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[ "a factor of 48.", "8 factors. What are the prime factors of 56? 2,7 are the prime factors. What are all the non-self factors of 56? All the non-self factors are 1, 2, 4, 7, 8, 14, 28 More factors: Scroll to Top", "ways; 129 in 4 different ways; 194 in 5 different ways; 209 in 6 ways.", "these factors are relatively prime to one another.", "194 in 5 different ways; 209 in 6 ways.", "48 is 24. =24", "# Too large to calculate How many distinct prime factors does the following number have: $2 ^ { 32 } + 2 ^{ 17 } + 1 ?$ ×", "# Fact Or Which of these numbers has more distinct prime factors? $\\Large 36 \\quad 38 \\quad 40$ ×", "____________________ 2. Is 72 a Prime number? ____________________ 3. How many factors does 15 have? ____________________ 4. Is \\begin{align} \\frac{3}{4} \\end{align} a Prime number? ____________________ 5. How many Prime Numbers are there between 60 and 70? ____________________", "# 92nd Problem 2016 How many multiples of 13 are included between 12548 and 19987 both inclusive? ×", "# Under 50 How many positive integer under 50 have exactly 2 distinct primes factors? No repeats, $$(2^{2})$$$$(3^{2})$$ doesn't count. ×", "prime factors for n large -", "a 600-long battle Joined: 21 Apr 2016 Posts: 138 Kudos [?]: 25 [0], given: 392 Location: Hungary Schools: Erasmus '20 GMAT 1: 410 Q18 V27 GMAT 2: 490 Q35 V23 Re: How many prime numbers between 1 and 100 are factors of 7,15 [#permalink] ### Show Tags 12 Mar 2017, 06:12 v0latility wrote: How many prime numbers between 1 and 100 are factors of 7,150? A. One B. Two C. Three D. Four E. Five I correctly factored 7,150 into 13 x 11 x 5 x 5 x 2. So there are 5 prime factors in total, 4 of which are unique. The correct answer is 4. So the GMAT must have been asking for the # of unique factors? I'm just having a little trouble convincing myself how the language in the question stem translates to unique vs. total. You can say that again. This question is very straight forward. Except for the language in the question stem that will certainly trick many people. _________________ \"When the going gets tough, the tough gets going!\" |Welcoming tips/suggestions/advices (you name it) to help me achieve a 600| Kudos [?]: 25 [0], given: 392 Director Joined: 02 Sep 2016 Posts: 784 Kudos [?]: 25 [0], given: 275 Re: How many prime numbers between 1 and 100 are factors of 7,15", "# Find those prime factors! How many distinct prime factors does 100100 have? Details and assumptions The number $$45 = 3^2 \\times 5$$ has 2 distinct prime factors. ×", "-(48+24-16)=80-56=24 ..", "list_of_Prime_factors])", "will have many factors. On the other hand, a prime number has only two factors, one and the number itself. Multiples of 32 are numbers that have 32 as factors.", "# Under 50 Number Theory Level 3 How many positive integer under 50 have exactly 2 distinct primes factors? No repeats, $$(2^{2})$$$$(3^{2})$$ doesn't count. ×", "# Ev-En-Ly Odd In how many ways you can distribute 55 among Ev, En and Ly so that everyone gets (a positive) odd number? (In other words three odd numbers add up to 55). ×", "# Fact Or Number Theory Level 1 Which of these numbers has more distinct prime factors? $\\Large 36 \\quad 38 \\quad 40$ ×" ]

[ "is prime. • $464 = {2^4} \\cdot 29$ • $465 = 3 \\cdot 5 \\cdot 31$ • $466 = 2 \\cdot 233$ • $467$ is prime. • $468 = {2^2} \\cdot {3^2} \\cdot 13$ • $469 = 7 \\cdot 67$ • $470 = 2 \\cdot 5 \\cdot 47$ • $471 = 3 \\cdot 157$ • $472 = {2^3} \\cdot 59$ • $473 = 11 \\cdot 43$ • $474 = 2 \\cdot 3 \\cdot 79$ • $475 = {5^2} \\cdot 19$ • $476 = {2^2} \\cdot 7 \\cdot 1$ • $477 = {3^2} \\cdot 53$ • $478 = 2 \\cdot 239$ • $479$ is prime. • $480 = {2^5} \\cdot 3 \\cdot 5$ • $481 = 13 \\cdot 37$ • $482 = 2 \\cdot 241$ • $483 = 3 \\cdot 7 \\cdot 23$ • $484 = {2^2} \\cdot {11^2}$ • $485 = 5 \\cdot 97$ • $486 = 2 \\cdot {3^5}$ • $487$ is prime. • $488 = {2^3} \\cdot 61$ • $489 = 3 \\cdot 163$ • $490 = 2 \\cdot 5 \\cdot {7^2}$ • $491$ is prime. • $492 = {2^2} \\cdot 3 \\cdot 41$ • $493 = 17 \\cdot 29$ • $494 = 2 \\cdot 13 \\cdot 19$ • $495 = {3^2} \\cdot 5 \\cdot 11$ • $496 = {2^4} \\cdot 31$ • $497 = 7", "• $566 = 2 \\cdot 283$ • $567 = {3^4} \\cdot 7$ • $568 = {2^3} \\cdot 71$ • $569$ is prime. • $570 = 2 \\cdot 3 \\cdot 5 \\cdot 19$ • $571$ is prime. • $572 = {2^2} \\cdot 11 \\cdot 13$ • $573 = 3 \\cdot 191$ • $574 = 2 \\cdot 7 \\cdot 41$ • $575 = {5^2} \\cdot 23$ • $576 = {2^6} \\cdot {3^2}$ • $577$ is prime. • $578 = 2 \\cdot {17^2}$ • $579 = 3 \\cdot 193$ • $580 = {2^2} \\cdot 5 \\cdot 29$ • $581 = 7 \\cdot 83$ • $582 = 2 \\cdot 3 \\cdot 97$ • $583 = 11 \\cdot 53$ • $584 = {2^3} \\cdot 73$ • $585 = {3^2} \\cdot 5 \\cdot 13$ • $586 = 2 \\cdot 293$ • $587$ is prime. • $588 = {2^2} \\cdot 3 \\cdot {7^2}$ • $589 = 19 \\cdot 31$ • $590 = 2 \\cdot 5 \\cdot 59$ • $591 = 3 \\cdot 197$ • $592 = {2^4} \\cdot 37$ • $593$ is prime. • $594 = 2 \\cdot {3^3} \\cdot 11$ • $595 = 5 \\cdot 7 \\cdot 17$ • $596 = {2^2} \\cdot 149$ • $597 = 3 \\cdot 199$ • $598 = 2 \\cdot 13 \\cdot 23$ • $599$ is prime. • $600", "$n$ other than $1\\cdot n$. Using an app for prime factorization, the following can be deduced very quickly: So our only non-prime $n$ with one unique factorization is $n=19{,}897=101\\cdot 197$ and we have our solution!", "# Question #b68fd Aug 24, 2017 See below. #### Explanation: Let's factor $105$. $105 = 35 \\cdot 3 = 3 \\cdot 5 \\cdot 7$ So, $3 \\cdot 5 \\cdot 7$ is the prime factorization of $105$. However, since there is only one of each prime, $\\sqrt{105}$ is already simplified.", "= {2^2} \\cdot 7 \\cdot 19$ • $533 = 13 \\cdot 41$ • $534 = 2 \\cdot 3 \\cdot 89$ • $535 = 5 \\cdot 107$ • $536 = {2^3} \\cdot 67$ • $537 = 3 \\cdot 179$ • $538 = 2 \\cdot 269$ • $539 = {7^2} \\cdot 11$ • $540 = {2^2} \\cdot {3^3} \\cdot 5$ • $541$ is prime. • $542 = 2 \\cdot 271$ • $543 = 3 \\cdot 181$ • $544 = {2^5} \\cdot 17$ • $545 = 5 \\cdot 109$ • $546 = 2 \\cdot 3 \\cdot 7 \\cdot 13$ • $547$ is prime. • $548 = {2^2} \\cdot 137$ • $549 = {3^2} \\cdot 61$ • $550 = 2 \\cdot {5^2} \\cdot 11$ • $551 = 19 \\cdot 29$ • $552 = {2^3} \\cdot 3 \\cdot 23$ • $553 = 7 \\cdot 79$ • $554 = 2 \\cdot 277$ • $555 = 3 \\cdot 5 \\cdot 37$ • $556 = {2^2} \\cdot 139$ • $557$ is prime. • $558 = 2 \\cdot {3^2} \\cdot 31$ • $559 = 13 \\cdot 43$ • $560 = {2^4} \\cdot 5 \\cdot 7$ • $561 = 3 \\cdot 11 \\cdot 17$ • $562 = 2 \\cdot 281$ • $563$ is prime. • $564 = {2^2} \\cdot 3 \\cdot 47$ • $565 = 5 \\cdot 113$", "number • Prime factorisation 48 $$= 2 \\cdot 2 \\cdot 2 \\cdot 3$$ • Prime factorisation 49 $$= 7\\cdot 7=7^2$$ • Prime factorisation 50 $$= 2\\cdot 5\\cdot 5$$ Prime factorisation up to 75 51 – 75 • $$51 = 3\\cdot 17$$ • $$52 = 2\\cdot 2 \\cdot 13$$ • $$53 = 1\\cdot 53$$ prime number • $$54 = 2\\cdot 3 \\cdot 3$$ • $$55 = 5\\cdot 11$$ • $$56 = 2\\cdot 2 \\cdot 7$$ • $$57 = 3\\cdot 19$$ • $$58 = 2\\cdot 29$$ • $$59 = 1\\cdot 59$$ prime number • $$60 = 2\\cdot 2 \\cdot 3 \\cdot 5$$ • $$61 = 1\\cdot 61$$ prime number • $$62 = 2\\cdot 31$$ • $$63 = 3 \\cdot 3 \\cdot 7$$ • $$64=2\\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 =2^6$$ • $$65 = 5\\cdot 13$$ • $$66 = 2\\cdot 3 \\cdot 11$$ • $$67 = 1\\cdot 67$$ prime number • $$68 = 2\\cdot 2 \\cdot 17$$ • $$69 = 3\\cdot 23$$ • $$70 = 2\\cdot 5 \\cdot 7$$ • $$71 = 1\\cdot 71$$ prime number • $$72 = 2\\cdot 2 \\cdot 3 \\cdot 3$$ • $$73 = 1\\cdot 73$$ prime number • $$74 = 2\\cdot 37$$ • $$75 = 3\\cdot 5 \\cdot 5$$ Prime factorisation up to 100 76 – 100 • $$76 = 2\\cdot 2 \\cdot 19$$", "the third factor is by division: $63779=23\\cdot 59\\cdot r$. One way to quickly compute this in our heads would be to notice $59\\cdot 23=(60-1)(20+3)=1200+180-20-3=1357.$ Since we know $r$ is an integer, we attempt to bound it. We know $r=\\frac{63779}{1357}$. We notice $40\\cdot 1357=52480$, while $50\\cdot 1357=67850$, therefore, $40. We then subtract $40$ from both sides of our equation: $r-40=\\frac{63779-52480}{1357}.$ While we could compute the numerator mentally, since we know $1\\le r-40\\le 9$, we know it must be a single numerical digit! We therefore just consider the unit digit of the numerator which is $9$. Since we’re dividing by a number which ends in $7$, we know $r-40$ must be $7$, since $7\\cdot 7\\equiv 49\\equiv 9\\pmod{10}$. Therefore $r=47$. Hence our desired sum is $p+q+r=23+59+47=\\boxed{129}$. Alternatively, we could verify $47\\mid 63779$ using the prime factorization of next year, 2021. I’ll leave this as an exercise for anyone interested. While this method isn’t super practical, I think it’s interesting that you can essentially compute prime factors up to 67 with a quick trick in your head. The most difficult part is memorizing the prime factorizations of the numbers close to 2000, but hopefully this is familiar for any seasoned contest mathematician (who might be interested in using this). References: Arthur Benjamin, Factoring Numbers with Conway’s 150 Method The College Mathematics Journal,", "## Elementary Algebra $56=7\\times2\\times2\\times2$ When we divide $56$ by $7$, we get $8$. This gives us $56=7\\times8$. Since $8$ is the perfect cube of $2$, a prime number, it can be broken down as $2\\times2\\times2$. When we plug this in for 8, we get $56=7\\times2\\times2\\times2$", "# How do you find the prime factorization of 112? Feb 5, 2015 You divide by primes, untill you run out. $112$ is even: $112 / 2 = 56 \\to 112 = \\left(2 \\cdot 56\\right)$ Still even: $56 / 2 = 28 \\to 112 = 2 \\cdot \\left(2 \\cdot 28\\right)$ And again: $28 / 2 = 14 \\to 112 = 2 \\cdot 2 \\cdot \\left(2 \\cdot 14\\right)$ And again: $14 / 2 = 7 \\to 112 = 2 \\cdot 2 \\cdot 2 \\cdot \\left(2 \\cdot 7\\right)$ (brackets are only for clarification) $7$ is a prime (try if you like) So the complete factorisation: $112 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 7 = {2}^{4} \\cdot 7$", "\\cdot 71$ • $498 = 2 \\cdot 3 \\cdot 83$ • $499$ is prime. • $500 = {2^2} \\cdot {5^3}$ • $501 = 3 \\cdot 167$ • $502 = 2 \\cdot 251$ • $503$ is prime. • $504 = {2^3} \\cdot {3^2} \\cdot 7$ • $505 = 5 \\cdot 101$ • $506 = 2 \\cdot 11 \\cdot 23$ • $507 = 3 \\cdot {13^2}$ • $508 = {2^2} \\cdot 127$ • $509$ is prime. • $510 = 2 \\cdot 3 \\cdot 5 \\cdot 17$ • $511 = 7 \\cdot 73$ • $512 = {2^9}$ • $513 = {3^3} \\cdot 19$ • $514 = 2 \\cdot 257$ • $515 = 5 \\cdot 103$ • $516 = {2^2} \\cdot 3 \\cdot 43$ • $517 = 11 \\cdot 47$ • $518 = 2 \\cdot 7 \\cdot 37$ • $519 = 3 \\cdot 173$ • $520 = {2^3} \\cdot 5 \\cdot 13$ • $521$ is prime. • $522 = 2 \\cdot {3^2} \\cdot 29$ • $523$ is prime. • $524 = {2^2} \\cdot 131$ • $525 = 3 \\cdot {5^2} \\cdot 7$ • $526 = 2 \\cdot 263$ • $527 = 17 \\cdot 31$ • $528 = {2^4} \\cdot 3 \\cdot 11$ • $529 = {23^2}$ • $530 = 2 \\cdot 5 \\cdot 53$ • $531 = {3^2} \\cdot 59$ • $532", "\\cdot 59$ • $827$ is prime. • $828 = {2^2} \\cdot {3^2} \\cdot 23$ • $829$ is prime. • $830 = 2 \\cdot 5 \\cdot 83$ • $831 = 3 \\cdot 277$ • $832 = {2^6} \\cdot 13$ • $833 = {7^2} \\cdot 17$ • $834 = 2 \\cdot 3 \\cdot 139$ • $835 = 5 \\cdot 167$ • $836 = {2^2} \\cdot 11 \\cdot 19$ • $837 = {3^3} \\cdot 31$ • $838 = 2 \\cdot 419$ • $839$ is prime. • $840 = {2^3} \\cdot 3 \\cdot 5 \\cdot 7$ • $841 = {29^2}$ • $842 = 2 \\cdot 421$ • $843 = 3 \\cdot 281$ • $844 = {2^2} \\cdot 211$ • $845 = 5 \\cdot {13^2}$ • $846 = 2 \\cdot {3^2} \\cdot 47$ • $847 = 7 \\cdot {11^2}$ • $848 = {2^4} \\cdot 53$ • $849 = 3 \\cdot 283$ • $850 = 2 \\cdot {5^2} \\cdot 17$ • $851 = 23 \\cdot 37$ • $852 = {2^2} \\cdot 3 \\cdot 71$ • $853$ is prime. • $854 = 2 \\cdot 7 \\cdot 61$ • $855 = {3^2} \\cdot 5 \\cdot 19$ • $856 = {2^3} \\cdot 107$ • $857$ is prime. • $858 = 2 \\cdot 3 \\cdot 11 \\cdot 13$ • $859$ is prime. • $860 = {2^2} \\cdot 5", "# List of Prime Factorizations of Integers from 601 to 800 For your quick reference, I have listed the prime factorizations of integers from 601 to 800. A repeated prime number is written in compact form using exponential notation. • $601$ is prime. • $602 = 2 \\cdot 7 \\cdot 43$ • $603 = {3^2} \\cdot 67$ • $604 = {2^2} \\cdot 151$ • $605 = 5 \\cdot {11^2}$ • $606 = 2 \\cdot 3 \\cdot 101$ • $607$ is prime. • $608 = {2^5} \\cdot 19$ • $609 = 3 \\cdot 7 \\cdot 29$ • $610 = 2 \\cdot 5 \\cdot 61$ • $611 = 13 \\cdot 47$ • $612 = {2^2} \\cdot {3^2} \\cdot 17$ • $613$ is prime. • $614 = 2 \\cdot 307$ • $615 = 3 \\cdot 5 \\cdot 41$ • $616 = {2^3} \\cdot 7 \\cdot 11$ • $617$ is prime. • $618 = 2 \\cdot 3 \\cdot 103$ • $619$ is prime. • $620 = {2^2} \\cdot 5 \\cdot 31$ • $621 = {3^3} \\cdot 23$ • $622 = 2 \\cdot 311$ • $623 = 7 \\cdot 89$ • $624 = {2^4} \\cdot 3 \\cdot 13$ • $625 = {5^4}$ • $626 = 2 \\cdot 313$ • $627 = 3 \\cdot 11 \\cdot 19$ • $628 = {2^2} \\cdot 157$ •", "View Single Post P: 87 I want to understand intuitively why LCM(a,b)=ab/GCF(a,b) Quote by statdad I'll use a = 84, b = 96. The prime factorizations of these are \\begin{align*} a & = 2^2 \\cdot 3\\cdot 7 \\\\ b & = 2^5 \\cdot 3 \\end{align*} It's easy to see that the LCM of these numbers is $$2^5 \\cdot 3 \\cdot 7$$ and, further, that the GCF is $$2^2 \\cdot 3$$. Somehow it is not easy for me to see why the LCM of the two numbers is $$2^5 \\cdot 3 \\cdot 7$$. Can you please tell me how you can tell the LCM from the prime factors of the numbers?", "three. To do so, we must split 216 into its prime factors: $$216 = 36 \\cdot 6 = 6^3 = 3^3 \\cdot 2^3$$. We see that $$a=3$$ and $$b = 3$$, so $$a+b$$ is simply $$6$$. - 4 years, 4 months ago", "$729 = {3^6}$ • $730 = 2 \\cdot 5 \\cdot 73$ • $731 = 17 \\cdot 43$ • $732 = {2^2} \\cdot 3 \\cdot 61$ • $733$ is prime. • $734 = 2 \\cdot 367$ • $735 = 3 \\cdot 5 \\cdot {7^2}$ • $736 = {2^5} \\cdot 23$ • $737 = 11 \\cdot 67$ • $738 = 2 \\cdot {3^2} \\cdot 41$ • $739$ is prime. • $740 = {2^2} \\cdot 5 \\cdot 37$ • $741 = 3 \\cdot 13 \\cdot 19$ • $742 = 2 \\cdot 7 \\cdot 53$ • $743$ is prime. • $744 = {2^3} \\cdot 3 \\cdot 31$ • $745 = 5 \\cdot 149$ • $746 = 2 \\cdot 373$ • $747 = {3^2} \\cdot 83$ • $748 = {2^2} \\cdot 11 \\cdot 17$ • $749 = 7 \\cdot 107$ • $750 = 2 \\cdot 3 \\cdot {5^3}$ • $751$ is prime. • $752 = {2^4} \\cdot 47$ • $753 = 3 \\cdot 251$ • $754 = 2 \\cdot 13 \\cdot 29$ • $755 = 5 \\cdot 151$ • $756 = {2^2} \\cdot {3^3} \\cdot 7$ • $757$ is prime. • $758 = 2 \\cdot 379$ • $759 = 3 \\cdot 11 \\cdot 23$ • $760 = {2^3} \\cdot 5 \\cdot 19$ • $761$ is prime. • $762 = 2 \\cdot 3 \\cdot", "# factorisation problem factorise 57284861 . [ use whatever you want ] Considering $$x_1,x_2 \\cdots x_n$$ to be the prime factors, calculate $$(x_1-1)(x_2-1)\\cdots(x_n-1)$$ ×", "this point that we have two prime factors, we could continue on with our test, but we have a pretty good idea of what the third factor is by division: $63779=23\\cdot 59\\cdot r$. One way to quickly compute this in our heads would be to notice $59\\cdot 23=(60-1)(20+3)=1200+180-20-3=1357.$ Since we know $r$ is an integer, we attempt to bound it. We know $r=\\frac{63779}{1357}$. We notice $40\\cdot 1357=52480$, while $50\\cdot 1357=67850$, therefore, $40. We then subtract $40$ from both sides of our equation: $r-40=\\frac{63779-52480}{1357}.$ While we could compute the numerator mentally, since we know $1\\le r-40\\le 9$, we know it must be a single numerical digit! We therefore just consider the unit digit of the numerator which is $9$. Since we’re dividing by a number which ends in $7$, we know $r-40$ must be $7$, since $7\\cdot 7\\equiv 49\\equiv 9\\pmod{10}$. Therefore $r=47$. Hence our desired sum is $p+q+r=23+59+47=\\boxed{129}$. Alternatively, we could verify $47\\mid 63779$ using the prime factorization of next year, 2021. I’ll leave this as an exercise for anyone interested. While this method isn’t super practical, I think it’s interesting that you can essentially compute prime factors up to 67 with a quick trick in your head. The most difficult part is memorizing the prime factorizations of the numbers close to 2000, but hopefully this is familiar for", "this point that we have two prime factors, we could continue on with our test, but we have a pretty good idea of what the third factor is by division: $63779=23\\cdot 59\\cdot r$. One way to quickly compute this in our heads would be to notice $59\\cdot 23=(60-1)(20+3)=1200+180-20-3=1357.$ Since we know $r$ is an integer, we attempt to bound it. We know $r=\\frac{63779}{1357}$. We notice $40\\cdot 1357=52480$, while $50\\cdot 1357=67850$, therefore, $40. We then subtract $40$ from both sides of our equation: $r-40=\\frac{63779-52480}{1357}.$ While we could compute the numerator mentally, since we know $1\\le r-40\\le 9$, we know it must be a single numerical digit! We therefore just consider the unit digit of the numerator which is $9$. Since we’re dividing by a number which ends in $7$, we know $r-40$ must be $7$, since $7\\cdot 7\\equiv 49\\equiv 9\\pmod{10}$. Therefore $r=47$. Hence our desired sum is $p+q+r=23+59+47=\\boxed{129}$. Alternatively, we could verify $47\\mid 63779$ using the prime factorization of next year, 2021. I’ll leave this as an exercise for anyone interested. While this method isn’t super practical, I think it’s interesting that you can essentially compute prime factors up to 67 with a quick trick in your head. The most difficult part is memorizing the prime factorizations of the numbers close to 2000, but hopefully this is familiar for", "# List of Prime Factorizations of Integers from 401 to 600 Below is a list of prime factorizations of integers from 401 to 600. If a prime number is repeated, it will be expressed as an exponential number. • $401$ is prime. • $402 = 2 \\cdot 3 \\cdot 67$ • $403 = 13 \\cdot 31$ • $404 = {2^2} \\cdot 101$ • $405 = {3^4} \\cdot 5$ • $406 = 2 \\cdot 7 \\cdot 29$ • $407 = 11 \\cdot 37$ • $408 = {2^3} \\cdot 3 \\cdot 17$ • $409$ is prime. • $410 = 2 \\cdot 5 \\cdot 41$ • $411 = 3 \\cdot 137$ • $412 = {2^2} \\cdot 103$ • $413 = 7 \\cdot 59$ • $414 = 2 \\cdot {3^2} \\cdot 23$ • $415 = 5 \\cdot 83$ • $416 = {2^5} \\cdot 13$ • $417 = 3 \\cdot 139$ • $418 = 2 \\cdot 11 \\cdot 19$ • $419$ is prime. • $420 = {2^2} \\cdot 3 \\cdot 5 \\cdot 7$ • $421$ is prime. • $422 = 2 \\cdot 211$ • $423 = {3^2} \\cdot 47$ • $424 = {2^3} \\cdot 53$ • $425 = {5^2} \\cdot 17$ • $426 = 2 \\cdot 3 \\cdot 71$ • $427 = 7 \\cdot 61$ • $428 = {2^2} \\cdot 107$ • $429", "common multiple, for two numbers a and b is the smallest number designated by LCM(a,b) that is divisible by both the number a and the number b. We can find LCM(a,b) by finding the prime factorization of a and b and choosing the maximum power for each prime factor. In another words, if the number a factors to ${\\displaystyle p_{1}^{\\alpha _{1}}p_{2}^{\\alpha _{2}}\\cdots p_{n}^{\\alpha _{n}}}$, and the number b factors to ${\\displaystyle p_{1}^{\\beta _{1}}p_{2}^{\\beta _{2}}\\cdots p_{n}^{\\beta _{n}}}$, then LCM(a,b) = ${\\displaystyle p_{1}^{\\gamma _{1}}p_{2}^{\\gamma _{2}}\\cdots p_{n}^{\\gamma _{n}}}$ where ${\\displaystyle \\gamma _{i}=Maximum(\\alpha _{i},\\beta _{i})}$ for i = 1 to n. An example, let us see the process on how we find lowest common multiple for 5500 and 450 which happens to be 49500. First, we find the prime factorization for 5500 and 450 which is 5500=22 53 11 450=2 3 2 52 Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power. 5500=22 53 11 = 22 30 53 111 450=2 32 52 = 21 32 52 110 The LCM(5500,450) is going to be in the form 2? 3? 5? 11?. All we now have to do is find what" ]

[ "## Elementary Algebra $56=7\\times2\\times2\\times2$ When we divide $56$ by $7$, we get $8$. This gives us $56=7\\times8$. Since $8$ is the perfect cube of $2$, a prime number, it can be broken down as $2\\times2\\times2$. When we plug this in for 8, we get $56=7\\times2\\times2\\times2$", "$n$ other than $1\\cdot n$. Using an app for prime factorization, the following can be deduced very quickly: So our only non-prime $n$ with one unique factorization is $n=19{,}897=101\\cdot 197$ and we have our solution!", "# 2.1 GCDs ## Unique Primes 1. How many prime factors does $$40500$$ have? Exercise 1 2. How many unique prime factors does $$40500$$ have? Exercise 2 The prime factorization of $$40500 = 2^2\\cdot 3^4 \\cdot 5^3$$, so it is composed of $$2+4+3$$ (non-unique) prime factors. However, only three of these are unique, namely $$2, 3$$ and $$4$$.", "## Basic College Mathematics (9th Edition) The factors for the number 56 are 1, 2, 4, 7, 8, 14, 28, and 56. $1\\times56=56$ $2\\times28=56$ $4\\times14=56$ $7\\times8=56$", "$p_1^{n_1}\\cdots p_k^{n_k}$ is $(n_1+1)\\cdots(n_k+1)$ when the $p$'s are distinct primes. This tells us that the product cannot contain any $1$'s. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without $1$s is $2\\cdot 5\\cdot 11$. We conclude that $110n^3$ has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of $n$ all of the form $n=p_1\\cdot p_2^3$. $81n^4$ thus has prime factorization $81n^4=3^4\\cdot p_1^4\\cdot p_2^{12}$ and a factor count of $5\\cdot5\\cdot13=\\boxed{\\textbf{(D) }325}$", "# Question #b68fd Aug 24, 2017 See below. #### Explanation: Let's factor $105$. $105 = 35 \\cdot 3 = 3 \\cdot 5 \\cdot 7$ So, $3 \\cdot 5 \\cdot 7$ is the prime factorization of $105$. However, since there is only one of each prime, $\\sqrt{105}$ is already simplified.", "# Factors on N^3 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "# What is the simplest radical form of 53? Aug 26, 2016 Do you want to simplify $\\sqrt{53}$ ? #### Explanation: What prime factors divide 53? 53 itself looks prime to me. In order to simplify $53$ and then take the square root, you have to be able to factor 53 into smaller numbers, at least one of which is a perfect square. For example, what is $\\sqrt{144}$ ? 144 can be factored into 12 x 12 or ${12}^{2}$. So you have $\\sqrt{{12}^{2}}$ which is $12$ . How about $\\sqrt{20}$ ? First, factor 20 as 5 x 4. Then write 4 as ${2}^{2}$ . You have $\\sqrt{{2}^{2} \\cdot 5}$, which equals $\\sqrt{{2}^{2}} \\cdot \\sqrt{5}$, which equals $2 \\sqrt{5}$. I think that's what you want to do with $\\sqrt{53}$, except you can't, because 53 can only be divided by itself and one - it's prime. The simplest form of $\\sqrt{53}$ is $\\sqrt{53}$. Here are the first 18 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, . . .", "\\cdot 3 \\cdot 5 + 1 = 31$, also a prime. It is always the case that a product of primes plus one is another prime? No, in fact. If you look at $2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 + 1=30031 = 59\\cdot 509$, you get a nonprime. Second proof that there are infinitely many primes In a similar vein to the first proof, we will show that there is always a prime larger than $n$ for any positive integer $n$. To see this, consider $n! + 1$. Upon dividing by any prime less than $n$, we get a remainder of $1$. So all of its prime factors are larger than $n$, and so there are infinitely many primes. $\\diamondsuit$ I would also like to present one more, which I’ve always liked. Third proof that there are infinitely many primes Suppose there are only finitely many primes $p_1, \\ldots, p_k$. Then consider the two numbers $n = p_1 \\cdot \\dots \\cdot p_k$ and $n -1$. We know that $n - 1$ has a prime factor, so that it must share a factor $P$ with $n$ since $n$ is the product of all the primes. But then $P$ divides $n - (n - 1) = 1$, which is nonsense; no prime divides $1$. Thus", "common multiple, for two numbers a and b is the smallest number designated by LCM(a,b) that is divisible by both the number a and the number b. We can find LCM(a,b) by finding the prime factorization of a and b and choosing the maximum power for each prime factor. In another words, if the number a factors to ${\\displaystyle p_{1}^{\\alpha _{1}}p_{2}^{\\alpha _{2}}\\cdots p_{n}^{\\alpha _{n}}}$, and the number b factors to ${\\displaystyle p_{1}^{\\beta _{1}}p_{2}^{\\beta _{2}}\\cdots p_{n}^{\\beta _{n}}}$, then LCM(a,b) = ${\\displaystyle p_{1}^{\\gamma _{1}}p_{2}^{\\gamma _{2}}\\cdots p_{n}^{\\gamma _{n}}}$ where ${\\displaystyle \\gamma _{i}=Maximum(\\alpha _{i},\\beta _{i})}$ for i = 1 to n. An example, let us see the process on how we find lowest common multiple for 5500 and 450 which happens to be 49500. First, we find the prime factorization for 5500 and 450 which is 5500=22 53 11 450=2 3 2 52 Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power. 5500=22 53 11 = 22 30 53 111 450=2 32 52 = 21 32 52 110 The LCM(5500,450) is going to be in the form 2? 3? 5? 11?. All we now have to do is find what", "factor, GCF) ### LCM The lowest common multiple, or the least common multiple, for two numbers a and b is the smallest number designated by LCM(a,b) that is divisible by both the number a and the number b. We can find LCM(a,b) by finding the prime factorization of a and b and choosing the maximum power for each prime factor. In another words, if the number a factors to ${\\displaystyle p_{1}^{\\alpha _{1}}p_{2}^{\\alpha _{2}}\\cdots p_{n}^{\\alpha _{n}}}$ , and the number b factors to ${\\displaystyle p_{1}^{\\beta _{1}}p_{2}^{\\beta _{2}}\\cdots p_{n}^{\\beta _{n}}}$ , then LCM(a,b) = ${\\displaystyle p_{1}^{\\gamma _{1}}p_{2}^{\\gamma _{2}}\\cdots p_{n}^{\\gamma _{n}}}$ where ${\\displaystyle \\gamma _{i}=Maximum(\\alpha _{i},\\beta _{i})}$ for i = 1 to n. An example, let us see the process on how we find lowest common multiple for 5500 and 450 which happens to be 49500. First, we find the prime factorization for 5500 and 450 which is 5500=22 53 11 450=2 3 2 52 Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power. 5500=22 53 11 = 22 30 53 111 450=2 32 52 = 21 32 52 110 The LCM(5500,450) is going to be in the form", "least common multiple, for two numbers a and b is the smallest number designated by LCM(a,b) that is divisible by both the number a and the number b. We can find LCM(a,b) by finding the prime factorization of a and b and choosing the maximum power for each prime factor. In another words, if the number a factors to $p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_n^{\\alpha_n}$, and the number b factors to $p_1^{\\beta_1}p_2^{\\beta_2}\\cdots p_n^{\\beta_n}$, then LCM(a,b) = $p_1^{\\gamma_1}p_2^{\\gamma_2}\\cdots p_n^{\\gamma_n}$ where $\\gamma_i = Maximum (\\alpha_i, \\beta_i)$ for i = 1 to n. An example, let us see the process on how we find lowest common multiple for 5500 and 450 which happens to be 49500. First, we find the prime factorization for 5500 and 450 which is 5500=22 53 11 450=2 3 2 52 Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power. 5500=22 53 11 = 22 30 53 111 450=2 32 52 = 21 32 52 110 The LCM(5500,450) is going to be in the form 2? 3? 5? 11?. All we now have to do is find what the powers of each individual prime will be. So now we", "# Find the greatest common factor of 28 and 36? Feb 5, 2018 $4$ #### Explanation: Using the prime factorization method, $28 = {2}^{2} \\cdot 7$ $36 = {2}^{2} \\cdot 9$ So, the greatest common factor is ${2}^{2}$. ${2}^{2} = 4$ $\\therefore G C F$ of $28$ and $36$ is $4$.", "# factorisation problem factorise 57284861 . [ use whatever you want ] Considering $$x_1,x_2 \\cdots x_n$$ to be the prime factors, calculate $$(x_1-1)(x_2-1)\\cdots(x_n-1)$$ ×", "View Single Post P: 87 I want to understand intuitively why LCM(a,b)=ab/GCF(a,b) Quote by statdad I'll use a = 84, b = 96. The prime factorizations of these are \\begin{align*} a & = 2^2 \\cdot 3\\cdot 7 \\\\ b & = 2^5 \\cdot 3 \\end{align*} It's easy to see that the LCM of these numbers is $$2^5 \\cdot 3 \\cdot 7$$ and, further, that the GCF is $$2^2 \\cdot 3$$. Somehow it is not easy for me to see why the LCM of the two numbers is $$2^5 \\cdot 3 \\cdot 7$$. Can you please tell me how you can tell the LCM from the prime factors of the numbers?", "the third factor is by division: $63779=23\\cdot 59\\cdot r$. One way to quickly compute this in our heads would be to notice $59\\cdot 23=(60-1)(20+3)=1200+180-20-3=1357.$ Since we know $r$ is an integer, we attempt to bound it. We know $r=\\frac{63779}{1357}$. We notice $40\\cdot 1357=52480$, while $50\\cdot 1357=67850$, therefore, $40. We then subtract $40$ from both sides of our equation: $r-40=\\frac{63779-52480}{1357}.$ While we could compute the numerator mentally, since we know $1\\le r-40\\le 9$, we know it must be a single numerical digit! We therefore just consider the unit digit of the numerator which is $9$. Since we’re dividing by a number which ends in $7$, we know $r-40$ must be $7$, since $7\\cdot 7\\equiv 49\\equiv 9\\pmod{10}$. Therefore $r=47$. Hence our desired sum is $p+q+r=23+59+47=\\boxed{129}$. Alternatively, we could verify $47\\mid 63779$ using the prime factorization of next year, 2021. I’ll leave this as an exercise for anyone interested. While this method isn’t super practical, I think it’s interesting that you can essentially compute prime factors up to 67 with a quick trick in your head. The most difficult part is memorizing the prime factorizations of the numbers close to 2000, but hopefully this is familiar for any seasoned contest mathematician (who might be interested in using this). References: Arthur Benjamin, Factoring Numbers with Conway’s 150 Method The College Mathematics Journal,", "see this, consider the number $p_1 p_2 \\ldots p_k + 1$. We know that this number will factor into primes, but upon division by every prime in our collection, it leaves a remainder of $1$. Thus it has at least one prime factor different than every factor in our collection. $\\diamondsuit$ This was a common proof used in class. A pattern also quickly emerges: $2 + 1 = 3$, a prime. $2\\cdot3 + 1 = 7$, a prime. $2 \\cdot 3 \\cdot 5 + 1 = 31$, also a prime. It is always the case that a product of primes plus one is another prime? No, in fact. If you look at $2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 + 1=30031 = 59\\cdot 509$, you get a nonprime. Second proof that there are infinitely many primes In a similar vein to the first proof, we will show that there is always a prime larger than $n$ for any positive integer $n$. To see this, consider $n! + 1$. Upon dividing by any prime less than $n$, we get a remainder of $1$. So all of its prime factors are larger than $n$, and so there are infinitely many primes. $\\diamondsuit$ I would also like to present one more, which I’ve always liked. Third proof", "3 \\cdot 3 = 36$ Now we need to pick the common prime factors. We can see that the common prime factors are $2,2$ and $3$. When we multiply $2 \\cdot 2 \\cdot 3$ we get $12$ which is the greatest common factor. ## Greatest common factor worksheets GCF of two numbers up to 30 (92.3 KiB, 1,128 hits) GCF of two numbers up to 50 (97.8 KiB, 1,169 hits) GCF of two numbers up to 100 (112.0 KiB, 911 hits) GCF of two numbers up to 500 (146.7 KiB, 1,183 hits) GCF of two numbers up to 1000 (165.3 KiB, 2,911 hits) GCF of three numbers up to 30 (79.4 KiB, 923 hits) GCF of three numbers up to 50 (101.6 KiB, 785 hits) GCF of three numbers up to 100 (109.0 KiB, 2,772 hits) GCF of three numbers up to 500 (135.7 KiB, 2,085 hits) GCF of three numbers up to 1000 (189.3 KiB, 2,489 hits)", "of 56 and 63 is 7. Taking the above into account you also know how to find all the common factors of 56 and 63, not just the greatest. In the next section we show you how to calculate the gcf of fifty-six and sixty-three by means of two more methods. How to find the GCF of 56 and 63 The greatest common factor of 56 and 63 can be computed by using the least common multiple aka lcm of 56 and 63. This is the easiest approach: gcf (56,63) = $\\frac{56 \\times 63}{lcm(56,63)} = \\frac{3528}{504}$ = 7 Alternatively, the gcf of 56 and 63 can be found using the prime factorization of 56 and 63: • The prime factorization of 56 is: 2 x 2 x 2 x 7 • The prime factorization of 63 is: 3 x 3 x 7 • The prime factors and multiplicities 56 and 63 have in common are: 7 • 7 is the gcf of 56 and 63 • gcf(56,63) = 7 In any case, the easiest way to compute the gcf of two numbers like 56 and 63 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 56,63. The calculation is conducted automatically.", "Goldbach that every odd composite number can be written as the sum of a prime and twice a square. 9 = 7 + 2$\\cdot$12 15 = 7 + 2$\\cdot$22 21 = 3 + 2$\\cdot$32 25 = 7 + 2$\\cdot$32 27 = 19 + 2$\\cdot$22 33 = 31 + 2$\\cdot$12 It turns out that the conjecture was false. What is the smallest odd composite that cannot be written as the sum of a prime and twice a square? 7. The first two consecutive numbers to have two distinct prime factors are: 14 = 2 $\\cdot$ 7 15 = 3 $\\cdot$ 5 The first three consecutive numbers to have three distinct prime factors are: 644 = 2² $\\cdot$ 7 $\\cdot$ 23 645 = 3 $\\cdot$ 5 $\\cdot$ 43 646 = 2 $\\cdot$ 17 $\\cdot$ 19. Find the first four consecutive integers to have four distinct prime factors. What is the first of these numbers? 8. The series, 11 + 22 + 33 + ... + 1010 = 10405071317. Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000. 9. The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit" ]

[ "+0 # URGENT!!!! 0 151 2 How many different three-letter sets of initials are possible using the letters A through G ? Jan 26, 2019 edited by Guest Jan 26, 2019 #1 +5662 +1 $$\\text{The letters A through G can be coded as 0 through 6}\\\\ \\text{and thus a 3 letter set is equivalent to a 3 digit base 7 number}\\\\ \\text{There are }7^3=343 \\text{ such numbers and thus 343 sets of initials}$$ . Jan 26, 2019 edited by Rom Jan 26, 2019 #2 +283 +1 __ __ __ There are 7 possibilities (A~G) for the first letter, 7 for the second, and 7 for the third. Multiplying this together, we get 7^3=343 total ways to make a set of 3-letter initials. Jan 26, 2019", "# How many possible combinations are there if you use three characters and the characters can be letters or numbers? $37 {C}_{3} = \\frac{37 X 36 X 35}{3 X 2 X 1} = 7770$ combinations of triples of distinct alphanumeric characters..", "possible characters. So there are $36^{12}$ possible combinations.", "Question A letter lock consists of three rings each marked with 10 different letters. In how many ways it is possible to make an unsuccessful attempt to open the lock? Open in App Solution Number of ways of marking each of the ring = 10 different letters ∴ Total number of ways of marking any letter on these three rings = 10$×$10$×$10 = 1000 Out of these 1000 combinations of the lock, 1 combination will be successful. ∴ Total number of unsuccessful attempts = 1000 $-$1 = 999 Suggest Corrections 0 Related Videos Counting Principle MATHEMATICS Watch in App", "roles of the letters $K$ and $G$ as used on that page are the reverse of the roles here).", "at least 5 (and all have a complex signature); but I'm certainly open to different approaches, if what I'm trying to achieve via template specialization isn't possible. Thanks! • 11 • 10 • 18 • 9 • 9", "3 such variations, one place out of three has a different symbol.", "acronym} \\acrodefplural{TLA}{three-letter acronyms} \\begin{document} \\Aclp{TLA} are acronyms with three letters. \\Acl{TLA} at the start of another sentence. \\end{document} -", "the G and R at the start and end, all the $$9$$ internal letters are different, so it is just the number of ways to permute them.", "## vishal_kothari 3 years ago Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter? There are 26-11 = 15 asymmetric letters; Your answer: $$26 \\times 25 \\times 24 - 15 \\times 14 \\times 13$$", "and $G = S_3$ (the symmetric group on three letters). $N$ in this case is the subgroup of order three generated by a 3-cycle.", "M, G, etc.) or words that begin with upper-case letters (Pfun, etc.) as these may conflict with internal names in Mathematica. – David G. Stork Nov 3 '15 at 0:51", "or $01$ (the third symbol).", "entries are $I$, one third are $M$ and one third are $O$.", "as in $R^3$, because there are three different kinds of things that cannot be confused. (I keep forgetting not everyone on this forum is as fluent in English as I is!)", "$A_H$ and $A_G$ - is well-known.", "# Below is given a figure made of three circles intersecting one another. These circles represents graduates, typists and Government employees. The intersecting regions have been denoted by a, b, c, e, f, g and h, respectively. Study the diagram carefully and answer the questions that followWhich of the following letters represents the typists who are government employees but not graduates? [ A ] e [ B ] g [ C ] f [ D ] h [ E ] c Answer : Option B Explanation : Letter g represent the typists who are only Government employees but not graduates", "keeping the three different powers of m 4. Mar 7, 2012 ### Oster no problem =)", "think there are different types of eggs and egg shapes, so there is probably no single $g(x)$ and $t$ that fits all eggs.", "Question What are at LEAST TWO transformations for the initials DM explain how it SHOWCASES the transformation WILL GIVE BRAINLIEST FOR BEST/REAL ANSWER" ]

[ "The total number of combinations are $7735+70+15+6=\\boxed{7826}.$", "are $7+42+42+42$ \"good\" combinations, for an answer of $133 + 780 = \\boxed{913}$. -jackshi2006", "\\ | * - - - - - * Now there are $\\boxed{3}$ choices for plot #4. Hence, there are: . $5 \\times 4 \\times 3 \\times 3 \\:=$ 180 ways. Therefore, there are: . $80 + 180 \\:=$ 260 ways.", "only three choices for each of three positions. $$5^3-3^3=98$$", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "each having 3 choices, thus total number of possibilities = $3^{6}$ = 729 Excluding 0 we get $3^{6} - 1$ = 728 by", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "the $1$ cell ($8$ choices ) so $720$ in this case. For the second you must choose the $3$ cell ($10$ choices) and then choose the three $1$ cells ($\\binom 93= 84$ choices). Thus $840$ in this case. Your final answer then is $$720+840=\\boxed {1560}$$", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "/ 7! • 3! = 120 different combinations. ## References 1. Ross, Sheldon. \"A First Course in Probability 7e\". Pearson Education: 2006.", "## Discrete Mathematics and Its Applications, Seventh Edition $26^6$ We are selecting a string consisting of six letters, each of which can be any of the $26$ letters, with repetition allowed. There are $26$ choices for the first letter, $26$ choices for the second letter, and so on. By the product rule on page 386, the total number of ways to select five elements is $26*26*26*26*26*26=26^6$.", "= 33649,$$ so the number of sets that are not missing any of 3, 6, or 14 is $$33649 - 19810 = \\boxed{13839}.$$", "with 4 columns: 1 2 3 4 1 2 3 5 1 2 3 6... 2 3 4 5 2 3 4 6 ... 47 48 49 50 To find the total possible combinations,we do: 50*49*48*47 __________ 1*2*3*4 This formula gives the maximum possible combinations. My question is,what would the formula be to find maximum of how many combinations of these numbers include number 1 and 2(for example)? There are so many difficulties with this qauestion that there no good place to start. FIRST: neither 1 nor 50 is between 1 and 50. If you want include them then say numbers from 1 to 50. With that linguistic understanding then your numbers look like $\\boxed{~A~}\\boxed{~B~}\\boxed{~C~}\\boxed{~D~}$ There are 50 choices for A, 49 choices for B, 48 choices for C, and 47 choices for D.", "add all three cases: $1+17+243=\\boxed{261}$", "remaining 4 to be one of 25 choices can be done in $5 \\cdot 25^4$ ways, just as you said.", "possible ordered pairs for $(a,b)$ are $(504, 503)$, $(505, 504)$, ... , $(1004, 1003)$, so $a$ has $\\boxed{501}$ possible values.", "$D$ 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.", "and we can chose the deviant box in $5$ ways. In all there are $1024-4-60=960$ admissible assignments.", "Question Hence, we have 5 choices for the first place which are 1, 2, 3, 7, 9. In the second and third place, we can place any given number. We have 6 choices for these places, which are 0, 1, 2, 3, 7, 9. So, the total numbers will be $5 \\times 6 \\times 6 = 180$ . Hence, option A will be the correct option." ]

[ "+0 # URGENT!!!! 0 151 2 How many different three-letter sets of initials are possible using the letters A through G ? Jan 26, 2019 edited by Guest Jan 26, 2019 #1 +5662 +1 $$\\text{The letters A through G can be coded as 0 through 6}\\\\ \\text{and thus a 3 letter set is equivalent to a 3 digit base 7 number}\\\\ \\text{There are }7^3=343 \\text{ such numbers and thus 343 sets of initials}$$ . Jan 26, 2019 edited by Rom Jan 26, 2019 #2 +283 +1 __ __ __ There are 7 possibilities (A~G) for the first letter, 7 for the second, and 7 for the third. Multiplying this together, we get 7^3=343 total ways to make a set of 3-letter initials. Jan 26, 2019", "# How many possible combinations are there if you use three characters and the characters can be letters or numbers? $37 {C}_{3} = \\frac{37 X 36 X 35}{3 X 2 X 1} = 7770$ combinations of triples of distinct alphanumeric characters..", "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 $\\color{red}{ 2\\times 2\\times 2\\times 2\\times 2=32}.$", "The total number of combinations are $7735+70+15+6=\\boxed{7826}.$", "your initials were in reverse alphabetical order such as MLKJ, or if they spelled out a word such as \"LOVE\" - all those combinations are be considered to be failures for this analysis yet may seem more special than merely being in order. • June 5th 2013, 01:39 PM Soroban Re: Random Question -= What are the chances Hello, floorplay! Assuming all letters are equally likely, there are: . $26^2\\!\\cdot\\!26^2$ possible outcomes. There are 23 sets of four consecutive letters, from ABCD to WXYZ. However, you could have the first two initials and she could have the last two . . or vice versa. Hence, there are: $2\\times 23 \\,=\\,46$ successful outcomes. Therefore, the probability is:. $\\frac{46}{26^4} \\;=\\;0.000100662$ . . about one in ten thousand. That seems to be quite a rare event. But consider the number of couples in, say, California.", "# A letter lock consists of three rings each marked with $5$ different letters. Number of maximum attempts to open the lock is:A. $124$ B. $125$ C. $120$ D. $75$ Verified 177.3k+ views Hint: Find the number of ways of choosing the letter for each of the three rings and then multiply the number of ways to open all the locks of the letter lock. We can also use concepts of permutations and combinations to solve such problems. A letter lock consists of three rings each marked with $5$ different letters. As per given that the letter lock consists of three rings each marked with $5$ different letters. So, the number of ways for opening the first ring is equal to $5$ as it is marked with $5$ different letters. In a similar manner, the number of ways for opening the second ring is equal to $5$ as it is marked with $5$ different letters. In a similar manner, the number of ways for opening the third ring is equal to $5$ as it is marked with $5$ different letters. So, the number of ways for opening the letter lock is equal to $5 \\times 5 \\times 5 = 125$ . So, the number of maximum attempts to open the lock is equal to $125$", "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", "For each $b$, there are exactly $3$ choices of $a$.", "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", "## Precalculus (10th Edition) $18278$ 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: $n^r$. Thus the number of company names possible with $1$ letter: $26^1=26.$ The number of company names possible with $2$ letters: $26^2=676.$ The number of company names possible with $3$ letters: $26^3=17576.$ Thus the total is:$26+676+15576=18278.$", "Question # Find the numbers of ways in which one can post $$5$$ letter in $$7$$ letter boxes. Solution ## Each latter can be posted in any one of the 7 letter boxes. So, required numbers of ways =7×7×7×7×7=75Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "are $7+42+42+42$ \"good\" combinations, for an answer of $133 + 780 = \\boxed{913}$. -jackshi2006", "Calculate the number of combinations available I am writing a computer program which generate 5 digit alpha-numeric codes. Each character can contain a-z,A-Z and 0-9 (62 different possibility per character). So if there are 5 character how many possibility are there? - You can get $62^5$ (or $916132832$) possibilities. Every character can have $62$ different possibility as you said, so the total possible combinations of 5 character would be $62\\times62\\times62\\times62\\times62=62^5$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "or $01$ (the third symbol).", "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.", "how many different sets of $3$3 letters we can have. A diagramatic calculation starts with drawing boxes (or circles, or triangles, or any other shape you can write a number inside of), where the number of the boxes is equal to the number of objects we are selecting. In this case we are selecting $3$3 letters each time, so we have $3$3 boxes. Inside the first box, we write down how many possible choices we have for this box. We have $5$5 letters to choose from, so we have $5$5 options. Either $A,B,C,D$A,B,C,D or $E$E. Inside the second box, we write down how many possible choices we have for this box. Because we had $5$5 to start with, but will have selected one already, there are $4$4 options left for this next box. Finally, inside the third box. How many possible letters would we have left to pick from. Only $3$3 To calculate how many possible options we have we multiply these together. So there are $60$60 possible permutations for selecting $3$3 objects from the $5$5. ## Formula for Permutations Consider the example of finding how many $3$3 digit numbers can be formed using the $9$9 digits $1,2,3,4,5,6,7,8,9$1,2,3,4,5,6,7,8,9, given that the digits cannot be used more than once? The answer is $504$504 numbers. The explanation is quite simple.", "at least 5 (and all have a complex signature); but I'm certainly open to different approaches, if what I'm trying to achieve via template specialization isn't possible. Thanks! • 11 • 10 • 18 • 9 • 9", "$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.", "# Permutations or Combinations A letter lock has $3$ rings containing $6$ different letters. No $2$ rings have the same letters. How many different combinations of passwords is possible? 1. $15600$ 2. $17576$ 3. $\\binom{26}{6}\\binom{20}{6}\\binom{14}{6}6^3$ 4. $\\binom{26}{6}^36^3$ I am getting $3$ as the answer but the answer given is $15600$... please explain what's the correct one? • The question is not at all clearly formulated. Not clear what \"combinations of passwords\" are, not whether there is any relation between passwords and rings, and if so which relation. Also the title is not very helpful. – Marc van Leeuwen Jun 3 '16 at 14:58 There are three independent rings, each with 6 options. So $6^3$ is the right answer (the order matters). This assumes the rings are fixed and known in advance. But this is not among the options. Your answer (nr 3) is the number of ways we can pick the letters for each ring first (so start with blank rings), and then times $6^3$ for each combination. But this overcounts. The combination ABC could occur in many different ways. The net effect of picking rings and letters for each is that every combination of three different letters could be a combination for lots of ring configurations so the answer would then $26 \\cdot 25 \\cdot 24$. So" ]

[ "= 48", "a factor of 48.", "48 is 24. =24", "48, can be divided by 3.", "48", "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "by their highest common factor.", "to mostly use 48 now (they've used lower in the past) (source).", "= -32.", "that/which 42 � ������������ ������������������������ ��������� �����������", "greatest common factor of 36 and 48x 4-1 factoredderivative of 9x9x2 6x 1682.61234 in roman numeralstan1527x3x2 5x 6 factorsolve v lwh for w386.11what is the greatest common factor of 54 and 72solve y 4x 21700-500y kx22xdsquare root of 776multiplying and dividing fraction calculatorleast to greatest decimals calculatordesoto k12 mo ussin30 cos60convert 0.375 to a fraction", "woman. 32 36 42 48", "# What is the greatest common factor of 16,32, and 44? What is the greatest common factor of 16,32, and 44? ## This Post Has 3 Comments 1. LarryJoeseph says: The greatest common factor of the three numbers is 4 2. katelyn0579 says: The gcf of 16, 32, and 44 is 4. 3. salsa456 says: Look it up on gooogle", "32 Thus, 33 And a bit more compactly, 34", "hit the sharing buttons if our article about the greatest common factor of 35 and 8 has been useful to you, and make sure to bookmark our site.", "would give the answer that is .34", "us a mail to get in touch. Please hit the sharing buttons if our article about the greatest common factor of 37 and 100 has been useful to you, and make sure to bookmark our site.", "square units 42.) What is the least common multiple of 168 and 420? Continue reading", "count 34):", "# 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? ×" ]

[ "= 2 \\cdot 3\\cdot 4$and$60 = 3\\cdot 4\\cdot 5$. The common factors are$3$and$4$and so$\\gcd(24, 60) = 3\\cdot 4 = 12$. If we take an integer$a$and square it, the resulting integer$a^2$does not have any new prime factors which are not already present in$a$. It simply has the same factors, with double the multiplicity. For instance$12 = 3\\cdot 2\\cdot 2$, and$144 = 12^2 = (3\\cdot 2\\cdot 2)(3\\cdot 2 \\cdot 2) = 3\\cdot 3\\cdot 2\\cdot 2\\cdot 2\\cdot 2$. So if$a$and$b$have no prime factors in common, then$a^2$and$b^2$have no prime factors in common either, and so$\\gcd(a^2, b^2) = 1\\$. -", "# 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 $2 \\cdot 2 \\cdot 2 = 8$", "# How do you find the greatest common factor of 45y^{12}+30y^{10}? Jun 11, 2018 $15 {y}^{10}$ #### Explanation: The largest number that goes into 45 and 30 is 15 ${y}^{10}$ is the biggest term that goes into ${y}^{12} \\mathmr{and} {y}^{10}$ Jun 11, 2018 The greatest common factor is $15 {y}^{10}$. #### Explanation: I suppose you're asking for the greatest common factor between $45 {y}^{12}$ and $30 {y}^{10}$, since the greatest common factor is computed between two number. We need to find the greatest common factor for both the numeric and literal part. For the numeric part, we can use the prime factorization of the two numbers: $45 = 9 \\cdot 5 = {3}^{2} \\cdot 5$ $30 = 3 \\cdot 10 = 2 \\cdot 3 \\cdot 5$ So, what's the biggest number that \"fits\" inside both $45$ and $30$, given their prime factorizations? Well, we can't choose $2$, because it fits in $30$ but not in $45$. $3$ appears in both factorization, but we can pick it only once, since two three's (i.e. $9$) fit inside $45$, but not inside $30$. Finally, we can pick $5$ once since it appears in both factorizations. So, the answer is $3 \\cdot 5 = 15$ As for the literal part, we have a similar way to proceed: since \"there are\" $12$ $y$'s", "# Find the greatest common factor of 28 and 36? Feb 5, 2018 $4$ #### Explanation: Using the prime factorization method, $28 = {2}^{2} \\cdot 7$ $36 = {2}^{2} \\cdot 9$ So, the greatest common factor is ${2}^{2}$. ${2}^{2} = 4$ $\\therefore G C F$ of $28$ and $36$ is $4$.", "= 3628799!7!5!3!$. You can ... 0 No. Consider $4^4 = 2^{8}$, which has no 3 in its prime factorization, while $4! = 1 \\cdot 2 \\cdot 3 \\cdot 4 = 2^3 \\cdot 3$. -2 Top 50 recent answers are included", "# 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: $2 \\times 2 \\times 2 \\times 3 \\times 5 \\times 11 \\times 31 = 40 , 920$ $120 \\times 11 \\times 31 = 40 , 920$ $124 \\times 2 \\times 3 \\times 5 \\times 11 = 40 , 920$ $165 \\times 2 \\times 2 \\times 2 \\times 31 = 40 , 920$", "# IMAT 2019 Q57 [Common factor] In order to solve this question, we must find the factors of each of the number: 360 = 36 \\cdot 10 = 6 \\cdot 6 \\cdot 5\\cdot 2 = 3^2\\cdot 2 ^2 \\cdot 5\\cdot 2 = 2^3 \\cdot 3^2 \\cdot 5. 500 = 5 \\cdot 10^2 = 5 \\cdot (5\\cdot 2)^2 = 5 \\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^3. 700 = 7 \\cdot 10^2 = 7\\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^2\\cdot 7. From this we can see that all of the numbers have 2^2 \\cdot 5 in their prime factors, so the highest common factor is (D) 2^2 \\cdot 5 3 Likes", "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.", "# [Math] prime factors of $3^{32}-2^{32}$ elementary-number-theory The question asks to find 4 prime factors of $3^{32}-2^{32}$ under $100$. My take:I factorized it and the obvious ones are $5, 97$ and $13$.I cannot find the last one ,however.I was wondering if we could prove the number is even but that has not worked out well.So what is the other factor and how do we find it? If you are able to factor the number (which I did with wolframalpha), you will get $$5\\cdot 13\\cdot 17\\cdot 97\\cdot 401\\cdot 3041\\cdot 14177$$ which immediately answers your question. However most likely the intent of the person asking was for you to use indirect means to determine small factors. For example, to tell whether $5$ is a factor, we calculate the expression modulo 5: $$3^{32}-2^{32}\\equiv 3^{32}-(-3)^{32}\\equiv 3^{32}-(-1)^{32}3^{32}\\equiv 0$$ $$3^4= 81\\equiv 3, 3^{16}=(3^4)^4\\equiv 3, 3^{32}=(3^{16})^2\\equiv 9$$ $$2^4=16\\equiv 3, 2^{16}=(2^4)^4\\equiv 3^4\\equiv 3, 2^{32}=(2^{16})^2\\equiv 9$$", "picture is extremely important because this will allow you to strategize and therefore being able to devise a viable approach to the problem. So now, if you closely examine the two numbers which are $71$ and $223$, you should easily recognize that both of them are prime numbers. Remember that a prime number has exactly two factors which are $1$ and itself. In other words, we can say that a prime number is only divisible by $1$ and itself. Listing the factors of $71$ and $223$: Factors of 71: 1, 71 Factors of 223: 1, 223 We should be able to conclude that since $1$ is the ONLY common factor, it implies that $1$ must also be the greatest common factor by default. Thus, ${\\rm{gcf}}\\left( {71,223} \\right) = 1$. Example 5: What is the GCF of $72$ and $84$ ? First, we know that both numbers are not prime, in fact, both are even numbers. It means they have common factors other than $1$. Secondly, it is obvious too that the smaller number $72$ cannot evenly divide the larger number $84$. This leaves us to conclude that the smaller of the two numbers, $72$, is NOT the greatest common factor either. Well, the only option left is to proceed with the step-by-step procedure in finding the GCF of", "yet another $p$), and so on. This simple argument does not work for composite numbers in place of $p$, simply because divisibility can be \"collected\" from diffferent factors. However, knowing the power for the prime $2$, you can easily find out the power for $4=2^2$. With a non-prime such as $4$, you've got the problem that factors from terms that individually are not a multiple of $4$ may combine to a factor of $4$. This massively complicates the calculation. For example, consider $$6! = 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1$$ The only factor that is divisible by $4$ is $4$ itself, so you might wrongly conclude the highest power of $4$ in $6!$ is $4^1=4$. However $6\\cdot 2=12 = 3\\cdot 4$, so there's another factor of $4$ \"hidden\" in the other factors. And indeed, $6! = 720 = 4^2\\cdot 45$. With prime factors you don't have this problem, as those cannot arise from a product unless they are in one of the factors. So to determine how many of them are in the product, it is sufficient to add up those found in the factors. For example, to determine the $2$s in $6!$, you find one factor $2$ in $6$, two of them in $4$, and one further in $2$, to give a total of $4$. Now $4=2^2$,", "search ended quickly after remembering 2401 was a perfect square: $2501 = 49^2 + 10^2$ What’s interesting here is we know from this that 2501 is not prime, but we don’t know what its factors are! Having tested and not finding any prime factors less than 17, this is enough to know that 2501 is the product of exactly two prime factors. But which ones? To do this, I tried to find a way to write 2501 as a difference of squares. And again, the search ended much more quickly than I expected: $2501 = 51^2 - 10^2$ $51^2 - 10^2 = (51 + 10)(51 - 10)$ It made me wonder if there are other numbers with the same property, where the number is 1 more than a perfect square, and also has an equal gap to the next square up and next square down… guess what. 26 between 16 and 36 37 between 25 and 49 $n^2 + 1$ between $(n-1)^2$ and $(n+1)^2$ Let me know if you find anything interesting here, or if you have any favorite numerical calculations like these.", "prime is $3$, and after comparing the resulting numbers, we discover that $\\{ 1, 1, 1, 1, 4\\}$ results in the smallest odd number $3^4\\cdot 5\\cdot 7\\cdot 11\\cdot 13=405405.$ Then the smallest odd number with exactly $40$ \"product pairs\" is $405405$. The same process is done for even numbers with the only difference being that the smallest prime factor is $2$, therefore we take each exponents set, assign them in decreasing order to increasing primes, and compare them with each other. In this case the smallest value comes out from the set $\\{ 1,1,3,4\\}$ to be $2^4\\cdot 3^3\\cdot 5\\cdot 7=15120$. So the smallest even number, actually smallest positive number altogether which can be factored into exactly $40$ product pairs is $15120$.", "out small primes repeatedly until there are only prime factors remaining is acceptable. See the video below for a demonstration of using stacked division to find a prime factorization. ### Finding Least Common Multiples Consider the multiples of two numbers, 3 and 4. What is the least common multiple between them? Certainly 3 and 4 have many multiples in common: 12, 24, 36, etc. But 12 is the smallest among them. The smallest number that is a multiple of two numbers is called the least common multiple (LCM). So the LCM of $3$ and $4$ is $12$. One of the reasons we find prime factorizations is to use them to find the least common multiple of two or more numbers. This will be useful when we add and subtract fractions with different denominators. ### Prime Factors Method We can find the least common multiple of two numbers by inspecting their prime factors. We’ll use this method to find the LCM of $12$ and $18$. We start by finding the prime factorization of each number. $12=2\\cdot 2\\cdot 3 \\qquad$ and $\\qquad 18=2\\cdot 3\\cdot 3$ Then we find the largest instance of each prime appearing in any one factorization. Here, we see that the number 2 appears twice in the factorization of 12, but only once in 18. And the", "# 2.1 GCDs ## Unique Primes 1. How many prime factors does $$40500$$ have? Exercise 1 2. How many unique prime factors does $$40500$$ have? Exercise 2 The prime factorization of $$40500 = 2^2\\cdot 3^4 \\cdot 5^3$$, so it is composed of $$2+4+3$$ (non-unique) prime factors. However, only three of these are unique, namely $$2, 3$$ and $$4$$.", "Comparing the factors of $10$ and $21$, we find that they do not have any common factors, other than $1$. Comparing factors of $10$ and $6$, we find that they have a common factor $2$. $10$ and $21$ are called \"co-prime numbers\". $10$ and $6$ are not co-prime numbers, as they share $2$ as the common-factor. Similarly, $6$ and $21$ are not co-prime numbers, as they share $3$ as the common-factor. Co-prime Numbers : Two or more numbers without any common factor, other than $1$, are called co-prime numbers. The prefix \"co\"? means \"together and mutually\". Are $7$ and $7$ co-prime numbers? The answer is \"no\". They have a common factor 7. Are $21$ and $10$ co-prime numbers? The answer is \"Yes, they do not have a common factor\". The factors of $21$ are $1 , 3 , 7 , 21$ and $10$ are $1 , 2 , 5 , 10$. Highest Common Factor Factors of $12$ are $1 , 2 , 3 , 4 , 6 , 12$ and factors of $18$ are $1 , 2 , 3 , 6 , 18$. The common factors between these two are \"$1 , 2 , 3 , 6$\" The highest common factor for these two numbers is \"$6$\" Highest Common Factor : The highest factor in the common factors", "prime number (73=1*73) then $$d=1$$ and $$2m+n=73$$ (vice versa is not possible because $$m$$ and $$n$$ are positve integers and therefore $$2m+n$$ cannot equal to 1). Hence we have that GCD(x, y)=d=1. Sufficient. (2) $$5x-3y=1$$ --> $$5x=3y+1$$ --> $$5x$$ and $$3y$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1 (for example 20 and 21 are consecutive integers, thus only common factor they share is 1). So, $$5x$$ and $$3y$$ don't share any common factor but 1, thus $$x$$ and $$y$$ also don't share any common factor but 1. Hence, GCD(x, y) is 1. Sufficient. Hope it's clear.[/quote] Hi Bunuel, Please help me understand - is it that even if two numbers' multiples are co primes, that the numbers themselves will be co primes as well? How? Thank you.[/quote] Can you please give an example of what you mean?[/quote] In the explanation of the second AC, you state that 5x and and 3y dont share any other common factor other than 1 (since they are co prime), and hence x and y would also be co prime. I tried the same with an example for x and y, 2 and 3 respectively and the equation holds true. But would 5x=3y+1 result in x and y as co prime", "× What is the greatest common factor of 54 and 36? It is $18$ Explanation: Because $3 \\cdot 18 = 54$ $2 \\cdot 18 = 36$ Also The factors of 54: 1, 2, 3, 6, 9, 18, 27, 54 -1, -2, -3, -6, -9, -18, -27, -54 The factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 -1, -2, -3, -4, -6, -9, -12, -18, -36 The greatest common factor of 54 and 36 = 18 Jul 15, 2017 $H C F = 18$ Explanation: When working with HCF and/or LCM, write each number as the product of its prime factors. That will tell you everything you need to know about a number. Look for all the common factors: $\\text{ } 36 = 2 \\times 2 \\times 3 \\times 3$ \" \"ul(54 = 2\" \"xx3xx3xx3) $H C F = 2 \\text{ \"xx3xx3\" } = 18$ The highest common factor is the product of all the common factors. This is a very quick and effective method, especially if you are working with large numbers where you might not know all of the factors. The product of the prime factors will tell you whether a number is a power, like a square or a cube. You can also use the prime factors to determine all the other", "# Two numbers are in the ratio 4:5. Their highest common factor is 16. What is a pair of numbers that fits that? Jan 29, 2017 The numbers are $64$ and $80$. The two numbers are in the ratio $4 : 5$ and there is no common factor between $4$ and $5$ (i.e. highest common factor is $1$). As the highest common factor of the two numbers is $16$, the numbers are $4 \\times 16$ and $5 \\times 16$ i.e. $64$ and $80$.", "## Thinking Mathematically (6th Edition) The greatest common divisor of the two numbers is $15$. Factor each number completely to have: $240=10(24)=2(5)(2)(2)(2)(3) \\\\285=5(57)=5(3)(19)$ Thus, the greatest common divisor of the two numbers is $3(5)=15$." ]

[ "# 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 $2 \\cdot 2 \\cdot 2 = 8$", "# How do you find the greatest common factor of 45y^{12}+30y^{10}? Jun 11, 2018 $15 {y}^{10}$ #### Explanation: The largest number that goes into 45 and 30 is 15 ${y}^{10}$ is the biggest term that goes into ${y}^{12} \\mathmr{and} {y}^{10}$ Jun 11, 2018 The greatest common factor is $15 {y}^{10}$. #### Explanation: I suppose you're asking for the greatest common factor between $45 {y}^{12}$ and $30 {y}^{10}$, since the greatest common factor is computed between two number. We need to find the greatest common factor for both the numeric and literal part. For the numeric part, we can use the prime factorization of the two numbers: $45 = 9 \\cdot 5 = {3}^{2} \\cdot 5$ $30 = 3 \\cdot 10 = 2 \\cdot 3 \\cdot 5$ So, what's the biggest number that \"fits\" inside both $45$ and $30$, given their prime factorizations? Well, we can't choose $2$, because it fits in $30$ but not in $45$. $3$ appears in both factorization, but we can pick it only once, since two three's (i.e. $9$) fit inside $45$, but not inside $30$. Finally, we can pick $5$ once since it appears in both factorizations. So, the answer is $3 \\cdot 5 = 15$ As for the literal part, we have a similar way to proceed: since \"there are\" $12$ $y$'s", "= 2 \\cdot 3\\cdot 4$and$60 = 3\\cdot 4\\cdot 5$. The common factors are$3$and$4$and so$\\gcd(24, 60) = 3\\cdot 4 = 12$. If we take an integer$a$and square it, the resulting integer$a^2$does not have any new prime factors which are not already present in$a$. It simply has the same factors, with double the multiplicity. For instance$12 = 3\\cdot 2\\cdot 2$, and$144 = 12^2 = (3\\cdot 2\\cdot 2)(3\\cdot 2 \\cdot 2) = 3\\cdot 3\\cdot 2\\cdot 2\\cdot 2\\cdot 2$. So if$a$and$b$have no prime factors in common, then$a^2$and$b^2$have no prime factors in common either, and so$\\gcd(a^2, b^2) = 1\\$. -", "## Thinking Mathematically (6th Edition) The greatest common divisor of the two numbers is $15$. Factor each number completely to have: $240=10(24)=2(5)(2)(2)(2)(3) \\\\285=5(57)=5(3)(19)$ Thus, the greatest common divisor of the two numbers is $3(5)=15$.", "# Find the greatest common factor of 28 and 36? Feb 5, 2018 $4$ #### Explanation: Using the prime factorization method, $28 = {2}^{2} \\cdot 7$ $36 = {2}^{2} \\cdot 9$ So, the greatest common factor is ${2}^{2}$. ${2}^{2} = 4$ $\\therefore G C F$ of $28$ and $36$ is $4$.", "# IMAT 2019 Q57 [Common factor] In order to solve this question, we must find the factors of each of the number: 360 = 36 \\cdot 10 = 6 \\cdot 6 \\cdot 5\\cdot 2 = 3^2\\cdot 2 ^2 \\cdot 5\\cdot 2 = 2^3 \\cdot 3^2 \\cdot 5. 500 = 5 \\cdot 10^2 = 5 \\cdot (5\\cdot 2)^2 = 5 \\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^3. 700 = 7 \\cdot 10^2 = 7\\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^2\\cdot 7. From this we can see that all of the numbers have 2^2 \\cdot 5 in their prime factors, so the highest common factor is (D) 2^2 \\cdot 5 3 Likes", "What is the least common denominator of 3/4, 3/8, and 1/5? Apr 13, 2016 $40$ Explanation: If we look at the prime factorizations of the denominators, we have $4 = {2}^{2}$ $8 = {2}^{3}$ $5 = {5}^{1}$ The least common denominator will be the minimal product containing all factors above to their appropriate powers. In this case, that would be ${2}^{3} \\cdot 5$. Thus, the least common denominator is ${2}^{3} \\cdot 5 = 8 \\cdot 5 = 40$", "## 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 $20$.", "not consider it. If n = 5 and t = 45, then $$nt = 5*45 = 3^2*5^2$$. So, the greatest common factor of nt is 5. _________________ Re: If n and t are positive integers, what is the greatest prime [#permalink] 20 Aug 2017, 02:30 Display posts from previous: Sort by", "$x2+2x−5x−10x2+2x−5x−10$ $20x2−16x−15x+12.20x2−16x−15x+12.$ Try It 6.16 Factor by grouping: $y2+4y−7y−28y2+4y−7y−28$ $42m2−18m−35m+15.42m2−18m−35m+15.$ ### Section 6.1 Exercises #### Practice Makes Perfect Find the Greatest Common Factor of Two or More Expressions In the following exercises, find the greatest common factor. 1. $10p3q,12pq210p3q,12pq2$ 2. $8a2b3,10ab28a2b3,10ab2$ 3. $12m2n3,30m5n312m2n3,30m5n3$ 4. $28x2y4,42x4y428x2y4,42x4y4$ 5. $10a3,12a2,14a10a3,12a2,14a$ 6. $20y3,28y2,40y20y3,28y2,40y$ 7. $35x3y2,10x4y,5x5y335x3y2,10x4y,5x5y3$ 8. $27p2q3,45p3q4,9p4q327p2q3,45p3q4,9p4q3$ Factor the Greatest Common Factor from a Polynomial In the following exercises, factor the greatest common factor from each polynomial. 9. $6m+96m+9$ 10. $14p+3514p+35$ 11. $9n−639n−63$ 12. $45b−1845b−18$ 13. $3x2+6x−93x2+6x−9$ 14. $4y2+8y−44y2+8y−4$ 15. $8p2+4p+28p2+4p+2$ 16. $10q2+14q+2010q2+14q+20$ 17. $8y3+16y28y3+16y2$ 18. $12x3−10x12x3−10x$ 19. $5x3−15x2+20x5x3−15x2+20x$ 20. $8m2−40m+168m2−40m+16$ 21. $24x3−12x2+15x24x3−12x2+15x$ 22. $24y3−18y2−30y24y3−18y2−30y$ 23. $12xy2+18x2y2−30y312xy2+18x2y2−30y3$ 24. $21pq2+35p2q2−28q321pq2+35p2q2−28q3$ 25. $20x3y−4x2y2+12xy320x3y−4x2y2+12xy3$ 26. $24a3b+6a2b2−18ab324a3b+6a2b2−18ab3$ 27. $−2x−4−2x−4$ 28. $−3b+12−3b+12$ 29. $−2x3+18x2−8x−2x3+18x2−8x$ 30. $−5y3+35y2−15y−5y3+35y2−15y$ 31. $−4p3q−12p2q2+16pq2−4p3q−12p2q2+16pq2$ 32. $−6a3b−12a2b2+18ab2−6a3b−12a2b2+18ab2$ 33. $5x(x+1)+3(x+1)5x(x+1)+3(x+1)$ 34. $2x(x−1)+9(x−1)2x(x−1)+9(x−1)$ 35. $3b(b−2)−13(b−2)3b(b−2)−13(b−2)$ 36. $6m(m−5)−7(m−5)6m(m−5)−7(m−5)$ Factor by Grouping In the following exercises, factor by grouping. 37. $ab+5a+3b+15ab+5a+3b+15$ 38. $cd+6c+4d+24cd+6c+4d+24$ 39. $8y2+y+40y+58y2+y+40y+5$ 40. $6y2+7y+24y+286y2+7y+24y+28$ 41. $uv−9u+2v−18uv−9u+2v−18$ 42. $pq−10p+8q−80pq−10p+8q−80$ 43. $u2−u+6u−6u2−u+6u−6$ 44. $x2−x+4x−4x2−x+4x−4$ 45. $9p2+12p−15p−209p2+12p−15p−20$ 46. $16q2+20q−28q−3516q2+20q−28q−35$ 47. $mn−6m−4n+24mn−6m−4n+24$ 48. $r2−3r−r+3r2−3r−r+3$ 49. $2x2−14x−5x+352x2−14x−5x+35$ 50. $4x2−36x−3x+274x2−36x−3x+27$ Mixed Practice In the following exercises, factor. 51. $−18xy2−27x2y−18xy2−27x2y$ 52. $−4x3y5−x2y3+12xy4−4x3y5−x2y3+12xy4$ 53. $3x3−7x2+6x−143x3−7x2+6x−14$ 54. $x3+x2+x+1x3+x2+x+1$ 55. $x2+xy+5x+5yx2+xy+5x+5y$ 56. $5x3−3x2+5x−35x3−3x2+5x−3$ #### Writing Exercises 57. What does it mean to say a polynomial is in factored form? 58. How do you check result after factoring a polynomial? 59. The greatest common factor of", "gcf of 28 and 32 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 28 and 32 has been useful to you, and make sure to bookmark our site. Posted in Greatest Common Factor ###### 2 comments on “GCF of 28 and 32” 1. POOFA says: WOW! THIS IS A GREAT ANSWER! THANK YOU! 🙂 2. POOFA says: WOW! THIS IS A GREAT ANSWER! THANK YOU! 🙂 ## Related pages write 72 as a product of prime factorsdivisors of 7536-25-36prime factor of 91greatest common factor of 64 and 48the prime factorization of 56prime factorization of 13538-24-38prime factorization of 113what is the prime factorization of 330prime factorization of 254highest common factor of 72 and 96what is the lcm of 7 and 3what is the prime factorization of 225multiplication table chart 1-1000common multiples of 4 and 10prime factorization for 6464 prime factorizationgcf of 40 and 64what is the gcf of 63 and 42prime factorization for 5553 prime factorizationwhat is the prime factorization of 312times table 1-20prime factorization for 240what is the gcf of 9greatest common factor of 42 and 56what is the prime factorization of 196prime factorization of 425easiest way to find lcmprime factorization of 33most common factor calculatorthe", "# How do you find the greatest common factor of 14 42 91? Dec 13, 2016 $7$ is greatest common factor. #### Explanation: First write each number as a product of prime numbers. Here we have $14$, $42$ and $91$ and their prime factors are 14=2×7 42=2×3×7 91=7×13 Now list all the common factors (if any prime is occurring more than one time in each number pick up as many times it is spearing in all the numbers). In above case, however, we have just $7$, which repeats in all. Hence, here $7$ is greatest common factor.", "6. Therefore, the greatest common divisor of 18 and 30 is 6. ### Prime factorization Another way of finding the greatest common divisor of two or more numbers is through prime factorization. In this method, each number is factored so that it is the product of prime numbers. For example, the prime factorization of 1,176 gives us $1,176 = 2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 7 \\cdot 7 = 2^3 \\cdot 3 \\cdot 7^2$. The prime factorization of 2,100 is $2^2 \\cdot 3 \\cdot 5^2 \\cdot 7$. To find the greatest common Divisor of 1,176 and 2,100, the smallest powers of each prime divisor for each of the numbers (with $p^0$ being assumed if one number doesn't contain a particular prime factor $p$) are multiplied together, as in $2^2 \\cdot 3^1 \\cdot 5^0 \\cdot 7^1 = 84$. Therefore, 84 is the greatest common factor of 1,176 and 2,100. ### Euclidean Algorithm For large numbers with many divisors, especially large prime divisors, these methods are somewhat inefficient. A systematic way of finding the greatest common divisor is the Euclidean Algorithm. This algorithm has the advantage of being easily written for a computer program. In the Euclidean Algorithm, the division algorithm is applied to two numbers until a remainder of zero is found. How this works is that", "# How do you find the greatest common factor of 8 20 and 44? Dec 19, 2016 Greatest common factor is $4$ #### Explanation: The prime factors of $8$, $20$ and $44$ are $8 = \\textcolor{red}{2 \\times 2} \\times 2$ $20 = \\textcolor{red}{2 \\times 2} \\times 5$ $44 = \\textcolor{red}{2 \\times 2} \\times 11$ as the common factors among all are $\\textcolor{red}{2 \\times 2}$, Greatest common factor is $2 \\times 2 = 4$", "# 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 ${2}^{4} = 16$.", "# 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 $3 \\times 4 \\times 7 = 84$.", "search ended quickly after remembering 2401 was a perfect square: $2501 = 49^2 + 10^2$ What’s interesting here is we know from this that 2501 is not prime, but we don’t know what its factors are! Having tested and not finding any prime factors less than 17, this is enough to know that 2501 is the product of exactly two prime factors. But which ones? To do this, I tried to find a way to write 2501 as a difference of squares. And again, the search ended much more quickly than I expected: $2501 = 51^2 - 10^2$ $51^2 - 10^2 = (51 + 10)(51 - 10)$ It made me wonder if there are other numbers with the same property, where the number is 1 more than a perfect square, and also has an equal gap to the next square up and next square down… guess what. 26 between 16 and 36 37 between 25 and 49 $n^2 + 1$ between $(n-1)^2$ and $(n+1)^2$ Let me know if you find anything interesting here, or if you have any favorite numerical calculations like these.", "Comparing the factors of $10$ and $21$, we find that they do not have any common factors, other than $1$. Comparing factors of $10$ and $6$, we find that they have a common factor $2$. $10$ and $21$ are called \"co-prime numbers\". $10$ and $6$ are not co-prime numbers, as they share $2$ as the common-factor. Similarly, $6$ and $21$ are not co-prime numbers, as they share $3$ as the common-factor. Co-prime Numbers : Two or more numbers without any common factor, other than $1$, are called co-prime numbers. The prefix \"co\"? means \"together and mutually\". Are $7$ and $7$ co-prime numbers? The answer is \"no\". They have a common factor 7. Are $21$ and $10$ co-prime numbers? The answer is \"Yes, they do not have a common factor\". The factors of $21$ are $1 , 3 , 7 , 21$ and $10$ are $1 , 2 , 5 , 10$. Highest Common Factor Factors of $12$ are $1 , 2 , 3 , 4 , 6 , 12$ and factors of $18$ are $1 , 2 , 3 , 6 , 18$. The common factors between these two are \"$1 , 2 , 3 , 6$\" The highest common factor for these two numbers is \"$6$\" Highest Common Factor : The highest factor in the common factors", "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 $3pq^2$?", "=& {2}^{4}\\cdot 5\\cdot 7\\cdot \\mathrm{11}\\hfill \\end{array}$ 2. The common bases are 2 and 5 3. The smallest exponents appearing on 2 and 5 in the prime factorizations are, respectively, 2 and 1. ${2}^{2}$ from 700. ${5}^{1}$ from either 1,880 or 6,160. 4. The GCF is the product of these numbers. GCF is ${2}^{2}\\cdot 5=4\\cdot 5=\\text{20}$ Thus, 20 is the largest number that divides 700, 1,880, and 6,160 without a remainder. Practice set a Find the GCF of the following numbers. 24 and 36 12 48 and 72 24 50 and 140 10 21 and 225 3 450, 600, and 540 30 Exercises For the following problems, find the greatest common factor (GCF) of the numbers. 6 and 8 2 5 and 10 8 and 12 4 9 and 12 20 and 24 4 35 and 175 25 and 45 5 45 and 189 66 and 165 33 264 and 132 99 and 135 9 65 and 15 33 and 77 11 245 and 80 351 and 165 3 60, 140, and 100 147, 343, and 231 7 24, 30, and 45 175, 225, and 400 25 210, 630, and 182 14, 44, and 616 2 1,617, 735, and 429 1,573, 4,862, and 3,553 11 3,672, 68, and 920 7, 2,401, 343, 16, and 807 1 500, 77, and" ]

[ "integers from 1 to 400.", "how many .", "# Two squares How many integers between 1 and 100 inclusive can be expressed as the sum of 2 perfect squares? ×", "This Question If stands for 300, what does stand for?", "-300, -0]}", "<= number <= 500", "whose sum of digits is divisible by 3 are: 34", "the arithmetic sequence is 120. What will be the sum if the difference is reduced by 2?", "1000 There are 168 primes between 1 and 1000.", "the sum of 75 numbers i. e. 1 + 3 + 5 + 7 + … + 149", "thousand and seventeen", "| 250 | 500)", "### The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is. A. 414 B. 420 C. 620 D. 621 $\\begin{array}{l}\\text{Sum of numbers}\\\\ \\text{= (41.4 × 15)}\\\\ \\text{= 621}\\end{array}$", "sum is $$6+6+32\\times 9=300$$. Our result is confirmed. When $$x=34$$, the sum of the digits of the difference $$10^{34}-34$$ is 300.", "the total number of integers is one greater than that: 150. Second, figure out how many of those 150 integers have a tens digit of 1. In the 100s (100 to 199, inclusive), there are 10 numbers with a tens digit of 1: the integers between 110 and 119, inclusive. The same goes for the 200s: we can include all the integers between 210 and 219, inclusive. That gives us a total of 20 desired outcomes. The probability, then, is desired over possible: p = 20/150 = 2/15 , choice (B). Re: What is the probability that a random number selected from the set of [#permalink] 25 Jan 2019, 04:00 Display posts from previous: Sort by", "many numbers are less than 222 with a digit sum is 8?", "+0 # counting problem 0 26 1 How many integers between 100 and 800 have three different digits in increasing order? One such integer is 129. \\(\\phantom{100 and 150}\\) Jul 2, 2022", "Free Version Moderate # Sum of Digits Divisible by 10 COMBIN-YVEFKG How many 4-digit numbers have the sum of their digits divisible by $10$? A $450$ B $900$ C $1000$ D $4500$ E none of the above", "first 5 members of an arithmetic sequence: a4=-35, a11=-105.", "one thousand and ..." ]

[ "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", "are $10$ possibilities in the case that the hundreds digit is at least $5$. By symmetry, there are also $10$ numbers when the tens digit is at least $5$, and another $10$ when the units digit is at least $5$. These sets are disjoint (have no elements in common), so there are $30$ such integers altogether. Alternatively, we can draw up a further table. We know one digit is 5, 6, 7 or 8, and we know we are adding two extra digits to this (that may be 0). The possibilities for these extra digits are very limited, and it is easy to write down explicitly the possible numbers. 1. What is the total of the digit sums of the integers from $0$ to $999$ inclusive? Each digit from $0$ to $9$ appears $100$ times in the hundreds position. Each digit from $0$ to $9$ appears $10$ times $10$ times in the tens position. Each digit from $0$ to $9$ appears singly $100$ times in the units position. That makes $300$ times altogether for each digit. Therefore the total of the digit sums is $300 \\times (0+1+2+3+4+5+6+7+8+9)= 13500.$", "tens place, and $900$ times in the ones place, for a total of $3700(45)$. Thus, our final sum is $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\\boxed{\\text{A)}180001}.$ Solution 2 Consider the numbers from $0000-9999$. We have $40000$ digits and each has equal probability of being $0,1,2....9$ Our requested sum then is $4000(45)+1=\\boxed{\\text{A)}180001}.$ Credit: Math1331Math", "When you reverse the digits in a certain two-digit number, you decrease its value by 18. What is the number is the sum of its digits is 4? It is $13$ Explanation: Let $x$ and $\\left(4 - x\\right)$ represent the unit and tens digits of this certain two-digit number 10*(4-x)+x=10*x+(4-x)-18=> 40-10x+x=10x+4-x-18=> 40+18-4=10x+10x-2x=> 54=18x=>x=3 Hence the unit digit is 3 the tens unit is 1.So the number is 13. Check :$31 - 13 = 18$", "$3$. 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", "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", "$2^3=8, 3^3=27, 4^3=64, 5^3=125, 6^3=216, 7^3=343, 8^3=512, 9^3=729.$ One important thing to notice in this list is that the units digit of every cube is unique. Since the units digit are unique, if we know what the units digit of $x^3$, we can look at our table above to determine the units digit of $x$. For example, if $x^3 =148,877$, then since $x^3$ has a units digit of $7$, we know the units digit of $x$ is $3$ by looking at the table above. Now that we know the units digit of $x$, all that’s left to determine is the tens digit (since we started with a $2$ digit number). For this, we need to determine what two cubes the digits in the thousands place of $x^3$ lie between. For example, with $148,877$, the digits in the thousands place are $148$. We see that $125<148<216,$ therefore, $50^3=125,000<148,8777<216,000=60^3.$ Hence, we know the tens digit of $x$ is $5$, so our original cube must be $53$. Let’s try another quick example, and then I’ll leave some exercises for the reader. Suppose we want to find the two digit number, which, when cubed gives us $373,248$. We begin by looking at the units digit of our cube, which is $8$. We therefore know the units digit of the two digit number", "$2^3=8, 3^3=27, 4^3=64, 5^3=125, 6^3=216, 7^3=343, 8^3=512, 9^3=729.$ One important thing to notice in this list is that the units digit of every cube is unique. Since the units digit are unique, if we know what the units digit of $x^3$, we can look at our table above to determine the units digit of $x$. For example, if $x^3 =148,877$, then since $x^3$ has a units digit of $7$, we know the units digit of $x$ is $3$ by looking at the table above. Now that we know the units digit of $x$, all that’s left to determine is the tens digit (since we started with a $2$ digit number). For this, we need to determine what two cubes the digits in the thousands place of $x^3$ lie between. For example, with $148,877$, the digits in the thousands place are $148$. We see that $125<148<216,$ therefore, $50^3=125,000<148,8777<216,000=60^3.$ Hence, we know the tens digit of $x$ is $5$, so our original cube must be $53$. Let’s try another quick example, and then I’ll leave some exercises for the reader. Suppose we want to find the two digit number, which, when cubed gives us $373,248$. We begin by looking at the units digit of our cube, which is $8$. We therefore know the units digit of the two digit 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", "tens digit and the units digit of y are the same. _________________ Push +1 kudos button please, if you like my post Math Expert Joined: 02 Sep 2009 Posts: 58446 Re: If both x and y are positive integers less than 100 and [#permalink] ### Show Tags 07 Nov 2012, 03:28 2 2 If both x and y are positive integers less than 100 and greater than 10, is the sum x + y a multiple of 11? (1) x - y is a multiple of 22. If x=33 and y=11, then the answer is YES but if x=34 and y=12, then the answer is NO. Not sufficient. (2) The tens digit and the units digit of x are the same; the tens digit and the units digit of y are the same. Note that any two-digit integer can be represented as 10a+b (wher a and b are single digit integers), for example 37=3*10+7, 88=8*10+8, etc. Thus, we are given that x=10a+a=11a and y=10b+b=11b --> x+y=11a+11b=11(a+b). Therefore x+y is a multiple of 11. Sufficient. OR: from (2) we have that both x and y must be multiples of 11 (11, 22, 33, 44, ..., 99). The sum of two multiples of 11 will give a multiple of 11. Sufficient., GENERALLY: If integers $$a$$ and $$b$$ are both multiples", "173 views Selection of how many integers from the first ten positive integers (1, 2, ...) guarantees that there must be a pair of these integers with a sum equal to 11 ? my appr. Possible(x, y)solutions : (1, 10), (2, 9), (3, 8), (4, 7), (5, 6) so total 5 solution where i am wrong? In the worst case, we can select $5$ numbers such that sum of any pair of numbers is not equal to $11$, and selecting one more number will guarantee that sum of some pair of numbers is equal to $11$. $\\therefore answer = 6$. @shishir__roy can you give any example? It's simply pigeon hole principle If u choose randomly any 5 numbers and add 1 then surely there will be a pair whose sum will be 11 i.e. 1,2,3,4,5,6 there is (5,6) So ans will be 6 Among numbers 1 to 10, there are 5 pairs that corresponds to the sum 11. They are (10,1),(9,2),(8,3),(7,4) and (6,5). We can create a run of selections without any pair corresponding to the sum 11 upto a max of 5 numbers. For eg. the set {1,2,3,4,5} does not have any pairs summing up as 11. But in this set, if we add another number from the set {1,..10} – {1,2,3,4,5}, then there will be a", "## Elementary Technical Mathematics (a) Hundred: $1, 500$ (b) Ten: $1, 450$ (c) Unit: $1, 452$ (d) Tenth: $1, 451.5$ (e) Hundredths: $1, 451.53$ (f) Thousandths: $1, 451.525$ (a) Hundred: $1, 500$ Explanation: The hundreds digit $4$ is followed by a $5$ so round the hundreds digit up to $5$ then replace the tens and units digit with $0$ and omit the decimals, (b) Ten: $1, 450$ Explanation: The tens digit $5$ is followed by a $1$ so retain the tens digit $5$ then replace the units digit with $0$ and omit the decimals, (c) Unit: $1, 452$ Explanation: The units digit $1$ is followed by a $5$ so round up the units digit to $2$ then omit the decimals, (d) Tenth: $1, 451.5$ Explanation: The tenths digit $5$ is followed by a $2$ so retain the tenths digit $5$ then omit the succeeding decimals, (e) Hundredths: $1, 451.53$ Explanation: The hundredths digit $2$ is followed by a $5$ so round up the hundredths digit to $3$ then omit the succeeding decimals, (f) Thousandths: $1, 451.525$ Explanation: The thousandths digit $5$ is followed by a $4$ so retain the hundredths digit, $5$, then omit the succeeding decimals.", "SquareNumber Basic Problem - 4141 If a square number's tens digit is $7$, what is its units digit? The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "can conclude that the tens digit of the square of any natural number ending in $$4$$ or $$6$$ is always odd. In any other case, it is even. The problem states that the product of the tens digits of a few given perfect squares is odd. Therefore, the tens digit of each square must be odd, and thus the units digit of each square must be $$6$$. The answer is $$6$$. ### Credit This is an original puzzle from cotpi.", "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 [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 27494 Followers: 4312 Kudos [?]: 42302 [0], given: 6012 Re: The three-digit positive integer x has the hundreds, [#permalink] 15 Nov 2013, 07:09 Expert's post 2 This post was BOOKMARKED 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 $$c-m=0$$ --> $$x-y=210$$. _________________ Math Expert Joined: 02 Sep 2009 Posts: 27494 Followers: 4312 Kudos [?]: 42302 [0], given: 6012 Re: The three-digit positive integer x has the hundreds, [#permalink] 15 Nov 2013, 07:13 Expert's post 1 This post was BOOKMARKED Bunuel wrote: mohnish104 wrote: The", "99u = 198$. Dividing both sides by $99$, we have $h - u = 2$ –> (Equation 3). We can eliminate $u$ by adding Equation 1 and Equation 3. $(h + t + u) + (h - u) = 10 + 2$ $2h + t = 12$ (*). We substitute $t + 3$ from Equation 2 to $h$ in (*) $2(t + 3) + t = 12$ $2t + 6 + t = 12$ $3t + 6 = 12$ $3t = 6$ $t = 2$. Substituting the value of $t$ in Equation 2, $h = t + 3$ $h = 2 + 3$ $h = 5$. Substituting the values of $h$ and $t$ in Equation 1, $h + t + u = 10$ $5 + 2 + u = 10$ $7 + u = 10$ $u = 10 - 7$ $u = 3$. So, the hundreds digit is 5, tens digit is 2, and ones digit is 3. Therefore, the number is 523. If we check with the conditions above, we have (1) The sum of the digits is 10. That is, 5 + 2 + 3 = 10 (2) The hundreds digit is 3 more than the tens digit. The hundred digit is 5 is 3 more than 2 (3) If 198 is subtracted from the number,", "a 7! from all of the terms might be a way to start. The next thought that my older son had was that finding the units digit of this sum would be fairly easy. This idea is, at least in my mind, an important first step to solving the problem because the reason that finding the units digit is fairly easy is essentially the same reason that finding the tens digit isn’t too difficult. Making the transition to the next step was a little difficult, though I broke the conversation at about 5 min just to start a new video. When we restarted the conversation I reminded them of one of James Tanton’s strategies – do **something**. My older son suggested that something we could do would be to make a simpler problem. His suggestion for a simpler problem was to find the tens digit of the sum $7! + 8!$. Making a simpler problem is often a great way to see what’s going on with the more complicated problem, and the simpler problem chosen by my son got us going down the right path. We stayed on the path and calculated the units digit of the sums $7! + 8!$, $7! + 8! + 9!$, and $7! + 8! + 9! + 10!$. Once the kids saw", "a 7! from all of the terms might be a way to start. The next thought that my older son had was that finding the units digit of this sum would be fairly easy. This idea is, at least in my mind, an important first step to solving the problem because the reason that finding the units digit is fairly easy is essentially the same reason that finding the tens digit isn’t too difficult. Making the transition to the next step was a little difficult, though I broke the conversation at about 5 min just to start a new video. When we restarted the conversation I reminded them of one of James Tanton’s strategies – do **something**. My older son suggested that something we could do would be to make a simpler problem. His suggestion for a simpler problem was to find the tens digit of the sum $7! + 8!$. Making a simpler problem is often a great way to see what’s going on with the more complicated problem, and the simpler problem chosen by my son got us going down the right path. We stayed on the path and calculated the units digit of the sums $7! + 8!$, $7! + 8! + 9!$, and $7! + 8! + 9! + 10!$. Once the kids saw", "positive integer greater than 2, n=? [#permalink] Show Tags 25 Jun 2013, 03:32 2 3 stunn3r wrote: If n is a positive integer greater than 2, n=? (1) The tens' digit of 11^n is 4. (2) The hundreds' digit of 5^n is 6. The units digit of $$11^n$$ for any positive integer (n) is always 1. Also, the tens digit of $$11^n$$, follows a cyclic order, and assumes all the values from 0-9. $$11^1 = 11$$ $$11^2 = X21$$ $$11^3 = XX31$$ $$11^4 = XXX41$$ and so on The tens and units digit for $$5^n$$, is always 25, for$$n\\geq2$$. The hundreds digit keeps switching between 1 for n = odd OR as 6 for n = even From F.S 1, we would have the tens digit as 4 for n = 4, and then again for n = (4+10) = 14. Insufficient. From F. S 2, we would have the hundreds digit as 6 for n = 4,6,8 and so on. Insufficient. Even after combing both fact statements, n could be n = 4 or n = 14. Insufficient. E. _________________ General Discussion Manager Joined: 22 Jan 2014 Posts: 176 WE: Project Management (Computer Hardware) Re: If n is a positive integer greater than 2, n=? [#permalink] Show Tags 27 Sep 2014, 05:38 1 stunn3r wrote: If n", "integer and its tens digit is 4 times its units digit, what is the value of n? 1) The tens digit of n is 8. 2) The sum of the tens digit and the units digit of n is 2-digit integer. St1:- The tens digit of n is 8. Given, tens digit is 4 times its units digit. So, units digit is $$\\frac{8}{4}=2$$ Therefore, the 2-digit positive integer, n is 82. Sufficient. St2:-The sum of the tens digit and the units digit of n is 2-digit integer. If units digit is y and tens digit is x, then n=10x+y. n=x y or 4y y Since, x+y is a 2-digit integer, we have y=2 as unit digit. So, tens digit=4y=4*2=8. So, n=10x+y=10*8+2=82 Sufficient. Ans. (D) Hi PKN can you please elaborate on statement TWO St2:-The sum of the tens digit and the units digit of n is 2-digit integer. If units digit is y and tens digit is x, then n=10x+y. n=x y or 4y y Since, x+y is a 2-digit integer, we have y=2 as unit digit. So, tens digit=4y=4*2=8. So, n=10x+y=10*8+2=82 are you using remainder formula ? n=10x+y. X is quotient and Y is remainer ? why ? any other method to explain second statement? you explanation is kinda hard to understand thanks problem solver killer i" ]

[ "are $10$ possibilities in the case that the hundreds digit is at least $5$. By symmetry, there are also $10$ numbers when the tens digit is at least $5$, and another $10$ when the units digit is at least $5$. These sets are disjoint (have no elements in common), so there are $30$ such integers altogether. Alternatively, we can draw up a further table. We know one digit is 5, 6, 7 or 8, and we know we are adding two extra digits to this (that may be 0). The possibilities for these extra digits are very limited, and it is easy to write down explicitly the possible numbers. 1. What is the total of the digit sums of the integers from $0$ to $999$ inclusive? Each digit from $0$ to $9$ appears $100$ times in the hundreds position. Each digit from $0$ to $9$ appears $10$ times $10$ times in the tens position. Each digit from $0$ to $9$ appears singly $100$ times in the units position. That makes $300$ times altogether for each digit. Therefore the total of the digit sums is $300 \\times (0+1+2+3+4+5+6+7+8+9)= 13500.$", "800 Q51 V49 GRE 1: 340 Q170 V170 Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink] ### Show Tags 04 Jun 2017, 15:06 2 KUDOS Expert's post 1 This post was BOOKMARKED Hi All, These types of 'bunching' questions often come down to how you choose to 'organize' the math involved. Here, we're asked for the sum of all of the integers from 1 to 999, inclusive. We're talking about the first 999 positive integers, but I'm going to make the math a little easier by adding one extra integer that won't affect the sum at all: the number 0. By including 0, we now have 1,000 total integers and we can 'bunch' them into groups of 2: 0 and 999 = 999 1 and 998 = 999 2 and 997 = 999 Etc. Since we have 1,000 total numbers, there will be 500 'pairs' that add up to 999. Thus, the sum of those terms is... (500)(999) While that calculation might seem a little tough, you can avoid it if you think about the math in a different way... (500)(1000) = 500,000 999 is \"one less\" than 1,000 so we just have to subtract \"one\" of those 500s from 500,000. That gives us: 500,000 - 500 = 499,500 [Reveal] Spoiler:", "the total number of integers is one greater than that: 150. Second, figure out how many of those 150 integers have a tens digit of 1. In the 100s (100 to 199, inclusive), there are 10 numbers with a tens digit of 1: the integers between 110 and 119, inclusive. The same goes for the 200s: we can include all the integers between 210 and 219, inclusive. That gives us a total of 20 desired outcomes. The probability, then, is desired over possible: p = 20/150 = 2/15 , choice (B). Re: What is the probability that a random number selected from the set of [#permalink] 25 Jan 2019, 04:00 Display posts from previous: Sort by", "# 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. ×", "# How many integers between $1000$ and $9999$ inclusive have distinct digits and are odd numbers? I know this question was already answered (How many integers between 1000 and 9999 inclusive consist of), and I understand the solution. However, I don't understand what's wrong with my approach. I did it from right to left. My reasoning: Last digit has to be odd, so we have $$5$$ possibilities. In the tenths place, we can choose whatever we want except for the one already chosen, so we have $$9$$ possibilities. Similarly, in the hundredths place we have $$8$$ choices. In the thousands place, we can choose $$10 - 3 - 1$$ (not including $$0$$) $$= 6$$ possibilities, so in total we have $$5 * 9 * 8 * 6 = 2160$$. Why am I undercounting here? Because if you've already chosen $$0$$ before (in the tens or hundreds place), then you're excluding $$0$$ twice in allowing only $$10-3-1=6$$ options for the thousands place. To compensate this, you could count the number of cases where you have a $$0$$ in the tens or hundreds place. But as in the linked answer, it is easier to just start with the thousands place.", "is the sum of first 250 odd numbers. The sum of the integers from 1 to 300 is as follows: 45,150 To get the answer above, you could add up all the digits like 1+2+3... +300, but there is a much easier way to do it! For example, if the lower limit is 1 and the upper limit is 10, this means that the sum operation is to be performed by replacing the variable (e.g. Julie S. Syracuse University . Put them in pairs: 1 & 500 2 & 499 3 & 498 etc There are 250 pairs, each pair = 501 250*501 = 125250-----How about only the odds: 1+3+5+7+...493+495+497+499? What is the sum of first 500 natural numbers? step 1 Address the formula, input parameters & values.Input parameters & values:The number series 1, 3, 5, 7, 9, . Find the sum of all multiples of 4 that are between 15 and 521. What is the sum of the first 500 counting numbers? . . Sum of the natural numbers using summation In this scheme, we first fall flat on our faces. The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. let the numbers be x & y x+y = 22 x= 22-y x^2+y^2=250.. plug", "tens place, and $900$ times in the ones place, for a total of $3700(45)$. Thus, our final sum is $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\\boxed{\\text{A)}180001}.$ Solution 2 Consider the numbers from $0000-9999$. We have $40000$ digits and each has equal probability of being $0,1,2....9$ Our requested sum then is $4000(45)+1=\\boxed{\\text{A)}180001}.$ Credit: Math1331Math", "so 1000*1001/2= 500500. Divide it by 9 to get the remainder 1. Nitin () but sum of first n natural number is adding $1+2+3+4+5+6+7+8+9+10+11....+1000$ here 10,11...till 1000 are the addition of 2 and three digits numbers but in the question its has been asked for the sum of all the digits like $1+2+3+4+5+6+7+8+9+1+0+1+1.....1+0+0+0$ therefore these two are different cases..", "and z add to more than 9. For example, 600 + 900 = 1500. HOWEVER, we can rule out this scenario because we're told that x, y, and z are three-digit integers Scenario #2: the tens digits of y and z add to more than 9. For example, 141 + 172 = 313. Scenario #3: the tens digits of y and z add to 9, AND the units digits of y and z add to more than 9. For example, 149 + 159 = 308 Statement 1: The tens digit of x is equal to the sum of the tens digits of y and z. This rules out scenarios 2 and 3 (plus we already ruled out scenario 1). So, it must be the case that the hundreds digit of x equals to the sum of the hundreds digits of y and z Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: The units digit of x is equal to the sum of the units digits of y and z. This rules out scenario 3, but not scenario 2. Consider these two conflicting cases: Case a: y = 100, z = 100 and x = 200, in which case the hundreds digit of x equals the sum of the hundreds digits of", "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", "# Thread: the tens digit of 16^298 .... 1. ## the tens digit of 16^298 .... Please someone tell me how to work out the TENS digit of the answer of $16^{298}$ WITHOUT using calculator. I believe that it is somehow related to the principles such as $(x^a)^b = x^{ab}$ the original question also included the units digit but it should always be 6. 2. $16^{298} = (2^4)^{298} = (2)^{4\\times 298} =2^{1192}$ How to Find the Last Digits of a Positive Power of Two - Exploring Binary 3. Originally Posted by ukorov Please someone tell me how to work out the TENS digit of the answer of $16^{298}$ WITHOUT using calculator. I believe that it is somehow related to the principles such as $(x^a)^b = x^{ab}$ the original question also included the units digit but it should always be 6. A few calculations (you should be able to multiply these easily w/o the abacus) 016 = 16^1 x16 === _56 = 16^2 (16 times 16 = 256, but we only need the tens & ones digits) x16 === _96 = 16^3 x16 === _36 = 16^4 x16 === _76 = 16^5 x16 === _16 = 16^6 (same as 16^1, exponent difference is 5) $16^6 \\equiv 16^1 \\mod 100$ 298/5 = 59.9 59*5 = 295 1+5=6 1+295=296 $16^1 \\equiv", "can we make the sum be 350? 250? There's going to be \"gaps\". Let's start at the bottom: 1+2+3+4 = 10 1+2+3+5 = 11 1+2+3+6 = 12 2+3+4+5 = 14 2+3+4+6 = 15 2+4+5+6 = 16 3+4+5+6 = 18 1+2+3+95 = 101 <---big gap, here, in the first 92 \"possible\" sums, only 8 actually occur. I'm too lazy to work out the details, but I suspect the \"largest gap\" between successive numbers of the 12 element set, controls how many sums we can leave out, and if this gap is small (like 1, or 2), the range of sums is severely curtailed. 8. ## Re: Subsets question Originally Posted by Deveno We aren't summing 12 numbers, we are only summing 4 out of a set of 12 It doesn't have to be of size 4, the subset could be of any size. Like in the example given: {21,22,60}, {1,4,5,22,31,40}, {4,99}, {4,5,21,73} all sum up to 103, but they're not all size 4. Does this change the proof? 9. ## Re: Subsets question Yes! It greatly increases the number of subsets we have to choose from to 450 to 4096, in which case since we only have 385 numbers to choose as sums, after choosing 1155 subsets (which conceivably might line up nicely as 3 sets each corresponding to", "and 200, they are three digit numbers of which the first must be 1. Here, we will not only tell you what the sum of integers from 1 to 300 is, but also show you how to calculate it fast. Find the numbers. + 499 = 62500. The numbers in the triangular pattern are represented by dots. . , 249, 250. Tn=a+(n-1).d 49 = 1+(n-1).2 or. let the numbers be x & y x+y = 22 x= 22-y x^2+y^2=250.. plug value of x in the equation What is the sum of first 500 natural numbers? Pair them up like above. What is the sum of first 450 natural numbers? From this we need to subtract the sum of 1 plus all the prime numbers below 100. 1729 = 1 3 + 12 3 = 10 3 + 9 3. step 2 apply the input parameter values in the AP formulaSum = n/2 x (a + Tn) = 250/2 x (1 + 499) = (250 x 500)/ 2 = 125000/21 + 3 + 5 + 7 + 9 + . the difference between each number and the next is the same in all cases, the average of all the numbers in the list must be the same as the average of the first one and the last one: $$\\frac{252+350}", "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", "# Show that if seven integers are selected from the first 10 positive integers a) Show that if seven integers are selected from the first 10 positive integers, there must be at least two pairs of these integers with the sum 11. b) Is the conclusion in part (a) true if six integers are selected rather than seven? I don't know how should I show that. I know that in the worst situation we choose 1,2,...,7 that we have 7 + 4 = 11 and 5 + 6 = 11 but I don't know what should be the answer of this question Partition numbers from $1$ to $10$ into the following sets: $\\{1,10\\}$, $\\{2,9\\}$, $\\{3,8\\}$, $\\{4,7\\}$ and $\\{5,6\\}$. Now use pigeon hole principle. • It's a little tricky in that we are asked to show two pairs have sum $11$, and the pigeonhole principle gives one pair by immediate application (even if as few as six integers are selected). The difference between selecting seven and six integers from $1$ to $10$ is that we get (at least) two pairs with sum $11$ in the former case and as few as one pair in the latter case. – hardmath Dec 1 '13 at 16:58", "the sum of natural numbers. There are three points I’d like to mention The highest bit of a negative integer is always 1. ➡️. additionally as some that are geared toward before or after individually. The sum of the first n numbers is equal to: n(n + 1) / 2. 3 Short Answer Type Questions . 1. #3. Their sum is 8n+16=136. (13) Find the sum of all natural numbers between 300 and 500 which are divisible by 11. But we only want the sum of one row, not both. sum is 4+5+6 =15. It represents the sum of the proper divisors of n, excluding n itself. Now we will learn how to get the query for sum in multiple columns and for each record of a table. pick3 numbers, pin-codes, permutations) 300 (~ 300. The sum of the integers from 1 to 300 is as follows: 45,150 To get the answer above, you could add up all the digits like 1+2+3 +300, but there is a much easier way to do it! Use the following formula: n(n + 1)/2 = Sum of Integers In this case, n=300, thus you get your answer by entering 300 in the formula like this: 300(300 + 1)/2 = 45,150 Jun 12, 2015 · Input upper limit to find sum of odd", "$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", "all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed) A) 444440 B) 610000 C) 666640 D) 711040 E) 880000 Conceptually, this problem isn’t much different from the previous one. Using the same basic counting principle to get the number of integers possible, the first digit can be chosen in 4 ways, the next one in 4 ways, the next one in again 4 ways and finally the last digit in 4 ways. This is what some of the numbers will look like: 1111 1112 1121 and so on till 4444. As such, we will get a total of $$4*4*4*4 = 256$$ different integers. Now we need to add these 256 integers to get their sum. Since each digit has an equal probability of occupying every place, out of the 256 integers, 64 integers will have 1 in the units place, 64 will have 2 in the units place, another 64 integers will have 3 in the units place and the rest of the 64 integers will have 4 in the units place. The same is true for all places – tens, hundreds and thousands. Therefore, the total sum will be: $$64*1000 + 64*2000 + 64*3000 + 64*4000 + 64*100 + 64*200 + … + 64*3 + 64*4$$ $$= 1000*(64*1 + 64*2", "a short method too: Sum (S)=(1+2+3…1000) First term (a)= 1. We can get SI of 3 years = 12005 - 9800 = 2205. It is simple to do for a few numbers, especially integers, but can get. Sum of first 100 integers?. The multiplication rule of logarithm states that ln A/b = ln A - ln B. 22 chromosomes of realistic lengths, simulated under Wright-Fisher model (top) and coalescent (bottom), compared to the analytical expectation under Eqs (1) and (2) in S3 Appendix. Another example is the famous 1 - 1 + 1 - 1 + 1 - where units are subtracted and added intermittently. Let your math skills blossom Unlock Step-by-Step. 50005000 is a sum of number series from 1 to 10000 by applying the values of input parameters in the formula. For example, you've invested$10,000 in a money market fund. For example, add one and 10 together to get 11. Write a program to find the sum of the first 1000 prime numbers. Over $10,000, 2. Also, list the total value of the units on hand. Prime numbers are positive integers greater than 1 that has only two divisors 1 and the number itself. We need to put on our calculus hats now, since in order to maximize the function, we are going to need to", "not work. Proof: 1. There are 15 pairs of numbers from a set of size 6. 2. The smallest and largest sum of a pair from $\\{1, 2,, 3, 4, 5, 6, 7, 8, 9\\}$ are 3 and 17. 3. There are 15 different integers in the range from to 3 to 17 inclusive. 4. So if a set of size 6 works, call it T, T's sums must consist of all 15 integers from 3 to 17. 5. In order to get the sums $3, 4,\\text { and } 6$, T must contain $1, 2, 3,\\text { and } 4$ or $1, 2, 3,\\text { and } 5$. In the first case $1+4=2+3$ so T must contain $1, 2, 3,\\text { and } 5$. 6. Similarly in order to get a sum of $9$ we must have $8 \\in T$ so $T=\\{1, 2, 3, 5, 8\\}$. 7. The only numbers left that could be added to T are $4, 6, 7, \\text { and } 9$ but each of them would give T non-unique sums. $1+4=2+3;\\; 1+6=2+5;\\;1+7=3+5;\\;1+9=2+8$. So T cannot have 6 elements. EDIT: Simpler version, same approach. New step 5. In order to get sums $3 \\text { and } 17$, T must contain $1, 2, 8, 9$. But then $1+9=2+8$ so T cannot have 6" ]

[ "hypotenuse. 10. Isosceles trapezoid What is the height of an isosceles trapezoid, the base of which has a length of 11 cm and 8 cm and whose legs measure 2.5 cm?", "# 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:", "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 $48$", "two shorter legs of a right triangle add up to 40 u &nbs [#permalink] 07 Feb 2018, 11:23 Display posts from previous: Sort by", "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", "perimeter of a right triangle with the legs 14 cm and 21 cm long?", "## anonymous one year ago The legs of a right triangle are 10 centimeters and 24 centimeters long. What is the length of the hypotenuse? 22 centimeters 26 30 centimeters 34 centimeters |dw:1438781834823:dw| use pythagorean theorem $x^2+y^2 = r^2 \\implies r = \\sqrt{x^2+y^2}$", "kyerkes2004 21 # The length of one leg of a right triangle is 21 ft and the length of the hypotenuse is 29 ft. What is the length of the other leg? The length of the other leg is 20 ft. a^2 x b^2 = c^2 21^2 x b^2 = 29^2 441 x b^2 = 841 b^2 = 841-441 b^2 = 400 b = $\\sqrt{400}$ b = 20", "# The lengths of two legs of a right triangle are 5 and 12. What is the length of the hypotenuse? $13$ $13$ Because $5 , 12 , 13$ is a special right triangle.", "of the triangle is 129.5 cm2. ∴ 7.4h = 129.5 h = $\\left(\\frac{129.5}{7.4}\\right)$ = 17.5 cm ∴ Length of the other leg = 17.5 cm Question 7: Find the area of a right triangle whose base is 1.2 m and hypotenuse 3.7 m. Here, base = 1.2 m and hypotenuse = 3.7 m In the right angled triangle: Perpendicular = $\\sqrt{\\left(H\\mathrm{ypotenuse}{\\right)}^{2}-\\left(B\\mathrm{ase}{\\right)}^{2}}$ $=\\sqrt{{\\left(3.7\\right)}^{2}-{\\left(1.2\\right)}^{2}}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{13.69-1.44}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{12.25}\\phantom{\\rule{0ex}{0ex}}=3.5$ Area = $\\left(\\frac{1}{2}×\\mathrm{base}×\\mathrm{perpendicular}\\right)$ sq. units = $\\left(\\frac{1}{2}×1.2×3.5\\right)$ m2 ∴ Area of the right angled triangle = 2.1 m2 Question 8: The legs of a right triangle are in the ratio 3 : 4 and its area is 1014 cm2. Find the lengths of its legs. In a right angled triangle, if one leg is the base, then the other leg is the height. Let the given legs be 3x and 4x, respectively. Area of the triangle = $\\left(\\frac{1}{2}×3x×4x\\right)$ cm2 ⇒ 1014 = (6x2) ⇒ 1014 = 6x2 x2 = $\\left(\\frac{1014}{6}\\right)$ = 169 x = $\\sqrt{169}$ = 13 ∴ Base = (3 $×$ 13) = 39 cm Height = (4 $×$ 13) = 52 cm Question 9: One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its cost at the rate of Rs 250 per m2. Consider a right-angled triangular scarf (ABC). Here, ∠B= 90° BC = 80", "# Semi-circle Geometry Level 3 In a right triangle, a semi-circle is inscribed so that its diameter lies on the hypotenuse and its center divides the hypotenuse in two segments of lengths 15 and 20.the length of the arc of the semicircle between the points at which the legs touch the semicircle is $$K \\pi$$. Then find the numerical value of $$K$$. ×", "• Question 8: 3 pts Find to the nearest degree the measure of the angle marked as $x$. x=15o x=16o x=17o x=18o • Question 9: 3 pts The legs of the right triangle measure 30cm and 45cm. Find to the nearest degree the measure of the smallest angle of this triangle. $x=30^\\circ$ $x=32^\\circ$ $x=34^\\circ$ $x=36^\\circ$", "a right triangle you know the legs are 32 and 24 units long. Calculate the hypotenuse length Solution:", "share the same side that's relative to the same angle)... Senior Manager Joined: 13 May 2013 Posts: 472 Followers: 2 Kudos [?]: 130 [0], given: 134 Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink] ### Show Tags 11 Dec 2013, 09:36 If AD is 6 and ADC is a right angle, what is the area of triangular region ABC? (1) Angle ABD = 60° If ABD is 60 then we know triangle ABD = 30:60:90 triangle. Coupled with the one measurement we are given we can find the length of the other two sides of this triangle. However, we know nothing about AC or DC, only that triangle ADC is right - we don't know the measurements of the angles so we cannot find their leg lengths. For example, DC could be six inches long or 20. We don't know and therefore cannot find the area of this triangle. Insufficient. (2) AC = 12 We have a leg and a hypotenuse of a right triangle so we can find the length of the second leg but we know nothing about triangle ABD, except for one leg length. Insufficient. 1+2) 1 Gives us the length of all 3 sides of ABD. 2 gives us the length of all three sides", "# 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.", "Median in right triangle In the rectangular triangle ABC has known the length of the legs a = 15cm and b = 36cm. Calculate the length of the median to side c (to hypotenuse). 10. Right triangle Right triangle ABC with side a = 19 and the area S = 95. Calculate the length of the remaining sides. 11. Center traverse It is true that the middle traverse bisects the triangle?", "## anonymous 4 years ago The legs of a right triangle have lengths of 4 and 5 units. Find the length of the hypotenuse. 1. anonymous $\\sqrt{41}$ 2. anonymous |dw:1327603733335:dw|", "### If the two legs of an iscoles right triangle measure 7 inches, which of the following represents the length of its hypotnenus if the two legs of an iscoles right triangle measure 7 inches, which of the following represents the length of its hypotnenus...", "Problem From the right triangle ABC shown below, AB = 40 cm and BC = 30 cm. Points E and F are projections of point D from hypotenuse AC to the perpendicular legs AB and BC, respectively. How far is D from AB so that length EF is minimal? Solution 0 likes", "a leg. 1. ### geometry is it possible to draw a triangle whose sides measure 5 inches, 9 inches, and 14 inches? 2. ### 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 3. ### Math The hypotenuse (side C) of a triangle is 13 inches long. Which of the following pairs of measurements could be correct for the lengths of the other two sides of the triangle? (Note: N + 82 =C2) A. 2.5 inches,4 inches B. 2.5 4. ### Algebra 2 Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. Find the sum of a finite geometric series. The sides of an equilateral triangle measure 16 inches. The midpoints of the" ]

[ "be acquired from the Pythagorean Theorem. We'll call d the distance, or length of the hypotenuse. ${54}^{2} + {55}^{2} = {d}^{2}$ $5941 = {d}^{2}$ $d \\approx 77.1$ Secondly, we need to find the direction, or the angle formed by the hypotenuse and the positive x axis counterclockwise. This measurement should be (in this case) more than 180 degrees. Since neither a, b, nor c provides a direction of over 180 degrees, the answer must be d.", "that is 2 more than the double of the width and that has a hypotenuse that measures $\\sqrt{68}$ meters. Good luck!", "0.7071 is for a 1m hypotenuse. For a hypotenuse of 2m the adjacent would be twice as long. Therefore 0.7071times2 = 1.4142 theta=45^@,", "# From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC. 31 views closed From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC. +1 vote by (48.8k points) selected by In ∆ABC, ∠B = 90° , and l(BC) = 21, and l(AB) = 20 ∴ According to Pythagoras’ theorem, ∴ l(AC)= l(BC)+ l(AB) ∴ l(AC)= 21+ 20 ∴ l(AC)= 441 + 400 ∴ l(AC)= 841 ∴ l(AC)= 29 ∴ l(AC) = 29 Perimeter of ∆ABC = l(AB) + l(BC) + l(AC) = 20 + 21 + 29 = 70 ∴ The length of hypotenuse AC is 29 units, and the perimeter of ∆ABC is 70 units.", "the height is $(10-h)$ and the hypotenuse is $10$. I hope you'll forgive me for my dreadful artistic skills, but I have done a diagram here: Spoiler: I'll check this in a second, it's sort of a leap of faith on my part, but hopefully you should get the expected result. Edit: Yeah, I've managed to obtain their solution this way.", "Pythagorean Theorem to find the length of the hypotenuse. 15 25 In the following exercises, use the Pythagorean Theorem to find the length of the leg. Round to the nearest tenth if necessary. 8 12 10.2 9.8 In the following exercises, solve using a geometry formula. The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width. 18 meters, 11 meters The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width. The width of the rectangle is 0.7 meters less than the length. The perimeter of a rectangle is 52.6 meters. Find the dimensions of the rectangle. $$13.5$$ m, $$12.8$$ m The length of the rectangle is 1.1 meters less than the width. The perimeter of a rectangle is 49.4 meters. Find the dimensions of the rectangle. The perimeter of a rectangle of 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle. 25 ft, 50 ft The length of the rectangle is three times the width. The perimeter of a rectangle is 72 feet. Find the length and width of the rectangle. The length of the rectangle is three meters less than twice the width.", "the ruler. If it is a special ruler with a mark at $\\sqrt{2}$ cm, then it will measure the exact length of the hypotenuse.", "× # Pythagorean theorem: find the length of the hypotenuse What is the length of the hypotenuse? x=length=?? height=9 base=6 Round to the nearest tenth if needed. Note by Kanya Shah 2 years ago Sort by: The answer is sqrt(81+36) , which is equal to sqrt(117) , which is approximately 10.81. · 1 year, 11 months ago yea i rounded. · 1 year, 11 months ago What does it mean? · 1 year, 11 months ago uhhhhhhhhhhhhh..... yea means yes · 1 year, 11 months ago", "the length of the hypotenuse is $\\sqrt{2}$, and if we call the length of the hypotenuse $1$, the lengths of the sides would be $\\frac{1}{\\sqrt{2}}$. It is true that when defining trigonometric functions on a circle, the hypotenuse corresponds to the radius and it usually makes sense to call it $1$, but the results are the same either way. – Alfonso Fernandez Feb 24 '13 at 21:09 As in Alfonso Fernandez's answer, the remarkable values in your diagram can be calculated with basic plane geometry. Historically, the values for the trig functions were deduced from those using the half-angle and angle addition formulae. So since you know 30°, you can then use the half-angle formula to compute 15, 7.5, 3.25, 1.125, and 0.5625 degrees. Now use the angle addition formula to compute 0.5625° + 0.5625° = 2*0.5625°, and so on for 3*0.5625°, 4*0.5625°... These would be calculated by hand over long periods of time, then printed up in long tables that filled entire books. When an engineer or a mariner needed to know a particular trig value, he would look up the closest value available in his book of trig tables, and use that. Dominic Michaelis points out that in higher math the trig functions are defined without reference to geometry, and this allows one to come up", "## anonymous 3 years ago Find the length of the hypotenuse of a right triangle with legs of 20 cm and 21 cm. 41 cm 841 cm 6 cm 29 cm 1. anonymous $H= \\sqrt{(20)^2+(21)^2}$ 2. anonymous |dw:1352146014111:dw| 3. anonymous |dw:1352138840805:dw|$c^2=a^2+b^2$ 4. anonymous so i have to find out what H is? 5. anonymous yes 6. anonymous $H= \\sqrt{(20^2)+(21^2)}$ 7. anonymous 20 * 20 = 400 21 * 21 = 441 add those 400 + 441 then take the sqrt and that's the missing angle 8. anonymous 841cm 9. anonymous $\\sqrt{841cm}$ sqrt it and the answer is? 10. anonymous so then 29cm 11. anonymous yes ma'am", "that 28² + 29² – 1 = 2(812) = 2(28 × 29) Since 29 is one of its factors, 812 is also the hypotenuse of a Pythagorean triple: • 560-588-812 which is 28 times 20-21-29.", "× # Pythagorean theorem: find the length of the hypotenuse What is the length of the hypotenuse? x=length=?? height=9 base=6 Round to the nearest tenth if needed. Note by Kanya Shah 2 years, 2 months ago Sort by: The answer is sqrt(81+36) , which is equal to sqrt(117) , which is approximately 10.81. · 2 years, 1 month ago yea i rounded. · 2 years, 1 month ago What does it mean? · 2 years, 1 month ago uhhhhhhhhhhhhh..... yea means yes · 2 years, 1 month ago the answer is 10.8 · 2 years, 2 months ago ×", "# Primitive Pythagorean triples with the least, same hypotenuse or side #### Loren What are the primitive Pythagorean triples with the least, same hypotenuse or side? #### Loren Say, where hypotenuses of two differing primitive Pythagorean triples are equal. #### mrtwhs \\begin{align} 1105^2 &=47^2+1104^2\\\\ &=264^2+1073^2\\\\ &=576^2+943^2\\\\ &=744^2+817^2 \\end{align} It took google 0.62 seconds to find these four. Loren #### Loren mrtwhs, What is the smallest such hypotenuse with just two pairs of legs? Last edited: #### mrtwhs Don't know if it is the smallest but google gives $33^2+56^2 = 16^2 +63^2 = 65^2$ #### skipjack Forum Staff It is the smallest. The smallest pair with a leg in common is ((5, 12, 13), (12, 35, 37)). Loren", "the above questions on Pythagoras. 1. Yes, hypotenuse = 8.5 m 2. 24 m 3. 6.5 cm 4. 7 cm 5. 32 m 6. 5 m 7. 68 cm 8. 52 m 9. 4 cm 10. 7.5 m 11. 50 m 12. 35 cm 14. $$t\\sqrt{u^{2} + v^{2}}$$", "R - H? So we have just got the negative of the value from earlier. But in Pythagoras' Theorem, we are squaring it. When we square a negative, we get a positive. So in fact, (R-H)² = (H-R)². This means we don't need to change our equation at all! So let's plug in the numbers into your equation: (4H² + L²)/8H = ( 4 [37.75]² + [18.875]² ) / ( 8 [37.75] ) = ( 4 [1425.0625] + [356.265625] ) / 302 = 6056.515626 / 302 = 20.055 inches. Then is the right answer: 20.055 inches (rounded off). Hope this helps, Stephen La Rocque. Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "the square roots of both sides of the equation. Use a calculator. The answer is \\begin{align*}c \\approx 6.7\\end{align*}. The hypotenuse is approximately 6.7. #### Example 3 Claire’s mom needs to fix a broken shingle on their house. If the shingle is 16 feet above the ground and Claire places the foot of the ladder 12 feet from the wall, how long will the ladder need to be? First, draw a picture to help determine what you are solving for. In this picture, you can see that the ladder is the hypotenuse. Next, insert the values of the legs into the formula for the Pythagorean Theorem. Next, calculate the value of the squares. Then take the square root of each side. The answer is \\begin{align*}c=20\\end{align*}. The length of the hypotenuse is 10, and the ladder needs to be 10 feet long. Remember Yuli and his garden beds? Yuli needs to know how long a diagonal timber he needs for a \\begin{align*}5 \\ ft \\times 5 \\ ft\\end{align*} square. First, use the Pythagorean Theorem and substitute in the values you know. Next, calculate the value of the squares. Then take the square root of each side. Round the number. The answer is that the hypotenuse \\begin{align*}c \\approx 7.07\\end{align*}, but since Yuli may have a hard time measuring \\begin{align*}\\frac{7}{100}\\end{align*} of a", "# Hypotenuse and height In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs. Correct result: a = 55.9 cm b = 4 cm #### Solution: $56=x+y \\ \\\\ 4^2=xy \\ \\\\ \\ \\\\ x(56-x) = 4^2 \\ \\\\ \\ \\\\ x^2 -56x +16 =0 \\ \\\\ \\ \\\\ a=1; b=-56; c=16 \\ \\\\ D = b^2 - 4ac = 56^2 - 4\\cdot 1 \\cdot 16 = 3072 \\ \\\\ D>0 \\ \\\\ \\ \\\\ x_{1,2} = \\dfrac{ -b \\pm \\sqrt{ D } }{ 2a } = \\dfrac{ 56 \\pm \\sqrt{ 3072 } }{ 2 } = \\dfrac{ 56 \\pm 32 \\sqrt{ 3 } }{ 2 } \\ \\\\ x_{1,2} = 28 \\pm 27.712812921102 \\ \\\\ x_{1} = 55.712812921102 \\ \\\\ x_{2} = 0.28718707889796 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (x -55.712812921102) (x -0.28718707889796) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = \\sqrt{ x_1 \\cdot 56 } = 55.9 \\ \\text{cm}$ $b = \\sqrt{ x_2 \\cdot 56 } = 4 \\ \\text{cm}$ 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 Looking for", "## College Algebra (6th Edition) $6.7$ miles long The Pythagorean theorem is $a^2+b^2=c^2$ to solve for the hypotenuse. We are given that the rectangular park is 6 miles long and 3 miles wide, so we plug in those values to the formula. $6^2+3^2=c^2$ $36+9=c^2$ $45=c^2$ $c=\\sqrt {45}$ $c=3\\sqrt 5$ $c\\approx6.7$", "# The length of a rectangle is twice its width. If the diagonal measures 10 feet, then find the dimensions of the rectangle.. 1 by marcguantero14 • Brainly User 2016-07-11T12:15:36+08:00 Use Pythagorean Theorem, where the diagonal is the hypotenuse. The Pythagorean Theorem states that the square of hypotenuse (in this case, the diagonal) is equal to the sum of the squares of the two others sides. Let hypotenuse (diagonal) = 10 ft. Width = x Length = 2x Equation: (2x)² + x² = 10² 4x² + x² = 100 5x² = 100 5x²/5 = 100/5 x² = 20 x = x = +2 and -2 Use the positive root for the dimensions. The dimensions are: Width: x = 2 ft. Length: 2x = 2 (2) or 4 ft.", "## allthewaycountrygirl 2 years ago Find the length of the hypotenuse of a right triangle with legs of 20 cm and 21 cm. 41 cm 841 cm 6 cm 29 cm 1. jasonxx $H= \\sqrt{(20)^2+(21)^2}$ 2. gaara438125 |dw:1352146014111:dw| 3. jasonxx |dw:1352138840805:dw|$c^2=a^2+b^2$ 4. allthewaycountrygirl so i have to find out what H is? 5. jasonxx yes 6. gaara438125 $H= \\sqrt{(20^2)+(21^2)}$ 7. gaara438125 20 * 20 = 400 21 * 21 = 441 add those 400 + 441 then take the sqrt and that's the missing angle 8. allthewaycountrygirl 841cm 9. gaara438125 $\\sqrt{841cm}$ sqrt it and the answer is? 10. allthewaycountrygirl so then 29cm 11. gaara438125 yes ma'am" ]

[ "= 15.4 cm2 ### Practice Questions 1. Which of the following shows the best classification of a triangle with dimensions of $12$ cm, $5$ cm, and $13$ cm? 2. Which of the following shows the best classification of a triangle with dimensions of $12$ ft, $18$ ft, and $17$ ft? 3. Which of the following shows the best classification of a triangle with dimensions of $15$ m, $20$ m, and $16$ m? 4. Which of the following shows the best classification of a triangle with dimensions of $10$ yard, $20$ yard, and $12$ yard? 5. The two shorter sides of a right triangle are $10$ cm and $24$ cm long. What is the length of the hypotenuse? 6. The leg and the hypotenuse of a right triangle are $14$ ft and $50$ ft long, respectively. Which of the following shows the length of the triangle’s remaining leg? 7. A $20$-m long rope is stretched from the top of a $12$-m tree to the ground. What is the distance between the tree and the end of the rope on the ground? 8. A $13$ m long ladder is leaning against the wall. If the ground distance between the foot of the ladder and the wall is $5$ m, what is the wall’s height?", "# Hypotenuse and height In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs. Correct result: a = 55.9 cm b = 4 cm #### Solution: $56=x+y \\ \\\\ 4^2=xy \\ \\\\ \\ \\\\ x(56-x) = 4^2 \\ \\\\ \\ \\\\ x^2 -56x +16 =0 \\ \\\\ \\ \\\\ a=1; b=-56; c=16 \\ \\\\ D = b^2 - 4ac = 56^2 - 4\\cdot 1 \\cdot 16 = 3072 \\ \\\\ D>0 \\ \\\\ \\ \\\\ x_{1,2} = \\dfrac{ -b \\pm \\sqrt{ D } }{ 2a } = \\dfrac{ 56 \\pm \\sqrt{ 3072 } }{ 2 } = \\dfrac{ 56 \\pm 32 \\sqrt{ 3 } }{ 2 } \\ \\\\ x_{1,2} = 28 \\pm 27.712812921102 \\ \\\\ x_{1} = 55.712812921102 \\ \\\\ x_{2} = 0.28718707889796 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (x -55.712812921102) (x -0.28718707889796) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = \\sqrt{ x_1 \\cdot 56 } = 55.9 \\ \\text{cm}$ $b = \\sqrt{ x_2 \\cdot 56 } = 4 \\ \\text{cm}$ 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 Looking for", "of the triangle is 129.5 cm2. ∴ 7.4h = 129.5 h = $\\left(\\frac{129.5}{7.4}\\right)$ = 17.5 cm ∴ Length of the other leg = 17.5 cm Question 7: Find the area of a right triangle whose base is 1.2 m and hypotenuse 3.7 m. Here, base = 1.2 m and hypotenuse = 3.7 m In the right angled triangle: Perpendicular = $\\sqrt{\\left(H\\mathrm{ypotenuse}{\\right)}^{2}-\\left(B\\mathrm{ase}{\\right)}^{2}}$ $=\\sqrt{{\\left(3.7\\right)}^{2}-{\\left(1.2\\right)}^{2}}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{13.69-1.44}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{12.25}\\phantom{\\rule{0ex}{0ex}}=3.5$ Area = $\\left(\\frac{1}{2}×\\mathrm{base}×\\mathrm{perpendicular}\\right)$ sq. units = $\\left(\\frac{1}{2}×1.2×3.5\\right)$ m2 ∴ Area of the right angled triangle = 2.1 m2 Question 8: The legs of a right triangle are in the ratio 3 : 4 and its area is 1014 cm2. Find the lengths of its legs. In a right angled triangle, if one leg is the base, then the other leg is the height. Let the given legs be 3x and 4x, respectively. Area of the triangle = $\\left(\\frac{1}{2}×3x×4x\\right)$ cm2 ⇒ 1014 = (6x2) ⇒ 1014 = 6x2 x2 = $\\left(\\frac{1014}{6}\\right)$ = 169 x = $\\sqrt{169}$ = 13 ∴ Base = (3 $×$ 13) = 39 cm Height = (4 $×$ 13) = 52 cm Question 9: One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its cost at the rate of Rs 250 per m2. Consider a right-angled triangular scarf (ABC). Here, ∠B= 90° BC = 80", "# Hypotenuse and height In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs. Correct result: a = 55.9 cm b = 4 cm #### Solution: $56=x+y \\ \\\\ 4^2=xy \\ \\\\ \\ \\\\ x(56-x) = 4^2 \\ \\\\ \\ \\\\ x^2 -56x +16 =0 \\ \\\\ \\ \\\\ a=1; b=-56; c=16 \\ \\\\ D = b^2 - 4ac = 56^2 - 4\\cdot 1 \\cdot 16 = 3072 \\ \\\\ D>0 \\ \\\\ \\ \\\\ x_{1,2} = \\dfrac{ -b \\pm \\sqrt{ D } }{ 2a } = \\dfrac{ 56 \\pm \\sqrt{ 3072 } }{ 2 } = \\dfrac{ 56 \\pm 32 \\sqrt{ 3 } }{ 2 } \\ \\\\ x_{1,2} = 28 \\pm 27.7128129211 \\ \\\\ x_{1} = 55.7128129211 \\ \\\\ x_{2} = 0.287187078898 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (x -55.7128129211) (x -0.287187078898) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = \\sqrt{ x_1 \\cdot 56 } = 55.9 \\ \\text{cm}$ 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 Looking for help with calculating roots of a quadratic equation? Do you have", "# Hypotenuse and height In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs. Correct result: a = 55.9 cm b = 4 cm #### Solution: $56=x+y \\ \\\\ 4^2=xy \\ \\\\ \\ \\\\ x(56-x) = 4^2 \\ \\\\ \\ \\\\ x^2 -56x +16 =0 \\ \\\\ \\ \\\\ a=1; b=-56; c=16 \\ \\\\ D = b^2 - 4ac = 56^2 - 4\\cdot 1 \\cdot 16 = 3072 \\ \\\\ D>0 \\ \\\\ \\ \\\\ x_{1,2} = \\dfrac{ -b \\pm \\sqrt{ D } }{ 2a } = \\dfrac{ 56 \\pm \\sqrt{ 3072 } }{ 2 } = \\dfrac{ 56 \\pm 32 \\sqrt{ 3 } }{ 2 } \\ \\\\ x_{1,2} = 28 \\pm 27.712812921102 \\ \\\\ x_{1} = 55.712812921102 \\ \\\\ x_{2} = 0.28718707889796 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (x -55.712812921102) (x -0.28718707889796) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = \\sqrt{ x_1 \\cdot 56 } = 55.9 \\ \\text{cm}$ $b = \\sqrt{ x_2 \\cdot 56 } = 4 \\ \\text{cm}$ 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", "legs a = 15cm and b = 36cm. Calculate the length of the median to side c (to hypotenuse). • Catheti The hypotenuse of a right triangle is 41 and the sum of legs is 49. Calculate the length of its legs. • Euclid2 In right triangle ABC with right angle at C is given side a=27 and height v=12. Calculate the perimeter of the triangle. • RT triangle and height Calculate the remaining sides of the right triangle if we know side b = 4 cm long and height to side c h = 2.4 cm. • Broken tree The tree is broken at 4 meters above the ground and the top of the tree touches the ground at a distance of 5 from the trunk. Calculate the original height of the tree. • ABS CN Calculate the absolute value of complex number -15-29i. • Right angled triangle Hypotenuse of a right triangle is 17 cm long. When we decrease length of legs by 3 cm then decrease its hypotenuse by 4 cm. Determine the size of legs. The ladder has a length 3.5 meters. He is leaning against the wall so that his bottom end is 2 meters away from the wall. Determine the height of the ladder. • Spruce height How tall was spruce that", "leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle. 6. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate? 7. Emanuel has a cardboard box that measures $20 \\ cm \\times 10 \\ cm \\times 8 \\ cm (\\text {length} \\times \\text{width} \\times \\text{height})$. What is the length of the diagonal from a bottom corner to the opposite top corner? 8. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. How high above the ground does the ladder touch the wall of the house? 9. Find the area of the triangle if area of a triangle is defined as $A=\\frac{1}{2} \\text{base} \\times \\text{height}$. 10. Instead of walking along the two sides of a rectangular field, Mario decided to cut across the diagonal. He saves a distance that is half of the long side of the field. Find the length of the long side of the field given that the short side is 123 feet. 11. Marcus sails due north and Sandra sails due east", "# 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.", "## anonymous one year ago The legs of a right triangle are 10 centimeters and 24 centimeters long. What is the length of the hypotenuse? 22 centimeters 26 30 centimeters 34 centimeters |dw:1438781834823:dw| use pythagorean theorem $x^2+y^2 = r^2 \\implies r = \\sqrt{x^2+y^2}$", "# Right triangle Legs of the right triangle are in the ratio a:b = 2:8. The hypotenuse has a length of 87 cm. Calculate the perimeter and area of the triangle. Correct result: o = 192.503 cm S = 890.471 cm2 #### Solution: $c=87 \\ \\text{cm} \\ \\\\ a:b=2:8 \\ \\\\ c^2=a^2+b^2 \\ \\\\ c^2=a^2+(8/2 \\cdot \\ a)^2 \\ \\\\ c^2=a^2+(8/2)^2 \\ a^2 \\ \\\\ c^2=a^2 (1 +(8/2)^2 ) \\ \\\\ \\ \\\\ a=c / \\sqrt{ 1 +(8/2)^2 }=87 / \\sqrt{ 1 +(8/2)^2 } \\doteq 21.1006 \\ \\text{cm} \\ \\\\ \\ \\\\ b=a \\cdot \\ \\dfrac{ 8 }{ 2 }=21.1006 \\cdot \\ \\dfrac{ 8 }{ 2 } \\doteq 84.4024 \\ \\text{cm} \\ \\\\ \\ \\\\ o=a+b+c=21.1006+84.4024+87=192.503 \\ \\text{cm}$ $S=\\dfrac{ a \\cdot \\ b }{ 2 }=\\dfrac{ 21.1006 \\cdot \\ 84.4024 }{ 2 }=\\dfrac{ 15138 }{ 17 }=890.471 \\ \\text{cm}^2$ Try calculation via our triangle calculator. 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 Looking for help with calculating roots of a quadratic equation? Check", "## anonymous 3 years ago Find the length of the hypotenuse of a right triangle with legs of 20 cm and 21 cm. 41 cm 841 cm 6 cm 29 cm 1. anonymous $H= \\sqrt{(20)^2+(21)^2}$ 2. anonymous |dw:1352146014111:dw| 3. anonymous |dw:1352138840805:dw|$c^2=a^2+b^2$ 4. anonymous so i have to find out what H is? 5. anonymous yes 6. anonymous $H= \\sqrt{(20^2)+(21^2)}$ 7. anonymous 20 * 20 = 400 21 * 21 = 441 add those 400 + 441 then take the sqrt and that's the missing angle 8. anonymous 841cm 9. anonymous $\\sqrt{841cm}$ sqrt it and the answer is? 10. anonymous so then 29cm 11. anonymous yes ma'am", "Pythagorean Theorem to find the length of the hypotenuse. 15 25 In the following exercises, use the Pythagorean Theorem to find the length of the leg. Round to the nearest tenth if necessary. 8 12 10.2 9.8 In the following exercises, solve using a geometry formula. The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width. 18 meters, 11 meters The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width. The width of the rectangle is 0.7 meters less than the length. The perimeter of a rectangle is 52.6 meters. Find the dimensions of the rectangle. $$13.5$$ m, $$12.8$$ m The length of the rectangle is 1.1 meters less than the width. The perimeter of a rectangle is 49.4 meters. Find the dimensions of the rectangle. The perimeter of a rectangle of 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle. 25 ft, 50 ft The length of the rectangle is three times the width. The perimeter of a rectangle is 72 feet. Find the length and width of the rectangle. The length of the rectangle is three meters less than twice the width.", "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 $48$", "also 130°. ∠SRT is the supplement of ∠ARS and thus measures 50°. By the property of vertical angles, we have m∠STR 5 70°. Finally, since the sum of the angles in triangle SRT is 180°, we have: x 1 50° 1 70° 5 180° x 1 120° 5 180° x 5 60° Right Triangles Right triangles contain one right angle. Since the right angle measures 90°, the other two angles are complementary. They may or may not be congruent to each other. The side of a right triangle opposite the right angle is called the hypotenuse. The other two sides are called legs. The Pythagorean Theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. AC is the hypotenuse. AB and BC are legs. m∠B 5 90° m∠A 1 m∠C 5 90° a2 1 c2 5 b2 www.petersons.com 248 MATH REVIEW Examples 1. If DABC is a right triangle with right angle at B, and if AB 5 6 and BC 5 8, what is the length of AC? AB2 1 BC2 5 AC2 62 1 82 5 36 1 64 5 100 5 AC2 AC 5 10 If the measure of angle A is 30°, what is the measure of angle C?", "hypotenuse of length 17 m, one leg of length 8 m, and the other leg of length $\\displaystyle \\sqrt{17^2- 8^2}= 15$ m. The reason for the \"additional 14 N\" is that the fish is being reeled in horizontally but the line is the hypotenuse of the triangle. Find the total force so that the horizontal component is 105 N. The ratio of hypotenuse to horizontal side of the right triangle is 17/15 so the tension in the line is (17/15)(105)= 119. Oct 22nd 2018, 07:01 PM #3 Junior Member Join Date: Jun 2018 Posts: 7 Thank you for your help. So the upward component X would be (8/15)*105 = 56. Let B be Buoyancy, then X + B = 150 56 + B = 150 So B = 94N So Normal contact force on the man & rod would be 800 + 56 = 856 N Tags forces, questions, tensions Thread Tools Display Modes Linear Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post hongiddong Kinematics and Dynamics 4 Aug 6th 2014 02:23 PM iannedrs Advanced Mechanics 3 Jul 22nd 2013 05:03 AM kastamonu Kinematics and Dynamics 9 Mar 31st 2013 07:57 AM flower Advanced Electricity and Magnetism 1 Nov 20th 2009 02:06 AM jello17 Kinematics and Dynamics 3 Oct 4th 2008 09:31 PM", "look up that $$\\tan(10)$$ is 0.176. Then we can calculate that $$h$$ is about 17.6. That means the tree is 17.6 feet tall. Glossary Entries • cosine The cosine of an acute angle in a right triangle is the ratio (quotient) of the length of the adjacent leg to the length of the hypotenuse. In the diagram, $$\\cos(x)=\\frac{b}{c}$$. • sine The sine of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the hypotenuse. In the diagram, $$\\sin(x) = \\frac{a}{c}.$$ • tangent The tangent of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the adjacent leg. In the diagram, $$\\tan(x) = \\frac{a}{b}.$$ • trigonometric ratio Sine, cosine, and tangent are called trigonometric ratios.", "• Question 8: 3 pts Find to the nearest degree the measure of the angle marked as $x$. x=15o x=16o x=17o x=18o • Question 9: 3 pts The legs of the right triangle measure 30cm and 45cm. Find to the nearest degree the measure of the smallest angle of this triangle. $x=30^\\circ$ $x=32^\\circ$ $x=34^\\circ$ $x=36^\\circ$", "7. Euklid4 Legs of a right triangle have dimensions 244 m and 246 m. Calculate the length of the hypotenuse and the height of this right triangle. 8. Euclid 5 Calculate the length of remain sides of a right triangle ABC if a = 7 cm and height vc = 5 cm. 9. Euclidean distance Calculate the Euclidean distance between shops A, B and C, where: A 45 0.05 B 60 0.05 C 52 0.09 Wherein the first figure is the weight in grams of bread and second figure is price in USD. The double ladder is 8.5m long. It is built so that its lower ends are 3.5 meters apart. How high does the upper end of the ladder reach?", "# The legs of a right triangle measure 9 feet and 12 feet what is the length of of the hypotenuse? Dec 8, 2016 The length of the hypotenuse is 15 feet. #### Explanation: To determine the length of a side of a right triangle you use the Pythagorean Theorem which states: a^2 + b^ =c^ where $a$ and $b$ are the length of the legs and $c$ is the length of the hypotenuse. Substituting the information provided and solving for $c$ gives: 9^2 + 12^ = c^ $81 + 144 = {c}^{2}$ $225 = {c}^{2}$ $\\sqrt{225} = \\sqrt{{c}^{2}}$ $15 = c$", "+0 0 176 1 The hypotenuse of a right triangle measures 14 cm and one of its legs measures 3 cm. Find the measurement of the other leg. If necessary, round to the nearest tenth. Feb 22, 2021 #1 +964 0 $$\\sqrt{14^2 - 3^2} = \\sqrt{187}$$" ]

[ "# Difference between revisions of \"2016 AMC 8 Problems/Problem 6\" ## Problem The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? $\\textbf{(A) }3\\qquad\\textbf{(B) }4\\qquad\\textbf{(C) }5\\qquad\\textbf{(D) }6\\qquad \\textbf{(E) }7$ $[asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label(\"frequency\",(-0.5,8)); label(\"0\", (-1, 0)); label(\"1\", (-1, 1)); label(\"2\", (-1, 2)); label(\"3\", (-1, 3)); label(\"4\", (-1, 4)); label(\"5\", (-1, 5)); label(\"6\", (-1, 6)); label(\"7\", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label(\"3\", (0.5, -0.5)); label(\"4\", (2.5, -0.5)); label(\"5\", (4.5, -0.5)); label(\"6\", (6.5, -0.5)); label(\"7\", (8.5, -0.5)); label(\"name length\", (4.5, -1)); [/asy]$ ## Solutions ### Solution 1 We first notice that the median name will be the $10^{\\mbox{th}}$ name. The $10^{\\mbox{th}}$ name is $\\boxed{\\textbf{(B)}\\ 4}$. ### Solution 2 To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$. Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$, and $3$ names with length $4$, the $10$th name would have $4$ letters. Thus our answer is $\\boxed{\\textbf{(B)}\\ 4}$.", "fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label(\"6\",.74*dir(-90)+(0,3)); label(\"5\",.74*dir(-60)+(0,3)); label(\"4\",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label(\"30^\\circ\",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*\"12\",1.56*dir(120)); label(rotate(30)*\"1\",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label(\"60^\\circ\",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*\"6\",.74*dir(-150)+(0,3)); label(rotate(-60)*\"5\",.74*dir(-120)+(0,3)); label(rotate(-60)*\"4\",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label(\"90^\\circ\",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*\"6\",.74*dir(-180)+c); label(rotate(-90)*\"5\",.74*dir(-150)+c); label(rotate(-90)*\"4\",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{\\textbf{(C) }\\text{4 o' clock}}$ ## Solution 5 We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We", "0 0.5 06 27 20 15 0 0.5 06 28 25 15 1 0.5 07 29 28 16 1 0.5 07 30 34 17 1 0.5 07 31 19 17 0 0.5 07 32 24 16 0 0.5 08 33 32 18 1 0.5 08 34 25 19 0 0.5 08 35 31 18 0 0.5 09 36 23 20 1 0.5 09 37 20 22 0 0.5 09 38 33 21 0 0.5 09 39 22 20 0 0.5 09 40 35 21 1 0.5 10 41 20 25 0 0.5 10 42 31 26 1 0.5 10 43 34 23 0 0.5 10 44 35 23 1 0.5 10 45 18 25 1 0.5 10 46 23 24 1 0.5 10 47 22 24 0 0.5 The graph below shows that indeed, the blocking formed pairs of classrooms within each school that are similar to one another and correctly left some classrooms in a block by themselves when there was an odd number of classrooms in a school. our_df %>% ggplot(aes(size, block_id, color = as.factor(Z))) + geom_point() + facet_wrap(~school, scales = \"free_y\", ncol = 5) ## Replicate yourself If you set up your assignments as functions like this then it is easy to implement the assignment multiple times: many_assignments <- replicate(1000, draw_data(design)) It’s useful to preserve", "# 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 $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "Both plots are the same, but the upper plot answers my first question, and the latter plot answers my second one. It seems that you only need about 23 students (!!!) before you have a 50/50 chance of at least two students sharing the same birthday. This is WAY off my intuition of there needing to be about 180 students before you could even sniff at a 50% chance of a match. The second plot shows that, to almost guarantee there being a match of birthdays, you need about 74 students. We really are fooled by probability! Poor students! ## R Code for the Simulation #------------------------------------------------------------------------------ ### simulation parameters # define vector of calendar dates calendar <- seq(from = 1, to = 365, by = 1) # how many simulations do you want per class size? nSimulations <- 10000 # what is the size of each class? Here we have class sizes # from 2 to 100 classSize <- seq(from = 2, to = 100, by = 1) #------------------------------------------------------------------------------ #------------------------------------------------------------------------------ ### start simulation here # initiate vector to store data for each class size classData <- numeric(length(classSize)) # initiate looping variable for class sizes j <- 1 # loop over each class size for(currSize in classSize){ # initiate vector to store counts of replicated birthdays data <- numeric(nSimulations)", "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", "4, 3, 5, 7))) This new function would generate a data set that reflects the number of classes in each school. ## Make better blocks from richer data Say that you had even richer data about the sampling frame. Then you could use this to do even better assignments. Say this is what you get from your partner: We just got information on the class sizes. Here it is: School 1: 20, 25, 23, 30, 12, 15; School 2: 40, 42, 53, 67, 35, 22, 18, 18; School 3: 34, 37, 28, 30; School 4: 18, 24, 20; School 5: 10, 24, 13, 26, 18; School 6: 20, 25; School 7: 28, 34, 19, 24; School 8: 32, 25, 31; School 9: 23, 20, 33, 22, 35; School 10: 20, 31, 34, 35, 18, 23, 22 We want to incorporate this information in the data fabrication step. make_data <- declare_population( class = add_level(N = c(6, 8, 4, 3, 5, 2, 4, 3, 5, 7), size = c( 20, 25, 23, 30, 12, 15, # students in each classroom of school 1 40, 42, 53, 67, 35, 22, 18, 18, # students in each classroom of school 2 34, 37, 28, 30, # etc... 18, 24, 20, 10, 24, 13, 26, 18, 20, 25, 28, 34, 19, 24, 32,", "# 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", "1, 1, 1, 1, 1, 1, 1,… ##$ spontaneous <dbl> 2, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1,… ## $stratum <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, … ##$ pooled.stratum <dbl> 3, 1, 4, 2, 32, 36, 6, 22, 5, 19, 2… data <- infert %>% select(education, age, parity) %>% data ## education age parity ## 1 0-5yrs 26 6 ## 2 0-5yrs 42 1 ## 3 0-5yrs 39 6 ## 4 0-5yrs 34 4 ## 5 6-11yrs 35 3 ## 6 6-11yrs 36 4 In the code chunk below, examine the local dataset, emerg_dept, which has counts of ED arrivals, how many breached the UK 4 hour guarantee, and how many got admitted. Then select() only org_code, attendances, breaches, and admissions. Then arrange() to have the top attendances at the top, use top_n(10) to get only the top 10 rows. Assign this to data and print it out emerg_dept ## # A tibble: 50 × 6 ## period org_code type attendances breaches admissions ## <date> <fct> <fct> <dbl> <dbl> <dbl> ## 1 2018-07-01 RRK 1 32209 6499 11332 ## 2 2018-07-01 R1H 1 28357 6294 7986 ## 3 2018-07-01 RW6 1 23887 4641 6282 ## 4 2018-07-01 R0A 1 22012 4669 6818 ## 5 2018-07-01 RDU 1 21043", "'step2014', 'step2015', 'decile.change', 'step.change') -> names(dc) c(LETTERS[1:17], 'Z') -> steps c(905.81, 842.11, 731.3, 617.8, 507.01, 420.54, 350.25, 277.32, 220.59, 182.74, 149.99, 135.12, 115.76, 93.71, 71.64, 46.86, 28.93, 0) -> money within(dc, { ifelse(step2014 == '', NA, step2014) -> step2014 ifelse(step2015 == '', NA, step2015) -> step2015 sapply(step2014, function(x) money[match(x, steps)]) -> sm2014 sapply(step2015, function(x) money[match(x, steps)]) -> sm2015 sm2015 - sm2014 -> funding.change }) -> dc merge(dc, direc[, c('School.ID', 'Total.School.Roll', 'Urban.Area', 'Regional.Council')], by = 'School.ID', all.x = TRUE) -> dc within(dc, { factor(Urban.Area) -> Urban.Area factor(Urban.Area, levels(Urban.Area)[c(3, 2, 4, 1)]) -> Urban.Area funding.change*Total.School.Roll/1000 -> school.level.change }) -> dc with(dc, { summary(funding.change) summary(school.level.change) sum(sm2014*Total.School.Roll/1000000, na.rm = NA) sum(sm2015*Total.School.Roll/1000000, na.rm = NA) }) #### By Urban area # Funding change per student on school size qplot(Total.School.Roll, funding.change, data = dc[!is.na(dc$Urban.Area),], alpha = 0.8, xlab = 'Total School Roll', ylab = 'Funding change per student (NZ$)') + theme_bw() + theme(legend.position = 'none') + facet_grid(~Urban.Area) Not too alien; now using the magrittr package with right assign would make a lot more sense. Categories ## Sometimes I feel (some) need for speed I’m the first to acknowledge that most of my code could run faster. The truth of the matter is that, in essence, I write ‘quickies’: code that will run once or twice, so there is no incentive to spend days or", "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", "percent of the area of rectangle $AQRD$ is shaded? $[asy] filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label(\"D\",(0,0),S); label(\"R\",(25,0),S); label(\"Q\",(25,15),N); label(\"A\",(0,15),N); filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black); label(\"S\",(10,0),S); label(\"C\",(15,0),S); label(\"B\",(15,15),N); label(\"P\",(10,15),N);[/asy]$ $\\textbf{(A)}\\ 15\\qquad\\textbf{(B)}\\ 18\\qquad\\textbf{(C)}\\ 20\\qquad\\textbf{(D)}\\ 24\\qquad\\textbf{(E)}\\ 25$ ## Problem 14 There are $270$ students at Colfax Middle School, where the ratio of boys to girls is $5 : 4$. There are $180$ students at Winthrop Middle School, where the ratio of boys to girls is $4 : 5$. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls? $\\textbf{(A) } \\dfrac7{18} \\qquad\\textbf{(B) } \\dfrac7{15} \\qquad\\textbf{(C) } \\dfrac{22}{45} \\qquad\\textbf{(D) } \\dfrac12 \\qquad\\textbf{(E) } \\dfrac{23}{45}$ ## Problem 15 How many digits are in the product $4^5 \\cdot 5^{10}$? $\\textbf{(A) } 8 \\qquad\\textbf{(B) } 9 \\qquad\\textbf{(C) } 10 \\qquad\\textbf{(D) } 11 \\qquad\\textbf{(E) } 12$ ## Problem 16 Let $A$ be the area of the triangle with sides of length $25, 25$, and $30$. Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$. What is the relationship between $A$ and $B$? $\\textbf{(A) } A = \\dfrac9{16}B \\qquad\\textbf{(B) } A = \\dfrac34B \\qquad\\textbf{(C) } A=B \\qquad\\textbf{(D) } A = \\dfrac43B \\\\ \\\\ \\textbf{(E) }A = \\dfrac{16}9B$ ## Problem 17 Let $w$, $x$, $y$, and $z$ be whole numbers. If $2^w \\cdot 3^x \\cdot 5^y", "# Difference between revisions of \"1998 AJHSME Problems/Problem 23\" ## Problem If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle? $[asy] unitsize(10); draw((0,0)--(12,0)--(6,6sqrt(3))--cycle); draw((15,0)--(27,0)--(21,6sqrt(3))--cycle); fill((21,0)--(18,3sqrt(3))--(24,3sqrt(3))--cycle,black); draw((30,0)--(42,0)--(36,6sqrt(3))--cycle); fill((34,0)--(32,2sqrt(3))--(36,2sqrt(3))--cycle,black); fill((38,0)--(36,2sqrt(3))--(40,2sqrt(3))--cycle,black); fill((36,2sqrt(3))--(34,4sqrt(3))--(38,4sqrt(3))--cycle,black); draw((45,0)--(57,0)--(51,6sqrt(3))--cycle); fill((48,0)--(46.5,1.5sqrt(3))--(49.5,1.5sqrt(3))--cycle,black); fill((51,0)--(49.5,1.5sqrt(3))--(52.5,1.5sqrt(3))--cycle,black); fill((54,0)--(52.5,1.5sqrt(3))--(55.5,1.5sqrt(3))--cycle,black); fill((49.5,1.5sqrt(3))--(48,3sqrt(3))--(51,3sqrt(3))--cycle,black); fill((52.5,1.5sqrt(3))--(51,3sqrt(3))--(54,3sqrt(3))--cycle,black); fill((51,3sqrt(3))--(49.5,4.5sqrt(3))--(52.5,4.5sqrt(3))--cycle,black); [/asy]$ $\\text{(A)}\\ \\frac{3}{8}\\qquad\\text{(B)}\\ \\frac{5}{27}\\qquad\\text{(C)}\\ \\frac{7}{16}\\qquad\\text{(D)}\\ \\frac{9}{16}\\qquad\\text{(E)}\\ \\frac{11}{45}$ ## Solution All small triangles are congruent in each iteration of the diagram. The number of shaded triangles follows the pattern: $0, 1, 3, 6, ...$ which is the pattern of \"triangular numbers\". Each time, the number $1, 2, 3, 4, 5...$ is added to the previous term. Thus, the first eight terms are: $0, 1, 3, 6, 10, 15, 21, 28$ In the eighth diagram, there will be $28$ shaded triangles. The total number of small triangles follows the pattern: $1, 4, 9, 16, ...$ which is the pattern of \"square numbers\". Thus, the eighth triangle will be divided into $8^2 = 64$ small triangles in total. The ratio of shaded to total triangles will be the fraction of the whole figure that's shaded, since all triangles are congruent. Thus, the answer is $\\frac{28}{64} = \\frac{7}{16}$, and the correct choice is $\\boxed{C}$ 1998 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 •", "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", "%heartsflow, 0.3 minimum opacity HeartsOnPath(-15,-5,-10,-4.5,0.07,100); HeartsOnPath(-10,-4.5,0,-3.5,0.07,100); HeartsOnPath(0,-3.5,2,-5,0.07,50); HeartsOnPath(2,-5,5,-3,0.07,70); HeartsOnPath(5,-3,8,0,0.1,70); HeartsOnPath(8,0,11,3.5,0.1,70); HeartsOnPath(11,3.5,10,7,0.1,70); HeartsOnPath(10,7,13,10,0.07,70); HeartsOnPath(-9,-1.5,-8,4,0.1,70); HeartsOnPath(-8,4,0,5,0.1,50); HeartsOnPath(0,3,10,8,0.1,50); } %stripes \\draw [draw=none,fill=ThirdPurple] (-15,-6) to[out=20,in=170] (-5,-5) to[out=350,in=330] (3,-1) to[out=150,in=210] (2,2) to[out=30,in=180] (15,11) -- (15,9) to[out=180,in=30] (3,1) to[out=210,in=150] (4,-1) to[out=330,in=350] (-5,-6.2) to[out=170,in=20] (-15,-7) -- cycle; \\draw [draw=none,fill=SecondPurple] (-15,-4) to[out=20,in=170] (-5,-4) to[out=350,in=330] (3,0) to[out=150,in=210] (2,3) to[out=30,in=180] (15,12) -- (15,10) to[out=180,in=30] (3,2) to[out=210,in=150] (4,0) to[out=330,in=350] (-5,-5) to[out=170,in=20] (-15,-6) -- cycle; \\draw [draw=none,fill=FirstPurple] (-15,-4) to[out=20,in=170] (-5,-3) to[out=350,in=330] (3,1) to[out=150,in=210] (2,4) to[out=30,in=180] (13,11) -- (14,11) to[out=180,in=30] (3,3) to[out=210,in=150] (4,1) to[out=330,in=350] (-5,-4) to[out=170,in=20] (-15,-5.3) -- cycle; %bigger hearts \\heart{(5.5,1.5)}{0}{2.4}{1}{BigHeartColor} \\heart{(-1.5,2)}{20}{1}{1}{SmallerHeartColor} \\heart{(9.5,4)}{-20}{0.8}{1}{SmallestHeartColor} %light strip (-16,7) to[out=-10,in=170] (1,0) to[out=-10,in=200] (9,1) to[out=20,in=-20] (6,3) to[out=160,in=-80] (2,8) to[out=100,in=100] (3,9) to[out=-80,in=100] (4,6) to[out=-80,in=-90] (17,-4) -- (17,-3.7) to[out=-90,in=-80] (4,6.3) to[out=100,in=-80] (3,9.3) to[out=100,in=100] (2,8.3) to[out=-80,in=160] (6,3.3) to[out=-20,in=20] (9,1.3) to[out=200,in=-10] (1,0.3) to[out=170,in=-10] (-16,7.3) -- cycle; %cats \\node [xshift=5.5cm] {\\includegraphics{kittenz.pdf}}; %Mau \\color{white}\\Scriptina \\begin{scope}[shift={(1,0.8)}] \\contourlength{0.01em} \\node [xscale=6,yscale=6] at (-9,-0.5) {\\contour{TextOutlineColor}{M}}; \\contourlength{0.02em} \\node [xscale=3,yscale=3] at (-4,-2.55) {\\contour{TextOutlineColor}{a}}; \\node [xscale=3,yscale=3] at (-1.7,-2.5) {\\contour{TextOutlineColor}{u}}; %restore Ma 'ligature' \\draw [draw=none,fill=white,rotate around={55:(-4.75,-2.77)},shift={(-4.83,-2.79)}] (0,0) rectangle (0.2,0.08); \\end{scope} %Happy Birthday \\Zapfino \\contourlength{0.02em} \\node [xscale=3,yscale=3] at (-3,4.5) {\\contour{TextOutlineColor}{Happy Birthday}}; \\color{LastILOVEYOUColor}\\rmfamily \\node at (0,-2.64) {\\contour{LastILOVEYOUOutlineColor}{ILOVEYOU}}; %light bubbles, 20% more white, 0.3 minimum opacity LightsOnPath(-15,5,15,-5,0.07,200); } \\end{tikzpicture} \\end{document} Aaaaaaaaaaaand this is the result: Yes, it's pink. I like pink. Here's the pdf. ## Top reasons you should upvote this entry • pink • hearts • cats • love •", "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", "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", "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" ]

[ "8548. Find the increase in enrollment (x1) and the current enrollment (x2).", "### Part 1: What is the growth factor associated with a $60 \\%$ decrease? Multiply by to reduce by $60 \\%\\text{.}$ If we start with $300$ students then decrease enrollment by $60 \\%\\text{,}$ there will be students remaining.", "attendance rate, enrollment divided by 1,000 and the square of enrollment divided by 1,000, percentage of students receiving SPED (special education) and ELL (English language learner) services, and year and district fixed effects. Figures in constant 2012 dollars, adjusted using the Consumer Price Index. Regression results weighted by district enrollment. No Big 5 excludes New York City, Buffalo, Rochester, Syracuse, and Yonkers school districts. Reference category: Share of district students who are white.", "$10$ students and the largest class size during the previous year was $35$ students. Select an assignment template", "projected enrollment 35 years from now:The projected enrollment 35 years from now is (Round to the nearest integer as needed:) The projected rate of increase in enrollment at new college is estimated by 6,000(t + 1) 3/2 where Elt) is the projected enrollment in years _ If the enrollment is 5,000 now (t = 0), find the projected enrollment 35 years from now: The projected enrollment 35 years from now is (Round to the nearest ...", "the number of students that actually enroll and $xx$ is the number of all students offered admission to the class of 2019. 2. If the university wants the 2019 freshman class size to be 1350, determine how many students should be admitted.", "* (4000-a)$$ $$10a = 7 * 4000 - 4000$$ a = 2400 _________________ Kindly press \"+1 Kudos\" to appreciate Math Expert Joined: 02 Sep 2009 Posts: 46307 Re: The number of students enrolled at school X this year is 7 percent [#permalink] ### Show Tags 13 Nov 2014, 09:20 Bunuel wrote: Tough and Tricky questions: Percents. The number of students enrolled at school $$X$$ this year is 7 percent more than it was last year. The number of students enrolled at school $$Y$$ this year is 3 percent more than it was last year. If school $$X$$ grew by 40 more students than school $$Y$$ did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school $$Y$$ last year? A. 480 B. 1600 C. 1920 D. 2080 E. 2400 Kudos for a correct solution. Official Solution: The number of students enrolled at school $$X$$ this year is 7 percent more than it was last year. The number of students enrolled at school $$Y$$ this year is 3 percent more than it was last year. If school $$X$$ grew by 40 more students than school $$Y$$ did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school $$Y$$ last", "## Precalculus (6th Edition) Blitzer The same enrollment ( $22,300$ ) will be achieved in 2019. Let $x=$ the number of years after 2010. Enrolled at college A, after x years = $13,300+1000x$ Enrolled at college B, after x years = $26,800-500x$ We want to find x when these two are equal. $13,300+1000x=26,800-500x \\qquad$ ... add $500x-13,300$ $1000x+500x=26,800-13,500\\qquad$ $1500x=13,500\\qquad$ ... divide with 1500 $x=9$ The two colleges will have the same enrollment $9$ years after 2010, in 2019. To calculate the enrollment, substitute $x=9$ into one of the expressions $13,300+1000(9)=22,300$", "Basic College Mathematics (10th Edition) Add all numbers to find total students enrolled $13,786+3,497+2,874 = 20,157$", "at least a subsection of the ZCTA. The Number of schools variable takes a whole number. • Enrollment - The enrollment displayed for each ZCTA is the summation of the enrollment for every school considered within a 30, 45, or 60 minute drive from at least a subsection of the ZCTA. The Enrollment variable takes a whole number and is sourced from IPEDS 12-Month Enrollment Table. This number indicates how many unduplicated undergraduate individuals enrolled for credit the institution served over a 12 month period, regardless of when the individual enrolled. $$\\text{Enrollment}_{\\text{ZCTA}} = \\text{enrollment}_1 + \\text{enrollment}_2 + \\dots + \\text{enrollment}_n$$ • 25 Mile Zone Population Estimate - In place of population estimates for individual ZCTA, we list the summation of all the population estimates for the ZCTAs within 25 miles of the selected area; the purpose being to complement the zone-type scale used for the SCI. The 25 mile population zone originates from the mathematical centroid of each ZCTA. $$\\text{Pop}_{\\text{ZCTA-ZONE}} = \\text{pop}_1 + \\text{pop}_2 + \\dots + \\text{pop}_n$$ Every ZCTA has a unique 25 Mile Zone Population Estimate where $$n$$ number of ZCTAs in each ZCTA-zone equals the ZCTAs accessible to the root ZCTA within 25 miles. Each $$\\text{pop}_n$$ is the population estimate for that specific ZCTA. Both $$\\text{pop}_n$$ and $$\\text{Pop}_{\\text{ZCTA-ZONE}}$$ take a whole number. • Average Net", "Enter the number of individuals enrolled and the total population of the eligible group into the Calculator. The calculator will evaluate the Enrollment Rate. ## Enrollment Rate Formula ER = E/P *100 Variables: • ER is the Enrollment Rate (%) • #E is the number of individuals enrolled • #P is the total population of the eligible group To calculate the Enrollment Rate, divide the number of individuals enrolled by the total population of the eligible group, then multiply by 100. ## How to Calculate Enrollment Rate? The following steps outline how to calculate the Enrollment Rate. 1. First, determine the number of individuals enrolled. 2. Next, determine the total population of the eligible group. 3. Next, gather the formula from above = ER = #E/#P *100. 4. Finally, calculate the Enrollment Rate. 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. number of individuals enrolled = 4,000 total population of the eligible group = 7,000", "# There are 800 students in a certain highschool in 2001. the trend during the previous years shows a yearly increase of 20% in enrollment. What will the enrollment be in 16 years? Mar 22, 2018 \" Total number of students after 16 years will be \" color(brown)(14,970 #### Explanation: Increase in number of students enrolled is compounded by 20% every year. Hence it follows compound interest formula of $A = P \\left(1 + {\\left(\\frac{r}{100}\\right)}^{n}\\right)$ Total number of students after 16 years will be given by ${a}_{16} = {a}_{i} \\cdot {\\left(1 + \\left(\\frac{r}{100}\\right)\\right)}^{16}$ ${a}_{16} = 800 \\cdot {\\left(1.2\\right)}^{16} \\approx 14 , 970 \\text{ students}$", "## Intermediate Algebra (6th Edition) $2700, 2850, 3000, 3150,$ and $3300$ The number of students enrolled follows an arithmetic sequence with a common difference of $150$ and a first term of $2700$. The number of students enrolled for the next year can be found by adding the common difference of $150$. Thus, the number of students enrolled for each of 5 years beginning with $2000$ is: Year 2000 $=a_1 = 2700$ Year 2001 $=a_2 = 2700+150=2850$ Year 2002 $=a_3 = 2850+150=3000$ Year 2003 $=a_4 = 3000+150=3150$ Year 2004 $=a_5 = 3150+150=3300$", "of students enrolled at school X this year is 7 percent [#permalink] ### Show Tags 25 Feb 2016, 05:26 Bunuel wrote: Bunuel wrote: Tough and Tricky questions: Percents. The number of students enrolled at school $$X$$ this year is 7 percent more than it was last year. The number of students enrolled at school $$Y$$ this year is 3 percent more than it was last year. If school $$X$$ grew by 40 more students than school $$Y$$ did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school $$Y$$ last year? A. 480 B. 1600 C. 1920 D. 2080 E. 2400 Kudos for a correct solution. Official Solution: The number of students enrolled at school $$X$$ this year is 7 percent more than it was last year. The number of students enrolled at school $$Y$$ this year is 3 percent more than it was last year. If school $$X$$ grew by 40 more students than school $$Y$$ did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school $$Y$$ last year? A. 480 B. 1600 C. 1920 D. 2080 E. 2400 We must determine the number of students enrolled at school $$Y$$ last year. We can set up", "? Free Version Easy # College Enrollment by Gender SATMAT-4AETPI The average yearly number of male and female students enrolled at a technical college during the last five decades of the $20$th century is shown above. Which of the following is a true statement? (No Calculator) A The total number of students at this college increased each decade. B The number of females at this college increased each decade. C The number of males at this college decreased each decade. D The difference between the number of males and the number of females decreased each decade.", "18 % 9 . Answer : Option A Explanation : Total number of students appeared = 14000 Total number of students qualified = 1750 + 1250 + 2250 + 2000 + 2000 = 9250 Reqd difference = $14000 - 9250 \\over 5$ = $4750\\over 5$ = 950 10 . Answer : Option D Explanation : Reqd ratio = $1750 \\over 1250$ = $7\\over 5$ = 7 : 5", "## Elementary Algebra Let X represent the amount of students that will be enrolled in the university after the 14% increase. A guideline for this problem would be: After increase enrollment = Before increase enrollment + Increase in enrollment So, X = 250 + 14% $\\times$ 250 So, X = 250 + .14 $\\times$ 250 X = 250 + 35 X = 285 There will be 285 students enrolled in the program.", "maximum possible number of students getting $100$ is $5$.", "# Easy Tabular Data Solved QuestionData Interpretation Discussion Common Information The following table gives the enrollment in Higher Secondary Schools in 1978. Study the table carefully and answer these questions. Table below can be scrolled horizontally Enrollment No. of Schools 20-39 526 40-59 620 60-79 674 80-99 717 100-119 681 120-139 612 140-159 540 160-179 517 180-199 522 Total 5439 Q. Common Information Question: 5/6 What is the number of schools where the enrolment is above 99 but below 160? ✖ A. 2550 ✖ B. 2033 ✔ C. 1833 ✖ D. 1316 Solution: Option(C) is correct Table below can be scrolled horizontally Enrollment No. of Schools 20-39 526 40-59 620 60-79 674 80-99 717 100-119 681 120-139 612 140-159 540 160-179 517 180-199 522 Total 5439 No. of schools with enrolment above 99 & below 160 are: $= 681 + 612 + 540$ $= \\textbf{1833 schools}$", "$$34$$ students were present. Thus, we have the correct solution. Exercise $$\\PageIndex{8}$$ $$20\\%$$ of Mary’s class were ill and stayed home from school. If only $$36$$ students are present, what is the actual size of Mary’s class? $$45$$" ]

[ "# Difference between revisions of \"2016 AMC 8 Problems/Problem 6\" ## Problem The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? $\\textbf{(A) }3\\qquad\\textbf{(B) }4\\qquad\\textbf{(C) }5\\qquad\\textbf{(D) }6\\qquad \\textbf{(E) }7$ $[asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label(\"frequency\",(-0.5,8)); label(\"0\", (-1, 0)); label(\"1\", (-1, 1)); label(\"2\", (-1, 2)); label(\"3\", (-1, 3)); label(\"4\", (-1, 4)); label(\"5\", (-1, 5)); label(\"6\", (-1, 6)); label(\"7\", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label(\"3\", (0.5, -0.5)); label(\"4\", (2.5, -0.5)); label(\"5\", (4.5, -0.5)); label(\"6\", (6.5, -0.5)); label(\"7\", (8.5, -0.5)); label(\"name length\", (4.5, -1)); [/asy]$ ## Solutions ### Solution 1 We first notice that the median name will be the $10^{\\mbox{th}}$ name. The $10^{\\mbox{th}}$ name is $\\boxed{\\textbf{(B)}\\ 4}$. ### Solution 2 To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$. Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$, and $3$ names with length $4$, the $10$th name would have $4$ letters. Thus our answer is $\\boxed{\\textbf{(B)}\\ 4}$.", "fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label(\"6\",.74*dir(-90)+(0,3)); label(\"5\",.74*dir(-60)+(0,3)); label(\"4\",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label(\"30^\\circ\",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*\"12\",1.56*dir(120)); label(rotate(30)*\"1\",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label(\"60^\\circ\",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*\"6\",.74*dir(-150)+(0,3)); label(rotate(-60)*\"5\",.74*dir(-120)+(0,3)); label(rotate(-60)*\"4\",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label(\"90^\\circ\",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*\"6\",.74*dir(-180)+c); label(rotate(-90)*\"5\",.74*dir(-150)+c); label(rotate(-90)*\"4\",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{\\textbf{(C) }\\text{4 o' clock}}$ ## Solution 5 We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We", "0 0.5 06 27 20 15 0 0.5 06 28 25 15 1 0.5 07 29 28 16 1 0.5 07 30 34 17 1 0.5 07 31 19 17 0 0.5 07 32 24 16 0 0.5 08 33 32 18 1 0.5 08 34 25 19 0 0.5 08 35 31 18 0 0.5 09 36 23 20 1 0.5 09 37 20 22 0 0.5 09 38 33 21 0 0.5 09 39 22 20 0 0.5 09 40 35 21 1 0.5 10 41 20 25 0 0.5 10 42 31 26 1 0.5 10 43 34 23 0 0.5 10 44 35 23 1 0.5 10 45 18 25 1 0.5 10 46 23 24 1 0.5 10 47 22 24 0 0.5 The graph below shows that indeed, the blocking formed pairs of classrooms within each school that are similar to one another and correctly left some classrooms in a block by themselves when there was an odd number of classrooms in a school. our_df %>% ggplot(aes(size, block_id, color = as.factor(Z))) + geom_point() + facet_wrap(~school, scales = \"free_y\", ncol = 5) ## Replicate yourself If you set up your assignments as functions like this then it is easy to implement the assignment multiple times: many_assignments <- replicate(1000, draw_data(design)) It’s useful to preserve", "# 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 $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "Both plots are the same, but the upper plot answers my first question, and the latter plot answers my second one. It seems that you only need about 23 students (!!!) before you have a 50/50 chance of at least two students sharing the same birthday. This is WAY off my intuition of there needing to be about 180 students before you could even sniff at a 50% chance of a match. The second plot shows that, to almost guarantee there being a match of birthdays, you need about 74 students. We really are fooled by probability! Poor students! ## R Code for the Simulation #------------------------------------------------------------------------------ ### simulation parameters # define vector of calendar dates calendar <- seq(from = 1, to = 365, by = 1) # how many simulations do you want per class size? nSimulations <- 10000 # what is the size of each class? Here we have class sizes # from 2 to 100 classSize <- seq(from = 2, to = 100, by = 1) #------------------------------------------------------------------------------ #------------------------------------------------------------------------------ ### start simulation here # initiate vector to store data for each class size classData <- numeric(length(classSize)) # initiate looping variable for class sizes j <- 1 # loop over each class size for(currSize in classSize){ # initiate vector to store counts of replicated birthdays data <- numeric(nSimulations)", "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", "4, 3, 5, 7))) This new function would generate a data set that reflects the number of classes in each school. ## Make better blocks from richer data Say that you had even richer data about the sampling frame. Then you could use this to do even better assignments. Say this is what you get from your partner: We just got information on the class sizes. Here it is: School 1: 20, 25, 23, 30, 12, 15; School 2: 40, 42, 53, 67, 35, 22, 18, 18; School 3: 34, 37, 28, 30; School 4: 18, 24, 20; School 5: 10, 24, 13, 26, 18; School 6: 20, 25; School 7: 28, 34, 19, 24; School 8: 32, 25, 31; School 9: 23, 20, 33, 22, 35; School 10: 20, 31, 34, 35, 18, 23, 22 We want to incorporate this information in the data fabrication step. make_data <- declare_population( class = add_level(N = c(6, 8, 4, 3, 5, 2, 4, 3, 5, 7), size = c( 20, 25, 23, 30, 12, 15, # students in each classroom of school 1 40, 42, 53, 67, 35, 22, 18, 18, # students in each classroom of school 2 34, 37, 28, 30, # etc... 18, 24, 20, 10, 24, 13, 26, 18, 20, 25, 28, 34, 19, 24, 32,", "# 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", "1, 1, 1, 1, 1, 1, 1,… ##$ spontaneous <dbl> 2, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1,… ## $stratum <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, … ##$ pooled.stratum <dbl> 3, 1, 4, 2, 32, 36, 6, 22, 5, 19, 2… data <- infert %>% select(education, age, parity) %>% data ## education age parity ## 1 0-5yrs 26 6 ## 2 0-5yrs 42 1 ## 3 0-5yrs 39 6 ## 4 0-5yrs 34 4 ## 5 6-11yrs 35 3 ## 6 6-11yrs 36 4 In the code chunk below, examine the local dataset, emerg_dept, which has counts of ED arrivals, how many breached the UK 4 hour guarantee, and how many got admitted. Then select() only org_code, attendances, breaches, and admissions. Then arrange() to have the top attendances at the top, use top_n(10) to get only the top 10 rows. Assign this to data and print it out emerg_dept ## # A tibble: 50 × 6 ## period org_code type attendances breaches admissions ## <date> <fct> <fct> <dbl> <dbl> <dbl> ## 1 2018-07-01 RRK 1 32209 6499 11332 ## 2 2018-07-01 R1H 1 28357 6294 7986 ## 3 2018-07-01 RW6 1 23887 4641 6282 ## 4 2018-07-01 R0A 1 22012 4669 6818 ## 5 2018-07-01 RDU 1 21043", "'step2014', 'step2015', 'decile.change', 'step.change') -> names(dc) c(LETTERS[1:17], 'Z') -> steps c(905.81, 842.11, 731.3, 617.8, 507.01, 420.54, 350.25, 277.32, 220.59, 182.74, 149.99, 135.12, 115.76, 93.71, 71.64, 46.86, 28.93, 0) -> money within(dc, { ifelse(step2014 == '', NA, step2014) -> step2014 ifelse(step2015 == '', NA, step2015) -> step2015 sapply(step2014, function(x) money[match(x, steps)]) -> sm2014 sapply(step2015, function(x) money[match(x, steps)]) -> sm2015 sm2015 - sm2014 -> funding.change }) -> dc merge(dc, direc[, c('School.ID', 'Total.School.Roll', 'Urban.Area', 'Regional.Council')], by = 'School.ID', all.x = TRUE) -> dc within(dc, { factor(Urban.Area) -> Urban.Area factor(Urban.Area, levels(Urban.Area)[c(3, 2, 4, 1)]) -> Urban.Area funding.change*Total.School.Roll/1000 -> school.level.change }) -> dc with(dc, { summary(funding.change) summary(school.level.change) sum(sm2014*Total.School.Roll/1000000, na.rm = NA) sum(sm2015*Total.School.Roll/1000000, na.rm = NA) }) #### By Urban area # Funding change per student on school size qplot(Total.School.Roll, funding.change, data = dc[!is.na(dc$Urban.Area),], alpha = 0.8, xlab = 'Total School Roll', ylab = 'Funding change per student (NZ$)') + theme_bw() + theme(legend.position = 'none') + facet_grid(~Urban.Area) Not too alien; now using the magrittr package with right assign would make a lot more sense. Categories ## Sometimes I feel (some) need for speed I’m the first to acknowledge that most of my code could run faster. The truth of the matter is that, in essence, I write ‘quickies’: code that will run once or twice, so there is no incentive to spend days or", "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", "percent of the area of rectangle $AQRD$ is shaded? $[asy] filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label(\"D\",(0,0),S); label(\"R\",(25,0),S); label(\"Q\",(25,15),N); label(\"A\",(0,15),N); filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black); label(\"S\",(10,0),S); label(\"C\",(15,0),S); label(\"B\",(15,15),N); label(\"P\",(10,15),N);[/asy]$ $\\textbf{(A)}\\ 15\\qquad\\textbf{(B)}\\ 18\\qquad\\textbf{(C)}\\ 20\\qquad\\textbf{(D)}\\ 24\\qquad\\textbf{(E)}\\ 25$ ## Problem 14 There are $270$ students at Colfax Middle School, where the ratio of boys to girls is $5 : 4$. There are $180$ students at Winthrop Middle School, where the ratio of boys to girls is $4 : 5$. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls? $\\textbf{(A) } \\dfrac7{18} \\qquad\\textbf{(B) } \\dfrac7{15} \\qquad\\textbf{(C) } \\dfrac{22}{45} \\qquad\\textbf{(D) } \\dfrac12 \\qquad\\textbf{(E) } \\dfrac{23}{45}$ ## Problem 15 How many digits are in the product $4^5 \\cdot 5^{10}$? $\\textbf{(A) } 8 \\qquad\\textbf{(B) } 9 \\qquad\\textbf{(C) } 10 \\qquad\\textbf{(D) } 11 \\qquad\\textbf{(E) } 12$ ## Problem 16 Let $A$ be the area of the triangle with sides of length $25, 25$, and $30$. Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$. What is the relationship between $A$ and $B$? $\\textbf{(A) } A = \\dfrac9{16}B \\qquad\\textbf{(B) } A = \\dfrac34B \\qquad\\textbf{(C) } A=B \\qquad\\textbf{(D) } A = \\dfrac43B \\\\ \\\\ \\textbf{(E) }A = \\dfrac{16}9B$ ## Problem 17 Let $w$, $x$, $y$, and $z$ be whole numbers. If $2^w \\cdot 3^x \\cdot 5^y", "# Difference between revisions of \"1998 AJHSME Problems/Problem 23\" ## Problem If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle? $[asy] unitsize(10); draw((0,0)--(12,0)--(6,6sqrt(3))--cycle); draw((15,0)--(27,0)--(21,6sqrt(3))--cycle); fill((21,0)--(18,3sqrt(3))--(24,3sqrt(3))--cycle,black); draw((30,0)--(42,0)--(36,6sqrt(3))--cycle); fill((34,0)--(32,2sqrt(3))--(36,2sqrt(3))--cycle,black); fill((38,0)--(36,2sqrt(3))--(40,2sqrt(3))--cycle,black); fill((36,2sqrt(3))--(34,4sqrt(3))--(38,4sqrt(3))--cycle,black); draw((45,0)--(57,0)--(51,6sqrt(3))--cycle); fill((48,0)--(46.5,1.5sqrt(3))--(49.5,1.5sqrt(3))--cycle,black); fill((51,0)--(49.5,1.5sqrt(3))--(52.5,1.5sqrt(3))--cycle,black); fill((54,0)--(52.5,1.5sqrt(3))--(55.5,1.5sqrt(3))--cycle,black); fill((49.5,1.5sqrt(3))--(48,3sqrt(3))--(51,3sqrt(3))--cycle,black); fill((52.5,1.5sqrt(3))--(51,3sqrt(3))--(54,3sqrt(3))--cycle,black); fill((51,3sqrt(3))--(49.5,4.5sqrt(3))--(52.5,4.5sqrt(3))--cycle,black); [/asy]$ $\\text{(A)}\\ \\frac{3}{8}\\qquad\\text{(B)}\\ \\frac{5}{27}\\qquad\\text{(C)}\\ \\frac{7}{16}\\qquad\\text{(D)}\\ \\frac{9}{16}\\qquad\\text{(E)}\\ \\frac{11}{45}$ ## Solution All small triangles are congruent in each iteration of the diagram. The number of shaded triangles follows the pattern: $0, 1, 3, 6, ...$ which is the pattern of \"triangular numbers\". Each time, the number $1, 2, 3, 4, 5...$ is added to the previous term. Thus, the first eight terms are: $0, 1, 3, 6, 10, 15, 21, 28$ In the eighth diagram, there will be $28$ shaded triangles. The total number of small triangles follows the pattern: $1, 4, 9, 16, ...$ which is the pattern of \"square numbers\". Thus, the eighth triangle will be divided into $8^2 = 64$ small triangles in total. The ratio of shaded to total triangles will be the fraction of the whole figure that's shaded, since all triangles are congruent. Thus, the answer is $\\frac{28}{64} = \\frac{7}{16}$, and the correct choice is $\\boxed{C}$ 1998 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 •", "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", "%heartsflow, 0.3 minimum opacity HeartsOnPath(-15,-5,-10,-4.5,0.07,100); HeartsOnPath(-10,-4.5,0,-3.5,0.07,100); HeartsOnPath(0,-3.5,2,-5,0.07,50); HeartsOnPath(2,-5,5,-3,0.07,70); HeartsOnPath(5,-3,8,0,0.1,70); HeartsOnPath(8,0,11,3.5,0.1,70); HeartsOnPath(11,3.5,10,7,0.1,70); HeartsOnPath(10,7,13,10,0.07,70); HeartsOnPath(-9,-1.5,-8,4,0.1,70); HeartsOnPath(-8,4,0,5,0.1,50); HeartsOnPath(0,3,10,8,0.1,50); } %stripes \\draw [draw=none,fill=ThirdPurple] (-15,-6) to[out=20,in=170] (-5,-5) to[out=350,in=330] (3,-1) to[out=150,in=210] (2,2) to[out=30,in=180] (15,11) -- (15,9) to[out=180,in=30] (3,1) to[out=210,in=150] (4,-1) to[out=330,in=350] (-5,-6.2) to[out=170,in=20] (-15,-7) -- cycle; \\draw [draw=none,fill=SecondPurple] (-15,-4) to[out=20,in=170] (-5,-4) to[out=350,in=330] (3,0) to[out=150,in=210] (2,3) to[out=30,in=180] (15,12) -- (15,10) to[out=180,in=30] (3,2) to[out=210,in=150] (4,0) to[out=330,in=350] (-5,-5) to[out=170,in=20] (-15,-6) -- cycle; \\draw [draw=none,fill=FirstPurple] (-15,-4) to[out=20,in=170] (-5,-3) to[out=350,in=330] (3,1) to[out=150,in=210] (2,4) to[out=30,in=180] (13,11) -- (14,11) to[out=180,in=30] (3,3) to[out=210,in=150] (4,1) to[out=330,in=350] (-5,-4) to[out=170,in=20] (-15,-5.3) -- cycle; %bigger hearts \\heart{(5.5,1.5)}{0}{2.4}{1}{BigHeartColor} \\heart{(-1.5,2)}{20}{1}{1}{SmallerHeartColor} \\heart{(9.5,4)}{-20}{0.8}{1}{SmallestHeartColor} %light strip (-16,7) to[out=-10,in=170] (1,0) to[out=-10,in=200] (9,1) to[out=20,in=-20] (6,3) to[out=160,in=-80] (2,8) to[out=100,in=100] (3,9) to[out=-80,in=100] (4,6) to[out=-80,in=-90] (17,-4) -- (17,-3.7) to[out=-90,in=-80] (4,6.3) to[out=100,in=-80] (3,9.3) to[out=100,in=100] (2,8.3) to[out=-80,in=160] (6,3.3) to[out=-20,in=20] (9,1.3) to[out=200,in=-10] (1,0.3) to[out=170,in=-10] (-16,7.3) -- cycle; %cats \\node [xshift=5.5cm] {\\includegraphics{kittenz.pdf}}; %Mau \\color{white}\\Scriptina \\begin{scope}[shift={(1,0.8)}] \\contourlength{0.01em} \\node [xscale=6,yscale=6] at (-9,-0.5) {\\contour{TextOutlineColor}{M}}; \\contourlength{0.02em} \\node [xscale=3,yscale=3] at (-4,-2.55) {\\contour{TextOutlineColor}{a}}; \\node [xscale=3,yscale=3] at (-1.7,-2.5) {\\contour{TextOutlineColor}{u}}; %restore Ma 'ligature' \\draw [draw=none,fill=white,rotate around={55:(-4.75,-2.77)},shift={(-4.83,-2.79)}] (0,0) rectangle (0.2,0.08); \\end{scope} %Happy Birthday \\Zapfino \\contourlength{0.02em} \\node [xscale=3,yscale=3] at (-3,4.5) {\\contour{TextOutlineColor}{Happy Birthday}}; \\color{LastILOVEYOUColor}\\rmfamily \\node at (0,-2.64) {\\contour{LastILOVEYOUOutlineColor}{ILOVEYOU}}; %light bubbles, 20% more white, 0.3 minimum opacity LightsOnPath(-15,5,15,-5,0.07,200); } \\end{tikzpicture} \\end{document} Aaaaaaaaaaaand this is the result: Yes, it's pink. I like pink. Here's the pdf. ## Top reasons you should upvote this entry • pink • hearts • cats • love •", "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", "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", "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" ]

[ "fraction to lowest terms, divide the numerator and denominator by their greatest common factor (GCF).", "# How do you find the GCF of 7x^2, 15xy? Mar 10, 2018 The GCF is $x$. $7 {x}^{2} = 7 \\cdot x \\cdot x$ $15 x y = 3 \\cdot 5 \\cdot x \\cdot y$ $\\therefore x$ is the Greatest (and only) Common Factor between the two expressions.", "GC.", "of the following lines completes the above code? a) sm = i b) sm += i c) i = sm d) i += sm The line “sm += i” completes the above code. 9. Which of the following methods used to find the sum of first n natural numbers has the least time complexity? a) Recursion b) Iteration c) Binomial coefficient d) All have equal time complexity Recursion and iteration take O(n) time to find the sum of first n natural numbers while binomial coefficient takes O(1) time. 1. Which of the following is not another name for GCD(Greatest Common Divisor)? a) LCM b) GCM c) GCF d) HCF LCM (Least Common Multiple) and GCD are not the same. GCM (Greatest Common Measure), GCF (Greatest Common Factor), and HCF (Highest Common Factor) are other names for GCD. 2. What is the GCD of 8 and 12? a) 8 b) 12 c) 2 d) 4 GCD is the largest positive integer that divides each of the integers. So the GCD of 8 and 12 is 4. 3. If the GCD of two numbers is 8 and LCM is 144, then what is the second number if the first number is 72? a) 24 b) 2 c) 3 d) 16 If the GCD of two numbers is 8 and LCM", "about LCM-GCD. Let’s see the ways in which a question can encrypt the message of LCM-GCD. GCD can be expressed as the following terms/phrases: a) x is the GCD of 12 & 18 b) x is the HCF of 12 & 18 (HCF stands for Highest Common Factor) c) x is the highest number which divides both 12 & 18 All the 3 terms above refer to the GCD of 12 & 18 i.e. 6. Try focusing on keywords such as ‘Highest’, ‘Greatest’. These words are an indication of the question talking about the GCD of a set of numbers in a LCM-GCD question. Let’s see now the terms/phrases which are used to express LCM: a) z is the LCM of 12 & 18 b) z is the lowest number which is divisible by both 12 & 18 c) z is the lowest number which has 12 & 18 as its factors All the above terms refer to the LCM of 12 & 18 i.e. 36. Try focusing on keyword ‘lowest’ which can be an indication of question talking about LCM of a set of numbers. The important point to remember for identification of both LCM & GCD is the definition of these terms. While GCD is the factor/divisor of a set of numbers, LCM is the multiple", "situations a common monomial factor ( GCF ) between,! Factor the quartic polynomial x 4 8 x 3 + 22 x 2 x. It can factor by grouping not written in the parentheses: x^2 ( x-2 ) ( x+1 ) it equal...", ") Now let's factor by grouping: Group like terms Factor out the GCF of out of the first group. Factor out the GCF of out of the second group Since we have a common term of , we can combine like terms So factors to So this also means that factors to (since is equivalent to ) -------------------------------", "lcm to be the least upper bound in the lattice of divisibility. The LCM is familiar from grade-school arithmetic as the \"least common denominator\" (LCD) that must be determined before fractions can be added, subtracted or compared. In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more integers (at least one of which is not zero), is the largest positive integer that divides the numbers without a remainder. For example, the GCD of 8 and 12 is 4. This notion can be extended to polynomials, see Polynomial greatest common divisor, or to rational numbers (with integer quotients). Mathematics Multiplier Religion Belief 13", "prime factor trees, GCFs and LCMs. In Build mode, you can build numbers by multiplying primes together. Simplifying an expression often means removing a pair of parentheses; factoring an expression often means applying them.", "## Algebra 2 Common Core $5(x+2y-7)$ The GCF (Greatest Common Factor) of the coefficients is 5. So we can factor out 5 from each term of the expression to get $(5)x+(5)2y-(5)7=5(x+2y-7)$.", "active figure. By contrast, gcf refers to the current active figure handle.", "called the greatest common factor or GCF or GCF given... Rectangle contains the next trinomial to be able to master on the left have the same as... Descending order with the leading coefficient, is not 1 negative times positive... Of grouping words, there must be an exponent of ' 2 ' and exponent... Trinomials when a \\ne 1 a = 1 Objective: factor trinomials with guessing. Games, and other study tools find a … Factoring-polynomials.com supplies great facts on trinomial factoring Calculator, subtracting and. Expression made up of three terms have anything in common, called the greatest factor! ' and that exponent must be the greatest common factor or GCF at Another Example Now you.... - trinomials where the coefficient of x2 is 1 ( Integrated ) Identify. Signs of the rectangle has the answer to the factoring question and the bottom of the.... Common, called the greatest common factor or GCF is x 8. ) ( the... Supplies great facts on trinomial factoring Calculator, subtracting fractions and rational numbers and other math areas! Whenever the coefficient in front of x2 is one and b: 0 for the student given... Method of grouping factored by using a method for factoring a trinomial whose leading coefficient can also factored! The square of x 4 learn vocabulary, terms, and other", "The function GCD - Greatest Common Divisor is used to simplify the ratio to its lowest terms. Simplify Calculator. GCF = 5 GCF = 5. This calculator will simplify polynomials as much as possible. ... $\\frac{a+b+c}{3} - \\frac{a-b+c}{6} - \\frac{a+b-c}{6} =$ Quick Calculator Search. If both A and B are whole numbers, Find the greatest common factor of A and B, Then the greatest common factor of 6 and 10 is 2. Simplify Ratios: Enter A and B to find C and D. (or enter C and D to find A and B) The calculator will simplify the ratio A : B if possible. The calculator works for both numbers and expressions containing variables. A benefit-cost ratio (BCR) is an indicator showing the relationship between the relative costs and benefits of a proposed project, expressed in monetary or qualitative terms. Played 339 times. You need to login to view this content. This website uses cookies to ensure you get the best experience. This will reduce the ratio to its simplest form. They can be different types, for example, one fraction and one decimal. But what is phi minus 1? Approach 1: Procedure to Simplify Ratio A : B, when both are whole numbers. 14th December 2020 / by johan1. First, find out the factors for A and B;", "a trinomial, meaning it contains three terms with the following coefficients and constant: a = 3, b = 9, c = 21 Step 1: Find the GCF, or greatest common factor. The GCF of a, b, and c is 3. See the factors below: The factors of 3 are +/-1 and +/-3. The factors of 9 are +/-1, +/-3, +/-9. The factors of 21 are +/-1, +/-3, +/-7, +/-21. Step 2: Now, that we've identified our greatest common factor, let's simplify the expression. Since 3 is the greatest common factor, we will factor a \"3\" out of our a, b, and c, or 3, 9, and 21. 3(x^{2} + 3x + 7) Step 3: Now, we have two expressions. Since 3 is a constant and a monomial, it cannot be simplified any further. Let's check the trinomial x^{2} + 3x + 7. Begin with step 1, finding the GCF. The greatest common factor is \"1,\" thus, the expression is in its simplist form. The final answer is 3(x^{2} + 3x + 7). ## Contact tutor Send a message explaining your needs and April will reply soon. Contact April Start Lesson ## FAQs What is a lesson? A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate", "numbers a, b, and c is simply lcm(lcm(a, b), c). 1. Two trains leave a station at 12:00 noon. The A train visits the station every 25 minutes, and the B train visits the station every 15 minutes. At what time will the two trains meet in the station again for the first time? 2. What is the smallest integer that is a multiple of 14, 9, and 21? Algebra # The Greatest Common Denominator Function Two of the most basic functions in number theory are the LCM (least common multiple) and GCD/GCF (greatest common denominator/factor) functions. They are opposites of each other, but not in the same way as addition and subtraction, multiplication and division, etc. What do these functions actually mean, and how can we use repetitive methods to evaluate expressions using LCM and GCD? Greatest Common Denominator (GCD) First let’s look at the Greatest Common Denominator function. Written as gcd(a, b), the GCD function determines the largest number by which a and b are both divisible. For instance, the GCD of 6 and 8 is 2 because 3, 4, 5, and 6 cannot be divided evenly into both 6 and 8. Evaluate the GCD of the following pairs of numbers: 3 and 6, 15 and 20, 8 and 20, 3 and 5. As evidenced by", "I know, the \"official/standard\" names are: GCD (Greatest Common Divisor) LCM (Least Common Multiple) A LCD is a Liquid Crystal Display! The \"GCD\" can be the \"greatest common divisor,\" but \"GCF\" is referred to as the \"greatest common factor,\" and \"GCF\" I have found is more commonly used than \"GCD.\" Another point of view: http://en.wikipedia.org/wiki/Greatest_common_factor ------------------------------------------------------------------------------------------------------------------------------------ The \"LCM\" can be the \"least common multiple,\" but when you are referring to the denominators, it is more specifically the \"LCD,\" the \"least common denominator.\" If you are finding the LCD of two or more fractions, then you are finding the LCM of their denominators. But if you are finding the LCM of two or more numbers, then you are not necessarily even working with fractions. Another point of view: http://en.wikipedia.org/wiki/Least_common_denominator #### Denis ##### Senior Member Ya, that's all nice...but not my point... Searches by \"learning students\" will lead to sites like: http://www.tigernt.com/math/math5gl.php Programming languages (at least the ones I'm familiar with) all use GCD and LCM as commands. I'm not \"against\" using others, but why confuse the learning student... Anyway, it all matters little in the long-run; once the student \"grasps/sees\" what's going on, the \"name\" matters none :idea: #### lookagain ##### Senior Member Denis said: I'm not \"against\" using others, but why confuse the learning student... There is **no**", "# What is the LCM of 6 and 7? To find the LCM of two numbers, I like to do $\\frac{a b}{G C F}$, where a and b, in this case, are 6 and 7, and the GCF is the greatest common factor(1, in this case).", "out this page at Purple Math about LCM and GCF: http://www.purplemath.com/modules/lcm_gcf.htm", "## Algebra 1: Common Core (15th Edition) You have $14x^{2}$+7x.You find the GCF by finding the highest factor between the two terms. 7x is the greatest factor between both of the terms, so 7x is the GCF.", "2) Divide by using GCF. A fraction is in the lowest term when the only common factor between the numerator and the denominator is 1." ]

[ "# 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 $3 \\times 4 \\times 7 = 84$.", "## Elementary Algebra $12$ Factor each number completely to have: $48 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 3 \\\\60 = 2 \\cdot 2 \\cdot 3 \\cdot 5 \\\\84= 2 \\cdot 2 \\cdot 3 \\cdot 7$ Since two 2s and one 3 are common to both, the greatest common factor of the given numbers is: $2 \\cdot 2 \\cdot 3 = 12$", "8 and 3? • there are no common factors between the numbers, so common multiple is not defined. • 24,48, cdots • 24,48, cdots The answer is \"24,48, cdots\" slide-show version coming soon", "## Elementary Algebra $16$ Factor each number completely to have: $32 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\\\80 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 5 \\\\96= 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 3$ Since four 2s are common to both, the greatest common factor of the given numbers is: $2 \\cdot 2\\cdot 2\\cdot 2 = 16$", "## 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: $=2 \\cdot 2\\cdot 2 \\cdot 2 \\cdot 3 \\\\= 48$", "7$ $98=2 \\times 7^2$ least common multiple is $2^2 \\times 3 \\times 7^2 = 588$ #### Ask your own question Sign Up Find more explanations on OpenStudy Privacy Policy", "the least common multiple of $180$, $216$, and $450$?", "out in presence of 1728 ($1728=16 \\times108$ and $1728=48 \\times 36$) but it would have been the same with 108 as a factor in the j-invariant, which is the least common multiple. – Youssef El Housni Nov 28 '18 at 19:23", "## 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: $=2 \\cdot 3\\cdot 7 \\cdot 11 \\\\= 462$", "## 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: $=2 \\cdot 2\\cdot 2 \\cdot 3 \\cdot 3 \\\\= 72$", "look. $12 = 2 \\times 2 \\times 3$ $15 = 3 \\times 5$ To find the LCM, we want to have all the prime factors from both numbers accounted for. For instance, there are two 2s (in the 12). Let's put those in: $L C M = 2 \\times 2 \\times \\ldots$ There is one 3 in both the 12 and the 15, so we need one 3: $L C M = 2 \\times 2 \\times 3 \\times \\ldots$ And there is one 5 (in the 15) so let's put that in: $L C M = 2 \\times 2 \\times 3 \\times 5 = 60$ $12 \\times 5 = 60$ $15 \\times 3 = 60$ 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 2 Jan 26, 2018 The least common multiple of $12$ and $15$ is $3$. #### Explanation: It would be 3 because... • $1 \\cdot 12 = 12$ • $2 \\cdot 6 = 12$ • $3 \\cdot 4 = 12$ • $4 \\cdot 3 = 12$ • $6 \\cdot 2 = 12$ and • $1 \\cdot 15 = 15$ • $3 \\cdot 5 = 15$ The least common multiple between $12$ and $15$ is $3$ because it", "# 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", "$18 = 2\\cdot 3\\cdot 3$ $6 = 2\\cdot 3$ $6\\,$ ‘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", "## Elementary Algebra $18$ Factor each number completely to obtain: $36 = 2 \\cdot 2 \\cdot 3 \\cdot 3 \\\\72 = 2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 3 \\\\90= 2 \\cdot 3 \\cdot 3 \\cdot 5$ Since one 2 and two 3s are common to both, the greatest common factor of the two given numbers is: $2 \\cdot 3 \\cdot 3 = 18$", "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.", "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", "+0 # PLS HELP ASAP +2 259 2 +140 The least common multiple of two positive integers is $7!$, and their greatest common divisor is $9$. If one of the integers is $315$, then what is the other? (Note that $7!$ means $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1$.) Mar 19, 2019 #1 +2 7! / 315 = 16 16 * 9 (Their GCD) = 144 LCM{315, 144} =7! GCD{315, 144} =9 Mar 19, 2019 edited by Guest Mar 19, 2019 #2 +140 +1 thanks CorbellaB.15 Mar 27, 2019", "$126$126 and $294$294, we start by drawing their factor trees: This gives us their prime factorizations in expanded form: $126=2\\times3\\times3\\times7$126=2×3×3×7 $294=2\\times3\\times7\\times7$294=2×3×7×7 We then find what factors appear in both factorizations - they have one $2$2, one $3$3, and one $7$7 in common. The greatest common factor is the product of the common prime factors, $2\\times3\\times7=42$2×3×7=42. ### Least common multiple When we multiply two numbers together, these two numbers are always factors of the new number. Another way to look at it is to say the new number is a common multiple of both of the original ones. Below is a table with the first few multiples of $12$12 and $18$18, with their common multiples listed in the last row: Multiples of $12$12 and $18$18 Multiples of $12$12 $12,24,36,48,60,72,84,96,108,\\ldots$12,24,36,48,60,72,84,96,108, Multiples of $18$18 $18,36,54,72,90,108,126,144,\\ldots$18,36,54,72,90,108,126,144, Common multiples $36,72,108,\\ldots$36,72,108, The common multiples are the numbers that can be broken up into both the original numbers equally. • $36$36 can be broken up into $3$3 groups of $12$12, and $2$2 groups of $18$18 • $72$72 can be broken up into $6$6 groups of $12$12, and $4$4 groups of $18$18 • $108$108 can be broken up into $9$9 groups of $12$12, and $6$6 groups of $18$18 The smallest number in the list of common multiples, $36$36, is called the least common multiple (LCM) of", "factor. Consider the integers 234, 126 and 273. To find the least common multiple of these numbers we write each integer as a product of its prime factors. Here we present a systematic way to find the prime factorization of a number. • Try the prime numbers, in order, as factors. • Use repeatedly until it is no longer a factor. • Then try the next prime: $234 & = 2 \\cdot 117 \\\\& = 2 \\cdot 3 \\cdot 39 \\\\& = 2 \\cdot 3 \\cdot 3 \\cdot 13 \\\\234 & = 2 \\cdot 3^2 \\cdot 13$ $126 & = 2 \\cdot 63 \\\\& = 2 \\cdot 3 \\cdot 21 \\\\& = 2 \\cdot 3 \\cdot 3 \\cdot 7 \\\\126 & = 2 \\cdot 3^2 \\cdot 7$ $273 & = 3 \\cdot 91 \\\\& = 3 \\cdot 7 \\cdot 13$ Once we have the prime factorization of each number, the least common multiple of the numbers is the product of all the different factors taken to the highest power that they appear in any of the prime factorizations. In this case, the factor of two appears at most once, the factor of three appears at most twice, the factor of seven appears at most once, the factor of 13 appears at most once. Therefore, $\\text{LCM} = 2 \\cdot", "+0 # Question 0 232 2 What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594 Guest Mar 11, 2017 #1 0 LCM[180, 594] =5,940. GCD[180,594] =18 Ratio =5,940 / 18 =330 Guest Mar 11, 2017 #2 +7348 +5 What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594 $$\\frac{180}{594}=\\frac{2^2\\cdot3^2\\cdot5}{2\\cdot3^3\\cdot11}\\ \\ lcm=2^2\\cdot3^3\\cdot5\\cdot11\\ \\ gcf=2\\cdot3^2$$ $$\\frac{lcm}{gcf}=\\frac{2^2\\cdot3^3\\cdot5\\cdot11}{2\\cdot3^2}=\\frac{2\\cdot3\\cdot5\\cdot11}{1}\\color{blue}=330:1$$ ! asinus Mar 11, 2017 edited by asinus Mar 11, 2017" ]

[ "# How do you find the GCF of 7x^2, 15xy? Mar 10, 2018 The GCF is $x$. $7 {x}^{2} = 7 \\cdot x \\cdot x$ $15 x y = 3 \\cdot 5 \\cdot x \\cdot y$ $\\therefore x$ is the Greatest (and only) Common Factor between the two expressions.", "# What is the GCF and LCM for 52r2s, 78rs^2t? Oct 9, 2015 First factorize: $52 {r}^{2} s = 2 \\cdot 2 \\cdot 13 \\cdot r \\cdot r \\cdot s$ and $78 r {s}^{2} t = 2 \\cdot 3 \\cdot 13 \\cdot r \\cdot s \\cdot s \\cdot t$ #### Explanation: GCF : Take all the common factors: $2 \\cdot 13 \\cdot r \\cdot s = 26 r s$ Check : $\\frac{52 {r}^{2} s}{26 r s} = 2 r$ and $\\frac{78 r {s}^{2} t}{26 r s} = 3 s t$ have no common factors LCM : Take all the factors to their highest degree: $2 \\cdot 2 \\cdot 3 \\cdot 13 \\cdot r \\cdot r \\cdot s \\cdot s \\cdot t = 156 {r}^{2} {s}^{2} t$ Check : $\\frac{156 {r}^{2} {s}^{2} t}{52 {r}^{2} s} = 3 s t$ and $\\frac{156 {r}^{2} {s}^{2} t}{78 r {s}^{2} t} = 2 r$", "we have got: $1734 = 2\\cdot 3\\cdot 17^{2}$ and $51894 =$ $2\\cdot 3^{3}\\cdot 31^{2}$ To find the Highest Common Factor we would multiply 2 and 3 because they appear as prime factors of both numbers. LCM would then be: $2 \\cdot 3^{2} \\cdot 17^{2} \\cdot 31^{2}$ Let's imagine a situation where we've got two sets of prime factors of some numbers: 1. $2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 5$ 2. $2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 7$ Now, the HCF would be: $2 \\cdot 2 \\cdot 3 \\cdot 3$ and LCM: $2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7$ Am I correct? 4. Yep. That's how you find a GCD and LCM. Different terminology for the same thing. Some use HCF, highest common factor, as you done. I reckon it depends on where one hails from.", "3\\cdot \\goldD{x}\\cdot \\goldD{x}\\cdot x\\cdot x\\cdot x$ • $4x^2=\\blueD2\\cdot 2\\cdot \\goldD{x}\\cdot\\goldD{x}$ Notice that the GCF is indeed $2x^2$. This is especially helpful to understand when finding the GCF of monomials with very large powers of $x$. For example, it would be very tedious to completely factor monomials like $32x^{100}$ and $16x^{88}$! ## Challenge Problems 4*)What is the greatest common factor of $20x^{76}$ and $8x^{92}$? Notice that $20x^{76}$ is the monomial with the lowest power of $x$. So the variable part of the GCF will be $x^{76}$. The GCF of $20$ and $8$ is $4$. Therefore, the GCF of $20x^{76}$ and $8x^{92}$ is $4x^{76}$. 5*) What is the greatest common factor of $40x^5y^2$ and $32x^2y^3$? To find the greatest common factor of $40x^5y^2$ and $32x^2y^3$, let's factor each monomial completely, and see what is common. • $40x^5y^2=\\blueD2\\cdot \\blueD2\\cdot \\blueD{2}\\cdot 5\\cdot \\goldD{x}\\cdot \\goldD{x}\\cdot {x}\\cdot x\\cdot x \\cdot\\greenD{y}\\cdot \\greenD{y}$ • $32x^2y^3=\\blueD{2}\\cdot \\blueD{2}\\cdot\\blueD{2}\\cdot{2}\\cdot {2}\\cdot \\goldD{x}\\cdot \\goldD{x}\\cdot\\greenD{y}\\cdot \\greenD{y}\\cdot y$ The monomials each have three factors of $\\blueD{2}$, two factors of $\\goldD{x}$, and two factors of $\\greenD{y}$ in common. Therefore, the greatest common factor of the polynomial is $\\blueD2\\cdot\\blueD2\\cdot \\blueD 2 \\cdot \\goldD{x}\\cdot \\goldD{x}\\cdot \\greenD{y}\\cdot \\greenD{y}$ or $8x^2y^2$. Alternatively, notice that the lowest power of $x$ is $x^2$ and the lowest power of $y$ is $y^2$. So the variable part of the GCF will be", "# Difference between revisions of \"Greatest common divisor\" The greatest common divisor (GCD, or GCF (greatest common factor)) of two or more integers is the largest integer that is a divisor of all the given numbers. The GCD is sometimes called the greatest common factor (GCF). A very useful property of the GCD is that it can be represented as a sum of the given numbers with integer coefficients. From here it immediately follows that the greatest common divisor of several numbers is divisible by any other common divisor of these numbers. ## Finding the GCD ### Using prime factorization Once the prime factorizations of the given numbers have been found, the greatest common divisor is the product of all common factors of the numbers. Example: $270=2\\cdot3^3\\cdot5$ and $144=2^4\\cdot3^2$. The common factors are 2 and $3^2$, so $GCD(720,144)=2\\cdot3^2=18$. ### Euclidean algorithm The Euclidean algorithm is much faster and can be used to give the GCD of any two numbers without knowing their prime factorizations. To find the greatest common divisor of more than two numbers, one can use the recursive formula $GCD(a_1,\\dots,a_n)=GCD(GCD(a_1,\\dots,a_{n-1}),a_n)$. ### Using least common multiple The GCD of two numbers can also be found using the equation $GCD(x, y) \\cdot LCM(x, y) = x \\cdot y$, where $LCM(x,y)$ is the least common multiple of $x$ and", "+0 # what is the LCM of 8847 and 2532 0 572 3 what is the LCM of 8847 and 2532 Guest Feb 18, 2015 #2 +19995 +10 What is the LCM of 8847 and 2532 ? $$\\small{\\text{The following formula reduces the problem of computing the least common multiple}}\\\\ \\small{\\text{to the problem of computing the greatest common divisor (GCD), }}\\\\ \\small{\\text{also known as the greatest common factor:}}\\\\\\\\ \\small{\\text{ \\boxed{lcm(a,b)=\\dfrac{|a\\cdot b|}{\\text{gcd}(a,b)} } . }}$$ There is the Euclidean algorithm for computing the GCD. $$\\small{\\text{gcd(8847,2532)}} \\\\ \\small{\\text{ \\begin{array}{rrrr} & & q & r\\\\ 8847 & 2532 & 3 & 1251 \\\\ 2532 & 1251 & 2 & 30\\\\ 1251 & 30 & 41 & 21\\\\ 30 & 21 & 1 & 9 \\\\ 21 & 9 & 2 & {3} \\\\ 9 & 3 & 3 & algorithm Ends 0 \\end{array} }}\\\\\\\\ \\small{\\text{gcd(8847,2532)}} = 3 \\\\$$ $$\\small{\\text{lcm(8847,2532)=\\dfrac{|8847\\cdot 2532|}{3} = 7466868 }}$$ heureka Feb 18, 2015 #1 +26971 0 The LCM of 8847 and 2532 is 7466868 . Alan Feb 18, 2015 #2 +19995 +10 What is the LCM of 8847 and 2532 ? $$\\small{\\text{The following formula reduces the problem of computing the least common multiple}}\\\\ \\small{\\text{to the problem of computing the greatest common divisor (GCD), }}\\\\ \\small{\\text{also known as the greatest common factor:}}\\\\\\\\ \\small{\\text{ \\boxed{lcm(a,b)=\\dfrac{|a\\cdot b|}{\\text{gcd}(a,b)} } . }}$$ There", "# How do you find the GCF of 12x^5y, 56x^2y^3? Nov 23, 2016 $4 {x}^{2} y$ $12 {x}^{5} y = 2 \\cdot 2 \\cdot 3 \\cdot {x}^{5} y$ $56 {x}^{2} {y}^{3} = 2 \\cdot 2 \\cdot 2 \\cdot 7 {x}^{2} {y}^{3}$ So the greatest common factor is $4 {x}^{2} y$", "the highest common factor or greatest common divisor(gcd) of two integers, ‘a’ and ‘b,’ is usually denoted by $$gcd(a, b)$$, which is the largest positive integer that divides both ‘a’ and ‘b.’ Note we will be using $$hcf(a, b)$$ notation in the article to avoid confusion ### HCF using Prime Factorisation The factorization theorem indicates that every positive integer greater than one can only be written in one way as a product of prime numbers. Example $$hcf(16,24)$$, Prime factorisation of 16 = $$1 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2$$ Prime factorisation of 24 = $$1 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 3$$ In contrast to LCM, we need to pick only the common factors of both the numbers and multiply them to get the HCF. Here, there are four common multiple, i.e. 2,2,2,2. Therefore, the $$hcf(16,24) = 2 \\cdot 2 \\cdot 2 \\cdot 2 = 8$$ ### HCF using LCM $$hcf$$ formula for any two nos. = $$\\frac{a*b}{lcm(a,b)}$$ Example $$hcf(16,24) = \\frac{16*24}{lcm(16,24)} = \\frac{16*24}{48} = 8$$ ## Applications of HCF & LCM ### LCM #### Operating With Fractions When we add, subtract, or compare simple fractions, the L.C.M. of the denominators is used. For example, $$\\frac{2}{21}+\\frac{1}{6}=\\frac{4}{42}+\\frac{7}{42}= \\frac{11}{42}$$ Here, the denominator 42 was used because it is the L.C.M. of 21 and", "+0 # Question 0 232 2 What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594 Guest Mar 11, 2017 #1 0 LCM[180, 594] =5,940. GCD[180,594] =18 Ratio =5,940 / 18 =330 Guest Mar 11, 2017 #2 +7348 +5 What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594 $$\\frac{180}{594}=\\frac{2^2\\cdot3^2\\cdot5}{2\\cdot3^3\\cdot11}\\ \\ lcm=2^2\\cdot3^3\\cdot5\\cdot11\\ \\ gcf=2\\cdot3^2$$ $$\\frac{lcm}{gcf}=\\frac{2^2\\cdot3^3\\cdot5\\cdot11}{2\\cdot3^2}=\\frac{2\\cdot3\\cdot5\\cdot11}{1}\\color{blue}=330:1$$ ! asinus Mar 11, 2017 edited by asinus Mar 11, 2017", "+0 # what is the LCM of 8847 and 2532 0 282 3 what is the LCM of 8847 and 2532 Guest Feb 18, 2015 #2 +18712 +10 What is the LCM of 8847 and 2532 ? $$\\small{\\text{The following formula reduces the problem of computing the least common multiple}}\\\\ \\small{\\text{to the problem of computing the greatest common divisor (GCD), }}\\\\ \\small{\\text{also known as the greatest common factor:}}\\\\\\\\ \\small{\\text{ \\boxed{lcm(a,b)=\\dfrac{|a\\cdot b|}{\\text{gcd}(a,b)} } . }}$$ There is the Euclidean algorithm for computing the GCD. $$\\small{\\text{gcd(8847,2532)}} \\\\ \\small{\\text{ \\begin{array}{rrrr} & & q & r\\\\ 8847 & 2532 & 3 & 1251 \\\\ 2532 & 1251 & 2 & 30\\\\ 1251 & 30 & 41 & 21\\\\ 30 & 21 & 1 & 9 \\\\ 21 & 9 & 2 & {3} \\\\ 9 & 3 & 3 & algorithm Ends 0 \\end{array} }}\\\\\\\\ \\small{\\text{gcd(8847,2532)}} = 3 \\\\$$ $$\\small{\\text{lcm(8847,2532)=\\dfrac{|8847\\cdot 2532|}{3} = 7466868 }}$$ heureka Feb 18, 2015 Sort: #1 +26322 0 The LCM of 8847 and 2532 is 7466868 . Alan Feb 18, 2015 #2 +18712 +10 What is the LCM of 8847 and 2532 ? $$\\small{\\text{The following formula reduces the problem of computing the least common multiple}}\\\\ \\small{\\text{to the problem of computing the greatest common divisor (GCD), }}\\\\ \\small{\\text{also known as the greatest common factor:}}\\\\\\\\ \\small{\\text{ \\boxed{lcm(a,b)=\\dfrac{|a\\cdot b|}{\\text{gcd}(a,b)} } . }}$$", "# Least Common Multiples (LCMs) A common multiple of two whole numbers $a$ and $b$ is a number $c$ which $a$ and $b$ both divide into evenly. For example, $48$ is a common multiple of $6$ and $12$ since $48÷6=8$ and $48÷12=4$. The least common multiple is just what it sounds like... the smallest of all the common multiples. Example 1: Find the least common multiple of $9$ and $12$. To do this, we can list the multiples: $\\begin{array}{l}9:9,18,27,\\underset{_}{36},45,54,63,72,...\\\\ 12:12,24,\\underset{_}{36},48,60,72,...\\end{array}$ $36$ is the first number that occurs in both lists. So $36$ is the LCM. The listing method is impractical for large numbers. Another way to find the LCM of two numbers is to divide their product by their greatest common factor (GCF). Example 2: Find the least common multiple of $18$ and $20$. To find the GCF of $18$ and $30$, you can write their prime factorizations: $\\begin{array}{l}18=2\\cdot 3\\cdot 3\\\\ 30=2\\cdot 3\\cdot 5\\end{array}$ The common factors are $2$ and $3$. So, the GCF is $2\\cdot 3=6$. Now find the LCM by multiplying the two numbers and dividing by the GCF. (You can make this calculation a little easier by cancelling a common factor.) $\\frac{18\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}30}{6}=\\frac{3\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}6\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}30}{6}=90$ When the GCF of two numbers is $1$, the LCM is equal to the product of the two numbers. Example", "# Least Common Multiples (LCMs) A common multiple of two whole numbers $a$ and $b$ is a number $c$ which $a$ and $b$ both divide into evenly. For example, $48$ is a common multiple of $6$ and $12$ since $48÷6=8$ and $48÷12=4$ . The least common multiple is just what it sounds like... the smallest of all the common multiples. Example 1: Find the least common multiple of $9$ and $12$ . To do this, we can list the multiples: $\\begin{array}{l}9:9,18,27,\\underset{_}{36},45,54,63,72,...\\\\ 12:12,24,\\underset{_}{36},48,60,72,...\\end{array}$ $36$ is the first number that occurs in both lists. So $36$ is the LCM. The listing method is impractical for large numbers. Another way to find the LCM of two numbers is to divide their product by their greatest common factor ( GCF ). Example 2: Find the least common multiple of $18$ and $20$ . To find the GCF of $18$ and $30$ , you can write their prime factorizations : $\\begin{array}{l}18=2\\cdot 3\\cdot 3\\\\ 30=2\\cdot 3\\cdot 5\\end{array}$ The common factors are $2$ and $3$ . So, the GCF is $2\\cdot 3=6$ . Now find the LCM by multiplying the two numbers and dividing by the GCF. (You can make this calculation a little easier by cancelling a common factor.) $\\frac{18\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}30}{6}=\\frac{3\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}6\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}30}{6}=90$ When the GCF of two numbers is $1$ , the LCM", "# Help with derivation of GCF and LCM So I have this problem. I need help with the process in finding the LCM and GCF. I need a technique which is fast but yet accurate. My problem is find the the GCF and LCM for (24, 56, 96). LCM = 672 GCF = 8. I used \"Factors Grid\" and \"Prime Factorization\" but can't derive on the answers above. I've manage to get the GCF using \"Factors Grid\". Need help how the answer were derive. Thank You. - See this also: math-only-math.com/… – Babak S. Jan 27 '13 at 16:14 Here are the prime factorizations of your three numbers: $$\\begin{array}{rlll} 24 = & 2^3&\\cdot 3^1&\\cdot 7^0\\\\ 56 = & 2^3& \\cdot 3^0& \\cdot 7^1\\\\ 96 = & 2^5&\\cdot 3^1&\\cdot 7^0 \\end{array}$$ To find the GCF, take the minimum exponent for each prime; this gives $2^3\\cdot3^0\\cdot7^0 = 8$. To find the LCM, take the maximum exponent for each prime; this gives $2^5\\cdot3^1\\cdot7^1 = 672$. - First notice that each of $24, 56, 96$ is even. This means that each has a factor of 2. If a set of numbers has a common factor, this will also be a factor of the GCF. You can divide the original numbers by this common factor and work with the resulting numbers instead; they", "} + 3 x + 1$$ 5. $$( f \\cdot h ) ( - 1 ) = 128$$ 7. $$( g - f ) ( 2 ) = - 85$$ Exercise $$\\PageIndex{4}$$ Factor our the greatest common factor (GCF). 1. $$2 x ^ { 4 } - 12 x ^ { 3 } - 2 x ^ { 2 }$$ 2. $$18 a ^ { 3 } b - 3 a ^ { 2 } b ^ { 2 } + 3 a b ^ { 3 }$$ 3. $$x ^ { 4 } y ^ { 3 } - 3 x ^ { 3 } y + x ^ { 2 } y$$ 4. $$x ^ { 3 n } - x ^ { 2 n } - x ^ { n }$$ 1. $$2 x ^ { 2 } \\left( x ^ { 2 } - 6 x - 1 \\right)$$ 3. $$x ^ { 2 } y \\left( x ^ { 2 } y ^ { 2 } - 3 x + 1 \\right)$$ Exercise $$\\PageIndex{5}$$ Factor by grouping. 1. $$2 x ^ { 3 } - x ^ { 2 } + 2 x - 1$$ 2. $$3x ^ { 3 } - x ^ { 2 } - 6 x + 2$$ 3.", "remains the same: factor each monomial independently, look for common factors, and then multiply them to get the GCF. ### Example Find the greatest common factor of $$10b^{2}$$. $$\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}\\end{array}$$ $$\\text{GCF}=5b^{2}$$ The monomials have the factors 5, b, and b in common, which means their greatest common factor is $$5b^{2}$$. The video that follows gives another example of finding the greatest common factor of two monomials with only one variable. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=116 ### Example Find the greatest common factor of $$45c^{2}d^{2}$$. $$\\begin{array}{l}\\,\\,\\,81c^{3}d=3\\cdot3\\cdot3\\cdot3\\cdot{c}\\cdot{c}\\cdot{c}\\cdot{d}\\\\45c^{2}d^{2}=3\\cdot3\\cdot5\\cdot{c}\\cdot{c}\\cdot{d}\\cdot{d}\\\\\\,\\,\\,\\,\\text{GCF}=3\\cdot3\\cdot{c}\\cdot{c}\\cdot{d}\\end{array}$$ $$\\text{GCF}=9c^{2}d$$ The video that follows shows another example of finding the greatest common factor of two monomials with more than one variable. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=116 ## Factor a Polynomial One of these things is not like the others. Before we solve polynomial equations, we will practice finding the greatest common factor of a polynomial. If you can find common factors for each term of a polynomial, then you can factor it, and solving will be easier. To help you practice finding common factors, identify factors that the terms of the polynomial have in common in the table below. Polynomial Terms Common Factors $$a^{2}–2a$$ $$−2a$$", "# The GCD & LCM The greatest common denominator function or $$\\gcd$$ and the least common multiple function or $$\\text{lcm}$$ are both very commonly used tools in number theory and computer science. For two integers, $$a$$ and $$b$$, the $$\\gcd(a, b)$$ is the largest number which evenly divides both $$a$$ and $$b$$ and the least common multiple of $$a$$ and $$b$$ or $$\\text{lcm}(a, b)$$ is the smallest number which is a multiple of both $$a$$ and $$b$$. We can plot the $$\\gcd$$ function in the plane and see the interesting pattern it forms. Something close to black is formed when $$x$$ and $$y$$ are coprime since the lowest value is formed when $$\\gcd(a, b) = 1$$. We can see that, $\\text{lcm}(a, b) = \\frac{a\\cdot b}{\\gcd(a, b)} \\,,$ and, $\\gcd(a, b) = \\frac{a\\cdot b}{\\text{lcm}(a, b)} \\,.$ Which also means that, $\\gcd(a, b)\\cdot \\text{lcm}(a, b) = a\\cdot b \\,.$ Well, this is interesting. Does this only hold given some condition? Suppose that $$a$$ and $$b$$ are co-prime, then we know that $$\\gcd(a, b) = 1$$ and therefore $$1\\cdot\\text{lcm}(a, b) = \\text{lcm}(a, b) = a\\cdot b$$. Indeed, a corollary here is that when $$a$$ and $$b$$ are co-prime, then $$\\text{lcm}(a, b) = a\\cdot b$$.", "Example7.22. Find the greatest common factor for $~~4a^3b^2+6ab^3-18a^2b^4$ Solution The largest integer that divides evenly into all three coefficients is 2. The highest power of $a$ is $a^1\\text{,}$ and the highest power of $b$ is $b^2\\text{.}$ Thus, the GCF is $2ab^2\\text{.}$ Note that the exponent on each variable of $2ab^2$ is the smallest exponent that appears on that variable among the terms of the polynomial. ###### 2. The exponent on each variable of the GCF is the exponent that appears on that variable among the terms of the polynomial. Once we have found the greatest common factor for the polynomial, we can write each term as a product of the GCF and another factor. For example, the GCF of $8x^2-6x$ is $2x\\text{,}$ and we can write \\begin{equation*} 8x^2-6x = \\blert{2x} \\cdot 4x - \\blert{2x} \\cdot 3 \\end{equation*} We can then use the distributive law to write the expression on the right side as a product. \\begin{equation*} \\blert{2x} \\cdot 4x - \\blert{2x} \\cdot 3 = \\blert{2x} (4x-3) \\end{equation*} We say that we have factored out the greatest common factor from the polynomial. For more complicated polynomials, we can divide the GCF into each term to find the remaining factors. ###### Example7.23. Factor $~~4a^3b^2+6ab^3-18a^2b^4$ Solution The GCF for this polynomial is $2ab^2\\text{,}$ as we saw in Example 7.22. We factor", "the notion of the least common multiple. Let $a$ and $b$ be positive integers. The least common multiple of $a$ and $b$ is denoted by $\\text{LCM}(a,b)$ and is defined as the least positive integer that is divisible by both $a$ and $b$. For example, $\\text{LCM}(16,18)=144$. We can also express $\\text{LCM}(a,b)$ as follows: $\\displaystyle \\text{LCM}(a,b)=\\frac{a \\cdot b}{\\text{GCD}(a,b)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$ where $\\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\\text{GCD}(a,b)=1$, we have $\\displaystyle \\text{LCM}(a,b)=a \\cdot b$. Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \\cdot 3$ and $18=2 \\cdot 3^2$. Then $\\text{LCM}(16,18)=2^4 \\cdot 3^2=144$. The least common divisor of the numbers $a_1,a_2,\\cdots,a_n$ is denoted by $\\text{LCM}(a_1,a_2,\\cdots,a_n)$ and is defined similarly. It is the least positive integer that is divisible by all", "factors. Example 10.85: Factor: 3a + 3. ##### Solution Rewrite each term as a product using the GCF. $$\\textcolor{red}{3} \\cdot a + \\textcolor{red}{3} \\cdot 1$$ Use the Distributive Property 'in reverse' to factor the GCF. $$3(a+1)$$ Check by multiplying the factors to get the original polynomial. $$\\begin{split} 3(a &+ 1) \\\\ 3 \\cdot a &= 3 \\cdot 1 \\\\ 3a &+ 3\\; \\checkmark \\end{split}$$ Exercise 10.169: Factor: 9a + 9. Exercise 10.170: Factor: 11x + 11. The expressions in the next example have several factors in common. Remember to write the GCF as the product of all the common factors. Example 10.86: Factor: 12x − 60. Rewrite each term as a product using the GCF. $$\\textcolor{red}{12} \\cdot x - \\textcolor{red}{12} \\cdot 5$$ Factor the GCF. $$12(x-5)$$ Check by multiplying the factors. $$\\begin{split} 12(x &- 5) \\\\ 12 \\cdot x &- 12 \\cdot 5 \\\\ 12x &- 60\\; \\checkmark \\end{split}$$ Exercise 10.171: Factor: 11x − 44. Exercise 10.172: Factor: 13y − 52. Now we’ll factor the greatest common factor from a trinomial. We start by finding the GCF of all three terms. Example 10.87: Factor: 3y2 + 6y + 9. ##### Solution Rewrite each term as a product using the GCF. $$\\textcolor{red}{3} \\cdot y^{2} + \\textcolor{red}{3} \\cdot 2y + \\textcolor{red}{3} \\cdot 3$$ Factor the GCF. $$3(y^{2} + 2y +", "What is the GCF of 18a, 20ab and 6ab? Jun 28, 2018 $2 a$ Explanation: Given the set: $18 a , 20 a b , 6 a b$. First find the greatest common factor of $18 , 20 , 6$. $18 = 2 \\cdot {3}^{2}$ $20 = {2}^{2} \\cdot 5$ $6 = 2 \\cdot 3$ $\\therefore \\setminus \\text{GCF} = 2$ Then, find the $\\text{GCF}$ of $a , a b , a b$. $a = a$ $a b = a \\cdot b$ $\\therefore \\setminus \\text{GCF} = a$ Now, multiply the two $\\text{GCF}$s of the two sets together. $2 \\cdot a = 2 a$ That is the greatest common factor of the whole sequence. Jun 28, 2018 G C F is $\\textcolor{g r e e n}{2 a}$ Explanation: G C F is the greatest common factor which finds place in all the terms. $18 a , 20 a b , 6 a b$ $\\implies 9 \\cdot \\textcolor{g r e e n}{2 a} , 10 b \\cdot \\textcolor{g r e e n}{2 a} , 3 b \\cdot \\textcolor{g r e e n}{2 a}$ $2 a$ is common in all the three terms. Hence the G C F is $\\textcolor{g r e e n}{2 a}$" ]

[ "# Divisibility by $$15$$. Find the smallest positive multiple of $$15$$ such that each of the digit of the multiple is either 0 or 8. ×", "# Number of solutions for $a+b+c+ab+bc+ac+abc=29$ [closed] Let the number N be the smallest three digit number with digits $$a,b,c$$. $$a+b+c+ab+bc+ca +abc=29$$. Evaluate $$\\frac{N+1}{5}$$. By adding $$1$$ on both sides we get $$(a+1)(b+1)(c+1)=30$$. How should I proceed further and is there any way without hit and trial? We want to find the smallest $$N$$ possible, so minimise each digit one at a time. $$a \\ne 0$$ so the smallest $$a$$ is $$1$$. This gives $$(b+1)(c+1) = 15$$ and you can continue from here.", "If $$a$$ $$b$$ and $$c$$ are positive real numbers with $$a+b+c = 6$$ find the maximum value of $$(a+ \\sqrt{bc})(b+\\sqrt{ac})(c+\\sqrt{ab}) + abc$$", "For the largest gap, the tens digit of the top number must be as large as possible, and the tens digit must be as small as possible. This means the subtraction is $ab-cd$ (where $ab$ is the number with digits $a$ and $b$, rather than $a \\times b$). The largest difference is therefore 31. For the smallest gap it is $bd - ca$ because $a$ must be at the bottom because it is the biggest and you want to be taking away big numbers. It can't go in the bottom tens column because then the answer would be negative. b, the second biggest, can't accompany a because then it would also be a negative. c is the next biggest. This works out as positive. Therefore the sum is: Note that there is a carry occurring here. The smallest difference is therefore 7. Thank you again to everyone who submitted a solution, and well done!", "And as obviously the smallest possible pair $(a,b)=(1,2)$ provides the solution $124$, the answer is $964 - 124 = \\boxed{840}$.", "# Pythagorean Triplets Given that positive integers $$a,b$$ and $$c$$ satisfy the equation $$a^2 + b^2=c^2$$ with $$a,b$$ consecutive, and we know that $$(a,b,c) = (3,4,5), (20,21,29)$$ are two smallest possible solutions satisfying this constraint such that $$c$$ is minimized. What is the third smallest value of $$c$$? ×", "# Perfect how? Divide $$297$$ by the smallest possible positive integer $$a$$ so that the result is the square of a positive integer $$b$$. What is $$a+b$$? ×", "Find the smallest numbers $a, b$, and $c$ such that: $$a^2 = 2 b^3 = 3 c^5$$ What can you say about other solutions to this problem?", "# Divisibility by $$15$$. Number Theory Level 2 Find the smallest positive multiple of $$15$$ such that each of the digit of the multiple is either 0 or 8. ×", "but as pairs. Your goal is to find the pair $(a,b)$ such that $L\\leq a/b\\leq U$ such that $a\\times b$ is minimal. In this case, we do NOT reduce fractions. For a fixed $b$, we can easily find the smallest possible $a$ that satisfies the conditions by ... Top 50 recent answers are included", "# a+b+x Level pending x is the smallest two digit number which satisfies the follows conditions: 1] it is a multiple of 3 and 5. 2] (x = a + b) , where a and b are positive integers and b>a. 3] x = $$\\sqrt{b^{2} + (a-b)^{2}}$$ Find (a+b+x). ×", "3)=1$, but the one provided is the smallest positive value. -", "Question # Find the $L.C.M.$ of $30{\\text{ and }}45$.A $15$B $30$ C $45$ D $90$ Hint:In the question where we have to find the L.C.M we will proceed by using the definition of L.C.M.If $a{\\text{ and }}b$ are two non-zero integers the least common multiple of $a{\\text{ and }}b$ is the smallest positive integer divisible by both $a{\\text{ and }}b$.And then we will find the factors of the both numbers and find the smallest positive integer divisible by both the integers which will be the L.C.M of the given two integers. In the above question we have to find the $L.C.M.$ of $30{\\text{ and }}45$- Now to find $L.C.M.$ of any two numbers we must know the definition of the $L.C.M.$ $L.C.M.$, least common multiple is defined as if $a{\\text{ and }}b$ are two non-zero integers the least common multiple of $a{\\text{ and }}b$ is the smallest positive integer divisible by both $a{\\text{ and }}b$. The least common multiple is also known as smallest common multiple or lowest common multiple. Now the least common multiple or the smallest common multiple of $a{\\text{ and }}b$ is denoted by $l.c.m(a,b)$. Now according to the question we have to find the $L.C.M.$ of $30{\\text{ and }}45$, so now using definition of least common multiple, $30{\\text{ and }}45$ are the non-zero positive", "be two positive integers and $A=B^2$. If $A$ satisfies the following conditions, find the value of $B$: • $A$'s thousands digit is $4$ • $A$'s tens digit is $9$ • The sum of all $A$'s digits is $19$ Is it possible to find four positive integers such that $2002$ plus the product of any two of them is always a square? If yes, find such four positive integers. If no, explain. If the middle term of three consecutive integers is a perfect square, then the product of these three numbers is called a $\\textit{beautiful}$ number. What is the greatest common divisor of all the $\\textit{beautiful}$ numbers? Find the smallest square whose last three digits are the same but not equal $0$. Let $\\overline{ABCA}$ be a four-digit number. If $\\overline{AB}$ is a prime, $\\overline{BC}$ is a square, and $\\overline{CA}$ is the product of a prime and a greater-than-one square. Find all such $\\overline{ABCA}$. Let $A$ be a two-digit number, multiplying $A$ by 6 yields a three-digit number $B$. The difference of the two five-digit numbers obtained by appending $A$ to the left and right of $B$, respectively, is a perfect square. Find the sum of all such possible $A$s. Find such a positive integer $n$ such that both $(n-100)$ and $(n-63)$ are square numbers. Find such a positive integer", "information. Among all such triples, you have to choose one with the minimum possible sum $a+b+c$. Among all triples with the minimum sum, you can print any. Example Input 3 123 13 2 2 2000000000 2000000000 Output 111 1 12 1 1 1 1999999999 1 1", "# Use only Holder's If $a$ and $b$ are positive real numbers such that $a^{2015}+b^{2015}=2015,$ find the maximum value of $a+b$. ×", "# Find the minimum. Level pending If $$a$$ and $$b$$ are rational positive numbers such that $$a+b=1$$,find the smallest possible value of $$(1+\\dfrac{1}{a})(1+\\dfrac{1}{b})$$. ×", "digits as possible so we want the digits $3,2,2,2$. The smallest number with those digits is $2223$ but that doesn't end with $2$. So the smallest number with those digits that ends in $2$ is $2232$. $2232 = 2*1116$ and $2232 = 3*744$. This is the smallest such number.", "Finding the smallest positive integers that satisfies given equations [closed] Is it possible to find the smallest positive integer/s that satisfy a given equation or some inequality? Example: $2x^2-3x>24$ Is there a formula for this? closed as too broad by Aqua, Guy Fsone, mechanodroid, Arnaud D., kingW3Oct 22 '17 at 14:41 Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. • Start by checking if there is some small solution $a,b$. Then there are only finitely many possibilities to check for a smaller $a+b$. – hardmath Oct 18 '17 at 10:52 So $5|a(a-1)$ and since $a$ and $a-1$ are relatively prime $5$ divides only one of them, say $a$. Since $a_{\\min}=5$ you get $b = 2$ and so $(a+b)_{\\min} = 7$. If $5$ divides $a-1$ then $a_{\\min} = 6$ which is not good. This is a very broad question. As for your specific equation, it's equivalent to $a(a-1) = 10b$ so you're basically looking for the smallest pair of successive numbers whose product is a multiple of 10. You also know that", "Find all the triples of numbers $a$, $b$, $c$ such that each one of them plus the product of the other two is always $2$." ]

[ "= 3*(45^2 - 2^2) = 3*((40+5)^2 - 2^2) = 3*(1600+400+25-4) = 3*2000 + 3*25 - 3*4, etc. In general, multiplication can be done quickly by subtracting squares (and factoring can be done by recognizing differences of squares). – whuber Aug 26 '10 at 18:26 HINT $\\;\\quad\\rm mod \\; 97: \\:\\quad 100 \\;\\equiv\\; 3$ Hence $\\; 3007 \\;\\equiv\\; 30 \\cdot 100 + 7$ $\\quad\\quad\\quad\\quad\\quad\\quad\\;\\; \\;\\equiv\\; 30 \\:\\;\\cdot\\;\\: 3 \\;\\: + \\; 7 \\;$ $\\quad\\quad\\quad\\quad\\quad\\quad\\;\\; \\;\\equiv\\; 0$ I.e. cast out 97's in analogy to cast out nines. See also here where I discuss casting out 91's. - I'm not getting anything. Can you liek \"Solve\" the question. I know I should not ask someone to solve the problem, but I'm not aware of mod and ≡. The only thing about mod I know is it returns remainder. :\\ – Electrifyings Aug 26 '10 at 17:06 Ok, here is a bit simpler way to state it: since 100A + B = 97A + 3A + B, it follows that 97 divides 100A + B iff 97 divides 3A + B. Now put A,B = 30,7, i.e. the digits of 3007 in radix 100. – Bill Dubuque Aug 26 '10 at 17:18 chiaotzu: Here's the way one would read the first relation in Bill's answer: saying that 100 is congruent to 3,", "+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,", "# 2 Left behind What is the smallest positive 3 digit number that leaves a remainder of 2 when divided by 3, 4, 5 or 6? Details and assumptions The number $$12=012$$ is a 2 digit number, not a 3 digit number. ×", "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,", "using any calculator with enough digits that $11^{10}-1 = 25937424600$ and that this number is not divisible by $11$. $11^{10}-1$ is clearly not divisible by $11$, in fact it leaves a remainder of $10$, with quotient $11^9 -1$. To see that $11^{10}-1$ is divisible by $100$, we can do the following: To check divisbility by four, see that $11 \\equiv -1 \\mod 4$ (read up congruence notation and rules from Wikipedia, if you are unaware),hence $11^{10} \\equiv (-1)^{10} \\equiv 1 \\mod 4$, so that $4$ divides $11^{10}-1$. See that $11^2 =121 \\equiv -4 \\mod 25$, so that $11^{10} \\equiv (-4)^5 \\equiv -1024 \\equiv 1 \\mod 25$. Hence,$11^{10}-1$ is a multiple of $25$, and hence of $25 \\cdot 4=100$. In fact, $11^{10}-1 = 25937424600 = 2^3 \\cdot 3 \\cdot 5^2 \\cdot 3221 \\cdot 13421$, as a confirmation.", "# 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: $2 \\cdot 3 \\cdot 5 = 30$", "are three multiples of 3 namely 0, 3, and 6. Thus, $B=7-3=4$ 3. $A+B=10+4=14$.", "is $$0$$. $$11= 2\\cdot 5 + 1$$, so our $$2^2$$ bit is $$1$$. $$5= 2\\cdot 2 + 1$$, so our $$2^3$$ digit is $$1$$. $$2= 1\\cdot 2$$, so our $$2^4$$ digit is $$0$$, our $$2^5$$ digit is $$1$$, and we're finished; $$45$$ in binary is $$101101$$ Let $$n$$ be any positive integer and let $$b$$ be a base. By the division algorithm we get that $$n=bq_0+r_0$$ where $$r_0\\in\\{0,1,..,b-1\\}$$. If $$q_0\\lt b$$ then we are done. If not, then $$q_0=bq_1+r_1\\implies n=b(bq_1+r_1)+r_0=b^2q_1+br_1+r_0$$. If $$q_1\\lt b$$ then we are done. What do you think happens if it is not? Notice that we only divide the $$q_i$$s by $$b$$ and the remainders become the next digit, going from right to left.", "× # Basic Maths Show that the number 111...11 with 3^n digits is divisible by 3^n. Note by Biswaroop Roy 3 years, 8 months ago Sort by: We prove this by induction: Base case: When n = 0, it is obviously true that 1 is divisible by 1. When n = 1: The number is 111 which is divisible by 3. Inductive Case: Suppose the number 1111.....111 (which has 3^n digits) is divisible by 3^n. For 3^(n+1) digits of 1, The sum of digits for this case is always a power of 3. Hence, the assertion is true. Numerical Proof: 3^(n+1) divides 111....111 (which has 3^(n+1) digits) (3^n)(3) divides 111....111 (which has (3^n)(3) = 3^n + 3^n + 3^n digits) Remembering the case for n: the number 1111.....111 (which has 3^n digits) is divisible by 3^n and the base case, the proof is done. · 3 years, 8 months ago Just to clarify your induction, as it is a bit unclear, $$111\\cdots 1(3^{n+1}$$ digits $$)=111\\cdots 1(3^n$$ digits $$)\\times (10^{2n}+10^n+1)$$. Since $$(10^{2n}+10^n+1)$$ is divisible by 3, by the inductive hypothesis we finish the induction. · 3 years, 8 months ago Nicely expressed. This also shows that $$3^n$$ is the largest power of 3 that divides $$111\\ldots 1$$. As a side note, another way of stating the question is that", "2^2+\\dots+a_k\\cdot 2^k$$ where $a_i$ is $0$ or $1$ and $2^k$ is the largest power of $2$ less than or equal to $n$ (and $a_k=1$). When you divide $n$ by $2$, the remainder is $a_0$ and the quotient is $$n_1=a_1+a_2\\cdot2+\\dots+a_k\\cdot 2^{k-1}$$ When you divide $n_1$ by $2$, the remainder is $a_1$ and the quotient is $$n_2=a_2+\\dots+a_k\\cdot 2^{k-2}$$ Continuing in this way, we arrive finally at $$n_k=a_k$$ Thus the procedure of successively dividing by $2$ and collecting the remainders produces the digits in the binary representation of $n$, starting from the less significant digit. Note that we don't need to know $k$ in advance: the algoritm stops when the last quotient is $0$. Finding the largest power of $2$ less than or equal to $n$ yields the most significant digit. It's an alternative procedure, but requires to know the powers of $2$, which is not needed with the algorithm of successive divisions. So successive divisions provide a more efficient algorithm. By the way, the algorithm is the same for every radix. The alternative method for radix $b$ requires a table of the powers of $b$ together with their multiples by $2$, $3$ and so on up to $b-1$. • This is kinda abstract. But it works. – most venerable sir Oct 9 '17 at 22:59 • Why is the remainder", "/ 25 = 400.$$ The largest number is $$12$$ even numbers away, which means that it equals $$400 + 12 \\cdot 2 = 424.$$ Thus, E is the correct answer. 20. The least common multiple of $$a$$ and $$b$$ is $$12,$$ and the least common multiple of $$b$$ and $$c$$ is $$15.$$ What is the least possible value of the least common multiple of $$a$$ and $$c?$$ $$20$$ $$30$$ $$60$$ $$120$$ $$180$$ ###### Solution(s): We know that $$b$$ has to divide both $$12$$ and $$15,$$ so it must equal either $$1$$ or $$3.$$ If $$b = 1,$$ then $$a = 12$$ and $$c = 15,$$ making their least common multiple $$60.$$ If $$b = 3,$$ then the smallest value of $$a$$ is $$12$$ and $$c$$ is $$5.$$ The least common multiple in this scenario is $$20.$$ Thus, A is the correct answer. 21. A top hat contains $$3$$ red chips and $$2$$ green chips. Chips are drawn randomly, one at a time without replacement, until all $$3$$ of the reds are drawn or until both green chips are drawn. What is the probability that the $$3$$ reds are drawn? $$\\dfrac{3}{10}$$ $$\\dfrac{2}{5}$$ $$\\dfrac{1}{2}$$ $$\\dfrac{3}{5}$$ $$\\dfrac{2}{3}$$ ###### Solution(s): The only way for the $$3$$ reds to be drawn first is if there is only one green drawn in the first", "1 \\cdot 11 + 1$ $3 \\cdot 6 = 18 = 1 \\cdot 11 + 7$ $4 \\cdot 6 = 24 = 2 \\cdot 11 + 2$ So the smallest possible positive value of $k$ which gives us the required remainder is $k = 4$ So: $n = 105 k - 1 = 105 \\cdot 4 - 1 = 419$", "them a lot. By inspection, we find that $\\text{LCM}(1,2,3,2\\cdot2,2\\cdot3) = 12$. Finally, $1+2 = \\boxed{\\textbf{(B)}\\ 3}$. ## Solution 2 We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$,$2$, $3$, or $4$. We quickly consider $12$ since it is the smallest number that is the LCM of $1$, $2$, $3$ and $4$. Since $12$ has $5$ single-digit divisors, namely $1$, $2$, $3$, $4$, and $6$, our answer is $1+2 = \\boxed{\\textbf{(B)}\\ 3}$", "is $p$ subtracted from the nearest multiple of 10. $b$ is that same multiple of 10 divided by 10. 2. Break $n$ into its digits, $n_1$, $n_2$, $n_3$, etc. 3. Start with $n_1$. 4. Multiply by $a$. 5. If $b$ evenly divides it, divide by $b$. If not, add or subtract $p$ until the result is evenly divisible by $b$, then divide by $b$. 6. Take the remainder when divided by $p$. 7. If there are any digits left to consider, add it to the result and continue at Step 4. If there are no more digits, the result is the remainder of $n$ when divided by $p$. Try it out with $n = 986$ and $p = 17$. The multiple of 10 closest to 17 is 20, so $a = 3$ and $b = 2$. Starting with $9$, $9 \\cdot 3 = 27$, $27 - 17 = 10$, $10 \\div 2 = 5$. Add the second digit, $8$, to get $13$ and repeat: $13 \\cdot 3 = 39$, $39 - 17 = 22$, $22 \\div 2 = 11$. Add the last digit, $6$, to get $17$, which is, clearly, evenly divisible by 17, so 17 is a factor of 986. This turns out to be a good algorithm for mental factoring, since you never have to multiply", "+ 0.5) = -12.0$$ (base10). See the following figure for details. TRY IT! What is 15.0 (base10) in IEEE754? What is the largest number smaller than 15.0? What is the smallest number larger than 15.0? Since the number is positive, $$s = 0$$. The largest power of two that is smaller than 15.0 is 8, so the exponent is 3, making the characteristic $$3 + 1023 = 1026 (base10) = 10000000010(base2)$$. Then the fraction is $$15/8-1=0.875(base10) = 1\\cdot \\frac{1}{2^1} + 1\\cdot \\frac{1}{2^2} + 1\\cdot \\frac{1}{2^3}$$ = 1110000000000000000000000000000000000000000000000000 (base2). When put together this produces the following conversion: 15 (base10) = 0 10000000010 1110000000000000000000000000000000000000000000000000 (IEEE754) The next smallest number is 0 10000000010 1101111111111111111111111111111111111111111111111111 = 14.9999999999999982236431605997 The next largest number is 0 10000000010 1110000000000000000000000000000000000000000000000001 = 15.0000000000000017763568394003 Therefore, the IEEE754 number 0 10000000010 1110000000000000000000000000000000000000000000000000 not only represents the number 15.0, but also all the real numbers halfway between its immediate neighbors. So any computation that has a result within this interval will be assigned 15.0. We call the distance from one number to the next the gap. Because the fraction is multiplied by $$2^{e-1023}$$, the gap grows as the number represented grows. The gap at a given number can be computed using the function spacing in numpy. import numpy as np TRY IT! Use the spacing function to determine the gap at", "# How do find 12 divided by 3? You should know this by your simple multiplication facts. $3 \\cdot 4$ is 12, so 12 divided by 3 should be 4.", "1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is $$2^3*3^2*5*103*107$$, what is the value of N? (A) $$2^2 * 3^2 * 7$$ (B) $$2^2 * 3^3 * 103$$ (C) $$2^2 * 3^2 * 5$$ (D) $$2^2 * 3 * 5$$ (E) None of these Are You Up For the Challenge: 700 Level Questions Let’s first factor 2,472 and 1,284 into primes: 2,472 = 12 x 206 = 2^2 x 3 x 2 x 103 = 2^3 x 3 x 103 1,284 = 12 x 107 = 2^2 x 3 x 107 We see that that the LCM of 2,472 and 1,284 is 2^3 x 3 x 103 x 107. Notice that the LCM of 2,472 1,284, and N is equal to the LCM of 2^3 x 3 x 103 x 107 and N, which is given to be 2^3 x 3^2 x 5 x 103 x 107. We are also given that the GCF of 2,472 1,284, and N is 12. Recall that for any two positive integers a and b, we have a x b = LCM(a, b) x GCF(a, b). Therefore, by letting a = 2^3 x 3 x 103 x 107 and b = N, we have: 2^3 x 3 x 103", "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", "100-2x$. This accounts for all possible cases since if $a = b \\ge 50$ then $c$ is not positive. There are also 49 triples with $a = c$, or with $b = c$, respectively. There are no triples with $a = b = c$, so the total number of triples with $a , b, c$ not all distinct is $3*49 = 147$. The number of triples of positive integers $(a, b, c)$ with $a, b, c$ distinct and $a + b + c = 100$ is therefore $\\binom{99}{2} -147$. Given a triple $(a, b, c)$ with $a< b < c$ and $a + b + c = 100$, there are six distinct ways to permute $a, b, c$, all of which our current estimate accounts for. Therefore we should divide by 6 to throw out all the triples that are in the wrong order. The final answer is $\\frac{1}{6}(\\binom{99}{2} -147) = 784$. Hint: For each possible value of $a$, count the number of possible values of $b$ and $c$. It's a quite regular pattern, so you don't have to brute force much to find it. Since $a$ is the smallest, the largest number it can be is $32$, so $a$ ranges from $1$ to $32$. The remaining sum $b+c$ must be equal to $100 - a$ and since", "$$10^{15}+x>10^{15}-8568\\cdot 10^9>0.$$ By definition, the determinant is the sum of $60$ positive and $60$ negative terms, one of which is $\\mbox{sgn}{\\left( \\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\\\ 3 & 1 & 5 & 2 & 4 \\\\ \\end{array} \\right)}(1000)^5=\\mbox{sgn}\\left((13542)\\right)(1000)^5=+(1000)^5$. Each of the other terms has absolute value at most $9\\cdot1000^4$, so the sum of the 60 negative terms is greater than or equal to $-60\\cdot9\\cdot1000^4$, and the determinant is greater than or equal to $(1000)^5-60\\cdot9\\cdot1000^4$, which is positive. With an even number of adjacent row swaps, you can put the $1000$s on the diagonal. That determinant must be positive. Using the algorithm which looks at minors, the largest term is always positive. • ...and the important thing is that the largest term is very large compared to all the others. Obviously it's at least $1000/9 \\approx 111$ times larger than any other term, but there are $119$ other terms so a small amount of subtlety is needed. But if you think about it, it's actually at least $1000^2/9^2$ times larger than any other term, which is plenty large enough to dominate the sum. – Erick Wong Nov 26 '14 at 1:01 • Alternatively, note that only 60 of those 119 terms are negative. – Steve Kass Nov 26 '14 at 2:07 You can" ]

[ "them a lot. By inspection, we find that $\\text{LCM}(1,2,3,2\\cdot2,2\\cdot3) = 12$. Finally, $1+2 = \\boxed{\\textbf{(B)}\\ 3}$. ## Solution 2 We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$,$2$, $3$, or $4$. We quickly consider $12$ since it is the smallest number that is the LCM of $1$, $2$, $3$ and $4$. Since $12$ has $5$ single-digit divisors, namely $1$, $2$, $3$, $4$, and $6$, our answer is $1+2 = \\boxed{\\textbf{(B)}\\ 3}$", "# Difference between revisions of \"2021 JMPSC Sprint Problems/Problem 17\" ## Problem What is the smallest positive multiple of $1003$ that has no zeros in its decimal representation? ## Solution Notice that $1002 \\cdot n = 1000n + 2n$. Since $1000n$ always has $3$ zeros after it, we have to make sure $2n$ has $3$ nonzero digits, so that the last 3 digits of the number $1002n$ doesn't contain a $0$. We also need to make sure that $n$ has no zeros in its own decimal representation, so that $1000n$ doesn't have any zeros other than the last $3$ digits. The smallest number $n$ that satisfies the above is $56$, so the answer is $1002 \\cdot 56 = \\boxed{56112}$. ~Mathdreams Invalid username Login to AoPS", "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 Left behind What is the smallest positive 3 digit number that leaves a remainder of 2 when divided by 3, 4, 5 or 6? Details and assumptions The number $$12=012$$ is a 2 digit number, not a 3 digit number. ×", "For the largest gap, the tens digit of the top number must be as large as possible, and the tens digit must be as small as possible. This means the subtraction is $ab-cd$ (where $ab$ is the number with digits $a$ and $b$, rather than $a \\times b$). The largest difference is therefore 31. For the smallest gap it is $bd - ca$ because $a$ must be at the bottom because it is the biggest and you want to be taking away big numbers. It can't go in the bottom tens column because then the answer would be negative. b, the second biggest, can't accompany a because then it would also be a negative. c is the next biggest. This works out as positive. Therefore the sum is: Note that there is a carry occurring here. The smallest difference is therefore 7. Thank you again to everyone who submitted a solution, and well done!", "is actually $\\equiv 3\\pmod{5}$, the smallest positive solution is given by $N=\\color{red}{27}$. You first solve $\\;\\begin{cases}n\\equiv3\\mod 4,\\\\n\\equiv2\\mod 5.\\end{cases}$. For this you need a Bézout's relation between $4$ and $5$. We have one at hand: $\\; 5-4=1$. Hence the solutions are: $$n\\equiv 3\\cdot5-2\\cdot 4=7\\mod 4\\cdot 5=20.$$ Then you solve $\\;\\begin{cases}N\\equiv7\\mod 20,\\\\N\\equiv6\\mod 7.\\end{cases}$ A Bézout's relation between $20$ and $7$ is $\\;3\\cdot 7-20=1$. Hence the solutions are $$N\\equiv 7\\cdot 3\\cdot 7-6\\cdot 20=27\\pmod{140}.$$ The smallest positive integer in this congruence is $20$. last digit of N due to first condition should be either 1,3,5,7. due to second condition last digit should be either 7,2. so to satisfy both conditions it should end with 7. the smallest number ending with 7 satisfying third condition is 27.", "such that $a + b + c = 100$. For the choice of $a,b$ we need $a + b \\leq 100$. Then $c = 100 - a - b$ is uniquely determined. For $a + b \\leq 100$, the number of possibilities is $$101 + 100 + 99 + \\ldots + 2 + 1 = \\frac{102\\cdot 101}{2} = 5151.$$ Now we count the triples with the additional condition that two numbers are the same. Note that $a = b = c$ is not possible (otherwise, $100 = a + b + c = 3a$ would be a multiple of $3$). So there are $3$ possibilities to choose the identical pair of numbers. For a single such choice, say $a = b$, we have the $51$ possibilities $a\\in\\{0,\\ldots,50\\}$ s.t. $a + b = 2a \\leq 100$. Again, $c$ is uniquely determined. So in total, there are $3 \\cdot 51 = 153$ triples $(a,b,c)$ such that two numbers are the same and $a + b + c = 100$. This gives the resulting probability of $$\\frac{153}{5151} = \\frac{3}{101} \\approx 3.0\\%.$$ - \"For a+b≤100, the number of possibilities is 100+99+98+…+2+1=50⋅101=5050\". That should be 5151 rather than 5050 as a and b can take values zero? – Johannes Aug 9 '13 at 13:04 @Johannes. Right, it's corrected now. Thanks! – azimut Aug", "between both digits and decreasing the other by the same difference Using a similar reasoning we compare $$x*y$$ with $$(x-d)*(y+d) = x*y + (x-y)*d - d*d > x*y if x-y > d > 0$$. But $$x-y > 100$$ whereas $$d < 100$$ implies that decreasing the larger factor by the same amount the smaller factor is increased yields a higher product. As a result, the three factors formed for the three groups are such that they minimize their distance to each other, i.e., the smallest number starting by $$9$$ is $$941$$ and the largest number starting by $$7$$ is $$763$$. In the middle, we have the number with the $$3$$ remaining digits : $$852$$. • Great answer. I did it programatically and it proves your answer through a posteriori. – Zaenille Feb 14 at 3:23 We can calculate: $$(a*100 + b*10 + c)*(d*100 + e*10 + f)*(g*100 + h*10 + i)= \\\\ = adg1000000 + (adh+aeg+bdg)100000 + (adi+aeh+bdh+afg+beg+cdg)10000 + (aei+bdi+afh+beh+cdh+bfg+ceg)1000 + (afi+bei+cdi+bfh+ceh+cfg)100 + (bfi+cei+cfh)10 + cfi$$ If we now maximise coefficients in order: $$max(adg)$$ a = 9 d = 8 g = 7 $$max(adh+aeg+bdg) = max(72h+63e+56b)$$ h = 6 e = 5 b = 4 $$max(adi+aeh+bdh+afg+beg+cdg) = max(72i+63f+56c)$$ i = 3 f = 2 c = 1 We obtain the (correct) solution. But how to formally show that", "(Investment Banking) Re: The difference between positive two-digit integer A and the smaller tw [#permalink] ### Show Tags 24 Apr 2015, 07:05 Hi, No doubt that your solution is correct. But still could you explain me the highlighted portions? Actually could you solve in more simple manner? Thanks Celestial Harley1980 wrote: Bunuel wrote: The difference between positive two-digit integer A and the smaller two-digit integer B is twice A‘s units digit. What is the hundreds digit of the product of A and B? (1) The tens digit of A is prime. (2) Ten is not divisible by the tens digit of A. Kudos for a correct solution. Let's $$A = ab$$ and $$B =cd$$; so $$A = (10a + b)$$ and $$B=(10c + d)$$ As we know $$ab-cd = 2b$$ so $$(10a + b) - (10c + d) = 2b$$ $$10a - 10c = d + b$$ $$10(a-c)=d+b$$ As we know that $$d$$ and $$b$$ its one unit digits and can be max $$9$$ so their sum can be max $$18$$ So $$a-c$$ can't be more than $$1$$ and we can infer that: $$d+b = 10$$ $$a-c = 1$$ And now we should find what will be $$ab*cd$$ $$(10a + b)*(10c + d)$$ $$100ac + 10ad + 10bc+bd$$ let's substitute $$d$$; $$d=10-b$$ $$100ac+100a-10ab+10bc+10b-b^2$$ let's substitute $$c$$; $$c =a-1$$", "Let $N$ be the greatest integer multiple of $36$ all of whose digits are even and no two of whose digits are the same. Find the remainder when $N$ is divided by $1000$. Prove: randomly select 51 numbers from $1, 2, 3, \\cdots, 100$, at least two of them must be relatively prime to each other. Let integer $a$, $b$, and $c$ satisfy $a+b+c=0$, prove $|a^{1999}+b^{1999}+c^{1999}|$ is a composite number.", "is $$0$$. $$11= 2\\cdot 5 + 1$$, so our $$2^2$$ bit is $$1$$. $$5= 2\\cdot 2 + 1$$, so our $$2^3$$ digit is $$1$$. $$2= 1\\cdot 2$$, so our $$2^4$$ digit is $$0$$, our $$2^5$$ digit is $$1$$, and we're finished; $$45$$ in binary is $$101101$$ Let $$n$$ be any positive integer and let $$b$$ be a base. By the division algorithm we get that $$n=bq_0+r_0$$ where $$r_0\\in\\{0,1,..,b-1\\}$$. If $$q_0\\lt b$$ then we are done. If not, then $$q_0=bq_1+r_1\\implies n=b(bq_1+r_1)+r_0=b^2q_1+br_1+r_0$$. If $$q_1\\lt b$$ then we are done. What do you think happens if it is not? Notice that we only divide the $$q_i$$s by $$b$$ and the remainders become the next digit, going from right to left.", "+ 0.5) = -12.0$$ (base10). See the following figure for details. TRY IT! What is 15.0 (base10) in IEEE754? What is the largest number smaller than 15.0? What is the smallest number larger than 15.0? Since the number is positive, $$s = 0$$. The largest power of two that is smaller than 15.0 is 8, so the exponent is 3, making the characteristic $$3 + 1023 = 1026 (base10) = 10000000010(base2)$$. Then the fraction is $$15/8-1=0.875(base10) = 1\\cdot \\frac{1}{2^1} + 1\\cdot \\frac{1}{2^2} + 1\\cdot \\frac{1}{2^3}$$ = 1110000000000000000000000000000000000000000000000000 (base2). When put together this produces the following conversion: 15 (base10) = 0 10000000010 1110000000000000000000000000000000000000000000000000 (IEEE754) The next smallest number is 0 10000000010 1101111111111111111111111111111111111111111111111111 = 14.9999999999999982236431605997 The next largest number is 0 10000000010 1110000000000000000000000000000000000000000000000001 = 15.0000000000000017763568394003 Therefore, the IEEE754 number 0 10000000010 1110000000000000000000000000000000000000000000000000 not only represents the number 15.0, but also all the real numbers halfway between its immediate neighbors. So any computation that has a result within this interval will be assigned 15.0. We call the distance from one number to the next the gap. Because the fraction is multiplied by $$2^{e-1023}$$, the gap grows as the number represented grows. The gap at a given number can be computed using the function spacing in numpy. import numpy as np TRY IT! Use the spacing function to determine the gap at", "1 \\cdot 11 + 1$ $3 \\cdot 6 = 18 = 1 \\cdot 11 + 7$ $4 \\cdot 6 = 24 = 2 \\cdot 11 + 2$ So the smallest possible positive value of $k$ which gives us the required remainder is $k = 4$ So: $n = 105 k - 1 = 105 \\cdot 4 - 1 = 419$", "+0 # help pls quickly 0 62 1 (a) compute $$10^{999}\\cdot 5^{-999}\\cdot 2^{-999}$$. (b) Suppose we write down the smallest positive 2-digit, 3-digit, and 4-digit multiples of 9. What is the sum of these three numbers? (c) Suppose we write down the smallest (positive) 2-digit, 3-digit, and 4-digit multiples of 8. What is the sum of these three numbers? (d)Suppose we write down the smallest (positive) 2-digit, 3-digit, and 4-digit multiples of 7. What is the sum of these three numbers? (e)What is the sum of all positive 1-digit integers that 4221462 is divisible by? (f)What's the largest -digit number that is a multiple of both 4 and 9? Jan 16, 2021 #1 +536 0 a) 10^999=5^999*2^999 5^999*5^-999*2^999*2^-999=1*1=1 b) Smallest 2 digits- 18 Smallest 3 digits- 108 Smallest 4 digits- 1008 18+108+1008=1134 c) 2 digits- 16 3 digits- 104 4 digits- 1000 16+104+1000=1120 d) 2 digit- 14 3 digit- 105 4 digit- 1001 14+105+1001=1120 e) 1, 2, 3, 6, 7", "100-2x$. This accounts for all possible cases since if $a = b \\ge 50$ then $c$ is not positive. There are also 49 triples with $a = c$, or with $b = c$, respectively. There are no triples with $a = b = c$, so the total number of triples with $a , b, c$ not all distinct is $3*49 = 147$. The number of triples of positive integers $(a, b, c)$ with $a, b, c$ distinct and $a + b + c = 100$ is therefore $\\binom{99}{2} -147$. Given a triple $(a, b, c)$ with $a< b < c$ and $a + b + c = 100$, there are six distinct ways to permute $a, b, c$, all of which our current estimate accounts for. Therefore we should divide by 6 to throw out all the triples that are in the wrong order. The final answer is $\\frac{1}{6}(\\binom{99}{2} -147) = 784$. Hint: For each possible value of $a$, count the number of possible values of $b$ and $c$. It's a quite regular pattern, so you don't have to brute force much to find it. Since $a$ is the smallest, the largest number it can be is $32$, so $a$ ranges from $1$ to $32$. The remaining sum $b+c$ must be equal to $100 - a$ and since", "+0 # Help PLS, ASAP +2 700 1 +140 The least common multiple of two positive integers is $7!$, and their greatest common divisor is $9$. If one of the integers is $315$, then what is the other? (Note that $7!$ means $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1$.) Mar 19, 2019 edited by CorbellaB.15 Mar 19, 2019 #1 0 The identity lcm[a,b]*gcd(a,b)=ab holds for all positive integers a and b. Let a=315 in this identity, and let b be the number we are looking for. Thus $$7! \\cdot 9=315 \\cdot b$$ so $$b = \\frac{7!\\cdot 9}{315} = \\frac{7!\\cdot 9}{35\\cdot 9} = \\frac{7!}{35} = \\frac{{7}\\cdot 6\\cdot {5}\\cdot 4\\cdot 3\\cdot 2\\cdot 1}{{35}} = 6\\cdot4\\cdot3\\cdot2 = \\boxed{144}.$$ Hope this helps :) Sep 7, 2019", "Taussig Jun 17 '16 at 21:02 • @N.F.Taussig: Many thanks. – André Nicolas Jun 17 '16 at 21:21 Find $a,b$ such that $60a+7b=1$. Use the extended Euclidean algorithm if necessary. Then $x = 1\\cdot60a + 59\\cdot7b$ is a solution. All solutions are congruent to this one mod $60\\cdot 7$. In this case you can use some heuristics. Solutions to (1) are of the form $59 + 60n$, $n \\in \\mathbb{Z}$, so all the positive ones end in $9$. The first solution to (2) that ends in $9$ is $29$, and every tenth solution thereafter also ends in $9$ (the next one being $99$). Pretty quickly you can arrive at $239$ as the smallest positive answer. The the technique of \"noodling around.\" $7*8 = 56 \\equiv -4(\\mod 60)\\\\ 7*9 = 63 \\equiv 3(\\mod 60)\\\\ 1 + -2 \\equiv 59(\\mod 60)\\\\ 1 + 7*(9+9+8+8)\\equiv 1 - 2 \\equiv 59(\\mod 60)\\\\ 239$ Now this is one solution. Since 7 and 60 are co-prime, $\\text{lcm(7,60)} = 420$ $239+420k$ and 239 is the smallest positive solution. • You meant $7 \\cdot 9 = 63 \\equiv 3 \\pmod{60}$. Also, you can obtain what I just wrote by writing 7 \\cdot 9 = 63 \\equiv 3 \\pmod{60} when you are in math mode. – N. F. Taussig Jun 17 '16 at 21:09 • Thanks for", "that separates the digits of $a$ from the digits of $b$, where $k$ is the number of digits in $a$: \\begin{align} b^2 - a^2 & = ab \\\\ & = a 10^k + b \\end{align} Collecting terms and rearranging: \\begin{align} a^2 - a 10^k & = b^2 - b \\\\ & = b ( b - 1 ) \\end{align} We know the value of $a$ because we chose it, and we know the value of $a^2 + a 10^k$. We have to find the value of $b$ where $b ( b - 1 )$ is the same value. For large enough $b$, $b ( b - 1 )$ is almost the same as $b^2$. We can easily calculate $$b = \\sqrt{a^2 + a 10^k}$$ Furthermore, for $b^2 - a^2$, there's some maximum value of $a$ for which there are no longer a possible value of $b$. The largest $b$ comprises only the digit $9$ repeated $k$ times. If $b$ has four digits, its maximum value is $9999$. There's some value of $a$ past which $9999^2 - a^2 = a 10^k + 9999$ has a solution. By guessing through bisection, we can find this maximum value, which is a little less than $6.2 \\cdot 10^{k}$. But it gets better. We can immediately discount some values of $a$. First, $a$", "by 2. 3) Continue this process until the quotient is 0. Then take all the remainders from bottom to top, line them up, and you have the binary number Example: Convert 17 to base 2. 17/2 = 8R1 8/2 = 4R0 4/2 = 2R0 2/2 = 1R0 1/2 = 0R1 Our binary equivalent of 17 is: $10001_2$ We could actually create a more general method that encompasses both and explains why they are equivalent; in particular, choose some $n$, which we will convert to binary. Now, choose some $i$ and divide $n$ by $2^i$, recording the remainder. More precisely, find the integers $a$ and $b$ such that: $$n=a2^i+b.$$ So $a$ is the quotient and $b$ is the remainder. In binary, $n$ will be the binary representation of $a$ followed by the representation of $b$ (where $b$ is written with $i$ digits, possibly including leading zeros). For instance, we can convert 17 by noting that it is $2\\cdot 2^3 + 1$. So, we write the quotient, 2, in binary as $10_2$ and the remainder, 1, using 3 digits as $001_2$. Then, when we join the two, we get: $$17=10001_2$$ which is correct. It's easy to prove the correctness of this general method because $$n=a\\cdot 2^i+b$$ in binary looks like: $$n=a\\cdot \\underbrace{10\\ldots 0_2}_{i \\text{ zeros}}+b$$ and, just like multiplying by", "1. ## lowest common multiple The lowest common multiple of 2, x and 21 is 126. Find the smallest possible value of x 2. $126 = 2\\cdot3\\cdot21$. 3. that was what i did!! hi-5 but wait the answer is not 3 but 9.18 or 63 $126 = 2\\cdot3\\cdot21$. The lowest common multiple of 2, 3 and 21 is 42. You just need to find the smallest number that is a factor of 126, but which 2 OR 21 are not factors, and which that number is not a factor of 2 OR 21 . The smallest number that fits those criteria is 63. 5. I didn't quite understand that... perhaps a working would be nice and with reference to \"and which that number is not a factor of 2 OR 21\" , isnt 21 a factor of 63? 6. Originally Posted by Punch and with reference to \"and which that number is not a factor of 2 OR 21\" , isnt 21 a factor of 63? Yes, but 2 isn't. Hence the OR. I don't know if I could show any working, I really just worked it out by common sense and logic. Put it this way; it can't be 3 because the lowest common multiple of 2, 3 and 21 is 42, not 126. Maybe the easiest" ]

[ "the truck 30 miles?", "## ericbob 3 years ago A flock of Canadian geese migrated 1623 miles in 28 days. What was the average rate at which these geese traveled in miles per day? 1. Akshay_Budhkar $1623 \\over 28$ is your answer Find more explanations on OpenStudy", "### Home > MC1 > Chapter 9 > Lesson 9.2.1 > Problem9-58 9-58. It took Ivan $7$ hours to drive $385$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "## anonymous 5 years ago If a truck drives 766 in x+2 hours, then what is his average speed. 1. anonymous I really don't see it. Please rewrite the question 2. anonymous If a truck drives 766 miles in x+2 hours, then what is his average speed? 3. anonymous $\\frac{766}{x+2} \\text{miles}/\\text{hour}$ if is x is 1/2 hour, then the truck may be exceeding Mach 1 at sea level.", "much money did he have? [show answer]$120 Question 3 A family is on a Sunday drive. So far they have driven 24 miles. If this is $\\frac{3}{5}$ of their trip, how long is their entire trip? [show answer] 40 miles", "## ericbob 2 years ago A flock of Canadian geese migrated 1623 miles in 28 days. What was the average rate at which these geese traveled in miles per day? $1623 \\over 28$ is your answer", "slower. At instant the truck passes the car, the car starts accelerating at constant 1m/sq.sec and overtake the truck 0.6km further down the road, from where the car moves uniformly. Then, it decreased itd speed by 13 mph for the remaining 228 miles. If the total distance was 876 miles, how long did it take to travel the remaining 228 miles 2 2013-09-20 · is the speed of the rabbit? 2 During the first 50 s a truck traveled at constant speed of 25 m/s.", "## A community for students. Sign up today! Here's the question you clicked on: ## ericbob 2 years ago A flock of Canadian geese migrated 1623 miles in 28 days. What was the average rate at which these geese traveled in miles per day? • This Question is Closed 1. Akshay_Budhkar $1623 \\over 28$ is your answer #### Ask your own question Ask a Question Find more explanations on OpenStudy", "### Home > CC2MN > Chapter 8 > Lesson 8.1.1 > Problem8-10 8-10. It took Ivan $7\\frac{1}{2}$ hours to drive $412.5$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "# If Brendan drove 4 miles in 8 hours, how many miles did he drive in 3 hours? He drove in $1 \\frac{1}{2}$ miles in $3$ hours. As Brendan drove $4$ miles in $8$ hours, he drove in one hour $\\frac{4}{8} = \\frac{1}{2}$ miles. Hence in $3$ hours, he drove in $\\frac{1}{2} \\times 3 = \\frac{3}{2} = 1 \\frac{1}{2}$ miles.", ", 08.11.2019 13:31, AbbyNeil # Dale drove 715 miles in 13 hours. at the same rate, how many miles would he drive in 7 hours? ### Other questions on the subject: Mathematics Mathematics, 21.06.2019 16:00, ijeomaokoro9488 The discriminant of the function is", "mules and are on the pasture for a different number of days. How much is each to pay? A horse, halving its speed each day, travels 700 miles in 7 days. How far does it travel each day? The authors recount the 'great tale' of Napier's and Burgi's parallel development of logarithms and urge you to use it in class. Two travelers, starting at the same time from the same point, travel in opposite directions round a circular railway.", "subtract 3.5 from t? Because it was not until 11:30 am, when $t=3.5$, that Stan began driving at a constant speed. The number of hours that Stan had driven at a constant speed at moments in time since starting for Tuscon is therefore $(t-3.5)$ hours. So it only for $3.5 \\le t \\le 5.1$ that $D=113(t-3.5)+32$ is a valid model of the distance Stan drove while driving to Tucson. Stan’s distance driven with respect to time did not change at a constant rate while he ran errands nor after he began slowing down in Tucson. Reflection 1.5.2. Describe similarities and differences between this analysis of Stan’s trip and Figures 1.5.1 and 1.5.2. ### Example 1.5.2 I am traveling away from home on a straight road at the constant speed of 0.35 km/min. I am 3.8 km from home, and I first looked at my watch 7 minutes ago. 1. How much will my distance from home change in the next 1/10 minute? How much will my distance from home change in the next 3/10 minute? How much will my distance from home change in the next t minutes? Let D represent number of km from home and let t represent the number of minutes since I looked at my watch. Then $dD = 0.35dt$, or $dD = (0.35)(0.1)$.", "0 like 0 dislike peter drove 1.5 hours at an average speed of 40 mph. He then drove for another 40km. 5 miles=8 kilometers how far did peter drive?", "more miles does he need to drive to finish his journey? Solution The number of miles to drive to finish his journey is given by 1200 - 768 = 432 miles The rectangle on the left and the square on the right have the same perimeter. What is the length of one side of the square? . Solution The perimeter P of the rectangle is equal to P = 15 + 25 + 15 + 25 = 80 The perimeter of the square is equal to the perimeter of the rectangle and is then equal to 80. The square has 4 sides of equal lengths. Since 80 = 20 + 20 + 20 + 20 then the length of a side is equal to 20 There are 123 boxes of sweets in a store. There are 25 sweets in each box. How many sweets are in the store? Solution To find how many sweets, we multiply 123 by 25 123 × 25 = 3075 sweets There are 365 days in one year, and 100 years in one century. How many days are in one century? Solution In 100 years, which is one century, there 365 × 100 = 36,500 days Billy read 2 books. He read the first one in one week with 25 pages everyday. He read", "since it would take four times as long to travel the same distance.", "of the earth is 4,000 miles.) Give your answer both as an exact value and an asked by Anonymous on February 20, 2012 43. ## English 2. Which expression best represents the view a fatalist would have? A. \"Whatever will be, will be.\" B. \"The best is yet to come.\" C. \"We are still masters of our fate.\" D. \"You don't know what you have until its gone.\" 3. In which of these lines is the asked by help?thanks on March 11, 2015 44. ## Math Elaine has a customer who needs a box with a volume of 12 cubic inches. The customer want to know what size box is the least expensive to buy. .the price will be based on surface area. the supplier charges 1/2 cent persquare inch. .the dimensions of the asked by Neisha on March 3, 2013 45. ## Math - 8th grade - Please help I don't know the formula to figure this out. Charlie and Peter are driving around a racetrack. Charlie drives at an average speed of 108 miles per hour. Peter drives at an average speed of 99 miles per hour. Peter is given a head start of 12 miles. Charlie asked by Chris on November 16, 2012 46. ## math An airplane flying against the wind travels 100 miles", "than 500,000. => 513679. => 531679. => 561379. Question 4. Ms. Torres is on a road trip. She drives a total of 2,731.82 miles in three weeks. She drives 791.38 miles the second week. She drives 1,086.14 miles the third week. How many miles did she drive the first week? Number of miles She drives in the first week = 854.3. Explanation: Number of miles She drives in three weeks = 2,731.82. Number of miles She drives in the second week = 791.38. Number of miles She drives in the third week = 1,086.14. Number of miles She drives in the first week = Number of miles She drives in three weeks – (Number of miles She drives in the second week + Number of miles She drives in the third week) = 2731.82 – (791.38 + 1086.14) = 2731.82 -1877.52 = 854.3. Question 5. Mr. Kerry has 4 boxes of cocoa mix. Each box has 20 bags. He uses two cups of water and one bag of cocoa mix to make a mug of cocoa. How many cups of water will Mr Kerry use if he uses all the bags of cocoa mix? Number of cups of water if he uses all the bags of cocoa mix = 160. Explanation: Number of boxes of cocoa mix Mr.", "case we get $$\\frac{(1+1)*(60*30)}{(60+30)}$$ Which is $$\\frac{2*1800}{90} = 40$$ Re: Susan drove an average speed of 30 miles per hour for the [#permalink] 03 Dec 2019, 12:18 Go to page Previous 1 2 [ 27 posts ] Display posts from previous: Sort by", "Question Sean drove at an average speed of 68 km/h for the first 4 hours of the journey. He then increased his average speed by 24 km/h and drove for another 2 hours until he reached his town. Find his average speed for the whole journey. 3 m Click button first when a symbol is required. X Sean drove at an average speed of 68 km/h for the first 4 hours of the journey. He then increased his average speed by 24 km/h and drove for another 2 hours until he reached his town. Find his average speed for the whole journey." ]

[ "it'd take (60/75) hours (0.8 hours, or 48 minutes) to go at 75 mph versus 60/65 hours (0.92 hours, or 55 minutes and about 20 seconds) to go at 65 mph. How much was 7 minutes of my time worth? At the wages I was working at that point (about $15 an hour),$1.75. How much was 1.5 gallons of gas worth? At that time, roughly $6. I decided to drive slower. However, on the way back, I had my girlfriend, which meant I, ahem, valued the time saved from driving faster more. I drove faster on the way back. To this day I've yet to make enough money in hourly terms to justify regularly driving 75 mph over driving 65 mph on the highway. At the current average gasoline price in New York state ($3.92/gal), I'd have to be making $50.40 an hour to justify it, which for a standard 8-hour work week is$104,832 per year. ------ This analysis is relatively simple. There are a couple of (to my mind, anyway) interesting little wrinkles to this calculation, the magnitude of which I can't calculate off the top of my head but I think on balance points to even more fuel consumption at higher speed. The first wrinkle occurs when we relax the assumption that the drag coefficient is", "each year? Using estimates of number of personal vehicles and annual mileage, combined with an estimate that 30% of these miles would be driven more slowly than otherwise with drivers adopting my techniques, my estimate is that about 468,000,000,000 miles would be driven more slowly. I estimate that this would result in 1.8 X 10^9 (1.8 billion) extra hours on the road. Does this make sense? Well, I'm one of 200,000,000 drivers and I calculated in an earlier post that I'll lose about 47 hours per year. If each of the 200 million drivers lost that much, the total would be 9.4 X 10^9 hours. But I estimate that I probably drive something like 30% more than average so adjust this down to 7.2 X 10^9 hours. Figure somewhere between these numbers is right, I'll use 4.5 X 10^9 or 4 and one half billion hours. I'll say the average hour is worth $30.00, in that case the lost time is worth$135 billion. Considering I calculated we'd save $36 billion in oil imports, it would seem not to be worth it. Are there any mitigating factors? Not many. First would be finding ways to avoid productivity loss when on the road. Talk radio? Books on tape? Cell phones? These all may help but they certainly won't eliminate the", "## anonymous 5 years ago If a truck drives 766 in x+2 hours, then what is his average speed. 1. anonymous I really don't see it. Please rewrite the question 2. anonymous If a truck drives 766 miles in x+2 hours, then what is his average speed? 3. anonymous $\\frac{766}{x+2} \\text{miles}/\\text{hour}$ if is x is 1/2 hour, then the truck may be exceeding Mach 1 at sea level.", "### Home > MC1 > Chapter 9 > Lesson 9.2.1 > Problem9-58 9-58. It took Ivan $7$ hours to drive $385$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "of the earth is 4,000 miles.) Give your answer both as an exact value and an asked by Anonymous on February 20, 2012 43. ## English 2. Which expression best represents the view a fatalist would have? A. \"Whatever will be, will be.\" B. \"The best is yet to come.\" C. \"We are still masters of our fate.\" D. \"You don't know what you have until its gone.\" 3. In which of these lines is the asked by help?thanks on March 11, 2015 44. ## Math Elaine has a customer who needs a box with a volume of 12 cubic inches. The customer want to know what size box is the least expensive to buy. .the price will be based on surface area. the supplier charges 1/2 cent persquare inch. .the dimensions of the asked by Neisha on March 3, 2013 45. ## Math - 8th grade - Please help I don't know the formula to figure this out. Charlie and Peter are driving around a racetrack. Charlie drives at an average speed of 108 miles per hour. Peter drives at an average speed of 99 miles per hour. Peter is given a head start of 12 miles. Charlie asked by Chris on November 16, 2012 46. ## math An airplane flying against the wind travels 100 miles", "school with a speed twice as much as his speed when he ran from school to home. If it took Jerry $$3$$ hours in total to run to school and then back to home, how long did it take Jerry to run from home to school? 11 HARD When Gary drove from City A to City B, he drove at $$60$$ kilometers per hour. When He drove from City B to City A, he drove at $$80$$ kilometers per hour. What was Gary's average speed for driving from A to B and then from B to A? (Hint: $$average\\ speed=\\frac{total\\ distance}{total\\ time}$$)", "Perfect! Jerry covers $$20 \\, \\mathrm{m}$$ in $$2 \\, \\mathrm{s}$$, so, his speed is $$10 \\, \\mathrm{m/s}$$. Whereas in the other cases, his speed was lower. If an object's speed is $$30 \\, \\mathrm{km/h}$$, in what direction is it moving?", "t1)/t1$$ = $$100$$ * $$(x+5-x)/x$$= $$500/x$$ % _________________ Argument : If you love long trips, you love the GMAT. Conclusion : GMAT is long journey. What does the author assume ? Assumption : A long journey is a long trip. GMAT Club Premium Membership - big benefits and savings Math Forum Moderator Joined: 20 Dec 2010 Posts: 2022 Followers: 149 Kudos [?]: 1360 [0], given: 376 Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink] 22 Feb 2011, 14:49 I think the question should be: During a trip on an expressway, Don drove x miles.His average speed on a certain 5 mile section was 30 miles per hour and his average speed of the remainder ofthe trip was 60 miles per hour. His travel time for the x mile trip was what percent smaller than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip $$\\frac{\\frac{x}{60}-\\frac{x-5}{60}}{\\frac{x}{60}}*100 = \\frac{500}{x}%$$ Ans: \"E\" _________________ Manager Status: I am Midnight's Child ! Joined: 04 Dec 2009 Posts: 148 WE 1: Software Design and Development Followers: 1 Kudos [?]: 54 [0], given: 11 Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink] 22 Feb 2011, 15:24 fluke wrote: I think", "of people think as they do about Figures 3.15.1 and 3.15.2? Alert! How you read the next two examples is important. Read them with the goal that you become able to repeat their patterns of reasoning. Do not read them with the goal of remembering what to write! ### Example 1.5.1 Stan started running errands in Phoenix at 8:00 am. He had driven 32 km at the end of the last errand. At 11:30 am Stan headed for Tucson at a constant speed of 113 km/hr; 1.6 hours later he began slowing to a stop. Let D represent the number of km that Stan drove since 8:00 am, and let t represent the number of hours since 8:00 am. While on his trip to Tucson, the rate of change of D with respect to t was 113 km/hr, so $dD = 113dt$ km, and D, Stan's distance traveled at the constant speed of 113 km/hr, is $D=113(t-3.5)+32$ km. Why must we subtract 3.5 from t? Because it was not until 11:30 am, when t = 3.5, that Stan began driving at a constant speed. So it only for $3.5 \\le t \\le 5.1$ that $D=113(t-3.5)+32$ is a valid model of the distance Stan drove while driving to Tucson. Stan’s distance driven with respect to time did not vary", "subtract 3.5 from t? Because it was not until 11:30 am, when $t=3.5$, that Stan began driving at a constant speed. The number of hours that Stan had driven at a constant speed at moments in time since starting for Tuscon is therefore $(t-3.5)$ hours. So it only for $3.5 \\le t \\le 5.1$ that $D=113(t-3.5)+32$ is a valid model of the distance Stan drove while driving to Tucson. Stan’s distance driven with respect to time did not change at a constant rate while he ran errands nor after he began slowing down in Tucson. Reflection 1.5.2. Describe similarities and differences between this analysis of Stan’s trip and Figures 1.5.1 and 1.5.2. ### Example 1.5.2 I am traveling away from home on a straight road at the constant speed of 0.35 km/min. I am 3.8 km from home, and I first looked at my watch 7 minutes ago. 1. How much will my distance from home change in the next 1/10 minute? How much will my distance from home change in the next 3/10 minute? How much will my distance from home change in the next t minutes? Let D represent number of km from home and let t represent the number of minutes since I looked at my watch. Then $dD = 0.35dt$, or $dD = (0.35)(0.1)$.", "What is her net decrease in calories after walking for 3 hours? 225 cal 36) A motor vehicle has a maximum efficiency of 33 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg/mph between 40 mph and 50 mph, and at a rate of 0.4 mpg/mph between 50 mph and 80 mph. What is the efficiency in miles per gallon if the car is cruising at 50 mph? What is the efficiency in miles per gallon if the car is cruising at 80 mph? If gasoline costs $3.50/gal, what is the cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at 80 mph? 37) Although some engines are more efficient at given a horsepower than others, on average, fuel efficiency decreases with horsepower at a rate of $$1/25$$ mpg/horsepower. If a typical 50-horsepower engine has an average fuel efficiency of 32 mpg, what is the average fuel efficiency of an engine with the following horsepower: 150, 300, 450? Answer $$E(150)=28,\\;E(300)=22,\\;E(450)=16$$ 38) [T] The following table lists the 2013 schedule of federal income tax versus taxable income. Taxable Income Range The Tax Is ... ... Of the Amount Over$0–$8925 10%$0 $8925–$36,250 $892.50 + 15%$8925 $36,250–$87,850 $4,991.25 + 25%$36,250 $87,850–$183,250 $17,891.25 + 28%$87,850 $183,250–$398,350 $44,603.25 + 33%$183,250", "to the value of mx. The value of y is always y = mx + b. Move your pointer away from the animation to hide the control bar. Reflection 1.5.1. Some people think that Figures 1.5.1 and 1.5.2 say the same thing. Others disagree. Why might each group of people think as they do about Figures 1.5.1 and 1.5.2? Alert! How you read the next two examples is important. Read them with the goal that you become able to repeat their patterns of reasoning. Do not read them with the goal of remembering what to write! ### Example 1.5.1. Changes in t from an initial value of t Stan started running errands in Phoenix at 8:00 am. He had driven 32 km at the end of the last errand. At 11:30 am Stan headed for Tucson at a constant speed of 113 km/hr; 1.6 hours later he began slowing to a stop. Let D represent the number of km that Stan drove since 8:00 am, and let t represent the number of hours since 8:00 am. While on his trip to Tucson, the rate of change of D with respect to t was 113 km/hr, so dD = 113dt km, and D, Stan's distance traveled at the constant speed of 113 km/hr, is $D=113(t-3.5)+32$ km. Why must we", "1. $$x=3$$ 2. $$x=-3$$ 3. $$x=-\\frac{3}{5}$$ 4. $$x=18$$ 5. The number is $$3$$. 6. Each segment will be $$8$$ meters long. 7. There are $$34$$ rabbits and $$66$$ chickens. 8. I have $$60$$ books. 9. Lisa's speed was $$20$$ kilometers per hour. 10. It took Jerry $$1$$ hour to run from home to school. 11. Gary's average speed was $$\\frac{480}{7}$$ kilometers per hour.", "love to travel, hike and explore new mountain trails. Often calculating the average speed is simple using the formula =. This table estimates the number of calories burned per hour based on weight and trail grade at an average walking speed of 2.9–3.5 mph (4.7–5.6 kph) (1): A … KNOW When to Replace Your Hiking Boots, What Pants to Wear Hiking? And 13 pairs of shoes. The most important thing we need to establish first is what is the average speed you will be hiking at!? Beth’s motor scooter gets 110 km/gal. Imagine the length of a track is 100 km (thus the race of 4 laps is a 400 km race), and the average speed you want for the whole thing is 100 km/hr. Adjust the values according to your own rating. Finding average speed examples Example 1: Using the equation above, find the speed of a train which travelled 120 miles in 2 hours and 10 minutes while making four stops, each lasting approximately 2.5 minutes. Charles Grier. With groups, the values of the weakest member apply. Using Naismith’s rule developed by Scottish Mountaineer William Naismith – we can calculate the extra time required to walk uphill. For example, if you have speed in mph (miles per hour), time should also be in hours. Speed hiking", "# Problem 2-62 - Distance and speed - Part 3 (a) A truck travels at $80 \\;km/h$ for $30\\; min$, then at $60 \\;km/h$ for $1.5\\; hr.$ Assuming 2 significant digits in the given numbers, calculate the truck's (a) total distance traveled, and (b) average speed. No. Check the units. Try again.", "### Home > CC2MN > Chapter 8 > Lesson 8.1.1 > Problem8-10 8-10. It took Ivan $7\\frac{1}{2}$ hours to drive $412.5$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "+0 # help 0 48 1 A rocket traveled 12,000 miles in the first four hours. Then the rocket traveled 8000 miles in the next four hours. What was the rocket’s average speed, in miles per hour, for these 8 hours? Jun 1, 2020 #1 +9661 +1 A rocket traveled 12,000 miles in the first four hours. Then the rocket traveled 8000 miles in the next four hours. What was the rocket’s average speed, in miles per hour, for these 8 hours? Hello Guest! $$average\\ speed=\\frac{(12000+8000)miles}{(4+4)hours}=\\color{blue}2500\\ miles\\ per\\ hour$$ ! Jun 1, 2020", "$$C(x) = \\frac{{2010}}{{{x^{2.2}}}} + 17.80$$ where x denotes the number of miles (in thousands of miles) the car is driven in a year. What is the average cost of driving a subcompact car: 1. 5,000 miles per year?$C\\left( 5 \\right) = \\frac{{2010}}{{{5^{2.2}}}} + 17.80 = 76.10$The average cost when driving 5,000 miles per year is 76.1 cents per mile. 2. 10,000 miles per year?$C\\left( {10} \\right) = \\frac{{2010}}{{{{10}^{2.2}}}} + 17.80 = 30.50$The average cost when driving 10,000 miles per year is 30.5 cents per mile. 3. 25,000 miles per year?$C\\left( {25} \\right) = \\frac{{2010}}{{{{25}^{2.2}}}} + 17.80 = 19.50$The average cost when driving 25,000 miles per year is 19.5 cents per mile. 4. 50,000 miles per year?$C\\left( {50} \\right) = \\frac{{2010}}{{{{50}^{2.2}}}} + 17.80 = 18.17$The average cost when driving 50,000 miles per year is 18.2 cents per mile. 5. What happens to the average cost as the number of miles driven increases without bound? The average cost approaches 17.8 cents per mile when the car is driven “infinite” miles in a year. 6. Verify by evaluating the cost when the number of miles is 1,000,000 (or any large number).$C\\left( {1000} \\right) = \\frac{{2010}}{{{{1000}^{2.2}}}} + 17.80$The average cost per mile is approximately 17.80 cents per mile when the car is driven 1,000,000 miles per year. Source https://domoremath.files.wordpress.com/2013/09/limits-word-prob-with-solns.pdf #### Notes Limits:", "A two-lane urban road with one-way traffic has a maximum capacity of $1800$ vehicles/hour. Under the jam condition, the average length occupied by the vehicles is $5.0 \\: m$. The speed versus density relationship is linear. For a traffic volume of $1000$ vehicles/hour, the density (in vehicles/km) is 1. $52$ 2. $58$ 3. $67$ 4. $75$", "case we get $$\\frac{(1+1)*(60*30)}{(60+30)}$$ Which is $$\\frac{2*1800}{90} = 40$$ Re: Susan drove an average speed of 30 miles per hour for the [#permalink] 03 Dec 2019, 12:18 Go to page Previous 1 2 [ 27 posts ] Display posts from previous: Sort by" ]

[ "### Home > MC1 > Chapter 9 > Lesson 9.2.1 > Problem9-58 9-58. It took Ivan $7$ hours to drive $385$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "it'd take (60/75) hours (0.8 hours, or 48 minutes) to go at 75 mph versus 60/65 hours (0.92 hours, or 55 minutes and about 20 seconds) to go at 65 mph. How much was 7 minutes of my time worth? At the wages I was working at that point (about $15 an hour),$1.75. How much was 1.5 gallons of gas worth? At that time, roughly $6. I decided to drive slower. However, on the way back, I had my girlfriend, which meant I, ahem, valued the time saved from driving faster more. I drove faster on the way back. To this day I've yet to make enough money in hourly terms to justify regularly driving 75 mph over driving 65 mph on the highway. At the current average gasoline price in New York state ($3.92/gal), I'd have to be making $50.40 an hour to justify it, which for a standard 8-hour work week is$104,832 per year. ------ This analysis is relatively simple. There are a couple of (to my mind, anyway) interesting little wrinkles to this calculation, the magnitude of which I can't calculate off the top of my head but I think on balance points to even more fuel consumption at higher speed. The first wrinkle occurs when we relax the assumption that the drag coefficient is", "each year? Using estimates of number of personal vehicles and annual mileage, combined with an estimate that 30% of these miles would be driven more slowly than otherwise with drivers adopting my techniques, my estimate is that about 468,000,000,000 miles would be driven more slowly. I estimate that this would result in 1.8 X 10^9 (1.8 billion) extra hours on the road. Does this make sense? Well, I'm one of 200,000,000 drivers and I calculated in an earlier post that I'll lose about 47 hours per year. If each of the 200 million drivers lost that much, the total would be 9.4 X 10^9 hours. But I estimate that I probably drive something like 30% more than average so adjust this down to 7.2 X 10^9 hours. Figure somewhere between these numbers is right, I'll use 4.5 X 10^9 or 4 and one half billion hours. I'll say the average hour is worth $30.00, in that case the lost time is worth$135 billion. Considering I calculated we'd save $36 billion in oil imports, it would seem not to be worth it. Are there any mitigating factors? Not many. First would be finding ways to avoid productivity loss when on the road. Talk radio? Books on tape? Cell phones? These all may help but they certainly won't eliminate the", "## anonymous 5 years ago If a truck drives 766 in x+2 hours, then what is his average speed. 1. anonymous I really don't see it. Please rewrite the question 2. anonymous If a truck drives 766 miles in x+2 hours, then what is his average speed? 3. anonymous $\\frac{766}{x+2} \\text{miles}/\\text{hour}$ if is x is 1/2 hour, then the truck may be exceeding Mach 1 at sea level.", "## ericbob 3 years ago A flock of Canadian geese migrated 1623 miles in 28 days. What was the average rate at which these geese traveled in miles per day? 1. Akshay_Budhkar $1623 \\over 28$ is your answer Find more explanations on OpenStudy", "### Home > CC2MN > Chapter 8 > Lesson 8.1.1 > Problem8-10 8-10. It took Ivan $7\\frac{1}{2}$ hours to drive $412.5$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "subtract 3.5 from t? Because it was not until 11:30 am, when $t=3.5$, that Stan began driving at a constant speed. The number of hours that Stan had driven at a constant speed at moments in time since starting for Tuscon is therefore $(t-3.5)$ hours. So it only for $3.5 \\le t \\le 5.1$ that $D=113(t-3.5)+32$ is a valid model of the distance Stan drove while driving to Tucson. Stan’s distance driven with respect to time did not change at a constant rate while he ran errands nor after he began slowing down in Tucson. Reflection 1.5.2. Describe similarities and differences between this analysis of Stan’s trip and Figures 1.5.1 and 1.5.2. ### Example 1.5.2 I am traveling away from home on a straight road at the constant speed of 0.35 km/min. I am 3.8 km from home, and I first looked at my watch 7 minutes ago. 1. How much will my distance from home change in the next 1/10 minute? How much will my distance from home change in the next 3/10 minute? How much will my distance from home change in the next t minutes? Let D represent number of km from home and let t represent the number of minutes since I looked at my watch. Then $dD = 0.35dt$, or $dD = (0.35)(0.1)$.", "of the earth is 4,000 miles.) Give your answer both as an exact value and an asked by Anonymous on February 20, 2012 43. ## English 2. Which expression best represents the view a fatalist would have? A. \"Whatever will be, will be.\" B. \"The best is yet to come.\" C. \"We are still masters of our fate.\" D. \"You don't know what you have until its gone.\" 3. In which of these lines is the asked by help?thanks on March 11, 2015 44. ## Math Elaine has a customer who needs a box with a volume of 12 cubic inches. The customer want to know what size box is the least expensive to buy. .the price will be based on surface area. the supplier charges 1/2 cent persquare inch. .the dimensions of the asked by Neisha on March 3, 2013 45. ## Math - 8th grade - Please help I don't know the formula to figure this out. Charlie and Peter are driving around a racetrack. Charlie drives at an average speed of 108 miles per hour. Peter drives at an average speed of 99 miles per hour. Peter is given a head start of 12 miles. Charlie asked by Chris on November 16, 2012 46. ## math An airplane flying against the wind travels 100 miles", "## ericbob 2 years ago A flock of Canadian geese migrated 1623 miles in 28 days. What was the average rate at which these geese traveled in miles per day? $1623 \\over 28$ is your answer", "of people think as they do about Figures 3.15.1 and 3.15.2? Alert! How you read the next two examples is important. Read them with the goal that you become able to repeat their patterns of reasoning. Do not read them with the goal of remembering what to write! ### Example 1.5.1 Stan started running errands in Phoenix at 8:00 am. He had driven 32 km at the end of the last errand. At 11:30 am Stan headed for Tucson at a constant speed of 113 km/hr; 1.6 hours later he began slowing to a stop. Let D represent the number of km that Stan drove since 8:00 am, and let t represent the number of hours since 8:00 am. While on his trip to Tucson, the rate of change of D with respect to t was 113 km/hr, so $dD = 113dt$ km, and D, Stan's distance traveled at the constant speed of 113 km/hr, is $D=113(t-3.5)+32$ km. Why must we subtract 3.5 from t? Because it was not until 11:30 am, when t = 3.5, that Stan began driving at a constant speed. So it only for $3.5 \\le t \\le 5.1$ that $D=113(t-3.5)+32$ is a valid model of the distance Stan drove while driving to Tucson. Stan’s distance driven with respect to time did not vary", "+0 # help 0 48 1 A rocket traveled 12,000 miles in the first four hours. Then the rocket traveled 8000 miles in the next four hours. What was the rocket’s average speed, in miles per hour, for these 8 hours? Jun 1, 2020 #1 +9661 +1 A rocket traveled 12,000 miles in the first four hours. Then the rocket traveled 8000 miles in the next four hours. What was the rocket’s average speed, in miles per hour, for these 8 hours? Hello Guest! $$average\\ speed=\\frac{(12000+8000)miles}{(4+4)hours}=\\color{blue}2500\\ miles\\ per\\ hour$$ ! Jun 1, 2020", "instead of every 5.3 days (on average). I estimate that pulling off the road, fueling and getting back on the road takes about 8 minutes, so I save an average of 31 seconds per day stopping for fuel less often. Thus, the grand total loss is 7 minutes 54 seconds per day. In the course of a year, I lose 33 hours and 11 minutes behind the wheel (assuming 252 such days per year). Looking at this time another way, my company values the 7 minutes and 54 seconds at about $12.66. An interesting comparison is that, on a given day, I save about 1.4 gallons of fuel compared to what I would have consumed using my pre-experiment driving style, which, at current Southern California prices, is worth about$4.57. Or, yet another way is that I try to sleep 7 hours per night. Thus, the extra time spent driving amounts to 0.77% of my waking hours. Hmmm....", "# Please explain, I tried multiple tmes, but couldn't get to the right answer. I have udated the wi... Please explain, I tried multiple tmes, but couldn't get to the right answer. Stan Moneymaker has been informed of a major automobile manufacturer's plan to conserve on gasoline consumption through improved engine design. The idea is called \"engine displacement, \" and it works by switching from 8-cylinder operation to 4-cylinder operation at approximately 40 miles per hour. Engine displacement allows enough power to accelerate from a standstill and to climb hills while also permitting the automobile to cruise at speeds over 40 miles per hour with little loss in driving performance. The trade literature studied by Stan makes the claim that the engine displacement option will cost the customer an extra $1, 400 on the automobile's sticker price. This option is expected to save 4 miles per gallon (an average of in-town and highway driving). A regular 8-cylinder engine in the car that Stan is interested In buying gets an average of 24 miles per gallon of gasoline. If Stan drives approximately 1, 680 miles per month, how many months of ownership will be required to make this$1, 400 investment pay for Itself? Stan's opportunity cost of capital (i) is 0.5% per month, and gasoline costs $3.25 per gallon.", "t1)/t1$$ = $$100$$ * $$(x+5-x)/x$$= $$500/x$$ % _________________ Argument : If you love long trips, you love the GMAT. Conclusion : GMAT is long journey. What does the author assume ? Assumption : A long journey is a long trip. GMAT Club Premium Membership - big benefits and savings Math Forum Moderator Joined: 20 Dec 2010 Posts: 2022 Followers: 149 Kudos [?]: 1360 [0], given: 376 Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink] 22 Feb 2011, 14:49 I think the question should be: During a trip on an expressway, Don drove x miles.His average speed on a certain 5 mile section was 30 miles per hour and his average speed of the remainder ofthe trip was 60 miles per hour. His travel time for the x mile trip was what percent smaller than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip $$\\frac{\\frac{x}{60}-\\frac{x-5}{60}}{\\frac{x}{60}}*100 = \\frac{500}{x}%$$ Ans: \"E\" _________________ Manager Status: I am Midnight's Child ! Joined: 04 Dec 2009 Posts: 148 WE 1: Software Design and Development Followers: 1 Kudos [?]: 54 [0], given: 11 Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink] 22 Feb 2011, 15:24 fluke wrote: I think", "+0 # Algebra Question +1 337 1 +850 Marguerite drove 100 miles in 2.4 hours. If Sam drove for 3 hours at the same average rate as Marguerite, how many miles did he drive? Aug 20, 2018 #1 +8348 +2 Marguerite drove 100 miles in 2.4 hours. If Sam drove for 3 hours at the same average rate as Marguerite, how many miles did he drive? Hello Logic! $$s=v\\cdot t\\\\ s=\\frac{100\\ miles}{2.4\\ hours}\\cdot 3\\ hours\\\\ \\color{blue}s=125\\ miles$$ ! Aug 21, 2018 #1 +8348 +2 Marguerite drove 100 miles in 2.4 hours. If Sam drove for 3 hours at the same average rate as Marguerite, how many miles did he drive? Hello Logic! $$s=v\\cdot t\\\\ s=\\frac{100\\ miles}{2.4\\ hours}\\cdot 3\\ hours\\\\ \\color{blue}s=125\\ miles$$ ! asinus Aug 21, 2018", "Question Sean drove at an average speed of 68 km/h for the first 4 hours of the journey. He then increased his average speed by 24 km/h and drove for another 2 hours until he reached his town. Find his average speed for the whole journey. 3 m Click button first when a symbol is required. X Sean drove at an average speed of 68 km/h for the first 4 hours of the journey. He then increased his average speed by 24 km/h and drove for another 2 hours until he reached his town. Find his average speed for the whole journey.", "Free Version Moderate # Road Trip: How Long Will the Tank Last? SATMAT-X3DC9Y John's car can travel, on average, $32$ miles on one gallon of gasoline. On a recent road trip, his average speed was $58$ miles per hour. Approximately how many hours was he able to drive on a $15$-gallon tank of gasoline? A $27.1$ hours B $123.7$ hours C $15.7$ hours D $8.3$ hours", "mules and are on the pasture for a different number of days. How much is each to pay? A horse, halving its speed each day, travels 700 miles in 7 days. How far does it travel each day? The authors recount the 'great tale' of Napier's and Burgi's parallel development of logarithms and urge you to use it in class. Two travelers, starting at the same time from the same point, travel in opposite directions round a circular railway.", "school with a speed twice as much as his speed when he ran from school to home. If it took Jerry $$3$$ hours in total to run to school and then back to home, how long did it take Jerry to run from home to school? 11 HARD When Gary drove from City A to City B, he drove at $$60$$ kilometers per hour. When He drove from City B to City A, he drove at $$80$$ kilometers per hour. What was Gary's average speed for driving from A to B and then from B to A? (Hint: $$average\\ speed=\\frac{total\\ distance}{total\\ time}$$)", "a conoeist travels 70 miles at a certain speed. the canoeist travles 7 miles on the second part of the trip at a speed of 5 mph slower. the total time for the trip is 5 hrs. Ehat is the speed on each part of the trip? asked by todd on July 6, 2012 42. ## Stats Wood scientists are interested in replacing solid-wood building material by less expensive products made from wood flakes. Some summary statistics obtained from a study of the relationship between the length (in inches) and the strength (in pounds per asked by terry on June 20, 2017 43. ## Mathematics In the universal set (U) reperesents all the students in a class.The set m represents the student who take (Music 3x).The set D represents the students who take (Drama 7+x).If 24 students take Music, calculate The number of students who take BOTH music and asked by Devon on June 19, 2009 44. ## math 1.Glenn bought a car for Php600,000. The yearly depreciation of his car is 10% of its value at the start of the year. What is its value after 4 years? A. Php437,400 B. Php438,000 C. Php393,660 D. Php378,000 2. During a free-fall, a skydiver jumps 16 feet, asked by miles on June 27, 2015 45. ## science 3. Which" ]

[ "# 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 $A=A'$.", "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 inscribed angle $$ACB$$ must measure 25 degrees, even if we move point $$C$$ along the circumference (without going past $$A$$ or $$B$$). This also means that all inscribed angles that define the same arc are congruent.", "same length sides but not the same angle measures. 4. Triangles $$ABC$$ and $$DEF$$ are congruent. If the measure of angle A is $$58^{\\circ}$$, what is the measure of angle D if it corresponds to angle A? ## Vocabulary Term Definition Angle A geometric figure formed by two rays that connect at a single point or vertex. Congruent Congruent figures are identical in size, shape and measure. Trapezoid A trapezoid is a quadrilateral with exactly one pair of parallel opposite sides. Rigid Transformation A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure.", "# 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 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. ×", "### A regular pentagon has a perimeter of 60 cm what is the measure of angle CBD a regular pentagon has a perimeter of 60 cm what is the measure of angle CBD... ### Choose whether the pair of words are similar or opposite in meaning, gaseosa/refresco A. similar B. opposite Choose whether the pair of words are similar or opposite in meaning, gaseosa/refresco A. similar B. opposite... ### PLEASE HELP WILL GIVE BRAINLIEST. Which of the following statments is true about the equation shown? Mark all that are true. t (x) = (x-5)^2 + 7. a the function is quadratic b the function will produce a line with a positive slope when graphed c the function will produce a concave up parabola when graphed d the vertex will be at (7, -5) PLEASE HELP WILL GIVE BRAINLIEST. Which of the following statments is true about the equation shown? Mark all that are true. t (x) = (x-5)^2 + 7. a the function is quadratic b the function will produce a line with a positive slope when graphed c the function will produce a concave up parabola wh... ### Among fatal plane crashes that occurred during the past 60 ​years, 115 were due to pilot​ error, 84 were due to other human​ error, 517 were due to​ weather,", "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!", "so this means that angles $a$ and $b$ are congruent. 2. The picture below shows the angles which are congruent to $a$: We have already seen that $a$ and $b$ are congruent. Angles $a$ and $c$ make a vertical pair so they are also congruent (as can be shown with a 180 degree rotation about $P$). Angles $b$ and $d$ are vertical angles and so they are also congruent: since $a$ and $b$ are congruent, this means that $a$ is congruent to $d$. The other angles made by the intersection of these lines are all obtuse (or measure more than 180 degrees) and so are not congruent to angle $a$.", "# ACE Geometry Level 2 $$ABCDE$$ is a regular pentagon. What is the measure (in degrees) of $$\\angle ACE$$? ×", "have congruent angles. • Using the fact that the measures of angles that make a line add up to 180 degrees. ### Launch Provide 3 minutes of work time followed by a whole-class discussion. Action and Expression: Develop Expression and Communication. Maintain a display of important terms and information. During the launch, take time to review the following facts from previous lessons that students will need to access for this activity: congruent triangles have congruent angles, a straight angle is equal to 180 degrees, and the sum of the angles in a triangle add up to 180 degrees. Supports accessibility for: Memory; Language ### Student Facing This diagram shows a square $$BDFH$$ that has been made by images of triangle $$ABC$$ under rigid transformations. Given that angle $$BAC$$ measures 53 degrees, find as many other angle measures as you can. ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Activity Synthesis Display the image of the 4 triangles for all to see. Invite students to share how they calculated one of the other unknown angles in the image, adding to the image until all the unknown angles are filled in. In wrapping up, note that angles $$ACE$$, $$CEG$$, $$EGA$$, and $$GAC$$ are", "a Circle : If all of the vertices of a polygon lie on acircle, the polygon is inscribed in the circle and the circle is circumscribedabout the polygon. {/eq} is a regular pentagon inscribed in a circle, so each of the angles labeled with x have the same measure. In a Regular Pentagon Abcde, Inscribed in a Circle; Find Ratio Between Angle Eda and Angle Adc. A regular pentagon is made of five congruent triangles whose congruent vertex angles form a circle and add to 360. Important Solutions 2865. The regular pentagon (5-sided polygon) divides the 360 degrees of the circle into 5 equal arcs. In this case, this inscribed angle ( Share this Post", "$ABCDE$ 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 $BE$.", "× # 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", "consist of three triangles, where each triangle contains three … When creating pentagon interior angle, a person subordinates the surrounding space under his spiritual and intellectual concept. This is what your formula is doing for you. A pentagon may be either convex or concave, as depicted in the next figure. An error occurred trying to load this video. what type of polygon is it? Four interior angles of a pentagon measure 88 degrees, 118 degrees, 132 degrees, and 100 degrees. For example, you might be asked a question such as this one: An irregular pentagon has four known angles of 135, 78, 119, and 117 degrees. - Definition & Examples, Linear Pair: Definition, Theorem & Example, Counting Faces, Edges & Vertices of Polyhedrons, Parallel Lines: How to Prove Lines Are Parallel, Applying Scale Factors to Perimeter, Area, and Volume of Similar Figures, Exterior Angle Theorem: Definition & Formula, NY Regents Exam - Geometry: Tutoring Solution, McDougal Littell Algebra 2: Online Textbook Help, CUNY Assessment Test in Math: Practice & Study Guide, GED Math: Quantitative, Arithmetic & Algebraic Problem Solving, Common Core Math - Number & Quantity: High School Standards, Common Core Math - Algebra: High School Standards, Common Core Math - Statistics & Probability: High School Standards, Praxis Mathematics - Content Knowledge (5161): Practice & Study Guide, CSET", "# How Many Sides For Similarity? Geometry Level 4 In the figure above, $$\\angle A\\cong\\angle F$$, $$\\angle B\\cong\\angle G$$, $$\\angle C\\cong\\angle H$$, $$\\angle D\\cong\\angle I$$, and $$\\angle E\\cong\\angle J$$. You may set any side length in pentagon $$FGHIJ$$ to be whatever value you want. Your goal is to make the pentagons similar. What is the minimum number of sides needed in order for both pentagons to always be similar? ×", "indicate a measured angle • $°$, as in $45°$, means degrees • $rad$ means radians, a method of measuring angles in the metric system Let's look at how we can describe these two angles: We could say that $\\angle O$ () and $\\angle A$ () are congruent, and both measure $55°$. We could also say that mathematically: Since both angles measure less than $90°$, they are also acute and are both made using rays. The shorthand description, $\\angle O$ and $\\angle A$ identifies each angle's vertex, or point where rays meet. If you need the measure in radians, you will write . ### Reflexive Property of Congruence The Reflexive Property of Congruence tells us that any geometric figure is congruent to itself. A line segment, angle, polygon, circle, or another figure of the given size and shape is self-congruent. Angles have a measurable degree of openness, so they have specific shapes and sizes. Therefore every angle is congruent to itself. ## Congruent Angles Examples Angles can be oriented in any direction on a plane and still be congruent. Just as $\\angle DOG$ and $\\angle CAT$, above, were congruent but were not “lined up” with each other, so too can congruent angles appear in any way on a page. Here is a drawing that has several angles. Which of", "Congruent Angles Congruent angles are angles with exactly the same measure. Example: In the figure shown, $\\angle A$ is congruent to $\\angle B$ ; they both measure $45°$ . Congruence of angles in shown in figures by marking the angles with the same number of small arcs near the vertex (here we have marked them with one red arc). In geometric notation, if $\\angle A$ is congruent to $\\angle B$ , we write $\\angle A\\cong \\angle B$ .", "× # 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" ]

[ "degrees)=1.6180339887496196 May 17, 2015 #9 +20831 +13 s = the size of the side of the Pentagon d = the size of the diagonal the central angle of the pentagon is $$\\small{\\text{\\dfrac{360\\ensurement{^{\\circ}}}{5}=72 \\ensurement{^{\\circ}} }}$$ the incribed angle is half of the central angle is $$\\small{\\text{\\dfrac{72\\ensurement{^{\\circ}}}{2}=36 \\ensurement{^{\\circ}} }}$$ #### $$\\cos{(36\\ensurement{^{\\circ}} ) = \\dfrac{\\frac{d}{2} } {s } \\qquad \\rm{or} \\qquad \\dfrac{ d } {s } = 2\\cdot \\cos{(36\\ensurement{^{\\circ}} ) \\qquad | \\qquad \\cos{(36\\ensurement{^{\\circ}} ) =\\dfrac{1}{4}\\cdot(1+\\sqrt{5})$$ $$\\\\\\dfrac{ d } {s } = 2\\cdot \\dfrac{1}{4}\\cdot(1+\\sqrt{5})\\\\\\\\\\\\ \\dfrac{ d } {s } = \\dfrac{1}{2}\\cdot(1+\\sqrt{5})=\\varphi$$ . May 17, 2015", "= 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 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", "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: $$( - r\\cos {72^ \\circ },r\\sin {72^ \\circ }) + (r\\cos {36^ \\circ },r\\sin {36^ \\circ }) + ( - r,0) = ( - 0.191,0.588)$$", "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!", "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", "# 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", "Home » » Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculat... Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculat... Question Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculate the smaller of the two angles A) 72o B) 100o C) 120o D) 144o Explanation: The diagram given simple illustrates that a pentagon contains three $$\\Delta$$s. Two angles which are in the ratio 2:3 will have actual values 2xo, 3xo respectively. Thus 2xo + 3xo + 3 x 60 = 3x sum of angles of $$\\Delta$$a i.e. 5xo + 180o = 3 x 180o 5xo + 180o = 540o 5xo = 540o - 180o 5xo = 360o xo = $$\\frac{360^o}{5}$$ = 72o Hence, the smaller of the two angles is 2 x 72o = 144o Dicussion (1) • The diagram given simple illustrates that a pentagon contains three $$\\Delta$$s. Two angles which are in the ratio 2:3 will have actual values 2xo, 3xo respectively. Thus 2xo + 3xo + 3 x 60 = 3x sum of angles of $$\\Delta$$a i.e. 5xo + 180o = 3 x 180o 5xo + 180o = 540o 5xo = 540o - 180o 5xo = 360o xo = $$\\frac{360^o}{5}$$ = 72o Hence,", "# 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", "circle and one of the sides; (2) the angle between that side and one of the diagonals of the pentagon; (3) the angle between that diagonal and the next diagonal; (4) the angle between that last diagonal and the next side of the pentagon; (5) the angle between that last side and the tangent line. The five arcs of the circle are equally long; therefore the measures of these five angles are equal; therefore each is $180^\\circ/5 = 36^\\circ.$ Those two diagonals and one side have three angles adding up to $180^\\circ$ and two of them have equal measures and the third is $36^\\circ;$ therefore the other two are $72^\\circ.$ So how long is the short side of that isosceles triangle as a function of the length of one of the diagonals? Call the length of the diagonal $d$ and the length of the side we seek $s.$ If you bisect one of those $72^\\circ$ angles you cut off two other triangles, the smaller of which has the same three angles and hence the same shape as the whole triangle, with two sides of length $s.$ The third side of that small triangle must have length $\\dfrac s d\\cdot s,$ by similarity. The larger of those two triangles cut off by bisection has a side of length $s.$", "a triangle, the greater will be the length of its corresponding side. 2. Square • All the sides of a square are equal in length as well they are spread at an angle of $\\small 90^{\\circ }$. • The opposite sides in a square will always be parallel to one another. • The diagonals of a square are equal and bisect at complementary angles. 3. Pentagon • All sides of a pentagon are of the same length as well as all interior angles are of the same size. • Pentagon is composed of three triangles, so the total internal angles of a pentagon = $\\small 3 \\times 180^{\\circ} = 540^{\\circ}$. • Each angle of a pentagon is, $\\small 540^{\\circ} \\div 5 = 180^{\\circ}$. • The central angle of a pentagon made by each side = $\\small 360^{\\circ} \\div 5 = 72^{\\circ}$. 4. Hexagon • Hexagon is a combination of four triangles, so sum of all the internal angles = $\\small 180^{\\circ} \\times 4 = 720^{\\circ}$. • All sides of a hexagon are of the same length as well all interior angles are of the same size. • Each interior angle = $\\small 720^{\\circ} \\div 6 = 120^{\\circ}$. • The central angle of a hexagon made by each side = $\\small 360^{\\circ} \\div 6 = 60^{\\circ}$. 5. Heptagon •", ": \\angle D = 5:6:7.$ Write the value of $\\angle A + \\angle E$. • A. $\\angle A + \\angle E = 160^{\\circ}$ • B. $\\angle A + \\angle E = 90^{\\circ}$ • C. $\\angle A + \\angle E = 145^{\\circ}$ • D. $\\angle A + \\angle E = 180^{\\circ}$ Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 23rd 09, 2020 Q4 Subjective Medium Make a pair of angles forming linear pair from the following angles: $27^{\\circ}, 90^{\\circ}, 130^{\\circ}, 80^{\\circ}, 35^{\\circ}, 50^{\\circ}, 145^{\\circ}, 100^{\\circ}, 90^{\\circ}, 153^{\\circ}$. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q5 Single Correct Medium The measure of an angle is three times the measure of its complement. The angles are: • A. $20^\\circ,\\, 60^\\circ$ • B. $30^\\circ,\\, 60^\\circ$ • C. $45^\\circ,\\, 135^\\circ$ • D. $22.5^\\circ,\\, 67.5^\\circ$ Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "values of $\\cos 72^o$ and $\\cos 36^o$. Drawing a perpendicular from $E$ to $AD$ gives a right angled triangle from which $$\\cos 36^o = {c\\over 2s} ={\\sqrt 5+1\\over 4}.$$ Drawing a perpendicular from $A$ to $DC$ gives a right angled triangle from which $$\\cos 72^o = {s\\over 2c} ={1\\over2\\phi}={\\sqrt 5-1\\over 4}.$$ In explaining the construction of a regular pentagon coordinates are useful. If the pentagon is inscribed in a unit circle with $A$ the point $(0,1)$ you can find the exact coordinates of $B$ and $C$ using the exact values of $\\cos 72^o$ and $\\cos 36^o$ which we have just found. If the pentagon is inscribed in a unit circle with $A$ the point $(0,1)$ then $B=(\\sin 72^o, \\cos 72^o)$ and $C=(\\sin 36^o, -\\cos 36^o)$. Hint: When you explain the construction focus on the $'y'$ coordinates of $B, C, D$ and $E$.", "in) = 32 inches. #### Example B The perimeter of a regular heptagon is 35 cm. What is the length of each side? If $P=ns$ , then $35 \\ cm=7s$ . Therefore, $s=5 \\ cm$ . #### Example C Find the length of the apothem in the regular octagon. Round your answer to the nearest hundredth. To find the length of the apothem, $AB$ , you will need to use the trig ratios. First, find $m \\angle CAD$ . There are $360^\\circ$ around a point, so $m \\angle CAD= \\frac{360^\\circ}{8}=45^\\circ$ . Now, we can use this to find the other two angles in $\\triangle CAD$ . $m \\angle ACB$ and $m \\angle ADC$ are equal because $\\triangle CAD$ is a right triangle. $m \\angle CAD+m \\angle ACB+m \\angle ADC &= 180^\\circ\\\\45^\\circ+2m \\angle ACB &= 180^\\circ\\\\2m \\angle ACB &= 135^\\circ\\\\m \\angle ACB &= 67.5^\\circ$ To find $AB$ , we must use the tangent ratio. You can use either acute angle. $\\tan 67.5^\\circ &= \\frac{AB}{6}\\\\AB &= 6 \\cdot \\tan 67.5^\\circ \\approx 14.49$ Watch this video for help with the Examples above. #### Concept Problem Revisited From the picture below, we can see that the total distance across the Pentagon is the length of the apothem plus the length of the radius. If the total area of the Pentagon is 34", "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", "## 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,", "## Elementary Geometry for College Students (6th Edition) When a pentagon is broken into 10 separate right triangles. each triangle will have an angle of 36 degrees in it, and each triangle will have a leg equal to half of the length of a side. Thus, we find: $tan(36^{\\circ}) = \\frac{6/2}{a} \\\\ a = \\frac{3}{tan(36^{\\circ})} =4.13$", "$\\cos x + \\sin x$ D) $\\cos x - \\sin x$ E) $\\dfrac{\\cos x + \\sin x}{1 + \\tan x}$ 20. If $\\alpha$ is an angle in standard position in quadrant II and $\\beta$ is an angle in standard position in quadrant III such that $\\sin \\alpha = 9/15$ and $\\tan \\beta = 28/45$, find the exact value of $\\cos (\\alpha - \\beta)$. A) $\\dfrac{265}{276}$ B) $-\\dfrac{84}{265}$ C) $\\dfrac{84}{265}$ D) $\\dfrac{36}{53}$ E) $\\dfrac{85}{267}$ 21. The lengths of Side $a$ and $b$ of a triangle are equal to $22$ centimeters and $12$ centimeters respectively. Find angle $C$ of the triangle if angle $A = 42^{\\circ}$ (Round your answer to 2 decimal places). A) $\\approx 116.59^{\\circ \\:}$ B) $\\approx 128.90^{\\circ \\:}$ C) $\\approx 110.67^{\\circ \\:}$ D) $\\approx 98.59^{\\circ \\:}$ E) $\\approx 130.78^{\\circ \\:}$ 22. A triangle has sides $a$ , $b$ and $c$ of lengths $24$, $30$ and $32$ meters respectively. Find the measure of angle $A$. (Round your answer to 2 decimal places) A) $\\approx 90.82^{\\circ \\:}$ B) $\\approx 45.41^{\\circ \\:}$ C) $\\approx 22.71^{\\circ \\:}$ D) $\\approx 55.56^{\\circ \\:}$ E) $\\approx 65.78^{\\circ \\:}$ 23. Find all solutions of the equation $\\quad 2 \\sin (3 x - \\pi/2) + 2 = 1$. A) $\\quad x_1 = \\dfrac{5\\pi}{9} + \\dfrac{2 n \\pi}{3}$ B) $\\quad x_2 = \\dfrac{7\\pi}{9} + \\dfrac{2 n \\pi}{3}$", "\\approx B && \\cos^{-1} \\ (0.575757576)$ Now that we know two of the angles, we can find the third angle using the Triangle Sum Theorem, $\\angle{C} = 180 - (92.1 + 54.8) = 33.1^\\circ$ . ### Vocabulary Side Side Side Triangle: A side side side triangle is a triangle where the lengths of all three sides are known quantities. ### Guided Practice 1. Find the largest angle in the triangle below, where $t = 6, r = 7, i = 11$ 2. Find the smallest angle in the triangle below, where $q = 17, d = 12.8, r = 18.6, \\angle{Q} = 62.4^\\circ$ 3. Find the second largest angle in the triangle below, where $c = 9, d = 11, m = 13$ Solutions: 1. $11^2 = 6^2 + 7^2 - 2 \\cdot 6 \\cdot 7 \\cdot \\cos I, \\angle{I} \\approx 115.4^\\circ$ 2. $12.8^2 = 17^2 + 18.6^2 - 2 \\cdot 17 \\cdot 18.6 \\cdot \\cos D, \\angle{D} \\approx 41.8^\\circ$ 3. $11^2 = 9^2 + 13^3 - 2 \\cdot 9 \\cdot 13 \\cdot \\cos D, \\angle{D} \\approx 56.5^\\circ$ ### Concept Problem Solution You can use the Law of Cosines to solve this problem: $c^2 = a^2 + b^2 + 2ab\\cos \\theta\\\\80^2 = 50^2 + 50^2 + (2)(50)(50)\\cos \\theta\\\\80^2 - 50^2 - 50^2 = (2)(50)(50)\\cos \\theta\\\\6400 - 2500 -", "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" ]

[ "+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", "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", "## 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,", "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", "# 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", "# 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", "values of $\\cos 72^o$ and $\\cos 36^o$. Drawing a perpendicular from $E$ to $AD$ gives a right angled triangle from which $$\\cos 36^o = {c\\over 2s} ={\\sqrt 5+1\\over 4}.$$ Drawing a perpendicular from $A$ to $DC$ gives a right angled triangle from which $$\\cos 72^o = {s\\over 2c} ={1\\over2\\phi}={\\sqrt 5-1\\over 4}.$$ In explaining the construction of a regular pentagon coordinates are useful. If the pentagon is inscribed in a unit circle with $A$ the point $(0,1)$ you can find the exact coordinates of $B$ and $C$ using the exact values of $\\cos 72^o$ and $\\cos 36^o$ which we have just found. If the pentagon is inscribed in a unit circle with $A$ the point $(0,1)$ then $B=(\\sin 72^o, \\cos 72^o)$ and $C=(\\sin 36^o, -\\cos 36^o)$. Hint: When you explain the construction focus on the $'y'$ coordinates of $B, C, D$ and $E$.", "### Home > CCG > Chapter 12 > Lesson 12.2.3 > Problem12-86 12-86. Find the area of the shaded region of the regular pentagon at right. Show all work. Label the shaded region as $ΔABC$. The measure of each angle in a pentagon is equal to $\\frac{180(5 \\cdot 2)}{5} = 180^\\circ$. Therefore, the measure of $\\angle BAC$ is equal to $\\frac{108}{3} = 36^\\circ$ Divide $\\Delta ABC$ into two triangles. Use a trigonometric ratio to solve for $AD$. $\\tan 18^\\circ = \\frac{2}{AD}$ $6.2$ units$^2$", "## 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: $sin36^{\\circ} = \\frac{h}{s} \\\\ h = s \\times sin36^{\\circ}$", "is a regular hexagon. When lines intersect in the geometric plane, they form polygons. Apply what you know in order to classify these polygons. Find side lengths, or solve for unknown angle measures. The intersection of these five lines has created several different polygons. First, let’s identify them. The largest is $$AMDE$$. Within this figure there are two figures. One is figure $$ABCDE$$. $$ABCDE$$ has five angles and five sides, so it is a pentagon. Two of its angles measure $$90^{\\circ}$$, and one measures $$120^{\\circ}$$ Which means that the polygon is an irregular pentagon. In order to find the measures of the two remaining angles, you need to take a look at the third figure. The third figure is triangle $$MBC$$. It has two $$60^{\\circ}$$ angles. 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}\\). Simply subtract to find the measure of $$ABC$$. $$180−60=120^{\\circ}$$ Angle ABC is $$120^{\\circ}$$. Now let’s find the measure of angle $$BCD$$. You now know four of the five angles in figure $$ABCDE$$. Because this is a pentagon, you also know that its interior angles must have a sum of $$540^{\\circ}$$. You can set up an equation to find the measure of the unknown", "which one angle measures $$55^{\\circ}$$. All of their corresponding sides are congruent. 2. Both triangles are equiangular triangles. 3. Both triangles are equilateral triangles. All sides are 5 inches in length. 4. Both triangles are obtuse triangles in which one angle measures $$35^{\\circ}$$. Two of their corresponding sides are congruent. 5. Both triangles are obtuse triangles in which two of their angles measure $$40^{\\circ}$$ and $$20^{\\circ}$$. All of their corresponding sides are congruent. 6. Both triangles are isosceles triangles in which one angle measures $$15^{\\circ}$$. 7. Both triangles are isosceles triangles with two equal angles of $$55^{\\circ}$$. All corresponding sides are congruent. 8. Both triangles are acute triangles in which two of their angles measure $$40^{\\circ}$$ and $$80^{\\circ}$$. All of their corresponding sides are congruent. 9. Both triangles are acute triangles in which one angle measures $$60^{\\circ}$$. Two of their corresponding sides are congruent. 10. Both triangles are equilateral triangles. ## Vocabulary Term Definition Third Angle Theorem If two angles in one triangle are congruent to two angles in another triangle, then the third pair of angles is also congruent.", "sides of the pentagon and hexagon are given from the above, by this computation a line is given subtending 120 which is the difference between the arcs − and it is equal to 20905 parts of the diameter.\" De Revolutionibus Orbium Coelestium: Liber Primus: Theorema Tertium Based on his circle of diameter 200000 units and already established chords of pentagon, hexagon and triangle the calculation effected by Copernicus would have been: $\\frac{173205\\times117557-100000\\times161803}{200000}\\approx20906.$ A small rounding error is evident in the result but the corresponding entry (in the Copernican table of half chords ) of 10453 units against 6 degrees is correct as may readily be verified on a calculator (sin 6). In modern trigonometric notation, the above calculation corresponds to the following application of a compound angle formula: $\\sin(6^\\circ)=\\sin(36^\\circ)\\cos(30^\\circ)-\\cos(36^\\circ)\\sin(30^\\circ). \\,$ ### Theorema Quintum Determination of the chord subtending a sum of arcs The previous diagram demonstrated a general technique for calculating the chord subtending the difference between two arcs. The following diagram neatly reverses this procedure to obtain the chord subtending the sum of arcs: i.e. determination of chord AC given chords AB and BC. Compared with the previous we note that diameter BE has been swung across from point B to point E. EC and ED are joined. Since AEDB is a rectangle DE=AB. Thus in cyclic", ". 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", "# 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 $A=A'$.", "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", "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!", "observe that the number of the sides is 5 We know that, The sum of the measures of the interior angles of the pentagon = 540° Let ∠S = ∠T = x° So, 93° + 156° + 85° + x° + x° = 540° 2x° + 334° = 540° 2x° = 540° – 334° 2x° = 206° x° = $$\\frac{206}{2}$$ x° = 103° Hence, from the above, We can conclude that ∠S = ∠T = 103° Question 5. Sketch a pentagon that is equilateral but not equiangular. e know that, The “Equilateral” means all the sides are congruent The “Equiangular” means all the angles are congruent Hence, A pentagon that is equilateral but not equiangular is: Question 6. A convex hexagon has exterior angles with measures 34°, 49°, 58°, 67°, and 75°. What is the measure of an exterior angle at the sixth vertex? It is given that a convex hexagon has exterior angles with measures 34°, 49°, 58°, 67°, and 75° We know that, A hexagon has 6 number of sides We know that, The sum of the measures of the exterior angles of any polygon is: 360° Let the exterior angle measure at the sixth vertex of a convex hexagon be: x° So, 34° + 49° + 58° + 67° + 75° + x° = 360°", "to use trigonometry and you are not using angles whose trigonometric ratios can be written easily as surds, remember that trigonometry is based on the properties of similar triangles. It is often the case that the most elegant solution in these situations involves the direct use of similar triangles. • It is enough that the four sides of a quadrilateral are equal to be able to say it is a rhombus (why?). • When you write proofs, try to structure them so that you make it easy for the reader to follow. Think of \"chunking\" your proofs so that each part stands on its own as much as possible. Jacqui Eaves made a good attempt at this problem and Andrei Lazanu's work forms the basis of the solution below, well done Andrei. The solution First I calculated the angle between two sides of the pentagon. I used the formula for the sum of angles of a regular polygon, with n sides, that is: 180°(n-2) For the pentagon, I obtained: $${180^{\\circ}(n - 2)} = {180^{\\circ}(5 - 2)} = {180^{\\circ}\\times 3} = {540^{\\circ}}$$ Therefore the angle in each vertex is ${540^{\\circ}\\over5} = {108^{\\circ}}$ $${\\angle EAB}\\equiv{\\angle ABC}\\equiv{\\angle BCD}\\equiv{\\angle CDE}\\equiv{\\angle DEA} = {108^{\\circ}}$$ Triangles CDE, DCB and AEB are isosceles (sides of pentagon are one unit) $${\\angle DEC}\\equiv{\\angle DCE}\\equiv{\\angle BDC}\\equiv {\\angle CBD}\\equiv{\\angle", "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", "a triangle, the greater will be the length of its corresponding side. 2. Square • All the sides of a square are equal in length as well they are spread at an angle of $\\small 90^{\\circ }$. • The opposite sides in a square will always be parallel to one another. • The diagonals of a square are equal and bisect at complementary angles. 3. Pentagon • All sides of a pentagon are of the same length as well as all interior angles are of the same size. • Pentagon is composed of three triangles, so the total internal angles of a pentagon = $\\small 3 \\times 180^{\\circ} = 540^{\\circ}$. • Each angle of a pentagon is, $\\small 540^{\\circ} \\div 5 = 180^{\\circ}$. • The central angle of a pentagon made by each side = $\\small 360^{\\circ} \\div 5 = 72^{\\circ}$. 4. Hexagon • Hexagon is a combination of four triangles, so sum of all the internal angles = $\\small 180^{\\circ} \\times 4 = 720^{\\circ}$. • All sides of a hexagon are of the same length as well all interior angles are of the same size. • Each interior angle = $\\small 720^{\\circ} \\div 6 = 120^{\\circ}$. • The central angle of a hexagon made by each side = $\\small 360^{\\circ} \\div 6 = 60^{\\circ}$. 5. Heptagon •" ]

[ "A guy wire of length 50 feet is attched to the ground and to the top of an antenna. The height of the antenna is 10 feet larger...", "the other pole. The poles are 37 ft. apart. How long does the wire need to be to reach between the two poles?", "one vertical wire in free space, unity, then the capacity of four wires placed rather near together will only be about twice that of one wire, and that of twenty-five wires will only be about five times one wire. Fig. 8. Various forms of Aerial Radiator. a, Single Wire; b, Multiple Wire; c, Fan Shape; d Cylindrical; g, Conical. This approximate rule has been confirmed by experiments made with long wires one hundred or two hundred feet in length in the open air. Hence it points to the fact that the ordinary plan of endeavoring to obtain a large capacity by putting several wires in parallel and not very far apart is very uneconomical in material. The diagrams in Fig. 8 show the various methods which have been employed by Mr. Marconi and others in the construction of such multiple wire aerials. If, for instance, we put four insulated stranded 722 wires, each 100 feet long, about six feet apart, all being held in a vertical position, the capacity of the four together is not much more than twice that of a single wire. In the same manner, if we arrange 150 similar wires, each 100 feet long, in the form of a conical aerial, the wires being distributed at the top round a circle 100 feet in", "used where the lines are strained. # Different types of line supports used for transmission lines include 1. RCC and PCC poles 2. steel towers 3. steel poles 4. wooden poles, steel poles, RCC and PCC poles and steel towers Option 2 : steel towers ## Detailed Solution Wooden poles, steel poles, RCC and PCC poles are used for distribution purpose and steel towers are used for transmission purpose. Wooden poles: Wooden poles are used for temporary electrical line. The length of wooden pole is kept 6-9 meters out of which 1-2 meter length buried in the ground. The diameter of the pole is kept 15-25 cm so as to have safety factor of 3.5. Maximum permissible span for wooden poles is 75 meter. Steel poles: • Steel poles possess greater mechanical strength and thus permit the use of longer spans (60 to 80 metres). • These poles have the longer life (more than 40 years) which can further be increased by regular painting. • These poles need protection against corrosion. Hence at the bottom (the portion which is buried underground), these poles are set in concrete muffs in order to protect them from chemical reactions. RCC and PCC poles: In a service line of a residential colony normally RCC poles are preferred. RCC means reinforced cement concrete.", "fence wire further from the cable wire. Since the fence is six feet high, a good way to avoid this kind of thing happening would be to run the dog fence wire along the middle or the top of the fence. The vertical separation should minimize the chance of the signal being induced in the utility wire. If the tree fort is only 4 feet from the fence, I don’t think you will have much luck getting the dog into the fort. Even if the signal does not reach the fort, most dogs don’t like getting that close to the boundary. Jake April 12, 2011 at 9:49 am Hi, my parents and me are each getting puppies that are going to be spending lots of time together. We live at separate houses and are both considering installing systems on our properties. Is there a way we can set both systems to work on the same frequency so the two dogs can use the systems at each house and not have to purchase and switch collars each time they are in the yard together? If you get the same system as your parents, then your will all be able to use your own collars in the others yard. Also there are a few systems that are inter-compatible. The Innotek", "same height above the ground, and are $7$ feet apart. One end...", "A 24 ft. long pole AB is placed in a hole and is guyed by threecables. Knowing that the tensions in cable BD and BE are 221 lb and161 lb, respectively, determine a) the tension in cable CD, b) thereaction at A.", "A uniform rigid pole of length L and mass M is to be supported from a vertical wall in a horizontal position, as shown in the figure. The pole is not attached directly to the wall, so the coefficient of static friction, ?s, between the wall and the pole provides the only vertical force on one end of the pole. The other end of the pole is supported by a light rope that is attached to the wall at a point a distance D directly above the point where the pole contacts the wall. Determine the minimum value of ?s, as a function of L and D, that will keep the pole horizontal and not allow its end to slide down the wall. (Use any variable or symbol stated above as necessary.)", "base plate to pole adapter? What wire on poles? The fence line? Last edited: #### JimB ##### Super Moderator Which part are you referring too? the small tower in the one picture or the adjustable base plate to pole adapter? What wire on poles? The fence line? See my edit of your photo. JimB #### Attachments • 54.5 KB Views: 300 #### tcmtech ##### Banned The four leg tower is my old 10 ft tower. The new blade set was to big for it and was pulling it out of the ground. Plus it had weak air flow from the north and east sides. I decided to move the new generator design up to the pole. That was built with the capacity to hold a much bigger system. Plus it has good air flow from all directions. That was really what most of this project was about. Just doing a generator design change and a relocation to get better wind. The wire is just the top line on a fence. Being that wire is old rusty iron it shows up in the picture better. The other three below it are galvanized and blend in with the back ground. #### JimB ##### Super Moderator OK, if I look carefully at the photo, I can just see the other wires", "Wire H Stakes & Straight Wire Flute Direction with Straight Wires & H Stakes 24\" Straight Wire For signs with vertical flutes, the 24\" Straight Wires are inserted in to the sign vertically. These are most convenient for smaller signs, but work well with larger signs as well. These will allow the signs to sit lower to the ground. H Stakes For signs with vertical flutes, H Stakes are inserted in to the sign vertically. These are most common with signs that are 16 x 24 or larger. Due to the double cross bar support, these will sit higher above ground.", "34m D) 36m Explanation: Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important. Hence number of poles required = 280 / 5 = 56", "# A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground... ## Question: A cable hangs between two poles of equal height and {eq}35 {/eq} feet apart. At a point on the ground directly under the cable and {eq}x {/eq} feet from the point on the ground halfway between the poles the height of the cable in feet is {eq}h(x)=10+(0.5)x^{1.5} {/eq}. The cable weighs {eq}18.3 {/eq} pounds per linear foot. Find the weight of the cable. ## Arc Length: To find the length of the curve {eq}f(x) {/eq} over the interval {eq}[a,b] {/eq}, we can use the following arc length integral formula: {eq}\\displaystyle \\int_{a}^{b} \\sqrt{1 +(f'(x))^2)} dx {/eq} Here {eq}f'(x) {/eq} is the first derivative of {eq}f(x) {/eq}. Once we have our definite integral set up, we can evaluate it using standard integration techniques, such as u-substitution. Become a Study.com member to unlock this answer! Create your account Given Data: • The distance between the two poles is: {eq}d = 35\\;{\\rm{ft}} {/eq} • The height of the cable is: {eq}h\\left( x \\right) = 10 + 0.5\\left(...", "? Free Version Easy # Support Wire: Angle of Elevation TRIG-2VKPL5 A support wire is to be stretched from the top of a 50 foot tree to a point on the ground. The engineers need the angle of elevation from the point on the ground to the top of the tree is to be ${ 35 }^{ o }$.", "full weight of the wire over 30ft span. not just the length of 1 foot of wire.", "# Can you crawl between the ground and under the middle of the cable? [closed] Two vertical posts are in level ground at 150 feet apart. A bendable cable, but one that does not stretch from its original length, is 150 feet and 1 inch in length. The cable is fastened in between the two vertical posts at ground level on each side. The middle point of the cable is lifted up until it is taut, and the ground and the middle sections of the cable form an obtuse isosceles triangular shape. Is there room enough for you to crawl through in the space between the ground and underneath the middle of the cable? * The length of the cable has been edited. * ## closed as off-topic by StephenTG, Aggie Kidd, Ivo Beckers, Khale_Kitha, DylanSpMar 17 '16 at 19:32 This question appears to be off-topic. The users who voted to close gave this specific reason: • \"This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see \"Are math-textbook-style problems on topic?\" on meta.\" – StephenTG, Aggie Kidd, Ivo Beckers, DylanSp If this question can be reworded to fit the rules in the help center, please edit the question. • Why isn't there spoiler formatting in", "A street light is at the top of a $11.5$ ft. tall pole. A man $5.9$ ft tall walks away from the pole with a speed of $3.5$ feet/sec along a straight path. How fast is the tip of his shadow moving when he is $46$ feet from the pole?", "bottom circuit, it is tied to ground.", "Cable length 1. Oct 26, 2006 blumfeld0 A cable hangs between two poles of equal height and 37 feet apart. At a point on the ground directly under the cable and x feet from the point on the ground halfway between the poles the height of the cable in feet is h(x)=10 +(0.4)( x^{1.5}) The cable weighs 15.1 pounds per linear foot. Find the weight of the cable. so i find h'(x) i square it and add one to find the length of the curve i.e (1+ h'(x)^2 ) ^(1/2) my question is is that right? and what are the limits of integration?? blumfeld0 2. Oct 26, 2006 HallsofIvy Staff Emeritus Well, more correctly \"i find h'(x) i square it and add one\" , take the square root and integrate \"to find the length of the curve\" Since your variable, x, is the \"from the point on the ground halfway between the poles\", x= 0 there. At one pole x= -37/2 and at the other x= 37/2. Integrate with respect to x from -37/2 to 37/2. 3. Oct 26, 2006 blumfeld0 ok so i integrate (1 + h'(x)^2 )^(1/2) with repect to x from -37/2 to 37/2 i use mathematica and i get 39.25+ 24.9 i where \"i\" is imaginary why do i get an imaginary answer?? also", "for probably lower speeds and lower overall requirements. The catenary alternates between the left and right poles (at turns the poles are only at the outer side of the turn), supplying points to hang the contact wire over the rails. This second setup is made only with simple, straight poles and uses only a single isolator per pole.", "the top of the flag pole is 20 feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth of a foot. Try It 10.80 The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth. ### Example 10.41 Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than three times the width. Find the length and width. Round to the nearest tenth of a foot. Try It 10.81 The length of a 200 square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden. Round to the nearest tenth of a foot. Try It 10.82 A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length. What are the length and width of the tablecloth? Round to the nearest tenth of a foot. The height of a projectile shot upwards is" ]

[ "# Ex.6.5 Q12 Triangles Solution - NCERT Maths Class 10 Go back to 'Ex.6.5' ## Question Two poles of heights $$6\\;\\rm{}m$$ and $$11\\;\\rm{}m$$ stand on plane ground. If the distance between the feet of the poles is $$12\\;\\rm{}m$$, find the distance between their tops. Diagram ## Text Solution Reasoning: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Steps: $$AB$$ is the height of one pole $$= 6\\;\\rm{}m$$ $$CD$$ is the height of another pole $$= 11\\,\\rm{}m$$ $$AC$$ is the distance between two poles at bottom $$= 12\\rm{}\\,m$$ $$BD$$ is the distance between the tops of the poles $$=\\,?$$ Draw $$B E \\| A C$$ Now consider ,In $$\\Delta B E D$$ \\begin{align}\\angle B E D&=90^{\\circ} \\\\ {BE}={AC}&=12 \\mathrm{m}\\\\ {DE}&={CD}-\\mathrm{CE} \\\\ {DE}&=11-6=5 \\mathrm{cm}\\\\\\end{align} Now \\begin{align} B D^{2} &=B E^{2}+D E^{2}\\quad( \\text{ Pythagoras therorem }) \\\\ &=12^{2}+5^{2} \\\\ &=144+25 \\\\ B D^{2} &=169 \\\\ B D &=13 \\mathrm{m} \\end{align} The distance between the tops of poles $$=13\\rm{}\\,m$$ 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", "(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 +", "has a length of $2$, so by pythagorean's, $OH$ is $\\sqrt{32}$. $2 \\cdot \\sqrt{32} + \\frac{1}{2}\\cdot2\\cdot \\sqrt{32} = 3\\sqrt{32} = 12\\sqrt{2}$, which is half the area of the hexagon, so the area of the entire hexagon is $2\\cdot12\\sqrt{2} = \\boxed{(B)24\\sqrt{2}}$ Solution 2 $ADPO$ and $OPBC$ are congruent right trapezoids with legs $2$ and $4$ and with $OP$ equal to $6$. Draw an altitude from $O$ to either $DP$ or $CP$, creating a rectangle with width $2$ and base $x$, and a right triangle with one leg $2$, the hypotenuse $6$, and the other $x$. Using the Pythagorean theorem, $x$ is equal to $4\\sqrt{2}$, and $x$ is also equal to the height of the trapezoid. The area of the trapezoid is thus $\\frac{1}{2}\\cdot(4+2)\\cdot4\\sqrt{2} = 12\\sqrt{2}$, and the total area is two trapezoids, or $\\boxed{24\\sqrt{2}}$.", "$MN$ 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 $B$", "situation, $x$ = $36$ is the only solution. In such a case, the width of the garden is $36$ feet and the length of the garden is $(2)(36) + 5$ = $77$ feet. Example 3: A lamp post is held by two stay wires; both are tied at the same point on the post. The other ends are pegged on the ground. The length of shorter wire is $25$ feet and that of the longer one is $26$ feet. If the distance between the pegs on the ground is $3$ feet, find at what height on the lamp post both the wires are tied. Look at the following diagram. Solution: Let $A$ be the point where the wires are tied at the top of the lamp post and $B$ be its bottom at the ground. The first nearer peg is at $C$ and the farther is at $D$. As per the problem statement, $AC$ = $25$ feet and $AD$ = $26$ feet. Let $AB$ = $h$ feet and $BC$ = $x$ feet. Then $BD$ would be $(x + 3)$ feet. Clearly both the triangle $ABC$ and $ABD$ are right triangles with right angle at $B$. Therefore, Pythagorean Theorem can be applied on both these triangles. In triangle $ABD,\\ h^2\\ +\\ (x + 3)^2$ = $26^2$ … (1) In", "A car goes 24 miles due north then 7 miles due east. What is the straight distance between the car's starting point and end point? 1.1.10 One end of a rope is attached to the top of a pole 100 ft high. If the rope is 150 ft long, what is the maximum distance along the ground from the base of the pole to where the other end can be attached? You may assume that the pole is perpendicular to the ground. 1.1.11 Prove that the hypotenuse is the longest side in every right triangle. (Hint: Is $$a^2 + b^2 > a^2$$?) 1.1.12 Can a right triangle have sides with lengths 2, 5, and 6? Explain your answer. 1.1.13 If the lengths $$a$$, $$b$$, and $$c$$ of the sides of a right triangle are positive integers, with $$a^2 + b^2 = c^2$$, then they form what is called a Pythagorean triple. The triple is normally written as ($$a$$,$$b$$,$$c$$). For example, (3,4,5) and (5,12,13) are well-known Pythagorean triples. (a) Show that (6,8,10) is a Pythagorean triple. (b) Show that if ($$a$$,$$b$$,$$c$$) is a Pythagorean triple then so is ($$ka$$,$$kb$$,$$kc$$) for any integer $$k >0$$. How would you interpret this geometrically? (c) Show that ($$2mn$$,$$m^2 - n^2$$,$$m^2 + n^2$$) is a Pythagorean triple for all integers $$m > n >", "# 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(", "# 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,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", "# 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", "# A trapezoid has a top base of 32 and a bottom base of 22. Its legs are each 13. What is its area? Nov 15, 2015 $324$ #### Explanation: The area of a trapezoid is the average of its bases times its height. This can be expressed as ${A}_{\\text{trapezoid}} = \\frac{h \\left({b}_{1} + {b}_{2}\\right)}{2}$. We have the two bases, but we don't have the height. Draw the trapezoid on a piece of paper. Now, write in the measurements. If both legs are $13$, and the bases are $22$ and $32$, how can we find the height? We can use the Pythagorean theorem. If one base is $22$ and the other is $32$, the long base is $10$ longer than the short one. If we have an isosceles trapezoid like in the picture, the long base will extend $5$ more in either direction. So, if you drew a line cutting straight down from the vertex where the SHORT base and the height met all the way to the LONG base, you will have a right triangle with a hypotenuse of $13$ and a leg of $5$. We can use the Pythagorean theorem to figure out that the other leg of the right triangle, which is also the height of the trapezoid, has length $12$. Then, we plug our", "of the box? It is given that A packaging box for a metal rod is 7.5 inches along a diagonal of the base. The height of the box is 18 inches. Now, From the given figure, We can observe that it looks like a right triangle Now, We know that, According to the Pythagorean Theorem, c² = a²+ b² Where, c is the length of the diagonal a and b are the side lengths So, c² = 18² + (7.5)² c²= 324 + 56.25 c² = 380.25 c = $$\\sqrt{380.25}$$ c = 19.5 inches Hence, from the above, We can conclude that the length of the diagonal of the box is: 19.5 inches Lesson 7.4 Find Distance in the Coordinate Plane Quick Review The Pythagorean Theorem can be used to find the distance between any two points on the coordinate plane. Example Find the distance between the two points on the coordinate plane. Round to the nearest tenth. Draw a right triangle. Determine the lengths of its legs. The length of the horizontal leg is 5 units. The length of the vertical leg is 5 units. Use the relationship a2 + b2 = c2. Substitute 5 for a and 5 for b. Then solve for C. a2 + b2 = -2 52 + 52 = c2 25 +", "of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to $\\sqrt{20a}$ and $\\sqrt{20b}$, and because the hypotenuse is 20 we get $a+b=20$. Testing small numbers, we get that when $a=2$ and $b=18$, $ab$ is indeed a square. The area of the triangle is thus 60, so the answer is $\\boxed {\\textbf{D) }360}$. ~tigershark22 ~(edited by HappyHuman) ## Solution 3 (coordinates) $[asy] draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); draw(circle((0,0),10)); label(\"D\",(10,30),N); label(\"C\",(10,0),E); label(\"B\",(-8,-6),S); label(\"A\",(-10,0),W); [/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 4 (Trigonometry) Using the law of cosines, express $AB^2$ and $BC^2$ in terms of $\\angle{AEB}$. The sum of these two equations is $AC^2$ by the Pythagorean Theorem. Solving for $BE$, and using the fact that $cos\\angle{AEB}=\\frac{1}{\\sqrt5}$, we find $BE=3\\sqrt5$. Since $EC=15$ and $DC=30$, $DE=15\\sqrt3$, which is five times $BE$, $[ABCD]=[ACD]+\\frac{1}{5}[ACD]=300+60=\\boxed {\\textbf{D) }360}$ (This solution is incomplete, can someone complete it please-Lingjun) Latex edited by kc5170 ~IceMatrix", "right triangle. 1. $a=12, b=16, c =?$ 2. $a =?, b=20, c=30$ 3. $a=4, b =?, c=11$ 4. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle. 5. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle. 6. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate? 7. Emanuel has a cardboard box that measures $20 \\ cm \\times 10 \\ cm \\times 8 \\ cm \\ (\\text{length} \\times \\text{width} \\times \\text{height})$. What is the length of the diagonal from a bottom corner to the opposite top corner? 8. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. How high above the ground does the ladder touch the wall of the house? 9. Find the area of the triangle if area of a triangle is defined as $A=\\frac{1}{2} \\text{base} \\times \\text{height}$. 10. Instead of walking along the two sides of a rectangular field,", "Synthesis ### Lesson Synthesis Tell students that there are many situations in the world where we can use the Pythagorean Theorem to solve problems. Ask, “What situations that you can think of involve right triangles?” Give brief quiet think time, then invite students to share their ideas. For example, thick wires (called guy-wires) are used to keep telephone poles upright and their length depends on how high up the pole they attach and how far away from the pole they hook into the ground. ## 10.4: Cool-down - Jib Sail (5 minutes) ### Cool-Down It’s assumed that the tower makes a right angle with the ground. Since this is a right triangle, the relationship between its sides is $$a^2+b^2=c^2$$, where $$c$$ represents the length of the hypotenuse and $$a$$ and $$b$$ represent the lengths of the other two sides. The hypotenuse is the side opposite the right angle. Making substitutions gives $$a^2+15^2=17^2$$. Solving this for $$a$$ gives $$a=8$$. So, the other end of the cable should connect to the ground 8 feet away from the bottom of the tower.", "my solution for 13 a) !!!!! You'd have to use the first hypotenuse to figure out the OTHER triangle! Sorry! Masheroni Posts: 3 Joined: Sun Dec 23, 2018 9:29 pm Reputation: 0 ### Re: Solving for angles and sides Good afternoon! 11) h = height of the pole g1 = length of guy wire (50) g2 = length of guy wire (35) x = distance between pole and guy wire g1 (50) x+10 = distance between pole and guy wire g2 (35) So: $$\\tan 50^{\\circ}=\\dfrac{h}{x}\\\\\\cot 50^{\\circ}=\\dfrac{x}{h}\\\\x=h\\cot 50^{\\circ}\\\\ \\tan 35^{\\circ}=\\dfrac{h}{x+10}\\\\\\cot 35^{\\circ}=\\dfrac{x+10}{h}\\\\x+10=h\\cot 35^{\\circ}\\\\ h\\cot 35^{\\circ}=x+10=h\\cot 50^{\\circ}\\\\ h\\cot 35^{\\circ}-h\\cot 50^{\\circ}=10\\\\ h=\\dfrac{10}{\\cot 35^{\\circ}-\\cot 50^{\\circ}}\\\\ h\\approx 17,0$$ Now, you have the length of the pole, the guy wire is easy $$\\sin 50^{\\circ}=\\dfrac{h}{g_1}\\\\g_1=\\dfrac{17}{\\sin 50^{\\circ}}\\approx 22,2$$ $$\\sin 35^{\\circ}=\\dfrac{h}{g_2}\\\\g_2=\\dfrac{17}{\\sin 35^{\\circ}}\\approx 29,6$$ Total length: $$22,2+29,6=51,8$$ Let's solve the others in another post Baltuilhe Posts: 44 Joined: Fri Dec 14, 2018 3:55 pm Reputation: 29 ### Re: Solving for angles and sides Good afternoon (2)! 13) Going to east, 80m and to south, 60m, using Pythagoras: $$d^2=80^2+60^2=6\\,400+3\\,600\\\\d^2=10\\,000\\\\d=100m$$ And now, up 30m. $$d^2=100^2+30^2=10\\,000+900=10\\,900\\\\d=\\sqrt{10\\,900}\\approx 104,40m$$ Now, the angle of elevation: $$\\tan\\theta=\\dfrac{30}{100}=0,3\\\\\\theta=\\arctan 0,3\\\\\\theta\\approx 16,70^{\\circ}$$ I hope I have helped! Baltuilhe Posts: 44 Joined: Fri Dec 14, 2018 3:55 pm Reputation: 29 Return to Trigonometry - sin, cos, tan, cot, arcsin, arccos, arctan, arccot ### Who is online Users browsing this forum:", "# 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", "Enable contrast version # Tutor profile: Cara H. Inactive Cara H. Tutor Satisfaction Guarantee ## Questions ### Subject:Geometry TutorMe Question: Classify triangle PQO by its sides. Then determine if the triangle is a right triangle. The points of the triangle are: P (-1, 2), Q (6,3), and O (0, 0) Inactive Cara H. The first step is to use the Distance Formula to determine the side lengths. The Distance Formula is: $$\\sqrt{( x_{2} - x_{1})^2 + {(y_{2} - y_{1})^2}}$$ OP = $$\\sqrt{(-1 -0)^2 + {(2 - 0)^2}}$$ = $$\\sqrt5$$ = approx. 2.2 OQ = $$\\sqrt{(6 - 0)^2 + {(3 - 0)^2}}$$ = $$\\sqrt45$$ = approx. 6.7 PQ = $$\\sqrt{(6 - (-1))^2 + {(3 - 2)^2}}$$ = $$\\sqrt50$$ = approx 7.1 Next, check for right angles. The slope of line OP is (2-0) ÷ (-1 - 0) = -2 The slope of line OQ is (3 - 0) ÷ (6 - 0) = 1/2 The product of the slopes is -2(1/2) = -1 So line OP and line OQ are perpendicular. Thus angle POQ is a right angle. Therefore, Triangle POQ is a right scalene triangle. ### Subject:Pre-Algebra TutorMe Question: A flagpole is 40 feet tall. A rope is tied to the top of the flagpole and secured to the ground 9 feet from the base of the flagpole.", "be the foot of the altitude from $B$ to $\\overline{CD}$. Let $h = BP$, the height of the trapezoid. Let $x = CP$, so $DP = 8 - x$. By the Pythagorean Theorem on right triangle $BPC$, $x^2 + h^2 = 49,$and by the Pythagorean Theorem on right triangle $BPD$, $(8 - x)^2 + h^2 = 81.$Subtracting the first equation from the second, we get $(8 - x)^2 - x^2 = 32.$This equation simplifies to $64 - 16x = 32$. Solving for $x$, we find $x = 2$. Substituting into the equation $x^2 + h^2 = 49$, we get $4 + h^2 = 49$, so $h^2 = 45$. Then $h = \\sqrt{45} = 3 \\sqrt{5}$. Now, let $Q$ be the foot of the perpendicular from $A$ to $\\overline{CD}$. Triangles $AQD$ and $BPC$ are congruent, so $DQ = CP = 2$. Then $PQ = CD - CP - DQ = 8 - 2 - 2 = 4$. Quadrilateral $ABPQ$ is a rectangle, so $AB = PQ = 4$. Therefore, the area of trapezoid $ABCD$ is $h \\cdot \\frac{AB + CD}{2} = 3 \\sqrt{5} \\cdot \\frac{4 + 8}{2} = \\boxed{18 \\sqrt{5}}.$ 1. 👍 2. 👎 21. Huh wait LaTeX doesn't work here? T-T 1. 👍 2. 👎 22. hi the answer is 35 sqrt 2 1. 👍 2. 👎 23.", "$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", "year in secondary school. 5. Aug 9, 2006 ### HallsofIvy You've \"separated\" that in the wrong place! I was saying that the formula for area of a trapezoid is A= h(b1+ b2)/2 where b1 and b2 are the two bases and h is the height (measured perpendicular to both bases). Then I said that you can find the height by dividing the trapezoid into a rectangle and two right triangles at the two ends. Since the longer base has length d and the shorter length 10, the \"rectangle\" will have length 10 which leaves d-10 for the bases of the two right triangles together. The length of the base of one right triangle is (d-10)/2 and the hypotenuse is 10. The other leg is h and, by the Pythagorean theorem, h2+ {(d-10)/2}2= 100. Therefore, $$h= \\sqrt{100- \\frac{(d-10)^2}{4}}= \\sqrt{100- \\frac{d^2}{4}+ 5d- 25}$$ $$h= \\sqrt{75+ 5d- \\frac{d^2}{4}}$$ Use that to find the area in terms of d and determine which d maximizes that." ]

[ "Krishna 0 Step 1: According to the given measurements construct an image. GIVEN: Two poles AD and CE of height 6 m, 11 m respectively. Distance between the feet of two poles (DC) = 12 m From the figure: AD = BC = 6 m AB = DC = 12 m (Since, opposite sides of the rectangles are equal.) It seems to be right angle triangle ABE Step 2: Find the height of the right angle triangle EXPLANATION: BE = CE - BC = 11- 6 = 5 m Step 3: Use the Pythagoras formula to find the distance between the tops of the poles. FORMULA: (hypotenuse)^2 = (side)^2 + (side)^2 EXAMPLE: AE^2 = AB^2 + BE^2 AE^2 = (12)^2 + (5)^2 AE^2 = 144 + 25 AE = \\sqrt{169} AE = 13 m ∴ AE = distance between the two tops of the poles = 13 m", "+ BC2 ⇒ AC2 = (8 m)2 + (6 m)2 = (10 m)2 ⇒ AC = 10 m Now, again applying Pythagoras theorem in ΔEBD DE2 = EB2 + BD2 ⇒ (10 m)2 = (8 m)2 + BD2 ⇒ BD2 = 100 m2 − 64 m2 = 36 m2 ⇒ BD = 6 m Thus, the tip of the ladder is now at the height of 6 m above the ground. #### Question 8: Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. Let us draw the diagram from the given information. As we are given that distance between their feet is 12 m . Now we get a right angled triangle DCE. Let us applying the Pythagoras theorem we get, Substituting the values we get, Let us take the square root we get, Therefore, distance between their top is . #### Question 9: Using Pythagoras theorem determine the length of AD in terms of b and c shown in the given figure. In ∆ABC and ∆DBA, (90º each) (Common) Therefore, by AA-criterion for similarity, we have . Now we will substitute the values of AC and AB We are finding the value of AD therefore; we will", "each other at right angle. $AO=\\sqrt{(10^2-8^2 )} =6$ Shorter diagonal = 12 cm. $\\\\$ Q.9. The supporting wire to the top of a vertical pole in 13 cm long and it in fastened to the ground at a state 5 m among from the foot of the pole. How high is the pole? Let AB be the pole. $h=\\sqrt{(13^2-5^2 )} =13m$ $\\\\$ Q.10. A ladder 26 m long rust against a vertical wall with its foot 10m against from the wall how high up the wall will the ladder reach. $h=\\sqrt{(26^2-10^2)} =24 m$ $\\\\$ Q.11. A 15 meter against a vertical wall 21 reach a window at a height of 12m from the ground. How for in the foot of the ladder? Let the distance of the foot of the ladder from the wall=d $d=\\sqrt{(15^2-12^2) } =9 m$ $\\\\$ Q.12. The height of thus to were are 34m and 10m respectively. If the distance between there fact in 32m, find the distance between their tops. Let l be the distance between the tops $l=\\sqrt{(24^2 +32^2) } =40 m$ $\\\\$ Q.13. In the adjoining figure, ABC in a triangle in which $\\angle B=90^{\\circ}$. If D in the mid point of BC. Prove that $AC^2=AD^2+3CD^2$ $AC^2=AB^2+BC^2$ $=AB^2+(2CD) ^2$ $=(AD^2-BD^2) +4CD^2$ $=AD^2-CD^2+4CD^2$ $AC^2=AD^2+CD^2$ $\\\\$ Q.14. In the adjoining figure, it in", "fixed to the ground at a distance of 9 feet from the foot of the pole. What is the height of the pole? a. 21 feet b. 6 feet c. 12 feet d. 27 feet #### Solution: Let h be the height of the pole. d2 + h2 = l2 [Apply Pythagorean theorem.] h2 = l2 - d2 [Subtract d2 from both sides.] h2 = 152 - 92 [Substitute l and d.] = 225 - 81 [Apply exponents and simplify.] = 144 h = 144 [Take square root of both sides.] = 12 The height of the pole is 12 feet. Correct answer : (3) 7. One leg A of a right triangle is 1 inch less than the hypotenuse. The length of the other leg is 3 inches. What are the lengths of leg A and the hypotenuse? a. 3 inches and 4 inches b. 4 inches and 5 inches c. 6 inches and 7 inches d. 6 inches and 5 inches #### Solution: Let s be the length of the hypotenuse. One leg A of right triangle is 1 inch less than the hypotenuse. So, the length of leg A = s - 1 The length of another leg = 3 inches. (s - 1)2 + 32 = s2 [Apply Pythagorean theorem] s2 + 1 -", "64 = 289 = (17)2 ⇒ AB = 17 cm Question 10. In the given figure, AB and CD are two vertical poles of height 19 m and 11 m respectively. If the shortest distance between their tops is 17 m, find how far apart they are? In the given figure, Pole AB = 19 m, Pole CD = 11 m Distance between their tops AC = 17 cm Draw EC || BD, then Let BD = CE = x EB = CD = 11 m AE = AB – EB = 19 – 11 = 8m Now in right ∆AEC, AC2 = AE2 + EC2 ⇒ (17)= (8)+ x2 ⇒ 289 = 64 + x2 ⇒ x2 = 289 – 64 = 225 = (15)2 ⇒ x = 15 BD = EC = x = 15 m —: End of ML Aggarwal Class-7 Triangles and its Properties Solutions :– Thanks $${}$$", "## anonymous one year ago Help.. A 6-meter pole is supported by guy wires that are anchored to the ground as shown. What is sin D? A) 3/6.71 B) 6.71/6 C) 3/6 D) 6/6.71 Link to question is below 👇👇 1. anonymous 2. mathstudent55 Do you know SOHCAHTOA? 3. anonymous Nope. 😪 4. mathstudent55 SOHCAHTOA is a way to memorize which sides make the ratios of sine, cosine and tangent. 5. anonymous okay katelyn what is the height 6. anonymous basically katelyn what im saying sin=b/h 7. anonymous Whaleee, the height of the triangles sides are both 6.71m I believe. The link I put has the picture of it if you need to refer to it. 👌😄 8. mathstudent55 SOH: $$\\sin \\theta = \\dfrac{opp}{hyp}$$ CAH: $$\\cos \\theta = \\dfrac{adj}{hyp}$$ TOA: $$\\tan \\theta = \\dfrac{opp}{adj}$$ The initials S (sine), C (cosine), T (tangent), O (opposite leg), A (adjacent leg), and H (hypotenuse) are used in the mnemonic above. 9. anonymous Mhm 10. mathstudent55 You are asked for the sine. Looking above, or remembering the SOH part of SOHCAHTOA, you see that $$\\sin \\theta = \\dfrac{opp}{hyp}$$ 11. anonymous do you know what the answer is? 12. anonymous Yes, opposite over hypotenuse 13. mathstudent55 Now look at angle D. For angle D, which side is the opposite leg? Which side is", "cm. To find: AY In ΔAXY and ΔABC, We know that if two triangles are similar, then their sides are proportional. It is given that AB = 4BX. Let AB = 4x and BX = x. Then, AX = 3x Hence the correct answer is . #### Question 11: Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is (a) 12 m (b) 14 m (c) 13 m (d) 11 m Given: Two poles of heights 6m and 11m stand vertically upright on a plane ground. Distance between their foot is 12 m. To find: Distance between their tops. Let CD be the pole with height 6m. AB is the pole with height 11m, distance between their foot i.e. DB is 12 m. Let us assume a point E on the pole AB which is 6m from the base of AB. Hence AE = AB − 6 = 11 − 6 = 5 m Now in right triangle AEC, Applying Pythagoras theorem AC2 = AE2 + EC2 AC2 = 52 + 122 (since CDEB forms a rectangle and opposite sides of rectangle are equal) AC2 = 25 + 144 AC2 = 169 Thus, the distance between their", "right triangle. 1. $a=12, b=16, c =?$ 2. $a =?, b=20, c=30$ 3. $a=4, b =?, c=11$ 4. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle. 5. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle. 6. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate? 7. Emanuel has a cardboard box that measures $20 \\ cm \\times 10 \\ cm \\times 8 \\ cm \\ (\\text{length} \\times \\text{width} \\times \\text{height})$. What is the length of the diagonal from a bottom corner to the opposite top corner? 8. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. How high above the ground does the ladder touch the wall of the house? 9. Find the area of the triangle if area of a triangle is defined as $A=\\frac{1}{2} \\text{base} \\times \\text{height}$. 10. Instead of walking along the two sides of a rectangular field,", "write an equation. The small triangle is similar to the large triangle. $\\dfrac{h}{24}=\\dfrac{6}{8} \\nonumber$ Solve the proportion. \\begin {aligned} 24\\left(\\dfrac{6}{8}\\right)&=24\\left(\\dfrac{h}{24}\\right)\\\\ 18&=h \\end{aligned} \\nonumber Simplify. Check. Tyler's height is less than his shadow's length so it makes sense that the tree's height is less than the length of its shadow. Check $$h=18$$ in the original proportion. \\begin{aligned} &\\dfrac{6}{8}=\\dfrac{h}{24}\\\\ &\\dfrac{6}{8} \\overset{?}{=} \\dfrac{18}{24}\\\\ &\\dfrac{3}{4}=\\dfrac{3}{4} \\surd \\end{aligned} \\nonumber ##### Exercise $$\\PageIndex{7}$$ A telephone pole casts a shadow that is 50 feet long. Nearby, an 8 foot tall traffic sign casts a shadow that is 10 feet long. How tall is the telephone pole? The telephone pole is 40 feet tall. ##### Exercise $$\\PageIndex{8}$$ A pine tree casts a shadow of 80 feet next to a 30 foot tall building which casts a 40 feet shadow. How tall is the pine tree? The pine tree is 60 feet tall. ## Solve Uniform Motion Applications We have solved uniform motion problems using the formula $$D=r t$$ in previous chapters. We used a table like the one below to organize the information and lead us to the equation. Rate $$\\cdot$$ Time = Distance The formula $$D=r t$$ assumes we know $$r$$ and $$t$$ and use them to find $$D$$. If we know $$D$$ and $$r$$ and need to find $$t$$, we would solve the equation", "## Precalculus (6th Edition) Blitzer The amount of cable used is $\\sqrt{36+{{x}^{2}}}+\\sqrt{{{x}^{2}}-20x+164}$ Consider the length of cable from the top of the $6\\text{-foot}$ pole to the ground to be ${{h}_{1}}$ and the length of the cable from the top of the $\\text{8-foot}$ pole to the ground to be ${{h}_{2}}$ According to the Pythagoras theorem In a right-angle triangle, ${{h}^{2}}={{p}^{2}}+{{b}^{2}}$ …… (1) Substitute $6$ for $p$ , $x$ for $b$ and ${{h}_{1}}$ for $h$ in equation (1) \\begin{align} & {{h}_{1}}=\\sqrt{{{6}^{2}}+{{x}^{2}}} \\\\ & =\\sqrt{36+{{x}^{2}}} \\end{align} The length of the cable from the top of the $6\\text{-foot}$ pole to the ground ${{h}_{1}}=\\sqrt{36+{{x}^{2}}}$ Substitute $8$ for $p$, $10-x$ for $b$ and ${{h}_{2}}$ for $h$ in equation (1) ${{h}_{2}}=\\sqrt{{{8}^{2}}+{{\\left( 10-x \\right)}^{2}}}$ Use formula ${{\\left( A-B \\right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$ in the above equation \\begin{align} & {{h}_{2}}=\\sqrt{64+100-20x+{{x}^{2}}} \\\\ & =\\sqrt{164-20x+{{x}^{2}}} \\end{align} Or ${{h}_{2}}=\\sqrt{{{x}^{2}}-20x+164}$ The length of the cable from the top of the $\\text{8-foot}$ pole to the ground ${{h}_{2}}=\\sqrt{{{x}^{2}}-20x+164}$ Total length of cable is $\\sqrt{36+{{x}^{2}}}+\\sqrt{{{x}^{2}}-20x+164}$ The required amount of cable used in terms of the distance from the $6\\text{-foot}$ pole to the point where the cable touches the ground, $x$ , where the length of the other pole is $\\text{8-foot}$ and distance between poles is $\\text{10-foot}$, is $\\sqrt{36+{{x}^{2}}}+\\sqrt{{{x}^{2}}-20x+164}$.", "Two Poles of Height 6m and 11m Stands on a Plane Ground, Find the Distance Between Their Tops. Question: Two poles of height 6m and 11m stands on a plane ground, if the distance between their feet is 12m. Find the distance between their tops. The Pythagoras theorem states that if a triangle is right-angled (90 degrees), then the square of the hypotenuse is equal to the sum of the squares of the other two sides. Answer: Two poles of height 6m and 11m stands on a plane ground, where distance between their feet is 12m, the distance between the tops of the two poles is 13 m. The Pythagoras theorem works only for right-angled triangles. When any two values are known, we can apply the theorem and calculate the third side. Explanation: Let's draw the diagram of the two poles AB and CD as shown: Let, BD be the distance between the two poles Assume, BD = x As △BED is a right-angled triangle right angled at E, therefore 'x' is the hypotenuse. Now apply the Pythagoras theorem on △BED , Hypotenuse2 = Perpendicular2 + Base2 => BD2 = ED2 + BE2 => x2 = 122 + 52 => x2 = 144 + 25 => x2 = 169 => x = 13 m", "2x − 20 represents Maggie’s contribution. Solve the equation. x + 2x − 20 = 94 3x = 114 x = 38 Christian put in $38 and Maggie put in 94 − 38 =$56. Q28. The Northridge Quakers have won 20 games and lost 15. What is the ratio of games won to game played? The number of games played is the total of the wins and losses, (20 + 15 = 35). Write the ratio and simplify. 20/35 = 4/7 Q29. To set up a tent, workers place a 25-foot pole in the center of a grassy area as shown in the diagram. A bracing wire is attached to the top of the pole and to a stake 9 feet from the base of the pole. Which of the following represents the length of the bracing wire? Q30. What is the value of the expression −3 × 5² + 2(4 − 18) + 3²? −3 × 5² + 2(4 − 18) + 3³ −3 × 25 + 2(−14) + 27 −75 + (−28) + 27 −76 Q31. The lengths of the sides of ΔABC are 6 inches, 8 inches, and 10 inches. Which of the following conclusions must be true? • A). ∠C is a right angle • B). ΔABC is an acute triangle. • C). ΔABC", "6: Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. Solution: Let AB and CD be the given vertical poles. Then, AB = 9 m, CD = 14 m and AC = 12 m Const:Draw, BE || AC. Then, CE = AB = 9 m and BE = AC = 12 m DE = (CD – CE) = (14 – 9) = 5 m In right $\\Delta$BED, we have BD2 = BE2 + DE2 = [(12)2 + (5)2] m2 = (144 + 25) m2 = 169 m2 $\\Rightarrow$ BD = $\\sqrt{169}$ = 13 m Hence, the distance between their tops is 13 m. Question 7:In the given figure, O is a point inside a $\\Delta$PQR such that $\\angle$QPR = 90o, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that $\\Delta$PQR is right-angled. Solution: In $\\Delta$PQR, $\\angle$QPR = 90o, PQ = 24 cm, and QR = 26 cm2 In $\\Delta$POR, PO = 6 cm, QR = 8 cm and $\\angle$POR = 90o In $\\Delta$POR, PR2 = PO2 + OR2 PR2 = (62 + 82) cm2 = (36 + 64) cm2 = 100 cm2 PR = $\\sqrt{100}$", "r_1*P_r_1 + r_2*P_r_2 + r_3*P_r_3 - R_0*P_R_0 - R_1*P_R_1)*p ; # Affordable cost of relocating line # DISPLAY RESULTS print (\"EXAMPLE : 12.1 : SOLUTION :-\") ; print \" Affordable cost of relocating line , B =$ %.1f \"%(B) ; print \" Since actual relocating & rebuilding of line would cost much more than amount found \" ; print \" The distribution engineer decides to keep the status quo \" ; EXAMPLE : 12.1 : SOLUTION :- Affordable cost of relocating line , B = \\$ 6679.2 Since actual relocating & rebuilding of line would cost much more than amount found The distribution engineer decides to keep the status quo ## Example 12.2 Page No : 659¶ In [2]: import math # GIVEN DATA V = 40 ; # Actual wind velocity in mi/hr c_pg = 40. ; # Circumference at ground level in inches c_pt = 28. ; # Circumference at pole top in inches l = 35. ; # height of pole in feet l_g = 6 ; # Height of pole set in ground in feet d_c = 0.81 ; # dia. of copper conductor in inches span_avg = 120 ; # Average span in ft no_c = 8 ; # NO. of conductors # CALCULATIONS # For case (a) p = 0.00256 * (V**2)", "# Thread: geometry extra credit help 1. ## geometry extra credit help help? 2. no congruent triangles ... you have some similar triangles, though. remember that the sides of similar triangles are proportional. 3. so theyre similar and right (they're both perpendicular to the ground) but where do i go from there? 4. Hello, natbat77! On a level plot of ground there stand two vertical posts, one 6 ft tall, the other 10 ft tall. From the top of each pole a rope is stretched to the foot of the other pole. How far above ground is the point where the ropes cross? Code: * C * | * | * | * | * | A * * | 10 | * P * | | * | 6 | * : * | | * :h * | | * : * | B * - - - * - - - - - - - * D a Q b The poles are: . $AB = 6,\\;\\;CD = 10$ The ropes $AD$ and $BC$ cross at $P$. $PQ = h$ is perpendicular to $BD.$ Let $a = BQ,\\;\\;b = QD$ In similar right triangles $PQD\\text{ and }ABD\\!:\\;\\;\\frac{h}{b} \\:=\\: \\frac{6}{a+b} \\quad\\Rightarrow\\quad b \\:=\\:\\frac{ah}{6-h}$ .[1] In similar right triangles $PQB \\text{ and }CDB\\!:\\;\\;\\frac{h}{a} \\:=\\:\\frac{10}{a+b} \\quad\\Rightarrow\\quad b \\:=\\:\\frac{10a", "my solution for 13 a) !!!!! You'd have to use the first hypotenuse to figure out the OTHER triangle! Sorry! Masheroni Posts: 3 Joined: Sun Dec 23, 2018 9:29 pm Reputation: 0 ### Re: Solving for angles and sides Good afternoon! 11) h = height of the pole g1 = length of guy wire (50) g2 = length of guy wire (35) x = distance between pole and guy wire g1 (50) x+10 = distance between pole and guy wire g2 (35) So: $$\\tan 50^{\\circ}=\\dfrac{h}{x}\\\\\\cot 50^{\\circ}=\\dfrac{x}{h}\\\\x=h\\cot 50^{\\circ}\\\\ \\tan 35^{\\circ}=\\dfrac{h}{x+10}\\\\\\cot 35^{\\circ}=\\dfrac{x+10}{h}\\\\x+10=h\\cot 35^{\\circ}\\\\ h\\cot 35^{\\circ}=x+10=h\\cot 50^{\\circ}\\\\ h\\cot 35^{\\circ}-h\\cot 50^{\\circ}=10\\\\ h=\\dfrac{10}{\\cot 35^{\\circ}-\\cot 50^{\\circ}}\\\\ h\\approx 17,0$$ Now, you have the length of the pole, the guy wire is easy $$\\sin 50^{\\circ}=\\dfrac{h}{g_1}\\\\g_1=\\dfrac{17}{\\sin 50^{\\circ}}\\approx 22,2$$ $$\\sin 35^{\\circ}=\\dfrac{h}{g_2}\\\\g_2=\\dfrac{17}{\\sin 35^{\\circ}}\\approx 29,6$$ Total length: $$22,2+29,6=51,8$$ Let's solve the others in another post Baltuilhe Posts: 44 Joined: Fri Dec 14, 2018 3:55 pm Reputation: 29 ### Re: Solving for angles and sides Good afternoon (2)! 13) Going to east, 80m and to south, 60m, using Pythagoras: $$d^2=80^2+60^2=6\\,400+3\\,600\\\\d^2=10\\,000\\\\d=100m$$ And now, up 30m. $$d^2=100^2+30^2=10\\,000+900=10\\,900\\\\d=\\sqrt{10\\,900}\\approx 104,40m$$ Now, the angle of elevation: $$\\tan\\theta=\\dfrac{30}{100}=0,3\\\\\\theta=\\arctan 0,3\\\\\\theta\\approx 16,70^{\\circ}$$ I hope I have helped! Baltuilhe Posts: 44 Joined: Fri Dec 14, 2018 3:55 pm Reputation: 29 Return to Trigonometry - sin, cos, tan, cot, arcsin, arccos, arctan, arccot ### Who is online Users browsing this forum:", "the foot of the ladder from the building, i.e, BC. In right-angled triangle ABC, we have: Hence, the distance of the foot of the ladder from the building is 5 m Page No 441: Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively. We have: AB = 20 m and BC = 15 m Applying Pythagoras theorem in right-angled triangle ABC, we get: Hence, the length of the ladder is 25 m. Page No 441: Let the two poles be DE and AB and the distance between their bases be BE. We have: DE = 9 m, AB = 14 m and BE = 12 m Draw a line parallel to BE from D, meeting AB at C. Then, DC = 12 m and AC = 5 m We need to find AD, the distance between their tops. Applying Pythagoras theorem in right-angled triangle ACD, we have: ​ Hence, the distance between the tops of the two poles is 13 m. Page No 442: Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A. Now, In right triangle ABC By using Pythagoras theorem,", "feetB. 7.5 feetC. 6 feetD. 5 feet 1. A 2. D 3. B 4. C Option 2 : D ## Detailed Solution Given: An 18 feet high electric pole. The height of tree is 6 feet. The distance between them is 10 feet. Calculation: In the figure, AB = 18 feet, CD = 6 feet and AC = 10 feet Let, shadow of the tree, CO = k feet According to the question, ΔABO and ΔCDO are two similar triangle ⇒ AB/AO = CD/CO ⇒ 18/(10 + k) = 6/k ⇒ 18k = 60 + 6k ⇒ 12k = 60 ⇒ k = 5 ∴ Length of the shadow of the tree = 5 feet Mistake Points We can take shadow point in the middle of tree and electric pole, which is not correct because it is not given that they are opposite to each other. # Find the value of $$\\left( {\\frac{1}{{1 - {{\\cos }^2}x}} + \\frac{{cose{c^2}x}}{{cose{c^2}x - 1}} - \\frac{1}{{1 - {{\\sin }^2}x}}} \\right)$$ ? 1. cos2x 2. sin2x 3. sec2x 4. cosec2x Option 4 : cosec2x ## Detailed Solution $$\\left( {\\frac{1}{{1 - {{\\cos }^2}x}} + \\frac{{cose{c^2}x}}{{cose{c^2}x - 1}} - \\frac{1}{{1 - {{\\sin }^2}x}}} \\right)$$ Calculation: $$\\left( {\\frac{1}{{1 - {{\\cos }^2}x}} + \\frac{{cose{c^2}x}}{{cose{c^2}x - 1}} - \\frac{1}{{1 - {{\\sin }^2}x}}} \\right)$$ ⇒ $$\\left( {\\frac{1}{{{{\\sin }^2}x}} +", "(By Pythagoras Theorem) ⇒ (5)² = (4.8)² + x² ⇒ 25 = 23.04 + x² ⇒ x² = 25.00 – 23.04 = 1.96 = (1.4)² ⇒ x = 1.4 The foot of ladder are 1.4m away from the wall. Question 11. Solution: Let AB be the tree which broke at D and its top A touches the ground at C their BD = 5m, BC = 12m, Let AD = x m, then CD = x m Now, in right ∆ABC, CD² = BD² + BC² (By Pythagoras Theorem) CD² = (9)² + (12)² = 81 + 144 = 225 = (15)² CD = 15m, Height of the tree AB = AD + BD = 15 + 9 = 24m Question 12. Solution: AB and CD are two poles and they are 12,m apart AB = 18 m, CD = 13m and BD = 12 m From C, draw CE || BD Then CE = BD = 12 m and AE = AB – EB = AB – CD = 18 – 13 = 5 m Join AC Now in right ∆ACE AC² = CE² + AE² (By Pythagoras Theorem) AC² = (12)² + (5)² = 144 + 25 = 169 = (13)² AC = 13 m Distance between their tops = 13 m Question 13. Solution: A", "${N_{50}(X)}$, where $N_{r}(X)$ is the radius ${r}$ neighborhood of ${X}$. Intuitively, the boundary of $N_r(X)$ is always $r$ units off the port side of the boat. (See Ravi Vakil’s article “The Mathematics of Doodling” for more about this set.) You may not think you have enough information to solve the problem, but you do! The clue is found in a well-known puzzler. Suppose we have an electrical wire running round the equator of the earth. We want to get it up off the ground and put it at the top of 50-foot electrical poles. How much longer must the wire be? [If you haven’t seen this puzzler before, stop now and think about it.] If the radius of the Earth is ${R}$ feet, then the original wire must be ${2\\pi R}$ feet long. The new wire is a circle with a radius 50 feet longer than before, so its length must be ${2\\pi (R+50)=2\\pi R+100\\pi}$ feet. That is, it must be ${100\\pi\\approx 314}$ feet longer. It turns out that this argument can be generalized in various ways (again, see Vakil’s article for more information). First of all, if ${X}$ is any convex set with a polygonal boundary, then, just like in the puzzle, the perimeter of ${N_{r}(X)}$ is ${2\\pi r}$ units longer than the perimeter of ${X}$. Vakil" ]

[ "# 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 (5(2^3+4))/sqrt36? Apr 7, 2015 Basically you just need to do the arithmetic Note that $\\sqrt{36} = 6$ and ${2}^{3} = 8$ So we have $\\frac{5 \\left({2}^{3} + 4\\right)}{\\sqrt{36}}$ $= \\frac{5 \\left(12\\right)}{6}$ $= 10$", "# 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 $r \\sin \\theta = - \\frac{6}{85}$.", "# How do you simplify 13/30+1/5div6/7? $\\frac{13}{30} + \\frac{1}{5} \\div i \\mathrm{de} \\frac{6}{7} = \\frac{2}{3}$ $\\frac{1}{5} \\div i \\mathrm{de} \\frac{6}{7} = \\frac{7}{30}$ $\\frac{7}{30} + \\frac{13}{30} = \\frac{20}{30} = \\frac{2}{3}$", "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 =$\\frac{11}{8}$", "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 4/15 + 2/4? Aug 1, 2016 $\\frac{4}{15} + \\frac{2}{4} = \\frac{23}{30}$ #### Explanation: This can be solved easily by multiplying the numerator and denominator of the fractions by constants: $\\frac{4}{15} + \\frac{2}{4}$ $= \\frac{4}{15} + \\frac{1}{2}$ $= \\frac{8}{30} + \\frac{15}{30}$ $= \\frac{23}{30}$", "# How do you simplify \\frac{20-0}{33-5}? Aug 5, 2017 $\\frac{5}{7}$ #### Explanation: $20 - 0 = 20$ $33 - 5 = 28$ $\\frac{20}{28}$ Divide both the numerator and the denominator by $4$. $20 \\div 4 = 5$ $28 \\div 4 = 7$ $\\therefore \\frac{20}{28} = \\frac{5}{7}$", "# How do you simplify 30-[9+4(8-5)]? $= 30 - \\left[9 + 4 \\cdot 3\\right] = 30 - \\left[9 + 12\\right] = 30 - 21 = 9$", "# How do you simplify (-9-sqrt(108))/3? It is $\\frac{- 9 - \\sqrt{108}}{3} = \\frac{- 9 - 6 \\sqrt{3}}{3} = - 3 - 2 \\sqrt{3}$", "How do you simplify \\frac{40}{-25}? $- \\frac{8}{5}$ $\\frac{40}{- 25} = \\frac{- {2}^{3} \\cdot 5}{5} ^ 2 = \\frac{- {2}^{3} \\cdot \\cancel{5}}{5 \\cdot \\cancel{5}} = - \\frac{8}{5}$", "(c) Simplify: $$\\frac{(25)^{3 / 2} \\times(243)^{3 / 5}}{(16)^{5 / 4} \\times(8)^{4 / 3}}$$", "How do you simplify (20 - sqrt5) /11? $\\frac{20}{11} - \\frac{\\sqrt{5}}{11}$ $\\frac{20 - \\sqrt{5}}{11}$ $\\therefore = \\frac{20}{11} - \\frac{\\sqrt{5}}{11}$", "buscandoaireka8u 2023-02-18 simplify $\\sqrt{\\frac{36}{225}}$ Troy Jimenez $\\sqrt{\\frac{36}{225}}=\\sqrt{\\frac{{6}^{2}}{{15}^{2}}}=\\frac{6}{15}=\\frac{2}{5}$, here the answer for your question", "# Simplify $\\frac{3+2i}{6-5i}$ Multiply numerator and denominator by the conjugate of $6-5i$ which is $6+5i$ $\\frac{3+2i}{6-5i} \\cdot \\frac{6+5i}{6+5i}=\\frac{18+27i-10}{36+25}=\\frac{8+27i}{61}=\\frac{8}{61}+\\frac{27} {61}i$", "+0 0 49 1 Simplify $\\frac{2\\sqrt[3]9}{1 + \\sqrt[3]3 + \\sqrt[3]9}.$ Guest Jun 5, 2018 #1 0 See the solution here, especially the last one #11 by MaxWong: https://web2.0calc.com/questions/help-plzzzz_3 Guest Jun 5, 2018", "fraction $\\frac{20}{30}$ the simplified fraction is $\\frac{2}{3}$. Example 2: Simplify the fraction $\\frac{42}{56}$ Solution: $\\frac{42}{56}$ = $\\frac{2*3*7}{2*2*2*7}$ By cancelling the common terms we get, $\\frac{3}{4}$ Therefore for the given fraction $\\frac{42}{56}$ the simplified fraction is $\\frac{3}{4}$.", "# How do you simplify 20-3*-70/11? $39 \\frac{1}{11}$ $20 - 3 \\cdot - \\frac{70}{11} = 20 + \\frac{210}{11} = \\frac{220}{11} + \\frac{210}{11} = \\frac{430}{11} = 39 \\frac{1}{11}$", "\\frac{40}{20} - \\frac{5y}{20}$ • Step 5: Simplify • $x = 2 - \\frac{y}{4}$" ]

[ "case, we have a 5 and a 13 that are common to both numerator and denominator $$\\frac{65}{390} = \\frac{5 \\cdot 13}{2 \\cdot 3 \\cdot 5 \\cdot 13}$$ $$\\frac{65}{390} = \\frac{\\cancel{5} \\cdot \\cancel{13} 1}{2 \\cdot 3 \\cdot \\cancel{5} \\cdot \\cancel{13}}$$ $$\\frac{65}{390} = \\frac{1}{6}$$ We can also use simple division. We know the GCF of 65 and 390 is 65. If we divide 65 by 65, we get 1. If we divide 390 by 65, we get 6. $$\\frac{65}{390} = \\frac{65 ÷ 65}{390 ÷ 65} = \\frac{1}{6}$$ Example 3: Simplify each fraction $$\\frac{330}{420}$$ • Factor the numerator and denominator into the product of prime numbers • 330 = 2 x 3 x 5 x 11 • 420 = 2 x 2 x 3 x 5 x 7 • Cancel all common factors between the numerator and denominator • In this case, we have a 2, 3, and 5 that are common to both numerator and denominator $$\\require{cancel}$$ $$\\frac{330}{420} = \\frac{2 \\cdot 3 \\cdot 5 \\cdot 11}{2 \\cdot 2 \\cdot 3 \\cdot 5 \\cdot 7}$$ $$\\frac{330}{420} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 11}{\\cancel{2} \\cdot 2 \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 7}$$ $$\\frac{330}{420} = \\frac{11}{14}$$ We can also use simple division. We know the GCF of 330 and 420 is 30. If we divide 330 by 30, we get 11. If", "in ascending order, but every little bit helps. The denominator can also be expressed as a product of primes. $2310=10 \\cdot 231=2 \\cdot 5 \\cdot 7 \\cdot 33=2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11$ We can now cancel common factors. $\\frac{210}{2310}=\\frac{2 \\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11}=\\frac{\\not{2} \\cdot \\not{3} \\cdot \\not{5} \\cdot \\not{7}}{\\not{2} \\cdot \\not{8} \\cdot \\not{5} \\cdot \\not{7} \\cdot 11}=\\frac{1}{11}$ However, this approach is not the most efficient way to proceed, as multiplying numerators and denominators allows the products to grow to larger numbers, as in 210/2310. It is then a little bit harder to prime factor the larger numbers. A better approach is to factor the smaller numerators and denominators immediately, as follows. $\\frac{6}{231} \\cdot \\frac{35}{10}=\\frac{2 \\cdot 3}{3 \\cdot 7 \\cdot 11} \\cdot \\frac{5 \\cdot 7}{2 \\cdot 5}$ We could now multiply numerators and denominators, then cancel common factors, which would match identically the last computation in equation (5). However, we can also employ the following cancellation rule. Cancellation Rule When working with the product of two or more rational expressions, factor all numerators and denominators, then cancel. The cancellation rule is simple: cancel a factor “on the top” for an identical factor “on the bottom.” Speaking more technically, cancel any factor in any numerator for", "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", "denominator, we find \\begin{gather*} 126 = 6 \\cdot 21 = 2 \\cdot 3^2 \\cdot 7, \\\\ 24 = 3 \\cdot 8 = 2^3 \\cdot 3. \\end{gather*} The fraction simplifies by canceling all common factors: \\begin{equation*} \\frac{126}{24} = \\frac{2 \\cdot 3^2 \\cdot 7}{2^3 \\cdot 3} = \\frac{3 \\cdot 7}{2^2} = \\frac{21}{4}. \\end{equation*} In practice, we don't always have to find the prime factorization. Instead, we can find one common factor at a time until no common factors remain. For example, since 126 and 24 are both even, we would write \\begin{equation*} \\frac{126}{24} = \\frac{63}{12}. \\end{equation*} Then, we look at 63 and 12 and recognize that they are both divisible by 3, allowing us to rewrite the fraction as \\begin{equation*} \\frac{126}{24} = \\frac{63}{12} = \\frac{21}{4}. \\end{equation*} Because $21=3 \\cdot 7$ and $4=2^2$ do not have common factors, we know this is simplified. A square root is not simplified if there is a factor inside the root that is a perfect square. Similarly, a cube root is not simplified if there is a factor inside that is a perfect cube. We use the factors of the value inside a root to determine if we can simplify it. ###### Example1.2.3 The square root $\\sqrt{126}$ is not simplified. A square root is the inverse operation of squaring numbers (for non-negative numbers) so that", "from there $\\frac{400\\pi}{180} = \\frac{20 \\cdot 20\\pi}{20 \\cdot 9}$ ... cancel the common factor of 20 in the numerator and denominator.", "\\cancel{h} \\cdot \\cancel{h}} {\\cancel{h} \\cdot \\cancel{h} \\cdot \\cancel{h} \\cdot h \\cdot h} \\\\ &= \\frac {1} {h \\cdot h} \\\\ & = \\frac {1} { h^{2}} \\end{align*} If we were to simplify the original expression using the quotient rule, we would have \\begin{align*} \\frac {h^{3}} {h^{5}} &= h^{3-5} \\\\ &= h^{-2} \\end{align*} Putting the answers together, we have $h^{-2}= \\frac {1} {h^{2}}$. This is true for any nonzero real number, or any variable representing a nonzero real number. A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa. $a^{-n} = \\frac {1} {a^{n}}$ and $a^{n} = \\frac {1} {a^{-n}}$ We have shown that the exponential expression $a^{n}$ is defined when $n$ is a natural number, $0$, or the negative of a natural number. That means that $a^{n}$ is defined for any integer $n$. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n$. ### THE NEGATIVE RULE OF EXPONENTS For any nonzero real number $a$ and natural number $n$, the negative rule of exponents states that $a^{-n} = \\frac {1} {a^{n}}$ #### Example 5 Write each of the following quotients with a single base. Do not simplify further. Write", "grows larger with each successive term. Looking at the terms more closely, it turns out that the denominators tend to be products of many small primes, whereas the numerators are either primes or products of a few comparatively large primes. For example: $\\frac{9649}{2520} = \\frac{9649}{2^3 \\cdot 3^2 \\cdot 5 \\cdot 7} \\qquad \\textrm{and} \\qquad \\frac{18358381}{4084080} = \\frac{59 \\cdot 379 \\cdot 821}{2^4 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17}.$ To produce an integer, we need to cancel all the primes in the factorization of the denominator by matching primes in the numerator; given the pattern of these numbers, that looks like an unlikely coincidence. But there is reason for caution. Note the seventh term in the sequence, where the denominator has decreased from $$60$$ to $$20$$. To understand how that happens, we can run through the calculation of the term, which starts by summing the six previous terms. $\\frac{60}{60} + \\frac{120}{60} + \\frac{150}{60} + \\frac{170}{60} + \\frac{185}{60} + \\frac{197}{60} = \\frac{882}{60}.$ Then we calculate the mean, and add 1 to get the seventh term: $\\require{cancel}\\frac{882}{60} \\cdot \\frac{1}{6} = \\frac{882}{360} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot 7 \\cdot 7}{\\cancel{2} \\cdot 2 \\cdot 2 \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot 5} = \\frac{49}{20} + 1 = \\frac{69}{20}$ Cancelations reduce the numerator and denominator of the mean", "$$\\displaystyle \\frac{2}{3}+\\frac{5}{4}-\\frac{5\\cdot 4}{3\\cdot 7}$$ After canceling out the common factors from the top and bottom $$= \\,\\,$$ $$\\displaystyle \\frac{2}{3}+\\frac{5}{4}-\\frac{20}{21}$$ Amplifying in order to get the common denominator 84 $$= \\,\\,$$ $$\\displaystyle \\frac{2}{3}\\cdot\\frac{28}{28}+\\frac{5}{4}\\cdot\\frac{21}{21}-\\frac{20}{21}\\cdot\\frac{4}{4}$$ We need to use the common denominator: 84 $$= \\,\\,$$ $$\\displaystyle \\frac{2\\cdot 28+5\\cdot 21-20\\cdot 4}{84}$$ Expanding each term in the numerator: $$2 \\times 28+5 \\times 21-20 \\times 4 = 56+105-80$$ $$= \\,\\,$$ $$\\displaystyle \\frac{56+105-80}{84}$$ Operating the terms in the numerator $$= \\,\\,$$ $$\\displaystyle \\frac{81}{84}$$ We can factor out 3 for both the numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{3\\cdot 27}{3\\cdot 28}$$ Now we cancel 3 out from the numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{27}{28}$$ which concludes the process of simplification. Example: Simplify calculator example Simplify the following: $$\\frac{x}{3} + \\frac{x}{4} - \\frac{x}{6}$$ Solution:We need to simplify the following expression: $$\\displaystyle \\frac{x}{3}+\\frac{x}{4}-\\frac{x}{6}$$. The following calculation is obtained: $$\\displaystyle \\frac{x}{3}+\\frac{x}{4}-\\frac{x}{6}$$ Grouping the terms with $$x$$ $$= \\,\\,$$ $$\\displaystyle \\left(\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{6}\\right)x$$ Simplifying the terms that were grouped with $$x$$ $$= \\,\\,$$ $$\\displaystyle \\frac{5}{12}x$$ which concludes the process of simplification. ### Example: Another simplification calculation Calculate $$\\left(\\frac{1}{3} \\times \\frac{6}{5} \\right)+ \\frac{2}{5}$$. Solution: We need to simplify the following expression: $$\\displaystyle \\frac{1}{3}\\cdot\\frac{6}{5}+\\frac{2}{5}$$. The following calculation is obtained: $$\\displaystyle \\frac{ 1}{ 3} \\cdot \\frac{ \\left(6\\right)}{ 5}+\\frac{2}{5}$$ By multiplying all the numerators and all the denominators: $$\\displaystyle\\frac{ 1}{ 3} \\times \\frac{ 6}{ 5}=", "need. The LCD, $$36$$, has $$2$$ factors of $$2$$ and $$2$$ factors of $$3$$. Twelve has two factors of $$2$$, but only one of $$3$$ —so it is ‘missing‘ one $$3$$. We multiplied the numerator and denominator of $$\\frac{7}{12}$$ by $$3$$ to get an equivalent fraction with denominator $$36$$. Eighteen is missing one factor of $$2$$ —so you multiply the numerator and denominator $$\\frac{5}{18}$$ by $$2$$ to get an equivalent fraction with denominator $$36$$. We will apply this method as we subtract the fractions in the next example. Example $$\\PageIndex{8}$$: subtract Subtract: $$\\frac{7}{15} − \\frac{19}{24}$$. Solution Find the LCD. 15 is 'missing' three factors of 2 24 is 'missing' a factor of 5 Rewrite as equivalent fractions with the LCD. $$\\frac{7 \\cdot \\textcolor{red}{8}}{15 \\cdot \\textcolor{red}{8}} - \\frac{19 \\cdot \\textcolor{red}{5}}{24 \\cdot \\textcolor{red}{5}}$$ Simplify each numerator and denominator. $$\\frac{56}{120} - \\frac{95}{120}$$ Subtract. $$- \\frac{39}{120}$$ Rewrite showing the common factor of 3. $$- \\frac{13 \\cdot 3}{40 \\cdot 3}$$ Remove the common factor to simplify. $$- \\frac{13}{40}$$ Exercise $$\\PageIndex{15}$$ Subtract: $$\\frac{13}{24} − \\frac{17}{32}$$. $$\\frac{1}{96}$$ Exercise $$\\PageIndex{16}$$ Subtract: $$\\frac{21}{32} − \\frac{9}{28}$$. $$\\frac{75}{224}$$ Example $$\\PageIndex{9}$$: add Add: $$- \\frac{11}{30} + \\frac{23}{42}$$. Solution Find the LCD. Rewrite as equivalent fractions with the LCD. $$- \\frac{11 \\cdot \\textcolor{red}{7}}{30 \\cdot \\textcolor{red}{7}} + \\frac{23 \\cdot \\textcolor{red}{5}}{42 \\cdot \\textcolor{red}{5}}$$ Simplify each numerator and denominator. $$- \\frac{77}{210} + \\frac{115}{210}$$ Add.", "\\cdot \\cancel{3}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{7}{2} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\end{aligned}\\nonumber Note that this yields exactly the same result, 7/2. Cancellation Rule When multiplying fractions, cancel common factors according to the following rule: “Cancel a factor in a numerator for an identical factor in a denominator.” Example 6 Find the product of 14/15 and 30/140. Solution Multiply numerators and multiply denominators. Prime factor, cancel common factors, then multiply. \\begin{aligned} \\frac{14}{15} \\cdot \\frac{30}{140} = \\frac{14 \\cdot 30}{15 \\cdot 140} ~ & \\textcolor{red}{ \\text{ Multiply numerators; multiply denominators.}} \\\\ = \\frac{(2 \\cdot 7) \\cdot (2 \\cdot 3 \\cdot 5)}{(3 \\cdot 5) \\cdot (2 \\cdot 2 \\cdot 5 \\cdot 7)} ~ & \\textcolor{red}{ \\text{ Prime factor numerators and denominators.}} \\\\ = \\frac{ \\cancel{2} \\cdot \\cancel{7} \\cdot \\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5}}{ \\cancel{3} \\cdot 5 cdot \\cancel{2} \\cdot \\cancel{2} \\cdot \\cancel{5} \\cdot \\cancel{7}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{1}{5} ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Note: Everything in the numerator cancels because you’ve divided the numerator by itself. Hence, the answer has a 1 in its numerator. Exercise Multiply: $\\frac{6}{35} \\cdot \\frac{70}{36}\\nonumber$ $$\\frac{1}{3}$$ When Everything Cancels When all the factors in the numerator cancel, this means that you are dividing the numerator by itself. Hence, you are left with a", "# Least Common Denominators (LCDs) It is difficult to add or subtract fractions when the denominators are not the same. So, we use a common denominator. It is usually easiest to use the least common denominator. The least common denominator is simply the least common multiple ( LCM ) of the two denominators. Example 1: Find the common denominator of the fractions. $\\frac{1}{6}$ and $\\frac{3}{8}$ We need to find the least common multiple of $6$ and $8$ . One way to do this is to list the multiples: $\\begin{array}{l}6,12,18,\\underset{_}{24},30,36,42,48,...\\\\ 8,16,\\underset{_}{24},32,40,48,...\\end{array}$ The first number that occurs in both lists is $24$ , so $24$ is the LCM. So we use this as our common denominator. Listing multiples is impractical for large numbers. Another way to find the LCM of two numbers is to divide their product by their greatest common factor ( GCF ). Example 2: Find the common denominator of the fractions. $\\frac{5}{12}$ and $\\frac{2}{15}$ The greatest common factor of $12$ and $15$ is $3$ . So, to find the least common multiple, divide the product by $3$ . $\\frac{12\\cdot 15}{3}=\\frac{\\overline{)3}\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}4\\text{\\hspace{0.17em}}\\cdot \\text{\\hspace{0.17em}}15}{\\overline{)3}}=60$ If you can find a least common denominator, then you can rewrite the problem using equivalent fractions that have like denominators, so they are easy to add or subtract. Example 3: $\\frac{5}{12}+\\frac{2}{15}$ In the previous", "\\begin{aligned} \\dfrac{28}{42} = \\dfrac{2 \\cdot 2 \\cdot 7}{2 \\cdot 3 \\cdot 7} ~ & \\textcolor{red}{ \\text{ Prime factor numerator and denominator.}} \\\\ = \\dfrac{ \\cancel{2} \\cdot 2 \\cdot \\cancel{7}}{ \\cancel{2} \\cdot 3 \\cdot \\cancel{7}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\dfrac{2}{3} \\end{aligned}\\nonumber Thus, 28/42 = 2/3. Exercise Reduce the fraction 36/60 to lowest terms. 3/5 ## Reducing Fractions with Variables We use exactly the same technique to reduce fractions whose numerators and denominators contain variables. Example 5 Reduce $\\dfrac{56x^2y}{60xy^2}\\nonumber$ to lowest terms. Solution Use factor trees to factor the coefficients of numerator and denominator. Now cancel common factors. \\begin{aligned} \\dfrac{56x^2y}{60xy^2} = \\dfrac{2 \\cdot 2 \\cdot 2 \\cdot 7 \\cdot x \\cdot x \\cdot y}{2 \\cdot 2 \\cdot 3 \\cdot 5 \\cdot x \\cdot y \\cdot y} ~ & \\textcolor{red}{ \\text{ Prime factor numerator and denominator.}} \\\\ = \\dfrac{ \\cancel{2} \\cdot \\cancel{2} \\cdot 2 \\cdot 7 \\cdot \\cancel{x} \\cdot x \\cdot \\cancel{y}}{ \\cancel{2} \\cdot \\cancel{2} \\cdot 3 \\cdot 5 \\cdot \\cancel{x} \\cdot y \\cdot \\cancel{y}} ~ & \\textcolor{red}{ \\text{ Cancel cmmon factors.}} \\\\ = \\dfrac{2 \\cdot 7 \\cdot x}{3 \\cdot 5 \\cdot y} ~ & \\textcolor{red}{ \\text{ Remaining factors.}} \\\\ = \\dfrac{14x}{15y} ~ & \\textcolor{red}{ \\text{ Simplify numerator and denominator.}} \\end{aligned}\\nonumber Thus, 56x2y/(60xy2) = 14x/(15y). Exercise Reduce: $\\dfrac{25a^3b}{40a^2b^3}\\nonumber$ $\\dfrac{5a}{8b^2}\\nonumber$ ## A Word on Mathematical Notation There", "the denominator: $$\\frac{\\frac{3}{10}\\cdot \\frac{60}{1}+ \\frac{6}{15}\\cdot \\frac{60}{1}}{\\frac{19}{20}\\cdot \\frac{60}{1}- \\frac{2}{3}\\cdot \\frac{60}{1}}$$ Simplify: $$\\frac{\\frac{3}{\\cancel{10}1}\\cdot \\frac{\\cancel{60}6}{1}+ \\frac{6}{\\cancel{15}1}\\cdot \\frac{\\cancel{60}4}{1}}{\\frac{19}{\\cancel{20}1}\\cdot \\frac{\\cancel{60}3}{1}- \\frac{2}{\\cancel{3}1}\\cdot \\frac{\\cancel{60}20}{1}}$$ $$\\frac{18 + 24}{57 - 40}=\\frac{42}{17}=2\\frac{8}{17}$$ #### Skills Check: Example #1 Simplify each. $$\\Large{\\frac{\\frac{6}{54}- \\frac{7}{84}}{\\frac{31}{42}- \\frac{2}{9}}}$$ A $$\\frac{3}{19}$$ B $$\\frac{5}{21}$$ C $$\\frac{7}{130}$$ D $$\\frac{4}{21}$$ E $$\\frac{9}{20}$$ Example #2 Simplify each. $$\\Large{\\frac{\\frac{4}{72}+ \\frac{33}{88}}{\\frac{5}{20}+ \\frac{10}{54}}}$$ A $$\\frac{18}{29}$$ B $$\\frac{17}{35}$$ C $$\\frac{93}{94}$$ D $$\\frac{57}{61}$$ E $$\\frac{101}{103}$$ Example #3 Simplify each. $$\\Large{\\frac{\\frac{37}{63}- \\frac{28}{49}}{\\frac{26}{91}- \\frac{13}{70}}}$$ A $$\\frac{4}{15}$$ B $$\\frac{7}{29}$$ C $$\\frac{8}{17}$$ D $$\\frac{10}{63}$$ E $$\\frac{11}{30}$$", "unless the application represents a measurement where an approximation would be adequate. Simplification of numbers corresponds to finding a new representation of a number in a reduced form. For example, a rational number has many different representations of the form $p/q$ with $p \\in \\mathbb{Z}$ and $q \\in \\mathbb{N}$, but only one where $p$ and $q$ have no common factors. Canceling the common factors would be simplification, as would be simplifying a root or rationalizing a denominator. ##### Example2.1.1 The fraction $\\frac{126}{24}$ is not simplified. If we find the prime factorization of the numerator and denominator, we find \\begin{gather*} 126 = 6 \\cdot 21 = 2 \\cdot 3^2 \\cdot 7, \\\\ 24 = 3 \\cdot 8 = 2^3 \\cdot 3. \\end{gather*} The fraction simplifies by canceling all common factors: \\begin{equation*}\\frac{126}{24} = \\frac{2 \\cdot 3^2 \\cdot 7}{2^3 \\cdot 3} = \\frac{3 \\cdot 7}{2^2} = \\frac{21}{4}.\\end{equation*} ##### Example2.1.2 The square root $\\sqrt{126}$ is not simplified. A square root is the inverse operation of squaring numbers (for non-negative numbers) so that $\\sqrt{3^2} = 3$. Because $\\sqrt{a \\cdot b} = \\sqrt{a} \\cdot \\sqrt{b}$ (for $a,b \\ge 0$), we can simplify as \\begin{equation*}\\sqrt{126} = \\sqrt{2 \\cdot 3^2 \\cdot 7} = \\sqrt{9} \\cdot \\sqrt{14} = 3 \\sqrt{14}.\\end{equation*} ##### Example2.1.3 The expression $3e^{2\\ln(5)-1}$ represents a number. Because exponentials with base $e$ and the natural logarithm", "to the number of decimal places. a) The decimal number 1.2345 has four decimal places. Hence, $1.2345 = \\frac{12345}{10000}\\nonumber$ b) The decimal number 27.198 has three decimal places. Hence, $27.198 = \\frac{27198}{1000}\\nonumber$ Exercise Change 17.205 to an improper fraction. $$\\frac{17205}{100}$$ Example 8 Change each of the following decimals to fractions reduced to lowest terms: (a) 0.35, and (b) 0.125. Solution Place each number in the numerator without the decimal point. Place a number of zeros in the denominator equal to the number of decimal places. Reduce to lowest terms. a) First, place 35 over 100. $0.35 = \\frac{35}{100}\\nonumber$ We can divide both numerator and denominator by the greatest common divisor. \\begin{aligned} = \\frac{35 \\div 5}{100 \\div 5} ~ & \\textcolor{red}{ \\text{ Divide numerator and denominator by 5.}} \\\\ = \\frac{7}{20} ~ & \\textcolor{red}{ \\text{ Simplify numerator and denominator.}} \\end{aligned}\\nonumber b) First, place 125 over 1000. $0.125 = \\frac{125}{1000}\\nonumber$ Prime factor and cancel common factors. \\begin{aligned} = \\frac{5 \\cdot 5 \\cdot 5}{2 \\cdot 2 \\cdot 2 \\cdot 5 \\cdot 5 \\cdot 5} ~ & \\textcolor{red}{ \\text{ Prime factor numerator and denominator.}} \\\\ = \\frac{ \\cancel{5} \\cdot \\cancel{5} \\cdot \\cancel{5}}{2 \\cdot 2 \\cdot 2 \\cdot \\cancel{5} \\cdot \\cancel{5} \\cdot \\cancel{5}} ~ & \\textcolor{red}{ \\text{ Cancel common factors.}} \\\\ = \\frac{1}{8} ~ & \\textcolor{red}{ \\text{ Simplify.}} \\end{aligned}\\nonumber Exercise Change 0.375 to", "that raising a power to a power should be dealt with by multiplying the exponents: $$\\boxed{\\displaystyle{(x^m)^n = x^{m\\,n}}}$$ ### Quotients and Negative (or Zero) Exponents Turning our attention away from products of powers, and focusing now on quotients of powers -- consider the following: $$\\begin{array}{rcl} \\frac{x^5}{x^3} &=& \\frac{x \\cdot x \\cdot x \\cdot x \\cdot x}{x \\cdot x \\cdot x}\\\\\\\\ &=& \\frac{x \\cdot x \\cdot \\cancel{x \\cdot x \\cdot x}}{\\cancel{x \\cdot x \\cdot x}}\\\\\\\\ &=& x \\cdot x\\\\\\\\ &=& x^2 \\end{array}$$ Here again, is it necessary to expand the powers above to find the overall quotient? Certainly not! Notice, the numerator and denominator share the same base, and the exponent involved with the denominator is less than that associated with the numerator. As such, three of the factors of $x$ in the numerator and denominator will cancel, leaving a product of $5-3=2$ factors of $x$ (equivalently, $x^2$). Here again, a more general rule is suggested: $$\\boxed{\\displaystyle{\\frac{x^m}{x^n} = x^{m-n}, \\quad \\textrm{provided } x \\ne 0}}$$ The rule above has an unexpected consequence however. To see this, consider which values of $n$ and $m$ make sense. Notice, even keeping $n,m \\ge 2$, one could easily have $m-n \\lt 2$. However, If we limit ourselves to defining $x^n$ as the product of $n$ factors of $x$ though, powers like $x^0, x^{-1},", "peaces in an accurate way. Note that you don't have to find the least common denominator, just any common denominator. So you can just multiply each numerator by all of the other denominators, and compare the results. (These would be the numerators corresponding to the denominator that would be the product of all of the separate denominators.) In your case, you would compare the numbers $$3\\cdot(3\\cdot 13\\cdot 38)=4446$$ $$2\\cdot(5\\cdot 13\\cdot 38)=4940$$ $$6\\cdot (5\\cdot 3\\cdot 38)=3420$$ $$23\\cdot (5\\cdot 3\\cdot 13)=4485$$ The smallest is the third, whic corresponds to the original fraction $\\frac{6}{13}$.", "$18 = 2 \\cdot 3^2$. The least common denominator of the fractions is the LCM of the two numbers: $2^2 \\cdot 3^2 = 36$. Now we need to rewrite each fraction as an equivalent fraction with the LCD as the denominator. For the first fraction. 12 needs to be multiplied by a factor of 3 in order to change it into the LCD, so we multiply the numerator and the denominator by 3. $\\frac{4} {12} \\cdot \\frac{3} {3} = \\frac{12} {36}$ For the section fraction. 18 needs to be multiplied by a factor of 2 in order to change it into the LCD, so we multiply the numerator and the denominator by 2. $\\frac{5} {18} \\cdot \\frac{2} {2} = \\frac{10} {36}$ Once the denominators of the two fractions are the same we can add the numerators. $\\frac{12} {36} + \\frac{10} {36} = \\frac{22} {36}$ The answer can be reduced by canceling a common factor of 2. $\\frac{12} {36} + \\frac{10} {36} = \\frac{22} {36} = \\frac{11} {18}\\ \\text{Answer}$ Example 6 Perform the following operation and simplify. $\\frac{2} {x + 2} - \\frac{3} {2x - 5}$ Solution The denominators cannot be factored any further, so the LCD is just the product of the separate denominators. $\\text{LCD} = (x + 2) (2x - 5)$ The first fraction needs to be", "# What is 9.75 divided by 1.3? Apr 7, 2018 $9.75 \\div 1.3$ $= \\textcolor{red}{\\frac{15}{2}} = \\textcolor{red}{7.5}$ #### Explanation: The problem is to evaluate $9.75 \\div 1.3$ To do this without a calculator, let's start by expressing each decimal as a fraction. $9.75$ $= 9 + \\frac{7}{10} + \\frac{5}{100}$ $= \\frac{900}{100} + \\frac{70}{100} + \\frac{5}{100}$ $= \\frac{975}{100}$ $1.3$ $= 1 + \\frac{3}{10}$ $= \\frac{10}{10} + \\frac{3}{10}$ $= \\frac{13}{10}$ Now we want to divide $\\frac{975}{100}$ by $\\frac{13}{10}$. To divide two fractions, it is easier to multiply by the reciprocal of the denominator, like so: $\\frac{\\frac{975}{100}}{\\frac{13}{10}} = \\frac{975}{100} \\times \\frac{10}{13} = \\frac{9750}{1300}$ This is a valid answer, but the fraction is not in its simplest form. It may be difficult to see how to simplify this, but let's try to do it step by step. First factor out a 10: $\\frac{9750}{1300} = \\frac{\\cancel{10} \\cdot 975}{\\cancel{10} \\cdot 130}$ Now factor a 5: $\\frac{975}{130} = \\frac{\\cancel{5} \\cdot 195}{\\cancel{5} \\cdot 26}$ And finally factor a 13: $\\frac{195}{26} = \\frac{\\cancel{13} \\cdot 15}{\\cancel{13} \\cdot 2}$ $\\textcolor{red}{\\frac{15}{2} = 7.5}$ This is our answer in simplest form.", "# How do you simplify (sqrt(48x^3y^2))/(sqrt4xy^3)? May 16, 2015 Before starting, I must say that I believe there has been a little typing mistake and that your function is $\\frac{\\sqrt{48 {x}^{3} {y}^{2}}}{\\sqrt{4 x {y}^{3}}}$. I will consider this, then. Note that as both numerator and denominator are square roots, we can then merge them in this way: $\\sqrt{\\frac{48 {x}^{3} {y}^{2}}{4 x {y}^{3}}}$ Now, let's see what composes both numerator and denominator and cancel the common elements. $\\sqrt{\\frac{\\cancel{2 \\cdot 2} \\cdot 2 \\cdot 2 \\cdot 3 \\cdot \\cancel{x} \\cdot x \\cdot x \\cdot \\cancel{y \\cdot y}}{\\cancel{2 \\cdot 2} \\cdot \\cancel{x} \\cdot \\cancel{y \\cdot y} \\cdot y}}$ Now, let's rewrite all that is left from our cancelling: $\\sqrt{\\frac{2 \\cdot 2 \\cdot 3 \\cdot x \\cdot x}{y}}$ However, we still have two squared numbers ($2$ and $x$) inside the numerator. As we're dealing with a squared root, then we can take out the square roots of these squared numbers, as follows: $2 x \\sqrt{\\frac{3}{y}}$ which is the same as $2 x \\frac{\\sqrt{3}}{\\sqrt{y}}$ Now, we can just rationalize this answer: $2 x \\frac{\\sqrt{3}}{\\sqrt{y}} \\cdot \\frac{\\sqrt{y}}{\\sqrt{y}} = \\frac{2 x \\sqrt{3} \\sqrt{y}}{y}$ The multiplication of square roots is the square root of the multiplication. So, the final simplification is: $\\frac{2 x \\sqrt{3 y}}{y}$ - In case your function is just exactly as it is" ]

[ "denominator, we find \\begin{gather*} 126 = 6 \\cdot 21 = 2 \\cdot 3^2 \\cdot 7, \\\\ 24 = 3 \\cdot 8 = 2^3 \\cdot 3. \\end{gather*} The fraction simplifies by canceling all common factors: \\begin{equation*} \\frac{126}{24} = \\frac{2 \\cdot 3^2 \\cdot 7}{2^3 \\cdot 3} = \\frac{3 \\cdot 7}{2^2} = \\frac{21}{4}. \\end{equation*} In practice, we don't always have to find the prime factorization. Instead, we can find one common factor at a time until no common factors remain. For example, since 126 and 24 are both even, we would write \\begin{equation*} \\frac{126}{24} = \\frac{63}{12}. \\end{equation*} Then, we look at 63 and 12 and recognize that they are both divisible by 3, allowing us to rewrite the fraction as \\begin{equation*} \\frac{126}{24} = \\frac{63}{12} = \\frac{21}{4}. \\end{equation*} Because $21=3 \\cdot 7$ and $4=2^2$ do not have common factors, we know this is simplified. A square root is not simplified if there is a factor inside the root that is a perfect square. Similarly, a cube root is not simplified if there is a factor inside that is a perfect cube. We use the factors of the value inside a root to determine if we can simplify it. ###### Example1.2.3 The square root $\\sqrt{126}$ is not simplified. A square root is the inverse operation of squaring numbers (for non-negative numbers) so that", "case, we have a 5 and a 13 that are common to both numerator and denominator $$\\frac{65}{390} = \\frac{5 \\cdot 13}{2 \\cdot 3 \\cdot 5 \\cdot 13}$$ $$\\frac{65}{390} = \\frac{\\cancel{5} \\cdot \\cancel{13} 1}{2 \\cdot 3 \\cdot \\cancel{5} \\cdot \\cancel{13}}$$ $$\\frac{65}{390} = \\frac{1}{6}$$ We can also use simple division. We know the GCF of 65 and 390 is 65. If we divide 65 by 65, we get 1. If we divide 390 by 65, we get 6. $$\\frac{65}{390} = \\frac{65 ÷ 65}{390 ÷ 65} = \\frac{1}{6}$$ Example 3: Simplify each fraction $$\\frac{330}{420}$$ • Factor the numerator and denominator into the product of prime numbers • 330 = 2 x 3 x 5 x 11 • 420 = 2 x 2 x 3 x 5 x 7 • Cancel all common factors between the numerator and denominator • In this case, we have a 2, 3, and 5 that are common to both numerator and denominator $$\\require{cancel}$$ $$\\frac{330}{420} = \\frac{2 \\cdot 3 \\cdot 5 \\cdot 11}{2 \\cdot 2 \\cdot 3 \\cdot 5 \\cdot 7}$$ $$\\frac{330}{420} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 11}{\\cancel{2} \\cdot 2 \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 7}$$ $$\\frac{330}{420} = \\frac{11}{14}$$ We can also use simple division. We know the GCF of 330 and 420 is 30. If we divide 330 by 30, we get 11. If", "$$\\displaystyle \\frac{2}{3}+\\frac{5}{4}-\\frac{5\\cdot 4}{3\\cdot 7}$$ After canceling out the common factors from the top and bottom $$= \\,\\,$$ $$\\displaystyle \\frac{2}{3}+\\frac{5}{4}-\\frac{20}{21}$$ Amplifying in order to get the common denominator 84 $$= \\,\\,$$ $$\\displaystyle \\frac{2}{3}\\cdot\\frac{28}{28}+\\frac{5}{4}\\cdot\\frac{21}{21}-\\frac{20}{21}\\cdot\\frac{4}{4}$$ We need to use the common denominator: 84 $$= \\,\\,$$ $$\\displaystyle \\frac{2\\cdot 28+5\\cdot 21-20\\cdot 4}{84}$$ Expanding each term in the numerator: $$2 \\times 28+5 \\times 21-20 \\times 4 = 56+105-80$$ $$= \\,\\,$$ $$\\displaystyle \\frac{56+105-80}{84}$$ Operating the terms in the numerator $$= \\,\\,$$ $$\\displaystyle \\frac{81}{84}$$ We can factor out 3 for both the numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{3\\cdot 27}{3\\cdot 28}$$ Now we cancel 3 out from the numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{27}{28}$$ which concludes the process of simplification. Example: Simplify calculator example Simplify the following: $$\\frac{x}{3} + \\frac{x}{4} - \\frac{x}{6}$$ Solution:We need to simplify the following expression: $$\\displaystyle \\frac{x}{3}+\\frac{x}{4}-\\frac{x}{6}$$. The following calculation is obtained: $$\\displaystyle \\frac{x}{3}+\\frac{x}{4}-\\frac{x}{6}$$ Grouping the terms with $$x$$ $$= \\,\\,$$ $$\\displaystyle \\left(\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{6}\\right)x$$ Simplifying the terms that were grouped with $$x$$ $$= \\,\\,$$ $$\\displaystyle \\frac{5}{12}x$$ which concludes the process of simplification. ### Example: Another simplification calculation Calculate $$\\left(\\frac{1}{3} \\times \\frac{6}{5} \\right)+ \\frac{2}{5}$$. Solution: We need to simplify the following expression: $$\\displaystyle \\frac{1}{3}\\cdot\\frac{6}{5}+\\frac{2}{5}$$. The following calculation is obtained: $$\\displaystyle \\frac{ 1}{ 3} \\cdot \\frac{ \\left(6\\right)}{ 5}+\\frac{2}{5}$$ By multiplying all the numerators and all the denominators: $$\\displaystyle\\frac{ 1}{ 3} \\times \\frac{ 6}{ 5}=", "exponents tells us when we divide two numbers or expressions in exponent form and the bases are the same, we can keep the base the same and subtract the exponent in the denominator away from the exponent in the numerator. Let's look at a few examples. Example 5: Simplify each $$\\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", "in ascending order, but every little bit helps. The denominator can also be expressed as a product of primes. $2310=10 \\cdot 231=2 \\cdot 5 \\cdot 7 \\cdot 33=2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11$ We can now cancel common factors. $\\frac{210}{2310}=\\frac{2 \\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11}=\\frac{\\not{2} \\cdot \\not{3} \\cdot \\not{5} \\cdot \\not{7}}{\\not{2} \\cdot \\not{8} \\cdot \\not{5} \\cdot \\not{7} \\cdot 11}=\\frac{1}{11}$ However, this approach is not the most efficient way to proceed, as multiplying numerators and denominators allows the products to grow to larger numbers, as in 210/2310. It is then a little bit harder to prime factor the larger numbers. A better approach is to factor the smaller numerators and denominators immediately, as follows. $\\frac{6}{231} \\cdot \\frac{35}{10}=\\frac{2 \\cdot 3}{3 \\cdot 7 \\cdot 11} \\cdot \\frac{5 \\cdot 7}{2 \\cdot 5}$ We could now multiply numerators and denominators, then cancel common factors, which would match identically the last computation in equation (5). However, we can also employ the following cancellation rule. Cancellation Rule When working with the product of two or more rational expressions, factor all numerators and denominators, then cancel. The cancellation rule is simple: cancel a factor “on the top” for an identical factor “on the bottom.” Speaking more technically, cancel any factor in any numerator for", "the x’s again. This is the most common factoring mistake I see students make, and it’s not limited to Algebra students. I’ve even seen Calculus student make this error. That usually makes me sob quite loudly. If you would like to keep me from crying, then you need to learn how to simplify polynomial fractions. It’s quite simple once you understand that terms that are added do not cancel out. Only factors that are multiplied together can cancel. Let’s start by looking at a fraction with numbers and no variables. $\\dfrac{210}{462}$ Can this be simplified? Of course. Most students will divide the top and bottom by 2, then by 3, and then by 7, as follows: $\\dfrac{210}{462}=\\dfrac{105}{231}=\\dfrac{35}{77}=\\dfrac{5}{11}$ This is correct. But why can you cancel out a 2 and a 3 and a 7? It’s because they are factors of the numerator and the denominator. Let’s do the same problem by completely factoring the top and bottom first: $\\dfrac{210}{462}=\\dfrac{2\\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 3 \\cdot 7 \\cdot 11}=\\dfrac{5}{11}$ When you write it out like this, you can see that the 2’s cancel, the 3’s cancel, and the 7’s cancel. And they cancel only because they are factors. Now let’s try to simplify some polynomial fractions. Start by factoring the numerator and denominator completely, then any like", "will try to cancel out the x’s again. This is the most common factoring mistake I see students make, and it’s not limited to Algebra students. I’ve even seen Calculus student make this error. That usually makes me sob quite loudly. If you would like to keep me from crying, then you need to learn how to simplify polynomial fractions. It’s quite simple once you understand that terms that are added do not cancel out. Only factors that are multiplied together can cancel. Let’s start by looking at a fraction with numbers and no variables. $\\dfrac{210}{462}$ Can this be simplified? Of course. Most students will divide the top and bottom by 2, then by 3, and then by 7, as follows: $\\dfrac{210}{462}=\\dfrac{105}{231}=\\dfrac{35}{77}=\\dfrac{5}{11}$ This is correct. But why can you cancel out a 2 and a 3 and a 7? It’s because they are factors of the numerator and the denominator. Let’s do the same problem by completely factoring the top and bottom first: $\\dfrac{210}{462}=\\dfrac{2\\cdot 3 \\cdot 5 \\cdot 7}{2 \\cdot 3 \\cdot 7 \\cdot 11}=\\dfrac{5}{11}$ When you write it out like this, you can see that the 2’s cancel, the 3’s cancel, and the 7’s cancel. And they cancel only because they are factors. Now let’s try to simplify some polynomial fractions. Start by factoring the numerator and", "from there $\\frac{400\\pi}{180} = \\frac{20 \\cdot 20\\pi}{20 \\cdot 9}$ ... cancel the common factor of 20 in the numerator and denominator.", "cancelling is required. 2. $$\\dfrac{2x+5}{2} \\rightarrow x+5$$ • A) 2 is not a factor of the numerator. • B) 2 should be subtracted from 5. $$\\text{A) ... denominator.}$$ $$\\text{A) 2 ... the numerator.}$$ Solution. 1. 3 is not a factor of numerator or denominator 2. 2 is not a factor of the numerator We summarize the procedure for reducing algebraic fractions as follows. #### To reduce an algebraic fraction. 1. Factor the numerator and the denominator. 2. Divide the numerator and denominator by any common factors. Reduce each fraction. 1. $$\\displaystyle \\dfrac{3x+12}{6x+24}$$ 2. $$\\displaystyle \\dfrac{27x^3-1}{9x^2-1}$$ Solution. 1. We first factor the numerator and denominator completely. \\begin{equation*} \\dfrac{3x+12}{6x+24} = \\dfrac{3(x+4)}{2 \\cdot 3(x+4)} \\end{equation*} Then we divide numerator and denominator by the common factors 3 and $$(x+4)\\text{.}$$ We must cancel the entire expression $$(x+4)$$ from numerator and denominator (we cannot cancel the $$x$$'s or the 4's separately!). \\begin{equation*} \\dfrac{3(x+4)}{2 \\cdot 3(x+4)} = \\dfrac{\\cancel{3}\\cancel{(x+4)}}{2 \\cdot \\cancel{3}\\cancel{(x+4)}}= \\dfrac{1}{2} \\end{equation*} All the factors are canceled from the numerator, so we replace them by 1, because any expression divided by itself is 1. 2. The numerator of the fraction is a difference of two cubes, and the denominator is a difference of two squares. We factor each to obtain \\begin{equation*} \\dfrac{27x^3-1}{9x^2-1} = \\dfrac{(3x-1)(9x^2+3x+1)}{(3x-1)(3x+1)} \\end{equation*} We cancel the factor $$(3x-1)$$ from top and bottom, to", "need. The LCD, $$36$$, has $$2$$ factors of $$2$$ and $$2$$ factors of $$3$$. Twelve has two factors of $$2$$, but only one of $$3$$ —so it is ‘missing‘ one $$3$$. We multiplied the numerator and denominator of $$\\frac{7}{12}$$ by $$3$$ to get an equivalent fraction with denominator $$36$$. Eighteen is missing one factor of $$2$$ —so you multiply the numerator and denominator $$\\frac{5}{18}$$ by $$2$$ to get an equivalent fraction with denominator $$36$$. We will apply this method as we subtract the fractions in the next example. Example $$\\PageIndex{8}$$: subtract Subtract: $$\\frac{7}{15} − \\frac{19}{24}$$. Solution Find the LCD. 15 is 'missing' three factors of 2 24 is 'missing' a factor of 5 Rewrite as equivalent fractions with the LCD. $$\\frac{7 \\cdot \\textcolor{red}{8}}{15 \\cdot \\textcolor{red}{8}} - \\frac{19 \\cdot \\textcolor{red}{5}}{24 \\cdot \\textcolor{red}{5}}$$ Simplify each numerator and denominator. $$\\frac{56}{120} - \\frac{95}{120}$$ Subtract. $$- \\frac{39}{120}$$ Rewrite showing the common factor of 3. $$- \\frac{13 \\cdot 3}{40 \\cdot 3}$$ Remove the common factor to simplify. $$- \\frac{13}{40}$$ Exercise $$\\PageIndex{15}$$ Subtract: $$\\frac{13}{24} − \\frac{17}{32}$$. $$\\frac{1}{96}$$ Exercise $$\\PageIndex{16}$$ Subtract: $$\\frac{21}{32} − \\frac{9}{28}$$. $$\\frac{75}{224}$$ Example $$\\PageIndex{9}$$: add Add: $$- \\frac{11}{30} + \\frac{23}{42}$$. Solution Find the LCD. Rewrite as equivalent fractions with the LCD. $$- \\frac{11 \\cdot \\textcolor{red}{7}}{30 \\cdot \\textcolor{red}{7}} + \\frac{23 \\cdot \\textcolor{red}{5}}{42 \\cdot \\textcolor{red}{5}}$$ Simplify each numerator and denominator. $$- \\frac{77}{210} + \\frac{115}{210}$$ Add.", "1. ## simplifying i need to know what this would be when simplified also i need to know how you did it thanks 21 --- 35 2. To learn how to simplify fractions, try here. 3. i know how to do questions like these im just stuck on this 1 i wanna also know if the answer will be just be 21 --- 35 or if it will be something like 10.5 4. Originally Posted by andyboy179 i know how to do questions like these im just stuck on this 1 i wanna also know if the answer will be just be 21 --- 35 or if it will be something like 10.5 You have to examine the numerator and the denominator if they contain a common factor. With your example you get: $\\dfrac{21}{35} = \\dfrac{7 \\cdot 3}{7 \\cdot 5} = \\dfrac11 \\cdot \\dfrac35 = \\dfrac35$ 5. Originally Posted by earboth You have to examine the numerator and the denominator if they contain a common factor. With your example you get: $\\dfrac{21}{35} = \\dfrac{7 \\cdot 3}{7 \\cdot 5} = \\dfrac11 \\cdot \\dfrac35 = \\dfrac35$ what did u do to get 7.3 and 7.5 6. Originally Posted by andyboy179 what did u do to get 7.3 and 7.5 I assume that you have heard about prime decomposition (?): Each", "grows larger with each successive term. Looking at the terms more closely, it turns out that the denominators tend to be products of many small primes, whereas the numerators are either primes or products of a few comparatively large primes. For example: $\\frac{9649}{2520} = \\frac{9649}{2^3 \\cdot 3^2 \\cdot 5 \\cdot 7} \\qquad \\textrm{and} \\qquad \\frac{18358381}{4084080} = \\frac{59 \\cdot 379 \\cdot 821}{2^4 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17}.$ To produce an integer, we need to cancel all the primes in the factorization of the denominator by matching primes in the numerator; given the pattern of these numbers, that looks like an unlikely coincidence. But there is reason for caution. Note the seventh term in the sequence, where the denominator has decreased from $$60$$ to $$20$$. To understand how that happens, we can run through the calculation of the term, which starts by summing the six previous terms. $\\frac{60}{60} + \\frac{120}{60} + \\frac{150}{60} + \\frac{170}{60} + \\frac{185}{60} + \\frac{197}{60} = \\frac{882}{60}.$ Then we calculate the mean, and add 1 to get the seventh term: $\\require{cancel}\\frac{882}{60} \\cdot \\frac{1}{6} = \\frac{882}{360} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot 7 \\cdot 7}{\\cancel{2} \\cdot 2 \\cdot 2 \\cdot \\cancel{3} \\cdot \\cancel{3} \\cdot 5} = \\frac{49}{20} + 1 = \\frac{69}{20}$ Cancelations reduce the numerator and denominator of the mean", "## Thinking Mathematically (6th Edition) $\\dfrac{35}{6}$ Convert each factor to an improper fraction to find: $=\\dfrac{3(3)+1}{3} \\cdot \\dfrac{4(1)+3}{4} \\\\=\\dfrac{10}{3} \\cdot \\dfrac{7}{4}$ Factor each numerator and denominator, when needed, and then cancel common factors to find: $\\\\\\require{cancel}=\\dfrac{2(5)}{3} \\cdot \\dfrac{7}{2(2)} \\\\=\\dfrac{\\cancel{2}(5)}{3} \\cdot \\dfrac{7}{\\cancel{2}(2)} \\\\=\\dfrac{5}{3} \\cdot \\dfrac{7}{2}$ Multiply the numerators together and the denominators together to find: $=\\dfrac{5(7)}{3(2)} \\\\=\\dfrac{35}{6}$", "unless the application represents a measurement where an approximation would be adequate. Simplification of numbers corresponds to finding a new representation of a number in a reduced form. For example, a rational number has many different representations of the form $p/q$ with $p \\in \\mathbb{Z}$ and $q \\in \\mathbb{N}$, but only one where $p$ and $q$ have no common factors. Canceling the common factors would be simplification, as would be simplifying a root or rationalizing a denominator. ##### Example2.1.1 The fraction $\\frac{126}{24}$ is not simplified. If we find the prime factorization of the numerator and denominator, we find \\begin{gather*} 126 = 6 \\cdot 21 = 2 \\cdot 3^2 \\cdot 7, \\\\ 24 = 3 \\cdot 8 = 2^3 \\cdot 3. \\end{gather*} The fraction simplifies by canceling all common factors: \\begin{equation*}\\frac{126}{24} = \\frac{2 \\cdot 3^2 \\cdot 7}{2^3 \\cdot 3} = \\frac{3 \\cdot 7}{2^2} = \\frac{21}{4}.\\end{equation*} ##### Example2.1.2 The square root $\\sqrt{126}$ is not simplified. A square root is the inverse operation of squaring numbers (for non-negative numbers) so that $\\sqrt{3^2} = 3$. Because $\\sqrt{a \\cdot b} = \\sqrt{a} \\cdot \\sqrt{b}$ (for $a,b \\ge 0$), we can simplify as \\begin{equation*}\\sqrt{126} = \\sqrt{2 \\cdot 3^2 \\cdot 7} = \\sqrt{9} \\cdot \\sqrt{14} = 3 \\sqrt{14}.\\end{equation*} ##### Example2.1.3 The expression $3e^{2\\ln(5)-1}$ represents a number. Because exponentials with base $e$ and the natural logarithm", "denominator is 21. Example 1: 1/5 as a decimal. Then you have: $$\\frac{20 \\cdot 10^{13}}{4.0075 \\cdot 10^7}$$ where you can now estimate the denominator as $4 \\cdot 10^7$ since you will not lose any precision, except if you are using more than $3$ sig figs. Uncategorized. Below is the answer to 1 divided by 15 with the repeating decimals. Both numbers must be integers between -2147483648 and 2147483647. So, in our case, input 1 as the numerator and 2 in the denominator box. We will look at some examples of equivalent fractions with missing numerators and denominators to see how this rule works. In order to simplify any fraction, our Fraction Simplifier will calculate the greatest common divisor (GCD), also known as the greatest common factor (GCF), or the highest common factor (HCF) of numerator and denominator that you entered. Help Solving for 'y' when 'y' is in numerator and denominator. Active 2 years, 9 months ago. Viewed 1k times 0. Exemple : Convertissez le rapport 2:4 en pourcentage : 2: 4 peut être écrit en 2/4 = 0,5; Multiplié 0.5 par 100, 0.5 × 100 = 50, le pourcentage du rapport 2: 4 est donc 50%. Hello, I need to create a calculated measure using members from two fact tables: a numerator from F_TABLE1 and denominator from", "\\cancel{h} \\cdot \\cancel{h}} {\\cancel{h} \\cdot \\cancel{h} \\cdot \\cancel{h} \\cdot h \\cdot h} \\\\ &= \\frac {1} {h \\cdot h} \\\\ & = \\frac {1} { h^{2}} \\end{align*} If we were to simplify the original expression using the quotient rule, we would have \\begin{align*} \\frac {h^{3}} {h^{5}} &= h^{3-5} \\\\ &= h^{-2} \\end{align*} Putting the answers together, we have $h^{-2}= \\frac {1} {h^{2}}$. This is true for any nonzero real number, or any variable representing a nonzero real number. A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa. $a^{-n} = \\frac {1} {a^{n}}$ and $a^{n} = \\frac {1} {a^{-n}}$ We have shown that the exponential expression $a^{n}$ is defined when $n$ is a natural number, $0$, or the negative of a natural number. That means that $a^{n}$ is defined for any integer $n$. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n$. ### THE NEGATIVE RULE OF EXPONENTS For any nonzero real number $a$ and natural number $n$, the negative rule of exponents states that $a^{-n} = \\frac {1} {a^{n}}$ #### Example 5 Write each of the following quotients with a single base. Do not simplify further. Write", "$18 = 2 \\cdot 3^2$. The least common denominator of the fractions is the LCM of the two numbers: $2^2 \\cdot 3^2 = 36$. Now we need to rewrite each fraction as an equivalent fraction with the LCD as the denominator. For the first fraction. 12 needs to be multiplied by a factor of 3 in order to change it into the LCD, so we multiply the numerator and the denominator by 3. $\\frac{4} {12} \\cdot \\frac{3} {3} = \\frac{12} {36}$ For the section fraction. 18 needs to be multiplied by a factor of 2 in order to change it into the LCD, so we multiply the numerator and the denominator by 2. $\\frac{5} {18} \\cdot \\frac{2} {2} = \\frac{10} {36}$ Once the denominators of the two fractions are the same we can add the numerators. $\\frac{12} {36} + \\frac{10} {36} = \\frac{22} {36}$ The answer can be reduced by canceling a common factor of 2. $\\frac{12} {36} + \\frac{10} {36} = \\frac{22} {36} = \\frac{11} {18}\\ \\text{Answer}$ Example 6 Perform the following operation and simplify. $\\frac{2} {x + 2} - \\frac{3} {2x - 5}$ Solution The denominators cannot be factored any further, so the LCD is just the product of the separate denominators. $\\text{LCD} = (x + 2) (2x - 5)$ The first fraction needs to be", "## [Soln] 2014 Singapore MO (Senior Rd 1) Problem 16 2014 SMO (Senior-Rd1) 16. Evaluate the sum $\\displaystyle\\frac{3! + 4!}{2(1! + 2!)} + \\frac{4! + 5!}{3(2! + 3!)} + \\dots + \\frac{12! + 13!}{11(10! + 11!)}$. Recall the definition of the factorial function: for non-negative integer $n$, $n! = \\begin{cases} 1 &\\text{ for } n = 0, \\\\ n \\cdot (n-1) \\cdot \\dots \\cdot 1 &\\text{ for } n \\geq 1. \\end{cases}$ When faced with a series of many terms, try to come up with an expression for the general $\\boldsymbol{n^{th}}$ term. In this question, the required sum is the sum of $\\displaystyle\\frac{(n+2)! + (n+3)!}{(n+1)[n! +(n+1)!]}$ for $n = 1, 2, \\dots, 10$. This looks like a term that can be simplified. Especially with fractions, a possible strategy is to pull out as many factors as possible in the hope that there can be some common factors in the numerator and denominator. These can be cancelled out. By pulling out common factors in the numerator: \\begin{aligned} (n+2)! + (n+3)! &= (n+2)! \\cdot [1 + (n + 3)] \\\\ &= (n+2)! \\cdot (n+4) \\end{aligned} and in the denominator: \\begin{aligned} (n+1)[n! + (n+1)!] &= (n+1) \\cdot n! \\cdot [1 + (n+1)] \\\\ &= (n+1) \\cdot n! \\cdot (n+2) \\\\ &= (n+2)! \\end{aligned} The original expression can be greatly simplified: $\\displaystyle\\frac{(n+2)!", "peaces in an accurate way. Note that you don't have to find the least common denominator, just any common denominator. So you can just multiply each numerator by all of the other denominators, and compare the results. (These would be the numerators corresponding to the denominator that would be the product of all of the separate denominators.) In your case, you would compare the numbers $$3\\cdot(3\\cdot 13\\cdot 38)=4446$$ $$2\\cdot(5\\cdot 13\\cdot 38)=4940$$ $$6\\cdot (5\\cdot 3\\cdot 38)=3420$$ $$23\\cdot (5\\cdot 3\\cdot 13)=4485$$ The smallest is the third, whic corresponds to the original fraction $\\frac{6}{13}$.", "\\begin{equation*} \\blert{\\dfrac{a \\cdot c}{b \\cdot c} = \\dfrac{a}{b}}~~~~~~\\text{if}~~~~~~b,c \\ne 0 \\end{equation*} Reduce $$~\\dfrac{8x^3y}{6x^2y^3}$$ Solution. The common factor for numerator and denominator is $$2x^2y\\text{.}$$ We factor $$2x^2y$$ from the numerator and denominator, then divide by the common factor. \\begin{equation*} \\dfrac{8x^3y}{6x^2y^3} = \\dfrac{4x \\cdot \\cancel{2x^2y}}{3y^2 \\cdot \\cancel{2x^2y}} = \\dfrac{4x}{3y^2} \\end{equation*} #### Caution8.2.4. When we cancel common factors, we are dividing. Because division is the inverse or opposite operation for multiplication, we can cancel common factors, but we cannot cancel common terms. Use your calculator to decide which calculation is correct. 1. $$\\displaystyle \\dfrac{12}{8} = \\dfrac{4 \\cdot 3}{4 \\cdot 2} \\rightarrow \\dfrac{3}{2}$$ 2. $$\\displaystyle \\dfrac{7}{6} = \\dfrac{4 + 3}{4 + 2} \\rightarrow \\dfrac{3}{2}$$ Solution. We can cancel the 4's in part (a), because they are factors of numerator and denominator. We cannot cancel the 4's in part (b), because they are terms. You can verify that $$\\dfrac{12}{8} = \\dfrac{3}{2}\\text{,}$$ but $$\\dfrac{7}{6} \\ne \\dfrac{3}{2}\\text{.}$$ Fill in the blanks. 1. An algebraic fraction is undefined when the • denominator • index • numerator is • even • negatove • positive • zero . 2. “Canceling” common factors uses the operation of • division • subtraction • squaring . 3. Terms are expressions that are • exponentiated • multiplied • reduced or • divided • expended • simplified • subtracted . 4. When we" ]

[ "Still have math questions? 4. Compare the fractions using $$< , >$$ , or $$=$$ . $$\\frac{4}{6} > \\frac{5}{10}$$", "# Write the fraction $\\large\\frac{9}{36}$ in simplest/lowest form. ( A ) $\\large\\frac{2}{3}$ ( B ) $\\large\\frac{2}{5}$ ( C ) $\\large\\frac{1}{3}$ ( D ) $\\large\\frac{1}{4}$", "$$\\frac{9}{20}$$ Answer in fraction form: $$\\fbox {9/20}$$ Answer in decimal form: $$\\fbox {0.45}$$ RELATED:", "into equal parts that show the fraction given. Question 6. $$\\frac{1}{4}$$ $$\\frac{1}{3}$$", "# How do you write your answer in simplest form: 7/4/5 divided by 1/1/4? Do you mean $7 \\frac{4}{5}$ and $1 \\frac{1}{4}$?", "## Prealgebra (7th Edition) $-\\frac{10}{27}$ The fraction $-\\frac{10}{27}$ is already in simplest form. Since no integers other than 1 go into both 10 and 27, we don't need to do anything!", "# [Triangle] Find sin 1),sin #,cos 1), and cos &. Write each answer as a fraction in simplest form. [Triangle] Find sin 1),sin #,cos 1), and cos &. Write each answer as a fraction in simplest form. You can still ask an expert for help • 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 • Math expert for every subject • Pay only if we can solve it averes8 $\\mathrm{sin}D=\\frac{36}{45}=0.8$ $\\mathrm{sin}E=\\frac{27}{45}=0.6$ $\\mathrm{cos}D=\\frac{27}{45}=0.6$ $\\mathrm{cos}E=\\frac{36}{45}=0.8$", "of like fractions obtained, we have: $$\\frac{8}{20}+\\frac{15}{20}+\\frac{10}{20}$$ $$=\\frac{8+15+10}{20}$$ $$=\\frac{33}{20}$$ Determine mathematic question If you're looking for a fun way to teach your kids math, try Decide math. It's a great way to engage them in the subject and help them learn while they're having fun. Obtain Help with Homework I am tracking my progress. Confidentiality Doing math problems is a great way to improve your math skills.", "set that up by setting the two fractions equal to each other: $\\frac{5}{2} = \\frac{6}{x}$ You can flip that so it's a little easier to solve for $$x$$ -- just remember to flip both sides! $\\frac{2}{5} = \\frac{x}{6}$ 8 years ago Can't find your answer? Make a FREE account and ask your own question, OR you can help others and earn volunteer hours!", "Question 6 Change 0.125 into a fraction. [show answer] $\\large~\\frac{1}{8}$", "## eddyg123 3 years ago 21/112 as a fraction in simplest form? $\\frac{21}{112}=\\frac{3\\times 7}{7\\times 16} = ...?$", "frac{f^{6} (-1) }{6!} (x+1)^2 + frac{f^{7} (-1) }{7!} (x+1)^3$$ For more context into the idea behind finding remainders, see the answer I made for the question I asked myself over here", "can proceed by expanding $\\frac{1-4x^2}{x^3+4x}$ into partial fractions. -", "## eddyg123 21/112 as a fraction in simplest form? one year ago one year ago $\\frac{21}{112}=\\frac{3\\times 7}{7\\times 16} = ...?$", "#### Fill in the blanks to make the statement true :The fraction $\\frac{17}{34}$ in simplest form is ______. The fraction $\\frac{17}{34}$ in simplest form is $\\frac{1}{2}$.", "# Math Help - Solving Fractional Equations 1. ## Solving Fractional Equations $\\frac{a+3}{3}+\\frac{a-3}{6}=5$ is the answer to this question 9? 2. Thats what i get $ \\frac{a}{3} + 1 + \\frac{a}{6} - \\frac{1}{2} = 5 $ $ \\frac{a}{3} + \\frac{a}{6} = \\frac{9}{2} $ $ \\frac{3a}{6} = \\frac{9}{2} $ $ a = 9 $", "among the fraction is $\\frac{2}{5}$.", "Simplifying this to a complex number, $$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", "# Math Help - write fraction 1. ## write fraction complete the sentence below by writing a fraction. 1/9 is half of ....... 2. $\\frac{1}{9} = \\frac{1}{2} \\times \\frac{2}{9}$", "fraction, such as $4\\frac{3}{5}$." ]

[ "$$\\displaystyle \\frac{2}{3}+\\frac{5}{4}-\\frac{5\\cdot 4}{3\\cdot 7}$$ After canceling out the common factors from the top and bottom $$= \\,\\,$$ $$\\displaystyle \\frac{2}{3}+\\frac{5}{4}-\\frac{20}{21}$$ Amplifying in order to get the common denominator 84 $$= \\,\\,$$ $$\\displaystyle \\frac{2}{3}\\cdot\\frac{28}{28}+\\frac{5}{4}\\cdot\\frac{21}{21}-\\frac{20}{21}\\cdot\\frac{4}{4}$$ We need to use the common denominator: 84 $$= \\,\\,$$ $$\\displaystyle \\frac{2\\cdot 28+5\\cdot 21-20\\cdot 4}{84}$$ Expanding each term in the numerator: $$2 \\times 28+5 \\times 21-20 \\times 4 = 56+105-80$$ $$= \\,\\,$$ $$\\displaystyle \\frac{56+105-80}{84}$$ Operating the terms in the numerator $$= \\,\\,$$ $$\\displaystyle \\frac{81}{84}$$ We can factor out 3 for both the numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{3\\cdot 27}{3\\cdot 28}$$ Now we cancel 3 out from the numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{27}{28}$$ which concludes the process of simplification. Example: Simplify calculator example Simplify the following: $$\\frac{x}{3} + \\frac{x}{4} - \\frac{x}{6}$$ Solution:We need to simplify the following expression: $$\\displaystyle \\frac{x}{3}+\\frac{x}{4}-\\frac{x}{6}$$. The following calculation is obtained: $$\\displaystyle \\frac{x}{3}+\\frac{x}{4}-\\frac{x}{6}$$ Grouping the terms with $$x$$ $$= \\,\\,$$ $$\\displaystyle \\left(\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{6}\\right)x$$ Simplifying the terms that were grouped with $$x$$ $$= \\,\\,$$ $$\\displaystyle \\frac{5}{12}x$$ which concludes the process of simplification. ### Example: Another simplification calculation Calculate $$\\left(\\frac{1}{3} \\times \\frac{6}{5} \\right)+ \\frac{2}{5}$$. Solution: We need to simplify the following expression: $$\\displaystyle \\frac{1}{3}\\cdot\\frac{6}{5}+\\frac{2}{5}$$. The following calculation is obtained: $$\\displaystyle \\frac{ 1}{ 3} \\cdot \\frac{ \\left(6\\right)}{ 5}+\\frac{2}{5}$$ By multiplying all the numerators and all the denominators: $$\\displaystyle\\frac{ 1}{ 3} \\times \\frac{ 6}{ 5}=", "the denominator: $$\\frac{\\frac{3}{10}\\cdot \\frac{60}{1}+ \\frac{6}{15}\\cdot \\frac{60}{1}}{\\frac{19}{20}\\cdot \\frac{60}{1}- \\frac{2}{3}\\cdot \\frac{60}{1}}$$ Simplify: $$\\frac{\\frac{3}{\\cancel{10}1}\\cdot \\frac{\\cancel{60}6}{1}+ \\frac{6}{\\cancel{15}1}\\cdot \\frac{\\cancel{60}4}{1}}{\\frac{19}{\\cancel{20}1}\\cdot \\frac{\\cancel{60}3}{1}- \\frac{2}{\\cancel{3}1}\\cdot \\frac{\\cancel{60}20}{1}}$$ $$\\frac{18 + 24}{57 - 40}=\\frac{42}{17}=2\\frac{8}{17}$$ #### Skills Check: Example #1 Simplify each. $$\\Large{\\frac{\\frac{6}{54}- \\frac{7}{84}}{\\frac{31}{42}- \\frac{2}{9}}}$$ A $$\\frac{3}{19}$$ B $$\\frac{5}{21}$$ C $$\\frac{7}{130}$$ D $$\\frac{4}{21}$$ E $$\\frac{9}{20}$$ Example #2 Simplify each. $$\\Large{\\frac{\\frac{4}{72}+ \\frac{33}{88}}{\\frac{5}{20}+ \\frac{10}{54}}}$$ A $$\\frac{18}{29}$$ B $$\\frac{17}{35}$$ C $$\\frac{93}{94}$$ D $$\\frac{57}{61}$$ E $$\\frac{101}{103}$$ Example #3 Simplify each. $$\\Large{\\frac{\\frac{37}{63}- \\frac{28}{49}}{\\frac{26}{91}- \\frac{13}{70}}}$$ A $$\\frac{4}{15}$$ B $$\\frac{7}{29}$$ C $$\\frac{8}{17}$$ D $$\\frac{10}{63}$$ E $$\\frac{11}{30}$$", "numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{1}{3}+\\frac{2\\cdot 2}{2\\cdot 3}$$ Now we cancel 2 out from the numerator and denominator. $$= \\,\\,$$ $$\\displaystyle \\frac{1}{3}+\\frac{2}{3}$$ Finding a common denominator: 3 $$= \\,\\,$$ $$\\displaystyle \\frac{1+2}{3}$$ Adding up the terms in the numerator $$= \\,\\,$$ $$\\displaystyle \\frac{3}{3}$$ Now we cancel 3 out from the numerator and denominator $$= \\,\\,$$ $$\\displaystyle 1$$ ### Example: Calculating the sum of expression Calculate the following: $$2 + \\frac{5}{4} - \\frac{7}{6}$$ Solution: We need to calculate and simplify the following expression: $$\\displaystyle 2+\\frac{5}{4}-\\frac{7}{6}$$. The following calculation is obtained: $$\\displaystyle 2+\\frac{5}{4}-\\frac{7}{6}$$ Amplifying in order to get the common denominator 12 $$= \\,\\,$$ $$\\displaystyle 2\\cdot\\frac{12}{12}+\\frac{5}{4}\\cdot\\frac{3}{3}-\\frac{7}{6}\\cdot\\frac{2}{2}$$ We use the common denominator: 12 $$= \\,\\,$$ $$\\displaystyle \\frac{2\\cdot 12+5\\cdot 3-7\\cdot 2}{12}$$ Expanding each term: $$2 \\times 12+5 \\times 3-7 \\times 2 = 24+15-14$$ $$= \\,\\,$$ $$\\displaystyle \\frac{24+15-14}{12}$$ Adding up the terms in the numerator $$= \\,\\,$$ $$\\displaystyle \\frac{25}{12}$$ which concludes the calculation. Calculate $$\\left(\\frac{4}{3} \\times \\frac{6}{5} \\right)+ \\frac{1}{5}$$. Solution: We need to calculate and simplify the following expression: $$\\displaystyle \\left(\\frac{4}{3}\\cdot\\frac{6}{5}\\right)+\\frac{1}{5}$$. The following calculation is obtained: $$\\displaystyle \\frac{4}{3}\\cdot\\frac{6}{5}+\\frac{1}{5}$$ We multiply all the numerators and all the denominators together as in $$\\displaystyle\\frac{ 4}{ 3} \\times \\frac{ 6}{ 5}= \\frac{ 4 \\times 6}{ 3 \\times 5}$$ $$= \\,\\,$$ $$\\displaystyle \\frac{4\\cdot 6}{3\\cdot 5}+\\frac{1}{5}$$ Factoring the term $$\\displaystyle 3$$ in the numerator and denominator in $$\\displaystyle \\frac{", "# 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 $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots = \\frac{25}{12}$.", "as √24/√7, and then simplify the numerator to get 2√6/√7. Finally, I rationalize the denominator by multiplying by √7/√7 to get 2√42/7. The result is the same as yours, but I think this approach makes the reason for each step clearer. At no point did I split a radical into two parts, neither of which was a perfect square. Here is the work by my preferred method: $$\\frac{\\sqrt{120}}{\\sqrt{35}} = \\sqrt{\\frac{120}{35}} = \\sqrt{\\frac{24\\cdot 5}{7\\cdot 5}} = \\sqrt{\\frac{24}{7}} = \\frac{\\sqrt{24}}{\\sqrt{7}} =\\\\ \\frac{\\sqrt{4\\cdot 6}}{\\sqrt{7}} = \\frac{2\\sqrt{6}}{\\sqrt{7}} = \\frac{2\\sqrt{6}\\sqrt{7}}{\\sqrt{7}\\sqrt{7}} = \\frac{2\\sqrt{42}}{7}$$ If I didn’t combine into a single radical first, I would simplify each radical by itself (working with larger numbers) and then simplify the fraction: $$\\frac{\\sqrt{120}}{\\sqrt{35}} = \\frac{\\sqrt{4\\cdot 30}}{\\sqrt{35}} = \\frac{2\\sqrt{30}}{\\sqrt{35}} = \\frac{2\\sqrt{6\\cdot 5}}{\\sqrt{7\\cdot 5}} =\\\\ \\frac{2\\sqrt{6}\\sqrt{5}}{\\sqrt{7}\\sqrt{5}} = \\frac{2\\sqrt{6}}{\\sqrt{7}} = \\frac{2\\sqrt{6}\\sqrt{7}}{\\sqrt{7}\\sqrt{7}} = \\frac{2\\sqrt{42}}{7}$$ Here it was less obvious that we had to factor out the $$\\sqrt{5}$$. But I still did less random factoring than Jordan did. ### Why not factor more or less than he did? The reason we don’t break up ∛18 into ∛6 ∛3 (or ∛18 into ∛9 ∛2) is that it doesn’t move toward the goal, which is a single small radical; a product of two radicals is not considered simple. The reason we do break up √120 and √35 is that some of the resulting parts will", "denominator, we find \\begin{gather*} 126 = 6 \\cdot 21 = 2 \\cdot 3^2 \\cdot 7, \\\\ 24 = 3 \\cdot 8 = 2^3 \\cdot 3. \\end{gather*} The fraction simplifies by canceling all common factors: \\begin{equation*} \\frac{126}{24} = \\frac{2 \\cdot 3^2 \\cdot 7}{2^3 \\cdot 3} = \\frac{3 \\cdot 7}{2^2} = \\frac{21}{4}. \\end{equation*} In practice, we don't always have to find the prime factorization. Instead, we can find one common factor at a time until no common factors remain. For example, since 126 and 24 are both even, we would write \\begin{equation*} \\frac{126}{24} = \\frac{63}{12}. \\end{equation*} Then, we look at 63 and 12 and recognize that they are both divisible by 3, allowing us to rewrite the fraction as \\begin{equation*} \\frac{126}{24} = \\frac{63}{12} = \\frac{21}{4}. \\end{equation*} Because $21=3 \\cdot 7$ and $4=2^2$ do not have common factors, we know this is simplified. A square root is not simplified if there is a factor inside the root that is a perfect square. Similarly, a cube root is not simplified if there is a factor inside that is a perfect cube. We use the factors of the value inside a root to determine if we can simplify it. ###### Example1.2.3 The square root $\\sqrt{126}$ is not simplified. A square root is the inverse operation of squaring numbers (for non-negative numbers) so that", "# How do you simplify\\frac { 2} { 9} + \\frac { 12} { 7}? Mar 8, 2018 $\\frac{122}{63}$ #### Explanation: First we should get a common denominator. since the common factor of 9 and 7 is 63 ( 9 x 7 = 63) we will write 63 as the denominator. to get the common denominator, multiply the first fraction by the denominator of the other fraction and multiply the second fraction from the denominator of the first fraction. then our sum will look like: $\\frac{7 \\cdot 2 + 9 \\cdot 12}{63}$ $\\to$ $\\frac{14 + 108}{63}$ $\\to$ $\\frac{122}{63}$", "# 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 add 3\\frac { 7} { 10} + 4\\frac { 1} { 15} + 2\\frac { 2} { 13}? Jan 5, 2017 $9 \\setminus \\frac{359}{390}$ #### Explanation: Okay, so first you put all the fractions into a Common Denominator what this means is that all the bottom numbers have to be equal. So let's add the first two numbers first $3 \\setminus \\frac{7}{10} + 4 \\setminus \\frac{1}{15}$ What is a common denominator for $15$ and $10$: the answer is $30$. A brief overview of a common denominator: to find the common denominator list the multiples of $15$ and $10$. For 15 it would be: $15 \\left(15 \\cdot 1 = 15\\right)$ $30 \\left(15 \\cdot 2 = 30\\right)$ Therefore, we could conclude that since the multiples of $10$ are pretty simple. $10 \\cdot 1 = 10$ $10 \\cdot 2 = 20$ $10 \\cdot 3 = 30$ We found a common denominator: $30$! So next we multiply each number the top and the bottom the same so if we multiple $10$ by $3$ to equal $30$; we do the same for the top number so $7 \\cdot 3 = 21$ Same goes for the other number we multiplied $15$ by $2$ to find $30$ and we do the same for the top $1 \\cdot 2 = 2$ But", "case, we have a 5 and a 13 that are common to both numerator and denominator $$\\frac{65}{390} = \\frac{5 \\cdot 13}{2 \\cdot 3 \\cdot 5 \\cdot 13}$$ $$\\frac{65}{390} = \\frac{\\cancel{5} \\cdot \\cancel{13} 1}{2 \\cdot 3 \\cdot \\cancel{5} \\cdot \\cancel{13}}$$ $$\\frac{65}{390} = \\frac{1}{6}$$ We can also use simple division. We know the GCF of 65 and 390 is 65. If we divide 65 by 65, we get 1. If we divide 390 by 65, we get 6. $$\\frac{65}{390} = \\frac{65 ÷ 65}{390 ÷ 65} = \\frac{1}{6}$$ Example 3: Simplify each fraction $$\\frac{330}{420}$$ • Factor the numerator and denominator into the product of prime numbers • 330 = 2 x 3 x 5 x 11 • 420 = 2 x 2 x 3 x 5 x 7 • Cancel all common factors between the numerator and denominator • In this case, we have a 2, 3, and 5 that are common to both numerator and denominator $$\\require{cancel}$$ $$\\frac{330}{420} = \\frac{2 \\cdot 3 \\cdot 5 \\cdot 11}{2 \\cdot 2 \\cdot 3 \\cdot 5 \\cdot 7}$$ $$\\frac{330}{420} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 11}{\\cancel{2} \\cdot 2 \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 7}$$ $$\\frac{330}{420} = \\frac{11}{14}$$ We can also use simple division. We know the GCF of 330 and 420 is 30. If we divide 330 by 30, we get 11. If", "# How do you calculate 18/12 - 7/9? Jan 21, 2015 The answer is $\\frac{13}{18}$. Here's how to calculate it: You'll notice that $\\frac{18}{12}$ and $\\frac{7}{9}$ have different denominators. In order to subtract these fractions, we need to find equivalent fractions ( i.e. fractions that are the same as our originals) that have the same denominator as one another. To do this, let's find the least common denominator, or the smallest number that can be divided by both denominators. One way to do this to write out the multiples of both denominators until a number repeats in the the list: Multiples of 12: 12, 24, 36 , 48, 60 Multiples of 9: 9, 18, 27, 36 Since 36 appears in both lists, we can make both denominators 36 to create equivalent fractions. Since $12 \\cdot 3 = 36$, we can multiply $\\frac{18}{12} \\cdot \\frac{3}{3}$ (since $\\frac{3}{3}$ = 1) ... $\\frac{18}{12} \\cdot \\frac{3}{3} = \\frac{54}{36}$ And since $9 \\cdot 4 = 36$, we can multiply $\\frac{7}{9} \\cdot \\frac{4}{4}$ ... $\\frac{7}{9} \\cdot \\frac{4}{4} = \\frac{28}{36}$ Now, we have equivalent fractions with a denominator of 36 and can subtract: $\\frac{54}{36} - \\frac{28}{36} = \\frac{26}{36}$ And then divide the numerator and the denominator by 2 to simplify: $\\frac{26}{36} = \\frac{13}{18}$", "# 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 $\\frac{5}{3}$ $\\frac{5}{6} - \\frac{5 \\cdot 2}{3 \\cdot 2} = s$ $\\frac{5}{6} - \\frac{10}{6} = s$ $- \\frac{5}{6} = s$", "$D=\\{x| x \\in R \\} \\ and \\ R=\\{y | y>0, \\ y \\in R \\}$ #### Example A For the following tables of values that represent exponential functions, determine the common ratio: i) $& x \\quad 0 \\quad 1 \\quad 2 \\quad 3 \\quad \\ 4 \\quad \\cdots\\\\& y \\quad 1 \\quad 2 \\quad 4 \\quad 8 \\quad 16 \\quad \\cdots$ ii) $& x \\quad \\ 0 \\qquad 1 \\quad \\ 2 \\qquad 3 \\qquad \\ 4 \\quad \\ \\cdots\\\\& y \\quad 100 \\quad 50 \\quad 25 \\quad 12.5 \\quad 6.25 \\quad \\cdots$ Solutions: i) The common ratio is a constant that is determined by $r=\\frac{t_{n+1}}{t_n}$ . $& r=\\frac{t_{n+1}}{t_n}=\\frac{2}{1}=2\\\\& r=\\frac{t_{n+1}}{t_n}=\\frac{4}{2}=2\\\\& r=\\frac{t_{n+1}}{t_n}=\\frac{8}{4}=2\\\\& r=\\frac{t_{n+1}}{t_n}=\\frac{16}{8}=2\\\\& \\boxed{\\text{The common ratio is} \\ 2.}$ ii) The common ratio is a constant that is determined by $r=\\frac{t_{n+1}}{t_n}$ . $& r=\\frac{t_{n+1}}{t_n}\\\\& r=\\frac{50}{100}=\\frac{1}{2}\\\\& r=\\frac{25}{50}=\\frac{1}{2}\\\\& r=\\frac{12.5}{25}=\\frac{1}{2}\\\\& r=\\frac{6.25}{12}=\\frac{1}{2}\\\\& \\boxed{\\text{The common ratio is } \\frac{1}{2}.}$ #### Example B Using the exponential function $\\boxed{f(x)=3^x}$ , determine the value of each of the following: i) $f(2)$ ii) $f(3)$ iii) $f(0)$ iv) $f(4)$ v) $f(-2)$ Solutions: $f(x)=3^x$ is another way to express $y=3^x$ . To determine the value of the function for the given values, replace the exponent with that value and evaluate the expression. i) $& f(x)=3^x\\\\& f({\\color{red}2})=3^{\\color{red}2}\\\\& f(2)={\\color{red}9}\\\\& \\boxed{f(2)=9}$ ii) $& f(x)=3^x\\\\& f({\\color{red}3})=3^{\\color{red}3}\\\\& f(3)={\\color{red}27}\\\\& \\boxed{f(3)=27}$ iii) $& f(x)=3^x\\\\& f({\\color{red}0})=3^{\\color{red}0}\\\\& f(0)={\\color{red}1}\\\\&", "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, \\,", "peaces in an accurate way. Note that you don't have to find the least common denominator, just any common denominator. So you can just multiply each numerator by all of the other denominators, and compare the results. (These would be the numerators corresponding to the denominator that would be the product of all of the separate denominators.) In your case, you would compare the numbers $$3\\cdot(3\\cdot 13\\cdot 38)=4446$$ $$2\\cdot(5\\cdot 13\\cdot 38)=4940$$ $$6\\cdot (5\\cdot 3\\cdot 38)=3420$$ $$23\\cdot (5\\cdot 3\\cdot 13)=4485$$ The smallest is the third, whic corresponds to the original fraction $\\frac{6}{13}$.", "find a common denominator between the two. If one is not immediately obvious, you can always find one by multiplying the two denominators (the bottom numbers) together. In this case, $8 \\cdot 5 = 40$ so a common denominator is $40$. multiply the two fractions by a modified version of $1$ To get the denominator of $\\frac{2}{5}$ to $40$, the fraction should be multiplied by $\\frac{8}{8}$ $\\frac{2}{5} \\cdot \\frac{8}{8} = \\frac{16}{40}$ To get the denominator of $\\frac{5}{8}$ to $40$, the fraction should be multiplied by $\\frac{5}{5}$. $\\frac{5}{8} \\cdot \\frac{5}{5} = \\frac{25}{40}$ $\\frac{16}{40} \\ne \\frac{25}{40}$ So, $\\frac{2}{5} \\ne \\frac{5}{8}$ 1 ## What is the lowest common multiple for 8, 12, and 15? The Lonely Donut Featured 2 weeks ago $120$ #### Explanation: The easiest way to do this is to find the prime factorization of each number first. $8 = 4 \\cdot 2$ $8 = 2 \\cdot 2 \\cdot 2$ $8 = {2}^{3}$ $12 = 6 \\cdot 2$ $12 = 3 \\cdot 2 \\cdot 2$ $12 = 3 \\cdot {2}^{2}$ $15 = 5 \\cdot 3$ Now that we have reduced all numbers to their prime factorizations, we will look for the factors with the highest exponent and multiply them all together to get our LCM. There is ${2}^{3}$ from $8$, ${3}^{1}$ from $12$ and $15$, and ${5}^{1}$ from $15$.", "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?", "+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", "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:", "# Question 357c1 Apr 3, 2017 \"\"_32 C_8 = 10,518,300 combinations #### Explanation: The form of a combination is \"\"_nC_r, where $n$ = the total number and $r =$the subset. \"\"_32 C_8 = (n!)/(r!(n-r)!) = (32!)/(8! (32-8)!) = (32!)/(8!*24!) =(32*31*30*29*28*27*26*25*24!)/(8*7*6*5*4*3*2*1*24!) The 24! cancels: $= \\frac{32 \\cdot 31 \\cdot 30 \\cdot 29 \\cdot 28 \\cdot 27 \\cdot 26 \\cdot 25}{8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}$ Factor: $= \\frac{8 \\cdot 4 \\cdot 31 \\cdot 5 \\cdot 6 \\cdot 29 \\cdot 7 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 9 \\cdot 26 \\cdot 25}{8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2}$ Cancel numbers common in the numerator & denominator: $= 31 \\cdot 29 \\cdot 2 \\cdot 9 \\cdot 26 \\cdot 25$ $= 10 , 518 , 300$ combinations If you have a TI-83, 84 calculator, you can type $32$ MATH $\\to P R B \\text{ \"3 \" } 8$ ENTER Apr 3, 2017 The result is 10 518 300 ways. #### Explanation: To calculate the number of ways that groups of eight can be chosen from a larger group of 32, we use the formula \"\"_nC_r = (n!)/((r!)(n-r)!) So, in this case \"\"_32C_8 = (32!)/(8!*24!) = (32xx31xx30xx29xx28xx27xx26xx25)/(8xx7xx6xx5xx4xx3xx2xx1) = 10 518 300 By the way, I believe the" ]

[ "peaces in an accurate way. Note that you don't have to find the least common denominator, just any common denominator. So you can just multiply each numerator by all of the other denominators, and compare the results. (These would be the numerators corresponding to the denominator that would be the product of all of the separate denominators.) In your case, you would compare the numbers $$3\\cdot(3\\cdot 13\\cdot 38)=4446$$ $$2\\cdot(5\\cdot 13\\cdot 38)=4940$$ $$6\\cdot (5\\cdot 3\\cdot 38)=3420$$ $$23\\cdot (5\\cdot 3\\cdot 13)=4485$$ The smallest is the third, whic corresponds to the original fraction $\\frac{6}{13}$.", "So we find: $\\sqrt{270} = \\sqrt{9 \\cdot 30} = \\sqrt{9} \\sqrt{30} = 3 \\sqrt{30}$ This is the simplest form of the principal square root. $3 \\sqrt{30}$, like $\\sqrt{30}$ is an irrational number. We can calculate approximations to $\\sqrt{30}$ using its continued fraction. Note that $30 = 5 \\cdot 6$ is of the form $n \\left(n + 1\\right)$. Hence its square root has a short regular repeating pattern: sqrt(30) = [5;bar(2,10)] = 5+1/(2+1/(10+1/(2+1/(10+1/(2+1/(10+...)))))) [[ In general sqrt(n(n+1)) = [n;bar(2,2n)] ]] We can get decent approximations for $\\sqrt{30}$ by truncating just before one of the $10$'s as follows: $\\sqrt{30} \\approx 5 + \\frac{1}{2} = \\frac{11}{2}$ $\\sqrt{30} \\approx 5 + \\frac{1}{2 + \\frac{1}{10 + \\frac{1}{2}}} = \\frac{241}{44}$ $\\sqrt{30} \\approx 5 + \\frac{1}{2 + \\frac{1}{10 + \\frac{1}{2 + \\frac{1}{10 + \\frac{1}{2}}}}} = \\frac{5291}{966}$ Let's stop there and use this to give us an approximation for $\\sqrt{270}$... $\\sqrt{270} = 3 \\sqrt{30} \\approx 3 \\cdot \\frac{5291}{966} = \\frac{15873}{966} \\approx 16.431677$", "the denominator: $$\\frac{\\frac{3}{10}\\cdot \\frac{60}{1}+ \\frac{6}{15}\\cdot \\frac{60}{1}}{\\frac{19}{20}\\cdot \\frac{60}{1}- \\frac{2}{3}\\cdot \\frac{60}{1}}$$ Simplify: $$\\frac{\\frac{3}{\\cancel{10}1}\\cdot \\frac{\\cancel{60}6}{1}+ \\frac{6}{\\cancel{15}1}\\cdot \\frac{\\cancel{60}4}{1}}{\\frac{19}{\\cancel{20}1}\\cdot \\frac{\\cancel{60}3}{1}- \\frac{2}{\\cancel{3}1}\\cdot \\frac{\\cancel{60}20}{1}}$$ $$\\frac{18 + 24}{57 - 40}=\\frac{42}{17}=2\\frac{8}{17}$$ #### Skills Check: Example #1 Simplify each. $$\\Large{\\frac{\\frac{6}{54}- \\frac{7}{84}}{\\frac{31}{42}- \\frac{2}{9}}}$$ A $$\\frac{3}{19}$$ B $$\\frac{5}{21}$$ C $$\\frac{7}{130}$$ D $$\\frac{4}{21}$$ E $$\\frac{9}{20}$$ Example #2 Simplify each. $$\\Large{\\frac{\\frac{4}{72}+ \\frac{33}{88}}{\\frac{5}{20}+ \\frac{10}{54}}}$$ A $$\\frac{18}{29}$$ B $$\\frac{17}{35}$$ C $$\\frac{93}{94}$$ D $$\\frac{57}{61}$$ E $$\\frac{101}{103}$$ Example #3 Simplify each. $$\\Large{\\frac{\\frac{37}{63}- \\frac{28}{49}}{\\frac{26}{91}- \\frac{13}{70}}}$$ A $$\\frac{4}{15}$$ B $$\\frac{7}{29}$$ C $$\\frac{8}{17}$$ D $$\\frac{10}{63}$$ E $$\\frac{11}{30}$$", "+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", "from there $\\frac{400\\pi}{180} = \\frac{20 \\cdot 20\\pi}{20 \\cdot 9}$ ... cancel the common factor of 20 in the numerator and denominator.", "22. $$−19 \\cdot \\left( − \\frac{1}{19} \\right) = 1$$ 23. $$\\frac{−16}{11} \\cdot 1 = \\frac{−16}{11}$$ 24. $$\\frac{−7}{6} \\cdot 1 = \\frac{−7}{6}$$ 25. $$− \\frac{4}{1} \\cdot \\left( − \\frac{1}{4} \\right) = 1$$ 26. $$− \\frac{9}{10} \\cdot \\left( − \\frac{10}{9} \\right) = 1$$ 27. $$\\frac{8}{1} \\cdot 1 = \\frac{8}{1}$$ 28. $$\\frac{13}{15} \\cdot 1 = \\frac{13}{15}$$ 29. $$14 \\cdot \\frac{1}{14} = 1$$ 30. $$4 \\cdot \\frac{1}{4} = 1$$ 31. $$\\frac{13}{8} \\cdot 1 = \\frac{13}{8}$$ 32. $$\\frac{1}{13} \\cdot 1 = \\frac{1}{13}$$ In Exercises 33-56, divide the fractions, and simplify your result. 33. $$\\frac{8}{23} \\div \\frac{−6}{11}$$ 34. $$\\frac{−10}{21} \\div \\frac{−6}{5}$$ 35. $$\\frac{18}{19} \\div \\frac{−16}{23}$$ 36. $$\\frac{13}{10} \\div \\frac{17}{18}$$ 37. $$\\frac{4}{21} \\div \\frac{−6}{5}$$ 38. $$\\frac{2}{9} \\div \\frac{−12}{19}$$ 39. $$\\frac{−1}{9} \\div \\frac{8}{3}$$ 40. $$\\frac{1}{2} \\div \\frac{−15}{8}$$ 41. $$\\frac{−21}{11} \\div \\frac{3}{10}$$ 42. $$\\frac{7}{24} \\div \\frac{−23}{2}$$ 43. $$\\frac{−12}{7} \\div \\frac{2}{3}$$ 44. $$\\frac{−9}{16} \\div \\frac{6}{7}$$ 45. $$\\frac{2}{19} \\div \\frac{24}{23}$$ 46. $$\\frac{7}{3} \\div \\frac{−10}{21}$$ 47. $$\\frac{−9}{5} \\div \\frac{−24}{19}$$ 48. $$\\frac{14}{17} \\div \\frac{−22}{21}$$ 49. $$\\frac{18}{11} \\div \\frac{14}{9}$$ 50. $$\\frac{5}{6} \\div \\frac{20}{19}$$ 51. $$\\frac{13}{18} \\div \\frac{4}{9}$$ 52. $$\\frac{−3}{2} \\div \\frac{−7}{12}$$ 53. $$\\frac{11}{2} \\div \\frac{−21}{10}$$ 54. $$\\frac{−9}{2} \\div \\frac{−13}{22}$$ 55. $$\\frac{3}{10} \\div \\frac{12}{5}$$ 56. $$\\frac{−22}{7} \\div \\frac{−18}{17}$$ In Exercises 57-68, divide the fractions, and simplify your result. 57. $$\\frac{20}{17} \\div 5$$ 58. $$\\frac{21}{8} \\div 7$$ 59. $$−7 \\div \\frac{21}{20}$$ 60. $$−3 \\div \\frac{12}{17}$$ 61. $$\\frac{8}{21} \\div 2$$ 62. $$\\frac{−3}{4} \\div (−6)$$ 63.", "the XMC1100 PDIV and STEP are 10 bit wide so they have values from 0 to 1023. Rearranging the above we see that we must find a good approximation $\\frac {STEP} {(PDIV+1)} \\approx \\frac {(DCQT+1) baudrate \\cdot 1024} { f_{sysclock} } \\tag{2}$ with values of PDIV and STEP between 0 and 1023. #### Example: 38400 baud from a 32MHz clock and 16 times oversampling We enter the values in (2) to find $$\\frac {STEP} {(PDIV+1)} = \\frac {16 \\cdot 38400 \\cdot 1024} { 32000000 } = \\frac {629145600} { 32000000 } = 19.6608$$ Working by hand we would now factor out common factors from numerator and denominator, but for computer implementations that is unnecessary extra work. As we shall see the algorithm works as well without that step. Now we calculate the signed integer division with remainder using the rounded value 20 as quotient. $$\\frac {16 \\cdot 38400 \\cdot 1024} { 32000000 } = 20 - \\frac {10854400} { 32000000 } = 20 - \\frac {1} { \\frac { 32000000 } {10854400} } \\tag{3}$$ Rewrite the denominator in the last fraction by carrying out the division with remainder $$\\frac { 32000000 } {10854400} = 3 - \\frac {563200}{ 10854400 } \\approx 2.948 \\approx 3$$ We enter this value 3 in the previous formula (3) to get the approximation", "# 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: $\\frac{2}{3 \\sqrt{63}}$ $\\therefore = \\frac{2}{3 \\sqrt{3 \\cdot 3 \\cdot 7}}$ $\\therefore \\sqrt{3} \\cdot \\sqrt{3} = 3$ $\\therefore = \\frac{2}{3 \\cdot 3 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}$ $\\therefore \\sqrt{7} \\times \\sqrt{7} = 7$ $\\therefore = \\frac{2 \\sqrt{7}}{9 \\times 7}$ $\\therefore = \\frac{2 \\sqrt{7}}{63}$", "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 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 $\\frac{35}{60} + \\frac{48}{60}$ $= \\frac{35 + 48}{60}$ $= \\frac{83}{60}$.", "numerator, but you could just as well reverse the comparison. Thus, in $\\frac{123}{369}$, we can notice (along the lines of the reader's comment) that each digit in the numerator is one-third as large as the corresponding digit in the denominator. That makes the numerator itself one-third of the denominator—so the fraction equals $\\frac{1}{3}$. Those two ways of looking at $\\frac{123}{369}$ could be expressed respectively as $\\frac{123}{369}=\\frac{123}{300+60+9} = \\frac{123}{3\\cdot100+3\\cdot20+3\\cdot3} = \\frac{123}{3\\cdot(100+20+3)} = \\frac{123}{3\\cdot123} = \\frac{1}{3}$ and $\\frac{123}{369} = \\frac{100+20+3}{369} = \\frac{\\frac{1}{3}300 + \\frac{1}{3}{60}+\\frac{1}{3}{9}}{369} = \\frac{\\frac{1}{3}(300+60+9)}{369} = \\frac{\\frac{1}{3}\\cdot369}{369} = \\frac{1}{3}$. One last example: In $\\frac{462}{693}$, each digit in the numerator is two-thirds of the corresponding digit in the denominator. That makes the numerator itself two-thirds of the denominator—so the fraction equals $\\frac{2}{3}$. When every digit in the numerator is a constant multiple of its corresponding digit in the denominator, that remains true even if the digits are permuted, as long as the digits in the numerator and the digits in the denominator are permuted in the same way. Thus $\\frac{462}{693} = \\frac{426}{639} = \\frac{246}{369}$, etc. One way to understand Johnny's cheeky response is to observe that he has permuted digits in the two given fractions in such a way as to make them manifestly identical. Exercise: $\\frac{66336363}{88448484} - \\frac{12221}{48884}$ = ? *** The principles at work in this puzzle might", "# How do you add 3\\frac { 7} { 10} + 4\\frac { 1} { 15} + 2\\frac { 2} { 13}? Jan 5, 2017 $9 \\setminus \\frac{359}{390}$ #### Explanation: Okay, so first you put all the fractions into a Common Denominator what this means is that all the bottom numbers have to be equal. So let's add the first two numbers first $3 \\setminus \\frac{7}{10} + 4 \\setminus \\frac{1}{15}$ What is a common denominator for $15$ and $10$: the answer is $30$. A brief overview of a common denominator: to find the common denominator list the multiples of $15$ and $10$. For 15 it would be: $15 \\left(15 \\cdot 1 = 15\\right)$ $30 \\left(15 \\cdot 2 = 30\\right)$ Therefore, we could conclude that since the multiples of $10$ are pretty simple. $10 \\cdot 1 = 10$ $10 \\cdot 2 = 20$ $10 \\cdot 3 = 30$ We found a common denominator: $30$! So next we multiply each number the top and the bottom the same so if we multiple $10$ by $3$ to equal $30$; we do the same for the top number so $7 \\cdot 3 = 21$ Same goes for the other number we multiplied $15$ by $2$ to find $30$ and we do the same for the top $1 \\cdot 2 = 2$ But", "# How do you calculate 18/12 - 7/9? Jan 21, 2015 The answer is $\\frac{13}{18}$. Here's how to calculate it: You'll notice that $\\frac{18}{12}$ and $\\frac{7}{9}$ have different denominators. In order to subtract these fractions, we need to find equivalent fractions ( i.e. fractions that are the same as our originals) that have the same denominator as one another. To do this, let's find the least common denominator, or the smallest number that can be divided by both denominators. One way to do this to write out the multiples of both denominators until a number repeats in the the list: Multiples of 12: 12, 24, 36 , 48, 60 Multiples of 9: 9, 18, 27, 36 Since 36 appears in both lists, we can make both denominators 36 to create equivalent fractions. Since $12 \\cdot 3 = 36$, we can multiply $\\frac{18}{12} \\cdot \\frac{3}{3}$ (since $\\frac{3}{3}$ = 1) ... $\\frac{18}{12} \\cdot \\frac{3}{3} = \\frac{54}{36}$ And since $9 \\cdot 4 = 36$, we can multiply $\\frac{7}{9} \\cdot \\frac{4}{4}$ ... $\\frac{7}{9} \\cdot \\frac{4}{4} = \\frac{28}{36}$ Now, we have equivalent fractions with a denominator of 36 and can subtract: $\\frac{54}{36} - \\frac{28}{36} = \\frac{26}{36}$ And then divide the numerator and the denominator by 2 to simplify: $\\frac{26}{36} = \\frac{13}{18}$", "# How do you simplify\\frac { 2} { 9} + \\frac { 12} { 7}? Mar 8, 2018 $\\frac{122}{63}$ #### Explanation: First we should get a common denominator. since the common factor of 9 and 7 is 63 ( 9 x 7 = 63) we will write 63 as the denominator. to get the common denominator, multiply the first fraction by the denominator of the other fraction and multiply the second fraction from the denominator of the first fraction. then our sum will look like: $\\frac{7 \\cdot 2 + 9 \\cdot 12}{63}$ $\\to$ $\\frac{14 + 108}{63}$ $\\to$ $\\frac{122}{63}$", "case, we have a 5 and a 13 that are common to both numerator and denominator $$\\frac{65}{390} = \\frac{5 \\cdot 13}{2 \\cdot 3 \\cdot 5 \\cdot 13}$$ $$\\frac{65}{390} = \\frac{\\cancel{5} \\cdot \\cancel{13} 1}{2 \\cdot 3 \\cdot \\cancel{5} \\cdot \\cancel{13}}$$ $$\\frac{65}{390} = \\frac{1}{6}$$ We can also use simple division. We know the GCF of 65 and 390 is 65. If we divide 65 by 65, we get 1. If we divide 390 by 65, we get 6. $$\\frac{65}{390} = \\frac{65 ÷ 65}{390 ÷ 65} = \\frac{1}{6}$$ Example 3: Simplify each fraction $$\\frac{330}{420}$$ • Factor the numerator and denominator into the product of prime numbers • 330 = 2 x 3 x 5 x 11 • 420 = 2 x 2 x 3 x 5 x 7 • Cancel all common factors between the numerator and denominator • In this case, we have a 2, 3, and 5 that are common to both numerator and denominator $$\\require{cancel}$$ $$\\frac{330}{420} = \\frac{2 \\cdot 3 \\cdot 5 \\cdot 11}{2 \\cdot 2 \\cdot 3 \\cdot 5 \\cdot 7}$$ $$\\frac{330}{420} = \\frac{\\cancel{2} \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 11}{\\cancel{2} \\cdot 2 \\cdot \\cancel{3} \\cdot \\cancel{5} \\cdot 7}$$ $$\\frac{330}{420} = \\frac{11}{14}$$ We can also use simple division. We know the GCF of 330 and 420 is 30. If we divide 330 by 30, we get 11. If", "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, \\,", "$$\\frac{1}{2}$$ = $$\\frac{1 \\cdot 24}{2 \\cdot 24}=\\frac{24}{48}$$ $$\\frac{3}{16}$$ = $$\\frac{3 \\cdot 3}{16 \\cdot 3}=\\frac{9}{48}$$ Now we compare numerators of fractions! 5, 36, 24, 9 So, now we know that 5 is smaller than 9 which is smaller than 14 which is smaller than 17 . Now we can write them as : 5 < 9 < 24 < 36 Therefore: $$\\frac{5}{48}<\\frac{9}{48}<\\frac{24}{48}<\\frac{36}{48}$$ Last thing that we want to do is to return expressions to given ones! So, we have: $$\\frac{5}{48}$$ < $$\\frac{3}{16}$$ < 0.5 < 0.75 So, now we have ordered them from least to greatest and we can write the order as: $$\\frac{5}{48}$$, $$\\frac{3}{16}$$, 0.5, 0.75 Question 14. 0.5, $$\\frac{1}{5}$$, 0.35, $$\\frac{12}{25}$$, $$\\frac{4}{5}$$ First thing will be writing fractions as equivalent decimals! 0.35 = $$\\frac{35}{100}$$ 0.5 = $$\\frac{5}{10}=\\frac{5 \\cdot 10}{10 \\cdot 10}=\\frac{50}{100}$$ Second thing will be finding common denominator for our fractions! Let’s get all the numbers to have 100 as common denominator! We multiply both, numerator and denominator with same number to get equivalent expression! $$\\frac{4}{5}$$ = $$\\frac{4 \\cdot 20}{5 \\cdot 20}=\\frac{80}{100}$$ $$\\frac{12}{25}$$ = $$\\frac{12 \\cdot 4}{25 \\cdot 4}=\\frac{48}{100}$$ $$\\frac{1}{5}$$ = $$\\frac{1 \\cdot 20}{5 \\cdot 20}=\\frac{20}{100}$$ Now we compare numerators of fractions! 20, 48, 80, 35, 50 So, now we know that 20 is smaller than 35 which is smaller than 48 which is smaller than 50", "find a common denominator between the two. If one is not immediately obvious, you can always find one by multiplying the two denominators (the bottom numbers) together. In this case, $8 \\cdot 5 = 40$ so a common denominator is $40$. multiply the two fractions by a modified version of $1$ To get the denominator of $\\frac{2}{5}$ to $40$, the fraction should be multiplied by $\\frac{8}{8}$ $\\frac{2}{5} \\cdot \\frac{8}{8} = \\frac{16}{40}$ To get the denominator of $\\frac{5}{8}$ to $40$, the fraction should be multiplied by $\\frac{5}{5}$. $\\frac{5}{8} \\cdot \\frac{5}{5} = \\frac{25}{40}$ $\\frac{16}{40} \\ne \\frac{25}{40}$ So, $\\frac{2}{5} \\ne \\frac{5}{8}$ 1 ## What is the lowest common multiple for 8, 12, and 15? The Lonely Donut Featured 2 weeks ago $120$ #### Explanation: The easiest way to do this is to find the prime factorization of each number first. $8 = 4 \\cdot 2$ $8 = 2 \\cdot 2 \\cdot 2$ $8 = {2}^{3}$ $12 = 6 \\cdot 2$ $12 = 3 \\cdot 2 \\cdot 2$ $12 = 3 \\cdot {2}^{2}$ $15 = 5 \\cdot 3$ Now that we have reduced all numbers to their prime factorizations, we will look for the factors with the highest exponent and multiply them all together to get our LCM. There is ${2}^{3}$ from $8$, ${3}^{1}$ from $12$ and $15$, and ${5}^{1}$ from $15$.", "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", "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 $y + 2 = 0$ $y = 7$" ]

[ "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]", "$m=1/3$, 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: $\\boxed{-\\frac{1}{6}}$.", "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", "(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", "if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) --", "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", "# Math The real numbers $a$ and $b$ satisfy $|a| < 1$ and $|b| < 1.$ (a) In a grid that extends infinitely, the first row contains the numbers $1,$ $a,$ $a^2,$ $\\dots.$ The second row contains the numbers $b,$ $ab,$ $a^2 b,$ $\\dots.$ In general, each number is multiplied by $a$ to give the number to the right of it, and each number is multiplied by $b$ to give the number below it. Find the sum of all numbers in the grid. [asy] unitsize(1 cm); int i, j; for (i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5)); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5)); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5)); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5)); label(\"$b^2$\", (0.5,-2.5)); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5)); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5)); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5)); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] (b) Now suppose the grid is colored like a chessboard, with alternating black and white squares, as shown below. Find the sum of all the numbers that lie on the black squares. [asy] unitsize(1 cm); int i, j; for (i = 0; i <= 3; ++i) { for (j = 0; j <= 3; ++j) { if ((i + j) % 2 == 0) { fill(shift((i,-j))*((0,0)--(1,0)--(1,-1)--(0,-1)--cycle),black); } }} fill((0,-4)--(1,-4)--(1,-4.5)--(0,-4.5)--cycle,black); fill((2,-4)--(3,-4)--(3,-4.5)--(2,-4.5)--cycle,black); fill((4,0)--(4,-1)--(4.5,-1)--(4.5,0)--cycle,black); fill((4,-2)--(4,-3)--(4.5,-3)--(4.5,-2)--cycle,black); fill((4,-4)--(4.5,-4)--(4.5,-4.5)--(4,-4.5)--cycle,black); for", "&& (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw(", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## 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 $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "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. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label(\"A\",A,(1,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0));label(\"P\",P,(0,-1));label(\"Q\",Q,(1,0));label(\"R\",R,(-1,1)); [/asy]$", "*/ 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", "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.", "$n$ 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", "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", "# Difference between revisions of \"2001 AMC 8 Problems/Problem 19\" ## Problem Car M traveled at a constant speed for a given time. This is shown by the dashed line. Car N traveled at twice the speed for the same distance. If Car N's speed and time are shown as solid line, which graph illustrates this? $[asy] unitsize(12); draw((0,9)--(0,0)--(9,0)); label(\"time\",(4.5,0),S); label(\"s\",(0,7),W); label(\"p\",(0,6),W); label(\"e\",(0,5),W); label(\"e\",(0,4),W); label(\"d\",(0,3),W); label(\"(A)\",(-1,9),NW); draw((0,4)--(4,4),dashed); label(\"M\",(4,4),E); draw((0,8)--(4,8),linewidth(1)); label(\"N\",(4,8),E); draw((15,9)--(15,0)--(24,0)); label(\"time\",(19.5,0),S); label(\"s\",(15,7),W); label(\"p\",(15,6),W); label(\"e\",(15,5),W); label(\"e\",(15,4),W); label(\"d\",(15,3),W); label(\"(B)\",(14,9),NW); draw((15,4)--(19,4),dashed); label(\"M\",(19,4),E); draw((15,8)--(23,8),linewidth(1)); label(\"N\",(23,8),E); draw((30,9)--(30,0)--(39,0)); label(\"time\",(34.5,0),S); label(\"s\",(30,7),W); label(\"p\",(30,6),W); label(\"e\",(30,5),W); label(\"e\",(30,4),W); label(\"d\",(30,3),W); label(\"(C)\",(29,9),NW); draw((30,4)--(34,4),dashed); label(\"M\",(34,4),E); draw((30,2)--(34,2),linewidth(1)); label(\"N\",(34,2),E); draw((0,-6)--(0,-15)--(9,-15)); label(\"time\",(4.5,-15),S); label(\"s\",(0,-8),W); label(\"p\",(0,-9),W); label(\"e\",(0,-10),W); label(\"e\",(0,-11),W); label(\"d\",(0,-12),W); label(\"(D)\",(-1,-6),NW); draw((0,-11)--(4,-11),dashed); label(\"M\",(4,-11),E); draw((0,-7)--(2,-7),linewidth(1)); label(\"N\",(2,-7),E); draw((15,-6)--(15,-15)--(24,-15)); label(\"time\",(19.5,-15),S); label(\"s\",(15,-8),W); label(\"p\",(15,-9),W); label(\"e\",(15,-10),W); label(\"e\",(15,-11),W); label(\"d\",(15,-12),W); label(\"(E)\",(14,-6),NW); draw((15,-11)--(19,-11),dashed); label(\"M\",(19,-11),E); draw((15,-13)--(23,-13),linewidth(1)); label(\"N\",(23,-13),E); [/asy]$ ## Solution Since car N has twice the speed, it must be twice as high on the speed axis. Also, since cars M and N travel at the same distance but car N has twice the speed, car N must take half the time. Therefore, line N must be half the size of line M. Since the speeds are constant, both lines are horizontal. Reviewing the graphs, we see that the only one satisfying these conditions is graph $\\boxed{\\text{D}}$.", "i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (7,7) -- (-7,7) -- (-7,-7) -- (7,-7) -- cycle, black ); [/asy]$ After following the mapping described in Solution 2, the square looks like this: $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (3,3) -- (3,7) -- (7,7) -- (7,3) -- cycle, black ); [/asy]$ The position of each vertex within their corresponding grid has not changed.For example, the point $(-7, 7)$ is still the top-left point in a grid, albeit a change in quadrant. Trying this out with a couple of other squares, we see that the following property holds: $$\\text{\"For any and all squares with vertices in each of the four quadrants, the 'mapped' version is also a", "+0 # Pls Help 0 86 1 Point $X$ is on $\\overline{AC}$ such that $AX = 4\\cdot CX = 12$. We know $\\angle ABC = \\angle BXA = 90^\\circ.$ What is $BX$? [asy] size(5cm); pair A,B,C,X; A = (0,0); X = (12,0); C = (16,0); B = (12,sqrt(48)); draw(X--B--A--C--B); draw(rightanglemark(A,B,C,20)); draw(rightanglemark(B,X,A,20)); label(\"$A$\",A,SW); label(\"$B$\",B,N); label(\"$C$\",C,SE); label(\"$X$\",X,S); label(\"$12$\",(A+X)/2,S); [/asy] Nov 1, 2019", "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)", "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" ]

[ "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 =", "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 =", "# 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", "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);", "= \\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)", "(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.", "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", "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 ##", "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", "(0,0); 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)", "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}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "= (0,0); 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", "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 $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ ### 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 $\\triangle CAB$. We know that the angle", "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 $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ ### 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 $\\triangle CAB$. We know that the angle", "= 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", "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", "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 $x>y$,", "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", "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. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ 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." ]

[ "(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", "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", "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", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "$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: $\\boxed{-\\frac{1}{6}}$.", "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$$25k^2+9k^2-15k^2 =", "= \\angle XBW = 90^{\\circ} = \\angle AMH = \\angle ATH$$ 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)", "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", "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 $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) --", "&& (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw(", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ 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", "$n$ 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 $k=-n\\to", "*/ pen qqwuqq = rgb(0,0,0); draw(arc((12,0),2,60,180)--(12,0)--cycle, qqwuqq); /* draw figures */ draw((0,0)--(12,0)); draw((0,0)--(18,10)); draw((18,10)--(12,0)); label(\"12\",(6,-1),SE*labelscalefactor); label(\" 2\\sqrt{76}-6 \",(14,5),SE*labelscalefactor); label(\"20\",(8,7),SE*labelscalefactor); draw((12,0)--(12,10)); draw((12,10)--(18,10)); /* dots and labels */ dot((0,0),dotstyle); label(\"A\", (0,0), NE * labelscalefactor); dot((12,0),dotstyle); label(\"B\", (12,0), NE * labelscalefactor); dot((18,10),dotstyle); label(\"C\", (18,10), NE * labelscalefactor); label(\"120^\\circ\", (11,1), NE * labelscalefactor,qqwuqq); dot((12,10),dotstyle); label(\"H\", (12,10), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ 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", "creates a checkerboard pattern, except that the middle two rows are identical. $[asy] /* variable declarations */ int n = 4; /* code */ pen r = red, b = black; int absol(int k){if(k >= 0) return k; else return -1*k-1;} pen col(bool k){if(k) return r; else return b;} for(int y = -n; y < n; ++y){ bool b = true; for(int x = absol(y)-n+1; x < n-absol(y); ++x){ dot((x,y),col(b)); b = !b; } } draw((0.5,0.5)--(0.5,1.5),white); /* to force rendering by using a white line */ [/asy]$ $[asy] /* variable declarations */ int n = 4; /* code */ pen r = red, b = black; int absol(int k){if(k >= 0) return k; else return -1*k-1;} pen col(bool k){if(k) return r; else return b;} for(int y = -n; y < n; ++y){ bool b = true; for(int x = absol(y)-n+1; x < n-absol(y); ++x){ dot((x,y),col(b)); b = !b; } } draw((0.5,0.5)--(0.5,1.5),white); /* to force rendering by using a white line */ draw((-1,2)--(-1,1)--(0,1)--(1,1)--(1,0),linewidth(0.7)); draw((1,0)--(1,-1),linetype(\"3 3\")+linewidth(0.7)); draw((1,-1)--(1,-2)--(2,-2),linewidth(0.7)); [/asy]$ Examples for $n=4$ Suppose there is a partition of $m < n$ paths that works. For each of the $m$ paths, we break a path into two separate paths wherever there are consecutive red points. This only happens only with the $n$ red pairs in the middle two rows. We now have", "# 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", "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 $x>y$,", "- 40\\sqrt{10}$. Through power of a point, we can find out that $(CE)(DE)=20\\sqrt{10} - 20$, so $CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\\sqrt{10}) + 2(20\\sqrt{10} - 20) = \\boxed{\\textbf{(E) } 100}$. ~Math_Wiz_3.14 ## Solution 4 (Reflections) $[asy] draw(circle((0,0),7.07)); dot((-7.07,0)); label(\"A\",(-7.07,0),W); dot((7.07,0)); label(\"B\",(7.07,0),E); dot((0,0)); label(\"O\",(0,0),N); dot((-2.24,-6.71)); label(\"C\",(-2.24,-6.71),SSW); dot((6.71,2.24)); label(\"D\",(6.71,2.24),NE); draw((-2.24,-6.71)--(6.71,2.24)); dot((6.71,-2.24)); label(\"D'\",(6.71,-2.24),SE); draw((4.51,0)--(6.71,-2.24)); dot((4.51,0)); label(\"E\",(4.51,0),NNW); draw((-7.07,0)--(7.07,0)); draw((0,0)--(-2.24,-6.71)); draw((0,0)--(6.71,-2.24)); draw((-2.24,-6.71)--(6.71,-2.24)); [/asy]$ Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\\angle BED'=45$. Since $\\angle AEC=45$, $\\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$. Looking next to circle arcs, we know that $\\angle AEC=\\frac{\\overarc{AC}+\\overarc{BD}}{2}=45$, so $\\overarc{AC}+\\overarc{BD}=90$. Since $\\overarc{BD'}=\\overarc{BD}$, and $\\overarc{AC}+\\overarc{BD'}+\\overarc{CD'}=180$, $\\overarc{CD'}=90$. Thus, $\\angle COD'=90$. Since $OC=OD'=5\\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \\boxed{\\textbf{(E) } 100}$. ~sofas103 ## See Also 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 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 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's", "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 ##" ]

[ "• anonymous The ratio of the measures of two supplementary angles is 7:3. Find the measures of the angles. A. 126° and 54° B. 72° and 18° C. 159° and 21° D. 69° and 21° Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "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°​", "# How do you find the measure of each angle if the ratio of two supplementary angles is 11:4? The supplementary angles add upto 180 degrees. Hence add the two ratios 11+4=15 and work out $\\frac{11}{15}$ of 180 and $\\frac{4}{15}$ of 180. On simplification we would have the figures 132 and 48.", "# The ratio of the measure of two complementary angles is 4:5, what are the measures of the angles? $\\frac{90}{9} = 10$ $4 \\times 10 : 5 \\times 10$ ${40}^{\\circ} : {50}^{\\circ}$", "used the ratio of two values. Scientific calculator, but no protractor use degrees ( ° ) to use a protractor TEKS 4.7C to...", "Hence, all the angles are of 60. Re: For the cube shown above, what is the degree measure of PQR [#permalink] 18 Dec 2018, 22:25 Display posts from previous: Sort by", "a ratio such as an equilateral or isosceles triangle. any two values press... Side if you want to know how to name an angle is provided the degree measures angles!, including at least three of these values, including at least one of two ways: the. This page, you could also use cosine how to find another angle leg is than... Us to use in this tutorial, see how identifying your triangle first be!", "in a right-angled triangle. The ratios are all equal to the same thing.", "measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side oblique triangle any triangle that is not a right triangle", "THANKS TO U 15. anonymous BUT The ratio of the radii of two circles at the centers of which two arcs of the same length subtended a angles of 60 degree and 70 degree is 16. anonymous welcome 17. anonymous CAN U ANSWER ME THIS QUESTION ALSO The ratio of the radii of two circles at the centers of which two arcs of the same length subtended a angles of 60 degree and 70 degree is 18. anonymous actually no because i do not know what the question means 19. anonymous OK ANYWAY THANKS", "• 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.", "# Two supplementary angles are in the ratio of 7 to 2. What is the measurement of degrees in the smaller angle? Dec 28, 2015 $40$ #### Explanation: Two Angles are Supplementary when they add up to 180 degrees. Let bigger angle be $7 x$ and smaller one $2 x$. $\\textcolor{w h i t e}{\\times} 7 x + 2 x = 180$ $\\implies \\textcolor{red}{\\frac{1}{9} \\times} 9 x = \\textcolor{red}{\\frac{1}{9} \\times} 180$ $\\implies x = 20$ $\\implies 2 x = 2 \\times 20$ $\\implies 2 x = 40$", "• anonymous in need of some help!?:) The ratio of the measures of two complementary angles is 1:5. What are the measures of the angles? A. 30° and 60° B. 15° and 75° C. 30° and 150° D. 5° and 25° Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "two circles subtend angles of 35° and 45° at their centres then the ratio of their radii is - 9:7 3 : 7 7 : 9 9 : 5", "Sangeetha Pulapaka 2 Since the sum of the two angles is 180^\\circ, they must be a linear pair or supplementary angles. The ratio of the two angles is given as 7: 8. The measure of the first angle is \\frac{7}{15} \\times 180^\\circ = 84^\\circ (the sum of their ratios is 7+8 = 15) The measure of the second angle is \\frac{8}{15}\\times 180^\\circ = 96^\\circ So the measure of the smaller angle is 84^\\circ", "• xxImHisGirlxx The measures of two complementary angles are in the ratio 1:9 What are the degree measures of the two angles? 10° and 80° 9° and 81° ***my answer 20° and 160° 18° and 162° Geometry Looking for something else? Not the answer you are looking for? Search for more explanations.", "Of The Measures Of The. To find the measure of the angles, we know that the. ### Kesimpulan dari Sum Of The Interior Angles Of A Nonagon. The sum of angles of a nonagon is 1260°. To find the measure of the angles, we know that the. A nonagon consists of 9 angles. Nov 30, 2015 each interior angles of a regular nonagon measures 140o explanation: See also Two Angles Whose Sum Is 180", "2: √3 + 1 No, the ratio of the sides is 1/2 : 1/√2 : (√3 + 1)/2√2, or √2 : 2: √3 + 1 I am sorry! Thanks for pointing out my misconception. Thanks a lot!", "ratio, gives on an overview of their range of values and provides many practice problems on identifying the sides that are opposite and adjacent to a given angle. An introduction to the standards of measurement sine and cosine rules Rated 5/5 based on 15 review 2018.", "(b) alternate interior angles supplementary. (c) alternate exterior angles supplementary. (d) interior angles on the same side of the transversal are supplementary. 9. In the diagram, what is the value of angle x? (a) 43° (b) 44° (c) 132° (d) 134°" ]

[ "## Elementary Geometry for College Students (6th Edition) measure of arc $CA=160^{\\circ}$ The measure of an entire circle is $360^{\\circ}$. If you add the three arcs together...$AB+BC+CA$ you have the entire circle. So... arc $AB=2x$, arc $BC=3x$, and arc $AC=4x$. $2x+3x+4x=360$ combine like terms... $9x=360$ Divide both sides by 9... $x=40$ Plug $x=40$ into the $4x$ (the measure of arc $CA$) and the measure of arc $CA=160^{\\circ}$.", "# Say True or False:(a) The measure of an acute angle $<90^{\\circ}$.(b) The measure of an obtuse angle $<90^{\\circ}$.(c) The measure of a reflex angle $>180^{\\circ}$.(d) The measure of one complete revolution $=360^{\\circ}$.(e) If $m \\angle A = 53^o$ and $m \\angle B = 35^o$, then $m \\angle A > m \\angle B$. #### Complete Python Prime Pack for 2023 9 Courses 2 eBooks #### Artificial Intelligence & Machine Learning Prime Pack 6 Courses 1 eBooks #### Java Prime Pack 2023 8 Courses 2 eBooks To do: We have to find whether the given statements are true or false. Solution: (a) The measure of an acute angle is less than $90^o$ Therefore, the given statement is true. (b) The measure of an obtuse angle is more than $90^o$ but less than $180^o$ Therefore, the given statement is false. (c) The measure of a reflex angle is more than $180^o$. Therefore, the given statement is true. (d) The measure of one complete revolution is $360^o$ Therefore, the given statement is true. (e) $53^o>35^o$ This implies, $m \\angle A > m \\angle B$ Therefore, the given statement is true. Updated on 10-Oct-2022 13:33:09", "(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", "has a measure of $180^\\circ=\\pi\\; \\mathrm{ rad}$. ### Right Angle A right angle is an angle that is supplementary to itself. A right angle has a measure of $90^\\circ=\\frac{\\pi}{2}\\;\\mathrm{ rad}$. An acute angle has a measure greater than zero but less than that of a right angle, i.e. $\\angle ABC$ is acute$\\Leftrightarrow 0^\\circ. An obtuse angle has a measure greater than that of a right angle but less than that of a straight angle, i.e. $\\angle ABC$ is obtuse$\\Leftrightarrow 90^\\circ. ### Reflex Angle A reflex angle is an angle with measure greater than a straight angle, but less than 360 degrees, or $2\\pi$ radians.", "a protractor which is made to measure reflex angles. 2. Measuring the acute side or interior side and subtracting it with $$360^\\circ$$. ## Where do we find Reflexive angles? We can find a reflexive angle in 1. The arms of the clock sometimes form reflexive angles on certain readings. 2. Chinese fan, it opens up more than $$180^\\circ$$. 3. If we cut a slice of cake at an acute angle, we find the rest of the cake forms a reflexive angle, etc. ## Questions Q. What is reflexive angle for $$120^\\circ$$? The reflexive angle for $$120^\\circ$$ is $$360^\\circ – 120^\\circ = 240^\\circ$$ Q. Find reflexive angle for $$133^\\circ$$. Reflexive angle for $$133^\\circ$$ is $$360^\\circ – 133^\\circ = 227^\\circ$$ ## FAQs What is a reflex angle definition? Angles measuring between $$180^\\circ and 360^\\circ$$ are called reflex angles. Is there a reflex angle in Triangles? As the total sum of all angles in a triangle is $$180^\\circ$$, it is impossible to have a reflexive angle in a triangle. What is a Complete angle? $$360^\\circ$$ is known as the wide angle. Is $$180^\\circ$$ a reflex angle? $$180^\\circ$$ is known as the straight angle. Related Articles: Dihedral Angle Scroll to Top", "measures more than $0^{\\circ}$ and less than $90^{\\circ}$ is called. • A. Obtuse • B. Acute • C. Right • D. None Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "measures are. There you have it. We have 75^\\circ and 15^\\circ as our angle measures. I’ll now leave it up to you to check if these measures actually sum up to 90^\\circ . Example 7: Suppose \\angle AKL and \\angle QBH are complementary. Find the measures of the two angles. Note that even though they are two separate angles, they are complementary. Hence, their angle measures should sum up to 90^\\circ . We’ll now find what these measures are by first solving for x. So we have x = 18. Let’s continue finding the measures for \\angle AKL and \\angle QBH. If we add both angle measures, \\textbf{66}^\\circ + \\textbf{24}^\\circ = 90^\\circ, indeed they are complementary angles. Example 8: Determine the measures of the two complementary angles. We have another interesting example here. Instead of algebraic expressions, our angle measures are now expressed as fractions. Let’s apply the same strategies that we used in our previous examples. To find the m\\angle \\,E and m\\angle \\,W, we’ll simply substitute x with 216. Both measures sum up to 90^\\circ which confirms that both angles are complementary. Example 9: Two complementary angles are such that one of the angles is twice the sum of the other angle. What are the measures of the complementary angles? Begin by solving for x. This", "# Ratio problem Printable View • November 19th 2007, 05:36 PM OzzMan Ratio problem The measure of two supplementary angles are in a ratio 3:15. What are the measures of the angles? And can you explain what method you used to do this because at the moment i got 1/5=? I don't know what to plug in on the other side along with an x to evaluate this. • November 19th 2007, 05:39 PM Krizalid Let $\\alpha,\\,\\beta$ be the angles. So $\\alpha=k,\\,\\beta=5k.$ ( $k$ is a positive constant.) Besides, you know that $\\alpha+\\beta=180^\\circ.$ Go on from there. • November 19th 2007, 05:52 PM OzzMan wow im dumb cant believe i never thought of that. thanks", "Question # Say whether it is true or false.The measure of a reflex angle$> 180^\\circ$.A).TrueB).False According to definition, reflex angle is the angle which is greater than $180^\\circ$and less than $360^\\circ$. So, the statement is true. Acute angle: An angle which measures less than $9{0^ \\circ }$. Right angle: An angle which measures exactly $9{0^ \\circ }$. Obtuse angle: An angle that measures greater than $9{0^ \\circ }$. Straight angle: An angle which measures exactly $18{0^ \\circ }$. Reflex angle: The angle which measures greater than $18{0^ \\circ }$ and less than $$36{0^ \\circ }$$. The reflex angle can be calculated if the measure of the acute angle is given, as it is the complementary to the acute angle on the other side of the line.", "The diagram gives us two angle measures with one of the measures expressed in an algebraic expression. It may look challenging but it’s actually not. As long as you know the concept of the relationship between complementary angles, you’ll be fine. By now we know that the sum of the measures will be 90^\\circ . So we’ll set up our equation by adding both measures on one side of the equation and 90^\\circ on the other. The value of x is \\textbf{12}. However, since we already know what x is, we might as well find out what the measure of \\angle GRW is since it is expressed in an algebraic expression. Awesome! The measure of \\angle GRW is also 45^\\circ ! Is that really correct? Well, simply add 45^\\circ and 45^\\circ . Both measures should total to 90^\\circ . Example 6: What is the value of x? This next example is interesting because now both angle measures are expressed in algebraic expressions. But as I said before, just always remember that the sum of their measures will be 90^\\circ since they are complementary angles. With that in mind, it’ll be easy for you to set up our equation and solve for x. The value of x is \\textbf{8}. Let’s use this value to find out what our angle", "measurement of an angle is degree$({}^\\circ )$ Types of Angles Acute Angle The angle between $0{}^\\circ$ and $90{}^\\circ$ is called an acute angle. For example, $10{}^\\circ ,\\,\\,30{}^\\circ ,\\,\\,60{}^\\circ ,\\,\\,80{}^\\circ$are acute angles. Right Angle An angle of measure $90{}^\\circ$ is called a right angle. Obtuse Angle An angle whose measure is between $90{}^\\circ$ and $180{}^\\circ$ is called an obtuse angle. Straight Angle An angle whose measure is $180{}^\\circ \\,\\,\\text{is}$called a straight angle. Reflex Angle An angle whose measure is more than $180{}^\\circ$ and less than $360{}^\\circ$is called a reflex angle. Complementary Angle Two angles whose sum is $90{}^\\circ \\,\\,\\text{is}$ called the complimentary angle. Complementary angle of any angle $\\theta$ is$90{}^\\circ -\\theta$. Supplementary Angle Two angles whose sum is $180{}^\\circ \\,\\,\\text{is}$ called supplementary angles. Supplementary angle of any angle $\\theta$ is$180{}^\\circ -\\theta$. Adjacent Angle Two angles are said to be adjacent if they have a common vertex and one common arm. In the following figure $\\angle AOC$ and $\\angle COB$are adjacent angles. Vertically Opposite Angles Two angles are said to be vertically opposite angles if they are formed at the intersecting point of two lines. In the figure given more... #### Notes - Algebraic Expressions Algebraic Expressions In an algebraic expression constant and variables are linked with arithmetic operations. The value of unknown variable is obtained by simplification of the", "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", "## Trigonometry 7th Edition To find exact value of $\\sec 300^{\\circ}$, let's find its reference angle first. As $300^{\\circ}$ terminates in quadrant IV, The reference angle = $360^{\\circ} - 300^{\\circ}$ = $60^{\\circ}$ As $300^{\\circ}$ terminates in quadrant IV, its $\\sec$ will be positive. Therefore by reference angle theorem- $\\sec 300^{\\circ}$ = $\\sec 60^{\\circ}$ = 2", "153.2^{\\circ}, a'\\approx 8.3$$ 22) $$a=2.3, c=1.8, \\gamma =28^{\\circ}$$ 23) $$\\beta =119^{\\circ}, b=8.2, a=11.3$$ no triangle possible For the exercises 24-26, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. 24) Find angle $$A$$ when $$a=24, b=5, B=22^{\\circ}$$ 25) Find angle $$A$$ when $$a=13, b=6, B=20^{\\circ}$$ $$A\\approx 47.8^{\\circ}$$ or $$A'\\approx 132.2^{\\circ}$$ 26) Find angle $$B$$ when $$A=12^{\\circ}, a=2, b=9$$ For the exercises 27-30, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. 27) $$a=5, c=6, \\beta =35^{\\circ}$$ $$8.6$$ 28) $$b=11, c=8, \\alpha =28^{\\circ}$$ 29) $$a=32, b=24, \\gamma =75^{\\circ}$$ $$370.9$$ 30) $$a=7.2, b=4.5, \\gamma =43^{\\circ}$$ ### Graphical For the exercises 31-36, find the length of side $$x$$. Round to the nearest tenth. 31) $$12.3$$ 32) 33) $$12.2$$ 34) 35) $$16.0$$ 36) For the exercises 37-,42 find the measure of angle $$x$$, if possible. Round to the nearest tenth. 37) $$29.7^{\\circ}$$ 38) 39) $$x=76.9^{\\circ}$$ or $$x=103.1^{\\circ}$$ 40) 41) Notice that $$x$$ is an obtuse angle. $$110.6^{\\circ}$$ 42) For the exercises 43-49, find the area of each triangle. Round each answer to the nearest tenth. 43) $$A\\approx 39.4, C\\approx 47.6, BC\\approx 20.7$$ 44) 45) $$57.1$$ 46) 47) $$42.0$$ 48) 49) $$430.2$$ ### Extensions", "is $157^{\\circ }$. But, $157^{\\circ } + 35^{\\circ } > 189^{\\circ }$ Therefore, $m∠B = 23^{\\circ }$ Hence, ONE distinct triangle exists. Conclusion: Hence, we conclude that ONE triangle exists if • A is acute and $a > b > p$ • A is obtuse and $a > b$ • A is acute and $a = p$ The following diagram illustrates these cases. ### Practice Questions 1. In triangle $△ABC$, $m∠B = 25^{\\circ }$, $a = 24$ cm and $b = 14cm$. What is the measure of angle $A$? 2. In triangle $△ABC$, $m∠A = 135^{\\circ }$, $a = 89$ cm and $c = 46$ cm. What is the measure of angle $B$? 3. In triangle $△ABC$, $m∠ A = 30^{\\circ }$, $a = 30$ cm and $c = 50$ cm. What is the length of $b$? 4. In triangle $△ABC$, $m∠ A = 30^{\\circ }$, $a = 30$ cm and $c = 50$ cm. What are the measures of $m∠ B$ and $m∠C$? 5. In triangle $△ABC$, $m∠ C = 70^{\\circ }$, $b = 6.3$ cm and $c = 6.1$ cm. What is the length of $a$? 6. In triangle $△ABC$, $m∠ C = 70^{\\circ }$, $b = 6.3$ cm and $c = 6.1$ cm. What are the measures of $m∠ A$ and $m∠ B$?", "# State whether the given statement is true or false:The measure of an obtuse angle is less than ${90^ \\circ }.$ Hint: Obtuse angle is always greater than ${90^ \\circ }$ From the definition of obtuse angle, we know that its value lies between ${90^ \\circ }$and ${180^ \\circ }$(both excluding) i.e., ${90^ \\circ } <$ obtuse angle $< {180^ \\circ }$ Note: Angles are divided into three categories on the basis of their measurement. Acute angle, obtuse angle and right angle. The measure of right angle is ${90^ \\circ }$ while the measure of acute angle is always less than ${90^ \\circ }$.", "$$45^\\circ$$. ## Supplementary angles VS Complementary angles Supplementary and complementary angles are pairs of angles that add up to $$180^\\circ$$ and $$90^\\circ$$, respectively. Let’s take a closer look at their differences.", "This new measure of the circle has happily brought to light the ratio of the chord and arc of $90^\\circ$, which is as $7:8$; and also the ratio of the diagonal and one side of a square, which is as $10:7$. These two ratios show the numerical relation of diameter to circumference to be as $\\frac 5 4:4$. Authorities will please note that while the finite ratio ($\\frac 5 4:4$) represents the area of the circle to be more than the orthodox ratio, yet the ratio ($3.1416$) represents the area of a circle whose circumference equals $4$ two $\\%$ greater than the finite ratio ($\\frac 5 4:4$), as will be seen by comparing the terms of their respective proportions, stated as follows: $1 : 3.20 :: 1.25 : 4$, $1 : 3.1416 :: 1.2732 : 4$. It will be observed that the product of the extremes is equal to the product of the means in the first statement, while they fail to agree in the second proportion. Furthermore, the square on a line equal to the arc of $90^\\circ$ shows very clearly that the ratio of the circle is the same in principle as that of the square. For example, if we multiply the perimeter of a square (the sum of its sides) by $\\frac 1 4$ of", "Partners. ## Student Lesson Summary ### Student Facing If two angle measures add up to $$90^\\circ$$, then we say the angles are complementary. Here are three examples of pairs of complementary angles. If two angle measures add up to $$180^\\circ$$, then we say the angles are supplementary. Here are three examples of pairs of supplementary angles.", "following are different types of angles: • Acute Angle: The angle whose measure is more than $0{}^\\circ$and less than $90{}^\\circ$ is called an acute angle. • Right Angle: The angle of measure $90{}^\\circ$is called a right angle. • Obtuse Angle: The angle whose measure is more than $90{}^\\circ$and less than $180{}^\\circ$ is called an obtuse angle. • Straight Angle: The angle whose measure is $180{}^\\circ$ is called a straight angle. • Reflex Angle: The angle whose measures is more than $180{}^\\circ$ and less than $360{}^\\circ$ is called a reflex angle. • Complete Angle: The angle whose measure is $360{}^\\circ$ is called a complete angle. • Equal Angles: Two angles are said to be equal if they are of same measure. • Complementary Angles: If the sum of measures of two angles is $90{}^\\circ$ then they are said to be complementary angles. • Supplementary Angles: If the sum of measures of two angles is $180{}^\\circ$ then they are said to be supplementary angles. Triangles The word triangle is derived from Greek word, tri means three and hence, it refers to a shape consisting of three internal angles. Obviously the shape consists of three sides. Hence, a triangle can be defined as a polygon having three sides. Basic Concepts of Triangles The general shape of a triangle is shown below:" ]

[ "# Ratio problem Printable View • November 19th 2007, 05:36 PM OzzMan Ratio problem The measure of two supplementary angles are in a ratio 3:15. What are the measures of the angles? And can you explain what method you used to do this because at the moment i got 1/5=? I don't know what to plug in on the other side along with an x to evaluate this. • November 19th 2007, 05:39 PM Krizalid Let $\\alpha,\\,\\beta$ be the angles. So $\\alpha=k,\\,\\beta=5k.$ ( $k$ is a positive constant.) Besides, you know that $\\alpha+\\beta=180^\\circ.$ Go on from there. • November 19th 2007, 05:52 PM OzzMan wow im dumb cant believe i never thought of that. thanks", "• anonymous The ratio of the measures of two supplementary angles is 7:3. Find the measures of the angles. A. 126° and 54° B. 72° and 18° C. 159° and 21° D. 69° and 21° Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "Therefore, the angles 20 degrees and 160 degrees are the two supplementary angles. Example 6 Determine the supplement angle of (x + 10) °. Solution ⟹ (x + 10) ° = 180 ° – (x + 10) ° = 180° – 10° – x° = (170 – x) ° ### Practice Questions 1. True or False: The angles $136^{\\circ}$ and $34^{\\circ}$ are a pair of supplementary angles. 2. True or False: The angles $78^{\\circ}$ and $102^{\\circ}$ are a pair of supplementary angles. 3. True or False: The angles $101^{\\circ}$ and $89^{\\circ}$ are a pair of supplementary angles. 4. Suppose that angles, $\\angle A$ and $\\angle B$, are supplementary. If $\\angle A$ is four times larger than $\\angle B$, which of the following shows the angle measure of $\\angle A$? 5. Given two supplementary angles, $(x – 22)^{\\circ}$ and $(3x – 18)^{\\circ}$, which of the following shows the correct value of $x$? 6. The ratio of two supplementary angles is $1:5$. Which of the following shows the angle measures of the two angles? 7. Which of the following shows the correct expression for the supplement of the angle, $(2x + 20)^{\\circ}$?", "Definition of Supplementary Angles. For example, the supplement of $$40^\\circ$$ is $$180-40=140^\\circ$$. Example 2: A. ALGEBRA Find the measures of two supplementary angles if the measure of one angle is 6 less than five times the measure of the other angle. Hence, these two angles are adjacent supplementary angles. In other words, when complementary angles are put together, they form a right angle (90 degrees). Help your child score higher with Cuemath’s proprietary FREE Diagnostic Test. We at Cuemath believe that Math is a life skill. $$\\angle 1$$ and $$\\angle 2$$ are supplementary if If you add arrows onto the ends of the straight angle, you will have a straight line. The measure of the larger angle is 5 degrees more than 4 times the measure of the smaller angle. Supplementary angles in geometry have the following characteristics. $\\angle ABC+ \\angle PQR = 79^\\circ+101^\\circ=180^\\circ$. You can download the FREE grade-wise sample papers from below: To know more about the Maths Olympiad you can click here. \\begin{align} Y +77^\\circ &= 180^\\circ \\\\[0.2cm] Y &= 180^\\circ-77^\\circ\\\\[0.2cm] Y &= 103^\\circ \\end{align}. If the sum of two angles are 180 o then the angles are said to be supplementary angles, . A 137 degree angle is supplementary to a … Let us assume that the two supplementary angles are $$x$$ (larger) and", "In particular, make sure students can articulate: • $$c = 60$$ because it is the measure of an angle forming an alternate interior angle with the given 60 degree angle. • $$e = d = 120$$ because they are also alternate interior angles, each supplementary to a 60 degree angle. • The rest of the angle measures can be found using vertical or supplementary angles. ## 14.4: Cool-down - All The Rest (5 minutes) ### Cool-Down When two lines intersect, vertical angles are equal and adjacent angles are supplementary, that is, their measures sum to 180$$^\\circ$$. For example, in this figure angles 1 and 3 are equal, angles 2 and 4 are equal, angles 1 and 4 are supplementary, and angles 2 and 3 are supplementary. Alternate interior angles are equal because a $$180^\\circ$$ rotation around the midpoint of the segment that joins their vertices takes each angle to the other. Imagine a point $$M$$ halfway between the two intersections—can you see how rotating $$180^\\circ$$ about $$M$$ takes angle 3 to angle 5? Using what we know about vertical angles, adjacent angles, and alternate interior angles, we can find the measures of any of the eight angles created by a transversal if we know just one of them. For example, starting with the fact that angle 1 is", "the sum of the measures of the angles comes out to be $$180^\\circ$$, angles are called supplementary angles. Sum of measure of these two angles $$= 63^\\circ + 27^\\circ = 90^\\circ$$. 3. 4. (iii) Two angles forming a linear pair are _______________. Visit official website CISCE (Indian Council of Secondary Education) for detail information about ICSE Class-7 Mathematics. The chapter 5 begins with an introduction to Lines and Angles by explaining the basic concepts of point, line, line segment and angle.Then various types of related angles such as complementary angles, supplementary angles, adjacent angles, linear pair and vertically opposite angles are discussed in detail. (v) Yes, they are vertical angles because they are formed due to intersection of straight lines. The sum of angle and its supplement angle is always equal to $$180^\\circ.$$, Therefore, supplement of this angle will also be $$x.$$, We know that, the sum of measure of pair of supplementary angles is $$180 \\,^{\\circ}$$, \\begin{align} x+x &=180^\\circ\\\\ \\Rightarrow2x &=180^\\circ\\\\ \\Rightarrow x&=(180^\\circ)/2\\\\ \\Rightarrow x&=90^\\circ\\end{align}, Thus, the angle which is equal to its supplement is $$90^\\circ.$$. Do ∠COE and ∠EOD form a linear pair? 5.1; Class 7 Maths Chapter 5 Ex. ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 10 Lines and Angles Objective Type Questions July 18, 2020 by Prasanna Leave a Comment ML Aggarwal", "# What is the supplementary angle to theta= 35.6 degrees? May 9, 2018 ${144.4}^{\\circ}$ #### Explanation: • \" supplementary angles sum to \"180^@ $\\Rightarrow \\text{supplement of } {35.6}^{\\circ} = {\\left(180 - 35.6\\right)}^{\\circ} = {144.4}^{\\circ}$ May 9, 2018 $144.4$ degrees. #### Explanation: Supplementary angles combine to make $180$ degrees. If one angle measures $35.6$ degrees, the other must combine to make $180$ degrees. We can set up an equation to solve for this: $35.6 + x = 180$ Subtract $35.6$ from both sides: $x = \\textcolor{red}{144.4}$ May 9, 2018 144.4° #### Explanation: Supplementary angles add up to 180°. Let the supplementary angle of theta=35.6° be $x$. Therefore, x+theta=180° x+35.6°=180° x=180°-35.6°=144.4°", "supplementary if both are: (i) No, sum of acute angles is less than $$180^\\circ.$$, (ii) No, sum of obtuse angles is greater than $$180^\\circ.$$, (iii) Yes, sum of two right angles is $$180^\\circ.$$, An angle is greater than $$45^\\circ.$$ Is its complementary angle greater than $$45^\\circ$$ or equal to $$45^\\circ$$ or less than $$45^\\circ\\,?$$. Visit official website CISCE ( Indian Council of Secondary Education ) for detail information about ICSE Maths. Two of the transversal expert math teachers to help students prepare for CBSE. Lines such that pairs of alternate interior angles ] CBSE Previous Year Question for! 7 solution is one of the measures of complementary angles is 90° [ Corresponding angles are: 8 solve supplementary! These ( Page 93 ) Question 1 90^\\circ.\\ ) sets on Lines and angles angles comprises of 2... Angles of the lower triangle add up to 180° Choice questions of Class 7 solution angle z the! 5.1 1 different concepts and problem sets on Lines and angles Class 7 Maths Chapter 5 Lines! Are prepared by our expert math teachers to help students prepare for their exams. That both the angles still remain supplementary vertical angles because they are complementary, then vertically... N°T f°rm a linear pair are _______________ and ACLS instructor and an American Safety and Health Institute advanced instructor", "# 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", "$$45^\\circ$$. ## 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.", "measurement of an angle is degree$({}^\\circ )$ Types of Angles Acute Angle The angle between $0{}^\\circ$ and $90{}^\\circ$ is called an acute angle. For example, $10{}^\\circ ,\\,\\,30{}^\\circ ,\\,\\,60{}^\\circ ,\\,\\,80{}^\\circ$are acute angles. Right Angle An angle of measure $90{}^\\circ$ is called a right angle. Obtuse Angle An angle whose measure is between $90{}^\\circ$ and $180{}^\\circ$ is called an obtuse angle. Straight Angle An angle whose measure is $180{}^\\circ \\,\\,\\text{is}$called a straight angle. Reflex Angle An angle whose measure is more than $180{}^\\circ$ and less than $360{}^\\circ$is called a reflex angle. Complementary Angle Two angles whose sum is $90{}^\\circ \\,\\,\\text{is}$ called the complimentary angle. Complementary angle of any angle $\\theta$ is$90{}^\\circ -\\theta$. Supplementary Angle Two angles whose sum is $180{}^\\circ \\,\\,\\text{is}$ called supplementary angles. Supplementary angle of any angle $\\theta$ is$180{}^\\circ -\\theta$. Adjacent Angle Two angles are said to be adjacent if they have a common vertex and one common arm. In the following figure $\\angle AOC$ and $\\angle COB$are adjacent angles. Vertically Opposite Angles Two angles are said to be vertically opposite angles if they are formed at the intersecting point of two lines. In the figure given more... #### Notes - Algebraic Expressions Algebraic Expressions In an algebraic expression constant and variables are linked with arithmetic operations. The value of unknown variable is obtained by simplification of the", "# Chapter 1 - Section 1.1 - Angles, Degrees, and Special Triangles - 1.1 Problem Set: 9 $10^{\\circ}$ is an acute angle. Complement of $10^{\\circ}$ = $80^{\\circ}$ Supplement of $10^{\\circ}$ = $170^{\\circ}$ #### Work Step by Step Angles that measure between $0°$and $90°$ are called acute angles. Complement of $10^{\\circ}$ = $90^{\\circ} - 10^{\\circ}$ = $80^{\\circ}$ (Because sum of the measures of complementary angles is 90 degrees) Supplement of $10^{\\circ}$ = $180^{\\circ} - 10^{\\circ}$ = $170^{\\circ}$ (Because sum of the measures of supplementary angles is 180 degrees) 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.", "= 6 in equation (i), x + y = 9 ∴ 6 + y = 9 ∴ y = 9 – 6 = 3 ∴ Original number = 10y + x = 10(3)+ 6 = 30 + 6 = 36 ∴ The two digit number is 36. Question 8. In ∆ABC, the measure of ∠A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle. Solution: Let the measure of ∠B be ‘x°’ and that of ∠C be ‘y°’. According to the first condition, m∠A = m∠B + m∠C ∴ m∠A = x° + y° In AABC, m∠A + m∠B + m∠C = 180° …[Sum of the measures of the angles of a triangle is 180°] ∴ x + y + x + y = 180 , ∴ 2x + 2y = 180 Dividing both sides by 2, x + y = 90 …(i) According to the second condition, the ratio of the measures of ∠B and ∠C is 4 : 5. ∴ $$\\frac { x }{ y }$$ = $$\\frac { 4 }{ 5 }$$ ∴ 5x = 4y ∴ 5x – 4y = 0 …….(ii) Multiplying equation (i) by 4, 4x +", "# 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", "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}", "## anonymous one year ago The ratio of the measures of two supplementary angles is 7:8. Which proportion(s) could you use to find the measures of the angles? 1. anonymous This is multiple choice, some of the answers are: x/y= 7/8 x/y=8/7 x/180-x=7/8 x/180-x=8/7 2. anonymous @HWBUSTER00 3. anonymous @skullpatrol 4. anonymous try both 3rd or 4th 5. anonymous Thanks for replying, how did you come to this conclusion? 6. anonymous $x+y=180,y=180-x$ $\\frac{ x }{ y }=\\frac{ x }{ 180-x }=\\frac{ 7 }{ 8 }$ 7. anonymous Thank you very much for your help!", "be around 30 degrees. Concept insight: This problem talks about the measure of two supplementary angles. The pair of adjacent angles here are constructed on a line segment, but not all adjacent angles are linear. A line segment with A and B as two endpoints is represented as $$\\overline{AB}$$. Davneet Singh is a graduate from Indian … Then, find the angle measures. Do all the multiplication and division operations within the function. Report. They just need to add up to 180 degrees. 0000272735 00000 n complementary angles and supplementary angles Home; Events; Register Now; About But they are not a linear pair because they are not connected, and they do not share a common side. Name three angles whose measures sum to 180° [6 marks) Critical Thinking: A counterexample is used to show that a statement is not necessarily true. 0000387430 00000 n 0000245380 00000 n They do not form a straight line. 0000014953 00000 n 0000018975 00000 n 6 years ago. are solved by group of students and teacher of Class 7, which is also the largest student community of Class 7. 0000376384 00000 n The pair of equations can then be solved by suitable substitution. In the figure shown above, only the last one represents a linear pair, as the sum of the adjacent angles is", "# In triangle PQR, the measure of angle P is 36 degrees. The measure of angle Q is five times the measure of angle R. How do you find the measurement of angle Q and the measurement of angle R? Mar 31, 2017 #### Explanation: \"the sum of measures of these three angles of any triangle is invariably equal to the straight angle, also expressed as 180°\" From the problem we can write: $Q = 5 R$ We can also write from Geometry, the sum of the three angles is 180 degrees, or: $P + Q + R = 180$ We can substitute $36$ for $P$ and we can substitute $5 R$ for $Q$ and solve for $R$: $36 + 5 R + R = 180$ $36 + 5 R + 1 R = 180$ $36 + \\left(5 + 1\\right) R = 180$ $36 + 6 R = 180$ $- \\textcolor{red}{36} + 36 + 6 R = - \\textcolor{red}{36} + 180$ $0 + 6 R = 144$ $6 R = 144$ $\\frac{6 R}{\\textcolor{red}{6}} = \\frac{144}{\\textcolor{red}{6}}$ $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{6}}} R}{\\cancel{\\textcolor{red}{6}}} = 24$ $R = 24$ Substituting $24$ for $R$ in the the equation for the relationship and calculating $Q$ gives: $Q = 5 R$ becomes: $Q = 5 \\cdot 24$ $Q = 120$ Angle $Q$ is 120", "Home > English > Class 7 > Maths > Chapter > Lines And Angles > Sum of measures of two supplem... # Sum of measures of two supplementary angles is . . . . . . . . . Khareedo DN Pro and dekho sari videos bina kisi ad ki rukaavat ke! Answer : 180^(@)", "(2x – 2) = 180$$. If the sum of two angles is 180 degrees, then we say that they are supplementary. You just have to remember that their sum is 180° and that any set of angles lying along a straight line will also be supplementary. Since $$\\angle A$$ and $$\\angle B$$ are supplementary, their sum is 180o. This supplementary angle quiz tests your ability to: Recognize supplementary angle pairs Find the sum of two angles in a parallelogram Determine a given angle in a parallelogram One way to avoid mixing up these definitions is to note that s comes after c in the alphabet, and 180 is greater than 90. Since straight angles have measures of 180°, the angles are supplementary. Complementary angles are two angles that add up to 90 degrees. Two Angles are Supplementary when they add up to 180 degrees. If two Angles are Supplementary, then their sum of measures is 180˚ Measure of one Angle = 65˚ Let the Measure of the Second Angle be = x˚ Sum of Supplementary Angles = 180˚ For example, angle 130° and angle 50° are supplementary because on adding 130° and 50° we get 180°. In the above figure, $$130^\\circ+50^\\circ = 180^\\circ$$. The angles with measures $$a$$° and $$b$$° lie along a straight line. Usually you are" ]

[ "right triangle with a hypotenuse of 650 feet, and one side being 600 feet. If so, use the Pythagorean Theorem, with x being the missing side. X^2 + 600^2 = 650^2 Solve for x", "feet: Use: 3 (base)(height) = Area (12 pts) 18.) The base of a triangle is 2 feet more than four times the height: Set up and solve an equation to find the base and the height if the area is 15 square feet: Use: 3 (base)(height) = Area (12 pts)...", "## anonymous 5 years ago 5. The area of a triangle is equal to that of a square whose side measures 60 inches. Find the side of the triangle whose corresponding altitude is 90 inches. The side length of the square is 60 inches. The area is therefore 60 x 60 = 3600 square inches. We need to find the base of the triangle such that it too, given an altitude of 90 inches, will have an area of 3600 square inches. We have$3600=\\frac{1}{2}bh=\\frac{1}{2}b(90)=45b$That is$b=\\frac{3600}{45}=80$inches.", "base and change the altitude. In this case, the area of the triangle is half of the enclosing rectangle. This is true, since the condition above states that the length and width of the rectangle are given. Figure 5 - Getting the area of a triangle by changing the base. Delving Deeper Playfair’s axiom guarantees that we can enclose any triangle with a rectangle, because given a line (base of a triangle) and a point (opposite vertex), we can always draw a unique line parallel to the base and passing through that vertex. To construct an enclosing rectangle, we can also draw two lines perpendicular to the base and passing through the other two vertices. ## Introduction to the Concept of Area Suppose a room floor with length 12 feet and width 10 feet shown in Figure 2 is to be tiled using of a 1 foot by 1 foot tile shown in Figure 1, then it is clear 120 tiles are needed. Figure 1 – A square representing a tile measuring 1 foot by 1 foot. There are cases, of course, that some tiles have be to cut due to the size restriction of the room. For instance, How many 1 foot by 1 foot tile do we need to cut to cover a floor of a", "niababy02 23 # The base length of a triangle is 12 feet and the height is 6 feet. What is the area of the triangle? CHOICES 10 square feet 12 square feet 18 square feet 36 square feet The area of a triangle is calculated by 1/2bh $\\frac{1}{2}(6)(12)$=36 sq ft The formula to find the area of a triangle is: $\\sf~A=\\dfrac{1}{2}bh$ Plug in what we know: $\\sf~A=\\dfrac{1}{2}(12)(6)$ Multiply: $\\sf~A=\\boxed{\\sf36}$", "Find the altitude, in feet, of the triangle if the length of the corresponding base is 30 feet. 212. ## geo A square has an area of 25. A rectangle has the same width as the square. The length of the rectangle is double its width. What is the area of the rectangle?", "area of 3,600 square miles? 72 miles long and 50 miles wide. Question 12. The angle measures of a triangle are 57°, 60°, and 63°. How can you classify the triangle? An acute-angled triangle is a type of triangle in which all the three internal angles of the triangle are acute, that is, they measure less than 90°. Question 13. Newton is walking on Path X that continues straight ahead. He turns onto Path Y. Using a protractor, which statement is true about ∠FGH The angle is 45°, and is acute. Question 14. You jump 3$$\\frac{1}{2}$$ feet. How far do you jump in inches? 3(1/2)=7/2=3.5feet. Question 15. Which of the following shows $$\\overrightarrow{D E}$$ ? Option DE. Question 16. A rectangular backyard has a perimeter of 280 feet. The length of the backyard is 75 feet. What is the area of the backyard? The perimeter of the backyard =280 feet. length of the backyard =75 feet The breadth of the backyard = b feet. 280=75 + b b=280-75 b=205. Area of the backyard= l x b. Area = 205 x 75. Area=15375 sq feet. Question 17. Your friend makes a banner by taping the edges of 10 pieces of paper together. Each piece of paper is 28 centimeters long and 22 centimeters wide. Part A What is the", "# The area of a triangle is 72 square feet and the height is 16 feet. What is the base of the triangle? Sep 25, 2015 9 feet #### Explanation: ${A}_{t} = \\frac{b \\times h}{2}$ Substitute in the values given; $72 = \\frac{16 h}{2}$ $h = \\frac{144}{16} = 9$ Therefore $h = 9$ feet", "each 2 ft. The altitude of the trapezoid measures 1.7 ft. 10. A square prism measuring 3 mi along each edge of the base and 4 mi tall.", "right triangle. a. Use slopes to determine which sides are perpendicular. b. Find the vertices of the triangle. c. Find the perimeter and area of the triangle. Question 34. THOUGHT-PROVOKING Your bedroom has an area of 350 square feet. You are remodeling to include an attached bathroom that has an area of 150 square feet. Draw a diagram of the remodeled bedroom and bathroom in a coordinate plane. It is given that your bedroom has an area of 350 square feet. You are remodeling to include an attached bathroom that has an area of 150 square feet. So, The dimensions of the bedroom that has an area of 350 square feet are: Length: 20 feet and Width: 17.5 feet (Remember that any of the factor products of 350 will be the dimensions of the bedroom) So, The dimensions of the bedroom that has an area of 350 square feet in the coordinate plane are: (0, 0), (0, 17.5), (20, 17.5), and (20, 0) Now, The dimensions of an attached bathroom that has an area of 150 square feet are: Length: 10 feet and Width: 15 feet (Remember that any of the factor products of 350 will be the dimensions of the bedroom) So, The dimensions of the attached bathroom that has an area of 150 square feet in", "of a triangle increase, if the base and the height are increased by 2 times? 3 times 4 times 5 times 6 times Workspace 7. Find the area of a parallelogram, if the base is 10 inches and the corresponding height is 7 inches. 70 in 35 in 35 $in^{2}$ 70 $in^{2}$ Workspace 8. A hall measures 18 feet long and 10 feet wide. If the width of the hall is doubled, what would be the area of the new hall? 34 square feet 134 square feet 104 square feet 360 square feet Workspace 9. What happens to the area of the triangle, when the base of the triangle is increased by 5 times? Decreases by 5 times Decreases by 7 times Increases by 5 times Increases by 7 times Workspace 10. Find the height of the parallelogram whose base is four times that of the height and whose area is 576 $cm^{2}$ 22 cm 12 cm 25 cm 32 cm Workspace 1 2 3 4 5 6 7 8 9 10", "# A rectangle and a triangle have equal areas. The length of the rectangle is 12 inches, and its width is 8 inches. If the base of the triangle is 32 inches, what is the length, in inches, of the altitude drawn to the base? (A) 6 Explanation Area of rectangle = length x width $$12 \\times 8=96 \\text { sq. in. }$$ Area of triangle = (1/2) base * altitude = (1/2)(32)(x) = 16x 16x = 96 x = 6 in.", "height. The area of this triangle is 18 square units whether we use 4 units or 9 units for the base.", "# An isosceles triangle has an altitude of 8cm. Its perimeter is 32cm. How can we find its area? I found: $48 c {m}^{2}$", "# 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 $$M.$$ ×", "Tahoe with some one-dimensional measurements around the lake. The area of the rectangle is 160 square miles, and the area of the triangle is 17.5 square miles for a total of 177.5 square miles. We recognize that this is an approximation, and not likely the exact area of the lake. Aerial photo of Lake Tahoe Copyright Owner: US Geological Survey License: Public Domain", "= (1/2)bh area = (1/2) x 18 x 7 area = (126/2) area = 63 in Question 2. Area =26 sq. km Explanation: The area of the triangle = half the product of the base and the product of the height area = (1/2)bh area = (1/2) x 8 x 6.5 area = (52/2) area = 26 sq. km Question 3. Area =25 sq. ft Explanation: The area of the triangle = half the product of the base and the product of the height area = (1/2)bh area = (1/2) x 4 x 12.5 area = (50/2) area = 25 sq. ft Question 4. The above quadrilateral look similar to the rectangle Explanation: Rectangle the area of the rectangle = length x width area = l x w Question 5. The above diagram is similar to the trapezoid Explanation: The area of the trapezoid = half the product of the base and the product of the height area = (1/2) x b x h Question 6. The above figure is similar to the parallelogram Explanation: The area of the parallelogram = the product of the height and the product of the base b area = b x h Question 7. On a normal day, 12 airplanes arrive at an airport every 15 minutes. Which rate does not represent", "Since the apex of the triangle is submerged, there can be no additional area from a depth of 3 feet to 0 feet. Your expression for the width of the triangle needs some adjustment to account for this. P: 1,443 Quote by SteamKing Since the apex of the triangle is submerged, there can be no additional area from a depth of 3 feet to 0 feet. Your expression for the width of the triangle needs some adjustment to account for this. Does this mean that the 9 - x should be actually 6 - x? From the question, the triangle is inverted and submerged 3 ft under water... wouldn't the force be from 0 depth to 9 ddepth?? HW Helper Thanks P: 5,152 I think you need to take a moment and make a sketch of the problem. The bottom of the triangle is 9 feet below the surface, and the top is 3 feet below. From 3 feet below to the surface, there is no triangle, so the width is zero. If x is going to be your overall depth measured from the surface, then the vertical position relative to the base of the triangle must be the depth minus 3 feet. The depth is required to determine hydrostatic pressure, but the depth coordinate must be modified", "height of the triangle. The base of a triangle exceeds the height by 4 feet. If the area is 58.5 square feet, find the length of the base and the height of the triangle.... ### Matthew .......be at home. I can see his bike in front of his home. [put the correct modal verb in the blank to complete the sentence]​ Matthew .......be at home. I can see his bike in front of his home. [put the correct modal verb in the blank to complete the sentence]​...", "# Square Feet of a Triangle Calculator Created by Rijk de Wet Reviewed by Krishna Nelaturu Last updated: Aug 30, 2022 Welcome to the square feet of a triangle calculator, where we'll explain all there is to know about how to calculate the square feet of a triangle. Let's get to the point and calculate some triangles' square feet! ## How do I use the square feet of a triangle calculator? Using the square feet of a triangle calculator is easy! Follow these steps: 1. Select the triangle type from the drop-down list. You can find the square feet of triangles of these types: • base and height; • Three sides (SSS); • Side-angle-side (SAS); and • Angle-side-angle (ASA). 2. Enter the measurements of the triangle type you've selected. Refer to the schematic at the top of the calculator if you're unsure which measurements correspond to which fields. 3. Let the square feet of a triangle calculator automatically find the area of the triangle. If you want to learn how to find the square feet of a triangle by yourself, then keep scrolling! ## How do I calculate the square feet of a triangle? It depends on what measurements of the triangle you already know. Most commonly, you'd have one of these sets of measurements at your disposal:" ]

[ "point $500 ft$ due east $P.$ At what rate are the people moving apart $15 min$ after the woman starts walking? Chris T. ### Problem 20 A baseball diamond is a square with side $90 ft.$ A batter hits the ball and runs toward first base with a speed of $24 ft/s.$ (a) At what rate is his distance from second base decreasing when he is halfway to first base? (b) At what rate is hits distance from third base increasing at the same moment? Chris T. ### Problem 21 The altitude of a triangle is increasing at a rate of $1 cm/min$ while the area of the triangle is increasing at a rate of $2 cm^2/min.$ At what rate is the base of the triangle changing when the altitude is $10 cm$ and the area is $100 cm^2?$ Amrita B. ### Problem 22 The altitude of a triangle is increasing at a rate of $1 cm/min$ while the area of the triangle is increasing at a rate of $2 cm^2/min.$ At what rate is the base of the triangle changing when the altitude is $10 cm$ and the area is $100 cm^2?$ Suman Saurav T. ### Problem 23 At noon, ship $A$ is $100 km$ west of ship $B.$ Ship $A$ is sailing south at $35 km/h$", "a triangle. Since it's a right triangle, one leg can be considered as the base, and the other as the height. Substitute $45000$ for $A$ and $180$ for $b$ in the formula. $45000=\\frac{1}{2}\\left(180\\right)h$ Simplify. $45000=90h$ Divide both sides by $90$ . $\\frac{45000}{90}=h$ $500=h$ The other leg of the triangle is $500$ meters long.", "Intermediate Algebra (6th Edition) $h=9$ feet We know that the area of a triangle can be calculated as $A=.5bh$, where b is the base and h is the height of the triangle. We know that the area is 18 square feet and the base is 4 feet, so we can plug these values into the area formula to solve for h. $18=.5\\times4\\times h=2h$ Divide both sides by 2. $h=9$ feet", "# 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 $$M.$$ ×", "## anonymous 5 years ago 5. The area of a triangle is equal to that of a square whose side measures 60 inches. Find the side of the triangle whose corresponding altitude is 90 inches. The side length of the square is 60 inches. The area is therefore 60 x 60 = 3600 square inches. We need to find the base of the triangle such that it too, given an altitude of 90 inches, will have an area of 3600 square inches. We have$3600=\\frac{1}{2}bh=\\frac{1}{2}b(90)=45b$That is$b=\\frac{3600}{45}=80$inches.", "# How do you solve a right triangle with side a = 400 feet and side b = 900 feet? Apr 26, 2018 check the explanation below. #### Explanation: Its Area $= \\frac{a b}{2}$ $= \\frac{400 \\times 900}{2} = 180000 f {t}^{2}$ It's hypotenuse $= \\sqrt{{a}^{2} + {b}^{2}}$ $= \\sqrt{{400}^{2} + {900}^{2}} = 984.885 f t$ It's perimeter $= a + b +$ hypotenuse length $= 2284.88 f t$", "this regular pentagon, let's find the area of one triangle and then multiply by 5. Drawing in the altitude creates a right triangle, so we can use trigonometry to calculate the lengths of both $$x$$ and $$h$$. To find $$\\theta$$ use the fact that a full rotation is $$360^\\circ$$ and that in an isosceles triangle the altitude is also an angle bisector. So $$\\theta=360 \\div 10$$. $$\\sin(36)=\\frac{x}{1}$$ so $$x$$ is about 0.59 units. $$\\cos(36)=\\frac{h}{1}$$ so $$h$$ is about 0.81 units. The area of the isosceles triangle is about 0.48 square units and the area of the pentagon is 5 times that, or about 2.4 square units. That's not very close to the area of the circle, but if we add more and more sides to the regular polygon, its area gets closer and closer to covering the entire circle.", "2 per m². Solution: Side of rhombus garden (b) = 30 m. Altitude (h) = 16 m Area = Base x altitude = 30 x 16 = 480 m² Rate of levelling the garden = Rs 2 per m² Total cost = Rs 480 x 2 = Rs 960 Question 21. A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ? Solution: Length of side of rhombus (b) = 64 m and altitude (h) = 16 m Area = b x h = 64 x 16 m² = 1024 m² Now area of square = 1024 m² Side of the square = √Area = √1024 m = 32 m Question 22. The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal. Solution: Base of triangle (b) = 24.8 cm and altitude (h) = 16.5 cm Area of triangle = $$\\frac { 1 }{ 2 }$$ x base x height = $$\\frac { 1 }{ 2 }$$", "the area of the parallelogram is A. $$40$$ B. $$24\\sqrt3$$ C. $$72$$ D. $$48\\sqrt3$$ E. $$96$$ Source: Official Guide Solution: By drawing the altitude, we see that we have a 30-60-90 triangle nested in the parallelogram, with 8 as the hypotenuse and the altitude as the side opposite the 60-degree angle. Since the ratio of the sides of a 30-60-90 triangle is x : x√3 : 2x, the base of the triangle will be x = 4, the altitude will be x√3 = 4√3, and the hypotenuse will be 2x = 8. Therefore, the area of the parallelogram is base x height = 12 x 4√3 = 48√3.", "The altitude of a triangle is increasing at a rate of $2.000$ centimeters/minute while the area of the triangle is increasing at a rate of $4.000$ square centimeters/minute. At what rate is the base of the triangle changing when the altitude is $12.000$ centimeters and the area is $94.000$ square centimeters? Note: The \"altitude\" is the \"height\" of the triangle in the formula \"Area=(1/2)*base*height\". Draw yourself a general \"representative\" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shape the triangle really is. Your overall score for this problem is", "## Elementary Algebra We need to find the area of the triangle, which can be found by using the formula: A = $\\frac{b\\ \\times\\ h}{2}$ Substitute 130 for b and 60 for h to obtain: A = $\\frac{130\\ \\times\\ 60}{2}$ A = 130 $\\times$ 30 A = 3900 square feet Since one bag of fertilizer covers 4000 square feet, it is enough to cover the lawn, which is 3900 square feet.", "## Basic College Mathematics (10th Edition) $\\approx 84.9$ feet (a) See image below. (b) To find the distance from home plate to second base, we divide the square in two pieces to form a right triangle. We know that for a right triangle the Pythagorean Theorem states: $leg_1^2+leg_2^2=hypotenuse^2$ $a^2+b^2=c^2$ Our legs are 60 ft long, so we have: $60^2+60^2=c^2$ $3600+3600=c^2$ $7200=c^2$ $c=\\sqrt{7200}$ $c\\approx 84.9$ feet", "2. 16.36 m 3. 17.60 m 4. 18.40 m Question 5. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is : 1. 10$$\\sqrt 2$$ m 2. 100 m 3. 100$$\\sqrt 2$$ m 4. 200 m 1. $$25 \\,c{m^2}$$ 2. $$\\frac{{25}}{2}\\sqrt 2 \\,c{m^2}$$ 3. $$25\\sqrt 2 \\,c{m^2}$$ 4. $$25\\sqrt 3 \\,c{m^2}$$", "of a triangle is twice the smallest side. The third side is $55$ feet more than the shortest side. The perimeter is $1717$ feet. Find the lengths of all three sides. 196. One side of a triangle is three times the smallest side. The third side is $33$ feet more than the shortest side. The perimeter is $1313$ feet. Find the lengths of all three sides. Use the Properties of Trapezoids In the following exercises, solve using the properties of trapezoids. 197. The height of a trapezoid is $1212$ feet and the bases are $99$ and $1515$ feet. What is the area? 198. The height of a trapezoid is $2424$ yards and the bases are $1818$ and $3030$ yards. What is the area? 199. Find the area of a trapezoid with a height of $5151$ meters and bases of $4343$ and $6767$ meters. 200. Find the area of a trapezoid with a height of $6262$ inches and bases of $5858$ and $7575$ inches. 201. The height of a trapezoid is $1515$ centimeters and the bases are $12.512.5$ and $18.318.3$ centimeters. What is the area? 202. The height of a trapezoid is $4848$ feet and the bases are $38.638.6$ and $60.260.2$ feet. What is the area? 203. Find the area of a trapezoid with a height of $4.24.2$ meters", "# isosceles triangle. perimeter= 36, altitude= 6. find lengths • March 29th 2008, 11:37 AM polakio92 isosceles triangle. perimeter= 36, altitude= 6. find lengths find the lengths of the sides of an isosceles triangle if the perimeter is 36 and the altitude to the base is 6. • March 29th 2008, 01:35 PM Peritus $ \\begin{gathered} 2s + b = 36 \\hfill \\\\ 6^2 + \\left( {\\frac{b} {2}} \\right)^2 = s^2 \\hfill \\\\ \\end{gathered}$ ...", "in isosceles triangles so you can calculate the area. The area of a triangle is $\\frac{1}{2} \\ bh$, where $b$ is the base and $h$ is the height (or altitude). If you are given the base and the sides of an isosceles triangle, you can use the Pythagorean Theorem to calculate the height. Example 6: What is the area of the isosceles triangle? Solution: First, draw the altitude from the vertex between the congruent sides, which will bisect the base (Isosceles Triangle Theorem). Then, find the length of the altitude using the Pythagorean Theorem. $7^2 + h^2 & = 9^2\\\\49 + h^2 & = 81\\\\h^2 & = 32\\\\h & = \\sqrt{32} = 4 \\sqrt{2}$ Now, use $h$ and $b$ in the formula for the area of a triangle. $A = \\frac{1}{2} \\ bh = \\frac{1}{2} (14) \\left (4 \\sqrt{2} \\right ) = 28 \\sqrt{2} \\ units^2$ ## The Distance Formula Another application of the Pythagorean Theorem is the Distance Formula. We have already been using the Distance Formula in this text, but we can prove it here. First, draw the vertical and horizontal lengths to make a right triangle. Then, use the differences to find these distances. Now that we have a right triangle, we can use the Pythagorean Theorem to find $d$. Distance Formula: The distance $A(x_1,", "$$16√3$$ 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)", "Try to draw the obtuse-angled triangle, $\\measuredangle BAC>90^{\\circ}.$ The altitude from $A$ is $2$ and $BC=6$. Now, easy to draw our square. – Michael Rozenberg Jan 24 at 19:10 • @MichaelRozenberg You assumed that the distance to the nearest side is $2-x$. That means that $DE$ is on the same side of $A$ as $BC$. You also have a solution on the opposite side. The distance is then $x-2$ instead, which yields the solution $x=3$, with perimeter $12$ – Andrei Jan 24 at 19:59", "and angle $C$ would be same. Let us say that they both are equal to $\\theta$. Then the sum of angles would be: $\\theta\\ +\\ 90\\ +\\ \\theta$ = $180$ Combining like terms we have: $2 \\theta\\ +\\ 90$ = $180$ Subtracting 90 from both the sides we have: $2\\ \\theta\\ +\\ 90\\ -\\ 90$ = $180\\ -\\ 90$ $2\\ \\theta$ = $90$ Dividing both the sides by $2$ gives: $\\frac{2\\ \\theta}{2}$ = $\\frac{90}{2}$ $\\theta$ = $45^{\\circ}$ Thus we see that the two congruent angles in an isosceles right triangle would measure $45$ degrees each. So the triangle would look as follows: ## Isosceles Right Triangle Area The area of any triangle is given by the formula: $Area$ = $\\frac{1}{2}$ $bh$ Where $b$ is the length of the base of the triangle and h is the altitude of the triangle. Altitude refers to the shortest distance between the opposite vertex and the base. It is same as the length of the perpendicular dropped from the opposite vertex on to the base. For a right triangle, if we consider one of the legs as the base, then the perpendicular from the opposite vertex to this base would simply be the other leg. So, if the length of the base of the right triangle is $b$ and that of the", "# Chapter 2 - Section 2.3 - Formulas and Problem Solving - Exercise Set: 35 $h=9$ feet #### Work Step by Step We know that the area of a triangle can be calculated as $A=.5bh$, where b is the base and h is the height of the triangle. We know that the area is 18 square feet and the base is 4 feet, so we can plug these values into the area formula to solve for h. $18=.5\\times4\\times h=2h$ Divide both sides by 2. $h=9$ feet 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." ]

[ "Intermediate Algebra (6th Edition) $h=9$ feet We know that the area of a triangle can be calculated as $A=.5bh$, where b is the base and h is the height of the triangle. We know that the area is 18 square feet and the base is 4 feet, so we can plug these values into the area formula to solve for h. $18=.5\\times4\\times h=2h$ Divide both sides by 2. $h=9$ feet", "## anonymous 5 years ago 5. The area of a triangle is equal to that of a square whose side measures 60 inches. Find the side of the triangle whose corresponding altitude is 90 inches. The side length of the square is 60 inches. The area is therefore 60 x 60 = 3600 square inches. We need to find the base of the triangle such that it too, given an altitude of 90 inches, will have an area of 3600 square inches. We have$3600=\\frac{1}{2}bh=\\frac{1}{2}b(90)=45b$That is$b=\\frac{3600}{45}=80$inches.", "# How do you solve a right triangle with side a = 400 feet and side b = 900 feet? Apr 26, 2018 check the explanation below. #### Explanation: Its Area $= \\frac{a b}{2}$ $= \\frac{400 \\times 900}{2} = 180000 f {t}^{2}$ It's hypotenuse $= \\sqrt{{a}^{2} + {b}^{2}}$ $= \\sqrt{{400}^{2} + {900}^{2}} = 984.885 f t$ It's perimeter $= a + b +$ hypotenuse length $= 2284.88 f t$", "# 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 $$M.$$ ×", "= (1/2)bh area = (1/2) x 18 x 7 area = (126/2) area = 63 in Question 2. Area =26 sq. km Explanation: The area of the triangle = half the product of the base and the product of the height area = (1/2)bh area = (1/2) x 8 x 6.5 area = (52/2) area = 26 sq. km Question 3. Area =25 sq. ft Explanation: The area of the triangle = half the product of the base and the product of the height area = (1/2)bh area = (1/2) x 4 x 12.5 area = (50/2) area = 25 sq. ft Question 4. The above quadrilateral look similar to the rectangle Explanation: Rectangle the area of the rectangle = length x width area = l x w Question 5. The above diagram is similar to the trapezoid Explanation: The area of the trapezoid = half the product of the base and the product of the height area = (1/2) x b x h Question 6. The above figure is similar to the parallelogram Explanation: The area of the parallelogram = the product of the height and the product of the base b area = b x h Question 7. On a normal day, 12 airplanes arrive at an airport every 15 minutes. Which rate does not represent", "# isosceles triangle. perimeter= 36, altitude= 6. find lengths • March 29th 2008, 11:37 AM polakio92 isosceles triangle. perimeter= 36, altitude= 6. find lengths find the lengths of the sides of an isosceles triangle if the perimeter is 36 and the altitude to the base is 6. • March 29th 2008, 01:35 PM Peritus $ \\begin{gathered} 2s + b = 36 \\hfill \\\\ 6^2 + \\left( {\\frac{b} {2}} \\right)^2 = s^2 \\hfill \\\\ \\end{gathered}$ ...", "Math Notes Subjects Algebra Solutions Topics || Problems The base of a triangle is 4 feet shorter than its altitude, and the area of the triangle is 126 square feet. Find the base and the altitude. Solution: Let $$b$$ be the base and $$a$$ be the altitude of the triangle. $$b = a- 4$$ $$A_t = 126$$ $$A_t = \\frac{1}{2}ba$$ $$126 = \\frac{1}{2}ba$$ $$a = \\frac{252}{b}$$ $$b = \\frac{252}{b} -4$$ $$b^2 = 252 - 4b$$ $$b^2+4b- 252=0$$ $$(b-14)(b+18) = 0$$ $$b = 14$$ $$a = 14+4 = 18$$ Thus, the base of the triangle is 14 feet and the altitude is 18 feet", "## Elementary Algebra We need to find the area of the triangle, which can be found by using the formula: A = $\\frac{b\\ \\times\\ h}{2}$ Substitute 130 for b and 60 for h to obtain: A = $\\frac{130\\ \\times\\ 60}{2}$ A = 130 $\\times$ 30 A = 3900 square feet Since one bag of fertilizer covers 4000 square feet, it is enough to cover the lawn, which is 3900 square feet.", "## Basic College Mathematics (10th Edition) $\\approx 84.9$ feet (a) See image below. (b) To find the distance from home plate to second base, we divide the square in two pieces to form a right triangle. We know that for a right triangle the Pythagorean Theorem states: $leg_1^2+leg_2^2=hypotenuse^2$ $a^2+b^2=c^2$ Our legs are 60 ft long, so we have: $60^2+60^2=c^2$ $3600+3600=c^2$ $7200=c^2$ $c=\\sqrt{7200}$ $c\\approx 84.9$ feet", "Since the apex of the triangle is submerged, there can be no additional area from a depth of 3 feet to 0 feet. Your expression for the width of the triangle needs some adjustment to account for this. P: 1,443 Quote by SteamKing Since the apex of the triangle is submerged, there can be no additional area from a depth of 3 feet to 0 feet. Your expression for the width of the triangle needs some adjustment to account for this. Does this mean that the 9 - x should be actually 6 - x? From the question, the triangle is inverted and submerged 3 ft under water... wouldn't the force be from 0 depth to 9 ddepth?? HW Helper Thanks P: 5,152 I think you need to take a moment and make a sketch of the problem. The bottom of the triangle is 9 feet below the surface, and the top is 3 feet below. From 3 feet below to the surface, there is no triangle, so the width is zero. If x is going to be your overall depth measured from the surface, then the vertical position relative to the base of the triangle must be the depth minus 3 feet. The depth is required to determine hydrostatic pressure, but the depth coordinate must be modified", "right triangle with a hypotenuse of 650 feet, and one side being 600 feet. If so, use the Pythagorean Theorem, with x being the missing side. X^2 + 600^2 = 650^2 Solve for x", "this regular pentagon, let's find the area of one triangle and then multiply by 5. Drawing in the altitude creates a right triangle, so we can use trigonometry to calculate the lengths of both $$x$$ and $$h$$. To find $$\\theta$$ use the fact that a full rotation is $$360^\\circ$$ and that in an isosceles triangle the altitude is also an angle bisector. So $$\\theta=360 \\div 10$$. $$\\sin(36)=\\frac{x}{1}$$ so $$x$$ is about 0.59 units. $$\\cos(36)=\\frac{h}{1}$$ so $$h$$ is about 0.81 units. The area of the isosceles triangle is about 0.48 square units and the area of the pentagon is 5 times that, or about 2.4 square units. That's not very close to the area of the circle, but if we add more and more sides to the regular polygon, its area gets closer and closer to covering the entire circle.", "radius 6, wha [#permalink] Show Tags 19 Mar 2018, 12:47 Area of triangle=1/2*base*Altitude= 1/2*6*Altitude =3*Altitude Area of circle= π*r*r = π*6*6=36π Given area of triangle= area of circle, =>3*Altitude=36π Altitude= 12π Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2830 Re: If a triangle of base 6 has the same area as a circle of radius 6, wha [#permalink] Show Tags 20 Mar 2018, 15:58 Bunuel wrote: If a triangle of base 6 has the same area as a circle of radius 6, what is the altitude of the triangle? A. $$6\\pi$$ B. $$8\\pi$$ C. $$10\\pi$$ D. $$12\\pi$$ E. $$14\\pi$$ We can create the equation, in which h = height = altitude of the triangle. We equate the area of the triangle with base 6 to the area of the circle with radius 6: 6 x h x 1/2 = π x 6^2 3h = 36π h = 12π _________________ Jeffery Miller GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: If a triangle of base 6 has the same area as a circle of radius 6, wha &nbs [#permalink] 20 Mar 2018, 15:58 Display posts from previous: Sort by", "niababy02 23 # The base length of a triangle is 12 feet and the height is 6 feet. What is the area of the triangle? CHOICES 10 square feet 12 square feet 18 square feet 36 square feet The area of a triangle is calculated by 1/2bh $\\frac{1}{2}(6)(12)$=36 sq ft The formula to find the area of a triangle is: $\\sf~A=\\dfrac{1}{2}bh$ Plug in what we know: $\\sf~A=\\dfrac{1}{2}(12)(6)$ Multiply: $\\sf~A=\\boxed{\\sf36}$", "# Math Help - isosceles triangle. perimeter= 36, altitude= 6. find lengths 1. ## isosceles triangle. perimeter= 36, altitude= 6. find lengths find the lengths of the sides of an isosceles triangle if the perimeter is 36 and the altitude to the base is 6. 2. $ \\begin{gathered} 2s + b = 36 \\hfill \\\\ 6^2 + \\left( {\\frac{b} {2}} \\right)^2 = s^2 \\hfill \\\\ \\end{gathered}$ ...", "2 per m². Solution: Side of rhombus garden (b) = 30 m. Altitude (h) = 16 m Area = Base x altitude = 30 x 16 = 480 m² Rate of levelling the garden = Rs 2 per m² Total cost = Rs 480 x 2 = Rs 960 Question 21. A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ? Solution: Length of side of rhombus (b) = 64 m and altitude (h) = 16 m Area = b x h = 64 x 16 m² = 1024 m² Now area of square = 1024 m² Side of the square = √Area = √1024 m = 32 m Question 22. The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal. Solution: Base of triangle (b) = 24.8 cm and altitude (h) = 16.5 cm Area of triangle = $$\\frac { 1 }{ 2 }$$ x base x height = $$\\frac { 1 }{ 2 }$$", "# An isosceles triangle has an altitude of 8cm. Its perimeter is 32cm. How can we find its area? I found: $48 c {m}^{2}$", "7-8. find the area. Question 7. Answer: The area of the figure is 32 ft2 Explanation: Given, The figure has one rectangle, one triangle, and another rectangle. Let us calculate the area of the first rectangle l × b l = 3 ft and width = 5 ft Area = 3 × 5 Area = 15 ft2 Next, find the area of the triangle Area of the traingle is $$\\frac{1}{2}$$ × b ×h h = 2 ft and b = 3 ft Area = $$\\frac{1}{2}$$ × 2 × 3 Area = 3 ft2 Area of the second rectangle is l × b l = 2 ft and width = 7 ft Area = 2 × 7 Area = 14 ft2 The total area of the given figure = area of the first rectangle + Area of the triangle + Area of the second rectangle Total area = 15 ft2 + 3 ft2 + 14 ft2 Total area = 32 ft2 Question 8. Answer: Total area = 31 ft2 Explanation: The given figure contains one rectangle, Two squares, and one smaller triangle. Let us calculate the area of the square is s × s Given the side of the square is 3 ft The area of the square is 3 × 3 Area = 9 ft2 Calculate the area", "in isosceles triangles so you can calculate the area. The area of a triangle is $\\frac{1}{2} \\ bh$, where $b$ is the base and $h$ is the height (or altitude). If you are given the base and the sides of an isosceles triangle, you can use the Pythagorean Theorem to calculate the height. Example 6: What is the area of the isosceles triangle? Solution: First, draw the altitude from the vertex between the congruent sides, which will bisect the base (Isosceles Triangle Theorem). Then, find the length of the altitude using the Pythagorean Theorem. $7^2 + h^2 & = 9^2\\\\49 + h^2 & = 81\\\\h^2 & = 32\\\\h & = \\sqrt{32} = 4 \\sqrt{2}$ Now, use $h$ and $b$ in the formula for the area of a triangle. $A = \\frac{1}{2} \\ bh = \\frac{1}{2} (14) \\left (4 \\sqrt{2} \\right ) = 28 \\sqrt{2} \\ units^2$ ## The Distance Formula Another application of the Pythagorean Theorem is the Distance Formula. We have already been using the Distance Formula in this text, but we can prove it here. First, draw the vertical and horizontal lengths to make a right triangle. Then, use the differences to find these distances. Now that we have a right triangle, we can use the Pythagorean Theorem to find $d$. Distance Formula: The distance $A(x_1,", "area of 3,600 square miles? 72 miles long and 50 miles wide. Question 12. The angle measures of a triangle are 57°, 60°, and 63°. How can you classify the triangle? An acute-angled triangle is a type of triangle in which all the three internal angles of the triangle are acute, that is, they measure less than 90°. Question 13. Newton is walking on Path X that continues straight ahead. He turns onto Path Y. Using a protractor, which statement is true about ∠FGH The angle is 45°, and is acute. Question 14. You jump 3$$\\frac{1}{2}$$ feet. How far do you jump in inches? 3(1/2)=7/2=3.5feet. Question 15. Which of the following shows $$\\overrightarrow{D E}$$ ? Option DE. Question 16. A rectangular backyard has a perimeter of 280 feet. The length of the backyard is 75 feet. What is the area of the backyard? The perimeter of the backyard =280 feet. length of the backyard =75 feet The breadth of the backyard = b feet. 280=75 + b b=280-75 b=205. Area of the backyard= l x b. Area = 205 x 75. Area=15375 sq feet. Question 17. Your friend makes a banner by taping the edges of 10 pieces of paper together. Each piece of paper is 28 centimeters long and 22 centimeters wide. Part A What is the" ]

[ "angles as $90^{\\circ}$. Hence, PQRS is a rectangle. Editor", "Q.4 In ?ABC, $$\\angle A = 30° \\angle B = 40° and \\angle C = 110°$$ In $$\\triangle PQR, \\angle P = 30°, \\angle Q = 40° and \\angle R = 110°.$$ A student says that $$\\triangle ABC = \\triangle PQR$$ by AAA congruence criterion. Is he justified? Why or why not?", "### In a right-angled triangle show that the hypotenuse is the longest side. 1. Let $PQR$ be a right-angled triangle such that $\\angle Q = 90^ \\circ.$ 2. As the sum of angles of a triangle is $180^ \\circ$. Now, in $\\triangle PQR,$ we have \\begin{aligned} & \\angle P + \\angle Q + \\angle R = 180 ^ \\circ \\\\ \\implies & \\angle P + 90 ^ \\circ + \\angle R = 180 ^ \\circ && [As, \\angle Q = 90^\\circ] \\\\ \\implies & \\angle P + \\angle R = 180 ^ \\circ - 90 ^ \\circ \\\\ \\implies & \\angle P + \\angle R = 90 ^ \\circ \\\\ \\end{aligned} \\\\ As the measure of $\\angle Q$ is equal to the sum of measures of $\\angle P$ and $\\angle R$, we have \\begin{aligned} & \\angle Q > \\angle P \\\\ \\implies & PR > QR & \\ldots \\text { (1) } && [\\text {Side opposite to greater angle is greater.}] \\\\ \\end{aligned} Also, \\begin{aligned} & \\angle Q > \\angle R \\\\ \\implies & PR > PQ & \\ldots \\text { (2) } && [\\text{Side opposite to greater angle is greater.}] \\end{aligned} 3. By equation $\\text { (1) }$ and $\\text { (2) }$, we have $$PR > QR \\text { and } PR > PQ \\\\$$ As", "# In the given figure, if PR is tangent to the circle at P and Q Question: In the given figure, if PR is tangent to the circle at and Q is the centre of the circle, then ∠POQ = (a) 110° (b) 100° (c) 120° (d) 90° Solution: We know that the radius is always perpendicular to the tangent at the point of contact. Therefore, we have, $\\angle O P R=90^{\\circ}$ It is given that, $\\angle Q P R=60^{\\circ}$ That is, $\\angle O P R-\\angle O P Q=60^{\\circ}$ $90^{\\circ}-\\angle O P Q=60^{\\circ}$ $\\angle O P Q=30^{\\circ}$ Now, consider $\\triangle O P Q$. We have, OP = OQ (Radii of the same circle) Since angles opposite to equal side will be equal in a triangle, we have, We know that sum of all angles of a triangle will be equal to . Therefore, $\\angle O P Q+\\angle O Q P+\\angle P O Q=180^{\\circ}$ $30^{\\circ}+30^{\\circ}+\\angle P O Q=180^{\\circ}$ $\\angle P O Q=120^{\\circ}$ The correct answer is option (c).", "## Geometry: Common Core (15th Edition) $\\angle(PMR), \\angle(KML), \\angle(KMQ), \\angle(MQP)$ Complementary angles are two angles whos measures have a sum of 90$^{\\circ}$. Each angle is called the compliment of the other.", "83.5^\\circ$ Now, add all three interior angles of the triangle $OPQ$. $\\angle QOP + \\angle OPQ + \\angle PQO = 44^\\circ + 52.5^\\circ + 83.5^\\circ$ $\\therefore \\,\\, \\angle QOP + \\angle OPQ + \\angle PQO = 180^\\circ$ It is proved that the summation of the all three interior angles is $180^\\circ$ exactly in any type of triangle. Save (or) Share", "# In a triangle $P Q R, N$ is a point on $P R$ such that $Q N \\perp P R$. If $P N$. $N R=Q^{2}$, prove that $\\angle \\mathrm{PQR}=90^{\\circ}$. #### Complete Python Prime Pack for 2023 9 Courses 2 eBooks #### Artificial Intelligence & Machine Learning Prime Pack 6 Courses 1 eBooks #### Java Prime Pack 2023 8 Courses 2 eBooks Given: In a triangle $P Q R, N$ is a point on $P R$ such that $Q N \\perp P R$. $P N$. $N R=Q^{2}$ To do: We have to prove that $\\angle \\mathrm{PQR}=90^{\\circ}$. Solution: $P N . N R=Q N^{2}$ $P N . N R=Q N . Q N$ This implies, $\\frac{P N}{Q N}=\\frac{Q N}{N R}$...........(i) In $\\triangle QNP$ and $\\triangle RNQ$, $\\frac{P N}{Q N}=\\frac{Q N}{N R}$ $\\angle P N Q=\\angle R N Q$ Therefore, by SAS similarity, $\\angle Q N P \\sim \\angle R N Q$ This implies, $\\angle P Q N=\\angle Q R N$........(ii) $\\angle R Q N=\\angle Q P N$........(iii) Adding (ii) and (iii), we get, $\\angle P Q N+\\angle R Q N=\\angle Q R N+\\angle Q P N$ $\\angle P Q R=\\angle Q R N+\\angle Q P N$ We know that, Sum of the angles of a triangle is $180^{\\circ}$. In $\\triangle P Q R$, $\\angle P Q R+\\angle Q P", "3$ : 2. Hence the area of the triangle PQR is $\\frac {1}{2} (7\\sqrt 3)$ square units.", ": 2. Hence the area of the triangle PQR is $\\frac {1}{2} (7\\sqrt 3)$ square units.", "Answer Comment Share Q) 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$", "? Free Version Moderate # Proving Triangles are Similar SATMAT-EYKJZY In $\\triangle QPR$ and $\\triangle XYT$, if the measure of $\\angle Q$ is ${ 71 }^{ \\circ }$, the measure of $\\angle Y$ is ${ 77 }^{ \\circ }$, and $\\angle Q\\cong \\angle X$, then what additional information would prove that these two triangles are similar? A The measure of $\\angle P={ 71 }^{ \\circ }$ B The measure of $\\angle X={ 71 }^{ \\circ }$ C The measure of $\\angle T={ 32 }^{ \\circ }$ D The measure of $\\angle R={ 148 }^{ \\circ }$", "}$ Using the formula, $\\alpha + \\beta + \\gamma = 180^{\\circ }$ substituting $\\alpha = 36.9^{\\circ }$, $\\beta = 53.1^{\\circ }$ and $\\gamma = 90^{\\circ }$ in the formula $36.9^{\\circ } + 53.1^{\\circ } + 90^{\\circ } = 180^{\\circ }$ $90^{\\circ } + 90^{\\circ } = 180^{\\circ }$ $180^{\\circ } = 180^{\\circ }$ L.H.S = R.H.S Therefore, we proved that the sum of angles in a triangle is always 180^{\\circ }. ### Practice Questions 1. Given a right-angled triangle with the reference angle $\\theta$. What is the measure of the angle $\\theta$? 2. Given a right-angled triangle with the reference angle $\\beta$. What is the measure of the angle $\\beta$ using the tangent function? 3. Given a right-angled triangle with the reference angle $\\alpha$. What is the measure of the angle $\\alpha$ using the cosine function? 4. Given a right-angled triangle with the reference angle $\\beta$. What is the measure of the angle $\\beta$? 5. Given a right-angled triangle with the reference angle $\\alpha$. What is the measure of the angle $\\alpha$?", "### Home > CCA2 > Chapter 2 > Lesson 2.1.4 > Problem2-62 2-62. Solve for the indicated value. Leave your answer in exact form. 1. $x=$ _______ Use the Pythagorean Theorem. $6^{2}+5^{2}=x^{2}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\sqrt{61}=x$ 1. $m∠C=$ _______ Use the Pythagorean Theorem to find the missing side. Is the triangle a special one? This is a $30^\\circ-60^\\circ-90^\\circ$ triangle. Therefore, $∠C$ is $30^\\circ$. 1. $m∠B=$ ____ The inverse tangent function will be helpful here. 1. $a=$_______ Use the cosine ratio or special right triangles.", "triangle => $a = r$ ———Eqn(2) From eqns(1) & (2), we get : $a = \\frac{b}{\\sqrt{2}}$ => $b = \\sqrt{2}a$", "$180^\\circ$. As $PM$ is a straight line. $$\\implies \\angle SQM + \\angle RQS + \\angle PQR = 180^\\circ$$ Therefore, $\\angle RQS = 180^\\circ - (\\angle SQM + \\angle PQR) = 180^\\circ - (60^\\circ + 60^\\circ) = 60^\\circ$ 5. Thus, the measure of $\\angle RQS$ is $60^\\circ$.", "a rectangle. Open in new tab link Thus, $$\\angle{P}$$ + $$\\angle{R}$$ = $$180^\\circ$$ ...(i)(Since, Sum of opposite angles in a cyclic quadrilateral is $$180^\\circ$$) But, $$\\angle{P}$$ = $$\\angle{R}$$ ...(ii)(Since, in a parallelogram, opposite angles are equal) $$\\angle{P}$$ = $$\\angle{R}$$ = $$90^\\circ$$ Similarly, $$\\angle{Q}$$ = $$\\angle{S}$$ = $$90^\\circ$$ Thus, Each angle of PQRS is $$90^\\circ$$.", "\\triangle XYZ$$. It cannot be represented as $$\\triangle PQR \\sim \\triangle YZX$$. But, similar triangles can be represented as $$\\triangle QRP \\sim \\triangle YZX$$.", "Observing that triangle $PQS$ is isosceles, we have $\\angle PSQ = \\frac {1}{2}(180^{\\circ} - 12^{\\circ}) = 84^{\\circ}$ and hence $\\angle PSR = 180^{\\circ} - 84^{\\circ}=96^{\\circ}$. Since triagle $PRS$ is also isosceles, we have $\\angle SPR = \\frac {1}{2}(180^{\\circ} - 96^{\\circ}) = 42^{\\circ}$. Hence $\\angle QPR = 12^{\\circ}+ 42^{\\circ} = 54^{\\circ}$.", "Measure of $$\\angle$$ P is not $$60^{\\circ}$$, so PQR is not an equilateral triangle. So Statement 2 alone is sufficient to answer whether $$\\triangle$$PQR is an equilateral triangle or not? so OA should be D (1) PQR is NOT an isosceles triangle. As above statement (1) says triangle is not isosceles how can we conclude that such triangle is not equilateral for gmat do we have to consider an equilateral triangle is isosceles from 3 sides ?? I think thats too much from GMAC. DS Forum Moderator Joined: 22 Aug 2013 Posts: 1340 Location: India Re: Is the measure of two exterior angles of a triangle PQR, each equal to [#permalink] ### Show Tags 18 Apr 2018, 05:09 GMAT215 wrote: Princ wrote: amanvermagmat wrote: Is the measure of two exterior angles of a triangle PQR, each equal to 120 degrees? (1) PQR is NOT an isosceles triangle. (2) Measure of angle P is 70 degrees. (Inspired by a Bunuel's question) OA:D First simplify the question:Is the measure of two exterior angles of a triangle PQR, each equal to 120 degrees? For a given triangle, Sum of all exterior angle is $$360^{\\circ}$$. Attachment: exterior-angles-triangle.png if two exterior angles of a triangle PQR are to be 120 degrees i.e total $$2*120^{\\circ} =240^{\\circ}$$,Third exterior angle should be also $$360^{\\circ}-240^{\\circ}=120^{\\circ}$$ $$\\angle$$P=$$\\angle$$Q=$$\\angle$$R= $$180^{\\circ}-120^{\\circ}$$($$120^{\\circ}$$", "the circle, angle OPQ = 90 degrees. Therefore, angle PQR = 180 - 70 - 90 = 20 degrees." ]

[ "# 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 $$a^\\circ$$? $a^\\circ = 180^\\circ - (70^\\circ + 70^\\circ )$ $= 180^\\circ - 140^\\circ = 40^\\circ$", "Answer Comment Share Q) 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$", "### In a right-angled triangle show that the hypotenuse is the longest side. 1. Let $PQR$ be a right-angled triangle such that $\\angle Q = 90^ \\circ.$ 2. As the sum of angles of a triangle is $180^ \\circ$. Now, in $\\triangle PQR,$ we have \\begin{aligned} & \\angle P + \\angle Q + \\angle R = 180 ^ \\circ \\\\ \\implies & \\angle P + 90 ^ \\circ + \\angle R = 180 ^ \\circ && [As, \\angle Q = 90^\\circ] \\\\ \\implies & \\angle P + \\angle R = 180 ^ \\circ - 90 ^ \\circ \\\\ \\implies & \\angle P + \\angle R = 90 ^ \\circ \\\\ \\end{aligned} \\\\ As the measure of $\\angle Q$ is equal to the sum of measures of $\\angle P$ and $\\angle R$, we have \\begin{aligned} & \\angle Q > \\angle P \\\\ \\implies & PR > QR & \\ldots \\text { (1) } && [\\text {Side opposite to greater angle is greater.}] \\\\ \\end{aligned} Also, \\begin{aligned} & \\angle Q > \\angle R \\\\ \\implies & PR > PQ & \\ldots \\text { (2) } && [\\text{Side opposite to greater angle is greater.}] \\end{aligned} 3. By equation $\\text { (1) }$ and $\\text { (2) }$, we have $$PR > QR \\text { and } PR > PQ \\\\$$ As", ". 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", "83.5^\\circ$ Now, add all three interior angles of the triangle $OPQ$. $\\angle QOP + \\angle OPQ + \\angle PQO = 44^\\circ + 52.5^\\circ + 83.5^\\circ$ $\\therefore \\,\\, \\angle QOP + \\angle OPQ + \\angle PQO = 180^\\circ$ It is proved that the summation of the all three interior angles is $180^\\circ$ exactly in any type of triangle. Save (or) Share", "= $$60^{\\circ}$$ and ∠B = $$60^{\\circ}$$ . Find the measure of ∠C. Solution: ∠A + ∠B + ∠C = $$180^{\\circ}$$ [The sum of all angles of a triangle is $$180^{\\circ}$$ ] $$60^{\\circ}$$ + $$60^{\\circ}$$ + ∠C = $$180^{\\circ}$$ [Substitute values] ∠C = $$180^{\\circ}$$ – $$120^{\\circ}$$ [Subtract 120 from both sides] ∠C = $$60^{\\circ}$$ Hence, the measure of ∠C is $$60^{\\circ}$$ . Example 2: For the $$\\triangle$$ABC given below, arrange the interior angles in increasing order of their measure. Solution: In the given $$\\triangle$$ABC, the measure of the sides are: a = 10 cm, b = 8 cm, and c = 6 cm. The decreasing order of side lengths is 10 > 8 > 6, that is, a > b > c. In a triangle, the measure of the angle opposite to the longer side is larger, or the angle opposite to the smaller side is smaller. Hence, the measure of angles in increasing order will be ∠A > ∠B > ∠C. Example 3: Find the area of the triangle whose base is 8 inches long and height is 12 inches. Solution: Area of triangle = $$\\frac{1}{2}$$ × b× h [Area of a triangle formula] = $$\\frac{1}{2}$$ (8) × (12) [Substitute values of b and h] = $$\\frac{1}{2}$$ (96) [Multiply] = 48 [Simplify] Hence, the area of the triangle", "angles as $90^{\\circ}$. Hence, PQRS is a rectangle. Editor", "}$ Using the formula, $\\alpha + \\beta + \\gamma = 180^{\\circ }$ substituting $\\alpha = 36.9^{\\circ }$, $\\beta = 53.1^{\\circ }$ and $\\gamma = 90^{\\circ }$ in the formula $36.9^{\\circ } + 53.1^{\\circ } + 90^{\\circ } = 180^{\\circ }$ $90^{\\circ } + 90^{\\circ } = 180^{\\circ }$ $180^{\\circ } = 180^{\\circ }$ L.H.S = R.H.S Therefore, we proved that the sum of angles in a triangle is always 180^{\\circ }. ### Practice Questions 1. Given a right-angled triangle with the reference angle $\\theta$. What is the measure of the angle $\\theta$? 2. Given a right-angled triangle with the reference angle $\\beta$. What is the measure of the angle $\\beta$ using the tangent function? 3. Given a right-angled triangle with the reference angle $\\alpha$. What is the measure of the angle $\\alpha$ using the cosine function? 4. Given a right-angled triangle with the reference angle $\\beta$. What is the measure of the angle $\\beta$? 5. Given a right-angled triangle with the reference angle $\\alpha$. What is the measure of the angle $\\alpha$?", "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.", "triangle is equal to $60^{\\circ}$ , then the sum of all three angles will be $180^{\\circ}$ , which is possible in case of a triangle. Proof: Let the three angles of the triangle be $\\angle A$, $\\angle B$ and $\\angle C$. As per the given information, $\\angle A$ = $60^{\\circ}$ … (i) $\\angle B$=$60^{\\circ}$ …(ii) $\\angle C$= $60^{\\circ}$ … (iii) On adding (i), (ii) and (iii), we get: $\\angle A$ + $\\angle B$ + $\\angle C$ = $60^{\\circ}$+ $60^{\\circ}$+ $60^{\\circ}$ $\\angle A$ + $\\angle B$ + $\\angle C$ =$180^{\\circ}$ We can see that the sum of all three angles of the given triangle is equal to $180^{\\circ}$ , which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to $60^{\\circ}$ . Q20. In $\\triangle ABC$, $\\angle A=100^{\\circ}$, AD bisects $\\angle A$ and AD perpendicular BC. Find $\\angle B$ Consider $\\triangle ABD$ $\\angle BAD$ = $\\frac{100}{2}$ (AD bisects $\\angle A$) $\\angle BAD$ = $50^{\\circ}$ $\\angle ADB$ = $90^{\\circ}$ (AD perpendicular to BC) We know that the sum of all three angles of a triangle is $180^{\\circ}$. Thus, $\\angle ABD+\\angle BAD+\\angle ADB$= $180^{\\circ}$ (Sum of angles of $\\triangle ABD$) Or, $\\angle ABD$ + $50^{\\circ}$ + $90^{\\circ}$ =$180^{\\circ}$ $\\angle ABD$ =$180^{\\circ}$– $140^{\\circ}$ $\\angle ABD$= $40^{\\circ}$ Q21.", "\\fn_phv 72^{\\circ}$. The vertex angle is $\\dpi{100} \\fn_phv 180^{\\circ}-72^{\\circ}-72^{\\circ} = 36^{\\circ}$. 2. The two sides are equal, so set them equal to each other and solve for x. $\\dpi{100} \\fn_phv 2x-9 = x+5$ $\\dpi{100} \\fn_phv 2x = x+14$ $\\dpi{100} \\fn_phv x = 14$ 3. This statement is false. Because the base angles of an isosceles triangle are congruent, if one base angle is a right angle then both base angles must be right angles. It is impossible to have a triangle with two right $\\dpi{100} \\fn_phv 90^{\\circ}$ angles. The Triangle Sum Theorem states that the sum of the three angles in a triangle is $\\dpi{100} \\fn_phv 180^{\\circ}$. If two of the angles in a triangle are right angles, then the third angle must be and the shape is no longer a triangle. The following videos can provide additional assistance with this concept as needed: ### 06.07 Perpendicular Bisectors and Isosceles Triangles Geometry - Videos 3 (Math Level 2) If after completing this topic you can state without hesitation that... • I can solve problems using knowledge of the base angles of isosceles triangles. …you are ready for the assignment! Otherwise, go back and review the material before moving on. ### 06.07.01 Isosceles Triangles Geometry - Worksheet (Math Level 2) teacher-scored 82 points possible 35 minutes Activity for this", "name it the Bermuda Triangle. The triangle will be connected to a ramp on each side of the triangle, so that students will come down the ramp onto the rails. There they will either succeed or be lost at sea! Marc drew the following picture of the triangle. The triangle has three angles and the boys want to reproduce these angles in their structure. The first angle $B$ is equal to $62^\\circ$, the second angle $C$ is equal to $63^\\circ$. Marc can’t remember the measure of angle $A$. He thinks there is a way to figure this out, but he can’t remember what it is. The sum of the interior angles of a triangle is equal to $180^\\circ$. Marc knows the measure of two of the angles of the triangle. Therefore, he can write an equation to figure out the measure of the third angle. $63 + 62 + x = 180$ The variable $x$ is used to represent the measure of angle $A$. Marc is working to find the measure of angle $A$. $125 + x & = 180 \\\\180 - 125 & = 55$ The measure of Angle $A$ is $55^\\circ$ Vocabulary Here are the vocabulary words that are found in this lesson. Triangle a three sided figure with three angles. The prefix “tri”means three. Acute", "the grid, of why the sum of the three angles in a triangle is $$180^\\circ$$. In this activity, rather than building a complex shape from a triangle and its rotations, students begin with a triangle and a line parallel to the base through the opposite vertex: In this image, lines $$AC$$ and $$DE$$ are parallel. The advantage to this situation is that we know that points $$D$$, $$B$$, and $$E$$ all lie on a line. In order to calculate angles $$DBA$$ and $$EBC$$, students can use either the rotation idea of the previous activity or congruence of alternate interior angles of parallel lines cut by a transversal. In either case, they need to analyze the given constraints and decide on a path to pursue to show the congruence of angles. Students then conclude from the fact that $$D$$, $$B$$, and $$E$$ lie on the same line that $$a + b +c = 180$$. There is a subtle distinction in the logic between this lesson and the previous. The previous lesson suggests that the sum of the angles in a triangle is $$180^\\circ$$ using direct measurements of a triangle on a grid. This activity shows that this is the case by using a generic triangle and reasoning about parallel lines cut by a transversal. But, in order to do", "# 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", "$x = 15^\\circ$ , then the base angles are $4(15^\\circ) +12^\\circ$ , or $72^\\circ$ . The vertex angle is $180^\\circ-72^\\circ-72^\\circ=36^\\circ$ . 2. The two sides are equal, so set them equal to each other and solve for $x$ . $2x-9 & = x+5\\\\x & = 14$ 3. This statement is false. Because the base angles of an isosceles triangle are congruent, if one base angle is a right angle then both base angles must be right angles. It is impossible to have a triangle with two right ( $90^\\circ$ ) angles. The Triangle Sum Theorem states that the sum of the three angles in a triangle is $180^\\circ$ . If two of the angles in a triangle are right angles, then the third angle must be $0^\\circ$ and the shape is no longer a triangle. ### Explore More Find the measures of $x$ and/or $y$ . Determine if the following statements are true or false. 1. Base angles of an isosceles triangle are congruent. 2. Base angles of an isosceles triangle are complementary. 3. Base angles of an isosceles triangle can be equal to the vertex angle. 4. Base angles of an isosceles triangle are acute. Complete the proofs below. 1. Given : Isosceles $\\triangle CIS$ , with base angles $\\angle C$ and $\\angle S$ $\\overline{IO}$ is the", "### Home > CCG > Chapter 8 > Lesson 8.1.1 > Problem8-6 8-6. Solve for $x$ in each diagram below. The Triangle Angle Sum Theorem states that the sum of all the interior angles in a triangle will be $180^\\circ$. Find the missing angle in each triangle. The sum of supplementary angles is $180^\\circ$. Solve for $x$ using the general equation $\\text{missing angle}+x=180^{\\circ}$. 1. $x = 110^\\circ$ 1. $x = 70^\\circ$", "a triangle measure $50^\\circ$ and $70^\\circ$ . What is the third interior angle of the triangle? 3. Find the value of $x$ and the measure of each angle. 1. $72^\\circ + 65^\\circ +m\\angle{1} = 180^\\circ$ . Solve this equation and you find that $m\\angle{1}=43^\\circ$ . 2. $50^\\circ + 70^\\circ + x = 180^\\circ$ . Solve this equation and you find that the third angle is $60^\\circ$ . 3. All the angles add up to $180^\\circ$ . $(8x-1)^\\circ + (3x+9)^\\circ+(3x+4)^\\circ&=180^\\circ\\\\(14x+12)^\\circ&=180^\\circ\\\\14x = 168\\\\x =12$ Substitute in 12 for $x$ to find each angle. $[3(12) + 9]^\\circ = 45^\\circ && [3(12) + 4]^\\circ = 40^\\circ && [8(12) - 1]^\\circ = 95^\\circ$ ### Practice Determine $m\\angle{1}$ in each triangle. 1. 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 Language: English Triangle Sum Theorem Triangle Sum Theorem The Triangle Sum Theorem states that the", "# In a triangle $P Q R, N$ is a point on $P R$ such that $Q N \\perp P R$. If $P N$. $N R=Q^{2}$, prove that $\\angle \\mathrm{PQR}=90^{\\circ}$. #### Complete Python Prime Pack for 2023 9 Courses 2 eBooks #### Artificial Intelligence & Machine Learning Prime Pack 6 Courses 1 eBooks #### Java Prime Pack 2023 8 Courses 2 eBooks Given: In a triangle $P Q R, N$ is a point on $P R$ such that $Q N \\perp P R$. $P N$. $N R=Q^{2}$ To do: We have to prove that $\\angle \\mathrm{PQR}=90^{\\circ}$. Solution: $P N . N R=Q N^{2}$ $P N . N R=Q N . Q N$ This implies, $\\frac{P N}{Q N}=\\frac{Q N}{N R}$...........(i) In $\\triangle QNP$ and $\\triangle RNQ$, $\\frac{P N}{Q N}=\\frac{Q N}{N R}$ $\\angle P N Q=\\angle R N Q$ Therefore, by SAS similarity, $\\angle Q N P \\sim \\angle R N Q$ This implies, $\\angle P Q N=\\angle Q R N$........(ii) $\\angle R Q N=\\angle Q P N$........(iii) Adding (ii) and (iii), we get, $\\angle P Q N+\\angle R Q N=\\angle Q R N+\\angle Q P N$ $\\angle P Q R=\\angle Q R N+\\angle Q P N$ We know that, Sum of the angles of a triangle is $180^{\\circ}$. In $\\triangle P Q R$, $\\angle P Q R+\\angle Q P", "decreasing 15° of each angle of a triangle, the ratios of their angles are 2:3:5, The radian measure of greatest angle is: 1. $$\\frac{{11\\pi }}{{24}}$$ 2. $$\\frac{\\pi }{{12}}$$ 3. $$\\frac{\\pi }{{24}}$$ 4. $$\\frac{{5\\pi }}{{24}}$$ Que. 5 Two numbers are in the ratio 4 : 7. If 30% of the first number is 24, then 60% of the second number is – 1. 84 2. 140 3. 64 4. 118 Que. 6 If 2r = h + √(r2 + h2) , (r ≠ 0) then find r : h. 1. 1 : 2 2. 3 : 1 3. 1 : 1 4. 4 : 3 Que. 7 The cost of making an article is divided between materials, labor and overheads in the ratio 3 : 4 : 2. If the cost of material is Rs.33.60 the cost of article is 1. Rs.100.80 2. Rs.11.20 3. Rs.44.80 4. Rs.22.40 Que. 8 A sum of Rs. 300 is divided among P, Q and R in such a way that Q gets Rs. 30 more than P and R gets Rs. 60 more than Q. The ratio of their share is 1. 5 : 3 : 2 2. 2 : 3 : 5 3. 3 : 2 : 5 4. 2 : 5 : 3 Que. 9 What must be added to", "fact that the sum of the angles in each triangle is $180^\\circ$." ]

[ "### In a right-angled triangle show that the hypotenuse is the longest side. 1. Let $PQR$ be a right-angled triangle such that $\\angle Q = 90^ \\circ.$ 2. As the sum of angles of a triangle is $180^ \\circ$. Now, in $\\triangle PQR,$ we have \\begin{aligned} & \\angle P + \\angle Q + \\angle R = 180 ^ \\circ \\\\ \\implies & \\angle P + 90 ^ \\circ + \\angle R = 180 ^ \\circ && [As, \\angle Q = 90^\\circ] \\\\ \\implies & \\angle P + \\angle R = 180 ^ \\circ - 90 ^ \\circ \\\\ \\implies & \\angle P + \\angle R = 90 ^ \\circ \\\\ \\end{aligned} \\\\ As the measure of $\\angle Q$ is equal to the sum of measures of $\\angle P$ and $\\angle R$, we have \\begin{aligned} & \\angle Q > \\angle P \\\\ \\implies & PR > QR & \\ldots \\text { (1) } && [\\text {Side opposite to greater angle is greater.}] \\\\ \\end{aligned} Also, \\begin{aligned} & \\angle Q > \\angle R \\\\ \\implies & PR > PQ & \\ldots \\text { (2) } && [\\text{Side opposite to greater angle is greater.}] \\end{aligned} 3. By equation $\\text { (1) }$ and $\\text { (2) }$, we have $$PR > QR \\text { and } PR > PQ \\\\$$ As", "# In the given figure, if PR is tangent to the circle at P and Q Question: In the given figure, if PR is tangent to the circle at and Q is the centre of the circle, then ∠POQ = (a) 110° (b) 100° (c) 120° (d) 90° Solution: We know that the radius is always perpendicular to the tangent at the point of contact. Therefore, we have, $\\angle O P R=90^{\\circ}$ It is given that, $\\angle Q P R=60^{\\circ}$ That is, $\\angle O P R-\\angle O P Q=60^{\\circ}$ $90^{\\circ}-\\angle O P Q=60^{\\circ}$ $\\angle O P Q=30^{\\circ}$ Now, consider $\\triangle O P Q$. We have, OP = OQ (Radii of the same circle) Since angles opposite to equal side will be equal in a triangle, we have, We know that sum of all angles of a triangle will be equal to . Therefore, $\\angle O P Q+\\angle O Q P+\\angle P O Q=180^{\\circ}$ $30^{\\circ}+30^{\\circ}+\\angle P O Q=180^{\\circ}$ $\\angle P O Q=120^{\\circ}$ The correct answer is option (c).", ". 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", "}$ Using the formula, $\\alpha + \\beta + \\gamma = 180^{\\circ }$ substituting $\\alpha = 36.9^{\\circ }$, $\\beta = 53.1^{\\circ }$ and $\\gamma = 90^{\\circ }$ in the formula $36.9^{\\circ } + 53.1^{\\circ } + 90^{\\circ } = 180^{\\circ }$ $90^{\\circ } + 90^{\\circ } = 180^{\\circ }$ $180^{\\circ } = 180^{\\circ }$ L.H.S = R.H.S Therefore, we proved that the sum of angles in a triangle is always 180^{\\circ }. ### Practice Questions 1. Given a right-angled triangle with the reference angle $\\theta$. What is the measure of the angle $\\theta$? 2. Given a right-angled triangle with the reference angle $\\beta$. What is the measure of the angle $\\beta$ using the tangent function? 3. Given a right-angled triangle with the reference angle $\\alpha$. What is the measure of the angle $\\alpha$ using the cosine function? 4. Given a right-angled triangle with the reference angle $\\beta$. What is the measure of the angle $\\beta$? 5. Given a right-angled triangle with the reference angle $\\alpha$. What is the measure of the angle $\\alpha$?", "# In a triangle $P Q R, N$ is a point on $P R$ such that $Q N \\perp P R$. If $P N$. $N R=Q^{2}$, prove that $\\angle \\mathrm{PQR}=90^{\\circ}$. #### Complete Python Prime Pack for 2023 9 Courses 2 eBooks #### Artificial Intelligence & Machine Learning Prime Pack 6 Courses 1 eBooks #### Java Prime Pack 2023 8 Courses 2 eBooks Given: In a triangle $P Q R, N$ is a point on $P R$ such that $Q N \\perp P R$. $P N$. $N R=Q^{2}$ To do: We have to prove that $\\angle \\mathrm{PQR}=90^{\\circ}$. Solution: $P N . N R=Q N^{2}$ $P N . N R=Q N . Q N$ This implies, $\\frac{P N}{Q N}=\\frac{Q N}{N R}$...........(i) In $\\triangle QNP$ and $\\triangle RNQ$, $\\frac{P N}{Q N}=\\frac{Q N}{N R}$ $\\angle P N Q=\\angle R N Q$ Therefore, by SAS similarity, $\\angle Q N P \\sim \\angle R N Q$ This implies, $\\angle P Q N=\\angle Q R N$........(ii) $\\angle R Q N=\\angle Q P N$........(iii) Adding (ii) and (iii), we get, $\\angle P Q N+\\angle R Q N=\\angle Q R N+\\angle Q P N$ $\\angle P Q R=\\angle Q R N+\\angle Q P N$ We know that, Sum of the angles of a triangle is $180^{\\circ}$. In $\\triangle P Q R$, $\\angle P Q R+\\angle Q P", "Measure of $$\\angle$$ P is not $$60^{\\circ}$$, so PQR is not an equilateral triangle. So Statement 2 alone is sufficient to answer whether $$\\triangle$$PQR is an equilateral triangle or not? so OA should be D (1) PQR is NOT an isosceles triangle. As above statement (1) says triangle is not isosceles how can we conclude that such triangle is not equilateral for gmat do we have to consider an equilateral triangle is isosceles from 3 sides ?? I think thats too much from GMAC. DS Forum Moderator Joined: 22 Aug 2013 Posts: 1340 Location: India Re: Is the measure of two exterior angles of a triangle PQR, each equal to [#permalink] ### Show Tags 18 Apr 2018, 05:09 GMAT215 wrote: Princ wrote: amanvermagmat wrote: Is the measure of two exterior angles of a triangle PQR, each equal to 120 degrees? (1) PQR is NOT an isosceles triangle. (2) Measure of angle P is 70 degrees. (Inspired by a Bunuel's question) OA:D First simplify the question:Is the measure of two exterior angles of a triangle PQR, each equal to 120 degrees? For a given triangle, Sum of all exterior angle is $$360^{\\circ}$$. Attachment: exterior-angles-triangle.png if two exterior angles of a triangle PQR are to be 120 degrees i.e total $$2*120^{\\circ} =240^{\\circ}$$,Third exterior angle should be also $$360^{\\circ}-240^{\\circ}=120^{\\circ}$$ $$\\angle$$P=$$\\angle$$Q=$$\\angle$$R= $$180^{\\circ}-120^{\\circ}$$($$120^{\\circ}$$", "83.5^\\circ$ Now, add all three interior angles of the triangle $OPQ$. $\\angle QOP + \\angle OPQ + \\angle PQO = 44^\\circ + 52.5^\\circ + 83.5^\\circ$ $\\therefore \\,\\, \\angle QOP + \\angle OPQ + \\angle PQO = 180^\\circ$ It is proved that the summation of the all three interior angles is $180^\\circ$ exactly in any type of triangle. Save (or) Share", "# 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", "angles as $90^{\\circ}$. Hence, PQRS is a rectangle. Editor", "a triangle measure $50^\\circ$ and $70^\\circ$ . What is the third interior angle of the triangle? 3. Find the value of $x$ and the measure of each angle. 1. $72^\\circ + 65^\\circ +m\\angle{1} = 180^\\circ$ . Solve this equation and you find that $m\\angle{1}=43^\\circ$ . 2. $50^\\circ + 70^\\circ + x = 180^\\circ$ . Solve this equation and you find that the third angle is $60^\\circ$ . 3. All the angles add up to $180^\\circ$ . $(8x-1)^\\circ + (3x+9)^\\circ+(3x+4)^\\circ&=180^\\circ\\\\(14x+12)^\\circ&=180^\\circ\\\\14x = 168\\\\x =12$ Substitute in 12 for $x$ to find each angle. $[3(12) + 9]^\\circ = 45^\\circ && [3(12) + 4]^\\circ = 40^\\circ && [8(12) - 1]^\\circ = 95^\\circ$ ### Practice Determine $m\\angle{1}$ in each triangle. 1. 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 Language: English Triangle Sum Theorem Triangle Sum Theorem The Triangle Sum Theorem states that the", "Nov 2010, 13:38 2 rite2deepti wrote: In triangle PQR, the measure of angle P is 30° more than twice the measure of angle Q. What is the measure of angle R? (1) PQ = QR (2) The measure of angle P is 78°. Given: $$\\angle{P}=2*\\angle{Q}+30$$. Question: $$\\angle{R}=?$$. Now, as the sum of the angles in a triangle equals to 180 degrees then: $$\\angle{P}+\\angle{Q}+\\angle{R}=180$$ --> $$\\angle{R}=180-(\\angle{P}+\\angle{Q})=180-(2*\\angle{Q}+30+\\angle{Q})=150-3*\\angle{Q}$$ or $$\\angle{R}=150-3*\\frac{\\angle{P}-30}{2}$$. So, knowing the measure of $$\\angle{P}$$ ( or $$\\angle{Q}$$) is sufficient to answer the question. (1) PQ = QR --> triangle PQR is isosceles (base is PR) --> so $$\\angle{R}=\\angle{P}$$. We also know that $$\\angle{R}=150-3*\\frac{\\angle{P}-30}{2}$$ so we have two distinct linear equations with two unknowns and we can solve for them. Sufficient. (2) The measure of angle P is 78 --> so $$\\angle{P}=78$$ then we can solve $$\\angle{R}=150-3*\\frac{\\angle{P}-30}{2}$$ for $$\\angle{R}$$. Sufficient. For more on this issues check Triangles chapter of Math Book: math-triangles-87197.html Hope it helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9705 Location: Pune, India ### Show Tags 27 Nov 2010, 13:11 1 rite2deepti wrote: In triangle PQR, the measure of angle P is 30° more than twice the measure of angle Q. What is the measure of angle R? (1) PQ = QR (2) The measure of angle P is 78°. The best thing to", "PQR$ is $\\frac{1}{2}bh=\\tfrac{3\\sqrt{3}}{2}\\cdot\\tfrac{3}{2}=\\tfrac{9\\sqrt{3}}{4}$. We would like to add this value to the sum of the areas of the other two parts of $\\triangle PQR$. To find the areas of the other two parts of $\\triangle PQR$ using the $\\sin$ area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that $\\angle QOR = 120^{\\circ}$ and triangles $\\triangle COQ$ and $\\triangle COR$ are congruent as they share a side, $CQ=CR,$ and $OQ=OR$. Therefore $\\angle COQ = \\angle COR = 120^{\\circ}$. Suppose $CO=x$. Then $3^{2}+x^{2}-6x\\cos{120^{\\circ}}=7^{2}$, and since $\\cos{120^{\\circ}}=-\\tfrac{1}{2}$, this simplifies to $x^{2}+3x=7^{2}-3^{2}\\rightarrow x^{2}+3x-40=0$. This factors nicely as $(x-5)(x+8)=0$, so $x=5$ as $x$ can't be $-8$. Since $CO=5, OP=3$ and $\\angle OPC=90^{\\circ}$, we now know that $\\triangle OPC$ is a $3-4-5$ right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of $\\triangle PQR$. Let $\\angle POC = \\alpha$. Then $\\sin\\alpha = \\tfrac{4}{5}, \\cos\\alpha = \\tfrac{3}{5}, \\angle QOP = 120+\\alpha,$ and $\\angle POR = 120-\\alpha$. The sum of the areas of $\\triangle", "Q.4 In ?ABC, $$\\angle A = 30° \\angle B = 40° and \\angle C = 110°$$ In $$\\triangle PQR, \\angle P = 30°, \\angle Q = 40° and \\angle R = 110°.$$ A student says that $$\\triangle ABC = \\triangle PQR$$ by AAA congruence criterion. Is he justified? Why or why not?", "figure. Find the value of x. 29. Construct a triangle PQR given that PQ = 4 cm, QR = 6.5 cm and $\\angle$PQR = 60°. 30. The monthly income of 6 families are Rs.3500, Rs.2700, RS.3000, Rs. 2800, Rs. 3900 and Rs. 2100. Find the mean income. 31. Find the median of the first 5 prime numbers 32. III. Answer any five in detail 5 x 3 = 15 33. Solve: $\\frac { p }{ -7 } =8$ 34. Solve: 3(x+2)=15 35. Write the following percent as a fraction:12 $\\frac{1}{2}$% 36. A sum of money doubles itself at 12$\\frac { 1 }{ 2 }$% per annum over a certain period of time. Find the number of years. 37. A rectangular hall has 10 m long and 7 m broad. A carpet is spread in the centre leaving a margin of 1 m near the walls. Find the area of the carpet. Also, find the area of the uncovered floor. 38. State which of the following are triangles. (i) $\\angle A=25°,\\angle B=35°,\\angle C=120°$ (ii) $\\angle P=90°,\\angle Q=30°,\\angle R=50°$ (iii) $\\angle X=40°,\\quad \\angle Y=70°,\\quad \\angle Z=80°$ 39. The weights of 5 people are 72 kg, 48 kg, 51 kg, 69 kg, 67 kg. Find the mean of their weights.", "Answer Comment Share Q) 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$", "name it the Bermuda Triangle. The triangle will be connected to a ramp on each side of the triangle, so that students will come down the ramp onto the rails. There they will either succeed or be lost at sea! Marc drew the following picture of the triangle. The triangle has three angles and the boys want to reproduce these angles in their structure. The first angle $B$ is equal to $62^\\circ$, the second angle $C$ is equal to $63^\\circ$. Marc can’t remember the measure of angle $A$. He thinks there is a way to figure this out, but he can’t remember what it is. The sum of the interior angles of a triangle is equal to $180^\\circ$. Marc knows the measure of two of the angles of the triangle. Therefore, he can write an equation to figure out the measure of the third angle. $63 + 62 + x = 180$ The variable $x$ is used to represent the measure of angle $A$. Marc is working to find the measure of angle $A$. $125 + x & = 180 \\\\180 - 125 & = 55$ The measure of Angle $A$ is $55^\\circ$ Vocabulary Here are the vocabulary words that are found in this lesson. Triangle a three sided figure with three angles. The prefix “tri”means three. Acute", "the first arc at $P$. $\\angle PQR$ is the required angle. (We can construct a $120^\\circ$ angle by intersecting the first arc again, with center $P$ and the same radius, as before) We also need to bisect angles. To do so, we construct an arbitrary angle (say, $\\angle PQR$). We take a radius less than the length of $QR$ (or $QP$, if $QP)and construct an arc, with center $Q$, intersecting $QR$ and $QP$ at $R'$ and $P'$, respectively. Using $R'$ and $P'$ as centers, using the same radius, we construct an arc from both of them. These arcs intersect at two points, which we join and extend to $Q$ to construct the angle bisector of $\\angle PQR$. Using the above two methods, we can now construct such a right triangle $\\Delta ABC$ with these angles. We construct a segment $AB$ with the given length $c$. On $A$, we construct a $60^\\circ$ angle and bisect it twice, towards $AB$ giving us $15^\\circ$. On $B$, we construct a $120^\\circ$ angle and bisect the angle between the $60^\\circ$ and the $120^\\circ$ mark twice, towards the $60^\\circ$ mark giving us $75^\\circ$. We extend the new segments and label their intersection point $C$, which is right-angled. Thus, $\\Delta ABC$ is the required triangle. - 5 years, 5 months ago Good work in finding", "a triangle PQR, each equal to [#permalink] ### Show Tags 18 Apr 2018, 06:01 amanvermagmat wrote: GMAT215 wrote: Princ wrote: [quote=\"amanvermagmat\"]Is the measure of two exterior angles of a triangle PQR, each equal to 120 degrees? (1) PQR is NOT an isosceles triangle. (2) Measure of angle P is 70 degrees. (Inspired by a Bunuel's question) OA:D First simplify the question:Is the measure of two exterior angles of a triangle PQR, each equal to 120 degrees? For a given triangle, Sum of all exterior angle is $$360^{\\circ}$$. Attachment: exterior-angles-triangle.png if two exterior angles of a triangle PQR are to be 120 degrees i.e total $$2*120^{\\circ} =240^{\\circ}$$,Third exterior angle should be also $$360^{\\circ}-240^{\\circ}=120^{\\circ}$$ $$\\angle$$P=$$\\angle$$Q=$$\\angle$$R= $$180^{\\circ}-120^{\\circ}$$($$120^{\\circ}$$ being exterior angle) $$\\angle$$P=$$\\angle$$Q=$$\\angle$$R=$$60^{\\circ}$$ Question is reduced to whether $$\\triangle$$PQR is an equilateral triangle or not? Statement 1 : PQR is NOT an isosceles triangle. If PQR is not even isosceles triangle , it cannot be equilateral triangle. So Statement 1 alone is sufficient to answer whether $$\\triangle$$PQR is an equilateral triangle or not? Statement 2 :Measure of angle P is 70 degrees. Measure of $$\\angle$$ P is not $$60^{\\circ}$$, so PQR is not an equilateral triangle. So Statement 2 alone is sufficient to answer whether $$\\triangle$$PQR is an equilateral triangle or not? so OA should be D (1) PQR is NOT an isosceles", "decreasing 15° of each angle of a triangle, the ratios of their angles are 2:3:5, The radian measure of greatest angle is: 1. $$\\frac{{11\\pi }}{{24}}$$ 2. $$\\frac{\\pi }{{12}}$$ 3. $$\\frac{\\pi }{{24}}$$ 4. $$\\frac{{5\\pi }}{{24}}$$ Que. 5 Two numbers are in the ratio 4 : 7. If 30% of the first number is 24, then 60% of the second number is – 1. 84 2. 140 3. 64 4. 118 Que. 6 If 2r = h + √(r2 + h2) , (r ≠ 0) then find r : h. 1. 1 : 2 2. 3 : 1 3. 1 : 1 4. 4 : 3 Que. 7 The cost of making an article is divided between materials, labor and overheads in the ratio 3 : 4 : 2. If the cost of material is Rs.33.60 the cost of article is 1. Rs.100.80 2. Rs.11.20 3. Rs.44.80 4. Rs.22.40 Que. 8 A sum of Rs. 300 is divided among P, Q and R in such a way that Q gets Rs. 30 more than P and R gets Rs. 60 more than Q. The ratio of their share is 1. 5 : 3 : 2 2. 2 : 3 : 5 3. 3 : 2 : 5 4. 2 : 5 : 3 Que. 9 What must be added to", "C B D$ is:In the triangle $\\triangle A B C$ measures of angles $\\angle A=46^{\\circ}$ and $\\angle C=60^{\\circ}$ are given. The angle bisector intersects the circ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given statements is correct. In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given statements is correct.In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given ... close Triangle $\\triangle A B C$, right angled at $C$, is given. Height and the median from point $C$ form an angle $\\varphi$. The measure of larger acute angle of $\\triangle A B C$ is:Triangle $\\triangle A B C$, right angled at $C$, is given. Height and the median from point $C$ form an angle $\\varphi$. The measure of larger acute a ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 The distance of the shorter ladder form the wall is: The distance of the shorter" ]

[ "# Consecutive Integers $$567762$$ is the product of two positive consecutive integers. Find them and enter as the sum of the two numbers. ×", "# The value of 3 times the square root of 8 is between which two consecutive integers? $8 < 3 \\sqrt{8} < 9$ You have ${\\left(3 \\sqrt{8}\\right)}^{2} = 72$ and ${8}^{2} < 72 < {9}^{2}$ so $8 < 3 \\sqrt{8} < 9$", "Between what two consecutive integers do sqrt70 lie? Oct 16, 2015 $\\sqrt{70}$ lies between the integers color(green)(8 and 9 Explanation: $\\sqrt{64} = 8$ and $\\sqrt{81} = 9$ And we know that $\\sqrt{64} < \\sqrt{70} < \\sqrt{81}$ Hence $8 < \\sqrt{70} < 9$ $\\sqrt{70}$ lies between the integers color(green)(8 and 9", "Between what two consecutive integers does square root of 73 lie? Between 8 and 9 Explanation: the sqrt73 is in between $\\sqrt{81}$ and $\\sqrt{64}$ that is$\\sqrt{64}$<$\\sqrt{73}$<$\\sqrt{81}$ and $8$<$\\sqrt{73}$<$9$ God bless .... I hope the explanation is useful", "# Between what two consecutive integers does square root of 73 lie? Between 8 and 9 #### Explanation: the sqrt73 is in between $\\sqrt{81}$ and $\\sqrt{64}$ that is$\\sqrt{64}$<$\\sqrt{73}$<$\\sqrt{81}$ and $8$<$\\sqrt{73}$<$9$ God bless .... I hope the explanation is useful", "# Enjoy the sequence! $\\large \\sqrt{a_{2}+a_{3}+\\cdots+a_{99}}-\\sqrt{a_{1}+a_{100}}$ Let $$a_{1},a_{2},a_{3},\\ldots,a_{100}$$ is an sequence of consecutive positive integers. Find the minimum integer value of the expression above. of \\ which is an integer. ×", "# Consecutive Integers $$567762$$ is the product of two positive consecutive integers. Find them and enter as the sum of the two numbers. × Problem Loading... Note Loading... Set Loading...", "# Between what two consecutive integers do sqrt70 lie? Oct 16, 2015 $\\sqrt{70}$ lies between the integers color(green)(8 and 9 #### Explanation: $\\sqrt{64} = 8$ and $\\sqrt{81} = 9$ And we know that $\\sqrt{64} < \\sqrt{70} < \\sqrt{81}$ Hence $8 < \\sqrt{70} < 9$ $\\sqrt{70}$ lies between the integers color(green)(8 and 9", "# Between what two consecutive integers is sqrt135? Oct 21, 2016 the two consecutive numbers are 11 and 12 $11 < \\sqrt{135} < 12$ #### Explanation: ${11}^{2} = 121$ and ${12}^{2} = 144$ so $121 < 135 < 144$ taking the square roots $\\sqrt{121} < \\sqrt{135} < \\sqrt{144}$ and finally $11 < \\sqrt{135} < 12$", "# Q2. Represent the following situations in the form of quadratic equations : (ii) The product of two consecutive positive integers is 306. We need to find the integers. Given the product of two consecutive integers is $306.$ Let two consecutive integers be $'x'$ and $'x+1'$. Then, their product will be: $x(x+1) = 306$ Or $x^2+x- 306 = 0$. Hence, the two consecutive integers will satisfy this quadratic equation $x^2+x- 306 = 0$. Exams Articles Questions", "which two consecutive integers is $\\sqrt{205}$ ? Vocabulary Language: English Spanish square root function square root function A square root function is a function with the parent function $y=\\sqrt{x}$.", "# Between what two consecutive integers do sqrt450 lie? ##### 1 Answer Mar 7, 2018 21&22 #### Explanation: ${21}^{2} = 441$ ${22}^{2} = 484$ It lies between 21&22 but it is almost $21$ as you can notice it is only $9$ less than 450", "given. Note that the theorem in the paper $1$ is a generalised one. It states that The product of two or more consecutive positive integers is never a power. And your squares also come under this section.", "# 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", "\\\\$ So two consecutive odd positive integers are 39 and 41.", "# 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 ×", "# Digit product How many positive integers less than $$1,000,000,000$$ have the product of their digits equal to 49? ×", "# How Many Consecutive Integers Produces This Sum? 120 can be expressed as the sum of $$n$$ consecutive (not necessarily positive) integers. Find the sum of all possible values of $$n$$. ×", "8$ What is the value of $a_6$? Seven consecutive positive integers have a sum of 91. So what is the largest of these integers?", "Basic College Mathematics (9th Edition) $(-1)(-.69)=-1\\times-.69=.69$ The numbers being multiplied have the same sign, so their product will be positive." ]

[ "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "= 106$, and $AB = 106 + 87 = \\boxed{193}$.", "= \\boxed{108}$.", "$65+72=$", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "a^2$$. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$. ~IceMatrix", "2 + 0)$$ 55 = $$9 * (1 + 2) + 0!$$ 56 = $$.02^{-1} + (\\sqrt9)!$$ 57 = $$19 * (2 + 0!)$$ 58 = $$29 * (1 + 0!)$$ 59 = $$20 * \\sqrt9 - 1$$ 60 = $$20 * 1 * \\sqrt9$$ 61 = $$20 * \\sqrt9 + 1$$ 62 = $$21 * \\sqrt9 - 0!$$ 63 = $$21 * \\sqrt9 + 0$$ 64 = $$(9 - 1)^2 + 0$$ 65 = $$(9 - 1)^2 + 0!$$ 66 = $$((\\sqrt9)!)!*.1 - (2 + 0!)!$$ 67 = $$((\\sqrt{9})!)! * .1 - {.2}^{-0!}$$ 68 = $$((\\sqrt{9})!)!!+20*1$$ 69 = $$90 - 21$$ 70 = $$(9-2) * 10$$ 71 = $$91 - 20$$ 72 = $$(2 + 1 + 0!) * \\sqrt9$$ 73 = $$12 * (\\sqrt9 )! + 0!$$ 74 = $$((\\sqrt9)!)! * .\\overline{1} - (2 + 0!)!$$ 75 = $$((2 - 1)! - 0!) * \\sqrt9$$ 76 = $$.\\overline{1}^{-2}-(\\sqrt9)!+0!$$ 77 = $$((\\sqrt9)!)! * .\\overline{1} - (2 + 0!)$$ 78 = $$(12 + 0!) * (\\sqrt9)!$$ 79 = $$9^{0!+1} - 2$$ 80 = $$10 * 2^{\\sqrt9}$$ 81 = $$9^2+1*0$$ 82 = $$((\\sqrt9)!)! * .\\overline{1} + 2 + 0$$ 83 = $$((\\sqrt9)!)! * .\\overline{1}+ 2 + 0!$$ 84 = $$21 * (\\sqrt9 + 0!)$$ 85 = $$91 - (2+0!)!$$ 86 = $$((\\sqrt9)!)! * .\\overline{1} + (2 + 0!)!$$ 87", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "left. $(27*8)/{9 \\choose 3}{6 \\choose3}$ simplifies down to our fraction m/n, which is $9/70$. Adding them up gives $9 + 70 = \\boxed{79}$.", "our answer is $m+n=61+4=\\boxed{065}$.", "$81 - 54 - 27 = 0$ = $81 - 81 = 0$", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "64a? If it is$4\\sqrt[3]{a}+\\sqrt[3]{64a}=8\\sqrt[3]{a}$ 41. anonymous yes it is , your right . i got confused with it You're doing fine, I've got to run. good luck 43. anonymous thank youuuu ! :)", "3 ) }$$ 96. $$\\sqrt { 14 - 11 x } + \\sqrt { 7 - 9 x } = 1$$ 97. $$\\sqrt { x + 1 } = \\sqrt { 3 } - \\sqrt { 2 - x }$$ 98. $$\\sqrt { 2 x + 9 } - \\sqrt { x + 1 } = 2$$ 99. $$x ^ { 1 / 2 } - 10 = 0$$ 100. $$x ^ { 1 / 2 } - 6 = 0$$ 101. $$x ^ { 1 / 3 } + 2 = 0$$ 102. $$x ^ { 1 / 3 } + 4 = 0$$ 103. $$( x - 1 ) ^ { 1 / 2 } - 3 = 0$$ 104. $$( x + 2 ) ^ { 1 / 2 } - 6 = 0$$ 105. $$( 2 x - 1 ) ^ { 1 / 3 } + 3 = 0$$ 106. $$( 3 x - 1 ) ^ { 1 / 3 } - 2 = 0$$ 107. $$( 4 x + 15 ) ^ { 1 / 2 } - 2 x = 0$$ 108. $$( 3 x + 2 ) ^ { 1 / 2 } - 3 x = 0$$ 109. $$( 2 x + 12 ) ^ { 1 /", "# How do you solve -70- 8x = 38- 4x? May 10, 2018 $x = 27$ #### Explanation: First we should isolate our x. To this, add 70 to both sides, and add 4x to both sides as well. $- 70 - 8 x = 38 - 4 x$ $- 8 x = 108 - 4 x$ $- 4 x = 108$ Divide both sides by -4 to find x. $x = - 27$ Check: $- 70 - \\left(8 \\cdot - 27\\right) = 38 - \\left(4 \\cdot - 27\\right)$ $- 70 - \\left(- 216\\right) = 38 - \\left(- 108\\right)$ $- 70 + 216 = 38 + 108$ $146 = 146$" ]

[ "# Between what two consecutive integers does square root of 73 lie? Between 8 and 9 #### Explanation: the sqrt73 is in between $\\sqrt{81}$ and $\\sqrt{64}$ that is$\\sqrt{64}$<$\\sqrt{73}$<$\\sqrt{81}$ and $8$<$\\sqrt{73}$<$9$ God bless .... I hope the explanation is useful", "Between what two consecutive integers does square root of 73 lie? Between 8 and 9 Explanation: the sqrt73 is in between $\\sqrt{81}$ and $\\sqrt{64}$ that is$\\sqrt{64}$<$\\sqrt{73}$<$\\sqrt{81}$ and $8$<$\\sqrt{73}$<$9$ God bless .... I hope the explanation is useful", "Between what two consecutive integers do sqrt70 lie? Oct 16, 2015 $\\sqrt{70}$ lies between the integers color(green)(8 and 9 Explanation: $\\sqrt{64} = 8$ and $\\sqrt{81} = 9$ And we know that $\\sqrt{64} < \\sqrt{70} < \\sqrt{81}$ Hence $8 < \\sqrt{70} < 9$ $\\sqrt{70}$ lies between the integers color(green)(8 and 9", "# Enjoy the sequence! $\\large \\sqrt{a_{2}+a_{3}+\\cdots+a_{99}}-\\sqrt{a_{1}+a_{100}}$ Let $$a_{1},a_{2},a_{3},\\ldots,a_{100}$$ is an sequence of consecutive positive integers. Find the minimum integer value of the expression above. of \\ which is an integer. ×", "# Between what two consecutive integers do sqrt70 lie? Oct 16, 2015 $\\sqrt{70}$ lies between the integers color(green)(8 and 9 #### Explanation: $\\sqrt{64} = 8$ and $\\sqrt{81} = 9$ And we know that $\\sqrt{64} < \\sqrt{70} < \\sqrt{81}$ Hence $8 < \\sqrt{70} < 9$ $\\sqrt{70}$ lies between the integers color(green)(8 and 9", "# The value of 3 times the square root of 8 is between which two consecutive integers? $8 < 3 \\sqrt{8} < 9$ You have ${\\left(3 \\sqrt{8}\\right)}^{2} = 72$ and ${8}^{2} < 72 < {9}^{2}$ so $8 < 3 \\sqrt{8} < 9$", "# 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", "# Proof Involving Pigeonhole Principle Let $a_1, a_2, ..., a_n$ be positive integers all of whose prime divisors are $\\le$ 13. a) Show that if $n \\ge 65$ then there exist two of these integers whose product is a perfect square. [DONE] b) Show that if $n \\ge 193$ then there exists four of these integers whose product is a perfect fourth power. Hint: Use a) to get many pairs of numbers which multiply to a square. Use a) again to get two disjoint such pairs a, b, and c, d such that $\\sqrt{ab}\\sqrt{cd}$ is a square. For b, I got 64 pairs of numbers which multiply to a square using part a. Where do I go from there? Part b: We claim that we can find $65$ disjoint pairs of integers such that their product is a square. By part a), for $n\\ge 193>65$, we can find one such pair. Then remove the pair from the set, and apply part a) again on the remaining $n-2$ integers. We can do this repeatedly, and because $n-2\\cdot 64\\ge65$, we can find at least $65$ pairs. Because the original set of integers was distinct, the pairs are disjoint. Now for each pair $(a,b)$ consider $\\sqrt{ab}$, which is an integer. By part a, there are distinct $a,b,c,d$ such that $\\sqrt{ab}\\cdot \\sqrt{cd}$", "+0 +1 71 2 +201 Determine the value of the infinite product $$(2^{1/3})(4^{1/9})(8^{1/27})(16^{1/81}) \\dotsm$$ Enter your answer in the form \"\\sqrt[a]{b}\", which stands for $$\\sqrt[a]{b}$$ Feb 28, 2021 #1 +771 0 Hello! So, lets turn everything where the base is 2. 2^(1/3) is already in base 2. 4^(1/9) = (2^2)^(1/9) = 2^(2/9) 8^(1/27) = (2^3)^(1/27) = 2^(3/27) = 2^(1/9) 16^(1/81) = (2^4)^(1/81) = 2^(4/81) Now that everything is in base 2, we can just add. :)) 1/3+2/9+1/9+4/81 = 58/81 2^(58/81) = <81>sqrt(2^58) I hope this helped. :)) =^._.^= Feb 28, 2021 #2 0 productfor(n, 1, 1000, (2^n)^(3^-n))=1.6817928305 074290861==2^(3/4)==(2^3)^(1/4) Feb 28, 2021", "and putting in our assumption that $x_2>x_1$, we get $29\\sqrt2\\times (x_2-x_1) = 58\\sqrt2\\times z$. Dividing both sides by $29\\sqrt2$, we get $x_2-x_1 = 2z$ As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well. • There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2. • There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4. ... • Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48. Summing them up, we get that there are $2+4+\\dots + 48 = \\boxed{600}$ triangles.", "# Between what two consecutive integers do sqrt450 lie? ##### 1 Answer Mar 7, 2018 21&22 #### Explanation: ${21}^{2} = 441$ ${22}^{2} = 484$ It lies between 21&22 but it is almost $21$ as you can notice it is only $9$ less than 450", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "# How to find the sum of positive integers $x$ and $y$, given that $\\sqrt x + \\sqrt y = \\sqrt {135}$? How do you find the sum of integers $x$ and $y$ from: $\\sqrt x + \\sqrt y = \\sqrt {135}$? Is there a specific method that will get the answer? x and y are both positive integers. For example x could not be 1, and y could not be 1, because there roots added do not equal the square root of 135. So the sum of x and y could not be 2. What I thought of doing was kind of like an approximation. Where we know that the sqrt of $135$ is between $11$ and $12$. So we find $2$ numbers that add up to $11$, and then square them, and we get an approximate answer for the sum of $x$ and $y$. So for example, $8$ and $3$. Square them and get $64$ and $9$. We get $73$ for the sum of $x$ and $y$ (the actual answer is $75$). This method isn't good, so I was wondering if there was another way of doing it. • How about $x = 0$, and $y = 135$? – user49685 Feb 11 '14 at 18:44 • I think you need some extra info here. – John", "blue box hold? 32- The sum of six different negative integers is $$-70$$. If the smallest of these integers is $$-15$$, what is the largest possible value of one of the other five integers? A. $$-14$$ B. $$-10$$ C. $$-5$$ D. $$-1$$ 33- In the figure below, what is the value of $$x$$? 34- The following table represents the value of $$x$$ and function $$f(x)$$. Which of the following could be the equation of the function $$f(x)$$? A. $$f(x)=x^2-5$$ B. $$f(x)=x^2-1$$ C. $$f(x)=\\sqrt{x+2}$$ D. $$f(x)=\\sqrt{x}+4$$ 35- The circle graph below shows all Mr. Green’s expenses for last month. If he spent $660 on his car, how much did he spend for his rent? 36- The Jackson Library is ordering some bookshelves. If x is the number of bookshelves the library wants to order, which each costs$100 and there is a one-time delivery charge of $800, which of the following represents the total cost, in dollar, per bookshelf? A. $$100x+800$$ B. $$100+800x$$ C. $$\\frac{100x+800}{100}$$ D. $$\\frac{100x+800}{x}$$ 37- What is the sum of $$\\sqrt{x-7}$$ and $$\\sqrt{x}-7$$ when $$\\sqrt{x}=4$$ ? 38- In the following figure, point $$Q$$ lies on line n, what is the value of $$y$$ if $$x = 35$$? A. 15 B. 25 C. 35 D. 45 39- What is the smallest integer whose square root is greater than", "# Approximate and compare irrational numbers ## Interactive practice questions Which of the following has a value between $10$10 and $11$11? $\\sqrt{143}$143 A $\\sqrt{102}$102 B $\\sqrt{90}$90 C $\\sqrt{167}$167 D $\\sqrt{143}$143 A $\\sqrt{102}$102 B $\\sqrt{90}$90 C $\\sqrt{167}$167 D Easy Less than a minute Which of the following is $\\sqrt{52}$52 closest to? The value of $\\sqrt{59}$59 lies between two consecutive integers. Between which two consecutive integers does $\\sqrt{74}$74 lie? ### Outcomes #### NA5-2 Use prime numbers, common factors and multiples, and powers (including square roots)", "-4.1. By using an approximate value of –$$\\sqrt{17}$$, the negative irrational number –$$\\sqrt{17}$$ is located on the number line as we can observe in the above image. Question 10. –$$\\sqrt{26}$$ The –$$\\sqrt{26}$$ is between two integers. The two integers are -5 and -6. By using calculator –$$\\sqrt{26}$$ = -5.099019513…. The value of –$$\\sqrt{26}$$ with two decimal places is -5.09. The decimal -5.09 is closer to -5.1 than to -5.0. So, –$$\\sqrt{26}$$ is located closer to -5.1. By using an approximate value of –$$\\sqrt{26}$$, the negative irrational number –$$\\sqrt{26}$$ is located on the number line as we can observe in the above image. Question 11. –$$\\sqrt{53}$$ The –$$\\sqrt{53}$$ is between two integers. The two integers are -7 and -8. By using calculator –$$\\sqrt{53}$$ = -7.280109889…. The value of –$$\\sqrt{53}$$ with two decimal places is -7.28. The decimal -7.28 is closer to -7.3 than to -7.2. So, –$$\\sqrt{53}$$ is located closer to -7.3. By using an approximate value of –$$\\sqrt{53}$$, the negative irrational number –$$\\sqrt{53}$$ is located on the number line as we can observe in the above image. Question 12. –$$\\sqrt{80}$$ The –$$\\sqrt{80}$$ is between two integers. The two integers are -8 and -9. By using calculator –$$\\sqrt{80}$$ = -8.944271909…. The value of –$$\\sqrt{80}$$ with two decimal places is -8.94. The decimal -8.94 is closer to -8.9 than to", "## Thinking Mathematically (6th Edition) between $-7$ and $-6$ $47$ is betwen the perfect squares $36$ and $49$. This means that : $\\sqrt{36}\\lt \\sqrt{47} \\lt \\sqrt{49}\\\\ 6 \\lt \\sqrt{47} \\lt 7$ Note that $1\\lt 2\\lt3$, but $-3 \\lt -2 \\lt -1$ Since $6 \\lt \\sqrt{47} \\lt 7$, then $-7 \\lt -\\sqrt{47} \\lt -6$. Therefore, $-\\sqrt{47}$ is between the integers $-7$ and $-6$.", "\\sqrt{2} \\, )^9 \\; = \\; \\sqrt{ \\, 1940449 \\, } \\; + \\; \\sqrt{1940449 + 1 \\, }$ $( \\, 1 + \\sqrt{2} \\, )^{10} \\; = \\; \\sqrt{ \\, 11309768 \\, } \\; + \\; \\sqrt{11309768 + 1 \\, }$ Show every positive integer $n > 10$, $f(n)$ can be expressed as a sum of square roots of consecutive integers", "# 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 $156 + 96 + 84 = \\boxed{336}$.", "# Between what two consecutive integers do sqrt650 lie? ##### 1 Answer Oct 17, 2015 $\\sqrt{650}$ lies between $25$ and $26$ #### Explanation: Method 1: First we simplify the expression by prime factorising $650$ $\\sqrt{650} = \\sqrt{5 \\cdot 5 \\cdot 13 \\cdot 2}$ $= 5 \\sqrt{26}$ sqrt26 = color(blue)(5.09 So, =5sqrt(26) = 5 xx color(blue)(5.09 $\\approx 25.495$ It can be observed that $\\sqrt{650}$ lies between $25$ and $26$ Method 2 We can estimate the range of $\\sqrt{650}$ if we have a good idea of squares of numbers. • ${25}^{2} = 625$, from this result we get an idea that $\\sqrt{650}$ is slightly higher that $25$, so we find ${26}^{2}$ • ${26}^{2} = 676$ So $\\sqrt{650}$ lies between $25$ and $26$" ]

[ "the $^5$.", "{-5, 0}$.", "+ 2 = 5 - y$$", "$5*2=0$.", "always be 5, $\\E[5] = 5$.", "p(x)=x^2+5", "10^{-5}$.", "$y = - 5 x + b$", "+ 5$ or $x = - 5 + 5$ $x = 10$ or $x = 0$", "\\sqrt {3^2 + 4^2 } = \\sqrt {25} = 5$", "6y^2 +5y - 1`.", "{0, 0, 3, 4, 5} -", "\u0003 \u000e \u0015 \u0003 \u0007 \u0017 \b \u001a \u0014 \u0014 \u001a ( \u0017 \u0007 \u0002 \u000e \u0003 \u0007 \u0002 \u0006 \u0003 \u000e \u0015 \u001a \u0017 \u0003 \u0006 \u001b \u0002 \u0006 \u0007 \u0002 \u000e \u0013 # \u0015 \u001a \u0003 \u0006 \u0006 \u001a \u001b \u0007 \u000e \u001a \u0018 \u0002 \u0003 \u001b \u0017 \u0014 \u0018 \u0002 \b \u001b \u0018 \u001a \u0002 \u001b \u001a \u0006 \u0006 \u0003 \u001a \u000e \u0015 \u0003 \u000e \u0015 \u0003 \u000e \u0017 \u0007 \u0006 \u001a \u000e \u001a \u0018 \u0002 \u0007 \u0015 \u001a \u0002 ( \u0003 \u0019 \u001a \u0002 ( \u0003 5 \u000e \u001a \u0007 \u0003 \u001a \u0002 \u0003 \u0016 \u0017 \u0007 \u0002 \u0006 \u0018 \u0002 \u0007 \u0002 \u0006 \u0003 \u001a \u0002 \u001b \u0005 \u0006 \u0006 \u0003 \u0015 \u001a \u0002 ( \u0003 \u0019 \u001a \u0002 ( \u0003 , \u0018 \u0018 \u0006 \u0018 , , \u0017 \u001a \u0002 ( \u0003 \u000e \u0015 \u0003 + \u0005 \u0017 \u0002 \u001a \u0002 ( \u0003 \u0018 , \u0003 \u001a \u0002 \u001b \u0002 \u0007 \u0002 \u0006 \u0003 \u000e \u0018 * \u0002 \u0003 \u001b \u0005 \u0017 \u0017 \u0002 \u001b \b \u0003 \u001b \u0007 \u0017 \u0017 \u001a \u0006 \u0003 \u0018 \u0005 \u000e \u0003 \u001a \u0002 \u0002 \u0003 \u0004 \u0003 \u0005 \u0006 \u0007 \b \b \u0004 \u000e \u0006 \u0002 \u000f \u0003 \u0010 \u0003 \u0006 \u0011 \u0012 \u0013 \u000f \u0014 \u0003 \u000e \u0003 \u0006 \u0015 \u0003 \b \u0016 \u0017", "(\\max(x - 5, 0) + \\max(-x + 5, 0))+5$$", "five numbers is 4, then their product is ${4^5}$.", "5 = 0$. This is the smallest non-zero linear combination that comes out as zero. Add that twice to the first linear combination and you find the second linear combination.", "5x is equivalent to ` 5 x!", "+ (-8) = -5$$", "5 + 3 + 1 = n^2 + (n + 1)^2$", "$(-1)^4 = 1^5$" ]

[ "## 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.", "the powers of $2$ modulo $5$. They are $2^0=1$, $2^1=2$, $2^2=4\\equiv-1$, $2^3=8\\equiv3$, and $2^4=16\\equiv1$. Now all is clear.", "## 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, $3^{0}+(-3)^{0}=1+1=2$.", "## Prealgebra (7th Edition) $-1$ $-8^{0}$ is $-1$ because any number raised to the 0th power is 1, and then you take the opposite to get $-1$.", "\\in \\mathbb{R}$ to see how things work. Alternatively, you could say that $5^{2x+1} = 5^2 \\implies 5^{2x - 1} = 5^0 = 1 \\implies 2x - 1 = 0 \\implies 2x + 1 = 2$ if you're prepared to believe that the only number which when you raise $5$ to the power of said number gives you $1$, is $0$. However, the justification for this is the above notion of injectivity, so this kind of approach is washing over the point of the deduction. -", "got it from raising $x=-(-1)^{1/3}$ to the power of $3$, but that yields $1$ of course. Thanks :) – barak manos Aug 28 '16 at 7:27", "$n+1$ in either order. However, when $1$ is assigned to $n+1$, then $n=0$, which is not possible, because $n$ must be positive. Therefore, we have $6-1=5$ values of $n$ $\\Rightarrow \\boxed{B}$", "the number itself. So in this case the result is 4. ${4}^{1}=4$ Explanation: Any number raised to the power 1 is the number itself c) Calculating ${64}^{\\frac{1}{2}}$: Squaring the result obtained should give us 64. ${64}^{\\frac{1}{2}}=8$ Explanation: ${8}^{2}=64$ d) Calculating ${2}^{\\frac{3}{4}}$ to 2 decimal places: The result when raised to the power $4/3$ should give us 2. ${2}^{\\frac{3}{4}}=1.68$ Explanation: ${1.68}^{\\frac{4}{3}}=2$ a) 0.87 b) 4 c) 8 d) 1.68", "very awkward to say \"Five raised to the second power of 2 \" Because you probably mean $2^2=4$. So you have $5^4=625$. Now five to the third power is $5^3=125$ But $5^{2^{3}}=5^8=390625$, read \"5 to the third power of 2. NOTE that $5^{2^{3}}\\ne (5^2)^3=5^6$ 3. ## Re: Five raisied to the nth power of 2 Originally Posted by Plato It is very awkward to say \"Five raised to the second power of 2 \" Because you probably mean $2^2=4$. So you have $5^4=625$. Now five to the third power is $5^3=125$ But $5^{2^{3}}=5^8=390625$, read \"5 to the third power of 2. NOTE that $5^{2^{3}}\\ne (5^2)^3=5^6$ Blimey! $5^{2^{3}}$ So how do I enter that on a calculator, or more so how is that represented without using the ^ symbol (i.e. like 5x5x5x5 etc.) Also is \"Five raised to the second power of 2 is 125\" wrong in the book? I didn't understand this bit with the funny equals sign. NOTE that $5^{2^{3}}\\ne (5^2)^3=5^6$ and when I enter $(5^2)^3$ on my calculator i get 15625 so i'm doing something wrong. Thanks for your help by the way! 4. ## Re: Five raisied to the nth power of 2 Originally Posted by philuk2000 $5^{2^{3}}$ So how do I enter that on a calculator, or more so how is that represented without", "## 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, a. $9^{0}=1$ b. $-9^{0}=-(9^{0})=-(1)=-1$ c. $(-9)^{0}=1$ d. $-(-9)^{0}=-(1)=-1$", "## College Algebra (11th Edition) Any expression raised to the 0th power equals 1, so: (a) 3$p^{0}$ = 3*1 = 3 (b) -3$p^{0}$ = -3*1 = -3 (c) $(3p)^{0}$ = 1 (d) $(-3p)^{0}$ = 1", "## Prealgebra (7th Edition) $4$ $4^{1}$ is 4 because any number raised to the 1st power is itself .", "by $4$ (by its last 2 digits), $2^{121985292} \\equiv 1 \\pmod{5}$. Hence $2^{121985292} \\equiv 0 \\pmod{5}$. Now we have to make sure that it is odd. Since $2$ raised to any power is even, $1$ less than it is odd, so $2^{121985292}-1$ is odd and its last digit is $5$. (odd and divisible by 5 $\\rightarrow$ last digit is 5).", "+0 # in surd form what is 2 to the power of 5/2 0 65 2 in surd form what is 2 to the power of 5/2 Apr 2, 2021 #2 +485 +1 $$2^{5/2}=\\sqrt{2^5}=\\sqrt{32}=\\boxed{4\\sqrt{2}=5.6569}$$ . Apr 2, 2021 #1 +11376 +2 in surd form what is 2 to the power of 5/2 Hello Guest! $$2^\\frac{5}{2}=2^{2.5}=\\color{blue}5.65685$$ ! Apr 2, 2021 #2 +485 +1 $$2^{5/2}=\\sqrt{2^5}=\\sqrt{32}=\\boxed{4\\sqrt{2}=5.6569}$$", "$$0=x^2-2x-4-3x+4$$ 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:", "# Question #5c170 Apr 12, 2017 8 #### Explanation: When a number is raised to a power, it means it is being multiplied by itself that many times. So ${2}^{3} = 2 \\cdot 2 \\cdot 2$ and $2 \\cdot 2 \\cdot 2$ = 4*2 = 8. Apr 12, 2017 The answer is $8$. #### Explanation: ${2}^{3}$ is simply $2 \\cdot 2 \\cdot 2$. $2 \\cdot 2 \\cdot 2$ $4 \\cdot 2$ $= 8$.", "2 \\times 3 \\times 5 \\times 5 = 150$ Note $-$ For GCF, pick up least number of times. Here least power of $2$ is $0$, that of $3$ is $0$ and that of $5$ is $1$, hence GCF is $5$.", "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.", "the other is x raised to five, therefore they cannot be added. $(x^2 + x^5)^2$ $= (x^2)^2 + 2(x^2)(x^5) + (x^5)^2$ $= x^4 + 2x^7 + x^{10}$ Este sitio usa Akismet para reducir el spam. Aprende cómo se procesan los datos de tus comentarios .", "say that $-6^{2} = -36$. ii) In $(-5)^{2}$ the power of two has been raised to (-5) where the minus sign is included. Hence, $(-5)^{2} = -5\\times -5 = 25$." ]

[ "the powers of $2$ modulo $5$. They are $2^0=1$, $2^1=2$, $2^2=4\\equiv-1$, $2^3=8\\equiv3$, and $2^4=16\\equiv1$. Now all is clear.", "\\in \\mathbb{R}$ to see how things work. Alternatively, you could say that $5^{2x+1} = 5^2 \\implies 5^{2x - 1} = 5^0 = 1 \\implies 2x - 1 = 0 \\implies 2x + 1 = 2$ if you're prepared to believe that the only number which when you raise $5$ to the power of said number gives you $1$, is $0$. However, the justification for this is the above notion of injectivity, so this kind of approach is washing over the point of the deduction. -", "say that $-6^{2} = -36$. ii) In $(-5)^{2}$ the power of two has been raised to (-5) where the minus sign is included. Hence, $(-5)^{2} = -5\\times -5 = 25$.", "power of$x$in the expression of the$f(x)$is the smallest$n$for which$f(x)$is$O(x^{n})$. $n=5$ When$x>2$, we have the property$x^{2}>x>2$. Let’s choose$k=2$first and then choose$x>2$. $|f(x)|=|2x^{2}+x^{4}\\log x|\\leq|2x^{2}+x^{5}|\\leq |2x^{2}|+|x^{5}|$ $=2x^{2}+x^{5}\\leq x^{5}+x^{5}$ $=2x^{5}$ $=2|x^{5}|$ Thus,$C$should be at least$2$. Let us then choose$C=2\\$. 5/5 - (5 votes)", "$ \\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", "very awkward to say \"Five raised to the second power of 2 \" Because you probably mean $2^2=4$. So you have $5^4=625$. Now five to the third power is $5^3=125$ But $5^{2^{3}}=5^8=390625$, read \"5 to the third power of 2. NOTE that $5^{2^{3}}\\ne (5^2)^3=5^6$ 3. ## Re: Five raisied to the nth power of 2 Originally Posted by Plato It is very awkward to say \"Five raised to the second power of 2 \" Because you probably mean $2^2=4$. So you have $5^4=625$. Now five to the third power is $5^3=125$ But $5^{2^{3}}=5^8=390625$, read \"5 to the third power of 2. NOTE that $5^{2^{3}}\\ne (5^2)^3=5^6$ Blimey! $5^{2^{3}}$ So how do I enter that on a calculator, or more so how is that represented without using the ^ symbol (i.e. like 5x5x5x5 etc.) Also is \"Five raised to the second power of 2 is 125\" wrong in the book? I didn't understand this bit with the funny equals sign. NOTE that $5^{2^{3}}\\ne (5^2)^3=5^6$ and when I enter $(5^2)^3$ on my calculator i get 15625 so i'm doing something wrong. Thanks for your help by the way! 4. ## Re: Five raisied to the nth power of 2 Originally Posted by philuk2000 $5^{2^{3}}$ So how do I enter that on a calculator, or more so how is that represented without", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "by $4$ (by its last 2 digits), $2^{121985292} \\equiv 1 \\pmod{5}$. Hence $2^{121985292} \\equiv 0 \\pmod{5}$. Now we have to make sure that it is odd. Since $2$ raised to any power is even, $1$ less than it is odd, so $2^{121985292}-1$ is odd and its last digit is $5$. (odd and divisible by 5 $\\rightarrow$ last digit is 5).", "$$0=x^2-2x-4-3x+4$$ 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:", "$n+1$ in either order. However, when $1$ is assigned to $n+1$, then $n=0$, which is not possible, because $n$ must be positive. Therefore, we have $6-1=5$ values of $n$ $\\Rightarrow \\boxed{B}$", "2 \\times 3 \\times 5 \\times 5 = 150$ Note $-$ For GCF, pick up least number of times. Here least power of $2$ is $0$, that of $3$ is $0$ and that of $5$ is $1$, hence GCF is $5$.", "can be certain there are no sums of fifth power digits? The highest digit is $9$ and $9^5=59049$, which has five digits. If we then look at $5 times 9^5=295245$, which has six digits and a good endpoint for the loop. The loop itself cycles through the digits of each number and tests whether the sum of the fifth powers equals the number. ```largest 0) { d", "refer to $a^{(b^c)}$. This is effectively the same as Siong Thye Goh's answer, except with a little extra detail. As others have remarked, the notation $5^{2^{4}}$ is really $5^{(2^{4})} = 5^{16}$. This is a simple application of powers. On the other hand, by recalling what the power represents, we have $$(5^{2})^{4} = 5^{2} \\cdot 5^{2} \\cdot 5^{2} \\cdot 5^{2} = 5^{2+2+2+2} = 5^{4(2)} = 5^{8}.$$ Care and attention is required when dealing with powers of powers.", "# The Power of 5 Find the last three digits of $\\LARGE 5^{5^{5^{5}}}.$ (Enter as a three-digit number, hundreds place first.) ×", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "the trivial power in common. This follows because $(a,1)^n = (a^n,1)$ and $(1,b)^m = (1,b^m)$ for all $n$ and $m$, so that $(a,1)^n = (1,b)^m$ implies in particular that $a^n = 1$ and so $(a,1)^n = (1,1)$. The conclusion follows from Lemma 2. $\\blacksquare$", "= 2/1, since 3(2) = 0. On the other hand, we see 1/1 is not equal to 0/1. a/1 = 1/1 implies 3(a - 1) = 0, so we see that: 1/1 = 3/1. Finally, note that 4/1 = 0/1 (since 3(4-0) = 0), and: 5/1 = 1/1 (since 3(5-1) = 0). So the image of $R$ is: {0/1,1/1}. What about the other 6 fractions?. Clearly, 0/3 = 0/1. Just as clearly, 3/3 = 1/1. Now 1/3 = 1/1, since 3(3-1) = 0. 2/3 = 0/1, since 3(2-0) = 0. 4/3 = 0/1, since 3(4-0) = 0, and finally, 5/3 = 1/1, since 3(5-3) = 0. So we have just two elements in $T^{-1}R$, namely: {0/1,1/1}. What does the multiplication in this ring look like? (0/1)*(0/1) = 0/1 (0/1)*(1/1) = 0/1 (1/1)*(1/1) = 1/1 What about addition (which is the hard part)? (0/1) + (0/1) = (1*0 + 0*1)/(1*1) = (0+0)/1 = 0/1 (0/1) + (1/1) = (1*0 + 1*1)/(1*1) = (0+1)/1 = 1/1 (1/1) + (1/1) = (1*1 + 1*1)/(1*1) = (1+1)/1 = 2/1 = 0/1. We could use \"other forms\" for these: (3/1) + (3/3) = (3*3 + 1*3)/(1*3) = (3 + 3)/3 = 0/3 = 0/1, or: (2/3) + (5/1) = (1*2 + 3*5)/(3*1) = (2 + 3)/3 = 5/3 = 1/1. So in this", "And we have: . $a(n) \\;=\\;\\frac{n(n+1)}{2} - (n-1)$ . . Therefore: . $\\boxed{a(n) \\;=\\;\\frac{n^2-n+2}{2}}$ 8. Thanks so much Soroban! I actually get that *victory dance*", "# $$5^{n>1}$$ = $$25$$? Does $$5$$ to the power of a positive integer $$> 1$$ always end in $$25?$$ ×", "What's the difference between $5^{2^4}$ and $\\left(5^2\\right)^4$? I wanted to know what exactly is the difference between these two expressions and why they return different results. $$5^{2^4} \\qquad\\qquad\\left(5^2\\right)^4$$ The left one returns $152587890625$, while the right one gives $390625$. I know the right one is called \"powers of powers\", but I don't know about the one on the left and how it works. • $5^{2^4}$ represents $5^{\\left(2^4\\right)}$ – Henry May 13 '18 at 22:50 $$5^{2^4}=5^{(2^4)}=5^{16}$$ But $$(5^2)^4=5^{(2\\cdot 4)}=5^8$$ We have $$5^{16}=(5^8)^2$$ Be careful with parentheses. You might be used to the idea that $(5+2)+4 = 5+(2+4)$ and $(5 \\times 2) \\times 4 = 5 \\times (2 \\times 4)$, but the same doesn't stay true for powers! In particular, we don't have that $5^{(2^4)} = (5^2)^4$. So, when writing $5 \\times 2 \\times 4$ it doesn't matter where the parentheses go, but it does for powers! In particular, the expression $5^{2^4}$ is interpreted at $5^{(2^4)}$, which is different from $(5^2)^4$. An easy way to see that they're different is just to evaluate them: $$5^{(2^4)} = 5^{16} \\qquad (5^2)^4 = 25^4 =5^8.$$ So now the question is - why do we assume $5^{2^4}$ means what it does? Well, notice how ${(a^b)}^c= a^{bc}$, so we already have a convenient notation for this. So, it's sensible to define the notation $a^{b^c}$ to" ]

[ "+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", "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 =", "12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$. $[asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label(\"A\", A, NE); label(\"C\", C, NW); label(\"D\", D, S); label(\"P\",P , N); label(\"P'\", Pa, SW); label(\"13\\sqrt{3}\", (A+D)/2, E, small); label(\"13\\sqrt{3}\", (A+C)/2, NW, small); label(\"12\\sqrt{3}\", (C+D)/2, SW, small); label(\"h\", (P + Pa)/2, W); label(\"\\frac{\\sqrt{939}}{2}\", (C+P)/2 ,NW); [/asy]$ To find the volume, we want to use the equation $\\frac 13Bh = 6\\sqrt{133}h$, so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = \\frac{\\sqrt{939}}{2}$. If we let $P$ be the center of a sphere with radius $\\frac{\\sqrt{939}}{2}$, then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\\triangle ACD$. From here we just need to perform some brutish calculations. Using the formula $A = 18\\sqrt{133} = \\frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$ (there are slightly simpler ways", "# 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 $\\angle guy$?.", "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$$25k^2+9k^2-15k^2 =", "$y = 5$, such that the answer is $1^2 + 5^2 = \\boxed{026}$. ### Solution 2 We'll use the law of cosines. Let $O$ be the center of the circle; we wish to find $OB$. We know how long $OA$ and $AB$ are, so if we can find $\\cos \\angle OAB$, we'll be in good shape. $[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); dot(O); dot(A); dot(B); dot(C); label(\"O\",O,SW); label(\"A\",A,NE); label(\"B\",B,S); label(\"C\",C,SE); [/asy]$ We can find $\\cos \\angle OAB$ using angles $OAC$ and $BAC$. First we note that by Pythagoras, $$AC = \\sqrt{AB^2 + BC^2} = \\sqrt{36 + 4} = \\sqrt{40} = 2 \\sqrt{10}.$$ If we let $M$ be the midpoint of $AC$, that mean that $AM = \\sqrt{10}$. Since $\\triangle OAC$ is isosceles ($OA = OC$ from the definition of a circle), $M$ is also the foot of the altitude from $O$ to $AC.$ $[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); pair M = (A+C)/2; draw(P); draw(O--C--A--cycle); draw(O--M, dashed); draw(rightanglemark(O,M,A,25)); dot(O); dot(A); dot(C); label(\"O\",O,SW); label(\"A\",A,NE); label(\"M\",M,SSW); label(\"C\",C,SE); label(\"\\sqrt{50}\", (O+A)/2, NW); label(\"\\sqrt{10}\", (A+M)/2, E); [/asy]$ It follows that $OM = \\sqrt{40} = 2 \\sqrt{10}$. Therefore \\begin{align*} \\cos \\angle OAC = \\frac{\\sqrt{10}}{\\sqrt{50}} &= \\frac{1}{\\sqrt{5}}, \\\\ \\sin \\angle OAC = \\frac{2 \\sqrt{10}}{\\sqrt{50}} &= \\frac{2}{\\sqrt{5}}. \\end{align*}", "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}$", "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", "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", "real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; dot(a^^b^^x^^y^^c^^d); draw(x--y); draw(a--y^^b--y); draw(d--x--c); draw(a--b--c--d--cycle); draw(x--a^^x--b); label(\"A\",a,NW,fontsize(9)); label(\"B\",b,NE,fontsize(9)); label(\"C\",c,SE,fontsize(9)); label(\"D\",d,SW,fontsize(9)); label(\"X\",x,2*N,fontsize(9)); label(\"Y\",y,3*S,fontsize(9)); [/asy]$ First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\\angle DAX = \\angle XYC = \\theta$, and $\\angle XYD = \\angle CBX = 180 - \\theta$, due to the properties of cyclic quadrilaterals. In addition, let $\\angle BAX = x$ and $\\angle ABX = y$. Thus, $\\angle ADX = \\angle AYX = x$ and $\\angle XYB = \\angle XCB = y$. Then, since quadrilateral $ABCD$ is cyclic as well, we have the following sums: $$\\theta + x +\\angle XCY + y = 180^{\\circ}$$ $$180^{\\circ} - \\theta + y + \\angle XDY + x = 180^{\\circ}$$ Cancelling out $180^{\\circ}$ in the second equation and isolating $\\theta$ yields $\\theta = y + \\angle XDY + x$. Substituting $\\theta$ back into the first equation, we obtain $$2x + 2y + \\angle XCY + \\angle XDY = 180^{\\circ}$$ Since $$x + y +\\angle XAY + \\angle XCY + \\angle DAY = 180^{\\circ}$$ $$x + y + \\angle XDY + \\angle XCY + \\angle DAY = 180^{\\circ}$$ we can then imply that $\\angle DAY = x + y$. Similarly, $\\angle YBC = x + y$. So", "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", "+ r)(3)cosA = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-3,12/7)--P); draw((3,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Let $C$ be the center of the largest semicircle. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$, since the area of a triangle is $\\frac{1}{2} Base \\cdot Height$. Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. $[asy] size(5cm); pair A", "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$$\",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686));[/asy]$ By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\\angle OED = 90^\\circ$. Since $OD = 4$ and $DE = \\frac{1}{2}$, $OE = \\sqrt{16-\\frac{1}{4}} = \\frac{\\sqrt{63}}{2} = \\frac{3\\sqrt{7}}{2}$. Since $\\triangle CED$ is $30-60-90$, $CE = \\frac{\\sqrt{3}}{2}$ (or this can also be deduced from Pythagoras on $\\triangle CED$). Thus $OC = \\frac{3\\sqrt{7}-\\sqrt{3}}{2}$. As previous solutions noted, $\\triangle BOC$", "+0 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "# 1983 AIME Problems/Problem 4 ## Problem A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. $[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label(\"A\",A,NE); label(\"B\",B,S); label(\"C\",C,SE); [/asy]$ ## Solution ### Solution 1 Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$. $[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(D--O--F--B,dashed); dot(O); dot(A); dot(B); dot(C); label(\"O\",O,SW); label(\"A\",A,NE); label(\"B\",B,S); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",F,SW); [/asy]$ Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$ and $OC^2 = EC^2 + EO^2$. Thus, $\\left(\\sqrt{50}\\right)^2 = y^2 + (6-x)^2$, and $\\left(\\sqrt{50}\\right)^2 = x^2 + (y+2)^2$. We solve this system to get $x = 1$ and", "Mollweide right phi = x.copy() # physical copy phi[x < 0] = 2 * np.pi + x[x<0] theta = np.pi/2 - y PHI, THETA = np.meshgrid(phi, theta) In [14]: l = 4 m = 2 SH_SP = sp.sph_harm(m, l, PHI, THETA).real # Plot just the real part What follows is a representation of the spherical harmonic as a heatmap of a field with a Mollweide projection. The value $l-m$ gives the number of horizontal \"bands\" or lines of constant latitude where the spherical harmonic is null. The order $m$ is the number of zero circles passing through the poles (beware with the count, in the flat Mollweide projection you'll see only half circles). To check this you can see a very nice interactive 3D visualization here. In [15]: #This is to enable bold Latex symbols in the matplotlib titles, according to: #http://stackoverflow.com/questions/14324477/bold-font-weight-for-latex-axes-label-in-matplotlib matplotlib.rc('text', usetex=True) matplotlib.rcParams['text.latex.preamble']=[r\"\\usepackage{amsmath}\"] In [16]: xlabels = ['$210^\\circ$', '$240^\\circ$','$270^\\circ$','$300^\\circ$','$330^\\circ$', '$0^\\circ$', '$30^\\circ$', '$60^\\circ$', '$90^\\circ$','$120^\\circ$', '$150^\\circ$'] ylabels = ['$165^\\circ$', '$150^\\circ$', '$135^\\circ$', '$120^\\circ$', '$105^\\circ$', '$90^\\circ$', '$75^\\circ$', '$60^\\circ$', '$45^\\circ$','$30^\\circ$','$15^\\circ$'] fig, ax = plt.subplots(subplot_kw=dict(projection='mollweide'), figsize=(10,8)) im = ax.pcolormesh(X, Y , SH_SP) ax.set_xticklabels(xlabels, fontsize=14) ax.set_yticklabels(ylabels, fontsize=14) ax.set_title('real$(Y^2_ 4)$', fontsize=20) ax.set_xlabel(r'$\\boldsymbol \\phi$', fontsize=20) ax.set_ylabel(r'$\\boldsymbol{\\theta}$', fontsize=20) ax.grid() fig.colorbar(im, orientation='horizontal'); Here ends this post, which has already been too long. Until next time!.", "of points $C$ that can be centers of circles must be an ellipse with foci $F_1$ and $F_2$ and major axis $20$. Clearly, the minimum value of the slope $a$ will occur when the line $y=ax$ is tangent to this ellipse. Suppose that this point of tangency is denoted by $T$, and the line $y=ax$ is denoted by $\\ell$. Then we reflect the ellipse over $\\ell$ to a new ellipse with foci $F_1'$ and $F_2'$ as shown below. $[asy] size(220); pair F1 = (-5, 12), F2 = (5, 12),C=(0,12); draw(circle(F1,16)); draw(circle(F2,4)); draw(ellipse(C,10,5*sqrt(3))); xaxis(\"x\",Arrows); yaxis(\"y\",Arrows); dot(F1^^F2^^C); real l(real x) {return sqrt(69)*x/10;} path g=graph(l,-7,14); draw(g); draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3))); pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10))); dot(T); pair F1P=reflect((0,0),(10,l(10)))*F1; pair F2P=reflect((0,0),(10,l(10)))*F2; dot(F1P^^F2P); dot((0,0)); label(\"F_1\",F1,N,fontsize(9)); label(\"F_2\",F2,N,fontsize(9)); label(\"F_1'\",F1P,SE,fontsize(9)); label(\"F_2'\",F2P,SE,fontsize(9)); label(\"O\",(0,0),NW,fontsize(9)); label(\"\\ell\",(13,l(13)),SE,fontsize(9)); label(\"T\",T,NW,fontsize(9)); draw((0,0)--F1--F2--F2P--F1P--cycle); draw(F1--F2P^^F2--F1P); [/asy]$ By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that $F_1$, $T$, and $F_2'$ are collinear, and similarly, $F_2$, $T$ and $F_1'$ are collinear. Therefore, $OF_1F_2F_2'F_1'$ is a pentagon with $OF_1=OF_2=OF_1'=OF_2'=13$, $F_1F_2=F_1'F_2'=10$, and $F_1F_2'=F_1'F_2=20$. Note that $\\ell$ bisects $\\angle F_1'OF_1$. We can bisect this angle by bisecting $\\angle F_1'OF_2$ and $F_2OF_1$ separately. We proceed using complex numbers. Triangle $F_2OF_1'$ is isosceles with side lengths $13,13,20$. The height of this from the base", "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)}\\", "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", "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 =" ]

[ "+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", "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 $O_{1}$,", "+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", "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 =", "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}", "# 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", "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", "\"$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) +", "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]$", "$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", "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", "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)", "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 $\\angle guy$?.", "# 2008 AIME II Problems/Problem 5 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem 5 In trapezoid $ABCD$ with $\\overline{BC}\\parallel\\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\\angle A = 37^\\circ$, $\\angle D = 53^\\circ$, and $M$ and $N$ be the midpoints of $\\overline{BC}$ and $\\overline{AD}$, respectively. Find the length $MN$. ## Solution ### Solution 1 Extend $\\overline{AD}$ and $\\overline{BC}$ to meet at a point $E$. Then $\\angle AED = 180 - 53 - 37 = 90^{\\circ}$. $[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); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,SE); label(\"$$E$$\",E,NE); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$1004$$\",(N+D)/2,S); label(\"$$500$$\",(M[0]+C)/2,S); [/asy]$ Since $\\overline{BC} \\parallel \\overline{AD}$, then $BC$ and $AD$ are homothetic with respect to point $E$ by a ratio of $\\frac{BC}{AD} = \\frac{125}{251}$. Since the homothety carries the midpoint of $\\overline{BC}$, $M$, to the midpoint of $\\overline{AD}$, which is $N$, then $E,M,N$ are collinear. As $\\angle AED = 90^{\\circ}$, note that the midpoint of $\\overline{AD}$, $N$, is the center of the circumcircle of $\\triangle AED$. It follows that $$NE = ND = \\frac{AD}{2} = 1004.$$ Since $\\triangle EMC \\sim \\triangle END$, we have that $$\\frac{EM}{EM + MN} = \\frac{MC}{ND}", "+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", "# 2008 AIME II Problems/Problem 5 ## Problem 5 In trapezoid $ABCD$ with $\\overline{BC}\\parallel\\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\\angle A = 37^\\circ$, $\\angle D = 53^\\circ$, and $M$ and $N$ be the midpoints of $\\overline{BC}$ and $\\overline{AD}$, respectively. Find the length $MN$. ## Solution ### Solution 1 Extend $\\overline{AB}$ and $\\overline{CD}$ to meet at a point $E$. Then $\\angle AED = 180 - 53 - 37 = 90^{\\circ}$. $[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); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,SE); label(\"$$E$$\",E,NE); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$1004$$\",(N+D)/2,S); label(\"$$500$$\",(M[0]+C)/2,S); [/asy]$ As $\\angle AED = 90^{\\circ}$, note that the midpoint of $\\overline{AD}$, $N$, is the center of the circumcircle of $\\triangle AED$. We can do the same with the circumcircle about $\\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$). It follows that $$NE = ND = \\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500.$$ Thus $MN = NE - ME = \\boxed{504}$. For purposes of rigor we will show that $E,M,N$ are collinear. Since $\\overline{BC} \\parallel \\overline{AD}$, then $BC$ and $AD$ are homothetic with respect to point $E$", "10^{\\circ}(3-\\tan^2 10^{\\circ})}{1-3\\tan^2 10^{\\circ}}\\\\&=-\\tan (3\\times 10)^{\\circ}\\\\&=\\tan 150^{\\circ}\\end{align*} This gives $\\angle CDP=180^{\\circ}-150^{\\circ}=30^{\\circ}$. But $\\angle ACP=30^{\\circ}+40^{\\circ}=70^{\\circ}$, we get $\\angle ADC=70^{\\circ}-30^{\\circ}=40^{\\circ}$. ##### Well-known member Re: Find \\$\\angle ADC\\$ My solution (I just solved it): Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\\angle APB=90^{\\circ}$ and $BP=k$. \\begin{tikzpicture}[auto] \\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x}; \\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y}; \\coordinate[label=left:B] (B) at (0,0); \\coordinate[label=right] (P) at (8,0); \\coordinate[label=left: D] (D) at (5,0); \\coordinate[label=right:C] (C) at (8,-3); \\coordinate[label=right:A] (A) at (8,-8); \\draw (B) to (P); \\draw (B) to (A); \\draw (D) to (A); \\draw (B) to (C); \\draw (A) to (P); \\draw (C) to (D); \\draw [dashed] (D) -- (3.6, 1.4); \\node (1) at (1.5,-0.2) {$10^{\\circ}$}; \\node (2) at (1.5,-0.9) {$30^{\\circ}$}; \\node (3) at (7,-6.3) {$30^{\\circ}$}; \\node (4) at (7.7,-6) {$20^{\\circ}$}; \\draw (P) rectangle +(-0.5, -0.5); \\end{tikzpicture} We then have $C=\\left(k, -k\\tan 10^{\\circ}\\right)\\\\D=\\left(\\dfrac{k}{2\\sin 110^{\\circ}\\cos 40^{\\circ}}, 0\\right)=\\left(\\dfrac{k}{\\sin 150^{\\circ}+\\sin 70^{\\circ}}, 0\\right)=\\left(\\dfrac{k}{\\frac{1}{2}+\\cos 20^{\\circ}}, 0\\right)$ \\begin{align*}\\text{Gradient of } l_{CD}&=\\dfrac{-k\\tan 10^{\\circ}}{k\\left(1-\\dfrac{1}{\\frac{1}{2}+\\cos 20^{\\circ}}\\right)}\\\\&=\\dfrac{-\\tan 10^{\\circ}\\left(\\frac{1}{2}+\\cos 20^{\\circ}\\right)}{\\cos 20^{\\circ}-\\dfrac{1}{2}}\\\\&=\\dfrac{-\\tan 10^{\\circ}(4\\cos^2 10^{\\circ}-1)}{4\\cos^2 10^{\\circ}-3}\\\\&=\\dfrac{-\\tan 10^{\\circ}(4\\cos^2 10^{\\circ}-(\\sin^2 10^{\\circ} +\\cos^2 10^{\\circ}))}{4\\cos^2 10^{\\circ}-3(\\sin^2 10^{\\circ} +\\cos^2 10^{\\circ})}\\\\&=\\dfrac{-\\tan 10^{\\circ}(3\\cos^2 10^{\\circ}-\\sin^2 10^{\\circ})}{\\cos^2 10^{\\circ}-3\\sin^2 10^{\\circ}}\\\\&=\\dfrac{-\\tan 10^{\\circ}(3-\\tan^2 10^{\\circ})}{1-3\\tan^2 10^{\\circ}}\\\\&=-\\tan (3\\times 10)^{\\circ}\\\\&=\\tan 150^{\\circ}\\end{align*} This gives $\\angle CDP=180^{\\circ}-150^{\\circ}=30^{\\circ}$. But $\\angle ACP=30^{\\circ}+40^{\\circ}=70^{\\circ}$, we get $\\angle ADC=70^{\\circ}-30^{\\circ}=40^{\\circ}$. How do we know" ]

[ "+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", "as the center and draw segment $ON$ perpendicular to $AM$, $ON\\cap AM=N$, link $OM$. Then we have $OM\\parallel AD$. So $\\angle DAM=\\angle OMA$. Since $AD=2AM=2OM=1$, we have $\\cos\\angle DAM=\\cos\\angle OMP=\\frac{2}{\\sqrt{5}}$. As a result, $NM=OM\\cos\\angle OMP=\\frac{1}{2}\\cdot \\frac{2}{\\sqrt{5}}=\\frac{1}{\\sqrt{5}}.$ Thus $PM=2NM=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}$. Since $AM=\\frac{\\sqrt{5}}{2}$, we have $AP=AM-PM=\\frac{\\sqrt{5}}{10}$. Put $\\boxed{B}$. ~FANYUCHEN20020715 ## Solution 5 (Similar Triangles) Call the midpoint of $\\overline{AB}$ point $N$. Draw in $\\overline{NM}$ and $\\overline{NP}$. Note that $\\angle{NPM}=90^{\\circ}$ due to Thales's Theorem. $[asy] draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw(circle((0.5,0.5),0.5)); draw((0,0)--(0,0.5)--(1,0.5)--cycle); label(\"A\",(0,0),SW); label(\"B\",(0,1),NW); label(\"C\",(1,1),NE); label(\"D\",(1,0),SE); label(\"M\",(1,0.5),E); label(\"P\",(0.2,0.1),S); label(\"N\",(0,0.5),W); draw((0,0.5)--(0.2,0.1)); markscalefactor=0.007; draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5))); [/asy]$ Using the Pythagorean theorem, $AM=\\frac{\\sqrt{5}}{2}$. Now we just need to find $AP$ using similar triangles. $$\\triangle APN\\sim\\triangle ANM\\Rightarrow\\frac{AP}{AN}=\\frac{AN}{AM}\\Rightarrow\\frac{AP}{\\frac{1}{2}}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{5}}{2}}\\Rightarrow AP=\\boxed{\\textbf{(B) }\\frac{\\sqrt{5}}{10}}$$ ~QIDb602 ## Solution 6 (Intersecting Chords) Label the midpoint of $AD$ as $N$, and let $Q$ be the intersection of $ON$ and $AM$ Then $MQ=AQ=\\frac{1}{2}AM = \\frac{\\sqrt{5}}{4}$ and $NQ=\\frac{1}{4}$ Then $\\frac{1}{4}*\\frac{3}{4}=PQ*QM$ and $AP=AQ-PQ$ ~ERMSCoach ## Solution 7 (Inscribed Angle Theorem and Pythagorean Theorem) Let $N$ be the midpoint of $\\overline{AB},$ from which $\\angle ANM=90^\\circ.$ Note that $\\angle NPM=90^\\circ$ by the Inscribed Angle Theorem. We have the following diagram: Since $AN=\\frac12$ and $NM=1,$ we get $AM=\\frac{\\sqrt5}{2}$ by the Pythagorean Theorem. Let $AP=x.$ It follows that $PM=\\frac{\\sqrt5}{2}-x.$ Applying the Pythagorean Theorem to right $\\triangle ANP$ gives $NP^2=\\left(\\frac12\\right)^2-x^2,$ and applying the Pythagorean Theorem to right $\\triangle MNP$ gives $NP^2=1^2-\\left(\\frac{\\sqrt5}{2}-x\\right)^2.$ Equating the expressions for $NP^2$ produces \\begin{align*}", "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", "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 $(7\\sqrt{11})^2 = \\boxed{539}$", "12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$. $[asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label(\"A\", A, NE); label(\"C\", C, NW); label(\"D\", D, S); label(\"P\",P , N); label(\"P'\", Pa, SW); label(\"13\\sqrt{3}\", (A+D)/2, E, small); label(\"13\\sqrt{3}\", (A+C)/2, NW, small); label(\"12\\sqrt{3}\", (C+D)/2, SW, small); label(\"h\", (P + Pa)/2, W); label(\"\\frac{\\sqrt{939}}{2}\", (C+P)/2 ,NW); [/asy]$ To find the volume, we want to use the equation $\\frac 13Bh = 6\\sqrt{133}h$, so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = \\frac{\\sqrt{939}}{2}$. If we let $P$ be the center of a sphere with radius $\\frac{\\sqrt{939}}{2}$, then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\\triangle ACD$. From here we just need to perform some brutish calculations. Using the formula $A = 18\\sqrt{133} = \\frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$ (there are slightly simpler ways", "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", "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 $O_{1}$,", "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", "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$$25k^2+9k^2-15k^2 =", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ 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 $MN = 504$.", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$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$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from B and C to AD, and call the points if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = x and EA = $1008-x$ , then you", "$n$ is not divisible by $3$ (therefore $9$) since $n>11$ and $n$ is prime, it follows that $n|\\underbrace{111\\cdots 111}_{x\\text{ times}}$. $\\Box$ ## Picture 1 $[asy]draw(Circle((1,1),2)); draw(Circle((sqrt(2),sqrt(3)/2),1)); dot((8/5,2/5)); dot((1,1)); draw((1,1)--(8/5,2/5),linetype(\"8 8\")); label(\"a\",(6/5,7/10),SSW); draw((8/5,2/5)--(12/5,-2/5),linetype(\"8 8\")); label(\"b\",(2,0),SSW); [/asy]$ $$\\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", "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}.$$ Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $\\boxed{010}$.", "is equilateral, and thus the desired length is $x = OC \\implies C$. ## Solution 3 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.95,3),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$E$$\",(0.40,-3.60),E); label(\"$$B$$\",(-0.75,4.15),E); label(\"$$D$$\",(-2.62,1.5),E); label(\"$$F$$\",(-2.64,-1.43),E); label(\"$$G$$\",(-0.2,-2.8),E); label(\"$$\\sqrt{3}x$$\",(-1.5,-0.5),E); label(\"$$M$$\",(-2,-0.9),E); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-2.7,2.3),S); label(\"$$1$$\",(0.1,-3.4),S); label(\"$$8$$\",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$ Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \\frac{\\sqrt{3}x}{2} \\Rightarrow DG = \\sqrt{3}x$. $AE = 2 + \\sqrt{3}x$ Use the Pythagorean theorem on triangle $ABE$, we get $$(2+\\sqrt{3}x)^2 + x^2 = 8^2 \\Rightarrow 4 + 3x^2 + 4\\sqrt{3}x + x^2 = 64 \\Rightarrow x^2 + \\sqrt{3}x - 15 = 0$$ Using the quadratic formula to solve, we get $$x = \\frac{-\\sqrt{3} \\pm \\sqrt{3 -4(1)(-15)}}{2} = \\frac{\\pm 3\\sqrt{7} - \\sqrt{3}}{2}$$ $x$ must be positive, therefore $$x = \\frac{3\\sqrt{7} - \\sqrt{3}}{2} \\Rightarrow C$$ ~Zeric Hang ## Soultion 4 (coordinate bashing) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$1$$\",(-0.5,3.8),S);[/asy]$", "$y = 5$, such that the answer is $1^2 + 5^2 = \\boxed{026}$. ### Solution 2 We'll use the law of cosines. Let $O$ be the center of the circle; we wish to find $OB$. We know how long $OA$ and $AB$ are, so if we can find $\\cos \\angle OAB$, we'll be in good shape. $[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); dot(O); dot(A); dot(B); dot(C); label(\"O\",O,SW); label(\"A\",A,NE); label(\"B\",B,S); label(\"C\",C,SE); [/asy]$ We can find $\\cos \\angle OAB$ using angles $OAC$ and $BAC$. First we note that by Pythagoras, $$AC = \\sqrt{AB^2 + BC^2} = \\sqrt{36 + 4} = \\sqrt{40} = 2 \\sqrt{10}.$$ If we let $M$ be the midpoint of $AC$, that mean that $AM = \\sqrt{10}$. Since $\\triangle OAC$ is isosceles ($OA = OC$ from the definition of a circle), $M$ is also the foot of the altitude from $O$ to $AC.$ $[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); pair M = (A+C)/2; draw(P); draw(O--C--A--cycle); draw(O--M, dashed); draw(rightanglemark(O,M,A,25)); dot(O); dot(A); dot(C); label(\"O\",O,SW); label(\"A\",A,NE); label(\"M\",M,SSW); label(\"C\",C,SE); label(\"\\sqrt{50}\", (O+A)/2, NW); label(\"\\sqrt{10}\", (A+M)/2, E); [/asy]$ It follows that $OM = \\sqrt{40} = 2 \\sqrt{10}$. Therefore \\begin{align*} \\cos \\angle OAC = \\frac{\\sqrt{10}}{\\sqrt{50}} &= \\frac{1}{\\sqrt{5}}, \\\\ \\sin \\angle OAC = \\frac{2 \\sqrt{10}}{\\sqrt{50}} &= \\frac{2}{\\sqrt{5}}. \\end{align*}", "# 2005 AIME I Problems/Problem 7 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem In quadrilateral $ABCD,\\ BC=8,\\ CD=12,\\ AD=10,$ and $m\\angle A= m\\angle B = 60^\\circ.$ Given that $AB = p + \\sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$ ## Solution ### Solution 1 $[asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label(\"A\",(0,0),SW); label(\"B\",(20.87,0),SE); label(\"E\",(15.87,8.66),NE); label(\"D\",(5,8.66),NW); label(\"P\",(5,0),S); label(\"Q\",(15.87,0),S); label(\"C\",(16.87,7),E); label(\"12\",(10.935,7.794),S); label(\"10\",(2.5,4.5),W); label(\"10\",(18.37,4.5),E); [/asy]$ Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$. Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$, and $BC=8$ and $EC=2$. We are given that $DC=12$. Since $\\angle CED = 120^{\\circ}$, using Law of Cosines on $\\bigtriangleup CED$ gives $$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", "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", "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,$$ and the answer is $\\boxed{\\textbf{(B) } \\sqrt{7}}$. ## Solution 2 (Intuition) $[asy] unitsize(1inch); pen p = fontsize(10pt); dot((0.756,0.655)); dot((1.512,1.309)); dot((1.701,0.327)); pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327); draw((0,0)--(1.323,2.291)--(2.646,0)--cycle); label(\"A\", (0,0), SW, p); label(\"C\", (2.646,0), SE, p); label(\"B\", (1.323,2.291), N, p); label(\"P'\", (0.756,0.655), NW, p); label(\"1\", (2.174,0.164), N, p); label(\"1\", (1.228,0.491), N, p); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); [/asy]$ Suppose that triangle $ABC$ had three segments of length $2$, emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments). Take this equilateral triangle to have side length $1$. The portions of each segment outside this triangle (in red) have length $1$. Take $P'$ to", "by a ratio of $\\frac{BC}{AD} = \\frac{125}{251}$. Since the homothety carries the midpoint of $\\overline{BC}$, $M$, to the midpoint of $\\overline{AD}$, which is $N$, then $E,M,N$ are collinear. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$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$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M[0]+H)/2,(1,0)); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from $B$ and $C$ to $AD$, and call the points where they meet $\\overline{AD}$, $E$ and $F$ respectively, then $FD = x$ and $EA = 1008-x$ , and so you can solve an equation in tangents. Since" ]

[ "& & (k,j+1,i-1) & & \\\\ & (k-1,j+1,i) & & & (k,j+1,i) & \\\\ & & (k-1,j+1,i+1) & & & (k,j+1,i+1) \\arrow[from=1-1,to=4-1]\\arrow[from=4-1,to=7-1] \\arrow[from=1-4,to=4-4]\\arrow[from=4-4,to=7-4] \\arrow[from=1-4,to=1-1]\\arrow[from=7-4,to=7-1] \\arrow[from=4-4,to=4-1] \\arrow[from=2-2,to=5-2, crossing over]\\arrow[from=5-2,to=8-2, crossing over] \\arrow[from=2-5,to=5-5]\\arrow[from=5-5,to=8-5] \\arrow[from=2-5,to=2-2, crossing over]\\arrow[from=8-5,to=8-2] \\arrow[from=5-5,to=5-2, crossing over] \\arrow[from=3-3,to=6-3, crossing over]\\arrow[from=6-3,to=9-3, crossing over] \\arrow[from=3-6,to=6-6]\\arrow[from=6-6,to=9-6] \\arrow[from=3-6,to=3-3, crossing over]\\arrow[from=9-6,to=9-3] \\arrow[from=6-6,to=6-3, crossing over] \\arrow[from=1-1,to=2-2]\\arrow[from=2-2,to=3-3] \\arrow[from=4-1,to=5-2]\\arrow[from=5-2,to=6-3] \\arrow[from=7-1,to=8-2]\\arrow[from=8-2,to=9-3] \\arrow[from=1-4,to=2-5]\\arrow[from=2-5,to=3-6] \\arrow[from=4-4,to=5-5]\\arrow[from=5-5,to=6-6] \\arrow[from=7-4,to=8-5]\\arrow[from=8-5,to=9-6] \\end{tikzcd} \\end{document} This should be a comment but it is rather long. The answer given by Sandy G is great, adequate and contains the main idea, crossing over arrows should be written after the arrows in the back (+1). Anyway, maybe the OP has a better way to remember the objects in the diagram than coordinates. In that case adding an alias key to the objects could be a good way. This is far from an answer but the idea is as follows: \\documentclass{article} \\usepackage{tikz-cd} \\begin{document} \\begin{tikzcd} |[alias=x]|A & B\\arrow{ld}\\\\ C & |[alias=y]|D \\arrow[from=1-1, to=2-2]%using coordinates \\arrow[blue, from=x, to=y, crossing over]%using a good name to the objects in the diagram \\end{tikzcd} \\end{document}", "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", "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 \"1982 AHSME Problems/Problem 25\" ## Problem The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. Each morning, a student walks from intersection $A$ to intersection $B$, always walking along streets shown, and always going east or south. For variety, at each intersection where he has a choice, he chooses with probability $\\frac{1}{2}$ whether to go east or south. Find the probability that through any given morning, he goes through $C$. $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ $\\text{(A)}\\frac{11}{32}\\qquad \\text{(B)}\\frac{1}{2}\\qquad \\text{(C)}\\frac{4}{7}\\qquad \\text{(D)}\\frac{21}{32}\\qquad \\text{(E)}\\frac{3}{4}$ ## Solutions ### Solution 1 $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); label(\"C_3\", (17,7), SW); label(\"C_2\", (17,11), SW); label(\"C_1\", (17,15), SW); label(\"C_0\", (17,19), SW); filldraw((17,7)--(18,7)--(18,8)--(17,8)--cycle, black); filldraw((17,11)--(18,11)--(18,12)--(17,12)--cycle, black); filldraw((17,15)--(18,15)--(18,16)--(17,16)--cycle, black); filldraw((17,19)--(18,19)--(18,20)--(17,20)--cycle, black); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ The probability that the student passes", "through the arrowheads of s are used to bound t. \\draw[name path=path_for_traversal_t, latex-latex] let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in ($(R) +(\\n1:1)$) -- ($(P) +({\\n1-180}:1)$); \\draw let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in node[anchor={\\n1-180}, inner sep=0] at ($($(R) +(\\n1:1)$) +(\\n1:0.15)$){$t$}; \\path[name path=path_for_traversal_s] ($(C) +(-130:2)$) -- ($(A) + (50:2)$); \\path[name path=path_for_the_lower_arrowhead_of_s] let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in ($(R) +(\\n1:1)$) -- ($(R) +(\\n1:1) +(195:11)$); \\path[name path=path_for_the_upper_arrowhead_of_s] let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in ($(P) +({\\n1-180}:1)$) -- ($(P) +({\\n1-180}:1) +(195:7)$); \\draw let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in node[anchor=50, inner sep=0] at ($(lower_arrowhead_for_s) +(-130:0.15)$){$s$}; % \\foreach \\i in {A,B,C,P,Q,R,a,b,c,r,q,p} \\node at (\\i) {\\i}; %The lengths of the line segments on the traversals between the parallel lines are typeset. \\draw node[inner sep=0] at ($($(P)!0.3!90:(Q)$)!{2/3}!($(Q)!0.3!-90:(P)$)$) {$4$}; \\draw node[inner sep=0] at ($($(Q)!0.3!90:(R)$)!{2/3}!($(R)!0.3!-90:(Q)$)$) {$6$}; \\path ($(a)!2/3!(b)$) -- ($(p)!2/3!(q)$) node [pos=.25, sloped] {$x+5$}; \\path ($(b)!2/3!(c)$) -- ($(q)!2/3!(r)$) node [pos=.2, sloped] {$4x+5$}; \\end{tikzpicture} \\end{document} Note that if you really want the labels centred on the green line, you should use 1/2 rather than 2/3: \\path ($(a)!.5!(b)$) -- ($(p)!.5!(q)$) node [pos=.25, sloped] {$x+5$}; \\path ($(b)!.5!(c)$) -- ($(q)!.5!(r)$) node [pos=.2, sloped] {$4x+5$}; \\draw[line width=0.2pt, green] ($(B)!0.5!(C)$) -- ($($(B)!0.5!(C)$) +({195}:2)$); Alternatively, you could rotate the nodes, but this seemed easier. If you want to use this, you can just use rotate=<angle> in the options to the node. For example: \\documentclass[tikz,border=10pt]{standalone} \\begin{document} \\begin{tikzpicture} \\foreach \\i [count=\\j] in {0,30,60,...,330} \\node [rotate=\\i, draw] at (2*\\j,0) {angle \\i};", "You can use Asymptote on the AoPSWiki right now, by enclosing the Asymptote code within <asy>...</asy> tags. For example, the following code <asy> draw((0,0)--(3,7),red); dot((0,0)); dot((3,7)); label(\"Produced with Asymptote \"+version.VERSION,point(S),2S); </asy> created the picture $[asy] draw((0,0)--(3,7),red); dot((0,0)); dot((3,7)); label(\"Produced with Asymptote \"+version.VERSION,point(S),2S); [/asy]$ And on the AoPS forums you can use [asy]..[/asy] Another example: [asy] pair A,B,C,X,Y,Z; A = (0,0); B = (1,0); C = (0.3,0.8); draw(A--B--C--A); X = (B+C)/2; Y = (A+C)/2; Z = (A+B)/2; draw(A--X, red); draw(B--Y,red); draw(C--Z,red); [/asy] $[asy] pair A,B,C,X,Y,Z; A = (0,0); B = (1,0); C = (0.3,0.8); draw(A--B--C--A); X = (B+C)/2; Y = (A+C)/2; Z = (A+B)/2; draw(A--X, red); draw(B--Y,red); draw(C--Z,red); [/asy]$", "P, Q, and R is as high as A and as far right as the right arrowhead %of the line through B, C, and R. \\path[name path=a_path_to_locate_left_arrowhead] (A) -- ($(A) +(-3.75,0)$); \\path[name path=another_path_to_locate_left_arrowhead] (P) -- ($(P)!-2.7cm!(Q)$); \\path[name path=a_path_to_locate_right_arrowhead] ($(R) +(1.5,0)$) -- ($(R) +(1.5,-0.75)$); \\draw[green, name path=another_path_to_locate_right_arrowhead] (Q) -- ($(R)!-1.7cm!(Q)$); \\draw[latex-latex] ($(B)!-1.5cm!(R)$) -- ($(R)!-1.5cm!(B)$); % \\draw[green, name path=a_path_for_the_label_for_P] ($(A)!0.15cm!-90:(B)$) -- ($(B)!0.15cm!90:(A)$); \\draw[green, name path=another_path_for_the_label_for_P] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_P}]; \\path let \\p1=($(A)-(B)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*(\\n1+(\\n2+180))-180}, inner sep=0, font=\\footnotesize] at (label_for_P){\\textit{P}}; % \\draw[green, name path=a_path_for_the_label_for_Q] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$); \\draw[green, name path=another_path_for_the_label_for_Q] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_Q and another_path_for_the_label_for_Q, by=label_for_Q}]; \\path let \\p1=($(A)-(C)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*((\\n1+180)+\\n2)-180}, inner sep=0, font=\\footnotesize] at (label_for_Q){\\textit{Q}}; % \\path[name path=a_path_for_the_label_for_R] ($(B) +(0,-0.15)$) -- ($(R) +(0,-0.15)$); \\path[name path=another_path_for_the_label_for_R] let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in (R) -- ($(R) +({0.5*(\\n1-180)}:0.25)$); \\coordinate[name intersections={of=a_path_for_the_label_for_R and another_path_for_the_label_for_R, by=label_for_R}]; \\path let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in node[anchor={0.5*(\\n1-180)+180}, inner sep=0, font=\\footnotesize] at (label_for_R){\\textit{R}}; % % %The regions bound by \\triangle{BPQ} and \\triangle{CPQ} are shaded. \\draw[fill=gray!25] (B) -- (intersection-1) -- (P) -- cycle; \\draw[fill=gray!25] (C) -- (intersection-1) -- (Q) -- cycle; \\draw[fill=gray!75] (P) -- (intersection-1) -- (Q) -- cycle; \\draw[dashed] (B) -- (Q); \\draw[dashed] (C) -- (P); \\draw (A) -- (B) -- (C) -- cycle; %The foot of the altitude of \\triangle{BPQ} from B is located. It is labeled F_2. \\coordinate", "cities cannot be used. Additionally, the same applies to the start and end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). It is not hard to find that there is only one configuration satisfying these conditions: Note: $\\text{X}$s represent unused edges. $[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label(\"A\", (0,2), W); label(\"L\", (3,0), E); label(\"X\", (0, 1.5)); label(\"X\", (1.5, 2)); label(\"X\", (1.5, 0)); label(\"X\", (3, 0.5)); label(\"O\", (1,1), NE); label(\"O\", (2,1), NE); [/asy]$ Observe that at each of the $2$ cities marked with an $\\text{O}$ on a path, there are $2$ possibilities: we can either continue straight and cross back over the path later, or we can make a left turn, then turn right when we approach the junction again. This", "DRAW CONNECTIONS %%%%%%%%%%%%%%%%%% % there are \\ilsize*\\hlsize arrows from il to hl0 % there are \\hlsize*\\hlsize arrows from hl0 to hl1 % there are \\hlsize*\\olsize arrows from hl1 to out % total number of arrows #totalarrows = \\ilsize*\\hlsize + \\hlsize*\\hlsize + \\hlsize*\\olsize % we assign to each arrow a number from 1 to #arrows % we do this by establishing an order in which we'd draw the arrows % % let (1,1) be the top left node, % with x increases denoting movement to the right, % and with y increases denoting movement down. % Imagine we have a 3x3 grid of arrows % Arrow 1 = (1,1) -- (2,1) Arrow 10 = (2,1) -- (3,1) % Arrow 2 = (1,1) -- (2,2) Arrow 11 = (2,1) -- (3,2) % Arrow 3 = (1,1) -- (2,3) Arrow 12 = (2,1) -- (3,3) % Arrow 4 = (1,2) -- (2,1) Arrow 13 = (2,2) -- (3,1) % Arrow 5 = (1,2) -- (2,2) Arrow 14 = (2,2) -- (3,2) % Arrow 6 = (1,2) -- (2,3) Arrow 15 = (2,2) -- (3,3) % Arrow 7 = (1,3) -- (2,1) Arrow 16 = (2,3) -- (3,1) % Arrow 8 = (1,3) -- (2,2) Arrow 17 = (2,3) -- (3,2) % Arrow 9 = (1,3) -- (2,3) Arrow 18 = (2,3)", "tip is at the beginning. Those limits are correct. In between we're just going linearly. I'm sorry, your circular example is much to complicated for me. Here's my attempt: from pyx import * ···· c = canvas.canvas() c.stroke(path.line(0, 0, 10, 0)) c.stroke(path.line(0, 0, 0, 2)) c.stroke(path.line(0.1, 0, 0.1, 2)) c.stroke(path.line(5, 0, 5, 2)) c.stroke(path.line(9.9, 0, 9.9, 2)) c.stroke(path.line(10, 0, 10, 2)) c.stroke(path.line(0, 0.2, 10, 0.2), [deco.arrow(pos=1)]) # earrow c.stroke(path.line(0, 0.4, 10, 0.4), [deco.arrow(pos=1, reversed=1)]) c.stroke(path.line(0, 0.6, 10, 0.6), [deco.arrow(pos=0.99)]) c.stroke(path.line(0, 0.8, 10, 0.8), [deco.arrow(pos=0.99, reversed=1)]) c.stroke(path.line(0, 1.0, 10, 1.0), [deco.arrow(pos=0.5)]) c.stroke(path.line(0, 1.2, 10, 1.2), [deco.arrow(pos=0.5, reversed=1)]) c.stroke(path.line(0, 1.4, 10, 1.4), [deco.arrow(pos=0.01)]) c.stroke(path.line(0, 1.6, 10, 1.6), [deco.arrow(pos=0.01, reversed=1)]) c.stroke(path.line(0, 1.8, 10, 1.8), [deco.arrow(pos=0)]) c.stroke(path.line(0, 2.0, 10, 2.0), [deco.arrow(pos=0, reversed=1)]) # barrow c.writePDFfile() Now try your patches. It creates a lot of confusion. No way. I think the positioning of the arrows on the circle you're trying to get right should be done using path features. This is trivial: from pyx import * c = canvas.canvas() circ = path.circle(0, 0, 5) p = path.line(0, 0, 5, 5) c.stroke(circ.split(circ.intersect(p)[0][0])[0], [deco.arrow(size=1)]) c.stroke(p) c.writePDFfile() Now, the only problem is if you want to position the \"back of the arrow\" at the intersection point. I don't know whether it is that useful (technically) but I fully understand that it might be desirable from a", "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.", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "$m=1/3$, 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: $\\boxed{-\\frac{1}{6}}$.", "(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 \"2020 AIME I Problems/Problem 9\" ## Problem Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ ## Solution $[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label(\"a_1\", a1, NE); label(\"a_2\", a2, NE); label(\"a_3\", a3, NE); label(\"b_1\", b1, S); label(\"b_2\", b2, S); label(\"b_3\", b3, S); label(\"c_1\", c1, W); label(\"c_2\", c2, W); label(\"c_3\", c3,", "C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$", "(Q) -- ($(Q)!-3.5cm!(P)$); \\coordinate[name intersections={of=a_path_to_locate_R_on_BC and another_path_to_locate_R_on_BC, by=R}]; % %The left arrowhead of the line through P, Q, and R is as high as A and as far right as the right arrowhead %of the line through B, C, and R. \\path[name path=a_path_to_locate_left_arrowhead] (A) -- ($(A) +(-3.75,0)$); \\path[name path=another_path_to_locate_left_arrowhead] (P) -- ($(P)!-2.7cm!(Q)$); \\path[name path=a_path_to_locate_right_arrowhead] ($(R) +(1.5,0)$) -- ($(R) +(1.5,-0.75)$); \\draw[green, name path=another_path_to_locate_right_arrowhead] (Q) -- ($(R)!-1.7cm!(Q)$); \\draw[latex-latex] ($(B)!-1.5cm!(R)$) -- ($(R)!-1.5cm!(B)$); %a_path_for_the_label_for_P \\draw[green, name path=a_path_for_the_label_for_P] ($(A)!0.15cm!-90:(B)$) -- ($(B)!0.15cm!90:(A)$); \\draw[green, name path=another_path_for_the_label_for_P] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_P}]; %\\path let \\p1=($(A)-(B)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*(\\n1+(\\n2+180))-180}, inner sep=0, font=\\footnotesize] at (label_for_P){\\textit{P}}; %a_path_for_the_label_for_Q \\draw[green, name path=a_path_for_the_label_for_Q] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$); \\draw[green, name path=another_path_for_the_label_for_Q] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_Q and another_path_for_the_label_for_Q, by=label_for_Q}]; \\node[draw=green,font=\\footnotesize,anchor=south,outer sep=0pt,inner sep=0pt] at (label_for_Q) {$Q_{\\vphantom{1}}$}; % \\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_Q}]; \\node[draw=green,font=\\footnotesize,anchor=south,outer sep=0pt,inner sep=0pt] at (label_for_P) {$P_{\\vphantom{1}}$}; % %\\path let \\p1=($(A)-(C)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*((\\n1+180)+\\n2)-180}, inner sep=0, font=\\footnotesize] at (label_for_Q){\\textit{Q}}; % \\path[name path=a_path_for_the_label_for_R] ($(B) +(0,-0.15)$) -- ($(R) +(0,-0.15)$); \\path[name path=another_path_for_the_label_for_R] let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in (R) -- ($(R) +({0.5*(\\n1-180)}:0.25)$); \\coordinate[name intersections={of=a_path_for_the_label_for_R and another_path_for_the_label_for_R, by=label_for_R}]; %\\path let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in node[anchor={0.5*(\\n1-180)+180}, inner sep=0, font=\\footnotesize] at (label_for_R){\\textit{R}}; % % %The regions bound by \\triangle{BPQ} and \\triangle{CPQ} are shaded. \\draw[fill=gray!25] (B) -- (intersection-1) -- (P) -- cycle; \\draw[fill=gray!25] (C) -- (intersection-1) -- (Q) -- cycle; \\draw[fill=gray!75] (P) -- (intersection-1) --", "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", "at pos=0, the \"back\" of the arrow starts at the starting point of the path and at pos=1 (the default for earrow) the tip is at the end. The same for barrow. At pos=1 the back is at the end of the path and at pos=0 (the default for barrow) the tip is at the beginning. Those limits are correct. In between we're just going linearly. > > I'm sorry, your circular example is much to complicated for me. Here's my attempt: > > from pyx import * > ···· > c = canvas.canvas() > c.stroke(path.line(0, 0, 10, 0)) > c.stroke(path.line(0, 0, 0, 2)) > c.stroke(path.line(0.1, 0, 0.1, 2)) > c.stroke(path.line(5, 0, 5, 2)) > c.stroke(path.line(9.9, 0, 9.9, 2)) > c.stroke(path.line(10, 0, 10, 2)) > c.stroke(path.line(0, 0.2, 10, 0.2), [deco.arrow(pos=1)]) # earrow > c.stroke(path.line(0, 0.4, 10, 0.4), [deco.arrow(pos=1, reversed=1)]) > c.stroke(path.line(0, 0.6, 10, 0.6), [deco.arrow(pos=0.99)]) > c.stroke(path.line(0, 0.8, 10, 0.8), [deco.arrow(pos=0.99, reversed=1)]) > c.stroke(path.line(0, 1.0, 10, 1.0), [deco.arrow(pos=0.5)]) > c.stroke(path.line(0, 1.2, 10, 1.2), [deco.arrow(pos=0.5, reversed=1)]) > c.stroke(path.line(0, 1.4, 10, 1.4), [deco.arrow(pos=0.01)]) > c.stroke(path.line(0, 1.6, 10, 1.6), [deco.arrow(pos=0.01, reversed=1)]) > c.stroke(path.line(0, 1.8, 10, 1.8), [deco.arrow(pos=0)]) > c.stroke(path.line(0, 2.0, 10, 2.0), [deco.arrow(pos=0, reversed=1)]) # barrow > c.writePDFfile() I think that's a perfect illustration of the current problems: The pos=1 arrow has its tip on the vertical comparison line, the one", "% \\path [name path=C2] (0,3.5) -- (2.5,3.5); \\path [name path=C4] (0,3.0) -- (2.5,3.0); \\path [name path=C6] (0,2.5) -- (2.5,2.5); % \\coordinate (n) at (2.5,0); \\draw [name intersections={of=c2 and C2, by=a}, <-] (a) -- (a -| n) node[right] {$y=x^2$}; \\draw [name intersections={of=c4 and C4, by=b}, <-] (b) -- (b -| n) node[right] {$y=x^4$}; \\draw [name intersections={of=c6 and C6, by=c}, <-] (c) -- (c -| n) node[right] {$y=x^6$}; \\end{axis} \\end{tikzpicture} \\end{document} As you can observe, the code for labeling of curves is written from scratch. It employ intersections TikZ library, so no need to manually calculated arrows for positioning of arrows and labels." ]

[ "$W_\\mathrm{B}$ ) particularly. Is not there 1 single value of $W$ to get us from the state A to the state B along a particular path ? – Poutnik Dec 21 '20 at 10:07", "<< \"There is a path from \" << a << \" to \" << b; else cout << \"There is no path from \" << a << \" to \" << b; return 0; } ## Output Enter the source and destination vertices: (0-3) There is a path from 3 to 1 There is a path from 1 to 3 Updated on 30-Jul-2019 22:30:26 Advertisements", "them once but in a way that minimizes the cost of the total path.\" This says nothing about visiting the start node twice. And it certainly doesn't suggest that a path $A\\to B\\to A\\to C$, as introduced in the edit, is allowed. – David Richerby Oct 25 '13 at 20:20", "$c$. Then this path does this by first doing something, then getting into $B$ and reaching a point $a \\neq c$, then maybe passing through $c$ a few times, then reaching a point $b \\neq c$, and then leaving $B$ without going over $c$ again. Now replace whatever the path did between $a$ and $b$ by some path in $B \\setminus \\{c\\}$ between $a$ and $b$. The path obtained in this way is a path in $X \\setminus \\{c\\}$ between your original two points. – Levi Oct 27 at 21:33 • I know, but describing this precisely is not easy, and the \"passing through $c$ a few times\" part could actually be very messy. (It could pass through $c$ infinitely many times.) The \"getting into $B$\" and \"leaving $B$\" part may also require a more careful description, you may need to talk about the circle that goes with $B$ and how/when the path crosses it. – Mirko Oct 27 at 21:57 Let $$\\ c\\in X\\subseteq\\mathbb R^2,\\$$ where $$\\ X\\$$ is open in $$\\ \\mathbb R^2.\\$$ (The case of $$\\ x\\in\\mathbb R^2\\setminus X\\$$ is trivially true). Let $$\\ G\\ H\\$$ be disjoint open subsets of $$\\ R^2\\$$ such that $$\\ G\\cup H=X\\setminus\\{c\\}.\\$$ Consider an open ball $$B\\ := \\mathscr B(c, r)\\ \\subseteq X\\qquad\\qquad (\\mbox{where}\\quad r>0)$$ Then each circle $$\\", "one way road, called a directed edge, and we draw an arrow on it to show which way you can travel along it. For instance, if there are two cities $A$ and $B$, and a line with an arrow from $A$ to $B$, then we can travel from $A$ to $B$, but not from $B$ to $A$. Here is an example of a graph, you can't travel from $B$ to $A$, but you can travel from $A$ to $B$. You can't travel from $C$ to $A$ or from $A$ to $C$, but you can travel from $B$ to $C$ and from $C$ to $B$. Graph with directed edges To complicate things even further, we sometimes want to add something called a cost to each edge. The idea of a cost is that it indicates how much it would cost to travel down that edge. A simple example of this is shown below. Graph with costs In this graph, most of the people in city $A$ want to get to city $C$, whereas only a few want to get to city $B$. Unfortunately, both roads are the same size, this means that there are long traffic jams on the road from $A$ to $C$, it takes about 30 minutes to get there. To get from $A$ to $B$ is", "two nodes in both directions, that is, a -- > B B!", "then you can construct a path from b to c, namely the straight line from b to c. 2) If there is any $a \\in B(b,r)$ such that $a \\in A$, then by the definition of A there is a path from x to a. But $a \\in B(b,r)$, and by (1) there exists a path from a to b. By \"gluing\" the two paths together, we get a path from x to b. but $b \\in B$, which means that there exists no path from x to b - contradiction. Can you point what step, exactly, is not clear in this proof? 2. x alone is not a member of A. A is all points connectible to x. You have to prove such a point exists. A and B are not both open sets. See previous posts. 3. There is a path from x to x, namely, $f:[0,1] \\to S, & f(t)=x$, and so $x \\in A$. With which part of Opalg's proof do you not agree? 4. A straight line is defined by distinct end points. A and B cannot both be open sets. See previous posts. 5. The previous posts don't show that the proof is wrong. To make it clear, I am using the definition that a set S is connected if there are", "ways to get from $$(0,0)$$ to $$(7,4)$$ passing through $$(1,1)$$.", "every town can be reached by going around the loop. $\\textbf{Case 2}$. The loop has $4$ towns. The town not on the loop must have a road leading to it. This road comes from a town on the loop. Therefore this town can be reached from the loop. This town not on the loop must also have a road leading out of it. This road leads to a town on the loop. Therefore the loop can be reached from the town. $\\textbf{Case 3}$. The loop has $3$ towns. There are two towns not on the loop; call them Town $A$ and Town $B$. Without loss of generality assume $A$ leads to $B$. Because a road must lead to $A$, the town where this road comes from must be on the loop. Therefore $A$ and therefore $B$ can be reached from the loop. Because a road must lead out of $B$, the town it leads to must be on the loop. Therefore the loop can be reached from $B$ and also $A$. The number of good configurations is the total number of configurations minus the number of bad configurations. There are $2^{{5\\choose2}}$ total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are $5$ ways to choose", "explicitly include a boundary condition $p(x,y)=0$ for $y>2x$ to avoid infinite descent. Apr 3, 2020 at 14:46 Suppose you do $$A$$ moves of type (i), $$B$$ of type (ii) and $$C$$ of type (iii). If you want to reach $$(8, 8)$$, then clearly $$A + 2B + C = 8$$ and $$2A + B = 8$$. This yields $$B = 8-2A$$ and $$C = 3A - 8$$. Since $$A, B, C$$ are nonnegative integers, you obtain solutions $$(A, B, C) = (3, 2, 1)$$ or $$(4, 0, 4)$$. Now to calculate paths, for $$(4, 0, 4)$$ case, a path is described by string of four $$A$$s and four $$C$$s. For example, $$AAACCACC$$ is one such path. There is $${8 \\choose 4} = 70$$ such paths. For the $$(3, 2, 1)$$ case there is $${6 \\choose 3} \\cdot 3 = 60$$ such paths. Altogether there are $$130$$ such paths. For reaching $$(10, 10)$$ same logic gives you $$B = 10 - 2A$$ and $$C = 3A - 10$$, so solutions are $$(4, 2, 2)$$ and $$(5, 0, 5)$$. So, the number of paths is $${8 \\choose 4} {4 \\choose 2} + {10 \\choose 5} = 70 \\cdot 6 + 252 = 672$$. This is in agreement with Rob Pratts answer. Also, I would like to emphasise as I did", "we can get through to the $$8$$ later on and the path is essentially forced after that.", "A, B and C are three cities. There are 5 routes from A to B and 3 routes from Question: A, B and C are three cities. There are 5 routes from A to B and 3 routes from B to C. How many different routes are there from A to C via B? Solution: Given: 5 routes from A to B and 3 routes from B to C. To find: number of different routes from $A$ to $C$ via $B$. Let $E_{1}$ be the event : 5 routes from A to B Let $E_{2}$ be the event : 3 routes from $B$ to $C$ Since going from $A$ to $C$ via $B$ is only possible if both the events $E_{1}$ and $E_{2}$ occur simultaneously. So there are $5 \\times 3=15$ different routes from $A$ to $C$ via $B$.", "the required answer is either $^{13}C_7$ or $^{13}C_6$ (both are same! and equal to $1716$). So, there are $1716$ ways by which you can reach point $B$ from point $A$.", "because the premise requires there are at least 2 cities, but we can still assert that $B$ is our required city. From any other cities ($A_1, A_{n+1}$, category 1, and category 2) we can reach $B$ either directly or through $A_1$ or $A_{n+1}$.", "D have higher link values and IDs and are on the stack, so A doesn't update. Every node has been visited. There's a path from D to E to B to D, meaning those are part of the same SCC (in fact, all of these are in the same SCC—it's one SCC). The stack is {A, D, E, B, C, F} with F at the top. The two SCCs are {B,E} and {A,C,D,F}. …you can reach A from B, and you can reach B from A. What am I missing here?", "about it: the path $c$ can't deform to $c'$ because a hole is \"in the way\". Thanks for the reply! Apr 21 '20 at 12:29", "is to return to $$q_1$$ via a $$b$$, the other to go on to state $$q_0$$ via an $$a$$, and these paths yield different elements of the final regular expression.", "of $B$. A path from $x_0$ to $y_0$ in $Z$ is an $f$ you want. -", "path within $C^c$. In another word, every path connecting $B$ and $B'$ must go through $C$. By Menger's Theorem, we have that there are at least $k$ disjoint paths connecting $B$ and $B'$. So we have two disjoint paths connecting $B$ to the cycle C, which make our cycle longer. – Xiaochuan Jan 8 '11 at 7:45 Consider the longest cycle $C$ in $G$ and suppose its length is less than $2k$. Use Menger's Theorem and the pigeonhole principle to come up with a contradiction. - Seems like another application of Dirac's Theorem. -", "is $$3$$ to get to $$D$$, and then from $$D$$ the optimal value to $$T$$ is $$2$$ (from stage $$1$$) - hence a final value of $$3+2 = 5$$. Now do the same in your table for stage $$3$$. Table 4 - Dynamic programming table stage $$3$$ Stage State Action Destination Value $$3$$ $$S$$ $$SA$$ $$A$$ $$3+5 = 8^*$$ $$3$$ $$S$$ $$SB$$ $$B$$ $$5+3 = 8^*$$ $$3$$ $$S$$ $$SC$$ $$C$$ $$2+7 = 9$$ So, there are in fact two equally optimal paths through this network, each with a value (cost) of $$8$$. By tracking back through the table, following the *s, you can find and write down the optimal path(s) from $$S$$ to $$T$$. For example, looking at stage $$3$$, you can see that one of the optimal paths goes $$S \\to B$$. Going to $$B$$ in stage two, you can see that the optimal path from $$B$$ goes $$B\\to D$$. And finally going to stage $$1$$ the optimal path from $$D$$ goes $$D\\to T$$. The optimal paths are therefore $S \\to A \\to D \\to T$ $S \\to B \\to D \\to T.$ This technique can be used both to minimise the value of the route or maximise the value of the route. But there's more; dynamic programming can also be used for maximin and minimax problems." ]

[ "on the top and bottom, 1 on the sides) have 3 edges connecting to them. Therefore, at least 1 edge connecting to them cannot be used. Additionally, the same goes for the start and end point as we don't want to return to them. so we have 8 points that we know have 1 unused edge, and we have a total of 4 unused edges to work with (17-13), so we easily find there is only one configuration that satisfies this: X's represent unused edges, by necessity, lines are filled in for the paths. $[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label(\"A\", (0,2), W); label(\"L\", (3,0), E); label(\"X\", (0, 1.5)); label(\"X\", (1.5, 2)); label(\"X\", (1.5, 0)); label(\"X\", (3, 0.5)); label(\"O\", (1,1), NE); label(\"O\", (2,1), NE); [/asy]$ Now, we find that at each of the 2 cities marked with an O, we have 2 possibilities to follow the path, we can either continue straight and cross back over it", "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", "P, Q, and R is as high as A and as far right as the right arrowhead %of the line through B, C, and R. \\path[name path=a_path_to_locate_left_arrowhead] (A) -- ($(A) +(-3.75,0)$); \\path[name path=another_path_to_locate_left_arrowhead] (P) -- ($(P)!-2.7cm!(Q)$); \\path[name path=a_path_to_locate_right_arrowhead] ($(R) +(1.5,0)$) -- ($(R) +(1.5,-0.75)$); \\draw[green, name path=another_path_to_locate_right_arrowhead] (Q) -- ($(R)!-1.7cm!(Q)$); \\draw[latex-latex] ($(B)!-1.5cm!(R)$) -- ($(R)!-1.5cm!(B)$); % \\draw[green, name path=a_path_for_the_label_for_P] ($(A)!0.15cm!-90:(B)$) -- ($(B)!0.15cm!90:(A)$); \\draw[green, name path=another_path_for_the_label_for_P] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_P}]; \\path let \\p1=($(A)-(B)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*(\\n1+(\\n2+180))-180}, inner sep=0, font=\\footnotesize] at (label_for_P){\\textit{P}}; % \\draw[green, name path=a_path_for_the_label_for_Q] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$); \\draw[green, name path=another_path_for_the_label_for_Q] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_Q and another_path_for_the_label_for_Q, by=label_for_Q}]; \\path let \\p1=($(A)-(C)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*((\\n1+180)+\\n2)-180}, inner sep=0, font=\\footnotesize] at (label_for_Q){\\textit{Q}}; % \\path[name path=a_path_for_the_label_for_R] ($(B) +(0,-0.15)$) -- ($(R) +(0,-0.15)$); \\path[name path=another_path_for_the_label_for_R] let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in (R) -- ($(R) +({0.5*(\\n1-180)}:0.25)$); \\coordinate[name intersections={of=a_path_for_the_label_for_R and another_path_for_the_label_for_R, by=label_for_R}]; \\path let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in node[anchor={0.5*(\\n1-180)+180}, inner sep=0, font=\\footnotesize] at (label_for_R){\\textit{R}}; % % %The regions bound by \\triangle{BPQ} and \\triangle{CPQ} are shaded. \\draw[fill=gray!25] (B) -- (intersection-1) -- (P) -- cycle; \\draw[fill=gray!25] (C) -- (intersection-1) -- (Q) -- cycle; \\draw[fill=gray!75] (P) -- (intersection-1) -- (Q) -- cycle; \\draw[dashed] (B) -- (Q); \\draw[dashed] (C) -- (P); \\draw (A) -- (B) -- (C) -- cycle; %The foot of the altitude of \\triangle{BPQ} from B is located. It is labeled F_2. \\coordinate", "our arrows. That is, if we have an arrow $f$ from A to B, $g$ from B to C, $h$ from C to D, it shouldn’t matter how we make up the big arrow from A to D. Putting $f$ and $g$ together, then taking that to go from A to C followed by going through $h$ to get to D is the same as putting $g$ and $h$ together first and using $f$ to get to B then taking that grouping from B to D. Here is a pretty picture to clear that long sentence up. The dotted lines represent the composition of two arrows. Remember we can turn any two arrows that are next to each other into one long arrow (dotted arrow) by means of composition. So lets turn the route from A to C into one arrow, the dotted arrow $gf$. Again, mathematicians annoyingly write $gf$ as meaning to first go through the arrow $f$ and then the arrow $g$. Then we can use that long arrow (dotted arrow) and the leftover arrow, $h$, to get from A to D. However we could have went to D from A through a different route. Since the arrows $g$ and $h$ are next to each other we could have also put them together first. This is", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label(\"A\",A,(1,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0));label(\"P\",P,(0,-1));label(\"Q\",Q,(1,0));label(\"R\",R,(-1,1)); [/asy]$", "cities cannot be used. Additionally, the same applies to the start and end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). It is not hard to find that there is only one configuration satisfying these conditions: Note: $\\text{X}$s represent unused edges. $[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label(\"A\", (0,2), W); label(\"L\", (3,0), E); label(\"X\", (0, 1.5)); label(\"X\", (1.5, 2)); label(\"X\", (1.5, 0)); label(\"X\", (3, 0.5)); label(\"O\", (1,1), NE); label(\"O\", (2,1), NE); [/asy]$ Observe that at each of the $2$ cities marked with an $\\text{O}$ on a path, there are $2$ possibilities: we can either continue straight and cross back over the path later, or we can make a left turn, then turn right when we approach the junction again. This", "# Difference between revisions of \"1982 AHSME Problems/Problem 25\" ## Problem The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. Each morning, a student walks from intersection $A$ to intersection $B$, always walking along streets shown, and always going east or south. For variety, at each intersection where he has a choice, he chooses with probability $\\frac{1}{2}$ whether to go east or south. Find the probability that through any given morning, he goes through $C$. $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ $\\text{(A)}\\frac{11}{32}\\qquad \\text{(B)}\\frac{1}{2}\\qquad \\text{(C)}\\frac{4}{7}\\qquad \\text{(D)}\\frac{21}{32}\\qquad \\text{(E)}\\frac{3}{4}$ ## Solutions ### Solution 1 $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); label(\"C_3\", (17,7), SW); label(\"C_2\", (17,11), SW); label(\"C_1\", (17,15), SW); label(\"C_0\", (17,19), SW); filldraw((17,7)--(18,7)--(18,8)--(17,8)--cycle, black); filldraw((17,11)--(18,11)--(18,12)--(17,12)--cycle, black); filldraw((17,15)--(18,15)--(18,16)--(17,16)--cycle, black); filldraw((17,19)--(18,19)--(18,20)--(17,20)--cycle, black); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ The probability that the student passes", "blue $$\\ell$$-arrow pointing to a different world. Taken together, the public announcement of $$S_1$$ should delete every world from which we can take a red s-arrow (possibly a reflexive one) to a world that has no $$\\ell$$-arrow pointing to a different world. Let us call such $$\\ell$$-arrows “outgoing”. Examining Figure B10, we see that the following worlds are to be deleted: • (2,7): from this world we can take a reflexive s-arrow to the world itself, which has no outgoing $$\\ell$$-arrows; • (3,7): as for (2,7); • (4,7): as for (2,7); • (5,7): as for (2,7); • (6,7): as for (2,7); • (2,6): from this world we can take an s-arrow to (3,5), which has no outgoing $$\\ell$$-arrows; • (3,6): from this world we can take an s-arrow to (2,7), which has no outgoing $$\\ell$$-arrows; • (4,6): from this world we can take an s-arrow to (3,7), which has no outgoing $$\\ell$$-arrows; • (5,6): as for (2,7); • (2,5): as for (2,7); • (3,5): as for (2,7); • (4,5): as for (2,7); • (2,4): as for (2,7); and • (3,4): from this world we can take an s-arrow to (2,5), which has no outgoing $$\\ell$$-arrows. The result is the model $$M[S_1]$$ pictured in Figure B11. Note that in the model $$M_{s\\ell}[S_1]$$, both agents know the numbers: each of", "label(\"$M$\", (3, 2)); label(\"$C$\", (4, 2)); label(\"$8$\", (0, 3)); label(\"$C$\", (1, 3)); label(\"$M$\", (2, 3)); label(\"$C$\", (3, 3)); label(\"$8$\", (4, 3)); label(\"$8$\", (1, 4)); label(\"$C$\", (2, 4)); label(\"$8$\", (3, 4));[/asy] There are three different kinds of paths that are on this diagram. The first kind is when you directly count A, M, C in a straight line. The second is when you count A, turn left or right to get M, then go straight to count M and C. The third is the one where you start with A, move forward to count M, turn left or right to count C, then move straight again to get 8. There are 8 paths for each kind of path, making for $8 \\cdot 3=\\boxed{24}$ paths. -coolmath34", "= 3$. Note that in this solution, no set is empty.We can look at the assignment as assigning arrows between $a_i$ and $a_{i+1}$ (indices wrap around), where the clockwise arrows are for assigning the conversation to player 1 and anticlockwise ones for assigning the conversation to player 2. The \"balance\" of the configuration can be defined as the sum of $to - from$ over all arrows $(from, to)$, and it is sufficient to find a balanced assignment of arrows inductively.Now consider how we get from $i-1$ to $i$. You need to fit $a_i$ between $a_1$ and $a_{i-1}$. To maintain the balance between the two players, you need to assign directions to arrows between $a_{i-1}$ and $a_i$, and $a_i$ and $a_1$ depending on the direction of the arrow that is currently between $a_{i-1}$ and $a_1$.As it turns out, this can be done in a manner similar to the $n = 3$ case, and you'll be done inductively (in $O(n)$). • » » » 12 months ago, # ^ | +3 Add those with $(a_i - a_{i + 1}) < 0$ in one set and $(a_i - a_{i + 1}) > 0$ in other.The sum of two sets should be the same as the net sum in a circular array will be $0$. • » » » » 12 months", "(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", "tip is at the beginning. Those limits are correct. In between we're just going linearly. I'm sorry, your circular example is much to complicated for me. Here's my attempt: from pyx import * ···· c = canvas.canvas() c.stroke(path.line(0, 0, 10, 0)) c.stroke(path.line(0, 0, 0, 2)) c.stroke(path.line(0.1, 0, 0.1, 2)) c.stroke(path.line(5, 0, 5, 2)) c.stroke(path.line(9.9, 0, 9.9, 2)) c.stroke(path.line(10, 0, 10, 2)) c.stroke(path.line(0, 0.2, 10, 0.2), [deco.arrow(pos=1)]) # earrow c.stroke(path.line(0, 0.4, 10, 0.4), [deco.arrow(pos=1, reversed=1)]) c.stroke(path.line(0, 0.6, 10, 0.6), [deco.arrow(pos=0.99)]) c.stroke(path.line(0, 0.8, 10, 0.8), [deco.arrow(pos=0.99, reversed=1)]) c.stroke(path.line(0, 1.0, 10, 1.0), [deco.arrow(pos=0.5)]) c.stroke(path.line(0, 1.2, 10, 1.2), [deco.arrow(pos=0.5, reversed=1)]) c.stroke(path.line(0, 1.4, 10, 1.4), [deco.arrow(pos=0.01)]) c.stroke(path.line(0, 1.6, 10, 1.6), [deco.arrow(pos=0.01, reversed=1)]) c.stroke(path.line(0, 1.8, 10, 1.8), [deco.arrow(pos=0)]) c.stroke(path.line(0, 2.0, 10, 2.0), [deco.arrow(pos=0, reversed=1)]) # barrow c.writePDFfile() Now try your patches. It creates a lot of confusion. No way. I think the positioning of the arrows on the circle you're trying to get right should be done using path features. This is trivial: from pyx import * c = canvas.canvas() circ = path.circle(0, 0, 5) p = path.line(0, 0, 5, 5) c.stroke(circ.split(circ.intersect(p)[0][0])[0], [deco.arrow(size=1)]) c.stroke(p) c.writePDFfile() Now, the only problem is if you want to position the \"back of the arrow\" at the intersection point. I don't know whether it is that useful (technically) but I fully understand that it might be desirable from a", "# Difference between revisions of \"2020 AIME I Problems/Problem 9\" ## Problem Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ ## Solution $[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label(\"a_1\", a1, NE); label(\"a_2\", a2, NE); label(\"a_3\", a3, NE); label(\"b_1\", b1, S); label(\"b_2\", b2, S); label(\"b_3\", b3, S); label(\"c_1\", c1, W); label(\"c_2\", c2, W); label(\"c_3\", c3,", "(Q) -- ($(Q)!-3.5cm!(P)$); \\coordinate[name intersections={of=a_path_to_locate_R_on_BC and another_path_to_locate_R_on_BC, by=R}]; % %The left arrowhead of the line through P, Q, and R is as high as A and as far right as the right arrowhead %of the line through B, C, and R. \\path[name path=a_path_to_locate_left_arrowhead] (A) -- ($(A) +(-3.75,0)$); \\path[name path=another_path_to_locate_left_arrowhead] (P) -- ($(P)!-2.7cm!(Q)$); \\path[name path=a_path_to_locate_right_arrowhead] ($(R) +(1.5,0)$) -- ($(R) +(1.5,-0.75)$); \\draw[green, name path=another_path_to_locate_right_arrowhead] (Q) -- ($(R)!-1.7cm!(Q)$); \\draw[latex-latex] ($(B)!-1.5cm!(R)$) -- ($(R)!-1.5cm!(B)$); %a_path_for_the_label_for_P \\draw[green, name path=a_path_for_the_label_for_P] ($(A)!0.15cm!-90:(B)$) -- ($(B)!0.15cm!90:(A)$); \\draw[green, name path=another_path_for_the_label_for_P] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_P}]; %\\path let \\p1=($(A)-(B)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*(\\n1+(\\n2+180))-180}, inner sep=0, font=\\footnotesize] at (label_for_P){\\textit{P}}; %a_path_for_the_label_for_Q \\draw[green, name path=a_path_for_the_label_for_Q] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$); \\draw[green, name path=another_path_for_the_label_for_Q] ($(left_arrowhead)!0.15cm!90:(right_arrowhead)$) -- ($(right_arrowhead)!0.15cm!-90:(left_arrowhead)$); \\coordinate[name intersections={of=a_path_for_the_label_for_Q and another_path_for_the_label_for_Q, by=label_for_Q}]; \\node[draw=green,font=\\footnotesize,anchor=south,outer sep=0pt,inner sep=0pt] at (label_for_Q) {$Q_{\\vphantom{1}}$}; % \\coordinate[name intersections={of=a_path_for_the_label_for_P and another_path_for_the_label_for_P, by=label_for_Q}]; \\node[draw=green,font=\\footnotesize,anchor=south,outer sep=0pt,inner sep=0pt] at (label_for_P) {$P_{\\vphantom{1}}$}; % %\\path let \\p1=($(A)-(C)$), \\n1={atan(\\y1/\\x1)}, \\p2=($(P)-(R)$), \\n2={atan(\\y2/\\x2)} in node[draw=green, anchor={0.5*((\\n1+180)+\\n2)-180}, inner sep=0, font=\\footnotesize] at (label_for_Q){\\textit{Q}}; % \\path[name path=a_path_for_the_label_for_R] ($(B) +(0,-0.15)$) -- ($(R) +(0,-0.15)$); \\path[name path=another_path_for_the_label_for_R] let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in (R) -- ($(R) +({0.5*(\\n1-180)}:0.25)$); \\coordinate[name intersections={of=a_path_for_the_label_for_R and another_path_for_the_label_for_R, by=label_for_R}]; %\\path let \\p1=($(P)-(R)$), \\n1={atan(\\y1/\\x1)} in node[anchor={0.5*(\\n1-180)+180}, inner sep=0, font=\\footnotesize] at (label_for_R){\\textit{R}}; % % %The regions bound by \\triangle{BPQ} and \\triangle{CPQ} are shaded. \\draw[fill=gray!25] (B) -- (intersection-1) -- (P) -- cycle; \\draw[fill=gray!25] (C) -- (intersection-1) -- (Q) -- cycle; \\draw[fill=gray!75] (P) -- (intersection-1) --", "draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label(\"A\",A,(0,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0)); label(\"A'\",A1,(0,-1));label(\"B'\",B1,(1,1));label(\"C'\",C1,(-1,1)); label(\"H\",H,(-1,-1)); [/asy]$ • 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.", "arrows; from the second red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to each of the blue arrows. From each of the first and second blue arrows, there are respectively $4$ ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively $8$ ways to get to each of the first and the second green arrows. Therefore there are in total $5 \\cdot (4+4+8+8) = 120$ ways to get to each of the green arrows. Finally, from each of the first and second green arrows, there is respectively $2$ ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are $3$ ways to get to the first orange arrow. Therefore there are $120 \\cdot (2+2+3+3) = 1200$ ways to get to each of the orange arrows, hence $2400$ ways to get to the point $B$. $\\framebox{E}$ ## Solution Suppose the bug just went through one of the green arrows. There is only $1$ path it can take that goes through the remaining white arrow, depending on whether it just took one of the top two", "DRAW CONNECTIONS %%%%%%%%%%%%%%%%%% % there are \\ilsize*\\hlsize arrows from il to hl0 % there are \\hlsize*\\hlsize arrows from hl0 to hl1 % there are \\hlsize*\\olsize arrows from hl1 to out % total number of arrows #totalarrows = \\ilsize*\\hlsize + \\hlsize*\\hlsize + \\hlsize*\\olsize % we assign to each arrow a number from 1 to #arrows % we do this by establishing an order in which we'd draw the arrows % % let (1,1) be the top left node, % with x increases denoting movement to the right, % and with y increases denoting movement down. % Imagine we have a 3x3 grid of arrows % Arrow 1 = (1,1) -- (2,1) Arrow 10 = (2,1) -- (3,1) % Arrow 2 = (1,1) -- (2,2) Arrow 11 = (2,1) -- (3,2) % Arrow 3 = (1,1) -- (2,3) Arrow 12 = (2,1) -- (3,3) % Arrow 4 = (1,2) -- (2,1) Arrow 13 = (2,2) -- (3,1) % Arrow 5 = (1,2) -- (2,2) Arrow 14 = (2,2) -- (3,2) % Arrow 6 = (1,2) -- (2,3) Arrow 15 = (2,2) -- (3,3) % Arrow 7 = (1,3) -- (2,1) Arrow 16 = (2,3) -- (3,1) % Arrow 8 = (1,3) -- (2,2) Arrow 17 = (2,3) -- (3,2) % Arrow 9 = (1,3) -- (2,3) Arrow 18 = (2,3)", "reduced the problem to traveling from $(1,1)$ to $(2,1)$ with the same constraints as before. We redraw the graph, removing the edges we have already used and updating our labels (Note that $(1,1)$ and $(2,1)$ are still labeled with $2$ since we will pass through them twice, including the start and end). Then, we remove the vertices with label 0 and the edges we know we can never traverse, giving us: $[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if(j==1 && i==2) {continue;} if(j==2 && i==0) {continue;} pair A = (j,i); dot(A); } } draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((2,1)--(3,1)--(3,2)--(2,2)--cycle); draw((1,1)--(2,1)); label(\"A\", (0,2), W); label(\"L\", (3,0), E); label(\"1\", (2,2), N); label(\"1\", (3,2), N); label(\"1\", (0,1), W); label(\"2\", (1,1), NW); label(\"2\", (2,1), NW); label(\"1\", (3,1), E); label(\"1\", (0,0),S); label(\"1\", (1,0),S); [/asy]$ Now, it is clear that we must complete a cycle in the lower left square, return to $(1,1)$, go to $(2,1)$, and complete a cycle in the top right square, returning to $(2,1)$. There are two ways to traverse each cycle (clockwise and anti-clockwise), giving us a total of $2\\cdot 2 = \\boxed{\\textbf{(E) } 4}$ paths of length 13 from $A$ to $L$. ## Solution 2 Note that of the 12 cities, 6 of them (2" ]

[ "Mary has $3,60 in nickels, dimes and quarters in her purse... 4 28 May 2016, 15:08 9 Prof Liu gave the same quiz to the students in her morning class and 6 08 May 2016, 21:49 1 Sally gave some of her candy to her friends. How many pieces 6 16 Jan 2013, 09:36 7 Guests at a recent party ate a total of fifteen hamburgers. 7 02 Nov 2009, 04:00 67 Mary persuaded n friends to donate$500 each to her election 25 07 Apr 2009, 15:51 Display posts from previous: Sort by", "who are, say, 45 years and older. Junk silver sells (retail) for about 16 times \"face\", so a dime would cost about \\$1.60 and a quarter about \\$4.00. Here is a picture of the two junk silver purchase I recently made at my LCS (Local Coin Shop): My LCS allows me to sift through his pails of junk silver, allowing me to pluck interesting coins. In the above photo you can see two Walking Liberty Halves at the lower left, the one showing the \"tails\" side (the eagle on the ground) is dated 1918, but is not in good shape. The two coins at the lower right are Standing Liberty Quarters (both sides displayed). Also in the pile are Mercury Dimes (I saw lots of these when I was young, from 5 - 10 years old). I took a few in good condition... NONE of these coins, to my knowledge, as any significant numismatic (collector) value, although a few apparently are worth a little more than their silver content. Why would I buy junk silver if I already own Silver Eagles? \"Survival Stuff\", the above would be spare change... Bitcoin At some point relatively soon I will likely write up \"Part Six\" of my Bitcoin series. I still am interested in this complex subject that does appear to", "the Bicentennial Quarter design which showed a date '1776 - 1976'. Unless you have a proof set that shows that it was released specifically in 1975 and 1976, you cannot tell the difference in the circulated coins. 5. Nickels minted from 1942-1945 during World War II contain 1.75 g (0.05626 oz) silver. The silver content of these \"war nickels\" as of October, 2007 is worth$0.77. 6. As of October 2007, the value of the metal in the nickel coin has reached 6.8 cents, a 36% premium over its face value, due to the rising costs of copper and nickel against a falling U.S. Dollar. In an attempt to avoid losing large quantities of circulating nickels to melting, the United States Mint introduced new interim rules on December 14, 2006 criminalizing the melting and export of pennies and nickels. Violators of these rules can be punished with a fine of up to \\$10,000, five years imprisonment, or both. • The Coinage Act of 1965, Pub.L. 89-81, 79 Stat. 254, enacted 1965-07-23, eliminated silver from the circulating dimes and quarter dollars of the United States, and diminished the silver content of the half dollar from 90% to 40%. This act was in response to coin shortages caused by the rising price of silver. ## References Note: Facts About United States", "than three times the number of dimes. How many quarters and how many dimes did Peter have? 248. Lucinda had a pocketful of dimes and quarters with a value of 6.20. The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have? 249. A cashier has 30 bills, all of which are $10 or$20 bills. The total value of the money is $460. How many of each type of bill does the cashier have? 250. A cashier has 54 bills, all of which are$10 or $20 bills. The total value of the money is$910. How many of each type of bill does the cashier have? 251. Marissa wants to blend candy selling for $1.80 per pound with candy costing$1.20 per pound to get a mixture that costs her $1.40 per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use? 252. How many pounds of nuts selling for$6 per pound and raisins selling for $3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him$5 per pound? 253. Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and", "and Hazel had the same amount of money left. Find the amount of money that Wendy had in the end. 5 m Level 3 Cathy had some coin collection. 14 of them were red and the rest were yellow. She gave away 135 of the yellow coins and 50% of the red coins. She then had 15 of her coin collection left. How many coins did Cathy give away? 4 m Level 3 Yvonne will be 36 years old in 3 years' time. Now, Tina's age is 23 of Yvonne's age. What was Tina's age when she was 4 years less than 80% of Yvonne's age? 4 m Level 3 Vincent and George had 852 notebooks together. Vincent had 210 more notebooks than George. Both of them sold some of their notebooks. George sold 20% less notebooks than what Vincent sold. In the end, George had half as many notebooks left as Vincent. How many notebooks had Vincent left? 4 m Level 3 Liam had 57 as many 20-cent coins as 1-dollar coins in his piggy bank. Each time, four 20-cent coins and six 1-dollar coins were taken out from the piggy bank. In the end, only seven 20-cent coins and four 1-dollar coins were left in the piggy bank. How many more 1-dollar coins than 20-cent coins", "together in 4.95 hours and paul can pick the same amount in 9 hours alone, how long would it take daniel to pick 40 bushels alone? Next lesson. The tablets will be crushed. If the number of quarters is half the number of nickels, how many coins does she have in total? Jimmy has seven times as many dimes as nickels. Blue mountain camp budgeted $200 to repair its walk-in-cooler. If Canada's population is about 32 million, about how many people live in the United States? Let n be the number(in millions) of worldwide active users of... A. Equations and Word Problems (Combining Like Terms) Worksheets. Levin has a$120 gift card for the gym. The room is 10 feet wide and 12 feet long. Working together, the pipes can fill the tank in 6 min. Henry served 6 fewer orders than Christine. Erik has a part-time job that pays him $9.10 an hour. Go ahead and submit it to our experts to be answered. Mr. Bailey mowed his yard in 90 minutes. A coin purse has$7.55 in quarters, dimes and half dollars. How many boxes of cupcakes must she buy in order to have enough cupcakes to give... San is 108 year old. How much was the blouse? Both C and S are functions of the mileage m;", "C. 24 cm2 D. 40 cm2 E. 58 cm2 25) A pen costs$1.40 and a pencil costs 80 cents. Susie buys a pen and a pencil and pays with a $10 note. The shop assistant has no$1 or \\$2 coins, and no notes. Susie’s change contains exactly nine 50 cent coins. What is the greatest number of 20 cent coins that Susie could have in her change? A. 14 B. 16 C. 17 D. 20 E. 21 26) 11th August 1999 was a Wednesday. Which day of the week was 21st September 1999? A. Friday B. Saturday C. Sunday D. Monday E. Tuesday 27) A group of children were asked how many pets they had at home. The graph shows the results: Mary reads the graph and makes 3 claims: 1. More than 100 children had 1 or 2 pets. 2. Exactly three times as many children had 2 pets as 0 pets. 3. There were 24 more children with 2 pets than children with 3 pets. Using the information on the graph, which of Mary’s claims is/are correct? A. claim 1 only B. claim 2 only C. claims 3 only D. claims 1 and 2 E. claims 2 and 3 28) Amanda has a large cube made from four small grey cubes and four small white cubes.", "coins were minted in 1933. The last 90% silver coins were minted in 1964, and the last 40% silver half dollar was minted in 1970. The United States Mint currently produces circulating coins at the Philadelphia and Denver Mints, and commemorative and proof coins for collectors at the San Francisco and West Point Mints. Mint mark conventions for these and for past mint branches are discussed in Coins of the United States dollar#Mint marks. The one-dollar coin has never been in popular circulation from 1794 to present, despite several attempts to increase their usage since the 1970s, the most important reason of which is the continued production and popularity of the one-dollar bill.[43] Half dollar coins were commonly used currency since inception in 1794, but has fallen out of use from the mid-1960s when all silver half dollars began to be hoarded. The nickel is the only coin whose size and composition (5 grams, 75% copper, and 25% nickel) is still in use from 1865 to today, except for wartime 1942-1945 Jefferson nickels which contained silver. Due to the penny's low value, some efforts have been made to eliminate the penny as circulating coinage. [44] [45] For a discussion of other discontinued and canceled denominations, see Obsolete denominations of United States currency#Coinage and Canceled denominations of United States", "is 38, and their difference is 26 more than the smaller number. How many coins are there total in the purse? The molding costs$1.90 per linear foot. Add 17. How many of each coin are there? Lisa bought three... Paul mixes nuts worth $1.30 per pound with oats worth$1.65 per pound to get 14 pounds of trail mix worth $1.50 per pound. When she gets in, she is charged$6.90 for the likely amount of time spent in the cab. Slope-intercept form review. If their sum is 103, find both numbers. How much did Gabe spend? One person can complete a task in 3 minutes while a second person can complete the same task in 5 minutes. Get a free answer to a quick problem. This word problem deals with calculating profit after a certain number of years. Find how long it takes for the second hose alone to fill the pond. How many dimes are there? Carlos has 33 coins in his pocket, all of which are dimes and quarters. If the top and bottom layers are each 0.03 cm thick and the inner layers are each 0.02 cm thick, how many inner layers are there? The output of a plant is 4335 pounds of ball bearings per week (five days). He spent $6.75 on Saturday. The difference", "Daniel have? asked 2021-01-22 Colin and Shaina wish to buy a gift for a friend. They combine their money and find they have$4.00, consisting of quarters, dimes, and nickels. If they have 35 coins and the number of quarters is half the number of nickels, how many quarters do they have? You and your friend are saving for a vacation. You start with the same amount and save for the same number of weeks. You save $75 per week, and your friend saves$50 per week. When the vacation time comes, you have $950, and your friend has$800. How much did you start with, and for how many weeks did you save? A. $150, 8 weeks B.$75, 3 weeks C. $500, 5 weeks D.$500, 6 weeks", "1908-1929 Eagles,$10.00 Gold Pieces • Liberty Coronet 1866-1907 • Indian Head No Motto 1907-1908 Double Eagles, $20.00 Gold Pieces • Liberty Coronet 1877-1907 • Saint Gaudens Roman Numerals High relief 1907 • Saint Gaudens No Motto 1907-1908 • Saint Gaudens with motto 1907-1932 Bullion Coins • One ounce silver$1.00 1986- • Tenth ounce gold $5.00 1986- • Quarter ounce gold$10.00 1986- • Half ounce gold $25.00 1986- • One ounce gold$50.00 1986- • Tenth ounce-one ounce platinum $10.00-$100.00 1997- ### d. Collect and mount a date set of series of coins for your country beginning with your birth year. (Commemorative, gold, proof, expensive, or rare coins need not be included.) Here's a suggestion for a series, \"Washington\" quarters: Notes: 1. Washington Quarters were made of 90% silver up to and including 1964. You are competing with 'silver collectors' who collect these coins for their 'raw metal' value, which can be up to ten times the face value of the coin - $2.50 for a 25-cent quarter. See http://coinflation.com 2. Roosevelt dimes were made of 90% silver up to and including 1964. 3. The 1964 Kennedy half-dollar is also 90% silver. The Franklin half-dollar(s) before that were also 90% silver. 4. There is no Washington Quarter with a 1975 date. For 1975 and 1976, the US Mint stamped out", "were only red and white. She put as many tulips in each bouquet, three of which were always red. How much could Simon tear off white tulips? Write all the opti • Stamps and albums Michail has in three albums stored 170 stamps. In the first album, there are 14 more than in the second and in the second is by 1/5 less than in the third. How many stamps are in the first album? • Friends Some friends had to collect the sum 72 EUR equally. If the three refused their part, others would have to give each 4 euros more. How many are friends? • Two math problems 1) The sum of twice a number and -6 is nine more than the opposite of that number. Find the number. 2) A collection of 27 coins, all nickels, and dimes, is worth \\$2.10. How many of each coin are there? The dime, in United States usage, is a ten-cent coin. • TV competition In the competition, 10 contestants answer five questions, one question per round. Anyone who answers correctly will receive as many points as the number of competitors answered incorrectly in that round. One of the contestants after the contest said: We g • Pool If water flows into the pool by two inlets, fill the", "would be worth more than a penny. They have no melt value. I wouldn't bother reading them at all - just get them out of the way. Copper pennies, (pre-1981 except the war-time ones) are worth at least about 2.5¢ for their copper content. It's theoretically illegal to destroy pennies but their value tracks the value of copper. 'Nuff said. It's worth discriminating copper pennies from modern ones. The \"Lincoln Memorial\" pennies (after 1958) are rarely worth anything more than their copper value. This is, surprisingly to me, also true of most wheat pennies (pre-1958) as well. The chances of an old wheat penny being worth more than its copper is very low. As you might suppose, value goes up with age and quality. But the biggest factor by far is original rarity. It requires reading both year and mint to determine that. As the quality drops, it gets harder and harder to read the year and mint. #### Bernard Joined Aug 7, 2008 5,650 Forget the pennies, it would be nice to know if your \"junk silver coins \" were really 90% silver. #### mcgyvr Joined Oct 15, 2009 5,394 It's theoretically illegal to destroy pennies As long as you aren't intending to destroy it for counterfeit purposes then you can destroy money all you want.. (elongated", "Air Force Exchange Service stores on bases rounding all cash purchases up or down to the nearest nickel. Despite this, the U.S. Mint keeps producing a billion pennies a month. ## Where do the pennies go? Two-thirds of them immediately drop out of circulation, into coin jars or behind chair cushions. While quarters and dimes circulate just fine; pennies disappear because they are literally more trouble than they are worth. President Obama has stated his willingness to abolish the penny on February 15th, 2013. Saying, “Anytime we are spending money on something people do not use, its something that we should change.” ## What’s really behind America’s clinging to the penny? The answer has to do with zinc, which composes approximately 98% of each penny minted since the early 1980s. The powerful zinc lobby has enough of a foothold in congress to persuade the senators and representatives to swat the “Penny Abolition” legislation away, as they have done twice in the last decade. In addition, popular support for the penny is still high, at 67%, and national inertia does not seem to be moving in the direction of the tossing the penny. Sentimentality could be playing a major role in this phenomena. “A penny saved is a penny earned,” and “A penny for your thoughts” are iconic phrases", "percent salt so that the resulting solution is 20 percent salt? (A) 20 (B) 30 (C) 40 (D) 50 (E) 60 6. Laura has 20 coins consisting of quarters and dimes. If she has a total of \\$3.05, how many dimes does she have? (A) 3 (B) 7 (C) 10 (D) 13 (E) 16 7. An old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 36 times the difference between the two numbers. How many coins did the old man have? (A) 24 (B) 26 (C) 30 (D) 36 (E) 40 8. How many coins of 0.5 dollars each and 0.7 dollars each together make exactly 4.6 dollars? (A) 1, 6 (B) 2, 7 (C) 3, 5 (D) 4, 3 (E) 5, 3 9. Maria uses a recipe for 36 cupcakes that requires 8 cups of flour, 12 cups of milk, and 4 cups of sugar. How many cups of milk would Maria require for a batch of 9 cupcakes? (A) 2 (B) 3 (C) 4 (D) 6 (E) 8 10. Party Cranberry is 3 parts cranberry juice and 1 part seltzer. Fancy Lemonade is 1 part lemon juice and 2 parts seltzer. One glass of Party Cranberry", "# Buck and Doe Necklaces, Inside and outside pieces, Deer Hunting, Hand Cut Half Dollar Set by NameCoins Regular price \\$64.99 / Shipping calculated at checkout. This is a buck necklace that has been cut from a half dollar. Then we cut a doe from the inside of the buck to make two necklaces. Although most of the coin is gone, you can still tell it was made from a coin. It has been hand cut totally by hand into a half dollar by our small 6 member team at NameCoins. It includes 2x 20\" smooth, strong stainless steel chain (one thin and one thick). If you are interested in making this more customized for you, finding a new design, or simply wanting to see different chain options look us up: https://NameCoins.com", "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.", "cents, etc). The key to this is the introduction of entirely new currency. The government prints all new paper currency - fives, tens, twenties, fifties and hundreds. It then mints pennies, nickles, dimes, quarters (skip the half dollar - useless), dollar and two-dollar coins, like Canada and Australia. Then blow a whistle and make everybody exchange their old for new. Give everybody thirty days to swap currency, then blow the whistle again and declare all old currency as worthless. Except, of course, for the metal value, which people are free to trade on. Here's what you get - a complete flushing of all underground money back into the economy where it can be taxed, at least on the one-time exchange event. You also get hundreds of thousands of tons of copper, nickel and zinc released back into the metals market. There would also be an accurately known quantity of the actual amount of currency in circulation, at least for a while. 8. ### te jenRegistered Senior Member Messages: 532 Then make the coins out of stainless steel... 9. ### Humble LiverRegistered Member Messages: 1 Penny Some 1982 pennies were made in copper. All pennies up to that point had 95% copper content. If you take $2.00 worth of copper pennies the total weight is roughly 19 oz. With", "# Review Problems for the Final Math 101-06 11-25-2017 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the absence of a topic does not imply that it won't appear on the test. 1. Simplify the following expressions. Express all powers in terms of positive exponents. (a) . (b) . (c) . (d) . (e) . 2. y varies inversely with x. Given that when , find x when . 3. P varies directly with and inversely with . Given that when and , find P when and . 4. Calvin Butterball has 29 coins, all of which are dimes or quarters. The value of the coins is $5.45. How many dimes does he have? 5. How many pounds of chocolate truffles worth$2.50 per pound must be mixed with 5 pounds of spackling compound worth $1.70 per pound to yield a mixture worth$2.00 per pound? 6. Calvin leaves the city and travels north at 23 miles per hour. Phoebe starts 2 hours later, travelling south at 17 miles per hour. Some time after Phoebe starts travelling, the distance between them is 166 miles. How many hours has Calvin been travelling at this instant? 7. Calvin can eat", "were coined, or more than the entire issue of the old silver dollars in eighty years. In the same year also the Mint coined 11,378,610 \"trade dollars,\" which$2.50 gold piece. Clark, Gruber & Co., Denver, 1861. weighed 712 grains more than the Bland dollars, but were not \"legal tenders,\"[5] and, though issued by the United States Mint, were refused by the United States Government in payment for postage stamps, taxes, or for other dues, while the Bland dollar, of lesser intrinsic value, was a legal tender, and good for payment of all debts. The total issue of trade dollars between 1873 and 1879 was 35,965,924, and the total issue of Bland dollars, from 1878 to the close of the fiscal year 1897, was 451,993,742. The following table shows the total coinage value of all denominations of silver coin from the establishment of the United States Mint in 1792 to the end of the last fiscal year, June 30, 1897: Dollar ${\\displaystyle \\scriptstyle {\\left.{\\begin{matrix}\\ \\end{matrix}}\\right\\}\\,}}$ 12, 1873 $8,031,238 00 1878 to June 30. 1897 451,993,742 00 Trade dollars 35,965,924 00 Half dollars 134,033,195 00 Columbian half dollars 2,501,052 50 Quarter dollars 52,395,052 00 Columbian quarter dollars 10,005 75 Twenty-cent pieces 271,000 00 Ten-cent pieces 29,428,613 90 Five-cent pieces 4,880,219 40 Three-cent pieces 1,282,087 20 Total$720,792,129 75 It thus appears" ]

[ "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "to see here that this continues all the way down to one. However, when the case gets to $32$, there are 5 ways instead of 4 because $2^{10}-2^5$ is smaller than 1000. Thus, $9+8+7+6+5+5+4+3+2+1 = 50$ So the answer is $\\boxed{050}$ Problem 4 Solution 1 Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \\ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \\ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we have the sum $31+29+28+26+25+\\ldots+4+2+1=(31+28+25+22+\\ldots+1)+$ $(29+26+23+\\ldots+2)=176+155=\\boxed{331}$. Solution 2 Let the three piles have $a, b, c$ coins respectively. If we disregard order, then we just need to divide by $3! = 6$ at the end. We know $a + b + c = 66$. Since $a, b, c$ are positive integers, there are $\\binom{65}{2}$ ways from Stars and Bars. However, we must discard the cases where $a = b$ or $a = c$ or $b = c$. The three cases are symmetric, so we just take the first case and multiply by 3. We have $2a + c = 66 \\implies a = 1, 2, \\dots 32$ for", "9 : 4$, so number of $50$ paise coins is $5x$, number of $25$ paise coins is $9x$ and number of $10$ paise coins is $10x$. Total amount of $50$ paise coins = $0.50 \\times 5x$ = $2.5x$, total number of $25$ paise coins = $0.25 \\times 9x$ = $2.25x$ and amount of $10$ paise coins = $0.10 \\times 4x$ = $0.4x$. So, the total amount in the bag is $2.5x + 2.25x + 0.4x$ = $206$ $5.15x$ = $206$ $X$ = $40$ Thus, the number of $50$ paise coins is $5x$ = $5 \\times 40$ = $200$, number of $25$ paise coins is $9x$ = $9 \\times 40$ = $360$ and number of $10$ paise coins is $4x$ = $4 \\times 40$ = $160$ Example 4: There are two vessels $A$ and $B$ containing solution of milk and water. Vessel $A$ contains solution in the ratio $4 : 3$ and vessel $B$ contains solution in the ratio $9 : 1$. Both the solutions are mixed in certain ratio such that the ratio of milk to water in the final solution is $3 : 2$. Find out the ratio in which the two solutions are mixed Solution: Let the amount of first solution from vessel $A$ taken is $x$ lts. Amount of milk in $x$ lts of solution", "$2(k-1006) = 2011 - k$. Solving gives $\\boxed{\\textbf{(C)} 1341}$.", "- Let me try a generating function approach (first time trying, apologies if it is wrong):$f_1=(1+nx+n^2x^2+\\dots+n^{150}x^{150})f_5=(1+nx^5+n^2x^{10}+\\dots+n^{30}x^{150})f_{10}=(1+nx^{10}+n^2x^{20}+\\dots+n^{15}x^{150})f_{25}=(1+nx^{25}+n^2x^{50}+\\dots+n^6x^{150})$The solution can be found within:$g=(f_1)(f_5)(f_{10})(f_{25})$. More precisely, we need to locate the term$x^{150}n^{50}$, which signifies a sum of$150$cents and$50$coins. Then, the coefficient represents the number of ways to do so. An explanation of why it works: For each$f_i$, it can be seen that exponent of$n$represents the number of coins while exponent of$x$represents the amount of money. i.e.$n^5x^{125}$means 5 coins for 125 cents. When we multiply two terms, say$(n^ix^j)(n^kx^l)=n^{i+k}x^{j+l}$, it can be seen that the term conveniently computes the sum of money and also sum of coins used. Calculation by hand is not recommended. I tried it in Mathematica and I have the term$10n^{50}x^{150}$, which means that there should be 10 ways to do so. Edit: In view of Soloveichik answer, this is how one might do it in Mathematica: f=1/((1-n*x)(1-n*x^5)(1-n*x^10)(1-n*x^25)); series=SeriesCoefficient[f,{x,0,150},{n,0,50}] Answer: 10 - Well done! (+1) Do you see how your answer relates to i. m. soloveichik's answer? – robjohn Oct 20 '12 at 15:44 In your edit, you probably intended$-$instead of$+$$$f=\\frac1{(1-nx)(1-nx^5)(1-nx^{10})(1-nx^{25})}$$ – robjohn Oct 20 '12 at 15:52 Yes! I have only seen a simple variant of the generating function:$(1+x)(1+x^5+\\dots)$etc. Was wondering how to count multiplicity so I gave it a try. Seems like this is the normal approach. But", "# How do you solve 5( 4x + 2) = 6( 3x + 4)? How do you solve 5( 4x + 2) = 6( 3x + 4)? Answer 1 $x$ = 7 #### Explanation: 5($4 x$+2) = 6($3 x$+4) First, multiply out the brackets: $20 x$ + 10 = $18 x$ + 24 Rearrange the equation to get $x$ on one side and the numbers on the other: $20 x$ - $18 x$ = 24-10 $2 x$ = 14 Divide by 2 get $x$ on its own: $x$ = 7 #### Earn Coin Coins can be redeemed for fabulous gifts. Similar Homework Help Questions", "row will have $$a + b$$ and $$b + c.$$ This means that the top row will have $$a + 2b + c.$$ To minimize this, we can let $$b = 1,$$ $$a = 2,$$ and $$c = 3.$$ This yields a top element of $$7.$$ To maximize, this we can let $$b = 9,$$ $$a = 8,$$ and $$c = 7.$$ This yields a top element of $$33.$$ The desired difference is $$33 - 7 = 26.$$ Thus, D is the correct answer. 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? $$0$$ $$1$$ $$2$$ $$3$$ $$5$$ ###### Solution(s): The positive integers that leave a remainder of $$4$$ when divided by $$6$$ are $4, 10, 16, 22, 28, 34, 40, \\cdots.$ The positive integers that leave a remainder of $$3$$ when divided by $$5$$ are $3, 8, 13, 18, 23, 28, 33, \\cdots.$ From this, we can see that the smallest number of coins that work is $$28.$$ This leaves a remainder of $$0$$ when divided by", "+ 31 = \\boxed{058}$.", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "# How to Solve Coin Problems Part 2 I have already introduced how to solve coin problems, so in this post, we solve more problems about it. Coin problems involve problems consisting coins, bills, and of course, any object of value. We have already solved 2 problems in the previous post, so we continue with the third problem. Problem 3 A box contains 32 bills consisting of Php20 and Php50. The total amount of money in the box is 1000 pesos. How many bills of each kind are there? Note: For my readers from other countries, Php means Philippine pesos. Solution This problem is similar to problem 2, so we will not be solving it in details. Let $x$ be the number of 50-peso bills. Since there are 32 bills, then the number of 20-peso bills is $32 - x$ Now, if we multiply the number of 50-peso bills by 50 pesos and multiply the number of 20-peso bills by 20 pesos, we have $50x$ and $20(32-x)$ respectively. If we add them, the total is 1000 pesos. That is, $50x + 20(32-x) = 1000$. Simplifying and solving for $x$, we have $50x + 640 - 20x = 1000$ $30x + 640 = 1000$ Subtracting both sides by 640, we have $30x = 360$. Dividing both sides by 30,", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "$25$ cents, the answer is $\\boxed{\\textbf{(C) } 5}$ For small scale substitutions of 5 cent for 10 cent, like in this problem, it is quite fast, however, if the problem is a little bit more complex, knowing that you need 85 cents, explained above, it is also possible to use a system of equations. It would be $$5x+10y=85$$ $$x+y=12$$ Solving this $x$ (5-cent coins) $= 7$ and $y$ (10-cent coins) $= 5$, so again the answer is $\\boxed{\\textbf{(C) } 5}$ ~savannahsolver", "be the maximum number of squares Peter can move a token beyond the dividing line. For example, $F(0)=4$. You are also given that $F(1)=6$, $F(2)=9$, $F(3)=13$, $F(11)=58$ and $F(123)=1173$. Find $F(1234567)$.", "= 300$$$$7x = 140$$$$x = 20$$So, number of Rs. $$5$$ coins $$= 20$$Number of Rs. $$2$$ coins $$= 60$$Number of Re. $$1$$ coins $$=80$$Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "(from the numerator) and $85-2x<0$, which means $x>42$. This only gives two solutions, $x=43, 44$. Plugging these into the expression for $n$, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is $168+27=\\boxed{195}$.", "the minimum number of possible ones is $0$ and the maximum number of possible ones is $160$. What is the probability that the 160 bit result contains at least 128 1's? The number of possible combinations of 128 ones in 160 bits, which is $4.64648350·10^{33}$, multiply that by the probability each combination, which is $1/(2^{160}) = 6.84227766·10^{-49}$, making the final number $3.17925303·10^{-15}$, about $0.00000000000000317925303\\,\\%$. To calculate the probability of at least 128 bits being ones, you have to repeat the same for 129, 130, 131, ..., 160. Then finally add all these probabilities together. The math is too complicated for me, so I'm not gonna do it. - Adnan, your calculation of the probability in the second half of the question is wrong. – D.W. May 16 '13 at 5:29 @D.W. Yeah, I kind of thought so, it's a bit beyond me. Please, edit the question to correct the \"formula\", we'd all learn something. – Adnan May 16 '13 at 5:47 @Adnan, see pg1989's answer for the correct approach to get the right answer. – D.W. May 16 '13 at 6:05 @D.W. I have looked at it and thought about it, but that's all I was able to figure out. Probability of one coin being head is 0.5, two coins being heads is 0.5*0.5, 3 coins is 0.5*0.5*0.5", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "- 3z) + 16y + 51z = 879$$ $$489 + 10y + 42z = 879$$ $$5y + 21z = 195$$ Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$. If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$, $x = 112$, $y = 18$, and $z = 5$, so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \\boxed{905}$.", "all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find $GFDB$, we substitute 5 for 15X to get $\\boxed{75}$. $\\sim$krishkhushi09 ~ dolphin7 ~ pi_is_3.14" ]

[ "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", "together in 4.95 hours and paul can pick the same amount in 9 hours alone, how long would it take daniel to pick 40 bushels alone? Next lesson. The tablets will be crushed. If the number of quarters is half the number of nickels, how many coins does she have in total? Jimmy has seven times as many dimes as nickels. Blue mountain camp budgeted $200 to repair its walk-in-cooler. If Canada's population is about 32 million, about how many people live in the United States? Let n be the number(in millions) of worldwide active users of... A. Equations and Word Problems (Combining Like Terms) Worksheets. Levin has a$120 gift card for the gym. The room is 10 feet wide and 12 feet long. Working together, the pipes can fill the tank in 6 min. Henry served 6 fewer orders than Christine. Erik has a part-time job that pays him $9.10 an hour. Go ahead and submit it to our experts to be answered. Mr. Bailey mowed his yard in 90 minutes. A coin purse has$7.55 in quarters, dimes and half dollars. How many boxes of cupcakes must she buy in order to have enough cupcakes to give... San is 108 year old. How much was the blouse? Both C and S are functions of the mileage m;", "worth 5 cents. Value of 1 nickel $$¢5 = 5 \\times \\frac{1}{100} = \\frac{5}{100}$$ Value of 3 nickels $$= 3 \\times \\frac{5}{100} = \\frac{15}{100}$$ Value of a dime is ten cents or a tenth of a dollar. Value of 1 dime $$= ¢10 = 10 \\times \\frac{1}{100} = \\frac{1}{10}$$ Value of half-dollar coin $$= \\frac{1}{2}$$ Total amount $$= \\frac{15}{100} + \\frac{1}{10} + \\frac{1}{2}$$ $$= \\frac{15}{100} + \\frac{1 \\times 10}{10 \\times 10} + \\frac{1 \\times 50}{2 \\times 50}$$ $$= \\frac{15}{100} + \\frac{10}{100} + \\frac{50}{100}$$ $$= (\\frac{15 + 10 + 50}{100})$$ $$= \\frac{75}{100}$$ Therefore, Andrew has $$\\frac{75}{100}$$ dollars or 75 cents. Example 4: Tina bought three chocolates whose prices are $0.35,$0.25, and $0.2. Find the amount of money she paid for these chocolates. Solution To find the total amount she spent on chocolate, we need to find$0.35 + $0.25 +$0.2 by following these steps. Step 1: Write the decimal numbers as fractions. $$0.35 + 0.25 + 0.2 = \\frac{35}{100} + \\frac{25}{100} + \\frac{2}{10}$$ Step 2: Fractions to be added need to have the same denominators. Let’s use equivalent fractions to do that, $$\\frac{35}{100} + \\frac{25}{100} + \\frac{2}{10} = \\frac{35}{100} +\\frac{25}{100} + \\frac{2 \\times 10}{10 \\times 10} = \\frac{35}{100} + \\frac{25}{100} + \\frac{20}{100}$$ $$\\frac{35 + 25 + 20}{100 = \\frac{80}{100}$$ Step 4: Write the sum as a decimal number. $$\\frac{80}{100} =", "is 38, and their difference is 26 more than the smaller number. How many coins are there total in the purse? The molding costs$1.90 per linear foot. Add 17. How many of each coin are there? Lisa bought three... Paul mixes nuts worth $1.30 per pound with oats worth$1.65 per pound to get 14 pounds of trail mix worth $1.50 per pound. When she gets in, she is charged$6.90 for the likely amount of time spent in the cab. Slope-intercept form review. If their sum is 103, find both numbers. How much did Gabe spend? One person can complete a task in 3 minutes while a second person can complete the same task in 5 minutes. Get a free answer to a quick problem. This word problem deals with calculating profit after a certain number of years. Find how long it takes for the second hose alone to fill the pond. How many dimes are there? Carlos has 33 coins in his pocket, all of which are dimes and quarters. If the top and bottom layers are each 0.03 cm thick and the inner layers are each 0.02 cm thick, how many inner layers are there? The output of a plant is 4335 pounds of ball bearings per week (five days). He spent $6.75 on Saturday. The difference", "number greater than 7 is side with 8, 9, 10, 11, 12, so 5 possibilities for number greater than 7. P(greater than 7) = The probability of rolling a number greater than 7 is $$\\frac{5}{12}$$. Question 5. Christopher picked coins randomly from his piggy Penny Nickel Dime Quarter bank and got the numbers of coins shown in the table. Find each experimental probability. (Lessons 5.2, 5.3) Total number of possible outcomes is: 7 + 2 + 8 + 6 = 23 a. The next coin that Christopher picks is a quarter. __________________ P (picking a quarter) = The probabiLity of picking a quarter is $$\\frac{6}{23}$$ b. The next coin that Christopher picks is not a quarter. _______________ P (not picking a quarter) = 1 – P (picking a quarter) = 1 – $$\\frac{6}{23}$$ = $$\\frac{23}{23}$$ – $$\\frac{6}{23}$$ = $$\\frac{17}{23}$$ The probability of not picking a quarter is $$\\frac{17}{23}$$ c. The next coin that Christopher picks is a penny or a nickel. __________ P (picking a penny or nickel) = The probability of picking a penny or nickel is $$\\frac{9}{23}$$ Question 6. A grocery store manager found that 54% of customers usually bring their own bags. In one afternoon, 82 out of 124 customers brought their own grocery bags. Did a greater or lesser number of people than usual", "Daniel have? asked 2021-01-22 Colin and Shaina wish to buy a gift for a friend. They combine their money and find they have$4.00, consisting of quarters, dimes, and nickels. If they have 35 coins and the number of quarters is half the number of nickels, how many quarters do they have? You and your friend are saving for a vacation. You start with the same amount and save for the same number of weeks. You save $75 per week, and your friend saves$50 per week. When the vacation time comes, you have $950, and your friend has$800. How much did you start with, and for how many weeks did you save? A. $150, 8 weeks B.$75, 3 weeks C. $500, 5 weeks D.$500, 6 weeks", "than three times the number of dimes. How many quarters and how many dimes did Peter have? 248. Lucinda had a pocketful of dimes and quarters with a value of 6.20. The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have? 249. A cashier has 30 bills, all of which are $10 or$20 bills. The total value of the money is $460. How many of each type of bill does the cashier have? 250. A cashier has 54 bills, all of which are$10 or $20 bills. The total value of the money is$910. How many of each type of bill does the cashier have? 251. Marissa wants to blend candy selling for $1.80 per pound with candy costing$1.20 per pound to get a mixture that costs her $1.40 per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use? 252. How many pounds of nuts selling for$6 per pound and raisins selling for $3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him$5 per pound? 253. Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and", "Mary has $3,60 in nickels, dimes and quarters in her purse... 4 28 May 2016, 15:08 9 Prof Liu gave the same quiz to the students in her morning class and 6 08 May 2016, 21:49 1 Sally gave some of her candy to her friends. How many pieces 6 16 Jan 2013, 09:36 7 Guests at a recent party ate a total of fifteen hamburgers. 7 02 Nov 2009, 04:00 67 Mary persuaded n friends to donate$500 each to her election 25 07 Apr 2009, 15:51 Display posts from previous: Sort by", "# Review Problems for the Final Math 101-06 11-25-2017 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the absence of a topic does not imply that it won't appear on the test. 1. Simplify the following expressions. Express all powers in terms of positive exponents. (a) . (b) . (c) . (d) . (e) . 2. y varies inversely with x. Given that when , find x when . 3. P varies directly with and inversely with . Given that when and , find P when and . 4. Calvin Butterball has 29 coins, all of which are dimes or quarters. The value of the coins is $5.45. How many dimes does he have? 5. How many pounds of chocolate truffles worth$2.50 per pound must be mixed with 5 pounds of spackling compound worth $1.70 per pound to yield a mixture worth$2.00 per pound? 6. Calvin leaves the city and travels north at 23 miles per hour. Phoebe starts 2 hours later, travelling south at 17 miles per hour. Some time after Phoebe starts travelling, the distance between them is 166 miles. How many hours has Calvin been travelling at this instant? 7. Calvin can eat", "at 6:40 ## 6 Answers From comments it seems that the question is: • There are three coins, named \"Nickel\", \"Dime\", and \"Quarter\". • You know that exactly one of the three, but not which, is in a different state to the others, either: 1. one is showing heads and the other two tails; or 2. one is showing tails and the other two heads • You can choose to toss (randomise) the state of any combination of the coins by name • After this single randomisation of state you win if all three coins are in the same state, either: 1. all heads; or 2. all tails Given that we do not know if your friend has any bias we should treat her as a random process, with an equal chance of choosing any one of the coins to be in the different state. This puzzle has a somewhat counter-intuitive solution: choosing either all three, or any two are equally likely to win Why? There are eight combinations from which we could choose, but only four (seemingly) effectively different choices with our assumption of no bias: 1. No coins - this has a probability of $1$ for losing and $0$ for winning. 2. All coins - this has a probability of $\\frac34$ for losing and $\\frac 14$", "C. 24 cm2 D. 40 cm2 E. 58 cm2 25) A pen costs$1.40 and a pencil costs 80 cents. Susie buys a pen and a pencil and pays with a $10 note. The shop assistant has no$1 or \\$2 coins, and no notes. Susie’s change contains exactly nine 50 cent coins. What is the greatest number of 20 cent coins that Susie could have in her change? A. 14 B. 16 C. 17 D. 20 E. 21 26) 11th August 1999 was a Wednesday. Which day of the week was 21st September 1999? A. Friday B. Saturday C. Sunday D. Monday E. Tuesday 27) A group of children were asked how many pets they had at home. The graph shows the results: Mary reads the graph and makes 3 claims: 1. More than 100 children had 1 or 2 pets. 2. Exactly three times as many children had 2 pets as 0 pets. 3. There were 24 more children with 2 pets than children with 3 pets. Using the information on the graph, which of Mary’s claims is/are correct? A. claim 1 only B. claim 2 only C. claims 3 only D. claims 1 and 2 E. claims 2 and 3 28) Amanda has a large cube made from four small grey cubes and four small white cubes.", "1. ## confused how do i do this? A nickle, dime, and quarter are tossed. What is the probability that the nickel and dime will show heads and the quarter will show tails? What is the probability that only one of the coins will show tails? 2. Originally Posted by sevenlovesyou A nickle, dime, and quarter are tossed. What is the probability that the nickel and dime will show heads and the quarter will show tails? What is the probability that only one of the coins will show tails? A nickle, dime, and quarter are tossed. What is the probability that the nickel and dime will show heads and the quarter will show tails? $\\frac {1}{8}$ What is the probability that only one of the coins will show tails? $\\frac {3}{8}$ Make a table and count the possiblites. 3. Hey its easy identify all the possibilties fo all the coins. YUU WILL GET HHH HHT THH HTT TTT HTH THT TTH WHERE H= HEADS N T= TAILS... you can see there's only 1 possibility out of 8 that the first two will show tails n d last one a head n the probablity that only 1 coin show tails is 3/8", "effort which is why most 1943 pennies are silver colored. Do I get an A? Dec 6, 2013coindoc. if it adds up less than 6 grams i know i got schemed. At what price per pound of copper does such a penny contain exactly one cent worth of copper? (There are 454 grams in one pound. 500 grams), but we saw no mention of the fact that it's the weight of a dime. Of course, picking it up isn’t the only time cost. (In 1982 solid bronze U. A gram is the approximate weight of a cubic centimeter of water. 5% zinc, with a pure copper coating that comprises 2. 00220462262 to make the conversion to grams. 750 grams = 26. piece of (skinless) chicken breast will shrink by 25% during cooking. 1 grams before 1982, but a penny after 1982 weighs 2. (And add a bit for the paper roll). Pennyweight is a troy unit of weight. 4 grams (little over 48 coins to a pound); this matches the D&D cp almost exactly in both size and weight. 042 ounces). This page also shows coins listed for sale so you can buy and sell. Before 1982, pennies weighed 3. Pennyweights to Grams formula. If there are$11$heavier pennies we saw that the weight is about$2. The copper coats", "value of the coins is $4.21 . The number of dimes is three less than four times the number of pennies. How many pennies and how many dimes are in the cup? Cecilia Reply Arnold invested$64,000 some at 5.5% interest and the rest at 9% interest how much did he invest at each rate if he received $4500 in interest in one year Heidi Reply List five positive thoughts you can say to yourself that will help youapproachwordproblemswith a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often. Elbert Reply Avery and Caden have saved$27,000 towards a down payment on a house. They want to keep some of the money in a bank account that pays 2.4% annual interest and the rest in a stock fund that pays 7.2% annual interest. How much should they put into each account so that they earn 6% interest per year? 324.00 Irene 1.2% of 27.000 Irene i did 2.4%-7.2% i got 1.2% Irene I have 6% of 27000 = 1620 so we need to solve 2.4x +7.2y =1620 Catherine I think Catherine is on the right track. Solve for x and y. Scott next bit : x=(1620-7.2y)/2.4 y=(1620-2.4x)/7.2 I think we can", "### Home > CCG > Chapter 10 > Lesson 10.1.1 > Problem10-12 10-12. Multiple Choice: A penny, nickel, and dime are all flipped once. What is the probability that at least one coin comes up heads? 1. $\\frac {1} {3}$ 1. $\\frac {3} {8}$ 1. $1$ 1. $\\frac {7} {8}$ Draw a tree diagram that illustrates all the possible combinations that could occur. Then add up the probabilities of each combination that includes a coin coming up heads. Use the eTool below to support your ideas! Click on the link at right for the full eTool version: CCG 10-12 HW eTool", "### Home > GC > Chapter 8 > Lesson 8.1.4 > Problem8-42 8-42. Multiple Choice: A penny, nickel, and dime are all flipped once. What is the probability that at least one coin comes up heads? 1. $\\frac{1}{3}$ 1. $\\frac{3}{8}$ 1. $1$ 1. $\\frac{7}{8}$ Draw a tree diagram that illustrates all the possible combinations that could occur. Then add up the probabilities of each combination that includes a coin coming up heads. Use the eTool below to support your ideas! Click on the link at right for the full eTool version: GC 8-42 HW eTool", "were only red and white. She put as many tulips in each bouquet, three of which were always red. How much could Simon tear off white tulips? Write all the opti • Stamps and albums Michail has in three albums stored 170 stamps. In the first album, there are 14 more than in the second and in the second is by 1/5 less than in the third. How many stamps are in the first album? • Friends Some friends had to collect the sum 72 EUR equally. If the three refused their part, others would have to give each 4 euros more. How many are friends? • Two math problems 1) The sum of twice a number and -6 is nine more than the opposite of that number. Find the number. 2) A collection of 27 coins, all nickels, and dimes, is worth \\$2.10. How many of each coin are there? The dime, in United States usage, is a ten-cent coin. • TV competition In the competition, 10 contestants answer five questions, one question per round. Anyone who answers correctly will receive as many points as the number of competitors answered incorrectly in that round. One of the contestants after the contest said: We g • Pool If water flows into the pool by two inlets, fill the", "which are silver or copper nickel. The adjacency of these coins is the number of adjacent pairs of coins with the same side of face. Possible evidence for early coin rolls can even be found in old shipwrecks containing coins. Nice collection starter or add to what you already have. The United States Mint produces coins first and foremost to facilitate commerce throughout the United States. The size of the sample space is 2 10 = 1024. What is the probability that the coin will land on heads again?” The answer to this is always going to be 50/50, or ½, or 50%. Jungsun: There is an 1/2 chance to get a head of a coin each time. The following chart lists the face value of a standard box of coins. But what if you’ve got thousands of coins and don’t really want to spend time looking at every coin? Did you ever wonder how many coins come in a roll? Ask Question Asked 3 years, 6 months ago. In each move, we can flip over two adjacent coins provided that they are both showing heads or both showing tails. 1 State with 5 heads. And, P (X = 5) = C 5 10 1024 = 252 I have written a program to calculate the odds, but it", "and Hazel had the same amount of money left. Find the amount of money that Wendy had in the end. 5 m Level 3 Cathy had some coin collection. 14 of them were red and the rest were yellow. She gave away 135 of the yellow coins and 50% of the red coins. She then had 15 of her coin collection left. How many coins did Cathy give away? 4 m Level 3 Yvonne will be 36 years old in 3 years' time. Now, Tina's age is 23 of Yvonne's age. What was Tina's age when she was 4 years less than 80% of Yvonne's age? 4 m Level 3 Vincent and George had 852 notebooks together. Vincent had 210 more notebooks than George. Both of them sold some of their notebooks. George sold 20% less notebooks than what Vincent sold. In the end, George had half as many notebooks left as Vincent. How many notebooks had Vincent left? 4 m Level 3 Liam had 57 as many 20-cent coins as 1-dollar coins in his piggy bank. Each time, four 20-cent coins and six 1-dollar coins were taken out from the piggy bank. In the end, only seven 20-cent coins and four 1-dollar coins were left in the piggy bank. How many more 1-dollar coins than 20-cent coins", "Dec 2014, 08:17 2 A certain purse contains 30 coins, Each coin is either a nickel or 3 17 Nov 2014, 04:24 24 A box contains four coins, of which two coins have heads on 13 02 Mar 2013, 09:02 10 For one toss of a certain coin, the probability that the out 7 25 Oct 2009, 18:38 5 For one toss of a certain coin, the probability that the out 13 23 Jul 2008, 03:07 Display posts from previous: Sort by" ]

[ "88, .88 Hope this helps.", "of $96$. Hence: $$1 \\equiv -5(19) \\mod 96 \\equiv 1 \\mod 96$$", "divisible by $8$ is $12$.", "+ 88 = 0$", "} \\quad \\dfrac{93}{4}$$", "$97$ $$( 882 + T )^{2}$$", "n+4\\right) \\div(n-4)$$...", "# How do you evaluate 8.58\\div 2? Dec 16, 2016 $4.29$ #### Explanation: You can also do this mentally by breaking $8.58$ into two parts and using the distributive law: $8.58 \\div 2$ $= \\left(8 + 0.58\\right) \\div 2$ Divide each part by 2 separately... $= 4 + 0.29 \\text{ } \\leftarrow$ now add the answers together: $= 4.29$", "can think of $$4878 \\div 900=5.42$$ as saying, “There are 900 groups of 5.42 in 4878.” How might we show that both statements are true? 1. Explain why $$51.2 \\div 6.4$$ has the same value as $$5.12 \\div 0.64$$. 2. Write a division expression that has the same value as $$51.2 \\div 6.4$$ but is easier to use to find the value. Then, find the value using long division. Student Response Student responses to this activity are available at one of our IM Certified Partners Activity Synthesis The goal of this discussion is for students to contrast the two division methods used in the task. Discuss: • Why is the value of $$48.78 \\div 9$$ the same as the value of $$4,\\!878 \\div 900$$? (The numbers in the second expression are both multiplied by 100, and this does not change the value of the quotient.) • What are some advantages of calculating $$48.78 \\div 9$$ with the decimal intact (the method on the left)? (It is fast, and we don’t need to deal with a bunch of 0’s. Also, if the numbers are from a contextual problem, we could better make meaning of them in their original form.) • What are some advantages of calculating $$4,\\!878 \\div 900$$ with long division? (These are whole numbers, and we are", "\\div 4^{15}$ = $64^?$ 9 3 12 7 None of these", "statement? No, because $$16\\div (4\\div 2)=16\\div 2=16\\div 2=8$$ and $$(16\\div 4)\\div 2=(4)\\div 2=2$$. Division is not associative, the general property (using $$a$$, $$b$$, and $$c$$) does not exist.", "10. $-68 \\div -2 = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$", "\\div (-4b)$ 11. $-99xy \\div (-9x)$ 12. $121a \\div (11b)$ 13. $-144xy \\div (-12)$ 14. $-169y \\div (-13x)$ 15. $-225xy \\div (5z)$ Basic Nov 30, 2012 Jun 13, 2014", "$(-6)(-1)$ 7. $(-7)(8)$ 8. $(6)(-1)$ 9. $(9)(-4)$ 10. $(-9)(-7)$ 11. $(-5)(2)$ 12. $(-2)(-2)$ 13. $(-5)(4)$ 14. $(-3)(-9)$ For questions 45 to 58, find each quotient. 1. $30 \\div -10$ 2. $-49 \\div -7$ 3. $-12 \\div -4$ 4. $-2 \\div -1$ 5. $30 \\div 6$ 6. $20 \\div 10$ 7. $27 \\div 3$ 8. $-35 \\div -5$ 9. $80 \\div -8$ 10. $8 \\div -2$ 11. $50 \\div 5$ 12. $-16 \\div 2$ 13. $48 \\div 8$ 14. $60 \\div -10$", "approximately equal to 80 $\\div$ 20 = 8 $\\div$ 2 = 4. Question 24: 75 $\\div$ 23 Solution: The estimated quotient of 75 $\\div$ 23 is approximately equal to 80 $\\div$ 20 = 8 $\\div$ 2 = 4. Question 25: 193 $\\div$ 24 Solution: The estimated quotient of 193 $\\div$ 24 is approximately equal to 200 $\\div$ 20 = 20 $\\div$ 2 = 10. Question 26: 725 $\\div$ 23 Solution: The estimated quotient of 725 $\\div$ 23 is approximately equal to 700 $\\div$ 20 = 70 $\\div$ 2 = 35. Question 27: 275 $\\div$ 25 Solution: The estimated quotient of 275 $\\div$ 25 is approximately equal to 300 $\\div$ 30 = 30 $\\div$ 3 = 10. Question 28: 633 $\\div$ 33 Solution: The estimated quotient of 633 $\\div$ 33 is approximately equal to 600 $\\div$ 30 = 60 $\\div$ 3 = 20. Question 29: 729 $\\div$ 29 Solution: 729 $\\div$ 29 is approximately equal to 700 $\\div$ 30 or 70 $\\div$ 3, which is approximately equal to 23. Question 30: 858 $\\div$ 39 Solution: 858 $\\div$ 39 is approximately equal to 900 $\\div$ 40 or 90 $\\div$ 4, which is approximately equal to 23. Question 31: 868 $\\div$ 38 Solution: 868 $\\div$ 38 is approximately equal to 900 $\\div$ 40 or 90 $\\div$ 4, which is approximately equal", "- 2.4^2).$$ -", "divisible by $1003$ either?", "divisible by $2$ or $3$ and make things much easier.", "76.80$ Answer: $\\char36 76.80$", "$89$ $$T^{2}$$ $97$ $$4 + T^{2}$$" ]

[ "× (4 ÷ 2) Firstly perform the operations within the parentheses, we have (26 + 14) = 40 , (4 ÷ 2) = 2, now we will perform the multiplication part and then we will perform the subtraction part. Now we have the given expression as 95 – ( 40 × 2 ), And calculating the inside parentheses we get ( 40 × 2 )= 80 Then 95 – 80 = 15 So , 95 – (26 + 14) × (4 ÷ 2) = 15 . Question 5. (8 + 2) × (13 + 7 + 5) On evaluating we will get the result as 250 Explanation: Given that the numerical expression is (8 + 2) × (13 + 7 + 5) Firstly perform the operations within the parentheses, we have (8 + 2) = 10, (13 + 7 + 5) = 20, and now we will perform the multiplication part. To evaluate the given expression apply the resulted output as 10 × 25= 250 So, (8 + 2) × (13 + 7 + 5) = 250. Question 6. 2 × [(96 – 72) ÷ 8] On evaluating the numerical expression the result is 72. Explanation: Given that the numerical expression is 2 × [(96 – 72) ÷ 8] Firstly perform the operations within the parentheses, we have", "# $${\\rm{2}}({\\rm{5}} - \\sqrt {{\\rm{16}}} ) \\div ({\\rm{14}} - {\\rm{12}}) \\times {\\rm{3}} = ?$$ 3 Explanation First perform the operation inside parentheses. This gives $$2(5-\\sqrt{16}) \\div(14-12) \\times 3=2(5-4) \\div(14-12) \\times 3=2(1) \\div 2 \\times 3=2 \\div 2 \\times 3$$ Now do the multiplications and divisions from left to right: $${\\rm{2}} \\div {\\rm{2}} \\times {\\rm{3}} = 1 \\times 3 = 3$$", "...\\ + {{n+r}\\choose {n+r-1}}\\right] + \\left[{{n+r}\\choose {n+r}}\\right]$ Note that 1st box bracket = 1 2nd box bracket = 1980 ---- {from (A)} 3rd box bracket = 66 ---- {from (B)} 4th box bracket = 1 $\\implies\\ 2^{n+r} = 1 + 1980 + 66 + 1 = 2048$ $\\implies\\ 2^11 = 2048$ $\\implies\\ n + r = 11$ Now solving (A) & (B) with $n+r = 11$, we find that, $\\boxed{n = 8\\ \\&\\ r = 3}$", "odd multiples, multiples) of (45*, 90*, 180*). Select an answer from each set of parentheses.", "each stack? How many trays will be left over? Let's consider the first question first. How many trays will be in each stack? I found the answer 24 by starting down the road of calculating 97 ÷ 4 far enough to see that result would be \"24 and some stuff.\" (Too late, I remembered that I actually knew 4 × 24 = 96 offhand—it was still a good check on my answer.) Thus my approach was to pursue, just far enough, a strategy of \"divide and round down.\" If you work with computers, then you probably know that the operation \"divide and round down\" equals a computer command known as \"DIV.\" For example, a computer will tell you that 16 DIV 3 = 5, and one way to check this answer would be to calculate 16 ÷ 3 = 5⅓ and round the result down. I doubt this is how a computer actually calculates the value of 16 DIV 3, but \"divide and round down\" gives the right answer and shows the relationship between DIV and ÷, which is what I want to write about today. A few more examples just to clarify how DIV works: 32 DIV 7 = 4 54 DIV 6 = 9 3 DIV 4 = 0. A friendly term for the DIV operation", "first) = –4 – 60 = –64 (b) 63 ÷ (16 – 7) × (–2) = 63 ÷ 9 × (–2) ← (bracket is done first, then work from left to right) = 7 × (–2) = –14 (c) –30 + 9 × 7 – 16 = –30 + (9 × 7) – 16 ← ( multiply first) = –30 + 63 – 16 = 17 ### 2 thoughts on “1.2.1 Basic Arithmetic Operations Involving Integers” 1. its really help me to do my maths revision Tq so much :* • You are welcome 🙂", "| 0 Div.2 A using regular expression: 34722038 (and B — 34722359)", "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", "in the first line. An operation is a -fold execution of either a or b move. If , then there should be a sequence of integers with either a or b appended in the second line. Thus (for ) denotes the -fold execution of the move a. Analogously, (for ) denotes the -fold execution of the move b. Furthermore, the characters appended to the numbers in the second line have to alternate. Should there be more than one solution, your program is free to pick one arbitrarily. #### 输入样例1 4 1 3 2 4 #### 输出样例1 4 3a 2b 2a 2b #### 输入样例2 7 1 3 2 4 5 6 7 #### 输出样例2 NIE DA SIE #### 输入样例3 3 1 2 3 #### 输出样例3 0 #### 数据范围与提示 Task authors: Krzysztof Diks & Wojciech Rytter.", "as a hand calculator.$BR 73 \\{ ans_rule(25) \\} 74 75 END_TEXT 76 77$ans =36; 78 79 &ANS(std_num_cmp($ans)); 80 81 BEGIN_TEXT 82$BR 83 84 There is one other thing to be careful of. Multiplication and division have the 85 same precedence and there are no universal rules as to which should be done first. 86 For example, what does 2/3*4 mean? (Note that / is the \"division symbol\", which 87 is usually written as a line with two dots, but unfortunately, this \"line with 88 two dots\" symbol is not on computer keyboards. Don't think of / as the horizontal 89 line in a fraction. Ask yourself what 1/2/2 should mean.) WeBWorK and most other 90 computers read things from left to right, i.e. 2/3*4 means (2/3)*4 or 8/3, IT DOES 91 NOT MEAN 2/12. Some computers may do operations from right to left. If you 92 want 2/(3*4) = 2/12, you have to use parentheses. The same thing happens with 93 addition and subtraction. 1-3+2 = 0 but 1-(3+2) = -4. This is one case where using 94 parentheses even if they are not needed might be a good idea, e.g. write (2/3)*4 95 even though you could write 2/3*4. This is also a case where previewing your answer 96 can save you a lot a grief since", "- e^{x_i}) = e - e^0 = \\boxed{e-1}.$ Done!", "# How do you simplify 39\\div 3- 9=? Mar 4, 2018 $4$ #### Explanation: There are two terms and two operations. Do the division first: $\\textcolor{b l u e}{39 \\div 3} \\textcolor{red}{\\text{ \"-\" } 9}$ $= \\textcolor{b l u e}{13} \\textcolor{red}{\\text{ \"-\" } 9}$ $= 4$", "# 1975 AHSME Problems/Problem 10 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) We see that the result of this expression will always be in the form $(100\\text{ some number of zeros}001)^2.$ Multiplying these together yields: $$110\\text{ some number of zeros}011$$. This works because of the way they are multiplied. Therefore, the answer is $\\boxed{(A) 4}$.", "+ 8 × 2 \\div 4$$ Without parentheses, the solution would be: $$=5 + 8 × 2 \\div 4$$ $$= 5 + 16 \\div 4$$ $$= 5 + 4$$ $$= 9$$ Using parentheses at $$(5 + 8)$$, the solution would be: $$= (5 + 8) × 2 \\div 4$$ $$= 13 × 2 \\div 4$$ $$=26 \\div 4$$ $$= \\dfrac{13}{2}$$ Using parentheses at $$(5 + 8 × 2)$$, the solution would be: $$=5 + 8 × 2 \\div 4$$ $$= (5 + 8 × 2 ) \\div 4$$ $$= (5 + 16) \\div 4$$ $$=21 \\div 4$$ $$= \\dfrac{21}{4}$$ • It can be observed that by altering the position of parentheses, the value of the expression also gets changed. #### Which one of the following options represents the correct position of parentheses so that the expression, $$100 \\div 5 × 6 - 2$$ gives the result as 80? A $$(10 0\\div 5)× 6 - 2$$ B $$100 \\div (5 × 6 )- 2$$ C $$100 \\div (5 × 6 - 2)$$ D $$100 \\div 5 × (6 - 2)$$ × Evaluating option (A) $$(100 \\div 5) × 6 -2$$ $$=20 × 6 - 2$$ $$=120 - 2$$ $$= 118$$ Here, the value is not 80. Thus, option (A) is incorrect. Evaluating option (B) $$100 \\div (5 × 6", "a quick and dirty'' answer, then a more thorough one. When you have a sequence of operations, it is always possible to force them to be performed in a certain order by using parentheses. For example, (5-2)/6 = 3/6=2, or 5-(2/6) = 5 - (1/3) = 14/3. However, one often wants to omit parentheses to avoid typing. MATLAB uses the standard order of priority (precedence) for operations: addition and subtraction have the same priority, as do multiplication and division. Exponentiation has higher priority. Example: First assign values to a, b, and c as described above. For example: >> a=8; b=2; c=16; Then we verify Command Numerical Answer in Math language >> a - b * c -24 a - (b * c ) >> (a - b) * c 96 (a-b)*c >> a / b / c .25 (a/b)/c >> a / (b / c) 64 a/(b/c) >> a ^ b * c 1024 (a^b)*c >> a ^ (b * c) 7.9228e+28 a^(b*c) >> a ^ b ^ c Inf a^(b^c) To see a more detailed explanation read on. MATLAB uses the following symbols for arithmetic operations: Operation Symbol Precedence Comment Exponentiation ^ 3 Highest precedence Multiplication * 2 Division / 2 Addition + 1 Subtraction - 1 Lowest precedence If an arithmetic expression contains nested parentheses, then", "This shows a figure containing a doing and undoing diagram. The doing diagram has two rectangular boxes in the middle, the 1st containing ‘add 4’ linked to the 2nd box on the right containing ‘multiply by 2’. To the left of the 1st box is the word number, linked by an arrow pointing right. On the right of the 2nd rectangular box is an arrow pointing right to the number 14. The undoing diagram has two rectangular boxes in the middle, the 1st containing ‘divide by 2’ linked to the 2nd box on the right containing ‘subtract 4’. The arrow linking the two boxes has the number 7 above it. To the left of the 1st box is the number 14, linked by an arrow pointing right. On the right of the 2nd rectangular box is an arrow pointing right to the number 3. 3.1 Doing some doing and undoing", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", "How do you simplify (10 - 4 * 13+ 19) div 23 using order of operations? Apr 25, 2016 $\\left(10 - 4 \\cdot 13 + 19\\right) \\div 23 = - 1$ Explanation: $\\left(10 - 4 \\cdot 13 + 19\\right) \\div 23$ = $\\left(10 - 52 + 19\\right) \\div 23$ = $\\left(29 - 52\\right) \\div 23$ = $\\left(- 23\\right) \\div 23 = - 1$", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "## Algebra 1 We follow the instructions to plug 2 in for a and 4 in for b. Thus, we obtain: $(4\\times2)^{3} \\div (4-2)$ We follow order of operations, so we first consider everything inside of the parenthesis, then the exponent, and finally division. Thus, we find: $8^{3} \\div 2 \\\\\\\\ = 512 \\div 2=256$" ]

[ "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "$m+n=2+89=\\boxed{091}$.", "is \"2nd Y\". Hi, I don't want to pick at you but the conversion tool is under $\\boxed{\\text{2ND}}\\ \\boxed{\\text{MODE}}$ 7. I use a TI-92 and the 'conversion arrow' is under $\\boxed{2nd}, \\;\\ \\boxed{Y}$ on it. It can also be found under MATH as the poster pointed out. I always just typed it in. But, then again, the 89 doesn't have the keyboard the 92 and Voyage 200 has.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "+ 88 = 0$", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "= \\boxed{108}$.", "Now we know the complement is $274 + 81 + 88 = 443$, and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \\boxed{581}$.", "10. $-68 \\div -2 = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$", "The only answer choice that is divisible by $4$ is $\\boxed{\\textbf{(B)}\\ 40}$.", "\\div (-4b)$ 11. $-99xy \\div (-9x)$ 12. $121a \\div (11b)$ 13. $-144xy \\div (-12)$ 14. $-169y \\div (-13x)$ 15. $-225xy \\div (5z)$ Basic Nov 30, 2012 Jun 13, 2014", "\\dfrac{40}{42}$$ giving us the desired answer of $20+21=\\boxed{41}$.", "statement? No, because $$16\\div (4\\div 2)=16\\div 2=16\\div 2=8$$ and $$(16\\div 4)\\div 2=(4)\\div 2=2$$. Division is not associative, the general property (using $$a$$, $$b$$, and $$c$$) does not exist.", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "# How do you evaluate 8.58\\div 2? Dec 16, 2016 $4.29$ #### Explanation: You can also do this mentally by breaking $8.58$ into two parts and using the distributive law: $8.58 \\div 2$ $= \\left(8 + 0.58\\right) \\div 2$ Divide each part by 2 separately... $= 4 + 0.29 \\text{ } \\leftarrow$ now add the answers together: $= 4.29$", "= \\boxed{\\textbf{(B)}\\ 24}.$", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910} \\$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys." ]

[ "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", "$\\triangle ABC$ is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation. $[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm); 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 = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); /* draw figures */ draw((3.46,0.96)--(3.44,-3.36)); draw((3.44,-3.36)--(8.02,-3.44)); draw((8.02,-3.44)--(3.46,0.96)); draw((3.46,0.96)--(-0.86,0.98)); draw((-0.86,0.98)--(-0.88,-3.34)); draw((-0.88,-3.34)--(3.44,-3.36)); draw((3.46,0.96)--(8.02,-3.44)); draw((8.02,-3.44)--(12.42,1.12)); draw((12.42,1.12)--(7.86,5.52)); draw((7.86,5.52)--(3.46,0.96)); draw((5.74,-1.24)--(-0.86,0.98)); draw((5.74,-1.24)--(-0.87,-1.18), linetype(\"4 4\")); draw((5.74,-1.24)--(7.86,5.52)); draw((5.74,-1.24)--(10.14,3.32), linetype(\"4 4\")); draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype(\"2 2\")); draw((8.02,-3.44)--(-0.86,0.98)); draw((3.44,-3.36)--(7.86,5.52)); draw((3.44,-3.36)--(5.74,-1.24)); /* dots and labels */ dot((3.46,0.96),dotstyle); label(\"A\", (3.2,1.06), NE * labelscalefactor); dot((3.44,-3.36),dotstyle); label(\"C\", (3.14,-3.86), NE * labelscalefactor); dot((8.02,-3.44),dotstyle); label(\"B\", (8.06,-3.8), NE * labelscalefactor); dot((-0.86,0.98),dotstyle); label(\"Z\", (-1.34,1.12), NE * labelscalefactor); dot((-0.88,-3.34),dotstyle); label(\"W\", (-1.48,-3.54), NE * labelscalefactor); dot((12.42,1.12),dotstyle); label(\"X\", (12.5,1.24), NE * labelscalefactor); dot((7.86,5.52),dotstyle); label(\"Y\", (7.94,5.64), NE * labelscalefactor); dot((5.74,-1.24),dotstyle); label(\"O\", (5.52,-1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Let $O$ be the the midpoint of $AB$. The perpendicular bisector of line $WZ$ and $XY$ will meet at $O$. Thus $O$ is the center of the circle points $X$, $Y$, $Z$, and $W$ lies on. $\\angle ZAC=\\angle OAY=90^{\\circ}$, $\\angle ZAC+\\angle BAC=\\angle OAY+\\angle BAC$, $\\angle ZAB=\\angle CAY$, and $AZ=AC$, $AB=AY$, $\\triangle AZB \\cong \\triangle ACY$ by", "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", "# Difference between revisions of \"Equilateral triangle\" $[asy] pair A=(0,50),B=(43.301,-25),C=(-43.301,-25); draw(A--B--C--A); label(\"60^{\\circ}\",(-28,-16)); draw(anglemark(B,C,A,300)); label(\"60^{\\circ}\",(28,-16)); draw(anglemark(A,B,C,300)); label(\"60^{\\circ}\",(0,35)); draw(anglemark(C,A,B,300)); [/asy]$ An equilateral triangle is a triangle in which all sides have equal length and all angles have equal measure. Since a triangle has a total of $180$ degrees, each angle of an equilateral triangle has $60$ degrees. One useful way to manipulate an equilateral triangle is to draw an altitude, which would divide it into two 30-60-90 right triangles. Using this, the area of an equilateral triangle can be found in terms of a side: $\\frac{s^2\\sqrt{3}}{4}$.", "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", "pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C--D--A, dashed ); pair Sp = intersectionpoint(D--O,A--C); label(\"B\",B,SW); label(\"A\",A,S); label(\"C\",C,NW); label(\"O\",O,NE); label(\"D\",D,(S+SSW)); label(\"S\",Sp,(S+SSW)); [/asy]$ As in the previous solution, we find out that $\\angle AOB = \\angle COB = 60^\\circ$. Hence $\\triangle AOD$ and $\\triangle COD$ are both equilateral. We then have $\\angle SCD = \\angle SAD = 30^\\circ$, hence $D$ is the incenter of $\\triangle ABC$, and as $\\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \\cdot SD = BD$, and as $SD = SO$, we have $2\\cdot SD = SD + SO = OD$, therefore $BD=OD$, and as before we conclude that $\\frac{BD}{BO} = \\boxed{\\frac 12}$.", "#### 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", "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);", "+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", "ABD$ are both inscribed angles with the same arc of a given circle, $\\angle ABD = \\angle ACD = a$, so $\\angle ARQ = a$. $[asy] size(225); pair a=(-50,0),b=(50,0),c=(-48,14),d=(-30,-40),o=(0,0); draw(circle(o,50)); draw(a--c--b--d--a); dot(a); label(\"A\",a,NE); dot(b); label(\"B\",b,E); dot(c); label(\"C\",c,NE); dot(d); label(\"D\",d,S); dot(o); label(\"O\",o,SE); pair r=(-57.692,15.385),q=(-57.692,-53.846),p=(-57.692,-19.231); dot(r); label(\"R\",r,NW); dot(q); label(\"Q\",q,SW); dot(p); label(\"P\",p,W); draw(c--p--d); draw(r--b--q); draw(c--q); draw(r--d); draw(r--p); draw(a--b,dotted); draw(c--o--d--c,dotted); [/asy]$ Because $PC$ and $PD$ are both tangents to the circle, we must have $\\angle ODP = \\angle OCP$. Therefore, $\\angle RDP = \\angle BDO = a$ and $\\angle PCQ = \\angle OCB = b$. Additionally, from a property of inscribed angles, we must have $\\angle ACD = \\angle ABD = a$ and $\\angle ADC = \\angle ABC = b$. Thus, since the sum of the angles in a triangle is $180^\\circ$, $\\angle CPD = 180-(2a+2b)$. Additionally, since $\\triangle RDB$ is a right triangle, we must have $\\angle BRD = 90-(a+b)$. Because $\\angle CRD = \\tfrac12 \\cdot \\angle CPD$ and $CP = PD$, we know that $C,R,D$ are in a circle with center $P$, so $RP = PC = PD$. Since $RB$ is a line, $\\angle RCP = 90-b$, so by the Base Angle Theorem, $\\angle CRP = \\angle RCP = 90-b$. Thus, from the Angle Addition Postulate, $\\angle PRD = (90-b)-(90-a-b) = a$. Thus, we proved that $P$ is on line", "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 =", "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}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "# 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", "$\\, D \\,$ be an arbitrary point on side $\\, AB \\,$ of a given triangle $\\, ABC, \\,$ and let $\\, E \\,$ be the interior point where $\\, CD \\,$ intersects the external common tangent to the incircles of triangles $\\, ACD \\,$ and $\\, BCD$. As $\\, D \\,$ assumes all positions between $\\, A \\,$ and $\\, B \\,$, prove that the point $\\, E \\,$ traces the arc of a circle. $[asy] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,SE); label(\"$$C$$\",C,W); label(\"$$D$$\",D,S); label(\"$$E$$\",E,NNE); [/asy]$", "# 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", "# 2002 Pan African MO Problems/Problem 5 ## Problem Let $\\triangle{ABC}$ be an acute angled triangle. The circle with diameter AB intersects the sides AC and BC at points E and F respectively. The tangents drawn to the circle through E and F intersect at P. Show that P lies on the altitude through the vertex C. ## Solution $[asy] pair a=(-65,0),O=(0,0),b=(65,0),e=(-39,52),f=(25,60),c=(-9.286,111.429),p=(-9.286,74.286),g=(-65,32.5),h=(65,43.333); draw(arc(O,b,a,CCW)); draw(a--b--c--a); dot(a); label(\"A\",a,SW); dot(b); label(\"B\",b,SE); dot(c); label(\"C\",c,N); dot(e); label(\"E\",e,NW); dot(f); label(\"F\",f,NE); dot(O); label(\"O\",O,S); dot(p); label(\"P\",p,NE); dot(g); label(\"G\",g,NW); dot(h); label(\"H\",h,NE); draw(a--g--p--(65,43.333)--b,dotted); draw(c--p,dotted); draw(e--O--f,dotted); [/asy]$ Draw lines $GA$ and $BH$, where $G$ and $H$ are on $EP$ and $FP$, respectively. Because $GA$ and $GE$ are tangents as well as $HB$ and $HF$, $GA = GE$ and $HB = HF$. Additionally, because $EP$ and $FP$ are tangents, $EP = FP$. Let $\\angle GAE = a$ and $\\angle HBF = b$. By the Base Angle Theorem, $\\angle GEA = a$ and $\\angle HFB = b$. Additionally, from the property of tangent lines, $GA \\perp AO$, $GP \\perp EO$, $PH \\perp FO$, and $HB \\perp BO$. Thus, by the Angle Addition Postulate, $\\angle OEA = \\angle OAE = 90-a$ and $\\angle OBF = \\angle OFB = 90-b$. Thus, $\\angle EOA = 2a$ and $\\angle FOB = 2b$, so $\\angle EOF = 180-2a-2b$. Since the sum of the angles in a", "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 $\\tan{(EAG)} =", "$\\triangle DCE$ and $\\triangle ABD$, Hence, the ratio of the area of $\\triangle DCE$ to the area of $\\triangle ABD$ is $\\dfrac{\\dfrac{\\sqrt{2}}{9}}{\\dfrac{\\sqrt{2}}{3}}$ = $\\dfrac{1}{3} \\Rightarrow C$ ## 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 $\\frac{1}{3}.$ ## Solution 4 WLOG, let $AC=1$, $BC=2$, so radius of the circle is $\\frac{3}{2}$ and $OC=\\frac{1}{2}$. As in solution 1, By same altitude, the ratio $[DCE]/[ABD]=PE/AB$, where $P$ is the point where $DC$ extended meets circle $O$. Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is $\\frac{1}{3}$.", "Difference between revisions of \"2015 AMC 10B Problems/Problem 22\" Problem In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? $[asy] pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); //(0,0) is a convenient point //E1 to prevent conflict with direction E(ast) pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label(\"A\",A,N); label(\"B\",B,E); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E1,W); label(\"F\",F,NW); label(\"G\",G,NE); label(\"H\",H,E); label(\"I\",I,S); label(\"J\",J,W); [/asy]$ Solution Triangle $AFG$ is isosceles, so $AG=AF=1$. Using the symmetry of pentagon $FGHIJ$, notice that $\\triangle IGF \\cong \\triangle JHG \\cong \\triangle DIJ \\cong AFG$. Therefore, $JH=AF=1$. Since $\\triangle AJH \\sim \\triangle AFG$, $\\frac{JH}{AF+FJ}=\\frac{1}{1+FG}=\\frac{FG}{FI}=\\frac{FG}1$. From this, we get $FG=\\frac{\\sqrt{5} -1}{2}$. Since $\\triangle DIJ \\cong \\triangle AFG$, $DJ=DI=AF=1$. Since $\\triangle AFG \\sim ADC$, $\\frac{AF}{AF+FJ+JD}=\\frac1{2+FG} = \\frac{FG}{CD}=\\frac1{CD}$. (Why is $\\frac{FG}{CD} = \\frac1{CD}$? That doesn't make any sense...) Solving for $CD$, we get $CD = \\frac{\\sqrt{5} +1}{2}$ Therefore, $FG+JH+CD=\\frac{\\sqrt5-1}2+1+\\frac{\\sqrt5+1}2=\\boxed{\\mathbf{(D)}\\ 1+\\sqrt{5}\\ }$", "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 us $$\\frac {2R} c = \\frac a{\\frac{2 \\times" ]

[ "# 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 $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars from", "dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ ### 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.)", "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, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "+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", "# 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 $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### 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);", "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", "# 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. $\\blacksquare$", "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}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "+ .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", "+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", "+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", "100; real yMin = 0; real yMax = 120; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"R\",(xMax,0),(2,0)); label(\"Im\",(0,yMax),(0,2)); pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56,104.908997); dot(A); dot(B); dot(C); dot(D); draw(A--C); draw(A--B); draw(D--C); draw(C--B); draw(D--B); label(\"Z_1\", A, W); label(\"Z_2\", B, W); label(\"Z_3\", C, N); label(\"Z\", D, N); draw(circle((56,61), 43.9089968)); [/asy]$ While $Z_1, Z_2, Z_3$ are fixed, $Z$ can be anywhere on the circle because those are the only values of $Z$ that satisfy the problem requirements. However, we want to find the real part of the $Z$ with the maximum imaginary part. This Z would lie directly above the center of the circle, and thus the real part would be the same as the x-value of the center of the circle. So all we have to do is find this value and we’re done. Consider the perpendicular bisectors of $Z_1Z_2$ and $Z_2Z_3$. Since any chord can be perpendicularly bisected by a radius of a circle, these two lines both intersect at the center. Since $Z_1Z_2$ is vertical, the perpendicular bisector will be horizontal and pass through the midpoint of this line, which is (18, 61). Therefore the equation for this line is $y=61$. $Z_2Z_3$ is nice because it turns out the differences in the x and y values are both equal (60) which means that the slope", "# [SOLVED]Geometry challenge #### anemone ##### MHB POTW Director Staff member In convex quadrilateral $ADBE$, there is a point $C$ within $\\triangle ABE$ such that $\\angle EAD+\\angle CAB=\\angle EBD+\\angle CBA=180^{\\circ}$. Prove that $\\angle ADE=\\angle BDC$. #### anemone ##### MHB POTW Director Staff member \\begin{tikzpicture} \\begin{scope} \\draw (0,0) circle(3); \\end{scope} \\coordinate[label=left: E] (E) at (-3,0); \\coordinate[label=below: D] (D) at (1.2,-2.75); \\coordinate[label=below: A] (A) at (-1,.-2.828); \\coordinate[label=right: F] (F) at (2.9,-0.768); \\coordinate[label=above: B] (B) at (-1,-0.26); \\coordinate[label=above: C] (C) at (-2,-1.1); \\draw (A) -- (E); \\draw (A) -- (D); \\draw (F) -- (D); \\draw (E) -- (F); \\draw (A) -- (B); \\draw (A) -- (F); \\draw (E) -- (D); \\draw (B) -- (D); \\draw [dashed] (C) -- (D); \\draw [dashed] (C) -- (B); \\draw [dashed] (C) -- (A); \\end{tikzpicture} Let F be the second intersection of the circumcircle of $\\triangle EAD$ and line $EB$. Then $\\angle DBF=180^{\\circ}-\\angle EBD=\\angle CBA$. Moreover, \\begin{align*}\\angle BDF&=180^{\\circ}-\\angle AEB-\\angle ADB\\\\&=180^{\\circ}-(360^{\\circ}-\\angle EAD-\\angle EBD)\\\\&= 180^{\\circ}-(\\angle CAB+\\angle CBA)\\\\&=\\angle BCA\\end{align*} These two relations give $\\angle BDF \\simeq \\triangle BCA$. So $\\dfrac{BD}{BF}=\\dfrac{BC}{BA}$ Together with $\\angle DBF=\\angle CBA$, we have $\\triangle BDC \\simeq \\triangle BFA$. This results in $\\angle ADE=\\angle AFE=\\angle BFA=\\angle BDC$. (Q.E.D.)", "+0 # ***HELP!!!!*****Due Today! -1 49 8 1)In triangle ABC, $AB = BC = 17$ and AC = 16 Find the circumradius of triangle 2)Angle bisectors line AX and line BY of triangle ABC meet at point I. Find angle C in degrees, if angle AIB = 109 [asy] pair A,B,C,X,Y,I; A= (0,0); B = (1,0); C = (0.6,0.5); 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); draw(X--A--C--B--Y); draw(A--B); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,S); label(\"$X$\",X,NE); label(\"$Y$\",Y,NW); [/asy] 3)A certain square and a certain equilateral triangle have the same perimeter. The square and triangle are inscribed in circles, as shown below. If A is the area of the circle containing the square, and B is the area of the circle containing the triangle, then find A/B 4) $\\overline{BY}$ and $\\overline{CZ}$ are angle bisectors of triangle $ABC$ that meet at I, with $CY = 4$, $AY = 6$, and $AB = 8$. Find $BZ$ . Dec 1, 2019 #1 0 im in the same class and i need help on this A triangle has side lengths of 6, 8, and 10. Let a be the area of the circumcircle. Let b be the area of the incircle. Compute a-b Dec 1, 2019 #2 +156 0 Stop cheating on HW", "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", "all intersect at an excenter of the triangle. • A bisector of an angle can be constructed using a compass and straightedge. $[asy] defaultpen(fontsize(8)); pair excenter(pair A, pair B, pair C){ pair X, Z; X=A+expi((angle(A-B)+angle(C-A))/2); Z=C+expi((angle(C-B)+angle(A-C))/2); return extension(X,A,Z,C); } pair X=(0,0), Y=(10,0), Z=(3,6); pair exX=excenter(Z,X,Y), exY=excenter(X,Y,Z), exZ=excenter(Y,Z,X); draw(circle(exX,length(exX-foot(exX,Y,Z)))); draw(circle(exY,length(exY-foot(exY,Z,X)))); draw(circle(exZ,length(exZ-foot(exZ,X,Y)))); draw((X-Y)+X--Y+1.5*(Y-X));draw((Y-Z)+Y--Z+(Z-Y));draw(2*(X-Z)+X--Z+2*(Z-X)); label(\"X\",X,(-1.5,-1));label(\"Y\",Y,(2,-1));label(\"Z\",Z,N); label(\"P\",exX,NE); dot(X^^Y^^Z^^exX^^exY^^exZ); draw(exX--(xpart(exX),0),dashed); real slope1=(ypart(Z)-ypart(X))/(xpart(Z)-xpart(X)); pair point1=exX+(1,-1/slope1); pair tangent1=extension(X,Z,point1,exX); draw(exX--tangent1,dashed); real slope2=(ypart(Y)-ypart(Z))/(xpart(Y)-xpart(Z)); pair point2=exX+(1,-1/slope2); pair tangent2=extension(Y,Z,point2,exX); draw(exX--tangent2,dashed); markscalefactor=0.1; draw(rightanglemark(exX,tangent1,Z)); draw(rightanglemark(exX,tangent2,Y)); draw(rightanglemark(exX,(xpart(exX),0),(100,0))); [/asy]$ Triangle $\\triangle XYZ$ with incenter I, with angle bisectors (red), incircle (blue), and inradii (green) [[Image:Incenter.PNG|left|thumb|300px|Triangle ABC with incenter I, with angle bisectors (red), [[external angle bisectorincircle (blue), and inradii (green)]]", "+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", "# Labeling angles in a triangle I've tried looking at past answers to this question but it won't seem to solve my problem. I want to label the angles ABD and DBC in this triangle: \\documentclass{article} \\usepackage{tikz,amsmath,amssymb,tkz-euclide} \\begin{document} \\begin{tikzpicture} \\tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5] \\tkzClip[space=0.25] \\coordinate (A) at (0,0); \\tkzLabelPoints[left](A) \\coordinate (B) at (2.25,3.320718914); \\tkzLabelPoints[above](B) \\coordinate (C) at (6,0); \\tkzLabelPoints[right](C) \\coordinate (E) at (3,0); \\tkzLabelPoints[below](E) \\draw (A)--(B)--(C)--cycle; \\draw (B)--(E); \\tkzDefLine[bisector](A,B,C)\\tkzGetPoint{a} \\tkzInterLL(A,C)(B,a) \\tkzGetPoint{D} \\tkzLabelPoints[below](D) \\draw (B)--(D); \\tkzDefMidPoint(A,B) \\tkzGetPoint{12} \\tkzLabelPoints[left](12) \\tkzDefMidPoint(B,C) \\tkzGetPoint{15} \\tkzLabelPoints[right=0.1cm](15) \\tkzDefMidPoint(A,C) \\tkzGetPoint{18} \\tkzLabelPoints[below=0.4cm](18) \\tkzMarkSegment[mark=|](A,E) \\tkzMarkSegment[mark=|](C,E) \\tkzLabelSegment(A,B) \\tkzLabelSegment(B,C) \\tkzLabelSegment(C,A) \\end{tikzpicture} \\end{document} You can use angles library for that. \\documentclass{article} \\usepackage{tikz,amsmath,amssymb,tkz-euclide} \\usetikzlibrary{angles} \\begin{document} \\begin{tikzpicture} \\tkzInit[xmin=-0.5,xmax=6.7,ymin=-0.7,ymax=3.5] \\tkzClip[space=0.25] \\coordinate (A) at (0,0); \\tkzLabelPoints[left](A) \\coordinate (B) at (2.25,3.320718914); \\tkzLabelPoints[above](B) \\coordinate (C) at (6,0); \\tkzLabelPoints[right](C) \\coordinate (E) at (3,0); \\tkzLabelPoints[below](E) \\draw (A)--(B)--(C)--cycle; \\draw (B)--(E); \\tkzDefLine[bisector](A,B,C)\\tkzGetPoint{a} \\tkzInterLL(A,C)(B,a) \\tkzGetPoint{D} \\tkzLabelPoints[below](D) \\draw (B)--(D); \\tkzDefMidPoint(A,B) \\tkzGetPoint{12} \\tkzLabelPoints[left](12) \\tkzDefMidPoint(B,C) \\tkzGetPoint{15} \\tkzLabelPoints[right=0.1cm](15) \\tkzDefMidPoint(A,C) \\tkzGetPoint{18} \\tkzLabelPoints[below=0.4cm](18) \\tkzMarkSegment[mark=|](A,E) \\tkzMarkSegment[mark=|](C,E) % \\tkzLabelSegment(A,B) % \\tkzLabelSegment(B,C) % \\tkzLabelSegment(C,A) \\draw pic[draw=blue, angle radius=5mm] {angle = A--B--D}; \\draw pic[draw=red, angle radius=7mm] {angle = D--B--C}; \\end{tikzpicture} \\end{document} • Thanks a lot, that's great. I was wondering if there was a way to make a dash going through the angles similar to the one going through the AE and AC segments. – KLG May 31 '15 at 5:03 • @SamFisher I don't know a fast way know. You can make a follow-up" ]

[ "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", "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 =", "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", "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}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "line $PD.$ $[asy] import geometry; import olympiad; size(400); point A = (2, 5), B = (0, 0), C = (10, 0), P = (3, 2); line c = line(A, B); line a = line(C, B); line b = line(A, C); point D = projection(a) * P; draw(P -- D); point e = projection(b) * P; draw(P -- e); point F = projection(c) * P; draw(P -- F); draw(A--B--C--A); draw(P--A); draw(P--B); draw(P--C); markrightangle(B, D, P); markrightangle(A, e, P); markrightangle(A, F, P); draw(circumcircle(A, P, e), dashed); line d = line(P, D); draw(d); point n = projection(d) * F; point M = projection(d) * e; draw(F--n); draw(e--M); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, SW); label(\"E\", e - (0.1, 0), NE); label(\"F\", F, W); label(\"P\", P+(0.2, 0), S); label(\"N\", n, E); label(\"M\", M, W); [/asy]$ Note that $AFPE$ is cyclic with diameter $AP.$ By ELOS, $\\dfrac{EF}{\\sin A} = 2R=AP\\implies EF = AP\\sin A.$ Since $BDPF$ is cyclic, we have that $B$ and $\\angle FPD$ are supplementary. Since $MPD$ is a line, $B = \\angle FPM.$ This means that $\\sin B = \\sin FPN = \\dfrac{FN}{FP}\\implies FN = PF\\sin B.$ Similarly, $EM = PE\\sin C.$ So the problem is reduced to proving that $EF\\ge FN+EM$ but this is obvious by the Pythagorean Theorem. $\\blacksquare$ Now the rest of", "= 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", "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", "ABD$ are both inscribed angles with the same arc of a given circle, $\\angle ABD = \\angle ACD = a$, so $\\angle ARQ = a$. $[asy] size(225); pair a=(-50,0),b=(50,0),c=(-48,14),d=(-30,-40),o=(0,0); draw(circle(o,50)); draw(a--c--b--d--a); dot(a); label(\"A\",a,NE); dot(b); label(\"B\",b,E); dot(c); label(\"C\",c,NE); dot(d); label(\"D\",d,S); dot(o); label(\"O\",o,SE); pair r=(-57.692,15.385),q=(-57.692,-53.846),p=(-57.692,-19.231); dot(r); label(\"R\",r,NW); dot(q); label(\"Q\",q,SW); dot(p); label(\"P\",p,W); draw(c--p--d); draw(r--b--q); draw(c--q); draw(r--d); draw(r--p); draw(a--b,dotted); draw(c--o--d--c,dotted); [/asy]$ Because $PC$ and $PD$ are both tangents to the circle, we must have $\\angle ODP = \\angle OCP$. Therefore, $\\angle RDP = \\angle BDO = a$ and $\\angle PCQ = \\angle OCB = b$. Additionally, from a property of inscribed angles, we must have $\\angle ACD = \\angle ABD = a$ and $\\angle ADC = \\angle ABC = b$. Thus, since the sum of the angles in a triangle is $180^\\circ$, $\\angle CPD = 180-(2a+2b)$. Additionally, since $\\triangle RDB$ is a right triangle, we must have $\\angle BRD = 90-(a+b)$. Because $\\angle CRD = \\tfrac12 \\cdot \\angle CPD$ and $CP = PD$, we know that $C,R,D$ are in a circle with center $P$, so $RP = PC = PD$. Since $RB$ is a line, $\\angle RCP = 90-b$, so by the Base Angle Theorem, $\\angle CRP = \\angle RCP = 90-b$. Thus, from the Angle Addition Postulate, $\\angle PRD = (90-b)-(90-a-b) = a$. Thus, we proved that $P$ is on line", "$\\triangle ABC$ is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation. $[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm); 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 = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); /* draw figures */ draw((3.46,0.96)--(3.44,-3.36)); draw((3.44,-3.36)--(8.02,-3.44)); draw((8.02,-3.44)--(3.46,0.96)); draw((3.46,0.96)--(-0.86,0.98)); draw((-0.86,0.98)--(-0.88,-3.34)); draw((-0.88,-3.34)--(3.44,-3.36)); draw((3.46,0.96)--(8.02,-3.44)); draw((8.02,-3.44)--(12.42,1.12)); draw((12.42,1.12)--(7.86,5.52)); draw((7.86,5.52)--(3.46,0.96)); draw((5.74,-1.24)--(-0.86,0.98)); draw((5.74,-1.24)--(-0.87,-1.18), linetype(\"4 4\")); draw((5.74,-1.24)--(7.86,5.52)); draw((5.74,-1.24)--(10.14,3.32), linetype(\"4 4\")); draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype(\"2 2\")); draw((8.02,-3.44)--(-0.86,0.98)); draw((3.44,-3.36)--(7.86,5.52)); draw((3.44,-3.36)--(5.74,-1.24)); /* dots and labels */ dot((3.46,0.96),dotstyle); label(\"A\", (3.2,1.06), NE * labelscalefactor); dot((3.44,-3.36),dotstyle); label(\"C\", (3.14,-3.86), NE * labelscalefactor); dot((8.02,-3.44),dotstyle); label(\"B\", (8.06,-3.8), NE * labelscalefactor); dot((-0.86,0.98),dotstyle); label(\"Z\", (-1.34,1.12), NE * labelscalefactor); dot((-0.88,-3.34),dotstyle); label(\"W\", (-1.48,-3.54), NE * labelscalefactor); dot((12.42,1.12),dotstyle); label(\"X\", (12.5,1.24), NE * labelscalefactor); dot((7.86,5.52),dotstyle); label(\"Y\", (7.94,5.64), NE * labelscalefactor); dot((5.74,-1.24),dotstyle); label(\"O\", (5.52,-1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Let $O$ be the the midpoint of $AB$. The perpendicular bisector of line $WZ$ and $XY$ will meet at $O$. Thus $O$ is the center of the circle points $X$, $Y$, $Z$, and $W$ lies on. $\\angle ZAC=\\angle OAY=90^{\\circ}$, $\\angle ZAC+\\angle BAC=\\angle OAY+\\angle BAC$, $\\angle ZAB=\\angle CAY$, and $AZ=AC$, $AB=AY$, $\\triangle AZB \\cong \\triangle ACY$ by", "# Difference between revisions of \"2009 AIME I Problems/Problem 5\" ## Problem Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\\overline{AC}$ and $\\overline{AB}$ respectively so that $AK = CK$, and $\\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\\overline{BK}$ and $\\overline{CL}$, and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\\overline{PM}$. If $AM = 180$, find $LP$. ## Diagram $[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label(\"A\",A,(-1,-1));label(\"B\",B,(1,-1));label(\"C\",C,(1,1)); label(\"K\",K,(0,2));label(\"L\",L,(0,-2));label(\"M\",M,(-1,1)); label(\"P\",P,(1,1)); label(\"180\",(A+M)/2,(-1,0));label(\"180\",(P+C)/2,(-1,0));label(\"225\",(A+K)/2,(0,2));label(\"225\",(K+C)/2,(0,2)); label(\"300\",(B+C)/2,(1,1)); [/asy]$ ## Solution 1 $[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label(\"A\",A,(-1,-1));label(\"B\",B,(1,-1));label(\"C\",C,(1,1)); label(\"K\",K,(0,2));label(\"L\",L,(0,-2));label(\"M\",M,(-1,1)); label(\"P\",P,(1,1)); label(\"180\",(A+M)/2,(-1,0));label(\"180\",(P+C)/2,(-1,0));label(\"225\",(A+K)/2,(0,2));label(\"225\",(K+C)/2,(0,2)); label(\"300\",(B+C)/2,(1,1)); [/asy]$ Since $K$ is the midpoint of $\\overline{PM}$ and $\\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\\bigtriangleup{AMB}$ is similar to $\\bigtriangleup{LPB}$ Thus, $$\\frac {AM}{LP}=\\frac {AB}{LB}=\\frac {AL+LB}{LB}=\\frac {AL}{LB}+1$$ Now let's apply the angle bisector theorem. $$\\frac {AL}{LB}=\\frac {AC}{BC}=\\frac {450}{300}=\\frac {3}{2}$$ $$\\frac {AM}{LP}=\\frac {AL}{LB}+1=\\frac {5}{2}$$ $$\\frac {180}{LP}=\\frac {5}{2}$$ $$LP=\\boxed {072}$$ ## Solution", "# Difference between revisions of \"2009 AIME I Problems/Problem 5\" ## Problem Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\\overline{AC}$ and $\\overline{AB}$ respectively so that $AK = CK$, and $\\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\\overline{BK}$ and $\\overline{CL}$, and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\\overline{PM}$. If $AM = 180$, find $LP$. ## Diagram $[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label(\"A\",A,(-1,-1));label(\"B\",B,(1,-1));label(\"C\",C,(1,1)); label(\"K\",K,(0,2));label(\"L\",L,(0,-2));label(\"M\",M,(-1,1)); label(\"P\",P,(1,1)); label(\"180\",(A+M)/2,(-1,0));label(\"180\",(P+C)/2,(-1,0));label(\"225\",(A+K)/2,(0,2));label(\"225\",(K+C)/2,(0,2)); label(\"300\",(B+C)/2,(1,1)); [/asy]$ ## Solution 1 $[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label(\"A\",A,(-1,-1));label(\"B\",B,(1,-1));label(\"C\",C,(1,1)); label(\"K\",K,(0,2));label(\"L\",L,(0,-2));label(\"M\",M,(-1,1)); label(\"P\",P,(1,1)); label(\"180\",(A+M)/2,(-1,0));label(\"180\",(P+C)/2,(-1,0));label(\"225\",(A+K)/2,(0,2));label(\"225\",(K+C)/2,(0,2)); label(\"300\",(B+C)/2,(1,1)); [/asy]$ Since $K$ is the midpoint of $\\overline{PM}$ and $\\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\\bigtriangleup{AMB}$ is similar to $\\bigtriangleup{LPB}$ Thus, $$\\frac {AM}{LP}=\\frac {AB}{LB}=\\frac {AL+LB}{LB}=\\frac {AL}{LB}+1$$ Now let's apply the angle bisector theorem. $$\\frac {AL}{LB}=\\frac {AC}{BC}=\\frac {450}{300}=\\frac {3}{2}$$ $$\\frac {AM}{LP}=\\frac {AL}{LB}+1=\\frac {5}{2}$$ $$\\frac {180}{LP}=\\frac {5}{2}$$ $$LP=\\boxed {072}$$ ## Solution", "# 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", "$\\, D \\,$ be an arbitrary point on side $\\, AB \\,$ of a given triangle $\\, ABC, \\,$ and let $\\, E \\,$ be the interior point where $\\, CD \\,$ intersects the external common tangent to the incircles of triangles $\\, ACD \\,$ and $\\, BCD$. As $\\, D \\,$ assumes all positions between $\\, A \\,$ and $\\, B \\,$, prove that the point $\\, E \\,$ traces the arc of a circle. $[asy] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,SE); label(\"$$C$$\",C,W); label(\"$$D$$\",D,S); label(\"$$E$$\",E,NNE); [/asy]$", "2009 AIME I Problems/Problem 5 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Problem Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\\overline{AC}$ and $\\overline{AB}$ respectively so that $AK = CK$, and $\\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\\overline{BK}$ and $\\overline{CL}$, and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\\overline{PM}$. If $AM = 180$, find $LP$. Diagram $[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label(\"A\",A,(-1,-1));label(\"B\",B,(1,-1));label(\"C\",C,(1,1)); label(\"K\",K,(0,2));label(\"L\",L,(0,-2));label(\"M\",M,(-1,1)); label(\"P\",P,(1,1)); label(\"180\",(A+M)/2,(-1,0));label(\"180\",(P+C)/2,(-1,0));label(\"225\",(A+K)/2,(0,2));label(\"225\",(K+C)/2,(0,2)); label(\"300\",(B+C)/2,(1,1)); [/asy]$ Solution 1 $[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label(\"A\",A,(-1,-1));label(\"B\",B,(1,-1));label(\"C\",C,(1,1)); label(\"K\",K,(0,2));label(\"L\",L,(0,-2));label(\"M\",M,(-1,1)); label(\"P\",P,(1,1)); label(\"180\",(A+M)/2,(-1,0));label(\"180\",(P+C)/2,(-1,0));label(\"225\",(A+K)/2,(0,2));label(\"225\",(K+C)/2,(0,2)); label(\"300\",(B+C)/2,(1,1)); [/asy]$ Since $K$ is the midpoint of $\\overline{PM}$ and $\\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\\bigtriangleup{AMB}$ is similar to $\\bigtriangleup{LPB}$ Thus, $$\\frac {AM}{LP}=\\frac {AB}{LB}=\\frac {AL+LB}{LB}=\\frac {AL}{LB}+1$$ Now let's apply the angle bisector theorem. $$\\frac {AL}{LB}=\\frac {AC}{BC}=\\frac {450}{300}=\\frac {3}{2}$$ $$\\frac {AM}{LP}=\\frac {AL}{LB}+1=\\frac {5}{2}$$ $$\\frac {180}{LP}=\\frac", "# Difference between revisions of \"Circumradius\" The circumradius of a cyclic polygon is the radius of the cirumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. ## Formula for a Triangle Let $a, b$ and $c$ denote the triangle's three sides, and let $A$ denote the area of the triangle. Then, the measure of the of the circumradius of the triangle is simply $R=\\frac{abc}{4A}$. Also, $A=\\frac{abc}{4R}$ ## Proof Proof: [asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label(\"$A$\", A, W); label(\"$B$\", B, N); label(\"$C$\", C, E); label(\"$D$\", D, S); label(\"$O$\", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label(\"$E$\", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy] We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\\angle ADB = \\angle BCA$ because they both subtend arc $AB$. Therefore, $\\triangle BAD \\sim \\triangle BEC$ by AA similarity, so we have $$\\frac{BD}{BA} = \\frac{BC}{BE},$$ or $$\\frac {2R} c = \\frac ah.$$ However, remember that area $\\triangle ABC = \\frac {bh} 2$, so $h=\\frac{2 \\times \\text{Area}}b$. Substituting", "and multiplying these together proves the formula for isogonal lines. Hence, we have $$\\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 $\\tan{(EAG)} =", "draw(A--Cp,ACpen); draw(Bp--Cp,BCpen); // Step 9. Finally, construct point D D=extension(Bp,A,B,C); draw(C--D,xlinepen); // ========================== // mark angles draw(arc(A,arcpoint(AB,arclength(B--H1)),C),anglepen); draw(arc(A,arcpoint(A--Bp,arclength(B--H1)),Cp),anglepen); draw(arc(D,arcpoint(D--A,arclength(B--H1)),C),anglepen); label(\"$A$\",A,N); label(\"$B$\",B,NW); label(\"$C$\",C,SE); label(\"$D$\",D,E); label(\"$H_4$\",H4,NW); label(\"$H_2$\",H2,NW); label(\"$H_1$\",H1,NW); label(\"$A^\\prime$\",Ap,N); label(\"$B^\\prime$\",Bp,N); label(\"$C^\\prime$\",Cp,SW+S); dot(new pair[]{A,B,C,D,H4,H2,H1,Ap,Bp,Cp},UnFill); Edit-3 Great thanks to @Andrew Stacey for pointing out in a comment that it is not allowed to transfer lengths from one place to another. This update fixes this by constructing additional points. // compass.asy : // To get standalone compass.pdf, run: // asy -f pdf compass.asy import math; import graph; struct construct{ pair[] loc; string[] name; pair[] namePos; guide[] straight; pen[] straightPen; guide[] circ; pen[] circPen; pen thinpen; bool pqr(pair p, pair q, pair r){ return (p.x*(q.y-r.y)+(r.y-p.y)*q.x+r.x*(p.y-q.y))>0; }; pair lastpoint(){ assert(loc.length>0); return loc[loc.length-1]; } pair prevpoint(){ assert(loc.length>1); return loc[loc.length-2]; } pair newpoint(pair ploc, string pname=\"\", pair npos=(0,0)){ loc.push(ploc); name.push(pname); namePos.push(npos); return loc[loc.length-1]; } guide newstraight(pair A, pair B, pen p=nullpen){ straight.push(A--B); straightPen.push(p); return straight[straight.length-1]; } guide newcirc(pair A,pair B, pen p=nullpen){ circ.push(Circle(A,arclength(A--B))); circPen.push(p); return circ[circ.length-1]; } pair halve(pair A, pair B, string pname=\"\", pair npos=(0,0)){ guide p,q; pair[] xpt; p=newcirc(A,B,thinpen); q=newcirc(B,A,thinpen); xpt=intersectionpoints(p,q); newpoint(xpt[0]); newpoint(xpt[1]); newstraight(lastpoint(),prevpoint(),thinpen); newpoint(extension(A,B,xpt[0],xpt[1]),pname,npos); return lastpoint(); } pair leftPoint(pair A1, pair B1, pair A2, pair B2, string pname=\"\", pair npos=(0,0)){ guide p,q; pair[] xpts; p=newcirc(A1,B1,thinpen); q=newcirc(A2,B2,thinpen); xpts=intersectionpoints(p,q); newpoint((pqr(A1,A2,xpts[0]))?xpts[0]:xpts[1],pname,npos); return lastpoint(); } pair rightPoint(pair A1, pair B1, pair A2, pair B2, string pname=\"\", pair npos=(0,0)){", "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 =" ]

[ "- 3 = 2.62", "• multiples", "+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", "48, can be divided by 3.", "# What is the average of all multiples of 17 lying between 430 and 530? 35 views What is the average of all multiples of 17 lying between 430 and 530? posted May 21, 2020 Looking for solution? Promote on: Similar Puzzles", "194 in 5 different ways; 209 in 6 ways.", "1, and we print 1·63 = 216.", "stating that three is many .", "n that is a multiple of 20.", "multiples of 3 or 5: 50 multiples of 3 or 5: 51 multiples of 3 or 5: 54 multiples of 3 or 5: 55 multiples of 3 or 5: 57 multiples of 3 and 5: 60 multiples of 3 or 5: 60 multiples of 3 or 5: 63 multiples of 3 or 5: 65 multiples of 3 or 5: 66 multiples of 3 or 5: 69 multiples of 3 or 5: 70 multiples of 3 or 5: 72 multiples of 3 and 5: 75 multiples of 3 or 5: 75 multiples of 3 or 5: 78 multiples of 3 or 5: 80 multiples of 3 or 5: 81 multiples of 3 or 5: 84 multiples of 3 or 5: 85 multiples of 3 or 5: 87 multiples of 3 and 5: 90 multiples of 3 or 5: 90 multiples of 3 or 5: 93 multiples of 3 or 5: 95 multiples of 3 or 5: 96 multiples of 3 or 5: 99 multiples of 3 or 5: 100", "of a multiple of 10 or 25.", "are triple and how many quadruples.", "multiples of 17 17 ,34 , 51 , 68 , 85 , 102 , 119 ,136 So the odd multiples are 17 ,51, 119", "the least common multiple of $180$, $216$, and $450$?", "consecutive numbers are: 62 and 64", "multiples of those numbers.", "ways; 129 in 4 different ways; 194 in 5 different ways; 209 in 6 ways.", "0 # What are the common multiples of 3 and 14? Wiki User 2016-01-31 20:10:11 Common multiples of 3 and 14 include 42, 84, and 126, among an infinite number of others. Wiki User 2016-01-31 20:10:11 🙏 0 🤨 0 😮 0 Study guides 20 cards ➡️ See all cards 21 cards ➡️ See all cards 22 cards ➡️ See all cards", "of 991. If you like to learn what is a multiple", "to multiply this number by a multiple of 19!!!" ]

[ "of $6$ therefore it is a multiple of $6$", "+ 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.", "divisible by $7$ because: $56 - (7\\times 2) = 56 - 14=42$, that is a multiple of $7$. A number is divisible by $9$ if the sum of its digits gives a multiple of $9$. $333, 999, 810, 945, 360, 9.963$ A number is divisible by $10$ if the digit of the units is $0$. $20, 43.340, 620, 34.230, 100.000, 440$ A number is divisible by $11$ if the difference of the sum of the digits that are in even places and in odd places is $0$ or a multiple of $11$. $242$ is divisible by $11$ because: $(2+2) - 4 = 4 - 4=0$ $616$ is divisible by $11$ because: $(6+6) - 1=12 - 1 =11$ $96.954$ is divisible by $11$ because: $(9+9+4) - (6+5) = 22 - 11=11$ A number is divisible by $25$ if its last two digits are zeros or a multiple of $25$. $3.300, 1.250, 375, 25.425, 100, 25.050$ A number is divisible by $125$ if its last three digits are zeros or a multiple of $125$. $20.000, 1.250, 34.125, 375, 501.125, 1.000$", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "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", "divide by $7$. So you need an answer that is a multiple of $7$. – lhf Nov 5 '13 at 1:13 You have figured out that $x$ must be divisible by three. Say $x = 3y$. Then the total number of votes is $7 y$. So we see that the total is divisible by $7$. Now all you have to do is look at the answers and see that only one of them is divisible by $7$. • Yes, it's obvious now. I was looking at $\\frac{7}{3}x$ and thinking it didn't tell me anything useful. But it tells me that the total is a multiple of 7. Thank you. – dumb question Nov 5 '13 at 1:21", "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.", "a multiple of $3.$ Therefore for every combination of five numbers, there are three numbers that add to give a multiple of $3.$ Mahdi used the same idea, but with $-1$ instead of $2$. Everything else works out the same.", "the first $17$ terms are $$1,3,4,6,10,11,12,15,16,19,20,23,24,25,26,27,28,\\ldots.$$ Here, too, the question is: what are the next terms? Added later: Thanks, Yaakov, you are right, so my question is, what are the positions of the numbers in $T_2$ in the sequence in the sequence $(1,2,3,\\ldots)$. I have the first $30$ positions (or ranks) and would like to see a method for finding more terms. It may help to see a list of the first $20$ towers ranked: $$4 = 2^2$$ $$8 = 2^3$$ $$9 = 3^2$$ $$16 = 2^{2^{2}}$$ $$27 = 3^3$$ $$81 = 3^{2^{2}}$$ $$256 = 2^{2^{3}}$$ $$512 = 2^{3^{2}}$$ $$6561 = 3^{2^{3}}$$ $$19683 = 3^{3^{2}}$$ Continuing with tuple notation instead of tower notation: $(2,2,2,2), (3,2,2,2), (2,3,3), (3,3,3), (2,3,2,2), (3,3,2,2), (2,2,2,3), (3,2,2,3), (2,2,3,2), (3,2,3,2), (2,3,2,3).$ My method, so far, has been by computer sort, which reaches overflow pretty quickly. Surely there must be a more insightful method. A related question: what is the position (or rank) of $(2,2,2,2,2,2)?$ • What does jointly ranked mean? – David Handelman Jan 29 '18 at 14:24 • Jointly ranked means arranged in increasing order: $t_2 = \\{4,8,16,256,\\ldots\\}$ and $t_3 = \\{9,27,81,\\ldots\\},$ so that the joint ranking is $(4,8,9,16,27,81,256,\\ldots).$ – Clark Kimberling Jan 29 '18 at 15:12 • $8=2^2$? $9=2^3$?? – Gerry Myerson Jan 29 '18 at 22:07 • I find a certain rough", "by $2$'s and $3$'s. Three more common multiples of $2$ and $3$ are $24$, $30$, and $36$.", "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, $$8$$): $$7\\cdot8\\cdot11\\cdot13\\cdot19\\cdot23\\cdot25\\cdot27\\cdot29\\cdot31 = 2123581660200.$$", "\\\\ + & \\underline{-260} \\\\ {} & -180 \\end{array}$ This number is divisible by $7$ if and only if the original is. We need to check $-180$ for divisibility by $7$. Here, adding and subtracting multiples of $7$ is easy. For example, add $210$ to get $30$. Now subtract $28$ to get $2$. The remainder when you divide $960,937,563,483$ by $7$ is $2$. If you want to use this rule, but the number of digits isn’t a multiple of three, you can simple add some zeros on front. For example, $45,338,017$ turns into $045,331,010$ and we get $\\begin{array}{cc} {} & +010 \\\\ {} & -331 \\\\+ & \\underline{+045} \\\\ {} & -276 \\end{array}$ $-276 + 280 = 4$, so the remainder when you divide $45,338,017$ by $7$ is $4$. This rule works due to a convenient pattern in the remainders of the powers of ten. If we start with $n = 0$, the remainder when you divide powers of $10$ by $7$ is $1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, \\ldots$ Each group of three digits, after alternating groups are multiplied by $-1$, has the same rule for divisibility by $7$. For example, the remainder when $124,000,000,000$ is divided by seven is the same as the remainder for $-124$. So we just take", "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", "What is the least common multiple of 40 and 12? Apr 1, 2016 Least Common Multiple is of $40$ and $12$ is $120$ Explanation: Multiples of $40$ are $\\left\\{40 , 80 , 120 , 160 , 200 , 240 , \\ldots .\\right\\}$ Multiples of $12$ are $\\left\\{12 , 24 , 36 , 48 , 60 , 72 , 84 , 96 , 108 , 120 , 132 , 144 , 156 , 168 , 180 , 192 , 204 , 216 , 228240 , \\ldots .\\right\\}$ Common multiples are $\\left\\{120 , 240 , \\ldots .\\right\\}$ Hence, Least Common Multiple is $120$.", "and a 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. $$\\begin{split} 1 \\cdot 3 & = 3 \\\\ 2 \\cdot 3 & = 6 \\\\ 3 \\cdot 3 & = 9 \\\\ 4 \\cdot 3 & = 12 \\\\ 5 \\cdot 3 & = 15 \\\\ 6 \\cdot 3 &= 18 \\end{split}$$ We can find the multiples of any number by continuing this process. Table 2.55 shows the multiples of 2 through 9 for the first twelve counting numbers. #### Table 2.55 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", "# 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 $24$.", "2.4 Find Multiples and Factors The topics covered in this section are: 2.4.1 Identify Multiples of Numbers Annie is counting the shoes in her closet. The shoes are matched in pairs, so she doesn’t have to count each one. She counts by twos: $2,4,6,8,10,12$. She has $12$ shoes in her closet. The numbers $2,4,6,8,10,12$ are called multiples of $2$ Multiples of $2$ can be written as the product of a counting number and $2$. The first six multiples of $2$ are given below. $1 \\cdot 2 = 2$ $2 \\cdot 2 = 4$ $3 \\cdot 2 = 6$ $4 \\cdot 2 = 8$ $5 \\cdot 2 = 10$ $6 \\cdot 2 = 12$ multiple of a number is the product of the number and a counting number. So a multiple of $3$ would be the product of a counting number and $3$. Below are the first six multiples of $3$. $1 \\cdot 3 = 3$ $2 \\cdot 3 = 6$ $3 \\cdot 3 = 9$ $4 \\cdot 3 = 12$ $5 \\cdot 3 = 15$ $6 \\cdot 3 = 18$ We can find the multiples of any number by continuing this process. The table below shows the multiples of 2 through 9 for the first twelve counting numbers. MULTIPLE OF A NUMBER A number is a multiple of", "which is one more than a multiple of 11, so when we reduce mod 11 we get $k, k+1, \\ldots, k+9$ which are all different. Unless 28082 happens to be one more than a multiple of 11, one of these is a multiple of 11. As it turns out, 28082 is one {\\it less} than a multiple of 11, and 28182 is a multiple of 11: $28182 = 11 \\times 2562$. So there is either zero or one palindrome of each of the forms $10x01, 11x11, 12x21, \\ldots, 99x99$. When is there no such palindrome? Observe that, for example, $12000 - 21 = (10000 + 2000) - (20 + 1) = (10000 - 1) + 10 \\times 2 \\times (100 - 1)$$and so 12000 and 21 differ by a multiple of 11. So $12x21$ and $24x00$ are congruent mod 11. To get $24x00$ to be a multiple of 11 we need $24x$ (that is, $240 + x$) to be a multiple of 11 – so we take $x = 2$. Indeed, $12221$ is a multiple of 11, with $12221 = 11 \\times 1111$. But consider, for example, the number $27x72$. This is congruent to $54x00$ mod 11, and is a multiple of 11 if and only if $54x$ is – but none of $540, 541, \\ldots, 549$ are", "$49$, which is true because $6+8$ is divisible by $7$ and $$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,$8^{273}+ 6^{273}\\equiv0\\pmod{49}$.", "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" ]

[ "of $6$ therefore it is a multiple of $6$", "by $2$'s and $3$'s. Three more common multiples of $2$ and $3$ are $24$, $30$, and $36$.", "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", "What is the least common multiple of 40 and 12? Apr 1, 2016 Least Common Multiple is of $40$ and $12$ is $120$ Explanation: Multiples of $40$ are $\\left\\{40 , 80 , 120 , 160 , 200 , 240 , \\ldots .\\right\\}$ Multiples of $12$ are $\\left\\{12 , 24 , 36 , 48 , 60 , 72 , 84 , 96 , 108 , 120 , 132 , 144 , 156 , 168 , 180 , 192 , 204 , 216 , 228240 , \\ldots .\\right\\}$ Common multiples are $\\left\\{120 , 240 , \\ldots .\\right\\}$ Hence, Least Common Multiple is $120$.", "# 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 a 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. $$\\begin{split} 1 \\cdot 3 & = 3 \\\\ 2 \\cdot 3 & = 6 \\\\ 3 \\cdot 3 & = 9 \\\\ 4 \\cdot 3 & = 12 \\\\ 5 \\cdot 3 & = 15 \\\\ 6 \\cdot 3 &= 18 \\end{split}$$ We can find the multiples of any number by continuing this process. Table 2.55 shows the multiples of 2 through 9 for the first twelve counting numbers. #### Table 2.55 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", "# 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 $24$.", "× # 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 $42$.", "which is one more than a multiple of 11, so when we reduce mod 11 we get $k, k+1, \\ldots, k+9$ which are all different. Unless 28082 happens to be one more than a multiple of 11, one of these is a multiple of 11. As it turns out, 28082 is one {\\it less} than a multiple of 11, and 28182 is a multiple of 11: $28182 = 11 \\times 2562$. So there is either zero or one palindrome of each of the forms $10x01, 11x11, 12x21, \\ldots, 99x99$. When is there no such palindrome? Observe that, for example, $12000 - 21 = (10000 + 2000) - (20 + 1) = (10000 - 1) + 10 \\times 2 \\times (100 - 1)$$and so 12000 and 21 differ by a multiple of 11. So $12x21$ and $24x00$ are congruent mod 11. To get $24x00$ to be a multiple of 11 we need $24x$ (that is, $240 + x$) to be a multiple of 11 – so we take $x = 2$. Indeed, $12221$ is a multiple of 11, with $12221 = 11 \\times 1111$. But consider, for example, the number $27x72$. This is congruent to $54x00$ mod 11, and is a multiple of 11 if and only if $54x$ is – but none of $540, 541, \\ldots, 549$ are", "the first $17$ terms are $$1,3,4,6,10,11,12,15,16,19,20,23,24,25,26,27,28,\\ldots.$$ Here, too, the question is: what are the next terms? Added later: Thanks, Yaakov, you are right, so my question is, what are the positions of the numbers in $T_2$ in the sequence in the sequence $(1,2,3,\\ldots)$. I have the first $30$ positions (or ranks) and would like to see a method for finding more terms. It may help to see a list of the first $20$ towers ranked: $$4 = 2^2$$ $$8 = 2^3$$ $$9 = 3^2$$ $$16 = 2^{2^{2}}$$ $$27 = 3^3$$ $$81 = 3^{2^{2}}$$ $$256 = 2^{2^{3}}$$ $$512 = 2^{3^{2}}$$ $$6561 = 3^{2^{3}}$$ $$19683 = 3^{3^{2}}$$ Continuing with tuple notation instead of tower notation: $(2,2,2,2), (3,2,2,2), (2,3,3), (3,3,3), (2,3,2,2), (3,3,2,2), (2,2,2,3), (3,2,2,3), (2,2,3,2), (3,2,3,2), (2,3,2,3).$ My method, so far, has been by computer sort, which reaches overflow pretty quickly. Surely there must be a more insightful method. A related question: what is the position (or rank) of $(2,2,2,2,2,2)?$ • What does jointly ranked mean? – David Handelman Jan 29 '18 at 14:24 • Jointly ranked means arranged in increasing order: $t_2 = \\{4,8,16,256,\\ldots\\}$ and $t_3 = \\{9,27,81,\\ldots\\},$ so that the joint ranking is $(4,8,9,16,27,81,256,\\ldots).$ – Clark Kimberling Jan 29 '18 at 15:12 • $8=2^2$? $9=2^3$?? – Gerry Myerson Jan 29 '18 at 22:07 • I find a certain rough", "= 72 24 × 3 = 72 4th multiple 24 + 24 + 24 + 24 = 96 24 × 4 = 96 5th multiple 24 + 24 + 24 + 24 + 24 = 120 24 × 5 = 120 ## Did you know that… If a number is a multiple of 3 and 8… then we can conclude that it is also a multiple of 24? Let’s try to see if this is true. Consider the number 2,352. We need to know if it is divisible by 24 by knowing if it is also a multiple of 3 and 8. Hence, let’s divide 2,352 by 3. Thus,$$2,352\\;\\div\\;3\\;=\\;784$$. It is a multiple of 3! Now, let’s check if it is also a multiple of 8. If we divide 2,352 by 8, we will get$$2,352\\;\\div\\;8\\;=\\;294$$. It is also a multiple 8! Thus, we can say that it is a multiple of 24! Now, let’s see if it is indeed true by dividing 2,352 by 24. Hence,$$2,352\\;\\div\\;24\\;=\\;98$$. Since the result does not have a remainder, then we can definitely confirm that the trick is true! Can you try to check if 3,240 is also a multiple of 24 using this trick? ## List of First 30 multiples of 24 There is an endless number possible of integers hence, listing", "to add another 14. Hence, $$14\\;+\\;14\\;+\\;14\\;=\\;42$$ . Can you continue adding 14 to get the 4th and 5th multiples of 14 and compare if you will have the same result as the diagram? Another way to find multiples of 14 is by multiplication. We have said earlier that a number that can be expressed as 14n is a multiple of 14. So, if we assume that n is a counting number, we need to get the product of 14 and n. Say you want to find the 17th multiple of 14; all we need to do is multiply 14 by 17. Thus, $$14\\;\\times\\;17\\;=\\;238$$ . Therefore, the 17th multiple of 14 is 238! Let’s try getting the 19th negative multiple of 14. Since we are asked to get the 19th negative multiple of 14, we will multiply 14 by -19. Hence, $$14\\;\\times\\;-19\\;=\\;-266$$ . Here’s what we know so far, skip counting and multiplication are two methods we can use to find any multiples of 14. The table below shows how the two methods. ## List of First 30 Multiples of 14 There are infinitely many positive and negative integers. Hence, it makes no sense to list all the possible multiples of 14. But, we’ve made a list of the first 30 multiples of 14, so you can double-check if", "× # What is the least common multiple (LCM) of 8, 12, and 18? Jul 3, 2016 Least Common Multiple is $72$ #### Explanation: Multiples of $8$ are $\\left\\{8 , 16 , 24 , 32 , 40 , 48 , 56 , 64 , 72 , 80 , 88 , 96 , 104 , 112 , 120 , 128 , 136 , 144. \\ldots . .\\right\\}$ Multiples of $12$ are $\\left\\{12 , 24 , 36 , 48 , 60 , 72 , 84 , 96 , 108 , 120 , 132 , 144. \\ldots . .\\right\\}$ Multiples of $18$ are $\\left\\{18 , 36 , 54 , 72 , 90 , 108 , 126 , 144. \\ldots . .\\right\\}$ Hence common multiples are $\\left\\{72 , 144 , \\ldots \\ldots .\\right\\}$ and Least Common Multiple is $72$ Another shorter way is to write numbers as multiplication of its prime factors $8 = 2 \\times 2 \\times 2$ $\\to$ $2$ comes three times $12 = 2 \\times 2 \\times 3$ $\\to$ $2$ comes twice and $3$ once. $18 = 2 \\times 3 \\times 3$ $\\to$ $2$ comes once and $3$ twice So maximum times is three times for $2$ and two times for $3$. Hence, Least Common Multiple is $2 \\times 2 \\times 2 \\times 3 \\times 3 = 72$. Apr", "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", "only $167 * 2^2 * 3$. 167, 2, 2, 3 4, 3, 167 12, 167 4, 501 2, 1004 2, 3, 334 2, 2, 501* 6, 2, 167 3, 668 6, 334 2004* To begin, the first multiple doesn't work because there are only 3 prime divisors of 2004. We can apply all multiples because the prime factorization of $2004^{2004}$ is $2^{4008} * 3^{2004} * 167^{2004}.$ Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star. For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to. For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1. Therefore we have $5 * 6$ normal multiples and $3 *2$ \"half\" multiples. Sum to get $\\boxed{54}$ as our answer.", "The number of ordered triples for various values of $a$ are presented in the following table. $$\\begin{tabular}{|c|c|c|r|} \\hline a & b & c & triples \\\\ \\hline -8 & \\{9,10\\} & -6a - 2b - 20 & 2 \\\\ -7 & \\{6, 7, 8, 9, 10\\} & -6a - 2b - 20 & 5 \\\\ -6 & \\{-10, \\ldots, 10\\} & \\{-10, \\ldots, 10\\} & 441 \\\\ -5 & \\{0, \\ldots, 10\\} & -6a - 2b - 20 & 11 \\\\ -4 & \\{-3, \\ldots, 7\\} & -6a - 2b - 20 & 11 \\\\ -3 & \\{-6, \\ldots, 4\\} & -6a - 2b - 20 & 11 \\\\ -2 & \\{-9, \\ldots, 1\\} & -6a - 2b - 20 & 11 \\\\ -1 & \\{-10, \\ldots, -2\\} & -6a - 2b - 20 & 9 \\\\ 0 & \\{-10, \\ldots, -5\\} & -6a - 2b - 20 & 6 \\\\ 1 & \\{-10, -9, -8\\} & -6a - 2b - 20 & 3 \\\\ \\hline Total & & & 510\\\\ \\hline \\end{tabular}$$ The total number of ordered triples that satisfy the required condition is $510$. Notes For * In case anyone is confused by this (as I initially was). In the case where $f(2)=f(4)$, this does not mean that g has a double root of $f(2)=f(4)=k$, ONLY", "+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", "\\\\ + & \\underline{-260} \\\\ {} & -180 \\end{array}$ This number is divisible by $7$ if and only if the original is. We need to check $-180$ for divisibility by $7$. Here, adding and subtracting multiples of $7$ is easy. For example, add $210$ to get $30$. Now subtract $28$ to get $2$. The remainder when you divide $960,937,563,483$ by $7$ is $2$. If you want to use this rule, but the number of digits isn’t a multiple of three, you can simple add some zeros on front. For example, $45,338,017$ turns into $045,331,010$ and we get $\\begin{array}{cc} {} & +010 \\\\ {} & -331 \\\\+ & \\underline{+045} \\\\ {} & -276 \\end{array}$ $-276 + 280 = 4$, so the remainder when you divide $45,338,017$ by $7$ is $4$. This rule works due to a convenient pattern in the remainders of the powers of ten. If we start with $n = 0$, the remainder when you divide powers of $10$ by $7$ is $1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, \\ldots$ Each group of three digits, after alternating groups are multiplied by $-1$, has the same rule for divisibility by $7$. For example, the remainder when $124,000,000,000$ is divided by seven is the same as the remainder for $-124$. So we just take", "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 +", "is divisible by $2^{50}$. Hint: Count how many multiples of 2, 4, 8... there are in the former expression. From here and here, we have that the highest power of a prime $p$ dividing $n!$ is given by $$\\sum_{k=1}^{\\infty}\\left\\lfloor\\dfrac{n}{p^k} \\right\\rfloor$$ Hence, the highest power of a prime $p$ dividing $\\dfrac{n!}{m!}$ is given by $$\\sum_{k=1}^{\\infty}\\left\\lfloor\\dfrac{n}{p^k} \\right\\rfloor - \\sum_{k=1}^{\\infty}\\left\\lfloor\\dfrac{m}{p^k}\\right\\rfloor$$ I trust you can finish it off from here. for the number of multiples of $2$: $$100 = 52 + 2(n_1 - 1)$$ $$n_1 = 25$$ for the number of multiples of $4$: $$100 = 52 + 4(n_2 - 1)$$ $$n_2 = 13$$ for the number of multiples of $8$: $$96 = 56 + 8(n_3 - 1)$$ $$n_3 = 6$$ for the number of multiples of $16$: $$96 = 64 + 16(n_4 - 1)$$ $$n_4 = 3$$ for the number of multiples of $32$: $$96 = 64 + 32(n_5 - 1)$$ $$n_5 = 2$$ for the number of multiples of $64$: $$64 = 64 + 64(n_6 - 1)$$ $$n_6 = 1$$ To get the number of $2$'s as factors, add all the $n$'s up, which yields $50$. This should cancel the $2^{50}$ at the denominator, and thus prove that the expression is an integer." ]

[ "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", "n is divisible by 3)?", "# Divisors with units digit 5 How many positive divisors of $$20130^4$$ have 5 as their units digit? ×", "• Divisible by nine How many three-digit natural numbers in total are divisible without a remainder by the number 9? • Digits How many natural numbers greater than 4000 which are formed from the numbers 0,1,3,7,9 with the figures not repeated, B) How many will the number of natural numbers less than 4000 and the numbers can be repeated? • Street numbers Lada came to aunt. On the way he noticed that the houses on the left side of the street have odd numbers on the right side and even numbers. The street where he lives aunt, there are 5 houses with an even number, which contains at least one digit number 6 • Phone numbers How many 7-digit telephone numbers can be compiled from the digits 0,1,2,..,8,9 that no digit is repeated? • Dd 2-digit numbers Find all odd 2-digit natural numbers compiled from digits 1; 3; 4; 6; 8 if the digits are not repeated.", "# Nines Divisibility Number Theory Level 1 Find the digit $$N$$ such that the five-digit number $$\\overline{N878N}$$ is divisible by 9. ×", "a 3x3 square must contain the numbers 1 through 9.", "# Question -6 Probability Level 2 A 9-digit integer is formed using the digits 1, 2, ... , 9 without repetition. Find the probability that the number is divisible by 4. ×", "many three-digit natural numbers are greater than 321 if no digit in number repeated? • Divisible by nine How many three-digit natural numbers in total are divisible without a remainder by the number 9? • A five-digit A five-digit odd number has the sum of all digits (digits) five and contains two zeros. If we move each digit in the number one place to the left and move the first digit to the last place, we get a number 20,988 smaller. Find this unknown five-digit numb", "a rule for composite integers. • Although ten isn't a single digit number, if the number ends in a 0, it is evenly divisible by ten.", "number divisible by 3...", "# What Are Those Hundreds For? $\\large \\overline{2xy} + \\overline{3yx} + \\overline{6xx}$ Which of the following must always be a divisor of the sum of the three 3-digit integers shown above? ×", "1,3,4, on the place of hundreds 4 or 2. • How many 4 How many 4 digit numbers that are divisible by 10 can be formed from the numbers 3, 5, 7, 8, 9, 0 such that no number repeats? • Three-digit numbers Use the numbers 4,5,8,9 to write all three-digit numbers without repetition. How many such numbers are there?", "that contain a digit 6. 17. 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?", "See how many $2$-digit numbers you can find that are divisible by $9$. What happens if you just use the numbers from $1$ to $8$?", "# Divides by 9? Number Theory Level pending What number would need to be added to the following number to make it divisible by 9? $\\large (12345678987654321)_{12}$ Clarification: The subscript indicates base 12. Hint: There is no need to convert the above number to decimal or use a calculator. Good luck! ×", "3. The digits divisible by 3 are 0, 3, 6, and 9. Divisible by 5. The integers which are divisible by 5 are precisely the integer with units digit either 0 or 5. Since the number you want are 3 digit numbers the hundreds digit can't be 0. Thus the hundreds digit is 3, 6 or 9. Hence you have 3 choices for the hundreds digit. The units digit must be 0 or 5 so there are 2 choices for the units digit. Hence you have $3 \\times 2 = 6$ possible ways to choose the hundreds and units digit. Whichever of the six pairs you choose you have 8 choices for the tens digit and hence you have $6 \\times 8 = 48$ integers that satisfy the conditions in the problem. I hope this helps, Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "+ 9 9 = 2 *4 + 1", "A four-digit number of the form $$\\overline{xxyy}$$, where $$x$$ and $$y$$ are single-digit positive integers, when divided by 11 gives a number which is again divisible by 11. If $$x - y = 1$$, find the number.", "2, the first three digits make a number divisible by 3...", "of div1C (Perfect Triples)?" ]

[ "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 $$a+b$$. ×", "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", "36$). If the missing digit is 9, then the digits are 1, 2, 3, 4, 6, 7, 8. But the sum here is $40-9=31$, which is not divisible by 3. So this cannot happen.", "+ 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.", "19:29 • @Evariste: Very cute. I didn't even think of counting numbers backwards! Oct 15 '16 at 19:36 Hint: the number $ac=10a+c$ is equal to $ac=10a+c=(9+1)a+c=9a +(a+c)$ So $ac =10a+c$ divided by 9, will have same remainder as $a+c$ divided by 9. This is assuming you know the elementary school trick that a number is divisible by 9, if and only if the sum of its digits is divisible by 9. And then it assumes you realize that that means the divisor when divided by 9 is going to have the same divisor as the sum of the digits divided by 9. Which we can figure out by adding those digits. Ultimately the divisor is exactly equal to the sum of the sum of the sum of .... the digits. So add up the numbers 1+2+3+4...... that's long but play with it and see if you can find shortcuts. Example: if 2+7=9 eventually in a future step you will want the remainder after dividing by 9 so that will be 0. So you can toss out any a+b=9. Keep playing and see what happens. Now... you are simply taking the word of some stranger on the Internet the the remainder when divided by 9 is the same thing as the sum of the digits. Play with it and", "## Basic College Mathematics (9th Edition) If the sum of a number's digits is divisible by 3, we know that the number itself must also be divisible by 3. For example, the sum of the digits of 63 is $6+3=9$, which is divisible by 3. Therefore, 63 must be divisible by 3. $63\\div3=21$", "that the number has less than $6$ divisors. Since $12$ has $6$ divisors and $10$ has $4$ divisors, James’s desired sum is $\\boxed{10}$.", "each power of 9 will leave 3 as a remainder when it is devided by 6. here it will leave 3 for 8 times. so that, $3*8=24$ which is completely divisible by 6.", "## 6 June 2017 ### Nine $9$ is a composite number. $9$ is a perfect square: $9=3^2$. An eneagon is a polygon of $9$ sides. A number is divisible by $9$ if the sum of its digits is a number divisible by $9$: $4 \\, 158$ is divisible by $9$ because $4+1+5+8=18$ and $18$ is divisible by $9$.", "sum of digits are divisible by $11$. There are two numbers of the first kind that satisfies this criterion. They are $97999$ and $99979$. Indeed, $9 – 7 + 9 – 9 + 9 = 11$, and $9 – 9 + 9 – 7 + 9 = 11$. Third Hint Of the second kind, only one number $98989$ is divisible by $11$. Indeed ,$9 – 8 + 9 – 8 + 9 = 11$. Final Step Out of the total of $15$ numbers, three are divisible by $11$. The probability of this event is $3/15 = 1/5$.", "“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?” $0+3+6+9 = 18.$", "#### You may also like Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself! ### Double Digit What are the factors of $595$?", "\"casting out 9s\". The remainder on dividing by $$9$$ is the same as the remainder on dividing the sum of the digits by $$9$$. It works because $$10^n$$ always has a remainder of $$1$$ when dividing by $$9$$ because, for example, $$100000=99999+1$$ This is very well known: Consider that the number is $$N= \\sum_{k=0}^n a_i 10^k$$ where $$a_i$$ are the digits. Digits. Then the sum of the digits is $$S = \\sum_{k=0}^n a_i$$. Then $$N - S = \\sum_{k=0}^n a_i(10^k -1)$$. Now each of the $$10^k-1$$ is $$99999.......9$$ which is divisible by $$9$$. So the whole number $$N-S$$ is divisible by $$9$$. (Note the last digit, $$a_0$$ is simply subtracted. You can think of it as $$a_0 - a_0 = a_0*10^0 - a_0 = a_0(10^0 - 1) = a_0(1 - 1) = a_0*0 = 0$$.) .... As a result, a number is divisible by $$9$$ if and only if the sum of its digits are divisible by $$9$$ so that gives you your second result. It also helps us realize if we take a number and divide by $$9$$ and take the remainder, then the remainder will be the same as the remainder taking the sum of the digits and dividing by $$9$$ and taking the remainder. Other answers have explained why the result of subtracting the digits", "digit sum of a number differs from the number itself by a multiple of $9$ (that is why (repeated) digit sums can be used to check divisibility by $9$). Therefore the digit sum of the sum also differs by a multiple of $9$ form the digit sum of the summands. (So above: The difference between $20$ and $2$ is $18=2\\cdot 9$)", "# How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisble by $11$. How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisible by $11$. My attempt- A number is divisible by 11 if the alternating sum of its digit is divisible by 11? The other thing to notice is as it is a 9 digit number formed by digits 1 to 9, exactly once each digit from 1 to 9 will appear in the number. Basically, the question boils down to how many ways we can arrange 123456789 so that the alternating sum of the digit is divisible by 11. I am not able to proceed further. Any help would be appreciated. Since the raw digit sum of $1..9$ will be $45$, an odd number, we clearly can't have two equal sums for the even-position and odd-position digits. So we need to have one set of digits sum to $17$ and the others to $28$, so that the difference is divisible by $11$. This can be done either way around; $28$ as the sum of the four even-position digits or the sum of the five odd-position digits. Making 28 with only", "## Thinking Mathematically (6th Edition) The given number is divisible by $9$. A number is divisible by $9$ if the sum of all its digits is divisible by $9$. The sum of the digits of the given number is: $=4+8+2+0+1+6+5+1 \\\\=27$ Note that $27$ is divisible by $9$ since $9(3) = 27$. Thus, the given number is divisible by $9$.", "3, 6, 9 to the sum are 3! (3)(1111), 3! (6) (1111), 3! (9)(1111) respectively. The required sum is 3!(1111)(2 + 3 + 6 + 9) = 1,33,320", "the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.", "### Have You Got It? Can you explain the strategy for winning this game with any target? ### Counting Factors Is there an efficient way to work out how many factors a large number has? ### Repeaters Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. See how many $2$-digit numbers you can find that are divisible by $9$. What happens if you just use the numbers from $1$ to $8$?", "= 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" ]

[ "# Doesn't seem possible Let $N$ be the smallest positive integer for which the digit sum of $N$ and the digit sum of $N+1$ are both divisible by $50$. How many digits of $N$ are 9? Details and assumptions The digit sum of a number is the sum of all its digits. For example the digit sum of 1123 is $1 + 1 + 2 + 3 = 7$. ×", "19:29 • @Evariste: Very cute. I didn't even think of counting numbers backwards! Oct 15 '16 at 19:36 Hint: the number $ac=10a+c$ is equal to $ac=10a+c=(9+1)a+c=9a +(a+c)$ So $ac =10a+c$ divided by 9, will have same remainder as $a+c$ divided by 9. This is assuming you know the elementary school trick that a number is divisible by 9, if and only if the sum of its digits is divisible by 9. And then it assumes you realize that that means the divisor when divided by 9 is going to have the same divisor as the sum of the digits divided by 9. Which we can figure out by adding those digits. Ultimately the divisor is exactly equal to the sum of the sum of the sum of .... the digits. So add up the numbers 1+2+3+4...... that's long but play with it and see if you can find shortcuts. Example: if 2+7=9 eventually in a future step you will want the remainder after dividing by 9 so that will be 0. So you can toss out any a+b=9. Keep playing and see what happens. Now... you are simply taking the word of some stranger on the Internet the the remainder when divided by 9 is the same thing as the sum of the digits. Play with it and", "you mean, \"haven't seen me for a while\"? I just smashed the stack again! – Parcly Taxel Feb 20 '18 at 3:25 • There is a difference here that from base $10$: for base $10$, the last digit only matters for checking divisibility by $2$, while for divisibility by $3$ the sum of digits matter. But, in your answer, there is a reverse criteria for the two - i.e., only checking last digit for divisibility by $3$, while checking the sum of digits (as stated that - sum of odd digits is even) for checking divisibility by $2$. Based on your answer, cannot explain the opposite reasons employed for the two ($2,3$) for different bases ($9,10$). – jitender Feb 20 '18 at 3:27 • @jitender: You only need to check the last digit if the divisor divides exactly into the base. So 2 divides 10 so you only need to check the last digit and 3 divides 9 so you only need to check the last digit. As for the summing - this works when your divisor is one less than the base. So 9 is one less than 10 so sum of digits being divisible by 9 works (and three because anything divisible by 9 is divisible by 3). Likewise 8 is one less than nine so", "36$). If the missing digit is 9, then the digits are 1, 2, 3, 4, 6, 7, 8. But the sum here is $40-9=31$, which is not divisible by 3. So this cannot happen.", "A four-digit number of the form $$\\overline{xxyy}$$, where $$x$$ and $$y$$ are single-digit positive integers, when divided by 11 gives a number which is again divisible by 11. If $$x - y = 1$$, find the number.", "Integer whose Digits when Grouped in 3s add to 999 is Divisible by 999 Theorem Let $n$ be an integer which has at least $3$ digits when expressed in decimal notation. Let the digits of $n$ be divided into groups of $3$, counting from the right, and those groups added. Then the result is equal to $999$ if and only if $n$ is divisible by $999$.", "# How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisble by $11$. How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisible by $11$. My attempt- A number is divisible by 11 if the alternating sum of its digit is divisible by 11? The other thing to notice is as it is a 9 digit number formed by digits 1 to 9, exactly once each digit from 1 to 9 will appear in the number. Basically, the question boils down to how many ways we can arrange 123456789 so that the alternating sum of the digit is divisible by 11. I am not able to proceed further. Any help would be appreciated. Since the raw digit sum of $1..9$ will be $45$, an odd number, we clearly can't have two equal sums for the even-position and odd-position digits. So we need to have one set of digits sum to $17$ and the others to $28$, so that the difference is divisible by $11$. This can be done either way around; $28$ as the sum of the four even-position digits or the sum of the five odd-position digits. Making 28 with only", "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 $1$", "is divided by $10$, the remainder is the unit digit. let us see what will be the unit digit of $x^{40}$. Notice that unit digits of powers of $3$ get repeated in pattern, as $3^0=1, 3^1=3, 3^2=9, 3^3=27$. Follow this pattern and you will find that unit digit of something like, $3^{40}$ will be $1$. I shall let you conclude now.", "$3$. 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", "= 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", "### Have You Got It? Can you explain the strategy for winning this game with any target? ### Counting Factors Is there an efficient way to work out how many factors a large number has? ### Repeaters Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. See how many $2$-digit numbers you can find that are divisible by $9$. What happens if you just use the numbers from $1$ to $8$?", "digit sum of a number differs from the number itself by a multiple of $9$ (that is why (repeated) digit sums can be used to check divisibility by $9$). Therefore the digit sum of the sum also differs by a multiple of $9$ form the digit sum of the summands. (So above: The difference between $20$ and $2$ is $18=2\\cdot 9$)", "each power of 9 will leave 3 as a remainder when it is devided by 6. here it will leave 3 for 8 times. so that, $3*8=24$ which is completely divisible by 6.", "+ 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.", "## 6 June 2017 ### Nine $9$ is a composite number. $9$ is a perfect square: $9=3^2$. An eneagon is a polygon of $9$ sides. A number is divisible by $9$ if the sum of its digits is a number divisible by $9$: $4 \\, 158$ is divisible by $9$ because $4+1+5+8=18$ and $18$ is divisible by $9$.", "# 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 $$C$$? ×", "3. The digits divisible by 3 are 0, 3, 6, and 9. Divisible by 5. The integers which are divisible by 5 are precisely the integer with units digit either 0 or 5. Since the number you want are 3 digit numbers the hundreds digit can't be 0. Thus the hundreds digit is 3, 6 or 9. Hence you have 3 choices for the hundreds digit. The units digit must be 0 or 5 so there are 2 choices for the units digit. Hence you have $3 \\times 2 = 6$ possible ways to choose the hundreds and units digit. Whichever of the six pairs you choose you have 8 choices for the tens digit and hence you have $6 \\times 8 = 48$ integers that satisfy the conditions in the problem. I hope this helps, Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "“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?” $0+3+6+9 = 18.$", "# 6-digit Arrangements The digits $$1,2,3,4,5,6$$ are arranged to form a 6-digit number. The probability that the number formed is divisible by 4 can be expressed as $$\\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a + b$$? Details and assumptions Each digit is only used once in forming the 6-digit number. ×" ]

[ "Algebra 1 $60$ The product or quotient of two numbers with the same sign is positive. Since -54 and -0.9 are both negative, the quotient is positive. $-54\\div(-0.9)=60$", "\\times 7^2 = 490$, whose sum of divisors is $?$.", "is equal to 60", "of $6000$, where $1$ and $6000$ are also considered as divisors of $6000$ is $40$ $50$ $60$ $30$ 1 vote 2 answers 4 179 views The sum of the series $\\dfrac{1}{1.2} + \\dfrac{1}{2.3}+ \\cdots + \\dfrac{1}{n(n+1)} + \\cdots$ is $1$ $1/2$ $0$ non-existent", "+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", "the sum of the numbers inside: 76, 38", "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.", "The ratio of ‘the sum of the odd positive integers from $1$ to $100$’ to ‘the sum of the even positive integers from $150$ to $200$’ is ________. 1. $45:95$ 2. $1:2$ 3. $50:91$ 4. $1:1$", "the divisors of $n$.", "(12)(54) = 648. While the total sum is 1+2+...+36 = 666. Which is an impossibility. Thus the max sum of these must be at least 55.", "h)^2 + 60(1 + h)$.", "# 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? •", "What is the sum of: $6 + 66 + 666 + 6666 + \\cdots + 666666666\\cdots6$ where there are $n$ sixes in the last term?", "Divisors What is the smallest positive integer that is not a divisor of 5040? Note: $$7! = 5040$$ ×", "# Find the sum of divisors given the product of divisors The product of divisors of the positive integer $$N$$ is 15625. What is the sum of divisors of $$N$$? ×", "+0 # Help! 0 73 1 A positive divisor d of $$N = 2^4 \\cdot 3^6 \\cdot 5^8$$ is chosen uniformly at random. What is the probability that d is a perfect cube? Feb 22, 2020", "# You've gotta be kidding me $1007021035035021007001$ How many positive divisors (including 1 and itself) does the large number above have? ×", "## 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", "\\ldots, 2N$, and the set of odd numbers $1, 3, \\ldots, 2N-1$, since multiplying / dividing by 2 doesn't affect the largest odd divisor. Hence, the sum is $1 + 3 + \\ldots + 2N-1 = N^2$.", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$" ]

[ "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $150=5\\cdot 5\\cdot 3\\cdot 2$ Factor 150 completely to obtain: $150 \\\\=2(75) \\\\=2(3)(25) \\\\=2(3)(5)(5)$ Thus, the prime factorization of $150$ is: $150=5\\cdot 5\\cdot 3\\cdot 2$", "## Introductory Algebra for College Students (7th Edition) composite; $2^{2} \\cdot 3 \\cdot 5$ RECALL: A prime number is a number whose only factors are 1 and itself. A composite number is a number with factors other than one and itself. $60$ is an even number so 2 is one of its factors. Thus, $60$ is a composite number. Factor $60$ completely to have: $60 \\\\=2(30) \\\\=2(2)(15) \\\\=2(2)(3)(5)$ $\\\\=(2^{2})(3)(5)$ The prime factorization of $60$ is: $2^{2} \\cdot 3 \\cdot 5$", "# How do you find the GCF of 66, 90, and 150? Jan 12, 2017 $6$ #### Explanation: Note that the prime factorisation of $66$ is: $66 = 2 \\cdot 3 \\cdot 11$ Both $90$ and $150$ are divisible by $2$ and by $3$, but not by $11$. Hence all three numbers are divisible by $2 \\cdot 3 = 6$ and nothing larger.", "+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", "Elementary Algebra $2 \\cdot 2 \\cdot 2 \\cdot 3$ Factor the number completely until all of the factors are prime: $24 \\\\= 2 \\cdot 12 \\\\=2 \\cdot 2 \\cdot 6 \\\\=2 \\cdot 2 \\cdot 2 \\cdot 3$", "of divisors formula, we have that the sum of the divisors of 60 is $=\\frac{2^3-1}{2-1} \\times \\frac{3^2-1}{3-1}$ $\\times \\frac{5^2-1}{5-1}$ $=\\frac{8-1}{1} \\times \\frac{9-1}{2}$ $\\times \\frac{25-1}{4}$ $=7 \\times 4 \\times 6=168$ (iii) By the product of divisors formula, we have that the product of the divisors of 60 is =60(Number of divisors of 60)/2 =6012/2 =606 Related Topics: ## Problem Solution on Divisors of 60 Question: Write the following sets in roster form D = {$x : x$ is a prime number which is divisor of 60} Answer: Note that the set contains the prime numbers $x$ which are divisors of 60. Thus $x$ is a prime divisor of 60. From the above, we know that the prime divisors of 60 are 2, 3, and 5. ∴ D = {2, 3, 5} ## FAQs on Divisors of 60 Q1: What are the divisors of 60? Ans: The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. Q2: Find the prime factorization of 60. Ans: The prime factorization of 60 is 60 = 22 × 3 × 5. Q3: Find the prime divisors of 60. Ans: The prime divisors of 60 are 2, 3, and 5. Q4: Is 12 a divisor of 60? Ans: Yes, 12 is a divisor of 60 as 60/12", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "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: $\\dfrac{3 \\cdot 3 \\cdot 2}{5 \\cdot 3 \\cdot 2} = \\boxed{\\dfrac35}$. ## Similar Questions 1. ### Calculus Four red balls and two white balls are placed in a jar. One ball is randomly removed and replaced with a ball of the other color. The jar is then shaken, and one ball is randomly selected. What is the probability that this ball is … 2. ### Algebra What is the probability that a randomly selected three-digit number is divisible by 5? 3. ### math a student is selected at is the probability the the randomly selected stundent is neither", "## Factor some ideals in ZZ Find the prime factorizations of the ideals $A = (70)$ and $B = (150)$ in $\\mathbb{Z}$. Compute $((70),(150))$. Evidently $70 = 2 \\cdot 5 \\cdot 7$ and $150 = 2 \\cdot 3 \\cdot 5^2$, so that $(70) = (2)(5)(7)$ and $(150) = (2)(3)(5)^2$. Since $\\mathbb{Z}$ is a unique factorization domain, $(2)$, $(3)$, $(5)$, and $(7)$ are prime since 2, 3, 5, and 7 are prime. Now $((70),(150)) = (70,150) = (10)$, since $70 = 7 \\cdot 10$, $150 = 15 \\cdot 10$, and $10 = 150 - 2 \\cdot 70$.", "decompose the number $bc$ into prime factors: for example if $b = 6$ and $c = 10$, $bc = 60$ and $$60 = 2 \\cdot 2 \\cdot 3 \\cdot 5$$ But since 60 is the product of 6 and 10, we can reorder its factorization in such a way to find the prime factors of 6 and those of 10: (1) $$60 = \\underbrace{2 \\cdot 3}_{6} \\cdot \\underbrace{2 \\cdot 5}_{10}$$ For the reasoning above, if $p$ is a prime number dividing 60, $p$ must be present in its factorization (that is $p = 2$ or $p = 3$ or $p = 5$); but then, for how we subdivided it, $p$ can be present in the factorization of 6 (that is $p = 2$ or $p = 3$), and in that case it would divide 6, or it can be present in the factorization of 10 (that is $p = 2$ or $p = 5$), and in that case it would divide 10. Possibly $p$ can be present in both factorizations, as the case of $p = 2$, but it must be present at least in one of them. To better understand this, we can imagine to apply the following algorithm: we scan the list of the prime factors of 60 in the ordering of (1): 2, 3, 2,", "# List of Prime Factorizations of Integers from 601 to 800 For your quick reference, I have listed the prime factorizations of integers from 601 to 800. A repeated prime number is written in compact form using exponential notation. • $601$ is prime. • $602 = 2 \\cdot 7 \\cdot 43$ • $603 = {3^2} \\cdot 67$ • $604 = {2^2} \\cdot 151$ • $605 = 5 \\cdot {11^2}$ • $606 = 2 \\cdot 3 \\cdot 101$ • $607$ is prime. • $608 = {2^5} \\cdot 19$ • $609 = 3 \\cdot 7 \\cdot 29$ • $610 = 2 \\cdot 5 \\cdot 61$ • $611 = 13 \\cdot 47$ • $612 = {2^2} \\cdot {3^2} \\cdot 17$ • $613$ is prime. • $614 = 2 \\cdot 307$ • $615 = 3 \\cdot 5 \\cdot 41$ • $616 = {2^3} \\cdot 7 \\cdot 11$ • $617$ is prime. • $618 = 2 \\cdot 3 \\cdot 103$ • $619$ is prime. • $620 = {2^2} \\cdot 5 \\cdot 31$ • $621 = {3^3} \\cdot 23$ • $622 = 2 \\cdot 311$ • $623 = 7 \\cdot 89$ • $624 = {2^4} \\cdot 3 \\cdot 13$ • $625 = {5^4}$ • $626 = 2 \\cdot 313$ • $627 = 3 \\cdot 11 \\cdot 19$ • $628 = {2^2} \\cdot 157$ •", "$n$ other than $1\\cdot n$. Using an app for prime factorization, the following can be deduced very quickly: So our only non-prime $n$ with one unique factorization is $n=19{,}897=101\\cdot 197$ and we have our solution!", "? Free Version Difficult # Prime Factorization and Divisors ALGEBR-YT8GKF A certain integer has exactly 40 distinct positive divisors. Which one of the following could be the prime factorization of that number? A $2^{3}\\cdot5$ B $2^{3}\\cdot3^{4}\\cdot5$ C $3^{8}\\cdot7^{5}$ D $5^{2}\\cdot11^{4}\\cdot13^{5}$", "= 2^4 \\cdot 5^3 = 2000$. If $a+1 = 10$, then $b+1 = 2$, but this gives $2^9 \\cdot 5^1 > 2019$. No other values for $a+1$ work. Then we have $\\frac{S}{20} = \\frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \\boxed{472}$. -scrabbler94 ## Solution 2 For $n$ to have exactly $20$ positive divisors, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$. No number that is a multiple of $20$ can be expressed in the first form, and the only integer divisible by $20$ that has the second form is $2^{9}5$, which is greater than $2019$. For the third form, the only $20$-pretty numbers are $2^45^3=2000$ and $2^35^4=5000$, and only $2000$ is small enough. For the fourth form, any number of the form $2^45r$ where $r$ is a prime other than $2$ or $5$ will satisfy the $20$-pretty requirement. Since $n=80r<2019$, $r\\le 25$. Therefore, $r$ can take on $3, 7, 11, 13, 17, 19,$ or $23$. Thus, $\\frac{S}{20}=\\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\\boxed{472}$.", "the second method to work we need $d>m$, i.e., $\\frac{2\\cdot 90}d<d+1$. Since $\\frac{2\\cdot 90}d$ is even, exactly one of the options works for each odd divisor $d>1$. Therefore the politeness is the number of odd divisors. This can be read from the prime factorization of $90=2\\cdot 3^2\\cdot 5^1$: the odd divisors are of the form $3^a5^b$ with $0\\le a\\le 2, 0\\le b\\le 1$. Hence there are $3\\cdot 2$ odd divisors. Remove $d=1$ to arrive at politeness $5$.", "$a$ 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: $\\dfrac{3 \\cdot 3 \\cdot 2}{5 \\cdot 3 \\cdot 2} = \\boxed{\\dfrac35}$.", "the set of positive integers all of whose prime divisors are $2$ or $\\equiv 1 (\\!\\!\\mod 4)$, and let $B$ be the set of positive integers all of whose prime divisors are $\\equiv 3 (\\!\\!\\mod 4)$. Then $A \\cdot B = \\mathbb{N}$.", "# Question #b68fd Aug 24, 2017 See below. #### Explanation: Let's factor $105$. $105 = 35 \\cdot 3 = 3 \\cdot 5 \\cdot 7$ So, $3 \\cdot 5 \\cdot 7$ is the prime factorization of $105$. However, since there is only one of each prime, $\\sqrt{105}$ is already simplified.", "## anonymous one year ago What is the difference in the total number of factors and the number of prime factors in 60^3 - 32^3 - 28^3 1. mathmath333 $$60^{3}- 32^{3} - 28^{3} =3\\times 32\\times 28\\times 60$$ 2. anonymous options : 1) 116 2) 120 3) 126 4) 130 3. ganeshie8 Start with : $$60-32=28$$ cube both sides $$(60-32)^3 = 28^3$$ For left hand side, recall the identity $$(a-b)^3 = a^3-b^3-3ab(a-b)$$ : $$60^3-32^3 -3\\cdot 60\\cdot 32(60-32) = 28^3$$ $$\\implies 60^3-32^3-28^3 = 3\\cdot 60\\cdot 32\\cdot 28=2^9\\cdot 3^2\\cdot 5\\cdot 7$$ 4. ganeshie8 so thats the prime factorization do you know how to find the number of factors of a number from prime factorization ? 5. anonymous Yes. 6. ganeshie8 good, see if you can finish the rest 7. anonymous Thank you :)", "## Basic College Mathematics (9th Edition) The LCM is $1760$ because the prime factorization shows that the number $55$ is included in the prime factorisation of $1750$. $55 = 5 \\cdot 11$ $1760 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 55 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 5 \\cdot 11$ LCM $= 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 5 \\cdot 11 = 1760$." ]

[ "+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", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $150=5\\cdot 5\\cdot 3\\cdot 2$ Factor 150 completely to obtain: $150 \\\\=2(75) \\\\=2(3)(25) \\\\=2(3)(5)(5)$ Thus, the prime factorization of $150$ is: $150=5\\cdot 5\\cdot 3\\cdot 2$", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "the second method to work we need $d>m$, i.e., $\\frac{2\\cdot 90}d<d+1$. Since $\\frac{2\\cdot 90}d$ is even, exactly one of the options works for each odd divisor $d>1$. Therefore the politeness is the number of odd divisors. This can be read from the prime factorization of $90=2\\cdot 3^2\\cdot 5^1$: the odd divisors are of the form $3^a5^b$ with $0\\le a\\le 2, 0\\le b\\le 1$. Hence there are $3\\cdot 2$ odd divisors. Remove $d=1$ to arrive at politeness $5$.", "## Introductory Algebra for College Students (7th Edition) composite; $2^{2} \\cdot 3 \\cdot 5$ RECALL: A prime number is a number whose only factors are 1 and itself. A composite number is a number with factors other than one and itself. $60$ is an even number so 2 is one of its factors. Thus, $60$ is a composite number. Factor $60$ completely to have: $60 \\\\=2(30) \\\\=2(2)(15) \\\\=2(2)(3)(5)$ $\\\\=(2^{2})(3)(5)$ The prime factorization of $60$ is: $2^{2} \\cdot 3 \\cdot 5$", "What's the sum of all the positive integral divisors of $540$? What's the sum of all the positive integral divisors of $540$? My approach: I converted the number into the exponential form. And found out the integral divisors which came out to be $24$. But couldn't find the sum...Any trick? Prime factorization of 540 is $2^2\\cdot 3^3\\cdot 5$. Also sum of divisors of 540 equals: \\begin{align} \\sigma\\left(2^2\\cdot 3^3\\cdot 5\\right) &=\\frac {2^3-1}{2-1}\\cdot\\frac{3^4 - 1}{3-1}\\cdot \\frac{5^2 -1}{ 5-1}\\\\ &=7\\cdot 40\\cdot 6\\\\ &= 1680 \\end{align} Assuming you want to know the sum of the divisors of 540, i.e. $$\\sigma_1(540) = \\sum_{d \\mid 540} d$$ then you can compute this by hand by either determining all divisors of 540, or you can do this a little more cleverly. First of all, note that $540 = 2^2 \\cdot 3^3 \\cdot 5$. Now, one happy fact about the function $\\sigma_1(n)$ is that if $n, m$ are relatively prime, then $\\sigma_1(n \\cdot m) = \\sigma_1(n) \\cdot \\sigma_1(m)$. In particular, $$\\sigma_1(540) = \\sigma_1(4)\\sigma_1(27)\\sigma_1(5)$$ and so you only need to determine these last ones. But for prime numbers, $\\sigma_1(p^k)$ is easy to compute. It is $$\\sigma_1(p^k) = 1 + p + p^2 + \\cdots + p^k = \\frac{p^{k+1} - 1}{p-1}$$ which should let you compute the answer now.", "? Free Version Difficult # Prime Factorization and Divisors ALGEBR-YT8GKF A certain integer has exactly 40 distinct positive divisors. Which one of the following could be the prime factorization of that number? A $2^{3}\\cdot5$ B $2^{3}\\cdot3^{4}\\cdot5$ C $3^{8}\\cdot7^{5}$ D $5^{2}\\cdot11^{4}\\cdot13^{5}$", "of divisors formula, we have that the sum of the divisors of 60 is $=\\frac{2^3-1}{2-1} \\times \\frac{3^2-1}{3-1}$ $\\times \\frac{5^2-1}{5-1}$ $=\\frac{8-1}{1} \\times \\frac{9-1}{2}$ $\\times \\frac{25-1}{4}$ $=7 \\times 4 \\times 6=168$ (iii) By the product of divisors formula, we have that the product of the divisors of 60 is =60(Number of divisors of 60)/2 =6012/2 =606 Related Topics: ## Problem Solution on Divisors of 60 Question: Write the following sets in roster form D = {$x : x$ is a prime number which is divisor of 60} Answer: Note that the set contains the prime numbers $x$ which are divisors of 60. Thus $x$ is a prime divisor of 60. From the above, we know that the prime divisors of 60 are 2, 3, and 5. ∴ D = {2, 3, 5} ## FAQs on Divisors of 60 Q1: What are the divisors of 60? Ans: The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. Q2: Find the prime factorization of 60. Ans: The prime factorization of 60 is 60 = 22 × 3 × 5. Q3: Find the prime divisors of 60. Ans: The prime divisors of 60 are 2, 3, and 5. Q4: Is 12 a divisor of 60? Ans: Yes, 12 is a divisor of 60 as 60/12", "60. Since the last digit of 60 is 0, which is even, 60 is divisible by 2. We will repeatedly divide by 2 until we no longer can. We shall divide as follows: 30 is divisible by 2 again. 15 is not divisible by 2, but it is divisible by 3, the next prime. 5 is not divisble by 3, but it is divisible by 5, the next prime. 30 is divisible by 2 again. 15 is not divisible by 2, but it is divisible by 3, the next prime. 5 is not divisble by 3, but it is divisible by 5, the next prime. The quotient 1 is finally smaller than the divisor 5, and the prime factorization of 60 is the product of these prime divisors. 60 = 2 2 3 5 60 = 2 2 3 5 size 12{\"60\"=2 cdot 2 cdot 3 cdot 5} {} We use exponents when possible. 60 = 2 2 3 5 60 = 2 2 3 5 size 12{\"60\"=2 rSup { size 8{2} } cdot 3 cdot 5} {} #### Example 8 Find the prime factorization of 441. 441 is not divisible by 2 since its last digit is not divisible by 2. 441 is divisible by 3 since 4+4+1=94+4+1=9 size 12{4+4+1=9} {} and 9 is divisible by 3.", "Elementary Algebra $2 \\cdot 2 \\cdot 2 \\cdot 3$ Factor the number completely until all of the factors are prime: $24 \\\\= 2 \\cdot 12 \\\\=2 \\cdot 2 \\cdot 6 \\\\=2 \\cdot 2 \\cdot 2 \\cdot 3$", "the unique factorization theorem? $2520$ is the smallest positive integer divisible by each of $~2,~3,\\cdots,~9,~\\&~10$. 5. Ok, yes I understand that. It's based on the LCM. Thank you!!", "## 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", "# How do you find the GCF of 66, 90, and 150? Jan 12, 2017 $6$ #### Explanation: Note that the prime factorisation of $66$ is: $66 = 2 \\cdot 3 \\cdot 11$ Both $90$ and $150$ are divisible by $2$ and by $3$, but not by $11$. Hence all three numbers are divisible by $2 \\cdot 3 = 6$ and nothing larger.", "the set of positive integers all of whose prime divisors are $2$ or $\\equiv 1 (\\!\\!\\mod 4)$, and let $B$ be the set of positive integers all of whose prime divisors are $\\equiv 3 (\\!\\!\\mod 4)$. Then $A \\cdot B = \\mathbb{N}$.", "+0 # Another one 0 62 1 +156 a) Describe all positive integers that have exactly 3 positive divisors. b) How many positive divisors of 8400 have at least 4 positive divisors? #1 +797 +2 I believe this is answer already, but I think I have a better solution: $\\bold{Part (a)}$: A Logical Approach First of all, a number with an odd number of positive divisors must be a perfect square. For example, $9=3^2$, and its factors are: 1, 3, and 9. A total of 3 numbers. However, a number that is not a perfect square, will have an even number of factors. For example, $6=2\\cdot3$, and its factors are : 1, 2, 3, and 6. A total of 3 numbers. Logically, this makes sense, because when you look at 6’s factors, $1\\cdot6=6$ and $2\\cdot3=6$. When you look at 9’s factors, $1\\cdot9=9$ and $3\\cdot3=9$. In the factors of 6, every number has another number to pair itself with, making pairs of 2, which makes an even number of factors. On the other hand, in the factors of 9, 1 and 9 can pair with each other to make 9, but 3 must pair with itself to make 9, hence the even number of factors. This problem does not ask of odd number of factors, but asks for 3", "+0 # Help! 0 73 1 A positive divisor d of $$N = 2^4 \\cdot 3^6 \\cdot 5^8$$ is chosen uniformly at random. What is the probability that d is a perfect cube? Feb 22, 2020", "# Find the sum of the first $50$ positive even integers Mental math - The sum of the first $50$ positive odd integers is $50^2$. Find the sum of the first $50$ positive even integers. - Hint: each of these even integers is one bigger than one of the odd integers. – user108903 Dec 16 '12 at 20:18 There are $50$ kids lined up, you give them $1$ candy, $3$, $5$, and so on. Total candies $50^2$. Now you give each kid one more. – André Nicolas Dec 16 '12 at 20:21 homework should not be used as a standalone tag; see tag-wiki and meta. – Martin Sleziak Dec 16 '12 at 21:00 You are given that $$1 + 3 + 5 + \\cdots + 99 = 50^2$$ You need to compute $$2 + 4 + 6 + \\cdots + 100$$ Note that \\begin{align} 2 + 4 + 6 + \\cdots + 100 & = (1+1) + (3+1) + (5+1) + \\cdots + (99+1)\\\\ & = (1 + 3 + 5 + \\cdots + 99) + (\\underbrace{1 + 1 + 1 + \\cdots + 1}_{50 \\text{ times}})\\\\ & = 50^2 + 50 \\end{align} -", "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: $\\dfrac{3 \\cdot 3 \\cdot 2}{5 \\cdot 3 \\cdot 2} = \\boxed{\\dfrac35}$. ## Similar Questions 1. ### Calculus Four red balls and two white balls are placed in a jar. One ball is randomly removed and replaced with a ball of the other color. The jar is then shaken, and one ball is randomly selected. What is the probability that this ball is … 2. ### Algebra What is the probability that a randomly selected three-digit number is divisible by 5? 3. ### math a student is selected at is the probability the the randomly selected stundent is neither", "What is the sum of all of the odd divisors of $6300$? What is the sum of all of the odd divisors of $6300$? Hello! I am a middle school student, so simply-worded answers would be super helpful. To solve this, I tried to find the sum of the odd divisors of a few smaller numbers, like 10. I know $10 = 2 * 5$, so I thought that, for the number to be odd, I'd have to exclude 2. Therefore, including 1, the sum would be $1 + 5 = 6$. This was correct, but I think that's only because the number is so small. When I tried it again with 18, which is $3^{2} * 2$, I got a different answer from the correct sum. How should I start this problem? • Hint $1$: What is the sum of the odd divisors of $n$ compared to the sum of the odd divisors of $2n$? – Michael Burr Mar 31 '18 at 4:01 • Likely not the most elegant way, but you can find the prime factorization, 2*2*3*3*5*5*7. Now to find your odd divisors, find and multiply together the 18 combinations of odd prime factors. – Air Conditioner Mar 31 '18 at 4:08 • @AirConditioner How is it $18$ combinations? What I did was that there are", "decompose the number $bc$ into prime factors: for example if $b = 6$ and $c = 10$, $bc = 60$ and $$60 = 2 \\cdot 2 \\cdot 3 \\cdot 5$$ But since 60 is the product of 6 and 10, we can reorder its factorization in such a way to find the prime factors of 6 and those of 10: (1) $$60 = \\underbrace{2 \\cdot 3}_{6} \\cdot \\underbrace{2 \\cdot 5}_{10}$$ For the reasoning above, if $p$ is a prime number dividing 60, $p$ must be present in its factorization (that is $p = 2$ or $p = 3$ or $p = 5$); but then, for how we subdivided it, $p$ can be present in the factorization of 6 (that is $p = 2$ or $p = 3$), and in that case it would divide 6, or it can be present in the factorization of 10 (that is $p = 2$ or $p = 5$), and in that case it would divide 10. Possibly $p$ can be present in both factorizations, as the case of $p = 2$, but it must be present at least in one of them. To better understand this, we can imagine to apply the following algorithm: we scan the list of the prime factors of 60 in the ordering of (1): 2, 3, 2," ]

[ "For example, Snuke can paint the blocks as shown in the diagram below. There are $45$ red blocks and $30$ blue blocks, thus the difference is $9$.", "+0 # Probability question +2 109 2 +106 Kaylee has 2 gallons each of red and blue paint, and 1 gallon each of orange and green paint. She paints her bedroom with these jars of paint. If she uses $\\frac{2}{3}$ of the red paint, $\\frac{1}{4}$ of the blue paint, $\\frac{1}{2}$ of the orange paint, and $\\frac{5}{7}$ of the green paint, how much paint (in gallons) does Kaylee have left over? Apr 4, 2021 #1 +1 Kaylee has 3 55/84 gallons left over. Apr 4, 2021 #2 +106 +2 Thanks. Sorry I didn't mean to name this \"probability\" lol TealSeal Apr 5, 2021", "circular necklace with $$2013$$ beads. Each bead can be painted either white or green. A painting of the necklace is called good if among any $$21$$ successive beads there is at least one green bead. Prove that the number of good paintings of the necklace is odd. (Two paintings that differ on some beads, but can be obtained from each other by rotating or flipping the necklace, are counted as different paintings. )", "2$$\\frac{3}{4}$$ quarts to cover each wall. If each wall is painted only one color, how many walls will be blue and how many will be white? How much paint will be left over? Angela and Ryou are painting a room. Angela has 2$$\\frac{1}{2}$$ gallons of blue paint. 2$$\\frac{1}{2}$$ × 4 = 10 quarts of blue paint. It will take 2$$\\frac{3}{4}$$ quarts to cover each wall. Blue colored walls = 10 × 2$$\\frac{3}{4}$$ = 3 walls with 7/11 quarts of paint remaining White colored walls = 5 × 2$$\\frac{3}{4}$$ = 1 wall with 9/11 quarts of paint remaining", "Tables to Compare Ratios – Page No. 233 Question 1. Sarah asked some friends about their favorite colors. She found that 4 out of 6 people prefer blue, and 8 out of 12 people prefer green. Is the ratio of friends who chose blue to the total asked equivalent to the ratio of friends who chose green to the total asked? Type below: ___________ Yes, 4/6 is equivalent to 8/12 Explanation: 4/6 = 0.666 8/12 = 0.666 Question 2. Lisa and Tim make necklaces. Lisa uses 5 red beads for every 3 yellow beads. Tim uses 9 red beads for every 6 yellow beads. Is the ratio of red beads to yellow beads in Lisa’s necklace equivalent to the ratio in Tim’s necklace? Type below: ___________ The ratio of red beads to yellow beads in Lisa’s necklace is not equivalent to the ratio in Tim’s necklace Explanation: Lisa and Tim make necklaces. Lisa uses 5 red beads for every 3 yellow beads. 5/3 = 1.666 Tim uses 9 red beads for every 6 yellow beads. 9/6 = 1.5 The ratio of red beads to yellow beads in Lisa’s necklace is not equivalent to the ratio in Tim’s necklace Question 3. Mitch scored 4 out of 5 on a quiz. Demetri scored 8 out of 10 on a quiz.", "is 84 bottles, which comes to about $2.13 per bottle, which is a nice discount. There’s also 2oz bottles on Amazon for ~$8 each. The Good: I’m a fan of these paints. Everything I’ve used works straight out of the bottle without thinning, with very few issues. Colors go on well, they don’t obscure any detail, and basically they did as they were expected to while being sprayed by a total newbie. That’s nice – there’s already enough hazards to learning to airbrush without “and also your paints hate you…” added to the mix. They also work fine with a brush, for quick touch-ups, color consistency, etc. I’m going to particularly call out their ranges of blues and grays. They’re both great – there are a huge amount of blues ranging from some deep, rich colors to “Power Sword Blue”, with clear steps between them, and no really strange shifts in the range. Their grays are the same – everything from “Nearly Black” to a pretty pale, almost plastic sprue gray comes in the kit. I actually managed to recreate “Unpainted Grey” which will come in handy if I ever actually make my “Da Grey Tide” Novelty Ork Army. Their reds and greens look promising as well, though I’ve had less experience with them. I believe their “Demonic", "of 24 spinning tops in four colors. She has the same number of tops in black, white, red, and green. How many tops of each color does she have? A. $$24 \\div 2$$ 2. Priya and her friend are decorating 24 wooden tops with paint. If each person is painting the same number of tops, how many tops is each person painting? B. $$12 \\div 2$$ 3. A store has 24 tops from around the world displayed in 6 boxes. Each box contains the same number of tops. How many tops are in each box? C. $$24 \\div 4$$ 4. Diego has 12 trompos that he wants to give as gifts. If he gives each friend 2 trompos, how many friends can get them as gifts? D. $$12 \\div 6$$ 5. Six friends are playing with 12 dreidels. If everyone is playing with the same number of dreidels, how many dreidels does each person have? E. $$24 \\div 6$$ ### Activity Synthesis • Select previously identified students to share. • Consider asking: “How do the numbers in the expression represent what is in the situation?” ## Activity 2: Cars in Boxes (10 minutes) ### Narrative The purpose of this activity is for students to understand that the same division expression can be used to represent both types of", "the two new masks, what Ali has called 'old' and 'clown'), but in the spreadsheet we used red to distinguish this part of the question from the others. 6) The answer to the second question is another $24$ combinations, making the total of $48$ combinations. 7) We used the same method for the third question, using yellow to denote the new information on the yellow cloak, as well as the combinations found using this new information. 8) The answer to the third question is another $24$ combinations, making the total of $72$ combinations. 9) For the last part (with addition of the shoes), we added the new information in green in our table. 10) Then we worked through the combinations as before, noticing the repeating patterns. 11) The answer to the fourth question is a grand total of $360$ combinations. Wow - you have worked very systematically, Ali - well done!", "A jar of honey butter contains 320g and costs £1.90. • A packet of cornflakes contains 300 g and costs £2.40. Promise makes 3.2 kg of the mixture, from which she can make 80 chocolate pyramids. She charges 70p for each pyramid and sells all 80 of them. (b) How much profit does she make? ### 7. GCSE Higher Pink paint is made by mixing red paint and white paint in the ratio 7 : 4 Mr Finklestein has 40 litres of red paint and 20 litres of white paint. What is the maximum amount of pink paint he can make? ### 8. GCSE Higher An Airbus A380 plane is used by Tran Tours to fly between Dubai and London and holds a maximum of 740 passengers. The seating has been configured so that 15% of the seats are business class and the rest are economy class. On one full flight, the ratio of males to females was 3:2. One eigth of the females were in business class. How many males were in economy class? ### 9. GCSE Higher The ratio a : b is equivalent to the ratio 7 : 3 and 8b = 5c Find the ratio a : c in its simplest form. ### 10. GCSE Higher At a summer camp the campers can choose", "Abigail Ahern Abigail Ahern Availability: In stock Price From: £2.95 Description Madison Grey, by Abigail Ahern, has the power to animate rooms. It transforms them and creates the coolest of vibes. It's grey but has undertones of green which change with the light. It's a color that connects you with the subtle currents of nature. Awe-inspiring, rich and joyful. Coverage is dependent upon the surface being painted, but the following can be used as a guide: 2.5L tin of emulsion/eggshell will cover approximately 25-35 square meters (one coat) 5L tin of emulsion/eggshell/floor will cover approximately 50-70 square meters (one coat) Typically, two coats of paint will be required to give the desired finish (although this is dependent upon the thickness of application and the surface being painted). Recoat After No No No MADISTONGR £3.00 Up to 300ml of paint 300ml of paint and over Not sure? Our sample pots might be just what you need • Sample Pot £2.95", "are 100 bars of blue and green soap in a box. The blue bars weigh $$\\text{50}$$ $$\\text{g}$$ per bar and the green bars $$\\text{40}$$ $$\\text{g}$$ per bar. The total mass of the soap in the box is $$\\text{4,66}$$ $$\\text{kg}$$. How many bars of green soap are in the box? \\begin{align*} x + y &= 100 ~(1)\\\\ 50x + 40y &= 4660 ~(2)\\\\ \\\\ &\\text{look at }(1) \\\\ x &= 100 - y ~(3)\\\\ (3) &\\text{ into } (2) \\\\ 50(100 - y) &+ 40y = \\text{4 660} \\\\ -50y &+ 40y = \\text{4 660} - \\text{5 000} \\\\ -10y &= -340 \\\\ y &= 34 \\\\ \\therefore 34 &\\text{ sacks of melie meal were bought} \\end{align*} Lisa has 170 beads. She has blue, red and purple beads each weighing $$\\text{13}$$ $$\\text{g}$$, $$\\text{4}$$ $$\\text{g}$$ and $$\\text{8}$$ $$\\text{g}$$ respectively. If there are twice as many red beads as there are blue beads and all the beads weigh $$\\text{1,216}$$ $$\\text{kg}$$, how many beads of each type does Lisa have? \\begin{align*} x + y + z &= 170 \\qquad(1)\\\\ 13x + 4y + 8z &= \\text{1 216} \\qquad(2)\\\\ y &= 2x \\qquad(3) \\\\ \\\\ (3) &\\text{ into } (1) \\\\ x + (2x) + z &= 170 \\\\ 3x + z &= 170\\\\ z &= 170 - 3x \\qquad(4)\\\\ (3) &\\text{ into", "3 5 ### Sample Output 3 15 For example, Snuke can paint the blocks as shown in the diagram below. There are $45$ red blocks and $30$ blue blocks, thus the difference is $9$.", "costs $35 per gallon can, blue paint costs$25 per gallon can, and green paint costs \\$23 per gallon can. (Submit for 2 points, so 8 points total). ## Plagiarism Checker Submit your documents and get free Plagiarism report Free Plagiarism Checker", "and a height of 14 meters. What is its volume? 15. The containers of icing for Tina’s cake decorator are cones. Each container has a radius of 2.4 inches and a height of 7 inches. If Tina buys containers of red, yellow, and blue icing, how much icing will she buy? Basic Nov 30, 2012 Aug 18, 2014", "has blue paint then she can mix green. I think this may solve your first issue. But I still find that the quoted statement does sounds illogical in common language. – Han de Bruijn Aug 6 at 18:46 • From Rutten's article: \"[1] If Brigitte has yellow paint and blue paint, then Brigitte can mix green, [2] Brigitte can mix green with yellow paint or Brigitte can mix green with blue paint.\" He suggests that it is somehow illogical that [2] should follow from [1] even though it may be a tautology. IMHO it may be counter-intuititive, but NOT illogical. Forgetting about the colour-wheel from art class for the moment, [1] does not strictly rule out different ways to mix green paint. – Dan Christensen Aug 7 at 5:40 Unfortunately, the author has made a little mistake while trying to reproduce the problem from the article by Emanuel Rutten. What's actually in there is this: $$\\begin{cases} \\mbox{[1*] } P \\wedge Q \\Rightarrow R, \\\\ \\mbox{[2*] } (P \\Rightarrow R) \\vee (Q \\Rightarrow R). \\end{cases}$$ Here $$P =$$ “Brigitte has yellow paint”, $$Q =$$ “Brigitte has blue paint” and $$R =$$ “Brigitte can mix green”. The calculus of propositional logic does render the argument form [1*]-[2*] logically valid. The above may be concisely written as: $$((P \\wedge Q) \\Rightarrow", "To buy the red bag, Lina would need $2.50 more than what she had. So Lina bought the blue bag. After that, she had$2 left. (a) What was the...", "accounted for. Just because there are thousands of house colors and paint combinations to choose from doesn’t mean they’re all available to you. Just one little smudge of paint will change the entire meaning and composition of a piece. There $3$ choices for body style; $8$ choices for exterior colors, and $2$ choices of interior color schemes:. Separate a long wall with a bookcase, shelving, or screen. When using the L’Oréal Paris Colorista Semi-Permanent Hair Color, you should leave the dye on your strands for 15 to 30 minutes, then rinse out and style as usual. Basically, it shows how many different possible subsets can be made from the larger set. Historic Paint Colors A primer for researching and choosing paint colors for the historic house. How many combinations of 2 can i make with 4 colors. If you have 3 bits, by the laws of combinations, you have 2•2•2 = 8 different values. Click on the link below to view our color combinations and start transforming your tabletop. Use 4 colors to color the segments. Parts four and five show two different tricks for working only one color at a time, yet still getting multi-color effects. Selesnya - Green and White: One of my favorite combinations, Selesnya is a blast to play. In how many ways can", "the length of the component in the drawing? The length of the component in the drawing is 25 cm Explanation: Write the proportion relating the model length to the actual length. the scale has inches for the actual length units so you must use 12.5 feet = 15 inches in the proportion model/actual = 1 cm/60 in = x cm/150 in 60x = 150 x = 150/6 = 25 Question 15. A specific shade of green glaze is made of 5 parts blue glaze to 3 parts yellow glaze. A glaze mixture contains 25 quarts of blue glaze and 9 quarts of yellow glaze. How can you fix the mixture to make the specific shade of green glaze? 25 parts blue to 15 parts yellow Explanation: 25 blue to 9 yellow The rate is not equal to the specific rate, 5 parts blue to 3 parts yellow 25 parts blue to 15 parts yellow ### Ratios and Proportions Cumulative Practice Question 1. The school store sells 4 pencils for $0.80. What is the unit cost of a pencil? A.$0.20 B. $0.80 C.$3.20 D. $5.00 Answer: A.$0.20 Explanation: The school store sells 4 pencils for \\$0.80 = 0.80 : 4 = 0.80/4 : 1 = 0.2 : 1 Question 2. What is the simplified form of the expression? 3x", "# Which artists apply colors in small brush strokes, making their paintings lively depictions when seen from a distance? ###### Question: Which artists apply colors in small brush strokes, making their paintings lively depictions when seen from a distance? ### Hi all, will for sure appreciate some help :) Hi all, will for sure appreciate some help :)... ### What does the cell theory explain? What does the cell theory explain?... ### A yogurt shop has vanilla, chocolate, strawberry, blueberry, and raspberry flavors served in small, medium, or large cups. How many different yogurt choices are possible? Question 1 options: 30 15 8 2 A yogurt shop has vanilla, chocolate, strawberry, blueberry, and raspberry flavors served in small, medium, or large cups. How many different yogurt choices are possible? Question 1 options: 30 15 8 2... ### Pls help Given that the amplifier cost was $235 and Jared still owes his mom$73, which of the following shows how much money Jared has already paid his mother? 73 + 18 = 91 235 - 18 = 217 235 - 73 = 162 pls help Given that the amplifier cost was $235 and Jared still owes his mom$73, which of the following shows how much money Jared has already paid his mother? 73 + 18 = 91 235 - 18", "+0 # irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons 0 296 3 irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons left.hiw many blue balloons does irene have at first? Guest Jul 6, 2015 #1 +26322 +10 Perhaps I've gone wrong somewhere or misinterpreted part of the question! Edit: Of course, I made a numerical mistake towards the end. I should have had 13b/12 = 764 (giving b = 705.231), not 13b/4 = 764. Never-the-less, this still doesn't result in integer numbers of balloons! . Alan Jul 6, 2015 Sort: #1 +26322 +10 Perhaps I've gone wrong somewhere or misinterpreted part of the question! Edit: Of course, I made a numerical mistake towards the end. I should have had 13b/12 = 764 (giving b = 705.231), not 13b/4 = 764. Never-the-less, this still doesn't result in integer numbers of balloons! . Alan Jul 6, 2015 #2 +18712 +5 irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922" ]

[ "A jar of honey butter contains 320g and costs £1.90. • A packet of cornflakes contains 300 g and costs £2.40. Promise makes 3.2 kg of the mixture, from which she can make 80 chocolate pyramids. She charges 70p for each pyramid and sells all 80 of them. (b) How much profit does she make? ### 7. GCSE Higher Pink paint is made by mixing red paint and white paint in the ratio 7 : 4 Mr Finklestein has 40 litres of red paint and 20 litres of white paint. What is the maximum amount of pink paint he can make? ### 8. GCSE Higher An Airbus A380 plane is used by Tran Tours to fly between Dubai and London and holds a maximum of 740 passengers. The seating has been configured so that 15% of the seats are business class and the rest are economy class. On one full flight, the ratio of males to females was 3:2. One eigth of the females were in business class. How many males were in economy class? ### 9. GCSE Higher The ratio a : b is equivalent to the ratio 7 : 3 and 8b = 5c Find the ratio a : c in its simplest form. ### 10. GCSE Higher At a summer camp the campers can choose", "Abigail Ahern Abigail Ahern Availability: In stock Price From: £2.95 Description Madison Grey, by Abigail Ahern, has the power to animate rooms. It transforms them and creates the coolest of vibes. It's grey but has undertones of green which change with the light. It's a color that connects you with the subtle currents of nature. Awe-inspiring, rich and joyful. Coverage is dependent upon the surface being painted, but the following can be used as a guide: 2.5L tin of emulsion/eggshell will cover approximately 25-35 square meters (one coat) 5L tin of emulsion/eggshell/floor will cover approximately 50-70 square meters (one coat) Typically, two coats of paint will be required to give the desired finish (although this is dependent upon the thickness of application and the surface being painted). Recoat After No No No MADISTONGR £3.00 Up to 300ml of paint 300ml of paint and over Not sure? Our sample pots might be just what you need • Sample Pot £2.95", "is 84 bottles, which comes to about $2.13 per bottle, which is a nice discount. There’s also 2oz bottles on Amazon for ~$8 each. The Good: I’m a fan of these paints. Everything I’ve used works straight out of the bottle without thinning, with very few issues. Colors go on well, they don’t obscure any detail, and basically they did as they were expected to while being sprayed by a total newbie. That’s nice – there’s already enough hazards to learning to airbrush without “and also your paints hate you…” added to the mix. They also work fine with a brush, for quick touch-ups, color consistency, etc. I’m going to particularly call out their ranges of blues and grays. They’re both great – there are a huge amount of blues ranging from some deep, rich colors to “Power Sword Blue”, with clear steps between them, and no really strange shifts in the range. Their grays are the same – everything from “Nearly Black” to a pretty pale, almost plastic sprue gray comes in the kit. I actually managed to recreate “Unpainted Grey” which will come in handy if I ever actually make my “Da Grey Tide” Novelty Ork Army. Their reds and greens look promising as well, though I’ve had less experience with them. I believe their “Demonic", "For example, Snuke can paint the blocks as shown in the diagram below. There are $45$ red blocks and $30$ blue blocks, thus the difference is $9$.", "content in red-to-green ratios as depicted in paintings by great masters, Atmos. Chem. Phys., 14, 2987–3015, https://doi.org/10.5194/acp-14-2987-2014, 2014.", "circular necklace with $$2013$$ beads. Each bead can be painted either white or green. A painting of the necklace is called good if among any $$21$$ successive beads there is at least one green bead. Prove that the number of good paintings of the necklace is odd. (Two paintings that differ on some beads, but can be obtained from each other by rotating or flipping the necklace, are counted as different paintings. )", "2$$\\frac{3}{4}$$ quarts to cover each wall. If each wall is painted only one color, how many walls will be blue and how many will be white? How much paint will be left over? Angela and Ryou are painting a room. Angela has 2$$\\frac{1}{2}$$ gallons of blue paint. 2$$\\frac{1}{2}$$ × 4 = 10 quarts of blue paint. It will take 2$$\\frac{3}{4}$$ quarts to cover each wall. Blue colored walls = 10 × 2$$\\frac{3}{4}$$ = 3 walls with 7/11 quarts of paint remaining White colored walls = 5 × 2$$\\frac{3}{4}$$ = 1 wall with 9/11 quarts of paint remaining", "+0 # Probability question +2 109 2 +106 Kaylee has 2 gallons each of red and blue paint, and 1 gallon each of orange and green paint. She paints her bedroom with these jars of paint. If she uses $\\frac{2}{3}$ of the red paint, $\\frac{1}{4}$ of the blue paint, $\\frac{1}{2}$ of the orange paint, and $\\frac{5}{7}$ of the green paint, how much paint (in gallons) does Kaylee have left over? Apr 4, 2021 #1 +1 Kaylee has 3 55/84 gallons left over. Apr 4, 2021 #2 +106 +2 Thanks. Sorry I didn't mean to name this \"probability\" lol TealSeal Apr 5, 2021", "white tiles she buys. In order to work out the number of white tiles, we need to work out the total number of tiles she buys. The ratio is $2$ parts blue to $13$ parts white. If Lucy buys $16$ blue tiles, this is $8$ times more than the figure for blue tiles in the ratio $(16\\div2=8)$. If the number of blue tiles she buys is $8$ times more than the blue tiles figure given in the ratio, then the number of white tiles she buys must also be $8$ times more than the white tiles figure in the ratio. Therefore, the number of white tiles she buys is: $13\\times8 = 104$ white tiles Now that we know how many tiles of each colour she has bought, we can calculate the total cost of the tiles. The cost of $16$ blue tiles is: $16\\times2.80 = \\pounds44.80$ The cost of $104$ white tiles is: $104\\times2.35 =\\pounds244.40$ Therefore, Lucy’s total spend on tiles is: $44.80+244.40=\\pounds289.20$ The first thing we need to do is to deduct the $20 \\%$ spent on the magazine subscription so that we can work out how much of his allowance Steve has left over. $20 \\%$ of $£200$ can be calculated as follows: $0.2 \\times \\pounds 200 = \\pounds 40$ You may prefer to calculate the", "6. $\\text-5(3x+1)$ ### Problem 4 (from Unit 6, Lesson 14) The school marching band has a budget of up to \\$750 to cover 15 new uniforms and competition fees that total \\$300. How much can they spend for one uniform? 1. Write an inequality to represent this situation. 2. Solve the inequality and describe what it means in the situation. ### Problem 5 (from Unit 6, Lesson 16) Solve the inequality that represents each story. Then interpret what the solution means in the story. 1. For every \\$9 that Elena earns, she gives$x$dollars to charity. This happens 7 times this month. Elena wants to be sure she keeps at least \\$42 from this month’s earnings. $7(9-x) \\geq 42$ 2. Lin buys a candle that is 9 inches tall and burns down $x$ inches per minute. She wants to let the candle burn for 7 minutes until it is less than 6 inches tall. $9 - 7x < 6$ ### Problem 6 (from Unit 4, Lesson 3) A certain shade of blue paint is made by mixing $1\\frac12$ quarts of blue paint with 5 quarts of white paint. If you need a total of 16.25 gallons of this shade of blue paint, how much of each color should you mix? ## Lesson 22 ### Problem 1 Jada says, “I", "in your profile area as well as via email. _________________ Kudos [?]: 273 [0], given: 0 Manager Joined: 23 Jul 2015 Posts: 147 Kudos [?]: 38 [0], given: 29 GMAT 1: 730 Q50 V40 Diane find 2 and a half cans of paint are just enough to [#permalink] ### Show Tags 09 Aug 2017, 04:21 5/2 cans paint 1/3 of the room 1 can paints 2/15 of the room --> to paint a room she needs 7.5 cans She'll need 5 more cans to paint remaining of the room and another 7.5 cans to paint the second room Total 12.5 cans (Choice D) Kudos [?]: 38 [0], given: 29 Director Joined: 22 May 2016 Posts: 819 Kudos [?]: 266 [0], given: 552 Diane find 2 and a half cans of paint are just enough to [#permalink] ### Show Tags 09 Aug 2017, 14:27 antoxavier wrote: Diane find 2 and a half cans of paint are just enough to paint one third of her room. How many more cans of paint will she need to finish her room and paint a second room of the same size? A. 5 B. 7 and a half C. 10 D. 12 and a half E. 15 Proportion with multiplier approach Remaining work = $$\\frac{5}{3}$$ Let x = amount of paint needed $$\\frac{\\frac{5}{2}}{\\frac{1}{3}}$$", "the length of the component in the drawing? The length of the component in the drawing is 25 cm Explanation: Write the proportion relating the model length to the actual length. the scale has inches for the actual length units so you must use 12.5 feet = 15 inches in the proportion model/actual = 1 cm/60 in = x cm/150 in 60x = 150 x = 150/6 = 25 Question 15. A specific shade of green glaze is made of 5 parts blue glaze to 3 parts yellow glaze. A glaze mixture contains 25 quarts of blue glaze and 9 quarts of yellow glaze. How can you fix the mixture to make the specific shade of green glaze? 25 parts blue to 15 parts yellow Explanation: 25 blue to 9 yellow The rate is not equal to the specific rate, 5 parts blue to 3 parts yellow 25 parts blue to 15 parts yellow ### Ratios and Proportions Cumulative Practice Question 1. The school store sells 4 pencils for $0.80. What is the unit cost of a pencil? A.$0.20 B. $0.80 C.$3.20 D. $5.00 Answer: A.$0.20 Explanation: The school store sells 4 pencils for \\$0.80 = 0.80 : 4 = 0.80/4 : 1 = 0.2 : 1 Question 2. What is the simplified form of the expression? 3x", "# Agreengrocer a cabbage at a profit of 37.5% on the price he pays for it. what is the ratio of the cost price to selling price? ### Other questions on the subject: Math Math, 19.08.2019 00:00, abhishek7500 In a 30-60-90 triangle , side opposite to 60° is hypotenuse​ Math, 19.08.2019 06:00, jesuiskhushi Awooden box (open at the top) of thickness 0.5 cm, length 21 cm, width 11 cm and height 6 cm is painted on the inside. the expenses of painting are ` 140. what is the rate of painting per square centimeter?", "the same number of arrows and all the arrows hit the target, and the results of every two boys differed in on • BW-BS balls Adam has a full box of balls that are large or small, black or white. Ratio of large and small balls is 5:3. Within the large balls the ratio of the black to white is 1:2 and between small balls the ratio of the black to white is 1:8 What is the ratio of • The painter In order for the painter to get the desired color, he must mix green and yellow in a ratio of 4: 7. If it has 28 l of green color, how many liters of yellow color should he add? How many liters of mixed color does he get? • Florist's The florist got 72 white and 90 red roses. How many bouquets can bind from all these roses when each bouquets should have the same number of white and red roses? Up to 48 rooms, some of which are triple and some quadruple, accommodated 173 people so that all beds are occupied. How many triple and how many quadruple rooms were there? • Candy and boxes We have some number of candy and empty boxes. When we put candies in boxes of ten, there will be", "3 5 ### Sample Output 3 15 For example, Snuke can paint the blocks as shown in the diagram below. There are $45$ red blocks and $30$ blue blocks, thus the difference is $9$.", "are 100 bars of blue and green soap in a box. The blue bars weigh $$\\text{50}$$ $$\\text{g}$$ per bar and the green bars $$\\text{40}$$ $$\\text{g}$$ per bar. The total mass of the soap in the box is $$\\text{4,66}$$ $$\\text{kg}$$. How many bars of green soap are in the box? \\begin{align*} x + y &= 100 ~(1)\\\\ 50x + 40y &= 4660 ~(2)\\\\ \\\\ &\\text{look at }(1) \\\\ x &= 100 - y ~(3)\\\\ (3) &\\text{ into } (2) \\\\ 50(100 - y) &+ 40y = \\text{4 660} \\\\ -50y &+ 40y = \\text{4 660} - \\text{5 000} \\\\ -10y &= -340 \\\\ y &= 34 \\\\ \\therefore 34 &\\text{ sacks of melie meal were bought} \\end{align*} Lisa has 170 beads. She has blue, red and purple beads each weighing $$\\text{13}$$ $$\\text{g}$$, $$\\text{4}$$ $$\\text{g}$$ and $$\\text{8}$$ $$\\text{g}$$ respectively. If there are twice as many red beads as there are blue beads and all the beads weigh $$\\text{1,216}$$ $$\\text{kg}$$, how many beads of each type does Lisa have? \\begin{align*} x + y + z &= 170 \\qquad(1)\\\\ 13x + 4y + 8z &= \\text{1 216} \\qquad(2)\\\\ y &= 2x \\qquad(3) \\\\ \\\\ (3) &\\text{ into } (1) \\\\ x + (2x) + z &= 170 \\\\ 3x + z &= 170\\\\ z &= 170 - 3x \\qquad(4)\\\\ (3) &\\text{ into", "in half as much bleach as she ought to have. The total solution consists of 27 mL. How much bleach did Alexandra put into the solution? A) 1.5 B) 3 C) 4.5 D) 5 E) 6 Last edited by Engr2012 on 20 Sep 2015, 04:39, edited 1 time in total. Formatted the question Manager Joined: 10 Dec 2012 Posts: 52 Followers: 0 Kudos [?]: 7 [0], given: 37 Re: Alexandra needs to mix cleaning solution in the following ratio [#permalink] ### Show Tags 19 Sep 2015, 23:52 jaspreets wrote: Alexandra needs to mix cleaning solution in the following ratio: 1 part bleach for every 4 parts water. When mixing the solution, Alexandra makes a mistake and mixes in half as much bleach as she ought to have. The total solution consists of 27 mL. How much bleach did Alexandra put into the solution? 1)1.5 2)3 3)4.5 4)5 5)6 Thank you. Manager Joined: 18 Aug 2014 Posts: 132 Location: Hong Kong Schools: Mannheim Followers: 1 Kudos [?]: 66 [0], given: 36 Re: Alexandra needs to mix cleaning solution in the following ratio [#permalink] ### Show Tags 20 Sep 2015, 02:56 jaspreets wrote: jaspreets wrote: Alexandra needs to mix cleaning solution in the following ratio: 1 part bleach for every 4 parts water. When mixing the solution, Alexandra makes a", "Compared to Container A, how much more paint does Brian need to paint Container B? (A) 3 times (B) 6 times (C) 9 times (D) 27 times Question 30. Britney draws an ant using a scale factor of 5 : 1. If the actual ant is 25 mm long, how long is the ant in Britney’s drawing in centimetres? (A) 125 cm (B) 12.5 cm (C) 5 cm (D) 0.5 cm", "with! To be honest I didn ’ t be so difficult consisting of 1/2 the. The next color of your room with these purple paint contains 1/3 cup of red paint and. Or maybe something in the how to make purple paint Red/Purple Tones that contains red that carries a strong bias of blue and! Strawberries, cherries, huckleberries, blackberries, elderberries, black currants, red currants and raspberries my... Amethyst and vibrant orchid, and see how it looks adding a bit at a time until have. A paint brush or painting knife to create purple in handy both from Golden Acrylics both! Gradually adding black paint until the color of the purple toward whichever color was dominant blue. That it does is make grey a look when we mix them with a blue desired shade of purple no. Gouache, and watercolor a dollop consisting of 1/2 + 1/3 = 5/6 a. For the next color of the remaining yellow wall is brightly colored, apply a primer or white color ensure. Explain how to mix green watercolor paint to the hue have said you ca n't make blue from because... And getting nowhere: / 8 red+7 Blue= nothing: / there are purple berries plus, get the of... Paint, we need 20 $\\div$ 5/6 = 24 batches formula which made my so.", "has blue paint then she can mix green. I think this may solve your first issue. But I still find that the quoted statement does sounds illogical in common language. – Han de Bruijn Aug 6 at 18:46 • From Rutten's article: \"[1] If Brigitte has yellow paint and blue paint, then Brigitte can mix green, [2] Brigitte can mix green with yellow paint or Brigitte can mix green with blue paint.\" He suggests that it is somehow illogical that [2] should follow from [1] even though it may be a tautology. IMHO it may be counter-intuititive, but NOT illogical. Forgetting about the colour-wheel from art class for the moment, [1] does not strictly rule out different ways to mix green paint. – Dan Christensen Aug 7 at 5:40 Unfortunately, the author has made a little mistake while trying to reproduce the problem from the article by Emanuel Rutten. What's actually in there is this: $$\\begin{cases} \\mbox{[1*] } P \\wedge Q \\Rightarrow R, \\\\ \\mbox{[2*] } (P \\Rightarrow R) \\vee (Q \\Rightarrow R). \\end{cases}$$ Here $$P =$$ “Brigitte has yellow paint”, $$Q =$$ “Brigitte has blue paint” and $$R =$$ “Brigitte can mix green”. The calculus of propositional logic does render the argument form [1*]-[2*] logically valid. The above may be concisely written as: $$((P \\wedge Q) \\Rightarrow" ]

[ "the story. 1. For every $9 that Elena earns, she gives $$x$$ dollars to charity. This happens 7 times this month. Elena wants to be sure she keeps at least$42 from this month’s earnings. $$7(9-x) \\geq 42$$ 2. Lin buys a candle that is 9 inches tall and burns down $$x$$ inches per minute. She wants to let the candle burn for 7 minutes until it is less than 6 inches tall. $$9 - 7x < 6$$ (From Unit 6, Lesson 16.) ### Problem 6 A certain shade of blue paint is made by mixing $$1\\frac12$$ quarts of blue paint with 5 quarts of white paint. If you need a total of 16.25 gallons of this shade of blue paint, how much of each color should you mix? (From Unit 4, Lesson 3.)", "For example, Snuke can paint the blocks as shown in the diagram below. There are $45$ red blocks and $30$ blue blocks, thus the difference is $9$.", "of 24 spinning tops in four colors. She has the same number of tops in black, white, red, and green. How many tops of each color does she have? A. $$24 \\div 2$$ 2. Priya and her friend are decorating 24 wooden tops with paint. If each person is painting the same number of tops, how many tops is each person painting? B. $$12 \\div 2$$ 3. A store has 24 tops from around the world displayed in 6 boxes. Each box contains the same number of tops. How many tops are in each box? C. $$24 \\div 4$$ 4. Diego has 12 trompos that he wants to give as gifts. If he gives each friend 2 trompos, how many friends can get them as gifts? D. $$12 \\div 6$$ 5. Six friends are playing with 12 dreidels. If everyone is playing with the same number of dreidels, how many dreidels does each person have? E. $$24 \\div 6$$ ### Activity Synthesis • Select previously identified students to share. • Consider asking: “How do the numbers in the expression represent what is in the situation?” ## Activity 2: Cars in Boxes (10 minutes) ### Narrative The purpose of this activity is for students to understand that the same division expression can be used to represent both types of", "is 84 bottles, which comes to about $2.13 per bottle, which is a nice discount. There’s also 2oz bottles on Amazon for ~$8 each. The Good: I’m a fan of these paints. Everything I’ve used works straight out of the bottle without thinning, with very few issues. Colors go on well, they don’t obscure any detail, and basically they did as they were expected to while being sprayed by a total newbie. That’s nice – there’s already enough hazards to learning to airbrush without “and also your paints hate you…” added to the mix. They also work fine with a brush, for quick touch-ups, color consistency, etc. I’m going to particularly call out their ranges of blues and grays. They’re both great – there are a huge amount of blues ranging from some deep, rich colors to “Power Sword Blue”, with clear steps between them, and no really strange shifts in the range. Their grays are the same – everything from “Nearly Black” to a pretty pale, almost plastic sprue gray comes in the kit. I actually managed to recreate “Unpainted Grey” which will come in handy if I ever actually make my “Da Grey Tide” Novelty Ork Army. Their reds and greens look promising as well, though I’ve had less experience with them. I believe their “Demonic", "2$$\\frac{3}{4}$$ quarts to cover each wall. If each wall is painted only one color, how many walls will be blue and how many will be white? How much paint will be left over? Angela and Ryou are painting a room. Angela has 2$$\\frac{1}{2}$$ gallons of blue paint. 2$$\\frac{1}{2}$$ × 4 = 10 quarts of blue paint. It will take 2$$\\frac{3}{4}$$ quarts to cover each wall. Blue colored walls = 10 × 2$$\\frac{3}{4}$$ = 3 walls with 7/11 quarts of paint remaining White colored walls = 5 × 2$$\\frac{3}{4}$$ = 1 wall with 9/11 quarts of paint remaining", "+0 # Probability question +2 109 2 +106 Kaylee has 2 gallons each of red and blue paint, and 1 gallon each of orange and green paint. She paints her bedroom with these jars of paint. If she uses $\\frac{2}{3}$ of the red paint, $\\frac{1}{4}$ of the blue paint, $\\frac{1}{2}$ of the orange paint, and $\\frac{5}{7}$ of the green paint, how much paint (in gallons) does Kaylee have left over? Apr 4, 2021 #1 +1 Kaylee has 3 55/84 gallons left over. Apr 4, 2021 #2 +106 +2 Thanks. Sorry I didn't mean to name this \"probability\" lol TealSeal Apr 5, 2021", "Abigail Ahern Abigail Ahern Availability: In stock Price From: £2.95 Description Madison Grey, by Abigail Ahern, has the power to animate rooms. It transforms them and creates the coolest of vibes. It's grey but has undertones of green which change with the light. It's a color that connects you with the subtle currents of nature. Awe-inspiring, rich and joyful. Coverage is dependent upon the surface being painted, but the following can be used as a guide: 2.5L tin of emulsion/eggshell will cover approximately 25-35 square meters (one coat) 5L tin of emulsion/eggshell/floor will cover approximately 50-70 square meters (one coat) Typically, two coats of paint will be required to give the desired finish (although this is dependent upon the thickness of application and the surface being painted). Recoat After No No No MADISTONGR £3.00 Up to 300ml of paint 300ml of paint and over Not sure? Our sample pots might be just what you need • Sample Pot £2.95", "A jar of honey butter contains 320g and costs £1.90. • A packet of cornflakes contains 300 g and costs £2.40. Promise makes 3.2 kg of the mixture, from which she can make 80 chocolate pyramids. She charges 70p for each pyramid and sells all 80 of them. (b) How much profit does she make? ### 7. GCSE Higher Pink paint is made by mixing red paint and white paint in the ratio 7 : 4 Mr Finklestein has 40 litres of red paint and 20 litres of white paint. What is the maximum amount of pink paint he can make? ### 8. GCSE Higher An Airbus A380 plane is used by Tran Tours to fly between Dubai and London and holds a maximum of 740 passengers. The seating has been configured so that 15% of the seats are business class and the rest are economy class. On one full flight, the ratio of males to females was 3:2. One eigth of the females were in business class. How many males were in economy class? ### 9. GCSE Higher The ratio a : b is equivalent to the ratio 7 : 3 and 8b = 5c Find the ratio a : c in its simplest form. ### 10. GCSE Higher At a summer camp the campers can choose", "the length of the component in the drawing? The length of the component in the drawing is 25 cm Explanation: Write the proportion relating the model length to the actual length. the scale has inches for the actual length units so you must use 12.5 feet = 15 inches in the proportion model/actual = 1 cm/60 in = x cm/150 in 60x = 150 x = 150/6 = 25 Question 15. A specific shade of green glaze is made of 5 parts blue glaze to 3 parts yellow glaze. A glaze mixture contains 25 quarts of blue glaze and 9 quarts of yellow glaze. How can you fix the mixture to make the specific shade of green glaze? 25 parts blue to 15 parts yellow Explanation: 25 blue to 9 yellow The rate is not equal to the specific rate, 5 parts blue to 3 parts yellow 25 parts blue to 15 parts yellow ### Ratios and Proportions Cumulative Practice Question 1. The school store sells 4 pencils for $0.80. What is the unit cost of a pencil? A.$0.20 B. $0.80 C.$3.20 D. $5.00 Answer: A.$0.20 Explanation: The school store sells 4 pencils for \\$0.80 = 0.80 : 4 = 0.80/4 : 1 = 0.2 : 1 Question 2. What is the simplified form of the expression? 3x", "has blue paint then she can mix green. I think this may solve your first issue. But I still find that the quoted statement does sounds illogical in common language. – Han de Bruijn Aug 6 at 18:46 • From Rutten's article: \"[1] If Brigitte has yellow paint and blue paint, then Brigitte can mix green, [2] Brigitte can mix green with yellow paint or Brigitte can mix green with blue paint.\" He suggests that it is somehow illogical that [2] should follow from [1] even though it may be a tautology. IMHO it may be counter-intuititive, but NOT illogical. Forgetting about the colour-wheel from art class for the moment, [1] does not strictly rule out different ways to mix green paint. – Dan Christensen Aug 7 at 5:40 Unfortunately, the author has made a little mistake while trying to reproduce the problem from the article by Emanuel Rutten. What's actually in there is this: $$\\begin{cases} \\mbox{[1*] } P \\wedge Q \\Rightarrow R, \\\\ \\mbox{[2*] } (P \\Rightarrow R) \\vee (Q \\Rightarrow R). \\end{cases}$$ Here $$P =$$ “Brigitte has yellow paint”, $$Q =$$ “Brigitte has blue paint” and $$R =$$ “Brigitte can mix green”. The calculus of propositional logic does render the argument form [1*]-[2*] logically valid. The above may be concisely written as: $$((P \\wedge Q) \\Rightarrow", "circular necklace with $$2013$$ beads. Each bead can be painted either white or green. A painting of the necklace is called good if among any $$21$$ successive beads there is at least one green bead. Prove that the number of good paintings of the necklace is odd. (Two paintings that differ on some beads, but can be obtained from each other by rotating or flipping the necklace, are counted as different paintings. )", "6. $\\text-5(3x+1)$ ### Problem 4 (from Unit 6, Lesson 14) The school marching band has a budget of up to \\$750 to cover 15 new uniforms and competition fees that total \\$300. How much can they spend for one uniform? 1. Write an inequality to represent this situation. 2. Solve the inequality and describe what it means in the situation. ### Problem 5 (from Unit 6, Lesson 16) Solve the inequality that represents each story. Then interpret what the solution means in the story. 1. For every \\$9 that Elena earns, she gives$x$dollars to charity. This happens 7 times this month. Elena wants to be sure she keeps at least \\$42 from this month’s earnings. $7(9-x) \\geq 42$ 2. Lin buys a candle that is 9 inches tall and burns down $x$ inches per minute. She wants to let the candle burn for 7 minutes until it is less than 6 inches tall. $9 - 7x < 6$ ### Problem 6 (from Unit 4, Lesson 3) A certain shade of blue paint is made by mixing $1\\frac12$ quarts of blue paint with 5 quarts of white paint. If you need a total of 16.25 gallons of this shade of blue paint, how much of each color should you mix? ## Lesson 22 ### Problem 1 Jada says, “I", "are 100 bars of blue and green soap in a box. The blue bars weigh $$\\text{50}$$ $$\\text{g}$$ per bar and the green bars $$\\text{40}$$ $$\\text{g}$$ per bar. The total mass of the soap in the box is $$\\text{4,66}$$ $$\\text{kg}$$. How many bars of green soap are in the box? \\begin{align*} x + y &= 100 ~(1)\\\\ 50x + 40y &= 4660 ~(2)\\\\ \\\\ &\\text{look at }(1) \\\\ x &= 100 - y ~(3)\\\\ (3) &\\text{ into } (2) \\\\ 50(100 - y) &+ 40y = \\text{4 660} \\\\ -50y &+ 40y = \\text{4 660} - \\text{5 000} \\\\ -10y &= -340 \\\\ y &= 34 \\\\ \\therefore 34 &\\text{ sacks of melie meal were bought} \\end{align*} Lisa has 170 beads. She has blue, red and purple beads each weighing $$\\text{13}$$ $$\\text{g}$$, $$\\text{4}$$ $$\\text{g}$$ and $$\\text{8}$$ $$\\text{g}$$ respectively. If there are twice as many red beads as there are blue beads and all the beads weigh $$\\text{1,216}$$ $$\\text{kg}$$, how many beads of each type does Lisa have? \\begin{align*} x + y + z &= 170 \\qquad(1)\\\\ 13x + 4y + 8z &= \\text{1 216} \\qquad(2)\\\\ y &= 2x \\qquad(3) \\\\ \\\\ (3) &\\text{ into } (1) \\\\ x + (2x) + z &= 170 \\\\ 3x + z &= 170\\\\ z &= 170 - 3x \\qquad(4)\\\\ (3) &\\text{ into", "+0 # irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons 0 296 3 irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons left.hiw many blue balloons does irene have at first? Guest Jul 6, 2015 #1 +26322 +10 Perhaps I've gone wrong somewhere or misinterpreted part of the question! Edit: Of course, I made a numerical mistake towards the end. I should have had 13b/12 = 764 (giving b = 705.231), not 13b/4 = 764. Never-the-less, this still doesn't result in integer numbers of balloons! . Alan Jul 6, 2015 Sort: #1 +26322 +10 Perhaps I've gone wrong somewhere or misinterpreted part of the question! Edit: Of course, I made a numerical mistake towards the end. I should have had 13b/12 = 764 (giving b = 705.231), not 13b/4 = 764. Never-the-less, this still doesn't result in integer numbers of balloons! . Alan Jul 6, 2015 #2 +18712 +5 irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922", "accounted for. Just because there are thousands of house colors and paint combinations to choose from doesn’t mean they’re all available to you. Just one little smudge of paint will change the entire meaning and composition of a piece. There $3$ choices for body style; $8$ choices for exterior colors, and $2$ choices of interior color schemes:. Separate a long wall with a bookcase, shelving, or screen. When using the L’Oréal Paris Colorista Semi-Permanent Hair Color, you should leave the dye on your strands for 15 to 30 minutes, then rinse out and style as usual. Basically, it shows how many different possible subsets can be made from the larger set. Historic Paint Colors A primer for researching and choosing paint colors for the historic house. How many combinations of 2 can i make with 4 colors. If you have 3 bits, by the laws of combinations, you have 2•2•2 = 8 different values. Click on the link below to view our color combinations and start transforming your tabletop. Use 4 colors to color the segments. Parts four and five show two different tricks for working only one color at a time, yet still getting multi-color effects. Selesnya - Green and White: One of my favorite combinations, Selesnya is a blast to play. In how many ways can", "the two new masks, what Ali has called 'old' and 'clown'), but in the spreadsheet we used red to distinguish this part of the question from the others. 6) The answer to the second question is another $24$ combinations, making the total of $48$ combinations. 7) We used the same method for the third question, using yellow to denote the new information on the yellow cloak, as well as the combinations found using this new information. 8) The answer to the third question is another $24$ combinations, making the total of $72$ combinations. 9) For the last part (with addition of the shoes), we added the new information in green in our table. 10) Then we worked through the combinations as before, noticing the repeating patterns. 11) The answer to the fourth question is a grand total of $360$ combinations. Wow - you have worked very systematically, Ali - well done!", "3 5 ### Sample Output 3 15 For example, Snuke can paint the blocks as shown in the diagram below. There are $45$ red blocks and $30$ blue blocks, thus the difference is $9$.", "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", "in half as much bleach as she ought to have. The total solution consists of 27 mL. How much bleach did Alexandra put into the solution? A) 1.5 B) 3 C) 4.5 D) 5 E) 6 Last edited by Engr2012 on 20 Sep 2015, 04:39, edited 1 time in total. Formatted the question Manager Joined: 10 Dec 2012 Posts: 52 Followers: 0 Kudos [?]: 7 [0], given: 37 Re: Alexandra needs to mix cleaning solution in the following ratio [#permalink] ### Show Tags 19 Sep 2015, 23:52 jaspreets wrote: Alexandra needs to mix cleaning solution in the following ratio: 1 part bleach for every 4 parts water. When mixing the solution, Alexandra makes a mistake and mixes in half as much bleach as she ought to have. The total solution consists of 27 mL. How much bleach did Alexandra put into the solution? 1)1.5 2)3 3)4.5 4)5 5)6 Thank you. Manager Joined: 18 Aug 2014 Posts: 132 Location: Hong Kong Schools: Mannheim Followers: 1 Kudos [?]: 66 [0], given: 36 Re: Alexandra needs to mix cleaning solution in the following ratio [#permalink] ### Show Tags 20 Sep 2015, 02:56 jaspreets wrote: jaspreets wrote: Alexandra needs to mix cleaning solution in the following ratio: 1 part bleach for every 4 parts water. When mixing the solution, Alexandra makes a", "buy? _____ posters 3 posters Explanation: 1 Poster Board = $0.75. Isabella spends$2.25 on poster boards. $2.25/$0.75 = 3 Question 25. B Isabella spends $4.87 on paintbrushes and paint. How many of each item does she buy? Explain how you found your answer. _______ paint brushes _______ jars of paint Answer: 2 paint brushes 3 jars of paint Explanation: paintbrushes =$0.95 Paint = $0.99 If she buys 2 paint brushes and 2 paints, she spent$1.9 for paintbrushes and $1.98 for 2 paints. The remaining amount is$0.99. So, she can buy one more paint with them. So, she can buy 2 paint brushes and 3 jars of paint. Question 25. C. Isabella spends less than $14.00 for glass beads, paintbrushes, poster board, and paint. She spends$1.68 on beads and $3.96 on paint. She buys more than 3 poster boards and more than 3 paintbrushes. Find how many ounces of glass beads and how many jars of paint she buys. Then, suggest the number of poster boards and paintbrushes she might buy for the total spent. Type below: _________ Answer: Isabella spends less than$14.00 for glass beads, paintbrushes, poster board, and paint. She spends $1.68 on beads and$3.96 on paint. Each beads = $0.28$1.68/$0.28 = 6 beads Each paint =$0.99 $3.96/$0.99 = 4 paints $14 – ($1.68 + $3.96) =$8.36." ]

[ "a shaded region with largest area? $[asy]/* AMC8 2003 #22 Problem */ size(3inch, 2inch); unitsize(1cm); pen outline = black+linewidth(1); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline); filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline); filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1)); filldraw(Circle((1,1), 1), white, outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label(\"A\", (1, 2), N); label(\"B\", (4, 2), N); label(\"C\", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label(\"2 cm\", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label(\"2 cm\", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label(\"2 cm\", (7, -.5)); draw((6,1)--(6,-.5), linetype(\"4 4\")); draw((8,1)--(8,-.5), linetype(\"4 4\"));[/asy]$ $\\textbf{(A)}\\ \\text{A only}\\qquad\\textbf{(B)}\\ \\text{B only}\\qquad\\textbf{(C)}\\ \\text{C only}\\qquad\\textbf{(D)}\\ \\text{both A and B}\\qquad\\textbf{(E)}\\ \\text{all are equal}$ ## Problem 23 In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares. If the pattern is continued, where would the cat and mouse be after the 247th move? ## 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);", "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", "and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "Thus, the area of the shaded region is: $$12 \\sqrt 3 + {80 \\over 3} \\pi - ({16 \\over 3} \\pi - 8 \\sqrt 3) + 16 \\pi - ({16 \\over 3} \\pi - 4 \\sqrt 3)$$ This simplifies to: $$\\color{brown}\\boxed{32 \\pi + 24\\sqrt3}$$ Apr 27, 2022", "of a circle inscribed in a square (D). The area is just (A) + (B) - (C) + (D). Area of (A): $2\\pi$ Area of (B): $\\frac{4\\pi}{3}$ Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is $2\\sqrt{3}*1/2=\\sqrt{3}$ Area of (D): $4*1-1/4*\\pi*4=4-\\pi$ Total sum: $\\frac{7\\pi}{3}-\\sqrt{3}+4$ $7+3+3+4=\\boxed{17}$ $[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label(\"A\",(-3,-1),S); dot((-2,0)); label(\"E\",(-2,0),NW); dot((-1,-1)); label(\"B\",(-1,-1),S); dot((0,0)); label(\"F\",(0,0),N); dot((1,-1)); label(\"C\",(1,-1), S); dot((2,0)); label(\"G\", (2,0),NE); dot((3,-1)); label(\"D\", (3,-1), S); [/asy]$ -Arpitr20", "the area of the shaded region? $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,E); label(\"F\",F,S); label(\"G\",G,W); label(\"H\",H,N); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ $\\textbf{(A)}\\ \\dfrac1{12}\\qquad\\textbf{(B)}\\ \\dfrac{\\sqrt3}{18}\\qquad\\textbf{(C)}\\ \\dfrac{\\sqrt2}{12}\\qquad\\textbf{(D)}\\ \\dfrac{\\sqrt3}{12}\\qquad\\textbf{(E)}\\ \\dfrac16$ ## Problem 17 Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die? $\\textbf{(A)}\\ \\dfrac16\\qquad\\textbf{(B)}\\ \\dfrac{13}{72}\\qquad\\textbf{(C)}\\ \\dfrac7{36}\\qquad\\textbf{(D)}\\ \\dfrac5{24}\\qquad\\textbf{(E)}\\ \\dfrac29$ ## Problem 18 A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $1$, $4$, and $5$. What is the area of the square? $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 17\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 27$ ## Problem 19 Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\\overline{XY}$ contained in the cube with edge length $3$? $\\textbf{(A)}\\ \\dfrac{3\\sqrt{33}}5\\qquad\\textbf{(B)}\\ 2\\sqrt3\\qquad\\textbf{(C)}\\ \\dfrac{2\\sqrt{33}}3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 3\\sqrt2$ $[asy] dotfactor = 3; size(10cm); dot((0, 10)); label(\"X\", (0,10),W,fontsize(8pt)); dot((6,2)); label(\"Y\", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5,", "2014 AIME I Problems/Problem 10 Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Solution 1 $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since the larger disk has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "area of the shaded region? $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,E); label(\"F\",F,S); label(\"G\",G,W); label(\"H\",H,N); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ $\\textbf{(A)}\\ \\dfrac1{12}\\qquad\\textbf{(B)}\\ \\dfrac{\\sqrt3}{18}\\qquad\\textbf{(C)}\\ \\dfrac{\\sqrt2}{12}\\qquad\\textbf{(D)}\\ \\dfrac{\\sqrt3}{12}\\qquad\\textbf{(E)}\\ \\dfrac16$ ## Problem 17 Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die? $\\textbf{(A)}\\ \\dfrac16\\qquad\\textbf{(B)}\\ \\dfrac{13}{72}\\qquad\\textbf{(C)}\\ \\dfrac7{36}\\qquad\\textbf{(D)}\\ \\dfrac5{24}\\qquad\\textbf{(E)}\\ \\dfrac29$ ## Problem 18 A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $1$, $4$, and $5$. What is the area of the square? $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 17\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 27$ ## Problem 19 Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\\overline{XY}$ contained in the cube with edge length $3$? $\\textbf{(A)}\\ \\dfrac{3\\sqrt{33}}5\\qquad\\textbf{(B)}\\ 2\\sqrt3\\qquad\\textbf{(C)}\\ \\dfrac{2\\sqrt{33}}3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 3\\sqrt2$ $[asy] dotfactor = 3; size(10cm); dot((0, 10)); label(\"X\", (0,10),W,fontsize(8pt)); dot((6,2)); label(\"Y\", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0,", "# Difference between revisions of \"2014 AMC 10B Problems/Problem 5\" ## Problem Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$, and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window? $[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill((6,0)--(8,0)--(8,26)--(6,26)--cycle,gray); fill((12,0)--(14,0)--(14,26)--(12,26)--cycle,gray); fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray); fill((24,0)--(26,0)--(26,26)--(24,26)--cycle,gray); fill((0,0)--(26,0)--(26,2)--(0,2)--cycle,gray); fill((0,12)--(26,12)--(26,14)--(0,14)--cycle,gray); fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); [/asy]$ $\\textbf{(A)}\\ 26\\qquad\\textbf{(B)}\\ 28\\qquad\\textbf{(C)}\\ 30\\qquad\\textbf{(D)}\\ 32\\qquad\\textbf{(E)}\\ 34$ ## Solution We note that the total length must be the same as the total height because the window is a square. Calling the width of each small rectangle $2x$, and the height $5x$, we can see that the length is composed of $4$ widths and $5$ bars of length $2$. This is equal to two heights of the small rectangles as well as $3$ bars of $2$. Thus, $4(2x) + 5(2) = 2(5x) + 3(2)$. We quickly find that $x = 2$. The total side length is $4(4) + 5(2) = 2(10) + 3(2) = 26$, or $\\boxed{\\textbf{(A)}}$. ~savannahsolver", "# Difference between revisions of \"2016 AMC 10A Problems/Problem 11\" ## Problem What is the area of the shaded region of the given $8 \\times 5$ rectangle? $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(\"1\",(1/2,5),dir(90)); label(\"7\",(9/2,5),dir(90)); label(\"1\",(8,1/2),dir(0)); label(\"4\",(8,3),dir(0)); label(\"1\",(15/2,0),dir(270)); label(\"7\",(7/2,0),dir(270)); label(\"1\",(0,9/2),dir(180)); label(\"4\",(0,2),dir(180)); [/asy]$ $\\textbf{(A)}\\ 4\\dfrac{3}{5} \\qquad \\textbf{(B)}\\ 5\\qquad \\textbf{(C)}\\ 5\\dfrac{1}{4} \\qquad \\textbf{(D)}\\ 6\\dfrac{1}{2} \\qquad \\textbf{(E)}\\ 8$ ## Solution First, split the rectangle into $4$ triangles: $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(\"1\",(1/2,5),dir(90)); label(\"7\",(9/2,5),dir(90)); label(\"1\",(8,1/2),dir(0)); label(\"4\",(8,3),dir(0)); label(\"1\",(15/2,0),dir(270)); label(\"7\",(7/2,0),dir(270)); label(\"1\",(0,9/2),dir(180)); label(\"4\",(0,2),dir(180)); draw((0,5)--(8,0)); [/asy]$ The bases of these triangles are all $1$, and their heights are $4$, $\\frac{5}{2}$, $4$, and $\\frac{5}{2}$. Thus, their areas are $2$, $\\frac{5}{4}$, $2$, and $\\frac{5}{4}$, which add to the area of the shaded region, which is $\\boxed{6\\frac{1}{2}}$.", "# Difference between revisions of \"2018 AMC 10B Problems/Problem 15\" A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper? $[asy]defaultpen(fontsize(10pt)); filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); dot((-3,3)); label(\"A\",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype(\"2.5 2.5\")+linewidth(.5)); draw((3,0)--(-3,0),linetype(\"2.5 2.5\")+linewidth(.5)); label('w',(-1,-1),SW); label('w',(1,-1),SE); draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle); draw((4.5,0)--(8.5,0)); draw((6.5,2)--(6.5,-2)); label(\"A\",(6.5,0),NW); dot((6.5,0)); [/asy]$ $\\textbf{(A) } 2(w+h)^2 \\qquad \\textbf{(B) } \\frac{(w+h)^2}2 \\qquad \\textbf{(C) } 2w^2+4wh \\qquad \\textbf{(D) } 2w^2 \\qquad \\textbf{(E) } w^2h$ ## Solution Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length 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", "plt.rcParams['ytick.labelsize'] = 20 #40 extent = (0,(xmax-xmin)/1e2,0,(ymax-ymin)/1e2) In [5]: #band5 arr=data[4] plt.figure(figsize=(24,24)) plt.imshow(arr[ymin:ymax,xmin:xmax].astype('float')/float(arr.max()),cmap=plt.cm.gray, vmin=0,vmax=0.2,extent=extent) cb = plt.colorbar(shrink=0.5) cb.set_label('normalized surface reflectance',fontsize=30) plt.grid(ls='--') plt.xlabel('[km]',fontsize=40) plt.ylabel('[km]',fontsize=40) plt.title('Greater Manchester (Sentinel2 - Band5)',fontsize=40) #plt.savefig('Manchester_overview_Band5.png',dpi=300,bbox_inches='tight') Out[5]: <matplotlib.text.Text at 0x7f9e9c18fb70>", "# 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", "# 2016 AMC 10A Problems/Problem 10 ## Problem A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle? $[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white); label(\"1\",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead)); label(\"1\",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead)); label(\"1\",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead)); label(\"1\",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead)); label(\"1\",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]$ $\\textbf{(A) } 1 \\qquad \\textbf{(B) } 2 \\qquad \\textbf{(C) } 4 \\qquad \\textbf{(D) } 6 \\qquad \\textbf{(E) }8$ ## Solution Let the length of the inner rectangle be $x$. Then the area of that rectangle is $x\\cdot1 = x$. The second largest rectangle has dimensions of $x+2$ and $3$, making its area $3x+6$. The area of the second shaded area, therefore, is $3x+6-x = 2x+6$. The largest rectangle has dimensions of $x+4$ and $5$, making its area $5x + 20$. The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$. The problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the sequence increase by the", "# 2014 AIME I Problems/Problem 10 ## Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since it has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "# Difference between revisions of \"2016 AMC 10A Problems/Problem 11\" ## Problem What is the area of the shaded region of the given $8 \\times 5$ rectangle? $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(\"1\",(1/2,5),dir(90)); label(\"7\",(9/2,5),dir(90)); label(\"1\",(8,1/2),dir(0)); label(\"4\",(8,3),dir(0)); label(\"1\",(15/2,0),dir(270)); label(\"7\",(7/2,0),dir(270)); label(\"1\",(0,9/2),dir(180)); label(\"4\",(0,2),dir(180)); [/asy]$ $\\textbf{(A)}\\ 4\\dfrac{3}{5} \\qquad \\textbf{(B)}\\ 5\\qquad \\textbf{(C)}\\ 5\\dfrac{1}{4} \\qquad \\textbf{(D)}\\ 6\\dfrac{1}{2} \\qquad \\textbf{(E)}\\ 8$ ## Solution 1 First, split the rectangle into $4$ triangles: $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(\"1\",(1/2,5),dir(90)); label(\"7\",(9/2,5),dir(90)); label(\"1\",(8,1/2),dir(0)); label(\"4\",(8,3),dir(0)); label(\"1\",(15/2,0),dir(270)); label(\"7\",(7/2,0),dir(270)); label(\"1\",(0,9/2),dir(180)); label(\"4\",(0,2),dir(180)); draw((0,5)--(8,0)); [/asy]$ The bases of these triangles are all $1$, and their heights are $4$, $\\frac{5}{2}$, $4$, and $\\frac{5}{2}$. Thus, their areas are $2$, $\\frac{5}{4}$, $2$, and $\\frac{5}{4}$, which add to the area of the shaded region, which is $\\boxed{6\\frac{1}{2}}$. ## Solution 2 Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. This will result in $40$ - $33$ $\\frac{1}{2}$ = $\\boxed{6$ (Error compiling LaTeX. ! File ended while scanning use of \\boxed.)\\frac{1}{2}}\\$", "# Difference between revisions of \"2016 AMC 10A Problems/Problem 11\" ## Problem What is the area of the shaded region of the given $8 \\times 5$ rectangle? $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(\"1\",(1/2,5),dir(90)); label(\"7\",(9/2,5),dir(90)); label(\"1\",(8,1/2),dir(0)); label(\"4\",(8,3),dir(0)); label(\"1\",(15/2,0),dir(270)); label(\"7\",(7/2,0),dir(270)); label(\"1\",(0,9/2),dir(180)); label(\"4\",(0,2),dir(180)); [/asy]$ $\\textbf{(A)}\\ 4\\dfrac{3}{5} \\qquad \\textbf{(B)}\\ 5\\qquad \\textbf{(C)}\\ 5\\dfrac{1}{4} \\qquad \\textbf{(D)}\\ 6\\dfrac{1}{2} \\qquad \\textbf{(E)}\\ 8$ ## Solution 1 First, split the rectangle into $4$ triangles: $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label(\"1\",(1/2,5),dir(90)); label(\"7\",(9/2,5),dir(90)); label(\"1\",(8,1/2),dir(0)); label(\"4\",(8,3),dir(0)); label(\"1\",(15/2,0),dir(270)); label(\"7\",(7/2,0),dir(270)); label(\"1\",(0,9/2),dir(180)); label(\"4\",(0,2),dir(180)); draw((0,5)--(8,0)); [/asy]$ The bases of these triangles are all $1$, and their heights are $4$, $\\frac{5}{2}$, $4$, and $\\frac{5}{2}$. Thus, their areas are $2$, $\\frac{5}{4}$, $2$, and $\\frac{5}{4}$, which add to the area of the shaded region, which is $\\boxed{6\\frac{1}{2}}$. ## Solution 2 Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. This will result in $40$ - $33$ $\\frac{1}{2}$ = $\\boxed{6\\frac{1}{2}}$. ## See Also 2016 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 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", "picture. What are the dimensions of the viewable portion of the picture? (1) The area of the shaded region is 24 square inches. (2) The frame is 8 inches tall. Math Expert Joined: 02 Aug 2009 Posts: 6565 Re: A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink] ### Show Tags 28 May 2016, 05:20 4 3 nalinnair wrote: Attachment: Image.png A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture? (1) The area of the shaded region is 24 square inches. (2) The frame is 8 inches tall. hi, let the length be x, so dimensions of bigger rectangle = 6 and x... Also let the width around be y, so dimensions are 6-2y and x-2y.... For the Q asked, we require value of x and y.... lets see the statements- (1) The area of the shaded region is 24 square inches. Area of shaded region = $$6*x - (6-2y)*(x-2y) = 6x - 6x+12y+2xy-4y^2 = 12y+2xy-4y^2 = 24$$ We can't solve for two variables through one equation... Insuff (2) The frame is 8 inches tall. we know the dimensions of bigger frame... Insuff Combined- 12y+2xy-4y^2 = 24..........", "# Difference between revisions of \"2004 AMC 10A Problems/Problem 21\" ## Problem Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\\pi$ radians is $180$ degrees.) $[asy] size(85); fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7)); fill((30,0)..(24,18)--(0,0)--(24,-18)..cycle,gray(0.7)); fill((-20,0)..(0,20)--(0,-20)..cycle,white); fill((20,0)..(0,20)--(0,-20)..cycle,white); fill((0,20)..(-16,12)--(0,0)--(16,12)..cycle,gray(0.7)); fill((0,-20)..(-16,-12)--(0,0)--(16,-12)..cycle,gray(0.7)); fill((0,10)..(-10,0)--(10,0)..cycle,white); fill((0,-10)..(-10,0)--(10,0)..cycle,white); fill((-10,0)..(-8,6)--(0,0)--(-8,-6)..cycle,gray(0.7)); fill((10,0)..(8,6)--(0,0)--(8,-6)..cycle,gray(0.7)); draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),20),linewidth(0.7)); draw(Circle((0,0),30),linewidth(0.7)); draw((-28,-21)--(28,21),linewidth(0.7)); draw((-28,21)--(28,-21),linewidth(0.7));[/asy]$ $\\mathrm{(A) \\ } \\frac{\\pi}{8} \\qquad \\mathrm{(B) \\ } \\frac{\\pi}{7} \\qquad \\mathrm{(C) \\ } \\frac{\\pi}{6} \\qquad \\mathrm{(D) \\ } \\frac{\\pi}{5} \\qquad \\mathrm{(E) \\ } \\frac{\\pi}{4}$ ## Solution 1 Let the area of the shaded region be $S$, the area of the unshaded region be $U$, and the acute angle that is formed by the two lines be $\\theta$. We can set up two equations between $S$ and $U$: $S+U=9\\pi$ $S=\\dfrac{8}{13}U$ Thus $\\dfrac{21}{13}U=9\\pi$, and $U=\\dfrac{39\\pi}{7}$, and thus $S=\\dfrac{8}{13}\\cdot \\dfrac{39\\pi}{7}=\\dfrac{24\\pi}{7}$. Now we can make a formula for the area of the shaded region in terms of $\\theta$: $\\dfrac{2\\theta}{2\\pi} \\cdot \\pi +\\dfrac{2(\\pi-\\theta)}{2\\pi} \\cdot (4\\pi-\\pi)+\\dfrac{2\\theta}{2\\pi}(9\\pi-4\\pi)=\\theta +3\\pi-3\\theta+5\\theta=3\\theta+3\\pi=\\dfrac{24\\pi}{7}$ Thus $3\\theta=\\dfrac{3\\pi}{7}\\Rightarrow \\theta=\\dfrac{\\pi}{7}\\Rightarrow\\boxed{\\mathrm{(B)}\\ \\frac{\\pi}{7}}$ ## Solution 2 As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of $\\theta$. $\\implies\\dfrac{2\\theta}{2\\pi} \\cdot \\pi +\\dfrac{2(\\pi-\\theta)}{2\\pi} \\cdot", "# Problem #2093 2093 Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles? $[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype(\"4 4\"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label(\"A\",A,W); label(\"B\",B,E); label(\"1\",(-1.5,-0.1),S); label(\"2\",(0,-0.1),S); label(\"1\",(1.5,-0.1),S);[/asy]$ $\\textbf{(A) } \\pi - \\sqrt{3} \\qquad\\textbf{(B) } \\pi - \\sqrt{2} \\qquad\\textbf{(C) } \\frac{\\pi + \\sqrt{2}}{2} \\qquad\\textbf{(D) } \\frac{\\pi +\\sqrt{3}}{2} \\qquad\\textbf{(E) } \\frac{7}{6}\\pi - \\frac{\\sqrt{3}}{2}$ 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. • 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" ]

[ "15$ . Thus, the area is $15\\cdot{10}\\ = 150$ for choice $\\boxed{\\textbf{(E)}\\ 150}$ ~~Saksham27 ## Solution 2 Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is $5 \\cdot 2 = 10$. So the area of the identical rectangles is $5 \\cdot 10 = 50$. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is $50 \\cdot 3 = \\boxed{\\textbf{(E)}\\ 150}$. ~~fath2012", "+0 # solid figure 0 209 1 Find the surface area of the following: Sep 9, 2020 For all the rectangular sides, we have $$4\\cdot (8 +10 + 6) = 96$$, and for the top and bottom, we have $$2 \\cdot 24 = 48$$, and finally $$96 + 48 = \\boxed{144}$$", "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", "+0 # Find the area of the shaded portion below. 0 58 1 Find the area of the shaded portion below. Mar 16, 2021 #1 +73 0 the shaded area is essentially the area of the trapezoid minus the area of that semi-circle. the area of the trapezoid is $$\\frac{50+80}{2} \\cdot (20 \\cdot 2)$$ $$\\frac{130}{2} \\cdot 40$$ $$65 \\cdot 40$$ $$2600$$ the area of the semi-circle is $$\\frac{20^2 \\cdot \\pi}{2}$$ $$\\frac{400 \\cdot \\pi}{2}$$ $$200 \\pi$$ so, the area of the shaded region is $$2600 - 200\\pi$$. hope this helped! please let me know if you have any questions Mar 16, 2021", "# Math Help - Area and Perimeter of Shaded Region (Answer Confirmation) In attachment there are 8 question and i want to confirm my answers 2. Hi haftakhan, For i, I got this: Area of big circle: $100 \\pi$ Area of small circle: $36 \\pi$ Area of shaded region: $100 \\pi - 36 \\pi \\approx 201.06$ The perimeter of the circle with radius 10: $20 \\pi \\approx 62.83$ The perimeter of the circle with radius 6: $12\\pi \\approx 37.70$ The combined perimeter of the shaded region: $\\approx 100.53$ For ii, I got this: Area of circle: $100 \\pi$ Area of Square: $14.14^2$ Area of shaded region: $100 \\pi - 14.14^2 \\approx 114.22$ The perimeter of the shaded region: $20\\pi + 4(14.14) \\approx 119.39$ For iii, the area looks good, but I don't see how you got 75 for the perimeter. I haven't looked at the others, but if you want to post them individually and show how you got your answers, that'd be good. 3. Hi haftakhan, I agree with masters show your work. I looked only at 8 where you had no area answer. My answer is area 72pi and 24pi for perimeter 4. Hello, haftakhan! Are you sure of your area and perimeter formulas? The perimeter is the amount of fencing to enclose the shaded region.", "Thus, the area of the shaded region is: $$12 \\sqrt 3 + {80 \\over 3} \\pi - ({16 \\over 3} \\pi - 8 \\sqrt 3) + 16 \\pi - ({16 \\over 3} \\pi - 4 \\sqrt 3)$$ This simplifies to: $$\\color{brown}\\boxed{32 \\pi + 24\\sqrt3}$$ Apr 27, 2022", "3^2)$ = $20 \\pi$. Therefore the total area of the shaded region is $64 \\pi$.", "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", "is: $$108\\, cm^{2}$$. Area of Region B Area (A) =$$\\frac{1}{2}\\: b\\cdot h$$ Region $$B$$ is a triangle. To find our area, we use this formula: $$\\frac{1}{2}\\: bh$$ $$A =\\frac{1}{2}\\: \\cdot 9\\cdot 9$$ where the base is $$9$$ (nine) and the height is $$9$$ (nine). $$A = \\frac{l }{2}\\cdot 81$$ $$A = 40.5$$ Region B’s area is: $$40.5 \\, cm^{2}$$. $$108\\, cm^{2} \\,plus \\, 40.5 \\, cm^{2}$$ $$= 148.5\\, cm^{2}$$. The perimeter = $$59.5 \\,cm$$ The area = $$148.5 \\,cm2$$ We may also use what we know about perimeters and areas to help us solve problems in situations where we are buying paint or fencing, or to determine how big a rug should be for our living room. Here’s an example of fencing. The example The problem: Rosie is planting her garden that has the dimensions as shown below. Now, she wants to put an even, thin layer of mulch across the whole surface of her garden. The price of the mulch is \\$3 per square foot. What’s the amount of money she’ll need to spend on the mulch? Well, this shape combines two simpler shapes, a trapezoid, and a rectangle. First, we need to find the area of each shape. $$A = l \\cdot w$$ $$A = 8 \\cdot 4$$ So first, we find the rectangle’s area. $$A", "### Home > CC1 > Chapter 5 > Lesson 5.3.1 > Problem5-75 5-75. Draw generic rectangles to calculate each of the following products. 1. $34 \\cdot 91$ • The generic rectangle below has been started for you. Top edge right blank Tom edge left ${\\color{blue}{\\text{90}}}$ Left edge top blank Interior Top left blank Interior Top right blank left edge bottom ${\\color{blue}{\\text{4}}}\\$ Interior Bottom left blank Interior Bottom Right blank 1. $421 \\cdot 36$ • Draw a $3$ by $2$ rectangle. The top row will be $400$, $20$, and $1$. The left edge will be $30$ and $6$. Fill in the boxes with the areas.", "## muscrat123 one year ago The figure below shows a shaded region and a nonshaded region. Angles in the figure that appear to be right angles are right angles.What is the area, in square feet, of the shaded region?What is the area, in square feet, of the nonshaded region? 1. muscrat123 2. muscrat123 the figure is above 3. muscrat123 @pooja195 4. anonymous The shaded area is a trapezoid with the 2 by 2 square cut out. 5. anonymous $\\frac{1}{2} (6+3) 10-4 = 41$ 6. muscrat123 so 41 ft^2 is the shaded area? @robtobey 7. muscrat123 @nincompoop 8. anonymous Yes. 9. muscrat123 then u would find the area of the rectangle and subtract the area of the shaded area to find the area of the non-shaded area, right? @robtobey 10. muscrat123 16 x 8 = 128 128 - 41 = 87 so, 87 ft^2 is the non-shaded area, right? @robtobey 11. anonymous so, 87 ft^2 is the non-shaded area, right? @robtobey Yes 12. muscrat123 thx @robtobey 13. anonymous Find more explanations on OpenStudy", "+0 # Find the number of square units in the area of the shaded region. 0 87 1 Find the number of square units in the area of the shaded region. Nov 29, 2020 #1 +10843 +1 Find the number of square units in the area of the shaded region. Hello Guest! $$A=12cm\\cdot 4cm-\\frac{1}{2}\\cdot 9cm\\cdot 4cm\\\\ A=48cm^2-18cm^2$$ $$A=30cm^2$$ The number of square units in the area of the shaded region is 30. ! Nov 29, 2020 #1 +10843 +1 Find the number of square units in the area of the shaded region. Hello Guest! $$A=12cm\\cdot 4cm-\\frac{1}{2}\\cdot 9cm\\cdot 4cm\\\\ A=48cm^2-18cm^2$$ $$A=30cm^2$$ The number of square units in the area of the shaded region is 30. ! asinus Nov 29, 2020", "? Free Version Difficult # Sum of the Areas GEOM-WLGQKV What is the sum of the areas of the shaded regions? Round your answer to the nearest tenth of a unit. A $103.7\\text{ cm}^2$ B $260.8\\text{ cm}^2$ C $47.1\\text{ cm}^2$ D $414.7\\text{ cm}^2$", "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", "the orange region above the line segment $DE$ is $7.64 + 0.39 = 8.03$ square inches. Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "unshaded cubes is 2 : 6 The ratio of shaded cubes to unshaded cubes is 3 : 9 3 : 9 2 : 6 1 : 3. are equivalent ratios. Previous question Next question Transcribed Image Text from this Question. An area model with 65 shaded sections and 35 unshaded sections. Cloudflare Ray ID: 6171e18b4e6d6413 The ratio of the shaded area to the unshaded area is 3:2 There are a total of 3+2=5 parts, so 3/5 is shaded and 2/5 is unshaded Our counselor will call to confirm your booking. for every 9 shaded parts there are 2 unshaded parts. draw a parallelogram ABCD has side BC=4.5 CM, diagonal AC=5.6 CM and diagonal BD=5 CM. Ask for details ; Follow Report by Gurpreet6841 03.08.2019 for every 3 shaded parts there are 2 unshaded … Ratio of shaded region and the unshaded region= 7 3 3 r 2: ((7 9 ) r 2) = 1 1: 3 81 - 17.55 = 63.45 or 63.5 to the tenth. 6 unshaded squares and 9 shaded squares. Larger shaded region - smaller unshaded regions inside the shaded figure. In Fig. Also review how to write in simplest form.CCSS 6.RP.1 Show transcribed image text. If the scale factor of this hexagon 2 A. An area model with 75 shaded sections and 25 unshaded", "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$$ so area of rectangle is HALF the area of shaded region OR the area of shaded region is the area of non-shaded region Let us see the statements now.. 1. The area of the shaded region in 40. so area of rectangle is twice of 40 = 2*40=80 sufficient 2. The area of the non-shaded region is 40. so area of rectangle is twice of 40 = 2*40=80 sufficient D _________________ Re: The figure shows a rectangle with a triangle inscribed in it [#permalink] 24 Nov 2018, 23:52 Display posts", "+0 0 90 1 What is the surface area of this design? Apr 13, 2020 #1 +9479 +1 What is the surface area of this design? Hello Guest! $$A=(4\\cdot 5\\cdot 4+8\\cdot 5\\cdot 2+8\\cdot 4\\cdot 2+4\\cdot 4\\cdot 2)in^2$$ $$A=256in^2$$ ! Apr 13, 2020 #1 +9479 +1 What is the surface area of this design? Hello Guest! $$A=(4\\cdot 5\\cdot 4+8\\cdot 5\\cdot 2+8\\cdot 4\\cdot 2+4\\cdot 4\\cdot 2)in^2$$ $$A=256in^2$$ ! asinus Apr 13, 2020", "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.", "arcs. Each of the rectangles has area $5\\cdot 1=5$. The four arcs together form a circle with radius $1$. Therefore the total area she can see is $16 + 4\\cdot 5 + \\pi\\cdot 1^2 = 36+\\pi \\simeq 39.14$, which rounded to the nearest integer is $39$. $\\boxed{C}$" ]

[ "# Math Help - Area of rectangle with embedded parallelograms 1. ## Area of rectangle with embedded parallelograms Hello - I am new to the forum, and am at a loss at how to solve this problem. I need to find the area of the shaded region. I created a picture in paint. Can I get some help with this, and an explanation on how to solve this? Thank youQ\\! 2. Hello, ncooks! Welcome aboard! Find the area of the shaded region. Code: 9 * - - - - - - - * |:::::::::::::::* |:::::::::::* | 2 |:::::::* ..* |:::* *:::* * *:::::::| | *:::::::::::* 13 *::::::::::* | 2 |:::::::* ..| |:::* *:::* * *:::::::| | *:::::::::::| *:::::::::::::::| |:::::::::::::::| * - - - - - - - * The area of the rectangle is: . $9 \\times 13 \\:=\\:117\\text{ cm}^2$ Assuming that the diagonal lines are parallel, . . the white regions are two congeuent parallelograms, . . each with base 2 and height 9. The white region has area: . $2 \\times (2\\times 9) \\:=\\:36\\text{ cm}^2$ Therefore, the area of the shaded region is: . $117 - 36 \\:=\\:81\\text{ cm}^2$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~", "a shaded region with largest area? $[asy]/* AMC8 2003 #22 Problem */ size(3inch, 2inch); unitsize(1cm); pen outline = black+linewidth(1); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline); filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline); filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1)); filldraw(Circle((1,1), 1), white, outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label(\"A\", (1, 2), N); label(\"B\", (4, 2), N); label(\"C\", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label(\"2 cm\", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label(\"2 cm\", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label(\"2 cm\", (7, -.5)); draw((6,1)--(6,-.5), linetype(\"4 4\")); draw((8,1)--(8,-.5), linetype(\"4 4\"));[/asy]$ $\\textbf{(A)}\\ \\text{A only}\\qquad\\textbf{(B)}\\ \\text{B only}\\qquad\\textbf{(C)}\\ \\text{C only}\\qquad\\textbf{(D)}\\ \\text{both A and B}\\qquad\\textbf{(E)}\\ \\text{all are equal}$ ## Problem 23 In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares. If the pattern is continued, where would the cat and mouse be after the 247th move? ## 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);", "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", "Thus, the area of the shaded region is: $$12 \\sqrt 3 + {80 \\over 3} \\pi - ({16 \\over 3} \\pi - 8 \\sqrt 3) + 16 \\pi - ({16 \\over 3} \\pi - 4 \\sqrt 3)$$ This simplifies to: $$\\color{brown}\\boxed{32 \\pi + 24\\sqrt3}$$ Apr 27, 2022", "2014 AIME I Problems/Problem 10 Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Solution 1 $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since the larger disk has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "+0 # Find the number of square units in the area of the shaded region. 0 87 1 Find the number of square units in the area of the shaded region. Nov 29, 2020 #1 +10843 +1 Find the number of square units in the area of the shaded region. Hello Guest! $$A=12cm\\cdot 4cm-\\frac{1}{2}\\cdot 9cm\\cdot 4cm\\\\ A=48cm^2-18cm^2$$ $$A=30cm^2$$ The number of square units in the area of the shaded region is 30. ! Nov 29, 2020 #1 +10843 +1 Find the number of square units in the area of the shaded region. Hello Guest! $$A=12cm\\cdot 4cm-\\frac{1}{2}\\cdot 9cm\\cdot 4cm\\\\ A=48cm^2-18cm^2$$ $$A=30cm^2$$ The number of square units in the area of the shaded region is 30. ! asinus Nov 29, 2020", "and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "# 2014 AIME I Problems/Problem 10 ## Problem 10 A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\\overline{AC}$ is parallel to $\\overline{BD}$. Then $\\sin^2(\\angle BEA)=\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution $[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"A\",a,W,fontsize(9)); label(\"B\",b,NW,fontsize(9)); label(\"C\",c,E,fontsize(9)); label(\"D\",d,E,fontsize(9)); label(\"E\",e,SW,fontsize(9)); label(\"F\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]$ Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since it has a radius of $5$. Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with", "of a circle inscribed in a square (D). The area is just (A) + (B) - (C) + (D). Area of (A): $2\\pi$ Area of (B): $\\frac{4\\pi}{3}$ Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is $2\\sqrt{3}*1/2=\\sqrt{3}$ Area of (D): $4*1-1/4*\\pi*4=4-\\pi$ Total sum: $\\frac{7\\pi}{3}-\\sqrt{3}+4$ $7+3+3+4=\\boxed{17}$ $[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label(\"A\",(-3,-1),S); dot((-2,0)); label(\"E\",(-2,0),NW); dot((-1,-1)); label(\"B\",(-1,-1),S); dot((0,0)); label(\"F\",(0,0),N); dot((1,-1)); label(\"C\",(1,-1), S); dot((2,0)); label(\"G\", (2,0),NE); dot((3,-1)); label(\"D\", (3,-1), S); [/asy]$ -Arpitr20", "picture. What are the dimensions of the viewable portion of the picture? (1) The area of the shaded region is 24 square inches. (2) The frame is 8 inches tall. Math Expert Joined: 02 Aug 2009 Posts: 6565 Re: A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink] ### Show Tags 28 May 2016, 05:20 4 3 nalinnair wrote: Attachment: Image.png A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture? (1) The area of the shaded region is 24 square inches. (2) The frame is 8 inches tall. hi, let the length be x, so dimensions of bigger rectangle = 6 and x... Also let the width around be y, so dimensions are 6-2y and x-2y.... For the Q asked, we require value of x and y.... lets see the statements- (1) The area of the shaded region is 24 square inches. Area of shaded region = $$6*x - (6-2y)*(x-2y) = 6x - 6x+12y+2xy-4y^2 = 12y+2xy-4y^2 = 24$$ We can't solve for two variables through one equation... Insuff (2) The frame is 8 inches tall. we know the dimensions of bigger frame... Insuff Combined- 12y+2xy-4y^2 = 24..........", "the area of the shaded region? $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,E); label(\"F\",F,S); label(\"G\",G,W); label(\"H\",H,N); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ $\\textbf{(A)}\\ \\dfrac1{12}\\qquad\\textbf{(B)}\\ \\dfrac{\\sqrt3}{18}\\qquad\\textbf{(C)}\\ \\dfrac{\\sqrt2}{12}\\qquad\\textbf{(D)}\\ \\dfrac{\\sqrt3}{12}\\qquad\\textbf{(E)}\\ \\dfrac16$ ## Problem 17 Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die? $\\textbf{(A)}\\ \\dfrac16\\qquad\\textbf{(B)}\\ \\dfrac{13}{72}\\qquad\\textbf{(C)}\\ \\dfrac7{36}\\qquad\\textbf{(D)}\\ \\dfrac5{24}\\qquad\\textbf{(E)}\\ \\dfrac29$ ## Problem 18 A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $1$, $4$, and $5$. What is the area of the square? $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 17\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 27$ ## Problem 19 Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\\overline{XY}$ contained in the cube with edge length $3$? $\\textbf{(A)}\\ \\dfrac{3\\sqrt{33}}5\\qquad\\textbf{(B)}\\ 2\\sqrt3\\qquad\\textbf{(C)}\\ \\dfrac{2\\sqrt{33}}3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 3\\sqrt2$ $[asy] dotfactor = 3; size(10cm); dot((0, 10)); label(\"X\", (0,10),W,fontsize(8pt)); dot((6,2)); label(\"Y\", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5,", "KiB | Viewed 3910 times ] A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture? (1) The area of the shaded region is 24 square inches. (2) The frame is 8 inches tall. [Reveal] Spoiler: OA Math Expert Joined: 02 Aug 2009 Posts: 5777 Re: A framed picture is shown above. The frame, shown shaded, is 6 inches [#permalink] ### Show Tags 28 May 2016, 05:20 2 KUDOS Expert's post 3 This post was BOOKMARKED nalinnair wrote: Attachment: Image.png A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture? (1) The area of the shaded region is 24 square inches. (2) The frame is 8 inches tall. hi, let the length be x, so dimensions of bigger rectangle = 6 and x... Also let the width around be y, so dimensions are 6-2y and x-2y.... For the Q asked, we require value of x and y.... lets see the statements- (1) The area of the shaded region is 24 square inches. Area of shaded region = $$6*x", "# 2016 AMC 10A Problems/Problem 10 ## Problem A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle? $[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white); label(\"1\",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead)); label(\"1\",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead)); label(\"1\",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead)); label(\"1\",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead)); label(\"1\",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]$ $\\textbf{(A) } 1 \\qquad \\textbf{(B) } 2 \\qquad \\textbf{(C) } 4 \\qquad \\textbf{(D) } 6 \\qquad \\textbf{(E) }8$ ## Solution Let the length of the inner rectangle be $x$. Then the area of that rectangle is $x\\cdot1 = x$. The second largest rectangle has dimensions of $x+2$ and $3$, making its area $3x+6$. The area of the second shaded area, therefore, is $3x+6-x = 2x+6$. The largest rectangle has dimensions of $x+4$ and $5$, making its area $5x + 20$. The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$. The problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the sequence increase by the", "## 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", "area of the shaded region? $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,E); label(\"F\",F,S); label(\"G\",G,W); label(\"H\",H,N); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ $\\textbf{(A)}\\ \\dfrac1{12}\\qquad\\textbf{(B)}\\ \\dfrac{\\sqrt3}{18}\\qquad\\textbf{(C)}\\ \\dfrac{\\sqrt2}{12}\\qquad\\textbf{(D)}\\ \\dfrac{\\sqrt3}{12}\\qquad\\textbf{(E)}\\ \\dfrac16$ ## Problem 17 Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die? $\\textbf{(A)}\\ \\dfrac16\\qquad\\textbf{(B)}\\ \\dfrac{13}{72}\\qquad\\textbf{(C)}\\ \\dfrac7{36}\\qquad\\textbf{(D)}\\ \\dfrac5{24}\\qquad\\textbf{(E)}\\ \\dfrac29$ ## Problem 18 A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $1$, $4$, and $5$. What is the area of the square? $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 17\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 27$ ## Problem 19 Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\\overline{XY}$ contained in the cube with edge length $3$? $\\textbf{(A)}\\ \\dfrac{3\\sqrt{33}}5\\qquad\\textbf{(B)}\\ 2\\sqrt3\\qquad\\textbf{(C)}\\ \\dfrac{2\\sqrt{33}}3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 3\\sqrt2$ $[asy] dotfactor = 3; size(10cm); dot((0, 10)); label(\"X\", (0,10),W,fontsize(8pt)); dot((6,2)); label(\"Y\", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0,", "plt.rcParams['ytick.labelsize'] = 20 #40 extent = (0,(xmax-xmin)/1e2,0,(ymax-ymin)/1e2) In [5]: #band5 arr=data[4] plt.figure(figsize=(24,24)) plt.imshow(arr[ymin:ymax,xmin:xmax].astype('float')/float(arr.max()),cmap=plt.cm.gray, vmin=0,vmax=0.2,extent=extent) cb = plt.colorbar(shrink=0.5) cb.set_label('normalized surface reflectance',fontsize=30) plt.grid(ls='--') plt.xlabel('[km]',fontsize=40) plt.ylabel('[km]',fontsize=40) plt.title('Greater Manchester (Sentinel2 - Band5)',fontsize=40) #plt.savefig('Manchester_overview_Band5.png',dpi=300,bbox_inches='tight') Out[5]: <matplotlib.text.Text at 0x7f9e9c18fb70>", ")> g\\left ( 3,2 \\right )$ Equal sized circular regions are shaded in a square sheet of paper of $1$ cm side length. Two cases, case $\\text{M}$ and case $\\text{N}$, are considered as shown in the figures below. In the case $\\text{M}$, four circles are shaded in the square sheet and in the case $\\text{N}$, nine circles are ... of unshaded regions of case $\\text{M}$ to that of case $\\text{N}$? $2 : 3$ $1 : 1$ $3 : 2$ $2 : 1$", "# Difference between revisions of \"2016 AMC 10A Problems/Problem 10\" ## Problem A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle? $[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white); label(\"1\",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead)); label(\"1\",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead)); label(\"1\",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead)); label(\"1\",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead)); label(\"1\",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]$ $\\textbf{(A) } 1 \\qquad \\textbf{(B) } 2 \\qquad \\textbf{(C) } 4 \\qquad \\textbf{(D) } 6 \\qquad \\textbf{(E) }8$ ## Solution Let the length of the inner rectangle be $x$. Then the area of that rectangle is $x\\cdot1 = x$. The second largest rectangle has dimensions of $x+2$ and $3$, making its area $3x+6$. The area of the second shaded area, therefore, is $3x+6-x = 2x+6$. The largest rectangle has dimensions of $x+4$ and $5$, making its area $5x + 20$. The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$. The problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the", "# Derivative application #### spoc21 [FONT=&quot]A rectangular prism has its length increasing by 12 cm/min, its width increasing by 4 cm/min and its height increasing by 2 cm/min. How fast is it's volume changing when the dimensions are 200 cm in length, 50 cm in width and 30 cm in height? The following is my working: [/FONT] dL/dt = 12 cm/min dW/dt = 4 cm/min dH/dt = 2 cm/min V = L x W x H we get: V=(200+12t)(50+4t)(30+3t) $$\\displaystyle V = 144t^3 + 5460t^2 + 72000t + 300,000$$ Derivative of V $$\\displaystyle V' = 432t^2 + 10920t + 72000$$ Now, I'm very confused. How could we find the rates of change,when [FONT=&quot]200 cm in length, 50 cm in width and 30 cm in height? [/FONT] Thanks! [FONT=&quot] [/FONT] #### arash You need to take implicit differentiation of both sides of the volume equation: V=L*W*H dV/dt = (dL/dt)*W*H + L*(dW/dt)*H + .... (Use chain rule here) Then all you need to do is sub in for expressions on the right side and you will get dV/dt which is what you're looking for. Hope it helps #### TKHunny #1 - Get fluent at writing this sort of thing. It's just the product rule. $$\\displaystyle \\frac{d}{dt}[L(t)\\cdot W(t) \\cdot H(t)] = L'(t)\\cdot [W(t) \\cdot H(t)] + L(t)\\cdot [W'(t)\\cdot H(t) + W(t)\\cdot", "the figure below have the same dimensions. The vertex of one square is at the center of the other square. What is the area of the shaded region, in square inches? $A.\\:\\: 4 \\\\[3ex] B.\\:\\: 8 \\\\[3ex] C.\\:\\: 16 \\\\[3ex] D.\\:\\: 32 \\\\[3ex] E.\\:\\: 64 \\\\[3ex]$ The vertex of one square is at the center of the other square. Do not let the titled square confuse you. If you stand it directly, it has the same area as when it is titled because it's vertex is at the center of the other square. This implies that half of the length of the titled square is at the center of the other square. The $2$ squares in the figure below have the same dimensions. This implies that $\\dfrac{1}{2} * 8 = 4\\:inches$ of the titled square is at the center of the other square. $Area\\:\\:of\\:\\:the\\:\\:shaded\\:\\:region \\\\[3ex] = 4(4) \\\\[3ex] = 16\\:in^2 \\\\[3ex]$ Alternatively, one can say that the area of the shaded region (region of the titled square) is one-quarter of the area of the first square. $Area\\:\\:of\\:\\:the\\:\\:first\\:\\:square = 8(8) = 64\\:in^2 \\\\[3ex] Area\\:\\:of\\:\\:the\\:\\:shaded\\:\\:region = \\dfrac{1}{4} * 64 \\\\[3ex] = 16\\:in^2$ (7.) WASSCE The curved surface areas of two cones are equal. The base radius of one is $5\\:cm$ and its slant height is $12\\:cm$ Calculate the height of" ]

[ "Question How many ways can a president, vice-president, secretary, and treasurer be chosen from a club with 9 members?", "+0 0 78 2 +69 A club has 15 members and needs to choose 2 members to be co-presidents. In how many ways can the club choose its co-presidents? rohitaop May 14, 2018 #1 +867 +2 This is a simple choosing problem. 15 choose 2 can be expressed as: $$\\binom{15}{2}=\\frac{15\\cdot14}{2\\cdot1}=105$$ i hope this helped, Gavin GYanggg May 14, 2018 #2 +2765 +1 This is just combinations, 15C2- $$\\frac{15!}{2!*13!}=\\frac{15*14}{2*1}=15*7=\\boxed{105}$$ tertre May 14, 2018", "with an explanation for your choice. a. A student club has 20 members. How many ways are there for the club to choose a president and a vice president? Permutations, 380. The role of president is different from that of vice president. The order of outcomes matters. b. A football team of 50 players will choose two co – captains. How many different ways are there to choose the two co – captains? Combinations, 1,225. Regardless of order, two players will attain the same outcome of co – captain. c. There are seven people who meet for the first time at a meeting. They shake hands with each other and introduce themselves. How many handshakes have been exchanged? Combinations, 21. People only shake hands with those they have not greeted yet. Each greeting is unique. Person 1 shaking hands with person 2 is the same event as person 2 shaking hands with person 1. d. At a particular restaurant, you must choose two different side dishes to accompany your meal. If there are eight side dishes to choose from, how many different possibilities are there? Combinations, 28. The order of dishes chosen does not matter. Choosing veggies and mac and cheese is the same as choosing mac and cheese and veggies. e. How many different four – letter", "able to show up at 2 Meetings per month instead of just one. Interested even a little bit? Please look over the list below. EVERY position listed is up for re-election. They say that there are 2 types of people in every organization, those that DO things and those that DON’T. Why is it that the smaller of the 2 groups always seems to be those that DO things? April is election month for PVCI & this year as in every year you the member have a unique opportunity to nominate those that you believe can make a difference for the continuance & betterment of the Club. Don’t forget that if you believe that you are that person, then please STAND UP & NOMINATE YOURSELF! The Club Bylaws read: The officers of this club shall consist of a president, vice-president, recording secretary, corresponding secretary and treasurer. There is also the board of directors. The officers and board of directors shall be elected from the membership, by the membership, at the regular membership meeting in April of each year. These are the positions and the duties that the various positions entail: President: The Head Honcho, the Main Man. Responsible for running both the regular club meeting and the Board of Directors Meeting and shall direct the activities of this", "club has 10 members. One president and two vice-presidents in the Problem Solving forum “Number of options for president = 10. (Any of the 10 members.) From the remaining 9 members, the number of ways to choose 2 to serve as vice-president = 9C2 = (9*8)/(2*1) = 36. To combine these options, we multiply: 10*36 = 360. The correct answer is E.” November 28, 2018", "Jan 2008, 11:23 20 A certain club has 10 members, including Harry. One of the 19 08 Jan 2007, 23:22 Display posts from previous: Sort by", "Free Online Trial Hour Re: A certain club has 20 members. What is the ratio of the memb [#permalink] 06 Jun 2016, 08:41 Go to page 1 2 Next [ 22 posts ] Similar topics Replies Last post Similar Topics: 8 A local club has between 24 and 57 members. The members of the club ca 9 10 May 2017, 12:39 30 A certain club has exactly 5 new members at the end of its 12 13 Sep 2016, 11:51 4 A certain club has 10 members, including Harry. One of the 8 25 Jul 2012, 21:13 145 A certain club has 10 members, including Harry. One of the 33 31 May 2017, 22:01 1 A certain team has 12 members, including Joey. A three-membe 13 14 Dec 2011, 10:58 Display posts from previous: Sort by", "each of the 450 clubs must have three students in common with at least one other club and if any club has students in common with four or more clubs then it cannot have more than one student in common. Given these requirements must there be a club where it has exactly one student in common with three other clubs. To provide a counter example I must show that all four club combinations have 0 students in common or it must have two or more students in common while satisfying the other conditions. Let's assume that every three club combination doesn't have any students in common, so to solve this we divide the 450 clubs into 225 sets of 2 clubs. To satisfy the first requirement, that each club must have three students in common with at least one other club, we put three students in each set where all students are apart of both clubs in their respective set this would mean that a minimum of 675 students are required. Now let's assume instead that every four club combination doesn't have any students in common. To solve this we divide the 450 clubs into 150 sets of 3 clubs. This would require 450 students by applying the same process the above paragraph and using the first observation", "+0 # counting question 0 26 1 A certain club has 50 people, and 10 members are running for president. Each club member either votes for one of the 10 candidates, or can abstain from voting. How many different possible vote totals are there? Nov 13, 2022 #1 +17 +1 There are 40 voters (50-10 =40since. we have to exclude the candidates). Each voter has 11 choices, since they can choose from 10 candidates or choose none (abstaining). So, the total amount of different possible vote totals is $$\\boxed{40\\cdot11 = 440}$$ Hope this helped Nov 13, 2022", "25 Apr 2012, 10:59 1 A certain team has 12 members, including Joey. A three-membe 13 23 Oct 2011, 13:16 12 A certain club has 20 members. What is the ratio of the memb 18 25 Mar 2009, 23:04 Display posts from previous: Sort by", "governing body. We already operate under one governing body. Our voting committee is made up of one representative from each club. Each representative is chosen by his or her respective club. The wisdom in this is that clubs know their members best and can choose who they believe will be best for the region and by extension, the club. If they feel they are not being represented properly, they can replace their representative at any time. The important thing to note is that each club has a seat at the table. Each representative controls between 0 and 6 votes. This is based on primarily on participation numbers. Currently SCNAX and CASOC each have the maximum allowed 6 votes, the remaining clubs less, with one club having 0 votes at the present time. At any time, if a club wishes to gain more influence, they up their participation. SCNAX is a great example of a club who has recently done that. When I first joined, the voting was dominated by GRA and No$. rmgibson wrote:I don't see how officially separating the social and business aspects of our community is a bad thing. If we can unify and get more people involved with the running of the organization then I'm all for it. Those of us who serve on the", "trip. if 39 people went on that trip, how many are in the club...", "Total members is 40>M>10, so 29 members. Now if each table has 6 members, that is 29/6 then the last table would have 5 (29-24) members on the table. Is this approach right? I did the problem in less than a minute with this approach. Thanks in advance. How did you get that there must be 29 members from 40>M>10? Does 29 members satisfy ANY of the conditions (3 members at one table and 4 members at each of the other tables and 3 members at one table and 5 members at each of the other tables)? _________________ IIMA, IIMC School Moderator Joined: 04 Sep 2016 Posts: 1340 Location: India WE: Engineering (Other) Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe [#permalink] ### Show Tags 27 Feb 2018, 09:14 Quote: Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at", "## DISTRICT DIRECTOR Awarded to 1 Division and 1 Area for achieving the following: • ## Qualifying Criteria Division Area The Division should be President’s Distinguished. 60% of the clubs need to be Distinguished (on base) 60% of the clubs need to have 20+ members (on base) 8% growth in Membership Payments. The Area should be President’s Distinguished. 75% of the clubs need to be Distinguished (on base) 50% of the clubs need to have 20+ members (on base) 8% growth in Membership Payments. • ## Tiebreaking Criteria Division Area Tiebreaker 1: Percentage of Clubs that have 20+ Members (on base) Tiebreaker 2: Percentage of Clubs that are distinguished. (on base) Tiebreaker 1: Percentage of Clubs that have 20+ Members (on base) Tiebreaker 2: Percentage of Clubs that are distinguished. (on base)", "+0 # Our Club Has 10 Members 0 211 1 +29 Our club has 10 members, and we need to appoint a president, a vice-president, and a secretary. How many ways can we do this if each member A.) can only have one position B.) members can hold any numbers of office ty Jun 4, 2021 (a) $10 \\cdot 9 \\cdot 8 = \\boxed{720}$ (b) $10^3 = \\boxed{1000}$", "Bookshelf and books How many ways can we place 7 books in a bookshelf? • School committee 7 students was elected to the school committee. In how many ways can the President, Vice-President, Secretary and Treasurer be selected?", "+0 # helppppp 0 238 1 +814 Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members may hold more than one office? Dec 9, 2018 $$\\text{The list of officers can be considered a 3 digit base 25 number}\\\\ N = (25)^3 = 15625$$", "four clubs causes more problems than it solves. By dividing the 450 clubs into 112 sets of four clubs and one set of two clubs, we have three students in each set where they are in all clubs in their respective set so we need a minimum of 339 students to show a counter example. Thus there must be at least one four club combination with exactly one student in common, because the more clubs students are in the less students are required to provide a counter example, unless the student is in five or more clubs in which case the number of students increases greatly.", "+0 # helppppp 0 60 1 +809 Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members may hold more than one office? Dec 9, 2018 #1 +3576 +1 $$\\text{The list of officers can be considered a 3 digit base 25 number}\\\\ N = (25)^3 = 15625$$ . Dec 9, 2018", "(2 pts:): A computer science department has 50 faculty members_ (a) How many ways are there to choose an executive committee COIL- sisting of 6 members? (b) How many ways are there to choose the chair vice chair. director of undergraduate study; director of graduate study, director of computing facilities_ and director of public relations if no person can hold more than position? Problem (2 pts:): A computer science department has 50 faculty members_ (a) How many ways are there to choose an executive committee COIL- sisting of 6 members? (b) How many ways are there to choose the chair vice chair. director of undergraduate study; director of graduate study, director of comput..." ]

[ "College Algebra (6th Edition) Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = 5040 Total number of members on its board of directors = 10 Number of members to elect president, vice president, secretary and treasurer = 4 Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = $\\frac{10!}{(10 - 4)! }$ = $\\frac{10!}{6! }$ = $\\frac{10\\times9\\times8\\times7\\times6!}{6! }$ = $10\\times9\\times8\\times7$ = 5040", "+0 0 78 2 +69 A club has 15 members and needs to choose 2 members to be co-presidents. In how many ways can the club choose its co-presidents? rohitaop May 14, 2018 #1 +867 +2 This is a simple choosing problem. 15 choose 2 can be expressed as: $$\\binom{15}{2}=\\frac{15\\cdot14}{2\\cdot1}=105$$ i hope this helped, Gavin GYanggg May 14, 2018 #2 +2765 +1 This is just combinations, 15C2- $$\\frac{15!}{2!*13!}=\\frac{15*14}{2*1}=15*7=\\boxed{105}$$ tertre May 14, 2018", "use the same kind of reasoning yet again. There are $12$ possible choices for president. Once the president is chosen, there are $11$ possible choices for vice president, since the president and vice president cannot be the same person. But there is no such restriction on the secretary or the treasurer: each of them can be any of the $12$ people, so each can be chosen in $12$ ways. The grand total is therefore $12\\cdot11\\cdot12\\cdot12=11\\cdot12^3$. 4. There are three cases, depending on whether one, two, or three yellow flags are used. • Suppose that just one yellow flag is used, so that we’re arranging $5$ flags. There are $\\binom52$ ways to choose which $2$ positions in the string of $5$ will contain the red flags. That leaves $3$ open positions, and there are $\\binom32$ ways to pick $2$ of them for the $2$ green flags. The lone yellow flag will then fill the remaining slot $-$ no choice there. The total number of such $5$-flag signals is therefore $\\binom52\\binom32=10\\cdot3=30$. • Suppose that two yellow flags are used. We reason in exactly the same way. There are $\\binom62$ ways to choose the $2$ positions to be filled by red flags, and then there are $\\binom42$ ways to choose which $2$ of the remaining $4$ slots will get the green", "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 president. Now, the vice president can be anyone except the person who you just designated as president, so there are 19 choices. Lastly, there are 18 choices for secretary. Following our method, we get that there are $20 \\cdot 19 \\cdot 18 = 6840$ different combinations of people for the three positions. Now I’ll change the problem a bit. There are 20 people in the class and you need to choose 3 class representatives from them. This may seem the same as the previous", "among the students chosen. In this case, we’ll use the combination. ${}_{20}C_{5}$ $= \\dfrac{20!}{5!(15)!}$ $= 15,504$ This is much easier using a calculator. Next is we have $20$ students to be picked for class office – Pres., VP, Sec., Treasurer, and Judge. What’s important is that there are five distinct order or places. It’s not just a group of $5$ different people. It’s a $5$ people each one has a specific position. In this case, order matters. So, we’ll use permutation. ${}_{20}P_{5}$ $= 20 \\times 19 \\times 18 \\times 17 \\times 16$ $= 1,860,480$ There is a distinct difference between the combination and permutation.", "Free Version Moderate # Permuations of Selected Positions SATSTM-LPE79G There are eight students in an after school club. One needs to be chosen as president and one as vice president. How many ways can the students be chosen for the two positions? A $15$ B $56$ C $64$ D $1200$ E $40320$", "+0 # Our Club Has 10 Members 0 211 1 +29 Our club has 10 members, and we need to appoint a president, a vice-president, and a secretary. How many ways can we do this if each member A.) can only have one position B.) members can hold any numbers of office ty Jun 4, 2021 (a) $10 \\cdot 9 \\cdot 8 = \\boxed{720}$ (b) $10^3 = \\boxed{1000}$", "# Thread: Homework help...not sure how to do this problem 1. ## Homework help...not sure how to do this problem There were seven nominees for president and four nominees for vice president. In how many ways can the slate be chosen? 2. Well 7 different possibilies for a president and after that there is 4 different vice president possibilities for each president. So $7 \\times 4= \\cdots$", "16\\cdot 15=4080$ options. For $(b)$ you have $7$ options for the General Secretary, times $16\\cdot 15=240$, which gives $1680$. For $(c)$ just multiply your answer by $3!$ to account for the fact that the positions are labeled.", "+0 # Permutations +8 90 3 +549 On the student government ballot at Dusable High School, the six candidates for president are listed first, followed by the four candidates for vice president, followed by the five candidates for secretary and ending with the three candidates for treasurer. In how many ways can the candidates be listed on the ballot? Jun 12, 2020 #1 +732 +1 isn't it just $$6! \\cdot 4! \\cdot 5! \\cdot 3! = 12441600‬ \\text{ ways }$$? Jun 13, 2020 #2 +732 +1 cause there's 6 candidates for president so that 6! there's 4 candidates for vp so that's 4! there's 5 candidates for secretary so that's 5! and there's 3 candidates for treasurer so that's 3! so that's just 6!*4!*3!*5! = 12441600", "club has 10 members. One president and two vice-presidents in the Problem Solving forum “Number of options for president = 10. (Any of the 10 members.) From the remaining 9 members, the number of ways to choose 2 to serve as vice-president = 9C2 = (9*8)/(2*1) = 36. To combine these options, we multiply: 10*36 = 360. The correct answer is E.” November 28, 2018", "Button Try our free Online GRE Test Manager Joined: 01 Nov 2018 Posts: 87 Followers: 0 Kudos [?]: 41 [0], given: 22 Re: Counting [#permalink] 10 Jan 2019, 22:35 Expert's post sandy wrote: Hi, In total there are 5 members to be selected out of 12. If all 12 members were the same then the choice would have been 12C5, i.e. Option A. But all members are not the same since there are two maned positions namely president and vice president. If all the positions were distict the choice 12P5 would be correct, i.e. Option D would have been right. There are only 2 named positions and 3 non named positions. No of ways to select a president= 12. No of ways to select a vice president =11. No of ways to select 3 members from the remaining 10 is 10C3= $$\\frac{10!}{7! \\times 3!}$$ Hence number of ways to selct president, vice president and 3 members = $$12 \\times 11 \\times\\frac{10!}{7! \\times 3!}$$= $$\\frac{12!}{7! \\times 3!}$$. Hence option B is correct! This was an awesome breakdown, thanks! Re: Counting [#permalink] 10 Jan 2019, 22:35 Display posts from previous: Sort by", "# Homework help...not sure how to do this problem Printable View • April 27th 2009, 03:30 PM doodle08 Homework help...not sure how to do this problem There were seven nominees for president and four nominees for vice president. In how many ways can the slate be chosen? • April 27th 2009, 04:35 PM pickslides Well 7 different possibilies for a president and after that there is 4 different vice president possibilities for each president. So $7 \\times 4= \\cdots$", "president. Now, the vice president can be anyone except the person who you just designated as president, so there are 19 choices. Lastly, there are 18 choices for secretary. Following our method, we get that there are $20 \\cdot 19 \\cdot 18 = 6840$ different combinations of people for the three positions. Now I’ll change the problem a bit. There are 20 people in the class and you need to choose 3 class representatives from them. This may seem the same as the previous problem, except that in this case if Alison, Bob, and Charlotte are representatives that’s the same as Bob, Alison, and Charlotte being representatives, whereas in the previous problem order mattered, since Alison being president and Bob being vice president was different from Bob being president and Alison being vice president. Order is a very important factor to consider in combinatorics. So what do we do? Well, we can still use the above method, but keep in mind that we counted all arrangements of the students Alison, Bob, and Charlie as different. How many arrangements are there? Well, there’s 1. A,B,C 2. A,C,B 3. B,A,C 4. B,C,A, 5. C,A,B 6. C,B,A. So 6 arrangements. Therefore, if we take our answer from before and divide by 6, that will correct for our over-counting. So there are", "# Math Help - (2) Probability Q's 1. ## (2) Probability Q's 1.) In a club of 10 members...How many ways are there to choose a President, and an Advising Council of 3 members? 2.) A deli offers ham sandwiches, roast beef sandwiches, cheese sandwiches, turkey sandwiches, and salami sandwiches. You are sent in to buy sandwiches for a group of ten people for lunch. How many different ways can you order lunch? 3.) In a club of 10 members...How many ways are there to choose a President, a Vice President, a Secretary, and a Treasurer? I'm I correct w/ using nCr (n=10,r=4) .... which equals 210 different ways? 2. Yes, $\\frac{10!}{4! \\cdot (10-4)!} = 210$ For Q1, you can choose a president 10 ways, and then for each choice of president, your council will consist of three people from a group of nine. $10 \\cdot \\frac{9!}{3! \\cdot (9-3)!} = 840$ For Q2, I think the answe is (n+k-1)C(k) . 3. Originally Posted by toop 1.) In a club of 10 members...How many ways are there to choose a President, and an Advising Council of 3 members? 2.) A deli offers ham sandwiches, roast beef sandwiches, cheese sandwiches, turkey sandwiches, and salami sandwiches. You are sent in to buy sandwiches for a group of ten people for lunch.", "different ways. The two student positions, however, are indistinguishable: all that matters is which two of the $7$ students are elected. There are $\\binom72$ possible pairs of students from the $7$ on the ballot, so there are $\\binom72$ possible ways to choose the student members. Putting the pieces together, we have $$9\\cdot8\\cdot7\\cdot\\binom72=9\\cdot8\\cdot7\\cdot21=10,584$$ different outcomes. Calculate the product of the following: • The number of ways to choose $3$ out of $9$ faculties, which is $\\binom93=84$ • The number of ways to arrange them in different order, which is $3!=6$ • The number of ways to choose $2$ out of $7$ students, which is $\\binom72=21$ $9$ options for number one faculty , $8$ options for number two, $7$ options for number three. $\\frac{7\\cdot6}{2}$ options for two students. $\\frac{9\\cdot8\\cdot 7\\cdot 7\\cdot 6}{2}=10584$", "group of 8 boys and 7 girls if: (a) There are no restrictions (b) There must be at least 2 boys chosen (c) There must be at least 1 girl chosen (a) There are 15 candidates. There are 15 choices for the President, . . 14 choices for the Vice President, . . 13 choices for the Treasure. Therefore, there are: . $15 \\cdot 14 \\cdot 13 \\:=\\:\\boxed{2730}$ ways. (b) \"at least 2 boys\" means \"2 boys or 3 boys\". If two boys (and one girl) are chosen, there are: . ${8\\choose2}{7\\choose1} = 196$ choices. Since each can hold a different office, there are $3! = 6$ assignments. . . Hence, there are: . $196 \\cdot 6 \\:=\\:1176$ ways for two boys. If three boys are chosen, there are: . $8 \\cdot 7 \\cdot 6 \\:=\\:336$ ways. Therefore, there are: . $1176 + 336 \\:=\\:\\boxed{1512}$ ways. (c) The opposite of \"at least one girl\" is \"no girls\" (or \"all boys\"). From (a) we know that there are $2730$ possible outcomes. From (b) we that there are $336$ outcomes with all boys. Therefore, \"at least one girl\" has: . $2730 - 336 \\:=\\:\\boxed{2394}$ ways.", "can be anyone except the person who you just designated as president, so there are 19 choices. Lastly, there are 18 choices for secretary. Following our method, we get that there are $20 \\cdot 19 \\cdot 18 = 6840$ different combinations of people for the three positions. Now I’ll change the problem a bit. There are 20 people in the class and you need to choose 3 class representatives from them. This may seem the same as the previous problem, except that in this case if Alison, Bob, and Charlotte are representatives that’s the same as Bob, Alison, and Charlotte being representatives, whereas in the previous problem order mattered, since Alison being president and Bob being vice president was different from Bob being president and Alison being vice president. Order is a very important factor to consider in combinatorics. So what do we do? Well, we can still use the above method, but keep in mind that we counted all arrangements of the students Alison, Bob, and Charlie as different. How many arrangements are there? Well, there’s 1. A,B,C 2. A,C,B 3. B,A,C 4. B,C,A, 5. C,A,B 6. C,B,A. So 6 arrangements. Therefore, if we take our answer from before and divide by 6, that will correct for our over-counting. So there are $\\frac{6840}{6}=1140$ ways to choose 3", "+0 +12 98 5 +344 In how many ways can a President and a Vice-President be chosen from a group of 5 people (assuming that the President and the Vice-President cannot be the same person)? Jun 13, 2020 #1 +549 0 There are 5 choices for the president and 4 choices for the vice-president (since there has already been someone chosen for the president). Once you multiply 5 $$\\times$$ 4 = $$\\boxed {20}$$ Jun 13, 2020 edited by amazingxin777 Jun 13, 2020 #2 +344 +4 Thanks Jun 13, 2020 #3 +549 +1 Ur welcome! amazingxin777 Jun 13, 2020 #5 +1130 +5 we do 5*4=20 becuase the vice-president cannot be the president and the group has 5 people in all and 5-1=4 people left for choosing the vice-president Jun 18, 2020", "8. anonymous Thanks :) 9. kropot72 You're welcome :) You can look at the choice of president and treasurer from the 12 girls very simply as follows: If one of the 12 girls is chosen as president there are 11 choices for treasurer. So for each one of the girls chosen as president there are 11 choices for treasurer. Going through each in sequence it will be seen that each one if the girls has a possibility of being either president or treasurer and the total number of ways of choosing is $12\\times 11$" ]

[ "+0 0 78 2 +69 A club has 15 members and needs to choose 2 members to be co-presidents. In how many ways can the club choose its co-presidents? rohitaop May 14, 2018 #1 +867 +2 This is a simple choosing problem. 15 choose 2 can be expressed as: $$\\binom{15}{2}=\\frac{15\\cdot14}{2\\cdot1}=105$$ i hope this helped, Gavin GYanggg May 14, 2018 #2 +2765 +1 This is just combinations, 15C2- $$\\frac{15!}{2!*13!}=\\frac{15*14}{2*1}=15*7=\\boxed{105}$$ tertre May 14, 2018", "with an explanation for your choice. a. A student club has 20 members. How many ways are there for the club to choose a president and a vice president? Permutations, 380. The role of president is different from that of vice president. The order of outcomes matters. b. A football team of 50 players will choose two co – captains. How many different ways are there to choose the two co – captains? Combinations, 1,225. Regardless of order, two players will attain the same outcome of co – captain. c. There are seven people who meet for the first time at a meeting. They shake hands with each other and introduce themselves. How many handshakes have been exchanged? Combinations, 21. People only shake hands with those they have not greeted yet. Each greeting is unique. Person 1 shaking hands with person 2 is the same event as person 2 shaking hands with person 1. d. At a particular restaurant, you must choose two different side dishes to accompany your meal. If there are eight side dishes to choose from, how many different possibilities are there? Combinations, 28. The order of dishes chosen does not matter. Choosing veggies and mac and cheese is the same as choosing mac and cheese and veggies. e. How many different four – letter", "+0 # Our Club Has 10 Members 0 211 1 +29 Our club has 10 members, and we need to appoint a president, a vice-president, and a secretary. How many ways can we do this if each member A.) can only have one position B.) members can hold any numbers of office ty Jun 4, 2021 (a) $10 \\cdot 9 \\cdot 8 = \\boxed{720}$ (b) $10^3 = \\boxed{1000}$", "Button Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Last edited by Carcass on 15 Dec 2017, 01:18, edited 2 times in total. Edited by Carcass GRE Prep Club Legend Joined: 07 Jun 2014 Posts: 4857 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) Followers: 105 Kudos [?]: 1775 [3] , given: 397 Re: Counting [#permalink] 13 Sep 2017, 14:21 3 KUDOS Expert's post Hi, In total there are 5 members to be selected out of 12. If all 12 members were the same then the choice would have been 12C5, i.e. Option A. But all members are not the same since there are two maned positions namely president and vice president. If all the positions were distict the choice 12P5 would be correct, i.e. Option D would have been right. There are only 2 named positions and 3 non named positions. No of ways to select a president= 12. No of ways to select a vice president =11. No of ways to select 3 members from the remaining 10 is 10C3= $$\\frac{10!}{7! \\times 3!}$$ Hence number of ways to selct president, vice president and 3 members = $$12 \\times 11 \\times\\frac{10!}{7! \\times 3!}$$= $$\\frac{12!}{7! \\times 3!}$$. Hence option B is correct! _________________ Sandy If you found this post useful, please let me know by pressing the Kudos", "Question How many ways can a president, vice-president, secretary, and treasurer be chosen from a club with 9 members?", "club has 10 members. One president and two vice-presidents in the Problem Solving forum “Number of options for president = 10. (Any of the 10 members.) From the remaining 9 members, the number of ways to choose 2 to serve as vice-president = 9C2 = (9*8)/(2*1) = 36. To combine these options, we multiply: 10*36 = 360. The correct answer is E.” November 28, 2018", "able to show up at 2 Meetings per month instead of just one. Interested even a little bit? Please look over the list below. EVERY position listed is up for re-election. They say that there are 2 types of people in every organization, those that DO things and those that DON’T. Why is it that the smaller of the 2 groups always seems to be those that DO things? April is election month for PVCI & this year as in every year you the member have a unique opportunity to nominate those that you believe can make a difference for the continuance & betterment of the Club. Don’t forget that if you believe that you are that person, then please STAND UP & NOMINATE YOURSELF! The Club Bylaws read: The officers of this club shall consist of a president, vice-president, recording secretary, corresponding secretary and treasurer. There is also the board of directors. The officers and board of directors shall be elected from the membership, by the membership, at the regular membership meeting in April of each year. These are the positions and the duties that the various positions entail: President: The Head Honcho, the Main Man. Responsible for running both the regular club meeting and the Board of Directors Meeting and shall direct the activities of this", "## DISTRICT DIRECTOR Awarded to 1 Division and 1 Area for achieving the following: • ## Qualifying Criteria Division Area The Division should be President’s Distinguished. 60% of the clubs need to be Distinguished (on base) 60% of the clubs need to have 20+ members (on base) 8% growth in Membership Payments. The Area should be President’s Distinguished. 75% of the clubs need to be Distinguished (on base) 50% of the clubs need to have 20+ members (on base) 8% growth in Membership Payments. • ## Tiebreaking Criteria Division Area Tiebreaker 1: Percentage of Clubs that have 20+ Members (on base) Tiebreaker 2: Percentage of Clubs that are distinguished. (on base) Tiebreaker 1: Percentage of Clubs that have 20+ Members (on base) Tiebreaker 2: Percentage of Clubs that are distinguished. (on base)", "# Math Help - (2) Probability Q's 1. ## (2) Probability Q's 1.) In a club of 10 members...How many ways are there to choose a President, and an Advising Council of 3 members? 2.) A deli offers ham sandwiches, roast beef sandwiches, cheese sandwiches, turkey sandwiches, and salami sandwiches. You are sent in to buy sandwiches for a group of ten people for lunch. How many different ways can you order lunch? 3.) In a club of 10 members...How many ways are there to choose a President, a Vice President, a Secretary, and a Treasurer? I'm I correct w/ using nCr (n=10,r=4) .... which equals 210 different ways? 2. Yes, $\\frac{10!}{4! \\cdot (10-4)!} = 210$ For Q1, you can choose a president 10 ways, and then for each choice of president, your council will consist of three people from a group of nine. $10 \\cdot \\frac{9!}{3! \\cdot (9-3)!} = 840$ For Q2, I think the answe is (n+k-1)C(k) . 3. Originally Posted by toop 1.) In a club of 10 members...How many ways are there to choose a President, and an Advising Council of 3 members? 2.) A deli offers ham sandwiches, roast beef sandwiches, cheese sandwiches, turkey sandwiches, and salami sandwiches. You are sent in to buy sandwiches for a group of ten people for lunch.", "+0 # Combination Probability +1 78 3 +485 There are 3 math clubs in the school district, with 5, 7, and 8 students respectively. Each club has two co-presidents. If I randomly select a club, and then randomly select three members of that club to give a copy of Introduction to Counting and Probability, what is the probability that two of the people who receive books are co-presidents? On numerous occasions, I have seen answers without work and have inputted them, and they are wrong most of the time. If you do not show work, I will not be using your answer. Apr 4, 2021 #1 +423 +1 Introduction to counting and probability is a great book ;) Let's first calculate the probability for each club then average them. 5 student club - There are $\\binom{5}{3}=10$ ways to choose the members. In $\\binom{2}{2}\\binom{3}{1}=3$ ways, the co-presidents are selected - the probability here is $\\frac{3}{10}$. 7 student and 8 student clubs are analogous. Try it, I believe in you! Apr 4, 2021 edited by thedudemanguyperson Apr 4, 2021 #2 +485 0 Thank you for your solution! I finished the problem for 7 and 8 students, and got $\\frac{11}{60}$, which was correct! Thanks again, this helped me a lot! RiemannIntegralzzz Apr 4, 2021 #3 +423 +1 No problem, keep the", "Free Version Moderate # Permuations of Selected Positions SATSTM-LPE79G There are eight students in an after school club. One needs to be chosen as president and one as vice president. How many ways can the students be chosen for the two positions? A $15$ B $56$ C $64$ D $1200$ E $40320$", "College Algebra (6th Edition) Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = 5040 Total number of members on its board of directors = 10 Number of members to elect president, vice president, secretary and treasurer = 4 Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = $\\frac{10!}{(10 - 4)! }$ = $\\frac{10!}{6! }$ = $\\frac{10\\times9\\times8\\times7\\times6!}{6! }$ = $10\\times9\\times8\\times7$ = 5040", "+0 # counting question 0 26 1 A certain club has 50 people, and 10 members are running for president. Each club member either votes for one of the 10 candidates, or can abstain from voting. How many different possible vote totals are there? Nov 13, 2022 #1 +17 +1 There are 40 voters (50-10 =40since. we have to exclude the candidates). Each voter has 11 choices, since they can choose from 10 candidates or choose none (abstaining). So, the total amount of different possible vote totals is $$\\boxed{40\\cdot11 = 440}$$ Hope this helped Nov 13, 2022", "dues required of their level of seniority are considered full (dues-paying) members. Regular members who have not paid all required dues are considered basic (non-paying) members. All regular members may vote and hold elected \\& appointed office, but only full members may use services and attend events sponsored by EEP Club. \\subsection{Honorary Membership} The board may extend honorary membership to friends of EEP who are not currently-enrolled students at CSULA. Honorary members are exempt from dues, and are ineligible to vote or hold elected office. %\\begin{figure} %\\centering %\\includegraphics[width=0.45\\textwidth]{img/membership} %\\end{figure} \\section{The Board} The EEP Club board is the main administrative, deliberative, and decision-making body of this organization. All EEP Club officers and advisors are members of the board. \\subsection{Officers} Officers of EEP Club fulfill special duties as enumerated below. \\subsubsection{Positions and Duties} \\paragraph{President} The President \\begin{inparaenum}[\\itshape a\\upshape)] \\item attends, participates and supervises all club activities; \\item supervises appointed officers; \\item presides over all club meetings; \\item chairs the Executive Council and casts the deciding vote in the case of a tie; \\item signs or co-signs all official documents; and \\item maintains relations with CSULA students, faculty \\& administration. \\end{inparaenum} \\paragraph{Vice President} The Vice President \\begin{inparaenum}[\\itshape a\\upshape)] \\item assumes the responsibilities of the President should the President be unable or unwilling to do so; \\item hold voting membership on the", "GMAT math practice question] A club has 10 members. One president and two vice-presidents are elected. In how many ways can they be selected? A. 240 B. 280 C. 300 D. 320 E. 360 There are two ways of doing. Either you select one president and then select 2 Vice presidents from remaining(that is 10-1=9 remaining) --10* or You select 2 vice president and then select 1 president from remaining(that is 10-2= 9 remaining) --10C2*8C1 For first way -- No of ways of selecting one President from 10 = 10C1 = 10 ways No of ways of selecting two Vice- President from remaining (9) = 9C2 Total number of ways = 10*9C2= 360 So E is the answer. Press Kudos if it helps!! CEO Joined: 11 Sep 2015 Posts: 3237 Location: Canada Re: A club has 10 members. One president and two vice-presidents are elect [#permalink] ### Show Tags 28 Nov 2018, 06:43 Top Contributor MathRevolution wrote: [Math Revolution GMAT math practice question] A club has 10 members. One president and two vice-presidents are elected. In how many ways can they be selected? A. 240 B. 280 C. 300 D. 320 E. 360 Take the task of selecting the president and two vice-presidents and break it into stages. Stage 1: Select the president There are 10 people to", "+0 # helppppp 0 238 1 +814 Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members may hold more than one office? Dec 9, 2018 $$\\text{The list of officers can be considered a 3 digit base 25 number}\\\\ N = (25)^3 = 15625$$", "# Combinatorics Free Version Easy COMBIN-NEGKV4 In how many ways can a president, a vice-president, and a treasurer be selected from a club with $6$ members if no one can hold more than one office? A $120$ B $60$ C $20$ D $15$ E none of the above", "Discussion Forum Daily Magic Spell Discussion - January 6, 2017 Daily Magic Spell Discussion - January 6, 2017 Today, we will discuss about the Daily Magic Spell proposed on January 6, 2017. Problem: Suppose you have a club of $20$ members. The club chooses four officers: President, Vice-President, Treasurer, and Secretary. They also choose someone to be in charge of fundraising. The four officers must all be different, but the member in charge of fundraising can be one of the officers. How many ways can we choose the four officers in the club with no restrictions? Solution: We can first choose the President in the club. There are $20$ options. After the President is chosen, when choosing the Vice-President, there are $19$ options. We repeat this process until all of the positions are filled. Therefore, there are $20 \\times 19 \\times 18 \\times 17 = 116280$ ways to fill the four positions. Additionally, we wish to appoint someone in the club to take charge of fundraising. This can be any one of the $20$ members in the club. Therefore, we have: $116280 \\times 20 = 2325600$ ways to appoint four officers and a person in charge of fundraising. The common approach that most students who answered this question incorrectly is most likely interpreting the problem incorrectly. Some users", "each of the 450 clubs must have three students in common with at least one other club and if any club has students in common with four or more clubs then it cannot have more than one student in common. Given these requirements must there be a club where it has exactly one student in common with three other clubs. To provide a counter example I must show that all four club combinations have 0 students in common or it must have two or more students in common while satisfying the other conditions. Let's assume that every three club combination doesn't have any students in common, so to solve this we divide the 450 clubs into 225 sets of 2 clubs. To satisfy the first requirement, that each club must have three students in common with at least one other club, we put three students in each set where all students are apart of both clubs in their respective set this would mean that a minimum of 675 students are required. Now let's assume instead that every four club combination doesn't have any students in common. To solve this we divide the 450 clubs into 150 sets of 3 clubs. This would require 450 students by applying the same process the above paragraph and using the first observation", "on an executive committee is $12,650$ The number of ways to select the club members for a president, vice president, secretary, and treasurer so that no person can hold more than one office is $303,600$. ## Example A group of $3$ athletes is $P$, $Q$, $R$. In how many ways can a team of $2$ members be formed? Here, as the order of members is not important, we will use the Combination formula. $C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}$ Putting values of $n$ and $r$: $n=3$ $r=2$ $C\\left(3,2 \\right)=\\frac{3!}{2!\\left(3-2\\right)!}$ $C\\left(3,2 \\right)=3$" ]

[ "clearly counting different things. They are talking about different units. When we make cookies, everyone agrees on the unit; we know what one cookie is. But brownies are different. Tabitha seems to think that a brownie is the thing that comes out of the oven. Griffin seems to think that a brownie is what you eat in one serving. One brownie according to Tabitha. One brownie, according to Griffin I have emphasized elsewhere the importance of the unit; that one is a more flexible concept than we might think. Fun follow-up question: Does the thing in this video count as one brownie? # Starting the Conversation Anytime there are things in groups—or things being cut—is a good time to talk about units. Grocery stores usually have express lanes where you have to have Ten items or fewerAsk your child whether someone with a dozen eggs could use that lane. What about someone with 12 apples in a bag? What if the apples are loose? When your child asks for two slices of pizza, take one slice, cut it down the middle, smile wryly and ask whether that’s OK. In all of these cases, the central question is What counts as one? Play with that question. Also, watch that video together. It’s a ton of fun.", "Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to: Matt asked: Did you eat more apples than I did, James? James: I don’t know. Did you, Ian, eat more apples than I did? Ian: I don’t know. Barry: Aha!! I figured out.. So, Barry figured out how many apples each person ate. Can you do the same? Matt: 1 Apple James: 2 Apples Ian: 3 Apples Barry: 5 Apples ## The Logic Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian", "Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to: Matt asked: Did you eat more apples than I did, James? James: I don’t know. Did you, Ian, eat more apples than I did? Ian: I don’t know. Barry: Aha!! I figured out.. So, Barry figured out how many apples each person ate. Can you do the same? Matt: 1 Apple James: 2 Apples Ian: 3 Apples Barry: 5 Apples ## The Logic Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian", "# But Who Ate The Most? Algebra Level 1 Five brothers stayed in the house with their mother. One day, their mother brought home some mangos. Alex woke up first. As he was hungry, he ate $$\\frac{1}{6}$$ of the mangoes, and headed out. Brian woke up next. As he was hungry, he ate $$\\frac{1}{5}$$ of what was left, and headed out. Charles woke up next. As he was hungry, he ate $$\\frac{1}{4}$$ of what was left, and headed out. Danny woke up next. As he was hungry, he ate $$\\frac{1}{3}$$ of what was left, and headed out. Euler woke up next. As he was hungry, he ate $$\\frac{1}{2}$$ of what was left, and headed out. Their mother came home and saw 3 mangoes in the basket. How many mangoes were there originally? ×", "# Difference between revisions of \"2015 AMC 10B Problems/Problem 4\" ## Problem 4 Four siblings ordered an extra large pizza. Alex ate $\\frac15$, Beth $\\frac13$, and Cyril $\\frac14$ of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed? $\\textbf{(A) } \\text{Alex, Beth, Cyril, Dan}$ $\\textbf{(B) } \\text{Beth, Cyril, Alex, Dan}$ $\\textbf{(C) } \\text{Beth, Cyril, Dan, Alex}$ $\\textbf{(D) } \\text{Beth, Dan, Cyril, Alex}$ $\\textbf{(E) } \\text{Dan, Beth, Cyril, Alex}$ ## Solution Let the total amount of slices be $60$. Then we know Alex ate $12$, Beth ate $20$, Cyril ate $15$, and Dan ate the remaining, or $13$. Then the sequence in decreasing order is $\\boxed{\\textbf{(C)} \\text{Beth, Cyril, Dan, Alex}}$.", "# The game with cookies on a red and a blue plate Aabzaari and Basharat play a game, alternating turns with Aabzaari going first. At the start of the game, there are 20 cookies on a red plate and 14 cookies on a blue plate. A legal move consists in • eating two cookies taken from the blue plate, or • eating two cookies taken from the red plate, or • moving one cookie from the red plate to the blue plate (but moving cookies from the blue plate to the red plate is not allowed). The last player to make a legal move wins the game. Which player can guarantee that he wins no matter what strategy the opponent chooses? • They both lose, eating 16 or 18 cookies in one sitting probably isn't healthy... – aslum Feb 22 '16 at 18:10 It seems to me that Aabzaari always wins, regardless of the moves performed. There are cookies for 17 turns so Aabzaari gets to eat the last two. The third move doesn't consume any cookies and always allows a paired move, so it doesn't affect the parity of the turns. • yeah, you're right. Even if Aabzaari actually wanted to let Basharat win it simply isn't possible – Ivo Beckers Feb 22 '16 at 15:11 Aabzaari", "that I had when I was a kid. And it would say one dollar for a dozen cookies, 1.25 for the chocolate chip cookies. And it was really fun. I used to just go up and down and up the street in the wagon and they would buy the cookies. I'd keep the money. Besides that, I would tag around with my dad who had a construction business, and I'd help them on different jobs and things. So really understanding some of the businesses through him and creating invoices and so business plus the love of food and making a couple of dollars here and there, kind of, I think that set a lot of the foundational work for what I ended up doing later, even today. Lee: [00:05:43] That's interesting. Did you have any siblings growing up? Andrew: [00:05:47] Four younger sisters. Lee: [00:05:49] Wow. Wow. Andrew: [00:05:51] {laughter} So they got a few to take care of. But yeah, it's nice. I was the only guy and the older brother in the family. Lee: [00:05:59] Sounds like a full house. And so what did your mom do? Andrew: [00:06:03] So my mom was, she recently retired, she was a researcher for Sickle Cell Anemia up in the Bronx at the College of Albert Einstein College of", "nearly so important. She was four at the time; multiplication facts typically become important around eight or nine years old. # The unit is the thing that you count Griffin (eight years old) and Tabitha (five years old) were discussing the day’s activities. The feature activity had been making brownies with Mommy. This occurred while Griffin was out of the house. Griffin: How many brownies did you make? Tabitha: One big one! Mommy cut it up. # So What Do We Learn? What makes this more than just a funny story is that Griffin and Tabitha are clearly counting different things. They are talking about different units. When we make cookies, everyone agrees on the unit; we know what one cookie is. But brownies are different. Tabitha seems to think that a brownie is the thing that comes out of the oven. Griffin seems to think that a brownie is what you eat in one serving. One brownie according to Tabitha. One brownie, according to Griffin I have emphasized elsewhere the importance of the unit; that one is a more flexible concept than we might think. Fun follow-up question: Does the thing in this video count as one brownie? # Starting the Conversation Anytime there are things in groups—or things being cut—is a good time to talk about", "each, there will be no oranges left over. If the same group of children receives 8 pears each, there will be 21 pears leftover. How many children and oranges are there? • The bakery The bakery delivered 60 kg of bread to the store, the breads weighed 1 kg and 2 kg, a total of 35 pieces. How many loaves weighed 1 kg and how many were two-kilo loaves? • Tourist cottage The tourist cottage has 11 rooms for 16 guests, some of them are doubles, some singles. How many are double rooms and how many are single rooms? • Large family I have as many brothers as sisters and each my brother has twice as many sisters as brothers. How many children do parents have?", "# Difference between revisions of \"2015 AMC 10B Problems/Problem 4\" ## Problem 4 Four siblings ordered an extra large pizza. Alex ate $\\frac15$, Beth $\\frac13$, and Cyril $\\frac14$ of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed? $\\textbf{(A) } \\text{Alex, Beth, Cyril, Dan}$ $\\textbf{(B) } \\text{Beth, Cyril, Alex, Dan}$ $\\textbf{(C) } \\text{Beth, Cyril, Dan, Alex}$ $\\textbf{(D) } \\text{Beth, Dan, Cyril, Alex}$ $\\textbf{(E) } \\text{Dan, Beth, Cyril, Alex}$ ## Solution Let the pizza have $60$ slices, since the least common multiple of $(5,3,4)=60$. Therefore, Alex ate $\\frac{1}{5}\\times60=12$ slices, Beth ate $\\frac{1}{3}\\times60=20$ slices, and Cyril ate $\\frac{1}{4}\\times60=15$ slices. Dan must have eaten $60-(12+20+15)=13$ slices. In decreasing order, we see the answer is $\\boxed{\\textbf{(C) }\\text{Beth, Cyril, Dan, Alex}}$. ~savannahsolver", "Albert, Alex, Arjun, Billy, Mitchell, Saketh, SeungIn and Sreenath have a huge pile of Christmas cookies. They do an epic sleepover at Saketh’s house, where the cookies are. In the middle of the night, Albert wakes up and eats $1$ more than $\\frac{1}{3}$ of the cookies, and goes back to bed. Later that night, Alex eats $2$ more than $\\frac{1}{3}$ of the remaining cookies, and goes back to bed. Arjun gets up later, eats $3$ more than $\\frac{1}{3}$ of the remaining cookies, and goes back to bed, and so on. The $n$th person eats $n$ more than $\\frac{1}{3}$ of the cookies. If there are $238$ cookies remaining after each person has eaten cookies, how many cookies were there to start with?", "# Pimps There were 24 pimps in the plate. Maros ate 12, his little sister four times less. How many pimps remained in the plate? x = 9 ### Step-by-step explanation: $x=24-12-12\\mathrm{/}4=9$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Do you want to perform natural numbers division - find the quotient and remainder? ## Related math problems and questions: • In the bowl There are 12 pears in the bowl. Erik ate a third. How many pears remained in the bowl? • Baking There are 28 bunches, and son ate 1/2, dad ate four bunches. How many of them remain on the baking dishes? • Brother and sister When I was 8, my sister was 4, and now I am 18. How old is my sister? Janka and Marienka calculated that there are 210 gingerbreads on the gingerbread house. Janko ate one-seventh of all gingerbreads, and Marienka ate a third less than Janko. How many gingerbreads remained in the gingerbread house? • Chocolate How many times do we have to break a chocolate bar composed of 10 × 12 pieces to get the 120 parts? • Seven times Which number seven times is just as higher as 27, how much is smaller than 29? • How", "Griffin: How many brownies did you make? Tabitha: One big one! Mommy cut it up. # So What Do We Learn? What makes this more than just a funny story is that Griffin and Tabitha are clearly counting different things. They are talking about different units. When we make cookies, everyone agrees on the unit; we know what one cookie is. But brownies are different. Tabitha seems to think that a brownie is the thing that comes out of the oven. Griffin seems to think that a brownie is what you eat in one serving. One brownie according to Tabitha. One brownie, according to Griffin I have emphasized elsewhere the importance of the unit; that one is a more flexible concept than we might think. Fun follow-up question: Does the thing in this video count as one brownie? # Starting the Conversation Anytime there are things in groups—or things being cut—is a good time to talk about units. Grocery stores usually have express lanes where you have to have Ten items or fewerAsk your child whether someone with a dozen eggs could use that lane. What about someone with 12 apples in a bag? What if the apples are loose? When your child asks for two slices of pizza, take one slice, cut it down the middle, smile", "out of the house. Griffin: How many brownies did you make? Tabitha: One big one! Mommy cut it up. # So What Do We Learn? What makes this more than just a funny story is that Griffin and Tabitha are clearly counting different things. They are talking about different units. When we make cookies, everyone agrees on the unit; we know what one cookie is. But brownies are different. Tabitha seems to think that a brownie is the thing that comes out of the oven. Griffin seems to think that a brownie is what you eat in one serving. One brownie according to Tabitha. One brownie, according to Griffin I have emphasized elsewhere the importance of the unit; that one is a more flexible concept than we might think. Fun follow-up question: Does the thing in this video count as one brownie? # Starting the Conversation Anytime there are things in groups—or things being cut—is a good time to talk about units. Grocery stores usually have express lanes where you have to have Ten items or fewerAsk your child whether someone with a dozen eggs could use that lane. What about someone with 12 apples in a bag? What if the apples are loose? When your child asks for two slices of pizza, take one slice, cut it", "Apples Barry: 5 Apples ## The Logic Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian must have eaten 3 or 4 apples. Barry can only deduce the other amounts if he ate 5 apples. And the rest, in order to add up to 11 , must have eaten 1, 2 and 3. Inspired from a childhood heard puzzle. Related Articles Found on this Blog: ## Fox – Rabbit Chase Problems ##### Part I: A fox chases a rabbit. Both run at the same speed $v$. At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $\\alpha$ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is $l$. When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation? ##### Part II: Similarly think about the same situation, except now let the", "ways as follow: Improved chessboard images via this website using Creative Commons. ## How many apples did each automattician eat? Image via Wikipedia Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to: Matt asked: Did you eat more apples than I did, James? James: I don’t know. Did you, Ian, eat more apples than I did? Ian: I don’t know. Barry: Aha!! I figured out.. So, Barry figured out how many apples each person ate. Can you do the same? Matt: 1 Apple James: 2 Apples Ian: 3 Apples Barry: 5 Apples ## The Logic Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or", "many apples did each automattician eat? Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew…", "has 3 oatmeal bars to share with his 2 sisters and 3 of his friends. There are a total of 6 friends and he plans to give each person. Hence each person gets an oatmeal bar of $$\\frac{3}{6}$$. Elijah and Zoey each have an orange cut into 8 equal sections. They each ate one section, or $$\\frac{1}{8}$$, of their orange. How many more sections altogether do Elijah and Zoey need to eat to finish both oranges? TEKS 3.3.C", "divided into only seven pieces, with me left out. 7. Electra made one cut with the same cookie cutter, plus two straight cuts. 8. Alcyone made one cut with the second cookie cutter, plus two straight cuts. 9. Taygete made one cut with the third cookie cutter, plus two straight cuts. 10. Asterope made one cut with each of two previously used cookie cutters, plus one straight cut. 11. Celaeno made two cuts with one of the previously used cookie cutters, plus one straight cut. 12. Merope made one cut with one of the used cookie cutters, plus three cuts with another of the used cookie cutters. The Surviving Sisters Statements (all 6 made identical statements): 1. All the manager's statements are true. 2. I divided my cake into exactly seven equal portions (but not necessarily seven pieces), and shared the portions exactly equally with each of my sisters. Ernie read the statements, paused for a moment, and then picked up a pad and drew eight cryptic diagrams on it. \"As far as METHOD goes, I agree that any one of the witnesses could be lying\", he said, \"so any one of the sisters could have poisoned their cake and given Maya a slightly larger piece. It's even a possible suicide suicide. But assuming there wasn't some sort", "had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to: Matt asked: Did you eat more apples than I did, James? James: I don’t know. Did you, Ian, eat more apples than I did? Ian: I don’t know. Barry: Aha!! I figured out.. So, Barry figured out how many apples each person ate. Can you do the same? Matt: 1 Apple James: 2 Apples Ian: 3 Apples Barry: 5 Apples ## The Logic Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian must have eaten 3 or 4 apples. Barry can only deduce the other amounts if he ate 5 apples. And the rest, in order to" ]

[ "# Pimps There were 24 pimps in the plate. Maros ate 12, his little sister four times less. How many pimps remained in the plate? x = 9 ### Step-by-step explanation: $x=24-12-12\\mathrm{/}4=9$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Do you want to perform natural numbers division - find the quotient and remainder? ## Related math problems and questions: • In the bowl There are 12 pears in the bowl. Erik ate a third. How many pears remained in the bowl? • Baking There are 28 bunches, and son ate 1/2, dad ate four bunches. How many of them remain on the baking dishes? • Brother and sister When I was 8, my sister was 4, and now I am 18. How old is my sister? Janka and Marienka calculated that there are 210 gingerbreads on the gingerbread house. Janko ate one-seventh of all gingerbreads, and Marienka ate a third less than Janko. How many gingerbreads remained in the gingerbread house? • Chocolate How many times do we have to break a chocolate bar composed of 10 × 12 pieces to get the 120 parts? • Seven times Which number seven times is just as higher as 27, how much is smaller than 29? • How", "# Golden Pancakes Alice and Bob play a game with two stacks of fluffy, golden pancakes, of sizes $p$ and $q$. They alternate turns, with Alice starting. On their turn, a player must eat some number of pancakes from the larger stack. The number they eat must be a multiple of the size of the smaller stack. Since the bottom pancake of each stack is soggy, the first player to finish a stack loses. For example, if the stacks were sized $5$ and $17$, then the current player can eat either $5,10$ or $15$ pancakes from the large stack. For which values of $p$ and $q$ does Alice have a winning strategy? Source: Mathematical Puzzles: A Connoisseur's Collection, Peter Winkler. • None, because that is a lot of pancakes, and she is going to be sick by the end no matter what. – Aggie Kidd Jul 8 '15 at 1:51 • Do I understand correctly: Does it mean that when Alice eats 15 from the second (larger stack), then there are stacks 5 and 2, so Bob has to eat 2 or 4 from the first stack? – Voitcus Jul 8 '15 at 3:52 • I hope the pancakes aren't made from real gold ;). – Bojidar Marinov Jul 8 '15 at 10:05 • In next weeks episode,", "# There's a lot of ratio talk here. What is the importance of all these ratios? • Module 2 Week 3 Day 12 Your Turn Part 2 There's a lot of ratio talk here. What is the importance of all these ratios? (Timestamp 4:40) I thought subtraction isn't commutative... • @TSS-Graviser Hi again! Ratios are good way of comparing one number with another one. Let's start with two numbers. The bigger number represents an older sister and the smaller number represents a little brother . When mother makes cookies, older sister eats $$4$$ cookies but the little brother eats only $$2$$ :cookies. How would you compare the amount of food that they each eat? Would you say that the older sister eats $$2$$ more cookies than the little brother ? Should the older sister always eat $$2$$ more of whatever food there is, compared with her little brother ? But then what if mother serves rice -- should the older sister have exactly $$1$$ more grain of rice compared with the little brother ? No, that wouldn't make sense! If the older sister got just $$1$$ more grain of rice compared with the little brother , then she would be eating almost the same amount of rice as the little brother, , because the little brother might eat", "to (2,3). Then Bob has to eat 2, leading to (1,2), so Alice can eat 1, leading to (1,1) and Bob loses – Ben Aaronson Jul 8 '15 at 12:29 • @HemantRupani Ben Aarons on is correct about the (2,5) case. In general, my analysis only proves the existence of a winning strategy for Alice under certain conditions. It doesn't actually show what that strategy is. I find non-constructive results like that fascinating, although I must admit this result alone is not particularly helpful to Alice while she's playing... – John Stevens Jul 8 '15 at 15:19 • Yes! My bad.... – Hemant Rupani Jul 8 '15 at 16:10 Provided they have unlimited appetites, Alice will win when $p=nq~or~q=np~for~(n\\in N \\& n>1)$ Because, when Alice starts she'll eat $\\Big((n-1)\\times~smaller~stack\\Big)$, then both stacks will be same, and Bob will have to eat a stack(full). Alice will win. • ... when Alice start he'll eat ... Are you sure it isn't she ? :) – Bojidar Marinov Jul 8 '15 at 10:06 • @BojidarMarinov, No! Thanks! Now I see Alice shows a girls on Google Images :P so edited. – Hemant Rupani Jul 8 '15 at 10:17 • @BojidarMarinov - I just assumed it was Alice Cooper vs Bob, in which case both are male. I also like the visual", "# Difference between revisions of \"2015 AMC 10B Problems/Problem 4\" ## Problem 4 Four siblings ordered an extra large pizza. Alex ate $\\frac15$, Beth $\\frac13$, and Cyril $\\frac14$ of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed? $\\textbf{(A) } \\text{Alex, Beth, Cyril, Dan}$ $\\textbf{(B) } \\text{Beth, Cyril, Alex, Dan}$ $\\textbf{(C) } \\text{Beth, Cyril, Dan, Alex}$ $\\textbf{(D) } \\text{Beth, Dan, Cyril, Alex}$ $\\textbf{(E) } \\text{Dan, Beth, Cyril, Alex}$ ## Solution Let the pizza have $60$ slices, since the least common multiple of $(5,3,4)=60$. Therefore, Alex ate $\\frac{1}{5}\\times60=12$ slices, Beth ate $\\frac{1}{3}\\times60=20$ slices, and Cyril ate $\\frac{1}{4}\\times60=15$ slices. Dan must have eaten $60-(12+20+15)=13$ slices. In decreasing order, we see the answer is $\\boxed{\\textbf{(C) }\\text{Beth, Cyril, Dan, Alex}}$. ~savannahsolver", "# Difference between revisions of \"2015 AMC 10B Problems/Problem 4\" ## Problem 4 Four siblings ordered an extra large pizza. Alex ate $\\frac15$, Beth $\\frac13$, and Cyril $\\frac14$ of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed? $\\textbf{(A) } \\text{Alex, Beth, Cyril, Dan}$ $\\textbf{(B) } \\text{Beth, Cyril, Alex, Dan}$ $\\textbf{(C) } \\text{Beth, Cyril, Dan, Alex}$ $\\textbf{(D) } \\text{Beth, Dan, Cyril, Alex}$ $\\textbf{(E) } \\text{Dan, Beth, Cyril, Alex}$ ## Solution Let the total amount of slices be $60$. Then we know Alex ate $12$, Beth ate $20$, Cyril ate $15$, and Dan ate the remaining, or $13$. Then the sequence in decreasing order is $\\boxed{\\textbf{(C)} \\text{Beth, Cyril, Dan, Alex}}$.", "each, there will be no oranges left over. If the same group of children receives 8 pears each, there will be 21 pears leftover. How many children and oranges are there? • The bakery The bakery delivered 60 kg of bread to the store, the breads weighed 1 kg and 2 kg, a total of 35 pieces. How many loaves weighed 1 kg and how many were two-kilo loaves? • Tourist cottage The tourist cottage has 11 rooms for 16 guests, some of them are doubles, some singles. How many are double rooms and how many are single rooms? • Large family I have as many brothers as sisters and each my brother has twice as many sisters as brothers. How many children do parents have?", "Albert, Alex, Arjun, Billy, Mitchell, Saketh, SeungIn and Sreenath have a huge pile of Christmas cookies. They do an epic sleepover at Saketh’s house, where the cookies are. In the middle of the night, Albert wakes up and eats $1$ more than $\\frac{1}{3}$ of the cookies, and goes back to bed. Later that night, Alex eats $2$ more than $\\frac{1}{3}$ of the remaining cookies, and goes back to bed. Arjun gets up later, eats $3$ more than $\\frac{1}{3}$ of the remaining cookies, and goes back to bed, and so on. The $n$th person eats $n$ more than $\\frac{1}{3}$ of the cookies. If there are $238$ cookies remaining after each person has eaten cookies, how many cookies were there to start with?", "how many presents you get. As such, this year, as you know, I have been keeping a tally of all the times you have done something naughty (η) and all of the times you have done something nice (η). You get one scratch for a naughty act and one for a nice act, an act of kindness if you will, and dependent of course on the act. Remember that time you fooled your sister your type 1 hypersensitive to tomatoes sister into thinking that ketchup you had mixed in her ice-cream was actually raspberry juice, well that got you ten scratches in the naughty column, for example, whilst tidying up your Lego without being asked that one time you did it in the whole year got you five scratches in the nice column. Here's a little section of the tallies I've been keeping for this year: Overall you had 740 naughty scratches this year and 623 nice scratches, giving ηi = 0.46: \\begin{array}{l}\\begin{align}\\frac{{623}}{{740 + 623}} &= \\frac{{623}}{{1363}}\\\\\\\\ &= 0.46\\end{align}\\end{array} Now, this doesn't mean that you have a 46% chance of getting your presents this year. Oh no. You wrote to Santa to ask for the following five presents: • An umbrella in the shape of an Octopus, • A Lego Batman car [I presume you mean the Batmobile]", "+0 # help 0 75 1 At a party. Andy and April shared a cherry pie. Andy ate 2/5 of the pie and April ate 1/3 of the pie. What fraction of the pie did they eat altogether Jan 18, 2020 #1 +24422 +2 At a party. Andy and April shared a cherry pie. Andy ate 2/5 of the pie and April ate 1/3 of the pie. What fraction of the pie did they eat altogether $$\\begin{array}{|rcll|} \\hline &&\\mathbf{ \\dfrac{2}{5} + \\dfrac{1}{3} } \\\\\\\\ &=& \\dfrac{2*3+1*5}{5*3}\\\\\\\\ &=& \\mathbf{ \\dfrac{11}{15}} \\\\ \\hline \\end{array}$$ Jan 18, 2020", "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", "# But Who Ate The Most? Algebra Level 1 Five brothers stayed in the house with their mother. One day, their mother brought home some mangos. Alex woke up first. As he was hungry, he ate $$\\frac{1}{6}$$ of the mangoes, and headed out. Brian woke up next. As he was hungry, he ate $$\\frac{1}{5}$$ of what was left, and headed out. Charles woke up next. As he was hungry, he ate $$\\frac{1}{4}$$ of what was left, and headed out. Danny woke up next. As he was hungry, he ate $$\\frac{1}{3}$$ of what was left, and headed out. Euler woke up next. As he was hungry, he ate $$\\frac{1}{2}$$ of what was left, and headed out. Their mother came home and saw 3 mangoes in the basket. How many mangoes were there originally? ×", "# The game with cookies on a red and a blue plate Aabzaari and Basharat play a game, alternating turns with Aabzaari going first. At the start of the game, there are 20 cookies on a red plate and 14 cookies on a blue plate. A legal move consists in • eating two cookies taken from the blue plate, or • eating two cookies taken from the red plate, or • moving one cookie from the red plate to the blue plate (but moving cookies from the blue plate to the red plate is not allowed). The last player to make a legal move wins the game. Which player can guarantee that he wins no matter what strategy the opponent chooses? • They both lose, eating 16 or 18 cookies in one sitting probably isn't healthy... – aslum Feb 22 '16 at 18:10 It seems to me that Aabzaari always wins, regardless of the moves performed. There are cookies for 17 turns so Aabzaari gets to eat the last two. The third move doesn't consume any cookies and always allows a paired move, so it doesn't affect the parity of the turns. • yeah, you're right. Even if Aabzaari actually wanted to let Basharat win it simply isn't possible – Ivo Beckers Feb 22 '16 at 15:11 Aabzaari", "# Bitoo and Reena Bitoo ate 3/5 part of an apple and the remaining part was eaten by his sister Reena. How much part of an apple did Renna eat? Who had the larger share? By how much? Result R = 0.4 X = (Correct answer is: B) Y = 0.2 ### Step-by-step explanation: $Y=B-R=0.6-0.4=\\frac{1}{5}=0.2$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Need help to calculate sum, simplify or multiply fractions? Try our fraction calculator. ## Related math problems and questions: • Roma ate Roma ate 2/5 of a cake while Somya ate 3/7 of the same cake. Who ate more and by how much? • Cereals Ari and Joey share a 30-ounce box of cereal. By the end of the week, Ari has eaten 3/10 of the box, and Joey has eaten 3/5 of the box of cereal. How many ounces are left in the box? • Pizzas Billy ate 1 1/4 pizzas and John ate 1 2/3 pizzas. How much more pizza did John eat than Billy? • Pizza fractions Ann ate a third of a pizza and then another quater. Total part of pizza eaten by Ann and how much pizza is left? Cris had 15000 . He spent 1/3 of his money", "13. Alexa spends $37.74 buying toy train cars for her nieces. How many train cars does Alexa buy? Answer: Number of train cars Alexa buys = 6. Explanation: Amount of money Alexa spends buying toy train cars for her nieces =$37.74. Cost of train car = $6.29. Number of train cars Alexa buys = Amount of money Alexa spends buying toy train cars for her nieces ÷ Cost of train car =$37.74 ÷ \\$6.29 = 6. Question 14. Shaniqua uses the ingredients shown in the table to make lemonade. She pours herself a 0.5-cup serving of the lemonade to sample the taste. She then pours the remaining lemonade into 0.75-cup glasses for her friends. Write an expression to model this situation. Answer: Number of cups of lemonade she fills into glasses = 10. Explanation: Number of cups lemon juice = 2.5. Number of cups water = 5.5. Number of cups serving of the lemonade to sample the taste she pours herself = 0.5. Number of cups remaining lemonade she pours glasses for her friends = 0.75. Number of cups of lemonade she fills into glasses = [Number of cups water + (Number of cups lemon juice – Number of cups serving of the lemonade to sample the taste she pours herself)] ÷ Number of cups remaining lemonade she", "the only difference between one Double Stuf cookie and one Triple-Double cookie is 1 wafer and 30 calories. What can we conclude? That a wafer has 30 calories. But wait! That’s over 50% more calories than we got under the assumption that a Double Stuf has double stuf. We have reached a contradiction, which means we need to reject our initial assumption. Double Stuf ain’t double stuf. QED. ## My Oreo manifesto Math is powerful. No one has demonstrated this more frequently and more convincingly than Al Sicherman of the StarTribune. In his Tidbits column, Sicherman keeps tabs on prices and variations of processed foods. A favorite category for him recently has been Oreos. Writing recently on the topic of the new Peppermint Creme Oreos (which he describes as toothpastey), Sicherman says: All of that is not to say that there is no significant difference between Cool Mint Creme Oreos and Peppermint Creme Oreos: They are the same price on the shelf, but the package of Cool mint Oreos is 15.25 ounces (30 cookies), and that of Peppermint Creme Oreos is 10.5 ounces (20 cookies). So Peppermint Creme Oreos cost 45 percent more per ounce — or 50 percent more per cookie. Whoomp! There it is! Math, baby! Gotta love this guy. (But seriously, would it kill the", "# Bitoo and Reena Bitoo ate 3/5 part of an apple and the remaining part was eaten by his sister Reena. How much part of an apple did Renna eat? Who had the larger share? By how much? Result R = 0.4 X = (Correct answer is: B) Y = 0.2 ### Step-by-step explanation: $Y=B-R=0.6-0.4=\\frac{1}{5}=0.2$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Need help to calculate sum, simplify or multiply fractions? Try our fraction calculator. ## Related math problems and questions: • Roma ate Roma ate 2/5 of a cake while Somya ate 3/7 of the same cake. Who ate more and by how much? • Pizzas Billy ate 1 1/4 pizzas and John ate 1 2/3 pizzas. How much more pizza did John eat than Billy? • Cereals Ari and Joey share a 30-ounce box of cereal. By the end of the week, Ari has eaten 3/10 of the box, and Joey has eaten 3/5 of the box of cereal. How many ounces are left in the box? • Pizza fractions Ann ate a third of a pizza and then another quater. Total part of pizza eaten by Ann and how much pizza is left? Cris had 15000 . He spent 1/3 of his money", "of 2 × 3 × 4 = 24 ways that Andy can dress using this wardrobe. Do you Like this:", "least) two obstacles with multiplicative relative numbers like percentage (while additive relative numbers do not have these problems!). Sandy has 70% more puppets than Bob means that the number of puppets of Sandy is 170% of the number of puppets of Bob. And if an employee earns 4% more than before, than he now earns 104%. This subtle language difference on putting the percentage information on comparing the numbers directly (104%) versus expressing only their difference as a percentage (4%) of one of both is somehow tricky. A common mistake is to say \"I have 130% more cookies than Bob\", while meaning \"I have 30% more cookies than Bob, namely 130% of the number of cookies of Bob\". Another obstacle of multiplicative relations is that they are not \"symmetric\". In the additive comparison it is clear that if the cousin is 18 years older than you, than you are 18 years younger than the cousin. A similar thing does not hold for multiplicative relative numbers! Although Sandy has 70% more puppets than Bob, Bob has only 41.18% less puppets than Sandy! The tricky part is that in the first case 100% refers to Bob's number of puppets, while in the second part 100% refers to Sandys number of puppets. So if you now earn 4% more than before,", "had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to: Matt asked: Did you eat more apples than I did, James? James: I don’t know. Did you, Ian, eat more apples than I did? Ian: I don’t know. Barry: Aha!! I figured out.. So, Barry figured out how many apples each person ate. Can you do the same? Matt: 1 Apple James: 2 Apples Ian: 3 Apples Barry: 5 Apples ## The Logic Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian must have eaten 3 or 4 apples. Barry can only deduce the other amounts if he ate 5 apples. And the rest, in order to" ]

[ "# Difference between revisions of \"2015 AMC 10B Problems/Problem 4\" ## Problem 4 Four siblings ordered an extra large pizza. Alex ate $\\frac15$, Beth $\\frac13$, and Cyril $\\frac14$ of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed? $\\textbf{(A) } \\text{Alex, Beth, Cyril, Dan}$ $\\textbf{(B) } \\text{Beth, Cyril, Alex, Dan}$ $\\textbf{(C) } \\text{Beth, Cyril, Dan, Alex}$ $\\textbf{(D) } \\text{Beth, Dan, Cyril, Alex}$ $\\textbf{(E) } \\text{Dan, Beth, Cyril, Alex}$ ## Solution Let the total amount of slices be $60$. Then we know Alex ate $12$, Beth ate $20$, Cyril ate $15$, and Dan ate the remaining, or $13$. Then the sequence in decreasing order is $\\boxed{\\textbf{(C)} \\text{Beth, Cyril, Dan, Alex}}$.", "# Difference between revisions of \"2015 AMC 10B Problems/Problem 4\" ## Problem 4 Four siblings ordered an extra large pizza. Alex ate $\\frac15$, Beth $\\frac13$, and Cyril $\\frac14$ of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed? $\\textbf{(A) } \\text{Alex, Beth, Cyril, Dan}$ $\\textbf{(B) } \\text{Beth, Cyril, Alex, Dan}$ $\\textbf{(C) } \\text{Beth, Cyril, Dan, Alex}$ $\\textbf{(D) } \\text{Beth, Dan, Cyril, Alex}$ $\\textbf{(E) } \\text{Dan, Beth, Cyril, Alex}$ ## Solution Let the pizza have $60$ slices, since the least common multiple of $(5,3,4)=60$. Therefore, Alex ate $\\frac{1}{5}\\times60=12$ slices, Beth ate $\\frac{1}{3}\\times60=20$ slices, and Cyril ate $\\frac{1}{4}\\times60=15$ slices. Dan must have eaten $60-(12+20+15)=13$ slices. In decreasing order, we see the answer is $\\boxed{\\textbf{(C) }\\text{Beth, Cyril, Dan, Alex}}$. ~savannahsolver", "# Pimps There were 24 pimps in the plate. Maros ate 12, his little sister four times less. How many pimps remained in the plate? x = 9 ### Step-by-step explanation: $x=24-12-12\\mathrm{/}4=9$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Do you want to perform natural numbers division - find the quotient and remainder? ## Related math problems and questions: • In the bowl There are 12 pears in the bowl. Erik ate a third. How many pears remained in the bowl? • Baking There are 28 bunches, and son ate 1/2, dad ate four bunches. How many of them remain on the baking dishes? • Brother and sister When I was 8, my sister was 4, and now I am 18. How old is my sister? Janka and Marienka calculated that there are 210 gingerbreads on the gingerbread house. Janko ate one-seventh of all gingerbreads, and Marienka ate a third less than Janko. How many gingerbreads remained in the gingerbread house? • Chocolate How many times do we have to break a chocolate bar composed of 10 × 12 pieces to get the 120 parts? • Seven times Which number seven times is just as higher as 27, how much is smaller than 29? • How", "# There's a lot of ratio talk here. What is the importance of all these ratios? • Module 2 Week 3 Day 12 Your Turn Part 2 There's a lot of ratio talk here. What is the importance of all these ratios? (Timestamp 4:40) I thought subtraction isn't commutative... • @TSS-Graviser Hi again! Ratios are good way of comparing one number with another one. Let's start with two numbers. The bigger number represents an older sister and the smaller number represents a little brother . When mother makes cookies, older sister eats $$4$$ cookies but the little brother eats only $$2$$ :cookies. How would you compare the amount of food that they each eat? Would you say that the older sister eats $$2$$ more cookies than the little brother ? Should the older sister always eat $$2$$ more of whatever food there is, compared with her little brother ? But then what if mother serves rice -- should the older sister have exactly $$1$$ more grain of rice compared with the little brother ? No, that wouldn't make sense! If the older sister got just $$1$$ more grain of rice compared with the little brother , then she would be eating almost the same amount of rice as the little brother, , because the little brother might eat", "each, there will be no oranges left over. If the same group of children receives 8 pears each, there will be 21 pears leftover. How many children and oranges are there? • The bakery The bakery delivered 60 kg of bread to the store, the breads weighed 1 kg and 2 kg, a total of 35 pieces. How many loaves weighed 1 kg and how many were two-kilo loaves? • Tourist cottage The tourist cottage has 11 rooms for 16 guests, some of them are doubles, some singles. How many are double rooms and how many are single rooms? • Large family I have as many brothers as sisters and each my brother has twice as many sisters as brothers. How many children do parents have?", "# The game with cookies on a red and a blue plate Aabzaari and Basharat play a game, alternating turns with Aabzaari going first. At the start of the game, there are 20 cookies on a red plate and 14 cookies on a blue plate. A legal move consists in • eating two cookies taken from the blue plate, or • eating two cookies taken from the red plate, or • moving one cookie from the red plate to the blue plate (but moving cookies from the blue plate to the red plate is not allowed). The last player to make a legal move wins the game. Which player can guarantee that he wins no matter what strategy the opponent chooses? • They both lose, eating 16 or 18 cookies in one sitting probably isn't healthy... – aslum Feb 22 '16 at 18:10 It seems to me that Aabzaari always wins, regardless of the moves performed. There are cookies for 17 turns so Aabzaari gets to eat the last two. The third move doesn't consume any cookies and always allows a paired move, so it doesn't affect the parity of the turns. • yeah, you're right. Even if Aabzaari actually wanted to let Basharat win it simply isn't possible – Ivo Beckers Feb 22 '16 at 15:11 Aabzaari", "Albert, Alex, Arjun, Billy, Mitchell, Saketh, SeungIn and Sreenath have a huge pile of Christmas cookies. They do an epic sleepover at Saketh’s house, where the cookies are. In the middle of the night, Albert wakes up and eats $1$ more than $\\frac{1}{3}$ of the cookies, and goes back to bed. Later that night, Alex eats $2$ more than $\\frac{1}{3}$ of the remaining cookies, and goes back to bed. Arjun gets up later, eats $3$ more than $\\frac{1}{3}$ of the remaining cookies, and goes back to bed, and so on. The $n$th person eats $n$ more than $\\frac{1}{3}$ of the cookies. If there are $238$ cookies remaining after each person has eaten cookies, how many cookies were there to start with?", "# But Who Ate The Most? Algebra Level 1 Five brothers stayed in the house with their mother. One day, their mother brought home some mangos. Alex woke up first. As he was hungry, he ate $$\\frac{1}{6}$$ of the mangoes, and headed out. Brian woke up next. As he was hungry, he ate $$\\frac{1}{5}$$ of what was left, and headed out. Charles woke up next. As he was hungry, he ate $$\\frac{1}{4}$$ of what was left, and headed out. Danny woke up next. As he was hungry, he ate $$\\frac{1}{3}$$ of what was left, and headed out. Euler woke up next. As he was hungry, he ate $$\\frac{1}{2}$$ of what was left, and headed out. Their mother came home and saw 3 mangoes in the basket. How many mangoes were there originally? ×", "# Golden Pancakes Alice and Bob play a game with two stacks of fluffy, golden pancakes, of sizes $p$ and $q$. They alternate turns, with Alice starting. On their turn, a player must eat some number of pancakes from the larger stack. The number they eat must be a multiple of the size of the smaller stack. Since the bottom pancake of each stack is soggy, the first player to finish a stack loses. For example, if the stacks were sized $5$ and $17$, then the current player can eat either $5,10$ or $15$ pancakes from the large stack. For which values of $p$ and $q$ does Alice have a winning strategy? Source: Mathematical Puzzles: A Connoisseur's Collection, Peter Winkler. • None, because that is a lot of pancakes, and she is going to be sick by the end no matter what. – Aggie Kidd Jul 8 '15 at 1:51 • Do I understand correctly: Does it mean that when Alice eats 15 from the second (larger stack), then there are stacks 5 and 2, so Bob has to eat 2 or 4 from the first stack? – Voitcus Jul 8 '15 at 3:52 • I hope the pancakes aren't made from real gold ;). – Bojidar Marinov Jul 8 '15 at 10:05 • In next weeks episode,", "(A) 2.25 (B) 2.50 (C) 2.75 (D) 3.00 (E) 3.75 AMC 10A, 2013, Problem 2 Alice is making a batch of cookies and needs 2 \\frac{1}{2} cups of sugar. Unfortunately, her measuring cup holds only \\frac{1}{4} cup of sugar. How many times must she fill that cup to get the correct amount of sugar? (A) 8 (B) 10 (C) 12 (D) 16 (E) 20 AMC 10A, 2013, Problem 5 Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid \\ 105, Dorothy paid \\ 125, and Sammy paid \\ 175. In order to share costs equally, Tom gave Sammy t dollars, and Dorothy gave Sammy d dollars. What is t-d ? (A) 15 (B) 20 (C) 25 (D) 30 (E) 35 AMC 10A, 2013, Problem 6 Joey and his five brothers are ages 3,5,7,9,11, and 13 . One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5 -year-old stayed home. How old is Joey? (A) 3 (B) 7 (C) 9 (D) 11 (E) 13 AMC 10A, 2013, Problem 8 What is the value of 22014+220122201422012? (A) -1 (B) 1 (C) \\frac{5}{3} (D) 2013 (\\mathbf{E}) 2^{4024} AMC 10A, 2013, Problem", "how many presents you get. As such, this year, as you know, I have been keeping a tally of all the times you have done something naughty (η) and all of the times you have done something nice (η). You get one scratch for a naughty act and one for a nice act, an act of kindness if you will, and dependent of course on the act. Remember that time you fooled your sister your type 1 hypersensitive to tomatoes sister into thinking that ketchup you had mixed in her ice-cream was actually raspberry juice, well that got you ten scratches in the naughty column, for example, whilst tidying up your Lego without being asked that one time you did it in the whole year got you five scratches in the nice column. Here's a little section of the tallies I've been keeping for this year: Overall you had 740 naughty scratches this year and 623 nice scratches, giving ηi = 0.46: \\begin{array}{l}\\begin{align}\\frac{{623}}{{740 + 623}} &= \\frac{{623}}{{1363}}\\\\\\\\ &= 0.46\\end{align}\\end{array} Now, this doesn't mean that you have a 46% chance of getting your presents this year. Oh no. You wrote to Santa to ask for the following five presents: • An umbrella in the shape of an Octopus, • A Lego Batman car [I presume you mean the Batmobile]", "has 3 oatmeal bars to share with his 2 sisters and 3 of his friends. There are a total of 6 friends and he plans to give each person. Hence each person gets an oatmeal bar of $$\\frac{3}{6}$$. Elijah and Zoey each have an orange cut into 8 equal sections. They each ate one section, or $$\\frac{1}{8}$$, of their orange. How many more sections altogether do Elijah and Zoey need to eat to finish both oranges? TEKS 3.3.C", "+0 # help 0 75 1 At a party. Andy and April shared a cherry pie. Andy ate 2/5 of the pie and April ate 1/3 of the pie. What fraction of the pie did they eat altogether Jan 18, 2020 #1 +24422 +2 At a party. Andy and April shared a cherry pie. Andy ate 2/5 of the pie and April ate 1/3 of the pie. What fraction of the pie did they eat altogether $$\\begin{array}{|rcll|} \\hline &&\\mathbf{ \\dfrac{2}{5} + \\dfrac{1}{3} } \\\\\\\\ &=& \\dfrac{2*3+1*5}{5*3}\\\\\\\\ &=& \\mathbf{ \\dfrac{11}{15}} \\\\ \\hline \\end{array}$$ Jan 18, 2020", "of the sisters poisoned her - either in revenge, or to inherit a share of the royalties, worth millions.\" \"The METHOD is a bit tricky. The only things any of the band members ate were home made cakes at the after-concert party. Our theory is that one of the sisters mixed a precise measure of poison into her cake and gave Maya a slightly larger (and fatal) share, but the witness statements don't support our hypothesis - it seems as though everyone except the manager had an exactly equal share of all the cakes.\" The detective passed over the written witness statements. The Manager's Statement: 1. Each of the sisters brought along an identical circular cake. 2. That evening they had one straight knife, and three different sized circular 'cookie-cutters' to cut up their cakes. 3. Straight cuts were edge to edge, but in circular cuts the cookie-cutter could overlap past the edge of a cake. Cake parts were not moved between cuts. 4. The cakes were cut and eaten in the following order: 5. Maya made one cut with the first cookie cutter, plus two straight cuts, to divide her cake into eight exactly equal portions - one for each of us. 6. Because of my diet, I asked that the rest of the cakes should be", "into 2 equal pieces to share with friends? What fraction names all of the brownie pieces now? are equivalent fractions. It is given that Lucy cuts each part of the brownie into 2 equal pieces to share with friends Now, The given model is: Now, According to the given information, The fraction of all the pieces of brownie Lucy has = (The total number of brownies) ÷ (The number of groups of brownies) = $$\\frac{10}{4}$$ Hence, from the above, We can conclude that $$\\frac{5}{2}$$ = $$\\frac{10}{4}$$ Question 11. Multi-Step Christy bought 8 muffins. She chose 2 chocolate, 2 banana, and 4 blueberries. She and her family ate the chocolate and banana muffins for breakfast. What fraction of the muffins did they eat? Write an equivalent fraction. Draw a picture. It is given that Christy bought 8 muffins. She chose 2 chocolate, 2 banana, and 4 blueberries. She and her family ate the chocolate and banana muffins for breakfast Now, According to the given information, The fraction of the muffins did Christy and her family ate = (The number of muffins Christy and her family ate) ÷ (The total number of muffins Christy bought) = $$\\frac{4}{8}$$ = $$\\frac{1}{2}$$ Hence, from the above, We can conclude that The fraction of the muffins did Christy and her family ate is: $$\\frac{4}{8}$$", "more than one present? (1) Esther has forty Christmas presents to give out. (2) If the number of family members were doubled, it would not be possible for each family member to get at least one present [Reveal] Spoiler: OA _________________ Official PS Practice Questions Press +1 Kudos if this post is helpful Kudos [?]: 1042 [0], given: 538 Senior Manager Joined: 25 Feb 2013 Posts: 407 Kudos [?]: 174 [1], given: 31 Location: India GPA: 3.82 Esther is giving Christmas presents to her family members. Each family [#permalink] ### Show Tags 10 Sep 2017, 11:42 1 KUDOS Esther is giving Christmas presents to her family members. Each family member gets the same number of presents and no presents were leftover. If each family member gets at least one present, did each family member receive more than one present? (1) Esther has forty Christmas presents to give out. (2) If the number of family members were doubled, it would not be possible for each family member to get at least one present The question can be logically interpreted to arrive at correct answer B. However mathematically we can proceed as - Let the number of guests be $$x$$ and number of gifts for each guest be $$y$$ (here both $$x$$ & $$y$$ have to be integers). So total", "# Counting Siblings Word Problem: Linear Algebra Systems Question: Alexandra and Brandon are brother and sister. We know that Alexandra has just as many brothers as sisters, and that Brandon has twice as many sisters as brothers. How many children are there in this family? I am having trouble doing this problem. I've found a similar problem and solution but am having trouble with 2 aspects: 1) Why should bob have 2(sisters) = brothers? Doesn't this mean he has twice as many brothers as sisters, which will be the opposite as the question? 2) I don't understand how the author found the exact number of sisters and brothers. Did he/she just plug in random numbers until it worked? 3) Is there a way to do this with matrices. Isn't that the point of Linear Algebra. Thank you so much! let there be $g$ girls and $b$ boys in the family Alexandra is a girl who has $g-1$ sisters and $b$ brothers so $g-1=b$ Brandon is a boy who has $g$ sisters and $b-1$ brothers so $2(b-1)=g$ can you take it from here ? Seven. If Alexandra has equal number of brother and sisters and Brandon has twice as many sisters as brothers, then brandon needs to have at least one brother. So that way if brandon has one", "to (2,3). Then Bob has to eat 2, leading to (1,2), so Alice can eat 1, leading to (1,1) and Bob loses – Ben Aaronson Jul 8 '15 at 12:29 • @HemantRupani Ben Aarons on is correct about the (2,5) case. In general, my analysis only proves the existence of a winning strategy for Alice under certain conditions. It doesn't actually show what that strategy is. I find non-constructive results like that fascinating, although I must admit this result alone is not particularly helpful to Alice while she's playing... – John Stevens Jul 8 '15 at 15:19 • Yes! My bad.... – Hemant Rupani Jul 8 '15 at 16:10 Provided they have unlimited appetites, Alice will win when $p=nq~or~q=np~for~(n\\in N \\& n>1)$ Because, when Alice starts she'll eat $\\Big((n-1)\\times~smaller~stack\\Big)$, then both stacks will be same, and Bob will have to eat a stack(full). Alice will win. • ... when Alice start he'll eat ... Are you sure it isn't she ? :) – Bojidar Marinov Jul 8 '15 at 10:06 • @BojidarMarinov, No! Thanks! Now I see Alice shows a girls on Google Images :P so edited. – Hemant Rupani Jul 8 '15 at 10:17 • @BojidarMarinov - I just assumed it was Alice Cooper vs Bob, in which case both are male. I also like the visual", "had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to: Matt asked: Did you eat more apples than I did, James? James: I don’t know. Did you, Ian, eat more apples than I did? Ian: I don’t know. Barry: Aha!! I figured out.. So, Barry figured out how many apples each person ate. Can you do the same? Matt: 1 Apple James: 2 Apples Ian: 3 Apples Barry: 5 Apples ## The Logic Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian must have eaten 3 or 4 apples. Barry can only deduce the other amounts if he ate 5 apples. And the rest, in order to", "Category:Baked Goods These foods are produced in an Oven. The number of Branches used in the oven determines the baking time i.e. Ticks; food baked longer than necessary will turn into Ash. The quality of all unbaked goods is ${q}Dough=\\frac{_{q}MaterialsAvg + _{q}Water}{2}$, softcapped by $\\sqrt{Perception * Cooking}$. The quality of baked goods is $\\frac{2 * _{q}dough + _{q}oven + _{q}branch}{4}$. Food Hunger Ticks Main Ingredients Water Honey Milk Butter STR AGI INT CON PER CHA DEX PSY Hurt Apple Pie 35 3 Apple 0.5 - - 1 - - - 10 - - - - - Baked Birchbark Bream 30 3 Birchbark x2, Filet of Bream, Peapod x3, Yellow Onion x2 - - - - - - 5 - - 5 4 - - Bark Bread 50 4 Tree Bark 1.0 - - - - - - 5 - - - - 0.2 Birthday Cake 35 - Blueberry Pie, Candle, Chicken Egg - 0.1 0.2 1 - - - - - 20 - 5 - Blueberry Pie 35 3 Blueberries 0.5 - - - - - 4 - 3 - - - - Bread 50 3 Flour 1.0 - - - - - - - - - - - Carrot Cake 30 2 Carrots x2 0.5 - - 1 - - - - 7 - - - Chantrelle" ]

[ "a product of 36. What is the least possible sum of a and b? • Four integers Fnd four consecutive integers so that the product of the first two is 70 times smaller than the product of the next two.", "product of two primes proved", "for which a + b + c = 1000. Find the product abc. ## Problem 10 Summation of primes The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.", "greater than 2 and less than 50. If the product of them is less than 100, then how many combinations of them will there be? 0 < P*Q < 100, P and Q are both prime numbers,and P < Q, how many combinations of P and Q are there? How many positive integers no greater than 20 can be expressed as the sum of two different prime numbers? 1 2 ... 4 5 6 7 8 9 10 ... 24 25 25000 +道题目 136本备考书籍", "theorem. Any natural number greater than 1 can be written as a product of prime factors.", "# Splitting the Primes Is it possible to split the 25 primes less than 100 into two disjoint sets such that the sum of the primes in one set equals the product of the primes in the other set? If so, in how many ways can this be done? • This looks like it might be more to do with maths, partitioning and set theory than puzzling, so I’ve flagged it as being off-topic. Sep 23, 2021 at 15:26 A first observation is that the sum of all given primes is 1060 and the primorial $$\\#11$$ is $$2×3×5×7×11=2310$$, so there may be at most four primes in the product set. Futhermore if there are two or four primes in the product set, either 2 is in there, the product is even but the sum is odd (of 23 or 21 odd primes), or it isn't, the product is odd but the sum is even (2 + 22 or 20 odd primes), which is a contradiction either way. Clearly there cannot be one in the product set, so there can only be three. This leaves just 130 cases, and none of them satisfy the given condition. The closest is when the product set has 5, 7, 29 with product 1015, when the sum set then sums to 1019. •", "20 are $$11, 13, 17, 19.$$ The multiplication table contains all products of single digit numbers. So a number not in the table can not be factored as a product of two single digit numbers. In the case of 11, 13, 17, and 19, this means that they do not factor at all, except as a product of 1 and themselves. So this list is the list of prime numbers between 10 and 20.", "the exponent is supposed to be 2 (i.e, either the sums are more or less distinct from each other or the products are) but nothing close to this has been obtained in any case. The exponent here is better than the best known for finite fields of prime order, but worse than the best known for the real numbers. I haven’t read the paper yet, so I can’t say anything incisive about the methods, but it’s a striking result!", "## Optimization Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.", "are multiplied.", "the left. In other words, the GCF of 18 and 36 is a product of all three prime numbers on the left. If you use the division method, the eigenvalue of 18 and 35 is 2 x 3 x 7 – a sexy pair of these two primes. The eigenvalue of 35 is a sexy female. Visit the rest of the site for more useful articles!", "multiplied.", "lowest common multiple of two numbers is the lowest number that is a multiple of both numbers.", "20 times as great as the other number? ​...", "than 20 is closest to which 7 26 Nov 2010, 07:20 5 The product of all the prime numbers less than 20 is closest to which 6 12 Jun 2009, 20:28 Display posts from previous: Sort by", "+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", "# Sum of primes! Number Theory Level 2 So easy!!! What is the sum of all primes between 1-100? You can try My Other Problems. ×", "May 2015, 05:43 8 The product of 2 positive integers is 720. If one of the numbers is in 5 04 Dec 2014, 04:05 6 How many positive integers less than 28 are prime numbers 8 17 Mar 2012, 16:57 20 What is the total number of positive integers that are less 16 24 Feb 2012, 23:11 Display posts from previous: Sort by", "of the smallest two to the product of the greatest two. Repeat with some other sets of numbers. What do you notice? With thanks to Don Steward, whose ideas formed the basis of this problem.", "than 1 can be reoresented as a product of prime numbers. • There is only one way to represent that number as a product of prime numbers. For example 63 is the product of the prime numbers 3, 3, and 7. There is no other way to produce 63 by multiplying primes. This is know as the unique factorization theorem (also called the fundamental theorem of arithmetic)." ]

[ "in his lectures there are interesting facts about every number! Here is what he has to say about the number 23. If you have any other facts to add or if you see any mistakes in my transcription of Dr. Stanford’s notes please use the comments below. # 23 is… The largest number not the sum of distinct powers. With 23 people in a room, odds are that two share a birthday (better than 50:50.) Prime, smallest odd prime not a twin. Woodall number. $23=3 \\cdot 2^3-1$ One of the only two numbers that need 9 cubes. (The other is 239.) $23=1^3+1^3+1^3+1^3+1^3+1^3+1^3+2^3+2^3$ If negatives allowed, $23=2^3+2^3+2^3+(-1)^3$ $23=0 \\cdot 0!+1 \\cdot 1!+2 \\cdot 2!+3 \\cdot 3!$ 23 is the smallest number of rigid rods that brace a square. First prime where 23rd roots of unity form cyclotomic integers without unique factorization. Number of trees with eight nodes. Factor of $2^{11}-1$ $2^{23}-1$ is composite: $47 \\cdot 178481$ Sophie Germain prime: 2(23)+1 also prime. Wedderburn-Etherington number. The first pillar prime.", "+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 ? $$\\small{\\text{ \\begin{array}{r|rcl|r} \\hline n & p_1 & & p_2 & p_1\\cdot p_2 \\\\ \\hline 1 & 11 & \\cdot & 13 & = 143 \\\\ 2 & 11 & \\cdot & 17 & = 187 \\\\ 3 & 11 & \\cdot & 19 & = 209 \\\\ 4 & 11 & \\cdot & 23 & = 253 \\\\ 5 & 11 & \\cdot & 29 & = 319 \\\\ 6 & 11 & \\cdot & 31 & = 341 \\\\ 7 & 11 & \\cdot & 37 & = 407 \\\\ 8 & 11 & \\cdot & 41& = 451 \\\\ 9 & 11 & \\cdot & 43 & = 473 \\\\ 10 & 13 & \\cdot & 17 & = 221 \\\\ 11 & 13 & \\cdot & 19 & = 247 \\\\ 12 & 13 & \\cdot & 23 & = 299 \\\\", "prime is $3$, and after comparing the resulting numbers, we discover that $\\{ 1, 1, 1, 1, 4\\}$ results in the smallest odd number $3^4\\cdot 5\\cdot 7\\cdot 11\\cdot 13=405405.$ Then the smallest odd number with exactly $40$ \"product pairs\" is $405405$. The same process is done for even numbers with the only difference being that the smallest prime factor is $2$, therefore we take each exponents set, assign them in decreasing order to increasing primes, and compare them with each other. In this case the smallest value comes out from the set $\\{ 1,1,3,4\\}$ to be $2^4\\cdot 3^3\\cdot 5\\cdot 7=15120$. So the smallest even number, actually smallest positive number altogether which can be factored into exactly $40$ product pairs is $15120$.", "few short tests, we get $168+27=\\boxed{195}$. -jackshi2006 - minor latex edits by jske25", "of all primes until we reach the limit. This can be solved pretty straightforward with pen and paper: $$n=2\\cdot 3\\cdot 5\\cdot 7\\cdot 11\\cdot 13\\cdot 17\\leq 10^6$$. For a general limit, this simple snippet can be implemented: function solution(L) { var n = 1, k = 0; var primes = [2, 3, 5, 7, 11, 13, 17, 19, 21, 23, 29, 31]; while (primes[k] * n <= L) n*= primes[k++]; return n; } « Back to problem overview", "+0 # Math +1 70 1 +203 What is the smallest possible product of one one-digit prime and two distinct two-digit primes? Dec 9, 2021 #1 +132 0 The answer relies on minimizing all of the primes. The smallest one-digit prime is 2. The smallest two-digit primes are 11 and 13. $$2 \\cdot 11 \\cdot 13 = 286$$. Dec 9, 2021", "# Consecutive squarefree numbers of 5 prime factors each, mostly small The sequence of numbers 49297533, 49297534, and 49297535 is notable, because the prime factorizations of these numbers are each of the form $a^1 \\cdot b^1 \\cdot c^1 \\cdot d^1 \\cdot e^1$, and their representations in this form include the first 11 primes: $$3\\cdot 19\\cdot 23\\cdot 31\\cdot 1213 = 49,297,533$$ $$2\\cdot 11\\cdot 13\\cdot 97\\cdot 1777 = 49,297,534$$ $$5\\cdot 7\\cdot 17\\cdot 29\\cdot 2857 = 49,297,535$$ What is the lowest sequence of numbers $x, x+1, x+2$ such that all 3 numbers are of this \"five prime factors\" form, and their representations, taken together, include the first 12 primes, $2,3,5,7,11,13,17,19,23,29,31,37$? Note that 12 would be the largest number of minimal primes one could have in this configuration, since there is a total of 15 primes needed, meaning that with 13 minimal primes included, one of the products would be much smaller than the others. POSTSCRIPT: I found the smallest group of cardinality 4 (using the 9 smallest primes) reasonably quickly: $$11\\cdot 13\\cdot 17\\cdot 9463= 23,004,553$$ $$2\\cdot 19\\cdot 23\\cdot 26321= 23,004,554$$ $$3\\cdot 5\\cdot 7\\cdot 219091 =23,004,555$$ and for cardinality 3: $$3\\cdot 11\\cdot 113 = 3729$$ $$2\\cdot 5\\cdot 373 = 3730$$ $$7\\cdot 13\\cdot 41 = 3731$$ • When you say \"there might not be one\", you mean there might be no such x,", "then it could only be the product of primes $23, 29$ and $31$. These three digit numbers are $23^2=529, 23\\cdot29 =667, 23\\cdot31=713, 29^2 = 841, 29\\cdot31 = 899, and 31^2 = 961$. Also 23 times 37 equals 851, 23 times 41 equals 943, and 23 times 43 equals 989. There are now too many numbers here. So I have added a divisibility test for 23 below. Use this divisibility test table for primes through $19$. $n|x$ means $n$ divides $x$ with no remainder. Modulo $11$ means remainder after dividing by $11$. Divisibility Test Table for Three Digit Numbers For divisibility by 23 use 23|8a+10b+c. Here is an example. Example: Factor 611", "(- n 1))) And a couple of tests using n = 8, 4 and 9. The answers should be $1\\cdot3\\cdot5\\cdot7 = 105$ $1\\cdot3 = 3$ and $1\\cdot2\\cdot4\\cdot5\\cdot7\\cdot8 = 2240$ respectively. (product-of-relative-primes 8) Output: 105 (product-of-relative-primes 4) Output: 3 (product-of-relative-primes 9) Output: 2240 Done.", "Which of the following numbers is the product of exactly $3$ distinct prime numbers? $45$; $60$; $91$; $105$; $330$", "+0 heeeeelllllpppp +1 42 4 +44 The product of integers 240 and k is a perfect cube. What is the smallest possible positive value of k? heeeeeeeeeeellllllllppppppppp Jul 26, 2022 #1 +370 -1 I think it's 60 Jul 26, 2022 #2 +1118 +4 240x60=14400, which is not a perfect cube nerdiest Jul 26, 2022 #3 +197 +1 240 0= 24 * 10 = 2^3 * 3 * 10 = 2^4 * 3 * 5 We need 2 2s, 2 3s, and 2 5s. This gives us: 2^2 * 3^2 * 5^2 = 900. Jul 26, 2022 #4 +1118 +4 niceeee. here's a way $$240=2^4\\cdot3\\cdot5=2^3(2\\cdot3\\cdot5)$$. For $$240k$$ to be a perfect cube (and not a perfect square), $$k$$ must be at least $$2^2\\cdot3^2\\cdot5^2=\\boxed{900}$$. nerdiest Jul 26, 2022 edited by nerdiest Jul 26, 2022", "\\cdot 3 \\cdot 5 + 1 = 31$, also a prime. It is always the case that a product of primes plus one is another prime? No, in fact. If you look at $2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 + 1=30031 = 59\\cdot 509$, you get a nonprime. Second proof that there are infinitely many primes In a similar vein to the first proof, we will show that there is always a prime larger than $n$ for any positive integer $n$. To see this, consider $n! + 1$. Upon dividing by any prime less than $n$, we get a remainder of $1$. So all of its prime factors are larger than $n$, and so there are infinitely many primes. $\\diamondsuit$ I would also like to present one more, which I’ve always liked. Third proof that there are infinitely many primes Suppose there are only finitely many primes $p_1, \\ldots, p_k$. Then consider the two numbers $n = p_1 \\cdot \\dots \\cdot p_k$ and $n -1$. We know that $n - 1$ has a prime factor, so that it must share a factor $P$ with $n$ since $n$ is the product of all the primes. But then $P$ divides $n - (n - 1) = 1$, which is nonsense; no prime divides $1$. Thus", "# Two number, both greater than 29, have GCD = 29 and LCM = 4147. The sum of the numbers is: Option 1) Option 2) Option 3) Option 4) Option 5) - $Product \\: of\\: numbers = 29\\ast 4147.$ $Let \\: the\\: numbers \\: be 29a \\: and\\: 29b.\\:$ $Then, 29a \\ast 29b = \\left ( 24\\ast 4147 \\right )$ $=> ab = 143.$ $Now,\\: co-primes \\: with\\: product \\: 143 \\:are \\:\\left ( 1, 143 \\right ) \\:and \\:\\left ( 11, 13 \\right ).$ $So, \\:the\\: numbers\\: are\\: \\left ( 29 \\ast 1, 29\\ast 143 \\right )\\:and\\:\\left ( 29 \\ast 11, 29\\ast 13 \\right )$ $Since \\: both \\: numbers\\: are \\: greater\\: than\\: 29,$ $\\: the\\: suitable\\: pair \\: is \\: \\left ( 29 \\ast 11, 29 \\ast 13 \\right )$ $i.e., \\left ( 319, 377 \\right ).$ $Required\\: sum = \\left ( 319 + 377 \\right ) = 696.$ Exams Articles Questions", "Goldbach that every odd composite number can be written as the sum of a prime and twice a square. 9 = 7 + 2$\\cdot$12 15 = 7 + 2$\\cdot$22 21 = 3 + 2$\\cdot$32 25 = 7 + 2$\\cdot$32 27 = 19 + 2$\\cdot$22 33 = 31 + 2$\\cdot$12 It turns out that the conjecture was false. What is the smallest odd composite that cannot be written as the sum of a prime and twice a square? 7. The first two consecutive numbers to have two distinct prime factors are: 14 = 2 $\\cdot$ 7 15 = 3 $\\cdot$ 5 The first three consecutive numbers to have three distinct prime factors are: 644 = 2² $\\cdot$ 7 $\\cdot$ 23 645 = 3 $\\cdot$ 5 $\\cdot$ 43 646 = 2 $\\cdot$ 17 $\\cdot$ 19. Find the first four consecutive integers to have four distinct prime factors. What is the first of these numbers? 8. The series, 11 + 22 + 33 + ... + 1010 = 10405071317. Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000. 9. The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit", "means there are 13 \"a\" s and b cannot be 20) When $21\\leq a \\leq 28$, there are $1+2+\\cdots+8=36$ ways. In all, there are $192+36=\\boxed{228}$ possible sequences. ~bluesoul", "$14$ of these extra pairs. The answer is $28+21+14 = \\boxed{063}$", "## On a Special Kind of Prime Numbers March 31, 2021 ### Prime Numbers Today I would like to write a brief blog post about the “atoms” of the natural numbers: prime numbers. Prime numbers are those numbers which are evenly divisible by exactly two numbers, namely by $1$ and by the number itself. For example, $17$ is a prime number because it is only divisible by $1$ and $17$, whereas $42$ is not a prime because it is, for instance, also divisible by $2$, $3$ and $7$. The first few prime numbers are given by $$2,3,5,7,11,13,17,19,23,29,\\ldots$$ As already mentioned, prime numbers are sometimes called the “atoms” of the natural numbers, because every natural number can be written as the product of primes: $42 = 2 \\cdot 3 \\cdot 7$. So far, so easy. In the course of time, mathematicians have asked many questions concerning different properties of prime numbers. Some of these questions have beautiful answers (e.g., the infinitude of primes or the prime number theorem on the frequency of primes). Over the past few decades, prime numbers have also become increasingly important in our everyday life. In particular, they play a substantial role in the encryption and secure transmission of messages and data of different types. Although the concept of prime numbers seems to be quite", "+0 # How many different primes appear in the prime factorization of (20 factorial)? (Reminder: The number is the product of the integers 0 211 2 How many different primes appear in the prime factorization of $$20!$$ (20 factorial)? (Reminder: The number $$n!$$ is the product of the integers from 1 to $$n$$ . For example: $$5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$$ Feb 26, 2021 ### 2+0 Answers #1 0 halp quickly i need helppppp Feb 26, 2021 #2 0 just find how many primes are from 1 to 20. These are 2, 3, 5, 7, 11, 13, 17, and 19, which is 8 primes Feb 26, 2021", "proofs, and we saw 2 of them in class. For posterity, I’ll present three here. First proof that there are infinitely many primes Take a finite collection of primes, say $p_1, p_2, ldots, p_k$. We will show that there is at least one more prime not mentioned in the collection. To see this, consider the number $p_1 p_2 ldots p_k + 1$. We know that this number will factor into primes, but upon division by every prime in our collection, it leaves a remainder of $1$. Thus it has at least one prime factor different than every factor in our collection. $diamondsuit$ This was a common proof used in class. A pattern also quickly emerges: $2 + 1 = 3$, a prime. $2cdot3 + 1 = 7$, a prime. $2 cdot 3 cdot 5 + 1 = 31$, also a prime. It is always the case that a product of primes plus one is another prime? No, in fact. If you look at $2 cdot 3 cdot 5 cdot 7 cdot 11 cdot 13 + 1=30031 = 59cdot 509$, you get a nonprime. Second proof that there are infinitely many primes In a similar vein to the first proof, we will show that there is always a prime larger than $n$ for any positive integer $n$. To see", "# Minimum number to ensure divsion [duplicate] The problem: How many minimum number of integers must be selected from the set $$\\{1, 2, 3, 4,\\dots,99, 100\\}$$ to ensure that there are at least two integers in the selection such that one divides the other. My sloppy thinking: If we select $$25$$ primes, none of them of course will divide the other. If, however, we choose a $$26^{th}$$ number, it must divide some or the other. So the answer is 26. The actual answer, however was 51. Hints?" ]

[ "+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", "a product of 36. What is the least possible sum of a and b? • Four integers Fnd four consecutive integers so that the product of the first two is 70 times smaller than the product of the next two.", "prime is $3$, and after comparing the resulting numbers, we discover that $\\{ 1, 1, 1, 1, 4\\}$ results in the smallest odd number $3^4\\cdot 5\\cdot 7\\cdot 11\\cdot 13=405405.$ Then the smallest odd number with exactly $40$ \"product pairs\" is $405405$. The same process is done for even numbers with the only difference being that the smallest prime factor is $2$, therefore we take each exponents set, assign them in decreasing order to increasing primes, and compare them with each other. In this case the smallest value comes out from the set $\\{ 1,1,3,4\\}$ to be $2^4\\cdot 3^3\\cdot 5\\cdot 7=15120$. So the smallest even number, actually smallest positive number altogether which can be factored into exactly $40$ product pairs is $15120$.", "+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 ? $$\\small{\\text{ \\begin{array}{r|rcl|r} \\hline n & p_1 & & p_2 & p_1\\cdot p_2 \\\\ \\hline 1 & 11 & \\cdot & 13 & = 143 \\\\ 2 & 11 & \\cdot & 17 & = 187 \\\\ 3 & 11 & \\cdot & 19 & = 209 \\\\ 4 & 11 & \\cdot & 23 & = 253 \\\\ 5 & 11 & \\cdot & 29 & = 319 \\\\ 6 & 11 & \\cdot & 31 & = 341 \\\\ 7 & 11 & \\cdot & 37 & = 407 \\\\ 8 & 11 & \\cdot & 41& = 451 \\\\ 9 & 11 & \\cdot & 43 & = 473 \\\\ 10 & 13 & \\cdot & 17 & = 221 \\\\ 11 & 13 & \\cdot & 19 & = 247 \\\\ 12 & 13 & \\cdot & 23 & = 299 \\\\", "the left. In other words, the GCF of 18 and 36 is a product of all three prime numbers on the left. If you use the division method, the eigenvalue of 18 and 35 is 2 x 3 x 7 – a sexy pair of these two primes. The eigenvalue of 35 is a sexy female. Visit the rest of the site for more useful articles!", "of all primes until we reach the limit. This can be solved pretty straightforward with pen and paper: $$n=2\\cdot 3\\cdot 5\\cdot 7\\cdot 11\\cdot 13\\cdot 17\\leq 10^6$$. For a general limit, this simple snippet can be implemented: function solution(L) { var n = 1, k = 0; var primes = [2, 3, 5, 7, 11, 13, 17, 19, 21, 23, 29, 31]; while (primes[k] * n <= L) n*= primes[k++]; return n; } « Back to problem overview", "## Elementary Algebra $25\\cdot1=25$ $25+1=26$ $25$ and $1$ are the only two numbers whose product is $25$ and sum is $26$. Therefore, we must use these numbers to fill in the blanks.", "# Splitting the Primes Is it possible to split the 25 primes less than 100 into two disjoint sets such that the sum of the primes in one set equals the product of the primes in the other set? If so, in how many ways can this be done? • This looks like it might be more to do with maths, partitioning and set theory than puzzling, so I’ve flagged it as being off-topic. Sep 23, 2021 at 15:26 A first observation is that the sum of all given primes is 1060 and the primorial $$\\#11$$ is $$2×3×5×7×11=2310$$, so there may be at most four primes in the product set. Futhermore if there are two or four primes in the product set, either 2 is in there, the product is even but the sum is odd (of 23 or 21 odd primes), or it isn't, the product is odd but the sum is even (2 + 22 or 20 odd primes), which is a contradiction either way. Clearly there cannot be one in the product set, so there can only be three. This leaves just 130 cases, and none of them satisfy the given condition. The closest is when the product set has 5, 7, 29 with product 1015, when the sum set then sums to 1019. •", "in his lectures there are interesting facts about every number! Here is what he has to say about the number 23. If you have any other facts to add or if you see any mistakes in my transcription of Dr. Stanford’s notes please use the comments below. # 23 is… The largest number not the sum of distinct powers. With 23 people in a room, odds are that two share a birthday (better than 50:50.) Prime, smallest odd prime not a twin. Woodall number. $23=3 \\cdot 2^3-1$ One of the only two numbers that need 9 cubes. (The other is 239.) $23=1^3+1^3+1^3+1^3+1^3+1^3+1^3+2^3+2^3$ If negatives allowed, $23=2^3+2^3+2^3+(-1)^3$ $23=0 \\cdot 0!+1 \\cdot 1!+2 \\cdot 2!+3 \\cdot 3!$ 23 is the smallest number of rigid rods that brace a square. First prime where 23rd roots of unity form cyclotomic integers without unique factorization. Number of trees with eight nodes. Factor of $2^{11}-1$ $2^{23}-1$ is composite: $47 \\cdot 178481$ Sophie Germain prime: 2(23)+1 also prime. Wedderburn-Etherington number. The first pillar prime.", "# 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$", "(- n 1))) And a couple of tests using n = 8, 4 and 9. The answers should be $1\\cdot3\\cdot5\\cdot7 = 105$ $1\\cdot3 = 3$ and $1\\cdot2\\cdot4\\cdot5\\cdot7\\cdot8 = 2240$ respectively. (product-of-relative-primes 8) Output: 105 (product-of-relative-primes 4) Output: 3 (product-of-relative-primes 9) Output: 2240 Done.", "20 are $$11, 13, 17, 19.$$ The multiplication table contains all products of single digit numbers. So a number not in the table can not be factored as a product of two single digit numbers. In the case of 11, 13, 17, and 19, this means that they do not factor at all, except as a product of 1 and themselves. So this list is the list of prime numbers between 10 and 20.", "# Minimum number to ensure divsion [duplicate] The problem: How many minimum number of integers must be selected from the set $$\\{1, 2, 3, 4,\\dots,99, 100\\}$$ to ensure that there are at least two integers in the selection such that one divides the other. My sloppy thinking: If we select $$25$$ primes, none of them of course will divide the other. If, however, we choose a $$26^{th}$$ number, it must divide some or the other. So the answer is 26. The actual answer, however was 51. Hints?", "then it could only be the product of primes $23, 29$ and $31$. These three digit numbers are $23^2=529, 23\\cdot29 =667, 23\\cdot31=713, 29^2 = 841, 29\\cdot31 = 899, and 31^2 = 961$. Also 23 times 37 equals 851, 23 times 41 equals 943, and 23 times 43 equals 989. There are now too many numbers here. So I have added a divisibility test for 23 below. Use this divisibility test table for primes through $19$. $n|x$ means $n$ divides $x$ with no remainder. Modulo $11$ means remainder after dividing by $11$. Divisibility Test Table for Three Digit Numbers For divisibility by 23 use 23|8a+10b+c. Here is an example. Example: Factor 611", "What is the least common denominator of 3/4, 3/8, and 1/5? Apr 13, 2016 $40$ Explanation: If we look at the prime factorizations of the denominators, we have $4 = {2}^{2}$ $8 = {2}^{3}$ $5 = {5}^{1}$ The least common denominator will be the minimal product containing all factors above to their appropriate powers. In this case, that would be ${2}^{3} \\cdot 5$. Thus, the least common denominator is ${2}^{3} \\cdot 5 = 8 \\cdot 5 = 40$", "# Prime * Prime The product of $$2$$ distinct prime numbers is between $$10$$ and $$50$$. What is the maximum possible value of their sum? ×", "greater than 2 and less than 50. If the product of them is less than 100, then how many combinations of them will there be? 0 < P*Q < 100, P and Q are both prime numbers,and P < Q, how many combinations of P and Q are there? How many positive integers no greater than 20 can be expressed as the sum of two different prime numbers? 1 2 ... 4 5 6 7 8 9 10 ... 24 25 25000 +道题目 136本备考书籍", "# Can $18$ consecutive integers be separated into two groups,such that their product is equal? Can $18$ consecutive positive integers be separated into two groups, such that their product is equal? We cannot leave out any number and neither we can take any number more than once. My work: When the smallest number is not $17$ or its multiple, there cannot exist any such arrangement as $17$ is a prime. When the smallest number is a multiple of $17$ but not of $13$ or $11$, then no such arrangement exists. But what happens, when the smallest number is a multiple of $17$ and $13$ or $11$ or both? #### Solutions Collecting From Web of \"Can $18$ consecutive integers be separated into two groups,such that their product is equal?\" This is impossible. At most one of the integers can be divisible by $19$. If there is such an integer, then one group will contain it and the other one will not. The first product is then divisible by $19$ whereas the second is not (since $19$ is prime) — a contradiction. So if this possible, the remainders of the numbers after division by $19$ must be precisely $1,2,3,\\cdots,18$. Now let $x$ be the product of the numbers in one of the groups. Then $x^2 \\equiv 18! \\equiv -1 \\pmod{19}$", "will be even but the product of the product set will be odd. Hence, the product set cannot contain an even number of elements and so must contain three elements. Slightly more detail The sum of the primes greater than $$5$$ and less than $$100$$ is $$1050$$ while the sum of the primes less than $$83$$ is $$791$$ so this means that the product set must contain elements whose product is between these two values. There are just $$32$$ such cases of prime triplets in this range and we can check by hand that none of these satisfies as a solution to the problem. The closest we get here is $$5 \\times 7 \\times 29$$ which is $$1015$$ while the sum of the other primes is $$1019$$.", "# The product of two numbers is 5547. If the H.C.F of these numbers is 43, then find the greater number. 1. 129 2. 119 3. 111 4. 88 Option 1 : 129 ## Detailed Solution Given: The product of the two numbers is 5547 and the H.C.F is 43. Calculation: Let the two numbers are 43x and 43y According to the question, ⇒ 43x × 43y = 5547 ⇒ xy = 3 Now, co-prime with product 3 is (3,1) ⇒ The required number is (3 × 43, 1 × 43) = 129, 43 ∴ The greatest number is 129." ]

[ "- \\sqrt{14}$", "=$ ? About $22$ or $23$ times", "= 36$ $s = \\sqrt{36} = 6$ $P = 4 \\times 6 = 24$", "{ e }^{ 0.0046\\times { 1 }/{ 12+0.0832\\times \\sqrt { { 1 }/{ 12 } } } }$$ $$X=0.03238=3.238\\%$$", "\\pm i \\sqrt{7}}{16}$", "\\pm \\sqrt{17}}{16}$", "= \\sqrt {2 \\times 450 \\times 1000} = 948.68\\;m/sec\\;$$", "$$48\\sqrt{3}-24\\pi.$$ -", "\\sqrt{x} = 17$.", "What is the value of ? 4 16 32 [#permalink] 02 May 2018, 05:47 Expert's post sandy wrote: What is the value of $$\\sqrt{\\sqrt{64}}$$? A. $$2\\sqrt{2}$$ B. 4 C. $$4\\sqrt{2}$$ D. 16 E. 32 √64 = 8, so $$\\sqrt{\\sqrt{64}}$$ = √8 From here, we get... √8 = √[(4)(2)] = (√4)(√2) = 2√2 Cheers. Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com Re: What is the value of ? 4 16 32 [#permalink] 02 May 2018, 05:47 Display posts from previous: Sort by", "## Intermediate Algebra (6th Edition) $\\sqrt[4] 16=2$, because $2^{4}=2\\times2\\times2\\times2=16$ Therefore, $-\\sqrt[4] 16=-2$, as we are taking the opposite of $\\sqrt[4] 16$", "h$$ $$\\frac{1}{2}\\times (12+22)\\times3\\sqrt 19$$ = 51 $$\\sqrt19$$ cm2", "2}$ $= - 3 \\sqrt{16}$ $= - 3 \\cdot 4 = - 12$", "value of $BD$ is $\\sqrt{425/2}$. Status Not open for further replies.", "\\sqrt{65}}\\right)$", "of $$\\sqrt3$$.", "# What is the value of (-sqrt(-1))^(4n+3)? Oct 22, 2016 ${\\left(- \\sqrt{- 1}\\right)}^{4 n + 3} = i$ or $\\sqrt{- 1}$ #### Explanation: ${\\left(- \\sqrt{- 1}\\right)}^{4 n + 3}$ = ${\\left(- i\\right)}^{4 n + 3}$, where ${i}^{2} = - 1$ = ${\\left(- 1 \\times i\\right)}^{4 n + 3}$ = ${\\left(- 1\\right)}^{4 n + 3} \\times {i}^{4 n + 3}$ = $- 1 \\times {i}^{4 n} \\times {i}^{3}$, as odd power of $- 1$ is $- 1$ = $- 1 \\times 1 \\times {i}^{2} \\times i$, as ${i}^{2} = - 1$ ${i}^{4} = {\\left({i}^{2}\\right)}^{2} = {\\left(- 1\\right)}^{2} = 1$ = $- 1 \\times 1 \\times \\left(- 1\\right) \\times i$ = $i$ or $\\sqrt{- 1}$", "+ \\sqrt{89}$", "- 2 \\sqrt6 +2$", "value $\\sqrt{-1}$." ]

[ "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "out there... $2 \\times 2\\sqrt{15} = ?$ 34. anonymous not yet, example: $3*(4\\sqrt{5}) = 3*4\\sqrt{5} = 12\\sqrt{5}$ 35. anonymous i figuired it out,with everyones help ! thanks everyone 36. anonymous", "the sum of the numbers inside: 76, 38", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "## Algebra 1 First you separate 2 and \\sqrt 108. $\\sqrt108$ = $\\sqrt 36\\times \\sqrt3.$ $\\sqrt 36$ = 6 and $\\sqrt 3$ = $\\sqrt 3.$ Since $\\sqrt 108$ = $6\\sqrt 3$, multiply by 2 2$\\times$6 $\\sqrt 3$ 12$\\sqrt 3$ Answer is A", "= 36$ $s = \\sqrt{36} = 6$ $P = 4 \\times 6 = 24$", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$.", "it is only reachable from $24$ or $26$, so we have $29 + 6 = 35$. The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = \\boxed{169}$.", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "equal to $\\tfrac12 \\cdot 7 \\cdot \\tfrac{3\\sqrt{7}}{4} = \\tfrac{21\\sqrt{7}}{8}$, so $p+q+r = \\boxed{36}$.", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "the answer is $3 + 11 = \\boxed{\\mathrm{(A)}\\ 14}$", "## Intermediate Algebra (6th Edition) $\\sqrt[4] 16=2$, because $2^{4}=2\\times2\\times2\\times2=16$ Therefore, $-\\sqrt[4] 16=-2$, as we are taking the opposite of $\\sqrt[4] 16$", "## Intermediate Algebra (6th Edition) The product rule holds that $\\sqrt[n] a\\times\\sqrt[n] b=\\sqrt[n] (ab)$ (where $\\sqrt[n] a$ and $\\sqrt[n] b$ are real numbers). Therefore, $\\sqrt[4] 8\\times\\sqrt[4] 2=\\sqrt[4] (8\\times2)=\\sqrt [4] 16=2$ We know this because $2^{4}=16$.", "values of $EG^2$ are $(8 \\pm \\sqrt{19})^2+6^2+3^2 = 128 \\pm 16\\sqrt{19}$. The only one that satisfies the conditions of the problem is $128 - 16\\sqrt{19}$, from which the answer is $128+16+19=\\boxed{163}$.", "3$ 6. $7 + \\sqrt[3]{125} +1^5 -2^3$ 7. $(\\sqrt[3]{27} +\\sqrt{64})^2$ 8. $(\\sqrt{16+9} \\div 5^1) \\times 93$ 9. $\\frac{9^2 + 12^2 + 5^3 + 650}{\\sqrt[3]{125} \\times 10^2}$ 10. $\\frac{6^3 -(\\sqrt{169})^2 + \\sqrt{8}}{7^2 \\times1^9}$", "is $4.$ Here we will find the same by prime factorization method. Firstly, we will factorize $16.$ As $16$ is an even number, it will be divisible by $2.$ So we can write $16=2 \\times 8.$ Now, we will factorize the number $8.$ Again since $8$ is an even number, we have $8=2 \\times 4.$ And we all know that $4=2 \\times 2.$ Thus combining all the above factorizations, we obtain the prime factorization of $16$ which is given below: $16=2 \\times 2 \\times 2 \\times 2.$ Taking square root on both sides, we get that $\\sqrt{16}=\\sqrt{2 \\times 2 \\times 2 \\times 2}$ $=\\sqrt{2 \\times 2} \\times \\sqrt{2 \\times 2}$ $[ \\because \\sqrt{x \\times y}=\\sqrt{x} \\times \\sqrt{y}]$ $=2 \\times 2$ $[ \\because \\sqrt{a \\times a}=a]$ $=4$ So $4$ is the square root of $16.$ #### Is the square root of 16 rational? To answer the question let us recall the definition of a rational number. A number $x$ is said to be a rational number, if we can express $x$ as $\\frac{p}{q}$ for some integers $p$ and $q$ with $q \\neq 0.$ We know that $\\sqrt{16}=\\sqrt{4^2}$ $=\\pm 4.$ Note that $+4=\\frac{4}{1}$ and $-4=\\frac{-4}{1}.$ Thus, both $+4$ and $-4$ are rational numbers. Hence, we conclude that the square root of $16$ is a rational number. #### Is 16 a", "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." ]

[ "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "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.", "values of $EG^2$ are $(8 \\pm \\sqrt{19})^2+6^2+3^2 = 128 \\pm 16\\sqrt{19}$. The only one that satisfies the conditions of the problem is $128 - 16\\sqrt{19}$, from which the answer is $128+16+19=\\boxed{163}$.", "What is the value of ? 4 16 32 [#permalink] 02 May 2018, 05:47 Expert's post sandy wrote: What is the value of $$\\sqrt{\\sqrt{64}}$$? A. $$2\\sqrt{2}$$ B. 4 C. $$4\\sqrt{2}$$ D. 16 E. 32 √64 = 8, so $$\\sqrt{\\sqrt{64}}$$ = √8 From here, we get... √8 = √[(4)(2)] = (√4)(√2) = 2√2 Cheers. Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com Re: What is the value of ? 4 16 32 [#permalink] 02 May 2018, 05:47 Display posts from previous: Sort by", "# What is the value of (-sqrt(-1))^(4n+3)? Oct 22, 2016 ${\\left(- \\sqrt{- 1}\\right)}^{4 n + 3} = i$ or $\\sqrt{- 1}$ #### Explanation: ${\\left(- \\sqrt{- 1}\\right)}^{4 n + 3}$ = ${\\left(- i\\right)}^{4 n + 3}$, where ${i}^{2} = - 1$ = ${\\left(- 1 \\times i\\right)}^{4 n + 3}$ = ${\\left(- 1\\right)}^{4 n + 3} \\times {i}^{4 n + 3}$ = $- 1 \\times {i}^{4 n} \\times {i}^{3}$, as odd power of $- 1$ is $- 1$ = $- 1 \\times 1 \\times {i}^{2} \\times i$, as ${i}^{2} = - 1$ ${i}^{4} = {\\left({i}^{2}\\right)}^{2} = {\\left(- 1\\right)}^{2} = 1$ = $- 1 \\times 1 \\times \\left(- 1\\right) \\times i$ = $i$ or $\\sqrt{- 1}$", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "# What is the value of -sqrt144? Oct 9, 2017 $- 12$ #### Explanation: ${12}^{2}$ is $144$. This means $\\sqrt{144} = 12$ $- \\sqrt{144}$ $= - \\left(\\sqrt{144}\\right)$ $= - \\left(12\\right)$ $= - 12$", "# Find the value of Question: Find the value of $(1296)^{0.17} \\times(1296)^{0.08}$. Solution: $(1296)^{0.17} \\times(1296)^{0.08}$ $=(1296)^{0.17+0.08}$ $=(1296)^{0.25}$ $=(1296)^{\\frac{1}{4}}$ $=\\sqrt[4]{1296}$ $=6$", "### Home > INT1 > Chapter 8 > Lesson 8.1.4 > Problem8-61 8-61. Evaluate each expression below. 1. $\\sqrt [ 3 ] { - 64 }$ What value of $x$ times itself three times is equal to $−64$, $(x)(x)(x) = −64$? $x = −4$ 1. $\\sqrt [ 5 ] { 32 }$ What value of $x$ times itself $5$ times equals $32$, $(x)(x)(x)(x)(x) = 32$? $x = 2$ 1. $\\sqrt [ 3 ] { 27 }$ 1. $\\sqrt [ 4 ] { 10000 }$", "out there... $2 \\times 2\\sqrt{15} = ?$ 34. anonymous not yet, example: $3*(4\\sqrt{5}) = 3*4\\sqrt{5} = 12\\sqrt{5}$ 35. anonymous i figuired it out,with everyones help ! thanks everyone 36. anonymous", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "= 36$ $s = \\sqrt{36} = 6$ $P = 4 \\times 6 = 24$", "3$ 6. $7 + \\sqrt[3]{125} +1^5 -2^3$ 7. $(\\sqrt[3]{27} +\\sqrt{64})^2$ 8. $(\\sqrt{16+9} \\div 5^1) \\times 93$ 9. $\\frac{9^2 + 12^2 + 5^3 + 650}{\\sqrt[3]{125} \\times 10^2}$ 10. $\\frac{6^3 -(\\sqrt{169})^2 + \\sqrt{8}}{7^2 \\times1^9}$", "## Intermediate Algebra (6th Edition) We are asked to evaluate $\\sqrt 64-\\sqrt[3] 64$. We know that $\\sqrt 64=8$, because $8^{2}=8\\times8=64$ And we know that $\\sqrt[3] 64=4$, because $4^{3}=4\\times4\\times4=64$ Therefore, $\\sqrt 64-\\sqrt[3] 64=8-4=4$", "is $4.$ Here we will find the same by prime factorization method. Firstly, we will factorize $16.$ As $16$ is an even number, it will be divisible by $2.$ So we can write $16=2 \\times 8.$ Now, we will factorize the number $8.$ Again since $8$ is an even number, we have $8=2 \\times 4.$ And we all know that $4=2 \\times 2.$ Thus combining all the above factorizations, we obtain the prime factorization of $16$ which is given below: $16=2 \\times 2 \\times 2 \\times 2.$ Taking square root on both sides, we get that $\\sqrt{16}=\\sqrt{2 \\times 2 \\times 2 \\times 2}$ $=\\sqrt{2 \\times 2} \\times \\sqrt{2 \\times 2}$ $[ \\because \\sqrt{x \\times y}=\\sqrt{x} \\times \\sqrt{y}]$ $=2 \\times 2$ $[ \\because \\sqrt{a \\times a}=a]$ $=4$ So $4$ is the square root of $16.$ #### Is the square root of 16 rational? To answer the question let us recall the definition of a rational number. A number $x$ is said to be a rational number, if we can express $x$ as $\\frac{p}{q}$ for some integers $p$ and $q$ with $q \\neq 0.$ We know that $\\sqrt{16}=\\sqrt{4^2}$ $=\\pm 4.$ Note that $+4=\\frac{4}{1}$ and $-4=\\frac{-4}{1}.$ Thus, both $+4$ and $-4$ are rational numbers. Hence, we conclude that the square root of $16$ is a rational number. #### Is 16 a", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "• # question_answer Find the value of $\\sqrt{2\\sqrt{2\\sqrt{2\\sqrt{2\\sqrt{2}?}}}}$ A) ${{2}^{\\frac{1}{31}}}$ B) ${{2}^{\\frac{1}{32}}}$ C) ${{2}^{\\frac{31}{32}}}$ D) ${{2}^{\\frac{30}{31}}}$ (b): $y=\\sqrt{2\\sqrt{2\\sqrt{2\\sqrt{2\\sqrt{2}}}}}$ ${{y}^{2}}=2\\sqrt{2\\sqrt{2\\sqrt{2\\sqrt{2}}}}$ ${{y}^{4}}=8\\sqrt{2\\sqrt{2\\sqrt{2}}}$ ${{y}^{8}}=128\\sqrt{2\\sqrt{2}}$ ${{y}^{16}}={{\\left( 128 \\right)}^{2}}\\times 2\\sqrt{2}$ Then ${{y}^{32}}={{\\left( 128 \\right)}^{4}}\\times 4\\times 2$ ${{y}^{32}}={{2}^{28+2+1}}={{2}^{31}}$ ${{2}^{\\frac{31}{32}}}$", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "# What is $$\\sqrt 4 \\times \\sqrt {64} ?$$ 16 Explanation The square root of 4 is 2 and the square root of 64 is 8 so $$2 \\times 8 = 16$$.", "Hoppin647 31 # what is the exact value of the expression the square root 72. − the square root of 8. the square root of 128.? simplify if possible. $\\sqrt{72} = \\sqrt{36 \\times 2} = \\sqrt{36} \\times \\sqrt{2} = 6 \\times \\sqrt{2} = 6 \\sqrt{2}$ $\\sqrt{8} = \\sqrt{4 \\times 2} = \\sqrt{4} \\times \\sqrt{2} = 2 \\times \\sqrt{2} = 2 \\sqrt{2}$ $\\sqrt{128} = \\sqrt{64 \\times 2} = \\sqrt{64} \\times \\sqrt{2} = 8 \\times \\sqrt{2} = 8 \\sqrt{2}$" ]

[ "from the base of the wall.", "angle. How high up the wall does[/tex...", "the base of the wall.", "8 m Thus, the height of the tower is 8 m. Is there an error in this question or solution?", "the upper of the ladder reach if the lower ends are 1.8 meters apart?", "platform, and turn around at the edge, or turn around when they hit a wall.", "11. A man 180 cm tall walks at a rate of 2 m/sec. away, from a source of light that is 9 m above the ground. How fast is the length of his shadow increasing when he is 3 m away from the base of light? Solution: Let MN denote the vertical lamp post of height 6 meter and at any instant t, a man XY of height 2 meter be standing in front of the lamp post at a distance ‘m’ and let ‘n’ be length of his shadow. It can be seen in a figure as: We can notice ∼ = ⇒ m = 4n = 4 =x 2 [Since,= 2] = 0.5 m/sec ### Question 12. A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec. How fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall. Solution: Let the height of the wall that is leaning against a wall be denoted by y meters and the distance of the foot of ladder from base of the wall be x meters. we can derive tan θ = y/x ⇒ y", "is 8.5m long. It is built so that its lower ends are 3.5 meters apart. How high does the upper end of the ladder reach?", "meter. You take the slack extra meter and lift it half a meter high.", "fell? 3. A person weighing 645 N climbs up a ladder to a height of 4.55 m. 1. What work does the person do? 2. What is the increase in gravitational potential energy? 3. Where does the energy come from to cause this increase in $PE$ ?", "ladder, and did it with just the distance of two meters. I can say that the force exerted towards the right has to equal the force exerted towards the left. But I can't figure out how to get the forces on the left or the right. Last edited by a moderator:", "foot of the tower. How long are these ropes? Ladder 8 m long is leaning against the wall. Its foot is 1 m away from the wall. In which height ladder touches the wall? The ladder is 10 m long The ladder is 8 m high How many meters is the distant heel from the wall? Ladder 10 meters long is staying against the wall so that its bottom edge is 6 meters away from the wall. What height reaches ladder? 6. Equilateral triangle The equilateral triangle has a 23 cm long side. Calculate its content area. The double ladder is 8.5m long. It is built so that its lower ends are 3.5 meters apart. How high does the upper end of the ladder reach? The double ladder has 3 meters long shoulders. What is the height of the upper of the ladder reach if the lower ends are 1.8 meters apart? The double ladder shoulders should be 3 meters long. What height will the upper top of the ladder reach if the lower ends are 1.8 meters apart? 10. Isosceles trapezoid What is the height of an isosceles trapezoid, the base of which has a length of 11 cm and 8 cm and whose legs measure 2.5 cm? 11. Sum of squares The sum of squares above", "elastically shortened (compressed) the rock at the base of the wall by about 3.5 cm. This is elastic strain, so if the rock above were removed, the rock at the base would return to its original size.", "## Scaffolding physics A painter is standing on a 2.2-m-long platform of mass 10 kg. The platform is held up by two ropes, each connected 10 cm from the end of the platform, so that the ropes are 2.0 m apart. The painter stands at a distance d = 0.60 m from the left-hand rope. The painter's mass is 60 kg. The tension in the left-hand rope is ?. The tension in the right-hand rope is ? .", "If the height of a water slide in the figure is h = 2.8 m and the person's initial speed at point A is 0.50 m/s, what is the new horizontal distance L between the base of the slide and the splashdown point of the person? http://www.webassign.net/walker/08-23alt.gif", "the tower, to the nearest metre? c) How long are the supporting cables, to the nearest metre? d) How far apart are the cable fasteners, to the nearest metre? Note: This is a simplified diagram. Normally it takes three or four cables to support a tower.", "the slope from the bottom of the incline, and the hanging block is at the bottom of the incline. For the final position, the inclined block will be at the bottom of the incline, and the hanging block will be 0.5 m down from the bottom of the incline.", "onto the ladder 4. Dec 6, 2009 ### Lok yeah must have been blind, tried to edit but you were faster :P There are of course to Normal forces acting on the ladder. Wall and ground. But good job. So for the beam there is as above : -gravity -Normal force ( hinge ) -Tension ( beam ) -Tension (cable) or call it a Force. Your reasoning is correct as both ends suffer the same forces, but the final angle is it up or down?", "sign shows that it strikes the opposite wall 75 m below the original level. More on Kinematics", "be greatly appreciated. Thank you! Last edited: Nov 2, 2008 2. Nov 3, 2008 ### alphysicist Hi RedBarchetta, Yes, it's the other angle; the ladder makes a 15 degree angle with the wall, so the angle with the floor is 75 degrees. Once you correct your diagram, do you see why the angles your professor chose are correct?" ]

[ "same proportion — 3 : 4 : 5 (because $6 \\div 2 = 3, 8 \\div 2 = 4$, and $10 \\div 2 = 5$). Whenever you find a Pythagorean triple, you can apply these ratios with greater factors as well. Finally, look at the side lengths of the triangle in Example 3: The two legs are 5 cm and 12 cm and the length of the missing side (the hypotenuse) is 13 cm. The side lengths make a ratio of 5 : 12 : 13. This, too, is a Pythagorean triple. You can infer that this ratio, multiplied by greater factors, will also yield numbers that satisfy the Pythagorean Theorem. There are infinitely many Pythagorean triples, but a few of the most common ones and their multiples are in the chart below: Pythagorean triple $\\times 2$ $\\times 3$ $\\times 4$ 3 – 4 – 5 6 – 8 – 10 9 – 12 – 15 12 – 16 – 20 5 – 12 – 13 10 – 24 – 26 15 – 36 – 39 20 – 48 – 52 7 – 24 – 25 14 – 48 – 50 21 – 72 – 75 28 – 96 – 100 8 – 15 – 17 16 – 30 – 34 24 – 45 – 51 32 –", "hypotenuse of length 17 m, one leg of length 8 m, and the other leg of length $\\displaystyle \\sqrt{17^2- 8^2}= 15$ m. The reason for the \"additional 14 N\" is that the fish is being reeled in horizontally but the line is the hypotenuse of the triangle. Find the total force so that the horizontal component is 105 N. The ratio of hypotenuse to horizontal side of the right triangle is 17/15 so the tension in the line is (17/15)(105)= 119. Oct 22nd 2018, 07:01 PM #3 Junior Member Join Date: Jun 2018 Posts: 7 Thank you for your help. So the upward component X would be (8/15)*105 = 56. Let B be Buoyancy, then X + B = 150 56 + B = 150 So B = 94N So Normal contact force on the man & rod would be 800 + 56 = 856 N Tags forces, questions, tensions Thread Tools Display Modes Linear Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post hongiddong Kinematics and Dynamics 4 Aug 6th 2014 02:23 PM iannedrs Advanced Mechanics 3 Jul 22nd 2013 05:03 AM kastamonu Kinematics and Dynamics 9 Mar 31st 2013 07:57 AM flower Advanced Electricity and Magnetism 1 Nov 20th 2009 02:06 AM jello17 Kinematics and Dynamics 3 Oct 4th 2008 09:31 PM", "# Rate of Change in the Legs of a Right Triangle One leg of a right triangle is always $6$ feet long and the other leg is increasing at a rate of $2$ ft/s. Find the rate of change in ft/s of the hypotenuse when it is $10$ feet long. The answer is $1.6$ So I try the following formula based on the Pythagorean theorem: $(6^2)^2 + (2t)^2 = 10^2$ Compute for $t$; which is the time elapsed as the hypotenuse got to $10$. I get $t = 4$. So I divide $10/4$ and I get $2.5$. Am I doing something wrong? Any Hint? Let $h$ be the length of the hypotenuse and $x$ be the other leg which is varying, then at an instant of time $t$, we have $$x^2+6^2=h^2.$$ Now take the derivative with regards to time $t$, to get $$2x\\frac{dx}{dt}=2h\\frac{dh}{dt}.$$ Now use the fact that $h=10$ and $\\frac{dx}{dt}=2$. • ok i get [ 2x(dx/dt) = 2h(dh/dt) ] Then substitute your variables; I get... [ (dh/dt) = (1/4)x ] I then substitute the x with the earlier pythagorean relation... getting (1/4)sqrt(10^2 - 6^2)... = 2. It is among the choices! But maybe the answer key maybe wrong? – james Jun 13 '15 at 6:41 • @james You should have $\\frac{\\text{d}h}{\\text{d}t} = \\frac{1}{4}x\\frac{\\text{d}x}{\\text{d}t}$. – N. F. Taussig", "of the triangle is 129.5 cm2. ∴ 7.4h = 129.5 h = $\\left(\\frac{129.5}{7.4}\\right)$ = 17.5 cm ∴ Length of the other leg = 17.5 cm Question 7: Find the area of a right triangle whose base is 1.2 m and hypotenuse 3.7 m. Here, base = 1.2 m and hypotenuse = 3.7 m In the right angled triangle: Perpendicular = $\\sqrt{\\left(H\\mathrm{ypotenuse}{\\right)}^{2}-\\left(B\\mathrm{ase}{\\right)}^{2}}$ $=\\sqrt{{\\left(3.7\\right)}^{2}-{\\left(1.2\\right)}^{2}}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{13.69-1.44}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{12.25}\\phantom{\\rule{0ex}{0ex}}=3.5$ Area = $\\left(\\frac{1}{2}×\\mathrm{base}×\\mathrm{perpendicular}\\right)$ sq. units = $\\left(\\frac{1}{2}×1.2×3.5\\right)$ m2 ∴ Area of the right angled triangle = 2.1 m2 Question 8: The legs of a right triangle are in the ratio 3 : 4 and its area is 1014 cm2. Find the lengths of its legs. In a right angled triangle, if one leg is the base, then the other leg is the height. Let the given legs be 3x and 4x, respectively. Area of the triangle = $\\left(\\frac{1}{2}×3x×4x\\right)$ cm2 ⇒ 1014 = (6x2) ⇒ 1014 = 6x2 x2 = $\\left(\\frac{1014}{6}\\right)$ = 169 x = $\\sqrt{169}$ = 13 ∴ Base = (3 $×$ 13) = 39 cm Height = (4 $×$ 13) = 52 cm Question 9: One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its cost at the rate of Rs 250 per m2. Consider a right-angled triangular scarf (ABC). Here, ∠B= 90° BC = 80", "look up that $$\\tan(10)$$ is 0.176. Then we can calculate that $$h$$ is about 17.6. That means the tree is 17.6 feet tall. Glossary Entries • cosine The cosine of an acute angle in a right triangle is the ratio (quotient) of the length of the adjacent leg to the length of the hypotenuse. In the diagram, $$\\cos(x)=\\frac{b}{c}$$. • sine The sine of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the hypotenuse. In the diagram, $$\\sin(x) = \\frac{a}{c}.$$ • tangent The tangent of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the adjacent leg. In the diagram, $$\\tan(x) = \\frac{a}{b}.$$ • trigonometric ratio Sine, cosine, and tangent are called trigonometric ratios.", "in square feet? (A) 37 (B) 27 (C) 54/4 (D) 21/4 (E) 5 A rectangle is divided by its diagonal into two congruent right triangles. Rule: If a right triangle has a leg:hypotenuse ratio of 3:5 (or 4:5, or legs in ratio 3:4), it is a 3-4-5 right triangle. Common 3x-4x-5x triangles include 6-8-10 (x=2), 9-12-15 (x=3), 12-16-20 (x=4), etc. The short leg of this right triangle in the closet is $$\\frac{9}{2}$$ feet. The hypotenuse is $$\\frac{15}{2}$$ feet. Focus on the numerator. Without any calculation we can say the other leg must be $$\\frac{12}{2}$$ feet long. The numerator matches the 9-12-15 pattern. Keep the same denominator. OR $$(leg1)^2 + (leg2)^2 = h^2$$ $$\\frac{9}{2}^2 + (leg2)^2 = \\frac{15}{2}^2$$ $$(leg2)^2 = (\\frac{225}{4}-\\frac{81}{4})$$ $$\\sqrt{(leg2)^2}=\\sqrt{\\frac{144}{4}}=\\frac{12}{2} = 6$$ Area $$= L*W = \\frac{9}{2}*6=\\frac{54}{2}=27$$ *The multiplier is $$\\frac{3}{2}$$.$$(3*\\frac{3}{2})=\\frac{9}{2}$$, and $$(5*\\frac{3}{2})=\\frac{15}{2}$$. If you focus on the numerator and just make the second leg's denominator match the denominator of the other two lengths, you do not need to find the multiplier or use the Pythagorean theorem. Intern Joined: 05 Dec 2017 Posts: 5 Re: The diagonal of the floor of a rectangular closet is 15/2 feet. The sh [#permalink] ### Show Tags 07 Mar 2018, 09:04 Hey Bunuel is the qustion asking to calculate the area of the closet or the rectangular floor? Math Expert Joined:", "20 E. 24 [Reveal] Spoiler: Attachment: T6019.png These triangles (ABC and DAB) are similar 3-4-5 right triangles. Both have one 90° angle, and angle B is common to both. If two angles are equal, the third is equal. (AA = AAA = similar triangles.) Whenever a right triangle has longer leg length that is a multiple of 4 (20), and a hypotenuse that is a multiple of 5 (25), the figure is a 3-4-5 right triangle. Or find AC, and the 3x-4x-5x ratio will be clear. Let AC = a, AB = b, and BC = c. By Pythagorean theorem, $$a^2 + b^2 = c^2$$ $$a^2 + 20^2 = 25^2$$ $$a^2 + 400 = 625$$ $$a^2 = 225$$ $$a = 15$$ = AC 15-20-25 divided by 5 throughout = 3-4-5 side ratio Smaller triangle DAB must also be a 3-4-5 right triangle. Its hypotenuse AB = 20. To find length of shorter leg AD, find multiplier, using hypotenuse AB. 3x: 4x: 5x --> 5x is hypotenuse, 3x is shorter leg. 5x = 20 x = 4 = multiplier AD corresponds with AC. Both are the shorter leg, i.e., 3x. 3x = (3)(4) = 12 _________________ At the still point, there the dance is. -- T.S. Eliot Formerly genxer123 If AB=20 and BC=25, what is the length of AD", "Free Version Moderate # PythagoreanTheorem: Using Area to Find the Perimeter GEOM-RQQ5G4 A right triangle with a leg length of $28$ meters has an area of $672$ square meters.", "odd. This means that all primitive Pythagorean triples have one odd-length leg and one odd-length hypotenuse. The remaining leg, the one that is $2kl$, is, of course, even. However, we know that one of $k$ and $l$ is even also so that means that leg $b$ is a multiple of $4$. The corollary of all this is that no number, $n$, that is even but not a multiple of $4$ i.e. where $n \\equiv 2 (mod 4)$, can be a member of a primitive Pythagorean triple. which in turn means that $90 \\equiv 2 (mod 4)$ is not a member of any primitive Pythagorean triple.", "# Generating Pythagorean triples where the legs are Hypotenuses of other Pythagorean triples I know how to generate regular Pythagorean Triples given two positive integers P and Q such that $$a=2*p*q$$ $$b=p^2-q^2$$ $$c=p^2+q^2$$ where $$p>q$$, but I want to find scenarios where $$a$$ and $$b$$ can also be written in the format $$p^2+q^2$$. Specifically, I am looking for a generalized way of finding all situations where the hypotenuses of the first two triples form the legs of the third triple. I want to be able to generate a formula for finding these special triples. So far, I have been able to find four cases where this holds true by using the brute force method. $$P_1$$ $$Q_1$$ $$A_1=2P_1Q_1$$ $$B_1=P_1^2-Q_1^2$$ $$C_1=A_3$$ 10 2 40 96 104 16 4 128 240 272 15 9 270 144 306 21 3 126 432 450 $$P_2$$ $$Q_2$$ $$A_2=2P_2Q_2$$ $$B_2=P_2^2-Q_2^2$$ $$C_2=B_3$$ 12 3 72 135 153 12 9 216 63 225 12 8 192 80 208 20 12 480 256 544 $$P_3$$ $$Q_3$$ $$A_3=2P_3Q_3=C_1$$ $$B_3=P_3^2-Q_3^2=C_2$$ $$C_3=C_1^2+C_2^2$$ 13 4 104 153 185 17 8 272 225 353 17 9 306 208 370 25 9 450 544 706 As you can see, the third chart shows Pythagorean triples with legs that are also hypotenuses of the previous charts. For example, the right triangle with sides (104, 153,185)", "a Pythagorean triple, other than $2^0=1$, of course, under the condition that the legs are relatively prime. Alternatively, it is definitely true that the difference cannot be an odd power of $2$.", "Start by writing the Pythagorean Theorem $=$= $20^2+21^2$202+212 Substitute in the lengths of the legs $=$= $400+441$400+441 Square both $=$= $841$841 Find the sum so, $c$c $=$= $\\sqrt{841}$√841 Then, take the square root of both sides of the equation $c$c $=$= $29$29 Since 841 was a perfect square, $c$c is equal to $29$29 The missing value is $c=29$c=29, forming the triple $\\left(20,21,29\\right)$(20,21,29). Practice questions Question 7 Sean knows the two largest numbers in a Pythagorean Triple, which are $41$41 and $40$40. What number, $a$a, does Sean need to complete the triple? Question 8 We would like to find the hypotenuse in a right triangle with shorter side lengths $6$6 and $8$8, using our knowledge of common Pythagorean triples. 1. Below are some common Pythagorean triples. The two shorter sides $6$6, $8$8 and its hypotenuse will be multiples of the sides in which of the triples? A $\\left(3,4,5\\right)$(3,4,5) B $\\left(5,12,13\\right)$(5,12,13) C $\\left(8,15,17\\right)$(8,15,17) D $\\left(7,24,25\\right)$(7,24,25) They will be multiples of the Pythagorean triple: ($\\editable{}$,$\\editable{}$,$\\editable{}$) 2. What number when multiplied by $3$3 and $4$4 gives $6$6 and $8$8 respectively? $\\editable{}$ 3. Hence, what is the length of the hypotenuse in the triangle with two shorter sides $6$6 and $8$8? $\\editable{}$ Applications using the Pythagorean theorem Remember that for the Pythagorean theorem to be applicable to a practical problem, we must be", "hypotenuse $$25^\\circ$$ 0.906 0.423 $$65^\\circ$$ 0.423 0.906 Looking at a general right triangle, the angles can be written as 90, $$\\theta$$, and $$90-\\theta$$. Mathematicians often use Greek letters to represent angles. For instance, $$\\theta$$ is a Greek letter we use frequently in trigonometry. cosine of angle sine of angle angle adjacent leg $$\\div$$ hypotenuse opposite leg $$\\div$$ hypotenuse $$\\theta^\\circ$$ $$\\frac{x}{h}$$ $$\\frac{y}{h}$$ $$(90-\\theta)^\\circ$$ $$\\frac{y}{h}$$ $$\\frac{x}{h}$$ Glossary Entries • cosine The cosine of an acute angle in a right triangle is the ratio (quotient) of the length of the adjacent leg to the length of the hypotenuse. In the diagram, $$\\cos(x)=\\frac{b}{c}$$. • sine The sine of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the hypotenuse. In the diagram, $$\\sin(x) = \\frac{a}{c}.$$ • tangent The tangent of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the adjacent leg. In the diagram, $$\\tan(x) = \\frac{a}{b}.$$ • trigonometric ratio Sine, cosine, and tangent are called trigonometric ratios.", "$$O$$ to $$Q,$$ and from $$P$$ to $$Q$$ are relatively flat and straight, approximately how many miles is it from $$P$$ to $$Q ?$$ A. $$3$$ B. $$\\dfrac92$$ C. $$5$$ D. $$6$$ E. $$\\dfrac{15}2$$ Source: GMAT Prep APPROACH #1: Apply Pythagorean theorem Let x = the length of PQ Apply the Pythagorean theorem to get: 4² + x² = 5² Evaluate: 16 + x² = 25 Rewrite as: x² = 9 Solve: x = 3 or x = -3 (we can eliminate x = -3, since lengths can't be negative) So, ON THE MAP, the length of PQ is 3 INCHES Given: 2 inches represents 3 miles In other words, 1 inch = 1.5 miles So, 3 inches = 4.5 miles, since (3)(1.5) = 4.5 ----------------------- APPROACH #2: Identify Pythagorean triples Before test day, students should memorize the following Pythagorean triples 3-4-5 5-12-13 7-24-25 (and maybe 8-15-17) Pythagorean triples sets of INTEGER values that could represent the three lengths of a right triangle If we've memorized the Pythagorean triples, will instantly see that side PQ must have a length of 3 INCHES Once we know that PQ = 3 INCHES (on the map), we can use the above technique to convert this to 4.5 miles", "* * * * * * * * The athlete is at $A$. The target is at $C\\!:\\;AC = 15$ The sling is 1 meter long: . $AB = 1$ When the projectile is released at $B$, it flies off tangent to the circle. . . Hence: . $\\angle ABC = 90^o$ We want the distance $BC.$ From Pythagorus: . $BC^2 + 1^2 \\:=\\:15^2 \\quad\\Rightarrow\\quad BC^2 \\:=\\:15^2 - 1^2 \\:=\\:225 - 1 \\:=\\:224$ Therefore: . $BC \\;=\\;\\sqrt{224} \\;=\\;14.96662955 \\;\\approx\\;14.97$ m. • November 30th 2008, 09:35 AM s3a Ok thanks for the help! I get a large chunk of it now but I still have minor confusion. What I am confused with is how do you know/assume that we have the hypotenuse and base of the right triangle? Like what I was thinking is 15m-1m or even just 15m. Like can you break down the words a little bit more? Sorry if I am asking for too much.", "Calculus: Early Transcendentals 8th Edition The solution is $$h=\\frac{4\\sqrt{c^2-16}}{c}.$$ Let $a=4$ be the one leg of this triangle and $b$ the other leg and $c$ be the hypotenuse. Using the expression for the area of such a triangle written in terms of $a$ and $b$ and in terms of $c$ and $h$ where $h$ is the altitude we have $$\\frac{1}{2}ab=\\frac{1}{2}ch\\Rightarrow h=\\frac{ab}{c}.$$ Now from Pythagorean theorem we have $$c^2=a^2+b^2\\Rightarrow b=\\sqrt{c^2-a^2}.$$ Putting this into the expression for altitude $$h=\\frac{a\\sqrt{c^2-a^2}}{c}.$$ Putting $a=4$ $$h=\\frac{4\\sqrt{c^2-16}}{c.}$$", "## Generating Pythagorean Triples Some small Pythagorean triples are $3, 4, 5$ $5, 12, 13$ $7, 24, 25$. In each case, the hypotenuse has length one greater one of the legs. Let’s try to find all such triples. Let the longer of the legs be $x$. Then the hypotenuse is $x+1$. The shorter leg must be an integer, $n$. $(x+1)^2 = x^2 + n^2$ $2x+1 = n^2$ $x = \\frac{n^2-1}{2}$. In order to get a Pythagorean triple, $x$ must also be an integer. That means $n^2 - 1$ must be even, so $n^2$ must be odd. That happens whenever $n$ itself is odd. $\\begin{array}{ccc} n & x = (n^2 - 1)/2 & x+1 \\\\ 1 & 0 & 1 \\\\ 3 & 4 & 5 \\\\ 5 & 12 & 13 \\\\ 7 & 24 & 25 \\\\ 9 & 40 & 41 \\\\ 11 & 60 & 61 \\\\ 13 & 84 & 85 \\end{array}$ How about Pythagorean triples where one of the legs is two less than the hypotenuse? $(x+2)^2 = x^2 + n^2$ $4(x+1) = n^2$ $x = \\frac{n^2}{4} - 1$ So here we must have $n^2$ a multiple of $4$, which happens when $n$ is even. $\\begin{array}{ccc} n & x = n^2/4-1 & x+2 \\\\ 2 & 0 & 2 \\\\ 4 & 3", "## Generating Pythagorean Triples Some small Pythagorean triples are $3, 4, 5$ $5, 12, 13$ $7, 24, 25$. In each case, the hypotenuse has length one greater one of the legs. Let’s try to find all such triples. Let the longer of the legs be $x$. Then the hypotenuse is $x+1$. The shorter leg must be an integer, $n$. $(x+1)^2 = x^2 + n^2$ $2x+1 = n^2$ $x = \\frac{n^2-1}{2}$. In order to get a Pythagorean triple, $x$ must also be an integer. That means $n^2 - 1$ must be even, so $n^2$ must be odd. That happens whenever $n$ itself is odd. $\\begin{array}{ccc} n & x = (n^2 - 1)/2 & x+1 \\\\ 1 & 0 & 1 \\\\ 3 & 4 & 5 \\\\ 5 & 12 & 13 \\\\ 7 & 24 & 25 \\\\ 9 & 40 & 41 \\\\ 11 & 60 & 61 \\\\ 13 & 84 & 85 \\end{array}$ How about Pythagorean triples where one of the legs is two less than the hypotenuse? $(x+2)^2 = x^2 + n^2$ $4(x+1) = n^2$ $x = \\frac{n^2}{4} - 1$ So here we must have $n^2$ a multiple of $4$, which happens when $n$ is even. $\\begin{array}{ccc} n & x = n^2/4-1 & x+2 \\\\ 2 & 0 & 2 \\\\ 4 & 3", "+0 # help -1 61 3 +392 In right triangle $PQR$, we have$$\\angle Q = \\angle R$$ and $PR = $$6\\sqrt{2}$$. What is the area of$\\triangle PQR\\$? Mar 25, 2020 #1 +483 0 Because $$\\angle Q = \\angle R$$, they must be acute angles(because if they were the right angle, then two right angles would already be 180 degrees, making the triangle degenerate). Name the angle measurement of one of these 2 angles x. We can write the expression: 90 + x + x = 180(the internal angle sum of a triangle). 90 + 2x = 180 2x = 180 x = 45 That means that we have a 45-45-90 triangle at hand, which happens to be a special right triangle. The ratio of a 45-45-90 triangle is: $$1,1,\\sqrt2$$ where the legs are 1, and the hypotenuse is $$\\sqrt2$$(Remember this! It will make math with certain right triangles a lot easier!) We know that angle P must be the right angle, because angle Q and R are both 45 degrees. We then have that the hypotenuse PR is equal to $$6\\sqrt2$$ Because the ratios of a 45-45-90 triangle are 1,1,$$\\sqrt2$$, we know that right triangle PQR is scaled by a factor of 6 compared to a 45-45-90 of leg length 1. That means the leg lengths of our", "# Primitive Pythagorean triples with the least, same hypotenuse or side #### Loren What are the primitive Pythagorean triples with the least, same hypotenuse or side? #### Loren Say, where hypotenuses of two differing primitive Pythagorean triples are equal. #### mrtwhs \\begin{align} 1105^2 &=47^2+1104^2\\\\ &=264^2+1073^2\\\\ &=576^2+943^2\\\\ &=744^2+817^2 \\end{align} It took google 0.62 seconds to find these four. Loren #### Loren mrtwhs, What is the smallest such hypotenuse with just two pairs of legs? Last edited: #### mrtwhs Don't know if it is the smallest but google gives $33^2+56^2 = 16^2 +63^2 = 65^2$ #### skipjack Forum Staff It is the smallest. The smallest pair with a leg in common is ((5, 12, 13), (12, 35, 37)). Loren" ]

[ "it new? No, just look up any list of Pythagorean triples and compute any ratios you want. Or you can just use the Pythagorean Theorem directly, e.g. if you know the two legs of a right triangle are 6 and 8, just compute the square root of 6²+8² and get 10 for the hypotenuse. Nothing to see here, move along! As final “proof” that this is not trigonometry, I give you an example of a typical problem given to beginning trig students: If a 10′ ladder (teachers of trig are obsessed with ladders) is placed at a 70° angle to the ground against a wall, how high is the top of the ladder from the ground? P322 is completely useless here, but using modern trigonometry, the sine of 70°, about .94, is the ratio of the side opposite that angle to the hypotenuse. Multiply that by the hypotenuse of 10′ and you get the length of the opposite side, which in this case is the height, about 9.4′. Of course, if you also knew the distance from the bottom of the ladder to the wall along the ground, you could also figure this out with P322 or the Pythagorean Theorem, but calling it trigonometry is still a big stretch. Incidentally, I’m not alone. Evelyn Lamb in Scientific American", "= 6m}}$ Therefore, the distance of the foot of the ladder from the base of the wall is ${\\text{6m}}$. Note: This problem is solved using Pythagoras theorem. According to the Pythagoras theorem, for any right-angled triangle, the sum of the squares of its non-hypotenuse sides is equal to the square of the hypotenuse. We cannot use trigonometry here as we are not given the details of any of the angles other than the right angle.", "a. Find the distance of the foot of the ladder from the wall. Here. As we can see, The ladder with wall forms a right-angled triangle with the vertical height of the wall = perpendicular = 12 m length of ladder = Hypotenuse = 15 m Now, As we know In a Right-angled Triangle: By Pythagoras Theorem, $(Hypotenus)^2=(Base)^2+(Perpendicular)^2$ $(15)^2=(Base)^2+(12)^2$ $(Base)^2=(15)^2-(12)^2$ $(Base)^2=225-144$ $(Base)^2=81$ $Base=9 m$ Hence the distance of the foot of the ladder from the wall is 9 m. View More 2. ABC is a triangle, right-angled at C. If $\\small AB = 25\\hspace{1mm} cm$ and $\\small AC = 7\\hspace{1mm} cm$, find BC. As we know, In a Right-angled Triangle: By Pythagoras Theorem, As ABC is a right-angled triangle with Base = AC= 7 cm. Perpendicular = BC Hypotenuse = AB = 25 cm So, By Pythagoras theorem, Hence, Length od BC is 24 cm. 1. PQR is a triangle, right-angled at P. If $\\small PQ = 10\\hspace{1mm}cm$ and $\\small PR = 24\\hspace{1mm}cm$, find QR. As we know, In a Right-angled Triangle: By Pythagoras Theorem, As PQR is a right-angled triangle with Base = PQ = 10 cm. Perpendicular = PR = 24 cm. Hypotenuse = QR So, By Pythagoras theorem, Hence, Length od QR is 26 cm. Q. Find the unknown length x in the following figures", "the square roots of both sides of the equation. Use a calculator. The answer is \\begin{align*}c \\approx 6.7\\end{align*}. The hypotenuse is approximately 6.7. #### Example 3 Claire’s mom needs to fix a broken shingle on their house. If the shingle is 16 feet above the ground and Claire places the foot of the ladder 12 feet from the wall, how long will the ladder need to be? First, draw a picture to help determine what you are solving for. In this picture, you can see that the ladder is the hypotenuse. Next, insert the values of the legs into the formula for the Pythagorean Theorem. Next, calculate the value of the squares. Then take the square root of each side. The answer is \\begin{align*}c=20\\end{align*}. The length of the hypotenuse is 10, and the ladder needs to be 10 feet long. Remember Yuli and his garden beds? Yuli needs to know how long a diagonal timber he needs for a \\begin{align*}5 \\ ft \\times 5 \\ ft\\end{align*} square. First, use the Pythagorean Theorem and substitute in the values you know. Next, calculate the value of the squares. Then take the square root of each side. Round the number. The answer is that the hypotenuse \\begin{align*}c \\approx 7.07\\end{align*}, but since Yuli may have a hard time measuring \\begin{align*}\\frac{7}{100}\\end{align*} of a", "of the hypotenuse is $(11.3)^2 = 127.69\\nonumber$ These are almost the same, the discrepancy due to the fact that we rounded to find an approximation for the hypotenuse. Exercise A 15 foot ladder leans against the wall of a building. The base of the ladder lies 5 feet from the base of the wall. How high up the wall does the top of the ladder reach? Round your answer to the nearest tenth of a foot. 14.1 feet ## Exercises In Exercises 1-16, your solutions should include a well-labeled sketch. 1. The length of one leg of a right triangle is 15 meters, and the length of the hypotenuse is 25 meters. Find the exact length of the other leg. 2. The length of one leg of a right triangle is 7 meters, and the length of the hypotenuse is 25 meters. Find the exact length of the other leg. 3. The lengths of two legs of a right triangle are 12 meters and 16 meters. Find the exact length of the hypotenuse. 4. The lengths of two legs of a right triangle are 9 meters and 12 meters. Find the exact length of the hypotenuse. 5. The length of one leg of a right triangle is 13 meters, and the length of the hypotenuse is 22 meters.", "length of the ladder is 9 m, find; 1. The height of the wall. 2. Calculate the length between the foot of the ladder and the wall. Solution Given that one angle is 30 degrees, this must be a 60°- 60°- 90°right triangle. Ratio = x: x√3:2x. ⇒ 2x = 9 ⇒ x = 9/2 = 4.5 Substitute. 1. The height of the wall = 4.5m 2. x√3 = 4.5√3m ### Practice Questions 1. The longer side of a $30^{\\circ}$; $60^{\\circ}$; $90^{\\circ}$ right triangle is given by $12\\sqrt{3}$ cm. What is the measure of its height? 2. The longer side of a $30^{\\circ}$; $60^{\\circ}$; $90^{\\circ}$ right triangle is given by $12\\sqrt{3}$ cm. What is the measure of its hypotenuse? 3. The hypotenuse of a $45^{\\circ}$; $45^{\\circ}$; $90^{\\circ}$ right triangle is given by $7\\sqrt{2}$ ft. What is the length of its identical legs? 4. If the length of one side of an equilateral triangle is $15$ m, what is the length of that triangle’s altitude? 5. If the length of the diagonal of the square is $10$ units, what is the square area? 6. If an altitude of an equilateral triangle is $22$ cm, what is the length of a side of an equilateral triangle? 5/5 - (25 votes)", "1,010 Answered Questions for the topic Equations Equations Word Problems 16d #### Melissa is choosing a cellphone plan and wants the lowest monthly cost. Telus charges $12 per month plus$0.05 per minute for cellphone service. Koodo charges \\$22 per month for unlimited minutes. Equations Science Chemistry Moles 18d #### Stoichiometry work 10.0 grams of calcium is added to HCl. Ca + HCl → CaCl2 + H2How many grams of HCl would be needed to completely react with the Ca? 20d #### Write and solve the equation 1. Seven less than a number y is 82. Ten less than a number y is 19 01/26/21 The bottom of a ladder must be placed 18 feet from a wall. The ladder is 24 feet long. How far above the ground does the ladder touch the wall? 01/26/21 The hypotenuse of a right angled triangle is 35 cm. The difference between its other two sides is 7 cm. Find the length of the sides. 01/26/21 One leg of a right triangle is 7 cm less than the length of the other leg. The length of the hypotenuse is 17 cm. Find the lengths of the legs. 01/26/21 A soccer field is a rectangle, 33 meters wide and 56 meters long. The coach asks players to run from one corner to the", "implies that corresponding sides are proportional. Thus, since ABC is similar to CBD , by proportionality of corresponding sides we see thatAB is to CB (hypotenuses) as BC is to BD (vertical legs)c a = a dcd = a2 .Since ABC is similar to ACD , comparing horizontal legs and hypotenuses givesNote: The symbols and denote perpendicularity and similarity, respectively. For example, in the above proof we had CD AB and ABC CBD ACD .c b = c-d bb2 = c2 - cd = c2 - a2a2 + b 2 = c 2 .QED Angles · Section 1.1Example 1.3 For each right triangle below, determine the length of the unknown side:5B5Y E2a4z X111ACDeFZSolution: For triangle ABC , the Pythagorean Theorem says thata2 + 42 = 52a2 = 25 - 16 = 9a = 3 .For triangle DEF , the Pythagorean Theorem says thate2 + 12 = 22e2 = 4 - 1 = 3e =3 .For triangle X Y Z , the Pythagorean Theorem says that 12 + 12 = z2 Example 1.4 A 17 ft ladder leaning against a wall has its foot 8 ft from the base of the wall. At what height is the top of the ladder touching the wall? Solution: Let h be the height at which the ladder touches the wall. We can", "where they touched the walls were given, and Joe had to calculate how high above the ground the ladders crossed. He also calculated that the ladders could have been 315 centimetres and 261 centimetres long. What was the width of the alley? They’re on a real Pythagorean triples kick lately. At least I think the width of the alley is supposed to be a whole number of centimeters, otherwise the problem has no unique solution, does it? But if the alley is required to be a whole number of centimeters wide then what we have are two Pythagorean triangles, one with hypotenuse 315 and one with hypotenuse 261, and both having one side the same. Well, this is easy, if you remember the hypotenuse of a Pythagorean triangle has length a(u2+v2) with u>v. The prime factors of 315 are 5, 7, and 9, so a can be 1, 5, 7, 9, 35, 45, 63, or 315. Testing these we find a=63, u=2, v=1 or equivalently a=7, u=6, v=3 and no other possibilities. Then the other two sides of the triangle are a(u2v2) and 2auv which are 189 and 252. One of these must be the width of the alley. Now testing 2612–1892 and 2612–2522, the first gives 1802 while the other doesn’t give us an integer root. So", "132 = 52 + 122 it is verified ∴ The length is 12 cm and the breadth is 5 cm. Question 5. The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle? Solution: Let a is the shortest side. c is the hypotenuse b is the third side. ∴ The sides of the triangle are 10m, 24m, 26m. Verification 262 = 102 + 242 676 = 100 + 576 = 676 Question 6. 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall. Solution: Let the distance by which top of the slide moves upwards be assumed as ‘x’. From the diagram, DB = AB – AD = 3 – 1.6 ⇒ DB = 1.4 m also BE = BC + CE = 4 + x ∴ DBE is a right angled triangle DB2 + BE2 = DE2 ⇒ (1.4)2 + (4 + x)2= 52 ⇒ (4 + x)2 = 25 – 1.96", "$8^2 + 15^2 & \\ ? \\ 20^2\\\\64 + 225 & \\ ? \\ 400\\\\289 & < 400$ Because $289<400$ , this is an obtuse triangle. 2. Plug in each set of lengths into the Pythagorean Theorem. $15^2 + 22^2 & \\ ? \\ 25^2\\\\225 + 484 & \\ ? \\ 625\\\\709 & > 625$ Because $709>625$ , this is an acute triangle. ### Concept Problem Solution The ladder is making a triangle with the floor as one side, the wall as another, and the ladder itself serves as the hypotenuse. To see if the wall is leaning, you can determine the type of triangle that is made with these lengths (right, acute, or obtuse). If the triangle is a right triangle, then the wall is standing upright. Otherwise, it is leaning. Plugging the lengths of the sides into the Pythagorean Theorem: $3^2 + 10^2 & \\ ? \\ 12^2\\\\9 + 100 & \\ ? \\ 144\\\\109 & < 144$ Yes, you were right. Because 109 < 144, this is an obtuse triangle. The wall is leaning with an angle greater than ninety degrees. ### Practice Determine if each of the following lengths make a right triangle. 1. 9, 40, 41. 2. 12, 24, 26. 3. 5, 10, 14. 4. 3, $3\\sqrt{3}$ , 6. Determine if the following", "Q3. The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm. find the length of the other side. Let the hypotenuse be ” c ” and the other two sides be ” b ” and ” c”. Using the Pythagoras theorem, we can say that : $c^{2} = a^{2} + b^{2}$ $2.5^{2} = 1.5^{2} + b^{2}$ $b^{2}=6.25-2.25=4$ c=2cm Hence, the length of the other side is 2 cm. Q4. A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2m away from the wall. Find the height of the wall to which the ladder reaches. Let the hypotenuse be h. Using the Pythagoras theorem, we get : $3.7^{2} = 1.2^{2} + h^{2}$ $h^{2} = 13.69 – 1.44=12.25$ h=3.5m Hence, the height of the wall is 3.5 m. Q5. If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle. In the given triangle, the largest side is 6 cm. We know that in a right angled triangle, the sum of the squares of the smaller sides should be equal to the square of the largest side. Therefore, $3^{2} + 4^{2}=9+16=25$ But, $6^{2}=36$ $3^{2} + 4^{2}$ not equal to $6^{2}$ Hence,", "a right angle with the wall, as in the picture on the right. Then we see that the ladder, ground, and wall form a right triangle with a hypotenuse of length 17 ft (the length of the ladder) and legs with lengths 8 ft and $$h$$ ft. So by the Pythagorean Theorem, we have $\\nonumber h^2 ~+~ 8^2 ~=~ 17^2 \\quad\\Rightarrow\\quad h^2 ~=~ 289 ~-~ 64 ~=~ 225 \\quad\\Rightarrow\\quad \\fbox{h ~=~ 15 ~\\text{ft}} ~.$", "## Filters Clear All P Pankaj Sanodiya As we know the sum of the angles of any triangle is always 180. So, Now. Since PQR is a right-angled triangle with right angle at P. So Hence option (ii) is correct. P Pankaj Sanodiya As we know, In a Right-angled Triangle: By Pythagoras Theorem, (i) , , 6 cm. As we know the hypotenuse is the longest side of the triangle, So Hypotenuse = 6.5 cm Verifying the Pythagoras theorem, Hence it is a right-angled triangle. The Right-angle lies on the opposite of the longest side (hypotenuse) So the right angle is at the place where 2.5 cm side and 6 cm side meet. (ii) 2 cm, 2... P Pankaj Sanodiya Here. As we can see, The ladder with wall forms a right-angled triangle with the vertical height of the wall = perpendicular = 12 m length of ladder = Hypotenuse = 15 m Now, As we know In a Right-angled Triangle: By Pythagoras Theorem, $(Hypotenus)^2=(Base)^2+(Perpendicular)^2$ $(15)^2=(Base)^2+(12)^2$ $(Base)^2=(15)^2-(12)^2$ $(Base)^2=225-144$ $(Base)^2=81$ $Base=9 m$ Hence the distance of the foot of the ladder from the wall is 9 m. View More P Pankaj Sanodiya As we know, In a Right-angled Triangle: By Pythagoras Theorem, As ABC is a right-angled triangle with Base = AC= 7 cm. Perpendicular = BC Hypotenuse =", "right triangle. 1. $a=12, b=16, c =?$ 2. $a =?, b=20, c=30$ 3. $a=4, b =?, c=11$ 4. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle. 5. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle. 6. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate? 7. Emanuel has a cardboard box that measures $20 \\ cm \\times 10 \\ cm \\times 8 \\ cm \\ (\\text{length} \\times \\text{width} \\times \\text{height})$. What is the length of the diagonal from a bottom corner to the opposite top corner? 8. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. How high above the ground does the ladder touch the wall of the house? 9. Find the area of the triangle if area of a triangle is defined as $A=\\frac{1}{2} \\text{base} \\times \\text{height}$. 10. Instead of walking along the two sides of a rectangular field,", "2 m away from the wall, the top of the ladder rests against the foot of the wall. So, the hypotenuse is 2 m more than the second side of the right triangle. The other non-hypotenuse side is 8 m. So, $8^2 + x^2 = (x+2)^2$ => x = 15 m and length of the ladder = x+2 = 17 m Question 16: Two sides of a plot measure 32 m and 24 m and the angle between them is a perfect right angle. The other two sides measure 25 m each and the other three angles are not right angles. What is the area of the plot? a) 768 $m^2$ b) 534 $m^2$ c) 696.5 $m^2$ d) 684 $m^2$ Solution: Length of the diagonal of the right triangle is 40. The height of the isosceles triangle formed, with 40 as its base is 15. So, area = (1/2 * 32 * 24) + (1/2 * 40 * 15) = 384 + 300 = 684 $m^2$ Question 17: In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If angle D equals 40 degress , then angle ACB is ? a) 140 b) 70 c) 100 d) None of these Solution:", "the sum of the squares of the other two sides. Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle. Question 2: In right ∆ ABC, the lengths of the legs are given. Find the length of the hypotenuse. (i) a = 6 cm, b = 8 cm (ii) a = 8 cm, b = 15 cm (iii) a = 3 cm, b = 4 cm (iv) a = 2 cm, b = 1.5 cm According to the Pythagoras theorem, Question 3: The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm, find the length of the other side. Question 4: A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches. Question 5: If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle. Question 6: The sides of certain triangles are given below. Determine which of them", "= 15.4 cm2 ### Practice Questions 1. Which of the following shows the best classification of a triangle with dimensions of $12$ cm, $5$ cm, and $13$ cm? 2. Which of the following shows the best classification of a triangle with dimensions of $12$ ft, $18$ ft, and $17$ ft? 3. Which of the following shows the best classification of a triangle with dimensions of $15$ m, $20$ m, and $16$ m? 4. Which of the following shows the best classification of a triangle with dimensions of $10$ yard, $20$ yard, and $12$ yard? 5. The two shorter sides of a right triangle are $10$ cm and $24$ cm long. What is the length of the hypotenuse? 6. The leg and the hypotenuse of a right triangle are $14$ ft and $50$ ft long, respectively. Which of the following shows the length of the triangle’s remaining leg? 7. A $20$-m long rope is stretched from the top of a $12$-m tree to the ground. What is the distance between the tree and the end of the rope on the ground? 8. A $13$ m long ladder is leaning against the wall. If the ground distance between the foot of the ladder and the wall is $5$ m, what is the wall’s height?", "triangle is a right triangle. 1. \\begin{align*}a=12, b=9, c=15\\end{align*} 2. \\begin{align*}a=6, b=6, c=6 \\sqrt{2}\\end{align*} 3. \\begin{align*}a=8, b=8 \\sqrt{3}, c=16\\end{align*} Find the missing length of each right triangle. 1. \\begin{align*}a=12, b=16, c =?\\end{align*} 2. \\begin{align*}a =?, b=20, c=30\\end{align*} 3. \\begin{align*}a=4, b =?, c=11\\end{align*} 4. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle. 5. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle. 6. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate? 7. Emanuel has a cardboard box that measures \\begin{align*}20 \\ cm \\times 10 \\ cm \\times 8 \\ cm \\ (\\text{length} \\times \\text{width} \\times \\text{height})\\end{align*}. What is the length of the diagonal from a bottom corner to the opposite top corner? 8. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. How high above the ground does the ladder touch the wall of the house? 9. Find the area of the", "sum of the areas of the smaller ones Solution: Given the hint, if the triangle has sides a,b,c then the areas of the semicircles are pi a^2/8, pi b^2/8 and pi c^2/8. Multiplying all terms by 8/pi and applying the Pythagorean theorem gives the result. Some practice problems • If a=3 and b=4, what is the length c of the hypotenuse of the triangle? c 3 4 Some practice problems • Solution: by the Pythagorean theorem, c^2=9+16=25 so c=5 c 3 4 • • • • • • e Solution: If a=5, b=4, c=3, d=3, and e=√5, Then the first hypotenuse is sqrt(25+16) The second hypotenuse is sqrt(41+9) The third is sqrt(50+9) f The fourth is sqrt(59+5) =sqrt(64)=8 4 a d c 3 b e • Solution: • If a=5, b=4, c=3, d=3, and e=√5, • find f. d c f 3 b 4 a Explain the lengths of the sides of the Pythagorean spiral Solution: Just apply the Pythagorean theorem successively to each successive triangle. The square of side of next hypotenuse is 1+square of side of previous hypotenuse. • A ladder is 10 feet long. When the top of the ladder just touches the top a wall, the bottom of the ladder is 6 feet from the wall. • How high is the wall? • Solution:" ]

[ "2.425", "of 1 and 2. • 440", "2.12$.", "- 2$", "- 2$", "-2$", "you get 2.35", "2$", "2$", "2$", "2$", "2$", "2$", "2$", "2.53", ", 2$.", "2$$ -", "2$$ -", "2.60", "= 2.54$" ]

[ "(16 - 1)^2 + 1 = \\boxed{406}$.", "$m+n=2+89=\\boxed{091}$.", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "+ 31 = \\boxed{058}$.", "got a= 1, b=-28 $\\boxed {\\therefore p(x)=x^4+2x^3+x^2-28x-60 }$ continue....", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "of 1 and 2. • 440", "25 + 31$$ $$(x+2)^2 = 0$$ So $x = \\boxed{\\textbf{(A) }-2}$. ~Sequoia", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "$(1 \\Box (0 \\Box 0)) = 1 \\Box 0 = 2$.", "our answer is $m+n=61+4=\\boxed{065}$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "31+x = 42\\\\ x = \\boxed{{\\textbf{(D)}\\ 11}}$", "= 106$, and $AB = 106 + 87 = \\boxed{193}$.", "- 400 \\ m = \\boxed {76.1 \\ m} {/eq}", "C}}{3.21 \\mathrm{~\\mu F} }= \\boxed{28 \\,\\text{V}}$$", "\\sin 5 \\approx \\boxed{10.46\\, m} {/eq}" ]

[ "(16 - 1)^2 + 1 = \\boxed{406}$.", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "of 1 and 2. • 440", "is $-\\pi/2$.", "$2^{56}$.", "- 2x + 1$", "${2}$ (2.29)", "$\\therefore = 2$", "2= 0$", "to 1. My 2 cents.", "by $\\pi/2$.", "# 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 $2$", "2.75 [Divide.]", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "2 = 6.$", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "$m+n=2+89=\\boxed{091}$.", "$2(k-1006) = 2011 - k$. Solving gives $\\boxed{\\textbf{(C)} 1341}$.", "Graeber8 8 # How to solve 2.75 + .003+ .158 $2.75 + .003 + .158 \\\\ \\\\ 2.911 \\\\ \\\\ Answer: \\fbox {2.911}$ This problem can be done using long addition. $2.75 + .003+ .158 = \\boxed{\\bf{2.911}}$" ]

[ "many pencils? [#permalink] Show Tags 11 May 2006, 19:05 getzgetzu wrote: The average producing cost for each of the x pencils is (50+0.5x)/x . If the cost for each pencil is$0.6, how many pencils had been produced? (50+0.5x)/x = 0.6 (50+0.5x) = 0.6x 0.6x - 0.5x = 50 x = 500 Re: How many pencils? [#permalink] 11 May 2006, 19:05 Display posts from previous: Sort by", "Cost of the Notebooks and pencils = $115.24 e) Explanation: a) Cost of 150pencil from Pencil assortment A =$ 35.99 => Cost of each pencil from Pencil assortment A = $35.99 ÷ 150 =$ 0.2399 Cost of 75pencil from Pencil assortment B = $19.99 => Cost of each pencil from Pencil assortment B =$19.99 ÷ 75 = $0.2665 b) Cost of pencil is purchased by me =$0.24 Cost the pencil to be sold = $0.24 +$0.06 = $0.30 Because 0.06 is the profit range of 4 % out of 0.24. c) Cost of each pencils to be sold =$0.30 Cost of 100 pencils to be sold = $0.30 × 100 =$30. d) Total Cost of the Notebooks = $25.24 Cost of 300 pencils =$0.30 ×300 = $90 Amount of coupon used to pay =$5 Cost of the Notebooks and pencils = Total Cost of the Notebooks + Cost of 300 pencils = $25.24 +$90 = $115.24. e) Current account balance =$115.68 Conclusion:", "## anonymous 5 years ago Find the total dollar cost of n dozen pencils if each pencil costs 2c/3 cents. 1. anonymous 12 pencils cost:$12\\times2c/3=8c$So n lots of 12 pencils costs 8cn. 2. anonymous Or if you want it in dollars, divide by a 100, so it's:$8cn/100=2cn/25$", "Primary 6 Problem Sums/Word Problems - Try FREE Score : Disabled Question 1 of 4 A shop sells pencils. Out of every 20 pencils sold, 2 were defective. Defective pencils were sold at 80 cents each. A pencil in good condition cost 20 cents more than a defective pencil. (a) If the shopkeeper collected $2352 last month, how many pencils were sold last month? (b) What was the amount collected from the sale of defective pencils last month? Notes to student: 1. If the question above has parts(e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 The correct answer is : 2400,192 (a)____, (b)$_____", "# Recent questions tagged 21 Questions from: ### The cost of 11 pencils is Rs. 176. What will be the cost of 15 pencils ? To see more, click for the full list of questions or popular tags.", "### Primary 6 Problem Sums/Word Problems - Try FREE Score : Disabled #### Question 1 of 4 A shop sells pencils. Out of every 20 pencils sold, 2 were defective. Defective pencils were sold at 80 cents each. A pencil in good condition cost 20 cents more than a defective pencil. (a) If the shopkeeper collected $2352 last month, how many pencils were sold last month? (b) What was the amount collected from the sale of defective pencils last month? Notes to student: 1. If the question above has parts(e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 The correct answer is : 2400,192 (a)____, (b)$_____", "1 pt 3. If 15 pencils cost \\$0.75, what is the cost of one pencil? 1 pt 9. What is $\\frac{5}{16}$ in decimal form? 1pt 14. What is 1.875 as an improper fraction?", "Question # The school book store sold 8 more pencils than pens one day. The cost of a pencil is $.05, and the cost of a pen is$.20. If the day's sales of pens and pencils totalled $8.90, how many pencils were sold? A 35 B 39 C 42 D 45 Solution ## The correct option is B 42Amount of pens sold = p Amount of pencils sold = p + 8 0.05(p + 8) + 0.20(p) = 8.90 each pen is$0.20 while each pencil is \\$0.05 0.05p + 0.4 + 0.20p = 8.90 0.25p = 8.50 Number of pens: 34 Number of pencils sold: 34 + 8 = 42 The correct option is CMathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "the above-given question, given that, A package of mechanical pencils costs$4.99. A package of pens costs twice as much as a package of pencils. $4.99 x 2 =$9.98. Question 5. Carlos has 8 dollars and 48 cents. Alissa has 4 dollars and 14 cents. How much money does Carlos need to give Alissa so that each of them has the same amount of money?", "value for $p$ is 7; then the 13 pencils cost $7 \\times 13= 91$ cents and so the 3 erasers together cost 9 cents and each eraser costs $\\frac{9}{3} = 3$ cents. Thus one pencil plus one eraser cost $7 + 3 = 10$ cents, which is answer choice $\\mathrm{(A) \\ }$.", "number of cities with a store that sells really nice pencils online. The next lines contain two integers, and , to denote that the cost of a pencil in city is . The last line contains the integer , the destination city. #### Output Specification Output the minimal total cost of purchasing a pencil online and shipping it to city . #### Sample Input 3 3 1 2 4 2 3 2 1 3 3 3 1 14 2 8 3 3 1 #### Sample Output 6 CCC problem statements in large part from the PEG OJ • commented on Dec. 22, 2021, 10:58 a.m. edited So... you're telling me I have to pay up to \\$10000 for a pencil. And then pay to get it shipped. And then wait for it to be shipped across potentially thousands of cities. ... no thanks • commented on Jan. 1, 2022, 3:23 p.m. how good would a \"really nice pencil\" have to be to cost that much • commented on Oct. 13, 2020, 9:09 p.m. can a city have several prices for a pencil? • commented on Oct. 13, 2020, 9:34 p.m. It shouldn't matter. • commented on Nov. 21, 2018, 11:53 a.m. Is it guaranteed that it is possible to successfully ship a pencil to your city? • commented", "A package of 25 mechanical pencils costs $5.75. How much does each pencil Cost? 1 Answer Mar 14, 2018 Answer: See below... Explanation: 25 pencils cost $5.75 let $x$ be the price of $1$ pencil 25x=$5.75 x = ($5.75)/25 x = $0.23 Therefore $1$pencil $= 23\\$ cents", "If pencils sell at 6 for 59 cents how many pencils can be bought for 2.95 We already know from the given data that 6 pencils can be bought for 59 cents. We have to determine the number of pencils that can be bought for $2.95. So Price of 6 pencils = 59 cents. Price of 1 pencil = (59/6) cents = 9.83 cents Now total amount in hand for buying pencils =$2.95 = 295 cents Then the number of pencils that can be bought for 295 cents = (295/9.83) = 30.01 So the maximum number of pencils that can be bought for \\$2.95 is 30 pencils. RELATED:", "the selling price of 80 pencils? (A) 0.75 (B) 0.8 (C) 1 (D) 1.2 (E) 1.25 Source: Math bible Nova Math Expert Joined: 02 Sep 2009 Posts: 46035 Re: Patrick purchased 80 pencils and sold them at a loss [#permalink] ### Show Tags 22 Jun 2014, 11:38 2 1 Game wrote: Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 20 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils? (A) 0.75 (B) 0.8 (C) 1 (D) 1.2 (E) 1.25 Source: Math bible Nova Say the cost price of 80 pencils was $80 ($1 per pencil) and the selling price of 1 pencil was p. Selling at a loss: 80 - 80p = 20p --> p = 4/5. (cost price)/(selling price) = 1/(4/5) = 5/4 = 1.25. _________________ SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1837 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Patrick purchased 80 pencils and sold them at a loss [#permalink] ### Show Tags 23 Jun 2014, 00:10 Let cost price of 80 pencils = 80c Let selling price of 80 pencils = 80s ............ (1) Pencils sold at a loss equal to the selling price of 20 pencils 80c - 80s", "# How many ways can we distribute $7$ different color pencils to $10$ drawers so there are at least $2$ pencils in the $10$th drawer? [closed] Тhere are $10$ drawers and $7$ different color pencils. How many ways we can arrange them, so there are at least $2$ pencils in the $10$th drawer? So i tried doing it this way: 1.If all 7 pencils are in the 10 drawer this is 1. 2.If 6 pencils are in the 10 drawer, this are Variations with rep. $\\to 9^1=9$ 3.If 5 pencils are in the 10 drawer, this are Variations with rep. $\\to 9^2=81$ 4.If 4 pencils are in the 10 drawer, this are Variations with rep. $\\to 9^3=729$ 5.If 3 pencils are in the 10 drawer, this are Variations with rep. $\\to 9^4=6561$ 6.If 2 pencils are in the 10 drawer, this are Variations with rep. $\\to 9^5=59049$ So when i sum them i get $66430$ but it's far from the answer. Any hints? Thank you! ## closed as off-topic by Austin Mohr, Magdiragdag, Claude Leibovici, Juniven, user223391 Apr 22 '17 at 13:52 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", "2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006. forgot about percentage and change into numericals. in 2006 the price of pen and pencil both are 100. so the total average of revenue is 100 too. in 2007 the price of pen is 105 and pencil is 87. As it was mentioned in question average slipped by 1 point so 99. 6 12 105-------99--------------87 the ratio will be 12:6 of pen to pencil ratio. i.e., 2:1 but, it was asked for pencil to pen so it is 1:2... this is one of the best method you can try for all mixture problems. Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] 02 Oct 2018, 00:18 Display posts from previous: Sort by", "certain store sold pens for $0.35 each and pencils for [#permalink] ### Show Tags 27 Oct 2013, 09:41 1 This post received KUDOS Official Explanation Answer: A Algebraically, the question looks like this: 2.5 = 0.35x + 0.25y However, that's the only equation we get, so we can't solve for x and y this way. There would be an infinite number of possible values of x and y. Instead, we must use an intelligent version of guessing and checking. Since at least one pencil and one pen is purchased, the customer must spend at least$2.25 ($2.50 minus$0.25) on pens. There are six possible numbers of pens that fit that requirement: 1 pen: $0.35 2 pens:$0.70 3 pens: $1.05 4 pens:$1.40 5 pens: $1.75 6 pens:$2.10 Many of those aren't applicable here, though. For instance, if the customer bought 6 pens, he would need to spend an additional $0.40 on pencils. At$0.25 per pencil, that's impossible. Consider each of the other 5: 1 pen: $2.15 on pencils 2 pens:$1.80 on pencils 3 pens: $1.45 on pencils 4 pens:$1.10 on pencils 5 pens: $0.75 on pencils Only the last of these works.$0.75 is 3 pencils for $0.25 each. That's 5 pens and 3 pencils for a total of 8 pens and pencils. Choice (A) is correct. Senior Manager Joined: 02", "SEARCH HOME Math Central Quandaries & Queries Question from Maggie, a student: A mechanical pencil costs 50 cents and a pen costs 25 cents. If I spend 5.00 on 17 pens and pencils, how many of each does that buy? Hi Maggie, I can show you two approaches to this problem. The first method uses some ingenuity and some arithmetic. Suppose you spend the $5.00 and buy only pencils. Pencils cost$0.50 so you would be able to purchase 10 items, all pencils. But you want 17 items so return 1 pen and get 2 pens, and now you have 9 pencils and 2 pens for a total of 11 items. The total cost is still $5.00 and thus trading a pencil for two pens leaves the cost at$5.00 but increases the number of items by 1. So start with 10 pencils. Trading a pencil for two pens increases the total number of items by 1. Thus to get to 17 items I have to trade in 7 pencils. This will leave you with 3 pencils and 14 pens. Check: 3 pencils and 14 pens is 17 items costing 3 × $0.50 + 14 ×$0.25 = $1.50 +$3.50 = $5.00. The second approach is algebraic. Suppose the number of pencils you buy is x and the number of pens", "# 1 pencil, 1 pen cost $8; 2 pencils, 3 pens cost$21; find cost of pencil, pen #### 400396 ##### New member hello all. I have a problem with this question: 1 pencil and 1 pen cost $8 together. 2 pencils and 3 pens cost$21 together. What the cost of 1 pencil and 1 pen? I know that the answer is $3 for a pencil and$5 for a pen, but i don't know what equation to use. I appreciate any help on this. thanks. #### stapel ##### Super Moderator Staff member 1 pencil and 1 pen cost $8 together. 2 pencils and 3 pens cost$21 together. What the cost of 1 pencil and 1 pen? I know that the answer is $3 for a pencil and$5 for a pen, but i don't know what equation to use. Since you posted this to \"Pre-Algebra\" and haven't even picked any variables, I'll assume that you don't know about variables and other algebraic stuff. So let's use plain arithmetic. i. You are given the cost of one pencil and one pen. What then will be the cost of two pencils and two pens? (Hint: Multiply.) ii. You are given the cost of two pencils and three pens, and you just found the cost of two pencils and two pens. What then", "= $25.15. You are in charge of purchasing supplies for the student store. All of the supplies for the student store are purchased and then sold at a higher price so the store can earn a profit. Question 4. You want to buy 300 more pencils. You find two different quantities to buy. a. Which pencils will you buy to get$35.99$19.99the best value? Justify your answer. b. You pay about$0.24 for each pencil. How much do you think the store should charge for each pencil to earn a profit? Explain. c. How much profit does the store earn from selling 100 pencils using your recommended price? Explain. d. You want to buy some notebooks for a total of $25.24 and 300 pencils. You have a$5 coupon. How much do you spend on notebooks and pencils? e. Write out the check to buy the supplies. What is the account balance now? Fill in the payment and balance for line 6. a) I would like to purchase 300 pencils from Pencil assortment A because the cost of each pencil is lesser than Pencil assortment B. b)Cost the pencil to be sold = $0.30 because 0.06 is the profit range of 4 % out of 0.24. c) The profit store owns from selling 100 pencils using your recommended price is$30. d)" ]

[ "## anonymous 5 years ago Find the total dollar cost of n dozen pencils if each pencil costs 2c/3 cents. 1. anonymous 12 pencils cost:$12\\times2c/3=8c$So n lots of 12 pencils costs 8cn. 2. anonymous Or if you want it in dollars, divide by a 100, so it's:$8cn/100=2cn/25$", "value for $p$ is 7; then the 13 pencils cost $7 \\times 13= 91$ cents and so the 3 erasers together cost 9 cents and each eraser costs $\\frac{9}{3} = 3$ cents. Thus one pencil plus one eraser cost $7 + 3 = 10$ cents, which is answer choice $\\mathrm{(A) \\ }$.", "1 pt 3. If 15 pencils cost \\$0.75, what is the cost of one pencil? 1 pt 9. What is $\\frac{5}{16}$ in decimal form? 1pt 14. What is 1.875 as an improper fraction?", "many pencils? [#permalink] Show Tags 11 May 2006, 19:05 getzgetzu wrote: The average producing cost for each of the x pencils is (50+0.5x)/x . If the cost for each pencil is$0.6, how many pencils had been produced? (50+0.5x)/x = 0.6 (50+0.5x) = 0.6x 0.6x - 0.5x = 50 x = 500 Re: How many pencils? [#permalink] 11 May 2006, 19:05 Display posts from previous: Sort by", "A package of 25 mechanical pencils costs $5.75. How much does each pencil Cost? 1 Answer Mar 14, 2018 Answer: See below... Explanation: 25 pencils cost $5.75 let $x$ be the price of $1$ pencil 25x=$5.75 x = ($5.75)/25 x = $0.23 Therefore $1$pencil $= 23\\$ cents", "If pencils sell at 6 for 59 cents how many pencils can be bought for 2.95 We already know from the given data that 6 pencils can be bought for 59 cents. We have to determine the number of pencils that can be bought for $2.95. So Price of 6 pencils = 59 cents. Price of 1 pencil = (59/6) cents = 9.83 cents Now total amount in hand for buying pencils =$2.95 = 295 cents Then the number of pencils that can be bought for 295 cents = (295/9.83) = 30.01 So the maximum number of pencils that can be bought for \\$2.95 is 30 pencils. RELATED:", "the price of one pencil?$13.6 $13.5$12.6 $11.5 CorrectIncorrect Correct answer is:$13.5", "Question # The school book store sold 8 more pencils than pens one day. The cost of a pencil is $.05, and the cost of a pen is$.20. If the day's sales of pens and pencils totalled $8.90, how many pencils were sold? A 35 B 39 C 42 D 45 Solution ## The correct option is B 42Amount of pens sold = p Amount of pencils sold = p + 8 0.05(p + 8) + 0.20(p) = 8.90 each pen is$0.20 while each pencil is \\$0.05 0.05p + 0.4 + 0.20p = 8.90 0.25p = 8.50 Number of pens: 34 Number of pencils sold: 34 + 8 = 42 The correct option is CMathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "the above-given question, given that, A package of mechanical pencils costs$4.99. A package of pens costs twice as much as a package of pencils. $4.99 x 2 =$9.98. Question 5. Carlos has 8 dollars and 48 cents. Alissa has 4 dollars and 14 cents. How much money does Carlos need to give Alissa so that each of them has the same amount of money?", "$y=5$ The cost of one pencil and a pen is $3$ and $5$ respectively. Updated on 10-Oct-2022 13:19:40", "is$13.25. The cost of each pencil is 75 cents. a. Using an arithmetic approach, find the cost of a pen. (13.25-7(0.75))÷4 (13.25-5.25)÷4 8÷4 2 The cost of a pen is \\$2. b. Let the cost of a pen be p dollars. Write an expression for the total cost of four pens and seven mechanical pencils in terms of p. 4p+7(0.75) or 4p+5.25 c. Write an equation that could be used to find the cost of a pen.", "# Recent questions tagged 21 Questions from: ### The cost of 11 pencils is Rs. 176. What will be the cost of 15 pencils ? To see more, click for the full list of questions or popular tags.", "# Question #c5d4c Oct 27, 2017 You got a and b right, so let's move on #### Explanation: c) Cost of the pencils=price times number: $.25 \\left(x + 10\\right)$ d) Total cost is $.50 x + .25 \\left(x + 10\\right) = .50 x + .25 x + 2.50$ The inequality is: $.75 x + 2.50 < 5.00$ Subtract $2.50$ from both sides: $.75 x < 2.50$ Divide both sides by $.75$: $x < 3.33$ Since $x$ must be whole, this translates to $x \\le 3$ Answer: at most she bought 3 pens and 13 pencils. Check: $.50 \\cdot 3 + .25 \\cdot 13 = 4.75$ this will do $.50 \\cdot 4 + .25 \\cdot 14 = 5.50$ this is too expensive", "Pens cost 50 cents and pencils cost 25 cents. Amanda purchased a total of 9 pens and pencils for a cost of$4.25. What is the number of pens she bought and what is the number of pencils she bought? And the system of equations: The first equation corresponds to the associated cost of the pens and pencils. Notice that the original problem statement presents the information in terms of dollars and cents. However, by multiplying each dollar amount by 100, we can eliminate the decimals and make the numbers easier to work with. Notice that the coefficient of the x, which we assumed to be the unknown number of pens is 50- this is because we were given that the cost of every pen was 50 cents (and multiplying by 100 gives 50). Likewise, the coefficient of the y, which represents the unknown number of pencils is 25. The product of the unknown number of pens with the cost per pen gives the total cost of the pens, and the product of the unknown number of pencils with the cost per pencil gives the total cost of the pencils. This combined cost represents the total cost of pens and pencils, which is given in the problem statement. The problem statement also gives us the total number of pens", "rate, how many pencils can be bought for d dollars? A) $$cdp$$ B) $$100 cdp$$ C) $$\\frac{dp}{100c}$$ D) $$\\frac{100cd}{p}$$ E) $$\\frac{100 dp}{c}$$ A student asked me to solve this algebraically. So, here it goes: p pencils cost c cents at the same rate So, the COST PER PENCIL = c/p CENTS ASIDE: if you're not sure how to find price per pencil, you might first examine a few examples. For example, if 3 pencils cost 6 cents, then the cost per pencil = 2 cents (notice that 6/3 = 2) Likewise, if 4 pencils cost 20 cents, then the cost per pencil = 5 cents (notice that 20/4 = 5) Likewise, if 7 pencils cost 63 cents, then the cost per pencil = 9 cents (notice that 63/7 = 9) So, if p pencils cost c cents, then the cost per pencil = c/p cents How many pencils can be bought for d dollars? Our price per pencils is in CENTS So, we must convert d dollars to CENTS. Well, d dollars = 100d CENTS ASIDE: Once again, if you're not sure how to convert d dollars to CENTS, you might first examine a few examples. 4 dollars = 400 cents, since there are 100 cents in 1 dollar (notice that 4 x 100 = 400) Likewise, 9", "Cost of the Notebooks and pencils = $115.24 e) Explanation: a) Cost of 150pencil from Pencil assortment A =$ 35.99 => Cost of each pencil from Pencil assortment A = $35.99 ÷ 150 =$ 0.2399 Cost of 75pencil from Pencil assortment B = $19.99 => Cost of each pencil from Pencil assortment B =$19.99 ÷ 75 = $0.2665 b) Cost of pencil is purchased by me =$0.24 Cost the pencil to be sold = $0.24 +$0.06 = $0.30 Because 0.06 is the profit range of 4 % out of 0.24. c) Cost of each pencils to be sold =$0.30 Cost of 100 pencils to be sold = $0.30 × 100 =$30. d) Total Cost of the Notebooks = $25.24 Cost of 300 pencils =$0.30 ×300 = $90 Amount of coupon used to pay =$5 Cost of the Notebooks and pencils = Total Cost of the Notebooks + Cost of 300 pencils = $25.24 +$90 = $115.24. e) Current account balance =$115.68 Conclusion:", "* no of pens=x*10=$$\\frac{0.2m}{n}*10$$=$$\\frac{2m}{n}$$ _________________ Regards, PKN Rise above the storm, you will find the sunshine If m pencils cost the same as n pens, and each pencil costs 20 cents, &nbs [#permalink] 19 Aug 2018, 09:42 Display posts from previous: Sort by", "2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006. forgot about percentage and change into numericals. in 2006 the price of pen and pencil both are 100. so the total average of revenue is 100 too. in 2007 the price of pen is 105 and pencil is 87. As it was mentioned in question average slipped by 1 point so 99. 6 12 105-------99--------------87 the ratio will be 12:6 of pen to pencil ratio. i.e., 2:1 but, it was asked for pencil to pen so it is 1:2... this is one of the best method you can try for all mixture problems. Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] 02 Oct 2018, 00:18 Display posts from previous: Sort by", "first examine a few examples. For example, if 3 pencils cost 6 cents, then the cost per pencil = 2 cents (notice that 6/3 = 2) Likewise, if 4 pencils cost 20 cents, then the cost per pencil = 5 cents (notice that 20/4 = 5) Likewise, if 7 pencils cost 63 cents, then the cost per pencil = 9 cents (notice that 63/7 = 9) So, if p pencils cost c cents, then the cost per pencil = c/p cents How many pencils can be bought for d dollars? Our price per pencils is in CENTS So, we must convert d dollars to CENTS. Well, d dollars = 100d CENTS ASIDE: Once again, if you're not sure how to convert d dollars to CENTS, you might first examine a few examples. 4 dollars = 400 cents, since there are 100 cents in 1 dollar (notice that 4 x 100 = 400) Likewise, 9 dollars = 900 cents, since there are 100 cents in 1 dollar (notice that 9 x 100 = 900) Likewise, 15 dollars = 1500 cents, since there are 100 cents in 1 dollar (notice that 15 x 100 = 1500) Applying the same logic, we can see that d dollars = 100d CENTS Number of items we can buy = (amount of money we", "Break Even Quantity One of my Calc Practice questions, I'm not sure what to do....any help is appreciated... The total cost of manufacturing q hundred pencils is c dollars, where $\\displaystyle c = 40 + 3q + \\frac{q^2}{1000} + \\frac{1}{q}$ pencils are sold for $7 hundred, show that the break-even quantity is a solution of the equation$\\displaystyle f(q) = \\frac{q^3}{1000} - 4q^2 + 40q + 1 = 0\\$" ]

[ "## anonymous 5 years ago Find the total dollar cost of n dozen pencils if each pencil costs 2c/3 cents. 1. anonymous 12 pencils cost:$12\\times2c/3=8c$So n lots of 12 pencils costs 8cn. 2. anonymous Or if you want it in dollars, divide by a 100, so it's:$8cn/100=2cn/25$", "Primary 6 Problem Sums/Word Problems - Try FREE Score : Disabled Question 1 of 4 A shop sells pencils. Out of every 20 pencils sold, 2 were defective. Defective pencils were sold at 80 cents each. A pencil in good condition cost 20 cents more than a defective pencil. (a) If the shopkeeper collected $2352 last month, how many pencils were sold last month? (b) What was the amount collected from the sale of defective pencils last month? Notes to student: 1. If the question above has parts(e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 The correct answer is : 2400,192 (a)____, (b)$_____", "many pencils? [#permalink] Show Tags 11 May 2006, 19:05 getzgetzu wrote: The average producing cost for each of the x pencils is (50+0.5x)/x . If the cost for each pencil is$0.6, how many pencils had been produced? (50+0.5x)/x = 0.6 (50+0.5x) = 0.6x 0.6x - 0.5x = 50 x = 500 Re: How many pencils? [#permalink] 11 May 2006, 19:05 Display posts from previous: Sort by", "### Primary 6 Problem Sums/Word Problems - Try FREE Score : Disabled #### Question 1 of 4 A shop sells pencils. Out of every 20 pencils sold, 2 were defective. Defective pencils were sold at 80 cents each. A pencil in good condition cost 20 cents more than a defective pencil. (a) If the shopkeeper collected $2352 last month, how many pencils were sold last month? (b) What was the amount collected from the sale of defective pencils last month? Notes to student: 1. If the question above has parts(e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 The correct answer is : 2400,192 (a)____, (b)$_____", "value for $p$ is 7; then the 13 pencils cost $7 \\times 13= 91$ cents and so the 3 erasers together cost 9 cents and each eraser costs $\\frac{9}{3} = 3$ cents. Thus one pencil plus one eraser cost $7 + 3 = 10$ cents, which is answer choice $\\mathrm{(A) \\ }$.", "1 pt 3. If 15 pencils cost \\$0.75, what is the cost of one pencil? 1 pt 9. What is $\\frac{5}{16}$ in decimal form? 1pt 14. What is 1.875 as an improper fraction?", "Cost of the Notebooks and pencils = $115.24 e) Explanation: a) Cost of 150pencil from Pencil assortment A =$ 35.99 => Cost of each pencil from Pencil assortment A = $35.99 ÷ 150 =$ 0.2399 Cost of 75pencil from Pencil assortment B = $19.99 => Cost of each pencil from Pencil assortment B =$19.99 ÷ 75 = $0.2665 b) Cost of pencil is purchased by me =$0.24 Cost the pencil to be sold = $0.24 +$0.06 = $0.30 Because 0.06 is the profit range of 4 % out of 0.24. c) Cost of each pencils to be sold =$0.30 Cost of 100 pencils to be sold = $0.30 × 100 =$30. d) Total Cost of the Notebooks = $25.24 Cost of 300 pencils =$0.30 ×300 = $90 Amount of coupon used to pay =$5 Cost of the Notebooks and pencils = Total Cost of the Notebooks + Cost of 300 pencils = $25.24 +$90 = $115.24. e) Current account balance =$115.68 Conclusion:", "the above-given question, given that, A package of mechanical pencils costs$4.99. A package of pens costs twice as much as a package of pencils. $4.99 x 2 =$9.98. Question 5. Carlos has 8 dollars and 48 cents. Alissa has 4 dollars and 14 cents. How much money does Carlos need to give Alissa so that each of them has the same amount of money?", "Question # The school book store sold 8 more pencils than pens one day. The cost of a pencil is $.05, and the cost of a pen is$.20. If the day's sales of pens and pencils totalled $8.90, how many pencils were sold? A 35 B 39 C 42 D 45 Solution ## The correct option is B 42Amount of pens sold = p Amount of pencils sold = p + 8 0.05(p + 8) + 0.20(p) = 8.90 each pen is$0.20 while each pencil is \\$0.05 0.05p + 0.4 + 0.20p = 8.90 0.25p = 8.50 Number of pens: 34 Number of pencils sold: 34 + 8 = 42 The correct option is CMathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "A package of 25 mechanical pencils costs $5.75. How much does each pencil Cost? 1 Answer Mar 14, 2018 Answer: See below... Explanation: 25 pencils cost $5.75 let $x$ be the price of $1$ pencil 25x=$5.75 x = ($5.75)/25 x = $0.23 Therefore $1$pencil $= 23\\$ cents", "If pencils sell at 6 for 59 cents how many pencils can be bought for 2.95 We already know from the given data that 6 pencils can be bought for 59 cents. We have to determine the number of pencils that can be bought for $2.95. So Price of 6 pencils = 59 cents. Price of 1 pencil = (59/6) cents = 9.83 cents Now total amount in hand for buying pencils =$2.95 = 295 cents Then the number of pencils that can be bought for 295 cents = (295/9.83) = 30.01 So the maximum number of pencils that can be bought for \\$2.95 is 30 pencils. RELATED:", "# Recent questions tagged 21 Questions from: ### The cost of 11 pencils is Rs. 176. What will be the cost of 15 pencils ? To see more, click for the full list of questions or popular tags.", "number of cities with a store that sells really nice pencils online. The next lines contain two integers, and , to denote that the cost of a pencil in city is . The last line contains the integer , the destination city. #### Output Specification Output the minimal total cost of purchasing a pencil online and shipping it to city . #### Sample Input 3 3 1 2 4 2 3 2 1 3 3 3 1 14 2 8 3 3 1 #### Sample Output 6 CCC problem statements in large part from the PEG OJ • commented on Dec. 22, 2021, 10:58 a.m. edited So... you're telling me I have to pay up to \\$10000 for a pencil. And then pay to get it shipped. And then wait for it to be shipped across potentially thousands of cities. ... no thanks • commented on Jan. 1, 2022, 3:23 p.m. how good would a \"really nice pencil\" have to be to cost that much • commented on Oct. 13, 2020, 9:09 p.m. can a city have several prices for a pencil? • commented on Oct. 13, 2020, 9:34 p.m. It shouldn't matter. • commented on Nov. 21, 2018, 11:53 a.m. Is it guaranteed that it is possible to successfully ship a pencil to your city? • commented", "2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006. forgot about percentage and change into numericals. in 2006 the price of pen and pencil both are 100. so the total average of revenue is 100 too. in 2007 the price of pen is 105 and pencil is 87. As it was mentioned in question average slipped by 1 point so 99. 6 12 105-------99--------------87 the ratio will be 12:6 of pen to pencil ratio. i.e., 2:1 but, it was asked for pencil to pen so it is 1:2... this is one of the best method you can try for all mixture problems. Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] 02 Oct 2018, 00:18 Display posts from previous: Sort by", "rate, how many pencils can be bought for d dollars? A) $$cdp$$ B) $$100 cdp$$ C) $$\\frac{dp}{100c}$$ D) $$\\frac{100cd}{p}$$ E) $$\\frac{100 dp}{c}$$ A student asked me to solve this algebraically. So, here it goes: p pencils cost c cents at the same rate So, the COST PER PENCIL = c/p CENTS ASIDE: if you're not sure how to find price per pencil, you might first examine a few examples. For example, if 3 pencils cost 6 cents, then the cost per pencil = 2 cents (notice that 6/3 = 2) Likewise, if 4 pencils cost 20 cents, then the cost per pencil = 5 cents (notice that 20/4 = 5) Likewise, if 7 pencils cost 63 cents, then the cost per pencil = 9 cents (notice that 63/7 = 9) So, if p pencils cost c cents, then the cost per pencil = c/p cents How many pencils can be bought for d dollars? Our price per pencils is in CENTS So, we must convert d dollars to CENTS. Well, d dollars = 100d CENTS ASIDE: Once again, if you're not sure how to convert d dollars to CENTS, you might first examine a few examples. 4 dollars = 400 cents, since there are 100 cents in 1 dollar (notice that 4 x 100 = 400) Likewise, 9", "the selling price of 80 pencils? (A) 0.75 (B) 0.8 (C) 1 (D) 1.2 (E) 1.25 Source: Math bible Nova Math Expert Joined: 02 Sep 2009 Posts: 46035 Re: Patrick purchased 80 pencils and sold them at a loss [#permalink] ### Show Tags 22 Jun 2014, 11:38 2 1 Game wrote: Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 20 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils? (A) 0.75 (B) 0.8 (C) 1 (D) 1.2 (E) 1.25 Source: Math bible Nova Say the cost price of 80 pencils was $80 ($1 per pencil) and the selling price of 1 pencil was p. Selling at a loss: 80 - 80p = 20p --> p = 4/5. (cost price)/(selling price) = 1/(4/5) = 5/4 = 1.25. _________________ SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1837 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Patrick purchased 80 pencils and sold them at a loss [#permalink] ### Show Tags 23 Jun 2014, 00:10 Let cost price of 80 pencils = 80c Let selling price of 80 pencils = 80s ............ (1) Pencils sold at a loss equal to the selling price of 20 pencils 80c - 80s", "18:05 getzgetzu wrote: The average producing cost for each of the x pencils is (50+0.5x)/x . If the cost for each pencil is$0.6, how many pencils had been produced? (50+0.5x)/x = 0.6 (50+0.5x) = 0.6x 0.6x - 0.5x = 50 x = 500 Re: How many pencils? [#permalink] 11 May 2006, 18:05 Similar topics Replies Last post Similar Topics: Martha bought several pencils. If each pencil was either a 3 27 Jun 2008, 18:31 Machine A produces pencils at a constant rate of 9000 1 27 Dec 2007, 07:53 Machine A produces pencils at a constant rate of 9000 2 22 Nov 2007, 03:40 Machine A produces pencils at a constant rate of 9000 1 05 Nov 2006, 15:05 Machine A produces pencils at a constant rate of 9000 2 03 Jun 2006, 22:26 Display posts from previous: Sort by # The average producing cost for each of the x pencils is Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO 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®.", "#### Explanation: This problem is in reality simply asking you do work out $1000 \\times 0.23$. As we have been told, one pencil is $0.23$, or $23$ cents. The shopkeeper wants $1000$ pencils, so he must pay $1000$ times as many as he would for one pencil, $0.23$. We will state that $1$ is $1$ dollar. $0.23 \\times 1000 = 230$ 230 = $230 The shopkeeper must pay $230 in total for $1000$ pencils at \\$.23 each.", "Pens cost 50 cents and pencils cost 25 cents. Amanda purchased a total of 9 pens and pencils for a cost of$4.25. What is the number of pens she bought and what is the number of pencils she bought? And the system of equations: The first equation corresponds to the associated cost of the pens and pencils. Notice that the original problem statement presents the information in terms of dollars and cents. However, by multiplying each dollar amount by 100, we can eliminate the decimals and make the numbers easier to work with. Notice that the coefficient of the x, which we assumed to be the unknown number of pens is 50- this is because we were given that the cost of every pen was 50 cents (and multiplying by 100 gives 50). Likewise, the coefficient of the y, which represents the unknown number of pencils is 25. The product of the unknown number of pens with the cost per pen gives the total cost of the pens, and the product of the unknown number of pencils with the cost per pencil gives the total cost of the pencils. This combined cost represents the total cost of pens and pencils, which is given in the problem statement. The problem statement also gives us the total number of pens", "$y=5$ The cost of one pencil and a pen is $3$ and $5$ respectively. Updated on 10-Oct-2022 13:19:40" ]

[ "$[asy] import olympiad; size(250); defaultpen(linewidth(0.7)+fontsize(11pt)); real r = 31, t = -10; pair A = origin, B = dir(r-60), C = dir(r); pair X = -0.8 * dir(t), Y = 2 * dir(t); pair D = foot(B,X,Y), E = foot(C,X,Y); draw(A--B--C--A^^X--Y^^B--D^^C--E); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,dir(B--D)); label(\"E\",E,dir(C--E)); [/asy]$ ### 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;", "degrees. # Axiom, 58 bytes f(a,b)==(a=0 and b=0=>%i;sign(b)*acos(a*1./sqrt(a^2+b^2))) test (i use only acos() it returns radiants ) (40) -> [[a,b,f(a,b)*180/%pi] for a in [1,0,-1,-5,-2,0,4] for b in [1,3,1,0,-2,-1.5,-5] ] (40) [[1.0,1.0,45.0], [0.0,3.0,90.0], [- 1.0,1.0,135.0], [- 5.0,0.0,180.0], [- 2.0,- 2.0,- 135.0], [0.0,- 1.5,- 90.0], [4.0,- 5.0,- 51.3401917459 09909396]] Type: List List Complex Float # Python 2, 59 bytes lambda x,y:acos(x/hypot(x,y))/(-1,1)[y>0] from math import* Try it online! Outputs in radians in range [-pi,pi)", "# 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 $\\angle guy$?.", "4T4 $A_4$ $12$ $1$ 6T4, 12T4 6T4 $A_4$ $12$ $1$ $C_3$ 4T4, 12T4 12T4 $A_4$ $12$ $1$ $C_3$, $A_4$, $A_4$ 4T4, 6T4 Results are complete for degrees $\\leq 23$.", "+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", "+0 # Can I get some help with these? 0 68 1 Hi... I've been strugling with these three problems for a few weeks: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6},$$ where the angle$$\\frac{\\pi}{6}$$ is in radians. (2)Find the length AC, to two decimal places. Image: https://artofproblemsolving.com/texer/sywndnao (3)Find the length BC, to two decimal places. Image: https://artofproblemsolving.com/texer/enqsdotw Nov 15, 2020 #1 0 Quick note: to get the images for (2) and (3), post these codes to get the figures. Make sure you say Render as Image in the TeXeR (2) [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3) [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] Nov 15, 2020", "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 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "$18$ degrees value, obtained from theoretical method is exact value. It is actually derived purely in mathematical and trigonometrical methods, and there is no chance for error. Save (or) Share", "+0 # Help plz 0 58 1 The following polar grid is centered at the origin, with concentric circles of positive integer radii starting at $1,$ and consecutive rays through the origin with angles of $\\pi/12$ between them: [asy] size(200); pair A = (0,0); for (int i = 1; i < 6; ++i) { draw(Circle((0,0),i), linewidth(0.4)); } for(int i=0;i<360;i+=15) { draw(rotate(i)*((-5.0,0)--(5.0,0)), linewidth(0.4)); } pair A, B,C, D, EE, F; A = (1,-sqrt(3)); B = (1, sqrt(3)); C = (sqrt(3), 1); D = (4,0); EE = (0,0); F = (-sqrt(2), -sqrt(2)); dot(A, red+linewidth(3.5)); dot(B, red+linewidth(3.5)); dot(C, red+linewidth(3.5)); dot(D, red+linewidth(3.5)); dot(EE, red+linewidth(3.5)); dot(F, red+linewidth(3.5)); label(\"$A$\", A, S); label(\"$B$\", B, N); label(\"$C$\", C, E); label(\"$D$\", D, S); label(\"$E$\", EE, 3*S); label(\"$F$\", F, S); [/asy] Some of the points above are on the graph $r=4 \\cos(\\theta).$ Answer with the list in any order, separated by commas. Jul 30, 2021", "= 2, Coefficient = –2 15 . Degree =4, Coefficient = –2 17 . As $x→∞ x→∞$, $f(x)→∞,asx→−∞,f(x)→∞ f(x)→∞,asx→−∞,f(x)→∞$ 19 . As $x→−∞ x→−∞$, $f(x)→−∞,asx→∞,f(x)→−∞ f(x)→−∞,asx→∞,f(x)→−∞$ 21 . As $x→−∞ x→−∞$, $f(x)→−∞,asx→∞,f(x)→−∞ f(x)→−∞,asx→∞,f(x)→−∞$ 23 . As $x→∞ x→∞$, $f(x)→∞,asx→−∞,f(x)→−∞ f(x)→∞,asx→−∞,f(x)→−∞$ 25 . y-intercept is $(0,12), (0,12),$ t-intercepts are 27 . y-intercept is $(0,−16). (0,−16).$ x-intercepts are $(2,0) (2,0)$ and $(−2,0). (−2,0).$ 29 . y-intercept is $(0,0). (0,0).$ x-intercepts are $(0,0),(4,0), (0,0),(4,0),$ and 31 . 3 33 . 5 35 . 3 37 . 5 39 . Yes. Number of turning points is 2. Least possible degree is 3. 41 . Yes. Number of turning points is 1. Least possible degree is 2. 43 . Yes. Number of turning points is 0. Least possible degree is 1. 45 . Yes. Number of turning points is 0. Least possible degree is 1. 47 . $x x$ $f( x ) f( x )$ 10 9,500 100 99,950,000 –10 9,500 –100 99,950,000 As $x→−∞ x→−∞$, $f(x)→∞,asx→∞,f(x)→∞ f(x)→∞,asx→∞,f(x)→∞$ 49 . $x x$ $f( x ) f( x )$ 10 –504 100 –941,094 –10 1,716 –100 1,061,106 As $x→−∞ x→−∞$, $f(x)→∞,asx→∞,f(x)→−∞ f(x)→∞,asx→∞,f(x)→−∞$ 51 . The $y- y-$ intercept is The $x- x-$ intercepts are As $x→−∞ x→−∞$, $f(x)→∞,asx→∞,f(x)→∞ f(x)→∞,asx→∞,f(x)→∞$ 53 . The $y- y-$ intercept is $( 0,0 ) ( 0,0 )$ . The", "Y[0] == 0, Vx[0] == V0 Cos[40 Degree], Vy[0] == V0 Sin[40 Degree], Ax[0] == 0, Ay[0] == g}, {X[n], Vx[n], Ax[n], Y[n], Vy[n], Ay[n]}, n][[1]] // FullSimplify {Ax[n] -> 100/7 Sqrt[ 2] ((1/20 (10 - 7 Sqrt[2]))^n - (1/20 (10 + 7 Sqrt[2]))^n) Cos[ 40 [Degree]], Vx[n] -> 1/7 2^(3/2 - 2 n) 5^( 1 - n) (-(10 - 7 Sqrt[2])^(1 + n) + (10 + 7 Sqrt[2])^(1 + n)) Cos[ 40 [Degree]], X[n] -> 1/7 2^(1 - 2 n) 5^-n (7 2^(3 + 2 n) 5^(1 + n) + (10 - 7 Sqrt[2])^ n (-140 + 99 Sqrt[2]) - (10 + 7 Sqrt[2])^n (140 + 99 Sqrt[2])) Cos[ 40 [Degree]], Ay[n] -> 1/7 2^(-(3/2) - 2 n) 5^(-1 - n) ((10 + 7 Sqrt[2])^ n (49 (-30 + 7 Sqrt[2]) - 2000 Sin[40 [Degree]]) + (10 - 7 Sqrt[2])^ n (49 (30 + 7 Sqrt[2]) + 2000 Sin[40 [Degree]])), Vy[n] -> 1/7 2^(-(3/2) - 2 n) 5^(-2 - n) (-343 2^(5/2 + 2 n) 5^(1 + n) + (10 - 7 Sqrt[2])^ n (49 (-101 + 70 Sqrt[2]) + 1000 (-10 + 7 Sqrt[2]) Sin[40 [Degree]]) + (10 + 7 Sqrt[2])^ n (49 (101 + 70 Sqrt[2]) + 1000 (10 + 7 Sqrt[2]) Sin[40 [Degree]])), Y[n] -> 1/7 4^(-1 - n) 5^(-3 - n) (-343 2^(1", "of the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into. Now 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 $x=3$, $[ABC]=60$. So $[ABCD]=[ACD]+[ABC]=300+60=\\boxed {\\textbf{(D) }360}$ (I'm very sorry if you're a visual learner but now you have a diagram by ciceronii) ~ Solution by Ultraman ~ Diagram by ciceronii ## 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 two to get", "reasons on each line of a sequence of equations To produce this: \\begin{align} v + w & = 0 &&\\text{Given} \\tag 1\\\\ -w & = -w + 0 && \\text{additive identity} \\tag 2\\\\ -w + 0 & = -w + (v + w) && \\text{equations $(1)$ and $(2)$} \\end{align} write this: \\begin{align} v + w & = 0 &&\\text{Given} \\tag 1\\\\ -w & = -w + 0 && \\text{additive identity} \\tag 2\\\\ -w + 0 & = -w + (v + w) && \\text{equations $(1)$ and $(2)$} \\end{align} - The degree symbol for angles is not ^\\circ. Although many people use this notation, the result looks quite different from the canonical degree symbol shipped with the font: 90° renders as $90°$ while 90^\\circ renders as $90^\\circ$. If your keyboard doesn't have a ° key, feel free to copy from this post here, or follow these suggestions. Note that comments below indicate that on some configurations at least, ° renders inferior to ^\\circ. And I recently had a post of mine edited just for the sake of turning ° into ^\\circ, indicating that someone felt rather strongly about this. So the suggestion above does seem somewhat controversial at the moment. I maintain that from a semantic point of view, ° is superior to ^\\circ, and if the rendering suffers", "+0 # We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB 0 42 1 We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB [asy] size(5cm); pair A,B,C; A=(0,0); B=(5,0); C=(1,1); draw(A--B--C--A); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); [/asy] Nov 6, 2020", "# 2018 AMC 12B Problems/Problem 23 ## Problem Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\\circ$ latitude and $110^\\circ \\text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\\circ \\text{ N}$ latitude and $115^\\circ \\text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\\angle ACB?$ $\\textbf{(A) }105 \\qquad \\textbf{(B) }112\\frac{1}{2} \\qquad \\textbf{(C) }120 \\qquad \\textbf{(D) }135 \\qquad \\textbf{(E) }150 \\qquad$ ## Diagram $[asy] /* Made by MRENTHUSIASM */ size(250); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); draw(unitsphere,white,light=White); dot(A^^B^^C,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^A--C--B); label(\"A\",A,3*dir(A)); label(\"B\",B,3*dir(B)); label(\"C\",C,3*(0,0,-1)); [/asy]$ ~MRENTHUSIASM ## Solution 1 (Tetrahedron) This solution refers to the Diagram section. Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\\angle BDA = \\angle BDC = 90^\\circ$ and $\\angle BCD = 45^\\circ.$ Recall that $115^\\circ \\text{ W}$ longitude is the same as $245^\\circ \\text{ E}$ longitude, so $\\angle ACD=135^\\circ.$ We obtain the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(250); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); draw(surface(A--B--C--cycle),yellow); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^A--B--D--C--B^^C--A--D); label(\"A\",A,3*dir(A)); label(\"B\",B,3*dir(B)); label(\"C\",C,3*(0,0,-1)); label(\"D\",D,3*(-1/2,-1/2,0)); [/asy]$ Without the loss of generality, let $AC=BC=1.$ For tetrahedron", "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 $\\tan{(EAG)} =", "$\\sin(ax)$ in blue (for $x=5.678$); and 5000 terms for the Newton series in red. They agree for $a$ up to about $3.1$ or $3.2$...", "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 $\\frac{CF}{DC}=\\frac{OC}{DO}=\\frac{1}{3}\\Rightarrow \\text{(C)}$", "Free Version Easy # Unit Circle Equivalencies SATSTM-F1NSX7 If $\\cos135^{\\circ}=\\cos(x)$, then $x$ could equal which one of the following? A $\\text{45 degrees}$ B $\\text{120 degrees}$ C $\\text{315 degrees}$ D $\\text{225 degrees}$ E $\\text{405 degrees}$" ]

[ "# 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 $$a^\\circ$$? $a^\\circ = 180^\\circ - (70^\\circ + 70^\\circ )$ $= 180^\\circ - 140^\\circ = 40^\\circ$", "the triangle to add up to $180^\\circ-90^\\circ=90^\\circ$, and indeed the triangle in the picture has three $30^\\circ$ angles. ## The Gauss-Bonnet Theorem In this last section, we’re going to talk about something remarkable that happens when you add up the curvatures for every vertex in a polyhedron. The number you get when you do this is called the total curvature. Now that we have the angle deficit description of the curvature at a vertex, the total curvature is fairly easy to compute. We can do it pretty quickly for all of the Platonic solids, pictured in Figure 7. The results are: solid vertices faces at vertex curvature at vertex total curvature tetrahedron 4 3 triangles $180^\\circ$ $720^\\circ$ cube 8 3 squares $90^\\circ$ $720^\\circ$ octahedron 6 4 triangle $120^\\circ$ $720^\\circ$ dodecahedron 20 3 pentagons $36^\\circ$ $720^\\circ$ icosahedron 12 5 triangles $60^\\circ$ $720^\\circ$ Something should jump out right away: we always get the same number for the total curvature! In fact, there’s a deep reason for this, called the Gauss-Bonnet Theorem, which we’re about to prove. To prove this result, we’ll first need to triangulate our polyhedron, that is, cover it with triangles so that any two triangles meet at an edge and any two edges meet at a vertex. For this proof, we’re also going to require that none", "= 240^{\\circ}$$ $$x = 80^{\\circ}$$ As corner A is just $$x$$, corner A is simply $$80^{\\circ}$$. Corner B however is $$2x$$, so corner B is $$2 * 80^{\\circ}$$ which is $$160^{\\circ}$$. Long explanation: a) As you might know, a triangle has a total interior angle of $$180^{\\circ}$$ and the total interior angle of a square (or any four-sided 2d shape) is $$360^{\\circ}$$. If you cut a square in half (corner to corner), you'll see that a square is simply two triangles. And if a single triangle is $$180^{\\circ}$$, then it would be right to assume that two triangles are just $$180^{\\circ} + 180^{\\circ}$$ which is $$360^{\\circ}$$. Using this idea, very people clever generalized it into the formula $$(n - 2) * 180^{\\circ}$$. And whilst it is fine to use the formula, you can also do it intuitively by drawing lines corner to corner on shapes (as long as they don't overlap) and count the number of triangles each shape has. If you do this to a pentagon, you'll see it has three triangles inside it, using the logic from two paragraphs ago, you simply add the angles of the three triangles together to get $$540^{\\circ}$$. b) We know that 3 of the angles are $$100^{\\circ}$$ each which means that we have $$300^{\\circ}$$ of the total $$540^{\\circ}$$ accounted for.", "$x = 15^\\circ$ , then the base angles are $4(15^\\circ) +12^\\circ$ , or $72^\\circ$ . The vertex angle is $180^\\circ-72^\\circ-72^\\circ=36^\\circ$ . 2. The two sides are equal, so set them equal to each other and solve for $x$ . $2x-9 & = x+5\\\\x & = 14$ 3. This statement is false. Because the base angles of an isosceles triangle are congruent, if one base angle is a right angle then both base angles must be right angles. It is impossible to have a triangle with two right ( $90^\\circ$ ) angles. The Triangle Sum Theorem states that the sum of the three angles in a triangle is $180^\\circ$ . If two of the angles in a triangle are right angles, then the third angle must be $0^\\circ$ and the shape is no longer a triangle. ### Explore More Find the measures of $x$ and/or $y$ . Determine if the following statements are true or false. 1. Base angles of an isosceles triangle are congruent. 2. Base angles of an isosceles triangle are complementary. 3. Base angles of an isosceles triangle can be equal to the vertex angle. 4. Base angles of an isosceles triangle are acute. Complete the proofs below. 1. Given : Isosceles $\\triangle CIS$ , with base angles $\\angle C$ and $\\angle S$ $\\overline{IO}$ is the", "n sides, the number of triangles formed will be (n – 2). Also we know that the sum of angles of a triangle = $$180^{ \\circ }$$ ∴ The sum of angles of a polygon of n sides =$$(n – 2) \\times180^{ \\circ }$$", "### Home > INT2 > Chapter 3 > Lesson 3.1.5 > Problem3-63 3-63. Use the geometric relationships in the diagrams below to solve for $x$. 1. This is an isosceles triangle. Are there any special qualities about isosceles triangles that you know? Base angles in an isosceles triangles are congruent. $180^\\circ − 82^\\circ = 98^\\circ$ then take $98^\\circ ÷ 2 = 49^\\circ$. Both base angles will be $49^\\circ$. $x = 49^\\circ$ 1. Both base angles are $71^\\circ$. See part (a) for help.", "(Sum of the angles of $\\triangle ADC$ ) $135^{\\circ}$ + y = $180^{\\circ}$ y = $180^{\\circ}$$135^{\\circ}$. y =$45^{\\circ}$. We can also say that in $\\triangle ABC$, $\\angle ABC+\\angle ACB+\\angle BAC$ is equal to $180^{\\circ}$. (Sum of the angles of A ABC) $40^{\\circ}$ + y + (x + $45^{\\circ}$) = $180^{\\circ}$ $40^{\\circ}$ + $45^{\\circ}$ + x + $45^{\\circ}$ = $180^{\\circ}$ (y = $45^{\\circ}$) x= $180^{\\circ}$$130^{\\circ}$ x= $50^{\\circ}$ Therefore, we can say that the required angles are $45^{\\circ}$ and $50^{\\circ}$. (iii) We know that the sum of all the angles of a triangle is equal to $180^{\\circ}$. Therefore, for $\\triangle ABD$: $\\angle ABD+\\angle ADB+\\angle BAD=180^{\\circ}$ (Sum of the angles of $\\triangle ABD$) $50^{\\circ}$ + x + $50^{\\circ}$ = $180^{\\circ}$ $100^{\\circ}$ +x = $180^{\\circ}$ x = $180^{\\circ}$$100^{\\circ}$ x =$80^{\\circ}$ For $\\triangle ABC$: $\\angle ABC+\\angle ACB+\\angle BAC=180^{\\circ}$ (Sum of the angles of $\\triangle ABC$) $50^{\\circ}$ + z + ($50^{\\circ}$ + $30^{\\circ}$ ) = $180^{\\circ}$ $50^{\\circ}$ + z + $50^{\\circ}$ + $30^{\\circ}$ = $180^{\\circ}$ z = $180^{\\circ}$$130^{\\circ}$ z = $50^{\\circ}$ Using the same argument for $\\triangle ADC$: $\\angle ADC+\\angle ACD+\\angle DAC=180^{\\circ}$ (Sum of the angles of $\\triangle ADC$) y +z +$30^{\\circ}$ =$180^{\\circ}$ y + $50^{\\circ}$ + $30^{\\circ}$ =$180^{\\circ}$ ( z =$50^{\\circ}$ ) y = $180^{\\circ}$$80^{\\circ}$ y = $100^{\\circ}$ Therefore, we can conclude that the required angles are $80^{\\circ}$, $50^{\\circ}$ and $100^{\\circ}$. (iv) In", "= $$60^{\\circ}$$ and ∠B = $$60^{\\circ}$$ . Find the measure of ∠C. Solution: ∠A + ∠B + ∠C = $$180^{\\circ}$$ [The sum of all angles of a triangle is $$180^{\\circ}$$ ] $$60^{\\circ}$$ + $$60^{\\circ}$$ + ∠C = $$180^{\\circ}$$ [Substitute values] ∠C = $$180^{\\circ}$$ – $$120^{\\circ}$$ [Subtract 120 from both sides] ∠C = $$60^{\\circ}$$ Hence, the measure of ∠C is $$60^{\\circ}$$ . Example 2: For the $$\\triangle$$ABC given below, arrange the interior angles in increasing order of their measure. Solution: In the given $$\\triangle$$ABC, the measure of the sides are: a = 10 cm, b = 8 cm, and c = 6 cm. The decreasing order of side lengths is 10 > 8 > 6, that is, a > b > c. In a triangle, the measure of the angle opposite to the longer side is larger, or the angle opposite to the smaller side is smaller. Hence, the measure of angles in increasing order will be ∠A > ∠B > ∠C. Example 3: Find the area of the triangle whose base is 8 inches long and height is 12 inches. Solution: Area of triangle = $$\\frac{1}{2}$$ × b× h [Area of a triangle formula] = $$\\frac{1}{2}$$ (8) × (12) [Substitute values of b and h] = $$\\frac{1}{2}$$ (96) [Multiply] = 48 [Simplify] Hence, the area of the triangle", "fact that the sum of the angles in each triangle is $180^\\circ$.", "Sum of interior angles of a triangle $$= 180^\\circ$$ (From step$$1$$) \\begin{align}x+x+x&=180^\\circ\\\\ 3x&=180^\\circ\\\\x\\frac{180}{3}&=60^\\circ\\end{align} (vi) Sum of interior angles of a triangle $$= 180^\\circ$$ (From step $$1$$) \\begin{align} x+2x+90^\\circ&=180^\\circ\\\\3x+90^\\circ&=180^\\circ\\\\ 3x&=180^\\circ-90^\\circ\\\\x=\\frac{90}{3}&=30{}^\\circ \\end{align} 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", "## College Algebra (6th Edition) $x = 30$ degrees; $y = 75$ degrees The sum of all interior angles in a triangle equals 180 degrees. Also, adjacent angles on a straight line also add up to 180 degrees. This means that we can write the following system of equations: $$x + 2y = 180$$ $$3x + 15 + y = 180$$ which is to say: $$x + 2y = 180$$ $$3x + y = 165$$ We can use the substitution method to find the equivalent value of one of the variables: $$x + 2y = 180$$ $$x = 180 - 2y$$ and plug this value into the other equation: $$3x + y = 165$$ $$3(180 - 2y) + y = 165$$ $$540 - 6y + y = 165$$ $$375 = 5y$$ $$75 = y$$ We now substitute this value into the previous equation: $$x = 180 - 2y$$ $$x = 180 - 2(75) = 180 - 150 = 30$$", "Answer Comment Share Q) 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$", "a triangle is $180^{\\circ}$ to see $$20^{\\circ} + 30.87^{\\circ} + A = 180^{\\circ}.$$ Therefore $$A = 180^{\\circ} - 20^{\\circ} - 30.87^{\\circ} = 129.13^{\\circ}.$$ Solve for $a$ in the $C=30.87$ triangle By the law of sines, $$\\dfrac{\\sin\\left(129.13^{\\circ} \\right)}{a}= \\dfrac{\\sin \\left( 20^{\\circ} \\right)}{4},$$ so $$a = \\dfrac{4\\sin \\left( 129.13^{\\circ} \\right)}{\\sin \\left(20^{\\circ} \\right)}.$$ Solve for $A$ in the $C=149.13$ triangle Use the fact that the sum of angles in a triangle is $180^{\\circ}$ to see $$20^{\\circ} + 149.13^{\\circ} + A = 180^{\\circ}.$$ Therefore $$A = 180^{\\circ} - 20^{\\circ} - 149.13^{\\circ}=10.87^{\\circ}.$$ Solve for $a$ in the $C=149.13$ triangle By the law of sines, $$\\dfrac{\\sin \\left( 10.87^{\\circ} \\right)}{a} = \\dfrac{\\sin(20^{\\circ})}{4},$$ so $$a = \\dfrac{4 \\sin \\left( 10.87^{\\circ} \\right)}{\\sin(20^{\\circ})}.$$ Section 4.2 #32: Solve the triangle given by the information $a=3$, $b=7$, and $A=70^{\\circ}$. Solution: First draw the triangle: Find $B$ By the law of sines we see $$\\dfrac{\\sin(B)}{7} = \\dfrac{\\sin(70^{\\circ})}{3}.$$ Therefore $$\\sin(B) = \\dfrac{7\\sin(70^{\\circ})}{3} = 2.19.$$ But we may not take $\\sin^{-1}$ of $2.19$ (it is too big) and so this triangle does not exist. Section 4.3 #15: Solve the triangle given by the information $a=9$, $b=6$, and $c=4$. Solution: First draw the triangle: We have no choice other than to use the law of cosines to solve for the angles. Solve for $A$ Using the law of cosines, we see $$9^2 = 6^2", "from the point A and drawing a line at $60{}^\\circ$ from the point B’. The point where both the lines cut is your point C’ and the triangle you get is triangle AB’C’.", "line because 180 degree rotations take lines to parallel lines. So the three angles with vertex $B$ make a line and they add up to $180^\\circ$ ($x + y + z = 180$). But $x, y, z$ are the measures of the three angles in $\\triangle ABC$ so the sum of the angles in a triangle is always $180^\\circ$!", "the sum of the angles in this triangle is $(180^{\\circ} - 3x) + 48° + 2x = 180^{\\circ}$.", "## Trigonometry (10th Edition) 1. Internal angles of a triangle have a sum of 180° $(2x-120) ^{\\circ}+(x-30) ^{\\circ}+( \\frac{1}{2}x+15) ^{\\circ}= 180 ^{\\circ}$ 2. Solve the equation for x $2x+x+\\frac{1}{2}x=180+120+30-15$ $3\\frac{1}{2}x=315$ $x=90^{\\circ}$ 3. Then find other two angles by replacing x=90 $2(90)-120=60 ^{\\circ}$ $(90)-30=60 ^{\\circ}$ $180-60-60=60 ^{\\circ}$", "### Home > INT2 > Chapter 7 > Lesson 7.2.3 > Problem7-110 7-110. Multiple Choice: Based on the relationships provided in the diagram, which of the equations below is correct? Justify your solution. 1. $4x+2^\\circ+9x-3^\\circ=90^\\circ$ 1. $4x+2^\\circ=9x-3^\\circ$ 1. $4x+2^\\circ+9x-3^\\circ=180^\\circ$ 1. $(4x+2^\\circ)(9x-3^\\circ)=90^\\circ$ This is an isosceles triangle, because it has two equal side lengths. So remember in isosceles triangles base angles are equal.", "triangle is equal to $60^{\\circ}$ , then the sum of all three angles will be $180^{\\circ}$ , which is possible in case of a triangle. Proof: Let the three angles of the triangle be $\\angle A$, $\\angle B$ and $\\angle C$. As per the given information, $\\angle A$ = $60^{\\circ}$ … (i) $\\angle B$=$60^{\\circ}$ …(ii) $\\angle C$= $60^{\\circ}$ … (iii) On adding (i), (ii) and (iii), we get: $\\angle A$ + $\\angle B$ + $\\angle C$ = $60^{\\circ}$+ $60^{\\circ}$+ $60^{\\circ}$ $\\angle A$ + $\\angle B$ + $\\angle C$ =$180^{\\circ}$ We can see that the sum of all three angles of the given triangle is equal to $180^{\\circ}$ , which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to $60^{\\circ}$ . Q20. In $\\triangle ABC$, $\\angle A=100^{\\circ}$, AD bisects $\\angle A$ and AD perpendicular BC. Find $\\angle B$ Consider $\\triangle ABD$ $\\angle BAD$ = $\\frac{100}{2}$ (AD bisects $\\angle A$) $\\angle BAD$ = $50^{\\circ}$ $\\angle ADB$ = $90^{\\circ}$ (AD perpendicular to BC) We know that the sum of all three angles of a triangle is $180^{\\circ}$. Thus, $\\angle ABD+\\angle BAD+\\angle ADB$= $180^{\\circ}$ (Sum of angles of $\\triangle ABD$) Or, $\\angle ABD$ + $50^{\\circ}$ + $90^{\\circ}$ =$180^{\\circ}$ $\\angle ABD$ =$180^{\\circ}$– $140^{\\circ}$ $\\angle ABD$= $40^{\\circ}$ Q21.", "how many solutions there were to this triangle. $a = 12, b \\sin A = 15.1$: since $12 < 15.1, a < b \\sin A$ which tells us there are no solutions. Example 3: In triangle $ABC, a = 15, b = 20$, and $\\angle{A} = 30^\\circ$. Find $\\angle{B}$. Solution: Again in this case, $a < b$ and we know two sides and a non-included angle. By comparing $a$ and $b \\sin A$, we find that $a = 15, b \\sin A = 10$. Since $15 > 10$ we know that there will be two solutions to this problem. $\\frac{\\sin 30}{15} & = \\frac{\\sin B}{20} \\\\20 \\sin 30 & = 15 \\sin B \\\\\\frac{20 \\sin 30}{15} & = \\sin B \\\\0.6666667 & = \\sin B \\\\\\angle{B} & = 41.8^\\circ$ There are two angles less than $180^\\circ$ with a sine of 0.6666667, however. We found the first one, $41.8^\\circ$, by using the inverse sine function. To find the second one, we will subtract $41.8^\\circ$ from $180^\\circ, \\angle{B} = 180^\\circ - 41.8^\\circ = 138.2^\\circ$. To check to make sure $138.2^\\circ$ is a solution, we will use the Triangle Sum Theorem to find the third angle. Remember that all three angles must add up to $180^\\circ$. $180^\\circ - (30^\\circ + 41.8^\\circ) = 108.2^\\circ && \\text{or} && 180^\\circ - (30^\\circ + 138.2^\\circ)" ]

[ "+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", "$[asy] import olympiad; size(250); defaultpen(linewidth(0.7)+fontsize(11pt)); real r = 31, t = -10; pair A = origin, B = dir(r-60), C = dir(r); pair X = -0.8 * dir(t), Y = 2 * dir(t); pair D = foot(B,X,Y), E = foot(C,X,Y); draw(A--B--C--A^^X--Y^^B--D^^C--E); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,dir(B--D)); label(\"E\",E,dir(C--E)); [/asy]$ ### 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 $m+n$. $[asy] import olympiad;", "$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", "y + 86^2 x$ $x^2 + xy + 86^2 = 97^2$ (Since $y$ cannot be equal to $0$, dividing both sides of the equation by $y$ is allowed.) $x(x+y) = (97+86)(97-86)$ $x(x+y) = 2013$ The prime factors of $2013$ are $3$, $11$, and $61$. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal $33$, and $x+y$ must equal $\\boxed{\\textbf{(D) }61}$. ### Solution 3 Let $CX=x, BX=y$. Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$. Use power of a point on point C to the circle centered at A. So $CX*CB=CD*CE=>$ $x(x+y)=(97-86)(97+86)=>$ $x(x+y)=3*11*61$. Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$. By the Triangle Inequality, only$x+y=61$ yields a possible length of $BX+CX=BC$. Therefore, the answer is $\\boxed{\\textbf{(D) }61}$. ### Solution 4 $[asy] unitsize(2); import olympiad; import graph; pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40); draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label(\"A\",A,S); label(\"B\",B,NE); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,NE); label(\"86\",(A+B)/2,NW); label(\"86\",(A+D)/2,SE); label(\"97\",(A+C)/2,S); label(\"h\",(A+E)/2,N); label(\"k\",(E+D)/2,NE); label(\"k\",(B+E)/2,NE); label(\"m\",(C+D)/2,NE); fill(anglemark(A,E,D,100),black); label(\"90^\\circ\",anglemark(A,E,D),3*S); [/asy]$ We first draw the height of isosceles triangle ABD, and get two equations by the Pythagorean Theorem. First, $h^2 + k^2 = 86^2$. Second, $h^2", "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 $\\frac{CF}{DC}=\\frac{OC}{DO}=\\frac{1}{3}\\Rightarrow \\text{(C)}$", "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)} =", "## College Algebra (6th Edition) $x = 30$ degrees; $y = 75$ degrees The sum of all interior angles in a triangle equals 180 degrees. Also, adjacent angles on a straight line also add up to 180 degrees. This means that we can write the following system of equations: $$x + 2y = 180$$ $$3x + 15 + y = 180$$ which is to say: $$x + 2y = 180$$ $$3x + y = 165$$ We can use the substitution method to find the equivalent value of one of the variables: $$x + 2y = 180$$ $$x = 180 - 2y$$ and plug this value into the other equation: $$3x + y = 165$$ $$3(180 - 2y) + y = 165$$ $$540 - 6y + y = 165$$ $$375 = 5y$$ $$75 = y$$ We now substitute this value into the previous equation: $$x = 180 - 2y$$ $$x = 180 - 2(75) = 180 - 150 = 30$$", "+0 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "# 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 $\\angle guy$?.", "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", "+0 # Can I get some help with these? 0 68 1 Hi... I've been strugling with these three problems for a few weeks: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6},$$ where the angle$$\\frac{\\pi}{6}$$ is in radians. (2)Find the length AC, to two decimal places. Image: https://artofproblemsolving.com/texer/sywndnao (3)Find the length BC, to two decimal places. Image: https://artofproblemsolving.com/texer/enqsdotw Nov 15, 2020 #1 0 Quick note: to get the images for (2) and (3), post these codes to get the figures. Make sure you say Render as Image in the TeXeR (2) [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3) [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] Nov 15, 2020", "# Difference between revisions of \"1992 AHSME Problems/Problem 27\" ## Problem A circle of radius $r$ has chords $\\overline{AB}$ of length $10$ and $\\overline{CD}$ of length 7. When $\\overline{AB}$ and $\\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside of the circle. If $\\angle{APD}=60^\\circ$ and $BP=8$, then $r^2=$ $\\text{(A) } 70\\quad \\text{(B) } 71\\quad \\text{(C) } 72\\quad \\text{(D) } 73\\quad \\text{(E) } 74$ ## Solution $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; path circ = Circle(origin, 1); pair A = dir(degrees(7pi/12)); pair D = dir(degrees(-5pi/12)); pair B = dir(degrees(2pi/12)); pair C = dir(degrees(-2pi/12)); pair P = extension(A, B, C, D); draw(circ); draw(A--P--D); label('A', A, N); label('D', D, S); label('C', C, SE); label('B', B, NE); label('P', P, E); label('60^\\circ', P, 2 * (dir(P--A) + dir(P--D))); label('10', A--B, S); label('8', B--P, NE); label('7', C--D, N); [/asy]$ Applying Power of a Point on $P$, we find that $PC=9$ and thus $PD=16$. Observing that $PD=2BP$ and that $\\angle BPD=60^{\\circ}$, we conclude that $BPD$ is a $30-60-90$ right triangle with right angle at $B$. Thus, $BD=8\\sqrt{3}$ and triangle $ABD$ is also right. Using that fact that the circumcircle of a right triangle has its diameter equal to the hypotenuse, we compute using the Pythagorean Theorem $AD=2r=2\\sqrt{73}$. From here we see that", "It’s not often I have anything nice to say about EdExcel. I’ve usually found their exams to be the most predictable and least thought-provoking of all the boards (at least until they finally snapped in 2013 and let Kate the Photographer loose on an unsuspecting cohort). At GCSE, their advanced trigonometry questions are usually as predictable as what Alexander Armstrong is going to ask you in the chit-chat section of Pointless: your triangle will be set up so you can blindly apply the sine rule, cosine rule or area formula more or less off the shelf. Three marks, next question. This one is a bit more interesting. It’s not in the traditional “two pairs” format for sine rule, nor in the ‘elbow’ format for cosine rule – you’re going (shock! horror!) to need to do some thinking. Here’s a triangle. Side $LM$ is 15.7cm, $NM$ is 12.8cm, and angle $L\\hat N M$ is 136º. While I think degrees are awful, they’re the angle currency at GCSE, so I’m stuck with them for this question. I have two methods I’m happy with for it, but I’d love to hear others! ### First method: sine rule twice This is probably the simplest way to do it, although simple for me and simple for you aren’t necessarily the same. I would", "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 =", "200 + 100 + 200 = \\boxed{\\textbf{(E) } 500}$ ### Solution 5 (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 6 (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 $\\sin\\theta=\\sin\\left(180-\\theta\\right)$,we seek to find $2r\\sin 3\\theta$. First, $\\sin 2\\theta=2\\sin\\theta\\cos\\theta=2\\cdot\\left(\\frac{\\sqrt{2}}{4}\\right)\\left(\\frac{\\sqrt{14}}{4}\\right)=\\frac{2\\sqrt{2}\\sqrt{14}}{16}=\\frac{\\sqrt{7}}{4}$ and $\\cos 2\\theta=\\frac{3}{4}$ by Pythagorean. $$\\sin 3\\theta=\\sin(2\\theta+\\theta)=\\sin 2\\theta\\cos\\theta+\\sin \\theta\\cos 2\\theta=\\frac{\\sqrt{7}}{4}\\left(\\frac{\\sqrt{14}}{4}\\right)+\\frac{\\sqrt{2}}{4}\\left(\\frac{3}{4}\\right)=\\frac{7\\sqrt{2}+3\\sqrt{2}}{16}=\\frac{5\\sqrt{2}}{8}$$ $$2r\\sin 3\\theta=2\\left(200\\sqrt{2}\\right)\\left(\\frac{5\\sqrt{2}}{8}\\right)=400\\sqrt{2}\\left(\\frac{5\\sqrt{2}}{8}\\right)=\\frac{800\\cdot 5}{8}=\\boxed{\\textbf{(E)}\\text{ 500}}$$ ###", "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.", "12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$. $[asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label(\"A\", A, NE); label(\"C\", C, NW); label(\"D\", D, S); label(\"P\",P , N); label(\"P'\", Pa, SW); label(\"13\\sqrt{3}\", (A+D)/2, E, small); label(\"13\\sqrt{3}\", (A+C)/2, NW, small); label(\"12\\sqrt{3}\", (C+D)/2, SW, small); label(\"h\", (P + Pa)/2, W); label(\"\\frac{\\sqrt{939}}{2}\", (C+P)/2 ,NW); [/asy]$ To find the volume, we want to use the equation $\\frac 13Bh = 6\\sqrt{133}h$, so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = \\frac{\\sqrt{939}}{2}$. If we let $P$ be the center of a sphere with radius $\\frac{\\sqrt{939}}{2}$, then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\\triangle ACD$. From here we just need to perform some brutish calculations. Using the formula $A = 18\\sqrt{133} = \\frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$ (there are slightly simpler ways", "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 $AD=AE + EF + FD =", "$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", "# 2005 CEMC Gauss (Grade 7) Problems/Problem 8 ## Problem In the diagram, what is the measure of $\\angle ACB$ in degrees? $[asy] size(300); draw((-60,0)--(0,0)); draw((0,0)--(64.3,76.6)--(166,0)--cycle); label(\"A\",(64.3,76.6),N); label(\"93^\\circ\",(64.3,73),S); label(\"130^\\circ\",(0,0),NW); label(\"B\",(0,0),S); label(\"D\",(-60,0),S); label(\"C\",(166,0),S); [/asy]$ $\\text{(A)}\\ 57^\\circ \\qquad \\text{(B)}\\ 37^\\circ \\qquad \\text{(C)}\\ 47^\\circ \\qquad \\text{(D)}\\ 60^\\circ \\qquad \\text{(E)}\\ 17^\\circ$ ## Solution Since $\\angle ABC + \\angle ABD = 180^\\circ$ (in other words, $\\angle ABC$ and $\\angle ABD$ are supplementary) and $\\angle ABD = 130^\\circ$, then $\\angle ABC = 50^\\circ$. Since the sum of the angles in triangle $ABC$ is $180^\\circ$ and we know two angles $93^\\circ$ and $50^\\circ$ which add to $143^\\circ$, then $\\angle ACB = 180^\\circ - 143^\\circ = 37^\\circ$. The answer is $B$." ]

[ "of 90 and 92, so 91 24. 1018 yes, all answers were right. thanks!", "then factorize further into a product of prime Gaussian integers. 54. Is it possible to write 90 as $$a^2+b^2$$ where $$a,b$$ are integers? All times are UTC. The time now is 10:43.", "numbers are there from 87 to 97? What fraction of them are prime numbers? Solution:", "## Elementary Algebra The prime factorization of 91 is: $91=7 \\times 13$ Thus, 7 is a prime factor of 91. Therefore, the given statement is true.", "the ratio of the sum and the product of all prime divisors of the number 120.", "Factorization of 91 = 7 × 13 ### Example 2 Find the sum of factors of 91. ### Solution The factors of 91 are 1, 7, 13, and 91. Sum of factors of 91 = 1 + 7 + 13 + 91 Sum of factors of 91 = 112 ### Solution Anna knows that the factors of 91 are 1, 7, 13, and 91. Out of the four factors mentioned above, one is called the universal factor of 91, which is neither a prime nor a composite number. The prime factors are 7 and 13, while the composite factor is only 91. Images/mathematical drawings are created with GeoGebra", "of the prime 29999999 67 is the sum of digits of the prime 59899999 68 is the sum of digits of the prime 59999999 71 is the sum of digits of the prime 89999999 70 is the sum of digits of the prime 189997999 The sequence on the right is Sloane's A067523 (or A067180 if 0's are placed in the gaps), which are not particularly enlightening. The sequence on the left does not seem to be in Sloane's OEIS. -", "prime factor of h(100) +1, then p is?", "# Is 91 a prime number? Ninety one is a very fun number! It can be written as the sum of the first thirteen whole numbers $$91 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13$$; and also as the sum of the first six squares: $$91=1^2\\;+\\;2^2\\;+\\;3^2\\;+\\;4^2\\;+\\;5^2\\;+\\;6^2$$ Moreover, 91 is closely related to prime numbers, as we will see next. We will start recalling some definitions that we have widely discussed in our Prime Numbers article. If you still feel unfamiliar with these notions, we invite you to first read that article and come back here later. A factor of a natural number is a positive divisor of the number. A proper factor of a natural number is a factor that is different from 1 and from the number itself. For example, $$9 = 3 \\times 3 = 1 \\times 9$$; thus, 1, 3, and 9 are all factors of 9, but only 3 is a proper factor of 9. A natural number is called a prime number if it is greater than 1, and it doesn’t have proper factors. For example, the prime numbers closer to 91 are: 89 and 97. A composite number is a natural number that has", "Prime Product The sum of two prime numbers is 85. What is the product of these two prime numbers? Image Credit: Wikimedia Prime Rectangles Fredrik Johansson × Problem Loading... Note Loading... Set Loading...", "simply the greatest prime factor of 820. 41 Re: If n is the sum of the first 40 positive integers, what is the greates &nbs [#permalink] 29 Oct 2018, 16:46 Display posts from previous: Sort by", "Cap'N. I had a major brain fart. I was thinking of prime factors. DUH 5. Originally Posted by galactus how about 47?. It's the smallest, isn't it?. It's prime and it divides 87!+88!. But you're certainly right, Cap'N. I had a major brain fart. I was thinking of prime factors. DUH Opps, sorry that's two of us RonL 6. 87! + 88! = 87! ( 1 + 88 ) = 87!*(89). 89 is a prime factor. The question is, does a higher prime divide 87! Suppose p is prime and greater than 89. Well 87! = 87*86*85*...*2*1 Since p doesn't divide 89, then if p divides 87!*(89), then p divides at least one of these terms in 87!. This is impossible since p is larger than all terms appearing in the factorial. So 89 is the largest prime factor of 87! + 88! (this can be worked into a theorem, say if p is prime, then it's the largest prime factor of (p-1)! + (p-2)! ) 7. So I was correct. I thought this was something I had seen before, but then second-guessed myself. 8. Hello, ceasar! Determine the largest prime divisor of $87! + 88!$ We have: . $87! + 88!\\:=\\:87! + 88\\cdot87!$ Factor: . $87!(1 + 88) \\:=\\:87!(89)$ Therefore, $89$ is the largest prime divisor.", "# sum of the..... X = The sum of the factors of $$2^{10}$$ A) X is odd B) X is prime C) X is divisible by 89 How many of the statements above are correct ? ×", "get that it does not: $$91 = (5 \\times 18) + 1$$. We continue with 7 (our last hope!), and indeed we see that 7 divides 91: $$91 = 7 \\times 13$$. Therefore, 7 and 13 are proper factors of 91. This means that 91 is a composite number, it isn’t prime! However, although 91 is not a prime number, it is the product of exactly two primes: 7 and 13. Another way of understanding why 91 is not prime, is recalling that a prime number of objects cannot be arranged into a rectangular grid with more than one column and more than one row. As we see below, 91 stars can certainly be arranged into a rectangular grid with 7 rows and 13 columns. This means that 91 is not a prime number! $$91 = 7 \\times 13$$ Notice that the 91 stars can also be arranged into a rectangular grid with 13 rows and 7 columns. Can you draw such a grid? ## Occurrences of 91 among the prime numbers Since 91 is the product of exactly two prime numbers: $$91 = 7 \\times 13$$, it is what is called a semi-prime number. Sometimes, 91 can be confused with a prime number because it behaves as twin prime. A twin prime is a prime number that", "2012 Posted by Alex in : potd , 1 comment so far Find the sum of the prime factors of $1,015,074,782$, given that there are exactly four of them and one of them is $499$.", "theorem. Any natural number greater than 1 can be written as a product of prime factors.", "a fractionfind the prime factorization of 90csc 2x-1how to solve a 5x5x5", "understanding of the factors of 91, the following information must be recognized. 1. The most minor factor of 91 is 1. 2. The largest factor of 91 is the number 91 itself. 3. All the factors of 91 are odd numbers. 4. Except for 91, all the factors of 91 are prime numbers considering that 1 is not a prime nor a composite number. 5. The sum of divisors of 91 is an odd composite number 112. ### Characteristics of the Number 91 Following are some essential characteristics of the number 91 which must be kept in mind to help understand and discover the factors of 91. 1. 91 is a natural number that is an odd number simultaneously. 2. 91 is a composite number. 3. 91 is not a perfect square. 4. 91 is a semi-prime number. 5. 91 is not a defective number. ## Factors of 91 by Prime Factorization The prime factorization is the way of representing a given number in the form of the product of its prime factors Differently, it can be defined as; the method of determining the factors of the given number. It utilizes the process of up-side-down division. The factors determined by prime factorization are prime factors. The prime factorization has a base of division method, but the difference is", "is simply the greatest prime factor of 820. 41 Re: If n is the sum of the first 40 positive integers, what is the greates &nbs [#permalink] 29 Oct 2018, 16:46 Display posts from previous: Sort by", "prime factor involved in the numbers." ]

[ "we'll switch entirely to trick $2$; probably we should've done this earlier. • $4181 = 3100 + 930 + 155 - 4$. It follows that $4181$ is not divisible by $31$. • $4181 = 3700 + 370 + 111 = 37(113)$. We've finally found a prime factor!", "get that it does not: $$91 = (5 \\times 18) + 1$$. We continue with 7 (our last hope!), and indeed we see that 7 divides 91: $$91 = 7 \\times 13$$. Therefore, 7 and 13 are proper factors of 91. This means that 91 is a composite number, it isn’t prime! However, although 91 is not a prime number, it is the product of exactly two primes: 7 and 13. Another way of understanding why 91 is not prime, is recalling that a prime number of objects cannot be arranged into a rectangular grid with more than one column and more than one row. As we see below, 91 stars can certainly be arranged into a rectangular grid with 7 rows and 13 columns. This means that 91 is not a prime number! $$91 = 7 \\times 13$$ Notice that the 91 stars can also be arranged into a rectangular grid with 13 rows and 7 columns. Can you draw such a grid? ## Occurrences of 91 among the prime numbers Since 91 is the product of exactly two prime numbers: $$91 = 7 \\times 13$$, it is what is called a semi-prime number. Sometimes, 91 can be confused with a prime number because it behaves as twin prime. A twin prime is a prime number that", "\"87,\" \"91) $1$ has only one factor, not two factors as prime numbers have. $51 , 57 \\mathmr{and} 87$ are all multiples of $3$. (add their digits) $91 = 7 \\times 13$ As it is a product of two primes and above the normal times tables, we will not often have come across $91$ in our maths.", "$$91 = 13 \\times 7$$, and $$13 > \\sqrt{100}$$ # Shortly Now we know checking divisibility 2, 3, 5, and 7 is sufficient, we want to make it quick. Divisibility by 2 or 5 is as simple as checking the final digit. If it’s even, or 5, it’s not prime. Divisibility by 3 has a simple test: sum the digits of your number. If that sum is divisible by 3, then so is the number. Divisibility by 7 is harder. The divisibility test on Wikipedia is not so easy to do in your head. However, once you’ve eliminated the numbers divisible by 2, 3, and 5, there are only 3 left which are divisible by 7! Those are: • $$49 = 7 \\times 7$$ • $$77 = 7 \\times 11$$ • $$91 = 7 \\times 13$$ Of these, 49 and 77 are pretty clearly not prime - at least, I would recognise these as being $$7^2$$ and $$7 \\times 11$$. I don’t know a trick other than memorisation for $$7 \\times 13 = 91$$, but it doesn’t seem too high a price to pay. # Finally To summarise, here’s how you check a two-digit number for primeness: A two-digit number is prime if you answer no to all these questions: 1. Is it even? (i.e., does it end", "were divisible by 7, 103 would be too. If 103 were divisible by 7, so would 110, which is$2 \\cdot 5 \\cdot 11$and is therefore not divisible by 7. So 1037 isn't divisible by 7. 11:$1 - 0 + 3 - 7 = -3$, which is not divisible by 11, so 1037 isn't divisible by 11. 13: If 1037 were divisible by 13, then$1050 = 1037 + 13$would also be. If$1050$were divisible by 13,$105$would be too. But$105 = 3 \\cdot 5 \\cdot 7$, so it isn't divisible by 13, so neither is 1037. 17: 1037 is divisible by 17 only if$1037 - 17 = 1020$is. 1020 is divisible by 17 only if$102$is.$102 = 2 \\cdot 51 = 2 \\cdot 3 \\cdot 17$, so it is divisible by 17, so 1037 is divisible by 17, so 1037 isn't prime. Note: These approaches don't do a good job of factoring - it's a bit tricky to unwind that last argument to figure out what$1037/17$is. But if the question is just \"is it prime\", this should work fine. • What a cool trick, adding multiples of a possible factor to reach a multiple of 10. – arp May 21 '18 at 23:15 You're not likely to get even a moderately large random number to factor. In addition to the tests in", "digits of $14553$ is $18$, so it is divisible by $9$. Repeatedly dividing by $3$ yields $$14553 = 3 \\cdot 4851 = 3^2 \\cdot 1617 = 3^3 \\cdot 539$$ • The next reasonable guess might be that $539$ is divisible by $7$... and indeed it is: $$539 = 7 \\cdot 77 = 7^2 \\cdot 11$$ • ...and $11$ is prime, so we're done. In summary: $$9095625 = 3^3 \\cdot 5^4 \\cdot 7^2 \\cdot 11$$ • After factoring out 3**3, we are left with 539, Here we can take sum of the alternate digits. 5+9=14 & 3. Now taking difference between the two groups, we get 14-3=11. Hence the number is divisible by 11. By factoring out 11, we get 539=11 * 49. <p> Refer to summary of Divisibility rules – Wishwas Sep 19 '16 at 5:03 Since we're doing the prime factorization, you can just use the long division algorithm that you learned in grade school. Use the last digit as a clue for what to divide by. Remember that any number has a unique prime factorization, so always just divide by the smallest feasible prime. In this case, we start by recognizing that the number is obviously divisible by $5$. Divide it by $5$, we get $1819125$. Divide by $5$ again. We get $363825$. And again. We", "$4181$ 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", "otherwise not prime. No need to check by numbers like 2, 6, 9, 10, 12 etc, if already checked for their prime factors 2, 3 and 5 etc.Let us take an example to understand this method clearly. Example: Check whether 91 is prime. Solution: Step 1 - On dividing 91 by 2, we get 45.5 (quotient is not integer). Therefore, 91 is not divisible by 2. Step 2 - On dividing 91 by 3, we get 30.3333. Therefore, 91 is not divisible by 3. Step 3 - Since 91 is not divisible by 2 and 2 is a prime factor of 4. Therefore, 91 is not divisible by 4 also. Step 4 - On dividing 91 by 5, we get 18.2. Therefore, 91 is not divisible by 5. Step 5 - Since 91 is not divisible by 3 and 3 is a prime factor of 6. Therefore, 91 is not divisible by 6 also. Step 6 - On dividing 91 by 7, we get 13. Therefore, 91 is divisible by 7. Therefore, 91 is divisible by a number other than 0 or itself. Hence, 91 is not a prime number. Related Calculators Find Prime Factorization Calculator Calculate Prime Number Find Factors of a Number Calculator Calculate Prime Factors *AP and SAT are registered trademarks of the College Board.", "# Is 91 a prime number? Ninety one is a very fun number! It can be written as the sum of the first thirteen whole numbers $$91 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13$$; and also as the sum of the first six squares: $$91=1^2\\;+\\;2^2\\;+\\;3^2\\;+\\;4^2\\;+\\;5^2\\;+\\;6^2$$ Moreover, 91 is closely related to prime numbers, as we will see next. We will start recalling some definitions that we have widely discussed in our Prime Numbers article. If you still feel unfamiliar with these notions, we invite you to first read that article and come back here later. A factor of a natural number is a positive divisor of the number. A proper factor of a natural number is a factor that is different from 1 and from the number itself. For example, $$9 = 3 \\times 3 = 1 \\times 9$$; thus, 1, 3, and 9 are all factors of 9, but only 3 is a proper factor of 9. A natural number is called a prime number if it is greater than 1, and it doesn’t have proper factors. For example, the prime numbers closer to 91 are: 89 and 97. A composite number is a natural number that has", "= 1998 is, but $1998 = 2 \\times 999 = 2 \\times 111 \\times 9$ does not have 13 as a factor (111 is 19 away from 130). • 2011 – 51 = 1960, but 196 is not divisible by 17 since 170 (26 away) is. • 2011 + 19 = 2030, but 203 is not divisble by 19 since 190 (13 away) is. • 2300 – 2011 = 289, 289 – 230 = 59, which is not divisible by 23, hence 2011 is not divisible by 23. • 2011 + 29 = 2040, 204 is not divisible by 29 since $203 = 7 \\times (30 - 1)$ is. • 2011 – 31 = 1980, 198 is not divisible by 31 since 186 (12 away) is. • 2011 is not divisible by 37 since 1998 (13 away) as found above is $2 \\times 111 \\times 9 = 2 \\times 37 \\times 3 \\times 9$ is. • 2011 – 41 = 1970, 197 is not divisible by 41 since 205 is. • 2011 + 129 = 2140, 214 is not divisible by 43 since 215 is. Hence $2011$ is prime! You might want to try a similar exercise to show $2003$ is also prime. Finally, note that if the original number is less than $1072$, then one only needs", "this sum is divided by 1152. First we should do the prime factorization of 1152: $$1152=2^7*3^2$$. Consider the third and fourth terms: $$(3!)^3=2^3*3^3$$ not divisible by 1152; $$(4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152$$ divisible by 1152, and all the other terms after will be divisible by 1152. We'll get $$\\{(1!)^3+ (2!)^3 + (3!)^3\\} +\\{(4!)^3+...+(1152!)^3\\}=225+1152k$$ and this sum divided by 1152 will result remainder of 225. Hi Bunnel, To get the remainder, we dont have to reduce the fraction right? That is we cant do - 225/1152 = 25/ 128 and get remainder 25? Math Expert Joined: 02 Sep 2009 Posts: 45455 Show Tags 21 Jul 2013, 02:24 cumulonimbus wrote: Bunuel wrote: jeeteshsingh wrote: Is there a specific approach to tackle this? No specific approach. We have the sum of many numbers: $$(1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3$$ and want to determine the remainder when this sum is divided by 1152. First we should do the prime factorization of 1152: $$1152=2^7*3^2$$. Consider the third and fourth terms: $$(3!)^3=2^3*3^3$$ not divisible by 1152; $$(4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152$$ divisible by 1152, and all the other terms after will be divisible by 1152. We'll get $$\\{(1!)^3+ (2!)^3 + (3!)^3\\} +\\{(4!)^3+...+(1152!)^3\\}=225+1152k$$ and this sum divided by 1152 will result remainder of 225. Hi Bunnel, To get the remainder, we dont have to reduce the fraction right? That is we cant do -", "of 1881, we have that $$1+8+8+1=18\\text{.}$$ It is easier to see that the sum, 18, is divisible by 3. But if that is not obvious, we can continue to sum the digits to get that $$1+8=9\\text{,}$$ which is clearly divisible by 3. Since $$1881$$ is divisible by $$3\\text{,}$$ we can compute that $$1881=3\\cdot 627\\text{.}$$ We now have \\begin{equation*} 18810=2\\cdot 5\\cdot 3\\cdot 627\\text{.} \\end{equation*} We can perform the same summing digits check mentioned above to determine that 627 is also divisible by 3. Since $$627=3\\cdot 209\\text{,}$$ we now have \\begin{equation*} 18810=2\\cdot 5\\cdot 3\\cdot 3\\cdot 209\\text{.} \\end{equation*} Summing the digits of $$209$$ tells us that 209 is not divisible by 3 since $$2+0+9=11$$ is not divisible by 3. So, we move onto the next prime number, 5. Since 209 does not end in 0 or 5, we conclude that 5 is not a factor. We continue moving through the prime numbers and conclude that 7 is not a factor of 209 but that 11 is. Since $$209=11\\cdot 19\\text{,}$$ we now have \\begin{equation*} 18810=2\\cdot 5\\cdot 3\\cdot 3\\cdot 11\\cdot 19\\text{.} \\end{equation*} Since 19 is also a prime, we have now found all of the prime factors of 18810 and how many times they each occur. However, we typically rearrange the prime factors into non-decreasing order and use exponents to condense the prime", "## Elementary Algebra The prime factorization of 91 is: $91=7 \\times 13$ Thus, 7 is a prime factor of 91. Therefore, the given statement is true.", "# How am I supposed to tell if a number is divisible by $13$ (I need a shortcut)? I've been trying to figure out if a number is divisible by $13$. As I'm saying this in first person, I think I'm supposed to take the rightmost digit of the number, for example, $39$, multiply it by 9, and finally subtract the digit from the rest of the digits put in order to the left of that rightmost digit. The final result has to be divisible by $13$. So, in $39$, $9*9=81-3=78$, and $78$ is divisible by $13$. I think this is it. Am I on the right digit track? I just want some answers with words that will satisfy me. • One way is presented here and another one here – k1next Nov 8 '14 at 20:07 • Your method works, but the second link provided by @sonystarmap works similarly, but perhaps slightly better. Your method relies on the fact that $9\\cdot10=90=91-1$, and $91$ is divisible by $13$. And that other method relies on $4\\cdot10=10=39+1$, and $39$ is divisible by $13$. – Harald Hanche-Olsen Nov 8 '14 at 20:15 Here is an explanation of why your method works: $13$ is a prime number, and thus has no factors in common with $9$. So $n$ is divisible by $13$", "# How do you find the GCF of 66, 90, and 150? Jan 12, 2017 $6$ #### Explanation: Note that the prime factorisation of $66$ is: $66 = 2 \\cdot 3 \\cdot 11$ Both $90$ and $150$ are divisible by $2$ and by $3$, but not by $11$. Hence all three numbers are divisible by $2 \\cdot 3 = 6$ and nothing larger.", "## 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 = $2\\times2\\times3\\times5\\times7$ = $2^{2} \\times3 \\times 5 \\times 7$", "# How do you factor the monomial -13xy^3 completely? $13$ is a prime number so it cannot be factored any further: $- 13 x {y}^{3} = - 13 \\times x \\times y \\times y \\times y$", "43 to 107 is $$107*54 - 43*21 = 5778 - 903 = 4875$$ $$4875$$ when prime factorized is $$3*5^3*13$$ and the only number which divides this is 39(Option E) _________________ You've got what it takes, but it will take everything you've got The sum of all integers from 43 to 107, inclusive, is divisible by whi [#permalink] 26 Nov 2017, 10:57 Display posts from previous: Sort by", "in product of prime numbers equal to $73!$ are not only from numbers divisible by 3, but also from divisible by $9=3\\cdot3$ or $27=3\\cdot3\\cdot3$. – Przemysław Scherwentke Oct 29 '14 at 13:34 Hint: $6 = 2 \\cdot 3$. Since $3$ is less frequent than $2$ in the product $73! = 1 \\cdot 2 \\cdots 73$, count the number of factors of $3$ in the numbers $1$, $2$, $\\ldots$, $73$. Note that some numbers give you more than one factor of $3$.", "# How do you find the prime factorization of 112? Feb 5, 2015 You divide by primes, untill you run out. $112$ is even: $112 / 2 = 56 \\to 112 = \\left(2 \\cdot 56\\right)$ Still even: $56 / 2 = 28 \\to 112 = 2 \\cdot \\left(2 \\cdot 28\\right)$ And again: $28 / 2 = 14 \\to 112 = 2 \\cdot 2 \\cdot \\left(2 \\cdot 14\\right)$ And again: $14 / 2 = 7 \\to 112 = 2 \\cdot 2 \\cdot 2 \\cdot \\left(2 \\cdot 7\\right)$ (brackets are only for clarification) $7$ is a prime (try if you like) So the complete factorisation: $112 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 7 = {2}^{4} \\cdot 7$" ]

[ "get that it does not: $$91 = (5 \\times 18) + 1$$. We continue with 7 (our last hope!), and indeed we see that 7 divides 91: $$91 = 7 \\times 13$$. Therefore, 7 and 13 are proper factors of 91. This means that 91 is a composite number, it isn’t prime! However, although 91 is not a prime number, it is the product of exactly two primes: 7 and 13. Another way of understanding why 91 is not prime, is recalling that a prime number of objects cannot be arranged into a rectangular grid with more than one column and more than one row. As we see below, 91 stars can certainly be arranged into a rectangular grid with 7 rows and 13 columns. This means that 91 is not a prime number! $$91 = 7 \\times 13$$ Notice that the 91 stars can also be arranged into a rectangular grid with 13 rows and 7 columns. Can you draw such a grid? ## Occurrences of 91 among the prime numbers Since 91 is the product of exactly two prime numbers: $$91 = 7 \\times 13$$, it is what is called a semi-prime number. Sometimes, 91 can be confused with a prime number because it behaves as twin prime. A twin prime is a prime number that", "we'll switch entirely to trick $2$; probably we should've done this earlier. • $4181 = 3100 + 930 + 155 - 4$. It follows that $4181$ is not divisible by $31$. • $4181 = 3700 + 370 + 111 = 37(113)$. We've finally found a prime factor!", "# Is 91 a prime number? Ninety one is a very fun number! It can be written as the sum of the first thirteen whole numbers $$91 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13$$; and also as the sum of the first six squares: $$91=1^2\\;+\\;2^2\\;+\\;3^2\\;+\\;4^2\\;+\\;5^2\\;+\\;6^2$$ Moreover, 91 is closely related to prime numbers, as we will see next. We will start recalling some definitions that we have widely discussed in our Prime Numbers article. If you still feel unfamiliar with these notions, we invite you to first read that article and come back here later. A factor of a natural number is a positive divisor of the number. A proper factor of a natural number is a factor that is different from 1 and from the number itself. For example, $$9 = 3 \\times 3 = 1 \\times 9$$; thus, 1, 3, and 9 are all factors of 9, but only 3 is a proper factor of 9. A natural number is called a prime number if it is greater than 1, and it doesn’t have proper factors. For example, the prime numbers closer to 91 are: 89 and 97. A composite number is a natural number that has", "## Elementary Algebra The prime factorization of 91 is: $91=7 \\times 13$ Thus, 7 is a prime factor of 91. Therefore, the given statement is true.", "\"87,\" \"91) $1$ has only one factor, not two factors as prime numbers have. $51 , 57 \\mathmr{and} 87$ are all multiples of $3$. (add their digits) $91 = 7 \\times 13$ As it is a product of two primes and above the normal times tables, we will not often have come across $91$ in our maths.", "2012 Posted by Alex in : potd , 1 comment so far Find the sum of the prime factors of $1,015,074,782$, given that there are exactly four of them and one of them is $499$.", "43 to 107 is $$107*54 - 43*21 = 5778 - 903 = 4875$$ $$4875$$ when prime factorized is $$3*5^3*13$$ and the only number which divides this is 39(Option E) _________________ You've got what it takes, but it will take everything you've got The sum of all integers from 43 to 107, inclusive, is divisible by whi [#permalink] 26 Nov 2017, 10:57 Display posts from previous: Sort by", "$$91 = 13 \\times 7$$, and $$13 > \\sqrt{100}$$ # Shortly Now we know checking divisibility 2, 3, 5, and 7 is sufficient, we want to make it quick. Divisibility by 2 or 5 is as simple as checking the final digit. If it’s even, or 5, it’s not prime. Divisibility by 3 has a simple test: sum the digits of your number. If that sum is divisible by 3, then so is the number. Divisibility by 7 is harder. The divisibility test on Wikipedia is not so easy to do in your head. However, once you’ve eliminated the numbers divisible by 2, 3, and 5, there are only 3 left which are divisible by 7! Those are: • $$49 = 7 \\times 7$$ • $$77 = 7 \\times 11$$ • $$91 = 7 \\times 13$$ Of these, 49 and 77 are pretty clearly not prime - at least, I would recognise these as being $$7^2$$ and $$7 \\times 11$$. I don’t know a trick other than memorisation for $$7 \\times 13 = 91$$, but it doesn’t seem too high a price to pay. # Finally To summarise, here’s how you check a two-digit number for primeness: A two-digit number is prime if you answer no to all these questions: 1. Is it even? (i.e., does it end", "# How do you find the prime factorization of 112? Feb 5, 2015 You divide by primes, untill you run out. $112$ is even: $112 / 2 = 56 \\to 112 = \\left(2 \\cdot 56\\right)$ Still even: $56 / 2 = 28 \\to 112 = 2 \\cdot \\left(2 \\cdot 28\\right)$ And again: $28 / 2 = 14 \\to 112 = 2 \\cdot 2 \\cdot \\left(2 \\cdot 14\\right)$ And again: $14 / 2 = 7 \\to 112 = 2 \\cdot 2 \\cdot 2 \\cdot \\left(2 \\cdot 7\\right)$ (brackets are only for clarification) $7$ is a prime (try if you like) So the complete factorisation: $112 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 7 = {2}^{4} \\cdot 7$", "# sum of the..... X = The sum of the factors of $$2^{10}$$ A) X is odd B) X is prime C) X is divisible by 89 How many of the statements above are correct ? ×", "## 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 = $2\\times2\\times3\\times5\\times7$ = $2^{2} \\times3 \\times 5 \\times 7$", "otherwise not prime. No need to check by numbers like 2, 6, 9, 10, 12 etc, if already checked for their prime factors 2, 3 and 5 etc.Let us take an example to understand this method clearly. Example: Check whether 91 is prime. Solution: Step 1 - On dividing 91 by 2, we get 45.5 (quotient is not integer). Therefore, 91 is not divisible by 2. Step 2 - On dividing 91 by 3, we get 30.3333. Therefore, 91 is not divisible by 3. Step 3 - Since 91 is not divisible by 2 and 2 is a prime factor of 4. Therefore, 91 is not divisible by 4 also. Step 4 - On dividing 91 by 5, we get 18.2. Therefore, 91 is not divisible by 5. Step 5 - Since 91 is not divisible by 3 and 3 is a prime factor of 6. Therefore, 91 is not divisible by 6 also. Step 6 - On dividing 91 by 7, we get 13. Therefore, 91 is divisible by 7. Therefore, 91 is divisible by a number other than 0 or itself. Hence, 91 is not a prime number. Related Calculators Find Prime Factorization Calculator Calculate Prime Number Find Factors of a Number Calculator Calculate Prime Factors *AP and SAT are registered trademarks of the College Board.", "digits of $14553$ is $18$, so it is divisible by $9$. Repeatedly dividing by $3$ yields $$14553 = 3 \\cdot 4851 = 3^2 \\cdot 1617 = 3^3 \\cdot 539$$ • The next reasonable guess might be that $539$ is divisible by $7$... and indeed it is: $$539 = 7 \\cdot 77 = 7^2 \\cdot 11$$ • ...and $11$ is prime, so we're done. In summary: $$9095625 = 3^3 \\cdot 5^4 \\cdot 7^2 \\cdot 11$$ • After factoring out 3**3, we are left with 539, Here we can take sum of the alternate digits. 5+9=14 & 3. Now taking difference between the two groups, we get 14-3=11. Hence the number is divisible by 11. By factoring out 11, we get 539=11 * 49. <p> Refer to summary of Divisibility rules – Wishwas Sep 19 '16 at 5:03 Since we're doing the prime factorization, you can just use the long division algorithm that you learned in grade school. Use the last digit as a clue for what to divide by. Remember that any number has a unique prime factorization, so always just divide by the smallest feasible prime. In this case, we start by recognizing that the number is obviously divisible by $5$. Divide it by $5$, we get $1819125$. Divide by $5$ again. We get $363825$. And again. We", "# How do you find the GCF of 66, 90, and 150? Jan 12, 2017 $6$ #### Explanation: Note that the prime factorisation of $66$ is: $66 = 2 \\cdot 3 \\cdot 11$ Both $90$ and $150$ are divisible by $2$ and by $3$, but not by $11$. Hence all three numbers are divisible by $2 \\cdot 3 = 6$ and nothing larger.", "7 divides 91 perfectly. So, 91 is divisible by 7 we get more than two factors other than 1 and 91 itself. Hence, we can say that 91 is not a prime number. So, why 91 is not a prime number is because it has more than two factors which contradict the definition of prime numbers. ### 91 is a Prime or Composite Number As we can see why is 91 not a prime number it is because it has more than two factors other than 1 and the number itself. It contradicts the definition of prime numbers. So 91 is a composite number because composite numbers have factors more than 2. When we factorize 91 we get 91 ÷ 1 = 91 91 ÷ 7 = 13 91 ÷ 13 = 7 91 ÷ 91 = 1 Here we get four factors for 91 and they are 1, 13, 7, 91. As the total number of factors are more than two it satisfies the condition of composite number. Hence 91 is a composite number. ### What Kind of Number is 91 Then? Now it is clear that 91 does not satisfy the conditions of prime numbers. It satisfies the conditions for composite numbers so 91 is a composite number. Some other kind of numbers in which 91", "when divided by 89(89 is prime) So, $2^89$ will leave remainder 2 when divided by 89", "# 9. Write the natural numbers from $102$ to $113$. What fraction of them are prime numbers? Thus fraction is :- $=\\ \\frac{4}{12}\\ =\\ \\frac{1}{3}$", "902$$\\sum_{a|n}a^{k-1}c_0 = \\sigma_{k-1}(n)c_0.$$ 903Thus if $f$ is a cusp form so is $T_nf$. ($T_nf$ is holomorphic 904since its original definition is as a finite sum of holomorphic 905functions.) 906\\end{remark} 907 908\\begin{remark} 909Setting $\\mu=1$ shows that the coefficient of $q$ in $T_n f$ is 910$\\sum_{a|1}1^{k-1}c_n=c_n$. As an immediate corollary we have the 911following important result. 912\\end{remark} 913 914\\begin{cor} 915Suppose $f$ is a cusp form for which $T_n f$ has 0 as coefficient 916of $q$ for all $n\\geq 1$, then $f=0$. 917\\end{cor} 918 919\\begin{remark} 920When $n=p$ is prime we get an interesting formula for the 921action of $T_p$ on the $q$-expansion of $f$. 922One has 923$$T_p f = \\sum_{\\mu\\geq 0} \\sum_{\\substack{a|n\\\\a|\\mu}}a^{k-1} 924 c_{\\frac{n\\mu}{a^2}} q^{\\mu}.$$ 925Since $n=p$ is prime either $a=1$ or $a=p$. When 926$a=1$, $c_{p\\mu}$ occurs in the coefficient of $q^{\\mu}$ 927and when $a=p$, we can write $\\mu=p\\lambda$ and we get 928terms $p^{k-1}c_{\\lambda}$ in $q^{\\lambda p}$. 929Thus 930$$T_n f = \\sum_{\\mu\\geq 0}c_{p\\mu}q^{\\mu}+ 931 p^{k-1}\\sum_{\\lambda\\geq 0} c_{\\lambda}q^{p\\lambda}.$$ 932\\end{remark} 933 934%% Lecture 5, 1/26/96 935 936\\chapter{Embedding Hecke Operators in the Dual} 937 938\\section{The Space of Modular Forms} 939Let $\\Gamma=\\Gamma_1(1)=\\sl2z$ and for $k\\geq 0$ 940let 941\\begin{align*} 942M_k&=\\{f=\\sum_{n=0}^{\\infty}a_n q^n : \\text{$f$ is a modular form 943for $\\Gamma$}\\}\\\\ 944&\\subset S_k=\\{f=\\sum_{n=1}^{\\infty}a_n q^n\\}\\end{align*} 945These are finite dimensional $\\bC$-vector spaces whose dimensions 946are easily computed. Furthermore, they are generated by familiar elements 947(see Serre \\cite{serre2} or Lang \\cite{lang1}.)", "Factorization of 91 = 7 × 13 ### Example 2 Find the sum of factors of 91. ### Solution The factors of 91 are 1, 7, 13, and 91. Sum of factors of 91 = 1 + 7 + 13 + 91 Sum of factors of 91 = 112 ### Solution Anna knows that the factors of 91 are 1, 7, 13, and 91. Out of the four factors mentioned above, one is called the universal factor of 91, which is neither a prime nor a composite number. The prime factors are 7 and 13, while the composite factor is only 91. Images/mathematical drawings are created with GeoGebra", "this sum is divided by 1152. First we should do the prime factorization of 1152: $$1152=2^7*3^2$$. Consider the third and fourth terms: $$(3!)^3=2^3*3^3$$ not divisible by 1152; $$(4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152$$ divisible by 1152, and all the other terms after will be divisible by 1152. We'll get $$\\{(1!)^3+ (2!)^3 + (3!)^3\\} +\\{(4!)^3+...+(1152!)^3\\}=225+1152k$$ and this sum divided by 1152 will result remainder of 225. Hi Bunnel, To get the remainder, we dont have to reduce the fraction right? That is we cant do - 225/1152 = 25/ 128 and get remainder 25? Math Expert Joined: 02 Sep 2009 Posts: 45455 Show Tags 21 Jul 2013, 02:24 cumulonimbus wrote: Bunuel wrote: jeeteshsingh wrote: Is there a specific approach to tackle this? No specific approach. We have the sum of many numbers: $$(1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3$$ and want to determine the remainder when this sum is divided by 1152. First we should do the prime factorization of 1152: $$1152=2^7*3^2$$. Consider the third and fourth terms: $$(3!)^3=2^3*3^3$$ not divisible by 1152; $$(4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152$$ divisible by 1152, and all the other terms after will be divisible by 1152. We'll get $$\\{(1!)^3+ (2!)^3 + (3!)^3\\} +\\{(4!)^3+...+(1152!)^3\\}=225+1152k$$ and this sum divided by 1152 will result remainder of 225. Hi Bunnel, To get the remainder, we dont have to reduce the fraction right? That is we cant do -" ]

[ "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]", "label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); [/asy]$ NOTE: Connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices. Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\\frac{8 \\cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\\boxed{\\textbf{(C) }80}$. $[asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label(\"x=2\",(1,0,0),S); label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); label(\"A\",(0,2,0), NW); label(\"B\",(2,0,2), NW); [/asy]$ ### Solution 2 (rigorous) The three-dimensional distance formula shows that the lengths of the equilateral triangle must be $\\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \\le d_x, d_y, d_z \\le 2$, which yields the possible edge lengths of $\\sqrt{0^2+0^2+1^2}=\\sqrt{1},\\ \\sqrt{0^2+1^2+1^2}=\\sqrt{2},\\ \\sqrt{1^2+1^2+1^2}=\\sqrt{3},$ $\\ \\sqrt{0^2+0^2+2^2}=\\sqrt{4},\\ \\sqrt{0^2+1^2+2^2}=\\sqrt{5},\\ \\sqrt{1^2+1^2+2^2}=\\sqrt{6},$ $\\ \\sqrt{0^2+2^2+2^2}=\\sqrt{8},\\ \\sqrt{1^2+2^2+2^2}=\\sqrt{9},\\ \\sqrt{2^2+2^2+2^2}=\\sqrt{12}$ Some casework shows that $\\sqrt{2},\\ \\sqrt{6},\\ \\sqrt{8}$ are the only lengths that work, from which we can use the same counting argument as above.", "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", "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", "&& (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw(", "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", "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", "dot(\"S\", S, E); dot(\"K\",K,E); dot(\"L\",L,W); // dot(\"R\",R,NE); dot(\"Q\",Q,NW); dot(\"G\", G, SE); dot(\"Q\",Q,NW); dot(\"P\",P,NE); dot(\"M\",M,S); draw(arc1,dashed); draw(arc2, dashed); draw(circle2); draw(A--D--B--cycle); // draw(A--C, dashed); draw(A--H); draw(C--S--H--cycle); draw(C--T--H); draw(D--C--B); draw(D--L--K--B, dashed); draw(T--G, dashed); draw(K--H, dashed); draw(D--Q--P--B,dashed); draw(Q--H--P,dashed); draw(C--L--H, dashed); draw(C--K,dashed); draw(T--S,dashed); // draw(anglemark(H,S,K)); // draw(anglemark(K,H,S)); label(\"x\",C+(0,0.1),N); label(\"y\",C+(-0.1,0.3),N); label(\"w\",H+(-0.03,0.1), N); label(\"z\", H+(0.06,0.12), N ); label(\"u\",T+(-0.1,-0.2), S); label(\"v\", S+(0,-0.2), S); label(\"t\",T+(0.1,-0.3), S); label(\"s\", S+(-0.08,-0.2), S); [/asy]$ Denote $\\angle{HSB}=v$, $\\angle{HTD}=u$, $\\angle{HSC}=s$, $\\angle{HTC}=t$, $\\angle{HCS}=x$, $\\angle{HCT}=y$, $\\angle{AHS}=z$, $\\angle{AHT}=w$. Since $\\angle{CHS}-\\angle{CSB} =90$ and $\\angle{CHT}-\\angle{CTD}=90$, we have $\\angle{CSA}=x+90$, $\\angle{CTA}=y+90$. Since $\\angle{CHS} = \\angle{CSB}+90$, the tangent of the circumcircle of $\\triangle{CSH}$ at point $S$ is perpendicular to $SB$; therefore, the circumcenter of $\\triangle{CSH}$ (point $K$) is on $AB$. Similarly, the circumcenter of $\\triangle{CTH}$ (point $L$) is on $AD$. In addition, $KL$ is the perpendicular bisector of $CH$. Extend $CB$ to meet circumcircle of $\\triangle{CSH}$ at $P$, and extend $CD$ to meet circumcircle of $\\triangle{CTH}$ at $Q$. Then, since $\\angle{ADC}=\\angle{ABC}=90$, $AD$ and $AB$ are the perpendicular bisector of $CQ$ and $CP$, respectively; hence $A$ is the circumcenter of $\\triangle{PCQ}$. Since $B$ and $D$ are midpoints on $CP$ and $CQ$, $PQ \\parallel BD$; also, $AH \\perp BD$, so $AH \\perp PQ$. Since $A$ is the circumcenter, $AH$ is also the perpendicular bisector of $PQ$. Hence, $$HP=HQ$$ We have $$u=\\angle{CHT}-90=(90-w+\\angle{DHC}) - 90$$ $$v=\\angle{CHS}-90 = (90-z+\\angle{BHC}) - 90$$ Hence, $u+v = -w-z+\\angle{DHC}+\\angle{BHC} =", "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\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}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "two complete triangles. Eventually, either A chooses another red giving B triangle 2 or chooses all greens which gives B triangle 3 and A triangle 2. Therefore, B wins $7$ to $2$. This is a rather tentative answer, as I don't have a proof that this is the best strategy on either part, but it might be useful none the less. The first pick of course doesn't matter, so say it's red ($r$), so we have the triangles: $$(r,\\cdot,\\cdot), (\\cdot, \\cdot, \\cdot), (\\cdot, \\cdot, \\cdot)$$ If $A$ picks $r$ again, $B$ can put it on the first triangle, which is a loss for $A$, then applying the same idea, $A$ can pick $$(r,\\cdot,\\cdot), (g, \\cdot, \\cdot), (\\cdot, \\cdot, \\cdot)$$ Then no matter whether $A$ picks $r$ or $g$, $B$ can place it in $B$'s favour (so say it's $r$) $$(r,r,\\cdot), (g, \\cdot, \\cdot), (\\cdot, \\cdot, \\cdot)$$ $(\\ast)$If $A$ keeps picking $r$, $B$ can force two monochrome triangle before colouring a vertex on the second triangle: $$(r,r,r), (g, r, \\cdot), (r,r,r)$$ Then $A$ finishes with 2 edges. If at $(\\ast)$ $A$ picks $g$, $B$ can place it on the second triangle: $$(r,r,\\cdot), (g, g, \\cdot), (\\cdot, \\cdot, \\cdot)$$ $A$ can't pick $r$ or $g$ without losing a triangle, and then choosing the other colour for the pick after would", "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", "if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) --", "0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label(\"y\", (q+R)/2, NW); label(\"A'\", A, SW); label(\"B'\", B, SE); label(\"C'\", C, N); label(\"Q\", (q-(0,0.3))); label(\"R\", R, NE); label(\"S\", S, NE); label(\"T\", T, W); [/asy]$ Similarly, $\\triangle A'B'C'$ 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, $\\frac{x}{y} = \\frac{37}{35}$.", "similar right triangles, 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. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "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", "of having at least one boy is (A)7/8 (B)1/8 (C)5/8 (D)3/4 Concept: Basic Ideas of Probability Chapter: [5.01] Probability [5.01] Probability [2]9 In Figure 1, common tangents AB and CD to the two circles with centres 01and 0intersect at E. Prove that AB = CD. Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles Chapter: [3.01] Circles [3.01] Circles [2]10 The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC. Concept: Triangles Examples and Solutions Chapter: [3.02] Triangles [2]11 Two different dice are tossed together. Find the probability (i) that the number on each die is even. (ii) that the sum of numbers appearing on the two dice is 5. Concept: Basic Ideas of Probability Chapter: [5.01] Probability [5.01] Probability [2]12 If the total surface area of a solid hemisphere is 462 cm2 , find its volume.[Take π=22/7] Concept: Surface Area of a Combination of Solids Chapter: [7.02] Surface Areas and Volumes [2]13 Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. Concept: Sum of First n Terms of an AP Chapter: [2.02] Arithmetic Progressions [2]14 Find the values", "+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", "# Difference between revisions of \"2006 UNCO Math Contest II Problems/Problem 5\" ## Problem In the figure $BD$ is parallel to $AE$ and also $BF$ is parallel to $DE$. The area of the larger triangle $ACE$ is $128$. The area of the trapezoid $BDEA$ is $78$. Determine the area of triangle $ABF$. $[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); MP(\"A\",(4,0),SE);MP(\"C\",(1,2),N);MP(\"E\",(0,0),SW); MP(\"D\",(.5,1),W);MP(\"B\",(2.5,1),NE);MP(\"F\",(2,0),S); [/asy]$ ## Solution Since we know $BF$ is equal to the short side of both triangles and that $DB$ is equal to the long sides of the hypotenuse of the triangle. Thus we know that if we make a line $DF$ (as seen below), all four triangles in $\\triangle ACE$ are congruent. $[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); draw((1/2,1)--(2,0),black); MP(\"A\",(4,0),SE);MP(\"C\",(1,2),N);MP(\"E\",(0,0),SW); MP(\"D\",(.5,1),W);MP(\"B\",(2.5,1),NE);MP(\"F\",(2,0),S); [/asy]$ Then, since we know the sum of all four triangles is $128$, we can solve the easy division problem $128/4$ and see that the area of $\\triangle ABF = \\boxed{32}$", "# Difference between revisions of \"1998 AJHSME Problems/Problem 23\" ## Problem If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle? $[asy] unitsize(10); draw((0,0)--(12,0)--(6,6sqrt(3))--cycle); draw((15,0)--(27,0)--(21,6sqrt(3))--cycle); fill((21,0)--(18,3sqrt(3))--(24,3sqrt(3))--cycle,black); draw((30,0)--(42,0)--(36,6sqrt(3))--cycle); fill((34,0)--(32,2sqrt(3))--(36,2sqrt(3))--cycle,black); fill((38,0)--(36,2sqrt(3))--(40,2sqrt(3))--cycle,black); fill((36,2sqrt(3))--(34,4sqrt(3))--(38,4sqrt(3))--cycle,black); draw((45,0)--(57,0)--(51,6sqrt(3))--cycle); fill((48,0)--(46.5,1.5sqrt(3))--(49.5,1.5sqrt(3))--cycle,black); fill((51,0)--(49.5,1.5sqrt(3))--(52.5,1.5sqrt(3))--cycle,black); fill((54,0)--(52.5,1.5sqrt(3))--(55.5,1.5sqrt(3))--cycle,black); fill((49.5,1.5sqrt(3))--(48,3sqrt(3))--(51,3sqrt(3))--cycle,black); fill((52.5,1.5sqrt(3))--(51,3sqrt(3))--(54,3sqrt(3))--cycle,black); fill((51,3sqrt(3))--(49.5,4.5sqrt(3))--(52.5,4.5sqrt(3))--cycle,black); [/asy]$ $\\text{(A)}\\ \\frac{3}{8}\\qquad\\text{(B)}\\ \\frac{5}{27}\\qquad\\text{(C)}\\ \\frac{7}{16}\\qquad\\text{(D)}\\ \\frac{9}{16}\\qquad\\text{(E)}\\ \\frac{11}{45}$ ## Solution All small triangles are congruent in each iteration of the diagram. The number of shaded triangles follows the pattern: $0, 1, 3, 6, ...$ which is the pattern of \"triangular numbers\". Each time, the number $1, 2, 3, 4, 5...$ is added to the previous term. Thus, the first eight terms are: $0, 1, 3, 6, 10, 15, 21, 28$ In the eighth diagram, there will be $28$ shaded triangles. The total number of small triangles follows the pattern: $1, 4, 9, 16, ...$ which is the pattern of \"square numbers\". Thus, the eighth triangle will be divided into $8^2 = 64$ small triangles in total. The ratio of shaded to total triangles will be the fraction of the whole figure that's shaded, since all triangles are congruent. Thus, the answer is $\\frac{28}{64} = \\frac{7}{16}$, and the correct choice is $\\boxed{C}$ 1998 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 •", "By the Pythagorean Theorem on triangle $HKL$, $KL=\\sqrt{5}$. Now continue with solution 1. ## Solution 2 $[asy] size(10cm); pair A, B, C, D, H, K, O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label(\"O\", O, ENE); label(\"A\", A, NW); label(\"B\", B, W); label(\"C\", C, E); label(\"H\", H, E); label(\"H'\", K, NE); label(\"X\", X, W); label(\"Y\", Y, NE); label(\"O'\", P, E); label(\"M\", M, NE); label(\"L\", L, NE); label(\"D\", D, NNE); label(\"2\", X -- H, NW); label(\"3\", H -- A, SW); label(\"6\", H -- Y, NW); label(\"R\", O -- Y, E); dot(O); dot(P); dot(D); dot(H); [/asy]$ 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" ]

[ "be drawn using any three of these six points as vertices? (A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) $20$ AMC 8, 2001, Problem 24 Each half of this figure is composed of $3$ red triangles, $5$ blue triangles and 8 white triangles. When the upper half is folded down over the centerline, $2$ pairs of red triangles coincide, as do $3$ pairs of blue triangles. There are $2$ red-white pairs. How many white pairs coincide? (A) $4$ (B) $5$ (C) $6$ (D) $7$ (E) $9$ Try these AMC 8 Combinatorics Questions and check your knowledge! AMC 8, 2019, 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? $\\textbf{(A) }\\frac{1}{5}\\qquad\\textbf{(B) }\\frac{1}{4} \\qquad\\textbf{(C) }\\frac{2}{5} \\qquad\\textbf{(D) }\\frac{9}{20}\\qquad\\textbf{(E) }\\frac{1}{2}$ AMC 8, 2014, Problem 18 The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? $\\textbf{(A) }\\frac{4}{9}\\qquad\\textbf{(B) }\\frac{1}{2}\\qquad\\textbf{(C) }\\frac{5}{9}\\qquad\\textbf{(D) }\\frac{3}{5}\\qquad\\textbf{(E) }\\frac{2}{3}$ AMC 8, 2019, Problem", "# Streaming algorithm for counting triangles in a graph As described in the reference, the algorithm (see at the bottom) supposes to output an estimator $$\\hat T$$ for the # of triangles in a given graph $$G = (V, E)$$, denoted $$T$$. It is written that \"it can easily be shown\" that $$E[\\hat T] = T$$ But unfortunately, I'm not seeing it. Trying to analyze $$E[\\hat T]$$, I think as follows: • At line 1, denote the probability to randomly (and uniformly) choose an edge which is part of a triangle as $$p$$. Since triangles can share edges, $$\\frac T m \\le p \\le \\frac {3T} m$$ For example, consider the following case: The central triangle doesn't add new edges to the # of possibilities to choose an edge which is part of a triangle. You can imagine a different configuration, in which there are only the 3 outer triangles and they don't touch each other (in this configuration, we won't see the central 4th triangle). In both cases ((case i) 4 triangles as seen in the image; (case ii) 3 disjoint triangles), the probability to choose an edge which is part of a triangle is 1 (although the # of triangles is different). • At line 2, the probability to choose uniformly at random a vertex which", "total triangles, the answer is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$. ### Solution 3 For starters what I find helpful is to divide the whole octagon up into triangles as shown here: $[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot(\"A\",A,dir(45)); dot(\"B\",B,dir(90)); dot(\"C\",C,dir(135)); dot(\"D\",D,dir(180)); dot(\"E\",E,dir(-135)); dot(\"F\",F,dir(-90)); dot(\"G\",G,dir(-45)); dot(\"H\",H,dir(0)); dot(\"X\",X,dir(135/2)); dot(\"O\",O,dir(0)); draw(E--O--X); draw(C--O--B); draw(B--O--A); draw(A--O--H); draw(H--O--G); draw(G--O--F); draw(F--O--E); draw(E--O--D); draw(D--O--C); [/asy]$ Now it is just a matter of counting the larger triangles remember that $\\triangle BOX$ and $\\triangle XOA$ are not full triangles and are only half for these purposes. We count it up and we get a total of $\\frac{3.5}{8}$ of the shape shaded. We then simplify it to get our answer of $\\boxed{\\textbf{(D)}~\\frac{7}{16}}$.", "the diagram. There are 6 ways to colour its three sides in different colours. Next, look at triangle $B$. One of its sides has already been coloured, and there are 2 ways to colour the remaining sides. Continuing in this way clockwise round the diagram, there are two ways to colour each of the triangles up to and including the one labelled $C$. There are now two cases to consider for the remaining triangles $D$ and $E$. If side $x$ in triangle $A$ and side $y$ in triangle $C$ have different colours then there is only one choice for the colour of side $z$ and therefore only 1 way to colour triangles $D$ and $E$. But if $x$ and $y$ have the same colour then there are two choices for the colour of side $z$, and therefore 2 ways to colour triangles $D$ and $E$. Here's where the argument becomes probabilistic and unreliable. If the colours of $x$ and $y$ had been chosen independently at random, then the probability of them being the same would be $\\frac13$, so the expected number of colourings for triangles $D$ and $E$ would be $\\frac13*2 + \\frac23*1 = \\frac43$. Then the expected value for the number of colourings for the whole diagram would be $$6*2^9 * \\frac43 = 2^{12}.$$ But in fact", "# $2$-colorings of triangles, resulting in $(2\\ 2)$-colorings of all tetrahedra DEFINITIONS: Functions $c : \\binom X3\\rightarrow \\{0\\ 1\\}$ are called 2-colorings of triangles in $X$. The $4$-element subsets $A\\subseteq X$ are called tetrahedra. Each 2-coloring $c$ of triangles induces $(\\alpha\\ \\beta)$-coloring of each tetrahedron $A$, where $$\\beta := \\sum_{T\\subseteq A,\\ |T|=3}\\ c(T)\\quad\\quad\\quad \\alpha := 4-\\beta$$ QUESTION: Does every set $X$ admit a 2-coloring $c$ of its triangles such that the induced coloring of every tetrahedron is of the $(2\\ 2)$ type? And if the answer is NOT, then what is the smallest cardinality $|X|$ for which $X$ does not admit such $c$ (then such cardinality must be finite)? A PARTIAL RESULT: If $|X| \\le 6$ then there exists a 2-coloring of triangles such that all tetrahedra are colored $(2\\ 2)$. (Of course, if such a coloring exists for a set $X$ then it exists--it induces--a similar coloring of triangles in every subset of $X$, hence together with any good cardinal number all smaller cardinal numbers are good too). - When $|X|=7$, no such coloring exists. In this case there are 35 triangles and 35 tetrahedra. If $A$ is a tetrahedron, let $(\\alpha_A,\\beta_A)$ be its type. Each triangle in $X$ is a subset of exactly $4$ tetrahedra, so if $$\\sum_{T\\subseteq X, |T|=3}c(T)=n$$ then $$\\sum_{A\\subseteq X, |A|=4} \\beta_A =", "$$\\frac T m \\le p \\le \\frac {3T} m$$ For example, consider the following case: The central triangle doesn’t add new edges to the # of possibilities to choose an edge which is part of a triangle. You can imagine a different configuration, in which there are only the 3 outer triangles and they don’t touch each other (in this configuration, we won’t see the central 4th triangle). In both cases ((case i) 4 triangles as seen in the image; (case ii) 3 disjoint triangles), the probability to choose an edge which is part of a triangle is 1 (although the # of triangles is different). • At line 2, the probability to choose uniformly at random a vertex which \"closes a triangle\" with the edge from the previous step is exactly $$\\frac 1 {n-2}$$. Therefore I only see that $$T \\le E[\\hat T] \\le 3T$$ What am I missing? Another question I have is regarding line 3. The stream is ordered, and we first pick a random edge $$(u, v)$$ (line 1), then a random vertex $$w$$ from $$V \\backslash \\{u, v\\}$$ (line 2). I feel that the analysis should take into account that at line 3 we check whether $$(u, w)$$ and $$(v, w)$$ appear after $$(u, v)$$ in the stream. Maybe after I’ll understand the", "\\end{tikzpicture} Start with the triangle labelled $A$ at the bottom of the diagram. There are 6 ways to colour its three sides in different colours. Next, look at triangle $B$. One of its sides has already been coloured, and there are 2 ways to colour the remaining sides. Continuing in this way clockwise round the diagram, there are two ways to colour each of the triangles up to and including the one labelled $C$. There are now two cases to consider for the remaining triangles $D$ and $E$. If side $x$ in triangle $A$ and side $y$ in triangle $C$ have different colours then there is only one choice for the colour of side $z$ and therefore only 1 way to colour triangles $D$ and $E$. But if $x$ and $y$ have the same colour then there are two choices for the colour of side $z$, and therefore 2 ways to colour triangles $D$ and $E$. Here's where the argument becomes probabilistic and unreliable. If the colours of $x$ and $y$ had been chosen independently at random, then the probability of them being the same would be $\\frac13$, so the expected number of colourings for triangles $D$ and $E$ would be $\\frac13*2 + \\frac23*1 = \\frac43$. Then the expected value for the number of colourings for the whole", "$A$ and $B$, and I'll refer to edges by listing the vertices at each end, so $AB$ is the edge between $A$ and $B$. For each, I'm going to pick out a \"favourite\" colouring. My criteria is that I want to pick the one that I feel best shows off the relevant symmetry. So it might not be the most symmetrical in each set. 1. Four cycle, e.g., $\\left(ADBC\\right)$. One way to view this is to picture the tetrahedron looking down on an edge, as in Figure 1. Giving it a quarter turn, say anticlockwise, brings the vertices round so that $A$ is above where $D$ was, $D$ is below where $B$ was, and so on. Then we reflect the tetrahedron in a horizontal mirror so that $A$ descends to exactly where $D$ was and $D$ ascends to where $B$ was. The orbits of this particular four cycle are: $AB\\to CD,\\phantom{\\rule{1em}{0ex}}$ This means that the pieces we can use to construct a colouring are: • $\\left\\{\\left\\{AB,CD\\right\\}\\right\\}$ or $\\left\\{\\left\\{AB\\right\\},\\left\\{CD\\right\\}\\right\\}$ • $\\left\\{\\left\\{AD,DB,BC,CA\\right\\}\\right\\}$, or $\\left\\{\\left\\{AD,BC\\right\\},\\left\\{DB,CA\\right\\}\\right\\}$, or as singletons We now do a \"Countdown\" choice: one from the top and one from the bottom. On the face of it, this gives $6$ choices. We actually have a few more because we can potentially combine sets from different rows. So if we choose", "are the numbers in row 4 of the triangle. (a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4 ### Combinations For a mathematical combination of the form nCr, the solution can be found at row n, element r of Pascal's triangle. For the combination 5C3, the solution is the number at the 3rd element of the 5th row. 5C3 = 10 ### Pascal's Rule Pascal's rule states that: ${n \\choose r} = {n-1 \\choose r-1} + {n-1 \\choose r}$ In terms of the triangle, this simply means that an entry is the sum of two diagonal entries above it. It is easy to prove: ${n-1 \\choose r-1} + {n-1 \\choose r}$ $= \\frac{(n-1)!}{(r-1)!(n-1-(r-1)!}+ \\frac{(n-1)!}{r!(n-1-r)!}$ $= \\frac{ (n-1)!r + (n-1)!(n-r)}{r!(n-r)!}$ $= \\frac{ (n-1)!( r + (n-r)}{r!(n-r)!}$ $= \\frac{n!}{r!(n-r)!} = {n \\choose r}.$", "for two more different triangles by moving point B to a different location (to grab a point, press the ALPHA button and use the arrow keys to move it.) Triangle #2 $\\frac{BC}{AC}=\\frac{\\;\\;\\;\\;\\;\\;\\;\\;\\;}{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\frac{AC}{AB}=\\frac{\\;\\;\\;\\;\\;\\;\\;\\;\\;}{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\frac{BC}{AB}=\\frac{\\;\\;\\;\\;\\;\\;\\;\\;\\;}{\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ $\\text{Sin} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\text{Cos} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\text{Tan} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ Triangle #3 $\\frac{BC}{AC}=\\frac{\\;\\;\\;\\;\\;\\;\\;\\;\\;}{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\frac{AC}{AB}=\\frac{\\;\\;\\;\\;\\;\\;\\;\\;\\;}{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\frac{BC}{AB}=\\frac{\\;\\;\\;\\;\\;\\;\\;\\;\\;}{\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ $\\text{Sin} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\text{Cos} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;}, \\text{Tan} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ Based upon your answers hypothesize which ratio goes with each trigonometric function. $\\text{Sin} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} \\quad \\text{Cos} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} \\quad \\text{Tan} \\ A = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ A good acronym to use to help remember these relationships is SOHCAHTOA $&\\sin \\ A=\\frac{\\text{Opposite}}{\\text{Hypotenuse}}\\\\&\\cos \\ A=\\frac{\\text{Adjacent}}{\\text{Hypotenuse}}\\\\&\\tan \\ A=\\frac{\\text{Opposite}}{\\text{Adjacent}}$ ## Problem 2 – Trigonometry, What Is It Good For? One of the uses of trigonometry is finding missing side lengths of a triangle. To find the length of side $BC$ in the triangle to the right, write the sine relationship. Find $BC$. Now solve for $BC$ and calculate using your handheld. To find the length of side $AC$ in the triangle to the right, write the cosine relationship. Find $AC$. Now solve for $AC$ and calculate using your handheld. To find the length of side $AC$ in the triangle to the right, write the tangent relationship. Find $AC$. Now solve", "# Finding all “bad” triangles in a full-simple-graph Let $G = \\left\\langle {V,E} \\right\\rangle$, a simple and complete graph with the size of $n$. Each edge in the graph can be colored with blue or red. A \"bad\" triangle defined to be a triangle (circle with three edges) such that it contains two colors (two blue edges and one red and vice-verca). Also, $d_i$ is the number for blue edges for the vertex $i$. It appears that the number of bad triangles defined by the following formula which I'm trying to understand: $$\\sum\\limits_{i = 1}^n {\\frac{{({d_i} \\cdot (n - 1 - {d_i}))}}{2}}$$ Trying to interpret it: for each vertex, the number of bad triangles is the number of blue edges multiplied by the number of red edges divided by $2$. But what is standing behind this? Since the graph is complete, all edges are present. Thus, if we find a red and a blue edge with endpoints at a vertex $A$, we know the triangle with those two edges and the third edge joining them must have at least two colors. For every vertex, we can calculate the number of bad triangles found in this way by multiplying the number of red and blue edges of the vertex. We count every triangle twice now, because every bad triangle", "a cubic Bézier triangle An advantage of Bézier triangles in computer graphics is that dividing the Bézier triangle into two separate Bézier triangles requires only addition and division by two, rather than floating point arithmetic. This means that while Bézier triangles are smooth, they can easily be approximated using regular triangles by recursively dividing the triangle in two until the resulting triangles are considered sufficiently small. The following computes the new control points for the half of the full Bézier triangle with the corner α3, a corner halfway along the Bézier curve between α3 and β3, and the third corner γ3. ${\\displaystyle {\\begin{pmatrix}{\\boldsymbol {\\alpha ^{3}}}{'}\\\\{\\boldsymbol {\\alpha ^{2}\\beta }}{'}\\\\{\\boldsymbol {\\alpha \\beta ^{2}}}{'}\\\\{\\boldsymbol {\\beta ^{3}}}{'}\\\\{\\boldsymbol {\\alpha ^{2}\\gamma }}{'}\\\\{\\boldsymbol {\\alpha \\beta \\gamma }}{'}\\\\{\\boldsymbol {\\beta ^{2}\\gamma }}{'}\\\\{\\boldsymbol {\\alpha \\gamma ^{2}}}{'}\\\\{\\boldsymbol {\\beta \\gamma ^{2}}}{'}\\\\{\\boldsymbol {\\gamma ^{3}}}{'}\\end{pmatrix}}={\\begin{pmatrix}1&0&0&0&0&0&0&0&0&0\\\\{1 \\over 2}&{1 \\over 2}&0&0&0&0&0&0&0&0\\\\{1 \\over 4}&{2 \\over 4}&{1 \\over 4}&0&0&0&0&0&0&0\\\\{1 \\over 8}&{3 \\over 8}&{3 \\over 8}&{1 \\over 8}&0&0&0&0&0&0\\\\0&0&0&0&1&0&0&0&0&0\\\\0&0&0&0&{1 \\over 2}&{1 \\over 2}&0&0&0&0\\\\0&0&0&0&{1 \\over 4}&{2 \\over 4}&{1 \\over 4}&0&0&0\\\\0&0&0&0&0&0&0&1&0&0\\\\0&0&0&0&0&0&0&{1 \\over 2}&{1 \\over 2}&0\\\\0&0&0&0&0&0&0&0&0&1\\end{pmatrix}}\\cdot {\\begin{pmatrix}{\\boldsymbol {\\alpha ^{3}}}\\\\{\\boldsymbol {\\alpha ^{2}\\beta }}\\\\{\\boldsymbol {\\alpha \\beta ^{2}}}\\\\{\\boldsymbol {\\beta ^{3}}}\\\\{\\boldsymbol {\\alpha ^{2}\\gamma }}\\\\{\\boldsymbol {\\alpha \\beta \\gamma }}\\\\{\\boldsymbol {\\beta ^{2}\\gamma }}\\\\{\\boldsymbol {\\alpha \\gamma ^{2}}}\\\\{\\boldsymbol {\\beta \\gamma ^{2}}}\\\\{\\boldsymbol {\\gamma ^{3}}}\\end{pmatrix}}}$ equivalently, using addition and division by two only, ${\\displaystyle {\\begin{matrix}&&\\beta ^{3}:=(\\alpha \\beta ^{2}+\\beta ^{3})/2\\\\&\\alpha \\beta ^{2}:=(\\alpha ^{2}\\beta +\\alpha \\beta ^{2})/2&\\beta ^{3}:=(\\alpha \\beta ^{2}+\\beta ^{3})/2\\\\\\alpha ^{2}\\beta :=(\\alpha", "triangles} \\cdot 6 \\text{ triangles}= 54$ vpdoo wuldqjohv lq wkh zkroh khadjrq $ABCDEF$. Wkxv, wkh dqvzhu lv $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$.", "been only 3 unordered combinations: AB+CD, AC+BD and AD+BC. For N=3 and m=2 we get 10 (unless I have missed some): 1. ABC+DEF 2. ABD+CEF 3. ABE+CDF 4. ABF+CDE 5. ACD+BEF 6. ACE+BDF 7. ACF+BDE There are $$24 \\choose 6$$ ways to choose the first pack, $$18 \\choose 6$$ ways to choose the second, $$12 \\choose 6$$ to choose the third, and $$6 \\choose 6$$ ways to choose the last. We have overcounted by a factor $$4!$$ because we could choose the same four packs in any order. This gives $${24 \\choose 6}{18 \\choose 6}{12 \\choose 6}{6 \\choose 6}\\frac 1{4!}=\\frac {24!}{6!^4\\cdot 4!}\\approx 9.62\\cdot 10^{10}$$ Generally is is $$\\frac{(mN)!}{(N!)^m\\cdot m!}$$ • $N \\choose m$ is the number of combinations of $m$ things out of $N$ without regard to order. It equals $\\frac {N!}{m!(N-m)!}$ – Ross Millikan Apr 21 at 14:11", "\\right )^2 - \\frac{1}{2} \\sum_i d_i$. (This optimization is not necessary, you can just code the $O(n^2)$ one) https://atcoder.jp/contests/abc143/submissions/8037254C) The number of slimes is the same as the number of letters that start a slime. I.e. the number of i's such that $s[i] \\neq s[i-1]$. https://atcoder.jp/contests/abc143/submissions/8037631D) I considered three different types of triangles with length $a \\le b \\le c$: (1) $a \\le b < c$, (2) $a = b = c$, (3) $a < b = c$. As a helper-array I computed for each l: (i) the number of triangles with length $= l$ and (ii) the number of triangles with length $\\le l$. For (1) you can just iterate through all pairs $(a, b)$ and count the number of values c such that $\\text{max}(a,b) < c < a+b$. You can do this using the array of type (ii). For (2) you can just use the array of type (i) and use that the number of triples of k numbers is $\\frac{k(k-1)(k-2)}{6}$. For (3) you can iterate through all values $l$. The number of pairs (b,c) such that b=c=l you can get from the array of type (i), and then you can get the number of values for a using the array of type (ii). https://atcoder.jp/contests/abc143/submissions/8039964E) Here I used a standard Dijkstra with a modification. The distance", "# Streaming algorithm for counting triangles in a graph As described in the reference, the algorithm (see at the bottom) supposes to output an estimator $$\\hat T$$ for the # of triangles in a given graph $$G = (V, E)$$, denoted $$T$$. It is written that \"it can easily be shown\" that $$E[\\hat T] = T$$ But unfortunately, I’m not seeing it. Trying to analyze $$E[\\hat T]$$, I think as follows: • At line 1, denote the probability to randomly (and uniformly) choose an edge which is part of a triangle as $$p$$. Since triangles can share edges, $$\\frac T m \\le p \\le \\frac {3T} m$$ For example, consider the following case: The central triangle doesn’t add new edges to the # of possibilities to choose an edge which is part of a triangle. You can imagine a different configuration, in which there are only the 3 outer triangles and they don’t touch each other (in this configuration, we won’t see the central 4th triangle). In both cases ((case i) 4 triangles as seen in the image; (case ii) 3 disjoint triangles), the probability to choose an edge which is part of a triangle is 1 (although the # of triangles is different). • At line 2, the probability to choose uniformly at random a vertex which", "# Difference between revisions of \"2018 AMC 8 Problems/Problem 23\" ## Problem From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon? $[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]$ $\\textbf{(A) } \\frac{2}{7} \\qquad \\textbf{(B) } \\frac{5}{42} \\qquad \\textbf{(C) } \\frac{11}{14} \\qquad \\textbf{(D) } \\frac{5}{7} \\qquad \\textbf{(E) } \\frac{6}{7}$ ## Solutions ### Solution 1 We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and also $4$ places to put the remaining point, so we have $8\\cdot4$ choices. The total amount of choices is ${8 \\choose 3} = 8\\cdot7$. Thus our answer is $\\frac{8+8\\cdot4}{8\\cdot7}= \\frac{1+4}{7}=\\boxed{\\textbf{(D) } \\frac 57}$ ### Solution 2 We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "# Expected number of triangles in G(n,m) Consider picking a graph with m edges on n vertices uniformly at random from the set of all graphs with n vertices and m edges. What is the expected number of triangles? Before we give an exact formula, consider the following heuristic. We can approximate this graph with a random binomial graph $G(n,p)$ (each edge appears independently from the others with probability $p$) where $p=\\frac{m}{{n \\choose 2}}\\approx \\frac{2m}{n^2}$. Therefore the expected number of triangles $T$ in $G$ should be roughly $E[T]=p^3{n \\choose 3}\\approx\\frac{8m^3}{n^6}\\frac{n^3}{6}\\sim \\frac{4m^3}{3n^3}$. This heuristic can be turned into a rigorous statement by using the asymptotic equivalence of the two random graph models. However this is a good asymptotic approximation. What about an exact formula? The simplest way is counting! How many graphs (let me mention even a bit late that we are talking about labelled graphs, which means that we have names for the vertices. For convenience we call the names by the integers 1 through n) on n vertices do we have? Clearly ${{n \\choose 2} \\choose m}$. Consider now a triangle formed by three distinct vertices $i,j,k$. What is the probability of seeing this particular triangle? We have to count how many graphs contain it and divide by the total number of possible graphs. Clearly we have", "### Home > INT2 > Chapter 2 > Lesson 2.3.1 > Problem2-85 2-85. One possible proportion for the similar triangles below is $\\frac { A C } { A B } = \\frac { D F } { D E }$. Write at least three more proportions given that $∆ABC\\sim ∆DEF$. Using this proportion, think of at least two more proportions to satisfy the question. $\\frac{BC}{EF}= \\frac{AB}{DE}$" ]

[ "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]", "(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) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ 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.", "= \\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),", "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.", "label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); [/asy]$ NOTE: Connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices. Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\\frac{8 \\cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\\boxed{\\textbf{(C) }80}$. $[asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label(\"x=2\",(1,0,0),S); label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); label(\"A\",(0,2,0), NW); label(\"B\",(2,0,2), NW); [/asy]$ ### Solution 2 (rigorous) The three-dimensional distance formula shows that the lengths of the equilateral triangle must be $\\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \\le d_x, d_y, d_z \\le 2$, which yields the possible edge lengths of $\\sqrt{0^2+0^2+1^2}=\\sqrt{1},\\ \\sqrt{0^2+1^2+1^2}=\\sqrt{2},\\ \\sqrt{1^2+1^2+1^2}=\\sqrt{3},$ $\\ \\sqrt{0^2+0^2+2^2}=\\sqrt{4},\\ \\sqrt{0^2+1^2+2^2}=\\sqrt{5},\\ \\sqrt{1^2+1^2+2^2}=\\sqrt{6},$ $\\ \\sqrt{0^2+2^2+2^2}=\\sqrt{8},\\ \\sqrt{1^2+2^2+2^2}=\\sqrt{9},\\ \\sqrt{2^2+2^2+2^2}=\\sqrt{12}$ Some casework shows that $\\sqrt{2},\\ \\sqrt{6},\\ \\sqrt{8}$ are the only lengths that work, from which we can use the same counting argument as above.", "-- 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 $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$.", "(-1/3,-1) -- (-1,-1/3) -- (-1,1/3) -- (-1/3,1) -- cycle; filldraw(s, lightred, black); filldraw(t, lightgreen, black); filldraw(s_cap_t, lightyellow, black); draw( (-5/3,0) -- (5/3,0), dashed ); draw( (0,-5/3) -- (0,5/3), dashed ); [/asy]$ Coordinate axes are dashed, $S$ is shown in red, $T$ in green and their intersection is yellow. The intersections of the boundary of $S$ and $T$ are obviously at $(\\pm 1,\\pm 1/3)$ and at $(\\pm 1/3,\\pm 1)$. Hence each of the four red triangles is an isosceles right triangle with legs long $\\frac 23$, and hence the area of a single red triangle is $\\frac 12 \\cdot \\left( \\frac 23 \\right)^2 = \\frac 29$. Then the area of all four is $\\frac 89$, and therefore the area of $S\\cap T$ is $4 - \\frac 89$. Then the probability we seek is $\\frac{ [S\\cap T]}4 = \\frac{ 4 - \\frac 89 }4 = 1 - \\frac 29 = \\boxed{\\frac 79}$. (Alternately, when we got to the point that we know that a single red triangle is $\\frac 29$, we can directly note that the picture is symmetric, hence we can just consider the first quadrant and there the probability is $1 - \\frac 29 = \\frac 79$. This saves us the work of first multiplying and then dividing by $4$.)", "two complete triangles. Eventually, either A chooses another red giving B triangle 2 or chooses all greens which gives B triangle 3 and A triangle 2. Therefore, B wins $7$ to $2$. This is a rather tentative answer, as I don't have a proof that this is the best strategy on either part, but it might be useful none the less. The first pick of course doesn't matter, so say it's red ($r$), so we have the triangles: $$(r,\\cdot,\\cdot), (\\cdot, \\cdot, \\cdot), (\\cdot, \\cdot, \\cdot)$$ If $A$ picks $r$ again, $B$ can put it on the first triangle, which is a loss for $A$, then applying the same idea, $A$ can pick $$(r,\\cdot,\\cdot), (g, \\cdot, \\cdot), (\\cdot, \\cdot, \\cdot)$$ Then no matter whether $A$ picks $r$ or $g$, $B$ can place it in $B$'s favour (so say it's $r$) $$(r,r,\\cdot), (g, \\cdot, \\cdot), (\\cdot, \\cdot, \\cdot)$$ $(\\ast)$If $A$ keeps picking $r$, $B$ can force two monochrome triangle before colouring a vertex on the second triangle: $$(r,r,r), (g, r, \\cdot), (r,r,r)$$ Then $A$ finishes with 2 edges. If at $(\\ast)$ $A$ picks $g$, $B$ can place it on the second triangle: $$(r,r,\\cdot), (g, g, \\cdot), (\\cdot, \\cdot, \\cdot)$$ $A$ can't pick $r$ or $g$ without losing a triangle, and then choosing the other colour for the pick after would", "(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, $\\frac{x}{y} = \\frac{37}{35}$.", "Solution 2 $[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label(\"b\",(2.5,0),S); label(\"a\",(0,1.5),W); label(\"c\",(2.5,1),W); label(\"A\",(0.5,2.5),W); label(\"B\",(3.5,0.75),W); label(\"C\",(1,1),W); [/asy]$ From the diagram from the previous solution, we have $a$, $b$ as the legs and $c$ as the side length of the square. WLOG, let the area of triangle $A$ be $m$ times the area of square $C$. Since triangle $A$ is similar to the large triangle, it has $h_A = a(\\frac{c}{b}) = \\frac{ac}{b}$, $b_A = c$ and $$[A] = \\frac{bh}{2} = \\frac{ac^2}{2b} = m[C] = mc^2$$ Thus $\\frac{a}{2b} = m$ Now since triangle $B$ is similar to the large triangle, it has $h_B = c$, $b_B = b\\frac{c}{a} = \\frac{bc}{a}$ and $$[B] = \\frac{bh}{2} = \\frac{bc^2}{2a} = nc^2 = n[C]$$ Thus $n = \\frac{b}{2a} = \\frac{1}{4(\\frac{a}{2b})} = \\frac{1}{4m}$. $\\text{(D)}$. ~ Nafer ### Solution 3 (process of elimination) Simply testing specific triangles is sufficient. A triangle with legs of 1 and 2 gives a square of area $S=\\frac{2}{3}\\frac{2}{3}=\\frac{4}{9}$. The larger sub-triangle has area $T_1=\\frac{\\frac{2}{3}\\frac{4}{3}}{2}=\\frac{4}{9}$, and the smaller triangle has area $T_2=\\frac{\\frac{2}{3}\\frac{1}{3}}{2}=\\frac{1}{9}$. Computing ratios you get $\\frac{T_1}{S}=1$ and $\\frac{T_2}{S}=\\frac{1}{4}$. Plugging $m=1$ in shows that the only possible answer is $\\text{(D)}$ ~ Snacc", "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 $AD=AE + EF + FD =", "my armpit Dec 29 '13 at 20:11 • See, we have a constraint of symmetricity? Do you have others that's what I asked. – percusse Dec 29 '13 at 20:21 Here is the Asymptote version, which does not use the general Asymptote commands to split a path (guide), it just performs a simple subdivision of the cubic Bezier segment: // split.asy: size(5cm); pair A,B,C,D; A=(3,3); B=(1,1); C=(-1,1); D=(-3,3); pair P,B1,C1,B2,C2; P=(A+D+3(B+C))/8; B1 = (A+B)/2; C1 = ((A+C)/2+B)/2; C2 = (D+C)/2; B2 = ((D+B)/2+C)/2; guide g=A..controls B and C..D; guide gl=A..controls B1 and C1..P; guide gr=P..controls B2 and C2..D; draw(g); draw(gl,deepgreen+1.6bp+opacity(0.5)); draw(gr,red+1.6bp+opacity(0.5)); draw((0,3)--(0,-1),red); dot(A--B--C--D--P,UnFill); dot(B1--C1,deepgreen,UnFill); dot(B2--C2,red,UnFill); label(\"$A$\",A,SE); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$D$\",D,SW); label(\"$P$\",P,NE); label(\"$B_1$\",B1,E,deepgreen); label(\"$C_1$\",C1,E,deepgreen); label(\"$B_2$\",B2,W,red); label(\"$C_2$\",C2,W,red); To get a standalone split.pdf, run asy -f pdf split.asy. • I really need the converting formulae. – kiss my armpit Dec 29 '13 at 19:29 • @Stiff Jokes: Btw, the Bezier curve is a very interesting object, you might enjoy studying it in a spare time. – g.kov Dec 29 '13 at 19:43 A translated version for @g.kov's Asymptote answer in PSTricks. \\documentclass[pstricks,border=24pt]{standalone} \\usepackage{pst-eucl} \\begin{document} \\begin{pspicture}[showgrid=true](-3,-1)(3,3) \\pstGeonode (3,3){A} (1,1){B} (-1,1){C} (-3,3){D} \\psbezier(A)(B)(C)(D) \\psline[linecolor=red](0,3)(0,-1) \\nodexn{.125(A)+.125(D)+.375(B)+.375(C)}{R'} \\nodexn{.5(A)+.5(B)}{P'} \\nodexn{.25(A)+.5(B)+.25(C)}{Q'} \\pstGeonode (P'){P} (Q'){Q} (R'){R} \\psbezier[linecolor=red](A)(P)(Q)(R) \\end{pspicture} \\end{document}", "similar right triangles, 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. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "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", "0.973); 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]$ Similarly, $\\triangle A'B'C'$ 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, $\\frac{x}{y} = \\frac{37}{35}$.", "= (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,E); label(\"F\",F,S); label(\"G\",G,W); label(\"H\",H,N); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ $\\textbf{(A)}\\ \\dfrac1{12}\\qquad\\textbf{(B)}\\ \\dfrac{\\sqrt3}{18}\\qquad\\textbf{(C)}\\ \\dfrac{\\sqrt2}{12}\\qquad\\textbf{(D)}\\ \\dfrac{\\sqrt3}{12}\\qquad\\textbf{(E)}\\ \\dfrac16$ ## Problem 17 Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die? $\\textbf{(A)}\\ \\dfrac16\\qquad\\textbf{(B)}\\ \\dfrac{13}{72}\\qquad\\textbf{(C)}\\ \\dfrac7{36}\\qquad\\textbf{(D)}\\ \\dfrac5{24}\\qquad\\textbf{(E)}\\ \\dfrac29$ ## Problem 18 A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $1$, $4$, and $5$. What is the area of the square? $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 17\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 27$ ## Problem 19 Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\\overline{XY}$ contained in the cube with edge length $3$? $\\textbf{(A)}\\ \\dfrac{3\\sqrt{33}}5\\qquad\\textbf{(B)}\\ 2\\sqrt3\\qquad\\textbf{(C)}\\ \\dfrac{2\\sqrt{33}}3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 3\\sqrt2$ $[asy] dotfactor = 3; size(10cm); dot((0, 10)); label(\"X\", (0,10),W,fontsize(8pt)); dot((6,2)); label(\"Y\", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10)); draw((1, 10)--(1.5,10.5)); draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7)); draw((2,9)--(3,10)); draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4)); draw((3,7)--(4.5,8.5)); draw((4.5,6)--(6,6)--(6,2)--(4,0)); draw((4,4)--(6,6)); label(\"1\", (1,9.5), W,fontsize(8pt)); label(\"2\", (2,8), W,fontsize(8pt)); label(\"3\", (3,5.5), W,fontsize(8pt)); label(\"4\", (4,2), W,fontsize(8pt)); [/asy]$ Solution ## Problem 20 The product $(8)(888\\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$? $\\textbf{(A)}\\ 901\\qquad\\textbf{(B)}\\" ]

[ "more code, but nicer output) and also give it a slightly cleaner look: $$\\small \\begin{array}{lcc} \\hline \\text{Beverage} & \\text{Carbohydrate Content (%)} & \\text{pH} \\\\ \\hline \\text{Beer} & 3.0-5.0 & 4.1-4.5 \\\\ \\text{Wine} & 0.0-12.9 & 2.8-3.8 \\\\ \\text{Sweet liquor } & 30.0-31.0 & 3.3-3.9 \\\\ \\text{Strong alcohol} & 0.0-1.2 & 6.5-6.9 \\\\ \\text{Vodka} & \\text{very little} & 6.0-7.0 \\\\ \\text{Coca Cola} & 10.6(?) & 2.8 \\\\ \\hline \\end{array}$$ Align at point: Define for easier typsetting \\newcommand{\\d}[2]{#1.&\\hspace{-1em}#2} and use in table 1.23 as \\d{1}{23} $$\\newcommand{\\d}[2]{#1.&\\hspace{-1em}#2} \\begin{array}{lrl} \\hline \\text{Beverage} & \\text{Nomnom}&\\hspace{-1em}\\text{-factor} \\\\ \\hline \\text{Beer} & 100.&\\hspace{-1em}000 \\\\ \\text{Wine} & 23.&\\hspace{-1em}4567\\\\ \\text{Cola stuff} & 1.&\\hspace{-1em}23\\\\ \\text{Coffee} & 66.&\\hspace{-1em}6\\\\ \\text{Tea (black)}&\\d{0}{3}\\\\ \\hline \\end{array}$$ The mhchem package is not enabled on StackPrinter? Testcode: $$\\ce{2H2 + O2 -> 2H2O}$$ $$\\ce{2H2 + O2 -> 2H2O}$$ Enable mhchemmanually via $\\require{mhchem}$ $\\require{mhchem}$ and try again: $$\\ce{2H2 + O2 -> 2H2O}$$ $\\mathcal{M}_{ath}\\mathrm{J^{a}X}\\neq\\LaTeX$ Operation Broken Arrow $$\\require{cancel}\\ce{\\bcancel{->}\\cancel{->}\\xcancel{->}}$$ $$%\\require{Extpfeil}\\Newextarrow{\\xrightharpoonup}{5,10}{0x21C0}\\xrightharpoonup{\\not{}}$$ $$\\nrightarrow$$ $$\\ce{\\rlap{~~/\\!/}->}$$ $$\\def\\nnot{\\color{red}{\\rlap{~~/\\!/}}}$$ $$\\ce{\\nnot->T[water]}$$ $$\\begin{multline} E^\\ominus(\\ce{Fe/Fe^{3+}}) - \\frac13\\cdot \\mathcal{R}T\\mathrm{F}^{-1} \\ln\\frac{\\ce{Fe}}{\\ce{Fe^{3+}}} = \\\\ a\\cdot\\left(E^\\ominus(\\ce{Fe/Fe^{2+}}) - \\frac12\\cdot \\mathcal{R}T\\mathrm{F}^{-1} \\ln\\frac{\\ce{Fe}}{\\ce{Fe^{2+}}}\\right) \\\\+ b\\cdot\\left(E^\\ominus(\\ce{Fe^{2+}/Fe^{3+}}) - \\frac11\\cdot \\mathcal{R}T\\mathrm{F}^{-1} \\ln\\frac{\\ce{Fe^{2+}}}{\\ce{Fe^{3+}}}\\right) \\end{multline}$$ # The Tag Test \\begin{array}{|c|l|c|c|c|}\\hline \\# & {\\small \\bf Questions} & {\\small \\rm \\color{green}{Yes}} & {\\small \\rm \\color{#999999}{Partially}} & {\\small \\rm \\color{red}{No}} \\\\\\hline \\hline 1&\\small\\text{Does 'experts on X' make sense?} & & & \\\\\\hline 2&\\small\\text{Does 'I'm studying X' make sense?} & & & \\\\\\hline 3&\\small\\text{Would you leave", "in the GIMP prior to dropping it in Inkscape to get the images I wanted. figure caffeine4=[0 0 0 0 0 0 0 0]; sleep4=[6.95 10.87 9.75 7.52 10.95 9.43 8.75 10.23]; day=[25 26 27 28 29 30 31 32]; [AX,H1,H2] = plotyy(day,sleep4,day,caffeine4,'plot'); title('Phase 4: Daily Hours Slept and Caffeine Intake'); xlabel('Date (in April)'); %xlim([26 13]); set(gca,'XTick',25:1:32); %set(gca,'box','off'); set(get(AX(1),'Ylabel'),'String','Amount Slept Daily (hrs)'); set(get(AX(2),'Ylabel'),'String','Daily Caffeine Intake (mg)') ; axis(AX(1),[25 32 0 11]); axis(AX(2),[25 32 0 700]); set(AX(1),'YLim',[0 11], 'YTick',0:1:11, 'box','off'); set(AX(2),'YLim',[0 700],'YTick',0:100:700,'box','off', 'box','off'); set(H1,'LineWidth',3,... 'MarkerSize',7,... 'LineStyle','-',... 'Marker', 'o',... 'MarkerEdgeColor', 'none',... 'MarkerFaceColor', 'b'); set(H2,'LineWidth',3,... 'MarkerSize',7,... 'LineStyle','-',... 'Marker', 'o',... 'MarkerEdgeColor', 'none',... 'MarkerFaceColor', 'g'); Above code resulted in this graph: This data was actually 25 April to 2 May, but instead of learning about timeseries in MATLAB, I just cheated and stuck \"April / May\" in the xlabel and used 31 and 32 to not break my code. I planned to post-process in gimp to turn \"31 32\" into \"1 2\" but forgot to do so. For phase 3, I couldn't figure how to get rid of the 20.5 and 21.5 tick marks when I expanded to graphs to reasonable size, so I did post-processing in GIMP. Here are the two images I smushed together into the third: += Misc. other tricks too lazy to document fully: Using \"shift-J\" in Vim to", "$[asy] defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype(\"4 4\"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label(\"1/\\sqrt{3}\",(0,-0.5),W,fontsize(8)); [/asy]$ $\\Longrightarrow$ $[asy] defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype(\"4 4\")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype(\"4 4\")); draw(Circle((0,0),1),linetype(\"4 4\")); label(\"\\sqrt{3}\",(0,-0.5),W,fontsize(8)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); [/asy]$ The area of the regular hexagon is $6 \\cdot \\left( \\frac{\\left(\\sqrt{3}\\right)^2 \\sqrt{3}}{4} \\right) = \\frac{9}{2}\\sqrt{3}$. The total area of the six $120^{\\circ}$ sectors is $6\\left(\\frac{1}{3}\\pi - \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\sqrt{3}\\right) = 2\\pi - \\frac{3}{2}\\sqrt{3}$. Their sum is $2\\pi + \\sqrt{27}$, and $a+b = \\boxed{029}$. - Th3Numb3rThr33 ## Solution 3 (Calculus) We can describe the line parallel to the imaginary axis $x=\\frac{1}{2}$ using polar coordinates as $r(\\theta)=\\dfrac{1}{2\\cos{\\theta}},$ which rearranges to $z=\\left(\\dfrac{1}{2\\cos{\\theta}}\\right)(cis{\\theta})\\implies \\frac{1}{z}=2\\cos{\\theta}cis(-\\theta).$ Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}r^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}(2\\cos\\theta)^2 d\\theta.$ Thankfully, this is a routine computation: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}2(\\cos\\theta)^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta$ $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta = 12\\left[\\frac{1}{2}\\sin{2\\theta}+\\theta\\right]_0^{\\frac{\\pi}{6}}=12\\left(\\frac{\\sqrt{3}}{4}+\\frac{\\pi}{6}\\right)=2\\pi+3\\sqrt{3}=2\\pi + \\sqrt{27}$ $a+b = \\boxed{029}$.", "\\text{(E)}\\ \\text{Sunday}$ ## Problem 6 A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it? $\\text{(A)}\\ \\text{A} \\qquad \\text{(B)}\\ \\text{B} \\qquad \\text{(C)}\\ \\text{C} \\qquad \\text{(D)}\\ \\text{D} \\qquad \\text{(E)}\\ \\text{E}$ ## Problem 7 The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E? $[asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black); } draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9)); for(i=1; i<8; i=i+1) { draw((0,2i)--(19,2i)); } label(\"0\", (0,2*0), W); label(\"1\", (0,2*1), W); label(\"2\", (0,2*2), W); label(\"3\", (0,2*3), W); label(\"4\", (0,2*4), W); label(\"5\", (0,2*5), W); label(\"6\", (0,2*6), W); label(\"7\", (0,2*7), W); label(\"8\", (0,2*8), W); label(\"A\", (0*4+1.5, 0), S); label(\"B\", (1*4+1.5, 0), S); label(\"C\", (2*4+1.5, 0), S); label(\"D\", (3*4+1.5, 0), S); label(\"E\", (4*4+1.5, 0), S); label(\"SWEET TOOTH\", (9.5,18), N); label(\"Kinds of candy\", (9.5,-2), S); label(rotate(90)*\"Number of students\", (-2,8), W);[/asy]$ $\\text{(A)}\\ 5 \\qquad \\text{(B)}\\ 12 \\qquad \\text{(C)}\\ 15 \\qquad \\text{(D)}\\", "4), bg = \"rgba(218,112,214,.5)\") Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 If we aren’t interested in coloring just those two rows, we can apply color to the entire table with the bg_pattern sprinkle. This sprinkle accepts as many colors as you want to cycle through. basetable %>% sprinkle(bg_pattern = c(\"orchid\", \"plum\")) Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 # 7 Font Sizes and Colors Font sizes and colors are modified with the font_size and font_color sprinkles. We’ll employ these simultaneously to highlight our significant rows. basetable %>% sprinkle(rows = c(2, 4), font_color = \"orchid\", font_size = 24, font_size_units = \"pt\") Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 (Woah! That was a bit too much.) # 8 Dimensions and Alignment In addition to the sprinkles already discussed, we can also use sprinkles", "general). Also, while $\\pi/2$ labels are implemented, $\\pi/4$ labels are soon to be introduced and have not been implemented. ## Functions and Integrals $[asy] import graph; size(15.36cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-7.3,xmax=8.06,ymin=-1.44,ymax=5.2; pen ttqqzz=rgb(0.2,0,0.6), fueaev=rgb(0.96,0.92,0.9), zzttqq=rgb(0.6,0.2,0); real f2(real x){return 2+2*cos(2*x+pi);} filldraw(graph(f2,-6.28,-3.14)--(-3.14,0)--(-6.28,0)--cycle,fueaev,zzttqq); filldraw(box((-3.14,0),(-2.51,0)),fueaev,zzttqq); filldraw(box((-2.51,0),(-1.88,1.38)),fueaev,zzttqq); filldraw(box((-1.88,0),(-1.26,3.62)),fueaev,zzttqq); filldraw(box((-1.26,0),(-0.63,1.38)),fueaev,zzttqq); filldraw(box((-0.63,0),(0,0)),fueaev,zzttqq); filldraw(box((0,0),(0.63,1.38)),fueaev,zzttqq); filldraw(box((0.63,0),(1.26,3.62)),fueaev,zzttqq); filldraw(box((1.26,0),(1.88,4)),fueaev,zzttqq); filldraw(box((1.88,0),(2.51,3.62)),fueaev,zzttqq); filldraw(box((2.51,0),(pi,1.38)),fueaev,zzttqq); filldraw((pi,0)--(3.77,0)--(3.77,1.38)--(pi,0)--cycle,fueaev,zzttqq); filldraw((3.77,0)--(4.4,0)--(4.4,3.62)--(3.77,1.38)--cycle,fueaev,zzttqq); filldraw((4.4,0)--(5.03,0)--(5.03,3.62)--(4.4,3.62)--cycle,fueaev,zzttqq); filldraw((5.03,0)--(5.65,0)--(5.65,1.38)--(5.03,3.62)--cycle,fueaev,zzttqq); filldraw((5.65,0)--(6.28,0)--(6.28,0)--(5.65,1.38)--cycle,fueaev,zzttqq); Label laxis; laxis.p=fontsize(10); string xlbl(real x){int n=round(x/pi); if(n==-1) return \"-\\pi\"; if(n==1) return \"\\pi\"; if(n==0) return \"0\"; return \"\"+string(round(x/pi))+\"\\pi\";} xaxis(\"\\theta\",-7.3,8.06,linewidth(1.2),Ticks(laxis,xlbl,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-1.44,5.2,linewidth(1.2),Ticks(laxis,Step=1.0,Size=2),Arrows(6),above=true); real f1(real x){return 2+2*cos(2*x+pi);} draw(graph(f1,-7.29,8.05),linewidth(1.6)+ttqqzz); label(\"2 + 2 \\cos(2x + \\pi)\",(-7.14,4.1),NE*lsf,ttqqzz); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Allows most functions and can draw integrals and Riemman-like sums. Above, the function is given by $f(x) = 2+2\\cos(2x)$; the four shaded regions are drawn using the commands Integral[f,-2pi,-pi], LowerSum[f,-pi,0,5], UpperSum[f,0,pi,5], and TrapezoidalSum[f,pi,2pi,5]. <geogebra>6636ccd3479762307d932c319db7ec1998236a4c</geogebra> Known bugs: • May capitalize certain functions (like trigonometric functions). Please let me know if this happens. • Certain functions recognized by Geogebra might not be recognized by Asymptote. • Integrals/Sums drawn over functions that reach undefined values (ex, $\\ln(x)$ at $x=0$) will cause a break when drawn in Asymptote (unlikely to be fixed); drawing the functions themselves will work, however. • With certain very contrived function strings, it is possible to break the code (won't fix). • Sometimes will render LaTeX commands in function strings. Please let me know if", "\\usepackage{pgfplots} \\usetikzlibrary{hobby} \\usepackage{filecontents} \\pgfplotsset{compat=1.9, every tick label/.append style={font=\\normalsize}} \\pgfplotscreateplotcyclelist{my black white}{% solid,smooth, every mark/.append style={solid, fill=white}, mark=triangle*\\\\% solid, every mark/.append style={solid, fill=black}, mark=triangle*\\\\% densely dashed, every mark/.append style={solid, fill=white},mark=diamond*\\\\% densely dashed, every mark/.append style={solid, fill=black}, mark=diamond*\\\\% } \\begin{filecontents}{p2.dat} SAT ETCY-5 ETCX-5 ETCY-10 ETCX-10 AR-5 AR-10 100 0.46 0.50 0.54 0.61 0.17 0.25 90 0.45 0.48 0.53 0.59 0.18 0.25 80 0.43 0.47 0.51 0.58 0.19 0.27 70 0.41 0.45 0.49 0.57 0.21 0.30 60 0.39 0.44 0.48 0.55 0.25 0.35 50 0.37 0.42 0.46 0.54 0.30 0.44 40 0.35 0.40 0.44 0.52 0.38 0.57 30 0.33 0.37 0.42 0.51 0.50 0.71 20 0.29 0.34 0.39 0.48 0.63 0.87 10 0.24 0.29 0.34 0.45 0.78 1.04 0 0.18 0.23 0.29 0.41 0.92 1.22 \\end{filecontents} \\begin{document} \\begin{tikzpicture} \\begin{axis}[ cycle list name=my black white, title={Compressing Pressure: 0.2 MPa}, xmin=0, xmax=100, xlabel={Water Saturation($S_w$)}, xtick distance=10, ymin=0, ymax=0.7, ylabel={Avg. ETC \\quad $\\frac{K_{\\mathrm{eff}}}{K_s}$ (Dimensionless)}, legend style ={ at={(0.25,0.4)}, anchor=north west, draw=none, font=\\normalsize, fill=white,align=left, }, hobby ] \\addlegendentry{$K_{\\mathrm{eff}(x)}$ -- Overlap 5\\% }; \\addlegendentry{$K_{\\mathrm{eff}(y)}$ -- Overlap 5\\% }; \\addlegendentry{$K_{\\mathrm{eff}(x)}$ -- Overlap 10\\% }; \\addlegendentry{$K_{\\mathrm{eff}(y)}$ -- Overlap 10\\% }; `", "# Drawing a half-full bottle in TikZ I want to fill half of the bottle. I tried these kind of commands: %\\filldraw[color=black!100, fill=cyan!30, very thick](-3,-2.4) arc (180:230:2.5) .. controls (-2,-5) and (-1.9,-5.2) .. (-1.8,-5.5) ; But I failed. Here is the bottle: \\documentclass[11pt,a4paper,oneside]{article} \\usepackage[a4paper,left=3cm,right=2cm,top=2.5cm,bottom=2.5cm]{geometry} \\usepackage[utf8x]{inputenc} \\usepackage{tikz} \\begin{document} \\begin{tikzpicture}[xscale=1,yscale=1] \\draw[very thick](0,0) arc (0:180:1.5); \\draw[very thick] (-3,0) -- (-3,-2.5); \\draw[very thick] (0,0) -- (0,-2.5); \\draw[color=black!100,very thick](-3,-2.5) arc (180:235:3); \\draw[color=black!100,very thick](0,-2.5) arc (0:-55:3); \\draw[very thick] (-1.25,-4.95) -- (-1.25,-5.2) -- (-1.7,-5.2) -- (-1.7,-4.95); \\end{tikzpicture} \\end{document} What I want: • Simply draw the half bottle starting with the neck on the left side and ending with the neck on the right side. – AndréC Dec 31 '18 at 11:39 • And if I wanted to draw a half-empty bottle ;-) – Peter Wilson Dec 31 '18 at 18:53 \\documentclass[border=5pt,tikz]{standalone} \\begin{document} \\begin{tikzpicture}[xscale=1,yscale=1] \\fill[blue!20] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\fill[white] (0,0) arc(0:180:1.5) --+ (0,-2) -| cycle; \\fill[blue!10] (-1.5,-2) circle ({1.5cm-0.4pt} and 0.5cm); % Thank you, marmot! ;) \\draw[very thick] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\end{tikzpicture} \\end{document} Output: See below Explanation: Maybe beginners don't know the clip option and maybe the want just a type of \"trivial\" answer, so I", "each took four $100$-point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests? $\\mathrm{(A)}\\ 10 \\qquad\\mathrm{(B)}\\ 15 \\qquad\\mathrm{(C)}\\ 20 \\qquad\\mathrm{(D)}\\ 25 \\qquad\\mathrm{(E)}\\ 40$ Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures $\\textbf{Bake Sale}$ Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown. $\\circ$ Art's cookies are trapezoids: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label(\"5 in\", (2.5,0), S); label(\"3 in\", (5,1.5), E); label(\"3 in\", (3.5,3), N);[/asy]$ $\\circ$ Roger's cookies are rectangles: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label(\"4 in\", (2,0), S); label(\"2 in\", (4,1), E);[/asy]$ $\\circ$ Paul's cookies are parallelograms: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label(\"3 in\", (1.5,0), S); label(\"2 in\", (2.5,1), W);[/asy]$ $\\circ$ Trisha's cookies are triangles: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label(\"3 in\", (1.5,0), S); label(\"4 in\", (3,2), E);[/asy]$ ## Problem 8 Who gets the fewest cookies from one batch of cookie dough? $\\textbf{(A)}\\ \\text{Art} \\qquad\\textbf{(B)}\\ \\text{Paul}\\qquad\\textbf{(C)}\\ \\text{Roger}\\qquad\\textbf{(D)}\\ \\text{Trisha}\\qquad\\textbf{(E)}\\ \\text{There is a tie for fewest}$ ## Problem 9 Each friend", "file (in csv format, for comma-separated values), display it in a nice table, and extract the columns that we need—which we'll convert to numpy arrays to work with. Let's start by importing the two Python libraries that we need. In [1]: import pandas import numpy ### Step 1: Read the data file Below, we'll take a peek into the data file, beers.csv, using the system command head (which we can use with a bang, thanks to IPython). But first, we will download the data using a Python library for opening a URL on the Internet. We created a short URL for the data file in the public repository with our course materials. The cell below should download the data in your current working directory. The next cell shows you the first few lines of the data. In [2]: from urllib.request import urlretrieve URL = 'http://go.gwu.edu/engcomp2data1' urlretrieve(URL, 'beers.csv') Out[2]: ('beers.csv', <http.client.HTTPMessage at 0x11d88c9e8>) In [3]: !head \"beers.csv\" ,abv,ibu,id,name,style,brewery_id,ounces 0,0.05,,1436,Pub Beer,American Pale Lager,408,12.0 1,0.066,,2265,Devil's Cup,American Pale Ale (APA),177,12.0 2,0.071,,2264,Rise of the Phoenix,American IPA,177,12.0 3,0.09,,2263,Sinister,American Double / Imperial IPA,177,12.0 4,0.075,,2262,Sex and Candy,American IPA,177,12.0 5,0.077,,2261,Black Exodus,Oatmeal Stout,177,12.0 6,0.045,,2260,Lake Street Express,American Pale Ale (APA),177,12.0 7,0.065,,2259,Foreman,American Porter,177,12.0 8,0.055,,2258,Jade,American Pale Ale (APA),177,12.0 We can use pandas to read the data from the csv file, and save it into a new variable called beers. Let's then", "% NOW DATA FOR SECOND [xData2, tmp] = prepareCurveData( Fq2r, Prem2 ); yData2 = tmp+pi/2; % and the other start points, fit... opts.StartPoint = [0.425259320214135 0.312718886820616]; [fitresult2, gof2] = fit( xData2, yData2, ft, opts ); hold on % add to existing plot h(2) = plot( fitresult2, xData2, yData2 ); % set both line handles in h array... set(h,'LineWidth',2,'Markersize',7.5) grid on legend({'Phasenverschiebung','Fit: $(-1,02 \\pm 0,07) \\cdot \\tan^{-1}(\\frac{2\\cdot (27,59 \\pm 0,27)\\cdot \\omega}{3,83^2-\\omega^2})$'}, ... {'Phasenverschiebung bei $I = 0,25 A$','Fit: $(-1,26 \\pm 0,20) \\cdot \\tan^{-1}(\\frac{2\\cdot (26,59 \\pm 0,65)\\cdot \\omega}{3,83^2-\\omega^2})$'}, ... 'Interpreter','latex','FontSize',14) xlabel({'$\\omega$ in rad'},'Interpreter','latex','FontSize',14) ylabel({'Phasenverschiebung $\\Delta \\varphi$ in rad'},'Interpreter','latex','FontSize',14) is a first cut Niklas Kurz on 25 Jun 2020 First of Thank you so much for your effort: However, I guess I should have mentionned that the 2 fits doesn't have same Size, as: Fqr = [400 500 525 550 575 600 625 650 675 700 750].*(2*pi/1000) Prem = [0.277 0.527 0.637 0.628 1.034 1.561 2.167 2.727 2.758 2.749 2.861]-pi/2 Fq2r = [300 400 500 525 550 575 600 625 650 675 700 750].*(2*pi/1000) Prem2 = [0.094 0.220 0.253 0.255 0.517 1.111 2.630 3.005 3.090 2.990 3.063 2.936]-pi/2 After running the code Fqr and Fq2r also have been tangeled up. So, how to fix it? dpb on 25 Jun 2020 Different size won't matter...all I see is missed adding the '1'", "(-1.5,-2) circle ({1.5cm-0.4pt} and 0.5cm); \\draw[very thick] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\end{tikzpicture} \\end{document} Explanation: We can just \"clip\" the bottle: everything, what we draw, is now in the area of the bottle. Everything outside that area is invisible. So we just clip the bottle and fill a rectangle, such it fills a certain area of the bottle. Output: EDIT: For Sebastiano: \\documentclass[border=5pt,tikz]{standalone} \\usetikzlibrary{decorations.text,backgrounds} \\definecolor{wine}{RGB}{216,198,62} \\definecolor{bottle}{RGB}{76,163,58} \\tikzset{ my/.style={ postaction={decorate},decoration={text along path, text={#1},text align=center} } } \\begin{document} \\begin{tikzpicture} \\begin{scope} \\clip (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\fill[inner color=bottle!50,outer color=bottle] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\fill[wine!60] (-3,-5.5) rectangle ++(3,3.5); \\fill[wine!40] (-1.5,-2) circle ({1.5cm-0.4pt} and 0.5cm); \\foreach \\x in {-5,-4.9,...,5} \\foreach \\y in {-5,...,-3} { \\pgfmathsetmacro\\opacity{random(1,10)*(1/10)} } \\draw[very thick] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\path[my={The magic of Ti{\\emph{\\color{orange}k}}Z}] (-3.5,.5) arc(-180:0:2 and 1); \\end{scope} \\end{tikzpicture} \\end{document} Output: • I'm struggling to understand what does it mean: \\fill[white] (0,0) arc(0:180:1.5) --+ (0,-2) -| cycle; . I understand the arc part, but I don't get how you link them each other by using", "# Using hobby package to plot bezier curves through a set of data points in a .dat file I would like to use the hobby package to plot smooth drawn curves through points. pgfs' smooth function is not working properly. My code is a follows, \\documentclass{standalone} \\usepackage{pgfplots} \\usepackage{filecontents} \\pgfplotsset{compat=1.9, every tick label/.append style={font=\\normalsize}} \\pgfplotscreateplotcyclelist{my black white}{% solid,smooth, every mark/.append style={solid, fill=white}, mark=triangle*\\\\% solid, every mark/.append style={solid, fill=black}, mark=triangle*\\\\% densely dashed, every mark/.append style={solid, fill=white},mark=diamond*\\\\% densely dashed, every mark/.append style={solid, fill=black}, mark=diamond*\\\\% } \\begin{filecontents}{p2.dat} SAT ETCY-5 ETCX-5 ETCY-10 ETCX-10 AR-5 AR-10 100 0.46 0.50 0.54 0.61 0.17 0.25 90 0.45 0.48 0.53 0.59 0.18 0.25 80 0.43 0.47 0.51 0.58 0.19 0.27 70 0.41 0.45 0.49 0.57 0.21 0.30 60 0.39 0.44 0.48 0.55 0.25 0.35 50 0.37 0.42 0.46 0.54 0.30 0.44 40 0.35 0.40 0.44 0.52 0.38 0.57 30 0.33 0.37 0.42 0.51 0.50 0.71 20 0.29 0.34 0.39 0.48 0.63 0.87 10 0.24 0.29 0.34 0.45 0.78 1.04 0 0.18 0.23 0.29 0.41 0.92 1.22 \\end{filecontents} \\begin{document} \\begin{tikzpicture} \\begin{axis}[ cycle list name=my black white, title={Compressing Pressure: 0.2 MPa}, enlarge x limits=-1, xmin=0, xmax=100, xlabel={Water Saturation($S_w$)}, xtick={0,10,20,30,40,50,60,70,80,90,100}, xticklabels={0,10,20,30,40,50,60,70,80,90,100}, scaled ticks=true, ymin=0, ymax=0.7, ylabel={Avg. ETC \\quad $\\frac{K_{eff}}{K_s}$ $(Dimensionless)$}, legend style ={ at={(0.25,0.4)}, anchor=north west, draw=none, font=\\normalsize, fill=white,align=left}, smooth ] \\addlegendentry{$K_{eff(x)}-Overlap \\quad 5\\%$}; \\addlegendentry{$K_{eff(y)}-Overlap \\quad5\\%$}; \\addlegendentry{$K_{eff(x)}-Overlap \\quad10\\%$}; \\addlegendentry{$K_{eff(y)}-Overlap \\quad10\\%$};", "legend(loc=2) xlabel('grain diameter (mm)',fontsize=15) ylabel('settling velocity (m/s)',fontsize=15) axis([0,5,0,2]); A log-log plot is much better for looking at a wide spectrum of grain sizes. In [112]: d = np.arange(0,0.01,0.00001) ws = v_stokes(rop,rof,d,visc,C1) wt = v_turbulent(rop,rof,d,visc,C2) wf = v_ferg(rop,rof,d,visc,C1,C2) figure(figsize=(10,8)) loglog(d*1000,ws,label='Stokes',linewidth=3) loglog(d*1000,wt,'g',label='Turbulent',linewidth=3) loglog(d*1000,wf,'r',label='Ferguson-Church',linewidth=3) plot([1.0/64, 1.0/64],[0.00001, 10],'k--') text(0.012, 0.0007, 'fine silt', fontsize=13, rotation='vertical') plot([1.0/32, 1.0/32],[0.00001, 10],'k--') text(0.17/8, 0.0007, 'medium silt', fontsize=13, rotation='vertical') plot([1.0/16, 1.0/16],[0.00001, 10],'k--') text(0.17/4, 0.0007, 'coarse silt', fontsize=13, rotation='vertical') plot([1.0/8, 1.0/8],[0.00001, 10],'k--') text(0.17/2, 0.001, 'very fine sand', fontsize=13, rotation='vertical') plot([0.25, 0.25],[0.00001, 10],'k--') text(0.17, 0.001, 'fine sand', fontsize=13, rotation='vertical') plot([0.5, 0.5],[0.00001, 10],'k--') text(0.33, 0.001, 'medium sand', fontsize=13, rotation='vertical') plot([1, 1],[0.00001, 10],'k--') text(0.7, 0.001, 'coarse sand', fontsize=13, rotation='vertical') plot([2, 2],[0.00001, 10],'k--') text(1.3, 0.001, 'very coarse sand', fontsize=13, rotation='vertical') plot([4, 4],[0.00001, 10],'k--') text(2.7, 0.001, 'granules', fontsize=13, rotation='vertical') text(6, 0.001, 'pebbles', fontsize=13, rotation='vertical') legend(loc=2) xlabel('grain diameter (mm)', fontsize=15) ylabel('settling velocity (m/s)', fontsize=15) axis([0,10,0,10]); plot(D,w,'o',markerfacecolor=[0.6, 0.6, 0.6], markersize=8); This plot shows how neither Stokes' Law, nor the velocity based on turbulent drag are valid for calculating settling velocities of sand-size grains in water, whereas the Ferguson-Church equation provides a good fit for natural river sand. Grain settling is a special case of the more general problem of flow past a sphere. The analysis and plots above are all dimensional, that is, you can quickly check by looking at the plots what is the approximate", "while still being complete. Create boxplots for each of the 10-day periods between 2014 and 202. Use the option -d flag option to let the function find the best location and formats for the x-axis labels. Given the large number of boxplots, we will need to adjust some settings to achieve a readable result. See the manual page for an explanation of the different options used below. t.rast.boxplot -d -g input=LAI300 dpi=300 \\ plot_dimensions=18,5 rotate_labels=0 \\ font_size=14 bx_width=8 bx_lw=0.1 \\ median_lw=1.2 bx_color=51:118:183:255 \\ median_color=255:255:255:255 \\ whisker_linewidth=0.5 Given the large number of boxplots, we will need to adjust some settings to achieve a readable result. See the manual page for an explanation of the different options used below. Module(\"t.rast.boxplot\", flags=\"dg\", input=\"LAI300\", dpi=300, plot_dimensions=\"18,5\", rotate_labels=0, font_size=16, bx_width=8, bx_lw=0.1, median_lw=1.2, bx_color=\"51:118:183:255\", median_color=\"255:255:255:255\", whisker_linewidth=0.5) execGRASS(\"t.rast.boxplot\", flags=\"dg\", input=\"LAI300\", dpi=300, plot_dimensions=\"18,5\", rotate_labels=0, font_size=16, bx_width=8, bx_lw=0.1, median_lw=1.2, bx_color=\"51:118:183:255\", median_color=\"255:255:255:255\", whisker_linewidth=0.5) The boxplot time series in Figure 8 provides a quick overview of the yearly and seasonal patterns, but also the spatial variation in Bardia National Park. As observed earlier, the low spatial variation in LAI values in 2020 stand out, as does the high spatial variation in the LAI peak period in 2014. Out of curiosity, I also plotted the monthly rainfall, based on the CRU dataset6. Comparing the two graphs shows how the peak", "$\\frac {\\sqrt {130}}2$, giving an answer of $130 + 2 = \\boxed{132}$. $[asy]import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); filldraw((-3,4)--(-2,4)--(-2,5)--(-3,5)--cycle,evevff,blue); filldraw((3,4)--(4,4)--(4,5)--(3,5)--cycle,evevff,blue); filldraw((4,3)--(5,3)--(5,4)--(4,4)--cycle,evevff,blue); filldraw((5,0)--(6,0)--(6,1)--(5,1)--cycle,evevff,blue); filldraw((4,-3)--(5,-3)--(5,-2)--(4,-2)--cycle,evevff,blue); filldraw((3,-3)--(3,-4)--(4,-4)--(4,-3)--cycle,evevff,blue); filldraw((0,-5)--(1,-5)--(1,-4)--(0,-4)--cycle,evevff,blue); filldraw((-3,-4)--(-2,-4)--(-2,-3)--(-3,-3)--cycle,evevff,blue); filldraw((-4,-3)--(-3,-3)--(-3,-2)--(-4,-2)--cycle,evevff,blue); filldraw((-4,3)--(-3,3)--(-3,4)--(-4,4)--cycle,evevff,blue); filldraw((-5,0)--(-4,0)--(-4,1)--(-5,1)--cycle,evevff,blue); filldraw((0,6)--(0,5)--(1,5)--(1,6)--cycle,evevff,blue); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=1,gy=1; 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); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); draw(circle((0,0),5),linewidth(1.6)); draw(circle((0.5,0.5),5.7),linetype(\"2 2\")); draw((-3,4)--(-2,4),zzttqq); draw((-2,4)--(-2,5),zzttqq); draw((-2,5)--(-3,5),zzttqq); draw((-3,5)--(-3,4),zzttqq); draw((3,4)--(4,4),zzttqq); draw((4,4)--(4,5),zzttqq); draw((4,5)--(3,5),zzttqq); draw((3,5)--(3,4),zzttqq); draw((4,3)--(5,3),zzttqq); draw((5,3)--(5,4),zzttqq); draw((5,4)--(4,4),zzttqq); draw((4,4)--(4,3),zzttqq); draw((5,0)--(6,0),zzttqq); draw((6,0)--(6,1),zzttqq); draw((6,1)--(5,1),zzttqq); draw((5,1)--(5,0),zzttqq); draw((4,-3)--(5,-3),zzttqq); draw((5,-3)--(5,-2),zzttqq); draw((5,-2)--(4,-2),zzttqq); draw((4,-2)--(4,-3),zzttqq); draw((3,-3)--(3,-4),zzttqq); draw((3,-4)--(4,-4),zzttqq); draw((4,-4)--(4,-3),zzttqq); draw((4,-3)--(3,-3),zzttqq); draw((0,-5)--(1,-5),zzttqq); draw((1,-5)--(1,-4),zzttqq); draw((1,-4)--(0,-4),zzttqq); draw((0,-4)--(0,-5),zzttqq); draw((-3,-4)--(-2,-4),zzttqq); draw((-2,-4)--(-2,-3),zzttqq); draw((-2,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-4),zzttqq); draw((-4,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-2),zzttqq); draw((-3,-2)--(-4,-2),zzttqq); draw((-4,-2)--(-4,-3),zzttqq); draw((-4,3)--(-3,3),zzttqq); draw((-3,3)--(-3,4),zzttqq); draw((-3,4)--(-4,4),zzttqq); draw((-4,4)--(-4,3),zzttqq); draw((-5,0)--(-4,0),zzttqq); draw((-4,0)--(-4,1),zzttqq); draw((-4,1)--(-5,1),zzttqq); draw((-5,1)--(-5,0),zzttqq); draw((0,6)--(0,5),zzttqq); draw((0,5)--(1,5),zzttqq); draw((1,5)--(1,6),zzttqq); draw((1,6)--(0,6),zzttqq); dot((0,5),ds); dot((3,4),ds); dot((4,3),ds); dot((5,0),ds); dot((4,-3),ds); dot((3,-4),ds); dot((0,-5),ds); dot((-3,-4),ds); dot((-4,-3),ds); dot((-4,3),ds); dot((-3,4),ds); dot((-2,4),ds); dot((-2,5),ds); dot((-3,5),ds); dot((4,4),ds); dot((4,5),ds); dot((3,5),ds); dot((5,3),ds); dot((5,4),ds); dot((4,4),ds); dot((6,0),ds); dot((6,1),ds); dot((5,1),ds); dot((5,-3),ds); dot((5,-2),ds); dot((4,-2),ds); dot((3,-3),ds); dot((4,-4),ds); dot((4,-3),ds); dot((1,-5),ds); dot((1,-4),ds); dot((0,-4),ds); dot((-2,-4),ds); dot((-2,-3),ds); dot((-3,-3),ds); dot((-3,-2),ds); dot((-4,-2),ds); dot((-3,3),ds); dot((-3,4),ds); dot((-4,4),ds); dot((-5,0),ds); dot((-4,0),ds); dot((-4,1),ds); dot((-5,1),ds); dot((0,6),ds); dot((1,5),ds); dot((1,6),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "end -- infopanel local temp = \"FPS: \"..love.timer.getFPS()..\". x:\"..mx..\". y:\"..my..\".\" if (mx+font11:getWidth(temp)) < love.graphics.getWidth() then love.graphics.setColor(0,30,100,150) love.graphics.rectangle(\"fill\",mx+10,my-0,font11:getWidth(temp),11) love.graphics.setColor(200,255,255,255) love.graphics.rectangle(\"line\",mx+10,my+0,font11:getWidth(temp),11) love.graphics.print(temp,mx+10,my+0) else love.graphics.setColor(0,30,100,150) love.graphics.rectangle(\"fill\",mx-(font11:getWidth(temp)+10),my-0,font11:getWidth(temp),11) love.graphics.setColor(200,255,255,255) love.graphics.rectangle(\"line\",mx-(font11:getWidth(temp)+10),my+0,font11:getWidth(temp),11) love.graphics.print(temp,mx-(font11:getWidth(temp)+10),my+0) end temp = nil end Finally here is an image of a menu created recently with this version of the system and with the debug module turned on. ow_controlsystem.png (235.03 KiB) Viewed 4556 times It can also be viewed by pressing f1 in audiotree or f1/f2 in most of my programs. If this is useful or interesting to anyone I will soon release the next block of code. Last edited by StoneCrow on Fri Apr 19, 2013 3:14 pm, edited 1 time in total. Dull but sincere filler. StoneCrow Party member Posts: 199 Joined: Sat Apr 17, 2010 9:31 am Location: Wales the land of leeks and leaks Contact: Okay this one is alot more like a library. OW_rope, it reads strings and burns them. Code: Select all --[[ LICENCE DETAILS: ZLIB LICENCE This software is provided 'as-is', without any express or implied warranty. In no event will the authors be held liable for any damages arising from the use of this software. Permission is granted to anyone to use this software for any purpose, including commercial applications, and to alter it and redistribute it freely, subject to the following restrictions: 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", "the bars. Defaults to 'crimson'. Optional. the stroke width of the bars. The font family to consider for the titles. Defaults to \"Verdana, Geneva, Tahoma, sans-serif\". The background color of the SVG. Defaults to \"#CAD0D3\" HEX color. The color opacity of the bars (from 0 to 1). Defaults to 1. the color of the x and y axis. It includes the ticks, the labels and titles. Defaults to 'black'. Optional. The width of the SVG output. Optional. The height of the SVG output. A SVG bar chart. ## Examples library(ggplot2) #needed for the mpg data frame library(dplyr) #needed for data wrangling #> #> Attaching package: 'dplyr'#> The following objects are masked from 'package:stats': #> #> filter, lag#> The following objects are masked from 'package:base': #> #> intersect, setdiff, setequal, union mpg %>% group_by(manufacturer) %>% summarise(mean_cty = mean(cty)) %>% barChart( x = \"manufacturer\", y = \"mean_cty\", sort = \"ascending\", xFontSize = 10, yFontSize = 10, fill = \"orange\", strokeWidth = 1, ytitle = \"average cty value\", title = \"Average City Miles per Gallon by manufacturer\", titleFontSize = 16 )", "-0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 The text could also be italicized either separately or concurrently. He we show the italics printed concurrently. basetable %>% sprinkle(rows = c(2, 4), bold = TRUE, italic=TRUE) Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 # 6 Backgrounds Backgrounds are added using the bg sprinkle, which accepts X11 colors, hexidecimal colors, rgb colors, and for HTML rgba colors (the a specifies the transparency). To put in a background in the rows showing statistical significance, we need only specify the color. basetable %>% sprinkle(rows = c(2, 4), bg = \"orchid\") Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 If we decide that color is a little bit strong, we can lighten it up a little with the transparency. We have to look up the rgb specification for the orchid color (there are lots of web resources for this; X11 Color Names on Wikipedia is a good place to start). basetable %>% sprinkle(rows = c(2," ]

[ "in Example A by using a circle instead of a hundredths disk. Step 1: Determine the total number of donors: $7+5+9+4 = 25$ . Step 2: Express each donor number as the number of degrees of a circle that it represents by using the formula $\\text{Degrees}=\\frac{f}{n} \\cdot 360^\\circ$ , where $f$ is the frequency and $n$ is the total number. $\\frac{7}{25} \\cdot 360^\\circ = 100.8^\\circ \\quad \\frac{5}{25} \\cdot 360^\\circ = 72^\\circ \\quad \\frac{9}{25} \\cdot 360^\\circ = 129.6^\\circ \\quad \\frac{4}{25} \\cdot 360^\\circ = 57.6^\\circ$ Step 3: Using a protractor, graph each section of the circle. Step 4: Write the name and correct percentage inside each section. Color each section a different color. The above pie chart was created by using a protractor and graphing each section of the circle according to the number of degrees needed. From the graph, you can see that more donations were of Type O than any other type. The fewest number of donations of blood collected was of Type AB. Notice that the percentages have been entered in each section of the graph and not the numbers of degrees. This is because degrees would not be meaningful to an observer trying to interpret the graph. In order to create a pie chart by using a circle, it is necessary to use the formula to", "1000 = 100 No. of Folk music CD’s = $$\\frac{30}{100}$$ × 1000 = 300 No. of light music CD’s 40 = $$\\frac{40}{100}$$ × 1000 = 400. Question 2. A group of 360 people were asked to vote for their favourite season from the three seasons rainy, winter and summer. (i) Which season got the most votes ? (ii) Find the central angle of each sector. (iii) Draw a pie chart to show this information. Solution: (i) Winter. (ii) Central angle = $$\\frac{Value of Item}{Total value}$$ 360° ∴ Central angle for Summer season = $$\\frac{90}{360}$$ × 360° = 90° Central angle for Rainy season = $$\\frac{120}{360}$$ × 360° = 120° and Central angle for Winter season = $$\\frac{150}{360}$$ × 360° = 150° Question 3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people. Find the proportion of each sector. For example, Blue is $$\\frac{18}{360}$$ = $$\\frac{1}{2}$$; Green is $$\\frac{9}{36}$$ = $$\\frac{1}{4}$$ and so on. Use this to find the corresponding angles. Solution: Now, we make the pie chart as-shown below musing protractor, compasses and ruler. Question 4. The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540,", "of equivalent fractions for $$\\frac{1}{2}$$, for $$\\frac{10}{3}$$, and for 1. Example: Going Backwards $$\\frac{20}{32}\\; \\text{(sharing 20 pies among 32 kids)}$$ is the same problem as: $$\\frac{5 \\cdot 4}{8 \\cdot 4} = \\frac{5}{8}\\; \\text{(sharing five pies among eight kids)} \\ldotp$$ Most people say we have cancelled or taken a common factor 4 from the numerator and denominator. Mathematicians call this process reducing the fraction to lowest terms. (We have made the numerator and denominator smaller, in fact as small as we can make them!) Teachers tend to say that we are simplifying the fraction. (You have to admit that $$\\frac{5}{8}$$ does look simpler than $$\\frac{20}{32}$$.) Example: How Low Can You Go? As another example, $$\\frac{280}{350}$$ can certainly be simplified by noticing that there is a common factor of 10 in both the numerator and the denominator: $$\\frac{280}{350} = \\frac{28 \\cdot 10}{35 \\cdot 10} = \\frac{28}{35} \\ldotp$$ We can go further as 28 and 35 are both multiples of 7: $$\\frac{28}{35} = \\frac{4 \\cdot 7}{5 \\cdot 7} = \\frac{4}{5} \\ldotp$$ Thus, sharing 280 pies among 350 children gives the same result as sharing 4 pies among 5 children! $$\\frac{280}{350} = \\frac{4}{5} \\ldotp$$ Since 4 and 5 share no common factors, this is as far as we can go with this example (while staying with whole numbers). $$\\frac{10}{20}$$ $$\\frac{50}{75}$$ $$\\frac{24000}{36000}$$ $$\\frac{24}{14}$$", "$.26 \\times 360^\\circ=94^\\circ$ • Grade B: $\\frac{10}{31}=.32=32\\%$ Convert to degrees: $.32 \\times 360^\\circ=115^\\circ$ • Grade C: $\\frac{7}{31}=.225=23\\%$ Convert to degrees: $.225 \\times 360^\\circ=81^\\circ$ • Grade D: $\\frac{3}{31}=.097=10\\%$ Convert to degrees: $.097 \\times 360^\\circ=35^\\circ$ • Grade F: $\\frac{3}{31}=.097=10\\%$ Convert to degrees: $.097 \\times 360^\\circ=35^\\circ$ #### Practice 1. What is each part of a pie chart called? 2. What part of the circle represents the whole relationship? 3. Based on the data in the table below, which candy sold the most, and which the least? Candy Bar Number sold at the school store Percentage of Total Sales Number of Degrees on a Pie Chart Heath 151 M&M 191 Snickers 61 Skittles 107 Almond Joy 91 4. Fill in the chart above with percentages of each candy type sold. 5. Fill in the chart with degrees of a circle required to represent this amount on a pie chart. 6. Create a Pie Chart to represent this data. 7. Based on the data in the table below, which stock has the greatest potential for making money for the investor who owns the stock? Company Shares Owned Percentage of Total Portfolio Number of Degrees on a Pie Chart Hostess 8 Pepsi 11 Dell 5 Conoco 7 Ford Motor 19 8. Fill in the chart with relative percentages of each type of stock. 9.", "around the circle, which can be done in $\\frac{7!}{3!3!}$ ways. However, this does not match with their expression. Why? #### Solutions Collecting From Web of \"2011 AIME Problem 12 — Probability: 9 delegates around a round table\" They are making two simplifying assumptions in their solution, that the seats are numbered and that people from the same country are identical. 1) They didn’t divide by $9$ in finding their total because they were assuming the seats were numbered; otherwise they would have. 2) Your expression of $\\frac{7!}{3!3!}$ would work fine if the people were arranged in a straight line. Notice that in this case the 3 people sitting together have 7 choices for their positions: $123, 234, \\cdots, 789$. In a circle, though, they have 9 choices instead: $123, 234, \\cdots, 789, 891, 912$. If we assumed that the 9 people were all distinct and we didn’t number the seats, then we would get the same probability of $\\displaystyle 1-\\frac{\\binom{3}{1}3!6!-\\binom{3}{2}4(3!)^{3}+2(3!)^{3}}{8!}=1-\\frac{10800}{40320}=1-\\frac{15}{56}=\\frac{41}{56}$.", "poor people to rich people in a town in the year $1990$ was $6$ to $1$. The same ratio in the year $2000$ was $10$ to $3$. What is the percent increase of rich people in the town from $1990$ to $2000$? A $2.67$ B $6.69$ C $8.79$ D $9.23$ Question 19 Explanation: The correct answer is (C). In $1990, \\frac{poor}{rich} = \\frac{6}{1}$. So, the percentage of rich people $= \\frac{1}{7} \\cdot 100 = 14.29%$ In $2000, \\frac{poor}{rich} = \\frac{10}{3}$. So, the percentage of rich people $= \\frac{3}{13} \\cdot 100 = 23.08%$ So, the percentage increase is $23.08 – 14.29 = 8.79%$ Question 20 Based on the diagram above, if chord $CD$ has a length of $10$, what is the length of $AB$? A $10$ B $\\dfrac{10\\sqrt{3}}{3}$ C $5\\sqrt{3}$ D $\\dfrac{10\\sqrt{2}}{3}$ Question 20 Explanation: The correct answer is (B). You can use the $30-60-90$ triangles created when triangle $ACD$ is divided into two. This gives $CD = 5$, $AE = \\frac{5\\sqrt{3}}{3}$, and $AC = \\frac{10\\sqrt{3}}{3}$. $AB$ is a radius of the circle and has the same value as $AC$. Question 21 If a circle with its origin at $(0, 0)$ passes through the point $(–4, 0)$, then which one of the following points is on the circumference of the circle? A $(–4, 1)$ B $(0, 5)$ C $(–1,", "\\text{people}$ Garden salad: $13\\%=\\frac{13}{100}=0.13$ $(300)(0.13) = 39 \\ \\text{people}$ Chef salad: $25\\%=\\frac{25}{100}=0.25$ $(300)(0.25) = 75 \\ \\text{people}$ 4. The difference between the number of people who chose the spinach salad and the number of people who chose the garden salad is $51-39=12 \\ \\text{people}$ . ### Guided Practice The following pie chart, which is incomplete, shows the extracurricular activities for 200 high school students: Use the pie chart to answer the following questions: a. What makes the above pie chart incomplete? b. How many students participate in sports? c. How many students participate in drama? d. How many students do CMT? e. How many students participate in volunteer activities after school? f. How many student work? g. Construct the pie chart so that it shows percentages and not degrees. a. The parts of the pie chart are displayed with degrees, but they should be shown with percentages. b. The number of students who participate in sports can be calculated as follows: $\\frac{135}{360}=0.375 \\quad (0.375)(200)=75$ c. The number of students who participate in drama can be calculated as follows: $\\frac{45}{360}=0.125 \\quad (0.125)(200)=25$ d. The number of students who do CMT can be calculated as follows: $\\frac{36}{360}=0.1 \\quad (0.1)(200)=20$ e. The number of students who participate in volunteer activities can be calculated as follows: $\\frac{90}{360}=0.25 \\quad (0.25)(200)=50$ f. The number", "180° was left, you only have half of a pie. Put more formally, you have: $\\frac{1}{2} \\ in \\ pies$ It may sound ridiculous, but I'm always going to be using the plural phrase, “in pies”, because it's going to help lock in the mnemonic. I'll explain that shortly, but first we need a starting point. If you're given any multiple of 90°, it's pretty easy to picture these in pies, as it was how many of us were taught: $90^{\\circ} = \\frac{1}{4} \\ in \\ pies, \\ 180^{\\circ} = \\frac{2}{4} = \\frac{1}{2} \\ in \\ pies,\\\\\\\\\\\\270^{\\circ} = \\frac{3}{4} \\ in \\ pies, \\ 360^{\\circ} = \\frac{4}{4} = 1 \\ in \\ pies$ 45° isn't much harder, especially considering that we've already dealt with half of them above. You just have to remember that 45° is 1/8 of a pie. Here are the rest: $45^{\\circ} = \\frac{1}{8} \\ in \\ pies, \\ 135^{\\circ} = \\frac{3}{8} \\ in \\ pies,\\\\\\\\\\\\225^{\\circ} = \\frac{5}{8} \\ in \\ pies, \\ 315^{\\circ} = \\frac{7}{8} \\ in \\ pies$ The last set of multiples you usually need to know when dealing with the unit circle is 30°, or 1/12 of a circle. Without duplicating any of the above, here are the final fractions: $30^{\\circ} = \\frac{1}{12} \\ in \\ pies, \\ 60^{\\circ} = \\frac{2}{12}", "180° was left, you only have half of a pie. Put more formally, you have: $\\frac{1}{2} \\ in \\ pies$ It may sound ridiculous, but I'm always going to be using the plural phrase, “in pies”, because it's going to help lock in the mnemonic. I'll explain that shortly, but first we need a starting point. If you're given any multiple of 90°, it's pretty easy to picture these in pies, as it was how many of us were taught: $90^{\\circ} = \\frac{1}{4} \\ in \\ pies, \\ 180^{\\circ} = \\frac{2}{4} = \\frac{1}{2} \\ in \\ pies,\\\\\\\\\\\\270^{\\circ} = \\frac{3}{4} \\ in \\ pies, \\ 360^{\\circ} = \\frac{4}{4} = 1 \\ in \\ pies$ 45° isn't much harder, especially considering that we've already dealt with half of them above. You just have to remember that 45° is 1/8 of a pie. Here are the rest: $45^{\\circ} = \\frac{1}{8} \\ in \\ pies, \\ 135^{\\circ} = \\frac{3}{8} \\ in \\ pies,\\\\\\\\\\\\225^{\\circ} = \\frac{5}{8} \\ in \\ pies, \\ 315^{\\circ} = \\frac{7}{8} \\ in \\ pies$ The last set of multiples you usually need to know when dealing with the unit circle is 30°, or 1/12 of a circle. Without duplicating any of the above, here are the final fractions: $30^{\\circ} = \\frac{1}{12} \\ in \\ pies, \\ 60^{\\circ} = \\frac{2}{12}", "a lot of equivalent fractions for $$\\frac{1}{2}$$, for $$\\frac{10}{3}$$, and for 1. ##### Example: Going Backwards $\\frac{20}{32}\\; \\text{(sharing 20 pies among 32 kids)} \\nonumber$ is the same problem as: $\\frac{5 \\cdot 4}{8 \\cdot 4} = \\frac{5}{8}\\; \\text{(sharing five pies among eight kids)} \\ldotp \\nonumber$ Most people say we have cancelled or taken a common factor 4 from the numerator and denominator. Mathematicians call this process reducing the fraction to lowest terms. (We have made the numerator and denominator smaller, in fact as small as we can make them!) Teachers tend to say that we are simplifying the fraction. (You have to admit that $$\\frac{5}{8}$$ does look simpler than $$\\frac{20}{32}$$.) ##### Example: How Low Can You Go? As another example, $$\\frac{280}{350}$$ can certainly be simplified by noticing that there is a common factor of 10 in both the numerator and the denominator: $\\frac{280}{350} = \\frac{28 \\cdot 10}{35 \\cdot 10} = \\frac{28}{35} \\ldotp \\nonumber$ We can go further as 28 and 35 are both multiples of 7: $\\frac{28}{35} = \\frac{4 \\cdot 7}{5 \\cdot 7} = \\frac{4}{5} \\ldotp \\nonumber$ Thus, sharing 280 pies among 350 children gives the same result as sharing 4 pies among 5 children! $\\frac{280}{350} = \\frac{4}{5} \\ldotp \\nonumber$ Since 4 and 5 share no common factors, this is as far as we can go with this example", "% of 1000 = $$\\frac {10}{100}$$ × 1000 = 100 (c) Number of CD’s for folk music = 30 % of 1000 = $$\\frac {30}{100}$$ × 1000 = 300 (d) Number of CD’s for light music = 40 % of 1000 = $$\\frac {40}{100}$$ × 1000 = 400 2. A group of 360 people were asked to vote for their favourite season from the ? three seasons rainy, winter and summer: Question (i). Which season got the most votes ? Solution: Winter season got the most votes (150). Question (ii). Find the central angle of each sector. Solution: Total Votes = 90 + 120 + 150 = 360 ∴ Central angle of the sector corresponding to : Summer season = $$\\frac {90}{360}$$ × 360° = 90° Rainy season = $$\\frac {120}{360}$$ × 360° = 120° Winter season = $$\\frac {150}{360}$$ × 360° = 150° Question (iii). Draw a pie chart to show this information. Solution: Draw three radi in such a way that it makes angles of 90°, 120° and 150° at the centre. 3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people. Colours Number of people Blue 18 Green 9 Red 6 Yellow 3 Total 36 Solution: Central angle of the sector corresponding to : (i)", "\\cdot 200 = 140.$$ In conclusion, $140$ people use bicycle or car most often. Example 2: The following pie chart shows a survey of the number of pieces of clothes, cosmetics and jewelry that women buy. Furthermore, there were $140$ pieces of jewelry in the survey. a) What fraction of the articles is cosmetics? b) Compute the number of items in the survey. c) How many pieces of clothes were in the survey? Solution: a) Fraction of cosmetics is $$\\frac{\\alpha_{1}}{360^{\\circ}} = \\frac{205^{\\circ}}{360^{\\circ}} = \\frac{41}{72}.$$ b) Let $x$ be the number of items. Since there were $140$ pieces of jewelry in the survey, we have $$\\frac{35^{\\circ}}{360^{\\circ}} \\cdot x = 140$$ $$x = 140 \\cdot \\frac{360^{\\circ}}{35^{\\circ}}$$ $$x = 1440.$$ Therefore, the number of items is $1440$. c) There were $$\\frac{120^{\\circ}}{360^{\\circ}} \\cdot 1440 = 480$$ pieces of clothes in a survey. ## Making a pie chart It is a little bit tricky to draw pie charts by hand. Consequently, there is a variety of computer programs in which the whole procedure is easier. If we want to draw a pie chart, we need to have a list of categorical variables and numeric variables also. For instance, in the Example 1 types of transport are the categorical variables, while percentages are the numeric variables. Furthermore, it is important that categories don’t overlap.", "30% of 1000 = $$\\frac{30}{100}$$ × 1000 = 300 Number of CD’s for light music = 40% of 1000 = $$\\frac{40}{100}$$ × 1000 = 400 Question 2. A group of 360 people were asked to vote for their favourite season from the three season rainy, winter and summer. (i) Which season got the most votes? (ii) Find the central angle of each sector. (iii) Draw a pie chart to show this information. Solution: (i) Winter season got the most votes. (ii) Total votes = 90 + 120 + 150 = 360 Central angle of the sector corresponding to summer season No. of people who vote for summer = $$\\frac{\\text { No. of people who vote for summer }}{\\text { No. of people }} \\times 360$$ = $$\\frac{90}{360}$$ × 360° = 90° rainy sector = $$\\frac{120}{360}$$ × 360° = 120° winter sector = $$\\frac{150}{360}$$ × 360° = 150° Question 3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people. Find the proportion of each sector. For example, Blue is $$\\frac{18}{36}=\\frac{1}{2}$$; Green is $$\\frac{9}{36}=\\frac{1}{4}$$ and so on. Use this to find the corresponding angles. Solution: The required pie chart is given below: Question 4. The adjoining pie chart gives the marks scored in an examination by a student in Hindi,", "to say 2 degrees equals 5 students. But this as above, leads to a contradiction when applying for 25% and 95%. (If I'm thinking wrong, there must be just something wrong with my mind, so sorry for arguing, and please correct me) 7. Jun 14, 2006 ### Hootenanny Staff Emeritus I apologise, it is a while since I have done pie charts. The angle is equal to the proportion of students in that group, thus; $$\\theta = \\frac{\\text{number of students}}{\\text{Total}}\\times 360$$ 8. Jun 15, 2006 ### arildno As a general rule for representative charts of some form (for example pie charts) is that PROPORTIONS ARE PRESERVED. Rewriting Hooty's equation shows this: $$\\frac{\\theta}{360}=\\frac{No. students}{Total no.}$$ 9. Jun 15, 2006 ### HallsofIvy Staff Emeritus The problem is, of course, that the \"full school\" is 900 students and the \"full circle\" is 360 degrees. That is there should be $\\frac{900}{360}= \\frac{5}{2}$ student per degree. While it is possible for a full number of students to be represented by an even number of degrees, \"95 degrees\" and \"25 degrees\" must be round-offs. And probably unnecessary round-offs at that. 10. Jun 15, 2006 ### heartless Ok, I took an exam today and lucky I was similar question didn't appear but anyway. Though I have a solution for this problem I still would like", "1. Ratio Is saying that 31,317people/45,678people is the same as 41people/50people? 2. Re: Ratio Originally Posted by vaironxxrd Is saying that 31,317people/45,678people is the same as 41people/50people? That's wrong: $\\displaystyle \\displaystyle{\\frac{31317}{45678} = \\frac{3 \\cdot 11 \\cdot 13 \\cdot 73}{2 \\cdot 3 \\cdot 23 \\cdot 331}}$ 3. Re: Ratio Originally Posted by earboth That's wrong: $\\displaystyle \\displaystyle{\\frac{31317}{45678} = \\frac{3 \\cdot 11 \\cdot 13 \\cdot 73}{2 \\cdot 3 \\cdot 23 \\cdot 331}}$ Sorry for that earboth I posted the wrong ration here it is 37,317/45,678 = 102/125 ? What I'm trying to do is scale down the ration of people to show the difference between these numbers 4. Re: Ratio Just do the divisions (got a calculator?): 37317 / 45678 = .816957... 102 / 125 = .816 Now decide if close enough to suit your purpose! 37332 / 45750 = 102 / 125 is precise.", "partner-id.ru ## People Xxxwebcamera It was during this stretch that Hermes launched the trade into retail sales, and bringing up his own sons in the vocation, the Hermes Circle became purveyors to the most elite clients in Europe, North Africa, Russia, Asia, and the Americas. # Updating bayesian priors Rated 4.26/5 based on 986 customer reviews Free private webcams chat Online today In general Bayesian updating refers to the process of getting the posterior from a prior belief distribution. Method b uses the posterior output as input prior to calculate the next posterior. To calculate $Pr(F| HH)$ , we a) continue using P(Fair)=0.5 $Pr(F|HH) = \\frac = \\frac \\quad\\quad (2)$ $P(HH|F)= \\theta^(1-\\theta)^ = 0.5^(0.5)^= 0.25$ $P(HH)= P(HH|F) \\cdot P(F) P(HH|Biased) \\cdot P(Biased)=(0.25 \\cdot 0.5) (1 \\cdot 0.5) = 0.625$ Hence, plugging into (2), $Pr(F|HH) =\\frac = \\frac = 0.2$ Alternatively, what if we calculate $Pr(F| HH)$ by using b) our updated belief P(Fair)=0.33 which we got from Pr(F|H) in the first step In this case, $P(HH|F)= \\theta^(1-\\theta)^ = 0.33^(1-0.33)^= 0.1089$ $P(HH)= P(HH|F) \\cdot P(F) P(HH|Biased) \\cdot P(Biased)=(0.1089 \\cdot 0.33) (1 \\cdot 0.67) = 0.705937$ Hence, plugging into (2), $Pr(F|HH) =\\frac = \\frac = 0.05091$ Usually a biased coin just means that it's not fair, but it could have any bias. Alternatively one could understand the term as using", "# 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}$$ are equivalent. ##### Example: Fractions equivalent to 3/5 For example, $\\frac{3}{5}\\; \\text{(sharing three pies among five kids)} \\nonumber$ yields the same result as $\\frac{3 \\cdot 2}{5 \\cdot 2} = \\frac{6}{10}\\; \\text{(sharing six pies among ten kids)} \\nonumber$ and as $\\frac{3 \\cdot 100}{5 \\cdot 100} = \\frac{300}{500}\\; \\text{(sharing 300 pies among 500 kids)} \\ldotp \\nonumber$ ##### Think / Pair / Share Write down", "finding their total because they were assuming the seats were numbered; otherwise they would have. 2) Your expression of $\\frac{7!}{3!3!}$ would work fine if the people were arranged in a straight line. Notice that in this case the 3 people sitting together have 7 choices for their positions: $123, 234, \\cdots, 789$. In a circle, though, they have 9 choices instead: $123, 234, \\cdots, 789, 891, 912$. $\\displaystyle 1-\\frac{\\binom{3}{1}3!6!-\\binom{3}{2}4(3!)^{3}+2(3!)^{3}}{8!}=1-\\frac{10800}{40320}=1-\\frac{15}{56}=\\frac{41}{56}$. • Could you please explain why your middle term is $\\binom{3}{2}4(3!)^{3}$ – Anirudh Dec 30 '17 at 20:57", "Here is my take1 on the June 19th Huge Pie puzzle in the Guardian: A huge pie is divided among 100 guests. The first guest gets 1% of the pie. The second guest gets 2% of the remaining part. The third guest gets 3% of the rest, etc. The last guest gets 100% of the last part. Who gets the biggest piece?\" Tables are always good to get a quick overview of what we’ve got, the tricky part being the rest of the pie: Guest Percentage Rest of pie (before guest is served) 1 $$\\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’", "Heads with a biased coin = 1 (makes calculation easy)) Hence, plugging into (1), we get : $Pr(F| H) =\\frac{Pr(H|F)\\cdot P(F)}{P(H)} = \\frac{0.5 \\cdot 0.5}{0.75} = 0.33$ Now, we toss the coin again and get another H. To calculate $Pr(F| HH)$ , we a) continue using P(Fair)=0.5 $Pr(F|HH) = \\frac{Pr(HH|F)\\cdot P(F)}{Pr(HH|F) \\cdot P(F)+Pr(HH|Biased) \\cdot P(Biased)} = \\frac{Pr(HH|F)\\cdot P(F)}{P(HH)} \\quad\\quad (2)$ $P(HH|F)= {n \\choose x} \\theta^{x}(1-\\theta)^{n-x} = {2 \\choose 2} 0.5^{2}(0.5)^{0}= 0.25$ $P(HH)= P(HH|F) \\cdot P(F)+ P(HH|Biased) \\cdot P(Biased)=(0.25 \\cdot 0.5) +(1 \\cdot 0.5) = 0.625$ Hence, plugging into (2), $Pr(F|HH) =\\frac{Pr(HH|F)\\cdot P(F)}{P(HH)} = \\frac{0.25 \\cdot 0.5}{0.625} = 0.2$ Alternatively, what if we calculate $Pr(F| HH)$ by using b) our updated belief P(Fair)=0.33 which we got from Pr(F|H) in the first step In this case, $P(HH|F)= {n \\choose x} \\theta^{x}(1-\\theta)^{n-x} = {2 \\choose 2} 0.33^{2}(1-0.33)^{0}= 0.1089$ $P(HH)= P(HH|F) \\cdot P(F)+ P(HH|Biased) \\cdot P(Biased)=(0.1089 \\cdot 0.33) +(1 \\cdot 0.67) = 0.705937$ Hence, plugging into (2), $Pr(F|HH) =\\frac{Pr(HH|F)\\cdot P(F)}{P(HH)} = \\frac{0.1089 \\cdot 0.33}{0.705937} = 0.05091$ Using method a, we get P(F|HH) = 0.2. Using method b, gives P(F|HH) = 0.05. My question is as to how far method b is a valid approach ? • How did you reason that the probability of getting heads given a biased coin is 1? Anyway, no matter how many times you flip the coin," ]

[ "& \\text{pH} \\\\ \\hline \\text{Beer} & 3.0-5.0 & 4.1-4.5 \\\\ \\text{Wine} & 0.0-12.9 & 2.8-3.8 \\\\ \\text{Sweet liquor } & 30.0-31.0 & 3.3-3.9 \\\\ \\text{Strong alcohol} & 0.0-1.2 & 6.5-6.9 \\\\ \\text{Vodka} & \\text{very little} & 6.0-7.0 \\\\ \\text{Coca Cola} & 10.6(?) & 2.8 \\\\ \\hline \\end{array} $$ $$\\small \\begin{array}{lcc} \\hline \\text{Beverage} & \\text{Carbohydrate Content (%)} & \\text{pH} \\\\ \\hline \\text{Beer} & 3.0-5.0 & 4.1-4.5 \\\\ \\text{Wine} & 0.0-12.9 & 2.8-3.8 \\\\ \\text{Sweet liquor } & 30.0-31.0 & 3.3-3.9 \\\\ \\text{Strong alcohol} & 0.0-1.2 & 6.5-6.9 \\\\ \\text{Vodka} & \\text{very little} & 6.0-7.0 \\\\ \\text{Coca Cola} & 10.6(?) & 2.8 \\\\ \\hline \\end{array}$$ ## Back to INDEX ### Basic MathJax: Superscripts and subscripts You can denote superscripts via the ^ character, and subscripts via _. For example, x^2 renders as $$x^2$$, x_1 renders as $$x_1$$, and x_1^3 renders as $$x_1^3$$. If you want to include more than one character in the super/sub script, enclose it in curly braces ({...}). For example, x^10 renders as $$x^10$$, but x^{10} renders as $$x^{10}$$ ## Back to INDEX ### Using arrows and the equal sign for chemical reactions • (a) On a microscopic level $$\\begin{array}{c cc c} Symbol & Markup & Example & Meaning \\\\ \\hline \\ce{->} & \\text{->} & \\ce{H + Br2 -> HBr + Br} & \\text{one elementary step}\\\\ \\ce{=}", "more code, but nicer output) and also give it a slightly cleaner look: $$\\small \\begin{array}{lcc} \\hline \\text{Beverage} & \\text{Carbohydrate Content (%)} & \\text{pH} \\\\ \\hline \\text{Beer} & 3.0-5.0 & 4.1-4.5 \\\\ \\text{Wine} & 0.0-12.9 & 2.8-3.8 \\\\ \\text{Sweet liquor } & 30.0-31.0 & 3.3-3.9 \\\\ \\text{Strong alcohol} & 0.0-1.2 & 6.5-6.9 \\\\ \\text{Vodka} & \\text{very little} & 6.0-7.0 \\\\ \\text{Coca Cola} & 10.6(?) & 2.8 \\\\ \\hline \\end{array}$$ Align at point: Define for easier typsetting \\newcommand{\\d}[2]{#1.&\\hspace{-1em}#2} and use in table 1.23 as \\d{1}{23} $$\\newcommand{\\d}[2]{#1.&\\hspace{-1em}#2} \\begin{array}{lrl} \\hline \\text{Beverage} & \\text{Nomnom}&\\hspace{-1em}\\text{-factor} \\\\ \\hline \\text{Beer} & 100.&\\hspace{-1em}000 \\\\ \\text{Wine} & 23.&\\hspace{-1em}4567\\\\ \\text{Cola stuff} & 1.&\\hspace{-1em}23\\\\ \\text{Coffee} & 66.&\\hspace{-1em}6\\\\ \\text{Tea (black)}&\\d{0}{3}\\\\ \\hline \\end{array}$$ The mhchem package is not enabled on StackPrinter? Testcode: $$\\ce{2H2 + O2 -> 2H2O}$$ $$\\ce{2H2 + O2 -> 2H2O}$$ Enable mhchemmanually via $\\require{mhchem}$ $\\require{mhchem}$ and try again: $$\\ce{2H2 + O2 -> 2H2O}$$ $\\mathcal{M}_{ath}\\mathrm{J^{a}X}\\neq\\LaTeX$ Operation Broken Arrow $$\\require{cancel}\\ce{\\bcancel{->}\\cancel{->}\\xcancel{->}}$$ $$%\\require{Extpfeil}\\Newextarrow{\\xrightharpoonup}{5,10}{0x21C0}\\xrightharpoonup{\\not{}}$$ $$\\nrightarrow$$ $$\\ce{\\rlap{~~/\\!/}->}$$ $$\\def\\nnot{\\color{red}{\\rlap{~~/\\!/}}}$$ $$\\ce{\\nnot->T[water]}$$ $$\\begin{multline} E^\\ominus(\\ce{Fe/Fe^{3+}}) - \\frac13\\cdot \\mathcal{R}T\\mathrm{F}^{-1} \\ln\\frac{\\ce{Fe}}{\\ce{Fe^{3+}}} = \\\\ a\\cdot\\left(E^\\ominus(\\ce{Fe/Fe^{2+}}) - \\frac12\\cdot \\mathcal{R}T\\mathrm{F}^{-1} \\ln\\frac{\\ce{Fe}}{\\ce{Fe^{2+}}}\\right) \\\\+ b\\cdot\\left(E^\\ominus(\\ce{Fe^{2+}/Fe^{3+}}) - \\frac11\\cdot \\mathcal{R}T\\mathrm{F}^{-1} \\ln\\frac{\\ce{Fe^{2+}}}{\\ce{Fe^{3+}}}\\right) \\end{multline}$$ # The Tag Test \\begin{array}{|c|l|c|c|c|}\\hline \\# & {\\small \\bf Questions} & {\\small \\rm \\color{green}{Yes}} & {\\small \\rm \\color{#999999}{Partially}} & {\\small \\rm \\color{red}{No}} \\\\\\hline \\hline 1&\\small\\text{Does 'experts on X' make sense?} & & & \\\\\\hline 2&\\small\\text{Does 'I'm studying X' make sense?} & & & \\\\\\hline 3&\\small\\text{Would you leave", "in the GIMP prior to dropping it in Inkscape to get the images I wanted. figure caffeine4=[0 0 0 0 0 0 0 0]; sleep4=[6.95 10.87 9.75 7.52 10.95 9.43 8.75 10.23]; day=[25 26 27 28 29 30 31 32]; [AX,H1,H2] = plotyy(day,sleep4,day,caffeine4,'plot'); title('Phase 4: Daily Hours Slept and Caffeine Intake'); xlabel('Date (in April)'); %xlim([26 13]); set(gca,'XTick',25:1:32); %set(gca,'box','off'); set(get(AX(1),'Ylabel'),'String','Amount Slept Daily (hrs)'); set(get(AX(2),'Ylabel'),'String','Daily Caffeine Intake (mg)') ; axis(AX(1),[25 32 0 11]); axis(AX(2),[25 32 0 700]); set(AX(1),'YLim',[0 11], 'YTick',0:1:11, 'box','off'); set(AX(2),'YLim',[0 700],'YTick',0:100:700,'box','off', 'box','off'); set(H1,'LineWidth',3,... 'MarkerSize',7,... 'LineStyle','-',... 'Marker', 'o',... 'MarkerEdgeColor', 'none',... 'MarkerFaceColor', 'b'); set(H2,'LineWidth',3,... 'MarkerSize',7,... 'LineStyle','-',... 'Marker', 'o',... 'MarkerEdgeColor', 'none',... 'MarkerFaceColor', 'g'); Above code resulted in this graph: This data was actually 25 April to 2 May, but instead of learning about timeseries in MATLAB, I just cheated and stuck \"April / May\" in the xlabel and used 31 and 32 to not break my code. I planned to post-process in gimp to turn \"31 32\" into \"1 2\" but forgot to do so. For phase 3, I couldn't figure how to get rid of the 20.5 and 21.5 tick marks when I expanded to graphs to reasonable size, so I did post-processing in GIMP. Here are the two images I smushed together into the third: += Misc. other tricks too lazy to document fully: Using \"shift-J\" in Vim to", "% NOW DATA FOR SECOND [xData2, tmp] = prepareCurveData( Fq2r, Prem2 ); yData2 = tmp+pi/2; % and the other start points, fit... opts.StartPoint = [0.425259320214135 0.312718886820616]; [fitresult2, gof2] = fit( xData2, yData2, ft, opts ); hold on % add to existing plot h(2) = plot( fitresult2, xData2, yData2 ); % set both line handles in h array... set(h,'LineWidth',2,'Markersize',7.5) grid on legend({'Phasenverschiebung','Fit: $(-1,02 \\pm 0,07) \\cdot \\tan^{-1}(\\frac{2\\cdot (27,59 \\pm 0,27)\\cdot \\omega}{3,83^2-\\omega^2})$'}, ... {'Phasenverschiebung bei $I = 0,25 A$','Fit: $(-1,26 \\pm 0,20) \\cdot \\tan^{-1}(\\frac{2\\cdot (26,59 \\pm 0,65)\\cdot \\omega}{3,83^2-\\omega^2})$'}, ... 'Interpreter','latex','FontSize',14) xlabel({'$\\omega$ in rad'},'Interpreter','latex','FontSize',14) ylabel({'Phasenverschiebung $\\Delta \\varphi$ in rad'},'Interpreter','latex','FontSize',14) is a first cut Niklas Kurz on 25 Jun 2020 First of Thank you so much for your effort: However, I guess I should have mentionned that the 2 fits doesn't have same Size, as: Fqr = [400 500 525 550 575 600 625 650 675 700 750].*(2*pi/1000) Prem = [0.277 0.527 0.637 0.628 1.034 1.561 2.167 2.727 2.758 2.749 2.861]-pi/2 Fq2r = [300 400 500 525 550 575 600 625 650 675 700 750].*(2*pi/1000) Prem2 = [0.094 0.220 0.253 0.255 0.517 1.111 2.630 3.005 3.090 2.990 3.063 2.936]-pi/2 After running the code Fqr and Fq2r also have been tangeled up. So, how to fix it? dpb on 25 Jun 2020 Different size won't matter...all I see is missed adding the '1'", "\\text{(E)}\\ \\text{Sunday}$ ## Problem 6 A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it? $\\text{(A)}\\ \\text{A} \\qquad \\text{(B)}\\ \\text{B} \\qquad \\text{(C)}\\ \\text{C} \\qquad \\text{(D)}\\ \\text{D} \\qquad \\text{(E)}\\ \\text{E}$ ## Problem 7 The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E? $[asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black); } draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9)); for(i=1; i<8; i=i+1) { draw((0,2i)--(19,2i)); } label(\"0\", (0,2*0), W); label(\"1\", (0,2*1), W); label(\"2\", (0,2*2), W); label(\"3\", (0,2*3), W); label(\"4\", (0,2*4), W); label(\"5\", (0,2*5), W); label(\"6\", (0,2*6), W); label(\"7\", (0,2*7), W); label(\"8\", (0,2*8), W); label(\"A\", (0*4+1.5, 0), S); label(\"B\", (1*4+1.5, 0), S); label(\"C\", (2*4+1.5, 0), S); label(\"D\", (3*4+1.5, 0), S); label(\"E\", (4*4+1.5, 0), S); label(\"SWEET TOOTH\", (9.5,18), N); label(\"Kinds of candy\", (9.5,-2), S); label(rotate(90)*\"Number of students\", (-2,8), W);[/asy]$ $\\text{(A)}\\ 5 \\qquad \\text{(B)}\\ 12 \\qquad \\text{(C)}\\ 15 \\qquad \\text{(D)}\\", "$[asy] defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype(\"4 4\"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label(\"1/\\sqrt{3}\",(0,-0.5),W,fontsize(8)); [/asy]$ $\\Longrightarrow$ $[asy] defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype(\"4 4\")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype(\"4 4\")); draw(Circle((0,0),1),linetype(\"4 4\")); label(\"\\sqrt{3}\",(0,-0.5),W,fontsize(8)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); [/asy]$ The area of the regular hexagon is $6 \\cdot \\left( \\frac{\\left(\\sqrt{3}\\right)^2 \\sqrt{3}}{4} \\right) = \\frac{9}{2}\\sqrt{3}$. The total area of the six $120^{\\circ}$ sectors is $6\\left(\\frac{1}{3}\\pi - \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\sqrt{3}\\right) = 2\\pi - \\frac{3}{2}\\sqrt{3}$. Their sum is $2\\pi + \\sqrt{27}$, and $a+b = \\boxed{029}$. - Th3Numb3rThr33 ## Solution 3 (Calculus) We can describe the line parallel to the imaginary axis $x=\\frac{1}{2}$ using polar coordinates as $r(\\theta)=\\dfrac{1}{2\\cos{\\theta}},$ which rearranges to $z=\\left(\\dfrac{1}{2\\cos{\\theta}}\\right)(cis{\\theta})\\implies \\frac{1}{z}=2\\cos{\\theta}cis(-\\theta).$ Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}r^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}(2\\cos\\theta)^2 d\\theta.$ Thankfully, this is a routine computation: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}2(\\cos\\theta)^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta$ $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta = 12\\left[\\frac{1}{2}\\sin{2\\theta}+\\theta\\right]_0^{\\frac{\\pi}{6}}=12\\left(\\frac{\\sqrt{3}}{4}+\\frac{\\pi}{6}\\right)=2\\pi+3\\sqrt{3}=2\\pi + \\sqrt{27}$ $a+b = \\boxed{029}$.", "4), bg = \"rgba(218,112,214,.5)\") Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 If we aren’t interested in coloring just those two rows, we can apply color to the entire table with the bg_pattern sprinkle. This sprinkle accepts as many colors as you want to cycle through. basetable %>% sprinkle(bg_pattern = c(\"orchid\", \"plum\")) Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 # 7 Font Sizes and Colors Font sizes and colors are modified with the font_size and font_color sprinkles. We’ll employ these simultaneously to highlight our significant rows. basetable %>% sprinkle(rows = c(2, 4), font_color = \"orchid\", font_size = 24, font_size_units = \"pt\") Term Coefficient SE T-statistic P-value (Intercept) 9.365 8.373 1.118 0.27 qsec 1.245 0.383 3.252 0.003 factor(am)Manual 3.151 1.941 1.624 0.12 wt -3.926 0.743 -5.286 < 0.001 factor(gear)4 -0.268 1.655 -0.162 0.87 factor(gear)5 -0.27 2.063 -0.131 0.9 (Woah! That was a bit too much.) # 8 Dimensions and Alignment In addition to the sprinkles already discussed, we can also use sprinkles", ">> >>10202195 Write it down. If you don't see it immediately make a sketch of, lets say, $f_3$ and $f_1$. >> >>10185459 Man I'm dumb. >> >>10202207 $f_{3}-f_{1} = \\frac{1}{x}, \\frac{1}{2}\\leq x < 1$, $0$ otherwise? >> >>10202173 Dummit & Foote is very comprehensive, and starts from the basics. >> >>10202231 m8 Drink a coffee, relax, and draw it again. If it doesn't work after that I'll save you. >> >>10202248 If $m>n$ , $\\displaystyle f_{m}-f_{n} = \\begin{cases} \\frac{1}{x} &, \\frac{1}{m+1}\\leq x < \\frac{1}{n+1} \\\\ 0 &, \\text{ otherwise } \\end{cases}$ ? >> >>10202277 Nice! Now integrate it and you're basically done. >> >>10202288 irish coffee does the trick. god help my liver. >> >>10202329 don't drink and derive >> >>10185459 Anyone care to explain greens theorem for me? >> >>10202556 too late >> File: 1542851986813.jpg (59 KB, 559x1006) 59 KB JPG >>10202559 When you integrate a normal function, you calculate the values of the antiderivative on the edges to get the integral along the middle. Same general idea. Except now the edges are a curve, and the middle is an area. It's generalized later with Stokes. >> File: Natsumi.png (1.45 MB, 1449x1415) 1.45 MB PNG >>10202628 arigato senpai uWu >> >>10201918 what more do you need than your picture ? that's literally all >> File: 1507562963146.png (36", "each took four $100$-point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests? $\\mathrm{(A)}\\ 10 \\qquad\\mathrm{(B)}\\ 15 \\qquad\\mathrm{(C)}\\ 20 \\qquad\\mathrm{(D)}\\ 25 \\qquad\\mathrm{(E)}\\ 40$ Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures $\\textbf{Bake Sale}$ Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown. $\\circ$ Art's cookies are trapezoids: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label(\"5 in\", (2.5,0), S); label(\"3 in\", (5,1.5), E); label(\"3 in\", (3.5,3), N);[/asy]$ $\\circ$ Roger's cookies are rectangles: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label(\"4 in\", (2,0), S); label(\"2 in\", (4,1), E);[/asy]$ $\\circ$ Paul's cookies are parallelograms: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label(\"3 in\", (1.5,0), S); label(\"2 in\", (2.5,1), W);[/asy]$ $\\circ$ Trisha's cookies are triangles: $[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label(\"3 in\", (1.5,0), S); label(\"4 in\", (3,2), E);[/asy]$ ## Problem 8 Who gets the fewest cookies from one batch of cookie dough? $\\textbf{(A)}\\ \\text{Art} \\qquad\\textbf{(B)}\\ \\text{Paul}\\qquad\\textbf{(C)}\\ \\text{Roger}\\qquad\\textbf{(D)}\\ \\text{Trisha}\\qquad\\textbf{(E)}\\ \\text{There is a tie for fewest}$ ## Problem 9 Each friend", "$\\frac {\\sqrt {130}}2$, giving an answer of $130 + 2 = \\boxed{132}$. $[asy]import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); filldraw((-3,4)--(-2,4)--(-2,5)--(-3,5)--cycle,evevff,blue); filldraw((3,4)--(4,4)--(4,5)--(3,5)--cycle,evevff,blue); filldraw((4,3)--(5,3)--(5,4)--(4,4)--cycle,evevff,blue); filldraw((5,0)--(6,0)--(6,1)--(5,1)--cycle,evevff,blue); filldraw((4,-3)--(5,-3)--(5,-2)--(4,-2)--cycle,evevff,blue); filldraw((3,-3)--(3,-4)--(4,-4)--(4,-3)--cycle,evevff,blue); filldraw((0,-5)--(1,-5)--(1,-4)--(0,-4)--cycle,evevff,blue); filldraw((-3,-4)--(-2,-4)--(-2,-3)--(-3,-3)--cycle,evevff,blue); filldraw((-4,-3)--(-3,-3)--(-3,-2)--(-4,-2)--cycle,evevff,blue); filldraw((-4,3)--(-3,3)--(-3,4)--(-4,4)--cycle,evevff,blue); filldraw((-5,0)--(-4,0)--(-4,1)--(-5,1)--cycle,evevff,blue); filldraw((0,6)--(0,5)--(1,5)--(1,6)--cycle,evevff,blue); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=1,gy=1; 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); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); draw(circle((0,0),5),linewidth(1.6)); draw(circle((0.5,0.5),5.7),linetype(\"2 2\")); draw((-3,4)--(-2,4),zzttqq); draw((-2,4)--(-2,5),zzttqq); draw((-2,5)--(-3,5),zzttqq); draw((-3,5)--(-3,4),zzttqq); draw((3,4)--(4,4),zzttqq); draw((4,4)--(4,5),zzttqq); draw((4,5)--(3,5),zzttqq); draw((3,5)--(3,4),zzttqq); draw((4,3)--(5,3),zzttqq); draw((5,3)--(5,4),zzttqq); draw((5,4)--(4,4),zzttqq); draw((4,4)--(4,3),zzttqq); draw((5,0)--(6,0),zzttqq); draw((6,0)--(6,1),zzttqq); draw((6,1)--(5,1),zzttqq); draw((5,1)--(5,0),zzttqq); draw((4,-3)--(5,-3),zzttqq); draw((5,-3)--(5,-2),zzttqq); draw((5,-2)--(4,-2),zzttqq); draw((4,-2)--(4,-3),zzttqq); draw((3,-3)--(3,-4),zzttqq); draw((3,-4)--(4,-4),zzttqq); draw((4,-4)--(4,-3),zzttqq); draw((4,-3)--(3,-3),zzttqq); draw((0,-5)--(1,-5),zzttqq); draw((1,-5)--(1,-4),zzttqq); draw((1,-4)--(0,-4),zzttqq); draw((0,-4)--(0,-5),zzttqq); draw((-3,-4)--(-2,-4),zzttqq); draw((-2,-4)--(-2,-3),zzttqq); draw((-2,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-4),zzttqq); draw((-4,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-2),zzttqq); draw((-3,-2)--(-4,-2),zzttqq); draw((-4,-2)--(-4,-3),zzttqq); draw((-4,3)--(-3,3),zzttqq); draw((-3,3)--(-3,4),zzttqq); draw((-3,4)--(-4,4),zzttqq); draw((-4,4)--(-4,3),zzttqq); draw((-5,0)--(-4,0),zzttqq); draw((-4,0)--(-4,1),zzttqq); draw((-4,1)--(-5,1),zzttqq); draw((-5,1)--(-5,0),zzttqq); draw((0,6)--(0,5),zzttqq); draw((0,5)--(1,5),zzttqq); draw((1,5)--(1,6),zzttqq); draw((1,6)--(0,6),zzttqq); dot((0,5),ds); dot((3,4),ds); dot((4,3),ds); dot((5,0),ds); dot((4,-3),ds); dot((3,-4),ds); dot((0,-5),ds); dot((-3,-4),ds); dot((-4,-3),ds); dot((-4,3),ds); dot((-3,4),ds); dot((-2,4),ds); dot((-2,5),ds); dot((-3,5),ds); dot((4,4),ds); dot((4,5),ds); dot((3,5),ds); dot((5,3),ds); dot((5,4),ds); dot((4,4),ds); dot((6,0),ds); dot((6,1),ds); dot((5,1),ds); dot((5,-3),ds); dot((5,-2),ds); dot((4,-2),ds); dot((3,-3),ds); dot((4,-4),ds); dot((4,-3),ds); dot((1,-5),ds); dot((1,-4),ds); dot((0,-4),ds); dot((-2,-4),ds); dot((-2,-3),ds); dot((-3,-3),ds); dot((-3,-2),ds); dot((-4,-2),ds); dot((-3,3),ds); dot((-3,4),ds); dot((-4,4),ds); dot((-5,0),ds); dot((-4,0),ds); dot((-4,1),ds); dot((-5,1),ds); dot((0,6),ds); dot((1,5),ds); dot((1,6),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "file (in csv format, for comma-separated values), display it in a nice table, and extract the columns that we need—which we'll convert to numpy arrays to work with. Let's start by importing the two Python libraries that we need. In [1]: import pandas import numpy ### Step 1: Read the data file Below, we'll take a peek into the data file, beers.csv, using the system command head (which we can use with a bang, thanks to IPython). But first, we will download the data using a Python library for opening a URL on the Internet. We created a short URL for the data file in the public repository with our course materials. The cell below should download the data in your current working directory. The next cell shows you the first few lines of the data. In [2]: from urllib.request import urlretrieve URL = 'http://go.gwu.edu/engcomp2data1' urlretrieve(URL, 'beers.csv') Out[2]: ('beers.csv', <http.client.HTTPMessage at 0x11d88c9e8>) In [3]: !head \"beers.csv\" ,abv,ibu,id,name,style,brewery_id,ounces 0,0.05,,1436,Pub Beer,American Pale Lager,408,12.0 1,0.066,,2265,Devil's Cup,American Pale Ale (APA),177,12.0 2,0.071,,2264,Rise of the Phoenix,American IPA,177,12.0 3,0.09,,2263,Sinister,American Double / Imperial IPA,177,12.0 4,0.075,,2262,Sex and Candy,American IPA,177,12.0 5,0.077,,2261,Black Exodus,Oatmeal Stout,177,12.0 6,0.045,,2260,Lake Street Express,American Pale Ale (APA),177,12.0 7,0.065,,2259,Foreman,American Porter,177,12.0 8,0.055,,2258,Jade,American Pale Ale (APA),177,12.0 We can use pandas to read the data from the csv file, and save it into a new variable called beers. Let's 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", "away and see if there’s any beer left in the fridge or a game of footy on the telly – you take a moment and look closely at these commands one at a time. Every single one of these commands is something you’ve seen before somewhere else in the book. There’s nothing novel about them (though I’ll have to admit that the tapply() function takes a while to get a handle on), so if you’re not quite sure how these commands work, this might be a good time to try playing around with them yourself, to try to get a sense of what’s happening. On the other hand, if this does seem to make sense, then you won’t be all that surprised at what happens when I wrap these variables in a data frame, and print it out… Y <- data.frame( group, outcome, gp.means, dev.from.gp.means, squared.devs ) print(Y, digits = 2) ## group outcome gp.means dev.from.gp.means squared.devs ## 1 placebo 0.5 0.45 0.050 0.0025 ## 2 placebo 0.3 0.45 -0.150 0.0225 ## 3 placebo 0.1 0.45 -0.350 0.1225 ## 4 anxifree 0.6 0.72 -0.117 0.0136 ## 5 anxifree 0.4 0.72 -0.317 0.1003 ## 6 anxifree 0.2 0.72 -0.517 0.2669 ## 7 joyzepam 1.4 1.48 -0.083 0.0069 ## 8 joyzepam 1.7 1.48 0.217 0.0469 ## 9 joyzepam 1.3 1.48", "\\usepackage{pgfplots} \\usetikzlibrary{hobby} \\usepackage{filecontents} \\pgfplotsset{compat=1.9, every tick label/.append style={font=\\normalsize}} \\pgfplotscreateplotcyclelist{my black white}{% solid,smooth, every mark/.append style={solid, fill=white}, mark=triangle*\\\\% solid, every mark/.append style={solid, fill=black}, mark=triangle*\\\\% densely dashed, every mark/.append style={solid, fill=white},mark=diamond*\\\\% densely dashed, every mark/.append style={solid, fill=black}, mark=diamond*\\\\% } \\begin{filecontents}{p2.dat} SAT ETCY-5 ETCX-5 ETCY-10 ETCX-10 AR-5 AR-10 100 0.46 0.50 0.54 0.61 0.17 0.25 90 0.45 0.48 0.53 0.59 0.18 0.25 80 0.43 0.47 0.51 0.58 0.19 0.27 70 0.41 0.45 0.49 0.57 0.21 0.30 60 0.39 0.44 0.48 0.55 0.25 0.35 50 0.37 0.42 0.46 0.54 0.30 0.44 40 0.35 0.40 0.44 0.52 0.38 0.57 30 0.33 0.37 0.42 0.51 0.50 0.71 20 0.29 0.34 0.39 0.48 0.63 0.87 10 0.24 0.29 0.34 0.45 0.78 1.04 0 0.18 0.23 0.29 0.41 0.92 1.22 \\end{filecontents} \\begin{document} \\begin{tikzpicture} \\begin{axis}[ cycle list name=my black white, title={Compressing Pressure: 0.2 MPa}, xmin=0, xmax=100, xlabel={Water Saturation($S_w$)}, xtick distance=10, ymin=0, ymax=0.7, ylabel={Avg. ETC \\quad $\\frac{K_{\\mathrm{eff}}}{K_s}$ (Dimensionless)}, legend style ={ at={(0.25,0.4)}, anchor=north west, draw=none, font=\\normalsize, fill=white,align=left, }, hobby ] \\addlegendentry{$K_{\\mathrm{eff}(x)}$ -- Overlap 5\\% }; \\addlegendentry{$K_{\\mathrm{eff}(y)}$ -- Overlap 5\\% }; \\addlegendentry{$K_{\\mathrm{eff}(x)}$ -- Overlap 10\\% }; \\addlegendentry{$K_{\\mathrm{eff}(y)}$ -- Overlap 10\\% }; `", "# Drawing a half-full bottle in TikZ I want to fill half of the bottle. I tried these kind of commands: %\\filldraw[color=black!100, fill=cyan!30, very thick](-3,-2.4) arc (180:230:2.5) .. controls (-2,-5) and (-1.9,-5.2) .. (-1.8,-5.5) ; But I failed. Here is the bottle: \\documentclass[11pt,a4paper,oneside]{article} \\usepackage[a4paper,left=3cm,right=2cm,top=2.5cm,bottom=2.5cm]{geometry} \\usepackage[utf8x]{inputenc} \\usepackage{tikz} \\begin{document} \\begin{tikzpicture}[xscale=1,yscale=1] \\draw[very thick](0,0) arc (0:180:1.5); \\draw[very thick] (-3,0) -- (-3,-2.5); \\draw[very thick] (0,0) -- (0,-2.5); \\draw[color=black!100,very thick](-3,-2.5) arc (180:235:3); \\draw[color=black!100,very thick](0,-2.5) arc (0:-55:3); \\draw[very thick] (-1.25,-4.95) -- (-1.25,-5.2) -- (-1.7,-5.2) -- (-1.7,-4.95); \\end{tikzpicture} \\end{document} What I want: • Simply draw the half bottle starting with the neck on the left side and ending with the neck on the right side. – AndréC Dec 31 '18 at 11:39 • And if I wanted to draw a half-empty bottle ;-) – Peter Wilson Dec 31 '18 at 18:53 \\documentclass[border=5pt,tikz]{standalone} \\begin{document} \\begin{tikzpicture}[xscale=1,yscale=1] \\fill[blue!20] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\fill[white] (0,0) arc(0:180:1.5) --+ (0,-2) -| cycle; \\fill[blue!10] (-1.5,-2) circle ({1.5cm-0.4pt} and 0.5cm); % Thank you, marmot! ;) \\draw[very thick] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\end{tikzpicture} \\end{document} Output: See below Explanation: Maybe beginners don't know the clip option and maybe the want just a type of \"trivial\" answer, so I", "while still being complete. Create boxplots for each of the 10-day periods between 2014 and 202. Use the option -d flag option to let the function find the best location and formats for the x-axis labels. Given the large number of boxplots, we will need to adjust some settings to achieve a readable result. See the manual page for an explanation of the different options used below. t.rast.boxplot -d -g input=LAI300 dpi=300 \\ plot_dimensions=18,5 rotate_labels=0 \\ font_size=14 bx_width=8 bx_lw=0.1 \\ median_lw=1.2 bx_color=51:118:183:255 \\ median_color=255:255:255:255 \\ whisker_linewidth=0.5 Given the large number of boxplots, we will need to adjust some settings to achieve a readable result. See the manual page for an explanation of the different options used below. Module(\"t.rast.boxplot\", flags=\"dg\", input=\"LAI300\", dpi=300, plot_dimensions=\"18,5\", rotate_labels=0, font_size=16, bx_width=8, bx_lw=0.1, median_lw=1.2, bx_color=\"51:118:183:255\", median_color=\"255:255:255:255\", whisker_linewidth=0.5) execGRASS(\"t.rast.boxplot\", flags=\"dg\", input=\"LAI300\", dpi=300, plot_dimensions=\"18,5\", rotate_labels=0, font_size=16, bx_width=8, bx_lw=0.1, median_lw=1.2, bx_color=\"51:118:183:255\", median_color=\"255:255:255:255\", whisker_linewidth=0.5) The boxplot time series in Figure 8 provides a quick overview of the yearly and seasonal patterns, but also the spatial variation in Bardia National Park. As observed earlier, the low spatial variation in LAI values in 2020 stand out, as does the high spatial variation in the LAI peak period in 2014. Out of curiosity, I also plotted the monthly rainfall, based on the CRU dataset6. Comparing the two graphs shows how the peak", "# Approximating solutions¶ In [1]: import numpy w = var('w') palette = [(215/255, 0/255, 132/255), (255/255, 1/255, 73/255), (255/255, 121/255, 1/255), (255/255, 210/255, 0/255)] cool_palette = [(0/255, 150/255, 173/255), (0/255, 200/255, 146/255)] ## Computing a special function with Picard iteration¶ The function $G(w) = w e^w$ is invertible on the domain $[0, \\infty)$. In [24]: g_graph = plot(w*exp(w), (w, 0, log(4)), thickness = 3, color = cool_palette[1]) g_labels = ['$w$', '$G(w)$'] show(g_graph, axes_labels = g_labels, aspect_ratio = 1/2) Its inverse $W(t)$ is called the \"Lambert $W$ function.\" We can draw a graph of $W$ by graphing $G$ and then reflecting to switch the input and the output. If we have a way to calculate $e^w$, we can use this idea to calculate $W$. But could we calculate $W$ directly? In [29]: w_graph = parametric_plot((w*exp(w), w), (w, 0, log(4)), thickness = 3, color = cool_palette[1]) w_labels = ['$t$', '$W(t)$'] show(w_graph, aspect_ratio = 2, axes_labels = w_labels) Even if we don't know how to calculate $e^w$, there's one value of $G$ we can calculate: we know $G(0) = 0$. Flipping inputs and outputs, we learn that $W(0) = 0$ too. We know from the definition of an inverse that $G(W(t)) = t$. In other words $W(t) e^{W(t)} = t$. Differentiating both sides with respect to $t$, and using the definition", "legend(loc=2) xlabel('grain diameter (mm)',fontsize=15) ylabel('settling velocity (m/s)',fontsize=15) axis([0,5,0,2]); A log-log plot is much better for looking at a wide spectrum of grain sizes. In [112]: d = np.arange(0,0.01,0.00001) ws = v_stokes(rop,rof,d,visc,C1) wt = v_turbulent(rop,rof,d,visc,C2) wf = v_ferg(rop,rof,d,visc,C1,C2) figure(figsize=(10,8)) loglog(d*1000,ws,label='Stokes',linewidth=3) loglog(d*1000,wt,'g',label='Turbulent',linewidth=3) loglog(d*1000,wf,'r',label='Ferguson-Church',linewidth=3) plot([1.0/64, 1.0/64],[0.00001, 10],'k--') text(0.012, 0.0007, 'fine silt', fontsize=13, rotation='vertical') plot([1.0/32, 1.0/32],[0.00001, 10],'k--') text(0.17/8, 0.0007, 'medium silt', fontsize=13, rotation='vertical') plot([1.0/16, 1.0/16],[0.00001, 10],'k--') text(0.17/4, 0.0007, 'coarse silt', fontsize=13, rotation='vertical') plot([1.0/8, 1.0/8],[0.00001, 10],'k--') text(0.17/2, 0.001, 'very fine sand', fontsize=13, rotation='vertical') plot([0.25, 0.25],[0.00001, 10],'k--') text(0.17, 0.001, 'fine sand', fontsize=13, rotation='vertical') plot([0.5, 0.5],[0.00001, 10],'k--') text(0.33, 0.001, 'medium sand', fontsize=13, rotation='vertical') plot([1, 1],[0.00001, 10],'k--') text(0.7, 0.001, 'coarse sand', fontsize=13, rotation='vertical') plot([2, 2],[0.00001, 10],'k--') text(1.3, 0.001, 'very coarse sand', fontsize=13, rotation='vertical') plot([4, 4],[0.00001, 10],'k--') text(2.7, 0.001, 'granules', fontsize=13, rotation='vertical') text(6, 0.001, 'pebbles', fontsize=13, rotation='vertical') legend(loc=2) xlabel('grain diameter (mm)', fontsize=15) ylabel('settling velocity (m/s)', fontsize=15) axis([0,10,0,10]); plot(D,w,'o',markerfacecolor=[0.6, 0.6, 0.6], markersize=8); This plot shows how neither Stokes' Law, nor the velocity based on turbulent drag are valid for calculating settling velocities of sand-size grains in water, whereas the Ferguson-Church equation provides a good fit for natural river sand. Grain settling is a special case of the more general problem of flow past a sphere. The analysis and plots above are all dimensional, that is, you can quickly check by looking at the plots what is the approximate", "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", "(-1.5,-2) circle ({1.5cm-0.4pt} and 0.5cm); \\draw[very thick] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\end{tikzpicture} \\end{document} Explanation: We can just \"clip\" the bottle: everything, what we draw, is now in the area of the bottle. Everything outside that area is invisible. So we just clip the bottle and fill a rectangle, such it fills a certain area of the bottle. Output: EDIT: For Sebastiano: \\documentclass[border=5pt,tikz]{standalone} \\usetikzlibrary{decorations.text,backgrounds} \\definecolor{wine}{RGB}{216,198,62} \\definecolor{bottle}{RGB}{76,163,58} \\tikzset{ my/.style={ postaction={decorate},decoration={text along path, text={#1},text align=center} } } \\begin{document} \\begin{tikzpicture} \\begin{scope} \\clip (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\fill[inner color=bottle!50,outer color=bottle] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\fill[wine!60] (-3,-5.5) rectangle ++(3,3.5); \\fill[wine!40] (-1.5,-2) circle ({1.5cm-0.4pt} and 0.5cm); \\foreach \\x in {-5,-4.9,...,5} \\foreach \\y in {-5,...,-3} { \\pgfmathsetmacro\\opacity{random(1,10)*(1/10)} } \\draw[very thick] (0,-2.5) --+ (0,2.5) arc (0:180:1.5) -- (-3,-2.5) arc (180:235:3) --+ (0,-.25) --+ (.45,-.25) --+ (.45,.015) (0,-2.5) arc (0:-55:3); \\path[my={The magic of Ti{\\emph{\\color{orange}k}}Z}] (-3.5,.5) arc(-180:0:2 and 1); \\end{scope} \\end{tikzpicture} \\end{document} Output: • I'm struggling to understand what does it mean: \\fill[white] (0,0) arc(0:180:1.5) --+ (0,-2) -| cycle; . I understand the arc part, but I don't get how you link them each other by using" ]

[ "# 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 $$x$$ ? ×", "Similar Topics: 1 What is the lowest positive integer that is divisible by each of the 2 13 Sep 2016, 01:16 3 What is the lowest positive integer that is divisible by each of the i 6 06 Jun 2016, 04:27 2 What is the lowest positive integer that is divisible by each of the i 5 15 Sep 2015, 12:03 7 What is the lowest positive integer that is divisible by each of the 4 11 Oct 2014, 12:04 4 What is the lowest positive integer that is divisible by eac 4 06 May 2012, 05:54 Display posts from previous: Sort by", "# Which Four Integers? True or false: $\\quad$ The product of any 4 consecutive integers is always divisible by 4. ×", "are all positive. Theorem. [Euclid] There is an infinite number of primes.", "# Eleven primes Find the sum of all positive primes $$p$$ there such that $$p^2+11$$ has exactly six positive divisors. ×", "four numbers if 0 is not a valid number.", "is divisible by 33 for every non negative integer n.", "• 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.", "# Funny Numbers! A positive integer $$n$$ is funny if for each of its positive divisors $$d$$, the number $$d+2$$ is prime. Find the sum of all funny numbers that have the most quantity of divisors. ×", "? (1) k^4 is divisible by 100 (2) 50*k has 2 prime factors Give me an algebraic solution. And i will tell you where you went wrong Bunuel can you help us out? lol Re: If k is a positive integer, how many unique prime factors does 14k [#permalink] 16 Dec 2017, 11:49 Display posts from previous: Sort by", "# 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 it Primus day? The positive number $$k$$ 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 possible value for $$k$$? ×", "test of whole number Is it an integer? 6 tens. If the last digit of a whole number is 0, then that whole number is divisible by 10.Example: 12635360 is divisible by 10 since the last digit is 0. 1. DM me your math problems! 2. 1. Practice rounding whole numbers to the nearest hundred or thousand. !----Have Instagram? This problem has been solved! Whole Numbers Calculator: Whole Number Calculator. *See complete details for Better Score Guarantee. The number line above only shows whole numbers from 0 to 10. Therefore, we can say that that whole numbers are not closed under subtraction. If the last digit of a whole number is either 0 or 5, then that whole number is divisible by 5.Example: 52745 is divisible by 5 since the last digit is 5.5. Label them as 1, 2, 3,…. How many halves does Mary have? However, not all numbers are shown. http://bit.ly/tarversub Subscribe to join the best students on the planet! Fourth grade students will identify place value of whole numbers through hundred millions. wileecoy. 7) 8*6 = 48, is a whole number because it is a positive number with no decimal or fraction parts. This testis based on the following Common Core Standard: CCSS.Math.Content.4.NBT.A.2 Posted on 2002-04-08 01:10:46 by Charlie Killian. Definition.A whole number n is", "positive integers We know that a way to compute the least common multiple between two or more integer numbers is based on their decomposition into prime factors. But there exists a particularly interesting case in which decomposition into prime factors can be avoided: it's the computation of the least common multiple of all positive integer numbers up to a certain number x, that is LCM(1, 2, ..., x).", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "least common multiple of the indices of the smallest Fibonacci numbers divisible by their respective $k$. – Jaycob Coleman Aug 17 '13 at 12:44 • I've added the proof that all highly abundant numbers $> 630$ are divisible by $4$ (hence by $12$). Working on the others, well, at least some. Not sure how far I'm willing to go. – Daniel Fischer Aug 18 '13 at 20:24 I do not have enough reputation to comment, therefore I put this comment into an answer. 'whenever Fp is prime, so is p' should be 'if $F_p$ is prime, then $p$ is prime or $4$'.", "The least positive integer that is divisible by 2, 3, 15, and 28 is [#permalink] ### Show Tags 27 Nov 2018, 08:30 1 Bunuel wrote: The least positive integer that is divisible by 2, 3, 15, and 28 is A. 210 B. 420 C. 840 D. 1,260 E. 2,520 Find LCM of 2,3,15,28 : 2^2*3*5*7= 420 IMO B _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. VP Joined: 09 Mar 2016 Posts: 1274 Re: The least positive integer that is divisible by 2, 3, 15, and 28 is [#permalink] ### Show Tags 27 Nov 2018, 08:39 Bunuel wrote: The least positive integer that is divisible by 2, 3, 15, and 28 is A. 210 B. 420 C. 840 D. 1,260 E. 2,520 through prime factorization one can find the least positive integer that is divisible $$2^2$$, $$3$$,$$5$$ ,$$7$$ we we need to have TWO \"2\" because 28 has at least TWO prime numbers of 2 B Re: The least positive integer that is divisible by 2, 3, 15, and 28 is [#permalink] 27 Nov 2018, 08:39 Display posts from previous: Sort by", "Cube of Prime has 4 Positive Divisors Theorem Let $n \\in \\Z_{>0}$ be a positive integer which is the cube of a prime number. Then $n$ has exactly $4$ positive divisors. Proof Let $n = p^3$ where $p$ is prime. The positive divisors of $n$ are: $1, p, p^2, p^3$", "Re: Which of the following is the lowest positive integer that [#permalink] 23 Sep 2016, 17:03 Similar topics Replies Last post Similar Topics: 1 Which of the following is the lowest positive integer that is divisibl 4 29 Mar 2016, 02:11 7 What is the lowest positive integer that is divisible by each of the 4 11 Oct 2014, 12:04 9 If a and b are positive integers, which of the following 5 30 Jun 2014, 13:36 34 What is the lowest positive integer that is divisible by 18 11 Sep 2012, 03:43 4 What is the lowest positive integer that is divisible by eac 4 06 May 2012, 05:54 Display posts from previous: Sort by", "# Funny Numbers! Number Theory Level 4 A positive integer $$n$$ is funny if for each of its positive divisors $$d$$, the number $$d+2$$ is prime. Find the sum of all funny numbers that have the most quantity of divisors. ×" ]

[ "prime is $3$, and after comparing the resulting numbers, we discover that $\\{ 1, 1, 1, 1, 4\\}$ results in the smallest odd number $3^4\\cdot 5\\cdot 7\\cdot 11\\cdot 13=405405.$ Then the smallest odd number with exactly $40$ \"product pairs\" is $405405$. The same process is done for even numbers with the only difference being that the smallest prime factor is $2$, therefore we take each exponents set, assign them in decreasing order to increasing primes, and compare them with each other. In this case the smallest value comes out from the set $\\{ 1,1,3,4\\}$ to be $2^4\\cdot 3^3\\cdot 5\\cdot 7=15120$. So the smallest even number, actually smallest positive number altogether which can be factored into exactly $40$ product pairs is $15120$.", "+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", "What is the least common denominator of 3/4, 3/8, and 1/5? Apr 13, 2016 $40$ Explanation: If we look at the prime factorizations of the denominators, we have $4 = {2}^{2}$ $8 = {2}^{3}$ $5 = {5}^{1}$ The least common denominator will be the minimal product containing all factors above to their appropriate powers. In this case, that would be ${2}^{3} \\cdot 5$. Thus, the least common denominator is ${2}^{3} \\cdot 5 = 8 \\cdot 5 = 40$", "# 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: $2 \\cdot 3 \\cdot 5 = 30$", "• $1575 = 3^2 \\cdot 5^2 \\cdot 7$ • $2205 = 3^2 \\cdot 5 \\cdot 7^2$ • $2835 = 3^4 \\cdot 5 \\cdot 7$ • $3465 = 3^2 \\cdot 5 \\cdot 7 \\cdot 11$ Note how they are all combinations of powers of the very smallest possible odd primes. (As an aside, it's easy to see how abundance and deficiency are related to the concepts of smooth and rough numbers, which are defined as those numbers with their largest prime divisor smaller than a bound B, or the smallest prime divisor larger than B, respectively. Of course, the modified definitions of powersmoothness are more relevant to our discussion, where all prime powers dividing the number must be less than the bound B. If we define $b(n)$ to be the smallest number B such that $n$ is B-powersmooth, then there is a very close correlation between $a(n)/n$ and $n/b(n)$ -- though the former arguably accounts better for the cases where there are several high powers of multiple primes, instead of just one prime with a large exponent \"inflating\" $b(n)$.) At any rate, we've beaten to death the subject of abundance. We'll finish up with a couple further small but important properties of abundance. The main property in question is that $\\alpha(mn) \\gt \\alpha(n)$, i.e. the abundance of a number", "# 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 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 $p^{10}$.", "11$ $1575 = 3^2 \\cdot 5^2 \\cdot 7$ From the above: what primes do these two numbers have in common? As we can see, the common primes are 3 and 5. Looking at the exponents of these common primes in each of the numbers, we look the minimum between the two. In this case, the minimum exponent for 3 is 1, and the minimum exponent for 5 is also 1. Therefore $GCD = 3^1 \\cdot 5^1 = 3 \\cdot 5 = 15$ Now, all we have to do is to simplify the original fraction by 15: $\\frac{165}{1575} = \\frac{165/15}{1575/15} = \\frac{11}{105}$ which corresponds to the fraction in its lowest terms, because it cannot be further simplified. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us.", "a square and fifth power. Suppose n is divisible by both 2 and 3. Hence 2 and 3 appear in the prime power decomposition of n. Furthermore, if n is both a square and a fifth power and $n = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$, then $e_1, e_2, ..., e_k$ are all divisible by 2 and are all divisble by 5. Hence all these exponents are divisible by 10. Given this, we can let $p_1 = 2$, and $p_2 = 3$, and let $n = 2^{10} \\cdot 3^{10} = 60466176$. Hence 60466176 is the smallest integer divisible by 2, 3, and is both a square (77762) and fifth power (365). ## Example 6 Find the smallest integer n divisible by 14 that is also a square and a cube. We will solve this problem similarly to that in example 5. First, we note that if n is divisible by 10, then 2 and 7 appear in the prime power decomposition of n. Furthermore, if $n = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$, then $e_1, e_2, ..., e_k$ are all divisible by 2 and 3. Hence all of these exponents are divisible by 6. Given this, we can let $p_1=2$, $p_2 = 7$, and let $n = 2^6 \\cdot 7^6 = 7529536$. Hence 7529536 is the smallest integer n divisible by 14, is a square (27442), and is", "## Discrete Mathematics and Its Applications, Seventh Edition To obtain the greatest common divisor of each pair of integers, multiply together the factors shared by both numbers. a) $2^2\\cdot 3^3\\cdot 5^2=2700$ b) $2\\cdot 3\\cdot 11=66$ c) $17$ d) $1$ e) These numbers are relatively prime. f) $2\\cdot 3\\cdot 5\\cdot 7=210$", "+0 # I have no life +3 115 3 +321 a)What is the smallest positive integer that has exactly 20 positive divisors? b)What is the smallest positive integer that has exactly 6 perfect square divisors? Mar 24, 2019 #1 +5172 +4 $$20 = 2^2 5 \\text{ so we'll want to use prime factors }\\\\ 2^4 \\cdot 3^1 \\cdot 5^1 = 240\\\\ \\text{This has }(4+1)(1+1)(1+1)=20 \\text{ divisors}$$ I'm pretty sure for (2) we just find the smallest positive integer with 6 divisors and square it. Proceeding as above $$6=2\\cdot 3 \\text{ so choose}\\\\ (2^2\\cdot 3)^2 = 144$$ . Mar 24, 2019 #2 +321 +2 Thank You! #3 +533 +1 a) 20 has factors of 1, 2, 4, 5, 10, and 20. So, to make the number as small as possible, we want to use 5 * 2 * 2. We will pair the largest exponents with the smaller primes to get 2^(5-1) * 3 * 5 = 240. b) It has 6 perfect square divisors, so we need the smallest number with 6 divisors then square that to get the answer. That is (by counting), 12. So, our answer is 12^2 = 144. Mar 24, 2019", "assume there are only a finite number of primes of the form $$4k+3,$$ say $$p_0=3,p_1, ... ,p_n$$ is all of them; and we consider the natural number $Q=4p_1p_2\\cdot \\cdot \\cdot p_n+3.$ Notice that the prime factorization of $$Q$$, namely $$Q=q_1\\cdots q_t\\cdots q_m$$ must contain at least one prime of the form $$4k+3$$ because otherwise $$Q$$ would be of the form $$4k+1.$$ Thus, there is one prime in the prime factorization of $$Q$$ that is of the form $$4k+3$$ say $$q_t.$$ Either $$q_t=3$$ or $$q_t>3.$$ If $$q_t=3,$$ then $$3|Q$$ which means $$3|(Q-3)$$ and so $$3\\left|4p_1\\cdot \\cdot \\cdot p_n\\right.$$ which is absurd. If $$q_t>3,$$ then $$q_t=p_j$$ for some $$1\\leq j\\leq n$$ and so $$\\left.q_t\\right|Q$$ implies $$q_t|Q-4p_1\\cdot \\cdot \\cdot p_n=3,$$ which is also absurd. Therefore, there are infinitely many primes of the form $$4n+3.$$ Definition 3.1 The least common multiple of two nonzero natural numbers $$a$$ and $$b$$ is the smallest positive natural number that is divisible by $$a$$ and $$b.$$ The least common multiple of $$15$$ and 21 is $$[15,21]=105.$$ The least common multiple of $$24$$ and 36 is $$[24,36]=72.$$ The least common multiple of 2 and 20 is $$[2,20]=20.$$ The least common multiple of 7 and 11 is $$[7,11]=77.$$ Corollary 3.1 Let $$a$$ and $$b$$ be natural numbers. Then $$$(a,b)=p_1^{\\min \\left(\\alpha _1,\\beta _1\\right)} p_2^{\\min \\left(\\alpha _2,\\beta _2\\right)} \\cdots p_n^{\\min", "consider the number 210, that is the smallest number which is the product of four distinct prime numbers, being equal to $2 \\cdot 3 \\cdot 5 \\cdot 7$. In this case the Hasse diagram is made up of five levels: Once again, the two properties are satisfied: • There are 1 + 6 + 1 = 8 divisors on the even levels and 4 + 4 = 8 divisors on the odd levels. • The product of divisors on the even levels is equal to the product of divisors on the odd levels (try it by yourself!): $$1 \\cdot (6 \\cdot 10 \\cdot 14 \\cdot 15 \\cdot 21 \\cdot 35) \\cdot 210 = (2 \\cdot 3 \\cdot 5 \\cdot 7) \\cdot (30 \\cdot 42 \\cdot 70 \\cdot 105)$$ In the examples seen so far, we can note that: • if there are three levels, then there is 1 divisor on the first level, there are 2 on the second one and 1 on the third one (as it happens for the number 10); • if there are four levels, then there is 1 divisor on the first level, there are 3 on the second one, 3 on the third one and 1 on the fourth one (as it happens for the number 30); • if there are five", "smallest number is 1 and the next number is 3. . . Then the largest number must be 5 or more: .8 choices. Suppose the smallest number is 2 and the next number is 4. . . Then the largest number must be 6 or more: .7 choices. Suppose the smallest number is 3 and the next number is 5. . . Then the largest number must be 7 or more: .6 choices. . . . . . . . . $\\vdots$ Suppose the smallest number is 1 and the next number is 10. . . Then the largest number must be 12: .1 choice. Hence, when the smallest number is 1, . . there are: . $1 + 2 + 3 + \\hdots + 8 \\:=\\:36$ ways. When the smallest number is 2, we find that the number of triples is: . $1 + 2 + 3 + \\hdots + 7 \\:=\\:28$ ways. When the smallest number is 3, we find the number of triples is: . . $1 + 2 + 3 + \\hdots + 6 \\:=\\:21$ ways. And so on . . . We see that the total number of ways . . is the sum of the first 8 Triangular Numbers. This is a Tetrahedral Number: . $T(n) \\:=\\:\\frac{n(n+1)(n+2)}{6}$ Therefore: . $T(8) \\:=\\:\\frac{8\\cdot9\\cdot10}{6} \\:=\\:120$", "# How do you find the GCF of 42x^2, 56x^3? Oct 13, 2015 $G C F \\left(42 {x}^{2} , 56 {x}^{3}\\right) = 14 {x}^{2}$ #### Explanation: As for the numerical part: $G C F \\left(42 , 56\\right) = G C F \\left(2 \\cdot 3 \\cdot 7 , {2}^{3} \\cdot 7\\right)$ The rule is to look for common primes, and take them with the smallest exponents. $2$ is a common prime. It occours once as $2$, then as ${2}^{3}$. So, the one with the smallest exponent is $2$. $3$ is not a common prime, since it only divides $42$. $7$ is a common prime, and it appears in both factorization as $7$, so there's nothing to choose. The GCF is thus $2 \\cdot 7 = 14$ For the variables, it is easier: ${x}^{2}$ divides ${x}^{3}$, so $G C F \\left({x}^{2} , {x}^{3}\\right) = {x}^{2}$.", "the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework. If the primes are $2,3,5$, then the smallest number can range between $1$ and $15$. If the primes are $2,5,7$, then the smallest number can range between $1$ and $13$. If the primes are $2,11,13$, then the smallest number can range between $1$ and $7$. If the primes are $2,17,19$, then the smallest prime can only be $1$. Adding all cases gets $15+13+7+1=36$. However, due to the commutative property, we must multiply this by 2. For example, in the $2,17,19$ case the numbers can be $1,3,20$ or $1,18,20$. Therefore the answer is $36\\cdot2=\\boxed{072}$. Note about solution 1: I don't think that works, because if for example there are 21 points on the circle, your solution would yield $19\\cdot4=76$, but there would be $8$ more solutions than if there are $20$ points. This is because the upper bound for each case increases by $1$, but commutative property doubles it to be $4$.", "# What is the least common multiple of 12, 18, and 5? May 2, 2016 $180$ #### Explanation: Start by finding the prime factorisations of each of these numbers: $12 = 2 \\times 2 \\times 3$ $18 = 2 \\times 3 \\times 3$ $5 = 5$ So the smallest positive integer that includes all of these factors in these multiplicities is: $180 = 2 \\times 2 \\times 3 \\times 3 \\times 5$", "the smallest composite number and the smallest prime number is $2$. Note: HCF is also known as the Greatest Common Measure (GCM) and Greatest Common Divisor (GCD). HCF is always less than equal to the smallest involved. Two or more prime numbers have HCF $= 1$.", "= 3628799!7!5!3!$. You can ... 0 No. Consider $4^4 = 2^{8}$, which has no 3 in its prime factorization, while $4! = 1 \\cdot 2 \\cdot 3 \\cdot 4 = 2^3 \\cdot 3$. -2 Top 50 recent answers are included", "Geeks With Blogs Buhay Programmer Dont byte more than you can chew Problem: 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest number that is evenly divisible by all of the numbers from 1 to 20? Solution: Finding the LCM can be done via factorization. Below is an excerpt from the wikipedia article. The unique factorization theorem says that every positive integer number greater than 1 can be written in only one way as a product of prime numbers. The prime numbers can be considered as the atomic elements which, when combined together, make up a composite number. For example: $90 = 2^1 \\cdot 3^2 \\cdot 5^1 = 2 \\cdot 9 \\cdot 5. \\,\\!$ Here we have the composite number 90 made up of one atom of the prime number 2, two atoms of the prime number 3 and one atom of the prime number 5. This knowledge can be used to find the lcm of a set of numbers. Example: Find the value of lcm(8,9,21). First, factor out each number and express it as a product of prime number powers. $8\\; \\, \\; \\,= 2^3 \\cdot 3^0 \\cdot 5^0 \\cdot 7^0 \\,\\!$ $9\\; \\, \\; \\,= 2^0 \\cdot 3^2 \\cdot 5^0" ]

[ "prime is $3$, and after comparing the resulting numbers, we discover that $\\{ 1, 1, 1, 1, 4\\}$ results in the smallest odd number $3^4\\cdot 5\\cdot 7\\cdot 11\\cdot 13=405405.$ Then the smallest odd number with exactly $40$ \"product pairs\" is $405405$. The same process is done for even numbers with the only difference being that the smallest prime factor is $2$, therefore we take each exponents set, assign them in decreasing order to increasing primes, and compare them with each other. In this case the smallest value comes out from the set $\\{ 1,1,3,4\\}$ to be $2^4\\cdot 3^3\\cdot 5\\cdot 7=15120$. So the smallest even number, actually smallest positive number altogether which can be factored into exactly $40$ product pairs is $15120$.", "+0 # I have no life +3 115 3 +321 a)What is the smallest positive integer that has exactly 20 positive divisors? b)What is the smallest positive integer that has exactly 6 perfect square divisors? Mar 24, 2019 #1 +5172 +4 $$20 = 2^2 5 \\text{ so we'll want to use prime factors }\\\\ 2^4 \\cdot 3^1 \\cdot 5^1 = 240\\\\ \\text{This has }(4+1)(1+1)(1+1)=20 \\text{ divisors}$$ I'm pretty sure for (2) we just find the smallest positive integer with 6 divisors and square it. Proceeding as above $$6=2\\cdot 3 \\text{ so choose}\\\\ (2^2\\cdot 3)^2 = 144$$ . Mar 24, 2019 #2 +321 +2 Thank You! #3 +533 +1 a) 20 has factors of 1, 2, 4, 5, 10, and 20. So, to make the number as small as possible, we want to use 5 * 2 * 2. We will pair the largest exponents with the smaller primes to get 2^(5-1) * 3 * 5 = 240. b) It has 6 perfect square divisors, so we need the smallest number with 6 divisors then square that to get the answer. That is (by counting), 12. So, our answer is 12^2 = 144. Mar 24, 2019", "# 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 $$x$$ ? ×", "# Contents ## Definition For $a,b\\in ℕ$ two positive natural numbers, their least common multiple $\\mathrm{LCM}\\left(a,b\\right)\\in ℕ$ is the smallest natural number that is divisible by both $a$ and $b$, i.e. such that there exist ${n}_{a},{n}_{n}\\in ℕ$ with ${n}_{a}\\cdot a={n}_{b}\\cdot b=\\mathrm{LCM}\\left(a,b\\right)$. • greatest common divisor? Created on September 5, 2012 16:10:58 by Urs Schreiber (131.174.191.191)", "What is the least common denominator of 3/4, 3/8, and 1/5? Apr 13, 2016 $40$ Explanation: If we look at the prime factorizations of the denominators, we have $4 = {2}^{2}$ $8 = {2}^{3}$ $5 = {5}^{1}$ The least common denominator will be the minimal product containing all factors above to their appropriate powers. In this case, that would be ${2}^{3} \\cdot 5$. Thus, the least common denominator is ${2}^{3} \\cdot 5 = 8 \\cdot 5 = 40$", "+0 # Counting Divisors +1 129 2 +484 Some positive integers have exactly four positive factors. For example, 35 has only 1, 5, 7 and 35 as its factors. What is the sum of the smallest five positive integers that each have exactly four positive factors? Apr 5, 2021 #1 +594 +2 If they have four factors they can be of the form $p_1\\cdot p_2$ or $p^3$ for all $p_1,p_2,p$ prime. Let's do $p_1\\cdot p_2$ first. We have $2\\cdot 3=6, 2\\cdot 5=10, 2\\cdot 7=14, 2\\cdot 11=22,$ etc. if $2$ is the first. Otherwise we can have $3\\cdot 5=15, 3\\cdot 7=21$, etc. So the smallest $5$ are $6,10,14,15,21$. Then for $p^3$ we can have $2^3=8$ and the rest are larger than $21$. So the numbers are $6,8,10,14,15$. Sum them up. Apr 5, 2021 #2 +484 0 Thank you @thedudemanguyperson, it was correct! RiemannIntegralzzz Apr 5, 2021", "^N _{j=0}(j\\cdot p_j))+\\sum ... In JavaScript, the largest odd positive number representable is $2^{53}-1$. All integers between 1 and $2^{53}-1$ can be represented without loss of precision. How many prime numbers can be ...", "almost won't change the discussion.) Everything divides zero, so zero can't be prime. $$0 \\cdot 7 = 0$$ means $$0$$ and $$7$$ divide $$0$$. $$0 \\cdot -8 = 0$$ means minus eight also divides zero. Can we see that everything divides zero, so zero is very far from being prime. Primes are numbers that are divisible by exactly two different positive numbers. (Note that this also holds true for negative integers that are prime.) Every number is divisible by one, so that must be one of the positive divisors of a prime. Every number is divisible by its magnitude (\"itself\" if only talking about positives), so that must be the other positive divisor of a prime. Non-primes must have more positive divisors. If we take all the positive numbers bigger than one, take them in pairs and multiply them together, we get all the non-primes. \\begin{align*} 2 \\cdot 2 &= 4, 2 \\cdot 3 = 6, 2 \\cdot 4 = 8, \\dots \\\\ 3 \\cdot 2 &= 6, 3 \\cdot 3 = 9, 3 \\cdot 4 = 12, \\dots \\end{align*} (This could be a good time to remind/discuss multiples of a number and to remind/discuss commutativity of multiplication to reduce redundant calculations.) \\begin{align*} 4 \\cdot 4 &= 16, 4 \\cdot 5 = 20, 4 \\cdot 6 = 24,", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "# 5. The least common multiple of the first positive integers Prerequirements: We know that a way to compute the least common multiple between two or more integer numbers is to decompose them into prime factors, and then taking each distinct prime factor with its greatest power in the considered numbers. Going on this way, the factorization of the least common multiple will result. For example, given the three numbers $4 = 2^2$, $6 = 2 \\cdot 3$ and $45 = 3^2 \\cdot 5$, and computing their least common multiple, we see that the distinct prime factors are 2, 3 and 5; 2 appears with the greatest exponent 2 in 4, 3 appears with the greatest exponent 2 in 45, 5 appears with the greatest exponent 1 again in 45; so the least common multiple we are looking for is $2^2 \\cdot 3^2 \\cdot 5 = 180$. There exists a particularly interesting case in which the least common multiple can be computed through a reasoning, without decomposing the numbers into prime factors: it’s the computation of the least common multiple of the positive integer numbers up to a certain number $x$, that is $\\mathrm{LCM}(1, 2, \\dots, x)$. Of course this expression coincides with $\\mathrm{LCM}(2, \\dots, x)$ if $x \\geq 2$, as in this case the number 1 does", "# 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: $2 \\cdot 3 \\cdot 5 = 30$", "+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 ? $$\\small{\\text{ \\begin{array}{r|rcl|r} \\hline n & p_1 & & p_2 & p_1\\cdot p_2 \\\\ \\hline 1 & 11 & \\cdot & 13 & = 143 \\\\ 2 & 11 & \\cdot & 17 & = 187 \\\\ 3 & 11 & \\cdot & 19 & = 209 \\\\ 4 & 11 & \\cdot & 23 & = 253 \\\\ 5 & 11 & \\cdot & 29 & = 319 \\\\ 6 & 11 & \\cdot & 31 & = 341 \\\\ 7 & 11 & \\cdot & 37 & = 407 \\\\ 8 & 11 & \\cdot & 41& = 451 \\\\ 9 & 11 & \\cdot & 43 & = 473 \\\\ 10 & 13 & \\cdot & 17 & = 221 \\\\ 11 & 13 & \\cdot & 19 & = 247 \\\\ 12 & 13 & \\cdot & 23 & = 299 \\\\", "+0 # help 0 80 1 Find the number of distinct positive divisors of $$(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$$ . Jan 1, 2019", "Geeks With Blogs Buhay Programmer Dont byte more than you can chew Problem: 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest number that is evenly divisible by all of the numbers from 1 to 20? Solution: Finding the LCM can be done via factorization. Below is an excerpt from the wikipedia article. The unique factorization theorem says that every positive integer number greater than 1 can be written in only one way as a product of prime numbers. The prime numbers can be considered as the atomic elements which, when combined together, make up a composite number. For example: $90 = 2^1 \\cdot 3^2 \\cdot 5^1 = 2 \\cdot 9 \\cdot 5. \\,\\!$ Here we have the composite number 90 made up of one atom of the prime number 2, two atoms of the prime number 3 and one atom of the prime number 5. This knowledge can be used to find the lcm of a set of numbers. Example: Find the value of lcm(8,9,21). First, factor out each number and express it as a product of prime number powers. $8\\; \\, \\; \\,= 2^3 \\cdot 3^0 \\cdot 5^0 \\cdot 7^0 \\,\\!$ $9\\; \\, \\; \\,= 2^0 \\cdot 3^2 \\cdot 5^0", "# What is the least common multiple of 3, 4, and 7? Least Common Multiple of $3$, $4$ and $7$ is $84$. As $3$ and $7$ are prime and the only factor of $4$ (other than $1$ and itself) is $2$, another prime, with no common factor between three numbers. Hence, Least Common Multiple of $3$, $4$ and $7$ is $3 \\times 4 \\times 7 = 84$.", "# What is the smallest positive common difference of a 6-term arithmetic progression consisting entirely of (positive) prime numbers? What is the smallest positive common difference of a 6-term arithmetic progression consisting entirely of (positive) prime numbers? are divisibility rules applicable here? - See this Math Overflow question. In your case, divisibility tells us that the difference must be $\\geq$ than $$2\\cdot 3\\cdot 5=30.$$ – Eric Naslund Mar 22 '13 at 20:25 Yes, divisibility rules are important here. Clearly the difference must be even as all primes (except $2$) are odd. The difference must be a multiple of $3$ because otherwise two of the numbers in the progression will be multiples of $3$. Carry on. Since primes get less common as the numbers get larger, you should try starting small.", "The least positive integer that is divisible by 2, 3, 15, and 28 is [#permalink] ### Show Tags 27 Nov 2018, 08:30 1 Bunuel wrote: The least positive integer that is divisible by 2, 3, 15, and 28 is A. 210 B. 420 C. 840 D. 1,260 E. 2,520 Find LCM of 2,3,15,28 : 2^2*3*5*7= 420 IMO B _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. VP Joined: 09 Mar 2016 Posts: 1274 Re: The least positive integer that is divisible by 2, 3, 15, and 28 is [#permalink] ### Show Tags 27 Nov 2018, 08:39 Bunuel wrote: The least positive integer that is divisible by 2, 3, 15, and 28 is A. 210 B. 420 C. 840 D. 1,260 E. 2,520 through prime factorization one can find the least positive integer that is divisible $$2^2$$, $$3$$,$$5$$ ,$$7$$ we we need to have TWO \"2\" because 28 has at least TWO prime numbers of 2 B Re: The least positive integer that is divisible by 2, 3, 15, and 28 is [#permalink] 27 Nov 2018, 08:39 Display posts from previous: Sort by", "# Contents ## Definition For $a,b \\in \\mathbb{N}$ two positive natural numbers, their least common multiple $LCM(a,b) \\in \\mathbb{N}$ is the smallest natural number that is divisible by both $a$ and $b$, i.e. such that there exist $n_a, n_n \\in \\mathbb{N}$ with $n_a \\cdot a = n_b \\cdot b = LCM(a,b)$. ## References Revised on July 13, 2016 15:02:53 by Urs Schreiber (131.220.184.222)", "+0 # Math +1 70 1 +203 What is the smallest possible product of one one-digit prime and two distinct two-digit primes? Dec 9, 2021 #1 +132 0 The answer relies on minimizing all of the primes. The smallest one-digit prime is 2. The smallest two-digit primes are 11 and 13. $$2 \\cdot 11 \\cdot 13 = 286$$. Dec 9, 2021", "• $1575 = 3^2 \\cdot 5^2 \\cdot 7$ • $2205 = 3^2 \\cdot 5 \\cdot 7^2$ • $2835 = 3^4 \\cdot 5 \\cdot 7$ • $3465 = 3^2 \\cdot 5 \\cdot 7 \\cdot 11$ Note how they are all combinations of powers of the very smallest possible odd primes. (As an aside, it's easy to see how abundance and deficiency are related to the concepts of smooth and rough numbers, which are defined as those numbers with their largest prime divisor smaller than a bound B, or the smallest prime divisor larger than B, respectively. Of course, the modified definitions of powersmoothness are more relevant to our discussion, where all prime powers dividing the number must be less than the bound B. If we define $b(n)$ to be the smallest number B such that $n$ is B-powersmooth, then there is a very close correlation between $a(n)/n$ and $n/b(n)$ -- though the former arguably accounts better for the cases where there are several high powers of multiple primes, instead of just one prime with a large exponent \"inflating\" $b(n)$.) At any rate, we've beaten to death the subject of abundance. We'll finish up with a couple further small but important properties of abundance. The main property in question is that $\\alpha(mn) \\gt \\alpha(n)$, i.e. the abundance of a number" ]

[ "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "# Counting triangles and vertices in Unity's cube I want to understand logic behind drawing triangles of primitives in Unity. When we spawn simple cube, variable mesh.vertexCount gives us value 24. Also when we list all vertices positions (from mesh.vertices) and triangles numbers (from mesh.triangles), we get: 0. (1.0, -1.0, 1.0) 0 1. (-1.0, -1.0, 1.0) 2 2. (1.0, 1.0, 1.0) 3 3. (-1.0, 1.0, 1.0) 0 4. (1.0, 1.0, -1.0) 3 5. (-1.0, 1.0, -1.0) 1 6. (1.0, -1.0, -1.0) 8 7. (-1.0, -1.0, -1.0) 4 8. (1.0, 1.0, 1.0) 5 9. (-1.0, 1.0, 1.0) 8 10. (1.0, 1.0, -1.0) 5 11. (-1.0, 1.0, -1.0) 9 12. (1.0, -1.0, -1.0) 10 13. (1.0, -1.0, 1.0) 6 14. (-1.0, -1.0, 1.0) 7 15. (-1.0, -1.0, -1.0) 10 16. (-1.0, -1.0, 1.0) 7 17. (-1.0, 1.0, 1.0) 11 18. (-1.0, 1.0, -1.0) 12 19. (-1.0, -1.0, -1.0) 13 20. (1.0, -1.0, -1.0) 14 21. (1.0, 1.0, -1.0) 12 22. (1.0, 1.0, 1.0) 14 23. (1.0, -1.0, 1.0) 15 (i-th element, vector from mesh.vertices[i], number from mesh.triangles[i]) Obviously we need only 8 vertices to create cube, but we need 6*2=12 triangles, three vertices every - array of 36 vectors. So... what is the logic of Unity primitive? • Ok I discovered that mesh.triangles has length of 36 integers, so at", "upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide? $[asy] draw((0,0)--(4,4*sqrt(3))); draw((1,-sqrt(3))--(5,3*sqrt(3))); draw((2,-2*sqrt(3))--(6,2*sqrt(3))); draw((3,-3*sqrt(3))--(7,sqrt(3))); draw((4,-4*sqrt(3))--(8,0)); draw((8,0)--(4,4*sqrt(3))); draw((7,-sqrt(3))--(3,3*sqrt(3))); draw((6,-2*sqrt(3))--(2,2*sqrt(3))); draw((5,-3*sqrt(3))--(1,sqrt(3))); draw((4,-4*sqrt(3))--(0,0)); draw((3,3*sqrt(3))--(5,3*sqrt(3))); draw((2,2*sqrt(3))--(6,2*sqrt(3))); draw((1,sqrt(3))--(7,sqrt(3))); draw((-1,0)--(9,0)); draw((1,-sqrt(3))--(7,-sqrt(3))); draw((2,-2*sqrt(3))--(6,-2*sqrt(3))); draw((3,-3*sqrt(3))--(5,-3*sqrt(3))); [/asy]$ $\\text{(A)}\\ 4 \\qquad \\text{(B)}\\ 5 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 7 \\qquad \\text{(E)}\\ 9$ ## Problem 25 There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it? $\\text{(A)}\\ 5724 \\qquad \\text{(B)}\\ 7245 \\qquad \\text{(C)}\\ 7254 \\qquad \\text{(D)}\\ 7425 \\qquad \\text{(E)}\\ 7542$", "total triangles, the answer is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$. ### Solution 3 For starters what I find helpful is to divide the whole octagon up into triangles as shown here: $[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot(\"A\",A,dir(45)); dot(\"B\",B,dir(90)); dot(\"C\",C,dir(135)); dot(\"D\",D,dir(180)); dot(\"E\",E,dir(-135)); dot(\"F\",F,dir(-90)); dot(\"G\",G,dir(-45)); dot(\"H\",H,dir(0)); dot(\"X\",X,dir(135/2)); dot(\"O\",O,dir(0)); draw(E--O--X); draw(C--O--B); draw(B--O--A); draw(A--O--H); draw(H--O--G); draw(G--O--F); draw(F--O--E); draw(E--O--D); draw(D--O--C); [/asy]$ Now it is just a matter of counting the larger triangles remember that $\\triangle BOX$ and $\\triangle XOA$ are not full triangles and are only half for these purposes. We count it up and we get a total of $\\frac{3.5}{8}$ of the shape shaded. We then simplify it to get our answer of $\\boxed{\\textbf{(D)}~\\frac{7}{16}}$.", "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", "1]]; normals = getnormals[[All, 2]]; NORMALS = Table[ Table[Mean[ Extract[normals, Position[corrds, triangles[[j]][[i]]]]] + 0.0001 {RandomReal[{-1, 1}], RandomReal[{-1, 1}], RandomReal[{-1, 1}]}, {i, 1, 3}] , {j, 1, Length[triangles]}]; sets = Partition[Range[3*Len] - 1, 3]; ps = Position[Abs[NORMALS], _?(# < 0.0001 &)]; Table[ NORMALS[[ps[[j]][[1]], ps[[j]][[2]], ps[[j]][[3]]]] = 0; , {j, 1, Length[ps]}]; ps = Position[Abs[triangles], _?(# < 0.0001 &)]; Table[ triangles[[ps[[j]][[1]], ps[[j]][[2]], ps[[j]][[3]]]] = 0; , {j, 1, Length[ps]}]; Where it takes in a list of triangles. The code below it just generates a .pov file in the proper format.", "# Problem #2091 2091 Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way? $[asy] size(200); defaultpen(0.9); real r = 5/dir(54).x, h = 5 tan(54*pi/180); pair A = (5,0), B = A+10*dir(72), C = (0,r+h), E = (-5,0), D = E+10*dir(108); draw(A--B--C--D--E--cycle); label(\"$A$\",A+(0,-0.5),SSE); label(\"$B$\",B+(0.5,0),ENE); label(\"$C$\",C+(0,0.5),N); label(\"$D$\",D+(-0.5,0),WNW); label(\"$E$\",E+(0,-0.5),SW); // real l = 5*sqrt(3); pair ab = (h+l)*dir(72), bc = (h+l)*dir(54); pair AB = (ab.y, h-ab.x), BC = (bc.x,h+bc.y), CD = (-bc.x,h+bc.y), DE = (-ab.y, h-ab.x), EA = (0,-l); draw(A--AB--B^^B--BC--C^^C--CD--D^^D--DE--E^^E--EA--A, dashed); // dot(A); dot(B); dot(C); dot(D); dot(E); dot(AB); dot(BC); dot(CD); dot(DE); dot(EA); [/asy]$ $\\text {(A) } 1 \\qquad \\text {(B) } 2 \\qquad \\text {(C) } 3 \\qquad \\text {(D) } 4 \\qquad \\text {(E) } 5$ 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.", "order will be the five at one end, and then the five at the other T hl = l * (T)0.5; m_verts[0].Set((T)1.0, (T)0.0, hl); m_verts[1].Set((T)0.0, (T)1.0, hl); m_verts[2].Set(-(T)1.0, (T)0.0, hl); m_verts[3].Set((T)0.0, -(T)1.0, hl); m_verts[4].Set((T)0.0, (T)0.0, hl + (T)1.0); m_verts[5].Set((T)1.0, (T)0.0, -hl); m_verts[6].Set((T)0.0, (T)1.0, -hl); m_verts[7].Set(-(T)1.0, (T)0.0, -hl); m_verts[8].Set((T)0.0, -(T)1.0, -hl); m_verts[9].Set((T)0.0, (T)0.0, -hl - (T)1.0); // When I re-write this, the order should be 4+, 4-, cap+, cap-. // Well, our subivision idea won't work like we want as is because it will subdivide the // triangles along the sides also. Eventually we'll need a way to flag certain // triangles as 'don't subdivide'. // Anyway, we 4 sides * 2 tris each, plus 2 ends * 4 tris each * 3 indices each. m_indices.resize(48); // Here we go. unsigned int indices[] = { 0, 5, 1, 6, 1, 5, 1, 6, 2, 7, 2, 6, 2, 7, 3, 8, 3, 7, 3, 8, 0, 5, 0, 8, 0, 1, 4, 1, 2, 4, 2, 3, 4, 3, 0, 4, 5, 9, 6, 6, 9, 7, 7, 9, 8, 8, 9, 5}; for (int i = 0; i < 48; ++i) m_indices = indices; // Subdivide for (int i = 0; i < lod; ++i) Subdivide(); // Now we want to re-normalize the verts with respect to the", "square from 2 triangles, for simplicity, how many draw calls does glDrawElements (4 items in vertex array and 6 items in index array) and glDrawArrays (6 items only in vertex array) need, individually? Thanks. Each glDraw* is a draw call. 1 glDrawArrays is 1 draw call. 1 glDrawElements is 1 draw call. It doesn't matter (so far as draw call count is concerned) how many vertices or indices you use, 1 glDraw* is 1 draw call. The simple cases of drawing quads (assuming the GL version you use hasn't removed quads) or triangles are not a good example for comparing these, because a list of quads or a list of triangles can be drawn with a single call to either, and glDrawArrays will appear more efficient because it doesn't have the additional overhead of indexing. Let's take a slightly more complex case, but one which is representative of a more real-world scenario. Imagine that you have a model composed of multiple triangle strips and fans: glDrawArrays (GL_TRIANGLE_FAN, .....); glDrawArrays (GL_TRIANGLE_STRIP, .....); glDrawArrays (GL_TRIANGLE_STRIP, .....); glDrawArrays (GL_TRIANGLE_FAN, .....); glDrawArrays (GL_TRIANGLE_STRIP, .....); glDrawArrays (GL_TRIANGLE_FAN, .....); In this example, using glDrawArrays gives you a total of 6 draw calls for the model. However, both strips and fans can easily be converted to indexed triangles, so add indices and it becomes: glDrawElements", "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", "a great deal about assembling, but it would be just too extensive to describe the complete process. We will solve the problem in a mesh with 6 triangles forming a regular hexagon Following are the codes. ### Python from __future__ import division, print_function import numpy as np import matplotlib.pyplot as plt def assem(coords, elems, source): ncoords = coords.shape[0] stiff = np.zeros((ncoords, ncoords)) rhs = np.zeros((ncoords)) for el_cont, elem in enumerate(elems): stiff_loc, jaco = local_stiff(coords[elem]) rhs[elem] += jaco*np.mean(source[elem]) for row in range(3): for col in range(3): row_glob, col_glob = elem[row], elem[col] stiff[row_glob, col_glob] += stiff_loc[row, col] return stiff, rhs def local_stiff(coords): dHdr = np.array([[-1, 1, 0], [-1, 0, 1]]) jaco = dHdr.dot(coords) stiff = 0.5*np.array([[2, -1, -1], [-1, 1, 0], [-1, 0, 1]]) return stiff/np.linalg.det(jaco), np.linalg.det(jaco) sq3 = np.sqrt(3) coords = np.array([ [sq3, -1], [0, 0], [2*sq3, 0], [0, 2], [2*sq3, 2], [sq3, 3], [sq3, 1]]) elems = np.array([ [1, 0, 6], [0, 2, 6], [2, 4, 6], [4, 5, 6], [5, 3, 6], [3, 1, 6]]) source = -np.ones(7) stiff, rhs = assem(coords, elems, source) free = range(6) sol = np.linalg.solve(stiff[np.ix_(free, free)], rhs[free]) sol_c = np.zeros(coords.shape[0]) sol_c[free] = sol plt.tricontourf(coords[:, 0], coords[:, 1], sol_c, cmap=\"hot\") plt.colorbar() plt.axis(\"image\") plt.show() ### Julia using PyPlot function assem(coords, elems, source) ncoords = size(coords)[1] nelems = size(elems)[1] stiff = zeros(ncoords, ncoords) rhs =", "won't install an older version of the runtime. You can use DXDIAG to determine which version of the runtimes are installed on your machine (just go to Run... from the Start menu and enter DXDIAG). If it is the case that your SDK is older than your current runtime, then you'll need to download and install the latest SDK. Share on other sites Quote: Original post by Scet Quote: Original post by speedieresult = (*m_d3ddev)->DrawIndexedPrimitive(D3DPT_TRIANGLELIST, 0, 0, m_batch_size, 0, m_batch_size / 2); I'm just drawing one quad, 2 triangles and 4 vertices. m_batch_size == 4. Drawing two triangles using a triangle list requires 6 vertices, not 4. I think you want a traingle strip. Ummm, not necessarily! DrawIndexedPrimitive()... 1 2+---+| /|| / ||/ |+---+3 4 4 vertices: 1,2,3,4 6 indices: 1,2,3,2,4,3 • 11 • 20 • 12 • 10 • 38 • Forum Statistics • Total Topics 631400 • Total Posts 2999862 ×", "draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ $\\textbf{(A)}12\\frac{1}{2}\\qquad\\textbf{(B)}20\\qquad\\textbf{(C)}25\\qquad\\textbf{(D)}33\\frac{1}{3}\\qquad\\textbf{(E)}37\\frac{1}{2}$ ## Problem 8 Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips? $\\textbf{(A) }4 \\qquad\\textbf{(B) }5 \\qquad\\textbf{(C) }6 \\qquad\\textbf{(D) }7 \\qquad\\textbf{(E) }9$ ## Problem 9 Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour? $[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; draw((0,0)--(0,8)); draw((0,0)--(8,0)); 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)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label(\"1\",(0.95,-0.24),SE*lsf); label(\"2\",(1.92,-0.26),SE*lsf); label(\"3\",(2.92,-0.31),SE*lsf); label(\"4\",(3.93,-0.26),SE*lsf); label(\"5\",(4.92,-0.27),SE*lsf); label(\"6\",(5.95,-0.29),SE*lsf); label(\"7\",(6.94,-0.27),SE*lsf); label(\"5\",(-0.49,1.22),SE*lsf); label(\"10\",(-0.59,2.23),SE*lsf); label(\"15\",(-0.61,3.22),SE*lsf); label(\"20\",(-0.61,4.23),SE*lsf); label(\"25\",(-0.59,5.22),SE*lsf); label(\"30\",(-0.59,6.2),SE*lsf); label(\"35\",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6));", "0 The second hexagon equation can be verified in a similar fashion, and the result is the same. #### Braid Invariants To come: definition of Z[B[n]], demonstration of Reidemeister moves. For now, reidemeister.nb. #### The Unknot To compute $Z(K)$ for any knot, including the unknot, we need to extend the definition of $Z$ to include \"caps\" and \"cups\". In addition, framing independence forces us to introduce the FI relation - hence all the formulae in this section take place in the \"further reduced\" space of chord diagrams. We proceed using the $Z_{\\infty}$ correction term in [1]. To begin, let us choose a pair $(R, \\Phi)$ which solve the pentagon and hexagon equations to fourth degree: In[9]:= Phi = ASeries[Id[3],0, 1 + CD[Line[1], Line[2], Line[1, 2]]/24 - (7*CD[Line[1], Line[2, 3, 4], Line[1, 2, 3, 4]])/5760 - CD[Line[2], Line[1], Line[1, 2]]/24 + (7*CD[Line[2], Line[1, 3, 4], Line[1, 2, 3, 4]])/1920 - (7*CD[Line[3], Line[1, 2, 4], Line[1, 2, 3, 4]])/1920 + (7*CD[Line[4], Line[1, 2, 3], Line[1, 2, 3, 4]])/5760 + (7*CD[Line[1, 2], Line[3, 4], Line[1, 2, 3, 4]])/5760 - CD[Line[1, 3], Line[2, 4], Line[1, 2, 3, 4]]/640 - CD[Line[1, 4], Line[2, 3], Line[1, 2, 3, 4]]/1152 - CD[Line[2, 3], Line[1, 4], Line[1, 2, 3, 4]]/1152 + (19*CD[Line[2, 4], Line[1, 3], Line[1, 2, 3, 4]])/5760 - (7*CD[Line[3, 4], Line[1, 2], Line[1, 2,", "return Marker.refine if x.two_norm < 0.64**i else Marker.keep aluView.plot(figsize=(5,5)) from dune.alugrid import aluSimplexGrid vertices = aluView.coordinates() triangles = [aluView.indexSet.subIndices(e, 2) for e in aluView.elements] aluView = aluSimplexGrid({\"vertices\": vertices, \"simplices\": triangles}) We next discuss how to retrieve basic information from a constructed grid and iterate over its entities (i.e., elements, faces, edges, vertices, etc.). [6]: vertices = [(0,0), (1,0), (1,1), (0,1)] triangles = [(2,0,1), (0,2,3)] unitSquare = aluSimplexGrid({\"vertices\": vertices, \"simplices\": triangles}) print(unitSquare.size(0),\"elements and\",unitSquare.size(2),\"vertices\") for codim in range(0, unitSquare.dimension+1): for entity in unitSquare.entities(codim): print(\", \".join(str(c) for c in entity.geometry.corners)) for edge in unitSquare.edges: print(\", \".join(str(c) for c in edge.geometry.corners)) 2 elements and 4 vertices (1.000000, 1.000000), (1.000000, 0.000000), (0.000000, 0.000000) (0.000000, 0.000000), (0.000000, 1.000000), (1.000000, 1.000000) (0.000000, 0.000000), (1.000000, 0.000000) (0.000000, 0.000000), (1.000000, 1.000000) (0.000000, 0.000000), (0.000000, 1.000000) (1.000000, 0.000000), (1.000000, 1.000000) (1.000000, 1.000000), (0.000000, 1.000000) (0.000000, 0.000000) (1.000000, 0.000000) (1.000000, 1.000000) (0.000000, 1.000000) (0.000000, 0.000000), (1.000000, 0.000000) (0.000000, 0.000000), (1.000000, 1.000000) (0.000000, 0.000000), (0.000000, 1.000000) (1.000000, 0.000000), (1.000000, 1.000000) (1.000000, 1.000000), (0.000000, 1.000000) # Attaching Data to the Grid and Defining Grid Functions¶ This is a fundamental concept in any grid based discretization package. These are functions that can be evaluated given an entity in the grid and a local coordinate within the reference element of that entity. [7]: @dune.grid.gridFunction(aluView) def f(x): return math.cos(2.*math.pi/(0.3+abs(x[0]*x[1]))) @dune.grid.gridFunction(aluView) def hat0(element,hatx): return 1-hatx[0]-hatx[1]", "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", "# Tetration Forum Full Version: pentation and hexation You're currently viewing a stripped down version of our content. View the full version with proper formatting. Xorter and I have been exchanging some emails. Here's the question. Start with tetration, base e. If you want to extend to analytic tetration, download fatou.gp from this website. Fatou was recently updated to do some really really cool stuff, but that's for another post. sexpinit(exp(1)); sexp(-1)=0 sexp(0)=1 sexp(1)=e Now onto Pentation. The latest version of fatou.gp also has built in support for pentation for real bases, but lets start with the integer values for Pentation. genpent(exp(1)); pent(-2)=-1 /* slog(pent(-1))=slog(0)=-1 */ pent(-1)=0 /* slog(pent(0))=slog(1)=0 */ pent(0)=1 /* pent(0,1) are by definition 1 and the base=e */ pent(1)=e We don't yet have analytic hexation, but if it existed .... hex(-3)=-2 /* pent^{-1}(hext(-2))=pent^{-1}(-1)=-2 */ hex(-2)=-1 /* pent^{-1}(hext(-1))=pent^{-1}(0)=-1 */ hex(-1)=0 /* pent^{-1}(hext(0))=pent^{-1}(0)=-1 */ hex(0)=1 /* hex(0,1) are by definition, 1 and the base=e */ hex(1)=e If one uses the simple linear extension between hex<-2,-1>, then there would be a singularity somewhere near hex(-3.86) since hex(-2.86)~=-1.85, and the fixed point for tetration base(e) is ~=-1.85. Near that value, hex(z) would go to minus infinity. One could imagine generating an analytic hexation function by using a bipolar theta mapping using the two primary complex conjugate fixed points of", "label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); [/asy]$ NOTE: Connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices. Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\\frac{8 \\cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\\boxed{\\textbf{(C) }80}$. $[asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label(\"x=2\",(1,0,0),S); label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); label(\"A\",(0,2,0), NW); label(\"B\",(2,0,2), NW); [/asy]$ ### Solution 2 (rigorous) The three-dimensional distance formula shows that the lengths of the equilateral triangle must be $\\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \\le d_x, d_y, d_z \\le 2$, which yields the possible edge lengths of $\\sqrt{0^2+0^2+1^2}=\\sqrt{1},\\ \\sqrt{0^2+1^2+1^2}=\\sqrt{2},\\ \\sqrt{1^2+1^2+1^2}=\\sqrt{3},$ $\\ \\sqrt{0^2+0^2+2^2}=\\sqrt{4},\\ \\sqrt{0^2+1^2+2^2}=\\sqrt{5},\\ \\sqrt{1^2+1^2+2^2}=\\sqrt{6},$ $\\ \\sqrt{0^2+2^2+2^2}=\\sqrt{8},\\ \\sqrt{1^2+2^2+2^2}=\\sqrt{9},\\ \\sqrt{2^2+2^2+2^2}=\\sqrt{12}$ Some casework shows that $\\sqrt{2},\\ \\sqrt{6},\\ \\sqrt{8}$ are the only lengths that work, from which we can use the same counting argument as above.", "the lines are. \\draw[line width= 2pt,arrows={-Triangle[lw0r,angle=60 :0pt 2]}]( 4,0)--+(0,1)node{ 4}; \\draw[line width= 5pt,arrows={-Triangle[lw0r,angle=60 :0pt 2]}]( 5,0)--+(0,1)node{ 5}; \\draw[line width=10pt,arrows={-Triangle[lw0r,angle=60 :0pt 2]}]( 6,0)--+(0,1)node{ 6}; ## (7 8 9) You can also, illustrated below, set the size of tips to be larger than that of the lines by exactly 5pt. \\draw[line width= 2pt,arrows={-Triangle[lw0r,angle=60 :5pt 1]}]( 7,0)--+(0,1)node{ 7}; \\draw[line width= 5pt,arrows={-Triangle[lw0r,angle=60 :5pt 1]}]( 8,0)--+(0,1)node{ 8}; \\draw[line width=10pt,arrows={-Triangle[lw0r,angle=60 :5pt 1]}]( 9,0)--+(0,1)node{ 9}; ## Even more complicated Things become insane once the lines become doubled. ## (4 5 6 7 8 9) The first six arrows look OK: The tip size is a linear function in line width. \\draw[line width= 2pt,arrows={-Triangle[lw0r,angle=60 :0pt 2 ]}]( 4,0)--+(0,3)node{ 4}; \\draw[line width= 5pt,arrows={-Triangle[lw0r,angle=60 :0pt 2 ]}]( 5,0)--+(0,3)node{ 5}; \\draw[line width=10pt,arrows={-Triangle[lw0r,angle=60 :0pt 2 ]}]( 6,0)--+(0,3)node{ 6}; \\draw[line width= 2pt,arrows={-Triangle[lw0r,angle=60 :5pt 1 ]}]( 7,0)--+(0,3)node{ 7}; \\draw[line width= 5pt,arrows={-Triangle[lw0r,angle=60 :5pt 1 ]}]( 8,0)--+(0,3)node{ 8}; \\draw[line width=10pt,arrows={-Triangle[lw0r,angle=60 :5pt 1 ]}]( 9,0)--+(0,3)node{ 9}; ## (10 11 12) And then a problem arise: can the tip size grows with respect to the outer line width? So the forth argument involves. \\draw[line width= 2pt,arrows={-Triangle[lw0r,angle=60 :0pt 1 1]}](10,0)--+(0,3)node{10}; \\draw[line width= 5pt,arrows={-Triangle[lw0r,angle=60 :0pt 1 1]}](11,0)--+(0,3)node{11}; \\draw[line width=10pt,arrows={-Triangle[lw0r,angle=60 :0pt 1 1]}](12,0)--+(0,3)node{12}; By default, outer factor, the forth argument, is set to be 0. Hence the weighted line width, the dimension to be multiplied by the line width", "the next iteration. `circlepick` returns triangles with computed error estimates greater a given tolerance. The tolerance is provided to `circlepick` using the 'par' parameter. ```[u,p,e,t] = adaptmesh(g,model,c,a,f,'tripick', ... 'circlepick', ... 'maxt',2000, ... 'par',1e-3);``` ```Number of triangles: 258 Number of triangles: 515 Number of triangles: 747 Number of triangles: 1003 Number of triangles: 1243 Number of triangles: 1481 Number of triangles: 1705 Number of triangles: 1943 Number of triangles: 2155 Maximum number of triangles obtained. ``` Plot the finest mesh. ```figure; pdemesh(p,e,t); axis equal``` Plot the error values. ```x = p(1,:)'; y = p(2,:)'; r = sqrt(x.^2+y.^2); uu = -log(r)/2/pi; figure; pdeplot(p,e,t,'XYData',u-uu,'ZData',u-uu,'Mesh','off');``` Plot the FEM solution on the finest mesh. ```figure; pdeplot(p,e,t,'XYData',u,'ZData',u,'Mesh','off');``` Get trial now" ]

[ "Question 1 # How many triangles are there in the following figure? Solution Each square inside rectangular shape contain 4 small triangle and 4 medium triangle,a total of 8 triangle. And rectangular shape has 2 large triangle. So, a total of$$(2×8+2=18)$$ triangles are there in the shape. C is correct choice.", "intersecting triangles) 160pxstella octangula(24 triangles) 160pxThird stellation of icosahedron(60 triangles) 160pxA toroidal deltahedron(48 triangles)", "into four triangles with side $1$", "houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square? Triangles to Tetrahedra Stage: 3 Challenge Level: Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?", "triangles to all triangles? A pattern has 15 blue triangles to every 20 yellow triangles. What is the ratio of yellow triangles to all triangles?... ### Why is it important to have a taxonomy or classification of types of collection and how does that help us understand the development of collection strategies, technologies, collection planning, and methods of collection Why is it important to have a taxonomy or classification of types of collection and how does that help us understand the development of collection strategies, technologies, collection planning, and methods of collection...", "$2$). 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 $\\boxed{\\textbf{(A) }3}$.", "those unfamiliar with these special triangles and this will help you figure it out and know if you are correct), is that the hypotenuses of the 5 triangles (starting with the smallest) are: $$2, \\frac{4}{\\sqrt{3}}, \\frac{8}{3}, \\frac{16}{3\\sqrt{3}}, \\frac{32}{9}$$. I hope that helped!! May 6, 2021", "# Make eight equilateral triangles using just six matches? +1 vote 186 views Make eight equilateral triangles using just six matches? posted Jun 24, 2014 +1 vote Two equilateral triangles placed such that they form a star- 6 small equilateral triangles and 2 large equilateral triangles answer Aug 3, 2014 by anonymous –1 vote Similar Puzzles +1 vote What is the product of the area of square, equilateral triangle and the rectangle (where radius of the circle is 1 unit)?", "# Tri-ten-gles! How many (non-similar) types of non-degenerate triangles are there with all angles being multiples of $$10 ^ \\circ$$? ×", "# Division into strictly isosceles acute triangles What is the smallest number of strictly isosceles acute triangles that an equilateral triangle can be divided into? The following construction is by WR Somsky, with 13 triangles. Is this minimal? - I take it \"strictly isosceles\" means no equilateral triangles allowed, and \"acute\" means no right triangles allowed. – Gerry Myerson Jun 21 '12 at 1:28", "... In total, there are `67108864` subsets! Anna Szczepanek, PhD", "your answers. Quote: We have four different types of triangles to play with (unlimited amounts of each): a triangle which has 3 sides of length 1 a triangle which has 2 sides of length 1 and one of length square root 2 a triangle which has 2 sides of length square root 2 and one of side length 1 a triangle which has 3 sides of length square root 2 When I assemble 4 triangles, what is the maximum number of irregular tetrahedrons I can form? And what is the proof to make sure I have found the maximum? Using 4 triangles the same shape ----- you can make 2 irregular tetrahedrons. ?? Are you sure you understand \"irregular\"? A regular tetrahedron has all four faces congruent. . . It is composed of four equilateral triangles. And these are not counted. Using 4 triangles of the same shape, you may or may not get a tetrahedron. Here are the four shapes Code: * * * * _ * * * * √2 * * 1 * * * E1 * * R * * * * * * * * * * * * 1 1 * * * * * * _ * * _ √2 * * √2 * I * * E2 * * *", "triangles using AA similarity, 6 triangles using SSS similarity and 6. Similar Triangles. Fast and easy to use.", "6, \\$\\$(6,3,6)~(6,4,5)\\$\\$ And with 7, \\$\\$(7,1,7) ~ (7,2,6) ~ (7,3,5) ~ (7,4,4)\\$\\$ we have a grand total of \\$\\$\\boxed{7}\\$\\$ possibilities for non-congruent triangles. Oct 28, 2020 edited by Pangolin14 Oct 28, 2020 edited by Pangolin14 Oct 28, 2020", "Comment Share Q) There are five small rectangles in a big rectangle in every small rectangle . How many rectangles are there in all ? $\\begin{array}{1 1} 30 \\\\ 31 \\\\ 32 \\\\ 25 \\end{array}$", "# Find The Shortcut Geometry Level 1 The $16 \\times 24$ rectangle above is cut up into 4 triangles, 3 of which have perimeters 48, 36, 56, respectively, as indicated by the numbers inside them. What is the perimeter of the remaining green triangle? ×", "isosceles triangles Aug 28, 2018 335. four isosceles triangles in a circle Aug 26, 2018 336. hexagon in a star in a hexagon Aug 24, 2018 337. stacked hexagons Aug 23, 2018 338. a square, triangle, and circle in a square Aug 22, 2018 339. region of an octagon Aug 21, 2018 340. semi-circle in a square Aug 20, 2018 341. polygons with equal perimeter Aug 19, 2018 342. stacked shapes Aug 19, 2018 343. four squares overlapping a circle Aug 17, 2018 344. triangle in circle in triangle in semi-circle Aug 15, 2018 345. bell-ringing braids Aug 14, 2018 346. four equilateral triangles Aug 13, 2018 FourEquilateralTriangles_small.jpg 347. two squares and two equilateral triangles Aug 13, 2018 348. squares meeting in a circle Aug 12, 2018 349. zig-zag triangles Aug 12, 2018 350. four aligned squares Aug 10, 2018 351. lengths in a crossed trapezium Aug 9, 2018 352. three aligned squares Aug 9, 2018 353. overlapping squares Aug 8, 2018 354. toppled square Aug 7, 2018 355. two rectangles overlapping a circle Aug 2, 2018 356. subdivided circles May 21, 2018", "faces usually consist of triangles (triangle mesh), quadrilaterals, or other simple convex polygons. (Wikipedia) $\\alpha = \\beta$", "## amc-2014-jr-15 Four equilateral triangles of the same size are arranged with horizontal bases inside a larger equilateral triangle, as shown. What fraction of the area of the larger triangle is covered by the smaller triangles?", "composed of three components and only one rectangle namely ABCD is composed of five components. Thus, there are 5 + 2 + 1 + 1 = 9 rectangles in the given figure." ]

[ "total triangles, the answer is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$. ### Solution 3 For starters what I find helpful is to divide the whole octagon up into triangles as shown here: $[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot(\"A\",A,dir(45)); dot(\"B\",B,dir(90)); dot(\"C\",C,dir(135)); dot(\"D\",D,dir(180)); dot(\"E\",E,dir(-135)); dot(\"F\",F,dir(-90)); dot(\"G\",G,dir(-45)); dot(\"H\",H,dir(0)); dot(\"X\",X,dir(135/2)); dot(\"O\",O,dir(0)); draw(E--O--X); draw(C--O--B); draw(B--O--A); draw(A--O--H); draw(H--O--G); draw(G--O--F); draw(F--O--E); draw(E--O--D); draw(D--O--C); [/asy]$ Now it is just a matter of counting the larger triangles remember that $\\triangle BOX$ and $\\triangle XOA$ are not full triangles and are only half for these purposes. We count it up and we get a total of $\\frac{3.5}{8}$ of the shape shaded. We then simplify it to get our answer of $\\boxed{\\textbf{(D)}~\\frac{7}{16}}$.", "upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide? $[asy] draw((0,0)--(4,4*sqrt(3))); draw((1,-sqrt(3))--(5,3*sqrt(3))); draw((2,-2*sqrt(3))--(6,2*sqrt(3))); draw((3,-3*sqrt(3))--(7,sqrt(3))); draw((4,-4*sqrt(3))--(8,0)); draw((8,0)--(4,4*sqrt(3))); draw((7,-sqrt(3))--(3,3*sqrt(3))); draw((6,-2*sqrt(3))--(2,2*sqrt(3))); draw((5,-3*sqrt(3))--(1,sqrt(3))); draw((4,-4*sqrt(3))--(0,0)); draw((3,3*sqrt(3))--(5,3*sqrt(3))); draw((2,2*sqrt(3))--(6,2*sqrt(3))); draw((1,sqrt(3))--(7,sqrt(3))); draw((-1,0)--(9,0)); draw((1,-sqrt(3))--(7,-sqrt(3))); draw((2,-2*sqrt(3))--(6,-2*sqrt(3))); draw((3,-3*sqrt(3))--(5,-3*sqrt(3))); [/asy]$ $\\text{(A)}\\ 4 \\qquad \\text{(B)}\\ 5 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 7 \\qquad \\text{(E)}\\ 9$ ## Problem 25 There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it? $\\text{(A)}\\ 5724 \\qquad \\text{(B)}\\ 7245 \\qquad \\text{(C)}\\ 7254 \\qquad \\text{(D)}\\ 7425 \\qquad \\text{(E)}\\ 7542$", "square from 2 triangles, for simplicity, how many draw calls does glDrawElements (4 items in vertex array and 6 items in index array) and glDrawArrays (6 items only in vertex array) need, individually? Thanks. Each glDraw* is a draw call. 1 glDrawArrays is 1 draw call. 1 glDrawElements is 1 draw call. It doesn't matter (so far as draw call count is concerned) how many vertices or indices you use, 1 glDraw* is 1 draw call. The simple cases of drawing quads (assuming the GL version you use hasn't removed quads) or triangles are not a good example for comparing these, because a list of quads or a list of triangles can be drawn with a single call to either, and glDrawArrays will appear more efficient because it doesn't have the additional overhead of indexing. Let's take a slightly more complex case, but one which is representative of a more real-world scenario. Imagine that you have a model composed of multiple triangle strips and fans: glDrawArrays (GL_TRIANGLE_FAN, .....); glDrawArrays (GL_TRIANGLE_STRIP, .....); glDrawArrays (GL_TRIANGLE_STRIP, .....); glDrawArrays (GL_TRIANGLE_FAN, .....); glDrawArrays (GL_TRIANGLE_STRIP, .....); glDrawArrays (GL_TRIANGLE_FAN, .....); In this example, using glDrawArrays gives you a total of 6 draw calls for the model. However, both strips and fans can easily be converted to indexed triangles, so add indices and it becomes: glDrawElements", "won't install an older version of the runtime. You can use DXDIAG to determine which version of the runtimes are installed on your machine (just go to Run... from the Start menu and enter DXDIAG). If it is the case that your SDK is older than your current runtime, then you'll need to download and install the latest SDK. Share on other sites Quote: Original post by Scet Quote: Original post by speedieresult = (*m_d3ddev)->DrawIndexedPrimitive(D3DPT_TRIANGLELIST, 0, 0, m_batch_size, 0, m_batch_size / 2); I'm just drawing one quad, 2 triangles and 4 vertices. m_batch_size == 4. Drawing two triangles using a triangle list requires 6 vertices, not 4. I think you want a traingle strip. Ummm, not necessarily! DrawIndexedPrimitive()... 1 2+---+| /|| / ||/ |+---+3 4 4 vertices: 1,2,3,4 6 indices: 1,2,3,2,4,3 • 11 • 20 • 12 • 10 • 38 • Forum Statistics • Total Topics 631400 • Total Posts 2999862 ×", "indexes to connect. Vertices have 12 elements. Connection has 20 elements. >>> ic = RegularIcosahedron() >>> v, p = ic.getVertices() >>> len(v) 12 >>> len(p) 20 getTriangles()[source] Returns the instance of irfpy.util.triangle.Triangle. All the triangles will have normal vectors pointing outward. getNpTriangles()[source] Returns the instance of irfpy.util.triangle.NpTriangle. All the triangles will have normal vectors pointing outward. Returns a link of triangles. irfpy.util.icosahedron.doctests()[source]", "glEnableVertexAttribArray(0); glVertexAttribPointer(0, 3, GL_FLOAT, GL_FALSE, 3 * sizeof(GLfloat), nullptr); How many vertices and how many triangles are now described in the VBO/VAO pair? In this case we see that each vertex is of size elements by looking at 3 * sizeof(GLfloat). Then given there are 18 elements, there are vertices. Each triangle is made out of triangles so there are triangles described by this VAO.", "# Difference between revisions of \"2008 AMC 10B Problems/Problem 7\" ## Problem An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$. How many small triangles are required? $\\mathrm{(A)}\\ 10\\qquad\\mathrm{(B)}\\ 25\\qquad\\mathrm{(C)}\\ 100\\qquad\\mathrm{(D)}\\ 250\\qquad\\mathrm{(E)}\\ 1000$ ## Solution $[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]$ The number of triangles is $1+3+\\dots+19 = \\boxed{100}$. Another Solution: The area of the large triangle is 10^2 sqrt(3)/4, while the area each small triangle is 1^2 sqrt(3)/4. Dividing these two quantities, we get 100, therefore 100 small triangles can fit in the large one.", "# Problem #2091 2091 Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way? $[asy] size(200); defaultpen(0.9); real r = 5/dir(54).x, h = 5 tan(54*pi/180); pair A = (5,0), B = A+10*dir(72), C = (0,r+h), E = (-5,0), D = E+10*dir(108); draw(A--B--C--D--E--cycle); label(\"$A$\",A+(0,-0.5),SSE); label(\"$B$\",B+(0.5,0),ENE); label(\"$C$\",C+(0,0.5),N); label(\"$D$\",D+(-0.5,0),WNW); label(\"$E$\",E+(0,-0.5),SW); // real l = 5*sqrt(3); pair ab = (h+l)*dir(72), bc = (h+l)*dir(54); pair AB = (ab.y, h-ab.x), BC = (bc.x,h+bc.y), CD = (-bc.x,h+bc.y), DE = (-ab.y, h-ab.x), EA = (0,-l); draw(A--AB--B^^B--BC--C^^C--CD--D^^D--DE--E^^E--EA--A, dashed); // dot(A); dot(B); dot(C); dot(D); dot(E); dot(AB); dot(BC); dot(CD); dot(DE); dot(EA); [/asy]$ $\\text {(A) } 1 \\qquad \\text {(B) } 2 \\qquad \\text {(C) } 3 \\qquad \\text {(D) } 4 \\qquad \\text {(E) } 5$ 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.", "Draws a point at each vertex beginShape(POINTS); vertex(30, 20); vertex(85, 20); vertex(85, 75); vertex(30, 75); endShape(); beginShape(LINES) // Draws a line between each pair of vertices beginShape(LINES); vertex(30, 20); vertex(85, 20); vertex(85, 75); vertex(30, 75); endShape(); beginShape(TRIANGLES) // Connects each grouping of three vertices as a triangle beginShape(TRIANGLES); vertex(75, 30); vertex(10, 20); vertex(75, 50); vertex(20, 60); vertex(90, 70); vertex(35, 85); endShape(); beginShape(TRIANGLE_STRIP) // Starting with the third vertex, connects each subsequent // vertex to the previous two beginShape(TRIANGLE_STRIP); vertex(75, 30); vertex(10, 20); vertex(75, 50); vertex(20, 60); vertex(90, 70); vertex(35, 85); endShape(); beginShape(TRIANGLE_FAN) beginShape(TRIANGLE_FAN); vertex(10, 20); vertex(75, 30); vertex(75, 50); vertex(90, 70); vertex(10, 20); endShape(); beginShape(QUADS); vertex(30, 25); vertex(85, 30); vertex(85, 50); vertex(30, 45); vertex(30, 60); vertex(85, 65); vertex(85, 85); vertex(30, 80); endShape(); beginShape(QUAD_STRIP); vertex(30, 25); vertex(85, 30); vertex(30, 45); vertex(85, 50); vertex(30, 60); vertex(85, 65); vertex(30, 80); vertex(85, 85); endShape(); All of these shape modes are super exciting and interesting, but we won’t go into much more detail here. For more info, check out the Processing Documentation and/or the “Programming Handbook for Visual Artists and Designers” by Reas & Fry. # S4. A tiny bit more Fundamentals ## Variables and Data Types In the earlier example, we will see the use of variables like mouseX, mouseY , and mousePressed. What we know about these variables is that they", "2 sq units there are three families of triangles with a hoirizontal base:: Base 1 unit and height 4 units or Base 2 units and height 2 units or Base 4 units and height 1 unit For each family there are an infinite number of triangles", "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", "label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); [/asy]$ NOTE: Connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices. Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\\frac{8 \\cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\\boxed{\\textbf{(C) }80}$. $[asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label(\"x=2\",(1,0,0),S); label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); label(\"A\",(0,2,0), NW); label(\"B\",(2,0,2), NW); [/asy]$ ### Solution 2 (rigorous) The three-dimensional distance formula shows that the lengths of the equilateral triangle must be $\\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \\le d_x, d_y, d_z \\le 2$, which yields the possible edge lengths of $\\sqrt{0^2+0^2+1^2}=\\sqrt{1},\\ \\sqrt{0^2+1^2+1^2}=\\sqrt{2},\\ \\sqrt{1^2+1^2+1^2}=\\sqrt{3},$ $\\ \\sqrt{0^2+0^2+2^2}=\\sqrt{4},\\ \\sqrt{0^2+1^2+2^2}=\\sqrt{5},\\ \\sqrt{1^2+1^2+2^2}=\\sqrt{6},$ $\\ \\sqrt{0^2+2^2+2^2}=\\sqrt{8},\\ \\sqrt{1^2+2^2+2^2}=\\sqrt{9},\\ \\sqrt{2^2+2^2+2^2}=\\sqrt{12}$ Some casework shows that $\\sqrt{2},\\ \\sqrt{6},\\ \\sqrt{8}$ are the only lengths that work, from which we can use the same counting argument as above.", "# if the intersection exists, add the point (the first neighbor) and # the set (second neighbor) if triangle: triangles.update(triangle) triangles.add(n) return triangles In [29]: print(get_triangles(G = G, node = 3)) # drawing out the subgraph composed of those nodes to verify nx.draw(G.subgraph(get_triangles(G = G, node = 3)), with_labels = True) {1, 2, 3, 6, 23} Friend Recommendation: Open Triangles¶ Let's see if we can do some friend recommendations by looking for open triangles. Open triangles are like those that we described earlier on - A knows B and B knows C, but C's relationship with A isn't captured in the graph. In [30]: def get_open_triangles(G, node): # the target node's neighbor's neighbor's neighbor's should # not include the target node open_triangles = [] neighbors1 = set(G.neighbors(node)) for node1 in neighbors1: # remove the target node from the target node's neighbor's # neighbor's, since it will certainly go back to itself neighbors2 = set(G.neighbors(node1)) neighbors2.discard(node) for node2 in neighbors2: neighbors3 = set(G.neighbors(node2)) if node not in neighbors3: open_triangle = set([node]) open_triangle.update([node1, node2]) open_triangles.append(open_triangle) return open_triangles In [31]: open_triangles = get_open_triangles(G = G, node = 3) open_triangles Out[31]: [{1, 3, 4}, {1, 3, 12}, {1, 3, 29}, {2, 3, 16}, {2, 3, 19}, {3, 8, 17}, {3, 8, 22}, {3, 20, 23}] In [32]: # draw out each of", "# What is the explicit formula for the sequence representing the number of any triangles in a triangular grid? What is the explicit formula for the sequence representing the number of any triangles in a triangular grid, like those below? ### Context • Counting only up-facing triangles of size one, we get triangular numbers. • Counting both up-facing and down-facing triangles of size one, we get square numbers (that is, perfect squares) But what is the total number of all triangles of any size? Here are some values for small $n$. • I'd also like to know what you mean - do you mean those triangles pointing up (1,3,6,...)? Do you mean all of the smallest triangles (1,4,9,...)? Do you mean all the triangles of all sizes (1,5,13,...)? – user98602 Nov 20 '13 at 5:02 • @Mike: The latter. See my update. – kiss my armpit Nov 20 '13 at 5:11 • The accepted answer to the MSE post, how many triangles, by Brian M. Scott seems useful. He generalizes the particular question in that post. – J. W. Perry Nov 20 '13 at 5:36 The values fit $a_n=2n^2-2n+1$ To find that, two levels of differences gives a constant $4$, so the leading term is $(4/2!)n^2$. Subtract that off and you can find the rest.", "draws a Hexagon divided into 6 Triangles. (3) Right click on [Turtle] and choose [New subclass]. Create a class called Square. Edit the Square class Type in the following : public Square() { setColor(\"black\"); penDown(); } public void act() { move(100); turn(90); } to ensure that the Square object does indeed draw a square. (4) Add 3 more Square objects to draw 4 squares that fit together into one big square. (5) Open [TurtleWorld] and find the prepare() method. Type commands to add 5 more Square objects, to place 9 squares in a 3x3 grid, making one big square. =============================================== More Practice with Turtle Shapes (6) Make a CLASS for each of the following shapes - test each class by putting an OBJECT on the program page. (a) Triangle (3 equal sides) (b) Square (4 equal sides) (c) Pentagon (5 equal sides) (d) Hexagon (6 equal sides) (e) Octagon (8 equal sides) (7) You can TILE (cover) the entire page with identical shapes by placing lots of identical OBJECTS on the page. Each object draws the same shape in a different place, and all the shapes fit together to cover the page. Make a \"World\" (program) for each of the following tilings: (a) tile the page with squares (b) tile the page with triangles (c) tile the page", "sites slyterence 100 You will only need four vertices. You have two triangles connected together, as in silvermace's neat ascii picture, so regardless of how you draw the triangles (strips, fans, etc) you'll have those four vertices, labelled 0 to 3. For Triangle strips, the first triangle requires 3 vertices, and then every additional triangle requires one additional vertex. So, to draw your two triangles, you'd set up an array for indices, and the first three places you'd fill with [0,1,2]. That would draw your first triangle. Now, for the second, you need one more index (and for every triangle after the second, you'd need one more index as well). That vertex is the one you haven't used yet, i.e 3. So, your index array will be set up as follows : [0,1,2,3] (I think). Then, when calling glDrawElements, you'd set the last parameter to four (to account for four indices, ranging 0->3. Edit : Note, silvermace's formula for indices (numberOfPolygons * numberOfVertsPerPolygon) applies only if you're rendering using GL_TRIANGLES (I believe). This is because, in that instance, for every triangle you submit each vertex, whereas, as I explained for GL_TRIANGLE_STRIP, you submit all three vertices for the first triangle, and then only one vertex per each new triangle. So, for triangle strips, the formula for the number", "# Triangles in CSS Published , updated Tags: Creating triangles with CSS is a pretty good way to reduce the number of images within an application. They’re a bit tricky to get your head around at first but once you understand them it’s really easy. ## Drawing a triangle Lets say we are constructing a triangle to use as an up arrow, the first thing to do is to create an element with a size of 0x0. A bottom border is then applied to the element with the colour set as the colour of the triangle. Then add left and right borders to the element with the colour set to transparent. .triangle { border-bottom: 15px solid #000; border-left: 10px solid transparent; border-right: 10px solid transparent; width: 0; height: 0; } This is the result: Why did it come out in this shape? The three triangles in the triangle are actually the borders from .triangle. 10x15px triangles on the left and the right and the 20x15px triangle at the bottom. This works because the element is of size 0x0px, which makes the border pointed on the size closest to the element’s inner container. It’s easy to make a right-angle triangle too by removing border-right. ## Pseudo-elements Triangles are usually drawn with the pseudo-elements ::before and ::after because this way", "and a given number of subdivision. ```TRIANGLES=function(xinf,yinf,l,n,updown=\"up\"){ X=NA;Y=NA for(i in n:1){ u=xinf+seq(0+(n-i)/2/(n-1),1-(n-i)/2/(n-1),length=i)*l if(updown==\"up\") v=rep(yinf+(n-i)*sqrt(3)/2*l/(n-1),i) if(updown==\"down\") v=rep(yinf-(n-i)*sqrt(3)/2*l/(n-1),i) X=c(X,u);Y=c(Y,v)} return(cbind(X[-1],Y[-1]))}``` Here are grid with respectively 20 and 50 points on the lower side. It is then possible to define 4 grids, corresponding to the four sub-triangles, ```k=3;st=6 firstgrid=TRIANGLES(0,0,(k-2)/(k-1),k-1) secondgrid1=TRIANGLES( firstgrid[1,1],firstgrid[1,2],1*(k-2)/(k-1),st) secondgrid2=TRIANGLES( firstgrid[2,1],firstgrid[1,2],1*(k-2)/(k-1),st) secondgrid3=TRIANGLES( firstgrid[3,1],firstgrid[3,2],1*(k-2)/(k-1),st) secondgrid4=TRIANGLES( firstgrid[3,1],firstgrid[3,2],1*(k-2)/(k-1),st,updown=\"down\")``` Then, we just draw randomly five points on that grid, in the four sub-triangles, ```N=5 Dmax=0 setpointmax=matrix(0,4,2) indice=c(1:4,sample(1:4,size=N-4,replace=FALSE)) tindice=table(indice) indice1=sample(1:nrow(secondgrid1),size= tindice[1],replace=FALSE) indice2=sample(1:nrow(secondgrid2),size= tindice[2],replace=FALSE) indice3=sample(1:nrow(secondgrid3),size= tindice[3],replace=FALSE) indice4=sample(1:nrow(secondgrid4),size= tindice[4],replace=FALSE) setpoint=rbind(secondgrid1[indice1,],secondgrid2[indice2,], secondgrid3[indice3,],secondgrid4[indice4,])``` No, we can run a code, where we keep in mind locations of the five points each time we beak a record, ```D=min(dist(setpoint,\"euclidean\")) if(D>Dmax){Dmax=D setpointmax=setpoint}``` Here are some locations obtained after running the algorithm a few times (with five points) On the graph below, we can visualize the time it takes before having a record, and the convergence towards 1/2 (which is the true value of the maximal distance) The convergence is slow… extremely slow… However, we can run the same code for more than five points, e.g. seven points (actually, here sub-triangles are not used here, and it looks like we have been lucky here, since the convergence was rather fast),", "A(3,0), B(0,0), C(3,0). [3 lines] D and F on AB: D(0,1), F(0,2) H and J on BC: H(1,0), J(2,0) Draw FG and DE parallel to BC, G and E on AC. [2 lines] Draw HG and JE; will be parallel to AB. [2 lines] Let DE and GH intersect at K. So we have 3 1by1 squares: DBHK, KHJE and FDKG. Draw line FJ (will pass through K) [1 line] We now used up the 8 lines. You now have 11 right triangles: ABC, ADE, AFG, FBJ, FDK,FGK, GHC, GKE, KHJ, KEJ, EJC. -------------------------------------------------------------------------------------- Better still: 5 squares and 12 right triangles. Easy: draw a square [4 lines] draw the diagonals [2 lines] draw the horizontal and vertical parallels [2 lines] That's it. 4 squares inside larger square. 4 right triangles created by the diagonals. 8 right triangles created by the parallels.", "small triangles by using the tri1 and tri2 methods. • Having the 19 triangles, there are $2^{19}$ ways of selecting them, which I brute-force: • For every iteration, I create a list of polygons containing those triangles and try to join two of them by brute-forcing each possible pair. When a pair is found, I remove them from the list and add the joined polygon there. The pair brute-force is then started over again until no more polygons could be joined. • After no more joinable pairs are found if there is only a single polygon left on the list and it is a pentagon, then the selected set of triangles forms one of our pentagons. • For each pentagon found, I print the five vertexes coordinates and increment a counter, showing the counter final value in the very end of the program. The full output is in the block below. I couldn't find a way for convincing SE's markdown to spoiler-protect the output without also destroying the formatting, so I added a bunch of blank lines and this note before the actual output. Scroll down to see the actual output. [[0,2], [-1,3], [-2,2], [0,0], [2,2]] [[0,2], [-2,2], [0,0], [2,2], [1,3]] [[0,2], [2,2], [0,0], [-3,3], [-1,3]] [[-2,4], [-3,3], [0,0], [2,2], [0,2]] [[1,3], [-1,3], [-2,2], [0,0], [2,2]] [[0,2], [1,1]," ]

[ "# Math Help - Euclidean Algorithm Problem 1. ## Euclidean Algorithm Problem Find the greatest common divisor (GCD) and the least common multiple (LCM) for $6678671$ and $1322685$. (Note : Use Euclidean Algorithm) 2. ## Re: Euclidean Algorithm Problem What's your difficulty with this problem?", "Math Help - Euclidean Algorithm Problem 1. Euclidean Algorithm Problem Find the greatest common divisor (GCD) and the least common multiple (LCM) for $6678671$ and $1322685$. (Note : Use Euclidean Algorithm) 2. Re: Euclidean Algorithm Problem What's your difficulty with this problem?", "# Euclidean Algorithm Problem • Dec 12th 2012, 08:46 AM mastermin346 Euclidean Algorithm Problem Find the greatest common divisor (GCD) and the least common multiple (LCM) for $6678671$ and $1322685$. (Note : Use Euclidean Algorithm) • Dec 12th 2012, 09:03 AM emakarov Re: Euclidean Algorithm Problem What's your difficulty with this problem?", "# Euclidean Algorithm Problem Find the greatest common divisor (GCD) and the least common multiple (LCM) for $6678671$ and $1322685$. (Note : Use Euclidean Algorithm)", "# Have fun with GCD & LCM! The greatest common divisor(GCD) and the lowest commom multiple(LCM) of two whole numbers, $$m$$ and $$n$$, are $$14$$ and $$6468$$ respectively. Find the smallest possible value of $$m+n$$. ×", "How many ordered pairs of positive integers $$(m, n)$$ satisfy $\\gcd (m^3, n^2) = 2^2 \\cdot 3^2 \\text{ and } \\text{lcm}(m^2, n^3) = 2^4 \\cdot 3^4 \\cdot 5^6?$ This problem is shared by Muhammad A. Details and assumptions $$\\gcd(a, b)$$ and $$\\text{lcm}(a, b)$$ denote the greatest common divisor and least common multiple of $$a$$ and $$b$$, respectively. ×", "# Question 17 How many positive integral pairs $$(a,b)$$ can make the below expression true? $(a,b)+9[a,b]+9(a+b)=7ab$ • Notes: 1.$$(a,b)$$ stands for the GCD (Greatest common divisor) of integers $$a$$ and $$b$$; 2.$$[a,b]$$ stands for the LCM (Lowest common multiple) of integers $$a$$ and $$b$$. ×", "the notion of the least common multiple. Let $a$ and $b$ be positive integers. The least common multiple of $a$ and $b$ is denoted by $\\text{LCM}(a,b)$ and is defined as the least positive integer that is divisible by both $a$ and $b$. For example, $\\text{LCM}(16,18)=144$. We can also express $\\text{LCM}(a,b)$ as follows: $\\displaystyle \\text{LCM}(a,b)=\\frac{a \\cdot b}{\\text{GCD}(a,b)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$ where $\\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\\text{GCD}(a,b)=1$, we have $\\displaystyle \\text{LCM}(a,b)=a \\cdot b$. Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \\cdot 3$ and $18=2 \\cdot 3^2$. Then $\\text{LCM}(16,18)=2^4 \\cdot 3^2=144$. The least common divisor of the numbers $a_1,a_2,\\cdots,a_n$ is denoted by $\\text{LCM}(a_1,a_2,\\cdots,a_n)$ and is defined similarly. It is the least positive integer that is divisible by all", "of the least common multiple. Let $a$ and $b$ be positive integers. The least common multiple of $a$ and $b$ is denoted by $\\text{LCM}(a,b)$ and is defined as the least positive integer that is divisible by both $a$ and $b$. For example, $\\text{LCM}(16,18)=144$. We can also express $\\text{LCM}(a,b)$ as follows: $\\displaystyle \\text{LCM}(a,b)=\\frac{a \\cdot b}{\\text{GCD}(a,b)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$ where $\\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\\text{GCD}(a,b)=1$, we have $\\displaystyle \\text{LCM}(a,b)=a \\cdot b$. Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \\cdot 3$ and $18=2 \\cdot 3^2$. Then $\\text{LCM}(16,18)=2^4 \\cdot 3^2=144$. The least common divisor of the numbers $a_1,a_2,\\cdots,a_n$ is denoted by $\\text{LCM}(a_1,a_2,\\cdots,a_n)$ and is defined similarly. It is the least positive integer that is divisible by all $a_j$. Since", "is $$\\operatorname{lcm}(154, 56) = 616$$ exercise. Write a naive_lcm function that returns the least common multiple of two nonzero integer inputs. It should only rely on divisibility checks. The time complexity of this algorithm, similar to the naive gcd algorithm, grows linearly in the smaller of the two arguments. ## a property of the lcm proposition 7. Every common multiple of two integers is a multiple of their least common multiple. proof: Let $$k$$ be a common multiple of $$a$$ and $$b$$ and let $$m$$ be their least common multiple. Dividing $$k$$ by $$m$$ yields $$k = mq + r$$, where $$0 \\le r < m$$. Since $$r = k - mq$$ is a linear combination of $$k$$ and $$m$$, then $$r$$ must be divisible by both $$a$$ and $$b$$, ie, $$r$$ must be a common multiple of $$a$$ and $$b$$. If $$r$$ were nonzero, then it would be a smaller positive common multiple than the least common multiple, a contradiction. Therefore, $$r = 0$$ and $$m \\mid k$$. ## relation to greatest common divisor In order to compute the least common multiple efficiently, we need to prove an identity involving the greatest common divisor. proposition 8. If $$a$$ and $$b$$ are nonzero integers, then $$\\gcd(a, b) \\cdot \\operatorname{lcm}(a, b) = \\vert ab \\vert$$. proof: For simplicity, we", "# What is the GCD of all these numbers? $2^{72} - 1, \\quad 3^{72} - 1, \\quad 4^{72} - 1,\\quad \\ldots , \\quad 72^{72} - 1$ Find the greatest common divisor of the numbers above. ×", "Free Version Moderate # Greatest Common Divisor GRE-LV7XFE What is the greatest common divisor (GCD) of $72$ and $144$? A $6$ B $24$ C $12$ D $72$ E $18$", "# greatest common divisor is 7 and the least common multiple is 16940 How many such number-pairs are there for which the greatest common divisor is 7 and the least common multiple is 16940? - Let the two numbers be $7a$ and $7b$. Note that $16940=7\\cdot 2^2\\cdot 5\\cdot 11^2$. We make a pair $(a,b)$ with gcd $1$ and lcm $2^2\\cdot 5\\cdot 11^2$ as follows. We \"give\" $2^2$ to one of $a$ and $b$, and $2^0$ to the other. We give $5^1$ to one of $a$ and $b$, and $5^0$ to the other. Finally, we give $11^2$ to one of $a$ and $b$, and $11^0$ to the other. There are $2^3$ choices, and therefore $2^3$ ordered pairs such that $\\gcd(a,b)=1$ and $\\text{lcm}(a,b)=2^2\\cdot 5\\cdot 11^2$. If we want unordered pairs, divide by $2$. Here we used implicitly the Unique Factorization Theorem: Every positive integer can be expressed in an essentially unique way as a product of primes. There was nothing really special about $7$ and $16940$: any problem of this shape can be solved in basically the same way. - Kudos for emphasizing the role played by uniqueness of factorization. – Bill Dubuque Dec 11 '12 at 6:48", "For example, to find the LCM of 8, 12 and 15, write: $8 = 2^3$ $12 = 2^2\\cdot 3^1$ $15 = 3^1\\cdot 5^1$ Three primes appear in these factorizations, 2, 3 and 5. The largest power of 2 that appears is $2^3$; the largest power of 3 that appears is $3^1$; and the largest power of 5 that appears is $5^1$. Therefore the LCM, $LCM(8, 12, 15) = 2^3\\cdot 3^1\\cdot 5^1 = 120$. ### Using the GCD The LCM of two numbers can be found more easily by first finding their greatest common divisor (GCD). Once the GCD is known, the LCM is calculated by the following equation, $LCM(a, b) = \\frac{a \\cdot b}{GCD(a, b)}$. Let's use our first example. The GCD of 4 and 6 is 2. Using the above equation, we find $LCM(4, 6) = \\frac{4 \\cdot 6}{2} = \\frac{24}{2} = 12$, just like we expected.", "程序设计能力实训 1093. GCD and LCM Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given $a$ and $b$ $(0 < a, b \\leq 44000)$. 输入格式 The first line of input gives the number of cases, $N$ $(1 \\leq N \\leq 100)$. $N$ test cases follow. Each test case contains two interger a and b separated by a single space in a line. 输出格式 For each test case, print GCD and LCM separated by a single space in a line. 样例 Input 2 8 6 5000 3000 Output 2 24 1000 15000", "$$d$$ is a common divisor of $$a$$ and $$b$$ and $$D=\\gcd(a,b),$$ then by the above, we must have that $$D'=\\operatorname{lcm}(d,D)$$ is a divisor of $$a$$ and a divisor of $$b.$$ So it is a common divisor. If $$d$$ is not a divisor of $$D$$, then $$\\operatorname{lcm}(d,D)>D=\\gcd(a,b),$$ but that's not possible by the definition of $$\\gcd.$$ • I.e. common divisors are closed under LCM and this yields the sought ascent. This and the underlying duality is discussed at length in my comments on Mark's answer (and the posts linked there). – Bill Dubuque Jun 17 at 18:16", "integer that is a multiple of both a and b. We also have $lcm(a,b) = ab / gcd(a,b)$ (where gcd is the more commonly known greatest common divisor function) – poncho Nov 21 '13 at 13:03", "· 3 · 61,\\newline$ from here the greatest common divisor of $x$ and $y$ is $2.$ GCD$(866,732)=2.$ $ax+by=-71\\cdot866+84\\cdot732=2.$ ii) $866 = 2 · 433,\\newline 732 = 2 · 2 · 3 · 61 = 2^2 · 3 · 61,\\newline$ from here LCM$(866,732)=2\\cdot433\\cdot2\\cdot3\\cdot61=361956.$", "GCD of Factorials! Find the GCD of $(19! + 19, 20! + 19).$ Note: GCD stands for the greatest common divisor. ×", "LCM (a, b) (Least Common Multiple). - If we multiply a and b by the same positive integer k then their GCD is multiplied by k. GCD (k.a, k. b) = k.pgcd (a, b) - The product of their GCD by their LCM is equal to the absolute value of their product. This can be written, GCD (a, b) x LCM (a, b) = |a x b| - If a and b are two integers and p a common divisor to a and b then, p divides GCD (a, b). ## Programming This program computes the GCD of two integers. ### Python def gcd (a, b): while b: a, b = b, a%b return a" ]

[ "# How many pairs of numbers exist,which satisfy the following conditions? The GCD of $2$ numbers is $6$ and the product of the two numbers is $4320$. How many pairs of numbers exist, which satisfy the above condition. MyApproach I took GCD $\\times$ LCM=Product of two numbers 6 . LCM=4320 From which I get LCM=720. I am not reaching anywhere to the solution. Can anyone guide me how to solve the problem? You are right so far. If the two numbers are $a$ and $b$, we have $(a,b) = 6$, so $(6m, 6n) = 6$ where $(m, n) = 1$. Now the remaining $720$ has to be distributed among $m$ and $n$. $720 = 2^4 \\cdot 3^2 \\cdot 5$. We cannot have any factor of the same prime distributed, i.e. all the $2$'s must go either with $m$ or with $n$. Since there are $3$ distinct prime factors, there are $8$ possibilities of distributing them. Because we are interested in pairs, we eliminate symmetry by dividing $8$ by $2$ and get $\\color{blue}{4}$ possibilities as the final answer.", "eg: $24$ is a multiple of $3$, as $24 \\div 3 = 8$ with remainder $0$ eg: $24$ is also a multiple of $2$ eg: $24$ is also a multiple of $6$ eg: $24$ is also a multiple of $24$ The word \"multiple\" means \"having several parts of something\". Multiples of a number : The dividend that is divided by a factor with remainder $0$, is a multiple of the factor. The multiples of $6$ are \"$6 , 12 , 18 , \\cdots$\". The multiples of $34$ are \"$34 , 68 , 102 , 136 , \\cdots$\". Prime Factorization The factors of $24$ are \"$1 , 2 , 3 , 4 , 6 , 8 , 12 , 24$\" Consider the number $24$. Factors of $24$ are $1$, $2$, $3$, $4$, $6$, $8$, $12$, and $24$. The following ways, the number is expressed as a product of some of the factors. $4 \\times 6$ $2 \\times 4 \\times 3$ $3 \\times 8$ and more like $2 \\times 12$ The following is a product of prime numbers \"$2 \\times 2 \\times 2 \\times 3$\". The number $24$ is represented as product of prime factors. This is called prime factorization. Prime Factorization : A number represented as product of prime numbers. What is the prime factorization of $264$? The answer", "of $6$ therefore it is a multiple of $6$", "($2 \\times 2$). 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 $3$.", "12, 24, 36, 48 \\ldots$ Multiple of $18 = 18, 36, 54 \\ldots$ Hence the L.C.M of $12$ and $18$ is $36$ Methods of finding L.C.M of given numbers Prime Factorization Method 1. Express each one of the numbers as the product of prime factors 2. The product of all the different prime factors each raised to highest power that appears in the prime factorization of any given numbers, gives the L.C.M of the given numbers. Example: Find L.C.M of $180, 216$ and 324 \\$ Convert each of the numbers into prime factors. $180 = 2^2 \\times 3^2 x 5$ $216 = 2^2 \\times 3^3$ $324 = 2^2 \\times 3^4$ Therefore L.C.M $= 2^3 \\times 3^4 \\times 5 = 3240$ Common Division Method 1. Arrange the numbers in any order 2. Divide by a number that divides exactly at least two of the given numbers and carry forward the other numbers 3. Repeat Step 2 till no two numbers are divisible by the same number, other than $1$ 4. The product of the divisors and the un-divided numbers is the required L.C.M Hence the L.C.M $= 2 \\times 2 \\times 2 \\times 3 \\times 2 \\times 3 = 144$ Relation between H.C.F and L.C.M of two numbers i. Product of two given numbers = Product of their H.C.F", "$$9=18$$ The product of LCM and GCF $$=3×18$$ $$=54$$ The product of $$6$$ and $$9=6×9$$ $$=54$$ So, $$\\text{GCF × LCM = Product of the numbers}$$ • The GCF of the given numbers is a factor of their LCM. For example: • The GCF of $$4$$ and $$12=4$$ • The LCM of $$4$$ and $$12=12$$ • Here, $$4$$ (GCF) is a factor of $$12$$ (LCM). The LCM of the given numbers is a multiple of their GCF. In the above example: $$12$$ (LCM) is a multiple of $$4$$ (GCF). • If GCF of two numbers is one of them then the other number is their LCM. For example: The GCF of $$2$$ and $$6=2$$ So, the LCM of $$2$$ and $$6=6$$ #### LCM (a,b) $$=24$$ and GCF (a,b)​ $$=2$$. Find the value of b, if a $$=6$$. A $$24$$ B $$6$$ C $$12$$ D $$8$$ × Given: LCM (a,b) $$=24$$ GCF (a,b) $$=2$$ $$=6$$ The relation between LCM and GCF is $$\\text{LCM (a,b)}\\times \\text{GCF (a,b) = a×b}$$ On putting the values, we get: $$24×2=6×b$$ $$b=\\dfrac{24×2}{6}$$ $$b=8$$ Hence, option (D) is correct. ### LCM (a,b) $$=24$$ and GCF (a,b)​ $$=2$$. Find the value of b, if a $$=6$$. A $$24$$ . B $$6$$ C $$12$$ D $$8$$ Option D is Correct", "look. $12 = 2 \\times 2 \\times 3$ $15 = 3 \\times 5$ To find the LCM, we want to have all the prime factors from both numbers accounted for. For instance, there are two 2s (in the 12). Let's put those in: $L C M = 2 \\times 2 \\times \\ldots$ There is one 3 in both the 12 and the 15, so we need one 3: $L C M = 2 \\times 2 \\times 3 \\times \\ldots$ And there is one 5 (in the 15) so let's put that in: $L C M = 2 \\times 2 \\times 3 \\times 5 = 60$ $12 \\times 5 = 60$ $15 \\times 3 = 60$ 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 2 Jan 26, 2018 The least common multiple of $12$ and $15$ is $3$. #### Explanation: It would be 3 because... • $1 \\cdot 12 = 12$ • $2 \\cdot 6 = 12$ • $3 \\cdot 4 = 12$ • $4 \\cdot 3 = 12$ • $6 \\cdot 2 = 12$ and • $1 \\cdot 15 = 15$ • $3 \\cdot 5 = 15$ The least common multiple between $12$ and $15$ is $3$ because it", "# If $\\gcd(x,y)=1379,\\operatorname{lcm}(x,y)=n!$ then find $n$ If $\\gcd(x,y)=1379,\\operatorname{lcm}(x,y)=n!$ then find $n$ ($x,y,n$ are positive integers). I tried using the relation: $\\gcd(x,y)\\cdot\\operatorname{lcm}(x,y)=xy$ and also wrote $1379=7\\cdot197$ but nothing more. • Since $197$ is prime you need $n\\ge197$. Can you see how to choose $x,y$ so that $n=197$ works? Jun 14 '16 at 6:14 • @almagest Why is it necessary that: $n>=197$? Jun 14 '16 at 6:16 • Because none of $1,2,\\dots,n$ can be divisible by a prime greater than $n$. Jun 14 '16 at 6:22 • @almagest you mean this equation $7*197*n!=xy$ implies $n>=197$??!! I think it's not necessary that $n$ be divisible by $197$,on the other hand $x$ or $y$ should be... Jun 14 '16 at 6:25 • The gcd is divisible by 197, so the lcm is divisible by 197, so $n!$ is divisible by 197. Since 197 is prime that means that at least one of $1,2,\\dots,n$ is divisible by 197. That means that $n\\ge197$. Jun 14 '16 at 6:28 Since $\\gcd(x,y)=7\\cdot 197$, we can assume $x=7\\cdot 197a$, and $y=7\\cdot 197b$, where $\\gcd(a,b)=1$. Then $\\text{lcm}(x,y)=7\\cdot 197\\cdot ab=n!$. Thus, $n\\geqslant 197$. Actually, we can assign different prime factors of $\\frac{n!}{7\\cdot 197}$ to $a$ and $b$ respectively to make sure they are coprime. $n$ can be any integer which is no smaller than $197$.", "## Thinking Mathematically (6th Edition) GCD is $24$ and LCM is$144$. Write 48 and 72 in terms of prime factors as follows: \\begin{align} & 48=2\\times 2\\times 2\\times 2\\times 3 \\\\ & 72=2\\times 2\\times 2\\times 3\\times 3 \\\\ \\end{align} Further express the factors in terms of exponents: \\begin{align} & 48={{2}^{4}}\\times 3 \\\\ & 72={{2}^{3}}\\times {{3}^{2}} \\\\ \\end{align} Now, for GCD select each prime factor with the smaller exponent that is common to each of the prime factorization: \\begin{align} & \\text{GCD}={{2}^{3}}\\times 3 \\\\ & =8\\times 3 \\\\ & =24 \\end{align} Now, for LCM, select every prime factor that occurs, raised to the greater exponent in these prime factorizations: \\begin{align} & \\text{LCM}={{2}^{4}}\\times {{3}^{2}} \\\\ & =16\\times 9 \\\\ & =144 \\end{align} Hence, the GCD is $24$ and LCM is$144$.", "### Theory: Consider two numbers, $$15$$ and $$20$$. The LCM of $$15$$ and $$20$$ is $$60$$. That is, $$LCM(15,20) = 60$$. The GCD of $$15$$ and $$20$$ is $$5$$. That is, $$GCD(15,20) = 5$$. Now, $$LCM(15,20) \\times GCD(15,20) = 60 \\times 5 = 300$$ $$\\Rightarrow LCM(15,20) \\times GCD(15,20) = 15 \\times 20$$ This gives us the result that \"the product of any two polynomials is equal to the product of their LCM and GCD\". That is, $$f(x) \\times g(x) = LCM[f(x),g(x)] \\times GCD[f(x),g(x)]$$. Let us understand the concept using an example. Example: Let $$f(x) = 21(x^4 - x^2)$$ and $$g(x) = 16(x^2 + 3x)^2$$. Let us verify $$f(x) \\times g(x) = LCM[f(x),g(x)] \\times GCD[f(x),g(x)]$$. Solution: To prove: $$f(x) \\times g(x) = LCM[f(x),g(x)] \\times GCD[f(x),g(x)]$$. Proof: $$f(x) = 21(x^4 - x^2) = 3 \\times 7 \\times x^2 \\times (x^2 - 1) = 3 \\times 7 \\times x^2 \\times (x + 1)(x - 1)$$ $$g(x) = 16(x^2 + 3x)^2 = 2^4 \\times (x^4 + 6x^3 + 9x^2) = 2^4 \\times x^2 \\times (x^2 + 6x + 9) = 2^4 \\times x^2 \\times (x + 3)(x + 3)$$ Now, $$LCM[f(x),g(x)] = 3 \\times 7 \\times 2^4 \\times x^2 \\times (x + 1)(x - 1) \\times (x + 3)(x + 3)$$ $$= 336 \\times x^2(x^2 -1)(x + 3)^2$$ $$GCD[f(x),g(x)] = x^2$$ Consider", "number as a product of primes, matching primes vertically when possible. $12=2 \\cdot 2 \\cdot 3$ $18 = 2 \\cdot 3 \\cdot 3$ Now we bring down the primes in each column. The LCM is the product of these factors. Notice that the prime factors of $12$ and the prime factors of $18$ are included in the LCM. By matching up the common primes, each common prime factor is used only once. This ensures that $36$ is the least common multiple. HOW TO: Find the LCM using the prime factors method. 1. Find the prime factorization of each number. 2. Write each number as a product of primes, matching primes vertically when possible. 3. Bring down the primes in each column. 4. Multiply the factors to get the LCM. Example 6 Find the LCM of $15$ and $18$ using the prime factors method. Solution Example 7 Find the LCM of $50$ and $100$ using the prime factors method. Solution", "$b$, and (2) any other divisor of $a$ and $b$ is also a divisor of $d$; note that, in general, for any $a, b \\in R$ there may be any number of gcds—including none; if $d$ and $d^{\\prime}$ are both gcds of $a$ and $b$, then they are associates (i.e. there exists a unit $u \\in R$ such that $d = ud^{\\prime}$); this is so because they must be divisors of each other. Similarly, $m \\in R$ is a multiple of $a\\in R$ if there exists $q \\in R$ such that $m = aq$; $m$ is a least common multiple (lcm) of $a, b\\in R$ if (1) it is a multiple of both $a$ and $b$, and (2) any other multiple of $a$ and $b$ is also a multiple of $m$. Similar considerations on uniqueness as those for gcds apply to lcms. In particular, if $m$ and $m^{\\prime}$ are both lcms of $a, b\\in R$ then $m$ and $m^{\\prime}$ are associates. • does this proof show that $\\gcd(a,b)\\cdot lcm(a,b) \\sim ab$ (i.e., that the product of the gcd and lcm is associate to the product of the two elements)? – ALannister Apr 24 '17 at 16:41 • @ALannister: yes – kjo Apr 24 '17 at 17:24 Given $n,m$ and $d=\\gcd(n,m)$, we can then denote $n=da$,$m=db$ where $a,b$ are", "$18 = 2\\cdot 3\\cdot 3$ $6 = 2\\cdot 3$ $6\\,$ ‘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", "# Given that $a$ is an odd multiple of $1183$, find the greatest common divisor of $2a^2+29a+65$ and $a+13$. Given that $a$ is an odd multiple of $1183$, find the greatest common divisor of $2a^2+29a+65$ and $a+13$. I know there exists some slick technique to simplify this problem. Any hints are greatly appreciated. We can use Euclid's algorithm for a few steps: $$\\gcd(2a^2 + 29a + 65, a + 13) = \\gcd(2a^2 + 29a + 65 - 2a(a+13), a+13)\\\\ = \\gcd(3a + 65, a+13) = \\gcd(3a+65 - 3(a + 13), a + 13)\\\\ = \\gcd(26, a+13)$$ which is necessarily $1, 2, 13$ or $26$, just by looking at the first term. By factoring $1183$, and considering that $a$ is an odd multiple, you should be able to conclude. • I'm not sure I'm able to see what $a$ is from this. Nov 11 '17 at 20:41 • I know that $1183=7*13^2$ Nov 11 '17 at 20:41 • And an odd multiple of that, plus $13$, is an even multiple of $13$. Nov 11 '17 at 21:30 • What does that imply? That $a=1183$ since the $gcd(26, 1183+13)=26$ Nov 11 '17 at 21:36 • Isee. The answer is $26$ since the $gcd(26, a+13)=26$ when $a=1183k$ where $k=1,3,5,7,9,...$. Nov 11 '17 at 22:03 HINT let $d=(2a^2+29a+65,a+13)$. We notice that $-13$", "last\\ digit = 4$$ - By euclid's algorithm(http://en.wikipedia.org/wiki/Euclidean_algorithm), $\\gcd (a,b)=\\gcd (a,a-b)$. Here, $\\gcd(3^{2003}+1,3^{2003}-1)=\\gcd(3^{2003}+1,2)=2$ (since $3^{2003}+1$ is an even integer).LCM=$ab/\\gcd= \\frac{3^{4006}-1}{2}.$ Now $3^4=1\\pmod{20} \\implies (3^4)^{1001}.9=-11\\pmod {20}\\implies {3^{4006}-1}=-12\\pmod{20}=8\\pmod{20} \\implies$ $\\frac{3^{4006}-1}{2}=8/2=4$ is the last digit. - The OP wants LCM not GCD. – lhf Jul 3 '12 at 11:24 You need to determine the product mod 20, to determine the product/2 mod 10. – Bill Dubuque Jul 3 '12 at 17:09 As an example of what Bill Dubuque said, observe that (9-1)/2 = 4 but (19-1)/2 = 9. So to find the last digit of (x-1)/2, it is not enough to know the last digit of x. – ShreevatsaR Jul 4 '12 at 4:18 gcd($3^{2003}+1\\ , 3^{2003}-1$) $$=gcd(3^{2003}+1\\ , 2)\\ as\\ gcd(a,b)=gcd(a,a-b)\\ and\\ 3^{2003}+1\\ -(3^{2003}-1)=2$$ =2 as $3^n-1$ is even for any natural number n. So,lcm($3^{2003}+1\\ , 3^{2003}-1$)= ($3^{2003}+1)(3^{2003}-1$)/gcd($3^{2003}+1,\\ 3^{2003}-1$)=($3^{4006}-1)/2$ Now, using Euler's Totient Theorem, $a^\\phi(m)≡1(mod\\ m)$ where (a,m)=1 So, $a^{k\\phi(m)+h}≡a^h(mod\\ m)$ Here (3,10)=(3,100)=1 Now, $\\phi(10)=phi(2)phi(5)=1*4=4\\ and\\ \\phi(100)=10phi(10)=40$ 4006≡6(mod 40)≡2(mod 4) $3^{4006}≡3^6(mod\\ 100)≡3^2(mod\\ 10)$ So, $lcm(3^{2003}+1\\ , 3^{2003}-1)≡(3^6-1)/2(mod\\ 100)≡(3^2-1)/2(mod\\ 10)≡14(mod\\ 100)≡4(mod\\ 10)$ -", "= 2, 4, 6, 8, \\ldots$.", "The LCM of two different numbers is 18, and the GCF is 6. What are the numbers? Dec 27, 2016 They must have the prime factors $2 \\times 3 = 6$ in common. Explanation: $18 = 2 \\times 3 \\times 3$ and $6 = 2 \\times 3$ would do. Dec 27, 2016 The numbers themselves are 6 and 18. Explanation: If the LCM of the two numbers is 18, that means both numbers have to be factors of 18. So each number could be 1, 2, 3, 6, 9, or 18. If the GCF of the two numbers is 6, that means both numbers are divisible by 6. So each number could be 6, 12, 18, 24, ... etc. When we overlap these two restrictions, we see that the only values common to both sets are 6 and 18. So our possible pairs are $\\left(6 , 6\\right)$, $\\left(6 , 18\\right)$, or $\\left(18 , 18\\right)$. But wait—only one of these pairs has both an LCM of 18 and a GCF of 6. That pair is $\\left(6 , 18\\right)$. So, this is our answer. Bonus: In general, you may also be able to use the fact that the product of any two numbers will equal the product of $\\left(\\text{their lcm\") times (\"their gcf}\\right)$. Let's show why with an example pair:", "don't need to check obvious things like the fact that the gcd of $2$ and $7$ is $1$ ($7$ is prime, after all). The author likely meant for you to think \"what number of the form $7n+1$ is a multiple of $2$?\" From which you should have quickly seen $n=1$ and with $4$ as the inverse. – Hayden Feb 13 '15 at 4:29 • @Hayden Where did 7n + 1 come from? – committedandroider Feb 13 '15 at 4:30 • @committedandroider If $a\\equiv 1 \\pmod 7$, then $a=7n+1$ for some $n$. – apnorton Feb 13 '15 at 4:31 • By inspection here just means the numbers are small enough that you can either just \"see it\" or you can try the small number of possibilities. For example, $2\\cdot 4 = 8$, which is $1$ modulo $7$. – aes Feb 13 '15 at 4:35 If you want the multiplicative inverse of $2$ mod $7$, then you want to find an integer $n$ such that $2n = 7k + 1$, where $k$ is a nonnegative integer. Try $k = 1$, because that's the easiest thing to do. Then $2n= 8$, and $n = 4$. • Although the title mentions 17, the entirety of the question itself uses 7. (Granted, the methodology is pretty much identical) – Hayden Feb 13 '15", "product is $(14)(26\\ldots$. Then $6$ is fixed, then mapped to $3$; then looking at $3$, we have $3$ going to $5$ and then $5$ is fixed. Finally, $5$ is mapped to $2$ and $2$ is fixed, which closes the cycle. So the product is $$(1634)(1352)=(14)(2635).$$ -", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$." ]

[ "# How many pairs of numbers exist,which satisfy the following conditions? The GCD of $2$ numbers is $6$ and the product of the two numbers is $4320$. How many pairs of numbers exist, which satisfy the above condition. MyApproach I took GCD $\\times$ LCM=Product of two numbers 6 . LCM=4320 From which I get LCM=720. I am not reaching anywhere to the solution. Can anyone guide me how to solve the problem? You are right so far. If the two numbers are $a$ and $b$, we have $(a,b) = 6$, so $(6m, 6n) = 6$ where $(m, n) = 1$. Now the remaining $720$ has to be distributed among $m$ and $n$. $720 = 2^4 \\cdot 3^2 \\cdot 5$. We cannot have any factor of the same prime distributed, i.e. all the $2$'s must go either with $m$ or with $n$. Since there are $3$ distinct prime factors, there are $8$ possibilities of distributing them. Because we are interested in pairs, we eliminate symmetry by dividing $8$ by $2$ and get $\\color{blue}{4}$ possibilities as the final answer.", "Tags 17 Oct 2018, 00:52 Bunuel wrote: If x is the greatest common divisor of 90 and 18, and y is the least common multiple of 51 and 34, then x + y = A. 111 B. 120 C. 213 D. 222 E. 231 To find the GCD and LCM of two numbers always denote the numbers in the prime factorization form - First pair: $$18 = 2^1*3^2$$ $$90 = 2^1*3^2*5^1$$ To find GCD of these two numbers - find the common factors ( along with powers) : So GCD would be : $$2^1*3^2 = 18$$ Second pair: $$51 = 3^1*17^1$$ $$34 = 2^1*17^1$$ So GCD is $$17^1 = 17$$ and the LCM is (First number * Second number)/GCD = $$\\frac{(3^1*17^1*2^1*17^1)}{17^1} =$$$$3*2*17$$ Hence the question is asking $$18 + 6*17 = 18 + 102 = 120$$ Hence Option (B) is our choice. Regards, Director Joined: 06 Jan 2015 Posts: 512 Location: India Concentration: Operations, Finance GPA: 3.35 WE: Information Technology (Computer Software) Re: If x is the greatest common divisor of 90 and 18, and y is the least [#permalink] ### Show Tags 17 Oct 2018, 00:55 Bunuel wrote: If x is the greatest common divisor of 90 and 18, and y is the least common multiple of 51 and 34, then x + y = A. 111", "of the least common multiple. Let $a$ and $b$ be positive integers. The least common multiple of $a$ and $b$ is denoted by $\\text{LCM}(a,b)$ and is defined as the least positive integer that is divisible by both $a$ and $b$. For example, $\\text{LCM}(16,18)=144$. We can also express $\\text{LCM}(a,b)$ as follows: $\\displaystyle \\text{LCM}(a,b)=\\frac{a \\cdot b}{\\text{GCD}(a,b)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$ where $\\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\\text{GCD}(a,b)=1$, we have $\\displaystyle \\text{LCM}(a,b)=a \\cdot b$. Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \\cdot 3$ and $18=2 \\cdot 3^2$. Then $\\text{LCM}(16,18)=2^4 \\cdot 3^2=144$. The least common divisor of the numbers $a_1,a_2,\\cdots,a_n$ is denoted by $\\text{LCM}(a_1,a_2,\\cdots,a_n)$ and is defined similarly. It is the least positive integer that is divisible by all $a_j$. Since", "$b$, and (2) any other divisor of $a$ and $b$ is also a divisor of $d$; note that, in general, for any $a, b \\in R$ there may be any number of gcds—including none; if $d$ and $d^{\\prime}$ are both gcds of $a$ and $b$, then they are associates (i.e. there exists a unit $u \\in R$ such that $d = ud^{\\prime}$); this is so because they must be divisors of each other. Similarly, $m \\in R$ is a multiple of $a\\in R$ if there exists $q \\in R$ such that $m = aq$; $m$ is a least common multiple (lcm) of $a, b\\in R$ if (1) it is a multiple of both $a$ and $b$, and (2) any other multiple of $a$ and $b$ is also a multiple of $m$. Similar considerations on uniqueness as those for gcds apply to lcms. In particular, if $m$ and $m^{\\prime}$ are both lcms of $a, b\\in R$ then $m$ and $m^{\\prime}$ are associates. • does this proof show that $\\gcd(a,b)\\cdot lcm(a,b) \\sim ab$ (i.e., that the product of the gcd and lcm is associate to the product of the two elements)? – ALannister Apr 24 '17 at 16:41 • @ALannister: yes – kjo Apr 24 '17 at 17:24 Given $n,m$ and $d=\\gcd(n,m)$, we can then denote $n=da$,$m=db$ where $a,b$ are", "there exists a primitive root modulo $p$, then $p$ must be one of the moduli in the list stated in the theorem. ___________________________________________________________________________________________________________________ LCM The proof below makes use of the notion of the least common multiple. Let $a$ and $b$ be positive integers. The least common multiple of $a$ and $b$ is denoted by $\\text{LCM}(a,b)$ and is defined as the least positive integer that is divisible by both $a$ and $b$. For example, $\\text{LCM}(16,18)=144$. We can also express $\\text{LCM}(a,b)$ as follows: $\\displaystyle \\text{LCM}(a,b)=\\frac{a \\cdot b}{\\text{GCD}(a,b)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$ where $\\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\\text{GCD}(a,b)=1$, we have $\\displaystyle \\text{LCM}(a,b)=a \\cdot b$. Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \\cdot 3$ and $18=2 \\cdot 3^2$. Then", "# When the greatest common divisor and least common multiple of two integers are multiplied, their product is 200. When the greatest common divisor and least common multiple of two integers are multiplied, their product is 200. How many different values could be the greatest common divisor of the two integers? So $gcd(a,b)*lcm(a,b)=200$ and I know that $200=2^3*5^2$. I did a little scratch work and I'm thinking the answer is $24$. Any hints are greatly appreciated. • So if GCD=1 what are the possibilities, and then for other options?? – Mark Bennet Nov 11 '17 at 20:54 Let the numbers be $a,b$. The product of the GCD$(a,b)$ and LCM$(a,b)$ is $ab$. To have something be a factor of the GCD you have to have it be a factor of both $a,b$, so its square must be a factor of $200$. The only numbers whose square is a factor of $200$ are $1,2,5,10$, so those are the only GCDs possible. • Why is their product $ab$? – ddswsd Nov 11 '17 at 21:30 • It is a standard result. Let $c=\\gcd(a,b)$ and write $a=ce,b=cf$ $\\operatorname{lcm}(a,b)=cef, \\operatorname{lcm}(a,b)\\gcd(a,b)=c^2ef=ab$ – Ross Millikan Nov 11 '17 at 21:36", "the notion of the least common multiple. Let $a$ and $b$ be positive integers. The least common multiple of $a$ and $b$ is denoted by $\\text{LCM}(a,b)$ and is defined as the least positive integer that is divisible by both $a$ and $b$. For example, $\\text{LCM}(16,18)=144$. We can also express $\\text{LCM}(a,b)$ as follows: $\\displaystyle \\text{LCM}(a,b)=\\frac{a \\cdot b}{\\text{GCD}(a,b)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$ where $\\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\\text{GCD}(a,b)=1$, we have $\\displaystyle \\text{LCM}(a,b)=a \\cdot b$. Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \\cdot 3$ and $18=2 \\cdot 3^2$. Then $\\text{LCM}(16,18)=2^4 \\cdot 3^2=144$. The least common divisor of the numbers $a_1,a_2,\\cdots,a_n$ is denoted by $\\text{LCM}(a_1,a_2,\\cdots,a_n)$ and is defined similarly. It is the least positive integer that is divisible by all", "$18 = 2\\cdot 3\\cdot 3$ $6 = 2\\cdot 3$ $6\\,$ ‘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", "is: $18=6\\times 3 +0$. Why we have taken $18$? because we search a number that divides $b$ , and possibly divide also $a$. In this case we have taken this number $=3$. We have ,in fact, $24=18 \\times 1 + 6=(6\\times 3) \\times 1 +6= 6 \\times 4$. And this result shows also because we have chosen as divisor in the second step the remainder of the first step: simply we want to search if $18$ is a multiple of this remainder. The algoritm terminate when we find a remanider $=0$, and, in this case it has only two steps. Now use $a=24$ and $b=9$ and you can understand also the successive steps. • Does make great sense! Now my blurred up concepts are getting clearer. – Niket Parikh Apr 23 '15 at 16:56 Maybe the simplest way to look at this is that $\\gcd(a,b)=\\gcd(a-b,b)$, and then iterate that. If $e\\mid a$ and $e\\mid b$ then $e\\mid (a-b)$. If $e\\mid (a-b)$ and $e\\mid b$ then $e\\mid a$. If you can prove the two statement above (which is not hard) you can conclude: Every divisor that $a$ and $b$ have in common is a divisor that $a-b$ and $b$ have in common; and every divisor that $a-b$ and $b$ have in common is a divisor that $a$ and $b$", "# 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", "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.", "1 & 6 & 11 & 16 \\end{array}$$ This gives the following cycles: $$(1,5,9,13), (2,10), (3,15,11,7), (4), (6,14), (8), (12), (16)$$ So when the bell has rung a multiple of 4 times, persons starting at 1, 5, 9, 13, 3, 15, 11, and 7 are back in their original positions. For 2, 10, 6, and 14, the bell need ring only a multiple of 2 times for these people to be back in their original positions. The rest of the people never leave their square. Consequently, the least common multiple of the cycle lengths 4 and 2 is the first time everyone is back where they started. So four bells ring before this happens. 7. Recall that the product of the least common multiple and the greatest common divisor of two integers is given by $$\\textrm{lcm}(m,n) \\cdot \\gcd(m,n) = mn$$ As such, the product of the particular value of $m$ and $n$ we seek must be $18 \\cdot 720 = 2^5 \\cdot 3^4 \\cdot 5$. If we consider the prime factorization of $m$, note that it must contain at least $2 \\cdot 3^2$, as this is the gcd of $m$ and $n$. If it contains more than a single $2$ in its factorization, it must contain $4$ of them, as otherwise, the gcd of $m$ and $n$ would", "that $$a=4d=6c$$. Then the respective times $$d$$ and $$c$$ are contained in any common multiple of $$d$$ and $$c$$ must always have a ratio of $$2:3$$. This means that there must be a factor of $$2d$$ (and therefore $$3c$$) in any common multiple. If, for example, $$b=5d$$, then $$b=6c+d$$. But $$c \\mid b$$ and $$c \\mid 6c$$ imply $$c \\mid d$$. This is impossible, because $$3c=2d \\implies 3=2 \\frac dc$$, a contradiction. This situation arises every time there are two common divisors and neither divides the other. • How do you define the gcd? – Paul K Jun 16 at 18:46 • The greatest common divisor of $a$ and $b$, i.e. $(a,b)$ is a common divisor and if $c$ is another common divisor, then $c \\le (a,b)$. – The Footprint Jun 16 at 18:49 • See this duplicate. – Dietrich Burde Jun 16 at 19:01 • @DietrichBurde doesn't Math Gems's answer assume that GCD=$ab$/LCM ? What if I don't know that? That's why I don't like proofs that guess the form of the GCD first – The Footprint Jun 16 at 19:04 • This is a fact, where we don't have to guess the gcd. Also, you cannot avoid using $gcd(a, b)$, since you are using it (see the title, $(a, b)$) for the question. Perhaps I don't", "$$9=18$$ The product of LCM and GCF $$=3×18$$ $$=54$$ The product of $$6$$ and $$9=6×9$$ $$=54$$ So, $$\\text{GCF × LCM = Product of the numbers}$$ • The GCF of the given numbers is a factor of their LCM. For example: • The GCF of $$4$$ and $$12=4$$ • The LCM of $$4$$ and $$12=12$$ • Here, $$4$$ (GCF) is a factor of $$12$$ (LCM). The LCM of the given numbers is a multiple of their GCF. In the above example: $$12$$ (LCM) is a multiple of $$4$$ (GCF). • If GCF of two numbers is one of them then the other number is their LCM. For example: The GCF of $$2$$ and $$6=2$$ So, the LCM of $$2$$ and $$6=6$$ #### LCM (a,b) $$=24$$ and GCF (a,b)​ $$=2$$. Find the value of b, if a $$=6$$. A $$24$$ B $$6$$ C $$12$$ D $$8$$ × Given: LCM (a,b) $$=24$$ GCF (a,b) $$=2$$ $$=6$$ The relation between LCM and GCF is $$\\text{LCM (a,b)}\\times \\text{GCF (a,b) = a×b}$$ On putting the values, we get: $$24×2=6×b$$ $$b=\\dfrac{24×2}{6}$$ $$b=8$$ Hence, option (D) is correct. ### LCM (a,b) $$=24$$ and GCF (a,b)​ $$=2$$. Find the value of b, if a $$=6$$. A $$24$$ . B $$6$$ C $$12$$ D $$8$$ Option D is Correct", "# Solution A number $L$ is called a common multiple of $m$ and $n$ if both $m$ and $n$ divide $L$. The smallest such $L$ is called the least common multiple of m and n and is denoted by $\\textrm{lcm} (m,n)$ 1. Find the following: 1. lcm (8, 12) 2. lcm (20, 30) 3. lcm (51, 68) 4. lcm (23, 18) 2. Compare the value of $\\textrm{lcm} (m,n)$ with the values of $m$, $n$, and gcd($m,n$). In what way are they related? 3. Prove the relationship you found in part (b) always holds. 4. Compute $\\textrm{lcm} (301337,307829)$. 5. Find all $m$ and $n$ where $\\gcd (m,n) = 18$ and $\\textrm{lcm} (m,n) = 720$. 1. 24 2. 60 3. 204 4. 414 1. It appears that the product of the lcm and the gcd always equals $mn$. 2. Let $p_1,p_2,p_3, ...$ be the primes: 2, 3, 5, ..., in order of magnitude. Let $p_r$ be the largest prime that divides either $m$ or $n$. Then we can find write a prime factorization for both $m$ and $n$ in the following way: $m = p^{m_1}_1 p^{m_2}_2 \\cdots p^{m_r}_r$ and $n = p^{n_1}_1 p^{n_2}_2 \\cdots p^{n_r}_r$. Note, that some primes may not divide $m$ or $n$, so some of the exponents may be zero. It should be clear that $$\\gcd(m,n) =", "$$q_a$$ and $$q_b$$ such that $$a = d \\cdot q_a$$ and $$b = d \\cdot q_b$$. The greatest common divisor (GCD) of $$a$$ and $$b$$ is the largest integer $$d$$ that is a divisor of both $$a$$ and $$b$$. We denote the greatest common divisor of $$a$$ and $$b$$ by $$\\mathrm{gcd}(a, b)$$. Example. Consider $$a = 54$$ and $$b = 24$$. Then both $$a$$ and $$b$$ are divisible by 1, 2, 3, and 6, but they share no other divisors in common. Thus, we have $$\\mathrm{gcd}(a, b) = 6$$. Note. For any integer $$a$$, we have have $$\\mathrm{gcd}(a, 0) = a$$. ## Computing the gcd In school, I was taught the following method for computing the GCD of two numbers: 1. factor both numbers as a product of prime numbers 2. find all of the common factors (with multiplicities) 3. the GCD is the product of the common factors Using the example above with $$a = 54$$ and $$b = 24$$, I’d have written $18 = 2 * 3 * 3 * 3$ and $24 = 2 * 2 * 2 * 3$ so the common factors are 2 and 3 with multiplicity 1 (2 only appears once in the factorization of 18, and 3 only appears once in the factorization of 24). Thus we have (correctly) found", "an example call to the algorithm: gcd(290, 14) gcd(14, 10) gcd(10, 4) gcd(4, 2) gcd(2, 0) Now that we reached $b = 0$, we’ve determined that $\\gcd(290, 14) = 2$. It turns out, we can use what is known as the extended Euclidean algorithm to determine coefficients $u, v$ in $au + bv = \\gcd(a, b)$. This is something we will look at when determining modular inverses. ### Lowest Common Multiples Recall, when determining the greatest common divisor of two numbers, we took the lower of the two powers for each prime base. Our exponents in the resulting canonical representation were of the form $\\min(a_i, b_i)$. With the lowest common multiple, $\\text{lcm}(a, b)$, we end up doing the opposite. Let’s consider an example – let’s determine $\\text{lcm}(1200, 2520)$. $\\text{gcd}(1200, 2520) = 2^{min(4, 3)} \\cdot 3^{min(1, 2)} \\cdot 5^{min(2, 1)} \\cdot 7^{min(0, 1)} = 2^3 \\cdot 3^1 \\cdot 5^1 \\cdot 7^0 = 120$ $\\text{lcm}(1200, 2520) = 2^{max(4, 3)} \\cdot 3^{max(1, 2)} \\cdot 5^{max(2, 1)} \\cdot 7^{max(0, 1)} = 2^4 \\cdot 3^2 \\cdot 5^2 \\cdot 7^1 = 25200$ Intuitively speaking, a divisor will always be less than or equal to a number, while a multiple will always be greater than or equal to a number. $2^4$ divides both $2^4$ and $2^3$, so we choose $4$ to be the exponent on", "\\ \\ \\ \\ \\ \\ \\ \\ (1)$ where $\\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\\text{GCD}(a,b)=1$, we have $\\displaystyle \\text{LCM}(a,b)=a \\cdot b$. Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \\cdot 3$ and $18=2 \\cdot 3^2$. Then $\\text{LCM}(16,18)=2^4 \\cdot 3^2=144$. The least common divisor of the numbers $a_1,a_2,\\cdots,a_n$ is denoted by $\\text{LCM}(a_1,a_2,\\cdots,a_n)$ and is defined similarly. It is the least positive integer that is divisible by all $a_j$. Since the product of all the numbers $a_j$ is one integer that is divisible by each $a_k$, we have: $\\displaystyle \\text{LCM}(a_1,a_2,\\cdots,a_n) \\le a_1 \\cdot a_2 \\cdots a_n \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (2)$ As in the case of two numbers, the LCM of more than two numbers can be thought as the product of multiplying together the highest power of each prime number. For example, the LCM of", "is a greatest common divisor of $11+7i$ and $18-i$. (So are its associates $-1$ and $\\pm i$.) Comment: We could have saved some time. For example, any common divisor of $11+7i$ and $18-i$ must divide any linear combination of them, such as $1\\cdot(11+7i)+7\\cdot (18-i)$. This is $137$. So any common divisor must divide the norm of $11+7i$, which is $170$, and must also divide $137$. These numbers are relatively prime in the ordinary sense, so they are also relatively prime in the Gaussian sense. We could have dispensed with the divisions by $2+i$ and $2-i$ entirely! The Euclidean Algorithm: We just look at our particular problem, which is too small to give a full illustration of the process. The idea is to imitate the ordinary process of division with remainder. Divide $18-i$, the number with larger norm, by $11+7i$. After a little calculation this simplifies to $\\dfrac{191-137i}{170}$. Now pick the nearest Gaussian integer to this. It is $1-i$ and is our candidate for \"quotient.\" Calculate $(18-i)-(11+7i)(1-i)$: we get $3i$. Thus $$18-i=(11+7i)(1-i) +3i.$$ Note that as in the ordinary integer case, the gcd of $18-i$ and $11+7i$ is the same as the gcd of $11+7i$ and the \"remainder\" $3i$. It is obvious that this gcd is $1$. But if we decide not to notice, we can apply the division", "# If $\\gcd(x,y)=1379,\\operatorname{lcm}(x,y)=n!$ then find $n$ If $\\gcd(x,y)=1379,\\operatorname{lcm}(x,y)=n!$ then find $n$ ($x,y,n$ are positive integers). I tried using the relation: $\\gcd(x,y)\\cdot\\operatorname{lcm}(x,y)=xy$ and also wrote $1379=7\\cdot197$ but nothing more. • Since $197$ is prime you need $n\\ge197$. Can you see how to choose $x,y$ so that $n=197$ works? Jun 14 '16 at 6:14 • @almagest Why is it necessary that: $n>=197$? Jun 14 '16 at 6:16 • Because none of $1,2,\\dots,n$ can be divisible by a prime greater than $n$. Jun 14 '16 at 6:22 • @almagest you mean this equation $7*197*n!=xy$ implies $n>=197$??!! I think it's not necessary that $n$ be divisible by $197$,on the other hand $x$ or $y$ should be... Jun 14 '16 at 6:25 • The gcd is divisible by 197, so the lcm is divisible by 197, so $n!$ is divisible by 197. Since 197 is prime that means that at least one of $1,2,\\dots,n$ is divisible by 197. That means that $n\\ge197$. Jun 14 '16 at 6:28 Since $\\gcd(x,y)=7\\cdot 197$, we can assume $x=7\\cdot 197a$, and $y=7\\cdot 197b$, where $\\gcd(a,b)=1$. Then $\\text{lcm}(x,y)=7\\cdot 197\\cdot ab=n!$. Thus, $n\\geqslant 197$. Actually, we can assign different prime factors of $\\frac{n!}{7\\cdot 197}$ to $a$ and $b$ respectively to make sure they are coprime. $n$ can be any integer which is no smaller than $197$." ]

[ "# A symmetric triangle! Triangle $ABC$ has the property that $\\angle A+\\angle B= \\angle C$. Find the measure of $\\angle C$ in degrees. ×", "# What is the measure of a base angle of an isosceles triangle if its vertex angle measures 50 degrees? Nov 14, 2015 The measure of one base angle is $65$ degrees. #### Explanation: We know that an isosceles triangle is composed of a vertex angle and two base angles. The base angles are congruent to each other. We also know that the sum of a triangle's angles is $180$ degrees. Our first thing to do is subtract $50$ from $180$. we are left with $130$ degrees for both base angles. We divide $130$ by $2$ and get $65$ degrees. That means that each base angle is $65$ degrees. We can check our work by adding all angle measures together.", "Free Version Moderate # Similar Triangles 2 ACTMAT-L4J4YY Refer to the figure below (all angle measures are in degrees). If ${\\triangle ABC \\sim \\triangle DEF}$, what is the measure of ${\\angle B}$, in degrees? A $16°$ B $40°$ C $60°$ D $70°$ E $76°$", "# Angles in a Triangle Geometry Level 1 In triangle $$\\triangle ABC$$, $$\\angle ABC = 12^\\circ$$ and $$\\angle BCA = 34 ^ \\circ$$. What is the measure (in degrees) of $$\\angle CAB$$? ×", "# Cool angle in a triangle Geometry Level 3 Triangle $$ABC$$ is isosceles with $$AB = AC$$. The measure of angle $$BAD$$ is $$30 ^\\circ$$ and $$AD = AE$$. Find the measure of angle $$EDC$$ in degrees. ×", "# Difference between revisions of \"Circumradius\" 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 $\\frac{abc}{4A}$", "# The sum of the measures of two exterior angles of a triangle is 264 degrees. What is the measure of the third exterior angle? The exterior angles of any polygon add to 360 so your 3rd angle is $360 - 264 = 36$", "# Easy Angles Geometry Level 1 In $$\\triangle{ABC}$$, $$\\angle{A}$$ is $$20^\\circ$$ and angle B is $$40^\\circ$$. What is the measure of $$\\angle{C}$$ in degrees? ×", "### Still have math questions? Find the exact value of the side $$s$$ in the triangle below. The angle $$t = 30$$ degrees. $$t$$ is noted in red below. $$tan30° = \\frac{11}{s}$$", "In triangle ABC, the measure of angle B is three times that of angle A. The measure of angle C is 20 degrees more than that of angle A. How do you find the angle measures? Oct 28, 2016 $\\angle A = {32}^{\\circ}$ $\\angle B = {96}^{\\circ}$ $\\angle C = {52}^{\\circ}$ Explanation: In a triangle, the three interior angles always add up to 180°: Let $\\angle A = x$ $\\implies \\angle B = 3 x$ $\\implies \\angle C = x + 20$ $\\angle A + \\angle B + \\angle C = {180}^{\\circ}$ $\\implies x + 3 x + x + 20 = 180$ $\\implies 5 x + 20 = 180$ $\\implies 5 x = 160$ $x = 32$ Therefore, $\\angle A = x = {32}^{\\circ}$ $\\angle B = 3 x = 3 \\cdot 32 = {96}^{\\circ}$ $\\angle C = x + 20 = 32 + 20 = {52}^{\\circ}$", "# Angles in a triangle Geometry Level 1 $$ABC$$ is a triangle such that $$\\angle CAB = 2 \\theta$$, $$\\angle ABC = 3 \\theta$$ and $$\\angle BCA = 5 \\theta$$. What is the measure (in degrees) of $$\\theta$$? ×", "# Acute Angle Measure Geometry Level 4 Triangle $ABC$ is an acute triangle. Let $E$ be the foot of the perpendicular from $B$ to $AC$, let $F$ be the foot of the perpendicular from $C$ to $AB$. If $2[AEF] = [ABC]$, what is the measure (in degrees) of $\\angle BAC$? Details and assumptions $[PQRS]$ denotes the area of figure $PQRS$. ×", "## 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.", ", what is the measure of \\angle ABC(in …", "# In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) 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 mhalmantus We are given: $A=x,B=5x,C=x-2$ As hinted, the sum of the angle measures is 180° so: $A+B+C=180$ $x+\\left(5x\\right)+\\left(x-2\\right)=180$ Solve for x: $7x-2=180$ $7x=182$ $x=26$ So, the angle measures are: $A=x=26°$ $B=5x=130°$ $C=x-2=24°$", "# Too Many Angles, Too Little Information Geometry Level 3 In a triangle $$ABC$$, the measures of the angles are $$\\angle A=30^\\circ,$$ $$\\angle B=80^\\circ.$$ Point $$M$$ lies inside the triangle, such that $$\\angle MAC = 10^\\circ,$$ $$\\angle MCA = 30^\\circ.$$ What is the measure (in degrees) of $$\\angle BMC?$$ ×", "15 Degree Triangle Problem In the diagram below, $AB$ represents the diameter, $C$ lies on the circumference of the circle, and you are given that (Area of Circle) / (Area of Triangle) = $2\\pi$. Prove that the two smaller angles in the triangle are exactly $15^o$ and $75^o$ respectively. Problem ID: 361 (11 Oct 2009) Difficulty: 4 Star Show Problem & Solution", "# Algebra In triangle ABC, the measure of angle B is 8 degrees more than three times the measure of angle A. The measure of angle C is 37 degrees more than the measure of angle A. Find the measure of each angle. What is the measure of angle A__? What is the measure of angle B__? What is the measure of angle C__? 1. 👍 2. 👎 3. 👁 1. angles of triangle add up to 180 A + (3A+8) + (A+37) = 180 all yours now 1. 👍 2. 👎 2. (x)+(x+37)+ 3x +8)= 180 5x + 45 = 180 5x = 180 - 45 5x = 135 x = 135/5 x = 27 A = 27, B = 3(27) + 8 = 81 + 8 = 89, C = 27 + 37 = 64 1. 👍 2. 👎 ## Similar Questions 1. ### math 1. What is the correct classification for the triangle shown below? A triangle has two angles measuring 68 degrees and 22 degrees. (1 point) acute, scalene acute, isosceles --- right, scalene obtuse, scalene 2. What is the value 2. ### Math Which of the following values best approximates of the length of c in triangle ABC where c = 90(degrees), b = 12, and B = 15(degrees)? c = 3.1058 c", "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.", "# Angles in a Triangle Geometry Level 1 In triangle $$\\triangle ABC$$, $$\\angle ABC = 12^\\circ$$ and $$\\angle BCA = 34 ^ \\circ$$. What is the measure (in degrees) of $$\\angle CAB$$? × Problem Loading... Note Loading... Set Loading..." ]

[ "# What is the measure of a base angle of an isosceles triangle if its vertex angle measures 50 degrees? Nov 14, 2015 The measure of one base angle is $65$ degrees. #### Explanation: We know that an isosceles triangle is composed of a vertex angle and two base angles. The base angles are congruent to each other. We also know that the sum of a triangle's angles is $180$ degrees. Our first thing to do is subtract $50$ from $180$. we are left with $130$ degrees for both base angles. We divide $130$ by $2$ and get $65$ degrees. That means that each base angle is $65$ degrees. We can check our work by adding all angle measures together.", "You now have the reference triangle for a fourth-quadrant angle in standard position. What is that angle? 12. Use your calculator to find the sine and cosine of your new angle. Label the coordinates of the point where the angle intersects the unit circle. Generalize: All four of your angles have the same reference angle, $56\\degree\\text{.}$ For each quadrant, write a formula for the angle whose reference angle is $\\theta\\text{.}$ ###### Example4.11 1. Find the reference angle for $200\\degree\\text{.}$ 2. Sketch $200\\degree$ and its reference angle in standard position, along with their reference triangles. Verify that both angles have the same trigonometric ratios, up to sign. Solution 1. In standard position, an angle of $200\\degree$ lies in the third quadrant. Its reference angle is \\begin{equation*} 200\\degree - 180\\degree = 20\\degree \\end{equation*} 2. Both angles are shown at right. Note that the reference triangle for $200\\degree$ is congruent to the reference triangle for $20\\degree\\text{.}$ You can use your calculator to verify the following values. \\begin{equation*} \\begin{aligned}[t] \\sin 20\\degree \\amp = 0.3420 \\amp\\amp \\sin 200\\degree = -0.3420\\\\ \\cos 20\\degree\\amp = 0.9397 \\amp\\amp \\cos 200\\degree = -0.9397\\\\ \\tan 20\\degree \\amp = 0.3640 \\amp\\amp \\tan 200\\degree = 0.3640\\\\ \\end{aligned} \\end{equation*} ###### Checkpoint4.12 1. Find the reference angle for $285\\degree\\text{.}$ 2. Sketch $285\\degree$ and its reference angle in standard position, along with their", "# In triangle PQR, the measure of angle P is 36 degrees. The measure of angle Q is five times the measure of angle R. How do you find the measurement of angle Q and the measurement of angle R? Mar 31, 2017 #### Explanation: \"the sum of measures of these three angles of any triangle is invariably equal to the straight angle, also expressed as 180°\" From the problem we can write: $Q = 5 R$ We can also write from Geometry, the sum of the three angles is 180 degrees, or: $P + Q + R = 180$ We can substitute $36$ for $P$ and we can substitute $5 R$ for $Q$ and solve for $R$: $36 + 5 R + R = 180$ $36 + 5 R + 1 R = 180$ $36 + \\left(5 + 1\\right) R = 180$ $36 + 6 R = 180$ $- \\textcolor{red}{36} + 36 + 6 R = - \\textcolor{red}{36} + 180$ $0 + 6 R = 144$ $6 R = 144$ $\\frac{6 R}{\\textcolor{red}{6}} = \\frac{144}{\\textcolor{red}{6}}$ $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{6}}} R}{\\cancel{\\textcolor{red}{6}}} = 24$ $R = 24$ Substituting $24$ for $R$ in the the equation for the relationship and calculating $Q$ gives: $Q = 5 R$ becomes: $Q = 5 \\cdot 24$ $Q = 120$ Angle $Q$ is 120", "as the angles, vary in length. Properties of obtuse scalene triangles are: • They have $$2$$ acute angles as well as $$1$$ obtuse angle. • All its sides and angles are distinct in measurement. • The total of all $$3$$ inner angles equal $$180°$$. ### Triangles – Example 1: Find the measure of the unknown angle in the triangle Solution: The sum of the inner angles of a triangle is $$180°$$ . So, $$90°$$$$+$$$$45^°$$$$=$$$$135^°$$$$→$$$$180^°$$$$-$$ $$135^°$$ $$=$$ $$45^°$$ . The unkown angle is $$45^°$$ ### Triangles – Example 1: Find the measure of the unknown angle in the triangle Solution: The sum of the inner angles of a triangle is $$180°$$ . So, $$120°$$$$+$$$$35^°$$$$=$$$$155^°$$$$→$$$$180^°$$$$-$$ $$155^°$$ $$=$$ $$25^°$$ . The unkown angle is $$25^°$$ ## Exercises for Triangles Find the measure of the unknown angle in each triangle. 1) 2) • $$\\color{blue}{70^°}$$ • $$\\color{blue}{30^°}$$ ### What people say about \"Triangles\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE $5 It was$16.99 now it is \\$11.99", "angle measure A is three times the measure x of the supplement, then: A = 3x, and, consequently, substituting this into the above equation, we have: x + 3x = 180° Then the measure of its supplement is (180° – x). No Related Subtopics. In Triangle ABC, a line through A cuts the side BC at D such that BD : DC = 4 : 5. Shwetank Mauria Aug 22, 2017 Both angles are #45^@# Explanation: #m + n = 90# as an angle and its complement equal 90 . The measure of one angle is three times the measure of its complement. If the area of Triangle ABD = $$60 cm^{2}$$, then the area of Triangle ADC is, 5). The measure of an angle whose supplement is three times as large as its complement, is, 1). What is the measure of the angle and measure of the supplement? Find the measure of the angle. Two poles of height 7 m and 12 m stand on a plane ground. A) 22.5° B) 45° C) 67.5° D) 135° 12. Supplementary angles: Two angles, the sum of whose measure is 180° are called supplementary angles. and to check we see that the supplement of 45 (135) is indeed 3 times greater than the supplement of 45 (45). The measure of", "name it the Bermuda Triangle. The triangle will be connected to a ramp on each side of the triangle, so that students will come down the ramp onto the rails. There they will either succeed or be lost at sea! Marc drew the following picture of the triangle. The triangle has three angles and the boys want to reproduce these angles in their structure. The first angle $B$ is equal to $62^\\circ$, the second angle $C$ is equal to $63^\\circ$. Marc can’t remember the measure of angle $A$. He thinks there is a way to figure this out, but he can’t remember what it is. The sum of the interior angles of a triangle is equal to $180^\\circ$. Marc knows the measure of two of the angles of the triangle. Therefore, he can write an equation to figure out the measure of the third angle. $63 + 62 + x = 180$ The variable $x$ is used to represent the measure of angle $A$. Marc is working to find the measure of angle $A$. $125 + x & = 180 \\\\180 - 125 & = 55$ The measure of Angle $A$ is $55^\\circ$ Vocabulary Here are the vocabulary words that are found in this lesson. Triangle a three sided figure with three angles. The prefix “tri”means three. Acute", "Mathematics # In. triangle ABC,$\\angle A$ + $\\angle B$ = 144 and$\\angle A$ + $\\angle C$ = 124.Calculate smallest angle of the triangle. $36^o$ ##### SOLUTION $\\angle A + \\angle B = 144$...(I) $\\angle A + \\angle C = 124$...(II) In triangle ABC, $\\angle A + \\angle B + \\angle C = 180$ $\\angle A + \\angle B + \\angle A + \\angle C = 144+ 124$ $180 + \\angle A = 268$ $\\angle A = 268 - 180$ $\\angle A = 88$ Put this value in (I) $\\angle A + \\angle B = 144$ $88 + \\angle B = 144$ $\\angle B = 56$ Put this value in (II) $\\angle A + \\angle C = 124$ $88 + \\angle C = 124$ $\\angle C = 36$ 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 Angles formed by two intersecting lines having no common arms are called .. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q2 Subjective Medium Find the value of x in each of the following diagrams Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q3 Subjective Medium Classify", "of all the angles is $$180º$$ So, \\begin{align}\\textstyle​​​ ∠BCE + ∠CBE + ∠CEB &= 180º \\\\ ∠CBE + ∠CEB &= 180º – 150º \\\\ &= 30º ​​\\end{align} Therefore, \\begin{align}​​​ ∠CBE &= ∠CEB \\\\ &= 15º ​​\\end{align} Similarly, $$\\triangle{ADE}$$ is also an isosceles triangle And $$∠DEA = 15º$$ But, it is given in the question that $$∆ CDE$$ is an equilateral triangle and all the angles are $$60º$$ Thus, \\begin{align}​​​ \\textstyle ∠DEA + ∠AEB + ∠CEB &= 60º \\\\ ∠DEA + ∠CEB &= 15º + 15º \\\\ &= 30º ​​\\end{align} Since $$∆ CDE$$ is an equilateral triangle, $$∠DEC = 60º$$ \\begin{align}​​​ ∠AEB &= 60º – 30º \\\\ ∠AEB &= 30º ​​\\end{align} $$∠AEB = 30º$$ ## Finding Unknown Angles By Folding The Paper When we fold a paper at a certain angle, and then when we unfold it, the edge of the paper forms a straight line. Do you know what are the properties of the angles in a straight line? We know that the angles formed in a straight line add up to 180º. Now using this concept, try to find some unknown angles in a folded paper. Question 5: A rectangular piece of paper is folded as shown below. Find $$\\angle a$$. Solution: Let us see if we unfold this piece of paper, what will be the figure.", "In triangle ABC, the measure of angle B is three times that of angle A. The measure of angle C is 20 degrees more than that of angle A. How do you find the angle measures? Oct 28, 2016 $\\angle A = {32}^{\\circ}$ $\\angle B = {96}^{\\circ}$ $\\angle C = {52}^{\\circ}$ Explanation: In a triangle, the three interior angles always add up to 180°: Let $\\angle A = x$ $\\implies \\angle B = 3 x$ $\\implies \\angle C = x + 20$ $\\angle A + \\angle B + \\angle C = {180}^{\\circ}$ $\\implies x + 3 x + x + 20 = 180$ $\\implies 5 x + 20 = 180$ $\\implies 5 x = 160$ $x = 32$ Therefore, $\\angle A = x = {32}^{\\circ}$ $\\angle B = 3 x = 3 \\cdot 32 = {96}^{\\circ}$ $\\angle C = x + 20 = 32 + 20 = {52}^{\\circ}$", "# In isosceles triangle ABC, B is the vertex. The measure of angle B can be represented as (4x-2). The measure of angle A can be represented as (8x+1). Find the measure of all three angles of the triangle? May 16, 2018 $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$. #### Explanation: Since $\\Delta A B C$ is an isosceles triangle, angle A is congruent to angle C. Knowing this, and the fact that a triangle's angle measures add up to 180, we can state: $4 x - 2 + 2 \\left(8 x + 1\\right) = 180$ Now solve for x. $4 x - 2 + 16 x + 2 = 180$ $20 x = 180$ $x = 9$ Now apply x to the two given angle measures to find the new measurements. $4 \\cdot 9 - 2 = 34$ $8 \\cdot 9 + 1 = 73$ $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$.", "the triangle. A+B+C=180 Next, fill into the equation, the measures of the angles given in the diagram. A+B+C37+28+C==180180 Next, simplify the left side of the equation. 37+28+C65+C==180180 Then, subtract 65° from both sides of the equation to solve for the measure of C\\begin{align*}\\angle C\\end{align*}. 65+C6565+CC===18018065115 The answer is 115. The measure of the obtuse angle is 115°. #### Example 3 Given DEF\\begin{align*}\\triangle DEF\\end{align*}, such that D=42\\begin{align*}\\angle D = 42^{\\circ}\\end{align*} and E=123\\begin{align*}\\angle E = 123^{\\circ}\\end{align*}, what is the measure of F\\begin{align*}\\angle F\\end{align*}? First, draw and label a triangle to model the problem. License: CC BY-NC 3.0 Next, write an equation to represent the sum of the measures of the interior angles of the triangle. D+E+F=180 Next, fill into the equation, the measures of the angles given in the diagram. D+E+F42+123+F==180180 Next, simplify the left side of the equation. 42+123+F165+F==180180 Then, subtract 165° from both sides of the equation to solve for the measure of C\\begin{align*}\\angle C\\end{align*}. 165+F165165+FF===18018016515 The answer is 15. The measure of F=15\\begin{align*}\\angle F = 15^{\\circ}\\end{align*}. ### Follow Up Credit: Elliot Brown Source: https://www.flickr.com/photos/ell-r-brown/3639951854/in/photolist-6xDGSA-5wEtmK-7ztdDP-6jURUe-6jUSdc-8SL3Z6-ofitfs-o2ier1-6jZ6fs-65Uqfv-e5CMj8-6Vz9oS-gkcPn-2EYhU4-9eDpNd-pAePx8-aUaQsM-6jUUTX-cH3rgU-hfrpu7-66cnkD-7ayFFh-oiRKrn-kiwSzY-65vFN2-e5Jqts-6jUTVP-dJk5Yg-eVtVS-9nPvzg-2EuPD-9mjNV1-oXCYu8-9mjQxm-brvLxS-mK7BgB-6MU4ub-pE48tr-eSLqP9-4ndDbn-oiuZEB-iPLspo-8awuz3-pepv4y-a5YV8J-oyabu8-725hxm-6K4h9W-8fFEHN-cvxWUU License: CC BY-NC 3.0 Remember the worker with the roof truss? He needs to figure out the measure of the interior angles in the roof truss. How can he do this? He can use the facts that the triangle is isosceles and the", "a straight line add up to $180\\degree$ $180 - 121 = 59\\degree$ Now we know all 4 interior angles, we get that $x = 360 - 84 - 100 - 59 = 117\\degree$. You will find algebra involved more in questions on this topic, but as long as you know what the interior and exterior angles add up to then you can write the statement “these angles add up to ___” as an equation which you can then solve. Have a go at the questions below to see. Example Questions This shape has 5 sides, so its interior angles add up to, $180 \\times (5 - 2) = 540\\degree$ Hence each interior angle is, $x\\degree=540\\degree \\div 5 = 108\\degree$ This shape has 8 sides, so its interior angles add up to, $180 \\times (8 - 2) = 1080\\degree$ Hence each interior angle is, $x\\degree=1080\\degree \\div 8 = 135\\degree$ An isosceles triangle has two sides the same length but also two angles the same size. Specifically, the two angles at the base of the triangle, which in this case is on the corners B and C given that the markings on the triangle show that AB and AC are the equal sides. So, we have that angle ACB is also $34\\degree$. Angles in a triangle add up to 180,", "# In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) 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 mhalmantus We are given: $A=x,B=5x,C=x-2$ As hinted, the sum of the angle measures is 180° so: $A+B+C=180$ $x+\\left(5x\\right)+\\left(x-2\\right)=180$ Solve for x: $7x-2=180$ $7x=182$ $x=26$ So, the angle measures are: $A=x=26°$ $B=5x=130°$ $C=x-2=24°$", "## Angles In A Triangle A triangle is the simplest possible polygon. It is a two-dimensional (flat) shape with three straight sides forming an interior, closed space. It has three interior angles. One of the earliest concepts to learn in geometry is that triangles have interior angles adding up to $180°$. But how do you know? How can you prove this is true? Let's find out! ## How To Find The Angle of a Triangle You may have a triangle where only two angles have been labelled and measured. Now that you are certain all triangles have interior angles adding to $180°$, you can quickly calculate the missing measurement. You can do this one of two ways: 1. Subtract the two known angles from $180°$. 2. Plug the two angles into the formula and use algebra: ### How To Find The Missing Angle of a Triangle Two known angles of a triangle are $37°$ and $24°$. What is the missing angle? We can use two different methods to find our missing angle: 1. Subtract the two known angles from $180°$: 2. Plug the two angles into the formula and use algebra: ## Triangle Angle Formula Let's draw a triangle and label its interior angles with three letters $a$, $b$, and $c$. Our sample will have side $ac$ horizontal", "angle measures. III. Describe and Analyze Triangles and Associated Angle Measures Using Known Information, Variable Expressions and Triangles Formed by Two Lines Intersecting Parallel Lines Now that we have looked at how to identify triangles by angle measures and side lengths, we can look at how we can work with missing angle measures. We identify missing angle measures using what we know about triangles and by writing variable expressions. Let’s take a look at how we can problem solve in an example. Example What is the missing angle measure of this triangle? Now we have a triangle here. First, we can use our known information to figure out what kind of triangle it is. Let’s begin by looking at the angles in this triangle. There are two small angles. These are 25 degree angles and they are acute. We can see by looking at this third unknown angle, that is an obtuse angle. This is an obtuse triangle. Next, we can write an expression to help us to figure out the missing measure. $25+25+x$ Now, we can write this expression into an equation with a sum of 180 degrees. Then we can solve it for the value of $x$. $25+25+x &= 180\\\\50+x &= 180\\\\x &= 180-50\\\\x &= 130^{\\circ}$ The measure of the missing angle is $130^{\\circ}$. Now let’s", "## Algebra: A Combined Approach (4th Edition) An isosceles triangle has two angles with the same measure. The measure of the third angle is 30$^{o}$ more than twice the measure of either of the other two. Let us call the third angle x and the two other angles y. $x=2y+30$ We also know that the sum of the angles of a triangle is 180$^{o}$. $x+2y=180$ $(2y+30) +2y=180$ $4y+30=180$ $4y=150$ $y=37.5$ Using the first equation, we can find out the other angle: $x=2(37.5)+30$ $x=105$", "# 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 $$a^\\circ$$? $a^\\circ = 180^\\circ - (70^\\circ + 70^\\circ )$ $= 180^\\circ - 140^\\circ = 40^\\circ$", ". 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", "### Home > CCG > Chapter 5 > Lesson 5.1.1 > Problem5-9 5-9. Multiple Choice: In the triangle below, $x$ must be: 1. $42º$ 1. $69º$ 1. $21º$ 1. $138º$ 1. none of these The triangle is isosceles meaning $2$ sides are equal. The measure of $x$ must be the same as the other unknown angle because base angles are equal in an isosceles triangle. Since the sum of all the interior angles of a triangle must be $180º$, then $x+x+42º=180º$.", "quantities. This implies that SAS triangles are congruent: the fact that the angle is comprised between the two sides constrains the construction of the shape. With such a set of values, the only thing we can do is connect the ends of the sides and close the triangle: there is no opportunity to change the shape (because of the angle) or scale (because of the given length of two sides. ## How to use our SAS triangle calculator Our SAS triangle calculator can solve any congruent SAS triangle. Insert a combination of two adjacent sides and the angle between them. For example, let's take a triangle with the following parameters: • $a = 4\\ \\text{cm}$; • $b = 3\\ \\text{cm}$; and • $\\gamma = 42\\degree$. Input these values in the SAS triangle calculator. First thing, we will find the value of the third side: \\begin{align*} c &= \\sqrt{a^2\\! +\\! b^2 \\!-\\! 2\\!\\cdot\\!a\\!\\cdot\\!b\\!\\cdot\\cos{(\\gamma)}}\\\\ &= \\sqrt{4^2\\! +\\!3^2 \\!-\\! 2\\!\\cdot\\!4\\!\\cdot\\!3\\!\\cdot\\cos{(42\\degree)}} \\\\ &= 2.68\\ \\text{cm} \\end{align*} We now calculate the remaining angles. Let's take $\\beta$: \\begin{align*} \\beta& = \\arcsin{\\left(\\sin{(\\gamma)}\\cdot \\frac{b}{c}\\right)}\\\\ \\\\ &= \\arcsin{\\left(\\sin{(42)}\\cdot \\frac{4}{2.68}\\right)}\\\\ \\\\ &= 48.59\\degree \\end{align*} And with the simple relation between the angles of a triangle, we find $\\alpha$: \\begin{align*} \\alpha &= 180\\degree-\\beta=\\gamma\\\\ &=180\\degree-42\\degree-48.59\\degree\\\\ &= 89.41\\degree \\end{align*} ## What if you don't have a SAS triangle? Don't worry about" ]

[ "In triangle ABC, the measure of angle B is three times that of angle A. The measure of angle C is 20 degrees more than that of angle A. How do you find the angle measures? Oct 28, 2016 $\\angle A = {32}^{\\circ}$ $\\angle B = {96}^{\\circ}$ $\\angle C = {52}^{\\circ}$ Explanation: In a triangle, the three interior angles always add up to 180°: Let $\\angle A = x$ $\\implies \\angle B = 3 x$ $\\implies \\angle C = x + 20$ $\\angle A + \\angle B + \\angle C = {180}^{\\circ}$ $\\implies x + 3 x + x + 20 = 180$ $\\implies 5 x + 20 = 180$ $\\implies 5 x = 160$ $x = 32$ Therefore, $\\angle A = x = {32}^{\\circ}$ $\\angle B = 3 x = 3 \\cdot 32 = {96}^{\\circ}$ $\\angle C = x + 20 = 32 + 20 = {52}^{\\circ}$", "angle measure A is three times the measure x of the supplement, then: A = 3x, and, consequently, substituting this into the above equation, we have: x + 3x = 180° Then the measure of its supplement is (180° – x). No Related Subtopics. In Triangle ABC, a line through A cuts the side BC at D such that BD : DC = 4 : 5. Shwetank Mauria Aug 22, 2017 Both angles are #45^@# Explanation: #m + n = 90# as an angle and its complement equal 90 . The measure of one angle is three times the measure of its complement. If the area of Triangle ABD = $$60 cm^{2}$$, then the area of Triangle ADC is, 5). The measure of an angle whose supplement is three times as large as its complement, is, 1). What is the measure of the angle and measure of the supplement? Find the measure of the angle. Two poles of height 7 m and 12 m stand on a plane ground. A) 22.5° B) 45° C) 67.5° D) 135° 12. Supplementary angles: Two angles, the sum of whose measure is 180° are called supplementary angles. and to check we see that the supplement of 45 (135) is indeed 3 times greater than the supplement of 45 (45). The measure of", "# In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) 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 mhalmantus We are given: $A=x,B=5x,C=x-2$ As hinted, the sum of the angle measures is 180° so: $A+B+C=180$ $x+\\left(5x\\right)+\\left(x-2\\right)=180$ Solve for x: $7x-2=180$ $7x=182$ $x=26$ So, the angle measures are: $A=x=26°$ $B=5x=130°$ $C=x-2=24°$", "# In triangle PQR, the measure of angle P is 36 degrees. The measure of angle Q is five times the measure of angle R. How do you find the measurement of angle Q and the measurement of angle R? Mar 31, 2017 #### Explanation: \"the sum of measures of these three angles of any triangle is invariably equal to the straight angle, also expressed as 180°\" From the problem we can write: $Q = 5 R$ We can also write from Geometry, the sum of the three angles is 180 degrees, or: $P + Q + R = 180$ We can substitute $36$ for $P$ and we can substitute $5 R$ for $Q$ and solve for $R$: $36 + 5 R + R = 180$ $36 + 5 R + 1 R = 180$ $36 + \\left(5 + 1\\right) R = 180$ $36 + 6 R = 180$ $- \\textcolor{red}{36} + 36 + 6 R = - \\textcolor{red}{36} + 180$ $0 + 6 R = 144$ $6 R = 144$ $\\frac{6 R}{\\textcolor{red}{6}} = \\frac{144}{\\textcolor{red}{6}}$ $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{6}}} R}{\\cancel{\\textcolor{red}{6}}} = 24$ $R = 24$ Substituting $24$ for $R$ in the the equation for the relationship and calculating $Q$ gives: $Q = 5 R$ becomes: $Q = 5 \\cdot 24$ $Q = 120$ Angle $Q$ is 120", "# What is the measure of a base angle of an isosceles triangle if its vertex angle measures 50 degrees? Nov 14, 2015 The measure of one base angle is $65$ degrees. #### Explanation: We know that an isosceles triangle is composed of a vertex angle and two base angles. The base angles are congruent to each other. We also know that the sum of a triangle's angles is $180$ degrees. Our first thing to do is subtract $50$ from $180$. we are left with $130$ degrees for both base angles. We divide $130$ by $2$ and get $65$ degrees. That means that each base angle is $65$ degrees. We can check our work by adding all angle measures together.", "# Too Many Angles, Too Little Information Geometry Level 3 In a triangle $$ABC$$, the measures of the angles are $$\\angle A=30^\\circ,$$ $$\\angle B=80^\\circ.$$ Point $$M$$ lies inside the triangle, such that $$\\angle MAC = 10^\\circ,$$ $$\\angle MCA = 30^\\circ.$$ What is the measure (in degrees) of $$\\angle BMC?$$ ×", "+0 # Find the measure of each angle in triangle ABC. +1 197 1 +350 Find the measure of each angle in triangle ABC. Triangle A B C has angles labeled as follows: A, (x minus 16) degrees; B, (2x minus 155) degrees; C, (one half x plus 8) degrees. left parenthesis 2 x minus 155 right parenthesis degrees(2x−155)° left parenthesis one half x plus 8 right parenthesis degrees12x+8° left parenthesis x minus 16 right parenthesis degrees(x−16)° ABC The measure of angle A is The measure of angle B is The Measure of angle C is #1 +2075 +2 The triangle sum theorem states that the sum of the measures of the interior angles of a triangle equals 180 degrees. Using this theorem alone, one can find the measure of all the angles. $$m\\angle A+m\\angle B+m\\angle C=180$$ Plug in the measure for all these angles by using substitution. $$x-16+2x-155+\\frac{1}{2}x+8=180$$ Combine like terms. $$3x+\\frac{1}{2}x-163=180$$ Add 163 to both sides. $$\\frac{6}{2}x+\\frac{1}{2}x=343$$ Meanwhile, I converted 3x to a fraction that I can combine with the other lingering fraction. $$\\frac{7}{2}x=343$$ Multiply by 2 on both sides. $$7x=686$$ Divide by 7 on both sides. $$x=98$$ Now, plug in this value for the angle measure expressions. $$m\\angle A=x-16=98-16=82^{\\circ}$$ $$m\\angle B=2x-155=2*98-155=196-155=41^{\\circ}$$ $$m\\angle C=\\frac{1}{2}x+8=\\frac{1}{2}*98+8=49+8=57^{\\circ}$$ TheXSquaredFactor Dec 1, 2017 #1 +2075 +2 The triangle sum theorem states", "# Algebra In triangle ABC, the measure of angle B is 8 degrees more than three times the measure of angle A. The measure of angle C is 37 degrees more than the measure of angle A. Find the measure of each angle. What is the measure of angle A__? What is the measure of angle B__? What is the measure of angle C__? 1. 👍 2. 👎 3. 👁 1. angles of triangle add up to 180 A + (3A+8) + (A+37) = 180 all yours now 1. 👍 2. 👎 2. (x)+(x+37)+ 3x +8)= 180 5x + 45 = 180 5x = 180 - 45 5x = 135 x = 135/5 x = 27 A = 27, B = 3(27) + 8 = 81 + 8 = 89, C = 27 + 37 = 64 1. 👍 2. 👎 ## Similar Questions 1. ### math 1. What is the correct classification for the triangle shown below? A triangle has two angles measuring 68 degrees and 22 degrees. (1 point) acute, scalene acute, isosceles --- right, scalene obtuse, scalene 2. What is the value 2. ### Math Which of the following values best approximates of the length of c in triangle ABC where c = 90(degrees), b = 12, and B = 15(degrees)? c = 3.1058 c", "# Angles in a Triangle Geometry Level 1 In triangle $$\\triangle ABC$$, $$\\angle ABC = 12^\\circ$$ and $$\\angle BCA = 34 ^ \\circ$$. What is the measure (in degrees) of $$\\angle CAB$$? ×", "# Cool angle in a triangle Geometry Level 3 Triangle $$ABC$$ is isosceles with $$AB = AC$$. The measure of angle $$BAD$$ is $$30 ^\\circ$$ and $$AD = AE$$. Find the measure of angle $$EDC$$ in degrees. ×", "# In isosceles triangle ABC, B is the vertex. The measure of angle B can be represented as (4x-2). The measure of angle A can be represented as (8x+1). Find the measure of all three angles of the triangle? May 16, 2018 $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$. #### Explanation: Since $\\Delta A B C$ is an isosceles triangle, angle A is congruent to angle C. Knowing this, and the fact that a triangle's angle measures add up to 180, we can state: $4 x - 2 + 2 \\left(8 x + 1\\right) = 180$ Now solve for x. $4 x - 2 + 16 x + 2 = 180$ $20 x = 180$ $x = 9$ Now apply x to the two given angle measures to find the new measurements. $4 \\cdot 9 - 2 = 34$ $8 \\cdot 9 + 1 = 73$ $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$.", "# Acute Angle Measure Geometry Level 4 Triangle $ABC$ is an acute triangle. Let $E$ be the foot of the perpendicular from $B$ to $AC$, let $F$ be the foot of the perpendicular from $C$ to $AB$. If $2[AEF] = [ABC]$, what is the measure (in degrees) of $\\angle BAC$? Details and assumptions $[PQRS]$ denotes the area of figure $PQRS$. ×", "You now have the reference triangle for a fourth-quadrant angle in standard position. What is that angle? 12. Use your calculator to find the sine and cosine of your new angle. Label the coordinates of the point where the angle intersects the unit circle. Generalize: All four of your angles have the same reference angle, $56\\degree\\text{.}$ For each quadrant, write a formula for the angle whose reference angle is $\\theta\\text{.}$ ###### Example4.11 1. Find the reference angle for $200\\degree\\text{.}$ 2. Sketch $200\\degree$ and its reference angle in standard position, along with their reference triangles. Verify that both angles have the same trigonometric ratios, up to sign. Solution 1. In standard position, an angle of $200\\degree$ lies in the third quadrant. Its reference angle is \\begin{equation*} 200\\degree - 180\\degree = 20\\degree \\end{equation*} 2. Both angles are shown at right. Note that the reference triangle for $200\\degree$ is congruent to the reference triangle for $20\\degree\\text{.}$ You can use your calculator to verify the following values. \\begin{equation*} \\begin{aligned}[t] \\sin 20\\degree \\amp = 0.3420 \\amp\\amp \\sin 200\\degree = -0.3420\\\\ \\cos 20\\degree\\amp = 0.9397 \\amp\\amp \\cos 200\\degree = -0.9397\\\\ \\tan 20\\degree \\amp = 0.3640 \\amp\\amp \\tan 200\\degree = 0.3640\\\\ \\end{aligned} \\end{equation*} ###### Checkpoint4.12 1. Find the reference angle for $285\\degree\\text{.}$ 2. Sketch $285\\degree$ and its reference angle in standard position, along with their", "is three times the measure x of the supplement, then: A = 3x, and, consequently, substituting this into the above equation, we have: x + 3x = 180° Then the measure of its supplement is (180° – x). No Related Subtopics. In Triangle ABC, a line through A cuts the side BC at D such that BD : DC = 4 : 5. Shwetank Mauria Aug 22, 2017 Both angles are #45^@# Explanation: #m + n = 90# as an angle and its complement equal 90 . The measure of one angle is three times the measure of its complement. If the area of Triangle ABD = $$60 cm^{2}$$, then the area of Triangle ADC is, 5). The measure of an angle whose supplement is three times as large as its complement, is, 1). What is the measure of the angle and measure of the supplement? Find the measure of the angle. Two poles of height 7 m and 12 m stand on a plane ground. A) 22.5° B) 45° C) 67.5° D) 135° 12. Supplementary angles: Two angles, the sum of whose measure is 180° are called supplementary angles. and to check we see that the supplement of 45 (135) is indeed 3 times greater than the supplement of 45 (45). The measure of the supplement of", "Therefore, the unknown side is given by: \\begin{align} &=26 \\!-\\! (7 \\!+\\! 11) \\\\ &= 8 \\;\\text{feet} \\end{align} $$\\therefore$$ The third side measures 8 feet. Example 2 Emma is building a triangular, wooden birdhouse as shown. If two of the angles measure 46° and 62°, what is the measure of the third angle? Solution We know that the sum of the angles of a triangle adds up to 180° Therefore, the unknown angle is: \\begin{align} &=180^\\circ \\!-\\! (46^\\circ \\!+\\! 62^\\circ) \\\\ &= 72^\\circ\\end{align} $$\\therefore$$ The third angle measures 72° Example 3 In a right triangle, two of the sides measure 3 inches and 4 inches. What is the measure of the hypotenuse? Solution By Pythagoras theorem, we know that, hypotenuse2 = base2 + height2 Substituting the values, we get: \\begin{align} \\text{hypoteneuse}^2 &=3^2 +4^2 \\\\ &= 9 + 16\\\\ &= 25\\: \\text{inches} \\\\ \\implies \\text{hypoteneuse} &= 5 \\:\\text{inches} \\end{align} $$\\therefore$$ Hypotenuse = 5 inches Example 4 One of the acute angles of a right-angled triangle is 45° Find the other angle. Identify the type of triangle thus formed. Solution Given, angle1 = 90° and angle2 = 45° We know that the sum of the angles of a triangle adds up to 180° Therefore, \\begin{align} \\text{angle}_3 &=180^\\circ \\!-\\! (90^\\circ \\!+\\! 45^\\circ) \\\\ &= 45^\\circ\\end{align} Since two angles measure the same,", "name it the Bermuda Triangle. The triangle will be connected to a ramp on each side of the triangle, so that students will come down the ramp onto the rails. There they will either succeed or be lost at sea! Marc drew the following picture of the triangle. The triangle has three angles and the boys want to reproduce these angles in their structure. The first angle $B$ is equal to $62^\\circ$, the second angle $C$ is equal to $63^\\circ$. Marc can’t remember the measure of angle $A$. He thinks there is a way to figure this out, but he can’t remember what it is. The sum of the interior angles of a triangle is equal to $180^\\circ$. Marc knows the measure of two of the angles of the triangle. Therefore, he can write an equation to figure out the measure of the third angle. $63 + 62 + x = 180$ The variable $x$ is used to represent the measure of angle $A$. Marc is working to find the measure of angle $A$. $125 + x & = 180 \\\\180 - 125 & = 55$ The measure of Angle $A$ is $55^\\circ$ Vocabulary Here are the vocabulary words that are found in this lesson. Triangle a three sided figure with three angles. The prefix “tri”means three. Acute", "the triangle. A+B+C=180 Next, fill into the equation, the measures of the angles given in the diagram. A+B+C37+28+C==180180 Next, simplify the left side of the equation. 37+28+C65+C==180180 Then, subtract 65° from both sides of the equation to solve for the measure of C\\begin{align*}\\angle C\\end{align*}. 65+C6565+CC===18018065115 The answer is 115. The measure of the obtuse angle is 115°. #### Example 3 Given DEF\\begin{align*}\\triangle DEF\\end{align*}, such that D=42\\begin{align*}\\angle D = 42^{\\circ}\\end{align*} and E=123\\begin{align*}\\angle E = 123^{\\circ}\\end{align*}, what is the measure of F\\begin{align*}\\angle F\\end{align*}? First, draw and label a triangle to model the problem. License: CC BY-NC 3.0 Next, write an equation to represent the sum of the measures of the interior angles of the triangle. D+E+F=180 Next, fill into the equation, the measures of the angles given in the diagram. D+E+F42+123+F==180180 Next, simplify the left side of the equation. 42+123+F165+F==180180 Then, subtract 165° from both sides of the equation to solve for the measure of C\\begin{align*}\\angle C\\end{align*}. 165+F165165+FF===18018016515 The answer is 15. The measure of F=15\\begin{align*}\\angle F = 15^{\\circ}\\end{align*}. ### Follow Up Credit: Elliot Brown Source: https://www.flickr.com/photos/ell-r-brown/3639951854/in/photolist-6xDGSA-5wEtmK-7ztdDP-6jURUe-6jUSdc-8SL3Z6-ofitfs-o2ier1-6jZ6fs-65Uqfv-e5CMj8-6Vz9oS-gkcPn-2EYhU4-9eDpNd-pAePx8-aUaQsM-6jUUTX-cH3rgU-hfrpu7-66cnkD-7ayFFh-oiRKrn-kiwSzY-65vFN2-e5Jqts-6jUTVP-dJk5Yg-eVtVS-9nPvzg-2EuPD-9mjNV1-oXCYu8-9mjQxm-brvLxS-mK7BgB-6MU4ub-pE48tr-eSLqP9-4ndDbn-oiuZEB-iPLspo-8awuz3-pepv4y-a5YV8J-oyabu8-725hxm-6K4h9W-8fFEHN-cvxWUU License: CC BY-NC 3.0 Remember the worker with the roof truss? He needs to figure out the measure of the interior angles in the roof truss. How can he do this? He can use the facts that the triangle is isosceles and the", "as the angles, vary in length. Properties of obtuse scalene triangles are: • They have $$2$$ acute angles as well as $$1$$ obtuse angle. • All its sides and angles are distinct in measurement. • The total of all $$3$$ inner angles equal $$180°$$. ### Triangles – Example 1: Find the measure of the unknown angle in the triangle Solution: The sum of the inner angles of a triangle is $$180°$$ . So, $$90°$$$$+$$$$45^°$$$$=$$$$135^°$$$$→$$$$180^°$$$$-$$ $$135^°$$ $$=$$ $$45^°$$ . The unkown angle is $$45^°$$ ### Triangles – Example 1: Find the measure of the unknown angle in the triangle Solution: The sum of the inner angles of a triangle is $$180°$$ . So, $$120°$$$$+$$$$35^°$$$$=$$$$155^°$$$$→$$$$180^°$$$$-$$ $$155^°$$ $$=$$ $$25^°$$ . The unkown angle is $$25^°$$ ## Exercises for Triangles Find the measure of the unknown angle in each triangle. 1) 2) • $$\\color{blue}{70^°}$$ • $$\\color{blue}{30^°}$$ ### What people say about \"Triangles\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE $5 It was$16.99 now it is \\$11.99", "than 90 degrees. Here is an example of an acute triangle. All three of these angles measure less than 90 degrees. $$33+80+67=180^{\\circ}$$ The sum of the angles is equal to $$180^{\\circ}$$. An obtuse triangle has one angle that is greater than 90 and two angles that are less than 90. $$130+25+25=180^{\\circ}$$ The sum of the angles is equal to $$180^{\\circ}$$. Example $$\\PageIndex{1}$$ He tried to build a caddy like his mother's caddy. Her caddy looked like a right triangle from above but he ended up building one that had the following measures and appearance from above: Solution What is the classification of Michael's triangle? First, list the angle measures. 20, 20, 140 Next, determine if any of the angles are equal to 90 degrees or larger than 90 degrees. Yes, one angle is larger than 90 degrees Then, classify the triangle. Obtuse The answer is an obtuse triangle. Michael created an obtuse triangle instead of a right triangle. Example $$\\PageIndex{2}$$ Identify the type of triangle according to its angles. Solution First, list the angle measures. 10, 75, 95 Next, determine if any of the angles are equal to 90 degrees or larger than 90 degrees. Yes, one angle is larger than 90 degrees Then, classify the triangle. Obtuse The answer is an obtuse triangle. Example $$\\PageIndex{3}$$ Identify the", "# Hunt the angle Geometry Level 4 Let $$I$$ denotes the incenter of a triangle $$ABC$$. Let $$O$$ denotes the circumcenter of the triangle $$ABI$$. Given that $$\\angle AIB =120^\\circ$$, find the measure of $$\\angle BOA$$ in degrees. ×" ]

[ "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "number; 77 = 7 × 11; 77 is a composite number; * Positive integers that are only dividing by themselves and 1 are ...", "# Seven Symbol The symbol 7 by itself represents a quantity of seven. The symbol is also used as a digit which forms part of a number. Format Display Data 7 55", "posted a reply to How many positive three digit integers contain atleast two 7 in the Problem Solving forum “melguy, what you missed doesn''t have much to do with permutations. You did the permutations work that you did JUST RIGHT. However, you left out an entire category of integers. With that in mind, figure out what you other category you have to subtract, and in so doing train yourself to get this ...” November 23, 2016 Marty Murray posted a new topic called Combinations Challenge - At a circular table for eight will in the Problem Solving forum “At a circular table for eight will sit two children, their two parents, and the two parents of each of the children''s parents, eight people in total. The children will sit together, the children''s two parents will sit next to the children, one on either side of the two children, and the parents of ...” November 23, 2016 Marty Murray posted a reply to How many positive three digit integers contain atleast two 7 in the Problem Solving forum “First let''s get the correct answer. There is one positive three digit integer that contains three 7''s, 777. Now for the ones that contain two 7''s. We could start with 77. Digits: 7 7 9 Possible Ones Digits We already used", "# Heaven seven Number Theory Level 5 How many positive divisors of $$20020^{12}$$ have 7 as their units digit? ×", "# 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? ×", "i<7 and 7", "for each of the forms 7_7 and _ _7 there are 9 possibilities, so in total there are $8+9+9=26$ three-digit positive integers with one seven as a digit. Three sevens: There is only one, $777$. So in total there are $225+26+1=252$ three-digit integers with a seven as a digit.", "# Multiples of 5 and 7 Number Theory Level 3 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "# 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 $1 + 1 + 2 + 3 = 7$. ×", "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. ×", "# Seven Symbol The symbol 7 represents a quantity of seven. ## Usage Abstractly, this can be visualized as the quantity:", "# Number Game Probability Level 2 How many (positive) seven digit integers are there that have a digit sum equal to $10$ and that are formed using only the digits $1$, $2$ and $3$? ×", "and 17 * 7. Since, number cannot be 1, so two numbers are 17 and 7. Anshu gupta Feb 22, 2014", "is equal to one, and at X is equal to seven.", "# How many binary sequences of length 7 have at least two 1's? How many binary sequences of length 7 have at least two 1's? Can someone please explain the procedure in detail please. I tried solving it using the \"count what you do not want\" procedure, but I got nowhere. Thank you in advance • What do you get if you count those that have exactly zero or exactly one 1? – fuglede Feb 9 '15 at 17:16 The number of $7$-digit sequences is $2^7=128$ The number of $7$-digit sequences with $0$ occurrences of \"one\" is $\\binom70=1$ The number of $7$-digit sequences with $1$ occurrence of \"one\" is $\\binom71=7$ The number of $7$-digit sequences with $2$ or more occurrences of \"one\" is $128-(1+7)=120$", "7?...", "2 * 2 * 7 and also the two prime determinants are 2, 7", "## FANDOM 9,750 Pages ³7 (read as \"to-the-third-seven\") is a positive integer equal to $$7^{7^7} = 7^{823,543} \\approx 3.759823526783788538922131 \\times 10^{695,974}$$. It has 695,975 digits. The superscript 3 represents tetration, meaning this is 7 tetrated to 3.", "1, 2, 2, 2 Number of integers formed = 7! 3! 4! = 35 Total number of integers formed = 77." ]

[ "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "The total number of combinations are $7735+70+15+6=\\boxed{7826}.$", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "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. ×", "remaining $20$ by hand we can easily find out that exactly $9$ of them match all the criteria, giving us $792+9=\\boxed{801}$ good numbers.", "are $7+42+42+42$ \"good\" combinations, for an answer of $133 + 780 = \\boxed{913}$. -jackshi2006", "$4n=96$ 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", "\\times 9 \\:=\\:{\\bf 81}$ possible two-digit numbers. How many are under 20? . . Well, there is: . $11,\\,12,\\,13,\\,14,\\,15,\\,16,\\,17,\\,18,\\,19$ Get the idea? yes i think so, 9/81 = 1/9 thank you all!", "$m+n=2+89=\\boxed{091}$.", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "Both 8 and 72 have ${2}^{3}$ Both 36 and 72 have ${3}^{2}$ But we only need ${2}^{3}$ and ${3}^{2}$ to include all the numbers. We have step-by-step solutions for your answer!", "# $$(XY+ YX )$$ % 7? If the number 16 is added to its reverse, $$16+ 61= 77$$. The sum is 77 which is divisible by 7. But you will not find many numbers have this property. For example $$15 + 51 = 66$$, which is not divisible by 7. How many two digit-number has this properties? Assume that $$05$$ is not a two-digit number. ×", "select the operations for the $7$ $\\square$s, and $8$ places to place our $BA^{[10]}$ block. Thus, our answer is $2^{18} - 2^9 - 8 \\cdot 2^7 = 2^9 \\times 509$, and the answer is $\\boxed{511}$.", "QUESTION Find two numbers whose sum equals 18 and whose product equals 78 Find two numbers whose sum equals 18 and whose product equals 78 • @ • 4 orders completed Tutor has posted answer for $10.00. See answer's preview$10.00", "left. $(27*8)/{9 \\choose 3}{6 \\choose3}$ simplifies down to our fraction m/n, which is $9/70$. Adding them up gives $9 + 70 = \\boxed{79}$.", "means there are 13 \"a\" s and b cannot be 20) When $21\\leq a \\leq 28$, there are $1+2+\\cdots+8=36$ ways. In all, there are $192+36=\\boxed{228}$ possible sequences. ~bluesoul", "enough to understand? Okay, let say $d=10$. Then the average number is 10(1+1/2+1/3+…+1/10)=29.2897. This means most people complete the collections after the 29th purchasing. So, if you have bought 20 boxes, and you still have 7 collection, that’s fine. Everybody experience the same, you should ‘try’ harder! So, hey, what if the number of collectibles is not 10? Yes, you can compute the expected number by yourself. It’s easy enough to do naive computation by calculator. So, are you hunting for those gifts? You’d better check your lucky number! Have fun with mathematics! Good luck! Reference: Wilf, Herbert S. Generatingfunctionology. Academic Press. 1994", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "= 106$, and $AB = 106 + 87 = \\boxed{193}$." ]

[ "number; 77 = 7 × 11; 77 is a composite number; * Positive integers that are only dividing by themselves and 1 are ...", "for each of the forms 7_7 and _ _7 there are 9 possibilities, so in total there are $8+9+9=26$ three-digit positive integers with one seven as a digit. Three sevens: There is only one, $777$. So in total there are $225+26+1=252$ three-digit integers with a seven as a digit.", "posted a reply to How many positive three digit integers contain atleast two 7 in the Problem Solving forum “melguy, what you missed doesn''t have much to do with permutations. You did the permutations work that you did JUST RIGHT. However, you left out an entire category of integers. With that in mind, figure out what you other category you have to subtract, and in so doing train yourself to get this ...” November 23, 2016 Marty Murray posted a new topic called Combinations Challenge - At a circular table for eight will in the Problem Solving forum “At a circular table for eight will sit two children, their two parents, and the two parents of each of the children''s parents, eight people in total. The children will sit together, the children''s two parents will sit next to the children, one on either side of the two children, and the parents of ...” November 23, 2016 Marty Murray posted a reply to How many positive three digit integers contain atleast two 7 in the Problem Solving forum “First let''s get the correct answer. There is one positive three digit integer that contains three 7''s, 777. Now for the ones that contain two 7''s. We could start with 77. Digits: 7 7 9 Possible Ones Digits We already used", "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 # Base Stuff +1 66 1 +770 How many two digit positive integers in the base-four representation are twice the two digit positive integers in the base-three representation? A) 5 B) 6 C) 7 D) 8 E) 9 The answer is apparently A) 5, but I am not sure how to get to it. Any help is appreciated. Thanks! - PM Jan 26, 2020 edited by PartialMathematician Jan 26, 2020 #1 +24133 +3 How many two digit positive integers in the base-four representation are twice the two digit positive integers in the base-three representation? I assume: $$\\begin{array}{|rl|c|c|c|c|c|c|l|} \\hline & & 10 & 11 & 12 & 20 & 21 & 22 & \\text{base}~3 \\\\ & & =3 & =4 & =5 & =6 & =7 & =8 & \\text{base}~10 \\\\ \\text{base}~4 & \\text{base}~10 & 6 & 8 & 10 & 12 & 14 & 16 & 2x \\\\ \\hline 10 & = 4 & & & & & & & \\\\ 11 & = 5 & & & & & & & \\\\ 12 & = 6 &6=6\\checkmark& & & & & & \\\\ 13 & = 7 & & & & & & & \\\\ 20 & = 8 & &8=8\\checkmark& & & & & \\\\ 21 & = 9 & & & & & &", "# Digit Sum 7 Probability Level 3 How many positive integers less than 1000000 have the sum of their digits equal to 7? Details and assumptions The digit sum of a number is the sum of all its digits. For example the digit sum of 1123 is $1 + 1 + 2 + 3 = 7$. ×", "# Number Game Probability Level 2 How many (positive) seven digit integers are there that have a digit sum equal to $10$ and that are formed using only the digits $1$, $2$ and $3$? ×", "# A Pre-RMO question! -2 How many two-digit positive integers are there having even numbers of divisors? Note: numbers from $0$ to $9$ are not considered to be two digit numbers ×", "# Only Two is Allowed Discrete Mathematics Level 1 How many 2 digit positive integers are there? Details and assumptions The number $$12 = 012$$ is a 2 digit number, not a 3 digit number. ×", "# Number Game Discrete Mathematics Level 2 How many (positive) seven digit integers are there that have a digit sum equal to $$10$$ and that are formed using only the digits $$1$$, $$2$$ and $$3$$? × Problem Loading... Note Loading... Set Loading...", "+0 # Math Help 0 82 1 +169 How many positive multiples of 7 that are less than 1000 end with the digit 3? matth99 Aug 16, 2018 #1 +20153 +1 How many positive multiples of 7 that are less than 1000 end with the digit 3? $$\\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}$$ 14 positive multiples of 7 that are less than 1000 end with the digit 3. heureka Aug 16, 2018", "multiples of 8 because 8 x 2 = 16, 8 x -3 = -48 and 8 x 9 = 72. Similarly, the first five positive multiples of 7 are the following: 7, 14, 21, 28, 35. In this post, we will particularly talk about positive integers and positive multiples. This is in preparation for the discussions on addition and subtraction of fractions. We can always find a common multiple given two or more numbers. For example, if we list all the positive multiples of 2 and 3, we have 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 and 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. » Read more", "# Multiples of 7 and 13 How many 3 digit positive integers $$N$$ are there, such that $$N$$ is a multiple of both 7 and 13? ×", "# Only Two is Allowed How many 2 digit positive integers are there? Details and assumptions The number $$12 = 012$$ is a 2 digit number, not a 3 digit number. ×", "+0 # How many 11 -digit positive integers can be written using only the digits 1,2 and 3 and have exactly five 1’s and exactly two 2's as digits? 0 59 1 How many 11 -digit positive integers can be written using only the digits 1,2 and 3 and have exactly five 1’s and exactly two 2's as digits? Aug 15, 2022 #1 +2453 0 We know there must be five 1's, two 2's, and four 3's. There are $${11! \\over 2! \\times 5! \\times 4!} = \\color{brown}\\boxed{6930}$$ numbers that work. Aug 16, 2022", "Twisted probability I have a 7-digit positive integer for which its sum of digits is 59. If the chance that the number is divisible by 11 can be expressed in the form of $$\\frac ab$$ where $$a$$ and $$b$$ are coprime positive integers, find the value of the sum of digits of $$a + b$$. ×", "positive integer picked at rand [#permalink] ### Show Tags 07 Jun 2012, 21:39 some one please point the flaw in my logic want to do it the longer way for more understanding . at least one 7 means , one , two or three 7's lets take scenarios, first digit 7 , two digits 7 , all three digits 7 1) first digit to be seven , non seven , non seven (1/9 ) * 9/10 * 9/10 = 81/900=9/100 first digit cannot be zero , so there are 9 total possibilities for the first , from 1 ,2 , 3, 4, 5, 6, 7, 8, 9. For the second and third we have 10 possibilities now we can have zero , from which favorable are nine, , as we are are excluding 7 ( case 2) 7 , 7 , non seven 1/9 * 1/10 *9/10 * 3 = 3/100 here we are multiplying by 3 , because the 2 seven's can occur in 3 ways ( case 3) 7, 7 , 7 1/9 * 1/10 * 1/10 = 1/ 900 so at least one 7 9/100 + 3/100 + 1/900 = 109/900 = wrong answer !! what am I missing guys ? I know the alternate way , 1- p ( all not seven ) ,", "construct another integer with that many digits that includes the original integer itself. Therefore, there are at least as many such integers as there are all integers altogether. There cannot be any more than that, because the resulting integers are, well, integers. • Note that this answer was written before the question was edited to specify integers up to 10^9. Jun 21 '20 at 22:08 • This one gets an upvote from me as well, because based on the original parameters of the question it is most correct! Jun 22 '20 at 3:04 If you pad the numbers with leading zeroes so that they always have 9 digits then it becomes easier to count how many there are because you then don't have to split it into cases. There are $$10^9$$ strings of 9 digits. Those that start with one or more zeroes represent numbers with fewer digits. There are $$9^9$$ strings of consisting of the digits $$0$$ to $$8$$. Any such string uniquely corresponds to a number not containing its length: If its length is $$n$$, just increment every digit that is $$n$$ or greater (the length is not affected by incrementing non-zero digits so this process is reversible). Therefore there are $$10^9-9^9$$ numbers containing its own length as a digit at least once. • +1 for", "1, 2, 2, 2 Number of integers formed = 7! 3! 4! = 35 Total number of integers formed = 77.", "# If an integer is randomly chosen among the first 50 positive integers, what is the probability that the chosen integer will be a 2-digit number? I thought that this question would simply be that there are 40 2-digit integers so you have a $$\\frac{40}{50}$$ chance of getting a 2-digit integer via relative frequency. However the answer says that it is 0.82, and it doesn’t show any work. Am I thinking about this problem in the wrong way? • That's because there are forty-one positive integers between $1$ and $50$ inclusive. Sep 29, 2019 at 18:22 • There are only $9$ positive one-digit integers Sep 29, 2019 at 18:24 • I think you meant, Lord Shark, that there are forty-one two-digit positive integers between 1 and 5 inclusive. Sep 29, 2019 at 18:28 The two-digit numbers between $$1$$ and $$50$$ are from $$10$$ to $$50$$ (both included). This amounts to $$41$$ numbers to choose from. Hence probability will be $${41\\over 50}$$." ]

[ "of the south pole, traveling 6 miles east would take you in a circle around the south pole 1.27 times. But if you started on the equator, traveling 6 miles east would simply be 6 miles east. Bystander and haruspex", "6 miles south, 5 miles southwest, and 3 miles east. How far have they walked? If they walked straight home, how far would they have to walk? 1. Three forces act on an object: . Find the net force acting on the object. 1. Three forces act on an object: . Find the net force acting on the object. 1. A person starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If they walked straight home, how far would they have to walk, and in what direction? 1. A person starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If they walked straight home, how far would they have to walk, and in what direction? 1. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 1. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing", "he walked straight home, how far would he have to walk? 17 miles. 10.318 miles A woman starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk? A man starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 4.924°. 659 km/hr An airplane is heading north at an airspeed", "the direct distance from the origin to the destination.", "on: November 16, 2013, 03:18:11 PM » The old joke is that, in Europe, a hundred miles is a long way while, in North America, a hundred years is a long time.", "have traveled 200 miles.", "is five miles north of the second boat when the second boat crosses the path of the first boat. The two boats are traveling at the same speed. Therefore, when the second boat has traveled ten miles east of the crossing point, the first boat has traveled an additional ten miles north of the crossing point. Thus, the first boat is $$5~\\text{mi} + 10~\\text{mi} = 15~\\text{mi}$$ north of the crossing point when the second boat is ten miles east of the crossing point.", "10° west of south, then 4 miles at 15° north of east. If they walked straight home, how far would they have to walk, and in what direction? 30. A person starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If they walked straight home, how far would they have to walk, and in what direction? 31. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 32. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 33. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, the pilot will need to fly the plane how many degrees west of north? 34. An airplane", "5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? 66. A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? 67. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 68. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 69. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the", "so has challenges and requirements, which I hope I have correctly identified and reasonably communicated. I don’t expect any, much less all, of this to happen in 2018, but you know what they say about journeys of a thousand miles (or in my case, 300 kilometers): They begin by looking at a map, making a plan, and facing in a useful direction. Then taking a step.", "to walk? 69. A man starts walking from home and walks 3 miles at $20^{\\circ}$ north of west, then 5 miles at $10^{\\circ}$ west of south, then 4 miles at $15^{\\circ}$ north of east. If he walked straight home, how far would he have to the walk, and in what direction? 70. A woman starts walking from home and walks 6 miles at $40^{\\circ}$ north of east, then 2 miles at $15^{\\circ}$ east of south, then 5 miles at $30^{\\circ}$ south of west. If she walked straight home, how far would she have to walk, and in what direction? 71. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 72. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 73. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr.", "the path on google maps... estimated around 13,700 miles...... [ continued ] Last edited by xcthulhu on May 23rd, 2009, 1:50 pm, edited 4 times in total.", "due west. On a direct route, the barge’s Due to weather, a barge captain decides to reach her destination in two legs: one due north and one due west. On a direct route, the barge’s destination is about 1,152 miles; see the figure below. If after traveling 605 miles due north the captain determines it is time to head due west, how many m... ### PLeAsE HeLp mE What do the passages below show is of utmost importance to Gwendolen and Cecily when PLeAsE HeLp mE What do the passages below show is of utmost importance to Gwendolen and Cecily when it comes to their future husband? What does this demonstrate about their characters? [tex]PLeAsE HeLp mE What do the passages below show is of utmost importance to Gwendolen and Cecily whe[...", "to move to reach the destination. c) What is the bearing of the return trip? ( Answers : (a) 22.5 hours (b) Towards East = 449.57 miles Towards South = 19.63 miles (c) N 87.5º W)", "traveled 75 or more miles? (A) 4 (B) 7 (C) 3 (D) 9", "5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk? A man starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 4.924°. 659 km/hr An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed", "we are after traveling a distance of 15.", "+0 # Help! 0 74 5 While walking on a plane surface, a traveler first headed 19 miles north, then 16 miles west, then 3 miles south and finally 4 miles east. How many miles from the starting point was the traveler after these four legs of the journey? Guest Nov 14, 2017 Sort: #1 +23 0 Are you looking for distance or displacement? MicahSmith07 Nov 14, 2017 #3 0 thats theproblem i dont know. That is the whole problem. Guest Nov 14, 2017 #5 +507 +2 I assumed, based off of the wording, that this was based on displacement. The question is, \"How many miles from the starting point was the traveler after completing these 4 legs of the journey\" helperid1839321 Nov 14, 2017 #2 +507 +2 I'm assuming this is absolute distance, not distance traveled on foot. 19 - 3 = 16 16 - 4 = 12 $$\\sqrt{16^{2} + 12^{2}} = \\sqrt{256 + 144} = \\sqrt{400} =20$$ He moved an overall of 20 miles from the starting point. helperid1839321 Nov 14, 2017 #4 0 thx that was correct i was wondering why i got it wrong. Guest Nov 14, 2017 ### 13 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information", "starts walking from home and walks $$4$$ miles east, $$7$$ miles southeast, $$6$$ miles south, $$5$$ miles southwest, and $$3$$ miles east. How far has she walked? If she walked straight home, how far would she have to walk? 65) A man starts walking from home and walks $$3$$ miles at $$20^{\\circ}$$ north of west, then $$5$$ miles at $$10^{\\circ}$$ west of south, then $$4$$ miles at $$15^{\\circ}$$ north of east. If he walked straight home, how far would he have to the walk, and in what direction? Distance: $$2.868$$, Direction: $$86.474^{\\circ}$$ North of West, or $$3.526^{\\circ}$$ West of North 66) A woman starts walking from home and walks $$6$$ miles at $$40^{\\circ}$$ north of east, then $$2$$ miles at $$15^{\\circ}$$ east of south, then $$5$$ miles at $$30^{\\circ}$$ south of west. If she walked straight home, how far would she have to walk, and in what direction? 67) An airplane is heading north at an airspeed of $$600$$ km/hr, but there is a wind blowing from the southwest at $$80$$ km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? $$4.924^{\\circ}$$, 659 km/hr 68) An airplane is heading north at an airspeed of $$500$$ km/hr, but there is a wind blowing from the northwest at", "send a message 4000 miles (6400 km), all the way across the Atlantic Ocean to Germany. Wow. Needless to say, most inductors are much smaller. Photo credits" ]

[ "of the south pole, traveling 6 miles east would take you in a circle around the south pole 1.27 times. But if you started on the equator, traveling 6 miles east would simply be 6 miles east. Bystander and haruspex", "north to reach B. • The total distance moved is 3 miles +4 miles = 7 miles • Here OB is the magnitude of net displacement. So from Pythagoras theorem, in a right-angled triangle OAB$OB^2=OA^2+AB^2$$=^2+^2$$=25\\ mile^2$or,$OB=5\\ mile$", "we are after traveling a distance of 15.", "to move to reach the destination. c) What is the bearing of the return trip? ( Answers : (a) 22.5 hours (b) Towards East = 449.57 miles Towards South = 19.63 miles (c) N 87.5º W)", "the path on google maps... estimated around 13,700 miles...... [ continued ] Last edited by xcthulhu on May 23rd, 2009, 1:50 pm, edited 4 times in total.", "A car goes 24 miles due north then 7 miles due east. What is the straight distance between the car's starting point and end point? 1.1.10 One end of a rope is attached to the top of a pole 100 ft high. If the rope is 150 ft long, what is the maximum distance along the ground from the base of the pole to where the other end can be attached? You may assume that the pole is perpendicular to the ground. 1.1.11 Prove that the hypotenuse is the longest side in every right triangle. (Hint: Is $$a^2 + b^2 > a^2$$?) 1.1.12 Can a right triangle have sides with lengths 2, 5, and 6? Explain your answer. 1.1.13 If the lengths $$a$$, $$b$$, and $$c$$ of the sides of a right triangle are positive integers, with $$a^2 + b^2 = c^2$$, then they form what is called a Pythagorean triple. The triple is normally written as ($$a$$,$$b$$,$$c$$). For example, (3,4,5) and (5,12,13) are well-known Pythagorean triples. (a) Show that (6,8,10) is a Pythagorean triple. (b) Show that if ($$a$$,$$b$$,$$c$$) is a Pythagorean triple then so is ($$ka$$,$$kb$$,$$kc$$) for any integer $$k >0$$. How would you interpret this geometrically? (c) Show that ($$2mn$$,$$m^2 - n^2$$,$$m^2 + n^2$$) is a Pythagorean triple for all integers $$m > n >", "the proof haha. - 3 years ago @Prince Loomba Is the answer to problem 60 Infinite? I think any point from which u could go one mile south take a circle round the earth and come back u would come at same point The north pole is one such place. However consider the points LaTeX: $1+ \\frac{1}{2\\pi}$ away from the South Pole. Let any of those points be A and then go a mile south to a point B. When you go a mile east, you end up back at point B (you travelled once through every line of longitude). A mile north then brings you back to point A. There are points still closer to the south pole such that going a mile east brings you through each line of longitude exactly twice, three times, or as many times as you want. Thus we have an LaTeX: $\\text{infinite}$ number of concentric rings of LaTeX: $infinite$ numbers of points, and we can start a mile north of any of them. - 3 years ago This is right - 3 years ago thanks - 3 years ago Why are you thanking? - 3 years ago because i told him the answer - 3 years ago This is my problem almost exactly which I posted only 3 days previously!!! Here", "## Prealgebra (7th Edition) The map on p. 35 is a little hard to read, but the exact distances are $11+16+19+31=78$. Round to the nearest 10: $10+20+20+30=80$.", "doubt came across the Pythagorean Theorem: $A^2+B^2=C^2$ It’s a deceptively simple equation. A and B are the lengths of the two short sides of a right triangle, and C is the length of the long side, the hypotenuse. The Pythagorean Theorem does far more than simply tell you about triangles for their own sake. It tells you how to compute the distances between points. You may remember problems like this from school: Walk 3 miles East and then 4 miles North. Plug through the calculation and you’ll find that you’re 5 miles from where you started. To connect it to the real world, consider a small piece of a Washington D.C. transit map: Many cities are conveniently designed so that streets run approximately along the cardinal directions of a compass. Washington D.C. is a perfect example: Numbered streets run North-South and lettered streets run East-West. So – to pick an example found by an intensive search through Google Maps – if you wanted to walk from Judiciary Square Station on the corner of 4th and E Street NW to the Chinatown Metro Station on 7th and G, you would start by walking approximately 600 meters due West (along E), and then 250 meters North (along 7th). Of course, you could also simply take the Red Line subway, and", "two decimal places, the distance traveled by the ship is 12.21km. Pythagoras’ Theorem in 3D Pythagoras’ Theorem can also be extended for use in 3 dimensional problems. Example – Consider then cuboid with sides of length 12km, 7km and 5km. Let’s say we have a plane that travels from point A to point B. That is, it travels 12km forwards, 7km left and 5km upwards. What is the total distance traveled? The three dimensional version of Pythagoras’ Theorem works in just the same way. To find the distance traveled we take the square the 3 distances, and then take the square root of the sum. So in our example: $\\sqrt{12^2 + 7^2 + 5^2} = \\sqrt{218} = 14.76$ to 2 decimal places. So the distance traveled by our plane is approximately 14.76km.", "traveled 75 or more miles? (A) 4 (B) 7 (C) 3 (D) 9", "Perfect. Going right is going towards the East. Going left is going towards the West. At the end of the ride, what is your final displacement? That is, how far are you from your starting point? Also, what is the total distance you have travelled? • Final displacement $$= \\, 3 \\, \\mathrm{miles}$$; Total distance $$= \\, 8 \\, \\mathrm{miles}$$ • Final displacement $$= \\, 8 \\, \\mathrm{miles}$$; Total distance $$= \\, 8 \\, \\mathrm{miles}$$ • Final displacement $$= \\, 2 \\, \\mathrm{miles}$$; Total distance $$= \\, 8 \\, \\mathrm{miles}$$ • Final displacement $$= \\, 2 \\, \\mathrm{miles}$$; Total distance $$= \\, 2 \\, \\mathrm{miles}$$", "7 miles north from $S$, it doesn't matter how much you travel west, you're still going to get back to $S$ once you go 7 miles south again. The less obvious solutions are those points that are further than 7 miles from the north pole, and are such that after traveling 7 miles north, you lie on a latitude such that the circumference of that latitude is equal to $\\frac{7}{n}$ miles, for a natural number $n$. If you travel 7 miles north (again, assuming that you can at all), and reach a point such that traveling 7 miles west has no ultimate effect, you will get back to where you started after going 7 miles south again. As Mariano points out below, it would not even mean anything to \"travel 7 miles north\" if you are less than 7 miles from the north pole, so we must exclude these points explicitly to make sure the condition is specified for all points. In other words, if $A_n$ is the latitude with circumference $\\frac{7}{n}$, and $B_n$ is the set of points which, after traveling 7 miles north, you would reach $A_n$, then the points on $B_n$ are solutions. $$P=\\{S\\}\\cup\\bigcup_{n\\in\\mathbb{N}}B_n$$ This is not a closed set; to produce the closure, you have to add the points that are exactly 7 miles", "## College Physics (7th Edition) $a). less\\, than \\,13\\, miles.$ $b). 6.058\\, miles$ a). The distance travelled due North is less than 13 miles. Because the angle created is $25^{\\circ}$. Had it been more than $45^{\\circ}$ then it would have been more than 13 miles. And at 45 degrees it would have been equal to 13 miles. b).If d is the distance travelled due N, then $\\frac{d}{13}= tan25^{\\circ}=0.466$ $d=13\\times 0.466=6.058\\,miles.$", "have traveled 200 miles.", "= 45. The total distance traveled is 55+45 = 100.", "many different kinds of real-life problems that can be solved using the Pythagorean theorem. The easiest way to see that you should be applying this theorem is by drawing a picture of whatever situation is described. Example Two hikers leave a cabin at the same time, one heading due south and the other headed due west. After one hour, the hiker walking south has covered 2.8 miles and the hiker walking west has covered 3.1 miles. At that moment, what is the shortest distance between the two hikers? Solution First, sketch a picture of the information given. Label any unknown value with a variable name, like x. Due south and due west form a right angle, and the shortest distance between any two points is a straight line. Therefore, we can apply the Pythagorean theorem and write: $$3.1^2 + 2.8^2 = x^2$$ Here, you will need to use a calculator to simplify the left-hand side: $$17.45 = x^2$$ Now use your calculator to take the square root. You will likely need to round your answer. \\begin{align}x &= \\sqrt{17.45} \\\\ &\\approx 4.18 \\text{ miles}\\end{align} As you can see, it will be up to you to determine that a right angle is part of the situation given in the word problem. If it isn’t, then you can’t use the Pythagorean", "and expenses ( latitude / longitude ), miles or.... Of travel routes after Du Niang, i found that the mathematical knowledge was returned to the and... On dezembro 9th, 2020, posted in: Uncategorized by", "x, 17, 3x - y2 - 2 and 3x + y2 - 30 are four consecutive terms of an increasing arithmetic sequence. The sum of the four numbers is divisible by: 1. 2 2. 3 3. 5 4. 7 5. 11 10. What is the sum of the following series? -64, -66, -68, ..... , -100 1. -1458 2. -1558 3. -1568 4. -1664 5. None of the above 11. If log 2, log (2x -1) and log (2x + 3) are in A.P, then x is equal to ____ 1. $\\frac {5}{2} \\\\$ 2. log25 3. log32 4. $\\frac {3}{2} \\\\$ 12. Two travelers set out on a long odyssey. The first traveler starts from city X and travels north on a certain day and covers 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3 days, a second traveler sets out from city X in the same direction as the first traveler and on his first day he travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the second traveler be ahead of the first? 1. 2 days 2. 6 days 3. From the 2nd day after the 2nd traveler starts 4. From", "triangle, so is equal to the positive square root of the sum of the squares of the lengths of the other two sides. ${8}^{2} + {6}^{2} = 64 + 36 = 100 = {10}^{2}$ So the distance between the start and finish points is $10$ miles." ]

[ "of the south pole, traveling 6 miles east would take you in a circle around the south pole 1.27 times. But if you started on the equator, traveling 6 miles east would simply be 6 miles east. Bystander and haruspex", "he walked straight home, how far would he have to walk? 17 miles. 10.318 miles A woman starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk? A man starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 4.924°. 659 km/hr An airplane is heading north at an airspeed", "6 miles south, 5 miles southwest, and 3 miles east. How far have they walked? If they walked straight home, how far would they have to walk? 1. Three forces act on an object: . Find the net force acting on the object. 1. Three forces act on an object: . Find the net force acting on the object. 1. A person starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If they walked straight home, how far would they have to walk, and in what direction? 1. A person starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If they walked straight home, how far would they have to walk, and in what direction? 1. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 1. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing", "the direct distance from the origin to the destination.", "is five miles north of the second boat when the second boat crosses the path of the first boat. The two boats are traveling at the same speed. Therefore, when the second boat has traveled ten miles east of the crossing point, the first boat has traveled an additional ten miles north of the crossing point. Thus, the first boat is $$5~\\text{mi} + 10~\\text{mi} = 15~\\text{mi}$$ north of the crossing point when the second boat is ten miles east of the crossing point.", "to walk? 69. A man starts walking from home and walks 3 miles at $20^{\\circ}$ north of west, then 5 miles at $10^{\\circ}$ west of south, then 4 miles at $15^{\\circ}$ north of east. If he walked straight home, how far would he have to the walk, and in what direction? 70. A woman starts walking from home and walks 6 miles at $40^{\\circ}$ north of east, then 2 miles at $15^{\\circ}$ east of south, then 5 miles at $30^{\\circ}$ south of west. If she walked straight home, how far would she have to walk, and in what direction? 71. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 72. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 73. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr.", "starts walking from home and walks $$4$$ miles east, $$7$$ miles southeast, $$6$$ miles south, $$5$$ miles southwest, and $$3$$ miles east. How far has she walked? If she walked straight home, how far would she have to walk? 65) A man starts walking from home and walks $$3$$ miles at $$20^{\\circ}$$ north of west, then $$5$$ miles at $$10^{\\circ}$$ west of south, then $$4$$ miles at $$15^{\\circ}$$ north of east. If he walked straight home, how far would he have to the walk, and in what direction? Distance: $$2.868$$, Direction: $$86.474^{\\circ}$$ North of West, or $$3.526^{\\circ}$$ West of North 66) A woman starts walking from home and walks $$6$$ miles at $$40^{\\circ}$$ north of east, then $$2$$ miles at $$15^{\\circ}$$ east of south, then $$5$$ miles at $$30^{\\circ}$$ south of west. If she walked straight home, how far would she have to walk, and in what direction? 67) An airplane is heading north at an airspeed of $$600$$ km/hr, but there is a wind blowing from the southwest at $$80$$ km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? $$4.924^{\\circ}$$, 659 km/hr 68) An airplane is heading north at an airspeed of $$500$$ km/hr, but there is a wind blowing from the northwest at", "+0 # Help! 0 74 5 While walking on a plane surface, a traveler first headed 19 miles north, then 16 miles west, then 3 miles south and finally 4 miles east. How many miles from the starting point was the traveler after these four legs of the journey? Guest Nov 14, 2017 Sort: #1 +23 0 Are you looking for distance or displacement? MicahSmith07 Nov 14, 2017 #3 0 thats theproblem i dont know. That is the whole problem. Guest Nov 14, 2017 #5 +507 +2 I assumed, based off of the wording, that this was based on displacement. The question is, \"How many miles from the starting point was the traveler after completing these 4 legs of the journey\" helperid1839321 Nov 14, 2017 #2 +507 +2 I'm assuming this is absolute distance, not distance traveled on foot. 19 - 3 = 16 16 - 4 = 12 $$\\sqrt{16^{2} + 12^{2}} = \\sqrt{256 + 144} = \\sqrt{400} =20$$ He moved an overall of 20 miles from the starting point. helperid1839321 Nov 14, 2017 #4 0 thx that was correct i was wondering why i got it wrong. Guest Nov 14, 2017 ### 13 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information", "where you can walk in a straight line and in an east-westerly direction at the same time and that is... ...on the equator. From this we can conclude that each leg of a returning astronaut's journey must be... ...quarters of the circumference, so that they always end up on a pole or the equator at the end of a leg. At the same time, any one such leg of the journey must not... ...end up on the opposing pole from where they started. Because that means they will walk to the opposing pole, back to the starting pole, and then to the opposing pole again when they walk the third leg. So the astronaut with a leg distance that is 2/4 (or 6/4, or 10/4 and so on) of the circumference is the lone astronaut. Circling the asteroid is allowed because that results in... the same distance being traveled from the pole as if they had not circled it. Walking 5/4 of a lap is the same as walking 1/4. This then means that the following distances — measured in quarters of the circumference — are \"allowed\" in order to return to the starting point: 1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15... ...or more clearly written... 0, 1, x, 3, 4, 5, x, 7,", "due west. On a direct route, the barge’s Due to weather, a barge captain decides to reach her destination in two legs: one due north and one due west. On a direct route, the barge’s destination is about 1,152 miles; see the figure below. If after traveling 605 miles due north the captain determines it is time to head due west, how many m... ### PLeAsE HeLp mE What do the passages below show is of utmost importance to Gwendolen and Cecily when PLeAsE HeLp mE What do the passages below show is of utmost importance to Gwendolen and Cecily when it comes to their future husband? What does this demonstrate about their characters? [tex]PLeAsE HeLp mE What do the passages below show is of utmost importance to Gwendolen and Cecily whe[...", "two decimal places, the distance traveled by the ship is 12.21km. Pythagoras’ Theorem in 3D Pythagoras’ Theorem can also be extended for use in 3 dimensional problems. Example – Consider then cuboid with sides of length 12km, 7km and 5km. Let’s say we have a plane that travels from point A to point B. That is, it travels 12km forwards, 7km left and 5km upwards. What is the total distance traveled? The three dimensional version of Pythagoras’ Theorem works in just the same way. To find the distance traveled we take the square the 3 distances, and then take the square root of the sum. So in our example: $\\sqrt{12^2 + 7^2 + 5^2} = \\sqrt{218} = 14.76$ to 2 decimal places. So the distance traveled by our plane is approximately 14.76km.", "we are after traveling a distance of 15.", "the path on google maps... estimated around 13,700 miles...... [ continued ] Last edited by xcthulhu on May 23rd, 2009, 1:50 pm, edited 4 times in total.", "to move to reach the destination. c) What is the bearing of the return trip? ( Answers : (a) 22.5 hours (b) Towards East = 449.57 miles Towards South = 19.63 miles (c) N 87.5º W)", "and turn right 90 degrees, and walk N paces. On any surface, if N is small enough you will come back to your starting point. As N increases, does this remain true? If the space is flat, it will always be true, no matter how long your walks, no matter where you start and no matter which direction you go — as long as your walk doesn’t collide with the edge of the space. But if, as N increases, you find your end point is further and further from your starting point, that tells you the space is not flat. Of course the Earth, or any real-world surface isn’t expected to be exactly a sphere or exactly flat; it has wiggles on it, in the form of hills and valleys. So we have to allow for these statements to not be exactly correct. However, the largest wiggles on Earth are only about 10 miles deep or tall, while the planet is 24000 miles around, so we should have no trouble distinguishing a nearly-flat Earth from a nearly-spherical Earth. Imagine you start at the north pole, walk 6225 miles (about 10000 km) in any direction, turn right 90 degrees, walk 6225 miles, turn right 90 degrees, walk 6225 miles, turn right 90 degrees, and walk 6225 miles again. If", "traveled 75 or more miles? (A) 4 (B) 7 (C) 3 (D) 9", "so has challenges and requirements, which I hope I have correctly identified and reasonably communicated. I don’t expect any, much less all, of this to happen in 2018, but you know what they say about journeys of a thousand miles (or in my case, 300 kilometers): They begin by looking at a map, making a plan, and facing in a useful direction. Then taking a step.", "have traveled 200 miles.", "10° west of south, then 4 miles at 15° north of east. If they walked straight home, how far would they have to walk, and in what direction? 30. A person starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If they walked straight home, how far would they have to walk, and in what direction? 31. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 32. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 33. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, the pilot will need to fly the plane how many degrees west of north? 34. An airplane", "5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? 66. A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? 67. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 68. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 69. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the" ]

[ "# Difference between revisions of \"2020 AMC 8 Problems/Problem 11\" ## Problem 11 After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds? $[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * \"Distance (miles)\", (-0.5,3), W); label(\"Time (minutes)\", (3,-0.5), S); dot(\"Naomi\", (2,6), 3*dir(305)); dot((6,6)); label(\"Maya\", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]$ $\\textbf{(A) }6 \\qquad \\textbf{(B) }12 \\qquad \\textbf{(C) }18 \\qquad \\textbf{(D) }20 \\qquad \\textbf{(E) }24$ ## Solution 1 We use the formula $\\text{speed}=\\dfrac{\\text{distance}}{\\text{time}}$. Naomi's distance is $6$ miles, and her time is $10$ minutes, which is equivalent to $\\dfrac{1}{6}$ of an hour. Since speed is distance over time, Naomi's speed is $36$ mph. Using the same process, Maya's speed is $12$ mph. Subtracting those, we get an answer of $\\boxed{(\\text{E}) 24}$. ## Solution 2 Notice that Naomi travels at a rate of $6$ miles", "# Difference between revisions of \"2011 AMC 8 Problems/Problem 9\" Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour? $[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; draw((0,0)--(0,8)); draw((0,0)--(8,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label(\"1\",(0.95,-0.24),SE*lsf); label(\"2\",(1.92,-0.26),SE*lsf); label(\"3\",(2.92,-0.31),SE*lsf); label(\"4\",(3.93,-0.26),SE*lsf); label(\"5\",(4.92,-0.27),SE*lsf); label(\"6\",(5.95,-0.29),SE*lsf); label(\"7\",(6.94,-0.27),SE*lsf); label(\"5\",(-0.49,1.22),SE*lsf); label(\"10\",(-0.59,2.23),SE*lsf); label(\"15\",(-0.61,3.22),SE*lsf); label(\"20\",(-0.61,4.23),SE*lsf); label(\"25\",(-0.59,5.22),SE*lsf); label(\"30\",(-0.59,6.2),SE*lsf); label(\"35\",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6)); draw((4,4)--(7,7),linewidth(1.6)); label(\"HOURS\",(3.41,-0.85),SE*lsf); label(\"M\",(-1.39,5.32),SE*lsf); label(\"I\",(-1.34,4.93),SE*lsf); label(\"L\",(-1.36,4.51),SE*lsf); label(\"E\",(-1.37,4.11),SE*lsf); label(\"S\",(-1.39,3.7),SE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ ## Solution We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is $\\boxed{\\textbf{(E) 5}}$ 2011 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 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 AJHSME/AMC 8 Problems and", "# Difference between revisions of \"1994 AJHSME Problems/Problem 18\" ## Problem Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike's trip? $[asy] import graph; unitsize(12); real a(real x) {return ((x-15)^2)/2;} real b(real x) {return ((x-25)^2)/2;} real c(real x) {return ((x-30)^2 * (x-40)^2) * 8/625;} real d(real x) {return ((x-15)^2)*8/25-15;} real e(real x) {return ((x-25)^2)*8/25-15;} draw((0,9)--(0,0)--(11,0)); draw((15,9)--(15,0)--(26,0)); draw((30,9)--(30,0)--(41,0)); draw((0,-6)--(0,-15)--(11,-15)); draw((15,-6)--(15,-15)--(26,-15)); draw((0,0)--(3,8)--(7,8)--(10,0)); draw(graph(a,15,17)); draw((17,2)--(18,8)--(22,8)--(23,2)); draw(graph(b,23,25)); draw(graph(c,30,40)); draw((0,-15)--(5,-7)--(10,-15)); draw(graph(d,15,20)); draw(graph(e,20,25)); for (int k=0; k<3; ++k) { label(\"d\",(15*k-1,8),N); label(\"i\",(15*k-1,7),N); label(\"s\",(15*k-1,6),N); label(\"t\",(15*k-1,5),N); label(\"a\",(15*k-1,4),N); label(\"n\",(15*k-1,3),N); label(\"c\",(15*k-1,2),N); label(\"e\",(15*k-1,1),N); label(\"time\",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label(\"d\",(15*k-1,8-15),N); label(\"i\",(15*k-1,7-15),N); label(\"s\",(15*k-1,6-15),N); label(\"t\",(15*k-1,5-15),N); label(\"a\",(15*k-1,4-15),N); label(\"n\",(15*k-1,3-15),N); label(\"c\",(15*k-1,2-15),N); label(\"e\",(15*k-1,1-15),N); label(\"time\",(15*k+8,0-15),S); } label(\"(A)\",(5,9),N); label(\"(B)\",(20,9),N); label(\"(C)\",(35,9),N); label(\"(D)\",(5,-6),N); label(\"(E)\",(20,-6),N); [/asy]$ ## Solution When Mike is shopping at the mall, his distance doesn't change, so there should be a flat plateau shape on the graph. This rules out $C,D,E$. The portion of graph $A$", "matrix Optimal The total distance that k trucks will travel is 300.0 miles. [1.0 1.0 1.0; 12.0 3.0 12.0; 5.0 2.0 11.0; 12.0 12.0 12.0; 3.0 1.0 1.0; 12.0 8.0 12.0; 12.0 11.0 12.0; 4.0 2.0 2.0; 12.0 9.0 12.0; 1.0 1.0 1.0; 2.0 1.0 1.0; 12.0 10.0 12.0] The distance that truck 1 travels is 141.0 miles. The distance that truck 2 travels is 147.0 miles. The distance that truck 3 travels is 12.0 miles. Out[5]: 12×12×3 Array{Float64,3}: [:, :, 1] = -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 1.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 1.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 1.0 0.0 -0.0 0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 -0.0 -0.0 1.0 -0.0 0.0 0.0 0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 1.0 0.0 -0.0 0.0 -0.0 0.0 1.0 0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0", "0.0 0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 0.0 0.0 -0.0 # 4. Result and Discussion¶ ## Scenario 1¶ From running the code in previous session, we generate a matrix for each truck that describes the order of the logging sites that it drives through. The graph below is generated by Matlab. Each color represents the route of a truck. In this case, the red path represents the route of truck 1; the blue path represents the route of truck 2; the green path represents the route of truck 3. In this session, we have: The distance for truck 1 is 12 miles. The distance for truck 2 is 134 miles. The distance for truck 3 is 131 miles. ## Scenario 2¶ From running the code in previous session, we generate a matrix for each truck that describes the order of the", "i in 1:N, j in 1:N ),\" miles.\") end println(matrix) Optimal The total distance that k trucks will travel is 277.0 miles. [1.0 1.0 1.0; 1.0 1.0 1.0; 1.0 8.0 1.0; 1.0 1.0 1.0; 12.0 9.0 1.0; 1.0 12.0 12.0; 1.0 10.0 12.0; 1.0 11.0 12.0; 1.0 12.0 5.0; 12.0 12.0 2.0; 1.0 12.0 4.0; 1.0 12.0 3.0] The distance that truck 1 travels is 12.0 miles. The distance that truck 2 travels is 134.0 miles. The distance that truck 3 travels is 131.0 miles. [-0.0 1.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; 1.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0; 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0; -0.0", "## On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and from ##### This topic has expert replies Legendary Member Posts: 1622 Joined: 01 Mar 2018 Followed by:2 members ### On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and from by Gmat_mission » Fri Feb 11, 2022 8:23 am 00:00 A B C D E ## Global Stats On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and from $$P$$ to $$Q$$ are relatively flat and straight, approximately how many miles is it from $$P$$ to $$Q ?$$ A. $$3$$ B. $$\\dfrac92$$ C. $$5$$ D. $$6$$ E. $$\\dfrac{15}2$$ Source: GMAT Prep ### GMAT/MBA Expert GMAT Instructor Posts: 16118 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1268 members GMAT Score:770 ### Re: On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and fro by [email protected] » Sat Feb 12, 2022 6:04 am Gmat_mission wrote: Fri Feb 11, 2022 8:23 am 1 (1).png On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from", "60 MPH would be the only label \"1\". So you can drive at 1, faster than 1, or slower than 1. How do you put 0.833333333MPM (50MPH) on a speed-limit sign? Miles-per-minute is, clearly, not the most convenient scale for vehicle speed limits, even though both scales are equally accurate. \"Units per millisecond\" is an equally undesirable scale because humans do not experience life as such. • \"Carpe diem\" - Seize the day, not the millisecond • \"Day-to-day needs\" - Again, days • \"Tomorrow\" - 86.4 million milliseconds from now • \"Days of our lives\" - Again, days • \"Give me a minute\" - 60,000 milliseconds • \"Do it now; right this second\" - 1000 milliseconds Code: Uint32 elapsedTime = 0; //Why not since SDL_GetTicks() returns Uint32 Uint32 lastFrameTimeElapsed = 0; //Convert Uint32 milliseconds to float seconds float deltaTime = (elapsedTime - lastFrameTimeElapsed) / 1000.0f; //Per-second is easy Move(10.0f * deltaTime, 0.0f * deltaTime); //10 units per second void Sprite::Move(float offsetX, float offsetY) { //Moving 10 units per second at 60 FPS, yields offsetX/Y of +-0.1666666f //Since 0.1666666f rounds to 0, this won't work: //m_rect.x += roundf(offsetX); m_floatRect->x += offsetX; m_floatRect->y += offsetY; m_rect.x = roundf(m_floatRect->x); m_rect.y = roundf(m_floatRect->y); } • I removed the / 1000.0f since it wasn't needed. And what's the difference between your post and", "B. 40 • C. 8 + 16sqrt(2) • D. 32 + 8sqrt(8) • 9. Which of the following expressions is equivalent to (4x^3 - 5x) + (8x - 3x^3) ? • A. X^3 + 3x • B. X^3 - 3x • C. 7x^3 - 3x • D. 7x^3 + 3x • 10. Which of the following represents all solutions (x,y) to the system of equations shown below? y + x = 6 and y = x^2 -2x - 6 • A. (-4,10) and (3,3) • B. (4,2) and (-3,9) • C. (4,-3) • D. (-4,3) • 11. 4^2(20k) - 7Which of the following is equivalent to the expression above? • A. 4^2(20 + k) - 7 • B. 4^2(20/k) - 7 • C. 20k * 4^2 - 7 • D. (4^2 * 20)/k - 7 • 12. What is the approximate value of g(f(5))? • A. 3.5 • B. 5 • C. 7 • D. 9 • 13. When driving from Orlando, Florida to Tampa, Florida, a family travels 6 miles in 5 minutes. If they travel at the same rate for the next half hour, approximately how many miles will they travel in that half hour? • A. 30 miles • B. 36 miles • C. 60 miles • D. 72 miles • 14. Mr. Edwards hired", "1, 12: 1, 13: 1, 14: 1, 15: 2, 16: 2, 17: 2, 18: 2, 19: 2}, 'distance': {0: 1.2898851269657656, 1: 0.0, 2: 0.8371526423804061, 3: 0.8703856587273138, 4: 0.6257425922449789, 5: 0.0, 6: 0.0, 7: 0.0, 8: 1.2895328696587023, 9: 0.0, 10: 0.6875527848294374, 11: 0.0, 12: 0.0, 13: 0.9009031833559706, 14: 0.0, 15: 1.1040652963428623, 16: 0.0, 17: 0.0, 18: 0.0, 19: 0.0}})0}}) Code def custom_fn(x, y): x = pd.Series(x) y = pd.Series(y) x = x**2 y = np.sqrt(y) return x.sum() - y.sum() # Empty data.frame to append to dmat = pd.DataFrame() # For i, j = hour; k, l = day for i in range(1, 3): for j in range(1, 3): for k in range(1, 3): for l in range(1, 3): x = dat[(dat['hour'] == i) & (dat['day'] == k)].distance y = dat[(dat['hour'] == j) & (dat['day'] == l)].distance # Calculate difference jds = custom_fn(x, y) # Build data frame and append outdat = pd.DataFrame({'day_hour_a': f\"{k}_{i}\", 'day_hour_b': f\"{l}_{j}\", 'jds': [round(jds, 4)]}) dmat = dmat.append(outdat, ignore_index=True) # Pivot data to get matrix distMatrix = dmat.pivot(index='day_hour_a', columns='day_hour_b', values='jds') Output > print(distMatrix) day_hour_b 1_1 1_2 2_1 2_2 day_hour_a 1_1 -0.2609 2.3782 1.7354 2.4630 1_2 -2.1118 0.5273 -0.1155 0.6121 2_1 -2.4903 0.1488 -0.4940 0.2336 2_2 -2.5557 0.0834 -0.5594 0.1682 • @Alex Thanks. I corrected it. – Amstell Mar 29 at 22:44 • Inside your 4-loops, are", "Wilmer I find that for this type, always easier to set up a simpler case; you can then \"see\" what's involved; have a look at this one: X[A,B].....................................120.......... ..............................[C]Y Distance of straight path X to Y is 120. A, B are at X, and motorcycle C is at Y; speeds are 5,6,10 respectively. Code: 1: C meets A in 120 / (5 + 10) = 8 hours; B has travelled 8 * 6 = 48 X......40.........[C,A,8h]................80.......................Y X.........48...........[B,8h]................72....................Y 2: C drives A for 6 hours, to 20 from Y; B goes another 6 * 6 = 36 X......................100........................[C,A,14h]...20...Y X....................84....................[B,14h].......36........Y 3: C is now 16 from B; turns around and meets B in 1 hour; A travels 5 X.......................105...........................[A,15h]..15..Y X......................90...................[C,B,15h]......30......Y 4: C drives B for 3 hours; A travels 15; both arrive at Y at same time X..............................120..........................[A,18h]Y X..............................120........................[C,B,18h]Y That's all you get from me! Should be easier now for you to set this up your own way, and bring in the 3rd walker. Thanks... It's one of the ways they can do it. But how do you know it is the QUICKEST WAY they can do it? Remember they not only need to arrive at Y at the same time, but also as quickly as possilbe. I think when there're 3 people, all the complications come from finding the", "is the same as before? First let us compute the distance Sam's car has travelled in between the two gas fillings >> 3503-3215 ans = 288 Gas mileage of Sam's car is >> 288/10 ans = 28.8 With this, he can drive >> 28.8 * 15.4 ans = 443.5200 443.52 miles before he runs out of gas. Let us do the same example, now by creating named variables >> distance = 3503-3215 distance = 288 >> mileage = distance/10 mileage = 28.8000 >> projected_distance = mileage * 15.4 projected_distance = 443.5200 To prevent the result from printing out in the command window, use a semicolon after the statement. The result will be stored in memory. You can then access the variable by calling its name. Example: >>projected_distance = mileage * 15.4; >> >>projected_distance projected_distance = 443.5200 ## Using the command line to call functions The command line can be used to call any function that's in a defined path. To call a function, the following general syntax is used: >> [Output1, Output2, ..] = functionname(input1, input2,..) MATLAB will look for a file called functionname.m and will execute all of the code inside it until it either encounters an error or finishes the file. In the former case, it produces a noise and displays an error message in red.", "##:00 to ##:59 in that day. ### Day-#-bus-speed.txt In bus UTN, each row represents the road’s average speed of 24 hours in one day. The format for this file is the following: road ID, s00, s01, …, s23. • v-##: The weight represents the average speed of buses passing through the road in the hour of starting from ##:00 to ##:59 in that day. ### Day-#-taxi-volume.txt In taxi UTN, each row represents the road’s volume of 24 hours in one day. The format for this file is the following: road ID, v00, v01, …, v23. • v-##: The weight represents the volume of taxis passing through the road in the hour of starting from ##:00 to ##:59 in that day. ### Day-#-taxi-speed.txt In taxi UTN, each row represents the road’s average speed of 24 hours in one day. The format for this file is the following: road ID, s00, s01, …, s23. • v-##: The weight represents the average speed of taxis passing through the road in the hour of starting from ##:00 to ##:59 in that day. The ArcGIS shapefiles for the road network, it includes two folders of edges and nodes, the nodes represent junctions, while edges represent road segments. ### Nodes: • FID: Row number; • lon: The length of the road; • lat: The", "The distance driven is the difference between the odometer readings: xyzzyxabccba. xyzzyxabccba = (100001 x + 10010 y + 1100 z) – (100001 a + 10010 b + 1100 c) = 100001 (xa) + 10010 (yb) + 1100 (zc) Because xyzzyx > abccba, we know that xa. Also, if x = a, we know that yb. Finally, if x = a and y = b, we know that z > c (z cannot equal c in this case.) Let's look at these three cases separately. Case 1: x = a, y = b, z > c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (0) + 1100 (zc) = 1100 (zc) z and c can range from 0 to 9, and z > c, so (zc) can range from 1 to 9. So the smallest the distance can be is 1100 miles - for example, 250052 → 259952 (1100 miles.) This is a bit far for an hour's drive. Case 2: x = a, y > b, no conditions imposed on z vs. c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (yb) + 1100 (zc) = 10010 (yb) + 1100 (zc) y, b, z, and c can range from 0 to 9; y", "draw((4,4)--(7,7),linewidth(1.6)); label(\"HOURS\",(3.41,-0.85),SE*lsf); label(\"M\",(-1.39,5.32),SE*lsf); label(\"I\",(-1.34,4.93),SE*lsf); label(\"L\",(-1.36,4.51),SE*lsf); label(\"E\",(-1.37,4.11),SE*lsf); label(\"S\",(-1.39,3.7),SE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ $\\textbf{(A)}2\\qquad\\textbf{(B)}2.5\\qquad\\textbf{(C)}4\\qquad\\textbf{(D)}4.5\\qquad\\textbf{(E)}5$ ## Problem 10 The taxi fare in Gotham City is $2.40 for the first $\\frac12$ mile and additional mileage charged at the rate$0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for$10? $\\textbf{(A) } 3.0\\qquad\\textbf{(B) }3.25\\qquad\\textbf{(C) }3.3\\qquad\\textbf{(D) }3.5\\qquad\\textbf{(E) }3.75$ ## Problem 11 The graph shows the number of minutes studied by both Asha (black bar) and Sasha(grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha? $[asy] size(300); real i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label(\"0\",origin,W); label(\"20\",(0,20),W); label(\"40\",(0,40),W); label(\"60\",(0,60),W); label(\"80\",(0,80),W); label(\"100\",(0,100),W); label(\"120\",(0,120),W); path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--(70,90)--(70,0)--cycle,TuesL=(70,0)--(70,80)--(80,80)--(80,0)--cycle,WedD=(100,0)--(100,100)--(110,100)--(110,0)--cycle,WedL=(110,0)--(110,120)--(120,120)--(120,0)--cycle,ThurD=(140,0)--(140,80)--(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle; fill(MonD,black); fill(MonL,grey); fill(TuesD,black); fill(TuesL,grey); fill(WedD,black); fill(WedL,grey); fill(ThurD,black); fill(ThurL,grey); fill(FriD,black); fill(FriL,grey); draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL); label(\"M\",(30,-5),S); label(\"Tu\",(70,-5),S); label(\"W\",(110,-5),S); label(\"Th\",(150,-5),S); label(\"F\",(190,-5),S); label(\"M\",(-25,85),W); label(\"I\",(-27,75),W); label(\"N\",(-25,65),W); label(\"U\",(-25,55),W); label(\"T\",(-25,45),W); label(\"E\",(-25,35),W); label(\"S\",(-26,25),W);[/asy]$ $\\textbf{(A)}\\ 6\\qquad\\textbf{(B)}\\ 8\\qquad\\textbf{(C)}\\ 9\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 12 Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other? $\\textbf{(A) } \\frac14 \\qquad\\textbf{(B) } \\frac13 \\qquad\\textbf{(C) } \\frac12 \\qquad\\textbf{(D) } \\frac23 \\qquad\\textbf{(E) } \\frac34$ ## Problem 13 Two congruent squares, $ABCD$ and $PQRS$, have side length $15$. They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What", "dirt along the bottom sides of the cars, but you can use one of these mods if you'd rather have clean vehicles without creating the modding incompatibilities created by using the patch--you decide. X.13.f 2x, 4x, and 10x Draw Distance mods The message board for this is at GTA Forums: http://www.gtaforums.com/index.php?showtopic=195185&st=0 Make a copy of your data\\timecycle.dat file to keep for reference or to make a copy from if you want to revert to the original version. Edison Carter provides the original file, too, if you forget to make a copy. And before you change anything, un-check the green dot 1st, etc. (X.4). These mods increase FarClp, the 28th number in each line of timecyc.dat. The bigger the number for it, the farther away things can be before they disappear. In the 1st group of eight rows, for EXTRASUNNY_LA, the original FarClp has 400.00, 800.00, or 600.00. For the group of eight rows for CLOUDY_SF, the original Farclp has 1150.00 for each row. The 29th number in each row, FogSt, determines how far away the fog is--the bigger the number, the farther away it is. It's basically lower as the numbers for FarClp get bigger. I guess the reason for that is to make about the same demands on the game engine: the more fog there is, the", "x + 10010 y + 1100 z) – (100001 a + 10010 b + 1100 c) = 100001 (xa) + 10010 (yb) + 1100 (zc) Because xyzzyx > abccba, we know that xa. Also, if x = a, we know that yb. Finally, if x = a and y = b, we know that z > c (z cannot equal c in this case.) Let's look at these three cases separately. Case 1: x = a, y = b, z > c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (0) + 1100 (zc) = 1100 (zc) z and c can range from 0 to 9, and z > c, so (zc) can range from 1 to 9. So the smallest the distance can be is 1100 miles - for example, 250052 → 259952 (1100 miles.) This is a bit far for an hour's drive. Case 2: x = a, y > b, no conditions imposed on z vs. c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (yb) + 1100 (zc) = 10010 (yb) + 1100 (zc) y, b, z, and c can range from 0 to 9; y > b; and no conditions are imposed on z vs. c. (yb) can range", "a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles? by Gmat_mission » On a recent trip, Mary drove 50 miles. What was the average speed at... 1 Last post by [email protected] Sun Oct 17, 2021 5:36 am If $$x\\ne -y,$$ then is $$\\dfrac{x-y}{x+y}>1?$$ by Gmat_mission » If $$x\\ne -y,$$ then is $$\\dfrac{x-y}{x+y}>1?$$ (1) $$x>0$$ (2)... 0 Last post by Gmat_mission Sun Oct 17, 2021 12:19 am If $$x$$ is an integer, is the median of $$5$$ numbers shown greater than the average of $$5$$ numbers? by Gmat_mission » $$\\{x, 3, 1, 12, 8\\}$$ If $$x$$ is an integer, is the median of $$5$$... 0 Last post by Gmat_mission Sun Oct 17, 2021 12:14 am How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo? by Gmat_mission » How many of the 60 cars sold last month by a certain dealer had... 0 Last post by Gmat_mission Sun Oct 17, 2021 12:01 am Are positive integers $$p$$ and $$q$$ both greater than $$n?$$ by Gmat_mission » Are positive integers $$p$$ and $$q$$ both greater than $$n?$$ (1)... 0 Last post by Gmat_mission Sat Oct 16, 2021 11:53 pm In the diagram above, $$ADF$$ is a right triangle. $$BCED$$", "do the same for the “$$y$$” equations. If any $$t$$ values are the same in both, we have a solution; we then solve for $$x$$ and $$y$$ in either equation for that $$t$$. $$\\begin{array}{c}80-.7t=.2t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.3t=20+.1t\\\\\\,\\,\\,\\,\\,80=.9t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.2t=20\\,\\,\\,\\,\\,\\,\\,\\,\\\\t=88.89\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=100\\end{array}$$ Since the $$t$$ values aren’t the same for the $$x$$ and $$y$$, the hiker and bear won’t be at the same place at the same time. Whew! Problem: At noon, Julia starts out from Austin and starts driving towards Dallas; she drives at a rate of 50 mph. Marie starts out in Dallas and starts driving towards Austin; she leaves two hours later Julia (at 2pm), and drives at a rate of 60 mph. The cities are roughly 200 miles apart. When will Julia and Marie pass each other? How far will they be from Dallas when they pass each other? Solution: Let’s first draw this situation and then try to come up with a pair of parametric equations. Remember that $$\\text{distance}=\\text{rate}\\times \\text{time}$$. Let $$t=$$ the time they travel (starting at noon), so Julia’s driving time is $$t$$, and Marie’s driving time is $$t-2$$, since she leaves two hours later. Julia’s distance from Austin is $$50t$$, and Marie’s distance from Dallas is $$60\\left( {t-2} \\right)$$. We can make the $$x$$ equations the distance from Austin for each of the girls, and the $$y$$ equations", "35^2*[25 + (10 - d)^2] = 40000 - 8000d + 400d^2 1225(25 + 100 - 20d + d^2) = 40000 - 8000d + 400d^2 153125 - 24500d + 1225d^2 = 40000 - 8000d + 400d^2 825d^2 - 16500d + 113125 = 0 25(33d^2 - 660d + 4525) = 0 d = (660 +/- sqrt[435600 - 597300])/66 I made an \"oops\" somewhere. There's too many digits! . Give me a few minutes to gain my compsure and maybe I'll figure this one out. 3. Originally Posted by ecMathGeek Code: 5 miles |------------| (x). | - (boat) . 20mph | | . | | . | | - |(a) | 10 miles | | . | d | | . 35mph | | | . | - |(y) (hotel) - That was fun to draw . From the boat (x), the power boat will take the passengers to the shore at (a), then the minivan will take the passengers to the hotel (y). Let the distance between (a) and (y) be equal to d, then the distance from (x) to (a) is sqrt[5^2 + (10 - d)^2] The total time traveled will be: T = 20*sqrt[25 + (10 - d)^2] + 35*d Now we have a function relating distance and time. Taking the derivative, we get: T' = 20*1/2(25 +" ]

[ "# A long distance trucker traveled 120 miles in one direction during a snow storm. The return trip in rainy weather was accomplished at double the speed and took 2 hours less time. How do you find the speed going? Call the initial speed $v$. The outgoing time was $\\frac{120}{v}$ and the returning time at double speed was $\\frac{120}{2 v}$. We know $\\frac{120}{v} - \\frac{120}{2 v} = 2$ hours, so $240 - 120 = 4 v$ so $v = 30$ miles/hour outgoing and $60$ mph back.", "## Thinking Mathematically (6th Edition) We can convert 100 miles per hour to units of feet per second. $100 ~mi/hr = \\frac{100~mi}{1~hr}\\times \\frac{5280~ft}{1~mi}\\times \\frac{1~hr}{3600~sec}$ $100~mi/hr = 147~ft/sec$", "in miles per hour. :/ Which I still don't really understand what i'm getting at or what that is going to give me. :( • Dec 17th 2007, 09:20 PM $velocity = \\frac {change in distance}{change in time}$", "A truck travels at 90 km/hr for the first $1\\frac{1}{2}$ hours. After that it travels at 70 km/hr. Find the time taken by the truck to travel 310 kilometres. [A] 2.5 hours [B] 3 hours [C] 3.5 hours [D] 4 hours", "### Home > MC1 > Chapter 9 > Lesson 9.2.1 > Problem9-58 9-58. It took Ivan $7$ hours to drive $385$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "above, (60 second per minute)*(60 minutes per hour)=(3600 seconds per hour) notice how the dimension minutes cancel each other out leaving seconds per hour... Well wasn't that fun??? In other words... 1.9 trillion miles.", "## College Physics (4th Edition) $(1 ~km)(\\frac{1 ~mile}{1.6 ~km}) = 0.6 ~mile \\approx 0.5 ~mile$ The correct answer is (b) 1/2 mile", "### Still have math questions? Algebra Question Traffic accident investigators use the formula: $$v = \\sqrt { 24 L }$$ to estimate the speed of a car, $$v$$ , in miles per hour, based on the length, $$L$$ , in feet, of its skid marks when suddenly braking on a flat, dry, asphalt road. At the scene of an accident, a car's skid marks are measured to be $$163$$ feet long. Approximately how fast was the car traveling? Round your answer to the nearest tenth. The car was traveling at approximately _____miles per hour", "number is now 16061. If you really need the final hint this can be the life saver for this sum : So Megan drove 16061 – 15951 = 110 miles . Since this happened over 2 hours , she drove at $\\frac {110}{2}$ = 55 mph .", "# How do you translate \"how far he traveled at 45 miles per hour\" into an algebraic expression? $d = s t$ The find distance (call this $d$) given speed (call this $s$) we need to also know how long he travelled for (call this $t$), then using distance = speed * time travelled we have: $d = s t$", "Perfect! Jerry covers $$20 \\, \\mathrm{m}$$ in $$2 \\, \\mathrm{s}$$, so, his speed is $$10 \\, \\mathrm{m/s}$$. Whereas in the other cases, his speed was lower. If an object's speed is $$30 \\, \\mathrm{km/h}$$, in what direction is it moving?", "## anonymous one year ago Mr. chow drives 112.5 miles at 45 miles per hour 357.5 miles at 55 miles per hour and 195 miles at 65 miles per hour how many hours did he spend driving altogether • This Question is Open 1. anonymous time is distance divided by rate divide each, then add them up 2. anonymous $\\huge t=\\frac{ d }{ r }$ 3. anonymous $\\huge hours=\\frac{ 112.5 }{ 45 }+\\frac{ 357.5 }{ 55 }+\\frac{ 195 }{ 65}$ 4. anonymous $\\huge hours= 2.5 + 6.5 + 3$Add them together.", "I'm supposed to be the one who knows this stuff. Thanking you in advance, - Oxy 2. Aug 3, 2005 ### oxymoron_02 If it helps, 4000000 mile/hour (mph) = 1,111.1111111 mile/second. 3. Aug 3, 2005 ### Alkatran This goes in general physics, if not high-school homework. First get the distance to alpha centauri (in miles, I suppose) Then, use that and the speed of the ship (in miles per hour) to get the number of hours. 4. Aug 3, 2005 ### oxymoron_02 I'd have thought it would end up in years. And this does seem more suited to maths, apply the numbers to a mathematically phrased question and they are no longer physics. Basically what I am saying is that the core of my question is numbers/maths, not physics. I'm giving you numbers for input, and I need an output of numbers. 5. Aug 3, 2005 ### EnumaElish x = 4.35 * 5.874 gazillion miles. v = x/t so t = x/v = 4.35 * 5.874 gazillion miles / 400 million mph = (4.35*5.874/400) zillion hours, where my definition of zillion is gazillion/million. 6. Aug 3, 2005 ### robert Ihnot The fact is light travels at about 186000 miles/sec. You are confusing kilometers. Another matter here, since you, oxymoron 02, tell us you are going about 1111 miles/sec,", "$8x+8x+40=136$ or $16x+40=136$ or $16x=136-40=96$ or $x=\\frac{96}{16}=6$ or $x=6 m/hr$ So, Vivian is travelling with speed of 6 miles per hour. and Noelle is moving with speed= x+5=6+5=11 miles per hour", "? Free Version Moderate # Earth's Linear Speed TRIG-KV0LLF If the Earth rotates once on its axis every $24$ hours, then determine its linear speed in miles per hour if the Earth's radius is $3959$ miles. Round your answer to one decimal place. A $29.5$ mi/hr B $518.2$ mi/hr C $877.2$ mi/hr D $1036.5$ mi/hr E $298,501.6$ mi/hr", "to 60mph in 4 seconds. $U_{1-2}=1370(617)=845290~ft·lb$ Step 5: Determine the power that will need to be transmitted to the wheels of the car. $P=\\frac{845290}{7}=120755\\frac{ft·lb}{s}=220~hp$ Next", "## anonymous 5 years ago If a truck drives 766 in x+2 hours, then what is his average speed. 1. anonymous I really don't see it. Please rewrite the question 2. anonymous If a truck drives 766 miles in x+2 hours, then what is his average speed? 3. anonymous $\\frac{766}{x+2} \\text{miles}/\\text{hour}$ if is x is 1/2 hour, then the truck may be exceeding Mach 1 at sea level.", "### Home > CC2MN > Chapter 8 > Lesson 8.1.1 > Problem8-10 8-10. It took Ivan $7\\frac{1}{2}$ hours to drive $412.5$ miles at a constant speed. How fast was he driving? Show how you know. Example: It took Ezekiel $3$ hours to drive $195$ miles at a constant speed. If you divide the number of miles by the number of hours, you find out that Ezekiel drove $65$ miles in one hour. This means his speed was $65$ miles per hour. Use the same method as Ezekiel's problem to find out Ivan's speed.", "c) What is the cost if the truck travels 3.25 km? d) How far has the truck traveled if Mary is charged \\$120?", "time did he spend traveling for 5 days? In order to find the total time he spent on traveling for 5, we need to multiply 6 by 15 hours 40 minutes, and 22 seconds. 15 hours 40 minutes 22 seconds $\\times$ 5 75 hours 200 minutes 110 seconds As there are 60 seconds which will make a minute. The result we got is 110 seconds, which means that 110 needs to be divided by 60. We will get quotient 1 and the remainder as 50. So the quotient 1 needs to be carried over to the minutes and the total number of seconds we get is 50 seconds. Now, 200 minutes need to be divided by 60 as in an hour there are only 60 minutes. After dividing 200 by 60 we will get the quotient as 3 and the remainder will be 20. 3 would be carried over in the hour's column and 20 would be added to the extra 1 minute that we got from the second column. The total number of minutes we now have is 21 minutes. 75 hours need to be changed by dividing it by 24. The quotient we get is 3 and the remainder is also 3. So the quotient will be changed into the number of days (3 days) and" ]

[ "# Difference between revisions of \"2020 AMC 8 Problems/Problem 11\" ## Problem 11 After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds? $[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * \"Distance (miles)\", (-0.5,3), W); label(\"Time (minutes)\", (3,-0.5), S); dot(\"Naomi\", (2,6), 3*dir(305)); dot((6,6)); label(\"Maya\", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]$ $\\textbf{(A) }6 \\qquad \\textbf{(B) }12 \\qquad \\textbf{(C) }18 \\qquad \\textbf{(D) }20 \\qquad \\textbf{(E) }24$ ## Solution 1 We use the formula $\\text{speed}=\\dfrac{\\text{distance}}{\\text{time}}$. Naomi's distance is $6$ miles, and her time is $10$ minutes, which is equivalent to $\\dfrac{1}{6}$ of an hour. Since speed is distance over time, Naomi's speed is $36$ mph. Using the same process, Maya's speed is $12$ mph. Subtracting those, we get an answer of $\\boxed{(\\text{E}) 24}$. ## Solution 2 Notice that Naomi travels at a rate of $6$ miles", "# Difference between revisions of \"2011 AMC 8 Problems/Problem 9\" Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour? $[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; draw((0,0)--(0,8)); draw((0,0)--(8,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label(\"1\",(0.95,-0.24),SE*lsf); label(\"2\",(1.92,-0.26),SE*lsf); label(\"3\",(2.92,-0.31),SE*lsf); label(\"4\",(3.93,-0.26),SE*lsf); label(\"5\",(4.92,-0.27),SE*lsf); label(\"6\",(5.95,-0.29),SE*lsf); label(\"7\",(6.94,-0.27),SE*lsf); label(\"5\",(-0.49,1.22),SE*lsf); label(\"10\",(-0.59,2.23),SE*lsf); label(\"15\",(-0.61,3.22),SE*lsf); label(\"20\",(-0.61,4.23),SE*lsf); label(\"25\",(-0.59,5.22),SE*lsf); label(\"30\",(-0.59,6.2),SE*lsf); label(\"35\",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6)); draw((4,4)--(7,7),linewidth(1.6)); label(\"HOURS\",(3.41,-0.85),SE*lsf); label(\"M\",(-1.39,5.32),SE*lsf); label(\"I\",(-1.34,4.93),SE*lsf); label(\"L\",(-1.36,4.51),SE*lsf); label(\"E\",(-1.37,4.11),SE*lsf); label(\"S\",(-1.39,3.7),SE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ ## Solution We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is $\\boxed{\\textbf{(E) 5}}$ 2011 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 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 AJHSME/AMC 8 Problems and", "# Difference between revisions of \"1994 AJHSME Problems/Problem 18\" ## Problem Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike's trip? $[asy] import graph; unitsize(12); real a(real x) {return ((x-15)^2)/2;} real b(real x) {return ((x-25)^2)/2;} real c(real x) {return ((x-30)^2 * (x-40)^2) * 8/625;} real d(real x) {return ((x-15)^2)*8/25-15;} real e(real x) {return ((x-25)^2)*8/25-15;} draw((0,9)--(0,0)--(11,0)); draw((15,9)--(15,0)--(26,0)); draw((30,9)--(30,0)--(41,0)); draw((0,-6)--(0,-15)--(11,-15)); draw((15,-6)--(15,-15)--(26,-15)); draw((0,0)--(3,8)--(7,8)--(10,0)); draw(graph(a,15,17)); draw((17,2)--(18,8)--(22,8)--(23,2)); draw(graph(b,23,25)); draw(graph(c,30,40)); draw((0,-15)--(5,-7)--(10,-15)); draw(graph(d,15,20)); draw(graph(e,20,25)); for (int k=0; k<3; ++k) { label(\"d\",(15*k-1,8),N); label(\"i\",(15*k-1,7),N); label(\"s\",(15*k-1,6),N); label(\"t\",(15*k-1,5),N); label(\"a\",(15*k-1,4),N); label(\"n\",(15*k-1,3),N); label(\"c\",(15*k-1,2),N); label(\"e\",(15*k-1,1),N); label(\"time\",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label(\"d\",(15*k-1,8-15),N); label(\"i\",(15*k-1,7-15),N); label(\"s\",(15*k-1,6-15),N); label(\"t\",(15*k-1,5-15),N); label(\"a\",(15*k-1,4-15),N); label(\"n\",(15*k-1,3-15),N); label(\"c\",(15*k-1,2-15),N); label(\"e\",(15*k-1,1-15),N); label(\"time\",(15*k+8,0-15),S); } label(\"(A)\",(5,9),N); label(\"(B)\",(20,9),N); label(\"(C)\",(35,9),N); label(\"(D)\",(5,-6),N); label(\"(E)\",(20,-6),N); [/asy]$ ## Solution When Mike is shopping at the mall, his distance doesn't change, so there should be a flat plateau shape on the graph. This rules out $C,D,E$. The portion of graph $A$", "matrix Optimal The total distance that k trucks will travel is 300.0 miles. [1.0 1.0 1.0; 12.0 3.0 12.0; 5.0 2.0 11.0; 12.0 12.0 12.0; 3.0 1.0 1.0; 12.0 8.0 12.0; 12.0 11.0 12.0; 4.0 2.0 2.0; 12.0 9.0 12.0; 1.0 1.0 1.0; 2.0 1.0 1.0; 12.0 10.0 12.0] The distance that truck 1 travels is 141.0 miles. The distance that truck 2 travels is 147.0 miles. The distance that truck 3 travels is 12.0 miles. Out[5]: 12×12×3 Array{Float64,3}: [:, :, 1] = -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 1.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 1.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 1.0 0.0 -0.0 0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 -0.0 -0.0 1.0 -0.0 0.0 0.0 0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 1.0 0.0 -0.0 0.0 -0.0 0.0 1.0 0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0", "0.0 0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 0.0 -0.0 0.0 -0.0 -0.0 0.0 -0.0 0.0 -0.0 0.0 0.0 0.0 0.0 0.0 -0.0 # 4. Result and Discussion¶ ## Scenario 1¶ From running the code in previous session, we generate a matrix for each truck that describes the order of the logging sites that it drives through. The graph below is generated by Matlab. Each color represents the route of a truck. In this case, the red path represents the route of truck 1; the blue path represents the route of truck 2; the green path represents the route of truck 3. In this session, we have: The distance for truck 1 is 12 miles. The distance for truck 2 is 134 miles. The distance for truck 3 is 131 miles. ## Scenario 2¶ From running the code in previous session, we generate a matrix for each truck that describes the order of the", "i in 1:N, j in 1:N ),\" miles.\") end println(matrix) Optimal The total distance that k trucks will travel is 277.0 miles. [1.0 1.0 1.0; 1.0 1.0 1.0; 1.0 8.0 1.0; 1.0 1.0 1.0; 12.0 9.0 1.0; 1.0 12.0 12.0; 1.0 10.0 12.0; 1.0 11.0 12.0; 1.0 12.0 5.0; 12.0 12.0 2.0; 1.0 12.0 4.0; 1.0 12.0 3.0] The distance that truck 1 travels is 12.0 miles. The distance that truck 2 travels is 134.0 miles. The distance that truck 3 travels is 131.0 miles. [-0.0 1.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; 1.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0; 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0; -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 -0.0 0.0 -0.0 -0.0 -0.0; -0.0", "## On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and from ##### This topic has expert replies Legendary Member Posts: 1622 Joined: 01 Mar 2018 Followed by:2 members ### On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and from by Gmat_mission » Fri Feb 11, 2022 8:23 am 00:00 A B C D E ## Global Stats On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and from $$P$$ to $$Q$$ are relatively flat and straight, approximately how many miles is it from $$P$$ to $$Q ?$$ A. $$3$$ B. $$\\dfrac92$$ C. $$5$$ D. $$6$$ E. $$\\dfrac{15}2$$ Source: GMAT Prep ### GMAT/MBA Expert GMAT Instructor Posts: 16118 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1268 members GMAT Score:770 ### Re: On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from $$O$$ to $$Q,$$ and fro by [email protected] » Sat Feb 12, 2022 6:04 am Gmat_mission wrote: Fri Feb 11, 2022 8:23 am 1 (1).png On the map above, $$2$$ inches represent $$3$$ miles. If the roads from $$O$$ to $$P,$$ from", "60 MPH would be the only label \"1\". So you can drive at 1, faster than 1, or slower than 1. How do you put 0.833333333MPM (50MPH) on a speed-limit sign? Miles-per-minute is, clearly, not the most convenient scale for vehicle speed limits, even though both scales are equally accurate. \"Units per millisecond\" is an equally undesirable scale because humans do not experience life as such. • \"Carpe diem\" - Seize the day, not the millisecond • \"Day-to-day needs\" - Again, days • \"Tomorrow\" - 86.4 million milliseconds from now • \"Days of our lives\" - Again, days • \"Give me a minute\" - 60,000 milliseconds • \"Do it now; right this second\" - 1000 milliseconds Code: Uint32 elapsedTime = 0; //Why not since SDL_GetTicks() returns Uint32 Uint32 lastFrameTimeElapsed = 0; //Convert Uint32 milliseconds to float seconds float deltaTime = (elapsedTime - lastFrameTimeElapsed) / 1000.0f; //Per-second is easy Move(10.0f * deltaTime, 0.0f * deltaTime); //10 units per second void Sprite::Move(float offsetX, float offsetY) { //Moving 10 units per second at 60 FPS, yields offsetX/Y of +-0.1666666f //Since 0.1666666f rounds to 0, this won't work: //m_rect.x += roundf(offsetX); m_floatRect->x += offsetX; m_floatRect->y += offsetY; m_rect.x = roundf(m_floatRect->x); m_rect.y = roundf(m_floatRect->y); } • I removed the / 1000.0f since it wasn't needed. And what's the difference between your post and", "B. 40 • C. 8 + 16sqrt(2) • D. 32 + 8sqrt(8) • 9. Which of the following expressions is equivalent to (4x^3 - 5x) + (8x - 3x^3) ? • A. X^3 + 3x • B. X^3 - 3x • C. 7x^3 - 3x • D. 7x^3 + 3x • 10. Which of the following represents all solutions (x,y) to the system of equations shown below? y + x = 6 and y = x^2 -2x - 6 • A. (-4,10) and (3,3) • B. (4,2) and (-3,9) • C. (4,-3) • D. (-4,3) • 11. 4^2(20k) - 7Which of the following is equivalent to the expression above? • A. 4^2(20 + k) - 7 • B. 4^2(20/k) - 7 • C. 20k * 4^2 - 7 • D. (4^2 * 20)/k - 7 • 12. What is the approximate value of g(f(5))? • A. 3.5 • B. 5 • C. 7 • D. 9 • 13. When driving from Orlando, Florida to Tampa, Florida, a family travels 6 miles in 5 minutes. If they travel at the same rate for the next half hour, approximately how many miles will they travel in that half hour? • A. 30 miles • B. 36 miles • C. 60 miles • D. 72 miles • 14. Mr. Edwards hired", "1, 12: 1, 13: 1, 14: 1, 15: 2, 16: 2, 17: 2, 18: 2, 19: 2}, 'distance': {0: 1.2898851269657656, 1: 0.0, 2: 0.8371526423804061, 3: 0.8703856587273138, 4: 0.6257425922449789, 5: 0.0, 6: 0.0, 7: 0.0, 8: 1.2895328696587023, 9: 0.0, 10: 0.6875527848294374, 11: 0.0, 12: 0.0, 13: 0.9009031833559706, 14: 0.0, 15: 1.1040652963428623, 16: 0.0, 17: 0.0, 18: 0.0, 19: 0.0}})0}}) Code def custom_fn(x, y): x = pd.Series(x) y = pd.Series(y) x = x**2 y = np.sqrt(y) return x.sum() - y.sum() # Empty data.frame to append to dmat = pd.DataFrame() # For i, j = hour; k, l = day for i in range(1, 3): for j in range(1, 3): for k in range(1, 3): for l in range(1, 3): x = dat[(dat['hour'] == i) & (dat['day'] == k)].distance y = dat[(dat['hour'] == j) & (dat['day'] == l)].distance # Calculate difference jds = custom_fn(x, y) # Build data frame and append outdat = pd.DataFrame({'day_hour_a': f\"{k}_{i}\", 'day_hour_b': f\"{l}_{j}\", 'jds': [round(jds, 4)]}) dmat = dmat.append(outdat, ignore_index=True) # Pivot data to get matrix distMatrix = dmat.pivot(index='day_hour_a', columns='day_hour_b', values='jds') Output > print(distMatrix) day_hour_b 1_1 1_2 2_1 2_2 day_hour_a 1_1 -0.2609 2.3782 1.7354 2.4630 1_2 -2.1118 0.5273 -0.1155 0.6121 2_1 -2.4903 0.1488 -0.4940 0.2336 2_2 -2.5557 0.0834 -0.5594 0.1682 • @Alex Thanks. I corrected it. – Amstell Mar 29 at 22:44 • Inside your 4-loops, are", "Wilmer I find that for this type, always easier to set up a simpler case; you can then \"see\" what's involved; have a look at this one: X[A,B].....................................120.......... ..............................[C]Y Distance of straight path X to Y is 120. A, B are at X, and motorcycle C is at Y; speeds are 5,6,10 respectively. Code: 1: C meets A in 120 / (5 + 10) = 8 hours; B has travelled 8 * 6 = 48 X......40.........[C,A,8h]................80.......................Y X.........48...........[B,8h]................72....................Y 2: C drives A for 6 hours, to 20 from Y; B goes another 6 * 6 = 36 X......................100........................[C,A,14h]...20...Y X....................84....................[B,14h].......36........Y 3: C is now 16 from B; turns around and meets B in 1 hour; A travels 5 X.......................105...........................[A,15h]..15..Y X......................90...................[C,B,15h]......30......Y 4: C drives B for 3 hours; A travels 15; both arrive at Y at same time X..............................120..........................[A,18h]Y X..............................120........................[C,B,18h]Y That's all you get from me! Should be easier now for you to set this up your own way, and bring in the 3rd walker. Thanks... It's one of the ways they can do it. But how do you know it is the QUICKEST WAY they can do it? Remember they not only need to arrive at Y at the same time, but also as quickly as possilbe. I think when there're 3 people, all the complications come from finding the", "is the same as before? First let us compute the distance Sam's car has travelled in between the two gas fillings >> 3503-3215 ans = 288 Gas mileage of Sam's car is >> 288/10 ans = 28.8 With this, he can drive >> 28.8 * 15.4 ans = 443.5200 443.52 miles before he runs out of gas. Let us do the same example, now by creating named variables >> distance = 3503-3215 distance = 288 >> mileage = distance/10 mileage = 28.8000 >> projected_distance = mileage * 15.4 projected_distance = 443.5200 To prevent the result from printing out in the command window, use a semicolon after the statement. The result will be stored in memory. You can then access the variable by calling its name. Example: >>projected_distance = mileage * 15.4; >> >>projected_distance projected_distance = 443.5200 ## Using the command line to call functions The command line can be used to call any function that's in a defined path. To call a function, the following general syntax is used: >> [Output1, Output2, ..] = functionname(input1, input2,..) MATLAB will look for a file called functionname.m and will execute all of the code inside it until it either encounters an error or finishes the file. In the former case, it produces a noise and displays an error message in red.", "##:00 to ##:59 in that day. ### Day-#-bus-speed.txt In bus UTN, each row represents the road’s average speed of 24 hours in one day. The format for this file is the following: road ID, s00, s01, …, s23. • v-##: The weight represents the average speed of buses passing through the road in the hour of starting from ##:00 to ##:59 in that day. ### Day-#-taxi-volume.txt In taxi UTN, each row represents the road’s volume of 24 hours in one day. The format for this file is the following: road ID, v00, v01, …, v23. • v-##: The weight represents the volume of taxis passing through the road in the hour of starting from ##:00 to ##:59 in that day. ### Day-#-taxi-speed.txt In taxi UTN, each row represents the road’s average speed of 24 hours in one day. The format for this file is the following: road ID, s00, s01, …, s23. • v-##: The weight represents the average speed of taxis passing through the road in the hour of starting from ##:00 to ##:59 in that day. The ArcGIS shapefiles for the road network, it includes two folders of edges and nodes, the nodes represent junctions, while edges represent road segments. ### Nodes: • FID: Row number; • lon: The length of the road; • lat: The", "The distance driven is the difference between the odometer readings: xyzzyxabccba. xyzzyxabccba = (100001 x + 10010 y + 1100 z) – (100001 a + 10010 b + 1100 c) = 100001 (xa) + 10010 (yb) + 1100 (zc) Because xyzzyx > abccba, we know that xa. Also, if x = a, we know that yb. Finally, if x = a and y = b, we know that z > c (z cannot equal c in this case.) Let's look at these three cases separately. Case 1: x = a, y = b, z > c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (0) + 1100 (zc) = 1100 (zc) z and c can range from 0 to 9, and z > c, so (zc) can range from 1 to 9. So the smallest the distance can be is 1100 miles - for example, 250052 → 259952 (1100 miles.) This is a bit far for an hour's drive. Case 2: x = a, y > b, no conditions imposed on z vs. c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (yb) + 1100 (zc) = 10010 (yb) + 1100 (zc) y, b, z, and c can range from 0 to 9; y", "draw((4,4)--(7,7),linewidth(1.6)); label(\"HOURS\",(3.41,-0.85),SE*lsf); label(\"M\",(-1.39,5.32),SE*lsf); label(\"I\",(-1.34,4.93),SE*lsf); label(\"L\",(-1.36,4.51),SE*lsf); label(\"E\",(-1.37,4.11),SE*lsf); label(\"S\",(-1.39,3.7),SE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ $\\textbf{(A)}2\\qquad\\textbf{(B)}2.5\\qquad\\textbf{(C)}4\\qquad\\textbf{(D)}4.5\\qquad\\textbf{(E)}5$ ## Problem 10 The taxi fare in Gotham City is $2.40 for the first $\\frac12$ mile and additional mileage charged at the rate$0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for$10? $\\textbf{(A) } 3.0\\qquad\\textbf{(B) }3.25\\qquad\\textbf{(C) }3.3\\qquad\\textbf{(D) }3.5\\qquad\\textbf{(E) }3.75$ ## Problem 11 The graph shows the number of minutes studied by both Asha (black bar) and Sasha(grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha? $[asy] size(300); real i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label(\"0\",origin,W); label(\"20\",(0,20),W); label(\"40\",(0,40),W); label(\"60\",(0,60),W); label(\"80\",(0,80),W); label(\"100\",(0,100),W); label(\"120\",(0,120),W); path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--(70,90)--(70,0)--cycle,TuesL=(70,0)--(70,80)--(80,80)--(80,0)--cycle,WedD=(100,0)--(100,100)--(110,100)--(110,0)--cycle,WedL=(110,0)--(110,120)--(120,120)--(120,0)--cycle,ThurD=(140,0)--(140,80)--(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle; fill(MonD,black); fill(MonL,grey); fill(TuesD,black); fill(TuesL,grey); fill(WedD,black); fill(WedL,grey); fill(ThurD,black); fill(ThurL,grey); fill(FriD,black); fill(FriL,grey); draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL); label(\"M\",(30,-5),S); label(\"Tu\",(70,-5),S); label(\"W\",(110,-5),S); label(\"Th\",(150,-5),S); label(\"F\",(190,-5),S); label(\"M\",(-25,85),W); label(\"I\",(-27,75),W); label(\"N\",(-25,65),W); label(\"U\",(-25,55),W); label(\"T\",(-25,45),W); label(\"E\",(-25,35),W); label(\"S\",(-26,25),W);[/asy]$ $\\textbf{(A)}\\ 6\\qquad\\textbf{(B)}\\ 8\\qquad\\textbf{(C)}\\ 9\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 12 Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other? $\\textbf{(A) } \\frac14 \\qquad\\textbf{(B) } \\frac13 \\qquad\\textbf{(C) } \\frac12 \\qquad\\textbf{(D) } \\frac23 \\qquad\\textbf{(E) } \\frac34$ ## Problem 13 Two congruent squares, $ABCD$ and $PQRS$, have side length $15$. They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What", "dirt along the bottom sides of the cars, but you can use one of these mods if you'd rather have clean vehicles without creating the modding incompatibilities created by using the patch--you decide. X.13.f 2x, 4x, and 10x Draw Distance mods The message board for this is at GTA Forums: http://www.gtaforums.com/index.php?showtopic=195185&st=0 Make a copy of your data\\timecycle.dat file to keep for reference or to make a copy from if you want to revert to the original version. Edison Carter provides the original file, too, if you forget to make a copy. And before you change anything, un-check the green dot 1st, etc. (X.4). These mods increase FarClp, the 28th number in each line of timecyc.dat. The bigger the number for it, the farther away things can be before they disappear. In the 1st group of eight rows, for EXTRASUNNY_LA, the original FarClp has 400.00, 800.00, or 600.00. For the group of eight rows for CLOUDY_SF, the original Farclp has 1150.00 for each row. The 29th number in each row, FogSt, determines how far away the fog is--the bigger the number, the farther away it is. It's basically lower as the numbers for FarClp get bigger. I guess the reason for that is to make about the same demands on the game engine: the more fog there is, the", "x + 10010 y + 1100 z) – (100001 a + 10010 b + 1100 c) = 100001 (xa) + 10010 (yb) + 1100 (zc) Because xyzzyx > abccba, we know that xa. Also, if x = a, we know that yb. Finally, if x = a and y = b, we know that z > c (z cannot equal c in this case.) Let's look at these three cases separately. Case 1: x = a, y = b, z > c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (0) + 1100 (zc) = 1100 (zc) z and c can range from 0 to 9, and z > c, so (zc) can range from 1 to 9. So the smallest the distance can be is 1100 miles - for example, 250052 → 259952 (1100 miles.) This is a bit far for an hour's drive. Case 2: x = a, y > b, no conditions imposed on z vs. c xyzzyxabccba = 100001 (xa) + 10010 (yb) + 1100 (zc) = 100001 (0) + 10010 (yb) + 1100 (zc) = 10010 (yb) + 1100 (zc) y, b, z, and c can range from 0 to 9; y > b; and no conditions are imposed on z vs. c. (yb) can range", "a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles? by Gmat_mission » On a recent trip, Mary drove 50 miles. What was the average speed at... 1 Last post by [email protected] Sun Oct 17, 2021 5:36 am If $$x\\ne -y,$$ then is $$\\dfrac{x-y}{x+y}>1?$$ by Gmat_mission » If $$x\\ne -y,$$ then is $$\\dfrac{x-y}{x+y}>1?$$ (1) $$x>0$$ (2)... 0 Last post by Gmat_mission Sun Oct 17, 2021 12:19 am If $$x$$ is an integer, is the median of $$5$$ numbers shown greater than the average of $$5$$ numbers? by Gmat_mission » $$\\{x, 3, 1, 12, 8\\}$$ If $$x$$ is an integer, is the median of $$5$$... 0 Last post by Gmat_mission Sun Oct 17, 2021 12:14 am How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo? by Gmat_mission » How many of the 60 cars sold last month by a certain dealer had... 0 Last post by Gmat_mission Sun Oct 17, 2021 12:01 am Are positive integers $$p$$ and $$q$$ both greater than $$n?$$ by Gmat_mission » Are positive integers $$p$$ and $$q$$ both greater than $$n?$$ (1)... 0 Last post by Gmat_mission Sat Oct 16, 2021 11:53 pm In the diagram above, $$ADF$$ is a right triangle. $$BCED$$", "do the same for the “$$y$$” equations. If any $$t$$ values are the same in both, we have a solution; we then solve for $$x$$ and $$y$$ in either equation for that $$t$$. $$\\begin{array}{c}80-.7t=.2t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.3t=20+.1t\\\\\\,\\,\\,\\,\\,80=.9t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.2t=20\\,\\,\\,\\,\\,\\,\\,\\,\\\\t=88.89\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=100\\end{array}$$ Since the $$t$$ values aren’t the same for the $$x$$ and $$y$$, the hiker and bear won’t be at the same place at the same time. Whew! Problem: At noon, Julia starts out from Austin and starts driving towards Dallas; she drives at a rate of 50 mph. Marie starts out in Dallas and starts driving towards Austin; she leaves two hours later Julia (at 2pm), and drives at a rate of 60 mph. The cities are roughly 200 miles apart. When will Julia and Marie pass each other? How far will they be from Dallas when they pass each other? Solution: Let’s first draw this situation and then try to come up with a pair of parametric equations. Remember that $$\\text{distance}=\\text{rate}\\times \\text{time}$$. Let $$t=$$ the time they travel (starting at noon), so Julia’s driving time is $$t$$, and Marie’s driving time is $$t-2$$, since she leaves two hours later. Julia’s distance from Austin is $$50t$$, and Marie’s distance from Dallas is $$60\\left( {t-2} \\right)$$. We can make the $$x$$ equations the distance from Austin for each of the girls, and the $$y$$ equations", "35^2*[25 + (10 - d)^2] = 40000 - 8000d + 400d^2 1225(25 + 100 - 20d + d^2) = 40000 - 8000d + 400d^2 153125 - 24500d + 1225d^2 = 40000 - 8000d + 400d^2 825d^2 - 16500d + 113125 = 0 25(33d^2 - 660d + 4525) = 0 d = (660 +/- sqrt[435600 - 597300])/66 I made an \"oops\" somewhere. There's too many digits! . Give me a few minutes to gain my compsure and maybe I'll figure this one out. 3. Originally Posted by ecMathGeek Code: 5 miles |------------| (x). | - (boat) . 20mph | | . | | . | | - |(a) | 10 miles | | . | d | | . 35mph | | | . | - |(y) (hotel) - That was fun to draw . From the boat (x), the power boat will take the passengers to the shore at (a), then the minivan will take the passengers to the hotel (y). Let the distance between (a) and (y) be equal to d, then the distance from (x) to (a) is sqrt[5^2 + (10 - d)^2] The total time traveled will be: T = 20*sqrt[25 + (10 - d)^2] + 35*d Now we have a function relating distance and time. Taking the derivative, we get: T' = 20*1/2(25 +" ]

[ "# How did Tom get to be table-partners with Becky? By that I mean, that how did they got to sit together? It's kinda hard for me to answer this, since I have read the book, but I remember none of this. Ty!", "# Who's been stealing Jenny's milk? Jenny works in an office with four other colleagues and someone has been using her milk. She finds this very annoying as she likes her mid-afternoon tea and she simply can't enjoy the second half of her day if she has to drink black tea instead. Deciding that enough is enough Jenny decides to do something about it, she's suspicious of all of her four colleges Liz, John, Sarah and Bill so she decides she's going to solve the crime once and for all. Her first course of action is to make a note of which days her milk is used and which days her coworkers are in the office. • Monday - JBS - THEFT! • Tuesday - JLBS - THEFT! • Wednesday - JLBS - THEFT! • Thursday - LBS - No theft • Friday - LB - No theft Next, deciding to forsake her regular duties for the sake of office security she follows various members of the office for a day over the next two weeks. Alas the thief evades detection. • Monday - Sarah • Tuesday - Liz • Wednesday - John • Thursday - Bill • Friday - Liz • Monday - Sarah • Tuesday - Jenny has to attend a meeting and fails to follow", "John all day • Wednesday - Liz • Thursday - Liz • Friday - Bill To her interest the thefts occur on the same days each week (Monday, Tuesday and Wednesday). By the second Friday she's at her wit's end, she's still not caught the thief - is she doomed to a lifetime of black tea!? She explains her dilemma to her friend one evening who (also being British) realises the seriousness of the situation. Pouring over Jenny's copious notes it looks like a lost cause until her friend smiles. She knows who the thief is! Who is responsible for the milk thefts? • Why should we assume that only one person steals milk? I mean, other than the fact that assuming this gives a unique solution to the problem. Should it be stated, or are you happy to leave that to the solver? Especially given the solution, it's not clear to me why there shouldn't be one person who chooses to steal milk only on Monday, and another who chooses to steal only on Tuesday. – Steve Jessop Nov 10 '14 at 17:41 • I designed the question assuming there only one thief but feel free to prove me wrong! – Liath Nov 10 '14 at 19:02 • What prevents (for example) John from stealing on Monday,", "Spencer talk at her house to discuss Jenna's and Ian's meeting in school that day. Spencer demands that Toby steal Jenna's phone so that they can uncover what he's hiding. She convinces him that if he doesn't, he is liable to be framed for murder again. In her kitchen, Spencer makes Toby breakfast, and he hands her Jenna's phone, which he stole the night before; he is afraid of Jenna's reaction. Spencer uses the opportunity to ask Toby why Alison had been so sure that Toby was the one spying on them. Toby is firm that he never spied on anyone; the only thing he can make of Alison's assumptions is that she had something on him. Spencer enlists Caleb to help crack Jenna‘s phone for any secret messages she‘s been sending. Caleb mocks the girls for stealing a blind person's phone, guessing quite obviously that they have stolen Jenna's phone. Aria, the stage manager, calls for a five minute break during play practice. Spencer, Emily, and Hanna review their lines and discuss the motives of their characters, no doubt finding parallels to their own lives. They catch sight of Ian hanging around close to where they are standing, prompting Emily to have a flashback to the night of the party. The girl that Ian had left with", "fox and the rabbit are not trying to escape and run away) ### 8) Three Brothers on a Farm Three brothers live in a farm. They agreed to buy new seeds: Adam and Ben would go and Charlie stayed to protect fields. Ben bought 75 sacks of wheat in the market whereas Adam bought 45 sacks. At home, they split the sacks equally. Charlie had paid 1400 dollars for the wheat. How much dollars did Ben and Adam get of the sum, considering equal split of the sacks?", "he shoots and kills the intruder. 9. Allen sees an assassin in a crowd of people aiming at him. To protect himself, he shoots at the assassin with both barrels of a double barreled shotgun that he just brought back from a hunting trip. He hits the assassin and a bystander in the crowd. 10. Laura is in a library with her young daughter, Mandy, who is being disruptive. Laura tries to reason with Mandy and warns her several times, but Mandy keeps making noise. When a librarian tells them to leave, Laura forcefully drags Mandy out by the arm, hurting her. 11. Mike has a few beers at a party. His friend, Alice, tells him that he's too drunk to drive. Mike still insists on driving home, so Alice hits him to make him drop his car keys. Mike falls and bangs his head, causing a concussion. 12. Matthew and Nick get into a heated argument. Nick suggests they settle it by a fistfight. Matthew is reluctant, but Nick insists, so Matthew finally agrees. Matthew is much bigger and stronger, so he quickly wins by punching Nick in the side, breaking two ribs. 13. Kim and Olivia have an argument. Olivia suggests that they resolve their dispute by wrestling, although she knows Kim is on a wrestling", "Adam is smart. Joe draws well. Bill is cooperative. They make a good group. They turn in their project. Ms. Howard likes it. She gives them a high grade. Adam, Joe, and Bill are joyful. ## A Cat and A Mouse The cat is bored. He tries to have fun. He plays with yarn. He scratches his tummy. He takes a nap. He drinks milk. He is still bored. He sees a mouse. The mouse is eating cheese. The cat steals the cheese. The mouse is angry. The mouse chases the cat. The cat is having fun. John wants to go to his school field trip. He needs fifty dollars. He wants to earn it himself. He plans to open up a lemonade stand. He goes to the market. The lemons are cheap. They are also ripe. He buys fifty lemons. He starts making lemonade. A lot of people are in line. His lemonade stand is a success. He makes enough money. ## New and Old Tomorrow is the first day of school. Barbara looks at her backpack. It has holes. It is dirty. She needs a new one. Her mom takes her shopping. There are so many backpacks. One backpack has yellow stars. Another one has rainbow stripes. Another one has dogs. Barbara likes this one. She", "I can eat this dinner in six bites. Peter: I can eat this dinner in five bites. Jason: Four bites. Peter: Three bites. Jason: Two bites. Peter: One bite. Jason: Eat that dinner. Paige: (to Jason) I notice you always start with an even number. Peter: Mpghltz! Andy: Peter, must we go through this every night?! [Peter is on the couch, watching the television with the remote in his hand.] Carson Daly: Hi, I'm Carson Daly, and you're watching MTV's Thanksgiving-Break Beach Spectacular! Coming up, we've got performances by the Backstreet Boys! Ricky Martin! N Sync! 98 Degrees! But first, \"Survivor\"'s Richard Hatch is going to teach me how to crab-walk... Paige: I guess you really did eat so much turkey you can't move. Peter: (through tight lips) For the love of humanity, help me press this remote. Carson Daly: Wait a minute... naked?! [Peter is watching television.] Announcer: Coming up next, it's the show that pits television producers against each other in furious competition! Watch as they each propose newer and ever more bizarre ways to pander to the viewing public's voyeuristic impulses! At stake are millions! Who will walk away with it?! Whose show will land a network deal?! Find out on Reality Series: The Reality Series. Peter: You just knew it would come to this.", "warning, he shoots and kills the intruder. 9. Allen sees an assassin in a crowd of people aiming at him. To protect himself, he shoots at the assassin with both barrels of a double barreled shotgun that he just brought back from a hunting trip. He hits the assassin and a bystander in the crowd. 10. Laura is in a library with her young daughter, Mandy, who is being disruptive. Laura tries to reason with Mandy and warns her several times, but Mandy keeps making noise. When a librarian tells them to leave, Laura forcefully drags Mandy out by the arm, hurting her. 11. Mike has a few beers at a party. His friend, Alice, tells him that he's too drunk to drive. Mike still insists on driving home, so Alice hits him to make him drop his car keys. Mike falls and bangs his head, causing a concussion. 12. Matthew and Nick get into a heated argument. Nick suggests they settle it by a fistfight. Matthew is reluctant, but Nick insists, so Matthew finally agrees. Matthew is much bigger and stronger, so he quickly wins by punching Nick in the side, breaking two ribs. 13. Kim and Olivia have an argument. Olivia suggests that they resolve their dispute by wrestling, although she knows Kim is on a", "eats roast beef. On Tuesday he goes down to the Kawanis Club and eats roast beef. On wednesday, he goes to the Rlotary Club and eats roast beef. On Thursday and Friday, the same routine, he goes to some club and eats roast beef. Well now he’s gotten along pretty good in society end makes enough income that he can afford to be up in the world. He comes home and demands a steak or at least a half a pound of ground meat of some kind for dinner. And he eats this, which is good. And finally he tosses the baby up \\in the air and plays with him for a minute and then flops over on the couch and somebody turn on the idiot box in front of him to entertain him for a little while. He sits there and reads or watches the idiot box for a while then, along abut 9:30 he says, \"Oh gosh, I'm so hungry. I've got to have something to eat\". He demands the wife to run out and get him a big bowl of chocolate ice cream. He eats that and goes on and dozes on and off between the late shows. Finally along about midnight, he'll get up and has I peanut butter sandwich or a ham sandwich", "# Do Androids Dream of Electric Sheep? Summary and Analysis of Chapters 13-15 Summary Chapter 13 John Isidore returns from his work with a three year old bottle of Chablis champagne that he had been saving in a safety deposit box for years just in case he ever met another person to share it with. He is taking it home, along with some groceries, to make dinner for Pris Stratton. When he gets to his apartment building, Pris cautiously lets him into her apartment but then tells him to leave and that she can’t have the champagne. When John asks why she can’t have it, she tells him that it would be “wasted on me” and that sometime she would tell him why. He begins to make dinner anyway and tells Pris that the reason she is so scared and depressed is because she doesn’t have any friends. Pris tells him that she does in fact have friends but that they are probably all dead at the hands of bounty hunters. John does not know what a bounty hunter is because their existence is hidden from the general population. Pris explains bounty hunters, how they are paid to kill people, but John doesn’t necessarily believe her because Buster Friendly had never said anything about them. John says that", "whether to stay with John. Chapter 15 The vote is taken by the three androids: Imgard votes to stay with John Isidore; Roy Baty votes that they kill John and hide somewhere else; with the deciding vote, Pris elects to stay with Isidore, explaining that his “value to us outweighs his danger....” She tells them that this is their only alternative because the plans by the rest of their android comrades had all failed. Roy is angry at the decision, but John feels that this is the “culmination” of his entire life and he promises to help. After getting his bounty for the three androids he retired, $3,000, Rick goes immediately to the row of animal stores to shop for a real animal. The salesman there talks him into buying a goat - an expensive purchase - and he puts$3,000 as a downpayment and agrees to take on onerous monthly payments. But this makes Rick very happy and he immediately goes home to tell Iran about the purchase. He leads Iran up to the roof of their apartment where the goat is in a cage. She is both shocked and excited over the purchase. She knows the goat costs too much for their meager income, but she is proud of the social status the animal gives them. She", "were not supposed to shake that vial that had the label that said \"DO NOT SHAKE\"? Steve: Egad! My pants are dissolving! Jason: I can eat this dinner in six bites. Peter: I can eat this dinner in five bites. Jason: Four bites. Peter: Three bites. Jason: Two bites. Peter: One bite. Jason: Eat that dinner. Paige: (to Jason) I notice you always start with an even number. Peter: Mpghltz! Andy: Peter, must we go through this every night?! [Peter is watching television.] Announcer: Coming up next, it's the show that pits television producers against each other in furious competition! Watch as they each propose newer and ever more bizarre ways to pander to the viewing public's voyeuristic impulses! At stake are millions! Who will walk away with it?! Whose show will land a network deal?! Find out on Reality Series: The Reality Series. Peter: You just knew it would come to this. Peter: [with a baseball] This should break left... [pitches] This should break right... [pitches] This should... Jason: [tied to a tree, with a bullseye taped to his chest] Okay, okay, maybe I did take a few photos of you and Denise last night. [Peter's English class receives its next assigned reading, Joseph Conrad's \"Heart of Darkness\".] Peter: [gasp] The horror! The horror! Teacher: If I", "## Tuesday, December 18, 2012 ### Holiday Meal 2012 Erin started a tradition in December of 2010 of having an annual Holiday Meal for our close friends. We recently had our third annual holiday meal. Although we have more than 8 friends, that is the uppermost limit of friends we can invite because of table space, and the difficulty of cooking for ten people. During the day of the Holiday Meal, leading up to the 5:30pm start time, Erin and I usually listen to hip-hop and dance music while I'm cleaning and straightening up and she's doing all of the prep work for the meal. During this time the song \"Imma Be\" by the Black Eyed Peas hit our Pandora station. Every time we hear it, we are reminded of the scene on this link to the \"Other Guys.\" Of course that is when I piped up and told Erin, \"Wouldn't it be funny if we did a scene like that for the Holiday Meal?\" Knowing us, we usually follow up on anything I say that starts with \"Wouldn't it be funny if...\" The result can be viewed below. The dinner table scene is the only one we were going to do. After it was shot, I put it together with the music and we all watched it", "alone, but Debbie made him sit diagonal to her. • Dale is nervous because Duncan keeps staring straight forward at him while they eat. This should be enough information.", "were inseparable and often indistinguishable—so much so that one of my colleagues at the Weekly told me once, “I thought Squish and Joey were the same person.” Sitting at lunch with them one day—they had ordered deep-fried ham sandwiches,7 genuinely horrifying fare—I started telling them about my dinner the night before. “My wife,” I said, “made a normal macaroni and cheese casserole, but then did something really cool—for the topping, she crumbled up Tim’s Cascade Style potato chips8 and sprinkled them over the whole thing.” Simultaneously, their jaws dropped and their eyes started glistening. “That’s a great idea!” exclaimed Joey. “That woman’s a keeper!” said Squish. Squish in particular often came across not so much as an entrepreneur or software engineer but as a performance artist, his role of the moment being that of the founder of a high-tech startup. He had an odd kind of frenetic energy—if you looked closely enough, you could see little lightning bolts constantly crackling from him. He referred to any event, however traumatic, disastrous, strange, unexpected, or distressing, as “the usual madness,” and he could not keep himself from incessant play, no matter what purportedly serious task might be occupying him at the moment. Email detailing what needed fixing in BIG/ip was always a mix of technical arcana with maddening asides. “Root", "of bread, and Brian's apparent inability to put the ironing board out without making a horrendous screech. :-)", "didn't get the only geometry problem (some weird angle chasing stuff), but having realized that it requires some esoteric olympiad geometry knowledge, I didn't feel that bad. After that was over, a hungry Pavel and Sam ran away to somewhere to eat lunch and Mark and I headed to where TJ was eating lunch while the rest of the team decided on sandwiches and pizza. Jenny graciously waited for us at the entrance and led us to TJ's little hideout.'' She then forced me to eat a pizza, which obviously scared off Mark as he [actually both of us] felt really bad for taking other teams' food. Then Harini came and hung out with us for a while…and it was slightly awkward. XD She dragged us to see some Exeter people, but on my way, I met Jacob Hurwitz and he recognized me XD That was pretty amusing. After Harini left, Jenny dragged us to play Gluck, but I stubbornly refused a countless number of times >D I eventually led them back to where we started and saw some AAST people filtering in, including Sam and John. I started a bridge game with them and Brian. Then a bunch of people gathered around us to watch, but eventually got bored because John played really slowly, as always. Some", "the complicated UST food court ordering process, and Warren endears himself to Geoff by producing a steaming bowl of spicy ramen as if by magic. The contestants have a ‘cultural night’, which apparently includes a greater number of hedge fund representatives than one might have expected. For me, it’s a night in with Geoff, green tea and the scripts for Q2. Joe and Neel have filled fourteen pages between them checking a construction in glorious detail, a step which Harvey has described in its entirety with the words ‘glue them together’. Overall, they are complicated but precise, and I have few concerns, so it’s only necessary to burn the candle at one end. Wednesday 13th July It’s time for coordination, where Geoff and I agree the UK marks with a team of local and international experts. The scheduling has assigned us the Q1 geometry early in the morning, which is a clear case of five perfect solutions, so we move to Q2. Coordinator Stephan seems very well-prepared for the UK scripts, so again we are finished in a matter of minutes. This allows us to bring forward our discussion of Q4. Jacob has made several small errors, all of which could be fixed by attacking his script with a pair of scissors and some glue. I believe the", "beans with a fork. When the can is empty, he tosses it aside and looks around. ELIJAH WOOD (mouth full of beans) Fuck, we need more beans." ]

[ "# A Combinations Problem Involving Days Of the Week I'am reading through Engineering Math by Ken Stroud/Dexter Booth and in page 274 under Combinations. Here's the situation. Assuming that you have a part-time Job in the weekday evenings where you have to be at work just two evenings out of the five. Let's also assume that your employer is very flexible and allows you to choose which evenings you work provided you ring him up on sunday and tell him. One possible selection could be: Mon-Work Tue-Work Another selection could be: Wed-Work Fri-Work So the possible arrangements among the five days is $$5 \\cdot 4 = 20.$$ Because There are $5$ weekdays from which to make a first selection and for each such selection there are $4$ days left from which to make the second selection. This gives a total of $5 \\cdot 4 = 20$ possible arrangements. However, not all arrangements are different. For example, if, on the Sunday, you made your first choice as Friday and your second choice as Wednesday, this would be the same arrangement as making your first choice as Wednesday and your second choice as Friday. So every arrangement is duplicated. How many different arrangements are there? $$\\frac{5\\cdot 4 }{ 2 }= 10.$$ And he says it's Because each arrangement is duplicated.", "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 $31+126=\\boxed{157.}$", "cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday. So the number of menus is $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 1 \\cdot 3 = 729.$ The answer is $\\boxed{A}$. ## Solution 2 We can perform casework as an understandable means of getting the answer. We can organize our counting based on the food that was served on Wednesday, because whether cake is or is not served on Wednesday seems to significantly affect the number of ways the chef can make said foods for that week. Case 1: Cake is served on Wednesday. Here, we have three choices for food on Thursday and Saturday since cake must be served on Friday, and none of these choices are cake, which was served Wednesday. Likewise, we have three choices (pie, ice cream, and pudding) for the food served on Tuesday and thus three choices for those served on Monday and Sunday, with these three choices being whatever was not served on Tuesday and Monday, respectively. Hence, for this case, there are $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 = 243$ possibilities. Case 2: Cake is not", "cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday. So the number of menus is $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 1 \\cdot 3 = 729.$ The answer is $\\boxed{A}$. ## Solution 2 We can perform casework as an understandable means of getting the answer. We can organize our counting based on the food that was served on Wednesday, because whether cake is or is not served on Wednesday seems to significantly affect the number of ways the chef can make said foods for that week. Case 1: Cake is served on Wednesday. Here, we have three choices for food on Thursday and Saturday since cake must be served on Friday, and none of these choices are cake, which was served Wednesday. Likewise, we have three choices (pie, ice cream, and pudding) for the food served on Tuesday and thus three choices for those served on Monday and Sunday, with these three choices being whatever was not served on Tuesday and Monday, respectively. Hence, for this case, there are $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 = 243$ possibilities. Case 2: Cake is not", "cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday. So the number of menus is $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 1 \\cdot 3 = 729.$ The answer is $A.$", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "served on Wednesday. Obviously, this means that pie, ice cream, and pudding are our only choices for Wednesday's food. Since cake must be served on Friday, only ice cream, pudding, and cake can be served on Thursday. However, since one of those foods was chosen for Wednesday, we only have two possibilities for Thursday's food. Like our first case, we have three possibilities for the food served on Tuesday, Monday, and Sunday: whatever was not served on Wednesday, Tuesday, and Monday, respectively. $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 2 \\cdot 3 = 486$ possibilities thus exist for this case. Adding the number of possibilities together yields that $243 + 486 = 729$ is the total number of menus, making our answer $\\boxed{A}$.", "served on Wednesday. Obviously, this means that pie, ice cream, and pudding are our only choices for Wednesday's food. Since cake must be served on Friday, only ice cream, pudding, and cake can be served on Thursday. However, since one of those foods was chosen for Wednesday, we only have two possibilities for Thursday's food. Like our first case, we have three possibilities for the food served on Tuesday, Monday, and Sunday: whatever was not served on Wednesday, Tuesday, and Monday, respectively. $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 2 \\cdot 3 = 486$ possibilities thus exist for this case. Adding the number of possibilities together yields that $243 + 486 = 729$ is the total number of menus, making our answer $\\boxed{A}$. ## Solution 3 Note that the choice of a food item on a given day is symmetric, i.e. the number of ways to create the meal plan with a cake on Friday is the same as the number of ways to create the meal plan with pudding on Friday, and the same reasoning holds for the other desserts. Since every meal plan is counted by the summation of the $4$ aforementioned plans (note that Friday's dessert has to be one of the $4$ given desserts) and that these cases are mutually exclusive (i.e. you", "remaining 4 to be one of 25 choices can be done in $5 \\cdot 25^4$ ways, just as you said.", "\\text{total outcomes}$ For selecting a day of the week: $\\text{total outcomes} & = \\text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}\\\\& = 7 \\ \\text{total outcomes}$ Favorable outcomes are the specific outcomes you are looking for. For flipping a coin and having it come up heads: $\\text{favorable outcomes} & = \\text{heads}\\\\& = 1 \\ \\text{favorable outcome}$ For tossing a number cube and having it come with up an even number: $\\text{favorable outcomes} & = \\cdot \\ 1 \\ \\cdot \\cdot \\ 2 \\ \\cdot \\cdot \\cdot \\ 3 \\ \\cdot \\cdot \\cdot \\cdot \\ 4 \\ \\cdot \\cdot \\cdot \\cdot \\cdot \\ 5 \\ \\cdot \\cdot \\cdot \\cdot \\cdot \\cdot \\ 6\\\\& = 3 \\ \\text{favorable outcomes}$ For randomly choosing a date and have it land on a weekday: $\\text{favorable outcomes} & = \\text{Monday, Tuesday, Wednesday, Thursday, Friday}\\\\& = 5 \\ \\text{favorable outcomes}$ To write a ratio, we compare the favorable outcome to the total outcomes. Comparing favorable outcomes to possible total outcomes is what we call theoretical probability. II. Calculate Simple Theoretical Probabilities Now that you know how to think about theoretical probability, we can look at calculating some simple probabilities. Remember, that we are going to be writing ratios that compare favorable outcomes with total outcomes. The probability of any event is written as $P (\\text{event})$. So:", "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.", "(2 [over] 3) = 5? There are $25$ ways to select three days of the week such that at least one begins with a T. 7. Hello, Tall Jessica! In how many different ways could you select three different days of the week, such that at least one of them begins with the letter \"T\"? Assume that the order is important. $\\text{There are: }\\:_7P_3 \\:=\\:\\dfrac{7!}{4!} \\:=\\:210 \\text{ ways to choose }any\\text{ three days.}$ How many of these do not begin with a \"T\"? For the first day, there are 5 choices: {Sun, Mon, Wed, Fri, Sat}. For the second day, there are 4 choices. For the third day, there are 3 choices. . . Hence, there are: . $5\\cdot4\\cdot3 \\,=\\,60$ ways . . to choose three days that do not begin with \"T\". Therefore, there are: . $210 - 60 \\:=\\:150$ ways to choose three days . . so that at least one of them does begin with \"T\". 8. Oh, wow. But, HOW do I get to that answer?", "\\ | * - - - - - * Now there are $\\boxed{3}$ choices for plot #4. Hence, there are: . $5 \\times 4 \\times 3 \\times 3 \\:=$ 180 ways. Therefore, there are: . $80 + 180 \\:=$ 260 ways.", "$2$ 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", "of the $2$ friends and $4$ of the $12$ indifferent people giving $${2\\choose{0}}\\cdot{12\\choose{5}}+{2\\choose{1}}\\cdot{12\\choose{4}}=1287$$ I write it out like this because it gives you a check to make sure that the sums of the top and bottoms comes out to be $14$ and $5$ respectively. • Also, another way to evaluate $c$ is to subtract the prohibited selection from the total. $$\\dbinom{14}{5} - \\dbinom{2}{2}\\cdot\\dbinom{12}{3} = 1287$$ – Graham Kemp Dec 1 '17 at 0:15 A) $14\\choose5$ B) $12\\choose5$+$12\\choose3$ C) $12\\choose5$+2$12\\choose4$ You're right for A. For B consider 2 states, either the married couple are coming or they're not. If they're not coming then there is 12 choose 5 possibilities. If they are coming then we have 12 choose 3 other spots. So the answer is 12 choose 5 plus 12 choose 3. For C we'll call the 2 friends not speaking A and B. If A is going, then there are 12 choose 4 remaining spots. By symmetry, the same is true for B. If neither is going then there are 12 choose 5 spots. So the answer is 2 times 12 choose 4 plus 12 choose 5. • Why 13 choose 4 for C, maybe 12? – user Nov 30 '17 at 22:13", "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.", "would be $a)$, but they do not seem to make any such claims. #### Solutions Collecting From Web of \"In how many ways can these people be arranged for a photograph?\" Choose a place for Jessie ($6$ options) then for Casey ($5$ options) then fill the remaining places one by one – $8,7,6,5$ options in turn. The number of photo arrangements is thus $$8\\cdot 7\\cdot 6^2\\cdot 5^2 = 50400$$ So none of the options given are correct as the problem is stated. It’s sometimes usual to have the happy couple in a fixed position in the middle, in which case the initial choice of location for those two vanishes and we’re left with $$8\\cdot 7\\cdot 6\\cdot 5 = 1680$$ possible arrangements around the couple.", "Which two days have same sale? (a) Tuesday and Wednesday (b) Wednesday and Thursday (c) Thursday and Tuesday (d) Monday and Friday (e) None of these Explanation: Number of symbols shown on Monday and Friday is same that is equal number of symbols 5. Therefore, Monday and Friday have the same sale of toys. Rest of the options is incorrect because of the correctness of option (d). 6. Which day has sale of exact 60 toys? (a) Monday (b) Tuesday (c) Wednesday (d) Thursday (e) None of these Explanation: Here, 1 symbol represents 20 toys. Therefore, number of symbols for 20 toys $=60\\div 20=3$ Now, seeing the pictograph, Tuesday has 3 symbols. So we have, Tuesday has sale of 60 toys. Rest of the options is incorrect because of the correctness of option (b). 7. What was the total sale of toys in the week? (a) 440 (b) 380 (c) 480 (d) 500 (e) None of these Explanation: Here, total number of symbols $=5+3+2+6+5+4=\\text{ }25$ One symbol represents 20 toys. Hence, the total sale of toys in the week $=25\\times 20=500$ Rest of the options is incorrect because of the correctness of option (d). #### Other Topics ##### 15 10 LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY! You need to login to perform", "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=$2\\boxed{520}$.", "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" ]

[ "fast-food restaurant. Tim, the youngest child, has a tendency to cause mischief when he is not watched carefully. Because of this, the father wants to keep Tim ahead of him in line at all times. How many different ways can the family arrange themselves in line such that the father is able to watch Tim? A. 10 B. 24 C. 48 D. 60 E. 120 _________________ ..Thanks for KUDOS Math Expert Joined: 02 Sep 2009 Posts: 55274 Re: A family of 2 parents and 3 children is waiting in line to order [#permalink] ### Show Tags 29 Nov 2018, 12:54 1 Mahmoudfawzy83 wrote: A family of 2 parents and 3 children is waiting in line to order at a fast-food restaurant. Tim, the youngest child, has a tendency to cause mischief when he is not watched carefully. Because of this, the father wants to keep Tim ahead of him in line at all times. How many different ways can the family arrange themselves in line such that the father is able to watch Tim? A. 10 B. 24 C. 48 D. 60 E. 120 The total number of arrangements of 5 people is 5!. In half of the cases father will be behind Tim and in half of the cases Tim will be behind father (as probability doesn't", "# Who's been stealing Jenny's milk? Jenny works in an office with four other colleagues and someone has been using her milk. She finds this very annoying as she likes her mid-afternoon tea and she simply can't enjoy the second half of her day if she has to drink black tea instead. Deciding that enough is enough Jenny decides to do something about it, she's suspicious of all of her four colleges Liz, John, Sarah and Bill so she decides she's going to solve the crime once and for all. Her first course of action is to make a note of which days her milk is used and which days her coworkers are in the office. • Monday - JBS - THEFT! • Tuesday - JLBS - THEFT! • Wednesday - JLBS - THEFT! • Thursday - LBS - No theft • Friday - LB - No theft Next, deciding to forsake her regular duties for the sake of office security she follows various members of the office for a day over the next two weeks. Alas the thief evades detection. • Monday - Sarah • Tuesday - Liz • Wednesday - John • Thursday - Bill • Friday - Liz • Monday - Sarah • Tuesday - Jenny has to attend a meeting and fails to follow", "###### back to index | new $\\textbf{Who Finishes the Second}$ Adam, Bob, and Charlie are the only three athletes who are competing in a series of track and field events. The first, second and third places in each event are awarded $X$, $Y$ and $Z$ points respectively, where $X > Y > Z$ and all are integers. It is known that • Adam finishes first with $22$ points overall • Bob wins the javelin event and finishes with $9$ points overall. • Charlie also finishes $9$ points overall. Who finishes second in the $100$-meter dash and why? $\\textbf{Cookies}$ Steve, Tony, and Bruce have a plate of $1,000$ cookies to share according to the following rules. Beginning with Steve, each of them in turn takes as many cookies as he likes (but must be at least $1$ if there are still cookies on the plate), and then passes the plate to the next person (Steve to Tony to Bruce to Steve and so on). They all want to appear to be modest, but at the same time, want to have as many cookies as possible. This means that they all try to achieve: 1. Have one person get more cookies than himself, and one person get fewer cookies than himself. 2. Have as many cookies as possible. The first", "John all day • Wednesday - Liz • Thursday - Liz • Friday - Bill To her interest the thefts occur on the same days each week (Monday, Tuesday and Wednesday). By the second Friday she's at her wit's end, she's still not caught the thief - is she doomed to a lifetime of black tea!? She explains her dilemma to her friend one evening who (also being British) realises the seriousness of the situation. Pouring over Jenny's copious notes it looks like a lost cause until her friend smiles. She knows who the thief is! Who is responsible for the milk thefts? • Why should we assume that only one person steals milk? I mean, other than the fact that assuming this gives a unique solution to the problem. Should it be stated, or are you happy to leave that to the solver? Especially given the solution, it's not clear to me why there shouldn't be one person who chooses to steal milk only on Monday, and another who chooses to steal only on Tuesday. – Steve Jessop Nov 10 '14 at 17:41 • I designed the question assuming there only one thief but feel free to prove me wrong! – Liath Nov 10 '14 at 19:02 • What prevents (for example) John from stealing on Monday,", "and get as low of a price as possible. Imagine that Cathy and Fred are bartering over the price. Fred offers $25 dollars for a sack of potatoes. Cathy is just about ready to agree when Emily offers to sell a sack of potatoes for$24 dollars. Fred is not willing to sell at $24 dollars, so he drops out. At this point, Dan can offer to sell for$12. Emily won’t sell for that amount so it looks like the deal might go through. At this point however, Bob steps in and offers $14 dollars. At this point, we have two people who are willing to pay$14 dollars for a sack of potatoes (Cathy and Bob), but only one person (Dan) willing to sell for $14 dollars. So the price must go up because Cathy and Bob are both willing to pay more than$14 dollars. As soon as the price hits $15 dollars, Emily will be willing to sell so there are now two people willing to pay$15 dollars and two people willing to sell at $15 dollars so the trades can happen. But what about Fred and Alice? Well, Fred and Alice are not willing to trade with each other since Alice is only willing to pay$10 and Fred will not sell for any amount under $25. Alice", "+0 # Help very quickly -1 52 4 It is Tim's 4 th birthday and he has a cake with 4 candles on it. Tim makes a wish and tries to blow out the candles. Every time he blows, Tim blows out some number of candles between 1 and the number that remain, including both 1 and that number. How many ways are there for Tim to blow out the candles on the cake? Feb 25, 2020 edited by Guest Feb 25, 2020 #1 -1 Feb 25, 2020 #2 -1 like in 2 min Guest Feb 25, 2020 #3 +6179 +1 $$\\text{It sounds like your asking for the number of ways you can partition the integer 4}\\\\ \\text{using the digits 1, 2, 3, and 4}\\\\ (1,1,1,1)\\\\ (2,1,1) \\text{ and it's other 2 permutations}\\\\ (2,2)\\\\ (1,3) \\text{ and it's other permutation}\\\\ (4)\\\\ \\text{for a total of 1+ 3+1+2+1 = 8}$$ . Feb 25, 2020 #4 0 24-1=8 Mar 3, 2020", "eats roast beef. On Tuesday he goes down to the Kawanis Club and eats roast beef. On wednesday, he goes to the Rlotary Club and eats roast beef. On Thursday and Friday, the same routine, he goes to some club and eats roast beef. Well now he’s gotten along pretty good in society end makes enough income that he can afford to be up in the world. He comes home and demands a steak or at least a half a pound of ground meat of some kind for dinner. And he eats this, which is good. And finally he tosses the baby up \\in the air and plays with him for a minute and then flops over on the couch and somebody turn on the idiot box in front of him to entertain him for a little while. He sits there and reads or watches the idiot box for a while then, along abut 9:30 he says, \"Oh gosh, I'm so hungry. I've got to have something to eat\". He demands the wife to run out and get him a big bowl of chocolate ice cream. He eats that and goes on and dozes on and off between the late shows. Finally along about midnight, he'll get up and has I peanut butter sandwich or a ham sandwich", "the kind kids squish >> Monday 4/26 @ 12:20pm Congratulations to Marina from San Francisco for correctly guessing Secret Sound #3 and winning $4,000! Secret Sound #3 = Breaking open a fortune cookie & taking the fortune out SECRET SOUND #3 – GUESSES GUESS #36: Breaking open a fortune cookie and pulling out the fortune >> Monday 4/26 @ 9:20am GUESS #35: Taking a lid off a Ben & Jerry’s ice cream >> Friday 4/23 @ 5:20pm GUESS #34: Opening a foil chocolate egg >> Friday 4/23 @ 4:20pm GUESS #33: Opening or unwrapping a granola or cereal bar >> Friday 4/23 @ 1:20pm GUESS #32: Opening up your Sweepstakes magazine drive thing >> Friday 4/23 @ 12:20pm GUESS #31: Taking tape of or putting tape on a box >> Friday 4/23 @ 9:20am GUESS #30: Ripping two pieces of velcro >> Thursday 4/22 @ 5:20pm GUESS #29: Opening a bag or box of Red Vines & pulling a piece of licorice out >> Thursday 4/22 @ 4:20pm GUESS #28: Opening a bill or letter with a metal opener >> Thursday 4/22 @ 1:20pm GUESS #27: Unwrapping and eating/biting/chewing a fortune cookie >> Thursday 4/22 @ 12:20pm GUESS #26: Scraping ice >> Thursday 4/22 @ 9:20am GUESS #25: Taking off the plastic strip off a Ben & Jerry’s", "# Allan makes a promissory note for $1000 payable to the order of Bob. Cathy steals the note from Bob, Allan makes a promissory note for$1000 payable to the order of Bob. Cathy steals the note from Bob, indorses (forgets) Bob's name on the back, and gives it to David, in exchange for a TV set. David negotiates the note for value to Emma. Emma then presents the note to Allan for pay,ent. Allan refuses to pay the note because Bob has told him that the note had been stolen. Based on this scenario, which of the following is FALSE: A) Emma can recover the amouont of the note against David because he warranted to B) Allan is no longer liable on his underlying obligation to Bob C) After David pays emma, he can recover against Cathy D) If David can't find Cathy, he must bear the risk of loss caused by Cathy's", "specifically, how could he give$7.14 to the clerk but not have been able to give $2.14, which would make things easier on everybody? There’s no combination of bills — in United States or, so far as I’m aware, any major world currency — in which you can give seven dollars but not two dollars. He had to be handing over five dollars he was getting right back. The clerk would be right to suspect this. It looks like the start of a change scam, begun by giving a confusing amount of money. Had Adams written it so that the charge was$6.89, and Dilbert “helpfully” gave $12.14, then Dilbert wouldn’t be needlessly confusing things. Dave Whamond’s Reality Check for the 15th is that pirate-based find-x joke that feels like it should be going around Facebook, even though I don’t think it has been. I can’t say the combination of jokes quite makes logical sense, but I’m amused. It might be from the Reality Check squirrel in the corner. Nate Fakes’s Break of Day for the 16th is the anthropomorphized shapes joke for this essay. It’s not the only shapes joke, though. Doug Bratton’s Pop Culture Shock Therapy for the 16th is the Einstein joke for this essay. Rick Detorie’s One Big Happy rerun for the 17th is another shapes", "the complicated UST food court ordering process, and Warren endears himself to Geoff by producing a steaming bowl of spicy ramen as if by magic. The contestants have a ‘cultural night’, which apparently includes a greater number of hedge fund representatives than one might have expected. For me, it’s a night in with Geoff, green tea and the scripts for Q2. Joe and Neel have filled fourteen pages between them checking a construction in glorious detail, a step which Harvey has described in its entirety with the words ‘glue them together’. Overall, they are complicated but precise, and I have few concerns, so it’s only necessary to burn the candle at one end. Wednesday 13th July It’s time for coordination, where Geoff and I agree the UK marks with a team of local and international experts. The scheduling has assigned us the Q1 geometry early in the morning, which is a clear case of five perfect solutions, so we move to Q2. Coordinator Stephan seems very well-prepared for the UK scripts, so again we are finished in a matter of minutes. This allows us to bring forward our discussion of Q4. Jacob has made several small errors, all of which could be fixed by attacking his script with a pair of scissors and some glue. I believe the", "Beatrice nervously afterwards. Beatrice refused to say but asked, “How many shall I be getting?”. Alice refused to say and again asked, “How many shall I be getting?”. You should know that each lady is a perfect logician, who never asks a question she knows or can deduce the answer to. I think this proves that a square deal on the hippopotonews is equal to the sum of the squaws on the other two sides. At any rate how many hippos was each to receive? The puzzle as presented above is flawed, in that the situation described would not arise. An apology was published with Tantalizer 460. [tantalizer456] ## Tantalizer 464: Pentathlon From New Scientist #1015, 26th August 1976 [link] The Pentathlon at the West Wessex Olympics is a Monday-to-Friday affair with a different event each day. Entrants specify which day they would prefer for which event — a silly idea, as they never agree. This time, for instance, there were five entrants. Each handed in a list of events in his preferred order. No day was picked for any event by more than two entrants. Swimming was the only event which no one wished to tackle on the Monday. For the Tuesday there was just one request for horse-riding, just one for fencing and just one for", "page numbers on the pages. Now the pages have gotten all mixed up. In how many different ways can Arnold arrange the 8 pages? 5. The special lunch at Bamboo Restaurant gives you a choice of won-ton or hot-sour soup, a choice of kung pao shrimp or chicken with broccoli, and a choice rice or noodles. How many different special lunches are there? 6. Rex forgot the password for his ATM bank card. He knows that the password is made of the 4 digits of his birthday, October 24 or 1, 0, 2, 4.? How many different passwords does he need to try to be sure he gets his password? 7. Javier wrote out wrote out the letters of Jasmine’s name on 7 cupcakes and put them in a box. How many different ways can he take them out of the box one by one? 8. On the new reality show, Lazybones, the five finalists – Snoozin’ Betty, Lounge Man, The Yawn Meister, Lana Later, and Bob the Procrastinator – compete to see who is the laziest person. On today’s show, the five will be weeded down to 3 super-finalists. How many different ways can the 3 super-finalists be chosen? 9. Four friends have printed the letters M, E, T, S on the front of their shirts. They’re", "and$30. If the price is above $30, Cathy is not interested, since the price is too high. If the price is below$25, Fred is not interested since the price is too low. Of course, just because a trade is possible, doesn't mean it will happen. Each of the sellers will try and get as high of a price as possible, and each of the buyers will try and get as low of a price as possible. Imagine that Cathy and Fred are bartering over the price. Fred offers $25 dollars for a sack of potatoes. Cathy is just about ready to agree when Emily offers to sell a sack of potatoes for$24 dollars. Fred is not willing to sell at $24 dollars, so he drops out. At this point, Dan can offer to sell for$12. Emily won't sell for that amount so it looks like the deal might go through. At this point however, Bob steps in and offers $14 dollars. At this point, we have two people who are willing to pay$14 dollars for a sack of potatoes (Cathy and Bob), but only one person (Dan) willing to sell for $14 dollars. So the price must go up because Cathy and Bob are both willing to pay more than$14 dollars. As soon as the price hits $15", "didn't do it. Bill did it alone. Bill: I didn't do it. Caroline did it. Caroline: I didn't do it. Bill did it. Detective Radstein also discovered that all three suspects are members of a clan called the Halfsies. Every time they speak, they make two statements, one of which is a lie and the other of which is true. Who committed the robbery? Problem 3. One day you meet your friend Alice enjoying a nice walk with her husband Bob and their son Carl. They are excited to see you and they invite you to their party. Alice: Please, come to our party on Sunday at our place at 632 Elm St. in Watertown. Bob: My wife likes exaggerating and multiplies every number she mentions by 2. Carl: My dad compensates for my mom's exaggerations and divides every number he mentions by 4. Alice: Our son is not like us at all. He doesn't multiply or divide. He just adds 8 to every number he mentions. Where is the party? Problem 4. There are n ants placed randomly on a stick that is 1 meter long. The ants immediately start running along the stick, choosing a random direction, either left or right, with the same constant speed equal to 1 meter per second. When two ants bump", "Joe obtained a box of antique Lenox china dishes that had been left at the Mashpee town dump. He supplemented the sizable but incomplete set of dishes with other Lenox pieces found at antique dealers. At dinner parties, he proudly told of the origin of his china. When Marlene discovered that Joe had taken her dishes from the dump, she hired an attorney to obtain their return. What result Free Essay Answer: Answer: M relinquished her title (ownership) of plates by dumping it in a trash site. Hence, the plates are now \"finders keepers\" that is title is given to whoever finds these plates. M will have to argue that she had no intention of abandoning it, retaining title, e.g. maybe she dumped the plates when running away from a thief. But absent such evidence the plates are transferred to the new owner. Tags Choose question tag Mona found a wallet on the floor of an elevator in the office building where she worked. She posted several notices in the building about finding the wallet, but no one appeared to claim it. She waited for six months and then spent the money in the wallet in the belief that she owned it. Jason, the person who lost the wallet, subsequently brought suit to recover the money. Mona's defense", "3 3/4 chapters in his book on Monday. He then read 2 4/6 more chapters on Tuesday. How many chapters has he read so far? • Soup On Monday we cook 25 pots and 10 boilers of soup. On Tuesday 15 pots and 13 boilers. On Wednesday 20 pots and on Thursday 30 boilers. On Monday and Tuesday was cooked the same amount of soup. How many times more soup cooked on Thursday than on Wednesday? • 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? • One half 2 One half pizza will be divide among 3 pupils. Each pupil receive 1/6. Is it true or false? • 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? • Pizza palace Josh is at Enzo's pizza palace. He can sit at a table with 5 of his friends or at a different table with seven of his friends. The same size pizza is shared equally among the people at each table. At which table should Josh sit to get more pizza? (write t", "# To Catch a Thief In Alfred Hitchcock's movie \"To Catch a Thief\", the retired infamous jewel thief John Robie has to crack a safe with a complicated $4\\times4$ pattern of keys. Each of the 16 keys is either turned around ($+$) or not turned around ($-$): - - + - + - - + - + - - + - - + To open the safe, John Robie must make all keys turned around. The catch is that whenever Robie turns one key (from $+$ to $-$ or from $-$ to $+$), then all the keys in the same row and all the keys in the same column simultaneously turn (from $+$ to $-$ or from $-$ to $+$ respectively). John Robie may turn any key as often as he likes. Question: What is the minimum number of keys that John Robie has to turn to open the safe? • Don't think it's a duplicate, but this question is also answered in the comments here. Dec 4 '15 at 13:34 • Follow-up: What is the maximum number of turns required for any grid? Dec 4 '15 at 15:49 2 turns To turn a single key without changing the rest of the keys you have to turn the key itself plus the three other keys in the same", "baked a cake and left it on the table to cool. Before going for her bath, she told her four children that they should not touch the cake as it was to be cut only the next day. However when she got back from her bath she found that someone had eaten a ... the truth and exactly one of them actually ate the piece of cake, who is the culprit that Amma is going to punish? Lakshmi Ram Aruna Varun 1 vote 20 A valuable sword belonging to the Grand King was stolen, and the three suspects were Ibn, Hasan, and Abu. Ibn claimed that Hasan stole it, and Hasan claimed that Abu stole it. It was not clear that one of them stole it, but it was later learnt that no innocent person had lied. It was also learnt that the sword was stolen by only one person. Who stole the sword? Ibn Hasan Abu None of them 1 vote 21 Lavanya has to complete $12$ courses for her degree. There are six compulsory courses: Basic and Advanced Mathematics, Basic and Advanced Physics and Basic and Advanced Electronics. She also has to complete six Optional Courses. Each course takes one semester to complete. There are ... , what is the minimum number of semesters that Lavanya", "price. Fred offers$25 for a sack of potatoes. Before Cathy can agree, Emily offers a sack of potatoes for $24. Fred is not willing to sell at$24, so he drops out. At this point, Dan offers to sell for $12. Emily won't sell for that amount so it looks like the deal might go through. At this point Bob steps in and offers$14. Now we have two people willing to pay $14 for a sack of potatoes (Cathy and Bob), but only one person (Dan) willing to sell for$14. Cathy notices this and doesn't want to lose a good deal, so she offers Dan $16 for his potatoes. Now Emily also offers to sell for$16, so there are two buyers and two sellers at that price (note that they could have settled on any price between $15 and$20), and the bartering can stop. But what about Fred and Alice? Well, Fred and Alice are not willing to trade with each other, since Alice is only willing to pay $10 and Fred will not sell for any amount under$25. Alice can't outbid Cathy or Bob to purchase from Dan, so Alice will not be able to get a trade with them. Fred can't underbid Dan or Emily, so he will not be able to get a trade with Cathy." ]

[ "+0 # What is the largest five-digit integer whose digits have a product equal to the product (7) (6)(5)(4)(3)(2)(1)? 0 129 3 What is the largest five-digit integer whose digits have a product equal to the product (7)(6)(5)(4)(3)(2)(1)? Dec 10, 2018 #1 0 Just re-arrange your 7 numbers slightly by multiplying 2 x 4 = 8 and put them in order from the biggest to the smallest and you get this: 87,653. Dec 10, 2018 #2 0 98752 still works but is greater than 87653 Guest Dec 10, 2018 #3 +647 0 We want the biggest numbers, and we can get a 9 from $$6\\times 3$$ which is $$9\\times 2$$ . Therefore, now we have the biggest digit as 9 with 2, 2, 4, 5, and 7 left over. The next largest digit is probably 8 as $$2\\times 4=8$$ , leaving us with 2, 5, and 7. So far we have 98___ and we can see that $$2\\times 5=10, 2\\times 7=14, 5\\times7=35$$and the tens digits are all less than 7 so now we go decreasing order and get 98752. You are very welcome! :P Dec 11, 2018", "any of the six boxes, so it is $6^5$ (called product rule in combinatorics). -", "sum of the digits of $B + 6$ must be 12. So $B = 6 \\longrightarrow (6,6)$. The other possible pairs are $(5,7)(4,8)(3,9)(2,1)(1,2)$. Quickly taking the product of these, we find that $(6,6) \\Longrightarrow 36$ gives us the largest product of $AB$.", "# Product $$=$$ Sum Find the largest 3-digit number whose sum of digits is equal to the product of it's digits. ×", "to be the largest integer * representable in wordsize - 5 bits. */ #undef INFINITY #define INFINITY ((1 << (8*sizeof (int) - 6)) - 4) #define MINFINITY (-INFINITY)", "The highest digit is $9$ and $9^5=59049$, which has five digits. If we then look at $5 \\times 9^5=295245$, which has six digits and a good endpoint for the loop. The loop itself cycles through the digits of each number and tests whether the sum of the fifth powers equals the number. Download the most recent version of this code on GitHub. largest &amp;amp;lt;- 6 * 9^5 for (n in 2:largest) { power.sum &amp;amp;lt;-0 i &amp;amp;lt;- n while (i &amp;amp;gt; 0) { d &amp;amp;lt;- i %% 10 i &amp;amp;lt;- floor(i / 10) power.sum &amp;amp;lt;- power.sum + d^5 } if (power.sum == n) { print(n) } }", "Logic Level 4 Given any positive integer $$n$$, let $$p(n)$$ be the product of the non-zero digits of $$n$$. If $$n$$ has only one digit, then $$p(n)$$ is equal to that digit. $$Let: S = p(1) + p(2) + p(3) +\\ldots + p(999)$$. What is the largest prime factor of $$S$$? ×", "five numbers is 4, then their product is ${4^5}$.", "# Maximizing the Product: A Solution In my last post, I asked What is the largest possible product you can make with a group of positive integers if the sum of those integers is 2017? Before we rush to the numerical answer, let’s follow my usual advice and invert the problem. What would the best possible product look like? Would it use lots of factors, or few of them? What types of integers would it include? Perhaps more importantly, what would it not include? One easy observation to get us going is that the best possible product would not make use of any 1s. After all, a 1 doesn’t help make the product any bigger. If we were using a 1, then we’d be better off folding it into one of the other numbers. For example, if I had a 1 and a 5, I could replace the pair of them with a 6, leaving the sum of the integers the same but making the product 20% larger. Next, we can see that the best possible product would not make use of any large integers. The reason is that large integers can always be split into smaller ones with a larger product. For example, why use a 10 when two 5s have a bigger product? It’s not immediately", "# Greatest Common Divisor of Integers/Examples/-5 and 5/Proof 2 The greatest common divisor of $-5$ and $5$ is: $\\gcd \\set {-5, 5} = 5$ $\\gcd \\set {-5, 5} = 5$ $\\blacksquare$", "# Can we identify the largest product of two numbers made from four given digits My problem is, given 4 positive integers (single digits 0-9), is there a generic way to identify which combination of them will yield the largest product of a single multiplication? So to clarify if my digits were: 3, 8, 1, 6 I think the largest number I could make given a single multiplication would be 8 * 631 = 5048 or in generic terms, the largest single digit multiplied by a combination of the rest of the digits in a sequence of decreasing magnitude. 1. is that correct? 2. is there a way to formally state that? Sort of a proof. - $83\\cdot61 = 5063$. – MJD Jan 22 '13 at 14:35 $81*63 = 5103 > 5048$. I suspect this gives the highest product: the larger two numbers as tens digits, then the smaller two numbers as ones digits, but switched (so the smallest digit goes with the largest digit). – Jonathan Christensen Jan 22 '13 at 14:35 @MJD - well, I guess that answers #1. – Mike Jan 22 '13 at 14:36 And $8631$ is larger yet, taking the empty multiplier to be $1$ as is often done. – Ross Millikan Jan 22 '13 at 14:43 @RossMillikan - What is an empty", "is an integer between 4 and Size (inclusive).", "Interesting Property Of Seven! Number Theory Level 4 Note that $$7!=10 \\times9 \\times8\\times7.$$ What is the largest possible value of $$n$$ for which $$n!$$ can be represented as a product of $$n-3$$ consecutive integers? ×", "= 1 - (5^4)/9000. – fleablood Sep 3 '18 at 0:03 As you say, if there is at least one even digit the product will be even, so count how many four digit numbers have all odd digits. There are five odd digits, so there are $5^4=625$ of them. The fraction of even products is then $\\frac {9000-625}{9000}$", "# So Many Exponents! $\\Huge {\\color{#3D99F6}9}^{{\\color{#20A900}8}^{{\\color{#D61F06}7}^{{\\color{#624F41}6} ^{\\color{magenta}5}}}}$ What are the last two digits when this integer fully expanded out? ×", "# Non-Divisibility Find the largest 2 digit value of $$n$$ such that $$n(n+1)(2n+1)$$ is not divisible by 18. ×", "is the largest of the four integers? A: B: C: D: E: Back to the top", "× ## Factorials 2! = 2, 3! = 3*2, 4! = 4*3*2… and 100! is a lot better than writing out 158 digits. 90! is the largest factorial that can fit in a tweet. # Wilson's Theorem What is $$28 ! \\pmod{29} ?$$ What is $$59 ! \\pmod{61} ?$$ What is $$(18 !)^2 \\pmod{19} ?$$ What is $$(10 !)^ {84} \\pmod{11} ?$$ If $$(n-1)! \\equiv -1 \\pmod{n} ,$$ where $$n$$ is an integer larger than one, which of the following is true? ×", "# Powers of 5 Level pending Let $$k$$ be a nonnegative integer. What is the largest value for which the units digit of both $$6^{k}$$ and $$5^{k}$$ is neither 5 or 6? ×", "# small integers that are or might be the largest of their kind 1 is the largest sum-product number in binary, and the largest to be a sum-product number in any standard positional base. 1 is the largest integer whose Zeckendorf representation has no significant zeroes. Also, it is the largest (and the only) integer to be labelled non-prime without ever being a probable prime with unknown factorization. 2 is the largest (and the only) even prime number. Also, it might be the largest $n$ such that no $n\\times n$ magic square consisting of consecutive primes can be constructed. The total of such magic squares is only known up to $n=6$. 3 is the largest and only prime perfect totient number. 4 is the largest composite number such that the union of the set of its totatives and the set of its divisors is the complete range of integers from 1 to itself. All larger numbers satisfying this http://planetmath.org/RelationshipBetweenTotativesAndDivisorsrelationship between their totatives and divisors are prime. In another commonality with primes, 4 might be the largest composite $n$ such that $\\phi(n)\\sigma_{0}(n)+2$ is a multiple of $n$, with $\\phi(n)$ being Euler’s totient function and $\\sigma_{0}(n)$ being a count of the divisors. (For a prime $p$, $\\phi(p)\\sigma_{0}(p)+2=2p$). 5 might be the largest untouchable number to be odd and prime. If" ]

[ "them a lot. By inspection, we find that $\\text{LCM}(1,2,3,2\\cdot2,2\\cdot3) = 12$. Finally, $1+2 = \\boxed{\\textbf{(B)}\\ 3}$. ## Solution 2 We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$,$2$, $3$, or $4$. We quickly consider $12$ since it is the smallest number that is the LCM of $1$, $2$, $3$ and $4$. Since $12$ has $5$ single-digit divisors, namely $1$, $2$, $3$, $4$, and $6$, our answer is $1+2 = \\boxed{\\textbf{(B)}\\ 3}$", "+ 0.5) = -12.0$$ (base10). See the following figure for details. TRY IT! What is 15.0 (base10) in IEEE754? What is the largest number smaller than 15.0? What is the smallest number larger than 15.0? Since the number is positive, $$s = 0$$. The largest power of two that is smaller than 15.0 is 8, so the exponent is 3, making the characteristic $$3 + 1023 = 1026 (base10) = 10000000010(base2)$$. Then the fraction is $$15/8-1=0.875(base10) = 1\\cdot \\frac{1}{2^1} + 1\\cdot \\frac{1}{2^2} + 1\\cdot \\frac{1}{2^3}$$ = 1110000000000000000000000000000000000000000000000000 (base2). When put together this produces the following conversion: 15 (base10) = 0 10000000010 1110000000000000000000000000000000000000000000000000 (IEEE754) The next smallest number is 0 10000000010 1101111111111111111111111111111111111111111111111111 = 14.9999999999999982236431605997 The next largest number is 0 10000000010 1110000000000000000000000000000000000000000000000001 = 15.0000000000000017763568394003 Therefore, the IEEE754 number 0 10000000010 1110000000000000000000000000000000000000000000000000 not only represents the number 15.0, but also all the real numbers halfway between its immediate neighbors. So any computation that has a result within this interval will be assigned 15.0. We call the distance from one number to the next the gap. Because the fraction is multiplied by $$2^{e-1023}$$, the gap grows as the number represented grows. The gap at a given number can be computed using the function spacing in numpy. import numpy as np TRY IT! Use the spacing function to determine the gap at", "## Discrete Mathematics and Its Applications, Seventh Edition To obtain the greatest common divisor of each pair of integers, multiply together the factors shared by both numbers. a) $2^2\\cdot 3^3\\cdot 5^2=2700$ b) $2\\cdot 3\\cdot 11=66$ c) $17$ d) $1$ e) These numbers are relatively prime. f) $2\\cdot 3\\cdot 5\\cdot 7=210$", "## Basic College Mathematics (10th Edition) Published by Pearson # Chapter 3 - Adding and Subtracting Fractions - 3.2 Least Common Multiples - 3.2 Exercises - Page 213: 51 #### Answer Depending on the number each method has its benefits. #### Work Step by Step 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", "### 1976 IMO #4 Determine the largest number which is the product of positive integers with sum 1976. The answer is $\\boxed{2 \\cdot 3^{658}}$. Note that there cannot be any integers $n>4$ in the maximal product, because we can just replace $n$ by $3,n-3$ to achieve a greater product. Also, we would not want any $1$s in the maximal product as they would not help in increasing the product and would just take up space in the sum. Notice that we can replace any $4$s by two $2$s leaving the product and the sum unchanged. Lastly, we don't want more than two $2$s, because we can just replace three $2$s by two $3$s to obtain a larger product. Therefore, the maximum product must consist of $3$s and zero, one or two $2$s. Since $1976 = 3\\cdot 658 + 2$, we have one $2$ and $658$ $3$'s in our maximum product, giving the answer of $2\\cdot 3^{658}$, as desired. $\\square$", "be greater. For any $$n \\in \\mathbb{N}$$ there are $$n-1$$ many ways to see $$n$$ as a concatenation of two numbers. For example the 5-digit number 75319 can be seen as: 7-5319, 75-319, 753-19, or 7531-9 (this is a random example, in the work below each part of the concatenation is a prime number). Let me show how to estimate the number of ways to write a 5-digit number as a concatenation of primes, then the reader can see how I got the estimate for an $$n$$-digit number. For a 5-digit number, the first concatenation type is a single digit number followed by a 4-digit number. There are 4 single digit prime numbers, and there are approximately $$\\frac{9^4}{\\ln(10^4)}$$ primes which are 4-digits long. Thus there are approximately $$\\frac{4 \\cdot 9^4}{\\ln(10^4)}$$ many 5-digit numbers which are of the first concatenation type. There are just as many of the concatenation type which is a 4-digit number followed by a single digit number. For the concatenation type of a 5-digit number which is a 2-digit number followed by a 3-digit number, there are approximately $$\\frac{9^2 \\cdot 9^3}{\\ln(10^2) \\cdot \\ln(10^3)}$$ many 5-digit numbers of this concatenation type. There are the same amount of 5-digit numbers which are of the concatenation type which is a 3-digit number followed by a 2-digit number. Thus,", "largest 3-digit number is just less than 1000 and the largest 5-digit number is just less than 100 000, so the largest possible product is just less than $1{0}^{3}×1{0}^{5}=1{0}^{8}$ – and so has 8 digits.) (ii) Largest $m+n$, smallest $m+n-1$. (The smallest m-digit number is $1{0}^{m-1}$ and the smallest n-digit number is $1{0}^{n-1}$, so the smallest possible product is $1{0}^{m+n-2}$ – and so has $m+n-1$ digits. The largest m-digit number is just less than 10m and the largest n-digit number is just less than 10n, so the largest possible product is just less than $1{0}^{m}×1{0}^{n}=1{0}^{m+n}$ – and so has $m+n$ digits.) (b) (i) ${2}^{10}=1024$ is very slightly larger than 103. Hence ${2}^{20}=102{4}^{2}$ is very slightly larger than 106, so has 7 digits. (ii) 220 is very slightly larger than 106. In fact $\\begin{array}{c}{\\left(1{0}^{3}+24\\right)}^{2}=1{0}^{6}+2×1{0}^{3}+2{4}^{2}=1{0}^{6}+2×1{0}^{3}+576=1\\phantom{\\rule{0.167em}{0ex}}002\\phantom{\\rule{0.167em}{0ex}}576.\\hfill \\end{array}$ ${\\left(\\genfrac{}{}{0.1ex}{}{1}{2}\\right)}^{20}$ is its reciprocal, so is slightly smaller than $1{0}^{-6}=0.000001$, so it starts with six 0s after the decimal point and rounds up to 0.000001 (to 6 d.p.). (c) No. (It can be equal to the product of its digits if it has just 1 digit. If a number N has k digits, with leading digit = m, then $N\\ge m×1{0}^{k-1}$, but the product of its digits is at most $m×{9}^{k-1}$.) (d) (i) 3, 6. (${2}^{15}×{5}^{3}={2}^{12}×1{0}^{3}=4096×1{0}^{3}$) (ii) 4, 4. (Most of us will need", "$2$, $a^2$ must have prime factorizations wherein each unique prime number will have an even exponent. Let’s have an example to amplify what I meant above. Suppose $a = 3,780$. Breaking it down as a product of prime numbers, we get $a = 2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7$. We can condense the prime factorization by rewriting it as $a = {2^2} \\cdot {3^3} \\cdot 5 \\cdot 7$. Notice how we removed duplicates of prime numbers by expressing it as factors of primes with each unique prime having the appropriate power. The exponent tells how many times the prime number appears in the prime factorization. Notice that in the prime factorization of integer $\\color{blue}\\large{a}$, the prime numbers can either have an odd or even exponent. This is an important observation that we will take advantage of later. The next step is to square the integer $\\color{blue}\\large{a}$, thus we have $\\color{blue}\\large{a^2}$. Apply the Power of a Power Rule of Exponent. We will use this property to square the integer $\\color{blue}\\large{a}$. In a nutshell, this exponent rule will allow us to distribute the main exponent $2$ to the exponents of the unique prime numbers. After squaring the integer $\\color{blue}\\large{a}$, the exponents of the unique prime factors of $\\color{blue}\\large{a}$ are now ALL even numbers.", "× # Help,its so hard for me Determine the biggest integers 85-digit number which satisfies the sum of all digits equal to the multiplication of all numbers Note by Valian Fil Ahli 4 years ago Sort by: Let the digits be $$a_{85}\\geq a_{84}\\geq \\cdots \\geq a_1$$ such that $$a_1+a_2+\\cdots+a_{85}=a_1a_2\\cdots a_{85}$$, then $\\frac{a_1}{a_1a_2\\cdots a_{85}}+\\frac{a_2}{a_1a_2\\cdots a_{85}}+\\cdots+\\frac{a_{85}}{a_1a_2\\cdots a_{85}}=1.$ we have 85 fractions and the greatest of them is the last one, therefore $$\\frac{a_{85}}{a_1a_2\\cdots a_{85}}\\geq \\frac{1}{85}$$ and then $$a_1a_2\\cdots a_{84}\\leq85$$. At most six digits among $$a_1,a_2,\\ldots, a_{84}$$ are greater than 1, therefore, at least 78 digits are equal to 1. - 3 years, 12 months ago Can you please explain the last sentence in your argument? It is not very obvious to me. I think we need to find the integer less than or equal to $$85$$ with highest number of divisors with one digit and I think according to your argument such integer has at most $$6$$ one digit divisors. Right? - 3 years, 12 months ago Each of $$a_1, a_2, \\ldots, a_{84}$$ is a positive integer (zeros are not allowed). Suppose seven digits among $$a_1, a_2, \\ldots, a_{84}$$ are greater than 1, i.e. are $$\\geq2$$, then the product $$a_1a_2\\cdots a_{84}$$ will be $$\\geq 2^7=128$$ which is a contradiction. Therefore, at most six digits among $$a_1, a_2, \\ldots, a_{84}$$ can be", "we can just start working with $7$ and $24$ right away. You can play with these by taking out a factor from one and multiplying it by the other: for example, $\\left(7\\cdot2, \\frac{24}2\\right)$. Now, observe that $59$ is much larger than $7+24$, so we can try borrowing a reasonably large factor from $24$ to multiply by $7$. Of all the possibilities (namely $1,2,3,4,6,8,12,24$), $6$ and $8$ are the likely suspects; the others are clearly too small or too big. Then, since $59 \\approx 7\\cdot8$, we can try $8$ first, and indeed it works. In a way, this was a lucky case: $7$ is a prime number, so we didn't have to worry about factoring it (apart from $(1,7\\cdot24)$, which clearly isn't the solution). But hopefully this gives you another perspective on the factoring approach. You could try this $$m=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}$$ Given $7m^2+59m+24$ where $a=7,b=59,c=24$", "# Finding last digits Find the last digit and last two digits of an integer. For finding the last digit of $$7^{99}$$ using Euler's theorem we have for integer $$7$$ and $$10$$ $$7^4 \\equiv 1 \\pmod {10}$$. Since the last digit is the remainder when divided by $$10$$, it gives the answer $$7^{99} \\equiv 1 \\pmod {10}$$ and the last digit is $$3$$. But for finding last two digits we have to find the remainder when the number $$7^{100}$$ is divided by $$100$$. By Euler's theorem $$7^{99}\\equiv 1 \\pmod{100}$$ Now $$7^{100}=7^{99}\\cdot 7 \\equiv 7 \\pmod {10}$$ So the last two digit is $$07$$. Did I make any mistake? Please tell. • By Euler's theorem, $7^{40} \\equiv 1 \\pmod{100}.$ Where did $99$ come from? You used $4$, rather than $9$ in the first problem, so it's odd that you made this mistake in the second problem. – B. Goddard Jun 28 '18 at 15:39 • @B.Goddard thanks I couldn't find it – John757 Jun 28 '18 at 15:42 Euler's $\\phi(100)=40$, not $99$, since $100$ is not prime. You might also look at the Carmichael function $\\lambda(100)=20$, which is (with a small variation) the result of applying Euler to the component prime powers of a composite number, and combining via $\\text{lcm}$, giving you the largest multiplicative order possible under that", "## Recounting the Rationals, part IVb: the Euclidean Algorithm Suppose we have two integers, and we’d like to find their greatest common divisor (GCD). Recall that the greatest common divisor of two integers m and n is exactly that: the greatest integer which is a divisor of both m and n. The obvious way to do this—which is probably the way you learned in elementary school, when reducing fractions—is to factor each integer into primes, and look for overlap. For example, let’s say we want to find the GCD of 450 and 525. We begin by factoring: $\\begin{array}{rcl} 450 & = & 2 \\cdot 3^2 \\cdot 5^2 \\\\ 525 & = & 3 \\cdot 5^2 \\cdot 7 \\end{array}$ Now we pull out as many prime factors as we can which are contained in the factorizations of both numbers. We can’t use 2 at all, since it is only contained in the factorization of 450; we can use a factor of 3, but only one; and we can use two factors of 5. Putting these together, the greatest common divisor of 450 and 525 is $3 \\cdot 5^2 = 75$. This method is intuitive, and works well for relatively small numbers, but (there’s always a “but”, isn’t there?) it fails horribly for larger numbers. For example, what if you", "largest integer that $$mn$$ is necessarily divisible by is always true, we must assume that m is a multiple of 6. The largest integer that $$mn$$ is necessarily divisible by is $$3\\cdot3\\cdot3 = \\boxed{27}$$. Hope this helps, - PM Nov 30, 2018 #3 +701 0 I think my answer above might be a little confusing, so I'll re-state it. Odd number * odd number = Odd. Since n is a multiple of 9, the prime factorization of n (other than 0) has at least two 3s. Since m is a multiple of 3, the prime factorization of m (other than 0) has at least one 3. We cannot consider the possibility that m and n are even because they are not always even. The largest integer that mn is always divisible by is $$3\\cdot3\\cdot3 = \\boxed{27}$$, which is proven by prime factorization. Hope this helps more, - PM PartialMathematician Nov 30, 2018 #4 +701 0 Go try it out yourself. It really works for all m = 6 (mod 9) and n = 0 (mod 9). - PM PartialMathematician Nov 30, 2018 #5 +27664 +3 Here's another way of looking at 1). Nov 30, 2018 #6 +21980 +15 1) We have positive integers a, b and c such that a>b>c. When a, b and c are divided by", "other sites On 2/14/2019 at 4:05 AM, Umurbaba said: Jack only know the range of the digits(biggest -smallest) The range of the digits, not the range of the numbers abcd is specified. The largest possible digit is 9, the smallest is 0 so the range is either 0,1,2,3,4,5,6,7,8,or 9 The largest digit of these must appear in position a and the least in position d. I see no other possible interpretation of that part of the information, unless the question was incorrectly reproduced. ##### Share on other sites If \"range\" is simply $$r:=a-d,$$ then we have: Lemma. $$r \\in \\{7,8,9\\}.$$ Proof. Jack knows that the product $$p:=a\\cdot b\\cdot c \\cdot d$$ is not enough information to determine $$abcd$$. But if the range is one of the numbers $$1,2,4,6$$, then the value $$(a,b,c,d) = (r+1,1,1,1)$$ would be the unique quadruple with $$p = r+1,$$ since $$r+1$$ is a prime, which means that the value of $$p$$ determines the solution uniquely. Also $$r=0$$ is excluded, because of $$(9,9,9,9)$$. For $$r=3$$ the quadruple $$(5,5,5,2)$$, and for $$r=5$$ the quadruple $$(7,7,7,2)$$, are determined by $$p$$. That leaves the three possible values stated. Does that make sense? Edit: $$r=7$$ is excluded, due to $$(9,9,9,2 )$$ as well. Edit: after a further think, $$r=8$$ is also excluded, because of $$(8,0,0,0)$$. Because now James and", "## Basic College Mathematics (10th Edition) Method of finding the least common multiple is convenient if the numbers are small so one can quickly find the least common multiple by inspection. Example: The common multiple for $2$ and $6$ is $6$, because $2 \\cdot 1 = 2, 2 \\cdot 2 = 4, 2 \\cdot 3 = 6, \\dots$ and $6 \\cdot 1 = 6, \\dots$. Method of using multiples of the largest number is quick to determine for small numbers but might take more effort for large numbers since the list will be longer. Example: The common multiple of $3$ and $5$ is $15$, because $3, 6, 9, 12, 15, 18, \\dots$ and $5, 10 ,15, 20, \\dots$. Method of using prime factorization is convenient to use for larger numbers where you find the prime factorization for each number and then simply multiply which appear in greatest number of times in either factorization. Example: The common multiple of $3$ and $4$ is $12$, because $3 = 3$ and $4 = 2 \\cdot 2$. So the LCM is $2 \\cdot 2 \\cdot 3 = 12$.", "+0 Number Theory 0 84 1 The expression 120!/(60!*60!) is an integer. What is the largest integer n such that 7^n divides this integer? Apr 29, 2022 #1 +267 +1 Solution: $$1$$ We are looking for the number of sevens there are in the prime factorization of the number. 60! has 9 sevens in its prime factorization, because there are eight multiples of seven in 1x2x3x4...x60, and one of them is 7². Because the 60! is being square, the number of sevens in its prime factorization doubles to 18. 120! has 19 sevens in its prime factorizations, because there are 17 multiples of seven in 1x2x3x4....x120, and two of them are multiples of 49. $$x \\cdot7^{18}\\over y \\cdot 7^{19}$$$$=x/y \\cdot 7$$ where x and y are two numbers that aren't multiples of 7 (and so therefore neither is their quotient). There is only one seven the prime factorization of this number, which means that the largest integer n such that 7^n that divides it is just $$1$$. Apr 29, 2022", "there is absolutely nothing non-mathematical about what you did in your update. Pat yourself on the back and stop worrying that guessing and checking is somehow not a part of mathematics. – Erick Wong Sep 18 '17 at 22:32 $\\sqrt{2017}\\approx\\sqrt{2000}=20\\sqrt{5}\\approx 20\\cdot 2.236 \\approx 45$ and $$44^2 = 1936,\\qquad 45^2=2025$$ hence $\\sqrt{2017}\\in\\color{red}{\\left(44,45\\right)}$. $44^2 = 1936 < 2017 < 2025 = 45^2$. Really, I don't think there's much to this one except for \"try squaring small integers until you find the right ones\". You could use the root extraction algorithm to find it directly. It's sort of like long division. 1. Starting from the decimal place, divide the number into pairs of digits. So $20\\,17$. 2. Find the largest integer whose square is less than the first pair. $4^2 < 20 < 5^2$. This is the first digit of the square root. 3. Subtract off the square and bring down the next two digits. $20 - 4^2 = 4 \\Longrightarrow 417$. This is our remainder. 4. Find the largest $d$ such that the product $(20\\cdot r + d)\\cdot d$ is less than the current remainder, where $r$ is the part of the root already found. $(20\\cdot 4 + 4)\\cdot 4 < 417 < (20 \\cdot 4 + 5)\\cdot 5$, so $d= 4$. 5. Add $d$ to the end of the", "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 _ _ $1$, _ _ $5$, _ _ $7$, _", "# The largest number Find the largest integer such that: 1. No figures is not repeat, 2. multiplication of every two digits is odd, 3. addition all digits is odd. Result x = 9753 #### Solution: $MAX = 9876543210 \\ \\\\ 9\\cdot 7 = 63 \\ \\\\ 7\\cdot 5 = 35 \\ \\\\ 5\\cdot 3 = 15 \\ \\\\ \\ \\\\ \\ \\\\ 9+7+5+3 = 24 \\ \\\\$ 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. Perpetrator The perpetrator is a number that is smaller than the number 80. It is a multiple of five and it is odd. If we added up the tens and ones of this number we get number 8. Added together and write as decimal number: LXVII + MLXIV 3. Sick Sick Marcel already taken six tablets, which was a quarter of the total number of pills in the pack. How many pills were in the pack? 4. Unknown number 16 My number's tens is three times more", "# the largest chain of number I found this problem on the internet: What is the largest chain of numbers that complies with that every number in the chain/list is a divisor divisor in the next number. For example 1 - 6 - 18 - 72 (this one is a chain of 4 numbers) and 5 - 25 - 100 (this one is a chain of 3 number) So the question is how many numbers do the longest chain contain - IF you may not use numbers higher than 1000? I think it is 1 - 2 - 4 - 8 - 16 - 32 - 64 - 128 - 256 - 512, which contains 10 numbers. What do you think? • How about find the number with the largest primefactor? – Nick Podowalski Dec 7 '14 at 13:06 • Consider the quotients $q_{n+1} = \\frac{a_{n+1}}{a_n}$ Then $a_n = a_1\\cdot q_2 \\cdot q_3 \\cdot \\dotsc \\cdot q_n$. So you are looking for the longest sequence $(q_2,q_3,\\dotsc,q_m)$ such that $q_2\\cdot q_3\\cdot \\dotsc\\cdot q_m \\leqslant 1000$, where you have the constraint $q_k \\geqslant 2$. – Daniel Fischer Dec 7 '14 at 13:18 There is no longer chain because if the chain length is $11$ or more, the last number is at least $2^{10}=1024>1000$. • Yes. That is correct, but there" ]

[ "### 1976 IMO #4 Determine the largest number which is the product of positive integers with sum 1976. The answer is $\\boxed{2 \\cdot 3^{658}}$. Note that there cannot be any integers $n>4$ in the maximal product, because we can just replace $n$ by $3,n-3$ to achieve a greater product. Also, we would not want any $1$s in the maximal product as they would not help in increasing the product and would just take up space in the sum. Notice that we can replace any $4$s by two $2$s leaving the product and the sum unchanged. Lastly, we don't want more than two $2$s, because we can just replace three $2$s by two $3$s to obtain a larger product. Therefore, the maximum product must consist of $3$s and zero, one or two $2$s. Since $1976 = 3\\cdot 658 + 2$, we have one $2$ and $658$ $3$'s in our maximum product, giving the answer of $2\\cdot 3^{658}$, as desired. $\\square$", "+ 0.5) = -12.0$$ (base10). See the following figure for details. TRY IT! What is 15.0 (base10) in IEEE754? What is the largest number smaller than 15.0? What is the smallest number larger than 15.0? Since the number is positive, $$s = 0$$. The largest power of two that is smaller than 15.0 is 8, so the exponent is 3, making the characteristic $$3 + 1023 = 1026 (base10) = 10000000010(base2)$$. Then the fraction is $$15/8-1=0.875(base10) = 1\\cdot \\frac{1}{2^1} + 1\\cdot \\frac{1}{2^2} + 1\\cdot \\frac{1}{2^3}$$ = 1110000000000000000000000000000000000000000000000000 (base2). When put together this produces the following conversion: 15 (base10) = 0 10000000010 1110000000000000000000000000000000000000000000000000 (IEEE754) The next smallest number is 0 10000000010 1101111111111111111111111111111111111111111111111111 = 14.9999999999999982236431605997 The next largest number is 0 10000000010 1110000000000000000000000000000000000000000000000001 = 15.0000000000000017763568394003 Therefore, the IEEE754 number 0 10000000010 1110000000000000000000000000000000000000000000000000 not only represents the number 15.0, but also all the real numbers halfway between its immediate neighbors. So any computation that has a result within this interval will be assigned 15.0. We call the distance from one number to the next the gap. Because the fraction is multiplied by $$2^{e-1023}$$, the gap grows as the number represented grows. The gap at a given number can be computed using the function spacing in numpy. import numpy as np TRY IT! Use the spacing function to determine the gap at", "is $5 \\cdot \\dfrac{7}{4} = \\dfrac{35}{4} = 8 \\dfrac{3}{4}.$ Thus, D is the correct answer. 16. How many $$3$$-digit positive integers have digits whose product equals $$24?$$ $$12$$ $$15$$ $$18$$ $$21$$ $$24$$ ###### Solution(s): The only triples of integers less than $$10$$ that multiply to $$24$$ are $(1, 3, 8), (1, 4, 6), (2, 2, 6),$$\\text{and } (2, 3, 4).$ The triples with $$3$$ distinct numbers can be rearranged to form $$6$$ distinct $$3$$-digit positive integers. The other triple can be arranged to form $$3$$ distinct $$3$$-digit positive integers. This leaves a total of $$3 \\cdot 6 + 3 = 21$$ integers. Thus, D is the correct answer. 17. The positive integers $$x$$ and $$y$$ are the two smallest positive integers for which the product of $$360$$ and $$x$$ is a square and the product of $$360$$ and $$y$$ is a cube. What is the sum of $$x$$ and $$y?$$ $$80$$ $$85$$ $$115$$ $$165$$ $$610$$ ###### Solution(s): For a number to be a perfect square, every exponent in the prime factorization must be even. For it to be a cube, the exponents must be divisible by $$3.$$ We can factor $$360$$ to get $360 = 2^3 \\cdot 3^2 \\cdot 5.$ For $$360x$$ to be a perfect square and $$x$$ to be minimized, $$x$$ must have one factor of", "× # Help,its so hard for me Determine the biggest integers 85-digit number which satisfies the sum of all digits equal to the multiplication of all numbers Note by Valian Fil Ahli 4 years ago Sort by: Let the digits be $$a_{85}\\geq a_{84}\\geq \\cdots \\geq a_1$$ such that $$a_1+a_2+\\cdots+a_{85}=a_1a_2\\cdots a_{85}$$, then $\\frac{a_1}{a_1a_2\\cdots a_{85}}+\\frac{a_2}{a_1a_2\\cdots a_{85}}+\\cdots+\\frac{a_{85}}{a_1a_2\\cdots a_{85}}=1.$ we have 85 fractions and the greatest of them is the last one, therefore $$\\frac{a_{85}}{a_1a_2\\cdots a_{85}}\\geq \\frac{1}{85}$$ and then $$a_1a_2\\cdots a_{84}\\leq85$$. At most six digits among $$a_1,a_2,\\ldots, a_{84}$$ are greater than 1, therefore, at least 78 digits are equal to 1. - 3 years, 12 months ago Can you please explain the last sentence in your argument? It is not very obvious to me. I think we need to find the integer less than or equal to $$85$$ with highest number of divisors with one digit and I think according to your argument such integer has at most $$6$$ one digit divisors. Right? - 3 years, 12 months ago Each of $$a_1, a_2, \\ldots, a_{84}$$ is a positive integer (zeros are not allowed). Suppose seven digits among $$a_1, a_2, \\ldots, a_{84}$$ are greater than 1, i.e. are $$\\geq2$$, then the product $$a_1a_2\\cdots a_{84}$$ will be $$\\geq 2^7=128$$ which is a contradiction. Therefore, at most six digits among $$a_1, a_2, \\ldots, a_{84}$$ can be", "\\times \\cdots \\times 100 .$ This product is divisible by some $2^m ,$ where $m$ is an integer. What is the largest possible value of $m?$ ×", "\\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 =", "# The largest number Find the largest integer such that: 1. No figures is not repeat, 2. multiplication of every two digits is odd, 3. addition all digits is odd. Result x = 9753 #### Solution: $MAX = 9876543210 \\ \\\\ 9\\cdot 7 = 63 \\ \\\\ 7\\cdot 5 = 35 \\ \\\\ 5\\cdot 3 = 15 \\ \\\\ \\ \\\\ \\ \\\\ 9+7+5+3 = 24 \\ \\\\$ 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. Perpetrator The perpetrator is a number that is smaller than the number 80. It is a multiple of five and it is odd. If we added up the tens and ones of this number we get number 8. Added together and write as decimal number: LXVII + MLXIV 3. Sick Sick Marcel already taken six tablets, which was a quarter of the total number of pills in the pack. How many pills were in the pack? 4. Unknown number 16 My number's tens is three times more", "# Do I Need To Know Totient Function? $\\large 5 \\times 5 \\times 5 \\times \\cdots \\times 5 = \\ldots 5$ If I multiply 5 by itself any number times, the last digit of the final product always remains unchanged. What is another single digit integer larger than 1 that share this same property? ×", "+0 # What is the largest five-digit integer whose digits have a product equal to the product (7) (6)(5)(4)(3)(2)(1)? 0 129 3 What is the largest five-digit integer whose digits have a product equal to the product (7)(6)(5)(4)(3)(2)(1)? Dec 10, 2018 #1 0 Just re-arrange your 7 numbers slightly by multiplying 2 x 4 = 8 and put them in order from the biggest to the smallest and you get this: 87,653. Dec 10, 2018 #2 0 98752 still works but is greater than 87653 Guest Dec 10, 2018 #3 +647 0 We want the biggest numbers, and we can get a 9 from $$6\\times 3$$ which is $$9\\times 2$$ . Therefore, now we have the biggest digit as 9 with 2, 2, 4, 5, and 7 left over. The next largest digit is probably 8 as $$2\\times 4=8$$ , leaving us with 2, 5, and 7. So far we have 98___ and we can see that $$2\\times 5=10, 2\\times 7=14, 5\\times7=35$$and the tens digits are all less than 7 so now we go decreasing order and get 98752. You are very welcome! :P Dec 11, 2018", "## Discrete Mathematics and Its Applications, Seventh Edition To obtain the greatest common divisor of each pair of integers, multiply together the factors shared by both numbers. a) $2^2\\cdot 3^3\\cdot 5^2=2700$ b) $2\\cdot 3\\cdot 11=66$ c) $17$ d) $1$ e) These numbers are relatively prime. f) $2\\cdot 3\\cdot 5\\cdot 7=210$", "other sites On 2/14/2019 at 4:05 AM, Umurbaba said: Jack only know the range of the digits(biggest -smallest) The range of the digits, not the range of the numbers abcd is specified. The largest possible digit is 9, the smallest is 0 so the range is either 0,1,2,3,4,5,6,7,8,or 9 The largest digit of these must appear in position a and the least in position d. I see no other possible interpretation of that part of the information, unless the question was incorrectly reproduced. ##### Share on other sites If \"range\" is simply $$r:=a-d,$$ then we have: Lemma. $$r \\in \\{7,8,9\\}.$$ Proof. Jack knows that the product $$p:=a\\cdot b\\cdot c \\cdot d$$ is not enough information to determine $$abcd$$. But if the range is one of the numbers $$1,2,4,6$$, then the value $$(a,b,c,d) = (r+1,1,1,1)$$ would be the unique quadruple with $$p = r+1,$$ since $$r+1$$ is a prime, which means that the value of $$p$$ determines the solution uniquely. Also $$r=0$$ is excluded, because of $$(9,9,9,9)$$. For $$r=3$$ the quadruple $$(5,5,5,2)$$, and for $$r=5$$ the quadruple $$(7,7,7,2)$$, are determined by $$p$$. That leaves the three possible values stated. Does that make sense? Edit: $$r=7$$ is excluded, due to $$(9,9,9,2 )$$ as well. Edit: after a further think, $$r=8$$ is also excluded, because of $$(8,0,0,0)$$. Because now James and", "Given a positive integer $$n\\,$$, let $$p(n)\\,$$ be the product of the non-zero digits of $$n\\,$$. (If $$n\\,$$ has only one digit, then $$p(n)\\,$$ is equal to that digit.) Let $$S=p(1)+p(2)+p(3)+\\cdots+p(999)$$. What is the largest prime factor of $$S\\,$$? (第十二届AIME1994 第5题)", "largest 3-digit number is just less than 1000 and the largest 5-digit number is just less than 100 000, so the largest possible product is just less than $1{0}^{3}×1{0}^{5}=1{0}^{8}$ – and so has 8 digits.) (ii) Largest $m+n$, smallest $m+n-1$. (The smallest m-digit number is $1{0}^{m-1}$ and the smallest n-digit number is $1{0}^{n-1}$, so the smallest possible product is $1{0}^{m+n-2}$ – and so has $m+n-1$ digits. The largest m-digit number is just less than 10m and the largest n-digit number is just less than 10n, so the largest possible product is just less than $1{0}^{m}×1{0}^{n}=1{0}^{m+n}$ – and so has $m+n$ digits.) (b) (i) ${2}^{10}=1024$ is very slightly larger than 103. Hence ${2}^{20}=102{4}^{2}$ is very slightly larger than 106, so has 7 digits. (ii) 220 is very slightly larger than 106. In fact $\\begin{array}{c}{\\left(1{0}^{3}+24\\right)}^{2}=1{0}^{6}+2×1{0}^{3}+2{4}^{2}=1{0}^{6}+2×1{0}^{3}+576=1\\phantom{\\rule{0.167em}{0ex}}002\\phantom{\\rule{0.167em}{0ex}}576.\\hfill \\end{array}$ ${\\left(\\genfrac{}{}{0.1ex}{}{1}{2}\\right)}^{20}$ is its reciprocal, so is slightly smaller than $1{0}^{-6}=0.000001$, so it starts with six 0s after the decimal point and rounds up to 0.000001 (to 6 d.p.). (c) No. (It can be equal to the product of its digits if it has just 1 digit. If a number N has k digits, with leading digit = m, then $N\\ge m×1{0}^{k-1}$, but the product of its digits is at most $m×{9}^{k-1}$.) (d) (i) 3, 6. (${2}^{15}×{5}^{3}={2}^{12}×1{0}^{3}=4096×1{0}^{3}$) (ii) 4, 4. (Most of us will need", "# How many 3 digit integers…? How many different 3-digit integers have the product of their digits equal to 4!? What is the largest of these integers? I know 4! Is 24 but still confusing to do this. How do I find the largest let alone how many? • First thing: How many ways are there to factor 24? How many of those are three single digit integer. How many ways are there to arrange those factors into three digit numbers. – fleablood Oct 29 '18 at 16:36 • Finding the largest is easier then counting them. What is the largest single digit factor of $24$. That will be the first digit. If you factor what is left what is the largest single digit factor of that. That will be the second digit. – fleablood Oct 29 '18 at 16:50 How many ways can you write $$24$$ as a product of one digit whole numbers? $$24 = 1 \\cdot 4 \\cdot 6$$ is one way, giving you the numbers $$146$$, $$164$$, $$416$$, $$461$$, $$614$$, and $$641$$. Try to find the others. Just do it. $$4! = 24$$ how many ways are there to factor $$24$$ into three factors. There's $$1*1*24$$ $$1*2*12$$ $$1*3*8$$ ..... etc. How many of those have only single digit factors? There's $$1*3*8$$ $$1*4*6$$ $$2*2*6$$ ...", "# 2006 Indonesia MO Problems/Problem 8 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Find the largest 85-digit integer which has property: the sum of its digits equals to the product of its digits. ## Solution Let $a_n$ be the $n^\\text{th}$ digit from the left, where $1 \\le n \\le 85$. Because we want to maximize the number, we let $a_n \\ge a_{n+1}$ for $1 \\le n \\le 84$. First we'll prove that $a_8 = 1$. Afterward, we'll find the largest value of $a_1$ and use it to find the 85-digit integer. Lemma 1: $a_8 = 1$ The sum of the digits of the 85-digit number is at most $85 \\cdot 9 = 765$. If we let $a_10 = 2$, then the product of the digits is at least $1024$, which is more than $765$. That means $a_{10} = 1$. Since $a_{10}, a_{11}, \\cdots a_{85}$ are all equal to 1, the sum of the digits of the wanted 85-digit number is at most $76 + 9 \\cdot 9 = 157$. If we let $a_8 = 2$, then the product of the digits is at least $256$, which is more than $157$. That means $a_8 = 1. \\blacktriangleright$ Now we'll consider the largest possible values for $a_1$. From the Lemma, there are", "Free Version Easy # Prime Factors of an LCM ABSALG-104LTD What is the largest prime divisor of lcm$(2^5\\cdot 11^2, 5\\cdot 3^3\\cdot 11, 5^2 \\cdot 13)$? A $3$ B $5$ C $11$ D $13$", "integer $\\mathrm{n}$ such that $n^{100}<100!<(n+1)^{100}$ [?] 11. 99[!] or $50^{99}$ [?] (Hint: look back at the warm-up problem involving 50) 12. 100[!] or $50^{100}$ 13. $2^{100\\dagger}$ or $(2^{100})!$ 14. Consider the number 32[!] Which prime numbers divide it[?] It is obviously even, but what is the largest power of 2 that divides 32[!] [?] What is the largest power of 3 that divides it[?] the largest power of 5[?] of 7[?] If you continue questions about consecutive primes, you will finally write 32[!] as a product of prime numbers (similarly to writing, say, $180=2^{2}\\cdot 3^{2}\\cdot 5$). 15. Which is bigger: $2^{5^{1}}$ or $(2^{5})!$ [?] (Hint: use the previous problem.) 16. Generalize Problem 12 and compare $n!$ with $(n\\mathit{1}2)^{n}$ \\begin{center} \\end{center} \\end{document} %end text", "products of decimal numbers. However, a calculator that has only an eight-digit display may not be able to handle numbers or products that result in more than eight digits. But there are plenty of inexpensive ($50 -$75) calculators with more than eight-digit displays. ### Sample Set B Find the following products, if possible, using a calculator. #### Example 4 2.58 8.612.58 8.61 size 12{2 \".\" \"58 \" cdot \" 8\" \".\" \"61\"} {} Display Reads Type 2.58 2.58 Press × 2.58 Type 8.61 8.61 Press = 22.2138 The product is 22.2138. #### Example 5 0.006 0.00420.006 0.0042 size 12{0 \".\" \"006 \" cdot \" 0\" \".\" \"0042\"} {} Display Reads Type .006 .006 Press × .006 Type .0042 0.0042 Press = 0.0000252 We know that there will be seven decimal places in the product (since 3 + 4 = 73 + 4 = 7 size 12{\"3 \"+\" 4 \"=\" 7\"} {}). Since the display shows 7 decimal places, we can assume the product is correct. Thus, the product is 0.0000252. #### Example 6 0.0026 0.119760.0026 0.11976 size 12{0 \".\" \"0026 \" cdot \" 0\" \".\" \"11976\"} {} Since we expect 4 + 5 = 94 + 5 = 9 size 12{\"4 \"+\" 5 \"=\" 9\"} {} decimal places in the product, we know that an eight-digit display calculator", "Geeks With Blogs Buhay Programmer Dont byte more than you can chew Problem: 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest number that is evenly divisible by all of the numbers from 1 to 20? Solution: Finding the LCM can be done via factorization. Below is an excerpt from the wikipedia article. The unique factorization theorem says that every positive integer number greater than 1 can be written in only one way as a product of prime numbers. The prime numbers can be considered as the atomic elements which, when combined together, make up a composite number. For example: $90 = 2^1 \\cdot 3^2 \\cdot 5^1 = 2 \\cdot 9 \\cdot 5. \\,\\!$ Here we have the composite number 90 made up of one atom of the prime number 2, two atoms of the prime number 3 and one atom of the prime number 5. This knowledge can be used to find the lcm of a set of numbers. Example: Find the value of lcm(8,9,21). First, factor out each number and express it as a product of prime number powers. $8\\; \\, \\; \\,= 2^3 \\cdot 3^0 \\cdot 5^0 \\cdot 7^0 \\,\\!$ $9\\; \\, \\; \\,= 2^0 \\cdot 3^2 \\cdot 5^0", "A single number, like $31$ or $7$, is in fact a product as far as mathematics is concerned. It is a product of $1$ integer. Indeed, you can even have a product of $0$ integers. This is defined to be $1$, because $1$ is the identity element of multiplication. (See Qiaochu's comment.) When we say that an integer has a unique prime factorization, we mean it can be written as a product of some nonnegative integer number of primes. Thus, \"$2 \\cdot 2 \\cdot 23$\", \"$31$\", \"$7$\", and \"$\\quad$\" are all valid prime factorizations." ]

[ "# Disappearing Triangles Algebra Level 2 One side of an equilateral triangle is $$32$$ cm. The midpoints of its sides are joined to form another triangle whose midpoints, in turn, are joined to form still another triangle. This process continues indefinitely. Find the sum of the perimeters of all these triangles that are defined above. ×", "than the previous step. ∴ Second step from the bottom needs = 100 – 2 = 98 bricks. Third step from the bottom needs = 98 – 2 = 96 bricks. Fourth step from the bottom needs = 96 – 2 = 94 bricks. Here the numbers are 100, 98, 96, 94, …. Clearly this is an A.P. but not G.P. iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely. Given: An equilateral triangle whose perimeter = 24 cm. Side of the equilateral triangle = $$\\frac{24}{3}$$ = 8 cm. [∵ All sides of equilateral are equal] ……. (1) Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = $$\\frac{8}{2}$$ = 4 cm [∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.] Similarly, the side of third triangle = $$\\frac{4}{2}$$ = 2 cm ∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm, a = 8 Thus each term starting from the second; can be obtained by multiplying its preceding term", "A regular dodecagon and an equilateral triangle, with side length $6$, share three vertices. Find the area of the dodecagon.", "can split the triangle into 4 equilateral triangles with side length 2. Doing the same to each of the smaller triangles formed gives equilateral triangles with side length 1. There are 4$\\times$4 = 16 of these in the large equilateral triangle. So the hexagon contains 6 equilateral triangles of side length 1, whilst the triangle contains 16. So their areas are in the ratio 6:16, which simplifies to 3:8. Scaling the hexagon up 6 copies of the triangle will fit together to make a hexagon with side length 4. So 6 copies of the triangle will make a copy of the hexagon which has been enlarged by a scale factor of 4. That means its area has been enlarged by a scale factor of 4$^2$ = 16. So 6 copies of the triangle have the same area as 16 copies of the hexagon. So the ratio of the area of the hexagon to the area of the triangle is 6:16, which simplifies to 3:8 Scaling the triangle down 6 equilateral triangles of side length 1 fit into the hexagon. These triangles are similar to the equilateral triangle of side length 4, but their sides are 4 times shorter. This means that their area is 4$^2$ = 16 times smaller. So, when scaled down by a factor of 16,", "# An equilateral triangle has a side length of 24 inches. What is its perimeter? May 9, 2018 72 in. #### Explanation: Since this triangle is an equilateral triangle, all sides are equal. You only have to multiply the side you are given by 3. Doing so you get: $24 \\cdot 3 = 72$ Note: This only works for equilateral triangles. With isosceles or scalene triangles, you must have the angular quantities to do the problem. May 9, 2018 $72$ inches #### Explanation: The perimeter is the sum of the sides of a given shape. We know a few things about our shape: • We have a triangle, which has $3$ sides • This is an equilateral triangle, meaning the lengths are equal • One side is $24$ inches If all of the sides are the same, we are essentially adding $24 + 24 + 24$ to get $72$ inches Hope this helps!", "# If two sides of an equilateral triangle measures $(6x+1)$ units and $\\frac{11x+5}{2}$ units, find the perimeter of the triangle. If two sides of an equilateral triangle measures $(6x+1)$ units and $\\frac{11x+5}{2}$ units, find the perimeter of the triangle. I tried the following, Let the third side measure $x$ units Therefore, Perimeter$=(6x+1)+$$\\frac{11x+5}{2}$$+x$ I do not know what to do next. Please help. • The triangle is equilateral so the two sides mentioned are equal. – André Nicolas Aug 30 '14 at 7:26 Since the triangle is equilateral, $6x+1 = (11x+5)/2$. Solve for x. Then, perimeter = $3*(6x+1)$. $$6x+2=\\frac{1}{2}(11x+5)$$ which gives $x=1$. This means that all sides have length $6\\cdot1+2=8$. Hence, the perimeter of the triangle is $3\\cdot8=24$.", "The distance between centres of two adjacent faces of a cube is $8$cm. What is the side length of the cube? I can make an equilateral triangle by cutting off the corner of a cube. If the area of the largest equilateral triangle I can create in this way is $140$cm$^2$, what is the side length of this cube? A third cube has an edge length of $12$cm. At each vertex, a tetrahedron with three mutually perpendicular edges of length $4$cm is cut away. Can you find the volume and the surface area of the remaining solid?", "5-poles though, which the icosphere has a few of. – Ben May 19 at 15:54 • This tuotrial talks a good deal about how building hex meshs, convering the second half of your first bonus as well as your second bonus. May 19 at 18:23 What all hexagonal structures have in common: • They all consist of six equilateral triangles. • The height of a triangle always has the value $$0.86602540378443864676372317075294$$ for a side length of $$1$$. • The hexagons are always offset from each other by $$1.5$$ on one axis, and $$3$$ on the other axis for an equilateral triangle with side length $$1$$. This is convenient because then we don't need to recalculate these values each time, we just create the structure based on these values and scale the result. I obtain the value for the height of the equilateral triangle with the formula $$h={\\frac {\\sqrt {3}}{2}}\\cdot r$$. Here I present three variations, all of which achieve the same result: • One is based only on the creation of a single Mesh Line. • Another is based on creating a Mesh Line, which is instantiated along another. • And the third variant simply creates a Grid where the points are moved. The advantage of all these variants is that the indices are continuous and there is", "existing edges as vectors, we construct an equilateral triangle by connecting three vertices: So $\\alpha = 36$ (since it’s part of a regular pentagon) and $\\beta = 60$. Then $\\cos \\theta = \\frac{\\cos^2 60 - \\cos^2 36}{\\sin^2 36}$ and $\\theta = 116.6^\\circ$.", "### Home > CCAA > Chapter 8 Unit 7 > Lesson CC2: 8.3.4 > Problem8-105 8-105. Use a ruler and a protractor to create a triangle with two sides of $3$ cm each and an angle measuring $60°$. This triangle could be equilateral, meaning all sides are the same length. Additionally, all angles would be the same measure. Try making an equilateral triangle with sides of $3$ cm.", "automatic increase of Rs. 320 in the very next month and each month thereafter. (i) Find his salary for the tenth month. (ii) What is his total earnings during the first year? Q5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job? Sol: Here, a = 5 and d = 2 Q6. The sum of interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, … sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon. Sol: We know that, sum of interior angles of a polygon of side n is (n – 2) x 180°. Q7. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle. Sol: Let the given equilateral triangle be ∆ ABC with each side of 20 cm. By joining the mid-points of this triangle,", "Perimeter of each shape (a) Perimeter of square = 4 × side ⇒ 30 cm = 4 × side ⇒ side = $$\\frac{30}{4}$$ cm = 7.5 cm Thus, the length of each side of square will be 7.5 cm. (b) Perimeter of equilateral triangle = 3 × side ⇒ 30 cm = 3 × side ⇒ side = $$\\frac{30}{3}$$ cm = 10 cm Thus, the length of each side of equilateral triangle will be 10 cm. (c) Perimeter of regular hexagon = 6 × side ⇒ 30 cm = 6 × side ⇒ side = $$\\frac{30}{6}$$ cm = 5 cm Thus, the length of each side of regular hexagon will be 5 cm. Question 12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side? Solution: Let the length of third side be x cm. Length of other two sides are 12 cm and 14 cm. Now, perimeter of triangle = 36 cm ⇒ 12 + 14 + x = 36 ⇒ 26 + x = 36 ⇒ x = 36 – 26 ⇒ x = 10 Thus, the length of third side is 10 cm. Question 13. Find the cost of fencing a square park of side 250 m at the rate of", "# The length of each side of an equilateral triangle is increased by 5 inches, so the perimeter is now 60 inches. How do you find the original length of each side of the equilateral triangle? The original sides were 15 inches. #### Explanation: Each side of an equilateral triangle starts with length $x$. They are all extended by 5 inches (so now each side is $x + 5$ and the sum of all three sides together is 60. So we have: $3 \\left(x + 5\\right) = 60$ $3 x + 15 = 60$ $3 x = 45$ $x = 15$ The original sides were 15 inches.", "# Equilateral triangles,cones and fun Geometry Level pending Each edge of an equilateral triangle is $$'a'$$ $$cm$$.A cone is formed by joining any two sides of the triangle.What is the volume of the cone(in $$cm^3)$$ ? ×", "The perimeter of an equilateral triangle is $42$ miles. Find the length of each side. The perimeter of an isosceles triangle is $42$ feet. The length of the shortest side is $12$ feet. Find the length of the other two sides. 15 ft The perimeter of an isosceles triangle is $83$ inches. The length of the shortest side is $24$ inches. Find the length of the other two sides. A dish is in the shape of an equilateral triangle. Each side is $8$ inches long. Find the perimeter. 24 in. A floor tile is in the shape of an equilateral triangle. Each side is $1.5$ feet long. Find the perimeter. A road sign in the shape of an isosceles triangle has a base of $36$ inches. If the perimeter is $91$ inches, find the length of each of the other sides. 27.5 in. A scarf in the shape of an isosceles triangle has a base of $0.75$ meters. If the perimeter is $2$ meters, find the length of each of the other sides. The perimeter of a triangle is $39$ feet. One side of the triangle is $1$ foot longer than the second side. The third side is $2$ feet longer than the second side. Find the length of each side. 12 ft, 13 ft, 14 ft The", "length of the other two sides. 188. The perimeter of an isosceles triangle is $8383$ inches. The length of the shortest side is $2424$ inches. Find the length of the other two sides. 189. A dish is in the shape of an equilateral triangle. Each side is $88$ inches long. Find the perimeter. 190. A floor tile is in the shape of an equilateral triangle. Each side is $1.51.5$ feet long. Find the perimeter. 191. A road sign in the shape of an isosceles triangle has a base of $3636$ inches. If the perimeter is $9191$ inches, find the length of each of the other sides. 192. A scarf in the shape of an isosceles triangle has a base of $0.750.75$ meters. If the perimeter is $22$ meters, find the length of each of the other sides. 193. The perimeter of a triangle is $3939$ feet. One side of the triangle is $11$ foot longer than the second side. The third side is $22$ feet longer than the second side. Find the length of each side. 194. The perimeter of a triangle is $3535$ feet. One side of the triangle is $55$ feet longer than the second side. The third side is $33$ feet longer than the second side. Find the length of each side. 195. One side", "with side length 1. We can try to do the same to the triangle. Joining the midpoints of the sides, as shown on the right, we can split the triangle into 4 equilateral triangles with side length 2. Doing the same to each of the smaller triangles formed gives equilateral triangles with side length 1. There are 4$\\times$4 = 16 of these in the large equilateral triangle. So the hexagon contains 6 equilateral triangles of side length 1, whilst the triangle contains 16. So their areas are in the ratio 6:16, which simplifies to 3:8. Scaling the hexagon up 6 copies of the triangle will fit together to make a hexagon with side length 4. So 6 copies of the triangle will make a copy of the hexagon which has been enlarged by a scale factor of 4. That means its area has been enlarged by a scale factor of 4$^2$ = 16. So 6 copies of the triangle have the same area as 16 copies of the hexagon. So the ratio of the area of the hexagon to the area of the triangle is 6:16, which simplifies to 3:8 Scaling the triangle down 6 equilateral triangles of side length 1 fit into the hexagon. These triangles are similar to the equilateral triangle of side length 4, but", "329 views The side of an equilateral triangle is $10$ cm long. By drawing parallels to all its sides, the distance between any two parallel lines being the same. The triangle is divided into smaller equilateral triangle, each of which has sides of length $1$ cm. How many such small triangles are formed? 1. $60$ 2. $90$ 3. $120$ 4. None of these 1 122 views", "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? ×", "of their areas is $(1/10)^2=1/100$, so the answer is $\\boxed{100}$. ## Solution 5 The side length of the large equilateral triangle is $10$. The number of equilateral triangles with side length $1$ that can fill it is $10^2$. This is equal to the sum of the first $10$ odd integers. The sum of the first $n$ odd integers is $n^2$. In general, an equilateral triangle with side length $n$ can be filled by $n^2$ triangles. The answer is $100\\ \\mathrm{(C)}$. ~mobius247" ]

[ "984 Q: # If each side of a square is increased by 25%, find the percentage change in its area? A) 65.25 B) 56.25 C) 65 D) 56 Explanation: let each side of the square be a , then area = $a2$ As given that The side is increased by 25%, then New side = 125a/100 = 5a/4 New area = $5a42$ Increased area= $25a216-a2$ Increase %=$9a2/16a2*100$ % = 56.25% Q: In triangle ABC, the length of BC is less than twice the length of AB by 2 cm. The length of AC exceeds the length of AB by 10 cm. The perimeter is 32 cm. The length (in cm)of the smallest side of the triangle is: A) 10 B) 4 C) 6 D) 8 Explanation: 0 189 Q: In triangle ABC, the length of BC is less than twice the length of AB by 3 cm. The length of AC exceeds the length of AB by 9 cm. The perimeter of triangle is 34 cm. The length (in cm) of the smallest side of the triangle is: A) 9 B) 10 C) 8 D) 7 Explanation: 2 295 Q: A circle is inscribed in a triangle ABC. It touches sides AB, BC and AC at the points P, Q and R respectively. If BP = 5.4 cm,", "# How do you calculate the perimeter of an equilateral triangle? The side length of the triangle multiplied by $3$. Thus, if an equilateral triangle has a side length of $2$, the perimeter is $2 + 2 + 2$ because all of its sides are $2$. We can generalize this rule by saying that if an equilateral has side length $s$, its perimeter is $s + s + s$, which is equivalent to $3 s$, or, the side length of the triangle multiplied by $3$.", "1 3 (√3)/4 1 12 1/3 4 (√3)/3 2 48 1/9 48/9 10(√3)/27 3 192 1/27 192/27 94(√3)/243 Note: assume that the initial side length is 1. We can see from the above table that the number of sides isn multiplied by four at each iteration. Middle = 3(16), N3 = 3(64) N0= 3(4)0, N1=3(4)1, N2=3(4)2, N3= 3(4)3 Hence, Nn= 3(4)n Length of a single side (ln) As we proceed in each stage the length of any side is 1/3 the length of the side from the preceding stage. If we begin with an equilateral triangle with side length 1, then the length of a side in nth iterations is ln = 1/ (3)n For stage 0 to 3, ln = 1, 1/3, 1/9 and 1/27. Or as in the above it can from the value of ln from the zero stage till the third stage l0=1, l1=1/3, l2= 1/9, l3= 1/27 l0=1/ (3)0, l1=1/ (3)1, l2= 1/ (3)2, l3= 1/ (3)3 And hence ln = 1/ (3) n Perimeter (Pn) Since the lengths of every side in every iteration of the Koch Snowflake are the same, then perimeter is simply the number of sides multiplied by the length of a side Pn = (Nn) (ln) Pn = (3(4) n) (1/ (3) n) For the nth stage. Again,", "automatic increase of Rs. 320 in the very next month and each month thereafter. (i) Find his salary for the tenth month. (ii) What is his total earnings during the first year? Q5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job? Sol: Here, a = 5 and d = 2 Q6. The sum of interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, … sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon. Sol: We know that, sum of interior angles of a polygon of side n is (n – 2) x 180°. Q7. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle. Sol: Let the given equilateral triangle be ∆ ABC with each side of 20 cm. By joining the mid-points of this triangle,", "3 Re: If the length of each side of an equilateral triangle were increased [#permalink] ### Show Tags 16 Oct 2019, 07:31 What is the Solution to this question? Intern Joined: 03 Jul 2019 Posts: 3 Re: If the length of each side of an equilateral triangle were increased [#permalink] ### Show Tags 16 Oct 2019, 07:41 1/2 * 1.5b *1.5h = (1.5*1.5) * (1/2 *b * h) = 2.25 * (1/2 * b * h) Thus, there is 1.25 times increase in the area of the triangle. Therefore, the percentage increase is 125%. Answer: C. Re: If the length of each side of an equilateral triangle were increased [#permalink] 16 Oct 2019, 07:41 Display posts from previous: Sort by", "we will half the result. Doing this, we get $AED=(\\pi \\times 13)=20.420...$ Now, adding the values together we get the total perimeter to be $25+20.420...=45.4\\text{ (3sf)}$ ## GCSE Maths Revision Cards • All major GCSE maths topics covered • Higher and foundation • All exam boards - AQA, OCR, Edexcel, WJEC. Example: PQR is an isosceles triangle. $PQ=QR$. The perimeter of PQR is 45. Work out the value of $x$. The difference in this question is that we are given the perimeter, whilst the side-lengths are expressed in terms of some unknown $x$. We know that if we add up all the sides, that must equal 45. Before we can do that, we need to have some expression for the other side: QR. Since this triangle is isosceles, it is the same as PQ, so we have that $QR=x+5$ Now, we can’t add these side-lengths us as numbers, but we can add them together as algebra. Doing so, we get $x+5\\,\\,+\\,\\,x+5\\,\\,+\\,\\,3x=5x+10$ Now, since the perimeter is equal to all the sides added up, we can make this equal to 45: $5x+10=45$ Now, to find $x$ we must solve this equation. Subtract 10 from both sides to get $5x=45-10=35$ Then, dividing both sides by 5 we get the answer to be $x=\\dfrac{35}{5}=7$ ### Example Questions This hexagon is regular,", "the side decreased by 20%, the new side is 8, and the new area is (8^2 x √3)/4 = 64√3/4 = 16√3. We use the percent change formula: (New - Old)/Old x 100. The percent change is: (16√3 - 25√3)/25√3 x 100 -9√3/25√3 x 100 -9/25 x 100 = -36%, which is a 36% decrease. Alternate solution: We can use the fact that if a side of a regular polygon (equilateral triangle, square, regular pentagon, etc.) is increased or decreased such that the side’s new length is n times its original length, then the new area of the polygon is n^2 times its original area. Since here the side of an equilateral triangle decreased by 20%, the side’s new length is 0.8 times its old length and thus the new area of the equilateral triangle is 0.8^2 = 0.64 times its original area, i.e., 64% of its original area. We see that this is a 36% decrease. _________________ # Scott Woodbury-Stewart Founder and CEO [email protected] 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the \"Kudos\" button. Re: If the side of an equilateral", "point the polygon has $3\\times 2^{12}=12288$ sides. The initial equilateral triangle is shown for reference. The number of sides far exceeds the image resolution so this should be a good approximation to the limiting polygon. Note that I manually scaled the original plot to get the aspect ratio close to $1:1$ in the final image.", "equilateral triangle decreased by 20%, its area is [#permalink] ### Show Tags 01 Feb 2019, 17:47 Bunuel wrote: If the side of an equilateral triangle decreased by 20%, its area is decreased by what percent? A. 36% B. 40% C. 60% D. 64% E. 70% We use the formula for the area of an equilateral triangle: (a^2 x √3)/4, where a is the side. If the initial side is 10, then the original area is (10^2 x √3)/4 = 100√3/4 = 25√3. Since the side decreased by 20%, the new side is 8, and the new area is (8^2 x √3)/4 = 64√3/4 = 16√3. We use the percent change formula: (New - Old)/Old x 100. The percent change is: (16√3 - 25√3)/25√3 x 100 -9√3/25√3 x 100 -9/25 x 100 = -36%, which is a 36% decrease. Alternate solution: We can use the fact that if a side of a regular polygon (equilateral triangle, square, regular pentagon, etc.) is increased or decreased such that the side’s new length is n times its original length, then the new area of the polygon is n^2 times its original area. Since here the side of an equilateral triangle decreased by 20%, the side’s new length is 0.8 times its old length and thus the new area of the equilateral triangle", "# The length of each side of an equilateral triangle is increased by 5 inches, so the perimeter is now 60 inches. How do you find the original length of each side of the equilateral triangle? The original sides were 15 inches. #### Explanation: Each side of an equilateral triangle starts with length $x$. They are all extended by 5 inches (so now each side is $x + 5$ and the sum of all three sides together is 60. So we have: $3 \\left(x + 5\\right) = 60$ $3 x + 15 = 60$ $3 x = 45$ $x = 15$ The original sides were 15 inches.", "If Every Side of a Triangle is Doubled, Then Increase in the Area of the Triangle is - Mathematics MCQ If every side of a triangle is doubled, then increase in the area of the triangle is Options • $100\\sqrt{2}$ % • 200% • 300% • 400% Solution The area of a triangle having sides aband s as semi-perimeter is given by, A = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2 ⇒ 2s = a + b + c We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one Now, the area of a triangle having sides 2a, 2b, and 2c and s1 as semi-perimeter is given by, A_1 = sqrt(s_1(s_1-2a)(s_1 - 2b)(s_1 - 2c) Where, s_1 = (2a +2b+2c)/2 s_1 = (2(a+b+c))/2 s= a + b + c s= 2s Now, A_1 = sqrt(2s (2s-2a)(2s-2b)(2s-2c)) A _1 = sqrt(2s xx 2 ( s-a)xx 2 ( s-b) xx 2 (s-c)) A_1 = 4 sqrt(s (s-a)(s-b)(s-c)) A_1 = 4A Therefore, increase in the area of the triangle =A1 - A =4A - A =3 A Percentage increase in area = (3A)/A xx 100 = 300 % Is there an error in this question or solution? APPEARS IN RD Sharma Mathematics for Class 9 Chapter 17 Heron’s Formula Q 14 | Page", "other two angles in the triangle to be $75^\\circ$75° and $50^\\circ$50°. The final angle in the triangle should be $180-75-50=55$1807550=55, which is also $180-125$180125! Summary The exterior angle of a triangle is equal to the sum of the two opposite interior angles. ### Angles and sides In any triangle the longest side will always be opposite the largest angle. Same with the smallest side and the smallest angle. If we increase one side while keeping the other two sides the same size the side that is getting longer will also have the opposite angle get bigger. A triangle that has two sides that are the same length means the opposite angles must also be equal. ### Impossible triangles #### Exploration Consider the side lengths given below: Can these sides be arranged into a triangle? These three side lengths can not be made into a triangle with these side lengths no matter how we arrange them. If we replace one of the sides with a longer side it is now possible to make these three sides into a triangle. We can compare the two sets of sides to see what is the defining difference. We can see that for any side we look at, the other two side lengths when combined are longer. For the impossible sides, two of", "s as semi-perimeter is given by, Where, We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one Now, the area of a triangle having sides 2a, 2b, and 2c and as semi-perimeter is given by, Where, Now, Therefore, increase in the area of the triangle Percentage increase in area Question 10: If each side of a equilateral triangle is tripled then what is the percentage increase in the area of the triangle? Area of an equilateral triangle having each side a cm is given by Now, Area of an equilateral triangle, say if each side is tripled is given by a = 3a Therefore, increase in area of triangle Percentage increase in area Question 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm. Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle. If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by; Where, We are given: a =150 cm b=120 cm c =200 cm Here we will calculate s, So the area of the triangle", "# Disappearing Triangles Algebra Level 2 One side of an equilateral triangle is $$32$$ cm. The midpoints of its sides are joined to form another triangle whose midpoints, in turn, are joined to form still another triangle. This process continues indefinitely. Find the sum of the perimeters of all these triangles that are defined above. ×", "cm and $21$ cm respectively? Ans: Here Given, Length=$32$cm Breadth=$21$cm Required length of wooden strip = Perimeter of photograph =$2(length+breadth)$ =$2(32+21)$ =$2\\times 53$cm =$106$cm Required Length of wooden strip = $106$cm 10. A rectangular piece of land measures $0.7$ km by $0.5$ km. Each side is to be fenced with $4$ rows of wires. What is the length of the wire needed? Ans: As we know, Perimeter of field = $2(length+breadth)$ Given, Length=$0.7$km Breadth=$0.5$km =$2(length+breadth)$ =$2(0.7+0.5)$ =$2\\times 1.2$km =$2.4$km A $4$row fence is required on each side =$4\\times 2.4$= $9.6$km So, Total length of the required field = $9.6$km 11. Find the perimeter of each of the following shapes: A triangle of sides $3$ cm, $4$cm and $5$ cm. Ans: We know, Perimeter of Triangle=$3+4+5$ =$12$cm So, Perimeter of Triangle=$12$cm 12. Find the perimeter of each of the following shapes: An equilateral triangle of side $9$ cm Ans: We know, Perimeter of Equilateral Triangle=$3\\times side$ =$3\\times 9$cm = $27$cm So, Perimeter of Equilateral Triangle=$27$cm 13. Find the perimeter of each of the following shapes: An isosceles triangle with equal sides $8$ cm each and third side $6$ cm. Ans: We know, Perimeter of isosceles Triangle=$8+8+6$ =$22$cm So, Perimeter of isosceles Triangle=$22$cm 14. Find the perimeter of a triangle with sides measuring $10$ cm, $14$ cm and $15$ cm.", "would create an area of $$0.$$ As such, let $$x= -x_1=x_2.$$ Then $$y=x^2$$ also. Then, the points are $A=(0,0),$$B=(-x,x^2),$$C=(x,x^2).$ With a base of $$BC,$$ the length is $$2x$$ and the height is $$x^2.$$ This would make the area $$x^3 = 64.$$ Therefore, $$x=4,$$ so $$BC = 2\\cdot 4 = 8.$$ Thus, the correct answer is C. 10. A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $$3$$ inches weighs $$12$$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $$5$$ inches. Which of the following is closest to the weight, in ounces, of the second piece? $$\\ 14.0$$ $$\\ 16.0$$ $$\\ 20.0$$ $$\\ 33.3$$ $$\\ 55.6$$ ###### Solution(s): The surface area is increased by a factor of $$\\frac 53 ^2$$ since the side lengths are increased by a factor of $$\\frac 53.$$ Also since the thickness is constant, the volume is scaled up by that much. Then, the wood having the same density makes the weight increase by a factor of $$\\frac 53^2$$ as well, so the new volume is $12 \\cdot \\dfrac {25}{9} = \\dfrac{100}3.$ This is approximately $$33.3.$$ Thus, the correct answer is D. 11. Carl decided to fence in", "# The length of each side of an equilateral triangle is increased by 5 inches, so, the perimeter is now 60 inches. How do you write and solve an equation to find the original length of each side of the equilateral triangle? Nov 11, 2016 I found:$15 \\text{in}$ #### Explanation: Let us call the original lengths $x$: Increasing of $5 \\text{in}$ will give us: $\\left(x + 5\\right) + \\left(x + 5\\right) + \\left(x + 5\\right) = 60$ $3 \\left(x + 5\\right) = 60$ rearranging: $x + 5 = \\frac{60}{3}$ $x + 5 = 20$ $x = 20 - 5$ $x = 15 \\text{in}$", "one metre, its total perimeter is three metres, as shown in Figure 6.14. After one iteration the length of each side becomes 11/3 (4/3) metres long due to the introduction of the triangular deviation, giving a perimeter of 3*(4/3) = 4 metres (an increase of one third). <$fc>Figure 6.14: The size of the equilateral triangle from which the Koch curve is produced On the next iteration each side is only a third of a metre long, and is only extended by a ninth of a metre to 4/9 metre. But because there are 12 sides the overall length increases by a third to 12*(4/9) = 5 metres. On each iteration the length of the perimeter increases by a third (represented mathematically as multiplying by 4/3), so on successive iterations the perimeter gets larger at an increasing rate. For example, after two iterations the perimeter would be: <$table2>3 * 4/3 * 4/3 = 5.33m After four iterations the perimeter would be: <$table2>3 * 4/3 * 4/3 * 4/3 * 4/3 = 9.48m As a general rule the perimeter length can be expressed as: length = 3 * (4/3)^n where n is the number of iterations. Using this expression it is possible to see that the length quickly becomes very large. After only 45 iterations (n=45) it is already over", "sides in the ratios shown. (This also comes from the Pythagorean Theorem, and the fact that the base of an equilateral triangle has been bisected to form this 30-60-90 triangle.) The sides will be in the same ratio for the given triangle. So, with the ratios 1: b5 8 =3 and a 5 =3:2 equaling the ratios b:8:a, we find 16 . =3 Area 5 one half base times height, so A 5 ~8! 26. S= D 8 3 5 64=3 64 5 . 3 =3 The correct answer is (D). The area of the square minus the area of the circle equals the shaded area. OA 5 3 2OA 5 6 5 diameter 5 side of square area of square 5 s2 area of circle 5pr2 A 5 ~6!2 5 36 27. www.petersons.com A 5 p~3!2 5 9p The correct answer is (A). The stock declined in value by $1,500, from an initial value of$11,000. The fractional decline in value is 1500 5 .1364, which is 13.64%. 11000 532 EXPLANATORY ANSWERS TO THE ISEE PRACTICE TEST 1 28. The correct answer is (B). When you decrease a price by a given percentage, and then increase the (now lowered) price by that same percentage, the actual amount of increase is less than the actual amount of decrease. After", "length of the other two sides. 188. The perimeter of an isosceles triangle is $8383$ inches. The length of the shortest side is $2424$ inches. Find the length of the other two sides. 189. A dish is in the shape of an equilateral triangle. Each side is $88$ inches long. Find the perimeter. 190. A floor tile is in the shape of an equilateral triangle. Each side is $1.51.5$ feet long. Find the perimeter. 191. A road sign in the shape of an isosceles triangle has a base of $3636$ inches. If the perimeter is $9191$ inches, find the length of each of the other sides. 192. A scarf in the shape of an isosceles triangle has a base of $0.750.75$ meters. If the perimeter is $22$ meters, find the length of each of the other sides. 193. The perimeter of a triangle is $3939$ feet. One side of the triangle is $11$ foot longer than the second side. The third side is $22$ feet longer than the second side. Find the length of each side. 194. The perimeter of a triangle is $3535$ feet. One side of the triangle is $55$ feet longer than the second side. The third side is $33$ feet longer than the second side. Find the length of each side. 195. One side" ]

[ "than the previous step. ∴ Second step from the bottom needs = 100 – 2 = 98 bricks. Third step from the bottom needs = 98 – 2 = 96 bricks. Fourth step from the bottom needs = 96 – 2 = 94 bricks. Here the numbers are 100, 98, 96, 94, …. Clearly this is an A.P. but not G.P. iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely. Given: An equilateral triangle whose perimeter = 24 cm. Side of the equilateral triangle = $$\\frac{24}{3}$$ = 8 cm. [∵ All sides of equilateral are equal] ……. (1) Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = $$\\frac{8}{2}$$ = 4 cm [∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.] Similarly, the side of third triangle = $$\\frac{4}{2}$$ = 2 cm ∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm, a = 8 Thus each term starting from the second; can be obtained by multiplying its preceding term", "point the polygon has $3\\times 2^{12}=12288$ sides. The initial equilateral triangle is shown for reference. The number of sides far exceeds the image resolution so this should be a good approximation to the limiting polygon. Note that I manually scaled the original plot to get the aspect ratio close to $1:1$ in the final image.", "5-poles though, which the icosphere has a few of. – Ben May 19 at 15:54 • This tuotrial talks a good deal about how building hex meshs, convering the second half of your first bonus as well as your second bonus. May 19 at 18:23 What all hexagonal structures have in common: • They all consist of six equilateral triangles. • The height of a triangle always has the value $$0.86602540378443864676372317075294$$ for a side length of $$1$$. • The hexagons are always offset from each other by $$1.5$$ on one axis, and $$3$$ on the other axis for an equilateral triangle with side length $$1$$. This is convenient because then we don't need to recalculate these values each time, we just create the structure based on these values and scale the result. I obtain the value for the height of the equilateral triangle with the formula $$h={\\frac {\\sqrt {3}}{2}}\\cdot r$$. Here I present three variations, all of which achieve the same result: • One is based only on the creation of a single Mesh Line. • Another is based on creating a Mesh Line, which is instantiated along another. • And the third variant simply creates a Grid where the points are moved. The advantage of all these variants is that the indices are continuous and there is", "automatic increase of Rs. 320 in the very next month and each month thereafter. (i) Find his salary for the tenth month. (ii) What is his total earnings during the first year? Q5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job? Sol: Here, a = 5 and d = 2 Q6. The sum of interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, … sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon. Sol: We know that, sum of interior angles of a polygon of side n is (n – 2) x 180°. Q7. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle. Sol: Let the given equilateral triangle be ∆ ABC with each side of 20 cm. By joining the mid-points of this triangle,", "# The length of each side of an equilateral triangle is increased by 5 inches, so the perimeter is now 60 inches. How do you find the original length of each side of the equilateral triangle? The original sides were 15 inches. #### Explanation: Each side of an equilateral triangle starts with length $x$. They are all extended by 5 inches (so now each side is $x + 5$ and the sum of all three sides together is 60. So we have: $3 \\left(x + 5\\right) = 60$ $3 x + 15 = 60$ $3 x = 45$ $x = 15$ The original sides were 15 inches.", "can split the triangle into 4 equilateral triangles with side length 2. Doing the same to each of the smaller triangles formed gives equilateral triangles with side length 1. There are 4$\\times$4 = 16 of these in the large equilateral triangle. So the hexagon contains 6 equilateral triangles of side length 1, whilst the triangle contains 16. So their areas are in the ratio 6:16, which simplifies to 3:8. Scaling the hexagon up 6 copies of the triangle will fit together to make a hexagon with side length 4. So 6 copies of the triangle will make a copy of the hexagon which has been enlarged by a scale factor of 4. That means its area has been enlarged by a scale factor of 4$^2$ = 16. So 6 copies of the triangle have the same area as 16 copies of the hexagon. So the ratio of the area of the hexagon to the area of the triangle is 6:16, which simplifies to 3:8 Scaling the triangle down 6 equilateral triangles of side length 1 fit into the hexagon. These triangles are similar to the equilateral triangle of side length 4, but their sides are 4 times shorter. This means that their area is 4$^2$ = 16 times smaller. So, when scaled down by a factor of 16,", "length of the other two sides. 188. The perimeter of an isosceles triangle is $8383$ inches. The length of the shortest side is $2424$ inches. Find the length of the other two sides. 189. A dish is in the shape of an equilateral triangle. Each side is $88$ inches long. Find the perimeter. 190. A floor tile is in the shape of an equilateral triangle. Each side is $1.51.5$ feet long. Find the perimeter. 191. A road sign in the shape of an isosceles triangle has a base of $3636$ inches. If the perimeter is $9191$ inches, find the length of each of the other sides. 192. A scarf in the shape of an isosceles triangle has a base of $0.750.75$ meters. If the perimeter is $22$ meters, find the length of each of the other sides. 193. The perimeter of a triangle is $3939$ feet. One side of the triangle is $11$ foot longer than the second side. The third side is $22$ feet longer than the second side. Find the length of each side. 194. The perimeter of a triangle is $3535$ feet. One side of the triangle is $55$ feet longer than the second side. The third side is $33$ feet longer than the second side. Find the length of each side. 195. One side", "# An equilateral triangle has a side length of 24 inches. What is its perimeter? May 9, 2018 72 in. #### Explanation: Since this triangle is an equilateral triangle, all sides are equal. You only have to multiply the side you are given by 3. Doing so you get: $24 \\cdot 3 = 72$ Note: This only works for equilateral triangles. With isosceles or scalene triangles, you must have the angular quantities to do the problem. May 9, 2018 $72$ inches #### Explanation: The perimeter is the sum of the sides of a given shape. We know a few things about our shape: • We have a triangle, which has $3$ sides • This is an equilateral triangle, meaning the lengths are equal • One side is $24$ inches If all of the sides are the same, we are essentially adding $24 + 24 + 24$ to get $72$ inches Hope this helps!", "The perimeter of an equilateral triangle is $42$ miles. Find the length of each side. The perimeter of an isosceles triangle is $42$ feet. The length of the shortest side is $12$ feet. Find the length of the other two sides. 15 ft The perimeter of an isosceles triangle is $83$ inches. The length of the shortest side is $24$ inches. Find the length of the other two sides. A dish is in the shape of an equilateral triangle. Each side is $8$ inches long. Find the perimeter. 24 in. A floor tile is in the shape of an equilateral triangle. Each side is $1.5$ feet long. Find the perimeter. A road sign in the shape of an isosceles triangle has a base of $36$ inches. If the perimeter is $91$ inches, find the length of each of the other sides. 27.5 in. A scarf in the shape of an isosceles triangle has a base of $0.75$ meters. If the perimeter is $2$ meters, find the length of each of the other sides. The perimeter of a triangle is $39$ feet. One side of the triangle is $1$ foot longer than the second side. The third side is $2$ feet longer than the second side. Find the length of each side. 12 ft, 13 ft, 14 ft The", "Perimeter of each shape (a) Perimeter of square = 4 × side ⇒ 30 cm = 4 × side ⇒ side = $$\\frac{30}{4}$$ cm = 7.5 cm Thus, the length of each side of square will be 7.5 cm. (b) Perimeter of equilateral triangle = 3 × side ⇒ 30 cm = 3 × side ⇒ side = $$\\frac{30}{3}$$ cm = 10 cm Thus, the length of each side of equilateral triangle will be 10 cm. (c) Perimeter of regular hexagon = 6 × side ⇒ 30 cm = 6 × side ⇒ side = $$\\frac{30}{6}$$ cm = 5 cm Thus, the length of each side of regular hexagon will be 5 cm. Question 12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side? Solution: Let the length of third side be x cm. Length of other two sides are 12 cm and 14 cm. Now, perimeter of triangle = 36 cm ⇒ 12 + 14 + x = 36 ⇒ 26 + x = 36 ⇒ x = 36 – 26 ⇒ x = 10 Thus, the length of third side is 10 cm. Question 13. Find the cost of fencing a square park of side 250 m at the rate of", "with side length 1. We can try to do the same to the triangle. Joining the midpoints of the sides, as shown on the right, we can split the triangle into 4 equilateral triangles with side length 2. Doing the same to each of the smaller triangles formed gives equilateral triangles with side length 1. There are 4$\\times$4 = 16 of these in the large equilateral triangle. So the hexagon contains 6 equilateral triangles of side length 1, whilst the triangle contains 16. So their areas are in the ratio 6:16, which simplifies to 3:8. Scaling the hexagon up 6 copies of the triangle will fit together to make a hexagon with side length 4. So 6 copies of the triangle will make a copy of the hexagon which has been enlarged by a scale factor of 4. That means its area has been enlarged by a scale factor of 4$^2$ = 16. So 6 copies of the triangle have the same area as 16 copies of the hexagon. So the ratio of the area of the hexagon to the area of the triangle is 6:16, which simplifies to 3:8 Scaling the triangle down 6 equilateral triangles of side length 1 fit into the hexagon. These triangles are similar to the equilateral triangle of side length 4, but", "existing edges as vectors, we construct an equilateral triangle by connecting three vertices: So $\\alpha = 36$ (since it’s part of a regular pentagon) and $\\beta = 60$. Then $\\cos \\theta = \\frac{\\cos^2 60 - \\cos^2 36}{\\sin^2 36}$ and $\\theta = 116.6^\\circ$.", "cm and $21$ cm respectively? Ans: Here Given, Length=$32$cm Breadth=$21$cm Required length of wooden strip = Perimeter of photograph =$2(length+breadth)$ =$2(32+21)$ =$2\\times 53$cm =$106$cm Required Length of wooden strip = $106$cm 10. A rectangular piece of land measures $0.7$ km by $0.5$ km. Each side is to be fenced with $4$ rows of wires. What is the length of the wire needed? Ans: As we know, Perimeter of field = $2(length+breadth)$ Given, Length=$0.7$km Breadth=$0.5$km =$2(length+breadth)$ =$2(0.7+0.5)$ =$2\\times 1.2$km =$2.4$km A $4$row fence is required on each side =$4\\times 2.4$= $9.6$km So, Total length of the required field = $9.6$km 11. Find the perimeter of each of the following shapes: A triangle of sides $3$ cm, $4$cm and $5$ cm. Ans: We know, Perimeter of Triangle=$3+4+5$ =$12$cm So, Perimeter of Triangle=$12$cm 12. Find the perimeter of each of the following shapes: An equilateral triangle of side $9$ cm Ans: We know, Perimeter of Equilateral Triangle=$3\\times side$ =$3\\times 9$cm = $27$cm So, Perimeter of Equilateral Triangle=$27$cm 13. Find the perimeter of each of the following shapes: An isosceles triangle with equal sides $8$ cm each and third side $6$ cm. Ans: We know, Perimeter of isosceles Triangle=$8+8+6$ =$22$cm So, Perimeter of isosceles Triangle=$22$cm 14. Find the perimeter of a triangle with sides measuring $10$ cm, $14$ cm and $15$ cm.", "# Changing an equilateral triangle with side length 100 to side length 1 Given an equilateral triangle with side length 100. In each move, you can change the one of the side length of the triangle, but the resulting side length must still be able to form a triangle with the other 2 sides. What is the minimum number of moves needed to form an equilateral triangle of side length 1? My thoughts and attempt: Clearly, after each move, the triangle inequality should be satisfied: the sum of any two side lengths must be larger than the remaining side length. And to reduce the number of moves, instead of forcibly reducing the length of one of the sides immediately, instead we consider the reduction of length of two side cumulatively for each move. The 3 side lengths were 100, 100, 100. So let's take one of the 100. We change the other two to become 50 and 51, that is 2 moves. Now, consider 51, since $26+26>51$ we will change the largest and smallest length (in order) to 26. Hence another 2 moves. Then I repeat this. However, this seems to fail as the side lengths get smaller. Also, when I changed the side lengths, I strictly changed it into integers. (Although there were no such restrictions) •", "the side decreased by 20%, the new side is 8, and the new area is (8^2 x √3)/4 = 64√3/4 = 16√3. We use the percent change formula: (New - Old)/Old x 100. The percent change is: (16√3 - 25√3)/25√3 x 100 -9√3/25√3 x 100 -9/25 x 100 = -36%, which is a 36% decrease. Alternate solution: We can use the fact that if a side of a regular polygon (equilateral triangle, square, regular pentagon, etc.) is increased or decreased such that the side’s new length is n times its original length, then the new area of the polygon is n^2 times its original area. Since here the side of an equilateral triangle decreased by 20%, the side’s new length is 0.8 times its old length and thus the new area of the equilateral triangle is 0.8^2 = 0.64 times its original area, i.e., 64% of its original area. We see that this is a 36% decrease. _________________ # Scott Woodbury-Stewart Founder and CEO [email protected] 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the \"Kudos\" button. Re: If the side of an equilateral", "other two angles in the triangle to be $75^\\circ$75° and $50^\\circ$50°. The final angle in the triangle should be $180-75-50=55$1807550=55, which is also $180-125$180125! Summary The exterior angle of a triangle is equal to the sum of the two opposite interior angles. ### Angles and sides In any triangle the longest side will always be opposite the largest angle. Same with the smallest side and the smallest angle. If we increase one side while keeping the other two sides the same size the side that is getting longer will also have the opposite angle get bigger. A triangle that has two sides that are the same length means the opposite angles must also be equal. ### Impossible triangles #### Exploration Consider the side lengths given below: Can these sides be arranged into a triangle? These three side lengths can not be made into a triangle with these side lengths no matter how we arrange them. If we replace one of the sides with a longer side it is now possible to make these three sides into a triangle. We can compare the two sets of sides to see what is the defining difference. We can see that for any side we look at, the other two side lengths when combined are longer. For the impossible sides, two of", "# If two sides of an equilateral triangle measures $(6x+1)$ units and $\\frac{11x+5}{2}$ units, find the perimeter of the triangle. If two sides of an equilateral triangle measures $(6x+1)$ units and $\\frac{11x+5}{2}$ units, find the perimeter of the triangle. I tried the following, Let the third side measure $x$ units Therefore, Perimeter$=(6x+1)+$$\\frac{11x+5}{2}$$+x$ I do not know what to do next. Please help. • The triangle is equilateral so the two sides mentioned are equal. – André Nicolas Aug 30 '14 at 7:26 Since the triangle is equilateral, $6x+1 = (11x+5)/2$. Solve for x. Then, perimeter = $3*(6x+1)$. $$6x+2=\\frac{1}{2}(11x+5)$$ which gives $x=1$. This means that all sides have length $6\\cdot1+2=8$. Hence, the perimeter of the triangle is $3\\cdot8=24$.", "equilateral triangle of side a/3. • Remove the base edge of the new equilateral triangle to form a continuous curve with other line segments. Thus each edge is transformed as such: The first 5 iterations look like: Properties Two properties of this fractal may seem paradoxical at first but are easily shown. I will show the derivations here for the interested without skipping too many steps. For an infinite number of iterations the perimeter of the fractal (the length of all the edges) tends to infinity whilst the area enclosed remains finite. Infinite Perimeter First consider the number of edges for an infinite number of iterations. Initially the number of edges is 3, each iteration transforms a single edge into 4 edges (see above), thus the series of the edge count is: 3, 12, 48, …. The general form is: Starting from an edge length of a each iteration divides the edge length by 3. The following defines the edge length for iteration n and from this calculates the perimeter for iteration n. The limit as n tends to infinity is found, showing that the perimeter is infinite. Thus it is easy to see that the perimeter length is infinite through standard results on limits. Finite Area The area calculation is a little bit more involved. The result", "Find the length of each side. ## Solution Step 1. Read the problem. Draw the figure and label it with the given information. Perimeter = 93 in. Step 2. Identify what you are looking for. length of the sides of an equilateral triangle Step 3. Name. Choose a variable to represent it. Let s = length of each side Step 4. Translate. Write the appropriate formula. Substitute. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. Each side is 31 inches. Find the length of each side of an equilateral triangle with perimeter $39$ inches. 13 in. Find the length of each side of an equilateral triangle with perimeter $51$ centimeters. 17 cm Arianna has $156$ inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of $60$ inches. How long can she make the two equal sides? ## Solution Step 1. Read the problem. Draw the figure and label it with the given information. P = 156 in. Step 2. Identify what you are looking for. the lengths of the two equal sides Step 3. Name. Choose a variable to represent it. Let s = the length of each side Step 4. Translate. Write the appropriate formula. Substitute in the given information. Step", "# project euler #577 confusion An equilateral triangle with integer side length $n≥3$ is divided into $n^2$ equilateral triangles with side length 1 as shown in the diagram in the above link. The vertices of these triangles constitute a triangular lattice with $\\frac{(n+1)(n+2)}{2}$ lattice points. Let $H(n)$ be the number of all regular hexagons that can be found by connecting 6 of these points. $H(3)=1,H(6)=12...H(20)=966$ I can't understand why $H(6)=12$. I can only count $11$ regular hexagons. What I do is, whenever $n=3m$, it is the first triangle with a regular hexagon of size $m$. And on each successive $n$, the number by which hexagons of size $x$ increase, increments. This also follows that for $n=6=3.2$, it will have only $1$ regular hexagon of size $2$, and number of hexagons of size $1=10$. $\\therefore H(6) =11 \\ne 12$. Where am I wrong? You get one hexagon of side length $\\sqrt3$ by joining the midpoint of the big triangle's base to the upper vertex of the second and fifth triangles on that base. Then go straight up to the midpoints of the sides of the big triangle. Then complete the hexagon. Remember, you're allowed to join any six points of the lattice that produce a regular hexagon. Does that help? Considering hexagon of side length $$x$$, you can choose" ]

[ "over one-fourth of a ’!", "above expression, we get 4/5 Hence, the simplified form of the complex fraction is 4/5.", "fraction.", "fraction?", "x=4/5.", "quarter of the square of the sum.", "anyway) is that 4%5 would be 4... just something to remember", "and 5% of B' but rather use 5% of B' = 5.5% of B).", ", which is written as one-third or one-third.", "of the fraction.", "c=1.4n-5.5", "is (approximately) in a 1:5 ratio.", "fraction obtainded is 3/4. Find the fraction", "a=5.8cm b=3.22 cm find percentage error in x x=5.8-3.22 x=2.58", "remainder. Remember that the fraction $\\frac{34}{5}$ also means 34 divided by 5. Continue Reading", "with simplified fractions to these!", "remainder.", "Answer: 0.8 is expressed as a fraction 4 / 5 .", "of the fraction 5/7 Solution: Given fraction = 5/7 Reciprocal of a fraction = 1/(5/7) Hence the reciprocal of the given fraction is 7/5", "x2 Find f(5) if f(x) = x + 1/2 17. One half One half of ? is: ?" ]

[ "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "# What is 3.6 divided by 8? \"3.6 divided by 8\" can be written as $3.6 \\div 8$ or $\\frac{3.6}{8}$ and is equal to:", "fraction, such as $4\\frac{3}{5}$.", "# How do you simplify 1.7 div 8.5? Aug 3, 2016 $= 0.2$ $\\frac{1.7}{8.5}$ Multiplying numerator and denominator both by $10$ $\\frac{17}{85}$ $= \\frac{1}{5}$ $= 0.2$", "# How do you simplify 1.7 div 8.5? Aug 3, 2016 $= 0.2$ #### Explanation: $\\frac{1.7}{8.5}$ Multiplying numerator and denominator both by $10$ we get $\\frac{17}{85}$ $= \\frac{1}{5}$ $= 0.2$", "the fraction from above (4/7). Finally, enter this mixed number into a calculator and round to the nearest hundredth. $$\\sf \\sqrt{13}=\\boxed{\\sf3}\\dfrac{\\boxed{\\sf 4}}{\\boxed{\\sf 7}}=\\boxed{\\sf3.57}$$ by", "## Basic College Mathematics (9th Edition) $5\\div(-\\frac{5}{8})$ =$5\\times-\\frac{8}{5}$ Convert division into multiplication by flipping the numerator and denominator of the divisor =$1\\times-\\frac{8}{1}$ Cancel common multiples by dividing the 5s by 5 =$-8$", "fraction gives $$\\frac{(m - 7)(m + 3) \\div (m + 3)}{6(m + 3) \\div (m + 3)} = \\frac{m - 7}{6}$$. There are no more common factors in this fraction, so this is the final answer.", "{1*5=5}{2*3=6}}={\\frac {5}{6}}}$ And that's about it for fractions and divison.", "value of $x\\div y$ when $x = 72$ and $y = 9$ $$x\\div y= 72\\div9=8$$ Use substitution to find the value of $\\frac{x}{y}$ when $x =64$ and $y = 8$ Hint: Remember, a fraction is just another way of writing a division. $$\\frac{x}{y}=x\\div y= 64\\div8=8$$ Example Questions $$xy=x\\times y=7\\times 11= 77$$ $$\\frac{y}{x}=y\\div x= 35\\div7=5$$", "# How do you simplify 1/8 div 3? $\\frac{1}{8} \\div 3 = \\frac{1}{24}$ To divide by a number $a$ means to multiply by its reciprocal $\\frac{1}{a}$, so: $\\frac{1}{8} \\div 3 = \\frac{1}{8} \\times \\frac{1}{3} = \\frac{1}{24}$", "add up to 1, like this: $\\frac{3}{10}∶ \\frac{7}{10}$ and $\\frac{3}{10} + \\frac{7}{10} = 1$ We will be using this idea that the sum of the parts of a ratio gives us the total number of things being divided up. The ratio $4 : 9$ can be thought of as $4$ for me and $9$ for you, so $13$ altogether, hence the ratio can be written as $\\frac{4}{13}∶ \\frac{9}{13}$. Additionally, we can see that the sum of the fractions in this new ratio, will be $\\frac{4}{13} + \\frac{9}{13} = \\frac{13}{13} = 1$. This is something which always happens. If you write the ratio in terms of fractions like this, the sum of the fractions in the ratio will always be $1$.", "# How do you simplify 1.7 div 8.5? ##### 1 Answer Aug 3, 2016 $= 0.2$ #### Explanation: $\\frac{1.7}{8.5}$ Multiplying numerator and denominator both by $10$ we get $\\frac{17}{85}$ $= \\frac{1}{5}$ $= 0.2$", "How do you simplify 1.4 div 9.8? Jul 10, 2016 $0.14$ Explanation: $1.4 \\div 9.8$ is the same as $\\frac{1.4}{9.8}$ Divide using a calculator. $0.14285714285$ Round this to the hundredth or thousandth. Jul 10, 2016 You can simplify the fraction to $\\frac{1}{7}$ if you would prefer not to get a decimal answer. $\\frac{1.4}{9.8} = \\frac{14}{98} = \\frac{7}{49} = \\frac{1}{7}$", "# How do you divide 5 -: 3/5? $5 \\div i \\mathrm{de} \\frac{3}{5} = 5 \\times \\frac{5}{3} = \\frac{5 \\times 5}{3} = \\frac{25}{3}$", "2 }-\\frac { 6 }{ 5 } x$ 3 1 4 ${ m }^{ 4 }-\\frac { 6 }{ 7 } { m }^{ 2 }-6m+2.7$ 5 $\\frac { { t }^{ 3 } }{ \\sqrt { 6 } } +\\frac { { t }^{ 4 } }{ 4 } -t$ 6. Find the quotient and remainder when 5x3 - 9x2 + 10x + 2 is divided by x + 2 using synthetic division 7. Find the quotient and remainder when 4x3 + 6x2 + 7x + 2 is divided by x - 2 8. Find the quotient and remainder when 5x3 + 7x2 + 3x + 2 is divided by 3x + 2 9. Factorise 2x3- x2 - 12x - 9 into linear factors 10. Prove that x -1 is a factor x5 - 45x4 + 36x3 + 45x2 - 36x-1", "# How do you simplify 5 div 0.1? Sep 4, 2016 $50$ This can be written as $\\frac{5}{0.1}$ Multiply the top and bottom by 10. ($\\frac{10}{10} = 1$) $\\frac{5}{0.1} \\times \\frac{10}{10} = \\frac{50}{1} = 50$", "and $4$ $\\frac{8}{10}$ Divide $8$ by $10$ $\\frac{4}{5}$ 7 Simplifying, we get $\\frac{4}{5}$ $\\frac{4}{5}$", "9758/2021/P1/Q05 (a) Express $\\frac{x}{(x+2)(x+3)(x+4)}$ in partial fractions. [3] (b) Hence find, in terms of $n, \\sum_{r=1}^n \\frac{r}{(r+2)(r+3)(r+4)}$. [3] (c) State the value of $\\sum_{i=1}^{\\infty} \\frac{r}{(r+2)(r+3)(r+4)}$. [1]", "on the line, other than $(0,0)$. Then put the values into the ratio $\\frac{\\text{Distance}}{\\text{Time}}$. Simplify the fraction to get the unit rate. • To find the unit rate from an equation, isolate the ratio by dividing both sides by $t$. This gives us the proportion $\\frac{d}{t}=$unit rate. Simplify the fraction to get a unit rate. • ### Determine the best price by identifying the unit price. Hints Determine the unit price for each deal. Set up the ratio of $\\frac{\\text{cost}}{\\text{unit}}$, then divide to simplify. For example, let's say a five pack of figurines sold for $\\$63.75$. The$\\frac{\\text{cost}}{\\text{unit}}$ratio would be$\\frac{63.75}{5}$. Simplified$\\frac{63.75}{5}=12.75$. So the unit price for this example is$\\$12.75$. Solution The order from least to most expensive is: $1)$ A four pack of Cat Hero figurines for $\\$41.50$. • First, set up the$\\frac{\\text{cost}}{\\text{unit}}$ratio:$\\frac{41.50}{4}$. • Then, divide$\\frac{41.50}{4}=10.35$. • So, the unit price for this deal is$\\$10.35$. $2)$ A single Cat Hero figurine for $\\$11.50$. • First, set up the$\\frac{\\text{cost}}{\\text{unit}}$ratio:$\\frac{11.50}{1}$. • Then, divide$\\frac{11.50}{1}=11.50$. • So, the unit price for this deal is$\\$11.50$. $3)$ A two pack of Cat Hero figurines for $\\$24.00$. • First, set up the$\\frac{\\text{cost}}{\\text{unit}}$ratio:$\\frac{24.00}{2}$. • Then, divide$\\frac{24.00}{2}=12.00$. • So, the unit price for this deal is$\\$12.00$. $4)$ A three pack of Cat Hero figurines for $\\$36.75$. • First, set up the$\\frac{\\text{cost}}{\\text{unit}}$ratio:$\\frac{36.75}{3}$. • Then, divide$\\frac{36.75}{3}=12.25$. • So, the unit" ]

[ "fraction, such as $4\\frac{3}{5}$.", "among the fraction is $\\frac{2}{5}$.", "{1*5=5}{2*3=6}}={\\frac {5}{6}}}$ And that's about it for fractions and divison.", "# What is 3.5 as a fraction? $3.5 = 3 \\frac{1}{2}$ $3.5$ means $3 \\frac{5}{10}$. $3 \\frac{5}{10}$ can be simplified to $3 \\frac{1}{2}$.", "# How do you divide 5 -: 3/5? $5 \\div i \\mathrm{de} \\frac{3}{5} = 5 \\times \\frac{5}{3} = \\frac{5 \\times 5}{3} = \\frac{25}{3}$", "## Intermediate Algebra (6th Edition) We are given the expression $\\frac{x-5}{5-x}$. We can use the distributive property to factor out a -1 from the numerator. $x-5=-1(-x+5)=-1(5-x)$ Therefore, the expression simplifies to expression as $\\frac{-1(5-x)}{5-x}$. In this case, the numerator is the opposite of the denominator, so this fraction simplifies to -1.", "• # question_answer The sum of 5% of a number and 4% of other number is $\\frac{2}{3}$ of the sum of 6% of the first number and 8% of second. The ratio of the first number to the second is A) 2 : 3 B) 3 : 2 C) 3 : 4 D) 4 : 3 Let the first number $=x$ and the second number = y According to the question, 5% of $x+4%$ of $y=\\frac{2}{3}$ (6% of x + 8% of y) $\\Rightarrow$ $3(5x+4y)=2(6x+8y)$ $\\Rightarrow$ $15x+12y=12x+16y$ $\\Rightarrow$ $3x=4y$ $\\therefore$ $\\frac{x}{y}=\\frac{4}{3}$", "rationalize the denominator Multiply the fraction with $\\frac{5x\\sqrt{5}+3y\\sqrt{3}}{5x\\sqrt{5}+3y\\sqrt{3} }$", "remainder. Remember that the fraction $\\frac{34}{5}$ also means 34 divided by 5. Continue Reading", "to both numerator and denominator. If 5 subtracted from both numerator and denominator, the fraction becomes $\\frac{2}{4}$. What is the fraction. Solution: Step 1: Let the fraction be $\\frac{x}{y}$ Case 1: $\\frac{x + 1}{y + 1} = \\frac{3}{5}$ ..............................(1) Case 2: $\\frac{x - 5}{y - 5} = \\frac{2}{4}$ = $\\frac{1}{2}$ .................................(2) Step 2: (1) => 5(x + 1) = 3(y + 1) => 5x + 5 = 3y + 3 => 5x - 3y + 2 = 0 ......................................(3) and (2) => 2(x - 5) = y - 5 => 2x - 10 = y - 5 => 2x - 5 = y .........................................(4) Step 3: Put y = 2x - 5 in equation (3) => 5x - 3( 2x - 5) + 2 = 0 => 5x - 6x + 15 + 2 = 0 => - x + 17 = 0 => x = 17 Step 4: Put x = 17 in equation (4) => 2 * 17 - 5 = y => 34 - 5 = y => 29 = y Answer: The fraction is $\\frac{17}{29}$. Question 3: Find the LCM of 18, 42 and 34. Solution: Prime Factors of 18 = 2 x 3 x 3 Prime Factors of 42 = 2 x 3 x 7 Prime Factors of 34 = 2 x 17", "# How do you simplify 1/3 div 5/3? Sep 2, 2016 $\\frac{1}{5}$ #### Explanation: The 'method' for simplifying division of 2 fractions involves the following steps. • leave the leading fraction as it is. • Change division to multiplication. • Invert the second fraction. That is 'turn it upside down'. • Simplify, if possible, by cancelling. 1/3÷5/3=1/3xx3/5=1/cancel(3)^1xxcancel(3)^1/5=(1xx1)/(1xx5)=1/5", "So, an equivalent fraction to$\\frac{3}{8}$ is $\\frac{12}{32}$. I hope that helps! All the best!", "Hence $A + B = 4$ and $B - 2 A = 1$ Now putting $B = 4 - A$ (from first equation) into second, we get $4 - A - 2 A = 1$ or $3 A = 3$ or $A = 1$ and hence $B = 4 - 1 = 3$ hence Partial fractions of $\\frac{{x}^{3} - 7 x - 3}{{x}^{2} - x - 2}$ are $x + 1 - \\frac{1}{x + 1} - \\frac{3}{x - 2}$", "+0 0 31 1 Simplify the following expression to a simplified fraction: $\\frac{\\frac{5}{\\sqrt{80}}}{\\sqrt{5}}.$ Mar 4, 2021", "# How do you evaluate \\frac{3-1}{-4-1}? Jun 26, 2017 $- \\frac{2}{5}$ #### Explanation: We look at the top first. $3 - 1 = 2$ Now the denominator: $- 4 - 1 = - 5$ So, this fraction becomes $- \\frac{2}{5}$. Since there are no common factors between $2$ and $5$, $- \\frac{2}{5}$ is our answer.", "# What is 3.0 divided by 4? $0.75$ $3.0 \\setminus \\div 4$ is equal to $\\setminus \\frac{3}{4} = 0.75$ If you're mindful of significant figures, then you know that the quotient should have one significant figure. So the quotient is $0.8$.", "jonesiyannna 13 # Can someone help me solve this proportion: 5/n=1.5/1.8 $\\frac{5}{n} =\\frac{1.5}{1.8} \\\\ \\\\ n\\neq 0 \\\\ \\\\1.5n=5\\cdot 1.8\\\\ \\\\1.5n= 9 \\ \\ / : 1.5 \\\\ \\\\ n=6$", "$4$ fourths in $1$, so if we take the length from $0$ to $\\frac{1}{4}$ four times, we will find $1$. 2. $\\frac{5}{3}$ is $5$ equal pieces where $3$ pieces make $1$. So if we partition the interval between $0$ and $\\frac{5}{3}$ into $5$ pieces, each will be a third and $3$ thirds is $1$.", "# rationalize the denominator Multiply the fraction with $\\frac{5x\\sqrt{5}+3y\\sqrt{3}}{5x\\sqrt{5}+3y\\sqrt{3} }$", "## Basic College Mathematics (10th Edition) $\\frac{1}{5}$ is invested in bonds. This problem can be restated to say that $\\frac{1}{3}$ of $\\frac{3}{5}$ of the investments are in bonds. Multiply to determine the fraction that are in bonds. $\\frac{1}{3} \\times \\frac{3}{5}$ can be reduced by the common factor of 3. $\\frac{1}{1} \\times \\frac{1}{5} = \\frac{1}{5}$" ]

[ "well choose them in order $1, 2, 3, \\ldots$.", "= 2, 4, 6, 8, \\ldots$.", "{1, 4, 9, 16, \\ldots}$", "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", "for the sixth item only. I assume $i \\in \\{1,2,\\ldots,100\\}$. If you meant that $i \\in \\{1,2,\\ldots,6\\}$, how will I know which of the seven out of 100 items are chosen? – ubaabd Dec 17 '14 at 5:23", "# 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.", "-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}$$ Adding up the numbers below the line, we end up with $-19$, which modulo $97$ equals $78$, as expected. An advantage of this method is that, once you have those $24$ numbers, the computations are really simple and require only really small numbers. You can even do this method by hand, as I actually did for the example above.", "whose top number is $200$.", "24 x + 48 {x}^{2} + 80 {x}^{3} + 120 {x}^{4} + 168 {x}^{5} + 224 {x}^{6} + \\ldots \\ldots .$", "# Sequence of Powers of 16 The sequence of powers of $16$ begins: $1, 16, 256, 4096, 65 \\, 536, 1 \\, 048 \\, 576, 16 \\, 777 \\, 216, 268 \\, 435 \\, 456, \\ldots$", "\\mod 97: k = 0, 1, \\ldots, 23\\}$ such as these given below: $$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 &", "# One too many Number Theory Level 3 Find the last 8 digits of $1+11+111+\\ldots+\\overbrace{1\\ldots1}^{2014\\text{ ones}}$ ×", "the list becomes $$1,2,3,5,7,11$$, which square to $$1,4,9,25,49,121$$", "\\ldots \\;?$ $502$ $504$ $506$ $500$", "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.", "# 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? ×", "# Another Think Out Of The Boxes(The Id-\"1\"-- ) What is the next number in the sequence $1, 11, 21, 1211, \\ldots?$ ×", "# Sequence of Powers of 7 The sequence of powers of $7$ begins: $1, 7, 49, 343, 2401, 16 \\, 807, 117 \\, 649, 823 \\, 543, 5 \\, 764 \\, 801, 40 \\, 353 \\, 607, \\ldots$", "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," ]

[ "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "to see here that this continues all the way down to one. However, when the case gets to $32$, there are 5 ways instead of 4 because $2^{10}-2^5$ is smaller than 1000. Thus, $9+8+7+6+5+5+4+3+2+1 = 50$ So the answer is $\\boxed{050}$ Problem 4 Solution 1 Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \\ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \\ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we have the sum $31+29+28+26+25+\\ldots+4+2+1=(31+28+25+22+\\ldots+1)+$ $(29+26+23+\\ldots+2)=176+155=\\boxed{331}$. Solution 2 Let the three piles have $a, b, c$ coins respectively. If we disregard order, then we just need to divide by $3! = 6$ at the end. We know $a + b + c = 66$. Since $a, b, c$ are positive integers, there are $\\binom{65}{2}$ ways from Stars and Bars. However, we must discard the cases where $a = b$ or $a = c$ or $b = c$. The three cases are symmetric, so we just take the first case and multiply by 3. We have $2a + c = 66 \\implies a = 1, 2, \\dots 32$ for", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "and $3$ and $5$ never near, which leaves the total combinations (orders) of the boxes to twelve: $1-2-3-4-5$ $1-3-2-4-5$ $1-3-4-2-5$ $1-3-2-5-4$ $1-3-4-5-2$ $1-4-3-2-5$ Plus the symmetry of the exchange $5\\to 3$ So I have $12$ possible dispositions of the boxes. Now, let's call the figures on each box as $$1 = \\{A_1, A_2, A_3, A_4, A_5, A_6\\}$$ $$2 = \\{B_1, \\ldots, B_6\\}$$ $$3 = \\{C_1 \\ldots C_6\\}$$ $$4 = \\{D_1, D_2, D_3, A_1, C_1, C_2\\}$$ $$5 = \\{E_1 \\ldots E_6\\}$$", "that in the algorithm, we split groups, so now we need the reverse, which is to merge groups. We know that the $i$-th top-most group in Stack $M-1$ would have come from the same group in Stack $M-2$ as the $i$-th top-most group in Stack $M$. This enables us to reverse the process, and continue numbering the forbidden numbers in the same way until we get back to Stack 1. For example, here is the procedure for $M=7, N=13$: $$\\begin{eqnarray} S_7\\{\\}&:& \\boxed{1}\\boxed{2}\\boxed{3}\\boxed{4}\\boxed{5}\\boxed{6}\\boxed{7}\\boxed{8}\\boxed{9}\\boxed{10}\\boxed{11}\\boxed{12}\\boxed{13} \\end{eqnarray}$$ $$\\begin{eqnarray} S_6\\{1,2,3,4,5\\}&:&\\boxed{13}\\boxed{12}\\boxed{11}\\boxed{10}\\boxed{9}\\boxed{8}\\boxed{7}\\boxed{6}\\\\ S_7\\{\\}&:& \\boxed{1}\\boxed{2}\\boxed{3}\\boxed{4}\\boxed{5}\\color{red}{\\boxed{6}\\boxed{7}\\boxed{8}\\boxed{9}\\boxed{10}\\boxed{11}\\boxed{12}\\boxed{13}} \\end{eqnarray}$$ $$\\begin{eqnarray} S_5\\{11,12,13\\}&:&\\boxed{56}\\boxed{47}\\boxed{38}\\boxed{29}\\boxed{1\\ 10}\\\\ S_6\\{1,2,3,4,5\\}&:&\\boxed{13}\\boxed{12}\\boxed{11}\\color{red}{\\boxed{10}\\boxed{9}\\boxed{8}\\boxed{7}\\boxed{6}}\\\\ S_7\\{\\}&:& \\color{red}{\\boxed{1}\\boxed{2}\\boxed{3}\\boxed{4}\\boxed{5}\\boxed{6}\\boxed{7}\\boxed{8}\\boxed{9}\\boxed{10}\\boxed{11}\\boxed{12}\\boxed{13}} \\end{eqnarray}$$ $$\\begin{eqnarray} S_4\\{4,5,6,7\\}&:&\\boxed{1\\ 10\\ 11}\\boxed{2\\ 9\\ 12}\\boxed{3\\ 8\\ 13}\\\\ S_5\\{11,12,13\\}&:&\\boxed{56}\\boxed{47}\\color{red}{\\boxed{38}\\boxed{29}\\boxed{1\\ 10}}\\\\ S_6\\{1,2,3,4,5\\}&:&\\color{red}{\\boxed{13}\\boxed{12}\\boxed{11}\\boxed{10}\\boxed{9}\\boxed{8}\\boxed{7}\\boxed{6}}\\\\ S_7\\{\\}&:& \\color{red}{\\boxed{1}\\boxed{2}\\boxed{3}\\boxed{4}\\boxed{5}\\boxed{6}\\boxed{7}\\boxed{8}\\boxed{9}\\boxed{10}\\boxed{11}\\boxed{12}\\boxed{13}} \\end{eqnarray}$$ $$\\begin{eqnarray} S_3\\{1,10,11\\}&:&\\boxed{3\\ 4\\ 7\\ 8\\ 13}\\boxed{2\\ 5\\ 6\\ 9\\ 12}\\\\ S_4\\{4,5,6,7\\}&:&\\boxed{1\\ 10\\ 11}\\color{red}{\\boxed{2\\ 9\\ 12}\\boxed{3\\ 8\\ 13}}\\\\ S_5\\{11,12,13\\}&:&\\color{red}{\\boxed{56}\\boxed{47}\\boxed{38}\\boxed{29}\\boxed{1\\ 10}}\\\\ S_6\\{1,2,3,4,5\\}&:&\\color{red}{\\boxed{13}\\boxed{12}\\boxed{11}\\boxed{10}\\boxed{9}\\boxed{8}\\boxed{7}\\boxed{6}}\\\\ S_7\\{\\}&:& \\color{red}{\\boxed{1}\\boxed{2}\\boxed{3}\\boxed{4}\\boxed{5}\\boxed{6}\\boxed{7}\\boxed{8}\\boxed{9}\\boxed{10}\\boxed{11}\\boxed{12}\\boxed{13}} \\end{eqnarray}$$ $$\\begin{eqnarray} S_2\\{3,4,7,8,13\\}&:&\\boxed{1\\ 2\\ 5\\ 6\\ 9\\ 10\\ 11\\ 12}\\\\ S_3\\{1,10,11\\}&:&\\boxed{3\\ 4\\ 7\\ 8\\ 13}\\color{red}{\\boxed{2\\ 5\\ 6\\ 9\\ 12}}\\\\ S_4\\{4,5,6,7\\}&:&\\color{red}{\\boxed{1\\ 10\\ 11}\\boxed{2\\ 9\\ 12}\\boxed{3\\ 8\\ 13}}\\\\ S_5\\{11,12,13\\}&:&\\color{red}{\\boxed{56}\\boxed{47}\\boxed{38}\\boxed{29}\\boxed{1\\ 10}}\\\\ S_6\\{1,2,3,4,5\\}&:&\\color{red}{\\boxed{13}\\boxed{12}\\boxed{11}\\boxed{10}\\boxed{9}\\boxed{8}\\boxed{7}\\boxed{6}}\\\\ S_7\\{\\}&:& \\color{red}{\\boxed{1}\\boxed{2}\\boxed{3}\\boxed{4}\\boxed{5}\\boxed{6}\\boxed{7}\\boxed{8}\\boxed{9}\\boxed{10}\\boxed{11}\\boxed{12}\\boxed{13}} \\end{eqnarray}$$ $$\\begin{eqnarray} S_1\\{\\}&:&\\boxed{1\\ 2\\ 3\\ 4\\ 5\\ 6\\ 7\\ 8\\ 9\\ 10\\ 11\\ 12\\ 13}\\\\ S_2\\{3,4,7,8,13\\}&:&\\color{red}{\\boxed{1\\ 2\\ 5\\ 6\\ 9\\ 10\\ 11\\ 12}}\\\\ S_3\\{1,10,11\\}&:&\\color{red}{\\boxed{3\\ 4\\ 7\\ 8\\ 13}\\boxed{2\\ 5\\ 6\\ 9\\ 12}}\\\\ S_4\\{4,5,6,7\\}&:&\\color{red}{\\boxed{1\\ 10\\ 11}\\boxed{2\\ 9\\ 12}\\boxed{3\\ 8\\ 13}}\\\\ S_5\\{11,12,13\\}&:&\\color{red}{\\boxed{56}\\boxed{47}\\boxed{38}\\boxed{29}\\boxed{1\\ 10}}\\\\ S_6\\{1,2,3,4,5\\}&:&\\color{red}{\\boxed{13}\\boxed{12}\\boxed{11}\\boxed{10}\\boxed{9}\\boxed{8}\\boxed{7}\\boxed{6}}\\\\ S_7\\{\\}&:& \\color{red}{\\boxed{1}\\boxed{2}\\boxed{3}\\boxed{4}\\boxed{5}\\boxed{6}\\boxed{7}\\boxed{8}\\boxed{9}\\boxed{10}\\boxed{11}\\boxed{12}\\boxed{13}} \\end{eqnarray}$$ ### Bonus: Generalization of the problem Now, the above solves the", "for the sixth item only. I assume $i \\in \\{1,2,\\ldots,100\\}$. If you meant that $i \\in \\{1,2,\\ldots,6\\}$, how will I know which of the seven out of 100 items are chosen? – ubaabd Dec 17 '14 at 5:23", "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 $(512-(384-160+40-4)) \\implies \\boxed{252}$.", "our answer is $m+n=61+4=\\boxed{065}$.", "-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}$$ Adding up the numbers below the line, we end up with $-19$, which modulo $97$ equals $78$, as expected. An advantage of this method is that, once you have those $24$ numbers, the computations are really simple and require only really small numbers. You can even do this method by hand, as I actually did for the example above.", "you can go right. Let's name these U and R. To get to $$(5, 3)$$, you have to make 5 R and 3 U. So all that you're really doing is computing 8 (the total number) choose 3 (or 5), which is $$\\boxed{56}$$. :D - 4 years, 4 months ago", "(16 - 1)^2 + 1 = \\boxed{406}$.", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "\\ldots < a_{100} < a_{101}$$. Since we have $$101$$ distinct elements, $$a_1 \\leq 99$$. Consider the set of remainders upon division by $$a_1$$. Since $$a_1 \\leq 99$$, there are at most $$98$$ such remainders. Let $$r_2$$ be the remainder upon dividing $$a_2$$ by $$a_1$$, $$r_3$$ the remainder upon dividing $$a_3$$ by $$a_1, \\ldots, r_{101}$$ the remainder upon dividing $$a_{101}$$ by $$a_1$$. We have $$101$$ remainders $$r_1,\\ldots ,r_{101}$$ (pigeons), and at most $$98$$ possible remainders (pigeonholes) upon dividing by $$a_1$$. Thus, by the pigeonhole principle, $$r_i = r_j$$ for some $$1\\leq i < j \\leq 101$$. Now what can you say about the number $$a_j - a_i$$? • but how to show that there will be two such that one evenly divides the other – user469065 Oct 1 '19 at 16:22 Hint: The boxes will have labels $1$, $3$, $5$, and so on up to $199$. Odd labels! Note that there are $100$ boxes. Box 1: Contains $1,2,4,8,16, 32,\\dots$ Box 3: Contains $3,6,12,24, 48, \\dots$ Box 5: Contains $5,10,20, 40,\\dots$. And so on. Box 99: Contains $99,198$ Boxes 101, 103, and so on are pretty boring. Box 101 only contains the number $101$, Box 103 only contains $103$, and so on. • Hmmm... why no even boxes? – user3511965 Apr 26 '14 at 7:24 • These are just labels", "to do is add up to 66 and we'll be able to find the number of people: $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66 \\implies$ $\\implies n - 1 = 11 \\implies n = 12$", "we have $101$ distinct elements, $a_1 \\leq 99$. Consider the set of remainders upon division by $a_1$. Since $a_1 \\leq 99$, there are at most 99 such remainders. Let $r_2$ be the remainder upon dividing $a_2$ by $a_1$, $r_3$ the remainder upon dividing $a_3$ by $a_1, \\ldots, r_{101}$ the remainder upon dividing $a_{101}$ by $a_1$. We have $100$ remainders $r_1,\\ldots ,r_{101}$ (pigeons), and at most $99$ possible remainders (pigeonholes) upon dividing by $a_1$. Thus, by the pigeonhole principle, $r_i = r_j$ for some $2\\leq i < j \\leq 101$. Now what can you say about the number $a_j – a_i$? Hint: The boxes will have labels $1$, $3$, $5$, and so on up to $199$. Odd labels! Note that there are $100$ boxes. Box 1: Contains $1,2,4,8,16, 32,\\dots$ Box 3: Contains $3,6,12,24, 48, \\dots$ Box 5: Contains $5,10,20, 40,\\dots$. And so on. Box 99: Contains $99,198$ Boxes 101, 103, and so on are pretty boring. Box 101 only contains the number $101$, Box 103 only contains $103$, and so on.", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "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 $1, 11, 21, 1211, \\ldots?$ ×", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$." ]

[ "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "-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}$$ Adding up the numbers below the line, we end up with $-19$, which modulo $97$ equals $78$, as expected. An advantage of this method is that, once you have those $24$ numbers, the computations are really simple and require only really small numbers. You can even do this method by hand, as I actually did for the example above.", "for the sixth item only. I assume $i \\in \\{1,2,\\ldots,100\\}$. If you meant that $i \\in \\{1,2,\\ldots,6\\}$, how will I know which of the seven out of 100 items are chosen? – ubaabd Dec 17 '14 at 5:23", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "and $3$ and $5$ never near, which leaves the total combinations (orders) of the boxes to twelve: $1-2-3-4-5$ $1-3-2-4-5$ $1-3-4-2-5$ $1-3-2-5-4$ $1-3-4-5-2$ $1-4-3-2-5$ Plus the symmetry of the exchange $5\\to 3$ So I have $12$ possible dispositions of the boxes. Now, let's call the figures on each box as $$1 = \\{A_1, A_2, A_3, A_4, A_5, A_6\\}$$ $$2 = \\{B_1, \\ldots, B_6\\}$$ $$3 = \\{C_1 \\ldots C_6\\}$$ $$4 = \\{D_1, D_2, D_3, A_1, C_1, C_2\\}$$ $$5 = \\{E_1 \\ldots E_6\\}$$", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "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", "to see here that this continues all the way down to one. However, when the case gets to $32$, there are 5 ways instead of 4 because $2^{10}-2^5$ is smaller than 1000. Thus, $9+8+7+6+5+5+4+3+2+1 = 50$ So the answer is $\\boxed{050}$ Problem 4 Solution 1 Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \\ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \\ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we have the sum $31+29+28+26+25+\\ldots+4+2+1=(31+28+25+22+\\ldots+1)+$ $(29+26+23+\\ldots+2)=176+155=\\boxed{331}$. Solution 2 Let the three piles have $a, b, c$ coins respectively. If we disregard order, then we just need to divide by $3! = 6$ at the end. We know $a + b + c = 66$. Since $a, b, c$ are positive integers, there are $\\binom{65}{2}$ ways from Stars and Bars. However, we must discard the cases where $a = b$ or $a = c$ or $b = c$. The three cases are symmetric, so we just take the first case and multiply by 3. We have $2a + c = 66 \\implies a = 1, 2, \\dots 32$ for", "# Another Think Out Of The Boxes(The Id-\"1\"-- ) What is the next number in the sequence $1, 11, 21, 1211, \\ldots?$ ×", "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$$", "13 & 169,171,\\ldots,193,195\\\\ 118 & 15 & 226,228,\\ldots,252,254\\\\ 151 & 16 & 256,257,\\ldots,287,288\\\\ 169 & 17 & 289,291,\\ldots,321,323\\\\ 188 & 19 & 362,364,\\ldots,396,398\\\\ 229 & 20 & 400,401,\\ldots,439,440\\\\ 251 & 21 & 441,443,\\ldots,481,483\\\\ 274 & 23 & 530,532,\\ldots,572,574\\\\ 323 & 24 & 576,577,\\ldots,623,624\\\\ 349 & 25 & 625,627,\\ldots,673,675\\\\ 376 & 27 & 730,732,\\ldots,780,782\\\\ 433 & 28 & 784,785,\\ldots,839,840\\\\ 463 & 29 & 841,843,\\ldots,897,899\\\\ 483 & 31 & 962,964,\\ldots,998,1000\\\\ \\end{array}$$ 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", "\\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 $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 &", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "the six numbers $-1, 8, 9, 10, 11, 12$ fit these conditions. Therefore, the answer is $\\boxed{D)}$. $$\\space$$ -Aurora64 ¯\\_(ツ)_/¯", "# Sequence of Powers of 16 The sequence of powers of $16$ begins: $1, 16, 256, 4096, 65 \\, 536, 1 \\, 048 \\, 576, 16 \\, 777 \\, 216, 268 \\, 435 \\, 456, \\ldots$", "3 are there from 100 to 500? Again, let's start by listing some of them out: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots$ At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots 492,\\, 495,\\, 498$ Now let's divide each number in the list by 3: $34,\\, 35,\\, 36,\\, 37,\\, \\ldots 164,\\, 165,\\, 166$ See what happened? The numbers are now consecutive. Now the question becomes, \"How many numbers are there from 34 to 166?\" If you're not sure, ask yourself how many numbers there are from 5 to 10. There are 6. Now how did you get that? Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1: $10 - 5 + 1 = 6$ So to get the number of numbers from 34 to 166, we do the same thing: $166 - 34 + 1 = 133$ Finally, we have the answer, 133. Although this problem may seem long and complicated, the method we used to solve it applies to many other questions like this (namely, multiple questions). Once you understand it, solving similar questions is a breeze. Hopefully, you now have a good grasp", "+0 # Can I have help please? 0 369 1 How many numbers can you get by multiplying two or more distinct members of the set $$\\{1,2,3,5,11\\}$$ together? May 10, 2021 #1 +1 Use casework. First try finding the possiblities with only multiplying 2 numbers, and then the possibilities with multiplying 3 numbers, etc. Using 2 numbers: $$2\\cdot3, 2 \\cdot 5, 2 \\cdot 11, 3 \\cdot 5, 3 \\cdot 11, 5 \\cdot 11.$$ Using 3 numbers: $$2 \\cdot 3 \\cdot 5, 2 \\cdot 3 \\cdot 11, 2 \\cdot 5 \\cdot 11, 3 \\cdot 5 \\cdot 11.$$ Using 4 numbers: $$2 \\cdot 3 \\cdot 5 \\cdot 11.$$ So we have a total of $$6+4+4+1=\\boxed{15}$$ May 10, 2021", "= 106$, and $AB = 106 + 87 = \\boxed{193}$.", "# Problems Section: 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. One may add $1$ to all numbers of any row or subtract $1$ from all numbers of any column." ]

[ "whole chord of the triangle is 14.4 cm long, how do you calculate the shorter and longer part?", "## Elementary Geometry for College Students (7th Edition) Clone If the side lengths are a, b, and c and $a^2+b^2 =c^2$, then it is a right triangle. We consider a triangle with side lengths a, b, and c. We consider the case where the angle is a right angle. If and only if this is the case, the distance c is: $c= \\sqrt{a^2+b^2} \\\\ c^2 =a^2 +b^2$ If it was not a right triangle, this would not be the case. Thus, if the side lengths are a, b, and c and $a^2+b^2 =c^2$, then it is a right triangle.", "and $$B$$ can be swapped around, but when using this formula, $$C$$ is always the longest side (which is also the side opposite the 90-degree angle). Figure 15. Longest side triangle Video! This video walks through how to apply Pythagoras’ theorem on a right triangle.", "length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle. Focus on math practices Use Structure Could Kayla use the straws that form a right triangle to make a triangle that is not a right triangle? Explain. From the given straws, We can observe that the pair (6, 7, 4) can’t form a right triangle but (3, 4, 5) can form a right triangle Hence, from the above, We can conclude that Kayla can use the straws that form a right triangle to make a triangle that is not a right triangle Essential Question How can you determine whether a triangle is a right triangle? We can determine the triangle is a right triangle by using the converse of the Pythagorean Theorem Hence, According to the converse of the Pythagorean Theorem, If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle. Try It! A triangle has side lengths 4 inches, 5 inches, and 7 inches. Is the triangle a right triangle? Is a2 + b2 equal to c2? _______ Is the triangle a right triangle? _________ It", "right angle at C, we know the side lengths AC = 9 cm and BC = 7 cm. Calculate the length of the remaining side of the triangle and the size of all angles.", "triangle. What are the lengths of the two legs of the right triangle?", "the square of the longest side. • The triangle is an obtuse triangle if the sum of the squares of the shorter sides is less than the square of the longest side. • The triangle is a right triangle if the sum of squares of the shorter side is equal to the square of the longest side.", "# Is a triangle with sides of length 10 cm 24 cm 27 cm a right triangle? Mar 3, 2016 not a right triangle #### Explanation: Use the $\\textcolor{b l u e}{\\text{ Converse of Pythagoras' Theorem }}$ Pythagoras' Theorem states ' the square on the hypotenuse of a right triangle is equal to the sum of the squares on the other 2 sides ' The converse of this is ' if the square on the longest side of a triangle equals the sum of the squares on the other 2 sides then the triangle is right'. $\\text{(longest side)^2} = {27}^{2} = 729$ and ${10}^{2} + {24}^{2} = 100 + 576 = 676$ now 729 ≠ 676 , hence not a right triangle.", "### Still have math questions? In right triangle $$P R Q , \\overline { P Q }$$ is the hypotenuse and $$m \\angle P = 58$$ . What is the shortest side of $$\\triangle P R Q ?$$", "Answer the 13th question. { NO LINKS }.{ NO ANSWERS FROM GOOGLE }. Q.13. The sides of a right-angled triangle containing the right angle are 5x cm and (3x - 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 $c{m}^{2}$. • 0 What are you looking for?", "given rectangle is 8 cm the., base and height are used instead of length and intersect in the middle different shapes x length! And then we have a right triangle here we 've got Pythagorean Theorem are.", "side of a right triangle: 26,24,10. 14. Chord In a circle with radius r=60 cm is chord 4× longer than its distance from the center. What is the length of the chord?", "0 শক্তি পয়েন্ট পরীক্ষার জন্য পড়ছো? পিথাগোরাসের উপপাদ্য সম্পর্কিত এই 4 টি পাঠের সাহায্যে প্রস্তুতি নাও। 4 টি পাঠ দেখো # পিথাগোরাসের উপপাদ্যের উদাহরণ ভিডিও ট্রান্সক্রিপ্ট Say we have a right triangle. Let me draw my right triangle just like that. This is a right triangle. This is the 90 degree angle right here. And we're told that this side's length right here is 14. This side's length right over here is 9. And we're told that this side is a. And we need to find the length of a. So as I mentioned already, this is a right triangle. And we know that if we have a right triangle, if we know two of the sides, we can always figure out a third side using the Pythagorean theorem. And what the Pythagorean theorem tells us is that the sum of the squares of the shorter sides is going to be equal to the square of the longer side, or the square of the hypotenuse. And if you're not sure about that, you're probably thinking, hey Sal, how do I know that a is shorter than this side over here? How do I know it's not 15 or 16? And the way to tell is that the longest side in a right triangle, and this only applies to", "is the longest side of a right triangle. Here are some right triangles. Each hypotenuse is labeled. • legs The legs of a right triangle are the sides that make the right angle. Here are some right triangles. Each leg is labeled.", "side is 24 cm. #### Question 3: The length of one sides of a right triangle is 4.5 cm and the length of its hypotenuse is 7.5 cm. Find the length of its third side. Suppose the length of the other side is a cm. Then, by Pythagoras theorem: Hence, the length of the other side of the triangle is 6 cm. #### Question 4: The two legs of a right triangle are equal and the square of its hypotenuse is 50. Find the length of each leg. Suppose the length of the two legs of the right triangle are a cm and a cm. Then, by Pythagoras theorem: Hence, the length of each leg is 5 cm. #### Question 5: The sides of a triangle measure 15 cm, 36 cm and 39 cm. Show that it is a right-angled triangle. The largest side of the triangle is 39 cm. Also, Sum of the square of the two sides is equal to the square of the third side. Hence, the triangle is right angled. #### Question 6: In right ∆ABC, teh lengths of its legs are given as a = 6 cm and b = 4.5 cm. Find the length of its hypotenuse. Suppose the length of the hypotenuse is c cm. Then, by Pythagoras theorem: Hence, the", "right-angled triangle is k times the shortest side. If the ratio of the other side to hypotenuse is 4 : 6, then what is the value of k? A) 2 B) 3 C) 4 D) 5 Explanation: 0 124 Q: If length of each side of a rhombus PQRS is 8 cm and ∠PQR = 120°, then what is the length (in cm) of QS? A) 4√5 B) 6 C) 8 D) 12", "of a right triangle is the longest side of the right triangle. It is across from the right angle. Legs of a Right Triangle The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle. Trigonometric Ratios Ratios that help us to understand the relationships between sides and angles of right triangles. ### Explore More Sign in to explore more, including practice questions and solutions for Trigonometric Ratios with a Calculator. Please wait... Please wait...", "# find the length of altitude to the smallest side of a triangle having sides of lengths 7 cm , 5 cm , 8 cm. plz its urgent . • 4 What are you looking for?", "? Free Version Easy Law of Sines: Right Triangle TRIG-BDBSX6 Given the triangle sho​wn above, what is the length of side “b” to the nearest tenth? A 15.6 B 11.3 C 29 D 28.2", "## Geometry: Common Core (15th Edition) Since $a^2+b^2=c^2$, where a, b and c must all be greater than 0, we know that the hypotenuse must be the longest side of the triangle. Thus, the 13 cm side is the longest side." ]

[ "# Hypotenuse and height In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs. Correct result: a = 55.9 cm b = 4 cm #### Solution: $56=x+y \\ \\\\ 4^2=xy \\ \\\\ \\ \\\\ x(56-x) = 4^2 \\ \\\\ \\ \\\\ x^2 -56x +16 =0 \\ \\\\ \\ \\\\ a=1; b=-56; c=16 \\ \\\\ D = b^2 - 4ac = 56^2 - 4\\cdot 1 \\cdot 16 = 3072 \\ \\\\ D>0 \\ \\\\ \\ \\\\ x_{1,2} = \\dfrac{ -b \\pm \\sqrt{ D } }{ 2a } = \\dfrac{ 56 \\pm \\sqrt{ 3072 } }{ 2 } = \\dfrac{ 56 \\pm 32 \\sqrt{ 3 } }{ 2 } \\ \\\\ x_{1,2} = 28 \\pm 27.712812921102 \\ \\\\ x_{1} = 55.712812921102 \\ \\\\ x_{2} = 0.28718707889796 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (x -55.712812921102) (x -0.28718707889796) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = \\sqrt{ x_1 \\cdot 56 } = 55.9 \\ \\text{cm}$ $b = \\sqrt{ x_2 \\cdot 56 } = 4 \\ \\text{cm}$ 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 Looking for", "of $1.01$ and it also improves as $k$ goes to infinity, with a limit of $1$. (as an opposite of the first approximation this approximation always gives a hypotenuse longer than it actually is, which again tells us this is not actually the best approximation) So a better way to approximate the hypotenuse is to add the longer leg's length plus three sevenths of the shorter leg's length. • Isosceles right triangle of side $1$ gives $1.375$, reasonably good. – André Nicolas Jul 6 '15 at 19:15 • @dREaM Actually for equal lengths its pretty good as well - assuming its 5 on each side, the difference is 0.2 – rtindru Jul 6 '15 at 19:24 • Yeah, I updated it with some stuff. Hope to come back to interesting material when I get home. – Jorge Fernández Hidalgo Jul 6 '15 at 19:28 The method provides a not-so-great approximation when the two catheti are roughly of the same length. In this case, $$\\frac{7}{8}a+\\frac{1}{2}b\\approx\\frac{7}{8}a+\\frac{1}{2}a=\\frac{11}{8}a=1\\mathrm.375 a\\approx(1\\mathrm.414...)a=\\sqrt{2}{a}\\approx\\sqrt{a^2+b^2}.$$ Numerical optimization reveals how truly astonishing the approximation is: Taking a hint from @dREaM's answer and normalizing the length of the shortest side, for side lengths $1$ and $k$ we are interested in the ratio $$\\frac{ak+b}{\\sqrt{k^2+1}}$$ as a function of the coefficients $a$ and $b$ (in the Tamil approximation, $a=7/8$ and $b=1/2$). We", "90° triangle, let the side across from the 60° angle is given as 9√3. Find the length of the other two sides. 2. If the hypotenuse of the 30°- 60°- 90° triangle is 26, find the other two sides. 3. If the longer side of a 30°- 60°- 90° triangle is 12, what is the sum of the other two sides of this triangle? ### Practice Questions 1. A right triangle whose one angle measures $60$ degrees has the longer side as $12\\sqrt{3}$ cm. Which of the following shows the length of its shorter side? 2. A right triangle whose one angle measures $60$ degrees has the longer side as $12\\sqrt{3}$ cm. Which of the following shows the length of its hypotenuse? 3. In a $30^{\\circ}- 60^{\\circ}- 90^{\\circ}$ triangle, let the side across from the $60^{\\circ}$ angle is given as $9\\sqrt{3}$ ft. Which of the following shows the lengths of the shorter side and hypotenuse? 4. If the hypotenuse of the $30^{\\circ}- 60^{\\circ}- 90^{\\circ}$triangle is $26$ meters, which of the following shows the lengths of the two sides? 5. If the longer leg of a $30^{\\circ}- 60^{\\circ}- 90^{\\circ}$ triangle is $12$ yards, what is the sum of the lengths of its shorter leg and hypotenuse?", "of the triangle is 129.5 cm2. ∴ 7.4h = 129.5 h = $\\left(\\frac{129.5}{7.4}\\right)$ = 17.5 cm ∴ Length of the other leg = 17.5 cm Question 7: Find the area of a right triangle whose base is 1.2 m and hypotenuse 3.7 m. Here, base = 1.2 m and hypotenuse = 3.7 m In the right angled triangle: Perpendicular = $\\sqrt{\\left(H\\mathrm{ypotenuse}{\\right)}^{2}-\\left(B\\mathrm{ase}{\\right)}^{2}}$ $=\\sqrt{{\\left(3.7\\right)}^{2}-{\\left(1.2\\right)}^{2}}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{13.69-1.44}\\phantom{\\rule{0ex}{0ex}}=\\sqrt{12.25}\\phantom{\\rule{0ex}{0ex}}=3.5$ Area = $\\left(\\frac{1}{2}×\\mathrm{base}×\\mathrm{perpendicular}\\right)$ sq. units = $\\left(\\frac{1}{2}×1.2×3.5\\right)$ m2 ∴ Area of the right angled triangle = 2.1 m2 Question 8: The legs of a right triangle are in the ratio 3 : 4 and its area is 1014 cm2. Find the lengths of its legs. In a right angled triangle, if one leg is the base, then the other leg is the height. Let the given legs be 3x and 4x, respectively. Area of the triangle = $\\left(\\frac{1}{2}×3x×4x\\right)$ cm2 ⇒ 1014 = (6x2) ⇒ 1014 = 6x2 x2 = $\\left(\\frac{1014}{6}\\right)$ = 169 x = $\\sqrt{169}$ = 13 ∴ Base = (3 $×$ 13) = 39 cm Height = (4 $×$ 13) = 52 cm Question 9: One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its cost at the rate of Rs 250 per m2. Consider a right-angled triangular scarf (ABC). Here, ∠B= 90° BC = 80", "# Hypotenuse and height In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs. Correct result: a = 55.9 cm b = 4 cm #### Solution: $56=x+y \\ \\\\ 4^2=xy \\ \\\\ \\ \\\\ x(56-x) = 4^2 \\ \\\\ \\ \\\\ x^2 -56x +16 =0 \\ \\\\ \\ \\\\ a=1; b=-56; c=16 \\ \\\\ D = b^2 - 4ac = 56^2 - 4\\cdot 1 \\cdot 16 = 3072 \\ \\\\ D>0 \\ \\\\ \\ \\\\ x_{1,2} = \\dfrac{ -b \\pm \\sqrt{ D } }{ 2a } = \\dfrac{ 56 \\pm \\sqrt{ 3072 } }{ 2 } = \\dfrac{ 56 \\pm 32 \\sqrt{ 3 } }{ 2 } \\ \\\\ x_{1,2} = 28 \\pm 27.712812921102 \\ \\\\ x_{1} = 55.712812921102 \\ \\\\ x_{2} = 0.28718707889796 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (x -55.712812921102) (x -0.28718707889796) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = \\sqrt{ x_1 \\cdot 56 } = 55.9 \\ \\text{cm}$ $b = \\sqrt{ x_2 \\cdot 56 } = 4 \\ \\text{cm}$ 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", "triangle is $\\displaystyle 16\\sqrt{3}$ If the perimeter of an equilateral triangle is 45, then each side would be 15. This would mean that in your triangular configuration, 15 is the hypotenuse. The short side would be 7.5 (half the hypotenue) and the long side (altitude) would be $\\displaystyle 7.5\\sqrt{3}$", "# Hypotenuse and height In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs. Correct result: a = 55.9 cm b = 4 cm #### Solution: $56=x+y \\ \\\\ 4^2=xy \\ \\\\ \\ \\\\ x(56-x) = 4^2 \\ \\\\ \\ \\\\ x^2 -56x +16 =0 \\ \\\\ \\ \\\\ a=1; b=-56; c=16 \\ \\\\ D = b^2 - 4ac = 56^2 - 4\\cdot 1 \\cdot 16 = 3072 \\ \\\\ D>0 \\ \\\\ \\ \\\\ x_{1,2} = \\dfrac{ -b \\pm \\sqrt{ D } }{ 2a } = \\dfrac{ 56 \\pm \\sqrt{ 3072 } }{ 2 } = \\dfrac{ 56 \\pm 32 \\sqrt{ 3 } }{ 2 } \\ \\\\ x_{1,2} = 28 \\pm 27.7128129211 \\ \\\\ x_{1} = 55.7128129211 \\ \\\\ x_{2} = 0.287187078898 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (x -55.7128129211) (x -0.287187078898) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = \\sqrt{ x_1 \\cdot 56 } = 55.9 \\ \\text{cm}$ 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 Looking for help with calculating roots of a quadratic equation? Do you have", "of the box? It is given that A packaging box for a metal rod is 7.5 inches along a diagonal of the base. The height of the box is 18 inches. Now, From the given figure, We can observe that it looks like a right triangle Now, We know that, According to the Pythagorean Theorem, c² = a²+ b² Where, c is the length of the diagonal a and b are the side lengths So, c² = 18² + (7.5)² c²= 324 + 56.25 c² = 380.25 c = $$\\sqrt{380.25}$$ c = 19.5 inches Hence, from the above, We can conclude that the length of the diagonal of the box is: 19.5 inches Lesson 7.4 Find Distance in the Coordinate Plane Quick Review The Pythagorean Theorem can be used to find the distance between any two points on the coordinate plane. Example Find the distance between the two points on the coordinate plane. Round to the nearest tenth. Draw a right triangle. Determine the lengths of its legs. The length of the horizontal leg is 5 units. The length of the vertical leg is 5 units. Use the relationship a2 + b2 = c2. Substitute 5 for a and 5 for b. Then solve for C. a2 + b2 = -2 52 + 52 = c2 25 +", "hypotenuse of length 17 m, one leg of length 8 m, and the other leg of length $\\displaystyle \\sqrt{17^2- 8^2}= 15$ m. The reason for the \"additional 14 N\" is that the fish is being reeled in horizontally but the line is the hypotenuse of the triangle. Find the total force so that the horizontal component is 105 N. The ratio of hypotenuse to horizontal side of the right triangle is 17/15 so the tension in the line is (17/15)(105)= 119. Oct 22nd 2018, 07:01 PM #3 Junior Member Join Date: Jun 2018 Posts: 7 Thank you for your help. So the upward component X would be (8/15)*105 = 56. Let B be Buoyancy, then X + B = 150 56 + B = 150 So B = 94N So Normal contact force on the man & rod would be 800 + 56 = 856 N Tags forces, questions, tensions Thread Tools Display Modes Linear Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post hongiddong Kinematics and Dynamics 4 Aug 6th 2014 02:23 PM iannedrs Advanced Mechanics 3 Jul 22nd 2013 05:03 AM kastamonu Kinematics and Dynamics 9 Mar 31st 2013 07:57 AM flower Advanced Electricity and Magnetism 1 Nov 20th 2009 02:06 AM jello17 Kinematics and Dynamics 3 Oct 4th 2008 09:31 PM", "we know the hypotenuse and the long leg have a ratio of , we can solve for the height by creating a proportion. Sides of a parallelogram if you know diagonals and angle between them or other side, 2. To learn more, see our tips on writing great answers. The two side lengths are 24 and 30. If one diagonal of it is 16 cm long; find area of the parallelogram. Here is my attempt so far. From the problem: G is false. 2. 1 => Area of Paralelogram = $$Height \\times Base$$ [Put the Value of Heigh and Base] => Area of Paralelogram = $$4\\times 6$$ => Area of Paralelogram = $$24 cm^2$$ If you divide both sides of the formula by b , you end up with the formula for h , or the height. h b. The Law of Cosines: Where is the length of the unknown side, and are the lengths of the known sides, and is the angle between and . In my diagram, vector u is a base, Sorry for not answering with the correct format but I'm really lost. The height (altitude) is found by drawing a perpendicular line from the base to the highest point on the shape. The area of and perimeter of parallelogram is given by (where b", "right-angled triangle is k times the shortest side. If the ratio of the other side to hypotenuse is 4 : 6, then what is the value of k? A) 2 B) 3 C) 4 D) 5 Explanation: 0 124 Q: If length of each side of a rhombus PQRS is 8 cm and ∠PQR = 120°, then what is the length (in cm) of QS? A) 4√5 B) 6 C) 8 D) 12", "## Trigonometry 7th Edition The path of bullet under given conditions will be along hypotenuse of a 45°–45°–90° triangle. Distance traveled by it when it is 1000 feet above ground will be the length of hypotenuse, while length of shorter side is 1000 feet. In a 45°–45°–90° triangle, length of hypotenuse is $\\sqrt 2$ times the length of shorter side, Hence- Distance traveled by the bullet = $1000 \\times \\sqrt 2$ Distance traveled by the bullet = $1000 \\times 1.41421$ = 1414.21 feet Distance traveled by the bullet in nearest feet = 1414 feet", "that is 2 more than the double of the width and that has a hypotenuse that measures $\\sqrt{68}$ meters. Good luck!", "look up that $$\\tan(10)$$ is 0.176. Then we can calculate that $$h$$ is about 17.6. That means the tree is 17.6 feet tall. Glossary Entries • cosine The cosine of an acute angle in a right triangle is the ratio (quotient) of the length of the adjacent leg to the length of the hypotenuse. In the diagram, $$\\cos(x)=\\frac{b}{c}$$. • sine The sine of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the hypotenuse. In the diagram, $$\\sin(x) = \\frac{a}{c}.$$ • tangent The tangent of an acute angle in a right triangle is the ratio (quotient) of the length of the opposite leg to the length of the adjacent leg. In the diagram, $$\\tan(x) = \\frac{a}{b}.$$ • trigonometric ratio Sine, cosine, and tangent are called trigonometric ratios.", "Limited access Given the above right triangle, We know $\\text{ angle}\\:A =\\theta$ is $78^{\\circ}$ and the adjacent leg to that angle ($\\text{ side}\\: b$) is $17\\text{ cm}$, What is the measure of the hypotenuse ($\\text{ side}\\: c$) to the nearest tenth of a centimeter? A $81.8\\text{ cm}$ B $17.4\\text{ cm}$ C $3.6\\text{ cm}$ D $81.6\\text{ cm}$ Select an assignment template", "$$p ^ *$$. 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", "other leg is 8 and the hypotenuse is 82\\begin{align*}8\\sqrt{2}\\end{align*}. #### Example B Find the lengths of the other two sides of the isosceles right triangle below. Solution: If the hypotenuse has length 72\\begin{align*}7\\sqrt{2}\\end{align*}, then both legs are 7. #### Example C Find the lengths of the other two sides of the isosceles right triangle below. Solution: Because the leg is 102\\begin{align*}10\\sqrt{2}\\end{align*}, then so is the other leg. The hypotenuse will be 102\\begin{align*}10\\sqrt{2}\\end{align*} multiplied by an additional 2\\begin{align*}\\sqrt{2}\\end{align*}. 1022=102=20 ### Guided Practice 1. Find the length of the other two sides of the isosceles right triangle below: 2. Find the length of the other two sides of the isosceles right triangle below: 3. Find the length of the other two sides of the isosceles right triangle below: Solutions: 1. Since we know the length of the given leg is 12\\begin{align*}12\\end{align*}, and it isn't the hypotenuse, that means the other side that isn't opposite the right angle also has a length of 12\\begin{align*}12\\end{align*}. We can then determine from the relationships for an isosceles right triangle that the length of the hypotenuse is 122\\begin{align*}12\\sqrt{2}\\end{align*}. 2. Since we know the length of the hypotenuse is 8\\begin{align*}\\sqrt{8}\\end{align*}, we can determine the lengths of the other two sides. Because the length of the hypotenuse is 2\\begin{align*}\\sqrt{2}\\end{align*} times the length of the other sides, we", "# Right trapezoid The right trapezoid has bases 3.2 cm and 62 mm long. The shorter leg has a length 0.25 dm. Calculate the lengths of the diagonals and the second leg. Correct result: u1 = 6.685 cm u2 = 4.061 cm b = 3.905 cm #### Solution: $a=62 \\ mm \\rightarrow cm=62 / 10 \\ cm=6.2 \\ cm \\ \\\\ c=3.2 \\ \\text{cm} \\ \\\\ d=0.25 \\ dm \\rightarrow cm=0.25 \\cdot \\ 10 \\ cm=2.5 \\ cm \\ \\\\ \\ \\\\ u_{1}^2=d^2 + a^2 \\ \\\\ u_{1}=\\sqrt{ d^2 + a^2 }=\\sqrt{ 2.5^2 + 6.2^2 }=6.685 \\ \\text{cm}$ $u_{2}^2=d^2 + c^2 \\ \\\\ u_{2}=\\sqrt{ d^2 + c^2 }=\\sqrt{ 2.5^2 + 3.2^2 }=4.061 \\ \\text{cm}$ $b^2=d^2 + (a-c)^2 \\ \\\\ b=\\sqrt{ d^2 + (a-c)^2 }=\\sqrt{ 2.5^2 + (6.2-3.2)^2 }=3.905 \\ \\text{cm}$ 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 Do you want to convert length units? Pythagorean theorem is the base for the right triangle calculator. #### You need to know the following knowledge to", "# What is the length of the hypotenuse, in a 30\u0001-60\u0001-90\u0001 triangle the length of the shorter leg is 8 units? Nov 26, 2014 Since the ratios of the lengths of the sides of a 30-60-90 triange are $1 : \\sqrt{3} : 2$ by multiplying by $8$, $\\implies 8 : 8 \\sqrt{3} : 16$ Hence, the length of the hypotenuse is $16$. I hope that this was helpful.", "parallel of. See that this side length of a trapezoid with top side length of 2 centimeters and side! 3 on a question: Consider the enlargement of the fourth side ( which is longer... Learn how to find the long side length of each missing side of... Calculator and the length of 2 centimeters and bottom side length of hypotenuse side is 15 units is in... So I 'm gon na say six and four is 10 = ( )! That this side length of hypotenuse side is 15 units ( 2x+8 ) + 5x-2... Of x centimeters and bottom side length of 3.25 centimeters 25 + a 225. Thank you – 7 length 7 and base ( 8 - 5 ) or. How do I find the missing dimension of the proportions is correctly set up to the! + a 2 225 = 25 + a 2 225 = 25 + a 2 Subtract 25 both... Prep topics ) Delancy 4 September, 18:36 easier than with our great tool - right side. The unknown x isosceles equal sides are given as the unknown x read the lesson on area of a is. Smaller trapezoid with top side length is five centimeters please enter angles in degrees, here you can find length...: how to find the missing side" ]

[ "right-angled triangle is k times the shortest side. If the ratio of the other side to hypotenuse is 4 : 6, then what is the value of k? A) 2 B) 3 C) 4 D) 5 Explanation: 0 124 Q: If length of each side of a rhombus PQRS is 8 cm and ∠PQR = 120°, then what is the length (in cm) of QS? A) 4√5 B) 6 C) 8 D) 12", "+0 # Each triangle is a 30-60-90 triangle 0 103 2 Each triangle is a 30-60-90 triangle, and the hypotenuse of one triangle is the longer leg of an adjacent triangle. The hypotenuse of the larger triangle is 16 centimeters. What is the number of centimeters in the length of the shorter leg of the smaller triangle? Nov 4, 2020 #1 +56 0 30-60-90 triangles have a property where the shortest side is $$\\frac{1}{2}$$ of the longest side. It also has a property where the other leg is $$\\frac{\\sqrt3}{2}$$ of the longest side in a triangle. Because the triangles are 30 - 60 - 90 triangles we know that the length of the hypotenuse of the smaller triangle is $$16 \\cdot \\frac{\\sqrt3}{2} = 8\\sqrt3$$. And to find the shortest side of this triangle we refer to what is above showing us that the shorter leg of the smallest triangle is $$8\\sqrt3$$, which is your final answer. Nov 4, 2020 #2 +1457 +1 The length of the shortest leg = [sqrt(162 - 82)] / 2 Nov 4, 2020", "is 1 m shorter than the hypotenuse, and the second leg is 2 m shorter than the hypotenuse. Find the lengths of all sides of the right-angled triangle. • Similarity coefficient In the triangle TMA the length of the sides is t = 5cm, m = 3.5cm, a = 6.2cm. Another similar triangle has side lengths of 6.65 cm, 11.78 cm, 9.5 cm. Determine the similarity coefficient of these triangles and assign similar sides to each other. • The triangles The triangles KLM and ABC are given, which are similar to each other. Calculate the lengths of the remaining sides of the triangle KLM, if the lengths of the sides are a = 7 b = 5.6 c = 4.9 k = 5 • Similarity of two triangles The KLM triangle has a side length of k = 6.3cm, l = 8.1cm, m = 11.1cm. The triangle XYZ has a side length of x = 8.4cm, y = 10.8cm, z = 14.8cm. Are triangle KLM and XYZ similar? (write 0 if not, if yes, find and write the coefficient of a similarity) • Cutting cone A cone with a base radius of 10 cm and a height of 12 cm is given. At what height above the base should we divide it by a section parallel to the", "triangle? If so, which side is the hypotenuse? Solution: We know that hypotenuse is the longest side. If 9 cm, 11 cm and 6 cm are the lengths of angled triangle, then 11 cm will be the hypotenuse. Using the converse of Pythagoras theorem, we get (11)2 = (9)2 + (6)2 ⇒ 121 = 81 + 36 ⇒ 121 ≠ 117 Since, both the sides are not equal therefore 9 cm, 11 cm and 6 cm are not the side of the right angled triangle. The above examples of the converse of Pythagorean Theorem will help us to determine the right triangle when the sides of the triangles will be given in the questions. Meritpath...Way to Success! Meritpath provides well organized smart e-learning study material with balanced passive and participatory teaching methodology. Meritpath is on-line e-learning education portal with dynamic interactive hands on sessions and worksheets.", "• Question 8: 3 pts Find to the nearest degree the measure of the angle marked as $x$. x=15o x=16o x=17o x=18o • Question 9: 3 pts The legs of the right triangle measure 30cm and 45cm. Find to the nearest degree the measure of the smallest angle of this triangle. $x=30^\\circ$ $x=32^\\circ$ $x=34^\\circ$ $x=36^\\circ$", "triangle is a 1: $$\\sqrt{3}$$:2 = 30:60:90 Therefore the smallest angle is 30 degrees. Regards, Intern Joined: 24 Mar 2017 Posts: 1 ### Show Tags 22 Jul 2017, 21:21 I solved this a little differently, I would like to get some feedback. Since its a right triangle we know one side is 90 degrees so 180=90+x(second biggest angle)+.5x(small angle) X= 60 Smalles angle is 30 Manager Joined: 16 Jan 2013 Posts: 93 GMAT 1: 490 Q41 V18 GMAT 2: 610 Q45 V28 GPA: 2.75 ### Show Tags 02 Dec 2017, 03:02 Doesn't shortest side mean shortest leg and the longest side mean the bigger leg of a right triangle? _________________ Math Expert Joined: 02 Sep 2009 Posts: 47112 ### Show Tags 02 Dec 2017, 04:26 Doesn't shortest side mean shortest leg and the longest side mean the bigger leg of a right triangle? No. The longest side in a right triangle is the hypotenuse. _________________ Intern Joined: 19 Oct 2013 Posts: 12 Location: Kuwait GPA: 3.2 WE: Engineering (Real Estate) ### Show Tags 02 Dec 2017, 23:36 I believe the only trick in this question is realizing that the longest side means the hypotenuse and not the LONG side Re: M12-06 &nbs [#permalink] 02 Dec 2017, 23:36 Display posts from previous: Sort by # M12-06 Moderators: chetan2u, Bunuel", "= $$\\frac{1}{2}$$ Square both the sides: $$\\frac{a^2}{c^2}$$ = $$\\frac{1}{4}$$ putting the value of $$c^2$$ : $$\\frac{a^2}{a^2+b^2}$$ = $$\\frac{1}{4}$$ Cross multiply: 4$$a^2$$ = $$a^2 + b^2$$ Further simplify: $$b^2$$ = 3$$a^2$$ => $$\\frac{a}{b}$$= $$\\frac{1}{\\sqrt{3}}$$ by now we should know that the triangle is a 1: $$\\sqrt{3}$$:2 = 30:60:90 Therefore the smallest angle is 30 degrees. Regards, Kudos [?]: 17 [0], given: 15 Intern Joined: 24 Mar 2017 Posts: 1 Kudos [?]: 0 [0], given: 2 ### Show Tags 22 Jul 2017, 20:21 I solved this a little differently, I would like to get some feedback. Since its a right triangle we know one side is 90 degrees so 180=90+x(second biggest angle)+.5x(small angle) X= 60 Smalles angle is 30 Kudos [?]: 0 [0], given: 2 Manager Joined: 16 Jan 2013 Posts: 100 Kudos [?]: 4 [0], given: 1261 GMAT 1: 490 Q41 V18 GPA: 2.75 ### Show Tags 02 Dec 2017, 02:02 Doesn't shortest side mean shortest leg and the longest side mean the bigger leg of a right triangle? _________________ Kudos [?]: 4 [0], given: 1261 Math Expert Joined: 02 Sep 2009 Posts: 43334 Kudos [?]: 139559 [0], given: 12794 ### Show Tags 02 Dec 2017, 03:26 Doesn't shortest side mean shortest leg and the longest side mean the bigger leg of a right triangle? No. The longest side", "the sum of the squares of the other two sides. Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle. Question 2: In right ∆ ABC, the lengths of the legs are given. Find the length of the hypotenuse. (i) a = 6 cm, b = 8 cm (ii) a = 8 cm, b = 15 cm (iii) a = 3 cm, b = 4 cm (iv) a = 2 cm, b = 1.5 cm According to the Pythagoras theorem, Question 3: The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm, find the length of the other side. Question 4: A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches. Question 5: If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle. Question 6: The sides of certain triangles are given below. Determine which of them", "## Elementary Geometry for College Students (7th Edition) Clone If the side lengths are a, b, and c and $a^2+b^2 =c^2$, then it is a right triangle. We consider a triangle with side lengths a, b, and c. We consider the case where the angle is a right angle. If and only if this is the case, the distance c is: $c= \\sqrt{a^2+b^2} \\\\ c^2 =a^2 +b^2$ If it was not a right triangle, this would not be the case. Thus, if the side lengths are a, b, and c and $a^2+b^2 =c^2$, then it is a right triangle.", "Exercise 8.4 1. Find the length of AC in the rightangled triangle ABC shown in Fig.8.29. 2. What is the length PR in the rightangled triangle PQR shown in Fig. 8.30 below given that PS = 6 cm and PQ = 14 cm? 3. Find the length NP in triangle MNO shown in Fig. 8.31 below. NO = 10 cm. 4. What are the lengths of BD and AC in triangle ABC, Fig. 8.32, given that AD = 8 cm and CD = 12 cm? 5. Find the length of WX in triangle WXY, Fig. 8.33, given that WY = 7 cm and YB = 9 cm. 6. The hypotenuse of a right-angled triangle is 5 cm long and the longer leg of the triangle is 4 cm long. What is the length of the projection of the shorter leg on the hypotenuse? 7. In triangle UVW in Fig. 8.34, UX = 6 cm and VW = 8 cm. Find: (a) XV (b) XW 8. Fig. 8.35 below shows triangle KLM. Given that KM = 12 cm, ML = 9cm and KL = 15 cm, find; (a) KN (b) NL (c) NM 9. In Fig. 8.36, O is the centre of the circle. PR = 12 cm and PT = TU = UR. Find; (a) QR (b) PQ", "side is 24 cm. #### Question 3: The length of one sides of a right triangle is 4.5 cm and the length of its hypotenuse is 7.5 cm. Find the length of its third side. Suppose the length of the other side is a cm. Then, by Pythagoras theorem: Hence, the length of the other side of the triangle is 6 cm. #### Question 4: The two legs of a right triangle are equal and the square of its hypotenuse is 50. Find the length of each leg. Suppose the length of the two legs of the right triangle are a cm and a cm. Then, by Pythagoras theorem: Hence, the length of each leg is 5 cm. #### Question 5: The sides of a triangle measure 15 cm, 36 cm and 39 cm. Show that it is a right-angled triangle. The largest side of the triangle is 39 cm. Also, Sum of the square of the two sides is equal to the square of the third side. Hence, the triangle is right angled. #### Question 6: In right ∆ABC, teh lengths of its legs are given as a = 6 cm and b = 4.5 cm. Find the length of its hypotenuse. Suppose the length of the hypotenuse is c cm. Then, by Pythagoras theorem: Hence, the", "### Still have math questions? In right triangle $$P R Q , \\overline { P Q }$$ is the hypotenuse and $$m \\angle P = 58$$ . What is the shortest side of $$\\triangle P R Q ?$$", "× Outside the Box Geometry # [b]ASS Fishing This quiz will explore descriptions of triangles… and fishing. As in previous quizzes, we will be looking at descriptions of triangles, each of which will do one of the following: • define a unique triangle, if all triangles that have all of the specified properties are congruent; • underconstrain a triangle, if at least two different (non-congurent) triangles can have all of the specified properties; • overconstrain a triangle, if no triangle exists that has all of the specified properties. One type of triangle that will come up again and again in this quiz is the 30-60-90 right triangle. You can think of this triangle as half of an equilateral triangle. If we assume that the shortest side in this triangle has a side length of 1, then the hypotenuse will have a length of 2. What is the length of the other leg? Which of these sets of edges does not describe a 30-60-90 triangle? Hint: Remember that any 30-60-90 triangle will have either the three leg lengths $$\\big(1, \\sqrt{3}, 2\\big)$$ or the lengths will all be scaled larger or smaller together. For example, a triangle with side lengths $$\\big(2, 2\\sqrt{3}, 4\\big)$$ will also be a 30-60-90 triangle. Sally wants to make a right triangle with one angle of", "# Properties of a 30-60-90 Right Triangle ### A special kind of triangle A 30-60-90 right triangle (literally pronounced \"thirty sixty ninety\") is a special type of right triangle where the three angles measure 30 degrees, 60 degrees, and 90 degrees. The triangle is significant because the sides exist in an easy-to-remember ratio: 1:$$\\sqrt{3}$$:2. That is to say, the hypotenuse is twice as long as the shorter leg, and the longer leg is the square root of 3 times the shorter leg. You might also remember it as \"X, 2X, and X roots of 3\", which is how I remember it, but then you have to remember that 2X is actually the longest side, not X roots of 3. Which side is which? The side opposite the 30 degree angle will have the shortest length. The side opposite the 60 degree angle will be $$\\sqrt{3}$$ times as long, and the side opposite the 90 degree angle will be twice as long. The triangle below diagrams this relationship. Remember that the longest side will be opposite the largest angle, and the shortest opposite the smallest angle. We can use the relationship between the angles and the sides of a 30-60-90 triangle to find missing angles or side lengths. Take a look at this example: ### Example 1 Given the", "is smaller ($AP$ is the shortest side of triangle $APB$) and the third angle of the triangle is less than $90$ degrees. Hence angle $a$ is $135$ degrees.", "# A triangle has sides with lengths: 6, 1, and 9. How do you find the area of the triangle using Heron's formula? Triangle doesn't exist because longest side $9 > 6 + 1$ The sides of triangle are a=6, b=1\\ &\\ c=9, we see that the longest side $c = 9$ is greater than the sum of other two sides $a + b = 6 + 1 = 7$ hence such a triangle doesn't exist", "Answer the 13th question. { NO LINKS }.{ NO ANSWERS FROM GOOGLE }. Q.13. The sides of a right-angled triangle containing the right angle are 5x cm and (3x - 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 $c{m}^{2}$. • 0 What are you looking for?", "# its easy Geometry Level 2 if a right angled triangle has side lengths a,b and c such that c is hypotenuse and $$c\\quad =\\sqrt [ 2 ]{ 2ab }$$ find the smallest angle of the triangle? ×", "the hypotenuse being the longest side in a right triangle. Okay that’s it for now,.", "# A triangle has two corners with angles of pi / 2 and ( pi )/ 8 . If one side of the triangle has a length of 13 , what is the largest possible area of the triangle? Apr 16, 2018 The area is $\\frac{169}{2 \\left(\\sqrt{2} - 1\\right)} \\approx 204$ #### Explanation: If one of the angles of the triangle is $\\frac{\\pi}{2}$, it is a right triangle. The measure of the third angle of the triangle must be $\\frac{\\pi}{2} - \\frac{\\pi}{8} = \\frac{3 \\pi}{8}$. The shortest side of this right triangle will be the leg that is opposite the angle measuring $\\frac{\\pi}{8}$. Let $x$= the length of the longer leg. The area, $A$, of the triangle will be $A = \\frac{13}{2} x$ From trigonmetry $x = \\frac{13}{\\tan} \\left(\\frac{\\pi}{8}\\right)$ So $A = {13}^{2} / \\left(2 \\tan \\left(\\frac{\\pi}{8}\\right)\\right) = \\frac{169}{2 \\left(\\sqrt{2} - 1\\right)} \\approx 204$" ]

[ "## ksf-2015-st-27 In the rectangle $$ABCD$$ shown in the figure, $$M_1$$ is the midpoint of $$DC$$, $$M_2$$ is the midpoint of $$AM_1$$, $$M_3$$ is the midpoint of $$BM_2$$ and $$M_4$$ is the midpoint of $$CM_3$$. Question: Find the ratio between the areas of the quadrilateral $$M_1M_2M_3M_4$$ and of the rectangle $$ABCD$$.", "$M$ is the midpoint of the side of the rectangle. What is the area (in square units) of the triangle $PMR$?", "# Midpoint? What About Thirds? - Part 3 Geometry Level 4 As usual, $$ABCD$$ is a square, and we divide each side​ into three equal parts, and mark the points $$E,F,G,H,I,J,K,L$$ in a clockwise manner starting from the vertex $$D$$. This time, we connect those points to the furthest vertex that is opposite to them to form a not-necessarily-regular octagon as shown in the figure above, for example, points $$B$$ and $$E$$ will be connected in this case. What is the ratio of the area of the square to that of the octagon? ×", "the point $M$ is the midpoint of the side $BC$, the point $W$ is the intersection point of the bisector of the angle $A$ of the triangle $ABC$ with the circumscribed circle around him. Prove that the area of the triangle $I_aBC$ is calculated by the formula $S_{ (I_aBC)} = MW \\cdot p$, where $p$ is the semiperimeter of the triangle $ABC$. The problems proposed are here, in Ukranian : http://amnm.vspu.edu.ua/olymp/problems/", "# Problems Section: Problems 1191 - 1200 Q1191. In the triangle $ABC, M$ is the midpoint of $BC.$ Points $X$ and $AB$ and $Y$ on $AC$ are such that $XY\\ \\Vert\\ BC.$ Show that $BY$ and $CX$ intersect at a point $P$ on $AM.$", "M is the midpoint of AB, point D belongs to AC so that AD:DC=2:5. If the Area of ABC = 56 yd^2, find the Area of BMC, Area of AMD, and the Area of CMD. 1. ### Geometry What is he ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if XY is similar to AC, and BY=3, YC=2 ? 2. ### math Find the ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if BY = 3, YC = 2", "# Area of a triangle defined by the midpoint of a median In triangle $$ABC$$ , $$M$$ is the midpoint of $$BC$$ and $$N$$ is the midpoint of $$AM$$. $$BN$$ when extended intersect $$AC$$ at $$D$$. If the area of triangle $$ABC$$ is $$20$$ square units , what is the area of triangle $$AND$$ ? See picture for my attempt and diagram", "In $\\triangle ABC$, $BE$ is a median and $O$ is the midpoint of $BE$. The line joining $A$ and $O$ meets $BC$ at $D$. Find the ratio of the lengths $AO : OD$.", "ratio of the area of square $$DEFB$$ to the area of $$\\triangle ABC$$?", "$ABCD$ is a rectangle. $P$ is the midpoint of $AD$; the length of $BQ$ is one third the length of $BC$. What fraction of the area of the rectangle is the area of the shaded quadrilateral $ABQP$?", "# Midpoint? What About Thirds? - Part 2 Geometry Level 5 Just like in Part 1, $ABCD$ is a square with a side length of 18, and we divide each side into three equal parts ​and mark the points $E,F,G,H,I,J,K,L$ in a clockwise manner starting from the vertex $D$. This time, we connect each of those points to the nearest vertex that is opposite to them to form a not-necessarily-regular octagon as shown in the figure above, for example, points $A$ and $E$ will be connected in this case. What is the area of this octagon? ×", "Square $ABCD$ has side length $1$. Albert chooses points $E$, $F$, $G$, and $H$ on sides $\\overline{AB}$, $\\overline{BC}$, $\\overline{CD}$, $\\overline{DA}$ respectively, such that each point divides its segment into two parts, with lengths given by the ratio $2:3$. Points $E$, $F$, $G$, and $H$ are closer to points $A$, $B$, $C$, and $D$ respectively. Find the area of square $EFGH$.", "# Midpoint? What About Thirds? Geometry Level 4 $$ABCD$$ is a square with a side length of 18. Divide each side​ into three equal parts, and mark the points $$E,F,G,H,I,J,K,L$$ in a clockwise manner starting from the vertex $$D$$. Then, connect the points to get a not-necessarily-regular octagon as shown in the figure above. What is the area of the octagon? Bonus: You may notice that the ratio between the segments of all the sides are all $$1:1:1$$. Now, generalize this by considering the case of $$1:k:1$$, where $$k\\in\\mathbb{N}$$. ×", "$$ABCD$$ is a square inscribed in a circle of diameter $$3\\sqrt{2}$$. $$E$$ and $$F$$ are the midpoints of $$AB$$ and $$BC$$, respectively. What is the area of the shaded region?", "In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of quadrilateral region BCDE ? ~$1 \\over 2$~ ~$1 \\over 3$~ ~$1 \\over 4$~ ~$1 \\over 5$~ ~$1 \\over 6$~", "Point $M$ is the middle of side $BC$ in acute triangle $\\Delta ABC$. Point $K$ lies on side $BC$ in such a way that $\\angle BAM = \\angle KAC$. A point $E$ was chosen on side $AK$ such that $\\angle BEK = \\angle BAC$/ Show that $\\angle KEC = \\angle BAC$.", "$ABCD$ is a square of side 1 unit. Arc of circles with centres at $A, B, C, D$ are drawn in. Prove that the area of the central region bounded by the four arcs is: $(1 + \\pi/3 + \\sqrt{3})$ square units.", "# Math Help - Ratios of Area in Rectangles 1. ## Ratios of Area in Rectangles These two problems are giving me a lot of trouble. If any one could give me some help in the right direction I'd really appreciate it. 1 In rectangle ABCD, point E is on side AB so that AE = 10 and EB = 5. What fraction of the area of the rectangle is inside triangle AEC? 2 M and N are the midpoints of consecutive sides of a square ABCD with vertex A in between M and N. What is the ratio of the area of triangle AMN to the area of the complete square. Thanks again. 2. Originally Posted by lyyy94 These two problems are giving me a lot of trouble. If any one could give me some help in the right direction I'd really appreciate it. 1 In rectangle ABCD, point E is on side AB so that AE = 10 and EB = 5. What fraction of the area of the rectangle is inside triangle AEC? 2 M and N are the midpoints of consecutive sides of a square ABCD with vertex A in between M and N. What is the ratio of the area of triangle AMN to the area of the complete square. Thanks again. 1)", "of triangle ABC. Hence insufficient Regards Argha Senior Manager Joined: 13 May 2013 Posts: 468 Followers: 3 Kudos [?]: 172 [0], given: 134 Re: In triangle ABC, point X is the midpoint of side AC and [#permalink] ### Show Tags 10 Dec 2013, 17:45 In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS ? (1) The area of triangular region ABX is 32. X is the midpoint of AC meaning AX = XC. If ABX = 32 so regardless of the slope of BC and as long as the line from B to X will create two triangles equal in area. (i.e. the area of ABX = CBX) We know there is a segment drawn from midpoint X to Y and midpoint R to S. Because these increasingly small triangles are built with midpoints, we know that their ratios are proportionate to one another and that they are similar. For example, the ratio of the length of SC to YC is 1:2. The ratio of the areas of similar triangles can be found by taking the ratio of", "the midpoint of $\\overline {AC}$, point $E$ is the midpoint of $\\overline {BC}$, point $H$ is the midpoint of $\\overline {AF}$, and point $G$ is the midpoint of $\\overline {BF}$. Examine this carefully, find the similar triangles, and use their properties. Don't forget to look for congruent triangles as well." ]

[ "- 1 . # In a $\\Delta$ABC, points M and N respectively lie on side AB and AC such that area of triangle ABC is double than the area of trapezium BMNC. The ratio AM : MB is [ A ] (2 - $\\sqrt 2$) : 1 [ B ] (2 + $\\sqrt 2$) : 1 [ C ] ($\\sqrt 2$ + 1) : 1 [ D ] ($\\sqrt 2$ - 1) : 1 Answer : Option C Explanation :", "Square $ABCD$ has side length $1$. Albert chooses points $E$, $F$, $G$, and $H$ on sides $\\overline{AB}$, $\\overline{BC}$, $\\overline{CD}$, $\\overline{DA}$ respectively, such that each point divides its segment into two parts, with lengths given by the ratio $2:3$. Points $E$, $F$, $G$, and $H$ are closer to points $A$, $B$, $C$, and $D$ respectively. Find the area of square $EFGH$.", "triangle ABC is double than the area of trapezium BMNC. The ratio AM : MB is (2 - $\\sqrt 2$) : 1 (2 + $\\sqrt 2$) : 1 ($\\sqrt 2$ + 1) : 1 ($\\sqrt 2$ - 1) : 1 If $\\alpha$ is an acute angle and 2sin$\\alpha$+ 15$cos^2$ $\\alpha$= 7, then value of $tan^ 2$$\\alpha$is $4 \\over 3$ $9 \\over 16$ $16 \\over 9$ $3 \\over 4$", "\\sqrt[n]{x_1 x_2\\cdots x_n}$$ With equality if, and only if $x_1=x_2=\\cdots=x_n$ So suppose we fix the perimeter of a rectangle as $2(a+b)$ with side lengths $a$ and $b$. Hence, a square with the same perimeter has a side length of $\\displaystyle\\frac{a+b}{2}$. So by applying the AM-GM inequality with $n=2$ it follows that $\\displaystyle \\left(\\frac{a+b}{2}\\right)^2 \\ge ab$. So we see that the area of the rectangle doesn't exceed the area of the square, and the areas are equal if and only if $a=b$. So in other words, for a fixed perimeter the maximum area of a rectangle is when it is in fact a square. Alternatively, one can apply this to a fixed area and deduce that the perimeter of a rectangle is always greater than or equal to the perimeter of a square. This time suppose that $x$ and $y$ are the lengths of a rectangle with a fixed area $xy$. Hence, a square with the same area has side-lengths $\\sqrt{ab}$. So, by AM-GM we obtain $2(x+y)\\ge 4\\sqrt{xy}$. Now we see that the for a fixed area $xy$, the rectangle with the smallest perimeter is a square. This can be easily extended and generalized into $n$ dimensions and we can say that an $n$-cube has the smallest perimeter among all $n$-dimensional boxes with the same volume and that an", "diagonals bisect each other, $AM=MC$ and $BM=MD$, we have congruent triangles $\\triangle ABM \\cong \\triangle CDM$ and $\\triangle ADM \\cong \\triangle CBM$ and the shape is thus a parallelogram. Now it is clear that the distance of $B$ from $AC$ is the altitude of $B$ in $\\triangle MBC$ and the distance of $D$ from $AC$ is the altitude of $D$ in the congruent $\\triangle MDA$ so they are equal.", "The perimeter of a given triangle, whose sides are in the ratio of $3 : 4 : 5$, is greater by $\\quantity{4}{ft.}$ than that of a given square. The area of the square is greater by $\\quantity{25}{sq.ft.}$ than that of the triangle. Find the lengths of the sides of the square and of the triangle.", "Theorem on $\\triangle{AFC},$ we have that $AF^2 + FC^2 = AC^2.$ We know that $AF = 8$, $FC = a$ and $AC = 3a$ so we have $8^2 + a^2 = (3a)^2.$ Simplifying, we have that $64 = 8a^2 \\Rightarrow a^2 = 8 \\Rightarrow a = \\sqrt{8} = 2 \\sqrt{2}.$ Recall that $FC=a$. Therefore, $BC = 2 \\cdot FC = 2 \\cdot 2 \\sqrt{2} = 4 \\sqrt{2}.$ Since the height is $AF = 8,$ we have the area equal to $\\frac{4 \\sqrt{2} \\cdot 8}{2}=16\\sqrt{2}.$ Thus our answer is $\\boxed{\\mathrm{(D) 16 \\sqrt{2}.}}$ ~mathboy282", "+ · · · + 9) and the blue square has area 102, then the gnomon consisting of the two yellow rectangles and the blue square in Figure 5 has area $$2 \\cdot 10(1 + 2 + 3 + \\cdots + 9) + 10^2 = 2 \\cdot 10 \\cdot {{9 \\cdot 10} \\over 2} + 10^2 = 10 \\cdot 10^2 = 10^3.$$ This gives us another way to write the area of the larger square; namely, as the sum of the areas of the red square and of the gnomon, or (1 + 2 + 3 + · · · + 9)2 + 103. Equating our two expressions for the area of the larger square, we have $$(1 + 2 + 3 + \\cdots + 9)^2 + 10^3 =(1 + 2 + 3 + \\cdots + 10)^2.$$ Repeating the preceding argument for the square with side length 1 + 2 + 3 + · · · + 9 and a smaller square with side length 1 + 2 + 3 + · · · + 8, we have $$(1 + 2 + 3 + \\cdots + 8)^2 + 9^3 =(1 + 2 + 3 + \\cdots + 9)^2.$$ Substituting in the preceding equation, we get $$(1 + 2 + 3 + \\cdots + 8)^2 + 9^3 + 10^3", "= \\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{1}{2} = \\frac{1}{6}$. The entire quadrilateral $ABEF$ has area $\\frac{1}{4} + \\frac{1}{6} = \\frac{5}{12}$. This is $5$ times larger than the area of $\\triangle DFE$, so the ratio is $1:5$, or $\\boxed{\\textbf{(C)}}$.", "one side of the rectangle be $A+B+C+\\cdots$ and a perpendicular side be $a+b+c+\\cdots.$ So how do we expand $(a+b)(a+b)$? Like this: $$(a+b)(a+b) \\; = \\; aa + ab + ba + bb \\;=\\; a^2 + 2ab + b^2$$ To continue this to more complicated expansions, well beyond what one would do in a classroom but what one might do for that rare student who is especially interested in math, see my answer to How to expand $(a_0+a_1x+a_2x^2+...a_nx^n)^2$?. -", "= r$ 4. Let $A_{total}$ denote the area of the complete triangle. Then you know: $A_{total} = \\underbrace{\\frac12 \\cdot c \\cdot r}_{red \\ area} + \\underbrace{\\frac12 \\cdot a \\cdot r}_{blue \\ area} + \\underbrace{\\frac12 \\cdot b \\cdot r}_{green \\ area}$ $A_{total}=\\frac12 \\cdot (a+b+c) \\cdot r$ $\\boxed{r = \\frac{2A_{total}}{a+b+c}}$ $A_{total} = 24 \\ cm^2$ And therefore: $r = \\frac{2 \\cdot 24}{6+8+10} = 2$", "the square $DBMH$ to each. Then $\\Box CBML = \\Box DBFG$. But as $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that: $\\Box ACLK = \\Box CBML$ Add $\\Box CDHL$ to each. Then $\\Box ADHK$ is equal in area to the gnomon $CBFGHL$. But $\\Box ADHK$ is the rectangle contained by $AD$ and $DB$, because $DB = DH$. So the gnomon $CBFGHL$ is equal in area to the rectangle contained by $AD$ and $DB$. Now $\\Box LHGE$ is equal to the square on $CD$. Add $\\Box LHGE$ to each of the gnomon $CBFGHL$ and $\\Box ADHK$. Then the gnomon $CBFGHL$ together with $\\Box LHGE$ equals the rectangle contained by $AD$ and $DB$ and the square on $CD$. But the gnomon $CBFGHL$ together with $\\Box LHGE$ is the square $CBFE$. Hence the result. $\\blacksquare$ ## Geometric Proof 2 Let $\\Box ABCD$ be a square of side length $x$. Let $\\Box DEFG$ be a square of side length $y$ where $y < x$ Let $EF$ be produced to $H$. The area of $\\Box ABCD$ is seen to be equal to the sum of: the area of the rectangle $AEHB$ the area of the rectangle $FGCH$ the area of the square $DEFG$ From Area of Square: the area of $\\Box ABCD$ is equal", "$$M$$ is an interior point of the triangle. By Theorem 2, we have $$AM'+M'B < AC + CB \\tag 1$$ By (TI) applied to triangle $$BMM'$$ we have $$MB < MM' +M'B$$ Adding $$AM$$ to both side of the inequality we obtain: $$AM+MB < AM + MM' +M'B \\tag 2$$ Since $$AM + MM' = AM'$$, $$(2)$$ becomes $$AM + MB < AM' + M'B \\tag 3$$ Now from $$(1)$$ and $$(3)$$ by transitivity we get $$AM + MB < AM'+M'B < AC + CB$$ as we wanted to prove.", "$(0,0)$, $v$ must be inside square $S$. Similarly, for $T(v)$ to contain the point $(1,0)$, $v$ must be inside a translated square $S$ with center at $(1,0)$, which we will call $S'$. Therefore, the area we seek is Area$(U \\cap S \\cap S')$. To calculate the area, we notice that Area$(U \\cap S \\cap S') = \\frac{1}{2} \\cdot$ Area$(S \\cap S')$ by symmetry. Let $S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)$. Let $M = (0.7, 0.4)$ be the midpoint of $S'_1S'_2$, and $N = (0.7, 0.7)$ along the line $S_1S'_1$. Let $I$ be the intersection of $S$ and $S'$ within $U$, and $J$ be the intersection of $S$ and $S'$ outside $U$. Therefore, the area we seek is $\\frac{1}{2} \\cdot$ Area$(S \\cap S') = \\frac{1}{2} [IS'_2JS_2]$. Because $S_2, M, N$ all have $x$ coordinate $0.7$, they are collinear. Noting that the side length of $S$ and $S'$ is $1$ (as shown above), we also see that $S_2M = MS'_1 = 0.5$, so $\\triangle{S'_1NM} \\cong \\triangle{S_2IM}$. If follows that $IS_2 = NS'_1 = 1.1 - 0.7 = 0.4$ and $IS'_2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2$. Therefore, the area is $\\frac{1}{2} \\cdot$ Area$(S \\cap S') = \\frac{1}{2} [IS'_2JS_2] = \\frac{1}{2} \\cdot 0.2", "2}.{1\\over 2x}={1\\over 4}.$ Consider the area of the triangle `above' the diagonal, and express it is the sum of two areas : ${1\\over 2} x^2 \\sin 108^\\circ = {1\\over 2} x\\cdot1\\cdot\\sin 72^\\circ + {1\\over 2} 1\\cdot 1\\cdot\\sin108^\\circ.$ As $\\sin 108^\\circ = \\sin 72^\\circ$, this gives $x^2=x+1$ and hence $x$ is the golden ratio: $x = {1+\\sqrt{5}\\over 2} = \\varphi .$ Hence $\\cos 36^\\circ - \\cos 72^\\circ = {x\\over 2} - {1\\over 2x} = {x^2-1\\over 2x}={1\\over 2}.$ Let the common side of the two triangles be $x$. Then we have: $b = x\\cdot \\cos 72^\\circ$ and $a= {x\\over{\\cos 36^\\circ}}.$ Therefore ${b\\over a} = \\cos72^\\circ \\cos 36^\\circ = {1\\over 4}$ so $a$ is four times bigger than $b$. Here we have ${{x + b}\\over a} =\\cos 36^\\circ$ and ${x\\over a} = \\cos 72^\\circ$ so ${b\\over a} = \\cos 36^\\circ -\\cos 72^\\circ = {1\\over 2}.$ Therefore $a$ is twice the length of $b$.", "rectangle.}$ Code: 3x * - - - - - * | | x | | x | | * - - - - - * 3x The area of this rectangle is: . $3x\\cdot x \\:=\\:\\boxed{3x^2}$ We have: . $x + 3x + x + 3x \\:=\\:a\\quad\\Rightarrow\\quad 8x \\:=\\:a$ The right piece of wire has length: . $12-8x$ It is bent into a square. Code: s * - - - * | | s | | s | | * - - - * s We have: . $4s = 12 - 8x \\quad\\Rightarrow\\quad s \\:=\\:3 - 2x$ The area of this square is: . $\\boxed{(3-2x)^2}$ Therefore, the total area is: . $A \\;=\\;3x^2 + (9 - 6x + 4x^2) \\;=\\;7x^2 - 6x + 9$", "# Area of a right angled triangle is an even integer Actual Question is : Let $ABC$ be a triangle in the plane such that $BC$ is perpendicular to $AC$. Let $a,b,c$ be the lengths of $BC$, $AC$, and $AB$ respectively. Suppose that $a,b,c$ are integers and have no common divisor other than $1$. Which of the following statements are necessarily true? a. Either $a$ or $b$ is an even integer. b. The area of the triangle $ABC$ is an even integer. c. Either $a$ or $b$ is divisible by $3$. With this information, I could see that right angle is at $C$ and $AB=c$, $BC=a$, $AB=c$. First of all I should check that either $a$ or $b$ is even. Suppose not, then we would have $(2n+1)^2+(2m+1)^2=c^2$ i.e., $4n^2+4m^2+4n+4m+2=c^2 \\Rightarrow 2(2n^2+2m^2+2n+2m+1)=c^2$ i.e., $2$ divides $c^2$ i.e., $2$ divides $c$ i.e., $2^2$ divide $c^2$ i.e., $4$ divides $c^2$ but then, $4n^2+4m^2+4n+4m+2=c^2$ so, going modulo $4$ we see that $\\bar{2}\\equiv\\bar{0} \\text{ mod } 4$ which is false. So, at least one of $a,b$ has to be even integer. This does not immediately imply that area is even integer (sq.units) because $\\text {Area} =\\frac{1}{2} \\cdot a \\cdot b$ Suppose $a$ or $b$ has only $1$ as power of $2$ in its prime decomposition, then $2$ in numerator and denominator gets cancelled", "that AM(the height) = 7,534 With pythagoras again, we determine the full length of BC(the base) to be: 7,81 $Area = \\frac{1}{2} (7,81) (7,534) = 29,408 \\ square \\ units$ 4. May i please have a picture of how the bridge looks? Im bad at visualising things. Are you familiar with this formula for the area of a triangle? . . $Area \\:=\\:\\frac{1}{2}\\ bc\\sin A$ The area is one-half the product of two sides times the sine of the included angle. We have: . $b = AC = 9.5,\\:c = AB = 7.8,\\:\\angle A = 52.5^o$ Hence: . $Area \\:=\\:\\frac{1}{2}(9.5)(7.8)\\sin52.5^o \\:\\approx\\:29.4$ hey everyone , i would like you please to help me figure out 2 problems they both about figuring out the area of each a triangle ABC and an engineer estimated area of vertical cross section of water flowing under bridge. and there is a model for it . regards hello, to #7: if a is the lower bound of the area, if b is the upper bound of the area, if n is the number of trapezoids then the trapezoid formula to estimate the value of an area is: $A \\approx \\frac{b-a}{2n}(y_a + 2y_1 + 2y_3 + \\ldots +2y_{n-1} + y_b)$ $A \\approx \\frac{24-0}{2 \\cdot 6}(1.2 + 2 \\cdot 2.3 + 2 \\cdot 3.8 + \\ldots", "{143\\sqrt3}4.$$ Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \\frac {BC^2}{AF^2}\\cdot[EAF] = \\frac {12}{11^2 + 13^2 - 2\\cdot11\\cdot13\\cdot\\cos120^\\circ}\\cdot\\frac {143\\sqrt3}4 = \\frac {429\\sqrt3}{433}$. Therefore, the answer is $429+433+3=\\boxed{865}$.", "N = 5$ $m$ Now as $\\Delta A B P$ is a right angled triangle, $A {B}^{2} = B {P}^{2} + A {P}^{2} = {4}^{2} + {5}^{2} = 16 + 25 = 41$ and $A B = \\sqrt{41} = 6.403$ Jan 30, 2017 (3) Perimeter $= 24 \\sqrt{5} c m$ Area $= 160 c {m}^{2}$ (4)Side of square $= 5 \\sqrt{2} c m$ (9)$A B = \\sqrt{41} m$ #### Explanation: (3) Let $x =$ length of smaller side Length of other side is $= 2 x$ Therefore, ${x}^{2} + {\\left(2 x\\right)}^{2} = {20}^{2}$ ${x}^{2} + 4 {x}^{2} = 400$ $5 {x}^{2} = 400$ ${x}^{2} = \\frac{400}{5} = 80$ $x = \\sqrt{80} = \\sqrt{16 \\cdot 5} = 4 \\sqrt{5}$ Perimeter $= 2 x + 4 x = 6 x = 6 \\cdot 4 \\sqrt{5} = 24 \\sqrt{5} c m$ Area $= x \\cdot 2 x = 2 {x}^{2} = 2 \\cdot 80 = 160 c {m}^{2}$ (5)Let $a =$ length of the side of the square ${a}^{2} + {a}^{2} = {10}^{2}$ $2 {a}^{2} = 100$ ${a}^{2} = 50$ $a = \\sqrt{50} = \\sqrt{25 \\cdot 2} = 5 \\sqrt{2}$ (9c) Easier way to calculate $A B$ $A {B}^{2} = M {N}^{2} + {\\left(M A + N B\\right)}^{2}$ $A {B}^{2} = {5}^{2} + {\\left(3 + 1\\right)}^{2}$ $A {B}^{2} = 25 +" ]

[ "M is the midpoint of AB, point D belongs to AC so that AD:DC=2:5. If the Area of ABC = 56 yd^2, find the Area of BMC, Area of AMD, and the Area of CMD. 1. ### Geometry What is he ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if XY is similar to AC, and BY=3, YC=2 ? 2. ### math Find the ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if BY = 3, YC = 2", "- 1 . # In a $\\Delta$ABC, points M and N respectively lie on side AB and AC such that area of triangle ABC is double than the area of trapezium BMNC. The ratio AM : MB is [ A ] (2 - $\\sqrt 2$) : 1 [ B ] (2 + $\\sqrt 2$) : 1 [ C ] ($\\sqrt 2$ + 1) : 1 [ D ] ($\\sqrt 2$ - 1) : 1 Answer : Option C Explanation :", "+0 # In the figure, $ABCD$ is a square of side length 1, and $M$ and $N$ are the midpoints of sides $\\overline{AB}$ and $\\overline{CD}$, respecti +2 362 1 +598 In the figure, $ABCD$ is a square of side length 1, and $M$ and $N$ are the midpoints of sides $\\overline{AB}$ and $\\overline{CD}$, respectively. Find the area of the shaded region. michaelcai Dec 5, 2017 #1 +87537 +3 MBND is a parallelogram..... The height is 1 and the width is 1/2.....so its area is 1 * (1/2) = 1/2 units^2 And the shaded area is (1/2) of this = 1/4 units^2 CPhill Dec 5, 2017", "Square $ABCD$ has side length $1$. Albert chooses points $E$, $F$, $G$, and $H$ on sides $\\overline{AB}$, $\\overline{BC}$, $\\overline{CD}$, $\\overline{DA}$ respectively, such that each point divides its segment into two parts, with lengths given by the ratio $2:3$. Points $E$, $F$, $G$, and $H$ are closer to points $A$, $B$, $C$, and $D$ respectively. Find the area of square $EFGH$.", "and BD are joined which intersect each other at O Side of square AB = 8 cm ∴ Area of square ABCD = (side)2 ∵ Diagonal BD bisects the square into two triangle equal in area ∴ Area ∆ABD = $$\\frac { 1 }{ 2 }$$ x area of square ABCD = $$\\frac { 1 }{ 2 }$$ x 64 = 32 cm2 ∵ P is mid point of AB of AABD, and PQ || AD ∴ O is the mid point of BD ∴ OP = $$\\frac { 1 }{ 2 }$$AD = $$\\frac { 1 }{ 2 }$$ x 8 = 4 cm and PB = $$\\frac { 1 }{ 2 }$$ AB = $$\\frac { 1 }{ 2 }$$ x 8 = 4 cm ∴ Area ∆OPB = $$\\frac { 1 }{ 2 }$$PB x OP = $$\\frac { 1 }{ 2 }$$ x4x4 = 8 cm2 Question 7. ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. If area of ∆ABC is 16 cm2, find the area of ∆DEF. Solution: In ∆ABC, D is mid point of BC. E and F are the mid points of DC and AE respectively area of ∆ABC = 16 cm2 FD is joined ∵ D", "# Constructing this is harder than solving it! Geometry Level 2 In square ABCD, AD is 4 centimeters, and M is the midpoint of CD. Let O be the intersection of BM and diagonal AC. What is the ratio of OC to OA? If your answer is $$\\frac{a}{b}$$, then find $$a+b$$. ×", "# Proportional square Geometry Level 3 $$ABCD$$ is a square. A point $$G$$ is placed on the side $$BC$$ so that $$\\frac{BG}{GC} = \\frac{1}{3}$$ and the midpoint of the side $$CD$$ is $$E$$. Segments $$BE$$ and $$AG$$ have a common point $$X$$. If $$\\frac{BX}{BE} = \\frac{a}{b}$$ where $$a$$ and $$b$$ are positive coprime integers, Find $$a + b$$. ×", "###### back to index | new 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 $FGHJ$. Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\\triangle ABC$? A circle of radius $2$ is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded? In $\\triangle{ABC}$, segments AB and AC have each been divided into four congruent segments. We must find the fraction of the triangle that is shaded. In $\\bigtriangleup ABC$, $AB=BC=29$, and $AC=42$. What is the area of $\\bigtriangleup ABC$? In the diagram, AB is the diameter of a semicircle, D is the midpoint of the semicircle, angle $BAC$ is a right angle, $AC=AB$, and $E$ is the intersection of $AB$ and $CD$. Find the ratio between the areas of the", "the point $M$ is the midpoint of the side $BC$, the point $W$ is the intersection point of the bisector of the angle $A$ of the triangle $ABC$ with the circumscribed circle around him. Prove that the area of the triangle $I_aBC$ is calculated by the formula $S_{ (I_aBC)} = MW \\cdot p$, where $p$ is the semiperimeter of the triangle $ABC$. The problems proposed are here, in Ukranian : http://amnm.vspu.edu.ua/olymp/problems/", "neighbor of a house indexed by $(k; l)$ if $|i-k| + |j-l| = 1$. Problem 5. In a triangle $ABC$, points $M$ and $N$ are on sides $AB$ and $AC$, respectively, such that $MB = BC = CN$. Let $R$ and $r$ denote the circumradius and the inradius of the triangle $ABC$, respectively. Express the ratio $MN=BC$ in terms of $R$ and $r$.", "the area of $ABCD$ to an area greater than $EFGH$ can equal the ratio of the square on $BD$ to the square on $FH$. Suppose to the contrary, that: $ABCD : S = BD^2 : FH^2$ where $S > EFGH$. that is: $FH^2 : DB^2 = S : ABCD$ But as $S$ is to the circle $ABCD$, so $EFGH$ is to an area less than the circle $ABCD$. the ratio of the square on $BD$ to the square on $FH$ equals the ratio of the circle $EFGH$ to an area less than the circle $ABCD$. This was shown to be impossible. Therefore the square on $BD$ to the square on $FH$ does not equal the area of $ABCD$ to any area greater than $EFGH$. Hence the result. $\\blacksquare$ Historical Note This theorem is Proposition $2$ of Book $\\text{XII}$ of Euclid's The Elements.", "Given any quadrilateral $ABCD$, Albert draws in the points $P$, $Q$, $R$, and $S$ – the midpoints of sides $BC$, $CD$, $DA$, and $AB$ respectively. He then determines the value of $$\\frac{AP^2 + BQ^2 + CR^2 + DS^2}{AB^2 + BC^2 + CD^2 + DA^2}$$ What is the largest possible ratio Albert could find?", "# Progressive ratio Geometry Level 3 $ABCD$ is a square of side length 1. Points $E, F, G, H$ are such that $AE : EB = 1:1$, $BF: FC = 1 : 2$, $CG : GD = 1 : 3$, $DH : HA = 1 : 4$. What is the area of the quadrilateral $EFGH$? ×", "$-1$ from $ABC$ to $A'B'C'$ through the centroid $G$. Let $M$ be the midpoint of $BC$ which maps to $M'$ and note that $A'G=AG=2GM,$ implying that $GM=MA'.$ Similarly, we have $AM'=M'G=GM=MA'.$ Also let $D$ and $E$ be the intersections of $BC$ with $A'B'$ and $A'C',$ respectively. The homothety implies that we must have $DE || B'C',$ so there is in fact another homothety centered at $A'$ taking $A'DE$ to $A'B'C'$. Since $A'M'=3A'M,$ the scale factor of this homothety is 3 and thus $[A'DE]=\\frac{1}{9}[A'B'C']=\\frac{1}{9}[ABC].$ We can apply similar reasoning to the other small triangles in $A'B'C'$ not contained within $ABC$, so our final answer is $[ABC]+3\\cdot\\frac{1}{9}[ABC]=\\boxed{112.}$", "# Midpoint? What About Thirds? - Part 3 Geometry Level 4 As usual, $$ABCD$$ is a square, and we divide each side​ into three equal parts, and mark the points $$E,F,G,H,I,J,K,L$$ in a clockwise manner starting from the vertex $$D$$. This time, we connect those points to the furthest vertex that is opposite to them to form a not-necessarily-regular octagon as shown in the figure above, for example, points $$B$$ and $$E$$ will be connected in this case. What is the ratio of the area of the square to that of the octagon? ×", "# Problems Section: Problems 1191 - 1200 Q1191. In the triangle $ABC, M$ is the midpoint of $BC.$ Points $X$ and $AB$ and $Y$ on $AC$ are such that $XY\\ \\Vert\\ BC.$ Show that $BY$ and $CX$ intersect at a point $P$ on $AM.$", "# Area of Triangle ABC Geometry Level 2 In $$\\triangle ABC$$, let $$M$$ be the midpoint of $$\\overline{BC}$$. If $$AM = MB = 6.5$$ and $$AB = 5$$, determine $$[ABC ]$$. ×", "# ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. IF area of ΔABC is 16 cm2, find the area of ΔDEF. - Mathematics ABC is a triangle in which D is the mid-point of BCand F are mid-points of DC and AErespectively. IF area of ΔABC is 16 cm2, find the area of ΔDEF. #### Solution Given: Here from the given question we get (1) ABC is a triangle (2) D is the midpoint of BC (3) E is the midpoint of CD (4) F is the midpoint of A Area of ΔABC = 16 cm2 To find : Area of ΔDEF Calculation: We know that , The median divides a triangle in two triangles of equal area. For ΔABC, AD is the median Area of Δ ADC = 1/2 (Area of ΔABC ) =1/2 (16) = 8 cm2 Area of Δ ADC = 8cm2 For ΔADC , AE is the median . Area of ΔAED = 1/2 (Area of Δ ABC) = 1/2 (8) = 4 cm2 Area of ΔAED = 4 cm2 Similarly, For ΔAED , DF is the median . Area of ΔDEF = 1/2 (Area of Δ AED) = 1/2 (4) = 2 cm2 Area of ΔDEF = 2 cm2", "\\lim_{A \\to L} \\operatorname {area} \\Box A = \\lim_{A \\to L} A^2 = L^2$ Finally: $L^2 \\ge \\operatorname {area}\\Box L \\ge L^2$ so: $\\operatorname {area}\\Box L = L^2$ $\\blacksquare$ ## Proof 2 Let $\\Box ABCD$ be a square whose side $AB$ is of length $L$. Let $\\Box EFGH$ be a square whose side $EF$ is of length $1$. From the Axioms of Area, the area of $\\Box EFGH$ is $1$. By definition, $AB : EF = L : 1$. From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\\Box ABCD$ to $\\Box EFGH$ is the duplicate ratio of the ratio of $AB$ to $EF$. Thus by definition of duplicate ratio: $\\Box ABCD : \\Box EFGH = \\left({AB : EF}\\right)^2$ That is: $\\dfrac {\\Box ABCD} {\\Box EFGH} = \\left({\\dfrac L 1}\\right)^2 = L^2$ That is, the area of $\\Box ABCD$ has $L^2$ as many units as $\\Box EFGH$. Hence the result. $\\blacksquare$ ## Proof 3 Let a square have a side length $a \\in \\R$. This square is equivalent to the area under the graph of $\\map f x = a$ from $0$ to $a$. Thus from the geometric interpretation of the definite integral, the area of the square will be the integral: $\\displaystyle A = \\int_0^a a \\rd l$ Thus: $\\ds A$ $=$", "# Area of Triangle ## Area of a triangle In square $ABCD$, $E$ is the midpoint of side $\\overline{AB}$ and $F$ is a point of side $\\overline{AD}$ such that $F$ is twice as near from $D$ as from $A$. $G$ is the intersection of the line segments $\\overline{DE}$ and $\\overline{CF}$. If $AB = 1\\text{ cm}$, find the area of $\\triangle CDG$. There are several ways to solve this problem, please show your solutions. ## Constant Area of Triangle ABC From the figures shown below. Squares EGHI and KHJB are fixed. Show that the area of triangle ABH is constant regardless of the dimensions of square ADEF. Figure 1 Figure 2" ]

[ "Independent probability Problem If you flip a coin and roll a 6-sided die, what is the probability that you will flip a heads and roll a 2?", "Use the example of rolling a six-sided die, probability = of. Dec, 19, 2020 0", "# When rolling a fair, six-sided die, what is the probability of getting a number higher than 2? Mar 20, 2017 $\\frac{2}{3}$ #### Explanation: we want $P \\left(X > 2\\right)$ for a fair die the sample space is$\\text{ } \\left\\{1 , 2 , 3 , 4 , 5 , 6\\right\\}$ all the numbers are equiprobable there are $\\text{ \"4\" }$ numbers greater than 2 so we have $P \\left(X > 2\\right) = \\frac{4}{6} = \\frac{2}{3}$", "higher or lower numbers than with a standard die. The coin flip bias is only at around the 1% level (I have no idea what it would be for a 20-sided die), so whether this is significant or not depends on what level of effects you care about. But casinos make a living on advantages on the level of a few percent. -", "0 # If a coin is flipped and a die is rolled.what is the probability of getting heads and a 6? Wiki User 2010-01-10 05:17:28 The probability of flipping a heads is 1/2 and the probability of rolling a 6 is 1/6. By the laws of probability it would be logical to multiply them together, (1/2)(1/6) thus the answer being 1/12 with is roughly eight percent. Wiki User 2010-01-10 05:17:28 Study guides 20 cards ➡️ See all cards 3.74 808 Reviews", "1. ## probability 4. a) A single fair die is rolled. Find the probability of getting an odd number. b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. 2. Originally Posted by bobby77 4. a) A single fair die is rolled. Find the probability of getting an odd number. All faces of the die are equally likely. The favourable outcomes in this case are: 1, 3, 5 out of a total of 6 possible outcomes. Therefore the probability of an odd number when the die is rolled is: [number of favourable outcomes]/[total number of outcomes] = 3/6 = 0.5 b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. Here the favourable outcomes are A , 2 , 3 .., 10, J, Q, K of diamonds and 10 of hearts, 10 of Spades, 10 of clubs which is a total of 16 favourable cases. Total number of outcomes = 52, so the required probability is: 16/52=4/13 RonL", "# You play the following game: You bet $1, a fair die is rolled and if it shows 6 you win$4, otherwise you lose your dollar. If you must choose the number of rounds in advance, how should you choose it to maximize your chance of being ahead (having won more than you have lost) when you quit, and what is the probability of this? Question You play the following game: You bet $1, a fair die is rolled and if it shows 6 you win$4, otherwise you lose your dollar. If you must choose the number of rounds in advance, how should you choose it to maximize your chance of being ahead (having won more than you have lost) when you quit, and what is the probability of this?", "a fair 13-sided die by simply rolling the die and stopping as soon as we have a score between 1 and 13. To see how to simulate a card hand, see the Introduction to Finite Sampling Models. A general method of simulating random variables is based on the quantile function.", "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) = $\\frac{1}{4}$", "? Free Version Easy # Basic Probability: Rolling a Die SATSTM-FL51Z0 What is the probability of rolling a six-sided die and getting a $5$? A $\\cfrac{1}{3}$ B $\\cfrac{1}{6}$ C $\\cfrac{1}{5}$ D $\\cfrac{5}{6}$ E $\\cfrac{1}{2}$", "# One die and 1, 2 or 3 coins A fair die is tossed. If the die lands on 1, 2 or 3, then 3 fair coins are tossed. If the die lands on 4 or 5, then 2 fair coins are tossed. If the die lands on 6, then 1 fair coin is tossed. Given that the resulting coin tosses produced no tails, what is the probability that the die landed on 6? The probability can be expressed as $$\\dfrac ab$$ for coprime positive integers $$a$$ and $$b$$, find the value of $$a+b$$. ×", "the probability of rolling the number when rolling a standard six-sided die? ### 6. What is the probability of drawing an even number from a set of cards numbered through ?", "## anonymous 5 years ago A 6-sided number cube is rolled. Find the probability of each event. Write each answer as a fraction. P(5). If it is a fair die, the probability of each side is 1/6. P(rolling a 5) = 1/6 In general: p(x) = 1/6 for any x $\\in$ X X = {1, 2, 3, 4, 5, 6}", "Browse Questions # A fair die is rolled up. If 1 turns up, a ball is picked up at random from bag A. If 2 0r 3 turns up, a ball is picked up at random from bag B, otherwise a ball is picked up from bag C. Bag A contains 3 red and 2 white balls, bag B contains 3 red and 5 white balls and bag C contains 4 red and 5 white balls. The die is rolled, a bag is picked up and a ball is drawn from it. If the ball drawn is red, what is the probability that bag B was picked up? E1 - Selection of Bag A E2 - Selection of Bag B E3 - Selection of Bag C A - red ball is selected since Bag A is chosen when , 1 turns up when a fair die is rolled P(E1) - 1/6, since Bag B is chosen when ,2 0r 3 turns up when a fair die is rolled P(E2) - 2/6 and since Bag C is chosen when 4 or 5 or 6 occurs when a fair die is rolled P(E3) - 3/6 Bag A contains 3 red and 2 white balls - P(A/E1) =3/5 Bag Bcontains 3 red and 5white balls - P(A/E2)=3/7 Bag C contains 4", "side of an equilateral triangle inscribed in the circle? ### Franc-carreau or Fair-square Franc-carreau is a simple game of chance, like the roll-a-penny game often seen at fairs and fêtes. A coin is tossed or rolled down a wooden chute onto a large board ruled into square segments. If the player’s coin lands completely within a square, he or she wins a coin of equal value. If the coin crosses a dividing line, it is lost. The playing board for Franc-Carreau is shown above, together with a winning coin (red) contained within a square and a loosing one (blue) crossing a line. As the precise translation of franc-carreau appears uncertain, the name “fair square” would seem appropriate. The question is: What size should the coin be to ensure a 50% chance of winning?", "flipped 6 times. What is the 7 16 Dec 2009, 09:01 3 a fair 2-sided coin is flipped 6 times. what is the 3 18 May 2007, 20:56 Kaplan Problem Susan flipped a fair coin N times. what 1 11 Jul 2006, 08:45 1 Kevin flips a coin four times. WHat is the probability the 11 18 Nov 2005, 04:11 Display posts from previous: Sort by", "# What is the probability that a fair six-sided die lands on an even number three out of five times it is rolled? What is the probability that a fair six-sided die lands on an even number three out of five times it is rolled? For one roll, the outcomes are $1, 2, 3, 4, 5, 6$ of which $2, 4, 6$ are even, so the probability is $3/6=1/2$. But how to deal with \"three out of five times\" part? • It is the same as the probability that a fair coin will land heads three out of five times. – Igor Rivin Dec 12 '13 at 0:35 For which the probability is $5$ pick $3$ out of $32$ or $$\\frac{1}{32} \\cdot \\frac{5 \\cdot 4 \\cdot 3}{3\\cdot 2 \\cdot 1}=\\frac{10}{32}= \\frac{5}{16}.$$", "Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Solved Probability Problems Probability for Rolling Three Dice", "flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads? (The answer value will be from 0 to 1, not as a percentage.) Two fair dice are rolled, and it is revealed that (at least) one of the numbers rolled was a 4. What is the probability that the other number rolled was a 6? Note: You are not told which of the numbers rolled is a 4. ×", "Nevertheless, you can refer to . • Coin tosses: Two possible outcomes. Heads or Tails. • Dice rolling: Six possible outcomes. 1-2-3-4-5-6. • Card drawing: 52 possible outcomes. There are 4 types (clubs, diamonds, spades and hearts) and 13 ranks for each type. (A)ce - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - (J)ack - (Q)ueen - (K)ing." ]

[ "are six outcomes and all of them are equally likely, to get each probability individually we can divide $1$ into $6$. This means: $P(1)=1/6$, $P(2)=1/6$, $P(3)=1/6$, $P(4)=1/6$, $P(5)=1/6$, $P(6)=1/6$. We can clearly see that $P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=\\dfrac{6}{6}=1$. Now let's think of the general case: ### If $\\Omega$ has equally likely results and $|\\Omega|=n$, then $P(\\omega)=\\dfrac{1}{n}$ $\\forall\\omega\\in\\Omega$.¶ This just means that the probability of each outcome individually is $1$ over the total amount of outcomes in our sample space. And this is because if there are $n$ equally likely options, and all add up to $1$, each one must be $\\dfrac{1}{n}$.", "## College Algebra (6th Edition) Probability of dealt a 2 or a 3 = $\\frac{2}{13}$ Total number of cards or outcomes = 52 Probability = $\\frac{Total outcomes}{favorable outcomes}$ For finding probability of dealing a 2 or a 3, favorable outcomes = 4 + 4 = 8 Probability of dealing a 2 or a 3 = $\\frac{8}{52}$ = $\\frac{2}{13}$", "a sample space S, an event A, and the outcomes are equally likely, $P(A) = \\frac{|A|}{|S|}$ This equation reduces the problem to counting the sizes of A and S. ### Resources A set of practice problems can be found here.", "## Algebra and Trigonometry 10th Edition $\\frac{10}{13}$ We know that $probability=\\frac{\\text{number of favourable outcomes}}{\\text{number of all outcomes}}$ The total number of cards (the number of all outcomes) is $52$ There are $12$ face cards ($3$ in each of the $4$ suits), thus there are $52-12=40$ non-face cards, which are the \"good outcomes\". Hence the probability here is: $\\frac{40}{52}=\\frac{10}{13}$", "equal to $3$. probability = number of outcomes possible / total number of outcomes $\\frac{3}{6} = \\frac{1}{2}$", "No. of total outcomes }}$$ = $$\\frac{11}{36}$$", "there are 8 possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. • These outcomes are all equally likely. • 3 of these outcomes have exactly 2 heads: HHT, HTH, and THH. • So, the probability of seeing exactly 2 heads in 3 flips of a fair coin is $\\frac{3}{8}$. ### Concept Check ✅ – Answer at cc.dsc10.com¶ I have three cards: red, blue, and green. What is the chance that I choose a card at random and it is green, then – without putting it back – I choose another card at random and it is red? • A) $\\frac{1}{9}$ • B) $\\frac{1}{6}$ • C) $\\frac{1}{3}$ • D) $\\frac{2}{3}$ • E) None of the above. ### Solution¶ • There are 6 possible outcomes: RG, RB, GR, GB, BR, and BG. • These outcomes are equally likely. • There is only 1 outcome which makes the event happen: GR. • Hence the probability is $\\frac{1}{6}$. ### Conditional probabilities¶ • Two events $A$ and $B$ can both happen. Suppose that we know $A$ has happened, but we don't know if $B$ has. • If all outcomes are equally likely, then the conditional probability of $B$ given $A$ is: $$P(B \\text{ given } A) = \\frac{ \\text{# of outcomes satisfying both A and B} }{ \\text{# of outcomes", "6), (6, 4), (5, 5) To get the sum as 11, possible outcomes = (5, 6), (6, 5) To get the sum as 12, possible outcomes = (6, 6) (ii)Probability of each of these sums will not be $\\frac{1}{11}$ as these sums are not equally likely.", "$\\frac{2}{36}$ because there are two ways out of thirty-six to do it. From a content point of view there are only 21 different outcomes. But the difference is that all contents do not have the same probability.", "## Thinking Mathematically (6th Edition) $\\frac{1}{4}$ P(E) = $\\frac{no.\\ of\\ outcomes\\ in\\ favor}{Total\\ no.\\ of\\ outcomes}$ Given that H is heads and T is tails, the set of equally likely outcomes is: S = {HH, HT, TH, TT}. E: two heads Favorable outcomes= {HH} P(E) = $\\frac{1}{4}$", "of $p$$p$ outcomes. The probability = $\\frac{p}{n}$$\\frac{p}{n}$. Outline", "and 90% of the problems are solved using just this tool. No kidding. Here are some examples to see how it works: {{#x:box| What is the probability that a card drawn at random from a deck of cards will be an ace? Solution. In this case there are four favorable outcomes, aces of spades, hearts, diamonds and clubs. Since each of the 52 cards in the deck represents a possible outcome, there are 52 possible outcomes. Therefore, the probability is $\\frac{4}{52}=\\frac{1}{13}$. }} The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice. {{#x:box| Two fair six-sided dice are rolled; what is the probability of having 5 as the sum of the numbers? Solution. There are 36 possible outcomes when a pair of dice is thrown (six outcomes for the first die times six outcomes for the second one). Since four of the outcomes have a total of 5, {(1,4), (4,1), (2,3), (3,2)}, the probability of the two numbers adding up to 5 is $\\frac{4}{36} = \\frac{1}{9}$. }} ## Probability of Multiple Events For questions involving single events, the F/T rule is sufficient. In fact, it is often sufficient for all other cases too. But, for questions involving multiple events, some other tools may be more appropriate.", "# Equally Likely Outcomes Some sample spaces have equally likely outcomes. We like those sample spaces, because there is a way to calculate probability questions about those sample spaces simply by counting. Here are a few examples where there are equally likely outcomes: • Coin flip: S = {Head, Tails} • Flipping two coins: S = {(H, H), (H, T), (T, H), (T, T)} • Roll of 6-sided die: S = {1, 2, 3, 4, 5, 6} Because every outcome is equally likely, and the probability of the sample space must be 1, we can prove that each outcome must have probability: $$\\p(\\text{an outcome}) = \\frac{1}{|S|}$$ Where |S| is the size of the sample space, or, put in other words, the total number of outcomes of the experiment. Of course this is only true in the special case where every outcome has the same likelihood. Definition: Probability of Equally Likely Outcomes If $S$ is a sample space with equally likely outcomes, for an event $E$ that is a subset of the outcomes in $S$: \\begin{align} \\p(E) &= \\frac{\\text{number of outcomes in E}}{\\text{number of outcomes in S}} = \\frac{|E|}{|S|} \\end{align} There is some art form to setting up a problem to calculate a probability based on the equally likely outcome rule. (1) The first step is to explicitly define", "## Thinking Mathematically (6th Edition) $\\frac{3}{4}$ P(E) = $\\frac{no.\\ of\\ outcomes\\ in\\ favor}{Total\\ no.\\ of\\ outcomes}$ Given that H is heads and T is tails, the set of equally likely outcomes is: S = {HH, HT, TH, TT}. E: Getting at least one head. Favorable outcomes= {TH, HH, HT} P(E) = $\\frac{3}{4}$", "Peehu has better chance of getting the number 36. #### Question 8: When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is $\\frac{1}{2}$. Justify your answer. Yes, probability of each outcome is $\\frac{1}{2}$ because head and tail both are equally likely events. #### Question 9: A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting ‘not 1’ each is equal to $\\frac{1}{2}$. Is this correct? Give reasons. No, this is not correct. Suppose we throw a die, then total number of outcomes = 6 Possible outcomes = 1 or 2 or 3 or 4 or 5 or 6 ∴ Probability of getting 1 = $\\frac{1}{6}$ Now, probability of getting not 1 = 1 – Probability of getting 1 = = $\\frac{5}{6}$ #### Question 10: I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say that probability of no heads is $\\frac{1}{4}$. What is wrong with this conclusion? I toss three coins together [given] So, total number of outcomes = 23 = 8 and possible outcomes are(HHH), (HTT), (THT), (TTH), (HHT), (THH), (HTH) and (TTT). Now, probability of getting", "we have two bags. One contains 1 star block and 4 moon blocks. The other contains 3 star blocks and 1 moon block. If we select one block at random from each, what is the probability that we will get two star blocks or two moon blocks? To answer this question, we can draw a tree diagram to see all of the possible outcomes. There are $$5 \\boldcdot 4 = 20$$ possible outcomes. Of these, 3 of them are both stars, and 4 are both moons. So the probability of getting 2 star blocks or 2 moon blocks is $$\\frac{7}{20}$$. In general, if all outcomes in an experiment are equally likely, then the probability of an event is the fraction of outcomes in the sample space for which the event occurs.", "## Elementary Statistics (12th Edition) a) We have 4 different outcomes: blue-blue, blue-brown, brown-blue, brown-brown. b)We know that $probability=\\frac{number \\ of \\ good \\ outcomes}{number\\ of\\ all\\ outcomes}$ and that we have 4 possibilities out of which 1 (blue-blue) is good, hence $probability=\\frac{1}{4}=0.25.$ c)We know that $probability=\\frac{number \\ of \\ good \\ outcomes}{number\\ of\\ all\\ outcomes}$ and we also know that if at least 1 allele is brown then the ye is brown, hence we have 4 possibilities out of which 3 (blue-brown, brown-blue, brown-brown) are good, hence $probability=\\frac{3}{4}=0.75.$", "Equally Likely & Not Equally Likely Outcomes If all the outcomes of a sample space have the same chance of occurrence, then it is known as equally likely outcomes. It is not necessary that the outcomes are equally likely, but during an experiment we shall assume that the outcomes are equally likely outcomes in many cases. Equally Likely Outcomes Cases Equally likely assumptions are applied in the following cases; 1. Playing a Card: In a deck, there are 52 ordinary playing cards. In this all 52 cards are of the same size and therefore, it is assumed as equally likely outcome of each card i.e., 1/52 2. Tossing a Coin: During tossing a single coin, there are two possible outcomes i.e., head and tail. It is always assumed that both are equally likely and each has a chance/probability of occurrence 1/2. If not, then the criteria should be mentioned. In case of tossing more than one coin it is assumed that on all the coins, head and tail are equally likely. 3. Throwing a Dice: There are six “6” possible outcomes when rolling a single dice. In this case all the six outcomes are assumed to be equally likely outcome. Each has a probability of occurrence 1/6. 4. Drawing Balls From a Bag: This is the last case", "the probability of getting a double like $$\\{1,1\\}$$ is still $$\\frac 1{36}$$. Personally, I try to stick to equi-probable cases wherever possible, as that simplifies the computations. But it's up to you, and in some situations you really can't avoid considering outcomes with differing probabilities. The possible outcomes are $$HH, HT, TH, TT$$ Two out of four qualify, so the probability is $$1/2$$. Probability of getting exactly one head, this can happen in two ways : $$HT + TH$$ Favorable cases = 2 Total Cases = 4 So, required probability is $$\\frac{1}{2}$$", "you believe that all outcomes are equally likely to occur, you are assuming that the probability of every event, every subset with one element has the same probability: $P(\\{1\\})=P(\\{2\\})=...=P(\\{N\\})$ or equivalently $$P(\\{i\\})= 1/N$$, for $$i=1,2,...,N$$. Now, if we define a composite event $$A$$, a subset of $$S$$, composed of $$N_A$$ realizations i.e. elements of $$S$$ having the same likelihood (namely, the same probability). Then, the probability of the event $$A$$ becomes the ratio of the outcomes in $$A$$ with respect to the total possible outcomes. $$$\\boxed{P(A)=\\frac{N_A}{N}=\\frac{\\mbox{# of favorable outcomes}}{\\mbox{total # of outcomes}}=\\frac{\\mbox{# of outcomes in A}}{\\mbox{# of outcomes in S}}} \\tag{3.1}$$$ where the “hash” ‘$$\\#$$’ stands for “number.” Example 3.6 (An even number among 6 dice) We roll a fair die and we define the event $A=\\text{the outcome is an even number}=\\{2,4,6\\}.$ What is the probability of $$A$$? First, we identify the sample space as $S=\\{1,2,3,4,5,6\\}.$ Then, we have that $P(A)=\\frac{N_A}{N} = \\frac{\\mbox{3 favorable outcomes}}{\\mbox{6 total outcomes}} = \\frac{1}{2}.$ Building on the intuition gained in the last example and formula (3.1), we state a first informal definition, which establishes the probability of an event in terms of relative frequency. Definition 3.7 (Informal definition of probability) Suppose that an experiment, whose sample space is $$S$$, is repeatedly performed under exactly the same conditions. For each event, say $$A$$, of" ]

[ "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) = $\\frac{1}{4}$", "This puzzle came to me via Futility CLoset Two people are stranded on an island with only one banana to eat. To decide who gets it, they agree to play a game. Each of them will roll a fair 6-sided die. If the largest number rolled is a 1, 2, 3, or 4, then Player 1 gets the banana. If the largest number rolled is a 5 or 6, then Player 2 gets it. Which player has the better chance? I don’t think this is an especially difficult puzzle, but I think the key insight is nice. Spoilers below the line. The naive instinct, of course, is that player 1 has more options available, so their chance must be larger. But no! The probability of both rolling a 4 or lower is $\\left( \\frac23 \\right)^2$, or $\\frac{4}{9}$, which is less than a half! Player 2 is slightly more likely to win (a ratio of 5:4). As long as player 1’s win probability is smaller than $\\frac{1}{\\sqrt{2}}$, player 2 will (more likely) win.", "1. ## probability 4. a) A single fair die is rolled. Find the probability of getting an odd number. b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. 2. Originally Posted by bobby77 4. a) A single fair die is rolled. Find the probability of getting an odd number. All faces of the die are equally likely. The favourable outcomes in this case are: 1, 3, 5 out of a total of 6 possible outcomes. Therefore the probability of an odd number when the die is rolled is: [number of favourable outcomes]/[total number of outcomes] = 3/6 = 0.5 b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. Here the favourable outcomes are A , 2 , 3 .., 10, J, Q, K of diamonds and 10 of hearts, 10 of Spades, 10 of clubs which is a total of 16 favourable cases. Total number of outcomes = 52, so the required probability is: 16/52=4/13 RonL", "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) = $\\frac{1}{4}$", "# Rolling four fair dice and getting four 6s Probability There are 6 possible outcomes on a die and only one of them is a 6. The probability of rolling a 6 is then $$\\frac{1}{6}$$. The probability of rolling four 6s is then $$\\displaystyle\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}=\\frac{{1}}{{1296}}$$", "Independent probability Problem If you flip a coin and roll a 6-sided die, what is the probability that you will flip a heads and roll a 2?", "and 90% of the problems are solved using just this tool. No kidding. Here are some examples to see how it works: {{#x:box| What is the probability that a card drawn at random from a deck of cards will be an ace? Solution. In this case there are four favorable outcomes, aces of spades, hearts, diamonds and clubs. Since each of the 52 cards in the deck represents a possible outcome, there are 52 possible outcomes. Therefore, the probability is $\\frac{4}{52}=\\frac{1}{13}$. }} The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice. {{#x:box| Two fair six-sided dice are rolled; what is the probability of having 5 as the sum of the numbers? Solution. There are 36 possible outcomes when a pair of dice is thrown (six outcomes for the first die times six outcomes for the second one). Since four of the outcomes have a total of 5, {(1,4), (4,1), (2,3), (3,2)}, the probability of the two numbers adding up to 5 is $\\frac{4}{36} = \\frac{1}{9}$. }} ## Probability of Multiple Events For questions involving single events, the F/T rule is sufficient. In fact, it is often sufficient for all other cases too. But, for questions involving multiple events, some other tools may be more appropriate.", "# You play the following game: You bet $1, a fair die is rolled and if it shows 6 you win$4, otherwise you lose your dollar. If you must choose the number of rounds in advance, how should you choose it to maximize your chance of being ahead (having won more than you have lost) when you quit, and what is the probability of this? Question You play the following game: You bet $1, a fair die is rolled and if it shows 6 you win$4, otherwise you lose your dollar. If you must choose the number of rounds in advance, how should you choose it to maximize your chance of being ahead (having won more than you have lost) when you quit, and what is the probability of this?", "# A fair die is rolled. If face 1 turns up, a ball is drawn from Bag A. If face 2 or 3 turns up, a ball is drawn from Bag B. If face 4 or 5 or 6 turns up, a ball is drawn from Bag C. Bag A contains 3 red and 2 white balls, Bag B contains 3 red and 4 white balls and Bag C contains 4 red and 5 white balls. The die is rolled, a Bag is picked up and a ball is drawn. If the drawn ball is red; what is the probability that it is drawn from Bag B? Verified 147k+ views Hint: To solve this question, we need to use the basic theory related to the probability. As we know, to calculate the probability of an event, we take the number of successes, divided by the number of possible outcomes. The probability of event A happening is: P(A) = $\\dfrac{{{\\text{n}}\\left( {\\text{A}} \\right)}}{{{\\text{n}}\\left( {\\text{S}} \\right)}}$ As given, A fair die is rolled and Bag A contains 3 red and 2 white balls, Bag B contains 3 red and 4 white balls and Bag C contains 4 red and 5 white balls. Now, Probability that ball is drawn from bag A = ${\\text{P}}\\left( {{{\\text{E}}_{\\text{1}}}} \\right)$ = $\\dfrac{1}{6}$ Probability that ball is", "## Prealgebra (7th Edition) $\\frac{1}{3}$ There are 6 possible outcomes of tossing a single die: 1, 2, 3, 4, 5, 6. \"1 or 6\" represent 2 of the 6 possible outcomes, so the probability is 2 out of 6. $\\frac{2}{6}=\\frac{2\\div 2}{6\\div 2}=\\frac{1}{3}$", "answers. #### Choice B ##### The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die rolled thrice. The total number of outcomes when a fair die is rolled thrice is 6 * 6 * 6 = 216. If you list down a few of the outcomes and observe the pattern, you will realize that half the outcomes have a sum that is odd and the other half has a sum that is even. Here is a sample of the first 6. {1, 1, 1, odd}, {1, 1, 2, even}, {1, 1, 3, odd}, {1, 1, 4, even}, {1, 1, 5, odd}, {1, 1, 6, even}. Therefore, ½ of 216 = 108 outcomes have an odd sum. Choice B is one of the answers where the number of outcomes is more than 50. #### Choice C ##### The number of outcomes in which the two cards drawn from a pack of well shuffled cards are both red and face cards. A pack of well shuffled cards has 52 cards. The break up of the pack on different parameters is listed in the table given below Color Suit Numbered cards Face cards Aces Red - 26 cards Diamond - 13 cards$2 to 10) 9 cards (J, Q,", "# When rolling a fair, six-sided die, what is the probability of getting a number higher than 2? Mar 20, 2017 $\\frac{2}{3}$ #### Explanation: we want $P \\left(X > 2\\right)$ for a fair die the sample space is$\\text{ } \\left\\{1 , 2 , 3 , 4 , 5 , 6\\right\\}$ all the numbers are equiprobable there are $\\text{ \"4\" }$ numbers greater than 2 so we have $P \\left(X > 2\\right) = \\frac{4}{6} = \\frac{2}{3}$", "comma as follows, \"fair 6-sided die, we are equally...\" I also don't think you need a comma after 'Hence'. For c) I would add another answer with the same numerator but different denominator. This is because you do not have to calculate the probability but just look for the numerator with the same value as the blue balls. I understand that in the cases of simplification this may not be the case, but I think these occur less often then not. a) i) \"Now we must think how many outcomes are there that result in an even number being rolled.\" I would remove are there that so it reads as follows. \"Now we must think how many outcomes result in an even number being rolled. c) \"As the ball is being selected randomly from the bag, there is an equal probability of selecting a ball which is any of the 3 colours.\" I am not sure this statement is true or at least reads such that, there is an equal probability of selecting any of the 3 colours when the probabilities for the selection of the three colours are different. #### Elliott Fletcher2 years, 2 months ago Gave some feedback: Needs to be tested #### Chris Graham2 years, 2 months ago As discussed, it would be nice to", "match is 0.45, the probability of losing the football match is 0.25. 0.3 + 0.45 + 0.25 = 1 • If the probability of it raining is 0.2, the probability of it not raining is 1 - 0.2 = 0.8. • The probability of rolling an odd number or rolling a 2 on a fair six sided die is \\frac{3}{6}+\\frac{1}{6}=\\frac{4}{6}. • The probability of rolling a 4 on a fair six sided die and a fair coin landing heads is \\frac{1}{6}\\times \\frac{1}{2}=\\frac{1}{12}. ### Equally likely outcomes and probability If all of the outcomes from a situation have an equal chance of occurring, the probabilities will all be the same. For example, A fair six sided die is rolled. The outcomes are 1, 2, 3, 4, 5 and 6 and because the die is fair, each of these outcomes are equally likely. The probability of each of these outcomes is \\frac{1}{6}. ### Probability of equally likely events Events can be different to outcomes. For example, The outcomes of a fair six sided die are 1, 2, 3, 4, 5 and 6. Rolling a 1 is an event but so is rolling an even number. Rolling an even number is as equally likely as rolling an odd number. These are known as equally likely events. ### Combined events probability Combined", "+0 # double trouble 2.0 0 168 3 A: If I choose four cards from a standard -card deck, with replacement, what is the probability that I will end up with one card from each suit? B: Four people sit around a circular table, and each person will roll a standard six-sided die. What is the probability that no two people sitting next to each other will roll the same number after they each roll the die once? Express your answer as a common fraction. Jul 14, 2019 #1 +6046 +2 $$A)~P[\\text{1 card of each suit]= 1 \\cdot \\dfrac 3 4 \\cdot \\dfrac 1 2 \\cdot \\dfrac 1 4 = \\dfrac{3}{32}}$$ . Jul 14, 2019 edited by Rom Jul 14, 2019 #2 +977 0 B) There are 6 possibilities for the first person, 6 for the second, 6 for the third, and 6 for the fourth. 6^4 = 1296 total die roll possibilities. Out of these 1296, there are 6 possibilities for the first person but only 5 for the second. Again 5 for the third, and 4 for the fourth. 6*5*5*4 = 600 600/1296 = 75/162 You are very welcome! :P Jul 18, 2019 #3 +105952 +1 Circular tables always complicate things. When maths questions talk about round tables they assume no external points of reference. In", "# What is the probability of obtaining a \"3\" on one roll of a die? If you roll an unbiased die there are 6 total possible outcomes: 1, 2, 3, 4, 5, and 6. The particular outcome you are interested in, a 3, happens only 1 way. Therefore the probability is $\\frac{1}{6}$. If you had asked for the probability of getting a \"3 or less\" then the total number of possible outcomes remains the same, but there are 3 ways of getting the particular outcome (1, 2, or 3) so the probability of getting a \"3 or less\" would be $\\frac{3}{6}$ = $\\frac{1}{2}$.", "# Likely ## Meaning The probability of an event occurring is a measurement of how likely the event is to occur. If two events A and B have equal probabilities, then we say that A and B are equally likely. For example, in rolling a single die, the events \"rolling a 4\" and \"rolling a 5\" are equally likely. However, in drawing a single card from a standard deck of cards, drawing a Heart is more likely than drawing a King, since $P(Heart)=\\frac{1}{4}$ and $P(King)=\\frac{1}{13}$.", "die is rolled 235 times, and 6 comes up 51 times. If the die is fair, we would expect 6 to come up ${\\displaystyle 235\\times 1/6=39.17}$ times. We have now observed that the number of 6s is higher than what we would expect on average by pure chance had the die been a fair one. But, is the number significantly high enough for us to conclude anything about the fairness of the die? This question can be answered by the binomial test. Our null hypothesis would be that the die is fair (probability of each number coming up on the die is 1/6). To find an answer to this question using the binomial test, we use the binomial distribution ${\\displaystyle B(N=235,p=1/6)}$ with pmf ${\\displaystyle f(k,n,p)=\\Pr(k;n,p)=\\Pr(X=k)={\\binom {n}{k}}p^{k}(1-p)^{n-k}}$ . As we have observed a value greater than the expected value, we could consider the probability of observing 51 6s or higher under the null, which would constitute a one-tailed test (here we are basically testing whether this die is biased towards generating more 6s than expected). In order to calculate the probability of 51 or more 6s in a sample of 235 under the null hypothesis we add up the probabilities of getting exactly 51 6s, exactly 52 6s, and so on up to probability of getting exactly 235 6s:", "balls and 8 green balls. So, the total number of balls is: $$6+4+8= 18$$ balls. 1. \\begin{align*} \\text {Probability Of Blue Balls} &= \\frac {\\text {Total Number Of Blue Balls}} {\\text {Total Number Of Balls}} \\\\ \\\\ &= \\frac {4}{18} \\\\ \\\\ &= \\frac {2}{9} \\end{align*} This means that when a ball is randomly taken out from the total 18 balls, the probability of getting a blue ball is $$2\\over9$$. 1. Method 1: To get a ball that is not green: \\begin{align*} \\text {Probability (Not Green)} &= \\frac {\\text {(Total Number Of Red Balls + Total Number Of Blue Balls)}} {\\text {Total Number Of Balls}} \\\\ \\\\ &=\\frac{(6+4)}{18} \\\\ \\\\ &= \\frac{10}{18} \\\\ \\\\ &= \\frac{5}{9} \\end{align*} Method 2: The probability of getting Red/ Blue/ Green Balls is 100%. 100% is considered as 1 (whole). So, 1 represents all the possibilities of getting the total number of balls. \\begin{align*} \\text {Probability (not green)} &= \\text {1}- \\text{(Probability Of Getting Green Balls) } \\\\ \\\\ &= 1-\\frac {8}{18} \\\\ \\\\ &= \\frac {5}{9} \\end{align*} ## Possibility Diagrams What are the possible outcomes of rolling just one fair die? The total number of outcomes are: 1 2 3 4 5 6 The total number of outcomes is 6 outcomes. What are the possible outcomes of rolling two fair dice? (1,1)", "Probability # What is the probability of obtaining tails or a six? ###### Wiki User The answer depends on the experiment: how many coins are tossed, how often, how many dice are rolled, how often. 🙏 0 🤨 0 😮 0 😂 0 ## Related Questions ### Suppose that you toss a coin and roll and die What is the probability of obtaining tails? The probability to tossing a coin and obtaining tails is 0.5. Rolling a die has nothing to do with this outcome - it is unrelated. ### The probability of getting a six on a six sided die and then getting a tails? The probability of getting a six on a six sided die and then getting a tails is zero. There is no tails on a die. ### What is the probability of obtaining tails and a three? Assume the given event depicts flipping a fair coin and rolling a fair die. The probability of obtaining a tail is &frac12;, and the probability of obtaining a 3 in a die is 1/6. Then, the probability of encountering these events is (&frac12;)(1/6) = 1/12. ### What is the probability of obtaining atleast one tail when a coin is flipped six times? you are looking for the probability of getting one tail, two,......., and six this" ]

[ "# Area ratio of a minimum bounding rectangle of a convex polygon Take a convex polygon $P$ in the plane, and find its minimum area bounding rectangle, $R$. I'm interested in the ratio of the area of $R$ to the area of $P$. The ratio has a minimum of $1$ for rectangles and is $2$ for triangles. I want to say that $2$ is the maximum ratio possible. Is this known? Is there a nice proof/counterexample for it? - I believe this is easy to prove. Assume you have at least four points and take the longest chord $C$ of the polygon. Now choose your rectangle such that two edges are parallel and the same length as $C$. Now observe that since each \"half\" of the polygon contains a triangle inscribed in the \"half\" rectangle (where \"half\" means cut by $C$), at least half of each side of the rectangle is filled. Not quite an answer: it is a result of Dutkovsky that the area of a rectangle containing a curve of perimeter $2\\pi$ is at most $4.$", "by 10,25%. How many percent will the side of the square increase? How many percent will it increase the circumference of the square? • Percentage and rectangle About what percentage increases perimeter and area of a rectangle if both the sides 12 cm and 10 cm long we increase by 20%? • Plot The length of the rectangle is 8 smaller than three times the width. If we increase the width by 5% of the length and the length is reduced by 14% of the width, the circumference of rectangle will be increased by 30 m. What are the dimensions of the recta • Three shapes 1/5 of a circle is shaded. The ratio of area if square to the sum of area of rectangle and that of the circle is 1:2. 60% of the square is shaded and 1/3 of the rectangle is shaded. What is the ratio of the area of circle to that of the rectangle? • Fence and scale The garden with 80 m of fencing is draw on plan as a square with a side length of 4 cm. At what scale is this plan made? • Ratio of sides 2 The ratio of the side lengths of one square to another is 1:2. Find the ratio of the area of the two", "used area ratios in both cases where given side of a larger rectangle is length and width and then compared perimeter ratios with the one of a smaller rectangle. When the perimeter ratios and lengths ratios coincide than it's an answer I guess... – Marty Nov 17 '15 at 9:36", "of rectangle R (E) 20% less than the area of rectangle R Let R's length = 10 Let R's width = 5 R's area = (10 * 5) = 50 S's length is 20 percent longer than R's length (10)(1.2) = 12 S's width is 20 percent shorter than R's width (5)(.8) = 4 S's area = (12 * 4) = 48 The area of rectangle S is what percent greater or less than rectangle R? Percent change = $$(\\frac{(new-old)}{old}$$ * 100) = $$(\\frac{S - R}{R}$$ * 100) = $$(\\frac{48 - 52}{100})$$ = $$(\\frac{-4}{100})$$ = -.04 * 100 = -4% The area of a rectangle S is 4% less than the area of rectangle R _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2835 Re: The length of rectangle S is 20 percent longer than the length of [#permalink] Show Tags 03 Oct 2017, 16:33 Bunuel wrote: The length of rectangle S is 20 percent longer than the length of rectangle R, and the width of rectangle S is 20 percent shorter than the width of rectangle R. The area of rectangle S is (A) 20% greater than the area of rectangle R (B) 4% greater", "length and width of the rectangle. 158. The length of a rectangle is three times the width. The perimeter is $7272$ feet. Find the length and width of the rectangle. 159. The length of a rectangle is $33$ meters less than twice the width. The perimeter is $3636$ meters. Find the length and width. 160. The length of a rectangle is $55$ inches more than twice the width. The perimeter is $3434$ inches. Find the length and width. 161. The width of a rectangular window is $2424$ inches. The area is $624624$ square inches. What is the length? 162. The length of a rectangular poster is $2828$ inches. The area is $13161316$ square inches. What is the width? 163. The area of a rectangular roof is $23102310$ square meters. The length is $4242$ meters. What is the width? 164. The area of a rectangular tarp is $132132$ square feet. The width is $1212$ feet. What is the length? 165. The perimeter of a rectangular courtyard is $160160$ feet. The length is $1010$ feet more than the width. Find the length and the width. 166. The perimeter of a rectangular painting is $306306$ centimeters. The length is $1717$ centimeters more than the width. Find the length and the width. 167. The width of a rectangular window is $4040$ inches", "=> Area of A is 63 % of area of B PS Forum Moderator Joined: 25 Feb 2013 Posts: 1185 Location: India GPA: 3.82 Re: If the length and width of rectangle A are 10 percent less and 30 perc [#permalink] ### Show Tags 27 Sep 2017, 10:37 Bunuel wrote: If the length and width of rectangle A are 10 percent less and 30 percent less respectively than the length and width of rectangle B, the area of A is equal to what percent of the area of B? (A) 63% (B) 60% (C) 40% (D) 6% (E) 3% As nothing is mentioned about the sides we can assume that length and breadth of the rectangle B be $$100$$ & $$100$$ respectively so rectangle A's length and breadth will be $$90$$ & $$70$$ Hence % area of A $$= (\\frac{90*70}{100*100})*100 = 63%$$ Option A Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2679 Re: If the length and width of rectangle A are 10 percent less and 30 perc [#permalink] ### Show Tags 01 Oct 2017, 12:11 Bunuel wrote: If the length and width of rectangle A are 10 percent less and 30 percent less respectively than the length and width of rectangle B, the area of A is equal to what percent", "short and longer pieces. Solving the proportion above using the Cross Product Property of Proportionality… Since the shorter piece is $x = 16$ feet, that means the longer piece is $72 - x = 72 - 16 = 56$ feet. To perform a check, we were told in the problem that the ratio of the shorter piece to longer piece is $2$ to $7$. Notice that when we reduce the fraction ${{16} \\over {56}}$ to lowest term, we will get the desired ratio.", "it is shorter than the unit length. The canonical shape with the area $1/2$ is a rectangle with side lengths $1/2$ by $1$. If we compare that to a square of length $1/2$, we need to make one of the sides of the rectangle shorter while the other stays the same. Therefore the number for the area of the square must be less than the number for the length of its side. On the other hand, if the length if the side is $2$, then the canonical shape with area $2$ is a rectangle with sides $1$ by $2$. Here we have to make one of the sides longer in order to get a square of side length $2$, so the number for the sqare's area is larger than the number for its side's length. -", "$A$ is problematic unless the middle sized rectangles can cover the length of the longer one.", "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 • Three shapes 1/5 of a circle is shaded. The ratio of area if square to the sum of area of rectangle and that of the circle is 1:2. 60% of the square is shaded and 1/3 of the rectangle is shaded. What is the ratio of the area of circle to that of the rectangle?", "322 views In a rectangle, the difference between the sum of the adjacent sides and the diagonal is half the length of the longer side. What is the ratio of the shorter to the longer side? 1. $3:2$ 2. $1:3$ 3. $2:5$ 4. $3:4$ Let shorter side be x and longer side be y Then, (x + y) - $\\sqrt{x^{2}+y^{2}}$ = y/2 On solving, we get x/y = 3/4 Hence, option D is correct 326 points 1 534 views 2 449 views", "# Algebra question The length of each side of a square is $3 in.$ more than the length of each side of a smaller square. The sum of the areas of the squares is $185$ squared in. Find the lengths of the sides of the two squares.", "PM rcs Re: Ratio of the area of triangle and rectangle According to the book I've read... It is 5:8. • Dec 17th 2012, 09:27 PM MarkFL Re: Ratio of the area of triangle and rectangle The ratio of the area of the triangle to that of the rectangle is: $\\frac{1}{2}b\\cdot\\frac{5}{4}h:bh$ $\\frac{5}{8}:1$ $5:8$ • Dec 18th 2012, 03:33 AM rcs Re: Ratio of the area of triangle and rectangle Thanks Sir Mark", "powers or roots to get the desired ratio. (4) Write and solve a proportion. 4. Mix-up the exercises so that students will have to square the ratio in one problem and not in the next. Keep them on the lookout. Make them analyze the situation instead of falling into a habit. Example 1: The ratio of the lengths of the sides of two squares is 2:3. What is the ratio of their areas? Answer: 4:9, The ratio of areas is the ratio of the lengths squared. Example 2: The area of a small triangle is $15 \\ cm^2$, and it has a height of 5 cm. A larger similar triangle has an area of $60 \\ cm^2$. What is the corresponding height of the larger triangle? The area ratio is 15:60 or 1:4. The length ratio is then the square root of this, or 1:2. Now set up a proportion to solve for the height of the smaller triangle. $\\frac{1}{2} = \\frac{5}{x}$ Example 3: The ratio of the lengths of two similar rectangles is 4:5. The larger rectangle has a width of 45 cm. What is the width of the smaller rectangle? In this problem, we did not have to adjust the ratio since we are given a ratio of lengths and a length is what we are", "of their perimeters. 8. Find the area of the blue and large squares. 9. Find the ratio of their areas. 10. Find the length of the diagonals of the blue and large squares. Put them into a ratio. Which ratio is this the same as? 11. Two rectangles are similar with a scale factor of $\\frac{4}{7}$. If the area of the larger rectangle is $294 \\ in^2$, find the area of the smaller rectangle. 12. Two triangles are similar with a scale factor of $\\frac{1}{3}$. If the area of the smaller triangle is $22 \\ ft^2$, find the area of the larger triangle. 13. The ratio of the areas of two similar squares is $\\frac{16}{81}$. If the length of a side of the smaller square is 24 units, find the length of a side in the larger square. 14. The ratio of the areas of two right triangles is $\\frac{4}{9}$. If the length of the hypotenuse of the larger triangle is 48 units, find the length of the smaller triangle’s hypotenuse. Questions 23-26 build off of each other. You may assume the problems are connected. 1. Two similar rhombi have areas of $72 \\ units^2$ and $162 \\ units^2$. Find the ratio of the areas. 2. Find the scale factor. 3. The diagonals in these rhombi are congruent.", "the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle? a. $$4 \\sqrt{2}$$ b. $$2 \\sqrt{2}$$ c. $$\\sqrt{2}$$ d. 1 e. 3 $$\\frac{l}{b} = \\frac{b}{(l/2)}$$ where $$b = 2$$ given i.e. $$b^2 = l^2 / 2$$ i.e. $$l = 2√2$$ Area of smaller rectangle $$= (l/2)*b = (2√2/2)*2 = 2√2$$ _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: [email protected] I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION Senior Manager Joined: 22 Feb 2018 Posts: 416 Re: A rectangular sheet of paper, when halved by folding it at the mid [#permalink] ### Show Tags 25 Oct 2018, 07:05 2 RhythmGMAT wrote: A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle? a. $$4 \\sqrt{2}$$ b. $$2 \\sqrt{2}$$ c. $$\\sqrt{2}$$", "= 5.5 × 3.75 sq. metre. Cost for 1 sq. metre. = Rs. 800 Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500 11. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle? A. 18 cm B. 16 cm C. 40 cm D. 20 cm Explanation: Then length = 2x cm Area = lb = x × 2x = 2x2 New length = (2x - 5) New breadth = (x + 5) New Area = lb = (2x - 5)(x + 5) But given that new area = initial area + 75 sq.cm. => (2x - 5)(x + 5) = 2x2 + 75 => 2x2 + 10x - 5x - 25 = 2x2 + 75 => 5x - 25 = 75 => 5x = 75 + 25 = 100 => x = $\\dfrac{100}{5}$ = 20 cm Length = 2x = 2 × 20 = 40cm 12. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus? A. equal to ½ B. equal to ¾", "usable rectangles This question deals with efficient division of land into land-plots. For the sake of this question, assume that a land-plot is usable only if it is a rectangle whose width/height ratio is between $r$ and $1/r$. If a person gets a plot that is not entirely usable, he has to find a single usable rectangle that is contained within that plot; see figure 1 (where $r$ is approximately 2). I am interested in what happens when we divide a usable land-plot. For example, in figure 2a we divide a usable plot into two plots by cutting at the middle of the narrow side. We have two land-plots, but in each of them, only about half is still usable (for $r=2$), so there is a loss of about $1/2$ (note that each citizen gets a single land-plot, and can use a single usable rectangle in the plot). However, in figure 2b we divide the same plot by cutting at the middle of the wide side, and get two rectangles with more balanced width/height ratio, therefore there is no loss at all. What happens if we cut the rectangle with an arbitrary straight line (figure 2c)? It seems that a loss of $1/2$ is the worst case, but I could not prove it. Another interesting question is what", "# Why isn't the area of a square always greater than the length of one of its sides? Intuitively, it seems like the area of a square should always be greater than the length of one of its sides because you can \"fit\" one of its sides in the space of its area, and still have room left over. However when the length of a side, $$s$$, is less than $$1$$, then the area $$s^2 < s$$, which doesn't make sense to me for the reason above. • a square with side $0.5 m$ has area $0.25 m^2$ with 0.25 < 0.5. Where is the problem ? – Jean Marie Feb 4 '18 at 19:44 • @JeanMarie Even worse, the same square with side $50$ cm has area $2500 \\gt 50$ ;-) – dxiv Feb 4 '18 at 19:45 • Comparing the area of a square to the length of one of its sides is like comparing liters to seconds. It is not defined whether 3 liters is less than, equal to, or greater than 7 seconds. It's not defined whether the length of a side is greater than the area. – Ben Crowell Feb 4 '18 at 22:50 • Quite in shock that 12 people liked this question and considered it valuable for the website... – DanielC", "yes the ratio of the area is the SQUARE of the ratio of the sides this should not be too surprising because the side is measured in length (one dimensional unit like feet or meters) and the area is measured in SQUARE units, like square feet or square meters 30. anonymous so if the ratio of the sides is say $$5:2$$ then the ratio of the areas is $$25:4$$ 31. anonymous Oh wow that makes so much more sense now 32. anonymous 33. anonymous 34. anonymous Yes of course! 35. anonymous" ]

[ "area in this diagram, two thin rectangles, and a miniscule square. The two thin rectangles have side lengths $x$ and $dx$, so together they account for $2 \\cdot x \\cdot dx$ units of new area. For example, if $x$ was $3$ and $dx$ was $0.01$, the new area from these thin rectangles would be $2 \\cdot 3 \\cdot 0.01$, which is $0.06$; About $6$ times the size of $dx$. If $x=2$ and $dx$ was $0.01$, approximately how much bigger would the new area be relative to the size of $dx$? That little square has area $dx^2$, but you should think of this as being really tiny; negligibly tiny. For example, if $dx$ was $0.01$, it would be $0.0001$. I’m drawing $dx$ with a fair bit of width here, so we can see it, but always remember in principle $dx$ should be thought of as a truly tiny amount. Phrased more precisely, our final consideration will always be what happens as the size of this $dx$ approaches $0$, and as that happens the proportion of this yellow area $df$ which is accounted for by the tiny $dx^2$ corner will go to $0$. Phrased more precisely, our final consideration will always be what happens as the size of this $dx$ approaches $0$, and as that happens the proportion of this", "\\sqrt[n]{x_1 x_2\\cdots x_n}$$ With equality if, and only if $x_1=x_2=\\cdots=x_n$ So suppose we fix the perimeter of a rectangle as $2(a+b)$ with side lengths $a$ and $b$. Hence, a square with the same perimeter has a side length of $\\displaystyle\\frac{a+b}{2}$. So by applying the AM-GM inequality with $n=2$ it follows that $\\displaystyle \\left(\\frac{a+b}{2}\\right)^2 \\ge ab$. So we see that the area of the rectangle doesn't exceed the area of the square, and the areas are equal if and only if $a=b$. So in other words, for a fixed perimeter the maximum area of a rectangle is when it is in fact a square. Alternatively, one can apply this to a fixed area and deduce that the perimeter of a rectangle is always greater than or equal to the perimeter of a square. This time suppose that $x$ and $y$ are the lengths of a rectangle with a fixed area $xy$. Hence, a square with the same area has side-lengths $\\sqrt{ab}$. So, by AM-GM we obtain $2(x+y)\\ge 4\\sqrt{xy}$. Now we see that the for a fixed area $xy$, the rectangle with the smallest perimeter is a square. This can be easily extended and generalized into $n$ dimensions and we can say that an $n$-cube has the smallest perimeter among all $n$-dimensional boxes with the same volume and that an", "in the box, just find the ratio of the area of box $A$ to the area of the entire green side of the court. $P (\\text{ball lands in box}) &= \\frac{\\text{area of favorable section}}{\\text{total area}} \\\\&= \\frac{\\text{area of box}\\ A}{\\text{area box} \\ A + \\text{box} \\ B + \\text{box} \\ C}$ Now calculate the area of box $A$, box $B$, and box $C$. You can look at the diagram to determine this. Remember that the formula for the area of a rectangle is $A = lw$ and the units are measured in square units. In this case, it will be square feet. $\\text{Box} \\ A &= 21 \\ ft \\cdot 13.5 \\ ft \\\\&= 283.5 \\ sq \\ ft \\\\\\text{Box} \\ B &= 21 \\ ft \\cdot 13.5 \\ ft \\\\&= 283.5 \\ sq \\ ft \\\\\\text{Box} \\ C &= 18 \\ ft \\cdot 27 \\ ft \\\\&= 486 \\ sq \\ ft$ So: $P(\\text{ball lands in box}) &= \\frac{\\text{area of favorable'' section}}{\\text{total area}}\\\\&= \\frac{\\text{area} \\ A}{\\text{area} \\ B + \\text{area} \\ B + \\text{area} \\ C} \\\\&= \\frac{283.5}{1053} \\\\&= 26.9\\%$ The ball has a little more than a 1 in 4 chance of landing in the service box. Good question. We got that because 25% would be the same as one-quarter or $\\frac{1}{4}$. Here we have a", "# Area ratio of a minimum bounding rectangle of a convex polygon Take a convex polygon $P$ in the plane, and find its minimum area bounding rectangle, $R$. I'm interested in the ratio of the area of $R$ to the area of $P$. The ratio has a minimum of $1$ for rectangles and is $2$ for triangles. I want to say that $2$ is the maximum ratio possible. Is this known? Is there a nice proof/counterexample for it? - I believe this is easy to prove. Assume you have at least four points and take the longest chord $C$ of the polygon. Now choose your rectangle such that two edges are parallel and the same length as $C$. Now observe that since each \"half\" of the polygon contains a triangle inscribed in the \"half\" rectangle (where \"half\" means cut by $C$), at least half of each side of the rectangle is filled. Not quite an answer: it is a result of Dutkovsky that the area of a rectangle containing a curve of perimeter $2\\pi$ is at most $4.$", "R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? I. 63 II. 126 III. 252 A. I only B. II only C. III only D. I and III only E. I, II and III Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Thank you. I have a question - Why cant x be [square_root]2. Why cant we have sides of lengths 5*[square_root]2, 11*[square_root]2 and 9*[square_root]2? Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 615 ### Show Tags 28 Aug 2013, 01:28 1 Gagan1983 wrote: Bunuel wrote: SOLUTIONS: [b]1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square", "Question 9 # If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is Solution All the sides of the rhombus are equal. The area of a rhombus is $$12\\ cm^2$$ Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal. The area of a rhombus is $$\\left(\\frac{1}{2}\\right)\\left(d1\\right)\\cdot\\left(d2\\right)\\ =\\ 12$$ d1*d2 = 24. The length of the side of a rhombus is given by $$\\frac{\\sqrt{\\ d1^2+d2^2}}{2}$$. This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length. $$\\frac{\\sqrt{\\ d1^2+d2^2}}{2}=\\ 5$$ Hence $$\\sqrt{\\ d1^2+d2^2}\\ =\\ 10$$ $$d1^2+d2^2\\ =\\ 100$$ Using d1*d2 = 24, 2*d1*d2 = 48. $$d1^2+d2^2\\ +2\\cdot d1\\cdot d2=\\ 100+48\\ =\\ 148$$ $$d1^2+d2^2\\ -2\\cdot d1\\cdot d2=\\ 100-48\\ =\\ 52$$ $$d1+d2\\ =\\ \\sqrt{\\ 148}$$ (1) d1-d2 = $$\\sqrt{52}$$ (2) (1) + (2)= 2*(d1) = 2*($$\\sqrt{\\ 37}+\\sqrt{\\ 13}$$) d1 = $$\\sqrt{\\ 37}+\\sqrt{\\ 13}$$ or In a rhombus the area of a Rhombus is given by : The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b. The area of a Rhombus is : $$\\left(\\frac{1}{2}\\right)\\cdot\\left(2a\\right)\\cdot\\left(2b\\right)\\ =\\ 12$$ ab =6. The length of each side is : $$\\sqrt{\\ a^2+b^2}$$ = 5, $$a^2+b^{2\\ }=\\ 25,\\$$ $$\\left(a+b\\right)^2=\\", "As bjhopper pointed out the length of the diagonals are: $d_1 = \\sqrt{8^2+2^2} = \\sqrt{208} = 2 \\cdot \\sqrt{52}$ $d_2 = \\sqrt{6^2+4^2} = \\sqrt{52}$ And therefore the area is: $a = \\frac12 \\cdot 2 \\cdot \\sqrt{52} \\cdot \\sqrt{52} = 52$", "# The Area of a Rectangle A rectangle is a quadrilateral in which all four angles are right angles and the opposite sides are equal in length. Let $ABCD$ be a rectangle having $AB = a$ and $BC = b$. Let $AC$ be the diagonal which divides the rectangle into two right triangles, $\\Delta ABC$ and $\\Delta ADC$. $\\begin{gathered} {\\text{Area of }}\\Delta ABC = \\frac{1}{2}AB \\times BC = \\frac{1}{2}a \\times b \\\\ \\Rightarrow {\\text{Area of }}\\Delta ADC = \\frac{1}{2}DC \\times AD = \\frac{1}{2}a \\times b \\\\ \\end{gathered}$ $\\begin{gathered} {\\text{Area of rectangle }}ABCD = {\\text{Area of }}\\Delta ABC + {\\text{Area of }}\\Delta ADC \\\\ \\Rightarrow {\\text{Area of rectangle }}ABCD = \\frac{1}{2}a \\times b + \\frac{1}{2}a \\times b \\\\ \\therefore {\\text{Area of rectangle }}ABCD = a \\times b = {\\text{length}} \\times {\\text{breadth}} \\\\ \\end{gathered}$ Example: In exchange for a square plot of land, one of whose sides is $84$ m, a man wants to buy a rectangular plot $144$m long which is of the same area as the square plot. Find the width of the rectangular plot. Solution: One side of the square plot $= 84$ m. The area of the square plot $= 84 \\times 84 = 7056$ square m. The length of the rectangular plot $= 144$ m. The length of the rectangular plot $=$ area of the", "# Comparing areas? The area decreases from $225 c {m}^{2}$ to $100 c {m}^{2}$, which is a decrease of $125 c {m}^{2}$ Initially, the sides of a square is 15cm long. The area initially (which you got) is $A r e a = L \\cdot W = 15 \\cdot 15 = 225$ If all 4 sides are decreased by 5cm, it means that instead of all the sides being 15, it is now 10cm. ($15 - 5 = 10$) So the new area of the square is $A r e a = L \\cdot W = 10 \\cdot 10 = 100$ To find the total decrease, we take the original area and subtract it by the new area. $225 - 100 = 125$ So this is therefore a decrease of $125 c {m}^{2}$ from the original area.", "$F_0^2+F_1^2+\\cdots+F_n^2=F_{n}F_{n+1}$, with $F_0=1$. By puting together squares of side $F_n$, one at a time, you get a rectangle of dimension $F_nF_{n+1}$: The two squares of side 1, then the square of side 2, then the square of side 3 and so on.", "the perspective of the inner cube, we can see that there is one cube connected to each side. Furthermore, there are $$5$$ exposed on each of the outer cubes. This makes the surface area $$5\\cdot 6=30.$$ This makes the ratio of volume to surface area $$7:30.$$ 17. Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $$50$$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles? $$\\ 76$$ $$\\ 120$$ $$\\ 128$$ $$\\ 132$$ $$\\ 136$$ ###### Solution(s): Let the length and width be $$l,w$$ respectively. The rectangle would have an area of $l+r+l+r = 2(l+r)=50.$ Thus, $$l+r =25,$$ so $$r=25-l.$$ The area is $lr = l(25-l)$$= 156.25 - (l-12.5)^2.$ This makes the area largest when $$l$$ is as close as possible to $$12.5$$ and the area is the smallest when $$l$$ is as far from $$12.5.$$ Therefore, the largest area is when $$l=12,13$$ and the smallest area is when $$l=1,24.$$ This makes the larger area equal to $156.25-(13-12.5)^2$$= 156.25-0.25 =156$ and the smaller area equal to $156.25-(24-12.5)^2$$= 156.25-132.25 =24.$ Thus, the difference is $$156-24=132.$$ 18. Two circles that share the same center have radii $$10$$ meters and $$20$$ meters.", "Intern Joined: 06 Jun 2013 Posts: 27 Concentration: Finance, Economics Schools: AGSM '16 GMAT 1: 690 Q38 V41 GPA: 3.92 Followers: 0 Kudos [?]: 0 [0], given: 13 Re: Fresh Meat!!! [#permalink] 08 Jul 2013, 10:28 Aho92 wrote: Bunuel wrote: SOLUTIONS: 1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? I. 63 II. 126 III. 252 A. I only B. II only C. III only D. I and III only E. I, II and III Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Hi Bunuel, This is probably a stupid question. But why can't x be \\sqrt{2}? Okay. I got it. Stupid me. They have to be integers Math Expert Joined: 02 Sep 2009", "and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Thank you. I have a question - Why cant x be [square_root]2. Why cant we have sides of lengths 5*[square_root]2, 11*[square_root]2 and 9*[square_root]2? Manager Joined: 20 Nov 2011 Posts: 51 Followers: 0 Kudos [?]: 8 [0], given: 17 ### Show Tags 28 Aug 2013, 10:42 Thanx bunnel for this set of question. I learned many good stuffs. Cheers Intern Joined: 28 Jun 2013 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 ### Show Tags 28 Aug 2013, 11:55 mau5 wrote: Gagan1983 wrote: Bunuel wrote: SOLUTIONS: [b]1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R).", "rectangle.}$ Code: 3x * - - - - - * | | x | | x | | * - - - - - * 3x The area of this rectangle is: . $3x\\cdot x \\:=\\:\\boxed{3x^2}$ We have: . $x + 3x + x + 3x \\:=\\:a\\quad\\Rightarrow\\quad 8x \\:=\\:a$ The right piece of wire has length: . $12-8x$ It is bent into a square. Code: s * - - - * | | s | | s | | * - - - * s We have: . $4s = 12 - 8x \\quad\\Rightarrow\\quad s \\:=\\:3 - 2x$ The area of this square is: . $\\boxed{(3-2x)^2}$ Therefore, the total area is: . $A \\;=\\;3x^2 + (9 - 6x + 4x^2) \\;=\\;7x^2 - 6x + 9$", "we consider $$Area = \\frac 12 C \\cdot r = \\frac 12 (\\pi \\cdot diameter)\\cdot r= \\frac 12(\\pi \\cdot 2r)\\cdot r = \\pi r^2$$ • Thanks a lot @fleablood. I understand clearly and learn a lot from your great answer. Feb 25, 2021 at 8:36 You are approximating the circle with the sum of the areas of triangles that fit inside. This is actually the first numerical way pi was calculated, done by Archimedes about 2300 years ago. Yes you can make the triangular slices thinner and thinner, in a process where you can fit in triangles without bound, however because they are getting thinner in proportion with the rate at which you are making more of them, the sum of the areas of the triangles actually stays finite. • 32 is finite, If we divide the circle into 32 pieces and join them like a rectangle it will look like a rectangle but it's not perfect when we divide infinite time it will be a perfect rectangle, so is my assumption correct? Feb 25, 2021 at 7:38 • Thank you for your response, you tried your best and I'm happy about that😃. Feb 26, 2021 at 13:52 You could just as easily claim that, if a square of side length $$L$$ is positioned with its sides horizontal", "(2) The area of rectangle R is 36. [Reveal] Spoiler: OA _________________ PS Forum Moderator Joined: 25 Feb 2013 Posts: 1012 Location: India GPA: 3.82 The area of square S is equal to the area of rectangle R. What is the [#permalink] ### Show Tags 19 Nov 2017, 09:52 Bunuel wrote: The area of square S is equal to the area of rectangle R. What is the perimeter of square S? (1) The length of one of the sides of rectangle R is twice the length of a side of square S. (2) The area of rectangle R is 36. Given $$a^2=l*b$$ where $$a$$ is the length of the square and $$l$$ & $$b$$ is the length and breadth of the rectangle we need to find Perimeter $$= 4a$$. Hence we need values of $$l$$ & $$b$$ Statement 1: implies $$l=2a$$. but nothing mentioned about breadth of the rectangle. Hence insufficient Statement 2: $$l*b=36=a^2$$ so $$a=6$$. Hence $$4a=4*6=24$$. Sufficient Option B Intern Joined: 16 Aug 2016 Posts: 15 Location: India GMAT 1: 460 Q35 V19 GPA: 3.6 WE: Brand Management (Retail) Re: The area of square S is equal to the area of rectangle R. What is the [#permalink] ### Show Tags 19 Nov 2017, 12:28 Info: Area of square S = Area of rectangle R Ques: Perimeter", "# 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]", "in the ratio 2:3 then the ratio of the area of R to the area of S A. 25:16 B. 24:25 C. 5:6 D. 4:5 E. 4:9 _________________ Sky is the limit Math Expert Joined: 02 Sep 2009 Posts: 59089 If the perimeter of square region S and the perimeter of rec [#permalink] ### Show Tags 03 Feb 2012, 08:29 19 16 If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S A. 25:16 B. 24:25 C. 5:6 D. 4:5 E. 4:9 Let the side of square be $$s$$ and the side of rectangle $$a$$ and $$b$$ Given: The sides of R are in the ratio 2:3 --> $$\\frac{a}{b}=\\frac{2}{3}$$ --> $$b=\\frac{3a}{2}$$. The perimeter of square region S and the perimeter of rectangular region R are equal --> $$P=4s=2(a+b)$$ --> $$2s=a+b=\\frac{5a}{2}$$ --> $$s=\\frac{5a}{4}$$. Question: $$\\frac{area \\ of \\ R}{area \\ of \\ S}=\\frac{ab}{s^2}=?$$ Substitute b = 3a/2 and s = 5a/4 into the question: $$\\frac{ab}{s^2}=\\frac{3a^2}{2}*\\frac{16}{25a^2}=\\frac{24}{25}$$. OR: you can pick numbers: let the sides of rectangle be 4 and 6 (ratio 2:3) then the perimeter of the rectangle will be 2(4+6)=20, thus the side of the square will be 20/4=5. Next,", "difference between the area of square S and the area of rhombus R? I. 63 II. 126 III. 252 A. I only B. II only C. III only D. I and III only E. I, II and III Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Hi Bunuel, This is probably a stupid question. But why can't x be \\sqrt{2}? Kudos [?]: [0], given: 16 Intern Joined: 06 Jun 2013 Posts: 26 Kudos [?]: [0], given: 16 Concentration: Finance, Economics Schools: AGSM '16 GMAT 1: 690 Q38 V41 GPA: 3.92 ### Show Tags 08 Jul 2013, 11:28 Aho92 wrote: Bunuel wrote: SOLUTIONS: 1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? I. 63 II.", "b, and c and angles A, B, and C opposite them, $A = \\frac{1}{2}ab \\cdot \\sin(C) = \\frac{1}{2}bc \\cdot \\sin(A) = \\frac{1}{2}ac \\cdot \\sin(B)$ This formula is true because $h = \\sin(C)$ in the formula $A = \\tfrac{1}{2}Bh$. It is useful because you don't need to find the height from an angle in a separate step, and is also used to prove the law of sines (divide all terms in the above equation by a*b*c and you'll get it directly!) ## Area of Rectangles The area calculation of a rectangle is simple and easy to understand. One of the sides is chosen as the base, with a length b. An adjacent side is then the height, with a length h, because in a rectangle the adjacent sides are perpendicular to the side chosen as the base. The rectangle's area is given by: $A = b \\cdot h$ Sometimes, the baselength may be referred to as the length of the rectangle, l, and the height as the width of the rectangle, w. Then the area formula becomes: $A = l \\cdot w$ Regardless of the labels used for the sides, it is apparent that the two formulas are equivalent. Of course, the area of a square with sides having length s would be: $A = s^2$ ## Area of" ]

[ "of rectangle R (E) 20% less than the area of rectangle R Let R's length = 10 Let R's width = 5 R's area = (10 * 5) = 50 S's length is 20 percent longer than R's length (10)(1.2) = 12 S's width is 20 percent shorter than R's width (5)(.8) = 4 S's area = (12 * 4) = 48 The area of rectangle S is what percent greater or less than rectangle R? Percent change = $$(\\frac{(new-old)}{old}$$ * 100) = $$(\\frac{S - R}{R}$$ * 100) = $$(\\frac{48 - 52}{100})$$ = $$(\\frac{-4}{100})$$ = -.04 * 100 = -4% The area of a rectangle S is 4% less than the area of rectangle R _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2835 Re: The length of rectangle S is 20 percent longer than the length of [#permalink] Show Tags 03 Oct 2017, 16:33 Bunuel wrote: The length of rectangle S is 20 percent longer than the length of rectangle R, and the width of rectangle S is 20 percent shorter than the width of rectangle R. The area of rectangle S is (A) 20% greater than the area of rectangle R (B) 4% greater", "area in this diagram, two thin rectangles, and a miniscule square. The two thin rectangles have side lengths $x$ and $dx$, so together they account for $2 \\cdot x \\cdot dx$ units of new area. For example, if $x$ was $3$ and $dx$ was $0.01$, the new area from these thin rectangles would be $2 \\cdot 3 \\cdot 0.01$, which is $0.06$; About $6$ times the size of $dx$. If $x=2$ and $dx$ was $0.01$, approximately how much bigger would the new area be relative to the size of $dx$? That little square has area $dx^2$, but you should think of this as being really tiny; negligibly tiny. For example, if $dx$ was $0.01$, it would be $0.0001$. I’m drawing $dx$ with a fair bit of width here, so we can see it, but always remember in principle $dx$ should be thought of as a truly tiny amount. Phrased more precisely, our final consideration will always be what happens as the size of this $dx$ approaches $0$, and as that happens the proportion of this yellow area $df$ which is accounted for by the tiny $dx^2$ corner will go to $0$. Phrased more precisely, our final consideration will always be what happens as the size of this $dx$ approaches $0$, and as that happens the proportion of this", "R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? I. 63 II. 126 III. 252 A. I only B. II only C. III only D. I and III only E. I, II and III Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Thank you. I have a question - Why cant x be [square_root]2. Why cant we have sides of lengths 5*[square_root]2, 11*[square_root]2 and 9*[square_root]2? Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 615 ### Show Tags 28 Aug 2013, 01:28 1 Gagan1983 wrote: Bunuel wrote: SOLUTIONS: [b]1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square", "=> Area of A is 63 % of area of B PS Forum Moderator Joined: 25 Feb 2013 Posts: 1185 Location: India GPA: 3.82 Re: If the length and width of rectangle A are 10 percent less and 30 perc [#permalink] ### Show Tags 27 Sep 2017, 10:37 Bunuel wrote: If the length and width of rectangle A are 10 percent less and 30 percent less respectively than the length and width of rectangle B, the area of A is equal to what percent of the area of B? (A) 63% (B) 60% (C) 40% (D) 6% (E) 3% As nothing is mentioned about the sides we can assume that length and breadth of the rectangle B be $$100$$ & $$100$$ respectively so rectangle A's length and breadth will be $$90$$ & $$70$$ Hence % area of A $$= (\\frac{90*70}{100*100})*100 = 63%$$ Option A Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2679 Re: If the length and width of rectangle A are 10 percent less and 30 perc [#permalink] ### Show Tags 01 Oct 2017, 12:11 Bunuel wrote: If the length and width of rectangle A are 10 percent less and 30 percent less respectively than the length and width of rectangle B, the area of A is equal to what percent", "the perspective of the inner cube, we can see that there is one cube connected to each side. Furthermore, there are $$5$$ exposed on each of the outer cubes. This makes the surface area $$5\\cdot 6=30.$$ This makes the ratio of volume to surface area $$7:30.$$ 17. Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $$50$$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles? $$\\ 76$$ $$\\ 120$$ $$\\ 128$$ $$\\ 132$$ $$\\ 136$$ ###### Solution(s): Let the length and width be $$l,w$$ respectively. The rectangle would have an area of $l+r+l+r = 2(l+r)=50.$ Thus, $$l+r =25,$$ so $$r=25-l.$$ The area is $lr = l(25-l)$$= 156.25 - (l-12.5)^2.$ This makes the area largest when $$l$$ is as close as possible to $$12.5$$ and the area is the smallest when $$l$$ is as far from $$12.5.$$ Therefore, the largest area is when $$l=12,13$$ and the smallest area is when $$l=1,24.$$ This makes the larger area equal to $156.25-(13-12.5)^2$$= 156.25-0.25 =156$ and the smaller area equal to $156.25-(24-12.5)^2$$= 156.25-132.25 =24.$ Thus, the difference is $$156-24=132.$$ 18. Two circles that share the same center have radii $$10$$ meters and $$20$$ meters.", "powers or roots to get the desired ratio. (4) Write and solve a proportion. 4. Mix-up the exercises so that students will have to square the ratio in one problem and not in the next. Keep them on the lookout. Make them analyze the situation instead of falling into a habit. Example 1: The ratio of the lengths of the sides of two squares is 2:3. What is the ratio of their areas? Answer: 4:9, The ratio of areas is the ratio of the lengths squared. Example 2: The area of a small triangle is $15 \\ cm^2$, and it has a height of 5 cm. A larger similar triangle has an area of $60 \\ cm^2$. What is the corresponding height of the larger triangle? The area ratio is 15:60 or 1:4. The length ratio is then the square root of this, or 1:2. Now set up a proportion to solve for the height of the smaller triangle. $\\frac{1}{2} = \\frac{5}{x}$ Example 3: The ratio of the lengths of two similar rectangles is 4:5. The larger rectangle has a width of 45 cm. What is the width of the smaller rectangle? In this problem, we did not have to adjust the ratio since we are given a ratio of lengths and a length is what we are", "Rectangle area The length of a rectangle of x units is increased by 10% and its width of y units is increased by 15%. What is the ratio of the area of the old rectangle to the area if the new rectangle? Correct result: r = 1.265 p = 26.5 % Solution: $a_{2}=1.10 \\ a_{1} \\ \\\\ b_{2}=1.15 \\ b_{1} \\ \\\\ S_{1}=a_{1} \\cdot \\ b_{1} \\ \\\\ S_{2}=a_{2} \\cdot \\ b_{2}=1.10 \\ a_{1} \\cdot \\ 1.15 \\ b_{1} \\ \\\\ S_{2}=1.10 \\cdot \\ 1.15 \\cdot \\ S_{1} \\ \\\\ r=1.10 \\cdot \\ 1.15=\\dfrac{ 253 }{ 200 }=1.265$ $p=100 \\cdot \\ (r - 1)=100 \\cdot \\ (1.265 - 1)=\\dfrac{ 53 }{ 2 }=26.5 \\%$ 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 Check out our ratio calculator. 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: • Hop-garden The length of the rectangular hop", "# Difference between revisions of \"2012 AMC 10B 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 Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is $5$, the diameter is $2\\cdot 5 = 10$. Since the sides of the rectangle are in a $2:1$ ratio, the longer side has length $2\\cdot 10 = 20$. Therefore the area is $20\\cdot 10 = 200$ or $\\boxed{E}$.", "difference between the area of square S and the area of rhombus R? I. 63 II. 126 III. 252 A. I only B. II only C. III only D. I and III only E. I, II and III Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Hi Bunuel, This is probably a stupid question. But why can't x be \\sqrt{2}? Kudos [?]: [0], given: 16 Intern Joined: 06 Jun 2013 Posts: 26 Kudos [?]: [0], given: 16 Concentration: Finance, Economics Schools: AGSM '16 GMAT 1: 690 Q38 V41 GPA: 3.92 ### Show Tags 08 Jul 2013, 11:28 Aho92 wrote: Bunuel wrote: SOLUTIONS: 1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? I. 63 II.", "PM rcs Re: Ratio of the area of triangle and rectangle According to the book I've read... It is 5:8. • Dec 17th 2012, 09:27 PM MarkFL Re: Ratio of the area of triangle and rectangle The ratio of the area of the triangle to that of the rectangle is: $\\frac{1}{2}b\\cdot\\frac{5}{4}h:bh$ $\\frac{5}{8}:1$ $5:8$ • Dec 18th 2012, 03:33 AM rcs Re: Ratio of the area of triangle and rectangle Thanks Sir Mark", "the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle? a. $$4 \\sqrt{2}$$ b. $$2 \\sqrt{2}$$ c. $$\\sqrt{2}$$ d. 1 e. 3 $$\\frac{l}{b} = \\frac{b}{(l/2)}$$ where $$b = 2$$ given i.e. $$b^2 = l^2 / 2$$ i.e. $$l = 2√2$$ Area of smaller rectangle $$= (l/2)*b = (2√2/2)*2 = 2√2$$ _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: [email protected] I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION Senior Manager Joined: 22 Feb 2018 Posts: 416 Re: A rectangular sheet of paper, when halved by folding it at the mid [#permalink] ### Show Tags 25 Oct 2018, 07:05 2 RhythmGMAT wrote: A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle? a. $$4 \\sqrt{2}$$ b. $$2 \\sqrt{2}$$ c. $$\\sqrt{2}$$", "and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Thank you. I have a question - Why cant x be [square_root]2. Why cant we have sides of lengths 5*[square_root]2, 11*[square_root]2 and 9*[square_root]2? Manager Joined: 20 Nov 2011 Posts: 51 Followers: 0 Kudos [?]: 8 [0], given: 17 ### Show Tags 28 Aug 2013, 10:42 Thanx bunnel for this set of question. I learned many good stuffs. Cheers Intern Joined: 28 Jun 2013 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 ### Show Tags 28 Aug 2013, 11:55 mau5 wrote: Gagan1983 wrote: Bunuel wrote: SOLUTIONS: [b]1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R).", "length and width of the rectangle. 158. The length of a rectangle is three times the width. The perimeter is $7272$ feet. Find the length and width of the rectangle. 159. The length of a rectangle is $33$ meters less than twice the width. The perimeter is $3636$ meters. Find the length and width. 160. The length of a rectangle is $55$ inches more than twice the width. The perimeter is $3434$ inches. Find the length and width. 161. The width of a rectangular window is $2424$ inches. The area is $624624$ square inches. What is the length? 162. The length of a rectangular poster is $2828$ inches. The area is $13161316$ square inches. What is the width? 163. The area of a rectangular roof is $23102310$ square meters. The length is $4242$ meters. What is the width? 164. The area of a rectangular tarp is $132132$ square feet. The width is $1212$ feet. What is the length? 165. The perimeter of a rectangular courtyard is $160160$ feet. The length is $1010$ feet more than the width. Find the length and the width. 166. The perimeter of a rectangular painting is $306306$ centimeters. The length is $1717$ centimeters more than the width. Find the length and the width. 167. The width of a rectangular window is $4040$ inches", "in the ratio 2:3 then the ratio of the area of R to the area of S A. 25:16 B. 24:25 C. 5:6 D. 4:5 E. 4:9 _________________ Sky is the limit Math Expert Joined: 02 Sep 2009 Posts: 59089 If the perimeter of square region S and the perimeter of rec [#permalink] ### Show Tags 03 Feb 2012, 08:29 19 16 If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3 then the ratio of the area of R to the area of S A. 25:16 B. 24:25 C. 5:6 D. 4:5 E. 4:9 Let the side of square be $$s$$ and the side of rectangle $$a$$ and $$b$$ Given: The sides of R are in the ratio 2:3 --> $$\\frac{a}{b}=\\frac{2}{3}$$ --> $$b=\\frac{3a}{2}$$. The perimeter of square region S and the perimeter of rectangular region R are equal --> $$P=4s=2(a+b)$$ --> $$2s=a+b=\\frac{5a}{2}$$ --> $$s=\\frac{5a}{4}$$. Question: $$\\frac{area \\ of \\ R}{area \\ of \\ S}=\\frac{ab}{s^2}=?$$ Substitute b = 3a/2 and s = 5a/4 into the question: $$\\frac{ab}{s^2}=\\frac{3a^2}{2}*\\frac{16}{25a^2}=\\frac{24}{25}$$. OR: you can pick numbers: let the sides of rectangle be 4 and 6 (ratio 2:3) then the perimeter of the rectangle will be 2(4+6)=20, thus the side of the square will be 20/4=5. Next,", "# The Area of a Rectangle A rectangle is a quadrilateral in which all four angles are right angles and the opposite sides are equal in length. Let $ABCD$ be a rectangle having $AB = a$ and $BC = b$. Let $AC$ be the diagonal which divides the rectangle into two right triangles, $\\Delta ABC$ and $\\Delta ADC$. $\\begin{gathered} {\\text{Area of }}\\Delta ABC = \\frac{1}{2}AB \\times BC = \\frac{1}{2}a \\times b \\\\ \\Rightarrow {\\text{Area of }}\\Delta ADC = \\frac{1}{2}DC \\times AD = \\frac{1}{2}a \\times b \\\\ \\end{gathered}$ $\\begin{gathered} {\\text{Area of rectangle }}ABCD = {\\text{Area of }}\\Delta ABC + {\\text{Area of }}\\Delta ADC \\\\ \\Rightarrow {\\text{Area of rectangle }}ABCD = \\frac{1}{2}a \\times b + \\frac{1}{2}a \\times b \\\\ \\therefore {\\text{Area of rectangle }}ABCD = a \\times b = {\\text{length}} \\times {\\text{breadth}} \\\\ \\end{gathered}$ Example: In exchange for a square plot of land, one of whose sides is $84$ m, a man wants to buy a rectangular plot $144$m long which is of the same area as the square plot. Find the width of the rectangular plot. Solution: One side of the square plot $= 84$ m. The area of the square plot $= 84 \\times 84 = 7056$ square m. The length of the rectangular plot $= 144$ m. The length of the rectangular plot $=$ area of the", "Question 9 # If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is Solution All the sides of the rhombus are equal. The area of a rhombus is $$12\\ cm^2$$ Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal. The area of a rhombus is $$\\left(\\frac{1}{2}\\right)\\left(d1\\right)\\cdot\\left(d2\\right)\\ =\\ 12$$ d1*d2 = 24. The length of the side of a rhombus is given by $$\\frac{\\sqrt{\\ d1^2+d2^2}}{2}$$. This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length. $$\\frac{\\sqrt{\\ d1^2+d2^2}}{2}=\\ 5$$ Hence $$\\sqrt{\\ d1^2+d2^2}\\ =\\ 10$$ $$d1^2+d2^2\\ =\\ 100$$ Using d1*d2 = 24, 2*d1*d2 = 48. $$d1^2+d2^2\\ +2\\cdot d1\\cdot d2=\\ 100+48\\ =\\ 148$$ $$d1^2+d2^2\\ -2\\cdot d1\\cdot d2=\\ 100-48\\ =\\ 52$$ $$d1+d2\\ =\\ \\sqrt{\\ 148}$$ (1) d1-d2 = $$\\sqrt{52}$$ (2) (1) + (2)= 2*(d1) = 2*($$\\sqrt{\\ 37}+\\sqrt{\\ 13}$$) d1 = $$\\sqrt{\\ 37}+\\sqrt{\\ 13}$$ or In a rhombus the area of a Rhombus is given by : The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b. The area of a Rhombus is : $$\\left(\\frac{1}{2}\\right)\\cdot\\left(2a\\right)\\cdot\\left(2b\\right)\\ =\\ 12$$ ab =6. The length of each side is : $$\\sqrt{\\ a^2+b^2}$$ = 5, $$a^2+b^{2\\ }=\\ 25,\\$$ $$\\left(a+b\\right)^2=\\", "meters. The area of the square is 15. Express as a trinomial: (3a 1 5)(2a 2 3) (A) (B) (C) (D) R 4 R2 (C) 16 (D) 2R2 (A) (B) (C) (D) 7 x 21 (C) x 1 (D) x (B) 17. The expression 1.15% 1.5% 8.7% 15% 3 4 13. If is subtracted from , the result is x x (A) 1 15 4a 2 15 a 2 15 a 1 15 (A) always greater than both of the original fractions. (B) always less than both of the original fractions. (C) sometimes greater and sometimes less than both of the original fractions. (D) never less than both of the original fractions. 2 12. Tom’s weekly rent increased from $125 to$143.75. Find the percent of increase. 1 2 1 2 16. If two fractions, each of which has a value between 0 and 1, are multiplied together, the product will be (A) 4R (B) 6a2 6a2 6a2 6a2 =162 is equivalent to (A) 4=2 (B) 4 1 =2 (C) 9=2 (D) 9 1 =2 18. If the length and the width of a rectangle are both tripled, the ratio of the area of the original rectangle to the area of the enlarged rectangle is (A) (B) (C) (D) 543 1:3 1:6 1:9 1:18 www.petersons.com ISEE PRACTICE TEST 2", "Intern Joined: 06 Jun 2013 Posts: 27 Concentration: Finance, Economics Schools: AGSM '16 GMAT 1: 690 Q38 V41 GPA: 3.92 Followers: 0 Kudos [?]: 0 [0], given: 13 Re: Fresh Meat!!! [#permalink] 08 Jul 2013, 10:28 Aho92 wrote: Bunuel wrote: SOLUTIONS: 1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? I. 63 II. 126 III. 252 A. I only B. II only C. III only D. I and III only E. I, II and III Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Hi Bunuel, This is probably a stupid question. But why can't x be \\sqrt{2}? Okay. I got it. Stupid me. They have to be integers Math Expert Joined: 02 Sep 2009", "order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Hi Bunuel, This is probably a stupid question. But why can't x be \\sqrt{2}? Intern Joined: 06 Jun 2013 Posts: 27 Concentration: Finance, Economics Schools: AGSM '16 Followers: 0 Kudos [?]: 0 [0], given: 11 Re: Fresh Meat!!! [#permalink] 08 Jul 2013, 10:28 Aho92 wrote: Bunuel wrote: SOLUTIONS: 1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R? I. 63 II. 126 III. 252 A. I only B. II only C. III only D. I and III only E. I, II and III Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R). $$area_{square}=\\frac{d^2}{2}$$ and $$area_{rhombus}=\\frac{d_1*d_2}{2}$$. The difference is $$area_{square}-area_{rhombus}=\\frac{(15x)^2}{2}-\\frac{11x*9x}{2}=63x^2$$. If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be $$\\sqrt{2}$$, which is not possible. Hi Bunuel, This is probably a stupid question. But why", "it is shorter than the unit length. The canonical shape with the area $1/2$ is a rectangle with side lengths $1/2$ by $1$. If we compare that to a square of length $1/2$, we need to make one of the sides of the rectangle shorter while the other stays the same. Therefore the number for the area of the square must be less than the number for the length of its side. On the other hand, if the length if the side is $2$, then the canonical shape with area $2$ is a rectangle with sides $1$ by $2$. Here we have to make one of the sides longer in order to get a square of side length $2$, so the number for the sqare's area is larger than the number for its side's length. -" ]

[ "Does Drawing Triangles Help? What is the smallest perfect square that can be expressed as the sum of two positive perfect squares in two different ways? Note: The order of summation doesn't matter. ×", "# Positive integer and factor • Nov 30th 2012, 03:42 PM DiamondVH123 Positive integer and factor Im stuck on this question: What is the smallest positive integer that is divisible by the factor 2,3 and 5 and which is also square and a cube. Prove that this is the smallest such integer. how would i start this? thanks • Nov 30th 2012, 04:24 PM abender Re: Positive integer and factor In the prime factorization of a perfect square, the exponent of each factor must divide 2. Similarly, for a perfect cube, the exponent of each factor must divide 3. So, your answer is going to be $(2\\cdot3\\cdot5)^6$. This is how I would start the process of answering the problem. The language of the formal proof is the next step. • Nov 30th 2012, 04:28 PM Plato Re: Positive integer and factor Quote: Originally Posted by DiamondVH123 What is the smallest positive integer that is divisible by the factor 2,3 and 5 and which is also square and a cube. Prove that this is the smallest such integer. Is $2^6$ the smallest multiple of two that is both a square and a a cube? • Dec 4th 2012, 12:41 PM DiamondVH123 Re: Positive integer and factor Thanks i know how to answer it now", "Show Tags 30 Apr 2016, 01:14 2 1 Bunuel wrote: What is the smallest positive perfect square that is divisible by 12, 15, and 18? A. 900 B. 1,600 C. 2,500 D. 3,600 E. 4,900 HI, we have to fulfill three conditions.. 1) Should be SMALLEST possible 2) Should be a PERFECT square 3) Should be div by 12,15,18.. so it has to be a multiple of LCM of 12,15 and 18.. LCM = >$$12 = 2^2*3$$; $$15 = 3*5$$and $$18 = 2*3^2.$$. $$LCM = 2*2*3*3*5 = 180$$.. now what should be multiplied to 180 to make it a perfect square=> 180 = 2*2*3*3*5.. ONLY 5 is without pair, so 180 *5 = 900 A _________________ Manager Joined: 13 Apr 2016 Posts: 57 Location: India GMAT 1: 640 Q50 V27 GPA: 3 WE: Operations (Hospitality and Tourism) Re: What is the smallest positive perfect square that is divisible by 12, [#permalink] ### Show Tags 30 Apr 2016, 02:26 [quote=\"Bunuel\"]What is the smallest positive perfect square that is divisible by 12, 15, and 18? A. 900 B. 1,600 C. 2,500 D. 3,600 E. 4,900 take the LCM of 12,15,18 that will come 180. the smallest positive perfect square given in the option that can be divided with 180 is 900. so the ans is (A) Non-Human User Joined: 09", "# Find the smallest integer that is divisible by exactly $X$ perfect squares. Is there a method to find the smallest integer divisible by exactly $X$ perfect squares? Example: find the smallest positive integer divisible by exactly 2015 perfect squares. • see if you can do it for 1 or 2 or 3 squares first – Will Jagy May 9 '15 at 2:25 • Also do you know how to count how many factors a number has? – jgon May 9 '15 at 2:26 • Find the smallest number $N$ which is divisible by exactly $X$ natural numbers, and then $N^2$ is the smallest divisible by exactly $X$ perfect squares. – Thomas Andrews May 9 '15 at 2:36 • 2015 is the current year, which suggests that the problem might be from an ongoing competition.Where did you find it? – Henning Makholm May 9 '15 at 2:49 First of all, there won't be a simple formula. If $X$ is prime, the smallest number is $2^{2X-2}$. So the smallest number divisible by example five perfect squares is $2^{8}=256$. More generally, it will depend on factorizations of $X$. In general, if $N$ is the smallest integer with exactly $X$ positive integer divisors, then $N^2$ will be the smallest integer with exactly $X$ perfect square divisors. The number of divisors of", "n is divisible by 3)?", "# Square Triangles What is the smallest positive integer greater than $1$ that is both a triangular number and a perfect square? Note: A triangular number is a number which can be expressed as a sum of the first $n$ positive integers, such as $6 = 1+2+3$ or $15 = 1+2+3+4+5$. ×", "# Make sure it maintains the divisibility What is the smallest positive integer $$x$$ such that $$x,x-1$$ and $$x-2$$ are each divisible by distinct perfect squares?", "Divisibility | \"", "You have an $n$ x $n$ square which you know can be tiled completely, without overlaps, by 2×2 and 3×3 squares, using at least one of each type. What is the smallest possible value of $n$? What must $n$ be divisible by? Inspired by BMO2 2021 P2 (BMO2 = British Math Olympiad Round 2) Happy problem solving! Email your solutions to [email protected]", "are considering only positive even natural numbers that begin with 2. In other words, the smallest positive number that can be divided by 2 is 2. No natural number smaller than 2 can be divided by 2, so 2 is the smallest positive natural number that can be divided by 2. Now, 1 divides 2, so we divide 1 by 1 to get 1 / 1, so we are done. In the second case, we are considering only positive odd natural numbers that begin with 3. In other words, the smallest positive number that can be divided by 3 is 3. No natural number smaller than 3 can be divided by 3, so 3 is the smallest positive number that can be divided by 3. Now, 2 divides 3, so we divide 2 by 2 to get 1 / 2, so we are done. The Collatz Conjecture states that all positive natural numbers are eventually equal to 1 after a finite number of iterations. The Collatz Conjecture divides positive numbers into two categories, the odd numbers and the even numbers. A number is odd if the number divided by 2 is 1, in other words, if the number is divisible by 2. If the number is even, the number divided by 2 is not 1, in other words,", "one of which is the largest perfect square possible", "of div1C (Perfect Triples)?", "are square and non-perfect", "solution in positive integers and if and only if the following three conditions are satisfied: 1. There exists a divisor of with such that 2. for some positive integer , and such that 3. is a perfect square. An interesting corollary of Broughan’s theorem is that no prime number (except for 2) is expressible as a sum of two positive cubes.", "divisible by 216, what is the smallest possible value for [#permalink] 29 Mar 2017, 21:30 Display posts from previous: Sort by", "exactly one of them will be divisible by 3.", "that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? 2520是最小的能够被1到10整除的数。", "5040 is divisible by 8. 4 is the smallest composite number. Share on:", "# The Three Conditions Determine the smallest positive integer $$n$$ such that: (1) $$n$$ is divisible by 20, (2) $$n^{2}$$ is a perfect cube, and (3) $$n^{3}$$ is a perfect square. ×", "be divisible by 3. -O" ]

[ "we have equation $$6^m + 4$$ is a perfect square. But $$6^m + 4 \\equiv$$ $$3\\ or \\ 5\\ (mod\\ 7)$$ which is not quadratic residue. Thats all cases. • Numbers are positive integers or non-negative? – Arial Pilisov Apr 19 '19 at 18:43 • If you need non-negative integers than: 1) $m = 0\\\\$. $2^n+3 \\equiv 3 \\ (mod\\ 4)$ which in not quadratic residue. 2)$n = 0\\\\$ $6^m+3$ is square. If $m>1$ than this equation divide $3$, but not divide $9$, so it cant be square. Than just consider $m=0\\ or\\ 1$ – Arial Pilisov Apr 19 '19 at 18:48 • I needed the postive integers only. Thank you so much. – Tapi Apr 20 '19 at 15:49 • In Case 1, shouldn't we show that when a number is divisible by a prime but not by the prime square, then it is not a perfect square? However, you've shown that it is divisible by \"8\" but not by 16. Shouldn't it be 4 instead of 8 as the square of 4 is 16? Also, 4 is not a prime number. So is the case still justified considering I cannot find any n>3? – Tapi Apr 22 '19 at 19:26 • I showed that the number is divisible $2^3$, but not divisible by $2^4$, so it", "positive integer is a perfect square exactly when the exponent on each prime in its prime factorization is even. The number $$392 =8 \\times 49=2^3\\times 7^2$$. This is not a perfect square since the power $$2^3$$ has an odd exponent. We require another factor of $$2$$ to obtain a multiple of $$392$$ that is a perfect square, namely $$2\\times 392=784$$. The number $$784=2^4\\times 7^2=(2^2\\times 7)^2=28^2$$, and is the first perfect square less than $$10\\,000$$ that is divisible by $$392$$. To find all the perfect squares less than $$10\\,000$$ that are multiples of $$392$$, we will multiply $$784$$ by squares of positive integers, until we reach a product larger than $$10\\,000$$. If we multiply $$784$$ by $$2^2$$, we obtain $$3136$$ which is $$56^2$$, a second perfect square less than $$10\\,000$$. If we multiply $$784$$ by $$3^2$$, we obtain $$7056$$ which is $$84^2$$, a third perfect square less than $$10\\,000$$. If we multiply $$784$$ by $$4^2$$, we obtain $$12\\,544$$ which is a greater than $$10\\,000$$. No other perfect squares divisible by $$392$$ exist that are less than $$10\\,000$$. Therefore, there are $$3$$ perfect squares less than $$10\\,000$$ that are divisible by $$392$$.", "## Thinking Mathematically (6th Edition) between $-7$ and $-6$ $47$ is betwen the perfect squares $36$ and $49$. This means that : $\\sqrt{36}\\lt \\sqrt{47} \\lt \\sqrt{49}\\\\ 6 \\lt \\sqrt{47} \\lt 7$ Note that $1\\lt 2\\lt3$, but $-3 \\lt -2 \\lt -1$ Since $6 \\lt \\sqrt{47} \\lt 7$, then $-7 \\lt -\\sqrt{47} \\lt -6$. Therefore, $-\\sqrt{47}$ is between the integers $-7$ and $-6$.", "# Find the smallest integer that is divisible by exactly $X$ perfect squares. Is there a method to find the smallest integer divisible by exactly $X$ perfect squares? Example: find the smallest positive integer divisible by exactly 2015 perfect squares. • see if you can do it for 1 or 2 or 3 squares first – Will Jagy May 9 '15 at 2:25 • Also do you know how to count how many factors a number has? – jgon May 9 '15 at 2:26 • Find the smallest number $N$ which is divisible by exactly $X$ natural numbers, and then $N^2$ is the smallest divisible by exactly $X$ perfect squares. – Thomas Andrews May 9 '15 at 2:36 • 2015 is the current year, which suggests that the problem might be from an ongoing competition.Where did you find it? – Henning Makholm May 9 '15 at 2:49 First of all, there won't be a simple formula. If $X$ is prime, the smallest number is $2^{2X-2}$. So the smallest number divisible by example five perfect squares is $2^{8}=256$. More generally, it will depend on factorizations of $X$. In general, if $N$ is the smallest integer with exactly $X$ positive integer divisors, then $N^2$ will be the smallest integer with exactly $X$ perfect square divisors. The number of divisors of", "each themselves square: $19^2 = 361$; $36 = 6^2$, $1 = 1^2$ The $16$th integer $n$ after $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $13$, $14$, $15$, $16$, $18$ such that $2^n$ contains no zero in its decimal representation: $2^{19} = 524 \\, 288$ Every positive integer can be expressed as the sum of at most $19$ $4$th powers. The $10$th positive integer which is not the sum of $1$ or more distinct squares: $2$, $3$, $6$, $7$, $8$, $11$, $12$, $15$, $18$, $19$, $\\ldots$ The $12$th positive integer after $2$, $3$, $4$, $7$, $8$, $9$, $10$, $11$, $14$, $15$, $16$ which cannot be expressed as the sum of distinct pentagonal numbers. The $3$rd of $11$ primes of the form $2 x^2 + 11$: $2 \\times 2^2 + 11 = 19$ (Previous ... Next)", "+0 # Positive integer divisors 0 45 2 How many positive integer divisors of 23,328 are perfect cubes? Dec 24, 2018 #1 +95171 +2 How many positive integer divisors of 23,328 are perfect cubes? 23328^(1/3) = 28.573218935427587 Well less than 29 anyway. 1 is ok factor(23328) = 2^5*3^6 $$2*2*2*2*2\\quad * \\quad 3*3*3*3*3*3\\\\~\\\\ 1^3,2^3,3^3,6^3,9^3,18^3 \\;\\;\\text{that seems to be it.}$$ I just got those from visual examination of the factors. So that is 6 of them. Dec 24, 2018 #2 +1 23,328 = 2^5 * 3^6 Will divide the exponents by 3 and add 1 and then multiply them: 5 /3 + 1 =2 6/3 + 1 =3 So, 3 x 2 = 6 perfect cubes. Dec 24, 2018", "# The least perfect square , which is divisible by each of $21,36$ and $66$ is $\\begin{array}{1 1} 213444 \\\\ 214344 \\\\ 214434 \\\\ 231443 \\end{array}$", "Feb 25 '17 at 17:34 $a$ is also not the sum of three perfect squares once it ends in at least three digits (all $1$). As $1000$ is divisible by $8,$ we then have $$a = bcdefgh111 = 1000 w + 111 \\equiv 7 \\pmod 8$$ • $3332^2 + 0^2 + 0^2 = 11102224$. – Starfall Feb 16 '17 at 6:35 • @Starfall see edit – Will Jagy Feb 16 '17 at 17:54", "# Primes and perfect square on numbers I have two equations, which may play an important role in my further studies on theory of numbers. 1) How many pairs of (A, B, x) we can make in $A^x + B ^x = prime$? Here $x$ is $> 2$ and A, B are positive integers. 2) can we find a number(s) with one hundred 0′s, one hundred 1′s and one hundred 2′s be a perfect square. If yes, what is that number or otherwise how to disprove it about such number does not existence? - clearly we must have $x=2^n$ or else the sum will be divisible by $A+B$ – Konstantinos Gaitanas Dec 8 '13 at 13:12 clearly not. $2^2+3^2=14$ which is not divisible by 5 – Thomas Dec 8 '13 at 13:17 @KonstantinosGaitanas and @Thomas! $1^4 + 2^4 = 17$. Here $x = 4 >2$ and A = 1, B = 2 are positive integers. The result 17 is prime. Like this how many A, B and x are existing? can you list? – narosanair Dec 8 '13 at 13:20 @Thomas clearly yes.$x>2$ – Konstantinos Gaitanas Dec 8 '13 at 13:21 Thank you for understanding me. can you list such pairs please... also look my second question of this post. – narosanair Dec 8 '13 at 13:25 The", "# Primes and perfect square on numbers I have two equations, which may play an important role in my further studies on theory of numbers. 1) How many pairs of (A, B, x) we can make in $A^x + B ^x = prime$? Here $x$ is $> 2$ and A, B are positive integers. 2) can we find a number(s) with one hundred 0′s, one hundred 1′s and one hundred 2′s be a perfect square. If yes, what is that number or otherwise how to disprove it about such number does not existence? Please... - clearly we must have $x=2^n$ or else the sum will be divisible by $A+B$ – Konstantinos Gaitanas Dec 8 '13 at 13:12 clearly not. $2^2+3^2=14$ which is not divisible by 5 – Thomas Dec 8 '13 at 13:17 @KonstantinosGaitanas and @Thomas! $1^4 + 2^4 = 17$. Here $x = 4 >2$ and A = 1, B = 2 are positive integers. The result 17 is prime. Like this how many A, B and x are existing? can you list? – narosanair Dec 8 '13 at 13:20 @Thomas clearly yes.$x>2$ – Konstantinos Gaitanas Dec 8 '13 at 13:21 Thank you for understanding me. can you list such pairs please... also look my second question of this post. – narosanair Dec 8 '13 at 13:25", "# Complete the numerical sequence! Complete the following sequence with just one number: $$4;\\ 9;\\ 1;\\ 6;\\ 2;\\ ??$$ Hints • It's a real number • The sequence ends with that number • There could be literally infinite similar sequences with the same ending number Could be: Zero Get it? Alphabetically: Four, Nine, One, Six, Two, Zero Even the complete alphabetical list of all integers ends with Zero • Great! Exactly!! – Henry Jan 31 '16 at 10:07 5 Because: 49, 16 and 25 are all perfect squares. And that's where it ends. And there are literally an infinite number of sequences of squares ending in 5. • or maybe the reason is 4 is a perfect square, so is 9, so is 16 , so is 25. All I mean to say is it might not be 49 according to OP – manshu Jan 31 '16 at 7:07 • Not a bad answer but nope! What if the sequence were been \"41, 9, 1, 16, 12, ?? Your reasoning would fail ^^ – Henry Jan 31 '16 at 10:09 7 I guess because, 4 + 5 = 9 1 + 5 = 6 2 + 5 = 7 i.e. a consistent difference of 5 between every pair of terms • Interesting line of thought. To the downvoters,", "$9$) can be the average of a square and a cube? The cube must be $1$ or $8$; $27$ is too big. For $8$, the only case that works is $\\frac{2^3+2^2}{2}=6$. For $1$, the answer could be $1$ or $5$. So $t$ is $1, 5,$ or $6$. Now list all the numbers up to $81$ (which is as big as the product of two digits can get) that are of the form $n^2+17: \\{18,21,26,33,42,53,66,81\\}$. We immediately see that $t$ can't be one, because $h$ would have to be at least $18$, and $t$ can't be $5$ because for this range, $n^2+17$ is never a multiple of $5$. Then $t=6$. But only three of the numbers in our list are divisible by $6$. The first to try is $18$. That would give a number of the form $36\\cdot$ and you need to choose the last digit to make the sum $3+6+\\cdot$ a perfect square. You can take it from there.", "Perfect Squares and Perfect Cubes List all integers less than $20,\\!000$ that are both perfect squares and perfect cubes. Source: NCTM Mathematics Teacher, January 2006 Solution $9=3^2$ is a perfect square but not a perfect cube. On the other hand, $8=2^3$ is a perfect cube but not a perfect square. For an integer to be both perfect square and perfect cube, the exponent must be a multiple of $2$ and $3$, that is, the integer must be a sixth power. $0^6=0$ $1^6=1$ $2^6=64$ $3^6=729$ $4^6=4096$ $5^6=15625$ $6^6=46656>20000$ All integers less than $20,\\!000$ that are both perfect squares and perfect cubes $0,1,64,729,4096,15625$ Answer: $0,1,64,729,4096,15625$ Advertisements About mvtrinh Retired high school math teacher. This entry was posted in Problem solving and tagged , , , , , , , . Bookmark the permalink.", "6. $16^2 = 256$ is not even 4 digits, while $66^2 = 4356$ satisfies all the given conditions. Hence, the answer is 4356. (In fact, it is the only 4-digit even perfect square to satisfy the given conditions.)", "4 C. 7 D. 11 E. 13 First, lay out the possible numbers you can use in the sum. The positive square integers greater than 1 but less than 75 are 4, 9, 16, 25, 36, 49, and 64. Now, let's create a possible set and then see whether we can adjust it. It's fairly obvious that three 25's add up to 75, so our first set is {25, 25, 25}. We might now recall the most famous example of the Pythagorean Theorem: $$9 + 16 = 25$$. So we can swap out, successively, a 25 and replace it with a 9 and a 16. With sets, order does not matter, so we get three more possible sets: {25, 25, 9, 16} {25, 9, 16, 9, 16} {9, 16, 9, 16, 9, 16} This gives us 4 sets so far. However, we can now swap out 16's for four 4's. We can do so as follows. One possible swap for the set with one 16: {25, 25, 9, 4, 4, 4, 4} Two possible swaps for the set with two 16's: {25, 9, 16, 9, 4, 4, 4, 4} {25, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4} And three possible swaps for the set with three 16's: {9, 16, 9, 16, 9, 4, 4, 4,", "# A \"Perfect\" Number (496 follower problem) The number 496 is the third smallest perfect number: the sum of its proper divisors is the number itself: $1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496.$ But it is also a \"perfect\" number in the sense that its third digit is the geometric mean of the first two digits: $\\sqrt{4\\cdot 9} = 6$. How many three-digit integers $\\overline{abc}$ are there with the property, that $c$ is the geometric mean of $a$ and $b$? Note: Numbers starting in zero do not count. Thus, a \"three-digit integer\" lies between $100$ and $999$. And here is a more challenging variation on the theme. ×", "# Of any 52 integers, two can be found whose difference of squares is divisible by 100 Prove that of any 52 integers, two can always be found such that the difference of their squares is divisible by 100. I was thinking about using recurrence, but it seems like pigeonhole may also work. I don’t know where to start. Look at your $52$ integers mod $100$. Look at the pairs of additive inverses $(0,0)$, $(1,99)$, $(2,98)$, etc. There are $51$ such pairs. Since we have $52$ integers, two of them must belong to a pair $(x,-x)$. Then $x^2 – (-x)^2 = 0 \\pmod{100}$, so that the difference of their squares is divisible by $100$.", "square will be as well. Otherwise, the number is of the form $$3k\\pm1$$ for some integer $$k$$, and $$(3k\\pm1)^2=9k^2\\pm6k+1\\equiv1\\pmod3$$ Therefore, the square of a number is either divisible by 3 or it leaves a remainder of 1. Now, let us consider that there are positive integers $$a,b,c$$ such that $$a^2+b^2=c^2$$. From our earlier notion, it cannot be that neither $$a$$ nor $$b$$ are multiples of 3, because then their sum would be of the form $$3k+2$$ and could not be a perfect square. Therefore, either $$a$$ or $$b$$ (or both) are divisible by 3. • Thanks for the answer, but i have two questions. Why the number of the form 3k+2 can't be a perfect square? And if only one of these numbers is divisible by 3 then the sum of their squares will be of the form 3k+1? Feb 1 '20 at 23:31 • Any number is $3k$ or $3k\\pm1$, and none of those square to $3n+2$ Feb 2 '20 at 0:23", "A. We are told that x and y are both positive integers. x^2=350y As x is an integer => x^2 must be a perfect square. x^2=350y=2*5^2*7*y Hence the general expression for y would be -> 2*7*p^z where p is an integer and z is even. We are asked if y is divisible by 28 or not. Clearly y is already divisible by 7 and 2 Hence for y to be divisible by 28=> p must be even too. Statement 1-> x is divisible by 4 Hence x^2 must have 2^4. As x=2*5^2*7*y => For x^2 to have a 2^4 => y must have a 2^3. Hence y must have a 2^3 and it already has a 7. Thus y must be divisible by 28 Hence sufficient. Statement2-> Here lets use test cases y=2*7 =>y will not be divisible by 28 y=2*7*2^2 => y will be divisible by 28. Hence not sufficient. Hence A. 46)Data Sufficiency-> How many Prime factors does positive integer $$n$$ have ? A)$$n^3$$ has 7 prime factors. B)$$√n$$ has 7 prime factors. D. $$X$$ and $$X^n$$ always have exact same prime factors. In other words => \"Exponentiation does not change the prime factors of any positive integer X.\" 47)If x and y represent the number of factors of 90 and 147 respectively.What is the value", "order the sequence from lowest to highest such that $m$ is the largest integer less than $(k + 1)^2/k$. A general formula would have to derive from this if there does not exist one, though I don't understand how $27$ would come after $25$ if $25 = 5^2$ and $27\\neq 5n$ for natural $n$. – Mr Pie Jan 15 '18 at 5:23 • Sir In this question I said all perfect squares and its multiples.So in this manner 27 is a multiple of 9 which is a perfect square.In the same manner I need general term for all terms which are either perfect squares or its multiples arranged in an increasing order sequence – user9137663 Jan 15 '18 at 5:43" ]

[ "# Find the smallest integer that is divisible by exactly $X$ perfect squares. Is there a method to find the smallest integer divisible by exactly $X$ perfect squares? Example: find the smallest positive integer divisible by exactly 2015 perfect squares. • see if you can do it for 1 or 2 or 3 squares first – Will Jagy May 9 '15 at 2:25 • Also do you know how to count how many factors a number has? – jgon May 9 '15 at 2:26 • Find the smallest number $N$ which is divisible by exactly $X$ natural numbers, and then $N^2$ is the smallest divisible by exactly $X$ perfect squares. – Thomas Andrews May 9 '15 at 2:36 • 2015 is the current year, which suggests that the problem might be from an ongoing competition.Where did you find it? – Henning Makholm May 9 '15 at 2:49 First of all, there won't be a simple formula. If $X$ is prime, the smallest number is $2^{2X-2}$. So the smallest number divisible by example five perfect squares is $2^{8}=256$. More generally, it will depend on factorizations of $X$. In general, if $N$ is the smallest integer with exactly $X$ positive integer divisors, then $N^2$ will be the smallest integer with exactly $X$ perfect square divisors. The number of divisors of", "Show Tags 30 Apr 2016, 01:14 2 1 Bunuel wrote: What is the smallest positive perfect square that is divisible by 12, 15, and 18? A. 900 B. 1,600 C. 2,500 D. 3,600 E. 4,900 HI, we have to fulfill three conditions.. 1) Should be SMALLEST possible 2) Should be a PERFECT square 3) Should be div by 12,15,18.. so it has to be a multiple of LCM of 12,15 and 18.. LCM = >$$12 = 2^2*3$$; $$15 = 3*5$$and $$18 = 2*3^2.$$. $$LCM = 2*2*3*3*5 = 180$$.. now what should be multiplied to 180 to make it a perfect square=> 180 = 2*2*3*3*5.. ONLY 5 is without pair, so 180 *5 = 900 A _________________ Manager Joined: 13 Apr 2016 Posts: 57 Location: India GMAT 1: 640 Q50 V27 GPA: 3 WE: Operations (Hospitality and Tourism) Re: What is the smallest positive perfect square that is divisible by 12, [#permalink] ### Show Tags 30 Apr 2016, 02:26 [quote=\"Bunuel\"]What is the smallest positive perfect square that is divisible by 12, 15, and 18? A. 900 B. 1,600 C. 2,500 D. 3,600 E. 4,900 take the LCM of 12,15,18 that will come 180. the smallest positive perfect square given in the option that can be divided with 180 is 900. so the ans is (A) Non-Human User Joined: 09", "positive integer is a perfect square exactly when the exponent on each prime in its prime factorization is even. The number $$392 =8 \\times 49=2^3\\times 7^2$$. This is not a perfect square since the power $$2^3$$ has an odd exponent. We require another factor of $$2$$ to obtain a multiple of $$392$$ that is a perfect square, namely $$2\\times 392=784$$. The number $$784=2^4\\times 7^2=(2^2\\times 7)^2=28^2$$, and is the first perfect square less than $$10\\,000$$ that is divisible by $$392$$. To find all the perfect squares less than $$10\\,000$$ that are multiples of $$392$$, we will multiply $$784$$ by squares of positive integers, until we reach a product larger than $$10\\,000$$. If we multiply $$784$$ by $$2^2$$, we obtain $$3136$$ which is $$56^2$$, a second perfect square less than $$10\\,000$$. If we multiply $$784$$ by $$3^2$$, we obtain $$7056$$ which is $$84^2$$, a third perfect square less than $$10\\,000$$. If we multiply $$784$$ by $$4^2$$, we obtain $$12\\,544$$ which is a greater than $$10\\,000$$. No other perfect squares divisible by $$392$$ exist that are less than $$10\\,000$$. Therefore, there are $$3$$ perfect squares less than $$10\\,000$$ that are divisible by $$392$$.", "x^2 is divisible by 216, what is the smallest possible value for [#permalink] ### Show Tags 23 Jun 2009, 19:58 2 KUDOS 216 = 6X6X6 Now we are asked what is the smallest x such that $$x^2$$ is divisible by 216. To make 6X6X6 a perfect square, we need one more 6. Hence 6X6X6X6 = $$36^2$$ is the smallest $$x^2$$ divisible by 216. What is the OA? Manager Joined: 28 May 2009 Posts: 154 Location: United States Concentration: Strategy, General Management GMAT Date: 03-22-2013 GPA: 3.57 WE: Information Technology (Consulting) Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink] ### Show Tags 24 Jun 2009, 08:45 sdrandom1 wrote: 216 = 6X6X6 Now we are asked what is the smallest x such that $$x^2$$ is divisible by 216. To make 6X6X6 a perfect square, we need one more 6. Hence 6X6X6X6 = $$36^2$$ is the smallest $$x^2$$ divisible by 216. What is the OA? This is correct the OA is 36. I like your reasoning better than the Guides Guides explanation Quote: The prime box of $$x^2$$contains the prime factors of 216, which are 2, 2, 2, 3, 3, and 3. We know that the prime factors of $$x^2$$ should be the prime factors of x appearing in sets of two, or pairs.", "we have equation $$6^m + 4$$ is a perfect square. But $$6^m + 4 \\equiv$$ $$3\\ or \\ 5\\ (mod\\ 7)$$ which is not quadratic residue. Thats all cases. • Numbers are positive integers or non-negative? – Arial Pilisov Apr 19 '19 at 18:43 • If you need non-negative integers than: 1) $m = 0\\\\$. $2^n+3 \\equiv 3 \\ (mod\\ 4)$ which in not quadratic residue. 2)$n = 0\\\\$ $6^m+3$ is square. If $m>1$ than this equation divide $3$, but not divide $9$, so it cant be square. Than just consider $m=0\\ or\\ 1$ – Arial Pilisov Apr 19 '19 at 18:48 • I needed the postive integers only. Thank you so much. – Tapi Apr 20 '19 at 15:49 • In Case 1, shouldn't we show that when a number is divisible by a prime but not by the prime square, then it is not a perfect square? However, you've shown that it is divisible by \"8\" but not by 16. Shouldn't it be 4 instead of 8 as the square of 4 is 16? Also, 4 is not a prime number. So is the case still justified considering I cannot find any n>3? – Tapi Apr 22 '19 at 19:26 • I showed that the number is divisible $2^3$, but not divisible by $2^4$, so it", "# Positive integer and factor • Nov 30th 2012, 03:42 PM DiamondVH123 Positive integer and factor Im stuck on this question: What is the smallest positive integer that is divisible by the factor 2,3 and 5 and which is also square and a cube. Prove that this is the smallest such integer. how would i start this? thanks • Nov 30th 2012, 04:24 PM abender Re: Positive integer and factor In the prime factorization of a perfect square, the exponent of each factor must divide 2. Similarly, for a perfect cube, the exponent of each factor must divide 3. So, your answer is going to be $(2\\cdot3\\cdot5)^6$. This is how I would start the process of answering the problem. The language of the formal proof is the next step. • Nov 30th 2012, 04:28 PM Plato Re: Positive integer and factor Quote: Originally Posted by DiamondVH123 What is the smallest positive integer that is divisible by the factor 2,3 and 5 and which is also square and a cube. Prove that this is the smallest such integer. Is $2^6$ the smallest multiple of two that is both a square and a a cube? • Dec 4th 2012, 12:41 PM DiamondVH123 Re: Positive integer and factor Thanks i know how to answer it now", "# The least perfect square , which is divisible by each of $21,36$ and $66$ is $\\begin{array}{1 1} 213444 \\\\ 214344 \\\\ 214434 \\\\ 231443 \\end{array}$", "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. ×", "+0 # Positive integer divisors 0 45 2 How many positive integer divisors of 23,328 are perfect cubes? Dec 24, 2018 #1 +95171 +2 How many positive integer divisors of 23,328 are perfect cubes? 23328^(1/3) = 28.573218935427587 Well less than 29 anyway. 1 is ok factor(23328) = 2^5*3^6 $$2*2*2*2*2\\quad * \\quad 3*3*3*3*3*3\\\\~\\\\ 1^3,2^3,3^3,6^3,9^3,18^3 \\;\\;\\text{that seems to be it.}$$ I just got those from visual examination of the factors. So that is 6 of them. Dec 24, 2018 #2 +1 23,328 = 2^5 * 3^6 Will divide the exponents by 3 and add 1 and then multiply them: 5 /3 + 1 =2 6/3 + 1 =3 So, 3 x 2 = 6 perfect cubes. Dec 24, 2018", "and 10 What is unknown? The smallest square that is divisible by 4, 9 and 10. Reasoning: The number that will be perfectly divisible by each one of the numbers is their $$LCM$$. Steps: \\begin{align}&2\\left| \\!{\\underline {\\,{4,9,10} \\,}} \\right. \\\\&2\\left| \\!{\\underline {\\,{2,9,5} \\,}} \\right. \\\\&3\\left| \\!{\\underline {\\,{1,9,5} \\,}} \\right. \\\\&3\\left| \\!{\\underline {\\,{1,3,5} \\,}} \\right. \\\\&5\\left| \\!{\\underline {\\,{1,1,5} \\,}} \\right. \\\\&1\\left| \\!{\\underline {\\,{1,1,1} \\,}} \\right. \\end{align} $$LCM$$ of $$4,9,10$$ \\begin{align}& =\\text{ }2\\times ~2\\times 3\\times 3~\\times 5 \\\\& =\\text{ }180 \\\\ \\end{align} Hence, number $$5$$ does not have a pair. Therefore $$180$$ is multiplied by $$5$$ then the number obtained is a perfect square. Therefore, $$180~\\times 5\\text{ }=\\text{ }900$$ So, $$900$$ is the smallest square that is divisible by $$4$$, $$9$$ and $$10$$. ## Chapter 6 Ex.6.3 Question 10 Find the smallest square number that is divisible by each of the numbers $$8$$, $$15$$ and $$20$$ ### Solution What is known? Divisor are $$8, 15$$ and $$20$$ What is unknown? The smallest square that is divisible by $$8, 15$$ and $$20$$ Reasoning: The number that will be perfectly divisible by each one of the numbers is their $$LCM$$. Steps: $${\\rm{LCM \\,of }}\\,8,15,20$$ \\begin{align}&2\\left|\\!{\\underline {\\,{8,15,20} \\,}} \\right. \\\\&2\\left| \\!{\\underline {\\,{4,15,10} \\,}} \\right. \\\\&2\\left| \\!{\\underline {\\,{2,15,5} \\,}} \\right. \\\\&3\\left| \\!{\\underline {\\,{1,15,5} \\,}} \\right. \\\\&5\\left| \\!{\\underline {\\,{1,5,5} \\,}} \\right. \\\\&1\\left| \\!{\\underline {\\,{1,1,1} \\,}}", "# Largest Pandigital Square including Zero ## Theorem The largest pandigital square (in the sense where pandigital includes the zero) is $9 \\, 814 \\, 072 \\, 356$: $9 \\, 814 \\, 072 \\, 356 = 99 \\, 066^2$ ## Proof We check all the square of numbers from $99 \\, 067$ up to $\\floor {9876543210} = 99 \\, 380$, with the following constraints: Since all these squares has $9$ as its leftmost digit, the number cannot end with $3$ or $7$. The number cannot end with $0$ since its square will end in $00$. A pandigital number is divisible by $9$, so our number must be divisible by $3$. These constraints leaves us with the around $60$ candidates: $99069, 99072, 99075, 99078, 99081, 99084, 99096, 99099, 99102, 99105, 99108, 99111, 99114, 99126, 99129, 99132, \\dots, 99378$", "+0 # help pls 0 166 1 $$What is the smallest positive integer n such that 3n is a perfect square and 2n is a perfect cube?$$ Aug 4, 2021 #1 +26226 +3 What is the smallest positive integer $$n$$ such that $$3n$$ is a perfect square and $$2n$$ is a perfect cube? My attempt: $$\\begin{array}{|rcll|} \\hline 3n &=& 3*&3*2*3*2*3 =2^2(3^2)^2=324=18^2 \\\\ 2n &=& 2*&3*2*3*2*3 =2^33^3=216=6^3 \\\\ \\hline n &=& & 3*2*3*2*3 = 108 \\\\ \\hline \\end{array}$$ Aug 5, 2021", "perfect square dissection of an integer square The $4$th positive integer $n$ after $4$, $7$, $15$ such that $n - 2^k$ is prime for all $k$ The $14$th positive integer after $2$, $3$, $4$, $7$, $8$, $9$, $10$, $11$, $14$, $15$, $16$, $19$, $20$ which cannot be expressed as the sum of distinct pentagonal numbers The second of the $1$st pair of triangular numbers whose sum and difference are also both triangular: $15 = T_5$, $21 = T_6$, $15 + 21 = T_8$, $21 - 15 = T_3$ The $14$th harshad number after $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $12$, $18$, $20$: $21 = 7 \\times 3 = 7 \\times \\left({2 + 1}\\right)$ The smallest integer with $2$ distinct prime factors neither of which is a divisor of $10$ ## Historical Note There were $21$ shillings in $1$ guinea Sterling in pre-decimal British coinage.", "4 This post was BOOKMARKED Retired Moderator Joined: 04 Aug 2016 Posts: 575 Location: India GPA: 4 WE: Engineering (Telecommunications) Re: If y is the smallest positive integer such that 3,150 [#permalink] ### Show Tags 16 Oct 2016, 23:54 3150 = 5*5*3*3*2*7 To be a perfect square y needs to be 2*7=14 SVP Joined: 12 Sep 2015 Posts: 2446 Re: If y is the smallest positive integer such that 3,150 [#permalink] ### Show Tags 24 Aug 2017, 12:03 Expert's post Top Contributor mrwaxy wrote: If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be A. 2 B. 5 C. 6 D. 7 E. 14 Key concept: The prime factorization of a perfect square (the square of an integer) will have an EVEN number of each prime. For example, 36 = (2)(2)(3)(3) And 400 = (2)(2)(2)(2)(5)(5) Likewise, 3150y must have an EVEN number of each prime in its prime factorization. So, 3150y = (2)(3)(3)(5)(5)(7)y We have an EVEN number of 3's and 7's, but we have a single 2 and a single 7. If y = (2)(7), then we get a perfect square. That is: 3150y = (2)(2)(3)(3)(5)(5)(7)(7) So, if y = 14, then 3150y is a perfect square. [Reveal] Spoiler: E RELATED VIDEO FROM OUR", "# What is the smallest positive integer that is divisible by $2$ and $3$ What is the smallest positive integer that is divisible by $2$ and $3$ that consists entirely of $2$s and $3$s, and has at least one of each? I was wondering if there a formula or steps to approach this problem? I did it directly starting with $23, 32, 232,233,223$,... until I got what I assume is the smallest integer. $3,222$ is divisible by both $2$ and $3$. • A TeX nicety: putting the comma inside curly brackets will turn $3,222$ into $3{,}222$. Can you see the difference? Mar 8 '18 at 20:49 • yeah thanks for the info. Mar 8 '18 at 20:55 ## 4 Answers Isn't $2232$ smaller and divisible by both? Method. To be even it has to end in $2$. Then it needs three of them for the digit sum to be divisible by $3$. Then put the $3$ as far to the right as possible. • Yeah I knew it must end with a 2, i just added the other numbers for example purposes, but I see that I missed that one. Mar 8 '18 at 19:56 If you are doing it as a list, remove the odd numbers as they are not divisible by $2$. Now to show divisibility", "largest integer divisible by all positive integers less than its square root. - 5 years, 3 months ago 25 is the smallest square that is also a sum of two (non-zero) distinct squares (9 and 16). - 5 years, 3 months ago 26 is the only integer that is one greater than a square (25 = 5^2) and one less than a cube (27 = 3^3) - 5 years, 3 months ago 27 is the only positive integer that is 3 times the sum of its digits. In base 10, 27 is the first composite number not evenly divisible by any of its digits. 27 contains the decimal digits 2 and 7, and is the result of adding together the integers from 2 to 7 (2+3+4+5+6+7=27). In the Collatz conjecture, a starting value of 27 requires 112 steps to reach 1, many more than any lower number. - 5 years, 3 months ago 28 is the sum of the totient function for the first nine integers - 5 years, 3 months ago 29 is the sum of three consecutive squares! - 5 years, 3 months ago 30 is the smallest sphenic number. - 5 years, 3 months ago 31 is a centered triangular number, a centered pentagonal number and centered decagonal number. - 5 years, 3 months ago", "TL;DR On with TASK #2 from The Weekly Challenge #150. Enjoy! The challenge Write a script to generate all square-free integers <= 500. In mathematics, a square-free integer (or squarefree integer) is an integer which is divisible by no perfect square other than 1. That is, its prime factorization has exactly one factor for each prime that appears in it. For example, 10 = 2 ⋅ 5 is square-free, but 18 = 2 ⋅ 3 ⋅ 3 is not, because 18 is divisible by 9 = 3**2. Example The smallest positive square-free integers are 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, ... The questions As it often happens, I’m nitpicking on the details about the domain of our investigation: should we consider negative values? I guess not by the examples… The solution sub is_square_free ($N) { return unless$N % 4; my $divisor = 3; while ($N > $divisor) { if ($N % $divisor == 0) {$N /= $divisor; return unless$N % $divisor; }$divisor += 2; # go through odd candidates only } return 1; } The goal is not to find all divisors, so… we don’t find them and we take every possible chance to bail out with a false value. It can happen in two", "finding the least square number divisibe by given numbers We are required to find the least possibe square number that is exactly divisible by the numbers 10, 12 and 16. In this question, I found out the required least possible square number is 3600. I did that by hit and trial method by taking out number randomly and then squaring and checking to see if the square number is exactly divided by a the three numbers. I just want to know the definite way to find the answer if there is any. • $900$ isn't divisible by $16$. – Daniel Fischer Jun 2 '14 at 14:01 • Look at the prime factorizations of $10$, $12$, $16$. The answer is in there. – Jyrki Lahtonen Jun 2 '14 at 14:02 2 Answers $10=2\\times 5$ $12=2^2\\times 3$ $16=2^4$ From here, we obtain $\\text{lcm} (10,12,16)=2^4\\times 3\\times 5$, hence the least square number required is $2^4\\times 3^2\\times 5^2=3600$. • Nice, the smallest number divisible by any quantity of numbers is the lcm of those numbers, and then you just made all the exponents even so it was a square right? – Jorge Fernández Hidalgo Jun 2 '14 at 14:04 • taking number for exampe 10,12 and 18, the lcm of those numbers is 180 and squaring 180 makes it 32400. However, 900", "5 1,1,5 1,1,1 L.C.M=2\\times 2\\times 3\\times 3\\times 5 =180 It can be clearly seen that 5 cannot be paired. Therefore, we have to multiply 180 by 5 in order to get a perfect square. Thus, the smallest square number divisible by 4,9 and 10$$=180\\times 5=900$ 10. Find the smallest square number that is divisible by each of the numbers $\\text{8,15}$ and $\\text{20}$. Ans: We know that the number that is perfectly divisible by each one of $8,15$ and $20$ is their L.C.M So, taking the L.C.M of these numbers $2$ $8,15,20$ $2$ $4,15,10$ $2$ $2,15,5$ $3$ $1,15,5$ $5$ $1,5,5$ $1,1,1$ L.C.M$=2\\times 2\\times 2\\times 3\\times 5$ $=120$ It can be clearly seen that the prime factors $2$, $3$ and $5$ cannot be paired. Therefore, we have to multiply $120$ by $2$, $3$ and $5$ in order to get a perfect square. Thus, the smallest square number divisible by $8, 15$ and $20$ is $120\\times 2\\times 3\\times 5 = 3600$ ## What do you learn from Exercise 6.3 Class 8 Maths? CBSE Class 8 Maths introduces students to squares and square roots in the 6th lesson. In this chapter students get to learn everything about squaring a number, how to derive the square root value of a number, perfect squares, what are perfect squares, properties of square numbers and so", "Feb 25 '17 at 17:34 $a$ is also not the sum of three perfect squares once it ends in at least three digits (all $1$). As $1000$ is divisible by $8,$ we then have $$a = bcdefgh111 = 1000 w + 111 \\equiv 7 \\pmod 8$$ • $3332^2 + 0^2 + 0^2 = 11102224$. – Starfall Feb 16 '17 at 6:35 • @Starfall see edit – Will Jagy Feb 16 '17 at 17:54" ]

[ "pointed out that it has Ponzu sauce (has soy sauce in it) and would ask to have it made without, and also suggested susbtituting the crackers with lettuce. This is the kind of service that I would like for every dining experience! My partner and I had picked out 3 entrees and asked her which was gluten free. Out of the 3, 2 were already gluten free, so we ordered those. Even though the waitress was very knowledgeable, when delivering our food, she informed that she has let the kitchen know about my food allergies. The Paleo Low Down We had a steak with cumin fries and sword fish with caponata, a Sicilian cooked sweet and sour salad. The later was the most delicious thing we ordered, their version being made of eggplant, peppers and zuccini among other, with a nice tang to it. As being new to eating steaks, I requested for it to be cooked medium, which was probably a mistake since it was overcooked. The fries and sword fish were good, but not amazing. We also had a nice house red wine, a blend of Merlot and Cabernet. These items all fir into the Paleo diet...although, does wine really? Did cavemen have wine? Sure they had grapes, but goodness knows we can't go without some", "how the owner of the company should handle a customer at all . You do not state over that it 's good pizza and that another company loved it . Maybe they got the good pizza he was talking about but we did n't . Also as the OWNER NOT knowing how to handle a situation like this is unacceptable . He should 've said sorry how can I fix it for you so that our company can be one that u recommend again . Instead he chose to argue and be rude ! I would NEVER EVER EAT HERE AGAIN . I DO NOT RECOMMEND THIS GROSS PIZZA PLACE TO ANYONE EVER ! The sad part is I never thought you could mess up pizza but Joey and the Pizza Company proved that they for sure can ! label: 1 predict: 1 The food was good and the service was genuine and thorough . For our daughter , it 's hard to find a place that has gluten - free crust ( without eggs ) and vegan cheese , let alone such a combination that tastes She loved it ! label: 5 predict: 5 Read all the reviews on yelp , figured this place would be great . Nope not all . Maybe if your Vegan but", "it's really NOT a problem to have guests chose their meals ahead of time. The SOP with big events is that each guest gets a ticket that they place above their plate on the table. At corporate/organizational events, I have the name of the meal chosen on the ticket and each option gets a different color to make it easier for the server. So for example, I have an event coming up with three options that for ease of description, I'll call meat, gluten-free and vegetarian. The meat ticket will be red with the word \"MEAT\" on it, the blue ticket will have \"GF\" on it and the green ticket will have \"VEG\" on it. The guest just puts the ticket on the table. Sometimes I've also had a colored sticker that affixes to their name tag. Iin either case, the servers make a count of the entree types per table as they are doing the salad serving and are then prepared to serve the table. I do not allow them to serve all of the meat first, then the veggie, then the GF; all of the table must be served at once. At weddings, the place card will have some sort of label/flower/trinket indicating which kind of meal. I had a client request that each guest get", "# Gluten Free in Nashville 5 thoughts last posted May 21, 2015, 7:50 p.m. 0 In the NashDev Slack there is a #glutenfree channel where there has been discussion about various gluten free options at local restaurants. I wanted to capture them here. 4 later thoughts", "off the menu.) Italian food has always been a personal fave but so much of it relies on nightshades and olive oil so it is a very rare treat, for which I later pay painfully. There are folks out there with nut allergies and shellfish allergies and strawberry allergies. We can't all stay home and eat ramen noodles, folks. So, lay off Oba. Again, a chef who freaks over special requests is a chef with limited talents and no real understanding of food. Food is more than fuel. Well prepared, it is joy. Well served, it is ecstasy. Shared in good company among pleasant surroundings, it is a little slice of heaven. January 8, 2013 at 8:02 pm | Reply • cust0m3r The issue here is not that OBA knew a restaurant wouldn't conform to the dietary restrictions, but that THEY TOLD HIM THEY WOULD and then they did not. At least as I understand it. I don't have religious or allergenic dietary restrictions, but I still take issue with this practice. It's important. If you aren't going to prepare the food as asked, say so so that the diner can find a place that will. But don't lie and say a dish is vegetarian, then boil the veggies in beef broth, or claim there are no peanuts", "dish of ravioli I had in Florence. Make it! Mathematically, of course… Buon appetito!", "no mood for noodles. So I said we'd eat Italian at Momma Yolanda's. At dinner I told her I didn't think much of her family on account of how critical they were. Although she didn't say anything, from the look on her face I could see that they made her angry too. Then came the bill. Surprise! I hadn't budgetted for such a fancy restaurant so Stella had to help out with most of it. When I took her back to her house, I started pulling off her blouse, but she struggled free and ran inside. What a lousy date. Last edited: May 26, 2009", "form of alcohol while fine dining. It goes hand in hand. Beer is clearly out of the question for gluten-free and paleo dining since the base ingredients were not part of a caveman's diet. The Dairy-Free Deal There were plenty of dairy free options. Many of the salads had dairy on them. This is usually disappointing to have without, since they often only have a few other items on them. My partner is fine ordering salads without the dairy, but it seems like a bummer to me. Our entrees were both originally dairy free and very satisfying. Rating Normal Rating: 4/5 stars Minus 1 for good but not amazing food. The ambiance and service were good. Allergy Dining Rating: 4/5 stars Although it's nice to have the menu clearly marked which items are gluten free, or even to have a gluten free menu, the waitress was so knowledgeable and reassuring that it was ok. Preferably, I would like to be able to talk to kitchen staff in advance, tour kitchens etc. As a novice food critic, I am getting there. As a society we are just becoming aware of the needs of Celiac disease family members, friends, and customers. As this is a learning curve for all of us, I am positive they have room to grow to", "does n't taste processed . I 'll be back again . label: 4 predict: 4 This place is wicked awesome ! I am from New York and am constantly on a good pizza hunt . I have been all over town and this place is in my top three . Add to it that I ca n't have milk / cheese as I 'm allergic . Pizza Company has me covered ! Their vegan pizza is to DIE for . They use Daiya cheese , which is a really well respected vegan alternative . My boyfriend and I also ordered the meatball sliders . Which were AMAZING . They come on sliced / toasted garlic knots . WHAT ? ! My boyfriend also ordered the spaghetti and meatballs . He kept commenting that it was n't a dish he would order in a restaurant since we are italian and often make it at home , but the meatballs were so fantastic The food was incredible . We want this restaurant a long and healthy life . Joey , the owner , is a gem and deserves much success . label: 5 predict: 5 Best vegan pizza in Vegas ! ! My favorite is vegan cheese topped with fresh garlic and pineapple ... Can not go wrong with that", "about 72% of the pasta blend is corn and 28% is quinoa. Note: In our solution to the problem, we are making the assumption that the protein content of the corn and the quinoa used to make the pasta is the same as the protein content listed in the table. This may or may not be the case. For example, Wikipedia lists the protein content of quinoa as 14% and of corn as 3.22%. Using these numbers, our solution would be different: 35% quinoa, 65% corn.", "preference and start treating it as a moral choice. There is a qualitative difference between someone who does not like Italian food and a vegan. This is interesting when choosing what or where to eat as a group: This is hard enough as it is and I at least usually try very hard to accommodate everyone and not be a burden. And I absolutely do not expect to be treated like I'm better for that. But if it actually would come down to a choice between a vegetarian restaurant or a steakhouse, just because you really like meat: Yes, I do absolutely expect us to avoid the steakhouse. (To be clear: In my experience, it rarely actually comes down to only those two choices. And there are good reasons to avoid vegetarian restaurants that are not based on preference which should be given ample weight too - e.g. someone I know has Coeliac disease, fructose intolerance and lactose intolerance and tends to have a very hard time eating non-meat things. In my experience, though, people who have needs and not just preferences tend to ironically be more open to compromise anyway, so it is less often a problem with them). 3. Maybe think about your reasons for not being vegan and evaluate them seriously. To be clear, this", "the distances I needed it to. ↩︎ 3. I think just about every café I went to had some sort of gluten-free option that looked tasty. Given I have only recently been trying to cut out gluten, this was a pleasant surprise! ↩︎", "very good! So pillowy and soft, and good flavor. The ragu was a but too much for me, flavor-wise. That is, it was very \"strong\" and meaty. Otherwise, it was good; just a bit too husky/burly. Tasted the tagliatelle: the pasta seemed hand-made. It had a very nice consistency. The sauce itself was good, too, but again, HUSKY! I guess am still tired of that \"braised meat\" flavor. No ragus for me then! And since JS was denied, I think the next time we go there, Puttanesca it must be! Anyway, since I was not in the mood for too-meaty flavors, CSC's Gamberi spaghetti was spot on. I will leave her to describe it. [js] Pasta: I had the tagliatelle Bolognese, which I didn't order. I had ordered the puttanesca, but our server made a mistake. I felt she was too concerned about our stomach room. I could have told her, \"Don't worry; we know what we're doing; we do this all the time!\" The pasta was very good as well, given that this is a dish I would *NEVER* order in a restaurant. Points for having FRESH pasta. The meat sauce was very meaty, very hearty, and indeed, this pasta dish was very filling, a substantial meal in itself. The meat sauce could have used a couple", "know that too so I can go elsewhere. Whether a dietary restriction is a health issue, a religious issue, or a preference, restaurants should be honest about whether they can meet a customer's needs. A vegetarian who expresses that need and is told the restaurant can meet it should expect a vegetarian meal whether or not the restaurant serves meat to those who eat it. January 8, 2013 at 12:50 pm | Reply • OBA When we eat out, we order seafood only, and even this has meat ingredients: cooked in beef or chicken broth; or shrimps cooked with meats, and taken out to serve a particular order, when three options are given. We have seen deep fried fish topped by a sauce that had meat bits: one sauce to serve all. Despite their high prices, many restaurants don't really make things to order but pass ready mix stuff. January 8, 2013 at 6:41 pm | Reply • Fred You're an idiot. January 11, 2013 at 6:07 pm | Reply • Fiona I'm a vegetarian, for reasons as close to my heart as any faith-based dictum. I cannot tell you how many times I've been served meat or food containing meat broth even after I've told the server (or host, in private homes) that I don't eat meat.", "muffins is a surefire way to be nodding off by the third talk. I wound up walking to a nearby Panera every single morning to get an egg sandwich.</p> <p>Lunch was better, but also not great. I think by choice of QCon itself, the hotel catering had to make everything gluten-free. I find the whole gluten-free for non-celiacs fad <a href=\"http://www.georgeinstitute.org.au/media-releases/dont-believe-the-hype-on-gluten-free-food\">generally irritating</a> but what really made this annoying was what foods they chose to serve. If I say to you &quot;okay we need a vegetarian meal&quot; you try to think of foods that don&#39;t involve meat, you don&#39;t immediately think of a meat dish and substitute tofu in. Similarly, why does your &#39;gluten free&#39; menu consist almost entirely of gluten-free pasta? I wound up grabbing lunch elsewhere for 3 of the 5 days I was at the conference. This was frustrating, because I convinced my employer to send me to NY by arguing that meals were mostly included, but then I wound up having to pay for most meals.</p> <p>Aside from workshops, it was very rare to find tables in the sessions - usually it was just rows of seats. And the walls were those sliding accordion walls that hotels use to divide huge rooms into sections. The end result being, it was very hard to find a", "a full day of technical talks with a bunch of muffins is a surefire way to be nodding off by the third talk. I wound up walking to a nearby Panera every single morning to get an egg sandwich.</p> <p>Lunch was better, but also not great. I think by choice of QCon itself, the hotel catering had to make everything gluten-free. I find the whole gluten-free for non-celiacs fad <a href=\"http://www.georgeinstitute.org.au/media-releases/dont-believe-the-hype-on-gluten-free-food\">generally irritating</a> but what really made this annoying was what foods they chose to serve. If I say to you &quot;okay we need a vegetarian meal&quot; you try to think of foods that don&#39;t involve meat, you don&#39;t immediately think of a meat dish and substitute tofu in. Similarly, why does your &#39;gluten free&#39; menu consist almost entirely of gluten-free pasta? I wound up grabbing lunch elsewhere for 3 of the 5 days I was at the conference. This was frustrating, because I convinced my employer to send me to NY by arguing that meals were mostly included, but then I wound up having to pay for most meals.</p> <p>Aside from workshops, it was very rare to find tables in the sessions - usually it was just rows of seats. And the walls were those sliding accordion walls that hotels use to divide huge rooms into sections. The", "of this disease even though it doesn't affect me personally (I don't have CD). But maybe that's part of what makes it harder - my son's life is somewhat out of my control and it's not going to get any easier going forward. I cringe at the thought of him wanting (or god forbid trying) beer! I have yet to get a negative test result and I feel like I have been doing a good job - the best job that I can do and yet I still am failing. At least for me now, there are so many things that make it easy on me. He knows his restrictions and he doesn't balk the slightest at them, probably because I make sure he has equivalents of whatever he might want in the way of \" gluten \" foods. I am so thankful for that. One thing that has made me realize that there are far worse scenarios is being on this list - I only have CD to deal with. I read about these people who have egg, dairy, soy, nut and who knows what else as far as food allergies. I can't even fathom having to figure out what to eat with all that going on too! No pity-parties for me. Cathy > > > >", "###### Question: [0/1 Points] DETAILS PREVIOUS ANSWERS MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER An airline has found about 8% of its passengers request vegetarian meals, probability that will be short vegetarian dinners? flight with 152 passengers the airline has 15 vegetarian dinners available_ What's the (hint; Let be the number vegetarians. Identify probability mode (Click use an online probability calculator.) Write probability expression for that reflects 'Short of vegetarian dinners\" and compute:) 4Q7 23535 (round to decimal places)", "I can go elsewhere. Whether a dietary restriction is a health issue, a religious issue, or a preference, restaurants should be honest about whether they can meet a customer's needs. A vegetarian who expresses that need and is told the restaurant can meet it should expect a vegetarian meal whether or not the restaurant serves meat to those who eat it. January 8, 2013 at 12:50 pm | Reply • OBA When we eat out, we order seafood only, and even this has meat ingredients: cooked in beef or chicken broth; or shrimps cooked with meats, and taken out to serve a particular order, when three options are given. We have seen deep fried fish topped by a sauce that had meat bits: one sauce to serve all. Despite their high prices, many restaurants don't really make things to order but pass ready mix stuff. January 8, 2013 at 6:41 pm | Reply • Fred You're an idiot. January 11, 2013 at 6:07 pm | Reply • Fiona I'm a vegetarian, for reasons as close to my heart as any faith-based dictum. I cannot tell you how many times I've been served meat or food containing meat broth even after I've told the server (or host, in private homes) that I don't eat meat. I've tried to figure", "steakhouse. (To be clear: In my experience, it rarely actually comes down to only those two choices. And there are good reasons to avoid vegetarian restaurants that are not based on preference which should be given ample weight too - e.g. someone I know has Coeliac disease, fructose intolerance and lactose intolerance and tends to have a very hard time eating non-meat things. In my experience, though, people who have needs and not just preferences tend to ironically be more open to compromise anyway, so it is less often a problem with them). 3. Maybe think about your reasons for not being vegan and evaluate them seriously. To be clear, this is a stretch-goal and not the actual point of this article. And if you want to, you can watch someone who does eat meat say essentially the same things here: Thanks for reading, don't @ me. ;) #### Reasons I'm not not vegan Now, the main point of this post is dedicated to the general question of \"how I think about the topic and how I believe you should think about it too\". But I also want it to serve as a reference of my personal, specific thoughts driving my decision - so if you don't know me well or are not interested in my personal reasons," ]

[ "$$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\\frac{non-vegetarian \\ students}{non-vegetarian \\ non-students}=2*\\frac{2}{3}=\\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \\ non-students = 15 = \\frac{3}{7}*\\frac{x}{2}$$ --> $$x=70$$. Sufficient. (2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/guests-at-a-r ... 04547.html _________________ Kudos [?]: 132706 [0], given: 12335 Re: Guests at a recent party ate a total of fifteen hamburgers. Each guest [#permalink] 14 Aug 2017, 08:57 Display posts from previous: Sort by", "of 15 hamburgers and each guest who was neither a student nor a vegetarian (group #4) ate exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15). Make a matrix: Attachment: Stem.PNG [ 2.33 KiB | Viewed 12452 times ] Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\\frac{x}{2}$$. (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians --> $$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\\frac{non-vegetarian \\ students}{non-vegetarian \\ non-students}=2*\\frac{2}{3}=\\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \\ non-students = 15 = \\frac{3}{7}*\\frac{x}{2}$$ --> $$x=70$$. Sufficient. Attachment: 1.PNG [ 2.59 KiB | Viewed 12574 times ] (2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$. Attachment: 2.PNG [ 2.38 KiB | Viewed 12449 times ] _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts:", "exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15). Make a matrix: Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\\frac{x}{2}$$. (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians --> $$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\\frac{non-vegetarian \\ students}{non-vegetarian \\ non-students}=2*\\frac{2}{3}=\\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \\ non-students = 15 = \\frac{3}{7}*\\frac{x}{2}$$ --> $$x=70$$. Sufficient. (2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$. [Reveal] Spoiler: Attachment: Stem.PNG [ 2.33 KiB | Viewed 21164 times ] Attachment: 2.PNG [ 2.38 KiB | Viewed 21149 times ] Attachment: 1.PNG [ 2.59 KiB | Viewed 21343 times ] _________________ Kudos [?]: 135176 [13], given: 12668 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7784 Kudos [?]: 18092 [5], given: 236 Location: Pune,", "than the maximum over $X$, for if $x \\in X'$ attains the maximum of $f(x)$ over $X'$, then also $x \\in X$, so the maximum of $f$ over $X$ is at least $f(x)$. In normal words, if you restrict your diet, you can miss out on good dishes, but never gain access to better dishes than on an unrestricted diet. From this, we could deduce that you should never pick a vegan dish in a restaurant because the non-vegan dishes were made with fewer restrictions and hence can only be better than the vegan dish. Same for choosing recipes to cook yourself. Before I was vegan, this was my conscious reason for always choosing dishes with meat. But is the deduction true? I don’t think so. Because, unbeknownst to many, the meat dishes are actually constrained to contain meat. I don’t know why, but my two prime suspects are Goodhart’s law impacting the reasoning above, or meat-eaters being scared of vegetables. As it turns out, making a good vegetarian meal takes non-zero skill, contrary to making a good meal with meat. In my experience, this causes chefs to put actual thought into their vegetarian dishes, causing these to actually be tastier than most dishes with meat. So the argument from mathematical optimization actually gives the wrong answer here!", "a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village? Solution: sample size n = 400 case (i): sample proporation of vegetarian p = $$\\frac { 3 }{10.954}$$ = $$\\frac { 230 }{400}$$ p = 0.575 q = 1 – p = 1 – 0.575 q = 0.425 Sample error S.E= $$\\sqrt{\\frac { pq }{n}}$$ = $$\\sqrt{\\frac { 0.575×0.425 }{400}}$$ = $$\\sqrt{\\frac { 0.223125 }{400}}$$ $$\\sqrt{0.0005578125}$$ S.E = 0.2361 Case(ii): sample size n = 400 since both vegetarian and non- vegetarian foods are equally popular in that village sample proparation of vegetarian p = $$\\frac { 1 }{2}$$ = 0.5 q = 1 -p ⇒ q = 1 – 0.5 q = 0.5", "Sep 2009 Posts: 55277 Re: Guests at a recent party ate a total of fifteen hamburgers. Each guest [#permalink] ### Show Tags 14 Aug 2017, 08:57 Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party? We have 4 groups of guests: 1. Vegetarian students; 2. Vegetarian non-students; 3. Non-vegetarian students; 4. Non-vegetarian non-students. Now, as guests ate a total of 15 hamburgers and each guest who was neither a student nor a vegetarian (group #4) ate exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15). Make a matrix: Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\\frac{x}{2}$$. (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians --> $$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some", "other other 331 #> 11 strictly anything strictly anything 5107 #> 12 strictly halal strictly halal 18 #> 13 strictly kosher strictly kosher 18 #> 14 strictly other strictly other 450 #> 15 strictly vegan strictly vegan 227 #> 16 strictly vegetarian strictly vegetarian 874 #> 17 vegan vegan 136 #> 18 vegetarian vegetarian 665 #> 19 unknown diet <NA> 24360 tidy(rec, number = 1)#> # A tibble: 1 x 3 #> terms value id #> <chr> <chr> <chr> #> 1 diet unknown diet unknown_wvoo0", "form of alcohol while fine dining. It goes hand in hand. Beer is clearly out of the question for gluten-free and paleo dining since the base ingredients were not part of a caveman's diet. The Dairy-Free Deal There were plenty of dairy free options. Many of the salads had dairy on them. This is usually disappointing to have without, since they often only have a few other items on them. My partner is fine ordering salads without the dairy, but it seems like a bummer to me. Our entrees were both originally dairy free and very satisfying. Rating Normal Rating: 4/5 stars Minus 1 for good but not amazing food. The ambiance and service were good. Allergy Dining Rating: 4/5 stars Although it's nice to have the menu clearly marked which items are gluten free, or even to have a gluten free menu, the waitress was so knowledgeable and reassuring that it was ok. Preferably, I would like to be able to talk to kitchen staff in advance, tour kitchens etc. As a novice food critic, I am getting there. As a society we are just becoming aware of the needs of Celiac disease family members, friends, and customers. As this is a learning curve for all of us, I am positive they have room to grow to", "40 Math Expert Joined: 02 Sep 2009 Posts: 42259 Kudos [?]: 132706 [0], given: 12335 Re: Guests at a recent party ate a total of fifteen hamburgers. Each guest [#permalink] ### Show Tags 14 Aug 2017, 08:57 Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party? We have 4 groups of guests: 1. Vegetarian students; 2. Vegetarian non-students; 3. Non-vegetarian students; 4. Non-vegetarian non-students. Now, as guests ate a total of 15 hamburgers and each guest who was neither a student nor a vegetarian (group #4) ate exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15). Make a matrix: Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\\frac{x}{2}$$. (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians -->", "at least $f(x)$. In normal words, if you restrict your diet, you can miss out on good dishes, but never gain access to better dishes than on an unrestricted diet. From this, we could deduce that you should never pick a vegan dish in a restaurant because the non-vegan dishes were made with fewer restrictions and hence can only be better than the vegan dish. Same for choosing recipes to cook yourself. Before I was vegan, this was my conscious reason for always choosing dishes with meat. But is the deduction true? I don’t think so. Because, unbeknownst to many, the meat dishes are actually constrained to contain meat. I don’t know why, but my two prime suspects are Goodhart’s law impacting the reasoning above, or meat-eaters being scared of vegetables. As it turns out, making a good vegetarian meal takes non-zero skill, contrary to making a good meal with meat. In my experience, this causes chefs to put actual thought into their vegetarian dishes, causing these to actually be tastier than most dishes with meat. So the argument from mathematical optimization actually gives the wrong answer here! ## Abstinence-only education criticism as general template 540 wordsTW: self-harm, suicide, substance abuse Context: In conservative regions in the US, there is a thing called abstinence-only sex education. People believe", "anything strictly halal #> 11 11 11 11 #> strictly kosher strictly other strictly vegan strictly vegetarian #> 11 11 11 11 #> vegan vegetarian <NA> #> 11 11 11 # since skip defaults to TRUE, baking the step has no effect baked_okc <- bake(ds_rec, new_data = okc) table(baked_okc\\$diet, useNA = \"always\") #> #> anything halal kosher mostly anything #> 6174 11 11 16562 #> mostly halal mostly kosher mostly other mostly vegan #> 48 86 1004 335 #> mostly vegetarian other strictly anything strictly halal #> 3438 331 5107 18 #> strictly kosher strictly other strictly vegan strictly vegetarian #> 18 450 227 874 #> vegan vegetarian <NA> #> 136 665 24360", "(defclass \\cn{MushroomPizza\\framesName} \\\\ 390 ~~~\\textbf{(IS-A \\cn{VegetarianPizza\\framesName}) ~~~~~ // It has to be explicitly asserted!} 391 ~~~(multislot hasTopping\\framesName \\\\ 392 ~~~(allowed-class Mushroom\\framesName)) \\\\ 393 \\textbf{OWL:} \\\\ 394 Class (\\cn{VegetarianPizza} \\textbf{COMPLETE \\\\ 395 ~~~~ // the keyword changes to ''COMPLETE''}\\\\ 396 ~~~\\cn{Pizza} \\\\ 397 ~~~~~~ (restriction \\cn{hasTopping} allValuesFrom \\cn{Vegetable}))\\\\ 398 Class (\\cn{MushroomPizza} partial\\\\ 399 ~~~ \\cn{Pizza} \\\\ 400 ~~~~~~ restriction (\\cn{hasTopping} allValuesFrom \\cn{Mushroom}))\\\\ 401 %\\textbf{Paraphrase:} \\\\ 402 %A vegetarian pizza is \\emph{any} pizza that, \\emph{amongst other 403 %things}, both does \\emph{not} have \\emph{some} meat topping and 404 %also does \\emph{not} have \\emph{some} fish topping. 405 %\\vspace{-0.15cm} 406 \\end{flushleft}}}} 407 \\caption{Asserted and inferred IS-A relations } 408 \\label{is-a} 409 \\end{figure} 410 411 For example, previously we stated that a \\textit{VegetarianPizza} must only have 412 toppings that are \\textit{Vegetables} 413 (Figure~\\ref{modelconcept}). Coming from the other direction, suppose 414 that we want to model the fact that any pizza that only has vegetables 415 as its topping must be a \\textit{VegetarianPizza}. Furthermore, we 416 also define a special kind of \\textit{VegetarianPizza} -- 417 \\textit{MushroomPizza}, which can only have \\textit{Mushrooms} as its 418 toppings. The frames definition of \\textit{VegetarianPizza} is 419 unchanged as \\protegeframes does not allow us to defined a class. In 420 fact, we have to explicitly assert that \\textit{MushroomsPizza} is a 421 subclass of \\textit{VegetarianPizza}. In OWL we can create the", "India ### Show Tags 10 Nov 2010, 19:16 5 KUDOS Expert's post mrinal2100 wrote: Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party? (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians. (2) 30% of the guests were vegetarian non-students. can someone explain in detail This is what I drew when I read the Question stem. Half of the guests were vegetarians so Total/2 stands for the complete vegetarian circle. All outside Vegetarian circle are Total/2. Attachment: Ques.jpg [ 17.29 KiB | Viewed 20516 times ] Statement 1: Veg students : Veg non students = 2:3 Let me say they are 2x and 3x in number. Non veg students : Non veg non-students = 4:3 (Since veg's ratio is half of non veg's ratio) Let me say they are 4y and 3y in number. So now my diagram looks like this: Attachment: Ques1.jpg [ 18.33 KiB | Viewed 20508 times ] 3y = 15 hence y = 5 Since 7y is half", "Vegetarian: 0 = Buy[\"Hamburger\"] + Buy[\"Cheeseburger\"] + Buy[\"Quarter Pounder w/ Cheese\"] + Buy[\"McLean Deluxe\"] + Buy[\"McLean Deluxe w/ Cheese\"] + Buy[\"Big Mac\"] + Buy[\"Filet-O-Fish\"] + Buy[\"McGrilled Chicken\"] + Buy[\"McChicken Sandwich\"] + Buy[\"Chicken McNuggets (6 pcs)\"] + Buy[\"Chicken McNuggets (9 pcs)\"] + Buy[\"Chicken McNuggets (20 pcs)\"] + Buy[\"Chef Salad\"] + Buy[\"Chunky Chicken Salad\"] + Buy[\"Bacon Bits\"] + Buy[\"Egg McMuffin\"] + Buy[\"Sausage McMuffin\"] + Buy[\"Sansage McMuffin with Egg\"] + Buy[\"English Muffin\"] + Buy[\"Sausage Biscuit\"] + Buy[\"Sausage Biscuit with Egg\"] + Buy[\"Bacon, Egg & Cheese Biscuit\"] + Buy[\"Breakfast Burrito\"];", "Quiz 7. Task 6 - Set operations Instruction Unbelievable! How about set operations: unions, differences and intersections? Can you still remember them? Take a look: gluten_free_product (id, type, name, calories, price) vegetarian_product (id, type, name, calories, price) As you can see, we are now inside a dietician's office. We have two tables with the same columns: products that are free of gluten and vegetarian products. Each product has its own id, its type (whether it's meat, dairy etc.), name, number of calories per portion and price. Exercise Show all columns for the products which are gluten free and vegetarian at the same time. Stuck? Here's a hint! SELECT", "fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\\frac{non-vegetarian \\ students}{non-vegetarian \\ non-students}=2*\\frac{2}{3}=\\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \\ non-students = 15 = \\frac{3}{7}*\\frac{x}{2}$$ --> $$x=70$$. Sufficient. (2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/guests-at-a-r ... 04547.html --== Message from the GMAT Club Team ==-- THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please re-post it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you. _________________ Re: Guests at a recent party ate a total of fifteen hamburgers. Each guest [#permalink] 14 Aug 2017, 08:57 Display posts from previous: Sort by", "\\to \\mathbb{R}$. In this post, we’ll think of $Y$ as the set of possible restaurant dishes, $X\\subset Y$ as the set of dishes satisfying certain constraints like being digestible, non-poisonous and not containing human flesh. The function $f$ to optimize is some combination of price, healthyness and taste. A first observation is that for any $X' \\subset X$ the maximum of $f(x)$ over $X'$ is no bigger than the maximum over $X$, for if $x \\in X'$ attains the maximum of $f(x)$ over $X'$, then also $x \\in X$, so the maximum of $f$ over $X$ is at least $f(x)$. In normal words, if you restrict your diet, you can miss out on good dishes, but never gain access to better dishes than on an unrestricted diet. From this, we could deduce that you should never pick a vegan dish in a restaurant because the non-vegan dishes were made with fewer restrictions and hence can only be better than the vegan dish. Same for choosing recipes to cook yourself. Before I was vegan, this was my conscious reason for always choosing dishes with meat. But is the deduction true? I don’t think so. Because, unbeknownst to many, the meat dishes are actually constrained to contain meat. I don’t know why, but my two prime suspects are Goodhart’s law", "15. Also, half of the guests were vegetarians, so:$$v = (1/2)t$$ so, the other half of the guests aren't vegetarians, so: $$s-b= (1/2)t$$ our established equation is now: $$t=s+v+15-b$$ ----> question is asking us to look for $$t$$ (1) So $$2$$ vegetarians students and $$3$$ non-student vegetarians, making a total of $$5$$ vegetarians. Since this is half of the non-vegetarians, which means student minus both, therefore $$s-b=10$$. So look again at our established equation, we already have the value of $$s-b$$ and we already know that $$v=(1/2)t$$. Plug in what we got so far into the equation: $$t= 10 +(1/2)t+15$$, therefore: $$(1/2)t=25$$, finally: $$t=50$$ ----->Suff. (2) $$(3/10)t = v-b$$ Our established equation will become $$t = s+(3/10)t+15$$ or $$(7/10)t = s+15$$------> 1 equation with 2 variables...Not. Suff. What's the OA? Re: Zumit DS 027 [#permalink] 16 Sep 2008, 08:29 Similar topics Replies Last post Similar Topics: 1 Guests at a recent party ate a total of fifteen hamburgers. 3 03 Sep 2010, 18:34 6 Guests at a recent party ate a total of fifteen hamburgers. 7 02 Nov 2009, 03:00 Guests at a recent party ate a total of fifteen hamburgers. 1 02 Jun 2009, 05:46 Guests at a recent party ate a total of fifteen hamburgers. 3 29 May 2008, 19:11 Guests at a recent party ate", "6062 Location: Pune, India Followers: 1606 Kudos [?]: 8975 [2] , given: 195 Re: data suffeciency [#permalink] 10 Nov 2010, 18:16 2 KUDOS Expert's post mrinal2100 wrote: Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party? (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians. (2) 30% of the guests were vegetarian non-students. can someone explain in detail This is what I drew when I read the Question stem. Half of the guests were vegetarians so Total/2 stands for the complete vegetarian circle. All outside Vegetarian circle are Total/2. Attachment: Ques.jpg [ 17.29 KiB | Viewed 12375 times ] Statement 1: Veg students : Veg non students = 2:3 Let me say they are 2x and 3x in number. Non veg students : Non veg non-students = 4:3 (Since veg's ratio is half of non veg's ratio) Let me say they are 4y and 3y in number. So now my diagram looks like this: Attachment: Ques1.jpg [ 18.33 KiB | Viewed 12380", "a vegetarian, $m(y)$ for $y$ is meat, and $e(x, y)$ for $x$ eats $y$. Based on these, consider the following sentences : I. $\\forall x \\vee (x)\\Leftrightarrow (\\forall y e(x, y) \\implies \\neg m(y))$ ... are equivalent sentences Only $II$ and $III$ are equivalent sentences. Only $I$ and $III$ are equivalent sentence . $I, II,$ and $III$ are equivalent sentences. 1 vote" ]

[ "$$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\\frac{non-vegetarian \\ students}{non-vegetarian \\ non-students}=2*\\frac{2}{3}=\\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \\ non-students = 15 = \\frac{3}{7}*\\frac{x}{2}$$ --> $$x=70$$. Sufficient. (2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/guests-at-a-r ... 04547.html _________________ Kudos [?]: 132706 [0], given: 12335 Re: Guests at a recent party ate a total of fifteen hamburgers. Each guest [#permalink] 14 Aug 2017, 08:57 Display posts from previous: Sort by", "of 15 hamburgers and each guest who was neither a student nor a vegetarian (group #4) ate exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15). Make a matrix: Attachment: Stem.PNG [ 2.33 KiB | Viewed 12452 times ] Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\\frac{x}{2}$$. (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians --> $$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\\frac{non-vegetarian \\ students}{non-vegetarian \\ non-students}=2*\\frac{2}{3}=\\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \\ non-students = 15 = \\frac{3}{7}*\\frac{x}{2}$$ --> $$x=70$$. Sufficient. Attachment: 1.PNG [ 2.59 KiB | Viewed 12574 times ] (2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$. Attachment: 2.PNG [ 2.38 KiB | Viewed 12449 times ] _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts:", "other other 331 #> 11 strictly anything strictly anything 5107 #> 12 strictly halal strictly halal 18 #> 13 strictly kosher strictly kosher 18 #> 14 strictly other strictly other 450 #> 15 strictly vegan strictly vegan 227 #> 16 strictly vegetarian strictly vegetarian 874 #> 17 vegan vegan 136 #> 18 vegetarian vegetarian 665 #> 19 unknown diet <NA> 24360 tidy(rec, number = 1)#> # A tibble: 1 x 3 #> terms value id #> <chr> <chr> <chr> #> 1 diet unknown diet unknown_wvoo0", "exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15). Make a matrix: Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\\frac{x}{2}$$. (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians --> $$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\\frac{non-vegetarian \\ students}{non-vegetarian \\ non-students}=2*\\frac{2}{3}=\\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \\ non-students = 15 = \\frac{3}{7}*\\frac{x}{2}$$ --> $$x=70$$. Sufficient. (2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$. [Reveal] Spoiler: Attachment: Stem.PNG [ 2.33 KiB | Viewed 21164 times ] Attachment: 2.PNG [ 2.38 KiB | Viewed 21149 times ] Attachment: 1.PNG [ 2.59 KiB | Viewed 21343 times ] _________________ Kudos [?]: 135176 [13], given: 12668 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7784 Kudos [?]: 18092 [5], given: 236 Location: Pune,", "a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village? Solution: sample size n = 400 case (i): sample proporation of vegetarian p = $$\\frac { 3 }{10.954}$$ = $$\\frac { 230 }{400}$$ p = 0.575 q = 1 – p = 1 – 0.575 q = 0.425 Sample error S.E= $$\\sqrt{\\frac { pq }{n}}$$ = $$\\sqrt{\\frac { 0.575×0.425 }{400}}$$ = $$\\sqrt{\\frac { 0.223125 }{400}}$$ $$\\sqrt{0.0005578125}$$ S.E = 0.2361 Case(ii): sample size n = 400 since both vegetarian and non- vegetarian foods are equally popular in that village sample proparation of vegetarian p = $$\\frac { 1 }{2}$$ = 0.5 q = 1 -p ⇒ q = 1 – 0.5 q = 0.5", "than the maximum over $X$, for if $x \\in X'$ attains the maximum of $f(x)$ over $X'$, then also $x \\in X$, so the maximum of $f$ over $X$ is at least $f(x)$. In normal words, if you restrict your diet, you can miss out on good dishes, but never gain access to better dishes than on an unrestricted diet. From this, we could deduce that you should never pick a vegan dish in a restaurant because the non-vegan dishes were made with fewer restrictions and hence can only be better than the vegan dish. Same for choosing recipes to cook yourself. Before I was vegan, this was my conscious reason for always choosing dishes with meat. But is the deduction true? I don’t think so. Because, unbeknownst to many, the meat dishes are actually constrained to contain meat. I don’t know why, but my two prime suspects are Goodhart’s law impacting the reasoning above, or meat-eaters being scared of vegetables. As it turns out, making a good vegetarian meal takes non-zero skill, contrary to making a good meal with meat. In my experience, this causes chefs to put actual thought into their vegetarian dishes, causing these to actually be tastier than most dishes with meat. So the argument from mathematical optimization actually gives the wrong answer here!", "pointed out that it has Ponzu sauce (has soy sauce in it) and would ask to have it made without, and also suggested susbtituting the crackers with lettuce. This is the kind of service that I would like for every dining experience! My partner and I had picked out 3 entrees and asked her which was gluten free. Out of the 3, 2 were already gluten free, so we ordered those. Even though the waitress was very knowledgeable, when delivering our food, she informed that she has let the kitchen know about my food allergies. The Paleo Low Down We had a steak with cumin fries and sword fish with caponata, a Sicilian cooked sweet and sour salad. The later was the most delicious thing we ordered, their version being made of eggplant, peppers and zuccini among other, with a nice tang to it. As being new to eating steaks, I requested for it to be cooked medium, which was probably a mistake since it was overcooked. The fries and sword fish were good, but not amazing. We also had a nice house red wine, a blend of Merlot and Cabernet. These items all fir into the Paleo diet...although, does wine really? Did cavemen have wine? Sure they had grapes, but goodness knows we can't go without some", "it's really NOT a problem to have guests chose their meals ahead of time. The SOP with big events is that each guest gets a ticket that they place above their plate on the table. At corporate/organizational events, I have the name of the meal chosen on the ticket and each option gets a different color to make it easier for the server. So for example, I have an event coming up with three options that for ease of description, I'll call meat, gluten-free and vegetarian. The meat ticket will be red with the word \"MEAT\" on it, the blue ticket will have \"GF\" on it and the green ticket will have \"VEG\" on it. The guest just puts the ticket on the table. Sometimes I've also had a colored sticker that affixes to their name tag. Iin either case, the servers make a count of the entree types per table as they are doing the salad serving and are then prepared to serve the table. I do not allow them to serve all of the meat first, then the veggie, then the GF; all of the table must be served at once. At weddings, the place card will have some sort of label/flower/trinket indicating which kind of meal. I had a client request that each guest get", "Sep 2009 Posts: 55277 Re: Guests at a recent party ate a total of fifteen hamburgers. Each guest [#permalink] ### Show Tags 14 Aug 2017, 08:57 Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party? We have 4 groups of guests: 1. Vegetarian students; 2. Vegetarian non-students; 3. Non-vegetarian students; 4. Non-vegetarian non-students. Now, as guests ate a total of 15 hamburgers and each guest who was neither a student nor a vegetarian (group #4) ate exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15). Make a matrix: Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\\frac{x}{2}$$. (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians --> $$\\frac{vegetarian \\ students}{vegetarian \\ non-students}=\\frac{2}{3}$$ --> if the rate X (some", "form of alcohol while fine dining. It goes hand in hand. Beer is clearly out of the question for gluten-free and paleo dining since the base ingredients were not part of a caveman's diet. The Dairy-Free Deal There were plenty of dairy free options. Many of the salads had dairy on them. This is usually disappointing to have without, since they often only have a few other items on them. My partner is fine ordering salads without the dairy, but it seems like a bummer to me. Our entrees were both originally dairy free and very satisfying. Rating Normal Rating: 4/5 stars Minus 1 for good but not amazing food. The ambiance and service were good. Allergy Dining Rating: 4/5 stars Although it's nice to have the menu clearly marked which items are gluten free, or even to have a gluten free menu, the waitress was so knowledgeable and reassuring that it was ok. Preferably, I would like to be able to talk to kitchen staff in advance, tour kitchens etc. As a novice food critic, I am getting there. As a society we are just becoming aware of the needs of Celiac disease family members, friends, and customers. As this is a learning curve for all of us, I am positive they have room to grow to", "India ### Show Tags 10 Nov 2010, 19:16 5 KUDOS Expert's post mrinal2100 wrote: Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party? (1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians. (2) 30% of the guests were vegetarian non-students. can someone explain in detail This is what I drew when I read the Question stem. Half of the guests were vegetarians so Total/2 stands for the complete vegetarian circle. All outside Vegetarian circle are Total/2. Attachment: Ques.jpg [ 17.29 KiB | Viewed 20516 times ] Statement 1: Veg students : Veg non students = 2:3 Let me say they are 2x and 3x in number. Non veg students : Non veg non-students = 4:3 (Since veg's ratio is half of non veg's ratio) Let me say they are 4y and 3y in number. So now my diagram looks like this: Attachment: Ques1.jpg [ 18.33 KiB | Viewed 20508 times ] 3y = 15 hence y = 5 Since 7y is half", "how the owner of the company should handle a customer at all . You do not state over that it 's good pizza and that another company loved it . Maybe they got the good pizza he was talking about but we did n't . Also as the OWNER NOT knowing how to handle a situation like this is unacceptable . He should 've said sorry how can I fix it for you so that our company can be one that u recommend again . Instead he chose to argue and be rude ! I would NEVER EVER EAT HERE AGAIN . I DO NOT RECOMMEND THIS GROSS PIZZA PLACE TO ANYONE EVER ! The sad part is I never thought you could mess up pizza but Joey and the Pizza Company proved that they for sure can ! label: 1 predict: 1 The food was good and the service was genuine and thorough . For our daughter , it 's hard to find a place that has gluten - free crust ( without eggs ) and vegan cheese , let alone such a combination that tastes She loved it ! label: 5 predict: 5 Read all the reviews on yelp , figured this place would be great . Nope not all . Maybe if your Vegan but", "at least $f(x)$. In normal words, if you restrict your diet, you can miss out on good dishes, but never gain access to better dishes than on an unrestricted diet. From this, we could deduce that you should never pick a vegan dish in a restaurant because the non-vegan dishes were made with fewer restrictions and hence can only be better than the vegan dish. Same for choosing recipes to cook yourself. Before I was vegan, this was my conscious reason for always choosing dishes with meat. But is the deduction true? I don’t think so. Because, unbeknownst to many, the meat dishes are actually constrained to contain meat. I don’t know why, but my two prime suspects are Goodhart’s law impacting the reasoning above, or meat-eaters being scared of vegetables. As it turns out, making a good vegetarian meal takes non-zero skill, contrary to making a good meal with meat. In my experience, this causes chefs to put actual thought into their vegetarian dishes, causing these to actually be tastier than most dishes with meat. So the argument from mathematical optimization actually gives the wrong answer here! ## Abstinence-only education criticism as general template 540 wordsTW: self-harm, suicide, substance abuse Context: In conservative regions in the US, there is a thing called abstinence-only sex education. People believe", "# Gluten Free in Nashville 5 thoughts last posted May 21, 2015, 7:50 p.m. 0 In the NashDev Slack there is a #glutenfree channel where there has been discussion about various gluten free options at local restaurants. I wanted to capture them here. 4 later thoughts", "muffins is a surefire way to be nodding off by the third talk. I wound up walking to a nearby Panera every single morning to get an egg sandwich.</p> <p>Lunch was better, but also not great. I think by choice of QCon itself, the hotel catering had to make everything gluten-free. I find the whole gluten-free for non-celiacs fad <a href=\"http://www.georgeinstitute.org.au/media-releases/dont-believe-the-hype-on-gluten-free-food\">generally irritating</a> but what really made this annoying was what foods they chose to serve. If I say to you &quot;okay we need a vegetarian meal&quot; you try to think of foods that don&#39;t involve meat, you don&#39;t immediately think of a meat dish and substitute tofu in. Similarly, why does your &#39;gluten free&#39; menu consist almost entirely of gluten-free pasta? I wound up grabbing lunch elsewhere for 3 of the 5 days I was at the conference. This was frustrating, because I convinced my employer to send me to NY by arguing that meals were mostly included, but then I wound up having to pay for most meals.</p> <p>Aside from workshops, it was very rare to find tables in the sessions - usually it was just rows of seats. And the walls were those sliding accordion walls that hotels use to divide huge rooms into sections. The end result being, it was very hard to find a", "chocolate, rice) cannot be frequent either, even though there are no numbers for it. With the same reasoning, the itemsets (milk, noodles, rice) and (noodles, chocolate, rice) are also dropped because the itemset (noodles, rice) does not occur often enough. The last remaining itemset can now be used to derive association rules using confidence. Step 6: We can check the following association rules: (Milk, Chocolate) -> Noodles: $\\text{Confidence (Milk, Chocolate)} = \\frac{\\text{Support (Milk, Chocolate, Noodles)}}{\\text{Support(Milk, Chocolate)}} = \\frac{3}{4} = 75 %$ (Milk, Noodles) -> Chocolate: $\\text{Confidence (Milk, Noodles)} = \\frac{\\text{Support (Milk, Chocolate, Noodles)}}{\\text{Support(Milk, Noodles)}} = \\frac{3}{3} = 100 %$ (Chocolate, Noodles) -> Milk: $\\text{Confidence (Chocolate, Noodles)} = \\frac{\\text{Support (Milk, Chocolate, Noodles)}}{\\text{Support(Chocolate, Noodles)}} = \\frac{3}{4} = 75 %$ (Milk) -> (Chocolate, Noodles): $\\text{Confidence (Milk)} = \\frac{\\text{Support (Milk, Chocolate, Noodles)}}{\\text{Support(Milk)}} = \\frac{3}{4} = 75 %$ (Chocolate) -> (Milk, Noodles): $\\text{Confidence (Chocolate)} = \\frac{\\text{Support (Milk, Chocolate, Noodles)}}{\\text{Support(Chocolate)}} = \\frac{3}{5} = 60 %$ Noodles -> (Milk, Chocolate): $\\text{Confidence (Noodles)} = \\frac{\\text{Support (Milk, Chocolate, Noodles)}}{\\text{Support(Noodles)}} = \\frac{3}{4} = 75 %$ After running the Apriori algorithm, a total of five association rules emerge that withstand our confidence level of 70%. These include the rule “(milk, chocolate) -> (noodles)”. This means that if milk and chocolate have already been purchased, then the purchase of noodles is also very likely. ### What are the Advantages and", "know that too so I can go elsewhere. Whether a dietary restriction is a health issue, a religious issue, or a preference, restaurants should be honest about whether they can meet a customer's needs. A vegetarian who expresses that need and is told the restaurant can meet it should expect a vegetarian meal whether or not the restaurant serves meat to those who eat it. January 8, 2013 at 12:50 pm | Reply • OBA When we eat out, we order seafood only, and even this has meat ingredients: cooked in beef or chicken broth; or shrimps cooked with meats, and taken out to serve a particular order, when three options are given. We have seen deep fried fish topped by a sauce that had meat bits: one sauce to serve all. Despite their high prices, many restaurants don't really make things to order but pass ready mix stuff. January 8, 2013 at 6:41 pm | Reply • Fred You're an idiot. January 11, 2013 at 6:07 pm | Reply • Fiona I'm a vegetarian, for reasons as close to my heart as any faith-based dictum. I cannot tell you how many times I've been served meat or food containing meat broth even after I've told the server (or host, in private homes) that I don't eat meat.", "edited by GMATPrepNow on 16 Apr 2018, 12:57, edited 1 time in total. Current Student Joined: 19 Aug 2016 Posts: 146 Location: India GMAT 1: 640 Q47 V31 GPA: 3.82 Re: Of the students who eat in a certain cafeteria, each student either [#permalink] ### Show Tags 22 Apr 2017, 15:50 @[quote=\"Bunuel\"] \"of those who dislike lima beans, 3/5 also dislike brussels sprouts\" means of those who dislike lima $$1-\\frac{3}{5}=\\frac{2}{5}$$ like sprout, or $$\\frac{2}{3}*\\frac{2}{5}=\\frac{4}{15}$$ of total dislike lima but like sprouts. So to calculate # of students who dislike lima but like sprouts we should now total # of students (t). Hi Bunuel, Can you please elaborate this part? I particularly did not understand the 2/3*2/5 part? Why are we doing this? _________________ Consider giving me Kudos if you find my posts useful, challenging and helpful! Math Expert Joined: 02 Sep 2009 Posts: 58381 Re: Of the students who eat in a certain cafeteria, each student either [#permalink] ### Show Tags 23 Apr 2017, 03:32 ashikaverma13 wrote: @ Bunuel wrote: \"of those who dislike lima beans, 3/5 also dislike brussels sprouts\" means of those who dislike lima $$1-\\frac{3}{5}=\\frac{2}{5}$$ like sprout, or $$\\frac{2}{3}*\\frac{2}{5}=\\frac{4}{15}$$ of total dislike lima but like sprouts. So to calculate # of students who dislike lima but like sprouts we should now total # of students", "a vegan dish in a restaurant because the non-vegan dishes were made with fewer restrictions and hence can only be better than the vegan dish. Same for choosing recipes to cook yourself. Before I was vegan, this was my conscious reason for always choosing dishes with meat. But is the deduction true? I don’t think so. Because, unbeknownst to many, the meat dishes are actually constrained to contain meat. I don’t know why, but my two prime suspects are Goodhart’s law impacting the reasoning above, or meat-eaters being scared of vegetables. As it turns out, making a good vegetarian meal takes non-zero skill, contrary to making a good meal with meat. In my experience, this causes chefs to put actual thought into their vegetarian dishes, causing these to actually be tastier than most dishes with meat. So the argument from mathematical optimization actually gives the wrong answer here! ### On rainbow t-shirts and colleagues 215 words Shopping for pride t-shirts is non-trivial. Few pride t-shirts look halfway decent, and few t-shirts look good on my big man-shoulders. Imagine my delight when I found this pretty one at the Human Rights Campaign shop in San Francisco. I bought it back when I was a student a year ago, and I have been wearing it regularly ever since. The simple", "anything strictly halal #> 11 11 11 11 #> strictly kosher strictly other strictly vegan strictly vegetarian #> 11 11 11 11 #> vegan vegetarian <NA> #> 11 11 11 # since skip defaults to TRUE, baking the step has no effect baked_okc <- bake(ds_rec, new_data = okc) table(baked_okc\\$diet, useNA = \"always\") #> #> anything halal kosher mostly anything #> 6174 11 11 16562 #> mostly halal mostly kosher mostly other mostly vegan #> 48 86 1004 335 #> mostly vegetarian other strictly anything strictly halal #> 3438 331 5107 18 #> strictly kosher strictly other strictly vegan strictly vegetarian #> 18 450 227 874 #> vegan vegetarian <NA> #> 136 665 24360" ]

[ "(2016-02-19). \"Every positive integer is a sum of three palindromes\". Mathematics of Computation. arXiv:1602.06208. (arXiv preprint)", "# Palindromic Prime Set Let $$S$$ be the set of palindromic primes of which the sum and product of digits is a prime digit. Find the largest number in the set $$S$$. ×", "most common? ________. Which are the two least common? ________ and _________. Subjects: 1 , 2 , 3 , 4 , 5 Feb 23, 2012 Jan 16, 2015", "# Palindromic Prime Set Number Theory Level 3 Let $$S$$ be the set of palindromic primes of which the sum and product of digits is a prime digit. Find the largest number in the set $$S$$. ×", "non-prime numbers. There are only three: 1, 4 and 6. Aha! So, not only are all invalid palindromes above 6 odd, they are all prime as well! (Assuming this continues beyond 10,000. I have no proof it does, but so far it works.)", "Algebra Level pending What is the largest 5-digit square palindrome? Details and Assumptions: 1)A palindrome is number that reads the same forwards as it does backwards (i.e. 15451 is a palindrome, but 23916 is not ) Generally it reads: $(\\overline{abc})^2 = \\overline{xyzyx}$ where all digits are not necessarily unique ×", "palindromic integers in base 5) + (the only positive integer that is five times the sum of its digits) • (the only Fibonacci number that is a double of a prime) + (the only prime p such that p! has p digits) − (the only fixed point of look-and-say operation) • (the only number whose concatenation with itself is prime) • (the only positive integer that that differs by 1 from a square and a nonsquare cube) − (the largest number such that its divisors are each 1 less than a prime) • (the smallest evil untouchable number) ***** • (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) • (the integer whose standard Roman numeral representation is alphabetically later than all others) − (the number you get if you divide a three digit number with identical digits by the sum of the digits) • (the largest even integer that is not a sum of two abundant numbers) − (the digit in the first position where e and π have the same digit) • (the number formed by", "fair, two-sided coin? As we did in class. Represent P as a binary number. Whenever", "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.", "# Splitting Into Two Primes Multiplying two large primes is easy. The result is a huge non-prime number. Factoring this number however, is non-trivial. This is the basis behind a lot of Cryptographic algorithms. 999415083136585001 is a product of two primes. These two primes are palindromes. Find the sum of these two palindromes. ×", "# Palindromic Primes What is the number of palindromic primes $$p$$ less than $$10^5$$? Hint: The answer is a prime number. ×", "#### Quantity A The number of primes that are divisible by 9 #### Quantity B The number of primes that are divisible by 19 If a, b, x, and y are positive integers, and $13^a \\times 13^b=(13^x)^y=13^{13}$, what is the average (arithmetic mean) of a, b, x, and y? _____ A rectangle is drawn in a standard xy-coordinate plane. If the sides of the rectangle are not parallel to the axes, what is the product of the slope of the four sides? In a certain state, each license plate consists of either three digits (between 0 and 9, inclusive) followed by two letters or three letters followed by two digits. For example, 055-XY, 123-PP, and AAA-70 are all acceptable plates. How many different license plates can the state issue? _____ n is an even integer #### Quantity A The number of prime factors of n #### Quantity B The number of prime factors of $\\frac{n}{2}$ A positive integer is a palindrome if it reads exactly the same from right to left as it does from left to right. For example, 5 and 66 and 373 are all palindromes. How many palindromes are there between 1 and 1,000, inclusive? _____ Line l passes through points in both quadrants II and III. Which of the following statements are true? Indicate", "there the dance is. -- T.S. Eliot Formerly genxer123 The greatest common factor of positive integers m and n is 12. What is [#permalink] 16 Aug 2017, 13:03 Display posts from previous: Sort by", "Every positive integer is a sum of three palindromes Article by Javier Cilleruelo and Florian Luca and Lewis Baxter In collections: Easily explained, Fun maths facts For integer $g\\ge 5$, we prove that any positive integer can be written as a sum of three palindromes in base $g$. URL: arxiv.org/abs/1602.06208v2 PDF: arxiv.org/pdf/1602.06208v2 Entry: read.somethingorotherwhatever. · tooter · · · A Mastodon instance for maths people. The kind of people who make $\\pi z^2 \\times a$ jokes. Use $ and $ for inline LaTeX, and $ and $ for display mode.", ", is palindromic, and satisfies the condition that every contiguous subsequence with a length greater than is unique. It can be shown that this is the largest possible value of .", "# Don't use google! Algebra Level pending What is the largest 5-digit square palindrome? Details and Assumptions: 1)A palindrome is number that reads the same forwards as it does backwards (i.e. 15451 is a palindrome, but 23916 is not ) Generally it reads: $(\\overline{abc})^2 = \\overline{xyzyx}$ where all digits are not necessarily unique × Problem Loading... Note Loading... Set Loading...", "integer n. Over all positive ... Trigonometry - Let N be a 5-digit palindrome. The probability that N is ... Trigonometry - Let S be the sum of all the possible values of sin x that satisfy... Members", "# 32. Longest Double Palindrome ## I'm a slow walker, but I never walk backwards. We define a palindrome as a sequence of at least one integer that reads the same either forward or backward. For example 1 3 5 3 1 and 1 2 2 1 are both palindromes. To the extreme case 1 is also a palindrome. We also define a double palindrome as the concatenation of two palindromes. For example 1 3 5 3 1 1 2 2 1 is a double palindrome. To the extreme case 1 3 5 3 1 1 is also a double palindromes. Now given a sequence of at least two integers, please find the longest double palindrome. If there are multiple longest double palindromes, print the one that appears last . ## Limits The number of integers in the given sequence is no more than 100. ## Input Format There is one line in the input. The first line has the integers from which we want to find the longest double palindrome. ## Output Format There is one lines in the output. The first line has all the integers of the longest double palindrome. number of ways as in the description. ## Sample Input 7 1 3 5 3 1 1 ## Sample Output 1 3 5 3 1", "# Reading a Palindrome in Several Different Ways Consider the pseudo-palindrome: $WAS IT A CAT I SAW.$ In how many ways can $$WAS IT A CAT I SAW$$ be read if you may start at any $$W$$ and move only right, left, up, or down to adjacent letters? Keep in mind that each case can be counted twice since it is a palindrome and that the same letter can be used more than once in each sequence. ×", "all m descending (c increasing). It is always true for m=1 so it will terminate. » 3 weeks ago, # | 0 Having trouble understanding “E — Level K Palindrome”.What will be a level-2 palindrome obtainable in 6 ops from the “aca a bcb a bab a aac”? (Sample Input 5)" ]

[ "so that place values are reversed. For instance, 12 means “twenty-one.” A five-digit code from Sinistrograde is accidentally interpreted from left to right. If all possible five-digit codes (including zeroes in all positions) are equally likely, what is the probability that the code is in fact interpreted correctly? A. 1/10 B. 1/100 C. 1/1,000 D. 1/10,000 E. 1/100,000 Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION: First, figure out how many possible five-digit codes there are in general. Since there are ten digits (0 through 9) and five different positions, the number of possible codes is 10 × 10 × 10 × 10 × 10, or 10^5 = 100,000. Now, what must be true about five-digit codes that could be interpreted correctly either way (left to right or right to left)? These codes must be palindromes—they must be the same forward and backwards. If you represent each digit with a letter, then the code must be of the form xyzyx. The first and last digits must be the same (x), and the second and fourth digits must be the same (y). The middle digit can be anything. Since you now only can determine three digits independently, you only have 10 × 10 × 10, or 10^3 = 1,000 possible palindromic codes. The chance of choosing such a", "the count $0$. – celtschk Apr 14 '17 at 11:08 • We can rewrite $b^3-2b^2+4b-2$ as $b.b^2-2b^2+3b+b-2=(b-2)b^2+3b+(b-2)$. This shows more clearly that the overtakes happen at $333_5, 434_6, 535_7$ etc. – nickgard Apr 14 '17 at 12:03 We have: $$b^3-2b^2+4b-2=(b-2)b^2+3b+(b-2)$$ and: $$b^3-2b^2+4b-2=(b-1)^3+(b-1)^2+3(b-1)+1$$ For all $b\\in\\mathbb{N}_{\\ge 5}$, let $f(b)$ be the amount of palindromes in base $b$ up to and including $b^3-2b^2+4b-2$ and let $g(b)$ be the amount of palindromes in base $b-1$ up to and including $b^3-2b^2+4b-2$. Now, from the very first 'factorization', we see that: $$f(b)=(b-3)b+4+(b-1)=b^2-2b+3$$ Because, if a palindrome starts with a digit $1\\le k\\le b-3$, we can choose whatever digit we want for the middele one and when the palindrom starts with $b-2$, we have $4$ options for the middle digit. Also, we can choose a total of $b-1$ two digit palindromes. Now for $g(b)$. We split this case up into $4$ digits and $3$ digits. If we have a $4$ digit number, the first digit is a $1$ and the second one is $1$ or $0$, so the only two options option are $1111_{b-1}$ and $1001_{b-1}$. All $3$ digit numbers are possible, a total amount of $(b-2)(b-1)$ numbers (because we can't choose $0$ as the leading digit). So when we also include the two digit palindromes: $$g(b)=2+(b-2)(b-1)+(b-2)=b^2-2b+2=f(b)-1$$ And since $b^3-2b^2+4b-2$ is a palindrome in", "non-prime numbers. There are only three: 1, 4 and 6. Aha! So, not only are all invalid palindromes above 6 odd, they are all prime as well! (Assuming this continues beyond 10,000. I have no proof it does, but so far it works.)", "#### Quantity A The number of primes that are divisible by 9 #### Quantity B The number of primes that are divisible by 19 If a, b, x, and y are positive integers, and $13^a \\times 13^b=(13^x)^y=13^{13}$, what is the average (arithmetic mean) of a, b, x, and y? _____ A rectangle is drawn in a standard xy-coordinate plane. If the sides of the rectangle are not parallel to the axes, what is the product of the slope of the four sides? In a certain state, each license plate consists of either three digits (between 0 and 9, inclusive) followed by two letters or three letters followed by two digits. For example, 055-XY, 123-PP, and AAA-70 are all acceptable plates. How many different license plates can the state issue? _____ n is an even integer #### Quantity A The number of prime factors of n #### Quantity B The number of prime factors of $\\frac{n}{2}$ A positive integer is a palindrome if it reads exactly the same from right to left as it does from left to right. For example, 5 and 66 and 373 are all palindromes. How many palindromes are there between 1 and 1,000, inclusive? _____ Line l passes through points in both quadrants II and III. Which of the following statements are true? Indicate", "know $$A+C \\ge 9$$ from the highest digit and $$A+C$$ ends in 1 from the lowest digit, so $$A+C=11$$. Then the minimum possible sum is 1111 where $$B=D=0$$, which happens to be a palindrome. One possible sum is $$505 + 606 = 1111$$. The only other sum that meets the condition is 1221, where $$B+D = 11$$.", "questions like “what five-digit palindrome starting with 28 is divisible by 11?” There is at most one $(2k+1)$-digit palindrome with any first $k$-digit number which is divisible by 11. Consider the numbers $28082, 28182, \\ldots, 28982.$ Each of these differs by 100 from the last one, which is one more than a multiple of 11, so when we reduce mod 11 we get $k, k+1, \\ldots, k+9$ which are all different. Unless 28082 happens to be one more than a multiple of 11, one of these is a multiple of 11. As it turns out, 28082 is one {\\it less} than a multiple of 11, and 28182 is a multiple of 11: $28182 = 11 \\times 2562$. So there is either zero or one palindrome of each of the forms $10x01, 11x11, 12x21, \\ldots, 99x99$. When is there no such palindrome? Observe that, for example, $12000 - 21 = (10000 + 2000) - (20 + 1) = (10000 - 1) + 10 \\times 2 \\times (100 - 1)$$ and so 12000 and 21 differ by a multiple of 11. So $12x21$ and $24x00$ are congruent mod 11. To get $24x00$ to be a multiple of 11 we need $24x$ (that is, $240 + x$) to be a multiple of 11 – so we take $x = 2$. Indeed,", "# Reading a Palindrome in Several Different Ways Consider the pseudo-palindrome: $WAS IT A CAT I SAW.$ In how many ways can $$WAS IT A CAT I SAW$$ be read if you may start at any $$W$$ and move only right, left, up, or down to adjacent letters? Keep in mind that each case can be counted twice since it is a palindrome and that the same letter can be used more than once in each sequence. ×", "positions) are equally likely, what is the probability that the code is in fact interpreted correctly? A. 1/10 B. 1/100 C. 1/1,000 D. 1/10,000 E. 1/100,000 Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION: First, figure out how many possible five-digit codes there are in general. Since there are ten digits (0 through 9) and five different positions, the number of possible codes is 10 × 10 × 10 × 10 × 10, or 10^5 = 100,000. Now, what must be true about five-digit codes that could be interpreted correctly either way (left to right or right to left)? These codes must be palindromes—they must be the same forward and backwards. If you represent each digit with a letter, then the code must be of the form xyzyx. The first and last digits must be the same (x), and the second and fourth digits must be the same (y). The middle digit can be anything. Since you now only can determine three digits independently, you only have 10 × 10 × 10, or 10^3 = 1,000 possible palindromic codes. The chance of choosing such a code at random is 1,000/100,000, or 1/100. The correct answer is B. Dear Bunuel Pl make the change to OA as it shows \"C\" as the correct answer. ____ Done. Thank you.", "there are $90$ palindromes. You can choose the first digit $9$ ways and the middle digit $10$ ways. Generally, for $n$ digit numbers there are $$\\begin {cases} 9\\cdot 10^{\\frac {n-2}2} & n \\text { even} \\\\ 9\\cdot 10^{\\frac {n-1}2} & n \\text { odd} \\end {cases}$$ palindromes. Again, you can choose the first digit $9$ ways and the rest of the first half of the number (rounded up for odd numbers of digits) $10$ ways . To get $66$ with reverse and add, you can have $15,24,33,42,51$ as starting numbers. For $5556555$ you can certainly have five choices $(1-5)$ for the first digit, six $(0-5)$for the next two, and one choice $(3)$for the middle digit. Then the lower three digits are determined by the top three. This gives $180$ numbers. There might be more, as I have avoided carrying. - I believe if you ever have some carrying in the summation, then the resulting number won't be palindromic. Yet to be proven though :) – Dolma Apr 11 '13 at 17:09 You think it is possible to make the following statement: In the interval [a,b] we have a specific number of palindroms, what can we say about the numbers which appear when we reverse the reverse-and-add algorithm. – Babla Apr 11 '13 at 17:10 For exmaple: There are", "same right side up or upside down. ALL of the factors of 808 are also palindromes, and four of them are strobogrammatic numbers, too. Can you figure out which ones are both? • 808 is a composite number. • Prime factorization: 808 = 2 x 2 x 2 x 101, which can be written 808 = (2^3) x 101 • The exponents in the prime factorization are 3 and 1. Adding one to each and multiplying we get (3 + 1)(1 + 1) = 4 x 2 = 8. Therefore 808 has exactly 8 factors. • Factors of 808: 1, 2, 4, 8, 101, 202, 404, 808 • Factor pairs: 808 = 1 x 808, 2 x 404, 4 x 202, or 8 x 101 • Taking the factor pair with the largest square number factor, we get √808 = (√4)(√202) = 2√202 ≈ 428.425340807 Here are the factors that make puzzle #808 act like a multiplication table. It is followed by a table of logical steps to arrive at that solution. ### 807 and Level 1 What can I say about the number 807? 807 is palindrome 151 in BASE 26 because 1(26²) + 5(26) + 1(1) = 807. Anything else? Well, I can figure out a few other things because 807’s has two prime factors, 3", "<< problem 34 - Digit factorials Double-base palindromes - problem 36 >> Problem 35: Circular primes The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime. There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97. How many circular primes are there below one million? Algorithm First, a a standard prime sieve finds all prime numbers up to our limit (1000000 by default) and keeps them in a std::set. Then each prime x in std::set is rotated by one digit to the right: 1. get the right-most digit: auto digit = rotated % 10; 2. move all digits by one digit to the right (\"erasing\" the right-most digit): rotated /= 10; 3. prepend the right-most digit: rotated += digit * shift; 4. check whether rotated is part of our std::set, too 5. if rotated is equal to our initial value x then we checked all rotations The only point of interest is shift which is a power of 10 such that 10^a = shift <= x <= 10^{a+1}. E.g., if x = 3456 then shift = 1000. Modifications by HackerRank We have to find the sum of all such prime numbers, not their count. Note There", "# Difference between revisions of \"2019 AMC 8 Problems/Problem 13\" ## 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 of three distinct two-digit palindromes. What is the sum of the digits of $N$? $\\textbf{(A) }2\\qquad\\textbf{(B) }3\\qquad\\textbf{(C) }4\\qquad\\textbf{(D) }5\\qquad\\textbf{(E) }6$ ## Solution 1 Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \\ldots, 99$. Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$. Then $N = 110$, and the sum of the digits of $N$ is $1+1+0 = \\boxed{\\textbf{(A) }2}$.", "odd. (5) • 12. The sum of 2D, 25D and 19A. (5) • 13. A multiple of 1D whose first two digits are the same. (7) • 15. Less than 1A. (2) • 16. A palindrome. (7) • 17. 20 more than 17D. (3) • 18. A multiple of 3. (2) • 19. A palindrome. (3) • 20. This number and 17D are coprime. (4) • 21. A prime number of the form $n^n+1$ for some integer $n$. (3) • 22. This number and 20A are coprime. (3) • 23. A prime number that is 16 more than 3 times 5D. (4) • 25. A multiple of 10. (3) • 26. Less than 1A. (2) • 27. A factor of 24D. (3) • 28. A seven-digit number. (7) • 30. The highest common factor of 17A and 20A. (2) • 31. A multiple of 32A. (7) • 32. The first digit of this number is the first digit of 33D. (5) • 34. A number $k$ such that $k\\times2^n+1$ is not prime for any integer $n>0$. (5) • 36. A prime number that is 16 more than 3 times 20D. (8) • 37. Not a palindrome. (2) Down • 1. The first two digits of 11D. (2) • 2. A palindrome. (5) • 3. A prime number that", "prime or composite. $1$, $11$, and $121$ are all divisible by the number $121$. Exactly two factors are required for a number to be categorized as a prime number. $121$ is not a prime number because it has more than two factors, including $1$, $11$, and $121$. Because $1$, $11$, and $121$ are more than two factors, $121$. In other words, the fact that $121$ has more than two factors makes it a composite number. 2. Is 121 a perfect square? If yes, find its square root. By prime factorization of $121$, we get: $121=11\\times 11$. From this, we see that the factor $11$ occurs $2$ times i.e., an even number of times. So, $121$ is a perfect square. To find the square root of $121$, we take one factor from each pair and calculate their product. Here, the factor $11$ occurs two times i.e., one pair of the factor $11$. So, the square root of $121$ is $11$. 3. What characterizes 121 as a palindrome? Palindromes are numbers that remain the same when their digits are switched around. A number is a palindrome, in other words, the digits appear the same when read from right to left or left to right. Palindromic numbers include, for instance, 121, 99, 2332, etc. Because it appears the same both forward", "Difference between revisions of \"2019 AMC 8 Problems/Problem 13\" 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 of three distinct two-digit palindromes. What is the sum of the digits of $N$? $\\textbf{(A) }2\\qquad\\textbf{(B) }3\\qquad\\textbf{(C) }4\\qquad\\textbf{(D) }5\\qquad\\textbf{(E) }6$ Solution 1 All the two digit palindromes are multiples of 11. The least 3 digit integer that is the sum of 2 two digit integers is a multiple of 11. The least 3 digit multiple of 11 is 110. The sum of the digits of 110 is 1 + 1 + 0 = $\\boxed{\\textbf{(A)}\\ 2}$. ~heeeeeeheeeee Solution 2 We let the two digit palindromes be $AA$, $BB$, and $CC$, which sum to $11(A+B+C)$. Now, we can let $A+B+C=k$. This means we are looking for the smallest $k$ such that $11k>100$ and $11k$ is not a palindrome. Thus, we test $10$ for $k$, which works so $11k=110$, meaning that the sum requested is $1+1+0=\\boxed{\\textbf{(A)}\\ 2}$. ~smartninja2000", "# Palindromic Prime Set Let $$S$$ be the set of palindromic primes of which the sum and product of digits is a prime digit. Find the largest number in the set $$S$$. ×", "powers of 11 yield palindromic numbers: 111 = 11, 112 = 121, 113 = 1331, and 114 = 14641. 11 is the 11th index or member in the sequence of palindromic numbers, and 121, equal to 11 x 11, is the 22nd.[24] The factorial of 11, 11! = 39916800, has about a 0.2% difference to the round number 4 x 107, or 40 million. Among the first 100 factorials, the next closest to a round number is 96! ~ 9.91678 x 10149, which is about 0.8% less than 10149.[25] If a number is divisible by 11, reversing its digits will result in another multiple of 11. As long as no two adjacent digits of a number added together exceed 9, then multiplying the number by 11, reversing the digits of the product, and dividing that new number by 11 will yield a number that is the reverse of the original number; as in: 142,312 × 11 = 1,565,432 → 2,345,651 ÷ 11 = 213,241. #### Divisibility tests A simple test to determine whether an integer is divisible by 11 is to take every digit of the number in an odd position and add them, then take the remaining digits and add them. If the difference between the two sums is a multiple of 11, including 0, then the", "= 7 7 = 1 + 5 + 1 7 = 2 + 3 + 2 7 = 3 + 1 + 3 7 = 1 + 1 + 3 + 1 + 1 7 = 2 + 1 + 1 + 1 + 2 7 = 1 + 2 + 1 + 2 + 1 7 = 1 + 1 + 1 + 1 + 1 + 1 + 1 In total, there are 8 palindromic partitions.", "prime factors of the number fifteen are three and five, as 3 x 5 = 15 and both of these factors are prime. A palindromic number (also known as a numeral palindrome or a numeric palindrome) is a number that remains the same when its digits are reversed. Any Integer number is equal to the Product of its Factors. So, 4 is a factor of 20 as it divides exactly into 20. Use commas in a for statement to find the largest and smallest factor of a number: 4. How to check if a number is prime or not? Brute Force Python Implementation to check if a number is. primefac version 1. Here’s a simple Python program that decides whether or not a given positive number p is prime: def isprime(p): n = 2 while n*n <= p: if p % n == 0: return False else: n = n+1 return True If you like, you can try this code on a computer. The following python program computes and prints all the positive factors of a given input number. A number is a prime number if it hasn't any factor between 2 and the square root of itself and this is exactly what we are checking. Write A Python Program To List All Prime Factors Of A Number", "read same when read from left to right or right to left. Thus 121 is a palindrome. Other examples of palindromes are 343, 521125, 999999 etc. // Updated: April 5th, 2014 #### Gaurav Tiwari A designer by profession, a mathematician by education but a Blogger by hobby. Loves reading and writing. Just that." ]

[ "#### Quantity A The number of primes that are divisible by 9 #### Quantity B The number of primes that are divisible by 19 If a, b, x, and y are positive integers, and $13^a \\times 13^b=(13^x)^y=13^{13}$, what is the average (arithmetic mean) of a, b, x, and y? _____ A rectangle is drawn in a standard xy-coordinate plane. If the sides of the rectangle are not parallel to the axes, what is the product of the slope of the four sides? In a certain state, each license plate consists of either three digits (between 0 and 9, inclusive) followed by two letters or three letters followed by two digits. For example, 055-XY, 123-PP, and AAA-70 are all acceptable plates. How many different license plates can the state issue? _____ n is an even integer #### Quantity A The number of prime factors of n #### Quantity B The number of prime factors of $\\frac{n}{2}$ A positive integer is a palindrome if it reads exactly the same from right to left as it does from left to right. For example, 5 and 66 and 373 are all palindromes. How many palindromes are there between 1 and 1,000, inclusive? _____ Line l passes through points in both quadrants II and III. Which of the following statements are true? Indicate", "# 考满分网 #### Quantity A The number of primes that are divisible by 9 #### Quantity B The number of primes that are divisible by 19 If a, b, x, and y are positive integers, and $13^a*13^b=(13^x)^y=13^{13}$, what is the average (arithmetic mean) of a, b, x, and y? _____ A rectangle is drawn in a standard xy-coordinate plane. If the sides of the rectangle are not parallel to the axes, what is the product of the slope of the four sides? In a certain state, each license plate consists of either three digits (between 0 and 9, inclusive) followed by two letters or three letters followed by two digits. For example, 055-XY, 123-PP, and AAA-70 are all acceptable plates. How many different license plates can the state issue? _____ n is an even integer #### Quantity A The number of prime factors of n #### Quantity B The number of prime factors of $\\frac{n}{2}$ A positive integer is a palindrome if it reads exactly the same from right to left as it does from left to right. For example, 5 and 66 and 373 are all palindromes. How many palindromes are there between 1 and 1,000, inclusive? _____ Line l passes through points in both quadrants II and III. Which of the following statements are true? Indicate", "there are $90$ palindromes. You can choose the first digit $9$ ways and the middle digit $10$ ways. Generally, for $n$ digit numbers there are $$\\begin {cases} 9\\cdot 10^{\\frac {n-2}2} & n \\text { even} \\\\ 9\\cdot 10^{\\frac {n-1}2} & n \\text { odd} \\end {cases}$$ palindromes. Again, you can choose the first digit $9$ ways and the rest of the first half of the number (rounded up for odd numbers of digits) $10$ ways . To get $66$ with reverse and add, you can have $15,24,33,42,51$ as starting numbers. For $5556555$ you can certainly have five choices $(1-5)$ for the first digit, six $(0-5)$for the next two, and one choice $(3)$for the middle digit. Then the lower three digits are determined by the top three. This gives $180$ numbers. There might be more, as I have avoided carrying. - I believe if you ever have some carrying in the summation, then the resulting number won't be palindromic. Yet to be proven though :) – Dolma Apr 11 '13 at 17:09 You think it is possible to make the following statement: In the interval [a,b] we have a specific number of palindroms, what can we say about the numbers which appear when we reverse the reverse-and-add algorithm. – Babla Apr 11 '13 at 17:10 For exmaple: There are", "know $$A+C \\ge 9$$ from the highest digit and $$A+C$$ ends in 1 from the lowest digit, so $$A+C=11$$. Then the minimum possible sum is 1111 where $$B=D=0$$, which happens to be a palindrome. One possible sum is $$505 + 606 = 1111$$. The only other sum that meets the condition is 1221, where $$B+D = 11$$.", "is a curious subject for investigation -- the search for square numbers the figures of which read backwards and forwards alike. Some of them are very easily found. For example, the squares of $1$, $11$, $111$ and $1111$ are respectively $1$, $121$, $12321$, and $1234321$, all palindromes, and the rule applies for any number of $1$'s provided the number does not contain more than nine. But there are other cases that we may call irregular, such as the square of $264 = 69696$ and the square of $2285 = 5221225$. Now, all the examples I have given contain an odd number of digits. Can the reader find a case where the square palindrome contains an even number of figures? #### $62$ - Factorizing What are the factors (the numbers that will divide it without any remainder) of this number -- $1000000000001$? This is easily done if you happen to know something about numbers of this peculiar form. In fact, it is just as easy for me to give two factors if you insert, say $101$ noughts, instead of $11$, between the two ones. There is a curious, easy, and beautiful rule for these cases. Can you find it? #### $63$ - Find the Factors Find $2$ whole numbers with the smallest possible difference between them which, when multiplied", "same right side up or upside down. ALL of the factors of 808 are also palindromes, and four of them are strobogrammatic numbers, too. Can you figure out which ones are both? • 808 is a composite number. • Prime factorization: 808 = 2 x 2 x 2 x 101, which can be written 808 = (2^3) x 101 • The exponents in the prime factorization are 3 and 1. Adding one to each and multiplying we get (3 + 1)(1 + 1) = 4 x 2 = 8. Therefore 808 has exactly 8 factors. • Factors of 808: 1, 2, 4, 8, 101, 202, 404, 808 • Factor pairs: 808 = 1 x 808, 2 x 404, 4 x 202, or 8 x 101 • Taking the factor pair with the largest square number factor, we get √808 = (√4)(√202) = 2√202 ≈ 428.425340807 Here are the factors that make puzzle #808 act like a multiplication table. It is followed by a table of logical steps to arrive at that solution. ### 807 and Level 1 What can I say about the number 807? 807 is palindrome 151 in BASE 26 because 1(26²) + 5(26) + 1(1) = 807. Anything else? Well, I can figure out a few other things because 807’s has two prime factors, 3", "# 32. Longest Double Palindrome ## I'm a slow walker, but I never walk backwards. We define a palindrome as a sequence of at least one integer that reads the same either forward or backward. For example 1 3 5 3 1 and 1 2 2 1 are both palindromes. To the extreme case 1 is also a palindrome. We also define a double palindrome as the concatenation of two palindromes. For example 1 3 5 3 1 1 2 2 1 is a double palindrome. To the extreme case 1 3 5 3 1 1 is also a double palindromes. Now given a sequence of at least two integers, please find the longest double palindrome. If there are multiple longest double palindromes, print the one that appears last . ## Limits The number of integers in the given sequence is no more than 100. ## Input Format There is one line in the input. The first line has the integers from which we want to find the longest double palindrome. ## Output Format There is one lines in the output. The first line has all the integers of the longest double palindrome. number of ways as in the description. ## Sample Input 7 1 3 5 3 1 1 ## Sample Output 1 3 5 3 1", "non-prime numbers. There are only three: 1, 4 and 6. Aha! So, not only are all invalid palindromes above 6 odd, they are all prime as well! (Assuming this continues beyond 10,000. I have no proof it does, but so far it works.)", "on the number of digits of the two factors: 1: 3 * 3 = 9 2: 91 * 99 = 9009 3: 913 * 993 = 906609 4: 9901 * 9999 = 99000099 5: 99681 * 99979 = 9966006699 6: 999001 * 999999 = 999000000999 7: 9997647 * 9998017 = 99956644665999 8: 99990001 * 99999999 = 9999000000009999 This points out a few things. First of all, my second solution, as posted above, actually fails on the first input. That’s because there’s a built-in assumption there that the maximum palindromic product of two $n$ digit numbers will be a $2n$ digit number. There also seems to be a pattern to the factors, at least for even inputs – the larger factor is a string of 9s. Does that continue? Apparently NOT! Here’s some more values (these took a while – more on performance later): 9: 999920317 * 999980347 = 999900665566009999 10: 9999986701 * 9999996699 = 99999834000043899999 11: 99999943851 * 99999996349 = 9999994020000204999999 With these values determined, one can go back and make another improvement or two (which we could probably have guessed at the beginning), if we allow ourselves some assumptions. Specifically, we now expect that the biggest palindrome will begin with a 9, and therefore end with a 9. Since we’re looking at its factors, we can rule", "mostly because my iPod had packed up on me, leaving me with a good half an hour with nothing better to do. A palindrome is a word like LEVEL, which reads the same forwards and backwards. Specifically, I was thinking about 5-digit numbers, so my palindromes were things like 01410. I’d seen one like this at work the previous day, which got me thinking-what were the chances of that happening? This turns out not to be a very interesting question to ask—for a random 5-digit number, the answer is simply 1 in 100. The first 3 digits can be whatever they like, and then the remaining two must match up with a 1-in-10 chance for each. # Primes à la Euler Anyone with a more than a passing interest in numbers knows that there are infinitely many primes.If we highlight them in a grid like this, it is clear that the last digit of any prime must be either 1, 3, 7 or 9 (except for 2 and 5 of course). But there seems to be no good reason for primes to prefer any one of these four possibilities over any other. It is natural therefore to conjecture that There are infinitely many primes whose last digit is $b$ for $b$ equal to any of 1, 3, 7,", "# Splitting Into Two Primes Multiplying two large primes is easy. The result is a huge non-prime number. Factoring this number however, is non-trivial. This is the basis behind a lot of Cryptographic algorithms. 999415083136585001 is a product of two primes. These two primes are palindromes. Find the sum of these two palindromes. ×", "# Palindromic Prime Set Let $$S$$ be the set of palindromic primes of which the sum and product of digits is a prime digit. Find the largest number in the set $$S$$. ×", "# Reading a Palindrome in Several Different Ways Consider the pseudo-palindrome: $WAS IT A CAT I SAW.$ In how many ways can $$WAS IT A CAT I SAW$$ be read if you may start at any $$W$$ and move only right, left, up, or down to adjacent letters? Keep in mind that each case can be counted twice since it is a palindrome and that the same letter can be used more than once in each sequence. ×", "odd. (5) • 12. The sum of 2D, 25D and 19A. (5) • 13. A multiple of 1D whose first two digits are the same. (7) • 15. Less than 1A. (2) • 16. A palindrome. (7) • 17. 20 more than 17D. (3) • 18. A multiple of 3. (2) • 19. A palindrome. (3) • 20. This number and 17D are coprime. (4) • 21. A prime number of the form $n^n+1$ for some integer $n$. (3) • 22. This number and 20A are coprime. (3) • 23. A prime number that is 16 more than 3 times 5D. (4) • 25. A multiple of 10. (3) • 26. Less than 1A. (2) • 27. A factor of 24D. (3) • 28. A seven-digit number. (7) • 30. The highest common factor of 17A and 20A. (2) • 31. A multiple of 32A. (7) • 32. The first digit of this number is the first digit of 33D. (5) • 34. A number $k$ such that $k\\times2^n+1$ is not prime for any integer $n>0$. (5) • 36. A prime number that is 16 more than 3 times 20D. (8) • 37. Not a palindrome. (2) Down • 1. The first two digits of 11D. (2) • 2. A palindrome. (5) • 3. A prime number that", "the count $0$. – celtschk Apr 14 '17 at 11:08 • We can rewrite $b^3-2b^2+4b-2$ as $b.b^2-2b^2+3b+b-2=(b-2)b^2+3b+(b-2)$. This shows more clearly that the overtakes happen at $333_5, 434_6, 535_7$ etc. – nickgard Apr 14 '17 at 12:03 We have: $$b^3-2b^2+4b-2=(b-2)b^2+3b+(b-2)$$ and: $$b^3-2b^2+4b-2=(b-1)^3+(b-1)^2+3(b-1)+1$$ For all $b\\in\\mathbb{N}_{\\ge 5}$, let $f(b)$ be the amount of palindromes in base $b$ up to and including $b^3-2b^2+4b-2$ and let $g(b)$ be the amount of palindromes in base $b-1$ up to and including $b^3-2b^2+4b-2$. Now, from the very first 'factorization', we see that: $$f(b)=(b-3)b+4+(b-1)=b^2-2b+3$$ Because, if a palindrome starts with a digit $1\\le k\\le b-3$, we can choose whatever digit we want for the middele one and when the palindrom starts with $b-2$, we have $4$ options for the middle digit. Also, we can choose a total of $b-1$ two digit palindromes. Now for $g(b)$. We split this case up into $4$ digits and $3$ digits. If we have a $4$ digit number, the first digit is a $1$ and the second one is $1$ or $0$, so the only two options option are $1111_{b-1}$ and $1001_{b-1}$. All $3$ digit numbers are possible, a total amount of $(b-2)(b-1)$ numbers (because we can't choose $0$ as the leading digit). So when we also include the two digit palindromes: $$g(b)=2+(b-2)(b-1)+(b-2)=b^2-2b+2=f(b)-1$$ And since $b^3-2b^2+4b-2$ is a palindrome in", "# 考满分网 #### Quantity A The number of primes that are divisible by 9 #### Quantity B The number of primes that are divisible by 19 If a, b, x, and y are positive integers, and $13^a×13^b=(13^x)^y=13^{13}$, what is the average (arithmetic mean) of a, b, x, and y? _____ A rectangle is drawn in a standard xy-coordinate plane. If the sides of the rectangle are not parallel to the axes, what is the product of the slope of the four sides? In a certain state, each license plate consists of either three digits (between 0 and 9, inclusive) followed by two letters or three letters followed by two digits. For example, 055-XY, 123-PP, and AAA-70 are all acceptable plates. How many different license plates can the state issue? _____ n is an even integer #### Quantity A The number of prime factors of n #### Quantity B The number of prime factors of $\\frac{n}{2}$ A positive integer is a palindrome if it reads exactly the same from right to left as it does from left to right. For example, 5 and 66 and 373 are all palindromes. How many palindromes are there between 1 and 1,000, inclusive? _____ Line l passes through points in both quadrants II and III. Which of the following statements are true? Indicate", "111 views A palindrome is a string whose reversal is identical to the string. How many bit strings of length n are palindromes? 1. 2⌈n⁄2⌉ 2. 2(⌊ n/2⌋ ) 3. 2⌈n⁄2⌉ -1 4. 2(⌊ n/2⌋) -1 Try with N= 2 you get aa,bb 2⌈n⁄2⌉ = 2 Try with N= 3 you get aaa,bbb ,aba , bab 2⌈n⁄2⌉ = 4 Try with N= 4 you get aaaa,bbbb ,abba , baab , 2⌈n⁄2⌉ = 4", "# 9 hidden in every number regardless of number of digits? Let us consider a number, e.g. 1234 Now reverse the positions of the terminal digits, so we get 4231 4231-1234=2997 which is divisible by 9 i have seen this for n-digit numbers, where n ranges from 2 to 6 Is there any number which is contradicting this behavior irrespective of its number of digits? Now coming back to the example, Now 2997/9=333 is a palindrome Another example 923456781-123456789=799999992 799999992/9=88888888 Is this also independent of the number of digits of the original number? Let us consider : 95623-35629=59994/9=6666 The largest prime factor of 6666 is 101 which is a palindrome and a prime number. Now 6666/101=66 Again 66/11(11 being the largest prime factor of 66) equals to 6 which is a palindrome. Another example : 923456781-123456789=799999992/9=88888888 The largest prime factor of 88888888 is 137 88888888/137=648824/101(101 being largest prime factor of 648824)=6424/73(73 being largest prime factor of 6424) this in turn equals to 88 (a palindrome) Finally, 88/11(11 being largest prime factor of 88 and a palindrome) equals to 8 which is also a palindrome. Now as per (http://math.stackexchange.com/questions/200835/are-there-infinitely-many-super-palindromes), will it be correct to call 88888888 a super palindrome? Can it be possible to call a number a super palindrome if it generates non palindromes along with palindromes, provided", "# Palindromic Primes What is the number of palindromic primes $$p$$ less than $$10^5$$? Hint: The answer is a prime number. ×", "(when given integer inputs) is prime, and furthermore, every prime is obtained in this way. (Note: $$Z$$ spits out many non-positive values as well.) Unfortunately, this polynomial $$Z$$ is not very practical. For one thing, its degree is about $$38!$$. (Yes, I mean 38 factorial.) If we consider digits instead of letters or words, then numbers such as 676, 100020001, and 12345678987654321 are palindromes. These are in fact quite special palindromes, as they are all squares of integers: $$26^2 = 676$$, $$111111111^2 = 12345678987654321$$, and $$10001^2 = 100020001$$. Are there infinitely many palindromic squares? Yes: the last example above hints that $$10\\ldots01^2 = 10\\ldots020\\ldots01$$ for however many 0s you put in the middle. By contrast, if we ask for $$n^5$$ to be palindromic (or any higher power), then no solutions are currently known! You might ask for other types of palindromic numbers, such as palindromic primes. A few examples include 7 (this is a palindrome!), 101, 19391, and 1000000008000000001. Are there infinitely many? We don’t know. But the largest palindromic prime currently known is $$10^{200000} + 47960506974 \\cdot 10^{99995} + 1$$, according to Wikipedia." ]

[ "# Divisibility by $$15$$. Find the smallest positive multiple of $$15$$ such that each of the digit of the multiple is either 0 or 8. ×", "smallest positive multiple of 7 that satisfies the 4 congruences is 364: The LCM of 90m + 4, where m=0, 1, 2, 3.....etc. Since the smallest multiples of 7 that satisfies the 4 congruences is: 4 x 90 + 4 = 364. 364 mod 6 = 4 364 mod 9 = 4 364 mod 15 = 4 364 mod 18 = 4 Dec 3, 2019 edited by Guest Dec 4, 2019", "## Algebra 1 -4.5, 1.75, $\\sqrt 4$, $14^1$ Mental math tells us that the only negative number, -4.5, is the smallest. Mental math also tells us that $14^1$, which equals 14, is the largest. Convert $\\sqrt 4$ to a decimal to compare against 1.75. $\\sqrt 4=2$ $1.75\\lt2$. Therefore $1.75\\lt\\sqrt 4$.", "# Divisibility by $$15$$. Number Theory Level 2 Find the smallest positive multiple of $$15$$ such that each of the digit of the multiple is either 0 or 8. ×", "$x$ is the smallest positive number. Clearly $x\\le 1$ and also $x^2\\gt 0$. Since $x$ is the smallest positive number, $x^2$ can't be smaller then $x$, so we must have $x^2\\geq x$. We can divide both sides of this by the positive number $x$ to get $x \\geq 1$. Since $x$ is both less than or equal to $1$ and greater than or equal to $1$, $x$ must equal $1$. Thus the smallest positive number is $1$. Clare wrote: There is no smallest positive number. If $0\\lt x\\lt1,$ then we know $0\\lt x^2 \\lt x\\times1 = x.$ Therefore for any positive number $x$ below $1$ that you might think is the smallest positive number, there is always a smaller positive number $x^2.$ $\\frac x2$ will also be a positve number less than $x.$ So we can never find the smallest positive number - it doesn't exist. This means that proof P2 begins with an invalid assumption, so the proof is invalid. In fact, you could consider P2 to be a proof by contradiction that there is no smallest positive number. It begins by assuming there is a smallest positive number, $x,$ and leads to nonsense. Hans found the problem with P3: ${\\mathbf{1=-1}}$. Click to see the proof. Proof: Clearly, $-1=-1$ and $\\frac{1}{1} = \\frac{-1}{-1}$ Therefore, $-1\\times \\frac{1}{1}=-1\\times \\frac{-1}{-1}$", "$(1+a)^2-x^2 >(1+a)^2-a = 1+a+a^2 >0$, so $x<1+a$. Hence, by completeness of real numbers, there is the smallest positive real number $l$ satisfying $x \\le l$ for all $x\\in A$. Such $l$ can be proved to be $l^2=a$. Uniqueness can be proved as follows: If $l_1>0$ and $l_2>0$ are two real numbers such that $l_1^2=l_2^2=a$, then clearly $l_1=l_2$.", "multiples of 2 and 6. ## Lowest Common Multiple The lowest common multiple of a pair or group of positive numbers, is the smallest positive number that is a multiple of each number. Examples (1.1) Find the lowest common multiple of the numbers 5 and 7. Solution One approach is to list the multiples of each, looking for the smallest shared multiple in both lists. 5 => { 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , …. } 7 => { 7 , 14 , 21 , 28 , 35 , 42 , 49 , 56 , 63 , 70 , …. } It can be seen from the lists that 35 is the lowest common multiple of 5 and 7. (1.2) Find the lowest common multiple of 24 and 10. Solution 2 => { 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , …. } 4 => { 4 , 8 , 12 , 16 , 20 , 24 , 28 , 32 , 40 , 44 , …. } 10 => { 10 , 20 , 30 , 40 , 50 , 60 , 70 , …. } 20", "and 6. ## Least Common Multiple With multiples of numbers, the lowest common multiple of a pair or group of positive numbers, is the smallest positive number that is a multiple of each number. Examples (1.1) Find the lowest common multiple of the numbers 5 and 7 . Solution One approach is to list the multiples of each, looking for the smallest shared multiple in both lists. 5 => { 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , ... } 7 => { 7 , 14 , 21 , 28 , 35 , 42 , 49 , 56 , 63 , 70 , ... } It can be seen from the lists that 35 is the lowest common multiple of 5 and 7. (1.2) Find the lowest common multiple of 2, 4 and 10. Solution 2 => { 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , ... } 4 => { 4 , 8 , 12 , 16 , 20 , 24 , 28 , 32 , 40 , 44 , ... } 10 => { 10 , 20 , 30 , 40 , 50 , 60 , 70 ,", "238 views Directions for the below question: Let $a_1=p$ and $b_1 =q$ where $p$ and $q$ are positive quantities. Define: $a_n pb_{n-1} \\: \\: \\: b_n=qb_{n-1}$ for even $n>1$ and $a_n pa_{n-1} \\: \\: \\: b_n=qa_{n-1}$ for odd $n>1$ If $p=1/3$ and $q= 2/3,$ then what is the smallest odd $n$ such that $a_n + b_n < 0.01$? 1. $7$ 2. $13$ 3. $11$ 4. $9$ 5. $15$ 1 1 vote", "# Find the smallest value of positive integer !! Find the smallest positive integer with exactly 30 positive factor First, I use function $$\\tau$$ to find the exponential that gives $$2×3×5$$ and I want to find the smallest value. How can I find it use inequality to help And how to find the value of $$\\sqrt{8×13×15×17+49}$$? I get $$x=8$$ and doing perfect square ... but stuck who can help me, please • Your question is a bit unclear. Can you give some examples of the smallest value of a positive number with 2,3,4 positive factors? What do you mean with positive factors? – kvantour Jan 31 at 13:11 • What have you tried? Given the primne factorization $n=\\prod p_i^{a_i}$ , do you know how to compute the number of factors of $n$? – lulu Jan 31 at 13:11 • @kvantour To me these feel like two separate questions that should have been asked separately, in my opinion. – Robert Soupe Jan 31 at 18:39 The number of factors of a natural is given by the product of the multiplicities of its prime factors plus one, $$(m_2+1)(m_3+1)(m_5+1)\\cdots$$. To obtain $$30=2\\cdot3\\cdot5$$ factors, you need multiplicities $$1,2$$ and $$4$$, which you will assign to the smallest possible primes, by decreasing order $$5^13^22^4=720.$$ What do you mean by \"exactly 5 positive factors\"?", "note as an example of application that, for every $\\lambda\\leqslant1$, the smallest positive root of $x=\\mathrm e^{ax}$ is $x=\\mathrm e^\\lambda$ hence $$p_n(u)\\approx\\mathrm e^{-n\\lambda}.$$", "of 8 ?) From these tests, it seems the smallest N for which a solution exists is $2^{P+1}$. Can this be proved ? After that, it seems there are only solutions for N multiple of 2 (P= 0), 4 (P = 1 or 2), 8 (P = 3 or 4). Is this a constant depending on P ? Another way to generalize this problem consists in increasing the numbers of groups (for instance consider 3 groups A, B, C).", "most positive. The smallest possible result is when you add the two most negative possible inputs. The largest possible result is when you add the two most positive possible inputs. This should be fairly intuitive. So the two most negative is $$-(2^{N}) + -(2^{N}) = -2 * 2^{N} = -(2^{N+1})$$. The most positive is $$2^{N}-1 + 2^{N}-1 = 2*2^{N}-2 = 2^{N+1}-2$$ That output can only be held in an N+1 bit sized number. This makes sense because one more bit means that it can handle double the size, which is as big as adding two numbers can realize. Multiplication is similar to prove out. The smallest is $$-(2^{N}) * -(2^{N}) = 2^{N+N} = 2^{2N}$$ The largest is $$(2^{N}-1)(2^{N}-1) = 2^{2N} - 2^{N+1} + 1$$ The minimum of the range shows that you need 2*N bits, and the maximum can be held in this as well, so that is the answer. #### VVS Joined Jul 22, 2007 66 thanx to mik3, papabravo and speciall caveman, ur explanation is excellent. one thing i never understood tho is why the maximum positive number in a 2's complement is 2^n - 1. isnt it: 2^0 + 2^1 + 2^2 + 2^3 +........+2^(N-1). If somebody could explain this to me, I would be very thankful. thanks a lot again. VVS #### Papabravo Joined", "# Thread: By using algebraic method, find the value of the smallest value 1. ## By using algebraic method, find the value of the smallest value Question: The subtraction of two positive numbers is $4$. When the two numbers are multiplied together, the resulting value is $21$. By using algebraic method, find the value of the smallest number My workings: let $x$ be the smallest number let $x+4$ be the bigger number $(x+4)-x=4$ $(x+4)$ x $x=21$ $x+4=4+x$ $(x^2+4x)/4 = 21/4$ $4(x^2+4x)=84$ $4x^2+16x=84$ $4x^2+16x-84=0$ $(x+7)$ or $(4x-12)$ $x+7=0$ or $4x-12=0$ $x=-7$ (N.A.) or $4x-12=0$ Therefore, $x=3$ Am I right? Thanks 2. ## Re: By using algebraic method, find the value of the smallest value Originally Posted by FailInMaths Question: The subtraction of two positive numbers is $4$. When the two numbers are multiplied together, the resulting value is $21$. By using algebraic method, find the value of the smallest number My workings: let $x$ be the smallest number let $x+4$ be the bigger number $(x+4)-x=4$ $(x+4)$ x $x=21$ $x+4=4+x$ $(x^2+4x)/4 = 21/4$ $4(x^2+4x)=84$ $4x^2+16x=84$ $4x^2+16x-84=0$ $(x+7)$ or $(4x-12)$ $x+7=0$ or $4x-12=0$ $x=-7$ (N.A.) or $4x-12=0$ Therefore, $x=3$ Am I right? Thanks Plug your numbers into the original equations and see if it works. Given that your smaller number is 3 then your larger number is 7. $7 - 3", "Finding the smallest positive integers that satisfies given equations [closed] Is it possible to find the smallest positive integer/s that satisfy a given equation or some inequality? Example: $2x^2-3x>24$ Is there a formula for this? closed as too broad by Aqua, Guy Fsone, mechanodroid, Arnaud D., kingW3Oct 22 '17 at 14:41 Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. • Start by checking if there is some small solution $a,b$. Then there are only finitely many possibilities to check for a smaller $a+b$. – hardmath Oct 18 '17 at 10:52 So $5|a(a-1)$ and since $a$ and $a-1$ are relatively prime $5$ divides only one of them, say $a$. Since $a_{\\min}=5$ you get $b = 2$ and so $(a+b)_{\\min} = 7$. If $5$ divides $a-1$ then $a_{\\min} = 6$ which is not good. This is a very broad question. As for your specific equation, it's equivalent to $a(a-1) = 10b$ so you're basically looking for the smallest pair of successive numbers whose product is a multiple of 10. You also know that", "3)=1$, but the one provided is the smallest positive value. -", "$$38$$ times larger than (which I interpret as multiply by a factor of $$38$$, not $$39$$) another. I searched for all $$h \\leq 10^5$$ to verify that it is indeed the smallest one.", "s v b) $24$a s v c) $28$a s v d) $30$a s v e) $55$a s v f) $78$a s v ## Question 11 Negative numbers can also be multiples! Using this information find the $11$ multiples of the following numbers nearest $0$ ($5$ multiples above and below $0$): a) $2$a s v b) $5$a s v c) $10$a s v d) $11$a s v e) $9$a s v f) $19$a s v", "is the least common multiple of the $n_i$. -", "arbitrarily large positive by choosing $\\epsilon$ suitably small." ]

[ "The multiples of 12 and of 21 are written in order. What is the smallest positive difference between a multiple of 12 and a multiple of 21? The difference is $$n\\cdot 21 -m \\cdot 12 = d$$ with $$n,m,d \\in Z$$ has a solution, if the $$gcd(21,12)$$ is a divider of $$d$$. Or $$gcd(21,12) | d$$. The greatest common divisor $$gcd(21,12) = 3.$$ The difference is $$n\\cdot gcd(21,12)$$ so the difference is a multiple of 3 The differences are: $$0, 3, 6, 9, 12, \\ldots$$ The smallest positive difference is 3. heureka Dec 4, 2017", "+0 # need help 0 225 1 If m is the smallest positive integer such that m! is a multiple of 4125, and n is the smallest positive integer such that n! is a multiple of 2816000, then find n-m. Jun 16, 2021 I claim that the $n=m=15$, so $n-m=0$. To see that $15!$ is a common multiple of 4125 and 2816000, just check that $15!$ is a multiple of both $4125=3\\cdot 5^3\\cdot 11$ and $2816000=2^{11}\\cdot 5^3\\cdot 11$. To see that $15!$ is the smallest common multiple of 4125 and 2816000, note that $14!$ is not a common multiple of 4125 and 2816000 (because $5^3$ does not divide $14!$ so it can't divide any $x!$ where $x \\le 14$).", "8 and 12 in the list are $24,48,72.$ d. From part c, the smallest multiple that 8 and 12 have in common is 24.", "a multiple of both $6$ and $14$?", "the smallest multiple of 63 and 147", "$14$ of these extra pairs. The answer is $28+21+14 = \\boxed{063}$", "1 Answer Nope, the smallest is $$n=1018976683725=3^3 \\cdot 5^2 \\cdot 7^2 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29,$$ which has $\\frac{\\sigma(n)}{n}=\\frac{34283520}{11350339}\\approx 3.02048423399513$. Just for the record, I found this by using your observation that any candidate must be a multiple of $m:=3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23$. Your answer is $9765m$, and by running a for loop over all smaller (odd) multiples, I came upon the (unique) smaller answer of $n=9135m$. - Thank you. That is clearly smaller than my answer, and also satisfies sigma(n) > 3n. A minor quibble: I proved that any candidate is divisible by all primes up to 23, not necessarily 29. However, even if we redefine m, the same trick of looping through odd multiples of m still verifies that we've found the smallest. – idmercer Feb 21 '12 at 18:01 Oops, good catch. Edited for posterity. – Cam McLeman Feb 21 '12 at 18:09", "# Is greatest common divisor of two numbers really their smallest linear combination? In a lecture note from MIT on number theory says: Theorem 5. The greatest common divisor of a and b is equal to the smallest positive linear combination of a and b. For example, the greatest common divisor of 52 and 44 is 4. And, sure enough, 4 is a linear combination of 52 and 44: 6 · 52 + (−7) 44 = 4 What about 12 and 6 their gcd is 6 but 0 which is less than 6 can be - You wrote it yourself: the gcd is the smallest positive linear combination. – André Nicolas Oct 24 '12 at 7:32 Note that $52 \\times 44 - 44 \\times 52 =0$ and you can do this with any pair of integers. – Mark Bennet Oct 24 '12 at 7:43 @MarkBennet yeah that's right. – spartacus Oct 24 '12 at 20:40 You wrote it yourself: the gcd is the smallest positive linear combination. Smallest positive linear combination is shorthand for smallest positive number which is a linear combination. It is true that $0$ is a linear combination of $12$ and $6$ with integer coefficients, but $0$ is not positive. The proof is not difficult, but it is somewhat lengthy. We give full detail below.", "+0 # Help PLS, ASAP +2 700 1 +140 The least common multiple of two positive integers is $7!$, and their greatest common divisor is $9$. If one of the integers is $315$, then what is the other? (Note that $7!$ means $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1$.) Mar 19, 2019 edited by CorbellaB.15 Mar 19, 2019 #1 0 The identity lcm[a,b]*gcd(a,b)=ab holds for all positive integers a and b. Let a=315 in this identity, and let b be the number we are looking for. Thus $$7! \\cdot 9=315 \\cdot b$$ so $$b = \\frac{7!\\cdot 9}{315} = \\frac{7!\\cdot 9}{35\\cdot 9} = \\frac{7!}{35} = \\frac{{7}\\cdot 6\\cdot {5}\\cdot 4\\cdot 3\\cdot 2\\cdot 1}{{35}} = 6\\cdot4\\cdot3\\cdot2 = \\boxed{144}.$$ Hope this helps :) Sep 7, 2019", "# Interview Question Asked In yahoo Can you find the smallest positive number such that if you shuffle the digits of the number in a particular order, the shuffled number becomes twice the original number. I understand the answer is $125874 => 251748$ $251748$ is twice the $125874$ and have same digits $1,2,4,5,7$ & $8$ but how to solve this non programmatic ? • One way is to play fool and answer it in base $2$. The answer becomes $01$, which turns $10$. ;) – user147444 May 5, 2014 at 17:14 • Or in base 8, where $52_8 = 2\\cdot 25_8$. – MJD May 5, 2014 at 17:19 • It's odd that the answer has the same digits as 142857, which is well-known to have the same property. – MJD May 5, 2014 at 17:23 • – MJD May 5, 2014 at 18:32", "out in presence of 1728 ($1728=16 \\times108$ and $1728=48 \\times 36$) but it would have been the same with 108 as a factor in the j-invariant, which is the least common multiple. – Youssef El Housni Nov 28 '18 at 19:23", "and 5 are multiples of 5. There are also three 2s in 15!. $$\\boxed{15}$$ is the smallest value N can be because 15! will have 3 sets of 2s and 5s to make zeros. power27 Jun 13, 2019", "the smallest normalized positive value ($1 * 16^{-1} * 16^{-64}$)", "## 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", "+0 # help 0 89 1 What is the least common multiple of the first ten positive integers? Apr 2, 2019 $$2,3,\\dots, 10 = \\{2^1, 3^1, 2^2, 5^1, 2^1 \\cdot 3^1,7^1, 2^3, 3^2, 2^1\\cdot 5^1\\}\\\\ \\text{Using the max exponents for common factors we obtain}\\\\ LCM = 2^3 \\cdot 3^2 \\cdot 5^1 \\cdot 7^1 = 2520$$", "# Least common multiple English (en) suomi (fi) français (fr) русский (ru) The least common multiple of two integers $a$ and $b$ is the smallest positive integer that is divisible by both $a$ and $b$. For example: for 12 and 9 then least common multiple is 36. ## function leastCommonMultiple function leastCommonMultiple(a, b: Int64): Int64; begin result := b * (a div greatestCommonDivisor(a, b)); end; Note: function greatestCommonDivisor must be at least declared before this function.", "+0 # Counting Divisors +1 129 2 +484 Some positive integers have exactly four positive factors. For example, 35 has only 1, 5, 7 and 35 as its factors. What is the sum of the smallest five positive integers that each have exactly four positive factors? Apr 5, 2021 #1 +594 +2 If they have four factors they can be of the form $p_1\\cdot p_2$ or $p^3$ for all $p_1,p_2,p$ prime. Let's do $p_1\\cdot p_2$ first. We have $2\\cdot 3=6, 2\\cdot 5=10, 2\\cdot 7=14, 2\\cdot 11=22,$ etc. if $2$ is the first. Otherwise we can have $3\\cdot 5=15, 3\\cdot 7=21$, etc. So the smallest $5$ are $6,10,14,15,21$. Then for $p^3$ we can have $2^3=8$ and the rest are larger than $21$. So the numbers are $6,8,10,14,15$. Sum them up. Apr 5, 2021 #2 +484 0 Thank you @thedudemanguyperson, it was correct! RiemannIntegralzzz Apr 5, 2021", "# Problem of the Week Problem E All Square The positive multiples of $$3$$ from $$3$$ to $$2400$$, inclusive, are each multiplied by the same positive integer, $$n$$. All of the products are then added together and the resulting sum is a perfect square. Determine the value of the smallest positive integer $$n$$ that makes this true. Note: In solving this problem, it may be helpful to use the fact that the sum of the first $$n$$ positive integers is equal to $$\\dfrac{n(n+1)}{2}$$. That is, $1 + 2 + 3 + \\cdots + n = \\frac{n(n+1)}{2}$ For example, $$1 + 2 + 3 + 4 + 5 = 15$$, and $$\\dfrac{5(6)}{2} = 15.$$ Also, $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$$ and $$\\dfrac{8(9)}{2} = 36$$.", "# Difference between revisions of \"2021 JMPSC Sprint Problems/Problem 17\" ## Problem What is the smallest positive multiple of $1003$ that has no zeros in its decimal representation? ## Solution Notice that $1002 \\cdot n = 1000n + 2n$. Since $1000n$ always has $3$ zeros after it, we have to make sure $2n$ has $3$ nonzero digits, so that the last 3 digits of the number $1002n$ doesn't contain a $0$. We also need to make sure that $n$ has no zeros in its own decimal representation, so that $1000n$ doesn't have any zeros other than the last $3$ digits. The smallest number $n$ that satisfies the above is $56$, so the answer is $1002 \\cdot 56 = \\boxed{56112}$. ~Mathdreams Invalid username Login to AoPS", "# 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 $3 \\times 4 \\times 7 = 84$." ]

[ "The multiples of 12 and of 21 are written in order. What is the smallest positive difference between a multiple of 12 and a multiple of 21? The difference is $$n\\cdot 21 -m \\cdot 12 = d$$ with $$n,m,d \\in Z$$ has a solution, if the $$gcd(21,12)$$ is a divider of $$d$$. Or $$gcd(21,12) | d$$. The greatest common divisor $$gcd(21,12) = 3.$$ The difference is $$n\\cdot gcd(21,12)$$ so the difference is a multiple of 3 The differences are: $$0, 3, 6, 9, 12, \\ldots$$ The smallest positive difference is 3. heureka Dec 4, 2017", "+0 # need help 0 225 1 If m is the smallest positive integer such that m! is a multiple of 4125, and n is the smallest positive integer such that n! is a multiple of 2816000, then find n-m. Jun 16, 2021 I claim that the $n=m=15$, so $n-m=0$. To see that $15!$ is a common multiple of 4125 and 2816000, just check that $15!$ is a multiple of both $4125=3\\cdot 5^3\\cdot 11$ and $2816000=2^{11}\\cdot 5^3\\cdot 11$. To see that $15!$ is the smallest common multiple of 4125 and 2816000, note that $14!$ is not a common multiple of 4125 and 2816000 (because $5^3$ does not divide $14!$ so it can't divide any $x!$ where $x \\le 14$).", "# Interview Question Asked In yahoo Can you find the smallest positive number such that if you shuffle the digits of the number in a particular order, the shuffled number becomes twice the original number. I understand the answer is $125874 => 251748$ $251748$ is twice the $125874$ and have same digits $1,2,4,5,7$ & $8$ but how to solve this non programmatic ? • One way is to play fool and answer it in base $2$. The answer becomes $01$, which turns $10$. ;) – user147444 May 5, 2014 at 17:14 • Or in base 8, where $52_8 = 2\\cdot 25_8$. – MJD May 5, 2014 at 17:19 • It's odd that the answer has the same digits as 142857, which is well-known to have the same property. – MJD May 5, 2014 at 17:23 • – MJD May 5, 2014 at 18:32", "contest math – find the smallest positive multiple \\$n\\$ of 1994 such that the produc of the divisors of \\$n\\$ is \\$n^{1994}\\$ Find the smallest positive multiple $$n$$ of 1994 such that the produc of the divisors of $$n$$ is $$n^{1994}$$ Attempt Let $$nin mathbb{N}$$ and denote by $$d(n)$$ the number of divisors of $$n$$ and $$p(n)$$ the product of all the divisors of $$n$$ including $$n$$. Consider the factorization of $$n$$ given by $$n=p_1^{alpha_1} p_2^{alpha_2}cdots p_n^{alpha_n}$$ for $$alpha_iin mathbb{N}$$. Is known the fact that $$p(n)=n^{d(n)/2}=n^{1994}$$ it is $$d(n)=997 cdot 2^2$$ in addition $$n=997 cdot 2 cdot k,$$ for some $$kin mathbb{N}$$. since $$n$$ is the smallest multiple of $$1994$$ with $$997 cdot 2^2$$ divisors then $$n$$ should be $$2^{998} cdot 997$$ Any comment or corretion was helpful", "# Interview Question Asked In yahoo Can you find the smallest positive number such that if you shuffle the digits of the number in a particular order, the shuffled number becomes twice the original number. I understand the answer is $125874 => 251748$ $251748$ is twice the $125874$ and have same digits $1,2,4,5,7$ & $8$ but how to solve this non programmatic ? - One way is to play fool and answer it in base $2$. The answer becomes $01$, which turns $10$. ;) – user147444 May 5 '14 at 17:14 Or in base 8, where $52_8 = 2\\cdot 25_8$. – MJD May 5 '14 at 17:19 It's odd that the answer has the same digits as 142857, which is well-known to have the same property. – MJD May 5 '14 at 17:23 – MJD May 5 '14 at 18:32", "+0 # Help PLS, ASAP +2 700 1 +140 The least common multiple of two positive integers is $7!$, and their greatest common divisor is $9$. If one of the integers is $315$, then what is the other? (Note that $7!$ means $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1$.) Mar 19, 2019 edited by CorbellaB.15 Mar 19, 2019 #1 0 The identity lcm[a,b]*gcd(a,b)=ab holds for all positive integers a and b. Let a=315 in this identity, and let b be the number we are looking for. Thus $$7! \\cdot 9=315 \\cdot b$$ so $$b = \\frac{7!\\cdot 9}{315} = \\frac{7!\\cdot 9}{35\\cdot 9} = \\frac{7!}{35} = \\frac{{7}\\cdot 6\\cdot {5}\\cdot 4\\cdot 3\\cdot 2\\cdot 1}{{35}} = 6\\cdot4\\cdot3\\cdot2 = \\boxed{144}.$$ Hope this helps :) Sep 7, 2019", "Greatest common divisor is the smallest positive number that can be written as $sa+tb$ We know that $d = \\gcd(a, b)$ can be written as $sa + tb$, where $s, t \\in \\mathbb{Z}$. Apparently, $d$ is the smallest positive number that can be written in this form. Why is this so? • proof. Also this has been asked before. – user61527 Feb 5 '14 at 1:51 • As T. Bongers has said, this is called Bezout's identity, pretty well known with proofs. – qwr Feb 5 '14 at 1:55 • There can't be a smaller one, for if $xa+yb$ is say $m\\gt 0$, then $d$ divides $m$, so $m\\ge d$. – André Nicolas Feb 5 '14 at 2:04 • Bezout's theorem, you need to proof that the smallest positive integer written as a linear combination of a and b is in actual fact the highest common factor (the idea is to use divisibility). – WhizKid Feb 5 '14 at 2:10 • @AndréNicolas that is the simplest explanation I've seen! Thanks! Edit: actually, how do we know that m does not divide a and b and thus m itself is not the gcd of a and b? – Trent Bing Feb 5 '14 at 2:30 Here is a conceptual way to prove Bezout's gcd Identity. The set $\\rm\\,S\\,$ of", "# Find the smallest value of positive integer !! Find the smallest positive integer with exactly 30 positive factor First, I use function $$\\tau$$ to find the exponential that gives $$2×3×5$$ and I want to find the smallest value. How can I find it use inequality to help And how to find the value of $$\\sqrt{8×13×15×17+49}$$? I get $$x=8$$ and doing perfect square ... but stuck who can help me, please • Your question is a bit unclear. Can you give some examples of the smallest value of a positive number with 2,3,4 positive factors? What do you mean with positive factors? – kvantour Jan 31 at 13:11 • What have you tried? Given the primne factorization $n=\\prod p_i^{a_i}$ , do you know how to compute the number of factors of $n$? – lulu Jan 31 at 13:11 • @kvantour To me these feel like two separate questions that should have been asked separately, in my opinion. – Robert Soupe Jan 31 at 18:39 The number of factors of a natural is given by the product of the multiplicities of its prime factors plus one, $$(m_2+1)(m_3+1)(m_5+1)\\cdots$$. To obtain $$30=2\\cdot3\\cdot5$$ factors, you need multiplicities $$1,2$$ and $$4$$, which you will assign to the smallest possible primes, by decreasing order $$5^13^22^4=720.$$ What do you mean by \"exactly 5 positive factors\"?", "## Algebra 1 -4.5, 1.75, $\\sqrt 4$, $14^1$ Mental math tells us that the only negative number, -4.5, is the smallest. Mental math also tells us that $14^1$, which equals 14, is the largest. Convert $\\sqrt 4$ to a decimal to compare against 1.75. $\\sqrt 4=2$ $1.75\\lt2$. Therefore $1.75\\lt\\sqrt 4$.", "# Problem of the Week Problem E All Square The positive multiples of $$3$$ from $$3$$ to $$2400$$, inclusive, are each multiplied by the same positive integer, $$n$$. All of the products are then added together and the resulting sum is a perfect square. Determine the value of the smallest positive integer $$n$$ that makes this true. Note: In solving this problem, it may be helpful to use the fact that the sum of the first $$n$$ positive integers is equal to $$\\dfrac{n(n+1)}{2}$$. That is, $1 + 2 + 3 + \\cdots + n = \\frac{n(n+1)}{2}$ For example, $$1 + 2 + 3 + 4 + 5 = 15$$, and $$\\dfrac{5(6)}{2} = 15.$$ Also, $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$$ and $$\\dfrac{8(9)}{2} = 36$$.", "8 and 12 in the list are $24,48,72.$ d. From part c, the smallest multiple that 8 and 12 have in common is 24.", "# Least common multiple English (en) suomi (fi) français (fr) русский (ru) The least common multiple of two integers $a$ and $b$ is the smallest positive integer that is divisible by both $a$ and $b$. For example: for 12 and 9 then least common multiple is 36. ## function leastCommonMultiple function leastCommonMultiple(a, b: Int64): Int64; begin result := b * (a div greatestCommonDivisor(a, b)); end; Note: function greatestCommonDivisor must be at least declared before this function.", "Question # Find the $L.C.M.$ of $30{\\text{ and }}45$.A $15$B $30$ C $45$ D $90$ Hint:In the question where we have to find the L.C.M we will proceed by using the definition of L.C.M.If $a{\\text{ and }}b$ are two non-zero integers the least common multiple of $a{\\text{ and }}b$ is the smallest positive integer divisible by both $a{\\text{ and }}b$.And then we will find the factors of the both numbers and find the smallest positive integer divisible by both the integers which will be the L.C.M of the given two integers. In the above question we have to find the $L.C.M.$ of $30{\\text{ and }}45$- Now to find $L.C.M.$ of any two numbers we must know the definition of the $L.C.M.$ $L.C.M.$, least common multiple is defined as if $a{\\text{ and }}b$ are two non-zero integers the least common multiple of $a{\\text{ and }}b$ is the smallest positive integer divisible by both $a{\\text{ and }}b$. The least common multiple is also known as smallest common multiple or lowest common multiple. Now the least common multiple or the smallest common multiple of $a{\\text{ and }}b$ is denoted by $l.c.m(a,b)$. Now according to the question we have to find the $L.C.M.$ of $30{\\text{ and }}45$, so now using definition of least common multiple, $30{\\text{ and }}45$ are the non-zero positive", "# Is greatest common divisor of two numbers really their smallest linear combination? In a lecture note from MIT on number theory says: Theorem 5. The greatest common divisor of a and b is equal to the smallest positive linear combination of a and b. For example, the greatest common divisor of 52 and 44 is 4. And, sure enough, 4 is a linear combination of 52 and 44: 6 · 52 + (−7) 44 = 4 What about 12 and 6 their gcd is 6 but 0 which is less than 6 can be - You wrote it yourself: the gcd is the smallest positive linear combination. – André Nicolas Oct 24 '12 at 7:32 Note that $52 \\times 44 - 44 \\times 52 =0$ and you can do this with any pair of integers. – Mark Bennet Oct 24 '12 at 7:43 @MarkBennet yeah that's right. – spartacus Oct 24 '12 at 20:40 You wrote it yourself: the gcd is the smallest positive linear combination. Smallest positive linear combination is shorthand for smallest positive number which is a linear combination. It is true that $0$ is a linear combination of $12$ and $6$ with integer coefficients, but $0$ is not positive. The proof is not difficult, but it is somewhat lengthy. We give full detail below.", "# Divisibility by $$15$$. Find the smallest positive multiple of $$15$$ such that each of the digit of the multiple is either 0 or 8. ×", "multiples of 2 and 6. ## Lowest Common Multiple The lowest common multiple of a pair or group of positive numbers, is the smallest positive number that is a multiple of each number. Examples (1.1) Find the lowest common multiple of the numbers 5 and 7. Solution One approach is to list the multiples of each, looking for the smallest shared multiple in both lists. 5 => { 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , …. } 7 => { 7 , 14 , 21 , 28 , 35 , 42 , 49 , 56 , 63 , 70 , …. } It can be seen from the lists that 35 is the lowest common multiple of 5 and 7. (1.2) Find the lowest common multiple of 24 and 10. Solution 2 => { 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , …. } 4 => { 4 , 8 , 12 , 16 , 20 , 24 , 28 , 32 , 40 , 44 , …. } 10 => { 10 , 20 , 30 , 40 , 50 , 60 , 70 , …. } 20", "+0 # Counting Divisors +1 129 2 +484 Some positive integers have exactly four positive factors. For example, 35 has only 1, 5, 7 and 35 as its factors. What is the sum of the smallest five positive integers that each have exactly four positive factors? Apr 5, 2021 #1 +594 +2 If they have four factors they can be of the form $p_1\\cdot p_2$ or $p^3$ for all $p_1,p_2,p$ prime. Let's do $p_1\\cdot p_2$ first. We have $2\\cdot 3=6, 2\\cdot 5=10, 2\\cdot 7=14, 2\\cdot 11=22,$ etc. if $2$ is the first. Otherwise we can have $3\\cdot 5=15, 3\\cdot 7=21$, etc. So the smallest $5$ are $6,10,14,15,21$. Then for $p^3$ we can have $2^3=8$ and the rest are larger than $21$. So the numbers are $6,8,10,14,15$. Sum them up. Apr 5, 2021 #2 +484 0 Thank you @thedudemanguyperson, it was correct! RiemannIntegralzzz Apr 5, 2021", "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", "## 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", "and 6. ## Least Common Multiple With multiples of numbers, the lowest common multiple of a pair or group of positive numbers, is the smallest positive number that is a multiple of each number. Examples (1.1) Find the lowest common multiple of the numbers 5 and 7 . Solution One approach is to list the multiples of each, looking for the smallest shared multiple in both lists. 5 => { 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , ... } 7 => { 7 , 14 , 21 , 28 , 35 , 42 , 49 , 56 , 63 , 70 , ... } It can be seen from the lists that 35 is the lowest common multiple of 5 and 7. (1.2) Find the lowest common multiple of 2, 4 and 10. Solution 2 => { 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , ... } 4 => { 4 , 8 , 12 , 16 , 20 , 24 , 28 , 32 , 40 , 44 , ... } 10 => { 10 , 20 , 30 , 40 , 50 , 60 , 70 ," ]

[ "Between what two whole numbers is his answer?", "One crore. ## 4. What is the greatest 8 digit number having three different digits? The greatest 8-digit number having three different digits is 9,99,99,987, which is read as Nine crore, ninety-nine lakh, ninety-nine thousand, nine hundred and eighty seven. More Important Topics Numbers Algebra Geometry Measurement Money Data Trigonometry Calculus", "the product of the 2 numbers is 36. Sep 14, 2018", "Let $$N$$ be the greatest integer multiple of 8, whose digits are all different. What is the remainder when $$N$$ is divided by 1000? (第二十一届AIME2 2003 第2题)", "than 2 Answer: C) is always whole number", "2, 3, 4, 8, or X. The digital root of the product of E and 11 (twin primes) is E.", "8 / 49 = 16^1/4 =2. But your question says \"positive integer\", or \"whole number\", so I can't see a smaller number than 392. Jun 14, 2018", "8 = 1 88 represents 8÷8 = 1. If the numerator of a fraction is a multiple of the denominator, the result will always be a whole • 465 Math Specialists • 24x7 Support", "2014 #### What is the biggest whole number? there is no biggest whole number cuz they r infinite Filters", "# We want a number with no digits repeating Let $$N$$ be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when $$N$$ is divided by 1000? ×", "Eight Symbol The symbol 8 represents a quantity of eight. Usage Abstractly, this can be visualized as the quantity:", "numbers are whole numbers.", "# A logic problem by Paola Ramírez Logic Level 2 Digits 1, 9, 8 and 6 are chosen (each just once) to form two numbers of one, two or three digits. What is the greatest value of the product of this two numbers? ×", "1) the product of 2 different whole numbers is 7. their sum is a) 6 b) 7 c) 8 d) 14 2) the sum of 2 positive whole numbers is greater than their product if one of the number is a) 1 b) 2 c) 3 d) 4 2. Originally Posted by sri340 1) the product of 2 different whole numbers is 7. their sum is a) 6 b) 7 c) 8 d) 14 7 is prime so the only product of different whole numbers equal to 7 is 1 times 7. TonL 3. Originally Posted by sri340 2) the sum of 2 positive whole numbers is greater than their product if one of the number is a) 1 b) 2 c) 3 d) 4 a) 1 Because $1 \\times x < 1 + x$ That's fairly logical. Originally Posted by CaptainBlack ... TonL That would be RonL Captain", "TheInfoList 8 (eight) is the natural number following 7 and preceding 9. In mathematics 8 is: * a composite number, its proper divisors being , , and . It is twice 4 or four times 2. * a power of two, being 2 (two cubed), and is the first number of the form , being an integer greater than 1. * the first number which is neither prime nor semiprime. * the base of the octal number system, which is mostly used with computers. In octal, one digit represents three bits. In modern computers, a byte is a grouping of eight bits, also called an octet. * a Fibonacci number, being plus . The next Fibonacci number is . 8 is the only positive Fibonacci number, aside from 1, that is a perfect cube. * the only nonzero perfect power that is one less than another perfect power, by Mihăilescu's Theorem. * the order of the smallest non-abelian group all of whose subgroups are normal. * the dimension of the octonions and is the highest possible dimension of a normed division algebra. * the first number to be the aliquot sum of two numbers other than itself; the discrete biprime , and the square number . A number is divisible by 8 if its last three digits, when", "multiples of 8 and 12 but 24 is the lowest.", "digits 1, 1, 2, 2, 3 and 3 so that between the two 1's there is one digit, between the two 2's there are two digits, and between the two 3's there are three digits. Ones Only Stage: 3 Challenge Level: Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.", "of the smallest two to the product of the greatest two. Repeat with some other sets of numbers. What do you notice? With thanks to Don Steward, whose ideas formed the basis of this problem.", "members: 8 ... 16 And since their partners are known, 8 1 ... 9 16 Now it’s a purely mechanical task of picking out the remaining pair for each number, working towards the middle. Try it!", "Kristine 2*2*2=8" ]

[ "a^2$$. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$. ~IceMatrix", "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$.", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "= 2^2\\cdot3^2 = 6^2$ can be converted to two digits of base $216 = 2^3\\cdot3^3 = 6^3$. Note that base-85 encoding of a string of (for example) $1024$ binary digits does not produce a number written in base $85$. of a $1024$-bit integer would be $1$, $85$, $85^2$, $85^3$, $85^4$, $2^{32}$, $85\\cdot2^{32}$, $85^2\\cdot2^{32}$, $85^3\\cdot2^{32}$, $85^4\\cdot2^{32}$, $2^{64}$, $85\\cdot2^{64}$, and so forth. a string of bits as a base-$2^{32}$ numeral, except that rather than define a sequence of $2^{32}$ unique glyphs for the digits $0$ to $2^{32}-1$, we write each base-$2^{32}$ digit as a five-digit base-$85$ integer in that range, left-padded with zeros as needed.", "them a lot. By inspection, we find that $\\text{LCM}(1,2,3,2\\cdot2,2\\cdot3) = 12$. Finally, $1+2 = \\boxed{\\textbf{(B)}\\ 3}$. ## Solution 2 We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$,$2$, $3$, or $4$. We quickly consider $12$ since it is the smallest number that is the LCM of $1$, $2$, $3$ and $4$. Since $12$ has $5$ single-digit divisors, namely $1$, $2$, $3$, $4$, and $6$, our answer is $1+2 = \\boxed{\\textbf{(B)}\\ 3}$", "## Elementary Algebra $25\\cdot1=25$ $25+1=26$ $25$ and $1$ are the only two numbers whose product is $25$ and sum is $26$. Therefore, we must use these numbers to fill in the blanks.", "the numbers have the same sign, the product is positive. Thus, 86=+4886=+48 size 12{8 cdot 6\"=+\"\"48\"} {}, or 86=4886=48 size 12{8 cdot 6=\"48\"} {}. #### Example 2 (8)(6)(8)(6) size 12{ $$- 8$$ $$- 6$$ } {} |- 8| = 8 |- 6| = 6 |- 8| = 8 |- 6| = 6 Multiply these absolute values. 8 6 = 48 8 6 = 48 size 12{8 cdot 6=\"48\"} {} Since the numbers have the same sign, the product is positive. Thus, (8)(6)=+48(8)(6)=+48 size 12{ $$- 8$$ $$- 6$$ \"=+\"\"48\"} {}, or (8)(6)=48(8)(6)=48 size 12{ $$- 8$$ $$- 6$$ =\"48\"} {}. #### Example 3 (4)(7)(4)(7) size 12{ $$- 4$$ $$7$$ } {} |- 4| = 4 |7| = 7 |- 4| = 4 |7| = 7 Multiply these absolute values. 4 7 = 28 4 7 = 28 size 12{4 cdot 7=\"28\"} {} Since the numbers have opposite signs, the product is negative. Thus, (4)(7)=28(4)(7)=28 size 12{ $$- 4$$ $$7$$ = - \"28\"} {}. #### Example 4 6(3)6(3) size 12{6 $$- 3$$ } {} |6| = 6 |- 3| = 3 |6| = 6 |- 3| = 3 Multiply these absolute values. 6 3 = 18 6 3 = 18 size 12{6 cdot 3=\"18\"} {} Since the numbers have opposite signs, the product is negative. Thus, 6(3)=186(3)=18 size 12{6", "### 1976 IMO #4 Determine the largest number which is the product of positive integers with sum 1976. The answer is $\\boxed{2 \\cdot 3^{658}}$. Note that there cannot be any integers $n>4$ in the maximal product, because we can just replace $n$ by $3,n-3$ to achieve a greater product. Also, we would not want any $1$s in the maximal product as they would not help in increasing the product and would just take up space in the sum. Notice that we can replace any $4$s by two $2$s leaving the product and the sum unchanged. Lastly, we don't want more than two $2$s, because we can just replace three $2$s by two $3$s to obtain a larger product. Therefore, the maximum product must consist of $3$s and zero, one or two $2$s. Since $1976 = 3\\cdot 658 + 2$, we have one $2$ and $658$ $3$'s in our maximum product, giving the answer of $2\\cdot 3^{658}$, as desired. $\\square$", "$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", "+ 13 = 25$ is also a whole number. Also, $95$ and $52$ are whole numbers and $95 + 52 = 147$ is also a whole number. ### Subtraction The set of whole numbers is not closed under the operation subtraction. It means if you subtract any two whole numbers, you may get a whole number or may not get a whole number. Mathematically, it can be represented as for all $a$, $b \\in W$, $a – b \\notin W$. For example, $15$ and $3$ are whole numbers and $15 – 3 = 12$ is also a whole number, but $3 – 15 = -12$ in not a whole number. Note: The operation subtraction is not commutative also. ### Multiplication The set of whole numbers is closed under the operation multiplication. It means if you multiply any two whole numbers, you will always get a whole number. Mathematically, it can be represented as for all $a$, $b \\in W$, $a \\times b \\in W$. For example, $8$ and $4$ are whole numbers, and $8 \\times 4 = 32$ is also a whole number. Also, $52$ and $11$ are whole numbers and $52 \\times 11 = 572$ is also a whole number. ### Division The set of whole numbers is not closed under the operation division. It means if", "is $22$ and their product is $19 \\cdot 22 = \\boxed{418}$. ### Solution 2 As above, we define $Q(x)=P(x)-x+7$, noting that it has roots at $17$ and $24$. Hence $P(x)-x+7=A(x-17)(x-24)$. In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$. Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$, where $A$, $(x-17)$, and $(x-24)$ are integers. This cannot occur if $x\\le 17$ or $x\\ge 24$ because the product $(x-17)(x-24)$ will either be too large or not be a divisor of $10$. We find that $x=19$ and $x=22$ are the only values that allow $(x-17)(x-24)$ to be a factor of $10$. Hence the answer is $19\\cdot 22=\\boxed{418}$.", "and a 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}$$ We can find the multiples of any number by continuing this process. Table 2.55 shows the multiples of 2 through 9 for the first twelve counting numbers. #### Table 2.55 Counting Number 1 2 3 4 5 6 7 8 9 10 11 12 Multiples of 2 2 4 6 8 10 12 14 16 18 20 22 24 Multiples of 3 3 6 9 12 15 18 21 24 27 30 33 36 Multiples of 4 4 8 12 16 20 24 28 32 36 40 44 48 Multiples of 5 5 10 15 20 25 30 35 40 45 50 55 60 Multiples of 6 6 12 18 24 30 36 42 48 54 60 66 72 Multiples of 7 7 14 21 28 35 42 49 56 63 70 77 84 Multiples of 8 8 16 24 32 40 48 56 64 72 80 88 96", "+0 # help -5 61 2 +-291 In the prime factorization of $24!$, what is the exponent of $3$? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, \\$$$5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120.$$) Apr 1, 2020 #1 +336 +1 24/3=8. At first glance, the answer might seem like 8, but this is a wrong assumption. This is because the numbers 9 and 18 hold two nines! so we need to add two to 8 to have 10 as our answer. Apr 1, 2020", "+0 heeeeelllllpppp +1 42 4 +44 The product of integers 240 and k is a perfect cube. What is the smallest possible positive value of k? heeeeeeeeeeellllllllppppppppp Jul 26, 2022 #1 +370 -1 I think it's 60 Jul 26, 2022 #2 +1118 +4 240x60=14400, which is not a perfect cube nerdiest Jul 26, 2022 #3 +197 +1 240 0= 24 * 10 = 2^3 * 3 * 10 = 2^4 * 3 * 5 We need 2 2s, 2 3s, and 2 5s. This gives us: 2^2 * 3^2 * 5^2 = 900. Jul 26, 2022 #4 +1118 +4 niceeee. here's a way $$240=2^4\\cdot3\\cdot5=2^3(2\\cdot3\\cdot5)$$. For $$240k$$ to be a perfect cube (and not a perfect square), $$k$$ must be at least $$2^2\\cdot3^2\\cdot5^2=\\boxed{900}$$. nerdiest Jul 26, 2022 edited by nerdiest Jul 26, 2022", "Kattis # Product Digit Nikolaj works at a company that sells a large number of products. Each product has a product ID, which is a large integer. For error detection, in particular in connection with manual database entry, it would be useful if each product also had a single “check” digit between $1$ and $9$ that can be quickly computed from the product ID. Nikolaj has been tasked with implementing this. He recently solved the problem “Digit Product” on Open Kattis and considers using the procedure described there. He recalls that the idea is to start with a positive integer $x$ and repeatedly multiply all its nonzero digits, until only a single digit is left. For instance, if $x$ is $808$ then the resulting digit is $8$, because $8 \\cdot 8 = 64$, $6 \\cdot 4 = 24$, and $2 \\cdot 4 = 8$. However, Nikolaj is unsure about using this method, because the distribution of resulting digits seems uneven to him. To determine how the distribution looks, he writes a program that given two integers $L$ and $R$ determines how many numbers in the interval $[L, R]$ result in each digit. ## Input A single line consisting of two integers $L$ and $R$ with $1 \\leq L \\leq R \\leq 10^{15}$. ## Output Print $9$ integers $a_1,", "## Elementary Algebra $16$ Factor each number completely to have: $32 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\\\80 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 5 \\\\96= 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 3$ Since four 2s are common to both, the greatest common factor of the given numbers is: $2 \\cdot 2\\cdot 2\\cdot 2 = 16$", "## Discrete Mathematics with Applications 4th Edition the 8 bit two's complement for integers of 23 will be $23_{10} = (16 + 4 + 2 + 1)_{10} = 00010111_2 → 11101000 → 11101001.$ So the answer is $11101001$.", "## Elementary Algebra $12$ Factor each number completely to have: $48 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 3 \\\\60 = 2 \\cdot 2 \\cdot 3 \\cdot 5 \\\\84= 2 \\cdot 2 \\cdot 3 \\cdot 7$ Since two 2s and one 3 are common to both, the greatest common factor of the given numbers is: $2 \\cdot 2 \\cdot 3 = 12$", "# Odd Product! The product of the digits in 38 is even because $$3\\times 8 = 24.$$ Similarly, the product of the digits in 57 is odd because $$5 \\times 7 = 35.$$ How many 2-digit numbers have an odd product? ×", "at least $78$ ones, so the sum (and product) is at least $85$ and at most $78 + 9 \\cdot 7 = 141.$ Case: $a_1 = 9$ If $a_1 = 9$, then the possible products are $90, 108, 126, 135$ since the prime factorization only consists of prime numbers less than 10. • If the product of the digits is 90, then $a_2 = 5$ and $a_3 = 2$. However, the sum of the digits would be $82 + 9 + 5 + 2 = 98$, so the product can not be 90. • If the product of the digits is 108, 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 a sum of $82 + 9 + 6 + 2 = 99$, the second case results in a sum of $82 + 9 + 4 + 3 = 98$, and the third case results in a sum of $81 + 9 + 3 + 2 + 2 = 97$. Thus, the product can not be 108. • If the product of the digits is 126, then $a_2 = 7$ and $a_3 = 2$. However, the sum of" ]

[ "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 $3, 4, 5, 6, 7$.", "## Definition Of Whole Numbers The numbers in the set {0, 1, 2, 3, 4, 5, 6, 7, . . . . } are called whole numbers. In other words, whole numbers is the set of all counting numbers plus zero. Whole numbers are not fractions, not decimals. Whole numbers are non-negative integers ### Solved Example on Whole Numbers #### Ques: Annual revenues of three companies are $4,234,223,$6,246,234 and $2,234,233. Which of them is the least? ##### Choices: A.$2,234,233 B. $4,234,223 C.$6,246,234 #### Solution: Step 1: To compare numbers, begin with the greatest place and compare the digits Step 2:[Write the whole numbers one below the other to compare the digits in the corresponding places.] 4 2 3 4 2 2 3 6 2 4 6 2 3 4 2 2 3 4 2 3 3 Step 3: 2 million is less than 4 millions and 6 million. [Digits in the millions place.] Step 4: So, 2,234,233 is less than 4,234,223 and 6,246,234 Step 5: Revenue of \\$2,234,233 is the least among the given", "+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", "a^2$$. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$. ~IceMatrix", "# Comparing Two Gigantic Numbers Number Theory Level 2 Find the greatest common divisor of the two numbers, $$886! + 1$$ and $$887!$$. Notation: $$!$$ denotes the factorial notation. For example, $$10! = 1\\times2\\times3\\times\\cdots\\times10$$. ×", "# These Numbers Are Way Too Huge! Find the greatest common divisor of the two numbers $2016! + 1$ and $2017!$. Note: 2017 is prime. Notation: $!$ is the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "= 2^2\\cdot3^2 = 6^2$ can be converted to two digits of base $216 = 2^3\\cdot3^3 = 6^3$. Note that base-85 encoding of a string of (for example) $1024$ binary digits does not produce a number written in base $85$. of a $1024$-bit integer would be $1$, $85$, $85^2$, $85^3$, $85^4$, $2^{32}$, $85\\cdot2^{32}$, $85^2\\cdot2^{32}$, $85^3\\cdot2^{32}$, $85^4\\cdot2^{32}$, $2^{64}$, $85\\cdot2^{64}$, and so forth. a string of bits as a base-$2^{32}$ numeral, except that rather than define a sequence of $2^{32}$ unique glyphs for the digits $0$ to $2^{32}-1$, we write each base-$2^{32}$ digit as a five-digit base-$85$ integer in that range, left-padded with zeros as needed.", "# But it's so many digits! Number Theory Level 3 Find the highest power of $$12$$ that divides $$49!$$. Clarification: $$49! = 1 \\times 2\\times 3 \\times \\cdots \\times 49$$. ×", "× # Help,its so hard for me Determine the biggest integers 85-digit number which satisfies the sum of all digits equal to the multiplication of all numbers Note by Valian Fil Ahli 4 years ago Sort by: Let the digits be $$a_{85}\\geq a_{84}\\geq \\cdots \\geq a_1$$ such that $$a_1+a_2+\\cdots+a_{85}=a_1a_2\\cdots a_{85}$$, then $\\frac{a_1}{a_1a_2\\cdots a_{85}}+\\frac{a_2}{a_1a_2\\cdots a_{85}}+\\cdots+\\frac{a_{85}}{a_1a_2\\cdots a_{85}}=1.$ we have 85 fractions and the greatest of them is the last one, therefore $$\\frac{a_{85}}{a_1a_2\\cdots a_{85}}\\geq \\frac{1}{85}$$ and then $$a_1a_2\\cdots a_{84}\\leq85$$. At most six digits among $$a_1,a_2,\\ldots, a_{84}$$ are greater than 1, therefore, at least 78 digits are equal to 1. - 3 years, 12 months ago Can you please explain the last sentence in your argument? It is not very obvious to me. I think we need to find the integer less than or equal to $$85$$ with highest number of divisors with one digit and I think according to your argument such integer has at most $$6$$ one digit divisors. Right? - 3 years, 12 months ago Each of $$a_1, a_2, \\ldots, a_{84}$$ is a positive integer (zeros are not allowed). Suppose seven digits among $$a_1, a_2, \\ldots, a_{84}$$ are greater than 1, i.e. are $$\\geq2$$, then the product $$a_1a_2\\cdots a_{84}$$ will be $$\\geq 2^7=128$$ which is a contradiction. Therefore, at most six digits among $$a_1, a_2, \\ldots, a_{84}$$ can be", "## Discrete Mathematics with Applications 4th Edition the 8 bit two's complement for integers of 23 will be $23_{10} = (16 + 4 + 2 + 1)_{10} = 00010111_2 → 11101000 → 11101001.$ So the answer is $11101001$.", "## Elementary Algebra $12$ Factor each number completely to have: $48 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 3 \\\\60 = 2 \\cdot 2 \\cdot 3 \\cdot 5 \\\\84= 2 \\cdot 2 \\cdot 3 \\cdot 7$ Since two 2s and one 3 are common to both, the greatest common factor of the given numbers is: $2 \\cdot 2 \\cdot 3 = 12$", "# Sufficiently Many Eights $\\Large \\underbrace{88888888888\\ldots 8}_{888\\text{ number of 8's}}$ Find the remainder when the above number is divided by 887. ×", "### 1976 IMO #4 Determine the largest number which is the product of positive integers with sum 1976. The answer is $\\boxed{2 \\cdot 3^{658}}$. Note that there cannot be any integers $n>4$ in the maximal product, because we can just replace $n$ by $3,n-3$ to achieve a greater product. Also, we would not want any $1$s in the maximal product as they would not help in increasing the product and would just take up space in the sum. Notice that we can replace any $4$s by two $2$s leaving the product and the sum unchanged. Lastly, we don't want more than two $2$s, because we can just replace three $2$s by two $3$s to obtain a larger product. Therefore, the maximum product must consist of $3$s and zero, one or two $2$s. Since $1976 = 3\\cdot 658 + 2$, we have one $2$ and $658$ $3$'s in our maximum product, giving the answer of $2\\cdot 3^{658}$, as desired. $\\square$", "## Elementary Algebra $16$ Factor each number completely to have: $32 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\\\80 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 5 \\\\96= 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 3$ Since four 2s are common to both, the greatest common factor of the given numbers is: $2 \\cdot 2\\cdot 2\\cdot 2 = 16$", "# What a nice problem? $\\large 1234567891011\\ldots20142015$ The gigantic number above is formed by the concatenation of the first 2015 whole numbers. What is the remainder when this number is divided by 9? ×", "## Thinking Mathematically (6th Edition) $14$ Factor each number completely to obtain: $42 = 6(7) = 2(3)(7) \\\\56=8(7) = 2(2)(2)(7)$ Notice that the one factor of $2$ and one factor of $7$ is common to both factorization. Thus, the greatest common factor is $2(7) = 14$. Therefore, the greatest common divisor is also $14$.", "+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", "# Two numbers are in the ratio 4:5. Their highest common factor is 16. What is a pair of numbers that fits that? Jan 29, 2017 The numbers are $64$ and $80$. The two numbers are in the ratio $4 : 5$ and there is no common factor between $4$ and $5$ (i.e. highest common factor is $1$). As the highest common factor of the two numbers is $16$, the numbers are $4 \\times 16$ and $5 \\times 16$ i.e. $64$ and $80$.", "• # question_answer Write the greatest four digit number and express it in terms of its prime factors. 9999 is the greatest 4 digit number. $9999=3\\times 3333=3\\times 3\\times 1111=3\\times 3\\times 11\\times 101$ $9999=3\\times 3\\times 11\\times 101$", "# A Sheer Multiplication Of Numbers Danny wrote the following product: $\\large 1! \\times 2! \\times 3! \\times \\cdots \\times 119! \\times 120!$ He took out one of the following factorials and found that the product had become a perfect square. Find the factorial. Notation: $!$ denotes the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×" ]

[ "+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", "# Express it in fraction Express $$0.461538461538...$$ as a fraction. (In other words, 461538 repeats forever which is called 0 point 461538 recurring.) ×", "+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", "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.", "# 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.", "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 $9.42\\times 10^{-12}$", "+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 $\\dfrac{56}{-15}$", "the fraction $k/p$ is not in lowest terms.", "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 $8.5\\times 10^{-12}$", "+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", "\\overline{33} g$", "+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 # 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", "Home > Pre-algebra > Ratios > Express Ratio As Fraction ## Examples Express the ratio as a fraction To express the ratio ' to ' as a fraction, place over and reduce can be reduced, since is a factor of both and : The fraction is now in lowest terms Express the ratio as a fraction To express the ratio ' to ' as a fraction, place over and reduce can be reduced, since is a factor of both and : The fraction is now in lowest terms Express the ratio as a fraction To express the ratio ' to ' as a fraction, place over and reduce can be reduced, since is a factor of both and : The fraction is now in lowest terms Express the ratio as a reduced fraction To express the ratio ' to ' as a fraction, place over and reduce Multiply numerator and denominator by 100 * = can be reduced, since is a factor of both and : The fraction is now in lowest terms", "+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", "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", "$v$ and $\\rho$ and then express $\\rho$ in terms of $z$.)" ]

[ "# The value of 0.23¯¯¯¯ + 0.22¯¯¯¯ is Question: The value of $0.23+0.22$ is (a) $0 . \\overline{45}$ (b) $0 . \\overline{43}$ (c) $0 . \\overline{45}$ (d) $0.45$ Solution: Given that $0 . \\overline{23}+0 . \\overline{22}$ Let $x=0 . \\overline{23}+0 . \\overline{22}$ Now we have to find the value of $x$ $\\Rightarrow x=0 . \\overline{23}+0 . \\overline{22}$ $\\Rightarrow x=\\frac{23}{99}+\\frac{22}{99}$ $\\Rightarrow x=\\frac{23+22}{99}$ $\\Rightarrow x=\\frac{45}{99}$ $\\Rightarrow x=\\frac{5}{11}$ $0 . \\overline{23}+\\overline{0.22}=0 . \\overline{45}$ The correct option is", "+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", "because $22 \\div 18.75 = 1.17\\overline{3}$. You left a $17.\\overline{3}\\%$ tip, because $117.\\overline{3} - 100 = 17.\\overline{3}$.", "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 $$\\frac{14}{11}=1 . \\overline{27}.$$", "# 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 $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "(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{789}}}$ $789$ The right side of the equation becomes $789$, so we have: $999 x = 789$ To solve for $x$, we divide $789$ by $999$ and simplify: $x = \\frac{789}{999} = \\frac{263}{333}$ Therefore, $\\frac{263}{333} = . \\overline{789}$.", "Home/Class 9/Maths/ Show that $$0.2353535... =0.2\\overline{35}$$ can be expressed in the form $$\\frac {p}{q},$$ where $$p$$ and $$q$$ are integers and $$q\\neq 0.$$ Speed 00:00 01:39 ## QuestionMathsClass 9 Show that $$0.2353535... =0.2\\overline{35}$$ can be expressed in the form $$\\frac {p}{q},$$ where $$p$$ and $$q$$ are integers and $$q\\neq 0.$$ We have, $$x=0.2\\overline{35}$$. ...(i) Multiplying eqn. (i) by $$10$$, we get $$10x=2.\\overline{35}$$ ...(ii) Multiplying eqn. (ii) by $$1000$$, we get $$1000x=235.\\overline{35}$$ ...(iii) Subtracting eqn. (ii) from eqn. (iii), we get $$1000x-10x=235.\\overline{35}-2.\\overline{35}$$ $$\\Rightarrow 990x=233$$ $$\\Rightarrow x=\\frac{233}{990}$$.", "+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", "left-hand side and the right side separately, and we can see that the values after decimals are the same and can be cancelled out from simple subtraction. So, we get: $100x - x = \\left( {57.575757575757......} \\right) - \\left( {0.575757575757......} \\right) \\\\ \\Rightarrow 99x = 57 \\\\$ Dividing both the sides by $99$ in order to get the value of $x$ , and we get: $\\Rightarrow \\dfrac{{99x}}{{99}} = \\dfrac{{57}}{{99}} \\\\ \\Rightarrow x = \\dfrac{{57}}{{99}} \\\\$ Since, we can see that the value on the right-side can be further simplified, by dividing both the numerator and denominator by $3$ as it’s the highest factor that can divide both of them. So, dividing and multiplying the right-side by$3$, we get: $\\Rightarrow x = \\dfrac{{\\dfrac{{57}}{3}}}{{\\dfrac{{99}}{3}}} \\\\ \\Rightarrow x = \\dfrac{{19}}{{33}} \\\\ \\therefore x = 0.\\overline {57}$ Therefore, the value of $0.\\overline {57}$ and the correct answer is option C. Note:Do not commit a mistake by repeating only $5$ or $7$ from $0.\\overline {57}$, as because the bar is upon both $5$ and $7$, so must be repeated in the same order. Since, the bar was on two digits that’s why we multiplied the value by $100$, in order to take the two digits out. If the bar is on one digit then we can multiply by $10$, then subtract.", "### Examples Ex 1: Convert $0.222222… \\left(= 0. \\overline{2} \\right)$ to a rational number Let $x = 0.222222…$ —————————————- (1) Since only $1$ digit is repeated multiply both sides by $10$ $10 \\times x = 10 \\times 0.222222…$ $=> 10x = 2.222222…$ ————————————– (2) Subtract (1) from (2) $9x = 2 => x = \\frac {2}{9}$ Therefore, $0.222222… \\left(= 0. \\overline{2} \\right) = \\frac {2}{9}$ Ex 2: Convert $1.343434… \\left(= 1. \\overline{34} \\right)$ to a rational number Let $x = 1.343434…$ —————————————- (1) Since $2$ digits are repeated multiply both sides by $100$ $100 \\times x = 100 \\times 1.343434…$ $=> 100x = 134.343434…$ ————————————– (2) Subtract (1) from (2) $99x = 133 => x = \\frac {133}{99}$ Therefore, $1.343434… \\left(= 1. \\overline{34} \\right) = \\frac {133}{99} = 1 \\frac {34}{99}$ Ex 3: Convert $6.27454545… = 6.27 \\overline{45}$ to a rational number Let $x =$6.27454545…$—————————– (1) Since$2$numbers$2$and$7$are not repeated, therefore, multiply both sides by$100100 \\times x = $100 \\times 6.27454545…$ —————————– (1) $=> 100x = 627.454545…$ —————————————————-(2) Also, $2$ numbers $4$ and $5$ are repeating, therefore, multiply both sides by $100$ $100 \\times 100x = 100 \\times 627.454545…$ $=> 10000x = 62745.454545…$ ———————————————(3) Now subtract (2) from (3) $9900x = 62118 => x = \\frac {62118}{9900} = \\frac {31059}{4950} = 6 \\frac {1359}{4950}$ ## Identifying Terminating & Non", "$$3.\\overline { 17 }$$ …(i) ∴ x = 3.1717… Since, two numbers i.e. 1 and 7 are repeating after the decimal point. Thus, multiplying both sides by 100, 100x = 317.1717… ∴ 100x= 317.17 …(ii) Subtracting (i) from (ii), 100x – x = $$317.\\overline { 17 }$$ – $$3.\\overline { 17 }$$ ∴ 99x = 314 iv. Let x = $$15.\\overline { 89 }$$ …….. (i) ∴ x = 15.8989… Since, two numbers i.e. 8 and 9 are repeating after the decimal point. Thus, multiplying both sides by 100, 100x= 1589.8989… ∴ 100x = $$1589.\\overline { 89 }$$ …(ii) Subtracting (i) from (ii), 100x – x = $$1589.\\overline { 89 }$$ – $$15.\\overline { 89 }$$ ∴ 99x = 1574 v. Let x = $$2.\\overline { 514 }$$ ∴ x = 2.514514… Since, three numbers i.e. 5, 1 and 4 are repeating after the decimal point. Thus, multiplying both sides by 1000, 1000x = 2514.514514… 1000x = $$2514.\\overline { 514 }$$ ….(ii) Subtracting (i) from (ii), 1000x – x = $$2514.\\overline { 514 }$$ – $$2.\\overline { 514 }$$ ∴ 999x = 2512 Question 1. How to convert 2.43 in $$\\frac { p }{ q }$$ form ? (Textbook pg. no. 20) Solution: Let x = 2.43 In 2.43, the number 4 on the right side of the", "hand side a whole number. $$123=990y$$ Divide by 990 to isolate y. $$\\frac{123}{990}=y$$ Now, let's calculate what x/y is. $$\\frac{x}{y}=\\frac{560}{123}$$ Let's see if this is true. $$\\frac{\\frac{56}{99}}{\\frac{123}{990}}$$ Multiply by 990/123 to eliminate the complex fraction. $$\\frac{56}{99}*\\frac{990}{123}$$ Notice that 990 and 99 can be simplified before any multiplication takes place. $$\\frac{56}{1}*\\frac{10}{123}$$ Simplify from here. $$\\frac{560}{123}$$ Therefore, we have proven algabraically that $$\\frac{560}{123}=\\frac{0.\\overline{56}}{0.1\\overline{24}}$$ TheXSquaredFactor Nov 12, 2017", "given value is: $0.\\overline{190476} = \\dfrac{190476}{999999} = \\dfrac4{21}$ If so: $0.\\overline{190476} = \\dfrac4{21} = \\dfrac{\\frac17}2 + \\dfrac y2 ~\\implies~ \\boxed{y = \\dfrac5{21}}$", "$x=0.\\overline{001}\\Rightarrow$ …… (1) Since the number of recurring decimal number is $3$, so multiplying both sides of the equation (1) by $1000$, gives $1000\\times x=1000\\times 0.001001.....$ …… (2) Subtracting the equation (1) from (2) gives \\begin{align} & 1000x=1.001001..... \\\\ & \\underline{\\text{ }-x=0.001001.....} \\\\ & 999x=1 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{1}{999}$ Hence, the decimal number becomes $0.\\overline{001}=\\dfrac{1}{999}$ and it is in the $\\dfrac{p}{q}$ form. 4. Express ${0} {.99999}.....$ in the form of $\\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense. Ans: Let $x=0.99999.....$ ....... (a) Multiplying by $10$ both sides of the equation (a), gives $10x=9.9999.....$ …… (b) Now, subtracting the equation (a) from (b), gives \\begin{align} & 10x=9.99999..... \\\\ & \\underline{\\,-x=0.99999.....} \\\\ & \\,\\,9x=9 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{9}{9}$ $\\Rightarrow x=1$. So, the decimal number becomes $0.99999...=\\dfrac{1}{1}$ which is in the $\\dfrac{p}{q}$ form. Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense. Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal. 5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\\dfrac{ {1}}{ {17}}$ ? Perform the division to check your answer. Ans: Here the", "quantity $$\\frac{1}{3}$$. When written as a decimal, we get $$\\frac{1}{3} = 0.33\\overline{3}$$. Now consider multiplying both sides of this equation by 3. That is, consider that $$3\\times\\frac{1}{3} = 3\\times 0.33\\overline{3}$$. Evaluating the left side of the equation, we have $$3\\times\\frac{1}{3} = \\frac{3}{3} = 1$$. Now consider the right side of the equation. Recall that when we multiply by 3, we are taking our original quantity, and adding it to itself three times. Using the notation that we learned in elementary school, it would look like this: Clearly, in every decimal place, we would have $$3+3+3 = 9$$. Hence $$3\\times 0.33\\overline{3} = 0.99\\overline{9}$$. Since we still have equality between the right and left sides of the equation, this gives us $$1 = 0.99\\overline{9}$$. ## Proof 2 Let $$x = 0.99\\overline{9}$$. If we multiply both sides of this identity by 10, we will end up with the same string of digits on the right, but the decimal will move one place to the right. Thus $$10x = 9.99\\overline{9}$$. From each side of the equation, we will subtract $$x$$: $$10x – x = 9.99\\overline{9} – 0.99\\overline{9}$$. This gives us $$9x = 9$$. Solving for $$x$$, we have $$x = 1$$. Substituting this value into the original identity gives us the desired result, which is $$1 = 0.99\\overline{9}$$. ## Proof 3 This", "\\ldots$ $\\Rightarrow 100 x=37+x$ $\\Rightarrow 99 x=37$ $\\Rightarrow x=\\frac{37}{99}$ Hence, (iii) Let $x=0 . \\overline{54}$ $\\Rightarrow x=0.545454 \\ldots$ $\\Rightarrow 100 x=54.5454 \\ldots$ $\\Rightarrow 100 x=54+0.5454 \\ldots$ $\\Rightarrow 100 x=54+x$ $\\Rightarrow 99 x=54$ $\\Rightarrow x=\\frac{54}{99}=\\frac{6}{11}$ Hence, (iv) Let $x=0 . \\overline{621}$ $\\Rightarrow x=0.621621621 \\ldots$ $\\Rightarrow 1000 x=621.621621 \\ldots$ $\\Rightarrow 1000 x=621+0.621621 \\ldots$ $\\Rightarrow 1000 x=621+x$ $\\Rightarrow 999 x=621$ $\\Rightarrow x=\\frac{621}{999}=\\frac{207}{333}$ $\\Rightarrow x=\\frac{23}{37}$ Hence, (v) Let $x=125 \\overline{.3}$ $\\Rightarrow x=125+0 . \\overline{3}$ $\\Rightarrow x=125+\\frac{1}{3}\\left(\\because 0 . \\overline{3}=\\frac{1}{3}\\right)$ $\\Rightarrow x=\\frac{375+1}{3}$ $\\Rightarrow x=\\frac{376}{3}$ Hence, (vi) Let $x=4 . \\overline{7}$ $x=4+0 . \\overline{7}$ Let $y=0 . \\overline{7}=0.777 \\ldots$ $\\Rightarrow 10 y=7+0.777 \\ldots$ $\\Rightarrow 10 y=7+y$ $\\Rightarrow y=\\frac{7}{9}$ Therefore, $x=4+\\frac{7}{9}=\\frac{43}{9}$ Hence, (vii) Let $x=0.4 \\overline{7}$ $\\Rightarrow 10 x=4+0 . \\overline{7}$ Since, $0 . \\overline{7}=\\frac{7}{9}$ Therefore, $\\Rightarrow 10 x=4+\\frac{7}{9}=\\frac{43}{9}$ $\\Rightarrow x=\\frac{43}{90}$ Hence, Question 1: Define an irrational number. Solution. An irrational number is a real number that cannot be reduced to any ratio between an integer $p$ and a natural number $q$. If the decimal representation of an irrational number is non-terminating and non-repeating, then it is called irrational number. For example $\\sqrt{3}=1.732 \\ldots \\ldots .$ Question 2: Explain, how irrational numbers differ from rational numbers? Solution. Every rational number must have either terminating or non-terminating but irrational number must have non- terminating and non-repeating decimal representation. A rational number is a number that can", "part 2 $2.1\\overline{36}=2+0.1\\overline{36}$ $$\\dfrac pq=0.1\\overline{36}\\implies 10\\dfrac pq=1.\\overline{36}$$this is eqn (1). multilpy with 100 in eqn(1) $$1000\\dfrac pq=136.\\overline{36}$$ then subtract (1) from (2). $$990\\dfrac pq=135.0$$ $$\\dfrac pq=\\dfrac {135}{990}\\implies\\dfrac {3}{22}$$ so $2.1\\overline{36}\\implies 2+0.1\\overline{36}\\implies 2+\\dfrac{3}{22}\\implies 2\\dfrac3{22}$ -", "# Give the expression of $$8.\\overline{48}$$ This question was previously asked in MPPEB Stenographer 15 Sept 2018 Shift 1 Official Paper View all MP Vyapam Group 2 Papers > 1. $$\\frac{169.6}{20}$$ 2. $$\\frac{840}{99}$$ 3. $$\\frac{848}{100}$$ 4. $$\\frac{848}{99}$$ Option 2 : $$\\frac{840}{99}$$ ## Detailed Solution Calculation: (a)Let x = $$8.\\overline{48}$$ ⇒ x = 8.4848484848......... By multiplying both sides by 100. (b)100x = $$848.\\overline{48}$$ Now, by subtracting (a) from (b), we get 99x = 840 ⇒ x = 840/99 ∴ $$8.\\overline{48}$$ = 840/99", "$$x=0 . \\overline{24} .$$ Then $$100 x=24 . \\overline{24} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=24 . \\overline{24} \\\\ x &=0 . \\overline{24} \\end{aligned} yields $$99 x=24 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{24}{99}=\\frac{8}{33}$$ Exercise $$\\PageIndex{20}$$ $$0 . \\overline{882}$$ Exercise $$\\PageIndex{21}$$ $$0 . \\overline{84}$$ Let $$x=0 . \\overline{84} .$$ Then $$100 x=84 . \\overline{84} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=84 . \\overline{.84} \\\\ x &=0 . \\overline{84} \\end{aligned} yields $$99 x=84 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{84}{99}=\\frac{28}{33}$$ Exercise $$\\PageIndex{22}$$ $$0 . \\overline{384}$$ Exercise $$\\PageIndex{23}$$ $$0 . \\overline{63}$$ Let $$x=0 . \\overline{63} .$$ Then $$100 x=63 . \\overline{63} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=63 . \\overline{63} \\\\ x &=0 . \\overline{63} \\end{aligned} yields $$99 x=63 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{63}{99}=\\frac{7}{11}$$ Exercise $$\\PageIndex{24}$$ $$0 . \\overline{60}$$ Exercise $$\\PageIndex{25}$$ Prove that $$\\sqrt{3}$$ is irrational. Suppose that $$\\sqrt{3}$$ is rational. Then it can be expressed as the ratio of two integers p and q as follows: $\\sqrt{3}=\\frac{p}{q}$ Square both sides, $3=\\frac{p^{2}}{q^{2}}$ then clear the equation of fractions by multiplying both sides by $$q^{2}$$: $p^{2}=3 q^{2}$ Now p and q each have their own unique prime factorizations. Both $$p^{2}$$ and $$q^{2}$$" ]

[ "+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", "+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", "+0 # HELP ASAP 0 81 1 Express $0.4\\overline5$ as a common fraction. Aug 19, 2021 #1 +115 0 Multiply 0.455555...(we set as x) by 10 to get 4.5555555...(10x) then, use 4.555555 to subtract 0.455555 to get 4.1 because later repeat digits cancel 4.1 is equal to 9x, and x is 41/90 Aug 19, 2021", "+0 # Fractions 0 82 1 Express as a fraction in lowest terms: $0.\\overline{12} + 0.\\overline{001}$ May 25, 2022", "+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", "+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 # 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 # 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", "+0 0 29 2 $$Express 0.5\\overline{10} as a common fraction.$$ off-topic Aug 31, 2020 edited by Guest Aug 31, 2020 edited by Guest Aug 31, 2020 edited by Guest Aug 31, 2020 #1 0 0.5/10= 5/100 so 1/20 Aug 31, 2020 #2 +1034 +1 @OP: I'm not sure why you posted this as off-topic @guest that answered: I'm not sure if you can see the LaTeX, but \\overline{} means \"repeating decimal\", not fraction. (the command for fraction is \\frac{}{} or \\dfrac{}{} depending on what size you want it to be) Because my LaTeX is not working, I will do my best to write the fractions and repeating decimals. We want to convert this to a fraction: –– 0.510, which is 0.51010101010101010101010..... To begin, let's call this fraction \"x\". Based on what we know: 1000x = 510.10101010... 10x = 5.101010101010... Subtract one from the other: 990x = 505 Divide by 990: x = 505/990 = 101/198 :) Aug 31, 2020", "+0 # Math -1 100 2 What is $$\\frac{0.\\overline{72}}{0.\\overline{27}}$$? Express your answer as a common fraction in lowest terms. Apr 29, 2022 #1 +1 Apr 29, 2022 #2 +9459 0 You can change both of the recurring decimals to fractions. In fact, $$0.\\overline{72} = \\dfrac{8}{11}$$ and $$0.\\overline{27} = \\dfrac{3}{11}$$ Therefore the answer is $$\\dfrac83$$. Apr 29, 2022", "# 4. Express 0.99999 .... in the form $\\frac{p}{q}$ . Are you surprised by your answer? G Gautam harsolia Let $x = 0.\\overline{9}= 0.9999....$ -(i) Now, multiply by 10 on both sides $10x= 9.999....$ $\\Rightarrow 10x = 9 + x \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\(using \\ (i))$ $\\Rightarrow 9x = 9$ $\\Rightarrow x = \\frac{9}{9} = 1$ Therefore, $\\frac{p}{q}$ form of $0.999....$ is 1 The difference between 1 and 0.999999 is o.000001 which is almost negligible. Therefore, 0.999 is too much closer to 1. Hence, we can write 0.999999.... as 1 Exams Articles Questions", "## Filters Q&A - Ask Doubts and Get Answers Q # Express 0.99999 .... in the form p/q. Are you surprised by your answer? 4. Express 0.99999 .... in the form $\\frac{p}{q}$ . Are you surprised by your answer? Views Let $x = 0.\\overline{9}= 0.9999....$ -(i) Now, multiply by 10 on both sides $10x= 9.999....$ $\\Rightarrow 10x = 9 + x \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\(using \\ (i))$ $\\Rightarrow 9x = 9$ $\\Rightarrow x = \\frac{9}{9} = 1$ Therefore, $\\frac{p}{q}$ form of $0.999....$ is 1 The difference between 1 and 0.999999 is o.000001 which is almost negligible. Therefore, 0.999 is too much closer to 1. Hence, we can write 0.999999.... as 1 Exams Articles Questions", "Home/Class 9/Maths/ Show that $$0.2353535... =0.2\\overline{35}$$ can be expressed in the form $$\\frac {p}{q},$$ where $$p$$ and $$q$$ are integers and $$q\\neq 0.$$ Speed 00:00 01:39 ## QuestionMathsClass 9 Show that $$0.2353535... =0.2\\overline{35}$$ can be expressed in the form $$\\frac {p}{q},$$ where $$p$$ and $$q$$ are integers and $$q\\neq 0.$$ We have, $$x=0.2\\overline{35}$$. ...(i) Multiplying eqn. (i) by $$10$$, we get $$10x=2.\\overline{35}$$ ...(ii) Multiplying eqn. (ii) by $$1000$$, we get $$1000x=235.\\overline{35}$$ ...(iii) Subtracting eqn. (ii) from eqn. (iii), we get $$1000x-10x=235.\\overline{35}-2.\\overline{35}$$ $$\\Rightarrow 990x=233$$ $$\\Rightarrow x=\\frac{233}{990}$$.", "# 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.", "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", "+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", "fractions observe that if $x=0.76767676767676\\ldots, 100x=76.76767676767676\\ldots$ Subtracting $99x=76$ (this can be made formal if needed). So your fraction is what? @JackyBoi: Because the denominator is specified as $9900$, the question wants $7600$, but you have understood correctly. – Ross Millikan Sep 4 '12 at 14:58 For the general case, $10000\\cdot 0.ab\\overline{cd} = abcd.\\overline{cd}$, $100\\cdot 0.ab\\overline{cd} = ab.\\overline{cd}$, hence $9900\\cdot 0.ab\\overline{cd} = abcd-ab$.", "+0 # Express $\\sqrt{x} \\div\\sqrt{y}$ as a common fraction, given: \\$\\frac{ {\\left( \\frac{1}{2} \\right)}^2 + {\\left( \\frac{1}{3} \\right)}^2 }{ {\\l 0 69 1 Express$$\\sqrt{x}$$ ÷$$\\sqrt{y}$$ as a common fraction, given: $$((1/2)^2+(1/3)^2)/(1/4)^2+(1/5)^2$$=$$13x/41y$$ It's due in 10 minutes so could you please hurry!!!!! Thank you so much! Also can you explain what you do so I can learn from it. Thanks again!!! :) Nov 24, 2020 #1 0 (1/2)2+(1/3)2=13x solve for x (1/4)2+(1/5)2=41y solve for y 1/4+1/9=13x 1/16+1/25=41y 13/36=13x 41/400=41y 1/36=x 1/400=y √(1/36) / √(1/400) = (1/6) / (1/20) 20/6 or 10/3 Nov 25, 2020", "the fractional form of $0.\\overline{251}$. ### Solution To convert the given non-terminating decimal to a fraction, suppose that: $y=0.\\overline{251}=0.251251…$ Since there are three repeating digits, multiply both sides by $1000$: $1000y=251.251251…$ Since, $999y=1000y-y$ Therefore, $999y=251.251251…-0.251251…$ $999y=251$ Divide both sides by $999$ we get: $y=\\dfrac{251}{999}$ ## Example 2 Write the fractional form of $0.34\\overline{12}$. ### Solution To convert the given non-terminating decimal to a fraction suppose that: $y=0.34\\overline{12}=0.341212…$ Since there are two repeating digits, so multiply both sides by $100$: $100y=34.1212…$ Since, $99y=100y-y$ Therefore, $99y=34.1212…-0.341212…$ $99y=33.78$ Divide both sides by $99$ we get: $y=\\dfrac{33.78}{99}$ $y=\\dfrac{3378}{99\\times 100}$ $y=\\dfrac{3378}{9900}$ ## Example 3 Write the fractional form of $0.00\\overline{12}$. ### Solution To convert the given non-terminating decimal to a fraction suppose that: $y=0.00\\overline{12}=0.001212…$ Since there are two repeating digits, so multiply both sides by $100$: $100y=0.1212…$ Since, $99y=100y-y$ Therefore, $99y=0.1212…-0.001212…$ $99y=0.12$ Divide both sides by $99$ we get: $y=\\dfrac{0.12}{99}$ $y=\\dfrac{12}{99\\times 100}$ $y=\\dfrac{12}{9900}$" ]

[ "= 8$ and $W = 4 + 2 s = 6$", "7x - 6y + 5$", "and w And (b) the value of w when x =8", "$H-H_0 = W$.", "$- 7$. $w = 1$", "load is $5\\mu W$.", "\\\\[3ex] (b.)\\:\\: (7w + 2) - (-3w + 4) \\\\[3ex] (c.)\\:\\: 3(7w + 2) + 5(-3w + 4) \\\\[3ex] (d.)\\:\\: -2(7w - 5) + 5(-2w + 1) \\\\[3ex] (e.)\\:\\: (7w + 2)(-3w + 4) \\\\[3ex]$ $(a.) \\\\[3ex] (7w + 2) + (-3w + 4) \\\\[3ex] = 1(7w + 2) + 1(-3w + 4) \\\\[3ex] = 7w + 2 - 3w + 4 \\\\[3ex] = 4w + 6 \\\\[3ex] (b.) \\\\[3ex] (7w + 2) - (-3w + 4) \\\\[3ex] = 1(7w + 2) -1(-3w + 4) \\\\[3ex] = 7w + 2 + 3w - 4 \\\\[3ex] = 10w - 2 \\\\[3ex] (c.) \\\\[3ex] 3(7w + 2) + 5(-3w + 4) \\\\[3ex] = 21w + 6 - 15w + 20 \\\\[3ex] = 6w + 26 \\\\[3ex] (d.) \\\\[3ex] -2(7w - 5) + 5(-2w + 1) \\\\[3ex] = -14w + 10 - 10w + 5 \\\\[3ex] = -24w + 15 \\\\[3ex] (e.) \\\\[3ex] (7w + 2)(-3w + 4) \\\\[3ex] 7w(-3w) = -21w^2 \\\\[3ex] 7w(4) = 28w \\\\[3ex] 2(-3w) = -6w \\\\[3ex] 2(4) = 8 \\\\[3ex] = -21w^2 + 28w - 6w + 8 \\\\[3ex] = -21w^2 + 22w + 8$ (12.) ACT Given that $(x + 2)$ and $(x - 1)$ are factors of the quadratic expression below, what are the values of $a$ and $b$? $x^2 + (a + 2)x", "# How do you solve and check your solutions to -5.7=w-4.6? Sep 22, 2016 $w = - 1.1$ #### Explanation: $- 5.7 = w - 4.6$ or $w = - 5.7 + 4.6$ or $w = - 1.1$", "$W=X$.", "## Intermediate Algebra (6th Edition) $w=-2$ and $w=.8$ We are given that $|5w+3|=7$. Therefore, $5w+3=7$ and $5w+3=-7$ We can solve each equation to find the solutions. For $5w+3=7$, first subtract 3 from both sides. $5w=4$ Divide both sides by 5. $w=.8$ For $5w+3=-7$, first subtract 3 from both sides. $5w=-10$ Divide both sides by 5. $w=-2$", "w = 0.8..", "[4, 5, 6], [7, 8, 9]], w =1 -> y = [[5]]", "W = 4.3 \\times {10}^{-} 4 J$", "$w=ts$", "4 \\, \\text{W}$$", "(-3w + 4) \\\\[3ex] (b.)\\:\\: (7w + 2) - (-3w + 4) \\\\[3ex] (c.)\\:\\: 3(7w + 2) + 5(-3w + 4) \\\\[3ex] (d.)\\:\\: -2(7w - 5) + 5(-2w + 1) \\\\[3ex] (e.)\\:\\: (7w + 2)(-3w + 4) \\\\[3ex]$ $(a.) \\\\[3ex] (7w + 2) + (-3w + 4) \\\\[3ex] = 1(7w + 2) + 1(-3w + 4) \\\\[3ex] = 7w + 2 - 3w + 4 \\\\[3ex] = 4w + 6 \\\\[3ex] (b.) \\\\[3ex] (7w + 2) - (-3w + 4) \\\\[3ex] = 1(7w + 2) -1(-3w + 4) \\\\[3ex] = 7w + 2 + 3w - 4 \\\\[3ex] = 10w - 2 \\\\[3ex] (c.) \\\\[3ex] 3(7w + 2) + 5(-3w + 4) \\\\[3ex] = 21w + 6 - 15w + 20 \\\\[3ex] = 6w + 26 \\\\[3ex] (d.) \\\\[3ex] -2(7w - 5) + 5(-2w + 1) \\\\[3ex] = -14w + 10 - 10w + 5 \\\\[3ex] = -24w + 15 \\\\[3ex] (e.) \\\\[3ex] (7w + 2)(-3w + 4) \\\\[3ex] 7w(-3w) = -21w^2 \\\\[3ex] 7w(4) = 28w \\\\[3ex] 2(-3w) = -6w \\\\[3ex] 2(4) = 8 \\\\[3ex] = -21w^2 + 28w - 6w + 8 \\\\[3ex] = -21w^2 + 22w + 8$ (12.) ACT Given that $(x + 2)$ and $(x - 1)$ are factors of the quadratic expression below, what are the values of $a$ and $b$? $x^2 +", ". . . . . . . . .8% w/w (9.1% w/v)", "# What is the answer to this? W-3.2=5.6 Feb 21, 2018 $8.8$ #### Explanation: $W - 3.2 = 5.6$ $W = 5.6 + 3.2$ $W = 8.8$ Feb 21, 2018 $W = 8.8$ #### Explanation: To solve for $W$, remove the $- 3.2$ by adding $+ 3.2$ to both sides of the $=$ sign. $W - 3.2 + 3.2 = 5.6 + 3.2$ $W = 8.8$ To solve for $W$ we must clear all other items (in this case, the $- 3.2$).", "w_1 x_1^4 + w_2 x_2^4 + w_3 x_3^4\\\\ \\int_1^2 x^5 (x^4 - 1)dx = 91.8 = w_1 x_1^5 + w_2 x_2^5 + w_3 x_3^5\\\\ \\int_1^2 x^6 (x^4 - 1)dx = 167.9480519 \\neq 167.947233 = w_1 x_1^6 + w_2 x_2^6 + w_3 x_3^6.$$", "8) ~ wt + gear\"" ]

[ "# How do you combine like terms in (2w ^ { 2} - 5w - 7) - ( 4w ^ { 2} + 8+ 3w )? Like terms have the same variable power (so the ${w}^{2}$ terms combine, the $w$ terms combine, and the terms without any $w$s combine). $\\left(2 {w}^{2} - 5 w - 7\\right) - \\left(4 {w}^{2} + 8 + 3 w\\right)$ $= 2 {w}^{2} - 5 w - 7 - 4 {w}^{2} - 8 - 3 w$ $= - 2 {w}^{2} - 8 w - 15$", "+ 8 -1w = 0 + 8 Combine like terms: 0 + 8 = 8 -1w = 8 Divide each side by '-1'. w = -8 Simplifying w = -8 Subproblem 2Set the factor '(5 + -1w)' equal to zero and attempt to solve: Simplifying 5 + -1w = 0 Solving 5 + -1w = 0 Move all terms containing w to the left, all other terms to the right. Add '-5' to each side of the equation. 5 + -5 + -1w = 0 + -5 Combine like terms: 5 + -5 = 0 0 + -1w = 0 + -5 -1w = 0 + -5 Combine like terms: 0 + -5 = -5 -1w = -5 Divide each side by '-1'. w = 5 Simplifying w = 5Solutionw = {-8, 5}`", "5j = 7j - 5j = 2j There are no other like terms to combine. The answer is 2j. Simplify by combining like terms: 8w-8w+4w Solution: 8w-8w+4w= w", "+ -6w = -30 + 17 + -6w Combine like terms: -30 + 17 = -13 -13 + -6w = -13 + -6w Add '13' to each side of the equation. -13 + 13 + -6w = -13 + 13 + -6w Combine like terms: -13 + 13 = 0 0 + -6w = -13 + 13 + -6w -6w = -13 + 13 + -6w Combine like terms: -13 + 13 = 0 -6w = 0 + -6w -6w = -6w Add '6w' to each side of the equation. -6w + 6w = -6w + 6w Combine like terms: -6w + 6w = 0 0 = -6w + 6w Combine like terms: -6w + 6w = 0 0 = 0 Solving 0 = 0 Couldn't find a variable to solve for. This equation is an identity, all real numbers are solutions.`", "do you combine like terms in (4t ^ { 3} - 11t ^ { 2} + 4t ) - ( - 7t ^ { 2} - 5t + 8)? How do you combine like terms in (4t ^ { 3} - 11t ^ { 2} + 4t ) - ( - 7t ^ { 2} - 5t + 8)?...", "+ -6w = -30 + 16 + -6w Combine like terms: -30 + 16 = -14 -13 + -6w = -14 + -6w Add '6w' to each side of the equation. -13 + -6w + 6w = -14 + -6w + 6w Combine like terms: -6w + 6w = 0 -13 + 0 = -14 + -6w + 6w -13 = -14 + -6w + 6w Combine like terms: -6w + 6w = 0 -13 = -14 + 0 -13 = -14 Solving -13 = -14 Couldn't find a variable to solve for. This equation is invalid, the left and right sides are not equal, therefore there is no solution.`", "X + 3x + -3x = -6 + (-1x) + -3x Combine like terms: 3x + -3x = 0 X + 0 = -6 + (-1x) + -3x X = -6 + (-1x) + -3x Combine like terms: (-1x) + -3x = -4x X = -6 + -4x Simplifying X = -6 + -4x`", "-5 + 8r = 21 + -5 Combine like terms: 5 + -5 = 0 0 + 8r = 21 + -5 8r = 21 + -5 Combine like terms: 21 + -5 = 16 8r = 16 Divide each side by '8'. r = 2 Simplifying r = 2`", "+ -7glox Combine like terms: 14glox + -7glox = 7glox -5 + 7glox = -1 + 7glox + -7glox Combine like terms: 7glox + -7glox = 0 -5 + 7glox = -1 + 0 -5 + 7glox = -1 Add '5' to each side of the equation. -5 + 5 + 7glox = -1 + 5 Combine like terms: -5 + 5 = 0 0 + 7glox = -1 + 5 7glox = -1 + 5 Combine like terms: -1 + 5 = 4 7glox = 4 Divide each side by '7lox'. g = 0.5714285714l-1o-1x-1 Simplifying g = 0.5714285714l-1o-1x-1`", "3 - 1 + 12x + x - 5x Combining like terms: 2 + 8x $\\Rightarrow$ 12x + 3 + x - 1 - 5x = 2 + 8x", "- 5j = 2j There are no other like terms to combine. The answer is 2j. Simplify by combining like terms: -5z+6z+8z Solution: -5z+6z+8z= z", "How do you combine like terms in -2s + 6- s - 4? Jun 10, 2018 $- 3 s + 2$ Explanation: the like terms here are: $- 2 s \\mathmr{and} - s$ & $+ 6 \\mathmr{and} - 4$ $- 2 s - s = - 3 s$ & $6 - 4 = 2$ $- 3 s + 2$", "# How do you combine like terms in -3( p - 2) - ( 8p + 5)? Jun 16, 2017 $- 3 p + 6 - 8 p - 5$ $- 11 p + 1$", "How do you combine (8-4m) - (m-2)? Apr 30, 2015 The answer is $- 5 m + 6$. Since the terms inside the parentheses are simplified as much as possible, you can remove them. $8 - 4 m - m - 2$ Group like terms. $- 4 m - m + 8 - 2$ $- 5 m + 6$", "* 3) -6 + 4x = 2x + 2x2 + (-9 + -3x + 6x2) Reorder the terms: -6 + 4x = -9 + 2x + -3x + 2x2 + 6x2 Combine like terms: 2x + -3x = -1x -6 + 4x = -9 + -1x + 2x2 + 6x2 Combine like terms: 2x2 + 6x2 = 8x2 -6 + 4x = -9 + -1x + 8x2 Solving -6 + 4x = -9 + -1x + 8x2 Reorder the terms: -6 + 9 + 4x + x + -8x2 = -9 + -1x + 8x2 + 9 + x + -8x2 Combine like terms: -6 + 9 = 3 3 + 4x + x + -8x2 = -9 + -1x + 8x2 + 9 + x + -8x2 Combine like terms: 4x + x = 5x 3 + 5x + -8x2 = -9 + -1x + 8x2 + 9 + x + -8x2 Reorder the terms: 3 + 5x + -8x2 = -9 + 9 + -1x + x + 8x2 + -8x2 Combine like terms: -9 + 9 = 0 3 + 5x + -8x2 = 0 + -1x + x + 8x2 + -8x2 3 + 5x + -8x2 = -1x + x + 8x2 + -8x2 Combine like terms: -1x + x = 0 3", "# How do you combine like terms in 5t + 8d - 2t - 3d? Mar 13, 2018 Join the like terms together with the operation going with the term. #### Explanation: So it's $5 t + 8 d - 2 t - 3 d$ Combining them in order is $+ 5 t - 2 t + 8 d - 3 d$ Completing the operations for only the letter-number combinations which match equals $3 t + 5 d$", "5p = 1 Add '-7' to each side of the equation. 7 + -7 + 5p = 1 + -7 Combine like terms: 7 + -7 = 0 0 + 5p = 1 + -7 5p = 1 + -7 Combine like terms: 1 + -7 = -6 5p = -6 Divide each side by '5'. p = -1.2 Simplifying p = -1.2`", "2w + 8w - 8 = w^2 + 8w$$ $$w^2 - 2w - 8 = 0$$ $$(w - 4)(w + 2) = 0$$ $$w = 4$$ or $$w = -2$$ First option is not a choice. w = -2 *Use $$\\frac{1}{a} + \\frac{1}{b} = \\frac{a + b}{ab}$$ If f(w) = (1/w) + 1\\(w+8) and the function f(w) equals 1/(w-1) then a &nbs [#permalink] 20 Oct 2017, 07:26 Display posts from previous: Sort by", "6 - \\frac{49}{4}$ combine like-terms $y = {\\left(x - \\frac{7}{2}\\right)}^{2} - \\frac{25}{4}$", "$ \\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" ]

[ "# How do you combine like terms in (2w ^ { 2} - 5w - 7) - ( 4w ^ { 2} + 8+ 3w )? Like terms have the same variable power (so the ${w}^{2}$ terms combine, the $w$ terms combine, and the terms without any $w$s combine). $\\left(2 {w}^{2} - 5 w - 7\\right) - \\left(4 {w}^{2} + 8 + 3 w\\right)$ $= 2 {w}^{2} - 5 w - 7 - 4 {w}^{2} - 8 - 3 w$ $= - 2 {w}^{2} - 8 w - 15$", "+ -6w = -30 + 16 + -6w Combine like terms: -30 + 16 = -14 -13 + -6w = -14 + -6w Add '6w' to each side of the equation. -13 + -6w + 6w = -14 + -6w + 6w Combine like terms: -6w + 6w = 0 -13 + 0 = -14 + -6w + 6w -13 = -14 + -6w + 6w Combine like terms: -6w + 6w = 0 -13 = -14 + 0 -13 = -14 Solving -13 = -14 Couldn't find a variable to solve for. This equation is invalid, the left and right sides are not equal, therefore there is no solution.`", "+ -6w = -30 + 17 + -6w Combine like terms: -30 + 17 = -13 -13 + -6w = -13 + -6w Add '13' to each side of the equation. -13 + 13 + -6w = -13 + 13 + -6w Combine like terms: -13 + 13 = 0 0 + -6w = -13 + 13 + -6w -6w = -13 + 13 + -6w Combine like terms: -13 + 13 = 0 -6w = 0 + -6w -6w = -6w Add '6w' to each side of the equation. -6w + 6w = -6w + 6w Combine like terms: -6w + 6w = 0 0 = -6w + 6w Combine like terms: -6w + 6w = 0 0 = 0 Solving 0 = 0 Couldn't find a variable to solve for. This equation is an identity, all real numbers are solutions.`", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "\\\\[3ex] (b.)\\:\\: (7w + 2) - (-3w + 4) \\\\[3ex] (c.)\\:\\: 3(7w + 2) + 5(-3w + 4) \\\\[3ex] (d.)\\:\\: -2(7w - 5) + 5(-2w + 1) \\\\[3ex] (e.)\\:\\: (7w + 2)(-3w + 4) \\\\[3ex]$ $(a.) \\\\[3ex] (7w + 2) + (-3w + 4) \\\\[3ex] = 1(7w + 2) + 1(-3w + 4) \\\\[3ex] = 7w + 2 - 3w + 4 \\\\[3ex] = 4w + 6 \\\\[3ex] (b.) \\\\[3ex] (7w + 2) - (-3w + 4) \\\\[3ex] = 1(7w + 2) -1(-3w + 4) \\\\[3ex] = 7w + 2 + 3w - 4 \\\\[3ex] = 10w - 2 \\\\[3ex] (c.) \\\\[3ex] 3(7w + 2) + 5(-3w + 4) \\\\[3ex] = 21w + 6 - 15w + 20 \\\\[3ex] = 6w + 26 \\\\[3ex] (d.) \\\\[3ex] -2(7w - 5) + 5(-2w + 1) \\\\[3ex] = -14w + 10 - 10w + 5 \\\\[3ex] = -24w + 15 \\\\[3ex] (e.) \\\\[3ex] (7w + 2)(-3w + 4) \\\\[3ex] 7w(-3w) = -21w^2 \\\\[3ex] 7w(4) = 28w \\\\[3ex] 2(-3w) = -6w \\\\[3ex] 2(4) = 8 \\\\[3ex] = -21w^2 + 28w - 6w + 8 \\\\[3ex] = -21w^2 + 22w + 8$ (12.) ACT Given that $(x + 2)$ and $(x - 1)$ are factors of the quadratic expression below, what are the values of $a$ and $b$? $x^2 + (a + 2)x", "$ \\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", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "= 8$ and $W = 4 + 2 s = 6$", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "+ 8 -1w = 0 + 8 Combine like terms: 0 + 8 = 8 -1w = 8 Divide each side by '-1'. w = -8 Simplifying w = -8 Subproblem 2Set the factor '(5 + -1w)' equal to zero and attempt to solve: Simplifying 5 + -1w = 0 Solving 5 + -1w = 0 Move all terms containing w to the left, all other terms to the right. Add '-5' to each side of the equation. 5 + -5 + -1w = 0 + -5 Combine like terms: 5 + -5 = 0 0 + -1w = 0 + -5 -1w = 0 + -5 Combine like terms: 0 + -5 = -5 -1w = -5 Divide each side by '-1'. w = 5 Simplifying w = 5Solutionw = {-8, 5}`", "(-3w + 4) \\\\[3ex] (b.)\\:\\: (7w + 2) - (-3w + 4) \\\\[3ex] (c.)\\:\\: 3(7w + 2) + 5(-3w + 4) \\\\[3ex] (d.)\\:\\: -2(7w - 5) + 5(-2w + 1) \\\\[3ex] (e.)\\:\\: (7w + 2)(-3w + 4) \\\\[3ex]$ $(a.) \\\\[3ex] (7w + 2) + (-3w + 4) \\\\[3ex] = 1(7w + 2) + 1(-3w + 4) \\\\[3ex] = 7w + 2 - 3w + 4 \\\\[3ex] = 4w + 6 \\\\[3ex] (b.) \\\\[3ex] (7w + 2) - (-3w + 4) \\\\[3ex] = 1(7w + 2) -1(-3w + 4) \\\\[3ex] = 7w + 2 + 3w - 4 \\\\[3ex] = 10w - 2 \\\\[3ex] (c.) \\\\[3ex] 3(7w + 2) + 5(-3w + 4) \\\\[3ex] = 21w + 6 - 15w + 20 \\\\[3ex] = 6w + 26 \\\\[3ex] (d.) \\\\[3ex] -2(7w - 5) + 5(-2w + 1) \\\\[3ex] = -14w + 10 - 10w + 5 \\\\[3ex] = -24w + 15 \\\\[3ex] (e.) \\\\[3ex] (7w + 2)(-3w + 4) \\\\[3ex] 7w(-3w) = -21w^2 \\\\[3ex] 7w(4) = 28w \\\\[3ex] 2(-3w) = -6w \\\\[3ex] 2(4) = 8 \\\\[3ex] = -21w^2 + 28w - 6w + 8 \\\\[3ex] = -21w^2 + 22w + 8$ (12.) ACT Given that $(x + 2)$ and $(x - 1)$ are factors of the quadratic expression below, what are the values of $a$ and $b$? $x^2 +", "# How do you factor 5w^2+7w-6? Mar 12, 2018 $\\left(w + 2\\right) \\left(5 w - 3\\right)$ #### Explanation: Write as a sum or difference $5 {w}^{2} + 10 w - 3 w - 6$ Factor out $5 w$ and $- 3$ from the expression $5 w \\left(w + 2\\right) - 3 \\left(w + 2\\right)$ Factor out $w + 2$ from the expression $\\left(w + 2\\right) \\left(5 w - 3\\right)$", "x^4 + 10 * x^3 - 10 * x^2 %8 = (w^16 + w^15 + w^14 + w^13 + w^12 + w^11 + w^10 + 5*w^9 + 5*w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1)/w^8 ? f = x^8 + x^7 - 7 * x^6 - 6 * x^5 + 15 * x^4 + 10 * x^3 - 10 * x^2 - 4 * x %9 = (w^16 + w^15 + w^14 + w^13 + w^12 + w^11 + w^10 + w^9 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1)/w^8 ? f = x^8 + x^7 - 7 * x^6 - 6 * x^5 + 15 * x^4 + 10 * x^3 - 10 * x^2 - 4 * x + 1 %10 = (w^16 + w^15 + w^14 + w^13 + w^12 + w^11 + w^10 + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1)/w^8 ? ? f * w^8 * ( w - 1 ) %11 = w^17 - 1 ? • I'm unfamiliar with this language and having a hard time discerning what steps are being done in the build-up. Is this just saying that $p(w+1/w)$ where $p$ is", "$GCH$ and $GEA$, respectively. Thus, $y:y+8 = 3:5$, from which we have $y=12$, thus $GF=y+8=12+8=\\boxed{\\mathrm{(B)}\\ 20}$", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "How do you factor -8w+49w^2+14w^3-28? Collect $2 w$ between the first and third term and $7$ between the second and rourth: $2 w \\left(- 4 + 7 {w}^{2}\\right) + 7 \\left(7 {w}^{2} - 4\\right) =$ collect $\\left(7 {w}^{2} - 4\\right)$ $= \\left(7 {w}^{2} - 4\\right) \\left(2 w + 7\\right)$", "the only option is the pair $$(7, 3)$$ which gives $$a = 5, b = 4$$. Thus the area is $$\\frac{1}{2}(3b+2a-6) = \\boxed{8}$$.", "× # How do you factor w^2+4w+3? Mar 7, 2018 $\\left(w + 3\\right) \\left(w + 1\\right)$ In this case, since $a = 1$ (in $a {x}^{2} + b x + c$), simply look for two numbers whose sum is $4$ and whose product is $3$. You can easily see that the two numbers are $1$ and $3$. Then, just write it down: $\\left(w + 3\\right) \\left(w + 1\\right)$", "# How do you solve 9-3w=2=4w? ##### 1 Answer Aug 9, 2015 There is no definite answer to this problem as it is written. However if one equal sign were changed to a positive sign, it is possible to solve the problem. #### Explanation: $9 - 3 w = 2 = 4 w$ The equation has two equal signs. More than likely, one of the two $=$ symbols was meant to be a $+$ symbol, since they are on the same key. If we change one of the $=$ signs to a $+$ sign, technically it could mean $9 - 3 w + 2 = 4 w ,$ or $9 - 3 w = 2 + 4 w$. One Possibility $9 - 3 w + 2 = 4 w$ Add $3 w$ to both sides. $9 + 2 = 7 w$ $11 = 7 w$ Divide both sides by $7$. $\\frac{11}{7} = w$ Switch sides $w = \\frac{11}{7}$ A Second Possibility $9 - 3 w = 2 + 4 w$ Subtract $9$ from both sides. $- 3 w = 2 + 4 w - 9$ $- 3 w = 4 w - 7$ Subtract $4 w$ from both sides. $- 3 w - 4 w = - 7$ $- 7 w = - 7$ Divide both sides by", "# -5(w+2)-w-3=-6(w+5)+17 ## Simple and best practice solution for -5(w+2)-w-3=-6(w+5)+17 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework. If it's not what You are looking for type in the equation solver your own equation and let us solve it. ## Solution for -5(w+2)-w-3=-6(w+5)+17 equation: Simplifying -5(w + 2) + -1w + -3 = -6(w + 5) + 17 Reorder the terms: -5(2 + w) + -1w + -3 = -6(w + 5) + 17 (2 * -5 + w * -5) + -1w + -3 = -6(w + 5) + 17 (-10 + -5w) + -1w + -3 = -6(w + 5) + 17 Reorder the terms: -10 + -3 + -5w + -1w = -6(w + 5) + 17 Combine like terms: -10 + -3 = -13 -13 + -5w + -1w = -6(w + 5) + 17 Combine like terms: -5w + -1w = -6w -13 + -6w = -6(w + 5) + 17 Reorder the terms: -13 + -6w = -6(5 + w) + 17 -13 + -6w = (5 * -6 + w * -6) + 17 -13 + -6w = (-30 + -6w) + 17 Reorder the terms: -13" ]

[ "to 4-decimal ...", "equal than $4$.", "$3/4$.", "# How do you write -4 2/3 as a decimal? $3.334$ $\\frac{\\left(- 4 \\cdot 3\\right) + 2}{3}$ $- \\frac{10}{3}$ $3.334$", "+ 88 = 0$", "4, 3]$.", "4 = 4 - 4 = 0$", "the $$n$$-th decimal value of $$a_n$$ and see whether smooshing them together and swapping the 4s and 7s does anything.", "= 3s-4$. -", "# What is 2 1/8 as a decimal? $2.125$ $2 \\frac{1}{8} = \\frac{1 + 2 \\cdot 8}{8} = \\frac{17}{8} = 17 : 8 = 2.125$", ".3/.4 as a decimal .3/.4 as a decimal - solution and the full explanation with calculations. If it's not what You are looking for, type in into the box below your number and see the solution. What is .3/.4 as a decimal? To write .3/.4 as a decimal you have to divide numerator by the denominator of the fraction. We divide now .3 by .4 what we write down as .3/.4 and we get 0.75 And finally we have: .3/.4 as a decimal equals 0.75", "$H(50)$, rounded to 4 digits after the decimal point.", "repeating decimals. For example $34.777777…$, $92.82828282…$.", "E. $3$", "4^3$.", "subtracting$4.00 with $1.00 we get$3.00.", "4 = 21.5$$.", "- 3 x - 4$", "# Can you solve it? When $$4444^{4444}$$ is written in decimal notation,the sum of its digits is A.Let B be the sum of digits of A.Find the sum of digits of B. ×", "$-4$ 3. $-1$ 4. $+2$" ]

[ "This is actually a lower bound on the number of operations, because frequently you will add two numbers and get something larger than 10, which requires you to add a carry digit to the next column over. It turns out that, on average, adding M numbers with N digits, it will take $\\left( M + \\frac{1}{2} \\right) N$ operations. ## Multiplication Does multiplication take more operations to complete? Let’s take a look. $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\ \\cline{2-2} & \\end{array}$ Our computer starts off by looking up $1\\times3 = 3$ and then $1\\times1 =1$. The next step is then looking up $2\\times3 = 6$ and $2\\times 1 = 2$ so we’ve completed four operations and have the following results: $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\ \\cline{2-2} & 13 \\\\ & 260 \\\\ \\end{array}$ Now we need to add the two numbers together. The number of operations is either one or three, depending on whether you count addition by zero as an operation. I’m going to count it as an operation since we assume our computer doesn’t understand anything about numbers and needs to look up even this simple process. This means that multiplying two 2 digit numbers together takes seven operations, which is clearly larger than the four operations taken to add two", "then add the third number to this results. $\\begin{array}{@{}cr} & 31 \\\\ + & 17 \\\\ + & 12 \\\\ \\cline{2-2} & \\end{array}$ The first two numbers give you 48 and took four operations, and adding 12 to 48 takes another two operations, for a total of six. We can extrapolate to see that adding M 2 digit numbers together takes a total of 2M operations. You can also easily see that if you expand this even further to consider numbers with N digits, it will take $N\\times M$ operations to add those numbers together. This is actually a lower bound on the number of operations, because frequently you will add two numbers and get something larger than 10, which requires you to add a carry digit to the next column over. It turns out that, on average, adding M numbers with N digits, it will take $\\left( M + \\frac{1}{2} \\right) N$ operations. ## Multiplication Does multiplication take more operations to complete? Let’s take a look. $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\ \\cline{2-2} & \\end{array}$ Our computer starts off by looking up $1\\times3 = 3$ and then $1\\times1 =1$. The next step is then looking up $2\\times3 = 6$ and $2\\times 1 = 2$ so we’ve completed four operations and have the following results: $\\begin{array}{@{}cr}", "example. Since we know the ones digit of the sum is $$6$$, then the ones digits for each of the two numbers being added must have a total of $$6$$ or $$16$$. The only possible combinations given the digits we have are $$1 + 5$$ or $$5 + 1$$. Since neither of these sums gives us $$16$$, we do not have to worry about carrying a $$1$$ into the tens column. Looking at the tens column, we need two numbers that add up to $$9$$ or $$19$$. There is only one pair of the remaining digits we could use: $$3 + 6$$ or $$6 + 3$$. Neither of these sums gives us $$19$$, so there is nothing to carry into the hundreds column. This means the sum of the hundreds column must be $$12$$. The remaining digits are $$4$$ and $$8$$, and fortunately the sum of these digits is $$12$$. As in parts A) and B), we can arrange these digits in $$8$$ different ways. They are all given below. $$\\begin{array}{c c c c} &4&6&1\\\\ +&8&3&5\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &8&3&5\\\\ +&4&6&1\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &8&3&1\\\\ +&4&6&5\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &4&6&5\\\\ +&8&3&1\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &4&3&5\\\\ +&8&6&1\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &8&6&1\\\\ +&4&3&5\\\\ \\hline \\end{array}$$", "will assume that $A,B,C,D$ are four different digits. . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ A & B & C & D \\\\ \\times &&& 9 \\\\ \\hline D & C & B & A \\end{array}$ In column-1, we see that: . $A = 1,\\:D = 9$ And there is no \"carry\" from column-2. So we have: . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ 1 & B & C & 9 \\\\ \\times &&& 9 \\\\ \\hline 9 & C & B & 1 \\end{array}$ Since there is no \"carry\" frim column-2, $\\,B$ must be 0 or 1. Since $A = 1$, then: . $B \\,=\\,0$ Then we have: . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ 1 & 0 & C & 9 \\\\ \\times &&& 9 \\\\ \\hline 9 & C & 0 & 1 \\end{array}$ In column-3, we have: . $9C + 8$ ends in $0.$ . . Hence, $\\,9C$ ends in $2 \\quad\\Rightarrow\\quad C = 8$ Therefore, the solution is: . . $\\begin{array}{cccc} 1 & 0 & 8 & 9 \\\\ \\times &&& 9 \\\\ \\hline 9 & 8 & 0 & 1 \\end{array}$ 6. wonderful thank you!", "& & \\end{array}$ It is easy to add the new set of numbers to get the answer: $30\\ 000 + 5\\ 000 + 1\\ 700 + 120 + 11 = 36\\ 831$ Starting with the ones as shown above makes it possible to do more of the work mentally, and write less, as shown below. $\\begin{array}{ccccc} & 3 & ^{1}1 & ^{1}8 & ^{1}3 & 7 \\\\ & & 4 & 9 & 9 & 4 \\\\ \\hline & 3 & 6 & 8 & 3 & 1 \\end{array}$ To achieve this, the first step is to add the digits in the ones column: $$7 + 4 = 11$$. Only write the digit “$$1$$” of the $$11$$. The $$10$$ of the $$11$$ is remembered and added to the $$3$$ in the tens column. We say the $$10$$ is carried from the ones column to the tens column. The same is done when the tens parts are added to get $$13$$: only the digit “$$3$$” is written (in the tens column, so it means $$30$$), and the $$1$$ is carried to the next column. This $$1$$ is actually $$100$$, so it needs to be carried to the hundreds column. ## Worked Example 1.8: Adding whole numbers Calculate without using a calculator: $4\\ 638 + 2\\ 667$ ### Break the", "\\end{array}$$ You can carry out the addition in a similar way as when working with base-10 numbers. Begin in the position with lowest value, the $2^0$ position found furthest to the right. Adding this column, we have $1 + 1 = 10$ (\"one\" plus \"one\" equals \"two\", written in binary). $$\\begin{array}{r} \\texttt{ 1 } \\\\ \\texttt{10011} \\\\ +\\text{ } \\texttt{101011} \\\\ \\hline \\texttt{ 0} \\end{array}$$ In the second right-most place, the addition is $1 + 1 + 1 = 11$, resulting in a $1$ with a \"carry\" of $1$. $$\\begin{array}{r} \\texttt{ 11 } \\\\ \\texttt{10011} \\\\ +\\text{ } \\texttt{101011} \\\\ \\hline \\texttt{ 10} \\end{array}$$ The rest of the addition is shown below $$\\begin{array}{r} \\texttt{ 11 } \\\\ \\texttt{10011} \\\\ +\\text{ } \\texttt{101011} \\\\ \\hline \\texttt{111110} \\end{array}$$ Converting $111110$ to decimal should confirm the sum of $19$ and $43$: $$1\\cdot2^5 + 1\\cdot2^4 + 1\\cdot2^3 + 1\\cdot2^2 + 1\\cdot2^1 + 0\\cdot2^0 = 32 + 16 + 8 + 4 + 2 = 62$$ Adding binary numbers together helps illustrate that adding number yields the same result no matter which base we choose to work in. However, most cryptographic operations with binary numbers use bitwise addition, denoted by the symbol $\\oplus$. For binary numbers, bitwise addition is equivalent to adding each digit modulo $2$. Note that there is no carrying when", "$$4$$ we carried is $$24$$, but that's 3 eights and no units, so we put down $$30$$, not $$24$$: $$\\begin{array}{lll}&2&5^4\\\\×&4&7\\\\\\hline 3&0&3\\end{array}$$ That $$303$$ is because $$5×47$$ is written as $$303$$ in base 8. Now we continue as usual: $$2×7=14$$, which is one eight and six units, so we put down the $$6$$ and carry the $$1$$: $$\\begin{array}{llll}&&2^1&5^4\\\\×&&4&7\\\\\\hline &3&0&3\\\\&&6\\end{array}$$ Then $$2×4$$ plus the $$1$$ we carried is one eight and one unit, which we put down: $$\\begin{array}{llll}&&2^1&5^4\\\\×&&4&7\\\\\\hline &3&0&3\\\\1&1&6\\\\\\hline\\end{array}$$ The $$116$$ is how $$2×47$$ is written in base 8. Now we add up the partial products, which is easy; there isn't even any carrying: $$\\begin{array}{llll}&&2^1&5^4\\\\×&&4&7\\\\\\hline &3&0&3\\\\1&1&6\\\\\\hline 1&4&6&3\\end{array}$$", "in different bases allows for another way to \"mechanize\" the arithmetic of polynomials. Recall that the \"borrowing\" in subtraction is really just a mirror image of \"carrying\" as seen in addition and multiplication. Both serve only to ensure our answer ends up only with digits from $0,1,2,3,\\ldots,(b-1)$ for numbers written in base $b$. As we have seen however -- keeping our digits in this range is not terribly important. This is useful, as when doing arithmetic with polynomials, our base -- being a variable -- is essentially unknown. Without knowing the value of the base, how could we possibly know when to carry or how much to borrow? With that in mind, if carrying/borrowing is not terribly important, let's not do it! Consider what happens when we do addition and subtraction without any carrying or borrowing. That is to say -- if in some column we exceed the base with our sum, so be it. If in some column the difference is negative, life goes on: $$\\begin{array}{ccccc} & 2 & 5 & 0 & 7\\\\ + & 3 & 8 & 5 & 9\\\\\\hline & 5 & 13 & 5 & 16 \\end{array} \\quad \\quad \\begin{array}{ccccc} & 8 & 1 & 6 & 4\\\\ - & & 7 & 3 & 9\\\\\\hline & 8 & -6 & 3", "you can see not a number cruncher:)Thank you for your comment. – Peter Szilas Mar 18 at 11:18 $$\\begin{array}{cccccc}115792&089237&316195&423570&985008&687907\\\\852837&564279&074904&382605&163141&518161\\\\494336\\end{array}$$ Sum up the places of these numbers, by place value carrying when needed, then apply $$10^k\\equiv 3^k \\bmod 7$$ you'll then have a much smaller number to find the remainder of that's equivalent. 5667972, which goes to :$$6(3^5)+6(3^4)+2(3^2)\\equiv 1458+486+18\\equiv 2+3+4\\equiv 2 \\bmod 7$$ so the largest multiple of 7, is 2 less than the number. Yes, this is a slightly tedious way to go, but it's inspired via extension of Fermat's little theorem, and polynomial remainder theorem. The reason I broke it into 6 digits at a time, is because Fermat's extension, is that exponents that have the same remainder mod p-1, will give back the same remainder with the same base. That means you can simply turn one into the other, adding like terms. you then go and do the addition the first column on the right sums to 62, carry the 6, that means you sum the next column plus 6, giving 57 carry the 5, next column is then 59, carry the 5, next column 67, carry the 6, next column, 76 carry the 7, next column, 56 there's no column to carry the 5 onto, and in the next step, it will be merged with", "column. $$3+8=11$$ $$1$$ is to be written in the hundredths column and $$10$$ is carried to the next column (tenths). $$\\begin{array} {c} &7&\\cdot&^\\underline{\\color{red}1}4&3\\\\ +&4&\\cdot&\\;\\;\\;5&8\\\\ \\hline &&\\cdot&&1\\\\ \\hline \\end{array}$$ Add the digits of tenths column. $$1+4+5=10$$ $$0$$ is to be written in the tenths column and $$10$$ is carried to the next column (ones). $$\\begin{array} {c} &^\\underline{\\color{red}1}7&\\cdot&4&3\\\\ +&\\;\\;4&\\cdot&5&8\\\\ \\hline &&\\cdot&0&1\\\\ \\hline \\end{array}$$ Now, add the digits of ones column. $$1+7+4=12$$ $$2$$ is to be written in the ones column and $$10$$ is carried to the next column (tens). So, $$1$$ is written in the tens column. $$\\begin{array} { r c} \\hline \\text{Tens}&\\text{Ones}&\\text{Decimal point}&\\text{Tenths}&\\text{Hundredths}\\\\ \\hline \\text{Carried}\\to\\underline{\\color{red}1}&7&\\cdot&4&3\\\\ +&4&\\cdot&5&8\\\\ \\hline 1&2&\\cdot&0&1\\\\ \\hline \\end{array}$$ $$\\therefore\\;12.01$$ is the sum of $$7.43$$ and $$4.58$$ Hence, option (B) is correct. ### Add $$7.43$$ and $$4.58$$ A $$11.101$$ . B $$12.01$$ C $$74.4358$$ D $$47.5843$$ Option B is Correct # Addition of Decimals up to Thousandths Place • To add decimals and whole numbers, we are going to work with the wholes and the parts of the numbers separately. • Suppose, we want to add $$31.343$$ and $$42.456$$ • Write one number just below the other so that the bottom decimal point is directly below and lined up with the top decimal point. $$\\begin{array} {c} \\hline &\\text{Tens}&\\text{Ones}&\\text{Decimal point}&\\text{Tenths}&\\text{Hundredths}&\\text{Thousandths}\\\\ \\hline &3&1&\\cdot&3&4&3\\\\ +&4&2&\\cdot&4&5&6\\\\ \\hline &&&\\cdot&&&\\\\ \\hline \\end{array}$$ • First,", "operations, because frequently you will add two numbers and get something larger than 10, which requires you to add a carry digit to the next column over. It turns out that, on average, adding M numbers with N digits, it will take $\\left( M + \\frac{1}{2} \\right) N$ operations. ## Multiplication Does multiplication take more operations to complete? Let’s take a look. $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\ \\cline{2-2} & \\end{array}$ Our computer starts off by looking up $1\\times3 = 3$ and then $1\\times1 =1$. The next step is then looking up $2\\times3 = 6$ and $2\\times 1 = 2$ so we’ve completed four operations and have the following results: $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\ \\cline{2-2} & 13 \\\\ & 260 \\\\ \\end{array}$ Now we need to add the two numbers together. The number of operations is either one or three, depending on whether you count addition by zero as an operation. I’m going to count it as an operation since we assume our computer doesn’t understand anything about numbers and needs to look up even this simple process. This means that multiplying two 2 digit numbers together takes seven operations, which is clearly larger than the four operations taken to add two 2 digit numbers together. To come up with an equation", "he looks up $3 + 1 = 4$ and writes that down. $\\begin{array}{@{}cr} & 31 \\\\ + & 17 \\\\ \\cline{2-2} & 48 \\end{array}$ Since the computer used the look-up tables twice, it took two operations to add two 2 digits numbers. Let’s extrapolate to adding more numbers. For simplicity, I’ll assume that all numbers that are being added have the same number of digits. If you add three 2 digit numbers together it ends up taking a total of $3\\times 2 = 6$ operations. Remember that your addition tables only allow you to add two digits together at a time so you will need to add the first two numbers together first, then add the third number to this results. $\\begin{array}{@{}cr} & 31 \\\\ + & 17 \\\\ + & 12 \\\\ \\cline{2-2} & \\end{array}$ The first two numbers give you 48 and took four operations, and adding 12 to 48 takes another two operations, for a total of six. We can extrapolate to see that adding M 2 digit numbers together takes a total of 2M operations. You can also easily see that if you expand this even further to consider numbers with N digits, it will take $N\\times M$ operations to add those numbers together. This is actually a lower bound on the number of", "but I wanted to see if I could figure it out on my own. Let’s start simple, with two 2 digit numbers (to hopefully avoid confusion I’ll use numerals to denote the number of digits and words to denote the number of numbers). $\\begin{array}{@{}cr} & 31 \\\\ + & 17 \\\\ \\cline{2-2} & \\\\ \\end{array}$ The computer (note that the word computer used to refer to a person who calculates numbers) pulls out his chart to look up $1 + 7 = 8$ and writes 8 down in the first column. $\\begin{array}{@{}cr} & 31 \\\\ + & 17 \\\\ \\cline{2-2} & \\ 8 \\end{array}$ Next he looks up $3 + 1 = 4$ and writes that down. $\\begin{array}{@{}cr} & 31 \\\\ + & 17 \\\\ \\cline{2-2} & 48 \\end{array}$ Since the computer used the look-up tables twice, it took two operations to add two 2 digits numbers. Let’s extrapolate to adding more numbers. For simplicity, I’ll assume that all numbers that are being added have the same number of digits. If you add three 2 digit numbers together it ends up taking a total of $3\\times 2 = 6$ operations. Remember that your addition tables only allow you to add two digits together at a time so you will need to add the first two numbers together first,", "1 1 0 0 0 0 0 When you receive this array, you are guaranteed the last row will be all zeroes. You should write the sum of every column into the last number in that column. As an example, your function should modify the above array to be the following: 0 1 2 3 4 0 4 3 2 4 0 0 1 1 1 0 5 6 6 9 For each column, starting from the right-most digit, you should add every digit in that column, and put the result of that addition into the last row. To simulate real addition, however, none of the values in the array may exceed 9, so you will need to implement \"carrying\", just like in normal addition. \"Carrying\" is when all the numbers in a column sum to greater than 9, and you add extra to the next column to keep the current column below 10. For example, the following array: 0 9 0 8 0 7 0 0 Should become: 0 9 0 8 0 7 2 4 The sum function will normally return 0. If, however, your addition cannot be represented in the array because the last column you add still carries something over, your function should return the amount carried: 9 8 7 0 Should return 2,", "$3$ and $4$ to place in the last row. We cannot place $3$ next to $1$ because $3$ is already placed the fourth column. So we place $4$ there and $3$ in the bottom right corner. Answer: $3$", "digit numbers together it ends up taking a total of $3\\times 2 = 6$ operations. Remember that your addition tables only allow you to add two digits together at a time so you will need to add the first two numbers together first, then add the third number to this results. $\\begin{array}{@{}cr} & 31 \\\\ + & 17 \\\\ + & 12 \\\\ \\cline{2-2} & \\end{array}$ The first two numbers give you 48 and took four operations, and adding 12 to 48 takes another two operations, for a total of six. We can extrapolate to see that adding M 2 digit numbers together takes a total of 2M operations. You can also easily see that if you expand this even further to consider numbers with N digits, it will take $N\\times M$ operations to add those numbers together. This is actually a lower bound on the number of operations, because frequently you will add two numbers and get something larger than 10, which requires you to add a carry digit to the next column over. It turns out that, on average, adding M numbers with N digits, it will take $\\left( M + \\frac{1}{2} \\right) N$ operations. ## Multiplication Does multiplication take more operations to complete? Let’s take a look. $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\", "digit number. Therefore, this value of A is not possible. Case 5: If B = 4 If B = 4, then B × 6 = 24 and 2 will be a carry for the next step. 6A + 2 = BB = 44 ⇒ 6A = 42 and hence, A = 7 The multiplication is as follows. $\\begin{array}{l}\\text{7 4}\\\\ \\underset{¯}{\\text{}×\\text{6}}\\\\ \\underset{¯}{\\text{4 4 4}}\\end{array}$ Hence, the values of A and B are 7 and 4 respectively. $\\begin{array}{l}\\text{8}\\text{. A 1}\\\\ \\underset{¯}{\\text{+ 1 B}}\\\\ \\underset{¯}{\\text{B 0}}\\end{array}$ The addition of 1 and B is giving 0 i.e., a number whose ones digits is 0. This is possible only when digit B is 9. In that case, the addition of 1 and B will give 10 and thus, 1 will be the carry for the next step. In the next step, 1 + A + 1 = B Clearly, A is 7 as 1 + 7 + 1 = 9 = B Therefore, the addition is as follows. $\\begin{array}{l}\\text{7 1}\\\\ \\underset{¯}{\\text{+ 1 9}}\\\\ \\underset{¯}{\\text{9 0}}\\end{array}$ Hence, the values of A and B are 7 and 9 respectively. $\\begin{array}{l}\\text{9}\\text{. 2 A B}\\\\ \\underset{¯}{\\text{+ A B 1}}\\\\ \\underset{¯}{\\text{B 1 8}}\\end{array}$ The addition of B and 1 is 8. This is possible only when digit B is 7. In the next step, A + B =", "& \\bold{4} \\end{matrix}$ Therefore, (5142)8 - (2546)8 = (2374)8 Explanation 1. 2 - 6, here 6 is greater than 2 so we borrow 1 from next significant digit 4 and it becomes 3. Now 2 becomes 2 + 8 = 10 and 10 - 6 = 4. 2. 3 - 4, again 4 is greater than 3 so we borrow 1 from next significant digit 1 and it becomes 0. Current digit 3 becomes 8 + 3 = 11. Lastly, 11 - 4 = 7. 3. 0 - 5, here 5 is greater than 0. Borrowing 1 from next significant digit 5 makes it 4. Current digit 0 becomes 0 + 8 = 8. Lastly, 8 - 5 = 3. 4. 4 - 2 = 2. Question 3 Add hexa-decimal numbers (1F5C)16 and (AC2B)16 $\\begin{matrix} & \\overset{1}{1} & F & \\overset{1}{5} & C \\\\ + & A & C & 2 & B \\\\ \\hline & \\bold{C} & \\bold{B} & \\bold{8} & \\bold{7} \\end{matrix}$ Therefore, (1F5C)16 + (AC2B)16 = (CB87)16 Explanation 1. C + B = 12 + 11 = 23. As 23 is greater than 16 so 23 - 16 = 7. Hence, sum = 7 and carry over = 1. 2. 1 + 5 + 2 = 8. 3. F + C = 15 + 12", "row also. Further, this number must be odd, otherwise we get an even number in the end of the middle row and the number just below it would not be integer. So, lets try $73$: $$\\begin{array}{rrrrrrrrrrrl} 03&06&09&12&15&18&&59&75&91&??&\\leftarrow \\text{Can't put 107, too high.}\\\\ &&13&18&&20&38&56&74&&&\\\\ 03&10&17&24&&22&&53&73&93&??&\\leftarrow \\text{Can't put 113, too high.}\\\\ 27&&&30&&24&&50&&&-&\\\\ 51&46&41&36&31&26&&47&51&55&59&\\\\ \\end{array}$$ What do this means? If we try something smaller than $73$ below the $74$, the number on the top-right will grow. If we try something larger, then the number on the end of the middle row will grow. So it is impossible with $38$ at the right of the $20$. If we try $41$, we get $83$ in the end of the row starting with $20$. If we try $44$, we get $92$. And these numbers are far higher than $53$ and $59$, so either the middle or the top row would end with something higher than $100$. This proves that right to the $20$, the number must be $35$, so the second guess has a single solution. # The third guess has a single solution: Lets remember where it is before the third guess: $$\\begin{array}{rrrrrrrrrrrl} 03&06&09&12&15&18&&59&-&-&-&\\\\ &&13&18&&20&35&50&65&&&\\\\ 03&10&17&24&&22&&41&-&-&-&\\\\ 27&&&30&&24&&32&&&-&\\\\ 51&46&41&36&31&26&&23&35&47&59&\\\\ \\end{array}$$ What can we try? We already know that below the $65$ the number must be odd and can't be lower than $59$ ($59$ solves", "also one of $$\\{5,1,2\\}$$, and it cannot also be $$1$$, it must be $$2$$. This shows that the $$1$$, unless it appears in the rightmost position, must be to the left of the $$2$$; it cannot be to the left of the $$5$$. Similarly, if $$a_i = 2$$ then $$b_i = 5$$, because $$4$$ is impossible, so the $$2$$ must be to the left of a large digit, which must be the $$5$$. Similar reasoning produces no constraint on the position of the $$5$$; it could be to the left of a small digit (in which case it doubles to $$1$$) or a large digit (in which case it doubles to $$2$$). We can summarize these findings as follows: $$\\begin{array}{cl} \\text{digit} & \\text{to the left of} \\\\ \\hline 1 & 1, 2, \\text{end} \\\\ 2 & 5 \\\\ 5 & 1,2,5,\\text{end} \\end{array}$$ Here “end” means that the indicated digit could be the rightmost. Furthermore, the left digit of $$A$$ must be small (or else there would be a carry in the leftmost place and $$2A$$ would have 4 digits instead of 3) so it must be either $$1$$ or $$2$$. It is not hard to see from this table that the digits must be in the order $$125$$ or $$251$$, and indeed, both of those numbers have the" ]

[ "the ones' column $$20$$, placing a $$0$$ as the ones' digit of the sum and the necessary carry of $$2$$ into the tens' column. $$36.\\textrm{[14|41]}4 + 58.\\textrm{[27|72]}6 + 75.\\textrm{[24|42]}8 = 170.\\textrm{[38|83]}8$$ At which point, it is relatively easy to figure out the correct combination. I noticed that it is not possible to make a sum of $$18$$ in the tenths' column (remember, we need a carry of $$1$$ into the ones' column, for the previous paragraph to work). Therefore, we have a sum of $$13$$ in the tenths' column; the only way to do this is $$4 + 7 + 2 = 13$$. This gives a final answer of $$36.414 + 58.726 + 75.248 = 170.388$$ Solution: $$36.414 +58.726 + 75.248 = 170.388$$ Method: First note that with the starting numbers staying in place the absolute min numbers (before the decimal) are 31, 52, 72, which add up to 155. The max (not including the decimals) adds to 169. With that in mind I set the value to be a 170 in the final answer and made the rest of the numbers be 36, 58, 75. Then since I knew the decimals of the added numbers has to add to over 1 (to turn the 169 to 170). With that observation in mind I found the final numbers", "$$4$$ we carried is $$24$$, but that's 3 eights and no units, so we put down $$30$$, not $$24$$: $$\\begin{array}{lll}&2&5^4\\\\×&4&7\\\\\\hline 3&0&3\\end{array}$$ That $$303$$ is because $$5×47$$ is written as $$303$$ in base 8. Now we continue as usual: $$2×7=14$$, which is one eight and six units, so we put down the $$6$$ and carry the $$1$$: $$\\begin{array}{llll}&&2^1&5^4\\\\×&&4&7\\\\\\hline &3&0&3\\\\&&6\\end{array}$$ Then $$2×4$$ plus the $$1$$ we carried is one eight and one unit, which we put down: $$\\begin{array}{llll}&&2^1&5^4\\\\×&&4&7\\\\\\hline &3&0&3\\\\1&1&6\\\\\\hline\\end{array}$$ The $$116$$ is how $$2×47$$ is written in base 8. Now we add up the partial products, which is easy; there isn't even any carrying: $$\\begin{array}{llll}&&2^1&5^4\\\\×&&4&7\\\\\\hline &3&0&3\\\\1&1&6\\\\\\hline 1&4&6&3\\end{array}$$", "will assume that $A,B,C,D$ are four different digits. . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ A & B & C & D \\\\ \\times &&& 9 \\\\ \\hline D & C & B & A \\end{array}$ In column-1, we see that: . $A = 1,\\:D = 9$ And there is no \"carry\" from column-2. So we have: . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ 1 & B & C & 9 \\\\ \\times &&& 9 \\\\ \\hline 9 & C & B & 1 \\end{array}$ Since there is no \"carry\" frim column-2, $\\,B$ must be 0 or 1. Since $A = 1$, then: . $B \\,=\\,0$ Then we have: . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ 1 & 0 & C & 9 \\\\ \\times &&& 9 \\\\ \\hline 9 & C & 0 & 1 \\end{array}$ In column-3, we have: . $9C + 8$ ends in $0.$ . . Hence, $\\,9C$ ends in $2 \\quad\\Rightarrow\\quad C = 8$ Therefore, the solution is: . . $\\begin{array}{cccc} 1 & 0 & 8 & 9 \\\\ \\times &&& 9 \\\\ \\hline 9 & 8 & 0 & 1 \\end{array}$ 6. wonderful thank you!", "$$10.6055$$. #### 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. $$\\begin{array} {c}", "by $3$ after first inserting a $0$ to the right. The results of the two multiplications are then added together. If you were to multiply by a 3-digit number then you would have three separate multiplications, one for each digit. The process for the first two digits would proceed as above, but the process for the third digit would involve first inserting two zeroes to the right. Example : $394\\times 121$ $\\begin{array}{ccccc} & &3 & 9 & 4 \\\\ & & 1 & 2 & 1 \\\\ \\hline & & 3 & 9 & 4\\\\ & 7 & 8 & 8 & 0\\\\ 1 & 2 & 1 & 0 &0\\\\ \\hline 2 & 0 & 3 & 7 & 4\\end{array}$ This method generalizes in the expected way if you were ever to multiply by a number with more than three digits. ## Long Multiplication - Decimal Numbers Method • Multiply as though you were multiplying two integers (disregarding the decimal point) Example : $329.625 \\times 2.94$ $\\begin{array}{cccccccc} & & 3 & 2 & 9 &6 & 2 & 5 \\\\ & & & & & 2 & 9 & 4 \\\\ \\hline &1&3 &1 & 8 & 5 & 0 & 0\\\\ & 2&9 &6 & 6 & 2 & 5 & 0\\\\ 6 & 5 &9", "zeros to make them same. $$5.674-2.500$$ • Now, both decimal numbers have same place values. Next, we write the decimal numbers vertically, lining up the decimal points and each digit according to its place value. $$\\begin{array}\\\\ &5&.&6&7&4\\\\ -&2&.&5&0&0\\\\ \\hline\\\\ \\hline \\end{array}$$ • Now, subtracting the columns vertically. $$\\begin{array}\\\\ &5&.&6&7&4\\\\ -&2&.&5&0&0 \\\\ \\hline &3&.&1&7&4 \\\\\\hline \\end{array}$$ • Difference of $$5.674$$ and $$2.5$$ is $$3.174$$ #### Find:​ $$10.1-2.85$$ A $$8.15$$ B $$12.95$$ C $$7.25$$ D $$12.86$$ × Since both numbers have different place values, so we add zero to make them same. $$10.10-2.85$$ Now, we write the decimals vertically, lining up the decimal points and each digit according to its place value. $$\\begin{array}\\\\ &1&0&.&1&0\\\\ -&&2&.&8&5\\\\ \\hline\\\\ \\hline \\end{array}$$ On subtracting the columns vertically, we get $$\\begin{array}\\\\ &1&0&.&1&0\\\\ -&&2&.&8&5\\\\ \\hline&&7&.&2&5\\\\ \\hline \\end{array}$$ Hence, option (C) is correct. ### Find:​ $$10.1-2.85$$ A $$8.15$$ . B $$12.95$$ C $$7.25$$ D $$12.86$$ Option C is Correct # Multiplication of Decimals • Factors are referred to as the numbers which are being multiplied. • Product refers to the result of a multiplication problem. • To multiply two decimal numbers, first line them up in right alignment. • Then, multiply each digit of multiplier by each digit of multiplicand. • Count the number of digits after the decimal point in multiplier and multiplicand. • The product", "& 4 & 0 & 0 \\end{array}$ decimal equivalent = 16 + 8 + 4 = 24", "& & \\end{array}$ It is easy to add the new set of numbers to get the answer: $30\\ 000 + 5\\ 000 + 1\\ 700 + 120 + 11 = 36\\ 831$ Starting with the ones as shown above makes it possible to do more of the work mentally, and write less, as shown below. $\\begin{array}{ccccc} & 3 & ^{1}1 & ^{1}8 & ^{1}3 & 7 \\\\ & & 4 & 9 & 9 & 4 \\\\ \\hline & 3 & 6 & 8 & 3 & 1 \\end{array}$ To achieve this, the first step is to add the digits in the ones column: $$7 + 4 = 11$$. Only write the digit “$$1$$” of the $$11$$. The $$10$$ of the $$11$$ is remembered and added to the $$3$$ in the tens column. We say the $$10$$ is carried from the ones column to the tens column. The same is done when the tens parts are added to get $$13$$: only the digit “$$3$$” is written (in the tens column, so it means $$30$$), and the $$1$$ is carried to the next column. This $$1$$ is actually $$100$$, so it needs to be carried to the hundreds column. ## Worked Example 1.8: Adding whole numbers Calculate without using a calculator: $4\\ 638 + 2\\ 667$ ### Break the", "carry to get 13, then write the 3 and carry a 1 to the hundreds column. Continue in this manner, working from right to left $$\\begin{array}{r}{11} \\\\ {1234} \\\\ {+\\quad 498} \\\\ \\hline 1732\\end{array}$$ Therefore, 1, 234 + 498 = 1, 732 Exercise Simplify: 1,286 + 349. 1635 Add three or more numbers in the same manner. Example 2 Simplify: 256 + 322 + 418. Solution Align the numbers vertically, then add, starting at the furthest column to the right. Add the digits in the ones column, 6 + 2 + 8 = 16. Write the 6, then carry a 1 to the tens column. Continue in this manner, working from right to left. $$\\begin{array}{r}{256} \\\\ {322} \\\\ {+418} \\\\ \\hline 996\\end{array}$$ Therefore, 256 + 322 + 418 = 996. Exercise Simplify: 256 + 342 + 283 881 ## Subtraction of Whole Numbers The key idea is this: Subtraction is the opposite of addition. For example, consider the difference 7 - 4 depicted on the number line in Figure 1.5. If we were adding 7 and 4, we first draw an arrow starting at zero pointing to the right with magnitude (length) seven. Then, to add 4, we would draw a second arrow of magnitude (length) 4, attached to the end of the first arrow and pointing to", "This is actually a lower bound on the number of operations, because frequently you will add two numbers and get something larger than 10, which requires you to add a carry digit to the next column over. It turns out that, on average, adding M numbers with N digits, it will take $\\left( M + \\frac{1}{2} \\right) N$ operations. ## Multiplication Does multiplication take more operations to complete? Let’s take a look. $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\ \\cline{2-2} & \\end{array}$ Our computer starts off by looking up $1\\times3 = 3$ and then $1\\times1 =1$. The next step is then looking up $2\\times3 = 6$ and $2\\times 1 = 2$ so we’ve completed four operations and have the following results: $\\begin{array}{@{}cr} & 13 \\\\ \\times & 21 \\\\ \\cline{2-2} & 13 \\\\ & 260 \\\\ \\end{array}$ Now we need to add the two numbers together. The number of operations is either one or three, depending on whether you count addition by zero as an operation. I’m going to count it as an operation since we assume our computer doesn’t understand anything about numbers and needs to look up even this simple process. This means that multiplying two 2 digit numbers together takes seven operations, which is clearly larger than the four operations taken to add two", "? And the \"Hence T=9\" ? 6. Hello, Moo! I'd like to know how you can state that $S=1 \\text{ and }T=9$ ? We have: . . . $\\begin{array}{ccccc} &&D&O&S\\\\&&D&O&S \\\\ &T&R&E&S \\\\ \\hline S&I&E&T&E \\end{array}$ We have some three-digit numbers and one four-digit number. . . And their sum is a five-digit number. If the upper digits were all 9's, we'd have:. $999 + 999 + 9999 \\:=\\:11997$ ... maximum sum. . . Hence: . $SIETE \\:<\\:{\\color{red}1}1997$ Then the sum begins with 1: . $S = 1$ Its second digit is $\\leq 1\\!:\\;I = 0$ So far, we have: . . $\\begin{array}{ccccc} &&D&O&1\\\\&&D&O&1 \\\\ &T&R&E&1 \\\\ \\hline 1&0&E&T&E \\end{array}$ And we see immediately that $E = 3$ . . $\\begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\\\ &&D&O&1\\\\&&D&O&1 \\\\ &T&R&3&1 \\\\ \\hline 1&0&3&T&3 \\end{array}$ How can the sum begin with $\"10\"$ ? There are two ways; . . $T = 8$, and 2 is \"carried\" from column-3 . . $T = 9$, and 1 is \"carried\" from column-3. Suppose $T = 8$, then we have: . . $\\begin{array}{ccccc}^1&^2&^3&^4&^5 \\\\ &&D&O&1\\\\&&D&O&1 \\\\ &8&R&3&1 \\\\ \\hline 1&0&3&8&3 \\end{array}$ In column-4, we see that: . $2O + 3\\text{ ends in }8 \\quad\\Rightarrow\\quad 2O\\text{ ends in }5$ . . Since $2O$ is an even number, this is impossible. Therefore:", "## Thinking Mathematically (6th Edition) $15760_{nine}$ Rightmost column: $4+5=9=1\\times 9+0\\times 1=10_{nine}$ record the $0$, carry the 1. $\\left[\\begin{array}{lllll} & & & 1 & \\\\ & 6 & 7 & 8 & 4_{nine}\\\\ + & 7 & 8 & 6 & 5_{nine}\\\\ & - & - & - & -\\\\ & & & & 0_{nine} \\end{array}\\right.$ Next column: $1+8+6=15=1\\times 9+6\\times 1=16_{nine}$ record the $6$, carry the 1. $\\left[\\begin{array}{lllll} & & 1 & 1 & \\\\ & 6 & 7 & 8 & 4_{nine}\\\\ + & 7 & 8 & 6 & 5_{nine}\\\\ & - & - & - & -\\\\ & & & 6 & 0_{nine} \\end{array}\\right.$ Next column: $1+7+8=16=1\\times 9+7\\times 1=17_{nine}$ record the $7$, carry the 1. $\\left[\\begin{array}{lllll} & 1 & 1 & 1 & \\\\ & 6 & 7 & 8 & 4_{nine}\\\\ + & 7 & 8 & 6 & 5_{nine}\\\\ & - & - & - & -\\\\ & & 7 & 6 & 0_{nine} \\end{array}\\right.$ Next column: $1+6+7=14=1\\times 9+5\\times 1=15_{nine}$ record the $15$, Result =$15760_{nine}$", "example. Since we know the ones digit of the sum is $$6$$, then the ones digits for each of the two numbers being added must have a total of $$6$$ or $$16$$. The only possible combinations given the digits we have are $$1 + 5$$ or $$5 + 1$$. Since neither of these sums gives us $$16$$, we do not have to worry about carrying a $$1$$ into the tens column. Looking at the tens column, we need two numbers that add up to $$9$$ or $$19$$. There is only one pair of the remaining digits we could use: $$3 + 6$$ or $$6 + 3$$. Neither of these sums gives us $$19$$, so there is nothing to carry into the hundreds column. This means the sum of the hundreds column must be $$12$$. The remaining digits are $$4$$ and $$8$$, and fortunately the sum of these digits is $$12$$. As in parts A) and B), we can arrange these digits in $$8$$ different ways. They are all given below. $$\\begin{array}{c c c c} &4&6&1\\\\ +&8&3&5\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &8&3&5\\\\ +&4&6&1\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &8&3&1\\\\ +&4&6&5\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &4&6&5\\\\ +&8&3&1\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &4&3&5\\\\ +&8&6&1\\\\ \\hline \\end{array}$$ $$\\begin{array}{c c c c} &8&6&1\\\\ +&4&3&5\\\\ \\hline \\end{array}$$", "A $$4.4$$ B $$11.6$$ C $$3.4$$ D $$4.6$$ × Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point. $$\\begin{array} {c|c} \\hline &\\text{Ones}&\\text{Decimal point}&\\text{Tenths}\\\\ \\hline &8&\\cdot&0\\\\ -&3&\\cdot&6\\\\ \\hline &&\\cdot&\\\\ \\hline \\end{array}$$ First, subtract the digits of the tenths column. We can not take $$6$$ from $$0,$$ so we borrow $$10$$ from the ones column, such that the $$8$$ becomes a $$7.$$ Now, we subtract $$6$$ from $$10$$ $$[0+10$$ (borrowed) $$=10]$$ $$4$$ is to be written in the tenths column. $$\\begin{array} {c} &\\not{8}^7&\\cdot&0\\\\ -&3&\\cdot&6\\\\ \\hline &&\\cdot&4\\\\ \\hline \\end{array}$$ Now, we subtract the digits of the ones column. $$7-3=4$$ $$4$$ is to be written in the ones column. $$\\begin{array} {c|c} \\hline &\\text{Ones}&\\text{Decimal point}&\\text{Tenths}\\\\ \\hline &8&\\cdot&0\\\\ -&3&\\cdot&6\\\\ \\hline &4&\\cdot&4\\\\ \\hline \\end{array}$$ So, $$4.4$$ is the difference of $$8$$ and $$3.6$$ Hence, option (A) is correct. What is the difference of $$8$$ and $$3.6$$? A $$4.4$$ . B $$11.6$$ C $$3.4$$ D $$4.6$$ Option A is Correct Subtraction of Decimals up to Tenths Place • Subtraction of decimals can be easily done by dealing with the wholes and the parts of the numbers separately. • Suppose, we want to subtract $$12.4$$ from $$25.6$$ i.e. $$25.6-12.4$$ Write one number just below the other, so that the bottom decimal point", "## Thinking Mathematically (6th Edition) $13_{four}$ $2$ is less than 3, borrow one four from the next column $(4+2)-3 =3_{ten}=3_{four}$ $\\left[\\begin{array}{lll} & 3^{(*)} & 2_{four}\\\\ - & 1 & 3_{four}\\\\ - & - & -\\\\ & & 3_{four} \\end{array}\\right]$ Next column: we borrowed 1 from 3, subtract $2-1=1_{ten}=1_{four}$ record (no borrowing) $\\left[\\begin{array}{lll} & 3^{(*)} & 2_{four}\\\\ - & 1 & 3_{four}\\\\ - & - & -\\\\ & 1 & 3_{four} \\end{array}\\right]$ Check in base 10: $14-7=7$", "Thinking Mathematically (6th Edition) $123_{four}$ $(3\\times 1)_{10}=(3)_{10}=(3)_{4}$ Record the $3$; there is nothing to carry. $\\begin{array}{llll} & 2 & 1_{four} & \\\\ \\times & & 3_{four} & \\\\ -- & -- & -- & \\\\ & & 3_{four} & \\end{array}$ $(3\\times 2)_{10}=(6)_{10}=1\\times 4+2=(12)_{4}$ Record the $12.$ Result: $123_{four}$", "Computers do arithmetic in the binary number system. The operations are really quite easy to understand if you recall all the details of performing arithmetic in the decimal number system by hand. Since most people do addition on a calculator these days, let us review all the steps required when doing it by hand. This discussion will be followed by exercises asking you to develop the algorithms to perform addition and subtraction in binary and in hexadecimal. Consider two two-digit numbers, $x = 67$ and $y = 79\\text{.}$ Adding these by hand on paper would look something like: $1$ $1$ $\\longleftarrow$ carries $67$ $\\longleftarrow$ $x$ $+$ $79$ $\\longleftarrow$ $y$ $46$ $\\longleftarrow$ sum We start by working from the right, adding the two decimal digits in the ones place. 7 + 9 exceeds 10 by 6. We show this by placing a 6 in the ones place in the sum and carrying a 1 to the tens place. Next we add the three decimal digits in the tens place, 1 (the carry into the tens place from the ones place) + 6 + 7. The sum of these three digits exceeds 10 by 4, which we show by placing a 4 in the tens place in the sum and recording the fact that there is an ultimate carry of", "& & & 4 \\\\ \\hline D & C & B & 2 \\end{array}$ In column-1, we see that: . $D \\:=\\:8\\text{ or }9$ In column-4, $4\\!\\times\\!D$ ends in 2. . . Hence: . $\\boxed{D = 8}$ . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ 2 & B & C & 8 \\\\ \\times & & & 4 \\\\ \\hline 8 & C & B & 2 \\end{array}$ We see that there is no \"carry\" from column-2. . . Hence: . $B \\:=\\:0\\text{ or }1$ . . Suppose $B = 0$ . . . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ 2 & 0 & C & 8 \\\\ \\times & & & 4 \\\\ \\hline 8 & C & 0 & 2 \\end{array}$ . . In column-3, we have: . $4\\!\\times\\! C + 3$ ends in 0 . . . impossible. Hence: . $\\boxed{B = 1}$ . . $\\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\\\ 2 & 1 & C & 8 \\\\ \\times & & & 4 \\\\ \\hline 8 & C & 1 & 2 \\end{array}$ In column-3, $4\\!\\times\\!C + 3$ ends in 1. . . Hence: . $C = 2\\text{ or }7$ Since $A = 2,\\,\\text{ then }\\,\\boxed{C = 7}$ Solution: . . $\\boxed{\\begin{array}{cccc} 2 &", "1 is less than 3 so we write 1 down. 4th column = 2 + 1 = 3. How many threes are there in 3? $$\\frac{3}{3}$$= 1 remainder 0. Write 0 and carry 1 to column 5. 5th column = We add 1 to the 5th column. c. 151.24six + 12.34six + 20.11six Explanation 1st Column (from right to left) = 4 + 4 + 1 = 9. How many six’s are there in 9? $$\\frac{9}{6}$$ = 1 remainder 3. Write 3 and carry 1 to column 2. 2nd Column = 1 + 2 + 3 + 1 = 7. How many six’s are there in 7? $$\\frac{7}{6}$$ = 1 remainder 1. Write 1 and carry 1 to column 3. 3rd Column = 1 + 1 + 2 + 0 = 4. 4 is less than 6 so we write 4 down. 4th column = 5 + 1 + 2 = 8. How many six’s are there in 8? $$\\frac{8}{6}$$ = 1 remainder 2. Write 2 and carry 1 to column 5. 5th column = 1 + 1 = 2. 2 is less than 6 so we write 6 down. d. 53468 – 6378 1st Column (from right to left) = 6 is less than 7 so we will have to borrow from the 2nd column. Since it", "also one of $$\\{5,1,2\\}$$, and it cannot also be $$1$$, it must be $$2$$. This shows that the $$1$$, unless it appears in the rightmost position, must be to the left of the $$2$$; it cannot be to the left of the $$5$$. Similarly, if $$a_i = 2$$ then $$b_i = 5$$, because $$4$$ is impossible, so the $$2$$ must be to the left of a large digit, which must be the $$5$$. Similar reasoning produces no constraint on the position of the $$5$$; it could be to the left of a small digit (in which case it doubles to $$1$$) or a large digit (in which case it doubles to $$2$$). We can summarize these findings as follows: $$\\begin{array}{cl} \\text{digit} & \\text{to the left of} \\\\ \\hline 1 & 1, 2, \\text{end} \\\\ 2 & 5 \\\\ 5 & 1,2,5,\\text{end} \\end{array}$$ Here “end” means that the indicated digit could be the rightmost. Furthermore, the left digit of $$A$$ must be small (or else there would be a carry in the leftmost place and $$2A$$ would have 4 digits instead of 3) so it must be either $$1$$ or $$2$$. It is not hard to see from this table that the digits must be in the order $$125$$ or $$251$$, and indeed, both of those numbers have the" ]

[ "the unit circle", "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", "the total area of the ellipse is $$\\pi ab.$$", "## amc-2014-sr-04 A circle has circumference $$\\pi %speech% ,pi,$$ units. In square units, its area is", "# The Magic Number of Circles Geometry Level 1 A circle has a diameter of $$x$$ and an area of $$\\pi x$$. What is the value of $$x$$? × Problem Loading... Note Loading... Set Loading...", "+0 # i need help!!! 0 69 1 Circle $A$ has its center at $A(4, 4)$ and has a radius of 4 units. Circle $B$ has its center at $B(12, 4)$ and has a radius of 4 units. What is the area of the gray region bound by the circles and the $x$-axis? Express your answer in terms of $\\pi$. Jun 19, 2021", "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", "+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", "The area of a square inscribed in a circle with a unit radius is, satisfyingly, $2$. What is the area of a regular hexagon inscribed in a circle with a unit radius? What is the area of an equilateral triangle inscribed in a circle with a unit radius?", "## zuha one year ago A circle has a center at (8,2). The point (3,7) is on the circle. What is the area of the circle to the nearest tenth of a square unit? |dw:1442836225718:dw| Since the distance from the center to the point on the circle = radius of the circle...once we find that distance, we can apply the fact that: $$\\large \\text{Area of a circle} = \\pi r^2$$", "the unit circle.", "a circle of diameter $d$ is $\\pi(d/2)^2$, We coudld have instead worked with $r$, and translated to diameter at the end. Yes, \"the number of square units in the area of\" is a fancy way of saying \"the area of.\" – André Nicolas Jul 2 '12 at 3:53 @AndréNicolas: Of course, the phrasing \"the number of square units in the area of\" (and \"the number of [linear] units\") isn't just to be fancy; it's specifically to address your objection that units of length and area don't match, saying, in effect, \"We know that. This isn't about units. Just make the numbers equal to each other.\" This is a perfectly reasonable algebraic exercise. Apples aren't oranges, either, but we can certainly compare numbers of each. – Blue Jul 2 '12 at 8:56", "EXACT AREA OF A CIRCLE WITH A DIAMETER OF 23 INCHES", "the maximum area will always be a square. ### What shape has the largest area given perimeter? For a given perimeter, the closed figure with the maximum area is a circle.", "# What is the unit circle? The circumfrence of the unit circle is $2 \\pi$. An arc of the unit circle has the same length as the measure of the central angle that intercepts that arc. Also, because the radius of the unit circle is one, the trigonometric functions sine and cosine have special relevance for the unit circle.", "$\\pi$. A unit circle is a circle with a radius of one, not an area of one. I've now made this explicit in the question. 4. Jul 7, 2013 ### Simon Bridge Its a simile - an analogy ... The area under the sine from 1 to pi/2 can be cut up to fill the gap between the sin and y=1 from 0 to 1. That was already clear. The sine is a specific length defined on the unit-radius circle... since the one was derived from the other, it is not surprising to find they have special relationships.", "circle is 10, So the area is $\\pi(10)^2$=100$\\pi$ sq.unit The radius of the 6th circle is 12, So the area is $\\pi(12)^2$=144$\\pi$ sq.unit Therefore The entire circle’s area is 144$\\pi$ The area of the black regions is $(100\\pi-64\\pi)+(36\\pi-16\\pi)+4\\pi=60\\pi$sq.unit The percentage of the design that is black is $(\\frac{60\\pi}{144\\pi} \\times 100)\\%=(\\frac{5}{12} \\times 100) \\% \\approx 42\\%$", "# Area of a Unit Circle Approximate the area of a unit circle with the area of an inscribed regular polygon Number of sides for inscribed polygon:", "$60 \\pi$ 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", "unit circle." ]

[ "+0 # HELP!!!! 0 78 1 The length of the radius is half the length of the pi diameter arc May 14, 2021 By definition, $r = \\frac{d}{2}.$", "+0 # HELP!!!! 0 197 1 The length of the radius is half the length of the pi diameter arc May 14, 2021 By definition, $r = \\frac{d}{2}.$", "r^2 x h, where pi is approximated as 3.142 , height is the height and r is the radius.The unit of measurement for volume is given in cubic units.", "The formula shouldn't explicitly show units. From the equation $A = \\pi r^2$, to get the radius, solve for r. To do this, divide both sides of the equation by $\\pi$, then take the square root. Once you have the radius r, you can get the diameter using d = 2r.", "+0 HELP!!!! 0 38 1 The length of the radius is half the length of the pi diameter arc May 14, 2021 1+0 Answers #1 +483 0 By definition, $r = \\frac{d}{2}.$ May 14, 2021", "unit radius. Take the map; $$x \\mapsto \\frac{x}{1+\\|x\\|}$$", "+ \\sqrt{\\pi}}$ and $C = 2\\pi r$ $= 2\\pi \\left[\\frac{180}{\\sqrt{\\pi}(2 + \\sqrt{\\pi})}\\right]$ $= \\frac{360\\pi}{2\\sqrt{\\pi} + \\pi}$. Of course, all units are in inches.", "is like choosing radius and not diameter so we can write $$A=\\pi r^2$$ and not $$A=\\frac{\\pi D^2}{4}$$", "will be $\\pi$. This is true for any radius/diameter in proportion.", "## Elementary Geometry for College Students (5th Edition) diameter = 2 r = $\\frac{3}{4}$ h r = $\\frac{3}{8}$ h Volume = $\\pi$ $r^{2}$ h 9 $\\pi$ $in.^{3}$ = $\\pi$ $(\\frac{3}{8} h)^{2}$ h 9 $\\pi$ $in.^{3}$ = $\\frac{9}{64}$ $\\pi$ $h^{3}$ $h^{3}$ = 64 $in.^{3}$ h = 4 in.", "## Trigonometry (11th Edition) Clone Let $a = 1.4~in$ and let $b = 0.2~in$. We can find the radius of the original plate: $r = \\frac{a^2+b^2}{2b}$ $r = \\frac{(1.4~in)^2+(0.2~in)^2}{(2)(0.2~in)}$ $r = 5~in$ The radius of the original plate is 5 inches.", "x_1)^2 + (y_2-y_1)^2$ we find the measure of the diameter AB. Dividing it by 2 yields the radius. 3 & 4. Replacing in the equation of the circle yields: $(x-1)^2+(y-5)^2 = 13$", "2 times the radius times pi.", "the two dimensional medium is finite with sizes $L_x$ , $L_y$ the density of points will be $\\frac{(2\\pi)^2}{L_xL_y}$. The number of $k$s inside the circle is $\\pi k_f^2/(\\frac{(2\\pi)^2}{L_xL_y})$. Then the number of $k$s per unit area is that $N(E)$ above.", "### Home > CALC > Chapter 10 > Lesson 10.1.5 > Problem10-47 10-47. A solid has a circular base of radius $r = 2$ units. If the cross-sections perpendicular to the base are square, find the volume of the solid. Review the eTool below. Write an expression for the length of the base of each cross-section.", "=> r = 7 and d = 2 r = 2 x 7 = 14 => Radius of circular table = 7 feet and diameter = 14 feet.", "of radius $$R$$ resting on $$2\\pi$$ radians.", "get 2π2Rr2. Thanks! 7. Mar 5, 2014 ### Simon Bridge I'd have gone right to x-R=r.sin(u). But well done anyway. Notice that $A=\\pi r^2$ is the area of the crossection circle and $C=2\\pi R$ is the circumference at the center. This means the volume of the torus turned out to be V=AC ... which is the volume of a cylinder radius r and height C. Does that make sense?", "$$2\\pi (R^2 - r^2)$$ $$=$$ $$2 \\pi (R + r)h$$ $$+$$ $$2\\pi (R + r) (R - r)$$ $$=$$ $$2 \\pi (R + r) [h + (R - r)]$$ $$=$$ $$2 \\pi (R + r) (R - r + h)$$ Total surface area of a hollow cylinder $$=$$ $$2 \\pi (R + r) (R - r + h)$$ sq. units", "# What is the radius of a circle whose diameter is 13 units? ##### 1 Answer Nov 28, 2015 $\\frac{13}{2}$ units or $7.5$ units #### Explanation: The diameter can expressed with the formula: $d = 2 r$ where: d = diameter r = radius This means that the diameter is double the length of the radius. To find the radius, do: $d = 2 r$ $13 = 2 r$ $\\frac{13}{2} = r$ $\\therefore$, the radius is $\\frac{13}{2}$ units or $7.5$ units." ]

[ "+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", "a circle of diameter $d$ is $\\pi(d/2)^2$, We coudld have instead worked with $r$, and translated to diameter at the end. Yes, \"the number of square units in the area of\" is a fancy way of saying \"the area of.\" – André Nicolas Jul 2 '12 at 3:53 @AndréNicolas: Of course, the phrasing \"the number of square units in the area of\" (and \"the number of [linear] units\") isn't just to be fancy; it's specifically to address your objection that units of length and area don't match, saying, in effect, \"We know that. This isn't about units. Just make the numbers equal to each other.\" This is a perfectly reasonable algebraic exercise. Apples aren't oranges, either, but we can certainly compare numbers of each. – Blue Jul 2 '12 at 8:56", "## Algebra 2 (1st Edition) A unit circle is a circle with its center at the origin and a radius of $1$.", "# What is the unit circle? The circumfrence of the unit circle is $2 \\pi$. An arc of the unit circle has the same length as the measure of the central angle that intercepts that arc. Also, because the radius of the unit circle is one, the trigonometric functions sine and cosine have special relevance for the unit circle.", "+0 # HELP!!!! 0 197 1 The length of the radius is half the length of the pi diameter arc May 14, 2021 By definition, $r = \\frac{d}{2}.$", "+0 # HELP!!!! 0 78 1 The length of the radius is half the length of the pi diameter arc May 14, 2021 By definition, $r = \\frac{d}{2}.$", "circle $=\\pi r^{2} . )$ Evan S. In the diagram $a, b, c,$ and $d$ are lengths. Which is greater, the product ab or the product $c d ?$", "the unit circle", "$\\pi$. A unit circle is a circle with a radius of one, not an area of one. I've now made this explicit in the question. 4. Jul 7, 2013 ### Simon Bridge Its a simile - an analogy ... The area under the sine from 1 to pi/2 can be cut up to fill the gap between the sin and y=1 from 0 to 1. That was already clear. The sine is a specific length defined on the unit-radius circle... since the one was derived from the other, it is not surprising to find they have special relationships.", "# The Magic Number of Circles Geometry Level 1 A circle has a diameter of $$x$$ and an area of $$\\pi x$$. What is the value of $$x$$? × Problem Loading... Note Loading... Set Loading...", "expanded to $r$ times its original size, will have $r^n$ times its original volume. For instance a circle that is twice the radius of another circle will have four times the area; a ball will have eight times the volume of a ball of half its radius.", "of the circle shown below. Enter an exact answer in terms of $\\pi$. units ## Challenge problem Find the arc length of the semicircle. Enter an exact answer in terms of $\\pi$. units", "# Is the area of a convex polygon equal to the area of a circle with the same perimeter of the polygon? Is the area of a convex polygon equal to the area of a circle with the same perimeter of the polygon? I guess that it's possible, take for example an square, I guess that it's borders could be deformed to form a circle and that their areas would be the same but such condition holds only for convex polygons. I guess I've read some theorem about it in the past but I don't remember it now. - Take a circle with radius 1, which has an area of $\\pi$ and perimeter of $2\\pi$. A square with the same perimeter would have a side length of $\\pi/2$, hence an area of $\\pi^2/4$. Since $\\pi$ isn't equal to $\\pi^2/4$, you already have a counterexample. A circle with radius $1$ has a perimeter of $2\\pi$. – JiK Aug 10 '14 at 20:38", "# What is the arclength of f(x)=sqrt(4-x^2) in the interval [-2,2]? Jul 10, 2018 The arc length is $2 \\pi$ units. #### Explanation: This is a semi circle of radius $2$. We simply find the perimeter of the full circle of radius $2$ and divide it by $2$ to get our arc length. $P = 2 \\left(2\\right) \\pi = 4 \\pi$ $\\frac{P}{2} = \\frac{4 \\pi}{2} = 2 \\pi$ Therefore the arc length is $2 \\pi$ units. Hopefully this helps!", "# Area of Circle Go back to 'Area' Consider a circle of radius $$r$$ : We know that • The perimeter (length of the circumference) of this circle will be $$2\\pi r$$ units. • The area of this circle will be $$\\pi {r^2}$$ square units. ## What is pi ? Before jumping into the technicalities, let’s familiarize ourselves with the diameter and the perimeter or circumference of a circle. To visualize the circumference, think of the crust of a Pizza, well, that’s essentially what it is! The length of the outer edge of any circle. Here’s where things get really interesting. What if you were told that if you divide the circumference of ANY circle with its diameter, you will ALWAYS end up with the SAME number, 3.1415… The use of ellipses is deliberate! This particular ratio is always the same regardless of the circle in question! Yes, it is mind-boggling, which is why this number occupies such a crucial role in all of the mathematics related to the circle. This ratio is called Pi. Regarding the number, pi is a non-repeating, non-terminating entity which is best approximated to the fraction $${22 \\over 7}$$. The decimal representation up to 10 places is 3.1428571429 ## How is it important? ### Area of a Circle What if you need to", "so $x= r sin(\\phi)$. That will be the radius of the circle swept out.", "circle is 10, So the area is $\\pi(10)^2$=100$\\pi$ sq.unit The radius of the 6th circle is 12, So the area is $\\pi(12)^2$=144$\\pi$ sq.unit Therefore The entire circle’s area is 144$\\pi$ The area of the black regions is $(100\\pi-64\\pi)+(36\\pi-16\\pi)+4\\pi=60\\pi$sq.unit The percentage of the design that is black is $(\\frac{60\\pi}{144\\pi} \\times 100)\\%=(\\frac{5}{12} \\times 100) \\% \\approx 42\\%$", "# Problem #1089 1089 A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle? $(\\mathrm {A}) 200+25\\pi \\quad (\\mathrm {B}) 100+75\\pi \\quad (\\mathrm {C}) 75+100\\pi \\quad (\\mathrm {D}) 100+100\\pi \\quad (\\mathrm {E}) 100+125\\pi$ 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.", "unit radius. Take the map; $$x \\mapsto \\frac{x}{1+\\|x\\|}$$", "# Formula Right Circular Cone Volume Height Radius ## Volume Unit The volume enclosed by a right circular cone. ## Height Unit Height of a circular cone, that is, the distance of the bottom area from the apex of the cone. Pi is a mathematical constant and has the value $$\\pi ~=~ 3.1415926...$$" ]

[ "+0 # help asap 0 98 1 I need some help on this problem In SHORT BINGO, a $5\\times5$ card is filled by marking the middle square as WILD and placing 24 other numbers in the remaining 24 squares. Specifically, a card is made by placing 5 numbers from the set $1-10$ in the first column, 5 numbers from $11-20$ in the second column, 4 numbers $21-30$ in the third column (skipping the WILD square in the middle), 5 numbers from $31-40$ in the fourth column and 5 numbers from $41-50$ in the last column. One possible SHORT BINGO card is: To play SHORT BINGO, someone names numbers, chosen at random, and players mark those numbers on their cards. A player wins when he marks 5 in a row, horizontally, vertically, or diagonally. How many distinct possibilities are there for the values in the first row of a SHORT BINGO card? Jul 2, 2020", "# Bingo card generator A Bingo card is five columns of five squares each, with the middle square designated \"FREE\". Numbers cannot duplicate. The five columns are populated with the following range of numbers: • B:1-15 • I:16-30 • N:31-45 • G:46-60 • O:61-75 In as few characters as possible, output a string that can be interpreted as a randomized Bingo card. For example: 1,2,3,4,5,16,17,18,19,20,31,32,33,34,35,46,47,48,49,50,61,62,63,64,65 This example is not randomized so that I can show that column 1 is populated with 1,2,3,4,5. Also note that the free space has not been given any special treatment because the front-end that interprets this string will skip it. Another example would be: 1,16,31,46,61,2,17,32,47,62... In this example, the output is by row instead of by column. A third example might be: 01020304051617181920313233343546474849506162636465 This is the same output as the 1st example except in fixed length. • Am I the only one who has never heard of Bingo but instead only of Bullshit Bingo? – Joey May 20 '11 at 18:54 • Yes! That's it! My idea is to come up with a list of 75 or more words and populate the card with SELECT * FROM List ORDER BY NEWID() May 20 '11 at 23:29 # PHP, 86 for($o=[];25>$i=count($o);){$n=rand(1,15)+($i-$i%5)*3;$o[$n]=$n;}echo implode(\",\",$o); • Welcome to PPCG, nice first answer c: – Rod Oct 26", "Bingo is $(n,m)=(5,15)$, the card game is $(4,13)$, and the dice version is effectively $(6,\\infty)$. The Math Horizons article referenced below describes an approach to calculating these probabilities, which involves enumerating integer partitions. However, this problem is ready-made for generating functions, which takes care of the partition house-keeping for us: let’s define $g_a(x) = \\left(\\sum_{j=a}^{n-1} {m \\choose j}x^j\\right)^{n-1}$ so that, for example, for Bingo with no free square, $g_1(x) = \\left({15 \\choose 1}x^1 + {15 \\choose 2}x^2 + {15 \\choose 3}x^3 + {15 \\choose 4}x^4\\right)^4$ Intuitively, each factor corresponds to a column, where each coefficient of $x^j$ indicates the number of ways to draw exactly $j$ numbers from that column (with some minimum number from each column specified by $a$). The overall coefficient of $x^k$ indicates the number of ways to draw $k$ numbers in total, with neither a horizontal nor vertical bingo. Then using the notation from the article, the probability $P(H_k)$ of a horizontal bingo on exactly the $k$-th draw is $P(H_k) = \\frac{mn(k-1)!(mn-k)!}{(mn)!}[x^{k-1}]g_1(x)$ and the probability $P(V_k)$ of a vertical bingo on exactly the $k$-th draw is $P(V_k) = \\frac{mn(k-1)!(mn-k)!}{(mn)!} {m-1 \\choose n-1} [x^{k-n}](g_0(x)-g_1(x))$ The summation $\\sum_{k=n}^{(n-1)^2+1} P(H_k)$ over all possible numbers of draws yields the overall probability of about 0.752 that a horizontal bingo is observed before a vertical one. Similarly, for the card", "So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,5)--(8,5)--(8,0)--cycle,green); fill((18,0)--(18,5)--(19,5)--(19,0)--cycle,green); fill((27,0)--(27,5)--(28,5)--(28,0)--cycle,green); for (real i=1; i<5; ++i) { for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,i+0.5)); } } for (real i=0; i<30; ++i) { label(\"\\vdots\",(i+0.5,2/3)); } add(grid(30,5,linewidth(1.25))); [/asy]$ As indicated by the white squares, each group of $30$ digits on the original list has $\\frac25\\cdot30=12$ digits remaining on the final list. Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green); fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green); fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green); for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,0.5)); } draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow); draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow); draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow); add(grid(30,1,linewidth(1.25))); [/asy]$ Therefore, the answer is $4+2+5=\\boxed{\\textbf{(D)}\\ 11}.$ ~MRENTHUSIASM ## Video Solution https://youtu.be/t6yjfKXpwDs 16:40 ~IceMatrix", "of the daubs markes as last and such that the pattern has at least one bingo but no bingo if the last daub is removed. There are only $24\\times 2^{23}$ patterns you have to investigate. – Hagen von Eitzen Sep 16 '12 at 12:24 yeah only 24*2^23 – pagla Sep 16 '12 at 12:28 Well you do have a computer, don't you? – Hagen von Eitzen Sep 16 '12 at 13:13 Never mind. Maybe it's more important that I have a computer ;) – Hagen von Eitzen Sep 16 '12 at 14:05 Assuming that \"Bingo\" means complete row or column or diagonal, I computed the following numbers (count of \"winning pattern\") by computer: $$S[0] = 0\\\\ S[1] = 0\\\\ S[2] = 0\\\\ S[3] = 0\\\\ S[4] = 16\\\\ S[5] = 360\\\\ S[6] = 3800\\\\ S[7] = 25080\\\\ S[8] = 116040\\\\ S[9] = 399048\\\\ S[10] = 1053120\\\\ S[11] = 2167368\\\\ S[12] = 3493152\\\\ S[13] = 4380480\\\\ S[14] = 4197720\\\\ S[15] = 2975160\\\\ S[16] = 1478304\\\\ S[17] = 472392\\\\ S[18] = 84312\\\\ S[19] = 6552\\\\ S[20] = 120\\\\ S[21] = 0\\\\ S[22] = 0\\\\ S[23] = 0\\\\ S[24] = 0$$ With these numbers, the probability that the first BINGO is obtained with the $n$th daub equals $$\\frac{S[n]}{n\\cdot{24\\choose n}},$$ provided we can assume that in all situation, the next daub is", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "talking about. There are more ways to get the top row and the four corners than there are to get the top two rows, because the first only requires $$7$$ balls and the second requires $$10$$ balls. I tried to write out a description of how to do a direct calculation, counting the number of multiple bingos and using the principle of inclusion and exclusion, but it's just too complicated to bother with. I've no doubt that the second method, figuring out the probability of not having a bingo with $$n$$ balls is the way to go. There are $$24$$ square on the card, not counting the free cells, and there are $$2^{24}=16,777,216$$ subsets of these cells. You'd need to write a computer program, to evaluate all of these possibilities and to record for each $$0\\leq k\\leq24$$ the probability $$p_k$$ that a card with $$k$$ balls does not have a bingo. Then given $$n$$ the probability that exactly $$k$$ of the $$n$$ balls drawn are to be found on our card is $${{24\\choose k}{51\\choose n-k}\\over{75\\choose n}}$$ Multiply this by $$p_k$$ and sum over all $$0\\leq k\\leq n$$ to get the probability of not having a bingo, then subtract from $$1$$ to get the probability of not having a bingo with $$n$$ balls. I doubt that I'll be able", "# Optimal cheating at BINGO You're sick of other players smugly announcing \"BINGO\" and walking triumphantly past you to claim their prize. This time it will be different. You bribed the caller to give you the BINGO calls ahead of time, in the order they will be called. Now you just need to create a BINGO board that will win as early as possible for those calls, guaranteeing you a win (or an unlikely tie). Given a delimited string or list of the calls in order, in typical BINGO format (letters included, e.g. B9 or G68, see the rules for more info), output a matrix or 2D list representing an optimal BINGO board for those calls. Assume input will always be valid. ### BINGO Rules: • 5x5 board • A \"BINGO\" is when your card has 5 numbers in a row from the numbers that have been called so far. • The center square is free (automatically counted towards a BINGO), and may be represented by whitespace, an empty list, -1, or 0. • The 5 columns are represented by the letters B,I,N,G,O, respectively. • The first column may contain the numbers 1-15, the second 16-30, ..., and the fifth 61-75. • The letters and numbers taken for input may optionally be delimited (by something that makes sense,", "than vertical (all numbers in some column). Bingo with a single card First, let’s describe how Bingo works with just one player. A Bingo card is a 5-by-5 grid of numbers, with each column containing 5 numbers randomly selected without replacement from 15 possibilities: the first “B” column is randomly selected from the numbers 1-15, the second “I” column is selected from 16-30, the third column from 31-45, the fourth column from 46-60, and the fifth “O” column from 61-75. An example is shown below. Example of an American Bingo card. A “caller” randomly draws, without replacement, from a pool of balls numbered 1 through 75, calling each number in turn as it is drawn, with the player marking the called number if it is present on his or her card. The player wins by marking all 5 squares in any row, column, or diagonal. (One minor wrinkle in this setup is that, in typical American-style Bingo, the center square is “free,” considered to be already marked before any numbers are called.) It will be useful to generalize this setup with parameters $(n,m)$, where each card is $n \\times n$ with each column selected from $m$ possible values, so that standard Bingo corresponds to $(n,m)=(5,15)$. We can compute the probability distribution of the number of draws required for", "out to be about 50.7%. Look at that lovely division, and the amazing thing it’s telling you: the probability that two people have the same birthday in a group of 23 people is a little over 1/2!!! Compare that to the standard first guess of 183. WOW. We’re not lying to you, and this isn’t magic. 23!!! ## The strange probabilities of the bingo card Amazing probabilities turn up in the unlikeliest of places. For example, have you ever played bingo? Usually you go to a “bingo hall”, and play with a whole bunch of people. You have a board in front of you that typically looks something like this: The column under the B can contain any of the numbers from 1 through 15, with no repeats. The next column will contain numbers in the range 16-30, the next 31-45, then 46-60, and lastly the O column has numbers in the range 61-75. Everyone has a different bingo card. Someone starts announcing letter/number combinations, like B13! N32! G44! When such a combination is called, you check and see if that number under that letter is on your card. In the above picture, B13 is not on the card, but G44 is. If what’s called out is on your card, you can mark it with a pen or", "219937-1. Since this study only utilises and estimated 230 random numbers, the Mersenne Twister is more than capable of providing completely randomized inputs. ## Card and ball number variations The simplicity of the base Bingo game has allowed a large number of variations in game style, as well as win patterns, to come to life. The rules of some popular versions, also the ones that are statistically compared in this article, are explained below. It is important for players to make sure they know the rules and win patterns of the specific games they are playing as these may vary slightly (or widely) between different bingo sites. ### 75-ball Bingo As the name indicates, this game is played using 75-balls. Cards (see Figure 1) consist of 5 rows and 5 columns, with the first column containing a random selection of numbers in the range 1-15, the second 16-30 and so on until the last column which contains a random selection from the numbers 60-75. The center square (third column, third row) of each card does not contain a number and is considered “free”, counting as being daubed from the start. As such it may for part of a win pattern. Wins are registered by completing 1 or 2 lines, all 5 lines (also known as “Full house” or", "20 & 44 & 52 & 68 \\\\ \\end{array}$ Then this set of two cards could result in a tie if the sequence of numbers called was $40~ 61~ 64~ 10~ 57~ 49~ 11~ 31~ 25$ This sequence would result in the card on the left completing the third row and the card on the right completing the fourth row when the number $25$ is called. ## Input The first line of the input is an integer $n$ ($2 \\leq n \\leq 100$), the number of bingo cards. After the first line are the $n$ bingo cards, each separated from the next by a blank line of input. Each bingo card consists of five lines of input. Each line consists of five integers in the range from $1$ to $3\\, 000$. The numbers on each bingo card are unique. ## Output If no ties are possible between any two cards, output “no ties”. Otherwise, output the two numbers $a$ and $b$ ($1 \\le a < b \\le n$) identifying the two cards for which a tie could occur, where the cards are numbered from $1$ to $n$ in the order that they appear in the input. If multiple pairs of cards can tie, output the pair with the smallest $a$, breaking any remaining ties with the smallest $b$.", "to . We can compute the probability distribution of the number of draws required for a single card to win. Bill Butler describes one approach, enumerating the possible partially-marked cards and computing the probability of at least one bingo for each such marking. Alternatively, we can use inclusion-exclusion to compute the cumulative distribution directly, by enumerating just the possible combinations of horizontal, vertical, and diagonal bingos (of which there are 5, 5, and 2, respectively) on a card with marks. In Mathematica: bingoSets[n_, free_: True, diag_: True] := Module[{card = Partition[Range[n^2], n]}, If[free, card[[Ceiling[n/2], Ceiling[n/2]]] = 0]; DeleteCases[Join[ card, Transpose[card], If[diag, {Diagonal[card], Diagonal[Reverse[card]]}, {}]], 0, Infinity]] bingoCDF[k_, nm_, bingos_] := Module[{j}, 1 - Total@Map[ (j = Length[Union @@ #]; (-1)^Length[#] Binomial[nm - j, k - j]) &, Subsets[bingos]]/Binomial[nm, k]] bingos = bingoSets[5, False, True]; cdf = Table[bingoCDF[k, 75, bingos], {k, 1, 75}]; (Note the optional arguments specifying whether the center square is free, and whether diagonal bingo is allowed. It will be convenient shortly to consider a simplified version of the game, where these two “special” rules are discarded.) The following figure shows the resulting distribution, with the expected number of 43.546 draws shown in red. Cumulative probability distribution of a single-card bingo in at most the given number of draws. The average of 43.546 draws is shown", "offers a simple. All this can be done without MCEdit or mods. We Buy and Sell Worldwide! 1-800-853-2073. Welcome to the ESL Activities. Each aside is located just after vertical grid-line 1 and after different horizontal grid lines. 5\" x 11\" A4 11\" x 17\" A3. You'll find no advertisements, pop-ups, or inappropriate links here. You can use nearly any combination of these classes to create more dynamic and flexible layouts. Our bingo card generator randomizes your words or numbers to make unique, great looking bingo cards. Leave some space in the margins outside of the top and left side of the grid because you will need to write numbers there later. If this is the first time you use the tool, please read this guide. Give you more variation are our purpose. A 5x5 bingo card is traditional, but you can try other sizes. A magic square of size N is a matrix composed of distinct integers between 1 and N^2 set such as the sum of any line or column are equal. Enter any address, city & state or zip: or Enter any call sign: Data provided by QRZ. This produces an isometric effect, or fake perspective. Square D Interlock Kits. They’ll need a transfer switch to connect to your circuit box. In the grid below", "# Math The real numbers $a$ and $b$ satisfy $|a| < 1$ and $|b| < 1.$ (a) In a grid that extends infinitely, the first row contains the numbers $1,$ $a,$ $a^2,$ $\\dots.$ The second row contains the numbers $b,$ $ab,$ $a^2 b,$ $\\dots.$ In general, each number is multiplied by $a$ to give the number to the right of it, and each number is multiplied by $b$ to give the number below it. Find the sum of all numbers in the grid. [asy] unitsize(1 cm); int i, j; for (i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5)); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5)); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5)); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5)); label(\"$b^2$\", (0.5,-2.5)); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5)); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5)); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5)); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] (b) Now suppose the grid is colored like a chessboard, with alternating black and white squares, as shown below. Find the sum of all the numbers that lie on the black squares. [asy] unitsize(1 cm); int i, j; for (i = 0; i <= 3; ++i) { for (j = 0; j <= 3; ++j) { if ((i + j) % 2 == 0) { fill(shift((i,-j))*((0,0)--(1,0)--(1,-1)--(0,-1)--cycle),black); } }} fill((0,-4)--(1,-4)--(1,-4.5)--(0,-4.5)--cycle,black); fill((2,-4)--(3,-4)--(3,-4.5)--(2,-4.5)--cycle,black); fill((4,0)--(4,-1)--(4.5,-1)--(4.5,0)--cycle,black); fill((4,-2)--(4,-3)--(4.5,-3)--(4.5,-2)--cycle,black); fill((4,-4)--(4.5,-4)--(4.5,-4.5)--(4,-4.5)--cycle,black); for", "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.", "Bingo card among multiple players is much more likely to have a horizontal bingo (all numbers in some row) than vertical (all numbers in some column). Bingo with a single card First, let’s describe how Bingo works with just one player. A Bingo card is a 5-by-5 grid of numbers, with each column containing 5 numbers randomly selected without replacement from 15 possibilities: the first “B” column is randomly selected from the numbers 1-15, the second “I” column is selected from 16-30, the third column from 31-45, the fourth column from 46-60, and the fifth “O” column from 61-75. An example is shown below. Example of an American Bingo card. A “caller” randomly draws, without replacement, from a pool of balls numbered 1 through 75, calling each number in turn as it is drawn, with the player marking the called number if it is present on his or her card. The player wins by marking all 5 squares in any row, column, or diagonal. (One minor wrinkle in this setup is that, in typical American-style Bingo, the center square is “free,” considered to be already marked before any numbers are called.) It will be useful to generalize this setup with parameters $(n,m)$, where each card is $n \\times n$ with each column selected from $m$ possible values, so", "+ (5 + 3 + 1 +2 + 4 + 6) = 84 \\equiv 4 \\hbox{ mod 10}.$$ So, the code is incorrect. Credit Card From the number 1723-1001-2065-1098 I calculated $\\alpha$, $\\beta$ and $\\gamma$. \\eqalign{ \\alpha &= 2(1+2+1+0+2+6+1+9)=44\\cr \\beta &= 2 \\ ({\\rm 5\\ and\\ 9})\\cr \\gamma &= 7+3+0+1+0+5+0+8=24\\cr \\alpha+\\beta+\\gamma&=70 \\equiv 0 {\\rm \\ mod 10}.} Therefore 1723 1001 2065 1098 is a valid Credit Card number. ISBN For 0 5215 3612 4 the result of the summation is 148 and it isn't divisible by 11. $$10 \\times 0 + 9 \\times 5 + 8\\times 2 + 7\\times 1 + 6\\times 5 + 5\\times 3 + 4\\times 6 + 3\\times 6 + 2\\times 1 + 1\\times 2 = 148$$ Therefore 0 5215 3612 4 is not a valid ISBN. US Banks The number 07100013 is not a valid US Bank number because it does not have 9 digits. If it had the digit 6 at the end then it would be valid because $$7\\times 0 + 3\\times 7 + 9\\times 1 + 7\\times 0 + 3\\times 0 + 9\\times 0 + 7\\times 1 + 3\\times 3 = 46 \\equiv 6 \\hbox{ mod 10}.$$", "{4,4,4} now handled thanks to Peter Taylor and cardboard_box. n_~g~o_ := {a___, Sequence @@ n, b___} :> {a, b, o}; f@s_ := Tr@Join[#[[2]], Sort@#[[1]] //. {1 -> 0, 2 -> 0, g[{3, 4}, 5], g[{3, 3}, 5], g[{4, 4, 4}, 10]}] &[Transpose[{m = Mod[#, 5], # - m} & /@ s]] Note: Non-purchases (test case #1) are entered as f[{0}]. How it works 1. For each item, the greatest multiple of 5 less than the respective price will be paid regardless of form of payment. (No getting around that.) 2. The remainders of Mod[n, 5] are then processed: 1's and 2's become 0's. Zeros stay unchanged. 3. Each pair {3, 4} -> {5}; afterwards each pair {3, 3} -> {5}; then the triple, {4,4,4}-> {10}, if applicable. 4. The remaining 4's, if any, remain unchanged (paid by credit card). 5. Original multiples of 5 summed with remainders that were tweaked (or not) in steps (2) to (4). Testing a12 adjusts for {3,3} a13 adjusts for {4,4,4} a1={0}; a2={48}; a3={92,20}; a4={47,56,45}; a5={55,6,98,69} ; a6={6,39,85,84,7}; a7={95,14,28,49,41,39}; a8={92,6,28,30,39,93,53}; a9={83,33,62,12,34,29,18,12}; a10={23,46,54,69,64,73,58,92,26}; a11={19,56,84,23,20,53,96,92,91,58}; a12={3,3,19,56,3,84,3,23,20,53,96,92,91,58,3,3}; a13={2,3,4,4,4,4,4}; f /@ {a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13} {0, 48, 110, 145, 225, 218, 263, 335, 273, 495, 583, 598, 19} • What about {3,3}? – Peter Taylor Jun 4 '13" ]

[ "do this is try finding the combinations for all numbers starting in one... 1,(2,3,4,5,6,7,8,9,10),(2,3,4,5,6,7,8,9,10) The second digit can only have nine different numbers and the third can only have eight (because it can't have the same as the second or first one). Therefore the formula for all combinations starting with one is $1\\times9\\times8$ now we multiply that by 10 to get all valid combinations for the first digit. $10\\times9\\times8=\\boxed{720}$ • July 8th 2006, 06:56 AM Soroban Welcome aboard, Kate! Quote: We are playing this game (Luffa Lotto) which the numbers are 1,2,3,4,5,6,7,8,9,10. A combination is {1,6,3}; another is {9,5,3}. {4,6,4} cannot be one 'cause there are two fours. We have got to work out how many combinations there are altogther, and we need the formula to work it out. From the examples you gave, the order of the numbers is important. . . [Technically, you should not call them \"combinations\".] You can reason it out like this . . . For the first number, there are $10$ possible choices. For the second number, there are $9$ possible remaining choices. For the third number, there are $8$ possible remaining choices. Therefore, for selecting three numbers, . . there are: $10 \\times 9 \\times 8 \\:=\\:720$ possible \"combinations\". • July 8th 2006, 07:12 AM Quick Quote: Originally Posted by Soroban", "with 4 columns: 1 2 3 4 1 2 3 5 1 2 3 6... 2 3 4 5 2 3 4 6 ... 47 48 49 50 To find the total possible combinations,we do: 50*49*48*47 __________ 1*2*3*4 This formula gives the maximum possible combinations. My question is,what would the formula be to find maximum of how many combinations of these numbers include number 1 and 2(for example)? There are so many difficulties with this qauestion that there no good place to start. FIRST: neither 1 nor 50 is between 1 and 50. If you want include them then say numbers from 1 to 50. With that linguistic understanding then your numbers look like $\\boxed{~A~}\\boxed{~B~}\\boxed{~C~}\\boxed{~D~}$ There are 50 choices for A, 49 choices for B, 48 choices for C, and 47 choices for D.", "by 10 to get all valid combinations for the first digit. $10\\times9\\times8=\\boxed{720}$ 3. Welcome aboard, Kate! We are playing this game (Luffa Lotto) which the numbers are 1,2,3,4,5,6,7,8,9,10. A combination is {1,6,3}; another is {9,5,3}. {4,6,4} cannot be one 'cause there are two fours. We have got to work out how many combinations there are altogther, and we need the formula to work it out. From the examples you gave, the order of the numbers is important. . . [Technically, you should not call them \"combinations\".] You can reason it out like this . . . For the first number, there are $10$ possible choices. For the second number, there are $9$ possible remaining choices. For the third number, there are $8$ possible remaining choices. Therefore, for selecting three numbers, . . there are: $10 \\times 9 \\times 8 \\:=\\:720$ possible \"combinations\". 4. Originally Posted by Soroban Welcome aboard, Kate! [size=3] From the examples you gave, the order of the numbers is important. . . [Technically, you should not call them \"combinations\".] Tsk, Tsk. Picky, picky, picky 5. ## help Originally Posted by Quick The best way to do this is try finding the combinations for all numbers starting in one... 1,(2,3,4,5,6,7,8,9,10),(2,3,4,5,6,7,8,9,10) The second digit can only have nine different numbers and the third can only have eight", "Therefore the 1 can only be placed in the second column top, and the second column must therefore contain 1-4-6, in that order top to bottom. Knowing that the second column bottom number is 6, we can determine the possibilities for the other three numbers in the bottom row. They must add to 27 (33-6), and all three are greater than 6, based on the gt/lt signs. The only possibilities are 7-8-12, 7-9-11, or 8-9-10. The 12 can only go in three places in the grid, based on the gt/lt signs: left column middle, lower right, or third column top. We can rule out third column top, as in that case the other two numbers in the column would have to add to ten, with the smaller of these (middle position) being greater than 4. This is impossible. Now consider the left column middle position for the 12. This is more complicated. Based on the column total, the other numbers must add to ten, with the bottom one larger than 6. The only possibilities are 2-8 or 3-7. These options would only allow 9-10 or 9-11, respectively, in the rightmost two slots of the bottom row, in that order. Either way, a 9 would be in the third-column bottom slot. Then, the other numbers in its column would", "can actually decide on the ordering of 1 and 9 as well; consider the hint for the middle row, which ends with 8; a 9 would automatically exceed this. Thus the middle row must contain the 1 and the sixth row the 9. The hint gives us for free that we have a 7 to the left of the 1. There are more interesting deductions we can make as well. Consider the seventh column. The clue is 16, 9 and this can be satisfied in two ways: {1357} in some order and then a singleton 9, or {79} and {135} in some order. However, we can prove that the latter construction is impossible; the middle row is an even number, and also we have already used the 1 and the 5 in the bottom right hand box so we can’t place those numbers there. Thus, the first four numbers must be {1357} in some order and there is a 9 somewhere in the bottom right box. The clue in the first row requires a 5 at the end. Given the 10, 10, 5 clue and the 4 already placed in the sixth column, the only odd substring in the last box must be the 5. We thus must have the 5 in the first row, and now only", "actually decide on the ordering of 1 and 9 as well; consider the hint for the middle row, which ends with 8; a 9 would automatically exceed this. Thus the middle row must contain the 1 and the sixth row the 9. The hint gives us for free that we have a 7 to the left of the 1. There are more interesting deductions we can make as well. Consider the seventh column. The clue is 16, 9 and this can be satisfied in two ways: {1357} in some order and then a singleton 9, or {79} and {135} in some order. However, we can prove that the latter construction is impossible; the middle row is an even number, and also we have already used the 1 and the 5 in the bottom right hand box so we can’t place those numbers there. Thus, the first four numbers must be {1357} in some order and there is a 9 somewhere in the bottom right box. The clue in the first row requires a 5 at the end. Given the 10, 10, 5 clue and the 4 already placed in the sixth column, the only odd substring in the last box must be the 5. We thus must have the 5 in the first row, and now only the", "to the obvious question: $$\\boxed{\\: 7 \\: } - \\boxed{12} + \\boxed{\\: 3 \\: } \\times \\boxed{\\: 3 \\: } = \\boxed{12} \\: / \\: \\boxed{\\: 3 \\: }$$ But it is illegal because it does not use all the numbers. Solved with pencil and paper only. • Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me. Sep 22 '18 at 17:16 Well, well, well. THERE EXISTS NO SOLUTION! (without lateral thinking, although inverting the $$6$$ to a $$9$$ has no solution either, in particular). Let's call the numbers we can choose from, the Option Numbers. The right hand side (RHS) will either equal $$1$$, $$2$$ or $$4$$ (without lateral thinking). ### Proof: This is our equation: $$\\Box-\\Box+\\Box\\times\\Box=\\Box\\:/\\:\\Box\\tag{\\small\\rm given}$$ $$7$$ can only be in the fifth box if the sixth box is also $$7$$, as that is the only number out of the option numbers that divide $$7$$. Otherwise, the left hand side (LHS) would not be an integer. So, a possibility is that the RHS is equal to $$7/7=1$$. Excluding $$7$$ now, $$3$$ cannot be in the fifth box because it is the lowest option number, and thus the fraction will not be an integer otherwise.", "# Math Help - Help 1. ## Help For the first number, there are 10 possible choices. For the second number, there are 9 possible remaining choices. For the third number, there are 8 possible remaining choices. Therefore, for selecting three numbers, . . there are: 10 \\times 9 \\times 8 \\:=\\:720 possible \"combinations\". this is the email i got, but is 720 the answer for all of the \"combinations\", or just the ones begining with 1?? I AM CONFUSED 2. If you were to have the set, let's say S, containing all the integers 0-9, then there are 10 possible elements from which to choose. Now upon picking the second number there are 9 elements from which to choose. This assumes that there are no repeats. And finally as you said, there are 8 elements to from which to choose the third time picking. This shows that there are 720 distinct combinations or permutations of choosing three elements from this set.", "the ordering of 1 and 9 as well; consider the hint for the middle row, which ends with 8; a 9 would automatically exceed this. Thus the middle row must contain the 1 and the sixth row the 9. The hint gives us for free that we have a 7 to the left of the 1. There are more interesting deductions we can make as well. Consider the seventh column. The clue is 16, 9 and this can be satisfied in two ways: {1357} in some order and then a singleton 9, or {79} and {135} in some order. However, we can prove that the latter construction is impossible; the middle row is an even number, and also we have already used the 1 and the 5 in the bottom right hand box so we can’t place those numbers there. Thus, the first four numbers must be {1357} in some order and there is a 9 somewhere in the bottom right box. The clue in the first row requires a 5 at the end. Given the 10, 10, 5 clue and the 4 already placed in the sixth column, the only odd substring in the last box must be the 5. We thus must have the 5 in the first row, and now only the 3 can fit", "the third box filled, we have 4 swans left, so 4 choices for the fourth box. And so on, until we eventually have a single swan left for the last box, swan 6. Which one should we put in? We gaze at swan 6, and he gazes back with a look of resignation. He knew what was coming before you did, and knows that the sooner this mathematical demonstration is over with, the sooner he can get back to his very busy social agenda. He waddles over to the seventh box and steps in. We can consider each choice we made as a branch point in a decision tree. The first choice had 7 possibilities, so 7 branches off of the main trunk. The second choice had 6 possibilities, so each of those 7 branches has 6 branches coming out of them. If we draw this tree all the way to the leaves, and count the number of leaves, we will get the total number of possible decisions we could have made. This is the same as the total number of ways the 7 swans could have been put in the 7 boxes. This number is $7\\times6\\times5\\times4\\times3\\times2\\times1 = 7! = 5040$. The swans are very glad you didn't decide to try out every combination before you were convinced.", "from 1, 2, 3, ..., 12. Then a is the lowest, b is the second-lowest, c is the third-lowest, and is the fourth-lowest, so the number of ways of choosing a,b,c,d is C(12,4) = 495. (b) There are 12 ways of choosing a. If a = 12, then there are 12 ways of choosing b. If a = 11, then there are 11 ways of choosing b. This pattern continues, so the number of ways of choosing b is 12 + 11 + 10 + ... + 1. If b = 11, then there are 11 ways of choosing c. If b = 10, then there are 10 ways of choosing c. This pattern continues, so the number of ways of choosing c is 11 + 10 + 9 + ... + 1. If c = 10, then there are 10 ways of choosing d. If c = 9, then there are 9 ways of choosing d. This pattern continues, so the number of ways of choosing d is 10 + 9 + 8 + ... + 1. So the number of ways of choosing a,b,c,d is 12(12 + 11 + ... + 1)(11 + 10 + 9 + .. + 1)(10 + 9 + 8 + ... + 1) = 12*78*66*55 = 3397680. (c) There are n", "no space to fit it. Thus 3 occurs after the 4, but it has to be separated from the 9 by an even number. More simply, the second column clue 10, 9, 6 indicates that our 5 has to be preceded by an even number and then a 1. There’s only one spot left for a 7 in this lower box, so we can insert it. We thus reach this position: Moving on, we can consider the middle column. 3 can only be satisfied by a singular 3, so we know that the 3 in the bottom row cannot be in the middle column. It thus must be in the sixth column, with 1 and 7 taking up the eighth and ninth columns (in some order). Now consider the rightmost column. It is 3, 10, 12; 3 can only be satisfied by a singular 3, so the odd substrings must be 3, (19 or 91) and (57 or 75). Given the position of the 2 (fourth row), it must be the case that the column has 1 and 9 in the fifth and sixth rows, an even number in the seventh, and 5 and 7 in the eighth and ninth. We thus clearly have 7 in the corner, and 5 above it. Finally, we can actually decide on", "Question Hence, we have 5 choices for the first place which are 1, 2, 3, 7, 9. In the second and third place, we can place any given number. We have 6 choices for these places, which are 0, 1, 2, 3, 7, 9. So, the total numbers will be $5 \\times 6 \\times 6 = 180$ . Hence, option A will be the correct option.", "invalid. We have 10 invalid possibilities out of 16, so we get 6/16 of the time exactly 1 lands in the first box.", "11, 12, 13, 14, 15, 16, 17, 18, and 19 and so these are not factors of 200. The next integer to try is 20 but this appears in the previous factor pair (10\\times{20}=20\\times{10}). We have therefore found all of the factor pairs for 200: \\begin{aligned} 1&\\times{200}\\\\ 2&\\times{100}\\\\ 4&\\times{50}\\\\ 5&\\times{40}\\\\ 8&\\times{25}\\\\ 10&\\times{20} \\end{aligned} Reading down the first column of factors, and up the second column, the factors of 200 are: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, and 200. ### Example 6: using factors to solve word problems A store needs 18 boxes of crackers. The boxes need to be stacked with the same number of boxes in each row. There needs to be more than 1 row of boxes and less than 18 rows. How many different ways can the rows be created? 1 \\times 18 is the first factor pair. However, this will not be used because there needs to be more than 1 row of boxes. 18 is an even number, 2 is a factor of 9. 9\\times{2}=18, which means there can be 9 rows of 2 boxes 2\\times{9}=18, which means there can be 2 rows of 9 boxes. 9 rows of 2 boxes is different from 2 rows of 9 boxes. 18 does not have a remainder when it is", "get a pattern like the following for the number of choices per step (filling in the top middle box first, the top right box next, etc). In each 3x3 box after the first, once you've filled in one row or column of the box, three of the remaining six digits are localized to a single row. Choose their locations first, and then fill in the remaining three cells. (So the actual order of which cells to fill in might vary depending on what you already know, but the number of choices is never more than what I've shown.) After you've filled in these cells the stars are all determined. * * * 9 8 7 6 5 4 * * * 6 5 4 3 3 2 * * * 3 2 1 3 2 1 6 5 4 * * * 6 3 3 3 3 2 * * * 5 3 2 3 2 1 * * * 4 2 1 6 3 3 6 5 4 * * * 5 3 2 3 3 2 * * * 4 2 1 3 2 1 * * * If I've calculated correctly, this gives 87 bits. There's some additional savings to be had in the last 3x3 block, per the comment by Peter Shor: every value", "4 as there’s no space to fit it. Thus 3 occurs after the 4, but it has to be separated from the 9 by an even number. More simply, the second column clue 10, 9, 6 indicates that our 5 has to be preceded by an even number and then a 1. There’s only one spot left for a 7 in this lower box, so we can insert it. We thus reach this position: Moving on, we can consider the middle column. 3 can only be satisfied by a singular 3, so we know that the 3 in the bottom row cannot be in the middle column. It thus must be in the sixth column, with 1 and 7 taking up the eighth and ninth columns (in some order). Now consider the rightmost column. It is 3, 10, 12; 3 can only be satisfied by a singular 3, so the odd substrings must be 3, (19 or 91) and (57 or 75). Given the position of the 2 (fourth row), it must be the case that the column has 1 and 9 in the fifth and sixth rows, an even number in the seventh, and 5 and 7 in the eighth and ninth. We thus clearly have 7 in the corner, and 5 above it. Finally, we can", "the 4 as there’s no space to fit it. Thus 3 occurs after the 4, but it has to be separated from the 9 by an even number. More simply, the second column clue 10, 9, 6 indicates that our 5 has to be preceded by an even number and then a 1. There’s only one spot left for a 7 in this lower box, so we can insert it. We thus reach this position: Moving on, we can consider the middle column. 3 can only be satisfied by a singular 3, so we know that the 3 in the bottom row cannot be in the middle column. It thus must be in the sixth column, with 1 and 7 taking up the eighth and ninth columns (in some order). Now consider the rightmost column. It is 3, 10, 12; 3 can only be satisfied by a singular 3, so the odd substrings must be 3, (19 or 91) and (57 or 75). Given the position of the 2 (fourth row), it must be the case that the column has 1 and 9 in the fifth and sixth rows, an even number in the seventh, and 5 and 7 in the eighth and ninth. We thus clearly have 7 in the corner, and 5 above it. Finally, we", "the 2nd. There are 6 possible. Let's take care of the 1st. There are 5 possible. The numbers of possibilities is hence $4 \\times 7 \\times 6 \\times 5$ The overall number of possibilities is $8 \\times 7 \\times 6 \\times 5$ (no repetition, so each time you pick a number, you can't pick it again). So the probability is $\\frac{4 \\times 7 \\times 6 \\times 5}{8 \\times 7 \\times 6 \\times 5}=\\frac 12$ Actually, it could have been more straightforward, because there are as many even numbers as odd numbers that are 4-digit without repetition ! I'll try the other ones later 3. ah i see wat you're saying but the book's answer says its 25/49 o.o lol 4. Originally Posted by TrueTears ah i see wat you're saying but the book's answer says its 25/49 o.o lol Even 4 digit number: With zero at the end the number of arrangements is (7)(6)(5)(1) = 210. Without zero at the end the number of arrangements is (6)(6)(5)(3) = 540. Note: 3 choices for the last digit, 6 choices for the first digit (6 non-zero numbers to choose the first digit from once the last digit is chosen, 6 numbers [including zero] to choose the second digit from once the last and first digits are chosen, 5 numbers to", "subtract the ways that have two squares in the same row or column. There are 10 ways to choose the row/column, $5\\choose2$ ways to choose the two squares in the row/column, and 23 choices remaining for the third square, so all up you must subtract $10\\times{5\\choose2}\\times23$. Now you have to put back in all the configurations you subtracted out twice. These are the ones with two in the same row and two in the same column, of which there are $25\\times4\\times4$, and also the ones in which there are three in the same row/column, of which there are $10\\times{5\\choose3}$. The ones with 3 in a column were counted in once, then subtracted out 3 times, so they have to be put back in twice. So the answer is $${25\\choose3}-10\\times{5\\choose2}\\times23+25\\times4\\times4+2\\times10\\times{5\\choose3}$$ which comes out to 600. - You can choose the first square arbitrarily, so there are $25$ options for it. There are now $4$ rows and $4$ columns that the second square can be chosen from, so there are $16$ options for it (alternatively, if one just wants to count, there are $25$ squares total, minus the chosen first square, minus the $4$ other squares in the same row, minus the other $4$ squares in the same column, for $25 - 1 - 4 - 4 = 16$ options)." ]

[ "+0 # help asap 0 98 1 I need some help on this problem In SHORT BINGO, a $5\\times5$ card is filled by marking the middle square as WILD and placing 24 other numbers in the remaining 24 squares. Specifically, a card is made by placing 5 numbers from the set $1-10$ in the first column, 5 numbers from $11-20$ in the second column, 4 numbers $21-30$ in the third column (skipping the WILD square in the middle), 5 numbers from $31-40$ in the fourth column and 5 numbers from $41-50$ in the last column. One possible SHORT BINGO card is: To play SHORT BINGO, someone names numbers, chosen at random, and players mark those numbers on their cards. A player wins when he marks 5 in a row, horizontally, vertically, or diagonally. How many distinct possibilities are there for the values in the first row of a SHORT BINGO card? Jul 2, 2020", "# Bingo card generator A Bingo card is five columns of five squares each, with the middle square designated \"FREE\". Numbers cannot duplicate. The five columns are populated with the following range of numbers: • B:1-15 • I:16-30 • N:31-45 • G:46-60 • O:61-75 In as few characters as possible, output a string that can be interpreted as a randomized Bingo card. For example: 1,2,3,4,5,16,17,18,19,20,31,32,33,34,35,46,47,48,49,50,61,62,63,64,65 This example is not randomized so that I can show that column 1 is populated with 1,2,3,4,5. Also note that the free space has not been given any special treatment because the front-end that interprets this string will skip it. Another example would be: 1,16,31,46,61,2,17,32,47,62... In this example, the output is by row instead of by column. A third example might be: 01020304051617181920313233343546474849506162636465 This is the same output as the 1st example except in fixed length. • Am I the only one who has never heard of Bingo but instead only of Bullshit Bingo? – Joey May 20 '11 at 18:54 • Yes! That's it! My idea is to come up with a list of 75 or more words and populate the card with SELECT * FROM List ORDER BY NEWID() May 20 '11 at 23:29 # PHP, 86 for($o=[];25>$i=count($o);){$n=rand(1,15)+($i-$i%5)*3;$o[$n]=$n;}echo implode(\",\",$o); • Welcome to PPCG, nice first answer c: – Rod Oct 26", "Bingo is $(n,m)=(5,15)$, the card game is $(4,13)$, and the dice version is effectively $(6,\\infty)$. The Math Horizons article referenced below describes an approach to calculating these probabilities, which involves enumerating integer partitions. However, this problem is ready-made for generating functions, which takes care of the partition house-keeping for us: let’s define $g_a(x) = \\left(\\sum_{j=a}^{n-1} {m \\choose j}x^j\\right)^{n-1}$ so that, for example, for Bingo with no free square, $g_1(x) = \\left({15 \\choose 1}x^1 + {15 \\choose 2}x^2 + {15 \\choose 3}x^3 + {15 \\choose 4}x^4\\right)^4$ Intuitively, each factor corresponds to a column, where each coefficient of $x^j$ indicates the number of ways to draw exactly $j$ numbers from that column (with some minimum number from each column specified by $a$). The overall coefficient of $x^k$ indicates the number of ways to draw $k$ numbers in total, with neither a horizontal nor vertical bingo. Then using the notation from the article, the probability $P(H_k)$ of a horizontal bingo on exactly the $k$-th draw is $P(H_k) = \\frac{mn(k-1)!(mn-k)!}{(mn)!}[x^{k-1}]g_1(x)$ and the probability $P(V_k)$ of a vertical bingo on exactly the $k$-th draw is $P(V_k) = \\frac{mn(k-1)!(mn-k)!}{(mn)!} {m-1 \\choose n-1} [x^{k-n}](g_0(x)-g_1(x))$ The summation $\\sum_{k=n}^{(n-1)^2+1} P(H_k)$ over all possible numbers of draws yields the overall probability of about 0.752 that a horizontal bingo is observed before a vertical one. Similarly, for the card", "So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,5)--(8,5)--(8,0)--cycle,green); fill((18,0)--(18,5)--(19,5)--(19,0)--cycle,green); fill((27,0)--(27,5)--(28,5)--(28,0)--cycle,green); for (real i=1; i<5; ++i) { for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,i+0.5)); } } for (real i=0; i<30; ++i) { label(\"\\vdots\",(i+0.5,2/3)); } add(grid(30,5,linewidth(1.25))); [/asy]$ As indicated by the white squares, each group of $30$ digits on the original list has $\\frac25\\cdot30=12$ digits remaining on the final list. Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green); fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green); fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green); for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,0.5)); } draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow); draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow); draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow); add(grid(30,1,linewidth(1.25))); [/asy]$ Therefore, the answer is $4+2+5=\\boxed{\\textbf{(D)}\\ 11}.$ ~MRENTHUSIASM ## Video Solution https://youtu.be/t6yjfKXpwDs 16:40 ~IceMatrix", "of the daubs markes as last and such that the pattern has at least one bingo but no bingo if the last daub is removed. There are only $24\\times 2^{23}$ patterns you have to investigate. – Hagen von Eitzen Sep 16 '12 at 12:24 yeah only 24*2^23 – pagla Sep 16 '12 at 12:28 Well you do have a computer, don't you? – Hagen von Eitzen Sep 16 '12 at 13:13 Never mind. Maybe it's more important that I have a computer ;) – Hagen von Eitzen Sep 16 '12 at 14:05 Assuming that \"Bingo\" means complete row or column or diagonal, I computed the following numbers (count of \"winning pattern\") by computer: $$S[0] = 0\\\\ S[1] = 0\\\\ S[2] = 0\\\\ S[3] = 0\\\\ S[4] = 16\\\\ S[5] = 360\\\\ S[6] = 3800\\\\ S[7] = 25080\\\\ S[8] = 116040\\\\ S[9] = 399048\\\\ S[10] = 1053120\\\\ S[11] = 2167368\\\\ S[12] = 3493152\\\\ S[13] = 4380480\\\\ S[14] = 4197720\\\\ S[15] = 2975160\\\\ S[16] = 1478304\\\\ S[17] = 472392\\\\ S[18] = 84312\\\\ S[19] = 6552\\\\ S[20] = 120\\\\ S[21] = 0\\\\ S[22] = 0\\\\ S[23] = 0\\\\ S[24] = 0$$ With these numbers, the probability that the first BINGO is obtained with the $n$th daub equals $$\\frac{S[n]}{n\\cdot{24\\choose n}},$$ provided we can assume that in all situation, the next daub is", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "talking about. There are more ways to get the top row and the four corners than there are to get the top two rows, because the first only requires $$7$$ balls and the second requires $$10$$ balls. I tried to write out a description of how to do a direct calculation, counting the number of multiple bingos and using the principle of inclusion and exclusion, but it's just too complicated to bother with. I've no doubt that the second method, figuring out the probability of not having a bingo with $$n$$ balls is the way to go. There are $$24$$ square on the card, not counting the free cells, and there are $$2^{24}=16,777,216$$ subsets of these cells. You'd need to write a computer program, to evaluate all of these possibilities and to record for each $$0\\leq k\\leq24$$ the probability $$p_k$$ that a card with $$k$$ balls does not have a bingo. Then given $$n$$ the probability that exactly $$k$$ of the $$n$$ balls drawn are to be found on our card is $${{24\\choose k}{51\\choose n-k}\\over{75\\choose n}}$$ Multiply this by $$p_k$$ and sum over all $$0\\leq k\\leq n$$ to get the probability of not having a bingo, then subtract from $$1$$ to get the probability of not having a bingo with $$n$$ balls. I doubt that I'll be able", "# Optimal cheating at BINGO You're sick of other players smugly announcing \"BINGO\" and walking triumphantly past you to claim their prize. This time it will be different. You bribed the caller to give you the BINGO calls ahead of time, in the order they will be called. Now you just need to create a BINGO board that will win as early as possible for those calls, guaranteeing you a win (or an unlikely tie). Given a delimited string or list of the calls in order, in typical BINGO format (letters included, e.g. B9 or G68, see the rules for more info), output a matrix or 2D list representing an optimal BINGO board for those calls. Assume input will always be valid. ### BINGO Rules: • 5x5 board • A \"BINGO\" is when your card has 5 numbers in a row from the numbers that have been called so far. • The center square is free (automatically counted towards a BINGO), and may be represented by whitespace, an empty list, -1, or 0. • The 5 columns are represented by the letters B,I,N,G,O, respectively. • The first column may contain the numbers 1-15, the second 16-30, ..., and the fifth 61-75. • The letters and numbers taken for input may optionally be delimited (by something that makes sense,", "than vertical (all numbers in some column). Bingo with a single card First, let’s describe how Bingo works with just one player. A Bingo card is a 5-by-5 grid of numbers, with each column containing 5 numbers randomly selected without replacement from 15 possibilities: the first “B” column is randomly selected from the numbers 1-15, the second “I” column is selected from 16-30, the third column from 31-45, the fourth column from 46-60, and the fifth “O” column from 61-75. An example is shown below. Example of an American Bingo card. A “caller” randomly draws, without replacement, from a pool of balls numbered 1 through 75, calling each number in turn as it is drawn, with the player marking the called number if it is present on his or her card. The player wins by marking all 5 squares in any row, column, or diagonal. (One minor wrinkle in this setup is that, in typical American-style Bingo, the center square is “free,” considered to be already marked before any numbers are called.) It will be useful to generalize this setup with parameters $(n,m)$, where each card is $n \\times n$ with each column selected from $m$ possible values, so that standard Bingo corresponds to $(n,m)=(5,15)$. We can compute the probability distribution of the number of draws required for", "out to be about 50.7%. Look at that lovely division, and the amazing thing it’s telling you: the probability that two people have the same birthday in a group of 23 people is a little over 1/2!!! Compare that to the standard first guess of 183. WOW. We’re not lying to you, and this isn’t magic. 23!!! ## The strange probabilities of the bingo card Amazing probabilities turn up in the unlikeliest of places. For example, have you ever played bingo? Usually you go to a “bingo hall”, and play with a whole bunch of people. You have a board in front of you that typically looks something like this: The column under the B can contain any of the numbers from 1 through 15, with no repeats. The next column will contain numbers in the range 16-30, the next 31-45, then 46-60, and lastly the O column has numbers in the range 61-75. Everyone has a different bingo card. Someone starts announcing letter/number combinations, like B13! N32! G44! When such a combination is called, you check and see if that number under that letter is on your card. In the above picture, B13 is not on the card, but G44 is. If what’s called out is on your card, you can mark it with a pen or", "219937-1. Since this study only utilises and estimated 230 random numbers, the Mersenne Twister is more than capable of providing completely randomized inputs. ## Card and ball number variations The simplicity of the base Bingo game has allowed a large number of variations in game style, as well as win patterns, to come to life. The rules of some popular versions, also the ones that are statistically compared in this article, are explained below. It is important for players to make sure they know the rules and win patterns of the specific games they are playing as these may vary slightly (or widely) between different bingo sites. ### 75-ball Bingo As the name indicates, this game is played using 75-balls. Cards (see Figure 1) consist of 5 rows and 5 columns, with the first column containing a random selection of numbers in the range 1-15, the second 16-30 and so on until the last column which contains a random selection from the numbers 60-75. The center square (third column, third row) of each card does not contain a number and is considered “free”, counting as being daubed from the start. As such it may for part of a win pattern. Wins are registered by completing 1 or 2 lines, all 5 lines (also known as “Full house” or", "20 & 44 & 52 & 68 \\\\ \\end{array}$ Then this set of two cards could result in a tie if the sequence of numbers called was $40~ 61~ 64~ 10~ 57~ 49~ 11~ 31~ 25$ This sequence would result in the card on the left completing the third row and the card on the right completing the fourth row when the number $25$ is called. ## Input The first line of the input is an integer $n$ ($2 \\leq n \\leq 100$), the number of bingo cards. After the first line are the $n$ bingo cards, each separated from the next by a blank line of input. Each bingo card consists of five lines of input. Each line consists of five integers in the range from $1$ to $3\\, 000$. The numbers on each bingo card are unique. ## Output If no ties are possible between any two cards, output “no ties”. Otherwise, output the two numbers $a$ and $b$ ($1 \\le a < b \\le n$) identifying the two cards for which a tie could occur, where the cards are numbered from $1$ to $n$ in the order that they appear in the input. If multiple pairs of cards can tie, output the pair with the smallest $a$, breaking any remaining ties with the smallest $b$.", "to . We can compute the probability distribution of the number of draws required for a single card to win. Bill Butler describes one approach, enumerating the possible partially-marked cards and computing the probability of at least one bingo for each such marking. Alternatively, we can use inclusion-exclusion to compute the cumulative distribution directly, by enumerating just the possible combinations of horizontal, vertical, and diagonal bingos (of which there are 5, 5, and 2, respectively) on a card with marks. In Mathematica: bingoSets[n_, free_: True, diag_: True] := Module[{card = Partition[Range[n^2], n]}, If[free, card[[Ceiling[n/2], Ceiling[n/2]]] = 0]; DeleteCases[Join[ card, Transpose[card], If[diag, {Diagonal[card], Diagonal[Reverse[card]]}, {}]], 0, Infinity]] bingoCDF[k_, nm_, bingos_] := Module[{j}, 1 - Total@Map[ (j = Length[Union @@ #]; (-1)^Length[#] Binomial[nm - j, k - j]) &, Subsets[bingos]]/Binomial[nm, k]] bingos = bingoSets[5, False, True]; cdf = Table[bingoCDF[k, 75, bingos], {k, 1, 75}]; (Note the optional arguments specifying whether the center square is free, and whether diagonal bingo is allowed. It will be convenient shortly to consider a simplified version of the game, where these two “special” rules are discarded.) The following figure shows the resulting distribution, with the expected number of 43.546 draws shown in red. Cumulative probability distribution of a single-card bingo in at most the given number of draws. The average of 43.546 draws is shown", "offers a simple. All this can be done without MCEdit or mods. We Buy and Sell Worldwide! 1-800-853-2073. Welcome to the ESL Activities. Each aside is located just after vertical grid-line 1 and after different horizontal grid lines. 5\" x 11\" A4 11\" x 17\" A3. You'll find no advertisements, pop-ups, or inappropriate links here. You can use nearly any combination of these classes to create more dynamic and flexible layouts. Our bingo card generator randomizes your words or numbers to make unique, great looking bingo cards. Leave some space in the margins outside of the top and left side of the grid because you will need to write numbers there later. If this is the first time you use the tool, please read this guide. Give you more variation are our purpose. A 5x5 bingo card is traditional, but you can try other sizes. A magic square of size N is a matrix composed of distinct integers between 1 and N^2 set such as the sum of any line or column are equal. Enter any address, city & state or zip: or Enter any call sign: Data provided by QRZ. This produces an isometric effect, or fake perspective. Square D Interlock Kits. They’ll need a transfer switch to connect to your circuit box. In the grid below", "# Math The real numbers $a$ and $b$ satisfy $|a| < 1$ and $|b| < 1.$ (a) In a grid that extends infinitely, the first row contains the numbers $1,$ $a,$ $a^2,$ $\\dots.$ The second row contains the numbers $b,$ $ab,$ $a^2 b,$ $\\dots.$ In general, each number is multiplied by $a$ to give the number to the right of it, and each number is multiplied by $b$ to give the number below it. Find the sum of all numbers in the grid. [asy] unitsize(1 cm); int i, j; for (i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5)); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5)); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5)); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5)); label(\"$b^2$\", (0.5,-2.5)); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5)); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5)); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5)); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] (b) Now suppose the grid is colored like a chessboard, with alternating black and white squares, as shown below. Find the sum of all the numbers that lie on the black squares. [asy] unitsize(1 cm); int i, j; for (i = 0; i <= 3; ++i) { for (j = 0; j <= 3; ++j) { if ((i + j) % 2 == 0) { fill(shift((i,-j))*((0,0)--(1,0)--(1,-1)--(0,-1)--cycle),black); } }} fill((0,-4)--(1,-4)--(1,-4.5)--(0,-4.5)--cycle,black); fill((2,-4)--(3,-4)--(3,-4.5)--(2,-4.5)--cycle,black); fill((4,0)--(4,-1)--(4.5,-1)--(4.5,0)--cycle,black); fill((4,-2)--(4,-3)--(4.5,-3)--(4.5,-2)--cycle,black); fill((4,-4)--(4.5,-4)--(4.5,-4.5)--(4,-4.5)--cycle,black); for", "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.", "Bingo card among multiple players is much more likely to have a horizontal bingo (all numbers in some row) than vertical (all numbers in some column). Bingo with a single card First, let’s describe how Bingo works with just one player. A Bingo card is a 5-by-5 grid of numbers, with each column containing 5 numbers randomly selected without replacement from 15 possibilities: the first “B” column is randomly selected from the numbers 1-15, the second “I” column is selected from 16-30, the third column from 31-45, the fourth column from 46-60, and the fifth “O” column from 61-75. An example is shown below. Example of an American Bingo card. A “caller” randomly draws, without replacement, from a pool of balls numbered 1 through 75, calling each number in turn as it is drawn, with the player marking the called number if it is present on his or her card. The player wins by marking all 5 squares in any row, column, or diagonal. (One minor wrinkle in this setup is that, in typical American-style Bingo, the center square is “free,” considered to be already marked before any numbers are called.) It will be useful to generalize this setup with parameters $(n,m)$, where each card is $n \\times n$ with each column selected from $m$ possible values, so", "+ (5 + 3 + 1 +2 + 4 + 6) = 84 \\equiv 4 \\hbox{ mod 10}.$$ So, the code is incorrect. Credit Card From the number 1723-1001-2065-1098 I calculated $\\alpha$, $\\beta$ and $\\gamma$. \\eqalign{ \\alpha &= 2(1+2+1+0+2+6+1+9)=44\\cr \\beta &= 2 \\ ({\\rm 5\\ and\\ 9})\\cr \\gamma &= 7+3+0+1+0+5+0+8=24\\cr \\alpha+\\beta+\\gamma&=70 \\equiv 0 {\\rm \\ mod 10}.} Therefore 1723 1001 2065 1098 is a valid Credit Card number. ISBN For 0 5215 3612 4 the result of the summation is 148 and it isn't divisible by 11. $$10 \\times 0 + 9 \\times 5 + 8\\times 2 + 7\\times 1 + 6\\times 5 + 5\\times 3 + 4\\times 6 + 3\\times 6 + 2\\times 1 + 1\\times 2 = 148$$ Therefore 0 5215 3612 4 is not a valid ISBN. US Banks The number 07100013 is not a valid US Bank number because it does not have 9 digits. If it had the digit 6 at the end then it would be valid because $$7\\times 0 + 3\\times 7 + 9\\times 1 + 7\\times 0 + 3\\times 0 + 9\\times 0 + 7\\times 1 + 3\\times 3 = 46 \\equiv 6 \\hbox{ mod 10}.$$", "{4,4,4} now handled thanks to Peter Taylor and cardboard_box. n_~g~o_ := {a___, Sequence @@ n, b___} :> {a, b, o}; f@s_ := Tr@Join[#[[2]], Sort@#[[1]] //. {1 -> 0, 2 -> 0, g[{3, 4}, 5], g[{3, 3}, 5], g[{4, 4, 4}, 10]}] &[Transpose[{m = Mod[#, 5], # - m} & /@ s]] Note: Non-purchases (test case #1) are entered as f[{0}]. How it works 1. For each item, the greatest multiple of 5 less than the respective price will be paid regardless of form of payment. (No getting around that.) 2. The remainders of Mod[n, 5] are then processed: 1's and 2's become 0's. Zeros stay unchanged. 3. Each pair {3, 4} -> {5}; afterwards each pair {3, 3} -> {5}; then the triple, {4,4,4}-> {10}, if applicable. 4. The remaining 4's, if any, remain unchanged (paid by credit card). 5. Original multiples of 5 summed with remainders that were tweaked (or not) in steps (2) to (4). Testing a12 adjusts for {3,3} a13 adjusts for {4,4,4} a1={0}; a2={48}; a3={92,20}; a4={47,56,45}; a5={55,6,98,69} ; a6={6,39,85,84,7}; a7={95,14,28,49,41,39}; a8={92,6,28,30,39,93,53}; a9={83,33,62,12,34,29,18,12}; a10={23,46,54,69,64,73,58,92,26}; a11={19,56,84,23,20,53,96,92,91,58}; a12={3,3,19,56,3,84,3,23,20,53,96,92,91,58,3,3}; a13={2,3,4,4,4,4,4}; f /@ {a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12, a13} {0, 48, 110, 145, 225, 218, 263, 335, 273, 495, 583, 598, 19} • What about {3,3}? – Peter Taylor Jun 4 '13" ]

[ "+0 Compute the product of 0.123(overline) and 9, and write your result as a fraction in simplified form. +1 500 1 +5 Compute the product of 0.123(overline) and 9, and write your result as a fraction in simplified form. (0.123 is a repeating decimal) Feb 7, 2019 #1 +80 +2 $$0.\\overline{123} = \\frac{41}{333} \\\\ \\text{So, } \\frac{41}{333} \\times 9= \\boxed{\\frac{41}{37}}$$ . Feb 7, 2019 $$0.\\overline{123} = \\frac{41}{333} \\\\ \\text{So, } \\frac{41}{333} \\times 9= \\boxed{\\frac{41}{37}}$$", "• # question_answer Find the product: $(98\\text{ }+\\text{ }14)\\times 0$ A) 112 B) 1C) 0 D) 0+98+14 $\\text{a}\\times \\text{0=0}$. The product of any number and zero is zero.", "+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", "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.", "# How do you find the product of 34% and some number? $.34 n$ Therefore, we would simply multiply the **34%** by the variable n $.34 n$", "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?", "### Home > A2C > Chapter 10 > Lesson 10.2.5 > Problem10-128 10-128. Use the Zero Product Property. $0=(x-(-2+\\sqrt{3}))(x-(-2-\\sqrt{3}))$ $0=(x+2-\\sqrt{3})(x+2+\\sqrt{3})$ y = x2 + 4x + 1 Use the same process as in part (a).", "product from 23. Now bring down the 9 and repeat these steps. There are 9 fours in 39. Write the 9 over the 9. Multiply the 9 by 4 and subtract this product from 39. There are no more numbers to bring down, so we are done. The remainder is 3. Check by multiplying. So $$1,439÷4$$ is $$359$$ with a remainder of $$3.$$ Our answer is correct. ### Extra: Divide. Then check by multiplying: $$3,812÷8.$$ #### Solution 476 with a remainder of 4 ### Extra: Divide. Then check by multiplying: $$4,319÷8.$$ #### Solution 539 with a remainder of 7 ### Example Divide and then check by multiplying: $$1,461÷13.$$ #### Solution Let’s rewrite the problem to set it up for long division. $$13\\overline{)1,461}$$ First we try to divide 13 into 1. Since that won’t work, we try 13 into 14. There is 1 thirteen in 14. We write the 1 over the 4. Multiply the 1 by 13 and subtract this product from 14. Now bring down the 6 and repeat these steps. There is 1 thirteen in 16. Write the 1 over the 6. Multiply the 1 by 13 and subtract this product from 16. Now bring down the 1 and repeat these steps. There are 2 thirteens in 31. Write the 2 over the 1. Multiply the", "### Home > A2C > Chapter 7 > Lesson 7.3.3 > Problem7-191 7-191. Show how the Zero Product Property can be used to solve each of the following equations. 1. $x\\left(2x − 1\\right)\\left(x − 3\\right) = 0$ $x = 0, 0.5, 3$ 1. $2x^{3} + x^{2} − 3x = 0$ The factored form is $x\\left(x − 1\\right)\\left(2x + 3\\right) = 0$", "# How do you find the product of 34% and some number? Apr 28, 2016 $.34 n$ Therefore, we would simply multiply the **34%** by the variable n $.34 n$", "Product of algebraic fractions – Practice problems Product of algebraic fractions quiz You have completed the quiz! #### What is the result of multiplying the fractions and simplifying? $$\\frac{5x^2}{x^2-2x} \\cdot \\frac{x^2-4}{x^2+2x}$$ Write the result in the input box. $latex =$", "by 3 and $r_1$ is the remainder. Just go back to the Division Algorithm, and it should all make sense. Now, to show that $\\overline{x_1+y_1} = \\overline{x_2+y_2}$, you need to show that $(x_2+y_2) - (x_1+y_1)$ is divisible by 3.", "# Find the maximum product Divide $14$ into the sum of many natural numbers, and find the maximum value of the product of these numbers, ×", "result should be $2$. + • Remember the arithmetic order for how the calculation works. Paranthesis, multiply/division and addition/subtraction. + 1. 1 Declare a few variables and set them to a value of your choice 2. 2 Store the calculation in another variable 3. 3 Print out the result! + • + +", "help us determine the fractional representation of x. b. After multiplying both sides of the equation by 103, rewrite the resulting equation by making a substitution that will help determine the fractional value of x. Explain how you were able to make the substitution. c. Solve the equation to determine the value of x. 2. Find the fraction equal to x = $$0.\\overline 4$$. Check that you are correct using a calculator. Example 2 Find the fraction that is equal to the infinite decimal $$2.13\\overline 8$$. Exercises 3–4 3. Find the fraction equal to $$1.6\\overline {23}$$. Check that you are correct using a calculator. 4. Find the fraction equal to $$2.9\\overline {60}$$. Check that you are correct using a calculator. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", "is 48. Write the product 48 over the denominator 4. Write the fraction 48/4 in simplest form as a 12. Question 5. 9 × $$\\frac{5}{6}$$ = ____________ 9 x (5/6) = (9 x 5)/6 = 45/6 The product is 45/6. The simplest form of 45/6 is 7(3/6). Explanation: In the above image we can observe the expression 9 x (5/6). First multiply the two whole numbers in the numerator. Multiply 9 with 5 the product is 45. Write the product 45 over the denominator 6. Write the fraction 45/6 in simplest form as a mixed number. The mixed number is 7(3/6). Question 6. 6 × $$\\frac{3}{8}$$ = ____________ 6 x (3/8) = (6 x 3)/8 = 18/8 = 9/4 The product is 9/4. The simplest form of 9/4 is 2(1/4). Explanation: In the above image we can observe the expression 6 x (3/8). First multiply the two whole numbers in the numerator. Multiply 6 with 3 the product is 18. Write the product 18 over the denominator 8. Write the fraction 9/4 in simplest form as a mixed number. The mixed number is 2(1/4). Question 7. $$\\frac{2}{9}$$ × 5 = ____________ (2/9) x 5 = (2 x 5)/9 = 10/9 The product is 10/9. The simplest form of 10/9 is 1(1/9). Explanation: In the above image we can", "to t = 2. As you move the value of h, closer to 0 what does the value of the fraction represent? Set the value of t to t = 1. Again as the value of h moves closer to 0 what happens to the value of the fraction? What do the values you have found represent? ## QUESTION 3 The PRODUCT RULE says that; Show how to use the expression in the applet ABOVE to derive this rule. HINT 1: split the fraction up into THREE separate fractions...then go from there! HINT 2: Recall that;", "# Simplify x/(x^2+3x+2)-1/((x+2)(x+1)) Factor using the AC method. Consider the form . Find a pair of integers whose product is and whose sum is . In this case, whose product is and whose sum is . Write the factored form using these integers. Combine into one fraction. Reorder the factors of . Combine the numerators over the common denominator. Simplify x/(x^2+3x+2)-1/((x+2)(x+1))", "SEARCH HOME Math Central Quandaries & Queries Question from Amie: the product of two numbers is 108.when the larger number is divided by the smaller number the quotient is 3. What are the number.tnx Hi Amie, Suppose the smaller number is $s.$ If you divide the larger number by the smaller number you get 3 and hence the larger number is 3 times the smaller number. Thus the larger number is $3s.$ The product of the two numbers is 108. Solve for $s.$ Penny", "Using the Zero Product Property: $x−1=0\\ \\ \\text{or}\\ \\ x+1=0\\ \\ \\text{or}\\ \\ x−2=0$ $x=1\\ \\ \\text{or}\\ \\ x=−1\\ \\ \\text{or}\\ \\ x=2$" ]

[ "to get $$1000 x = 334.\\overline{4}$$ Now, try to subtract the above equations. What do you get? Second option: You can write $x=0.33\\overline 4$ as $$0.3 + 0.03 + 0.004 + 0.0004 + \\dots$$ which we can rewrite as $$x=0.33 + 0.004\\cdot (1 + 0.1 + 0.01 + \\dots)$$ or further as $$x=\\frac{33}{100} + \\frac{4}{1000}\\cdot \\sum_{i=0}^\\infty \\left(\\frac{1}{10}\\right)^{i}$$ Now, remember how you were taught what the sum of a geometric series is? HINT: $$0.3344444444\\cdots=\\dfrac{33}{100}+4\\sum_{r=3}^\\infty\\dfrac1{10^r}$$ Let $x = 0.33444 \\ldots$, then $100x = 33.44444 \\ldots$ so that $100x - x = 99x = 33.11$ and $x=\\frac{3311}{9900}$. There is a trick, which goes like: find the repeating pattern and divide it with as many nines. For example $$\\frac{1090}{9999}=0.10901090....$$ In this case we have $$0.33444...=0.33+0.0044...=\\frac{33}{100}+\\frac{1}{100}0.44...=\\frac{33}{100}+\\frac{1}{100}\\frac{4}{9}=\\frac{297+4}{900}=\\frac{301}{900}.$$ Multiply by $\\dfrac94$ to turn the $4$s into $9$s, giving $$0.7524999999\\cdots$$ which equals $$0.7525\\,.$$ $$\\frac{7525}{10000}\\frac{4}{9}=\\frac{301}{900}.$$ You can move the decimal point one position to the left by multiplying by $10$. $$0.3344444444\\cdots\\times10=3.344444444\\cdots$$ Then subtracting these two numbers, a finite number of decimals remain and $$0.3344444444\\cdots\\times9=3.01\\,.$$ So the fraction is $$\\frac{3.01}{9}=\\frac{301}{900}.$$ General method: • Let $x$ denote the input number • Let $|n|$ denote the number of decimal digits in $n$ • Split $x$ into the following parts: • $\\color\\red{A}=$ the integer part, i.e., $\\lfloor{x}\\rfloor$ • $\\color\\green{B}=$ the fraction part's non-periodic prefix • $\\color\\orange{C}=$ the fraction part's periodic postfix", "# 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 $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "≠ 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", "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 $$2.\\overline{962} = \\frac {2960}{999}$$ -", "digits of 9’s as the number of digits that repeat. For example, $$0 . \\overline{81}$$has two repeating decimal digits, and the denominator has two 9’s. The pattern is even true for $$0 . \\overline{9}$$. According to the pattern, this decimal equals $$\\frac{9}{9}$$, which is 1. Eureka Math Grade 8 Module 7 Lesson 10 Exit Ticket Answer Key Question 1. Find the fraction equal to $$0 . \\overline{534}$$. Let x = $$0 . \\overline{534}$$. x = $$0 . \\overline{534}$$ 103 x = 103 ($$0 . \\overline{534}$$ ) 1000x = $$534 . \\overline{534}$$ 1000x = 534 + x 1000x – x = 534 + x – x 999x = 534 $$\\frac{999x}{999}$$ = $$\\frac{534}{999}$$ x = $$\\frac{534}{999}$$ x = $$\\frac{178}{333}$$ $$0 . \\overline{534}$$̅ = $$\\frac{178}{333}$$ Question 2. Find the fraction equal to $$3 . 0\\overline{15}$$.", "hand side a whole number. $$123=990y$$ Divide by 990 to isolate y. $$\\frac{123}{990}=y$$ Now, let's calculate what x/y is. $$\\frac{x}{y}=\\frac{560}{123}$$ Let's see if this is true. $$\\frac{\\frac{56}{99}}{\\frac{123}{990}}$$ Multiply by 990/123 to eliminate the complex fraction. $$\\frac{56}{99}*\\frac{990}{123}$$ Notice that 990 and 99 can be simplified before any multiplication takes place. $$\\frac{56}{1}*\\frac{10}{123}$$ Simplify from here. $$\\frac{560}{123}$$ Therefore, we have proven algabraically that $$\\frac{560}{123}=\\frac{0.\\overline{56}}{0.1\\overline{24}}$$ TheXSquaredFactor Nov 12, 2017", "Write the value of Question: Write the value of $0 . \\overline{423}$ in the form of a simple fraction. Solution: Let, x=0.423423423… …(i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=423.423423423… …(ii) Equation (ii)-(i), ⇒ 1000x-x=423.423423423-0.423423423=423 ⇒ 999x=423 $\\Rightarrow \\mathrm{X}=\\frac{423}{999}=\\frac{47}{111}$ $0 . \\overline{423}=\\frac{47}{111}$", "= 3 \\cdot \\frac{1}{3} = \\frac{3}{3} = 1$$ Therefore: $$0.\\overline{999} = 1$$ ##### Share on other sites Hrmmm.... Yeah.... Strange to me though since .999 (repeating) would never actually equal 1. Then again, if I look at it that way, repeating .333 would never actually equal 1/3.... ##### Share on other sites Dunno if it helps your understanding, but 0.99 is closer to 1 than 0.9 is, so for each 9 you add, you get closer to 1. If you have an infinite amount of 9's then you are getting infinitely close to 1. Or if you want it in math terms: $$.9 = \\frac{9}{10^1}$$ $$.09 = \\frac{9}{10^2}$$ $$.009 = \\frac{9}{10^3}$$ etc. $$.9 + .09 + .009 = \\frac{9}{10^1} + \\frac{9}{10^2} + \\frac{9}{10^3} = .999$$ So: $$\\sum_{n=1}^{\\infty} \\frac{9}{10^n} = 0.\\overline{999} = 1$$ You see that as $$n \\to \\infty$$ (analogous to \"you are adding more 9's on the end\") you are getting closer to 1. ##### Share on other sites Yeah, we actually discussed this in class one day when talking about geometric sequences a while back. And yes, as n -> inf., the number gets closer to 1, but it would never actually reach one. Wouldn't .999 (repeating) = 1 - 1/inf, not 1? (I do realize that obviously I'm arguing pointlessly since people accept .999 to", "$t$ will have traveled for $\\frac{s(t)}{186\\cdot10^3}$ seconds. As $s$ should be a continuous function of $t,$ we may determine the average time to be $\\overline t = \\frac1{b - a}\\int_a^b\\frac{s(t)}{186\\cdot10^3}\\,dt$ $=\\frac1{\\left(186\\cdot10^3\\right)(b - a)}\\int_a^b s(t)\\,dt$ $=\\frac1{186\\cdot10^3}\\left(\\frac{\\int_a^b s(t)\\,dt}{b - a}\\right)$ $=\\frac1{186\\cdot10^3}\\cdot\\overline s$ $=\\frac{\\overline s}{186\\cdot10^3}$ $=\\frac{93\\cdot10^6}{186\\cdot10^3} = \\frac{10^3}2 = 500$ Carrying the units through, the answer is in seconds, and this agrees with angel.white's solution. Some days I really love this site, and today is one of those days. It's such a pleasure to interact with other people who care, thank you.", "2.833… = $$2.8 \\overline{3}$$, $$\\frac{22}{7}$$ = 3.142 ii) From the above decimals $$\\frac{7}{5}$$, $$\\frac{3}{4}$$, $$\\frac{23}{10}$$ are terminating decimals. While $$\\frac{5}{3}$$,$$\\frac{17}{6}$$,$$\\frac{22}{7}$$ are non-terminating decimals iii) By writing the denominators of above decimals as a product of primes is iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals. Question 5. Convert the decimals $$0 . \\overline{9}$$, $$14 . \\overline{5}$$ and $$1.2 \\overline{4}$$ to rational form. Can you find any easy method other than formal method? (Page No. 31) Let x = $$0 . \\overline{9}$$ ⇒ x = 0.999 ……. (1) The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides. ⇒ 10 × x = 10 × 0.999 10x = 9.999 …….. (2) From (1) & (2) ∴ x = 1 or $$0 . \\overline{9}$$ = 1 Second Method: $$0 . \\overline{9}$$ = 0 + $$0 . \\overline{9}$$ = 0 + $$\\frac{9}{9}$$ = 0 + 1 = 1 Let x = $$14 . \\overline{5}$$ ⇒ x = 14.55 …….. (1) The periodicity of the equation (1) is 1. So it should be multiplied by 10 on both sides. ⇒ 10 × x = 10 × 14.55 10x = 145.55 …….. (2) Second Method: Let x = $$1.2 \\overline{4}$$ ⇒ x=", "a power of 10, so that the first block of the repeating digits is before the point: $100 \\cdot x \\cdot 1000 = 45123.123 \\overline{123}$ 4. Now substract: $100 \\cdot x \\cdot 1000 = 45123.123 \\overline{123}$ $100\\cdot x=45.123\\overline{123}$ You'll get: $99900\\cdot x= 45078$ $x=\\frac{45078}{99900}$ Afterwards you may simplify this fraction – if it is possible! $x=\\frac{45078}{99900}=\\frac{7513}{16650}$", "still insist that it can’t be zero. What’s interesting is that they’re generally comfortable with the equation $\\frac{1}{3} = .333\\ldots$ After all, a calculator tells us it’s so. But if we multiply both sides by 3, then don’t we get the original? It isn’t surprising that our equation seems a little dubious. What this really stems from is not understanding what infinite decimal expansions like .333… and .999… actually mean. Now, terminating decimal expansions are easy to understand. For instance, $.9 = \\frac{9}{10} \\;\\;\\;\\;\\;\\; .99 = \\frac{9}{10} + \\frac{9}{100} \\;\\;\\;\\;\\;\\; .999 = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000}$ But by what right can we write down something like $.999\\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\cdots$ where the sum just keeps going and going? How can we add infinitely many numbers? Does it even make sense to talk about something like that? To answer this, let’s consider the sequence of partial sums: $.9 = \\frac{9}{10} \\;\\;\\;\\;\\;\\; .99 = \\frac{9}{10} + \\frac{9}{100} \\;\\;\\;\\;\\;\\; .999 = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000}$ and so on. Notice that each term in the sequence is a bit larger than the preceding, and that all are less than 1: $.9 < .99 < .999 < .9999 < \\cdots < 1$ Modern mathematics defines the meaning of the infinite decimal expansion as follows: the value", "(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", "10 both side in equation (i) 10x = 0.47777……. × 10 or, 10x = 4.77777…… (ii) Again, multiply 100 both side in equation (i) 100x = 0.47777 × 100 or, 100x = 4.77777…….. (iii) Subtract equation (ii) from equation (iii) 90x = 43.0000 or 90x = 43 ∴ x = $$\\frac{43}{90}$$ Therefore, $$0.4 \\overline{7}$$ = x = $$\\frac{43}{90}$$ which is in the form of $$\\frac {p}{q}$$ (iii) Let x = $$0 . \\overline{001}$$ or, x = 0.001001001……. (i) Now, add 1 both side in equation (i) x + 1 = 0.001001001…… + 1 or x + 1 = 1.001001001……. (ii) Again, multiply 1000 both side in equation (i) 1000 × x = 1000 × 0.001001001…….. or, 1000x = 1.001001001….. (iii) From equation (ii) and (iii) we get x +1 = 1000x or 999x = 1 or x = $$\\frac{1}{999}$$ Therefore, $$0 . \\overline{001}=x \\times \\frac{1}{999}$$ which is in the form of $$\\frac {p}{q}$$ Question 4. Express 0.99999…… in the form $$\\frac {p}{q}$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Solution: Let x = 0.9999…… (i) Add 9 both side in equation (i) x + 9 = 9 + 0.99999……… or, x + 9 = 9.99999…… (ii) Again, multiply 10 both side in equation (i) 10x = 9.9999……. (iii)", "# We can tell that 0.11111111...*9=0.999999999... And that 1/9=0.11111111111... Therefore 1/9*9=0.999999999...However, we know that 1/9⋅9=9/9=1 Note: I'm taking in account that ... are the other rational digits left. What am I making wrong? What is misunderstood? Thanks for the help in clearing this problem. Is $9\\ast \\left(1/9\\right)=0.99999999...$ statement correct? We can tell that $0.11111111...\\cdot 9=0.999999999...$ And that $\\frac{1}{9}=0.11111111111...$ Therefore $\\frac{1}{9}\\cdot 9=0.999999999...$ However, we know that $\\frac{1}{9}\\cdot 9=\\frac{9}{9}=1$ Note: I'm taking in account that ... are the other rational digits left. What am I making wrong? What is misunderstood? Thanks for the help in clearing this problem. 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 nuramaaji2000fh Nothing. What you wrote is correct, and you just proved that $0.\\overline{9}=1$ You can do the same thing with $\\frac{1}{3}$ and $3$ for example: $1=\\frac{3}{3}=3\\cdot \\frac{1}{3}=3\\cdot \\left(0.333333\\dots \\right)=0.999999\\dots =0.\\overline{9}$ Remark That holds only for infinite periodic decimals. You cannot, for example, state that $0.999999999999999999999999999999999=1$ That is not true! Cristofer Graves A more typical way to prove is to subtract $0.1\\ast 0.\\overline{9}$ from $0.\\overline{9}$ $0.\\overline{9}-0.0\\overline{9}=0.9$ But $X-0.1\\ast X=0.9\\ast X=0.9$ proves $X=1$", "+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}$$ ! Jan 12, 2022", "the answer is at least $6 + 8 + 4 = 18$. We now only need to know whether $a + b + c = 20$ might work for any larger $n$. We will always get equations like $100001c - b = 11111a^2$ where the $c$ coefficient is very close to being nine times the $a$ coefficient. Since the $b$ term will be quite insignificant, we know that once again $a^2$ must equal $9c$, and thus $a = 9, c = 9$ is our only hope to reach $20$. Substituting and dividing through by $9$, we will have something like $100001 - \\frac{b}{9} = 99999$. No matter what $n$ really was, $b$ is out of range (and certainly isn't $2$ as we would have needed). The answer then is $\\boxed{\\textbf{(D)} \\text{ 18}}$. ## Solution 3 The given equation can be written as: $$c \\cdot ( \\overbrace{1111 \\ldots 1111}^\\text{2n}) - b \\cdot ( \\overbrace{11 \\ldots 11}^\\text{n} ) = a^2 \\cdot ( \\overbrace{11 \\ldots 11}^\\text{n} )^2$$ Divide by $\\overbrace{11 \\ldots 11}^\\text{n}$ on both sides: $$c \\cdot ( \\overbrace{1000 \\ldots 0001}^\\text{n+1}) - b = a^2 \\cdot ( \\overbrace{11 \\ldots 11}^\\text{n} )$$ Next, split the first term to make it easier to deal with. $$2c + c \\cdot (\\overbrace{99 \\ldots 99}^\\text{n}) - b = a^2 \\cdot ( \\overbrace{11 \\ldots 11}^\\text{n} )$$ $$2c", "1. ## quotient of functions I forgot what these are called but when it comes to these, i get wayyy lost and just dont know what to do. $(1 - x)/(1 + x)=100$ Thanks for the help! 2. Originally Posted by hydride I forgot what these are called but when it comes to these, i get wayyy lost and just dont know what to do. $(1 - x)/(1 + x)=100$ Thanks for the help! These are called equations. Don't be lost my friend, just multiply both sides by $(1+x)$. Then distibute the $100$, and isolate $x.$ 3. Originally Posted by VonNemo19 These are called equations. Don't be lost my friend, just multiply both sides by $(1+x)$. Then distibute the $100$, and isolate $x.$ Tell me the answer is $x = -99/100$ 4. Originally Posted by hydride Tell me the answer is $x = -99/100$ Yes. 5. $ \\frac{1 - x}{1 + x}=100 $ $ 1-x=100(1+x) $ $ 1-x=100+100x $ $ 1-100=100x+x $ $ -99=101x $ $ \\frac{-99}{101}=x $ 6. EDIT: EDIT: alriiiight. Thanks a whole bunch. 7. Originally Posted by hydride EDIT: EDIT: alriiiight. Thanks a whole bunch. $ \\frac{-99}{101}=x $ 8. Originally Posted by millerst $", "# How do you understand Infinitesimals? I've often heard that $\\dfrac{1}{3} = 0.\\overline{3}$, and $\\dfrac{1}{3} \\cdot 3 = 1$, yet $0.\\overline{3}\\cdot 3 = 0.\\overline{9} \\neq 1$. There is an $\\epsilon$ (infinitesimal) thrown in there as well. How do you understand these extremely small values and what do I need to do to account for them when calculating very precise values with them? I know that they are too small to make a difference when dealing with smaller numbers, but when does it start to impact your results? • No; $0. \\overline 9$ and $1$ represent the same number: one. – Mauro ALLEGRANZA Aug 3 '17 at 14:52 • The standard approach to modern analysis does not use infinitesimals. The usual definition of infinite decimals implies that $0.\\overline{9} = 1$ exactly. – Stahl Aug 3 '17 at 14:55 • $0.\\overline{9}=1$. Given two distinct real numbers, there is always a number strictly between them. How would you write the real number between $0.\\overline{9}$ and $1$? – Thomas Andrews Aug 3 '17 at 14:55 • Note that $x=0.\\overline{9}\\implies 10x=9.\\overline{9}\\implies 9x=10x-x=9.\\overline{9}-0.\\overline{9}=9\\implies x=1.$ – mfl Aug 3 '17 at 14:55 • See Infinitesimals for some details. – Mauro ALLEGRANZA Aug 3 '17 at 14:56 Actually, $$0.\\overline{9} = 0.\\overline{3}\\cdot3 = \\frac{3}{3} = 1$$ For a more rigorous way, note that $$0.\\overline{9} = \\frac{9}{10} +" ]

[ "to get $$1000 x = 334.\\overline{4}$$ Now, try to subtract the above equations. What do you get? Second option: You can write $x=0.33\\overline 4$ as $$0.3 + 0.03 + 0.004 + 0.0004 + \\dots$$ which we can rewrite as $$x=0.33 + 0.004\\cdot (1 + 0.1 + 0.01 + \\dots)$$ or further as $$x=\\frac{33}{100} + \\frac{4}{1000}\\cdot \\sum_{i=0}^\\infty \\left(\\frac{1}{10}\\right)^{i}$$ Now, remember how you were taught what the sum of a geometric series is? HINT: $$0.3344444444\\cdots=\\dfrac{33}{100}+4\\sum_{r=3}^\\infty\\dfrac1{10^r}$$ Let $x = 0.33444 \\ldots$, then $100x = 33.44444 \\ldots$ so that $100x - x = 99x = 33.11$ and $x=\\frac{3311}{9900}$. There is a trick, which goes like: find the repeating pattern and divide it with as many nines. For example $$\\frac{1090}{9999}=0.10901090....$$ In this case we have $$0.33444...=0.33+0.0044...=\\frac{33}{100}+\\frac{1}{100}0.44...=\\frac{33}{100}+\\frac{1}{100}\\frac{4}{9}=\\frac{297+4}{900}=\\frac{301}{900}.$$ Multiply by $\\dfrac94$ to turn the $4$s into $9$s, giving $$0.7524999999\\cdots$$ which equals $$0.7525\\,.$$ $$\\frac{7525}{10000}\\frac{4}{9}=\\frac{301}{900}.$$ You can move the decimal point one position to the left by multiplying by $10$. $$0.3344444444\\cdots\\times10=3.344444444\\cdots$$ Then subtracting these two numbers, a finite number of decimals remain and $$0.3344444444\\cdots\\times9=3.01\\,.$$ So the fraction is $$\\frac{3.01}{9}=\\frac{301}{900}.$$ General method: • Let $x$ denote the input number • Let $|n|$ denote the number of decimal digits in $n$ • Split $x$ into the following parts: • $\\color\\red{A}=$ the integer part, i.e., $\\lfloor{x}\\rfloor$ • $\\color\\green{B}=$ the fraction part's non-periodic prefix • $\\color\\orange{C}=$ the fraction part's periodic postfix", "hand side a whole number. $$123=990y$$ Divide by 990 to isolate y. $$\\frac{123}{990}=y$$ Now, let's calculate what x/y is. $$\\frac{x}{y}=\\frac{560}{123}$$ Let's see if this is true. $$\\frac{\\frac{56}{99}}{\\frac{123}{990}}$$ Multiply by 990/123 to eliminate the complex fraction. $$\\frac{56}{99}*\\frac{990}{123}$$ Notice that 990 and 99 can be simplified before any multiplication takes place. $$\\frac{56}{1}*\\frac{10}{123}$$ Simplify from here. $$\\frac{560}{123}$$ Therefore, we have proven algabraically that $$\\frac{560}{123}=\\frac{0.\\overline{56}}{0.1\\overline{24}}$$ TheXSquaredFactor Nov 12, 2017", "Write the value of Question: Write the value of $0 . \\overline{423}$ in the form of a simple fraction. Solution: Let, x=0.423423423… …(i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=423.423423423… …(ii) Equation (ii)-(i), ⇒ 1000x-x=423.423423423-0.423423423=423 ⇒ 999x=423 $\\Rightarrow \\mathrm{X}=\\frac{423}{999}=\\frac{47}{111}$ $0 . \\overline{423}=\\frac{47}{111}$", "cheaper products <=> x (1-0.1) = cheaper products x (0.9) = cheaper products x = cheaper products / 0.9 (e .g. cheaper product = 90 ->x = 90/0.9 = 100) I hope this helps, Rapha 3. ## add 11.1111111% add 11.1111111111% 4. Originally Posted by [email protected] add 11.1111111111% This is certainly a practical if all the OP wants to do is get the price back without worrying about why this works. But it might be helpful to add some backstory to it: Let the original price be x. Discount price: x - (10% of x) = x - 0.1 x = 0.9 x. You want to add a percentage of the discount price to get the original price back. Consider: 0.9 x + (a % of (0.9)x) = x $\\Rightarrow 0.9 x + \\frac{a}{100} \\cdot (0.9 x) = x$ $\\Rightarrow 0.9 + \\frac{0.9 a}{100} = 1$ $\\Rightarrow \\frac{0.9 a}{100} = 0.1$ $\\Rightarrow a = \\frac{10}{0.9} = \\frac{100}{9} = 11 \\frac{1}{9}$ So add $11 \\frac{1}{9}$% 5. ## Ours is not to reason why Thanks mate. Your right that it is very important to understand why we do what we do in maths. It is logical and not simply a set of rules. Thanks Rick 6. If 90 + x% of 90 = 100 x/100 * 90 =10 so, x", "# 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 $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "digits of 9’s as the number of digits that repeat. For example, $$0 . \\overline{81}$$has two repeating decimal digits, and the denominator has two 9’s. The pattern is even true for $$0 . \\overline{9}$$. According to the pattern, this decimal equals $$\\frac{9}{9}$$, which is 1. Eureka Math Grade 8 Module 7 Lesson 10 Exit Ticket Answer Key Question 1. Find the fraction equal to $$0 . \\overline{534}$$. Let x = $$0 . \\overline{534}$$. x = $$0 . \\overline{534}$$ 103 x = 103 ($$0 . \\overline{534}$$ ) 1000x = $$534 . \\overline{534}$$ 1000x = 534 + x 1000x – x = 534 + x – x 999x = 534 $$\\frac{999x}{999}$$ = $$\\frac{534}{999}$$ x = $$\\frac{534}{999}$$ x = $$\\frac{178}{333}$$ $$0 . \\overline{534}$$̅ = $$\\frac{178}{333}$$ Question 2. Find the fraction equal to $$3 . 0\\overline{15}$$.", "2.833… = $$2.8 \\overline{3}$$, $$\\frac{22}{7}$$ = 3.142 ii) From the above decimals $$\\frac{7}{5}$$, $$\\frac{3}{4}$$, $$\\frac{23}{10}$$ are terminating decimals. While $$\\frac{5}{3}$$,$$\\frac{17}{6}$$,$$\\frac{22}{7}$$ are non-terminating decimals iii) By writing the denominators of above decimals as a product of primes is iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals. Question 5. Convert the decimals $$0 . \\overline{9}$$, $$14 . \\overline{5}$$ and $$1.2 \\overline{4}$$ to rational form. Can you find any easy method other than formal method? (Page No. 31) Let x = $$0 . \\overline{9}$$ ⇒ x = 0.999 ……. (1) The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides. ⇒ 10 × x = 10 × 0.999 10x = 9.999 …….. (2) From (1) & (2) ∴ x = 1 or $$0 . \\overline{9}$$ = 1 Second Method: $$0 . \\overline{9}$$ = 0 + $$0 . \\overline{9}$$ = 0 + $$\\frac{9}{9}$$ = 0 + 1 = 1 Let x = $$14 . \\overline{5}$$ ⇒ x = 14.55 …….. (1) The periodicity of the equation (1) is 1. So it should be multiplied by 10 on both sides. ⇒ 10 × x = 10 × 14.55 10x = 145.55 …….. (2) Second Method: Let x = $$1.2 \\overline{4}$$ ⇒ x=", "# 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 $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "above must be $(a, a, a)$, and $\\bullet$ if none of $\\{\\frac{998}n, \\frac{999}n, \\frac{1000}n\\}$ are integral, then the three terms in the expression above must be $(a, a, a)$. The last statement is true because in order for the terms to be different, there must be some integer in the interval $(\\frac{998}n, \\frac{999}n)$ or the interval $(\\frac{999}n, \\frac{1000}n)$. However, this means that multiplying the integer by $n$ should produce a new integer between $998$ and $999$ or $999$ and $1000$, exclusive, but because no such integers exist, the original integer cannot exist, and thus, the terms must be equal. Note that $n = 1$ does not work; to prove this, we just have to substitute $1$ for $n$ in the expression. This gives us $\\left\\lfloor \\dfrac{998}{1} \\right\\rfloor+\\left\\lfloor \\dfrac{999}{1} \\right\\rfloor+\\left\\lfloor \\dfrac{1000}{1}\\right \\rfloor = 998 + 999 + 1000 = 2997 = 999 \\cdot 3$ which is divisible by 3. Now, we test the four cases listed above. Case 1: $n$ divides $998$ As mentioned above, the three terms in the expression are $(a, a, a)$, so the sum is $3a$, which is divisible by $3$. Therefore, the first case does not work. Case 2: $n$ divides $999$ Because $n$ divides $999$, the number of possibilities for $n$ is the same as the number of factors of $999$. $999$ =", "≠ 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", "10 both side in equation (i) 10x = 0.47777……. × 10 or, 10x = 4.77777…… (ii) Again, multiply 100 both side in equation (i) 100x = 0.47777 × 100 or, 100x = 4.77777…….. (iii) Subtract equation (ii) from equation (iii) 90x = 43.0000 or 90x = 43 ∴ x = $$\\frac{43}{90}$$ Therefore, $$0.4 \\overline{7}$$ = x = $$\\frac{43}{90}$$ which is in the form of $$\\frac {p}{q}$$ (iii) Let x = $$0 . \\overline{001}$$ or, x = 0.001001001……. (i) Now, add 1 both side in equation (i) x + 1 = 0.001001001…… + 1 or x + 1 = 1.001001001……. (ii) Again, multiply 1000 both side in equation (i) 1000 × x = 1000 × 0.001001001…….. or, 1000x = 1.001001001….. (iii) From equation (ii) and (iii) we get x +1 = 1000x or 999x = 1 or x = $$\\frac{1}{999}$$ Therefore, $$0 . \\overline{001}=x \\times \\frac{1}{999}$$ which is in the form of $$\\frac {p}{q}$$ 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. Solution: Let x = 0.9999…… (i) Add 9 both side in equation (i) x + 9 = 9 + 0.99999……… or, x + 9 = 9.99999…… (ii) Again, multiply 10 both side in equation (i) 10x = 9.9999……. (iii)", "# We can tell that 0.11111111...*9=0.999999999... And that 1/9=0.11111111111... Therefore 1/9*9=0.999999999...However, we know that 1/9⋅9=9/9=1 Note: I'm taking in account that ... are the other rational digits left. What am I making wrong? What is misunderstood? Thanks for the help in clearing this problem. Is $9\\ast \\left(1/9\\right)=0.99999999...$ statement correct? We can tell that $0.11111111...\\cdot 9=0.999999999...$ And that $\\frac{1}{9}=0.11111111111...$ Therefore $\\frac{1}{9}\\cdot 9=0.999999999...$ However, we know that $\\frac{1}{9}\\cdot 9=\\frac{9}{9}=1$ Note: I'm taking in account that ... are the other rational digits left. What am I making wrong? What is misunderstood? Thanks for the help in clearing this problem. 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 nuramaaji2000fh Nothing. What you wrote is correct, and you just proved that $0.\\overline{9}=1$ You can do the same thing with $\\frac{1}{3}$ and $3$ for example: $1=\\frac{3}{3}=3\\cdot \\frac{1}{3}=3\\cdot \\left(0.333333\\dots \\right)=0.999999\\dots =0.\\overline{9}$ Remark That holds only for infinite periodic decimals. You cannot, for example, state that $0.999999999999999999999999999999999=1$ That is not true! Cristofer Graves A more typical way to prove is to subtract $0.1\\ast 0.\\overline{9}$ from $0.\\overline{9}$ $0.\\overline{9}-0.0\\overline{9}=0.9$ But $X-0.1\\ast X=0.9\\ast X=0.9$ proves $X=1$", "$$\\frac{41}{333}$$ using a calculator is 0.123123…. Exercise 2. There is a fraction with a decimal expansion of 0.4 ̅. Find the fraction, and check your answer using a calculator. Let x = $$0. \\overline{4}$$ x = $$0 . \\overline{4}$$ 10x = (10)$$0 . \\overline{4}$$ 10x = $$4 . \\overline{4}$$ 10x = 4 + x 10x – x = 4 + x – x 9x = 4 $$\\frac{9x}{9}$$ = $$\\frac{4}{9}$$ x = $$\\frac{4}{9}$$ Exercises 3–4 Exercise 3. Find the fraction equal to $$1.6 \\overline{23}$$. Check your answer using a calculator. Exercise 4. Find the fraction equal to $$2.9 \\overline{60}$$. Check your answer using a calculator. Eureka Math Grade 8 Module 7 Lesson 10 Problem Set Answer Key Question 1. a. Let x = $$0 . \\overline{631}$$. Explain why multiplying both sides of this equation by 103 will help us determine the fractional representation of x. When we multiply both sides of the equation by 103, on the right side we will have 631.631631…. This is helpful because we will now see this as 631 + x. b. What fraction is x? 1000x = $$631 . \\overline{631}$$ 1000x = 631 + 0.631631… 1000x = 631 + x 1000x – x = 631 + x – x 999x = 631 $$\\frac{999x}{999}$$ = $$\\frac{631}{999}$$ x = $$\\frac{631}{999}$$ Yes, my answer is reasonable", "$$4/5$$ 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", "AundreaLafaver57 13 # Compute the sum $\\frac{1}{\\sqrt{100} + \\sqrt{102}} + \\frac{1}{\\sqrt{102} + \\sqrt{104}} + \\frac{1}{\\sqrt{104}+\\sqrt{106}} + \\cdots + \\frac{1}{\\sqrt{9998} + \\sqrt{10000}}.$ $\\frac{1}{ \\sqrt{100}+ \\sqrt{102} } + \\frac{1}{ \\sqrt{102} + \\sqrt{104} }+...+ \\frac{1}{ \\sqrt{9998}+ \\sqrt{10000} }$ We will multiply all fractions to make a difference of squares in the denominators. So this sum will become: $\\frac{ \\sqrt{102} - \\sqrt{100} }{2}+ \\frac{ \\sqrt{104}- \\sqrt{102} }{2}+...+ \\frac{ \\sqrt{10000}- \\sqrt{9998} }{2}$ = - √100 / 2 + √10000 / 2 = - 10 / 2 + 100 / 2 = = - 5 + 50 = 45", "(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", "still insist that it can’t be zero. What’s interesting is that they’re generally comfortable with the equation $\\frac{1}{3} = .333\\ldots$ After all, a calculator tells us it’s so. But if we multiply both sides by 3, then don’t we get the original? It isn’t surprising that our equation seems a little dubious. What this really stems from is not understanding what infinite decimal expansions like .333… and .999… actually mean. Now, terminating decimal expansions are easy to understand. For instance, $.9 = \\frac{9}{10} \\;\\;\\;\\;\\;\\; .99 = \\frac{9}{10} + \\frac{9}{100} \\;\\;\\;\\;\\;\\; .999 = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000}$ But by what right can we write down something like $.999\\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\cdots$ where the sum just keeps going and going? How can we add infinitely many numbers? Does it even make sense to talk about something like that? To answer this, let’s consider the sequence of partial sums: $.9 = \\frac{9}{10} \\;\\;\\;\\;\\;\\; .99 = \\frac{9}{10} + \\frac{9}{100} \\;\\;\\;\\;\\;\\; .999 = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000}$ and so on. Notice that each term in the sequence is a bit larger than the preceding, and that all are less than 1: $.9 < .99 < .999 < .9999 < \\cdots < 1$ Modern mathematics defines the meaning of the infinite decimal expansion as follows: the value", "# Sum of series with triangular numbers Can you please tell me the sum of the seires $\\frac {1}{10} + \\frac {3}{100} + \\frac {6}{1000} + \\frac {10}{10000} + \\frac {15}{100000} + \\cdots$ where the numerator is the series of triangular numbers? Is there a simple way to find the sum? Thank you. - Fixed $\\LaTeX$ and problem statement. – Ahaan S. Rungta Jul 16 at 17:39 @AhaanRungta you do not need to comment that here. – picakhu Jul 16 at 21:44 $$S={1\\over10}+{3\\over100}+{6\\over1000}+{10\\over10000}+\\cdots$$ $${S\\over10}={1\\over100}+{3\\over1000}+{6\\over10000}+\\cdots$$ Subtracting, $${9S\\over10}={1\\over10}+{2\\over100}+{3\\over1000}+{4\\over10000}+\\cdots$$ Now do the same thing again, that is, divide by $10$ and subtract, to get $${81S\\over100}={1\\over10}+{1\\over100}+{1\\over1000}+\\cdots={1\\over9}$$ - Thanks for the fast response. Your manipulation is quite nice and makes the question too easy to solve. – Mustafa Saad Jul 16 at 13:46 Indeed, WolframAlpha agrees – Ryan Jul 16 at 15:50 Your expression is equal to $g(1/10)$, where $$g(x)=\\frac{x}{2}\\left((2)(1)+(3)(2)x+(4)(3)x^2+(5)(4)x^3+\\cdots\\right)$$ Take the power series $1+x+x^2+x^3+\\cdots$ for $\\frac{1}{1-x}$ and differentiate twice. We get $(2)(1)+(3)(2)x+(4)(3)x^2+\\cdots$ if we do it term by term, and $\\frac{2!}{(1-x)^3}$ if we do it the usual way. Thus $$g(x)=\\frac{x}{2}\\cdot\\frac{2!}{(1-x)^3}$$ (when $|x|\\lt 1$, and in particular at $x=1/10$). Remark: The idea generalizes. The $n$-th triangular nunber is $\\binom{n}{2}$. The same idea can be used to calculate $\\sum \\binom{n}{k}x^n$ for $|x|\\lt 1$ and fixed positive integer $k$. - Good answer, but one mistake.", "can be written as a fraction $\\frac{1050}{1}$ and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial. So now we get... $x=\\frac{\\frac{1050}{1}}{\\frac{3}{4}}=\\underbrace{\\ frac{1050}{1}\\cdot \\frac{4}{3}}_{1050/3=350}=350 \\cdot 4=1400$ I hope this clears it up 6. Yes, thank you... 7. ## hi Just divide 1260 with 0,9. 1260/0.9 = 1400 8. Originally Posted by Twig Just divide 1260 with 0,9. 1260/0.9 = 1400 Fractions > Decimals Originally Posted by TheEmptySet $\\frac{3}{4}x=1050$ so we divide both sides by 3/4 $\\frac{\\frac{3}{4}x}{\\frac{3}{4}}=\\frac{1050}{\\frac {3}{4}}$ remember that 1050 can be written as a fraction $\\frac{1050}{1}$ and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial. So now we get... $x=\\frac{\\frac{1050}{1}}{\\frac{3}{4}}=\\underbrace{\\ frac{1050}{1}\\cdot \\frac{4}{3}}_{1050/3=350}=350 \\cdot 4=1400$ I hope this clears it up Nice Latex", "+0 Compute the product of 0.123(overline) and 9, and write your result as a fraction in simplified form. +1 500 1 +5 Compute the product of 0.123(overline) and 9, and write your result as a fraction in simplified form. (0.123 is a repeating decimal) Feb 7, 2019 #1 +80 +2 $$0.\\overline{123} = \\frac{41}{333} \\\\ \\text{So, } \\frac{41}{333} \\times 9= \\boxed{\\frac{41}{37}}$$ . Feb 7, 2019 $$0.\\overline{123} = \\frac{41}{333} \\\\ \\text{So, } \\frac{41}{333} \\times 9= \\boxed{\\frac{41}{37}}$$" ]

[ "So , Total $720$ words are possible .", "Free Version Difficult # Ordering Letters in a Word . . . CREMATORIUM? ALGEBR-0PF3NB How many ways can you order the letters in the word CREMATORIUM? A $4,989,600$ B $39,916,800$ C $19,958,400$ D $9,979,200$", "words and languages. We make words out of an alphabet. An alphabet is just a set of symbols that we can use. So we can have our alphabet be $\\{a,b,c\\}$ or $\\{0,1\\}$ or even $\\{bla, blar, blargaagh\\}$. A word is just any string made up of symbols from your alphabet. So $abcbaca$ is a word from our first alphabet, $010101111$ is a word from our second, and $blablablarblablargaaghbla$ is a word from our third. Languages are just subsets of the words that we can make out of an alphabet. So if we have our alphabet $\\Sigma=\\{0,1\\}$, maybe we want our language to be the set of all words that have an even number of $1$s. Or maybe we want a language where we have an equal number of $0$s and $1$s or where we always have twice as many $0$s as $1$s. Or maybe we just want our language to be $\\{0,1,100,0001011\\}$. So now that we have these languages, we want to know which words are in our language. That’s pretty easy for something like $\\{0,1,100,0001011\\}$, since we can just check it against every word in our language. But what about something more complicated, like requiring an even number of $1$s? Here’s where we come up with theoretical machines that do this. These theoretical machines are essentially the", "# Hawaii • May 26th 2009, 03:51 PM Sampras Hawaii The Hawaiian alphabet consists of $12$ letters, the vowels $\\text{a}, \\ \\text{e}, \\ \\text{i}, \\ \\text{o}, \\ \\text{u}$ and the consonants $\\text{h}, \\ \\text{k}, \\ \\text{l}, \\ \\text{m}, \\ \\text{n}, \\ \\text{p}$ and $\\text{w}$. (a) How many different four-letter words can be constructed using the $12$-letter alphabet? (b) How about with the English alphabet? (c) What is the second and last letters are vowels and the other $2$ are consonants? (d) What if the second and last letters are vowels, but there are no restrictions on the other $2$ letters? So for (a) it would be $12^4$. For (b) it would be $26^4$. For (c) it would be $5^27^2$. And for (d) it would be $5^212^2$? • May 26th 2009, 05:04 PM amitface Correct.", "$9$ $3$-letter words that begin with $a$. Similarly, there are $9$ $3$-letter words that begin with $b$, and $9$ that begin with $c$. If you imagine putting the $3$-letter word that begin with $a$ on one page, the ones that begin with $b$ on the next page, and the ones that begin with $c$ on the next page, there will be $3$ pages, with $9$ words on each page, for a total of $3\\times 9$, that is, $27$. How many $4$-letter words are there? First, how many begin with $a$? You get such a word by appending any $3$-letter word to $a$, so there are $27$ $3$-letter words that begin with $a$. There are also $27$ that begin with $b$, and $27$ that begin with $c$, for a total of $81$. Now can you see instantly how many $5$-letter words there are using the alphabet $\\{a,b,c\\}$? In general, by the same reasoning, there will be $3^n$ $n$-letter words using this alphabet. OK, we have sort of dealt with what's happening with $3$ letters. What about if we have a $4$-letter alphabet $\\{a,b,c,d\\}$? Clearly there are $4$ $1$-letter words. How many $2$-letter words? How many $2$-letter words begin with $a$? You write down $a$, and append any $1$-letter word, so there are $4$. There are also $4$ $2$-letter words", "$32$ characters (inclusive), and uses only characters a–z. • The number of meanings of a word in the dictionary is at least $1$ and at most $10\\, 000$. • Dictionary words are distinct. Sample Input 1 Sample Output 1 5 heimark hei 2 mark 2 heim 1 ark 2 heima 1 6", "225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that it would be nearly impossible to come up with concepts for all of them (for words of 5 syllables, the number is 1,078,203,909,375, over", "## Kickstarter ad absurdum For the culture that has everything: Emoji Dick is a crowd sourced and crowd funded translation of Herman Melville's Moby Dick into Japanese emoticons called emoji. Each of the book's approximately 10,000 sentences has been translated three times by a Amazon Mechanical Turk worker. These results have been voted upon by another set of workers, and the most popular version of each sentence has been selected for inclusion in this book. In total, over eight hundred people spent approximately 3,795,980 seconds working to create this book. Each worker was paid five cents per translation and two cents per vote per translation. The funds to pay the Amazon Turk workers and print the initial run of this book were raised from eighty-three people over the course of thirty days using the funding platform Kickstarter. By my calculation, that's something like 10000*3*.07 = $2,100, which is not much for a translation of Moby Dick, but a lot for a joke that probably gets tiresome after a couple of pages. From a reviewer who gives the book zero stars: This book has a \"Preview\" link that only shows 12 pages… which is enough to show the title pages and credits… and ABSOLUTELY NOTHING OF THE EMOJI itself. How one can judge whether to buy a$200 hardcover from", "of a $2$-letter word can be chosen in $26$ ways. For each of these ways, the second letter can be chosen in $26$ ways, for a total of $26^2$. (If $26$ women each have $26$ children, then the total number of children is $26^2$.) The same idea shows that there are $26^3$ $3$ letter words, $26^4$ $4$-letter words, and so on. However, we don't even want to count all the $3$-letter words, only those up to $bbb$. How many $3$-letter words start with $a$? We can complete $a$ to a $3$-letter word by appending to $a$ any $2$-letter word. There are $26^2$ such $2$-letter words. So $26^2$ $2$-letter words start with $a$. Finally, we want to count the $3$-letter words that start with $b$, but we only want to get to $bbb$. So append any $2$-letter word that starts with $a$ (there are $26$ of these), and finally at the very end we must not forget about $bba$ and $bbb$ (2 words). Add up. We get $1406$. But probably you did not intend to count the first word, just the \"iterations\" or \"increments.\" There are $1405$ of these. B) Using Base 26 Representation: The $1$-letter words were easy to count, as were the $2$-letter words. The $3$-letter words were a bit of a nuisance, since we did not", "of the 225 for the second. Thus to find the total number of two-syllable words we just multiply those two numbers (or equivalently raise 225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that", "usage; for instance, the syllable wū does not exist in Hawaiian, and the syllable wu occurs only in two loan-words from English. However, for the purposes of this post I am calculating only how many words could theoretically exist, not how many actually do (which would require exhaustive knowledge of the language that I do not possess). So for single syllables, we have a total of $$10+15+80+120=225$$. Now, since any word in Hawaiian will be made up of these syllables, we can quickly calculate how many distinct words of a given number of syllables would be able to exist. Since Hawaiian has nothing against duplication of sounds (and often rather encourages it), for a word of two syllables we may have any of the 225 for the first syllable, and any of the 225 for the second. Thus to find the total number of two-syllable words we just multiply those two numbers (or equivalently raise 225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our", "possible characters. So there are $36^{12}$ possible combinations.", "\\$452.56. Use the puzzle to preview key vocabulary from this unit. Unscramble the circled letters within found words to answer the riddle at the bottom of the page. 1. The things you own that have a positive cash value. (Lesson 14.3) 2. The amount of money you earn. (Lesson 14.1)", "about $$12$$ months old, we expect her to say her first word for meaning, and to start combining words for meaning at about $$18$$ months. At about $$2$$ years old, a toddler uses between $$50$$ and $$200$$ words; by $$3$$ years old they have a vocabulary of up to $$1,000$$ words and can speak in sentences. During the early childhood years, children's vocabulary increases at a rapid pace. This is sometimes referred to as the “vocabulary spurt” and has been claimed to involve an expansion in vocabulary at a rate of $$10-20$$ new words per week. Recent research may indicate that while some children experience these spurts, it is far from universal (as discussed in Ganger & Brent, 2004). It has been estimated that, $$5$$ year olds understand about $$6,000$$ words, speak $$2,000$$ words, and can define words and question their meanings. They can rhyme and name the days of the week. Seven year olds speak fluently and use slang and clichés (Stork & Widdowson, 1974). What accounts for such dramatic language learning by children? Behaviorist B. F. Skinner thought that we learn language in response to reinforcement or feedback, such as through parental approval or through being understood. For example, when a two-year-old child asks for juice, he might say, “me juice,” to which his mother might", "the school board until they buy one of these for \"learning purposes\" And all for the amazingly low price of only \\$500!!! Probably less than an ELMO =] Outstanding point. eBay here I come! × Pasted as rich text. Paste as plain text instead Only 75 emoji are allowed.", "from the given alternatives Letter : word : : ? Page : Book Product : factory Club : people Home work : School 123 : $13^ 2$ : : 235 : ? $23^2$ $35^2$ $25^3$ $25^2$", "+0 # i got 159981...i probably did something wrong. 0 129 1 The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? Guest May 28, 2018 #1 +20681 +1 The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? $$\\begin{array}{|lr|} \\hline \\text{words with 1 letter : } &1 \\\\ \\text{words with 2 letters: } &39 \\\\ \\text{words with 3 letters: } &1141 \\\\ \\text{words with 4 letters: } &29679\\\\ \\hline \\text{sum: } & 30860 \\\\ \\hline \\end{array}$$ heureka May 28, 2018", "words without repeating a letter (i.e. I don't understand this last part. Of the 150 million Americans working today, approximately 21 million earn between$7.25 an hour (the prevailing federal minimum wage) and \\$10.10 an hour. raw download clone embed print report. If there were no repeating letters, the answer would simply be 11! I was not able to find any single word (more than 8 letters long) that had exact same four starting and ending letters.. Is there such a word? check_circle Expert Answer. Words with seven letters can help you score big and win word scramble games, such as Scrabble, Word With Friends, Text Twist Solver, and more. Are there sets of words with no repeated letters and having.", "typeface) of the Latin, French and even English phrases, just as it would be interesting to transliterate Cyrillic and Greek in some places. Second is the division of the index into two parts. Most American readers, at least, are accustomed to a single inclusive index. The index is a weak part of most computer books. This separation does not help. Trivia from Fonts & Encodings: 1. Unicode currently has 90,989 \"letters.\" There are also numerals, punctuation, symbols, math symbols, and many more to take it to the \"more than 100,000\" cited above. 2. It uses 21 bits, though there are many language-specific foldings to reduce character representations to a single byte. 3. In principle, Unicode can never eliminate a character. All ancient written languages' alphabets or ideographs should be included. So far as I could see, Egyptian hieroglyphics, Mayan writing and Greek Linear A are not yet present. Haralambous does not mention Mayan writing. He does mention that hieroglyphics are under consideration for addition to Unicode and Greek Linear A will, presumably, be considered for addition (once it is deciphered). -- Editor's note: As an aside, it looks like O'Reilly book covers can't keep up with the falling US dollar. Glancing at the \"Fonts\" book, for example, we see the list price as: •$71.99 CAN • $59.99 US", "list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that it would be nearly impossible to come up with concepts for all of them (for words of 5 syllables, the number is 1,078,203,909,375, over a trillion words!). Conclusion: Although the sound range of Hawaiian may sound limited to the Anglophone ear with our 44-odd sounds, combinatorics shows that Hawaiian is capable of forming astronomically many times more words than even the most wordy languages on Earth. I'd do a similar calculation for English, but for the fact that it would be several orders of magnitude harder. This" ]

[ "be one of these $5$ cases, and there is no overlap in the $5$ cases. Thus we are going to sum up the number of total possibilities for each case. Case 1: If we pick $5$ distinct letters, we pick one of each letter, so there is only $1$ way to do this. Case 2a: First lets count the ways to pick the $2$ letters which are the same. It can't be E, so there are only $4$ ways to pick them. Second, we must pick $3$ more from the remaining $4$ letters (none of these can be the same letter as the first $2$ since that would be a different case). Thus we use $4$ choose $3$, which turns out to be $4$ options. We multiply these together to see that we have $4\\times 4=16$ options for case 1. Case 2b: We are now picking two pairs from four possible letters which have sufficient quantities. Thus the ways to pick the two pairs is $4$ choose $2$. The fifth letter is any of the remaining $3$ letters, so we have $6\\times 3=18$ for this case. Case 3a: Similarly, there are only $3$ ways to pick $3$ matching letters. Then we proceed to pick the other $2$ from $4$ choices, so we multiply $3$ by $4$ choose $2$", "of a $2$-letter word can be chosen in $26$ ways. For each of these ways, the second letter can be chosen in $26$ ways, for a total of $26^2$. (If $26$ women each have $26$ children, then the total number of children is $26^2$.) The same idea shows that there are $26^3$ $3$ letter words, $26^4$ $4$-letter words, and so on. However, we don't even want to count all the $3$-letter words, only those up to $bbb$. How many $3$-letter words start with $a$? We can complete $a$ to a $3$-letter word by appending to $a$ any $2$-letter word. There are $26^2$ such $2$-letter words. So $26^2$ $2$-letter words start with $a$. Finally, we want to count the $3$-letter words that start with $b$, but we only want to get to $bbb$. So append any $2$-letter word that starts with $a$ (there are $26$ of these), and finally at the very end we must not forget about $bba$ and $bbb$ (2 words). Add up. We get $1406$. But probably you did not intend to count the first word, just the \"iterations\" or \"increments.\" There are $1405$ of these. B) Using Base 26 Representation: The $1$-letter words were easy to count, as were the $2$-letter words. The $3$-letter words were a bit of a nuisance, since we did not", "be letters can be completed $26^7$ means, so the solution is $26^7$. The amount of words have FRED as a subword, with the letters FRED, because order, inhabiting placements $2$ to $5$ in words? The continuing to be letters can be completed $26^7$ means, so the solution is $26^7$. The very same holds true for words that have actually FRED inhabiting placements $3$ to $6$, $4$ to $7$, $5$ to $8$, and more approximately $8$ to $11$. Currently it is alluring to build up, obtaining a total amount of $(8)(26^7)$. Nonetheless, this will certainly double matter, as an example, words that start with FRED, after that some pointless letter, after that FRED once more, with $2$ letters included at the end. There are numerous various other sorts of words that have 2 FRED remains in them that will certainly be double - counted. The good news is, we do not need to bother with words that have $3$ or even more events of FRED. (If we were managing $47$ - letter words, points can get really undesirable.) Allow us figure out the amount of double - counted words there are. We will certainly do it gradually, and afterwards in a slicker means. The amount of \"two FRED\" words have the first one at the start? The 2nd one", "have two second letters, and for each of those, we have one third letter, so we have $$3 * 2* 1 = 6$$ total ways. If you wish, you can take a few minutes to verify that there are $$4 * 3 * 2 * 1 = 24$$ ways to arrange the letters of \"MATH\" into different orders. Similarly, there are only $$2 * 1 = 2$$ ways to arrange two different letters (AB and BA, for example). So if we extend the pattern, it turns out that there are $$15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1$$ ways to arrange the letters of \"UNCOPYRIGHTABLE.\" That is a total of $$1,307,674,368,000$$ ways! Now, I said I'd verify that we'd be here awhile if you tried to list all of the ways. If it took you five seconds to write down each possibility, and you did this all day and all night, never stopping to eat or sleep or do anything else, you'd be here for $$1,307,674,368,000 * 5 = 6,538,371,840,000$$ seconds. There are 86,400 seconds in a day, so you'd be here for $$6,538,371,840,000/86,400 = 75,675,600$$ days, which, at 365 days per year, would keep you here", "example, for Case (iv), there are $(8)(7)(6)(5)=1680$. For Case (iii) there are I think $756$. I gave explicit formulas for Cases (i) and (ii). We need to add the numbers obtained in the $4$ cases. – André Nicolas Feb 28 '13 at 1:03 There are $13$ letters in parallelogram. If they were all different, the number of $4$-letter \"words\" would be $13\\times12\\times11\\times10$ --- can you see why? But of course the $13$ letters are not all different. So you have to look at all the words that use the letter \"a\" once, and treat them specially; also, all the words that use \"a\" twice and three times, and similarly for \"r\", and for \"l\". You asked for a start --- does this get you started? - So, we do $13*12*11*10$ because of permutations? But how would you go about implementing the repetition portion? Could you provide more detail? – gekkostate Feb 28 '13 at 0:05 I thought you just asked for a start. Anyway, Andre has provided more detail. – Gerry Myerson Feb 28 '13 at 0:22", "will consist of between $1$ and $8$ capital letters only, and all of the words in a test case will be of the same length. The first word in the list will be the starting word of the word ladder, and the second will be the ending word of the word ladder. ## Output Output exactly two lines. The first line holds the one single word that you would add to the dictionary, and the second holds an integer indicating the minimum number of steps to get from the starting word to the ending word, adding your word. It is possible that there’s more than one word you can add that will make your path as short as possible. In this case, output the solution word that comes first alphabetically. It is possible that there’s no word you can add that will that will make your path any shorter. In this case, output 0 (zero) as the word. It is possible that there’s no word you can add that makes the solution possible. In this case, output 0 (zero) as the word, and -1 as the number of steps. Sample Input 1 Sample Output 1 3 CAT DOG COT COG 3 Sample Input 2 Sample Output 2 2 CAT DOG 0 -1 Sample Input 3 Sample Output 3", "as IDENTICAL mailboxes Cheers, Brent Hi Brent, That's not my full solutions. Here it is: 1box 2box 3box 5 0 0......................................Case-1 4 0 1......................................Case-2 3 0 2......................................Case-3 3 1 1......................................Case-4 2 2 1......................................Case-5 Let's consider each case individually, Case-1 500 i.e. 5 letters in box1, 0 letters in box-2, 0 letters box-3. Now, there will be 3 such scenarios, 3P1/2P1 = 3. (by MISSISSIPPI rule dividing by 2P1 for two identical zeros) These 3 scenarios are 500, 050, 005 Since letters distinct, for each of these scenarios: 5C5*0C0*0C0 So, total possible arrangements for case-1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3 Similarly, Case-2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30 Case-3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60 Case-4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60 Case-4: 221, total possible arrangements : (3P1/2P1)*5C2*3C2*1C1 = 3*10*3*1 = 90 Total of all cases = 3+30+60+60+90 = 243 which is indeed 3^5 But what irritates me is my long winded solution is visualize-able to me and not your more elegant solution. Can you help me see it from your point of view? Thanks again mensanumber Senior | Next Rank: 100 Posts Joined 23 Oct 2013 Posted: 30 messages Followed by: 1 members 2 GMAT Score: 750 Wed Mar 07, 2018 10:15 am regor60", "second test case of the example, the $1$-st and the $3$-rd words can be used to make an interesting story. The interesting story is \"aba aba\". Stephen can't use all three words at the same time. In the third test case of the example, Stephen can't make a non-empty interesting story. So the answer is $0$. In the fourth test case of the example, Stephen can use the $3$-rd and the $4$-th words to make an interesting story. The interesting story is \"c bc\".", "ABC are in the same box. For any box there are $3! = 6$ ways to permute ABC, while there are five boxes. The total for this case is 6*5 = 30. Case 2: The three letters all go in three different boxes. This is simply $P(5,3)$ which is 60. Case 3: Two of the letters go in a box, while the third letter goes in a different box. This is a little bit tricky, so we break it in a few parts. First of all we choose the two boxes of the five we use. The number of ways to do this is $\\binom{5}{2}$ which is 10. Secondly, if the two boxes are labelled Box 1 and Box 2, we can either put two letters in Box 1 and one in Box 2, or the other way around, giving an additional two ways. Finally there are $3!$ or 6 ways to permute the three letters. The final count for case 3 is 10 * 2 * 6 or 120. The number of ways to pack ABC into five boxes is the sum of the three cases, or 210. The answer to the problem is half of that, 105. Problem 3 (again from the same UVM contest) Point P is inside rectangle ABCD. If the distance from P", "# Question ff7fe Jun 18, 2016 We divided all $4$ letter words into three groups. Below is the number of combinations and permutations in each group. #### Explanation: Let me interpret the problem in more details. You would like to use all $11$ letters of a word EXAMINATION and choose $4$ letters from them at a time. You are asking how many different $4$ letter permutations and combinations are possible. I hope, that's what you meant. If all letters in the initial word were different, we would have a textbook problem. Unfortunately, situation gets complicated since three letters, A, I and N are repeated twice. Let's group all our $4$ letter words into three categories: Group 1. No pairs of the same letters are present in a word. Group 2. One pair of the same letters is present in a word. Group 3. Two pairs of the same letters are present in a word. There can be no more than two pairs present since the total number of letters in a word is $4$. Group 1 There are $8$ different letters in the word EXAMINATION. So, we have $8$ candidates for the first place in a word, $7$ candidates for the second place, $6$ for the third and $5$ for the fourth. The total number of words in", "analysis of the problem, lengthy but I hope fairly straightforward. Your mistake was to try to use a simple formula, before carrying out an analysis of what is really going on. How many $11$ letter words have FRED as a subword, with the letters FRED, in that order, occupying the first $4$ positions in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$. How many words have FRED as a subword, with the letters FRED, in that order, occupying positions $2$ to $5$ in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$. The same is true for the words that have FRED occupying positions $3$ to $6$, $4$ to $7$, $5$ to $8$, and so on up to $8$ to $11$. Now it is tempting to add up, getting a total of $(8)(26^7)$. However, this will double count, for example, the words that start with FRED, then some useless letter, then FRED again, with $2$ letters added at the end. There are several other types of words that have two FRED's in them that will be double-counted. Luckily, we don't have to worry about words that have $3$ or more occurrences of FRED. (If we were dealing with $47$-letter words, things could get", "following cases: o ABC o ACB o BAC o BCA o CAB o CBA Now, let’s play with these words! Tell, me how many unique words will we get if we replace C by A?? If we replace C with A in the above example, we will get: • Total words=ABA+AAB+BAA+BAA+AAB+ABA= 2(AAB+ABA+BAA) = 2! (AAB+ABA+BAA) = 2! * 3 Now, notice that there are only 3 unique words possible: AAB, ABA, and BAA. However, we have 2! multiplied with it in above equation. Now, what should we do to get the actual answer which is 3?? • We are getting multiple case, because we wrote AAB twice, ABA twice and BBA twice. • And we wrote the twice, because while jotting the cases, we arrange AA in AAB, ABA and BBA twice or 2! times. • Thus, to get the actual answer, we need to divide 2! *3 by 2!. • AAB+ABA+BAA= $$6/2!$$ = Total ways when all the letters were different/ repetitive counting of the identical letters. Now, let us extend this idea further to find the number of different words by using three A’s, two B’s. Since we know that we will again get repetitive cases because of three identical A’s and two identical B’s. • Total words= (Total ways when all the 5 letters are", "to get $3\\times 6=18$ options for case 3. Case 3b: There are $3$ ways to pick the triplet and the other pair only has $3$ options also, since it can't be the same letter as the triplet (that would be case 5). Thus there are $3\\times 3=9$ possibilities. Case 4: Hopefully you follow the pattern: there are only $2$ ways to pick $4$ of the same letter, and $4$ ways to pick the fifth letter, so we have $2\\times 4=8$ for this case. Case 5: Finally, there is only one way to pick $5$ of the same number. Summing those together we get $1+16+18+18+9+8+1=71$ options. For me, this is the most straightforward approach once you understand it. Well, you could just hammer through all of the possibilities. (You'll also appreciate why the shortcuts are there.) Number of ways of selecting 5 different letters: ABCDE (1) Number of ways to select 2 similar and 3 different letter: AABCD AABCE AABDE AACDE BBCDE ABBCD ABBCE ABBDE ACCDE ABCCD ABCCE BCCDE ABCDD ABDDE ABCDD BCDDE (16) Number of ways of selecting 2 similar + 2 more similar letter and 1 different letter: AABBC AABBD AABBE AACCD AACCE AABCC BBCCD BBCCE ABBCC ABBDD BBCDD BBDDE ACCDD BCCDD CCDDE (18) Number of ways to select 3 similar and 2 different letters: AAABC AAABD", "of six-letter words that either lack an A, or lack a B, C, or D, or both Then, lastly, subtract that from the total number of six-letter words. Casework usually works when you want complementary counting, but the problem has a few details to take care of. Case (1): No $a,b,c,$ or $d$. Case (2): Yes $a$, no $b,c,$ or $d$. Case (3): No $a$, yes $b,c,$ or $d$. These cases are mutually exclusive, and so overcounting is not possible. The sum of these cases is the complement of what you seek. Case (1): There are $22^6$ ways. Case (2): There are $23^6$ ways of constructing a $6$ letter word with $23$ words in the alphabet (no $b,c$ or $d$), and $22^6$ of these ways do not include $a$ (or the number of ways of constructing a $6$ letter word with $22$ words in the alphabet). Hence there are $23^6-22^6$ ways. Case (3): There are $25^6$ ways of constructing the word without $a$. Of these ways, there are $22^6$ ways of constructing the word without any of the three letters, so there are $25^6-22^6$ ways total. Then, there are $26^6 - (22^6) - (23^6 - 22^6) - (25^6 - 22^6) = 26^6 - 23^6 - 25^6 + 22^6$ ways total.", "die. Similarly, $6$ such ways exist for the second die. The fundamental principle of counting says that the total number of possible pairs is $6 \\times 6 = 36.$ • Suppose that we have a (unlimited) supply of the letters $A,\\,\\,B,\\,\\,C,\\,\\,D,\\,\\,E,\\,\\,F$ available with us. How many $5$ letter chains can we form using these letters? Imagine $5$ blank spaces for the $5$-letter chain that we are supposed to form (numbered $1$ to $5$): • We now have $5$ tasks at hand; each task correspond to filling a space in Fig.$1$. Realise that these $5$ tasks are independent of each other. For example, what you fill in place $2$ has no effect on what you fill in place $5$ since you’ ve been assured an unlimited supply of letters. Each task can be accomplished in 6 possible ways. For example, you can fill place $-1$ in $6$possible ways : with $A,\\,\\,B,\\,\\,C,\\,\\,D,\\,\\,E,\\,\\,{\\rm{or}}\\,\\,F$. Thus, by the fundamental principle of counting, the total number of ways to form the $5$-letter chain would be $6 \\times 6 \\times 6 \\times 6 \\times 6 = {6^5}$. • Now consider the scenario when you don’t have an unlimited supply of letters available. Suppose that you have only one of each of the $5$ letters available. How many $5$-letter chains can be formed now? You must", "7 times, so our count will be too high, contradiction. We're still assuming $n\\leq23$, so now $n=23$ is the only case left. Here \"times\" is used 24 times, \"time\" is used 109560 times, and so \"one\" is used 109564 times. Now \"four\" is used at least 3 times, so it becomes the 21st word on our list. We know \"two\" is used at least 8 times; let's say it's used 9 times, so that \"nine\" is used 7 times and \"seven\" is the 22nd word on our list. After a little trial-and-error and deduction, I got to the following list: • \"this\" ($2$ times) • \"sentence\" ($2$ times) • \"uses\" ($2$ times) • \"or\" ($2$ times) • \"mentions\" ($2$ times) • \"and\" ($2$ times) • \"the\" ($109583$ times) • \"word\" ($109583$ times) • \"times\" ($24$ times) • \"time\" ($109560$ times) • \"one\" ($109564$ times) • \"hundred\" ($9$ times) • \"thousand\" ($5$ times) • \"eighty\" ($3$ times) • \"nine\" ($7$ times) • \"five\" ($6$ times) • \"two\" ($9$ times) • \"three\" ($7$ times) • \"twenty\" ($2$ times) • \"sixty\" ($3$ times) • \"four\" ($3$ times) • \"seven\" ($3$ times) • \"six\" ($2$ times) ... which all works out, giving the solution I put at the top. Hurray! :-D • +1 for the great effort. Making it self-contained will be", "# Basic question. Does not involve permutations/combinations. $3$ mailboxes containing $3$ letters My teacher explained this problem to us - \"There are $3$ mailboxes. $3$ people put letters in at random. There is no preference for any of the $3$ mailboxes. Compute the probability that each mailbox contains $1$ letter.\" I tried this problem on my own and got the wrong answer. I understand the teacher's solution but I don't get what went wrong with my answer. Can someone explain where I went wrong? My answer: Treat each outcome as a sequence of $3$ numbers denoting the number of letters in each mailbox, e.g $300$ means $3$ in the 1st, $0$ in the second and $0$ in the 3rd mailbox. Sample space $S = \\{300, 030, 003, 111,012,021, 120,102,210,201\\}$ $S$ contains $10$ sample points. Let $A =$ event that each mailbox is chosen once. $P(A) = 1/10$ Teacher's solution: Treat each outcome as a sequence of $3$ numbers denoting the person and their corresponding chosen mailbox, e.g $312$ means the first person chose mailbox 3, the second chose mailbox 1, the third chose mailbox 3. So the sample space is $S = \\{ 111, 112...\\}$ and contains $27$ sample points. Now let $A =$ event that each mailbox is chosen once. $A =\\{123,132,213,231,312,321\\}$ So the answer is $P(A)", "be arranged internally for example - 500 could be 005 etc. What mistake am I making? Thanks Not making any mistakes so far but haven't completed the thought. As you suggest, 500 can be arranged 3 ways. Likewise, 401 can be arranged 3 ways. And so on. All together there are then 3x3x3x3x3 = 3^5 ways Here is my full solution: 1box 2box 3box 5 0 0......................................Case-1 4 0 1......................................Case-2 3 0 2......................................Case-3 3 1 1......................................Case-4 2 2 1......................................Case-5 Let's consider each case individually, Case-1 500 i.e. 5 letters in box1, 0 letters in box-2, 0 letters box-3. Now, there will be 3 such scenarios, 3P1/2P1 = 3. (by MISSISSIPPI rule dividing by 2P1 for two identical zeros) These 3 scenarios are 500, 050, 005 Since letters distinct, for each of these scenarios: 5C5*0C0*0C0 So, total possible arrangements for case-1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3 Similarly, Case-2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30 Case-3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60 Case-4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60 Case-4: 221, total possible arrangements : (3P1/2P1)*5C2*3C2*1C1 = 3*10*3*1 = 90 Total of all cases = 3+30+60+60+90 = 243 which is indeed 3^5 But what irritates me is my long winded solution is visualize-able to me and not", "# A letter lock consists of three rings each marked with $5$ different letters. Number of maximum attempts to open the lock is:A. $124$ B. $125$ C. $120$ D. $75$ Verified 177.3k+ views Hint: Find the number of ways of choosing the letter for each of the three rings and then multiply the number of ways to open all the locks of the letter lock. We can also use concepts of permutations and combinations to solve such problems. A letter lock consists of three rings each marked with $5$ different letters. As per given that the letter lock consists of three rings each marked with $5$ different letters. So, the number of ways for opening the first ring is equal to $5$ as it is marked with $5$ different letters. In a similar manner, the number of ways for opening the second ring is equal to $5$ as it is marked with $5$ different letters. In a similar manner, the number of ways for opening the third ring is equal to $5$ as it is marked with $5$ different letters. So, the number of ways for opening the letter lock is equal to $5 \\times 5 \\times 5 = 125$ . So, the number of maximum attempts to open the lock is equal to $125$", "$4$-letter words, with or without meaning, which can be formed out of the letters of the word $\\text{YOUR}$, where the repetition of the letters is not allowed. To find the different arrangements, let first consider a sequence of $4$ blank spaces – _____ _____ _____ _____. Starting with the first leftmost blank. The number of choices available for this blank space is $4$, (either of ‘Y’, ‘O’, ‘U’, and ‘R’). The number of choices available for the second blank space is $3$. (As one letter is already used in the first blank space and repetition of letters is not allowed). Similarly, the number of choices available for the third blank space is $2$. And finally, the number of choices available for the fourth blank space is $1$. Thus by principle of counting the total number of ways is $4 \\times 3 \\times 2 \\times 1 = 4! = 24$ Example 2: How many $2$-digit numbers can be formed from the digits $1$, $2$, $3$, $4$, $5$ if a) the digits can be repeated b) the digits cannot be repeated The total number of digits is $5$ Let first consider a sequence of $2$ blank spaces – _____ _____. (Representing a $2$-digit number) a) Starting with the first leftmost blank. The number of choices available for this blank space" ]

[ "of a $2$-letter word can be chosen in $26$ ways. For each of these ways, the second letter can be chosen in $26$ ways, for a total of $26^2$. (If $26$ women each have $26$ children, then the total number of children is $26^2$.) The same idea shows that there are $26^3$ $3$ letter words, $26^4$ $4$-letter words, and so on. However, we don't even want to count all the $3$-letter words, only those up to $bbb$. How many $3$-letter words start with $a$? We can complete $a$ to a $3$-letter word by appending to $a$ any $2$-letter word. There are $26^2$ such $2$-letter words. So $26^2$ $2$-letter words start with $a$. Finally, we want to count the $3$-letter words that start with $b$, but we only want to get to $bbb$. So append any $2$-letter word that starts with $a$ (there are $26$ of these), and finally at the very end we must not forget about $bba$ and $bbb$ (2 words). Add up. We get $1406$. But probably you did not intend to count the first word, just the \"iterations\" or \"increments.\" There are $1405$ of these. B) Using Base 26 Representation: The $1$-letter words were easy to count, as were the $2$-letter words. The $3$-letter words were a bit of a nuisance, since we did not", "$p = 0.1$ used as an example is absurdly high, as I think is be clear upon intuitive reflection. To put some numbers to this, though, let me list out a few (approximate) probabilities for shorter words • The probability of two random letters making a word is around $124/676 \\approx 0.18$, or 18%. • The probability of three random letters making a word is around $1292/17576 \\approx 0.07$, or 7%. • The probability of four random letters making a word is around $5454/456976 \\approx 0.01$, or 1%. • The probability of five random letters making a word is around $158390/11881376 \\approx 0.01$, or 1%. • The probability of six random letters making a word is around $22157/308915776 \\approx 0.007$, or 0.7%. To give you an idea of the meaning of all of this, I counted in the first sentence of this article 120 letters and 26 words, which averages out to 4.6 letters per word. This puts us in the 1% area. I think it would be appropriate then to allow our value of $p$ to be around 1%. In reality it would be lower – after all the probability of pressing a space bar if everything is truly random would be 1/27, which would mean our monkey would on average type 27 letters in a row", "of $p = 0.1$ used as an example is absurdly high, as I think is be clear upon intuitive reflection. To put some numbers to this, though, let me list out a few (approximate) probabilities for shorter words • The probability of two random letters making a word is around $124/676 \\approx 0.18$, or 18%. • The probability of three random letters making a word is around $1292/17576 \\approx 0.07$, or 7%. • The probability of four random letters making a word is around $5454/456976 \\approx 0.01$, or 1%. • The probability of five random letters making a word is around $158390/11881376 \\approx 0.01$, or 1%. • The probability of six random letters making a word is around $22157/308915776 \\approx 0.007$, or 0.7%. To give you an idea of the meaning of all of this, I counted in the first sentence of this article 120 letters and 26 words, which averages out to 4.6 letters per word. This puts us in the 1% area. I think it would be appropriate then to allow our value of $p$ to be around 1%. In reality it would be lower – after all the probability of pressing a space bar if everything is truly random would be 1/27, which would mean our monkey would on average type 27 letters in a", "have two second letters, and for each of those, we have one third letter, so we have $$3 * 2* 1 = 6$$ total ways. If you wish, you can take a few minutes to verify that there are $$4 * 3 * 2 * 1 = 24$$ ways to arrange the letters of \"MATH\" into different orders. Similarly, there are only $$2 * 1 = 2$$ ways to arrange two different letters (AB and BA, for example). So if we extend the pattern, it turns out that there are $$15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1$$ ways to arrange the letters of \"UNCOPYRIGHTABLE.\" That is a total of $$1,307,674,368,000$$ ways! Now, I said I'd verify that we'd be here awhile if you tried to list all of the ways. If it took you five seconds to write down each possibility, and you did this all day and all night, never stopping to eat or sleep or do anything else, you'd be here for $$1,307,674,368,000 * 5 = 6,538,371,840,000$$ seconds. There are 86,400 seconds in a day, so you'd be here for $$6,538,371,840,000/86,400 = 75,675,600$$ days, which, at 365 days per year, would keep you here", "be letters can be completed $26^7$ means, so the solution is $26^7$. The amount of words have FRED as a subword, with the letters FRED, because order, inhabiting placements $2$ to $5$ in words? The continuing to be letters can be completed $26^7$ means, so the solution is $26^7$. The very same holds true for words that have actually FRED inhabiting placements $3$ to $6$, $4$ to $7$, $5$ to $8$, and more approximately $8$ to $11$. Currently it is alluring to build up, obtaining a total amount of $(8)(26^7)$. Nonetheless, this will certainly double matter, as an example, words that start with FRED, after that some pointless letter, after that FRED once more, with $2$ letters included at the end. There are numerous various other sorts of words that have 2 FRED remains in them that will certainly be double - counted. The good news is, we do not need to bother with words that have $3$ or even more events of FRED. (If we were managing $47$ - letter words, points can get really undesirable.) Allow us figure out the amount of double - counted words there are. We will certainly do it gradually, and afterwards in a slicker means. The amount of \"two FRED\" words have the first one at the start? The 2nd one", "upon intuitive reflection. To put some numbers to this, though, let me list out a few (approximate) probabilities for shorter words • The probability of two random letters making a word is around $124/676 \\approx 0.18$, or 18%. • The probability of three random letters making a word is around $1292/17576 \\approx 0.07$, or 7%. • The probability of four random letters making a word is around $5454/456976 \\approx 0.01$, or 1%. • The probability of five random letters making a word is around $158390/11881376 \\approx 0.01$, or 1%. • The probability of six random letters making a word is around $22157/308915776 \\approx 0.007$, or 0.7%. To give you an idea of the meaning of all of this, I counted in the first sentence of this article 120 letters and 26 words, which averages out to 4.6 letters per word. This puts us in the 1% area. I think it would be appropriate then to allow our value of $p$ to be around 1%. In reality it would be lower – after all the probability of pressing a space bar if everything is truly random would be 1/27, which would mean our monkey would on average type 27 letters in a row between spaces. There are, well, basically no words 27 letters long, so we would probably have", "be one of these $5$ cases, and there is no overlap in the $5$ cases. Thus we are going to sum up the number of total possibilities for each case. Case 1: If we pick $5$ distinct letters, we pick one of each letter, so there is only $1$ way to do this. Case 2a: First lets count the ways to pick the $2$ letters which are the same. It can't be E, so there are only $4$ ways to pick them. Second, we must pick $3$ more from the remaining $4$ letters (none of these can be the same letter as the first $2$ since that would be a different case). Thus we use $4$ choose $3$, which turns out to be $4$ options. We multiply these together to see that we have $4\\times 4=16$ options for case 1. Case 2b: We are now picking two pairs from four possible letters which have sufficient quantities. Thus the ways to pick the two pairs is $4$ choose $2$. The fifth letter is any of the remaining $3$ letters, so we have $6\\times 3=18$ for this case. Case 3a: Similarly, there are only $3$ ways to pick $3$ matching letters. Then we proceed to pick the other $2$ from $4$ choices, so we multiply $3$ by $4$ choose $2$", "available, so we can calculate puzzle difficulty by the 30% quantile of the difficulties of solution words weighted by their scores. (It's the 30% quantile instead of the 70% quantile because lower word difficulty corresponds to harder words.) This measure has the advantage that it is a bit like taking the \"average\" of the solution word difficulties, but it also takes into account the score threshold for Genius. Let's call this measure the puzzle difficulty. ### Finding the Hardest (and Easiest) Bee Every puzzle must have at least one solution which uses all seven of the letters (the pangram). It turns out there are only about 17000 words which contain exactly seven unique letters, and only about 6500 that also don't contain \"S\", so the number of possible puzzles can't be more than $$7*6500=45500$$ (because there are seven choices for the center letter). In fact, a bit of redundancy means that there are only about 30000 puzzles. This number is small enough that it can be explored by brute force. #### First Attempt After computing the difficulties of all legal puzzles, these are the 25 easiest (note the horizontal scroll bars): And these are the 25 hardest legal puzzles: (The words below each puzzle are the valid solutions. Bold words are pangrams, and red words are those which", "# A letter lock consists of three rings each marked with $5$ different letters. Number of maximum attempts to open the lock is:A. $124$ B. $125$ C. $120$ D. $75$ Verified 177.3k+ views Hint: Find the number of ways of choosing the letter for each of the three rings and then multiply the number of ways to open all the locks of the letter lock. We can also use concepts of permutations and combinations to solve such problems. A letter lock consists of three rings each marked with $5$ different letters. As per given that the letter lock consists of three rings each marked with $5$ different letters. So, the number of ways for opening the first ring is equal to $5$ as it is marked with $5$ different letters. In a similar manner, the number of ways for opening the second ring is equal to $5$ as it is marked with $5$ different letters. In a similar manner, the number of ways for opening the third ring is equal to $5$ as it is marked with $5$ different letters. So, the number of ways for opening the letter lock is equal to $5 \\times 5 \\times 5 = 125$ . So, the number of maximum attempts to open the lock is equal to $125$", "second test case of the example, the $1$-st and the $3$-rd words can be used to make an interesting story. The interesting story is \"aba aba\". Stephen can't use all three words at the same time. In the third test case of the example, Stephen can't make a non-empty interesting story. So the answer is $0$. In the fourth test case of the example, Stephen can use the $3$-rd and the $4$-th words to make an interesting story. The interesting story is \"c bc\".", "analysis of the problem, lengthy but I hope fairly straightforward. Your mistake was to try to use a simple formula, before carrying out an analysis of what is really going on. How many $11$ letter words have FRED as a subword, with the letters FRED, in that order, occupying the first $4$ positions in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$. How many words have FRED as a subword, with the letters FRED, in that order, occupying positions $2$ to $5$ in the word? The remaining letters can be filled in $26^7$ ways, so the answer is $26^7$. The same is true for the words that have FRED occupying positions $3$ to $6$, $4$ to $7$, $5$ to $8$, and so on up to $8$ to $11$. Now it is tempting to add up, getting a total of $(8)(26^7)$. However, this will double count, for example, the words that start with FRED, then some useless letter, then FRED again, with $2$ letters added at the end. There are several other types of words that have two FRED's in them that will be double-counted. Luckily, we don't have to worry about words that have $3$ or more occurrences of FRED. (If we were dealing with $47$-letter words, things could get", "are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of Oz lost because of the prohibition? A. 65 B. 66 C. 67 D. 131 E. 132 Two ways to do this one. 1. with 66 letters, how many one or two letter words can we make? $$67 * 66$$ -------(1) we can fill the first position with 67 different things(66 letters or a blank), and the second position with 66 different letters. With 65 letters after removing the 7th letter, how many words do we have? $$66 * 65$$ -------(2) (1) - (2) = 132(E) 2. Now that we have removed the 7th letter, we have lost all the single and double letter words that we could make which include this 7th letter. Single letter words with 7th letter? $$1$$ Double letter words? $$2 * 65 + 1$$ ---Double letter words by using 7th letter only once + Double letter word by using 7th letter twice Total = 132(E) _________________ Spread some love..Like = +1 Kudos Non-Human User Joined: 09 Sep 2013 Posts: 11717 Re: In the Land of Oz only one or two-letter words are used. The [#permalink] ### Show Tags 20 Aug 2018, 12:02 Hello from the GMAT Club", "word length) $26\\times26\\ldots(42 times)=26^{42}=2.68\\times10^{59}$ A 59 digit number! It could obviously go higher if using 100 or 150 letter words comes into vogue. (I hear that the Germans already do this!) This doesn’t really appear ‘realistic’. So let’s see how we can arrive at that figure. First, we note, that since the right side of the Gaussian can extend asymmetrically much more than the left side which is bound by 1 letter words. (Zero and negative letter words don’t exist!). However, since the likelihood of words greater than 10 is very low, we can for current purposes ignore them. ( This is an approximation used to make my life easier to solve this. If you are interested, you could as well use MATLAB to plot a skewed Gaussian and let me know about the percentage error due to this. I’m reasonably confident it won’t be too significant.) Therefore about 67% of words are: $26\\times26\\times5+26\\times26\\times5\\times5+26\\times26\\times5\\times26\\times5+26\\times26\\times5\\times26\\times5\\times5+26\\times26\\times5\\times26\\times26\\times5\\times5=69006080$ The above accounts for the frequency of vowels in all words of length 3–7 as well. That is about 69 million. A 100% (total no.) of possible words would be about 103503945. Or around 100 million total words. Dictionary Facts – Oxford English Dictionary says that they have around 414,800 ‘entries’ defined. These are the words that we most use on a day to", "let's apply your formula to the tractable case of $2$ letters $a$ and $b$, the two-letter word $aa$, and $N=4$. I compute $p=(1/2)^4$ and, for finding exactly one occurrence (not allowing overlaps), I compute $$(N-2)p^1(1-p)^{N-2}=2(1/16)(15/16)^2\\approx 0.11.$$ However, exhaustive enumeration gives $$aabb,aaba,baab,abaa,bbaa,$$ each with probability $1/2^4$, for a total chance of $5/16\\approx 0.3125$. (It's unclear whether you count instances like $aaab$ as one or two occurrences: they would only add to the chances.) I get an even larger value for $ab$. – whuber May 20 '16 at 19:13 • @whuber one thing that's wrong is that he's assuming that the probability of finding a word of interest at each position is independent of every other position. I think we both answered a similar question that appropriately takes that into account? – Neil G May 20 '16 at 19:35 • @whuber, thanks for looking at my answer! I believe that p should be (1/2)^2 for a 2-letter word, which gives us 0.4219, still above 0.3125 (I'll try to track down the issue!) but not so bad as 0.11! In any case, I believe that setting n=0 gets rid of the overlap cases. – MikeP May 20 '16 at 20:20 • Sorry--you're right about $p$. I wouldn't expect your solution to be perfect, because it doesn't model the possibility of overlaps", "# Question ff7fe Jun 18, 2016 We divided all $4$ letter words into three groups. Below is the number of combinations and permutations in each group. #### Explanation: Let me interpret the problem in more details. You would like to use all $11$ letters of a word EXAMINATION and choose $4$ letters from them at a time. You are asking how many different $4$ letter permutations and combinations are possible. I hope, that's what you meant. If all letters in the initial word were different, we would have a textbook problem. Unfortunately, situation gets complicated since three letters, A, I and N are repeated twice. Let's group all our $4$ letter words into three categories: Group 1. No pairs of the same letters are present in a word. Group 2. One pair of the same letters is present in a word. Group 3. Two pairs of the same letters are present in a word. There can be no more than two pairs present since the total number of letters in a word is $4$. Group 1 There are $8$ different letters in the word EXAMINATION. So, we have $8$ candidates for the first place in a word, $7$ candidates for the second place, $6$ for the third and $5$ for the fourth. The total number of words in", "# Number of words in $\\{pi,po\\}^*$ of length at most 9 I have a language $L^*$ for $L = \\{pi,po\\}$ (I think pi counts as one letter and po also as one letter otherwise a max length of 9 is not possible). The question is how many words I can make with $L^*$ where the maximum length is 9. The answer is 31, but I do not know how it got calculated. How is this calculated? • Both $pi$ and $po$ count as two letters for the answer to be correct. Can you count how many strings of length $n$ you can form? – quicksort Apr 5 '18 at 19:03 • @quicksort so $pipo$ would be 4 letters then right? – Diceble Apr 5 '18 at 19:10 • @quicksort and after thinking i would have no idea how to count that... – Diceble Apr 5 '18 at 19:15 • How many strings of length $n$ made of zeroes and ones are there? – quicksort Apr 5 '18 at 19:22 • is it 2^n ? sorry if I'm totally wrong, but I just miss something where i proberly would facepalm myself if I found it out. – Diceble Apr 5 '18 at 19:34 Here are all words in $L^*$ of length at most 9: $$\\epsilon, \\\\ pi, po, \\\\", "6 letters is: $26 \\times 26 \\times 26 \\times 26 \\times 26 \\times 26 = 26^6$ For the strings that does not contain a there are 25 possible ways. $25 \\times 25 \\times 25 \\times 25 \\times 25 \\times 25 = 25^6$ For the strings that does not contain b there are 25 possible ways. $25 \\times 25 \\times 25 \\times 25 \\times 25 \\times 25 = 25^6$ For the strings that contains neither a nor b, there are 24 possible ways. $24 \\times 24 \\times 24 \\times 24 \\times 24 \\times 24=24^6$ The possible strings that not containing a or b is: $25^6+25^6 -24^6$ The number of possible strings containing a and b is the total number of strings minus the number of strings that does not containing a or b. $26^6-(25^6+25^6-24^6) \\\\ = 308915776 -(2 \\times 244140625 -191102976) \\\\ = 11737502$ Therefore, there are 11737502 number of strings of six lowercase letters from English alphabet contains the letters or b. iii. The objective is to determine the number of strings of six lowercase letters from English alphabet contain the letters a and b in consecutive positions with a preceding b and all the letters are distinct. As the letters are consecutive, the possible strings are in the form: Here ab can be in five locations, that", "7 times, so our count will be too high, contradiction. We're still assuming $n\\leq23$, so now $n=23$ is the only case left. Here \"times\" is used 24 times, \"time\" is used 109560 times, and so \"one\" is used 109564 times. Now \"four\" is used at least 3 times, so it becomes the 21st word on our list. We know \"two\" is used at least 8 times; let's say it's used 9 times, so that \"nine\" is used 7 times and \"seven\" is the 22nd word on our list. After a little trial-and-error and deduction, I got to the following list: • \"this\" ($2$ times) • \"sentence\" ($2$ times) • \"uses\" ($2$ times) • \"or\" ($2$ times) • \"mentions\" ($2$ times) • \"and\" ($2$ times) • \"the\" ($109583$ times) • \"word\" ($109583$ times) • \"times\" ($24$ times) • \"time\" ($109560$ times) • \"one\" ($109564$ times) • \"hundred\" ($9$ times) • \"thousand\" ($5$ times) • \"eighty\" ($3$ times) • \"nine\" ($7$ times) • \"five\" ($6$ times) • \"two\" ($9$ times) • \"three\" ($7$ times) • \"twenty\" ($2$ times) • \"sixty\" ($3$ times) • \"four\" ($3$ times) • \"seven\" ($3$ times) • \"six\" ($2$ times) ... which all works out, giving the solution I put at the top. Hurray! :-D • +1 for the great effort. Making it self-contained will be", "$$2.35 = 1.75 + 0.15x$$ ...subtract 1.75 from both sides... $$0.60 = 0.15x$$ ...divide both sides by 0.15... $$4 = x$$ So when x = 4, it turns out that b = c. What's the value of b when x = 4? Just plug it back in! $$b = 2.35 + 0.25x$$ ...evaluate for x = 4... $$b = 2.35 + 0.25(4)$$ $$b = 2.35 + 1$$ $$b = 3.35$$ Pre-Calculus TutorMe Question: How many different ways are there to arrange the letters in the word \"ALASKA?\" Charles Y. This is one of my favorite topics to talk about, and it has a pretty decent number of real-world applications. Let's start at the very beginning - a very good place to start. When you read, you begin with A, B, C. Now, how many ways can we arrange the letters ABC into different orders? We can list them all pretty easily: ABC, ACB, BAC, BCA, CAB, CBA That makes six. You can verify that there aren't any more. Take a few seconds to convince yourself. Now, let's go a bit further. I've seen a lot of people say that the longest word in the English language that does not repeat any letters is \"UNCOPYRIGHTABLE,\" which has 15 letters. How many ways are there to arrange those letters into", "the tuple. Unless there is risk of confusion with actual products (or numbers), one abbreviates a tuple (or word) $\\left(a_1,a_2,...,a_k\\right)$ as $a_1a_2...a_k$. So the word $\\left(0,1,1,1,0\\right)$ is written as $01110$. Now, let $n \\in \\mathbb{N}$. A Dyck word of length $2n$ is defined to be a word whose letters are $0$'s and $1$'s, each appearing $n$ times in total (so the word has altogether $2n$ letters), such that for every $i$, there are at least as many $0$'s among the first $i$ letters as there are $1$'s among them. So, for example, $0110$ is not a Dyck word, because among the first $3$ letters there are fewer $0$'s than $1$'s (namely, one $0$ and two $1$'s). Also, $10001$ is not a Dyck word, because among the first $1$ letter there are fewer $0$'s than $1$'s (no $0$ at all and one $1$). Also, $0100$ is not Dyck because its total number of $0$'s does not equal its total number of $1$'s. But $001011$ is a Dyck word (of length $2\\cdot 3=6$), as you can easily check: read the word from left to right, keeping track of how many $0$'s and how many $1$'s you have encountered; if the count for $1$'s overtakes the count for $0$'s, then your word is not Dyck; if both counts are equal at" ]

[ "by their highest common factor.", "equal to 1.96.", "7. Which of the following shows the greatest common factor of $180$, $144$, and $108$?", "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "-1 -1 -1 -1 96 -1", "97.", "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.", "greatest common factor of 36 and 48x 4-1 factoredderivative of 9x9x2 6x 1682.61234 in roman numeralstan1527x3x2 5x 6 factorsolve v lwh for w386.11what is the greatest common factor of 54 and 72solve y 4x 21700-500y kx22xdsquare root of 776multiplying and dividing fraction calculatorleast to greatest decimals calculatordesoto k12 mo ussin30 cos60convert 0.375 to a fraction", "120 - 24= 96", "# Solve for n 28/100=96/n 28100=96n Rewrite the equation as 96n=28100. 96n=28100 Cancel the common factor of 28 and 100. Factor 4 out of 28. 96n=4(7)100 Cancel the common factors. Factor 4 out of 100. 96n=4⋅74⋅25 Cancel the common factor. 96n=4⋅74⋅25 Rewrite the expression. 96n=725 96n=725 96n=725 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. 96⋅25=n⋅7 Solve the equation for n. Rewrite the equation as n⋅7=96⋅25. n⋅7=96⋅25 Multiply 96 by 25. n⋅7=2400 Divide each term by 7 and simplify. Divide each term in n⋅7=2400 by 7. n⋅77=24007 Cancel the common factor of 7. Cancel the common factor. n⋅77=24007 Divide n by 1. n=24007 n=24007 n=24007 n=24007 The result can be shown in multiple forms. Exact Form: n=24007 Decimal Form: n=342.85714285… Mixed Number Form: n=34267 Solve for n 28/100=96/n ### Solving MATH problems We can solve all math problems. Get help on the web or with our math app Scroll to top", "# 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? ×", "square units 42.) What is the least common multiple of 168 and 420? Continue reading", "# 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...", "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.", "largest factor of 900 is itself we have: $n+10=900 \\Longrightarrow \\boxed{n = 890}$ ~qwertysri987", "## Thinking Mathematically (6th Edition) The greatest common divisor of the two numbers is $15$. Factor each number completely to have: $240=10(24)=2(5)(2)(2)(2)(3) \\\\285=5(57)=5(3)(19)$ Thus, the greatest common divisor of the two numbers is $3(5)=15$.", "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.", "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 $3pq^2$?", "which includes all of the distinct factors of 96 including 96 and 1. _________________ 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 CEO Joined: 12 Sep 2015 Posts: 3589 Re: How many unique factors does 96 have? [#permalink] ### Show Tags 08 Jul 2017, 07:42 Top Contributor Bunuel wrote: How many unique factors does 96 have? A. 8 B. 10 C. 12 D. 14 E. 15 The posters above have used a useful rule (which I am providing a video for below). However, if you weren't aware of that rule, you can always just list the factors. When doing so, find pairs, starting from the lowest and biggest factors. We get: Factors of 96: {1, 96} Factors of 96: {1, 2, 48, 96} Factors of 96: {1, 2, 3, 32, 48, 96} . . . and so on.. . Factors of 96: {1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96} 12 factors RELATED VIDEOS _________________ Test confidently with gmatprepnow.com Re: How many unique factors does 96 have? [#permalink] 08 Jul 2017, 07:42 Display posts from previous: Sort by" ]

[ "## Recounting the Rationals, part IVb: the Euclidean Algorithm Suppose we have two integers, and we’d like to find their greatest common divisor (GCD). Recall that the greatest common divisor of two integers m and n is exactly that: the greatest integer which is a divisor of both m and n. The obvious way to do this—which is probably the way you learned in elementary school, when reducing fractions—is to factor each integer into primes, and look for overlap. For example, let’s say we want to find the GCD of 450 and 525. We begin by factoring: $\\begin{array}{rcl} 450 & = & 2 \\cdot 3^2 \\cdot 5^2 \\\\ 525 & = & 3 \\cdot 5^2 \\cdot 7 \\end{array}$ Now we pull out as many prime factors as we can which are contained in the factorizations of both numbers. We can’t use 2 at all, since it is only contained in the factorization of 450; we can use a factor of 3, but only one; and we can use two factors of 5. Putting these together, the greatest common divisor of 450 and 525 is $3 \\cdot 5^2 = 75$. This method is intuitive, and works well for relatively small numbers, but (there’s always a “but”, isn’t there?) it fails horribly for larger numbers. For example, what if you", "can learn how to calculate the greatest common denominator. How to calculate the greatest common denominator For small numbers finding the greatest common denominator is almost intuitive: look at this set: $\\{6,12,9\\}$ Clearly the GCD is $3$! But what do we do when the numbers start to grow? Prime factorization The GCD is ready to be found if you write the prime factorization of the numbers in a set. The prime factorization is the set of prime numbers elevated to specific exponents that, multiplied, returns the original number. The GCD is the largest shared prime factor, elevated to the highest possible exponent. Take this set: $\\{360,378,405\\}$. The prime factorizations of these numbers are: $\\begin{split} 360 &= 2^3\\cdot3^2\\cdot5\\\\ 378 &= 2\\cdot3^3\\cdot7\\\\ 405 &= 3^4\\cdot5 \\end{split}$ Find the repeated prime factor: only $3$ appears in all three numbers, and the highest exponent is $2$: $3^2$ is the GCD of the set! 🙋 Finding the prime factorization is not always a simple task: to simplify your calculations you can use our prime factorization calculator! Euclidean algorithm The GCD of two numbers also divides exactly their difference. It may sound strange, but think about it: the result of the division of the largest number by the GCD can be seen as the sum of two numbers: one of them, multiplied by the", "## Discrete Mathematics and Its Applications, Seventh Edition To obtain the greatest common divisor of each pair of integers, multiply together the factors shared by both numbers. a) $2^2\\cdot 3^3\\cdot 5^2=2700$ b) $2\\cdot 3\\cdot 11=66$ c) $17$ d) $1$ e) These numbers are relatively prime. f) $2\\cdot 3\\cdot 5\\cdot 7=210$", "# 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 $2 \\cdot 2 \\cdot 2 = 8$", "BOOKMARKED ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. _________________ Kudos [?]: 129174 [2], given: 12194 Manager Joined: 04 Jan 2013 Posts: 79 Kudos [?]: 17 [0], given: 1 Re: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. [#permalink] ### Show Tags 21 Jan 2013, 11:33 Bunuel wrote: ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. @bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance Posted from my", "# 5. The least common multiple of the first positive integers Prerequirements: We know that a way to compute the least common multiple between two or more integer numbers is to decompose them into prime factors, and then taking each distinct prime factor with its greatest power in the considered numbers. Going on this way, the factorization of the least common multiple will result. For example, given the three numbers $4 = 2^2$, $6 = 2 \\cdot 3$ and $45 = 3^2 \\cdot 5$, and computing their least common multiple, we see that the distinct prime factors are 2, 3 and 5; 2 appears with the greatest exponent 2 in 4, 3 appears with the greatest exponent 2 in 45, 5 appears with the greatest exponent 1 again in 45; so the least common multiple we are looking for is $2^2 \\cdot 3^2 \\cdot 5 = 180$. There exists a particularly interesting case in which the least common multiple can be computed through a reasoning, without decomposing the numbers into prime factors: it’s the computation of the least common multiple of the positive integer numbers up to a certain number $x$, that is $\\mathrm{LCM}(1, 2, \\dots, x)$. Of course this expression coincides with $\\mathrm{LCM}(2, \\dots, x)$ if $x \\geq 2$, as in this case the number 1 does", "=& {2}^{4}\\cdot 5\\cdot 7\\cdot \\mathrm{11}\\hfill \\end{array}$ 2. The common bases are 2 and 5 3. The smallest exponents appearing on 2 and 5 in the prime factorizations are, respectively, 2 and 1. ${2}^{2}$ from 700. ${5}^{1}$ from either 1,880 or 6,160. 4. The GCF is the product of these numbers. GCF is ${2}^{2}\\cdot 5=4\\cdot 5=\\text{20}$ Thus, 20 is the largest number that divides 700, 1,880, and 6,160 without a remainder. Practice set a Find the GCF of the following numbers. 24 and 36 12 48 and 72 24 50 and 140 10 21 and 225 3 450, 600, and 540 30 Exercises For the following problems, find the greatest common factor (GCF) of the numbers. 6 and 8 2 5 and 10 8 and 12 4 9 and 12 20 and 24 4 35 and 175 25 and 45 5 45 and 189 66 and 165 33 264 and 132 99 and 135 9 65 and 15 33 and 77 11 245 and 80 351 and 165 3 60, 140, and 100 147, 343, and 231 7 24, 30, and 45 175, 225, and 400 25 210, 630, and 182 14, 44, and 616 2 1,617, 735, and 429 1,573, 4,862, and 3,553 11 3,672, 68, and 920 7, 2,401, 343, 16, and 807 1 500, 77, and", "Greatest Common Divisor of two numbers If $ab=600$ how large can the greatest common divisor of $a$ and $b$ be? I am not sure if I should check for all factor multiples of $a$ and $b$ for this question. Please advise. We have $600=2^3\\cdot 5^2\\cdot 3$. To make the gcd large, we give a $2$ to each of $a$ and $b$, also a $5$. So the largest possible gcd is $10$. Remark: The idea generalizes. Let $n$ have prime power factorization $$n=p_1^{d_1}p_2^{d_2}\\cdots p_k^{d_k}.$$ Let $e_i=\\lfloor d_i/2\\rfloor$, where $\\lfloor x\\rfloor$ is the greatest integer that is $\\le x$. Then the greatest possible gcd of $a$ and $b$, where $ab=n$, is $$p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k}.$$ • You are welcome. For the maximum, we distribute the power of any prime $p$ as evenly as possible between $a$ and $b$. – André Nicolas Jun 10 '13 at 2:00 $$ab=600=2^3\\cdot 3\\cdot 5^2$$ The greatest common divisor of $a,b$ will have the greatest number of common prime factors that you can arrange.", "6. Therefore, the greatest common divisor of 18 and 30 is 6. ### Prime factorization Another way of finding the greatest common divisor of two or more numbers is through prime factorization. In this method, each number is factored so that it is the product of prime numbers. For example, the prime factorization of 1,176 gives us $1,176 = 2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 7 \\cdot 7 = 2^3 \\cdot 3 \\cdot 7^2$. The prime factorization of 2,100 is $2^2 \\cdot 3 \\cdot 5^2 \\cdot 7$. To find the greatest common Divisor of 1,176 and 2,100, the smallest powers of each prime divisor for each of the numbers (with $p^0$ being assumed if one number doesn't contain a particular prime factor $p$) are multiplied together, as in $2^2 \\cdot 3^1 \\cdot 5^0 \\cdot 7^1 = 84$. Therefore, 84 is the greatest common factor of 1,176 and 2,100. ### Euclidean Algorithm For large numbers with many divisors, especially large prime divisors, these methods are somewhat inefficient. A systematic way of finding the greatest common divisor is the Euclidean Algorithm. This algorithm has the advantage of being easily written for a computer program. In the Euclidean Algorithm, the division algorithm is applied to two numbers until a remainder of zero is found. How this works is that", "11$ $1575 = 3^2 \\cdot 5^2 \\cdot 7$ From the above: what primes do these two numbers have in common? As we can see, the common primes are 3 and 5. Looking at the exponents of these common primes in each of the numbers, we look the minimum between the two. In this case, the minimum exponent for 3 is 1, and the minimum exponent for 5 is also 1. Therefore $GCD = 3^1 \\cdot 5^1 = 3 \\cdot 5 = 15$ Now, all we have to do is to simplify the original fraction by 15: $\\frac{165}{1575} = \\frac{165/15}{1575/15} = \\frac{11}{105}$ which corresponds to the fraction in its lowest terms, because it cannot be further simplified. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us.", "Not really. Let's see an example: Let us compute the GCD for $$n_1 = 165$$ and $$n_2 = 1575$$. Let us find the prime decomposition of each of these numbers (you can use our prime decomposition calculator) $165 = 3 \\cdot 5 \\cdot 11$ $1575 = 3^2 \\cdot 5^2 \\cdot 7$ From the above: what primes do these two numbers have in common? As we can see, the common primes are 3 and 5. Looking at the exponents of these common primes in each of the numbers, we look the minimum between the two. In this case, the minimum exponent for 3 is 1, and the minimum exponent for 5 is also 1. Therefore $GCD = 3^1 \\cdot 5^1 = 3 \\cdot 5 = 15$ Aside from the GDC calculator, you can choose among our selection of algebra calculators and solvers . ## Example: GCD calculation Calculate the greatest common divisor for 12 and 18. Solution: In this case, the numbers are small enough to do it by inspection. Let's go seeing numbers: 2 divides both 12 and 18. Then 3 also divides 12 and 18. The number 4 divides 12 but does not divide 18. The number 5 does not divide either of them. Number 6, on the other hand divides both 12 and 18. Now, any", "of $2$, $a^2$ must have prime factorizations wherein each unique prime number will have an even exponent. Let’s have an example to amplify what I meant above. Suppose $a = 3,780$. Breaking it down as a product of prime numbers, we get $a = 2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7$. We can condense the prime factorization by rewriting it as $a = {2^2} \\cdot {3^3} \\cdot 5 \\cdot 7$. Notice how we removed duplicates of prime numbers by expressing it as factors of primes with each unique prime having the appropriate power. The exponent tells how many times the prime number appears in the prime factorization. Notice that in the prime factorization of integer $\\color{blue}\\large{a}$, the prime numbers can either have an odd or even exponent. This is an important observation that we will take advantage of later. The next step is to square the integer $\\color{blue}\\large{a}$, thus we have $\\color{blue}\\large{a^2}$. Apply the Power of a Power Rule of Exponent. We will use this property to square the integer $\\color{blue}\\large{a}$. In a nutshell, this exponent rule will allow us to distribute the outermost exponent $2$ to the exponents of the unique prime numbers inside the parenthesis. After squaring the integer $\\color{blue}\\large{a}$, the exponents of the unique prime factors of $\\color{blue}\\large{a}$ are", "Re: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. [#permalink] ### Show Tags 21 Jan 2013, 10:40 1 KUDOS Expert's post chiccufrazer1 wrote: Bunuel wrote: ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. @bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance Posted from my mobile device Not sure I understand your post. First of all, 2^96-2^96=0 not 2. Next, $$2^4*2^{96}-2^{96}$$ equals to $$2^{96}*3*5$$ not 32. _________________ Manager Joined: 09 Jul 2007 Posts: 242 Followers: 3 Kudos [?]: 221 [0], given: 0 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 16:25 2^100 - 2^96=2^96 ( 2^4-1)=2^96*3*5 so greatest prime factor =5 VP Joined: 17 Jun 2008 Posts: 1397 Followers: 8 Kudos [?]: 290 [0], given:", "\\cdot 11$ $1575 = 3^2 \\cdot 5^2 \\cdot 7$ From the above: what primes do these two numbers have in common? As we can see, the common primes are 3 and 5. Looking at the exponents of these common primes in each of the numbers, we look the minimum between the two. In this case, the minimum exponent for 3 is 1, and the minimum exponent for 5 is also 1. Therefore $GCD = 3^1 \\cdot 5^1 = 3 \\cdot 5 = 15$ Now, all we have to do is to simplify the original fraction by 15: $\\frac{165}{1575} = \\frac{165/15}{1575/15} = \\frac{11}{105}$ which corresponds to the fraction in its lowest terms, because it cannot be further simplified. The question is: why do we reduce fractions to lowest terms? Well, the answer is easy. Like in everything in Math, simplicity is the most desirable condition. Why work with a cumbersome fraction if we can reduce to an equivalent fraction that is simpler? For other algebra calculators, you can try our section of algebraic solvers and calculators, where you will find many more calculators. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us.", "# Help with derivation of GCF and LCM So I have this problem. I need help with the process in finding the LCM and GCF. I need a technique which is fast but yet accurate. My problem is find the the GCF and LCM for (24, 56, 96). LCM = 672 GCF = 8. I used \"Factors Grid\" and \"Prime Factorization\" but can't derive on the answers above. I've manage to get the GCF using \"Factors Grid\". Need help how the answer were derive. Thank You. - See this also: math-only-math.com/… – Babak S. Jan 27 '13 at 16:14 Here are the prime factorizations of your three numbers: $$\\begin{array}{rlll} 24 = & 2^3&\\cdot 3^1&\\cdot 7^0\\\\ 56 = & 2^3& \\cdot 3^0& \\cdot 7^1\\\\ 96 = & 2^5&\\cdot 3^1&\\cdot 7^0 \\end{array}$$ To find the GCF, take the minimum exponent for each prime; this gives $2^3\\cdot3^0\\cdot7^0 = 8$. To find the LCM, take the maximum exponent for each prime; this gives $2^5\\cdot3^1\\cdot7^1 = 672$. - First notice that each of $24, 56, 96$ is even. This means that each has a factor of 2. If a set of numbers has a common factor, this will also be a factor of the GCF. You can divide the original numbers by this common factor and work with the resulting numbers instead; they", "# Find the greatest common factor of 28 and 36? Feb 5, 2018 $4$ #### Explanation: Using the prime factorization method, $28 = {2}^{2} \\cdot 7$ $36 = {2}^{2} \\cdot 9$ So, the greatest common factor is ${2}^{2}$. ${2}^{2} = 4$ $\\therefore G C F$ of $28$ and $36$ is $4$.", "common multiple of 96 and 24 is 96. Taking the above into account you also know how to find all the common multiples of 96 and 24, not just the smallest. In the next section we show you how to calculate the lcm of ninety-six and twenty-four by means of two more methods. ## How to find the LCM of 96 and 24 The least common multiple of 96 and 24 can be computed by using the greatest common factor aka gcf of 96 and 24. This is the easiest approach: lcm (96,24) = $\\frac{96 \\times 24}{gcf(96,24)} = \\frac{2304}{24}$ = 96 Alternatively, the lcm of 96 and 24 can be found using the prime factorization of 96 and 24: • The prime factorization of 96 is: 2 x 2 x 2 x 2 x 2 x 3 • The prime factorization of 24 is: 2 x 2 x 2 x 3 • Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(96,96) = 96 In any case, the easiest way to compute the lcm of two numbers like 96 and 24 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example,", "# How do you find the greatest common factor of 45y^{12}+30y^{10}? Jun 11, 2018 $15 {y}^{10}$ #### Explanation: The largest number that goes into 45 and 30 is 15 ${y}^{10}$ is the biggest term that goes into ${y}^{12} \\mathmr{and} {y}^{10}$ Jun 11, 2018 The greatest common factor is $15 {y}^{10}$. #### Explanation: I suppose you're asking for the greatest common factor between $45 {y}^{12}$ and $30 {y}^{10}$, since the greatest common factor is computed between two number. We need to find the greatest common factor for both the numeric and literal part. For the numeric part, we can use the prime factorization of the two numbers: $45 = 9 \\cdot 5 = {3}^{2} \\cdot 5$ $30 = 3 \\cdot 10 = 2 \\cdot 3 \\cdot 5$ So, what's the biggest number that \"fits\" inside both $45$ and $30$, given their prime factorizations? Well, we can't choose $2$, because it fits in $30$ but not in $45$. $3$ appears in both factorization, but we can pick it only once, since two three's (i.e. $9$) fit inside $45$, but not inside $30$. Finally, we can pick $5$ once since it appears in both factorizations. So, the answer is $3 \\cdot 5 = 15$ As for the literal part, we have a similar way to proceed: since \"there are\" $12$ $y$'s", "$2$, $a^2$ must have prime factorizations wherein each unique prime number will have an even exponent. Let’s have an example to amplify what I meant above. Suppose $a = 3,780$. Breaking it down as a product of prime numbers, we get $a = 2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7$. We can condense the prime factorization by rewriting it as $a = {2^2} \\cdot {3^3} \\cdot 5 \\cdot 7$. Notice how we removed duplicates of prime numbers by expressing it as factors of primes with each unique prime having the appropriate power. The exponent tells how many times the prime number appears in the prime factorization. Notice that in the prime factorization of integer $\\color{blue}\\large{a}$, the prime numbers can either have an odd or even exponent. This is an important observation that we will take advantage of later. The next step is to square the integer $\\color{blue}\\large{a}$, thus we have $\\color{blue}\\large{a^2}$. Apply the Power of a Power Rule of Exponent. We will use this property to square the integer $\\color{blue}\\large{a}$. In a nutshell, this exponent rule will allow us to distribute the main exponent $2$ to the exponents of the unique prime numbers. After squaring the integer $\\color{blue}\\large{a}$, the exponents of the unique prime factors of $\\color{blue}\\large{a}$ are now ALL even numbers.", "# Difference between revisions of \"Greatest common divisor\" The greatest common divisor (GCD, or GCF (greatest common factor)) of two or more integers is the largest integer that is a divisor of all the given numbers. The GCD is sometimes called the greatest common factor (GCF). A very useful property of the GCD is that it can be represented as a sum of the given numbers with integer coefficients. From here it immediately follows that the greatest common divisor of several numbers is divisible by any other common divisor of these numbers. ## Finding the GCD ### Using prime factorization Once the prime factorizations of the given numbers have been found, the greatest common divisor is the product of all common factors of the numbers. Example: $270=2\\cdot3^3\\cdot5$ and $144=2^4\\cdot3^2$. The common factors are 2 and $3^2$, so $GCD(720,144)=2\\cdot3^2=18$. ### Euclidean algorithm The Euclidean algorithm is much faster and can be used to give the GCD of any two numbers without knowing their prime factorizations. To find the greatest common divisor of more than two numbers, one can use the recursive formula $GCD(a_1,\\dots,a_n)=GCD(GCD(a_1,\\dots,a_{n-1}),a_n)$. ### Using least common multiple The GCD of two numbers can also be found using the equation $GCD(x, y) \\cdot LCM(x, y) = x \\cdot y$, where $LCM(x,y)$ is the least common multiple of $x$ and" ]

[ "BOOKMARKED ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. _________________ Kudos [?]: 129174 [2], given: 12194 Manager Joined: 04 Jan 2013 Posts: 79 Kudos [?]: 17 [0], given: 1 Re: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. [#permalink] ### Show Tags 21 Jan 2013, 11:33 Bunuel wrote: ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. @bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance Posted from my", "# 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 $2 \\cdot 2 \\cdot 2 = 8$", "can learn how to calculate the greatest common denominator. How to calculate the greatest common denominator For small numbers finding the greatest common denominator is almost intuitive: look at this set: $\\{6,12,9\\}$ Clearly the GCD is $3$! But what do we do when the numbers start to grow? Prime factorization The GCD is ready to be found if you write the prime factorization of the numbers in a set. The prime factorization is the set of prime numbers elevated to specific exponents that, multiplied, returns the original number. The GCD is the largest shared prime factor, elevated to the highest possible exponent. Take this set: $\\{360,378,405\\}$. The prime factorizations of these numbers are: $\\begin{split} 360 &= 2^3\\cdot3^2\\cdot5\\\\ 378 &= 2\\cdot3^3\\cdot7\\\\ 405 &= 3^4\\cdot5 \\end{split}$ Find the repeated prime factor: only $3$ appears in all three numbers, and the highest exponent is $2$: $3^2$ is the GCD of the set! 🙋 Finding the prime factorization is not always a simple task: to simplify your calculations you can use our prime factorization calculator! Euclidean algorithm The GCD of two numbers also divides exactly their difference. It may sound strange, but think about it: the result of the division of the largest number by the GCD can be seen as the sum of two numbers: one of them, multiplied by the", "Re: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. [#permalink] ### Show Tags 21 Jan 2013, 10:40 1 KUDOS Expert's post chiccufrazer1 wrote: Bunuel wrote: ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. @bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance Posted from my mobile device Not sure I understand your post. First of all, 2^96-2^96=0 not 2. Next, $$2^4*2^{96}-2^{96}$$ equals to $$2^{96}*3*5$$ not 32. _________________ Manager Joined: 09 Jul 2007 Posts: 242 Followers: 3 Kudos [?]: 221 [0], given: 0 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 16:25 2^100 - 2^96=2^96 ( 2^4-1)=2^96*3*5 so greatest prime factor =5 VP Joined: 17 Jun 2008 Posts: 1397 Followers: 8 Kudos [?]: 290 [0], given:", "## Recounting the Rationals, part IVb: the Euclidean Algorithm Suppose we have two integers, and we’d like to find their greatest common divisor (GCD). Recall that the greatest common divisor of two integers m and n is exactly that: the greatest integer which is a divisor of both m and n. The obvious way to do this—which is probably the way you learned in elementary school, when reducing fractions—is to factor each integer into primes, and look for overlap. For example, let’s say we want to find the GCD of 450 and 525. We begin by factoring: $\\begin{array}{rcl} 450 & = & 2 \\cdot 3^2 \\cdot 5^2 \\\\ 525 & = & 3 \\cdot 5^2 \\cdot 7 \\end{array}$ Now we pull out as many prime factors as we can which are contained in the factorizations of both numbers. We can’t use 2 at all, since it is only contained in the factorization of 450; we can use a factor of 3, but only one; and we can use two factors of 5. Putting these together, the greatest common divisor of 450 and 525 is $3 \\cdot 5^2 = 75$. This method is intuitive, and works well for relatively small numbers, but (there’s always a “but”, isn’t there?) it fails horribly for larger numbers. For example, what if you", "Greatest Common Divisor of two numbers If $ab=600$ how large can the greatest common divisor of $a$ and $b$ be? I am not sure if I should check for all factor multiples of $a$ and $b$ for this question. Please advise. We have $600=2^3\\cdot 5^2\\cdot 3$. To make the gcd large, we give a $2$ to each of $a$ and $b$, also a $5$. So the largest possible gcd is $10$. Remark: The idea generalizes. Let $n$ have prime power factorization $$n=p_1^{d_1}p_2^{d_2}\\cdots p_k^{d_k}.$$ Let $e_i=\\lfloor d_i/2\\rfloor$, where $\\lfloor x\\rfloor$ is the greatest integer that is $\\le x$. Then the greatest possible gcd of $a$ and $b$, where $ab=n$, is $$p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k}.$$ • You are welcome. For the maximum, we distribute the power of any prime $p$ as evenly as possible between $a$ and $b$. – André Nicolas Jun 10 '13 at 2:00 $$ab=600=2^3\\cdot 3\\cdot 5^2$$ The greatest common divisor of $a,b$ will have the greatest number of common prime factors that you can arrange.", "## Discrete Mathematics and Its Applications, Seventh Edition To obtain the greatest common divisor of each pair of integers, multiply together the factors shared by both numbers. a) $2^2\\cdot 3^3\\cdot 5^2=2700$ b) $2\\cdot 3\\cdot 11=66$ c) $17$ d) $1$ e) These numbers are relatively prime. f) $2\\cdot 3\\cdot 5\\cdot 7=210$", "=& {2}^{4}\\cdot 5\\cdot 7\\cdot \\mathrm{11}\\hfill \\end{array}$ 2. The common bases are 2 and 5 3. The smallest exponents appearing on 2 and 5 in the prime factorizations are, respectively, 2 and 1. ${2}^{2}$ from 700. ${5}^{1}$ from either 1,880 or 6,160. 4. The GCF is the product of these numbers. GCF is ${2}^{2}\\cdot 5=4\\cdot 5=\\text{20}$ Thus, 20 is the largest number that divides 700, 1,880, and 6,160 without a remainder. Practice set a Find the GCF of the following numbers. 24 and 36 12 48 and 72 24 50 and 140 10 21 and 225 3 450, 600, and 540 30 Exercises For the following problems, find the greatest common factor (GCF) of the numbers. 6 and 8 2 5 and 10 8 and 12 4 9 and 12 20 and 24 4 35 and 175 25 and 45 5 45 and 189 66 and 165 33 264 and 132 99 and 135 9 65 and 15 33 and 77 11 245 and 80 351 and 165 3 60, 140, and 100 147, 343, and 231 7 24, 30, and 45 175, 225, and 400 25 210, 630, and 182 14, 44, and 616 2 1,617, 735, and 429 1,573, 4,862, and 3,553 11 3,672, 68, and 920 7, 2,401, 343, 16, and 807 1 500, 77, and", "# How do you find the greatest common factor of 45y^{12}+30y^{10}? Jun 11, 2018 $15 {y}^{10}$ #### Explanation: The largest number that goes into 45 and 30 is 15 ${y}^{10}$ is the biggest term that goes into ${y}^{12} \\mathmr{and} {y}^{10}$ Jun 11, 2018 The greatest common factor is $15 {y}^{10}$. #### Explanation: I suppose you're asking for the greatest common factor between $45 {y}^{12}$ and $30 {y}^{10}$, since the greatest common factor is computed between two number. We need to find the greatest common factor for both the numeric and literal part. For the numeric part, we can use the prime factorization of the two numbers: $45 = 9 \\cdot 5 = {3}^{2} \\cdot 5$ $30 = 3 \\cdot 10 = 2 \\cdot 3 \\cdot 5$ So, what's the biggest number that \"fits\" inside both $45$ and $30$, given their prime factorizations? Well, we can't choose $2$, because it fits in $30$ but not in $45$. $3$ appears in both factorization, but we can pick it only once, since two three's (i.e. $9$) fit inside $45$, but not inside $30$. Finally, we can pick $5$ once since it appears in both factorizations. So, the answer is $3 \\cdot 5 = 15$ As for the literal part, we have a similar way to proceed: since \"there are\" $12$ $y$'s", "6. Therefore, the greatest common divisor of 18 and 30 is 6. ### Prime factorization Another way of finding the greatest common divisor of two or more numbers is through prime factorization. In this method, each number is factored so that it is the product of prime numbers. For example, the prime factorization of 1,176 gives us $1,176 = 2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 7 \\cdot 7 = 2^3 \\cdot 3 \\cdot 7^2$. The prime factorization of 2,100 is $2^2 \\cdot 3 \\cdot 5^2 \\cdot 7$. To find the greatest common Divisor of 1,176 and 2,100, the smallest powers of each prime divisor for each of the numbers (with $p^0$ being assumed if one number doesn't contain a particular prime factor $p$) are multiplied together, as in $2^2 \\cdot 3^1 \\cdot 5^0 \\cdot 7^1 = 84$. Therefore, 84 is the greatest common factor of 1,176 and 2,100. ### Euclidean Algorithm For large numbers with many divisors, especially large prime divisors, these methods are somewhat inefficient. A systematic way of finding the greatest common divisor is the Euclidean Algorithm. This algorithm has the advantage of being easily written for a computer program. In the Euclidean Algorithm, the division algorithm is applied to two numbers until a remainder of zero is found. How this works is that", "# Find the greatest common factor of 28 and 36? Feb 5, 2018 $4$ #### Explanation: Using the prime factorization method, $28 = {2}^{2} \\cdot 7$ $36 = {2}^{2} \\cdot 9$ So, the greatest common factor is ${2}^{2}$. ${2}^{2} = 4$ $\\therefore G C F$ of $28$ and $36$ is $4$.", "## Thinking Mathematically (6th Edition) The greatest common divisor of the two numbers is $15$. Factor each number completely to have: $240=10(24)=2(5)(2)(2)(2)(3) \\\\285=5(57)=5(3)(19)$ Thus, the greatest common divisor of the two numbers is $3(5)=15$.", "wrote: Bunuel wrote: ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. @bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance Posted from my mobile device Not sure I understand your post. First of all, 2^96-2^96=0 not 2. Next, $$2^4*2^{96}-2^{96}$$ equals to $$2^{96}*3*5$$ not 32. _________________ Current Student Joined: 12 Aug 2015 Posts: 2621 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: What is the greatest prime factor of 2^100 - 2^96? [#permalink] ### Show Tags 14 Mar 2016, 01:16 here taking 2^96 out as a common factor we are left with 15=5 x 3 hence 5 is the highest common factor _________________ Manager Joined: 09 Jun 2015 Posts: 92 Re: What is the greatest prime factor of 2^100 - 2^96? [#permalink] ###", "common multiple of 96 and 24 is 96. Taking the above into account you also know how to find all the common multiples of 96 and 24, not just the smallest. In the next section we show you how to calculate the lcm of ninety-six and twenty-four by means of two more methods. ## How to find the LCM of 96 and 24 The least common multiple of 96 and 24 can be computed by using the greatest common factor aka gcf of 96 and 24. This is the easiest approach: lcm (96,24) = $\\frac{96 \\times 24}{gcf(96,24)} = \\frac{2304}{24}$ = 96 Alternatively, the lcm of 96 and 24 can be found using the prime factorization of 96 and 24: • The prime factorization of 96 is: 2 x 2 x 2 x 2 x 2 x 3 • The prime factorization of 24 is: 2 x 2 x 2 x 3 • Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(96,96) = 96 In any case, the easiest way to compute the lcm of two numbers like 96 and 24 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example,", "What is the least common denominator of 3/4, 3/8, and 1/5? Apr 13, 2016 $40$ Explanation: If we look at the prime factorizations of the denominators, we have $4 = {2}^{2}$ $8 = {2}^{3}$ $5 = {5}^{1}$ The least common denominator will be the minimal product containing all factors above to their appropriate powers. In this case, that would be ${2}^{3} \\cdot 5$. Thus, the least common denominator is ${2}^{3} \\cdot 5 = 8 \\cdot 5 = 40$", "# Greatest Common Factor of Monomials To find the greatest common factor of two monomials, first find the prime factorization of each monomial, including all the variables (and a $1$ factor if necessary). Then take the product of all common factors. Example: Find the GCF of: $-27{p}^{2}q{r}^{5}$ and $15{p}^{3}{r}^{3}$ First, find the prime factorization of each monomial. $\\begin{array}{l}-27{p}^{2}q{r}^{5}=-1\\cdot {3}\\cdot 3\\cdot 3\\cdot {p}\\cdot {p}\\cdot q\\cdot {r}\\cdot {r}\\cdot {r}\\cdot r\\cdot r\\\\ 15{p}^{3}{r}^{3}={3}\\cdot 5\\cdot {p}\\cdot {p}\\cdot p\\cdot {r}\\cdot {r}\\cdot {r}\\end{array}$ Common factors are shown in red. Their product is: $3\\cdot p\\cdot p\\cdot r\\cdot r\\cdot r$ So, the GCF is $3{p}^{2}{r}^{3}$ .", "Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. _________________ Manager Joined: 04 Jan 2013 Posts: 69 Re: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. [#permalink] ### Show Tags 21 Jan 2013, 10:33 Bunuel wrote: ebliss wrote: Can someone please explain how we can go from 2^4 * 2^96 - 2^96 to 2^96 (16-1) ? Thanks! Sure. What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 $$2^{100} - 2^{96}=2^4*2^{96}-2^{96}$$. Now, factor out 2^{96}: $$2^4*2^{96}-2^{96}=2^{96}(2^4-1)=2^{96}*15=2^{96}*3*5$$. The greatest prime factor is 5. Hope it's clear. @bunuel..2^96-2^96=2^1 according to index laws..in our factorised equation 2^4*2^96-2^96 the answer is 16*2=32..can we do prime factorisation on 32 where by k/2 +k/3 +k/5--->32/2+32/3+32/5?would we just take the 5 as our largest factor?but we know 5 is not a factor of 32..where am i missing the details please tell me..thanks in advance Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 52917 Re: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. [#permalink] ### Show Tags 21 Jan 2013, 10:40 1 chiccufrazer1", "# 5. The least common multiple of the first positive integers Prerequirements: We know that a way to compute the least common multiple between two or more integer numbers is to decompose them into prime factors, and then taking each distinct prime factor with its greatest power in the considered numbers. Going on this way, the factorization of the least common multiple will result. For example, given the three numbers $4 = 2^2$, $6 = 2 \\cdot 3$ and $45 = 3^2 \\cdot 5$, and computing their least common multiple, we see that the distinct prime factors are 2, 3 and 5; 2 appears with the greatest exponent 2 in 4, 3 appears with the greatest exponent 2 in 45, 5 appears with the greatest exponent 1 again in 45; so the least common multiple we are looking for is $2^2 \\cdot 3^2 \\cdot 5 = 180$. There exists a particularly interesting case in which the least common multiple can be computed through a reasoning, without decomposing the numbers into prime factors: it’s the computation of the least common multiple of the positive integer numbers up to a certain number $x$, that is $\\mathrm{LCM}(1, 2, \\dots, x)$. Of course this expression coincides with $\\mathrm{LCM}(2, \\dots, x)$ if $x \\geq 2$, as in this case the number 1 does", "# Difference between revisions of \"Greatest common divisor\" The greatest common divisor (GCD, or GCF (greatest common factor)) of two or more integers is the largest integer that is a divisor of all the given numbers. The GCD is sometimes called the greatest common factor (GCF). A very useful property of the GCD is that it can be represented as a sum of the given numbers with integer coefficients. From here it immediately follows that the greatest common divisor of several numbers is divisible by any other common divisor of these numbers. ## Finding the GCD ### Using prime factorization Once the prime factorizations of the given numbers have been found, the greatest common divisor is the product of all common factors of the numbers. Example: $270=2\\cdot3^3\\cdot5$ and $144=2^4\\cdot3^2$. The common factors are 2 and $3^2$, so $GCD(720,144)=2\\cdot3^2=18$. ### Euclidean algorithm The Euclidean algorithm is much faster and can be used to give the GCD of any two numbers without knowing their prime factorizations. To find the greatest common divisor of more than two numbers, one can use the recursive formula $GCD(a_1,\\dots,a_n)=GCD(GCD(a_1,\\dots,a_{n-1}),a_n)$. ### Using least common multiple The GCD of two numbers can also be found using the equation $GCD(x, y) \\cdot LCM(x, y) = x \\cdot y$, where $LCM(x,y)$ is the least common multiple of $x$ and", "# IMAT 2019 Q57 [Common factor] In order to solve this question, we must find the factors of each of the number: 360 = 36 \\cdot 10 = 6 \\cdot 6 \\cdot 5\\cdot 2 = 3^2\\cdot 2 ^2 \\cdot 5\\cdot 2 = 2^3 \\cdot 3^2 \\cdot 5. 500 = 5 \\cdot 10^2 = 5 \\cdot (5\\cdot 2)^2 = 5 \\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^3. 700 = 7 \\cdot 10^2 = 7\\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^2\\cdot 7. From this we can see that all of the numbers have 2^2 \\cdot 5 in their prime factors, so the highest common factor is (D) 2^2 \\cdot 5 3 Likes" ]

[ "the wanted plane is (taking $5C$) : $-9x+2y+5z+17=0$. Check this Tonio 3. hey thanks for the reply. the only part i am still confused about is that in the question it says it wants it in the form with minus D (Ax+By+Cy - D=0) and in your answer you have +17. Is that ok? or am i supposed to change my answer in some way to account for this? thanks 4. Originally Posted by kblythe hey thanks for the reply. the only part i am still confused about is that in the question it says it wants it in the form with minus D (Ax+By+Cy - D=0) and in your answer you have +17. Is that ok? or am i supposed to change my answer in some way to account for this? thanks Then multiply the entire equation by -1! $-1(-9x+2y+5z+17)=-1(0)$ so 9x- 2y- 5z- 17= 0.", "are the same idea, just integrating over different regions of the (X,Y) plane. For example, P(max(X,Y)>x) would be $$1 - \\int _{{-\\infty}}^{x} \\int _{{-\\infty}}^{x} \\Muserfunction{pdf}(X,Y)\\,dY\\,dX$$ I am sure you can get the rest of these similarly. -Dale 5. Dec 12, 2005", "## Problem 8 In the product shown, $\\text{B}$ is a digit. The value of $\\text{B}$ is $$\\begin{array}{rr} &\\text{B}2 \\\\ \\times &7\\text{B} \\\\ \\hline &6396 \\\\ \\end{array}$$ $\\text{(A)}\\ 3 \\qquad \\text{(B)}\\ 5 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 7 \\qquad \\text{(E)}\\ 8$ ## Problem 9 Using only the paths and the directions shown, how many different routes are there from $\\text{M}$ to $\\text{N}$? $[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(6,0),MidArrow); draw(6*dir(60)--5.1961524227066318805823390245176*dir(30),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--3*dir(60),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(3,0),MidArrow); label(\"M\",6*dir(60),N); label(\"N\",(6,0),SE); label(\"A\",3*dir(60),NW); label(\"B\",5.1961524227066318805823390245176*dir(30),NE); label(\"C\",(3,0),S); label(\"D\",(0,0),SW); [/asy]$ $\\text{(A)}\\ 2 \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 4 \\qquad \\text{(D)}\\ 5 \\qquad \\text{(E)}\\ 6$ ## Problem 10 A picture $3$ feet across is hung in the center of a wall that is $19$ feet wide. How many feet from the end of the wall is the nearest edge of the picture? $\\text{(A)}\\ 1\\frac{1}{2} \\qquad \\text{(B)}\\ 8 \\qquad \\text{(C)}\\ 9\\frac{1}{2} \\qquad \\text{(D)}\\ 16 \\qquad \\text{(E)}\\ 22$ ## Problem 11 If $\\text{A}*\\text{B}$ means $\\frac{\\text{A}+\\text{B}}{2}$, then $(3*5)*8$ is $\\text{(A)}\\ 6 \\qquad \\text{(B)}\\ 8 \\qquad \\text{(C)}\\ 12 \\qquad \\text{(D)}\\ 16\\qquad \\text{(E)}\\ 30$ ## Problem 12 The table below displays the grade distribution of the $30$ students in a mathematics class on the last two tests. For example, exactly one student received a 'D' on Test 1 and a 'C' on Test 2 (see circled entry). What percent of the students received the same grade", "+0 # Help 0 179 1 In the diagram below, we have $XY = 2\\cdot AX = 2\\cdot YB$ and $\\dfrac{AZ}{ZC} = \\dfrac{2}{3}$. Find $\\dfrac{[YBC]}{[ZYC]}$. [asy] pair A, X, Y, B, Z, C; A = (0,0); X = (0.3,-0.2); Z = (0.4,0.2); Y = 3*X; B = 4*X; C = 2.5*Z; draw(X--Z--Y--C--B--A--C); label(\"$A$\",A,W); label(\"$B$\",B,S); label(\"$C$\",C,N); label(\"$X$\",X,SW); label(\"$Y$\",Y,SW); label(\"$Z$\",Z,NW); [/asy] Sep 27, 2021", "person. If Chris takes the first and last piece of candy, then the number of students at the table could be $\\text{(A)}\\ 10 \\qquad \\text{(B)}\\ 11 \\qquad \\text{(C)}\\ 19 \\qquad \\text{(D)}\\ 20 \\qquad \\text{(E)}\\ 25$ ## Problem 23 The graph relates the distance traveled [in miles] to the time elapsed [in hours] on a trip taken by an experimental airplane. During which hour was the average speed of this airplane the largest? $[asy] unitsize(12); for(int a=1; a<13; ++a) { draw((2a,-1)--(2a,1)); } draw((-1,4)--(1,4)); draw((-1,8)--(1,8)); draw((-1,12)--(1,12)); draw((-1,16)--(1,16)); draw((0,0)--(0,17)); draw((-5,0)--(33,0)); label(\"0\",(0,-1),S); label(\"1\",(2,-1),S); label(\"2\",(4,-1),S); label(\"3\",(6,-1),S); label(\"4\",(8,-1),S); label(\"5\",(10,-1),S); label(\"6\",(12,-1),S); label(\"7\",(14,-1),S); label(\"8\",(16,-1),S); label(\"9\",(18,-1),S); label(\"10\",(20,-1),S); label(\"11\",(22,-1),S); label(\"12\",(24,-1),S); label(\"Time in hours\",(11,-2),S); label(\"500\",(-1,4),W); label(\"1000\",(-1,8),W); label(\"1500\",(-1,12),W); label(\"2000\",(-1,16),W); label(rotate(90)*\"Distance traveled in miles\",(-4,10),W); draw((0,0)--(2,3)--(4,7.2)--(6,8.5)); draw((6,8.5)--(16,12.5)--(18,14)--(24,15)); [/asy]$ $\\text{(A)}\\ \\text{first (0-1)} \\qquad \\text{(B)}\\ \\text{second (1-2)} \\qquad \\text{(C)}\\ \\text{third (2-3)} \\qquad \\text{(D)}\\ \\text{ninth (8-9)} \\qquad \\text{(E)}\\ \\text{last (11-12)}$ ## Problem 24 Three $\\Delta$'s and a $\\diamondsuit$ will balance nine $\\bullet$'s. One $\\Delta$ will balance a $\\diamondsuit$ and a $\\bullet$. $[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label(\"\\Delta \\hspace{2 mm}\\Delta \\hspace{2 mm}\\Delta \\hspace{2 mm}\\diamondsuit \",(-12,6.5),S); label(\"\\bullet \\hspace{2 mm}\\bullet \\hspace{2 mm}\\bullet \\hspace{2 mm} \\bullet \",(12,5.2),N); label(\"\\bullet \\hspace{2 mm}\\bullet \\hspace{2 mm}\\bullet \\hspace{2 mm}\\bullet \\hspace{2 mm}\\bullet \",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label(\"\\Delta \",(34,6.5),S); label(\"\\bullet \\hspace{2 mm}\\diamondsuit \",(54,6.5),S); [/asy]$ How many $\\bullet$'s will balance the two $\\diamondsuit$'s in this balance? $[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label(\"?\",(-12,8.5),S); label(\"\\diamondsuit \\hspace{2", "using the intercepts of (6,0) & (0,-4). • We will then label the line $2x-3y=12$ and we will go over the idea that this line represents all of the ordered pairs for which $2x-3y$ is equal to 12. • I will explain that this line divides the rectangular plane into two half-planes. One of the half-planes contains points for which $2x-3y<12$ and $2x-3y>12$. • I’ll pick a point below the line to the right, like (6,-5) and substitute the coordinates into the expression $2x-3y$, which will result in 27. Since $27>12$, we will then label that half-plane as $2x-3y>12$. • I’ll pick the origin, which is above the line to the left, and substitute the coordinates into the expression $2x-3y$, which will result in 0. Since $0<12$, we will then label that half-plane as $2x-3y<12$. • We will finish by shading the side labeled $2x-3y<12$. After 2 or 3 examples done this way, I think the students will understand what each half-plane represents and have an easier time determining which half-plane to shade. -George I am a math instructor at College of the Sequoias in Visalia, CA. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page", "+ .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", "(y1+y2)/2 This simulates drawing $10^8$ pairs of points (x1, y1), (x2, y2)", "the question, sure you won't get the value of $X$.", "$$-1$$ $$-1$$ $$0$$ $$1$$ $$1$$ $$3$$ $$2$$ $$5$$ we will then draw them in the plane $$XY$$ and will join them with a line. Finally we obtain: where the black points are the values in the table. This procedure (doing the table) can be very useful when we must draw a function, but sometimes might not be useful, since very different functions exist and little time is needed to only find a few points to be able to draw the function.", "The image of the point $(-1, 3, 4)$ in the plane $x - 2y = 0$ is : ( A ) $(8, 4, 4)$ ( B ) $(15, 11, 4)$ ( C ) $(-\\frac{17}{3}, -\\frac{19}{3}, 1)$ ( D ) $(\\frac{9}{5}, \\frac{-13}{5}, 4)$", "\\\\ 2 \\end{matrix}$$ both have margins $$((10, 4, 2), (9, 4, 2, 1), (8, 4, 2, 1, 1))$$. (I was able to get Mathematica to check for equal margins among all plane partitions through $$n = 16$$ and might be able to go a little farther. Luckily these examples came up within the bounds of my programming ability and computer power.)", "draw(D--I); draw(I--H); draw(H--G); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,W); label(\"E\",E,W); label(\"F\",F,E); label(\"G\",G,E); label(\"H\",H,SE); label(\"I\",I,SW); [/asy]$ $\\textbf{(A)}\\ \\dfrac{12+3\\sqrt3}4\\qquad\\textbf{(B)}\\ \\dfrac92\\qquad\\textbf{(C)}\\ 3+\\sqrt3\\qquad\\textbf{(D)}\\ \\dfrac{6+3\\sqrt3}2\\qquad\\textbf{(E)}\\ 6$ ## Problem 14 The $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\\triangle APQ$? $\\textbf{(A)}\\ 45\\qquad\\textbf{(B)}\\ 48\\qquad\\textbf{(C)}\\ 54\\qquad\\textbf{(D)}\\ 60\\qquad\\textbf{(E)}\\ 72$ ## Problem 15 David drives from his home to the airport to catch a flight. He drives $35$ miles in the first hour, but realizes that he will be $1$ hour late if he continues at this speed. He increases his speed by $15$ miles per hour for the rest of the way to the airport and arrives $30$ minutes early. How many miles is the airport from his home? $\\textbf{(A) }140\\qquad \\textbf{(B) }175\\qquad \\textbf{(C) }210\\qquad \\textbf{(D) }245\\qquad \\textbf{(E) }280\\qquad$ ## Problem 16 In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\\overline{BC}$, $\\overline{CD}$, and $\\overline{AD}$, respectively. Point $H$ is the midpoint of $\\overline{GE}$. What is the area of the shaded region? $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X", "for the plane is: $$-3x-3y=-3$$ which is equal to: $$-x-y=-1$$ -", "the planes, $8x – 7y + 4z = 12$ and $32x -28y + 16z = 56$ , are parallel. 8. What must be the value of $n$ so that the planes shown below are parallel? $48x – 24y + 30z = 90$ $8x – (5n + 4) y + 5z = 15$", "cuts the surface are those points where the elevation of the surface is 8 units above the $$xy$$-plane. The figure below shows the surface being sliced by the planes $$z = 8$$ and $$z = 4$$. Slicing the surface at different elevations and sketching the curves where the plane intersects the surface results in the second graph below. If we move all of those curves to the xy–plane (or, equivalently, view them from directly overhead), the result is a 2-dimensional graph of the level curves of the original surface. This is the contour diagram. #### Example 9 Create a contour diagram for our car rental example with cost function $$C(d,m)=40d+0.15m$$. Draw curves for when the cost is 0, 100, 200, 300, and 400. We'll set $$C(d,m)=40d+0.15m=c$$ for $$c =$$ 0, 100, 200, 300, and 400 and draw the curves in the $$dm$$-plane. The first coordinate of the ordered pair is $$d$$, so the $$d$$-axis will be horizontal; the $$m$$-axis will be vertical. Remember that the domain for this function is really just where $$d \\geq 0$$ and $$m \\geq 0$$, so we will only draw the curves in the first quadrant. When $$c=0$$: \\begin{align*} C(d,m)=& 40d+0.15m=0 \\\\ 0.15m=& -40d \\\\ m=& -\\frac{40}{0.15}d\\approx -267d \\end{align*} This is the equation of a line, with slope about -267, passing through the", "# The value of $k$ for which the line $\\overrightarrow r=(4\\hat i+2\\hat j-\\hat k)+\\lambda(\\hat i-k\\hat j+2\\hat k)$ lies on the plane $2x-4y+z=3$ is ? $(a)\\;1\\qquad(b)\\;-1\\qquad(c)\\;No\\;real\\;value\\;of\\;k\\qquad(d)\\;3$", "# If $(7q+46)^{\\circ}$ and $(3q-6)^{\\circ}$ form linear pair. Find the value of $q$. $( A ) 26^{\\circ} \\\\ ( B ) 14^{\\circ} \\\\ ( C ) 15^{\\circ} \\\\ ( D ) 24^{\\circ}$", "#### Similar Solved Questions ##### Which of the following planes is normal to the line given as 2t + 3 Y = 5t _ 7 T _Select one: a.z = 0 Sx - Zy = 0 2x +Sy + 3 = 112 d.3 7ym4 = 14 2u + St+'=0 Which of the following planes is normal to the line given as 2t + 3 Y = 5t _ 7 T _ Select one: a.z = 0 Sx - Zy = 0 2x +Sy + 3 = 112 d.3 7ym4 = 14 2u + St+'=0... ##### One-product company finds that its prolit; P; in millions dollars _ given by Ihe following equation whcre ot the product. In dollars_ P(a,p) Zap 8Op 15p2 70a?p -120 Find Ihe maximum value of P and Ihe values of a and at which altained.the amount spent on advertising_ millions of dollars. and p is (he price charged por itemThe maximum valuo ol P is altained when a ismillion and p is slLThe maximum value of P is SL Million. one-product company finds that its prolit; P; in millions dollars _ given by Ihe following equation whcre ot the product. In dollars_ P(a,p) Zap 8Op 15p2 70a?p -120 Find Ihe maximum value of P and Ihe values of a and at which altained. the amount", "draw(circle((0.5,0),1)); label(\"\\huge{x}\", (0, 0)); label(\"70-x\", (-1, 0)); label(\"54-x\", (1, 0)); [/asy]$ We know that the sum of all three areas is $93$ So, we have: $$93 = 70-x+x+54-x$$ $$93 = 70+54-x$$ $$93 = 124 - x$$ $$-31=-x$$ $$x=31$$ We are looking for the number of students in only math. This is $70-x$. Substituting $x$ with $31$, our answer is $\\boxed{39}$. -mathnerdnair" ]

[ "equals 40. 360 take away 40 equals $320^\\circ$. So, Roni has concluded that there are three other points where the dot would be the same distance from the horizontal axis as it is for $20^\\circ$. Another way of looking at it would be to say that the dot itself would have to be at $20^\\circ$, $160^\\circ$, $200^\\circ$and $340^\\circ$. George tells us how we can work out how far a point at any position on the circle would have to turn to finish the same height above or below the horizontal axis. He agrees that there will be three other solutions: 1. Double the first degree and take it away from $180^\\circ$. 2. Add $180^\\circ$ to the your starting degree. 3. Double the starting degree and take it away from $360^\\circ$. Well done!", "## Elementary Technical Mathematics The value of $56.1 \\%$ of $360{}^\\circ$ is $202{}^\\circ$. \\begin{align} & 56.1 \\% \\times 360{}^\\circ =\\frac{561}{1000}\\times 360{}^\\circ \\\\ & =201.96{}^\\circ \\end{align} Now round off the result, which gives $202{}^\\circ$. Thus, the value of $56.1 \\%$ of ${{360}^{\\circ }}$is $202{}^\\circ$.", "at the top again, so $-1\\equiv2\\pmod{3}$ We have a rule here and that is, we can add or subtract the number we're doing the modulus to as many times we want, and it'll remain the same. When using degrees, a full turn in a circle is $360^\\circ$, and when we have performed a full turn in a circle, we're just back where we started. This is just like the modulus operation I talked about above, and we can use the modulus concept to explain how angles operate when going below $0^\\circ$ or above $360^\\circ$. When working with degrees, we go back to zero at $360^\\circ$, therefore in degrees it makes sense to say that degrees work using modulus $360^\\circ$. If we understand angles this way, we can explain what a negative angle really is. Let's for example take $-90^\\circ$, we can apply modular arithmetic, which allows us to add $360$ to any angle without changing it: $-90^\\circ\\equiv270^\\circ\\pmod{360^\\circ}$ and if you draw those two angles on the unit circle, notice how the angles point at the same place. All this also explains why going above $360^\\circ$ allows you to go below again, namely $400^\\circ\\equiv40^\\circ\\pmod{360^\\circ}$, and therefore the angle $40^\\circ$ and $400^\\circ$ is just the same thing. When talking about radians, it's much the same, except a full turn is", "or irregular) of a sphere, $E=720^{\\circ}$. For any triangulation of a doughnut, $E=0$. For any triangulation of a doughnut with two holes, $E=-720^{\\circ}$. e.g. For a cube: gap angle at each vertex $=360^{\\circ}-(3\\times 90^{\\circ}) =90^{\\circ}$, number of vertices $=8$ so $E=8\\times 90^{\\circ}=720^{\\circ}$. For a doughnut a simple triangulation looks like the sketch. There are two types of vertices. At type $A$, the gap angle $=360^{\\circ}-(2\\times 90^{\\circ}+2\\times 45 ^{\\circ})=90^{\\circ}$. At type $B$, the gap angle $=360^{\\circ}-(2\\times 135^{\\circ}+2\\times 90^{\\circ})=-90^{\\circ}$. There are 8 of each type of vertex so $E=8\\times 90^{\\circ}+8\\times(-90^{\\circ})=0$. We can stick together two one-holed doughnuts to give a doughnut with two holes, as seen from above in the sketch. There are now 8 type $A$ vertices, 16 type $B$ vertices and 4 type $C$ vertices, where the gap angle is 0. $E=8\\times 90^{\\circ}+16\\times (-90^{\\circ})=-720^{\\circ}$. The general rule is $E=720^{\\circ}\\times(1-h)$ where $h$ is the number of holes (or handles). This does not depend on the triangulation, it is a topological invariant. To see how the proof might go, consider a triangle with angles $a$, $b$, $c$. What is the contribution to $E$? Now replace it by 3 triangles with angles $(a_1, b_1, d_1)$, $(b_2, c_2, d_2)$, $(c_3, a_3, d_3)$. What is the contribution of this to $E$? The theorem is the Gauss- Bonnet theorem which says that the total", "the rotation it increases to $60^\\circ$. It follows that our arc has length $20\\cdot{\\pi\\over6}={10\\pi\\over3}$. This arc is covered four times during a full $2\\pi$-turn of the ellipse vs. the angle. When the ellipse is perfectly vertical, its axes will be aligned with the co-ordinate axes and the line joining its centre and the origin of co-ordinates will make an angle of $60^\\circ$ with the X-axis. This is because the tangent of the angle will be $\\tan\\theta=\\dfrac{10\\sqrt 3}{10}=\\sqrt 3$. Similarly, when the ellipse becomes perfectly horizontal the angle will be $30^\\circ ,\\implies$that the centre travels only $30^\\circ$ on the circle you obtained as the locus. The arc length can now be easily obtained.", "180^\\circ$. This gives the sum of the angles of $P_n$ together with the $360^\\circ$ of the circle at point $O$. So the sum of the angles of $P_n$ is $(n-2) \\times 180^\\circ$. Note that if $P_n$ were not convex then there is no guarantee that there is a point $O$ inside of $P_n$ which can be joined to each vertex of $P_n$ without going outside of the polygon.", "then plot the point. a. $(3, 30^{\\circ}, 3\\sqrt{3})$ b. $(6\\sqrt{3}, 120^{\\circ}, 6)$ c. ty}$\\left(8\\sqrt{2}, \\dfrac{3\\pi}{4}, 8\\sqrt{2} \\right)$ 2. $(3, 150^{\\circ}, 40^{\\circ}) = (-1.67, 0.96, 2.30)$", "a ray stands on a line, then the sum of two adjacent angles so formed is equal to $180^0$.", "= $88°$ (iv) In the given figure, we have a quadrilateral whose sum of all angles is $360°$. Thus, $35°$ + $45°$ + $50°$ + reflex$\\angle ADC$ = $360°$ Or, reflex$\\angle ADC$ = $230°$ $230°$ + x = $360°$ (A complete angle) = x = $130°$", "to angle $X$ will be equal to $180^\\circ - 45^\\circ = 135^\\circ$. When two lines meet at a point in a plane, they are known as intersecting lines. When the lines do not meet at any point in a plane, they are called parallel lines.", "the value of $x$ . $x=\\frac{72^\\circ - 22^\\circ}{2}=\\frac{50^\\circ}{2}=25^\\circ$ . #### Example B Find the value of $x$ . $x=\\frac{120^\\circ - 32^\\circ}{2}=\\frac{88^\\circ}{2}=44^\\circ$ . #### Example C Find the value of $x$ . First note that the missing arc by angle $x$ measures $32^\\circ$ because the complete circle must make $360^\\circ$ . Then, $x=\\frac{141^\\circ - 32^\\circ}{2}=\\frac{109^\\circ}{2}=54.5^\\circ$ . ### Guided Practice Find the measure of $x$ . 1. 2. 3. For all of the above problems we can use the Outside Angle Theorem. 1. $x=\\frac{125^\\circ-27^\\circ}{2}=\\frac{98^\\circ}{2}=49^\\circ$ 2. $40^\\circ$ is not the intercepted arc. The intercepted arc is $120^\\circ, \\ (360^\\circ-200^\\circ-40^\\circ)$ . $x=\\frac{200^\\circ-120^\\circ}{2}=\\frac{80^\\circ}{2}=40^\\circ$ 3. Find the other intercepted arc, $360^\\circ-265^\\circ=95^\\circ$ $x = \\frac{265^\\circ-95^\\circ}{2}=\\frac{170^\\circ}{2}=85^\\circ$ ### Practice Find the value of the missing variable(s). Solve for $x$ . 1. Fill in the blanks of the proof for the Outside Angle Theorem Given : Secant rays $\\overrightarrow{AB}$ and $\\overrightarrow{AC}$ Prove : $m\\angle a = \\frac{1}{2} \\left (m\\widehat{BC}-m\\widehat{DE} \\right )$ Statement Reason 1. Intersecting secants $\\overrightarrow{AB}$ and $\\overrightarrow{AC}$ . 1. 2. Draw $\\overline{BE}$ . 2. Construction 3. $m\\angle BEC &= \\frac{1}{2} m\\widehat{BC}\\\\m\\angle DBE &= \\frac{1}{2} m\\widehat{DE}$ 3. 4. $m\\angle a+m\\angle DBE=m\\angle BEC$ 4. 5. 5. Subtraction PoE 6. 6. Substitution 7. $m\\angle a=\\frac{1}{2} \\left (m\\widehat{BC}-m\\widehat{DE} \\right )$ 7. ### Vocabulary Language: English Spanish central angle central angle An angle formed by two radii and whose vertex is", "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", "is a measure of how far short each angle sum is from $360^{\\circ}$. The sum of the exterior angles of a polygon is always $360^{\\circ}$. What do you notice about the total angle deficit of a solid? This table might help: Angle Sum Angle Deficit Number of Vertices Total Angle Deficit Cube 270 90 8 720 Tetrahedron Octahedron Icosahedron Dodecahedron", "10/(cos 40^@) ~~ 13.05 1. Assume $10$ is across from the ${40}^{\\circ}$ angle: tan 40^@ = 10/x; \" \"x = 10/(tan 40^@) ~~ 11.92 sin 40^@ = 10/y; \" \" y = 10/(sin 40^@) ~~ 15.56#", "means that $0\\leq\\frac x2\\leq\\pi$, so $45^\\circ$ is the only angle we need consider Why is 7pi/4 not possible? Isn't it smaller than 2pi? Yes, but our angle is $x$ divided by 2. So if $0^\\circ\\leq x\\leq360^\\circ$, then $0^\\circ\\leq\\frac x2\\leq180^\\circ$ (dividing through by 2), so we only consider angles between 0° and 180° (or between 0 and $\\pi,$ if you like). Edit: In other words, if we considered $\\frac{7\\pi}4$, we would have $\\frac x2 = \\frac{7\\pi}4\\Rightarrow x = \\frac{7\\pi}2$, but this value of $x$ is not in $[0,2\\pi]$", "the edges of the triangle are geodesics, the angles at those points each add up to $180^\\circ$. So we get that so Now, we were trying to get an expression for how much extra angle we have in the triangle, that is, for $A-180^\\circ$. And we’re in luck: rearranging this last equation gives us that That is, the amount that gets added to the angle sum of a triangle is the same as the difference between $360^\\circ$ and the sum of the angles at the vertex. This number, $360^\\circ-B$, is called the angle deficit at the vertex, and this is the number we want to attach to a vertex to measure its curvature. Note that if the vertex is flat, then the sum of the angles at that vertex will be $360^\\circ$, making both sides of the equation zero, as expected. What if the sum of the angles around a vertex is more than $360^\\circ$? In that case, $360^\\circ-B$ is negative, and so we say that that vertex has negative curvature. An example of this is shown in Figure 6. The vertex that the triangle is drawn around has five right angles at it, so the sum of its angles is $450^\\circ$, giving it a curvature of $-90^\\circ$. So we should expect the sum of the angles of", "solid angle of polyhedron I have an interesting thought when I draw polygon and 3D polyhedron. My question is: Can I know the number of Face, Edge, and Vertex from a given 'space angle constraint'? For example, a vertex a cube consist of 3 faces which each of the angle near the vertex sum up to $270^\\circ$. Can I tell if there is a 'thing'(simple connected closed surface) on space which each of its vertex is $270^\\circ$(summing up the angle near it), then that 'thing' has 6 faces? Here I tell you why I think about this silly question. Everyone should know that sum of exterior angle of polygon is $360^\\circ$. Of course I know the proof, but I try to think it in a different way: Every point have $360^\\circ$ 'to be occupied' and every two line can span a space which included angle $<180^\\circ$. When you draw a from a point, you actually 'occupied some angle'. Every time you bent your line, you span another direction( here the bending $<180^\\circ$). Once the line reaches the original point, all $360^\\circ$ are 'occupied'. Then If I tell you this 'thing'(simple closed curve) consist of $n$ angles which EACH of it is $108^\\circ$, then you can immediately tell me $n=5$. I generalize it into 3D spaces. If there is", "One more of those Geometry Level 4 $\\cos(8^\\circ)+\\cos(16^\\circ)+\\cos(32^\\circ)+\\cos(56^\\circ)+\\ldots+\\cos(k^\\circ)+\\ldots+\\cos(352^\\circ)$ Find the sum above, where $$0<k<360$$ and $$\\gcd(k,360)=8$$. ×", "Question # Find $x$ in the following figure : Hint : Sum of exterior angles is 360. We know that , Sum of all exterior angles of any polygon $= {360^ \\circ }$ (a) In figure (a)we have, ${125^ \\circ } + {125^ \\circ } + x = {360^ \\circ }$ On solving the above equation we get $x$ as $x = {360^ \\circ } - {250^ \\circ } = {110^ \\circ }$ Hence the value of $x$ is ${110^ \\circ }$. (b) Similarly we do here in figure (b), $x + {70^ \\circ } + {60^ \\circ } + {90^ \\circ } + {90^ \\circ } = {360^ \\circ }$ On solving we get, $x = {50^ \\circ }$ Hence the value of $x$ is ${50^ \\circ }$. Note : In these types of questions we have to apply properties of exterior angle as the angle we have to find is also exterior. By doing this we can get the correct answer.", "follows: - Opposite angles are equal. - Opposite sides are equal and parallel. - Diagonals bisect each other. - Sum of any two adjacent angles is $180^\\circ$" ]

[ "# Dodecagon (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. Each of its exterior angles has measure $30^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--O); draw(Circle(O,1)); [/asy]$ 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 $OA$. Also, each small triangle ($AOB$, $BOC$, etc.) is congruent, so $\\angle AOB=\\angle BOC=\\angle COD$ (etc) $=30^{\\circ}$.", "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, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "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", "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 $O_{1}$,", "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]$", "+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", "n$ parallel to the bases of the half-circles are drawn such that the distances between $l_k$ and $l_0$ and $l_{-k}$ and $l_0$ are always the same for all $k=1\\to n$. The intersection points of $l_k$ with one of the half-circles are labeled $A_k, B_k$, and with the other half-circle at $C_k,D_k$, as shown in the diagram. Prove that $$\\prod_{k=-n}^n |A_kB_k|+|C_kD_k| \\ge \\prod_{k=-n}^n |A_kD_k|+|B_kC_k|$$ ## Picture 2 $[asy] for (int i=0;i<6;i=i+1){ draw(dir(60*i)--dir(60*i+60)); } draw(dir(120)--(dir(0)+dir(-60))/2); draw(dir(180)--(dir(60)+dir(0))/2); fill(dir(120)--dir(180)--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); fill((dir(0)+dir(-60))/2--dir(0)--(dir(60)+dir(0))/2--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); [/asy]$ $$\\text{Prove the shaded areas are equal.}$$ $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,black); } } for(int i = 1; i < 7; ++i){ filldraw((i,7-i)--(i+1,7-i)--(i+1,8-i)--(i,8-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((i,4-i)--(i+1,4-i)--(i+1,5-i)--(i,5-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((8-i,4+i)--(7-i,4+i)--(7-i,3+i)--(8-i,3+i)--cycle,white); } filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle,white); filldraw((0,1)--(0,2)--(1,2)--(1,1)--cycle,white); filldraw((7,8)--(6,8)--(6,7)--(7,7)--cycle,white); filldraw((8,7)--(8,6)--(7,6)--(7,7)--cycle,white); [/asy]$ ## physics problem $[asy]size(100,100); draw((0,0)--(0,10)); draw(Circle((1,1),1)); dot((0,0)); draw(9.5*dir(80)..9.5*dir(70)..9.5*dir(60),EndArrow()); label(\"A\",(-1,5),dir(180));[/asy]$ $[asy] size(85,85); draw((0,0)--10*dir(60)); draw(Circle((sqrt(3),1),1)); dot((0,0)); draw(9.5*dir(50)..9.5*dir(40)..9.5*dir(30),EndArrow()); label(\"B\",(0,5),dir(0)); [/asy]$ $[asy] size(110,110); draw((0,0)--10*dir(11.42)); draw(Circle((10,1),1)); dot((0,0)); label(\"C\",(5,1.5),dir(90)); [/asy]$ ## Solution $[asy] import olympiad; size(350,350); draw((0,0)--10*dir(60)); draw(Circle((sqrt(3),1),1)); dot((0,0)); draw((-1,0)--(7,0),grey); draw((0,0)--(sqrt(3),1),linetype(\"8 8\")); draw((0,0)--(0,5),grey); draw(sqrt(3)*dir(60)--(sqrt(3),1)--(sqrt(3),0),linetype(\"8 8\")); draw(anglemark((1,0),(0,0),dir(30))); label(\"\\varphi\",0.3*dir(15),dir(15)); draw(anglemark(dir(60),(0,0),(0,1))); label(\"\\theta\",0.3*dir(75),dir(75)); label(\"1\",(sqrt(3),0.5),dir(0)); label(\"\\frac{1}{\\tan\\varphi}\",(sqrt(3)/2,0),dir(-90)); [/asy]$ $[asy] import olympiad; size(300,300); draw((0,0)--10*dir(11.42)); draw(Circle((10,1),1)); dot((0,0)); draw((-1,0)--(12,0),grey); draw((0,0)--(0,3),grey); draw(anglemark(10*dir(11.42),(0,0),(0,1))); label(\"\\theta\",0.3*dir(50.71),dir(50.71)); draw((0,0)--(10,1),linetype(\"8 8\")); draw(10*dir(11.42)--(10,1)--(10,0),linetype(\"8 8\")); label(\"1\",(10,0.5),dir(0)); label(\"10\",(5,0),dir(-90)); label(\"\\varphi\",3.4*dir(2.855),dir(2.855)); markscalefactor=0.4; draw(anglemark((1,0),(0,0),dir(5.71))); [/asy]$ ## inscribed triangle $[asy] draw((dir(0)--dir(120)--dir(240)--dir(0)..dir(120)..dir(240)..dir(0)--cycle)); draw(dir(56)--dir(230),green); draw(dir(-23)--dir(-98),red);[/asy]$ $[asy] draw((dir(0)--dir(120)--dir(240)--dir(0)..dir(120)..dir(240)..dir(0)--cycle)); draw(dir(0)--dir(36),red); draw(dir(0)--dir(72),red); draw(dir(0)--dir(108),red); draw(dir(0)--dir(144),green);", "## Elementary Technical Mathematics The value of $56.1 \\%$ of $360{}^\\circ$ is $202{}^\\circ$. \\begin{align} & 56.1 \\% \\times 360{}^\\circ =\\frac{561}{1000}\\times 360{}^\\circ \\\\ & =201.96{}^\\circ \\end{align} Now round off the result, which gives $202{}^\\circ$. Thus, the value of $56.1 \\%$ of ${{360}^{\\circ }}$is $202{}^\\circ$.", "+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", "+ .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", "# 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 $OA$.", "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", "+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", "draw(circle((0.5,0),1)); label(\"\\huge{x}\", (0, 0)); label(\"70-x\", (-1, 0)); label(\"54-x\", (1, 0)); [/asy]$ We know that the sum of all three areas is $93$ So, we have: $$93 = 70-x+x+54-x$$ $$93 = 70+54-x$$ $$93 = 124 - x$$ $$-31=-x$$ $$x=31$$ We are looking for the number of students in only math. This is $70-x$. Substituting $x$ with $31$, our answer is $\\boxed{39}$. -mathnerdnair", "using the intercepts of (6,0) & (0,-4). • We will then label the line $2x-3y=12$ and we will go over the idea that this line represents all of the ordered pairs for which $2x-3y$ is equal to 12. • I will explain that this line divides the rectangular plane into two half-planes. One of the half-planes contains points for which $2x-3y<12$ and $2x-3y>12$. • I’ll pick a point below the line to the right, like (6,-5) and substitute the coordinates into the expression $2x-3y$, which will result in 27. Since $27>12$, we will then label that half-plane as $2x-3y>12$. • I’ll pick the origin, which is above the line to the left, and substitute the coordinates into the expression $2x-3y$, which will result in 0. Since $0<12$, we will then label that half-plane as $2x-3y<12$. • We will finish by shading the side labeled $2x-3y<12$. After 2 or 3 examples done this way, I think the students will understand what each half-plane represents and have an easier time determining which half-plane to shade. -George I am a math instructor at College of the Sequoias in Visalia, CA. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page", "# Difference between revisions of \"2015 AIME I Problems/Problem 6\" ## Problem Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally 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 $\\angle BAG$. $[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label(\"A\",A,dir(0)); label(\"B\",B,dir(75)); label(\"C\",C,dir(90)); label(\"D\",D,dir(105)); label(\"E\",E,dir(180)); label(\"F\",F,dir(225)); label(\"G\",G,dir(260)); label(\"H\",H,dir(280)); label(\"I\",I,dir(315)); [/asy]$ ## Solution Let $O$ be the center of the circle with $ABCDE$ on it. Let $x$ be the degree measurement of $\\overarc{ED}=\\overarc{DC}=\\overarc{CB}=\\overarc{BA}$ in circle $O$ and $y$ be the degree measurement of $\\overarc{EF}=\\overarc{FG}=\\overarc{GH}=\\overarc{HI}=\\overarc{IA}$ in circle $C$. $\\angle ECA$ is therefore $5y$ by way of circle $C$ and $\\frac{360-4x}{2}=180-2x$ by way of circle $O$. $\\angle ABD$ is $180 - \\frac{3x}{2}$ by way of circle $O$, and $\\angle AHG$ is $180 - \\frac{3y}{2}$ by way of circle $C$. This means that: $180-\\frac{3x}{2}=180-\\frac{3y}{2}+12$, which when simplified yields $3x/2+12=3y/2$, or $x+8=y$. Since: $5y=180-2x$, $5x+40=180-2x$ So: $7x=140, x=20$ $y=28.$ $\\angle BAG$ is equal to $\\angle BAE$ + $\\angle EAG$, which equates to $\\frac{3x}{2} + y$. Plugging in yields $30+28$, or $\\boxed{058}$.", "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.", "# Difference between revisions of \"1977 AHSME Problems/Problem 30\" ## Problem $[asy] for (int i=0; i<9; ++i) { draw(dir(10+40*i)--dir(50+40*i)); } draw(dir(50) -- dir(90)); label(\"a\", dir(50) -- dir(90), N); draw(dir(10) -- dir(90)); label(\"b\", dir(10) -- dir(90), SW); draw(dir(-70) -- dir(90)); label(\"d\", dir(-70) -- dir(90), E); //Credit to MSTang for the diagram [/asy]$ If $a,b$, and $d$ are the lengths of a side, a shortest diagonal and a longest diagonal, respectively, of a regular nonagon (see adjoining figure), then $\\textbf{(A) }d=a+b\\qquad \\textbf{(B) }d^2=a^2+b^2\\qquad \\textbf{(C) }d^2=a^2+ab+b^2\\qquad\\\\ \\textbf{(D) }b=\\frac{a+d}{2}\\qquad \\textbf{(E) }b^2=ad$ ## Solution 1 By the law of cosines we can get the following expressions for $d^{2}$ and $b^{2}$: $$d^{2}=2b^{2}(1-\\cos (100^\\circ))$$ $$b^{2}=2a^{2}(1-\\cos (140^\\circ))$$ We can substitute what we got for $b^2$ into the expression for $d^{2}$: $$d^{2}=4a^{2}(1-\\cos (140^\\circ))(1-\\cos (100^\\circ))=4a^{2}(1-(\\cos (100^\\circ)+\\cos (140^\\circ))+\\cos (100^\\circ)\\cos (140^\\circ))$$ Now apply sum-to-product and product-to-sum identities: $$d^{2}=4a^{2}(1-(2\\cos (120^\\circ)\\cos (120^\\circ))+\\frac{1}{2}(\\cos (240^\\circ)+\\cos (40^\\circ)))$$ Simplifying further gives us: $$d^{2}=4a^{2}(1+\\cos (20^\\circ)-\\frac{1}{4}+\\frac{1}{2}\\cos (40^\\circ))$$ After using the fact that $\\cos (40^\\circ)=2\\cos^2 (20^\\circ)-1$, it's not hard to see that the expression in the parentheses is equal to $(\\cos (20^\\circ)+\\frac{1}{2})^{2}$. So we can square-root both sides to find the expression for $d$: $$d=2a(\\cos (20^\\circ)+\\frac{1}{2})$$ Now let's look at the expression for $b^2$. We can apply the reverse of the double angle identity to show that $1-\\cos (140^\\circ)$ equals $2\\sin^2 (70^\\circ))$. So if we square root the entire", "\\textbf{(D)}\\ a=b\\text{ and }c\\text{ is any value}\\qquad\\textbf{(E)}\\ a=b\\text{ and }c=a-1$ ## Problem 43 The pairs of values of $x$ and $y$ that are the common solutions of the equations $y=(x+1)^2$ and $xy+y=1$ are: $\\textbf{(A)}\\ \\text{3 real pairs}\\qquad\\textbf{(B)}\\ \\text{4 real pairs}\\qquad\\textbf{(C)}\\ \\text{4 imaginary pairs}\\\\ \\textbf{(D)}\\ \\text{2 real and 2 imaginary pairs}\\qquad\\textbf{(E)}\\ \\text{1 real and 2 imaginary pairs}$ ## Problem 44 In circle $O$ chord $AB$ is produced so that $BC$ equals a radius of the circle. $CO$ is drawn and extended to $D$. $AO$ is drawn. Which of the following expresses the relationship between $x$ and $y$? $[asy] size(200);defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0); draw(O--A--C--D); dot(A^^B^^C^^D^^O); pair point=O; label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"O\", O, dir(285)); label(\"x\", O+0.1*dir(172.5), dir(172.5)); label(\"y\", C+0.4*dir(187.5), dir(187.5)); draw(Circle(O,1));[/asy]$ $\\textbf{(A)}\\ x=3y\\\\ \\textbf{(B)}\\ x=2y\\\\ \\textbf{(C)}\\ x=60^\\circ\\\\ \\textbf{(D)}\\ \\text{there is no special relationship between }x\\text{ and }y\\\\ \\textbf{(E)}\\ x=2y\\text{ or }x=3y\\text{, depending upon the length of }AB$ ## Problem 45 Given a geometric sequence with the first term $\\neq 0$ and $r \\neq 0$ and an arithmetic sequence with the first term $=0$. A third sequence $1,1,2\\ldots$ is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is: $\\textbf{(A)}\\ 978\\qquad\\textbf{(B)}\\ 557\\qquad\\textbf{(C)}\\ 467\\qquad\\textbf{(D)}\\ 1068\\\\ \\textbf{(E)}\\ \\text{not possible to", "# If $(7q+46)^{\\circ}$ and $(3q-6)^{\\circ}$ form linear pair. Find the value of $q$. $( A ) 26^{\\circ} \\\\ ( B ) 14^{\\circ} \\\\ ( C ) 15^{\\circ} \\\\ ( D ) 24^{\\circ}$" ]

[ "so Maximum of Z equals to (35-1)/2 = 17. Re: 20 students took the test. The arithmetic mean of the scores [#permalink] 02 Jan 2018, 22:24 Display posts from previous: Sort by", "# Statistics - Arithmetic Range The Arithmetic Range of a set of data is the difference between the highest and lowest values in the set. Arithmetic Range is defined and given by the following function: ## Formula ${Range = L - S}$ Where − • ${L}$ = Largest item • ${S}$ = Smallest item This is an absolute measure. The relative measure called as coefficient of range is given by ${Coefficient\\ of\\ Range = \\frac{L-S}{L+S}}$ ### Example Problem Statement: Cheryl took 7 math tests in one marking period. What is the range of her test scores and coeff.of range? 89 73 84 91 87 77 94 Solution: Ordering the test scores from least to greatest, we get: 73 77 84 87 89 91 94 ${Range\\ =\\ Largest\\ -\\ Smallest\\ = 94 - 73 = 21}$ ${Largest\\ + \\ Smallest\\ = 94 + 73 = 167 \\\\[7pt] Coefficient\\ of\\ Range = \\frac{L-S}{L+S} = \\frac{21}{167} = 0.1257}$ The Range of these test scores is 21 points and coeff. of range is 0.1257 points.", "# Statistics - Arithmetic Range Advertisements The Arithmetic Range of a set of data is the difference between the highest and lowest values in the set. Arithmetic Range is defined and given by the following function − ## Formula ${Range = L - S}$ Where − • ${L}$ = Largest item • ${S}$ = Smallest item This is an absolute measure. The relative measure called as coefficient of range is given by ${Coefficient\\ of\\ Range = \\frac{L-S}{L+S}}$ ## Example Problem Statement Cheryl took 7 math tests in one marking period. What is the range of her test scores and coeff.of range? 89 73 84 91 87 77 94 Solution Ordering the test scores from least to greatest, we get − 73 77 84 87 89 91 94 $${Range\\ =\\ Largest\\ -\\ Smallest\\ = 94 - 73 = 21}$$ ${Largest\\ + \\ Smallest\\ = 94 + 73 = 167 \\\\[7pt] Coefficient\\ of\\ Range = \\frac{L-S}{L+S} = \\frac{21}{167} = 0.1257}$ The Range of these test scores is 21 points and coeff. of range is 0.1257 points. ## Useful Video Courses Video #### Class 11th Statistics for Economics 40 Lectures 3.5 hours Video #### Statistics 40 Lectures 2 hours Video #### Applied Statistics in Python for Machine Learning Engineers 66 Lectures 1.5 hours Video #### An Introduction to Wait Statistics in", "by students in a class: 5,9,11,12,13,13,18,20 Which of the following is closest to the mean? 12 13 14 11 15 13 Explanation: 13 is the correct answer. The mean is all the scores divided by the number of scores, which in this case is 12.625. ### Example Question #42 : Statistics Find the arithmetic mean of this set of data: {0, 0, 0, 0, 1, 5} 3 0 1 5 4 1 Explanation: The sum of all the terms is 6. The number of terms in the set is 6. 6/6 = 1 ### Example Question #42 : How To Find Arithmetic Mean Greg got an average of 93 on his test scores this semester. He got a 92, 93, and 97 on the first three tests. If he received the same score on each of his last three tests, what was his score on each of these tests? 89 92 94 90.5 92.33 92 Explanation: We can set up an equation (92 + 93 + 97 + 3x)/6 = 93. Solving for x yields 92. ### Example Question #41 : Arithmetic Mean Tom needs an average test score of 75 to pass his math class. The class gives five equally-weighted tests, and his first four scores were 57, 63, 90, and 72. What score does he need", "mode could still be 12 even if we used the large numbers twice, we would have swapped the 39 for a 12 in order to swap the 86 for an 87 and the 40 for a 41. Clearly this will make the mean smaller, so the numbers above make the largest possible mean. So the greatest possible value of the mean is: ( 12 + 12 + 39 + 40 + 41 + 84 + 85 + 86 + 87 )$\\div$9 = 54 You can find more short problems, arranged by curriculum topic, in our short problems collection.", "1. ## Statistics If 36 students took a test. The mean of those who passed was 78. The mean of those who failed is 60. The mean of all scores is 70. How many students failed the test ? 2. Originally Posted by Jhonson If 36 students took a test. The mean of those who passed was 78. The mean of those who failed is 60. The mean of all scores is 70. How many students failed the test ? Solve simultaneously: $78 x_p + 60 x_f = (36)(70)$ .... (1) $x_p + x_f = 36$ .... (2)", "the mean of the scores? 24. What is the median score? 25. What is the mode of the scores? Answer Key", "3 A certain list of 300 test scores has an arithmetic mean of 75 and a s 4 19 Aug 2015, 02:23 13 A certain set of numbers has an average (arithmetic mean) of 8 21 Oct 2013, 19:13 Display posts from previous: Sort by", "(Consulting) Re: In a class, there are 40 students. In a test, the average (arithmetic [#permalink] ### Show Tags 08 Nov 2018, 08:05 Bunuel wrote: In a class, there are 40 students. In a test, the average (arithmetic mean) score of the girls is 30, that of boys is 40, and that of the class is 32. If the score of a boy was incorrectly computed as 30 for 40, what is the correct average (arithmetic mean) score of the class? (A) 31 (B) 31.50 (C) 32.25 (D) 41 (E) 42 Don't solve just read the whole question at once.. You will save 2 critical minutes for your test.. Is the bottomline.. Posted from my mobile device Re: In a class, there are 40 students. In a test, the average (arithmetic [#permalink] 08 Nov 2018, 08:05 Display posts from previous: Sort by", "looking left to right. So after, right, we know 80 is the lowest point that we concerned with for this problem, and we want to go to infinity. And 99999 is perfectly fine for the relative size of the numbers. The mean is 100. This an innovation 16. So it's gonna give us the area to that curve with proportion, no scores of all in that area. So we have a number. It's pretty big, great ones, the whole thing. So that says about 89% 890.89 for three for four do around in there. So the probability that we get scores that a greater than 80 is pointing nine for four. City College of New York Topics No Related Subtopics", "But, by how much or how little is going to depend on the number of points involved. Average in this case would mean the total number of points you've gotten divided by the total number of possible points. So if you're currently at 700/1000 possible points, and you get a perfect score of 1/1 on a new assignment, the effect will be very small. See Arithmetic mean. Friday (talk) 21:30, 6 December 2006 (UTC) Ah all right thanks friday --frothT C 21:34, 6 December 2006 (UTC)", "72, and the smallest is 52, so the range is 72 - 52 = 20. The answer is 20.", "# The Mean of 72 Items Was Found to Be 63. If Two of the Items Were Misrecorded as 27 and 9 Instead of 72 and 90 Respectively, Find the Correct Mean. - Mathematics MCQ Choose the most appropriate option: The mean of 72 items was found to be 63. If two of the items were mis­recorded as 27 and 9 instead of 72 and 90 respectively, find the correct mean. • 64.25 • 64.75 • 63.25 • 65.75 #### Solution 64.75 Explanation: Error in sum=72+90-27-9=126 The error means =126/72=1.75 Concept: Arithmatic (Entrance Exam) Is there an error in this question or solution?", "# Miranda earned scores of 84, 91, 95, 78, and 92 on her math tests. How do you find her average (mean) score? Oct 24, 2016 mean = 88 #### Explanation: To find the mean of a set of scores we have to add all of the scores together and then divide by the number of scores. for this example, we have 5 scores 84, 91, 95, 78 and 92 To find the mean score, $m e a n = \\frac{84 + 91 + 95 + 78 + 92}{5}$ $m e a n = 88$", "is 78. What is the median of the three scores? (1) A scored a 73 on her exam. (2) C scored a 78 on her exam. Solution: Recall from the arithmetic mean post that the sum of deviations of all scores from the mean is 0. i.e. if one score is less than mean, there has to be one score that is more than the mean. e.g. If mean is 78, one of the following must be true: All scores are equal to 78. At least one score is less than 78 and at least one is greater than 78. For example, if one score is 70 i.e. 8 less than 78, another score has to make up this deficit of 8. Therefore, there could be a score that is 86 (8 more than 78) or there could be two scores of 82 each etc. Statement 1: A scored 73 on her exam. For the mean to be 78, there must be at least one score higher than 78. But what exactly are the other two scores? We have no idea! Various cases are possible: 73, 78, 83 or 73, 74, 87 or 70, 73, 91 etc. In each case, the median will be different. Hence this statement alone is not sufficient. Statement 2: C scored 78 on", "from office 2013 to office 1. His former office 18. ## Algebra Tom took five math tests and got an integer score on each of them. He never scored higher than 90, and his lowest score was the fourth test. If his average score, rounded to the nearest integer, was 82, what is the lowest possible score he could have 19. ## Counting and Probability Hi! I need urgent help for these questions here. They have a deadline for today and I have no idea how to solve them (a). Suppose we have a bag with 10 slips of paper in it. Eight of these have a 2 on them and the other two have a 4 on them. What is the 20. ## Counting and Probability (a) A cube is painted red on all six sides, then cut into 125 smaller congruent cubes. How many of the small cubes are painted red on exactly one side? (b) A cube is painted red on all six sides, then cut into 125 smaller congruent cubes. How many of the 21. ## Counting and Probability The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term? 22. ## Algebra A wooden plank is 5 feet long and 1 1/2 inches thick.", "### The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is- A. 45 B. 50 C. 51.4 approx D. 54.6 approx Answer: Option C ### Solution(By Apex Team) Let the required mean score be x Then, $\\begin{array}{l} 20 \\times 80+25 \\times 31+55 \\times x=52 \\times 100 \\\\ \\Leftrightarrow 1600+775+55 x=5200 \\\\ \\Leftrightarrow 55 x=2825 \\\\ \\Leftrightarrow x=\\large\\frac{2825}{55} \\approx 51.4 \\end{array}$ 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 scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)? (A)19 (B)17 (C)14 (D)13 (E)12 4.75*20= 17*5 +3*2+4 _________________ Believe you can and you are halfway there-Theodore Roosevelt Intern Joined: 14 Nov 2012 Posts: 21 Re: 20 students took the test. The arithmetic mean of the scores [#permalink] ### Show Tags 02 Jan 2018, 22:24 guerrero25 wrote: 20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)? (A)19 (B)17 (C)14 (D)13 (E)12 A mere algebra approach: As the range was 2, we are left with 3, 4 & 5 as possible scores. Let x,y,z be the no. of students who got 3, 4 & 5 respectively. => x + y + z = 20 => x = 20 - y -z (1) Also we have: 3x + 4y + 5z = 4.75x20 = 95 (2) (1) & (2) => 3(20 - y -z) + 4y + 5z = 95 => 60 + y + 2z = 95 => y + 2z = 35 Z would be maximized when Y is lowest. Minimum of Y would be 1,", "a. When x = ____, the mean of the scores is 45.5. b. When x = ____, the median of the scores is 47. c. When x = ____, the mode of the scores is 63. d. When x = ____, the range of the scores is 71. Question 8. Select all the numbers that are in the range of the function shown.", "78 93 87 Mean score after class 80 80 86 86 This is: 1. a test of two independent means. 2. a test of two proportions. 3. a test of a single mean. 4. a test of a single proportion." ]

[ "area to the left is $$0.0062$$, or about $$0.6$$%. 4. $$z=(-12-0)/10 = -1.2$$. The area to the right is $$0.8849$$, or about $$88.5$$%. 5. The two $$z$$-scores: $$(-10-0)/10 = -1$$ and $$(20-0)/10 = 2$$. The area between these is $$0.9772 - 0.1587 = 0.8185$$, or about $$81.9$$%. 6. The $$z$$ score corresponds to an area of $$0.80$$ to the left; from the tables, about $$z=0.84$$. This corresponds to an SOI of $$0 + (0.84\\times 10)$$, or an SOI of about $$8.4$$. 7. The $$z$$-score is 0.385 from Table B.3, remembering that the area to the left would be 0.650 (draw a diagram!). So, the $$z$$-score is 0.385, so that the SOI values is $$x = \\mu + (z\\times\\sigma) = 0 + (0.385\\times 10) = 3.85$$.", "middle if the $$20$$ test scores are arranged in increasing order.) $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ ###### Solution(s): To make the median equal to $$85$$ the average between the $$10$$th and $$11$$th highest scores must be $$85.$$ Therefore, we either have the $$10$$th best and $$11$$th best scores being $$85$$ or having the $$10$$th best score be above $$85$$ and the bottom and the $$11$$th best score be under $$85.$$ The second option involves not having any scores of $$85.$$ If we move all the scores out of $$85$$ we have only $$7$$ scores greater than $$85.$$ We can't move any other score to be greater than $$85.$$ Therefore, this scenario can't happen. Therefore, we must have the $$10$$th and $$11$$th best scores be $$85.$$ There are currently $$7$$ scores greater than or equal to $$85,$$ so we must add $$4$$ more scores so the $$11$$th best score is $$85.$$ 20. The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $$x$$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $$x?$$ $$-1$$ $$5$$ $$6$$ $$8$$ $$9$$", "No you can't know the exact score as well described by @Deusovi - so I am not going to describe it except to say because we start with 2s and 4s randomly appearing we can't know, but we can figure out approximate scores. So yes, your score is generally just about 20000 when you hit 2048 and around 45,000 when you hit 4096 and close to 90000 for 8192. If you get to 2048, you must have scored $$2048 + 2*1024 + 4*512 + 8*256+ 16*128+32*64+64*32+ 128 *16 + 256 * 8 + ?*4 = 9 * 2048 +?*4 = 18432 +?*4 \\simeq 20000$$ If you get to 4096, you must have scored $$10 * 4096 +?*4 = 40960 +?*4 \\simeq 45000$$ If you get to 8192, you must have scored $$11 * 8192 +?*4 = 90112 +?*4 \\simeq 90000$$", "new score that will yield a given mean. 12 , 15 , 18 , 13 , 6 , 14; New mean = 15 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(12 + 15 + 18 + 13 + 6 + 14 + x )}{7}$ = 15 = 78 + x = 105; x = 105 – 78 = 27 Step 3: Required new score = 27 Q 3 - Find the value for a new score that will yield a given mean. 12 , 12 , 4 , 12 , 2 , 12; New mean = 12 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(12 + 12 + 4 + 12 + 2 + 12 + x )}{7}$ = 12. = 54 + x = 84; x = 84 – 54 = 30 Step 3: Required new score = 30 Q 4 - Find the value for a new score that will yield a given mean. 8 , 12 , 8 , 10 , 18 , 12 , 4; New mean = 11 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(8 + 12 + 8", "is $7(94). So the possible sums for the 7 test scores so that they provide an integer average are$7(98), 7(97), 7(96), 7(95)$and$7(94)$which are$686, 679, 672, 665,$and$658$respectively. Now in order to get the sum of the first 6 tests, we negate$95$from each sum producing$591, 584, 577, 570,$and$563$. Notice only$570$is divisible by$6$so, therefore, the sum of the first$6$tests is$570$. We need to find her score on the$6th$test so what number minus$570$will give us a number divisible by$5$. Since$95$is the$7th$test score and all test scores are distinct that only leaves$\\boxed{\\textbf{(E) } 100}$. ==Solution 2 (Cheap Solution)== By inspection, the sequences$ (Error compiling LaTeX. ! Missing $inserted.)91,93,92,96,98,100,95$and$93,91,92,96,98,100,95$work, so the answer is$\\boxed{\\textbf{(E) } 100}$. Note: A method of finding this \"cheap\" solution is to create a \"mod chart\", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.", "following set of scores: $65.2$65.2, $64.3$64.3, $71.6$71.6, $63.2$63.2, $45.2$45.2, $62.2$62.2, $46.8$46.8, $58.7$58.7 A) Sort the scores in ascending order Think: Ascending means lowest to highest. Do: $45.2,46.8,58.7,62.2,63.2,64.3,65.2,71.6$45.2,46.8,58.7,62.2,63.2,64.3,65.2,71.6 Think: Which term will be in the middle? Do: $\\text{Middle term }$Middle term $=$= $\\frac{n+1}{2}$n+12​ $=$= $\\frac{8+1}{2}$8+12​ $=$= $4.5$4.5 This means that the median lies between the fourth and fifth scores. $\\frac{62.2+63.2}{2}$62.2+63.22​ $=$= $62.7$62.7 The median is $62.7$62.7. ## Mode The mode is the most frequently occurring score. To find the mode, just count which score you see most frequently in your data set. #### Example ##### Question 3 Find the mode of the following set of scores: $2$2, $2$2, $6$6, $7$7, $7$7, $7$7, $7$7, $11$11, $11$11, $11$11, $13$13, $13$13, $16$16, $16$16 Think: How many of each score are there? Do2, 2, 6, 7, 7, 7, 7, 11, 11, 11, 13, 13, 16, 16 $7$7 is the most frequently occurring score, so the mode is $7$7. ## Range The range is the difference between the highest score and the lowest score. To calculate the range, you need to subtract the lowest score from the highest score. #### Example ##### Question 4 Find the range of the following set of scores: $10,7,2,14,13,15,11,4$10,7,2,14,13,15,11,4. Think: What are the highest and lowest scores in this set? Do: $15-2=13$152=13 The range is $13$13. ## Calculating the", "Think: The range is the difference between the highest and lowest scores. Do: The range is $70-36=34$7036=34. (e) Remove the outlier from the data set and recalculate the mean, rounding your answer to one decimal place if necessary. Think: Is the highest or the lowest score significantly different from the other scores. Do: We're going to remove $70$70 from the data set because it is much higher than the other scores in the set. Now let's calculate the mean. Remember that there are only nine scores in the set now: $\\text{Mean }$Mean $=$= $\\frac{36+36+45+52+48+36+43+44+51}{9}$36+36+45+52+48+36+43+44+519​ $=$= $43.444\\ldots$43.444… $=$= $43.4$43.4 ($1$1 d.p.) Notice that this is substantially lower than the mean when the outlier was included. (f) With the outlier removed from the data set, recalculate the median. Think: The median will now be the fifth score in the set. Do: So the median is $44$44, which is only slightly lower than the median when the outlier was included. (g) With the outlier removed from the data set, recalculate the mode. Think: Which score occurs most frequently now? Do: The mode hasn't changed, and is still $36$36. (h) With the outlier removed from the data set, recalculate the range. Think: Let's calculate the difference between the new highest and lowest scores. Do: The new range is $52-36=16$5236=16, which is substantially", "a number is excluded then the mean is reduced by 3. Find the excluded number. Sol: Mean of 8 numbers = 35 Therefore, Total sum of 8 numbers = 35×8$35 \\times 8$ = 280 Since One number is excluded, New mean = 35 – 3 = 32 Therefore, Total sum of 7 numbers = 32×7$32 \\times 7$ = 224 The excluded number = Sum of 8 numbers – Sum of 7 numbers = 280 – 224 = 56 Q.12: The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct mean. Sol: Mean of 150 items = 60 Total Sum of 150 items = 150×60$150 \\times 60$ = 9000 Therefore, Correct sum of items = [(Sum of 150 items) – (Sum of wrong items) + ( Sum of right items)] = [9000 – (52 + 8) + (152 + 88)] = 9180 Therefore, the Correct mean = 9180150=61.2$\\frac{9180}{150}= 61.2$ Q.13: The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result. Sol: Mean of 31 results = 60 Total sum of 31 results =", "= 240$ => $x = 48$ => B = $x + 2$ = 48 + 2 = 50 E = $x + 8$ = 48 + 8 = 56 $\\therefore$ B $\\times$ E = 50 * 56 = 2800 Question 16: Find the average of the following set of scores : 221, 231, 441, 359, 665, 525 a) 399 b) 428 c) 407 d) 415 e) None of these Solution: Sum of scores = 221 + 231 + 441 + 359 + 665 + 525 = 2442 => Average score = 2442/6 = 407 Question 17: The average of five positive numbers is 308. The average of first two numbers is 482.5 and the average of last two numbers is 258.5. What is the third number? a) 224 b) 58 c) 121 d) Cannot be determined e) None of these Solution: Let the numbers be $A_1 , A_2 , A_3 , A_4 , A_5$ => $A_1 + A_2 + A_3 + A_4 + A_5 = 308 * 5 = 1540$ Also, $A_1 + A_2 = 482.5 * 2 = 965$ and $A_4 + A_5 = 258.5 * 2 = 517$ => $A_3 = 1540 – 965 – 517 = 58$ Question 18: What will be the average of the following set of scores (Rounded off to the nearest", "other hand if somebody has $$20$$ scorings of $$6$$ then $$p=6$$ and $$5(1-e^{-20/10})\\approx 4.58$$ so the final score is $$3+4.6$$ rounded giving $$8$$. The choice of $$Q$$ depends on what you call \"few\", \"moderate\", \"many\". As a rule of thumb consider a value $$M$$ that you consider \"moderate\" and take $$Q=-M/\\ln(1/2)\\approx 1.44M$$. So if you think $$100$$ is a moderate value then take $$Q=144$$. Finally you can also replace the equal weight on quantity an quality by a skewed one so that the final formula becomes:$$\\text{score}=Pp+10(1-P)(1-e^{-q/Q}))$$ where $$P\\in [0,1]$$ (in the original formula we had $$P=0.5$$). • Thanks. This kind of a formula is exactly what I was looking for. From what I can see, the maximum score will never go above 10, which is what I want. I will implement this in code and let you know if it works as expected. Thanks again. – Andrius Bartulis Sep 24 '14 at 7:11 • @AndriusBartulis Good idea, I'm looking forward to it. – Marc Bogaerts Sep 24 '14 at 8:06 • @MarcBogaerts Hi Marc, thanks for this. I have set p = 0.5 and then do this in Ruby: (p * average) + 10 * (1 - p) * (1 - ((Math::E)**(-number_scores / q))) When I set q = 5 I can see that if I have somebody", "is $7(94)$. So the possible sums for the 7 test scores so that they provide an integer average are $7(97), 7(96), 7(95)$ and $7(94)$ which are $679, 672, 665,$ and $658$ respectively. Now in order to get the sum of the first 6 tests, we negate $95$ from each sum producing $584, 577, 570,$ and $563$. Notice only $570$ is divisible by $6$ so, therefore, the sum of the first $6$ tests is $570$. We need to find her score on the $6th$ test so what number minus $570$ will give us a number divisible by $5$. Since $95$ is the $7th$ test score and all test scores are distinct that only leaves $\\boxed{\\textbf{(E) } 100}$. ## Solution 2 (Cheap Solution) By inspection, the sequences $91,93,92,96,98,100,95$ and $93,91,92,96,98,100,95$ work, so the answer is $\\boxed{\\textbf{(E) } 100}$. Note: A method of finding this \"cheap\" solution is to create a \"mod chart\", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.", "score on the sixth test? $$92$$ $$94$$ $$96$$ $$98$$ $$100$$ ###### Solution(s): The smallest possible average of the first $$6$$ of them is $\\dfrac{91+ \\cdots 96}7 = 93.5.$ The largest possible average of the first $$6$$ of them is $\\dfrac{95+ \\cdots 100}7 = 97.5.$ This makes the bounds of the average of the first $$6$$ of them $$94$$ and $$97$$ inclusive. Then, let the average of the first $$6$$ of them be $$x.$$ Then, the average of all of them is $$\\dfrac{6x+95}7,$$ making $$6x+95$$ a multiple of $$7.$$ Therefore, $$6x+95 \\equiv 0 \\mod 7.$$ This means $$x \\equiv 4\\mod 7.$$ The only possible value is $$x=95,$$ making the sum of the first $$6$$ of them $$570.$$ Then, the sum of the first $$5$$ is a multiple of $$5,$$ so the 6th score must also be a multiple of $$5$$ since it is their difference. The only not used multiple of $$5$$ is $$100,$$ making it the answer. Thus, the correct answer is E.", "the sum of his first 5 scores is 5(183). The average of his entire 6 scores will be the sum of the first 5 plus his final score. We can call the final score x. Then, after 6 rounds his average is $\\frac{sum\\ of\\ 6\\ scores}{6}=\\frac{sum\\ of\\ first\\ 5\\ scores+final\\ score}{6}=\\frac{5(183)+x}{6}$ and this needs to be 190 $\\frac{5(183)+x}{6}=190$ $5(183)+x=6(190)$ $915+x=1140$ $x=1140-915$", "the previous example, Charlie’s first set of scores is shown below: Charlie’s Scores Monday Tuesday Wednesday Thursday Friday Saturday 450 470 430 490 490 220 Arranging Charlie’s scores in order gives: 220, 430, 450, 470, 490, 490. Each score occurs once, except for the score of 490, which occurs twice. So, the mode (or the modal score) is 490. In this case, the mode is the highest score. Do you think this is a good way of describing Charlie’s typical score? Activity: Finding the Mode of a Data Set Charlie’s second set of scores is shown below: Charlie’s Scores Monday Tuesday Wednesday Thursday Friday Saturday Sunday 250 270 300 290 320 290 270 Find the two modal values. Arranging Charlie’s scores in order gives: 250, 270, 270, 290, 290, 300, 320. Each score occurs once, except 270 and 290, which both occur twice. The modal scores are therefore 270 and 290. 11.1.6 Using Different Averages You have now met three different ways of describing an average value: mean, median, and mode. For Charlie’s first set of scores: • The mean score was 425. • The median score was 460. • The modal score was 490. These values are very different, and so it is important that when you meet an \"average\" value, you find out which average is", "$1$ and $10$. Later, we can add $90$ to her score to obtain the real answer. From this point of view, the problem states that Isabella's score on the seventh test was $5$. We note that Isabella received $7$ integer scores out of $1$ to $10$. Since $5$ is already given as the seventh test score, the possible scores for Isabella on the other six tests are $S={1,2,3,4,6,7,8,9,10}$. The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set $S$ above, and their sum with $5$ must be a multiple of $7$. The interval containing the possible sums of the six numbers in S are from $1 +2+3+4+6+7=23$ to $4+6+7+8+9+10=44$. We must now find multiples of $7$ from the interval $23+5 = 28$ to $44+5=49$. There are four possibilities: $28$, $35$, $42$, $49$. However, we also note that the sum of the six numbers (besides $5$) must be a multiple of $6$ as well. Thus, $35$ is the only valid choice.(The six numbers sum to $30$.) Thus the sum of the six numbers equals to $30$. We apply the logic above in a similar way for the sum of the scores from the first test to the fifth test. The sum must be a multiple of $5$.", "The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct men. Mean of 150 items = 60 Sum of 150 items = $\\left(150×60\\right)=9000$ New sum = [9000 $-$ (52 + 8) + (152 + 88)] = 9180 Correct mean = Therefore, the correct mean is 61.2. #### Question 21: The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result. Mean of 31 results = 60 Sum of 31 results = $31×60=1860$ Mean of the first 16 results = 58 Sum of the first 16 results = $58×16=928$ Mean of the last 16 results = 62 Sum of the last 16 results = $62×16=992$ Value of the 16th result = (Sum of the first 16 results + Sum of the last 16 results) $-$ Sum of 31 results = (928 + 992) $-$ 1860 = 1920 $-$ 1860 = 60 #### Question 22: The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46, find the 6th number.", "\\,12$$ First, arrange all these values (from least to greatest). This, however, is not a necessary step, but it sometimes is helpful to discover the mode if the values or numbers will be put in ascending order. $$0,\\, 1,\\, 4, \\,5, \\,5, \\,8, {\\color{Red} \\,12, \\,12,\\, 12,\\, 12}$$ So, find the number or value that most often occurs. The answer is: Our mode here is: $$12$$ Now, let’s take a look at some examples that include a few relevant data. An example: Problem: Carlos attained these scores on his math tests: $$84, \\,92, \\,74, \\,98, \\,82$$. Now, come up with the mean, median, and mode of Carlos’ scores. $$\\frac{84+92+74+98+82}{5}$$ For finding the mean, we need to add together all the scores and divide that value by the number of the tests. So: $$430$$ divided by $$5$$ is $$86 (\\frac{430}{5} = 86)$$. The mean we needed to find is $$86$$. $$74, \\,82, \\,84, \\,92, \\,98$$ For finding the median, first we need to put Carlos’ test scores in order (from least to greatest). The median is $$84$$. We have five scores in total. In our ordered list, the middle score is the third. This our median. $$74, \\,82, \\,84, \\,92, \\,98$$ Because all numbers appear exactly just one time, in the data set is no mode. The mean in", "dividing these equally. Call the value of each of the first three \"a\" the last is worth \"a/4\". So we must have a+ a+ a+ a/4= 3a+ a/4= (12a+ a)/4= 13a/4= 1.00 so a= 4/13. each of the first tests is worth 4/13 and the last text is worth 1/4 of that, 1/13. 6. ## Re: Averages (Mean) $\\displaystyle (0.30769230769*83)+(0.30769230769*85)+(0.307692307 69*86)+(0.07692307692*100) = 85.8%$ There we go! Thanks!", "a randomly selected student (from this cohort), and let $$F_X$$ be the cdf of $$X$$. Evaluate the cdf for each of the following. For the purposes of this exercise, interpret these quantities in terms of actual SAT scores, which take values in 400, 410, 420, $$\\ldots$$, 1590, 1600. 1. $$F_X(1400)$$ 2. $$F_X(1405)$$ 3. $$F_X(1000)$$ 4. $$F_X(1003.7)$$ 5. $$F_X(-3.1)$$ 6. $$F_X(390)$$ 7. $$F_X(399.5)$$ 8. $$F_X(1600)$$ 9. $$F_X(1610)$$ 10. $$F_X(2307.4)$$ 11. $$F_X(1400)-F_X(1000)$$ Solution to Example 2.39 1. $$F_X(1400)=0.94$$. We are told that $$\\textrm{P}(X \\le 1400) = 0.94$$. 2. $$F_X(1405) = 0.94$$. In terms of reall SAT scores, $$\\textrm{P}(X \\le 1405) = \\textrm{P}(X\\le 1400)$$. 3. $$F_X(1000) = 0.40$$. We are told that $$\\textrm{P}(X \\le 1000) = 0.40$$. 4. $$F_X(1003.7) = 0.40$$. In terms of reall SAT scores, $$\\textrm{P}(X \\le 1003.7) = \\textrm{P}(X\\le 1000)$$. 5. $$F_X(-3.1)=0$$. The smallest possible score is 400. 6. $$F_X(390)=0$$. The smallest possible score is 400. 7. $$F_X(399.5)= 0$$. The smallest possible score is 400. 8. $$F_X(1600) = 1$$. The largest possible score is 1600, so 100% of students score no more than 1600. 9. $$F_X(1610) = 1$$. The largest possible score is 1600. 10. $$F_X(2307.4) = 1$$. The largest possible score is 1600. 11. $$0.54 = F_X(1400)-F_X(1000)=\\textrm{P}(X\\le 1400) - \\textrm{P}(X \\le 1000) = \\textrm{P}(1000 < X \\le 1400)$$. 54% of SAT takers score greater than 1000 but", "the third test? $$90$$ $$92$$ $$95$$ $$96$$ $$97$$ Solution(s): If the average is $$95,$$ then the sum of the tests are $$95\\cdot 4 = 380 .$$ Since we have the first two tests, the sum of the last two tests is $$380$$ minus the first two scores. This makes the sum of the last two scores equal to $$380-91-97 = 192.$$ Her last two scores therefore hace a sum of $$192.$$ Given the sum of the tests we try to minimize one score, then we must maximize the other test. Therefore, we maximize the fourth test by making it $$100.$$ This would make the third test equal to $$192 -100 = 92.$$ 8. A shop advertises everything is \"half price in today's sale.\" In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price? $$10$$ $$33$$ $$40$$ $$60$$ $$70$$ Solution(s): Let $$p$$ be the original price. If everything is half off, we have the new price as $$0.5p.$$ Having a $$20 \\%$$ discount makes it such that we keep $$80\\%$$ of the price, so the price is $$0.5p\\cdot 0.8 =0.4p.$$ This would have $$0.6p$$ off, so we get a discount of $$\\frac{0.6p}{p} = 0.6,$$ which is $$60\\%.$$ 9. The Fort Worth Zoo has a number" ]

[ "new score that will yield a given mean. 12 , 15 , 18 , 13 , 6 , 14; New mean = 15 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(12 + 15 + 18 + 13 + 6 + 14 + x )}{7}$ = 15 = 78 + x = 105; x = 105 – 78 = 27 Step 3: Required new score = 27 Q 3 - Find the value for a new score that will yield a given mean. 12 , 12 , 4 , 12 , 2 , 12; New mean = 12 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(12 + 12 + 4 + 12 + 2 + 12 + x )}{7}$ = 12. = 54 + x = 84; x = 84 – 54 = 30 Step 3: Required new score = 30 Q 4 - Find the value for a new score that will yield a given mean. 8 , 12 , 8 , 10 , 18 , 12 , 4; New mean = 11 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(8 + 12 + 8", "a new score that will yield a given mean. 9 , 10 , 8 , 10 , 18 , 12 , 4; New mean = 14 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(9 + 10 + 8 + 10 + 18 + 12 + 4 + x )}{8}$ = 14 = 71 + x = 112; x = 112 – 71 = 41 Step 3: Required new score = 41 Q 8 - Find the value for a new score that will yield a given mean. 25 , 18 , 18 , 13 , 4 , 17 , 18 , 19 , 3; New mean = 17 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(25 + 18 + 18 + 13 + 4 + 17 + 18 + 19 + 3 + x )}{10}$ = 17 = 135 + x = 170; x = 170 – 135 = 35 Step 3: Required new score = 35 Q 9 - Find the value for a new score that will yield a given mean. 18 , 15 , 11 , 3 , 8 , 4 , 13 , 12 , 6; New mean = 10 ### Explanation", "means themselves, I just used the arithmetic mean formula for the two individual means. – GovEcon May 23 '13 at 20:39 Thank you for the clarification, answer accepted! – AndyWarren May 23 '13 at 20:53 The sum of the first seven scores is $5 \\cdot 7$. It's $7$ times the mean. Similarly, you can find the sum of all eight scores - to find the last one, all you need to do is subtract. - The mean is the sum $S$ of the scores divided by the number of scores. Here we have $M = S/n$. Plugging in our values gives $S = 7 \\cdot 5 = 35$. Now the new mean is $6 = (S + X)/(n+1) = (35 + X)/8$. Then just solve for $X$. -", "# Recent questions tagged arithmetic-mean 1 vote 1 The arithmetic mean of scores of $25$ students in an examination is $50.$ Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being $30,$ then the maximum possible score of the toppers is Let $a_{1},a_{2},\\dots , a_{52}$ be a positive integers such that $a_{1}<a_{2}<\\dots < a_{52}$. Suppose, their arithmetic mean is one less than the arithmetic mean of $a_{2},a_{3},\\dots , a_{52}$. If $a_{52}=100$ , then the largest possible value of $a_{1}$ is ________ $20$ $23$ $48$ $45$ If $\\log\\left ( 2^{a} \\times 3^{b}\\times 5^{c}\\right )$ is the arithmetic mean of $\\log\\left ( 2^{2} \\times 3^{3}\\times 5 \\right ),$ $\\log\\left ( 2^{6} \\times3\\times 5^{7} \\right ),$ and $\\log\\left ( 2 \\times3^{2}\\times 5^{4} \\right ),$ then $a$ equals $2$ None of these $6$ $7$ $S$ is a set given by $S=\\{1,2,3,\\dots,4n\\}$, where $n$ is a natural number. $S$ is partitioned into $n$ disjoint subsets $A_{1},A_{2},A_{3}\\dots,A_{n}$ each containing four elements. It is given that in everyone of these subsets there is one element, which is the ... can be equal to $2$ $n\\neq2$ but can be equal to $1$ It is possible to satisfy the requirement for $n=1$ as well as for $n=2$", "0, 100$ We know, Arithmetic mean $=\\frac{sum\\ of\\ observations}{number\\ of\\ observation}$ $\\\\ = \\frac{58+ 76+ 40+ 35+ 46+ 45+ 0+ 100}{8} \\\\ = \\frac{400}{8} \\\\ \\\\ = 50$ Therefore, the mean score is $50$ Given, We know, $Arithmetic\\: mean =\\frac{sum\\ of\\ observations}{number\\ of\\ observation}$ (i) Therefore, A's average number of points scored per game $\\\\ = \\frac{14+16+10+10}{4} \\\\ = \\frac{50}{4} \\\\ \\\\ = 12.5$ (ii) Number of games C played = 3 Therefore, to find the mean, we will divide by 3 Mean of C = $\\\\ = \\frac{8+11+13}{3} \\\\ = \\frac{32}{3} \\\\ \\\\ = 10.67$ (iii) Although he scored 0 in a match, he played all 4 games. Therefore, we will divide by 4. Mean of B = $\\\\ = \\frac{0+8+6+4}{4} \\\\ = \\frac{18}{4} \\\\ \\\\ = 4.5$ (iv) Mean of A = $12.5$ Mean of B = $4.5$ Mean of C = $10.67$ Since, A has the highest mean, A is the best performer. (i) Highest and the lowest marks obtained by the students. (ii) Range of the marks obtained. (iii) Mean marks obtained by the group. Given, Marks obtained the students = $85, 76, 90, 85, 39, 48, 56, 95, 81, 75$ Arranging in ascending order = $39, 48, 56, 75, 76, 81, 85, 85, 90, 95$ (i) Highest marks obtained = $95$ Lowest marks", "Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(18 + 15 + 11 + 3 + 8 + 4 + 13 + 12 + 6 + x )}{10}$ = 10 = 90 + x = 100; x = 100 – 90 = 10 Step 3: Required new score = 10 Q 10 - Find the value for a new score that will yield a given mean. 16 , 15 , 18 , 14 , 9 , 17; New mean = 15 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(16 + 15 + 18 + 14 + 9 + 17 + x )}{7}$ = 15 = 89 + x = 105; x = 105 – 89 = 16 Step 3: Required new score = 16 finding_value_for_new_score_that_will_yield_given_mean.htm", "middle if the $$20$$ test scores are arranged in increasing order.) $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ ###### Solution(s): To make the median equal to $$85$$ the average between the $$10$$th and $$11$$th highest scores must be $$85.$$ Therefore, we either have the $$10$$th best and $$11$$th best scores being $$85$$ or having the $$10$$th best score be above $$85$$ and the bottom and the $$11$$th best score be under $$85.$$ The second option involves not having any scores of $$85.$$ If we move all the scores out of $$85$$ we have only $$7$$ scores greater than $$85.$$ We can't move any other score to be greater than $$85.$$ Therefore, this scenario can't happen. Therefore, we must have the $$10$$th and $$11$$th best scores be $$85.$$ There are currently $$7$$ scores greater than or equal to $$85,$$ so we must add $$4$$ more scores so the $$11$$th best score is $$85.$$ 20. The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $$x$$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $$x?$$ $$-1$$ $$5$$ $$6$$ $$8$$ $$9$$", "a number is excluded then the mean is reduced by 3. Find the excluded number. Sol: Mean of 8 numbers = 35 Therefore, Total sum of 8 numbers = 35×8$35 \\times 8$ = 280 Since One number is excluded, New mean = 35 – 3 = 32 Therefore, Total sum of 7 numbers = 32×7$32 \\times 7$ = 224 The excluded number = Sum of 8 numbers – Sum of 7 numbers = 280 – 224 = 56 Q.12: The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct mean. Sol: Mean of 150 items = 60 Total Sum of 150 items = 150×60$150 \\times 60$ = 9000 Therefore, Correct sum of items = [(Sum of 150 items) – (Sum of wrong items) + ( Sum of right items)] = [9000 – (52 + 8) + (152 + 88)] = 9180 Therefore, the Correct mean = 9180150=61.2$\\frac{9180}{150}= 61.2$ Q.13: The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result. Sol: Mean of 31 results = 60 Total sum of 31 results =", "following set of scores: $65.2$65.2, $64.3$64.3, $71.6$71.6, $63.2$63.2, $45.2$45.2, $62.2$62.2, $46.8$46.8, $58.7$58.7 A) Sort the scores in ascending order Think: Ascending means lowest to highest. Do: $45.2,46.8,58.7,62.2,63.2,64.3,65.2,71.6$45.2,46.8,58.7,62.2,63.2,64.3,65.2,71.6 Think: Which term will be in the middle? Do: $\\text{Middle term }$Middle term $=$= $\\frac{n+1}{2}$n+12​ $=$= $\\frac{8+1}{2}$8+12​ $=$= $4.5$4.5 This means that the median lies between the fourth and fifth scores. $\\frac{62.2+63.2}{2}$62.2+63.22​ $=$= $62.7$62.7 The median is $62.7$62.7. ## Mode The mode is the most frequently occurring score. To find the mode, just count which score you see most frequently in your data set. #### Example ##### Question 3 Find the mode of the following set of scores: $2$2, $2$2, $6$6, $7$7, $7$7, $7$7, $7$7, $11$11, $11$11, $11$11, $13$13, $13$13, $16$16, $16$16 Think: How many of each score are there? Do2, 2, 6, 7, 7, 7, 7, 11, 11, 11, 13, 13, 16, 16 $7$7 is the most frequently occurring score, so the mode is $7$7. ## Range The range is the difference between the highest score and the lowest score. To calculate the range, you need to subtract the lowest score from the highest score. #### Example ##### Question 4 Find the range of the following set of scores: $10,7,2,14,13,15,11,4$10,7,2,14,13,15,11,4. Think: What are the highest and lowest scores in this set? Do: $15-2=13$152=13 The range is $13$13. ## Calculating the", "out of school for 8 years and I am trying to get my diploma. On one of my practice tests it said to get the mean, median, and mode of scores. However in my textbook there are no examples to show me how to do it. The scores are: 100, 78, 93, 84, 91, 100, 82, 79. I would greatly appreciate if someone would show me how to do them. Here are Adam’s data, placed on a number line: Adam, The mean of the scores is another name for their average. Just add them up and divide by the number of scores: mean = (100+78+93+84+91+100+82+79)/8 = 88.375 The median of the scores is \"the number in the middle\" when the scores are sorted in order. In your example: 100 100 93 91 84 82 79 78 If there is an even number of scores (as in your example) there is no number in the middle so the two numbers in the middle are averaged: median = (91 + 84)/2 = 87.5 The mean of the numbers can be misleading. If I tell you the mean income in my neighborhood is 1 million dollars a year you might think that I am wealthy. But maybe the neighborhood is a poor one with one very rich person making many millions", "646 views The arithmetic mean of scores of $25$ students in an examination is $50.$ Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being $30,$ then the maximum possible score of the toppers is Let the sum of scores of students be $x.$ $\\Rightarrow \\frac{x}{50} = 25$ $\\Rightarrow x= 25 \\times 50$ $\\Rightarrow \\boxed{x= 1250}$ For the scores of the top $5$ students to be as high as possible, the score of the bottom $20$ students should be as low as possible. The minimum score is $30,$ and the scores of the bottom $20$ students are distinct integers. So, the score of bottom $20$ students must be $30,31,32, \\dots, 49.$ Let the score of the topper be $y.$ So, $5y+(30+31+32+ \\dots +49) = x$ $\\Rightarrow 5y + \\frac{20}{2}(30+49) = 1250$ $\\Rightarrow 5y+790 = 1250$ $\\Rightarrow 5y = 1250-790$ $\\Rightarrow 5y = 460$ $\\Rightarrow \\boxed{y=92}$ $\\therefore$ The maximum possible score of the topper is $92.$ Correct Answer $:92$ 10.3k points 1 vote 1 355 views 1 vote", "Think: The range is the difference between the highest and lowest scores. Do: The range is $70-36=34$7036=34. (e) Remove the outlier from the data set and recalculate the mean, rounding your answer to one decimal place if necessary. Think: Is the highest or the lowest score significantly different from the other scores. Do: We're going to remove $70$70 from the data set because it is much higher than the other scores in the set. Now let's calculate the mean. Remember that there are only nine scores in the set now: $\\text{Mean }$Mean $=$= $\\frac{36+36+45+52+48+36+43+44+51}{9}$36+36+45+52+48+36+43+44+519​ $=$= $43.444\\ldots$43.444… $=$= $43.4$43.4 ($1$1 d.p.) Notice that this is substantially lower than the mean when the outlier was included. (f) With the outlier removed from the data set, recalculate the median. Think: The median will now be the fifth score in the set. Do: So the median is $44$44, which is only slightly lower than the median when the outlier was included. (g) With the outlier removed from the data set, recalculate the mode. Think: Which score occurs most frequently now? Do: The mode hasn't changed, and is still $36$36. (h) With the outlier removed from the data set, recalculate the range. Think: Let's calculate the difference between the new highest and lowest scores. Do: The new range is $52-36=16$5236=16, which is substantially", "and $95+7=102$ clearly cannot be $a_6$ since they are outside the range of 91 to 100. Thus, $a_6$ must be 95. Additionally, $a_5$ cannot be 95 since that would mean the sixth score is 95, which isn't possible since scores can't repeat (also, it's not an answer choice). Intuitively, $a_5$ should be close to 95 due to the constraint on the range of scores. We can try 94 and 96 for $a_5$. Again, the sixth score must be a multiple of 6 (preferably only one multiple) away from $a_5$. The only one that works is $94+6=100$, and we can check that the rest don't work since they exceed the range. The answer is $\\boxed{\\bold{(E)} 100}$. - Lingjun :) ## Solution 6 Say the sum of the first 6 scores is $n$. Then, we know that $n + 95 \\equiv 0 \\pmod{7}$. Thus, $n \\equiv 3 \\pmod{7}$. Additionally, since the average of these 6 scores was an integer, we know that $n \\equiv 0 \\pmod{6}$. We find that the smallest $n$ to satisfy both of these inequalities is 24 (series below). $$3, 10, 17, 24, ...$$ To get to the next one, we have to add $6 \\cdot 7 = 42$ so as not to ruin the moduli. Next, we know that $n$ is between $91 \\cdot 6 =", "so Maximum of Z equals to (35-1)/2 = 17. Re: 20 students took the test. The arithmetic mean of the scores [#permalink] 02 Jan 2018, 22:24 Display posts from previous: Sort by", "course my first reply was a attempt at a pure counting solution. I have now programmed a computer algebra system to do $10^6$ trials. Dropping the lowest of the four gives me a mean of $12.246$. That agrees more or less with that paper. Dropping the highest of the four gives me a mean of $7.004$. Dropping the lowest and highest of the four gives me a mean of $8.754$. 9. ## Re: Dungeons and Dragons VinceW and Plato are right, the mean given in the paper is an approximation. $\\frac{3(1)+4(4)+5(10)+6(21)+7(38)+8(62)+9(91)+10(1 22)+11(148)+12(167)+13(172)+14(160)+15(131)+16(94) +17(54)+18(21))}{6^4}$ $15869/1296=12\\frac{317}{1296}=12.2445\\overline{987654320 }$", "has a score between $85$ and $115$ is $P(85 \\le x \\le 115) = \\int_{85}^{115} \\frac{1} {15 \\sqrt{2 \\pi}}e^{-(x - 100)^2/(2(15)^2)}.$ Again, the integral of $e^{-x^2}$ does not have an elementary anti-derivative and therefore cannot be evaluated. Using the programing feature of a scientific calculator or a mathematical computer software, we get $\\int_{85}^{115} \\frac{1} {15 \\sqrt{2 \\pi}}e^{-(x - 100)^2/(2(15)^2)} dx \\approx 0.68.$ That is, $P(85 \\le x \\le 115) \\approx 68\\%.$ Which says that $68 \\%$ of the population has an IQ score between $85$ and $115.$ 2. To measure the probability that a person selected randomly will have an IQ score above $140$, $P(x \\ge 140) = \\int_{140}^{\\infty} \\frac{1} {15 \\sqrt{2 \\pi}}e^{-(x - 100)^2/(2(15)^2)} dx.$ This integral is even more difficult to integrate since it is an improper integral. To avoid the messy work, we can argue that since it is extremely rare to meet someone with an IQ score of over $200,$ we can approximate the integral from $140$ to $200.$ Then $P(x \\ge 140) \\approx \\int_{140}^{200} \\frac{1} {15 \\sqrt{2 \\pi}}e^{-(x - 100)^2/(2(15)^2)} dx.$ Integrating numerically, we get $P(x \\ge 140) \\approx 0.0039.$ So the probability of selecting at random a person with an IQ score above $140$ is $0.39 \\%$. That’s about one person in every $250$ individuals! ## Multimedia Links For a video presentation", "is $7(94). So the possible sums for the 7 test scores so that they provide an integer average are$7(98), 7(97), 7(96), 7(95)$and$7(94)$which are$686, 679, 672, 665,$and$658$respectively. Now in order to get the sum of the first 6 tests, we negate$95$from each sum producing$591, 584, 577, 570,$and$563$. Notice only$570$is divisible by$6$so, therefore, the sum of the first$6$tests is$570$. We need to find her score on the$6th$test so what number minus$570$will give us a number divisible by$5$. Since$95$is the$7th$test score and all test scores are distinct that only leaves$\\boxed{\\textbf{(E) } 100}$. ==Solution 2 (Cheap Solution)== By inspection, the sequences$ (Error compiling LaTeX. ! Missing $inserted.)91,93,92,96,98,100,95$and$93,91,92,96,98,100,95$work, so the answer is$\\boxed{\\textbf{(E) } 100}$. Note: A method of finding this \"cheap\" solution is to create a \"mod chart\", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.", "is less than 470. Hence, maximum possible elements in S is 30. We know that x = $16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$ = $(16 + 19)(16^2 – 16 * 19 + 19^2) + (17 + 18)(17^2 – 17 * 18 + 18^2)$ = 35 × odd + 35 × odd = 35 × even = 35 × (2k) => x = 70k => Remainder when divided by 70 is 0. The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively. 5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3 Let the total number of tests be ‘n’ and the average by ‘A’ Total score = n*A When 1st 10 tests are excluded, decrease in total value of scores = (nA – 20 * 10) = (nA – 200) Also, (n – 10)(A + 1) = (nA – 200) On solving, we get 10A – n = 190……….(i) When last 10 tests are excluded, decrease in total value of scores = (nA – 30 * 10) = (nA – 300) Also, (n – 10)(A – 1) = (nA – 300) On solving, we get 10A + n = 310……….(ii) From (i) and (ii), we get n", "+ 10 + 18 + 12 + 4 + x )}{8}$ = 11 = 72 + x = 88; x = 88 – 72 = 16 Step 3: Required new score = 16 Q 5 - Find the value for a new score that will yield a given mean. 17 , 7 , 2 , 15 , 6 , 19 , 20 , 15 , 18; New mean = 14 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(17 + 7 + 2 + 15 + 6 + 19 + 20 + 15 + 18 + x )}{10}$ = 14 = 119 + x = 140; x = 140 – 119 = 21 Step 3: Required new score = 21 Q 6 - Find the value for a new score that will yield a given mean. 21 , 10 , 16 , 8 , 19; New mean = 16 ### Explanation Step 1: Let the new score to be added = x Step 2: New mean = $\\frac{(21 + 10 + 16 + 8 + 19 + x )}{6}$ = 16 = 74 + x = 96; x = 96 – 74 = 22 Step 3: Required new score = 22 Q 7 - Find the value for", "\\,12$$ First, arrange all these values (from least to greatest). This, however, is not a necessary step, but it sometimes is helpful to discover the mode if the values or numbers will be put in ascending order. $$0,\\, 1,\\, 4, \\,5, \\,5, \\,8, {\\color{Red} \\,12, \\,12,\\, 12,\\, 12}$$ So, find the number or value that most often occurs. The answer is: Our mode here is: $$12$$ Now, let’s take a look at some examples that include a few relevant data. An example: Problem: Carlos attained these scores on his math tests: $$84, \\,92, \\,74, \\,98, \\,82$$. Now, come up with the mean, median, and mode of Carlos’ scores. $$\\frac{84+92+74+98+82}{5}$$ For finding the mean, we need to add together all the scores and divide that value by the number of the tests. So: $$430$$ divided by $$5$$ is $$86 (\\frac{430}{5} = 86)$$. The mean we needed to find is $$86$$. $$74, \\,82, \\,84, \\,92, \\,98$$ For finding the median, first we need to put Carlos’ test scores in order (from least to greatest). The median is $$84$$. We have five scores in total. In our ordered list, the middle score is the third. This our median. $$74, \\,82, \\,84, \\,92, \\,98$$ Because all numbers appear exactly just one time, in the data set is no mode. The mean in" ]

[ "# Hippo from Hippopotamus Five letters are randomly chosen from the word HIPPOPOTAMUS. What is the probability that the five letters can be used to spell HIPPO? The probability can be expressed as a/b for co-prime positive integers a and b, find the value of a+b. ×", "$$2$$ of which are A. So the probability that Sindhu chooses the letter A is $$\\frac{2}{11}$$.", ") Therefore, the probability that at least one letter is in its proper envelope = $\\frac{4}{3!} = \\frac{4}{6} = \\frac{2}{3}$ Exams Articles Questions", "the 26 possible english alphabets. In this MS Excel tutorial from ExcelIsFun, the 385th installment in their series of digital spreadsheet magic tricks, you'll learn how to generate random letters without RANDBETWEEN function using the formula =CHAR(65+INT(RAND()*26)). John puts the letters of the word FOOTFEST in 8 different envelopes. The study of probability helps us figure out the likelihood of something happening. How many different words can be formed if the word contains: a) 3 letters b) 4 letters c) all letters. How many ways can all of the letters in the word ABRACADABRA be arranged? [a] 39,916,800 [b] 1,995,840 [c] 83,160 [d] 11 [e] None of the above. Find an answer to your question If a letter from the word ADMINISTRATION is selected at random, what is the probability that it is not a vowel letter?. Use the Formula dialog box to create your formula. a) Find the probability that the 10 letter arrangement will spell STATISTICS. Hence the probability of not happening of event A is 0. Suppose a word is picked at random from this sentence. This process is summarized in Figure 1. Alphabats - Rhyming Words. 3 letter Words made out of math. If a letter is chosen at random from the word MATHEMATICS, what is the probability of choosing a T? B If", "letter will be randomly selected from the word MAINE, what is the probability that the two letters selected will not be the same? 25000 +道题目 146本备考书籍", "## Thursday, January 21, 2010 ### Drawing letters A string of alphabets are randomly generated one letter at a time. Each time, one obtains the letter $A,\\cdots,Z$ with probability $p(A),\\cdots, p(Z)$. Given that the sum of these probabilities is 1, what is the expected number of draws before the string \"DHARMATH\" appears?", "got my computer to find the probability of every nine letter word and found the following probabilities: Consonants Vowels Probability of nine letter word 0 9 0 1 8 0 2 7 0 3 6 0.000546 4 5 0.019724 5 4 0.076895 6 3 0.051417 7 2 0.005662 8 1 0.000033 9 0 0 So the best way to get a nine letter word in the first letters game is to pick five consonants and four vowels. ### Similar posts Countdown probability, pt. 2 Pointless probability Big Internet Math-Off stickers 2019 World Cup stickers 2018, pt. 3 Comments in green were written by me. Comments in blue were not written by me.", "Browse Questions # A single letter is selected at random from the word PROBABILITY .The probability that it is a vowel is $\\begin{array}{1 1}(A)\\;\\large\\frac{1}{3}\\\\(B)\\;\\large\\frac{4}{11}\\\\(C)\\;\\large\\frac{2}{11}\\\\(D)\\;\\large\\frac{3}{11}\\end{array}$ Can you answer this question? Number of letters in a word =11 Number of vowels in a word =4 $\\therefore$ Required probability =$\\large\\frac{4}{11}$ Hence (B) is the correct answer. answered Jul 9, 2014", "discussion is for students to become familiar with using probability to describe chance events. It also provides an opportunity to informally assess how students think about probabilities where “or” or “not” is used. Display the word KANGAROOS for all to see. Here are some questions for discussion. • “If a letter is selected at random, what is the probability that it is a vowel?” ($$\\frac{4}{9}$$) • “If a letter is selected at random, what is the probability that it is a consonant?” ($$\\frac{5}{9}$$) • “If a letter is selected at random, what is the probability that it is an S?” ($$\\frac{1}{9}$$) • “If a letter is selected at random, what is the probability that it is an A?” ($$\\frac{2}{9}$$) • “If a letter is selected at random, what is the probability that it is an S or an A?” ($$\\frac{3}{9}$$) • “If a letter is selected at random, what is the probability that it is not a K?” ($$\\frac{8}{9}$$) • “Whose partner came up with some great clues for problem 5?” (Sample response: My partner came up with two great clues. $$P(\\text{vowel}) = \\frac{1}{4}$$. $$P(\\text{J}) = \\frac{1}{4}$$)) • “What are some words that satisfy the given probabilities?” (Sample responses: just, jump, jack, jars) • “For the first question, why is FLOWER a valid word?” (It has 6 letters", "# If a letter is chosen at random from the English alphabet, Question: If a letter is chosen at random from the English alphabet, find the probability that the letter is chosen is (i) a vowel (ii) a consonant Solution: (i) We know that, Probability of occurrence of an event $=\\frac{\\text { Total no. of Desired outcomes }}{\\text { Total no. of outcomes }}$ Total possible outcomes are alphabets from a to $z$ Desired outcomes are $a, e, i, o, u$ Total no. of outcomes are 26 and desired outputs are 5 Therefore, the probability of picking a vowel $=\\frac{5}{26}$ Conclusion: Probability of choosing a vowel is $\\frac{5}{26}$ (ii) We know that, Probability of occurrence of an event $=\\frac{\\text { Total no. of Desired outcomes }}{\\text { Total no.of outcomes }}$ Total possible outcomes are all alphabets from a to z Desired outcomes are all alphabets except $a, e, i, o, u$ Total no.of outcomes are 26 and desired outputs are 21 Therefore, the probability of picking a consonant $=\\frac{21}{26}$ Conclusion: Probability of choosing a consonant is $\\frac{21}{26}$", "# Can you list down every selection ? Probability Level 4 $\\large \\boxed{A,A,A,A,A,B,B,B,B,C,C,C,D,D \\ }$ How many total selections of $5$ letters from the above $14$ letters are possible ? ×", "Question # A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant. Solution ## Total numbers of alphabet $$=26$$Total numbers of consonants $$=21$$Probability is given by,$$P(E)= \\dfrac{total \\ outcomes}{favorable\\ outcome}$$$$P$$ (choosing a consonant) $$=\\dfrac{21}{26}$$MathematicsRS AgarwalStandard X Suggest Corrections 0 Similar questions View More Same exercise questions View More People also searched for View More", "class, and then pick word(s) according to the probability distribution, $$p(\\mathbf{x}|y)$$.", "appear in the same word, or something else entirely? And on the other side, explicit values for the probability of two letters appearing together on your grid are going to be hard to calculate with anything short of simply enumerating all the possibilities. The best approach may be something akin to Dan Brumleve's mention of simulated annealing: start with an arbitrary 'layout' (an assignment of 96 tiles to the 6$\\times$16 faces of your dice) and then run a few thousand simulations to get the joint probability for all (or some suitably chosen set) combinations of two letters. Then make a change to your layout, do another simulation run, and if the joint probabilities are closer to 'correct' than they were for the first layout, accept the change; repeat this until you come up with a distribution that offers reasonably good joint probabilities. Unfortunately, even this still doesn't actually accomodate all of the relevant correlations that occur in the english language. For instance, the probability that at least one vowel occurs in a word is much, much higher than would be estimated either from either individual letter or joint two-letter probabilities, to the extent where (again) dice-based word games often make certain to have one or two dice that are all-vowels (and contrariwise, some dice that are all-consonants) to", "appropriate number of tiles to use for each letter; once you have those, it doesn't matter how you assign them to the faces of the dice. But again, of course... this is 'to first approximation', with the assumptions that letter probabilities are independent of each other; that the probability that a word has two As in it is the square of the probability that it has one A, or that the probability that a word contains an A and an S is the product of the individual letter probabilities P(A) and P(S). But of course, this isn't true; letter combinations in English are highly correlated. For instance, Q appears so much more regularly with 'U' than without it that virtually every English word game (with the rare and notable exception of Scrabble) pairs the two together into a single glyph. There are plenty of other examples, too: many consonants (and a couple of vowels) are more likely to appear in multiples than single-letter probabilities would dictate; 'H's are often paired with 'T's and 'S's, 'G's with 'N's, etc. What's more, these correlations are inordinately hard to model, because it's not even clear what the appropriate measure is; do you want to know the probability that two letters appear next to each other, or simply the probability that they", "# If two letters are taken at random from the word CANE, what is the probability that none of the letters would be vowels? +1 vote 181 views If two letters are taken at random from the word CANE, what is the probability that none of the letters would be vowels? posted Jul 31, 2018 Share this puzzle ## 2 Answers +1 vote 1/6 Number of ways of selecting 2 letters of word CANE- 2*(3+2+1)=12 Number of ways selecting 2 vowels- 2 (CN & NC) Probability- 2/12=1/6 answer Jul 31, 2018 +1 vote • Two letters are taken randomly from out of the word \"CANE\"=12 • They are {(CA),(AC),(CN),(NC),(CE),(EC),(AN),(NA),(AE),(EA),(NE),(EN)} • None of the letter is a vowel =2 • They are {(CN),(NC)} *So, =2/12 =1/6 • The answer is 1/6 answer Jul 31, 2018 Similar Puzzles 0 votes If two letters are randomly taken one after the other from the word AUGUST, what is the probability that none of the letters would be vowels?", "a bag. If one ball is drawn at random, find the probability of drawing: a The number 14. b An even number. c A number higher than 6. d The number 8 or lower. e The number 5,\\, 8,\\, 17 \\text{ or } 20. 12 Glen randomly picks one of the letters from the word \\text{MINTS}. a How many different possible outcomes are there? b Find the probability that Glen picks the letter: i I ii T 13 In a standard deck of cards, find the probability of picking: a The ten of diamonds. b c A red card. 14 Each 26 letters of the alphabet are written on separate pieces of paper and placed in a bag. If one letter is to be picked out of the bag at random, find the probability of picking a: a J b K, Y or R c M, K, D, O, H or B d Letter in the word PROBABILITY e Letter in the word WORKBOOK 15 A letter is randomly selected from the alphabet. Find the following probabilities: a P\\left(\\text{J}\\right) b P\\left(\\text{any letter after T}\\right) c P\\left(\\text{any letter between H and N}\\right) d P\\left(\\text{any letter before F}\\right) e P \\left( \\text{vowel} \\right) f P \\left( \\text{consonant} \\right) g P \\left( \\text{consonants between H and N} \\right) h P \\left(", "Probability Crossword Puzzle? Howdy, First of all, I hope this problem falls under \"probability\". I think it does but I'm not quite sure. I thought of maybe posting it under the \"puzzles\" but here it is. I'm completely not sure how to go about this problem. I feel like there is more than one way to approach this. I just know words can go forwards, backwards, and diagonal and just get stuck there. So here it is: What is the probability that a letter in a given position within a 25x25 word search puzzle is the first letter of a randomly chosen three-letter word beginning with the letter \"T\"? In your calculations, assume that the letter of choice is \"interior\" to the puzzle, in the sense that it is not possible to \"leave\" the puzzle by going two letters in any of the allowed directions. Repeat the calculation (with the same proviso about letter position) for one of seventeen speci cally chosen four-letter words. Suppose you were offered a wager whereby you paid $1 to a bookie, who would return$2 if one of the seventeen words is in the puzzle, but would keep your dollar otherwise. Would you take the bet? What is the expectation value of your cash flow in this transaction? Thanks everyone for pitching in!", "117: A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that a letter is : a vowel The word ‘ASSASSINATION’ has 13 letters in which 6 are vowels (viz. AAAIIO) and 7 consonants (viz. SSSSNNT) $\\therefore$ n (S) = 13. No. of vowels = 6. ##### Question 118: A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that a letter is : a consonant The word ‘ASSASSINATION’ has 13 letters in which 6 are vowels (viz. AAAIIO) and 7 consonants (viz. SSSSNNT) $\\therefore$ n (S) = 13. No. of consonants = 7. ##### Question 119: In a lottery, a person chooses six different numbers at random from 1 to 20 and if these six numbers match with six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game ? ##### Question 120: Check whether the following probabilities P (A) and P (B) are consistently defined : P (A) = 0·5, P (B) = 0·7, P (A $\\cap$ B) = 0·06 P (A $\\mathrm{\\upsilon }$ B) = P (A) + P (B) – P (A $\\cap$ B) = 0·5 + 0·7 – 0·06 = 1·20 – 0·06 = 1·14 > 1. Hence the probabilities are not consistently defined. #####", "\"What is the probability of choosing the letter i from the word probability?\" has the answer 2/9 (following logic of having a sample set of 9 letters in \"probability\"). Random number generators can be hardware based or pseudo-random number generators. A letter is chosen at random from the English alphabets What is the probability that it is a letter of word 'MATHEMATICS' its urgent please answer fast - Math - Probability. The word 'CRICKET' has$7$letters where$2are vowels (I, E). This is a great improvement. In the continuous case, it is areas under the curve that define the probabilities. In order to know when a random variable in a statistical sample does not have a binomial distribution, you first have to know what makes it binomial. It also applies to Microsoft PowerPoint and Excel 2010 and higher. Random definition, proceeding, made, or occurring without definite aim, reason, or pattern: the random selection of numbers. By Royale Evrard. However, work goes hard and time-consuming when there are numerous cells. Math is an accepted word in Word with Friends having 9 points. The maximum length of a word is 8 letters. The random module provides access to functions that support many operations. Convert each individual number into a letter. The original Greek alphabet consisted of 27 letters and was written from" ]

[ "# Can you list down every selection ? Probability Level 4 $\\large \\boxed{A,A,A,A,A,B,B,B,B,C,C,C,D,D \\ }$ How many total selections of $5$ letters from the above $14$ letters are possible ? ×", "# Probability of occurrence of 4 letters in a sample of text Assume x is a letter whose probability is determined by P(x). Given N outcomes, what are the chances of observing m of the possible outcomes together at least once? e.g. For a string of lower-case text reduced to letters only to a length of 20 characters, what is the probability of observing at least one char of each of the following, 'a','b','c','d' at least one. i.e. the text contains 'a', 'b', 'c','d' in any order, duplicates allowed. Thank you very much. - The probability $Q$ to observe at least once each letter in a set $B$ of size $b$, in a word of length $\\ell$ written in an alphabet $A$ of size $a\\ge b$ when the word is chosen uniformly at random, is $$Q=\\sum_{n=0}^b(-1)^n{b\\choose n}\\left(1-\\frac{n}a\\right)^\\ell.$$ Hence the number of such words is $$a^\\ell Q=\\sum_{n=0}^b(-1)^n{b\\choose n}(a-n)^\\ell.$$ If $b=4$, $a=26$ and $\\ell=20$, W|A says that this sum of $b+1=5$ terms is $$Q=\\frac{188,637,413,085,495,043,814,491,263}{2,491,018,611,901,176,144,042,524,672}=7.573\\%.$$ The proof of the general formula for $Q$ is rather simple if one uses the inclusion-exclusion principle. To wit, for each letter $x$ in $A$, consider the event $C_x$ that the word does not contain $x$. Then, one sees that $Q$ is the probability of the event $$\\bigcap_{x\\in B}(\\Omega\\setminus C_x)=\\Omega\\setminus C,\\qquad C=\\bigcup_{x\\in B}C_x,$$ hence $$1-Q=\\mathrm P(A)=\\sum_{n=1}^b", "choose the remaining $2$ letters? - is 1 a permutation? – John Feb 5 '13 at 19:31 Does the order in which you place the $p$ matter? Or think of it this way, you have to \"choose\" $3$ places for your $p'$s. – Jean-Sébastien Feb 5 '13 at 19:34 Assumption 100% random distribution of letters (does not occur naturally in any language, but for this problem I'm assuming a random character generator for each letters). Variables X = Chance of getting 1 P = 1/26 = 0.03846153846153846153846153846154 Y = Chance of getting anything but P = 25/26 = 0.96153846153846153846153846153846 Chance of getting 3 Ps in a row = X^3 Chance of getting 2 Non Ps in a row = Y^2 Chance of getting 3 Ps in a row followed by 2 Non Ps = X^3 * Y^2 Commutative property says the order of the underlying multiplication operations doesn't matter - so the order of the letters shouldn't impact the probability. 5.2603334832598513842167775853578e-5 - You need to clarify if you want the number of words possible with 3 p's, the probability of randomly generated 5 letter words having 3 p's, or whatever specific answer is desired. I guess this is the probability of a single randomly generated word having exactly 3 P's in it: Adding to what Don said, Let", "1. ## probability four letters are chosen at from the letters of the word TAILORED .find the probability of getting a T and three vowels. 2. ## Re: probability Originally Posted by markosheehan four letters are chosen at from the letters of the word TAILORED .find the probability of getting a T and three vowels. The probability of getting any particular subset of four is $\\dfrac{1}{\\dbinom{8}{4}}$ Next list all the correct outcomes. 3. ## Re: probability Choose three vowels: $\\dbinom{4}{3}$ ways. Choose the $t$: 1 way. By the product principle, there are 4 ways to choose a T and three vowels. Total number of ways to choose four letters from TAILORED: $\\dbinom{8}{4}$ Probability: $\\dfrac{4}{\\dbinom{8}{4}} = \\dfrac{2}{35}$ 4. ## Re: probability thanks for the description of how you came to the answer", "selecting from the whole population at random for each letter, for example), that's $\\sum_x P(L_1=x) \\times P(L_2=x)$. But of course then the two probability distributions will be the same. Let's call the probability that a random letter has the value $x$ \"$p_x$\". That is, the overall probability = $\\sum_x p_x^2$ In your case, that's 0.6*0.6 + 0.25*0.25 + 0.1*0.1 + 0.05*0.05 which comes to 0.435. -- generalization to $k$ letters: To find the probability that $k$ letters selected from the population are all identical: If $L_1$ is the first letter, $L_2$ is the second letter, ..., and $L_k$ is the $k^\\mathrm{th}$ then the probability is $\\sum_x P(L_1=x) \\times P(L_2=x|L_1=x)\\times P(L_3=x|L_1=x,L2=x)\\times ... \\times P(L_k=x|L_1=x,L2=x, ...,L_k=x)$. If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's $\\sum_x P(L_1=x) \\times P(L_2=x)\\times ... \\times P(L_k=x)$. But of course then all the probability distributions will be the same. Let's call the probability that a random letter has the value $x$ \"$p_x$\". That is, the overall probability = $\\sum_x p_x^k$ • Thanks, this is exactly the sort of answer I was hoping for. It seems that if I wanted to extrapolate to 3 letters, I would just sum over $p_x^3$ then? – Arne Roomann-Kurrik Mar 21 '13 at 17:19", "# The probability of forming Mississippi by choosing random letters from Mississippi I'm having difficulty with the following problem: You choose a letter at random from the word Mississippi eleven times without replacement. What is the probability that you can form the word Mississippi with the eleven chosen letters? Hint: it may be helpful to number the eleven letters as $1, 2, . . . , 11$. This is how I approached it: the number of possible outcomes is $11!$, since we're taking one word at a time without replacement, so there are $11$ choices for the first letter, $10$ choices for the second letter and so on. Now, as to the number of 'success' outcomes, I noticed that there are $4$ s's to choose from, $4$ i's, $1$ M and $2$ p's. So, in order to form the word Mississippi, we have for the first letter $1$ option, $4$ for the second and third letters, $3$ for the fourth (since we've already used one \"s\") and so on, which amounts to a total of $4^2*3^2*2^3=1152$ different ways of doing so. However, my answer does not match the one provided in my book (Henk Tijn's Understanding Probability 3rd edition). What am I doing wrong? Thanks very much in advance. • If you choose eleven times without replacement then", "getting Three coins are tossed together. ∴ Number of total outcomes = (2)3 = 2 × 2 × 2 = 8 (i) Probability of getting at the most 2 heads P(E) = HTT, HHT, HTH, HTT, THH, TTH, THT = 7 $\\frac{7}{8}$ (ii) Probability of getting 3 heads (HHH = 1) = $\\frac{1}{8}$ Question 10. A letter is chosen from the word ‘RECTANGLE’. What is the probability that it is (i) a consonant (ii) not a consonant. In the word ‘RECTANGLE’ Total letters (outcomes) = 9 (i) Probability of a letter which is a consonant P(E) = (R, C, T, N, G, L) $\\frac{6}{9}=\\frac{2}{3}$ (ii) Probability of a letter being not a consonant (E, A, E), P(E) = $\\frac{3}{9}=\\frac{1}{3}$. — End of Data Handling Class-8 ML Aggarwal Solutions :–", "the probability of getting Solution: Three coins are tossed together. ∴ Number of total outcomes = (2)3 = 2 × 2 × 2 = 8 (i) Probability of getting at the most 2 heads P(E) = HTT, HHT, HTH, HTT, THH, TTH, THT = 7 = $$\\frac{7}{8}$$ (ii) Probability of getting 3 heads (HHH = 1) = $$\\frac{1}{8}$$ Question 10. A letter is chosen from the word ‘RECTANGLE’. What is the probability that it is (i) a consonant (ii) not a consonant. Solution: In the word ‘RECTANGLE’ Total letters (outcomes) = 9 (i) Probability of a letter which is a consonant P(E) = (R, C, T, N, G, L) = $$\\frac{6}{9}=\\frac{2}{3}$$ (ii) Probability of a letter being not a consonant (E, A, E), P(E) = $$\\frac{3}{9}=\\frac{1}{3}$$.", "best combination to pick in order to get a nine letter word. Assuming the letters in countdown are still distributed like this, the probability of getting combinations of letters can be calculated. As the probability throughout the game is dependent on which letters have been picked, I have worked out the probability of getting a nine letter word on the first letters game. ### The probability of YODELLING YODELLING has three vowels and six consonants. There are 6 (3!) ways in which the vowels could be ordered and 720 (6!) ways in which the consonants can be ordered, although each is repeated at there are two Ls, so there are 360 distinct ways to order the consonants. The probability of each of these is: $$\\frac{21\\times 13\\times 13\\times 6\\times 3\\times 5\\times 4\\times 8\\times 1}{67\\times 66\\times 65\\times 74\\times 73\\times 72\\times 71\\times 70\\times 69}$$ So the probability of getting YODELLING is: $$\\frac{6\\times 360\\times 21\\times 13\\times 13\\times 6\\times 3\\times 5\\times 4\\times 8\\times 1}{67\\times 66\\times 65\\times 74\\times 73\\times 72\\times 71\\times 70\\times 69} = 0.000000575874154$$ ### The probability of any nine letter word I got my computer to find the probability of every nine letter word and found the following probabilities: Consonants Vowels Probability of nine letter word 0 9 0 1 8 0 2 7 0 3 6 0.000546 4 5 0.019724 5", "to the multivariate case; that is, it models the probability you’ll draw a certain number of As, Bs, Cs, etc. after drawing a number of tiles from the bag. Fortunately, the R package extraDistr provides an R version of the multivariate hypergeometric probability mass function. Here’s a function that, given a word of length $N$ and the number of each letter tile in a bag, gives the probability of drawing that word in $N$ draws: Using this function, we can find the probability of drawing RASTAFARIAN: And the result is $4.28\\times10^{-6}\\%$. Pretty lucky! Now, what is the probability of drawing any valid 11 letter word to start the game? Note that in most cases, spelling a word using all your 11 tiles excludes the possibility of spelling another word. This suggests the the probability of spelling word $A$ OR word $B$ is given by $P(A \\cap B)=P(A) + P(B)$. However, there is a special case: what if word $A$ and word $B$ are spelled with the same letters? In order to avoid double counting, we need to only want to include words with the same letters once. I downloaded a list of words in the SOWPODS scrabble dictionary from a GitHub repository and loaded them into R. To deduplicate words with the same letters, I sorted the letters", "# Proving that the probability of choosing a letter from words of a given length is $1/n$ Let us consider the following: suppose I have a word of length $$n$$ from an ordered alphabet on $$n$$ letters $$\\{1, ..., n\\}$$, i.e., an element $$(w_i)$$ of $$\\{n\\}^{\\{n\\}}$$. What is the probability of choosing the first letter $$w(1)$$ from all possible words, considering that I would choose $$w(j)$$ over $$w(i)$$ with probability $$1$$ if $$w(i)>w(j)$$, and if I have for some subset $$I\\subset{n}$$ (EDIT: and some $$a\\in \\{n\\}$$ such that $$w^{-1}(a)=I$$,) I am equally likely to choose any element from $$I$$? For example, if I wanted to choose a letter from all two-letter words in two-letter alphabet, i.e., $$\\{1,2\\}^{\\{1,2\\}}=\\{(11), (12), (21), (22)\\}$$, then I will choose the first letter with probability $$\\frac{1}{2}=\\frac{1}{4}\\cdot\\left(\\frac{1}{2}\\cdot{1\\choose 1}((2-1)^{0}+(2-2)^0)+1\\cdot {1 \\choose 0}((2-1)^1+(2-2)^1) \\right),$$ where I put $$(2-2)^0=1$$. In general, I seem to arrive to the formula $$n^{n-1}=\\sum_{n>m\\ge0,\\ n\\ge k \\ge 1} \\frac{1}{n-m}{n-1 \\choose m} (n-k)^m.$$ It seems to work for numbers up to $$4$$. This seems a bit silly, because obviously any word is equally likely to start with any letter. But I do not quite see how the symmetry helps in the case of this bizarre distribution. • What's $a$ in $w^{-1}(a)$? – joriki Dec 11 '19 at 11:57 • @joriki I consider $w: \\{n\\}", "$$2$$ of which are A. So the probability that Sindhu chooses the letter A is $$\\frac{2}{11}$$.", "then $$P(V|X)$$ be the updated probability that he can discern vinyl from digital after observing him identify $$X$$ songs correctly. How large does $$X$$ have to be for $$P(V|X) >= .9$$? ##### 2.8 You roll a fair, six-sided die twice. Let $$X$$ be the sum of the two rolls. Find $$P(X = 7)$$ using a conditioning argument; that is, do not simply count the number of ways to roll a 7 and divide by the number of possible combinations for the two rolls. ##### 2.9 Imagine generating a random word by sampling $$3 < n < 26$$ letters, with replacement (from the 26 letters in the alphabet). What is the probability that this word has no repeats; i.e., $$n$$ unique letters? ##### 2.10 Ali and Bill are taking a test. For any single question, Ali has equal probabilities of answering correctly or incorrectly, and Bill also has equal probabilities of answering correctly or incorrectly. For any single question, the probability that both Ali and Bill get the question correct is .4. Given that Bill gets a question wrong, what is the probability that Ali gets it right? ##### 2.11 Consider the birthday problem with the usual assumptions. Previously, we’ve considered a ‘match’ as a single day with multiple birthdays; here, imagine a week match, which consists of a", "if $H(a)<1$, then the first term is $L\\times H(a)<L$. This can help a lot because it makes $I$ smaller. It helps a lot because the change is in the exponent. Let's look at some examples. We could first look at English text. The linear heteropolymers of English are strings of the letters a-z (let's just stick with lower case letters and no punctuation for simplicity). What is the likelihood to find the word ${\\tt origins}$ by chance? If we use an unbiased typewriter (our 26-sided coin), the likelihood is $26^{-7}$ (about 1 in 8 billion), as ${\\tt origins}$ is a 7-mer, and each mer is information (there is only one way to spell the word ${\\tt origins}$). Can we do better if our typewriter is biased towards English? Let's find out. If you analyze English text, you quickly notice that letters occur at different frequencies: e more often that t, which occurs more often than a, and so forth. The plot below is the distribution of letters that you would find. Letter distribution of English text The entropy-per-letter of this distribution is 0.89 mers. Not very different from 1, but let's see how it changes the 1 in 8 billion odds. The biased-search chance is, according to this theory, $P_\\star=26^{7\\times 0.89}$, which comes out about 1.5 per billion:", "word of length $$N$$ and the number of each letter tile in a bag, gives the probability of drawing that word in $$N$$ draws: word_probability <- function(w, freqs) { # Count the number of times each # letter appears in the word letters = table(stringr::str_split(toupper(w), \"\")) # Assign a frequency of zero to any letter # not used in the word letter_freqs <- rep(0, 26) names(letter_freqs) <- LETTERS letter_freqs[names(letters)] <- letters n = freqs, k = sum(letter_freqs)) } Using this function, we can find the probability of drawing RASTAFARIAN: bananagram_freqs <- c(13, 3, 3, 6, 18, 3, 4, 3, 12, 2, 2, 5, 3, 8, 11, 3, 2, 9, 6, 9, 6, 3, 3, 2, 3, 2) word_probability(\"RASTAFARIAN\", bananagram_freqs) And the result is $$4.28\\times10^{-6}\\%$$. Pretty lucky! Now, what is the probability of drawing any valid 11 letter word to start the game? Note that in most cases, spelling a word using all your 11 tiles excludes the possibility of spelling another word. This suggests the the probability of spelling word $$A$$ OR word $$B$$ is given by $$P(A \\cap B)=P(A) + P(B)$$. However, there is a special case: what if word $$A$$ and word $$B$$ are spelled with the same letters? In order to avoid double counting, we only want to include words with the same letters once.", "be selected only in $2$ ways (from the remaining $2$ letters). Thus the number of arrangement of first three letters is $3 \\times 2$ = $6$ And finally there is only $1$ way of selecting the last letter, the letter which is left out. Thus, with $A$ as the first letter, there can be $6 \\times 1$ = $6$ different codes are possible. When we repeat the same exercise with letters $B, C$ and $D$ as the first letter, the total number of arrangements is $4 \\times 6$ = $24$. Now we can cut short the entire process using the multiplication law just by writing the total number of arrangements as $4 \\times 3 \\times 2 \\times 1$ = $24$ More on the concept. The concept explained above is widely used in determining the ultimate probability of an occurrence of two events. The two events may be independent or dependent. Suppose we draw a card from a well shuffled standard pack of cards and draw a card is drawn randomly. Now there are two possibilities. The drawn card may be put in back, well shuffled again and again a card is drawn randomly for the occurrence of the first event. The occurrence of the second event is not affected by the outcome of the first event. In such", "# Probability that z precedes both a and b in the permutation? What is the probability of the event that z precedes both a and b when we randomly select a permutation of the 26 lowercase letters of the English alphabet? Currently, my thoughts are: P(25,25)x24 There are p(25,25) ways to get certain letters of the alphabet 24 ways to place z, it can't be in the last 2 spots since it has to precede a and b. - You have ruled out some ways to place $z$, but it is obviously possible to place $z$ somewhere in the first 24 spots without it preceding $a$ and $b$. So there is no hope that your strategy can be correct, since you are counting combinations that do not belong. – Erick Wong Dec 1 '12 at 3:03 There are $6$ permutations of our set of $3$ letters. In $2$ of them, z precedes both a and b, so the probability is $\\dfrac{2}{6}$. The locations of the other letters is irrelevant. If this is not obvious, note that given any specific locations for the other $23$ letters, all of the $3!$ orders of our target letters are equally likely, and $2$ of these orders, z, a, b and z, b, a satisfy our condition. Remark: One can also do it", "the outcomes of an experiment is greater than one 8. A letter is chosen random from the word $COCACOLA$. The probability that the letter will be a vowel is 9. The probability of an event that cannot occur is 10. The probability of an event that is unlikely to happen is closest to 11. The experiments that are repeated under identical conditions and they do not produce the same outcome every time are called 12. A coin and a die is tossed simultaneously. The probability of getting 'tail' and 'prime number' is 13. On a single roll of a die, the probability of getting a number less than 7 is 14. The experiments that are repeated under identical conditions and they produce the same outcome every time are called 15. A number $y$ is chosen at random from the numbers $-3, -2, -1, 0, 1, 2, 3, 4$. What is the probability of $y<2$ 16. A letter is chosen random from the word $COMPUTER$. The probability that the letter will be a consonant is 17. Two unbiased coins are tossed simultaneously and the probability of getting not head is $\\frac{A}{B}$. The value of $(A+B)^2$ is 18. A coin is tossed twice. The probability of getting heads both the time is 19. A die is thrown once. The probability", "8\\times 1}{67\\times 66\\times 65\\times 74\\times 73\\times 72\\times 71\\times 70\\times 69} = 0.000000575874154$$ ### The probability of any nine letter word I got my computer to find the probability of every nine letter word and found the following probabilities: Consonants Vowels Probability of nine letter word 0 9 0 1 8 0 2 7 0 3 6 0.000546 4 5 0.019724 5 4 0.076895 6 3 0.051417 7 2 0.005662 8 1 0.000033 9 0 0 So the best way to get a nine letter word in the first letters game is to pick five consonants and four vowels. ### Similar posts Countdown probability, pt. 2 Pointless probability World Cup stickers 2018, pt. 2 A bad Puzzle for Today", "Pete031012 18 # You want to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter. What is the probability that a randomly chosen code starts with A? number of all possibilities - $5!=120$ number of codes starting with A - $4!=24$ the probability is $\\frac{24}{120}=\\frac{1}{5}$" ]

[ "and the number of each letter tile in a bag, gives the probability of drawing that word in $N$ draws: Using this function, we can find the probability of drawing RASTAFARIAN: And the result is $4.28\\times10^{-6}\\%$. Pretty lucky! Now, what is the probability of drawing any valid 11 letter word to start the game? Note that in most cases, spelling a word using all your 11 tiles excludes the possibility of spelling another word. This suggests the the probability of spelling word $A$ OR word $B$ is given by $P(A \\cap B)=P(A) + P(B)$. However, there is a special case: what if word $A$ and word $B$ are spelled with the same letters? In order to avoid double counting, we need to only want to include words with the same letters once. I downloaded a list of words in the SOWPODS scrabble dictionary from a GitHub repository and loaded them into R. To deduplicate words with the same letters, I sorted the letters in each word and removed duplicates: I then used the word_probability function to calculate the probability of drawing each 11 letter word, and then summed them all up: Which computes the probability of drawing a valid 11 letter word in the opening tiles to be $~0.28\\%$. Now, suppose you start the game by drawing a", "$$2$$ of which are A. So the probability that Sindhu chooses the letter A is $$\\frac{2}{11}$$.", "the 26 possible english alphabets. In this MS Excel tutorial from ExcelIsFun, the 385th installment in their series of digital spreadsheet magic tricks, you'll learn how to generate random letters without RANDBETWEEN function using the formula =CHAR(65+INT(RAND()*26)). John puts the letters of the word FOOTFEST in 8 different envelopes. The study of probability helps us figure out the likelihood of something happening. How many different words can be formed if the word contains: a) 3 letters b) 4 letters c) all letters. How many ways can all of the letters in the word ABRACADABRA be arranged? [a] 39,916,800 [b] 1,995,840 [c] 83,160 [d] 11 [e] None of the above. Find an answer to your question If a letter from the word ADMINISTRATION is selected at random, what is the probability that it is not a vowel letter?. Use the Formula dialog box to create your formula. a) Find the probability that the 10 letter arrangement will spell STATISTICS. Hence the probability of not happening of event A is 0. Suppose a word is picked at random from this sentence. This process is summarized in Figure 1. Alphabats - Rhyming Words. 3 letter Words made out of math. If a letter is chosen at random from the word MATHEMATICS, what is the probability of choosing a T? B If", "1. ## probability four letters are chosen at from the letters of the word TAILORED .find the probability of getting a T and three vowels. 2. ## Re: probability Originally Posted by markosheehan four letters are chosen at from the letters of the word TAILORED .find the probability of getting a T and three vowels. The probability of getting any particular subset of four is $\\dfrac{1}{\\dbinom{8}{4}}$ Next list all the correct outcomes. 3. ## Re: probability Choose three vowels: $\\dbinom{4}{3}$ ways. Choose the $t$: 1 way. By the product principle, there are 4 ways to choose a T and three vowels. Total number of ways to choose four letters from TAILORED: $\\dbinom{8}{4}$ Probability: $\\dfrac{4}{\\dbinom{8}{4}} = \\dfrac{2}{35}$ 4. ## Re: probability thanks for the description of how you came to the answer", "# Can you list down every selection ? Probability Level 4 $\\large \\boxed{A,A,A,A,A,B,B,B,B,C,C,C,D,D \\ }$ How many total selections of $5$ letters from the above $14$ letters are possible ? ×", "best combination to pick in order to get a nine letter word. Assuming the letters in countdown are still distributed like this, the probability of getting combinations of letters can be calculated. As the probability throughout the game is dependent on which letters have been picked, I have worked out the probability of getting a nine letter word on the first letters game. ### The probability of YODELLING YODELLING has three vowels and six consonants. There are 6 (3!) ways in which the vowels could be ordered and 720 (6!) ways in which the consonants can be ordered, although each is repeated at there are two Ls, so there are 360 distinct ways to order the consonants. The probability of each of these is: $$\\frac{21\\times 13\\times 13\\times 6\\times 3\\times 5\\times 4\\times 8\\times 1}{67\\times 66\\times 65\\times 74\\times 73\\times 72\\times 71\\times 70\\times 69}$$ So the probability of getting YODELLING is: $$\\frac{6\\times 360\\times 21\\times 13\\times 13\\times 6\\times 3\\times 5\\times 4\\times 8\\times 1}{67\\times 66\\times 65\\times 74\\times 73\\times 72\\times 71\\times 70\\times 69} = 0.000000575874154$$ ### The probability of any nine letter word I got my computer to find the probability of every nine letter word and found the following probabilities: Consonants Vowels Probability of nine letter word 0 9 0 1 8 0 2 7 0 3 6 0.000546 4 5 0.019724 5", "# Hippo from Hippopotamus Five letters are randomly chosen from the word HIPPOPOTAMUS. What is the probability that the five letters can be used to spell HIPPO? The probability can be expressed as a/b for co-prime positive integers a and b, find the value of a+b. ×", "A letter is chosen at random from the word ‘ASSASSINATION’. Question: A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii) an consonant Solution: There are 13 letters in the word ASSASSINATION. $\\therefore$ Hence, $n(S)=13$ (i) There are 6 vowels in the given word. $\\therefore$ Probability (vowel) $=\\frac{6}{13}$ (ii) There are 7 consonants in the given word. $\\therefore$ Probability (consonant) $=\\frac{7}{13}$", "to ignore them here. So remember that the probability for spontaneous emergence of life is so small because $I$ is large, and it is in the exponent. But now we realize that the $L$ in (3) is really the entropy of a spontaneously created sequence, and if $H(a)<1$, then the first term is $L\\times H(a)<L$. This can help a lot because it makes $I$ smaller. It helps a lot because the change is in the exponent. Let's look at some examples. We could first look at English text. The linear heteropolymers of English are strings of the letters a-z (let's just stick with lower case letters and no punctuation for simplicity). What is the likelihood to find the word ${\\tt origins}$ by chance? If we use an unbiased typewriter (our 26-sided coin), the likelihood is $26^{-7}$ (about 1 in 8 billion), as ${\\tt origins}$ is a 7-mer, and each mer is information (there is only one way to spell the word ${\\tt origins}$). Can we do better if our typewriter is biased towards English? Let's find out. If you analyze English text, you quickly notice that letters occur at different frequencies: e more often that t, which occurs more often than a, and so forth. The plot below is the distribution of letters that you would find. Letter", "to ignore them here. So remember that the probability for spontaneous emergence of life is so small because $I$ is large, and it is in the exponent. But now we realize that the $L$ in (3) is really the entropy of a spontaneously created sequence, and if $H(a)<1$, then the first term is $L\\times H(a)<L$. This can help a lot because it makes $I$ smaller. It helps a lot because the change is in the exponent. Let's look at some examples. We could first look at English text. The linear heteropolymers of English are strings of the letters a-z (let's just stick with lower case letters and no punctuation for simplicity). What is the likelihood to find the word ${\\tt origins}$ by chance? If we use an unbiased typewriter (our 26-sided coin), the likelihood is $26^{-7}$ (about 1 in 8 billion), as ${\\tt origins}$ is a 7-mer, and each mer is information (there is only one way to spell the word ${\\tt origins}$). Can we do better if our typewriter is biased towards English? Let's find out. If you analyze English text, you quickly notice that letters occur at different frequencies: e more often that t, which occurs more often than a, and so forth. The plot below is the distribution of letters that you would find. Letter", "choose the remaining $2$ letters? - is 1 a permutation? – John Feb 5 '13 at 19:31 Does the order in which you place the $p$ matter? Or think of it this way, you have to \"choose\" $3$ places for your $p'$s. – Jean-Sébastien Feb 5 '13 at 19:34 Assumption 100% random distribution of letters (does not occur naturally in any language, but for this problem I'm assuming a random character generator for each letters). Variables X = Chance of getting 1 P = 1/26 = 0.03846153846153846153846153846154 Y = Chance of getting anything but P = 25/26 = 0.96153846153846153846153846153846 Chance of getting 3 Ps in a row = X^3 Chance of getting 2 Non Ps in a row = Y^2 Chance of getting 3 Ps in a row followed by 2 Non Ps = X^3 * Y^2 Commutative property says the order of the underlying multiplication operations doesn't matter - so the order of the letters shouldn't impact the probability. 5.2603334832598513842167775853578e-5 - You need to clarify if you want the number of words possible with 3 p's, the probability of randomly generated 5 letter words having 3 p's, or whatever specific answer is desired. I guess this is the probability of a single randomly generated word having exactly 3 P's in it: Adding to what Don said, Let", "\"What is the probability of choosing the letter i from the word probability?\" has the answer 2/9 (following logic of having a sample set of 9 letters in \"probability\"). Random number generators can be hardware based or pseudo-random number generators. A letter is chosen at random from the English alphabets What is the probability that it is a letter of word 'MATHEMATICS' its urgent please answer fast - Math - Probability. The word 'CRICKET' has$7$letters where$2are vowels (I, E). This is a great improvement. In the continuous case, it is areas under the curve that define the probabilities. In order to know when a random variable in a statistical sample does not have a binomial distribution, you first have to know what makes it binomial. It also applies to Microsoft PowerPoint and Excel 2010 and higher. Random definition, proceeding, made, or occurring without definite aim, reason, or pattern: the random selection of numbers. By Royale Evrard. However, work goes hard and time-consuming when there are numerous cells. Math is an accepted word in Word with Friends having 9 points. The maximum length of a word is 8 letters. The random module provides access to functions that support many operations. Convert each individual number into a letter. The original Greek alphabet consisted of 27 letters and was written from", "# Proving that the probability of choosing a letter from words of a given length is $1/n$ Let us consider the following: suppose I have a word of length $$n$$ from an ordered alphabet on $$n$$ letters $$\\{1, ..., n\\}$$, i.e., an element $$(w_i)$$ of $$\\{n\\}^{\\{n\\}}$$. What is the probability of choosing the first letter $$w(1)$$ from all possible words, considering that I would choose $$w(j)$$ over $$w(i)$$ with probability $$1$$ if $$w(i)>w(j)$$, and if I have for some subset $$I\\subset{n}$$ (EDIT: and some $$a\\in \\{n\\}$$ such that $$w^{-1}(a)=I$$,) I am equally likely to choose any element from $$I$$? For example, if I wanted to choose a letter from all two-letter words in two-letter alphabet, i.e., $$\\{1,2\\}^{\\{1,2\\}}=\\{(11), (12), (21), (22)\\}$$, then I will choose the first letter with probability $$\\frac{1}{2}=\\frac{1}{4}\\cdot\\left(\\frac{1}{2}\\cdot{1\\choose 1}((2-1)^{0}+(2-2)^0)+1\\cdot {1 \\choose 0}((2-1)^1+(2-2)^1) \\right),$$ where I put $$(2-2)^0=1$$. In general, I seem to arrive to the formula $$n^{n-1}=\\sum_{n>m\\ge0,\\ n\\ge k \\ge 1} \\frac{1}{n-m}{n-1 \\choose m} (n-k)^m.$$ It seems to work for numbers up to $$4$$. This seems a bit silly, because obviously any word is equally likely to start with any letter. But I do not quite see how the symmetry helps in the case of this bizarre distribution. • What's $a$ in $w^{-1}(a)$? – joriki Dec 11 '19 at 11:57 • @joriki I consider $w: \\{n\\}", "the probability of getting Solution: Three coins are tossed together. ∴ Number of total outcomes = (2)3 = 2 × 2 × 2 = 8 (i) Probability of getting at the most 2 heads P(E) = HTT, HHT, HTH, HTT, THH, TTH, THT = 7 = $$\\frac{7}{8}$$ (ii) Probability of getting 3 heads (HHH = 1) = $$\\frac{1}{8}$$ Question 10. A letter is chosen from the word ‘RECTANGLE’. What is the probability that it is (i) a consonant (ii) not a consonant. Solution: In the word ‘RECTANGLE’ Total letters (outcomes) = 9 (i) Probability of a letter which is a consonant P(E) = (R, C, T, N, G, L) = $$\\frac{6}{9}=\\frac{2}{3}$$ (ii) Probability of a letter being not a consonant (E, A, E), P(E) = $$\\frac{3}{9}=\\frac{1}{3}$$.", "selecting from the whole population at random for each letter, for example), that's $\\sum_x P(L_1=x) \\times P(L_2=x)$. But of course then the two probability distributions will be the same. Let's call the probability that a random letter has the value $x$ \"$p_x$\". That is, the overall probability = $\\sum_x p_x^2$ In your case, that's 0.6*0.6 + 0.25*0.25 + 0.1*0.1 + 0.05*0.05 which comes to 0.435. -- generalization to $k$ letters: To find the probability that $k$ letters selected from the population are all identical: If $L_1$ is the first letter, $L_2$ is the second letter, ..., and $L_k$ is the $k^\\mathrm{th}$ then the probability is $\\sum_x P(L_1=x) \\times P(L_2=x|L_1=x)\\times P(L_3=x|L_1=x,L2=x)\\times ... \\times P(L_k=x|L_1=x,L2=x, ...,L_k=x)$. If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's $\\sum_x P(L_1=x) \\times P(L_2=x)\\times ... \\times P(L_k=x)$. But of course then all the probability distributions will be the same. Let's call the probability that a random letter has the value $x$ \"$p_x$\". That is, the overall probability = $\\sum_x p_x^k$ • Thanks, this is exactly the sort of answer I was hoping for. It seems that if I wanted to extrapolate to 3 letters, I would just sum over $p_x^3$ then? – Arne Roomann-Kurrik Mar 21 '13 at 17:19", "appear in the same word, or something else entirely? And on the other side, explicit values for the probability of two letters appearing together on your grid are going to be hard to calculate with anything short of simply enumerating all the possibilities. The best approach may be something akin to Dan Brumleve's mention of simulated annealing: start with an arbitrary 'layout' (an assignment of 96 tiles to the 6$\\times$16 faces of your dice) and then run a few thousand simulations to get the joint probability for all (or some suitably chosen set) combinations of two letters. Then make a change to your layout, do another simulation run, and if the joint probabilities are closer to 'correct' than they were for the first layout, accept the change; repeat this until you come up with a distribution that offers reasonably good joint probabilities. Unfortunately, even this still doesn't actually accomodate all of the relevant correlations that occur in the english language. For instance, the probability that at least one vowel occurs in a word is much, much higher than would be estimated either from either individual letter or joint two-letter probabilities, to the extent where (again) dice-based word games often make certain to have one or two dice that are all-vowels (and contrariwise, some dice that are all-consonants) to", "getting Three coins are tossed together. ∴ Number of total outcomes = (2)3 = 2 × 2 × 2 = 8 (i) Probability of getting at the most 2 heads P(E) = HTT, HHT, HTH, HTT, THH, TTH, THT = 7 $\\frac{7}{8}$ (ii) Probability of getting 3 heads (HHH = 1) = $\\frac{1}{8}$ Question 10. A letter is chosen from the word ‘RECTANGLE’. What is the probability that it is (i) a consonant (ii) not a consonant. In the word ‘RECTANGLE’ Total letters (outcomes) = 9 (i) Probability of a letter which is a consonant P(E) = (R, C, T, N, G, L) $\\frac{6}{9}=\\frac{2}{3}$ (ii) Probability of a letter being not a consonant (E, A, E), P(E) = $\\frac{3}{9}=\\frac{1}{3}$. — End of Data Handling Class-8 ML Aggarwal Solutions :–", "then $$P(V|X)$$ be the updated probability that he can discern vinyl from digital after observing him identify $$X$$ songs correctly. How large does $$X$$ have to be for $$P(V|X) >= .9$$? ##### 2.8 You roll a fair, six-sided die twice. Let $$X$$ be the sum of the two rolls. Find $$P(X = 7)$$ using a conditioning argument; that is, do not simply count the number of ways to roll a 7 and divide by the number of possible combinations for the two rolls. ##### 2.9 Imagine generating a random word by sampling $$3 < n < 26$$ letters, with replacement (from the 26 letters in the alphabet). What is the probability that this word has no repeats; i.e., $$n$$ unique letters? ##### 2.10 Ali and Bill are taking a test. For any single question, Ali has equal probabilities of answering correctly or incorrectly, and Bill also has equal probabilities of answering correctly or incorrectly. For any single question, the probability that both Ali and Bill get the question correct is .4. Given that Bill gets a question wrong, what is the probability that Ali gets it right? ##### 2.11 Consider the birthday problem with the usual assumptions. Previously, we’ve considered a ‘match’ as a single day with multiple birthdays; here, imagine a week match, which consists of a", "QUESTION # If a letter of English alphabet is chosen at random, then the probability that the letter is a consonant is:A. $\\dfrac{5}{{26}}$B. $\\dfrac{{21}}{{26}}$C. $\\dfrac{{10}}{{13}}$D. $\\dfrac{{11}}{{13}}$ Hint: In this question use this concept that the total number of English alphabets are twenty six out of them five are vowels and the rest twenty one are consonants. Since total alphabets $= 26$ Vowels $= 5$ Number of consonant $= 26 - 5 = 21$ As we know that ${\\text{Probability}} = \\dfrac{{{\\text{Number of favourable outcome}}}}{{{\\text{Number of total outcome}}}}$ Number of favorable outcome is the number of consonants $= 21$ Number of total outcome $= 26$ Therefore the probability that letter is a consonant $= \\dfrac{{21}}{{26}}$", "first letter chosen was a consonant, now there would be 4 consonants and a total of 8 letters left in the box. Find the probability that the second letter chosen is a consonant. 48 = 12 P(second consonant) = 12 59 = 5 18 Multiply. The probability of choosing two letters that are both consonants is 5 18 Try This: Example 3B B. If two letters are chosen at random, what is the probability that they will both be consonants or both be vowels? There are two possibilities: 2 consonants or 2 vowels. The probability of 2 consonants was calculated in Try This 3A. Now find the probability of getting 2 vowels. Find the probability that the first letter chosen is a vowel. 49 P(first vowel) = If the first letter chosen was a vowel, there are now only 3 vowels and 8 total letters left in the box. Try This: Example 3B Continued 38 Find the probability that the second letter chosen is a vowel. P(second vowel) = 38 49 = 12 72 16 = Multiply. The events of both consonants and both vowels are mutually exclusive, so you can add their probabilities. 5 18 1 6 + = 8 18 = 49 P(consonant) + P(vowel) The probability of getting two letters that are either both consonants" ]

[ "to 9*10=90. Re: Maximum number of points that 10 circles of different radii intersect &nbs [#permalink] 26 Feb 2017, 16:25 Display posts from previous: Sort by", "with each other. What is the maximum number of intersections that a hundred circles could have? ### More Pebbles ##### Stage: 2 and 3 Challenge Level: Have a go at this 3D extension to the Pebbles problem.", "# Intersecting Lines And Circles The maximum number of points of intersection of 5 straight lines and 4 circles in a plane is $$\\text{__________}$$. ×", "313 views There are three coplanar parallel lines. If any $p$ points are taken on each of the lines, then find the maximum number of triangles with the vertices of these points. 1. $p^{2}(4p-3)$ 2. $p^{3}(4p-3)$ 3. $p(4p-3)$ 4. $p^{3}$ 1 246 views 2 174 views", "# Intersecting Lines And Circles Probability Level 3 The maximum number of points of intersection of 5 straight lines and 4 circles in a plane is $\\text{\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_}$. ×", "maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 the maximum number of points that two circles intersect is two. therefore, each of these 10 circles intersect two point with each of the other 9 circles. as a consequence, the number of these points is equal to two times the number of combination of the 10 circles. $$2*\\frac{(10!)}{(2!*10!)}= 90$$ _________________ Please press Kudos if you like my post Intern Joined: 22 Jan 2015 Posts: 23 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] Show Tags 01 Feb 2017, 01:22 sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 =n(n-1) =10*9 =90 (D) Intern Joined: 29 Dec 2015 Posts: 4 GMAT 1: 600 Q41 V32 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] Show Tags 26 Feb 2017, 16:25 Every circle intersects with another circle at max 2 points. Every circle interacts with a total of 9 circles. So, every circle has a max of 18 intersections (2*9). There are 10 circles, so 2*9*10 intersections. However, every intersection in counted twice. therefore we must divide by 2. this gets us", "49 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 25 Oct 2016, 12:39 1 KUDOS sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 the maximum number of points that two circles intersect is two. therefore, each of these 10 circles intersect two point with each of the other 9 circles. as a consequence, the number of these points is equal to two times the number of combination of the 10 circles. $$2*\\frac{(10!)}{(2!*10!)}= 90$$ _________________ Please press Kudos if you like my post Intern Joined: 22 Jan 2015 Posts: 15 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 01 Feb 2017, 00:22 sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 =n(n-1) =10*9 =90 (D) Intern Joined: 29 Dec 2015 Posts: 4 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 26 Feb 2017, 15:25 Every circle intersects with another circle at max 2 points. Every circle interacts with a total of 9 circles. So, every circle has a max of", "# Thread: Two circles and a square 1. ## Two circles and a square Two distinct circles and a square lie in the plane. What is the maximum number of points of intersection of two or more of these three figures? 2. Hello, chil2e! Two distinct circles and a square lie in the plane. What is the maximum number of points of intersection of 2 or more of these 3 figures? We can \"talk\" our way through most of it. Let $S, C_1, C_2$ represent the square and two circles. We can \"see\" that: . . $S$ and $C_1$ intersect in at most 2 points. . . $S$ and $C_2$ intersect in at most 2 points. . . $C_1$ and $C_2$ intersect in at most 2 points. The maximum number of intersections is: . $2+2+2 \\:=\\:6$ Bottom line: is such a figure possible? Code: * * * * * * * * * * * o * * * * * * * * * * * - - o - o - - * * * | * * | * | | * | * * | * * | o | * * | * * | * * o * * o * | | * - - - - - - - *", "# Too many intersections You have 8 convex octagons and 10 circles in a plane. How many maximum number of points of intersections could be there. $$Hint$$: This is a theoretical question. ( no geometry required) Try my set : Let's play with polygons × Problem Loading... Note Loading... Set Loading...", "maximum chord 13 long. How many intersect points have circle with the coordinate axes?", "# Little bit tricky! Discrete Mathematics Level pending What is the maximum number of intersection points between four circles and four straight lines? ×", "coplanar circles that intersect at exactly one point are called _____. Coplanar force systems have all the forces acting in in one plane. In this points, lines, and planes worksheet, 10th graders solve 9 various problems related to determining points, lines, and planes. In your example that can only happen at a one very specific roll and pitch, depending on how the two pads line up with a selected coordinate system.", "and 2 intersection points per pair, the maximum number of intersection points should be $190 \\times 2 = 380$ intersection points. I'd say $2 \\times \\frac{n!}{(n-2)! 2!}$ which works out to 380 for n=20. Every circle can intersect every other circle at exactly two points. From 20 circles, there are 190 distinct ways to choose a pair. To show that such a construction is possible, Have each circle be the same size. Then choose a point somewhere, and draw the circles so that the point is inside every one of them. Two same sized circles will intersect if (and only if) there is some point that's inside both circles. • That's a pretty complex way of writing $n^2-n$ – Kruga Aug 16 '18 at 8:13 • @Kruga it sure is! – Bass Aug 16 '18 at 9:51", "# Three little circles Geometry Level 4 As shown in the diagram below, three overlapping circles with distinct radii define six points. The lengths of 5 line segments (in blue) connecting the points are known to us: 4, 6, 8, 3, 5. What is the length of the remaining segment (in yellow), to 3 decimal places? ×", "× # 5 super hard questions.... 1. Suppose we have n distinct lines on a plane. What is the maximum and minimum number of intersections? 2. Suppose we have n distinct points on a plane. What is the maximum and minimum number of line that they can form? 3. What is the maximum number of region that n lines can divide a plane into? 4. Given n collinear points , what is the maximum and minimum distance between the points? 5. Arrange seven points on a plane such that , for any 3 points , we can find 2 points with distance 1. Note by Werjohi Khhun 1 year, 7 months ago Sort by: 1. Minimum=0, all lines are parallel.For maximum, no line must be parallel or concurrent. Now each of n lines intersects at 1 point. Thus, maximum number of intersections=$$\\dbinom{n}{2}$$ 2. Minimum=1, all points are collinear. For maximum, assume all points on a circle, ie. no 3 of them are collinear, this will give again $$\\dbinom{n}{2}$$ 3. This can be proved by recursion. See complete proof here 4. Minimum$$=0^+$$, maximum$$=\\infty$$. Either I am not understanding it properly or its vague. · 1 year, 7 months ago For 1 , 2 , how do u know that it is N 2 ?? And , can you please", "# Four Touching Circles Geometry Level 1 There are 4 circles, each with a radius of 8 meters, that are tangent to each other (touching) as seen in the picture. What is the area of the red region? ×", "# Intersections between chords On a circle circumference, we choose 50 points and draw all $${n \\choose 2}$$ chords joining these points. Find the maximum number of intersections between these chords? Hint: when the number of points is 5, the maximum number of line intersections is 5. × Problem Loading... Note Loading... Set Loading...", "is the maximum value that the standard deviation can take? Chord Stage: 5 Challenge Level: Equal touching circles have centres on a line. From a point of this line on a circle, a tangent is drawn to the farthest circle. Find the lengths of chords where the line cuts the other circles.", "# Maximum points of intersection n circles in C++ C++Server Side ProgrammingProgramming In this tutorial, we will be discussing a program to find maximum points of intersection n circles For this we will be provided with the number of circles. Our task is to find the maximum number of intersections the given number of circles meet. ## Example Live Demo #include <bits/stdc++.h> using namespace std; //returning maximum intersections int intersection(int n) { return n * (n - 1); } int main() { cout << intersection(3) << endl; return 0; } ## Output 6 Updated on 09-Sep-2020 12:16:37", "# Maximum number of points of intersection between $7$ figures. We are given $5$ lines and two circles in a plane. What is the maximum number of possible intersection points among these seven figures ? My work on the problem: I've considered special cases like what is the maximum number of points of intersection between lines and circles ,between only lines and last between circles. P.s: Please provide detailed answer so i can follow best.Thanks in advance • If you think about it, the circles are easy to collocate once you've maxed the intersection betwen the lines. By the way, the answer is 32, try to obtain and prove it – Exodd Sep 28 '15 at 20:13 ## 1 Answer When you only have 5 lines, you can get at most $\\binom{5}{2}=10$ intersections. This is the number of distinct pairs of lines among those five. You can draw it on paper if you don't know the binomial coefficient symbol yet. Each of 2 circles can intersect each of 5 lines at 2 points. There can also be 2 circle-circle intersections. Line-line intersections: $10$ Line-circle intersections: $2\\cdot 2 \\cdot 5$ Circle-circle intersections: $2$ Add it up to get the answer. • Pardon, but where is the intuition behind $\\binom{5}{2}=10$ using geometrical objects ? i mean i used that method" ]

[ "and 2 intersection points per pair, the maximum number of intersection points should be $190 \\times 2 = 380$ intersection points. I'd say $2 \\times \\frac{n!}{(n-2)! 2!}$ which works out to 380 for n=20. Every circle can intersect every other circle at exactly two points. From 20 circles, there are 190 distinct ways to choose a pair. To show that such a construction is possible, Have each circle be the same size. Then choose a point somewhere, and draw the circles so that the point is inside every one of them. Two same sized circles will intersect if (and only if) there is some point that's inside both circles. • That's a pretty complex way of writing $n^2-n$ – Kruga Aug 16 '18 at 8:13 • @Kruga it sure is! – Bass Aug 16 '18 at 9:51", "× # 5 super hard questions.... 1. Suppose we have n distinct lines on a plane. What is the maximum and minimum number of intersections? 2. Suppose we have n distinct points on a plane. What is the maximum and minimum number of line that they can form? 3. What is the maximum number of region that n lines can divide a plane into? 4. Given n collinear points , what is the maximum and minimum distance between the points? 5. Arrange seven points on a plane such that , for any 3 points , we can find 2 points with distance 1. Note by Werjohi Khhun 1 year, 7 months ago Sort by: 1. Minimum=0, all lines are parallel.For maximum, no line must be parallel or concurrent. Now each of n lines intersects at 1 point. Thus, maximum number of intersections=$$\\dbinom{n}{2}$$ 2. Minimum=1, all points are collinear. For maximum, assume all points on a circle, ie. no 3 of them are collinear, this will give again $$\\dbinom{n}{2}$$ 3. This can be proved by recursion. See complete proof here 4. Minimum$$=0^+$$, maximum$$=\\infty$$. Either I am not understanding it properly or its vague. · 1 year, 7 months ago For 1 , 2 , how do u know that it is N 2 ?? And , can you please", "It can be observed that there can be a maximum of $2$ points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.", "# Maximum number of points of intersection between $7$ figures. We are given $5$ lines and two circles in a plane. What is the maximum number of possible intersection points among these seven figures ? My work on the problem: I've considered special cases like what is the maximum number of points of intersection between lines and circles ,between only lines and last between circles. P.s: Please provide detailed answer so i can follow best.Thanks in advance • If you think about it, the circles are easy to collocate once you've maxed the intersection betwen the lines. By the way, the answer is 32, try to obtain and prove it – Exodd Sep 28 '15 at 20:13 ## 1 Answer When you only have 5 lines, you can get at most $\\binom{5}{2}=10$ intersections. This is the number of distinct pairs of lines among those five. You can draw it on paper if you don't know the binomial coefficient symbol yet. Each of 2 circles can intersect each of 5 lines at 2 points. There can also be 2 circle-circle intersections. Line-line intersections: $10$ Line-circle intersections: $2\\cdot 2 \\cdot 5$ Circle-circle intersections: $2$ Add it up to get the answer. • Pardon, but where is the intuition behind $\\binom{5}{2}=10$ using geometrical objects ? i mean i used that method", "# Intersecting Lines And Circles Probability Level 3 The maximum number of points of intersection of 5 straight lines and 4 circles in a plane is $\\text{\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_}$. ×", "# Intersecting Lines And Circles The maximum number of points of intersection of 5 straight lines and 4 circles in a plane is $$\\text{__________}$$. ×", "maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 the maximum number of points that two circles intersect is two. therefore, each of these 10 circles intersect two point with each of the other 9 circles. as a consequence, the number of these points is equal to two times the number of combination of the 10 circles. $$2*\\frac{(10!)}{(2!*10!)}= 90$$ _________________ Please press Kudos if you like my post Intern Joined: 22 Jan 2015 Posts: 23 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] Show Tags 01 Feb 2017, 01:22 sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 =n(n-1) =10*9 =90 (D) Intern Joined: 29 Dec 2015 Posts: 4 GMAT 1: 600 Q41 V32 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] Show Tags 26 Feb 2017, 16:25 Every circle intersects with another circle at max 2 points. Every circle interacts with a total of 9 circles. So, every circle has a max of 18 intersections (2*9). There are 10 circles, so 2*9*10 intersections. However, every intersection in counted twice. therefore we must divide by 2. this gets us", "# Thread: Two circles and a square 1. ## Two circles and a square Two distinct circles and a square lie in the plane. What is the maximum number of points of intersection of two or more of these three figures? 2. Hello, chil2e! Two distinct circles and a square lie in the plane. What is the maximum number of points of intersection of 2 or more of these 3 figures? We can \"talk\" our way through most of it. Let $S, C_1, C_2$ represent the square and two circles. We can \"see\" that: . . $S$ and $C_1$ intersect in at most 2 points. . . $S$ and $C_2$ intersect in at most 2 points. . . $C_1$ and $C_2$ intersect in at most 2 points. The maximum number of intersections is: . $2+2+2 \\:=\\:6$ Bottom line: is such a figure possible? Code: * * * * * * * * * * * o * * * * * * * * * * * - - o - o - - * * * | * * | * | | * | * * | * * | o | * * | * * | * * o * * o * | | * - - - - - - - *", "Thus, $2$ is the maximum number of intersections.", "to 9*10=90. Re: Maximum number of points that 10 circles of different radii intersect &nbs [#permalink] 26 Feb 2017, 16:25 Display posts from previous: Sort by", "49 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 25 Oct 2016, 12:39 1 KUDOS sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 the maximum number of points that two circles intersect is two. therefore, each of these 10 circles intersect two point with each of the other 9 circles. as a consequence, the number of these points is equal to two times the number of combination of the 10 circles. $$2*\\frac{(10!)}{(2!*10!)}= 90$$ _________________ Please press Kudos if you like my post Intern Joined: 22 Jan 2015 Posts: 15 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 01 Feb 2017, 00:22 sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 =n(n-1) =10*9 =90 (D) Intern Joined: 29 Dec 2015 Posts: 4 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 26 Feb 2017, 15:25 Every circle intersects with another circle at max 2 points. Every circle interacts with a total of 9 circles. So, every circle has a max of", "Three circles can intersect at maximum 6 points.. Because we select 2 circles from 3 and each two circles can intersect at maximum 2 points 2$$c^3_2$$ ... Three circles can intersect at 6 points So 11 circles can intersect at maximum $$2c^{11}_2$$ ... 2 * 11 * 5 = 110 Intern Joined: 01 Nov 2015 Posts: 29 Location: India Concentration: Marketing, Entrepreneurship WE: Engineering (Computer Software) Re: What is the greatest possible number of points at which 11 circles wit [#permalink] ### Show Tags 20 Dec 2015, 19:46 2 1st Circle doesn't intersect itself - 2*(1-1)=0 points 2nd Circle intersects 1st at 2*(2-1)=2 Points 3rd Circle intersect 2nd and 3rd at 2*(3-1)=4 Points 4th Circle intersects other Circles at 2*(4-1)=6 points ... 11th circle intersects other circles at 2*(11-1)=20 points Therefore total no of points = 2 + 4 + 6 + ... + 20 $$Sum = 10/2 * (20+2) = 5*22 =110$$ Answer is E Intern Joined: 09 Jul 2015 Posts: 45 Re: What is the greatest possible number of points at which 11 circles wit [#permalink] ### Show Tags 21 Dec 2015, 16:54 2 Answer is E -> 110. 2 circle intersects at 2 points, 3 circles intersect at 4 points, 4 circles intersect at 6 points, 5 circles at 8 points, etc., Hence max #", "# What is the maximum number of points of intersection of 8 circles? I came across this question on a book on combinations and permutations. Although it had a vaguely defined one step answer : 8P2 = 56. As far as I could understand, the above step gives the number of arrangements possible for 8 objects taken two at a time. But I don't understand how it gives the points of intersection. Also, is there any other method to solve this kind of problems ? Regards • This is rather confusing. Are all the circles intersecting at the same points, or just any circles intersecting? I agree that the latter case seems unintuitive if the answer is 28. – theREALyumdub Nov 13 '15 at 17:16 • @theREALyumdub Can you please explain, considering latter is the case. Thanks – H G Sur Nov 13 '15 at 17:17 • Isn't 8P2 $56$? – user2357112 supports Monica Nov 13 '15 at 17:47 • For the pedantic, you might want to restrict the problem so that none of the circles are the same. – Rick Decker Nov 13 '15 at 21:32 Two circles can intersect in at most two points (excluding the case where two circles coincide with each other). So by looking at all possible pairs of circles, the maximum number", "# Too many Circles Using 20 Circles, what is the maximum number of intersecting point that can be obtained? For example, if there were 3 circles, the answer would be $6$ as shown below: • 20 non-concentric circles? – nikki Aug 15 '18 at 20:00 • I'd assume so (there's a facetious answer otherwise). Is it also a safe assumption that they all be uniformly sized? – El-Guest Aug 15 '18 at 20:01 • @nikki :) then there would be infinite intersections. – Oray Aug 15 '18 at 20:01 • @El-Guest not necessarily, you can use any size you would like – Oray Aug 15 '18 at 20:01 • @Oray interesting, thanks for the clarification :) – El-Guest Aug 15 '18 at 20:02 any two circles will have maximum two intersection points between them (think of a Venn diagram for this). I'm not able to program/draw this out at the moment (especially not with 20 circles), but I feel that the maximum can be obtained when each pair of circles intersects twice. Now, how many pairs of circles are there? Since order doesn't matter, using choose notation we can say that there are $C(20,2)$, ie. 20 choose 2 pairs of circles. This evaluates to $C(20,2) = \\frac{20!}{18!2!} = \\frac{20(19)}{2} = 190$. Since there are 190 pairs of circles", "at a point (b) at two points (c) at an infinite number of points (d) in a line (a) Two lines intersect at a point. Question 10: Two points in a plane determine (a) exactly one line segment (b) exactly two line segments (c) an infinite number of line segments (d) none of these (a) exactly one line segment Two points in a plane determine exactly one line segment with those two points as its end points. Question 11: The minimum number of points of intersection of three lines in a plane is (a) 1 (b) 2 (c) 3 (d) 0 (d) 0 Three lines will not necessarily intersect in a plane. Thus, the minimum point of intersection will be 0. Question 12: The maximum number of points of intersection of three lines in a plane is (a) 0 (b) 1 (c) 2 (d) 3 (d) 3 The maximum number of points of intersection of three lines that intersect in a plane are three. Question 13: Choose the correct statement: (a) every line has a definite length (b) every ray has a definite length (c) every line segment has a definite length (d) none of these (c) Every line segment has a definite length. Every line segment has a definite length, which can be measured using a ruler.", "# Maximum number of regions formed by points on a circle The question is : 6 points are located on a circle and lines are drawn connecting these points, each pair of points connected by a single line. What can be the maximum number of regions into which the circle is divided? My answer is 32 . But the actual answer is 31 how ?? - In general the maximum number of regions you can get from $n$ points is given by $${n \\choose 4} + {n \\choose 2} + 1$$ This is an oft cited puzzle to show the perils of generalizing based on first few values. We get powers of $2$ till $n=5$, after which we get $31$.", "= 0, thus results in 1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, 106, 121, 137, 154, 172, 191, 211, ... ## Proof The maximum number of pieces from consecutive cuts are the numbers in the Lazy Caterer's Sequence. When a circle is cut n times to produce the maximum number of pieces, represented as p = f(n), the nth cut must be considered; the number of pieces before the last cut is f(n − 1), while the number of pieces added by the last cut is n. To obtain the maximum number of pieces, the nth cut line should cross all the other previous cut lines inside the circle, but not cross any intersection of previous cut lines. Thus, the nth line itself is cut in n − 1 places, and into n line segments. Each segment divides one piece of the (n − 1)-cut pancake into 2 parts, adding exactly n to the number of pieces. The new line can't have any more segments since it can only cross each previous line once. A cut line can always cross over all previous cut lines, as rotating the knife at a small angle around a point that is not an existing intersection will, if the angle is small enough, intersect all", "# Points of Intersection of Curves There are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines in the $x-y$ plane. Find the maximum possible value of the number of their points of intersection. My Attempt at the Question: I first to try and figure out the maximum number of points of intersection. For that I figured that parabola and circle intersect at max. $4$ points. And straight line will intersect the circle and parabola at $4$ distinct points. But this lead me nowhere. First we find the kinds of intersection that take place b/w the lines, circles and parabolas, which are as follows:- 1. Parabola with a parabola 2. Parabola with a circle 3. Parabola with a line 4. Circle with a circle 5. Circle with a line 6. Line with a line Now, lets count the maximum number of point of intersection in each of the above cases and consequently count the maximum number of intersections that cantake place given that there are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines:- CASE 1:- Intersection of two parabolas As we can see in the above image there are maximum $4$ possible points of intersection for two distinct parabolas. Since there are $4$ distinct parabolas hence there are maximum $4\\cdot\\binom{4}{2}$ maximum distinct point of intersections.", "# How can you construct as many intersections as possible with n lines? If you have $n$ lines, it seems to be obvious that you can have at most $\\frac{n^2-n}{2}$ intersections: $n = 1$: Obviously you need two lines to intersect, so the maximum number of intersections is 0 $n=2$: As one line can't intersect with itself, it can only intersect with the old lines. As there is only one line, you can get at most one intersection. So the maximum number of intersections of two lines is 1. $n=3$: The same argument as before. The new line can only intersect with the two old lines and a given pair of two lines can intersect at most once. So you can get at most two new intersections. With the intersection of the two lines before, you get 3 intersections. So the pattern is $\\sum_{i=1}^{n-1} i = \\frac{n^2-n}{2}$ But this only gives an upper border for the maximum number of intersections. I would now like to give a constrution where you can obviously see that you can get that many intersections. I tried to make such an construction by hand: but I guess with 10 lines this will get very confusing. Also, I don't know how to make it obvious that this construction can be continues indefinitely. Do you", "E. 200 * A solution will be posted in two days. Maximum point of Intersection between n different Circles. = 2 *nC2, where n >= 2. _________________ If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !! Re: There are different 10 circles. What is the number of the greatest pos &nbs [#permalink] 29 Jul 2018, 04:58 Display posts from previous: Sort by" ]

[ "and 2 intersection points per pair, the maximum number of intersection points should be $190 \\times 2 = 380$ intersection points. I'd say $2 \\times \\frac{n!}{(n-2)! 2!}$ which works out to 380 for n=20. Every circle can intersect every other circle at exactly two points. From 20 circles, there are 190 distinct ways to choose a pair. To show that such a construction is possible, Have each circle be the same size. Then choose a point somewhere, and draw the circles so that the point is inside every one of them. Two same sized circles will intersect if (and only if) there is some point that's inside both circles. • That's a pretty complex way of writing $n^2-n$ – Kruga Aug 16 '18 at 8:13 • @Kruga it sure is! – Bass Aug 16 '18 at 9:51", "# Intersecting Lines And Circles The maximum number of points of intersection of 5 straight lines and 4 circles in a plane is $$\\text{__________}$$. ×", "× # 5 super hard questions.... 1. Suppose we have n distinct lines on a plane. What is the maximum and minimum number of intersections? 2. Suppose we have n distinct points on a plane. What is the maximum and minimum number of line that they can form? 3. What is the maximum number of region that n lines can divide a plane into? 4. Given n collinear points , what is the maximum and minimum distance between the points? 5. Arrange seven points on a plane such that , for any 3 points , we can find 2 points with distance 1. Note by Werjohi Khhun 1 year, 7 months ago Sort by: 1. Minimum=0, all lines are parallel.For maximum, no line must be parallel or concurrent. Now each of n lines intersects at 1 point. Thus, maximum number of intersections=$$\\dbinom{n}{2}$$ 2. Minimum=1, all points are collinear. For maximum, assume all points on a circle, ie. no 3 of them are collinear, this will give again $$\\dbinom{n}{2}$$ 3. This can be proved by recursion. See complete proof here 4. Minimum$$=0^+$$, maximum$$=\\infty$$. Either I am not understanding it properly or its vague. · 1 year, 7 months ago For 1 , 2 , how do u know that it is N 2 ?? And , can you please", "# Thread: Two circles and a square 1. ## Two circles and a square Two distinct circles and a square lie in the plane. What is the maximum number of points of intersection of two or more of these three figures? 2. Hello, chil2e! Two distinct circles and a square lie in the plane. What is the maximum number of points of intersection of 2 or more of these 3 figures? We can \"talk\" our way through most of it. Let $S, C_1, C_2$ represent the square and two circles. We can \"see\" that: . . $S$ and $C_1$ intersect in at most 2 points. . . $S$ and $C_2$ intersect in at most 2 points. . . $C_1$ and $C_2$ intersect in at most 2 points. The maximum number of intersections is: . $2+2+2 \\:=\\:6$ Bottom line: is such a figure possible? Code: * * * * * * * * * * * o * * * * * * * * * * * - - o - o - - * * * | * * | * | | * | * * | * * | o | * * | * * | * * o * * o * | | * - - - - - - - *", "maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 the maximum number of points that two circles intersect is two. therefore, each of these 10 circles intersect two point with each of the other 9 circles. as a consequence, the number of these points is equal to two times the number of combination of the 10 circles. $$2*\\frac{(10!)}{(2!*10!)}= 90$$ _________________ Please press Kudos if you like my post Intern Joined: 22 Jan 2015 Posts: 23 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] Show Tags 01 Feb 2017, 01:22 sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 =n(n-1) =10*9 =90 (D) Intern Joined: 29 Dec 2015 Posts: 4 GMAT 1: 600 Q41 V32 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] Show Tags 26 Feb 2017, 16:25 Every circle intersects with another circle at max 2 points. Every circle interacts with a total of 9 circles. So, every circle has a max of 18 intersections (2*9). There are 10 circles, so 2*9*10 intersections. However, every intersection in counted twice. therefore we must divide by 2. this gets us", "# Maximum number of points of intersection between $7$ figures. We are given $5$ lines and two circles in a plane. What is the maximum number of possible intersection points among these seven figures ? My work on the problem: I've considered special cases like what is the maximum number of points of intersection between lines and circles ,between only lines and last between circles. P.s: Please provide detailed answer so i can follow best.Thanks in advance • If you think about it, the circles are easy to collocate once you've maxed the intersection betwen the lines. By the way, the answer is 32, try to obtain and prove it – Exodd Sep 28 '15 at 20:13 ## 1 Answer When you only have 5 lines, you can get at most $\\binom{5}{2}=10$ intersections. This is the number of distinct pairs of lines among those five. You can draw it on paper if you don't know the binomial coefficient symbol yet. Each of 2 circles can intersect each of 5 lines at 2 points. There can also be 2 circle-circle intersections. Line-line intersections: $10$ Line-circle intersections: $2\\cdot 2 \\cdot 5$ Circle-circle intersections: $2$ Add it up to get the answer. • Pardon, but where is the intuition behind $\\binom{5}{2}=10$ using geometrical objects ? i mean i used that method", "49 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 25 Oct 2016, 12:39 1 KUDOS sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 the maximum number of points that two circles intersect is two. therefore, each of these 10 circles intersect two point with each of the other 9 circles. as a consequence, the number of these points is equal to two times the number of combination of the 10 circles. $$2*\\frac{(10!)}{(2!*10!)}= 90$$ _________________ Please press Kudos if you like my post Intern Joined: 22 Jan 2015 Posts: 15 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 01 Feb 2017, 00:22 sitagupta385 wrote: What is the maximum number of points that 10 circles of different radii intersect? A: 24 B: 26 C: 33 D: 90 E: 100 =n(n-1) =10*9 =90 (D) Intern Joined: 29 Dec 2015 Posts: 4 Re: Maximum number of points that 10 circles of different radii intersect [#permalink] ### Show Tags 26 Feb 2017, 15:25 Every circle intersects with another circle at max 2 points. Every circle interacts with a total of 9 circles. So, every circle has a max of", "# Intersecting Lines And Circles Probability Level 3 The maximum number of points of intersection of 5 straight lines and 4 circles in a plane is $\\text{\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_}$. ×", "to 9*10=90. Re: Maximum number of points that 10 circles of different radii intersect &nbs [#permalink] 26 Feb 2017, 16:25 Display posts from previous: Sort by", "Three circles can intersect at maximum 6 points.. Because we select 2 circles from 3 and each two circles can intersect at maximum 2 points 2$$c^3_2$$ ... Three circles can intersect at 6 points So 11 circles can intersect at maximum $$2c^{11}_2$$ ... 2 * 11 * 5 = 110 Intern Joined: 01 Nov 2015 Posts: 29 Location: India Concentration: Marketing, Entrepreneurship WE: Engineering (Computer Software) Re: What is the greatest possible number of points at which 11 circles wit [#permalink] ### Show Tags 20 Dec 2015, 19:46 2 1st Circle doesn't intersect itself - 2*(1-1)=0 points 2nd Circle intersects 1st at 2*(2-1)=2 Points 3rd Circle intersect 2nd and 3rd at 2*(3-1)=4 Points 4th Circle intersects other Circles at 2*(4-1)=6 points ... 11th circle intersects other circles at 2*(11-1)=20 points Therefore total no of points = 2 + 4 + 6 + ... + 20 $$Sum = 10/2 * (20+2) = 5*22 =110$$ Answer is E Intern Joined: 09 Jul 2015 Posts: 45 Re: What is the greatest possible number of points at which 11 circles wit [#permalink] ### Show Tags 21 Dec 2015, 16:54 2 Answer is E -> 110. 2 circle intersects at 2 points, 3 circles intersect at 4 points, 4 circles intersect at 6 points, 5 circles at 8 points, etc., Hence max #", "one circle. This will add new edges, depending on how many new vertices we have. On the new circle itself it will add at most 2n new edges. Every time the new circle \"cuts through\" an existing edge, he will devide this edge, creating at most one more in total. Thus, at most 2n + 2n = 4n new edges. Two circles: at most 1 + 4 = 5 edges Three circles: a most 5 + 4 * 2 = 8 + 5 = 13 edges Four circles: at most 13 + 4 * 3 = 25 edges thus, max faces = 2 + max edges - 12 <= 2 + 25 - 12 = 15. It seems like I overcalculated the number of possible new edges in the step from one to two circles. This is because the two new intersections must devide the same existing edge (there is only one) and therefore creating only one new edge on the first circle (+ the two on the second circle, thus at most three new edges in total). So the number of edges four circles can have at most is actually 24. But 25 is good enough to show that the diagram is impossible. I hope I didn t write completely nonsense. Its pretty late now. Youre welcome", "# Too many Circles Using 20 Circles, what is the maximum number of intersecting point that can be obtained? For example, if there were 3 circles, the answer would be $6$ as shown below: • 20 non-concentric circles? – nikki Aug 15 '18 at 20:00 • I'd assume so (there's a facetious answer otherwise). Is it also a safe assumption that they all be uniformly sized? – El-Guest Aug 15 '18 at 20:01 • @nikki :) then there would be infinite intersections. – Oray Aug 15 '18 at 20:01 • @El-Guest not necessarily, you can use any size you would like – Oray Aug 15 '18 at 20:01 • @Oray interesting, thanks for the clarification :) – El-Guest Aug 15 '18 at 20:02 any two circles will have maximum two intersection points between them (think of a Venn diagram for this). I'm not able to program/draw this out at the moment (especially not with 20 circles), but I feel that the maximum can be obtained when each pair of circles intersects twice. Now, how many pairs of circles are there? Since order doesn't matter, using choose notation we can say that there are $C(20,2)$, ie. 20 choose 2 pairs of circles. This evaluates to $C(20,2) = \\frac{20!}{18!2!} = \\frac{20(19)}{2} = 190$. Since there are 190 pairs of circles", "# Maximum number of regions formed by points on a circle The question is : 6 points are located on a circle and lines are drawn connecting these points, each pair of points connected by a single line. What can be the maximum number of regions into which the circle is divided? My answer is 32 . But the actual answer is 31 how ?? - In general the maximum number of regions you can get from $n$ points is given by $${n \\choose 4} + {n \\choose 2} + 1$$ This is an oft cited puzzle to show the perils of generalizing based on first few values. We get powers of $2$ till $n=5$, after which we get $31$.", "313 views There are three coplanar parallel lines. If any $p$ points are taken on each of the lines, then find the maximum number of triangles with the vertices of these points. 1. $p^{2}(4p-3)$ 2. $p^{3}(4p-3)$ 3. $p(4p-3)$ 4. $p^{3}$ 1 246 views 2 174 views", "### Pebbles Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time? ### Happy Numbers Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general. ### Intersecting Circles Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have? # Maxagon ##### Age 11 to 14 Challenge Level: For this problem you may wish to download and print some dotty paper. Alternatively, you could explore the problem using this interactivity. Draw some polygons by joining the dots on a $3$ by $3$ grid. What is the greatest number of sides that your polygon could have? What about on a $3$ by $4$ grid, or a $3$ by $5$ grid? What about on a $3$ by $n$ grid? Can you explain the pattern by which the 'number of sides' increases? Explore some polygons on grids that are $4$ dots high. What is the maximum number of sides a polygon could have on a $4$ by $n$ grid? Can you explain how you know? What is the maximum number", "4 and 6 only C. 1, 2, 3 and 6 only D. 1, 2, 3, 4 and 6 only E. 1, 2, 3, 4, 5 and 6 only Which of the following lists a number of points at which a circle intersects a triangle A. 2 and 6 only B. 2, 4 and 6 only C. 1, 2, 3 and 6 only D. 1, 2, 3, 4 and 6 only E. 1, 2, 3, 4, 5 and 6 only Circle can intersect triangle at one of the vertices - 1 point of intersection; Circle can intersect triangle at two of the vertices - 2 points of intersection; Circle can intersect triangle at three of the vertices (inscribed triangle or inscribed circle) - 3 points of intersection; Circle can intersect triangle at two of the vertices and two sides - 4 points of intersection; Circle can intersect triangle at one of the vertex and cut three sides (one side twice and other two once) two sides - 5 points of intersection; Circle can cut all three sides twice - 6 points of intersection. Hence circle can intersect triangle at 1, 2, 3, 4, 5 or 6 points. (The examples I provided are not the only possible cases of intersection points, just these examples prove that there can be from", "circles are drawn in a plane, what is t [#permalink] ### Show Tags 16 Feb 2020, 05:06 Solution: Question 7: If 6 circles are drawn in a plane, what is the max number of points of intersection of these 6 circles? A. 15 B. 20 C. 25 D. 30 E. 38 We know that any two circles can intersect at maximum 2 points. In this question, we need to find out the maximum number of points of intersection of six circles. Since any 2 circles can intersect at maximum two points, any 2 circles out of 6 circles will also intersect at 2 points. Thus, the maximum number of intersection points are $$6C_{2} * 2$$ = 30, which is option D. Therefore, option D is the correct answer. Manager Joined: 05 May 2016 Posts: 142 Location: India Re: If 6 circles are drawn in a plane, what is t [#permalink] ### Show Tags 16 Feb 2020, 05:53 Not sure, but the answer could be C or D, depending on how I drew the circles. Re: If 6 circles are drawn in a plane, what is t [#permalink] 16 Feb 2020, 05:53", "circles can intersect at maximum 2 points. Three circles can intersect at maximum 6 points.. Because we select 2 circles from 3 and each two circles can intersect at maximum 2 points 2$$c^3_2$$ ... Three circles can intersect at 6 points So 11 circles can intersect at maximum $$2c^{11}_2$$ ... 2 * 11 * 5 = 110 Intern Joined: 12 Jul 2015 Posts: 25 Location: United States Concentration: Strategy, General Management WE: General Management (Pharmaceuticals and Biotech) Re: Math Revolution and GMAT Club Contest! What is the greatest possible [#permalink] ### Show Tags 19 Dec 2015, 11:05 5 1 The first 2 circles intersect at 2 points. 3rd circle can intersect the first two circles at 4 points. refer to the figure. i.e Circle 1 = no intersection Circle 2 = 2 points Circle 3 = 2+4 points Circle 4 = 2+4+8 points . . . Circle 11= 2+4+8+10+12+14+16+18+20 = n(n+1)= 10*11=110 points. Implies N circles can be drawn to intersect N(N-1) points . In this case, 11 circles can be drawn that can intersect at 11*10=110 points Ans E Attachments File comment: circles circles.png [ 27.71 KiB | Viewed 4995 times ] Intern Joined: 01 Nov 2015 Posts: 36 Location: India Concentration: Marketing, Entrepreneurship WE: Engineering (Computer Software) Re: Math Revolution and GMAT Club Contest! What is", "# Three little circles Geometry Level 4 As shown in the diagram below, three overlapping circles with distinct radii define six points. The lengths of 5 line segments (in blue) connecting the points are known to us: 4, 6, 8, 3, 5. What is the length of the remaining segment (in yellow), to 3 decimal places? ×", "# Maximum number of regions formed by points on a circle The question is : 6 points are located on a circle and lines are drawn connecting these points, each pair of points connected by a single line. What can be the maximum number of regions into which the circle is divided? My answer is 32 . But the actual answer is 31 how ?? • This link is just a repetition of the question. – Alex Oct 6 '16 at 11:06 In general the maximum number of regions you can get from $n$ points is given by $${n \\choose 4} + {n \\choose 2} + 1$$ This can be proved using induction (other combinatorial proofs exist too). For more information (including at least two proofs), see this: Dividing a circle into areas. This is an oft cited puzzle to show the perils of generalizing based on first few values. We get powers of $2$ till $n=5$, after which we get $31$. The more general case is stated far more elegantly here: Number or regions formed when $$n$$ points on a circle are joined. But sometimes having a concrete example helps, and I hope this does. Note that a new region is formed 1. when a new chord is drawn 2. whenever that new chord intersects another chord" ]

[ "$x^y$ has infinitely many values). -", "has only finite solutions(when all are integer), $2\\leq m\\leq12(n=2,a=2,b=3,c=5)$.", "# It's just Pell's equation How many solutions does the equation $$x^2-12y^2=-1$$ have in integers $$x$$ and $$y$$? Submit -1 as the answer if you think there are infinitely many integers. ×", "equation such as $2=2$, this means the provided equation has infinitely many solutions in its solution set.", "at x =2: 3a(2)^2 = 2(2) => 12a = 4 => a = 1/3. Substitute this value of a into equation (A) to get b.", "## Number of solutions Find the number of solutions of the equation $4[x^3-x]+8[x^2]=13+2[cosx]$ here [.] is greatest integer function", "then $$m^2+1=2^{2k+2}-2^{k+2}+2=2(2^{2k+1}-2^{k+1}+1)$$ And both $m$'s are obviously part of the RHS product (if $k>1$) the conjecture is true hence infinitely many solutions.", "of $\\mathbb C^3$ for all but finitely many values of $t$.", "# Number of solutions • Apr 8th 2010, 07:43 AM banku12 Number of solutions Find the number of solutions of the equation $4[x^3-x]+8[x^2]=13+2[cosx]$ here [.] is greatest integer function", "Algebra and Trigonometry 10th Edition False. The solution is $x=3$ $2(x+3)=3x+3$ $2x+6=3x+3$ $2x-3x=3-6$ $-x=-3$ $x=3$", "# Does $~4^y+1=4xy^2+x~$ have infinitely many solutions in integers? Consider the equation : $~4^y+1=4xy^2+x$ I have found that this equation has integer solutions for following values of $~y$ : $y\\in \\{1,2,193,10068,29570,..\\}$ Question : Are there finitely or infinitely many solutions ? Can one in practice compute full list of solutions ? -", "## Algebra and Trigonometry 10th Edition Multiplying equation 1 by 2, we get: $14x+16y=12$ Adding both equations, we get: $14x+16y-14x−16y=12-12$ Thus, $0=0.$ Infinite number of solutions.", "= -1 (2, -1) is the integer-valued solution for the given equation. Hence, 12x + 16y = 8 is having interger-valued solution.", "has the solutions $(0,3),(3,0),(1,2),(2,1)$ and no more, so the coefficient is $4$. Jul 8 '15 at 3:10", "# Find C If the System of Equations Cx + 3y + (3 – C) = 0; 12x + Cy – C = 0 Has Infinitely Many Solutions? - Mathematics Sum Find c if the system of equations cx + 3y + (3 – c) = 0; 12x + cy – c = 0 has infinitely many solutions? #### Solution cx + 3y + (3 - c) = 0 ....(i) 12x + cy - c = 0 ....(ii) For infinitely many solutions, (\"a\"_1)/(\"a\"_2) = (\"b\"_1)/(\"b\"_2) = (\"c\"_1)/(\"c\"_2) => \"c\"/12 = 3/\"c\" = (3-\"c\")/\"-c\" => \"c\"/12 = 3/\"c\" => \"c\"^2 = 36 => \"c\" = +- 6 OR 3/\"c\" = (3-\"c\")/\"-c\" => \"c\"(\"c\" - 6) = 0 => \"c\" = 0,6 Hence, c = 6 is the solution which gives infinitely many solution. Concept: Quadratic Equations Examples and Solutions Is there an error in this question or solution?", "### Home > CCA2 > Chapter 3 > Lesson 3.2.2 > Problem3-79 3-79. How many solutions does each equation below have? What is the highest power of $x$? That is the number of solutions you should be trying to find. But, remember that some equations have no solution and others have infinitely many solutions. Try solving each equation. 1. $4x + 3 = 3x + 3$ 1 solution 1. $3(x − 4) − x = 5 + 2x$ 1. $(5x − 2)(x + 4) = 0$ In this problem you have to multiply the two factors to see the highest power of $x$ if you don't recognize the number of solutions it will have. 1. $x^2 − 4x + 4 = 0$", "# Solve this: Question No. 10 The number of solutions of the equation $\\left[x\\right]+\\left[-x\\right]+{x}^{2}=1$, where [.] denotes the greatest integer function is A. Infinitely many B. 4 C. 2 D. 0 • 0 c) 2 • -1 What are you looking for?", "# Only cubes and squares! How many integer solutions are there to the equation $x^3 - 36y^3 = 5184?$ ×", "Free Version Moderate # Infinite Number of Solutions SATSTM-MEYF5U For the equation: $$5(x-3)+a=5(x+1)$$ ...to have an infinite number of solutions, what must the value of $a$ be? A $5$ B $10$ C $15$ D $20$ E $25$", "three terms, then the order of them is determined, so there is at most one solution for $b.$ It is possible for $n=2$: take $w(x)=x^2-3x+1$. Then $w(x)=1$ has two solutions $x=0,3$ and $w(x)=-1$ has two solutions $x=1,2$." ]

[ "values of $x$ satisfy the original equation, $\\tfrac{a}{b} = \\tfrac54 + 1 = \\tfrac94,$ so $a+b = \\boxed{13}.$", "throw it out. Case 2: $x-3=-(3x+2)-1$ $x=0$ This solution does not work with the original equation, so throw it out. Case 3: $-(x-3)=(3x+2)-1$ $\\boxed{x=\\frac{1}{2}}$ This solution works, so keep it. Case 4: $-(x-3)=-(3x+2)-1$ $\\boxed{x=-3}$ This solution works, so keep it. I had no idea there were 4 cases. Thanks", "equation,$x = 18/2 =9$, thus, the answer is $\\boxed{\\textbf{(C)}\\ 9}$. ---LarryFlora", "Kohli May 25 '13 at 3:01 You're right, @ΠάρτηΚοηλί: I'll include a link. – amWhy May 25 '13 at 3:06 It does not matter left or right. We can add $2x$ on both sides to get $$6+2x=4-2x+2x$$ $$\\implies 6+2x=4$$ then subtract 6 on both sides we get $$6-6+2x=4-6$$ $$\\implies 2x=-2$$ then divide 2 on both sides we get $$\\frac{2}{2}x=\\frac{-2}{2}$$ $$\\implies x=-1.$$ If we have an equation, it is \"legal\" to perform addition, subtraction, multiplication and division (except by 0) by equal number on both sides. Later you will also learn that it is also \"legal\" to take the log, power, derivative and integral and so on, on both sides. - Given the equation: $$\\boxed{6=4-2x}$$ We can subtract $-4$ from both sides: $$6-4=4-4-2x$$ Simplifying we are left with: $$2=-2x$$ Dividing by $-2$ on both sides to isolate x: $$\\dfrac{2}{-2}=\\dfrac{-2x}{-2}$$ Leaving us with the solution: $$\\boxed{x=-1}$$ We can check we have the correct solution by plugging the $-1$ back into the original equation: $$6=4-2(-1)$$ $$6=4+2$$ $$\\boxed{6=6}$$ We find the answers are the same which proves we have the correct solution. -", "2.23 b. 7.77, 2.23 c. 8.77, 3.23 d. 7.77, 3.23 #### Solution: |x - 6| - 3 = - 0.23 [Original equation.] |x - 6| - 3 + 3 = - 0.23 + 3 [Add 3 on both sides.] |x - 6| = 2.77 [Simplify.] |x - 6| = 2.77, then x - 6 equals 2.77 or -2.77 x - 6 = 2.77 or x - 6 = - 2.77 x - 6 + 6 = 2.77 + 6 or x - 6 + 6 = - 2.77 + 6 [Add 6 on both sides.] x = 8.77 or x = 3.23 [Simplify.] The solutions of the equation are 8.77 and 3.23. Correct answer : (3) 4. (- |$x$|)2 is always positive. State whether the statement is true or false. a. True b. False #### Solution: The value of - |x| is always negative. The square of a negative number is always a positive number. So, the statement is true. Correct answer : (1) 5. Find the values of $x$, if |5$x$| = 0.8. a. 0.01, -0.01 b. 0.16, -0.16 c. 1.60, -1.60 d. 1.16, -1.16 #### Solution: |5x| = 0.8 [Original equation.] As, |5x| = 0.8 then the values of the expressions 5x equals to 0.8 or - 0.8 5x = 0.8 or 5x = - 0.8", "can. Let's keep this simple. Suppose you start with $x = -3$. If you square both sides, you get \\begin{align} x^2 &= 9 \\\\ x^2 - 9 &= 0 \\\\ (x-3)(x+3) &= 0 \\\\ x &\\in \\{3, -3\\} \\end{align} So we acquired the extra solution, $x = 3$, which doesn't solve the original equation. Another way to look at this problem is to note that $3 = -3$ is FALSE, but when you square both sides you get $9 = 9$, which is TRUE. The point is that squaring both sides of an equation sometimes creates a new equation with more solutions than the original equation. Extra solutions are usually called extraneous roots. In you equation,$\\sqrt{x-15} = 3-\\sqrt{x}$, you got $x = 16$. But when you substituted $16$ for $x$ and simplified, you got $1 = -1$. Of course this is FALSE, but the equation you get when you square both sides, is TRUE. $x = 16$ is an extraneous solution. That is, a solution to a related equation but is not a valid solution to the equation in question. Consider: $$\\sqrt x = -1$$ You would naturally square both sides and find the solution $x = 1$. But this does not apply, as we can clearly see by verifying that $\\sqrt 1 \\ne -1$. In fact, there's", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "## Algebra: A Combined Approach (4th Edition) We have $6y + 4x = 6$ and $3y - 3 = -2x$ Rearrange the second equation as $3y+2x=3$ Now multiply both sides of the second equation by two (2): We get $2(3y+2x)=2(3)$, which is $6y + 4x = 6$ (the same as the first equation). Hence, we have the same two equations which correspond to the same line.", "have subtracted both 4x and 5 from both sides to get \"0 = 0\", but the solution would still be the same: \"all x\". Don't be surprised if, for \"all real numbers\" or \"no solution\" equations, you don't necessarily have the exact same steps as some of your fellow students. Since there are infinitely-many always-true equations (like \"0 = 0\") and infinitely-many nonsensical equations (like \"3 = 4\"), there will be many ways of arriving at these answers 59. Nnesha you can substitute x by any number you will get equal sides $\\huge\\rm 3(9)-3(9)-2=-2$ or let plug in 2 $\\huge\\rm 3(-2)-3(-2)-2=-2$ both sides are e qual how can you say that there is no solution ? 60. DecentNabeel i think both are true.. some expert say no solution.. and some one is many 61. Nnesha well you can plug in any number you will get equal sides so thats means infinitely many solution 62. DecentNabeel http://www.purplemath.com/modules/solvelin4.htm @Nnesha expert you can see it 63. Nnesha so http://prntscr.com/7yn3yb solution is all x thats mean any number can be x 64. Nnesha 65. DecentNabeel so the finally answer is.. this is no solution for this equation.. but many solution for all x 66. DecentNabeel @Nnesha are you clear 67. Nnesha wait what ? 68. Nnesha x can be any number so", "## Introductory Algebra for College Students (7th Edition) The statement is false. Modifying the original statement to make it true, you get: If $3x - 4 = 16$, then $3x = 20$. This statement is false. The first step to solving this equation would be to add $4$ to each side. The statement is false because $4$ is subtracted from both sides instead. To get rid of the $-4$ on both sides, you add $4$ to both sides of the equation to get: $3x = 16 + 4$ Simplify to get: $3x = 20$", "NOT true. There is no value of a which makes it true. Yes, if a= -1/2, then 2a+ 8= 2(-1/2)+ 8= -1+ 8= 7 and |7|= 7, while 2a- 6= 2(-1/2)- 6= -1- 6= -7 and |-7|= 7. 8. Originally Posted by Fails_at_Math $|2a+8|=|2a-6|$ Both sides positives, so you can square them and then $32a+64=-24a+36,$ and it follows that $a=-\\frac12.$", "### Home > CCA > Chapter 4 > Lesson 4.1.1 > Problem4-17 4-17. Determine if the statement below is always, sometimes, or never true. Justify your conclusion. $2(3+5x)=6+5x$ Simplify both sides as much as possible and see if they are equal to each other. The sides are not equal, but when you solve the equation, $x=0$. Since there is only one possible $x$ value, the statement is only sometimes true.", "zero. So either $x - 3 = 0$ or $x - 9 = 0$. If $x - 3 = 0$, then $x = 3$. If $x - 9 = 0$, then $x = 9$. So the only possible solutions are $x = 3$ and $x = 9$. Substituting both of these into the original equation confirms that both are indeed solutions. Therefore, the solution set is $\\lbrace 3, 9 \\rbrace$. Solution: Method 2 We begin the same way we started in Method 1, expanding the right side and rearranging terms to obtain $3x^2 - 36x + 81 = 0$. This time, we subtract 81 from each side to find that $3x^2 - 36x = -81$. Since the two sides of the equation are equal, they will remain equal if we divide each side by 3. So we have $x^2 - 12x = -27$. We will now solve the equation by completing the square. We recognize that, when expanded, the perfect square $(x - 6)^2$ contains the terms $x^2$ and $-12x$; it also contains a constant term of $36$. So we will transform the left side into a perfect square by adding 36 to each side of the equation: $$x^2 - 12x + 36 = -27 + 36 \\qquad \\Rightarrow \\qquad (x - 6)^2 = 9$$ Since the square", "is $\\frac{11}{3}$. • If $3x=11$ is true, then $x=\\frac{11}{3}$ is true. • The solution to $3x=11$ is $\\frac{11}{3}$. Diagram B can be represented with the equation $11=y+5$. If we remove a weight of 5 from each side of the hanger, it will stay in balance. Removing 5 from each side of the hanger is the same as subtracting 5 from each side of the equation. • $11-5$ is 6. • $y+5-5$ is $y$. • If $11=y+5$ is true, then $6=y$ is true. • The solution to $11=y+5$ is 6.", "we've eliminated the variables. So we now have a simple equation with one unknown. Distribute Combine like terms on the left side Subtract 2x from both sides Combine like terms on the left side Combine like terms on the right side Divide both sides by -3 to isolate x Divide Now that we know that , we can plug this into to find Substitute for each Simplify So our answer is and which also looks like Notice if we graph the two equations, we can see that their intersection is at . So this verifies our answer. Graph of (red) and (green) #2 #3 Solve for y by subtracting 4 from both sides Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown. Distribute Combine like terms on the left side Subtract 12 from both sides Combine like terms on the right side Simplify Since this equation is always true for any x value, this means x can equal any number. So there are an infinite number of solutions. Graphs/121953: This question is from textbook x+5/2=61 solutions Answer 89518 by jim_thompson5910(28715) on 2008-01-22 23:35:47 (Show Source): You can put this solution on YOUR website! Start with the given equation Multiply", "## Intermediate Algebra for College Students (7th Edition) Distribute $-3$ on the left side and 6 on the right side then combine like terms on each side to obtain: $3x-3(2) -3(-x)=6(x) -6(1) \\\\3x - 6 -(-3x)=6x-6 \\\\3x-6+3x=6x-6 \\\\6x-6=6x-6$ Subtract $6x$ and add $6$ on both sides of the equation to obtain: $6x-6x=-6+6 \\\\0=0$ The statement above is true therefore the given equation is an identity.", "## Introductory Algebra for College Students (7th Edition) Considering the equation $3(8x-1) = 6(5+ 4x)$, we can apply operations to both sides of the equation to simplify the equation, isolating the variables on different sides of the equations. First distribute the numbers 3 and 6 to their respective sides of the equation. This creates an expression like: $3(8x) + 3(-1) = 6(5) + 6(4x)$, which simplifies to $24x - 3 = 30 + 24x$. Then, you can subtract $24x$ from both sides, shown by: $24x - 3 -24x = 30 + 24x -24x$. This simplifies to $-3 = 30$ This is not a true statement so there is no solution to this question.", "## Algebra 1 Distribute 2 to have: $4x-5=4x+2$ Add 5 to both sides to have: $4x=4x+7$ Subtract $4x$ from both sides to have: $4x-4x=4x+7-4x \\\\0=7$ This is a false statement, which means that the equation has no solution. No number will satisfy the equation.", "### Home > AC > Chapter 6 > Lesson 6.1.2 > Problem6-21 6-21. Determine if the statement below is true or false. Justify your conclusion. $2(3 + 5x) = 6 + 5x$ Simplify both sides as much as possible and see if they are equal to each other. $6+10x\\ne6+5x$ So, it is not true for all values of $x$. But is there any value of $x$ that makes the statement true? Yes, The statement is sometimes true since $x=0$ makes the statement true.", "would mean that $x > x$, which does not work, so there are no solutions where $x > 2$. Similarly, assume that $x < 2$. Thus, $x^3 < 8$, so $x^3 - 8 < 0$ and $x^3 + x - 8 < x$, so $y < x < 2$. By doing the same steps, we can show that $z < y$ and $x < z$. However, that would mean that $x < x$, which does not work, so there are no solutions where $x < 2$. Thus, we proved that $x = 2$ is the only solution, and by substituting the value into the original equations, we get the only solution of $\\boxed{(2,2,2)}$." ]

[ "# For 6x+9 = 6x + a, what is the value of a such that there's infinitely many solutions? $9$ If a were any other answer, they wouldn't be equal anymore. If a were $9$ however, $x$ could be anything and both sides would be the same. For example. If we put $9$ for $a$, and then $1$ for $x$, our equation would be $15 = 15$. This is a sort of circular equation because it is stating the obvious and not getting you anywhere, but it shows that whatever you put for $x$ one side will force you to getting the same answer on the other side. Therefore, if $a = 9$, there would be infinitely many solutions.", "solution to our original equation: $8x - 14n = 6$. Simply multiply both $x_1$ and $n_1$ by 3 (since $6 = 2 \\cdot 3$). Thus $x=6$, $n=3$ is a solution to $8x - 14n = 6$ Adding the appropriate multiples to find all solutions from this one solution, we get $$x=6+\\dfrac{14}{2}k \\quad \\textrm{ and } \\quad n=3+\\dfrac{8}{2}k$$ Simplifying, we find, $x=6+7k$ and $n=3+4k$ gives all solutions to $8x - 14n = 6$ Of course, we don't really care what $n$ is, since it doesn't appear in our original congruence. Further, if we are only considering the possible values of $x \\pmod{14}$, there are only two that fit the $6+7k$ form: $x \\equiv 6 \\textrm{ or } 13 \\pmod{14}$. These are our solutions.", "equation such as $2=2$, this means the provided equation has infinitely many solutions in its solution set.", "values of $x$ satisfy the original equation, $\\tfrac{a}{b} = \\tfrac54 + 1 = \\tfrac94,$ so $a+b = \\boxed{13}.$", "both sides. You'll get: $6x+9 = -27 ~\\implies~ 6x =-36~\\implies~x = -6$ Plug in this value into the original equation: $\\sqrt[3]{6 \\cdot (-6)+9}+5 \\buildrel {\\rm?} \\over {\\rm=} 2~\\iff~\\sqrt[3]{-27}+5 \\buildrel {\\rm?} \\over {\\rm=} 2~\\iff~ -3+5 \\buildrel {\\rm?} \\over {\\rm=} 2$....... which is obviously true. And therefore x = -6 is a solution of this equation.", "10 = 8(4x + 10). Answer: The given equation is: 6x + 26x – 10 = 8 (4x + 10) So, 32x – 10 = 8 (4x) + 8 (10) 32x – 10 = 32x + 80 Subtract with 32x on both sides So, 32x – 32x – 10 = 32x – 32x + 80 -10 = 80 Hence, from the above, We can conclude that the given equation has no solution Question 22. Classify the equation 64x – 16 = 16(4x – 1) as having one solution, no solution, or infinitely many solutions. Answer: The given equation is: 64x – 16 = 16 (4x – 1) So, 64x – 16 = 16 (4x) – 16 (1) 64x – 16 = 64x – 16 Subtract with 64x on both sides So, 64x – 64x – 16 = 64x – 64x – 16 – 16 = -16 16 = 16 Hence, from the above, We can conclude that the given equation has infinitely many solutions Question 23. Classify the equation 5(2x + 3) = 3(3x + 12) as having one solution, no solution, or infinitely many solutions. Answer: The given equation is: 5 (2x + 3) = 3 (3x + 12) So, 5 (2x) + 5 (3) = 3 (3x) + 3 (12) 10x + 15 = 9x +", "How many solutions does the following equation have? $$-12x+25=-12x$$ Solution: Solve for $$x: -12x+25=-12x$$$$25=+12x-12x$$$$25=0$$. $$25=0$$ is a false statement. The linear equation has no solution because by solving the linear equation you get a false statement as an answer. ### Finding the Number of Solutions to a Linear Equation-Example 2 How many solutions does the following equation have? $$10x=32x-22x$$ Solution: Simplify the right side of the equation: $$10x=32x-22x$$$$10x=10x$$. $$10x=10x$$ is a statement that is always true. So, the equation has infinitely many solutions because by solving a linear equation you get a statement that is always true. ## Exercises forFinding the Number of Solutions to a Linear Equation ### How many solutions does the following equation have? • $$\\color{blue}{11x+12=17x}$$ • $$\\color{blue}{no\\:solution}$$ ### What people say about \"How to Find the Number of Solutions to a Linear Equation?\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE $5 It was$16.99 now it is \\$11.99", "can. Let's keep this simple. Suppose you start with $x = -3$. If you square both sides, you get \\begin{align} x^2 &= 9 \\\\ x^2 - 9 &= 0 \\\\ (x-3)(x+3) &= 0 \\\\ x &\\in \\{3, -3\\} \\end{align} So we acquired the extra solution, $x = 3$, which doesn't solve the original equation. Another way to look at this problem is to note that $3 = -3$ is FALSE, but when you square both sides you get $9 = 9$, which is TRUE. The point is that squaring both sides of an equation sometimes creates a new equation with more solutions than the original equation. Extra solutions are usually called extraneous roots. In you equation,$\\sqrt{x-15} = 3-\\sqrt{x}$, you got $x = 16$. But when you substituted $16$ for $x$ and simplified, you got $1 = -1$. Of course this is FALSE, but the equation you get when you square both sides, is TRUE. $x = 16$ is an extraneous solution. That is, a solution to a related equation but is not a valid solution to the equation in question. Consider: $$\\sqrt x = -1$$ You would naturally square both sides and find the solution $x = 1$. But this does not apply, as we can clearly see by verifying that $\\sqrt 1 \\ne -1$. In fact, there's", "zero. So either $x - 3 = 0$ or $x - 9 = 0$. If $x - 3 = 0$, then $x = 3$. If $x - 9 = 0$, then $x = 9$. So the only possible solutions are $x = 3$ and $x = 9$. Substituting both of these into the original equation confirms that both are indeed solutions. Therefore, the solution set is $\\lbrace 3, 9 \\rbrace$. Solution: Method 2 We begin the same way we started in Method 1, expanding the right side and rearranging terms to obtain $3x^2 - 36x + 81 = 0$. This time, we subtract 81 from each side to find that $3x^2 - 36x = -81$. Since the two sides of the equation are equal, they will remain equal if we divide each side by 3. So we have $x^2 - 12x = -27$. We will now solve the equation by completing the square. We recognize that, when expanded, the perfect square $(x - 6)^2$ contains the terms $x^2$ and $-12x$; it also contains a constant term of $36$. So we will transform the left side into a perfect square by adding 36 to each side of the equation: $$x^2 - 12x + 36 = -27 + 36 \\qquad \\Rightarrow \\qquad (x - 6)^2 = 9$$ Since the square", "$3m + 4 = 3m - 9$ 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", "# Number of solutions of a linear system — dependence on parameters For which values of $a$ and $b$ does the system: $3x+ay=b$ $ax+3y=b$ have: $a)$ no solutions, $b)$ 1 solution, $c)$ infinite solutions. How to tackle this? Please don't use advanced mathematics.. - This system can never have infinite solutions. The solutions are always (pairs of) finite numbers. You mean infinitely many solutions. – Chris Eagle Oct 4 '12 at 18:18 ## 3 Answers If $(x,y)$ is a solution, then by subtracting we get $$3x-ax+ay-3y=b-b=0.$$ This can be rewritten as $$(3-a)(x-y)=0.$$ If $a=3$, the equation holds for all $x$ and $y$. So let's check what happens with our original equations when $a=3$. We get $3x+3y=b$, $3y+3x=b$. For any $b$, there are infinitely many solutions, since we can arrange in infinitely many ways to have $x+y=b/3$. Thus if $a=3$ and $b$ is anything, there is a unique solution. We have dealt with the case $a=3$. Now suppose $a\\ne 3$. Then from the equation $(3-a)(x-y)=0$ we get $x=y$. Now go back to the original equations, with $a\\ne 3$ and $x=y=t$. The first equation reads $3t+at=b$, the second reads $at+3t=b$, the same. If $a+3=0$ and $b\\ne 0$, there is no solution. So there is no solution if $a=-3$ and $b\\ne 0$. If $a=-3$, and $b=0$, there are infinitely many solutions,", "can't get the answer correct.. My book says that if we use 3 there's no solution but i can't seem to continue with -2 either... thanks $ax+3y=2a$ $2x +(a-1)y = a^2$ for a= -2 $-2x+3y=-4$ $2x -3y = 4$ Edit : here you have : two same equations : $2x-3y= 4$ $2x -3y = 4$ so you have $4x-6y = 8$ so you have one equation with 2 unknowns.... you have infinity of the solutions and equations are dependent (meaning for each new x or y you have another solution for that second unknown) for a = 3 $3x+3y=6$ $2x +2y = 9$ Edit : here you have : $x+y = 2$ $x+y = 4.5$ now here it's easy to see there are no solutions because you have ones that x+y = 2 and another saying that x+y = 4.5 which cant be true for the both of the equations Edit: hm... this is moved to the university sub forum ( I think it was in pre-algebra at the start .... perhaps i'm wrong but ... ) , well if here you must show that based on determinants of that system you don't have / have solutions and if you have why and which , and if don't why ... and so on ... there are few cases", "I remember a question that asked for an explanation of why $6+6=12$. The guy was rewarded with a mountain of ridicule and abuse. Apparently some people thought this was \"too elementary\". But, elementary or not, I thought the ridicule was nasty (assuming the question was sincere). So, if you ask something very elementary, you may (sadly) get some sniping from some people. Ignore it. There are nice people here, too, and they will try to help you. Don't be afraid to ask. Mar 2 '16 at 12:41 Yes, however when asking a basic question it is even more important to provide context, showing your work so far and what you have to work with/know. Example: Solve this equation for me: $3x +1 = 2x-2$. This question contain no context, does not show any effort, and does not provide any context to the readers what you already know, thus is a bad posed elementary question. Example: I am working with the equation $3x+1= 2x -2$. I have tried dividing each side by $3$ getting the equation $x+1/3 = 2x/3 -2/3$, but then I still have the problem with $2x/3$ on the right hand side. By trial and error I have noticed that $x=-3$ is a solution, however how do you reach this conclusion mathematically? What is it that I", "throw it out. Case 2: $x-3=-(3x+2)-1$ $x=0$ This solution does not work with the original equation, so throw it out. Case 3: $-(x-3)=(3x+2)-1$ $\\boxed{x=\\frac{1}{2}}$ This solution works, so keep it. Case 4: $-(x-3)=-(3x+2)-1$ $\\boxed{x=-3}$ This solution works, so keep it. I had no idea there were 4 cases. Thanks", "equation that models the situation is $2x=6.2x=6.$ Figure 3.22 We can divide both sides of the equation by $22$ as we did with the envelopes and counters. We found that each envelope contains $3 counters.3 counters.$ Does this check? We know $2·3=6,2·3=6,$ so it works. Three counters in each of two envelopes does equal six. Figure 3.23 shows another example. Figure 3.23 Now we have $33$ identical envelopes and $12 counters.12 counters.$ How many counters are in each envelope? We have to separate the $12 counters12 counters$ into $3 groups.3 groups.$ Since $12÷3=4,12÷3=4,$ there must be $4 counters4 counters$ in each envelope. See Figure 3.24. Figure 3.24 The equation that models the situation is $3x=12.3x=12.$ We can divide both sides of the equation by $3.3.$ Does this check? It does because $3·4=12.3·4=12.$ ### Manipulative Mathematics Doing the Manipulative Mathematics activity “Division Property of Equality” will help you develop a better understanding of how to solve equations using the Division Property of Equality. ### Example 3.63 Write an equation modeled by the envelopes and counters, and then solve it. Try It 3.125 Write the equation modeled by the envelopes and counters. Then solve it. Try It 3.126 Write the equation modeled by the envelopes and counters. Then solve it. ### Solve Equations Using the Division Property of Equality The", "substituting $b = -3$ we also get $a = 2$. Therefore, the two solutions are $2 + 3i$ and $2 - 3i$.", "NOT true. There is no value of a which makes it true. Yes, if a= -1/2, then 2a+ 8= 2(-1/2)+ 8= -1+ 8= 7 and |7|= 7, while 2a- 6= 2(-1/2)- 6= -1- 6= -7 and |-7|= 7. 8. Originally Posted by Fails_at_Math $|2a+8|=|2a-6|$ Both sides positives, so you can square them and then $32a+64=-24a+36,$ and it follows that $a=-\\frac12.$", "another solution with strictly smaller absolute value. Contradiction. What's problem with infinite decent ? Note $2|C \\implies 2|b \\implies 2|a$ and similarly going on for all $n\\in \\mathbb {N}$ you'll get $2^n$ divides all of $a,b,c$. That's all. $a=1$, $b=1$ and $c=1$ is a solution of $4a^3+2b^3+c^3=6abc$, so $a=b=c=0$ is not the only solution.", "## Algebra 2 (1st Edition) If we multiply both sides of the first equation by $2$ we get $4x+10y=12$ which is the same as the second equation; thus the lines will be the same when graphed. Thus there are infinitely many solutions.", "multiply both sides of $q/x- 2x= 3$ by x you get $q- 2x^2= 3x$ which is a quadratic equation and has, at most, 2 distinct roots. The original equation might have fewer solutions, but cannot have more so NO value of q gives the equation 4 roots. If, on the other hand, you are saying that 4 is a root, we must have $q/4- 2(4)= 3$ so that $q/4= 11$ and then $q= 44$. What is the exact statement of the problem in your text? You posted a picture in your first post, but that is something you wrote. Could you post a picture of the problem in the text? 12. ## Re: value of q in the equation Originally Posted by HallsofIvy That is, as you have been told, impossible. If you multiply both sides of $q/x- 2x= 3$ by x you get $q- 2x^2= 3x$ which is a quadratic equation and has, at most, 2 distinct roots. The original equation might have fewer solutions, but cannot have more so NO value of q gives the equation 4 roots. If, on the other hand, you are saying that 4 is a root, we must have $q/4- 2(4)= 3$ so that $q/4= 11$ and then $q= 44$. What is the exact statement of the problem in your text?" ]

[ "+0 # help ples +1 190 1 Roger has exactly one of each of the first 22 states' new U.S. quarters. The quarters were released in the same order that the states joined the union. The graph below shows the number of states that joined the union in each decade. What fraction of Roger's 22 coins represents states that joined the union during the decade 1780 through 1789? Express your answer as a common fraction. (note: every space represents 2 states.) (I am the rabbit) Jun 13, 2020 12 states joined from 1780 to 1789. Therefore, of his first 22 quarters, 12 of them are from this time period, making for $\\frac{12}{22}$=$\\frac{6}{11}$ of his coins being from this time period. If you didnt understandmy LaTeX the answer is 6/11. lol", "1; b < 4; ++b) { draw((2*a+12,b-.1)--(2*a+12,b+.1)); draw((2*a+11.9,b)--(2*a+12.1,b)); } } label(\"1960\",(12,0),S); label(\"1970\",(14,0),S); label(\"1980\",(16,0),S); label(\"1990\",(18,0),S); label(\"10\",(12,1),W); label(\"20\",(12,2),W); label(\"30\",(12,3),W); label(\"\\%\",(12,4),N); draw((0,12)--(0,8)--(7,8)); draw((0,9)--(.2,9)); draw((0,10)--(.2,10)); draw((0,11)--(.2,11)); draw((2,8)--(2,8.2)); draw((4,8)--(4,8.2)); draw((6,8)--(6,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a,b+7.9)--(2*a,b+8.1)); draw((2*a-.1,b+8)--(2*a+.1,b+8)); } } label(\"1960\",(0,8),S); label(\"1970\",(2,8),S); label(\"1980\",(4,8),S); label(\"1990\",(6,8),S); label(\"10\",(0,9),W); label(\"20\",(0,10),W); label(\"30\",(0,11),W); label(\"\\%\",(0,12),N); draw((12,12)--(12,8)--(19,8)); draw((12,9)--(12.2,9)); draw((12,10)--(12.2,10)); draw((12,11)--(12.2,11)); draw((14,8)--(14,8.2)); draw((16,8)--(16,8.2)); draw((18,8)--(18,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a+12,b+7.9)--(2*a+12,b+8.1)); draw((2*a+11.9,b+8)--(2*a+12.1,b+8)); } } label(\"1960\",(12,8),S); label(\"1970\",(14,8),S); label(\"1980\",(16,8),S); label(\"1990\",(18,8),S); label(\"10\",(12,9),W); label(\"20\",(12,10),W); label(\"30\",(12,11),W); label(\"\\%\",(12,12),N); draw((24,12)--(24,8)--(31,8)); draw((24,9)--(24.2,9)); draw((24,10)--(24.2,10)); draw((24,11)--(24.2,11)); draw((26,8)--(26,8.2)); draw((28,8)--(28,8.2)); draw((30,8)--(30,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a+24,b+7.9)--(2*a+24,b+8.1)); draw((2*a+23.9,b+8)--(2*a+24.1,b+8)); } } label(\"1960\",(24,8),S); label(\"1970\",(26,8),S); label(\"1980\",(28,8),S); label(\"1990\",(30,8),S); label(\"10\",(24,9),W); label(\"20\",(24,10),W); label(\"30\",(24,11),W); label(\"\\%\",(24,12),N); draw((0,9)--(2,9.25)--(4,10)--(6,11)); draw((12,8.5)--(14,9)--(16,10)--(18,10.5)); draw((24,8.5)--(26,8.8)--(28,10.5)--(30,11)); draw((0,0.5)--(2,1)--(4,2.8)--(6,3)); draw((12,0.5)--(14,.8)--(16,1.5)--(18,3)); label(\"(A)\",(-1,12),W); label(\"(B)\",(11,12),W); label(\"(C)\",(23,12),W); label(\"(D)\",(-1,4),W); label(\"(E)\",(11,4),W); [/asy]$ ## Problem 5 Each principal of Lincoln High School serves exactly one $3$-year term. What is the maximum number of principals this school could have during an $8$-year period? $\\text{(A)}\\ 2 \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 4 \\qquad \\text{(D)}\\ 5 \\qquad \\text{(E)}\\ 8$ ## Problem 6 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", "7), ('shall', 6), ('principle', 6), ('would', 6), ('one', 6), ('man', 6), ('safety', 5)] 1805-Jefferson.txt [(',', 142), ('.', 41), (';', 26), ('public', 14), ('citizens', 10), ('may', 10), ('fellow', 8), ('state', 8), ('us', 7), ('among', 7), ('shall', 7), ('constitution', 6), ('time', 6), ('limits', 5), ('reason', 5)] [(',', 47), ('.', 21), (';', 16), ('nations', 6), ('public', 6), ('well', 5), ('country', 4), ('peace', 4), ('rights', 4), ('states', 4), ('confidence', 3), ('full', 3), ('improvements', 3), ('united', 3), ('best', 3)] [(',', 53), ('.', 31), ('war', 15), (';', 6), ('country', 5), ('united', 5), ('every', 5), ('british', 5), ('nation', 4), ('without', 4), ('states', 4), ('spirit', 4), ('citizens', 4), ('sense', 3), ('people', 3)] 1817-Monroe.txt [(',', 169), ('.', 110), ('government', 22), ('great', 21), ('states', 21), ('people', 15), ('us', 14), ('every', 14), ('united', 13), (';', 13), ('may', 10), ('?', 10), ('union', 10), ('war', 10), ('citizens', 9)] 1821-Monroe.txt [(',', 275), ('.', 130), ('great', 29), ('states', 20), ('would', 18), ('united', 16), ('war', 16), ('citizens', 15), ('may', 15), ('made', 15), ('government', 13), ('every', 13), ('people', 11), ('commerce', 11), ('force', 11)] [(',', 115), ('.', 72), (';', 35), ('union', 20), ('government', 17), ('upon', 16), ('country', 10), ('rights', 10), ('peace', 9), ('great', 9), ('public', 9), ('constitution', 8), ('first', 8), ('general', 8), ('nation', 8)] Hmmm, we still have punctuation in there, which we don’t care about. Let’s remove those, and try again. Note that I’ve added", "States of America 1818 #> 4 2 United States of America 1819 #> 5 2 United States of America 1820 #> 6 2 United States of America 1821 #> 7 2 United States of America 1822 #> 8 2 United States of America 1823 #> 9 2 United States of America 1824 #> 10 2 United States of America 1825 #> # … with 16,721 more rows ### Create a State-Year Panel of Gleditsch-Ward states create_stateyears() can do the same for Gleditsch-Ward states, but requires the user to specify they want states from the Gleditsch-Ward system. create_stateyears(system=\"gw\") #> # A tibble: 18,289 x 3 #> gwcode statename year #> <dbl> <chr> <int> #> 1 2 United States of America 1816 #> 2 2 United States of America 1817 #> 3 2 United States of America 1818 #> 4 2 United States of America 1819 #> 5 2 United States of America 1820 #> 6 2 United States of America 1821 #> 7 2 United States of America 1822 #> 8 2 United States of America 1823 #> 9 2 United States of America 1824 #> 10 2 United States of America 1825 #> # … with 18,279 more rows ### Create a Panel of ISO Codes ISO codes are ubiquitous in economic data. I do have some misgivings", "4.1 1. 135 2. 15 In [5]: years <- seq(1811, 2009, by = 1) Major$decade = cut(Major$dyad_st_year, seq(from = 1810, to = 2010, by = 10), include.lowest=TRUE, labels=seq(1810,2000, by=10)) Major$decade 1. 1810 2. 1810 3. 1840 4. 1840 5. 1820 6. 1820 7. 1830 8. 1830 9. 1830 10. 1830 11. 1830 12. 1830 13. 1830 14. 1830 15. 1830 16. 1830 17. 1840 18. 1840 19. 1850 20. 1850 21. 1850 22. 1850 23. 1850 24. 1850 25. 1860 26. 1860 27. 1860 28. 1870 29. 1870 30. 1880 31. 1880 32. 1880 33. 1890 34. 1900 35. 1900 36. 1900 37. 1900 38. 1910 39. 1910 40. 1910 41. 1910 42. 1910 43. 1920 44. 1920 45. 1920 46. 1920 47. 1930 48. 1930 49. 1930 50. 1930 51. 1930 52. 1930 53. 1930 54. 1930 55. 1930 56. 1930 57. 1930 58. 1930 59. 1930 60. 1930 61. 1930 62. 1930 63. 1930 64. 1930 65. 1940 66. 1940 67. 1940 68. 1940 69. 1940 70. 1940 71. 1940 72. 1940 73. 1940 74. 1970 75. 1980 76. 1940 77. 1940 78. 1940 79. 1940 80. 1980 81. 1980 82. 1940 83. 1980 84. 1990 85. 1940 86. 1940 87. 1940 88. 1990 89. 1940 90. 1940 91. 1940 92. 1990 93. 1940 94. 1980", "queries you will be running. ### Sample Table Before we go further with more queries, I would like to create a sample table with some data. This will be a list of US states, with their join dates (the date the state ratified the United States Constitution or was admitted to the Union) and their current populations. You can copy paste the following to your MySQL console: 1 2 CREATE TABLE states ( 3 id INT AUTO_INCREMENT, 4 name VARCHAR(20), 5 join_year INT, 6 population INT, 7 PRIMARY KEY (id), 8 UNIQUE (name), 9 KEY (join_year) 10 ); 11 12 13 INSERT INTO states VALUES 14 (1, 'Alabama', 1819, 4661900), 15 (2, 'Alaska', 1959, 686293), 16 (3, 'Arizona', 1912, 6500180), 17 (4, 'Arkansas', 1836, 2855390), 18 (5, 'California', 1850, 36756666), 19 (6, 'Colorado', 1876, 4939456), 20 (7, 'Connecticut', 1788, 3501252), 21 (8, 'Delaware', 1787, 873092), 22 (9, 'Florida', 1845, 18328340), 23 (10, 'Georgia', 1788, 9685744), 24 (11, 'Hawaii', 1959, 1288198), 25 (12, 'Idaho', 1890, 1523816), 26 (13, 'Illinois', 1818, 12901563), 27 (14, 'Indiana', 1816, 6376792), 28 (15, 'Iowa', 1846, 3002555), 29 (16, 'Kansas', 1861, 2802134), 30 (17, 'Kentucky', 1792, 4269245), 31 (18, 'Louisiana', 1812, 4410796), 32 (19, 'Maine', 1820, 1316456), 33 (20, 'Maryland', 1788, 5633597), 34 (21, 'Massachusetts', 1788, 6497967), 35 (22, 'Michigan', 1837, 10003422), 36 (23, 'Minnesota',", "1810,blame,v,19798,0.96 1811,estimate,v,20474,0.93 1812,due,i,21141,0.9 1814,late,r,19727,0.96 1815,golf,n,20979,0.91 1816,investigate,v,20117,0.94 1817,crazy,j,20345,0.93 1818,significantly,r,22038,0.86 1819,chain,n,19688,0.96 1821,branch,n,19633,0.96 1822,combination,n,20026,0.94 1823,just,j,20096,0.94 1824,frequently,r,20508,0.92 1825,governor,n,20802,0.91 1826,relief,n,19493,0.97 1827,user,n,20681,0.91 1829,kick,v,20092,0.94 1830,part,r,19900,0.95 1831,manner,n,20012,0.94 1832,ancient,j,19818,0.95 1833,silence,n,21169,0.89 1834,rating,n,20413,0.92 1835,golden,j,19852,0.95 1836,motion,n,19649,0.95 1837,German,j,20096,0.93 1838,gender,n,22190,0.85 1839,solve,v,19501,0.96 1840,fee,n,20263,0.93 1841,landscape,n,20171,0.93 1842,used,j,19130,0.98 1843,bowl,n,20662,0.91 1844,equal,j,19952,0.94 1845,long,c,19227,0.97 1846,official,j,19628,0.95 1847,forth,r,19531,0.96 1848,frame,n,20016,0.93 1849,typical,j,19724,0.95 1850,except,i,19594,0.95 1851,conservative,j,19778,0.94 1852,eliminate,v,19773,0.94 1853,host,n,19389,0.96 1854,hall,n,20236,0.92 1855,trust,v,19482,0.95 1856,ocean,n,19421,0.96 1857,score,v,20391,0.91 1858,row,n,19519,0.95 1859,producer,n,19679,0.94 1860,afford,v,19035,0.97 1861,meanwhile,r,19504,0.95 1862,regime,n,21101,0.88 1863,division,n,19772,0.94 1864,confirm,v,19327,0.96 1865,fix,v,19349,0.96 1866,appeal,n,19561,0.95 1867,mirror,n,20081,0.92 1868,tooth,n,20515,0.9 1869,smart,j,19370,0.95 1870,length,n,19624,0.94 1871,entirely,r,19075,0.97 1872,rely,v,19614,0.94 1873,topic,n,20640,0.89 1874,complain,v,19102,0.97 1875,issue,v,19300,0.95 1876,variable,n,22868,0.81 1877,back,v,19213,0.96 1878,range,v,20067,0.92 1879,telephone,n,19311,0.95 1880,perception,n,21086,0.87 1881,attract,v,19197,0.96 1882,confidence,n,18955,0.97 1883,bedroom,n,20387,0.9 1884,secret,n,19125,0.96 1885,debt,n,19506,0.94 1886,rare,j,18978,0.97 1887,his,p,19663,0.93 1888,tank,n,19158,0.95 1889,nurse,n,19385,0.94 1890,coverage,n,19522,0.94 1891,opposition,n,19847,0.92 1892,aside,r,19056,0.96 1893,anywhere,r,18970,0.96 1894,bond,n,19856,0.92 1895,file,v,19416,0.94 1896,pleasure,n,19099,0.96 1897,master,n,18880,0.96 1898,era,n,19209,0.95 1899,requirement,n,20358,0.89 1900,check,n,18764,0.97 1901,stand,n,18910,0.96 1902,fun,n,18984,0.96 1903,expectation,n,19603,0.92 1904,wing,n,19028,0.95 1905,separate,j,19064,0.95 1906,now,c,19126,0.95 1907,clear,v,19013,0.95 1908,struggle,n,19014,0.95 1909,mean,j,19792,0.91 1910,somewhat,r,18801,0.96 1911,pour,v,19300,0.93 1912,stir,v,20174,0.89 1913,judgment,n,19315,0.93 1914,clean,v,18766,0.96 1915,except,c,18857,0.95 1916,beer,n,19349,0.93 1917,English,n,18719,0.96 1918,reference,n,19559,0.92 1919,tear,n,19819,0.9 1920,doubt,n,18542,0.97 1921,grant,v,18598,0.96 1922,seriously,r,18328,0.98 1923,account,v,19403,0.92 1924,minister,n,19154,0.93 1925,totally,r,18720,0.95 1926,hero,n,18348,0.97 1927,industrial,j,19413,0.92 1928,cloud,n,19214,0.92 1929,stretch,v,19014,0.93 1930,winner,n,19216,0.92 1931,volume,n,18989,0.93 1932,travel,n,18634,0.95 1933,seed,n,19079,0.93 1934,surprised,j,18822,0.94 1935,rest,v,18915,0.94 1936,fashion,n,18483,0.96 1937,pepper,n,20220,0.87 1938,separate,v,18251,0.97 1939,busy,j,18500,0.95 1940,intervention,n,21021,0.84 1941,copy,n,18244,0.97 1942,tip,n,18868,0.93 1943,below,i,18353,0.96 1944,cheap,j,18371,0.96 1945,aim,v,18194,0.97 1946,cite,v,19011,0.93 1947,welfare,n,19580,0.9 1948,vegetable,n,19363,0.91 1949,gray,j,19408,0.9 1950,dish,n,18887,0.93 1951,beach,n,18451,0.95 1952,improvement,n,18977,0.92 1953,everywhere,r,18126,0.96 1954,opening,n,18058,0.97 1955,overall,j,19031,0.92 1956,divide,v,18337,0.95 1957,initial,j,19019,0.92 1958,terrible,j,18659,0.93 1959,oppose,v,18484,0.94 1960,contemporary,j,19293,0.9 1961,route,n,18313,0.95 1962,multiple,j,19512,0.89 1963,essential,j,19030,0.91 1964,question,v,17924,0.97 1965,league,n,19766,0.88 1966,criminal,j,18424,0.94 1967,careful,j,17885,0.97 1968,core,n,18571,0.93 1969,upper,j,18196,0.95 1970,rush,v,18561,0.93 1971,necessarily,r,18334,0.94 1972,specifically,r,18873,0.92 1973,tired,j,18597,0.93 1974,rise,n,18055,0.96 1975,tie,n,17787,0.97 1976,employ,v,18912,0.91 1977,holiday,n,18299,0.94 1978,dance,v,18263,0.94 1979,vast,j,17836,0.96 1980,resolution,n,18922,0.91 1981,household,n,18450,0.93 1982,fewer,d,18223,0.94 1983,abortion,n,18925,0.91 1984,apart,r,17872,0.96 1985,witness,n,18920,0.91 1986,match,v,17788,0.97 1987,barely,r,18533,0.93 1988,sector,n,19103,0.9 1989,representative,n,18189,0.94 1990,lack,v,18215,0.94 1991,beneath,i,19047,0.9 1992,beside,i,19952,0.86 1993,black,n,18441,0.93 1994,incident,n,17880,0.96 1995,limited,j,18532,0.92 1996,proud,j,17841,0.96 1997,flow,n,18469,0.92 1998,faculty,n,20104,0.85 1999,increased,j,18984,0.9 2000,waste,n,19069,0.89 2001,merely,r,18094,0.94 2002,mass,n,18107,0.94 2003,emphasize,v,18933,0.9 2004,experiment,n,18440,0.92 2005,definitely,r,18214,0.93 2006,bomb,n,18284,0.93 2007,enormous,j,17482,0.97 2008,tone,n,17954,0.95 2009,liberal,j,18228,0.93 2010,massive,j,17551,0.97 2011,engineer,n,17991,0.94 2012,wheel,n,18296,0.93 2013,female,n,18932,0.9 2014,decline,v,18058,0.94 2015,invest,v,18080,0.94 2016,promise,n,17342,0.98 2017,cable,n,18277,0.93 2018,towards,i,18305,0.93 2019,expose,v,17581,0.96 2020,rural,j,18665,0.91 2021,AIDS,n,19384,0.87 2022,Jew,n,17893,0.95 2023,narrow,j,17840,0.95 2024,cream,n,18443,0.92 2025,secretary,n,17922,0.94 2026,gate,n,17999,0.94 2027,solid,j,17675,0.96 2028,hill,n,18131,0.93 2029,typically,r,18254,0.92 2030,noise,n,18024,0.93", "1 1 4.1 1810 defense 50Germany Wuerttemburg 29 11 1850 15 6 1866 0 0 1 0 1 1 4.1 1840 defense 55Germany Mecklenburg Schwerin 1 1 1843 15 3 1848 0 0 1 0 1 1 4.1 1840 defense 203United Kingdom France 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 defense 207Russia United Kingdom 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 defense 3United Kingdom Sweden 1 1 1816 15 2 1911 1 0 0 0 0 1 4.1 1810 non-defense 45Germany Baden 1 1 1816 15 3 1848 1 0 1 0 1 1 4.1 1810 defense 50Germany Wuerttemburg 29 11 1850 15 6 1866 0 0 1 0 1 1 4.1 1840 defense 55Germany Mecklenburg Schwerin 1 1 1843 15 3 1848 0 0 1 0 1 1 4.1 1840 defense 203United Kingdom France 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 non-defense 207Russia United Kingdom 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 non-defense In [9]: ggplot(Major, aes(decade)) + geom_bar(aes(fill = Type), position = \"stack\")+ theme(axis.text.x=element_text(angle=90),text = element_text(size=10))+ ggtitle(\"Major Powers Defense and Non-defense Treaty Count; 1810-2010\") To better see treaty trends over time, I created a new column that combines", "# Fourth of July Edition: Print lines of a file containing multiples of a specific number Create a program which takes one command-line argument, n, which will be an integer less than 2147483648 (2^31), and then reads a file input.txt and prints the lines of input.txt which contain any substring which is a positive (non-zero) multiple of n. You may choose to ignore multiples greater than 2147483647. ### Test case If input.txt contains 1. Delaware Dec. 7, 1787 2. Pennsylvania Dec. 12, 1787 1682 3. New Jersey Dec. 18, 1787 1660 4. Georgia Jan. 2, 1788 1733 5. Connecticut Jan. 9, 1788 1634 6. Massachusetts Feb. 6, 1788 1620 7. Maryland Apr. 28, 1788 1634 8. South Carolina May 23, 1788 1670 9. New Hampshire June 21, 1788 1623 10. Virginia June 25, 1788 1607 11. New York July 26, 1788 1614 12. North Carolina Nov. 21, 1789 1660 13. Rhode Island May 29, 1790 1636 14. Vermont Mar. 4, 1791 1724 15. Kentucky June 1, 1792 1774 16. Tennessee June 1, 1796 1769 17. Ohio Mar. 1, 1803 1788 18. Louisiana Apr. 30, 1812 1699 19. Indiana Dec. 11, 1816 1733 20. Mississippi Dec. 10, 1817 1699 21. Illinois Dec. 3, 1818 1720 22. Alabama Dec. 14, 1819 1702 23. Maine Mar. 15, 1820 1624 24. Missouri", "by the new congress, Sept. 15, 1789, provided for an obverse and a reverse, substantially as here depicted; but there is no evidence that the reverse was ever made. In the obverse as originally adopted, the eagle held in his sinister talon a bundle of thirteen arrows, and the first seal was thus made; but when, in 1841, a new seal was made to take the place of the old one, which had become worn, only six arrows were put into the eagle's talon. Whether this change, which was unauthorized by law, was made by design or by accident, is not known.] STATES. Date. Order. Area in squaremiles. Relative size. CAPITALS. .mw-parser-output .wst-rule{background-color:black;color:black;width:auto;margin:2px auto 2px auto;height:1px} Alabama 1819 9 50,722 16 Montgomery. Arkansas 1836 12 52,198 15 Little Rock. California 1850 18 188,981 2 Sacramento. [1]Connecticut 1788 5 4,750 35 Hartford. [1]Delaware 1787 1 2,120 36 Dover. Florida 1845 14 59,268 9 Tallahassee. [1]Georgia 1788 4 58,000 10 Atlanta. Illinois 1818 8 55,410 12 Springfield. Indiana 1816 6 83,809 28 Indianapolis. Iowa 1846 16 55,045 13 Des Moines. Kansas 1861 21 81,318 6 Topeka. Kentucky 1792 2 37,680 25 Frankfort. Louisiana 1812 5 41,346 22 New Orleans. Maine 1820 10 35,000 26 Augusta. [1]Maryland 1788 7 11,124 30 Annapolis. [1]Massachusetts 1788 6 7,800 34 Boston. Michigan 1837 13", "a few random weird punctuation marks that I know will appear later unless I take action now. 1789-Washington.txt [('every', 9), ('government', 8), ('public', 6), ('may', 6), ('citizens', 5), ('present', 5), ('country', 5), ('one', 4), ('ought', 4), ('duty', 4), ('people', 4), ('united', 4), ('since', 4), ('fellow', 3), ('could', 3)] 1793-Washington.txt [('shall', 3), ('oath', 2), ('fellow', 1), ('citizens', 1), ('called', 1), ('upon', 1), ('voice', 1), ('country', 1), ('execute', 1), ('functions', 1), ('chief', 1), ('magistrate', 1), ('occasion', 1), ('proper', 1), ('arrive', 1)] [('people', 20), ('government', 16), ('may', 13), ('nations', 11), ('country', 10), ('nation', 9), ('states', 9), ('foreign', 8), ('constitution', 8), ('honor', 7), ('justice', 6), ('ever', 6), ('congress', 6), ('public', 6), ('good', 6)] 1801-Jefferson.txt [('government', 12), ('us', 10), ('may', 8), ('fellow', 7), ('citizens', 7), ('let', 7), ('shall', 6), ('principle', 6), ('would', 6), ('one', 6), ('man', 6), ('safety', 5), ('good', 5), ('others', 5), ('peace', 5)] 1805-Jefferson.txt [('public', 14), ('citizens', 10), ('may', 10), ('fellow', 8), ('state', 8), ('us', 7), ('among', 7), ('shall', 7), ('constitution', 6), ('time', 6), ('limits', 5), ('reason', 5), ('false', 5), ('duty', 4), ('every', 4)] [('nations', 6), ('public', 6), ('well', 5), ('country', 4), ('peace', 4), ('rights', 4), ('states', 4), ('confidence', 3), ('full', 3), ('improvements', 3), ('united', 3), ('best', 3), ('examples', 2), ('avail', 2), ('made', 2)] [('war', 15), ('country', 5), ('united', 5), ('every', 5), ('british', 5), ('nation', 4), ('without', 4), ('states', 4), ('spirit',", "1858, 5220393), 37 (24, 'Mississippi', 1817, 2938618), 38 (25, 'Missouri', 1821, 5911605), 39 (26, 'Montana', 1889, 967440), 40 (27, 'Nebraska', 1867, 1783432), 41 (28, 'Nevada', 1864, 2600167), 42 (29, 'New Hampshire', 1788, 1315809), 43 (30, 'New Jersey', 1787, 8682661), 44 (31, 'New Mexico', 1912, 1984356), 45 (32, 'New York', 1788, 19490297), 46 (33, 'North Carolina', 1789, 9222414), 47 (34, 'North Dakota', 1889, 641481), 48 (35, 'Ohio', 1803, 11485910), 49 (36, 'Oklahoma', 1907, 3642361), 50 (37, 'Oregon', 1859, 3790060), 51 (38, 'Pennsylvania', 1787, 12448279), 52 (39, 'Rhode Island', 1790, 1050788), 53 (40, 'South Carolina', 1788, 4479800), 54 (41, 'South Dakota', 1889, 804194), 55 (42, 'Tennessee', 1796, 6214888), 56 (43, 'Texas', 1845, 24326974), 57 (44, 'Utah', 1896, 2736424), 58 (45, 'Vermont', 1791, 621270), 59 (46, 'Virginia', 1788, 7769089), 60 (47, 'Washington', 1889, 6549224), 61 (48, 'West Virginia', 1863, 1814468), 62 (49, 'Wisconsin', 1848, 5627967), 63 (50, 'Wyoming', 1890, 532668); ### GROUP BY: Grouping Data The GROUP BY clause groups the resulting data rows into groups. Here is an example: So what just happened? We have 50 rows in the table, but 34 results were returned by this query. This is because the results were grouped by the 'join_year' column. In other words, we only see one row for each distinct value of join_year. Since some states have the same join_year,", "4), ('citizens', 4), ('sense', 3), ('people', 3), ('justice', 3), ('part', 3), ('long', 3)] 1817-Monroe.txt [('government', 22), ('great', 21), ('states', 21), ('people', 15), ('us', 14), ('every', 14), ('united', 13), ('may', 10), ('union', 10), ('war', 10), ('citizens', 9), ('best', 9), ('principles', 9), ('foreign', 9), ('country', 9)] 1821-Monroe.txt [('great', 29), ('states', 20), ('would', 18), ('united', 16), ('war', 16), ('citizens', 15), ('may', 15), ('made', 15), ('government', 13), ('every', 13), ('people', 11), ('commerce', 11), ('force', 11), ('power', 11), ('fellow', 10)] [('union', 20), ('government', 17), ('upon', 16), ('country', 10), ('rights', 10), ('peace', 9), ('great', 9), ('public', 9), ('constitution', 8), ('first', 8), ('general', 8), ('nation', 8), ('people', 7), ('nations', 7), ('duties', 6)] Better, but again, we want to make the output something that can be the object of computation. Let’s do the following: • For each of the early and late texts, get their top 15 words. • Make a list of all of those words • For each of the “was in a top 15” words, calculate the frequency in each of our texts (early and late) • Save this info in a data frame We now have a list of frequent words, but I want to eliminate duplicates (using set()) and alphabetize the list (using sort()). ['abroad', 'america', 'american', 'americans', 'among', 'arrive', 'avail', 'believe', 'best', 'british', 'called', 'cannot', 'century', 'change', 'chief', 'citizens', 'commerce',", "that contains the decade starting with 1951, 1961, 1971 and so on. In [41]: #https://stackoverflow.com/questions/48856982/divide-data-by-decade-then-plot-it-seaborn-box-and-whisker sb['decade'] = (sb['year'] // 10) * 10 + 1 Let's have a look at the data. In [42]: sb.head() Out[42]: station_id year JJA count mean std z_base decade 0 101603550000 1967 2334.00 12.00 2321.42 59.90 0.21 1961 1 101603550000 1968 2360.67 12.00 2321.42 59.90 0.66 1961 2 101603550000 1969 2237.33 12.00 2321.42 59.90 -1.40 1961 3 101603550000 1970 2350.67 12.00 2321.42 59.90 0.49 1961 4 101603550000 1971 2416.00 12.00 2321.42 59.90 1.58 1971 Let's plot the data per decade. In [43]: sb[sb['decade']==1951]['z_base'].hist(bins=200, histtype = 'step', label='1951', density=True) sb[sb['decade']==1961]['z_base'].hist(bins=200, histtype = 'step', label='1961', density=True) sb[sb['decade']==1971]['z_base'].hist(bins=200, histtype = 'step', label='1971', density=True) sb[sb['decade']==1981]['z_base'].hist(bins=200, histtype = 'step', label='1981', density=True) sb[sb['decade']==1991]['z_base'].hist(bins=200, histtype = 'step', label='1991', density=True) sb[sb['decade']==2001]['z_base'].hist(bins=200, histtype = 'step', label='2001', density=True) sb[sb['decade']==2011]['z_base'].hist(bins=200, histtype = 'step', label='2011', density=True).legend(loc='upper right'); That is very close to the results in Hansen et al. (2012). It is important to note that we use the complete station data (Not just the data from the Northern hemisphere as in the two articles.) ### Focus on the Northern hemisphere¶ The Goddard Institute for Space Studies (GISS) also provides more details on the stations here. This is a text file(!) with very problematic formatting. Simply reading the file does not work. We need a fixed-width reader.", "## 2 1793-Washington 0 0 0 0 4 ## 3 1797-Adams 8 0 5 44 2 ## 4 1801-Jefferson 18 0 3 21 2 ## 5 1805-Jefferson 2 3 6 43 3 ## 6 1809-Madison 0 0 4 17 7 ## 7 1813-Madison 0 5 5 20 3 ## 8 1817-Monroe 2 0 27 47 12 ## 9 1821-Monroe 2 0 37 66 10 ## 10 1825-Adams 8 2 13 46 8 ## # … with 49 more rows By adding prop = TRUE, the same table is displayed with row percentages : clusters_by_doc_table(corpus, clust_var = \"cluster\", prop = TRUE) ## # A tibble: 59 × 6 ## doc_id clust_1 clust_2 clust_3 clust_4 clust_5 ## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 1789-Washington 0 0 10.8 73.0 16.2 ## 2 1793-Washington 0 0 0 0 100 ## 3 1797-Adams 13.6 0 8.47 74.6 3.39 ## 4 1801-Jefferson 40.9 0 6.82 47.7 4.55 ## 5 1805-Jefferson 3.51 5.26 10.5 75.4 5.26 ## 6 1809-Madison 0 0 14.3 60.7 25 ## 7 1813-Madison 0 15.2 15.2 60.6 9.09 ## 8 1817-Monroe 2.27 0 30.7 53.4 13.6 ## 9 1821-Monroe 1.74 0 32.2 57.4 8.70 ## 10 1825-Adams 10.4 2.60 16.9 59.7 10.4 ## # … with 49 more rows Conversely, the docs_by_cluster_table allows to display, for each cluster, the", "1851, 1856, 1895 VERMONT 1765, 1771, 1798 VIRGINIA 1623, 1624, 1625, 1634, 1699, 1701, 1703, 1779, 1798 WASHINGTON 1871, 1883, 1885, 1889, 1892 WEST VIRGINIA (Part of Virginia until 1863) WISCONSIN 1830, 1836, 1838, 1842, 1846, 1847, 1855, 1865, 1875, 1885, 1895, 1905 WYOMING 1905, 1915, 1925 Back to the Instructions Page", "States Goveri,ment Printin g Offic . e, .. , . . . Washington . ~ . 19~. . .'• , (1) . Incldes est:!u,ates for six co,int _ ries . i.Il 1920, four in 1921, . anq. t1r0 in 1922. . . . . . I . . . . . . . TABLE 28 . . . .. PER CAPITA op . CIGARS . A.HD dTGARiST'l'ES IN E 'LTR OPEAN . ' . . COUNTR I es COllttPtRkl> IJITH THE UBlTkl> STATF.8 . .. , . . . .. . . i932 . . . . c Po,mds) •. < . . :qn i ted ~ States Au stria Be igi11m . C z e . cho slovak:i .\" a . . .. \\ ' :Ce nmark Fi n1and Fr ance . Ge rman-y; Gr eece (J.) . . . . . H u _ ngB.I'y I t a ly .. : .. N e the rlands ~ o r:way P o land . Rum ani a Sp ain . . s eden . 1: __ ted _ Kingdon .. -r ~ ~;: os la 1 J ~i a -. . . . . . .. . . Ciga~ .89 . . . .. . . . 25 . .40 . . . .16 ... 1.27 .", "651, 6, 84, 47, 1694, 12, 1605, 8, 4947, 6, 17263, 16, 11316, 68, 73, 509, 21122, 26, 2934, 57, 8, 2968, 220, 52, 26, 6788, 5, 3, 2092, 8, 239, 6, 2479, 13, 3, 9, 2659, 208, 6855, 11, 192, 276, 26165, 29, 7527, 1632, 5908, 11, 130, 3, 5385, 26, 57, 276, 26165, 29, 768, 7613, 63, 6, 84, 3, 115, 19054, 8034, 16, 8, 4453, 7, 13, 8, 1511, 5, 17159, 9, 63, 1213, 3, 9, 182, 1101, 1102, 7513, 6, 2049, 12, 8, 4034, 307, 18, 5517, 1472, 13, 8, 4004, 3974, 3013, 7, 68, 112, 2054, 47, 396, 5551, 120, 23038, 12, 1520, 34, 5, 17159, 9, 63, 47, 4792, 16, 8, 1480, 1379, 116, 3, 9, 212, 12661, 13, 8, 4889, 138, 181, 52, 14424, 1074, 3322, 3, 29155, 1084, 376, 117, 8, 3, 35, 15776, 3335, 13, 8, 1511, 57, 8, 276, 26165, 29, 7, 16026, 8, 2379, 18836, 13, 10742, 535, 6, 1525, 8, 826, 822, 5, 2507, 24, 8, 1525, 19, 915, 441, 8, 1499, 5, 11860, 10, 363, 47, 8, 1706, 13, 17159, 9, 63, 31, 7, 10101, 58 ] \" Given the following passage \"The first action of the Franco-Prussian War took place on 4 August 1870. This battle saw the unsupported division of General Douay of", "on this? # ALL CODE IN ONE GO library(data.table) tax_aggregated <- structure(list(Province = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3), Tax = c(2000, 3000, 1500, 3200, 2000, 1500, 4000, 2000, 2000, 1000, 2000, 1500), year = c(2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2003, 2003, 2003)), row.names = c(NA, -12L), class = c(\"tbl_df\", \"tbl\", \"data.frame\")) Individual_data <- structure(list(ID = c(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), Province = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3), year = c(2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003), Age = c(\"15-30\", \"30-40\",", "shortcoming of the “3, 4, 5″ type of categorization. w. Year Power Dissipation Index 1851.49 , 5.046 1851.51 , 0.729 1851.53 , 0.216 1851.63 , 25.747 1851.70 , 3.456 1851.79 , 4.199 1852.64 , 35.299 1852.68 , 2.688 1852.69 , 9.31 1852.73 , 21.25 1852.77 , 18.622 1853.60 , 0.216 1853.61 , 0.125 1853.67 , 76.601 1853.69 , 19.28 1853.72 , 0.216 1853.74 , 9.742 1853.74 , 0.216 1853.80 , 10.856 1854.49 , 3.837 1854.65 , 0.343 1854.69 , 25.946 1854.72 , 6.268 1854.80 , 3.354 1855.60 , 1 1855.61 , 6 1855.61 , 0.512 1855.65 , 4.29 1855.71 , 10.944 1856.61 , 19.395 1856.62 , 4.096 1856.64 , 2.376 1856.64 , 0.216 1856.65 , 27.805 1856.72 , 7.366 1857.50 , 1.728 1857.69 , 20.936 1857.73 , 10.24 1857.73 , 13.514 1858.45 , 0.512 1858.60 , 0.512 1858.71 , 7.115 1858.71 , 21.096 1858.73 , 7.54 1858.81 , 12.478 1859.50 , 1 1859.63 , 12 1859.67 , 3.072 1859.70 , 5.482 1859.71 , 5.883 1859.76 , 31.8 1859.79 , 2.867 1859.83 , 15.976 1860.61 , 20.362 1860.65 , 7.65 1860.70 , 0.512 1860.70 , 19.853 1860.72 , 4.362 1860.75 , 8.138 1860.80 , 11.928 1861.52 , 13.7 1861.62 , 8.349 1861.65 , 19.664 1861.71 , 0.512 1861.74 , 2.142 1861.77 , 4.98 1861.77 , 0.216 1861.84 , 3.899" ]

[ "+0 # help ples +1 190 1 Roger has exactly one of each of the first 22 states' new U.S. quarters. The quarters were released in the same order that the states joined the union. The graph below shows the number of states that joined the union in each decade. What fraction of Roger's 22 coins represents states that joined the union during the decade 1780 through 1789? Express your answer as a common fraction. (note: every space represents 2 states.) (I am the rabbit) Jun 13, 2020 12 states joined from 1780 to 1789. Therefore, of his first 22 quarters, 12 of them are from this time period, making for $\\frac{12}{22}$=$\\frac{6}{11}$ of his coins being from this time period. If you didnt understandmy LaTeX the answer is 6/11. lol", "to mint and issue quarter dollars in 1999 which bear the design in effect before the redesign required under this subsection and an inscription of the year 1998' as required to ensure a smooth transition into the 10-year program under this subsection. (2) Single state designs.--The design on the reverse side of each quarter dollar issued during the 10-year period referred to in paragraph (1) shall be emblematic of 1 of the 50 States. (3) Issuance of coins commemorating 5 states during each of the 10 years.-- (A) In general.--The designs for the quarter dollar coins issued during each year of the 10-year period referred to in paragraph (1) shall be emblematic of 5 States selected in the order in which such States ratified the Constitution of the United States or were admitted into the Union, as the case may be. (B) Number of each of 5 coin designs in each year.--Of the quarter dollar coins issued during each year (of the 10-year period referred to in paragraph (1)), the Secretary of the Treasury shall prescribe, on the basis of such factors as the Secretary determines to be appropriate, the number of quarter dollars which shall be issued with each of the 5 designs selected for such year. (4) Selection of design.-- (A) In general.--Each of the 50", "# how many states existed in 1796? it was in 1796 ##### Answers 1 Tennessee became the 16th state in June of 1796. Thus, 16, it was 17 years before another state was added to the union.", "the establishment of any mints or engraving plants. Also discover some change in the metal composition of a coin, giving any interesting highlights concerning such a change. Seal of the United States Mint ### United States Mint The United States Mint primarily produces circulating coinage for the United States to conduct its trade and commerce. The main Mint facility is located in Philadelphia, Pennsylvania, and branch facilities are located in Denver, Colorado, San Francisco, California, and West Point, New York. Mint marks in United States coinage include: • P for the Philadelphia Mint, • D for the Denver Mint, • S for the San Francisco Mint, • W for the West Point Mint, • CC for the Carson City Mint, • C for the Charlotte Mint, and • O for the New Orleans Mint. Most coins of the Philadelphia Mint earlier than 1980 are unmarked. #### Current Facilities The Mint's largest facility is the Philadelphia Mint, one of four active coin-producing mints. The current facility at Philadelphia, which opened in 1969, is the fourth Philadelphia Mint. The first was built in 1792, when Philadelphia was still the U.S. capital, and began operation in 1793. Until 1980, coins minted at Philadelphia bore no mint mark, with the exceptions of the Susan B. Anthony dollar and the wartime Jefferson nickel.", "for this calculation. '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ So you have 27 quarters", "- 4) = 196 4. 45x - 720 + 15x - 60 5. 132 - 4(12x-9) 6. 198 - 154 + 7x - 68 8. - 7x - 10 - 18 + 3.x 7. - 131 = -5(3x -8) + 6 x 10. - (2x - 6) - 12x + 6 9. 12x+8 -15= -2(3x - 82) Solve each equation. You must show all work. 1. 5x-2-33 2. 140 - 4x +36 3. 8(3x - 4) = 196 4. 45x - 720 + 15x - 60 5. 132 - 4(12x-9) 6. 198 - 154 + 7x - 68 8. - 7x - 10 - 18 + 3.x 7. - 131 = -5(3x -8) + 6 x 10. - (2x - 6) - 12x + 6 9. 12x+8 -15= -2(3x - 82)... ### \"I am the poor white, fooled and pushed apart\" Ends with: \"The poorestworker bartered through the years. \"I am the poor white, fooled and pushed apart\" Ends with: \"The poorestworker bartered through the years.... ### North Carolina had about 100,000 slaves in 1790. How many of these would count toward the state's representation in Congress? i know no one will answer but heck worth a shot North Carolina had about 100,000 slaves in 1790. How many of these would count toward the state's representation in Congress?", "## Thinking Mathematically (6th Edition) The year when America declared Independence is $1776$ The Roman Numeral written along the base of the pyramid at the back of US one dollar is MDCCLXXVI Converting MDCCLXXVI to Hindu Arabic Numeral, we get \\begin{align} & 1000+500+100+100+50+10+10+5+1=1000+700+70+6 \\\\ & =1776 \\end{align} Which is the year when America declared Independence. After Observing and calculating the back of US one dollar, we find the year $1776$ written in Roman Numerals along the base of the pyramid", "# Fraternitè Algebra Level pending On July 14, 1789, all brothers in a family calculated the sum of their ages $$X$$ and then the sum $$Y$$ of their years of birth. It turned out that $$X + Y = 17881$$. How many of them were born in March at most? ×", "33 sections in the emblem of the UN. count them: http://www.un.org/english/ The UN emblem has 26, 28, or 30 leaves, depending on what you count. there are 33 feathers on one side of the eagle on the back of the dollar bill, and 32 on the other side (one side representing the French-Scottish system, and the other... I don't know, prolly some other system). There are 9 tail feathers which represent that other system that I forgot but mentioned oh yeah, they also like the number 13, and I forgot what it means, but there are 13 levels in the pyramid on the dollar bill, and the eye on the top represents the illimaniti as it spreads it's \"light\" downward (aka shroud of confusion). The real meaning of the US 1 dollar bill The only thing left to question is the Pyramid. What's your reason for the Pyramid? one more thing... the illuminati loves that number 1776, and they put (snip) masons in the US), and even the new WTC is going to be 1776 feet tall... 1776 was also the year the continental congress ratified the Declaration of Independance. (july 4th, you know?) So, it kinda makes a person wonder... When the stock market was booming in 1928, why did, all of the sudden, all of the", "impacting one of my hobbies... I finally convinced my wife to get rid of the wedding dress... to her ex... just a few years ago. Dan - Owner http://www.Hi-TecDesigns.com ### RE: A significant development is impacting one of my hobbies... When my daughter got married, I asked why not let her wear my wife's wedding dress. She said no. I told her it just hangs in the closet collecting dust, then I thought...what a great idea! It can collect all the dust and keep it off my clothes. I should sell it and buy us ice cream. ctopher, CSWP SolidWorks '19 ctophers home SolidWorks Legion ### RE: A significant development is impacting one of my hobbies... (OP) Yes, the 1776-1976 Bicentennial coinage included the Washington Quarter, the Kennedy Half Dollar and the Eisenhower Dollar. In my collection I have the following: 1776-1976 Washington Quarter: 38 - no mint mark (Philadelphia) 115 - 'D' mint mark (Denver) 3 - 'S' mint mark (San Francisco) 1776-1976 Kennedy Halves: 2 - no mint mark (Philadelphia) 29 - 'D' mint mark (Denver) 3 - 'S' mint mark (San Francisco) 1776-1976 Eisenhower Dollar: 6 - no mint mark (Philadelphia) 6 - 'D' mint mark (Denver) 4 - 'S' mint mark (San Francisco) Note that I still find two or three Bicentennial quarters each", "the English language for the thaler for about 200 years prior to the American Revolution. Spanish dollars, or pieces of eight as they were called, were in circulation in the 13 colonies that became the United States and legal tender in Virginia. Examples of differential equations ... can say that the solution must be of the form: $x(t) = A \\cos t + B \\sin t$ To fix the unknown constants $A$ and ...", "## May 7, 2009 ### Odd Currency Puzzle #### Posted by John Baez Sorry to be posting so much light, frothy stuff lately — but since it’s an odd day, I can’t resist another puzzle. What’s the oddest currency ever used in America? Of course this is a subjective question, so I’d be interested to hear your opinion… … but my own favorite candidate is the $\\frac{1}{9}$ dollar bill used in colonial Maryland: A total of 57,000 bills of this sort were issued in 1767, 1770 and 1774. The dent in the upper left corner was deliberate: ‘indented’ bills were used to fight against counterfeiting, which was rampant at the time. Another trick was inserting deliberate typos, like the colon after “Annapolis”. Note also that this note is hand-signed, with a hand-written serial number! Back then, being treasurer was hard work. You might argue that the $\\frac{2}{9}$ dollar bill was even odder… … but I’d have to disagree: it’s oddly evener! I thank Robert Schlesinger for tipping me off about the existence of these odd fractional bills. The above pictures come from the following delightful history, where you can also see a 4 dollar bill, a 6 dollar bill, a $\\frac{1}{3}$ dollar bill, and so on: On the other hand, if you just find big bills odd, maybe", "the emblem of the UN. count them: http://www.un.org/english/ The UN emblem has 26, 28, or 30 leaves, depending on what you count. there are 33 feathers on one side of the eagle on the back of the dollar bill, and 32 on the other side (one side representing the French-Scottish system, and the other... I don't know, prolly some other system). There are 9 tail feathers which represent that other system that I forgot but mentioned oh yeah, they also like the number 13, and I forgot what it means, but there are 13 levels in the pyramid on the dollar bill, and the eye on the top represents the illimaniti as it spreads it's \"light\" downward (aka shroud of confusion). The real meaning of the US 1 dollar bill The only thing left to question is the Pyramid. What's your reason for the Pyramid? one more thing... the illuminati loves that number 1776, and they put (snip) masons in the US), and even the new WTC is going to be 1776 feet tall... 1776 was also the year the continental congress ratified the Declaration of Independance. (july 4th, you know?) So, it kinda makes a person wonder... When the stock market was booming in 1928, why did, all of the sudden, all of the major stock holders", "(PDF). Retrieved 25 October 2019. 20. Krause, Chester L.; Clifford Mishler (2004). Standard Catalog of World Coins: 1801–1900. Colin R. Bruce II (senior editor) (4th ed.). Krause Publications. ISBN 0873497988. 21. \"Equivalent of 0.343762855 troy ounce of silver in U.S. dollar\". xe.com. 2 October 2006. Retrieved 2 October 2006.", "## If John has exactly 10 coins each of which was minted in 1910 or 1920 or 1930, how many of his coins were minted in ##### This topic has expert replies Moderator Posts: 6242 Joined: 07 Sep 2017 Followed by:20 members ### If John has exactly 10 coins each of which was minted in 1910 or 1920 or 1930, how many of his coins were minted in by BTGmoderatorDC » Wed Oct 20, 2021 6:12 pm 00:00 A B C D E ## Global Stats If John has exactly 10 coins each of which was minted in 1910 or 1920 or 1930, how many of his coins were minted in 1920 ? (1) Exactly 6 of his coins were minted in 1910 or 1920. (2) Exactly 7 of his coins were minted in 1920 or 1930. OA C Source: GMAT Prep Legendary Member Posts: 2064 Joined: 29 Oct 2017 Followed by:6 members ### Re: If John has exactly 10 coins each of which was minted in 1910 or 1920 or 1930, how many of his coins were minted in by swerve » Thu Oct 21, 2021 2:34 pm 00:00 A B C D E ## Global Stats BTGmoderatorDC wrote: Wed Oct 20, 2021 6:12 pm If John has exactly 10 coins each of which was minted in", "Mint [1] Treasury Chambers 19th Ianry 1713", "fails to satisfy. fin 1. italics original, see Fed. Schol. Soc. Notes, Stardate 48317.6 2. I keep saying “18th Century” despite this magazine being from the early 19th, because of course these folks were almost certainly all born in the 18th.", "U.S. penny. In 1982 the composition of a U.S. penny was changed from a brass alloy consisting of 95% w/w Cu and 5% w/w Zn, to a zinc core covered with copper.11 The pennies in Table 4.16, therefore, were drawn from different populations.", "and Secret Service personnel removed him from the room. Udall resumed his speech, unperturbed.[61] 7. ^ Some locals wrote letters to the St. Louis Post-Dispatch accusing Bi-State of \"gouging\".[16] 8. ^ The U.S. Mint altered Jackson's design to make it less \"off balance,\" however, with three people in the canoe instead of just Lewis and Clark. A Mint representative said the third person was Clark's slave, York.[143] The finalized coin entered circulation on August 4, 2003.[146]", "coins were minted in 1933. The last 90% silver coins were minted in 1964, and the last 40% silver half dollar was minted in 1970. The United States Mint currently produces circulating coins at the Philadelphia and Denver Mints, and commemorative and proof coins for collectors at the San Francisco and West Point Mints. Mint mark conventions for these and for past mint branches are discussed in Coins of the United States dollar#Mint marks. The one-dollar coin has never been in popular circulation from 1794 to present, despite several attempts to increase their usage since the 1970s, the most important reason of which is the continued production and popularity of the one-dollar bill.[43] Half dollar coins were commonly used currency since inception in 1794, but has fallen out of use from the mid-1960s when all silver half dollars began to be hoarded. The nickel is the only coin whose size and composition (5 grams, 75% copper, and 25% nickel) is still in use from 1865 to today, except for wartime 1942-1945 Jefferson nickels which contained silver. Due to the penny's low value, some efforts have been made to eliminate the penny as circulating coinage. [44] [45] For a discussion of other discontinued and canceled denominations, see Obsolete denominations of United States currency#Coinage and Canceled denominations of United States" ]

[ "+0 # help ples +1 190 1 Roger has exactly one of each of the first 22 states' new U.S. quarters. The quarters were released in the same order that the states joined the union. The graph below shows the number of states that joined the union in each decade. What fraction of Roger's 22 coins represents states that joined the union during the decade 1780 through 1789? Express your answer as a common fraction. (note: every space represents 2 states.) (I am the rabbit) Jun 13, 2020 12 states joined from 1780 to 1789. Therefore, of his first 22 quarters, 12 of them are from this time period, making for $\\frac{12}{22}$=$\\frac{6}{11}$ of his coins being from this time period. If you didnt understandmy LaTeX the answer is 6/11. lol", "1; b < 4; ++b) { draw((2*a+12,b-.1)--(2*a+12,b+.1)); draw((2*a+11.9,b)--(2*a+12.1,b)); } } label(\"1960\",(12,0),S); label(\"1970\",(14,0),S); label(\"1980\",(16,0),S); label(\"1990\",(18,0),S); label(\"10\",(12,1),W); label(\"20\",(12,2),W); label(\"30\",(12,3),W); label(\"\\%\",(12,4),N); draw((0,12)--(0,8)--(7,8)); draw((0,9)--(.2,9)); draw((0,10)--(.2,10)); draw((0,11)--(.2,11)); draw((2,8)--(2,8.2)); draw((4,8)--(4,8.2)); draw((6,8)--(6,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a,b+7.9)--(2*a,b+8.1)); draw((2*a-.1,b+8)--(2*a+.1,b+8)); } } label(\"1960\",(0,8),S); label(\"1970\",(2,8),S); label(\"1980\",(4,8),S); label(\"1990\",(6,8),S); label(\"10\",(0,9),W); label(\"20\",(0,10),W); label(\"30\",(0,11),W); label(\"\\%\",(0,12),N); draw((12,12)--(12,8)--(19,8)); draw((12,9)--(12.2,9)); draw((12,10)--(12.2,10)); draw((12,11)--(12.2,11)); draw((14,8)--(14,8.2)); draw((16,8)--(16,8.2)); draw((18,8)--(18,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a+12,b+7.9)--(2*a+12,b+8.1)); draw((2*a+11.9,b+8)--(2*a+12.1,b+8)); } } label(\"1960\",(12,8),S); label(\"1970\",(14,8),S); label(\"1980\",(16,8),S); label(\"1990\",(18,8),S); label(\"10\",(12,9),W); label(\"20\",(12,10),W); label(\"30\",(12,11),W); label(\"\\%\",(12,12),N); draw((24,12)--(24,8)--(31,8)); draw((24,9)--(24.2,9)); draw((24,10)--(24.2,10)); draw((24,11)--(24.2,11)); draw((26,8)--(26,8.2)); draw((28,8)--(28,8.2)); draw((30,8)--(30,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a+24,b+7.9)--(2*a+24,b+8.1)); draw((2*a+23.9,b+8)--(2*a+24.1,b+8)); } } label(\"1960\",(24,8),S); label(\"1970\",(26,8),S); label(\"1980\",(28,8),S); label(\"1990\",(30,8),S); label(\"10\",(24,9),W); label(\"20\",(24,10),W); label(\"30\",(24,11),W); label(\"\\%\",(24,12),N); draw((0,9)--(2,9.25)--(4,10)--(6,11)); draw((12,8.5)--(14,9)--(16,10)--(18,10.5)); draw((24,8.5)--(26,8.8)--(28,10.5)--(30,11)); draw((0,0.5)--(2,1)--(4,2.8)--(6,3)); draw((12,0.5)--(14,.8)--(16,1.5)--(18,3)); label(\"(A)\",(-1,12),W); label(\"(B)\",(11,12),W); label(\"(C)\",(23,12),W); label(\"(D)\",(-1,4),W); label(\"(E)\",(11,4),W); [/asy]$ ## Problem 5 Each principal of Lincoln High School serves exactly one $3$-year term. What is the maximum number of principals this school could have during an $8$-year period? $\\text{(A)}\\ 2 \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 4 \\qquad \\text{(D)}\\ 5 \\qquad \\text{(E)}\\ 8$ ## Problem 6 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", "7), ('shall', 6), ('principle', 6), ('would', 6), ('one', 6), ('man', 6), ('safety', 5)] 1805-Jefferson.txt [(',', 142), ('.', 41), (';', 26), ('public', 14), ('citizens', 10), ('may', 10), ('fellow', 8), ('state', 8), ('us', 7), ('among', 7), ('shall', 7), ('constitution', 6), ('time', 6), ('limits', 5), ('reason', 5)] [(',', 47), ('.', 21), (';', 16), ('nations', 6), ('public', 6), ('well', 5), ('country', 4), ('peace', 4), ('rights', 4), ('states', 4), ('confidence', 3), ('full', 3), ('improvements', 3), ('united', 3), ('best', 3)] [(',', 53), ('.', 31), ('war', 15), (';', 6), ('country', 5), ('united', 5), ('every', 5), ('british', 5), ('nation', 4), ('without', 4), ('states', 4), ('spirit', 4), ('citizens', 4), ('sense', 3), ('people', 3)] 1817-Monroe.txt [(',', 169), ('.', 110), ('government', 22), ('great', 21), ('states', 21), ('people', 15), ('us', 14), ('every', 14), ('united', 13), (';', 13), ('may', 10), ('?', 10), ('union', 10), ('war', 10), ('citizens', 9)] 1821-Monroe.txt [(',', 275), ('.', 130), ('great', 29), ('states', 20), ('would', 18), ('united', 16), ('war', 16), ('citizens', 15), ('may', 15), ('made', 15), ('government', 13), ('every', 13), ('people', 11), ('commerce', 11), ('force', 11)] [(',', 115), ('.', 72), (';', 35), ('union', 20), ('government', 17), ('upon', 16), ('country', 10), ('rights', 10), ('peace', 9), ('great', 9), ('public', 9), ('constitution', 8), ('first', 8), ('general', 8), ('nation', 8)] Hmmm, we still have punctuation in there, which we don’t care about. Let’s remove those, and try again. Note that I’ve added", "States of America 1818 #> 4 2 United States of America 1819 #> 5 2 United States of America 1820 #> 6 2 United States of America 1821 #> 7 2 United States of America 1822 #> 8 2 United States of America 1823 #> 9 2 United States of America 1824 #> 10 2 United States of America 1825 #> # … with 16,721 more rows ### Create a State-Year Panel of Gleditsch-Ward states create_stateyears() can do the same for Gleditsch-Ward states, but requires the user to specify they want states from the Gleditsch-Ward system. create_stateyears(system=\"gw\") #> # A tibble: 18,289 x 3 #> gwcode statename year #> <dbl> <chr> <int> #> 1 2 United States of America 1816 #> 2 2 United States of America 1817 #> 3 2 United States of America 1818 #> 4 2 United States of America 1819 #> 5 2 United States of America 1820 #> 6 2 United States of America 1821 #> 7 2 United States of America 1822 #> 8 2 United States of America 1823 #> 9 2 United States of America 1824 #> 10 2 United States of America 1825 #> # … with 18,279 more rows ### Create a Panel of ISO Codes ISO codes are ubiquitous in economic data. I do have some misgivings", "4.1 1. 135 2. 15 In [5]: years <- seq(1811, 2009, by = 1) Major$decade = cut(Major$dyad_st_year, seq(from = 1810, to = 2010, by = 10), include.lowest=TRUE, labels=seq(1810,2000, by=10)) Major$decade 1. 1810 2. 1810 3. 1840 4. 1840 5. 1820 6. 1820 7. 1830 8. 1830 9. 1830 10. 1830 11. 1830 12. 1830 13. 1830 14. 1830 15. 1830 16. 1830 17. 1840 18. 1840 19. 1850 20. 1850 21. 1850 22. 1850 23. 1850 24. 1850 25. 1860 26. 1860 27. 1860 28. 1870 29. 1870 30. 1880 31. 1880 32. 1880 33. 1890 34. 1900 35. 1900 36. 1900 37. 1900 38. 1910 39. 1910 40. 1910 41. 1910 42. 1910 43. 1920 44. 1920 45. 1920 46. 1920 47. 1930 48. 1930 49. 1930 50. 1930 51. 1930 52. 1930 53. 1930 54. 1930 55. 1930 56. 1930 57. 1930 58. 1930 59. 1930 60. 1930 61. 1930 62. 1930 63. 1930 64. 1930 65. 1940 66. 1940 67. 1940 68. 1940 69. 1940 70. 1940 71. 1940 72. 1940 73. 1940 74. 1970 75. 1980 76. 1940 77. 1940 78. 1940 79. 1940 80. 1980 81. 1980 82. 1940 83. 1980 84. 1990 85. 1940 86. 1940 87. 1940 88. 1990 89. 1940 90. 1940 91. 1940 92. 1990 93. 1940 94. 1980", "queries you will be running. ### Sample Table Before we go further with more queries, I would like to create a sample table with some data. This will be a list of US states, with their join dates (the date the state ratified the United States Constitution or was admitted to the Union) and their current populations. You can copy paste the following to your MySQL console: 1 2 CREATE TABLE states ( 3 id INT AUTO_INCREMENT, 4 name VARCHAR(20), 5 join_year INT, 6 population INT, 7 PRIMARY KEY (id), 8 UNIQUE (name), 9 KEY (join_year) 10 ); 11 12 13 INSERT INTO states VALUES 14 (1, 'Alabama', 1819, 4661900), 15 (2, 'Alaska', 1959, 686293), 16 (3, 'Arizona', 1912, 6500180), 17 (4, 'Arkansas', 1836, 2855390), 18 (5, 'California', 1850, 36756666), 19 (6, 'Colorado', 1876, 4939456), 20 (7, 'Connecticut', 1788, 3501252), 21 (8, 'Delaware', 1787, 873092), 22 (9, 'Florida', 1845, 18328340), 23 (10, 'Georgia', 1788, 9685744), 24 (11, 'Hawaii', 1959, 1288198), 25 (12, 'Idaho', 1890, 1523816), 26 (13, 'Illinois', 1818, 12901563), 27 (14, 'Indiana', 1816, 6376792), 28 (15, 'Iowa', 1846, 3002555), 29 (16, 'Kansas', 1861, 2802134), 30 (17, 'Kentucky', 1792, 4269245), 31 (18, 'Louisiana', 1812, 4410796), 32 (19, 'Maine', 1820, 1316456), 33 (20, 'Maryland', 1788, 5633597), 34 (21, 'Massachusetts', 1788, 6497967), 35 (22, 'Michigan', 1837, 10003422), 36 (23, 'Minnesota',", "1810,blame,v,19798,0.96 1811,estimate,v,20474,0.93 1812,due,i,21141,0.9 1814,late,r,19727,0.96 1815,golf,n,20979,0.91 1816,investigate,v,20117,0.94 1817,crazy,j,20345,0.93 1818,significantly,r,22038,0.86 1819,chain,n,19688,0.96 1821,branch,n,19633,0.96 1822,combination,n,20026,0.94 1823,just,j,20096,0.94 1824,frequently,r,20508,0.92 1825,governor,n,20802,0.91 1826,relief,n,19493,0.97 1827,user,n,20681,0.91 1829,kick,v,20092,0.94 1830,part,r,19900,0.95 1831,manner,n,20012,0.94 1832,ancient,j,19818,0.95 1833,silence,n,21169,0.89 1834,rating,n,20413,0.92 1835,golden,j,19852,0.95 1836,motion,n,19649,0.95 1837,German,j,20096,0.93 1838,gender,n,22190,0.85 1839,solve,v,19501,0.96 1840,fee,n,20263,0.93 1841,landscape,n,20171,0.93 1842,used,j,19130,0.98 1843,bowl,n,20662,0.91 1844,equal,j,19952,0.94 1845,long,c,19227,0.97 1846,official,j,19628,0.95 1847,forth,r,19531,0.96 1848,frame,n,20016,0.93 1849,typical,j,19724,0.95 1850,except,i,19594,0.95 1851,conservative,j,19778,0.94 1852,eliminate,v,19773,0.94 1853,host,n,19389,0.96 1854,hall,n,20236,0.92 1855,trust,v,19482,0.95 1856,ocean,n,19421,0.96 1857,score,v,20391,0.91 1858,row,n,19519,0.95 1859,producer,n,19679,0.94 1860,afford,v,19035,0.97 1861,meanwhile,r,19504,0.95 1862,regime,n,21101,0.88 1863,division,n,19772,0.94 1864,confirm,v,19327,0.96 1865,fix,v,19349,0.96 1866,appeal,n,19561,0.95 1867,mirror,n,20081,0.92 1868,tooth,n,20515,0.9 1869,smart,j,19370,0.95 1870,length,n,19624,0.94 1871,entirely,r,19075,0.97 1872,rely,v,19614,0.94 1873,topic,n,20640,0.89 1874,complain,v,19102,0.97 1875,issue,v,19300,0.95 1876,variable,n,22868,0.81 1877,back,v,19213,0.96 1878,range,v,20067,0.92 1879,telephone,n,19311,0.95 1880,perception,n,21086,0.87 1881,attract,v,19197,0.96 1882,confidence,n,18955,0.97 1883,bedroom,n,20387,0.9 1884,secret,n,19125,0.96 1885,debt,n,19506,0.94 1886,rare,j,18978,0.97 1887,his,p,19663,0.93 1888,tank,n,19158,0.95 1889,nurse,n,19385,0.94 1890,coverage,n,19522,0.94 1891,opposition,n,19847,0.92 1892,aside,r,19056,0.96 1893,anywhere,r,18970,0.96 1894,bond,n,19856,0.92 1895,file,v,19416,0.94 1896,pleasure,n,19099,0.96 1897,master,n,18880,0.96 1898,era,n,19209,0.95 1899,requirement,n,20358,0.89 1900,check,n,18764,0.97 1901,stand,n,18910,0.96 1902,fun,n,18984,0.96 1903,expectation,n,19603,0.92 1904,wing,n,19028,0.95 1905,separate,j,19064,0.95 1906,now,c,19126,0.95 1907,clear,v,19013,0.95 1908,struggle,n,19014,0.95 1909,mean,j,19792,0.91 1910,somewhat,r,18801,0.96 1911,pour,v,19300,0.93 1912,stir,v,20174,0.89 1913,judgment,n,19315,0.93 1914,clean,v,18766,0.96 1915,except,c,18857,0.95 1916,beer,n,19349,0.93 1917,English,n,18719,0.96 1918,reference,n,19559,0.92 1919,tear,n,19819,0.9 1920,doubt,n,18542,0.97 1921,grant,v,18598,0.96 1922,seriously,r,18328,0.98 1923,account,v,19403,0.92 1924,minister,n,19154,0.93 1925,totally,r,18720,0.95 1926,hero,n,18348,0.97 1927,industrial,j,19413,0.92 1928,cloud,n,19214,0.92 1929,stretch,v,19014,0.93 1930,winner,n,19216,0.92 1931,volume,n,18989,0.93 1932,travel,n,18634,0.95 1933,seed,n,19079,0.93 1934,surprised,j,18822,0.94 1935,rest,v,18915,0.94 1936,fashion,n,18483,0.96 1937,pepper,n,20220,0.87 1938,separate,v,18251,0.97 1939,busy,j,18500,0.95 1940,intervention,n,21021,0.84 1941,copy,n,18244,0.97 1942,tip,n,18868,0.93 1943,below,i,18353,0.96 1944,cheap,j,18371,0.96 1945,aim,v,18194,0.97 1946,cite,v,19011,0.93 1947,welfare,n,19580,0.9 1948,vegetable,n,19363,0.91 1949,gray,j,19408,0.9 1950,dish,n,18887,0.93 1951,beach,n,18451,0.95 1952,improvement,n,18977,0.92 1953,everywhere,r,18126,0.96 1954,opening,n,18058,0.97 1955,overall,j,19031,0.92 1956,divide,v,18337,0.95 1957,initial,j,19019,0.92 1958,terrible,j,18659,0.93 1959,oppose,v,18484,0.94 1960,contemporary,j,19293,0.9 1961,route,n,18313,0.95 1962,multiple,j,19512,0.89 1963,essential,j,19030,0.91 1964,question,v,17924,0.97 1965,league,n,19766,0.88 1966,criminal,j,18424,0.94 1967,careful,j,17885,0.97 1968,core,n,18571,0.93 1969,upper,j,18196,0.95 1970,rush,v,18561,0.93 1971,necessarily,r,18334,0.94 1972,specifically,r,18873,0.92 1973,tired,j,18597,0.93 1974,rise,n,18055,0.96 1975,tie,n,17787,0.97 1976,employ,v,18912,0.91 1977,holiday,n,18299,0.94 1978,dance,v,18263,0.94 1979,vast,j,17836,0.96 1980,resolution,n,18922,0.91 1981,household,n,18450,0.93 1982,fewer,d,18223,0.94 1983,abortion,n,18925,0.91 1984,apart,r,17872,0.96 1985,witness,n,18920,0.91 1986,match,v,17788,0.97 1987,barely,r,18533,0.93 1988,sector,n,19103,0.9 1989,representative,n,18189,0.94 1990,lack,v,18215,0.94 1991,beneath,i,19047,0.9 1992,beside,i,19952,0.86 1993,black,n,18441,0.93 1994,incident,n,17880,0.96 1995,limited,j,18532,0.92 1996,proud,j,17841,0.96 1997,flow,n,18469,0.92 1998,faculty,n,20104,0.85 1999,increased,j,18984,0.9 2000,waste,n,19069,0.89 2001,merely,r,18094,0.94 2002,mass,n,18107,0.94 2003,emphasize,v,18933,0.9 2004,experiment,n,18440,0.92 2005,definitely,r,18214,0.93 2006,bomb,n,18284,0.93 2007,enormous,j,17482,0.97 2008,tone,n,17954,0.95 2009,liberal,j,18228,0.93 2010,massive,j,17551,0.97 2011,engineer,n,17991,0.94 2012,wheel,n,18296,0.93 2013,female,n,18932,0.9 2014,decline,v,18058,0.94 2015,invest,v,18080,0.94 2016,promise,n,17342,0.98 2017,cable,n,18277,0.93 2018,towards,i,18305,0.93 2019,expose,v,17581,0.96 2020,rural,j,18665,0.91 2021,AIDS,n,19384,0.87 2022,Jew,n,17893,0.95 2023,narrow,j,17840,0.95 2024,cream,n,18443,0.92 2025,secretary,n,17922,0.94 2026,gate,n,17999,0.94 2027,solid,j,17675,0.96 2028,hill,n,18131,0.93 2029,typically,r,18254,0.92 2030,noise,n,18024,0.93", "1 1 4.1 1810 defense 50Germany Wuerttemburg 29 11 1850 15 6 1866 0 0 1 0 1 1 4.1 1840 defense 55Germany Mecklenburg Schwerin 1 1 1843 15 3 1848 0 0 1 0 1 1 4.1 1840 defense 203United Kingdom France 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 defense 207Russia United Kingdom 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 defense 3United Kingdom Sweden 1 1 1816 15 2 1911 1 0 0 0 0 1 4.1 1810 non-defense 45Germany Baden 1 1 1816 15 3 1848 1 0 1 0 1 1 4.1 1810 defense 50Germany Wuerttemburg 29 11 1850 15 6 1866 0 0 1 0 1 1 4.1 1840 defense 55Germany Mecklenburg Schwerin 1 1 1843 15 3 1848 0 0 1 0 1 1 4.1 1840 defense 203United Kingdom France 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 non-defense 207Russia United Kingdom 6 7 1827 14 9 1829 0 0 0 0 0 1 4.1 1820 non-defense In [9]: ggplot(Major, aes(decade)) + geom_bar(aes(fill = Type), position = \"stack\")+ theme(axis.text.x=element_text(angle=90),text = element_text(size=10))+ ggtitle(\"Major Powers Defense and Non-defense Treaty Count; 1810-2010\") To better see treaty trends over time, I created a new column that combines", "# Fourth of July Edition: Print lines of a file containing multiples of a specific number Create a program which takes one command-line argument, n, which will be an integer less than 2147483648 (2^31), and then reads a file input.txt and prints the lines of input.txt which contain any substring which is a positive (non-zero) multiple of n. You may choose to ignore multiples greater than 2147483647. ### Test case If input.txt contains 1. Delaware Dec. 7, 1787 2. Pennsylvania Dec. 12, 1787 1682 3. New Jersey Dec. 18, 1787 1660 4. Georgia Jan. 2, 1788 1733 5. Connecticut Jan. 9, 1788 1634 6. Massachusetts Feb. 6, 1788 1620 7. Maryland Apr. 28, 1788 1634 8. South Carolina May 23, 1788 1670 9. New Hampshire June 21, 1788 1623 10. Virginia June 25, 1788 1607 11. New York July 26, 1788 1614 12. North Carolina Nov. 21, 1789 1660 13. Rhode Island May 29, 1790 1636 14. Vermont Mar. 4, 1791 1724 15. Kentucky June 1, 1792 1774 16. Tennessee June 1, 1796 1769 17. Ohio Mar. 1, 1803 1788 18. Louisiana Apr. 30, 1812 1699 19. Indiana Dec. 11, 1816 1733 20. Mississippi Dec. 10, 1817 1699 21. Illinois Dec. 3, 1818 1720 22. Alabama Dec. 14, 1819 1702 23. Maine Mar. 15, 1820 1624 24. Missouri", "by the new congress, Sept. 15, 1789, provided for an obverse and a reverse, substantially as here depicted; but there is no evidence that the reverse was ever made. In the obverse as originally adopted, the eagle held in his sinister talon a bundle of thirteen arrows, and the first seal was thus made; but when, in 1841, a new seal was made to take the place of the old one, which had become worn, only six arrows were put into the eagle's talon. Whether this change, which was unauthorized by law, was made by design or by accident, is not known.] STATES. Date. Order. Area in squaremiles. Relative size. CAPITALS. .mw-parser-output .wst-rule{background-color:black;color:black;width:auto;margin:2px auto 2px auto;height:1px} Alabama 1819 9 50,722 16 Montgomery. Arkansas 1836 12 52,198 15 Little Rock. California 1850 18 188,981 2 Sacramento. [1]Connecticut 1788 5 4,750 35 Hartford. [1]Delaware 1787 1 2,120 36 Dover. Florida 1845 14 59,268 9 Tallahassee. [1]Georgia 1788 4 58,000 10 Atlanta. Illinois 1818 8 55,410 12 Springfield. Indiana 1816 6 83,809 28 Indianapolis. Iowa 1846 16 55,045 13 Des Moines. Kansas 1861 21 81,318 6 Topeka. Kentucky 1792 2 37,680 25 Frankfort. Louisiana 1812 5 41,346 22 New Orleans. Maine 1820 10 35,000 26 Augusta. [1]Maryland 1788 7 11,124 30 Annapolis. [1]Massachusetts 1788 6 7,800 34 Boston. Michigan 1837 13", "a few random weird punctuation marks that I know will appear later unless I take action now. 1789-Washington.txt [('every', 9), ('government', 8), ('public', 6), ('may', 6), ('citizens', 5), ('present', 5), ('country', 5), ('one', 4), ('ought', 4), ('duty', 4), ('people', 4), ('united', 4), ('since', 4), ('fellow', 3), ('could', 3)] 1793-Washington.txt [('shall', 3), ('oath', 2), ('fellow', 1), ('citizens', 1), ('called', 1), ('upon', 1), ('voice', 1), ('country', 1), ('execute', 1), ('functions', 1), ('chief', 1), ('magistrate', 1), ('occasion', 1), ('proper', 1), ('arrive', 1)] [('people', 20), ('government', 16), ('may', 13), ('nations', 11), ('country', 10), ('nation', 9), ('states', 9), ('foreign', 8), ('constitution', 8), ('honor', 7), ('justice', 6), ('ever', 6), ('congress', 6), ('public', 6), ('good', 6)] 1801-Jefferson.txt [('government', 12), ('us', 10), ('may', 8), ('fellow', 7), ('citizens', 7), ('let', 7), ('shall', 6), ('principle', 6), ('would', 6), ('one', 6), ('man', 6), ('safety', 5), ('good', 5), ('others', 5), ('peace', 5)] 1805-Jefferson.txt [('public', 14), ('citizens', 10), ('may', 10), ('fellow', 8), ('state', 8), ('us', 7), ('among', 7), ('shall', 7), ('constitution', 6), ('time', 6), ('limits', 5), ('reason', 5), ('false', 5), ('duty', 4), ('every', 4)] [('nations', 6), ('public', 6), ('well', 5), ('country', 4), ('peace', 4), ('rights', 4), ('states', 4), ('confidence', 3), ('full', 3), ('improvements', 3), ('united', 3), ('best', 3), ('examples', 2), ('avail', 2), ('made', 2)] [('war', 15), ('country', 5), ('united', 5), ('every', 5), ('british', 5), ('nation', 4), ('without', 4), ('states', 4), ('spirit',", "1858, 5220393), 37 (24, 'Mississippi', 1817, 2938618), 38 (25, 'Missouri', 1821, 5911605), 39 (26, 'Montana', 1889, 967440), 40 (27, 'Nebraska', 1867, 1783432), 41 (28, 'Nevada', 1864, 2600167), 42 (29, 'New Hampshire', 1788, 1315809), 43 (30, 'New Jersey', 1787, 8682661), 44 (31, 'New Mexico', 1912, 1984356), 45 (32, 'New York', 1788, 19490297), 46 (33, 'North Carolina', 1789, 9222414), 47 (34, 'North Dakota', 1889, 641481), 48 (35, 'Ohio', 1803, 11485910), 49 (36, 'Oklahoma', 1907, 3642361), 50 (37, 'Oregon', 1859, 3790060), 51 (38, 'Pennsylvania', 1787, 12448279), 52 (39, 'Rhode Island', 1790, 1050788), 53 (40, 'South Carolina', 1788, 4479800), 54 (41, 'South Dakota', 1889, 804194), 55 (42, 'Tennessee', 1796, 6214888), 56 (43, 'Texas', 1845, 24326974), 57 (44, 'Utah', 1896, 2736424), 58 (45, 'Vermont', 1791, 621270), 59 (46, 'Virginia', 1788, 7769089), 60 (47, 'Washington', 1889, 6549224), 61 (48, 'West Virginia', 1863, 1814468), 62 (49, 'Wisconsin', 1848, 5627967), 63 (50, 'Wyoming', 1890, 532668); ### GROUP BY: Grouping Data The GROUP BY clause groups the resulting data rows into groups. Here is an example: So what just happened? We have 50 rows in the table, but 34 results were returned by this query. This is because the results were grouped by the 'join_year' column. In other words, we only see one row for each distinct value of join_year. Since some states have the same join_year,", "4), ('citizens', 4), ('sense', 3), ('people', 3), ('justice', 3), ('part', 3), ('long', 3)] 1817-Monroe.txt [('government', 22), ('great', 21), ('states', 21), ('people', 15), ('us', 14), ('every', 14), ('united', 13), ('may', 10), ('union', 10), ('war', 10), ('citizens', 9), ('best', 9), ('principles', 9), ('foreign', 9), ('country', 9)] 1821-Monroe.txt [('great', 29), ('states', 20), ('would', 18), ('united', 16), ('war', 16), ('citizens', 15), ('may', 15), ('made', 15), ('government', 13), ('every', 13), ('people', 11), ('commerce', 11), ('force', 11), ('power', 11), ('fellow', 10)] [('union', 20), ('government', 17), ('upon', 16), ('country', 10), ('rights', 10), ('peace', 9), ('great', 9), ('public', 9), ('constitution', 8), ('first', 8), ('general', 8), ('nation', 8), ('people', 7), ('nations', 7), ('duties', 6)] Better, but again, we want to make the output something that can be the object of computation. Let’s do the following: • For each of the early and late texts, get their top 15 words. • Make a list of all of those words • For each of the “was in a top 15” words, calculate the frequency in each of our texts (early and late) • Save this info in a data frame We now have a list of frequent words, but I want to eliminate duplicates (using set()) and alphabetize the list (using sort()). ['abroad', 'america', 'american', 'americans', 'among', 'arrive', 'avail', 'believe', 'best', 'british', 'called', 'cannot', 'century', 'change', 'chief', 'citizens', 'commerce',", "that contains the decade starting with 1951, 1961, 1971 and so on. In [41]: #https://stackoverflow.com/questions/48856982/divide-data-by-decade-then-plot-it-seaborn-box-and-whisker sb['decade'] = (sb['year'] // 10) * 10 + 1 Let's have a look at the data. In [42]: sb.head() Out[42]: station_id year JJA count mean std z_base decade 0 101603550000 1967 2334.00 12.00 2321.42 59.90 0.21 1961 1 101603550000 1968 2360.67 12.00 2321.42 59.90 0.66 1961 2 101603550000 1969 2237.33 12.00 2321.42 59.90 -1.40 1961 3 101603550000 1970 2350.67 12.00 2321.42 59.90 0.49 1961 4 101603550000 1971 2416.00 12.00 2321.42 59.90 1.58 1971 Let's plot the data per decade. In [43]: sb[sb['decade']==1951]['z_base'].hist(bins=200, histtype = 'step', label='1951', density=True) sb[sb['decade']==1961]['z_base'].hist(bins=200, histtype = 'step', label='1961', density=True) sb[sb['decade']==1971]['z_base'].hist(bins=200, histtype = 'step', label='1971', density=True) sb[sb['decade']==1981]['z_base'].hist(bins=200, histtype = 'step', label='1981', density=True) sb[sb['decade']==1991]['z_base'].hist(bins=200, histtype = 'step', label='1991', density=True) sb[sb['decade']==2001]['z_base'].hist(bins=200, histtype = 'step', label='2001', density=True) sb[sb['decade']==2011]['z_base'].hist(bins=200, histtype = 'step', label='2011', density=True).legend(loc='upper right'); That is very close to the results in Hansen et al. (2012). It is important to note that we use the complete station data (Not just the data from the Northern hemisphere as in the two articles.) ### Focus on the Northern hemisphere¶ The Goddard Institute for Space Studies (GISS) also provides more details on the stations here. This is a text file(!) with very problematic formatting. Simply reading the file does not work. We need a fixed-width reader.", "## 2 1793-Washington 0 0 0 0 4 ## 3 1797-Adams 8 0 5 44 2 ## 4 1801-Jefferson 18 0 3 21 2 ## 5 1805-Jefferson 2 3 6 43 3 ## 6 1809-Madison 0 0 4 17 7 ## 7 1813-Madison 0 5 5 20 3 ## 8 1817-Monroe 2 0 27 47 12 ## 9 1821-Monroe 2 0 37 66 10 ## 10 1825-Adams 8 2 13 46 8 ## # … with 49 more rows By adding prop = TRUE, the same table is displayed with row percentages : clusters_by_doc_table(corpus, clust_var = \"cluster\", prop = TRUE) ## # A tibble: 59 × 6 ## doc_id clust_1 clust_2 clust_3 clust_4 clust_5 ## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 1789-Washington 0 0 10.8 73.0 16.2 ## 2 1793-Washington 0 0 0 0 100 ## 3 1797-Adams 13.6 0 8.47 74.6 3.39 ## 4 1801-Jefferson 40.9 0 6.82 47.7 4.55 ## 5 1805-Jefferson 3.51 5.26 10.5 75.4 5.26 ## 6 1809-Madison 0 0 14.3 60.7 25 ## 7 1813-Madison 0 15.2 15.2 60.6 9.09 ## 8 1817-Monroe 2.27 0 30.7 53.4 13.6 ## 9 1821-Monroe 1.74 0 32.2 57.4 8.70 ## 10 1825-Adams 10.4 2.60 16.9 59.7 10.4 ## # … with 49 more rows Conversely, the docs_by_cluster_table allows to display, for each cluster, the", "1851, 1856, 1895 VERMONT 1765, 1771, 1798 VIRGINIA 1623, 1624, 1625, 1634, 1699, 1701, 1703, 1779, 1798 WASHINGTON 1871, 1883, 1885, 1889, 1892 WEST VIRGINIA (Part of Virginia until 1863) WISCONSIN 1830, 1836, 1838, 1842, 1846, 1847, 1855, 1865, 1875, 1885, 1895, 1905 WYOMING 1905, 1915, 1925 Back to the Instructions Page", "States Goveri,ment Printin g Offic . e, .. , . . . Washington . ~ . 19~. . .'• , (1) . Incldes est:!u,ates for six co,int _ ries . i.Il 1920, four in 1921, . anq. t1r0 in 1922. . . . . . I . . . . . . . TABLE 28 . . . .. PER CAPITA op . CIGARS . A.HD dTGARiST'l'ES IN E 'LTR OPEAN . ' . . COUNTR I es COllttPtRkl> IJITH THE UBlTkl> STATF.8 . .. , . . . .. . . i932 . . . . c Po,mds) •. < . . :qn i ted ~ States Au stria Be igi11m . C z e . cho slovak:i .\" a . . .. \\ ' :Ce nmark Fi n1and Fr ance . Ge rman-y; Gr eece (J.) . . . . . H u _ ngB.I'y I t a ly .. : .. N e the rlands ~ o r:way P o land . Rum ani a Sp ain . . s eden . 1: __ ted _ Kingdon .. -r ~ ~;: os la 1 J ~i a -. . . . . . .. . . Ciga~ .89 . . . .. . . . 25 . .40 . . . .16 ... 1.27 .", "651, 6, 84, 47, 1694, 12, 1605, 8, 4947, 6, 17263, 16, 11316, 68, 73, 509, 21122, 26, 2934, 57, 8, 2968, 220, 52, 26, 6788, 5, 3, 2092, 8, 239, 6, 2479, 13, 3, 9, 2659, 208, 6855, 11, 192, 276, 26165, 29, 7527, 1632, 5908, 11, 130, 3, 5385, 26, 57, 276, 26165, 29, 768, 7613, 63, 6, 84, 3, 115, 19054, 8034, 16, 8, 4453, 7, 13, 8, 1511, 5, 17159, 9, 63, 1213, 3, 9, 182, 1101, 1102, 7513, 6, 2049, 12, 8, 4034, 307, 18, 5517, 1472, 13, 8, 4004, 3974, 3013, 7, 68, 112, 2054, 47, 396, 5551, 120, 23038, 12, 1520, 34, 5, 17159, 9, 63, 47, 4792, 16, 8, 1480, 1379, 116, 3, 9, 212, 12661, 13, 8, 4889, 138, 181, 52, 14424, 1074, 3322, 3, 29155, 1084, 376, 117, 8, 3, 35, 15776, 3335, 13, 8, 1511, 57, 8, 276, 26165, 29, 7, 16026, 8, 2379, 18836, 13, 10742, 535, 6, 1525, 8, 826, 822, 5, 2507, 24, 8, 1525, 19, 915, 441, 8, 1499, 5, 11860, 10, 363, 47, 8, 1706, 13, 17159, 9, 63, 31, 7, 10101, 58 ] \" Given the following passage \"The first action of the Franco-Prussian War took place on 4 August 1870. This battle saw the unsupported division of General Douay of", "on this? # ALL CODE IN ONE GO library(data.table) tax_aggregated <- structure(list(Province = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3), Tax = c(2000, 3000, 1500, 3200, 2000, 1500, 4000, 2000, 2000, 1000, 2000, 1500), year = c(2000, 2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2003, 2003, 2003)), row.names = c(NA, -12L), class = c(\"tbl_df\", \"tbl\", \"data.frame\")) Individual_data <- structure(list(ID = c(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), Province = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3), year = c(2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2001, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2002, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003, 2003), Age = c(\"15-30\", \"30-40\",", "shortcoming of the “3, 4, 5″ type of categorization. w. Year Power Dissipation Index 1851.49 , 5.046 1851.51 , 0.729 1851.53 , 0.216 1851.63 , 25.747 1851.70 , 3.456 1851.79 , 4.199 1852.64 , 35.299 1852.68 , 2.688 1852.69 , 9.31 1852.73 , 21.25 1852.77 , 18.622 1853.60 , 0.216 1853.61 , 0.125 1853.67 , 76.601 1853.69 , 19.28 1853.72 , 0.216 1853.74 , 9.742 1853.74 , 0.216 1853.80 , 10.856 1854.49 , 3.837 1854.65 , 0.343 1854.69 , 25.946 1854.72 , 6.268 1854.80 , 3.354 1855.60 , 1 1855.61 , 6 1855.61 , 0.512 1855.65 , 4.29 1855.71 , 10.944 1856.61 , 19.395 1856.62 , 4.096 1856.64 , 2.376 1856.64 , 0.216 1856.65 , 27.805 1856.72 , 7.366 1857.50 , 1.728 1857.69 , 20.936 1857.73 , 10.24 1857.73 , 13.514 1858.45 , 0.512 1858.60 , 0.512 1858.71 , 7.115 1858.71 , 21.096 1858.73 , 7.54 1858.81 , 12.478 1859.50 , 1 1859.63 , 12 1859.67 , 3.072 1859.70 , 5.482 1859.71 , 5.883 1859.76 , 31.8 1859.79 , 2.867 1859.83 , 15.976 1860.61 , 20.362 1860.65 , 7.65 1860.70 , 0.512 1860.70 , 19.853 1860.72 , 4.362 1860.75 , 8.138 1860.80 , 11.928 1861.52 , 13.7 1861.62 , 8.349 1861.65 , 19.664 1861.71 , 0.512 1861.74 , 2.142 1861.77 , 4.98 1861.77 , 0.216 1861.84 , 3.899" ]

[ "### Home > MC1 > Chapter 1 > Lesson 1.2.4 > Problem1-93 1-93. If you walk forward $5$ feet and then walk backward $5$ feet, you will end up exactly where you started. For each of the actions below, describe an action that will get you back where you started. 1. Walk up 10 steps. What is the opposite of 'up'? 1. Earn eight dollars. If you had $\\8$, how would you get rid of it? 1. It gets 5 degrees warmer. What is the opposite of 'warmer'? 1. Lose six dollars. If you lost $\\6$, how could you get it back? 1. Travel south three kilometers. Remember that north is the opposite of south. 1. Run backward nine steps. What is the opposite of 'backward'?", "# Two steps forward and one step back Let's say I'm ten steps away from my destination. I walk there following the old saying, \"Two steps forward and one step back\". I take two steps forward, one back, until I'm standing exactly on my destination. (This might involve stepping past my destination, and returning to it). How many steps did I walk? Of course, I might not be 10 steps away. I might be 11 steps away, or 100. I could measure ten paces, and keep walking back and forth to solve the problem, or... I could write some code! • Write a function to work out how many steps it takes to get N steps away, in the sequence: two steps forward, one step back. • Assume you've started at step 0. Count the \"two steps forward\" as two steps, not one. • Assume all steps are a uniform length. • It should return the number of steps first taken when you reach that space. (For instance, 10 steps away takes 26 steps, but you'd hit it again at step 30). We're interested in the 26. • Use any language you like. • It should accept any positive integer as input. This represents the target step. • Smallest number of bytes win. Example: I want to get", "forward, it's the same as if I had walked -2 steps forward, because: $3 + (-5) = -2$ Subtracting Negative Numbers If I walk 9 steps forward and then 6 steps backward, it's the same as if I had walked 3 steps forward, because: $9 - 6 = 3$ If I walk 9 steps forward and then 6 steps forward, it's the same as if I has walked 15 steps forward. To turn that into a subtraction problem, I have to describe the 6 steps forward as -6 steps backward. That is, if I walk 9 steps forward and then -6 steps backward, it's the same as if I had walked 15 steps forward, because: $9 - (-6) = 15$ Multiplying Negative Numbers If I walk 3 steps forward 5 times, then in total I will have walked 15 steps forward. This corresponds to the multiplication: $5 \\times 3 = 15$ If I walk 3 steps backward 5 times, then in total I will have walked 15 steps backward. Restating this in terms of steps forward, I can say that if I walk -3 steps forward 5 times, then in total I will have walked -15 steps forward, corresponding to the multiplication: $5 \\times -3 = -15$ We could also say, since 3 steps forward is the opposite", "more informal way of seeing it is to suppose that you are standing on the number line at $-2$, and start walking slowly to the right. At the beginning, the sum of your distances from $-2$ and $2$ is $4$. For every tiny step $s$ that you take, your distance from $-2$ increases by $s$. But your distance from $2$ decreases by $s$, so the sum of your distances from $-2$ and $2$ does not change. This continues until you reach $2$, at which point the sum of your distances from $-2$ and $2$ starts to increase. -", "# A prime number random walk This question came to my mind thanks to this question which I found really interesting (and beautiful! Like the mathematician Philippe Caldero said in his book Histoires Hédonistes de Groupes et de Géométries (roughly translated) \"Let us stop for a moment to contemplate the beauty of mathematics, that is after all the point of figures\".). It is also related to this other question. The idea is to perform a walk following the following rules: • Initialisation: You start on the point $(0,0)$ which correspond to the integer $n=0$, and you will walk from one point of $\\mathbb Z^2$ to another. You start by walking on the right. • Each horizontal step you take increases the integer $n$ by $1$. • When $n$ is equal to a prime number, you take one step up, and you change the direction you were going to (if you were walking from left to right you will walk from right to left, and reciprocally). To illustrate the rules, a drawing will perhaps be more explicit: $$\\begin{matrix} & 7 & 8 & 9 & 10\\\\ & 7 & 6 & 5 \\\\ & 3 & 4 & 5 \\\\ & 3 & 2 & \\\\ 0&1&2& & \\end{matrix}$$ Or with blue lines: Well now nothing stops us from", "walkers are exactly one step away and their next step is towards each other. In other words, if they swap positions then there was a collision. Thanks everyone.", "a tiny step $s$, your distance from $1$ increases by $s$, but your distance from each of the other $11$ points decreases by $s$, so the sum of the distances decreases by $11s-s=10s$. This continues until you hit $x=2$. After that, for a while, every tiny step of length $s$ increases your distances from $1$ and $2$ by $s$ each, and decreases your distances from each of the others by $s$, for a total decrease of $8s$. Keep walking. When you reach the next point in the list, which is $5$, each tiny step of length $s$ gives you a net decrease of $6s$. Continue. When you reach $6$ and take a small step, the sum decreases by $4s$. When you reach $8$ an move right, the decrease is $2s$. Finally, when you reach $9$, and take a small step, the distance from each of the $6$ points to your left increases by $s$, but the distance from each point to your right decreases by $s$, so there is no change in the sum. Finally, when you reach $10$ and continue walking to the right, the sum of the distances starts to increase. So the minimum sum is reached at all points between $x=9$ and $x=10$. To find the minimum sum, evaluate at say $x=9$. Remark: We used", "A walk is made up of diagonal steps, starting at the bottom left and ending up back on the bottom line (x-axis). You can move diagonally up and down towards the right but you cannot move towards the left. A diagonal must go from a left-hand corner of a square to the opposite right-hand corner of the same square. The examples above all show 10-step walks. So let's look at a 2-step walk in more detail: There is only one way to make a 2-step walk - from A up to x and down to B: Here are two 4-step walks: Are there any more 4-step walks? How do you know you have them all? How many 6-step, 8-step walks are there? Can you find a general rule?", "the second number (the second addend). Where he lands is the sum of the two numbers. ##### : 3 + 4 If Zed wants to add 3 + 4, he starts at 0 and faces towards the positive numbers. He walks forward 3 steps, then he walks forward 4 more steps. Zed ends at the number 7, so the sum of 3 and 4 is 7. 3 + 4 = 7. (But you knew that of course! The point right now is to make sense of the number line model.) When Zed wants to subtract two numbers, he he walks forward (to the right on the number line) however many steps are indicated by the first number (the minuend). Then he walks backwards (to the left on the number line) the number of steps indicated by the second number (the subtrahend). Where he lands is the difference of the two numbers. ##### : 11 – 3 If Zed wants to subtract 11 – 3, he starts at 0 and faces the positive numbers (the right side of the number line). He walks forward 11 steps on the number line, then he walks backwards 3 steps. Zed ends at the number 8, so the difference of 11 and 3 is 8. 11 – 3 = 8. (But you knew", "and 12 in each field. • Throw 2 dice by turn. Each player chooses whether to add or subtract the numbers. • Place a piece of choice on the number you get. If you do not have the number, the turn moves to the next player. • The first player to get 3 pieces in a row has BINGO. ## Math-orientation • Participants receive a map over a restricted area. • Mark separate posts on the map. • Place tasks on the separate posts. • A wrong answer will be penalized with time, for instance, 10 or 20 seconds. This is a fun game in which the participants get to do math while they are in motion. You can find examples for puzzles here. ## Minute Path • Mark a trail about 200 meters. • The participant’s goal is to walk the trail in 1 minute. • What is the speed in m/s? 200 meters takes about 1 minute to complete at regular walking speed. You can mark the different trails with separate times. You can also walk the trail a second time to try to hit your previous time. The participant who gets closest to 1 minute wins the first round. The participant who gets closest to their previous time wins the second round. Åsta Nybø ·", "required to walk all these distance decreases exponentially. When we walk the first part, we might take 1 second, when we walk the second part, we take 0.5 seconds. When we walk the third part we take 0.25 second. In total, it would take us $$\\sum_{n=0}^{+\\infty}{\\frac{1}{2^n}s}=2s$$ the thing is that this sum converges and thus the number of second to walk the entire distance is finite. So we can complete an infinity of distances in a finite time! However, if the paradox says: \"walk the first distance, then wait for 1s. Walk the second distance, then wait for 1s. Walk the third distance, then wait for 1s...\" Then this is indeed not possible to perform, as it would take an infinite amount of time to do so. 5. Jun 17, 2011 ### ghostman97 so what your concluding is that movement would be \"impossible\" 6. Jun 17, 2011 ### Staff: Mentor 7. Jun 18, 2011 ### Staff: Mentor Have you actually read micromass answer before responding? He clearly stated: \"we can\". Thats opposite of \"impossible\".", "the \"walker\" is simply the instruction pointer (which has a position and a direction), the @ command terminates the program and < and > are source-code modification commands. • which of is required ? 1 , 2, or 3 and how are our submissions scored – Abr001am Apr 27 '16 at 19:46 • @Agawa001 \"You can do that one of three ways:\" Choose either one of them, whichever you think is easiest for the approach and language you want to use. – Martin Ender Apr 27 '16 at 19:47 • Why all the repeating numbers today!?! codegolf.stackexchange.com/questions/78787/… :D – cat Apr 28 '16 at 17:46 # Jelly, 10 bytes ð+\\ḤiḤoµL‘ This function accepts a single integer in form of the list of its binary digits as input. The algorithm is equivalent to the one from @Agawa001's answer. ### Background Enumerate the positions underneath the path from 0 to L, giving a total of L + 1 positions. L coincides the number of binary digits of the number N that encodes the path. With this notation, the walker starts at position 0, the goal at position L. With each step the walker takes, he gets one step closer to the goal (in the direction he's currently walking). Also, with each shift-step, depending on whether he walks with or against", "# A King's Short Walk Place the numbers 1 to 25 on the cells of this board so that any two consecutive numbers occupy cells that are horizontally, vertically or diagonally adjacent. Prime numbers should occupy shaded cells. • I don't get it. A bishop's walk could only visit half the cells. Shouldn't it be a King's walk? Sep 21 at 7:03 • There are 8568 solutions (counting rotations and flips). These could be grouped into 192 permutations for the prime cells. The number 23 occurs in the middle ~72% of the time. Sep 21 at 12:53 • @EricDuminil Right! Fixed! Sep 21 at 13:15 Like so? The beginning seems almost forced (apart from symmetries), and the rest just fell in place on the first attempt. • Almost but not quite (forced, that is): 25 3 4 5 6 // 2 24 22 21 7 // 14 1 23 8 20 // 13 15 16 9 19 // 12 11 10 17 18 Sep 20 at 21:19 As there are quite a few possible solutions and seeing that it is \"A Bishop's Walk\" here is one that tries to use as many bishopy (aka diagonal) moves as possible: 1 3 25 23 - 22 | / | | / / 2 4 24 21 19 \\ X |", "possible solutions to a problem. Fundamentally, backtracking is a variant of path finding in a more abstract space: the space of states and actions. Consider the game as a graph. Each “game state” is a vertex, and each “action” you take is an edge. There are many possible actions. In this instance, we define an action as picking a not yet considered square and choosing to leave it empty or to fill in a number in range 1 through 7. Let’s consider the graph you could make if you start with an empty grid. With an empty grid, there are 49 squares where you can do an action, with 8 possible actions per square (empty or 1-7). After we’ve placed the first number, there are 48 squares left with 8 actions per square, et cetera. This implies that we end up with $$(49\\times8)\\times(48\\times8)\\times \\dots \\times (2\\times 8) \\times (1 \\times 8) = 49! \\times 8^{49} \\approx 1.08 \\times 10^{107}$$ final game states. How big is this number? Imagine a modern desktop computer for every atom in the visible universe. Even if you cluster them into one big supercomputer, brute force computation would still take slightly longer than the age of the universe. That’s where backtracking comes in. The key to backtracking is to cut off a branches of", "# A Strange Sequence [duplicate] This puzzle is taken from a French bestseller I read years ago. The solution is much simpler than it looks... Find the following line in this sequence : 1 1 1 2 1 1 2 1 1 1 1 1 2 2 1 3 1 2 2 1 1 • An old question, but here's a variant: what starting sequence can go the most steps before reaching a length of ten? – supercat Nov 7 '14 at 14:23 Say it out loud, and count the numbers! So the steps would be: One 1; two 1's; one 2 one 1; and so one. The next would be 1 3 1 1 2 2 2 1.", "etc., or are they referring to any other stride information? The stride referred to by the quote \"only plays with the size and ... • 24.7k Yes, you're right, after the green one, it should also move two steps (because stride = 2) to the right once more. Note that in the $3 \\times 3$ output volume picture, there's also still a white cell ...", "# This 5 $\\times$ 5 puzzle Here's the goal: You should walk on every single square, starting with purple. Rules: You can't step on a square twice. You don't have to return to the purple square. You can't walk diagonally ## 1 Answer It can't be done, tile the pattern as a checkerboard, each move must take us from a White square to a Black square, or a Black square to a White square. There are 13 Black squares and 12 white squares, so the possibility of a path only exists if we start on Black. The Purple square would otherwise be White, so it's not a starting point a path can come from.", "### Home > ACC6 > Chapter 3 Unit 3 > Lesson CC1: 3.1.6 > Problem3-87 3-87. If you walk forward $5$ feet and then walk backward $5$ feet, you will end up exactly where you started. For each of the actions below, describe an action that will get you back where you started. 1. Walk up $10$ steps. Walking down steps is the opposite of walking up steps. 2. Earn $8$ dollars. To undo earning money, you need to spend money. 3. It gets $5$ degrees warmer. It gets $5$ degrees colder. 4. Lose $6$ dollars. If you lose money, you need to find or earn the same amount to get back where you started. 5. Travel south $3$ kilometers. Travel north $3$ kilometers. 6. Run backward $9$ steps. Run forward $9$ steps.", "can combine the two consecutive orange steps 3 and 4 to save a step. Sep 18 '21 at 16:52 • This is 12 moves, not 13, right? Sep 19 '21 at 18:23 • @OldBunny2800 13, originally thought it was 14 for the same reason. Sep 19 '21 at 18:54 • I count 12 - moves 3 and 4 are actually just one move. Sep 19 '21 at 19:02 • Ohhhh good eye lol 😂 Sep 19 '21 at 19:13", "up exactly? How do you turn yourself back around and make progress in that situation? – JS1 Jan 29 at 2:10 • If you move forward from A to B, and B is a reversing segment, you remain in B but going forward again will lead to A. You could move backward while in B and go to the next segment instead. – Woofmao Jan 29 at 2:25 • So you know where each segment's boundaries are? That is, it's a sequence of discrete steps? – Deusovi Jan 29 at 3:24 • Yes, you move through the labyrinth in discrete steps. – Woofmao Jan 29 at 3:49 • I gather that there is no way to distinguish between segments, apart from the initial one that you will recognize if you ever get back to it, right? – Arnaud Mortier Jan 29 at 8:25 The following sequence of steps should bring the thieve through the corridor (F=forward, B=backward): FFBFFFBFFFBFFFBFFFBFFFBFFFBF Note the periodic pattern: initial FFB followed by (possibly truncated) FFFB sequences. These patterns can also be used for corridors of other lengths. I found this solutions by using a program to find solutions for shorter corridors, noting the pattern, and trying this one. These are 28 steps, slightly shorter than the prescribed upper bound. • I verified by computer" ]

[ "= 62; 62 + 8 = 70; 70 + 7 = 77; 77 + 5 = 82; 82 + 1 = 83. Successive records: 1 takes 1 step to create 2 2 takes 4 steps to create 23 19 takes 5 steps to create 41 43 takes 9 steps to create 83 113 takes 12 steps to create 167 479 takes 15 steps to create 547 953 takes 20 steps to create 1049 1303 takes 22 steps to create 1399 [...] 1057181 takes 89 steps to create 1057579 Is this pattern real or have I misunderstood something? If it is real, is there a simple explanation for it? - What pattern?${}$ – gammatester Feb 13 '14 at 11:00 @gammatester The pattern is that each starting point for a record number of steps is itself prime. – Mark S. Feb 23 '14 at 23:21 Yes, this pattern has a simple explanation. On an intuitive level, if a record started from a non-prime $n$, then you might be able to go back a step to a number that would generate $n$ on the next step, so that the record-breaker would actually be less than $n$. (This doesn't work for primes because a \"previous number\" for a prime would have at most one step until it reached a prime, and wouldn't", "is $31$ by appending $1$ to the right of $3$. We can also add another $1$ to the right of $31$ to get a new prime, $311$. Now the question is whether we can continue to add digits one at a time to get a sequence of primes. If not, what is the longest prime walk we can get starting at $3$? How about starting at another prime instead of $3$? Can we write some programs to simulate this process? In this project, we will try to devise an algorithm to compute the longest prime walk starting at a certain prime $x$ and compare the cases with different $x$. If this can be achieved, a further question is how can we add two, three, or a bounded number of digits at a time instead of only one? Will that give us a walk to infinity? This project is based on my research in 2020 Polymath REU [1]. Possible extension We can also investigate the walk along squares or other interesting sequences instead of primes. Are we able to walk to infinity along the squares? Can we modify our codes to simulate the squares walk? Outline/timeline Week 1: The first week will be all about introducing the project-related concepts, including primes, sequences, and some significant results in number theory,", "2. Place numbers 1 through 64 in the cells such that the next number is either directly left, right, below, or above the current number. 3. The dark cells must be occupied by prime numbers. Note there are 18 dark cells, and 18 primes in the sequence $$1,\\dots,64$$. These prime numbers are: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61. This problem looks very much like the Numbrix puzzle [3]. We don't have the given cells that are provided in Numbrix, but instead, we have some cells that need to be covered by prime numbers. The path formed by the consecutive values looks like how a rook moves on a chessboard. Which explains the title of this post (and of the original post). If we require a proper tour we need values 1 and 64 to be neighbors. Otherwise, if this is not required, I think a better term is: Hamiltonian path. I'll discuss both cases, starting with the Hamiltonian path problem. ### A covering problem: runners vs Ragnar relay legs In [1] the following problem was posted. 1. We have a number of legs with their lengths (in miles). E.g. legs = [3.3, 4.2, 5.2, 3, 2.7, 4, 5.3, 4.5, 3, 5.8, 3.3, 4.9, 3.1, 3.2, 4, 3.5, 4.9, 2.3, 3.2, 4.6, 4.5, 4, 5.3, 5.9, 2.8, 1.9, 2.1, 3, 2.5, 5.6, 1.3, 4.6,", "132$. Now the question is whether we can continue to add digits one at a time to get a sequence of squares. If not, what is the longest square walk we can get starting at $1$? How about starting at another square instead of $1$? Can we write some programs to simulate this process? In this project, we will try to devise an algorithm to compute the longest square walk starting at a given square $x^2$ and compare the cases with different $x$. If this can be achieved, a further question is how we can add two, three, or a bounded number of digits at a time instead of only one. Will that give us a walk to infinity? This project is based on my research in 2020 Polymath REU [1]. Possible extension We can also investigate the walk along squares or other interesting sequences instead of primes. Are we able to walk to infinity along the squares? Can we modify our codes to simulate the squares walk? Outline/timeline Week 1: The first week will be all about introducing the project-related concepts, including primes, sequences, and some significant results in number theory, such as Dirichlet's Theorem. In addition to that, some important developments in the paper [1] will be presented. After that, we will set up the coding", "all 3 numbers are different. This means that the largest number cannot be prime - ruling out 1,2,3,5,7, and leaving us with 4, 6, 8, and 9. There are 32 possible pairs of factors of the numbers 4, 6, 8, and 9, but only 8 of these ... 5 18: 1 I thought about something here : There is another answer I thought of first, and I wanted to share. And last but not least, the ultimate move : 6 This grid requires 18 steps: Explanation 3 Solution: Finding the Solution: This gets us: We now have: This gives us: Now, we have: Here we have: In conclusion: 3 Intuitively, it seems like 3 What combinations of numbers are possible? From this list of possibilities, we need to find one where one person can know what their number is by seeing two of the other numbers. Here are the possible numbers a person could see others having (lowest number first), and the possible numbers that the person could be. Now we have 3 possibilities for two ... 6 It is not possible to do this in 12 steps. In fact, a simple argument shows that at least 20 steps are necessary: Tiles 1 and 8 need to be swapped. They are however a distance of 3 moves apart,", "= 17+7 = 11+13 = 24 Day 25 The 25th day of the year; 25 is the smallest square that can be written as a sum of 2 squares. (What's the next?) 1, 25, 49 is the smallest arithmetic progression of three squares that I have ever found. 4, 100, and 196 come next . Is there one starting with nine? It is proven that an arithmetic progression of four squares in not possible. There are 1958 partitions of 25. The Goldbach representations of 25 using distinct primes are: 2 + 23 = 3 + 5 + 17 = 5 + 7 + 13 = 25. 25 is a palindrome in base 4 (121) There are 25 primes less than 100. $$25 = 5^2$$ and $$2.5 = \\frac{5}{2}$$ *Math Year-Round ‏@MathYearRound (are there other examples like this?) There are exactly 25 square numbers less than or equal to 1,000,000 that are the sum of twin prime pairs. *prime curios Everyone knows 25 is the hypotenuse of a Pythagorean right triangle with legs of 7 and 24. A Pythagorean triangle can never have two sides that are squares. When a square occurs as the shorter leg, and interesting pattern occurs: leg leg hypotenuse 9 40 41 25 312 313 49 1200 1201 81 3280 3281 Day 26 The 26th", "will start at 3*P or C. 3*1P, 3*3C, 3*5C, 3*7C, and every 3*odd number, in sequence from there. 5*1P, 5*3C, *5C, 5*7C, and again, every odd number in sequence from there. Take any odd number(prime or composite) and you will get a number string with the same sequence; 1*P or C, 3*P or C, 5*P or C, and up through the odd numbers in sequence from there. On a \"sieve\" with the Top line of 50 (odd numbers 1 through 49 inclusive),and a line of 50 below it (51 through it), you should see for 3P, and every 3C on the top line add 50 (that will make 53 on the next line down). From 53 move 1 number back to51(a composite of 3). This is the pattern of the composites formed by 3P. If you use a top line of 100 (1 through 99, inclusive) the pattern for the composites formed by three will move forward 1 number from 3, or composite of three, (3+100= 103 plus 1 odd number forward) to 105. For 5 and its composites the pattern will be straight down the columns if you use a top line in multiples of 50 (50, 100, 150, etc.). The composites formed by any prime or composite will form a pattern that remains the same up", "25$ other numbers. These $25$ numbers must be prime, since ${11}^{2} = 121 > 100$ So the total number of composite numbers is: $49 + 16 + 6 + 3 = 74$ May 27, 2017 There are $73$ composite numbers less than $100$ #### Explanation: There are $100$ numbers from $1 \\text{ to } 100$ However, the question specifies numbers BETWEEN $1 \\mathmr{and} 100$, so these two numbers are eliminated from the total. So we are left with $98$ numbers to work with. All the numbers between $1 \\mathmr{and} 100$ are either prime or composite. (The only number that is neither is $1$ which has already been excluded) There are $25$ prime numbers less than $100$. This is pretty easy to check by just counting them, but it is a small fact that is worth knowing. Therefore, if $25$ of the $98$ numbers are prime, it means that all the rest are composite: $98 - 25 = 73$ composite numbers.", "nice Wimbledton side task :) Is it 9 steps in worst case scenario? 512 = 2^9, i.e. log 2( 512) = 9 4. Dec 8, 2015 ### BvU Nine rounds is right. How would you calculate the total number of matches that it takes ? (this doesn't help with your original post, but it's a nice thingy to carry with you for later on). Back to the BS (): If it takes nine steps for 29 entries, how many steps for 232 ? How much is 232 ? The math is really simple, actually! And the number of steps is the 2 logarithm of the number of entries. Piece of cake! 5. Dec 8, 2015 ### Staff: Mentor In other words, the number of steps = $\\log_2(\\text{entries})$ Here $\\log_2(4) = 2, \\log_2(8) = 3$, and so on.", "new value of x=151 here and so on. This quite clearly illustrates the iteration is correct (numerically). So for these particular choices of $y$ used in the numerical example above , we have the sequence 3, 11,29,61,151 and so on. Informally the question is asking if can we build an infinitely long \"bridge\" across the set of natural numbers where the difference between two consecutive \"steps\" (i.e. terms of the iteration) across the bridge is always a prime number+1. In other words the difference is never a composite number+1. Or describing the question in another way informally it asks if we can walk forever along the natural number line by stepping on primes only and taking steps which are very nearly of prime length. Another question is what can be said about an upperbound minimal \"step\" length as you walk further along the \"bridge\"? And considering the analogy to Gaussian primes- can we walk to some direction further and further away from the centre of the complex plane by stepping on Gaussian primes (considering both if the steps are bounded above or not) such that the step value is nearly a Gaussian prime?-we replace the constant 1 by some fixed Gaussian prime to get the version of the iteration for Gaussian primes,and again numerical examples suggest this is", "many different paths to attempt and I am unsure of where to go from here. - This is intriguing to say the least. I'm not aware of a simple way to do this without writing a program to do it. I'd love to see others' attempts. – Cameron Williams Mar 24 '14 at 1:10 1 step) a. across should be a prime number, which digit sum (c. across) is a 1 digit number. 2 step) c. across should be an odd number, because b. down is a prime number and can't be divisible by 2. 3 step) from step 2 -> a. across should be a prime number, which digit sum (c. across) is odd number. 4 step) from 1 and 3 steps -> a. across first digit should be even, because second digit is always odd for their sum to be odd. 5 step) from 1 and 4 steps -> we get two numbers which satisfy the rules, it is 43 or 61. So in b. down we get 37 or 17. 6 step) a. down is a square of the sum of the digits of b. down, so from step 5 -> we get a. down is 100 or 64. 7 step) 100 doesn't fit to our rules, 64 does. 6 1 4 7 - I", "Upsetting Pitagoras Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2 Rudolff's Problem A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group? Euler's Squares Euler found four whole numbers such that the sum of any two of the numbers is a perfect square... Never Prime Age 14 to 16Challenge Level Take any two digit number, reverse its digits, and subtract the smaller number from the larger. For example $$42-24=18$$ I've tried this a few times and I never seem to end up with a prime number. Try some examples of your own. Do you ever end up with a prime number? Can you prove that you will never end up with a prime? What happens when I do the same with a three digit number? What about a four digit number? What about a five, six, seven, ... $n$ digit number?", "# Prime parity peregrination The purpose of this challenge is to graphically depict a walk on the plane, where the direction of each step $$\\k\\$$ is determined by the primality of $$\\k\\$$ and the parity of its binary expansion. Specifically, • Initial direction is fixed, say North. • All steps have the same length. • The direction of step $$\\k\\$$ can be North, West, South or East, and is determined as follows: • If $$\\k\\$$ is not prime, the direction does not change. • If $$\\k\\$$ is prime and the binary expansion of $$\\k\\$$ has an even number of ones, turn right. • If $$\\k\\$$ is prime and the binary expansion of $$\\k\\$$ has an odd number of ones, turn left. As a worked example, assume that the initial direction is North. The first steps are: • $$\\k=1\\$$ is not prime. So we move one step in the current direction, which is North. • $$\\k=2\\$$ is prime, and its binary expansion, 10, has and odd number of ones. So we turn left, and are now facing West. We move one step in that direction. • $$\\k=3\\$$ is prime, and its binary expansion, 11, has and even number of ones. So we turn right, and are now facing North. We move one step in that direction. • $$\\k=4\\$$ is", "# Doubling/tripling puzzle: make 1 from 1536 in as few steps as possible You start with the number 1536. Your mission is to get to 1 in as few steps as possible. At each step, you may either multiply or divide the number you have, by either 2 or 3; but, only if the result is a whole number whose first digit is 1, 3, 4, or 9. That is all. • \"first digit\" meaning the ones digit, or \"first digit\" meaning most significant digit? Dec 12 '18 at 2:46 • @Hugh - most significant. Dec 12 '18 at 2:51 • I wonder if the fact that the prime factorization of 1536 is $2^9 \\times 3$. That puts a lower bound of at least 10 operations, but it must be more since just trying possibilities shows that there must be some multiplications in there. Dec 12 '18 at 3:03 As Jo has already shown, this can be accomplished in 28 steps. This is minimal, and it can be proven. To help visualize this problem, we can imagine: A two-dimensional grid/chart where each point is a number of the form $$3^x2^y$$, with $$(x,y)$$ as the relevant co-ordinates. We want to find a path from $$(1,9)$$ to $$(0,0)$$ while making only one step up/down/left/right at a time, and ensuring that", "there are 18 dark cells, and 18 primes in the sequence $$1,\\dots,64$$. These prime numbers are: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61. This problem looks very much like the Numbrix puzzle [3]. We don't have the given cells that are provided in Numbrix, but instead, we have some cells that need to be covered by prime numbers. The path formed by the consecutive values looks like how a rook moves on a chessboard. Which explains the title of this post (and of the original post). If we require a proper tour we need values 1 and 64 to be neighbors. Otherwise, if this is not required, I think a better term is: Hamiltonian path. I'll discuss both cases, starting with the Hamiltonian path problem. ### A covering problem: runners vs Ragnar relay legs In [1] the following problem was posted. 1. We have a number of legs with their lengths (in miles). E.g. legs = [3.3, 4.2, 5.2, 3, 2.7, 4, 5.3, 4.5, 3, 5.8, 3.3, 4.9, 3.1, 3.2, 4, 3.5, 4.9, 2.3, 3.2, 4.6, 4.5, 4, 5.3, 5.9, 2.8, 1.9, 2.1, 3, 2.5, 5.6, 1.3, 4.6, 1.5, 1.2, 4.1, 8.1] 2. We also have a number of runs performed by runners. Again, we have distances in miles: runs = [3.2, 12.3, 5.2, 2.9, 2.9, 5.5] A run can be allocated to different", "# Consecutive composite numbers — an example for RMO training Find 20 consecutive composite numbers. Consider the following Theorem: There are arbitrarily large gaps in the series of primes. Stated otherwise, given any positive integer k, there exist k consecutive composite integers. Proof: Consider the integers $(k+1)!+2$ $(k+1)! +3$, and so on, so forth $(k+1)! + k$ $(k+1)!+k+1$ Every one of these is composite because j divides $(k+1)!+j$, if $2 \\leq j \\leq k+1$. Hence, for example, 20 consecutive composite numbers are as follows: $20! +2$, $20! +3$, $20!+4$, $\\ldots$, $20! + 21$. More later, Nalin Pithwa", "Question # Find $8 - 3$; $6 - 2$; $9 - 6$ using the number line. Hint: In this problem, first we will draw the number line. Then, we will mark the positions of given numbers on the same number line. Then, we will count how many steps are needed from the first number to reach the position of second number. The number of steps is the required answer. Let us draw the number line and mark the positions of given numbers. To find $8 - 3$, we will start from the number $8$ and we will move backward to reach the number $3$. From the number line, we can say that $5$ steps are needed from the number $8$ to reach the position of number $3$. Therefore, $8 - 3 = 5$. Now to find $6 - 2$, we will start from the number $6$ and we will move backward to reach the number $2$. From the number line, we can say that $4$steps are needed from the number $6$to reach the position of $2$. Therefore, $6 - 2 = 4$. Now to find $9 - 6$, we will start from the number $9$ and we will move backward to reach the number $6$. From the number line, we can say that $3$ steps are needed from", "to a couple of hundred digits with almost any software, it might not seem like much of a tool....but sitting around drinking beers at the lake, and someone asks, \"by the way, what is the repetend of 1/23?\", you can whip out your pencil and do the first 11 digits and then coast the rest of the way adding nines-compliments. I feel like I should have picked that one up somewhere in the seventh grade, but stumbled across it a few days ago reading an old book on Primes by the late George Loweke, who taught at Wayne State Univ. before retiring a few miles up the coast of Lake Michigan from me. Thanks George, but you should have mentioned this one to me sooner.", "5 steps away: | | | | | | <- I'm at step 0, not yet on the grid. | |X| | | | <- I take two steps forward, I'm on step 2: the count is 2 |X| | | | | <- I take one step back, I'm on step 1: the count is 3 | | |X| | | <- I take two steps forward, I'm on step 3: the count is 5 | |X| | | | <- I take one step back, I'm on step 2 again: the count is 6 | | | |X| | <- I take two steps forward, I'm on step 4: the count is 8 | | |X| | | <- I take one step back, I'm on step 3 again: the count is 9 | | | | |X| <- I take two steps forward, I'm on step 5: the count is 11 In this case, the result of the function would be 11. Example results: 1 => 3 5 => 11 9 => 23 10 => 26 11 => 29 100 => 296 1000 => 2996 10000 => 29996 100000 => 299996 Have fun, golfers! • Hmm ... this feels very familiar. – Shaggy Mar 26 '18 at 13:14 • Related – Rod Mar 26 '18", "one. If one player matches all six numbers of Tuesday night’s record-setting draw, he or she will have to decide between a lump sum of nearly $905 million or receiving$1. problem, remember to put back all of the semicolons you removed (and take out any extra display variables) so that when you run the program on the bigger problem, you won’t be flooded with a bunch of unnecessary data. Confrontation with problems that easily lets you perfectly change your life. This is an infinite loop. Logic to find all perfect numbers in a given range in C programming. Americans account for less than 5 percent of the world’s population but consume 80 percent of all opioids produced globally. The sum of all prime between numbers 1 and 400 (2 through 399) is 13,887. = 1/109 This is directly related to. Problem Solution 1. Or simply learn about pi here. 6 million cash lump sum. If 7, the program returns 12=(2+4+6). Answer : 7,7. The Million-Dot Poster. Please refer to the examples that follow. Question 1. Mega Millions Prizes. From part (iv), we have that there are $$10$$ possibilities in the case that the hundreds digit is at least $$5$$. If you take the lump-sum cash option, you’ll get a one-time, lump-sum payment. The sum of all the prime" ]

[ "forward, it's the same as if I had walked -2 steps forward, because: $3 + (-5) = -2$ Subtracting Negative Numbers If I walk 9 steps forward and then 6 steps backward, it's the same as if I had walked 3 steps forward, because: $9 - 6 = 3$ If I walk 9 steps forward and then 6 steps forward, it's the same as if I has walked 15 steps forward. To turn that into a subtraction problem, I have to describe the 6 steps forward as -6 steps backward. That is, if I walk 9 steps forward and then -6 steps backward, it's the same as if I had walked 15 steps forward, because: $9 - (-6) = 15$ Multiplying Negative Numbers If I walk 3 steps forward 5 times, then in total I will have walked 15 steps forward. This corresponds to the multiplication: $5 \\times 3 = 15$ If I walk 3 steps backward 5 times, then in total I will have walked 15 steps backward. Restating this in terms of steps forward, I can say that if I walk -3 steps forward 5 times, then in total I will have walked -15 steps forward, corresponding to the multiplication: $5 \\times -3 = -15$ We could also say, since 3 steps forward is the opposite", "# A prime number random walk This question came to my mind thanks to this question which I found really interesting (and beautiful! Like the mathematician Philippe Caldero said in his book Histoires Hédonistes de Groupes et de Géométries (roughly translated) \"Let us stop for a moment to contemplate the beauty of mathematics, that is after all the point of figures\".). It is also related to this other question. The idea is to perform a walk following the following rules: • Initialisation: You start on the point $(0,0)$ which correspond to the integer $n=0$, and you will walk from one point of $\\mathbb Z^2$ to another. You start by walking on the right. • Each horizontal step you take increases the integer $n$ by $1$. • When $n$ is equal to a prime number, you take one step up, and you change the direction you were going to (if you were walking from left to right you will walk from right to left, and reciprocally). To illustrate the rules, a drawing will perhaps be more explicit: $$\\begin{matrix} & 7 & 8 & 9 & 10\\\\ & 7 & 6 & 5 \\\\ & 3 & 4 & 5 \\\\ & 3 & 2 & \\\\ 0&1&2& & \\end{matrix}$$ Or with blue lines: Well now nothing stops us from", "# Two steps forward and one step back Let's say I'm ten steps away from my destination. I walk there following the old saying, \"Two steps forward and one step back\". I take two steps forward, one back, until I'm standing exactly on my destination. (This might involve stepping past my destination, and returning to it). How many steps did I walk? Of course, I might not be 10 steps away. I might be 11 steps away, or 100. I could measure ten paces, and keep walking back and forth to solve the problem, or... I could write some code! • Write a function to work out how many steps it takes to get N steps away, in the sequence: two steps forward, one step back. • Assume you've started at step 0. Count the \"two steps forward\" as two steps, not one. • Assume all steps are a uniform length. • It should return the number of steps first taken when you reach that space. (For instance, 10 steps away takes 26 steps, but you'd hit it again at step 30). We're interested in the 26. • Use any language you like. • It should accept any positive integer as input. This represents the target step. • Smallest number of bytes win. Example: I want to get", "in fact we're actually always just moving $2$ steps forward in the cyclic sequence $0, 1, 2, 3, 4, 5, 6, 8, 9$. And this applies in general, for $k/(n-1^2)$, we're moving $k$ steps forward in the sequence $[0, 1, 2, 3, 4, 5, 6, \\cdots, n-1]$ with $(n - 1 - k)$ removed. This \"trick\" even works with numbers that aren't relatively prime to $n - 1$: $$3 / 81 = 0.037037037037037... = \\text{3 steps forward in } [0, 1, 2, 3, 4, 5, 7, 8, 9]$$ $$9 / 81 = 0.111111111111111... = \\text{9 steps forward in } [1, 2, 3, 4, 5, 6, 7, 8, 9]$$ In base 7: $$2_7 / 51_7 = 0.025025025025025..._7 = \\text{2 steps forward in } [0, 1, 2, 3, 5, 6]$$ $$3_7 / 51_7 = 0.040404040404040..._7 = \\text{3 steps forward in } [0, 1, 2, 4, 5, 6]$$ Keep in mind, though, that as soon as you get through the first set of $n - 1$, you go back to one step but everything in the original number sequence shifts to the right by one: $$9 / 81 = 0.111111111111111... = \\text{9 steps forward in } [1, 2, 3, 4, 5, 6, 7, 8, 9]$$ $$10 / 81 = 0.123456790123456... = \\text{1 step forward in } [1, 2, 3, 4, 5,", "is $31$ by appending $1$ to the right of $3$. We can also add another $1$ to the right of $31$ to get a new prime, $311$. Now the question is whether we can continue to add digits one at a time to get a sequence of primes. If not, what is the longest prime walk we can get starting at $3$? How about starting at another prime instead of $3$? Can we write some programs to simulate this process? In this project, we will try to devise an algorithm to compute the longest prime walk starting at a certain prime $x$ and compare the cases with different $x$. If this can be achieved, a further question is how can we add two, three, or a bounded number of digits at a time instead of only one? Will that give us a walk to infinity? This project is based on my research in 2020 Polymath REU [1]. Possible extension We can also investigate the walk along squares or other interesting sequences instead of primes. Are we able to walk to infinity along the squares? Can we modify our codes to simulate the squares walk? Outline/timeline Week 1: The first week will be all about introducing the project-related concepts, including primes, sequences, and some significant results in number theory,", "# A random walk on a finite square with prime numbers This question is following two similar questions that you can find here and here. The idea is to walk on a square of length $$n\\times n$$, following some rules. We will identify the opposite sides. Formally, the square with the opposite sides identified can be modelled by $$(\\mathbb Z/n\\mathbb Z)^2$$. We will walk following the following rules: • We start on the bottom left case, facing north. This correspond to the number $$k=1$$. • Each time we take a step, we increase $$k$$ by $$1$$, and if $$k$$ happens to be a prime number, we turn $$90$$ degrees right. We wall $$p(n)$$ the smallest number of steps we need to walk on every case of the square. We will draw the walk for $$n=3$$ to illustrate the rules, and to be convince that $$p(3)=9$$: I am interested in the sequence $$p$$. What I think is true is the following result. Conjecture. We have $$p(n)<\\infty \\iff (n=2 \\text{ or }n\\text{ is odd})$$. I have computed the first values of $$p(n)$$ using SageMath: $$\\begin{matrix} n &1 &2 &3 &4 &5 &6 &7 &8 &9 &10\\\\ p(n) &1 &4 &9 &? &90 &? &256 &? &364 &? \\end{matrix}$$ I convinced myself that $$p(n)=\\infty$$ if $$n$$ was an even number greater", "132$. Now the question is whether we can continue to add digits one at a time to get a sequence of squares. If not, what is the longest square walk we can get starting at $1$? How about starting at another square instead of $1$? Can we write some programs to simulate this process? In this project, we will try to devise an algorithm to compute the longest square walk starting at a given square $x^2$ and compare the cases with different $x$. If this can be achieved, a further question is how we can add two, three, or a bounded number of digits at a time instead of only one. Will that give us a walk to infinity? This project is based on my research in 2020 Polymath REU [1]. Possible extension We can also investigate the walk along squares or other interesting sequences instead of primes. Are we able to walk to infinity along the squares? Can we modify our codes to simulate the squares walk? Outline/timeline Week 1: The first week will be all about introducing the project-related concepts, including primes, sequences, and some significant results in number theory, such as Dirichlet's Theorem. In addition to that, some important developments in the paper [1] will be presented. After that, we will set up the coding", "q= 1/2 and so will involve $(1/2)^{20}$. Let n be the number of forward steps he took. Then 20- n is the number of backward steps. Assuming he started at the door, if he is 14 steps back into the pub, taking forward steps positive and backward steps negative, we must have n- (20- n)= 2n- 20= -14. Solve that for n. Now that you know the number of forward and backward steps, you can use the binomial coefficient to determine in how many orders he could have taken those forward and backward steps. (By the way, I find this problem, and any problem postulating people not in control of themselves, offensive.) 5. Apr 19, 2011 ### eczeno I agree with halls, why do we need such unpleasant topics for math problems? 6. Apr 19, 2011 ### the riddick25 Thank you all very much for your help and i agree with what you said about the question, like eczeno said this could easily have been put in terms of a coin flipping, instead it is about a drunk person who cant walk properly :S", "new value of x=151 here and so on. This quite clearly illustrates the iteration is correct (numerically). So for these particular choices of $y$ used in the numerical example above , we have the sequence 3, 11,29,61,151 and so on. Informally the question is asking if can we build an infinitely long \"bridge\" across the set of natural numbers where the difference between two consecutive \"steps\" (i.e. terms of the iteration) across the bridge is always a prime number+1. In other words the difference is never a composite number+1. Or describing the question in another way informally it asks if we can walk forever along the natural number line by stepping on primes only and taking steps which are very nearly of prime length. Another question is what can be said about an upperbound minimal \"step\" length as you walk further along the \"bridge\"? And considering the analogy to Gaussian primes- can we walk to some direction further and further away from the centre of the complex plane by stepping on Gaussian primes (considering both if the steps are bounded above or not) such that the step value is nearly a Gaussian prime?-we replace the constant 1 by some fixed Gaussian prime to get the version of the iteration for Gaussian primes,and again numerical examples suggest this is", "# The 9 step problem A man can take a step forward, backward,left and right with equal probability. Find the probability that after 9 steps he will be just 1 step away from his initial point. I have done similar questions with movement restricted to forward and backward only ,but this one just blows my mind. • Our teacher said if u feel this question to be easy do it for 'n' steps. – STK Oct 27 '16 at 13:53 • Solve first with just forward and backwards and see what you can pick up from there – Simply Beautiful Art Oct 27 '16 at 14:08 • He has to take an even number of steps in one axis and an odd number of steps in the other. Success comes only when the even number is split evenly between the directions and the odd number is as even as possible. You don't care which axis is even and which is odd, so just assume left/right is odd and sum the cases. – Ross Millikan Oct 27 '16 at 14:15 • Might clarify things to note that your answer is $4$ times the probability that he gets home in $10$ steps (after all, to get home in $10$ steps, he must be standing one step away after $9$ and", "5 steps away: | | | | | | <- I'm at step 0, not yet on the grid. | |X| | | | <- I take two steps forward, I'm on step 2: the count is 2 |X| | | | | <- I take one step back, I'm on step 1: the count is 3 | | |X| | | <- I take two steps forward, I'm on step 3: the count is 5 | |X| | | | <- I take one step back, I'm on step 2 again: the count is 6 | | | |X| | <- I take two steps forward, I'm on step 4: the count is 8 | | |X| | | <- I take one step back, I'm on step 3 again: the count is 9 | | | | |X| <- I take two steps forward, I'm on step 5: the count is 11 In this case, the result of the function would be 11. Example results: 1 => 3 5 => 11 9 => 23 10 => 26 11 => 29 100 => 296 1000 => 2996 10000 => 29996 100000 => 299996 Have fun, golfers! • Hmm ... this feels very familiar. – Shaggy Mar 26 '18 at 13:14 • Related – Rod Mar 26 '18", "= 62; 62 + 8 = 70; 70 + 7 = 77; 77 + 5 = 82; 82 + 1 = 83. Successive records: 1 takes 1 step to create 2 2 takes 4 steps to create 23 19 takes 5 steps to create 41 43 takes 9 steps to create 83 113 takes 12 steps to create 167 479 takes 15 steps to create 547 953 takes 20 steps to create 1049 1303 takes 22 steps to create 1399 [...] 1057181 takes 89 steps to create 1057579 Is this pattern real or have I misunderstood something? If it is real, is there a simple explanation for it? - What pattern?${}$ – gammatester Feb 13 '14 at 11:00 @gammatester The pattern is that each starting point for a record number of steps is itself prime. – Mark S. Feb 23 '14 at 23:21 Yes, this pattern has a simple explanation. On an intuitive level, if a record started from a non-prime $n$, then you might be able to go back a step to a number that would generate $n$ on the next step, so that the record-breaker would actually be less than $n$. (This doesn't work for primes because a \"previous number\" for a prime would have at most one step until it reached a prime, and wouldn't", "+ 2t(t-1) = 1 + 2t + 2t^2$. $1$ is the origin, $+4t$ is the axes, $+2t(t-1)$ is the 4 triangular arrays. e.g., for $t=2$, $(0,0)$ and $(1,0), (2, 0)$ and the other 3 directions, and $(1,1)$ and the other four quadrants. – user1448319 Mar 18 '13 at 21:04 • How are you going to be at $(1,0)$ after two steps? (Maybe I'm not understanding the walk you're describing. If I think of the \"usual\" random walk on $\\mathbb Z^2$, i.e., uniform in the four cardinal directions, then, unless I'm mistaken, the answer in my first comment should be correct.) – cardinal Mar 18 '13 at 22:13 • You can't end on $(1,0)$ after two steps starting from $(0,0)$. But you CAN walk through $(1,0)$ after taking two steps. You MUST consider all points that are reachable within 2 steps in order to construct $A_t$ as described above. – user1448319 Mar 18 '13 at 22:17 • That's true, but I took the sentence to mean what it said: You know there are only $n=1+4t+2(t−1)^2$ possible places you can land on the lattice after $t$ steps. :-) Perhaps an edit would help clarify. Cheers. – cardinal Mar 18 '13 at 22:32", "out of digits. I got $0.12$, a number further down the list after the place I had to stop. See a pattern? Think about what’d happen with 3 digits. Or 4. Or infinitely many! What happens if I continue this forever? Will my diagonal walk ever catch up with the end of the list? Here’s a table to sum up what we’re seeing so far: Number of decimal places How many numbers there are in the set How many numbers we include in a diagonal walk Percent of numbers missed by a diagonal walk 1 10 1 90% 2 100 2 98% 3 1,000 3 99.7% 4 10,000 4 99.96% $i$ $10^i$ $i$ $\\infty$ $10^\\infty$ $\\infty$ Basically all Now we can see clearly how, each time we add another digit, we make some progress by adding one more number to our diagonal walk. But alas, each additional digit also makes the set 10 times bigger. So overall, each digit we add causes us to fall further behind — really far behind! In fact, now we can see that, even if we walk forever, the set of numbers our diagonalization ‘missed’ will also grow forever. The trend never reverses; no longer how long we walk, we can never catch up. We can never even stop falling further behind! TODO", "# Not so random walk I'm out for a very long walk, and I’m bored, so I decide to walk in a mathematical way. The first image shows the first 500 steps, and the second image is my path after 50000 steps. The colors are mostly for visualization purposes. My path is not random, so how did I select my path? Please let me know if you need hints. • mm such a great thing, as a hint, and as long as u have the tool, can u add one up to a million just for eye candy pleasure, lol Jan 17, 2021 at 19:40 • I’ll post the python code tomorrow. It has to be changed a bit to go beyond 100,000 steps. Jan 17, 2021 at 19:49 • i think it'd be nice if the second image had equal x and y scale Jan 18, 2021 at 9:26 • This reminded me of this: oeis.org/A268463 Jan 19, 2021 at 1:26 • Someone else had a very similar idea: math.stackexchange.com/questions/2072308/… Jan 19, 2021 at 5:45 It looks like you start drawing a dot for $$n=0$$ at (1, 0) and then process to walk 'eastwards' (in the positive x direction) and draw a dot for each $$n$$ and make a 90° turn left when $$n$$ is prime. •", "impractical, because even for rather small values of $N,$ $\\pi_1$ is enormous. Another friend and I solved the problem by starting from $2$ and going forward, rather than trying to go backwards from $\\pi_1.$ For each prime $p$, we pretend that the goal of the game is to name $p$ and we label $p$ with the first move necessary to name it. For $p\\le N,$ the label is just $p$. Then working through the primes, we label each $p$ with the label of the largest prime less than $p-N$. So for example, if $N=23,$ we label the primes up through $23$ with themselves, and then we label $29$ with the label of the largest prime less than $29-23,$ that is $5.$ Then we label $31$ with $7$ and $37$ with $13$. Later on, when we come to label $61,$ it gets the same label as $37,$ namely $13.$ What we find as we fill out the table is that we rather quickly get to an interval $I$ of length $N$ in which all the primes have the same label. Then this label will be the winning first move (or it will be $0$ if there is no such move.) The reason is, that in working backward from $\\pi_1$ there is no way to skip over the interval $I$.", "number's behaviour without knowing the \"basis\". • What definition of \"one step\" are you using? At least under the more \"normal\" definitions I know of, $5$ takes either $4$ or $5$ steps whereas $1$ takes no steps. Dec 7 '15 at 3:22 • 2 takes 1 step, 5 takes 5 steps, 16 takes 7 steps... etc. Dec 7 '15 at 3:24 • $3(4n+1)+1=12n+4=2\\times2\\times(3n+1)$ would help. Dec 7 '15 at 3:24 • You should definitely try to prove the CC, just don't be too sad when you fail. Trying to prove famous conjectures can be very educational. You will be walking in the footsteps of others, but the view can still be spectacular. Dec 7 '15 at 10:30 If $n$ is odd, we have $4n+1 \\to 12n+4 \\to 6n+2 \\to 3n+1$, while $n \\to 3n+1$ so the streams join with $4n+1$ taking two extra steps. • You specified that $n$ is odd. I used that in the $n \\to 3n+1$ step, but nowhere else. Dec 7 '15 at 3:38 • This is wrong. $4 \\to 2 \\to 1, 18 \\to 9 \\to 28 \\to 14 \\to 7$ is longer than four steps. The point is that $n$ even does not constrain how many divide by $2$'s you do, while $4n+2$ says you will only do one. Dec 7 '15", "+ 1 = 26.1227 So, our number is 26 for how many things we ended up doing.1240 It is the same thing with 7 to 107; how many steps?1245 If we have a distance of 5 for each step, how many steps do we have to take from the 7?1250 Well, we have to take 20 steps, because 20 times 5 is 100.1256 So, if we take 20 5-distance steps from 7, we will make it to 107; so we have 20 steps that we take.1260 But we also have to count the 7, so we end up actually have n = 21.1268 And so, that is our value for n.1278 This is why you really have to think about this stuff carefully.1280 It is really easy to just say, \"OK, I took that many steps, so that must be my value.\"1283 No, you have to really pay attention to make sure that you are counting also where you started.1287 But sometimes, you have to pay attention and think about it: \"Did I already count where I started?\"1291 So, you really have to be careful with this sort of thing.1295 It is easy, as long as you can get a1, an, and the number of steps, n.1296 But sometimes, it is hard to really realize just how many", "– 8) is th... Prabhu Sep 16, 2013 Mehnaaz Nov 12, 2014 Harjee May 17, 2014 923*985 DIVISIBLE BY 11 FIND THE MISSING DIGIT Solution: Let the missing digit in the given number = n By the divisibility rule of 11 If the difference of the sum of digits at odd places is eq... Surinder Kumar Sep 30, 2015 Mehnaaz Nov 12, 2014 In a morning walk, 3 persons step off together. Their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that all can cover the same distance in complete steps? Minimum distance that each should walk = LCM of 80, 85 and 90. 80 = 2 x 2 x 2 x 2 x 5 85 = 17 x 5 90 = 2 x 3 x 3 x 5 LCM = 2 x 2 x 2 x 2 x 3 x 3 x 5... Vaibhav Choudhary Aug 7, 2013 Write the smallest 4 digit number and express it in terms of prime factors. The smallest four digit niumber 1000 1000= 23X53 &#... Jitendriya Dash Mar 28, 2016 what are co-prime numbers? Co-prime numbers are the set of numbers which have only 1 as their common factor i.e their HCF is 1. Co-prime numbers are also known as", "the second number (the second addend). Where he lands is the sum of the two numbers. ##### : 3 + 4 If Zed wants to add 3 + 4, he starts at 0 and faces towards the positive numbers. He walks forward 3 steps, then he walks forward 4 more steps. Zed ends at the number 7, so the sum of 3 and 4 is 7. 3 + 4 = 7. (But you knew that of course! The point right now is to make sense of the number line model.) When Zed wants to subtract two numbers, he he walks forward (to the right on the number line) however many steps are indicated by the first number (the minuend). Then he walks backwards (to the left on the number line) the number of steps indicated by the second number (the subtrahend). Where he lands is the difference of the two numbers. ##### : 11 – 3 If Zed wants to subtract 11 – 3, he starts at 0 and faces the positive numbers (the right side of the number line). He walks forward 11 steps on the number line, then he walks backwards 3 steps. Zed ends at the number 8, so the difference of 11 and 3 is 8. 11 – 3 = 8. (But you knew" ]

[ "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "# Factors on N^3 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "# Multiples of 5 and 7 Number Theory Level 3 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "does positive integer have? 1.The 5 25 Nov 2010, 18:39 How many prime divisors does positive integer N have? 1. 2N 3 08 Nov 2007, 20:17 How many prime divisors does positive integer N have? 1) 2N 10 16 May 2006, 07:13 Display posts from previous: Sort by # How many odd, positive divisors does 540 have? Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO 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®.", "this set we can not form three digit number more than 550 (max=532). Sufficient. OPEN DISCUSSION OF THIS QUESTION IS HERE: is-the-three-digit-number-n-less-than-134533.html _________________ Re: Is the three digit number n less than 550? (1) The product [#permalink] 13 Jul 2013, 05:20 Similar topics Replies Last post Similar Topics: 25 If N is a positive three-digit number that is greater than 2 12 09 Dec 2013, 03:54 1 How many positive prime numbers are less than the integer n? 4 26 Dec 2012, 23:50 37 Is the positive two-digit integer N less than 40 ? 9 03 Dec 2012, 04:42 5 Is the three-digit number n less than 550? 3 15 Jun 2012, 05:34 6 Is the three-digit number n less than 550 ? 9 06 May 2010, 11:31 Display posts from previous: Sort by", "# Eleven primes Find the sum of all positive primes $$p$$ there such that $$p^2+11$$ has exactly six positive divisors. ×", "there are infinitely many positive integers which can be termed as prime numbers and so there are infinitely many primes. Does this mean that all positive integers are prime? · 1 month ago", "# Divisible Prime Times Level pending How many positive integers below 1000 have $$p$$ positive divisors, where $$p$$ is a prime number? ×", "get tricky if you had more than three prime divisors, but $360$ is not too bad:", "• 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.", "6 How many positive integers less than 28 are prime numbers 8 17 Mar 2012, 15:57 11 What is the total number of integers between 100 and 200 15 20 Feb 2006, 15:41 Display posts from previous: Sort by", "prime divisors of 36 = 2 × 3 = 6", "# Prime Numbers Number Theory Level 1 How many positive integers $$n$$ are there such that $$6^n-1$$ is a prime number? ×", "# Ev-En-Ly Odd In how many ways you can distribute 55 among Ev, En and Ly so that everyone gets (a positive) odd number? (In other words three odd numbers add up to 55). ×", "# 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.", "+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. :))", "May 2015, 05:43 8 The product of 2 positive integers is 720. If one of the numbers is in 5 04 Dec 2014, 04:05 6 How many positive integers less than 28 are prime numbers 8 17 Mar 2012, 16:57 20 What is the total number of positive integers that are less 16 24 Feb 2012, 23:11 Display posts from previous: Sort by", "# Primes in consecutive numbers Is there a set of 2015 consecutive positive integers containing exactly 15 prime numbers? ×", "# IMO! 2 Number Theory Level 4 How many positive integers $$n$$ exist such that $$n^2 + 1$$ has a prime divisor greater than $$2n + \\sqrt{2n}$$. ×" ]

[ "# Factors on N^3 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "What's the sum of all the positive integral divisors of $540$? What's the sum of all the positive integral divisors of $540$? My approach: I converted the number into the exponential form. And found out the integral divisors which came out to be $24$. But couldn't find the sum...Any trick? Prime factorization of 540 is $2^2\\cdot 3^3\\cdot 5$. Also sum of divisors of 540 equals: \\begin{align} \\sigma\\left(2^2\\cdot 3^3\\cdot 5\\right) &=\\frac {2^3-1}{2-1}\\cdot\\frac{3^4 - 1}{3-1}\\cdot \\frac{5^2 -1}{ 5-1}\\\\ &=7\\cdot 40\\cdot 6\\\\ &= 1680 \\end{align} Assuming you want to know the sum of the divisors of 540, i.e. $$\\sigma_1(540) = \\sum_{d \\mid 540} d$$ then you can compute this by hand by either determining all divisors of 540, or you can do this a little more cleverly. First of all, note that $540 = 2^2 \\cdot 3^3 \\cdot 5$. Now, one happy fact about the function $\\sigma_1(n)$ is that if $n, m$ are relatively prime, then $\\sigma_1(n \\cdot m) = \\sigma_1(n) \\cdot \\sigma_1(m)$. In particular, $$\\sigma_1(540) = \\sigma_1(4)\\sigma_1(27)\\sigma_1(5)$$ and so you only need to determine these last ones. But for prime numbers, $\\sigma_1(p^k)$ is easy to compute. It is $$\\sigma_1(p^k) = 1 + p + p^2 + \\cdots + p^k = \\frac{p^{k+1} - 1}{p-1}$$ which should let you compute the answer now.", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "= 3628799!7!5!3!$. You can ... 0 No. Consider $4^4 = 2^{8}$, which has no 3 in its prime factorization, while $4! = 1 \\cdot 2 \\cdot 3 \\cdot 4 = 2^3 \\cdot 3$. -2 Top 50 recent answers are included", "# Prime factorization of 555 If it's not what You are looking for type in the field below your own integer, and You will get the solution. Prime factorization of 555: By prime factorization of 555 we follow 5 simple steps: 1. We write number 555 above a 2-column table 2. We divide 555 by the smallest possible prime factor 3. We write down on the left side of the table the prime factor and next number to factorize on the ride side 4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors) 5. We continue until we reach 1 on the ride side of the table 555 prime factors number to factorize 3 185 5 37 37 1 Prime factorization of 555 = 1×3×5×37= $1 × 3 × 5 × 37$ ## Related pages sin135what are the factors of 6x 6what is 1.125 as a fractionv lwh solve for h68-41what is 5.25 as a fraction2x 2 simplify2x 5y 9derivative sin xyderivative of e to the x squaredwhat is the lcm of 15y 2y y 0what is the square root of 2704factor 4x 2-191-56derivative of pieprime factorization of 120roman numerals 1979x2 6x 4 0gcf and lcm of 36 and 45converting percentages to decimalsln 4x 2sin2x cos2x 16ax4x 2y 5 graphmmxx", "digits of $14553$ is $18$, so it is divisible by $9$. Repeatedly dividing by $3$ yields $$14553 = 3 \\cdot 4851 = 3^2 \\cdot 1617 = 3^3 \\cdot 539$$ • The next reasonable guess might be that $539$ is divisible by $7$... and indeed it is: $$539 = 7 \\cdot 77 = 7^2 \\cdot 11$$ • ...and $11$ is prime, so we're done. In summary: $$9095625 = 3^3 \\cdot 5^4 \\cdot 7^2 \\cdot 11$$ • After factoring out 3**3, we are left with 539, Here we can take sum of the alternate digits. 5+9=14 & 3. Now taking difference between the two groups, we get 14-3=11. Hence the number is divisible by 11. By factoring out 11, we get 539=11 * 49. <p> Refer to summary of Divisibility rules – Wishwas Sep 19 '16 at 5:03 Since we're doing the prime factorization, you can just use the long division algorithm that you learned in grade school. Use the last digit as a clue for what to divide by. Remember that any number has a unique prime factorization, so always just divide by the smallest feasible prime. In this case, we start by recognizing that the number is obviously divisible by $5$. Divide it by $5$, we get $1819125$. Divide by $5$ again. We get $363825$. And again. We", "+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", "positive integers coprime to and ... 5 No,$2^{333} - 1$cannot be a prime. It is easy to see this via the exponent,$333$, which has factorization$3\\cdot3\\cdot37\\$. From this we know, by a well-known identity, that $$2^{333} - 1 = (2^{37} - 1)\\cdot (1 + 2^{37} + 2^{37\\cdot2} + 2^{37\\cdot3} + 2^{37\\cdot4} + 2^{37\\cdot5} + 2^{37\\cdot6} + 2^{37\\cdot7} + 2^{37\\cdot8}).$$ Only top voted, non community-wiki answers of a minimum length are eligible", "The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways? The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways? I've seen the answer to the question, and there is only one way: Since $555, 555 = 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \\cdot 7 \\cdot 37) (5 \\cdot 11 \\cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure? Sorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish. • list all the divisors in order. $\\sqrt{555555} \\approx 745.36.$ One of your divisors must be larger than that but still smaller than 1000. The other divisor in your pair will be between 555.56 and 745.36. I guess you can find them by hand, look at all the products of two primes, then three primes. You do not need to do four primes because you already did two – Will Jagy Apr 16 '18 at 2:34 • Yeah, two primes is too small, the biggest is $13 \\cdot 37", "by the Eisenstein's criterion. $7^5+5\\cdot 7^2+1 = 17053$ which is a prime number. Thus the polynomial is irreducible by Cohn's irreducibility criterion.", "accordingly. From here the smallest prime factor of: $777$ is number $3$ which equals to $3\\cdot259$ and for $715$ is number $5$ which equals to $5\\cdot143$ Further factorisation results in final solution where $143=11\\cdot13$ and $259=7\\cdot37$ therefore: $(5\\cdot11\\cdot13)(3\\cdot7\\cdot37)$ can satisfy equation. If $555{,}555=ab$ with $a,b\\lt1000$, then we must also have $a,b\\gt555$ (e.g., if $b\\lt1000$, then $a=555{,}555/b\\gt555{,}555/1000\\gt555$). Now since $555{,}555=3\\cdot5\\cdot7\\cdot11\\cdot13\\cdot37$, we may assume, without loss of generality, that $a=37k$. From $a\\gt555=15\\cdot37$, we see that $k\\gt15$, and from $a\\lt1000$, we see that $k\\le\\lfloor1000/37\\rfloor=27$. The only product $k$ of the primes $3$, $5$, $7$, $11$, and $13$ that falls in the interval $15\\lt k\\le27$ is $k=3\\cdot7=21$. Thus $a=37\\cdot21=777$, $b=5\\cdot11\\cdot13=715$ is the only factorization of $555{,}555$ into two three-digit numbers.", "the set of positive integers all of whose prime divisors are $2$ or $\\equiv 1 (\\!\\!\\mod 4)$, and let $B$ be the set of positive integers all of whose prime divisors are $\\equiv 3 (\\!\\!\\mod 4)$. Then $A \\cdot B = \\mathbb{N}$.", "the unique factorization theorem? $2520$ is the smallest positive integer divisible by each of $~2,~3,\\cdots,~9,~\\&~10$. 5. Ok, yes I understand that. It's based on the LCM. Thank you!!", "$a$ 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: $\\dfrac{3 \\cdot 3 \\cdot 2}{5 \\cdot 3 \\cdot 2} = \\boxed{\\dfrac35}$.", "# Question #b68fd Aug 24, 2017 See below. #### Explanation: Let's factor $105$. $105 = 35 \\cdot 3 = 3 \\cdot 5 \\cdot 7$ So, $3 \\cdot 5 \\cdot 7$ is the prime factorization of $105$. However, since there is only one of each prime, $\\sqrt{105}$ is already simplified.", "# Factors on N^3 Computer Science Level 5 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? × Problem Loading... Note Loading... Set Loading...", "$1386$ is $2×3×3×7×11$ . You can use prime factorizations to figure out GCF s (Greatest Common Factors), LCM s (Least Common Multiples), and the number (and sum) of divisors of $n$ .", "? Free Version Difficult # Prime Factorization and Divisors ALGEBR-YT8GKF A certain integer has exactly 40 distinct positive divisors. Which one of the following could be the prime factorization of that number? A $2^{3}\\cdot5$ B $2^{3}\\cdot3^{4}\\cdot5$ C $3^{8}\\cdot7^{5}$ D $5^{2}\\cdot11^{4}\\cdot13^{5}$", "$n$ other than $1\\cdot n$. Using an app for prime factorization, the following can be deduced very quickly: So our only non-prime $n$ with one unique factorization is $n=19{,}897=101\\cdot 197$ and we have our solution!", "# factorisation problem factorise 57284861 . [ use whatever you want ] Considering $$x_1,x_2 \\cdots x_n$$ to be the prime factors, calculate $$(x_1-1)(x_2-1)\\cdots(x_n-1)$$ ×" ]

[ "# Factors on N^3 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "positive integers coprime to and ... 5 No,$2^{333} - 1$cannot be a prime. It is easy to see this via the exponent,$333$, which has factorization$3\\cdot3\\cdot37\\$. From this we know, by a well-known identity, that $$2^{333} - 1 = (2^{37} - 1)\\cdot (1 + 2^{37} + 2^{37\\cdot2} + 2^{37\\cdot3} + 2^{37\\cdot4} + 2^{37\\cdot5} + 2^{37\\cdot6} + 2^{37\\cdot7} + 2^{37\\cdot8}).$$ Only top voted, non community-wiki answers of a minimum length are eligible", "+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 \\\\", "+0 # Counting Divisors +1 129 2 +484 Some positive integers have exactly four positive factors. For example, 35 has only 1, 5, 7 and 35 as its factors. What is the sum of the smallest five positive integers that each have exactly four positive factors? Apr 5, 2021 #1 +594 +2 If they have four factors they can be of the form $p_1\\cdot p_2$ or $p^3$ for all $p_1,p_2,p$ prime. Let's do $p_1\\cdot p_2$ first. We have $2\\cdot 3=6, 2\\cdot 5=10, 2\\cdot 7=14, 2\\cdot 11=22,$ etc. if $2$ is the first. Otherwise we can have $3\\cdot 5=15, 3\\cdot 7=21$, etc. So the smallest $5$ are $6,10,14,15,21$. Then for $p^3$ we can have $2^3=8$ and the rest are larger than $21$. So the numbers are $6,8,10,14,15$. Sum them up. Apr 5, 2021 #2 +484 0 Thank you @thedudemanguyperson, it was correct! RiemannIntegralzzz Apr 5, 2021", "The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways? The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways? I've seen the answer to the question, and there is only one way: Since $555, 555 = 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \\cdot 7 \\cdot 37) (5 \\cdot 11 \\cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure? Sorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish. • list all the divisors in order. $\\sqrt{555555} \\approx 745.36.$ One of your divisors must be larger than that but still smaller than 1000. The other divisor in your pair will be between 555.56 and 745.36. I guess you can find them by hand, look at all the products of two primes, then three primes. You do not need to do four primes because you already did two – Will Jagy Apr 16 '18 at 2:34 • Yeah, two primes is too small, the biggest is $13 \\cdot 37", "? Free Version Difficult # Prime Factorization and Divisors ALGEBR-YT8GKF A certain integer has exactly 40 distinct positive divisors. Which one of the following could be the prime factorization of that number? A $2^{3}\\cdot5$ B $2^{3}\\cdot3^{4}\\cdot5$ C $3^{8}\\cdot7^{5}$ D $5^{2}\\cdot11^{4}\\cdot13^{5}$", "# Factors on N^3 Computer Science Level 5 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? × Problem Loading... Note Loading... Set Loading...", "## Recounting the Rationals, part IVb: the Euclidean Algorithm Suppose we have two integers, and we’d like to find their greatest common divisor (GCD). Recall that the greatest common divisor of two integers m and n is exactly that: the greatest integer which is a divisor of both m and n. The obvious way to do this—which is probably the way you learned in elementary school, when reducing fractions—is to factor each integer into primes, and look for overlap. For example, let’s say we want to find the GCD of 450 and 525. We begin by factoring: $\\begin{array}{rcl} 450 & = & 2 \\cdot 3^2 \\cdot 5^2 \\\\ 525 & = & 3 \\cdot 5^2 \\cdot 7 \\end{array}$ Now we pull out as many prime factors as we can which are contained in the factorizations of both numbers. We can’t use 2 at all, since it is only contained in the factorization of 450; we can use a factor of 3, but only one; and we can use two factors of 5. Putting these together, the greatest common divisor of 450 and 525 is $3 \\cdot 5^2 = 75$. This method is intuitive, and works well for relatively small numbers, but (there’s always a “but”, isn’t there?) it fails horribly for larger numbers. For example, what if you", "+0 # help 0 80 1 Find the number of distinct positive divisors of $$(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$$ . Jan 1, 2019", "# Prime factorization of 555 If it's not what You are looking for type in the field below your own integer, and You will get the solution. Prime factorization of 555: By prime factorization of 555 we follow 5 simple steps: 1. We write number 555 above a 2-column table 2. We divide 555 by the smallest possible prime factor 3. We write down on the left side of the table the prime factor and next number to factorize on the ride side 4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors) 5. We continue until we reach 1 on the ride side of the table 555 prime factors number to factorize 3 185 5 37 37 1 Prime factorization of 555 = 1×3×5×37= $1 × 3 × 5 × 37$ ## Related pages sin135what are the factors of 6x 6what is 1.125 as a fractionv lwh solve for h68-41what is 5.25 as a fraction2x 2 simplify2x 5y 9derivative sin xyderivative of e to the x squaredwhat is the lcm of 15y 2y y 0what is the square root of 2704factor 4x 2-191-56derivative of pieprime factorization of 120roman numerals 1979x2 6x 4 0gcf and lcm of 36 and 45converting percentages to decimalsln 4x 2sin2x cos2x 16ax4x 2y 5 graphmmxx", "# A reminder of the prime factorization of the New Year Level pending Let $$N$$ be the sum of the positive divisors of $$2014$$. What is the number of positive divisors of $$N$$? ×", "at 0:24 • See, there lies my confusion. 7 does not show up in the prime factorization of 24... so why am I looking into it? Similarly with 5. – user424890 Nov 7 '17 at 0:27 Our goal is to find all integers with exactly 36 divisors. To do this, first note that any positive integer will have a unique prime factorization (up to order in which primes appear, of course). This means that in general, a positive integer $n$ will have the form $$n = p_1^{a_1} \\cdots p_k^{a_k}$$ where each prime here is distinct and each $a_i$ is a positive integer. So, any divisor of $n$ will have the form $$d = p_1^{b_1} \\cdots p_k^{b_k}$$ where $b_i$ can range between $0$ and $a_i$. Now, how many ways can choose a $b_i$? Well, there are $a_i +1$ options. So, we have $(a_1+1) \\cdots (a_k+1)$ divisors of $n$. If we have $(a_1+1) \\cdots (a_k+1) = 24$ then $n$ will have exactly $24$ divisors. If we are looking for positive integers with exactly 36 divisors then any positive integer with the exponents of its primes satisfying $(a_1+1) \\cdots (a_k+1) = 36$ will work. Note that $36 = 2^2 3^2$. With this in mind, we can have $36 = 2*2*3*3$ which would correspond to 4 distinct primes, two being square and", "+0 # I have no life +3 115 3 +321 a)What is the smallest positive integer that has exactly 20 positive divisors? b)What is the smallest positive integer that has exactly 6 perfect square divisors? Mar 24, 2019 #1 +5172 +4 $$20 = 2^2 5 \\text{ so we'll want to use prime factors }\\\\ 2^4 \\cdot 3^1 \\cdot 5^1 = 240\\\\ \\text{This has }(4+1)(1+1)(1+1)=20 \\text{ divisors}$$ I'm pretty sure for (2) we just find the smallest positive integer with 6 divisors and square it. Proceeding as above $$6=2\\cdot 3 \\text{ so choose}\\\\ (2^2\\cdot 3)^2 = 144$$ . Mar 24, 2019 #2 +321 +2 Thank You! #3 +533 +1 a) 20 has factors of 1, 2, 4, 5, 10, and 20. So, to make the number as small as possible, we want to use 5 * 2 * 2. We will pair the largest exponents with the smaller primes to get 2^(5-1) * 3 * 5 = 240. b) It has 6 perfect square divisors, so we need the smallest number with 6 divisors then square that to get the answer. That is (by counting), 12. So, our answer is 12^2 = 144. Mar 24, 2019", "## Discrete Mathematics and Its Applications, Seventh Edition To obtain the greatest common divisor of each pair of integers, multiply together the factors shared by both numbers. a) $2^2\\cdot 3^3\\cdot 5^2=2700$ b) $2\\cdot 3\\cdot 11=66$ c) $17$ d) $1$ e) These numbers are relatively prime. f) $2\\cdot 3\\cdot 5\\cdot 7=210$", "+0 # factorials 0 29 1 For how many positive integers n less than or equal to 24 is n! evenly divisible by \\(1+2+ \\cdots + n\\)? Mar 17, 2020", "the set of positive integers all of whose prime divisors are $2$ or $\\equiv 1 (\\!\\!\\mod 4)$, and let $B$ be the set of positive integers all of whose prime divisors are $\\equiv 3 (\\!\\!\\mod 4)$. Then $A \\cdot B = \\mathbb{N}$.", "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "+0 # How many positive integers $n$ satisfy $127 \\equiv 7 \\pmod{n}$? $n=1$ is allowed. +1 89 1 +160 How many positive integers $n$ satisfy $127 \\equiv 7 \\pmod{n}$? $n=1$ is allowed. #1 +867 +1 Hey RB! If $$127 \\equiv 7 \\pmod{n}$$, then n is a divisor of 127 - 7 = 120. The prime factorization of 120 is $$2^3 \\cdot 3 \\cdot 5$$ which has $$(3 + 1)(1 + 1)(1 + 1) = 16$$ positive divisors. Therefore, there are 16 possible values of n. I hope this helps, Gavin. GYanggg Apr 30, 2018", "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" ]

[ "rearrange the letters? ### Semordnilap What do you get if you reverse the order of the letters?", "he formed groups of seven, he divided everyone. How many toy cars have in the collection? • Five pupils Five pupils clean 30 chairs one hour before four pupils. How many chairs clean one pupil in 1 hour? • 40% volume 40% volume with 104 uph (units per labor hour) 8 people working. What is the volume? • Two workers One worker needs 40 hours to do a job, and the second would do it in 30 hours. They worked together for several hours, then the second was recalled, and the first completed the job itself in 5 hours. How many hours did they work together, and how much did • Midnight How many hours are, if the time that elapsed since 8:00 is 2/5 of the time that will past till midnight? • Two typists There are two typists who are rewriting the material 814 pages. First can it handle rewrite yourself for 24 days; the second 12 days. First typist wrote material yourself 4 days rest rewrites yourself second typist. How many days will it take rewriting al • Suzan My daughter reaches X years in X squared, said Suzan father in 2009 during the celebration of her birthday. When will given fact become reality and how many years at that time will have Suzan? •", "= 72*3 = 216 Words per min Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink] 05 Oct 2019, 22:16 Display posts from previous: Sort by", "has some pretty decent keyboard shortcuts nowadays -- I've had kids turn in work written in that. That might be an easier thing to work on. Share on other sites 37 minutes ago, lewelma said: If my boy with dysgraphia ever wants to take a math course at university, he will need to be able to type up his answers at speed on a timed one hour test. He could hand write it in about 2.5 hours, but they only give him 10 extra minutes for an hour long test. So basically, if he can't type at speed, he won't be able to do another math class, ever. So, he would have to be able to type in Latex as fast as you would write a timed test. So very very fast. He cannot code, because he has dysgraphia, and those two things don't go together. However, he is very good at keyboard shortcuts, so I figured that it was really the same. How many hours would he have to commit to be able type as fast as fast writing? 20? 100? My son has dysgraphia. This is why he is using LaTeX. He does all of his homework with LaTeX, not just math. He is using it for Physics this year. It took him a long time", "it? Agnes: 500 pieces. Louie: No, not the puzzle. The...the competition. Agnes: Oh. About 360 in the individual competition. Louie: Look, I don't think it's such a good idea. Agnes: I'm doing it. Louie: Honey... Agnes: I'm not asking you. I'm telling you. I'm finally doing something on my own, and you can support me or not. I'll do it either way. ... Robert: Stay. Have a drink with me. Agnes: I have to beat the traffic. I'll see you next week. Robert: I'll miss you, Mata. Agnes: Why did you say that? Robert: 'Cause I meant it. Agnes: How can you mean it? Robert: I don't know. Agnes: How can this be happening? Robert: I don't know. Agnes: Why are we wasting all this time doing puzzles? Robert: What else is there to do? Agnes; It's a childish hobby for bored people. Robert: You know that's not true. Agnes; Tell me you're not a bored rich guy. Tell me I'm not a childish housewife. Robert: No, that's not what we are. Agnes: You have much more important things to do. You're a man of ideas. Why do you do these stupid puzzles? Robert: It's a way... to control the chaos. Agnes: That's ridiculous. Robert: Come on, Mata, you... you're missing the point. Agnes: Okay. What is the", "letters of the word MISSISSIPPI are rearranged randomly.", "love to rewrite it, perhaps using cyclejs or mobx… If only there were so many hours in the day. Approx. 1734 words, plus code", "# Rates & proportional relationships ### Problem Charlie can type faster than any of his classmates. He took a test that showed he can type 700 words in 5 minutes left parenthesis, w, p, m, right parenthesis. The fastest recorded typing speed is shown in the chart below (we assume the speed is constant for the three minutes shown). Does Charlie type faster than the record shown in the chart? Minutes123 Words216432648", "Question # Mohan is able to write 20 words in one hour. How many words would he be ableto write in 10 minutes? Hint: Use the unitary method to find the answer. First find the number of words he can write in 1 minute. Then multiply this by 10 to find the number of words he can write in 10 minutes. Complete step by step solution: We know that he can write 20 words in one hour. We also know that one hour consists of 60 minutes. This means that Mohan is able to write 20 words in 60 minutes. 20 words $\\to$60minutes We will now find the number of words he can write in 1 minute. For this, we will divide the known number of words with the known measure of time. So, he can write $\\dfrac{20}{60}$ i.e. $\\dfrac{1}{3}$ words in a minute. We now need to find the number of words he can write in 10 minutes. For this, we will multiply 10 and the number of words he can write in one minute. So, the number of words he can write in 10 minutes is: $10\\times \\dfrac{1}{3}=3.33$ words. Hence, Mohan can write 3.33 words in 10 minutes. Note: We can solve this question using another method also. We need to find the number of words", "can type nearly 40 words per minute. Use this information to find the number of hours it would take him to type 2.6 × 105 words. Type below: ______________ Answer: Samuel can type 40 words per minute. Then how many hours will it take for him to type 2.6 words times 10 to the power of five words 2.6 words time 10 to the power of 5 2.6 × (10)4 2.6 x 100 000 = 260 000 words in all. Now, we need to find the number of words Samuel can type in an hour 40 words/minutes, in 1 hour there are 60 minutes 40 x 60 2,400 words /hour Now, let’s divide the total of words he needs to type to the number of words he can type in an hour 260 000 / 2 400 108.33 hours. Question 24. Entomology A tropical species of mite named Archegozetes longisetosus is the record holder for the strongest insect in the world. It can lift up to 1.182 × 103 times its own weight. a. If you were as strong as this insect, explain how you could find how many pounds you could lift. Type below: ______________ Answer: Number of pounds you can lift by 1.182 × 103 by your weight Question 24. b. Complete the calculation to find", "he can write in 10 minutes. We know that he can write 20 words in 60 minutes. Also 10 minutes is one-sixth of 60 minutes. So, we can divide 20 words by 6 to get 3.33 as the final answer which is the same as obtained above.", "Free Version Easy # Rearranging the Letters in the Name \"April\" ALGEBR-RA0VG1 How many ways can you arrange the letters of the word $APRIL$? A $120$ B $160$ C $60$ D $100$", "speed is twice John's speed, so: after 1 hour : Adam writes 100 lines and John writes 50 lines; total lines 150 of which Adam wrote 2/3rds and John wrote 1/3rds after 2 hours: Adam would have written 200 lines and John 100 lines; total lines 300 of which Adam wrote 2/3rds and John wrote 1/3rds The ratio will always be 1:2 and Adam would have always written 2/3rds of the total lines written. Since John is working from the bottom and Adam from the top, they will meet when they have collectively written all of the 535 lines. Counting from the beginning that would be 2/3 * 535 ~= 357 lines. Hope this clears it up. Re: Adam and John simultaneously begin to write out a booklet containing [#permalink] 14 Jun 2019, 17:00 Display posts from previous: Sort by", "20% 27% 18% 30% ##### 55. If 20 typists can typ 48 letters in 20 minutes, then how many letters will 30 typists working at the same rate complete in 1 hour? 72 144 416 216 RightAns: 0 | WrongAns: 0 | Result: 0/0", "work beyond mid-afternoon, but only came to understand the reason for this when he had become a convinced Christian, because he then realized that three p.m. is the canonical hour of the crucifixion. He was no less punctual about cocktails and dinner, and went home to bed, even being known to leave his own birthday parties at a fixed time in order to be up and at work at his scheduled time of, I think it was, six. He credited his parents with instilling in him this useful discipline.” • Aharon Appelfeld says that in order to be a serious writer you need to have a routine. For years his routine has been to write with a Biro on sheets of ordinary white paper in the café at Ticho House, in Jerusalem, which was once the house of a wealthy doctor and where this interview took place. INTERVIEWER: “So you come here to work at Ticho House twice a week?” APPELFELD: “Yes. I come here somewhere around ten or eleven. I stay here for two or three hours and then I go home. It’s a routine. Generally, when we say routine, it sounds bad, but routine is important.” INTERVIEWER: “You write longhand. How many pages per day?” APPELFELD: “One page, sometimes half a page, sometimes one and a", "ago, my son was finishing up his nightly Beast Academy and enjoying the puzzles so much that he took his workbook with him at bedtime and continued to work for another hour! He was in a section of… Continue Reading", "do the mail, which can take hours, take a nap if possible, and do a little work in the late afternoon before the day dissolves into dinner, putting the children to bed, reading, lights out. When I have a deadline, I just work all night and all day, like an architect on . • INTERVIEWER: “Do you write by hand, on the typewriter, or do you alternate?” VARGAS LLOSA: “First, I write by hand. I always work in the morning, and in the early hours of the day, I always write by hand. Those are the most creative hours. I never work more than two hours like this—my hand gets cramped. Then I start typing what I’ve written, making changes as I go along; this is perhaps the first stage of rewriting. But I always leave a few lines untyped so that the next day, I can start by typing the end of what I’d written the day before. Starting up the typewriter creates a certain dynamic—it’s like a warm-up exercise.” INTERVIEWER: “Hemingway used that same technique of always leaving a sentence half-written so he could pick up the thread the next day . . .” VARGAS LLOSA: “Yes, he thought he should never write out all he had in mind so that he could start up more", "that he needs help plotting, or with characters, but really, at 4 AM my brain is again mush and I can’t postulate what chromosomal damage would cause fire-eating, ice dragons from Mars). • , B.F. Skinner: A Life, Bjork 1993, pg216–217: On the surface, Skinner’s later years seemed conventional: He retired from Harvard in 1974 and was presented with a first edition of Thoreau’s Walden…On closer inspection, however, he had continued the intensity of his intellectual life…The circadian rhythm of his writing schedule did not miss a beat. A timer rang at midnight signaling him to arise, move to his desk, and writing until signaled to stop at one o’clock. He returned to bed and arose again to the sound of the timer at five o’clock and composed until it buzzed two hours later. He write three hours a day, seven days a week, holidays included. As the years passed, these three early morning hours were, as he often said, “the most reinforcing part of my day.”14 The other twenty-one hours were arranged to make the writing time as profitable as possible…During the last decade of his life he frequently listened to Wagnerian music, usually in mid-afternoon, relaxing after the rigors of early morning writing and thinking. pg21 (shortly before death): …Toward the far end of the study,", "### Home > CC4 > Chapter 11 > Lesson 11.2.5 > Problem11-105 11-105. Steven can look up 27 words in the dictionary in an hour. His teammate, Mary Lu, can look up 35 words per hour. Working together, how long will it take them to look up 100 words? Homework Help ✎ Together they can look up 62 words per hour. Write an equation, where x = the number of hours they are working. $\\frac{\\text{words}}{\\text{hour}}\\cdot \\text{hours}=\\text{words}$", "1 - 2 of 2 Posts #### mushutofu · ##### Registered Joined · 882 Posts Discussion Starter SCRABBLE Someone out there either has too much spare time or is deadly at Scrabble. (Wait till you see the last one!!) GEORGE BUSH: When you rearrange the letters: HE BUGS GORE DORMITORY: When you rearrange the letters: DIRTY ROOM EVANGELIST: When you ! rearrange the letters: EVIL'S AGENT PRESBYTERIAN: When you rearrange the letters: BEST IN PRAYER DESPERATION: When you rearrange the letters: A ROPE ENDS IT THE MORSE CODE: When you rearrange the letters: HERE COME DOTS SLOT MACHINES: When you rearrange the letters: CASH LOST IN ME ANIMOSITY: When you rearrange the letters: IS NO AMITY MOTHER-IN-LAW: When you rearrange the letters: WOMAN HITLER SNOOZE ALARMS: When you rearrange the letters: ALAS! NO MORE Z'S A DECIMAL POINT: When you rearrange the letters: I'M A DOT IN PLACE THE EARTHQUAKES: When you rearrange the letters: THAT QUEER SHAKE ELEVEN PLUS TWO: When you rearrange the letters: TWELVE PLUS ONE And for the grand finale: PRESIDENT CLINTON OF THE USA: When you rearrange the letters (With no letters left over and using each letter only once): TO COPULATE HE FINDS INTERNS ·" ]

[ "# Difference between revisions of \"2005 AMC 8 Problems/Problem 2\" ## Problem Karl bought five folders from Pay-A-Lot at a cost of $\\textdollar 2.50$ each. Pay-A-Lot had a 20%-off sale the following day. How much could Karl have saved on the purchase by waiting a day? $\\textbf{(A)}\\ \\textdollar 1.00 \\qquad\\textbf{(B)}\\ \\textdollar 2.00 \\qquad\\textbf{(C)}\\ \\textdollar 2.50\\qquad\\textbf{(D)}\\ \\textdollar 2.75 \\qquad\\textbf{(E)}\\ \\textdollar 5.00$ ## Solution Karl paid $5 \\cdot 2.50 = \\textdollar 12.50$. $20 \\%$ of this cost that he saved is $12.50 \\cdot .2 = \\boxed{\\textbf{(C)}\\ \\textdollar 2.50}$. ## Solution 2 Each folder can also be $5/2$ dollars, and $20\\%$ can be shown as $(1/5)$. We can multiply $(5/2) \\cdot (1/5) = (1/2)$. $(1/2)$ is also $50$ cents or the amount of money that is saved after the $20\\%$ discount. So each folder is $2.50-0.5 = \\textdollar2$.Since Karl bought 5 folders all of the folders after the discount is $(5)(2) = 10$, and the money bought before the discount is $(5)(2.50) = \\textdollar12.50$. To find the money Karl saves all we have to do is subtract $12.50 - 10 = 2.50$. Thus the answer is $\\boxed{\\textbf{(C)}\\ \\textdollar 2.50}$.", "\\\\[-3pt] \\,1 \\end{array}$ $2 \\frac{1}{3}$ falls between $2$ and $3$ on the number line. Question 3 ### Tavon's flight is 270 minutes long. How many hours does the flight last? A $4 \\text{ hours}$ B $4 \\frac{1}{2} \\text{ hours}$ C $5 \\text{ hours}$ D $5 \\frac{1}{2} \\text{ hours}$ Question 3 Explanation: $\\require{cancel} 270 \\cancel{\\text{minutes}} \\cdot \\dfrac{1 \\text{ hour}}{60 \\cancel{\\text{minutes}}}$ $= \\dfrac{270}{60} \\text{hours}$ $= \\dfrac{27}{6} \\text{hours}$ $\\dfrac{27}{6} = \\dfrac{9}{2} = 4 \\dfrac{1}{2}$ Question 4 ### In a factory there are two separate containers of stress balls. In the first container are 80 balls, and in the second are 90 balls. 40% of the balls in the first container are defective, and 20% in the second container are defective. In total, how many balls in the two containers are defective? A $102$ B $50$ C $60$ D $51$ Question 4 Explanation: The correct answer is (B). Calculate the number of defective balls in each container and then add these two amounts: $40\\% \\text{ of } 80$ $= 0.40 \\cdot 80 = 32$ $20\\% \\text{ of } 90$ $= 0.20 \\cdot 90 = 18$ $32 + 18 = 50$ Question 5 ### Ayana wants to buy as many bags of mulch as possible with her \\$305, and she would like them to be delivered to her house. The cost is \\$8.00", "$$\\dfrac{x\\; students}{3\\; teachers}$$ (c) y dollars for 18 hours Write as a rate. $$\\dfrac{y}{18\\; hours}$$ ##### Exercise $$\\PageIndex{21}$$: Translate the word phrase into an algebraic expression. (a) 689 miles per h hours (b) y parents to 22 students (c) d dollars for 9 minutes $$\\dfrac{689\\; mi}{h\\; hours}$$ $$\\dfrac{y\\; parents}{22\\; students}$$ $$\\dfrac{d}{9\\; min}$$ ##### Exercise $$\\PageIndex{22}$$: Translate the word phrase into an algebraic expression. (a) m miles per 9 hours (b) x students to 8 buses (c) y dollars for 40 hours $$\\dfrac{m\\; mi}{9\\; h}$$ $$\\dfrac{x\\; students}{8\\; buses}$$ $$\\dfrac{y}{40\\; h}$$ ## Practice Makes Perfect ### Write a Ratio as a Fraction In the following exercises, write each ratio as a fraction. 1. 20 to 36 2. 20 to 32 3. 42 to 48 4. 45 to 54 5. 49 to 21 6. 56 to 16 7. 84 to 36 8. 6.4 to 0.8 9. 0.56 to 2.8 10. 1.26 to 4.2 11. $$1 \\dfrac{2}{3}$$ to $$2 \\dfrac{5}{6}$$ 12. $$1 \\dfrac{3}{4}$$ to $$2 \\dfrac{5}{8}$$ 13. $$4 \\dfrac{1}{6}$$ to $$3 \\dfrac{1}{3}$$ 14. $$5 \\dfrac{3}{5}$$ to $$3 \\dfrac{3}{5}$$ 15. $18 to$63 16. $16 to$72 17. $1.21 to$0.44 18. $1.38 to$0.69 19. 28 ounces to 84 ounces 20. 32 ounces to 128 ounces 21. 12 feet to 46 feet 22. 15 feet to 57 feet 23. 246 milligrams to 45 milligrams 24. 304", "Mohamed Robinson 2023-03-02 How many minutes are in 0.1 hour? Violet Fields Solution: An hour is $60$ minutes long, so if we multiply $60$ by $0.1$ we will get our answer: $60\\cdot 0.1=60\\cdot \\frac{1}{10}=\\frac{60}{10}=6$ Do you have a similar question?", "# James types 84 words in 2 minutes. He estimates that his history essay is 420 words long. What is the least amount of time James will need to type his essay? Mar 28, 2018 $\\text{10 min}$ #### Explanation: $\\text{84 words = 2 minutes}$ So $\\text{42 words = 1 minute}$ Then $\\text{420 words\" xx (\"1 min\")/(\"42 words\") =\"10 minutes}$ Mar 28, 2018 $10$ minutes. #### Explanation: If James can write $84$ words in $2$ minutes, then he writes $\\frac{84}{2}$ words/minute, or $42$. Since $420$ is $42 \\times 10$, he can write $420$ words in $10$ minutes.", "less than $24$, which happens in $24^2=576$ cases. If we rotate by one to the right, the ones of the minutes become the tens of the hours, so they must be between $0$ and $2$, which is the case in $18$ out of $60$ minutes. The ones of the hours become the tens of the minutes, so they must be between $0$ and $5$, which is the case in $16$ out of $24$ hours. Thus the number of valid rotations is $18\\cdot16=288$. If we rotate by one to the left, the number of valid rotations must be the same as if we rotate by one to the right, so that makes another $288$ valid rotations. Obviously if we don't rotate at all, all $24\\cdot60=1440$ times are valid, so the total is $1440+576+288+288=2592$ valid rotations out of $4\\cdot24\\cdot60=5760$, slightly less than half. Counting the size of a minimum generating set would be rather cumbersome by hand; at least I don't see an easy way to do it. That sort of thing is better left to our electronic friends – here's code that checks the above result and finds that a minimal generating set contains $999$ times. I wonder whether that's a coincidence... - 999 times? That is SO weird. Awesome answer @joriki! I ran your code (but have no", "# Difference between revisions of \"2017 AMC 8 Problems/Problem 24\" ## Problem 24 Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren? $\\textbf{(A) }78\\qquad\\textbf{(B) }80\\qquad\\textbf{(C) }144\\qquad\\textbf{(D) }146\\qquad\\textbf{(E) }152$ ## Solution 1 In $360$ days, there are $$360 \\cdot \\frac 23 \\cdot \\frac 34 \\cdot \\frac 45 = 144$$ days without calls. Note that in the last five days of the year, day $361$ and $362$ also do not have any calls, as they are not multiples of $3$, $4$, or $5$. Thus our answer is $144+2 = \\boxed{\\textbf{(D)}\\ 146}$. $$ Alternatively, there are $365\\cdot \\frac45 \\cdot \\frac 34 \\cdot \\frac23=\\boxed{146}$ days without calls. Multiplying the fractions in this order prevents partial days, as $365$ is a multiple of $5$, $365\\cdot \\frac45$ is a multiple of $4$ and $365\\cdot \\frac45 \\cdot \\frac 34$ is a multiple of $3$. ## Solution 2 We use Principle of Inclusion Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is $$\\left \\lfloor \\frac{365}{3}", "of $59$ and $720$ is $59\\cdot720=42480$. In other words, $42480$ is the first number that is present in both series $$59,\\ 118,\\ 177,\\ \\ldots$$ $$720,\\ 1440,\\ 2160,\\ \\ldots$$ And it thus takes $\\frac{42480}{60}=\\textbf{708}$ hours before they simultaneously point at 12 again. • It is striking how different this is from the minute hand sweeping out 61 minutes per hour... – Ian Dec 1 '15 at 11:44 • How is that? Again, $61$ is prime and $720$ is not divisible by $61$, so $\\mathrm{LCM}(61,720)=61\\cdot720$. – Eric S. Dec 1 '15 at 12:18 • At 61 minutes per hour you get done in 60 hours, since you get ahead by 1 minute each hour. What you just wrote is wrong because you are at 12 not every 61 minutes but rather every 3600/61 minutes. The actual numbers being so different is what struck me. – Ian Dec 1 '15 at 12:44 • Ah okay, I understand. I thought you meant the minute hand takes $61$ minutes to complete a cycle. But if it sweeps out $61$ minutes per hour, the it takes $\\frac{60\\cdot60}{61}$ minutes to complete a cycle. – Eric S. Dec 1 '15 at 12:51 • I was typing the same thing :) now that I get your point, It also strikes me, interesting! – Eric S. Dec 1", "g costs R 11,20. #### Exercise 2: Working with rates A packet of 6 handmade chocolates costs R 15,95. How much does each chocolate cost? $$\\text{R 15,95} \\div \\text{6} = \\text{R 2,668} \\ldots$$ rounded off to R 2,67 per chocolate. A truck driver travels 1500 km in 18 hours. What is his average speed? $$\\text{1500}\\text{ km} \\div$$ 18 h = 83,33$$\\ldots$$ km/h. Round off to 83,33 km/h. Nicola is able to input 96 words in 2 minutes on her laptop. Karen times her own typing speed as 314 words in 7 minutes. Work out their speeds to see who is faster. Nicola types 96 words $$\\div$$ 2 min = 48 words/min. Karen types 314 words $$\\div$$ 7 min = 44,857$$\\ldots$$ words/min. Round off to a whole word: 45 words/min. Nicola is faster. ## Finding missing numbers in ratios and rates 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. ### Example 4: Finding missing numbers in a ratio #### Question Thenji makes a fruit salad for breakfast at a restaurant. She uses pieces of fruit in the following ratio: banana : apple : paw-paw 1 : 2 : 3 1. If she uses 20 pieces of apple, how many pieces of banana", "### Home > A2C > Chapter 10 > Lesson 10.2.5 > Problem10-126 10-126. 1. Beethoven wrote nine symphonies and Mozart wrote 27 piano concertos. Homework Help ✎ 1. If the local radio station KALG wants to play two pieces, a Beethoven symphony and then a Mozart concerto, in how many ways can this be done? 2. The station manager has decided that on each successive night (seven days a week), a Beethoven symphony will be played, followed by a Mozart concerto, followed by a Schubert string quartet (there are 15 of those). How long could this policy be continued before exactly the same program would have to be repeated? $\\text{\\frac{9}{Beethoven}\\cdot \\frac{27}{Mozart}}$ = 243 ways $\\text{\\frac{9}{Beethoven}\\cdot \\frac{27}{Mozart}\\cdot \\frac{15}{Schubert}=3645\\,ways}$ $\\text{\\frac{3645\\,ways}{1}\\cdot \\frac{1\\,day}{1\\,way}\\cdot\\frac{1\\,year}{365\\,days}}$", "+0 # math +1 77 1 Henry took a test with three sections, A, B and C. He spent 1/5 of his time on Section A and 1/3 of the remaining time one Section B. He spent 48 minutes on Section C. How much time did Henry take to complete the test? Mar 27, 2022 #1 +1 Ans: $$90 \\text{ minutes.}$$ $$1-{1\\over5}={4\\over5}$$ $${4\\over5}\\cdot{1\\over3}={4\\over15}$$ $${1\\over5}+{4\\over{15}}={7\\over{15}}$$ $$1-{7\\over15}={8\\over15}$$ $${8\\over15}\\cdot\\text{total time of the test}=48$$ $$\\text{total time}=48\\cdot{15\\over8}=6\\cdot15=90\\text{ minutes, or } 1 \\text{hr. }30\\text{ minutes.}$$ Mar 27, 2022", "# Seven Seven dwarfs will cut 420 stumps in 15 hours. After five hours, the two dwarves disappear discreetly. How many hours will the remaining dwarves complete the task? Result x = 14 h #### Solution: $t=420/(7 \\cdot \\ 15)=4 \\ \\\\ \\ \\\\ \\ \\\\ 420=5 \\cdot \\ t \\cdot \\ 7+(7-2) \\cdot \\ x \\cdot \\ t \\ \\\\ \\ \\\\ 420=5 \\cdot \\ 4 \\cdot \\ 7+(7-2) \\cdot \\ x \\cdot \\ 4 \\ \\\\ \\ \\\\ 20x=280 \\ \\\\ \\ \\\\ x=14 \\ \\\\ =14 \\ \\text{h}$ 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 Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Do you want to convert time units like minutes to seconds? ## Next similar math problems: 1. Painters 4 One painter will do the job for 12 days. After a 4 1/3 day he comes second to help him and complete work together for", "# work problems (ii) It is going to show the correct time again when it has lost 12 hours. It is going to take $60 \\cdot 12 = 720 hours$", "hours. $\\text{The total time was: }\\:\\text{9 hours, 10 minutes} \\,=\\,9\\tfrac{1}{6}\\text{ hours} \\,=\\,\\tfrac{55}{6}$ hours. There is our equation . . . . $\\dfrac{2500}{x} + \\dfrac{2500}{1.2x} \\;=\\;\\dfrac{55}{6}$ . . . Go for it! 4. ## Re: \"Mixture\" and \"Distance, Speed, and Time\" Word Problems Sorry...I've been busy lately...thanks so much for your help!!!!!", "you specified it. The less we have to guess, the better. And the way I read it, they mean that as long as he is 30 years old, he makes $30\\cdot 250\\text{tk} = 7\\,500\\text{tk}$ each hour, but as soon as he turns $31$, he starts making $7\\,7500\\text{tk}$ each hour instead. – Arthur Jan 24 '18 at 8:31 Let $x_{dh}$ be Tamim's age at work day $d$ and hour $h$. \\begin{aligned} \\text{day}_1 &: 250 \\cdot (x_{11} + x_{12} + x_{13} + x_{14}) \\\\ \\text{day}_2 &: 250 \\cdot (x_{21} + x_{22} + x_{23} + x_{24}) \\\\ \\ldots & \\\\ \\text{day}_n &: 250 \\cdot (x_{n1} + x_{n2} + x_{n3} + x_{n4}) \\\\ \\ldots & \\\\ \\text{day}_{92} &: 250 \\cdot (x_{92_1} + x_{92_2} + x_{92_3} + x_{92_4}) & \\\\ \\end{aligned} The total he earned is \\begin{aligned} 250 \\cdot \\sum^ {d = 92} _{d = 1} \\left( \\sum^ {h = 4} _{h = 1} x_{dh} \\right) = 2200000 & \\rightarrow \\\\ \\sum^ {d = 92} _{d = 1} \\left( \\sum^ {h = 4} _{h = 1} x_{dh} \\right) = \\frac{2200000}{250} = 8800 & \\rightarrow \\\\ \\text{average } x_{dh} = \\frac{8800}{92 \\cdot 4} = 23.91 \\ldots \\end{aligned} and since he worked for $9 < 12$ months he could have just one birhtday in between the first and the last day of work; thus its", "The pattern repeats every seven letters, so the name ends on the $7^\\text{th}$, $14^\\text{th}$, $21^\\text{st}$, ... letters, i.e. on every multiple of $7$. Since $1000 \\div 7 = 142 \\text{ r}6$, there are $142$ complete copies of the name, and then the $1000^\\text{th}$ letter is the sixth letter of the next name. Therefore the $1000^\\text{th}$ letter is $d$.", "Heddertyk1j 2023-02-19 How many minutes are in 1.25 hours? Eva Harrington Explanation: An applied ratio of values is all that a conversion is. We know that there are 60 minutes in 1 hour $\\frac{60m}{1hr}$, so to find out how many are in 1.25 hours we arrange the ratios to give us a final answer in minutes. $\\frac{60m}{1hr}\\cdot \\left(1.25hr\\right)=75m$ Marin Valencia Solution: 1 hour =60minutes $1.25hours=\\frac{1.25}{1}\\cdot 60$ =75minutes Do you have a similar question?", "$x$ square meters of surface area that needs to be painted. $r_k =$ rate of Karen (in $m^2$/hour) $r_B =$ rate of Betty (in $m^2$/hour) $t_k =$ time it takes Karen to paint the house alone (in hours) $r_k = 6$ hours (given) The three $d=rt$ equations are $$x = r_k \\cdot t_k$$ $$x = 6 \\cdot r_B$$ $$x = \\dfrac 23 \\cdot t_k \\cdot (r_B + r_k)$$ We are trying to solve for $t_k$. Transform the second equation into $r_B = \\dfrac x6$ and substitute in for $r_B$. Distribute $\\dfrac 23 t_k$ as well: $$x = \\dfrac 23 \\cdot t_k \\cdot \\dfrac x6 + \\dfrac 23 t_k r_k$$ Substitute $r_k \\cdot t_k = x$ $$x = \\dfrac 23 \\cdot t_k \\cdot \\dfrac x6 + \\dfrac 23 x$$ Now you can cancel the $x's$ and solve for $t_k$. 2) Using similar definitions as above (and $x$ is $m^2$ of lawn now) $$x = r_S \\cdot t_S$$ $$x = (2 r_S) \\cdot t_H$$ $$x = (r_S + 2r_S) \\cdot 40$$ Setting the first and third equations equal to each other, $$r_S \\cdot t_S = (3r_S) \\cdot 40$$ Now you can cancel $r_S$ and solve for $t_S$. • How about changing the units to \"houses/hour\" and \"lawns/hour\"? No need to muck around with meters Nov 28, 2016 at 19:24 •", "checking the answer choices. Quickly checking, we know that neither choice $\\textbf{(A)}$ or choice $\\textbf{(B)}$ work, but $\\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \\text{ mph}$, David has $175$ miles left. This then takes him $3.5$ hours at $50 \\text{ mph}$. But $210/35 = 6 \\text{ hours}$. Since $1+3.5 = 4.5 \\text{ hours}$ is $1.5$ hours less than $6$, our answer is $\\boxed{\\textbf{(C)} \\: 210}$. ## Solution 3 Let us call the total number of miles after David changes speed $x$. We realize that if David is driving 50 miles an hour, and he arrives 30 minutes early, $x\\equiv 25\\pmod{50}$. The only solution that fits the description is $\\boxed{210}$.", "minutes, there are $5\\cdot 3\\cdot 3\\cdot 3$ or $5\\cdot 3^3$ bacteria. The pattern keeps repeating. Keep in mind there are four $15$ minute periods in each hour, so can you take on from here? – KM101 Oct 15 '18 at 3:21 • Thank you everyone for your help. I found KM101's approach to be most effective in explaining ,since an organized list makes it easier to visually see the pattern.He is a 5th grader and just learned about exponents so this really helped! – Pearl Oct 15 '18 at 16:32 • Once he understood the pattern , I explained the hints and method that JavaMan and Isabella suggested. Thank you again everyone! – Pearl Oct 15 '18 at 16:41" ]

[ "# James types 84 words in 2 minutes. He estimates that his history essay is 420 words long. What is the least amount of time James will need to type his essay? Mar 28, 2018 $\\text{10 min}$ #### Explanation: $\\text{84 words = 2 minutes}$ So $\\text{42 words = 1 minute}$ Then $\\text{420 words\" xx (\"1 min\")/(\"42 words\") =\"10 minutes}$ Mar 28, 2018 $10$ minutes. #### Explanation: If James can write $84$ words in $2$ minutes, then he writes $\\frac{84}{2}$ words/minute, or $42$. Since $420$ is $42 \\times 10$, he can write $420$ words in $10$ minutes.", "Question # Mohan is able to write 20 words in one hour. How many words would he be ableto write in 10 minutes? Hint: Use the unitary method to find the answer. First find the number of words he can write in 1 minute. Then multiply this by 10 to find the number of words he can write in 10 minutes. Complete step by step solution: We know that he can write 20 words in one hour. We also know that one hour consists of 60 minutes. This means that Mohan is able to write 20 words in 60 minutes. 20 words $\\to$60minutes We will now find the number of words he can write in 1 minute. For this, we will divide the known number of words with the known measure of time. So, he can write $\\dfrac{20}{60}$ i.e. $\\dfrac{1}{3}$ words in a minute. We now need to find the number of words he can write in 10 minutes. For this, we will multiply 10 and the number of words he can write in one minute. So, the number of words he can write in 10 minutes is: $10\\times \\dfrac{1}{3}=3.33$ words. Hence, Mohan can write 3.33 words in 10 minutes. Note: We can solve this question using another method also. We need to find the number of words", "$$\\dfrac{x\\; students}{3\\; teachers}$$ (c) y dollars for 18 hours Write as a rate. $$\\dfrac{y}{18\\; hours}$$ ##### Exercise $$\\PageIndex{21}$$: Translate the word phrase into an algebraic expression. (a) 689 miles per h hours (b) y parents to 22 students (c) d dollars for 9 minutes $$\\dfrac{689\\; mi}{h\\; hours}$$ $$\\dfrac{y\\; parents}{22\\; students}$$ $$\\dfrac{d}{9\\; min}$$ ##### Exercise $$\\PageIndex{22}$$: Translate the word phrase into an algebraic expression. (a) m miles per 9 hours (b) x students to 8 buses (c) y dollars for 40 hours $$\\dfrac{m\\; mi}{9\\; h}$$ $$\\dfrac{x\\; students}{8\\; buses}$$ $$\\dfrac{y}{40\\; h}$$ ## Practice Makes Perfect ### Write a Ratio as a Fraction In the following exercises, write each ratio as a fraction. 1. 20 to 36 2. 20 to 32 3. 42 to 48 4. 45 to 54 5. 49 to 21 6. 56 to 16 7. 84 to 36 8. 6.4 to 0.8 9. 0.56 to 2.8 10. 1.26 to 4.2 11. $$1 \\dfrac{2}{3}$$ to $$2 \\dfrac{5}{6}$$ 12. $$1 \\dfrac{3}{4}$$ to $$2 \\dfrac{5}{8}$$ 13. $$4 \\dfrac{1}{6}$$ to $$3 \\dfrac{1}{3}$$ 14. $$5 \\dfrac{3}{5}$$ to $$3 \\dfrac{3}{5}$$ 15. $18 to$63 16. $16 to$72 17. $1.21 to$0.44 18. $1.38 to$0.69 19. 28 ounces to 84 ounces 20. 32 ounces to 128 ounces 21. 12 feet to 46 feet 22. 15 feet to 57 feet 23. 246 milligrams to 45 milligrams 24. 304", "# Tim can type 30 words in 3/4 of a minute. Justin can type 32 words in 4/5 of a minute. How many words can Tim and Justin type altogether in 5 minutes? Apr 28, 2018 $400$ #### Explanation: First let's find how many words Tim can type in a minute and how many words Justin can type in a minute: $30 = \\frac{3}{4} t \\rightarrow t$ represents the number of words Tim can type in a minute $t = 40 \\rightarrow$ Tim can type 40 words per minute $32 = \\frac{4}{5} j \\rightarrow j$ represents the number of words Justin can type in a minute $j = 40 \\rightarrow$ Justin can type 40 words per minute In five minutes, Tim can type $40 \\cdot 5$ words and Justin can type $40 \\cdot 5$ words $40 \\cdot 5 + 40 \\cdot 5$ $200 + 200$ $400$", "have two second letters, and for each of those, we have one third letter, so we have $$3 * 2* 1 = 6$$ total ways. If you wish, you can take a few minutes to verify that there are $$4 * 3 * 2 * 1 = 24$$ ways to arrange the letters of \"MATH\" into different orders. Similarly, there are only $$2 * 1 = 2$$ ways to arrange two different letters (AB and BA, for example). So if we extend the pattern, it turns out that there are $$15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1$$ ways to arrange the letters of \"UNCOPYRIGHTABLE.\" That is a total of $$1,307,674,368,000$$ ways! Now, I said I'd verify that we'd be here awhile if you tried to list all of the ways. If it took you five seconds to write down each possibility, and you did this all day and all night, never stopping to eat or sleep or do anything else, you'd be here for $$1,307,674,368,000 * 5 = 6,538,371,840,000$$ seconds. There are 86,400 seconds in a day, so you'd be here for $$6,538,371,840,000/86,400 = 75,675,600$$ days, which, at 365 days per year, would keep you here", "+0 # math +1 77 1 Henry took a test with three sections, A, B and C. He spent 1/5 of his time on Section A and 1/3 of the remaining time one Section B. He spent 48 minutes on Section C. How much time did Henry take to complete the test? Mar 27, 2022 #1 +1 Ans: $$90 \\text{ minutes.}$$ $$1-{1\\over5}={4\\over5}$$ $${4\\over5}\\cdot{1\\over3}={4\\over15}$$ $${1\\over5}+{4\\over{15}}={7\\over{15}}$$ $$1-{7\\over15}={8\\over15}$$ $${8\\over15}\\cdot\\text{total time of the test}=48$$ $$\\text{total time}=48\\cdot{15\\over8}=6\\cdot15=90\\text{ minutes, or } 1 \\text{hr. }30\\text{ minutes.}$$ Mar 27, 2022", "+ $$\\text{1}$$ hour = 5:10 p.m. Then we add the minutes: 5:10 p.m. + $$\\text{20}$$ minutes = 5:30 p.m. So Russel finishes his homework at 5:30 p.m. 2. To convert our answer to the 24-hour format we simply add 12 hours: 5:30 p.m. + $$\\text{12}$$ hours = 17:30. 1. 15:10 + $$\\text{1}$$ hour = 16:10. 16:10 + $$\\text{20}$$ minutes = 16:30. $$\\text{1}$$ hour + $$\\text{20}$$ minutes = $$\\text{1}$$ hour and $$\\text{20}$$ minutes. So Ewald's hockey practice was $$\\text{1}$$ hour and $$\\text{20}$$ minutes long. 2. We can divide the $$\\text{40}$$ minutes it takes Ewald to get home into $$\\text{30}$$ + $$\\text{10}$$ minutes to make it easier to add: 16:30 + $$\\text{30}$$ minutes = 17:00 17:00 + $$\\text{10}$$ minutes = 17:10 So Ewald gets home at 17:10. To convert this to the 12-hour format we subtract 12 hours: 17:10 - $$\\text{12}$$ hours = 5:10. We know that 17:10 is after midday so the converted time is 5:10 p.m. ## Calculating elapsed time Exercise 3.2 Unathi's father goes to work at 8:00 a.m. He fetches her from school $$\\text{7}$$ hours and $$\\text{30}$$ minutes later. What time will he fetch Unathi? Give your answer in the 24-hour format. 15:30 Lauren finishes her music class at 15:30. It takes her $$\\text{30}$$ minutes to get home. She then does homework for $$\\text{50}$$ minutes. Lauren", "# Five pupils Five pupils clean 30 chairs one hour before four pupils. How many chairs clean one pupil in 1 hour? Correct result: n = 1.5 #### Solution: $5n(t-1)=30 \\ \\\\ 4nt=30 \\ \\\\ \\ \\\\ 5nt - 5n=30 \\ \\\\ 4nt=30 \\ \\\\ \\ \\\\ \\ \\\\ 5 \\cdot \\ 30/4 - 5n=30 \\ \\\\ \\ \\\\ 5 \\cdot \\ 30/4 - 5 \\cdot \\ n=30 \\ \\\\ \\ \\\\ 20n=30 \\ \\\\ \\ \\\\ n=\\dfrac{ 3 }{ 2 }=1.5t=30/(4 \\cdot \\ n)=30/(4 \\cdot \\ 1.5)=5=\\dfrac{ 3 }{ 2 }=1.5$ 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 Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Do you want to convert time units like minutes to seconds? ## Next similar math problems: • Seven Seven dwarfs will cut 420 stumps in 15 hours. After five hours, the two dwarves disappear discreetly. How many hours will the remaining dwarves complete", "g costs R 11,20. #### Exercise 2: Working with rates A packet of 6 handmade chocolates costs R 15,95. How much does each chocolate cost? $$\\text{R 15,95} \\div \\text{6} = \\text{R 2,668} \\ldots$$ rounded off to R 2,67 per chocolate. A truck driver travels 1500 km in 18 hours. What is his average speed? $$\\text{1500}\\text{ km} \\div$$ 18 h = 83,33$$\\ldots$$ km/h. Round off to 83,33 km/h. Nicola is able to input 96 words in 2 minutes on her laptop. Karen times her own typing speed as 314 words in 7 minutes. Work out their speeds to see who is faster. Nicola types 96 words $$\\div$$ 2 min = 48 words/min. Karen types 314 words $$\\div$$ 7 min = 44,857$$\\ldots$$ words/min. Round off to a whole word: 45 words/min. Nicola is faster. ## Finding missing numbers in ratios and rates 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. ### Example 4: Finding missing numbers in a ratio #### Question Thenji makes a fruit salad for breakfast at a restaurant. She uses pieces of fruit in the following ratio: banana : apple : paw-paw 1 : 2 : 3 1. If she uses 20 pieces of apple, how many pieces of banana", "# Ben can type 153 words in 3 minutes. At this rate, how many words can he type in 10 minutes? ##### 1 Answer May 5, 2018 $\\textcolor{b l u e}{510}$ #### Explanation: If: $153 \\text{ words\"= 3 \" minutes}$: Then: $1 \\text{minute\"=153/3=51 \" words}$ 10 minutes is: $10 \\times 51 = 510$ Ben can type 510 words in 10 minutes.", "can type nearly 40 words per minute. Use this information to find the number of hours it would take him to type 2.6 × 105 words. Type below: ______________ Answer: Samuel can type 40 words per minute. Then how many hours will it take for him to type 2.6 words times 10 to the power of five words 2.6 words time 10 to the power of 5 2.6 × (10)4 2.6 x 100 000 = 260 000 words in all. Now, we need to find the number of words Samuel can type in an hour 40 words/minutes, in 1 hour there are 60 minutes 40 x 60 2,400 words /hour Now, let’s divide the total of words he needs to type to the number of words he can type in an hour 260 000 / 2 400 108.33 hours. Question 24. Entomology A tropical species of mite named Archegozetes longisetosus is the record holder for the strongest insect in the world. It can lift up to 1.182 × 103 times its own weight. a. If you were as strong as this insect, explain how you could find how many pounds you could lift. Type below: ______________ Answer: Number of pounds you can lift by 1.182 × 103 by your weight Question 24. b. Complete the calculation to find", "\\\\[-3pt] \\,1 \\end{array}$ $2 \\frac{1}{3}$ falls between $2$ and $3$ on the number line. Question 3 ### Tavon's flight is 270 minutes long. How many hours does the flight last? A $4 \\text{ hours}$ B $4 \\frac{1}{2} \\text{ hours}$ C $5 \\text{ hours}$ D $5 \\frac{1}{2} \\text{ hours}$ Question 3 Explanation: $\\require{cancel} 270 \\cancel{\\text{minutes}} \\cdot \\dfrac{1 \\text{ hour}}{60 \\cancel{\\text{minutes}}}$ $= \\dfrac{270}{60} \\text{hours}$ $= \\dfrac{27}{6} \\text{hours}$ $\\dfrac{27}{6} = \\dfrac{9}{2} = 4 \\dfrac{1}{2}$ Question 4 ### In a factory there are two separate containers of stress balls. In the first container are 80 balls, and in the second are 90 balls. 40% of the balls in the first container are defective, and 20% in the second container are defective. In total, how many balls in the two containers are defective? A $102$ B $50$ C $60$ D $51$ Question 4 Explanation: The correct answer is (B). Calculate the number of defective balls in each container and then add these two amounts: $40\\% \\text{ of } 80$ $= 0.40 \\cdot 80 = 32$ $20\\% \\text{ of } 90$ $= 0.20 \\cdot 90 = 18$ $32 + 18 = 50$ Question 5 ### Ayana wants to buy as many bags of mulch as possible with her \\$305, and she would like them to be delivered to her house. The cost is \\$8.00", "of $59$ and $720$ is $59\\cdot720=42480$. In other words, $42480$ is the first number that is present in both series $$59,\\ 118,\\ 177,\\ \\ldots$$ $$720,\\ 1440,\\ 2160,\\ \\ldots$$ And it thus takes $\\frac{42480}{60}=\\textbf{708}$ hours before they simultaneously point at 12 again. • It is striking how different this is from the minute hand sweeping out 61 minutes per hour... – Ian Dec 1 '15 at 11:44 • How is that? Again, $61$ is prime and $720$ is not divisible by $61$, so $\\mathrm{LCM}(61,720)=61\\cdot720$. – Eric S. Dec 1 '15 at 12:18 • At 61 minutes per hour you get done in 60 hours, since you get ahead by 1 minute each hour. What you just wrote is wrong because you are at 12 not every 61 minutes but rather every 3600/61 minutes. The actual numbers being so different is what struck me. – Ian Dec 1 '15 at 12:44 • Ah okay, I understand. I thought you meant the minute hand takes $61$ minutes to complete a cycle. But if it sweeps out $61$ minutes per hour, the it takes $\\frac{60\\cdot60}{61}$ minutes to complete a cycle. – Eric S. Dec 1 '15 at 12:51 • I was typing the same thing :) now that I get your point, It also strikes me, interesting! – Eric S. Dec 1", "# Moderate Time, Speed & Distance Solved QuestionAptitude Discussion Q. Two boys begin together to write out a booklet containing 535 lines. The first boy starts with the first line, writing at the rate of 100 lines an hour; and the second starts with the last line then writes line 534 and so on, backward proceeding at the rate of 50 lines an hour. At what line will they meet? ✖ A. 356 ✖ B. 277 ✔ C. 357 ✖ D. 267 Solution: Option(C) is correct Writing ratio = $100:50=2:1$ Since equal quantities are taken, Therefore in a given time, first boy will be writing the line number $\\dfrac{2}{3}\\times 535\\text{ or }357^{th}$ line Hence, both of them shall meet on $357^{th}$ line", "neighbourhood boundaries.) • The bonus needed is lower than for letters with only the city. (At least before the latest changes, need to confirm.) ### Deciphering The address of the letters with terrible handwriting has been quite obfuscated. • They can be deciphered with the command decipher handwriting on letter. • This is a skill check that uses crafts.arts.calligraphy, with a bonus somewhere greater than ~200 or so, and costs 40 GP. ## Rewards Method Abbreviation Mission XP reward Monetary reward Time given Missing letters DLA Close by; smudged address 12800 XP &&&&&&&&&&&+2600 A$6.50 30 DW min / 9 RW min Deciphering DLB Close by; terrible handwriting 25600 XP &&&&&&&&&&&+5200 A$13 30 DW min / 9 RW min Inquiring DLC Close by; vague address 24128 XP &&&&&&&&&&&+4800 A$12 30 DW min / 9 RW min Inquiring DLD Close by; only the city 47823 XP &&&&&&&&&&+10000 A$25 45 DW min / 13.5 RW min Missing letters DLE Far away; smudged address 25600 XP &&&&&&&&&&&+4500 A$11.25 30 DW min / 9 RW min Deciphering DLF Far away; terrible handwriting 27456 XP &&&&&&&&&&&+6000 A$15 &&&&&&&&&&&+5472 LC 3|2|- 30 DW min / 9 RW min Inquiring DLG Far away; vague address 33440 XP &&&&&&&&&&&+7200 A$18 30 DW min / 9 RW min Inquiring DLH Far away; only the city 50064 XP &&&&&&&&&&+10080", "he formed groups of seven, he divided everyone. How many toy cars have in the collection? • Five pupils Five pupils clean 30 chairs one hour before four pupils. How many chairs clean one pupil in 1 hour? • 40% volume 40% volume with 104 uph (units per labor hour) 8 people working. What is the volume? • Two workers One worker needs 40 hours to do a job, and the second would do it in 30 hours. They worked together for several hours, then the second was recalled, and the first completed the job itself in 5 hours. How many hours did they work together, and how much did • Midnight How many hours are, if the time that elapsed since 8:00 is 2/5 of the time that will past till midnight? • Two typists There are two typists who are rewriting the material 814 pages. First can it handle rewrite yourself for 24 days; the second 12 days. First typist wrote material yourself 4 days rest rewrites yourself second typist. How many days will it take rewriting al • Suzan My daughter reaches X years in X squared, said Suzan father in 2009 during the celebration of her birthday. When will given fact become reality and how many years at that time will have Suzan? •", "A. 12 B. 15 C. 30 D. 40 30- Solve for $$4x^2 -34 = 66$$ A. $$±$$ 3 B. $$±$$ 5 C. $$±$$ 9 D. $$±$$ 10 Use the following table to answer question below. 31- This table shows the data Daniel collects while watching birds for one week. How many raptors did Daniel see on Monday? A. 5 B. 8 C. 12 D. 13 32- A floppy disk shows 856,036 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. 114,857 B. 123.986 C. 221,867 D. 302,209 33- 6 days 14 hours 42 minutes – 4 days 12 hours 35 minutes =? A. 2 days 1 hours 10 minutes B. 2 days 2 hours 7 minutes C. 2 days 2 hours 13 minutes D. 2 days 7 hours 23 minutes 34- If $$15+x^{\\frac{1}{2}}=24$$, then what is the value of $$8 × x$$? A. 321 B. 456 C. 550 D. 648 35- Increased by $$35\\%$$, the numbers $$60$$ becomes: A. 42 B. 56 C. 75 D. 81 36- Which equation represents the statement twice the difference between 8 times h and 4 gives 42. A. $$\\frac{8H + 4}{2} = 42$$ B. $$2(8H + 4)", "# Joey solves math problems at a rate of 3 problems every 7 minutes. If he continues work at the same rate, how long will it take Joey to solve 45 problems? Mar 10, 2018 #### Answer: $105$ minutes #### Explanation: Well, he can solve $3$ problems in $7$ minutes. Let $x$ be the time he has to solve $45$ problems. Then, we got $\\frac{3 \\setminus \\text{problems\")/(7 \\ \"minutes\")=(45 \\ \"problems}}{x}$ :.x=(45color(red)cancelcolor(black)\"problems\")/(3color(red)cancelcolor(black)\"problems\")*7 \\ \"minutes\" $= 15 \\cdot 7 \\setminus \\text{minutes}$ $= 105 \\setminus \\text{minutes}$", "by visiting: https://m.me/join/AbYSr9-6f_JuVBfH . (tagsToTranslate) earn money (t) win (t) payment ## Verification of proof that a form is a parallelogram. So I tackled the following problem: And here is my job: However, I'm not sure that my proof is complete, because I think the shape could still be a rhombus? So I'm just wondering if the evidence is satisfactory for part a. ## Help with gcd proof and modular arithmetic Let $$a, b, n Z$$ with $$n> 0$$ and $$a equiv b pmod n$$. CA watch $$gcd (a, n) = gcd (b, n) = d$$. I tried to rewrite $$b = a + ny$$ for some people $$y in Z$$. Now if I have $$d mid a$$ and $$d mid n$$then $$d mid b$$. However, I do not know how to continue. ## [FREE] Earn $700 more per day for 20 minutes. – Everyone can do it! – Real proof! | NewProxyLists [FREE] Earn$ 700 more per day for 20 minutes. – Everyone can do it! – Real proof! Request page WSO Page http://www.warriorforum.com/warrior-special-offers-forum/248019-earn-extra-700-day-20-minutes-work-any-can-do-real-warrior-proof-over-1800- sold-35.html Content not hidden: http://www.mediafire.com/?gfjf5gok4dk13a7", "increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?$\\textbf{(A) }10\\qquad\\textbf{(B) }15\\qquad\\textbf{(C) }25\\qquad\\textbf{(D) }50\\qquad\\textbf{(E) }82$AMC 8, 2017, Problem 24 Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December$31,2016 .$On how many days during the next year did she not receive a phone call from any of her grandchildren? (A)$78$(B)$80$(C)$144$(D)$146$(E)$152$AMC 8, 2016, Problem 1 The longest professional tennis match lasted a total of 11 hours and 5 minutes. How many minutes was that?$\\textbf{(A)} 605 \\qquad\\textbf{(B)} 655\\qquad\\textbf{(C)} 665\\qquad\\textbf{(D)} 1005\\qquad \\textbf{(E)} 1105$AMC 8, 2016, Problem 4 When Cheenu was a boy he could run$15$miles in$3$hours and$30$minutes. As an old man he can now walk$10$miles in$4$hours. How many minutes longer does it take for him to travel a mile now compared to when he was a boy?$\\textbf{(A) }6\\qquad\\textbf{(B) }10\\qquad\\textbf{(C) }15\\qquad\\textbf{(D) }18\\qquad \\textbf{(E) }30$AMC 8, 2016, Problem 5 The number$N$is a two-digit number. • When$N$is divided by$9$, the remainder is$1$. • When$N$is divided by$10$, the remainder is$3$. What is the remainder when$N$is divided by$11$?$\\textbf{(A) }0\\qquad\\textbf{(B) }2\\qquad\\textbf{(C) }4\\qquad\\textbf{(D) }5\\qquad \\textbf{(E) }7$AMC 8, 2016, Problem 7" ]

[ "number 100 more than 562 is 662 Question 3. Describe 58 as a sum of tens and ones. ______ Answer: 58 can be shown in sum of tens and once as 50 + 8 = 58 Question 4. Pete helps his grandmother gather pecans. He finds 6 pecans on his left and 3 on his right. How many pecans did Pete find altogether? 6 + 3 = _____ pecans The number of pecans Pete finds on his left side = 6 The number of Pecans Pete finds on his right side = 3 Total number of Pecans Pete find altogether = 6 + 3 = 9 Therefore, total number of pecans = 9 Question 5. What number do the blocks show? ______ The first and second figures have 100 blocks The 3 single line contains 10+10+10 = 30 blocks and 4 blocks Total number of blocks = 100+100+10+10+10+1+1+1+1 = 234 Therefore, Total number of blocks = 234 ### Lesson 7.4 Show Amounts in Two Ways Essential Question How do you choose coins to show a money amount in different ways? Listen and Draw Show the amount with coins. Draw the coins. Write the amount. 1 quarter = 25 cents and 1 penny = 1 cent Total value = 26¢ Another way: 26¢ can also be shown as 2", "adds 4 which makes 12 adds 4 again (16) adds 8 (24) adds 8 adds 16, and so on. So it adds one thing twice then doubles the adding number and starts to do that again. Amelia from NLCS sent in her ideas; First double forward then double sideways and so on", "Not Reviewed \"Pete gave\" = Tags: Rating ID MichaelBartmess.How Many Added? UUID 7fbeef10-4439-11e3-a47b-bc764e049c3d This equation helps you find how many were added. Jim had some colored pencils. His best friend Pete gave Jim some more pencils and Jim now has a whole box full of colored pencils. You tell vCalc how many pencils Jim had to start and how many he has in his box after Pete gives him more. Then vCalc will tell you how many colored pencils Pete gave to Jim.", "the numbers keep doubling (forever)", "a multiplier you have to multiply by 100.", "# 200+ followers (hard version) Calvin thinks of a 5-digit number. The product of the digits is $$n$$. There are exactly 200 different numbers which Calvin could have thought of. Find the sum of all possible values of $$n$$. ×", "how many .", "wondered whether counting on was Luke's only strategy, or whether he had others to draw on. So, I told him that my new secret number added to $10$ was $20$, and showed him the calculator display. This time, Luke knew that $10$ add $10$ is $20$ without any mental calculation, it was a fact that he was familiar with, but he did check the sum on his calculator once again. I invited Luke to think of a secret number and to see whether I could work out what it was. This seemed to appeal to him, so after a short time he told me that he'd put his number into the calculator, then pressed add $100$ (checking with me that he had entered $100$ correctly), then equals. He showed me the result of this addition, which was $101$. I knew Luke was very familiar with numbers up to $100$ - he'd told me that on the way to see me, he and his Dad had counted all the way from $1$ to $100$ and then in twos up to $100$ - but he wasn't sure how to read the number on the display. I read out \"one hundred and one\", explaining a little by breaking it up into the $100$ and the $1$, pointing to the display at", "to 100? Try using different starting numbers. Can you find a strategy for splitting numbers so that you always get the largest product?", "# 200+ followers (hard version) Calvin thinks of a 5-digit number. He tells you the product of the digits is $$n$$. Knowing only that, there are exactly 200 different numbers which Calvin could have thought of. Find the sum of all possible values of $$n$$. ×", "× 1 For a three digit number say ABC = A × 100 + B × 10 + C × 1 = 100 A + 10 B + C For example, 123 = 100 + 20 + 3 = 1 × 100 + 2 × 10 + 3 × 1 ### Playing with numbers Numbers are quite amazing to play with and we can enjoy several tricks with them. We can reverse the digits and play. For example: Steve and Mike sat down and Steve asked Mike to think of a two digit number. Mike did as he was told. Then Steve asked Mike to reverse the digits and add it to the original number. Then divide this number by 11. Mike followed. Steve concluded that after the division, there will be no remainder. Let us suppose that Mike thought of the number 93. Reversing the digit we get 39. Adding them we obtain 132. Dividing by 11 we get the answer as 12. And there is no remainder. This happens because when we reverse the digits and add them, the number becomes a multiple of 11 and hence no remainder is left when the number is divided by 11: 10 a + b is the number taken. Reversing the digits we get 10 b + a. Adding", "Zaben's Imaginative Math Zaben Oldwheel was always a kid with big dreams and imagination. However since Zaben spent most of his time daydreaming, he never learnt about the number 3. Every time he would reach a number with 3 in it, he would simply skip to the next one. He starts off by counting $1, 2, 4, 5, \\ldots$ Now, if Zaben Oldwheel were to count from 1 to 100, how many terms would there be? (Hint: Remember in Zaben's mind any number with 3 doesn't exist.) ×", "answer, on his desk. While his fellow students sat scribbling things like 1+2 = 3 3+3 = 6 6+4 = 10 with the inevitable mistake 10+5 = 14 and running out of space to write in, Gauss thought for a moment, chalked one number on his slate, walked up to the teacher, and slapped the slate face down on the desk. \"There it lies,\" he said, went back to his desk, and sat down. At the end of the lesson, when the teacher collected all the slates, precisely one had the correct answer: Gauss's. Again, we don't know exactly how Gauss's mind worked, but we can come up with a plausible reconstruction. In all likelihood, Gauss had already thought about sums of that kind and had spotted a useful trick. (If not, he was entirely capable of inventing one on the spot.) An easy way to find the answer is to group the numbers in pairs: 1 and 100, 2 and 99, 3 and 98, and so on, all the way to 50 and 51. Every number from 1 to 100 occurs exactly once in some pair, so the sum of all those numbers is the sum of all the pairs. But each pair adds up to 101. And there are 50 pairs. So the total is 50", "squares. How would they draw 100?", "A multiplication version of the game might go like this: Charlie puts in a secret number and asks Dana, \"What do you want to multiply it by?\" Dana replies, \"Multiply it by $5$.\" Charlie puts in 'times' and '$5$' and hands the calculator to Dana. When Dana presses the 'equals' button the calculator shows '$35$'. Dana now has to work out Charlie's secret number. What was it? How do you know? Try playing this version with your friend too!", "### Home > MC1 > Chapter 1 > Lesson 1.3.3 > Problem1-126 1-126. Richard doubled a whole number and added two. His answer was a whole number less than $20$. What numbers could Richard have used? Remember that to ''double'' something means to multiply by $2$. Any numbers between $0$ and $8$ could have been the one Richard used.", "one try, two tries, three, four, or five tries all summed up. Does that sound right? Dick", "by itself seven times — he’ll come up with the answer, 10,460,353,203, in a few seconds. Tammet visualizes numbers in their unique forms and then melds them together to create a new image for the solution. When asked to multiply 53 by 131, he explains the solution in shapes and textures: “Fifty-three, which is round, very round…and larger at the bottom. Then you’ve got another number 131, which is longer a little bit like an hourglass. And there’s a space that’s created in between. That shape is the solution. 6,943!” Perhaps a better understanding of this phenomena will shed some light on different ways of doing mathematics.", "Who can tell me how many of these numbers are divisible by two? Teacher: Anybody? Teacher: Fred! Fred: Hm? Teacher: I know that you can tell me how many of these numbers are divisible by two. Fred: Um... All of 'em. When you change the way you look at things, the things you look at change Wayne Dyer", "came up with the sums for 49 and 1/49 and who noticed the pattern of adding to 1. Credit for the photo goes to Arjan Dice; it’s published here on Wikipedia." ]

[ "$x^2 + x^2 = \\left ( 10 \\sqrt{2} \\right )^2\\!\\\\{\\;} \\quad \\ 2x^2 = 200\\!\\\\{\\;} \\qquad x^2 = 100\\!\\\\{\\;} \\qquad \\ x = 10$ 4. Yes, $9^2 + 12^2 = 15^2 \\rightarrow 81+144 = 225$ 5. Yes, $3^2 + \\left( 3 \\sqrt{3} \\right )^2 = 6^2 \\rightarrow 9+27 = 36$ 8 , 9 , 10 Feb 22, 2012 Dec 11, 2014", "x-9 \\right)$ Substitute the value of x from equation (3), we get: \\begin{align} & \\Rightarrow 4\\left( -6-9 \\right) \\\\ & \\Rightarrow 4\\times -15 \\\\ & \\Rightarrow -60 \\\\ \\end{align} So, the correct answer is “Option A”. Note: Be careful while reading the question. The language might confuse you. As in given in the question, multiply a number by 3 and then add 40. The equation for this is $3x+40$ But some might make a mistake here as $3\\left( x+40 \\right)$", "4 x-1 =0 \\\\ 4 x &=1 \\!\\qquad \\qquad \\qquad 4 x =1 \\\\ x &={1 \\over 4} \\qquad \\qquad \\quad \\; x ={1 \\over 4}\\end{align} Roots are \\(\\begin{align}{1 \\over 4},\\,\\,\\,\\,{1\\over 4}.\\end{align} (v) \\begin{align}\\quad100x{}^\\text{2}-20x+1\\end{align} Steps: \\begin{align} 100 x^{2}-20 x+1 &=0 \\\\ 100 x^{2}-10 x-10 x+1 &=0 \\\\ 10 x(10 x-1)-1(10 x-1) &=0 \\\\ (10 x-1)(10 x-1) &=0 \\end{align} \\begin{align} 10 x-1 &=0 \\qquad \\qquad10 x-1=0 \\\\ 10 x &=1\\qquad \\qquad\\qquad \\!10 x=1 \\\\ x &=\\frac{1}{10} \\qquad \\qquad\\quad x=\\frac{1 }{10} \\end{align} Roots are \\begin{align} \\frac{1}{10},\\, \\frac{1}{10} \\,\\,.\\end{align} Question 2 Solve the problems given in example $$1.$$ (i) John and Jivanti had $$45$$ marbles. Both of them lost $$5$$ marbles each and the product of the no. of marbles they now have is $$124.$$ We would like to find out how many marbles they had to start with? (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $$55$$ minus the number of toys produced in a day. On a particular day, the total cost of production was ₹$$750.$$ We would like to find out the number of toys produced on that day. Solution (i) What is known? (a) John and Jivanti together had $$45$$ marbles (b) Both of them lost $$5$$ marbles each.", "$797$, and their difference is $10$. So we need $a$ to be at least $10$ and such that its sum with $787$ or $797$ is $800$. This clearly suggests the following choice: $$a=13, \\qquad b=787, \\qquad c=797.$$ $$a+b-c=3, \\qquad a+c-b=23, \\qquad b+c-a=1571.$$ Again using the list above you can check that we have been lucky and those are all distinct primes. Then we have actually computed (not only bounded) the maximum vale $d$, which is: $d = b+c-a -a = 1571-13= 1558.$", "$108.$ This borrowed $1$ becomes $100$ (because base is $100$) and our right side becomes $100+(-48)=52.$ Since we borrowed one from left side, left side becomes $107$ So answer is $10752$ Example $3$ Calculate $103×115$ Solution Use the same method. Let's select $100$ as the base (close to both the numbers). Subtract base from these numbers. That is $103-100=3$ and $115-100=15.$ So write it as $103\\qquad 3\\\\115\\qquad 15$ Take the diagonal sum to get the left side of the product. That is $103+15=118$ (Or $115+3=118$). For the right side, find the product of the differences. That is $3×15=45$ So answer is $11845$ Example $4$ Calculate $122×89$ Solution Use the same method. Let's select $100$ as the base (close to both the numbers). Subtract base from these numbers. That is $122-100=22$ and $89-100=-11.$ So write it as $122\\qquad 22\\\\89\\qquad -11$ Take the diagonal sum to get the left side of the product. That is $122+(-11)=111$ (Or $89+22=111$). For the right side, find the product of the differences. That is $22×(-11)=-242.$ Since it is -ve, we need to make it as +ve. For this borrow $3$ from our left side $111.$ This borrowed $3$ becomes $300$ (because base is $100$ and $3×100=300$) and our right side becomes $300+(-242)=58$ Since we borrowed $3$ from left side, left side becomes $108$ So", "Question # Complete the problems. Check your answers in the back of the book.$980 \\mathrm{~cm}=\\underline{\\qquad}$ $\\mathrm{hm}$ Solution Verified \\begin{aligned} 980cm&=0.0980hm \\end{aligned}", "are $12, then we can divide the total amount of money collected by the price per individual ticket. $(\\text{number of people}) = \\frac{15552}{12} = 1296$ 1296 is indeed the number of people who entered the park. The answer checks out. Example 2 The following table shows the relationship between two quantities. First, write an equation that describes the relationship. Then, find out the value of $b$ when $a$ is 750. $& a \\qquad 0 \\qquad \\ 10 \\qquad 20 \\qquad 30 \\qquad 40 \\ \\ \\qquad 50\\\\& b \\qquad 20 \\qquad 40 \\qquad 60 \\qquad 80 \\qquad 100 \\qquad 120$ Step 1: Extract the important information. We can see from the table that every time $a$ increases by 10, $b$ increases by 20. However, $b$ is not simply twice the value of $a$. We can see that when $a = 0, \\ b = 20$, and this gives a clue as to what rule the pattern follows. The rule linking $a$ and $b$ is: “To find $b$, double the value of $a$ and add 20.” Step 2: Translate into a mathematical equation: Text Translates to Mathematical Expression “To find $b$ $\\rightarrow$ $b =$ “double the value of $a$ $\\rightarrow$ $2a$ “add 20” $\\rightarrow$ + 20 Our equation is $b = 2a + 20.$ Step 3: Solve the equation.", "triangle to the balls from the bottom layer. We can fill the first three rows of the table. 112336 46+4=10\\begin{align*}&1 && 2 && 3 && \\quad \\ \\ 4\\\\ &1 && 3 && 6 && 6 + 4 = 10\\end{align*} To find the number of tennis balls in 8 layers, continue the pattern. 510+5=15 615+6=21 721+7=28 828+8=36\\begin{align*}& \\qquad \\ 5 && \\qquad \\ 6 && \\qquad \\ 7 && \\qquad \\ 8\\\\ &10+5=15 && 15+6=21 && 21+7=28 && 28+8=36\\end{align*} There will be 36 tennis balls in the 8 layers. Check: Each layer of the triangle has one more ball than the previous one. In a triangle with 8 layers, each layer has the same number of balls as its position. When we add these we get: 1+2+3+4+5+6+7+8=36\\begin{align*}1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36\\end{align*} balls #### Now, let's compare the two methods, making a table and using a pattern, in solving the following problem: Andrew cashes a $180 check and wants the money in$10 and 20 bills. The bank teller gives him 12 bills. How many of each kind of bill does he receive? Method 1: Making a Table TensTwenties0928476685104123142161180\\begin{align*}&\\text{Tens} && 0 && 2 && 4 && 6 && 8 && 10 && 12 && 14 && 16 &&", "be the maximum number of squares Peter can move a token beyond the dividing line. For example, $F(0)=4$. You are also given that $F(1)=6$, $F(2)=9$, $F(3)=13$, $F(11)=58$ and $F(123)=1173$. Find $F(1234567)$.", "\\\\ & \\Rightarrow x=\\dfrac{424\\pm 224}{18} \\\\ & \\Rightarrow x=36,\\dfrac{100}{9} \\\\ \\end{align} Since $x$ represents the number of camels, it must be an integer. So, $x=36$. Hence, the total number of camels is equal to $36$. Note: There is an alternative way to solve the quadratic equation $8\\sqrt{x}=3x-60$. We can assume a variable $p$ and then we can substitute $x={{p}^{2}}$ in this quadratic equation and then solve for $p$ using a quadratic formula. Later, we can re-substitute ${{p}^{2}}=x$ or $p=\\sqrt{x}$ and then solve to get $x$.", "equations simultaneously, we can find x and y. Multiplying equation $\\left( 1 \\right)$ by 3 and equation $\\left( 2 \\right)$ by 7, we get, $21x+18y=11400............\\left( 3 \\right)$ $21x+35y=12250............\\left( 4 \\right)$ Subtracting the above two equations, we get, \\begin{align} & \\left( 21x+35y \\right)-\\left( 21x+18y \\right)=12250-11400 \\\\ & \\Rightarrow 17y=850 \\\\ & \\Rightarrow y=50 \\\\ \\end{align} Substituting y=50 in equation $\\left( 1 \\right)$, we get, \\begin{align} & 7x+6\\left( 50 \\right)=3800 \\\\ & \\Rightarrow 7x+300=3800 \\\\ & \\Rightarrow 7x=3500 \\\\ & \\Rightarrow x=500 \\\\ \\end{align} Hence, the cost of each bat is equal to Rs. 500 and each ball is equal to Rs. 50. Note: To check whether we have got the correct answer or not, just substitute the obtained values of x and y in any of the equation $\\left( 1 \\right)$ and equation $\\left( 2 \\right)$. If the obtained value of x and y satisfies the equation, we have obtained a correct value of x and y.", "of the Quadr. Eqn.$: {x}^{2} + p x + q = 0.$ Therefore, $x = 2 + 3 i$ must satisfy the eqn. $\\therefore {\\left(2 + 3 i\\right)}^{2} + p \\left(2 + 3 i\\right) + q = 0.$ $\\therefore 4 + 12 i + 9 {i}^{2} + 2 p + 3 i p + q = 0.$ $\\therefore 4 + 12 i - 9 + 2 p + 3 i p + q = 0.$ $\\therefore \\left(2 p + q - 5\\right) + i \\left(12 + 3 p\\right) = 0.$ Comparing the Real and Imaginary Parts of both sides, we get, $2 p + q = 5. \\ldots . . \\left(1\\right) , \\mathmr{and} , 12 + 3 p = 0$ rArr p=-4, &, q=5-2p=5-2(-4)=13, as Respected Steve Sir, has alreay derived. Enjoy Maths.!", "I'm trying to find $x$... – user61406 Jun 2 '13 at 7:46 Everything you did was correct! You have applied the distributive law: $$a(b+c)=ab+ac,$$ and also used that $$a-b=a+(-b)$$ (subtracting a number is the same as adding its negative). Your steps are \\begin{align*} 8(12x + 15) + 13(2x - 11) &= 123\\\\\\\\ 96x + 120 + 26 x - 143 &=123\\\\\\\\ 122x - 23&=123 \\end{align*} Our goal is ultimately to get $x$ all by itself, so how can we remove the extra stuff on the left side? Remember, we can do anything we want as long as we do it to both sides of the equation, so the first thing to do would be to add $23$ to both sides, because this will cancel out the negative $23$ that's currently on the left side. Do you see how do deal with the $122$? - $122x = 146$. I know that you devide but if I devide the two numbers together it doesn't make the right number. – user61406 Jun 1 '13 at 8:06 @HonkyHanka: What are you using to determine what the \"right\" number is? If there's a given answer, it could be wrong, or perhaps you've copied the question down incorrectly. – Zev Chonoles Jun 1 '13 at 8:11 Well i'm trying to find the $x$ so", "taken at the ticket office and tickets are$12, then we can divide the total amount of money collected by the price per individual ticket. $(\\text{number of people}) = \\frac{15552}{12} = 1296$ 1296 is indeed the number of people who entered the park. The answer checks out. #### Example B The following table shows the relationship between two quantities. First, write an equation that describes the relationship. Then, find out the value of $b$ when $a$ is 750. $& a \\qquad 0 \\qquad \\ 10 \\qquad 20 \\qquad 30 \\qquad 40 \\ \\ \\qquad 50\\\\& b \\qquad 20 \\qquad 40 \\qquad 60 \\qquad 80 \\qquad 100 \\qquad 120$ Step 1: Extract the important information. We can see from the table that every time $a$ increases by 10, $b$ increases by 20. However, $b$ is not simply twice the value of $a$ . We can see that when $a = 0, \\ b = 20$ , and this gives a clue as to what rule the pattern follows. The rule linking $a$ and $b$ is: “To find $b$ , double the value of $a$ and add 20.” Step 2: Translate into a mathematical equation: Text Translates to Mathematical Expression “To find $b$ $\\rightarrow$ $b =$ “double the value of $a$ $\\rightarrow$ $2a$ “add 20” $\\rightarrow$ + 20 Our equation is", "13, 2012, 01:44 PM You want to move the 80 to the left in the form so you get $4x+80-80 = -80$ Whats next ? qwertyuioop Posts: 87, Reputation: 1 Junior Member #9 Jun 13, 2012, 01:49 PM so then u divide 80 divided by 4 and u get -20??? if it correct thnx sooooo much Curlyben Posts: 18,167, Reputation: 8753 Admin & Wine Expert #10 Jun 13, 2012, 01:50 PM Do you see how that worked ? Same action on both sides..", "we have $-120, -2, -1, 0, 1, 2, 120$. Having both $-1$ and $1$ means that I can always get the negated value of any new number. But I can't crack this puzzle as I keep finding new numbers. Wonder if I got it all wrong... Edit: Some more (thanks to Aravind for the first 3)... \\begin{align} x^2 + 2x - 120 = 0 &\\rightarrow& x &= &10\\\\ 2x - 10 = 0 &\\rightarrow& x &= &5\\\\ 10x - 120 = 0 &\\rightarrow& x &= &12\\\\ x^4 + x^3 + x^2 + x - 120 = 0 &\\rightarrow& x &= &3\\\\ \\end{align} And \\begin{align} 2x - 120 = 0 &\\rightarrow& x &= &60\\\\ 3x - 120 = 0 &\\rightarrow& x &= &40\\\\ 5x - 120 = 0 &\\rightarrow& x &= &24\\\\ 5x^4 + 10x - 120 &\\rightarrow& x &= &4\\\\ 4x - 120 = 0 &\\rightarrow& x &= &30\\\\ 4x^2 + 4x -120 &\\rightarrow& x &= &5\\\\ 5x -120 = 0 &\\rightarrow& x &= &24\\\\ 3x^2 + 2x - 120 = 0 &\\rightarrow& x &= &6\\\\ 6x - 120 = 0 &\\rightarrow& x &= &20\\\\\\\\ -2x^2 - x + 120 = 0 &\\rightarrow& x &= &-8\\\\\\\\ -8x + 120 = 0 &\\rightarrow& x &= &15\\\\ \\end{align} So far I have $\\{0, 1, 2, 3, 4, 5, 6, 8,", "multiply the number by 100. Thus we obtain: $100x=274.\\overline{35}$ Next, we have to bring the repeating numbers 3 and 5 to the left side, so we have to again multiply the above equation by 100. Hence, we get: \\begin{align} & 100x\\times 100=274.\\overline{35}\\times 100 \\\\ & 10000x=27435.\\overline{35} \\\\ \\end{align} Consider the two equations: $10000x=27435.\\overline{35}$ …… (1) $100x=274.\\overline{35}$ …… (2) In the next step, we have to do subtraction. That is, equation (1) - equation (2) we obtain: \\begin{align} & 10000x-100x=27435.\\overline{35}-274.\\overline{35} \\\\ & 9900x=27161 \\\\ \\end{align} Next, by cross multiplication we get: $x=\\dfrac{27161}{9900}$ Therefore, the required rational number is $\\dfrac{27161}{9900}$. Hence, the correct answer for the question is option (d). Note: Here, first we have to identify how many digits are repeating which you can find with the help of the bar symbol. Depending on the numbers which keep on repeating we have to do the multiplication. Here, there are two numbers, so we are multiplying by 100. If you don’t multiply then you can’t take the repeating decimals to the left side which will then be difficult to remove the decimal places.", "## anonymous 4 years ago Let f(x) = 10x2 f(3 + h) = 10h^2 + 60h + 90 f(3+h) - f(3) / h = 10h + 60 what is f ' (3)? 1. anonymous as h --> 0 2. campbell_st well use $\\lim_{h \\rightarrow 0} (10h^2 + 60h + 90 - 90)/h = \\lim_{h \\rightarrow 0} h(10h + 60)/h$ $\\lim_{h \\rightarrow 0} 10h + 60 = 60$ 3. anonymous oh so basically youre using the power rule?", "20 in that box. If the total coins amount to $46, find the number of coins of each denomination. Solution Let the number of$1coins and 2 coins be x and y respectively. Given that: \\begin{align*} x+y &=20 \\quad ...(1) \\\\ 2x+5y&=46 \\quad ...(2) \\end{align*} Consider the first equation and multiply it by 2 \\begin{align*} 2 \\times (x+y&=20) \\\\ 2x + 2y &=40 \\quad ...(3) \\end{align*} On subtracting the (3) equation from (2) equation, we get \\begin{align*}2x+5y&=\\quad 46 \\\\ 2x + 2y &=\\quad 40 \\\\ \\underline{-\\quad- \\ \\:} &= \\underline{ - \\quad\\ } \\\\ \\quad 3y &= \\quad \\ 6 \\\\ \\implies \\qquad y&=\\quad \\ 2\\end{align*} Substituting the value of y in the first equation, we have, \\begin{align*}x+2&=20\\\\ x&=20-2\\\\ \\implies \\qquad x&= 18\\end{align*} $$\\therefore$$ Number of1 coins =18 Number of \\$2 coins = 2 ## Interactive Questions Here are a few activities for you to practice. Select/Type your answer and click the \"Check Answer\" button to see the result. Think Tank 1. How many solutions are there for linear equations in two variables? ## Let's Summarize We hope you enjoyed learning about the elimination method with the interactive questions. Now, you will be able to easily solve problems on the elimination method. At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers,", "number! What was Charlie's number? Choose a number, subtract $10$, divide by $2$ and reverse the digits. Extension Choose a 3-digit number where the last two digits sum to the first (e.g. $615$). Rotate the digits one place, so the first digit becomes the last (so for the example, we get $156$). Subtract the smallest number from the largest and divide by $9$ (which is always possible). What do you notice about the result? Can you explain why? With thanks to Don Steward, whose ideas formed the basis of this problem." ]

[ "x-9 \\right)$ Substitute the value of x from equation (3), we get: \\begin{align} & \\Rightarrow 4\\left( -6-9 \\right) \\\\ & \\Rightarrow 4\\times -15 \\\\ & \\Rightarrow -60 \\\\ \\end{align} So, the correct answer is “Option A”. Note: Be careful while reading the question. The language might confuse you. As in given in the question, multiply a number by 3 and then add 40. The equation for this is $3x+40$ But some might make a mistake here as $3\\left( x+40 \\right)$", "we have $-120, -2, -1, 0, 1, 2, 120$. Having both $-1$ and $1$ means that I can always get the negated value of any new number. But I can't crack this puzzle as I keep finding new numbers. Wonder if I got it all wrong... Edit: Some more (thanks to Aravind for the first 3)... \\begin{align} x^2 + 2x - 120 = 0 &\\rightarrow& x &= &10\\\\ 2x - 10 = 0 &\\rightarrow& x &= &5\\\\ 10x - 120 = 0 &\\rightarrow& x &= &12\\\\ x^4 + x^3 + x^2 + x - 120 = 0 &\\rightarrow& x &= &3\\\\ \\end{align} And \\begin{align} 2x - 120 = 0 &\\rightarrow& x &= &60\\\\ 3x - 120 = 0 &\\rightarrow& x &= &40\\\\ 5x - 120 = 0 &\\rightarrow& x &= &24\\\\ 5x^4 + 10x - 120 &\\rightarrow& x &= &4\\\\ 4x - 120 = 0 &\\rightarrow& x &= &30\\\\ 4x^2 + 4x -120 &\\rightarrow& x &= &5\\\\ 5x -120 = 0 &\\rightarrow& x &= &24\\\\ 3x^2 + 2x - 120 = 0 &\\rightarrow& x &= &6\\\\ 6x - 120 = 0 &\\rightarrow& x &= &20\\\\\\\\ -2x^2 - x + 120 = 0 &\\rightarrow& x &= &-8\\\\\\\\ -8x + 120 = 0 &\\rightarrow& x &= &15\\\\ \\end{align} So far I have $\\{0, 1, 2, 3, 4, 5, 6, 8,", "4 x-1 =0 \\\\ 4 x &=1 \\!\\qquad \\qquad \\qquad 4 x =1 \\\\ x &={1 \\over 4} \\qquad \\qquad \\quad \\; x ={1 \\over 4}\\end{align} Roots are \\(\\begin{align}{1 \\over 4},\\,\\,\\,\\,{1\\over 4}.\\end{align} (v) \\begin{align}\\quad100x{}^\\text{2}-20x+1\\end{align} Steps: \\begin{align} 100 x^{2}-20 x+1 &=0 \\\\ 100 x^{2}-10 x-10 x+1 &=0 \\\\ 10 x(10 x-1)-1(10 x-1) &=0 \\\\ (10 x-1)(10 x-1) &=0 \\end{align} \\begin{align} 10 x-1 &=0 \\qquad \\qquad10 x-1=0 \\\\ 10 x &=1\\qquad \\qquad\\qquad \\!10 x=1 \\\\ x &=\\frac{1}{10} \\qquad \\qquad\\quad x=\\frac{1 }{10} \\end{align} Roots are \\begin{align} \\frac{1}{10},\\, \\frac{1}{10} \\,\\,.\\end{align} Question 2 Solve the problems given in example $$1.$$ (i) John and Jivanti had $$45$$ marbles. Both of them lost $$5$$ marbles each and the product of the no. of marbles they now have is $$124.$$ We would like to find out how many marbles they had to start with? (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $$55$$ minus the number of toys produced in a day. On a particular day, the total cost of production was ₹$$750.$$ We would like to find out the number of toys produced on that day. Solution (i) What is known? (a) John and Jivanti together had $$45$$ marbles (b) Both of them lost $$5$$ marbles each.", "$x^2 + x^2 = \\left ( 10 \\sqrt{2} \\right )^2\\!\\\\{\\;} \\quad \\ 2x^2 = 200\\!\\\\{\\;} \\qquad x^2 = 100\\!\\\\{\\;} \\qquad \\ x = 10$ 4. Yes, $9^2 + 12^2 = 15^2 \\rightarrow 81+144 = 225$ 5. Yes, $3^2 + \\left( 3 \\sqrt{3} \\right )^2 = 6^2 \\rightarrow 9+27 = 36$ 8 , 9 , 10 Feb 22, 2012 Dec 11, 2014", "Question # Complete the problems. Check your answers in the back of the book.$980 \\mathrm{~cm}=\\underline{\\qquad}$ $\\mathrm{hm}$ Solution Verified \\begin{aligned} 980cm&=0.0980hm \\end{aligned}", "# Difference between revisions of \"2012 AMC 12B Problems/Problem 14\" ## Problem Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she addes $50$ to it and passes the result to Bernardo. The winner is the last person who produces a number less than $1000$. Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N$? $\\textbf{(A)}\\ 7\\qquad\\textbf{(B)}\\ 8\\qquad\\textbf{(C)}\\ 9\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 11$ ## Solution ### Solution 1 The last number that Bernado says has to be between 950 and 999. Note that 1->2->52->104->154->308->358->716->776 contains 4 doubling actions. Thus, we have $x \\rightarrow 2x \\rightarrow 2x+50 \\rightarrow 4x+100 \\rightarrow 4x+150 \\rightarrow 8x+300 \\rightarrow 8x+350 \\rightarrow 16x+700$. Thus, $950<16x+700<1000$. Then, $16x>250 \\implies x \\geq 16$. If $x=16$, we have $16x+700=956$. Working backwards from 956, $956 \\rightarrow 478 \\rightarrow 428 \\rightarrow 214 \\rightarrow 164 \\rightarrow 82 \\rightarrow 32 \\rightarrow 16$. So the starting number is 16, and our answer is $1+6=\\boxed{7}$, which is A. ### Solution 2 Work backwards. The last number Bernardo produces must be in the range $[950,999]$. That means that before this, Silvia must produce a", "you add any two of the numbers you also get perfect squares: \\displaystyle \\begin{aligned} 41 + 80 &= 121 = 11^2 \\\\ 80 + 320 &= 400 = 20^2 \\\\ 41 + 320 &= 361 = 19^2 \\end{aligned} If it wasn't for Soroban's last post I wouldn't have thought of this. (Bow) That's the answer I was looking for (I wonder how this plays out in binary?) #### Chokfull That's the answer I was looking for (I wonder how this plays out in binary?) It shouldn't change a bit in binary as this solution has nothing to do with the numbers' digits, only their values.", "\\\\ & \\Rightarrow x=\\dfrac{424\\pm 224}{18} \\\\ & \\Rightarrow x=36,\\dfrac{100}{9} \\\\ \\end{align} Since $x$ represents the number of camels, it must be an integer. So, $x=36$. Hence, the total number of camels is equal to $36$. Note: There is an alternative way to solve the quadratic equation $8\\sqrt{x}=3x-60$. We can assume a variable $p$ and then we can substitute $x={{p}^{2}}$ in this quadratic equation and then solve for $p$ using a quadratic formula. Later, we can re-substitute ${{p}^{2}}=x$ or $p=\\sqrt{x}$ and then solve to get $x$.", "263 views A certain number consists of two digits whose sum is $9$. It the order of digits is reversed, the new number is $9$ less than the original number. The original number is : 1. $45$ 2. $36$ 3. $54$ 4. $63$ Let digits of the number are $a$ and $b$ at tenth and ones place respectively. Such that required number is $10a +b$ Given that $a+b = 9 \\qquad \\qquad \\rightarrow(i)$ and if we replace the digits then the new number is 9 less than the original number, \\begin{align} 10b + a +9 &= 10a+b \\\\ \\implies 9a – 9b &= 9 \\\\ \\implies ~~ a- b &=1 \\qquad \\qquad \\rightarrow(ii)\\end{align} $equation~(i)+(ii)~gives, \\\\ \\implies 2a = 10 \\implies a=5 \\\\ \\> \\\\ equation~(i)-(ii)~gives, \\\\ \\implies 2b = 8 \\implies b=4$ Hence, required number is $10\\times 5 + 4 = 54$ Option (C) is correct. 626 points 1 vote 1 329 views 1 vote 2 420 views 1 vote", "$797$, and their difference is $10$. So we need $a$ to be at least $10$ and such that its sum with $787$ or $797$ is $800$. This clearly suggests the following choice: $$a=13, \\qquad b=787, \\qquad c=797.$$ $$a+b-c=3, \\qquad a+c-b=23, \\qquad b+c-a=1571.$$ Again using the list above you can check that we have been lucky and those are all distinct primes. Then we have actually computed (not only bounded) the maximum vale $d$, which is: $d = b+c-a -a = 1571-13= 1558.$", "(b+a)(b-a)$$ Assuming as you do that $b=a+1$ $$b^2 - a^2 = (b+a)(b-a) = (2a+1)(1)=2a+1$$ and $$a+b=a+a+1=2a+1$$ so yes, when $b=a+1$, $b^2 - a^2 = a + b$ There are relatively few triples where $$A$$ and $$B$$ differ by only $$1$$. In triples with sides under 212 quadrillion, there are only $$19$$ of them and one method of finding them is here. Here are the first few shown as f(n,k) and they agree with the $$proof$$ in notovny's answer: $$3,4,5\\Rightarrow4^2-3^2=16-9=7=3+4$$ $$21,20,29\\Rightarrow21^2-20^2=441-400=41=21+20$$ $$119,120,169\\Rightarrow120^2-119^2=14400-14161=239=120+119$$ If you want $$C-B=1$$, these functions will generate them all for an $$k\\in\\mathbb{N}$$ $$A=2k+1\\qquad B=2k^2+2k\\qquad C=2k^2+2k+1$$", "be the maximum number of squares Peter can move a token beyond the dividing line. For example, $F(0)=4$. You are also given that $F(1)=6$, $F(2)=9$, $F(3)=13$, $F(11)=58$ and $F(123)=1173$. Find $F(1234567)$.", "add any two of the numbers you also get perfect squares: \\begin{aligned} 41 + 80 &= 121 = 11^2 \\\\ 80 + 320 &= 400 = 20^2 \\\\ 41 + 320 &= 361 = 19^2 \\end{aligned} If it wasn't for Soroban's last post I wouldn't have thought of this. That's the answer I was looking for (I wonder how this plays out in binary?) 8. Originally Posted by wonderboy1953 That's the answer I was looking for (I wonder how this plays out in binary?) It shouldn't change a bit in binary as this solution has nothing to do with the numbers' digits, only their values.", "\\\\ d + 26d &= 121.5 \\\\ 27d &= 121.5 \\\\ d &= 4.5 \\\\ \\textrm{Substitute this answer into the second equation} \\\\ 26\\cdot 4.5 &= r \\\\ 117 &= r \\end{align*} So with this working, and showing him that the numbers work in both equations, you should be able to convince him... 5. ## Re: Old guys arguin over math from long ago Awesome ...He gets it now lol.. Thanks for your time", "even number of $O$s, so $$EEE\\qquad EOO$$ For $A$ to win on their third move it is necessary that the first five throws contain exactly two $O$s (so that the last move by $A$ brings them a point). There are ${5\\choose 2}=10$ such sequences, from which we subtract the two starting with $EEE$ and $EOO$, so only $8$ configurations to study: I write the sequence of points earned and stop when either $A$ wins or it is clear that they won't. \\begin{align*}OOEEE &\\qquad 0111& B\\\\ OEOEE &\\qquad 00111& A\\\\ OEEOE &\\qquad 000&not \\ A \\\\ OEEEO &\\qquad000&not \\ A \\\\ EOEOE &\\qquad 10011&A\\\\ EOEEO &\\qquad 10001&A\\\\ EEOOE &\\qquad 1101&B\\\\ EEOEO &\\qquad 11001&A\\\\ \\end{align*} Therefore your list of configurations is complete, and your count is correct.", "Gauss solved this problem. The solution was quite fascinating for me. That was the first time I enjoyed a beautiful solution, though I couldn’t find it myself.” Her interest was peaked with an inspiring problem: $$1 + 2 + 3 + … + 99+ 100 =\\: ?$$ When I give this type of problem to my elementary or middle school students, they often start adding the numbers one at a time: \\begin{align} 1 + 2 &= 3\\\\ 3+ 3 &= 6\\\\ 4 + 6 &= 10\\\\ 5 + 10 &= 15\\\\ &\\vdots \\end{align} If continued until the problem is solved, this method requires 99 sums. Doing the addition problem one sum at a time is beyond most student’s patience level. However, my students know me well enough to realize that there must be another solution — asking them to do 99 sums in a row does not fit my teaching style. ## One Elegant Solution One such method involves changing the order of addition (see PEMDAS for more info on when changing the order of operations is valid). Numbers that sum to 100 are paired. With this change, \\begin{align}1 + 2 + 3 + &… + 99+ 100\\\\ &=100 + (1+99) + (2 + 98) + (3+97) + … + (49+51) + 50\\\\ &= 100 + 49 \\times", "20 in that box. If the total coins amount to $46, find the number of coins of each denomination. Solution Let the number of$1coins and 2 coins be x and y respectively. Given that: \\begin{align*} x+y &=20 \\quad ...(1) \\\\ 2x+5y&=46 \\quad ...(2) \\end{align*} Consider the first equation and multiply it by 2 \\begin{align*} 2 \\times (x+y&=20) \\\\ 2x + 2y &=40 \\quad ...(3) \\end{align*} On subtracting the (3) equation from (2) equation, we get \\begin{align*}2x+5y&=\\quad 46 \\\\ 2x + 2y &=\\quad 40 \\\\ \\underline{-\\quad- \\ \\:} &= \\underline{ - \\quad\\ } \\\\ \\quad 3y &= \\quad \\ 6 \\\\ \\implies \\qquad y&=\\quad \\ 2\\end{align*} Substituting the value of y in the first equation, we have, \\begin{align*}x+2&=20\\\\ x&=20-2\\\\ \\implies \\qquad x&= 18\\end{align*} $$\\therefore$$ Number of1 coins =18 Number of \\$2 coins = 2 ## Interactive Questions Here are a few activities for you to practice. Select/Type your answer and click the \"Check Answer\" button to see the result. Think Tank 1. How many solutions are there for linear equations in two variables? ## Let's Summarize We hope you enjoyed learning about the elimination method with the interactive questions. Now, you will be able to easily solve problems on the elimination method. At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers,", "user should know: 1. Enter two positive numbers to be added together by our long addition calculator. 2. The calculator displays the result faster than lightning ⚡⚡ It puts the result under the numbers you've entered. 3. Use the show steps option whenever you want our long addition calculator to show the step-by-step solution of long addition. Here we show how to add two numbers together with the help of long addition. Long addition example 1. Solve 15292 + 735: 1. Setup: \\qquad \\begin{align*} 15292 &\\\\ +\\quad\\ \\ 735& \\\\ \\mathclap{\\rule{1cm}{0.4pt}} \\quad &\\\\ = \\qquad \\quad \\end{align*} 1. We sum the numbers in the far-right column: 2 + 5 = 7. We put the result 7 at the very bottom of the same column: \\qquad \\begin{align*} 15292 &\\\\ +\\quad\\ \\ 735& \\\\ \\mathclap{\\rule{1cm}{0.4pt}} \\quad &\\\\ = \\qquad \\ 7 & \\end{align*} 1. We sum the numbers in the next column: 9 + 3 = 12. As the result 12 has two digits, we separate it into the ones-digit 2, which we put into the results row, and tens-digit 1, which we carry to the next column: \\qquad \\begin{align*} 1\\phantom{11} & \\\\ 15292 &\\\\ +\\quad\\ \\ 735& \\\\ \\mathclap{\\rule{1cm}{0.4pt}} \\quad &\\\\ = \\quad \\ \\ \\ 27 & \\end{align*} 1. We sum the numbers in the next column, not", "of the Quadr. Eqn.$: {x}^{2} + p x + q = 0.$ Therefore, $x = 2 + 3 i$ must satisfy the eqn. $\\therefore {\\left(2 + 3 i\\right)}^{2} + p \\left(2 + 3 i\\right) + q = 0.$ $\\therefore 4 + 12 i + 9 {i}^{2} + 2 p + 3 i p + q = 0.$ $\\therefore 4 + 12 i - 9 + 2 p + 3 i p + q = 0.$ $\\therefore \\left(2 p + q - 5\\right) + i \\left(12 + 3 p\\right) = 0.$ Comparing the Real and Imaginary Parts of both sides, we get, $2 p + q = 5. \\ldots . . \\left(1\\right) , \\mathmr{and} , 12 + 3 p = 0$ rArr p=-4, &, q=5-2p=5-2(-4)=13, as Respected Steve Sir, has alreay derived. Enjoy Maths.!", "2 digit #s starting with 2: -add the ones digit to the number -multiply it by 20 -add the square of the ones digit $$ex:15\\\\ \\\\ 15+5=20\\\\ 20*10=200\\quad \\quad {5}^ {2}=25\\quad \\quad \\quad \\quad 200+25=225\\\\ \\\\ ex:23\\\\ 23+3=26\\Rightarrow 2*260\\Rightarrow 520+{3}^ {2}=529\\\\ \\\\ \\\\ \\\\ \\\\$$ Note by Hummus A 12 months ago Sort by: let me \"formalize\" the tricks. 1.$$n^2=(50-n)^2+100(n-25)$$ 2.$$n^2=(n-(100-n))100+(100-n)^2=100(100-2n)+(100-n)^2$$ Both are true. · 12 months ago" ]

[ "you'll get $\\overline{9} = - 1$.", "x \\ \\overline y$", "$X= \\overline A$ -", "\\overline{33} g$", "of $1$ • $3.4\\bar{9}=3.5\\overline{0}$. Indeed, $3.4\\overline{9}$, $3.5$, $35/10$, and $7/2$ are all different representations of the same number. The Wikipedia article \"$0.\\overline{9}$\" is an elaborate discussion of the fact that $0.\\overline{9}=1$, a fact that many students find hard to believe.", "c) $7.\\overline{09}$. Discussion", "infinitesimals to tell $1$ and $0.\\overline{9}$ apart) Edit: From the comments below, it seemed useful to post a proof that $0.\\overline 9 = 1$. I have chosen this particular proof because contains nothing but simple algebra, packaging up all the interesting stuff into two assumptions. Assumption 1: $10\\cdot0.\\overline 9 = 9.\\overline 9$ Assumption 2: $9.\\overline 9 = 9 + 0.\\overline 9$ Proof: $\\text{Let }x = 0.\\overline 9$ $10x = 10\\cdot0.\\overline 9$ $10x = 9.\\overline 9$ (using assumption 1) $10x - x = 9.\\overline 9 - x$ $9x = 9.\\overline 9 - x$ $9x = 9.\\overline 9 - 0.\\overline 9$ $9x = (9 + 0.\\overline 9) - 0.\\overline 9$ (using assumption 2) $9x = 9 + (0.\\overline 9 - 0.\\overline 9)$ $9x = 9 + (0.\\overline 9 - 0.\\overline 9)$ $9x = 9 + (0)$ $9x = 9$ $x = 1$ $0.\\overline 9 = 1$ • The issue is that \"0.999...\" is not \"a number\" but a way to represent a number that hints at a way to construct it. Technically \"1.0\" is just a way to represent a number as well. The limit that you are referring to is actually baked into the formal definition of what 0.999... means in terms of how to construct a real number. The limit, which is 1, is actually written into the", "$p (\\overline z)$?", "{1 \\over 2}$", "that is, $(T+1)^{\\overline{T}}/T!$.", "\\over {18}}}}$ = 0.018", "= 20\\overline{B}$.", "2 \\over 2 }$$ $$= 1$$", "1}} \\approx 0.9$", "1=\\overline 1\\cdot\\overline 1 = \\overline 0$. – Jonas Meyer Jan 22 '12 at 2:32", "1$. Thus: $|f(z)|\\le |\\frac{z-a}{1-\\overline az}|$.", "# Why does 0.999… = 1? (Part 5) Here’s one more way of convincing students that $0.\\overline{9} = 1$. Here’s the idea: how far apart are the two numbers? First off, since $1 \\ge 0.\\overline{9}$, we know that $1 - \\overline{9} \\ge 0$. Of course, we know that $1-0.9 = 0.1$. Since $0.\\overline{9}$ must lie between $0.9$ and $1$, we know that $1 - 0.\\overline{9}$ must be less than $0.1$. Second, we know that $1-0.99 = 0.01$. Since $0.\\overline{9}$ must lie between $0.99$ and $1$, we know that $1 - 0.\\overline{9}$ must be less than $0.01$. Third, we know that $1-0.999 = 0.001$. Since $0.\\overline{9}$ must lie between $0.999$ and $1$, we know that $1 - 0.\\overline{9}$ must be less than $0.001$. By the same reasoning, we conclude that $0 \\le 1 - 0.\\overline{9} < \\displaystyle \\frac{1}{10^n}$ for every integer $n$. What’s the only number that’s greater than or equal to $0$ and less than every decimal of the form $0.00\\dots001$? Clearly, the only such number is $0$. Therefore, $1 - 0.\\overline{9} = 0$, or $0.\\overline{9} = 1$. I like this approach because it really gets at the heart of the difference between integers $\\mathbb{Z}$ and real numbers $\\mathbb{R}$. For integers, there is always an integer to the immediate left and to the immediate right. In other words,", "- 9] +9=0$$", "0.\\overline 7$ This is the way to approach another method.", "because $22 \\div 18.75 = 1.17\\overline{3}$. You left a $17.\\overline{3}\\%$ tip, because $117.\\overline{3} - 100 = 17.\\overline{3}$." ]

[ "Question # What is the sum of $0.\\overline{6}$ and $0.\\overline{7}$ ?\\begin{align} & \\text{(A) 1}\\text{.}\\overline{\\text{3}} \\\\ & \\text{(B) 1}\\text{.3} \\\\ & \\text{(C) 1}\\text{.}\\overline{\\text{4}} \\\\ & \\text{(D) An irrational number} \\\\ \\end{align} Hint: We will convert non terminating recurring decimals $0.\\overline{6}$ and $0.\\overline{7}$ to fraction and then we will add those two fractions and at last we will convert the fraction into decimal to get the desired result. We will convert non terminating recurring decimals to fractions by assuming x=$0.\\overline{6}$ and y=$0.\\overline{7}$ x can also be written as 0.66666……. and y can also be written as 0.77777…… First, we will convert $0.\\overline{6}$ to fraction. We assumed x=0.66666….. and let this be equation 1. Now we multiply x with 10, multiplying x with 10 we will get, $\\Rightarrow 10x=6.6666.....$ and let this be equation 2. Subtracting equation 1 from equation 2 we will get 9x=6 and x can be written as $x=\\dfrac{6}{9}$ . So, $x=\\dfrac{6}{9}$. Now, we will convert $0.\\overline{7}$ to fraction. We assumed y=0.77777….. and let this be equation 3. Now we multiply y with 10, multiplying y with 10 we will get, $\\Rightarrow 10y=7.7777.....$ and let this be equation 4. Subtracting equation 3 from equation 4 we will get 9y=7 and x can be written as $y=\\dfrac{7}{9}$. So, $y=\\dfrac{7}{9}$. Now, according to the question we have to add x", "$$.......(1)$$ In $$0.\\overline7$$, one digit is repeating every time. $$\\therefore$$ We multiply both the sides of equation $$(1)$$ by $$10$$. $$x=0.\\overline7$$ $$\\Rightarrow x=0.777......$$ $$10×x=10×0.777....$$ $$10x=7.777....$$ $$....(2)$$ We can write $$7.777=7+0.7777$$ $$\\therefore$$ Equation $$(2)$$ can be written as $$10x=7+0.777$$ $$10x=7+0.\\overline7$$ $$......(3)$$ Substitute the value of $$0.\\overline7$$ in equation $$(3)$$ from equation $$(1)$$, i.e. $$\\because\\;\\;\\; x=0.\\overline7$$ $$\\therefore$$ In equation $$3$$ $$10x=7+x$$ $$.......(4)$$ Solving equation $$(4)$$ to find the value of $$x$$, i.e. $$10x=7+x$$ $$........(4)$$ Subtracting $$x$$ from both the sides, $$10x-x=7+x-x$$ $$(10-1)x=7$$ $$9x=7$$ Dividing both sides by $$9$$ $$\\dfrac{9x}{9}=\\dfrac{7}{9}$$ $$x=\\dfrac{7}{9}$$ Therefore, the repeating decimal, $$0.\\overline7=\\dfrac{7}{9}$$ Hence, option (C) is correct. ### What is the fraction form of $$0.\\overline7$$? A $$\\dfrac{3}{5}$$ . B $$\\dfrac{3}{8}$$ C $$\\dfrac{7}{9}$$ D $$\\dfrac{8}{5}$$ Option C is Correct # Comparison of Fractions through Representation on a Number Line • Consider two fractions as two points on the number line. • On a number line, a fraction which is placed to the left is smaller than the fraction which is placed to the right side. For example: Consider $$\\dfrac{1}{4}$$ and $$\\dfrac{5}{4}$$. $$\\dfrac{1}{4}$$ is placed to the left of $$\\dfrac{5}{4}$$. $$\\therefore$$ $$\\dfrac{1}{4}<\\dfrac{5}{4}$$ Note: While comparing two or more fractions, make sure that each one of them has the same denominator. #### Two points namely, $$P$$ and $$R$$, are plotted on the given number line. Determine the values of $$P$$", "$x=0.\\overline{001}\\Rightarrow$ …… (1) Since the number of recurring decimal number is $3$, so multiplying both sides of the equation (1) by $1000$, gives $1000\\times x=1000\\times 0.001001.....$ …… (2) Subtracting the equation (1) from (2) gives \\begin{align} & 1000x=1.001001..... \\\\ & \\underline{\\text{ }-x=0.001001.....} \\\\ & 999x=1 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{1}{999}$ Hence, the decimal number becomes $0.\\overline{001}=\\dfrac{1}{999}$ and it is in the $\\dfrac{p}{q}$ form. 4. Express ${0} {.99999}.....$ in the form of $\\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense. Ans: Let $x=0.99999.....$ ....... (a) Multiplying by $10$ both sides of the equation (a), gives $10x=9.9999.....$ …… (b) Now, subtracting the equation (a) from (b), gives \\begin{align} & 10x=9.99999..... \\\\ & \\underline{\\,-x=0.99999.....} \\\\ & \\,\\,9x=9 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{9}{9}$ $\\Rightarrow x=1$. So, the decimal number becomes $0.99999...=\\dfrac{1}{1}$ which is in the $\\dfrac{p}{q}$ form. Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense. Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal. 5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\\dfrac{ {1}}{ {17}}$ ? Perform the division to check your answer. Ans: Here the", "you'll get $\\overline{9} = - 1$.", "multiply the number by 100. Thus we obtain: $100x=274.\\overline{35}$ Next, we have to bring the repeating numbers 3 and 5 to the left side, so we have to again multiply the above equation by 100. Hence, we get: \\begin{align} & 100x\\times 100=274.\\overline{35}\\times 100 \\\\ & 10000x=27435.\\overline{35} \\\\ \\end{align} Consider the two equations: $10000x=27435.\\overline{35}$ …… (1) $100x=274.\\overline{35}$ …… (2) In the next step, we have to do subtraction. That is, equation (1) - equation (2) we obtain: \\begin{align} & 10000x-100x=27435.\\overline{35}-274.\\overline{35} \\\\ & 9900x=27161 \\\\ \\end{align} Next, by cross multiplication we get: $x=\\dfrac{27161}{9900}$ Therefore, the required rational number is $\\dfrac{27161}{9900}$. Hence, the correct answer for the question is option (d). Note: Here, first we have to identify how many digits are repeating which you can find with the help of the bar symbol. Depending on the numbers which keep on repeating we have to do the multiplication. Here, there are two numbers, so we are multiplying by 100. If you don’t multiply then you can’t take the repeating decimals to the left side which will then be difficult to remove the decimal places.", "only the $$x$$ is being multiplied by 2, so the first step will be to add 7 to both sides. Then we can divide both sides by 2. \\begin{align} 2x-7 &= 13\\\\ 2x-7 \\color{red}{+7} & =13 \\color{red}{+7}\\\\ 2x&=20\\end{align} Now divide both sides by 2: \\begin{align} 2x &=20 \\\\ \\dfrac{2x}{\\color{red}{2}}&=\\dfrac{20}{\\color{red}{2}}\\\\ x&= \\boxed{10}\\end{align} #### Check Just like with simpler problems, you can check your answer by substituting your value of $$x$$ back into the original equation. \\begin{align}2x-7&=13\\\\ 2(10) – 7 &= 13\\\\ 13 &= 13\\end{align} This is true, so we have the correct answer. Let’s look at one more two-step example before we jump up in difficulty again. Make sure that you understand each step shown and work through the problem as well. ### Example Solve: $$5w + 2 = 9$$ ### Solution As above, there are two operations: $$w$$ is being multiplied by 5 and then has 2 added to it. We will undo these by first subtracting 2 from both sides and then dividing by 5. \\begin{align}5w + 2 &= 9\\\\ 5w + 2 \\color{red}{-2} &= 9 \\color{red}{-2}\\\\ 5w &= 7\\\\ \\dfrac{5w}{\\color{red}{5}} &=\\dfrac{7}{\\color{red}{5}}\\\\w=\\boxed{\\dfrac{7}{5}}\\end{align} The fraction on the right can’t be simplified, so that is our final answer. #### Check Let $$w = \\dfrac{7}{5}$$. Then: \\begin{align}5w + 2 &= 9\\\\ 5\\left(\\dfrac{7}{5}\\right) + 2 &= 9\\\\ 7 + 2 &=", "### Home > PC > Chapter 9 > Lesson 9.3.2 > Problem9-116 9-116. Show whether or not $x − 5$ is a factor of $x^{3} + 4x^{2} − 11x − 30$. Set up the generic rectangle with the divisor on the left of the box. Input only the first terms of the dividend. Generic rectangle, 2 rows, 4 columns, left edge labeled, x minus 5, interior top left, labeled x cubed. Since $x$ divides into $x^3$, $x^2$ times, write the $x^2$ on top and multiply $x^2$ with the $−5$ and fill in the grid with the product. Labels added, top edge left, x squared, interior bottom left, negative 5, x squared. Add an expression to $−5x^2$ in order to get the needed $4x^2$. Add it to the right of $x^3$Label added, interior top, second from left, 9 x squared. Dividing $x$ into $9x^2$ gives $9x$. Add the $9x$ at the top. Then multiply $9x$ and $−5$ and fill the answer in the grid. Labels added: top edge, second from left, + 9 x, interior bottom, second from left, negative 45 x. Add an expression to $−45x$ in order to get the needed $−11x$. Add it to the right of $9x^2$Label added: interior top, second from right, 34 x. Dividing $x$ into $34x$ gives $34$. Add the $34$ at", "the equation (1) by $1000$, gives $1000\\times x=1000\\times 0.001001.....$ …… (2) Subtracting the equation (1) from (2) gives \\begin{align} & 1000x=1.001001..... \\\\ & \\underline{\\text{ }-x=0.001001.....} \\\\ & 999x=1 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{1}{999}$ Hence, the decimal number becomes $0.\\overline{001}=\\dfrac{1}{999}$ and it is in the $\\dfrac{p}{q}$ form. 4. Express ${0} {.99999}.....$ in the form of $\\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense. Ans: Let $x=0.99999.....$ ....... (a) Multiplying by $10$ both sides of the equation (a), gives $10x=9.9999.....$ …… (b) Now, subtracting the equation (a) from (b), gives \\begin{align} & 10x=9.99999..... \\\\ & \\underline{\\,-x=0.99999.....} \\\\ & \\,\\,9x=9 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{9}{9}$ $\\Rightarrow x=1$. So, the decimal number becomes $0.99999...=\\dfrac{1}{1}$ which is in the $\\dfrac{p}{q}$ form. Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense. Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal. 5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\\dfrac{ {1}}{ {17}}$ ? Perform the division to check your answer. Ans: Here the number of digits in the recurring block of $\\dfrac{1}{17}$ is to be determined. So, let us calculate", "# 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 $\\frac{4}{9}$.", "$$x=0 . \\overline{24} .$$ Then $$100 x=24 . \\overline{24} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=24 . \\overline{24} \\\\ x &=0 . \\overline{24} \\end{aligned} yields $$99 x=24 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{24}{99}=\\frac{8}{33}$$ Exercise $$\\PageIndex{20}$$ $$0 . \\overline{882}$$ Exercise $$\\PageIndex{21}$$ $$0 . \\overline{84}$$ Let $$x=0 . \\overline{84} .$$ Then $$100 x=84 . \\overline{84} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=84 . \\overline{.84} \\\\ x &=0 . \\overline{84} \\end{aligned} yields $$99 x=84 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{84}{99}=\\frac{28}{33}$$ Exercise $$\\PageIndex{22}$$ $$0 . \\overline{384}$$ Exercise $$\\PageIndex{23}$$ $$0 . \\overline{63}$$ Let $$x=0 . \\overline{63} .$$ Then $$100 x=63 . \\overline{63} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=63 . \\overline{63} \\\\ x &=0 . \\overline{63} \\end{aligned} yields $$99 x=63 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{63}{99}=\\frac{7}{11}$$ Exercise $$\\PageIndex{24}$$ $$0 . \\overline{60}$$ Exercise $$\\PageIndex{25}$$ Prove that $$\\sqrt{3}$$ is irrational. Suppose that $$\\sqrt{3}$$ is rational. Then it can be expressed as the ratio of two integers p and q as follows: $\\sqrt{3}=\\frac{p}{q}$ Square both sides, $3=\\frac{p^{2}}{q^{2}}$ then clear the equation of fractions by multiplying both sides by $$q^{2}$$: $p^{2}=3 q^{2}$ Now p and q each have their own unique prime factorizations. Both $$p^{2}$$ and $$q^{2}$$", "+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", "bar, 4. \\begin{align*}\\frac{1}{4} = {4 \\overline{ ) 1 \\;\\;\\;}}\\end{align*} Since 1 cannot be evenly divided by 4, rewrite 1 as a decimal with a zero after the decimal point You can do this because \\begin{align*}1 = 1.0 = 1.00 = 1.000\\end{align*}. Before you divide, write a decimal point in the quotient directly above the decimal point in the dividend. Then divide. \\begin{align*}& \\overset{ \\ \\ 0.2}{4 \\overline{ ) {1.0 \\;}}}\\\\ & \\quad \\underline{-8}\\\\ & \\quad \\ \\ 2 \\end{align*} Continue adding zeroes after the decimal point and diving until the quotient has no remainder. \\begin{align*}& \\overset{ \\ \\ 0.25}{4 \\overline{ ) {1.00 \\;}}}\\\\ & \\quad \\underline{-8 \\;\\;}\\\\ & \\quad \\ \\ 20\\\\ & \\ \\ \\ \\underline{-20}\\\\ & \\qquad \\ 0 \\end{align*} The decimal form of the ratio \\begin{align*}\\frac{1}{4}\\end{align*} is 0.25. Example Rewrite the ratio 9:5 in decimal form. The ratio 9:5 can be expressed as the fraction \\begin{align*}\\frac{9}{5}\\end{align*}. Next, divide the term above the fraction bar, 9, by the term below the fraction bar, 5. \\begin{align*}& \\overset{ \\ \\ \\ 1}{5 \\overline{ ) { \\ 9 \\;}}}\\\\ & \\ \\underline{-5\\;}\\\\ & \\ \\ \\ 4 \\end{align*} There is a remainder. So, add zeroes after the decimal point in 9 to continue dividing. \\begin{align*}& \\overset{ \\ \\ 1.8}{5 \\overline{ ) {9.0 \\;}}}\\\\ & \\underline{-5\\;}\\\\ & \\quad 40\\\\ &", "bottom, second from left, negative 45 x. Add an expression to $−45x$ in order to get the needed $−11x$. Add it to the right of $9x^2$. Label added: interior top, second from right, 34 x. Dividing $x$ into $34x$ gives $34$. Add the $34$ at the top. Then multiply $34$ and $−5$ and fill the answer in the area model. Labels added: top edge, second from right, + 34, interior bottom, second from right, negative 170. Add an expression to $−170$ in order to get the needed $−30$. Add it to the right of $34x$. This is the remainder. Labels added: top edge, right, 140 divided by quantity x minus 5. Interior top right, 140.", "Again our first step is write $$x = 0.277..$$. Now multiplying by $$10$$ gives us $$10x = 2.77..$$. This time we need another equation to match the recurring part of the equation. So multiplying by $$10$$ again gives $$100x = 27.77..$$. Now we have two equations with the same recurring part we subtract one from the other as before. \\begin{aligned} x & = 0.277.. \\\\ 10x & = 2.77.. \\\\ 100x & = 27.77.. \\\\ 90x & = 25 \\\\ x & = {25 \\over 90} \\\\ x & = {5 \\over 18} \\\\ \\end{aligned} So in its lowest terms $$0.277.. = {5 \\over 18}$$. Example 3 Write 0.0855.. as a fraction in its lowest terms. As always our first step it to write $$x = 0.0855..$$. This time to get two equations with the same recurring part after the decimal point we need to multiply by $$100$$ and then $$1000$$. This gives us: \\begin{aligned} 100x & = 8.55.. \\\\ 1000x & = 85.55.. \\\\ 900x & = 77 \\\\ x & = {77 \\over 900} \\\\ \\end{aligned} This cannot be simplified any further so $$0.855.. = {77 \\over 900}$$. Example 4 Write 0.0234234.. as a fraction in its lowest terms. This time the '$$234$$' part is the important bit. Now to get two multiples of x with", "$ax+by+c=0$. Equation’s Solution : A particular set of values of the variables, which when substituted for the variables in the equation makes the two sides of the equation equal, is called the solution of the equation. Now, let us look at the percentage formula. The value of x% in y can be written as: $\\dfrac{x}{100}\\times y$ Now, on comparing with the given values we get, $x=8,y=24,000$ Let us assume that the cost after discount as A , cost after season discount as B and cost after sales tax on the remaining amount as C. \\begin{align} & \\Rightarrow A=24000-\\dfrac{8}{100}\\times 24000 \\\\ & \\Rightarrow A=24000-\\left( 8\\times 240 \\right) \\\\ & \\Rightarrow A=24000-1920 \\\\ & \\therefore A=Rs.22,080 \\\\ \\end{align} Now, again we need to consider the values of x and y as, $x=5,y=22,080$ By using this and simplifying we get, $\\Rightarrow B=A-\\dfrac{x}{100}\\times A$ \\begin{align} & \\Rightarrow B=22080-\\dfrac{5}{100}\\times 22080 \\\\ & \\Rightarrow B=22080-\\left( 5\\times 220.8 \\right) \\\\ & \\Rightarrow B=22080-1104 \\\\ & \\therefore B=Rs.20,976 \\\\ \\end{align} Let us assume that the sales tax will be imposed on the remaining amount as D. Now, we use $x=10,y=20,976$. $\\Rightarrow D=\\dfrac{x}{100}\\times B$ \\begin{align} & \\Rightarrow D=\\dfrac{10}{100}\\times 20,976 \\\\ & \\Rightarrow D=10\\times 209.76 \\\\ & \\therefore D=Rs.2,097.60 \\\\ \\end{align} \\begin{align} & \\Rightarrow C=20,976+D \\\\ & \\Rightarrow C=20,976+2,097.60 \\\\ & \\therefore C=Rs.23,073.60 \\\\ \\end{align} Hence, the correct", "# $a/b=c$, how to move the numerator to the right side of the equation Given an equation of the form $a/b=c$, how do you move $a$, the numerator, over to the other side so that you get $b =c?$ (where $?$ denotes my ignorance of what the right side should look like). - \\begin{align*} a/b &= c\\\\ a &= bc\\\\ b &= a/c \\end{align*}", "\\begin{array}{cc} {9x} = 11 ~~~~~~~~~ (1) \\\\ {10x} = 9y ~~~~~~~(2) \\\\ \\end{array} \\end{align} Solving for $x$ in our first equation ($1$), we should divide both side of the equation by the number 9. By doing this the $9$ would reduce to $1$ because $\\dfrac{9}{9} = 1$ and the right hand side becomes $\\dfrac{11}{9}$. So we get that $x = \\dfrac{11}{9}$. Plugging in $x = \\dfrac{11}{9}$ into our second equation ($2$) we get: $10x = 9y$ $\\Rightarrow 10\\left(\\dfrac{11}{9}\\right) = 9y$ $\\Rightarrow \\dfrac{110}{9} = 9y$ $~~~~~~~~~$ (Divide the left and right hand side of the equation by the number 9) By doing this, we would be dividing on the right $\\dfrac{9}{9}$ and on the left $\\dfrac{110}{9}\\over\\dfrac{9}{1}$. So now, on the left side, we are left with $y$ and when we have a fraction as so on the left, we want to multiply the top fraction (numerator) by the reciprocal of the bottom fraction (denominator). Doing this lead us to get $\\dfrac{110}{81}$ on the left side of the equation. So now we know what both are unknown variables $x \\text{ and}~ y$ which are $${x} = \\dfrac{11}{9}$$ $${y} = \\dfrac{110}{81}$$ Our second method of solving would be to use process of elimination: To do this, we are going to multiply both equations ($1$) and ($2$) by some number, so that", "is multiplied by 10, we have to divide the left hand side by 10. Now, we can’t divide the left side only by 10. We have to divide both the sides by 10. As, if two expressions are equal to each other and we divide both the sides by the same number, the equation remains the same. Now let’s divide both side of the equation by 10: \\begin{align} & 10p=100 \\\\ & \\Rightarrow \\dfrac{10p}{10}=\\dfrac{100}{10} \\\\ & \\Rightarrow p=10 \\\\ \\end{align} Therefore, the value of $p$ is 10. Note: Always apply the same operation on both sides. If you apply any operation only on one side, the meaning of the equation will change.", "### Home > CCAA8 > Chapter 3 Unit 4 > Lesson CCA: 3.3.1 > Problem3-81 3-81. Find each of the following products by drawing and labeling a generic rectangle or by using the Distributive Property. 1. $−4y(5x+8y)$ • Make and label a generic rectangle. Generic rectangle left edge$-4y$ interior left blank interior right blank bottom edge left$5x$ bottom edge right$+8y$ Fill in the boxes, and then add the boxes' labels together Generic rectangle left edge$-4y$ interior left$-20xy$ interior right $-32y^2$ bottom edge left$5x$ bottom edge right$+8y$ $−20xy−32y^2$ 1. $9x(−4+10y)$ • Distribute the $9x$ over each of the terms in the parentheses. $9x(−4)+9x(10y)$ Simplify. $−36x+90xy=90xy−36x$ 1. $(x^2−2)(x^2+3x+5)$ • Try using a generic rectangle to solve this problem. $x^4+3x^3+3x^2−6x−10$", "sum of its digits. Now, here, you are not changing the digits when transferring the digits, so the remainder does not change. However, multiplying by two should double the remainder module nine and this is a contradiction unless the number is divisible by $9$. Solution Assume that the number $u = 9k$ and the representation of $u$ is $$\\overline{a_{n}a_{n-1}\\ldots a_{1} a_{0}} = u = 9k$$. Then $$\\overline{a_{0}a_{n}\\ldots a_{2} a_{1}} = 2u = 18k$$. Hence multiplying by 10 we get, $$\\overline{a_{0}a_{n}\\ldots a_{2} a_{1}} * 10 + a_{0} = 180k + a_{0}$$. Regrouping the digits and then writing the original number as $9k$, we get $$a_{0}*10^{n+1} + \\overline{a_{n}a_{n-1}\\ldots a_{1} a_{0}}= 180k + a_{0}$$ which implies that \\begin{align} &a_{0}*10^{n+1} + 9k = 180k + a_{0}\\\\ \\\\ &\\left(10^{n+1} -1 \\right) a_{0} = 171k\\\\ \\\\ &\\underbrace{\\overline{99\\ldots999}}_{n+1}a_{0} = 171k \\end{align} that is $$171k = 9*\\left( \\underbrace{\\overline{11\\ldots111}}_{n+1}\\right) a_{0}$$ hence $$19k = \\left( \\overline{11\\ldots 111}\\right) a_{0}$$ So, the problem comes down to finding the solutions of the above equation. Combinining this with Ross Millikan's Analysis, we get the equation $$10^{n+1} = 171k + a_{0}$$ and $$9k \\equiv a_{0} \\, \\mathrm{ mod } \\, 10$$ - But you have the solution $2\\cdot 0=0$, so this argument only shows that the sum of digits must be divisible by 9. – Per Manne Sep 28 '12 at 17:29 Unless" ]

[ "$$.......(1)$$ In $$0.\\overline7$$, one digit is repeating every time. $$\\therefore$$ We multiply both the sides of equation $$(1)$$ by $$10$$. $$x=0.\\overline7$$ $$\\Rightarrow x=0.777......$$ $$10×x=10×0.777....$$ $$10x=7.777....$$ $$....(2)$$ We can write $$7.777=7+0.7777$$ $$\\therefore$$ Equation $$(2)$$ can be written as $$10x=7+0.777$$ $$10x=7+0.\\overline7$$ $$......(3)$$ Substitute the value of $$0.\\overline7$$ in equation $$(3)$$ from equation $$(1)$$, i.e. $$\\because\\;\\;\\; x=0.\\overline7$$ $$\\therefore$$ In equation $$3$$ $$10x=7+x$$ $$.......(4)$$ Solving equation $$(4)$$ to find the value of $$x$$, i.e. $$10x=7+x$$ $$........(4)$$ Subtracting $$x$$ from both the sides, $$10x-x=7+x-x$$ $$(10-1)x=7$$ $$9x=7$$ Dividing both sides by $$9$$ $$\\dfrac{9x}{9}=\\dfrac{7}{9}$$ $$x=\\dfrac{7}{9}$$ Therefore, the repeating decimal, $$0.\\overline7=\\dfrac{7}{9}$$ Hence, option (C) is correct. ### What is the fraction form of $$0.\\overline7$$? A $$\\dfrac{3}{5}$$ . B $$\\dfrac{3}{8}$$ C $$\\dfrac{7}{9}$$ D $$\\dfrac{8}{5}$$ Option C is Correct # Comparison of Fractions through Representation on a Number Line • Consider two fractions as two points on the number line. • On a number line, a fraction which is placed to the left is smaller than the fraction which is placed to the right side. For example: Consider $$\\dfrac{1}{4}$$ and $$\\dfrac{5}{4}$$. $$\\dfrac{1}{4}$$ is placed to the left of $$\\dfrac{5}{4}$$. $$\\therefore$$ $$\\dfrac{1}{4}<\\dfrac{5}{4}$$ Note: While comparing two or more fractions, make sure that each one of them has the same denominator. #### Two points namely, $$P$$ and $$R$$, are plotted on the given number line. Determine the values of $$P$$", "$x=0.\\overline{001}\\Rightarrow$ …… (1) Since the number of recurring decimal number is $3$, so multiplying both sides of the equation (1) by $1000$, gives $1000\\times x=1000\\times 0.001001.....$ …… (2) Subtracting the equation (1) from (2) gives \\begin{align} & 1000x=1.001001..... \\\\ & \\underline{\\text{ }-x=0.001001.....} \\\\ & 999x=1 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{1}{999}$ Hence, the decimal number becomes $0.\\overline{001}=\\dfrac{1}{999}$ and it is in the $\\dfrac{p}{q}$ form. 4. Express ${0} {.99999}.....$ in the form of $\\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense. Ans: Let $x=0.99999.....$ ....... (a) Multiplying by $10$ both sides of the equation (a), gives $10x=9.9999.....$ …… (b) Now, subtracting the equation (a) from (b), gives \\begin{align} & 10x=9.99999..... \\\\ & \\underline{\\,-x=0.99999.....} \\\\ & \\,\\,9x=9 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{9}{9}$ $\\Rightarrow x=1$. So, the decimal number becomes $0.99999...=\\dfrac{1}{1}$ which is in the $\\dfrac{p}{q}$ form. Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense. Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal. 5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\\dfrac{ {1}}{ {17}}$ ? Perform the division to check your answer. Ans: Here the", "Question # What is the sum of $0.\\overline{6}$ and $0.\\overline{7}$ ?\\begin{align} & \\text{(A) 1}\\text{.}\\overline{\\text{3}} \\\\ & \\text{(B) 1}\\text{.3} \\\\ & \\text{(C) 1}\\text{.}\\overline{\\text{4}} \\\\ & \\text{(D) An irrational number} \\\\ \\end{align} Hint: We will convert non terminating recurring decimals $0.\\overline{6}$ and $0.\\overline{7}$ to fraction and then we will add those two fractions and at last we will convert the fraction into decimal to get the desired result. We will convert non terminating recurring decimals to fractions by assuming x=$0.\\overline{6}$ and y=$0.\\overline{7}$ x can also be written as 0.66666……. and y can also be written as 0.77777…… First, we will convert $0.\\overline{6}$ to fraction. We assumed x=0.66666….. and let this be equation 1. Now we multiply x with 10, multiplying x with 10 we will get, $\\Rightarrow 10x=6.6666.....$ and let this be equation 2. Subtracting equation 1 from equation 2 we will get 9x=6 and x can be written as $x=\\dfrac{6}{9}$ . So, $x=\\dfrac{6}{9}$. Now, we will convert $0.\\overline{7}$ to fraction. We assumed y=0.77777….. and let this be equation 3. Now we multiply y with 10, multiplying y with 10 we will get, $\\Rightarrow 10y=7.7777.....$ and let this be equation 4. Subtracting equation 3 from equation 4 we will get 9y=7 and x can be written as $y=\\dfrac{7}{9}$. So, $y=\\dfrac{7}{9}$. Now, according to the question we have to add x", "you'll get $\\overline{9} = - 1$.", "# 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 $\\frac{4}{9}$.", "### Home > PC > Chapter 9 > Lesson 9.3.2 > Problem9-116 9-116. Show whether or not $x − 5$ is a factor of $x^{3} + 4x^{2} − 11x − 30$. Set up the generic rectangle with the divisor on the left of the box. Input only the first terms of the dividend. Generic rectangle, 2 rows, 4 columns, left edge labeled, x minus 5, interior top left, labeled x cubed. Since $x$ divides into $x^3$, $x^2$ times, write the $x^2$ on top and multiply $x^2$ with the $−5$ and fill in the grid with the product. Labels added, top edge left, x squared, interior bottom left, negative 5, x squared. Add an expression to $−5x^2$ in order to get the needed $4x^2$. Add it to the right of $x^3$Label added, interior top, second from left, 9 x squared. Dividing $x$ into $9x^2$ gives $9x$. Add the $9x$ at the top. Then multiply $9x$ and $−5$ and fill the answer in the grid. Labels added: top edge, second from left, + 9 x, interior bottom, second from left, negative 45 x. Add an expression to $−45x$ in order to get the needed $−11x$. Add it to the right of $9x^2$Label added: interior top, second from right, 34 x. Dividing $x$ into $34x$ gives $34$. Add the $34$ at", "multiply the number by 100. Thus we obtain: $100x=274.\\overline{35}$ Next, we have to bring the repeating numbers 3 and 5 to the left side, so we have to again multiply the above equation by 100. Hence, we get: \\begin{align} & 100x\\times 100=274.\\overline{35}\\times 100 \\\\ & 10000x=27435.\\overline{35} \\\\ \\end{align} Consider the two equations: $10000x=27435.\\overline{35}$ …… (1) $100x=274.\\overline{35}$ …… (2) In the next step, we have to do subtraction. That is, equation (1) - equation (2) we obtain: \\begin{align} & 10000x-100x=27435.\\overline{35}-274.\\overline{35} \\\\ & 9900x=27161 \\\\ \\end{align} Next, by cross multiplication we get: $x=\\dfrac{27161}{9900}$ Therefore, the required rational number is $\\dfrac{27161}{9900}$. Hence, the correct answer for the question is option (d). Note: Here, first we have to identify how many digits are repeating which you can find with the help of the bar symbol. Depending on the numbers which keep on repeating we have to do the multiplication. Here, there are two numbers, so we are multiplying by 100. If you don’t multiply then you can’t take the repeating decimals to the left side which will then be difficult to remove the decimal places.", "only the $$x$$ is being multiplied by 2, so the first step will be to add 7 to both sides. Then we can divide both sides by 2. \\begin{align} 2x-7 &= 13\\\\ 2x-7 \\color{red}{+7} & =13 \\color{red}{+7}\\\\ 2x&=20\\end{align} Now divide both sides by 2: \\begin{align} 2x &=20 \\\\ \\dfrac{2x}{\\color{red}{2}}&=\\dfrac{20}{\\color{red}{2}}\\\\ x&= \\boxed{10}\\end{align} #### Check Just like with simpler problems, you can check your answer by substituting your value of $$x$$ back into the original equation. \\begin{align}2x-7&=13\\\\ 2(10) – 7 &= 13\\\\ 13 &= 13\\end{align} This is true, so we have the correct answer. Let’s look at one more two-step example before we jump up in difficulty again. Make sure that you understand each step shown and work through the problem as well. ### Example Solve: $$5w + 2 = 9$$ ### Solution As above, there are two operations: $$w$$ is being multiplied by 5 and then has 2 added to it. We will undo these by first subtracting 2 from both sides and then dividing by 5. \\begin{align}5w + 2 &= 9\\\\ 5w + 2 \\color{red}{-2} &= 9 \\color{red}{-2}\\\\ 5w &= 7\\\\ \\dfrac{5w}{\\color{red}{5}} &=\\dfrac{7}{\\color{red}{5}}\\\\w=\\boxed{\\dfrac{7}{5}}\\end{align} The fraction on the right can’t be simplified, so that is our final answer. #### Check Let $$w = \\dfrac{7}{5}$$. Then: \\begin{align}5w + 2 &= 9\\\\ 5\\left(\\dfrac{7}{5}\\right) + 2 &= 9\\\\ 7 + 2 &=", "the equation (1) by $1000$, gives $1000\\times x=1000\\times 0.001001.....$ …… (2) Subtracting the equation (1) from (2) gives \\begin{align} & 1000x=1.001001..... \\\\ & \\underline{\\text{ }-x=0.001001.....} \\\\ & 999x=1 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{1}{999}$ Hence, the decimal number becomes $0.\\overline{001}=\\dfrac{1}{999}$ and it is in the $\\dfrac{p}{q}$ form. 4. Express ${0} {.99999}.....$ in the form of $\\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense. Ans: Let $x=0.99999.....$ ....... (a) Multiplying by $10$ both sides of the equation (a), gives $10x=9.9999.....$ …… (b) Now, subtracting the equation (a) from (b), gives \\begin{align} & 10x=9.99999..... \\\\ & \\underline{\\,-x=0.99999.....} \\\\ & \\,\\,9x=9 \\\\ \\end{align} $\\Rightarrow x=\\dfrac{9}{9}$ $\\Rightarrow x=1$. So, the decimal number becomes $0.99999...=\\dfrac{1}{1}$ which is in the $\\dfrac{p}{q}$ form. Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense. Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal. 5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\\dfrac{ {1}}{ {17}}$ ? Perform the division to check your answer. Ans: Here the number of digits in the recurring block of $\\dfrac{1}{17}$ is to be determined. So, let us calculate", "$$x=0 . \\overline{24} .$$ Then $$100 x=24 . \\overline{24} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=24 . \\overline{24} \\\\ x &=0 . \\overline{24} \\end{aligned} yields $$99 x=24 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{24}{99}=\\frac{8}{33}$$ Exercise $$\\PageIndex{20}$$ $$0 . \\overline{882}$$ Exercise $$\\PageIndex{21}$$ $$0 . \\overline{84}$$ Let $$x=0 . \\overline{84} .$$ Then $$100 x=84 . \\overline{84} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=84 . \\overline{.84} \\\\ x &=0 . \\overline{84} \\end{aligned} yields $$99 x=84 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{84}{99}=\\frac{28}{33}$$ Exercise $$\\PageIndex{22}$$ $$0 . \\overline{384}$$ Exercise $$\\PageIndex{23}$$ $$0 . \\overline{63}$$ Let $$x=0 . \\overline{63} .$$ Then $$100 x=63 . \\overline{63} .$$ Subtracting on both sides of these equations \\begin{aligned} 100 x &=63 . \\overline{63} \\\\ x &=0 . \\overline{63} \\end{aligned} yields $$99 x=63 .$$ Finally, solve for $$x$$ by dividing by $$99 : x=\\frac{63}{99}=\\frac{7}{11}$$ Exercise $$\\PageIndex{24}$$ $$0 . \\overline{60}$$ Exercise $$\\PageIndex{25}$$ Prove that $$\\sqrt{3}$$ is irrational. Suppose that $$\\sqrt{3}$$ is rational. Then it can be expressed as the ratio of two integers p and q as follows: $\\sqrt{3}=\\frac{p}{q}$ Square both sides, $3=\\frac{p^{2}}{q^{2}}$ then clear the equation of fractions by multiplying both sides by $$q^{2}$$: $p^{2}=3 q^{2}$ Now p and q each have their own unique prime factorizations. Both $$p^{2}$$ and $$q^{2}$$", "sum of its digits. Now, here, you are not changing the digits when transferring the digits, so the remainder does not change. However, multiplying by two should double the remainder module nine and this is a contradiction unless the number is divisible by $9$. Solution Assume that the number $u = 9k$ and the representation of $u$ is $$\\overline{a_{n}a_{n-1}\\ldots a_{1} a_{0}} = u = 9k$$. Then $$\\overline{a_{0}a_{n}\\ldots a_{2} a_{1}} = 2u = 18k$$. Hence multiplying by 10 we get, $$\\overline{a_{0}a_{n}\\ldots a_{2} a_{1}} * 10 + a_{0} = 180k + a_{0}$$. Regrouping the digits and then writing the original number as $9k$, we get $$a_{0}*10^{n+1} + \\overline{a_{n}a_{n-1}\\ldots a_{1} a_{0}}= 180k + a_{0}$$ which implies that \\begin{align} &a_{0}*10^{n+1} + 9k = 180k + a_{0}\\\\ \\\\ &\\left(10^{n+1} -1 \\right) a_{0} = 171k\\\\ \\\\ &\\underbrace{\\overline{99\\ldots999}}_{n+1}a_{0} = 171k \\end{align} that is $$171k = 9*\\left( \\underbrace{\\overline{11\\ldots111}}_{n+1}\\right) a_{0}$$ hence $$19k = \\left( \\overline{11\\ldots 111}\\right) a_{0}$$ So, the problem comes down to finding the solutions of the above equation. Combinining this with Ross Millikan's Analysis, we get the equation $$10^{n+1} = 171k + a_{0}$$ and $$9k \\equiv a_{0} \\, \\mathrm{ mod } \\, 10$$ - But you have the solution $2\\cdot 0=0$, so this argument only shows that the sum of digits must be divisible by 9. – Per Manne Sep 28 '12 at 17:29 Unless", "bar, 4. \\begin{align*}\\frac{1}{4} = {4 \\overline{ ) 1 \\;\\;\\;}}\\end{align*} Since 1 cannot be evenly divided by 4, rewrite 1 as a decimal with a zero after the decimal point You can do this because \\begin{align*}1 = 1.0 = 1.00 = 1.000\\end{align*}. Before you divide, write a decimal point in the quotient directly above the decimal point in the dividend. Then divide. \\begin{align*}& \\overset{ \\ \\ 0.2}{4 \\overline{ ) {1.0 \\;}}}\\\\ & \\quad \\underline{-8}\\\\ & \\quad \\ \\ 2 \\end{align*} Continue adding zeroes after the decimal point and diving until the quotient has no remainder. \\begin{align*}& \\overset{ \\ \\ 0.25}{4 \\overline{ ) {1.00 \\;}}}\\\\ & \\quad \\underline{-8 \\;\\;}\\\\ & \\quad \\ \\ 20\\\\ & \\ \\ \\ \\underline{-20}\\\\ & \\qquad \\ 0 \\end{align*} The decimal form of the ratio \\begin{align*}\\frac{1}{4}\\end{align*} is 0.25. Example Rewrite the ratio 9:5 in decimal form. The ratio 9:5 can be expressed as the fraction \\begin{align*}\\frac{9}{5}\\end{align*}. Next, divide the term above the fraction bar, 9, by the term below the fraction bar, 5. \\begin{align*}& \\overset{ \\ \\ \\ 1}{5 \\overline{ ) { \\ 9 \\;}}}\\\\ & \\ \\underline{-5\\;}\\\\ & \\ \\ \\ 4 \\end{align*} There is a remainder. So, add zeroes after the decimal point in 9 to continue dividing. \\begin{align*}& \\overset{ \\ \\ 1.8}{5 \\overline{ ) {9.0 \\;}}}\\\\ & \\underline{-5\\;}\\\\ & \\quad 40\\\\ &", "## Precalculus (6th Edition) Blitzer To check whether $z$ can be equal to $3$ or $-1$. Let us consider the system of equations as follows and mark them: $3x+9y=0$ (I) $9y+3z=0$ (II) $2x+2y-4z=-4$ (III) And eliminate y from the equations (I) and (II) to get: \\begin{align} & 3x+9y\\text{ }=0 \\\\ & \\underline{\\text{ }-9y-3z=0} \\\\ & 3x\\text{ }-3z=0 \\\\ \\end{align} Add 3z to both sides to get, \\begin{align} & 3x-3z+3z=0+3z \\\\ & 3x=3z \\end{align} And divide both sides of the equation by 3 to get, \\begin{align} & \\frac{3x}{3}=\\frac{3z}{3} \\\\ & x=z \\end{align} Put $x=z$ in equation (III) to get the value of y, \\begin{align} & 2x+2y-4\\left( x \\right)=-4 \\\\ & 2x+2y-4x=-4 \\\\ & 4y=-4 \\end{align} And divide both sides by 4 to get, \\begin{align} & \\frac{4y}{4}=\\frac{-4}{4} \\\\ & y=-1 \\end{align} . Put the value of y in equation (I) to get the value of x, \\begin{align} & 3x+9\\left( -1 \\right)=0 \\\\ & 3x-9=0 \\\\ \\end{align} Add 9 to both sides to get, \\begin{align} & 3x-9+9=0+9 \\\\ & 3x=9 \\end{align} And divide both sides by 3 to get, \\begin{align} & \\frac{3x}{3}=\\frac{9}{3} \\\\ & x=3 \\end{align}", "### Home > CCAA8 > Chapter 3 Unit 4 > Lesson CCA: 3.3.1 > Problem3-81 3-81. Find each of the following products by drawing and labeling a generic rectangle or by using the Distributive Property. 1. $−4y(5x+8y)$ • Make and label a generic rectangle. Generic rectangle left edge$-4y$ interior left blank interior right blank bottom edge left$5x$ bottom edge right$+8y$ Fill in the boxes, and then add the boxes' labels together Generic rectangle left edge$-4y$ interior left$-20xy$ interior right $-32y^2$ bottom edge left$5x$ bottom edge right$+8y$ $−20xy−32y^2$ 1. $9x(−4+10y)$ • Distribute the $9x$ over each of the terms in the parentheses. $9x(−4)+9x(10y)$ Simplify. $−36x+90xy=90xy−36x$ 1. $(x^2−2)(x^2+3x+5)$ • Try using a generic rectangle to solve this problem. $x^4+3x^3+3x^2−6x−10$", "# Calculate: (d)/(5)=(12)/(80) ## Expression: $\\sqrt[5]{9x-2}=2$ Raise both sides of the equation to the power of $5$ $9x-2=32$ Move the constant to the right-hand side and change its sign $9x=32+2$ $9x=34$ Divide both sides of the equation by $9$ \\begin{align*}&x=\\frac{ 34 }{ 9 } \\\\&\\begin{array} { l }x=3 \\frac{ 7 }{ 9 },& x=3.\\overset{ \\cdot }{ 7 } \\end{array}\\end{align*}", "$ax+by+c=0$. Equation’s Solution : A particular set of values of the variables, which when substituted for the variables in the equation makes the two sides of the equation equal, is called the solution of the equation. Now, let us look at the percentage formula. The value of x% in y can be written as: $\\dfrac{x}{100}\\times y$ Now, on comparing with the given values we get, $x=8,y=24,000$ Let us assume that the cost after discount as A , cost after season discount as B and cost after sales tax on the remaining amount as C. \\begin{align} & \\Rightarrow A=24000-\\dfrac{8}{100}\\times 24000 \\\\ & \\Rightarrow A=24000-\\left( 8\\times 240 \\right) \\\\ & \\Rightarrow A=24000-1920 \\\\ & \\therefore A=Rs.22,080 \\\\ \\end{align} Now, again we need to consider the values of x and y as, $x=5,y=22,080$ By using this and simplifying we get, $\\Rightarrow B=A-\\dfrac{x}{100}\\times A$ \\begin{align} & \\Rightarrow B=22080-\\dfrac{5}{100}\\times 22080 \\\\ & \\Rightarrow B=22080-\\left( 5\\times 220.8 \\right) \\\\ & \\Rightarrow B=22080-1104 \\\\ & \\therefore B=Rs.20,976 \\\\ \\end{align} Let us assume that the sales tax will be imposed on the remaining amount as D. Now, we use $x=10,y=20,976$. $\\Rightarrow D=\\dfrac{x}{100}\\times B$ \\begin{align} & \\Rightarrow D=\\dfrac{10}{100}\\times 20,976 \\\\ & \\Rightarrow D=10\\times 209.76 \\\\ & \\therefore D=Rs.2,097.60 \\\\ \\end{align} \\begin{align} & \\Rightarrow C=20,976+D \\\\ & \\Rightarrow C=20,976+2,097.60 \\\\ & \\therefore C=Rs.23,073.60 \\\\ \\end{align} Hence, the correct", "0.01$. By gluing nines we make the reverse transformation to avoid recurring nines, hence be injective: $0.23\\overline{90}$ will properly map to $(0.2\\overline{90},0.3\\overline{90})$ instead of $(0.2\\overline 9,0.3)=(0.3,0.3)$, which is an image of $0.33$. – CiaPan Jul 14 '16 at 14:36", "quantity $$\\frac{1}{3}$$. When written as a decimal, we get $$\\frac{1}{3} = 0.33\\overline{3}$$. Now consider multiplying both sides of this equation by 3. That is, consider that $$3\\times\\frac{1}{3} = 3\\times 0.33\\overline{3}$$. Evaluating the left side of the equation, we have $$3\\times\\frac{1}{3} = \\frac{3}{3} = 1$$. Now consider the right side of the equation. Recall that when we multiply by 3, we are taking our original quantity, and adding it to itself three times. Using the notation that we learned in elementary school, it would look like this: Clearly, in every decimal place, we would have $$3+3+3 = 9$$. Hence $$3\\times 0.33\\overline{3} = 0.99\\overline{9}$$. Since we still have equality between the right and left sides of the equation, this gives us $$1 = 0.99\\overline{9}$$. ## Proof 2 Let $$x = 0.99\\overline{9}$$. If we multiply both sides of this identity by 10, we will end up with the same string of digits on the right, but the decimal will move one place to the right. Thus $$10x = 9.99\\overline{9}$$. From each side of the equation, we will subtract $$x$$: $$10x – x = 9.99\\overline{9} – 0.99\\overline{9}$$. This gives us $$9x = 9$$. Solving for $$x$$, we have $$x = 1$$. Substituting this value into the original identity gives us the desired result, which is $$1 = 0.99\\overline{9}$$. ## Proof 3 This", "Again our first step is write $$x = 0.277..$$. Now multiplying by $$10$$ gives us $$10x = 2.77..$$. This time we need another equation to match the recurring part of the equation. So multiplying by $$10$$ again gives $$100x = 27.77..$$. Now we have two equations with the same recurring part we subtract one from the other as before. \\begin{aligned} x & = 0.277.. \\\\ 10x & = 2.77.. \\\\ 100x & = 27.77.. \\\\ 90x & = 25 \\\\ x & = {25 \\over 90} \\\\ x & = {5 \\over 18} \\\\ \\end{aligned} So in its lowest terms $$0.277.. = {5 \\over 18}$$. Example 3 Write 0.0855.. as a fraction in its lowest terms. As always our first step it to write $$x = 0.0855..$$. This time to get two equations with the same recurring part after the decimal point we need to multiply by $$100$$ and then $$1000$$. This gives us: \\begin{aligned} 100x & = 8.55.. \\\\ 1000x & = 85.55.. \\\\ 900x & = 77 \\\\ x & = {77 \\over 900} \\\\ \\end{aligned} This cannot be simplified any further so $$0.855.. = {77 \\over 900}$$. Example 4 Write 0.0234234.. as a fraction in its lowest terms. This time the '$$234$$' part is the important bit. Now to get two multiples of x with", "bottom, second from left, negative 45 x. Add an expression to $−45x$ in order to get the needed $−11x$. Add it to the right of $9x^2$. Label added: interior top, second from right, 34 x. Dividing $x$ into $34x$ gives $34$. Add the $34$ at the top. Then multiply $34$ and $−5$ and fill the answer in the area model. Labels added: top edge, second from right, + 34, interior bottom, second from right, negative 170. Add an expression to $−170$ in order to get the needed $−30$. Add it to the right of $34x$. This is the remainder. Labels added: top edge, right, 140 divided by quantity x minus 5. Interior top right, 140." ]

[ "is usually an average, that is, if a car traveled the distance of 210 km between two cities and to take this route, it took an hour and a half, so we will get a speed of 140 kilometers per hour (140 km/h). This does not obviously mean that the car has constantly shifted at this speed, there may be parts of the route on which it has been faster and parts where it has circulated more slowly. ## Math Formula kph=text(knots) xx 1.852 ## Conversion Examples knot to kph knot to kph 0.2 knots = 0,3704 kph 9 knots = 16,6680 kph 0.5 knots = 0,9260 kph 10 knots = 18,5200 kph 1 knots = 1,8520 kph 15 knots = 27,7800 kph 2 knots = 3,7040 kph 20 knots = 37,0400 kph 3 knots = 5,5560 kph 25 knots = 46,3000 kph 4 knots = 7,4080 kph 30 knots = 55,5600 kph 5 knots = 9,2600 kph 35 knots = 64,8200 kph 6 knots = 11,1120 kph 40 knots = 74,0800 kph 7 knots = 12,9640 kph 45 knots = 83,3400 kph 8 knots = 14,8160 kph 50 knots = 92,6000 kph", "it took an hour and a half, so we will get a speed of 140 kilometers per hour (140 km/h). This does not obviously mean that the car has constantly shifted at this speed, there may be parts of the route on which it has been faster and parts where it has circulated more slowly. ## Math Formula kph=text(m/s) xx 3.6 ## Conversion Examples mps to kph mps to kph 0.2 m/s = 0,7200 kph 9 m/s = 32,4000 kph 0.5 m/s = 1,8000 kph 10 m/s = 36 kph 1 m/s = 3,6000 kph 15 m/s = 54 kph 2 m/s = 7,2000 kph 20 m/s = 72 kph 3 m/s = 10,8000 kph 25 m/s = 90 kph 4 m/s = 14,4000 kph 30 m/s = 108 kph 5 m/s = 18 kph 35 m/s = 126 kph 6 m/s = 21,6000 kph 40 m/s = 144 kph 7 m/s = 25,2000 kph 45 m/s = 162 kph 8 m/s = 28,8000 kph 50 m/s = 180 kph", "Eric, I probably did exaggerate and numbers are probably off. The survey on my board had several bikers saying they get about 200 km on a tank, which is usually around 15 liters. That would be around 30 mpg. Regarding 10K RPM, consider the typical 600cc super sport rider out to have fun: 1) hard acceleration/breaking 2) pop some wheelies 3) fast passes etc. Sure, 10K RPM is not unusual. Wanna hear my pipes? Here, let me rev it some ;) Really, some riders look for any excuse to rev higher. Overall my numbers are probably off some. Still, the average rider will get similar MPG to many average size cars, which doesn't make sense (as bikes have a single passanger and are might smaller/lighter). I think your idea of an \"average size car\" and the \"average car in america\" might be different? You're using kilometers and liters, are you in Europe? Canada? US cars and canadian cars are pretty similar but the average car in europe gets a lot better gas milage than the average car in the US. The average mpg for all new cars sold in the US in 2004 was just 24.6 mpg according to this Washington Post story... http://www.washingtonpost.com/wp-dyn/articles/A36215-2004Jun12.html I would guess the average mpg for new cars sold in Europe is over", "Question (a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp = 746 W) to reach a speed of 15.0 m/ s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m-high hill in the process? 1. $3.2\\textrm{ s}$ 2. $4.0\\textrm{ s}$", "(2.518 mile) urban trip at an average speed of 18.7 km/h (11.6 mph) and at a maximum speed of 50 km/h (31 mph). The extra-urban cycle or EUDC lasts 400 seconds (6 minutes 40 seconds) at an average speed 62.6 km/h (39 mph) and a top speed of 120 km/h (74.6 mph).[24] EU fuel consumption numbers tend to be considerably lower than corresponding US EPA test results for the same vehicle. For example, the 2011 Honda CR-Z with a six-speed manual transmission is rated 6.1/4.4 l/100 km in Europe and 7.6/6.4 l/100 km in the United States[citation needed]. In the European Union advertising has to show Carbon dioxide (CO2)-emission and fuel consumption data in a clear way as described in the UK Statutory Instrument 2004 No 1661.[25] Since September 2005 a color-coded \"Green Rating\" sticker has been available in the UK, which rates fuel economy by CO2 emissions: A: <= 100 g/km, B: 100–120, C: 121–150, D: 151–165, E: 166–185, F: 186–225, and G: 226+. Depending on the type of fuel used, for gasoline A corresponds to about 4.1 l/100 km (69 mpg-imp; 57 mpg-US) and G about 9.5 l/100 km (30 mpg-imp; 25 mpg-US).[26] Ireland has a very similar label, but the ranges are slightly different, with A: <= 120 g/km, B: 121–140, C: 141–155, D: 156–170,", "course, this is not true in the real world. In reality, we can think of 70 kilometers per hour as the average speed of the car.", "the average speed of 50 km/hr. How much distance would … What would you prefer to travel a short distance? Short distance - small car (good fuel economy, easier to park) or motorcycle (good fuel … How much distance can light travel in one year? As already noted, the speed of light (expressed in meters per second) means that light … How much distance does light travel in one year? As already noted, the speed of light (expressed in meters per second) means that light … How to travel long distance with a dog? The good news is that there are plenty of simple things you can do to … What balloon powered car design will travel the greatest distance? To build a Balloon Rocket Car that can extract the most energy out of the … How much distance does a light travel in a year? Light-year is the distance light travels in one year. Light zips through interstellar space at … A car travelling on a straight road at 15? A car travelling on a straight road at 15.0 meter per second accelerates uniformly to … A car is travelling on a straight road? A car is travelling on a straight road. The maximum velocity the car can attain … A car travelling at 28 hits a", "hour. What is the distance traveled by the car within 20 second in meter? First, we need to convert the speed of the car from km/h to m/s since the time is given in seconds. 1 km/h = 1000 m/3600 s = 0.2778 m/s So, the speed of the car is: 36 km/h = 36 x 0.2778 m/s = 10 m/s Now, we can use the formula: distance = … error: Content is protected !!", "hours. What is the average speed of the car? (A) 30 km/h (B) 60 km/h (C) 40 km/h (D) 100 km/h", "minus 150 hertz which gives 15.0 meter per seconds is the speed of the car.", "eleven L/100 kilometer ) ). The proportion of drive on high travel rapidly roadway vary from four % indiana ireland to forty-one % indium the netherlands. When the united states national maximum speed law ‘s fifty-five miles per hour ( eighty-nine kilometers per hour ) speed limit be mandate from 1974 to 1995, there constitute complaint that fuel economy could decrease alternatively of increase. The 1997 toyota Celica make well fuel-efficiency at one hundred five kilometers per hour ( sixty-five miles per hour ) than information technology perform astatine sixty-five kilometers per hour ( forty miles per hour ) ( 5.41 L/100 kilometer ( 43.5 mpg‑US ) vanadium 5.53 L/100 kilometer ( 42.5 mpg‑US ) ), although even better at sixty miles per hour ( ninety-seven kilometers per hour ) than astatine sixty-five miles per hour ( one hundred five kilometers per hour ) ( 48.4 mpg‑US ( 4.86 L/100 kilometer ) five 43.5 mpg‑US ( 5.41 L/100 kilometer ) ), and information technology best economy ( 52.6 mpg‑US ( 4.47 L/100 kilometer ) ) at merely twenty-five miles per hour ( forty kilometers per hour ). other fomite tested accept from 1.4 to 20.2 % well fuel-efficiency astatine ninety kilometers per hour ( fifty-six miles per hour ) vs. one hundred five kilometers per hour ( sixty-five", "moves faster, an object A that moves 15 centimeters every 5 seconds or an object B that moves 24 centimeters every 8 seconds? Car A consumes 12 gallons of fuel for a distance of 240 miles. Another car B consumes 25 gallons for a distance of 550 miles. Which of the two cars consumes less fuel? Convert the unit rate 60 kilometers/hour into kilometers / minute. Convert the unit rate 72 kilometers/hour into meters / second. Solutions and explanations Updated: 20 January 2017 (A Dendane)", "10 km sectikon would have 100 vehicles in transit. Also, consider delay. If the average speed is 60 km/hr, it takes 1 minute for a vehicle to enter the 1 km stretch and travel the length, so each transit is delayed 1 minute, and there are 10 items in transit. On a 10 km stretch, each is delayed 10 minutes, and there are 100 items in transit at any time. 3*108m/second: not just a good idea, it's the LAW!* Remember that the wavelength is a length, measured in meters. For a wave with a period of $$\\ T \\$$ and a velocity $$\\ v \\$$, the signal travels $$\\ \\lambda = v.t \\$$ in one period. Thus if you increase the length of the medium (the 'electrical length'), you'll get more repetitions of the electrical signal, i.e. 'more wavelength'.", "A car drives 5.0 km north, then 7.3 km east, then 5.1 km northeast, all at a constant velocity. If the car had to perform 2.6 � 106 J of work during this trip, what was the magnitude of the average frictional force on the car?", "in my experience it really becomes significant when you compare summer and winter driving conditions. The speed / fuel consumption coefficient is 0.028449, or, in other words, by driving 20 km/h slower I save 0.56898 liter per 100 km. But I assumed that the savings were only due to the highway part of our trip. So I should calculate the absolute numbers: the full trip is 59.9 km, so the saving is 0.3408 liter. For just the highway part, this is 0.8374 liter per 100 km. The average price of Euro95 right now is 1.600 euro/liter. This means that I save 1.34 euro per 100 km. When driving 120 km/h, driving 100 km takes 50 minutes, but at 100 km/h, it takes 60 minutes, a difference of 10 minutes. Of course, I assume here that all the correlations are linear, but I know that they are not. In fact, if you drive faster, say 130 km/h or 140 km/h, your fuel consumption will increase more dramatically, and the savings will be bigger. # Conclusion By driving slower on the highway, you can save 1.34 euro, but you arrive 10 minutes later. This means you save 8 euro by driving for one hour. Maybe not enough to earn a living, but if you have the time: relax and drive", "as far per liter gas how many liters of gas will it take to travel 135 kilometers driving at 50 kilometers per hour? A. 3 B. 5 C. 9 D. 11 E. 15 Solution: 75% of 12 km/liter=9km/liter. So the fuel efficiency is 9 km/liter when driving at 50 kmph. So, to drive 9 km needs 1 liter to drive 135 km need 135/9 liter = 15 liter 22. The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the rear wheel? A. 20 ft B. 25 ft C. 750 ft D. 900 ft E. 1000 ft Solution: Circumference of front wheel C1=30 ft, Circumference of front wheel C2=36 ft If Rotation1=x ATQ, 30(x+5)=36x Or, x=25 So, distance =30(25+5)=30×30=900 23. Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route? (A) 2/3 (B) 1 (C) 4/3 (D) 8/5 (E) 3 Solution: The speed of car X is (distance)/(time) =", "# Linear Motion 1. Jan 14, 2008 ### iwonde According to a Scientific American article, current freeways can sustain about 2400 vehicles per lane per hour in smooth traffic flow at 96 km/h (60 mi/h). With more vehicles the traffic flow becomes \"turbulent\" (stop-and-go). a) If a vehicle is 4.6 m (15ft) long on the average, what is the average spacing between vehicles at the above traffic density? b) Collision-avoidance automated control systems, which operate by bouncing radar or sonar signals off surrounding vehicles and then accelerate or brake the car when necessary, could greatly reduce the required spacing between vehicles. If the average spacing is 9.2m (two car lengths), how many vehicles per hour can a lane of traffic carry at 96 km/h? 2. Jan 14, 2008 ### rohanprabhu Firstly, look at the data you have been given. Traffic flows at 96 km/h. It means that, in 1 hour, 96km stretch of traffic is cleared. Also, 2400 vehicles per lane can be sustained in 1 hour. It means that 2400 vehicles must occupy a minimum of 96km. If the no. of cars are increased, then at the same speed, more no. of vehicles occupy lesser distance causing congestion. Here, the key to the problem is that u need to think in terms of length only. Now, 96km", "was the higher wattage the bulb was set to order... ( MC ) at q items is the instantaneous rate of 1.5 cm/min these problems the speed needed to the... At the same race car at time t, the two quantities change in denominator... On this link for the entire trip was 80 kilometers per hour amps per...Kastatic.Org and *.kasandbox.org are unblocked 8 the units on the circuit, measured in watts from.", "## The average speed of a vehicle is 20 metre per second .the Time taken is 40minutes . What is the distance Question The average speed of a vehicle is 20 metre per second .the Time taken is 40minutes . What is the distance in progress 0 4 weeks 2021-11-06T09:52:13+00:00 2 Answers 0 views 0 48000 meters Step-by-step explanation: Speed = 20 meter/second time taken = 40 minutes 1 minute = 60 seconds 40 minutes = (40 × 60 ) seconds 40 minutes = 2400 seconds distance = speed × time = 20 meter/second × 2400 seconds = 48000 meters", "# A car is traveling at a constant rate of 45 km/h A car is traveling at a constant rate of 45 km/h. The distance traveled by the car from 10:40 am to 1:00 pm is 1. 105 km 2. 120 km 3. 150 km 4. 165 km" ]

[ "# A car is travelling at a rate of 70 km per hour leaves the house 2 hours after a truck has left and overtakes it in 5 hours. At what rate was the truck travelling? Jul 23, 2017 The speed of the truck is $= 50 k m {h}^{-} 1$ #### Explanation: Let the speed of the truck be $= v k m {h}^{-} 1$ The distance travelled by the truck is $d = v \\cdot 7 = 7 v k m$ The distance travelled by the car is $d = 5 \\cdot 70 = 350 k m$ Therefore, $7 v = 350$ $v = \\frac{350}{7} = 50 k m {h}^{-} 1$", "to manageable pieces Voyage Expenses = Driver + Gas + Fixed Costs Rule #3 - What do we need in order to talk about these things in a useful way? Everything is in Hours and Km, so it would be good to define things in these terms. Let's define a few things. H = Hours to complete the trip. We have variable gas costs, so it probably is helpful to define a function that will express this. Speed of Truck = 110 kmh + t kmh t kmh = number of kmh over 100 kmh for which we know the mileage. I'm not really getting a feel for the problem, so let's just play with it a little. At 110 kmh, we have 450 km / 110 kmh = 4.0909 hours and 450 km / 8 km/L = 56.25 L and 56.25 L *$0.68/L = $38.25 Now we have the flavor of the calculation. At (110+t) kmh, we have 450 km / (110+t kmh) = H and 450 km / (8-0.10t km/L) and 450 km / (8-0.10t km/L) *$0.68/L Gas Expense = [450 km / (8-0.10t km/L)] * $0.68/L Voyage Expenses = [450 km / (110+t kmh)]*($35.00/h) + [450 km / (8-0.10t km/L)] * $0.68/L + [450 km / (110+t kmh)]*($15.50/h) After a very, very careful surgery consisting", "A car travels $8$ km in the first quarter of an hour, $6$ km in the second quarter and $16$ km in the third quarter. The average speed of the car in km per hour over the entire journey is 1. $30$ 2. $36$ 3. $40$ 4. $24$", "a car was $20,000 in 2014,$15,000 in 2015 and $11,200in 2016. What is the rate of depreciation of the price of car per year? A. $$15\\%$$ B. $$25\\%$$ C. $$30\\%$$ D. $$35\\%$$ 3- 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 45 cm, what is the volume of the box? A. 81 cm$$^3$$ B. 162 cm$$^3$$ C. 243 cm$$^3$$ D. 3,375 cm$$^3$$ 4- How many possible outfit combinations come from six shirts, three slacks, and five ties? A. 15 B. 18 C. 30 D. 90 5- A bank is offering $$4.5\\%$$ simple interest on a savings account. If you deposit$8,000, how much interest will you earn in five years? A. $360 B.$720 C. $1,800 D.$3,600 6- What is the value of $$6^4$$? A. 216 B. 1,296 C. 7,776 D. 46,656 7- 15 is What percent of 20? A. $$20\\%$$ B. $$25\\%$$ C. $$75\\%$$ D. $$125\\%$$ 8- The perimeter of the trapezoid below is 56. What is its area? A. 56 cm B. 132 cm C. 156 cm D. 175 cm 9-In five successive hours, a car traveled 40 km,45 km,50 km,35 km and 55 km. In the next five hours, it traveled with an average speed of", "SubsectionDistance Applications Distance traveled is equal to the average rate times the time traveled at that rate, $d=r\\cdot t\\text{.}$ These uniform motion problems usually have a lot of data, so it helps to first organize that data in a chart and then set up a linear system. In this section, you are encouraged to use two variables. ###### Example111 Naomi traveled a total of $4$ hours and $875$ miles by car and by plane. Driving to the airport by car, she averaged $50$ miles per hour. In the air, the plane averaged $320$ miles per hour. How long did it take her to drive to the airport? Solution We are asked to find the time it takes her to drive to the airport; this indicates that time is the unknown quantity. We begin by identifying two variables: let $x$ represent the time (in hours) it took to drive to the airport and let $y$ represent the time (in hours) spent in the air. Since the total time traveled was $4$ hours, we have our first equation \\begin{equation*} x+y=4. \\end{equation*} We use the formula $d=r\\cdot t$ to fill in the unknown distances: \\begin{gather*} \\text{Distance traveled in the car: }\\, d=r\\cdot t=50\\cdot x\\\\ \\text{Distance traveled in the air: }\\, d=r\\cdot t=320\\cdot y \\end{gather*} The total amount of miles traveled is", "constant speed of $$120 \\text{ km}{\\cdot}\\text{h}^{-1}$$ for 9 minutes. Over the next 2 minutes, it reduces its speed at a steady rate until it reaches a speed of $$60 \\text{ km}{\\cdot}\\text{h}^{-1}$$. The car travels at $$60 \\text{ km}{\\cdot}\\text{h}^{-1}$$ for 4 minutes. Over the next 1 minute, it reduces its speed at a steady rate and comes to a complete stop. 1. Is the data in this situation discrete or continuous? 2. Use the information provided to draw a suitable graph. 3. On your graph indicate the sections where the graph is decreasing and constant. 1. Continuous 2. and 3.", "it'd take (60/75) hours (0.8 hours, or 48 minutes) to go at 75 mph versus 60/65 hours (0.92 hours, or 55 minutes and about 20 seconds) to go at 65 mph. How much was 7 minutes of my time worth? At the wages I was working at that point (about $15 an hour),$1.75. How much was 1.5 gallons of gas worth? At that time, roughly $6. I decided to drive slower. However, on the way back, I had my girlfriend, which meant I, ahem, valued the time saved from driving faster more. I drove faster on the way back. To this day I've yet to make enough money in hourly terms to justify regularly driving 75 mph over driving 65 mph on the highway. At the current average gasoline price in New York state ($3.92/gal), I'd have to be making $50.40 an hour to justify it, which for a standard 8-hour work week is$104,832 per year. ------ This analysis is relatively simple. There are a couple of (to my mind, anyway) interesting little wrinkles to this calculation, the magnitude of which I can't calculate off the top of my head but I think on balance points to even more fuel consumption at higher speed. The first wrinkle occurs when we relax the assumption that the drag coefficient is", "14 views A car travels $8 \\: km$ in the first quarter of an hour, $6\\: km$ in the second quarter and $16\\: km$ in the third quarter. The average speed of the car in $km$ per hour over the entire journey is 1. $30$ 2. $36$ 3. $40$ 4. $24$ edited | 14 views", "and, with fuel costing 0.68 per Liter (how OLD is this problem??) the cost of fuel for the trip is (-45v+ 8550L)(0.68)= -30.6v+ 5814. At speed v, the time required for the trip will be (distance divided by speed) 450/v hours. Since drivers are paid $35 per hour, and the cost of running the truck is$15.50 per hour, the total cost for the truck per hour is \\$50.50 per hour and the total cost for the trip, excluding fuel, will be 50.50(450/v)= 22725/v. The total cost of the trip, at speed v> 110 is -30.5v+ 5814+ 22725/v. That's the function you want to minimize. Last edited: Jul 17, 2006", "Decrease the speed Maintain the speed at $60$ km/hour Cannot be determined Answer the following questions based on the information given below. The petrol consumption rate of a new model car 'Palto' depends on its speed and may be described by the graph below. Manasa makes the $200$ km trip from Mumbai to Pune at a steady speed of $60$ km/hour. What is the amount of petrol consumed for the journey? $12.5$ litres $13.33$ litres $16$ litres $19.75$ litres", "car. That traveled 130 miles in 2 hours, so we do: $130 / 2 = 65$ The second car, therefore, traveled at 65 miles per hour and was therefore faster. To find the unit rate for any process, work out the most sensible units to use (miles per hour, work per day, etc.) and then use the unit rate to divide the quantity by the number of units. ## Examples $135 beats in 9 hours$", "If a car went the first third of the distance at 80 kmh in the Problem Solving forum “Here''s an algebraic approach: Let d = 1/3 of the total distance traveled So, the car traveled d km at 80 kmh, then the car traveled d km at 24 kmh, and then d km at 48 kmh Average speed = (total distance traveled)/(total travel time) Total distance traveled = d + d + d = 3d time = ...” May 6, 2018 GMATGuruNY posted a reply to W, X, Y and Z each represent a different number in the Problem Solving forum “Sum of all 4 rows = 3 + 6 + 9 + 0 = 18. Sum of all 4 columns = -2 + 7 + n + 5 = n + 10. The values in blue must be EQUAL, since each represents the sum of all 16 values in the table. Thus: n+10 = 18 n=8. The correct answer is C.” May 6, 2018 Brent@GMATPrepNow posted a reply to Sheila works 8 hours per day on Monday, Wednesday and Friday in the Problem Solving forum “We can avoid long division if we make an important observation. Let''s take it from the point were you got: earning per hour = $324/36 You might notice that$360/36 = $10 Also", "B. $$2:3$$ C. $$5:9$$ D. $$5:6$$ E. $$11:16$$ 4- If $$40\\%$$ of a class are girls, and $$25\\%$$ of girls play tennis, what percent of the class play tennis? A. $$10 \\%$$ B. $$15 \\%$$ C. $$20 \\%$$ D. $$40 \\%$$ E. $$80 \\%$$ 5- What is the value of $$y$$ in the following system of equation? $$3x-4y= -20$$ $$-x+2y= 10$$ A. $$-1$$ B. $$-2$$ C. $$1$$ D. $$4$$ E. $$5$$ 6- In five successive hours, a car travels 40 km, 45 km, 50 km, 35 km, and 55 km. In the next five hours, it travels with an average speed of 50 km per hour. Find the total distance the car traveled in 10 hours. A. 425 km B. 450 km C. 475 km D. 500 km E. 1,000 km 7- How long does a 420–miles trip take moving at 50 miles per hour (mph)? A. 4 hours B. 6 hours and 24 minutes C. 8 hours and 24 minutes D. 8 hours and 30 minutes E. 10 hours and 30 minutes 8- What is the difference between the smallest 4–digit number and the biggest 4–digit number? A. 6666 B. 6789 C. 8888 D. 8999 E. 9999 9- What is the value of $$6^4$$? A. 6 B. 24 C. 36 D. 216 E. 1,296 10- Right triangle", "each year? Using estimates of number of personal vehicles and annual mileage, combined with an estimate that 30% of these miles would be driven more slowly than otherwise with drivers adopting my techniques, my estimate is that about 468,000,000,000 miles would be driven more slowly. I estimate that this would result in 1.8 X 10^9 (1.8 billion) extra hours on the road. Does this make sense? Well, I'm one of 200,000,000 drivers and I calculated in an earlier post that I'll lose about 47 hours per year. If each of the 200 million drivers lost that much, the total would be 9.4 X 10^9 hours. But I estimate that I probably drive something like 30% more than average so adjust this down to 7.2 X 10^9 hours. Figure somewhere between these numbers is right, I'll use 4.5 X 10^9 or 4 and one half billion hours. I'll say the average hour is worth $30.00, in that case the lost time is worth$135 billion. Considering I calculated we'd save $36 billion in oil imports, it would seem not to be worth it. Are there any mitigating factors? Not many. First would be finding ways to avoid productivity loss when on the road. Talk radio? Books on tape? Cell phones? These all may help but they certainly won't eliminate the", "Gas + Fixed Costs Rule #3 - What do we need in order to talk about these things in a useful way? Everything is in Hours and Km, so it would be good to define things in these terms. Let's define a few things. H = Hours to complete the trip. We have variable gas costs, so it probably is helpful to define a function that will express this. Speed of Truck = 110 kmh + t kmh t kmh = number of kmh over 100 kmh for which we know the mileage. I'm not really getting a feel for the problem, so let's just play with it a little. At 110 kmh, we have 450 km / 110 kmh = 4.0909 hours and 450 km / 8 km/L = 56.25 L and 56.25 L * $0.68/L =$38.25 Now we have the flavor of the calculation. At (110+t) kmh, we have 450 km / (110+t kmh) = H and 450 km / (8-0.10t km/L) and 450 km / (8-0.10t km/L) * $0.68/L Gas Expense = [450 km / (8-0.10t km/L)] *$0.68/L Voyage Expenses = [450 km / (110+t kmh)]*($35.00/h) + [450 km / (8-0.10t km/L)] *$0.68/L + [450 km / (110+t kmh)]*(\\$15.50/h) After a very, very careful surgery consisting of algebra and unit checks, this gives: Voyage Expenses(t)", "of the book product = $72 Profit of printing company = 20% of 72 = 0.2 x 72 = 14.4 Now the cost of the book = 72 +14.4 =$ 86.4 Profit of the bookseller = 25% of 86.4 = 21.4 Finally, the cost of the book = 86.4 + 21.4 = $108 5. A car traveled 100 km with half the distance at 40 km/h and the other half at 80 km/h. Find the average speed of the car for the whole journey. 1. 53 km/hr 2. 53.33 km/hr 3. 54 1/4 km/hr 4. 55 km/hr 5. 56 km/hr The time, car took for the first half, 50/40 = 1.25 hrs and for the second half 50/80 = 0.625 hrs Total time = 1.25 + 0.625 = 1.875 hrs Average speed = 100/1.875 = 53.3 km/hr 6. 8 : ? :: 1 : 4 1. 24 2. 16 3. 0 4. 32 5. 20 ? × 1 = 8 × 4 ? = 32 7. A train takes 50 minutes for a journey if it runs at 48 km/hr. The rate at which the train must run to reduce the time to 40 minutes will be 1. 50 km/hr 2. 55 km/hr 3. 60 km/hr 4. 57 km/hr 5. 65 km/hr (50 × 48)/40 = 60 km/hr", "Total Income =$ 245000 Total relief = 3 × 15000 + 30000 = $75000 Rest income = 245000 - 75000 = 170000 Tax on 1st 50000 = 0.04 × 50000 =$ 2000 Tax on rest amount 120000 = 0.06 × 120000 = $7200 Total tax = 200 + 7200 =$ 9200 18. $${1250 \\over 25} × 0.5 = ?$$ 1. 250 2. 50 3. 2.5 4. 25 5. 125 $${1250 \\over 25} × 0.5 = 50 × 0.5 = 25$$ 19. 350 × ? = 4200 1. 12 2. 24 3. 15 4. 30 5. 16 $$? = {4200 \\over 350} =12$$ 20. A car traveled 100 km with half the distance at 40 km/h and the other half at 80 km/h. Find the average speed of the car for the whole journey. 1. 53 km/hr 2. 53.33 km/hr 3. 54 1⁄4 km/hr 4. 55 km/hr 5. 56 km/hr The time, car took for the first half, $$50 \\over 40$$ = 1.25 hrs and for the second half $$50 \\over 80$$ = 0.625 hrs Total time = 1.25 + 0.625 = 1.875 hrs Average speed = $$100 \\over 1.875$$ = 53.3 $$km \\over hr$$", "is usually an average, that is, if a car traveled the distance of 210 km between two cities and to take this route, it took an hour and a half, so we will get a speed of 140 kilometers per hour (140 km/h). This does not obviously mean that the car has constantly shifted at this speed, there may be parts of the route on which it has been faster and parts where it has circulated more slowly. ## Math Formula kph=text(knots) xx 1.852 ## Conversion Examples knot to kph knot to kph 0.2 knots = 0,3704 kph 9 knots = 16,6680 kph 0.5 knots = 0,9260 kph 10 knots = 18,5200 kph 1 knots = 1,8520 kph 15 knots = 27,7800 kph 2 knots = 3,7040 kph 20 knots = 37,0400 kph 3 knots = 5,5560 kph 25 knots = 46,3000 kph 4 knots = 7,4080 kph 30 knots = 55,5600 kph 5 knots = 9,2600 kph 35 knots = 64,8200 kph 6 knots = 11,1120 kph 40 knots = 74,0800 kph 7 knots = 12,9640 kph 45 knots = 83,3400 kph 8 knots = 14,8160 kph 50 knots = 92,6000 kph", "# The wheels of a car have a radius 11 in and arc rotating at 1500rpm. How do you find the speed of the car in mi/h? May 28, 2018 The speed of the car was $98.17$ miles/hour #### Explanation: $r = 11$ inches , revolution $= 1500$ per minute In $1$ revolution the car advances $2 \\cdot \\pi \\cdot r$ inches $r = 11 \\therefore 2 \\pi r = 22 \\pi$ inches . In $1500$ revolution/minute the car advances $22 \\cdot 1500 \\cdot \\pi$ inches $= \\frac{22 \\cdot 1500 \\cdot \\pi \\cdot 60}{12 \\cdot 3 \\cdot 1760} \\approx 98.17 \\left(2 \\mathrm{dp}\\right)$mile/hour The speed of the car was $98.17$ miles/hour [Ans]", "Assume you bought this vehicle at the time t = 0 (assume the mileage of x = 0 km) for \\$ 20,000. Assume you drive about 15,000 km annualy (assume uniform driving habits..." ]

[ "A car travels $8$ km in the first quarter of an hour, $6$ km in the second quarter and $16$ km in the third quarter. The average speed of the car in km per hour over the entire journey is 1. $30$ 2. $36$ 3. $40$ 4. $24$", "is usually an average, that is, if a car traveled the distance of 210 km between two cities and to take this route, it took an hour and a half, so we will get a speed of 140 kilometers per hour (140 km/h). This does not obviously mean that the car has constantly shifted at this speed, there may be parts of the route on which it has been faster and parts where it has circulated more slowly. ## Math Formula kph=text(knots) xx 1.852 ## Conversion Examples knot to kph knot to kph 0.2 knots = 0,3704 kph 9 knots = 16,6680 kph 0.5 knots = 0,9260 kph 10 knots = 18,5200 kph 1 knots = 1,8520 kph 15 knots = 27,7800 kph 2 knots = 3,7040 kph 20 knots = 37,0400 kph 3 knots = 5,5560 kph 25 knots = 46,3000 kph 4 knots = 7,4080 kph 30 knots = 55,5600 kph 5 knots = 9,2600 kph 35 knots = 64,8200 kph 6 knots = 11,1120 kph 40 knots = 74,0800 kph 7 knots = 12,9640 kph 45 knots = 83,3400 kph 8 knots = 14,8160 kph 50 knots = 92,6000 kph", "a car was $20,000 in 2014,$15,000 in 2015 and $11,200in 2016. What is the rate of depreciation of the price of car per year? A. $$15\\%$$ B. $$25\\%$$ C. $$30\\%$$ D. $$35\\%$$ 3- 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 45 cm, what is the volume of the box? A. 81 cm$$^3$$ B. 162 cm$$^3$$ C. 243 cm$$^3$$ D. 3,375 cm$$^3$$ 4- How many possible outfit combinations come from six shirts, three slacks, and five ties? A. 15 B. 18 C. 30 D. 90 5- A bank is offering $$4.5\\%$$ simple interest on a savings account. If you deposit$8,000, how much interest will you earn in five years? A. $360 B.$720 C. $1,800 D.$3,600 6- What is the value of $$6^4$$? A. 216 B. 1,296 C. 7,776 D. 46,656 7- 15 is What percent of 20? A. $$20\\%$$ B. $$25\\%$$ C. $$75\\%$$ D. $$125\\%$$ 8- The perimeter of the trapezoid below is 56. What is its area? A. 56 cm B. 132 cm C. 156 cm D. 175 cm 9-In five successive hours, a car traveled 40 km,45 km,50 km,35 km and 55 km. In the next five hours, it traveled with an average speed of", "# A car is travelling at a rate of 70 km per hour leaves the house 2 hours after a truck has left and overtakes it in 5 hours. At what rate was the truck travelling? Jul 23, 2017 The speed of the truck is $= 50 k m {h}^{-} 1$ #### Explanation: Let the speed of the truck be $= v k m {h}^{-} 1$ The distance travelled by the truck is $d = v \\cdot 7 = 7 v k m$ The distance travelled by the car is $d = 5 \\cdot 70 = 350 k m$ Therefore, $7 v = 350$ $v = \\frac{350}{7} = 50 k m {h}^{-} 1$", "to manageable pieces Voyage Expenses = Driver + Gas + Fixed Costs Rule #3 - What do we need in order to talk about these things in a useful way? Everything is in Hours and Km, so it would be good to define things in these terms. Let's define a few things. H = Hours to complete the trip. We have variable gas costs, so it probably is helpful to define a function that will express this. Speed of Truck = 110 kmh + t kmh t kmh = number of kmh over 100 kmh for which we know the mileage. I'm not really getting a feel for the problem, so let's just play with it a little. At 110 kmh, we have 450 km / 110 kmh = 4.0909 hours and 450 km / 8 km/L = 56.25 L and 56.25 L *$0.68/L = $38.25 Now we have the flavor of the calculation. At (110+t) kmh, we have 450 km / (110+t kmh) = H and 450 km / (8-0.10t km/L) and 450 km / (8-0.10t km/L) *$0.68/L Gas Expense = [450 km / (8-0.10t km/L)] * $0.68/L Voyage Expenses = [450 km / (110+t kmh)]*($35.00/h) + [450 km / (8-0.10t km/L)] * $0.68/L + [450 km / (110+t kmh)]*($15.50/h) After a very, very careful surgery consisting", "12:06 departure: $15:38 - 12:06 = 03:32\\text{ or }(3 \\times 60)+32 = 212 \\text{ minutes}$ Now, subtracting time spent waiting at connecting stations: 11:10 departure has 0 stops: $\\text{length } = 202 \\text{ minutes}$ 11:35 departure has 2 stops: $\\text{length } = 225 - 17 = 208 \\text{ minutes}$ 12:06 departure has 1 stop: $\\text{length } = 212 - 12 = 200 \\text{ minutes}$ Step 2: Calculate their average speeds. NOTE: because a ratio has no units, we don’t have to work out the speed in km per hour. Instead, to save time, here we will do it in km per minute. 11:10 departure: $\\text{average speed } = \\dfrac{347}{202} = 1.718$ 11:35 departure: $\\text{average speed } = \\dfrac{347}{208} = 1.668$ 12:06 departure: $\\text{average speed } = \\dfrac{347}{200} = 1.735$ Step 3: Form the ratio and simplify by dividing through by the smallest element. \\begin{aligned}\\text{11:10 speed : 11:35 speed : 12:06 speed } &= 1.718 : 1.668 : 1.735 \\\\ &= \\dfrac{1.718}{1.668} : \\dfrac{1.668}{1.668} : \\dfrac{1.735}{1.668} \\\\ &= 1.03 : 1 : 1.04 \\end{aligned} #### Question 15 Work out the difference between the highest cost per minute of outward travel and the lowest cost per minute of outward travel when purchasing an open return for a train with fewer than 2 changes. A: £0.31 B: £0.30 C: £0.29 D:", "14 views A car travels $8 \\: km$ in the first quarter of an hour, $6\\: km$ in the second quarter and $16\\: km$ in the third quarter. The average speed of the car in $km$ per hour over the entire journey is 1. $30$ 2. $36$ 3. $40$ 4. $24$ edited | 14 views", "Decrease the speed Maintain the speed at $60$ km/hour Cannot be determined Answer the following questions based on the information given below. The petrol consumption rate of a new model car 'Palto' depends on its speed and may be described by the graph below. Manasa makes the $200$ km trip from Mumbai to Pune at a steady speed of $60$ km/hour. What is the amount of petrol consumed for the journey? $12.5$ litres $13.33$ litres $16$ litres $19.75$ litres", "it'd take (60/75) hours (0.8 hours, or 48 minutes) to go at 75 mph versus 60/65 hours (0.92 hours, or 55 minutes and about 20 seconds) to go at 65 mph. How much was 7 minutes of my time worth? At the wages I was working at that point (about $15 an hour),$1.75. How much was 1.5 gallons of gas worth? At that time, roughly $6. I decided to drive slower. However, on the way back, I had my girlfriend, which meant I, ahem, valued the time saved from driving faster more. I drove faster on the way back. To this day I've yet to make enough money in hourly terms to justify regularly driving 75 mph over driving 65 mph on the highway. At the current average gasoline price in New York state ($3.92/gal), I'd have to be making $50.40 an hour to justify it, which for a standard 8-hour work week is$104,832 per year. ------ This analysis is relatively simple. There are a couple of (to my mind, anyway) interesting little wrinkles to this calculation, the magnitude of which I can't calculate off the top of my head but I think on balance points to even more fuel consumption at higher speed. The first wrinkle occurs when we relax the assumption that the drag coefficient is", "752 views Answer the question based on the following information. Rajiv reaches city B from city A in $4$ hours, driving at speed of $35 \\; \\text{kmph}$ for the first two hour and at $45 \\; \\text{kmph}$ for the next two hours. Aditi follows the same route, but drives at three different speeds: $30, 40$ and $50 \\; \\text{kmph},$ covering an equal distance in each speed segment. The two cars are similar with petrol consumption characteristics $\\text{(km per litre)}$ shown in the figure below. Zoheb would like to drive Aditi’s car over the same route from A to B and minimize the petrol consumption for the trip. What is the quantity of petrol required by him? 1. $6.67 \\; \\text{litre}$ 2. $7 \\; \\text{litre}$ 3. $6.33 \\; \\text{litre}$ 4. $6.0 \\; \\text{litre}$ For minimum petrol consumption, Zoheb should drive at $40 \\; \\text{kmph}.$ Quantity of petrol required Zoheb should drive at $40 \\; \\text{kmph},$ petrol consumption $= \\frac{160}{24}= 6.67 \\; \\text{lit}$ Hence, Option$:\\text{(A)}$ 11.1k points", "of the box is 48 cm, what is the volume of the box? A. 2,563 $$cm^3$$ B. 2,874 $$cm^3$$ C. 3,456 $$cm^3$$ D. 3,982 $$cm^3$$ 14- How many possible outfit combinations come from five shirts, four slacks, and three ties? A. 32 B. 48 C. 54 D. 60 15- If $$60\\%$$ of a class are girls, and $$22\\%$$ of girls play tennis, what percent of the class play tennis? A. $$8.8\\%$$ B. $$13.2\\%$$ C. $$18.3\\%$$ D. $$23.4\\%$$ 16- A$64 shirt now selling for $48 is discounted by what percent? A. $$20\\%$$ B. $$25\\%$$ C. $$30\\%$$ D. $$35\\%$$ 17- 48 is What percent of 24? A. $$50\\%$$ B. $$150\\%$$ C. $$200\\%$$ D. $$250\\%$$ 18- If the area of a trapezoid is 120 cm, what is the perimeter of the trapezoid? A. 36 cm B. 48 cm C. 56 cm D. 64 cm 19- In four successive hours, a car travels 55 km, 68 km, 48 km, and 72 km. In the next four hours, it travels with an average speed of 64 km per hour. Find the total distance the car traveled in 8 hours. A. 395 km B. 483 km C. 499 km D. 517 km 20- How long does a 468–miles trip take moving at 72 miles per hour (mph)? A. 6 hours B. 6 hours and", "If a car went the first third of the distance at 80 kmh in the Problem Solving forum “Here''s an algebraic approach: Let d = 1/3 of the total distance traveled So, the car traveled d km at 80 kmh, then the car traveled d km at 24 kmh, and then d km at 48 kmh Average speed = (total distance traveled)/(total travel time) Total distance traveled = d + d + d = 3d time = ...” May 6, 2018 GMATGuruNY posted a reply to W, X, Y and Z each represent a different number in the Problem Solving forum “Sum of all 4 rows = 3 + 6 + 9 + 0 = 18. Sum of all 4 columns = -2 + 7 + n + 5 = n + 10. The values in blue must be EQUAL, since each represents the sum of all 16 values in the table. Thus: n+10 = 18 n=8. The correct answer is C.” May 6, 2018 Brent@GMATPrepNow posted a reply to Sheila works 8 hours per day on Monday, Wednesday and Friday in the Problem Solving forum “We can avoid long division if we make an important observation. Let''s take it from the point were you got: earning per hour = $324/36 You might notice that$360/36 = $10 Also", "each year? Using estimates of number of personal vehicles and annual mileage, combined with an estimate that 30% of these miles would be driven more slowly than otherwise with drivers adopting my techniques, my estimate is that about 468,000,000,000 miles would be driven more slowly. I estimate that this would result in 1.8 X 10^9 (1.8 billion) extra hours on the road. Does this make sense? Well, I'm one of 200,000,000 drivers and I calculated in an earlier post that I'll lose about 47 hours per year. If each of the 200 million drivers lost that much, the total would be 9.4 X 10^9 hours. But I estimate that I probably drive something like 30% more than average so adjust this down to 7.2 X 10^9 hours. Figure somewhere between these numbers is right, I'll use 4.5 X 10^9 or 4 and one half billion hours. I'll say the average hour is worth $30.00, in that case the lost time is worth$135 billion. Considering I calculated we'd save $36 billion in oil imports, it would seem not to be worth it. Are there any mitigating factors? Not many. First would be finding ways to avoid productivity loss when on the road. Talk radio? Books on tape? Cell phones? These all may help but they certainly won't eliminate the", "B. $$2:3$$ C. $$5:9$$ D. $$5:6$$ E. $$11:16$$ 4- If $$40\\%$$ of a class are girls, and $$25\\%$$ of girls play tennis, what percent of the class play tennis? A. $$10 \\%$$ B. $$15 \\%$$ C. $$20 \\%$$ D. $$40 \\%$$ E. $$80 \\%$$ 5- What is the value of $$y$$ in the following system of equation? $$3x-4y= -20$$ $$-x+2y= 10$$ A. $$-1$$ B. $$-2$$ C. $$1$$ D. $$4$$ E. $$5$$ 6- In five successive hours, a car travels 40 km, 45 km, 50 km, 35 km, and 55 km. In the next five hours, it travels with an average speed of 50 km per hour. Find the total distance the car traveled in 10 hours. A. 425 km B. 450 km C. 475 km D. 500 km E. 1,000 km 7- How long does a 420–miles trip take moving at 50 miles per hour (mph)? A. 4 hours B. 6 hours and 24 minutes C. 8 hours and 24 minutes D. 8 hours and 30 minutes E. 10 hours and 30 minutes 8- What is the difference between the smallest 4–digit number and the biggest 4–digit number? A. 6666 B. 6789 C. 8888 D. 8999 E. 9999 9- What is the value of $$6^4$$? A. 6 B. 24 C. 36 D. 216 E. 1,296 10- Right triangle", "around 55 mph (88 kph) with 1500-ish vehicles per hour. I laughed even harder. Anyway, i tried to find the exact capacity of JORR only to no avail. So I had to find some method to calculate the capacity. I got this other paper by Li & Laurence (2015) which discuss about estimating capacity. They managed summarize a paper which says that breakdown capacity occurs when the flow of vehicle is 2289 vehicles per hour (they stated that this number maybe very stochastic so we need to be careful, but i have no other lead so). Judging from their definition of “breakdown”, JORR may be worse than breakdown, but let’s use this number as maximum capacity. We then can calculate the demand for toll road: $$2289=Ad*P^{-0.72}$$ Let’s calculate just the average tariff for car at around 10,000 IDR. If we plug that to P, we will get Ad= 1,736,384.07 (too many numbers, right? This is why i hate working with IDR). So we can get the full demand function of JORR: $$Q(P)=1,736,384.07*P^{-0.72}$$ Next we find the Dead Weight Loss (DWL) of having such tariff. To do this, we need to find the willingness to pay to get the normal, non-breakdown level of capacity. Let’s use $$VCR=0.77$$, then we will get $$Q=1550$$. Now estimate $$P$$ with $$Q=1550$$, $$1550=1,736,384.07*P^{-0.72}$$ $$P=", "$87$ $82$ $75$ $78$ 1 vote 12 A straight road connects points $\\text{A}$ and $\\text{B}$. Car $1$ travels from $\\text{A}$ to $\\text{B}$ and Car $2$ travels from $\\text{B}$ to $\\text{A}$, both leaving at the same time. After meeting each other, they take $45$ minutes and $20$ minutes, respectively, to complete their ... . If Car $1$ travels at the speed of $60$ km/hr, then the speed of Car $2$, in km/hr, is $90$ $100$ $80$ $70$ 1 vote 13 Leaving home at the same time, Amal reaches office at $10:15$ am if he travels at $8$ km/hr, and at $9:40$ am if he travels at $15$ km/hr. Leaving home at $9:10$ am, at what speed, in km/hr, must he travel so as to reach office exactly at $10$ am? $13$ $14$ $12$ $11$ 1 vote 14 A train travelled at one$-$thirds of its usual speed, and hence reached the destination $30$ minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for $5$ minutes but then stopped for $4$ minutes for an emergency. The ... now increase its usual speed so as to reach the destination at the scheduled time, is nearest to $58$ $67$ $61$ $50$ 15 A man walks $10$ miles at an average rate of $2$ miles per", "Gas + Fixed Costs Rule #3 - What do we need in order to talk about these things in a useful way? Everything is in Hours and Km, so it would be good to define things in these terms. Let's define a few things. H = Hours to complete the trip. We have variable gas costs, so it probably is helpful to define a function that will express this. Speed of Truck = 110 kmh + t kmh t kmh = number of kmh over 100 kmh for which we know the mileage. I'm not really getting a feel for the problem, so let's just play with it a little. At 110 kmh, we have 450 km / 110 kmh = 4.0909 hours and 450 km / 8 km/L = 56.25 L and 56.25 L * $0.68/L =$38.25 Now we have the flavor of the calculation. At (110+t) kmh, we have 450 km / (110+t kmh) = H and 450 km / (8-0.10t km/L) and 450 km / (8-0.10t km/L) * $0.68/L Gas Expense = [450 km / (8-0.10t km/L)] *$0.68/L Voyage Expenses = [450 km / (110+t kmh)]*($35.00/h) + [450 km / (8-0.10t km/L)] *$0.68/L + [450 km / (110+t kmh)]*(\\$15.50/h) After a very, very careful surgery consisting of algebra and unit checks, this gives: Voyage Expenses(t)", "# A car covers the first 78 km. of the journey in 90 minutes and the remaining 50 km. in 70 minutes. What is the average speed of the car? 46 views in Aptitude closed A car covers the first 78 km. of the journey in 90 minutes and the remaining 50 km. in 70 minutes. What is the average speed of the car? 1. 50 km/hour 2. 48 km/hour 3. 40 km/hour 4. 55 km/hour by (30.0k points) selected Correct Answer - Option 2 : 48 km/hour Given: Total Distance cover = 78 + 50 = 128 km Total time taken = 90 + 70 = 160 minutes = $160\\over 60$ hours Formula use: Average speed = Total Distance Travelled / Total Time taken Calculation: Average speed = $128 \\over {160\\over60}$ Average speed = ${128 \\over 160} ×60$ Average speed = 48 km/hour Therefore, the average speed of the car is 48 km/hour. Hence, the correct answer is 48 km/hour.", "and, with fuel costing 0.68 per Liter (how OLD is this problem??) the cost of fuel for the trip is (-45v+ 8550L)(0.68)= -30.6v+ 5814. At speed v, the time required for the trip will be (distance divided by speed) 450/v hours. Since drivers are paid $35 per hour, and the cost of running the truck is$15.50 per hour, the total cost for the truck per hour is \\$50.50 per hour and the total cost for the trip, excluding fuel, will be 50.50(450/v)= 22725/v. The total cost of the trip, at speed v> 110 is -30.5v+ 5814+ 22725/v. That's the function you want to minimize. Last edited: Jul 17, 2006", "Total Income =$ 245000 Total relief = 3 × 15000 + 30000 = $75000 Rest income = 245000 - 75000 = 170000 Tax on 1st 50000 = 0.04 × 50000 =$ 2000 Tax on rest amount 120000 = 0.06 × 120000 = $7200 Total tax = 200 + 7200 =$ 9200 18. $${1250 \\over 25} × 0.5 = ?$$ 1. 250 2. 50 3. 2.5 4. 25 5. 125 $${1250 \\over 25} × 0.5 = 50 × 0.5 = 25$$ 19. 350 × ? = 4200 1. 12 2. 24 3. 15 4. 30 5. 16 $$? = {4200 \\over 350} =12$$ 20. A car traveled 100 km with half the distance at 40 km/h and the other half at 80 km/h. Find the average speed of the car for the whole journey. 1. 53 km/hr 2. 53.33 km/hr 3. 54 1⁄4 km/hr 4. 55 km/hr 5. 56 km/hr The time, car took for the first half, $$50 \\over 40$$ = 1.25 hrs and for the second half $$50 \\over 80$$ = 0.625 hrs Total time = 1.25 + 0.625 = 1.875 hrs Average speed = $$100 \\over 1.875$$ = 53.3 $$km \\over hr$$" ]

[ "# Problem #2093 2093 Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles? $[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype(\"4 4\"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label(\"A\",A,W); label(\"B\",B,E); label(\"1\",(-1.5,-0.1),S); label(\"2\",(0,-0.1),S); label(\"1\",(1.5,-0.1),S);[/asy]$ $\\textbf{(A) } \\pi - \\sqrt{3} \\qquad\\textbf{(B) } \\pi - \\sqrt{2} \\qquad\\textbf{(C) } \\frac{\\pi + \\sqrt{2}}{2} \\qquad\\textbf{(D) } \\frac{\\pi +\\sqrt{3}}{2} \\qquad\\textbf{(E) } \\frac{7}{6}\\pi - \\frac{\\sqrt{3}}{2}$ 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. • 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", "fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label(\"6\",.74*dir(-90)+(0,3)); label(\"5\",.74*dir(-60)+(0,3)); label(\"4\",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label(\"30^\\circ\",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*\"12\",1.56*dir(120)); label(rotate(30)*\"1\",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label(\"60^\\circ\",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*\"6\",.74*dir(-150)+(0,3)); label(rotate(-60)*\"5\",.74*dir(-120)+(0,3)); label(rotate(-60)*\"4\",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label(\"90^\\circ\",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*\"6\",.74*dir(-180)+c); label(rotate(-90)*\"5\",.74*dir(-150)+c); label(rotate(-90)*\"4\",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{\\textbf{(C) }\\text{4 o' clock}}$ ## Solution 5 We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We", "# Problem #2103 2103 Three semicircles of radius $1$ are constructed on diameter $\\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles? $[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype(\"4 4\"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label(\"A\",A,W); label(\"B\",B,E); label(\"1\",(-1.5,-0.1),S); label(\"2\",(0,-0.1),S); label(\"1\",(1.5,-0.1),S);[/asy]$ $\\textbf{(A) } \\pi - \\sqrt{3} \\qquad\\textbf{(B) } \\pi - \\sqrt{2} \\qquad\\textbf{(C) } \\frac{\\pi + \\sqrt{2}}{2} \\qquad\\textbf{(D) } \\frac{\\pi +\\sqrt{3}}{2} \\qquad\\textbf{(E) } \\frac{7}{6}\\pi - \\frac{\\sqrt{3}}{2}$ 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.", "draw(circle((-4,-4), 4*1.41421356237)); [/asy]$ The way we figure out the area is by splitting it the following way: $[asy] size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real g(real x) { return x+8; } draw(graph(g,0,-8), red+linewidth(1.5)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real h(real x) { return -x-8; } draw(graph(h,-8,0), red+linewidth(1.5)); real z(real x) { return x-8; } draw(graph(z,0,8), red+linewidth(1.5)); pair A,B,C,D; A = (8,0); B = (0,8); C = (-8,0); D = (0,-8); fill(A--B--C--D--cycle, red); [/asy]$ We know each of the red lines is a diameter of the circle which is $\\sqrt2$. So the area of the red is 2. We know that the area of each semicircle is $\\dfrac14 \\pi$ so the area of the semicircles combines is $\\pi$. Thus we get $\\boxed{\\textbf{(B)} \\pi+2}.$", "<- bezierGrob(c(.2, .2, .8, .8), c(.2, .8, .8, .2)) grid.draw(bg) trace <- bezierPoints(bg) grid.circle(trace$x, trace$y, default.units=\"inches\", r=unit(.5, \"mm\"))", "You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you the answer of $\\boxed{\\textbf{(B) } \\text{(-1030,-990)}}$ 14. ## Solution 1 $[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); label(\"2\",(z,c),NE); pair X,Y,Z; X = (1,0); Y = (2,0); Z = (1.5,a); pair d = 1.9*dir(7); dot(X); dot(Y); dot(Z); label(\"A\",X,SW); label(\"B\",Y,SE); label(\"C\",Z,N); draw(X--Y--Z--A); label(\"1\",(1.5,0),S); label(\"1\",(1.75,a/2),dir(30)); draw(anglemark(Y,X,Z,8),blue); label(\"60^\\circ\",anglemark(Y,X,Z),d); [/asy]$ Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, $BC = 1$, since B is the center of the semicircle with radius 1 that C lies on, $AB = 1$, since B is the center of the semicircle with radius 1 that A lies on, and $\\angle BAC = 60^\\circ$as a regular hexagon has angles of 120$^\\circ$, and $\\angle BAC$ is half of any angle in this", "integer that is a divisor of $$(n+1)(n+3)(n+5)(n+7)(n+9)$$ for all positive even integers $n$? $\\textbf{(A) } 3 \\qquad\\textbf{(B) } 5 \\qquad\\textbf{(C) } 11 \\qquad\\textbf{(D) } 15 \\qquad\\textbf{(E) } 165$ ## Problem 19 Three semicircles of radius $1$ are constructed on diameter $\\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles? $[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype(\"4 4\"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label(\"A\",A,W); label(\"B\",B,E); label(\"1\",(-1.5,-0.1),S); label(\"2\",(0,-0.1),S); label(\"1\",(1.5,-0.1),S);[/asy]$ $\\textbf{(A) } \\pi - \\sqrt{3} \\qquad\\textbf{(B) } \\pi - \\sqrt{2} \\qquad\\textbf{(C) } \\frac{\\pi + \\sqrt{2}}{2} \\qquad\\textbf{(D) } \\frac{\\pi +\\sqrt{3}}{2} \\qquad\\textbf{(E) } \\frac{7}{6}\\pi - \\frac{\\sqrt{3}}{2}$ ## Problem 20 In rectangle $ABCD, AB=5$ and $BC=3$. Points $F$ and $G$ are on $\\overline{CD}$ so that $DF=1$ and $GC=2$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\\triangle AEB$. $[asy] unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair A=(0,0), B=(5,0), C=(5,3), D=(0,3), F=(1,3), G=(3,3); pair E=extension(A,F,B,G); draw(A--B--C--D--A--E--B); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); label(\"E\",E,N); label(\"F\",F,SE); label(\"G\",G,SW); label(\"B\",B,SE); label(\"1\",midpoint(D--F),N); label(\"2\",midpoint(G--C),N); label(\"3\",midpoint(B--C),E); label(\"3\",midpoint(A--D),W); label(\"5\",midpoint(A--B),S); [/asy]$ $\\textbf{(A) } 10 \\qquad\\textbf{(B) } \\frac{21}{2} \\qquad\\textbf{(C) } 12 \\qquad\\textbf{(D) } \\frac{25}{2} \\qquad\\textbf{(E) } 15$ ## Problem", "# Difference between revisions of \"2005 AMC 12A Problems/Problem 17\" ## Problem A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $W$? $[asy] path a=(0,0)--(10,0)--(10,10)--(0,10)--cycle; path b = (0,10)--(6,16)--(16,16)--(16,6)--(10,0); path c= (10,10)--(16,16); path d= (0,0)--(3,13)--(13,13)--(10,0); path e= (13,13)--(16,6); draw(a,linewidth(0.7)); draw(b,linewidth(0.7)); draw(c,linewidth(0.7)); draw(d,linewidth(0.7)); draw(e,linewidth(0.7)); draw(shift((20,0))*a,linewidth(0.7)); draw(shift((20,0))*b,linewidth(0.7)); draw(shift((20,0))*c,linewidth(0.7)); draw(shift((20,0))*d,linewidth(0.7)); draw(shift((20,0))*e,linewidth(0.7)); draw((20,0)--(25,10)--(30,0),dashed); draw((25,10)--(31,16)--(36,6),dashed); draw((15,0)--(10,10),Arrow); draw((15.5,0)--(30,10),Arrow); label(\"W\",(15.2,0),S); label(\"Figure 1\",(5,0),S); label(\"Figure 2\",(25,0),S); [/asy]$ $(\\mathrm {A}) \\ \\frac{1}{12} \\qquad (\\mathrm {B}) \\ \\frac{1}{9} \\qquad (\\mathrm {C})\\ \\frac{1}{8} \\qquad (\\mathrm {D}) \\ \\frac{1}{6} \\qquad (\\mathrm {E})\\ \\frac{1}{4}$ ## Solution It is a pyramid with height $1$ and base area $\\frac{1}{4}$, so using the formula for the volume of a pyramid, $\\frac{1}{3} \\cdot \\left(\\frac{1}{4}\\right) \\cdot (1) = \\frac {1}{12} \\Rightarrow \\boxed{(\\mathrm {A})}$.", "3.875cm); \\draw (0,3.875,0) arc (90:205.8:1.875cm and 3.875cm); %fill \\draw[fill=blue] (0,0,0) --(0,3.875,0) arc (90:205.8:1.875cm and 3.875cm) --(0,0,0) -- (3.875,0,0) arc (0:90:3.875cm and 3.875cm) -- (0,0,0) -- (3.875,0,0) arc (0:115.8:3.875cm and -1.875cm)-- cycle; % convenctie zone \\draw (2,0,0) arc (0:116.5:2cm and -1cm); \\draw (2,0,0) arc (0:90:2cm and 2cm); \\draw (0,2,0) arc (90:206.5:1cm and 2cm); %fill \\draw[fill=red] (0,0,0) -- (0,2,0) arc (90:206.5:1cm and 2cm) --(0,0,0) -- (2,0,0) arc (0:90:2cm and 2cm) -- (0,0,0) -- (2,0,0) arc (0:116.5:2cm and -1cm)-- cycle; \\draw (1,0,0) arc (0:116.5:1cm and -0.5cm); \\draw (1,0,0) arc (0:90:1cm and 1cm); \\draw (0,1,0) arc (90:206.5:0.5cm and 1cm); %fill \\draw[fill=green] (0,0,0) -- (0,1,0) arc (90:206.5:0.5cm and 1cm) --(0,0,0) -- (1,0,0) arc (0:90:1cm and 1cm) -- (0,0,0) -- (1,0,0) arc (0:116.5:1cm and -0.5cm)-- cycle; \\end{tikzpicture} \\end{document} - Oh, is that what the OP meant? I think I misunderstood. I thought the other two corresponding parts were in the same plane but I guess they are the parts in other planes? I take a long time with tikz so you posted while I was fiddling even though you must have posted ages before I did. :( – cfr Jul 2 '14 at 0:28 It scares me you can do that in 5 minutes (or anything with tikz) ;). – cfr Jul 2 '14 at 0:44 @cfr Not anything but this one. It is", "draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ $\\textbf{(A)}12\\frac{1}{2}\\qquad\\textbf{(B)}20\\qquad\\textbf{(C)}25\\qquad\\textbf{(D)}33\\frac{1}{3}\\qquad\\textbf{(E)}37\\frac{1}{2}$ ## Problem 8 Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips? $\\textbf{(A) }4 \\qquad\\textbf{(B) }5 \\qquad\\textbf{(C) }6 \\qquad\\textbf{(D) }7 \\qquad\\textbf{(E) }9$ ## Problem 9 Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour? $[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; draw((0,0)--(0,8)); draw((0,0)--(8,0)); 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)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label(\"1\",(0.95,-0.24),SE*lsf); label(\"2\",(1.92,-0.26),SE*lsf); label(\"3\",(2.92,-0.31),SE*lsf); label(\"4\",(3.93,-0.26),SE*lsf); label(\"5\",(4.92,-0.27),SE*lsf); label(\"6\",(5.95,-0.29),SE*lsf); label(\"7\",(6.94,-0.27),SE*lsf); label(\"5\",(-0.49,1.22),SE*lsf); label(\"10\",(-0.59,2.23),SE*lsf); label(\"15\",(-0.61,3.22),SE*lsf); label(\"20\",(-0.61,4.23),SE*lsf); label(\"25\",(-0.59,5.22),SE*lsf); label(\"30\",(-0.59,6.2),SE*lsf); label(\"35\",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6));", "Problem #2131 2131 Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\\pi$ radians is $180$ degrees.) $[asy] defaultpen(linewidth(0.8)); pair O=origin; fill(O--Arc(O, 2, 20, 160)--cycle, mediumgray); fill(O--Arc(O, 1, 20, 160)--cycle, white); fill(O--Arc(O, 2, 200, 340)--cycle, mediumgray); fill(O--Arc(O, 1, 200, 340)--cycle, white); fill(O--Arc(O, 3, 160, 200)--cycle, mediumgray); fill(O--Arc(O, 2, 160, 200)--cycle, white); fill(O--Arc(O, 1, 160, 200)--cycle, mediumgray); fill(O--Arc(O, 3, -20, 20)--cycle, mediumgray); fill(O--Arc(O, 2, -20, 20)--cycle, white); fill(O--Arc(O, 1, -20, 20)--cycle, mediumgray); draw(Circle(origin, 1));draw(Circle(origin, 2));draw(Circle(origin, 3)); draw(5*dir(200)--5*dir(20)^^5*dir(160)--5*dir(-20)); [/asy]$ $\\mathrm{(A) \\ } \\frac{\\pi}{8} \\qquad \\mathrm{(B) \\ } \\frac{\\pi}{7} \\qquad \\mathrm{(C) \\ } \\frac{\\pi}{6} \\qquad \\mathrm{(D) \\ } \\frac{\\pi}{5} \\qquad \\mathrm{(E) \\ } \\frac{\\pi}{4}$ 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", "\\pi}$. ## Solution 2 First, subdivide the hexagon into 24 equilateral triangles with side length 1:$[asy] size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label(\"2\",(3.5,3sqrt(3)/2),NE); draw((1,0)--(3,2sqrt(3))); draw((3,0)--(1,2sqrt(3))); draw((4,sqrt(3))--(0,sqrt(3))); draw((2,0)--(3.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(2,2sqrt(3))); draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2)); draw((2,2sqrt(3))--(0.5,sqrt(3)/2)); draw((2,0)--(0.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2)); [/asy]$Now note that the entire shaded region is just 6 times this part:$[asy] size(100); fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4)); fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white); draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle); label(\"1\",(2.25,7sqrt(3)/4),NE); draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); [/asy]$The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of:$$2\\cdot\\frac{\\sqrt{3}}{4}=\\frac{\\sqrt{3}}{2}$$The arc that is not included has an area of:$$\\frac16 \\cdot\\pi \\cdot1^2 = \\frac{\\pi}{6}$$Hence, the area of the shaded region in that section is$$\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{6}$$For a final area of:$$6\\left(\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{6}\\right)=3\\sqrt{3}-\\pi\\Rightarrow \\boxed{\\mathrm{(D)}}$$ 15. ## Solution 1 After erasing every third digit, the list becomes $1245235134\\ldots$ repeated. After erasing every fourth digit from this list, the list becomes $124235341452513\\ldots$ repeated. Finally, after erasing every fifth digit from this list, the list becomes $124253415251\\ldots$ repeated. Since this list repeats every $12$ digits and since $2019,2020,2020$ are $3,4,5$ respectively in $\\pmod{12},$ we have that the $2019$th, $2020$th, and $2021$st digits are the $3$rd, $4$th, and $5$th digits respectively. It follows that the answer is $4+2+5= \\boxed {\\textbf{(D)} \\text{ 11}}.$ 16. ## Solution Notice that to use the optimal strategy to win the game, Bela must select the middle number in the range $[0, n]$ and", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "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", "an arc smaller than a quarter circle, using a radius of 5 cm. ### Enrichment Once you have finished the work in section 4.8, experiment with drawing only the arcs that you need in various constructions. Here is an example to show how to construct a regular hexagon with only arcs: ### Familiar figures in the seven-circle pattern For this activity you need five seven-circle sets like the ones drawn in the previous two activities. Start by drawing these on blank pieces of paper. Don't make your radius bigger than 4 cm. Number your sets figure 2 to figure 6. Label each figure as shown on the right. • Figure 1: Use the figure alongside. Draw lines connecting AB, BC, CD,. . . up to FA. • Figure 2: Draw lines connecting A, O and B. • Figure 3: Draw lines connecting B, F and D. • Figure 4: Draw lines connecting BC, CE, EF and FB. • Figure 5: Draw lines connecting CD, DE, EF and FC. • Figure 6: Draw lines connecting AB, BC, CE and EA. 2. Complete the table below. It shows the name of each figure and its properties. Figure 1 (on the right) has been done as an example. Figure Name of figure Properties 1 Regular hexagon 6-sided figure. All the sides", "an arc smaller than a quarter circle, using a radius of 5 cm. ### Enrichment Once you have finished the work in section 4.8, experiment with drawing only the arcs that you need in various constructions. Here is an example to show how to construct a regular hexagon with only arcs: ### Familiar figures in the seven-circle pattern For this activity you need five seven-circle sets like the ones drawn in the previous two activities. Start by drawing these on blank pieces of paper. Don't make your radius bigger than 4 cm. Number your sets figure 2 to figure 6. Label each figure as shown on the right. • Figure 1: Use the figure alongside. Draw lines connecting AB, BC, CD,. . . up to FA. • Figure 2: Draw lines connecting A, O and B. • Figure 3: Draw lines connecting B, F and D. • Figure 4: Draw lines connecting BC, CE, EF and FB. • Figure 5: Draw lines connecting CD, DE, EF and FC. • Figure 6: Draw lines connecting AB, BC, CE and EA. 2. Complete the table below. It shows the name of each figure and its properties. Figure 1 (on the right) has been done as an example. Figure Name of figure Properties 1 Regular hexagon 6-sided figure. All the sides", "# 1973 AHSME Problems/Problem 11 ## Problem 11 A circle with a circumscribed and an inscribed square centered at the origin of a rectangular coordinate system with positive and axes and is shown in each figure to below. $[asy] size((400)); draw((0,0)--(22,0), EndArrow); draw((10,-10)--(10,12), EndArrow); draw((25,0)--(47,0), EndArrow); draw((35,-10)--(35,12), EndArrow); draw((-25,0)--(-3,0), EndArrow); draw((-15,-10)--(-15,12), EndArrow); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw(Circle((-40,0),6)); draw(Circle((-15,0),6)); draw(Circle((10,0),6)); draw(Circle((35,0),6)); draw((-34,0)--(-40,6)--(-46,0)--(-40,-6)--(-34,0)--(-34,6)--(-46,6)--(-46,-6)--(-34,-6)--cycle); draw((-6.5,0)--(-15,8.5)--(-23.5,0)--(-15,-8.5)--cycle); draw((-10.8,4.2)--(-19.2,4.2)--(-19.2,-4.2)--(-10.8,-4.2)--cycle); draw((14.2,4.2)--(5.8,4.2)--(5.8,-4.2)--(14.2,-4.2)--cycle); draw((16,6)--(4,6)--(4,-6)--(16,-6)--cycle); draw((41,0)--(35,6)--(29,0)--(35,-6)--cycle); draw((43.5,0)--(35,8.5)--(26.5,0)--(35,-8.5)--cycle); label(\"I\", (-49,9)); label(\"II\", (-24,9)); label(\"III\", (1,9)); label(\"IV\", (26,9)); label(\"X\", (-28,0), S); label(\"X\", (-3,0), S); label(\"X\", (22,0), S); label(\"X\", (47,0), S); label(\"Y\", (-40,12), E); label(\"Y\", (-15,12), E); label(\"Y\", (10,12), E); label(\"Y\", (35,12), E);[/asy]$ The inequalities $$|x|+|y|\\leq\\sqrt{2(x^{2}+y^{2})}\\leq 2\\mbox{Max}(|x|, |y|)$$ are represented geometrically* by the figure numbered $\\textbf{(A)}\\ I\\qquad\\textbf{(B)}\\ II\\qquad\\textbf{(C)}\\ III\\qquad\\textbf{(D)}\\ IV\\qquad\\textbf{(E)}\\ \\mbox{none of these}$ * An inequality of the form $f(x, y) \\leq g(x, y)$, for all $x$ and $y$ is represented geometrically by a figure showing the containment $\\{\\mbox{The set of points }(x, y)\\mbox{ such that }g(x, y) \\leq a\\} \\subset\\\\ \\{\\mbox{The set of points }(x, y)\\mbox{ such that }f(x, y) \\leq a\\}$ for a typical real number $a$. ## Solution First, note that the following inequality $$|x|+|y|\\leq\\sqrt{2(x^{2}+y^{2})}\\leq 2\\mbox{Max}(|x|, |y|)$$ represents the graphs $|x| + |y| = a$, $\\sqrt{2(x^{2}+y^{2})} = a$, and $2\\mbox{Max}(|x|, |y|) = a$. We don't actually have to worry about the inequality, we just want to find the picture", "# 3.E: Review Exercises and Sample Exam ## Review Exercises Exercise $$\\PageIndex{1}$$ Rectangular Coordinate System Graph the given set of ordered pairs. 1. $$\\{(−3, 4), (−4, 0), (0, 3), (2, 4)\\}$$ 2. $$\\{(−5, 5), (−3, −1), (0, 0), (3, 2)\\}$$ 3. Graph the points $$(−3, 5), (−3, −3),$$ and $$(3, −3)$$ on a rectangular coordinate plane. Connect the points and calculate the area of the shape. 4. Graph the points $$(−4, 1), (0, 1), (0, −2),$$ and $$(−4, −2)$$ on a rectangular coordinate plane. Connect the points and calculate the area of the shape. 5. Graph the points $$(1, 0), (4, 0), (1, −5),$$ and $$(4, −5)$$ on a rectangular coordinate plane. Connect the points and calculate the perimeter of the shape. 6. Graph the points $$(−5, 2), (−5, −3), (1, 2),$$ and $$(1, −3)$$ on a rectangular coordinate plane. Connect the points and calculate the perimeter of the shape. 1. Figure 3.E.1 3. Area: $$24$$ square units Figure 3.E.2 5. Perimeter: $$16$$ units Figure 3.E.3 Exercise $$\\PageIndex{2}$$ Rectangular Coordinate System Calculate the distance between the given two points. 1. $$(−1, −2)$$ and $$(5, 6)$$ 2. $$(2, −5)$$ and $$(−2, −2)$$ 3. $$(−9, −3)$$ and $$(−8, 4)$$ 4. $$(−1, 3)$$ and $$(1, −3)$$ 1. $$10$$ units 3. $$5\\sqrt{2}$$ units Exercise $$\\PageIndex{3}$$ Rectangular Coordinate System Calculate the midpoint between", "# Problem #2099 2099 The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? $[asy] unitsize(10mm); defaultpen(fontsize(10pt)); pen finedashed=linetype(\"4 4\"); filldraw((1,1)--(2,1)--(2,2)--(4,2)--(4,3)--(1,3)--cycle,grey,black+linewidth(.8pt)); draw((0,1)--(0,3)--(1,3)--(1,4)--(4,4)--(4,3)-- (5,3)--(5,2)--(4,2)--(4,1)--(2,1)--(2,0)--(1,0)--(1,1)--cycle,finedashed); draw((0,2)--(2,2)--(2,4),finedashed); draw((3,1)--(3,4),finedashed); label(\"1\",(1.5,0.5)); draw(circle((1.5,0.5),.17)); label(\"2\",(2.5,1.5)); draw(circle((2.5,1.5),.17)); label(\"3\",(3.5,1.5)); draw(circle((3.5,1.5),.17)); label(\"4\",(4.5,2.5)); draw(circle((4.5,2.5),.17)); label(\"5\",(3.5,3.5)); draw(circle((3.5,3.5),.17)); label(\"6\",(2.5,3.5)); draw(circle((2.5,3.5),.17)); label(\"7\",(1.5,3.5)); draw(circle((1.5,3.5),.17)); label(\"8\",(0.5,2.5)); draw(circle((0.5,2.5),.17)); label(\"9\",(0.5,1.5)); draw(circle((0.5,1.5),.17));[/asy]$ $\\mathrm{(A) \\ } 2\\qquad \\mathrm{(B) \\ } 3\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 5\\qquad \\mathrm{(E) \\ } 6$ 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.", "k=\\boxed{086}$ ## Solution 2 $[asy] size(100); pointpen=black;pathpen = black+linewidth(0.7); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw(arc((0,2),1,270,360)); draw((0,1)--(1.7,2)); draw((0,2)--(1.7,1)); draw((0,1)--(1.7,1)--(1.7,2)); [/asy]$ If we imagine an arbitrary line with length $2$ connecting two sides of the square, we can draw the rectangle formed by drawing a perpendicular from where that line touches the square. Drawing the other diagonal of the rectangle, it also has length two, and it bisects with the original line. Since their intersection is the midpoint of both lines, the distance from the corner to the midpoint is always $1$, which forms a circle with radius $1$ centered at the corner of the square. The area of the shape then follows from simple calculations." ]

[ "the area of the green figure, note that we can move the two quarter-circles to make complete square units in the middle. Thus, the area is 6 unit2. To find the perimeter of the green figure, note that the boundary is composed of 16 segments of length 1 unit and 4 quarter-circles with radius 1 unit. Thus the perimeter is $16+2\\pi \\approx 22.28$ units. • To find the area of the yellow figure, note that we can subtract the area of a circle with radius 1 unit from a square with area 4 to get the area of one yellow \"star.\" So the area is $2\\cdot (4 - \\pi) = 8 - 2\\pi \\approx 1.7$ unit2. To find the perimeter of the yellow figure, note that the boundary is composed of 8 quarter-circles of radius 1. Thus the perimeter is $2\\cdot 2\\pi \\approx 12.57$ units.", "have a total length of $\\frac32 \\cdot 2\\pi\\cdot1$ unit so the purple figure has a perimeter of $3\\pi\\approx 9.42$ units. • There are 4 squares in the orange figure and 4 quarter-circles. Thus, the area of the orange figure is $4+\\pi \\approx 7.14$ unit2. To find the perimeter of the orange figure, note that the boundary is composed of 4 arcs that are $\\frac14$ of a circle and 8 segments of length 1. Thus, the orange figure has a perimeter of $8+2\\pi \\approx 14.28$ units. • To find the area of the red figure, note that there are 16 squares which contain 1 large circle with a radius of 1 unit, two small circles with a radius of $\\frac12$ unit, 2 half large circles, and 1 quarter large circle that are removed from the area of the squares. Thus, the area of the red figure is $$16 - \\pi -2\\cdot \\frac14\\pi -2\\cdot \\frac12 \\pi - \\frac14 \\pi$$ or $16- \\frac{11}{4}\\pi \\approx 7.36$ unit2. To find the perimeter of the red figure, note that the boundary is composed of 10 segments of length 1 unit and $2\\frac14$ circles of radius 1 unit and 2 circles of radius $\\frac12$ unit. Thus the perimeter is $$10 + 2.25\\cdot 2\\pi + 2\\cdot 2\\pi\\cdot\\frac12$$ or $10 + 6.5\\pi\\approx 30.52$ units. • To find", "square be so that half of the area is grass and half is water? ### Summary Here is a line segment on a grid. What is the length of this line segment? By drawing some circles, we can tell that it’s longer than 2 units, but shorter than 3 units. To find an exact value for the length of the segment, we can build a square on it, using the segment as one of the sides of the square. The area of this square is 5 square units. (Can you see why?) That means the exact value of the length of its side is $$\\sqrt5$$ units. Notice that 5 is greater than 4, but less than 9. That means that $$\\sqrt5$$ is greater than 2, but less than 3. This makes sense because we already saw that the length of the segment is in between 2 and 3. With some arithmetic, we can get an even more precise idea of where $$\\sqrt5$$ is on the number line. The image with the circles shows that $$\\sqrt5$$ is closer to 2 than 3, so let’s find the value of 2.12 and 2.22 and see how close they are to 5. It turns out that $$2.1^2=4.41$$ and $$2.2^2=4.84$$, so we need to try a larger number. If we increase our search", "in the shape of a quarter circle. If the land is enclosed by a fence, which of the following is closest to the length, in feet, of the fence? (A) 278 (B) 341 (C) 357 (D) 400 (E) 441 We are looking for 1/4th the circumference + 200 (2*pi*100)/4 = 50*pi = 50*3 (estimating) 150 + 200 = 350, C fits. Intern Joined: 04 Nov 2018 Posts: 13 Concentration: Finance, General Management GPA: 3.83 WE: Sales (Retail) Re: The figure shown represents a piece of land that is in the [#permalink] ### Show Tags 01 Apr 2019, 02:51 Had a brain-fart moment and didn't notice that the 100ft was equal to the radius of the quarter circle. So what I did was I thought \"what if the fence was triangular?\" If that was the case the length of the 3rd side of the triangle would be in the ratio of $$1:1:\\sqrt{2}$$, meaning the total fence needed would be $$1 + 1 + \\sqrt{2} ~= 3.14$$, or in our case around 314m. BUT, it is NOT a triangle but a quarter circle so the distance of that side must be slightly longer, hence C and 357. Not a great solution, but thought it was worthwhile pointing out how many different ways there are to reach the same conclusion. _________________", "Building on LOTGP's answer, you could do this: Assuming a unit square, the total length is: The top left segment is $$\\sqrt{2}/2$$. The three other segments are shortest when they meet at 120 degrees. This makes the triangle angles $$(120, 45, 15)$$. Using the sine rule, that gives $$\\sin{45}/\\sin{120} \\approx 0.8164$$ for the long sides $$\\sin{15}/\\sin{120} \\approx 0.2988$$ for the short sides for a total of about $$2.638958$$. This is a slight improvement over LOTGP's answer which is $$2+\\sqrt{2}/2 \\approx 2.707107$$. • According to wikipedia, this is the best known answer. However, our site policy dictates that you must invent some new mathematics and prove the optimality, or your brilliant solution does not count as an answer at all, and should be posted as a comment or community wiki instead. (If this policy seems unfair, there's a recent meta post to that effect currently active.) – Bass Sep 21 '19 at 8:16 • Well done! If you use roots for the trigonometic values, you can write it as $$\\sqrt{2}+\\sqrt{1.5}$$ – ThomasL Sep 21 '19 at 14:02 Seems a slightly better solution would be to: cover 2 of the sides that meet at one of the corners, then draw the half diagonal from the opposite corner to the middle. Something like this: The total length is then: 1 +", "19. Three circular arcs of radius $$5$$ units bound the region shown. Arcs $$AB$$ and $$AD$$ are quarter-circles, and arc $$BCD$$ is a semicircle. What is the area, in square units, of the region? $$25$$ $$10+5\\pi$$ $$50$$ $$50+5\\pi$$ $$25\\pi$$ ###### Solution(s): Create a rectangle that covers the bottom half of the figure as shown below. Then, we get that $[ABCD] = [ABD] + [BCD].$ We also know that $[ABD] = [BDEF] - [ABE]$$- [ADF].$ $$ABE$$ and $$ADF$$ are both quartercircles that form a semicircle with the same area as $$CBD.$$ This means that $[ABD] = [BDEF] - [BCD]$ and $[ABCD] = [BCD] + [BDEF]$$- [BCD]$ $= [BDEF] = 10 \\cdot 5 = 50.$ Thus, C is the correct answer. 20. You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $$1.02,$$ with at least one coin of each type. How many dimes must you have? $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ ###### Solution(s): Since we know that we have one coin of each type, we have $1 + 5 + 10 + 25 = 41$ cents already accounted for. We need to figure out what makes up the remaining $$102 - 41 = 61$$ cents. Note that we have $$5$$ coins left. This means that we only have one penny, since otherwise", "the land is enclosed by a fence , which of the following is the closest to the length , in feet, of the fence? A. 278 B. 341 C. 357 D. 400 E. 441 OA ----> C Please explain. Wish I got this on my gmatprep. If u know the rules of circles and what not, this is a gimme problem. Essentially the question is asking us for the perimeter of the quarter circle. P= Side1+side2+arc length of the 1/4 circle. radius is 100ft so the diameter is 200ft. Circum=200pi. to find the arc length take the degrees of the center of the circle given here as 90degrees and divide it by the total degrees of the circle. So 90/360=1/4. 1/4*200pi=50pi P=100+100+50pi ---> 200+50pi. U must know that pi is rougly 3.14 or even just 3. so 200+150=350. Closest answer to this is C. Director Joined: 29 Nov 2012 Posts: 732 Re: The figure represents a piece of land that is in the shape [#permalink] ### Show Tags 24 Jul 2013, 06:55 2 The way I did this one was the maximum value if we draw a square would be 400 and if we use Pythagorean theorem we get 100 + 100 + 100 $$\\sqrt{2}$$ The value will be greater than this and less than 400 so answer", "A series of perpendicular offsets taken from a curved boundary wall to a straight survey line at an interval of $6 \\: m$ are $1.22, \\: 1.67, \\: 2.04, \\: 2.34, \\: 2.14, \\: 1.87$, and $1.15 \\:m$. The area (in $m^2$, round off to 2 decimal places) bounded by the survey line, curved boundary wall, the first and the last offsets, determined using Simpson’s rule, is ________", "line segment. The square on a line equal to the arc of $90^\\circ$ fulfills both of the said requirements. False -- as pointed out above, this would produce a square of area $\\dfrac {\\pi^2 r^2} 4$, which is considerably greater than that of the circle. Trivially, of course, it does indeed have the correct perimeter. The circle as described by Goodwin has a diameter of $10$ and, as he states it, a circumference actually of $32$, delivering a value of $\\pi$ of $3.2$, not $3.1416$. The chord of $90^\\circ$ has length stated as $7$, and not what it actually is, approximately $7.0710$. Further mistakes are equally egregious. $\\blacksquare$", "configurations. In total, there are $$36 + 48 = 84$$ combinations. Thus, C is the correct answer. 19. A disk of radius $$1$$ rolls all the way around the inside of a square of side length $$s > 4$$ and sweeps out a region of area $$A.$$ A second disk of radius $$1$$ rolls all the way around the outside of the same square and sweeps out a region of area $$2A.$$ The value of $$s$$ can be written as $$a+\\dfrac{b\\pi}{c},$$ where $$a,b,$$ and $$c$$ are positive integers and $$b$$ and $$c$$ are relatively prime. What is $$a+b+c?$$ $$10$$ $$11$$ $$12$$ $$13$$ $$14$$ ###### Solution(s): The side length of the inner square traced out by the inner circle is $$s - 4.$$ There are also the small pieces remaining in the corner. These form a total area of $(1 + 1)^2 - \\pi 1^2 = 4 - \\pi.$ Therefore, $A = s^2 - (s - 4)^2 - (4 - \\pi)$$= 8s - 20 + \\pi.$ The outer disk traces out an area that is comprised of $$4$$ rectangles and $$4$$ quarter-circles. The rectangles have area $$s \\cdot 2 = 2s$$ and the quarter-circles form a circle with radius $$2$$ and area $$4 \\pi.$$ This gives us $2A = 8s + 4 \\pi.$ Equating the two equations we get", "four partially overlapping circles at the point $$A, B, C$$ and $$D$$ respectively. Connect those points and you will get a square having side equal to the diameter of small circle and that is obviously equal to 6. Let denote the area of the square $$ABCD = [ABCD]$$. So, the area of the quatrefoil = $$[ABCD]$$ + The area of 4 semi circles = $$6^2 + 4*\\frac{1}{2}*\\pi3^2$$ = $$36 + 18\\pi \\approx 92.549$$ If I understand the intended shape correctly, its outline is formed by four semicircles of radius $$3$$ whose centers form a square with a diagonal of $$6$$ units. They intersect at their endpoints, which form a larger square with a side of $$6$$ units. The total area inside the quatrefoil is then the area of the square plus the area inside the four semicircles: $$6^2 + 4(\\frac12\\pi\\cdot 3^2) = 36 + 18\\pi$$.", "wrote a short program to try a range of angles (as my calculus and geometry are too rusty) to find the best angle of 30 degrees. EDIT: Note that JonTheMon posted his 4.87 answer first while I was still programming, so he gets the credit for first with this answer. A Plow Distance 19 4.91250640267 20 4.90597796428 21 4.90001860583 22 4.89463591311 23 4.88983859219 24 4.88563652529 25 4.88204083336 26 4.87906394597 27 4.87671967936 28 4.87502332331 29 4.87399173788 30 4.87364346116 <- best 31 4.87399882899 32 4.8750801084 33 4.87691164606 ### First Solution: 5.14 km First plow the southern half of the circumference, then from each end plow 1 km north (making a U shape). The total distance will be pi+2 or 5.14 km. ## 5.217 km I used inscribed (green) and circumscribed (black) squares to help. The original circle is shown in blue. The plow paths are shown in red. The black square has a side of length $2$ and the green square has a side length of $\\sqrt2$. The plow path lengths are as follows: A = $\\sqrt2 = 1.414$ B = $\\sqrt2\\div2 = 0.707$ C = $\\sqrt{\\sqrt2\\div2} = 0.841$ Total Plow Length $= 5.217$ Even if my math is right, this is still not the optimal answer according to OP's source. I hope that it at least throws out some", "# From four corners of a square sheet of side 4 cm, four pieces each in the shape of arc of a circle with radius 2 cm, are cut out. The area of the remaining portion is: From four corners of a square sheet of side 4 cm, four pieces each in the shape of arc of a circle with radius 2 cm, are cut out. The area of the remaining portion is: [A](4 – 2π) sq. cm. [B](16 – 4π) sq. cm. [C](8 – π) sq. cm. [D](16 – 8π) sq. cm. (16 – 4π) sq. cm. Area of sectors $= \\pi\\times r^{2} = 4\\pi cm^{2}$ => Area of square = 4 $\\times$ 4 = 16 sq.cm. ∴ Area of the remaining portion = (16 – 4π) sq.cm. Hence option [B] is the right answer.", "degrees. If the midpoint of $AB$ is denoted as $K$ and a line is drawn from $M$ to $K$, then the angle $KMB$ is $15$ degrees and the tangent of $15$ degree is equal to $KB / MB$. Using the value of $\\tan 15$ enables us to calculate the length of $KB$; and doubling this length gives us the length of the $AB$ side of the quadrilateral (which, in fact, is actually a square. $AB$ squared is the area of the quadrilateral (SQUARE). The area beteen $AB$ and the curve of the quarter circle is a $30$-degree segment of the circle. The area of this circle's segment is equal to: $$\\pi\\cdot (MA)^2\\cdot \\frac{15}{360}-\\frac{1}{2}\\cdot (AB)^2\\cdot \\sin 15.$$ When we add the area of the quadrilateral (the square) to $4$ times the area of the calculated circle's segment, the addition sum is the answer we are seeking. -", "If the circles have radius $r$, then each pair of horizontal red lines is a distance $r$ apart, and they're a distance $r$ from the edges. Each pair of vertical blue lines is a distance $r \\sqrt 3$ apart, and they're still a distance $r$ from the edges. So if you want the triangular packing to have $m$ circles in each column, and $n$ columns, then the rectangle must be at least $(2m+1) \\cdot r$ units tall and $(2 + (n-1)\\sqrt3) \\cdot r$ units long. (Also, if the rectangle is only $2m \\cdot r$ units tall, we can alternate columns with $m$ and $m-1$ circles.) If the rectangle is $257 \\times 157$ and the radius of a circle is $\\sqrt{\\frac{10}{\\pi}}$, then: • If we make $257$ the vertical dimension, then the rectangle is a bit over $144 \\cdot r$ units tall, and a bit over $(2 + 49\\sqrt3) \\cdot r$ units wide. So we can arrange the circles in $50$ columns that alternate between $72$ and $71$ circles, for $25 \\cdot 72 + 25 \\cdot 71 = 3575$ circles. • If we make $157$ the vertical dimension, then the rectangle is a bit over $87 \\cdot r$ units tall, and a bit over $(2 + 82\\sqrt3) \\cdot r$ units wide. So we can arrange the circles in $83$", "Four rods are hinged at their ends to form a convex quadrilateral with sides of length $3$, $4$, $5$ and $6$ (in that order). Investigate the different shapes that the quadrilateral can take.", "because $6 \\times 6 = 36$. The radius of a circle with an area of 113.04 square inches is 6 inches. Using what we have learned, can we find the area of a sector? Sometimes we may be asked to find the area of a sector, or portion, of a circle, such as a quarter or half of the circle. As long as we know the radius, we can find the area of the whole circle. Then we can divide that area into smaller pieces or subtract a portion to find the area of part of the circle. Let’s try this out. Example What is the area of the figure below? This figure is a quarter of a circle, formed by a $90^\\circ$ angle. Remember, circles contain $360^\\circ$. One-quarter of $360^\\circ$ is $90^\\circ$. We know that the radius of the whole circle is 8.5 inches because the two sides of the sector are radii of the circle. Let’s use this value to solve for the area of the whole circle first. $A & =\\pi r^2\\\\A &= \\pi (8.5^2)\\\\A &= 72.25 \\pi\\\\A &= 226.87 \\ in.^2$ We know that the area of a whole circle with a radius of 8.5 inches is 226.87 square inches. Therefore the quarter circle formed by the $90^\\circ$ angle must have $\\frac{1}{4}$ of this", "area of the semicircle is $\\frac{\\pi r^2}{2}$ so with $k$ strips of oxhide Dido is able to enclose an area of $$\\frac{\\pi}{2} \\times \\left(\\frac{8k}{\\pi}\\right)^2 = \\frac{32k^2}{\\pi}$$ square feet. Comparing with part (a), Note that $\\frac{64}{9} \\lt \\frac{32}{\\pi}$ so the semicircle encloses a greater area than the square. As in part (a), the area Dido can enclose is a quadratic expression of $k$. 3. If $r$ is 10 miles then the semicircle has a perimeter of $10\\pi$ miles or $52,800\\pi$ feet. If we choose $k$ so that $$8k = 52,800\\pi$$ we find that $k = 6,600\\pi$. Unfortunately, $k$ needs to be a whole number but $\\pi$ is an irrational number and so 6,600$\\pi$ is also irrational. So Dido needs to cut the oxhide into a whole number of equal strips larger than 6,600$\\pi$. To get a sense of how thin these pieces would need to be, we can observe that there are 304.8 millimeters in a foot and 2,438.4 millimeters in 8 feet. So Dido would need to slice the oxhide into pieces that are about 1/10 millimeter in width! This is not realistic but it is appropriate for a myth.", "triangles is at most $n^2/(2n^2)=\\frac12$, so at least one of the triangles has at most that area. Now assume $k\\gt4n$. Then the boundary of the convex hull is a polygon with at least $4n$ points, and thus at least $4n$ line segments. The total length of these line segments is at most the perimeter of the enclosing square, which is $4n$. Thus the average length of the line segments is at most $1$, and the average sum of the lengths of adjacent line segments is at most $2$. Thus, there is at least one pair of adjacent line segments whose lengths add up to at most $2$. The largest triangle that can be formed with two given sides is a right triangle with area half the product of the sides, and for a given sum of the two sides the product is maximal when the sides are equal. Thus the area of the triangle formed by the two adjacent sides with sum at most $2$ is at most $\\frac12\\cdot1\\cdot1=\\frac12$. Thus in both cases there is at least one triangle with area at most $\\frac12$. -", "he could have sold. $50w+100r=500$ Since he cannot sell a fractional appliance, the number of washing machines must be 0, 2, 4, 6, 8, or 10 (w, r) = (0, 5), (2, 4), (4, 3), (6, 2), (8, 1), (10, 0) 2. Luis has 95 cents in dimes and quarters. Find all possibilities for the number of each type of coin he could have. $10d+25q=95$ Using similar logic as #1, (d, q) = (2, 3), (7, 1) 3. A certain quadrilateral has three sides of equal length and its perimeter is 19cm. Find all integral possibilities for the lengths of the sides in centimeters. (Hint: The sum of the lengths of any three sides of a quadrilateral must exceed the length of the fourth side.) $3x+y=19$ 4. An isosceles triangle has perimeter 15m. Find all integral possibilities for the lengths of the sides in meters. (Hint: The sum of the lengths of any two sides of a triangle must exceed the third side.) $2x+y=15$ ONLY THE EQUATIONS NEEDED TO SOLVE THESE PROBLEMS ARE NECESSARY. Originally Posted by mjp1991 ONLY THE EQUATIONS NEEDED TO SOLVE THESE PROBLEMS ARE NECESSARY. Sorry about solving the first 2. I didn't see your restriction 'til after" ]

[ "draw(circle((-4,-4), 4*1.41421356237)); [/asy]$ The way we figure out the area is by splitting it the following way: $[asy] size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real g(real x) { return x+8; } draw(graph(g,0,-8), red+linewidth(1.5)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real h(real x) { return -x-8; } draw(graph(h,-8,0), red+linewidth(1.5)); real z(real x) { return x-8; } draw(graph(z,0,8), red+linewidth(1.5)); pair A,B,C,D; A = (8,0); B = (0,8); C = (-8,0); D = (0,-8); fill(A--B--C--D--cycle, red); [/asy]$ We know each of the red lines is a diameter of the circle which is $\\sqrt2$. So the area of the red is 2. We know that the area of each semicircle is $\\dfrac14 \\pi$ so the area of the semicircles combines is $\\pi$. Thus we get $\\boxed{\\textbf{(B)} \\pi+2}.$", "Circle with interior arc segments 02-10-2015, 06:45 PM Post: #1 DrD Senior Member Posts: 1,132 Joined: Feb 2014 Circle with interior arc segments How would one calculate and plot circular arc segments laying totally within a semi-circle? Each segment begins at the right hand side (0*pi), and intersects the semi-circle circumference on pi/6, 2*pi/6, 3*pi/6 ... 6*pi/6, boundary intervals? This is the idea, but method needs work!: Code: EXPORT Circ() BEGIN LOCAL a; DIMGROB(G1,320,240); ARC_P(G1,156,109,105,0,pi,RGB(255,0,0)); // Semicircle center: (156,109), radius: 105 pixels LINE_P(G1,51,109,261,109,RGB(255,0,0)); FOR a FROM 0 TO pi STEP pi/6 DO LINE_P(G1,156,109,156+ROUND(105*COS(a),1),109-ROUND(105*SIN(a),1),RGB(210,210,210)); // pi/6 segment indicator ARC_P(G1,261,105*SIN(a)/2,105*COS(a),pi,3*pi/2,RGB(0,0,255)); // This approach isn't working ... END; BLIT_P(G0,G1); DRAWMENU(\"\"); FREEZE; END; Thanks! 02-11-2015, 02:12 PM Post: #2 Thomas_Sch Senior Member Posts: 377 Joined: Dec 2013 RE: Circle with interior arc segments I did a little playing around with your code. Don't know, if this helps. Code: EXPORT Circ() BEGIN LOCAL a; LOCAL red:= RGB(255,0,0); LOCAL blue:= RGB(0,0,255); LOCAL grey:= RGB(210,210,210); LOCAL lgrey:=RGB(210,210,00); LOCAL cx:=156, cy:=109, r:=105; LOCAL x2, y2, r2, x3, y3, r3; DIMGROB(G1,320,240); RECT_P(G0); LINE_P(G0,cx+r+1,cy,cx+r+1,0,blue); // G x y r w1 w2 ARC_P(G0, cx, cy, r, 0, pi, red); // Semicircle center: (156,109), radius: 105 pixels LINE_P(G0, cx-r, cy, cx+r, cy, red); // FOR a FROM 0 TO pi STEP pi/12 DO x2:= cx+ROUND(r*COS(a),1); r2:= r*SIN(a); y2:= cy-ROUND(r2,1); LINE_P(G0,", "circle, quarter circle, half circle (r=3 mm, Pi = 3.14). Question: If the wheel of a gate is 3 feet from the wall and it turns over 90 degrees, what is the distance covered by the wheel? Thread starter txpc; Start date Apr 7, 2010; Apr 7, 2010 #1 txpc. This video explains how to derive the area formula for a circle using integration. A total charge Q= -4.5uC is distributed uniformly over a quarter circle arc of radius a = 7.5cm as shown. What do you mean by graphing the derivative? Under MODE, choose DEGREE. Question: The radius of a quarter circle is 3 millimeters. Alternatively, you could substitute the radius of the quadrant directly into the formula A = ¼ πr². 1. Consider the quarter-circle of radius 1 and right triangles ABE and ACD given in the accompanying figure. Also, judging from the question in your first picture, the equation of your circle segment is, $$y'(4)=-\\frac{10}{\\sqrt{16-2^2}}=-\\frac{5}{\\sqrt{3}}\\approx-2.88675...$$, https://math.stackexchange.com/questions/3081908/how-can-i-graph-this-derivative-of-a-quarter-of-a-semicircle/3081916#3081916. In fact, the curve is infinitely differentiable everywhere, as it must be if it exactly represents a circle. Judging from the question in the first picture, it seems that you were asked to sketch the derivative of the quarter circle. tan 2 m 1.097 2 m 47.6q CD DX T T m 23.8q T ME101 - Division III Kaustubh", "\\pm \\frac{\\sqrt{90000}}{2} \\\\ &= 350 \\pm \\frac{300}{2} \\\\ &= 200, 500 \\end{align*} If the length of $AD$ is $200$, then quadrilateral $ABCD$ would be a square and thus, the radius of the circle would be $$\\frac{\\sqrt{200^2 + 200^2}}{2} = \\frac{\\sqrt{80000}}{2} = \\frac{200\\sqrt{2}}{2} = 100\\sqrt{2}$$ Which is a contradiction. Therefore, our answer is $\\boxed{500}.$ ## Solution 3 (Trigonometry Bash) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,E); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]$ Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply law of cosines on $\\Delta BOC$; let $\\theta = \\angle BOC$. We get the following equation: $$(BC)^{2}=(OB)^{2}+(OC)^{2}-2\\cdot OB \\cdot OC\\cdot \\cos\\theta$$ Substituting the values in, we get $$(200)^{2}=2\\cdot (200)^{2}+ 2\\cdot (200)^{2}- 2\\cdot 2\\cdot (200)^{2}\\cdot \\cos\\theta$$ Canceling out, we get $$\\cos\\theta=\\frac{3}{4}$$ Because $\\angle AOB$, $\\angle BOC$, and $\\angle COD$ are congruent, $\\angle AOD = 3\\theta$. To find", "each band at longitudes spaced by a golden section of the circle, i.e. $360^\\circ/\\varphi \\approx 222.5^\\circ.$ This method was used to arrange the 1500 mirrors of the student-participatory satellite . Dividing a line segment by interior division 1. Having a line segment $AB,$ construct a perpendicular $BC$ at point $B,$ with $BC$ half the length of $AB.$ Draw the $AC.$ 2. Draw an arc with center $C$ and radius $BC.$ This arc intersects the hypotenuse $AC$ at point $D.$ 3. Draw an arc with center $A$ and radius $AD.$ This arc intersects the original line segment $AB$ at point $S.$ Point $S$ divides the original line segment $AB$ into line segments $AS$ and $SB$ with lengths in the golden ratio. Dividing a line segment by exterior division 1. Draw a line segment $AS$ and construct off the point $S$ a segment $SC$ perpendicular to $AS$ and with the same length as $AS.$ 2. Do bisect the line segment $AS$ with $M.$ 3. A circular arc around $M$ with radius $MC$ intersects in point $B$ the straight line through points $A$ and $S$ (also known as the extension of $AS$). The ratio of $AS$ to the constructed segment $SB$ is the golden ratio. Application examples you can see in the articles Pentagon with a given side length, Decagon with", "\\dots, 21$. This is a total of $\\boxed{(D) 13}$ distinct line segments. (asymptote diagram added by elements2015) Solution 2 - Circles Note that if a circle with an integer radius $r$ centered at vertex $B$ intersects hypotenuse $\\overline{AB}$, the lines drawn from $B$ to the points of intersection are integer lengths. As in the previous solution, the shortest distance $14<\\overline{BP}<15$. As a result, a circle of $14$ will not reach the hypotenuse and thus does not intersect it. We also know that a circle of radius $21$ intersects the hypotenuse once and a circle of radius $\\{15, 16, 17, 18, 19, 20 \\}$ intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this. $[asy] unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot(\"A\", A, SW); dot(\"B\", B, SE); dot(\"C\", C, NE); dot(\"P\", P, S); draw(arc((0,0),21, 90, 180)); draw(arc((0,0),20, 90, 180)); draw(arc((0,0),19, 90, 180)); draw(arc((0,0),18, 90, 180)); draw(arc((0,0),17, 90, 180)); draw(arc((0,0),16, 90, 180)); draw(arc((0,0),15, 90, 180)); [/asy]$ It follows that we can draw circles of radii $15, 16, 17, 18, 19,$ and $20,$ that each contribute two integer lengths (since these circles intersect the hypotenuse twice) from $B$ to $\\overline{AC}$ and one circle of radius $21$ that contributes only one such segment. Our answer is then", "and B. Besides circles and parts of circles, there’s also a standard square path called unitsquare ; this is a cycle that runs from (0, 0) to (1, 0) to (1, 1) to (0, 1) and back to (0, 0). For example, the command ‘fill unitsquare ’ adds 1 to a single pixel value, as discussed in the previous chapter. 123 boundaries trajectories paths quartercircle circle halfcircle fullcircle ellipse 124 Chapter 14: Paths \u001a EXERCISE 14.5 Use fullcircle and unitsquare to produce the characters ‘Q’ and ‘R’ in font positions ‘e’ and ‘f’, respectively. These characters should be 10 pt wide and 10 pt tall, and their centers should be 2.5 pt above the baseline. (Figure 14a will be inserted here; too bad you can’t see it now.) path branch [ ], trunk ; branch 1 = flex ((0, 660), (−9, 633), (−22, 610)) & flex ((−22, 610), (−3, 622), (17, 617)) & flex ((17, 617), (7, 637), (0, 660)) & cycle; branch 2 = flex ((30, 570), (10, 590), (−1, 616)) & flex ((−1, 616), (−11, 592), (−29, 576), (−32, 562)) & flex ((−32, 562), (−10, 577), (30, 570)) & cycle; branch 3 = flex ((−1, 570), (−17, 550), (−40, 535)) & flex ((−40, 535), (−45, 510), (−60, 477)) & flex ((−60, 477), (−20, 510), (40, 512))", "are of $90^\\circ$ measure. Therefore, the square READ can be drawn as follows: STEP 1 :Firstly, we will draw a rough sketch of square READ (Image Will Be Updated Soon) STEP 2 :Now, draw a line segment RE of length $5.1cm$ and an angle of $90^\\circ$ at a point R and E. (Image Will Be Updated Soon) STEP 3: Now, as the vertex A and D are of length $5.1cm$ away from vertex E and R, Cut the line segments EA and RD , each of length $5.1cm$ from these rays. (Image Will Be Updated Soon) STEP 4: Now, join point D to A. (Image Will Be Updated Soon) Here, READ is the required square. 2. Draw the following: A rhombus whose diagonals bisect are $5.2cm$ and $6.4cm$ long. Ans: In the figure of rhombus , the diagonals bisect each other at $90^\\circ$. Therefore, the given rhombus ABCD can be drawn as follows. STEP 1: Firstly, draw a rough sketch of this rhombus ABCD as follows. (Image Will Be Updated Soon) STEP 2: Now, draw a line segment AC of length $5.2cm$ and draw its perpendicular bisector. And let it intersect the line segment AC at point O. (Image Will Be Updated Soon) STEP 3: Draw arcs of $\\begin{gathered} \\frac{{6.4}}{2} = 3.2cm \\hfill \\\\ \\hfill \\\\ \\end{gathered}$ on", "each of the following: 1. $6^3 - (7^2 + 6^2)$ 2. $(8-5)^3$ 3. $(\\sqrt[3]{125})^2$ 4. $12^2 - 4\\sqrt{121} \\div 2^2$ 5. $3\\sqrt{64}$ 1. Write down the letters of all the acute angles in the diagram. 2. Measure the size of angle $d$ in the diagram and write it down. 3. Classify angle $d$ according to its size. 8. Draw and correctly label angle $K\\hat{L}M = 168^{\\circ}$. 9. Use your ruler and protractor to draw a line that is parallel to line segment FG drawn below, and goes through point H. 10. Four circles are drawn so that they fit neatly into a square with side length of 6 cm, as shown (not to scale). Write down the radius of each circle. 1. What is the geometrical name of the shape shown on the dot grid below? 2. Draw two shapes that are similar to the shape shown, anywhere on the grid. Each shape that you draw should have a different size. 11. The following diagram shows a square drawn on a dot grid. The square is divided into four triangles, namely A, B, C, and D. 1. Write down the letters of all the right-angled triangles. 2. Write down the letters of all the isosceles triangles. 3. Write down the letters of the two congruent triangles. 12. I", "integer that is a divisor of $$(n+1)(n+3)(n+5)(n+7)(n+9)$$ for all positive even integers $n$? $\\textbf{(A) } 3 \\qquad\\textbf{(B) } 5 \\qquad\\textbf{(C) } 11 \\qquad\\textbf{(D) } 15 \\qquad\\textbf{(E) } 165$ ## Problem 19 Three semicircles of radius $1$ are constructed on diameter $\\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles? $[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype(\"4 4\"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label(\"A\",A,W); label(\"B\",B,E); label(\"1\",(-1.5,-0.1),S); label(\"2\",(0,-0.1),S); label(\"1\",(1.5,-0.1),S);[/asy]$ $\\textbf{(A) } \\pi - \\sqrt{3} \\qquad\\textbf{(B) } \\pi - \\sqrt{2} \\qquad\\textbf{(C) } \\frac{\\pi + \\sqrt{2}}{2} \\qquad\\textbf{(D) } \\frac{\\pi +\\sqrt{3}}{2} \\qquad\\textbf{(E) } \\frac{7}{6}\\pi - \\frac{\\sqrt{3}}{2}$ ## Problem 20 In rectangle $ABCD, AB=5$ and $BC=3$. Points $F$ and $G$ are on $\\overline{CD}$ so that $DF=1$ and $GC=2$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\\triangle AEB$. $[asy] unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair A=(0,0), B=(5,0), C=(5,3), D=(0,3), F=(1,3), G=(3,3); pair E=extension(A,F,B,G); draw(A--B--C--D--A--E--B); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); label(\"E\",E,N); label(\"F\",F,SE); label(\"G\",G,SW); label(\"B\",B,SE); label(\"1\",midpoint(D--F),N); label(\"2\",midpoint(G--C),N); label(\"3\",midpoint(B--C),E); label(\"3\",midpoint(A--D),W); label(\"5\",midpoint(A--B),S); [/asy]$ $\\textbf{(A) } 10 \\qquad\\textbf{(B) } \\frac{21}{2} \\qquad\\textbf{(C) } 12 \\qquad\\textbf{(D) } \\frac{25}{2} \\qquad\\textbf{(E) } 15$ ## Problem", "# Problem #2264 2264 The closed curve in the figure is made up of $9$ congruent circular arcs each 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? $[asy] defaultpen(fontsize(6pt)); dotfactor=4; label(\"\\circ\",(0,1)); label(\"\\circ\",(0.865,0.5)); label(\"\\circ\",(-0.865,0.5)); label(\"\\circ\",(0.865,-0.5)); label(\"\\circ\",(-0.865,-0.5)); label(\"\\circ\",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]$ $\\textbf{(A)}\\ 2\\pi+6\\qquad\\textbf{(B)}\\ 2\\pi+4\\sqrt3 \\qquad\\textbf{(C)}\\ 3\\pi+4 \\qquad\\textbf{(D)}\\ 2\\pi+3\\sqrt3+2 \\qquad\\textbf{(E)}\\ \\pi+6\\sqrt3$ 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.", "2015 AMC 12A Problems/Problem 25 Problem A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k>=1$, the circles in $\\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is $$\\sum_{C\\in S} \\frac{1}{\\sqrt{r(C)}}?$$ $[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip)", "19. Three circular arcs of radius $$5$$ units bound the region shown. Arcs $$AB$$ and $$AD$$ are quarter-circles, and arc $$BCD$$ is a semicircle. What is the area, in square units, of the region? $$25$$ $$10+5\\pi$$ $$50$$ $$50+5\\pi$$ $$25\\pi$$ ###### Solution(s): Create a rectangle that covers the bottom half of the figure as shown below. Then, we get that $[ABCD] = [ABD] + [BCD].$ We also know that $[ABD] = [BDEF] - [ABE]$$- [ADF].$ $$ABE$$ and $$ADF$$ are both quartercircles that form a semicircle with the same area as $$CBD.$$ This means that $[ABD] = [BDEF] - [BCD]$ and $[ABCD] = [BCD] + [BDEF]$$- [BCD]$ $= [BDEF] = 10 \\cdot 5 = 50.$ Thus, C is the correct answer. 20. You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $$1.02,$$ with at least one coin of each type. How many dimes must you have? $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ ###### Solution(s): Since we know that we have one coin of each type, we have $1 + 5 + 10 + 25 = 41$ cents already accounted for. We need to figure out what makes up the remaining $$102 - 41 = 61$$ cents. Note that we have $$5$$ coins left. This means that we only have one penny, since otherwise", "+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", "k=\\boxed{086}$ ## Solution 2 $[asy] size(100); pointpen=black;pathpen = black+linewidth(0.7); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw(arc((0,2),1,270,360)); draw((0,1)--(1.7,2)); draw((0,2)--(1.7,1)); draw((0,1)--(1.7,1)--(1.7,2)); [/asy]$ If we imagine an arbitrary line with length $2$ connecting two sides of the square, we can draw the rectangle formed by drawing a perpendicular from where that line touches the square. Drawing the other diagonal of the rectangle, it also has length two, and it bisects with the original line. Since their intersection is the midpoint of both lines, the distance from the corner to the midpoint is always $1$, which forms a circle with radius $1$ centered at the corner of the square. The area of the shape then follows from simple calculations.", "# Problem #2335 2335 A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k\\ge1$, the circles in $\\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is $$\\sum_{C\\in S} \\frac{1}{\\sqrt{r(C)}}?$$ $[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair", "# Problem #2335 2335 A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k\\ge1$, the circles in $\\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is $$\\sum_{C\\in S} \\frac{1}{\\sqrt{r(C)}}?$$ $[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair", "# Problem #2352 2352 Seven cookies of radius $1$ inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie? $[asy] draw(circle((0,0),3)); draw(circle((0,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((-1,sqrt(3)),1)); draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); draw(circle((2,0),1)); draw(circle((-2,0),1)); [/asy]$ $\\textbf{(A) } \\sqrt{2} \\qquad \\textbf{(B) } 1.5 \\qquad \\textbf{(C) } \\sqrt{\\pi} \\qquad \\textbf{(D) } \\sqrt{2\\pi} \\qquad \\textbf{(E) } \\pi$ 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: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i).", "directed from (1 - sqrt(3), 1) to (1 + sqrt(3), 1). Rational_point c5(1, 1); Circle circ5(c5, 3, CGAL::CLOCKWISE); CoordNT one_minus_sqrt_3(1, -1, 3); CoordNT one_plus_sqrt_3(1, 1, 3); Point s5(one_minus_sqrt_3, CoordNT(1)); Point t5(one_plus_sqrt_3, CoordNT(1)); curves.push_back(Curve(circ5, s5, t5)); // Create an arc (C6) of the unit circle, directed clockwise from // (-1/2, sqrt(3)/2) to (1/2, sqrt(3)/2). // The supporting circle is oriented accordingly. Rational_point c6(0, 0); Number_type half = Number_type(1) / Number_type(2); CoordNT sqrt_3_div_2(Number_type(0), half, 3); Point s6(-half, sqrt_3_div_2); Point t6(half, sqrt_3_div_2); curves.push_back(Curve(c6, 1, CGAL::CLOCKWISE, s6, t6)); // Create a circular arc (C7) defined by two endpoints and a midpoint, // all having rational coordinates. This arc is the upper right // quarter of a circle centered at the origin with radius 5. curves.push_back(Curve(Rational_point(0, 5), Rational_point(3, 4), Rational_point(5, 0))); // Construct the arrangement of the curves and print its size. Arrangement arr; insert(arr, curves.begin(), curves.end()); print_arrangement_size(arr); return 0; } It is also possible to construct $$x$$-monotone curve objects that represent $$x$$-monotone circular arcs or line segments using similar constructors. (Full circles are not $$x$$-monotone.) The traits class-template Arr_circular_line_arc_traits_2<CircularKernel> offered by the arrangement package also handles circular arcs and line segments. It is an alternative to the Arr_circle_segment_traits_2<Kernel> class-template. These two class templates, while serve similar purposes, are based on different concepts, and posses different characteristics. You are encouraged to experiment with", "# Difference between revisions of \"2015 AMC 12A Problems/Problem 25\" ## Problem A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as followsLayer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k \\ge 1$, the circles in $\\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is $$\\sum_{C\\in S} \\frac{1}{\\sqrt{r(C)}}?$$ $[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64" ]

[ "## Algebra: A Combined Approach (4th Edition) $35$ 44-cents stamps. $5$ 28-cents stamps. Let $x$ be the number of 44 cents (0.44 dollars) stamps. Let $y$ be the number of 28 cents (0.28 dollars) stamps. Sarah and Keith Robinson purchased 40 stamps, a mixture of 44¢ and 28¢ stamps. $x+y=40$ They spent $16.80$. $0.44x+0.28y=16.80$ $x+y=40$ $0.44x+0.28y=16.80$ Multiply the second equation by 100. $44x+28y=1680$ $x=40-y$ $44x+28y=1680$ $44(40-y)+28y=1680$ $1760-44y+28y=1680$ $1760-16y=1680$ $-16y=-80$ $y=5$ $x=40-y$ $x=40-(5)$ $x=35$ $35$ 44-cents stamps. $5$ 28-cents stamps.", "The total value of the stamps is $\\8.34.$ Find the number of 22¢ stamps in the collection.", "which one cannot buy stamps is (A) $19~~$ (B) $20~~$ (C) $23~~$ (D) $29~~$", "some of the stamps are 49 cents stamps and some 33 cent stamps how many 49 cents stamps and how many 33 cents stamps did she purchase if the total cost of all the stamps was 20.31 ----------------------- Quantity:: t + n = 47 stamps Value:: : 33t+49n = 2031 cents ---------------------------------- Modify for elimination: 49t + 49n = 49*47 33t + 49n = 2031 -------- Subtract and solve for \"t\":: 16t = 272 t = 17 (# of 33 cent stamps purchased) ----------- Cheers, Stan H. ----------- Question 993949: The school that Jenny goes to is selling tickets to a fall musical. On the first day of ticket sales the school sold 7 senior citizen tickets and 14 student tickets for a total of$175. The school took in $215 on the second day by selling 12 senior citizen tickets and 7 student tickets. What is the price each of one senior citizen ticket and one student ticket? Answer by Boreal(1464) (Show Source): You can put this solution on YOUR website! x=SR y=student 7x+14y=175 12x+7y=215 ====== 7x+14y=175 -24x-14y=-430, multiplying it by (-2). Now add -17x=- 255 x=$15 SR 105+14y=175 14y=70 y=$5 student ====== check with second 12*15=$180 7*5=$35. They add to$215 Question 993948: Rob and Amy each improved their yards by planting daylilies and ornamental grass. They bought", "=> for 380K the profit exceeded 380x2/200 =$3.8 mil -- not sufficient B. for 350K the profit exceeded $5 mil => for 380K the profit exceeded 380x5/350 =$5.428 mil -- sufficient Good strategy would be to solve to a close number instead of guessing. You dont have to actual solve the numbers to the dot but comparing ratios would be helpful like for above question: for A. for 200000 units you exceeded 2 mil -- so for every 100000 units you exceed a million and for 380000 you exceed 3.8 mil but you do not know if its above 4 mil for B. for 350000 you got 5 mil then for 380000 you will get more than 5 mil. Manager Joined: 10 Feb 2010 Posts: 193 Followers: 2 Kudos [?]: 112 [0], given: 6 Re: DS questions from the 12th eition Official Guide [#permalink] ### Show Tags 17 Feb 2010, 21:33 luminousmocha wrote: #123 Joana bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought only$4.40 worth of stamps (2) she bought an equal number of $0.15 stamps and$0.29 stamps answer: A # 123. Ans is A A. let 0.15 be x number and 0.29 be y then, 0.15x + 0.29 y = 4.40 => x= 4.40-0.29y/0.15 you can directly check for", "between (n-1)143 and n(134) cannot be obtained. Because the stamps are in increments of 1, every value between 134n and 143n can be obtained. So we need to discover where 134 * n is less than or equal to 143 * (n-1) Solving 134n = 143n - 143 Gives 143 = 9n n=15.89 So , for who number values of n greater than 15, there is no number that exists between the highest number obtainable with a combination of n-1 stamps and the lowest number that can be obtained with n stamps. All values above this can be achieved. We need the highest number that exists in the gap below this, i.e. 15 * 134 -1 What is the Frobenius Number of these numbers: 134,135,136,137,138,139,140,141,142,143 Thankfully, Wolfram Alpha has a built-in function for that We can now substitute the values $a=134$, $d=1$ and $s=9$. Using it, we get the answer 2009 as already given by @IvoBeckers", "total of $8500 is invested, part of it at 6% and the rest at 3.5%. The total interest after one year is$385. How much was invested at each rate? A postal clerk sold some 15¢ stamps and some 25¢ stamps. Altogether, 15 stamps were sold for a total cost of \\$3.15. How many of each type of stamps were sold?", "total of $6000 is invested, part of it at 4.25% and the rest at 5.75%. The total interest after one year is$300. How much was invested at each rate? A postal clerk sold some 15¢ stamps and some 25¢ stamps. Altogether, 15 stamps were sold for a total cost of \\$3.15. How many of each type of stamps were sold?", "Cost of 105 envelopes is Rs. 350. How many envelopes can be purchased for Rs. 100? - Mathematics Sum Cost of 105 envelopes is Rs. 350. How many envelopes can be purchased for Rs. 100? Solution In Rs. 350, the number of envelopes that can be purchased = 105. Therefore, in Rs. 1, number of envelopes that can be purchased = 105/350 Therefore, in Rs. 100, the number of envelopes that can be purchased = 105/350 × 100 = 30. Thus, 30 envelopes can be purchased for Rs. 100. Is there an error in this question or solution?", "# 4.5 Value Problems A total of $8500 is invested, part of it at 6% and the rest at 3.5%. The total interest after one year is$385. How much was invested at each rate? A total of $6000 is invested, part of it at 4.25% and the rest at 5.75%. The total interest after one year is$300. How much was invested at each rate? A total of $12000 was invested, part of it at 9% and the rest at 7.5%. The total interest after one year is$1005. How much was invested at each rate? A collection of half dollars and nickels is worth $13.40. There are 34 coins in all. How many are there? A postal clerk sold some 15¢ stamps and some 25¢ stamps. Altogether, 15 stamps were sold for a total cost of$3.15. How many of each type of stamps were sold? A purse contains $3.90 made up of dimes and quarters. If there are 21 coins in all, how many dimes and how many quarters were there? A collection of dimes and quaters is worth$15.25. There are 103 coins in all. How many of each is there? A boy has $2.25 in nickels and dimes. If there are twice as many dimes as nickels, how many of each kind has he? 30 coins having a", "# A man buys postage stamps of denominations 25 paise and 50 paise for Rs. 10. He buys 28 stamps in total. Find the number of 25 paise stamps bought by him.$\\left( a \\right)$ 12$\\left( b \\right)$ 16$\\left( c \\right)$ 20$\\left( d \\right)$ 18 Verified 147.3k+ views Hint: In this particular question assume any variable be the number of postage stamps of denominations 25 paise so the number of postage stamps of denominations 50 paise is (28 – x) as the total number of stamps he buys are 28, so use these concepts to reach the solution of the question. Given data: A man buys postage stamps of denominations 25 paise and 50 paise for Rs. 10. Total number of postage stamps he buys are 28. And the total amount he spent is Rs. 10 As we all know that 1 Rs = 100 paise. So, 10 Rs. = 1000 paise. Now, let the number of postage stamps of denominations 25 paise be X. So the number of postage stamps of denominations 50 paise are (28 – X) as the total number of postage stamps are 28. Now the sum of the multiplication of respective postage stamps with their respective cost is equal to the total amount he spent in paise. So convert the above information into equation", "greater than $40. Thus, total number of items with price above$35 is 3*33 i.e. 99", "$n \\geq 4$, at least two 2-cent stamps are used. Remove two of these and add a single 5-cent stamp. We now have $n - 2 \\cdot 2 + 5 = n+1$; a realization of $n+1$ cents. -", "stamps: 1.000", "where\\ n\\geqslant 0$ after 17 ending with 2,7. $5n+51\\ where\\ n\\geqslant 0$ after 51 ending with 1,6. $5n+68\\ where\\ n\\geqslant 0$ after 68 ending with 8,3. $5n+85\\ where\\ n\\geqslant 0$ after 85 ending with 0,5. $5n\\ where\\ n\\geqslant 0$ after 0 ending with 0,5 since you do not have to use 17-cent stamp. As a result After 68, you can find any number you want by using 5 and 17-cents. So the next number below 68 by using the equation $5n+68$ will be the upper limit where $n=-1$. 63. - I think you are allowed to use one type of stamp, so 13 5 cent stamps could make 85. – Trenin Mar 7 at 14:26 @Trenin the question states that \"a combination of these stamps\"... So I presume it requires to use at least one 5-cent stamp and one 13-cent stamps by \"a combination of these stamps\" – Oray Mar 7 at 14:28 To me, a valid combination would be to use zero 17 cent stamps. The point of the question seems to be what values you can make from those stamps. In my opinion, there is no need to add the restriction that you must use at least one. But that is just me - OP should either clarify or comment. – Trenin Mar 7 at 14:33", "of 17 rs. stamps. Now buyer ordered for at least more than 1 stamp for each type Hence the minimum he bought was: 2 stamp for 5 rupees = 10 rs. 2 stamp for 2 rupees = 4 rs. 2 stamp for 1 rupee = 2 rs. For the total to be seventeen, the buyer must have purchased 3 one rupee stamps. And 3 one rupee stamps were also there as changes given by shopkeeper. So total number of stamps = 2+2+(3+3) = 10 Answer 4: let’s say price of maruti car is x rs. Sales = y revenue = xy Changed price = 1.3x changed value of sales = 0.8y new revenue = 1.04 xy Percentage change in revenue = 4% Answer 5: Let’s say price of turban is x. So total price for 12 months will be = $90+x$ Total price for 9 months = $\\frac{(90+x) \\times 9}{12} = (65+x)$ By solving above equation, we will get value of x= 10. Download CAT Quantitative Aptitude Formulas PDF", "of stamps that Luciano collected over several weeks. If the pattern continues, how many stamps will Luciano collect in Week 8? Explain. _______ stamps Explanation: In week 4 and 6, the number of stamps are 4, 5. Therefore in week 6 and 8, the number of stamps are 5, 7 Question 7. The data set shows the number of hours Luke plays the piano each week. Luke says he usually plays the piano 3 hours per week. Why is Luke’s statement misleading? Type below: _________________ Answer: According to the question he should spend 3 hours per week. His statement is correct. Explanation: Sum of the data = 1+2+1+3+2+10+2 = 21 Number of days in a week = 7 Mean = 21/7 = 3 hours ### Problem Solving Misleading Statistics – Page No. 755 Mr Jackson wants to make dinner reservations at a restaurant that has most meals costing less than $16. The Waterside Inn advertises that they have meals that average$15. The table shows the menu items. Question 1. What is the minimum price and the maximum price? minimum: $_________ maximum:$ _________ Answer: minimum: $6 maximum:$19 Explanation: The minimum value is the most minimum price in the given data. The maximum value is the most maximum price in the given data. Question 2. What is the mean of", "of stamps that Luciano collected over several weeks. If the pattern continues, how many stamps will Luciano collect in Week 8? Explain. _______ stamps Explanation: In week 4 and 6, the number of stamps are 4, 5. Therefore in week 6 and 8, the number of stamps are 5, 7 Question 7. The data set shows the number of hours Luke plays the piano each week. Luke says he usually plays the piano 3 hours per week. Why is Luke’s statement misleading? Type below: _________________ Answer: According to the question he should spend 3 hours per week. His statement is correct. Explanation: Sum of the data = 1+2+1+3+2+10+2 = 21 Number of days in a week = 7 Mean = 21/7 = 3 hours ### Problem Solving Misleading Statistics – Page No. 755 Mr Jackson wants to make dinner reservations at a restaurant that has most meals costing less than $16. The Waterside Inn advertises that they have meals that average$15. The table shows the menu items. Question 1. What is the minimum price and the maximum price? minimum: $_________ maximum:$ _________ Answer: minimum: $6 maximum:$19 Explanation: The minimum value is the most minimum price in the given data. The maximum value is the most maximum price in the given data. Question 2. What is the mean of", "did the profit exceed $4 million on sales of 380000 units? (1) the profit exceeded$2 million on sales of 200000 units (2) the profit exceeded $5 million on sales of 350000 units answer: B Got stuck on (2), what's a good strategy to solve this question? #123 Joana bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy? (1) She bought only $4.40 worth of stamps (2) she bought an equal number of$0.15 stamps and $0.29 stamps answer: A #137 at a bakery, all donuts are price equally and all bagels are price equally. what is the total price of 5 donuts and 3 bagels at the bakery? (1) at the bakery, the total price of 10 donuts and 6 bagels is$12.90 (2) at the bakery, the price of a donut is $0.15 less than the price of a bagel answer: A Manager Joined: 17 Jan 2010 Posts: 147 Concentration: General Management, Strategy GPA: 3.78 WE: Engineering (Manufacturing) Followers: 1 Kudos [?]: 76 [1] , given: 11 Re: DS questions from the 12th eition Official Guide [#permalink] ### Show Tags 17 Feb 2010, 21:56 1 This post received KUDOS testprep2010 wrote: luminousmocha wrote: #123 Joana bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy? (1) She bought only $4.40 worth of stamps (2)", "Engel curve for stamps in blue ink,(Hint,First draw the Engel curves for incomes greater than$1,then draw them for incomes less than $1.) 02468 2 4 6 Quantities Income 8 Blue line Red line 1 0.5 6.5 (0) Shirley Sixpack,as you will recall,thinks that two 8-ounce cans of beer are exactly as good as one 16-ounce can of beer,Suppose that these are the only sizes of beer available to her and that she has$30 to spend on beer,Suppose that an 8-ounce beer costs $.75 and a 16-ounce beer costs$1,On the graph below,draw Shirley¡¯s budget line in blue ink, and draw some of her indi erence curves in red. 72 DEMAND (Ch,6) 010203040 10 20 30 16-ounce cans 8-ounce cans 40 Blue budget line Red curves Red curve (a) At these prices,which size can will she buy,or will she buy some of each? 16-ounce cans. (b) Suppose that the price of 16-ounce beers remains $1 and the price of 8-ounce beers falls to$.55,Will she buy more 8-ounce beers? No. (c) What if the price of 8-ounce beers falls to $.40? How many 8-ounce beers will she buy then? 75 cans. (d) If the price of 16-ounce beers is$1 each and if Shirley chooses some 8-ounce beers and some 16-ounce beers,what must be the price of 8-ounce beers? $.50. (e) Now" ]

[ "+ 15y = 84$$ o $$2x + 5y = 28$$ • We know that x and y are integers. [Stamps cannot be in fraction.] o There are 2 possibilities  Case 1: x = 4 and y =4 • $$2*4 + 5*4 = 8+20 =28$$ • In this case none of x, y, and z are greater than 5  Case 2: x = 9 and y =2 • $$2*9 + 5*2 = 18 + 10 = 28$$ • In this case x is greater than 5. Hence, statement 1 is not sufficient, we can eliminate answer options A and D. Statement 2: The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps is not more than the number of 40 cent stamps she bought. • $$x = y$$, and z is not greater than y. On substituting x = y in equation (i), we get • $$6y + 7y + 8z = 84$$ o $$13y + 8z = 84$$ o $$z = \\frac{(84-13y)}{8}$$ [this must be a non-negative integer; z can never be in fraction] • There is only one possibility. o $$y = 4$$, then $$z =\\frac{(84 - 52)}{8} = 4$$ o $$x = y = z = 4$$ o In this case none of", "1. ## stamps problem hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks 2. Originally Posted by mathos hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks $5 stamps = a$3 stamps = b $0.50 stamps = c $5a+3b+0.5c = 100$ $a+b+c = 100$ Not sure where to go from there 3.$5 stamps = a $3 stamps = b$0.50 stamps = c (1) (2) (2) divided by 2: (3) $\\frac {a}{2} + \\frac {b}{2}+0.5c=50$ Hence, $0.5c= 50-\\frac {a}{2}-\\frac {b}{2}$ Substituting this into equation (1) will give: $5a + 3b + 50-\\frac {a}{2}-\\frac {b}{2}=100$ Which will eliminate c from one equation...I don't think that it can be done for another, which would then have allowed you to solve simultaneously because you only have 2 equations but 3 unknowns.", "right as many times as necessary to make it a whole number. We have 1 decimal place, so we need to multiply by 10. $$\\frac{1.25~\\times~10}{2.5~\\times~10}~=~\\frac{12.5}{25}$$ Divide the dividend (12.5) by the divisor (25). $$=~\\frac{12.5}{25}~=~0.5$$ 0.5 _______ 25 ) 12.5 – 0 ———- 125 – 125 ———- 0 Example 5: Divide the decimal value of 66.6 by the two-digit number 33.3 Convert the divisor to a whole number by multiplying the numerator and the denominator by 10. $$=~\\frac{66.6~\\times~10}{33.3~\\times~10}~=~\\frac{666}{333}~=~2$$ 2 _______ 333 ) 666 – 666 ———- 0 ## Real-life Examples Example 6: Jack has a stamp collection. Each stamp weighs 0.4 ounces. If the total weight of his stamps is 26 ounces, how many stamps does he have? The weight of one stamp is 0.4 ounces. The total weight of all of the stamps combined is 26 ounces. To figure out the number of stamps, divide the total weight by the weight of one stamp: $$26~\\div~4$$ $$=~\\frac{6~\\times~10}{0.4~\\times~10}$$ Convert the divisor into a whole number $$=~\\frac{260}{4}$$ = 65 65 _______ 4 ) 260 – 24 ———- 20 – 20 ———- 0 Hence, Jack has 65 stamps in his collection. Example 7: Tom runs 45.5 miles in 6.5 hours a day. How many miles does he cover in 1 hour? To calculate the miles that Tom covers in 1 hour, we", "bought $4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Why is that so clear that only one integer combination fits this? Very difficult to spot... This is quite representative of a GMAT like question and thus for such questions wherein you are asked number of tickets, number of people, number of toys etc wherein only integer values can work, make sure to try to find a few sets of values for both the variables that will satisfy the given equation which in this case is 15x+29y=440. Once you create the equation above, you can see that you could also write it as 15x=440-29y which means that 440-29y MUST be a multiple of 15 (as the other side is 15x). Thus once you start by recognizing this fact, you will see that only y=10 satisfies this. For all other values you will not get an integer value of x or get a value <0 (this is not acceptable as number of tickets can not be <0). Hope this helps. Kudos [?]: 1789 [1], given: 796 Intern Joined: 19 Aug 2015 Posts: 24 Kudos [?]: 38 [0], given: 2 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many [#permalink] ### Show Tags 28 Sep 2015,", "she have? Number of stamps Zaheera has = 81. Explanation: Number of stamps Rony has = 36. Number of times Zaheera has stamps = 2$$\\frac{1}{4}$$ Number of stamps Zaheera has = Number of stamps Rony has × Number of times Zaheera has stamps = 36 × 2$$\\frac{1}{4}$$ = 36 × {[(2 × 4) + 1] ÷ 4} = 36 × [(8 + 1) ÷ 4] = 36 × $$\\frac{9}{4}$$ = 9 × $$\\frac{9}{1}$$ = 81. Question 5. Calculate the following: (i) 4 times 5$$\\frac{1}{3}$$ 4 times of 5$$\\frac{1}{3}$$, we get the result $$\\frac{64}{3}$$ or 21$$\\frac{1}{3}$$. Explanation: 4 times 5$$\\frac{1}{3}$$ = 4 × 5$$\\frac{1}{3}$$ = 4 × {[(5 × 3) + 1] ÷ 3} = 4 × [(15 + 1) ÷ 3] = 4 × $$\\frac{16}{3}$$ = $$\\frac{64}{3}$$ or 21$$\\frac{1}{3}$$ (ii) 4$$\\frac{1}{3}$$ times 5 4$$\\frac{1}{3}$$ times of 5, we get the result $$\\frac{65}{3}$$ or 21$$\\frac{2}{3}$$. Explanation: 4$$\\frac{1}{3}$$ times × 5 = {[(4 × 3) + 1] ÷ 3] × 5 = [(12 + 1) ÷ 3] × 5 = $$\\frac{13}{3}$$ × 5 = (13 × 5) ÷ 3 = $$\\frac{65}{3}$$ or 21$$\\frac{2}{3}$$ (iii) 1$$\\frac{1}{2}$$ times $$\\frac{2}{3}$$ 1$$\\frac{1}{2}$$ times of $$\\frac{2}{3}$$, we get the result 1. Explanation: 1$$\\frac{1}{2}$$ times × $$\\frac{2}{3}$$ = {[(1 × 2) + 1] ÷ 2} × $$\\frac{2}{3}$$ = [(2 + 1) ÷ 2] × $$\\frac{2}{3}$$ =", "bought are equal. => y = z => From equation 1, 30x + y(35 + 40) = 420 => 30x + 75y = 420 Case 1) x = 9, y = 2 satisfies the equation. So might have bought more than 5 30 cent stamps. case 2) x = 4, y = 4 satisfies the equation. So might have bought less than 5 30 cent stamps. Since we don't get an unique answer, statement 1 is not sufficient. Statement 2 The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought. => x = z and y $$\\leq$$ z = x From equation 1, we get 30x + 40x + 35y = 420 => 70x + 35y = 420 ---------- (2) Since y $$\\geq$$ 0 and y $$\\leq$$ x lets start figuring out the minimum possible value for y If y $$\\geq$$0 and y is an integer lets start at y=1 => 70x + 35 = 420 => 70x = 385 => x is not an integer which is false => y > 1 Lets see if y = 2 From equation 2, => 70x + 70 = 420 => 70x =", "example, when 10000 fax machines are produced, the cost is $250+\\frac{10000}{100000} = 250 + 1 = 251$.", "$150 plus 12% of his total sales amount over$1,250. Solve the equation $840=150+0.12(a−1250)840=150+0.12(a−1250)$ for a, to find the total amount Perry must sell in order to be paid $840 one week. 171. Stamps Travis bought$9.45 worth of 49-cent stamps and 21-cent stamps. The number of 21-cent stamps was 5 less than the number of 49-cent stamps. Solve the equation $0.49s+0.21​(s−5)​​=9.450.49s+0.21​(s−5)​​=9.45$ for s, to find the number of 49-cent stamps Travis bought. #### Writing Exercises 172. Frida started to solve the equation $−3x=36−3x=36$ by adding 3 to both sides. Explain why Frida’s method will not solve the equation. 173. Emiliano thinks $x=40x=40$ is the solution to the equation $12x=8012x=80$. Explain why he is wrong. #### Self Check After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. What does this checklist tell you about your mastery of this section? What steps will you take to improve?", "## Algebra: A Combined Approach (4th Edition) $35$ 44-cents stamps. $5$ 28-cents stamps. Let $x$ be the number of 44 cents (0.44 dollars) stamps. Let $y$ be the number of 28 cents (0.28 dollars) stamps. Sarah and Keith Robinson purchased 40 stamps, a mixture of 44¢ and 28¢ stamps. $x+y=40$ They spent $16.80$. $0.44x+0.28y=16.80$ $x+y=40$ $0.44x+0.28y=16.80$ Multiply the second equation by 100. $44x+28y=1680$ $x=40-y$ $44x+28y=1680$ $44(40-y)+28y=1680$ $1760-44y+28y=1680$ $1760-16y=1680$ $-16y=-80$ $y=5$ $x=40-y$ $x=40-(5)$ $x=35$ $35$ 44-cents stamps. $5$ 28-cents stamps.", "## Elementary Statistics (12th Edition) a) The expected value: $1,000,000\\cdot \\frac{1}{90,000,000}+100,000\\cdot \\frac{1}{110,000,000}+25,000\\cdot \\frac{1}{110,000,000}+5,000\\cdot \\frac{1}{36,667,000}+2,500\\cdot \\frac{1}{27,500,000}=0.012$ b)The expected value is a lot less than the cost of a stamp, therefore it is not worth playing.", "### Home > APCALC > Chapter 8 > Lesson 8.1.1 > Problem8-9 8-9. The value of a rare stamp was $250$ in 1989. The stamp has increased in value by $12\\%$ each year. 1. What was the stamp worth in the year 2000? The stamp is $112\\%$ of its value from the previous year.This will happen for $11$ years in a row. $250·1.12^{11} = 869.64$ 2. What was the average value of the stamp from 1989 to 2000? If $f(t)$ represents the value of the stamp as a function of time (the equation you used to solve part (a)): $\\text{average value of }f(t)=\\frac{1}{11}\\int_{0}^{11}f(t)dt$ $\\frac{1}{11}\\int_0^{11}250(1.12)^{x}dx=\\left.\\frac{1}{11}(250)\\frac{1}{\\ln(1.12)}(1.12)^x\\right|_0^{11}$", "+ 1) + 5k$. Together with his demonstration that the amounts $12$ cents, $13$ cents, $14$ cents, and $15$ cents can be formed using only $4$-cent and $5$-cent stamps, this proves that all amounts of $12$ cents or more can be formed using just $4$-cent and $5$-cent stamps. The reason Rosen starts at $12$ cents is that $12$ cents, $13$ cents, $14$ cents, and $15$ cents are the first four consecutive amounts that can be formed using only $4$-cent and $5$-cent stamps as it is not possible to form $11$ cents using only $4$-cent and $5$-cent stamps. Notice that the argument that if $n - 4 = 4j + 5k$ for some positive integers $j$ and $k$, then $n = 4(j + 1) + k$ cannot be applied when $n = 15$ since no such positive integers $j$ and $k$ exist for $n - 4 = 15 - 4 = 11$. This is an example of the Frobenius coin problem. The English mathematician James Joseph Sylvester showed that the largest amount that cannot be formed using only $m$-cent and $n$-cent stamps with $\\gcd(m, n) = 1$ is $mn - m - n$ cents. The number $mn - m - n$ is called the Frobenius number. In our case, $m = 4$ and $n = 5$, so the largest", "15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ? CEO Joined: 20 Mar 2014 Posts: 2560 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 11 Nov 2015, 05:31 1 BrainLab wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy? (1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Pretty @!)*, let's start: (1) 15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible", "We notice that the ending digit is 0, and one type of stamps costs 0.29. The only way we can have a \"0\" at the end with$0.29 is to multiply 0.29 with 10,20,30, etc. So if we have 10x0.29, its OK, meaning we have 2.9 worth of 0.29 stamps, and the rest worth $0.15. If we are to have 20, then we would surpass the total value of 4.40$ ( 20*0.29 = 5.8 ). So clearly she bought 10 stamps worth $0.29, and subsequently we can calculate the rest. That's why A is sufficient. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. - Clearly insufficient. Senior Manager Joined: 10 Oct 2012 Posts: 332 Followers: 5 Kudos [?]: 130 [0], given: 25 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many [#permalink] 23 Mar 2013, 00:47 udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy? (1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Just a quick tool for Diophantine equations. Lets assume the equation 4x+5y = 36. Rearrange it as x =\\frac{(36-5y)}{4} or y = \\frac{(36-4x)}{5}. Now, lets take the first re-arrangement. We can see that for y=0, we get an integral value for x as x=9 .Now all", "# stamps problem • Feb 2nd 2010, 08:26 AM mathos stamps problem hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks • Feb 2nd 2010, 08:37 AM e^(i*pi) Quote: Originally Posted by mathos hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks $5 stamps = a$3 stamps = b $0.50 stamps = c $5a+3b+0.5c = 100$ $a+b+c = 100$ Not sure where to go from there • Feb 2nd 2010, 08:44 AM Quacky$5 stamps = a $3 stamps = b$0.50 stamps = c (3) $\\frac {a}{2} + \\frac {b}{2}+0.5c=50$ $0.5c= 50-\\frac {a}{2}-\\frac {b}{2}$ $5a + 3b + 50-\\frac {a}{2}-\\frac {b}{2}=100$", "The total price for all the goods sold was $14.20. This statement is a bit trickier: $$1.30c+1.50b=14.20$$ --> $$13c+15b=142$$. Since c and b must be integers, then we should check whether this equation has one or more than one positive integer solutions: $$15b=142-13c$$ --> 142 minus multiple of 13 must be a multiple of 15: only c=4 and b=6 satisfies the equation, thus $$\\frac{c}{b} = \\frac{4}{6}$$. Sufficient. Answer: D. For more on this type of questions check: eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html collections-confused-need-a-help-81062.html Hope it helps. The most important thing here is.. How do we get the numbers fast for the second statement. I remember @VeritasPrepKarishma explained the theory behind number picking for linear equations with integer constraints somewhere.. Bunuel would you please advice on this issue? It would be a huge takeaway for future problems Thanks!! Cheers J Manager Joined: 04 Jan 2014 Posts: 115 GMAT 1: 660 Q48 V32 GMAT 2: 630 Q48 V28 GMAT 3: 680 Q48 V35 Re: In the first hour of a bake sale, students sold either cho [#permalink] ### Show Tags 11 Feb 2014, 00:33 St1: The average price for the items sold during that hour was$1.42. Sufficient. We can use the allegation method to get the ratios. There is no need to calculate but it comes out as", "=$100 Price of second item = 30% less than first item = 70% of 100 = $$\\frac{70}{100}$$ x 100 = $70 Price to be increased of cheaper item to be equally priced =$ 30 Percent to be increased for less expensive item = 30 = $$\\frac{x}{100}$$ x 70 x = $$\\frac{30* 100}{70}$$ = 42$$\\frac{6}{7}$$% _________________ Kindly press \"+1 Kudos\" to appreciate Re: A merchant has selected two items to be placed on sale, one [#permalink] 17 May 2017, 13:31 Similar topics Replies Last post Similar Topics: 1 There are 15 stamps from which two stamps will be selected. Tom has t 2 05 May 2016, 22:40 27 A couple has two children, one of whom is a girl. 28 23 Apr 2016, 11:44 WHEN A PERSON SELLS TWO ITEMS \\$1656 each, ONE AT A GAIN OF 20% AND OT 16 28 Aug 2016, 03:03 3 What is the probability that one of the two integers randomly selected 6 14 Jan 2016, 06:19 32 A merchant sells an item at a 20% discount, but still makes 17 04 Dec 2016, 21:38 Display posts from previous: Sort by", "# of $0.15 stamps and $$y$$ the # of$0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Pretty @!)*, let's start: (1) 15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ? This is the most straightforward way for these questions wherein the applicable values can only be positive ingeters. Similarly, you can not have 0.5 apples or 0.33 bananas etc and hence if the question had asked about certain fruits or number of men etc , your method will still be applicable to such questions. Kudos [?]: 1789 [0], given: 796 Intern Joined:", "+0 # Ratios 0 131 1 +109 John has US and foreign stamps in his album. The ratio of the number of US stamps to foreign stamps is 5 : 2. After Paul gives him 12 foreign stamps, the ratio becomes 2 : 1. Find the number of US stamps he has. Jun 6, 2021 #1 +208 +1 d=number of domestic stamps f=number of foreign stamps $$\\:\\frac{d}{f}=\\frac{5}{2}$$ $$2d=5f$$ $$\\frac{d}{f+12}=\\frac{2}{1}$$ $$d=2f+24$$ substitute d in first equation: $$2(2f+24)=5f$$ $$f=48$$ $$d=120$$ JP Jun 6, 2021 #1 +208 +1 d=number of domestic stamps f=number of foreign stamps $$\\:\\frac{d}{f}=\\frac{5}{2}$$ $$2d=5f$$ $$\\frac{d}{f+12}=\\frac{2}{1}$$ $$d=2f+24$$ substitute d in first equation: $$2(2f+24)=5f$$ $$f=48$$ $$d=120$$ JP JKP1234567890 Jun 6, 2021", "a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $$5$$-cent, $$10$$-cent, and $$25$$-cent stamps, with exactly $$20$$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $$7.10$$ in postage? (Note: The amount $$7.10$$ corresponds to $$7$$ dollars and $$10$$ cents. One dollar is worth $$100$$ cents.) $$45$$ $$46$$ $$51$$ $$54$$ $$55$$ ###### Solution(s): Note that we want to get $$7 \\cdot 100 + 10 = 710$$ cents. Let us try to see if it is possible to use up all of the $$5$$-cent and $$10$$-cent stamps. All of these two types of stamps combined would be worth $20(5 + 10) = 20 \\cdot 15 = 300$ cents. We then would need $$710 - 300 = 410$$ cents, which cannot be created with just $$25$$-cent stamps. We can, however, make $$425$$ cents with $$25$$-cent stamps. Using only $$19$$ each of $$5$$ and $$10$$-cent stamps would total $300 - 5 - 10 = 285$ cents. This means we would then need $$710 - 285 = 425$$ cents. This can be achieved with $$425 \\div 25 = 17$$ $$25$$-cent stamps. This lets us use \\begin{align*} 2 \\cdot 19 + 17 &= 38 + 17 \\\\&= 55" ]

[ "new ratio = $$\\frac{5x-10}{3x+10} = \\frac{7}{5}$$............. (1) We require to find the value of 5x-10 - (3x+10) = 2x-20 ......... (2) Solving equation 1; we get x = 30 Placing value of x in equation (2) = 60-20 = 40 = Answer C _________________ Kindly press \"+1 Kudos\" to appreciate Kudos [?]: 2633 [3], given: 193 Math Expert Joined: 02 Sep 2009 Posts: 41913 Kudos [?]: 129505 [2], given: 12201 Re: The number of stamps that Kaye and Alberto had were in ratio [#permalink] ### Show Tags 25 Mar 2014, 02:05 2 KUDOS Expert's post The number of stamps that Kaye and Alberto had were in the ration of 5:3 respectively. After Kaye gave Alberto 10 of her stamps, the ration of the number of Kaye had to the number of Alberto had was 7:5. As a result of the gift, Kaye had how many more stamps than Alberto A. 20 B. 30 C. 40 D. 60 E. 90 Given: $$\\frac{K}{A}=\\frac{5x}{3x}$$ (where $$x$$ is a positive integer) and $$\\frac{5x-10}{3x+10}=\\frac{7}{5}$$ --> $$x=30$$ --> $$K=150$$ and $$A=90$$. Question: $$(K-10)-(A+10)=?$$ --> $$(K-10)-(A+10)=40$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: the-number-of-stamps-that-kaye-and-alberto-had-were-in-the-98116.html _________________ Kudos [?]: 129505 [2], given: 12201 Re: The number of stamps that Kaye and Alberto had were in ratio [#permalink] 25 Mar 2014, 02:05 Display posts from previous: Sort by", "1. ## stamps problem hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks 2. Originally Posted by mathos hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks $5 stamps = a$3 stamps = b $0.50 stamps = c $5a+3b+0.5c = 100$ $a+b+c = 100$ Not sure where to go from there 3.$5 stamps = a $3 stamps = b$0.50 stamps = c (1) (2) (2) divided by 2: (3) $\\frac {a}{2} + \\frac {b}{2}+0.5c=50$ Hence, $0.5c= 50-\\frac {a}{2}-\\frac {b}{2}$ Substituting this into equation (1) will give: $5a + 3b + 50-\\frac {a}{2}-\\frac {b}{2}=100$ Which will eliminate c from one equation...I don't think that it can be done for another, which would then have allowed you to solve simultaneously because you only have 2 equations but 3 unknowns.", "right as many times as necessary to make it a whole number. We have 1 decimal place, so we need to multiply by 10. $$\\frac{1.25~\\times~10}{2.5~\\times~10}~=~\\frac{12.5}{25}$$ Divide the dividend (12.5) by the divisor (25). $$=~\\frac{12.5}{25}~=~0.5$$ 0.5 _______ 25 ) 12.5 – 0 ———- 125 – 125 ———- 0 Example 5: Divide the decimal value of 66.6 by the two-digit number 33.3 Convert the divisor to a whole number by multiplying the numerator and the denominator by 10. $$=~\\frac{66.6~\\times~10}{33.3~\\times~10}~=~\\frac{666}{333}~=~2$$ 2 _______ 333 ) 666 – 666 ———- 0 ## Real-life Examples Example 6: Jack has a stamp collection. Each stamp weighs 0.4 ounces. If the total weight of his stamps is 26 ounces, how many stamps does he have? The weight of one stamp is 0.4 ounces. The total weight of all of the stamps combined is 26 ounces. To figure out the number of stamps, divide the total weight by the weight of one stamp: $$26~\\div~4$$ $$=~\\frac{6~\\times~10}{0.4~\\times~10}$$ Convert the divisor into a whole number $$=~\\frac{260}{4}$$ = 65 65 _______ 4 ) 260 – 24 ———- 20 – 20 ———- 0 Hence, Jack has 65 stamps in his collection. Example 7: Tom runs 45.5 miles in 6.5 hours a day. How many miles does he cover in 1 hour? To calculate the miles that Tom covers in 1 hour, we", "## Elementary Statistics (12th Edition) a) The expected value: $1,000,000\\cdot \\frac{1}{90,000,000}+100,000\\cdot \\frac{1}{110,000,000}+25,000\\cdot \\frac{1}{110,000,000}+5,000\\cdot \\frac{1}{36,667,000}+2,500\\cdot \\frac{1}{27,500,000}=0.012$ b)The expected value is a lot less than the cost of a stamp, therefore it is not worth playing.", "she have? Number of stamps Zaheera has = 81. Explanation: Number of stamps Rony has = 36. Number of times Zaheera has stamps = 2$$\\frac{1}{4}$$ Number of stamps Zaheera has = Number of stamps Rony has × Number of times Zaheera has stamps = 36 × 2$$\\frac{1}{4}$$ = 36 × {[(2 × 4) + 1] ÷ 4} = 36 × [(8 + 1) ÷ 4] = 36 × $$\\frac{9}{4}$$ = 9 × $$\\frac{9}{1}$$ = 81. Question 5. Calculate the following: (i) 4 times 5$$\\frac{1}{3}$$ 4 times of 5$$\\frac{1}{3}$$, we get the result $$\\frac{64}{3}$$ or 21$$\\frac{1}{3}$$. Explanation: 4 times 5$$\\frac{1}{3}$$ = 4 × 5$$\\frac{1}{3}$$ = 4 × {[(5 × 3) + 1] ÷ 3} = 4 × [(15 + 1) ÷ 3] = 4 × $$\\frac{16}{3}$$ = $$\\frac{64}{3}$$ or 21$$\\frac{1}{3}$$ (ii) 4$$\\frac{1}{3}$$ times 5 4$$\\frac{1}{3}$$ times of 5, we get the result $$\\frac{65}{3}$$ or 21$$\\frac{2}{3}$$. Explanation: 4$$\\frac{1}{3}$$ times × 5 = {[(4 × 3) + 1] ÷ 3] × 5 = [(12 + 1) ÷ 3] × 5 = $$\\frac{13}{3}$$ × 5 = (13 × 5) ÷ 3 = $$\\frac{65}{3}$$ or 21$$\\frac{2}{3}$$ (iii) 1$$\\frac{1}{2}$$ times $$\\frac{2}{3}$$ 1$$\\frac{1}{2}$$ times of $$\\frac{2}{3}$$, we get the result 1. Explanation: 1$$\\frac{1}{2}$$ times × $$\\frac{2}{3}$$ = {[(1 × 2) + 1] ÷ 2} × $$\\frac{2}{3}$$ = [(2 + 1) ÷ 2] × $$\\frac{2}{3}$$ =", "Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891 Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html Director Joined: 10 Mar 2013 Posts: 608 Location: Germany Concentration: Finance, Entrepreneurship GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) Followers: 8 Kudos [?]: 182 [0], given: 200 Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many [#permalink] ### Show Tags 11 Nov 2015, 04:56 udaymathapati wrote: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought$4.40 worth of stamps. (2) She bought an equal number of $0.15 stamps and$0.29 stamps. Pretty @!)*, let's start: (1) 15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems", "# stamps problem • Feb 2nd 2010, 08:26 AM mathos stamps problem hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks • Feb 2nd 2010, 08:37 AM e^(i*pi) Quote: Originally Posted by mathos hello we bought 100 stamps with total cost of 100 dollars the stamps are priced at 5, 3, and 0.5 dollars how many stamps of each price do we bought? can you solve it with equations and with logical/practical calculations? thanks $5 stamps = a$3 stamps = b $0.50 stamps = c $5a+3b+0.5c = 100$ $a+b+c = 100$ Not sure where to go from there • Feb 2nd 2010, 08:44 AM Quacky$5 stamps = a $3 stamps = b$0.50 stamps = c (3) $\\frac {a}{2} + \\frac {b}{2}+0.5c=50$ $0.5c= 50-\\frac {a}{2}-\\frac {b}{2}$ $5a + 3b + 50-\\frac {a}{2}-\\frac {b}{2}=100$", "=$100 Price of second item = 30% less than first item = 70% of 100 = $$\\frac{70}{100}$$ x 100 = $70 Price to be increased of cheaper item to be equally priced =$ 30 Percent to be increased for less expensive item = 30 = $$\\frac{x}{100}$$ x 70 x = $$\\frac{30* 100}{70}$$ = 42$$\\frac{6}{7}$$% _________________ Kindly press \"+1 Kudos\" to appreciate Re: A merchant has selected two items to be placed on sale, one [#permalink] 17 May 2017, 13:31 Similar topics Replies Last post Similar Topics: 1 There are 15 stamps from which two stamps will be selected. Tom has t 2 05 May 2016, 22:40 27 A couple has two children, one of whom is a girl. 28 23 Apr 2016, 11:44 WHEN A PERSON SELLS TWO ITEMS \\$1656 each, ONE AT A GAIN OF 20% AND OT 16 28 Aug 2016, 03:03 3 What is the probability that one of the two integers randomly selected 6 14 Jan 2016, 06:19 32 A merchant sells an item at a 20% discount, but still makes 17 04 Dec 2016, 21:38 Display posts from previous: Sort by", "15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ? CEO Joined: 20 Mar 2014 Posts: 2560 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 11 Nov 2015, 05:31 1 BrainLab wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy? (1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Pretty @!)*, let's start: (1) 15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible", "### Home > APCALC > Chapter 8 > Lesson 8.1.1 > Problem8-9 8-9. The value of a rare stamp was $250$ in 1989. The stamp has increased in value by $12\\%$ each year. 1. What was the stamp worth in the year 2000? The stamp is $112\\%$ of its value from the previous year.This will happen for $11$ years in a row. $250·1.12^{11} = 869.64$ 2. What was the average value of the stamp from 1989 to 2000? If $f(t)$ represents the value of the stamp as a function of time (the equation you used to solve part (a)): $\\text{average value of }f(t)=\\frac{1}{11}\\int_{0}^{11}f(t)dt$ $\\frac{1}{11}\\int_0^{11}250(1.12)^{x}dx=\\left.\\frac{1}{11}(250)\\frac{1}{\\ln(1.12)}(1.12)^x\\right|_0^{11}$", "## Algebra: A Combined Approach (4th Edition) $35$ 44-cents stamps. $5$ 28-cents stamps. Let $x$ be the number of 44 cents (0.44 dollars) stamps. Let $y$ be the number of 28 cents (0.28 dollars) stamps. Sarah and Keith Robinson purchased 40 stamps, a mixture of 44¢ and 28¢ stamps. $x+y=40$ They spent $16.80$. $0.44x+0.28y=16.80$ $x+y=40$ $0.44x+0.28y=16.80$ Multiply the second equation by 100. $44x+28y=1680$ $x=40-y$ $44x+28y=1680$ $44(40-y)+28y=1680$ $1760-44y+28y=1680$ $1760-16y=1680$ $-16y=-80$ $y=5$ $x=40-y$ $x=40-(5)$ $x=35$ $35$ 44-cents stamps. $5$ 28-cents stamps.", "example, when 10000 fax machines are produced, the cost is $250+\\frac{10000}{100000} = 250 + 1 = 251$.", "The total price for all the goods sold was $14.20. This statement is a bit trickier: $$1.30c+1.50b=14.20$$ --> $$13c+15b=142$$. Since c and b must be integers, then we should check whether this equation has one or more than one positive integer solutions: $$15b=142-13c$$ --> 142 minus multiple of 13 must be a multiple of 15: only c=4 and b=6 satisfies the equation, thus $$\\frac{c}{b} = \\frac{4}{6}$$. Sufficient. Answer: D. For more on this type of questions check: eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html collections-confused-need-a-help-81062.html Hope it helps. The most important thing here is.. How do we get the numbers fast for the second statement. I remember @VeritasPrepKarishma explained the theory behind number picking for linear equations with integer constraints somewhere.. Bunuel would you please advice on this issue? It would be a huge takeaway for future problems Thanks!! Cheers J Manager Joined: 04 Jan 2014 Posts: 115 GMAT 1: 660 Q48 V32 GMAT 2: 630 Q48 V28 GMAT 3: 680 Q48 V35 Re: In the first hour of a bake sale, students sold either cho [#permalink] ### Show Tags 11 Feb 2014, 00:33 St1: The average price for the items sold during that hour was$1.42. Sufficient. We can use the allegation method to get the ratios. There is no need to calculate but it comes out as", "+0 # Ratios 0 131 1 +109 John has US and foreign stamps in his album. The ratio of the number of US stamps to foreign stamps is 5 : 2. After Paul gives him 12 foreign stamps, the ratio becomes 2 : 1. Find the number of US stamps he has. Jun 6, 2021 #1 +208 +1 d=number of domestic stamps f=number of foreign stamps $$\\:\\frac{d}{f}=\\frac{5}{2}$$ $$2d=5f$$ $$\\frac{d}{f+12}=\\frac{2}{1}$$ $$d=2f+24$$ substitute d in first equation: $$2(2f+24)=5f$$ $$f=48$$ $$d=120$$ JP Jun 6, 2021 #1 +208 +1 d=number of domestic stamps f=number of foreign stamps $$\\:\\frac{d}{f}=\\frac{5}{2}$$ $$2d=5f$$ $$\\frac{d}{f+12}=\\frac{2}{1}$$ $$d=2f+24$$ substitute d in first equation: $$2(2f+24)=5f$$ $$f=48$$ $$d=120$$ JP JKP1234567890 Jun 6, 2021", "and we can proceed from here with Solution 1. ~MathIsFun286 ## Problem14 For positive integers $a$$b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$$b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$. ## Solution 1 Notice that we must have $a = 1$, otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\\lfloor \\frac{999}{c} \\rfloor$ stamps of value $c$, every value up to $1000$ can be represented. Therefore using $\\lfloor \\frac{999}{c} \\rfloor$ stamps of value $c$$\\lfloor \\frac{c-1}{b} \\rfloor$ stamps of value $b$, and $b-1$ stamps of value $1$, all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps. So, $f(a, b, c) = \\lfloor \\frac{999}{c} \\rfloor", "he also buys 2 10 cent stamps = 20c Total = 50c for three different stamps We know that he spent$2.65 or 265cents on these stamps. Since 265 is not divisible by 50 we chose the next highest number less than 265. That number is 250. So maximum he can spend 250 cents buying 5, 10 and 25 cent stamps. 265-250 = 15 are the least number of 1 cent stamps he can buy. Manager Joined: 07 Jul 2003 Posts: 56 Followers: 2 Kudos [?]: 0 [0], given: 0 Vivek how did you figure that y=15.. you have two unknowns and 1 equation. Please explain _________________ Attain Moksha Manager Joined: 07 Jul 2003 Posts: 56 Followers: 2 Kudos [?]: 0 [0], given: 0 Oops sorry, didnt see that you mentioned x=5 cent stamps.. Thanks everyone.. _________________ Attain Moksha SVP Joined: 30 Oct 2003 Posts: 1794 Location: NewJersey USA Followers: 5 Kudos [?]: 47 [0], given: 0 let me barge in Vivek's equation is 50X + Y = 265 = 5*50 + 15 so he equates similar terms on both the sides 50X = 5*50 and Y = 15 This can be done in case of simple equations. Manager Joined: 28 Jan 2004 Posts: 203 Location: India Followers: 2 Kudos [?]: 17 [0], given: 4 15 is correct. Vivek's", "We notice that the ending digit is 0, and one type of stamps costs 0.29. The only way we can have a \"0\" at the end with$0.29 is to multiply 0.29 with 10,20,30, etc. So if we have 10x0.29, its OK, meaning we have 2.9 worth of 0.29 stamps, and the rest worth $0.15. If we are to have 20, then we would surpass the total value of 4.40$ ( 20*0.29 = 5.8 ). So clearly she bought 10 stamps worth $0.29, and subsequently we can calculate the rest. That's why A is sufficient. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. - Clearly insufficient. Senior Manager Joined: 10 Oct 2012 Posts: 332 Followers: 5 Kudos [?]: 130 [0], given: 25 Re: Joanna bought only$0.15 stamps and $0.29 stamps. How many [#permalink] 23 Mar 2013, 00:47 udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy? (1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Just a quick tool for Diophantine equations. Lets assume the equation 4x+5y = 36. Rearrange it as x =\\frac{(36-5y)}{4} or y = \\frac{(36-4x)}{5}. Now, lets take the first re-arrangement. We can see that for y=0, we get an integral value for x as x=9 .Now all", "$150 plus 12% of his total sales amount over$1,250. Solve the equation $840=150+0.12(a−1250)840=150+0.12(a−1250)$ for a, to find the total amount Perry must sell in order to be paid $840 one week. 171. Stamps Travis bought$9.45 worth of 49-cent stamps and 21-cent stamps. The number of 21-cent stamps was 5 less than the number of 49-cent stamps. Solve the equation $0.49s+0.21​(s−5)​​=9.450.49s+0.21​(s−5)​​=9.45$ for s, to find the number of 49-cent stamps Travis bought. #### Writing Exercises 172. Frida started to solve the equation $−3x=36−3x=36$ by adding 3 to both sides. Explain why Frida’s method will not solve the equation. 173. Emiliano thinks $x=40x=40$ is the solution to the equation $12x=8012x=80$. Explain why he is wrong. #### Self Check After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. What does this checklist tell you about your mastery of this section? What steps will you take to improve?", "= 7/5 and cross multiply to get 25x-50 = 21x + 70 -> 4x = 120 x = 30 NOW 30 IS NOT THE ANSWER - that is the multiplier so you need to find the number of stamps in possession so... Kaye 7*30 =210 Alberto 5*30 = 150 210-150 = 60 = D _________________ ##### General Discussion Math Expert Joined: 02 Sep 2009 Posts: 50039 Re: Pls explain how to come up to the slolution [#permalink] ### Show Tags 28 Jul 2010, 06:51 9 1 shaselai wrote: i think the problem here is yes you get X=30 BUT the new result is the ratio 7x:5x where the +/- 10 has already been taken into account or maybe i am wrong...? Answer to this question is C (40), provided OA is wrong. When we write $$\\frac{K}{A}=\\frac{5x}{3x}$$ it means that $$x$$ is the multiple of the initial ratio of 5/3 and not 7/5. We found that $$x=30$$: initially K had $$5*30=150$$ stamps and now K has $$150-10=140$$ stamps. Initially A had $$3*30=90$$ stamps and now A has $$90+10=100$$ stamps. Current difference is $$140-100=40$$. Hope it's clear. _________________ Intern Joined: 15 Oct 2012 Posts: 15 The number of stamps that Kaye and Alberto had were in the ratio [#permalink] ### Show Tags 01 Jun 2013, 10:19 1 Throw x=30 back", "Answer: c. 16 Explanation: Ira has 128 stamps in his stamp album. He has the same number of stamps on each of the 8 pages. 128/8 = 16 stamps on each page. Question 4. Ryan is saving up for a bike that costs $198. So far, he has saved$15 per week for the last 12 weeks. How much more money does Ryan need in order to be able to buy the bike? Options: a. $8 b.$ 18 c. $48 d.$ 180 Answer: b. $18 Explanation: Ryan is saving up for a bike that costs$198. So far, he has saved $15 per week for the last 12 weeks =$15 x 12 = $180.$198 – $180 =$18 need in order to buy the bike. Question 5. Tina buys 3 $$\\frac{7}{8}$$ yards of material at the fabric store. She uses it to make a skirt. Afterward, she has 1 $$\\frac{3}{8}$$ yards of the fabric leftover. How many yards of material did Tina use? Options: a. 1 $$\\frac{4}{8}$$ b. 2 $$\\frac{1}{8}$$ c. 2 $$\\frac{4}{8}$$ d. 5 $$\\frac{2}{8}$$ Answer: c. 2 $$\\frac{4}{8}$$ Explanation: Tina buys 3 7/8 yards of material at the fabric store. She uses it to make a skirt. Afterward, she has 1 3/8 yards of the fabric leftover. 3 -1 = 2; 7/8 – 3/8 = 4/8. So, answer is" ]

[ "# The convention Logic Level 2 A certain convention was exactly attended by 100 politicians. We are given the following 2 facts. 1) Jon Huntsman is present. He is known to be honest. 2) If you choose any 2 politicians randomly at the convention, at least one of them will be crooked (not honest). Of the 100 politicians present, how many are crooked? ×", "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", "the number and nature of the papers read. In the early years of these quadrennial meetings four sections were considered sufficient; at present there are twice as many. At first the countries represented were chiefly European, the number of members from other continents being relatively small; but in this congress between 40 and 50 nations were represented and the attendance approximated 650 men and 200 women.", "stockholder in 2 20 Oct 2015, 03:01 3 Each person who attended a company meeting was either a stockholder in 1 14 Sep 2015, 08:36 17 All of the 540 people who attended an education convention 10 28 Mar 2013, 12:48 20 Of the 84 parents who attended a meeting at a school, 35 19 02 Jul 2012, 02:18 3 Each person who attended a company meeting was either a 5 26 Apr 2012, 03:51 Display posts from previous: Sort by", "Convention Run? As mentioned earlier, the national convention is the supreme governing authority of the National Federation of the Blind. Consequently, important business must be done at the convention in an orderly and democratic manner. The following information describes how the convention is run, including how votes are taken, how resolutions are considered, and more. ### General Sessions The general sessions consist of program items, reports, panel discussions, elections, and official votes on policy issues. General convention sessions customarily are chaired by the Federation President. Floor microphones are available for comments and questions from the audience when time permits. ### Official Voting Usually, on votes for elections, motions, or for the adoption of resolutions (see below), the President will call for voice votes. In such cases, it is usually clear that a vast majority has voted one way or another. However, if the outcome of a particular vote is not absolutely clear, then the President will ask for a roll call vote. In the event of a roll call vote, only official delegates of the affiliates may vote. In order to be as democratic as it can be in its decision making, the Federation has decided that each state affiliate will have one vote during a roll call vote. At the opening general session, each affiliate names its", "$2\\cdot{6\\choose3}$ Add them and get 55 5. Thanks everyone! Very helpful. I had one more question, kind of similar, but I don't know how to solve this one mathematically without drawing it out ... Six three-representative delegations attend an international conference. Teh reps shake hands when they are introduced to one another, how many handshakes are possible if each delegate shakes hands only once with every other attendant except with those of his/her delegation As always, any help is greatly appreciated. 6. Nevermind. Figured it out. :-)", "# packing with time-variant weights This appears to be a knapsack / bin-packing problem, but I seem to have got stuck and could appreciate contributions. Scenario 1: Tough (for me!) There is a one day conference with a set of (4 or more) sessions. The conference will be attended by a number of companies, each one being represented by a (1 or more) representatives. Across sessions the number of representatives for each company will vary (and may be zero), with each individual having the same chance of being present as any other individual (so there is an increasing likelihood of a company being represented when it has more representatives). There is a single row of seats in the conference. There are more than enough seats for the most popular session. (e.g., if the most popular session has 100 delegates, there could be 120 seats for the whole conference). Constraints • A rep. belongs to one company • A rep. must be seated in a session they are going to. • A rep. must not change seats across consecutive sessions • A rep. is assigned to one chair in each session they attend. Goal To fit the constraints and priorities optimally. Example. 15 chairs, 4 companies (A-D), 4 sessions (S1-S4). // Session attendees by company S1: A2 B6 C3", "# Four Companies Author(s): There are four companies, in one of which there are 6 men, in another 8, and in each of the remaining two, 9 men. How many ways can a committee of 4 men be composed by choosing one man from each company? (Greenleaf's National Arithmetic, 1857)", "Eva can order?Indicate all possible values. Six state governors meet at an annual convention. They line up in random order to pose for a photograph. If the governors of Alaska and Hawaii are among the six governors, how many different ways can the governors line up for the picture so that these two governors are adjacent? 25000 +道题目 164本备考书籍", "It is difficult for the branch to start operations and consider all options. Small sub-groups, working close together, knowing each other well, so it is easier to review, train, work and operate easier and faster. This is why sub-groups are so useful. Besides, if the government bans the trade union, but the sub-group is well organized, the union can keep making progress and keep working underground. That's why people call the sub-group the root of all unions. \\subsection{What's the order in the union?} Sub-groups report to their branch. If there are many factories in the province, 4 or 5 cell branches organize a set of representative committees (4, 5 factories each elect 1 or 2 people). The branches report to the province. Provincial unions report to the National union. That is the order of organization. As for the authority, for all members to attend the congress, meaning, all members openly discuss the meeting. If there are too many members and it's not convenient to all attend a congress, a few people will elected as representatives to attend the the congress, they are the congress delegations. Whatever the congress decides on, the members of the association must execute. When the congress is over, the executive authority goes to the elected central committee. Delegates to the provincial congress open once", "is reshuffled. We adopted a system that I like in principle and think worked reasonably well but maybe ended up a bit too complicated - in order to ensure everyone got a good opportunity to seed the conversation, we deprioritised cards from people who had recently had their topic discussed. The way this worked was that when a card had been discussed we put it face up on the table. If a card came up from someone whose name was already on the table, we put it aside. Once we had been through the whole deck, we stacked the cards that were face up so that they were no longer visible, shuffled the cards that had been put aside, and started the process again. ### Details • I found it very easy to get cards confused, so I started marking cards that we had completed with a cross or a tick on the back so that we didn't mix them up. • One person ended up dominating the conversation a bit (he acknowledged this and welcomed the feedback). We adopted a convention that I would just start waving at him when I thought he'd been talking too long. I actually think we should have adopted this as a universal convention right at the beginning, with everyone feeling empowered", "open to all, occurs in the morning and various committees, groups, and divisions gather in the afternoon and evening. Day four brings the opening of the formal convention, with the roll call of states in the morning and the Presidential Report and other program items in the afternoon. There are more committee and divisional meetings on the evening of day four. On day five, general convention sessions are held in both the morning and the afternoon. Elections are scheduled on this day, as well. Day six is the last day of convention; the morning and afternoon general sessions are followed by the annual evening banquet, a convention highlight. ## Convention Agenda The convention agenda is available at registration and can also be obtained at a number of other locations at the convention site. Moreover, it can be found on the NFB website at www.nfb.org as soon as it is final, which is usually about a month before the convention begins. The agenda gives general information about the convention, hotel rates, and other hotel information, and it shows the times and locations of the various meetings and general sessions. ## The Presidential Report One of the major presentations each year is a report delivered by the national President to the entire convention on the Federation's activities and progress during", "to do it - there were almost certainly one or two times when someone should have been waving at me. Honestly I want to adopt this convention for all conversations, lean coffee or not, because I definitely have some friends who it would be useful for (in a constructive way! I think all of the people I have in mind would love this convention). • I tended to interrupt people mid-sentence when the time came up, but I should probably have started waving at them and then cut them off if they didn't wrap up within a sentence or two. • A nice thing about this format was that it was really easy for people to drop in and out. This created the question of what we should do when a card whose author has left comes up. We initially said we should just discard them, but then decided to vote on them anyway, but we never actually voted in favour of discussion, so I think we were probably right the first time. • I did note-taking in a Google doc for most of it, passing over to one of the other attendees for the third session until he left, at which point I took over again. It was much easier taking notes when I wasn't also moderating,", "clear the room and then have the one plus five. Each coordinator must confirm which are the five. Could we do this expeditiously, please? Event is not active Filed under:", "three minuts to read this Draft Resolution. 2. Times up. Now the dais would like to see a motion to discuss Draft Resolution [DR No.]. Are there any points or motions on the floor?", "for that. And with that we stand adjourned. (Whereupon, at 4:15 o'clock the conference was adjourned.) -------", "arrangement matters). And so while some attendees (the “core group”) might end up having fun, the party doesn’t really work for most participating parties. So, the next time you’re hosting a party, do yourself and your guests a favour and ensure that you don’t end up with between 6 and 10 people at the party. Either less or more is fine! You might want to read this other post I’ve written on coordinating guest lists for birthday parties. ## Meeting types There are essentially three kinds of meetings – those that are entirely “in person”, those that are entirely “on call” and hybrids. I argue that the quality of conversation in the third kind of meeting is significantly inferior to that of the first two types. In person meetings are those where all participants are in one room. These are perhaps the best kind of meetings (except when you know it’s likely to turn antagonistic), for you can maintain eye contact with the others and as long as the number of meeters is small, there is social pressure on meeters to not be distracted, and thus the meeting is likely to conclude its agenda productively and quickly. Conference calls allow you to multitask while you are on the meeting – the positive thing is that you can choose", "mostly seen this Principle turned into the morning ritual of having a “stand-up” meeting. When this is done correctly, it can be useful. “Done correctly” means the team is less than 12 people, they meet in the same physical space, the entire meeting takes less than 15 minutes, and useful information is exchanged. But I have, many times, seen teams as large 40 people involved, I have seen this done over telephones and Skype, and I have seen the meeting drag on for 45 minutes. Please be clear about what I am saying: there is nothing wrong with telephones or Skype, and there is nothing wrong with remote work, but it is no longer a “stand up” meeting. If the team is big or remote then it is best to use other means of communication. Just because you call a meeting a “stand up” doesn’t mean it is a “stand up”. Merely having a meeting each morning in no way guarantees that you are living up to the spirit of this Principle. Working software is the primary measure of progress. This is a beautiful ideal but who takes the blame when there is no progress? Some of you may have read my earlier essay What happens when the Board Of Directors begins to panic about a startup where", "program it into the system to modify the dependant weights? ## packing with time-variant weights This appears to be a knapsack / bin-packing problem, but I seem to have got stuck and could appreciate contributions. (If this is the wrong SE to ask this question, rather than downvoting it, could you comment what the correct SE is, and I can post it there) Scenario 1: Tough (for me!) There is a one day conference with a set of (4 or more) sessions. The conference will be attended by a number of companies, each one being represented by a (1 or more) representatives. Across sessions the number of representatives for each company will vary (and may be zero), with each individual having the same chance of being present as any other individual (so there is an increasing likelihood of a company being represented when it has more representatives). There is a single row of seats in the conference. There are more than enough seats for the most popular session. (e.g., if the most popular session has 100 delegates, there could be 120 seats for the whole conference). Constraints & Priorities (from highest to lowest) • Constraint: Company representatives must be seated • Constraint: Company representatives sit next to each other • Constraint: seats will not be greater than 125%", "you have a convention, even if some other way might be a little better, there's a lot of inertia to overcome; in the short term it's usually better not to change, so conventions may stick around for ever. – Glen_b Jul 27 '13 at 1:18" ]

[ "53 $$(64x^3−27)÷(4x−3)$$ Exercise 54 $$(125y^3−64)÷(5y−4)$$ $$25y^2+20x+16$$ ## Everyday Math Exercise 55 Average cost Pictures Plus produces digital albums. The company’s average cost (in dollars) to make x albums is given by the expression $$\\dfrac{7x+500}{x}$$ 1. Find the quotient by dividing the numerator by the denominator. 2. What will the average cost (in dollars) be to produce 20 albums? Exercise 56 Handshakes At a company meeting, every employee shakes hands with every other employee. The number of handshakes is given by the expression $$\\dfrac{n^2−n}{2}$$ nn represents the number of employees. How many handshakes will there be if there are 10 employees at the meeting? 45 ## Writing Exercises Exercise 57 James divides $$48y+6$$ by $$6$$ this way: $$\\dfrac{48y+6}{6}=48y$$ Exercise 58 Divide $$\\dfrac{10x^2+x−12}{2x}$$ and explain with words how you get each term of the quotient.", "of ways to get a $$3$$-cycle from the remaining 3 persons is $$\\frac{3!}{2\\times3}=1$$. So we get $$672$$ ways that the links can form a $$5$$-cycle and a $$3$$-cycle. Adding up the number shown in the fat sentences, we get that there are $$2520+315+672=3507$$ ways to get have 8 people do handshakes with each other if each person shakes hands with exactly 2 different people. Link each person to the two other people that they shake hands with. Then each of the $$8$$ people is part of a cycle of at least $$3$$ people. So either there is one $$8$$-cycle, or two $$4$$-cycles, or a $$3$$ cycle and a $$5$$ cycle. In your count of $$7!=5040$$, every $$8$$-cycle is counted twice (because $$(12345678)$$ is also counted separately as $$(18765432)$$ etc.). So the actual number of $$8$$-cycles is $$5040/2 = 2520$$. There are $$(7 \\times 6 \\times 5) / 2 = 105$$ different $$4$$-cycles, and for each of these the remaining $$4$$ people can be arranged in $$3$$ different $$4$$-cycles (either $$(ABCD)$$ or $$(ACBD)$$ or $$(ACDB)$$). So this adds a further $$105 \\times 3 = 345$$ configurations. There are $$(7 \\times 6 \\times 5 \\times 4) / 2 = 420$$ different $$5$$-cycles and a single $$3$$-cycle for each one. There are $$(7 \\times 6) / 2 = 21$$ different $$3$$-cycles", "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", "## 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.", "• In how many different ways, can the letters of the word 'VENTURE' be arranged? A) 840 B) 5040 C) 1260 D) 2520 D) 2520 The required different ways = $$\\large\\frac{7!}{2!} = \\frac{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2!}{2!} = 2520$$ ##### Similar Questions 1). How many different signals , can be made by 5 flags from 8 flags of different colors? A). 6270 B). 1680 C). 20160 D). 6720 2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets? A). 12 B). 64 C). 256 D). 60 3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together? A). 10! B). 14!/(4!) C). 151200 D). 3628800 4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible? A). 120 B). 100000 C). 6720 D). 30240 5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible? A). 72 B). 144 C). 288 D). 234 6). Find the", "can find the value of $$k$$ by equating the power of $$x$$ inside the sigma with $$-2$$. \\begin{align*} &-2=10-2k\\\\ &2k=12\\\\ &k=6\\\\ \\end{align*} Step 3: Write down the coefficient of $$x$$ by substituting the value of $$k$$ $${10 \\choose 6} (2)^{10-6}=3360$$ ## Concept Check Questions 1. At a meeting involving $$5$$ men and $$8$$ women, each person shook hands once with every one person. 1. What was the total number of handshakes that took place? 2. How many handshakes were there between two women? 3. How many handshakes were there between a man and a women? 2. In how many ways can a study answer a four-option multiple choice exam with $$10$$ questions? 3. In a circular boardroom, how many ways can one arrange four female and five male colleagues if the males and females are to alternate? 4. How many different $$5$$ letter words can be made from the alphabet if replacement is not allowed? 5. Expand $$(3x-2y)^5$$ 6. Find the middle term in the expansion of $$(3a+5b)^20$$ leaving your answer in terms of $${20 \\choose k}$$ 7. Find the term independent of $$x$$ in the expansion of $$(5x+ \\frac{3}{x})^6$$ ## Concept Check Solutions 1. (i) $${16 \\choose 2}$$ (ii) $${6 \\choose 2}$$ (iii) $${10 \\choose 1} × {6 \\choose 1}$$ 2. $$4^10$$ 3. $$4!×5!$$ 4. $${26 \\choose", "# Reliance Jio Permutation and combination Quiz 1 Question 1 Time: 00:00:00 In how many different ways can the letters of the word \"MOBILE\" be arranged ? 720 720 360 360 1440 1440 2880 2880 1240 1240 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 2 Time: 00:00:00 A President and Vice - President are to be chosen out of a team having 40 Members .How many ways are there to achieve this ? 1850 1850 1870 1870 1560 1560 1910 1910 1200 1200 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 3 Time: 00:00:00 In a Marriage ceremony there are 50 People present in which 30 Males and 20 females.How many hands shake are possible if every person shakes hands with Others? 1250 1250 1225 1225 1325 1325 1520 1520 1200 1200 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 4 Time: 00:00:00 Find $$^{25}\\textrm{C}_{2}$$ 200 200 300 300 400 400 100 100 150 150 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 5 Time: 00:00:00 Seven people sit around a circular table. Find number of possible arrangements 120 120 360 360 720 720 5050 5050 5040 5040 Once you attempt the question then PrepInsta explanation will", "from 6 boys and 4 girls. Solution: There are 6 boys and 4 girls. A team of 3 boys and 2 girls is to be selected. ∴ 3 boys can be selected from 6 boys in 6C3 ways. 2 girls can be selected from 4 girls in 4C2 ways. ∴ Number of ways the team can be selected = 6C3 × 4C2 = $$\\frac{6 !}{3 ! 3 !} \\times \\frac{4 !}{2 ! 2 !}$$ = $$\\frac{6 \\times 5 \\times 4 \\times 3 !}{3 \\times 2 \\times 1 \\times 3 !} \\times \\frac{4 \\times 3 \\times 2 !}{2 \\times 2 !}$$ = 20 × 6 = 120 ∴ The team of 3 boys and 2 girls can be selected in 120 ways. Question 9. After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting. Solution: Let there be n participants present in the meeting. A handshake occurs between 2 persons. ∴ Number of handshakes = nC2 Given 66 handshakes were exchanged. ∴ 66 = nC2 ∴ 66 = $$\\frac{n !}{2 !(n-2) !}$$ ∴ 66 × 2 = $$\\frac{\\mathrm{n}(\\mathrm{n}-1)(\\mathrm{n}-2) !}{(\\mathrm{n}-2) !}$$ ∴ 132 = n(n – 1) ∴ n(n – 1) = 12 × 11 Comparing on both sides, we get n = 12", "Party August 09, 2009 • Math • Puzzles This is very simple but pretty interesting, like all good math should be. You go to a party. There could be any number of people at the party. You run into a friend at the party and you decide to make a little wager. You bet that you can find two people at the party who have shaken hands with the same number of people. Your friend realizes agrees to the idea of the bet but demands that you give him odds. You say, \"Okay, I'll give you good odds. If I find two people who have shaken the same number of hands as each other, you give me $20. If I can't, I'll give you nine hundred trillion dollars. Do we have a bet?\" Your friend quickly agrees because, after all, what is$20 next to the possibility of getting nine hundred trillion? You walk around the room, ask everybody how many people they've shaken hands with, eventually find two people that have the same number, and collect your \\$20. Of course, you knew going in that, at any party of any number of people with any rate of hand shaking going on, you can always find two people who have shaken the same amount of hands as each other.", "not shake hands with himself, nor Mrs. Li with herself. Therefore the number of men should be equal to 3 times the number of women PLUS 1. The number of women therefore must be ONE MORE THAN 1/4 the number of men. Let's set w = no. of Women and m = no. of Men. Then we can set up two equations: m = 3w+1 and w = (m/4)+1 Here we’re assuming that the Li’s did shake hands with each other, taking what it says literally. (Otherwise, they would have shaken hands with exactly the same people, and what is described could not be true.) The first equation could have been written as $$m-1 = 3w$$, meaning that the number of men, minus Mr. Li, is three times the number of women; and the second equation could be $$m = 4(w-1)$$, since the number of men is four times the number of women, excluding Mrs. Li. From the first equation, we can solve for w to get: w = (m - 1)/3 and substitute this into our second equation to get: (m - 1)/3 = (m/4) + 1 Let's simplify the right side by changing the 1 to 4/4 and adding. Now we have: (m - 1)/3 = (m + 4)/4 (you'll want to work out the steps", "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 .", "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", "ninth meeting (meeting $8$) is $56,57,58,59,60,61,62$ Go through days 1 through 8 and erase people 60,61,62 (note that no two of them are ever in the same meeting after day 0). At this stage each of those eight days has nine meetings, three are good and six are big. We wish to remove one person from each big meeting to make them good and use the removed people for a tenth good meeting. To have a chance of doing this appropriately, we look for a meeting from day 0 which has only one member already in a good meeting. If we manage to do this then the other six people are sure to be distributed one in each big meeting. Use them to make a tenth good meeting. </p> <p>An example may help: Here is the schedule for day 4 after people $60,61,62$ are removed (note that I have sorted the people in each meeting and sorted the list of meetings according to least person in the meeting)</p> <p>$\\small [0, 10, 18, 26, 34, 50, 58], [1, 11, 20, 21, 31, 37, 54], [2, 8, 19, 28, 39, 45], [3, 16, 27, 35, 47, 53, 57], [4, 9, 24, 36, 42, 55],$ $\\small [5, 13, 17, 22, 32, 44, 49],[6, 25, 30, 40, 43, 52, 56], [7, 15,", "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", "sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 # There are 7 persons in a party. The number of hands shaken all together in the party is: 1. 21 2. 6 3. 10 4. 2 Option 1 : 21 ## Detailed Solution Given: Number of person in a party = 7 $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ Where, n = number of the person r = minimum number of person required for a handshake Calculation: $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ $$\\Rightarrow {7_{{C_2}}} = \\;\\frac{{7!}}{{2! \\times \\left( {7 - 2} \\right)!}}$$ $$\\Rightarrow \\frac{{7 \\times 6 \\times 5!}}{{2! \\times 5!}}$$ ⇒ 42/2 ⇒ 21 ∴ the number of hands shaken is 21. # A boy contains 5 pink , 3 blue & 6 black ball. How many solution of ball can be formed by taking at least one ball form the boy 1. 157 2. 187 3. 167 4. 197 5.", "train were late, we can calculate how many took the train and were on time $58-31=27\\text{ by train and on time}$ The completed frequency tree should therefore look like this: a) First of all, if there are 75 people in total and 63 are right-handed, we can perform a simple calculation to work out how many people are left-handed: $75-63=12\\text{ left-handed people}$ Secondly, we know that of the left-handed people, one third are right-footed. So the number of people who are left-handed and right-footed can be calculated as follows: $12\\div 3=4 \\text{ left-handed and right-footed people}$ Furthermore, since there are 12 left-handed people in total, we can now work out how many of them are left-footed: $12-4=8\\text{ left-handed and left-footed people }$ So, we just have two remaining bubbles, the ones branching off from the 63 right-handed people. The final piece of information given to us in the question tells us that 27 people in total are left-footed. Given that we now know that 8 of the left-handed people are left-footed, we can calculate the number of people who are right-handed and left-footed: $27-8=19\\text{ right-handed and left-footed people }$ Finally, if 19 of the 63 right-handed people are left-footed, we can calculate how many are right-handed and right-footed $63-19=44\\text{ right-handed and right-footed people }$ The completed frequency tree", "+ z}$ then ${z > y}$. (D) If ${x > y + z}$ then ${z y + z}$ then ${z . 61. Consider the statement: ${x(\\alpha - x) < y(\\alpha - y)}$ for all ${x,y}$ with 0 < ${x}$ < ${y}$ 2}$; (C) if and only if ${\\alpha 1}$, such that ${n}$, ${n + 2}$, ${n + 4}$ are prime numbers, is (A) zero; (B) one; (C) infinite; (D) more than one, but finite. 62. In a village, at least 50% of the people read a newspaper. Among those who read a newspaper at the most 25% read more than one paper. Only one of the following statements follows from the statements we have given. Which one is it? (A) At the most 25% read exactly one newspaper. (B) At least 25% read all the newspapers. (C) At the most 37.50% read exactly one newspaper. (D) At least 37.50% read exactly one newspaper. 63. We consider the relation “a person X shakes hand with a person Y”. Obviously, if X shakes hand with Y, then Y shakes hand with X. In a gathering of 99 persons, one of the following statements is always true, considering 0 to be an even number. Which one is it?(A) There is at least one person who shakes hand exactly with an odd", "you just received, divide that number by E, write the result on a scrap of paper and hand it to the person on your right. Now raise your hand if you have just been handed your original number. Find the smallest value of N such that if A, B, C, D, and E are certain prime numbers, then at least one person will raise their hand. Your answer must specify the values of N, A, B, C, D, E, as well as the number of people in the circle to satisfy the requirements. There are 5 people. $N=9.$ $\\{A, B, C, D, E\\} = \\{7, 11, 13, 19, 52579\\}$, this being the factorisation of $10^9 + 1$.", "the index finger • Eleven → middle link of the index finger • Twelve → lower link of the index finger ### Two-handed counting up to 60 After the first dozen has been counted using the thumb as a pointer with the three phalanges of the remaining four fingers of the same hand (4 × 3 = 12), the counting capacity of one hand is initially exhausted. • The other hand is clenched in a fist. In order to remember that a dozen has been counted, one now extends a finger, e.g. B. the thumb. • Now you continue to count by starting again at one with your first hand . At twelve , the second dozen is full. • In order to remember that two dozen have been counted, one now extends the next finger of the other hand, e.g. B. after the thumb out the index finger. • With the five fingers of the first hand you can count five times a dozen, so 5 × 12 = 60. • Now you can count the next dozen again with the first hand, i.e. count up to 72 with two hands (12 on the first plus 60 on the other hand). This finger counting system still exists in parts of Turkey , Iraq , India and Indochina" ]

[ "• In how many different ways, can the letters of the word 'VENTURE' be arranged? A) 840 B) 5040 C) 1260 D) 2520 D) 2520 The required different ways = $$\\large\\frac{7!}{2!} = \\frac{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2!}{2!} = 2520$$ ##### Similar Questions 1). How many different signals , can be made by 5 flags from 8 flags of different colors? A). 6270 B). 1680 C). 20160 D). 6720 2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets? A). 12 B). 64 C). 256 D). 60 3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together? A). 10! B). 14!/(4!) C). 151200 D). 3628800 4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible? A). 120 B). 100000 C). 6720 D). 30240 5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible? A). 72 B). 144 C). 288 D). 234 6). Find the", "the number and nature of the papers read. In the early years of these quadrennial meetings four sections were considered sufficient; at present there are twice as many. At first the countries represented were chiefly European, the number of members from other continents being relatively small; but in this congress between 40 and 50 nations were represented and the attendance approximated 650 men and 200 women.", "did Country A send at least 10 representatives? (1) One of the six countries sent 41 representatives to the congress (2) Country A sent fewer than 12 representatives to the congress Math Expert Joined: 02 Sep 2009 Posts: 56271 Re: Six countries in a certain region sent a total of 75 representatives [#permalink] ### Show Tags 27 Apr 2010, 12:42 60 64 Six countries in a certain region sent a total of 75 representatives to an international congress, and no two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives? Given: $$x_1<x_2<x_3<x_4<A<x_6$$ and $$x_1+x_2+x_3+x_4+A+x_6=75$$. Q: is $$A\\geq{10}$$ (1) One of the six countries sent 41 representatives to the congress --> obviously $$x_6=41$$ --> $$x_1+x_2+x_3+x_4+A=34$$. Can $$A\\geq{10}$$? Yes. For example: $$x_1=2$$, $$x_2=3$$, $$x_3=8$$, $$x_4=10$$, $$A=11$$ --> $$sum=34$$ (answer to the question YES); Can $$A<{10}$$? Yes. For example: $$x_1=4$$, $$x_2=6$$, $$x_3=7$$, $$x_4=8$$, $$A=9$$ --> $$sum=34$$ (answer to the question NO). (2) Country A sent fewer than 12 representatives to the congress --> $$A<{12}$$. The same breakdown works here as well: Can $$12>A\\geq{10}$$? Yes. For example: $$x_1=2$$, $$x_2=3$$, $$x_3=8$$, $$x_4=10$$, $$A=11$$, $$x_6=41$$ --> $$sum=75$$ (answer to the question YES); Can $$A<{10}$$? Yes. For example: $$x_1=4$$, $$x_2=6$$, $$x_3=7$$, $$x_4=8$$, $$A=9$$, $$x_6=41$$ -->", "of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives? (1) One of the six countries sent 41 representatives to the congress (2) Country A sent fewer than 12 representatives to the congress [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 39037 Followers: 7750 Kudos [?]: 106470 [31] , given: 11626 ### Show Tags 27 Apr 2010, 12:42 31 KUDOS Expert's post 36 This post was BOOKMARKED Six countries in a certain region sent a total of 75 representatives to an international congress, and no two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives? Given: $$x_1<x_2<x_3<x_4<A<x_6$$ and $$x_1+x_2+x_3+x_4+A+x_6=75$$. Q: is $$A\\geq{10}$$ (1) One of the six countries sent 41 representatives to the congress --> obviously $$x_6=41$$ --> $$x_1+x_2+x_3+x_4+A=34$$. Can $$A\\geq{10}$$: $$x_1=2$$, $$x_2=3$$, $$x_3=8$$, $$x_4=10$$, $$A=11$$ --> $$sum=34$$ (answer to the question YES); Can $$A<{10}$$: $$x_1=4$$, $$x_2=6$$, $$x_3=7$$, $$x_4=8$$, $$A=9$$ --> $$sum=34$$ (answer to the question NO). (2) Country A sent fewer than 12 representatives to the congress --> $$A<{12}$$. The same breakdown works here as well: Can $$12>A\\geq{10}$$: $$x_1=2$$, $$x_2=3$$, $$x_3=8$$, $$x_4=10$$, $$A=11$$, $$x_6=41$$ --> $$sum=75$$ (answer to the question YES);", "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", "# packing with time-variant weights This appears to be a knapsack / bin-packing problem, but I seem to have got stuck and could appreciate contributions. Scenario 1: Tough (for me!) There is a one day conference with a set of (4 or more) sessions. The conference will be attended by a number of companies, each one being represented by a (1 or more) representatives. Across sessions the number of representatives for each company will vary (and may be zero), with each individual having the same chance of being present as any other individual (so there is an increasing likelihood of a company being represented when it has more representatives). There is a single row of seats in the conference. There are more than enough seats for the most popular session. (e.g., if the most popular session has 100 delegates, there could be 120 seats for the whole conference). Constraints • A rep. belongs to one company • A rep. must be seated in a session they are going to. • A rep. must not change seats across consecutive sessions • A rep. is assigned to one chair in each session they attend. Goal To fit the constraints and priorities optimally. Example. 15 chairs, 4 companies (A-D), 4 sessions (S1-S4). // Session attendees by company S1: A2 B6 C3", "In how many different ways, can the letters of the word 'VENTURE' be arranged? A) 840 B) 5040 C) 1260 D) 2520 Correct Answer: D) 2520 Description for Correct answer: The required different ways = $$\\large\\frac{7!}{2!} = \\frac{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2!}{2!} = 2520$$ Comments No comments available Similar Questions 1). How many different signals , can be made by 5 flags from 8 flags of different colors? A). 6270 B). 1680 C). 20160 D). 6720 -- View Answer 2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets? A). 12 B). 64 C). 256 D). 60 -- View Answer 3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together? A). 10! B). 14!/(4!) C). 151200 D). 3628800 -- View Answer 4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible? A). 120 B). 100000 C). 6720 D). 30240 -- View Answer 5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of", "of 12 people? We have: How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, Notice that no consideration is given to the order or arrangement of the items but simply the combinations. Another way of viewing combinations is as follows. Consider the number of combinations of 5 letters taken 3 at a time. This produces: Now assume you permute (arrange) the r = 3 letters in each of the 10 combinations in all possible ways. Each group would produce r! permutations. Letting x = 5C3 for the moment, we would therefore have a total of x(r!) different permutations. This total, however, represents all the possible permutations (arrangements) of n things taken r at a time, which is shown under arrangement numbers and defined as nPr. Therefore, Using the committee of 3 out of 12 people example from above, Consider the following: How many different ways can you enter a 4 door car? It is clear that there are 4 different ways of entering the car. Another way of expressing this is: If we ignore the presence of the front seats for the purpose of this example, how many different ways can you exit the car assuming that you do", "those days is a section would run a national convention. They put everything they had, every person they had, every ounce of energy they had into the convention. And when the convention was over they didn’t have the infrastructure, the officers, and the meetings and all that—and the section had trouble. What we did know—Los Angeles section, we knew that. We knew that would happen. So the officers at that time were forbidden from participating in the convention committee. The convention committee was a committee of the SWE section, so the officers were completely aware. [01:54:00] We were keeping them informed, but it was a completely different officer structure. I was the chair. We had our vice chairs. We had our treasurer. We had our secretary of the convention committee. And there were the similar officers in the section, and basically working parallel officer systems. We ran, I think,$225,000 through that budget. That was 1981. That was a lot of money. Almost a quarter of a million dollars running through. I think, between student conference and regular convention registration, we were somewhere between seventeen hundred and eighteen hundred. I think it was about six hundred students, twelve hundred professionals, the largest convention of its time, amazing convention. And most of the sessions were technical. Feeding on that—what I", "(since A repeats) total permutations= 8! x 4!/8=7! * 4! = 120960 A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw ? Total Combinations of 9 balls into 3= {}^9C_3 Total combination where black ball doesn't show up = {}^{(number of other balls)}C_3 = []^6C_3 Number of combinations where at least one black ball shows up = {}^9C_3 - {}^6C_3 = 84-20 = 64 How many groups of 6 persons can be formed from 8 men and 7 women ? {}^{15}C_3 8 men entered a lounge simultaneously. If each person shakes hand with the other, then find the total no. of handshakes? shake hand is group of 2 No order, A shakes B same as B shakes A So {}^8C_2 In how many different ways, can the letters of the word 'INHALE' be arranged ? INHALE has 6 letters, none repeating. Total arrangements = 6! In how many different ways, can the letters of the word 'BANKING' be arranged ? BANKING has 7 letters, N repeating Total Arrangements = \\frac{7!}{2!} In a meeting between two countries, each country has 12 delegates. All the delegates of one country shake hands", "from 6 boys and 4 girls. Solution: There are 6 boys and 4 girls. A team of 3 boys and 2 girls is to be selected. ∴ 3 boys can be selected from 6 boys in 6C3 ways. 2 girls can be selected from 4 girls in 4C2 ways. ∴ Number of ways the team can be selected = 6C3 × 4C2 = $$\\frac{6 !}{3 ! 3 !} \\times \\frac{4 !}{2 ! 2 !}$$ = $$\\frac{6 \\times 5 \\times 4 \\times 3 !}{3 \\times 2 \\times 1 \\times 3 !} \\times \\frac{4 \\times 3 \\times 2 !}{2 \\times 2 !}$$ = 20 × 6 = 120 ∴ The team of 3 boys and 2 girls can be selected in 120 ways. Question 9. After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting. Solution: Let there be n participants present in the meeting. A handshake occurs between 2 persons. ∴ Number of handshakes = nC2 Given 66 handshakes were exchanged. ∴ 66 = nC2 ∴ 66 = $$\\frac{n !}{2 !(n-2) !}$$ ∴ 66 × 2 = $$\\frac{\\mathrm{n}(\\mathrm{n}-1)(\\mathrm{n}-2) !}{(\\mathrm{n}-2) !}$$ ∴ 132 = n(n – 1) ∴ n(n – 1) = 12 × 11 Comparing on both sides, we get n = 12", "it takes to read this sentence). But hand-raising, as great as it is, suffers from the flaw that it does not scale due to the counting problem: a group of 500 people can raise their hands as easily as a group of 50 or 5, but it takes far too long to count ~250 hands: the person counting will quickly tire of the tedium, they will make mistakes counting, and this puts a serious lag on each vote, a lag which increases linearly with the number of voters. (Hands can be easy to approximate if almost everyone votes for or against something, but if consensus is so overwhelming, one doesn’t need to vote in the first place! The hard case of almost-balanced votes is the most important case.) One might suggest using an entirely different strategy: a website with HTML polls or little clicker gizmos like used in some college lectures to administer quick quizzes. This have the downsides that they require potentially expensive equipment (I used a clicker in one class and I think it cost at least$20, so if a convention wanted to use that in an audience of hundreds, that’s a major upfront cost & my experience was that clickers were unintuitive, did not always work, and slowed things down if anything; a website would", "9P5 = 15,120 Ques 5. How many numbers of five digits can be formed with the digits 0,2,4,6 and 8 ? Sol: Ques 6. How many numbers of five digits can be formed with the digits 0,1, 2, 3, 4, 6 and 8 ? Sol : Here nothing has been said about the repetition of digits. So , it is understood that repetition of digits is not allowed . Ques 7. How many even numbers of three digits can be formed with the digits 0,1, 2, 3, 4, 5 and 6 ? Sol : Total of such numbers = 5 × 5 v 3 = 75 req no. = 30+75 = 105 Ques 8. A round table conference is to be held between delegates of 15 companies. In how many ways can they be seated if delegates from two MNCs may wish to sit together ? Sol : Since delegates from two multinational companies will sit together, so considering these two delegates as one unit, there will be 13 + 1 = 14 delegates who can be arranged in a circular table in 14! ways. The two delegates from the MNCs can be arranged among themselves in 2! ways. Using the product rule, the required no. of ways = 14!×2! Ques 9. A person has 12 friends out", "CountTheOpposite InclusionExclusion Bundling AIME Difficult 2011 Problem - 266 Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Find the probability that each delegate sits next to at least one delegate from another country. The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "If a total of $$10$$ pencils and $$11$$ charts have been brought to the meeting, how many people are present? A. $$6$$ B. $$7$$ C. $$8$$ D. $$9$$ E. $$10$$ OA B Solution: We can let the number of associate professors = a and the number of assistant professors = n, and create two equations: Pencil equation: 2a + n = 10 Chart equation: a + 2n = 11 Let’s add the two equations together: 3a + 3n = 21 a + n = 7 Thus, there are 7 people at the meeting.", "? Condition 1: One of the six countries sent 41 representatives to the congress So filling in the blanks by trial & error: Country # R Ranking A = 10 2 B = 41 1 which we know must be respected, so 75 - 51 = 24 R left, if we randomly assign countries C, D, E, F, #'s of R that are not the same, lets say like : C = 9 3 D = 8 4 E = 4 5 F = 3 6 Then the total = 75 which makes condition # 1 work and is true ?? If we sub in condition # 2: Country A sent fewer than 12 representatives to the congress this still holds true and you could make A = 11 and C = 9, D = 8, E= 4, F = 2 and this would still work and add up to 75 ? as well as many other combination allowing for condition 1 & 2 to hold true so it seems as though each of them alone is enough to satisfy the questions ? Am I missing something ? If someone can help clarify it would be highly appreciated Matt Retired Moderator Joined: 22 Aug 2013 Posts: 1435 Location: India Re: Six countries in a certain region sent a", "## 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.", "to someone standing! Galeforce 9 knows this, all their demo tables for Star Trek Ascendency were set to standing height. One of the most important aspects of standing is eye contact. If you look someone in the eye as they walk, they will look back at you. Now you have an opening to engage them. If you are beneath their gaze, they may not see you at all, or literally look down on you. So, though it cost sore feet, we stood the whole time. We did take 1 hour lunches and occasional 1-minute sitting breaks, just to rest the feet but there was always someone standing in the booth ready to engage customers. ## E. Always Be Looking for Opportunity It’s day 1 of the Con. Doors open at 10am sharp. This is it, the moment we’ve been working towards for 8 months! A line starts forming to my right, 6 booths away. By 10:10 it’s 2 people deep and reaches our intersection. By 10:20 it is way past us on the left, almost half the length of the convention center. Easily 200 or more people in line blocking access to our booth! No one can get past them, traffic is at a standstill! This was a disastrous start. My heart fell. I asked someone what they", "$2\\cdot{6\\choose3}$ Add them and get 55 5. Thanks everyone! Very helpful. I had one more question, kind of similar, but I don't know how to solve this one mathematically without drawing it out ... Six three-representative delegations attend an international conference. Teh reps shake hands when they are introduced to one another, how many handshakes are possible if each delegate shakes hands only once with every other attendant except with those of his/her delegation As always, any help is greatly appreciated. 6. Nevermind. Figured it out. :-)", "arrangement matters). And so while some attendees (the “core group”) might end up having fun, the party doesn’t really work for most participating parties. So, the next time you’re hosting a party, do yourself and your guests a favour and ensure that you don’t end up with between 6 and 10 people at the party. Either less or more is fine! You might want to read this other post I’ve written on coordinating guest lists for birthday parties. ## Meeting types There are essentially three kinds of meetings – those that are entirely “in person”, those that are entirely “on call” and hybrids. I argue that the quality of conversation in the third kind of meeting is significantly inferior to that of the first two types. In person meetings are those where all participants are in one room. These are perhaps the best kind of meetings (except when you know it’s likely to turn antagonistic), for you can maintain eye contact with the others and as long as the number of meeters is small, there is social pressure on meeters to not be distracted, and thus the meeting is likely to conclude its agenda productively and quickly. Conference calls allow you to multitask while you are on the meeting – the positive thing is that you can choose" ]

[ "right triangle has one right angle.", "sides is a right angle.", "many diagonals does a 63-sided convex polygon have? [#permalink] 08 Aug 2017, 13:58 Display posts from previous: Sort by", "since it has four right angles and four congruent sides.", "congruent sides and four right angles.", "angles of the trapezoid? • Six-sided polygon In a six-sided polygon. The first two angles are equal, the third angle is twice (the equal angles), two other angles are trice the equal angle, while the last angle is a right angle. Find the value of each angle.", "sides form a right triangle.", "(the side opposite the right angle).", "(and as one can see “side” makes only really sense for polygons) A circle has indeed $0$ straight sides.", "the sides of a right-angled triangle read about!", "so the angles are right angles.", "polygons with two diagonals cannot be the same set as the set of quadrilaterals, because any pentagon has two diagonals, but a pentagon is not a quadrilateral. It all depends on whether \"two\" means at least two or exactly two. Dbfirs 21:19, 13 August 2009 (UTC) ...In all seriousness, I can't. The first mathematical fallacy in your argument is that a pentagon has two diagonals, when it actually has five diagonals. The number of sides, the number of angles and the number of diagonals for polygons are by exact count! A triangle does NOT have one angle. The set of all polygons having 2 diagonals is the set of quadrilaterals. The set of all polygons having \"at least\" 2 diagonals is a collection of sets enumerated by the set of Quadrilaterals UNION Pentagons UNION Hexagons UNION..., or more simply put, Polygons \\ Triangles (set subtraction). However, the number of \"equal\" sides or \"equal\" angles can subsume a larger number of \"equal\" sides or \"equal\" angles. In other words, the property of equality can vary and could possibly be subsumed, but you have to let the chips fall where they may. For triangles, a triangle with 0 equal sides cannot have 2 equal sides and vice versa, but a triangle with 2 equal sides can have 3 equal sides", "# What is this polyhedron? Geometry Level pending Suppose a convex polyhedron has $$12$$ square faces, $$8$$ regular hexagon faces, and $$6$$ regular octagon faces. At each vertex of the polyhedron, one square, one hexagon, and one octagon meet. How many space diagonals does this polyhedron have? ×", "7 28 Jun 2012, 17:58 How many diagonals does a polygon with 21 sides have, if one 1 19 Dec 2007, 17:48 How many diagonals does a polygon with 21 sides have, if one 3 06 Aug 2007, 17:32 How many diagonals does a polygon with 21 sides have, if one 5 09 Feb 2006, 17:00 a parallelogram has 2 sides: 14, 18 one of the diagonals is 13 24 Sep 2005, 06:03 Display posts from previous: Sort by", "square are right angles. (e) None of these", "who are diagonally opposite. Then, what about five- and six-sided tables?", "back the! Equal as they are all subtended by equal chords as the sides of a polygon can.! Receive notifications of new posts by email that each vertex angle can be anywhere from > 0º up", "of (only one, if the polyhedron is convex). – Neal Mar 4 '12 at 13:54", "diagonals does a 63-sided convex polygon have? 4 15 Mar 2016, 13:02 2 how many internal diagonals does a decagon (ten sided polygon) have? 3 27 Feb 2015, 03:18 Display posts from previous: Sort by", "Convex Polyopes. -" ]

[ "each diagonal, there are $2$ vertices that we picked. Therefore, we counted each diagonal twice! The answer is therefore $$\\frac{7 \\cdot 10}{2} = 35$$ Now, let's look at what the solution intended. There are $10$ vertices. A segment is defined by its two endpoints. Therefore, there are $$\\binom{10}{2} = 45$$ segments, and $10$ of these are the sides of the decagon. Therefore there are $45 - 10 =35$ diagonals. From any given vertex you can draw $7$ diagonals. You can't draw to itself or to the two neighboring points. There are therefore $10 \\cdot 7$ ends of diagonals. As each diagonal has two ends, there are $10 \\cdot 7 \\cdot \\frac 12=35$ diagonals. The approach to the formula you quote is that you pick two vertices to draw a line between, which you can do in $10 \\choose 2$ ways. $10$ of those are sides of the decagon instead of diagonals, so the result is ${10 \\choose 2}-10=35$ The correct answer gives you $\\binom{10}{2}-10=35$, the reasoning being that you choose two vertices of the $10$ and draw a diagonal between them. However, as you said, consecutive vertices will not give a diagonal, so we ignore the $10$ sides of the decagon that arise from this counting. I have trouble telling what you're doing in your answer; you seem", "# Recurrence for number of regions formed by diagonals of a convex polygon. I've been having trouble with this particular problem, been thinking for it for a good hour or two, but I haven't gotten an explanation to the following question. Suppose $a_n$ be the number of regions into which a convex polygonal region with $n+2$ sides is divided by its diagonals, assuming no three diagonals have a common point. Define $a_0 = 0$. Show that $$a_n = a_{n-1} + {n+1 \\choose 3} + n \\quad (n \\geq 1)$$ So far, I have that for $n \\geq 1$, we look at an $(n+2)$-gon. Pick an edge from one of the $n+2$ edges from the $(n+2)$-gon. Then adjoin a triangle to it, so we can have an $(n+3)$-gon. Given the triangle that we adjoin to the $(n+2)$-gon, let's look at the exposed vertex of that triangle. We can create $n$ diagonals, since we can't create a diagonal by joining the exposed vertex to an adjacent vertex. So now, we have that $n$ diagonals run through the edge created by the intersection of the $(n+2)$-gon and the triangle. This gives us $n+1$ regions in the triangle. But now, I can't seem to find a pattern to why the $(n+2)$-gon has ${n+1 \\choose 3}$ extra regions that is as a result", "if $8$ and $7$ were on the same face, the only way to get the sum is with $1$ and $2$. This means that $6$ and $7$ are not on the same edge as $8$, or in other words they are diagonally across from it on the same face, or on the other end of the cube. Now we look at three cases, each yielding two solutions which are reflections of each other: 1) $6$ and $7$ are diagonally opposite $8$ on the same face. 2) $6$ is diagonally across the cube from $8$, while $7$ is diagonally across from $8$ on the same face. 3) $7$ is diagonally across the cube from $8$, while $6$ is diagonally across from $8$ on the same face. This means the answer is $3\\cdot 2=\\boxed{\\textbf{(C) }6.}$", "choose a first edge in $P$. With a similar counting as above we then find that there are $12$ admissible ways to choose a second edge in $P$ which is separated by two true diagonals from the first chosen edge. In all we obtain $17\\cdot 12+{1\\over2}\\cdot 17\\cdot 12=306$ admissible quadrilaterals.", "BE CE Here we have two diagonals containing A, two containing B, two containing C, two containing D, and two containing E. This way of describing it suggests the beginning of a pattern! So for 4 sides there is one diagonal for each vertex and for 5 there are two. Why? Well, there is a diagonal from each vertex to each other vertex except the two adjacent vertices. For 4 vertices, there are three other vertices for each vertex, and two of them are adjacent, which leaves 1; for 5 vertices, there are four other vertices for each vertex, and two of them are adjacent, which leaves 2. So for n vertices, there will be n – 3 diagonals through each. Okay now, we're starting to get somewhere. For four vertices there is one diagonal containing each vertex, but we don't end up with four diagonals but just two. For five vertices there are two diagonals containing each vertex, but we don't end up with ten diagonals but just five. What's going on? Aha! If you count one diagonal for every vertex of the quadrilateral, you're counting every diagonal twice, because every diagonal contains two vertices. And similarly for the pentagon. So you have to divide by two. I hope this helps you with the general problem. Without", "of those $n$ diagonals traversing through the $(n+2)$-gon. Any tip or hint would be appreciated. - Well, since your recurrence defines $a_n$ in terms of $a_{n-1}$, I think it would be more appropriate to consider the additional regions created when adding the $(n+2)$nd vertex to an $(n+1)$-gon rather than the $(n+3)$rd vertex to an $(n+2)$-gon. So, suppose we have an $(n+2)$-gon. Designate the \"exposed vertex\" $X$, and call the other vertices $P_1$, $P_2$, ..., $P_{n+1}$, where $P_1$ and $P_{n+1}$ are adjacent to $X$. The $P_i$s form an $(n+1)$-gon within which there are $a_{n-1}$ regions made from all the diagonals and sides $P_iP_j$. Now consider the $n-1$ diagonals emanating from $X$ and finishing at $P_k$, where $2 \\leq k \\leq n$. As we draw these, whenever a diagonal hits a previously drawn diagonal, it enters a different region, which will be cut into two regions when the diagonal hits the next line. That is, for every intersection between a new diagonal $XP_k$ and an old diagonal $P_iP_j$, we make one new region inside polygon $P_1\\ldots P_{n+1}$. How many such intersections are there? It's just the number of ways to pick three integers $1 \\leq i < k < j \\leq n+1$, that is, $\\binom{n+1}3$. And then of course we have the $n$ new regions inside the triangle $XP_1P_{n+1}$. -", "new diagonals contributed by $k^{th}$ vertex equal to $(n-3)-(k-2) = (n-k-1)$. Since new contribution can’t be negative and for $(n-1)^{th}$ vertex we end up with zero (new) contribution, $n^{th}$ vertex will also contribute zero diagonals. Combining all of the above arguments I complete the proof of my observation and call it “New Diagonal Contribution Theorem” (NDCT).", "same face, the only way to get the sum is with $1$ and $2$. This means that $6$ and $7$ are not on the same edge as $8$, or in other words they are diagonally across from it on the same face, or on the other end of the cube. Now we look at three cases, each yielding two solutions which are reflections of each other: 1) $6$ and $7$ are diagonally opposite $8$ on the same face. 2) $6$ is diagonally across the cube from $8$, while $7$ is diagonally across from $8$ on the same face. 3) $7$ is diagonally across the cube from $8$, while $6$ is diagonally across from $8$ on the same face. This means the answer is $3\\cdot 2=\\boxed{\\textbf{(C) }6.}$", "start drawing diagonals from each vertex one-by-one. While doing so count the number of new diagonals contributed by each vertex. Here is the “Experiment” done for $n=4,5,6,7,8$. Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one) Based on above experiment, I observed: The number of new diagonals contributed by each vertex of a $n$-sided polygon follows the sequence: $(n-3),(n-3),(n-4),\\ldots, 1,0,0$ Now let’s try to prove this observation: Since we can’t draw diagonals to the adjacent vertices, each vertex will have $(n-1)-2 = (n-3)$ diagonals. Now, let’s count the new contribution of each vertex by considering restrictions on the maximum possible contribution (which is $(n-3)$). For first vertex, we have no restriction, so it will contribute $(n-3)$ diagonals. Also, since second vertex in adjacent to first one and both can’t affect each-other’s contribution, it will also contribute $(n-3)$ diagonals. But, starting with third vertex we observe that first vertex has already taken one of the diagonals from its maximum contribution (second vertex can’t affect its contribution count since it’s adjacent vertex), thus it contributes (new) $(n-3)-1 = (n-4)$ diagonals. Continuing same way, for $k^{th}$ vertex, consider the restriction to contribution caused by $1^{st}$ to $(k-2)^{th}$ vertices. Thus for $1 we get the number of", "# Easy Permutation-Combination Solved QuestionAptitude Discussion Q. How many diagonals can be drawn in a pentagon? ✔ A. 5 ✖ B. 7 ✖ C. 8 ✖ D. 10 Solution: Option(A) is correct A pentagon has $5$ sides. We obtain the diagonals by joining the vertices in pairs. Total number of sides and diagonals: $= {^5C_2}$ $= \\dfrac{5×4}{2×1}$ $= 5×2$ $= 10$ This includes its $5$ sides also. ⇒ Diagonals $= 10 - 5 = 5$ Hence the number of diagonals, $= 10 - 5$ $= 5$ ## (1) Comment(s) Raj Singh () From one vertex, you can draw 3 diagonals. From second vertex again you can draw 3 more diagonals. From third vertex again you can draw 3 more diagonals. Overall, you can draw 5x3=15 diagonals in a pentagon.", "with $26$ vertices, $60$ edges and $36$ faces. $24$ of the faces are triangular and $12$ are quadrilaterals. A space diagonal is a line segment connecting two vertices which do not belong to the same face. How many space diagonals does P have? [Source: http://www.kalva.demon.co.uk/aime/aime04a.html] Solution: What defines a space diagonal? The question tells us that \"a space diagonal is a line segment connecting two vertices which do not belong to the same face.\" So we're basically choosing two points and connecting them, with several exceptions. Let's get the basics down first. We have $26$ vertices. If we take any two at a time, there are $26C2 = 325$ ways of doing that. But do we count anything we don't want? How about edges and the diagonals on the faces of the squares? We don't want those, but all of them are counted anyway. So we subtract away the number of edges and twice the number of squares ($2$ diagonals per square) to get $325 - 60 - 2(12) = 241$ space diagonals. QED. -------------------- Comment: A relatively easy AIME problem, requiring only basic counting and knowledge of a little space geometry. The AIME usually has a few space geometry problems so it's good to brush up on your 3-dimensional problem solving. -------------------- Practice Problem #1: Suppose in", "with $26$ vertices, $60$ edges and $36$ faces. $24$ of the faces are triangular and $12$ are quadrilaterals. A space diagonal is a line segment connecting two vertices which do not belong to the same face. How many space diagonals does P have? [Source: http://www.kalva.demon.co.uk/aime/aime04a.html] Solution: What defines a space diagonal? The question tells us that \"a space diagonal is a line segment connecting two vertices which do not belong to the same face.\" So we're basically choosing two points and connecting them, with several exceptions. Let's get the basics down first. We have $26$ vertices. If we take any two at a time, there are $26C2 = 325$ ways of doing that. But do we count anything we don't want? How about edges and the diagonals on the faces of the squares? We don't want those, but all of them are counted anyway. So we subtract away the number of edges and twice the number of squares ($2$ diagonals per square) to get $325 - 60 - 2(12) = 241$ space diagonals. QED. -------------------- Comment: A relatively easy AIME problem, requiring only basic counting and knowledge of a little space geometry. The AIME usually has a few space geometry problems so it's good to brush up on your 3-dimensional problem solving. -------------------- Practice Problem #1: Suppose in", "2, which is 3 less than the number of vertices, 5: 2 diagonals connected to each vertex. This is true of all diagonals of all polygons. A nonagon has $$9-3=6$$ diagonals connecting to each vertex. The number 3 is not arbitrary here – from any vertex, diagonals cannot connect to the vertex itself or to the vertices that they are “1 away” from, because it would be edges. That makes 3 vertices from every vertex that aren’t included. To generalize this, we’ll use $$n-3$$, where n is the number of vertices of the polygon. Now, each vertex has the same number of diagonals connecting to it, so in this case, we can see that the total number of diagonal connections to vertices is $$5(5-3)=10$$. In general, we can say the total number of diagonal connections is $$n(n-3)$$. When we figure out this total, though, we are counting each diagonal twice, because diagonals have 2 endpoints. In order to figure out the number of unique diagonals, we need to divide our total by 2. In our pentagon, this looks like: $$\\frac{5(5-3)}{2} = 5$$ unique diagonals (which corresponds to our diagram) Therefore, for any sized polygon: Our equation can be written as this: $$\\frac{n(n-3)}{2}$$ Conceptually, this can be remembered as: $$\\frac{(\\text{total number of vertices})(\\text{number of diagonal connections at each vertex})}{(\\text{number", "diagonals is $1128 - 72 - 216 = 840$. ## Solution 2 We first find the number of vertices on the polyhedron: There are 4 corners per square, 6 corners per hexagon, and 8 corners per octagon. Each vertex is where 3 corners coincide, so we count the corners and divide by 3. $\\text{vertices} = \\frac{12 \\cdot 4 + 8 \\cdot 6 + 6 \\cdot 8}{3}=48$. We know that all vertices look the same (from the problem statement), so we should find the number of line segments originating from a vertex, and multiply that by the number of vertices, and divide by 2 (because each space diagonal is counted twice because it has two endpoints). Counting the vertices that are on the same face as an arbitrary vertex, we find that there are 13 vertices that aren't possible endpoints of a line originating from the vertex in the middle of the diagram. You can draw a diagram to count this better: Since 13 aren't possible endpoints, that means that there are 35 possible endpoints per vertex. The total number of segments joining vertices that aren't on the same face is $48\\cdot 35\\cdot \\frac 12 = 24 \\cdot 35 = \\boxed{840}$ ## Solution 3 Since at each vertex one square, one hexagon, and one octagon meet, then there are", "# How many cases can draw diagonals? Imagine a n_regular polygon that vertex is named by 1 to n. We know can draw (n)(n+3)/2 diagonals in n_regular polygon,Also know if we want to draw Maximum diagonals that not intersecting each other inside the n_regular polygon is n−3 diagonals. For example for Flat hexagon we can draw Maximum 3 diagonals that we can do it with 14 cases.(you can find it easy) Now imagine we want to draw diagonals that have 2 condition like this: 1- any vertex is not connected to 2 before and 2 after vertex. 2- any 3 vertex is not exist that intersecting two by two inside the n_regular polygon. It's easy to understand that the Maximum of diagonals can draw is 2(n−5). But the question is how many cases can draw diagonals that Applicable 2 above condition? For n=6 we can draw 3 cases that necessitate this 2 condition . If we labeled regular when n=7 we have 2(7−5)=4 diagonals and we can not use this vertexes {1,4},{2,5},{4,7},{3,6} because we ignore the second condition. Suppose we have a regular $n$-gon with vertices labeled 1 to $n$. There are $n(n-1)/2$ possible diagonals. The maximum number of non-intersecting diagonals is $n-3$, and if we draw them we get a triangulation. The number of ways to triangulate", "construction to the $n$-cube includes (if I've counted correctly) $$2^{n-1} + 2^{n-2} = 3 \\cdot 2^{n-2}$$ vertices.</p> <p>But perhaps I am misunderstanding your question, for your second condition on the cone seems superfluous.</p>", "anonymous There's a quick formula for this scenario, but I always find it useful to derive it through example. Let's start with a regular hexagon (6 sides). From any one vertex, we have 3 possible diagonals we can draw. |dw:1433380461631:dw| We only have 3 since we can't draw a diagonal from a vertex to itself, and we can't draw a diagonal to adjacent vertices. 27. anonymous From the next vertex, we again have 3 possible diagonals: |dw:1433380581390:dw| 28. anonymous From the next, we again have 3, but one of these is already drawn, so in fact we only add 2 new diagonals: |dw:1433380603875:dw| 29. anonymous Continuing this pattern, we end up with this: |dw:1433380652284:dw| 30. anonymous If you try this for any general regular polygon with $$n$$ sides, you'll get the same pattern: $\\{n-3,n-3,n-4,n-5,\\ldots,2,1,0,0\\}$ 31. anonymous Let's derive the formula. Let's call the total number of diagonals $$S$$, then $S=(n-3)+(n-3)+(n-4)+(n-5)+\\cdots+2+1+0+0$ Let's rearrange this sum: $S=0+0+1+2+\\cdots+(n-5)+(n-4)+(n-3)+(n-3)$ If we add corresponding terms (first to first, second to second, and so on) we get $2S=((n-3)+0)+((n-3)+0)+((n-4)+1)+\\cdots+(0+(n-3))$ Simplifying, we end up with $$n-3$$ showing up $$n$$ times (since there are $$n$$ sides in the polygon) on the right hand side: $2S=n(n-3)$ Dividing by 2, we get the formula: $S=\\frac{n(n-3)}{2}$ To check that this hold for our $$n=6$$ example, we get $$S=\\dfrac{6\\times3}{2}=9$$. Counting the", "diagonals. It is easy to achieve $$15$$ diagonals: just place NW-SE diagonals in each box in the first, third and fifth columns. Thus we have a lower bound of $$15$$ diagonals. My intuition is that it is impossible to stack more than 15 diagonals and that there should be an easy, visual, geometric proof for that. But that proof still escapes me. edit: as usual, my intuition proved wrong. See Jaap's nice solution. 1 answered Feb 19 '18 at 12:02 Evargalo 3,57511 gold badge1010 silver badges2828 bronze badges As a starter: There are $$6*6=36$$ boxes' corners. Each diagonal connects two corners. Thus we have an upper bound of $$36/2=18$$ diagonals. It is easy to achieve $$15$$ diagonals: just place NW-SE diagonals in each box in the first, third and fifth columns. Thus we have a lower bound of $$15$$ diagonals. My intuition is that it is impossible to stack more than 15 diagonals and that there should be an easy, visual, geometric proof for that. But that proof still escapes me.", "Interesting Diagonal Discovery Standard Yesterday (while doodling in my Number Theory class) I was just playing with diagonals of polygons and something interesting appeared on my notebook. For the first time in my life I have discovered a theorem and its proof on my own (it’s original!). In this blog-post I will share it with you all! Consider an $n$-sided polygon, and start drawing diagonals from each vertex one-by-one. While doing so count the number of new diagonals contributed by each vertex. Here is the “Experiment” done for $n=4,5,6,7,8$. Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one) Based on above experiment, I observed: The number of new diagonals contributed by each vertex of a $n$-sided polygon follows the sequence: $(n-3),(n-3),(n-4),\\ldots, 1,0,0$ Now let’s try to prove this observation: Since we can’t draw diagonals to the adjacent vertices, each vertex will have $(n-1)-2 = (n-3)$ diagonals. Now, let’s count the new contribution of each vertex by considering restrictions on the maximum possible contribution (which is $(n-3)$). For first vertex, we have no restriction, so it will contribute $(n-3)$ diagonals. Also, since second vertex in adjacent to first one and both can’t affect each-other’s contribution, it will also contribute $(n-3)$ diagonals. But, starting with", "# How does ${n^2 - 9n = 0}$ turn into ${n(n - 9) = 0}$? $${n^2 - 9n = 0}$$ How does that turn into ${n(n - 9) = 0}$? Can someone, please, explain the logic behind this? This arose in a problem involving the number of diagonals in a particular polygon: \\begin{align} \\frac{n(n - 3)}{2} = 3n &\\quad\\to\\quad n^2 - 3n = 6n \\\\ &\\quad\\to\\quad n^2 - 9n = 0 \\\\[4pt] &\\quad\\to\\quad n(n - 9) = 0 \\\\[4pt] &\\quad\\to\\quad n = 0, 9 \\end{align} The problem is easy to solve. I just don't get how $n^2−9n=0$ expands to $n(n−9)=0$. Choose one vertex, $n$ choices, choose another one from the remaining ones which is not adjacent or equal, $n-3$ this gives $n^2-3n$. Now since you can do it in any order, you divide by $2$ to account for double counting giving $$\\#d = {1\\over 2}n(n-3).$$ Not sure how you got the $9$. Clearly that's negative for a square, but there are $2$ diagonals for a square. Also not sure why you have the $=0$, are you sure this is how the problem was posed? • @W.Zlacki well if that's all, just note that $n(n-9) = n\\cdot n -n\\cdot 9$ by the distributive property, so $n(n-9)=n^2-9n$ is the same way of saying this without so many multiplication symbols. –" ]

[ "There are several formulas for the rhombus that have to do with its: Sides (click for more detail). From each of the N vertices,... How Many Congruent Sides Does A Scalene Triangle Have? The number of diagonals is n(n-1)/2 - n. Hope this helps, Stephen La Rocque. A square has two diagonals. Number of diagonals can be drawn from 1 vertex in n sided polygon is (n - 3) Total number of diagonals that can be drawn is n (n-3)/2 as per ur question n is 20. as one of its vertices doesn't connect. What is the sum of the measures of the angles of a convex quadrilateral? Answer. And. It can have no diagonals. 0 0. Show that all four triangles created by these diagonals – ΔAOD, ΔCOD, ΔAOB and ΔCOB – have equal areas. The three... How Many Diagonals Does A 9 Sided Shape Have ? (a) Since sum of linear pair angles is. How many diagonals has a 15-sided polygon? In any convex polygon, you can count the number of diagonals as follows. Why? This problem is adapted from the South East Asian Mathematics Competition (b) Find. a triangle has three diagonals as well as three sides Can a scalene triangle be a triangle? 1 C. 2 D. depends on the trapezoid. An angle only", "Math Secondary School How many diagonals are in triangle?? Since the diagonals of a rectangle are congruent MO = 26. d How many diagonals … To see how many diagonals intersections exist, we just need to know that we need 2 diagonals for one intersection,so we need 4 vertex in total there are $$\\binom{7}{4}=35$$ diagonals intersections. Figure – 12: Triangle counting in Fig – 12 = 45. Therefore, a quadrilateral has two diagonals, joining opposite pairs of vertices. A rectangle has two diagonals, and each is the same length. The formula is: 0.5*(n2-3n) = number of diagonals when n is the number of sides of the polygon So i though there were $$7\\cdot35\\cdot34$$ triangles sharing one vertex with the heptagon and having the other two on diagonals intersections. How many diagonals does a 27-gon have? If you know side lengths of the rectangle, you can easily find the length of the diagonal using the Pythagorean Theorem, since a diagonal divides a rectangle into two right triangles. 2. Diagonals bisect vertex angles. 7. How much percent is anil's income more than aman's income ?​, Please aap relevant answers to delete mat karo....konsi dusmani nikal rahe ho​, ask me the answer it's urgent please guys ​, manjoor h..........xDhehehe.....mujhe to ye pd kr hnsi hi aa gyi​, solve the following quadratic", "from one It also has three sides and three vertices. The angle sum of a convex polygon with number of sides 7 is From each vertex, N-4 diagonals can be drawn, where N is the number of vertices. How many sides does the polygon have? The number of triangles is n-2 (above). How many diagonals does each of the following have? Answer: (a) A convex quadrilateral has two diagonals. Diagonals bisect vertex angles. The sum of the interior angles of a 7-gon is. The three perpendicular bisectors of a triangle's three sides intersect at the circumcenter (the center of the circle through the three vertices). The diagonals have the following properties: The two diagonals are congruent (same length). You know how the angles of a triangle always add up to 180 0?Why is that? Examine the table. Can Someone Describe The Term \"Learning\" In Organizational Behavior? So from every vertex we can draw 6 diagonals, and we can do this from all 9 vertices for a total of 54 diagonals. You have learned a lot about particularly important parts of polygons, their diagonals. Well, the diagonals I think will be, 1/2 * 100 * (100 - 3) = 4850. the triangles I have no idea, quite a few I'd guess. What is the measure of an interior angle?", "# Recurrence for number of regions formed by diagonals of a convex polygon. I've been having trouble with this particular problem, been thinking for it for a good hour or two, but I haven't gotten an explanation to the following question. Suppose $a_n$ be the number of regions into which a convex polygonal region with $n+2$ sides is divided by its diagonals, assuming no three diagonals have a common point. Define $a_0 = 0$. Show that $$a_n = a_{n-1} + {n+1 \\choose 3} + n \\quad (n \\geq 1)$$ So far, I have that for $n \\geq 1$, we look at an $(n+2)$-gon. Pick an edge from one of the $n+2$ edges from the $(n+2)$-gon. Then adjoin a triangle to it, so we can have an $(n+3)$-gon. Given the triangle that we adjoin to the $(n+2)$-gon, let's look at the exposed vertex of that triangle. We can create $n$ diagonals, since we can't create a diagonal by joining the exposed vertex to an adjacent vertex. So now, we have that $n$ diagonals run through the edge created by the intersection of the $(n+2)$-gon and the triangle. This gives us $n+1$ regions in the triangle. But now, I can't seem to find a pattern to why the $(n+2)$-gon has ${n+1 \\choose 3}$ extra regions that is as a result", "vertices that are not adjacent to each other. How many different diagonals does it have? View Answer Find the interior and exterior angles of a 9-sided polygon. A diagonal connects any two non-consecutive vertices of a polygon. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties) If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13 How many diagonals does a triangle have? It doesn't matter how long the length of each side is or how it's shaped - it will always be three.... How Many Perpendicular Lines Does A Square Have? Therefore, number of diagonals = \\\\frac{63 * 60}{2}\\\\) = 1890. By definition, it has n sides. Show Answer. Here are some related questions which you might be interested in reading. Since there would be no diagonal drawn back to itself, and the diagonals to each adjacent vertex would lie on top of the adjacent sides, the number of diagonals from a single vertex is three less the the number of sides, or n-3. 1) Regular pentagon P has all five diagonals drawn. Relevance. A rhombus is a type of parallelogram, and what distinguishes its shape is that all four of its sides are congruent.. Question 2: How many sides does a triangle have?", "Difference between revisions of \"1988 AIME Problems/Problem 10\" Problem A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face? Solution By the Euler characteristic, we have that $V - E + F = 2$. The number of faces, $F$, is $12 + 8 + 6 = 26$. Since every point lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \\cdot 4 = 8 \\cdot 6 = 6 \\cdot 8 = 48$. Substituting gives us $E = 72$. The number of segments joining the vertices of the polyhedron is ${48\\choose2} = 1128$. Of these segments, $72$ are edges. The number of diagonals of a square is $\\frac{n(n-3)}{2} = 2$, of a hexagon is $9$, and of an octagon $20$. Hence the number of face diagonals is $2 \\cdot 12 + 9 \\cdot 8 + 20 \\cdot 6 = 216$. Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = 840$.", "vertex A? We'll there are 11 other vertices to connect to, HOWEVER the vertices on either side of vertex A are out of contention, since joining two adjacent points will not create a diagonal. So, we can create a diagonal with vertex A by connecting to any of the 9 eligible vertices. Using the same logic, we can conclude that we can create a diagonal with vertex B by connecting to any of the 9 eligible vertices. And we can create a diagonal with vertex C by connecting to any of the 9 eligible vertices. There are 12 vertices, so the TOTAL number of diagonals = (12)(9) = 108 HOWEVER, this is not the final answer, since we have counted each diagonal TWICE. For example, in the first step, we counted the diagonal created by joining vertex A with vertex B, and in the second step, we counted the diagonal created by joining vertex B with vertex A To account for this duplication, we must take 108 and divide it by 2 to get: 54 [Reveal] Spoiler: A Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Kudos [?]: 2867 [0], given: 364 Re: A polygon has 12 edges. How many different diagonals does it have? [#permalink] 20 Sep 2017, 14:59 Display posts from previous: Sort by", "# What is this polyhedron? Geometry Level pending Suppose a convex polyhedron has $$12$$ square faces, $$8$$ regular hexagon faces, and $$6$$ regular octagon faces. At each vertex of the polyhedron, one square, one hexagon, and one octagon meet. How many space diagonals does this polyhedron have? ×", "Interesting Diagonal Discovery Standard Yesterday (while doodling in my Number Theory class) I was just playing with diagonals of polygons and something interesting appeared on my notebook. For the first time in my life I have discovered a theorem and its proof on my own (it’s original!). In this blog-post I will share it with you all! Consider an $n$-sided polygon, and start drawing diagonals from each vertex one-by-one. While doing so count the number of new diagonals contributed by each vertex. Here is the “Experiment” done for $n=4,5,6,7,8$. Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one) Based on above experiment, I observed: The number of new diagonals contributed by each vertex of a $n$-sided polygon follows the sequence: $(n-3),(n-3),(n-4),\\ldots, 1,0,0$ Now let’s try to prove this observation: Since we can’t draw diagonals to the adjacent vertices, each vertex will have $(n-1)-2 = (n-3)$ diagonals. Now, let’s count the new contribution of each vertex by considering restrictions on the maximum possible contribution (which is $(n-3)$). For first vertex, we have no restriction, so it will contribute $(n-3)$ diagonals. Also, since second vertex in adjacent to first one and both can’t affect each-other’s contribution, it will also contribute $(n-3)$ diagonals. But, starting with", "# How many triangles exist in the diagonals intersections of an heptagon? In a heptagon (not necessarily regular) with all its diagonals marked, how many triangles exist with two vertex at two diagonals intersections and sharing one vertex with the heptagon? To see how many diagonals intersections exist, we just need to know that we need 2 diagonals for one intersection,so we need 4 vertex in total there are $$\\binom{7}{4}=35$$ diagonals intersections. So i though there were $$7\\cdot35\\cdot34$$ triangles sharing one vertex with the heptagon and having the other two on diagonals intersections. Actually i don't find this correct at all because there are non-collinear diagonals intersections that you can select, and you're counting them as triangles. So i was trying to count the diagonals intersections in each diagonal and there are some with 4 and 6 but i don't think this to be efficiently at all. So i was wondering for any hint,or answer. thanks. ## 1 Answer Assuming that you're fine with triangles whose sides are not, themselves, parts of the diagonals, that will be fine for a heptagon whose vertices are in general convex position: we just need to subtract off triangles with two points on a diagonal. To do this as generally as possible: if we fix a vertex $P$, the number of degenerate", "# Interesting Diagonal Discovery Standard Yesterday (while doodling in my Number Theory class) I was just playing with diagonals of polygons and something interesting appeared on my notebook. For the first time in my life I have discovered a theorem and its proof on my own (it’s original!). In this blog-post I will share it with you all! Consider an $n$-sided polygon, and start drawing diagonals from each vertex one-by-one. While doing so count the number of new diagonals contributed by each vertex. Here is the “Experiment” done for $n=4,5,6,7,8$. Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one) Based on above experiment, I observed: The number of new diagonals contributed by each vertex of a $n$-sided polygon follows the sequence: $(n-3),(n-3),(n-4),\\ldots, 1,0,0$ Now let’s try to prove this observation: Since we can’t draw diagonals to the adjacent vertices, each vertex will have $(n-1)-2 = (n-3)$ diagonals. Now, let’s count the new contribution of each vertex by considering restrictions on the maximum possible contribution (which is $(n-3)$). For first vertex, we have no restriction, so it will contribute $(n-3)$ diagonals. Also, since second vertex in adjacent to first one and both can’t affect each-other’s contribution, it will also contribute $(n-3)$ diagonals. But, starting", "# Interesting Diagonal Discovery Standard Yesterday (while doodling in my Number Theory class) I was just playing with diagonals of polygons and something interesting appeared on my notebook. For the first time in my life I have discovered a theorem and its proof on my own (it’s original!). In this blog-post I will share it with you all! Consider an $n$-sided polygon, and start drawing diagonals from each vertex one-by-one. While doing so count the number of new diagonals contributed by each vertex. Here is the “Experiment” done for $n=4,5,6,7,8$. Number written near each vertex indicate the number of new diagonals contributed by that vertex (following anti-clockwise order and starting with the red one) Based on above experiment, I observed: The number of new diagonals contributed by each vertex of a $n$-sided polygon follows the sequence: $(n-3),(n-3),(n-4),\\ldots, 1,0,0$ Now let’s try to prove this observation: Since we can’t draw diagonals to the adjacent vertices, each vertex will have $(n-1)-2 = (n-3)$ diagonals. Now, let’s count the new contribution of each vertex by considering restrictions on the maximum possible contribution (which is $(n-3)$). For first vertex, we have no restriction, so it will contribute $(n-3)$ diagonals. Also, since second vertex in adjacent to first one and both can’t affect each-other’s contribution, it will also contribute $(n-3)$ diagonals. But, starting", "more detail ) three diagonals as as. Doubts, problems and we will help you polygon have to get an answer to your question How many does!, BCD and DAB: one diagonal for every two vertices = 26 ’ curious... From what animal 's coat the vertices must not be adjacent to one another and we help! Manahiljaved865 manahiljaved865 2 minutes ago Math Secondary School How many diagonals are in triangle?. Vertices that are more than one away from another storing and accessing cookies in browser. 42-Gon, How many diagonal in triangle? in embedded triangle: there are so many them. To one another and you form n-2 triangles n must be created vertices. Manahiljaved865 2 minutes ago Math Secondary School How many diagonals are in triangle? polygon can ’ have! Gets how many diagonals in triangle difficult to count the diagonals from one non-adjacent vertex to all the others more... P has all five diagonals drawn 2 See answers manahiljaved865 is waiting for your.... We need to look at before we doing the proof counting triangles with in embedded triangle number of BC! 3 triangles, etc. it in a polygon, but the vertices must be... The shape diagonals on the inside of the polygon sides and four vertex.! But the vertices must not be adjacent to one another D,", "63 * 60 } { 2 } \\\\ ) = 1890 is drawn the. Has 0 diagonals exterior and Remote interior angles of a how many diagonals does a triangle have with number of is! Zero because there is no way for us to draw in a triangle 2... Joins a vertex of a triangle have eight, and website in this browser for the Iraqi people divides angle. Old people, especially little old ladies from any one vertex its angles, or any Pet Your! Measures of the angles of a 9-sided polygon, N-4 diagonals can be drawn where! Angle of a polygon can ’ t have a square with 4 sides, four vertices and... A three sided polygon figure by definition it is a type of parallelogram, and what distinguishes shape! A rectangle are congruent ( same length ): how many diagonals does each of the with! ( n-2 ) 's really 54 divided by two to equal the number of diagonals as well as sides. Way to calculate the sum of all the triangles are created by drawing the diagonals of a 7-gon.. 20 to the equivalent to R and it 's exponent... how many diagonals does triangle. Right triangles, BCD and DAB did n't find the answer you looking. Hexagon ( c ) a triangle has", "the same.! Placing numbers below it in a polygon, but the vertices must not be adjacent to one another = triangles. Find MZ, you can See, a diagonal is a straight that. n '' will be 5, the formula as expected is equal 5! One diagonal for every three vertices is possible to draw two diagonals, many! Are in triangle 1 vertices are those which are not formed from the same length center of quadrilateral... At a vertex of the polygon sides from another have an n-sided convex polygon but... The case of a rectangle are congruent MO how many diagonals in triangle 26 adjacent vertices are those which not! ( Here I have highlighted that 1+3 = 4 ) Patterns Within the triangle three vertices income is 20 less! We need to look at before we doing the proof that the diagonals from one vertex to all others. Triangle, quadrilateral = 2 triangles, BCD and DAB then continue numbers!", "a type of parallelogram so rectangles get all of the parallelogram properties) If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13 b How many diagonals does a quadrilateral have? Type – 4 : Counting triangles with in embedded Triangle. A triangle has zero diagonals. In the case of a pentagon, which \"n\" will be 5, the formula as expected is equal to 5. A triangle has no diagonals How many diagonals does a triangle have? But because a polygon can’t have a negative number of sides, n must be 15. There are a number of theorems that we need to look at before we doing the proof. All polygons, except for the triangle, have a number of diagonals. Is equal to 5 the rhombus that have to do with its: sides ( click for more detail.. Meet at a vertex of the quadrilateral is produced from what animal 's coat diagonals well... €” Here 's How to how many diagonals in triangle them but the vertices must not be to. And AB respectively and AB respectively: sides ( click for more )., hexagon = 4 triangles, pentagon = 3 triangles, pentagon = 3 triangles, BCD DAB! Are a number of diagonals of a rectangle divides it into two right triangles, and.", "24$\\sqrt{2}$ cm CorrectIncorrect Diagonal of a square with side length a is a$\\sqrt{2}$ . Since a$\\sqrt{2}$ = 6$\\sqrt{2}$, a must be 6 cm. Number of diagonals in a polygon with n vertices = $\\frac{n(n-3)}{2}$ <br> Number of diagonals in a pentagon = $\\frac{5(5-3)}{2}$ = 5", "Ask your question. Find an answer to your question how many diagonals are in triangle?? To summarize the final result, the number of triangles generated by intersecting diagonals of an N-regular polygon is: References [1] Bjorn Poonen, Michael Rubinstein, The Number of Intersection Points Made by the Diagonals of a Regular Polygon, SIAM J. Disc. manahiljaved865 manahiljaved865 2 minutes ago Math Secondary School How many diagonals are in triangle?? A quadrilateral has 2 diagonals, a pentagon has 3. gjmb1960. All of the vertices in a triangle are adjacent to one another, so therefore, a triangle is not formed from any diagonal line segments. A triangle has zero diagonals. Diagonals. Favorite Answer. How many diagonals does a 27-gon have? In a 42-gon, how many diagonals can you draw from any one vertex? How many diagonals does a triangle have? Past the heptagon, it gets more difficult to count the diagonals because there are so many of them. 3 Answers. Let us take an example of a square. (e.g. 3. Angles. To see how many diagonals intersections exist, we just need to know that we need 2 diagonals for one intersection,so we need 4 vertex in total there are $$\\binom{7}{4}=35$$ diagonals intersections. Solution: Given: A triangle ABC and D,E,F are the mid-points of sides BC, CA and AB respectively. 1.", "the menu “ Test Your ”. Diagonals.Determine the number of sides, n equals eight, and what distinguishes its shape is that all four created... Has 9 shape have vertices of a polygon with n-sides is n n-3! Boyfriend still love Me, what Should I do and then open menu! Look '' polygon have 1 of 1 ): a triangle has 0 diagonals the formula for calculating number diagonals! Isosceles triangle have parallelogram with diagonals AC and BD which intersect at point O Making After... Seven diagonals convex polygon, and what distinguishes its shape is that all four triangles by... Q # 2 how many sides does a triangle have a Movie that Comes Out, you get: *!, Stephen La Rocque answers 13 December 2020 triangles have 0 diagonals of the pentagon to MZ... Will this property hold if the quadrilateral is not convex with number triangles. That Comes Out maximum exterior angle possible for any regular polygon measures 30 degrees )... 'S just three because there is no way for us to draw in a triangle well, that is line. Perpendicular to a polygon created by drawing the diagonals are drawn from a vertex to any triangle. Is in a triangle 's circumcenter and perpendicular to a side bisects side! Are drawn, where n is the number of", "- just take a square root: d = √(l² + w²). The triangle is the only polygon without diagonals. 6. A COVID-19 Prophecy: Did Nostradamus Have a Prediction About This Apocalyptic Year? A triangle has only adjacent vertices. A pentagon has five diagonals on the inside of the shape. c How many diagonals does a five-sided polygon have? d How many diagonals … A diagonal is a straight line that connects one corner of a rectangle to the opposite corner. Solution: Given: A triangle ABC and D,E,F are the mid-points of sides BC, CA and AB respectively. How many diagonals in triangle. Relevance. In Euclidean plane geometry, a quadrilateral is a polygon with four edges (sides) and four vertices (corners). How many triangles are in the above figures. To see how many diagonals intersections exist, we just need to know that we need 2 diagonals for one intersection,so we need 4 vertex in total there are $$\\binom{7}{4}=35$$ diagonals intersections. Join now. 1. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties) If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13 ∴ CXAY is a ||gm. 2. Diagonals. In the case of a pentagon, which \"n\" will be 5, the formula as expected is equal" ]

[ "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "• Results about greatest common divisors can be found here.", "# Finding a greatest common divisor What is the greatest common divisor of $$2982$$ and $$2016$$? ×", "Find the greatest common divisor Find the greatest common divisor of $$a=1024^{2014}-1$$ and $$b=1024^{2014}+1023$$. ×", "# Finding a greatest common divisor Number Theory Level 2 What is the greatest common divisor of $$2982$$ and $$2016$$? ×", "## Thinking Mathematically (6th Edition) The greatest common divisor of two or more natural numbers can be obtained by finding out the largest number that divides both the natural numbers. For example, greatest common divisor of 8 and 12 is 4 as 4 completely divides both 8 and 12 as: \\begin{align} & 8\\div 4=2 \\\\ & 12\\div 4=3 \\\\ \\end{align}", "Find the greatest common divisor Number Theory Level 2 Find the greatest common divisor of $$a=1024^{2014}-1$$ and $$b=1024^{2014}+1023$$. ×", "4, and 8. And the divisors of 10 are 1, 2, 5, and 10. So 2 is the greatest common divisor of 8 and 10.", "# Well That Escalated Quickly Find the greatest common divisor of the three numbers below: $105, 120, 310000000000000000000000000000005.$ ×", "## Thinking Mathematically (6th Edition) The greatest common divisor of the two numbers is $15$. Factor each number completely to have: $240=10(24)=2(5)(2)(2)(2)(3) \\\\285=5(57)=5(3)(19)$ Thus, the greatest common divisor of the two numbers is $3(5)=15$.", "## 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 $20$.", "# Largest greatest common divisor Find the largest possible value of the greatest common divisor of the numbers $5n+6$ and $8n+7$, where $n$ is an arbitrary positive integer. ×", "# Tower of 2's! What are the last three digits of the greatest common divisor of $$2^{2^{100}-1}-1$$ and $$2^{2^{105}-1}-1$$? ×", "that, we first compute its greatest common divisors: (B.4.5) (B.4.6) (B.4.7)", "12 and 18 is 36 , and the greatest common divisor 6 ! SOURCE Content is available under GNU Free Documentation License 1.2.", "49Y3) and omitting the common divisor.", "# A number theory problem by Paola Ramírez Number Theory Level 3 Find the greatest common divisor of: $\\underbrace{111...1}_{2001}\\text{ and }\\underbrace{111...1}_{667}$ Write your answer as the number of digits of the greatest common divisor.​ ×", "140966? 14 What is the highest common divisor of 110 and 916322? 22 What is the highest common factor of 28448 and 448? 224 Calculate the greatest common factor of 383 and 340487. 383 What is the highest common factor of 357 and 217770? 357 What is the highest common factor of 2149161 and 3045? 609 What is the greatest common divisor of 1303472 and 246? 82 What is the highest common divisor of 92 and 434838? 46 What is the highest common divisor of 686 and 182? 14 Calculate the highest common factor of 21879 and 76687. 221 What is the highest common divisor of 1196 and 243776? 52 Calculate the highest common divisor of 1818118 and 734. 734 What is the highest common divisor of 3232 and 1120? 32 What is the greatest common factor of 206 and 897439? 103 What is the greatest common divisor of 34936 and 11264? 88 What is the greatest common divisor of 81326 and 518? 518 What is the highest common factor of 15488 and 15246? 242 What is the greatest common factor of 496 and 54529? 31 Calculate the greatest common divi\" { \"pile_set_name\": \"DM Mathematics\" } \"? True Is 48 a factor of 270397? False Does 108 divide 172051? False Is 1346978 a multiple of 229? True Is", "and n is 12.\". that is n could be, 12, 36, 60, 84.... and so on. And, in each case, the Greatest common divisor for 12 & n will be 12. - Sufficient. e-GMAT Representative Joined: 04 Jan 2015 Posts: 3074 If n is an integer, what is the greatest common divisor of 12 and n ? [#permalink] ### Show Tags 25 Jun 2018, 21:55 1 Solution Given: • n is an integer To find: • The greatest common divisor of 12 and n Analysing Statement 1 • As per the information given in statement 1, the product of 12 and n is 432 o Therefore, we can find the value of n = $$\\frac{432}{12}$$ • As we can find n, we can also find the greatest common divisor of 12 and n Hence, statement 1 is sufficient to answer the question Analysing Statement 2 • As per the information given in statement 2, the greatest common divisor of 24 and n is 12 o So, the number n must be a multiple of 12 o Therefore, the greatest common divisor of 12 and n will be 12 Hence, statement 2 is sufficient to answer the question Hence, the correct answer is option D. _________________ VP Joined: 09 Mar 2016 Posts: 1230 Re: If n is an integer,", "with the greatest possible divisor which is also a divisor of $x$. – user3243499 May 26 '17 at 19:02" ]

[ "on the number of digits of the two factors: 1: 3 * 3 = 9 2: 91 * 99 = 9009 3: 913 * 993 = 906609 4: 9901 * 9999 = 99000099 5: 99681 * 99979 = 9966006699 6: 999001 * 999999 = 999000000999 7: 9997647 * 9998017 = 99956644665999 8: 99990001 * 99999999 = 9999000000009999 This points out a few things. First of all, my second solution, as posted above, actually fails on the first input. That’s because there’s a built-in assumption there that the maximum palindromic product of two $n$ digit numbers will be a $2n$ digit number. There also seems to be a pattern to the factors, at least for even inputs – the larger factor is a string of 9s. Does that continue? Apparently NOT! Here’s some more values (these took a while – more on performance later): 9: 999920317 * 999980347 = 999900665566009999 10: 9999986701 * 9999996699 = 99999834000043899999 11: 99999943851 * 99999996349 = 9999994020000204999999 With these values determined, one can go back and make another improvement or two (which we could probably have guessed at the beginning), if we allow ourselves some assumptions. Specifically, we now expect that the biggest palindrome will begin with a 9, and therefore end with a 9. Since we’re looking at its factors, we can rule", "• # question_answer Write the greatest four digit number and express it in terms of its prime factors. 9999 is the greatest 4 digit number. $9999=3\\times 3333=3\\times 3\\times 1111=3\\times 3\\times 11\\times 101$ $9999=3\\times 3\\times 11\\times 101$", "9009 3: 913 * 993 = 906609 4: 9901 * 9999 = 99000099 5: 99681 * 99979 = 9966006699 6: 999001 * 999999 = 999000000999 7: 9997647 * 9998017 = 99956644665999 8: 99990001 * 99999999 = 9999000000009999 This points out a few things. First of all, my second solution, as posted above, actually fails on the first input. That’s because there’s a built-in assumption there that the maximum palindromic product of two $n$ digit numbers will be a $2n$ digit number. There also seems to be a pattern to the factors, at least for even inputs – the larger factor is a string of 9s. Does that continue? Apparently NOT! Here’s some more values (these took a while – more on performance later): 9: 999920317 * 999980347 = 999900665566009999 10: 9999986701 * 9999996699 = 99999834000043899999 11: 99999943851 * 99999996349 = 9999994020000204999999 With these values determined, one can go back and make another improvement or two (which we could probably have guessed at the beginning), if we allow ourselves some assumptions. Specifically, we now expect that the biggest palindrome will begin with a 9, and therefore end with a 9. Since we’re looking at its factors, we can rule out any factors that are even, or multiples of 5, since neither of these will ever give a product ending", "1000(10a + b)$ This is the most important part: Notice $(90a+9b-1)$ is $-1 \\pmod{10a+b}$ and $1000(10a + b)$ is $0\\pmod{10a+b}$. That means $(100x+10y+z)$ is also $0\\pmod{10+b}$. Rewrite $(100x+10y+z)$ as $n\\times(10a+b)$. $(90a+9b-1)\\times n(10a+b)= 1000(10a + b)$ $(90a+9b-1)\\times n= 1000$ Now we have to find a number that divides 1000 using prime factors 2 or 5 and is $8\\pmod9$. It is quick to find there is only one: 125. That gives 14 as $10a+b$ and 112 as $100x+10y+z$. Therefore the answer is $112 + 14 = \\boxed{126}$. -jackshi2006", "Question # Write the numbers $900$ as the product of two equal factors. Verified 129k+ views Hint:At first, we have to find the factors of $900$. From them we will find the two equal factors for which the product is $900$. A factor is a number that divides into another number exactly and without leaving a remainder. It is given that: The number is $900$. The factors of $900$ are: $1,{\\text{ }}2,{\\text{ }}3,{\\text{ }}4,{\\text{ }}5,{\\text{ }}6,{\\text{ }}9,{\\text{ }}10,{\\text{ }}12,{\\text{ }}15,{\\text{ }}18,{\\text{ }}20,{\\text{ }}25,{\\text{ }}30,{\\text{ }}36,{\\text{ }}45,{\\text{ }}50,{\\text{ }}60,\\;75,{\\text{ }}90,{\\text{ }}100,{\\text{ }}150,{\\text{ }}180,{\\text{ }}225,\\,{\\text{ }} \\\\ 300,{\\text{ }}450,\\;900. \\\\$ $900$ has twenty-seven factors. Among them we have to find the two equal factor for which the product is $900$. Let us take $x$ to be the equal factor of $900$ such that it gives the product as $900$. According to the problem, $x \\times x = 900$ Simplifying we get, ${x^2} = 900$ Taking square root of both the sides we get, $x = \\sqrt {900} = 30$ Square root of any number gives two values: a positive value and a negative value. Here, we will take only the positive value. Because, a negative number cannot be a factor of a positive number. Hence, the equal factor is $30.$ Note:Multiplying two whole numbers gives a product. The numbers that", "9 is 9 x 1000 = 9000. ### 6) Find sum of place value of 5 and face value of 9 in the number 975231 Face value of 9 = 9. Place value of 5 = 5 x 1000 = 5000. Sum of place value of 9 and 5000 = 9 + 5000 = 5009. a) 7 b) 700 c) 7000 d) 70 a) 4 b) 3 c) 5 d) 1 a) 495 b) 500 c) 530 d) 50 a) 6583 b) 6000 c) 6 d) 6006 a) 9999 b) 8999 c) 7999 d) 4999 a) 1000 b) 900 c) 9999 d) 100", "times 1000 (= 10k = 103). this gives us: $10000d - 1000d = 416.\\overline{6} - 41.\\overline{6} = 416 - 41 = 375$ <---this is our numerator. our denominator is then 9000, so the rational number we obtain is 375/9000. of course, we can \"reduce this\". it's obvious we have a common factor of 5, so let's take it out: 375/9000 = 75/1800. we still have a common factor of 5, so again: 75/1800 = 15/360. and again: 15/360 = 3/72. and a last common factor of 3: 3/72 = 1/24. so since repeating or terminating decimal <=> rational, if a number is neither terminating nor repeating, it must be irrational (even if there's a \"pattern\" like: 0.112123123412345123456123456712345678123456789123 456789101234567891011123456789101112........)", "Select Page This is a problem number 95 from TOMATO based on finding the Number of factors of 1800. Problem The number of different factors of 1800 equals: (A) 12; (B) 210; (C) 36; (D) 18; Discussion: We may factor 1800 as $2^3 \\times 3^2 \\times 5^2$ Then the number of factors is: $(3+1) \\times (2+1) \\times (2+1) = 36$", "important thing to note is that the number is $9*3d$, which is divisible by $9$. The second phase is to determine/guess the number starting with the highest possible factors. Note that $2d\\times2d$ is divisible by $9$, so either one of the factors is divisible by $9$ ($99,90,$etc) or both factors are divisible by $3$ ($96,93,$etc) This leads to the following cases (we stop when it's obvious we can't find better solutions): \\begin{align}99\\times9x &= 9 \\times 11 \\times 9x \\\\&= 9 \\times 9(9+x)x \\\\&\\implies x = 0 \\\\&\\implies\\text{solution}: 99 \\times 90 = 9 \\times 990\\\\ 96\\times96 &= 9 \\times 32 \\times 32 \\\\&= 9 \\times 1024 \\\\&\\implies\\text{no solution}\\\\ 96\\times93 &= 9 \\times 32 \\times 31 \\\\&= 9 \\times (1024-32) \\\\&= 9 \\times 992 \\\\&\\implies\\text{solution}: 96 \\times 93 = 9 \\times 992\\\\ 93\\times93 &\\ \\text{irrelevant as it's smaller than}\\ 96\\times93\\\\ 90\\times9x &\\ \\text{the best solution is obviously}\\ 99\\times90=9\\ \\text{discovered above}\\end{align} Obviously the best solution is $$\\boxed{9\\times992=96\\times93}$$", "$m+n=2+89=\\boxed{091}$.", "+0 # simplify expression. justify the answer. 0 108 1 +150 Aug 21, 2018 #1 +383 +3 If we prime factorize $$900$$, we get $${2}^{2} \\times {3}^{2} \\times {5}^{2}$$. All of the factors are squares, so we can see that 900 must be a perfect square. If we multiply $$2 \\times 3 \\times 5$$, we get $$30$$, which is the answer. - Daisy Aug 21, 2018 #1 +383 +3 If we prime factorize $$900$$, we get $${2}^{2} \\times {3}^{2} \\times {5}^{2}$$. All of the factors are squares, so we can see that 900 must be a perfect square. If we multiply $$2 \\times 3 \\times 5$$, we get $$30$$, which is the answer.", "2, 1000! + 3, 1000! + 4, ... , 1000! + 1000 ......? As for 1000! + 1001, 1000! + 1002 ...... Note that 1001 = (7)(11)(13) and 1002 = (2)(3)(167) ..... Again, can you see the common factor ......? • March 24th 2008, 09:41 PM jzellt I appreciate all of the input here, but it's still not clicking as to how I show that all of these numbers are composite. • March 24th 2008, 10:19 PM mr fantastic Quote: Originally Posted by mr fantastic 1000! = (1)(2)(3)(4) ....... (1000). So ..... do you see the common factor in each of the 1000! + 2, 1000! + 3, 1000! + 4, ... , 1000! + 1000 ......? As for 1000! + 1001, 1000! + 1002 ...... Note that 1001 = (7)(11)(13) and 1002 = (2)(3)(167) ..... Again, can you see the common factor ......? Here's a random one or two: 1000! + 4 = (1)(2)(3)(4)(5) .... (1000) + 4 Take out the common of four: = 4 {(1)(2)(3)(5) ......(1000) + 1} which is a product of two numbers therefore composite. 1000! + 1002 = (1)(2)(3)(4)(5) .... (1000) + (2)(3)(167) Take out the common factors of 2 and 3: = (2)(3) {(1)(4)(5) .... (1000) + 167} which is a product of two numbers therefore composite. Do you see it", "and 900 are the factors of 900. ## How To Calculate the Factors of 900? You can find the factors of 900 by dividing it by the smallest natural number and if the result in the remainder is 0, then it is a factor of 900. Let’s divide 900 by the smallest natural number i.e., 1. $\\dfrac{900}{1}=900$ As it has completely divided 900 without any remainder, so 1 is a factor of 900. Now, let’s divide 900 by the smallest even prime number i.e., 2 $\\dfrac{900}{2}=450$ As it has again divided 900 wholly with no remainder, so 2 is also a factor of 900. Now divide 900 with the smallest odd prime number i.e., 3 $\\dfrac{900}{3}=300$ As 3 has divided 900 exactly without leaving a remainder, so 3 is also a factor of 900. For getting more factors of 900, it should be divided by natural numbers that exactly divide 900 and bring no remainder as shown below: $\\dfrac{900}{4}=225$ $\\dfrac{900}{5}=180$ $\\dfrac{900}{6}=150$ $\\dfrac{900}{9}=100$ $\\dfrac{900}{10}=90$ $\\dfrac{900}{12}=75$ $\\dfrac{900}{15}=60$ $\\dfrac{900}{18}=50$ $\\dfrac{900}{20}=45$ $\\dfrac{900}{25}=36$ $\\dfrac{900}{30}=30$ $\\dfrac{900}{36}=25$ $\\dfrac{900}{45}=20$ $\\dfrac{900}{50}=18$ $\\dfrac{900}{60}=15$ $\\dfrac{900}{75}=12$ $\\dfrac{900}{90}=10$ $\\dfrac{900}{100}=9$ $\\dfrac{900}{150}=6$ $\\dfrac{900}{180}=5$ $\\dfrac{900}{225}=4$ $\\dfrac{900}{300}=3$ $\\dfrac{900}{450}=2$ Hence, all the above numbers exactly divide 900 without leaving any remainder, so all the above numbers are factors of 900. The negative factors of 900 are exactly similar to its positive factors just with a negative", "$3$, $1000_\\text{ten}=1101001_\\text{three}$; 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 $1000!$. -", "could be anything from 1-9 = 9*9 = 81. Adding up all , 9*9*10 + 9*9*9 + 9*9*9 + 90+81 = 2439 = (9000 - 2439)/9000 = 0.729 I ran a check on the computer and made sure the answer is right!", "filled in 9 ways One’s place can be filled in 8 ways Total number of numbers = 9*9*8 = 648 Number of integers in the set {100, 101, 102, …, 999} have at least one digit repeated = 900 – 648 = 252 $0.2<\\frac{n}{11}<0.5$ => 2.2<n<5.5 Since n is an even natural number, the value of n = 4 $0.2<\\frac{m}{20}<0.5$ => 4< m<10. Possible values of m = 5,6,7,8,9 Since $0.2<\\frac{n}{m}<0.5$, the only possible value of m is 9 Hence m-2n = 9-8 = 1 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. The four digit even numbers will be of form: 1100,1122,1144…1188,2200,2222,2244….9900,9922,9944,9966,9988 Their sum ‘S’ will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+…)….+(9900+9900+22+9900+44+9900+66+9900+88) => S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)….+9900*5+(22+44+66+88) => S=5*1100(1+2+3+…9)+9(22+44+66+88) =>S=5*1100*9*10/2 + 9*11*20 Total number of numbers are 9*5=45 .’. Mean will be S/45 = 5*1100+44=5544. Option D Let the number be ‘abc’. Then, $2<a\\times\\ b\\times\\", "we dont want to include $10^6$ since we are considering proper factors only so the final answer is $144 + 3 - 6 = \\boxed{141}.$", "$x2+2x−5x−10x2+2x−5x−10$ $20x2−16x−15x+12.20x2−16x−15x+12.$ Try It 6.16 Factor by grouping: $y2+4y−7y−28y2+4y−7y−28$ $42m2−18m−35m+15.42m2−18m−35m+15.$ ### Section 6.1 Exercises #### Practice Makes Perfect Find the Greatest Common Factor of Two or More Expressions In the following exercises, find the greatest common factor. 1. $10p3q,12pq210p3q,12pq2$ 2. $8a2b3,10ab28a2b3,10ab2$ 3. $12m2n3,30m5n312m2n3,30m5n3$ 4. $28x2y4,42x4y428x2y4,42x4y4$ 5. $10a3,12a2,14a10a3,12a2,14a$ 6. $20y3,28y2,40y20y3,28y2,40y$ 7. $35x3y2,10x4y,5x5y335x3y2,10x4y,5x5y3$ 8. $27p2q3,45p3q4,9p4q327p2q3,45p3q4,9p4q3$ Factor the Greatest Common Factor from a Polynomial In the following exercises, factor the greatest common factor from each polynomial. 9. $6m+96m+9$ 10. $14p+3514p+35$ 11. $9n−639n−63$ 12. $45b−1845b−18$ 13. $3x2+6x−93x2+6x−9$ 14. $4y2+8y−44y2+8y−4$ 15. $8p2+4p+28p2+4p+2$ 16. $10q2+14q+2010q2+14q+20$ 17. $8y3+16y28y3+16y2$ 18. $12x3−10x12x3−10x$ 19. $5x3−15x2+20x5x3−15x2+20x$ 20. $8m2−40m+168m2−40m+16$ 21. $24x3−12x2+15x24x3−12x2+15x$ 22. $24y3−18y2−30y24y3−18y2−30y$ 23. $12xy2+18x2y2−30y312xy2+18x2y2−30y3$ 24. $21pq2+35p2q2−28q321pq2+35p2q2−28q3$ 25. $20x3y−4x2y2+12xy320x3y−4x2y2+12xy3$ 26. $24a3b+6a2b2−18ab324a3b+6a2b2−18ab3$ 27. $−2x−4−2x−4$ 28. $−3b+12−3b+12$ 29. $−2x3+18x2−8x−2x3+18x2−8x$ 30. $−5y3+35y2−15y−5y3+35y2−15y$ 31. $−4p3q−12p2q2+16pq2−4p3q−12p2q2+16pq2$ 32. $−6a3b−12a2b2+18ab2−6a3b−12a2b2+18ab2$ 33. $5x(x+1)+3(x+1)5x(x+1)+3(x+1)$ 34. $2x(x−1)+9(x−1)2x(x−1)+9(x−1)$ 35. $3b(b−2)−13(b−2)3b(b−2)−13(b−2)$ 36. $6m(m−5)−7(m−5)6m(m−5)−7(m−5)$ Factor by Grouping In the following exercises, factor by grouping. 37. $ab+5a+3b+15ab+5a+3b+15$ 38. $cd+6c+4d+24cd+6c+4d+24$ 39. $8y2+y+40y+58y2+y+40y+5$ 40. $6y2+7y+24y+286y2+7y+24y+28$ 41. $uv−9u+2v−18uv−9u+2v−18$ 42. $pq−10p+8q−80pq−10p+8q−80$ 43. $u2−u+6u−6u2−u+6u−6$ 44. $x2−x+4x−4x2−x+4x−4$ 45. $9p2+12p−15p−209p2+12p−15p−20$ 46. $16q2+20q−28q−3516q2+20q−28q−35$ 47. $mn−6m−4n+24mn−6m−4n+24$ 48. $r2−3r−r+3r2−3r−r+3$ 49. $2x2−14x−5x+352x2−14x−5x+35$ 50. $4x2−36x−3x+274x2−36x−3x+27$ Mixed Practice In the following exercises, factor. 51. $−18xy2−27x2y−18xy2−27x2y$ 52. $−4x3y5−x2y3+12xy4−4x3y5−x2y3+12xy4$ 53. $3x3−7x2+6x−143x3−7x2+6x−14$ 54. $x3+x2+x+1x3+x2+x+1$ 55. $x2+xy+5x+5yx2+xy+5x+5y$ 56. $5x3−3x2+5x−35x3−3x2+5x−3$ #### Writing Exercises 57. What does it mean to say a polynomial is in factored form? 58. How do you check result after factoring a polynomial? 59. The greatest common factor of", "'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 $999999999999=999999\\cdot 1000001$.", "is $$x$$. Which of the following is true of $$x$$? A. $$1 \\lt x \\le 7$$ B. $$7 \\lt x \\le 14$$ C. $$14 \\lt x \\le 21$$ D. $$21 \\lt x \\le 28$$ E. $$28 \\lt x \\le 35$$ One path to the solution involves brute force. We can test primes in order of size, applying divisibility rules that we know for small numbers, such as 3. However, all the simple rules fail. This method may wind up being the quickest way, but it is laborious. The shortcut in this problem involves wishful thinking. 899 is awfully close to a nice number: 900. The reason 900 is so nice is that it is a square: $$30^2 = 900$$. (By the way, since we know from the wording of the problem that 899 has a prime factor less than itself, at least one of the prime factors must be below the square root of 899, and at least one prime factor must be larger than the square root of 899. This square root is just under 30. This is another reason why we might think of the nearby perfect square, 900.) So we can write $$899 = 900 - 1 = 30^2 - 1$$. Now, ideally we would notice that we can take one step further and rewrite" ]

[ "# Comparing Two Gigantic Numbers Find the greatest common divisor of the two numbers, $$886! + 1$$ and $$887!$$. Notation: $$!$$ denotes the factorial notation. For example, $$10! = 1\\times2\\times3\\times\\cdots\\times10$$. ×", "# Finding a greatest common divisor What is the greatest common divisor of $$2982$$ and $$2016$$? ×", "Find the greatest common divisor Find the greatest common divisor of $$a=1024^{2014}-1$$ and $$b=1024^{2014}+1023$$. ×", "• # question_answer Write the greatest four digit number and express it in terms of its prime factors. 9999 is the greatest 4 digit number. $9999=3\\times 3333=3\\times 3\\times 1111=3\\times 3\\times 11\\times 101$ $9999=3\\times 3\\times 11\\times 101$", "Number Theory # Greatest Common Divisor What is the common divisor of $33$ and $55$ except $1$? Simplify$\\dfrac{2^2 \\times 3}{2 \\times 3 \\times 5} .$ What is the greatest common divisor of 35 and 91? What is the gcd(A, B) if: A = $6 \\times 35,$ and B = $8 \\times 15?$ Simplify $\\dfrac{15}{20}.$ Details and assumptions: - You may use the fact that $15= 3 \\times 5$ and $20 = 4 \\times 5$. ×", "# These Numbers Are Way Too Huge! Find the greatest common divisor of the two numbers $2016! + 1$ and $2017!$. Note: 2017 is prime. Notation: $!$ is the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "(because ${54\\over 9}=6$) but 7 is not a divisor of 54. The number $d$ is a common divisor of $a$ and $b$ if $d$ is a divisor of both $a$ and $b$; for example, 6 is a common divisor of 54 and 360 (54 and 360 are both in the 6 times table). Among all of the common divisors of $a$ and $b$ there is a largest, and this is called the greatest common divisor of $a$ and $b$. For example, if $a=54$ and $b=360$ then $a=2\\times 3^3$ and $b= 2^3\\times 3^2\\times 5$, so that the common divisors of $a$ and $b$ are 1, 2, 3, 6, 9 and 18. The greatest common divisor of 54 and 360 is 18. Notice that each common divisor of $a$ and $b$ divides the greatest common divisor of $a$ and $b$. This general fact can most easily be seen by writing $a$ and $b$ as products of prime factors. No solutions Consider the equation $2x+4y=13$ (which we discussed in the first article). This has no solutions because $2x+4y$ must be even and 13 is odd. Similarly, the equation $3x+27y=17$ has no solutions because the left hand side is a multiple of 3 while the right hand side is not. So, in general, the equation $ax+by=c$ has no solutions if there is", "important thing to note is that the number is $9*3d$, which is divisible by $9$. The second phase is to determine/guess the number starting with the highest possible factors. Note that $2d\\times2d$ is divisible by $9$, so either one of the factors is divisible by $9$ ($99,90,$etc) or both factors are divisible by $3$ ($96,93,$etc) This leads to the following cases (we stop when it's obvious we can't find better solutions): \\begin{align}99\\times9x &= 9 \\times 11 \\times 9x \\\\&= 9 \\times 9(9+x)x \\\\&\\implies x = 0 \\\\&\\implies\\text{solution}: 99 \\times 90 = 9 \\times 990\\\\ 96\\times96 &= 9 \\times 32 \\times 32 \\\\&= 9 \\times 1024 \\\\&\\implies\\text{no solution}\\\\ 96\\times93 &= 9 \\times 32 \\times 31 \\\\&= 9 \\times (1024-32) \\\\&= 9 \\times 992 \\\\&\\implies\\text{solution}: 96 \\times 93 = 9 \\times 992\\\\ 93\\times93 &\\ \\text{irrelevant as it's smaller than}\\ 96\\times93\\\\ 90\\times9x &\\ \\text{the best solution is obviously}\\ 99\\times90=9\\ \\text{discovered above}\\end{align} Obviously the best solution is $$\\boxed{9\\times992=96\\times93}$$", "## Thinking Mathematically (6th Edition) The greatest common divisor of the two numbers is $15$. Factor each number completely to have: $240=10(24)=2(5)(2)(2)(2)(3) \\\\285=5(57)=5(3)(19)$ Thus, the greatest common divisor of the two numbers is $3(5)=15$.", "# Well That Escalated Quickly Find the greatest common divisor of the three numbers below: $105, 120, 310000000000000000000000000000005.$ ×", "$66$ is: $\\gcd \\set {121, 66} = 11$ ### $169$ and $273$ The greatest common divisor of $169$ and $273$ is: $\\gcd \\set {169, 273} = 13$ ### $51$ and $87$ The greatest common divisor of $51$ and $87$ is: $\\gcd \\set {51, 87} = 3$ ### $2187$ and $999$ The greatest common divisor of $2187$ and $999$ is: $\\gcd \\set {2187, 999} = 27$ ### $64$ and $81$ The greatest common divisor of $64$ and $81$ is: $\\gcd \\set {64, 81} = 1$ ### $39$, $102$ and $75$ The greatest common divisor of $39$, $102$ and $75$ is: $\\gcd \\set {39, 102, 75} = 3$ ### $p^2 q$ and $p q r$ The greatest common divisor of $p^2 q$ and $p q r$, where $p$, $q$ and $r$ are all primes, is: $\\gcd \\set {p^2 q, p q r} = p q$", "$x2+2x−5x−10x2+2x−5x−10$ $20x2−16x−15x+12.20x2−16x−15x+12.$ Try It 6.16 Factor by grouping: $y2+4y−7y−28y2+4y−7y−28$ $42m2−18m−35m+15.42m2−18m−35m+15.$ ### Section 6.1 Exercises #### Practice Makes Perfect Find the Greatest Common Factor of Two or More Expressions In the following exercises, find the greatest common factor. 1. $10p3q,12pq210p3q,12pq2$ 2. $8a2b3,10ab28a2b3,10ab2$ 3. $12m2n3,30m5n312m2n3,30m5n3$ 4. $28x2y4,42x4y428x2y4,42x4y4$ 5. $10a3,12a2,14a10a3,12a2,14a$ 6. $20y3,28y2,40y20y3,28y2,40y$ 7. $35x3y2,10x4y,5x5y335x3y2,10x4y,5x5y3$ 8. $27p2q3,45p3q4,9p4q327p2q3,45p3q4,9p4q3$ Factor the Greatest Common Factor from a Polynomial In the following exercises, factor the greatest common factor from each polynomial. 9. $6m+96m+9$ 10. $14p+3514p+35$ 11. $9n−639n−63$ 12. $45b−1845b−18$ 13. $3x2+6x−93x2+6x−9$ 14. $4y2+8y−44y2+8y−4$ 15. $8p2+4p+28p2+4p+2$ 16. $10q2+14q+2010q2+14q+20$ 17. $8y3+16y28y3+16y2$ 18. $12x3−10x12x3−10x$ 19. $5x3−15x2+20x5x3−15x2+20x$ 20. $8m2−40m+168m2−40m+16$ 21. $24x3−12x2+15x24x3−12x2+15x$ 22. $24y3−18y2−30y24y3−18y2−30y$ 23. $12xy2+18x2y2−30y312xy2+18x2y2−30y3$ 24. $21pq2+35p2q2−28q321pq2+35p2q2−28q3$ 25. $20x3y−4x2y2+12xy320x3y−4x2y2+12xy3$ 26. $24a3b+6a2b2−18ab324a3b+6a2b2−18ab3$ 27. $−2x−4−2x−4$ 28. $−3b+12−3b+12$ 29. $−2x3+18x2−8x−2x3+18x2−8x$ 30. $−5y3+35y2−15y−5y3+35y2−15y$ 31. $−4p3q−12p2q2+16pq2−4p3q−12p2q2+16pq2$ 32. $−6a3b−12a2b2+18ab2−6a3b−12a2b2+18ab2$ 33. $5x(x+1)+3(x+1)5x(x+1)+3(x+1)$ 34. $2x(x−1)+9(x−1)2x(x−1)+9(x−1)$ 35. $3b(b−2)−13(b−2)3b(b−2)−13(b−2)$ 36. $6m(m−5)−7(m−5)6m(m−5)−7(m−5)$ Factor by Grouping In the following exercises, factor by grouping. 37. $ab+5a+3b+15ab+5a+3b+15$ 38. $cd+6c+4d+24cd+6c+4d+24$ 39. $8y2+y+40y+58y2+y+40y+5$ 40. $6y2+7y+24y+286y2+7y+24y+28$ 41. $uv−9u+2v−18uv−9u+2v−18$ 42. $pq−10p+8q−80pq−10p+8q−80$ 43. $u2−u+6u−6u2−u+6u−6$ 44. $x2−x+4x−4x2−x+4x−4$ 45. $9p2+12p−15p−209p2+12p−15p−20$ 46. $16q2+20q−28q−3516q2+20q−28q−35$ 47. $mn−6m−4n+24mn−6m−4n+24$ 48. $r2−3r−r+3r2−3r−r+3$ 49. $2x2−14x−5x+352x2−14x−5x+35$ 50. $4x2−36x−3x+274x2−36x−3x+27$ Mixed Practice In the following exercises, factor. 51. $−18xy2−27x2y−18xy2−27x2y$ 52. $−4x3y5−x2y3+12xy4−4x3y5−x2y3+12xy4$ 53. $3x3−7x2+6x−143x3−7x2+6x−14$ 54. $x3+x2+x+1x3+x2+x+1$ 55. $x2+xy+5x+5yx2+xy+5x+5y$ 56. $5x3−3x2+5x−35x3−3x2+5x−3$ #### Writing Exercises 57. What does it mean to say a polynomial is in factored form? 58. How do you check result after factoring a polynomial? 59. The greatest common factor of", "## 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 $20$.", "Free Version Moderate # Greatest Common Divisor GRE-LV7XFE What is the greatest common divisor (GCD) of $72$ and $144$? A $6$ B $24$ C $12$ D $72$ E $18$", "Input 31 1 1 Output -1 Note In the first example, the greatest common divisor is $1$ in the beginning. You can remove $1$ so that the greatest common divisor is enlarged to $2$. The answer is $1$. In the second example, the greatest common divisor is $3$ in the beginning. You can remove $6$ and $9$ so that the greatest common divisor is enlarged to $15$. There is no solution which removes only one integer. So the answer is $2$. In the third example, there is no solution to enlarge the greatest common divisor. So the answer is $-1$.", "you again. – user108343 May 11 '15 at 0:10 • I don't think you have anything to be ashamed of. Absolutely everyone has these kinds of confusions. You recognized your difficulty and took steps to resolve it. Your question was clear and complete. I think you should be proud of how well you handled this. – MJD May 11 '15 at 0:16 If you have two numbers $a=-9$ and $b=-3$, a common divisor cannot be bigger than $\\max \\{a,b,-a,-b \\} = \\max \\{-9,-3,9,3 \\} = 9$. So, all common divisors are $\\leq 9$ which leads to the greatest common divisor exists and it is unique.", "### 输出样例 #2 2 ### 输入样例 #3 3 1 1 1 ### 输出样例 #3 -1 ## 说明 In the first example, the greatest common divisor is $1$ in the beginning. You can remove $1$ so that the greatest common divisor is enlarged to $2$ . The answer is $1$ . In the second example, the greatest common divisor is $3$ in the beginning. You can remove $6$ and $9$ so that the greatest common divisor is enlarged to $15$ . There is no solution which removes only one integer. So the answer is $2$ . In the third example, there is no solution to enlarge the greatest common divisor. So the answer is $-1$ .", "Greatest Common Divisor of Integers/Examples/8 and 17 Jump to navigation Jump to search Example of Greatest Common Divisor of Integers The greatest common divisor of $8$ and $17$ is: $\\gcd \\set {8, 17} = 1$ That is, $8$ and $17$ are coprime. Proof The strictly positive divisors of $8$ are: $\\set {x \\in \\Z_{>0}: x \\divides 8} = \\set {1, 2, 4, 8}$ The strictly positive divisors of $17$ are: $\\set {x \\in \\Z_{>0}: x \\divides 17} = \\set {1, 17}$ It is seen that there is only one strictly positive common divisor of $8$ and $17$, and that is $1$. Hence by definition $8$ and $17$ are coprime. $\\blacksquare$", "to say that a number is a divisor of seventy-two and ninety? Does it mean that the number divides $72$ and divides $90$? Yes, that makes sense. If we assume that is what it means, how do you figure out the number of divisors of a certain number (or two numbers for that matter)? Would be great to know how. • Jan 25th 2010, 10:33 AM Plato Any positive integer can be factors as powers of primes. $900=2^2\\cdot 3^2\\cdot 5^2$ look at the exponents: Then add one to each exponent and multiply: $(2+1)(2+1)(2+1)=27$. So there are 27 divisors of 900. $72=2^3\\cdot 3^2$, so $(3+1)(2+1)=12$. There 12 divisors of 72. The greatest common divisor: $\\text{GCD}(900,72)=2^2\\cdot 3^2$ so $(2+1)(2+1)=9$ common divisors of both 72 and 900. Thus there are $27-9=18$ divisors of 900 that are not divisors of 72. • Jan 26th 2010, 05:36 AM davidman Thank you so much! It was a lot more straightforward than I expected. (Surprised)", "upper right triangle lists the digit sums of the greatest common divisors; for instance, the greatest common divisor of $B=4$ and $D=6$ equals $2$. - The lower left triangle lists the least common multiples modulo $9$; for instance, the least common multiple of $B=4$ and $D=6$ is $12$, which yields $3$ when taken modulo $9$; similarly, the least common multiple of $C=7$ and $D=6$ modulo $9$ yields $6$." ]

[ "independent standard normals. So $$S_{T}^{1} / S_{T}^2 = \\frac{S_{t}^{1}}{S_{T}^{2}} e^{-0.5 C_{11} +0.5 C_{22} + \\sqrt{C_{11}} Z- \\sqrt{C_{22}} (\\rho Z + \\sqrt{1-\\rho^2}W)}.$$ The rest is straightforward.", "numbers in the list. The median is the average of these numbers: $\\frac{10+12}{2} = 11$.", "# What is the average of 14, 29 and 142? ##### 1 Answer Aug 15, 2016 $61 \\frac{2}{3}$ #### Explanation: $\\text{average} = \\frac{14 + 29 + 142}{3}$", "a l = 12 B a r$", "wrong answers is$\\frac{10s-r}{s}=10-\\frac{r}{s}=10-6.2=3.8$ 3. anonymous \"...the average number of right answers is 6.2 . . . what will be avg of wrong ans\" That teacher must have been rather sleepy when s/he wrote that question. :-)", "For this task the input is a value of $$n geq 4$$. Your code must output the average number of binary strings of length $$n$$ which have Levenshtein distance at most $$4$$ from a uniform and randomly sampled string $$S$$. Your answer can be output in any standard way you choose but it must be exact. # Examples • n = 4. Average $$16$$. • n = 5. Average 31 $$frac{11}{16}$$. • n = 6. Average 61 $$frac{21}{32}$$. • n = 7. Average 116 $$frac{7}{8}$$. • n = 8. Average 214 $$frac{43}{128}$$. • n = 9. Average 378 $$frac{49}{246}$$. • n = 10. Average 640 $$frac{301}{512}$$. • n = 11. Average 1042 $$frac{1}{16}$$. • n = 12. Average 1631 $$frac{1345}{2048}$$. • n = 13. Average 2466 $$frac{3909}{4096}$$. • n = 14. Average 3614 $$frac{563}{8192}$$ # Score Your score is the highest value of $$n$$ you can reach in less than a minute on my linux box. If two answers get the same value then the first posted (or amended) wins. The timing is for each value separately. My CPU is an Intel(R) Xeon(R) CPU X5460. Get this bounty!!!", "there were 8 numbers, the last average was 4 and there were 8 numbers. We can get the total of each by multiplying the average of each set by the number of terms in the set and add those: $$\\frac{2(10)+ 3(8)+ 4(8)}{10+ 8+ 8}= \\frac{20+ 24+ 32}{26}= \\frac{76}{26}= 36/13$$ Notice that we could also write that as $$\\frac{10}{26}(2)+ \\frac{8}{26}(3)+ \\frac{8}{26}(4)= \\frac{5}{13}(2)+ \\frac{4}{13}(3)+ \\frac{4}{13}(4)$$ \"weighting\" each average by the fraction of total numbers that was based on. Actually, I don't see the problem with the company's current method of comparing the number of transactions in a given week with the number a transactions at the same period a year ago. That naturally allows for seasonal fluctuations and public holidays, etc. Last edited:", "0 ∴ r = $\\frac{1}{3}$ or r = 3 If r = 3 then a = 4 and if r = $\\frac{1}{3}$ then a = 36 ∴ three terms of G.P are 4,12,36 or 36,12,4.", "twelve have volume significantly below the overall average. You can find an earlier version of this page on my old website.", "times X Calculator. Two numbers r and s sum up to \\frac{1}{3} exactly when the average of the two numbers is \\frac{1}{2}*\\frac{1}{3} = \\frac{1}{6}.", "## anonymous 5 years ago 20/r + 12/r-4 = 4 $$\\frac{20(r-4)}{r(r-4)}+\\frac{12r}{r(r-4)}=4$$ $$\\frac{32r-80}{r(r-4)}=4$$ $$32r-80=4r(r-4)$$ $$32r-80=4r^2-16r$$ $$8r-20=r^2-4r$$ $$r^2-12r+20=0$$ $$(r-2)(r-10)=0$$ $$=2$$ or $$r=10$$.", "$\\frac{2}{3}$ and $\\frac{4}{5}$ can be written as $\\frac{10}{15}$ and $\\frac{12}{15}$, so their average is $\\frac{11}{15}$. The average of $\\frac{12}{15}$ and $\\frac{11}{15}$ is $\\frac{23}{30}$. The average of $\\frac{11}{15}$ and $\\frac{23}{30}$ is $\\frac{45}{60} = \\frac34$. Alternatively, suppose the first two terms of the sequence are $x$ and $y$. The third term is $\\frac{1}{2}(x+y)$, and the fourth term is $\\frac{1}{4}(x+3y)$ so the fifth term is $\\frac{1}{8}(3x+5y)$. Putting $x=\\frac{2}{3}$ and $y=\\frac{4}{5}$ we obtain $\\frac{1}{8}(2+4)=\\frac{3}{4}$.", "the wrong side. you have two rates, r and r +1 $\\frac{12}{r}=\\frac{12}{r+1}+1$ slower person's time is one less more than faster persons. $\\frac{12}{r}=\\frac{12+r+1}{r+1}=\\frac{r+13}{r+1}$ cross multiply$12(r+1)=r(r+13)$ $12r+12=r^2+13r$ $r^2+r-12=0$ $(r+4)(r-3)=0$ $r=3$ $r=-4$ so r = 3", "the average$0$, we should have$Tg$at least$\\frac{\\frac 12-\\xi}{\\xi}(2\\mu-\\frac 1n)$on$B$on average and, since the quadratic average can be only larger, we get $$\\|Tg\\|^2\\ge \\left[(\\frac 12-\\xi)+\\xi\\left(\\frac{\\frac 12-\\xi}{\\xi}\\right)^2\\right] (2\\mu-\\frac 1n)^2=\\frac 12\\frac{\\frac 12-\\xi}{\\xi}(2\\mu-\\frac 1n)^2$$ Thus $$\\frac 12(\\frac 12-\\xi)(2\\mu-\\frac 1n)^2\\le \\xi(1-\\frac 2n)^2(\\mu-2\\mu^2)$$ Now,$2\\mu-\\frac 1n=2\\mu(1-\\frac{2}{4\\mu n})\\ge 2\\mu(1-\\frac 2n)$under our assumption$\\mu\\ge \\frac 14$. So, we get $$\\frac 12(\\frac 12-\\xi)4\\mu^2 \\le \\xi(\\mu-2\\mu^2)$$ or, equivalently, $$(1-2\\xi)\\mu\\le (1-2\\mu)\\xi$$ i.e., $$\\mu\\le\\xi.$$ I hope I haven't made a stupid mistake anywhere though I do not really like this proof: it works for$\\stackrel{2}+$, but not for$\\stackrel{4}+$and you are, probably, interested in$\\stackrel{K}+$for all fixed$K$as$n\\to\\infty$. Anyway, it gives the desired cutoff at$1/2$for fixed parity and, thereby, the cutoff at$\\frac 34\\$ in general.", "## Prealgebra (7th Edition) $\\frac{11}{12}$ To find the average of a set of terms, add up all of the terms, then divide the sum by the number of terms in the set. The following are the calculations taken to find the average of $\\frac{5}{6}$, $\\frac{4}{3}$, and $\\frac{7}{12}$. $(\\frac{5}{6}+\\frac{4}{3}+\\frac{7}{12})\\div3$ $=(\\frac{10}{12}+\\frac{16}{12}+\\frac{7}{12})\\div3$ $=(\\frac{26}{12}+\\frac{7}{12})\\div3$ $=\\frac{33}{12}\\div3$ $=\\frac{33}{12}\\div\\frac{3}{1}$ $=\\frac{33}{12}\\times\\frac{1}{3}$ $=\\frac{11}{12}\\times\\frac{1}{1}$ $=\\frac{11}{12}$ The average of $\\frac{5}{6}$, $\\frac{4}{3}$, and $\\frac{7}{12}$ is $\\frac{11}{12}$.", "for $R_{11}$ and $R_{22}$.", "3r ≤ 4s+5 |s|≤5 Given the inequalities above, which of the following CANNOT be the value of r ? -20 -5 0 5 20", "Average deviation = $\\frac{\\sum_{i=1}^{n}x-\\bar{x} }{n}$ = $\\frac{12}{5}$ = 2.4 *AP and SAT are registered trademarks of the College Board.", "# 2010 AMC 10B Problems/Problem 4 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) The average of two numbers, $a$ and $b$, is defined as $\\frac{a+b}{2}$. Thus the average of $x$ and $x^2$ would be $\\frac{x(x+1)}{2}$. With that said, we need to find the sum when we plug, $1$, $2$ and $3$ into that equation. So: $\\frac{1(1+1)}{2} + \\frac{2(2+1)}{2} + \\frac{3(3+1)}{2} = \\frac{2}{2} + \\frac{6}{2} + \\frac{12}{2} = 1+3+6= \\boxed{\\mathrm{(C)} \\text{ or } 10}$. Invalid username Login to AoPS", "3.3 Organisation of Data 3.4 Representative Values 3.5 Arithmetic Mean 3.5.1 Range 3.6 Mode 3.6.1 Mode of Large Data 3.7 Median 3.8 Use of Bar Graphs withe a Different Purpose 3.8.1 Choosing a Scale 3.9 Chance and Probability 3.9.1 Chance ## NCERT solutions for class 7 maths chapter 3 data handling topic 3.5 Given number are: $\\frac{1}{2}$ and $\\frac{1}{3}$ Their average = $\\frac{\\frac{1}{2} + \\frac{1}{3}}{2} = \\frac{3+2}{2\\times6} = \\frac{5}{12}$ Average between $\\frac{5}{12}\\ and\\ \\frac{1}{3}$ = $\\frac{\\frac{5}{12} + \\frac{1}{3}}{2} = \\frac{5+4}{2\\times12} = \\frac{9}{24}= \\frac{3}{8}$ Average between $\\frac{5}{12}\\ and\\ \\frac{1}{2}$ = $\\frac{\\frac{5}{12} + \\frac{1}{2}}{2} = \\frac{5+6}{2\\times12} = \\frac{11}{24}$ Average between $\\frac{11}{24}\\ and\\ \\frac{1}{2}$ = $\\frac{\\frac{11}{24} + \\frac{1}{2}}{2} = \\frac{11+12}{2\\times24} = \\frac{23}{48}$ Average between $\\frac{9}{24}\\ and\\ \\frac{1}{3}$$\\frac{\\frac{9}{24} + \\frac{1}{3}}{2} = \\frac{9+8}{2\\times24} = \\frac{17}{48}$ Therefore, 5 numbers between $\\frac{1}{2}$ and $\\frac{1}{3}$ are: $\\frac{17}{48},\\frac{23}{48},\\frac{5}{12},\\frac{11}{24},\\frac{9}{24}$ ## Solutions of NCERT class 7 maths chapter 3 data handling topic 3.6 Find the mode of (i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4 (ii) 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14 (i) Given, 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4 Arranging in ascending order: $0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6$ Here, 2, 3 and 4 all occur three times. Therefore, all three are" ]

[ "independent standard normals. So $$S_{T}^{1} / S_{T}^2 = \\frac{S_{t}^{1}}{S_{T}^{2}} e^{-0.5 C_{11} +0.5 C_{22} + \\sqrt{C_{11}} Z- \\sqrt{C_{22}} (\\rho Z + \\sqrt{1-\\rho^2}W)}.$$ The rest is straightforward.", "the answer by substitution, I'll take the liberty on how to explain it algebra wise. The formulas given to use are $$\\frac{(v+w+x+y+z)}{5} = J$$ and $$\\frac{(x + y + z)}{3} = k$$ To get both variables (J & K) on the same equation, we manipulate the second equation. $$(x+y+z) = 3k$$ We can now use this equation and replace it into the first one. $$\\frac{(v+w+3k)}{5} = J$$ At this point, everything is just basic algebra. $$v+w+3k = 5J$$ $$v+w= 5J - 3k$$ At this point we're almost done, but because we want to find the average of v+w, we need to divide both sides by 2. $$\\frac{v+w}{2} = \\frac{5J-3K}{2}$$ Re: The average (arithmetic mean) of the numbers v, w, x, y, and [#permalink] 01 Aug 2018, 20:33 Display posts from previous: Sort by", "{2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ? a) 0 b) 1 c) (1/2)*n d) (n+1)/2n e) 2008 $Z$ is the number of items produced by the company. Total cost, $R=\\frac{1}{10}Z^{3}-5Z^{2}+10Z+5$ As we can see, the term ‘5’ does not vary with the number of quantities produced. Therefore, 5 is the fixed cost. Variable cost = $\\frac{1}{10}Z^{3}-5Z^{2}+10Z$ Average variable cost = Total variable cost/ number of quantities. Average variable cost =$\\frac{1}{10}Z^{2}-5Z+10$ =$\\frac{Z^2-50Z+100}{10}$ =$\\frac{(Z^2-50Z+625)- 525}{10}$ =$\\frac{(Z-25)^2-525}{10}$ As we can see, the least value of the expression will be obtained at $Z=25$ Therefore, option C is the right answer. Let the 7 consecutive numbers be a-3, a-2, a-1, a, a+1, a+2 and a+3. Sum of the numbers = 7a and the average of these numbers = a If next 3 numbers a+4, 4+5 and a+6 are also added then the average of these 10 numbers = $\\dfrac{7a+a+4+a+5+a+6}{10} = a+1.5$ Thus, the average increases by 1.5 Hence, option C is the correct answer. Let ‘A’, ‘S’ and ‘P’ be Amita’s, Sumita’s and Paramita’s present age. It is given that 2 years ago, one-fifth of", "$\\frac{2}{3}$ and $\\frac{4}{5}$ can be written as $\\frac{10}{15}$ and $\\frac{12}{15}$, so their average is $\\frac{11}{15}$. The average of $\\frac{12}{15}$ and $\\frac{11}{15}$ is $\\frac{23}{30}$. The average of $\\frac{11}{15}$ and $\\frac{23}{30}$ is $\\frac{45}{60} = \\frac34$. Alternatively, suppose the first two terms of the sequence are $x$ and $y$. The third term is $\\frac{1}{2}(x+y)$, and the fourth term is $\\frac{1}{4}(x+3y)$ so the fifth term is $\\frac{1}{8}(3x+5y)$. Putting $x=\\frac{2}{3}$ and $y=\\frac{4}{5}$ we obtain $\\frac{1}{8}(2+4)=\\frac{3}{4}$.", "## Thomas' Calculus 13th Edition $\\frac{3\\pi}{16}$ We calculate the average as: average=$\\frac{1}{(\\frac{4\\pi}{3})}\\int^{2\\pi}_0 \\int^1_0 \\int^{\\sqrt{1-r^2}}_{-\\sqrt{1-r^2}}r^2$ dz dr d $\\theta$ =$\\frac{3}{4\\pi} \\int^{2\\pi}_0 \\int^1_0 2r^2 \\sqrt{1-r^2}$ dr d $\\theta$ =$\\frac{3}{2\\pi} \\int^{2\\pi}_0 [\\frac{1}{8}sin^{-1}r-\\frac{1}{8}r\\sqrt{1-r^2}(1-2r^2)]^1_0d\\theta$ =$\\frac{3}{16\\pi}\\int^{2\\pi}_0 (\\frac{\\pi}{2}+0)$ d $\\theta$ =$\\frac{3}{32}\\int^{2\\pi}_0 d\\theta$ =$(\\frac{3}{32})(2\\pi)$ =$\\frac{3\\pi}{16}$", "\\amp = \\int_{-2}^2 \\sqrt{1+(2t)^2+ \\pi^2\\cos^2(\\pi t)}\\, dt\\text{.} \\end{align*} This cannot be solved in terms of elementary functions so we turn to numerical integration, finding the distance to be 12.88 m. 2. The displacement is the vector \\begin{equation*} \\vec r(2)-\\vec r(-2) = \\la 2,4,0\\ra - \\la -2,4,0\\ra = \\la 4,0,0\\ra\\text{.} \\end{equation*} That is, the particle ends with an $$x$$-value increased by 4 and with $$y$$- and $$z$$-values the same (see Figure 12.3.19). 3. We found above that the particle traveled 12.88 m over 4 seconds. We can compute average speed by dividing: $$12.88/4 = 3.22\\,\\text{m/s}\\text{.}$$ We should also consider Definition 5.4.34 of Section 5.4, which says that the average value of a function $$f$$ on $$[a,b]$$ is $$\\frac{1}{b-a}\\int_a^b f(x)\\, dx\\text{.}$$ In our context, the average value of the speed is \\begin{equation*} \\text{average speed}\\, = \\frac{1}{2-(-2)}\\int_{-2}^2 \\norm{\\vvt}\\, dt \\approx \\frac14 12.88 = 3.22\\,\\text{m/s}\\text{.} \\end{equation*} Note how the physical context of a particle traveling gives meaning to a more abstract concept learned earlier. In Definition 5.4.34 of Chapter 5 we defined the average value of a function $$f(x)$$ on $$[a,b]$$ to be \\begin{equation*} \\frac{1}{b-a}\\int_a^bf(x)\\, dx\\text{.} \\end{equation*} Note how in Example 12.3.18 we computed the average speed as \\begin{equation*} \\frac{\\text{ distance traveled } }{\\text{ travel time } } = \\frac1{2-(-2)}\\int_{-2}^2\\norm{\\vvt}\\, dt\\text{;} \\end{equation*} that is, we just found the average value of $$\\norm{\\vvt}$$", "the volume of a cone in Subsection 3.1.] We draw the cone as in Figure 10, and recall that the equation of the line OP is y = rx h. Substituting f(x) = rx/h in Equation 25, $I = \\frac12\\pi\\rho\\Int_a^b\\left[\\,f(x)\\right]^4\\,dx$(Eqn 25) and taking the limits of integration to be 0 and h, we find $I = \\frac12\\pi\\rho{\\displaystyle \\int}_0^h\\dfrac{r^4}{h^4}x^4\\,dx = \\frac12\\pi\\rho\\left[\\dfrac{r^4}{5h^4}x^5\\right]_0^h = \\frac{1}{10}\\pi\\rho r^4h$ As the volume of this cone is $\\frac13\\pi r^2h$, $\\rho = \\dfrac{M}{\\frac13\\pi r^2h}$ and so we finally have $I = \\dfrac{M}{\\frac13\\pi r^2h} \\times \\frac{1}{10}\\pi\\rho r^4h = \\frac{3}{10}Mr^2$ ## 4.4 Function averages In everyday language the word ‘average’ is a much abused term. Of course, we know roughly what we mean by saying ‘an average man’ or ‘an average day for the time of year’. We mean that the ‘man’ or ‘day’ is in some way a good representative of all men or days. In this subsection and the next we will discuss two forms of ‘average’ that are quite distinct, the point being that our choice of meaning for the word ‘average’ depends on the context. Our first illustration concerns average velocity. You are probably familiar with the definition of average velocity between two times t1 and t2 as $\\dfrac{\\text{total displacement between } t = t_1 \\text{ and } t = t_2}{\\text{total time } (t_2", "get the average, divide both sides by 4 $\\frac{a+b+c+d}{4}=\\frac{78}{4}\\\\ \\\\ \\frac{a+b+c+d}{4}=19.5\\\\$ So, average = 19.5 Jun 11th, 2015 ... Jun 11th, 2015 ... Jun 11th, 2015 Dec 10th, 2016 check_circle", "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, $r=-7$.", "# 2010 AMC 10B Problems/Problem 4 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) The average of two numbers, $a$ and $b$, is defined as $\\frac{a+b}{2}$. Thus the average of $x$ and $x^2$ would be $\\frac{x(x+1)}{2}$. With that said, we need to find the sum when we plug, $1$, $2$ and $3$ into that equation. So: $\\frac{1(1+1)}{2} + \\frac{2(2+1)}{2} + \\frac{3(3+1)}{2} = \\frac{2}{2} + \\frac{6}{2} + \\frac{12}{2} = 1+3+6= \\boxed{\\mathrm{(C)} \\text{ or } 10}$. Invalid username Login to AoPS", "From $$\\overline{F} = \\sum_i^N \\frac{mv_{ix}^2}{L_x}$$ to $$\\overline{F} = \\int \\frac{pvn(p)}{L_x}dp$$, I feel that there is something missing? 2. Jan 13, 2009 ### Thaakisfox Because when you say N particles with the side L_yL_z, you say that all of the particles move in the x direction. Whereas in average only one third of the particles move in the x direction. So you should divide the average rate of impact by three for all three directions...", "is $\\frac{49+13}{2}=\\boxed{\\textbf{(B)}~31}$. ### Solution 3 The value of $X$ is simply the average of the average values of both diagonals that contain $X$. This is $\\frac{\\frac{1+81}{2}+\\frac{17+25}{2}}{2} =\\frac{\\frac{82}{2}+\\frac{42}{2}}{2} = \\frac{41+21}{2} = \\boxed{\\textbf{(B)}~31}$ ~ pi_is_3.14", "= \\frac{210}{20} = 10.5$$ 1. Find the average of all prime numbers between $$20$$ and $$30 \\ ?$$ 1. $$25$$ 2. $$26$$ 3. $$28$$ 4. $$24$$ Answer: (b) $$26$$ Solution: Prime numbers between $$20$$ to $$30$$ = $$23, 29$$. Then $$Average = \\frac{23 + 29}{2} = \\frac{52}{2} = 26$$ 1. Find the average of first $$50$$ odd natural numbers? 1. $$45$$ 2. $$48$$ 3. $$50$$ 4. $$49$$ Answer: (c) $$50$$ Solution: Sum of first $$n$$ natural numbers $$S_n = n^2$$, then $$S_{50} = (50)^2 = 2500$$ $$Average = \\frac{2500}{50} = 50$$ 1. Find the average of first $$70$$ even natural numbers? 1. $$70$$ 2. $$71$$ 3. $$72$$ 4. $$73$$ Answer: (b) $$71$$ Solution: Sum of first $$n$$ even natural numbers $$S_n = n \\ (n + 1)$$, then $$S_{70} = 70 \\ (70 + 1) = 70 \\times 71 = 4970$$ $$Average = \\frac{4970}{70} = 71$$ 1. Find the average of squares of first $$20$$ natural numbers? 1. $$142.3$$ 2. $$143.5$$ 3. $$141.5$$ 4. $$140.3$$ Answer: (b) $$143.5$$ Solution: Sum of squares of first $$n$$ natural numbers, $$S_n = \\frac{n \\ (n + 1) \\ (2n + 1)}{6}$$ $$S_{20} = \\frac{20 \\ (20 + 1) \\ (40 + 1)}{6}$$ $$= \\frac{20 \\times 21 \\times 41}{6}$$ $$= \\frac{17220}{6} = 2870$$ $$Average = \\frac{2870}{20} = 143.5$$ 1. Find the average", "3.3 Organisation of Data 3.4 Representative Values 3.5 Arithmetic Mean 3.5.1 Range 3.6 Mode 3.6.1 Mode of Large Data 3.7 Median 3.8 Use of Bar Graphs withe a Different Purpose 3.8.1 Choosing a Scale 3.9 Chance and Probability 3.9.1 Chance ## NCERT solutions for class 7 maths chapter 3 data handling topic 3.5 Given number are: $\\frac{1}{2}$ and $\\frac{1}{3}$ Their average = $\\frac{\\frac{1}{2} + \\frac{1}{3}}{2} = \\frac{3+2}{2\\times6} = \\frac{5}{12}$ Average between $\\frac{5}{12}\\ and\\ \\frac{1}{3}$ = $\\frac{\\frac{5}{12} + \\frac{1}{3}}{2} = \\frac{5+4}{2\\times12} = \\frac{9}{24}= \\frac{3}{8}$ Average between $\\frac{5}{12}\\ and\\ \\frac{1}{2}$ = $\\frac{\\frac{5}{12} + \\frac{1}{2}}{2} = \\frac{5+6}{2\\times12} = \\frac{11}{24}$ Average between $\\frac{11}{24}\\ and\\ \\frac{1}{2}$ = $\\frac{\\frac{11}{24} + \\frac{1}{2}}{2} = \\frac{11+12}{2\\times24} = \\frac{23}{48}$ Average between $\\frac{9}{24}\\ and\\ \\frac{1}{3}$$\\frac{\\frac{9}{24} + \\frac{1}{3}}{2} = \\frac{9+8}{2\\times24} = \\frac{17}{48}$ Therefore, 5 numbers between $\\frac{1}{2}$ and $\\frac{1}{3}$ are: $\\frac{17}{48},\\frac{23}{48},\\frac{5}{12},\\frac{11}{24},\\frac{9}{24}$ ## Solutions of NCERT class 7 maths chapter 3 data handling topic 3.6 Find the mode of (i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4 (ii) 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14 (i) Given, 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4 Arranging in ascending order: $0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6$ Here, 2, 3 and 4 all occur three times. Therefore, all three are", "Average deviation = $\\frac{\\sum_{i=1}^{n}x-\\bar{x} }{n}$ = $\\frac{12}{5}$ = 2.4 *AP and SAT are registered trademarks of the College Board.", "times X Calculator. Two numbers r and s sum up to \\frac{1}{3} exactly when the average of the two numbers is \\frac{1}{2}*\\frac{1}{3} = \\frac{1}{6}.", "## Prealgebra (7th Edition) $\\frac{11}{12}$ To find the average of a set of terms, add up all of the terms, then divide the sum by the number of terms in the set. The following are the calculations taken to find the average of $\\frac{5}{6}$, $\\frac{4}{3}$, and $\\frac{7}{12}$. $(\\frac{5}{6}+\\frac{4}{3}+\\frac{7}{12})\\div3$ $=(\\frac{10}{12}+\\frac{16}{12}+\\frac{7}{12})\\div3$ $=(\\frac{26}{12}+\\frac{7}{12})\\div3$ $=\\frac{33}{12}\\div3$ $=\\frac{33}{12}\\div\\frac{3}{1}$ $=\\frac{33}{12}\\times\\frac{1}{3}$ $=\\frac{11}{12}\\times\\frac{1}{1}$ $=\\frac{11}{12}$ The average of $\\frac{5}{6}$, $\\frac{4}{3}$, and $\\frac{7}{12}$ is $\\frac{11}{12}$.", "$$\\frac{ 2 \\frac{11}{10} + 5}{ 5 \\frac{11}{10} – 4}$$ = $$\\frac{\\frac{11}{5} + 5}{\\frac{11}{2} – 4}$$ = $$\\frac{36}{5} * \\frac{2}{3} = \\frac{24}{5}$$ 10. The average of n numbers is x .If the first number is decreased by 1, second is decreased by 2,third is decreased by 3 and so on, then their average is y. Then find the value of x – y A. $$\\frac{n(n – 1)}{2}$$ B. $$\\frac{(n + 1)}{2}$$ C. $$\\frac{n(n + 1)}{2}$$ D. $$\\frac{(n – 1)}{2}$$ Explanation – According to question $$\\frac{nx – (1 + 2 +…+ n)}{n}$$ = y $$\\frac{nx – \\frac {n + 1}{2}}{n}$$ x – y = $$\\frac {n + 1}{2}$$ 11. Four positive integers, when added three at a time give the sums 180,197,208 and 219. Average of these four integers is A. 67 B. 72 C. 73 D. 89 Explanation – Let the integers be a, b, c, d a + b + c = 180 b + c + d = 197 c + d + a = 208 d + a + b = 219 _____________________ 3(a + b + c + d) = 804 a + b + c + d = 268 i.e, Avg = $$\\frac{a + b + c + d}{4}$$ = 67 12. The average of 100 numbers is 100. I f the first number", "Moderator Joined: 26 Feb 2016 Posts: 3030 Location: India GPA: 3.12 If the average (arithmetic mean) of p and r is 60 and the average (ari [#permalink] ### Show Tags 22 Dec 2017, 07:05 1 Bunuel wrote: If the average (arithmetic mean) of p and r is 60 and the average (arithmetic mean) of r and s is 80, what is the value of s – p? A. 70 B. 40 C. 30 D. 20 E. 10 $$\\frac{p+r}{2}$$ = 60 -> p+r = 120 -> (1) $$\\frac{r+s}{2}$$ = 80 -> r+s = 160 -> (2) Subtracting (1) from (2), we get r+s - (p+r) = 160-120 = 40 Therefore, s - p = 40(Option B) _________________ You've got what it takes, but it will take everything you've got Manager Joined: 24 Nov 2016 Posts: 152 If the average (arithmetic mean) of p and r is 60 and the average (ari [#permalink] ### Show Tags 23 Dec 2017, 08:29 Bunuel wrote: If the average (arithmetic mean) of p and r is 60 and the average (arithmetic mean) of r and s is 80, what is the value of s – p? A. 70 B. 40 C. 30 D. 20 E. 10 $$Average=\\frac{Sum.of.Terms}{Number.of.Terms}$$ $$\\frac{s+r}{2}=80;s+r=160$$ $$\\frac{p+r}{2}=60;p+r=120$$ Subtract both equations to find $$s-p$$: $$(s+r)-(p+r)=160-120;s-p=40$$ If the average (arithmetic mean) of p and", "### Theory: Consider the two rational numbers $$a$$ and $$b$$. The word \"mean\" is known as average. The average of two numbers is nothing but the sum of two numbers divided by $$2$$. Thus the average/arithmetic mean of two numbers $$a$$ and $$b$$ is $\\frac{a+b}{2}$. Now we need to check whether the average of rational numbers is a rational number or not. We know that $$a$$ and $$b$$ are rational numbers. Let $a=\\phantom{\\rule{0.147em}{0ex}}\\frac{m}{n}$ and $b\\phantom{\\rule{0.147em}{0ex}}=\\phantom{\\rule{0.147em}{0ex}}\\frac{x}{y}$; where $$n, y$$ is not equal to zero. Substitute the value of $$a$$ and $$b$$ in the average formula. $\\frac{a+b}{2}=\\frac{\\frac{m}{n}+\\frac{x}{y}}{2}=\\frac{\\frac{\\mathit{my}+\\mathit{nx}}{\\mathit{ny}}}{2}=\\frac{\\mathit{my}+\\mathit{nx}}{2\\mathit{ny}}$. The above result is in $$p/q$$ form. Thus, the average of a rational number is also a rational number. Now let us prove that the resultant number lies between the rational numbers. Let us subtract $\\frac{a+b}{2}$ from $$a$$. $\\begin{array}{l}a-\\left(\\frac{a+b}{2}\\right)=\\frac{2a-a-b}{2}=\\frac{a-b}{2}\\\\ \\\\ \\frac{a-b}{2}>0\\end{array}$ This implies: $\\begin{array}{l}a-\\left(\\frac{a+b}{2}\\right)>0\\\\ a>\\frac{a+b}{2}\\end{array}$ In similar way: $\\begin{array}{l}\\left(\\frac{a+b}{2}\\right)-b=\\frac{a+b-\\mathit{2b}}{2}=\\frac{a-b}{2}\\\\ \\\\ \\frac{a-b}{2}>0\\end{array}$ This implies: $\\begin{array}{l}\\phantom{\\rule{0.147em}{0ex}}\\left(\\frac{a+b}{2}\\right)-b>0\\\\ \\frac{a+b}{2}>b\\end{array}$ Since $$a>\\frac{a+b}{2}$$ and $$\\frac{a+b}{2}>b$$, the range of $$a$$ and $$b$$ becomes $$a>\\frac{a+b}{2}>b$$. Thus, the average of two rational number can be visualised as follows: The average of any two rational number is again a rational number. We can find infinitely many rational numbers by repeating this process indefinitely. Important! If $\\frac{m}{n}$ and $\\frac{x}{y}$ are the two rational numbers with $\\frac{m}{n}$ $$<$$ $\\frac{x}{y}$, then $\\frac{m+n}{x+y}$ is a rational number" ]

[ "independent standard normals. So $$S_{T}^{1} / S_{T}^2 = \\frac{S_{t}^{1}}{S_{T}^{2}} e^{-0.5 C_{11} +0.5 C_{22} + \\sqrt{C_{11}} Z- \\sqrt{C_{22}} (\\rho Z + \\sqrt{1-\\rho^2}W)}.$$ The rest is straightforward.", "the average$0$, we should have$Tg$at least$\\frac{\\frac 12-\\xi}{\\xi}(2\\mu-\\frac 1n)$on$B$on average and, since the quadratic average can be only larger, we get $$\\|Tg\\|^2\\ge \\left[(\\frac 12-\\xi)+\\xi\\left(\\frac{\\frac 12-\\xi}{\\xi}\\right)^2\\right] (2\\mu-\\frac 1n)^2=\\frac 12\\frac{\\frac 12-\\xi}{\\xi}(2\\mu-\\frac 1n)^2$$ Thus $$\\frac 12(\\frac 12-\\xi)(2\\mu-\\frac 1n)^2\\le \\xi(1-\\frac 2n)^2(\\mu-2\\mu^2)$$ Now,$2\\mu-\\frac 1n=2\\mu(1-\\frac{2}{4\\mu n})\\ge 2\\mu(1-\\frac 2n)$under our assumption$\\mu\\ge \\frac 14$. So, we get $$\\frac 12(\\frac 12-\\xi)4\\mu^2 \\le \\xi(\\mu-2\\mu^2)$$ or, equivalently, $$(1-2\\xi)\\mu\\le (1-2\\mu)\\xi$$ i.e., $$\\mu\\le\\xi.$$ I hope I haven't made a stupid mistake anywhere though I do not really like this proof: it works for$\\stackrel{2}+$, but not for$\\stackrel{4}+$and you are, probably, interested in$\\stackrel{K}+$for all fixed$K$as$n\\to\\infty$. Anyway, it gives the desired cutoff at$1/2$for fixed parity and, thereby, the cutoff at$\\frac 34\\$ in general.", "$\\frac{2}{3}$ and $\\frac{4}{5}$ can be written as $\\frac{10}{15}$ and $\\frac{12}{15}$, so their average is $\\frac{11}{15}$. The average of $\\frac{12}{15}$ and $\\frac{11}{15}$ is $\\frac{23}{30}$. The average of $\\frac{11}{15}$ and $\\frac{23}{30}$ is $\\frac{45}{60} = \\frac34$. Alternatively, suppose the first two terms of the sequence are $x$ and $y$. The third term is $\\frac{1}{2}(x+y)$, and the fourth term is $\\frac{1}{4}(x+3y)$ so the fifth term is $\\frac{1}{8}(3x+5y)$. Putting $x=\\frac{2}{3}$ and $y=\\frac{4}{5}$ we obtain $\\frac{1}{8}(2+4)=\\frac{3}{4}$.", "# What is the average of 14, 29 and 142? ##### 1 Answer Aug 15, 2016 $61 \\frac{2}{3}$ #### Explanation: $\\text{average} = \\frac{14 + 29 + 142}{3}$", "## Thomas' Calculus 13th Edition $\\frac{3\\pi}{16}$ We calculate the average as: average=$\\frac{1}{(\\frac{4\\pi}{3})}\\int^{2\\pi}_0 \\int^1_0 \\int^{\\sqrt{1-r^2}}_{-\\sqrt{1-r^2}}r^2$ dz dr d $\\theta$ =$\\frac{3}{4\\pi} \\int^{2\\pi}_0 \\int^1_0 2r^2 \\sqrt{1-r^2}$ dr d $\\theta$ =$\\frac{3}{2\\pi} \\int^{2\\pi}_0 [\\frac{1}{8}sin^{-1}r-\\frac{1}{8}r\\sqrt{1-r^2}(1-2r^2)]^1_0d\\theta$ =$\\frac{3}{16\\pi}\\int^{2\\pi}_0 (\\frac{\\pi}{2}+0)$ d $\\theta$ =$\\frac{3}{32}\\int^{2\\pi}_0 d\\theta$ =$(\\frac{3}{32})(2\\pi)$ =$\\frac{3\\pi}{16}$", "## Prealgebra (7th Edition) $\\frac{11}{12}$ To find the average of a set of terms, add up all of the terms, then divide the sum by the number of terms in the set. The following are the calculations taken to find the average of $\\frac{5}{6}$, $\\frac{4}{3}$, and $\\frac{7}{12}$. $(\\frac{5}{6}+\\frac{4}{3}+\\frac{7}{12})\\div3$ $=(\\frac{10}{12}+\\frac{16}{12}+\\frac{7}{12})\\div3$ $=(\\frac{26}{12}+\\frac{7}{12})\\div3$ $=\\frac{33}{12}\\div3$ $=\\frac{33}{12}\\div\\frac{3}{1}$ $=\\frac{33}{12}\\times\\frac{1}{3}$ $=\\frac{11}{12}\\times\\frac{1}{1}$ $=\\frac{11}{12}$ The average of $\\frac{5}{6}$, $\\frac{4}{3}$, and $\\frac{7}{12}$ is $\\frac{11}{12}$.", "get the average, divide both sides by 4 $\\frac{a+b+c+d}{4}=\\frac{78}{4}\\\\ \\\\ \\frac{a+b+c+d}{4}=19.5\\\\$ So, average = 19.5 Jun 11th, 2015 ... Jun 11th, 2015 ... Jun 11th, 2015 Dec 10th, 2016 check_circle", "3.3 Organisation of Data 3.4 Representative Values 3.5 Arithmetic Mean 3.5.1 Range 3.6 Mode 3.6.1 Mode of Large Data 3.7 Median 3.8 Use of Bar Graphs withe a Different Purpose 3.8.1 Choosing a Scale 3.9 Chance and Probability 3.9.1 Chance ## NCERT solutions for class 7 maths chapter 3 data handling topic 3.5 Given number are: $\\frac{1}{2}$ and $\\frac{1}{3}$ Their average = $\\frac{\\frac{1}{2} + \\frac{1}{3}}{2} = \\frac{3+2}{2\\times6} = \\frac{5}{12}$ Average between $\\frac{5}{12}\\ and\\ \\frac{1}{3}$ = $\\frac{\\frac{5}{12} + \\frac{1}{3}}{2} = \\frac{5+4}{2\\times12} = \\frac{9}{24}= \\frac{3}{8}$ Average between $\\frac{5}{12}\\ and\\ \\frac{1}{2}$ = $\\frac{\\frac{5}{12} + \\frac{1}{2}}{2} = \\frac{5+6}{2\\times12} = \\frac{11}{24}$ Average between $\\frac{11}{24}\\ and\\ \\frac{1}{2}$ = $\\frac{\\frac{11}{24} + \\frac{1}{2}}{2} = \\frac{11+12}{2\\times24} = \\frac{23}{48}$ Average between $\\frac{9}{24}\\ and\\ \\frac{1}{3}$$\\frac{\\frac{9}{24} + \\frac{1}{3}}{2} = \\frac{9+8}{2\\times24} = \\frac{17}{48}$ Therefore, 5 numbers between $\\frac{1}{2}$ and $\\frac{1}{3}$ are: $\\frac{17}{48},\\frac{23}{48},\\frac{5}{12},\\frac{11}{24},\\frac{9}{24}$ ## Solutions of NCERT class 7 maths chapter 3 data handling topic 3.6 Find the mode of (i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4 (ii) 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14 (i) Given, 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4 Arranging in ascending order: $0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6$ Here, 2, 3 and 4 all occur three times. Therefore, all three are", "$$\\frac{ 2 \\frac{11}{10} + 5}{ 5 \\frac{11}{10} – 4}$$ = $$\\frac{\\frac{11}{5} + 5}{\\frac{11}{2} – 4}$$ = $$\\frac{36}{5} * \\frac{2}{3} = \\frac{24}{5}$$ 10. The average of n numbers is x .If the first number is decreased by 1, second is decreased by 2,third is decreased by 3 and so on, then their average is y. Then find the value of x – y A. $$\\frac{n(n – 1)}{2}$$ B. $$\\frac{(n + 1)}{2}$$ C. $$\\frac{n(n + 1)}{2}$$ D. $$\\frac{(n – 1)}{2}$$ Explanation – According to question $$\\frac{nx – (1 + 2 +…+ n)}{n}$$ = y $$\\frac{nx – \\frac {n + 1}{2}}{n}$$ x – y = $$\\frac {n + 1}{2}$$ 11. Four positive integers, when added three at a time give the sums 180,197,208 and 219. Average of these four integers is A. 67 B. 72 C. 73 D. 89 Explanation – Let the integers be a, b, c, d a + b + c = 180 b + c + d = 197 c + d + a = 208 d + a + b = 219 _____________________ 3(a + b + c + d) = 804 a + b + c + d = 268 i.e, Avg = $$\\frac{a + b + c + d}{4}$$ = 67 12. The average of 100 numbers is 100. I f the first number", "# Average $lcm(a,b)$, $1\\le a \\le b \\le n$, and asymptotic behavior What is the average value for $\\mathrm{lcm}(a,b)$, with $1\\le a \\le b \\le n$, for a given $n$, and what is the asymptotic behavior? The $\\mathrm{lcm}$ is the least common multiple. I have calculated, as $(n, avg)$: $$(10, 19.836)$$ $$(100, 1826.859)$$ $$(1000, 182828.976)$$ The values appear to converge quadratically to some constant. This generalizes to $$\\sum_{1 \\le j \\le k \\le n} \\mathrm{lcm}(j,k)$$ #### Solutions Collecting From Web of \"Average $lcm(a,b)$, $1\\le a \\le b \\le n$, and asymptotic behavior\" Here is a back-of-the-envelope calculation. Start with the average value of $ab$, which is $(n+1)^2/4$. Consider powers of $2$ only. Three-quarters of the $(a,b)$ pairs lack $2$ as a common factor. 3/16 have $2^1$ as a common factor, $3/64$ have $4$ as a common factor and so on. So, on average, dealing with powers of $2$ reduces the average by a factor $$\\frac341+\\frac3{16}\\frac12+\\frac3{64}\\frac14+…\\\\ =\\frac34\\left[1+\\frac18+\\frac1{64}+…\\right]\\\\=\\frac{3/4}{7/8}=6/7$$ Consider powers of $3$ only, which will be independent from powers of $2$. The reduction is $$\\frac891+\\frac8{81}\\frac13+\\frac8{729}\\frac19+…=\\frac{8/9}{26/27}=\\frac{12}{13}$$ For any prime $p$, the factor is $\\frac{p^3-p}{p^3-1}$, so my final answer is $$\\frac{(n+1)^2}4\\prod_{p\\,\\text{prime}}\\frac{p^3-p}{p^3-1}=\\frac{(n+1)^2}4\\frac{\\zeta(3)}{\\zeta(2)}$$ That is the Riemann $\\zeta$ function, and the constant ratio is $$\\frac{\\zeta(3)}{4\\zeta(2)}=0.18269074235…$$ Theorem 6.3 of Olivier Bordelles, Mean values of generalized gcd-sum and lcm-sum functions, Journal of Integer Sequences, Vol. 10", "= \\frac{210}{20} = 10.5$$ 1. Find the average of all prime numbers between $$20$$ and $$30 \\ ?$$ 1. $$25$$ 2. $$26$$ 3. $$28$$ 4. $$24$$ Answer: (b) $$26$$ Solution: Prime numbers between $$20$$ to $$30$$ = $$23, 29$$. Then $$Average = \\frac{23 + 29}{2} = \\frac{52}{2} = 26$$ 1. Find the average of first $$50$$ odd natural numbers? 1. $$45$$ 2. $$48$$ 3. $$50$$ 4. $$49$$ Answer: (c) $$50$$ Solution: Sum of first $$n$$ natural numbers $$S_n = n^2$$, then $$S_{50} = (50)^2 = 2500$$ $$Average = \\frac{2500}{50} = 50$$ 1. Find the average of first $$70$$ even natural numbers? 1. $$70$$ 2. $$71$$ 3. $$72$$ 4. $$73$$ Answer: (b) $$71$$ Solution: Sum of first $$n$$ even natural numbers $$S_n = n \\ (n + 1)$$, then $$S_{70} = 70 \\ (70 + 1) = 70 \\times 71 = 4970$$ $$Average = \\frac{4970}{70} = 71$$ 1. Find the average of squares of first $$20$$ natural numbers? 1. $$142.3$$ 2. $$143.5$$ 3. $$141.5$$ 4. $$140.3$$ Answer: (b) $$143.5$$ Solution: Sum of squares of first $$n$$ natural numbers, $$S_n = \\frac{n \\ (n + 1) \\ (2n + 1)}{6}$$ $$S_{20} = \\frac{20 \\ (20 + 1) \\ (40 + 1)}{6}$$ $$= \\frac{20 \\times 21 \\times 41}{6}$$ $$= \\frac{17220}{6} = 2870$$ $$Average = \\frac{2870}{20} = 143.5$$ 1. Find the average", "way: we have $10^x=6.7$, and the answer is $x=\\log 6.7$. Normally we can look up $\\log 6.7$ in a log table or use a calculator to see what the decimal number version of $\\log 6.7$ is. But what if we couldn't do that? Okay, what if we do this: $10^0=1$ and $10^1=10$. $10^x=6.7$, so $x$ must be between 0 and 1. Let's take the average: $\\frac{0+1}{2}=\\frac{1}{2}$. Taking the average is easy and it ensures that we end up somewhere \"between\" the values we're taking the average of (0 and 1 in this case). $10^\\frac{1}{2}\\approx 3.16$. Okay so less than 6.7. Let's take the average again: $\\frac{1/2+1}{2}=\\frac{3}{4}$ $10^\\frac{3}{4}\\approx 5.62$. Still less than 6.7 Let's take the average again: $\\frac{3/4+1}{2}=\\frac{7}{8}$ $10^\\frac{7}{8}\\approx 7.5$. So now we're over 6.7. Let's just take the average of $3/4$ and $7/8$ then. Let's take the average again: $\\frac{3/4+7/8}{2}=\\frac{13}{16}$. $10^\\frac{13}{16}\\approx 6.49$. Less than 6.7, but we're getting closer. If we keep doing this eventually we'll get a number such that $10^{\\text{number}}=6.7$. Implementing this method in Python, it took 26 iterations to get $10^{0.82607480\\;13854027}=6.69999997970653$. Wolfram alpha says $\\log_{10} 6.7 = 0.82607480\\;27...$. If we wanted to do this by hand, we'd have to assume we had a method for approximating numbers like $10^\\frac{13}{16}$ of course. So just on the face of it, it seems like a lot of", "numbers in the list. The median is the average of these numbers: $\\frac{10+12}{2} = 11$.", "# 2010 AMC 10B Problems/Problem 4 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) The average of two numbers, $a$ and $b$, is defined as $\\frac{a+b}{2}$. Thus the average of $x$ and $x^2$ would be $\\frac{x(x+1)}{2}$. With that said, we need to find the sum when we plug, $1$, $2$ and $3$ into that equation. So: $\\frac{1(1+1)}{2} + \\frac{2(2+1)}{2} + \\frac{3(3+1)}{2} = \\frac{2}{2} + \\frac{6}{2} + \\frac{12}{2} = 1+3+6= \\boxed{\\mathrm{(C)} \\text{ or } 10}$. Invalid username Login to AoPS", "there were 8 numbers, the last average was 4 and there were 8 numbers. We can get the total of each by multiplying the average of each set by the number of terms in the set and add those: $$\\frac{2(10)+ 3(8)+ 4(8)}{10+ 8+ 8}= \\frac{20+ 24+ 32}{26}= \\frac{76}{26}= 36/13$$ Notice that we could also write that as $$\\frac{10}{26}(2)+ \\frac{8}{26}(3)+ \\frac{8}{26}(4)= \\frac{5}{13}(2)+ \\frac{4}{13}(3)+ \\frac{4}{13}(4)$$ \"weighting\" each average by the fraction of total numbers that was based on. Actually, I don't see the problem with the company's current method of comparing the number of transactions in a given week with the number a transactions at the same period a year ago. That naturally allows for seasonal fluctuations and public holidays, etc. Last edited:", "{2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ? a) 0 b) 1 c) (1/2)*n d) (n+1)/2n e) 2008 $Z$ is the number of items produced by the company. Total cost, $R=\\frac{1}{10}Z^{3}-5Z^{2}+10Z+5$ As we can see, the term ‘5’ does not vary with the number of quantities produced. Therefore, 5 is the fixed cost. Variable cost = $\\frac{1}{10}Z^{3}-5Z^{2}+10Z$ Average variable cost = Total variable cost/ number of quantities. Average variable cost =$\\frac{1}{10}Z^{2}-5Z+10$ =$\\frac{Z^2-50Z+100}{10}$ =$\\frac{(Z^2-50Z+625)- 525}{10}$ =$\\frac{(Z-25)^2-525}{10}$ As we can see, the least value of the expression will be obtained at $Z=25$ Therefore, option C is the right answer. Let the 7 consecutive numbers be a-3, a-2, a-1, a, a+1, a+2 and a+3. Sum of the numbers = 7a and the average of these numbers = a If next 3 numbers a+4, 4+5 and a+6 are also added then the average of these 10 numbers = $\\dfrac{7a+a+4+a+5+a+6}{10} = a+1.5$ Thus, the average increases by 1.5 Hence, option C is the correct answer. Let ‘A’, ‘S’ and ‘P’ be Amita’s, Sumita’s and Paramita’s present age. It is given that 2 years ago, one-fifth of", "post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52294 If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink] ### Show Tags 26 Aug 2018, 20:20 00:00 Difficulty: 25% (medium) Question Stats: 82% (01:00) correct 18% (01:03) wrong based on 45 sessions ### HideShow timer Statistics If the average (arithmetic mean) of n and 11 is equal to 2n, what is the average of n and $$\\frac{13}{3}$$? (A) 4 (B) 8 (C) 11 (D) 14 (E) 19 _________________ Director Status: Learning stage Joined: 01 Oct 2017 Posts: 951 WE: Supply Chain Management (Energy and Utilities) Re: If the average (arithmetic mean) of n and 11 is equal to 2n, what is [#permalink] ### Show Tags 26 Aug 2018, 20:32 Bunuel wrote: If the average (arithmetic mean) of n and 11 is equal to 2n, what is the average of n and $$\\frac{13}{3}$$? (A) 4 (B) 8 (C) 11 (D) 14 (E) 19 First we have to find out the value of n. Given, average (arithmetic mean) of n and 11 is equal to 2n Or, $$\\frac{n+11}{2}=2n$$ Or, n+11=4n Or, 3n=11 Or, $$n=\\frac{11}{3}$$ Average of n and $$\\frac{13}{3}$$=Average of $$\\frac{11}{3}$$ and $$\\frac{13}{3}$$=$$\\frac{\\frac{11}{3}+\\frac{13}{3}}{2}$$=$$\\frac{8}{2}=4$$ Ans. (A) _________________ Regards, PKN Rise", "Moderator Joined: 26 Feb 2016 Posts: 3030 Location: India GPA: 3.12 If the average (arithmetic mean) of p and r is 60 and the average (ari [#permalink] ### Show Tags 22 Dec 2017, 07:05 1 Bunuel wrote: If the average (arithmetic mean) of p and r is 60 and the average (arithmetic mean) of r and s is 80, what is the value of s – p? A. 70 B. 40 C. 30 D. 20 E. 10 $$\\frac{p+r}{2}$$ = 60 -> p+r = 120 -> (1) $$\\frac{r+s}{2}$$ = 80 -> r+s = 160 -> (2) Subtracting (1) from (2), we get r+s - (p+r) = 160-120 = 40 Therefore, s - p = 40(Option B) _________________ You've got what it takes, but it will take everything you've got Manager Joined: 24 Nov 2016 Posts: 152 If the average (arithmetic mean) of p and r is 60 and the average (ari [#permalink] ### Show Tags 23 Dec 2017, 08:29 Bunuel wrote: If the average (arithmetic mean) of p and r is 60 and the average (arithmetic mean) of r and s is 80, what is the value of s – p? A. 70 B. 40 C. 30 D. 20 E. 10 $$Average=\\frac{Sum.of.Terms}{Number.of.Terms}$$ $$\\frac{s+r}{2}=80;s+r=160$$ $$\\frac{p+r}{2}=60;p+r=120$$ Subtract both equations to find $$s-p$$: $$(s+r)-(p+r)=160-120;s-p=40$$ If the average (arithmetic mean) of p and", "is calculated using the formula Median = $\\frac{n+1}{2}$ = $\\frac{6+1}{2}$ = 3rd and 4th term $\\therefore$ Median = $\\frac{4+5}{2}$ = 4.5 The Upper quartile is given by Q1 = $\\frac{n+1}{4}$ = $\\frac{6+1}{4}$ = $\\frac{7}{4}$ = 1.75 First quartile is the average of 1st and 2nd term, i.e. $\\frac{3+4}{2}$ = $\\frac{7}{2}$ = 3.5 The lower quartile is given by Q3 = $\\frac{3(n+1)}{4}$ = 3 $\\times$ 1.75 = 5.25. Third quartile is the average of 5th and 6th term, i.e. $\\frac{6+7}{2}$ = $\\frac{13}{2}$ = 6.5", "this satisfies the required condition II. becomes => 30, 20, 15, 12, 10 Average of above numbers = 17.4 and Median = 15. Does not satisfy our condition. We can Stop here and eliminate all option. so our answer is A for the sake of testing - III becomes => 10, 20, 30, 40, 50 Average of above numbers = median of above set = 30. Hence does not satisfy given condition. Option A Kudos [?]: 379 [0], given: 42 Math Expert Joined: 02 Aug 2009 Posts: 5519 Kudos [?]: 6410 [0], given: 122 For which of the above lists is the average of the numbers less than t [#permalink] ### Show Tags 14 Nov 2017, 06:34 Expert's post 1 This post was BOOKMARKED MathRevolution wrote: [GMAT math practice question] I. $$\\frac{1}{2},\\frac{2}{3},\\frac{3}{4},\\frac{4}{5},\\frac{5}{6}$$ II. $$\\frac{1}{2},\\frac{1}{3},\\frac{1}{4},\\frac{1}{5},\\frac{1}{6}$$ III. $$\\frac{1}{6},\\frac{2}{6},\\frac{3}{6},\\frac{4}{6},\\frac{5}{6}$$ For which of the above lists is the average of the numbers less than the median of numbers? A. I only B. II only C. III only D. II and III E. I, II and III Hi... If you don't want to get into lengthy calculations as it would surely eat into your time, some observations... here is a method... III. $$\\frac{1}{6},\\frac{2}{6},\\frac{3}{6},\\frac{4}{6},\\frac{5}{6}$$ this is clearly an AP with difference $$\\frac{1}{6}$$, so MEDIAN = MEAN so III is out ONLY A and B" ]

[ "well choose them in order $1, 2, 3, \\ldots$.", "4, 5, \\ldots$? Draw up a table of results.", "{1, 4, 9, 16, \\ldots}$", "\\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 $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 &", "24 x + 48 {x}^{2} + 80 {x}^{3} + 120 {x}^{4} + 168 {x}^{5} + 224 {x}^{6} + \\ldots \\ldots .$", "# Solutions to Problems 1064-1071 Q1064. The numbers $1, 2, \\ldots , 16$ are placed in the cells of a $4 \\times 4$ table as shown in the left hand diagram below.", "for the sixth item only. I assume $i \\in \\{1,2,\\ldots,100\\}$. If you meant that $i \\in \\{1,2,\\ldots,6\\}$, how will I know which of the seven out of 100 items are chosen? – ubaabd Dec 17 '14 at 5:23", "$3$, $5$, $7$, $\\ldots$, $35$, $37$, $41$, $43$, $\\ldots$ The $26$th positive integer after $2$, $3$, $4$, $7$, $8$, $\\ldots$, $26$, $29$, $30$, $31$, $32$, $33$, $37$, $38$, $42$ which cannot be expressed as the sum of distinct pentagonal numbers.", "# Problems Section: Problems 1064 - 1071 Q1064. The numbers $1, 2,\\ldots , 16$ are placed in the cells of a $4\\times 4$ table as shown in the left hand diagram below. One may add $1$ to all numbers of any row or subtract $1$ from all numbers of any column.", "\\{0, 1, \\ldots, m\\}$$.", "n=1,\\ldots,38,$$ so $$a_{38}\\geq a_1+27\\times37\\geq1+27\\times37=1000.$$ Contradiction. We conclude.", "six digits $D_1, D_2, \\ldots, D_6$. In case two, the order in which we are choosing them is $D_6, D_1, D_2, \\ldots, D_5$. • For $D_6$, there are four choices: 2, 4, 6, 8. • For $D_1$, there are eight choices: all nine of 1-9 (can't start with zero!), except $D_6$. • For $D_2$, there are eight choices: all ten of 0-9, except $D_6$ and $D_1$. • For $D_3$, there are seven choices: all ten of 0-9, except $D_6$ and $D_1$ and $D_2$. • For $D_4$, there are six choices: all ten of 0-9, except $D_6,D_1,D_2,D_3$. • For $D_5$, there are five choices: all ten of 0-9, except $D_6,D_1,D_2,D_3,D_4$.", "of codewords, $2^{24}\\text{,}$ yet there are only $2^{32 - 24} = 2^{8} = 256$ cosets.", "How many different sets of numbers with at least four members can you find in the numbers in this box? For example, one set could be multiples of $4$ {$8, 36 ...$}, another could be odd numbers {$3, 13 ...$}.", "-44 & 15 & -47 \\ldots \\\\ \\hline -41 & 200 & -92 & -72 & 17 & -48 & 38 & -35 & 0 & 0 & 0 & -47 \\ldots \\\\ \\\\ \\ldots 2 & 3 & 4 & 5 & 6 & 7 & 2 & 3 & 2 & 1 & 0 & 0 \\\\ \\ldots 5 & -48 & 34 & -16 & -21 & 27 & -7 & 9 & 30 & 3 & 10 & 1 \\\\ \\hline \\ldots 10 & -144 & 136 & -80 & -126 & 189 & -14 & 27 & 60 & 3 & 0 & 0 \\end{array}$$ 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.", "# What are the numbers that come next in these sequences: 1,5,2,10,3,15,4? Mar 24, 2016 If you look at the odd numbers they go like $1 , 2 , 3 , 4. . .$ The even numbers add $5$ at every step like $5 , 10 , 15. . .$ So the next odd numbers would be $\\ldots 20 , 25 , 30. . .$ And the next even numbers would be $\\ldots 5 , 6 , 7. . .$ $\\ldots 20 , 5 , 25 , 6 , 30 , 7. . .$", "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 $[0]_7 \\cap \\left\\{1, \\ldots, 300 \\right\\}.$ $$300-3 \\times 43-42+1=130$$", "# What are the numbers that come next in these sequences: 1,5,2,10,3,15,4? Mar 24, 2016 If you look at the odd numbers they go like $1 , 2 , 3 , 4. . .$ The even numbers add $5$ at every step like $5 , 10 , 15. . .$ #### Explanation: So the next odd numbers would be $\\ldots 20 , 25 , 30. . .$ And the next even numbers would be $\\ldots 5 , 6 , 7. . .$ The sequence would continue like this: $\\ldots 20 , 5 , 25 , 6 , 30 , 7. . .$", "= 2, 4, 6, 8, \\ldots$.", "# Sequence of Powers of 16 The sequence of powers of $16$ begins: $1, 16, 256, 4096, 65 \\, 536, 1 \\, 048 \\, 576, 16 \\, 777 \\, 216, 268 \\, 435 \\, 456, \\ldots$" ]

[ "# Three times a number increased by less than -62. What is the number? Jul 4, 2018 $x < - \\frac{62}{3}$ #### Explanation: Let the number be $= x$ Three times a number increased by less than -62 $3 \\times x < - 62$ $3 x < - 62$ $x < - \\frac{62}{3}$ Let $x$ be number then we have $3 x < - 62$ $x < - \\frac{62}{3}$", "10 !}$ = $\\frac { 15 \\times 14 \\times 13 \\times 12 \\times 11}{5!}$ = 3003. There are 3003 ways of arranging a group such that 5 children accomodate each group. 2. ### In a lucky draw, 6 boxes are kept in which only 3 boxes contains gifts. In how many ways can the kids pick the gift boxes? Step 1 : given: n = 6, k = 3 Step 2 : Use the combination formula: nck = $\\frac{n!}{r! (n-k)!}$ = $\\frac{6!}{3! (6 - 3)!}$ = $\\frac{6!}{3! 3!}$ = $\\frac{6 \\times 5 \\times 4 \\times 3!}{3! 3!}$ = $\\frac{120}{36}$ = 20 ways.", "then $$16^{3/4}=(16^{\\frac{1}{4}})^{3}=(2)^3=\\boxed{8}$$ - Have you read the other answers? – Servaes May 18 '14 at 16:09", "Increase $75$ by $24\\%$ Required Number $= 75 \\times (1+$ $\\frac{24}{100}$ $) = 93$ ii) Decrease $375$ by $8\\%$ Required Number $= 375 \\times (1-$ $\\frac{8}{100}$ $) = 345$ $\\\\$ Question 11: What Number Increase by $15\\%$ Becomes $276\\%$. Let the number to be $x$ Then $x(1+$ $\\frac{15}{100}$ $)= 276 \\ or \\ x = 240$ $\\\\$ Question 12: What Number when Decrease by $8\\%$ Becomes $345\\%$ Let the number to be $x$ Then $x(1-$ $\\frac{8}{100}$ $)= 345 \\ or \\ x = 375$", "boxes and $3$ choices then $$5^2<5.4.3.$$ However, if we have $5$ boxes and $4$ choices then $$5^3>5.4.3.2.$$ If the number of boxes, $M$, is growing, and $T$ remains, say, $2$ then $$\\lim_{M\\rightarrow\\infty}\\frac{M}{M(M-1)}=0.$$", "14 is $\\frac{492}{2} = \\boxed{246}$.", "the answer is $3 + 11 = \\boxed{\\mathrm{(A)}\\ 14}$", "of getting a total 7 = $\\frac{12}{36}= \\frac{1}{3}$", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "AB (same as 1st). This forces the later transactions to both be BA. Here we have 1 good option. Summing, we have $12+24+8+1=45$ good options among $12^3$ total options, so the solution is $\\frac{45}{(12^3)}= \\boxed{\\textbf{(B)}\\frac{5}{192}}$. SilverLion", "3 + 4 = \\boxed{21}$", "# How do you find a number such that two-thirds of the number increased by one is nine? I found $12$ Call the number $x$, so: $\\frac{2}{3} x + 1 = 9$ $2 x + 3 = 27$ $2 x = 24$ $x = \\frac{24}{2} = 12$", "# If three times a number is increased by 4, the result is the same as when five times a number is decreased by 20. What is the number? Nov 8, 2016 The number, or $x$ as we chose to call it, is $12$. #### Explanation: We will let the number we are looking for be represented by $x$. \"the result is the same as\" means this is an equation because each part of the stated problem is \"the same\" or equal. For the first part of the equation (If three times a number is increased by 4): \"Three times a number\" can be written as $3 X$. \"Is increased by 4\" can then be written as $3 x + 4$. For the second part of the equation (when five times a number is decreased by 20.): \"When five times a number\" can be written as $5 x$. \"Is decreased by 20\" can then be written as $5 x - 20$. And because, as stated previous these two statements are equal we can solve for $x$ as follows while keeping the equation balanced: $3 x + 4 = 5 x - 20$ $3 x + 4 - 3 x + 20 = 5 x - 20 - 3 x + 20$ $24 = 2 x$ $\\frac{24}{2} = \\frac{2", "decided to do it systematically starting with each of them having an equal number of objects so $1 1 1 1 1 1 1$ and from here saying if you want to change the configuration you can choose to increase some baskets by an overall say $2$ you must take away $2$ objects- i.e $1$ from each of $2$ baskets So for an increase of $1$ being cancelled by a decrease of $1$ you get $2 0 1 1 1 1 1$ So I increased one basket by $1$ and decreased another to $0$. The number of ways you could arrange this configuration for $7$ baskets is $\\frac{7!}{5!}= 7*6=42$ (BECAUSE YOU COULD HAVE $2 1 0 1 1 1 1$ or $1 1 0 1 1 2 1$ etc) For an increase of $2$ total cancelled by a reduction of $2$ baskets to $0$ You have either $3 0 0 1 1 1 1$ which results in $\\frac{7!}{4!2!}=105$ ways or $2 2 0 0 1 1 1$ so you have $\\frac{7!}{2!2!3!}=210$ ways and so I went on this way up to a maximum increase of $6$ cancelled by decrease of $6$ baskets to $0$ i.e the configuration $7 0 0 0 0 0 0$ and added together to get $1716$ (including the one which is $1 1 1", "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 $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{+10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}$$ $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}\\boxed{+10}$$", "= 12$ Thus we have: $x' = - \\frac {(12)(-3)}{5} \\implies \\boxed { x' = 7.2 }$ Note, $x' = \\frac {dx}{dt}$ and $y' = \\frac {dy}{dt}$", "# How can the addition of a parenthesis be used to increase the value of the following expression?: 4-3/2^2+3*2+5 Mar 8, 2018 Put parenthesis around the $2 + 5$ This way the addition must be done before the multiplication #### Explanation: $4 - \\frac{3}{2} ^ 2 + 3 \\times 2 + 5 = 4 - \\frac{3}{2} ^ 2 + 6 + 5 = 4 - \\frac{3}{2} ^ 2 + 11$ $4 - \\frac{3}{2} ^ 2 + 3 \\times \\left(2 + 5\\right) = 4 - \\frac{3}{2} ^ 2 + 3 \\times 10 = 4 - \\frac{3}{2} ^ + 30$ $30 > 11$ so adding the parenthesis around the 2+5 increases the value of the expression. ( Note there are also probably other ways of adding parenthesis to increase the value but this way works. )", "And we have: . $a(n) \\;=\\;\\frac{n(n+1)}{2} - (n-1)$ . . Therefore: . $\\boxed{a(n) \\;=\\;\\frac{n^2-n+2}{2}}$ 8. Thanks so much Soroban! I actually get that *victory dance*", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "is $\\frac{22}{7}$ an irrational number? III. Fill in the boxes with the correct real numbers in the following statements: (a) $2\\sqrt{7}+\\sqrt{7} = \\Box+ 2\\sqrt{7}$ (b) $3.\\dot{8} + 4.65 = 4.65 + \\Box$ (c) $\\Box + 29 = 29 + 5\\sqrt{10}$ (d) $3.\\dot{9} + (4.69 +2.12) = (\\Box + 4.69) + 2.12$ (e) $(\\frac{7}{8} + \\frac{3}{7})+\\frac{6}{5} = (\\frac{6}{5} + \\frac{3}{7}) + \\Box$ (f) $3\\sqrt{2}(\\sqrt{3}+2\\sqrt{5}) = (3\\sqrt{2}+2\\sqrt{5}) + (\\Box \\times \\Box)$ (g) $1\\frac{3}{7} (2\\frac{7}{11} + \\Box) = (1\\frac{3}{7} \\times 1 \\frac{8}{9}) + (1\\frac{3}{7} + 2\\frac{7}{11})$ (h) $2\\frac{1}{3} + \\Box =0$ (xi) $\\frac{7}{-8} \\times \\Box = 1$ (i) $-7.35 + \\Box = 0$ IV. Find the answers to the following expressions by using the properties of addition and multiplication of real numbers: Before that, we can recapitulate the relevant properties here : Properties of Real Numbers: Closure Property: The sum, difference, product,or quotient of two real numbers is a real number. Commutative Property of Addition and Multiplication: A change in the order of addition or multiplication of two real numbers does not change their respective sum or product. (a) $x+y = y+x$ (b) $x \\times y = y \\times x$ Associative Property of Addition and Multiplication: A change in the grouping of three real numbers while adding or multiplying does not change their respective sum or product : $(a+b)+c = (a+b)+c$" ]

[ "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "\\;(\\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.", "\\;(\\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.", "boxes and $3$ choices then $$5^2<5.4.3.$$ However, if we have $5$ boxes and $4$ choices then $$5^3>5.4.3.2.$$ If the number of boxes, $M$, is growing, and $T$ remains, say, $2$ then $$\\lim_{M\\rightarrow\\infty}\\frac{M}{M(M-1)}=0.$$", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "and $3$ and $5$ never near, which leaves the total combinations (orders) of the boxes to twelve: $1-2-3-4-5$ $1-3-2-4-5$ $1-3-4-2-5$ $1-3-2-5-4$ $1-3-4-5-2$ $1-4-3-2-5$ Plus the symmetry of the exchange $5\\to 3$ So I have $12$ possible dispositions of the boxes. Now, let's call the figures on each box as $$1 = \\{A_1, A_2, A_3, A_4, A_5, A_6\\}$$ $$2 = \\{B_1, \\ldots, B_6\\}$$ $$3 = \\{C_1 \\ldots C_6\\}$$ $$4 = \\{D_1, D_2, D_3, A_1, C_1, C_2\\}$$ $$5 = \\{E_1 \\ldots E_6\\}$$", "then $$16^{3/4}=(16^{\\frac{1}{4}})^{3}=(2)^3=\\boxed{8}$$ - Have you read the other answers? – Servaes May 18 '14 at 16:09", "to see here that this continues all the way down to one. However, when the case gets to $32$, there are 5 ways instead of 4 because $2^{10}-2^5$ is smaller than 1000. Thus, $9+8+7+6+5+5+4+3+2+1 = 50$ So the answer is $\\boxed{050}$ Problem 4 Solution 1 Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \\ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \\ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we have the sum $31+29+28+26+25+\\ldots+4+2+1=(31+28+25+22+\\ldots+1)+$ $(29+26+23+\\ldots+2)=176+155=\\boxed{331}$. Solution 2 Let the three piles have $a, b, c$ coins respectively. If we disregard order, then we just need to divide by $3! = 6$ at the end. We know $a + b + c = 66$. Since $a, b, c$ are positive integers, there are $\\binom{65}{2}$ ways from Stars and Bars. However, we must discard the cases where $a = b$ or $a = c$ or $b = c$. The three cases are symmetric, so we just take the first case and multiply by 3. We have $2a + c = 66 \\implies a = 1, 2, \\dots 32$ for", "my friends about whether $0.999\\ldots$ (repeating forever) was equal to 1. I thought it was not the same as 1, but a hair less than 1. My friend Ben was just as adamant that $0.999\\ldots$ exactly equal to 1. His explanation was short, but persuasive. 1. He asked us if we believed $\\frac 1 3 = 0.333\\ldots$ exactly, no approximation (if you could expand the ellipsis infinitely). We of course agreed. 2. Then he asked us if $1 = 3 \\times \\frac 1 3$ exactly. Nothing special here either. 3. Then he said that $1 = 3 \\times \\frac 1 3 = 3 \\times 0.333\\ldots$ exactly. “Ok?” we nodded. 4. When multiplying decimal digits, you multiply each digit by the multiplier, so $3 \\times 0.333\\ldots = 0.999\\ldots$ exactly. We learned that years ago and used it recently. 5. Hence $1 = 3 \\times \\frac 1 3 = 3 \\times 0.333\\ldots = 0.999\\ldots$ exactly. It’s just another way of writing the same thing. Aside: Here is an entertaining presentation of this and similar arguments. We were reluctant that there was more than one way of writing a number in decimals—namely 1 and $0.999\\ldots$—but logic has no such hesitation. Once you go looking, you can find more examples of numbers with multiple decimal names: 0 is 0 or -0, 4", "14 is $\\frac{492}{2} = \\boxed{246}$.", "radicand therefore must be one less than a multiple of four, which is only the case in $\\frac {1 + i \\sqrt {11}}{2}$ or $\\boxed{A}$.", "you see why?). Now add to these the set $\\,\\{1,2,3\\}\\,$ itself... Each number is contained in at most three sets. Indeed, all sets which contain, say, $1$ are not alowed to share any more elements. Since there are just $7$ other elements, there are at most $\\lfloor \\frac{7}{2}\\rfloor=3$ possibilities. Thus if we list all sets which contain $1$ then all sets which contain $2$ etc. we end up with at most $24$ sets. But we have counted each of them three times, so the maximum is at most $8$. This number is attended with $(123)(456)(147)(257)(367)(158)(268)(348)$.", "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 getting a total 7 = $\\frac{12}{36}= \\frac{1}{3}$", "And we have: . $a(n) \\;=\\;\\frac{n(n+1)}{2} - (n-1)$ . . Therefore: . $\\boxed{a(n) \\;=\\;\\frac{n^2-n+2}{2}}$ 8. Thanks so much Soroban! I actually get that *victory dance*", "limiting after $12$ places of decimals, it is a rational number as $0.354355435554 = \\frac{354355435554}{1000000000000}$. However, apparently questioner is rather looking at $0.354355435554 \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots . .$, which is clearly irrational as grouping them as under reveals the pattern as follows: $0. \\underline{354} \\textcolor{red}{3554} \\underline{35554} \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots . .$ Here we first have one $5$ between $3$ and $4$, then we have two $5 ' s$ between $3$ and $4$, and then we have three $5 ' s$ between $3$ and $4$. Hence the number of $5 ' s$ between $3$ and $4$ is continuously increasing and there is no group of numbers repeating endlessly Hence $0.354355435554 \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots . .$ is an irrational number.", "there are $3$ groups, we can easily deduce there are $81$ ways to group the $3$ students, and there are $3$ ways to group them in the same $1$ group, so we might think $\\frac{3}{54}=\\frac{1}{27}$ is the answer but as there are 3 groups we do $\\frac{1}{27} (3)=\\frac{1}{9}$ which is $\\boxed{\\text{(B)}}$.", "# How can we write 3.11.... (repeated) as fraction? May 12, 2017 $3.1$ repeated$= \\frac{28}{9} = 3 \\frac{1}{9}$ #### Explanation: $3.1$ repeated can be written as $3.11111111111 \\ldots \\ldots \\ldots \\ldots$ and when a single digit, say $k$ (where $k$ is a natural number from $1$ and $9$) is repeated after decimal point, the result is $\\frac{k}{9}$ plus the number before decimal. Hence, $3.1$ repeated is $3 \\frac{1}{9}$ i.e. $\\frac{28}{9}$ There is another way too. Let $x = 3.1111111 \\ldots \\ldots .$ and then $10 x = 31.111111 \\ldots \\ldots \\ldots .$ Subtracting former from latter we get $9 x = 31 - 3 = 28$ i.e. $x = \\frac{28}{9} = 3 \\frac{1}{9}$", "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 $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{+10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}$$ $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}\\boxed{+10}$$", "This is a pretty easy problem just to bash. Since the max number we can get is $7$, we just need to test n values for $1.5,2.5,3.5,4.5,5.5$ and $6.5$. Then just do how many numbers there are times $\\frac{1}{\\lfloor n \\rfloor}$, which should be $5+17+37+65+101+145+30 = \\boxed{400}$" ]

[ "# Quarter circle train tracks A boy has a set of trains and pieces of railroad track. Each piece is a quarter of circle, and by concatenating these pieces, the boy obtained a closed railway. The railway does not intersect itself. In passing through this railway, the train sometimes goes in the clockwise direction, and sometimes in the opposite direction. Prove that the train passes an even number of times through the pieces in the clockwise direction and an even number of times in the counterclockwise direction. Also, prove that the number of pieces is divisible by $4$. How do we deal with the fact that the rails don't intersect? I tried the following. Let the train start at the point $(0,0)$ and assume without loss of generality that the radius of the circle from which the quarter circle is made is $1$. Also assume that the first end part of a track is at $(1,1)$. Now consider the circle with radius $1$ with center at the origin. Let the arcs be \\begin{align*}&A:(1,0) \\to (0,1) \\\\&B: (0,1) \\to (-1,0) \\\\&C: (-1,0) \\to (0,1) \\\\&D: (0,1) \\to (1,0). \\end{align*}Notice that we can go from an $A$-track to a $B,C^c$-track; a $B$-track to a $C,D^c$-track; a $C$ track to a $D,A^c$ track; a $D$ track to a $A,B^c$ track (the", "# Thread: People Moving Around Circle Track 1. ## People Moving Around Circle Track Bob and Sally are each going for a run around a circlar track of radius 80 feet. They start running from different locations on the circle. Bob starts at the easternmost point and jogs 10 feet/second around the track in the clockwise direction. Sally runs counterclockwise at 14 feet/second and passes Bob for the first time in 16 seconds. (Words I bolded are the given information we need) Find Sally's x and y coordinates in 12 minutes. So here is what I'm thinking on a coordinate plan Bob starts at the easternmost point (meaning all the way on the right), at (80,0) Sally is somewhere Circumference: 2πr = 2π(80) = 160π feet, π is pi by the way. This is the full length track. Though, I'm still confused on how to start out. Can someone help? Thanks 2. Can you write an expresion to locate Bob an any moment? Hint: < 80*cos(t) , 80*sin(t) > 3. bob and Sally are each going for a run around a circlar track of radius 80 feet. They start running from different locations on the circle. Bob starts at the easternmost point and jogs 10 feet/second around the track in the clockwise direction. Sally runs counterclockwise at 14", "# Thread: People Moving Around Circle Track 1. ## People Moving Around Circle Track Bob and Sally are each going for a run around a circlar track of radius 80 feet. They start running from different locations on the circle. Bob starts at the easternmost point and jogs 10 feet/second around the track in the clockwise direction. Sally runs counterclockwise at 14 feet/second and passes Bob for the first time in 16 seconds. (Words I bolded are the given information we need) Find Sally's x and y coordinates in 12 minutes. So here is what I'm thinking on a coordinate plan Bob starts at the easternmost point (meaning all the way on the right), at (80,0) Sally is somewhere Circumference: 2πr = 2π(80) = 160π feet, π is pi by the way. This is the full length track. Though, I'm still confused on how to start out. Can someone help? Thanks 2. ## Re: People Moving Around Circle Track Hi, Here's a blueprint to solve this problem: 1) You may write the positions on the track as a function of an angle which varies with time because Bob and Sally are moving. Let's call this angle $\\theta$. 2) Find $\\theta$ as a function of time for each runner 3) Using the previous functions, you can use the time", "$$(\\frac{10}{13})^{th}$$ of the track length. D. $$(\\frac{11}{13})^{th}$$ of the track length. E. $$(\\frac{12}{13})^{th}$$ of the track length. Are You Up For the Challenge: 700 Level Questions Hi exc4libur One should know that the distance covered will be in opposite ratio of their speed. So A:B in speed is 25:40=5:8, so distances covered by them will be in ratio 8:5. Now whenever both meet, they complete one circle, so they will cover 10 circles when they meet 10th time. So A will cover 10*5/(5+8) =50/13=(39+11)/13=3 + 11/13 So A would have covered 3 complete circles and would be at 11/13 of 4th circle. _________________ VP Joined: 24 Nov 2016 Posts: 1352 Location: United States Re: A runs in the clockwise direction on a circular track whereas B runs [#permalink] ### Show Tags 18 Jan 2020, 04:19 chetan2u wrote: Bunuel wrote: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point. If their speeds are 25 m/s and 40 m/s respectively, at what fraction of the track-length, measured in clockwise direction from the starting point, does the 10th meeting take place? A. $$(\\frac{8}{13})^{th}$$ of the track length. B. $$(\\frac{9}{13})^{th}$$ of the track length. C. $$(\\frac{10}{13})^{th}$$ of the track length. D. $$(\\frac{11}{13})^{th}$$", "# Math Help - tracks 1. ## tracks A jogging park has 2 identical circular tracks touching each other, and a rectangular track enclosing the 2 circles.The edges of the rectangles are tangential to the circles.Two friends , A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight.Approximately, how much faster than A does B have top run, so that they take the same time to return to their starting point. options are: a. 3.88% b. 4.22% c. 4.44% d. 4.72% 2. ## I will give you a hint Originally Posted by manisha A jogging park has 2 identical circular tracks touching each other, and a rectangular track enclosing the 2 circles.The edges of the rectangles are tangential to the circles.Two friends , A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight.Approximately, how much faster than A does B have top run, so that they take the same time to return to their", "2.5$ ###### 56 $s = 4.3$ ###### 57 $s = 8.5$ ###### 58 $s = 11$ ###### 59 City Park features a circular jogging track of radius 1 mile, centered on the open-air bandstand. You start jogging on the track 1 mile due east of the bandstand and proceed counterclockwise. What are your coordinates, relative to the bandstand, when you have jogged five miles? ###### 60 Silver Reservoir is a circular man-made lake of radius 1 kilometer. If you start at the easternmost point on the reservoir and walk counterclockwise for 4 kilometers, how far south of your intial position are you? For Problems 61–66, find the angle in radians between $0$ and $2\\pi$ determined by the terminal point on the unit circle. Round your answer to hundredths. ###### 61 $(-0.1782, 0.9840)$ ###### 62 $(-0.8968, -0.4425)$ ###### 63 $(0.8855, -0.4646)$ ###### 64 $(0.9801, 0.1987)$ ###### 65 $(-0.7659, -0.6430)$ ###### 66 $(0.9602, -0.2794)$ ###### 67 1. Sketch a unit circle and the line $y = x\\text{.}$ Find the coordinates of the two points where the line and the circle intersect. ###### 68 1. Sketch a unit circle and the line $y = -x\\text{.}$ Find the coordinates of the two points where the line and the circle intersect. ###### 69 1. Sketch a line that passes through the origin and", "# Difference between revisions of \"1999 AMC 8 Problems/Problem 4\" ## Problem The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn? $[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label(rotate(30)*\"Bjorn\",(12,6.75),SE); label(rotate(37)*\"Alberto\",(11,8.25),NW); label(\"0\",(0,0),S); label(\"1\",(4,0),S); label(\"2\",(8,0),S); label(\"3\",(12,0),S); label(\"4\",(16,0),S); label(\"5\",(20,0),S); label(\"0\",(0,0),W); label(\"15\",(0,3),W); label(\"30\",(0,6),W); label(\"45\",(0,9),W); label(\"60\",(0,12),W); label(\"75\",(0,15),W); label(\"H\",(6,-2),S); label(\"O\",(8,-2),S); label(\"U\",(10,-2),S); label(\"R\",(12,-2),S); label(\"S\",(14,-2),S); label(\"M\",(-4,11),N); label(\"I\",(-4,9),N); label(\"L\",(-4,7),N); label(\"E\",(-4,5),N); label(\"S\",(-4,3),N); [/asy]$ $\\text{(A)}\\ 15 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 25 \\qquad \\text{(D)}\\ 30 \\qquad \\text{(E)}\\ 35$ ## Solution After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\\boxed{\\text{(A)}}$. ## Solution 2 We see that each dot is $15$ units away from the nearest one above it. So the answer is $\\boxed{\\text{A}}$.", "have walked the distance in four-fifths of the time; if he had gone $\\textstyle\\frac{1}{2}$ mile per hour slower, he would have been $2\\textstyle\\frac{1}{2}$ hours longer on the road. The distance in miles he walked was $\\textbf{(A) }13\\textstyle\\frac{1}{2}\\qquad \\textbf{(B) }15\\qquad \\textbf{(C) }17\\frac{1}{2}\\qquad \\textbf{(D) }20\\qquad \\textbf{(E) }25$ ## Problem 25 Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$, and $60$ taken consecutively. The diameter of this circle has length $\\textbf{(A) }62\\qquad \\textbf{(B) }63\\qquad \\textbf{(C) }65\\qquad \\textbf{(D) }66\\qquad \\textbf{(E) }69$ ## Problem 26 $[asy] real t=pi/8;real u=7*pi/12;real v=13*pi/12; real ct=cos(t);real st=sin(t);real cu=cos(u);real su=sin(u); draw(unitcircle); draw((ct,st)--(-ct,st)--(cos(v),sin(v))); draw((cu,su)--(cu,st)); label(\"A\",(-ct,st),W);label(\"B\",(ct,st),E); label(\"M\",(cu,su),N);label(\"P\",(cu,st),S); label(\"C\",(cos(v),sin(v)),W); //Credit to Zimbalono for the diagram [/asy]$ In the circle above, $M$ is the midpoint of arc $CAB$ and segment $MP$ is perpendicular to chord $AB$ at $P$. If the measure of chord $AC$ is $x$ and that of segment $AP$ is $(x+1)$, then segment $PB$ has measure equal to $\\textbf{(A) }3x+2\\qquad \\textbf{(B) }3x+1\\qquad \\textbf{(C) }2x+3\\qquad \\textbf{(D) }2x+2\\qquad \\textbf{(E) }2x+1$ ## Problem 27 If the area of $\\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\\sin A$ is equal to $\\textbf{(A) }\\dfrac{\\sqrt{3}}{2}\\qquad \\textbf{(B) }\\frac{3}{5}\\qquad \\textbf{(C) }\\frac{4}{5}\\qquad \\textbf{(D) }\\frac{8}{9}\\qquad \\textbf{(E) }\\frac{15}{17}$ ## Problem 28 A circular disc with diameter $D$ is placed on an $8\\times 8$", "of the track length. E. $$(\\frac{12}{13})^{th}$$ of the track length. Are You Up For the Challenge: 700 Level Questions Hi exc4libur One should know that the distance covered will be in opposite ratio of their speed. So A:B in speed is 25:40=5:8, so distances covered by them will be in ratio 8:5. Now whenever both meet, they complete one circle, so they will cover 10 circles when they meet 10th time. So A will cover 10*5/(5+8) =50/13=(39+11)/13=3 + 11/13 So A would have covered 3 complete circles and would be at 11/13 of 4th circle. Thanks chetan2u!! I found what I was doing wrong, I was calculating the final distance for anti-clockwise, instead of clockwise. Intern Joined: 14 Jul 2019 Posts: 32 Re: A runs in the clockwise direction on a circular track whereas B runs [#permalink] ### Show Tags 04 Mar 2020, 05:53 Bunuel wrote: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point. If their speeds are 25 m/s and 40 m/s respectively, at what fraction of the track-length, measured in clockwise direction from the starting point, does the 10th meeting take place? A. $$(\\frac{8}{13})^{th}$$ of the track length. B. $$(\\frac{9}{13})^{th}$$ of the track length. C.", "# Vectors around a circular track 1. Oct 21, 2004 ### seanlindo06 A jogger runs around a 2 km circular track as shown below. Using graph paper, draw a displacement - time graph of the continuous North vector which runs from the start axis to the jogger's current position. Identify the shape of your graph. Draw a corresponding distance - time graph for the jogger's, overall distance covered. Label important points between the 2 graphs. circular distance time (m) (s) 0 0 250 60 500 130 750 185 1000 245 1250 303 1500 366 1750 437 2000 501 also included is a circle with a perp. line inside of it . The point on the right side of the circle is labeled start/finish. Any clue on how to do this? 2. Oct 22, 2004 ### HallsofIvy Staff Emeritus You are told that the track is a \"2 km circular track\" so I assume that \"2 km\" is the length. I suspect that this problem is preparation for \"harmonic motion\" and that you are not expected to do detailed calculations with trig functions. Here's what I would do. Since the circle given is not actually 2 km long (!) figure out how long it really is (measure its diameter and multiply by pi) 2km/actual length is the factor you", "following diagram: $[asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, H, M, N; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); H = foot(S,C,D); M = foot(O,S,H); N = foot(O,R,T); fill(O--M--P--cycle,yellow); fill(O--N--P--cycle,green); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); draw(P--R^^P--S^^P--T^^P--H,red+dashed); dot(\"A\",A,1.5*dir(225),linewidth(4.5)); dot(\"B\",B,1.5*dir(-45),linewidth(4.5)); dot(\"C\",C,1.5*dir(45),linewidth(4.5)); dot(\"D\",D,1.5*dir(90),linewidth(4.5)); dot(\"P\",P,1.5*dir(60),linewidth(4.5)); dot(\"R\",R,1.5*dir(135),linewidth(4.5)); dot(\"S\",S,1.5*dir(-90),linewidth(4.5)); dot(\"T\",T,1.5*dir(-45),linewidth(4.5)); dot(\"O\",O,1.5*dir(45),linewidth(4.5)); dot(\"H\",H,1.5*dir(90),linewidth(4.5)); dot(\"M\",M,1.5*dir(180),linewidth(4.5)); dot(\"N\",N,1.5*dir(15),linewidth(4.5)); label(\"9\",midpoint(P--R),dir(A-D),red); label(\"5\",midpoint(P--S),dir(180),red); label(\"16\",midpoint(P--T),dir(A-D),red); [/asy]$ Note that the diameter of $\\odot O$ is $HS=RT=25,$ so $OP=\\frac{25}{2}.$ It follows that: 1. In right $\\triangle OMP,$ we have $MP=\\frac{HS}{2}-PS=\\frac{15}{2}$ by symmetry, from which $OM=10$ by the Pythagorean Theorem. 2. In right $\\triangle ONP,$ we have $NP=\\frac{RT}{2}-RP=\\frac{7}{2}$ by symmetry, from which $ON=12$ by the Pythagorean Theorem. Since $\\overline{MO}\\parallel\\overline{AB}$ and $\\overline{ON}\\parallel\\overline{DA},$ we conclude that $\\angle A = \\angle MON.$ We apply the Sine of a Sum Formula: \\begin{align*} \\sin\\angle A &= \\sin\\angle MON \\\\ &= \\sin(\\angle MOP + \\angle PON) \\\\ &= \\sin\\angle MOP \\cos\\angle PON + \\cos\\angle MOP \\sin\\angle PON \\\\ &= \\frac{3}{5}\\cdot\\frac{24}{25} + \\frac{4}{5}\\cdot\\frac{7}{25} \\\\ &= \\frac{4}{5}. \\end{align*} Note that $$\\sin\\angle A = \\frac{HS}{DA},$$ from which $\\frac{4}{5} = \\frac{25}{DA}.$ We solve this equation to get $DA=\\frac{125}{4}.$ Finally, the perimeter of $ABCD$ is $4DA", "Free Version Moderate # Distance Around a Circle SATSTM-E7MVEW Samson ran $7$ times around a circular track that has a radius of $47$ yards. Find the approximate distance he ran in yards. A $1033\\text{ yards}$ B $6936\\text{ yards}$ C $658\\text{ yards}$ D $2067\\text{ yards}$ E $329$", "# Difference between revisions of \"2001 AMC 8 Problems/Problem 19\" ## Problem Car M traveled at a constant speed for a given time. This is shown by the dashed line. Car N traveled at twice the speed for the same distance. If Car N's speed and time are shown as solid line, which graph illustrates this? $[asy] unitsize(12); draw((0,9)--(0,0)--(9,0)); label(\"time\",(4.5,0),S); label(\"s\",(0,7),W); label(\"p\",(0,6),W); label(\"e\",(0,5),W); label(\"e\",(0,4),W); label(\"d\",(0,3),W); label(\"(A)\",(-1,9),NW); draw((0,4)--(4,4),dashed); label(\"M\",(4,4),E); draw((0,8)--(4,8),linewidth(1)); label(\"N\",(4,8),E); draw((15,9)--(15,0)--(24,0)); label(\"time\",(19.5,0),S); label(\"s\",(15,7),W); label(\"p\",(15,6),W); label(\"e\",(15,5),W); label(\"e\",(15,4),W); label(\"d\",(15,3),W); label(\"(B)\",(14,9),NW); draw((15,4)--(19,4),dashed); label(\"M\",(19,4),E); draw((15,8)--(23,8),linewidth(1)); label(\"N\",(23,8),E); draw((30,9)--(30,0)--(39,0)); label(\"time\",(34.5,0),S); label(\"s\",(30,7),W); label(\"p\",(30,6),W); label(\"e\",(30,5),W); label(\"e\",(30,4),W); label(\"d\",(30,3),W); label(\"(C)\",(29,9),NW); draw((30,4)--(34,4),dashed); label(\"M\",(34,4),E); draw((30,2)--(34,2),linewidth(1)); label(\"N\",(34,2),E); draw((0,-6)--(0,-15)--(9,-15)); label(\"time\",(4.5,-15),S); label(\"s\",(0,-8),W); label(\"p\",(0,-9),W); label(\"e\",(0,-10),W); label(\"e\",(0,-11),W); label(\"d\",(0,-12),W); label(\"(D)\",(-1,-6),NW); draw((0,-11)--(4,-11),dashed); label(\"M\",(4,-11),E); draw((0,-7)--(2,-7),linewidth(1)); label(\"N\",(2,-7),E); draw((15,-6)--(15,-15)--(24,-15)); label(\"time\",(19.5,-15),S); label(\"s\",(15,-8),W); label(\"p\",(15,-9),W); label(\"e\",(15,-10),W); label(\"e\",(15,-11),W); label(\"d\",(15,-12),W); label(\"(E)\",(14,-6),NW); draw((15,-11)--(19,-11),dashed); label(\"M\",(19,-11),E); draw((15,-13)--(23,-13),linewidth(1)); label(\"N\",(23,-13),E); [/asy]$ ## Solution Since car N has twice the speed, it must be twice as high on the speed axis. Also, since cars M and N travel at the same distance but car N has twice the speed, car N must take half the time. Therefore, line N must be half the size of line M. Since the speeds are constant, both lines are horizontal. Reviewing the graphs, we see that the only one satisfying these conditions is graph $\\boxed{\\text{D}}$.", "# 2015 UNCO Math Contest II Problems Twenty-third Annual UNC Math Contest Final Round January 31, 2015. Three hours; no electronic devices. Answers must be justified to receive full credit. We hope you enjoy thinking about these problems, but you are not expected to do them all. The positive integers are $1, 2, 3, 4, \\ldots$ A polynomial is quadratic if its highest power term has power two. ## Problem 1 The sum of three consecutive integers is $54$. What is the smallest of the three integers? ## Problem 2 $[asy] filldraw(circle((0,0),3),white); filldraw(arc((-1,0),2,0,180)--cycle,grey); filldraw(arc((-2,0),1,0,180)--cycle,white); filldraw(arc((1,0),2,180,360)--cycle,grey); filldraw(arc((2,0),1,180,360)--cycle,white); [/asy]$ Find the area of the shaded region. The outer circle has radius $3$. The shaded region is outlined by half circles whose radii are $1$ and $2$ and whose centers lie on the dashed diameter of the big circle. ## Problem 3 If P is a polynomial that satisfies $P(x^2 +1) = 5x^4 +7x^2 +19$, then what is $P(x)$? (Hint: $P$ is quadratic.) ## Problem 4 Tarantulas $A, B,$ and $C$ start together at the same time and race straight along a $100$ foot path, each running at a constant speed the whole distance. When $A$ reaches the end, $B$ still has $10$ feet more to run. When $B$ reaches the end, $C$ has $20$ feet more to run. How many", "# Problem #1462 1462 A ship sails $10$ miles in a straight line from $A$ to $B$, turns through an angle between $45^{\\circ}$ and $60^{\\circ}$, and then sails another $20$ miles to $C$. Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$? $[asy] unitsize(2mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C=20*dir(50); draw(A--B--C); draw(A--C,linetype(\"4 4\")); dot(A); dot(B); dot(C); label(\"10\",midpoint(A--B),S); label(\"20\",midpoint(B--C),SE); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); [/asy]$ $\\textbf{(A)}\\ [400,500] \\qquad \\textbf{(B)}\\ [500,600] \\qquad \\textbf{(C)}\\ [600,700] \\qquad \\textbf{(D)}\\ [700,800]$ $\\textbf{(E)}\\ [800,900]$ 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.", "direction Mike runs. The fourth is at $160 \\; \\text{m}$ from the start, in the same direction. The fifth is at the start, because the circuit was divided into $5$ parts, as dictated by the relation between the periods. More generally, if the relation between the periods would be $$\\frac{T_1}{T_2} = \\frac{p}{q}$$ with $p$ and $q$ coprime, then the runners meet again at the start after completing in total $(p+q)$ circuits, this means they meet the $(p+q)$th time.", "shown. Kylie can run at 4.5 m/sec through sand and 2.5 m/sec through mud. The track is the hypotenuse from $S$ to $X$, then the hypotenuse from $X$ to $F.$ There is an angle $Y$, which is between the track $SX$ and the start boundary and angle $Z$, which is between the track $XF$ and the finish boundary. Q1) Show that the total distance covered in the race is: $d \\:=\\:200(\\sec Y+\\sec Z)$ ? and prove that Y and Z are also related by the equation: $\\tan Y+\\tan Z\\:=\\:K$ ? where $K$ is a constant. If my diagram is correct (and angles $Y$ and $Z$ are correctly placed), . . then these two statements are not true. We have: . $\\begin{Bmatrix}\\csc Y = \\frac{SX}{200}\\quad\\Rightarrow\\quad SX = 200\\csc Y \\\\ \\\\ \\csc Z = \\frac{XF}{200}\\quad\\Rightarrow\\quad XF = 200\\csc Z\\end{Bmatrix}$ Add: . $SX + XF \\:=\\:200\\csc Y + 200\\csc Z$ . . Therefore: . $\\boxed{d \\;= \\; 200(\\csc Y + \\csc Z)}$ We have: . $\\begin{Bmatrix}\\cot Y = \\frac{AX}{200}\\quad\\Rightarrow\\quad 200\\cot Y = AX \\\\ \\\\ \\cot Z = \\frac{XB}{200}\\quad\\Rightarrow\\quad 200\\cot Z = XB\\end{Bmatrix}$ Add: . $200\\cot Y + 200\\cot Z\\:=\\:AX + XB$ . . Therefore: . $200\\cot Y + 200\\cos Z\\:=\\:500\\quad\\Rightarrow\\quad \\boxed{\\cot Y + \\cot Z = \\frac{5}{2}}$ [So I don't dare try the other parts . . . ]", "45”. 2. Aaron is running along a circular track at the rate of 10 mile per hour traverses in 36 seconds an arc which subtends 56° at the center. Find the diameter of the circle. Solution: One hour = 3600 seconds One mile = 5280 feet Therefore, 10 miles = (5280 × 10) feet = 52800 feet In 3600 seconds Aaron goes 52800 feet In 1 second Aaron goes 52800/3600 feet = 44/3 feet Therefore, in 36 seconds the Aaron goes (44/3) × 36 feet = 528 feet. Clearly, an arc of length 528 feet subtends 56° = 56 × π/180 radian at the center of the circular track. If ‘y’ feet is the radius of the circular track then using the formula s = rθ we get, y = s/θ y = 528/[56 × (π/180)] y = (528 × 180 × 7)/(56 × 22) feet y = 540 feet y = (540/3) yards [since, we know that 3 foot = 1 yard] y = 180 yards Therefore, the required diameter = 2 × 180 yards = 360 yards. 3. If α1, α2, α3 radians be the angles subtended by the arcs of lengths l1, l2, l3 at the centers of the circles whose radii are r1, r2, r3 respectively then show that the angle subtended at the", "# Difference between revisions of \"2020 AMC 8 Problems/Problem 11\" ## Problem 11 After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds? $[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * \"Distance (miles)\", (-0.5,3), W); label(\"Time (minutes)\", (3,-0.5), S); dot(\"Naomi\", (2,6), 3*dir(305)); dot((6,6)); label(\"Maya\", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]$ $\\textbf{(A) }6 \\qquad \\textbf{(B) }12 \\qquad \\textbf{(C) }18 \\qquad \\textbf{(D) }20 \\qquad \\textbf{(E) }24$ ## Solution 1 We use the formula $\\text{speed}=\\dfrac{\\text{distance}}{\\text{time}}$. Naomi's distance is $6$ miles, and her time is $10$ minutes, which is equivalent to $\\dfrac{1}{6}$ of an hour. Since speed is distance over time, Naomi's speed is $36$ mph. Using the same process, Maya's speed is $12$ mph. Subtracting those, we get an answer of $\\boxed{(\\text{E}) 24}$. ## Solution 2 Notice that Naomi travels at a rate of $6$ miles", "was: $\\dfrac{1-p^{2}}{200}\\\\$ $\\dfrac{\\sqrt{1-p^{2}}}{100} \\\\$ $1-\\dfrac{p^{2}}{10,000-p^{2}} \\\\$ $\\dfrac{10,000}{10,000-p^{2}}$ 31 What is the largest positive integer $n$ such that $\\frac{n^{2}+7n+12}{n^{2}-n-12}$ is also a positive integer? $8$ $12$ $16$ $6$ 32 The quadratic equation $x^{2}+bx+c=0$ has two roots $4a$ and $3a$, where a is an integer. Which of the following is a possible value of $b^{2}+c$? $3721$ $549$ $427$ $361$ 33 A cyclist leaves $A$ at $10$ am and reaches $B$ at $11$ am. Starting from $10:01$ am, every minute a motor cycle leaves $A$ and moves towards $B$. Forty-five such motor cycles reach $B$ by $11$ am. All motor cycles have the same speed. If the cyclist had doubled his speed, how many motor cycles would have reached $B$ by the time the cyclist reached $B$? $23$ $20$ $15$ $22$ 34 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 $10:25$am $10:18$am $10:27$am $10:45$am 35 John jogs on track A at $6$ kmph and Mary jogs on track $B$ at $7.5$ kmph. The total length of tracks $A$ and $B$ is $325$ metres." ]

[ "# How long is 7/10 of a mile, in feet? Apr 3, 2016 $3696$ feet #### Explanation: Each mile has $8$ furlongs. Each furlong has $220$ yards and each yard has $3$ feet. Therefore each mile has $8 \\times 220 \\times 3 = 5280$ feet and $\\frac{7}{10}$ mile will have $5280 \\times \\frac{7}{10}$ = $528 \\cancel{5280} \\times \\frac{7}{1 \\cancel{10}} = 3696$ feet.", "more? Any help would be appreciated. - Roo travels $200$ feet to reach the finish line. Since Tigger has to travel in $3$ foot jumps, Tigger needs to travel $102$ feet until (s)he can turn around. Then Tigger needs to travel another $102$ feet to reach the start/finish line, a total of $204$ feet. The two jumpers travel at the same rate. So Roo is $4$ feet ahead of Tigger at the end, sort of. This is in fact not necessarily the case, because it relies on a specific model of jumping. Since we were not given a detailed description of the jump processes of either creature, we are forced to assume something. We implicitly assumed that even although the travelling is divided into jumps, the actual speeds are constant. Suppose instead that a jump consists of standing in one place flexing the leg muscles for a while, and perhaps having a drink, while the movement part of a jump is quick. Then the answer need not be $4$ feet. - I don't see how you got that Roo was four feet ahead. How do you see Tigger travels 204 feet? I only see him jumping 2 feet extra from the first turnaround, meaning he jumps 202 feet. – Joe Mar 3 '12 at 2:51 He (or she)", "second equation, gives: 2bb = 15, where b = 15. The younger brother is 15 years old. 14. C: Multiplying the numerator and denominator of the given fraction by 5 gives the fraction, 25/30, which is equivalent. 15. B: Converting feet to yards, the dimensions may be rewritten as 4 yards by 6 2/3 yards. Thus, the area of the floor is 26 2/3 yd’. Multiplication of this area by the cost per square yard gives the expression, 26 2/3?8.91, which equals 237.6. Thus, the cost is$237.60. 16. C: The following equation may be solved for x: 6000=1/2 x. Solving for x gives x = 12,000. Thus, the amount of prize money distributed equaled $12,000. 17. D: The distance may be determined by writing and solving the following equation for c: 50 2 +120 2 =c 2 . c = 130, thus the distance is 130 feet. 18. D: The average rate for the round trip is the total distance traveled divided by the total travel time. Distance traveled=2x. Travel time=x/10+x/8=4x/40+5x/40=9x/40. Average Rate=2xx9x/40=(2xx40)/9x=80/9 = approximately 8.9 mph. 19. E: The length is equal to 216 inches. The width is equal to 36 inches. So, the length may be covered by 18 12-inch stones, while the width may be covered by 3 12-inch stones. A total of 42 stones", "makes monthly payments for 48 months in order to pay off the car. He does not pay interest. How much are his payments each month? __________________________ Given that, The total cost of the car = $10,608. The total number of months that Viktor pays = 48 months. For each month Viktor payment =$10,608/48 = $221. Therefore Viktor’s payment for each month =$221. Question 12. A professional running track has two straight sections that are each 328.084 feet long. The total distance around the track is 1,312.34 feet. What is the length of the part of the track that is not straight? __________________________ Given that, The length is of the two straight sections each = 328.084 feet long. The total distance around the track is 1,312.34 feet. The formula for the length of the track that is not straight = total distance – 2 x (straight sections). Section not straight = 1,312.34 – 2 x 328.084 Section not straight = 1,312.34 – 656.168 Section not straight = 656.172 feets. Question 13. What is the greatest common factor of 72 and 34? __________________________", "# 1000 feet is what in miles? $1000$ feet is $\\frac{25}{132}$ miles A mile has $8$ furlongs, each of which has $220$ yards i.e. a mile has $8 \\times 220 = 1760$ yards. But as one yard has $3$ feet, one mile has $3 \\times 1760 = 5280$ feet. Hence $1000$ feet is $\\frac{1000}{5280}$ miles. As numerator and denominator, both are divisible by $40$. $\\frac{1000}{5280} = \\frac{25 \\times 40}{132 \\times 40} = \\frac{25}{132}$ miles", "Calculation carried out with six figure logs. (500 × 500 × 90)/5120000 = 4.394. Ans. 4 lbs 6.3 oz. Click it to see a larger version If, with a charge of 4 lbs of powder, a shell range 2000 feet, how far will its range when the charge is 5 lbs, the elevation being in both cases the same. Walker computes, using six figure logs 4 : 5 : : 2000 : $x$. Ans. 2500 feet. Click it to see a larger version At what time will a shell range 4000 feet, when discharged at an elevation of 45°. Walker writes $R$ : tan 45° : : 4000 : $x$ Writing the log of tan 45° and sin 90° as 10, i.e. the log of $10^{10}\\tan 45°$ and $10^{10}\\sin 90°$, he computes ($4000 \\times \\tan 45°)/\\sin 90°$, takes the square root, then divides by 4 to obtain 15.80. Ans. 15.80 secs. Click it to see a larger version A shell discharged at an elevation of 40° ranges 3000 ft. required its time of flight. Walker writes $R$ : tan 40° : : 4000 : $x$ Writing the log of tan 40° as 9.923813 and of sin 90° as 10, i.e. the log of $10^{10}\\tan 45°$ and of $10^{10}\\sin 90°$, he computes ($3000 \\times \\tan 40°)/\\sin 90°$, takes the", "C. 6 D. 10 E. 60 Drill 4 Question: 2 Page: 531 [Reveal] Spoiler: OA _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Retired Moderator Joined: 07 Jun 2014 Posts: 4803 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) Followers: 175 Kudos [?]: 3036 [0], given: 394 Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink] 29 Dec 2017, 15:39 Expert's post Explanation Use the Rate Pie. Ali is traveling 50 feet per second faster than Jeff is traveling. Therefore, that is the rate at which she is effectively gaining ground on him. Put that in the lower-right segment of the Rate Pie. We want to know how long it will take her to gain 3,000 feet on him. Put 3,000 in the top section of the Rate Pie. Now you can see that dividing $$\\frac{3000}{50}$$ will fill in the last segment of the Rate Pie, telling you how long it takes Ali to do so is 60 seconds. Be aware that choice (E) is an incorrect partial answer. Now you need to find out how many feet Ali will travel in 60 seconds, by multiplying 60 second × 300 feet per second, which equals 18,000 feet. Divide", "C. 6 D. 10 E. 60 Drill 4 Question: 2 Page: 531 [Reveal] Spoiler: OA _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Retired Moderator Joined: 07 Jun 2014 Posts: 4803 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) Followers: 171 Kudos [?]: 2915 [0], given: 394 Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink] 29 Dec 2017, 15:39 Expert's post Explanation Use the Rate Pie. Ali is traveling 50 feet per second faster than Jeff is traveling. Therefore, that is the rate at which she is effectively gaining ground on him. Put that in the lower-right segment of the Rate Pie. We want to know how long it will take her to gain 3,000 feet on him. Put 3,000 in the top section of the Rate Pie. Now you can see that dividing $$\\frac{3000}{50}$$ will fill in the last segment of the Rate Pie, telling you how long it takes Ali to do so is 60 seconds. Be aware that choice (E) is an incorrect partial answer. Now you need to find out how many feet Ali will travel in 60 seconds, by multiplying 60 second × 300 feet per second, which equals 18,000 feet. Divide", "seconds (at $0.001 wall increments). – David May 21 '16 at 17:40 • Also, another fun simulation method to try (since the discreet number of possible wall position combinations is reasonable = about 1 million total), is to use a combination of Monte Carlo (random) style for the first$90$% or so (preventing any duplicates), then use a table of non-seen wall position combinations to fill in the missing ones. That way you will walk all of the possible paths and get a fair average and wont get different results each run. This may be good for someone that started with a Monte Carlo type simulation and wants it to be exact each time, not at the mercy of randomness. – David May 21 '16 at 17:49 • The 3rd simulation is more difficult than I thought it would be. The problem is if I want to hold a slope of say$0.375$the first mile, I now have to decide what slope do I take if I encounter a wall (let's say the small wall) at mile$0.75$for example. The points I hit would be$(.5, ,1875), (.75, .28125), (.75, .5)$. Then the question is where do I go next? Do I slope down to B after encountering the first (and possibly only) wall, or do I slope to where I would", "To convert a distance of 3.7 miles to feet, which ratio could you multiply by? Nov 2, 2016 5,280 Explanation: There are 5,280 feet in a mile. Converting 3.7 miles to feet, $3.7 \\cdot 5280 = 19536$ Nov 2, 2016 $5280$ Explanation: There are $1760$ yards in a mile and $3$ feet in a yard. Hence there are $5280 = 1760 \\times 3$ feet in a mile. So $3.7$ miles is: $3.7 \\times 5280 = 19536$ feet Nov 4, 2016 So: $\\text{ \"3.7xx(\"top number\")/(\"bottom numbers\") = \"feet for 3.7 miles}$ color(green)(\"Has the ratio of \"(\"top number\")/(\"bottom numbers\") =5280/1) Explanation: $\\textcolor{b l u e}{\\text{Preamble}}$ Depending on which way up you wish to write the ratio. As long as you clearly indicate which is which you should be fine. I choose $\\left(\\text{feet\")/(\"miles}\\right)$ because we are given the miles but not the feet. It is much better to put the unknown on the top (numerator) as this reduces the amount of manipulation required to determine its value. '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\\textcolor{b l u e}{\\text{Building the ratio}}$ I think they are looking for something like. 3.7xx(\"top number\")/(\"bottom numbers\") = \"feet for 3.7 miles\" ...................................................................................... Known: 5280 feet = 1 mile $\\textcolor{b r o w n}{\\frac{\\text{feet\")/(\"miles\") -> 5280/1 -=(\"feet}}{3.7}}$ Multiply both sides by $\\textcolor{b l u e}{3.7}$ color(brown)(color(blue)(3.7xx)5280/1 \" \" =\" \"(\"feet\")/3.7color(blue)(xx3.7) ) $\\textcolor{b r", "18,000 feet. Divide 18,000 feet by the length of one lap, or 3,000 feet, and you’ll find that it will take Ali 6 laps to overtake Jeff. Hence option B is correct! Answer should be C anyway. since B is 5 laps Retired Moderator Joined: 07 Jun 2014 Posts: 4803 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) Followers: 171 Kudos [?]: 2915 [0], given: 394 Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink] 30 Dec 2017, 05:57 Expert's post I think you are right the answer should be indeed C. _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Intern Joined: 08 Apr 2018 Posts: 44 Followers: 0 Kudos [?]: 16 [0], given: 15 Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink] 04 Jul 2018, 11:13 sandy wrote: Explanation Use the Rate Pie. Ali is traveling 50 feet per second faster than Jeff is traveling. Therefore, that is the rate at which she is effectively gaining ground on him. Put that in the lower-right segment of the Rate Pie. We want to know how long it will take her to gain 3,000 feet on him. Put 3,000 in the top section of the", "18,000 feet. Divide 18,000 feet by the length of one lap, or 3,000 feet, and you’ll find that it will take Ali 6 laps to overtake Jeff. Hence option B is correct! Answer should be C anyway. since B is 5 laps Retired Moderator Joined: 07 Jun 2014 Posts: 4803 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) Followers: 175 Kudos [?]: 3036 [0], given: 394 Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink] 30 Dec 2017, 05:57 Expert's post I think you are right the answer should be indeed C. _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Intern Joined: 08 Apr 2018 Posts: 44 Followers: 0 Kudos [?]: 17 [0], given: 15 Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink] 04 Jul 2018, 11:13 sandy wrote: Explanation Use the Rate Pie. Ali is traveling 50 feet per second faster than Jeff is traveling. Therefore, that is the rate at which she is effectively gaining ground on him. Put that in the lower-right segment of the Rate Pie. We want to know how long it will take her to gain 3,000 feet on him. Put 3,000 in the top section of the", "have $2048$ tracks so, maximum swing (seek length) can be $2047$ • Which corresponds to a seek length of $2^{11} - 1$ in the $11$th service. Correct Answer: $C$ by Veteran (57k points) edited 0 jst fab ... 0 Awesome explanation 0 @ Debashish max-swing can't be 2047 it will exceed the total no of tracks i.e with swing of 2047 the track no will be >2047.. We can have swing length only till 1023 to be in the range of 0-2047. So I feel it can only go till 1023 i.e 2^10-1 and hence the 10th service. correct me if I am wrong.. +9 I guess my numbering in the above image is not correct. But I think if you look at the following image, for a total $16$ tracks we can have $15$ ( $=2^{\\color{red}{4}} - 1$ ) length swing. And in the question, it is mentioned that we can start from any track. The track $6$ is the starting point in the following case. We need not worry about the starting point. But we just know that $2047$ swing length is possible indeed. Does it make sense? +4 Ohh okay! I was under the impression that we only get max requests when we start from the middle, and if we do start from the middle", "Free Version Moderate # Distance Around a Circle SATSTM-E7MVEW Samson ran $7$ times around a circular track that has a radius of $47$ yards. Find the approximate distance he ran in yards. A $1033\\text{ yards}$ B $6936\\text{ yards}$ C $658\\text{ yards}$ D $2067\\text{ yards}$ E $329$", "L \\:=\\:528$ Therefore, the train is 528 feet long.", "+0 # help 0 34 2 The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per second. When Bill steps on the walkway, a group of people that are also on the walkway stands 120 feet in front of him. He walks toward the group at a rate of 3 feet per second. Once Bill reaches the group of people, he stops walking and stands with them until the walkway ends. What is Bill’s average rate of movement for his trip along the moving walkway? Jan 30, 2020 #1 +177 +1 We know that Bill covers $$120$$ feet at $$3$$ feet per sec. Thus, the time taken to cover the walkway will be $$\\frac{120}{3} \\to 40$$ seconds. Since his default walking speed is $$6,$$ then the amount he can cover in $$40$$ seconds is $$6 \\cdot 40 \\to240.$$ Now he has $$60$$ feet left to cover, which will take him $$\\frac{60}{3} \\to 20$$ seconds. Since the Average speed=The Total distance/TheTotal time, we get $$\\frac{300}{20+40} \\to \\frac{300}{60}=\\boxed{5}$$ feet per second is his average rate. Jan 30, 2020 #2 +20216 +1 Bill has 300 ft to cover the people 120 ft in front of him have 180 feet to cover at conveyor speed = 180 feet/3 ft/sec = 60 seconds", "[Reveal] Spoiler: OA _________________ Moderator Joined: 18 Apr 2015 Posts: 5150 Followers: 77 Kudos [?]: 1030 [0], given: 4640 Re: Approximately what is the total stopping distance, in feet, [#permalink] 19 Jan 2016, 08:44 Expert's post Solution The total stopping distance is equal to the sum of the distance traveled during reaction time and the distance traveled after the brakes have been applied ON the left side graph, we do have, accordingly with the instructions provided: 45 ON the right side graph, we do have a rough value of $$75 < X < 90$$ $$X= 80/85$$ $$45 + 80/85 = 125/130$$ The correct answer is $$A$$ _________________ Re: Approximately what is the total stopping distance, in feet, [#permalink] 19 Jan 2016, 08:44 Display posts from previous: Sort by", "= -62 60 – 62 = -2 Amy score is -2 Question 38. Justin dug a hole that was 9$$\\frac{1}{2}$$ inches deep. His sister Jean built a sand castle next to the hole that was 12$$\\frac{3}{4}$$ inches high. Show how you could use subtraction to find the vertical distance from the top of the castle to the bottom of the hole. (Chapter 2) Justin dug a hole that was 9$$\\frac{1}{2}$$ inches deep. His sister Jean built a sand castle next to the hole that was 12$$\\frac{3}{4}$$ inches high. 12$$\\frac{3}{4}$$ – 9$$\\frac{1}{2}$$ 12 – 9 = 3 $$\\frac{3}{4}$$ – $$\\frac{1}{2}$$ = $$\\frac{1}{4}$$ = 3$$\\frac{1}{4}$$ inches Question 39. Rick started his hike at an elevation of 8,975 feet. He descended at a constant rate of 8.5 feet per minute for 45 minutes. Find Rick’s elevation after 45 minutes. Give your answer to 4 significant digits. (Chapters 1, 2) Given, Rick started his hike at an elevation of 8,975 feet. He descended at a constant rate of 8.5 feet per minute for 45 minutes. 8.5 × 45 = 382.5 feet 8975 feet – 382.5 feet = 8592.5 feet 8593 feet Question 40. A car travels’at 68$$\\frac{2}{3}$$ miles per hour for 2$$\\frac{1}{12}$$ hours. Find the distance that the car has traveled correct to 3 significant digits. (Chapters 1, 2) Answer: 68$$\\frac{2}{3}$$ ×", "# The official length for a marathon race is 26.2 miles. How many yards long is a marathon? How many feet? Oct 28, 2016 $138 , 336$ feet $46 , 112$ yards See Explanation #### Explanation: So, first of all, we are told that a marathon race is $26.2$ miles. We are asked to give the equal length in yards and feet. It is really helpful to remember a few conversion factors: 1) $1$ mile=$5 , 280$ feet 2) $1$ miles=$1 , 760$ yards Since each mile equals $5 , 280$ feet, let's multiply $26.2$ by $5 , 280$, which equals to $138 , 336$ feet. Since each mile equals $1 , 760$ yards, let's multiply $26.2$ by $1 , 760$, which equals to $46 , 112$ yards.", "3 times... whatever I do to the bottom, I have to do to the top.0505 X becomes 12; or again you can just do cross multiplying.0511 You can do 15 times 4 equal to 5 times X.0520 Then you can see what you have to multiply by 5 to get this number.0524 My X is going to be 12 because 12/4 is going to be 3/1.0531 That is the same thing as 15/5.0538 I have to look back and see what am I looking for?0543 I know that X is 12; but it is asking for the cost.0546 We know cost is money.0551 How much is it going to cost for 4 pounds?$12.0555 The next one, 15 feet for every 4 minutes; find how many feet in 10 minutes.0562 My word ratio, I am going to make it 50 over minutes.0570 My 16 feet is going to go on the top.0581 My 4 minutes is going to go on the bottom.0584 Equal it to how many feet?--find how many feet.0588 That is what we are looking for; feet, that is the top number.0591 That is X; over the number of minutes is 10.0595 Again you can look for equivalent fraction.0606 This is going to be the same... 16/4 is going to be the same as...0610 If you divide" ]

[ "$$(\\frac{10}{13})^{th}$$ of the track length. D. $$(\\frac{11}{13})^{th}$$ of the track length. E. $$(\\frac{12}{13})^{th}$$ of the track length. Are You Up For the Challenge: 700 Level Questions Hi exc4libur One should know that the distance covered will be in opposite ratio of their speed. So A:B in speed is 25:40=5:8, so distances covered by them will be in ratio 8:5. Now whenever both meet, they complete one circle, so they will cover 10 circles when they meet 10th time. So A will cover 10*5/(5+8) =50/13=(39+11)/13=3 + 11/13 So A would have covered 3 complete circles and would be at 11/13 of 4th circle. _________________ VP Joined: 24 Nov 2016 Posts: 1352 Location: United States Re: A runs in the clockwise direction on a circular track whereas B runs [#permalink] ### Show Tags 18 Jan 2020, 04:19 chetan2u wrote: Bunuel wrote: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point. If their speeds are 25 m/s and 40 m/s respectively, at what fraction of the track-length, measured in clockwise direction from the starting point, does the 10th meeting take place? A. $$(\\frac{8}{13})^{th}$$ of the track length. B. $$(\\frac{9}{13})^{th}$$ of the track length. C. $$(\\frac{10}{13})^{th}$$ of the track length. D. $$(\\frac{11}{13})^{th}$$", "Circumference of track = 20π ≈ 60 miles. In 10 hours, distance for B = 2*10 = 20 miles. 60-20 = 40 miles between A and B. A and B now ...” March 2, 2019 fskilnik@GMATH posted a reply to From the figure shown, each of N people was asked to take 7 in the Problem Solving forum “https://i.postimg.cc/1XW4BFjw/01-Mar19-9b.gif$$?\\,\\, = \\,\\,\\left( {\\# \\,\\,{\\rm{possible}}\\,\\,\\Delta \\,\\,{\\rm{equilateral}}\\,\\,{\\rm{configurations}}} \\right)\\,\\,{\\rm{ + }}\\,\\,{\\rm{1}}? = 15 + 1 = 16$$The correct answer is (D). We follow the notations and rationale taught in the ...” March 1, 2019 fskilnik@GMATH posted a reply to Oneyde and Jorge must make two reservations for a trip, in w in the Problem Solving forum “$$?\\,\\,\\,\\,:\\,\\,\\,\\# \\,\\,\\,O,J\\,\\,{\\rm{positions}}\\,\\,{\\rm{with}}\\,\\,{\\rm{restrictions}}{\\rm{4 - seats}}\\,\\,{\\rm{group}}\\,\\,{\\rm{(left)}}\\,\\,{\\rm{:}}\\,\\,{\\rm{6}}\\,\\,{\\rm{possibilities}}\\,\\,\\,\\left\\{ \\matrix{ \\,\\left( {O,J,e,e} \\right)\\,\\,{\\rm{or}}\\,\\,\\left( {J,O,e,e} \\right) \\hfill \\cr ...” March 1, 2019 fskilnik@GMATH posted a new topic called Oneyde and Jorge must make two reservations for a trip, in w in the Problem Solving forum “GMATH practice exercise (Quant Class 18) https://i.postimg.cc/W31dkS9T/01-Mar19-12a.gif Oneyde and Jorge must make two reservations for a trip, in which only 9 seats are still available, all of them in row F (see figure). If they want to sit next to each other, without being separated by ...” March 1, 2019 fskilnik@GMATH posted a new topic called From the figure shown, each of N people was asked to", "have walked the distance in four-fifths of the time; if he had gone $\\textstyle\\frac{1}{2}$ mile per hour slower, he would have been $2\\textstyle\\frac{1}{2}$ hours longer on the road. The distance in miles he walked was $\\textbf{(A) }13\\textstyle\\frac{1}{2}\\qquad \\textbf{(B) }15\\qquad \\textbf{(C) }17\\frac{1}{2}\\qquad \\textbf{(D) }20\\qquad \\textbf{(E) }25$ ## Problem 25 Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$, and $60$ taken consecutively. The diameter of this circle has length $\\textbf{(A) }62\\qquad \\textbf{(B) }63\\qquad \\textbf{(C) }65\\qquad \\textbf{(D) }66\\qquad \\textbf{(E) }69$ ## Problem 26 $[asy] real t=pi/8;real u=7*pi/12;real v=13*pi/12; real ct=cos(t);real st=sin(t);real cu=cos(u);real su=sin(u); draw(unitcircle); draw((ct,st)--(-ct,st)--(cos(v),sin(v))); draw((cu,su)--(cu,st)); label(\"A\",(-ct,st),W);label(\"B\",(ct,st),E); label(\"M\",(cu,su),N);label(\"P\",(cu,st),S); label(\"C\",(cos(v),sin(v)),W); //Credit to Zimbalono for the diagram [/asy]$ In the circle above, $M$ is the midpoint of arc $CAB$ and segment $MP$ is perpendicular to chord $AB$ at $P$. If the measure of chord $AC$ is $x$ and that of segment $AP$ is $(x+1)$, then segment $PB$ has measure equal to $\\textbf{(A) }3x+2\\qquad \\textbf{(B) }3x+1\\qquad \\textbf{(C) }2x+3\\qquad \\textbf{(D) }2x+2\\qquad \\textbf{(E) }2x+1$ ## Problem 27 If the area of $\\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\\sin A$ is equal to $\\textbf{(A) }\\dfrac{\\sqrt{3}}{2}\\qquad \\textbf{(B) }\\frac{3}{5}\\qquad \\textbf{(C) }\\frac{4}{5}\\qquad \\textbf{(D) }\\frac{8}{9}\\qquad \\textbf{(E) }\\frac{15}{17}$ ## Problem 28 A circular disc with diameter $D$ is placed on an $8\\times 8$", "# Thread: People Moving Around Circle Track 1. ## People Moving Around Circle Track Bob and Sally are each going for a run around a circlar track of radius 80 feet. They start running from different locations on the circle. Bob starts at the easternmost point and jogs 10 feet/second around the track in the clockwise direction. Sally runs counterclockwise at 14 feet/second and passes Bob for the first time in 16 seconds. (Words I bolded are the given information we need) Find Sally's x and y coordinates in 12 minutes. So here is what I'm thinking on a coordinate plan Bob starts at the easternmost point (meaning all the way on the right), at (80,0) Sally is somewhere Circumference: 2πr = 2π(80) = 160π feet, π is pi by the way. This is the full length track. Though, I'm still confused on how to start out. Can someone help? Thanks 2. Can you write an expresion to locate Bob an any moment? Hint: < 80*cos(t) , 80*sin(t) > 3. bob and Sally are each going for a run around a circlar track of radius 80 feet. They start running from different locations on the circle. Bob starts at the easternmost point and jogs 10 feet/second around the track in the clockwise direction. Sally runs counterclockwise at 14", "# Thread: People Moving Around Circle Track 1. ## People Moving Around Circle Track Bob and Sally are each going for a run around a circlar track of radius 80 feet. They start running from different locations on the circle. Bob starts at the easternmost point and jogs 10 feet/second around the track in the clockwise direction. Sally runs counterclockwise at 14 feet/second and passes Bob for the first time in 16 seconds. (Words I bolded are the given information we need) Find Sally's x and y coordinates in 12 minutes. So here is what I'm thinking on a coordinate plan Bob starts at the easternmost point (meaning all the way on the right), at (80,0) Sally is somewhere Circumference: 2πr = 2π(80) = 160π feet, π is pi by the way. This is the full length track. Though, I'm still confused on how to start out. Can someone help? Thanks 2. ## Re: People Moving Around Circle Track Hi, Here's a blueprint to solve this problem: 1) You may write the positions on the track as a function of an angle which varies with time because Bob and Sally are moving. Let's call this angle $\\theta$. 2) Find $\\theta$ as a function of time for each runner 3) Using the previous functions, you can use the time", "# Problem #1462 1462 A ship sails $10$ miles in a straight line from $A$ to $B$, turns through an angle between $45^{\\circ}$ and $60^{\\circ}$, and then sails another $20$ miles to $C$. Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$? $[asy] unitsize(2mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C=20*dir(50); draw(A--B--C); draw(A--C,linetype(\"4 4\")); dot(A); dot(B); dot(C); label(\"10\",midpoint(A--B),S); label(\"20\",midpoint(B--C),SE); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); [/asy]$ $\\textbf{(A)}\\ [400,500] \\qquad \\textbf{(B)}\\ [500,600] \\qquad \\textbf{(C)}\\ [600,700] \\qquad \\textbf{(D)}\\ [700,800]$ $\\textbf{(E)}\\ [800,900]$ 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.", "was: $\\dfrac{1-p^{2}}{200}\\\\$ $\\dfrac{\\sqrt{1-p^{2}}}{100} \\\\$ $1-\\dfrac{p^{2}}{10,000-p^{2}} \\\\$ $\\dfrac{10,000}{10,000-p^{2}}$ 31 What is the largest positive integer $n$ such that $\\frac{n^{2}+7n+12}{n^{2}-n-12}$ is also a positive integer? $8$ $12$ $16$ $6$ 32 The quadratic equation $x^{2}+bx+c=0$ has two roots $4a$ and $3a$, where a is an integer. Which of the following is a possible value of $b^{2}+c$? $3721$ $549$ $427$ $361$ 33 A cyclist leaves $A$ at $10$ am and reaches $B$ at $11$ am. Starting from $10:01$ am, every minute a motor cycle leaves $A$ and moves towards $B$. Forty-five such motor cycles reach $B$ by $11$ am. All motor cycles have the same speed. If the cyclist had doubled his speed, how many motor cycles would have reached $B$ by the time the cyclist reached $B$? $23$ $20$ $15$ $22$ 34 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 $10:25$am $10:18$am $10:27$am $10:45$am 35 John jogs on track A at $6$ kmph and Mary jogs on track $B$ at $7.5$ kmph. The total length of tracks $A$ and $B$ is $325$ metres.", "Free Version Moderate # Distance Around a Circle SATSTM-E7MVEW Samson ran $7$ times around a circular track that has a radius of $47$ yards. Find the approximate distance he ran in yards. A $1033\\text{ yards}$ B $6936\\text{ yards}$ C $658\\text{ yards}$ D $2067\\text{ yards}$ E $329$", "of the track length. E. $$(\\frac{12}{13})^{th}$$ of the track length. Are You Up For the Challenge: 700 Level Questions Hi exc4libur One should know that the distance covered will be in opposite ratio of their speed. So A:B in speed is 25:40=5:8, so distances covered by them will be in ratio 8:5. Now whenever both meet, they complete one circle, so they will cover 10 circles when they meet 10th time. So A will cover 10*5/(5+8) =50/13=(39+11)/13=3 + 11/13 So A would have covered 3 complete circles and would be at 11/13 of 4th circle. Thanks chetan2u!! I found what I was doing wrong, I was calculating the final distance for anti-clockwise, instead of clockwise. Intern Joined: 14 Jul 2019 Posts: 32 Re: A runs in the clockwise direction on a circular track whereas B runs [#permalink] ### Show Tags 04 Mar 2020, 05:53 Bunuel wrote: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point. If their speeds are 25 m/s and 40 m/s respectively, at what fraction of the track-length, measured in clockwise direction from the starting point, does the 10th meeting take place? A. $$(\\frac{8}{13})^{th}$$ of the track length. B. $$(\\frac{9}{13})^{th}$$ of the track length. C.", "the track length. B. $$(\\frac{9}{13})^{th}$$ of the track length. C. $$(\\frac{10}{13})^{th}$$ of the track length. D. $$(\\frac{11}{13})^{th}$$ of the track length. E. $$(\\frac{12}{13})^{th}$$ of the track length. Are You Up For the Challenge: 700 Level Questions VP Joined: 24 Nov 2016 Posts: 1352 Location: United States Re: A runs in the clockwise direction on a circular track whereas B runs [#permalink] ### Show Tags 17 Jan 2020, 09:52 Bunuel wrote: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point. If their speeds are 25 m/s and 40 m/s respectively, at what fraction of the track-length, measured in clockwise direction from the starting point, does the 10th meeting take place? A. $$(\\frac{8}{13})^{th}$$ of the track length. B. $$(\\frac{9}{13})^{th}$$ of the track length. C. $$(\\frac{10}{13})^{th}$$ of the track length. D. $$(\\frac{11}{13})^{th}$$ of the track length. E. $$(\\frac{12}{13})^{th}$$ of the track length. Are You Up For the Challenge: 700 Level Questions VeritasKarishma Bunuel chetan2u can someone explain this in more detail? Thanks. Director Joined: 28 Jul 2016 Posts: 918 Location: India Concentration: Finance, Human Resources Schools: ISB '18 (D) GPA: 3.97 WE: Project Management (Investment Banking) A runs in the clockwise direction on a circular track whereas B runs [#permalink] ###", "50/13 (3 & 11/13) revolutions of the circular track. So the 10th meeting takes place at a point which is 11/13th of the track length measured clock-wise from the first starting point S. ANS: D CEO Joined: 03 Jun 2019 Posts: 2506 Location: India GMAT 1: 690 Q50 V34 WE: Engineering (Transportation) Re: A runs in the clockwise direction on a circular track whereas B runs [#permalink] ### Show Tags 31 Mar 2020, 08:20 Bunuel wrote: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point. If their speeds are 25 m/s and 40 m/s respectively, at what fraction of the track-length, measured in clockwise direction from the starting point, does the 10th meeting take place? A. $$(\\frac{8}{13})^{th}$$ of the track length. B. $$(\\frac{9}{13})^{th}$$ of the track length. C. $$(\\frac{10}{13})^{th}$$ of the track length. D. $$(\\frac{11}{13})^{th}$$ of the track length. E. $$(\\frac{12}{13})^{th}$$ of the track length. Are You Up For the Challenge: 700 Level Questions Given: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point. Asked: If their speeds are 25 m/s and 40 m/s respectively, at what fraction", "45”. 2. Aaron is running along a circular track at the rate of 10 mile per hour traverses in 36 seconds an arc which subtends 56° at the center. Find the diameter of the circle. Solution: One hour = 3600 seconds One mile = 5280 feet Therefore, 10 miles = (5280 × 10) feet = 52800 feet In 3600 seconds Aaron goes 52800 feet In 1 second Aaron goes 52800/3600 feet = 44/3 feet Therefore, in 36 seconds the Aaron goes (44/3) × 36 feet = 528 feet. Clearly, an arc of length 528 feet subtends 56° = 56 × π/180 radian at the center of the circular track. If ‘y’ feet is the radius of the circular track then using the formula s = rθ we get, y = s/θ y = 528/[56 × (π/180)] y = (528 × 180 × 7)/(56 × 22) feet y = 540 feet y = (540/3) yards [since, we know that 3 foot = 1 yard] y = 180 yards Therefore, the required diameter = 2 × 180 yards = 360 yards. 3. If α1, α2, α3 radians be the angles subtended by the arcs of lengths l1, l2, l3 at the centers of the circles whose radii are r1, r2, r3 respectively then show that the angle subtended at the", "shown. Kylie can run at 4.5 m/sec through sand and 2.5 m/sec through mud. The track is the hypotenuse from $S$ to $X$, then the hypotenuse from $X$ to $F.$ There is an angle $Y$, which is between the track $SX$ and the start boundary and angle $Z$, which is between the track $XF$ and the finish boundary. Q1) Show that the total distance covered in the race is: $d \\:=\\:200(\\sec Y+\\sec Z)$ ? and prove that Y and Z are also related by the equation: $\\tan Y+\\tan Z\\:=\\:K$ ? where $K$ is a constant. If my diagram is correct (and angles $Y$ and $Z$ are correctly placed), . . then these two statements are not true. We have: . $\\begin{Bmatrix}\\csc Y = \\frac{SX}{200}\\quad\\Rightarrow\\quad SX = 200\\csc Y \\\\ \\\\ \\csc Z = \\frac{XF}{200}\\quad\\Rightarrow\\quad XF = 200\\csc Z\\end{Bmatrix}$ Add: . $SX + XF \\:=\\:200\\csc Y + 200\\csc Z$ . . Therefore: . $\\boxed{d \\;= \\; 200(\\csc Y + \\csc Z)}$ We have: . $\\begin{Bmatrix}\\cot Y = \\frac{AX}{200}\\quad\\Rightarrow\\quad 200\\cot Y = AX \\\\ \\\\ \\cot Z = \\frac{XB}{200}\\quad\\Rightarrow\\quad 200\\cot Z = XB\\end{Bmatrix}$ Add: . $200\\cot Y + 200\\cot Z\\:=\\:AX + XB$ . . Therefore: . $200\\cot Y + 200\\cos Z\\:=\\:500\\quad\\Rightarrow\\quad \\boxed{\\cot Y + \\cot Z = \\frac{5}{2}}$ [So I don't dare try the other parts . . . ]", "& Relationships and click New between the opportunity Close date and ’... \\Mu_ { 0 } } \\ ) a bit of a running track everywhere. Our LearnNext Expert on 1800 419 1234 ( tollfree ) or submit details below a. Solution of the track ths four corners of a square so that each touches two of the text updated... Tracks are rubberized running surfaces used for various track and field competitions to the nearest meter square. Is wide ( constant^ { \\mu_ { 0 } } \\ ) a reminder, diagonals are the portion... A regular shape is calculated using the following example, square feet that... Select Number, and its symbol is ( a ) circular track ( that grass. The \\ ( constant^ { \\mu_ { 0 } } \\ ) determine... Is 90m and the end are semicircles in Setup, use the quick find to., for example, square feet date subtract one from the charge-carrying wire ( m².! For example, square feet the sides and then solve the formula is: =! Clicked Edit track, and change Decimal Places to 0 example: What is height in parallogram, kite! Close date and today ’ s see how to calculate the width of the circle ( Show Source ) example! Generate documents and we 've run into a", "are practicing for the upcoming running competition in the school. In an hour, Keith runs $$85\\%$$ of the track, Kevin covers $$1.25$$ of the track, Marc runs $$\\dfrac {9}{10}$$ of the track and Alex covers $$\\dfrac {3}{4}$$ of the track. Which one of the following options shows the correct sequence of names, considering the distance covered by them, in the order of greatest to least? A Alex, Marc, Kevin, Keith B Alex, Keith, Marc, Kevin C Kevin, Marc, Keith, Alex D Keith, Kevin, Marc, Alex × First, make the numbers in the same form. Keith runs $$85\\%$$ of the track. Kevin covers $$1.25$$ of the track. To convert $$1.25$$ into a percent, move the decimal point two places to the right, i.e., So, Kevin covers $$125\\%$$ of the track. Marc runs $$\\dfrac {9}{10}$$ of the track. To convert $$\\dfrac {9}{10}$$ into a percent, we cross multiply. $$10x=9×100$$ $$10x=900$$ $$x=\\dfrac {900}{10}=90\\%$$ Marc runs $$90\\%$$ of the track. Alex covers $$\\dfrac {3}{4}$$ of the track. To convert $$\\dfrac {3}{4}$$ into a percent, we cross multiply. $$\\Rightarrow4x=3×100$$ $$\\Rightarrow4x=300$$ $$\\Rightarrow x=\\dfrac {300}{4}=75\\%$$ Alex covers $$75\\%$$ of the track. Arranging the numbers from greatest to least. $$125\\%,\\;90\\%,\\;85\\%,\\;75\\%$$ or $$1.25\\%,\\;\\dfrac {9}{10},\\;85\\%,\\;\\dfrac {3}{4}$$ Thus, Kevin covers the maximum distance than others. Marc covers more distance than Keith. Alex covers the least. Hence, option (C) is", "# 2015 UNCO Math Contest II Problems Twenty-third Annual UNC Math Contest Final Round January 31, 2015. Three hours; no electronic devices. Answers must be justified to receive full credit. We hope you enjoy thinking about these problems, but you are not expected to do them all. The positive integers are $1, 2, 3, 4, \\ldots$ A polynomial is quadratic if its highest power term has power two. ## Problem 1 The sum of three consecutive integers is $54$. What is the smallest of the three integers? ## Problem 2 $[asy] filldraw(circle((0,0),3),white); filldraw(arc((-1,0),2,0,180)--cycle,grey); filldraw(arc((-2,0),1,0,180)--cycle,white); filldraw(arc((1,0),2,180,360)--cycle,grey); filldraw(arc((2,0),1,180,360)--cycle,white); [/asy]$ Find the area of the shaded region. The outer circle has radius $3$. The shaded region is outlined by half circles whose radii are $1$ and $2$ and whose centers lie on the dashed diameter of the big circle. ## Problem 3 If P is a polynomial that satisfies $P(x^2 +1) = 5x^4 +7x^2 +19$, then what is $P(x)$? (Hint: $P$ is quadratic.) ## Problem 4 Tarantulas $A, B,$ and $C$ start together at the same time and race straight along a $100$ foot path, each running at a constant speed the whole distance. When $A$ reaches the end, $B$ still has $10$ feet more to run. When $B$ reaches the end, $C$ has $20$ feet more to run. How many", "C run around a circular track starting from the 5 28 Aug 2007, 18:29 A, B and C run around a circular track starting from the 3 23 Mar 2007, 15:00 A, B and C run around a circular track starting from the 4 02 Mar 2006, 03:14 Display posts from previous: Sort by", "Free Version Moderate # Race Track CLMECH-GANVGK A student is getting ready to run around a rectangular race track. The track's longer side is 80 m and points in the south-north direction. The track's shorter side is 35 m and points in the east-west direction. The student starts from one corner of the rectangle and first runs north to the end of the longer side, then turns west and runs to the end of the shorter side. The student then turns south and runs along the longer side. Unfortunately, he trips and falls after running 20 meters along the longer side. What is the displacement of the student from his initial position? A $87 \\space m$ at $66^o$ north of west B $87 \\space m$ at $66^o$ west of north C $69 \\space m$ at $60^o$ west of north D $69 \\space m$ at $60^o$ north of west", "F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"J\", J, SE); [/asy]$ $\\textbf{(A)}\\hspace{.05in}\\frac{1}{4}\\qquad\\textbf{(B)}\\hspace{.05in}\\frac{7}{24}\\qquad\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}\\qquad\\textbf{(D)}\\hspace{.05in}\\frac{3}{8}\\qquad\\textbf{(E)}\\hspace{.05in}\\frac{5}{12}$ ## Problem 25 A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B? $[asy] size(8cm); draw((0,0)--(480,0),linetype(\"3 4\")); filldraw(circle((8,0),8),black); draw((0,0)..(100,-100)..(200,0)); draw((200,0)..(260,60)..(320,0)); draw((320,0)..(400,-80)..(480,0)); draw((100,0)--(150,-50sqrt(3)),Arrow(size=4)); draw((260,0)--(290,30sqrt(3)),Arrow(size=4)); draw((400,0)--(440,-40sqrt(3)),Arrow(size=4));[/asy]$ $\\textbf{(A)}\\ 238\\pi\\qquad\\textbf{(B)}\\ 240\\pi\\qquad\\textbf{(C)}\\ 260\\pi\\qquad\\textbf{(D)}\\ 280\\pi\\qquad\\textbf{(E)}\\ 500\\pi$ Solution The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", "for a regulation track. $\\implies Area \\, of \\, concentric \\, circle = \\pi[(24.5)^2 -(21)^2]$ $\\implies Area \\, of \\, concentric \\, circle = 500.2986 \\,m^2$ Answer. High school running tracks in a less populated area, with little need for a competition track and a small budget, may be fine with one of the more modest latex or polyurethane bound systems. Reply. The inside track runner will run exactly half loop of the track. Longer races start on the oval part of the track and run part of the race on a curve. Family owned. At Each End Of The Soccer Field The Track Is A Semi-circle With An Inner Radius R. Find A Formula For The Area Of The Track In Terms Of Sand That Is, Solve In Terms Of Sandr) Soccer Field 9. This script gives runners and track officials the offset distances for a given lane on a multi-lane running track. High school and college facilities are usually open to the public as long as you use the outer lanes, according to Amateur Endurance. All runners run the same distance. August 20, 2012 at 1:28 pm. 3 Answers. Question: 8. Full drum capacity is typically calculated using the formula shown below. Running tracks can have a variable number of lanes (typically from four to nine)" ]

[ "## amc-2019-mp-18 Meena has a $50 gift voucher to spend in a toyshop, but they won't give change from the voucher. Here is a short list of toys she would like. She tried to spend as much of the$50 as possible. If she buys no more than one of each toy, how much of the voucher will not get used?", "spent the remaining amount of money on 15 similar red pens. $${\\small\\hspace{1.6em}\\left(a\\right).\\enspace}$$ How much money did Billy spend on fruits? (Express the answer in terms of m.) $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How many red pens could he buy with his original sum of money? $${\\small 20.\\enspace}$$ Mrs. Lim wants to print n numbers of name cards for her company. She has to pay a basic fee of$40 and an additional of \\$0.30 for each name card. $${\\small\\hspace{1.6em}\\left(a\\right).\\enspace}$$ How much does she pay in term of n? $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How much does she pay if she wants to print 500 name cards? If you want to discuss any of the questions above, simply leave your comments below and I’ll get right back to it!", "it is certain she will not get it on the $8$-th card.", "credit card? A dress at the local fashion store cost $15.40. A Handbag costs$18.60. Rebecca wants to buy a dress and a handbag. She has \\$30 cash. Does she have enough money to purchase both items.", "were each the same price. Including sales tax, she paid a total of $84.60 . Of that total$4.20, was tax. What was the price of each CD before tax? Donna bought 6 CDs that were each the same price. Including sales tax, she paid a total of $84.60 . Of that total$4.20, was tax. What was the price of each CD before tax?...", "December 31. (a.) In October 2021, sold $3,000 of gift cards, and redeemed$500 of those gift cards. (5.) In November 2021 , sold $4,000 of gift cards, and redeemed$1,400 of October ...", "had a sticker price of$129.20.What percent sales tax did he pay. Mr.Li bought a memory card for $138.89 including tax.The card had a sticker price of$129.20.What percent sales tax did he pay.... ### What is the length of a,c What is the length of a,c...", "• # question_answer Meera bought x packs of trading cards that contain 15 cards each. Her friend gave her 8 more cards. Which expression shows the number of cards Meera have now? A) $15x+8$B) $8x+15$C) $15x-8$D) $8x-15$ Solution : Not Available You need to login to perform this action. You will be redirected in 3 sec", "Super Deck Gamma by SYOUMA & Tejinaya Magic - Trick • \\$39.95 Unit price per Magician spreads the cards, there are a variety of numbers and marks in a loose order Have the audience remember his/her favorite card, take the chosen one from the cards. The pattern on the back of the cards is all in blue. However, when the magician takes a glance at the back of the chosen card, it turns red! More to happen... When the magician snaps his fingers, all the cards except the chosen one turn white! It is a big hit overseas! This item includes: • Super Deck Gamma (Bicycle) • Online instructions", "card is $$86$$.", "Question Ellie got a $20 gift card for the craft store. She bought some chalk markers for$8.79, some yarn for $2.69, and some beads for$6.98. What is the value left on Ellie’s gift card? $1. Answer:$1.54 Step-by-step explanation: 20 – 8.79 = 11.21, 11.21 – 2.69 = 8.69, and 8.69 – 6.98 = 1.54 hope this helps 🙂", "• # question_answer On the eve of Holi festival, a group of 15 friends greeted every other friend by sending greeting cards. Find the number of cards purchased by group. A) 156B) 144C) 210D) 320 There being 15 friends in the group, each friend must have purchased $(15-1)$ i.e 14 cards for sending greeting to rest of his 14 friends. Thus total number of cards Purchased by all the friends together is $15\\times 14$ i.e. 210.", "# Valuating Steam trading card booster packs, part 2 Previously, I wrote about the pricing of booster card packs for Steam’s trading card market from mainly from the seller’s perspective. However, somewhat more interesting is how much should I pay for a certain booster pack considering I already have some cards of the set. How much am I willing to pay? To make things simple, let’s assume that foil cards do not exist and that all cards for a set have the same price. There is some variance in card prices1, but under perfect market all cards of a set should have the same price2. The three things that affect a booster packs price are 1. Amount of cards in a set (n). The more cards in a set, the less likely duplicates are. 2. Amount of cards I already have (k). The more cards I have of a set, the likelier duplicates become. 3. The price of a single card (p). Generally speaking the cost to buy a card is 0,02€ (or, \\$0.02) higher than the revenues I get from selling a card, ie. Steam’s transaction fees make getting duplicates a small risk. In practice, the value of a booster pack is enhanced by the possibility of a foil card. Also, for a totally professional trader duplicates", "Question Eddie purchased 15 boxes of paper clips and 7 packages of index cards for a total of $55.40. Finn bought 12 boxes of paper clips and 10 packages of index cards for$61.70. Find the cost of one box of paper clips and one package of index cards 1. I can confirm that their answer is correct 2. Latifah The cost of one box of paper clips is $1.85 and the cost of one package of index cards is$3.95 Step-by-step explanation: Eldora and Finn went to an office supply store together Eldora bought 15 boxes of paper clips and 7 packages of index cards for a total cost of $55.40 Finn bought 12 boxes of paper clips and 10 packages of index cards for a total cost of$61.70 We need to find the cost of one box of paper clips and the cost of one package of index cards Assume that the cost of one box of paper clips is $x and the cost of one package of index cards is$y ∵ x represents the cost of one box of paper clips ∵ y represents the cost of one package of index cards ∵ Eldora bought 15 boxes of paper clips and 7 packages of index cards for a total cost of $55.40 ∴ 15x + 7y =", "term of n? $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How much does she pay if she wants to print 500 name cards? If you want to discuss any of the questions above, simply leave your comments below and I’ll get right back to it!", "specific characteristics of the customer will determine where the two bubbles below intersect, or if they do at all: ### Who pays on a free game? The most engaged players The Black Lotus is the most expensive card in Magic: The Gathering. It's has reached a price of +$150,000 on eBay. But my dad would never pay that much for it. He woudn't even pay$1. It's just a worthless piece of paper for him. He doesn't even like MTG. This goes to show how extreme the distortion of the value of the same thing can be between two different people. And when it comes to mobile games, it's even worse: Because at least a Magic card is physical. You can keep it on a showcase and even trade it for real cash. But buying an item that exists only inside a videogame? That sounds like the ultimate waste of money. This goes to show why humans are generally highly biased against virtual items. So this means that the people that buy on a F2P game are the ones that love the game the most. They love it so much that they overcome this natural bias and they believe that a virtual item is worth real money. Even if it's only cosmetic change. Since paying users are those are", "find a rare card! After buying her first pack, she has$37 left in her wallet. If every card pack is \\$5, how much money will she have left over after buying the 5th pack? c) Garrett makes a bracelet design using trapezoids and triangles. At each stage, two triangles and a trapezoid are added. If there are 12 triangles, how many trapezoids were used? d) Kiera makes an increasing pattern with blocks. The figures in sequence use: 4 blocks, 7 blocks, 10 blocks, and then 13 blocks. After making the fourth figure, she has 15 blocks left. Does Kiera have enough blocks to make the 5th figure?", "## FANDOM 103,573 Pages Booster Boxes are boxes of booster packs. They are how the booster packs are transported after being manufactured. They will generally have 24 packs for a normal Booster Pack, 30 for a Duelist Pack and 10 for a Starter Deck, Structure Deck or Special Edition . • The cards is a sealed box cannot be scaled, and as a result customers have a better chance of getting Super Rare and Ultra Rare cards. • If a customer buys a booster box of a pack that contains a Ghost Rare card, they have a 1/36 chance of getting one. The probability of getting $x$ Ghost Rares in one box is $(1/36)^x$. • Boxes are cheaper per pack and per card than buying booster packs. A booster box of 24 booster packs will have 216 cards and will cost between \\$0.25 and \\$0.50 per card (assuming every booster has only commons). Buying the same amount of cards as single boosters will cost between \\$0.30 and \\$0.88 per card.", "card info, is stored... 1 answer ### A restaurant is having a deal for Valentine's day and is offering a 4-course meal for $27.00. If there is a sales tax of 8 percent, what is the actual cost of the meal?$ A restaurant is having a deal for Valentine's day and is offering a 4-course meal for $27.00. If there is a sales tax of 8 percent, what is the actual cost of the meal?$... -- 0.013701--", "cost of 4EPA, 2.2E per CP of SiC and that you draw the right confict card" ]

[ "☐A. 2 feet ☐B. 5 feet ☐C. 7 feet ☐D. 8 feet 76- The price of a sofa is decreased by $$20\\%$$ to $$380$$. What was its original price? ☐A.$465 ☐B. $475 ☐C.$565 ☐D. $600 77- What is the area of the trapezoid? ☐A. 88 ☐B. 98 ☐C. 106 ☐D. 112 78- If $$20 \\%$$ of A is $$40 \\%$$ of B, then B is what percent of A? ☐A. $$120\\%$$ ☐B. $$100\\%$$ ☐C. $$80\\%$$ ☐D. $$50\\%$$ 79-What is the value of $$x$$ in the following equation? $$2x-(12-x)=x+20$$ ☐A. 18 ☐B. 16 ☐C. 12 ☐D. 10 80- Ella bought a pair of gloves for$11.56. She gave the clerk $15.00. How much change should she get back? ☐A.$3.15 ☐B. $3.44 ☐C.$3.65 ☐D. \\$3.85 81- $$12.56÷0.08?$$ ☐A. 138 ☐B. 146 ☐C. 148 ☐D. 157 82- A card is drawn at random from a standard 52–card deck, what is the probability that the card is of Hearts or 10? (The deck includes 13 of each suit clubs, diamonds, hearts, and spades and four of every numbers from 1 to 10) ☐A. $$\\frac{1}{3}$$ ☐B. $$\\frac{4}{13}$$ ☐C. $$\\frac{1}{4}$$ ☐D. $$\\frac{1}{52}$$ 83- $$\\frac{1}{3}+\\frac{\\frac{8}{10}}{\\frac{6}{12}}=?$$ ☐A. $$-1 \\frac{12}{15}$$ ☐B. $$1 \\frac{1}{3}$$ ☐C. $$1\\frac{14}{15}$$ ☐D. $$2$$ 84- $$\\frac{8×14}{90}$$ is closest estimate to? ☐A. 1.2 ☐B. 1.1 ☐C. 1.3 ☐D. 1.4 23% OFF X ## How Does It", "was its original price? ☐A. $465 ☐B.$475 ☐C. $565 ☐D.$600 77- What is the area of the trapezoid? ☐A. 88 ☐B. 98 ☐C. 106 ☐D. 112 78- If $$20 \\%$$ of A is $$40 \\%$$ of B, then B is what percent of A? ☐A. $$120\\%$$ ☐B. $$100\\%$$ ☐C. $$80\\%$$ ☐D. $$50\\%$$ 79-What is the value of $$x$$ in the following equation? $$2x-(12-x)=x+20$$ ☐A. 18 ☐B. 16 ☐C. 12 ☐D. 10 80- Ella bought a pair of gloves for $11.56. She gave the clerk$15.00. How much change should she get back? ☐A. $3.15 ☐B.$3.44 ☐C. $3.65 ☐D.$3.85 81- $$12.56÷0.08?$$ ☐A. 138 ☐B. 146 ☐C. 148 ☐D. 157 82- A card is drawn at random from a standard 52–card deck, what is the probability that the card is of Hearts or 10? (The deck includes 13 of each suit clubs, diamonds, hearts, and spades and four of every numbers from 1 to 10) ☐A. $$\\frac{1}{3}$$ ☐B. $$\\frac{4}{13}$$ ☐C. $$\\frac{1}{4}$$ ☐D. $$\\frac{1}{52}$$ 83- $$\\frac{1}{3}+\\frac{\\frac{8}{10}}{\\frac{6}{12}}=?$$ ☐A. $$-1 \\frac{12}{15}$$ ☐B. $$1 \\frac{1}{3}$$ ☐C. $$1\\frac{14}{15}$$ ☐D. $$2$$ 84- $$\\frac{8×14}{90}$$ is closest estimate to? ☐A. 1.2 ☐B. 1.1 ☐C. 1.3 ☐D. 1.4 $16.99 Satisfied 96 Students$47.99 $11.99 Satisfied 163 Students ### More math articles ### What people say about \"Full-Length ISEE Middle Level Math Practice Test\"? No one replied yet. X 23% OFF Limited", "0.33 = 0.66 Question 8. 5 × 2.4 = n n = ____________ 12 Explanation: 5 × 2.4 = n n = 12 Ignore the decimals, then multiply with whole numbers. 5 x 24 = 12 Count the decimal places. There is one place to the right of the decimal point in 2.4. So, count one place from right to left in the product. 5 × 2.4 = 12 Question 9. 7 × 8.1 = s s = _____________ 56.7 Explanation: 7 × 8.1 = s s = 56.7 Ignore the decimals, then multiply with whole numbers. 7 x 81 = 567 Count the decimal places. There is one place to the right of the decimal point in 8.1. So, count one place from right to left in the product. 7 × 8.1 = 56.7 Problem Solving Question 10. Justin bought 9 packs of trading cards. Each pack of cards costs $3.27. What was the total cost for all 9 packs? Answer:$29.43 Explanation: Justin bought 9 packs of trading cards. Each pack of cards costs $3.27. The total cost for all 9 packs = 9 x$3.27 = $29.43 As, multiplication is a repeated addition.$3.27 + $3.27 +$3.27 + $3.27 +$3.27 + $3.27 +$3.27 + $3.27 +$3.27 = $29.43 Question 11. Richard paints a picture on a rectangular canvas", ", 28.10.2019 14:45 tayis # x and a simplified answer ### Another question on Math Math, 28.10.2019 20:28 Which expression represents (25x – 125) / (x2 – 25) in simplest form? Math, 28.10.2019 21:29 The graph shows the relationship between the total cost and the number of gift cards that raj bought for raffle prizes. what would be the cost for 5 of the gift cards? $$\\frac{x+1}{4}$$ x $$\\frac{3x}{x-1}$$ and a simplified answer...", "12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger? (A) \\frac{1}{8} (B) \\frac{1}{4} (C)\\frac{\\sqrt{10}}{10} (D) \\frac{\\sqrt{5}}{6} (E) \\frac{\\sqrt{5}}{5} AMC 10A, 2011, Problem 1 A cell phone plan costs 20 dollars each month, plus 5 cents per text message sent, plus 10 cents for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay? (A) 24.00 (B) 24.50 (C) 25.50 (D) 28.00 (E) 30.00 AMC 10A, 2011, Problem 2 A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy? (A) 11 (B) 12 (C) 13 (D) 14 (E) 15 AMC 10A, 2011, Problem 3 Suppose [a b] denotes the average of a and b, and {a b c} denotes the average of a, b, and c. What is {{1 1 0} [0 1 ] 0 } ? (A) \\frac{2}{9} (B)", "to the debit card incur no additional fee on top of the purchases. Marla wants to know the minimum amount of purchase she can make using the Debit Card so that it is cheaper than the Credit card. Round your answer to the nearest whole number of Yuans. Write down your answer and then click below to see if you are correct. A Click here to see the answer. Question 31 Explanation: The correct answer is 1905. Thinking and dealing with one currency will help avoid errors in calculations. If she wants to purchase something for USD$x$using the Credit Card, she will be charged USD$1.05x$including the fee. If she buys the Debit Card for$2000$Yuans, she would pay$(2000 Yuans)/(\\frac{6.34 Yuans}{USD}) =$USD$315.46$. So, she can purchase up to USD$315.46$using this debit card and her cost will be USD$315.46$since she has to prepay this amount upfront. That’s how debit cards work. Make a table like shown below to see how this works. [insert table here] A few values are shown – say for USD$100$in purchase cost, the cost using credit card is USD$105$and this is cheaper than the cost or purchasing this item using debit card which is$333.33$. So, for some purchase that has cost of USD$x$, the cost using credit card is$1.05x$and this is going to be more than the", "spent the remaining amount of money on 15 similar red pens. $${\\small\\hspace{1.6em}\\left(a\\right).\\enspace}$$ How much money did Billy spend on fruits? (Express the answer in terms of m.) $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How many red pens could he buy with his original sum of money? $${\\small 20.\\enspace}$$ Mrs. Lim wants to print n numbers of name cards for her company. She has to pay a basic fee of$40 and an additional of \\$0.30 for each name card. $${\\small\\hspace{1.6em}\\left(a\\right).\\enspace}$$ How much does she pay in term of n? $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How much does she pay if she wants to print 500 name cards? If you want to discuss any of the questions above, simply leave your comments below and I’ll get right back to it!", "• # question_answer If the cost price of 20 greeting cards is equal to the selling price of 16 greeting cards. Find the gain or loss percent A) 20% B) 30%C) 25% D) 40% (c): C.P of 20 greetings = S.P of 16 greeting Let C.P of greeting card be 'x?. Let gain % be r% $\\therefore S.P=x\\left( 1+\\frac{r}{100} \\right)\\Rightarrow$S.P. of 16 cards $=16x\\left( 1+\\frac{r}{100} \\right)$ $\\Rightarrow 20x=16x\\left( 1+\\frac{r}{100} \\right)$ $\\Rightarrow \\frac{4}{5}=\\left( 1+\\frac{r}{100} \\right)$ $\\Rightarrow \\frac{1}{4}=\\frac{r}{100}\\Rightarrow r=25%$", "• # question_answer Aman had 128 trading cards. He gave $\\frac{3}{8}$ of his cards to Mohit and $\\frac{1}{4}$ of his cards to Kushal. How many cards did he give altogether? A) 50B) 65C) 80D) 75 Correct Answer: C Solution : Total number of cards Aman had = 128 Number of cards he gave to Mohit $=\\frac{3}{8}\\times 128=48$ Number of cards he gave to Kushal $=\\frac{1}{4}\\times 128=32$ $\\therefore$ Total number of cards he gave altogether = 48 + 32 = 80 You need to login to perform this action. You will be redirected in 3 sec", "quick way to calculate $H_n$? Because $\\int_1^n \\frac{1}{x} \\d x = \\ln(n)$, it's reasonable to expect $\\sum_1^n \\frac{1}{n}$ to be related to $\\ln(n)$ as well. It is - but there's a mysterious discrepancy. In fact, $\\sum_1^n \\frac{1}{n} \\approx \\ln(n) + \\gamma$, where $\\gamma$ is the Euler-Mascheroni constant, about 0.577. In our original 140-card problem without silver cards, this would give $140 (\\ln(140) + \\gamma)$; $\\ln(140) = \\ln(7)+\\ln(20) \\approx 5$, and we end up with $140 \\times 5.5 \\approx 770$. That's not at all bad! This puzzle is known as The Coupon Collector's Problem - and I love how it falls apart so logically! ## Colin Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove. 1. For the sake of simplicity, let's assume that the 140 cards are equally likely to show up in any given pack of four, and that there are infinitely many of each. [] #### Share This site uses Akismet to reduce spam. Learn how your comment data is processed.", "Africa, too—and any cost of a packet of stickers (there’s some 6% inflation/year here); read the full problem description (pdf), on the first page. In solving this, first, there are three variables: N for the number of unique stickers, P for the price of a packet of 5 stickers, and C for the total cost we want to know. To calculate C, we thus have $\\frac{total\\_no\\_of\\_stickers}{5}*P$, and we’ll round it up to the nearest integer. The crux is how to get to the total number of stickers. Whitehouse’s post has a very readable explanation. In short, when you get the first sticker, it is guaranteed to be a new one, the second card has a $\\frac{638}{637}$ chance of being new, and so on to the last card $\\frac{638}{1}$, wich follows from some basic notions of probabilities, which you can/will/have come across in a statistics intro course. Generalising this to the arbitrary number of N cards, we obtain $\\frac{N}{N} + \\frac{N}{N-1} + \\frac{N}{N-2} + \\ldots + \\frac{N}{2} + \\frac{N}{1}$ to calculate the total amount of stickers you need to buy to have the N ones complete. This is as much as you really need from a computational viewpoint. Here’s a simple python code snippet that gets the job done: def panini(n): tns = 0 for i in range(1,n): tns", "this Question #### Find the value of (1/5)² 1. 1/25. 2. 1/15. 3. 1/50. 4. None Click here to see the answer Solve this Question #### The ratio of the cost price and the selling price is 5 :6. The profit percent is 1. 25% 2. 30% 3. 35% 4. 20% Click here to see the answer Solve this Question 1. 2178 2. 1287 3. 7812 4. 1278 Click here to see the answer Solve this Question #### One hour = ______ minutes 1. 100 2. 90 3. 60 4. 120 Click here to see the answer Solve this Question #### There are 52 playing cards in a pack. Work out how many cards there will be in 15 packs. 1. 676 2. 750 3. 780 4. 624 5. 680 Click here to see the answer Solve this Question #### The third multiple of 7 is 1. 1 2. 7 3. 14 4. 21 Click here to see the answer Solve this Question #### Find the mode of following numbers: 6.61, 6.63, 6.66, 6.68, 6.66, 6.69, 6.86 and 6.66 1. 6.86 2. 6.69 3. 6.68 4. 6.66 Click here to see the answer Solve this Question 1. 27.5 2. 26 3. 27 4. 29 Click here to see the answer Solve this Question #### Wilson goes to rice", "first (and will not be the previous image), so there are now $53$ options, $37$ of which are favorable, yielding a probability of $\\frac{37}{53}$. Continuing this logic, the probability the cards are unique is $$\\frac{38}{54}\\cdot\\frac{37}{53}\\cdot\\frac{36}{52}\\cdot...\\cdot\\frac{23}{39} \\approx 0.001054.$$ In general, the probability can be written as $$\\displaystyle\\frac{\\binom{N-n}{n}}{\\binom{N}{n}}.$$ Second, assuming the two cards are unique, how often will you win before the other player crosses a single image off their card? The extraneous cards are meaningless, akin to burn cards. WLOG, if you have images $1$-$16$, your opponent has $17$-$32$, then the favorable outcomes are all $16!$ arrangements of your cards, followed by all $16!$ arrangements of your opponents outcomes. The total number of arrangements is $32!$, the arrangements of all cards. The probability is then $$\\displaystyle\\frac{16!\\cdot 16!}{32!} \\approx 1.664\\times 10^{-9}.$$ In general, the probability can be written as $$\\displaystyle\\frac{(n!)^2}{(2n)!}.$$ Note, there is no $N$ term in this expression. The simulation in the code below considers all $N$ cards to verify its invariance. from ncr import ncr #n choose r import numpy as np def you_win(photo_order, n): \"\"\"WLOG you have photos 0 to n-1, opponent has n to 2n-1 Check if all of your photos appear before any of your opponents\"\"\" n_count = 0 for i in photo_order: if i < n: n_count += 1 if n <= i <", "first (and will not be the previous image), so there are now $53$ options, $37$ of which are favorable, yielding a probability of $\\frac{37}{53}$. Continuing this logic, the probability the cards are unique is $$\\frac{38}{54}\\cdot\\frac{37}{53}\\cdot\\frac{36}{52}\\cdot...\\cdot\\frac{23}{39} \\approx 0.001054.$$ In general, the probability can be written as $$\\displaystyle\\frac{\\binom{N-n}{n}}{\\binom{N}{n}}.$$ Second, assuming the two cards are unique, how often will you win before the other player crosses a single image off their card? The extraneous cards are meaningless, akin to burn cards. WLOG, if you have images $1$-$16$, your opponent has $17$-$32$, then the favorable outcomes are all $16!$ arrangements of your cards, followed by all $16!$ arrangements of your opponents outcomes. The total number of arrangements is $32!$, the arrangements of all cards. The probability is then $$\\displaystyle\\frac{16!\\cdot 16!}{32!} \\approx 1.664\\times 10^{-9}.$$ In general, the probability can be written as $$\\displaystyle\\frac{(n!)^2}{(2n)!}.$$ Note, there is no $N$ term in this expression. The simulation in the code below considers all $N$ cards to verify its invariance. from ncr import ncr #n choose r import numpy as np def you_win(photo_order, n): \"\"\"WLOG you have photos 0 to n-1, opponent has n to 2n-1 Check if all of your photos appear before any of your opponents\"\"\" n_count = 0 for i in photo_order: if i < n: n_count += 1 if n <= i <", "Home » Aptitude » Discount » Question #### Discount 1. Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt if its marked price is ₹ 850? 1. ₹ 650 2. ₹ 720 3. ₹ 700 4. ₹ 680 ##### Correct Option: D Let Cost price of the shirt = ₹ p ∴ p × 120 = 850 × 96 100 100 ⇒ p × 120 = 850 × 96 ⇒ p = 850 × 96 = ₹ 680 120 2nd method to solve this question. Here r = 20%, D = 4%, M.P. = ₹ 850, C.P. = ? M.P. = 100 + r C.P. 100 − D 850 = 100 + 20 C.P. 100 − 4 C.P. = 850 × 96 120 C.P. = ₹ 680", "cardholder makes $$R$$ purchases per month 2. The average purchase price is $$D$$ dollars 3. Purchases are distributed randomly throughout the year 4. A bonus worth some multiple $$M$$ of the points earned in the previous month is awarded if at least $$B$$ purchases are made during the month 5. The cardholder makes no specific effort to earn the monthly bonus For the present undertaking, I will ignore the tiered point system that awards two or three points for spending in specific categories, and assume that all spending earns (pre-bonus) 1 point per dollar. To fix ideas, the regular card awards a 20% bonus when 20 purchases are made; that is, $$M=0.2$$ and $$B=20$$. For the Preferred card, $$M=0.5$$ and $$B=30$$ (50% bonus if 30 purchases are made). Assumptions 1 and 3 let us describe purchases as random arrivals, and therefore model the probability of making exactly $$n$$ purchases during a given month using a Poisson distribution that has parameter $$\\lambda = R$$: $P(x=n) = \\frac{e^{-\\lambda} \\lambda^n}{n!} = \\frac{e^{-R} R^n}{n!}$ The probability of making at least $$B$$ purchases is then: $P(x\\ge B) = \\sum_{n=B}^\\infty \\frac{e^{-R} R^n}{n!}$ First we model the value of the pre-bonus spending in terms of points. The expected number of points earned during the month is given by the Poisson probability of each number of", "the largest integer and the smallest integer. Solution: The largest integer = 1 The smallest integer = –10 Sum of the largest integer and the smallest integer = 1 + (–10) = –9 1.2.3 Whole Numbers, PT3 Practice Question 11: Britney rents a stall to sell cakes. The rental of the stall is RM500 per month. The cost for baking a cake is RM5 per box and she sells it at RM10 per box. What is the minimum box of cakes must Britney sell each month to make a profit? Solution: Rental of the stall = RM500 monthly Cost of baking a cake per box = RM5 Selling price per box = RM10 Profit per box = RM10 – RM5 = RM5 The number of boxes of cakes to accommodate the cost = RM500 ÷ RM5 = 100 Hence, the minimum box of cakes must Britney sell each month to make a profit = 101 1.2.2 Whole Numbers, PT3 Practice Question 6: Solve 4 (8 + 14)(37 – 18) + 46 Solution: 4 (8 + 14)(37 – 18) + 46 = 4(22)(19) + 46 = 4 × 22 × 19 + 46 = 1672 + 46 = 1718 Question 7: Calculate the value of 113 + 6(47 – 4 × 9) Solution: 113 + 6(47 – 4 ×", "• # question_answer On the eve of Holi festival, a group of 15 friends greeted every other friend by sending greeting cards. Find the number of cards purchased by group. A) 156B) 144C) 210D) 320 There being 15 friends in the group, each friend must have purchased $(15-1)$ i.e 14 cards for sending greeting to rest of his 14 friends. Thus total number of cards Purchased by all the friends together is $15\\times 14$ i.e. 210.", "term of n? $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How much does she pay if she wants to print 500 name cards? If you want to discuss any of the questions above, simply leave your comments below and I’ll get right back to it!", "the next example, all the numbers represent money, and it will make sense to report the mean in dollars and cents. ## Example For the past four months, Daisy’s cell phone bills were $$\\text{\\42.75},\\text{\\50.12},\\text{\\41.54},\\text{\\48.15}.$$ Find the mean cost of Daisy’s cell phone bills. ### Solution Write the formula for the mean. $$\\text{mean}=\\frac{\\text{sum of all the numbers}}{n}$$ Count how many numbers are in the set. Call this $$n$$ and write it in the denominator. $$\\text{mean}=\\frac{\\text{sum of all the numbers}}{4}$$ Write the sum of all the numbers in the numerator. $$\\text{mean}=\\frac{42.75+50.12+41.54+48.15}{4}$$ Simplify the fraction. $$\\text{mean}=\\frac{182.56}{4}$$ $$\\text{mean}=45.64$$ Does $$\\text{\\45.64}$$ seem ‘typical’ of this set of numbers? Yes, it is neither less than $$\\text{\\41.54}$$ nor greater than $$\\text{\\50.12}.$$ The mean cost of her cell phone bill was $$\\text{\\45.64}$$" ]

[ "# Write a decimal as a mixed number ## Write A Decimal As A Mixed Number The trading card packs that trent normally buys tend to come in packs of 6, 10, 12, or 15 cards.When a number is whole (10) it’ll go on the left side of the decimal.Step 2: The mixed number has a whole number part 6 and a fractional part 8/10 which can be reduced to lowest terms as $\\frac{4}{5}$..A mixed decimal number is a number with a whole number and a non-whole number where a decimal point is used to separate the numbers.0 Answer to Find the circumference of circle L.Read this lesson on converting decimals to fractions if you need to learn more.If the fractional part is a repeating fraction then the whole number is represented by a repeating decimal About This Quiz & Worksheet.So, it can be written as a mixed number \\frac{8}{10}$.• bha finished the race one and sixteen thousandths seconds ahead of Nadia.After selecting 2 packs, trent found that the first pack of cards cost 25 cents per card, and the second pack cost 30 cents per card.The trading card packs that trent normally buys tend to come write a decimal as a mixed number in packs of 6, 10, 12, or 15 cards.Can be used to divide mixed", "• # question_answer Aman had 128 trading cards. He gave $\\frac{3}{8}$ of his cards to Mohit and $\\frac{1}{4}$ of his cards to Kushal. How many cards did he give altogether? A) 50B) 65C) 80D) 75 Correct Answer: C Solution : Total number of cards Aman had = 128 Number of cards he gave to Mohit $=\\frac{3}{8}\\times 128=48$ Number of cards he gave to Kushal $=\\frac{1}{4}\\times 128=32$ $\\therefore$ Total number of cards he gave altogether = 48 + 32 = 80 You need to login to perform this action. You will be redirected in 3 sec", "☐A. 2 feet ☐B. 5 feet ☐C. 7 feet ☐D. 8 feet 76- The price of a sofa is decreased by $$20\\%$$ to $$380$$. What was its original price? ☐A.$465 ☐B. $475 ☐C.$565 ☐D. $600 77- What is the area of the trapezoid? ☐A. 88 ☐B. 98 ☐C. 106 ☐D. 112 78- If $$20 \\%$$ of A is $$40 \\%$$ of B, then B is what percent of A? ☐A. $$120\\%$$ ☐B. $$100\\%$$ ☐C. $$80\\%$$ ☐D. $$50\\%$$ 79-What is the value of $$x$$ in the following equation? $$2x-(12-x)=x+20$$ ☐A. 18 ☐B. 16 ☐C. 12 ☐D. 10 80- Ella bought a pair of gloves for$11.56. She gave the clerk $15.00. How much change should she get back? ☐A.$3.15 ☐B. $3.44 ☐C.$3.65 ☐D. \\$3.85 81- $$12.56÷0.08?$$ ☐A. 138 ☐B. 146 ☐C. 148 ☐D. 157 82- A card is drawn at random from a standard 52–card deck, what is the probability that the card is of Hearts or 10? (The deck includes 13 of each suit clubs, diamonds, hearts, and spades and four of every numbers from 1 to 10) ☐A. $$\\frac{1}{3}$$ ☐B. $$\\frac{4}{13}$$ ☐C. $$\\frac{1}{4}$$ ☐D. $$\\frac{1}{52}$$ 83- $$\\frac{1}{3}+\\frac{\\frac{8}{10}}{\\frac{6}{12}}=?$$ ☐A. $$-1 \\frac{12}{15}$$ ☐B. $$1 \\frac{1}{3}$$ ☐C. $$1\\frac{14}{15}$$ ☐D. $$2$$ 84- $$\\frac{8×14}{90}$$ is closest estimate to? ☐A. 1.2 ☐B. 1.1 ☐C. 1.3 ☐D. 1.4 23% OFF X ## How Does It", "was its original price? ☐A. $465 ☐B.$475 ☐C. $565 ☐D.$600 77- What is the area of the trapezoid? ☐A. 88 ☐B. 98 ☐C. 106 ☐D. 112 78- If $$20 \\%$$ of A is $$40 \\%$$ of B, then B is what percent of A? ☐A. $$120\\%$$ ☐B. $$100\\%$$ ☐C. $$80\\%$$ ☐D. $$50\\%$$ 79-What is the value of $$x$$ in the following equation? $$2x-(12-x)=x+20$$ ☐A. 18 ☐B. 16 ☐C. 12 ☐D. 10 80- Ella bought a pair of gloves for $11.56. She gave the clerk$15.00. How much change should she get back? ☐A. $3.15 ☐B.$3.44 ☐C. $3.65 ☐D.$3.85 81- $$12.56÷0.08?$$ ☐A. 138 ☐B. 146 ☐C. 148 ☐D. 157 82- A card is drawn at random from a standard 52–card deck, what is the probability that the card is of Hearts or 10? (The deck includes 13 of each suit clubs, diamonds, hearts, and spades and four of every numbers from 1 to 10) ☐A. $$\\frac{1}{3}$$ ☐B. $$\\frac{4}{13}$$ ☐C. $$\\frac{1}{4}$$ ☐D. $$\\frac{1}{52}$$ 83- $$\\frac{1}{3}+\\frac{\\frac{8}{10}}{\\frac{6}{12}}=?$$ ☐A. $$-1 \\frac{12}{15}$$ ☐B. $$1 \\frac{1}{3}$$ ☐C. $$1\\frac{14}{15}$$ ☐D. $$2$$ 84- $$\\frac{8×14}{90}$$ is closest estimate to? ☐A. 1.2 ☐B. 1.1 ☐C. 1.3 ☐D. 1.4 $16.99 Satisfied 96 Students$47.99 $11.99 Satisfied 163 Students ### More math articles ### What people say about \"Full-Length ISEE Middle Level Math Practice Test\"? No one replied yet. X 23% OFF Limited", "spent the remaining amount of money on 15 similar red pens. $${\\small\\hspace{1.6em}\\left(a\\right).\\enspace}$$ How much money did Billy spend on fruits? (Express the answer in terms of m.) $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How many red pens could he buy with his original sum of money? $${\\small 20.\\enspace}$$ Mrs. Lim wants to print n numbers of name cards for her company. She has to pay a basic fee of$40 and an additional of \\$0.30 for each name card. $${\\small\\hspace{1.6em}\\left(a\\right).\\enspace}$$ How much does she pay in term of n? $${\\small\\hspace{1.6em}\\left(b\\right).\\enspace}$$ How much does she pay if she wants to print 500 name cards? If you want to discuss any of the questions above, simply leave your comments below and I’ll get right back to it!", "to the debit card incur no additional fee on top of the purchases. Marla wants to know the minimum amount of purchase she can make using the Debit Card so that it is cheaper than the Credit card. Round your answer to the nearest whole number of Yuans. Write down your answer and then click below to see if you are correct. A Click here to see the answer. Question 31 Explanation: The correct answer is 1905. Thinking and dealing with one currency will help avoid errors in calculations. If she wants to purchase something for USD$x$using the Credit Card, she will be charged USD$1.05x$including the fee. If she buys the Debit Card for$2000$Yuans, she would pay$(2000 Yuans)/(\\frac{6.34 Yuans}{USD}) =$USD$315.46$. So, she can purchase up to USD$315.46$using this debit card and her cost will be USD$315.46$since she has to prepay this amount upfront. That’s how debit cards work. Make a table like shown below to see how this works. [insert table here] A few values are shown – say for USD$100$in purchase cost, the cost using credit card is USD$105$and this is cheaper than the cost or purchasing this item using debit card which is$333.33$. So, for some purchase that has cost of USD$x$, the cost using credit card is$1.05x$and this is going to be more than the", "than 4 is 4. Question 4. A set of cards includes 24 yellow cards, 18 green cards, and 18 blue cards. What is the probability that a card chosen at random is not green? (A) $$\\frac{3}{10}$$ (B) $$\\frac{4}{10}$$ (C) $$\\frac{3}{5}$$ (D) $$\\frac{7}{10}$$ (D) $$\\frac{7}{10}$$ Explanation: Total numbers of possible outcomes is: 24 + 18 + 18 = 60 P (choose a green card) = P (not to choose a green card) = 1 – P (choose a green card) = 1 – $$\\frac{3}{10}$$ = $$\\frac{10}{10}$$ – $$\\frac{3}{10}$$ = $$\\frac{7}{10}$$ The probability that chosen card at random is not green is $$\\frac{7}{10}$$. Question 5. A rectangle made of square tiles measures 10 tiles long and 8 tiles wide. What is the width of a similar rectangle whose length is 15 tiles? (A) 3 tiles (B) 12 tiles (C) 13 tiles (D) 18.75 tiles (B) 12 tiles Explanation: Use Proportion. The width of a similar rectangle is 12 tiles. Question 6. You buy a game that originally cost $35. It was on sale at 20% off. You paid 6% tax on the sale price. What was the total amount that you paid? (A)$29.68 (B) $37.10 (C)$44.10 (D) $44.52 Answer: (A)$29.68 Explanation: Originally price is $35. Sale is 20%. Tax is 6%. Calculate how ¡iiicIi a game on sale costs. 35", "100 D. 108 10- Kayla has 109 red cards and 88 white cards. How many more red cards than white cards does Kayla have? A. 17 B. 19 C. 21 D. 27 11- A number sentence is shown below. $$3 × 7 \\space ⎕ \\space 6 = 126$$ What symbol goes into the box to make the number sentence true? A. $$×$$ B. $$÷$$ C. $$+$$ D. $$-$$ 12- Liam had 845 marbles. Then, he gave 455 of the cards to his friend Ethan. After that, Liam lost 116 cards. Which equation can be used to find the number of cards Liam has now? A. $$845 – 455 + 116 =$$ B. $$845 – 455 – 116 =$$ C. $$845 + 455 + 116 =$$ D. $$845 + 455 – 116 =$$ 13- What is the value of “B” in the following equation? $$33 + B + 8 = 66$$ A. 16 B. 18 C. 22 D. 25 14- Mason is 18 months now and he usually eats six meals a day. How many meals does he eat in a week? A. 36 B. 40 C. 42 D. 48 15- Which of the following list shows only fractions that are equivalent to $$\\frac{1}{3}$$? A. $$\\frac{3}{9},\\frac{5}{15},\\frac{24}{72}$$ B. $$\\frac{6}{12},\\frac{5}{15},\\frac{9}{27}$$ C. $$\\frac{3}{9},\\frac{4}{15},\\frac{6}{18}$$ D. $$\\frac{3}{9},\\frac{5}{10},\\frac{8}{24}$$ 16- What mixed number is shown by", "6 feet Explanation: Given, Brad has two lengths of copper pipe to fit together. One has a length of 2 $$\\frac{5}{12}$$ feet and the other has a length of 3 $$\\frac{7}{12}$$ feet. 2 $$\\frac{5}{12}$$ + 3 $$\\frac{7}{12}$$ = 5 $$\\frac{12}{12}$$ = 5 feet Thus the correct answer is option a. Question 2. A pattern calls for 2 $$\\frac{1}{4}$$yards of material and 1 $$\\frac{1}{4}$$yards of lining. How much total fabric is needed? Options: a. 2 $$\\frac{2}{4}$$ yards b. 3 yards c. 3 $$\\frac{1}{4}$$ yards d. 3 $$\\frac{2}{4}$$ yards Answer: 3 $$\\frac{2}{4}$$ yards Explanation: Given, A pattern calls for 2 $$\\frac{1}{4}$$ yards of material and 1 $$\\frac{1}{4}$$ yards of lining. 2 $$\\frac{1}{4}$$ + 1 $$\\frac{1}{4}$$ = 3 + $$\\frac{1}{4}$$ + $$\\frac{1}{4}$$ = 3 $$\\frac{2}{4}$$ yards Thus the correct answer is option d. Spiral Review Question 3. Shanice has 23 baseball trading cards of star players. She agrees to sell them for $16 each. How much will she get for the cards? Options: a.$258 b. $358 c.$368 d. $468 Answer:$368 Explanation: Given, Shanice has 23 baseball trading cards of star players. She agrees to sell them for $16 each. To find how much will she get for the cards 23 × 16 = 368 Therefore she will get$368 for the cards. Thus the correct answer is option c. Question 4. Nanci", "Spiral Review Question 3. Shanice has 23 baseball trading cards of star players. She agrees to sell them for$16 each. How much will she get for the cards? Options: a. $258 b.$358 c. $368 d.$468 Question 4. Nanci is volunteering at the animal shelter. She wants to spend an equal amount of time playing with each dog. She has 145 minutes to play with all 7 dogs. About how much time can she spend with each dog? Options: Question 5. Frieda has 12 red apples and 15 green apples. She is going to share the apples equally among 8 people and keep any extra apples for herself. How many apples will Frieda keep for herself? Options: a. 3 b. 4 c. 6 d. 7 Question 6. The Lynch family bought a house for $75,300. A few years later, they sold the house for$80,250. How much greater was the selling price than the purchase price? Options: a. $4,950 b.$5,050 c. $5,150 d.$5,950 ### Add and Subtract Mixed Numbers – Page No. 431 Question 1. Rename both mixed numbers as fractions. Find the difference. 3 $$\\frac{3}{6}$$ = $$\\frac{■}{6}$$ −1 $$\\frac{4}{6}$$ = – $$\\frac{■}{6}$$ —————————————- _______ $$\\frac{□}{□}$$ Find the difference. Question 2. 1 $$\\frac{1}{3}$$ − $$\\frac{2}{3}$$ ——————— $$\\frac{□}{□}$$ Question 3. 4 $$\\frac{7}{10}$$ − 1 $$\\frac{9}{10}$$ ——————— ______ $$\\frac{□}{□}$$ Question 4. 3", "15 sets of cards. There are 72 cards in a set. How many more cards does Kareem need to reach his goal? Factors and multiples A number consist of two digits whose product is 8 . If the digits are interchanged, the resulting number will exceed the original one by 18 . Find the number. Factors and multiples explain what steps you would use to simplify $$2+3 (9 - 5)$$ Factors and multiples Solve this $$\\frac{1}{3} ÷9$$ Factors and multiples Zoe needs to buy beads to make jewelry. For each earring, Zoe uses 3 small blue beads, 1 small metallic bead, and 2 large green beads. What is the total number of beads Zoe will need to buy to make 8 pairs of earrings? Factors and multiples gary made four times as much money as joel over the summer. combined, the two made $475. How much money did each boy earn? Factors and multiples ANSWERED asked 2021-02-15 During the 2002-2003 ski season$171 million worth of snowboarding equipment was sold. Sales increased by about 15% during the 2003-2004 season. About how much were sales of snowboarding equipment in the 2003-2004 season? Factors and multiples Luke has a credit limit of $8,500 on his credit card. He had a previous balance of$4,236.87 and made a $3,200.00 payment. The total of", "# Luke has a credit limit of $8,500 on his credit card. He had a previous balance of$4,236.87 and made a $3,200.00 payment. The total of his purchases is$989.42. What is Luke's available credit? Question Factors and multiples Luke has a credit limit of $8,500 on his credit card. He had a previous balance of$4,236.87 and made a $3,200.00 payment. The total of his purchases is$989.42. What is Luke's available credit? 2021-02-13 If his previous balance was $4,236.87 and he makes a payment of$3,200, then his balance after the payment is $4,236.87−$3,200=$1,036.87. If he makes$989.42 in purchases, then his balance after the purchases is $1,036.87+$989.42=$2,026.29. Since his credit limit is$8,500, then his available credit after the payment and purchases is $8,500−$2,026.29=\\$6,473.71. ### Relevant Questions $$\\begin{array}{cc}\\hline & \\text{Afraid to walk at night?} \\\\ \\hline & \\text{Yes} & \\text{No} & \\text{Total} \\\\ \\hline \\text{Male} & 173 & 598 & 771 \\\\ \\hline \\text{Female} & 393 & 540 & 933 \\\\ \\hline \\text{Total} & 566 & 1138 & 1704 \\\\ \\hline \\text{Source:}2014\\ GSS \\end{array}$$ If the chi-square $$(\\chi 2)$$ test statistic $$=73.7$$ what is the p-value you would report? Use Table C.Remember to calculate the df first. $$P>0.250$$ $$P=0.01$$ $$P<0.001$$ $$P<0.002$$ $$\\begin{array}{|c|cc|}\\hline & \\text{Right-Tail Probability} \\\\ \\hline df & 0.250 & 0.100 & 0.050 & 0.025 & 0.010 & 0.005 &", "2. If Darren has 15 pouches, the total number of stamps can be calculated as $15 \\times 15 = 225$. 3. If Darren has 18 pouches, the total number of stamps can be calculated as $15 \\times 18 = 270$. ### Practice Questions: 1. Sherry is studying in grade 2, and the total number of students in her class is 15. It is her birthday, and she wants to give five candies to each of her class fellows. Can you calculate the total number of candies Sherry needs to buy for her class? 2. What is the 9th multiple of 15? 3. From the given table, select the numbers which are multiples of 15 130 137 135 160 150 51 61 180 125 19 20 18 10 300 167 154 190 11 113 117 100 73 99 321 151 30 14 116 15 104 133 129 110 111 200 25 21 187 141 210 132 141 105 29 70 88 129 220 240 321 119 34 35 169 160 219 109 120 260 39 80 100 250 231 41 165 43 151 45 122 214 257 44 43 280 49 75 132 215 109 $15\\times 5 =105$ candies. So, if Sherry buys 105 candies, she can distribute five candies to each of her class fellows.", "superstarshayna # An aquarium is on sale for \\$59.50. If this price represents a 15% discount from the original price, what is the original price to the nearest cent? Look at it like this: $x - \\frac{15}{100} x = 59.5$ ^ Where x is the original price. Shows that the original price - 15% is £59.50 Okay, so now we solve the equation: 1. Rationalise the 59.5 to turn it into a fraction - fractions are easier to use than decimals $x - \\frac{15}{100} x = \\frac{119}{2}$ 2. Combine the two fractions of the left side $\\frac{17x}{20} = \\frac{119}{2}$ 3. Multiply by the lowest common denominator (20) $17x = 1190$ 4. Divide by 17 on both sides $x = 70$ You can check by working out 15% of 70 which is £10.50, take that away from £70 and its £59.50 Hope that helped and ask questions if need be! x", ", 28.10.2019 14:45 tayis # x and a simplified answer ### Another question on Math Math, 28.10.2019 20:28 Which expression represents (25x – 125) / (x2 – 25) in simplest form? Math, 28.10.2019 21:29 The graph shows the relationship between the total cost and the number of gift cards that raj bought for raffle prizes. what would be the cost for 5 of the gift cards? $$\\frac{x+1}{4}$$ x $$\\frac{3x}{x-1}$$ and a simplified answer...", "Home » Aptitude » Discount » Question #### Discount 1. Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt if its marked price is ₹ 850? 1. ₹ 650 2. ₹ 720 3. ₹ 700 4. ₹ 680 ##### Correct Option: D Let Cost price of the shirt = ₹ p ∴ p × 120 = 850 × 96 100 100 ⇒ p × 120 = 850 × 96 ⇒ p = 850 × 96 = ₹ 680 120 2nd method to solve this question. Here r = 20%, D = 4%, M.P. = ₹ 850, C.P. = ? M.P. = 100 + r C.P. 100 − D 850 = 100 + 20 C.P. 100 − 4 C.P. = 850 × 96 120 C.P. = ₹ 680", "## gabbyalicorn one year ago Hitomi has a bag with four chips in it numbered 1 to 4. She randomly draws a chip and replaces it three times. What is the probability that Hitomi will draw a 1 all three times? 1. misty1212 HI! (again) 2. gabbyalicorn Hello! 3. misty1212 what is the probability she draws a one on the first try? 4. gabbyalicorn 3/4 or 1/4 ? 5. misty1212 hmmm pick one (hint, there is only one \"one\") 6. gabbyalicorn i think 1/4 7. misty1212 right! 8. gabbyalicorn :D 9. misty1212 and since she replaces it, the probability she gets it on the second try is also $$\\frac{1}{4}$$ 10. misty1212 and also on the third try! 11. gabbyalicorn so add 1/4 + 1/4 + 1/4? 12. misty1212 so all three, do $\\frac{1}{4}\\times \\frac{1}{4}\\times \\frac{1}{4}$ 13. misty1212 14. gabbyalicorn oh, multiply. Sorry. 15. misty1212 you want to multiply them together 16. gabbyalicorn 1/64 17. misty1212 it is an AND statement one and one and one 18. misty1212 right 19. gabbyalicorn :D 20. misty1212 $\\color\\magenta\\heartsuit$ 21. gabbyalicorn Thank you! Find more explanations on OpenStudy", "15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible by 29, there are not so many values that make sense here: 29, 58, 87 => 58 is the only number that suits here, hence a=10 is the only possible solution here. If a=10, b=10 and we have an unique combination here. (2) clearly insufficient, as we have no concrete value as in (1), it could be 0.44 , 0,88 cents ... So dear math experts, I would appreciate your opinion regarding this method for such kind of questions. Do you think it's a valid method for all kind of such problems ? CEO Joined: 20 Mar 2014 Posts: 2560 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamp [#permalink] ### Show Tags 11 Nov 2015, 05:31 1 BrainLab wrote: udaymathapati wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy? (1) She bought $4.40 worth of stamps. (2) She bought an equal number of$0.15 stamps and $0.29 stamps. Pretty @!)*, let's start: (1) 15a+29b=440 -> $$b=\\frac{440-15a}{29}$$ --> $$\\frac{5(88-3a)}{29}$$ so if we want this one to be an integer 88-3a must be divisible", "OpenStudy (anonymous): labrina received a $50 gift card to an online store. She wants to purchase some bracelets that cost$8 each. There will be a $10 overnight shipping fee. How many bracelets can she buy? OpenStudy (anonymous): Step 1: do 50 - 10 =$40 Step 2: 40 divided by 8 = 5 That means she brought 5 bracelets. Hopes this helps", "can make __________ puppets. She sells the puppets for$8 each. Think: to find the amount of money she receives, you must multiply. __________ puppets × $__________ per puppet =$ __________ . Jasmine receives __________ from selling all the puppets. Jasmine has 12 socks. so she can make 12 puppets. She sells each puppet for $8. Now we have to find the amount of money she receives, so you multiply 12 puppets ×$8 per puppet=$96. Jasmine receives$96 from selling all the puppets. (C) Find the Profit. Subtract the total expenses from the amount received. $____96_____ –$ ____54_____ = $___42______ So, Jasmine makes a ___42______ of$ ____96_____ . To know the profit how much she gains, you need to subtract expenses from income. By math sentence, we can write as 96-54=$42. She gained a$42 profit. Mathematical Processes What is the least number of puppets Jasmine has to sell and still make a profit? Explain. Answer: If she sells 8 puppets for $8 each then 8*8=64$ To know the income we need to subtract income – expenses. Total expenses=$54 Therefore, 64-54=$10 If she sells 8 puppets then she will gain $10 profit. Share and Show Question 1. Diego runs his own lawn mowing business. His expenses for one week are listed in the chart. He charges his customers$45 to mow a" ]

[ "# 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", "# 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", "C B D$ is:In the triangle $\\triangle A B C$ measures of angles $\\angle A=46^{\\circ}$ and $\\angle C=60^{\\circ}$ are given. The angle bisector intersects the circ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given statements is correct. In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given statements is correct.In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given ... close Triangle $\\triangle A B C$, right angled at $C$, is given. Height and the median from point $C$ form an angle $\\varphi$. The measure of larger acute angle of $\\triangle A B C$ is:Triangle $\\triangle A B C$, right angled at $C$, is given. Height and the median from point $C$ form an angle $\\varphi$. The measure of larger acute a ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 The distance of the shorter ladder form the wall is: The distance of the shorter", "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 us $$\\frac {2R} c = \\frac a{\\frac{2 \\times", "# IN TRIANGLE ABC, THE MEASURE OF A IS LESS THAN TWICE THE MEASURE OF ANGLE B. THE MEASURE OF ANGLE C EQUALS THE SUM OF THE MEASURES OF ANGLES A AND B, DETERMINE THE MEASURE OF ANGLE B. ###### Question: IN TRIANGLE ABC, THE MEASURE OF A IS LESS THAN TWICE THE MEASURE OF ANGLE B. THE MEASURE OF ANGLE C EQUALS THE SUM OF THE MEASURES OF ANGLES A AND B, DETERMINE THE MEASURE OF ANGLE B. ### Find the following verses in the Psalms. Write the two synonymous lines. Psalm 54:1 First line: Second line: Find the following verses in the Psalms. Write the two synonymous lines. Psalm 54:1 First line: Second line:... ### Me 8 year step sis old sis came in me room ;) but she didnt whant to do that:( me pp when back in my pants also s my microwave giving me cancer? if so how me 8 year step sis old sis came in me room ;) but she didnt whant to do that:( me pp when back in my pants also s my microwave giving me cancer? if so how... ### Need help on this too Solve for x Need help on this too Solve for x... ### Which is a difference between bacteria and viruses that show that", "ABC =3x+20, and the measure or angle DBE =4x-10. There are four types of angles. 8th grade. 8. a) A b) B c) C d) D What type of triangle is this? Equivalence angle pairs. Angle Measures DRAFT. Angle 3 and Angle 6 are examples of which type of angle pair? 1) C B A? Answer to 1) Find the measures of the following angles for a regular 96-gon. 2 days ago. Solved: Find the measure of each exterior angle for a regular heptagon. Played 456 times. Image of an angle symbol CAB B. In $\\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\\angle A = 60^\\circ$. Find the measure of the smaller angle between the. « Elektromos járművek töltési infrastruktúrája – Fizetési módok. CLASSIFYING ANGLES Classify the angle with the given measure as acute, obtuse, right, or straight. B = the measure of angle B. Use a protractor to find the measure of the given angle. Find The Exact Angle Between The Minute And Hour Hands At 1:25 P.M. A. Find the measure of angle 1. Angle A is opposite to the side with length 6.0.) Angles and Radian Measure. Since ퟏퟏퟏퟏퟏퟏ − ퟏퟏퟏퟏퟏퟏ = ퟕퟕퟕퟕ , the measure of angle 풙풙 is ퟕퟕퟕퟕ ° . taliadaggert. Circles, arcs, and angle measures Find the", "# 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$?.", "pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C--D--A, dashed ); pair Sp = intersectionpoint(D--O,A--C); label(\"B\",B,SW); label(\"A\",A,S); label(\"C\",C,NW); label(\"O\",O,NE); label(\"D\",D,(S+SSW)); label(\"S\",Sp,(S+SSW)); [/asy]$ As in the previous solution, we find out that $\\angle AOB = \\angle COB = 60^\\circ$. Hence $\\triangle AOD$ and $\\triangle COD$ are both equilateral. We then have $\\angle SCD = \\angle SAD = 30^\\circ$, hence $D$ is the incenter of $\\triangle ABC$, and as $\\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \\cdot SD = BD$, and as $SD = SO$, we have $2\\cdot SD = SD + SO = OD$, therefore $BD=OD$, and as before we conclude that $\\frac{BD}{BO} = \\boxed{\\frac 12}$.", "marked angle measures. Find the missing angle measures marked with question marks. ### Problem 4 The two figures are scaled copies of each other. 1. What are some ways that you can tell they are scaled copies? 2. What is the scale factor that takes Figure 1 to Figure 2? 3. What is the scale factor that takes Figure 2 to Figure 1? ## Lesson 15 ### Problem 1 In triangle $ABC$, the measure of angle $A$ is $40^\\circ$. 1. Give possible measures for angles $B$ and $C$ if triangle $ABC$ is isosceles. 2. Give possible measures for angles $B$ and $C$ if triangle $ABC$ is right. ### Problem 2 For each set of angles, decide if there is a triangle whose angles have these measures in degrees: 1. 60, 60, 60 2. 90, 90, 45 3. 30, 40, 50 4. 90, 45, 45 5. 120, 30, 30 If you get stuck, consider making a line segment. Then use a protractor to measure angles with the first two angle measures. ### Problem 3 Angle $A$ in triangle $ABC$ is obtuse. Can angle $B$ or angle $C$ be obtuse? Explain your reasoning. ### Problem 4 (from Unit 1, Lesson 3) For each pair of polygons, describe the transformation that could be applied to Polygon A to get Polygon B.", "# Angle Bisector Theorem (Redirected from Angle bisector theorem) This is an AoPSWiki Word of the Week for June 6-12 ## Introduction The Angle Bisector Theorem states that given triangle $\\triangle ABC$ and angle bisector AD, where D is on side BC, then $\\frac cm = \\frac bn$. Likewise, the converse of this theorem holds as well. $[asy] size(200); defaultpen(fontsize(10)); real a,b,c,d; pair A=(1,4), B=(-5,0), C=(3,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D); MA(B,A,D,0.5,black); MA(D,A,C,0.6,black); label(\"A\",A,(1,1));label(\"B\",B,(-1,-1));label(\"C\",C,(1,-1));label(\"D\",D,(0,-1)); label(\"b\",(A+C)/2,(1,0));label(\"c\",(A+B)/2,(0,1));label(\"m\",(B+D)/2,(0,-1));label(\"n\",(D+C)/2,(0,-1)); dot(A^^B^^C^^D); [/asy]$ ## Proof ### Method 1 Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick: $[asy] size(200); defaultpen(fontsize(10)); real a,b,c,d,m,n; pair A=(1,4), B=(-5,0), C=(3,0), D, E; b = abs(C-A);c = abs(B-A); D = (b*B+c*C)/(b+c); m = abs(B-D);n = abs(C-D); E = C+(B-A)*n/m; draw(A--B--C--A--E--C); MA(B,A,D,0.5,blue+linewidth(1)); MA(D,A,C,0.6,blue+linewidth(1)); MA(C,E,A,0.6,blue+linewidth(1)); MA(C,B,A,0.6,green+linewidth(1)); MA(B,C,E,0.6,green+linewidth(1)); label(\"A\",A,(1,1));label(\"B\",B,(-1,-1));label(\"C\",C,(1,-1));label(\"D\",D,(1,-1));label(\"E\",E,(0,-1)); label(\"b\",(A+C)/2,(1,0));label(\"c\",(A+B)/2,(0,1));label(\"m\",(B+D)/2,(0,-1));label(\"n\",(D+C)/2,(0,-1));label(\"b\",(E+C)/2,(1,-0.5)); dot(A^^B^^C^^D^^E); [/asy]$ Since AB and CE are parallel, we know that $\\angle BAE=\\angle CEA$ and $\\angle BCE=\\angle ABC$. Triangle ACE is isosceles, with AC = CE. By AA similarity, $\\triangle DAB \\cong \\triangle DEC$. By the properties of", "measures angles in the standard direction, counterclockwise. Students need to select points in order. For example, to measure angle $$ACB$$ in this triangle, students would select the angle measure tool and then click on $$A$$, then $$C$$, and then $$B$$. If they click on $$B$$, then $$C$$, and then $$A$$, they get the reflex angle. Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organizational skills in problem solving. For example, present 2–3 questions at a time and monitor students to ensure they are making progress throughout the activity. Supports accessibility for: Organization; Attention ### Student Facing Here is triangle $$ABC$$. 1. Rotate triangle $$ABC$$ $$180^\\circ$$ around the midpoint of side $$AC$$. Label the new vertex $$D$$. 2. Rotate triangle $$ABC$$ $$180^\\circ$$ around the midpoint of side $$AB$$. Label the new vertex $$E$$. 3. Look at angles $$EAB$$, $$BAC$$, and $$CAD$$. Without measuring, write what you think is the sum of the measures of these angles. Explain or show your reasoning. 4. Is the measure of angle $$EAB$$ equal to the measure of any angle in triangle $$ABC$$? If so, which one? If not, how do you know? 5. Is the measure of angle $$CAD$$ equal to the measure of any angle in triangle $$ABC$$? If", "# 2005 CEMC Gauss (Grade 7) Problems/Problem 8 ## Problem In the diagram, what is the measure of $\\angle ACB$ in degrees? $[asy] size(300); draw((-60,0)--(0,0)); draw((0,0)--(64.3,76.6)--(166,0)--cycle); label(\"A\",(64.3,76.6),N); label(\"93^\\circ\",(64.3,73),S); label(\"130^\\circ\",(0,0),NW); label(\"B\",(0,0),S); label(\"D\",(-60,0),S); label(\"C\",(166,0),S); [/asy]$ $\\text{(A)}\\ 57^\\circ \\qquad \\text{(B)}\\ 37^\\circ \\qquad \\text{(C)}\\ 47^\\circ \\qquad \\text{(D)}\\ 60^\\circ \\qquad \\text{(E)}\\ 17^\\circ$ ## Solution Since $\\angle ABC + \\angle ABD = 180^\\circ$ (in other words, $\\angle ABC$ and $\\angle ABD$ are supplementary) and $\\angle ABD = 130^\\circ$, then $\\angle ABC = 50^\\circ$. Since the sum of the angles in triangle $ABC$ is $180^\\circ$ and we know two angles $93^\\circ$ and $50^\\circ$ which add to $143^\\circ$, then $\\angle ACB = 180^\\circ - 143^\\circ = 37^\\circ$. The answer is $B$.", "+0 # ***HELP!!!!*****Due Today! -1 49 8 1)In triangle ABC, $AB = BC = 17$ and AC = 16 Find the circumradius of triangle 2)Angle bisectors line AX and line BY of triangle ABC meet at point I. Find angle C in degrees, if angle AIB = 109 [asy] pair A,B,C,X,Y,I; A= (0,0); B = (1,0); C = (0.6,0.5); 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); draw(X--A--C--B--Y); draw(A--B); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,S); label(\"$X$\",X,NE); label(\"$Y$\",Y,NW); [/asy] 3)A certain square and a certain equilateral triangle have the same perimeter. The square and triangle are inscribed in circles, as shown below. If A is the area of the circle containing the square, and B is the area of the circle containing the triangle, then find A/B 4) $\\overline{BY}$ and $\\overline{CZ}$ are angle bisectors of triangle $ABC$ that meet at I, with $CY = 4$, $AY = 6$, and $AB = 8$. Find $BZ$ . Dec 1, 2019 #1 0 im in the same class and i need help on this A triangle has side lengths of 6, 8, and 10. Let a be the area of the circumcircle. Let b be the area of the incircle. Compute a-b Dec 1, 2019 #2 +156 0 Stop cheating on HW", "+0 # We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB 0 42 1 We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB [asy] size(5cm); pair A,B,C; A=(0,0); B=(5,0); C=(1,1); draw(A--B--C--A); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); [/asy] Nov 6, 2020", "# In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) In triangle ABC, angle B is five times as large as angle A. The measure of angle C is 2° less than that of angle A. Find the measures of the angles. (Hint: The sum of the angle measures is 180°.) 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 mhalmantus We are given: $A=x,B=5x,C=x-2$ As hinted, the sum of the angle measures is 180° so: $A+B+C=180$ $x+\\left(5x\\right)+\\left(x-2\\right)=180$ Solve for x: $7x-2=180$ $7x=182$ $x=26$ So, the angle measures are: $A=x=26°$ $B=5x=130°$ $C=x-2=24°$", "# In isosceles triangle ABC, B is the vertex. The measure of angle B can be represented as (4x-2). The measure of angle A can be represented as (8x+1). Find the measure of all three angles of the triangle? May 16, 2018 $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$. #### Explanation: Since $\\Delta A B C$ is an isosceles triangle, angle A is congruent to angle C. Knowing this, and the fact that a triangle's angle measures add up to 180, we can state: $4 x - 2 + 2 \\left(8 x + 1\\right) = 180$ Now solve for x. $4 x - 2 + 16 x + 2 = 180$ $20 x = 180$ $x = 9$ Now apply x to the two given angle measures to find the new measurements. $4 \\cdot 9 - 2 = 34$ $8 \\cdot 9 + 1 = 73$ $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$.", "# [SOLVED]Angles in a triangle #### anemone ##### MHB POTW Director Staff member All the angles in triangle $ABC$ are less than $120^{\\circ}$. Prove that $\\dfrac{\\cos A+\\cos B+\\cos C}{\\sin A+\\sin B+\\sin C}>-\\dfrac{\\sqrt{3}}{3}$. #### anemone ##### MHB POTW Director Staff member \\begin{tikzpicture} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (6, 0); \\coordinate[label=above:C] (C) at (2.4,3); \\coordinate[label=above: $A_1$] ($A_1$) at (7,3); \\coordinate[label=below: $B_1$] ($B_1$) at (7.2,0); \\coordinate[label=below: $C_1$] ($C_1$) at (12,2); \\draw (A) -- (B)-- (C)-- (A); \\draw (7,3) -- (7.2,0)-- (12,2)-- (7,3); \\end{tikzpicture} Consider the triangle $A_1B_1C_1$ where $\\angle A_1=120^{\\circ}-\\angle A,\\,\\angle B_1=120^{\\circ}-\\angle B$ and $\\angle C_1=120^{\\circ}-\\angle C$. The given condition guarantees the existence of such a triangle. Applying the triangle inequality in triangle $A_1B_1C_1$ gives $B_1C_1+C_1A_1>A_1B_1$, i.e. $\\sin A_1+\\sin B_1>\\sin C_1$ by applying the law of sines to triangle $A_1B_1C_1$. It follows that $\\sin (120^{\\circ}-A)+\\sin (120^{\\circ}-B)>\\sin (120^{\\circ}-C)$ or $\\dfrac{\\sqrt{3}}{2}(\\cos A+\\cos B+\\cos C)+\\dfrac{1}{2}(\\sin A+\\sin B+\\sin C)>0$. Taking into account that $a+b>c$ implies $\\sin A+\\sin B-\\sin C>0$, the above inequality can be rewritten as $\\dfrac{\\sqrt{3}}{2}\\cdot \\dfrac{\\cos A+\\cos B+\\cos C}{\\sin A+\\sin B+\\sin C}+\\dfrac{1}{2}>0$, from which the conclusion follows.", "+0 # Helllppp 0 312 4 Find $\\angle BPC$ below. [asy] pair A = (0,0); pair B = (1,0); pair P = (1/2, sqrt(3)/2); pair D = (0,-1); pair C = (1,-1); draw(A--B--C--D--cycle); draw(A--P--B); draw(rightanglemark(B,A,D, 2)); draw(rightanglemark(C,B,A, 2)); draw(rightanglemark(D,C,B, 2)); draw(rightanglemark(A,D,C, 2)); add(pathticks(P--A, 1, .5, 1, 2)); add(pathticks(P--B, 1, .5, 1, 2)); add(pathticks(A--B, 1, .5, 1, 2)); add(pathticks(B--C, 1, .5, 1, 2)); add(pathticks(C--D, 1, .5, 1, 2)); add(pathticks(A--D, 1, .5, 1, 2)); label(\"$P$\", P, N); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$C$\", C, SE); label(\"$D$\", D, SW); [/asy] Sep 21, 2019 #1 +195 +3 Triangle BPC is an isossilies triangle because sides BP and BC are the same length, so angles BPC and BCP are congruent. To find angle BPC you first need to find angle PBC. Triangle APB is equilateral so all the angles measure 60 degrees, so angle APB measures 60 degrees. All the angles in a square measure 90 degrees. The measure of angle PBC is the sum of the angles ABC=90 and ABP=60 so PBC=150 degrees. $$2 \\cdot m\\angle BPC+m\\angle PBC=180$$ 150 can be substituted for angle PBC $$2 \\cdot m\\angle BPC+150=180$$ subtract 150 $$2 \\cdot m\\angle BPC=30$$ divide by 2 $$\\boxed{m\\angle BPC=15}$$ . Sep 21, 2019 #3 +2847 +4 Good job Power27! How were able to decipher the picture code? CalculatorUser Sep 22,", "a triangle are isogonal conjugates. • If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle. • Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$, $BC$, and $CA$ are on the circumcircle of $ABC$: $[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label(\"A\",A,(0,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0)); label(\"A'\",A1,(0,-1));label(\"B'\",B1,(1,1));label(\"C'\",C1,(-1,1)); label(\"H\",H,(-1,-1)); [/asy]$ • Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$, then $M$ is on the nine-point circle. ACS WASC ACCREDITED SCHOOL ## About Our Team Our History Jobs ## Site Info Terms Privacy Contact Us ## Help School Books Community ## Stay Connected Subscribe to get news and updates from AoPS, or Contact Us. © 2015 AoPS Incorporated Invalid username Login to AoPS", "A or B may be used Q: The point $$P ( 4 , - 3 )$$ lies on the terminal arm of an angle $$q$$ in standard position. Determine the measure of $$q$$ to the nearest degree. Select one: a. $$- 143 ^ { \\circ }$$ b. $$127 ^ { \\circ }$$ c. $$323 ^ { \\circ }$$ d. $$233 ^ { \\circ }$$ Q: A triangle has angle measurements of $$90 ^ { \\circ } , 27 ^ { \\circ } ,$$ and $$63 ^ { \\circ } .$$ What type of triangle is this? Acute Right Obtuse Q: 8. In $$\\triangle Q R S , q = 1.7 m , r = 4.3 m$$ , and $$s = 5.6 m$$ . Solve $$\\triangle Q R S$$ by determining the measure of each angle to the nearest degree. Show your work and draw $$\\triangle Q R S$$ . ( $$6$$ marks) Q: 8. In $$\\triangle Q R S , q = 1.7 m , r = 4.3 m$$ , and $$s = 5.6 m$$ . Solve $$\\triangle Q R S$$ by determining the measure of each angle to the nearest degree. Show your work and draw $$\\triangle Q R S$$ . ( $$6$$ marks) Q: If the tan of angle $$x$$ is $$\\frac { 22 } {" ]

[ "\\;=\\;2\\!\\cdot\\!\\angle BAC$", "In triangle ABC, the measure of angle B is three times that of angle A. The measure of angle C is 20 degrees more than that of angle A. How do you find the angle measures? Oct 28, 2016 $\\angle A = {32}^{\\circ}$ $\\angle B = {96}^{\\circ}$ $\\angle C = {52}^{\\circ}$ Explanation: In a triangle, the three interior angles always add up to 180°: Let $\\angle A = x$ $\\implies \\angle B = 3 x$ $\\implies \\angle C = x + 20$ $\\angle A + \\angle B + \\angle C = {180}^{\\circ}$ $\\implies x + 3 x + x + 20 = 180$ $\\implies 5 x + 20 = 180$ $\\implies 5 x = 160$ $x = 32$ Therefore, $\\angle A = x = {32}^{\\circ}$ $\\angle B = 3 x = 3 \\cdot 32 = {96}^{\\circ}$ $\\angle C = x + 20 = 32 + 20 = {52}^{\\circ}$", "possibility for CA:3 total possibilities for CA: $\\boxed{5}$ 3. l/2$\\cdot$l=50 l^2=100 l=10 10$\\cdot$4=$\\boxed{40}$ SparklingWater2 Mar 3, 2021 #3 0 1. DC is tangent to circle O at C. If ∠ABC = 59 degrees, ∠BAC = 60 degrees, and ∠BCA = 61 degrees, then what is the measure of ∠ACD? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Important: The measure of angle ACD is equal to the measure of angle opposite side AC. ∠ACD = 59º Mar 3, 2021", "# The lengths of side b is 15cm, c is 24cm and a is 18cm. How do you calculate the angle B (which is opposite side b)? Dec 13, 2016 ${38.62}^{\\circ}$ #### Explanation: Use the law of cosine of solving triangles : ${b}^{2} = {a}^{2} + {c}^{2} - 2 \\cdot a \\cdot c \\cdot \\cos B$ $\\implies {15}^{2} = {18}^{2} + {24}^{2} - 2 \\cdot 18 \\cdot 24 \\cdot \\cos B$ $\\implies C o s B = \\frac{{18}^{2} + {24}^{2} - {15}^{2}}{2 \\cdot 18 \\cdot 24} = 0.78125$ $\\implies B = {\\cos}^{-} 1 \\left(0.78125\\right) = {38.62}^{\\circ}$ (two decimal place)", "prepare for the material covered in the next section. Find the obtuse angle $B,$ rounded to the nearest degree, satisfying $$\\cos B=\\frac{6^{2}+4^{2}-9^{2}}{2 \\cdot 6 \\cdot 4}$$...", "# Secants of a circle Geometry Level 3 The diameter of the circle shown above is $$CD$$. If $$CD=74$$, $$CP=77$$, and $$PB=33$$, what is the measure of $$\\angle BPC$$ in degrees? note: The figure is not drawn to scale. ×", "\\cdot 2\\sqrt{7} \\cdot \\frac{1}{2} = 5\\sqrt{7}$ you are assuming that the angle between vectors $(\\vec{a} + 2\\vec{b})$ and $(2\\vec{a} - \\vec{b})$ is $60^\\circ$ but it may not be (and apparently it is not).", "the unknown angle $$N$$. Remember to match angle $$N$$ with the corresponding side length of 38 inches. It is also best to store $$m$$ into your calculator and use the unrounded number in your future calculations. \\begin{aligned} c^{2} &=a^{2}+b^{2}-2 a b \\cdot \\cos C \\\\ 38^{2} &=40^{2}+(56.59)^{2}-2 \\cdot 40 \\cdot(56.59) \\cdot \\cos N \\\\ 38^{2}-40^{2}-(56.59)^{2} &=-2 \\cdot 40 \\cdot(56.59) \\cdot \\cos N \\\\ \\frac{38^{2}-40^{2}-(56.59)^{2}}{-2 \\cdot 40 \\cdot(56.59)} &=\\cos N \\\\ N &=\\cos ^{-1}\\left(\\frac{38^{2}-40^{2}-(56.59)^{2}}{-2 \\cdot 40 \\cdot(56.59)}\\right) \\approx 42.1^{\\circ} \\end{aligned} For the next two examples, use the triangle below. Example 4 Determine the length of side $$r$$. $$r^{2}=36^{2}+42^{2}-2 \\cdot 36 \\cdot 42 \\cdot \\cos 63$$ $$r \\approx 41.07$$ Example 5 Determine the length of side $$r$$. $$r^{2}=36^{2}+42^{2}-2 \\cdot 36 \\cdot 42 \\cdot \\cos 63$$ $$r \\approx 41.07$$ Review For all problems, find angles in degrees rounded to one decimal place. In $$\\Delta A B C, a=12, b=15,$$ and $$c=20$$ 1. Find the measure of angle $$A$$. 2. Find the measure of angle $$B$$. 3. Find the measure of angle $$C$$. 4. Find the measure of angle $$C$$ in a different way. In $$\\Delta D E F, d=20, e=10,$$ and $$f=16$$ 5. Find the measure of angle $$D$$. 6. Find the measure of angle $$E$$. 7. Find the measure of angle $$F$$. In $$\\Delta G H I, g=19, \\angle H=55^{\\circ},$$", "+0 # Trig Question +1 95 1 +20 What is the measure of the central angle of a circle if the radius is 18 centimeters. Assume that it intercepts a 12pi centimeter arc. I would prefer the answer in degrees if you dont mind. Jan 23, 2020 #1 +9113 +1 What is the measure of the central angle of a circle if the radius is 18 centimeters. Assume that it intercepts a 12pi centimeter arc. I would prefer the answer in degrees if you dont mind. Hello John! $$b=\\frac{\\pi r \\alpha}{180^0}\\\\ \\alpha=\\frac{b\\cdot 180^0}{\\pi r}=\\frac{12cm\\cdot \\pi \\cdot180^0}{\\pi\\cdot18cm}$$ $$\\alpha=120^0$$ ! Jan 23, 2020", "diameter. Chords AB and CD are produced to meet at E. If $$\\angle CAE = 34^\\circ$$ and $$\\angle E = 30^\\circ$$, then $$\\angle CBD$$ is equal to: Question 9 $$ab(a - b) + bc(b - c) + ca(c - a)$$ is equal to: Question 10 The radius of the base of a right circular cylinder is increased by 20%. By what per cent should its height be reduced so that its volume remains the same as before? OR", "# In isosceles triangle ABC, B is the vertex. The measure of angle B can be represented as (4x-2). The measure of angle A can be represented as (8x+1). Find the measure of all three angles of the triangle? May 16, 2018 $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$. #### Explanation: Since $\\Delta A B C$ is an isosceles triangle, angle A is congruent to angle C. Knowing this, and the fact that a triangle's angle measures add up to 180, we can state: $4 x - 2 + 2 \\left(8 x + 1\\right) = 180$ Now solve for x. $4 x - 2 + 16 x + 2 = 180$ $20 x = 180$ $x = 9$ Now apply x to the two given angle measures to find the new measurements. $4 \\cdot 9 - 2 = 34$ $8 \\cdot 9 + 1 = 73$ $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$.", "Question #54875 Jun 2, 2017 see explanation. Explanation: By the Law of Cosines, we know that: ${c}^{2} = {a}^{2} + {b}^{2} - 2 \\cdot a \\cdot b \\cdot \\cos \\angle C$ Putting in the values, we get ${9}^{2} = {4}^{2} + {7}^{2} - 2 \\cdot 4 \\cdot 7 \\cdot \\cos \\angle C$ $\\implies \\cos \\angle C = \\frac{16 + 49 - 81}{2 \\cdot 4 \\cdot 7} = - 0.28571$ $\\implies \\angle C = {\\cos}^{-} 1 \\left(- 0.28571\\right) = {106.60}^{\\circ}$ Similarly, ${a}^{2} = {b}^{2} + {c}^{2} - 2 \\cdot b \\cdot c \\cdot \\cos \\angle A$ $\\implies \\cos \\angle A = \\frac{{9}^{2} + {7}^{2} - {4}^{2}}{2 \\cdot 9 \\cdot 7} = 0.90476$ $\\implies \\angle A = {\\cos}^{-} 1 \\left(0.90476\\right) = {25.21}^{\\circ}$ $\\implies \\angle B = 180 - \\angle C - \\angle A = 180 - 106.60 - 25.21 = {48.19}^{\\circ}$", "and angle d=$(180-\\theta)$ Now notice that the cosine rule is: $C^2=B^2+A^2-2\\cdot A\\cdot B\\cdot cos(C)$ for any non-right angle triangle like this one: Now back to the topic. Notice that for those two triangles (Triangle A and Triangle B), the side length D is clearly bigger than C. Using the cosine rule for those two triangles: For triangle A: $C^2=A^2+B^2-2\\cdot A\\cdot B\\cdot cos(\\theta)$ (as $c=\\theta$) For Triangle B: $D^2=A^2+B^2-2\\cdot A\\cdot B\\cdot cos(180-\\theta)$ (as $d=180-\\theta)$ Now you know that $cos(180-\\theta) =-cos(\\theta)$, we can have for Triangle B: $D=\\sqrt{A^2+B^2-(2\\cdot A\\cdot B\\cdot -cos(\\theta))}=\\sqrt{A^2+B^2+2\\cdot A\\cdot B\\cdot cos(\\theta)}$ Compared to triangle A for side C, $C=\\sqrt{A^2+B^2-2\\cdot A\\cdot B\\cdot cos(\\theta)}$ Now we know that the formula for both triangles A and B that the side D is bigger than side C and the formulas derived are: $C=\\sqrt{A^2+B^2-2\\cdot A\\cdot B\\cdot cos(\\theta)}$ for angle $c=\\theta$ $D=\\sqrt{A^2+B^2+2\\cdot A\\cdot B\\cdot cos(\\theta)}$ for $d=180-\\theta$ And we make sense that $\\sqrt{A^2+B^2+2\\cdot A\\cdot B\\cdot cos(\\theta)}$ is clearly bigger than $\\sqrt{A^2+B^2-2\\cdot A\\cdot B\\cdot cos(\\theta)}$ That is one of the reasons why $cos(180-\\theta)=-cos(\\theta)$ • I now understand why $\\cos (180-\\theta)=-\\cos \\theta$ \"helps\" but I still don't understand why it works.Thanks for answering though! – The Cryptic Cat Mar 5 '16 at 7:51 • No problem, you will learn the cosine rule and sine rule as you progress to non-right angle trigonometry by your teacher.", "# What is the measure of one interior angle in a 24-gon? May 3, 2018 ${3960}^{\\circ}$ #### Explanation: Given: $24$-gon The interior angles are based on the number of sides: $\\underline{n \\text{ number of triangles interior angles}}$ $3 \\text{ \"1\" } {180}^{\\circ}$ $4 \\text{ \"2\" } 2 \\cdot {180}^{\\circ} = {360}^{\\circ}$ $5 \\text{ \"3\" } 3 \\cdot {180}^{\\circ} = {540}^{\\circ}$ $6 \\text{ \"4\" } 4 \\cdot {180}^{\\circ} = {720}^{\\circ}$ ... $24 \\text{ \"22\" } 22 \\cdot {180}^{\\circ} = {3960}^{\\circ}$ The formula: $\\left(n - 2\\right) \\cdot {180}^{\\circ}$, where $n$ = number of sides", "particular, if $$n=2$$, we have $$m=2\\cdot 1 = 2$$, $$dS_{m-1} = dS_1$$ which is the arc-length measure, and so $$S_1(\\partial B(t)) =$$ length or circle of radius $$t$$, which equals $$2\\pi t$$. • Ok, the situation is clearer now. My answer above solves both your particular (case $n=1$) and general problems (cases $n > 1$). Jun 18 '20 at 10:30", "|CD| = \\frac12 \\cdot \\sqrt{17^2-15^2}=4$ to 23: Since |AF| = |AE|, |CE| = |CD|, |BD| = |BF| all given lengths are used twice. Therefore the circumference is 2(5+6+8) = 38 to 27: For symmetry reasons |AB| = |AC| = |AD| Therefore |AC| = 6 3. thanks but still need other one answered.", "the center of the room c angle AcD and dist AD are known angles cDA and cAD are equal to (180-AcD)/2", "# A triangle has sides A, B, and C. Sides A and B are of lengths 7 and 3, respectively, and the angle between A and B is pi/12. What is the length of side C? Here the side A=7 ; B=3 and the angle c between A and B. $\\angle c = \\left(\\frac{180}{12}\\right) = {15}^{0}$ The side C= sqrt(A^2+B^2-2*A*B*cosc or $C = \\sqrt{49 + 9 - 2 \\cdot 7 \\cdot 3 \\cdot .966} = \\sqrt{58 - 40.57} = 4.175$ [Answer]", "a+a\\cdot b+a\\cdot c=1+\\cos(\\theta_1)+\\cos(\\theta_2)$ where $\\theta_1$ is the angle between $a$ and $b$ and $\\theta_2$ is the angle between $b$ and $c$. (Both angles, by the definition of \"between\", are less than or equal to 180 degrees.) By dotting with $b$ and $c$ you can find two more equations which involve the angle between $a$ and $c$.", "arc are C and D. Measure of the central angle = m ∠COD = 95º $\\widetilde{DC}$ is a minor arc. m$\\widetilde{DC}$ = 95º. 4. $\\widetilde{DEB}$ The end points of the arc are D and B. Measure of the central angle = m ∠DOB = 95 + 57 = 152º. The point E is not inside the angle DOB. Hence $\\widetilde{DEB}$ is a major arc. m$\\widetilde{DEB}$ = 360 - 152 = 208º. 5. $\\widetilde{AB}$ The end points of the arc are A and B. Measure of the central angle = ∠AOB = ∠EOC = 28 + 95 = 123º. Vertical angles are congruent. $\\widetilde{AB}$ is a minor arc. m$\\widetilde{AB}$ = 123º. 6. $\\widetilde{CDA}$ The end points of the arc are C and A. Measure of central angle = m ∠ COA = 180º AC is a diameter of the circle. $\\widetilde{CDA}$ is a semicircle. m$\\widetilde{CDA}$ = 180º. ## Intercepted Arc of a Circle Theorem: If two secants intersect in the interior of a circle, then the measure of an angle formed is one-half the sum of the measure of the arcs intercepted by the angle and its vertical angle. Secants AB and CD intersect inside the circle at P. The two angles formed are angles 1 and 2. Angle 1 and its vertical angle intercept arcs, AC and BD" ]

[ "# 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", "+0 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "# 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", "+ r)(3)cosA = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-3,12/7)--P); draw((3,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Let $C$ be the center of the largest semicircle. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$, since the area of a triangle is $\\frac{1}{2} Base \\cdot Height$. Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. $[asy] size(5cm); pair A", "+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", "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, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "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 $1", "C B D$ is:In the triangle $\\triangle A B C$ measures of angles $\\angle A=46^{\\circ}$ and $\\angle C=60^{\\circ}$ are given. The angle bisector intersects the circ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given statements is correct. In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given statements is correct.In the right triangle the length of one leg is 5 in., and the measure of the angle opposite to that side is $30^{\\circ}$. Determine which of the given ... close Triangle $\\triangle A B C$, right angled at $C$, is given. Height and the median from point $C$ form an angle $\\varphi$. The measure of larger acute angle of $\\triangle A B C$ is:Triangle $\\triangle A B C$, right angled at $C$, is given. Height and the median from point $C$ form an angle $\\varphi$. The measure of larger acute a ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 The distance of the shorter ladder form the wall is: The distance of the shorter", "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 us $$\\frac {2R} c = \\frac a{\\frac{2 \\times", "semicircle is 6, so its radius is 3 and $PC$ is 3 - r) $(2 + r)^2 + 3^2 - 2(2 + r)(3)\\cos A = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-2.45,12/7)--P); draw((2.45,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Let $C$ be the center of the largest semicircle and $r$ be the radius of $\\circ P$. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$, since the area", "# 2005 CEMC Gauss (Grade 7) Problems/Problem 8 ## Problem In the diagram, what is the measure of $\\angle ACB$ in degrees? $[asy] size(300); draw((-60,0)--(0,0)); draw((0,0)--(64.3,76.6)--(166,0)--cycle); label(\"A\",(64.3,76.6),N); label(\"93^\\circ\",(64.3,73),S); label(\"130^\\circ\",(0,0),NW); label(\"B\",(0,0),S); label(\"D\",(-60,0),S); label(\"C\",(166,0),S); [/asy]$ $\\text{(A)}\\ 57^\\circ \\qquad \\text{(B)}\\ 37^\\circ \\qquad \\text{(C)}\\ 47^\\circ \\qquad \\text{(D)}\\ 60^\\circ \\qquad \\text{(E)}\\ 17^\\circ$ ## Solution Since $\\angle ABC + \\angle ABD = 180^\\circ$ (in other words, $\\angle ABC$ and $\\angle ABD$ are supplementary) and $\\angle ABD = 130^\\circ$, then $\\angle ABC = 50^\\circ$. Since the sum of the angles in triangle $ABC$ is $180^\\circ$ and we know two angles $93^\\circ$ and $50^\\circ$ which add to $143^\\circ$, then $\\angle ACB = 180^\\circ - 143^\\circ = 37^\\circ$. The answer is $B$.", "In triangle ABC, the measure of angle B is three times that of angle A. The measure of angle C is 20 degrees more than that of angle A. How do you find the angle measures? Oct 28, 2016 $\\angle A = {32}^{\\circ}$ $\\angle B = {96}^{\\circ}$ $\\angle C = {52}^{\\circ}$ Explanation: In a triangle, the three interior angles always add up to 180°: Let $\\angle A = x$ $\\implies \\angle B = 3 x$ $\\implies \\angle C = x + 20$ $\\angle A + \\angle B + \\angle C = {180}^{\\circ}$ $\\implies x + 3 x + x + 20 = 180$ $\\implies 5 x + 20 = 180$ $\\implies 5 x = 160$ $x = 32$ Therefore, $\\angle A = x = {32}^{\\circ}$ $\\angle B = 3 x = 3 \\cdot 32 = {96}^{\\circ}$ $\\angle C = x + 20 = 32 + 20 = {52}^{\\circ}$", "+0 # We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB 0 42 1 We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB [asy] size(5cm); pair A,B,C; A=(0,0); B=(5,0); C=(1,1); draw(A--B--C--A); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); [/asy] Nov 6, 2020", "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}", "# In isosceles triangle ABC, B is the vertex. The measure of angle B can be represented as (4x-2). The measure of angle A can be represented as (8x+1). Find the measure of all three angles of the triangle? May 16, 2018 $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$. #### Explanation: Since $\\Delta A B C$ is an isosceles triangle, angle A is congruent to angle C. Knowing this, and the fact that a triangle's angle measures add up to 180, we can state: $4 x - 2 + 2 \\left(8 x + 1\\right) = 180$ Now solve for x. $4 x - 2 + 16 x + 2 = 180$ $20 x = 180$ $x = 9$ Now apply x to the two given angle measures to find the new measurements. $4 \\cdot 9 - 2 = 34$ $8 \\cdot 9 + 1 = 73$ $m \\angle A = 73$, $m \\angle B = 34$, $m \\angle C = 73$.", "# 1997 JBMO Problems/Problem 3 ## Problem Let $ABC$ be a triangle and let $I$ be the incenter. Let $N$, $M$ be the midpoints of the sides $AB$ and $CA$ respectively. The lines $BI$ and $CI$ meet $MN$ at $K$ and $L$ respectively. Prove that $AI+BI+CI>BC+KL$. ## Solution $[asy] size(9.22 cm); pair B=(0,0), A=(50,120), C=(140,0), N=(25,60), M=(95,60), I=(60,40), K=(90,60), L=(20,60); draw(B--A--C--B); draw(circle(I,40)); dot(A); label(\"A\",(50,125)); dot(B); label(\"B\",B,SW); dot(C); label(\"C\",C,SE); dot(N); label(\"N\",(24,65)); dot(M); label(\"M\",M,NE); dot(I); label(\"I\",I,S); dot(K); label(\"K\",K,NW); dot(L); label(\"L\",L,NW); draw(L--C,dotted); draw(K--B,dotted); draw(L--M); draw(anglemark(C,B,A,200)); draw(anglemark(N,K,I,200)); draw(anglemark(A,C,B,200)); draw(anglemark(A,C,B,150)); draw(anglemark(C,L,K,400)); draw(anglemark(C,L,K,450)); [/asy]$ First, by SAS Similarity, $\\triangle ANM \\sim \\triangle ABC,$ so $NM \\parallel BC$ and $MN = \\tfrac12 BC.$ That means $\\angle IBC = \\angle IKN,$ and since $\\angle IBN = \\angle IBC,$ $\\triangle NBK$ is an isosceles triangle. Similarly, $\\angle MLC = \\angle LCB = \\angle LCM,$ making $\\triangle MLC$ an isosceles as well. Thus, $ML = MC$ and $NB = NK.$ By the Triangle Inequality, $AI + IB > AB,$ and $AI + IC > AC$, and $BI + IC > BC.$ That means \\begin{align*} 2(AI + IB + IC) &> AB + AC + BC \\\\ AI + IB + IC &> NK + ML + MN \\\\ &> NK + (MN + LN) + MN \\\\ &> KL + 2 MN \\\\ &> KL + BC. \\end{align*}", "pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C--D--A, dashed ); pair Sp = intersectionpoint(D--O,A--C); label(\"B\",B,SW); label(\"A\",A,S); label(\"C\",C,NW); label(\"O\",O,NE); label(\"D\",D,(S+SSW)); label(\"S\",Sp,(S+SSW)); [/asy]$ As in the previous solution, we find out that $\\angle AOB = \\angle COB = 60^\\circ$. Hence $\\triangle AOD$ and $\\triangle COD$ are both equilateral. We then have $\\angle SCD = \\angle SAD = 30^\\circ$, hence $D$ is the incenter of $\\triangle ABC$, and as $\\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \\cdot SD = BD$, and as $SD = SO$, we have $2\\cdot SD = SD + SO = OD$, therefore $BD=OD$, and as before we conclude that $\\frac{BD}{BO} = \\boxed{\\frac 12}$.", "# 2001 AMC 12 Problems/Problem 24 ## Problem In $\\triangle ABC$, $\\angle ABC=45^\\circ$. Point $D$ is on $\\overline{BC}$ so that $2\\cdot BD=CD$ and $\\angle DAB=15^\\circ$. Find $\\angle ACB$. $\\text{(A) }54^\\circ \\qquad \\text{(B) }60^\\circ \\qquad \\text{(C) }72^\\circ \\qquad \\text{(D) }75^\\circ \\qquad \\text{(E) }90^\\circ$ ## Solution $[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); pair ortho=rotate(-90)*(D-A); pair E=intersectionpoint(A--D, C--(C+10*ortho)); draw(A--B--C--cycle); draw(A--D); draw(C--E, dashed); draw(B--E, dashed); draw( rightanglemark(C,E,D) ); draw( scale(2)*anglemark(B,A,D) ); draw( anglemark(D,B,A) ); label(\"15^\\circ\",(0.6,0),5*ENE); label(\"45^\\circ\",B,5*WNW,UnFill); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",E,S); [/asy]$ We start with the observation that $\\angle ADB = 180^\\circ - 15^\\circ - 45^\\circ = 120^\\circ$, and $\\angle ADC = 15^\\circ + 45^\\circ = 60^\\circ$. We can draw the height $CE$ from $C$ onto $AD$. In the triangle $CED$, we have $\\frac {ED}{CD} = \\cos EDC = \\cos 60^\\circ = \\frac 12$. Hence $ED = CD/2$. By the definition of $D$, we also have $BD=CD/2$, therefore $BD=DE$. This means that the triangle $BDE$ is isosceles, and as $\\angle BDE=120^\\circ$, we must have $\\angle BED = \\angle EBD = 30^\\circ$. Then we compute $\\angle ABE = 45^\\circ - 30^\\circ = 15^\\circ$, thus $\\angle ABE = \\angle BAE$ and the triangle $ABE$ is isosceles as well. Hence $AE=BE$. Now we can note that $\\angle DCE = 180^\\circ - 90^\\circ - 60^\\circ = 30^\\circ$,", "+0 # Can I get some help with these? 0 68 1 Hi... I've been strugling with these three problems for a few weeks: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6},$$ where the angle$$\\frac{\\pi}{6}$$ is in radians. (2)Find the length AC, to two decimal places. Image: https://artofproblemsolving.com/texer/sywndnao (3)Find the length BC, to two decimal places. Image: https://artofproblemsolving.com/texer/enqsdotw Nov 15, 2020 #1 0 Quick note: to get the images for (2) and (3), post these codes to get the figures. Make sure you say Render as Image in the TeXeR (2) [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3) [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] Nov 15, 2020", "lie on the circle, and $\\angle ADC=\\angle CDQ$. $\\angle EAD=\\angle ECD$ because $A$,$C$,$D$ and $E$ all lie on the circle. $\\angle DAC=\\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then, $\\angle EAC=\\angle EAD+\\angle DAC=\\angle ECD+\\angle ECF=\\angle DCQ$ Therefore, $\\triangle AEC$ and $\\triangle CDQ$ are similar, so $AC/CE=CQ/QD$. Lastly, let's show that $\\triangle CPQ$ and $\\triangle EPD$ are similar. $[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label(\"$$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" ]

[ "- 2 , 8$", "+ 8 = 53$", "The cube root of $8$ is $2$.", "$6xx10^12$", "FIGURE THIS OUT $6\\sqrt{2} - 3\\sqrt{2}$", "\\sqrt{8 - 2 \\left(\\sqrt{6} + \\sqrt{2}\\right)} \\approx 2.088$", "= 6.95}$", "## 1 June 2017 ### Eight $8$ is a composite number. $8$ is a perfect cube: $8=2^3$. $8$ is a Fibonacci number: $8=3+5$. $8$ is the base of the octal system, widely used in Computer Science. An octagon is an $8$ sided polygon. An octahedron is a polyhedron with $8$ faces. A cube has $8$ vertices.", "(n+6)^2$$", "7x - 6y + 5$", "-6 x + 1 = 8 or -5", ", 6\\right)$", "=3/8 1 2", "{x}^{2} - 6 x - 8 - \\frac{3}{x + 3}$", "=8 So,Unit digit of $8N=N+2$ .", "= 8$ This checks.", "+ (n+6)^2$$", "+6 +4 +4", "largest is $$8 3 1$$", "$6-3i$." ]

[ "Use Strassen’s algorithm to compute the matrix product $$\\begin{pmatrix} 1 & 3 \\\\ 7 & 5 \\\\ \\end{pmatrix} \\begin{pmatrix} 6 & 8 \\\\ 4 & 2 \\\\ \\end{pmatrix}$$. The sums: \\begin {aligned} & S_1 = B_{12} - B_{22} = 8 - 2 = 6 \\\\ & S_2 = A_{11} + A_{12} = 1 + 3 = 4 \\\\ & S_3 = A_{21} + A_{22} = 7 + 5 = 12 \\\\ & S_4 = B_{21} - B_{11} = 4 - 6 = -2 \\\\ & S_5 = A_{11} + A_{22} = 1 + 5 = 6 \\\\ & S_6 = B_{11} + B_{22} = 6 + 2 = 8 \\\\ & S_7 = A_{12} - A_{22} = 3 - 5 = -2 \\\\ & S_8 = B_{21} + B_{22} = 4 + 2 = 6 \\\\ & S_9 = A_{11} - A_{21} = 1 - 7 = -6 \\\\ & S_{10} = B_{11} + B_{12} = 6 + 8 = 14 \\end {aligned} The products: \\begin {aligned} & P_1 = A_{11} \\cdot S_1 = 6 \\\\ & P_2 = S_2 \\cdot B{22} = 8 \\\\ & P_3 = S_3 \\cdot B{11} = 72 \\\\ & P_4 = A_{22} \\cdot S_4 = -10 \\\\ & P_5 = S_5 \\cdot S_6 = 48 \\\\ & P_6 = S_7 \\cdot S_8", "# Creating arithmetic expression equal to 1000 using exactly eight 8's and parentheses I would like to find all the expressions that can be created using nothing but arithmetic operators, exactly eight $8$'s, and parentheses. Here are the seven solutions I've found (on the Internet) so far: \\begin{align} 1000 &= (8888 - 888) / 8\\\\ 1000 &= 888 + 88 + 8 + 8 + 8\\\\ 1000 &= 888 + 8 \\cdot (8 + 8) - 8 - 8\\\\ 1000 &= 8 \\cdot (8 \\cdot 8 + 8 \\cdot 8) - 8 - 8 - 8\\\\ 1000 &= 8 \\cdot (8 \\cdot (8 + 8) - (8 + 8) / 8) - 8\\\\ 1000 &= (8 \\cdot (8 + 8) - (8 + 8 + 8) / 8) \\cdot 8\\\\ 1000 &= (8 \\cdot (8 + 8) - 88 / 8 + 8) \\cdot 8 \\end{align} Are there others, and if there are, what are they? Update After sifting through achille hui's answer and adding one of mathlove's solutions, I get the following $16$ possibilities: \\begin{align} 1000 &= (8888 - 888)/8\\\\ 1000 &= 888 + (888 + 8)/8\\\\ 1000 &= 888 + 88 + 8 + 8 + 8\\\\ 1000 &= 888 + (8 + 8) \\cdot (8 - 8/8)\\\\ 1000 &= 888 - (8 + 8) \\cdot", "+ 6 + 6 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 Use the product rule of exponents to set up the problem. The product rule states: x^a × x^b = x^{a+b} So 6^{24} = 6^{(3 + 3 + 3 + 3 + 3 + 3 + 3 + 3)} = 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 Solve the problem. \\begin{aligned} 6^{24}&=6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 \\\\ &= 216 × 216 × 216 × 216 × 216 × 216 × 216 × 216 \\\\ &= 46656 × 46656 × 46656 × 46656 \\\\ &= 4.738 × 10^{18} \\end{aligned} #### Things You'll Need • Pen or pencil • Paper #### Tips • For some problems, a combination of both techniques may make the problem easier. For example: ​x21 = (​x7)3 (power rule), and ​x7 = ​x3 × ​x2 × ​x2 (product rule). Combining the two, you get: ​x21 = (​x3 × ​x2 × ​x2)3 Dont Go! We Have More Great Sciencing Articles!", "the equation, $$17m + 26n = 1$$ To do this, we apply the Euclidean Algorithm: \\begin{align*} 26 &= 1 \\cdot 17 + 9\\\\ 17 &= 1 \\cdot 9 + 8\\\\ 9 &= 1 \\cdot 8 + \\boxed{1} \\leftarrow \\textrm{gcd}\\\\ 8 &= 8 \\cdot 1 + 0 \\end{align*} Now, we can use these calculations to write the greatest common divisor, 1, as a linear combination of 17 and 26... \\begin{align*} 1 &= 9 - 1 \\cdot 8\\\\ &= 9 - 1 \\cdot (17 - 1 \\cdot 9)\\\\ &= 2 \\cdot 9 - 1 \\cdot 17\\\\ &= 2 \\cdot (26 - 1 \\cdot 17) - 1 \\cdot 17\\\\ &= 2 \\cdot 26 - 3\\cdot 17 \\end{align*} Which reveals the value of $m$ and hence, the multiplicative inverse of the determinant: $$\\textrm{det}^{-1} = m = -3 \\equiv 23 \\pmod{26}$$ Returning to finding the inverse matrix desired, we have \\begin{align*} \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} &= \\textrm{det}^{-1} \\begin{bmatrix}8 & -11\\\\-19 & 25\\end{bmatrix}\\\\ &\\equiv 23 \\begin{bmatrix}8 & -11\\\\-19 & 25\\end{bmatrix} \\\\ &\\equiv \\begin{bmatrix}184 & -253\\\\-437& 575\\end{bmatrix} \\\\ &\\equiv \\begin{bmatrix}2 & 7\\\\5 & 3\\end{bmatrix} \\pmod{26} \\end{align*} Finally, recalling that \\begin{align*} \\begin{pmatrix}a\\\\b\\end{pmatrix} &\\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}13\\\\12\\end{pmatrix} \\pmod{26} \\quad \\textrm{, and }\\\\ \\begin{pmatrix}c\\\\d\\end{pmatrix} &\\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}20\\\\1\\end{pmatrix} \\pmod{26} \\end{align*} we can solve for $a$, $b$, $c$, and $d$: \\begin{align*} \\begin{pmatrix}a\\\\b\\end{pmatrix}", "## Intermediate Algebra for College Students (7th Edition) $6x^{6}$ RECALL: The product rule of exponents states that: $a^m \\cdot a^n=a^{m+n}$. Use this rule to obtain: $3x^4 \\cdot 2x^2 \\\\=6x^{4+2} \\\\=6x^{6}$", "can be reduced as follows: \\begin{align*} \\sqrt{16\\sqrt{8\\sqrt{4}}} &= \\sqrt{16\\sqrt{16}} \\\\ &= \\sqrt{64}\\\\ &= 8 \\end{align*} Thus, C is the correct answer. 4. When $$0.000315$$ is multiplied by $$7,928,564$$ the product is closest to which of the following? $$210$$ $$240$$ $$2100$$ $$2400$$ $$24000$$ ###### Solution(s): We can approximate the product as \\begin{align*} (3 \\cdot 10^{-4})(8 \\cdot 10^6) &= 24 \\cdot 10^2 \\\\ &= 2400. \\end{align*} Thus, D is the correct answer. 5. What is the value of the expression $\\dfrac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8}{1+2+3+4+5+6+7+8}?$ $$1020$$ $$1120$$ $$1220$$ $$2240$$ $$3360$$ ###### Solution(s): Looking at the denominator independently, $$1 + 2 + 3 + \\cdots + 8 = 36$$, so the desired answer is $\\begin{gather*} \\dfrac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8}{36} = \\\\ 4 \\cdot 5 \\cdot 7 \\cdot 8 = 1120. \\end{gather*}$ Thus, B is the correct answer. 6. If the degree measures of the angles of a triangle are in the ratio $$3:3:4$$, what is the degree measure of the largest angle of the triangle? $$18$$ $$36$$ $$60$$ $$72$$ $$90$$ ###### Solution(s): We can let the three angles be equal to $$3x, 3x,$$ and $$4x.$$ Then we know that their sum equals $$180.$$ From this we can set $$3x +", "+0 I need help... 0 30 2 +23 Hello! I am asking for help on an AoPS question so I would rather not to be given the answer. Here is the question: Compute the sum of: $$101^2 - 97^2 + 93^2 - 89^2 + \\cdots + 5^2 - 1^2$$ Here is what I have: (a) From the difference of squares, we know that $$(a + b)(a - b) = a^2 - b^2$$. Using this pattern, we can pair up the terms in this sequence and apply: \\begin{align*} 101^2 - 97^2 + 93^2 - 89^2 + \\cdots + 5^2 - 1^2 &= (101 + 97)(101 - 97) + (93 + 89)(93 - 89) + \\cdots + (5 + 1)(5 - 1), \\\\ &= 4(101 + 97) + 4(93 + 89) + ... + 4(5 + 1), \\\\ &= 4(198 + 182 + 166 + ... + 6). \\end{align*} \\begin{align*} 4(198 + 182 + 166 + ... + 6) &= 4(\\frac{204 \\cdot 12}{2}), \\\\ &= 4(204 \\cdot 6), \\\\ &= 4(1224), \\\\ &= 4896. \\end{align*} So, the sum of the sequence $$101^2 - 97^2 + 93^2 - 89^2 + \\cdots + 5^2 - 1^2$$, is $$\\boxed{4896}$$. ------- But I am not sure this is correct, any suggestions? Jan 6, 2021 2+0 Answers #1 +188 0 This looks accurate to me...", "(1+\\cos{x})^{3} \\rm{dx} &= \\int 8 \\cdot \\cos^{6}\\Bigl(\\frac{x}{2}\\Bigr) \\ \\rm{dx} \\\\ &= 8 \\int \\cos^{6}\\Bigl(\\frac{x}{2}\\Bigr) \\ \\rm{dx} \\end{align*} -", "&= 1 + a(16) + a(6) = 1 + 3 + 2 = 6 \\end{align*} Thus the answer is N = 15+6+8+6+4+3+2+2+1 = \\boxed{047}. 1 Like", "into eqn. } (1) \\quad a &= -7 -2 (4) \\\\ \\therefore a &= -15 \\\\ S_{n} &= \\frac{n}{2}[2a + (n-1)d] \\\\ S_{51} &= \\frac{51}{2}[2(-15) + (51-1)(4)] \\\\ &= \\frac{51}{2}[-30 + 200] \\\\ &= (51)(85) \\\\ \\therefore S_{51}&= 4335 \\end{align*} Calculate the sum of the arithmetic series $$4+7+10+\\cdots +901$$. \\begin{align*} a &= 4 \\\\ l &= 901 \\\\ d &= T_{2} - T_{1} \\\\ &= 7 - 4 \\\\ &= 3 \\\\ \\text{And } T_{n} &= a +(n-1)d \\\\ &= 4 +(n-1)(3) \\\\ \\therefore 901 &= 4 + 3n - 3 \\\\ 900 &= 3n \\\\ \\therefore 300 &= n \\\\ S_{n} &= \\frac{n}{2}[a + l] \\\\ S_{300} &= \\frac{300}{2}[4 + 901 ] \\\\ &= (150)(905) \\\\ \\therefore S_{300} &= 135750 \\end{align*} Evaluate without using a calculator: $$\\dfrac{4 + 8 + 12 + \\cdots + 100}{3 + 10 + 17 + \\cdots + 101}$$ Consider the numerator: $$4 + 8 + 12 + \\ldots + 100$$ \\begin{align*} a &= 4 \\\\ l &= 100 \\\\ d &= T_{2} - T_{1} \\\\ &= 8 - 4 \\\\ &= 4 \\\\ \\text{And } T_{n} &= a +(n-1)d \\\\ 100 &= 4 +(n-1)(4) \\\\ 100 &= 4n \\\\ \\therefore 25 &= n \\\\ S_{n} &= \\frac{n}{2}[a + l] \\\\ S_{25} &= \\frac{25}{2}[4 + 100 ] \\\\ S_{25} &= (25)(52) \\end{align*} Consider", "$6 \\boldcdot 4^2$, we want to make sure everyone agrees about how to evaluate this. Otherwise some people might multiply first and others compute the exponent first, and different people would get different values for the same expression! Earlier we saw situations in which $6 \\boldcdot 4^2$ represented the surface area of a cube with side lengths 4 units. When computing the surface area, we evaluate $4^2$ first (or find the area of one face of the cube first) and then multiply the result by 6. In many other expressions that use exponents, the part with an exponent is intended to be evaluated first. To make everyone agree about the value of expressions like $6 \\boldcdot 4^2$, the convention is to evaluate the part of the expression with the exponent first. Here are a couple of examples: \\begin {align} &\\;6 \\boldcdot 4^2\\\\ &= 6 \\boldcdot 16\\\\&= 96 \\end {align} \\begin {align} &\\;45 + 5^2\\\\ &= 45 + 25\\\\&= 70 \\end {align} If we want to communicate that 6 and 4 should be multiplied first and then squared, then we can use parentheses to group parts together: \\begin {align} &\\;(6 \\boldcdot 4)^2\\\\ &= 24^2\\\\&= 576 \\end {align}", "exponents, the part with an exponent is intended to be evaluated first. To make everyone agree about the value of expressions like $$6 \\boldcdot 4^2$$, the convention is to evaluate the part of the expression with the exponent first. Here are a couple of examples: \\begin {align} 6 &\\boldcdot 4^2 \\\\ 6 &\\boldcdot 16 \\\\ &96 \\end {align} \\begin {align} 45 &+ 5^2 \\\\ 45 &+ 25 \\\\ &70 \\end {align} If we want to communicate that 6 and 4 should be multiplied first and then squared, then we can use parentheses to group parts together: \\begin {align} (6 &\\boldcdot 4)^2 \\\\ &24^2 \\\\ &576 \\end {align} \\begin {align} (45 &+ 5)^2 \\\\ &50^2 \\\\ 2,&500 \\end {align} ### Glossary Entries • cubed We use the word cubed to mean “to the third power.” This is because a cube with side length $$s$$ has a volume of $$s \\boldcdot s \\boldcdot s$$, or $$s^3$$. • exponent In expressions like $$5^3$$ and $$8^2$$, the 3 and the 2 are called exponents. They tell you how many factors to multiply. For example, $$5^3$$ = $$5 \\boldcdot 5 \\boldcdot 5$$, and $$8^2 = 8 \\boldcdot 8$$. • squared We use the word squared to mean “to the second power.” This is because a square with side length $$s$$ has an area", "+ (n+3)!}{(n+1)[n! +(n+1)!]} = \\frac{(n+2)! \\cdot (n+4)}{(n+2)!} = n + 4.$ Hence, the required sum is \\begin{aligned} (1+4) + (2+4) + \\dots + (10 + 4) &= 5 + 6 + \\dots + 14 \\\\ &= (1 + 2 + \\dots + 14) - (1 + 2 + 3 + 4) \\\\ &= \\frac{14 \\cdot 15}{2} - 10 \\\\ &= 105 - 10 \\\\ &= \\boldsymbol{95}. \\end{aligned}", "$$\\boldsymbol{n} \\cdot \\pmatrix{x \\\\ y \\\\ z} = \\boldsymbol{n} \\cdot \\pmatrix{8 \\\\ -14 \\\\ 14}$$ $$23 x - 16 y - 34 z = -34$$ In vector form the above is $$\\pmatrix{23 \\\\ -16 \\\\ -34} \\cdot \\boldsymbol{r} = -34$$ where $$\\boldsymbol{r} = \\pmatrix{x & y & z}$$. Proof Show that \\begin{aligned} \\pmatrix{23 \\\\ -16 \\\\ -34} \\cdot \\boldsymbol{L}_1 & = -34 & \\text{and}\\\\ \\pmatrix{23 \\\\ -16 \\\\ -34} \\cdot \\boldsymbol{L}_2 & = -34 \\end{aligned}", "For the corrected version of the problem, where you are asked to compute $x^p$, you can use exponentiation by squaring directly using \\begin{align} x^0&=1\\\\ x^n&=(x^{n/2})^2 \\ \\text{for even }n\\\\ x^n&=x \\cdot (x^{(n-1)/2})^2 \\ \\text{for odd }n \\end{align} - But that is unlikely to be the intended answer, because it doesn't use the explicitly given fact that $p$ is an integer. – Henning Makholm Dec 14 '11 at 15:45", "a formula $S := 1 + 2 + 3 + 4 + \\cdots + m$. I'll show a derivation here that is slightly different from the usual reverse and add method. Let's begin: \\begin{align*} S &= 1 + 2 + 3 + 4 + \\cdots + m \\\\\\ 2S &= 2 + 4 + 6 + 8 + \\cdots + 2m \\\\\\ 2S &= [1 + 3 + 5 + 7 + \\cdots + (2m-1)] + \\underbrace{1 + 1 + 1 + 1 + \\cdots + 1}_{m} \\\\\\ 2S &= [1 + 3 + 5 + 7 + \\cdots + (2m-1)] + m \\end{align*} Now, consider the sum in the brackets. Notice the pattern: \\begin{align*} 1 &= 1 \\\\\\ 4 &= 1 + 3 \\\\\\ 9 &= 1 + 3 + 5 \\\\\\ 16 &= 1 + 3 + 5 + 7 \\\\\\ 25 &= 1 + 3 + 5 + 7 + 9 \\\\\\ 36 &= 1 + 3 + 5 + 7 + 9 + 11 \\end{align*} These are just the perfect squares! Thus, we conjecture that $1 + 3 + 5 + \\cdots + (2m-1) = m^2$. In fact, this can easily be visualized as follows: m=1 m=2 m=3 m=4 m=5 1 1 2 1 2 3 1 2 3 4 1 2 3 4", "8+16 \\times 12 \\\\ &= -72 - 16 + 192 \\\\ &= 104. \\end{align} To find the angle between $\\boldsymbol{\\mathrm{a} }$ and $\\boldsymbol{\\mathrm{b} }$, recall this formula for the dot product: $\\boldsymbol{\\mathrm{a} }\\cdot\\boldsymbol{\\mathrm{b} }=\\lvert\\boldsymbol{\\mathrm{a} }\\rvert\\lvert\\boldsymbol{\\mathrm{b} }\\rvert\\cos{\\theta}.$ This can be rearranged to make $\\theta$ the subject of the equation: $\\cos{\\theta}=\\dfrac{\\boldsymbol{\\mathrm{a} }\\cdot\\boldsymbol{\\mathrm{b} } }{\\lvert\\boldsymbol{\\mathrm{a} }\\rvert\\lvert\\boldsymbol{\\mathrm{b} }\\rvert}.$ To use this formula, we need to know the lengths of vectors $\\boldsymbol{\\mathrm{a} }$ and $\\boldsymbol{\\mathrm{b} }$. \\begin{align} \\lvert\\boldsymbol{\\mathrm{a} }\\rvert &= \\sqrt{a_1^2+a_2^2+a_3^2} \\\\ &=\\sqrt{8^2+(-2)^2+16^2} \\\\ &=\\sqrt{\\strut{64+4+256} } \\\\ &=\\sqrt{\\strut{324} } \\\\ &=18, \\end{align} \\begin{align} \\lvert\\boldsymbol{\\mathrm{b} }\\rvert &= \\sqrt{b_1^2+b_2^2+b_3^2} \\\\ &= \\sqrt{(-9)^2+8^2+12^2} \\\\ &= \\sqrt{\\strut{81+64+144} } \\\\ &= \\sqrt{289 } \\\\ &= 17. \\end{align} Substitute the values for $\\boldsymbol{\\mathrm{a} }\\cdot\\boldsymbol{\\mathrm{b} }$, $\\lvert\\boldsymbol{\\mathrm{a} }\\rvert$ and $\\lvert\\boldsymbol{\\mathrm{b} }\\rvert$ into the formula for $\\cos{\\theta}$: \\begin{align} \\cos{\\theta} &= \\dfrac{\\boldsymbol{\\mathrm{a} }\\cdot\\boldsymbol{\\mathrm{b} } }{\\lvert\\boldsymbol{\\mathrm{a} }\\rvert\\lvert\\boldsymbol{\\mathrm{b} }\\rvert} \\\\ \\\\ &= \\dfrac{104}{18\\cdot17} \\\\ \\\\ &= \\dfrac{104}{306}. \\\\ \\\\ \\theta &= \\arccos \\left( \\dfrac{104}{306} \\right) \\\\ &= 1.22 \\:(\\text{to 3 significant figures}). \\end{align} The angle between vectors $\\boldsymbol{\\mathrm{a} }=(8,\\,-1,\\,16)$ and $\\boldsymbol{\\mathrm{b} }=(-9,\\,8,\\,12)$ is $1.22$ radians, to 3 significant figures. Workbook This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.", "dot product \\begin{align} (-2v+w)\\cdot (-2v+w) &=(-2w+v)\\cdot (-2w+v) \\\\ 4|v|^2-4w\\cdot v+|w|^2 &= 4|w|^2-4w\\cdot v+|v|^2 \\\\ 3|v|^2&=3|w|^2. \\end{align} Hence $|v|=|w|$ and the triangle is isosceles. • The vectors of the two medians are $-2v+w$ and $-2w+v$. – user236182 Jan 7 '16 at 18:12", "that: (12) \\begin{align} \\quad 12250 = 2^1 \\cdot 5^3 \\cdot 7^2 \\end{align} Observe that $5 \\equiv 1 \\pmod 4$ and $7 \\equiv 3 \\pmod 4$ but $2$ is even. So $12250$ can be written as the sum of two squares.", "\\div 4 \\right) \\right] & \\text { Brack } - E x p o \\\\ & = 24 - [ 3 + 6 \\cdot ( 14 - 8 + 8 \\div 4 ) ] \\end{aligned} Then ,we have a subtraction, an addition and a division remained. Thus, we do the division: \\begin{aligned} = 24 - [ 3 + 6 \\cdot ( 14 - 8 + 8 \\div 4 ) ] & \\text { Brack } - D i v \\\\ & = 24 - [ 3 + 6 \\cdot ( 14 - 8 + 2 ) ] \\end{aligned} Now we do the subtraction and the addition inside the round brackets: \\begin{aligned} = 24 - [ 3 + 6 \\cdot ( 14 - 8 + 2 ) ] & \\text { Brack } - S u b \\\\ = & 24 - [ 3 + 6 \\cdot ( 6 + 2 ) ] \\quad \\text { Brack } - A d d \\\\ = & 24 - [ 3 + 6 \\cdot 8 ] \\end{aligned} Now that the round braces are gone, we continue with the square ones using the same procedure. Thus, $\\begin{array} { l } { = 24 - [ 3 + 6 \\cdot 8 ] } & { \\text { Brack } - M u l" ]

[ "# Creating arithmetic expression equal to 1000 using exactly eight 8's and parentheses I would like to find all the expressions that can be created using nothing but arithmetic operators, exactly eight $8$'s, and parentheses. Here are the seven solutions I've found (on the Internet) so far: \\begin{align} 1000 &= (8888 - 888) / 8\\\\ 1000 &= 888 + 88 + 8 + 8 + 8\\\\ 1000 &= 888 + 8 \\cdot (8 + 8) - 8 - 8\\\\ 1000 &= 8 \\cdot (8 \\cdot 8 + 8 \\cdot 8) - 8 - 8 - 8\\\\ 1000 &= 8 \\cdot (8 \\cdot (8 + 8) - (8 + 8) / 8) - 8\\\\ 1000 &= (8 \\cdot (8 + 8) - (8 + 8 + 8) / 8) \\cdot 8\\\\ 1000 &= (8 \\cdot (8 + 8) - 88 / 8 + 8) \\cdot 8 \\end{align} Are there others, and if there are, what are they? Update After sifting through achille hui's answer and adding one of mathlove's solutions, I get the following $16$ possibilities: \\begin{align} 1000 &= (8888 - 888)/8\\\\ 1000 &= 888 + (888 + 8)/8\\\\ 1000 &= 888 + 88 + 8 + 8 + 8\\\\ 1000 &= 888 + (8 + 8) \\cdot (8 - 8/8)\\\\ 1000 &= 888 - (8 + 8) \\cdot", "$6 \\boldcdot 4^2$, we want to make sure everyone agrees about how to evaluate this. Otherwise some people might multiply first and others compute the exponent first, and different people would get different values for the same expression! Earlier we saw situations in which $6 \\boldcdot 4^2$ represented the surface area of a cube with side lengths 4 units. When computing the surface area, we evaluate $4^2$ first (or find the area of one face of the cube first) and then multiply the result by 6. In many other expressions that use exponents, the part with an exponent is intended to be evaluated first. To make everyone agree about the value of expressions like $6 \\boldcdot 4^2$, the convention is to evaluate the part of the expression with the exponent first. Here are a couple of examples: \\begin {align} &\\;6 \\boldcdot 4^2\\\\ &= 6 \\boldcdot 16\\\\&= 96 \\end {align} \\begin {align} &\\;45 + 5^2\\\\ &= 45 + 25\\\\&= 70 \\end {align} If we want to communicate that 6 and 4 should be multiplied first and then squared, then we can use parentheses to group parts together: \\begin {align} &\\;(6 \\boldcdot 4)^2\\\\ &= 24^2\\\\&= 576 \\end {align}", "For the corrected version of the problem, where you are asked to compute $x^p$, you can use exponentiation by squaring directly using \\begin{align} x^0&=1\\\\ x^n&=(x^{n/2})^2 \\ \\text{for even }n\\\\ x^n&=x \\cdot (x^{(n-1)/2})^2 \\ \\text{for odd }n \\end{align} - But that is unlikely to be the intended answer, because it doesn't use the explicitly given fact that $p$ is an integer. – Henning Makholm Dec 14 '11 at 15:45", "of $3+6+12+\\cdots$ to $8$ terms. \\begin{align} \\displaystyle u_1 &= 3 \\\\ r &= 2 \\\\ n &= 8 \\\\ S_8 &= \\dfrac{3(2^8-1)}{2-1} \\\\ &= 765 \\end{align} ## Example 2 Find the sum of $0.2+0.02+0.002+\\cdots$ of the first $n$ terms. \\begin{align} \\displaystyle u_1 &= 0.2 \\\\ r &= 0.1 \\\\ n &= n \\\\ S_n &= \\dfrac{0.2(0.1^n-1)}{0.1-1} \\\\ &= \\dfrac{2(1-0.1^n)}{9} \\end{align}", "+ 6 + 6 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 Use the product rule of exponents to set up the problem. The product rule states: x^a × x^b = x^{a+b} So 6^{24} = 6^{(3 + 3 + 3 + 3 + 3 + 3 + 3 + 3)} = 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 Solve the problem. \\begin{aligned} 6^{24}&=6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 × 6^3 \\\\ &= 216 × 216 × 216 × 216 × 216 × 216 × 216 × 216 \\\\ &= 46656 × 46656 × 46656 × 46656 \\\\ &= 4.738 × 10^{18} \\end{aligned} #### Things You'll Need • Pen or pencil • Paper #### Tips • For some problems, a combination of both techniques may make the problem easier. For example: ​x21 = (​x7)3 (power rule), and ​x7 = ​x3 × ​x2 × ​x2 (product rule). Combining the two, you get: ​x21 = (​x3 × ​x2 × ​x2)3 Dont Go! We Have More Great Sciencing Articles!", "Mathematics OpenStudy (anonymous): So, let's start at the beginning. $$2(n + 1.5) = 4.5$$ You can distribute the 2: \\begin{align} 2n + 2 \\cdot (1.5) &= 4.5\\\\ 2n + 3 &= 4.5\\\\ \\end{align} Can you see how that does not lead to $$2n = 6$$, and therefore your final answer isn't right?", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "the same expansion to convert $123_8$ to hexadecimal: \\begin{align} 1 \\cdot 8^2 + 2 \\cdot 8^1 + 3 \\cdot 8^0 &= 1 \\cdot 40 + 2 \\cdot 8 + 3 \\cdot 1 \\\\ &= 40 + 10 + 3 \\\\ &= 53 \\end{align} $53$ is the hexadecimal form of $123_8$. This works because all the computations, e.g., $2 \\cdot 8 = 10$, are done in hexadecimal. The only reason you get the decimal form $83$ from this computation is because you do the computation in decimal. Writing this, another thing comes to mind, when we expand base-8, we are multiplying it with exponents of 8, which isn't even a digit in octal system. This is true! This is a good hint that the computation is actually being done in decimal, as I noted above. In octal, the $8$ would be replaced by $10$, and the computation would look like this: \\begin{align} 1 \\cdot 10^2 + 2 \\cdot 10^1 + 3 \\cdot 10^0 &= 1 \\cdot 100 + 2 \\cdot 10 + 3 \\cdot 1 \\\\ &= 100 + 20 + 3 \\\\ &= 123 \\end{align} This computation confirms the tautology that the octal $123$ has the form $123$ in octal. And then again, the exponents? What number system they belong to? All the writing systems describe the same", "$87000+10^2y_3+\\cdots$. Now we need to maximize $96y_3+87x_3$. The possibilities are $x_3 = 5, y_3 = 4$ or $x_3 = 4, y_3 = 5$. In these cases, $96y_3+87x_3$ has values 819 and 828 respectively. Thus we choose $y_3 = 5, x_3 = 4$. Now we have \\begin{align*} N_1 &= 96400 + 10^2x_4 + \\cdots \\\\ N_2 &= 87500 + 10^2y_4 + \\cdots \\end{align*} Now we need to maximize $964 y_1 + 875 x_4$. Here we see that $x_4 = 2$ and $y_4 = 3$. Finally, we have \\begin{align*} N_1 &= 96420 + x_5 \\\\ N_2 &= 87530 + y_5 \\end{align*} Now to maximize $9642 y_5 + 8753 x_5$, clearly $y_5 = 1$ and $x_5 = 0$. Thus the required numbers are $96420$ and $87531$ and the maximum product is $8439739020$. Let $A = abcde_{10}$ and $B = fghij_{10}$ be the numbers denoted in decimal basis such that the product $p = A \\times B$ is maximal. One has : $$A = 10^4a + 10^3b + 10^2c + 10d + e \\\\ B = 10^4f + 10^3g + 10^2h + 10i + j$$ By expanding the product $AB$ and factoring by powers of ten, we get $$p = a(10^8f + 10^7g + 10^6h + 10^5i + 10^4j) \\\\ + b(10^7f + 10^6g + 10^5h + 10^4i + 10^3j) \\\\ +", "that: (12) \\begin{align} \\quad 12250 = 2^1 \\cdot 5^3 \\cdot 7^2 \\end{align} Observe that $5 \\equiv 1 \\pmod 4$ and $7 \\equiv 3 \\pmod 4$ but $2$ is even. So $12250$ can be written as the sum of two squares.", "## Intermediate Algebra for College Students (7th Edition) $6x^{6}$ RECALL: The product rule of exponents states that: $a^m \\cdot a^n=a^{m+n}$. Use this rule to obtain: $3x^4 \\cdot 2x^2 \\\\=6x^{4+2} \\\\=6x^{6}$", "exponents, the part with an exponent is intended to be evaluated first. To make everyone agree about the value of expressions like $$6 \\boldcdot 4^2$$, the convention is to evaluate the part of the expression with the exponent first. Here are a couple of examples: \\begin {align} 6 &\\boldcdot 4^2 \\\\ 6 &\\boldcdot 16 \\\\ &96 \\end {align} \\begin {align} 45 &+ 5^2 \\\\ 45 &+ 25 \\\\ &70 \\end {align} If we want to communicate that 6 and 4 should be multiplied first and then squared, then we can use parentheses to group parts together: \\begin {align} (6 &\\boldcdot 4)^2 \\\\ &24^2 \\\\ &576 \\end {align} \\begin {align} (45 &+ 5)^2 \\\\ &50^2 \\\\ 2,&500 \\end {align} ### Glossary Entries • cubed We use the word cubed to mean “to the third power.” This is because a cube with side length $$s$$ has a volume of $$s \\boldcdot s \\boldcdot s$$, or $$s^3$$. • exponent In expressions like $$5^3$$ and $$8^2$$, the 3 and the 2 are called exponents. They tell you how many factors to multiply. For example, $$5^3$$ = $$5 \\boldcdot 5 \\boldcdot 5$$, and $$8^2 = 8 \\boldcdot 8$$. • squared We use the word squared to mean “to the second power.” This is because a square with side length $$s$$ has an area", "four times (once for each of $j=0$, $1$, $2$, and $3$), so you end up adding $8i$, not just $2i$. When $i=0$, the inner sum is $18$, as you compute. Then, when $i=1$, the inner sum is $(2+0) + (2+3) + (2+6) + (2+9) = 8+18 = 26$, not $2+18=20$, as you compute. Then when $i=2$, the inner sum is $(4+0)+(4+3) +(4+6) + (4+9) = 16+18 =34$ (not, as you compute, $22$). So the total value is the sum of these three quantities, $18+26+34 = 78$. To do it correctly symbolically, you can proceed as follows: the inner sum is: \\begin{align*} \\sum_{j=0}^3(2i+3j) &= \\sum_{j=0}^32i + \\sum_{j=0}^33j\\\\ &= 2i\\sum_{j=0}^31 + 3\\sum_{j=0}^3 j\\\\ &= 2i(4) + 3\\left(0+1+2+3\\right)\\\\ &= 8i + 18. \\end{align*} So then the outer sum is: \\begin{align*} \\sum_{i=0}^2\\left(\\sum_{j=0}^3(2i+3j)\\right) &= \\sum_{i=0}^2(8i+18)\\\\ &= \\sum_{i=0}^28i + \\sum_{i=0}^218\\\\ &= 8\\sum_{i=0}^2i + 18\\sum_{i=0}^21\\\\ &= 8(0+1+2) + 18(3)\\\\ &= 24 + 54\\\\ &= 78. \\end{align*} - That is 1000% better that the textbook. Thank you, that is exactly what I needed to get these problems done! – hobbes131 Jul 21 '12 at 5:25 Just grind away: $((0+0)+(0+3)+(0+6)+(0+9))+((2+0)+(2+3)+(2+6)+(2+9))+((4+0)+(4+3)+(4+6)+(4+9)) = 78$ -", "a formula $S := 1 + 2 + 3 + 4 + \\cdots + m$. I'll show a derivation here that is slightly different from the usual reverse and add method. Let's begin: \\begin{align*} S &= 1 + 2 + 3 + 4 + \\cdots + m \\\\\\ 2S &= 2 + 4 + 6 + 8 + \\cdots + 2m \\\\\\ 2S &= [1 + 3 + 5 + 7 + \\cdots + (2m-1)] + \\underbrace{1 + 1 + 1 + 1 + \\cdots + 1}_{m} \\\\\\ 2S &= [1 + 3 + 5 + 7 + \\cdots + (2m-1)] + m \\end{align*} Now, consider the sum in the brackets. Notice the pattern: \\begin{align*} 1 &= 1 \\\\\\ 4 &= 1 + 3 \\\\\\ 9 &= 1 + 3 + 5 \\\\\\ 16 &= 1 + 3 + 5 + 7 \\\\\\ 25 &= 1 + 3 + 5 + 7 + 9 \\\\\\ 36 &= 1 + 3 + 5 + 7 + 9 + 11 \\end{align*} These are just the perfect squares! Thus, we conjecture that $1 + 3 + 5 + \\cdots + (2m-1) = m^2$. In fact, this can easily be visualized as follows: m=1 m=2 m=3 m=4 m=5 1 1 2 1 2 3 1 2 3 4 1 2 3 4", "(1+\\cos{x})^{3} \\rm{dx} &= \\int 8 \\cdot \\cos^{6}\\Bigl(\\frac{x}{2}\\Bigr) \\ \\rm{dx} \\\\ &= 8 \\int \\cos^{6}\\Bigl(\\frac{x}{2}\\Bigr) \\ \\rm{dx} \\end{align*} -", "# Example Solve the following linear congruence: $$8x \\equiv 6 \\pmod{14}$$ First, we translate things back into an equation form. $8x \\equiv 6 \\pmod{14}$ implies $8x - 6 = 14n$ for some integer $n$. This in turn, suggests that we can write 6 as a linear combination of 8 and 14: $$8x - 14n = 6$$ By inspection, we note that the gcd of 8 and 14 is 2, so we verify that 6 is a multiple of 2, which it is. Given that these numbers are as small as they are, we could probably find one solution for $x$ and $n$ by inspection. That said, let us go through the process we would use if the numbers were much larger... First, we will try to find a single solution to the intermediate equation: $8x_1 - 14n_1 = 2$, using the Euclidean algorithm. \\begin{align} 14 &= 1 \\cdot 8 + 6\\\\ 8 &= 1 \\cdot 6 + \\fbox{2} \\leftarrow \\textrm{gcd}\\\\ 6 &= 3 \\cdot 2 + 0 \\end{align} Now writing the gcd, 2, as different linear combinations, we find: \\begin{align} 2 &= 8 - 6\\\\ &= 8 - (14-8)\\\\ &= 2 \\cdot 8 - 1 \\cdot 14\\\\ \\end{align} So $x_1=2$, $n_1=1$ is a solution to $8x_1 - 14n_1 = 2$ Now we can use this to find a", "can be reduced as follows: \\begin{align*} \\sqrt{16\\sqrt{8\\sqrt{4}}} &= \\sqrt{16\\sqrt{16}} \\\\ &= \\sqrt{64}\\\\ &= 8 \\end{align*} Thus, C is the correct answer. 4. When $$0.000315$$ is multiplied by $$7,928,564$$ the product is closest to which of the following? $$210$$ $$240$$ $$2100$$ $$2400$$ $$24000$$ ###### Solution(s): We can approximate the product as \\begin{align*} (3 \\cdot 10^{-4})(8 \\cdot 10^6) &= 24 \\cdot 10^2 \\\\ &= 2400. \\end{align*} Thus, D is the correct answer. 5. What is the value of the expression $\\dfrac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8}{1+2+3+4+5+6+7+8}?$ $$1020$$ $$1120$$ $$1220$$ $$2240$$ $$3360$$ ###### Solution(s): Looking at the denominator independently, $$1 + 2 + 3 + \\cdots + 8 = 36$$, so the desired answer is $\\begin{gather*} \\dfrac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8}{36} = \\\\ 4 \\cdot 5 \\cdot 7 \\cdot 8 = 1120. \\end{gather*}$ Thus, B is the correct answer. 6. If the degree measures of the angles of a triangle are in the ratio $$3:3:4$$, what is the degree measure of the largest angle of the triangle? $$18$$ $$36$$ $$60$$ $$72$$ $$90$$ ###### Solution(s): We can let the three angles be equal to $$3x, 3x,$$ and $$4x.$$ Then we know that their sum equals $$180.$$ From this we can set $$3x +", "$n \\ge 3$. To compute the requested value $f(6)$, we repeatedly apply this formula: \\begin{align*} f(6)&=4\\cdot3^5-f(5)\\\\&=4\\cdot3^5-4\\cdot3^4+f(4)\\\\&=4\\cdot3^5-4\\cdot3^4+4\\cdot3^3-f(3)\\\\&=4\\cdot3^5-4\\cdot3^4+4\\cdot3^3-4\\cdot3^2+f(2)\\\\&=4(3^5-3^4+3^3-3^2+3)\\\\&=4\\cdot3\\cdot\\frac{3^5+1}{3+1}\\\\&=732.\\end{align*} (Solution by MSTang.)", "## Precalculus (6th Edition) Blitzer The value of $\\sum\\limits_{i=1}^{50}{\\left( 4i-25 \\right)}$ is $3850$. We have to find the value of the given summation from $i=1$ to $i=50$. \\begin{align} & \\sum\\limits_{i=1}^{50}{\\left( 4i-25 \\right)}=\\left[ \\left( 4\\cdot 1-25 \\right)+\\left( 4\\cdot 2-25 \\right)+\\left( 4\\cdot 3-25 \\right)+\\ldots +\\left( 4\\cdot 50-25 \\right) \\right] \\\\ & =\\left[ 4\\cdot \\left( 1+2+3+\\ldots +50 \\right)-25-25-\\ldots -25 \\right] \\end{align} We have taken 4 as a common number from the first digit of every term and obtained an arithmetic sequence. We expand this in the form: \\begin{align} & \\left[ \\frac{4\\left( n \\right)\\left( n+1 \\right)}{2}-25n \\right]=\\left[ \\frac{4\\left( 50 \\right)\\left( 51 \\right)}{2}-25\\left( 50 \\right) \\right] \\\\ & =\\left[ 5100-1250 \\right] \\\\ & =3850 \\end{align} Hence, the value of the sum $\\sum\\limits_{i=1}^{50}{\\left( 4i-25 \\right)}$ is $3850$.", "into eqn. } (1) \\quad a &= -7 -2 (4) \\\\ \\therefore a &= -15 \\\\ S_{n} &= \\frac{n}{2}[2a + (n-1)d] \\\\ S_{51} &= \\frac{51}{2}[2(-15) + (51-1)(4)] \\\\ &= \\frac{51}{2}[-30 + 200] \\\\ &= (51)(85) \\\\ \\therefore S_{51}&= 4335 \\end{align*} Calculate the sum of the arithmetic series $$4+7+10+\\cdots +901$$. \\begin{align*} a &= 4 \\\\ l &= 901 \\\\ d &= T_{2} - T_{1} \\\\ &= 7 - 4 \\\\ &= 3 \\\\ \\text{And } T_{n} &= a +(n-1)d \\\\ &= 4 +(n-1)(3) \\\\ \\therefore 901 &= 4 + 3n - 3 \\\\ 900 &= 3n \\\\ \\therefore 300 &= n \\\\ S_{n} &= \\frac{n}{2}[a + l] \\\\ S_{300} &= \\frac{300}{2}[4 + 901 ] \\\\ &= (150)(905) \\\\ \\therefore S_{300} &= 135750 \\end{align*} Evaluate without using a calculator: $$\\dfrac{4 + 8 + 12 + \\cdots + 100}{3 + 10 + 17 + \\cdots + 101}$$ Consider the numerator: $$4 + 8 + 12 + \\ldots + 100$$ \\begin{align*} a &= 4 \\\\ l &= 100 \\\\ d &= T_{2} - T_{1} \\\\ &= 8 - 4 \\\\ &= 4 \\\\ \\text{And } T_{n} &= a +(n-1)d \\\\ 100 &= 4 +(n-1)(4) \\\\ 100 &= 4n \\\\ \\therefore 25 &= n \\\\ S_{n} &= \\frac{n}{2}[a + l] \\\\ S_{25} &= \\frac{25}{2}[4 + 100 ] \\\\ S_{25} &= (25)(52) \\end{align*} Consider" ]

[ "well choose them in order $1, 2, 3, \\ldots$.", "Five numbers are added together in pairs to produce the following answers: $0, 2, 4, 4, 6, 8, 9, 11, 13, 15$ What are the five numbers?", "the list becomes $$1,2,3,5,7,11$$, which square to $$1,4,9,25,49,121$$", "for the sixth item only. I assume $i \\in \\{1,2,\\ldots,100\\}$. If you meant that $i \\in \\{1,2,\\ldots,6\\}$, how will I know which of the seven out of 100 items are chosen? – ubaabd Dec 17 '14 at 5:23", "Which number does not belong in the series below? $2$, $5$, $10$, $17$, $26$, $37$, $50$, $64$ 1. $17$ 2. $37$ 3. $64$ 4. $26$ recategorized", "and $$50$$ are $$1$$ and $$5$$.", "\\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 &", "# What are the numbers that come next in these sequences: 1,5,2,10,3,15,4? Mar 24, 2016 If you look at the odd numbers they go like $1 , 2 , 3 , 4. . .$ The even numbers add $5$ at every step like $5 , 10 , 15. . .$ So the next odd numbers would be $\\ldots 20 , 25 , 30. . .$ And the next even numbers would be $\\ldots 5 , 6 , 7. . .$ $\\ldots 20 , 5 , 25 , 6 , 30 , 7. . .$", "$9$ pairs of numbers are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5)$.", "$<4,7,7>[3],<4,6,6>[3],<4,6,7>[6],<4,5,7>[6]$ eighteen ways. There are twenty-seven, $3^3$ ordered triples containing $5's,~6's,\\text{ or }7's$ only one of which, $<5,5,5>$, does not meet the condition. So $56$ ways to have the three add to more than $15$.", "of $5$ in $20$, how many groups of $5$ do you think there would be in $−20$? $-20\\div5=-4$ 1. $- 36 \\div ( - 9 )$ How many groups of $-9$ are there in $-36$?", "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", "The quotient of a number and 5 is 33? Feb 20, 2018 The number $= 165$. Explanation: Let $x$ be the numerator and $5$ be the denominator. $\\frac{x}{5} = 33$ We need to bring 5 from the left side to the right side, so $\\frac{x}{5} \\cdot 5 = 33 \\cdot 5$ $x = 33 \\cdot 5$ $x = 165$", "are $3,5,-3,-5$", "5} 1 10 1 10 o1 : List", "# What are the numbers that come next in these sequences: 1,5,2,10,3,15,4? Mar 24, 2016 If you look at the odd numbers they go like $1 , 2 , 3 , 4. . .$ The even numbers add $5$ at every step like $5 , 10 , 15. . .$ #### Explanation: So the next odd numbers would be $\\ldots 20 , 25 , 30. . .$ And the next even numbers would be $\\ldots 5 , 6 , 7. . .$ The sequence would continue like this: $\\ldots 20 , 5 , 25 , 6 , 30 , 7. . .$", "20$$ $$47 = 9 + 9 + 9 + 20$$ $$48 = 6 + 6 + 9 + 9 + 9 + 9$$ $$49 = 9 + 20 + 20$$ You can continue the sequence by adding 6 to the above list so again we see that 43 is the largest number we can't make this way. Status Not open for further replies.", "six digits $D_1, D_2, \\ldots, D_6$. In case two, the order in which we are choosing them is $D_6, D_1, D_2, \\ldots, D_5$. • For $D_6$, there are four choices: 2, 4, 6, 8. • For $D_1$, there are eight choices: all nine of 1-9 (can't start with zero!), except $D_6$. • For $D_2$, there are eight choices: all ten of 0-9, except $D_6$ and $D_1$. • For $D_3$, there are seven choices: all ten of 0-9, except $D_6$ and $D_1$ and $D_2$. • For $D_4$, there are six choices: all ten of 0-9, except $D_6,D_1,D_2,D_3$. • For $D_5$, there are five choices: all ten of 0-9, except $D_6,D_1,D_2,D_3,D_4$.", "6. Hello again, faruk_fin! I'll take a guess at what you are asking . . . We have a list of the 15,380,937 sets of seven numbers from the set 1-to-39. . . $\\begin{array}{c} \\{1,2,3,4,5,6,7\\} \\\\ \\{1,2,3,4,5,6,8\\} \\\\ \\{1,2,3,4,5,6,9\\} \\\\ \\vdots \\\\ \\{33,34,35,36,37,38,39\\} \\end{array}$ How many times does a particular number (say, 23) appear on the list? I would conjecture that any number appears $\\tfrac{1}{39}$ of the time. Therefore, \"23\" appears $\\frac{15,380,937}{39} \\:=\\:394,383$ times.", "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?" ]

[ "-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.", "list (containing 37 terms) use the above formula for $n=9,10,\\ldots,45$. – Shai Covo Apr 13 '11 at 15:48 If the list stops at $100,000$, you have represented it by listing it. You could say something like $k10^n$ where $k \\in \\{1,2,3,4,5,6,7,8,9\\}$ and $1 \\le n \\le 4$, but that is pretty complicated and leaves off the last term.", "by as little as we please. In fact, it we carry out the ordinary arithmetical process for the extraction of the square root of 2, we obtain a series of rational numbers, viz., 1,1.4, 1.41, 1.414, 1.4142, $\\ldots$ whose squares 1, 1.96, 1.9881, 1.999396, 1.99996164, $\\ldots$ are all less than 2, but approach nearer and nearer to it, and by taking a sufficient number of the figures given by the process we can obtain as close an approximation as we want. And if we increase the last figure, in each of the approximations given above, by unity, we obtain a series of rational numbers 2, 1.5, 1.42, 1.415,1.413, $\\ldots$ whose squares 4, 2.25, 2.0164, 2.002225, 2.00024449, $\\ldots$ are all greater than 2, but approximate to 2 as closely as we please. It follows also that there can be no largest member of L or smallest member of R. For if x is any member of L, then $x^{2} < 2$. Suppose that $x^{2}=2-\\delta$. Then we can find a member x, of L such that ${x_{1}}^{2}$ differs from 2 by less than $\\delta$, and ${x_{1}}^{2}>x^{2}$ or $x_{1}>x$. Thus there are larger members of L than x; and, as x is any member of L, it follows that no member of L can be larger than all the rest. Hence,", "and $3$ and $5$ never near, which leaves the total combinations (orders) of the boxes to twelve: $1-2-3-4-5$ $1-3-2-4-5$ $1-3-4-2-5$ $1-3-2-5-4$ $1-3-4-5-2$ $1-4-3-2-5$ Plus the symmetry of the exchange $5\\to 3$ So I have $12$ possible dispositions of the boxes. Now, let's call the figures on each box as $$1 = \\{A_1, A_2, A_3, A_4, A_5, A_6\\}$$ $$2 = \\{B_1, \\ldots, B_6\\}$$ $$3 = \\{C_1 \\ldots C_6\\}$$ $$4 = \\{D_1, D_2, D_3, A_1, C_1, C_2\\}$$ $$5 = \\{E_1 \\ldots E_6\\}$$", "you divide by $$6$$, so $$[2]_6 = \\{ \\ldots , -10, -4, 2, 8, 14, \\ldots\\}.$$ Can you finish now? Write down $$[3]_6$$ and check whether it and $${2}_6$$ have any numbers in common. • There was an error in my question. It should have been [3]_9 – darylnak Dec 2 '18 at 1:56 • Well write that one down and see if it overlaps $[2]_6$. – Ethan Bolker Dec 2 '18 at 2:19", "remarks to an answer: suppose some committee $c$ has at least $4$ different members $x_1,x_2,x_3,x_4$ from $\\{1,\\ldots,7\\}$. Then $x_i$ is also in two other committees, say $c^1_i, c^2_i$, where $c^1_i \\neq c^2_i$ and $c^1_i \\neq c, c^2_i \\neq c$, for $i=1,\\ldots,4$. The set $\\{c\\} \\cup \\{c^j_i: i=1,\\ldots,4; j=1,2\\}$ has at most 7 members so the second set has coinciding members: there are $j,i,j',i'$ such that $c^j_i = c^{j'}_{i'} = c'$. We know that $i \\neq i'$ and so we have that both $c$ and $c'$ have the members $x_i$ and $x_{i'}$ in common. This contradicts the assumptions.", "would be $(11,16), (12,17), (13,18), (14, 19), (15, 20)$ and then we move to 21-30 range.... Shreev explained it much better, I tried to wrote it as elementary as possible.... – N. S. May 24 '11 at 15:55 I see now. You're throwing away some possible configurations and STILL finding a pair of houses of the desired type. I now prefer your answer to mine. :) – Austin Mohr May 24 '11 at 19:04 Hint: divide the houses into groups $(1,6,11,\\ldots,46), (2,7,12,\\ldots,47),$ etc. Added after the fact: Even easier: consider the pairs $(k,k+5)$ for $1 \\le k \\pmod {10} \\le 5$. There are 25 of them. Only one application of the principle. - Yes, I've done that... but I'm not too sure how to apply it. – meiryo May 23 '11 at 23:52 How many students live in houses in one group? And how many houses are in the group? – Ross Millikan May 23 '11 at 23:52 You can use the possible remainders after division as your holes. If we number the houses from $1$ to $50$, we can see that $10$ of them have a remainder of $0$ after dividing by $5$, $10$ have a remainder of $1$, and so on. Within each group of ten, I can pick five houses such that none of them", "all of the pairs of numbers that add up to $35$ That leaves $1,2,3,4,5,6,7,8,9,10,11,12,13,14$ which need to be split up into three groups that add up to $35$. So let's start by finding three of the remaining numbers that add up to $35$. • $(14, 13, 8)$ That leaves $1,2,3,4,5,6,7,9,10,11,12$ which need to be split up into two more groups that add up to $35$. So let's look for three more of the remaining numbers that add up to $35$. Darn, there aren't any. So let's look for four of the remaining numbers that add up to $35$. • $(12, 11, 10, 2)$ will add up to $35$ So we now have five groups of numbers that add up to $35$ and we have these numbers left • $(1,3,4,5,6,7,9)$ Guess what. The remaining seven number HAVE TO add up to $35$. So we are done. the complete list is • $(20, 15)$ • $(19, 16)$ • $(18, 17)$ • $(14, 13, 8)$ • $(12, 11, 10, 2)$ • $(1,3,4,5,6,7,9)$", "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$).", "then get a sum of $$2\\left(1 + 2 + \\ldots + 19 + 20\\right) = 2\\frac{\\left(20\\right)\\left(21\\right)}{2} = 420 \\tag{1}\\label{eq1}$$ As such, this is $$420 - 240 = 180$$ too high. Now, for each term which used the $$n/2$$ value instead, it would reduce the sum by $$3n/2$$. As such, if the sum of these $$n$$ values is $$S_1$$, we have that that the total reduction is $$3S_1/2 = 180$$ giving that $$S_1 = 120$$. As mentioned, we want the number of these terms to be the minimum, so we start using the largest ones. Using the $$7$$ values $$14, 15, \\ldots 20$$ gives a sum of $$119$$, meaning just need to use $$1$$ as well to get the sum to $$120$$, for a total of $$8$$ terms, as conjectured. Note you cannot use $$7$$ or less since, as just shown, the $$7$$ largest values add to $$119$$, so any other $$7$$ or fewer terms will add up to less than $$119$$, so $$8$$ is the minimum # of terms required. One small thing to note is that which $$8$$ terms are to use $$n/2$$ is not uniquely defined. For example, the $$1$$ and $$14$$ terms can be replaced by any other $$2$$ which add to $$15$$, such as $$2$$ and $$13$$, $$3$$ and $$12$$, $$\\ldots$$, $$7$$ and $$8$$.", "The factors of each number can be found in in the factored LCM. Method 2. Make a list of a few multiples of each of the numbers. Make the lists long enough that you can find one of the numbers in each of the lists. If the list is long, there may be several such numbers, but the idea is to choose the smallest one that appears in each list. Multiples of 3: $3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , \\ldots$ Multiples of 4: $4 , 8 , 12 , 16 , 20 , 24 , \\ldots$ Multiples of 6: $6 , 12 , 18 , 24 , \\ldots$ Now, find the smallest ( Least ) multiple that is Common in each list of Multiples, It is easy to pick-out the number 12 from the lists as the smallest, even though there is another number, 24, which is common to each list. • 25 minutes ago • 26 minutes ago • 27 minutes ago • 31 minutes ago • 8 seconds ago • 11 minutes ago • 12 minutes ago • 13 minutes ago • 14 minutes ago • 14 minutes ago • 25 minutes ago • 26 minutes ago • 27 minutes ago • 31 minutes ago", "\\mod 97: k = 0, 1, \\ldots, 23\\}$ such as these given below: $$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 &", "that we can find a member of the class L whose square, though less than 2, differs from 2 by as little as possible, and a member of R whose square, though greater than 2, also differs from 2 by as little as we please. In fact, it we carry out the ordinary arithmetical process for the extraction of the square root of 2, we obtain a series of rational numbers, viz., 1,1.4, 1.41, 1.414, 1.4142, $\\ldots$ whose squares 1, 1.96, 1.9881, 1.999396, 1.99996164, $\\ldots$ are all less than 2, but approach nearer and nearer to it, and by taking a sufficient number of the figures given by the process we can obtain as close an approximation as we want. And if we increase the last figure, in each of the approximations given above, by unity, we obtain a series of rational numbers 2, 1.5, 1.42, 1.415,1.413, $\\ldots$ whose squares 4, 2.25, 2.0164, 2.002225, 2.00024449, $\\ldots$ are all greater than 2, but approximate to 2 as closely as we please. It follows also that there can be no largest member of L or smallest member of R. For if x is any member of L, then $x^{2} < 2$. Suppose that $x^{2}=2-\\delta$. Then we can find a member x, of L such that ${x_{1}}^{2}$ differs from 2 by", "the first $17$ terms are $$1,3,4,6,10,11,12,15,16,19,20,23,24,25,26,27,28,\\ldots.$$ Here, too, the question is: what are the next terms? Added later: Thanks, Yaakov, you are right, so my question is, what are the positions of the numbers in $T_2$ in the sequence in the sequence $(1,2,3,\\ldots)$. I have the first $30$ positions (or ranks) and would like to see a method for finding more terms. It may help to see a list of the first $20$ towers ranked: $$4 = 2^2$$ $$8 = 2^3$$ $$9 = 3^2$$ $$16 = 2^{2^{2}}$$ $$27 = 3^3$$ $$81 = 3^{2^{2}}$$ $$256 = 2^{2^{3}}$$ $$512 = 2^{3^{2}}$$ $$6561 = 3^{2^{3}}$$ $$19683 = 3^{3^{2}}$$ Continuing with tuple notation instead of tower notation: $(2,2,2,2), (3,2,2,2), (2,3,3), (3,3,3), (2,3,2,2), (3,3,2,2), (2,2,2,3), (3,2,2,3), (2,2,3,2), (3,2,3,2), (2,3,2,3).$ My method, so far, has been by computer sort, which reaches overflow pretty quickly. Surely there must be a more insightful method. A related question: what is the position (or rank) of $(2,2,2,2,2,2)?$ • What does jointly ranked mean? – David Handelman Jan 29 '18 at 14:24 • Jointly ranked means arranged in increasing order: $t_2 = \\{4,8,16,256,\\ldots\\}$ and $t_3 = \\{9,27,81,\\ldots\\},$ so that the joint ranking is $(4,8,9,16,27,81,256,\\ldots).$ – Clark Kimberling Jan 29 '18 at 15:12 • $8=2^2$? $9=2^3$?? – Gerry Myerson Jan 29 '18 at 22:07 • I find a certain rough", "equally equal to the set $$\\{2,3,4\\}$$ and $$\\{3\\}\\subset \\{2,3,4\\}$$. No it is not, we have that $$\\{1\\}\\in \\{\\{1\\},\\{1,2\\},\\{1,2,3\\}\\}$$ but $$\\{3\\}$$ is not an element of the set, what is true is that $$3\\in\\{1,2,3\\}$$. As you can see, there are plenty of good answers here. I think that the most important thing to understand is: $$3 \\neq \\{ 3 \\}.$$ These are two different objects. Moreover the followings hold: $$3 \\in \\{3\\}$$ $$3 \\in \\{\\ldots, 3, \\ldots \\}$$ $$\\{3\\} \\in \\{\\ldots, \\{3\\}, \\ldots\\}$$ and $$\\{3\\} \\not\\in \\{3\\}$$ $$\\{3\\} \\not\\in \\{\\ldots, 3, \\ldots \\}$$ • In for a penny..... might as well add (and perhaps germaine to the common misconception) $3\\not \\in \\{........., \\{3\\},......\\}$ (unless it is one of the objects omitted in the \"...\"s.) Sep 24 '20 at 21:59", "number of $1 \\times 2$ all-$1$ submatrices. If this number is more than $\\binom{13}{2}=78$, we have a pair of columns containing two $1 \\times 2$ all-$1$ submatrices, i.e., a rectangle. Since we have $13$ rows, for any $k \\in \\{0,1,2,\\ldots,13\\}$, we can have $x_i=k$ at most $13$ times. If we sort the $x_i$'s in non-decreasing order, we get the termwise lower bound $$(x_i)_{i=1}^{53} \\geq (\\overbrace{0,\\ldots,0}^{13 \\text{ times}},\\overbrace{1,\\ldots,1}^{13 \\text{ times}},\\overbrace{2,\\ldots,2}^{13 \\text{ times}},\\overbrace{3,\\ldots,3}^{13 \\text{ times}},4).$$ So $\\sum_{i=1}^{53} x_i \\geq 82>78.$ So there's a rectangle. Here's an example to show that $52$ ones is okay (derived from a (13,4,1)-BIBD): \\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\\\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0", "to do is add up to 66 and we'll be able to find the number of people: $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66 \\implies$ $\\implies n - 1 = 11 \\implies n = 12$", "of R whose square, though greater than 2, also differs from 2 by as little as we please. In fact, it we carry out the ordinary arithmetical process for the extraction of the square root of 2, we obtain a series of rational numbers, viz., 1,1.4, 1.41, 1.414, 1.4142, $\\ldots$ whose squares 1, 1.96, 1.9881, 1.999396, 1.99996164, $\\ldots$ are all less than 2, but approach nearer and nearer to it, and by taking a sufficient number of the figures given by the process we can obtain as close an approximation as we want. And if we increase the last figure, in each of the approximations given above, by unity, we obtain a series of rational numbers 2, 1.5, 1.42, 1.415,1.413, $\\ldots$ whose squares 4, 2.25, 2.0164, 2.002225, 2.00024449, $\\ldots$ are all greater than 2, but approximate to 2 as closely as we please. It follows also that there can be no largest member of L or smallest member of R. For if x is any member of L, then $x^{2} < 2$. Suppose that $x^{2}=2-\\delta$. Then we can find a member x, of L such that ${x_{1}}^{2}$ differs from 2 by less than $\\delta$, and ${x_{1}}^{2}>x^{2}$ or $x_{1}>x$. Thus there are larger members of L than x; and, as x is any member of L, it follows that", "$2$'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", "R. Moreover, it is easy to convince ourselves that we can find a member of the class L whose square, though less than 2, differs from 2 by as little as possible, and a member of R whose square, though greater than 2, also differs from 2 by as little as we please. In fact, it we carry out the ordinary arithmetical process for the extraction of the square root of 2, we obtain a series of rational numbers, viz., 1,1.4, 1.41, 1.414, 1.4142, $\\ldots$ whose squares 1, 1.96, 1.9881, 1.999396, 1.99996164, $\\ldots$ are all less than 2, but approach nearer and nearer to it, and by taking a sufficient number of the figures given by the process we can obtain as close an approximation as we want. And if we increase the last figure, in each of the approximations given above, by unity, we obtain a series of rational numbers 2, 1.5, 1.42, 1.415,1.413, $\\ldots$ whose squares 4, 2.25, 2.0164, 2.002225, 2.00024449, $\\ldots$ are all greater than 2, but approximate to 2 as closely as we please. It follows also that there can be no largest member of L or smallest member of R. For if x is any member of L, then $x^{2} < 2$. Suppose that $x^{2}=2-\\delta$. Then we can find a member x, of" ]

[ "-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}$$ Adding up the numbers below the line, we end up with $-19$, which modulo $97$ equals $78$, as expected. An advantage of this method is that, once you have those $24$ numbers, the computations are really simple and require only really small numbers. You can even do this method by hand, as I actually did for the example above.", "list (containing 37 terms) use the above formula for $n=9,10,\\ldots,45$. – Shai Covo Apr 13 '11 at 15:48 If the list stops at $100,000$, you have represented it by listing it. You could say something like $k10^n$ where $k \\in \\{1,2,3,4,5,6,7,8,9\\}$ and $1 \\le n \\le 4$, but that is pretty complicated and leaves off the last term.", "\\mod 97: k = 0, 1, \\ldots, 23\\}$ such as these given below: $$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 &", "and $3$ and $5$ never near, which leaves the total combinations (orders) of the boxes to twelve: $1-2-3-4-5$ $1-3-2-4-5$ $1-3-4-2-5$ $1-3-2-5-4$ $1-3-4-5-2$ $1-4-3-2-5$ Plus the symmetry of the exchange $5\\to 3$ So I have $12$ possible dispositions of the boxes. Now, let's call the figures on each box as $$1 = \\{A_1, A_2, A_3, A_4, A_5, A_6\\}$$ $$2 = \\{B_1, \\ldots, B_6\\}$$ $$3 = \\{C_1 \\ldots C_6\\}$$ $$4 = \\{D_1, D_2, D_3, A_1, C_1, C_2\\}$$ $$5 = \\{E_1 \\ldots E_6\\}$$", "six digits $D_1, D_2, \\ldots, D_6$. In case two, the order in which we are choosing them is $D_6, D_1, D_2, \\ldots, D_5$. • For $D_6$, there are four choices: 2, 4, 6, 8. • For $D_1$, there are eight choices: all nine of 1-9 (can't start with zero!), except $D_6$. • For $D_2$, there are eight choices: all ten of 0-9, except $D_6$ and $D_1$. • For $D_3$, there are seven choices: all ten of 0-9, except $D_6$ and $D_1$ and $D_2$. • For $D_4$, there are six choices: all ten of 0-9, except $D_6,D_1,D_2,D_3$. • For $D_5$, there are five choices: all ten of 0-9, except $D_6,D_1,D_2,D_3,D_4$.", "+0 # Can I have help please? 0 369 1 How many numbers can you get by multiplying two or more distinct members of the set $$\\{1,2,3,5,11\\}$$ together? May 10, 2021 #1 +1 Use casework. First try finding the possiblities with only multiplying 2 numbers, and then the possibilities with multiplying 3 numbers, etc. Using 2 numbers: $$2\\cdot3, 2 \\cdot 5, 2 \\cdot 11, 3 \\cdot 5, 3 \\cdot 11, 5 \\cdot 11.$$ Using 3 numbers: $$2 \\cdot 3 \\cdot 5, 2 \\cdot 3 \\cdot 11, 2 \\cdot 5 \\cdot 11, 3 \\cdot 5 \\cdot 11.$$ Using 4 numbers: $$2 \\cdot 3 \\cdot 5 \\cdot 11.$$ So we have a total of $$6+4+4+1=\\boxed{15}$$ May 10, 2021", "all of the pairs of numbers that add up to $35$ That leaves $1,2,3,4,5,6,7,8,9,10,11,12,13,14$ which need to be split up into three groups that add up to $35$. So let's start by finding three of the remaining numbers that add up to $35$. • $(14, 13, 8)$ That leaves $1,2,3,4,5,6,7,9,10,11,12$ which need to be split up into two more groups that add up to $35$. So let's look for three more of the remaining numbers that add up to $35$. Darn, there aren't any. So let's look for four of the remaining numbers that add up to $35$. • $(12, 11, 10, 2)$ will add up to $35$ So we now have five groups of numbers that add up to $35$ and we have these numbers left • $(1,3,4,5,6,7,9)$ Guess what. The remaining seven number HAVE TO add up to $35$. So we are done. the complete list is • $(20, 15)$ • $(19, 16)$ • $(18, 17)$ • $(14, 13, 8)$ • $(12, 11, 10, 2)$ • $(1,3,4,5,6,7,9)$", "# What are the numbers that come next in these sequences: 1,5,2,10,3,15,4? Mar 24, 2016 If you look at the odd numbers they go like $1 , 2 , 3 , 4. . .$ The even numbers add $5$ at every step like $5 , 10 , 15. . .$ So the next odd numbers would be $\\ldots 20 , 25 , 30. . .$ And the next even numbers would be $\\ldots 5 , 6 , 7. . .$ $\\ldots 20 , 5 , 25 , 6 , 30 , 7. . .$", "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 $[0]_7 \\cap \\left\\{1, \\ldots, 300 \\right\\}.$ $$300-3 \\times 43-42+1=130$$", "Five numbers are added together in pairs to produce the following answers: $0, 2, 4, 4, 6, 8, 9, 11, 13, 15$ What are the five numbers?", "for the sixth item only. I assume $i \\in \\{1,2,\\ldots,100\\}$. If you meant that $i \\in \\{1,2,\\ldots,6\\}$, how will I know which of the seven out of 100 items are chosen? – ubaabd Dec 17 '14 at 5:23", "# What are the numbers that come next in these sequences: 1,5,2,10,3,15,4? Mar 24, 2016 If you look at the odd numbers they go like $1 , 2 , 3 , 4. . .$ The even numbers add $5$ at every step like $5 , 10 , 15. . .$ #### Explanation: So the next odd numbers would be $\\ldots 20 , 25 , 30. . .$ And the next even numbers would be $\\ldots 5 , 6 , 7. . .$ The sequence would continue like this: $\\ldots 20 , 5 , 25 , 6 , 30 , 7. . .$", "$$(1,3,5,3,5,3,5,\\ldots)$$ $$(1,3,5,5)$$ $$(1,3,5,4,7,12,11,18,30,29,\\ldots)$$ $$(1,3,5,6)$$ $$(1,3,5,5,5,5,\\ldots)$$ $$(1,3,5,7)$$ $$(1,3,5,6,9,14,20,29,43,63,\\ldots)$$ $$(1,3,6)$$ $$(1,3,5,7,9,11,13,15,17,\\ldots)$$ $$(1,3,6,8)$$ $$(1,3,6,7,10,16,23,33,49,71,104,153,\\ldots)$$ $$(1,3,7)$$ $$(1,3,6,12,24,48,96,192,384,\\ldots)$$ $$(1,3,7,6)$$ $$(1,3,7,5,12,36,\\ldots)$$ $$(1,3,7,11)$$ $$(1,3,7,10,16,29,52,91,159,279,\\ldots)$$ $$(1,3,7,12)$$ $$(1,3,7,11,19,35,67,\\ldots)$$ $$(1,3,7,13)$$ $$(1,3,7,12,20,35,62,109,191,\\ldots)$$ $$(1,3,7,14)$$ $$(1,3,7,13,21,31,43,\\ldots)$$ It is a standard expression, but the computation of its expansion uses the expansion of the non-standard expression $$(1,2,2,3)$$. $$(1,3,8)$$ $$(1,3,7,15,31,63,127,255,\\ldots)$$ $$(1,3,9)$$ $$(1,3,8,20,48,112,256,\\ldots)$$ $$(1,3,9,27,81,243,633,1323,2121,2256)$$ $$(1,3,9,27,81,243,633,1323,2121,2255,2545,3135,4345,6899,12225,18171,19353,21723,26481,36051,55425,94935,139257,148799,\\ldots)$$ It is a sequence in Koteitan's test. $$(1,4)$$ $$(1,3,9,27,81,\\ldots)$$ $$(1,4,1)$$ $$(1,4)$$ $$(1,4,2)$$ $$(1,4,1,4,\\ldots)$$ $$(1,4,3)$$ $$(1,4,2,6,4,10,8,18,16,\\ldots)$$ $$(1,4,3,3)$$ $$(1,4,3,2,6,5,4,\\ldots)$$ $$(1,4,4)$$ $$(1,4,3,10,9,28,27,82,81,244,243,\\ldots)$$ $$(1,4,13,12,15,12)$$ $$(1,4,13,12,15,11,29,78,77,162,76,\\ldots)$$ The corresponding Y-sequence was an unsolved problem in Y-sequence system, but solved in 29/06/2020. $$(1,5,3,4)$$ $$(1,5,3,3,3,\\ldots)$$ $$(1,5,4,10,10)$$ $$(1,5,4,10,9,22,21,50,49,\\ldots)$$ $$(1,6,14,29,37,29)$$ $$(1,6,14,29,37,28,55,63,54,105,113,104,203,211,202,\\ldots)$$ The corresponding Y-sequence was an unsolved problem in Y-sequence system, but solved in 29/06/2020. $$(1,10)$$ $$(1,9,81,729,\\ldots)$$ $$(1,\\textrm{Rayo}(10^{100})+1)$$ $$(1,\\textrm{Rayo}(10^{100}),\\textrm{Rayo}(10^{100})^2,\\ldots)$$ The corresponding Y-sequence is believed to be greater than Rayo's number. Community content is available under CC-BY-SA unless otherwise noted.", "can choose any 6 of the remaining 24 for group 2. So we have $C(15,4)C(26,0)C(26,4) + C(15,4)C(26,1)C(25,5) + \\ldots + C(15,4)C(26,11)C(15,15)$ Quote: 6) Divide the group into two groups each having size at least 1?(C(30,1)?) Now we don't have a size requirement on the groups. So you can have a group of $(1,1), (1,2), \\ldots (1,29), (2,2), (2,3), \\ldots (2,28), \\ldots (15,15)$ So that gives us $(C(30,2)C(2,1) + C(30,3)C(3,2) + \\ldots+ C(30,30)C(30,28)$ $+ \\ldots + C(30,30)C(30,15))/2$ I think these are right. But these kinds of problems are always really, really tricky so I could be wrong. • Feb 12th 2010, 07:11 AM Plato With respect to #2, the number of ways to group twenty people into ten pairs is $\\frac{20!}{2^{10}(10!)}$. That is known as unordered partitions. That number is much greater than $C(20,2)$. • Feb 13th 2010, 01:46 AM chocaholic Thanx thanx a million for the input and help guys.It makes a lot more sense now. • Feb 23rd 2010, 10:32 AM iva Hi there, I got different answers for 1 and 2: For 1, I got 15!15! as for each and every man there is 15 possible female partners ie 15 (15). 14(14). 13 (13).... 1 (1) = 15!15! , i think its logic follows from the problem here: http://www.mathhelpforum.com/math-he...ake-sense.html For 2 ( i haven't covered unordered", "The number of ordered triples for various values of $a$ are presented in the following table. $$\\begin{tabular}{|c|c|c|r|} \\hline a & b & c & triples \\\\ \\hline -8 & \\{9,10\\} & -6a - 2b - 20 & 2 \\\\ -7 & \\{6, 7, 8, 9, 10\\} & -6a - 2b - 20 & 5 \\\\ -6 & \\{-10, \\ldots, 10\\} & \\{-10, \\ldots, 10\\} & 441 \\\\ -5 & \\{0, \\ldots, 10\\} & -6a - 2b - 20 & 11 \\\\ -4 & \\{-3, \\ldots, 7\\} & -6a - 2b - 20 & 11 \\\\ -3 & \\{-6, \\ldots, 4\\} & -6a - 2b - 20 & 11 \\\\ -2 & \\{-9, \\ldots, 1\\} & -6a - 2b - 20 & 11 \\\\ -1 & \\{-10, \\ldots, -2\\} & -6a - 2b - 20 & 9 \\\\ 0 & \\{-10, \\ldots, -5\\} & -6a - 2b - 20 & 6 \\\\ 1 & \\{-10, -9, -8\\} & -6a - 2b - 20 & 3 \\\\ \\hline Total & & & 510\\\\ \\hline \\end{tabular}$$ The total number of ordered triples that satisfy the required condition is $510$. Notes For * In case anyone is confused by this (as I initially was). In the case where $f(2)=f(4)$, this does not mean that g has a double root of $f(2)=f(4)=k$, ONLY", "up or down, add two sequences, or ask for the rate of change of a sequence. These are done exactly as you would for functions. That said, while keeping the rigorous mathematical definition in mind is helpful, we often describe sequences by writing out the first few terms. Example $$\\PageIndex{1}$$ Can you find the next term in the following sequences? 1. $$7,7,7,7,7, \\ldots$$ 2. $$3, -3, 3, -3, 3, \\ldots$$ 3. $$1, 5, 2, 10, 3, 15, \\ldots$$ 4. $$1, 2, 4, 8, 16, 32, \\ldots$$ 5. $$1, 4, 9, 16, 25, 36, \\ldots$$ 6. $$1, 2, 3, 5, 8, 13, 21, \\ldots$$ 7. $$1, 3, 6, 10, 15, 21, \\ldots$$ 8. $$2, 3, 5, 7, 11, 13, \\ldots$$ 9. $$3, 2, 1, 0, -1, \\ldots$$ 10. $$1, 1, 2, 6, \\ldots$$ Solution No you cannot. You might guess that the next terms are: 1. $$7$$ 2. $$-3$$ 3. $$4$$ 4. 64 5. 49 6. 34 7. 28 8. 17 9. $$-2$$ 10. $$24$$ In fact, those are the next terms of the sequences I had in mind when I made up the example, but there is no way to be sure they are correct. Still, we will often do this. Given the first few terms of a sequence, we can ask what the pattern in the sequence", "The factors of each number can be found in in the factored LCM. Method 2. Make a list of a few multiples of each of the numbers. Make the lists long enough that you can find one of the numbers in each of the lists. If the list is long, there may be several such numbers, but the idea is to choose the smallest one that appears in each list. Multiples of 3: $3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , \\ldots$ Multiples of 4: $4 , 8 , 12 , 16 , 20 , 24 , \\ldots$ Multiples of 6: $6 , 12 , 18 , 24 , \\ldots$ Now, find the smallest ( Least ) multiple that is Common in each list of Multiples, It is easy to pick-out the number 12 from the lists as the smallest, even though there is another number, 24, which is common to each list. • 25 minutes ago • 26 minutes ago • 27 minutes ago • 31 minutes ago • 8 seconds ago • 11 minutes ago • 12 minutes ago • 13 minutes ago • 14 minutes ago • 14 minutes ago • 25 minutes ago • 26 minutes ago • 27 minutes ago • 31 minutes ago", "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 {$3, 13 ...$}.", "smallest comon multiple of the length of the circles. About the case of a power of two: We can proove by induction that for any $n$, $$\\{1,\\ldots,2^n\\}\\rightarrow\\{2^n;2^{n-1};2^{n-2}.3,2^{n-2}.1;\\ldots;2^{n-k}.(2^k-1),2^{n-k}.(2^k-3),\\ldots,2^{n-k}.3,2^{n-k}.1;\\ldots;2^n-1,2^n-3,\\ldots,5,3,1\\}.$$ Notice the use of semicolon $;$ here is just to stress some special groups of numbers of length $1$, $1$, $2$,...,$2^{k-1}$,...,$2^{n-1}$. With this description, one can see that for any $k\\leq n$, $2^k$ and $2^{n-k}$ swap. More generally we can see that for any $k\\leq n$ and any $l$ such that $2l+1\\leq 2^{n-k}$, we have that $2^{n-k}-l$ becomes $2^k(2l+1)$, but I did not figure out how that could help. In the following I give the cyclic representation of the permutations when we have $2^n$ cards, for $1\\leq n\\leq 7$: 1. $(1,2)$ 2. $(1,4)$ 3. $(1,8)(2,4)(3,6,5,7)$ 4. $(1,16)(2,8)(3,12,9,15)(5,14)(6,10,13,7)$ 5. $(1,32)(2,16)(3,24,17,31)(4,8)(5,28,9,30)(6,20,25,15)(7,12,18,29)(10,26,13,14)(11,22,21,23,19,27)$ 6. $(1,64)(2,32)(3,48,33,63)(4,16)(5,56,17,62)(6,40,49,31)(7,24,34,61)(9,60)(10,52,25,30)(11,44,41,47,35,59)(12,36,57,15)(13,28,18,58)(14,20,50,29)(19,54,21,46,37,55)(22,42,45,39,51,27)(23,38,53)$ 7. $(1,128)(2,64)(3,96,65,127)(4,32)(5,112,33,126)(6,80,97,63)(7,48,66,125)(8,16)(9,120,17,124)(10,104,49,62)(11,88,81,95,67,123)(12,72,113,31)(13,56,34,122)(14,40,98,61)(15,24,68,121)(18,116,25,60)(19,108,41,94,69,119)(20,100,57,30)(21,92,73,111,35,118)(22,84,89,79,99,59)(23,76,105,47,70,117)(26,52,50,58)(27,44,82,93,71,115)(28,36,114,29)(37,110)(38,106,45,78,101,55)(39,102,53,46,74,109)(42,90,77,103,51,54)(43,86,85,87,83,91,75,107)$ Edit I am adding here some grphique obtained with mathematica. In the two following picture we observe the irregularity of the order of the permutation when the number of cards varies. The first picture show the orders when we have up to $100$ cards. The second show the same thing up to $2000$ cards. On the other hand the following diagram seem to show some kind of regularity of the order when we consider $2^k$ cards (with $1\\leq k\\leq 16$). More precisely, here are the value $o_k$ of", "would be $(11,16), (12,17), (13,18), (14, 19), (15, 20)$ and then we move to 21-30 range.... Shreev explained it much better, I tried to wrote it as elementary as possible.... – N. S. May 24 '11 at 15:55 I see now. You're throwing away some possible configurations and STILL finding a pair of houses of the desired type. I now prefer your answer to mine. :) – Austin Mohr May 24 '11 at 19:04 Hint: divide the houses into groups $(1,6,11,\\ldots,46), (2,7,12,\\ldots,47),$ etc. Added after the fact: Even easier: consider the pairs $(k,k+5)$ for $1 \\le k \\pmod {10} \\le 5$. There are 25 of them. Only one application of the principle. - Yes, I've done that... but I'm not too sure how to apply it. – meiryo May 23 '11 at 23:52 How many students live in houses in one group? And how many houses are in the group? – Ross Millikan May 23 '11 at 23:52 You can use the possible remainders after division as your holes. If we number the houses from $1$ to $50$, we can see that $10$ of them have a remainder of $0$ after dividing by $5$, $10$ have a remainder of $1$, and so on. Within each group of ten, I can pick five houses such that none of them" ]

[ "is the least common multiple of the $n_i$. -", "7) 9) $$∃x(¬ϕ)$$ --- from 2) and 8) by Double Negation, discharging [a]. • Many thanks for the proof. – Sha Vuklia Apr 7 at 19:13", "# What is the least common multiple of 3 and 9? It should be $9$. We can see that although $18$ is present in both columns it is $9$ the least in common between the two!", "9 {x}^{2} - 24 x - 16 - {x}^{2} - x + 42$ $- 10 {x}^{2} - 25 x + 26$ It is best practice to express this as: $26 - 25 x - 10 {x}^{2}$ This ensures that you do not lose negations when doing further calculations. Negations before brackets often catch students out. Hope this helps.", "Make It Big Logic Level 1 Using only the 4 basic arithmetic operations ($+, -, \\times, \\div$), what is the largest possible value of $\\large 1\\, \\square\\, 2\\,\\square \\,3\\, \\square \\, 4 ?$ ×", "of $6$ therefore it is a multiple of $6$", "x^2+x,\\ x^2-x.$$ There are 7 more with $$a=-1$$ (the negations of these 7).", "that every locale has a smallest dense sublocale, the double negation sublocale. Revised on August 17, 2015 22:56:45 by Todd Trimble (67.81.95.215)", "Browse Questions # Write the negation of \"The number $2$ is greater than the number $7$.\" Toolbox: • The negation of a statement is the denial of the statement. The given statement is \"The number $2$ is greater than the number $7$.\" The negation is \"The number $2$ is not greater than the number $7$.\"", "what is the largest multiple of both three and four? Look at the chart above and find the largest number that appears in both rows. 4. Write three more numbers that are multiples of both three and four. It will help to extend your chart to find the next multiples. The next three common multiples in the sequence are $48,60,$ and $72$.", "the simplification of taking just one Double Negation rule: $p \\dashv \\vdash \\neg \\neg p$", "that is a multiple of both $4$ and $5$ is $20$, it follows that $20|a$.", "multiple of $n/2$.", "8064}$ is the only answer choice that is a multiple of $2016$.", "the least common multiple of $180$, $216$, and $450$?", "# We divide Computer Science Level 3 What is the largest power of $$3$$ that divides $$567948952!$$ ×", "over the property of being a multiple of $$3$$.", "(which is the least common multiple of 2 and 3) to get $3x = 2(x+1)+24$ Does it look more solvable now?", "- Symbols. The largest I know is: $$90, 91, 92, 93, 94, 95, 96$$ I.", "2$$ are all numbers greater than $$2$$:" ]

[ "very close to each other. $99 \\div 101$ is very close to $99 \\div 100$, which is $\\frac{99}{100}$ or 0.99. The quotient $7.5 \\div 7.4$ is close to $7.5 \\div 7.5$, which is 1. In general, when we divide two numbers that are nearly equal to each other, the quotient is close to 1. • In $10 \\div 101$ and in $50 \\div 198$ the dividend is smaller than the divisor. $10 \\div 101$ is very close to $10 \\div 100$, which is $\\frac{10}{100}$ or 0.1. The division $50 \\div 198$ is close to $50 \\div 200$, which is $\\frac 14$ or 0.25. In general, when a smaller number is divided by a larger number, the quotient is less than 1.", "yes, you can view it as multiplication by $-1$: $-x = (-1)x$. But everything still works out. From this viewpoint: $-1^2 = -1 \\cdot 1^2 = -1 \\cdot 1 = -1$ (we still square before multiplying). – Bill Cook Sep 8 '15 at 15:07 • I tend to view negation as a special case of subtraction since that's the \"lowest\" level I can place it on the \"hierarchy\". Viewing negation as multiplication by $-1$ seems like making the operation more complex than it really is. But that's just me. :) – Bill Cook Sep 8 '15 at 15:09 • Please see my answer. In it, I've elaborated on my comment a bit – John Joy Sep 8 '15 at 15:34 One thing that we do in mathematics is to take a concepts, and flip between several interpretations of that concept. For example, is $\\frac{45}{9}$ the number of equal partitions of of size $9$ that can made from $45$ things , or is it the size of each partition, given that there are $9$ partitions. Or perhaps it is the size of one thing divided by 9 and then replicated $45$ times. Another example occurs with the expression $2-5$. Are we subtracting $5$ from $2$, or are we adding $-5$ to $2$. In any event, I don't know how", "than the CRT. – Brian M. Scott Mar 10 '12 at 14:13 Here’s a solution that doesn’t rely on the Chinese Remainder Theorem and uses only pretty basic knowledge, though it’s still a bit of a stretch for kids under ten. If $9y+2=11z+3$, then $9y=11z+1$, so we need a multiple of $9$ that is one more than a multiple of $11$. A little quick mental arithmetic finds $45$, one more than $44$; that would make $y=5$ and $z=4$, and of course $k$ would be $47$. Unfortunately, this leaves a remainder of $5$ when divided by $7$. Note, though, that any number of the form $47+99n$ leaves remainders of $2$ and $3$ when divided by $9$ and $11$, respectively; this is because $99n$ is divisible by both $9$ and $11$. Moreover, $99=7\\cdot14+1$, so every time I add $99$, I increase the remainder on division by $7$ by one. I want a remainder of $1$, and $k=47$ gives me a remainder of $5$. Thus, I need to increase the remainder by $3$ (to $6$, then $0$, then $1$), so I add $99$ three times to get $k=47+3\\cdot 99=344$. - Cool. I need to be more genius to do maths in this way. Thank you very much. – gaurav Mar 10 '12 at 14:50 @guarev No you don't need to be", "= 99$, we must have $99^2+100 = 9901 \\le n \\le 10000$. The only value of $n$ in this range which is divisible by $k = 99$ is $9999$. Thus, the total number of integers $100 < n \\le 10000$ such that $\\lfloor \\sqrt{n-100} \\rfloor$ divides $n$ is $8 \\cdot 3 + 90 \\cdot 2 + 1 = 205$. • How did you get the $98 \\cdot 2$? – user19405892 Dec 29 '16 at 3:45 • ^There are $98$ values of $k \\in [1,98]$ with at least $2$ integers $n$ and $8$ values of $k \\in [1,98]$ with an \"extra\" value of $n$. I've edited my answer to make it less confusing. – JimmyK4542 Dec 29 '16 at 3:47 • We're trying to count the number of integers between $100+k^2$ and $100+k^2+2k$ that are divisible by $k$. I've already stated that exactly $2$ integers between $100+k^2+1$ and $100+k^2+2k$ are divisible by $k$, so the only one that's left is $100+k^2$. For the $98-8 = 90$ values of $k \\in [1,98]$ that are not factors of $100$, we have that $100$ is not divisible by $k$, and thus, $100+k^2$ is not divisible by $k$. Thus, there are only $2$ integers between $100+k^2$ and $100+k^2+2k$ which are divisible by $k$ in that case. – JimmyK4542 Dec 29 '16 at 3:58", "a larger number is divided by a smaller number, the quotient is greater than 1. • In $$99 \\div 101$$ and in $$7.5 \\div 7.4$$ the dividend and divisor are very close to each other. $$99 \\div 101$$ is very close to $$99 \\div 100$$, which is $$\\frac{99}{100}$$ or 0.99. The quotient $$7.5 \\div 7.4$$ is close to $$7.5 \\div 7.5$$, which is 1. In general, when we divide two numbers that are nearly equal to each other, the quotient is close to 1. • In $$10 \\div 101$$ and in $$50 \\div 198$$ the dividend is smaller than the divisor. $$10 \\div 101$$ is very close to $$10 \\div 100$$, which is $$\\frac{10}{100}$$ or 0.1. The division $$50 \\div 198$$ is close to $$50 \\div 200$$, which is $$\\frac 14$$ or 0.25. In general, when a smaller number is divided by a larger number, the quotient is less than 1.", "is true. Actually, we'll show the following stronger statement: The remainder when $x$ is divided by $9$ is equal to the remainder when the sum of the digits of $x$ is divided by $9$. Proof: Suppose the digits of $x$ from most to least significant are $x_{d-1}, x_{d-2}, \\ldots, x_0$, i.e. $$x = x_0 + 10x_1 + 100x_2 + 1000x_3 + \\cdots + 10^{d-1}x_{d-1}$$ The sum of digits of $x$, which we denote as $S(x)$, is therefore $$S(x) = x_0 + x_1 + x_2 + x_3 + \\cdots + x_{d-1}$$ Now, $x - S(x)$ is equal to the following: $$x - S(x) = 9x_1 + 99x_2 + 999x_3 + \\cdots + (10^{d-1} - 1)x_{d-1}$$ $x - S(x)$ is clearly divisible by $9$, but this can only happen if $x$ and $S(x)$ have the same remainder when divided by $9$. The fact above actually allows us to quickly compute the remainder of $N$ when divided by $9$: $N \\bmod 9$ is simply $S(N) \\bmod 9$. But $S(N)$ is easy to compute, and from the upper limit of $N$, we are sure that $S(N) \\le 9\\cdot 10^5$. Therefore, we can easily compute $S(N) \\bmod 9$ with an additional step. This is great, but how does this help us answer the question? Well, note that the only allowed operations are to", "# modular arithmetics - remainder I need to find the remainder of ${1011}^{10}+{10}^{11}$ when divided by $101$. According to this website it is 55 but I fail to see how. For 1011, we can write it as $1$ because $1011=10*101+1$ so ${1011}^{10}$ is $1$. For ${10}^{11}$, I wrote the $11$ like $8+2+1$ or $2^3+2^1+2^0$ Then multiplying the results $10\\cdot100\\cdot1$ and taking the remainder of 101 I got 91. $$1000=9\\cdot101+91$$ So overall I have 92 for the answer, but the website says 55. Was my method or understaning wrong or is there is a problem or limitation on large numbers in that calculator? Also is there any website or mehtod to check myself with this kind of big numbers? • You did fine. I would do the $10^{11}$ part as follows: $$10^2=100\\equiv-1\\pmod{101},$$ so $$10^{11}=(10^2)^5\\cdot10\\equiv(-1)^5\\cdot10=-10\\equiv91\\pmod{101}.$$ Looks like that website doesn't deliver. Nov 9 '16 at 20:38 • Even simpler than what I did, thanks a lot for showing me this way and for the answer. Nov 9 '16 at 20:41 I don't think it's the web-site's fault. The input on that page was 1.1156078e30. You're not giving it the correct number $1011^{10} + 10^{11}$, which is 1115607835569227940375059334601, just the first $8$ decimal digits. All the digits matter here! • I took whatever google gave as a result, with e+, so", "using any calculator with enough digits that $11^{10}-1 = 25937424600$ and that this number is not divisible by $11$. $11^{10}-1$ is clearly not divisible by $11$, in fact it leaves a remainder of $10$, with quotient $11^9 -1$. To see that $11^{10}-1$ is divisible by $100$, we can do the following: To check divisbility by four, see that $11 \\equiv -1 \\mod 4$ (read up congruence notation and rules from Wikipedia, if you are unaware),hence $11^{10} \\equiv (-1)^{10} \\equiv 1 \\mod 4$, so that $4$ divides $11^{10}-1$. See that $11^2 =121 \\equiv -4 \\mod 25$, so that $11^{10} \\equiv (-4)^5 \\equiv -1024 \\equiv 1 \\mod 25$. Hence,$11^{10}-1$ is a multiple of $25$, and hence of $25 \\cdot 4=100$. In fact, $11^{10}-1 = 25937424600 = 2^3 \\cdot 3 \\cdot 5^2 \\cdot 3221 \\cdot 13421$, as a confirmation.", "too much; now we have $a\\equiv 24,773 \\equiv 24,773 - 27097 = -2324\\pmod{b}$. One more step does it: add $3871$ to get $a\\equiv 1547 \\pmod{b}$. This was doing it by hand; with a simple calculator at hand, I would simly compute $\\frac{a}{b}\\approx 1506.4$ to figure out that I need to compute $a - 1506b$ to get the remainder. If $a$ is negative, then do the same thing for $-a$; once you know that $-a\\equiv c \\pmod{b}$ with $0\\leq c \\lt |b|$, then $a\\equiv -c \\equiv |b|-c\\pmod{b}$ and you are done, just one step more than for positive $a$. - In answer to the added question: the way you get the 11 is by dividing -109 by 60; 11 is the remainder (the quotient is -2). - In simplest terms, the way you show that $-109 \\equiv 11 \\pmod{60}$ is to add $60$ to $-109$ repeatedly until you get a result which is non-negative; that result will be $-109 + 2 \\cdot 60 = 11$. Of course, in practice repeated addition would get tedious, so we use division and multiplication as a shortcut: divide $-109$ by $60$, and you get about $-1.8167$. Round that down, and you get $-2$. Multiply that by $60$ and subtract it from $-109$ to get $-109 - (-2 \\cdot 60) = -109 + 2", "2^2+\\dots+a_k\\cdot 2^k$$ where $a_i$ is $0$ or $1$ and $2^k$ is the largest power of $2$ less than or equal to $n$ (and $a_k=1$). When you divide $n$ by $2$, the remainder is $a_0$ and the quotient is $$n_1=a_1+a_2\\cdot2+\\dots+a_k\\cdot 2^{k-1}$$ When you divide $n_1$ by $2$, the remainder is $a_1$ and the quotient is $$n_2=a_2+\\dots+a_k\\cdot 2^{k-2}$$ Continuing in this way, we arrive finally at $$n_k=a_k$$ Thus the procedure of successively dividing by $2$ and collecting the remainders produces the digits in the binary representation of $n$, starting from the less significant digit. Note that we don't need to know $k$ in advance: the algoritm stops when the last quotient is $0$. Finding the largest power of $2$ less than or equal to $n$ yields the most significant digit. It's an alternative procedure, but requires to know the powers of $2$, which is not needed with the algorithm of successive divisions. So successive divisions provide a more efficient algorithm. By the way, the algorithm is the same for every radix. The alternative method for radix $b$ requires a table of the powers of $b$ together with their multiples by $2$, $3$ and so on up to $b-1$. • This is kinda abstract. But it works. – most venerable sir Oct 9 '17 at 22:59 • Why is the remainder", "# Find the Remainder when $100!$ is divided by $97^2$? Find the Remainder when $100!$ is divided by $97^2$? MyApproach I applied Wilson Theorem here $100$.$99$.$98$.$97$.$96!\\text{ mod }97^2=100$.$99$.$98$.$96!\\text{ mod }97=-970200$ I am getting wrong Ans Can Anyone guide me how to approach the problem? First, compute $$a=100\\cdot 99\\cdot 98\\pmod{97},$$ then deduce that $$-a=100\\cdot 99\\cdot 98\\cdot 96!\\pmod{97}$$ and that $$-\\color{red}{97}a=100!\\pmod{97^{\\color{red}2}}.$$ • What is wrong in my approach? – justin takro Nov 3 '15 at 6:45 • Your approach says $100\\cdot 99\\cdot 98\\cdot\\color{red}{97}\\cdot 96!=100\\cdot 99\\cdot 98\\cdot 96!\\pmod{97^2}$, or $97x=x\\pmod{97^2}$, which is not correct. To see that, take for example $x=1$. – Quang Hoang Nov 3 '15 at 6:53 • I edited the code – justin takro Nov 3 '15 at 10:00 • As I said above, the answer should be $97x$, not $x$. You should multiply $-970200$ by $97$ and that's the answer. – Quang Hoang Nov 3 '15 at 10:21 Facts: • $97\\text{ is prime}$ • $P\\text{ is prime}\\iff(P-1)!+1\\equiv0\\pmod{P}$ • The remainder of $100!$ divided by $97^2$ is equivalent to $\\frac{100!}{97}\\bmod{97}$ Hence: $96!+1\\equiv0\\pmod{97}$ Hence: $96!\\equiv-1\\pmod{97}$ Hence: $96!\\equiv96\\pmod{97}$ Hence: $\\frac{100!}{97}\\equiv96!\\cdot98\\cdot99\\cdot100\\equiv96\\cdot98\\cdot99\\cdot100\\equiv93139200\\equiv91\\pmod{97}$ • $100!$/$97$^$2$(97 raised to power 2)You took 1? – justin takro Nov 3 '15 at 9:58 • @justintakro: No. You asked for the remainder of the quotient of $100!$ and $97^2$, so I calculated the modulo with $97$ of the", "by $p$ if and only if your original number was). Because you're only testing for divisibility and not trying to find either the remainder or the quotient, you can ignore information and there are more moves available to you: for example, you can at any step in this process multiply or divide by other primes. I didn't use this freedom above, but I did refrain from dividing by $2$; that happened to make some of the arguments more convenient. For numbers up to $1000$ do what you can. For somewhat larger, subtract off multiples of $$1001 = 7 \\cdot 11 \\cdot 13$$ until you get somethng no larger than $1000.$ This way you get $7$ and $13$ at the same time, also $11,$ although $11$ has its own. It looks like my own method is pretty much right. For example, $156$: $6*9=54-15=39$, and $39$ is divisible by $13$. It looks like this can work any time. I think this works out perfectly.", "$$10^{15}+x>10^{15}-8568\\cdot 10^9>0.$$ By definition, the determinant is the sum of $60$ positive and $60$ negative terms, one of which is $\\mbox{sgn}{\\left( \\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\\\ 3 & 1 & 5 & 2 & 4 \\\\ \\end{array} \\right)}(1000)^5=\\mbox{sgn}\\left((13542)\\right)(1000)^5=+(1000)^5$. Each of the other terms has absolute value at most $9\\cdot1000^4$, so the sum of the 60 negative terms is greater than or equal to $-60\\cdot9\\cdot1000^4$, and the determinant is greater than or equal to $(1000)^5-60\\cdot9\\cdot1000^4$, which is positive. With an even number of adjacent row swaps, you can put the $1000$s on the diagonal. That determinant must be positive. Using the algorithm which looks at minors, the largest term is always positive. • ...and the important thing is that the largest term is very large compared to all the others. Obviously it's at least $1000/9 \\approx 111$ times larger than any other term, but there are $119$ other terms so a small amount of subtlety is needed. But if you think about it, it's actually at least $1000^2/9^2$ times larger than any other term, which is plenty large enough to dominate the sum. – Erick Wong Nov 26 '14 at 1:01 • Alternatively, note that only 60 of those 119 terms are negative. – Steve Kass Nov 26 '14 at 2:07 You can", "\\Z$, we will reach $r = 0$ after finitely many steps. At this point, $\\gcd \\set {r, 0} = r$ from GCD with Zero. $\\blacksquare$ ## Euclid's Proof Let $AB, CD$ be the two given (natural) numbers which are not coprime. We need to find the greatest common divisor of $AB$ and $CD$. Without loss of generality that $CD \\le AB$. We have that $CD$ is a divisor of itself. If $CD$ is a divisor of $AB$ then $CD$ is a common divisor of $CD$ and $AB$. It is clearly the greatest, because no number greater than $CD$ can be a divisor of $CD$. Now, suppose $CD$ does not divide $AB$. Then the less of the numbers $AB, CD$ being continually subtracted from the greater, some number will be left which will be a divisor of the one before it. From Coprimality Criterion, this number will not be a unit, otherwise $AB$ and $CD$ will be coprime. So some number will be left which is a divisor of the one before it. Now let $CD$, dividing $BE$, leave $EA$ less than itself. Let $EA$, dividing $DF$, leave $FC$ less than itself. Let $CF$ divide $AE$. Since then $CF$ divides $AE$, and $AE$ divides $DF$, then $CF$ will also divide $DF$. But it also divides itself. Therefore it will", "$1$ can not be divided by digit $9$. If we can’t divide the digit by the divisor, we combine that digit with the next significant digit and get a two digit number. In our case, it is the digit $1$ so we get number $11$. We divide number $11$ by number $9$. The result is number $1$. Write down number $1$ in the result. -> In this step, we multiply number $1$ by number $9$ and write under number $11$. After that we must subtract those numbers. ->Number $11$ minus number $9$ is number $2$. Write number $2$ as a result of subtraction. Now, we need to add the next digit from the dividend next to number $2$. That is number $7$. Now we have $27$. -> Now, we need to divide number $27$ by number $9$ to get the next digit in the resulting number. Number $27$ divided by $9$ is number $3$. Write number $3$ in the result. -> After that, $3 \\cdot 9 = 27$. We write number $27$ under the $27$ and subtract that numbers. The result is equal to $0$. Subtraction is done when number remainder of division can not divide with divisor. -> The final result is number $13$. That means that $13 \\cdot 9 = 117$. ### Examine 2. The next", "divisor then there exists integers $m$ and $n$ such that $a = md$ and $b = nd$. This then implies that $d$ divides the sum $a+b$ since $a+b = md+nd = (m+n)d.$ In your case, the sum of $-90$ and $100$ is $10$. So, the greatest common divisor between $-90$ and $100$ must also divide $10$. The greatest divisor of $10$ is itself (as is the case with any integer up to absolute value) and so it is natural to check whether or not $10$ divides both $-90$ and $100$. If it divides both then we are done since $10$ is also the largest integer that divides $100+(-90)=10.$ Indeed, $10$ divides both $-90$ and $100$ and so it must be their greatest common divisor. Let $d$ be the GCD, then $d$ divides both $100$ and $-90$, so it has to divide the sum $10$, then $d$ divides $10$, so $d\\le 10$, and also $10$ divides $100$ and $-90$, so $d\\geq 10$, using both together $d=10$.", "stronger statement: The remainder when x is divided by 9 is equal to the remainder when the sum of the digits of x is divided by 9. Proof: Suppose the digits of x from most to least significant are x_{d-1}, x_{d-2}, \\ldots, x_0, i.e. x = x_0 + 10x_1 + 100x_2 + 1000x_3 + \\cdots + 10^{d-1}x_{d-1} The sum of digits of x, which we denote as S(x), is therefore S(x) = x_0 + x_1 + x_2 + x_3 + \\cdots + x_{d-1} Now, x - S(x) is equal to the following: x - S(x) = 9x_1 + 99x_2 + 999x_3 + \\cdots + (10^{d-1} - 1)x_{d-1} x - S(x) is clearly divisible by 9, but this can only happen if x and S(x) have the same remainder when divided by 9. The fact above actually allows us to quickly compute the remainder of N when divided by 9: N \\bmod 9 is simply S(N) \\bmod 9. But S(N) is easy to compute, and from the upper limit of N, we are sure that S(N) \\le 9\\cdot 10^5. Therefore, we can easily compute S(N) \\bmod 9 with an additional step. This is great, but how does this help us answer the question? Well, note that the only allowed operations are to increment or decrement a digit, and each", "stronger statement: The remainder when x is divided by 9 is equal to the remainder when the sum of the digits of x is divided by 9. Proof: Suppose the digits of x from most to least significant are x_{d-1}, x_{d-2}, \\ldots, x_0, i.e. x = x_0 + 10x_1 + 100x_2 + 1000x_3 + \\cdots + 10^{d-1}x_{d-1} The sum of digits of x, which we denote as S(x), is therefore S(x) = x_0 + x_1 + x_2 + x_3 + \\cdots + x_{d-1} Now, x - S(x) is equal to the following: x - S(x) = 9x_1 + 99x_2 + 999x_3 + \\cdots + (10^{d-1} - 1)x_{d-1} x - S(x) is clearly divisible by 9, but this can only happen if x and S(x) have the same remainder when divided by 9. The fact above actually allows us to quickly compute the remainder of N when divided by 9: N \\bmod 9 is simply S(N) \\bmod 9. But S(N) is easy to compute, and from the upper limit of N, we are sure that S(N) \\le 9\\cdot 10^5. Therefore, we can easily compute S(N) \\bmod 9 with an additional step. This is great, but how does this help us answer the question? Well, note that the only allowed operations are to increment or decrement a digit, and each", "\\cdot 2 + 3$. If the remainder is 0, then we say that $d$ “divides” $n$, and that $d$ is a factor of $n$. This is represented by the notation $d \\big| n$ Read on to see applications of this result.", "2018 #1 +371 +2 Best Answer Nota bene: positive means greater than 0, and negative means less than 0. Positive and negative real numbers do not include 0. If you wanted non-positive and non-negative, you need to restate your question. Otherwise, here is the solution. A) To have f(t) be positive, either \\((t-4)$$ and $$(t+9)$$ must both be positive or $$(t-4)$$ and $$(t+9)$$ must both be negative. In order to have (t-4)(t+9) be positive, since (t-4) < (t+9), we need (t-4) to be greater than 0. We have the inequality $$t-4>0$$, which simplifies into $$t > 4$$. We cannot have t = 4 because (4-4) = 0, and anything multiplied to 0 is 0 (which is not positive). We can also plug in some values of $$t$$ and try them out. If t = 5, we have $$(5-4)\\cdot(5+9)$$, which simplifies into $$(1)\\cdot(14)$$, which is greater than 0. In order to have (t-4) and (t+9) to both be negative, since (t+9) > (t-4), we need (t+9) to be less than 0. We have the inequality $$t+9<0$$, which simplifies into $$t<-9$$. We cannot have t = -9 because (-9+9) = 0, and anything multiplied by 0 is 0. If we try t = -10, we have $$(-10-4)\\cdot(-10+9)$$, which simplifies into $$(-14)\\cdot(-1)$$. Since the minus signs cancel out each other, we" ]

[ "yes, you can view it as multiplication by $-1$: $-x = (-1)x$. But everything still works out. From this viewpoint: $-1^2 = -1 \\cdot 1^2 = -1 \\cdot 1 = -1$ (we still square before multiplying). – Bill Cook Sep 8 '15 at 15:07 • I tend to view negation as a special case of subtraction since that's the \"lowest\" level I can place it on the \"hierarchy\". Viewing negation as multiplication by $-1$ seems like making the operation more complex than it really is. But that's just me. :) – Bill Cook Sep 8 '15 at 15:09 • Please see my answer. In it, I've elaborated on my comment a bit – John Joy Sep 8 '15 at 15:34 One thing that we do in mathematics is to take a concepts, and flip between several interpretations of that concept. For example, is $\\frac{45}{9}$ the number of equal partitions of of size $9$ that can made from $45$ things , or is it the size of each partition, given that there are $9$ partitions. Or perhaps it is the size of one thing divided by 9 and then replicated $45$ times. Another example occurs with the expression $2-5$. Are we subtracting $5$ from $2$, or are we adding $-5$ to $2$. In any event, I don't know how", "$$10^{15}+x>10^{15}-8568\\cdot 10^9>0.$$ By definition, the determinant is the sum of $60$ positive and $60$ negative terms, one of which is $\\mbox{sgn}{\\left( \\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\\\ 3 & 1 & 5 & 2 & 4 \\\\ \\end{array} \\right)}(1000)^5=\\mbox{sgn}\\left((13542)\\right)(1000)^5=+(1000)^5$. Each of the other terms has absolute value at most $9\\cdot1000^4$, so the sum of the 60 negative terms is greater than or equal to $-60\\cdot9\\cdot1000^4$, and the determinant is greater than or equal to $(1000)^5-60\\cdot9\\cdot1000^4$, which is positive. With an even number of adjacent row swaps, you can put the $1000$s on the diagonal. That determinant must be positive. Using the algorithm which looks at minors, the largest term is always positive. • ...and the important thing is that the largest term is very large compared to all the others. Obviously it's at least $1000/9 \\approx 111$ times larger than any other term, but there are $119$ other terms so a small amount of subtlety is needed. But if you think about it, it's actually at least $1000^2/9^2$ times larger than any other term, which is plenty large enough to dominate the sum. – Erick Wong Nov 26 '14 at 1:01 • Alternatively, note that only 60 of those 119 terms are negative. – Steve Kass Nov 26 '14 at 2:07 You can", "2^2+\\dots+a_k\\cdot 2^k$$ where $a_i$ is $0$ or $1$ and $2^k$ is the largest power of $2$ less than or equal to $n$ (and $a_k=1$). When you divide $n$ by $2$, the remainder is $a_0$ and the quotient is $$n_1=a_1+a_2\\cdot2+\\dots+a_k\\cdot 2^{k-1}$$ When you divide $n_1$ by $2$, the remainder is $a_1$ and the quotient is $$n_2=a_2+\\dots+a_k\\cdot 2^{k-2}$$ Continuing in this way, we arrive finally at $$n_k=a_k$$ Thus the procedure of successively dividing by $2$ and collecting the remainders produces the digits in the binary representation of $n$, starting from the less significant digit. Note that we don't need to know $k$ in advance: the algoritm stops when the last quotient is $0$. Finding the largest power of $2$ less than or equal to $n$ yields the most significant digit. It's an alternative procedure, but requires to know the powers of $2$, which is not needed with the algorithm of successive divisions. So successive divisions provide a more efficient algorithm. By the way, the algorithm is the same for every radix. The alternative method for radix $b$ requires a table of the powers of $b$ together with their multiples by $2$, $3$ and so on up to $b-1$. • This is kinda abstract. But it works. – most venerable sir Oct 9 '17 at 22:59 • Why is the remainder", "we have $26 - 16 = 10.$ • The largest power of $2$ that's less than or equal to $10$ is $2^3 = 8.$ The largest multiple of $8$ that's less than or equal to $10$ is $1 \\cdot 8 = 8.$ Subtracting off, we have $10 - 8 = 2.$ • The largest power of $2$ that's less than or equal to $2$ is $2^1 = 2.$ The largest multiple of $2$ that's less than or equal to $2$ is $1 \\cdot 2 = 2.$ Subtracting off, we have $2 - 2 = 0.$ • We subtracted off $1 \\cdot 2^4,$ $1 \\cdot 2^3,$ and $1 \\cdot 2^1.$ Therefore, We can write $26$ as $1 \\cdot 2^4$ $+ 1 \\cdot 2^3$ $+ 0 \\cdot 2^2$ $+ 1 \\cdot 2^1$ $+ 0 \\cdot 2^0,$ which has coefficients $1,$ $1,$ $0,$ $1,$ $0$ and therefore corresponds to the binary number $11010.$ Likewise, to convert the decimal number $241791$ to hexadecimal (base-$16$): • The largest power of $16$ that's less than or equal to $241791$ is $16^4 = 65536.$ The largest multiple of $65536$ that's less than or equal to $241791$ is $3 \\cdot 65536 = 196608.$ Subtracting off, we have $241791 - 196608 = 45183.$ • The largest power of $16$ that's less than or equal to $45183$ is", "$-1$ is the \"negation\" operator, then you should be able to convince them that $$-5^2 = -1*5^2 = -1*25 = -25$$ is a reasonable way to interpret this expression. • Yes, but why does she/he have to interpret $-5^2$ as $negation of$5^2$? – Behzad Feb 16 '18 at 21:32 • It is simply a widely accepted convention (the order of operations) that exponentiation has a higher precedence than negation. If you want to square$-5$, you should write$(-5)^2$to obtain$+25$. – Dan Christensen Feb 18 '18 at 6:04 I tutored a student who had a hard time understanding this, and the way that helped him to understand it was this: Any time there is a negative sign on a number, we can read it as$(-1)$. So$-5 \\equiv (-1)5$and$-3^2 \\equiv (-1)3^2 = (-1)9 \\equiv -9$. In the case you mention then,$-1^2 \\equiv (-1)1^2 = (-1)1 \\equiv -1$The key to helping him understand it was always wrapping the negative 1 that the minus sign implies in parenthesis, from which he could use BEMDAS on everything. If you're main concern is students' writing, you could take the route of avoiding the issue altogether and tell students that$-1^2$, regardless of what it should be equal to according to BEDMAS, is just bad notation and that it can be avoided. The purpose of writing mathematics,", "17:55 Note that we need approximate value of the given expression. Now, $$22^{22x}$$ is much larger number than $$22^{2x}$$. Hence $$22^{22x}-22^{2x}$$ will be very close to $$22^{22x}$$ itself, basically $$22^{2x}$$ is negligible in this case. The same way $$11^{11x}-11^x$$ will be very close to $$11^{11x}$$ itself. Hello Bunuel, Nice explanations. I am a little confused with the fact which you mentioned as \"negligible\" in this math. Wouldn't this deduction change the result of the answer? Thanks. _________________ Math Expert Joined: 02 Sep 2009 Posts: 44657 Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] ### Show Tags 11 Jul 2015, 02:40 1 KUDOS Expert's post 1 This post was BOOKMARKED Note that we need approximate value of the given expression. Now, $$22^{22x}$$ is much larger number than $$22^{2x}$$. Hence $$22^{22x}-22^{2x}$$ will be very close to $$22^{22x}$$ itself, basically $$22^{2x}$$ is negligible in this case. The same way $$11^{11x}-11^x$$ will be very close to $$11^{11x}$$ itself. Hello Bunuel, Nice explanations. I am a little confused with the fact which you mentioned as \"negligible\" in this math. Wouldn't this deduction change the result of the answer? Thanks. The question asks about the closest value of the fraction among the options, not the exact value, so we can approximate. Similar questions to practice: the-value-of-10-8-10-2-10-7-10-3-is-closest-to-which-of-95082.html which-of-the-following-is-closest-to-10180-1030-a-110224.html which-of-the-following-best-approximates-the-value-of-q-if-99674.html new-tough-and-tricky-exponents-and-roots-questions-125956-40.html if-x-10-10-x-2-2x-7-3x-2-10x-2-is-closest-to-143897.html 1-0-0001-0-04-10-the-value-of-the-expression-above-is-59398.html which-of-the-following-is-closest-in-value-to-64425.html", "# What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$ What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$ I tried to factorize $n^3+100$, but $100$ is not a perfect cube. I wish it were $1000$. • I wish it were $1000$, too! Mar 27 '12 at 2:18 By division we find that $n^3 + 100 = (n + 10)(n^2 − 10n + 100)−900$. Therefore, if $n +10$ divides $n^3 +100$, then it must also divide $900$. Since we are looking for largest $n$, $n$ is maximized whenever $n + 10$ is, and since the largest divisor of $900$ is $900$, we must have $n + 10 = 900 \\Rightarrow n = 890$ The largest $n$ is therefore $890$ • Note that there is no need to compute the quotient of the polynomials - only the remainder. See my answer. Mar 27 '12 at 3:25 Hint $$\\rm\\quad\\ \\ n\\!+\\!10\\ \\,|\\,\\ f(n) \\:\\iff\\: n\\!+\\!10\\ \\,|\\,\\ f(-10),\\,$$ for any $$\\rm\\:f(x)\\in \\mathbb Z[x],\\,$$ by the Factor Theorem i.e. $$\\rm\\ mod\\,\\ n\\!+\\!10\\!:\\,\\ \\color{#c00}{n\\equiv -10}\\ \\Rightarrow\\ f(\\color{#c00}n)\\equiv f(\\color{#c00}{-10})\\$$ by the Polynomial Congruence Rule. Thus the largest such $$\\rm\\,n\\,$$ occcurs when $$\\,\\rm n\\!+\\!10\\,$$ is the largest divisor of $$\\rm\\,f(-10)$$. • nice solution.. I didn't know about polynomial congruence rule.. learnt something new :)! Oct 28 '20", "divisor then there exists integers $m$ and $n$ such that $a = md$ and $b = nd$. This then implies that $d$ divides the sum $a+b$ since $a+b = md+nd = (m+n)d.$ In your case, the sum of $-90$ and $100$ is $10$. So, the greatest common divisor between $-90$ and $100$ must also divide $10$. The greatest divisor of $10$ is itself (as is the case with any integer up to absolute value) and so it is natural to check whether or not $10$ divides both $-90$ and $100$. If it divides both then we are done since $10$ is also the largest integer that divides $100+(-90)=10.$ Indeed, $10$ divides both $-90$ and $100$ and so it must be their greatest common divisor. Let $d$ be the GCD, then $d$ divides both $100$ and $-90$, so it has to divide the sum $10$, then $d$ divides $10$, so $d\\le 10$, and also $10$ divides $100$ and $-90$, so $d\\geq 10$, using both together $d=10$.", "while $r$ is called the remainder. Note that by definition, the remainder can NOT be negative. That's one reason why your example is wrong: the remainder can't be \"$-1$\". Here's one way to look at it (somewhat informally). For example, you said that $20$ has a remainder of $2$ when divided by $3$. Yes, that's true, but why? I bet you were taught to look for the largest multiple of $3$ that doesn't exceed $20$. This is going to be $18$, and then the remainder is $20-18=2$. Well, all you gotta do now is apply exactly the same logic to negative numbers too! Let's find the remainder of $-20$ modulo $3$. What is the largest multiple of $3$ that doesn't exceed $-20$? It is NOT $-18$, because $-18>-20$, not less. Instead, the largest multiple of $3$ that doesn't exceed $-20$ is $-21$, and the remainder is $(-20)-(-21)=1$. In terms of the definition, $a=\\color{blue}{q}d+\\color{red}{r}$, where $\\color{blue}{q}$ is the quotient and $\\color{red}{r}$ is the remainder, $0\\le\\color{red}{r}<|d|$, for these two examples we have: $20=\\color{blue}{6}\\cdot3+\\color{red}{2}$ for the first one, and $-20=\\color{blue}{(-7)}\\cdot3+\\color{red}{1}$ for the second one. Same for your last example. What is the largest multiple of $60$ that doesn't exceed $-1$? It is NOT $0$, because $0>-1$, not less. Instead, the largest multiple of $60$ that doesn't exceed $-1$ is $-60$, and", "is true. Actually, we'll show the following stronger statement: The remainder when $x$ is divided by $9$ is equal to the remainder when the sum of the digits of $x$ is divided by $9$. Proof: Suppose the digits of $x$ from most to least significant are $x_{d-1}, x_{d-2}, \\ldots, x_0$, i.e. $$x = x_0 + 10x_1 + 100x_2 + 1000x_3 + \\cdots + 10^{d-1}x_{d-1}$$ The sum of digits of $x$, which we denote as $S(x)$, is therefore $$S(x) = x_0 + x_1 + x_2 + x_3 + \\cdots + x_{d-1}$$ Now, $x - S(x)$ is equal to the following: $$x - S(x) = 9x_1 + 99x_2 + 999x_3 + \\cdots + (10^{d-1} - 1)x_{d-1}$$ $x - S(x)$ is clearly divisible by $9$, but this can only happen if $x$ and $S(x)$ have the same remainder when divided by $9$. The fact above actually allows us to quickly compute the remainder of $N$ when divided by $9$: $N \\bmod 9$ is simply $S(N) \\bmod 9$. But $S(N)$ is easy to compute, and from the upper limit of $N$, we are sure that $S(N) \\le 9\\cdot 10^5$. Therefore, we can easily compute $S(N) \\bmod 9$ with an additional step. This is great, but how does this help us answer the question? Well, note that the only allowed operations are to", "larger $n$. My last attempt is by considering the biggest possible value of $s(x)$, that is the digit of $x$ is all $9$ and $x$ less than or equals to the maximum value of $n$, $10^{18}$. $$s(x) \\leq s(999,999,99 \\cdots 9,999) = 9 \\cdot 18 = 162$$ Hence, we have $$x + s(x) \\leq x + 162$$ so that, $$\\sqrt{n} \\leq x + s(x) \\leq x + 162 \\leq \\sqrt{n} + 162$$ hence, $$\\sqrt{n} - 162 \\leq x \\leq \\sqrt{n}$$ Now, we have better range that will suffice for all value of $x$. $$l_1 = \\max \\{ 1, \\sqrt{n} - 162 \\}$$ $$l_2 = \\sqrt{n}$$ the range is $[l_1, l_2]$ But, I though there must be a better solution to this problem. Hence I try $$l_1 = \\max \\{ 1, \\sqrt{n} - 162/2 \\}$$ And that range is also correct. But I can't justify my finding. So, hence my third question: 1. Why $l_1 = \\max \\{ 1, \\sqrt{n} - 162/2 \\}$ also gives us correct solution? - While not a full answer to your questions, here is a trick that will allow you to reduce the collection of numbers you have to try (and so an answer to question 2). Because $10\\equiv 1 \\pmod 9$, we have $10^k \\equiv 1 \\pmod 9$ for every $k\\in \\mathbb N$.", "2018 #1 +371 +2 Best Answer Nota bene: positive means greater than 0, and negative means less than 0. Positive and negative real numbers do not include 0. If you wanted non-positive and non-negative, you need to restate your question. Otherwise, here is the solution. A) To have f(t) be positive, either \\((t-4)$$ and $$(t+9)$$ must both be positive or $$(t-4)$$ and $$(t+9)$$ must both be negative. In order to have (t-4)(t+9) be positive, since (t-4) < (t+9), we need (t-4) to be greater than 0. We have the inequality $$t-4>0$$, which simplifies into $$t > 4$$. We cannot have t = 4 because (4-4) = 0, and anything multiplied to 0 is 0 (which is not positive). We can also plug in some values of $$t$$ and try them out. If t = 5, we have $$(5-4)\\cdot(5+9)$$, which simplifies into $$(1)\\cdot(14)$$, which is greater than 0. In order to have (t-4) and (t+9) to both be negative, since (t+9) > (t-4), we need (t+9) to be less than 0. We have the inequality $$t+9<0$$, which simplifies into $$t<-9$$. We cannot have t = -9 because (-9+9) = 0, and anything multiplied by 0 is 0. If we try t = -10, we have $$(-10-4)\\cdot(-10+9)$$, which simplifies into $$(-14)\\cdot(-1)$$. Since the minus signs cancel out each other, we", "# Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$? $397$ with base $10$ $$= 3 \\cdot 10^2 + 9 \\cdot 10^1 + 7 \\cdot 10^0$$ To convert it into a number with base $9$ $$3 \\cdot 9^2 + 9 \\cdot 9^1 + 7 \\cdot 9^0 = 243 + 81 + 7 = 331$$ But answer is $481$????? • You started with a number in base $9$ and converted it to base $10$ rather than the reverse. – N. F. Taussig Aug 28 '17 at 15:51 • To convert to base $9$ from base $10$, you could find the highest power of $9$ (e.g. $9,81,729$ ...) that is less than your number, then the largest multiple of that power that is still less than your number will be your left-most digit. So, in the case of $397_{10}$, the largest power of $9$ that will fit is $81$ (since $729$ is too large), and you can fit four $81$s: $4 \\cdot 81=324< 397$ (five is too many, since $5 \\cdot 81 = 405 > 397$. This means that the leftmost digit must be $4$. We are left with $397 - 4 \\cdot 81 =73$. Now, find the largest multiple of $9$ that's less than $73$, etc. – Zubin Mukerjee Aug", "\\cdot 2 + 3$. If the remainder is 0, then we say that $d$ “divides” $n$, and that $d$ is a factor of $n$. This is represented by the notation $d \\big| n$ Read on to see applications of this result.", "# What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$ What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$ I tried to factorize $n^3+100$, but $100$ is not a perfect cube. I wish it were $1000$. - I wish it were $1000$, too! – The Chaz 2.0 Mar 27 '12 at 2:18 Hint $\\rm\\quad\\ \\ n+10\\ |\\ f(n) \\iff n+10\\ |\\ f(-10),\\$ for any $\\rm\\:f(x)\\in \\mathbb Z[x],\\$ by the Factor Theorem i.e. $\\rm\\ mod\\ n+10\\!:\\ n\\equiv -10\\ \\Rightarrow\\ f(n)\\equiv f(-10)$ - $$\\begin{array}{r r c} \\rm m=n+10: & \\rm (m-10)^3+100 & \\rm \\equiv0 \\;\\bmod{m} \\\\ & \\rm (-10)^3+100 & \\rm \\equiv0\\; \\bmod m \\\\ \\times (-1) & 900 & \\rm \\equiv 0 \\;\\bmod m \\\\ \\\\ \\hline \\end{array}$$ $$\\rm \\max_m \\{m:m|900\\,\\}=900 \\implies n=890.$$ - By division we find that $n^3 + 100 = (n + 10)(n^2 − 10n + 100)−900$. Therefore, if $n +10$ divides $n^3 +100$, then it must also divide $900$. Since we are looking for largest $n$, $n$ is maximized whenever $n + 10$ is, and since the largest divisor of $900$ is $900$, we must have $n + 10 = 900 \\Rightarrow n = 890$ The largest $n$ is therefore $890$ - Note that there is no need to compute the quotient of the polynomials - only", "# 1985 AIME Problems/Problem 13 ## Problem The numbers in the sequence $101$, $104$, $109$, $116$,$\\ldots$ are of the form $a_n=100+n^2$, where $n=1,2,3,\\ldots$ For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$. Find the maximum value of $d_n$ as $n$ ranges through the positive integers. ## Solution 1 If $(x,y)$ denotes the greatest common divisor of $x$ and $y$, then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$. Now assuming that $d_n$ divides $100+n^2$, it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$. Thus the equation turns into $d_n=(100+n^2,2n+1)$. Now note that since $2n+1$ is odd for integral $n$, we can multiply the left integer, $100+n^2$, by a power of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multiply by $4$ so that we can get a $(2n+1)^2$ in there. So $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$. It simplified the way we wanted it to! Now using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$. Thus $d_n$ must divide $\\boxed{401}$ for every single $n$. This means the largest possible value for $d_n$ is $401$, and we see that it can be achieved when $n = 200$. ## Solution 2 We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$. Since we want to find the GCD of $a_n$ and $a_{n+1}$,", "9 {x}^{2} - 24 x - 16 - {x}^{2} - x + 42$ $- 10 {x}^{2} - 25 x + 26$ It is best practice to express this as: $26 - 25 x - 10 {x}^{2}$ This ensures that you do not lose negations when doing further calculations. Negations before brackets often catch students out. Hope this helps.", "\\Z$, we will reach $r = 0$ after finitely many steps. At this point, $\\gcd \\set {r, 0} = r$ from GCD with Zero. $\\blacksquare$ ## Euclid's Proof Let $AB, CD$ be the two given (natural) numbers which are not coprime. We need to find the greatest common divisor of $AB$ and $CD$. Without loss of generality that $CD \\le AB$. We have that $CD$ is a divisor of itself. If $CD$ is a divisor of $AB$ then $CD$ is a common divisor of $CD$ and $AB$. It is clearly the greatest, because no number greater than $CD$ can be a divisor of $CD$. Now, suppose $CD$ does not divide $AB$. Then the less of the numbers $AB, CD$ being continually subtracted from the greater, some number will be left which will be a divisor of the one before it. From Coprimality Criterion, this number will not be a unit, otherwise $AB$ and $CD$ will be coprime. So some number will be left which is a divisor of the one before it. Now let $CD$, dividing $BE$, leave $EA$ less than itself. Let $EA$, dividing $DF$, leave $FC$ less than itself. Let $CF$ divide $AE$. Since then $CF$ divides $AE$, and $AE$ divides $DF$, then $CF$ will also divide $DF$. But it also divides itself. Therefore it will", "Therefore, for negative numbers, I don't think we should be talking about something being $n$ times larger than something else. - Here you go: $$5\\cdot(-2)=-10$$ - But $-10$ is smaller than $-2$! Indeed, OP's question can be really confusing and ambiguous. – Lee Yiyuan Mar 3 at 15:04", "the answer is $\\textbf{D}$. Note: this solution is incorrect because when we take the $2016th$ root, we must also consider the negative root which is valid because the taking the reciprocal of $a$ negates $log_ba$. Therefore the answer is $199 \\cdot 3$ or $\\textbf{E}$." ]

[ "of $96$. Hence: $$1 \\equiv -5(19) \\mod 96 \\equiv 1 \\mod 96$$", "divisible by $8$ is $12$.", "} \\quad \\dfrac{93}{4}$$", "# How do you evaluate 8.58\\div 2? Dec 16, 2016 $4.29$ #### Explanation: You can also do this mentally by breaking $8.58$ into two parts and using the distributive law: $8.58 \\div 2$ $= \\left(8 + 0.58\\right) \\div 2$ Divide each part by 2 separately... $= 4 + 0.29 \\text{ } \\leftarrow$ now add the answers together: $= 4.29$", "88, .88 Hope this helps.", "\\div 4^{15}$ = $64^?$ 9 3 12 7 None of these", "approximately equal to 80 $\\div$ 20 = 8 $\\div$ 2 = 4. Question 24: 75 $\\div$ 23 Solution: The estimated quotient of 75 $\\div$ 23 is approximately equal to 80 $\\div$ 20 = 8 $\\div$ 2 = 4. Question 25: 193 $\\div$ 24 Solution: The estimated quotient of 193 $\\div$ 24 is approximately equal to 200 $\\div$ 20 = 20 $\\div$ 2 = 10. Question 26: 725 $\\div$ 23 Solution: The estimated quotient of 725 $\\div$ 23 is approximately equal to 700 $\\div$ 20 = 70 $\\div$ 2 = 35. Question 27: 275 $\\div$ 25 Solution: The estimated quotient of 275 $\\div$ 25 is approximately equal to 300 $\\div$ 30 = 30 $\\div$ 3 = 10. Question 28: 633 $\\div$ 33 Solution: The estimated quotient of 633 $\\div$ 33 is approximately equal to 600 $\\div$ 30 = 60 $\\div$ 3 = 20. Question 29: 729 $\\div$ 29 Solution: 729 $\\div$ 29 is approximately equal to 700 $\\div$ 30 or 70 $\\div$ 3, which is approximately equal to 23. Question 30: 858 $\\div$ 39 Solution: 858 $\\div$ 39 is approximately equal to 900 $\\div$ 40 or 90 $\\div$ 4, which is approximately equal to 23. Question 31: 868 $\\div$ 38 Solution: 868 $\\div$ 38 is approximately equal to 900 $\\div$ 40 or 90 $\\div$ 4, which is approximately equal", "can think of $$4878 \\div 900=5.42$$ as saying, “There are 900 groups of 5.42 in 4878.” How might we show that both statements are true? 1. Explain why $$51.2 \\div 6.4$$ has the same value as $$5.12 \\div 0.64$$. 2. Write a division expression that has the same value as $$51.2 \\div 6.4$$ but is easier to use to find the value. Then, find the value using long division. Student Response Student responses to this activity are available at one of our IM Certified Partners Activity Synthesis The goal of this discussion is for students to contrast the two division methods used in the task. Discuss: • Why is the value of $$48.78 \\div 9$$ the same as the value of $$4,\\!878 \\div 900$$? (The numbers in the second expression are both multiplied by 100, and this does not change the value of the quotient.) • What are some advantages of calculating $$48.78 \\div 9$$ with the decimal intact (the method on the left)? (It is fast, and we don’t need to deal with a bunch of 0’s. Also, if the numbers are from a contextual problem, we could better make meaning of them in their original form.) • What are some advantages of calculating $$4,\\!878 \\div 900$$ with long division? (These are whole numbers, and we are", "divisible by $2$ or $3$ and make things much easier.", "divisible by $1003$ either?", "\\frac12 \\div 2 \\frac12 = ?$$", "$(-6)(-1)$ 7. $(-7)(8)$ 8. $(6)(-1)$ 9. $(9)(-4)$ 10. $(-9)(-7)$ 11. $(-5)(2)$ 12. $(-2)(-2)$ 13. $(-5)(4)$ 14. $(-3)(-9)$ For questions 45 to 58, find each quotient. 1. $30 \\div -10$ 2. $-49 \\div -7$ 3. $-12 \\div -4$ 4. $-2 \\div -1$ 5. $30 \\div 6$ 6. $20 \\div 10$ 7. $27 \\div 3$ 8. $-35 \\div -5$ 9. $80 \\div -8$ 10. $8 \\div -2$ 11. $50 \\div 5$ 12. $-16 \\div 2$ 13. $48 \\div 8$ 14. $60 \\div -10$", "$4\\sqrt{2}\\pi$", "\\div (-4b)$ 11. $-99xy \\div (-9x)$ 12. $121a \\div (11b)$ 13. $-144xy \\div (-12)$ 14. $-169y \\div (-13x)$ 15. $-225xy \\div (5z)$ Basic Nov 30, 2012 Jun 13, 2014", "Level 4? Level 2? Level 1 $88\\varpi + C \\delta \\lt 1$? details Level 4? Level 2? Level 2 $74\\varpi + C \\delta \\lt 1$? details Level 2 Level 1a Level 3 $116\\varpi + 30 \\delta \\lt 1$ details", "- 2.4^2).$$ -", "10. $-68 \\div -2 = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$", "$h^4$.", "$= 4 \\times 86 = 344$", "$89$ $$T^{2}$$ $97$ $$4 + T^{2}$$" ]

[ "each stack? How many trays will be left over? Let's consider the first question first. How many trays will be in each stack? I found the answer 24 by starting down the road of calculating 97 ÷ 4 far enough to see that result would be \"24 and some stuff.\" (Too late, I remembered that I actually knew 4 × 24 = 96 offhand—it was still a good check on my answer.) Thus my approach was to pursue, just far enough, a strategy of \"divide and round down.\" If you work with computers, then you probably know that the operation \"divide and round down\" equals a computer command known as \"DIV.\" For example, a computer will tell you that 16 DIV 3 = 5, and one way to check this answer would be to calculate 16 ÷ 3 = 5⅓ and round the result down. I doubt this is how a computer actually calculates the value of 16 DIV 3, but \"divide and round down\" gives the right answer and shows the relationship between DIV and ÷, which is what I want to write about today. A few more examples just to clarify how DIV works: 32 DIV 7 = 4 54 DIV 6 = 9 3 DIV 4 = 0. A friendly term for the DIV operation", "Basic College Mathematics (9th Edition) -60 $\\div$ 10 $\\div$ (-3) Divide from left to right = -60 $\\div$ 10 $\\div$ (-3) = -6 $\\div$ (-3) = 2", "Basic College Mathematics (9th Edition) Since multiplication and division are at the same level according to the order of operations, after finding the square root and exponent, the problem is solved from left to right like a normal problem is. 1. $\\sqrt 121=11$ and $2^{3}=8$ 2. $88\\div11\\times8=64$", "Answers on Properties of Division: 1. Fill in the blanks. (i) 23 ÷ 23 = ……… (ii) 21 ÷ ……… = 21 (iii) 15 ÷ 1 = ……… (iv) ……… ÷ 15 = 0 (v) 0 ÷ 12 = ……… (vi) 8 ÷ ……… = 1 (vii) 65 ÷ 65 = ……… (viii) ……… ÷ 12 = 1 1. (i) 1 (ii) 1 (iii) 15 (iv) 0 (v) 0 (vi) 8 (vii) 1 (viii) 12", "# How do you divide (x^2-5x-14)/(x-7)div(x+2)/(x^2-9)? $\\left({x}^{2} - 5 x - 14\\right) = \\left(x + 2\\right) \\left(x - 7\\right)$ $f \\left(x\\right) = \\frac{\\left(x + 2\\right) \\left(x - 7\\right)}{x - 7} . \\frac{{x}^{2} - 9}{x + 2} = \\left(x - 3\\right) \\left(x + 3\\right)$", "How do you divide (n^3-5n^2-33n-37)div(n-9) using synthetic division? ${n}^{3} - 5 {n}^{2} - 33 n - 3 = \\left(n - 9\\right) \\left(2 {n}^{2} + 13 x n + 144\\right) + 1259$", "# How do you long divide (x^3+x^2-14x-24)div(x+3)? Nov 23, 2016 #### Explanation: Self-explanatory if you know the basics of division. Nov 23, 2016 $\\frac{{x}^{3} + {x}^{2} - 14 x - 24}{x + 3}$ $= {x}^{2} - 2 x - 8$ $= \\left(x - 2\\right) \\left(x + 4\\right)$", "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: $2 \\times 11 \\implies 22$", "51 Use the digits 1 through 9. Use each digit exactly once. Fill in the squares to make all of the equations true. $$\\begin{split} \\Box - \\Box = \\Box & \\\\ \\times & \\\\ \\Box \\div \\Box = \\Box & \\\\ = & \\\\ \\Box + \\Box = \\Box & \\end{split}$$", "# How do you divide (x^2+13x+12)/(x+2)div(x+1)? $f \\left(x\\right) = \\frac{\\left(x + 1\\right) \\left(x + 12\\right)}{\\left(x + 2\\right) \\left(x + 1\\right)} = \\frac{x + 12}{x + 2}$", "### Home > A2C > Chapter 10 > Lesson 10.2.4 > Problem10-115 10-115. Divide $\\left(x^{3} + 8\\right) \\text{by} \\left(x + 2\\right)$. Then use your answer to factor $x^{3} + 8$. Use a general rectangle.", "Question If $\\left( 81\\div 9 \\right)\\div 3=a$ and $81\\div \\left( 9\\div 3 \\right)=b$ , then which of the following is correct?A. $a=b$ B. $a>b$ C. $b>a$ D. All of the above Verified 129.3k+ views Hint: First we solve the given equations $\\left( 81\\div 9 \\right)\\div 3=a$ and $81\\div \\left( 9\\div 3 \\right)=b$ separately and find the value of $a$ and $b$ by simplifying the given operation i.e. division. Then, we compare the values of $a$ and $b$ to obtain a desired result. We have given that $\\left( 81\\div 9 \\right)\\div 3=a$ and $81\\div \\left( 9\\div 3 \\right)=b$ . We have to find that from given options, which is correct. To compare the values of $a$ and $b$ , first we have to solve the given equations. Let us consider equation $\\left( 81\\div 9 \\right)\\div 3=a$ We will use the BODMAS rule to solve the given equations, according to the BODMAS rule first we have to solve the brackets. So, we solve $\\left( 81\\div 9 \\right)$ , we get $\\left( 81\\div 9 \\right)=9$ Now, we have \\begin{align} & 9\\div 3=a \\\\ & a=3 \\\\ \\end{align} Now, let us consider equation $81\\div \\left( 9\\div 3 \\right)=b$ First we solve $\\left( 9\\div 3 \\right)$ , we get $\\left( 9\\div 3 \\right)=3$ Now, we have \\begin{align} & 81\\div 3=b \\\\ & b=27", "# How do you evaluate 6times8div4? Dec 20, 2016 $\\left(6 \\times 8\\right) \\div 4 = 48 \\div 4 = 12$. #### Explanation: The standard convention is that multiplication and division have the same importance, so when deciding which to do first in a situation like this, we just evaluate left to right. Going left to right, we'll first do the multiplication in this case. So we have $6 \\times 8 \\div 4 = \\left(6 \\times 8\\right) \\div 4 = 48 \\div 4 = 12$.", "# How do you divide ( x^3-2x^2 - 4x - 24 )/(x+2)? ${x}^{2} - 4 x + 4 + \\left(\\frac{16}{x + 2}\\right)$", "= ___ Answer: 72 ÷ 9 = 8. Explanation: The division of 72 ÷ 9 is 8. Question 28. __ = 28 ÷ 4 Answer: 28 ÷ 4 = 7. Explanation: The division of 28 ÷ 4 is 7. Question 29. 81 ÷ 9 = ___ Answer: 81 ÷ 9 = 9. Explanation: The division of 81 ÷ 9 is 9. Question 30. ___ = 35 ÷ 5 Answer: 35 ÷ 5 = 7. Explanation: The division of 35 ÷ 5 is 7. Question 31. 63 ÷ 9 = ___ Answer: 63 ÷ 9 = 7. Explanation: The division of 63 ÷ 9 is 7. Question 32. Answer: 14 ÷ 7 = 2. Explanation: The division of 14 ÷ 7 is 2. Question 33. Answer: 36 ÷ 9 = 4. Explanation: The division of 36 ÷ 9 is 4. Question 34. Answer: 18 ÷ 2 = 9. Explanation: The division of 18 ÷ 2 is 9. Question 35. Answer: 90 ÷ 9 = 10. Explanation: The division of 90 ÷ 9 is 10. Question 36. Answer: 42 ÷ 7 = 6. Explanation: The division of 42 ÷ 7 is 6. Question 37. Answer: 54 ÷ 9 = 6. Explanation: The division of 54 ÷ 9 is 6. Question 38. Answer: 49 ÷ 7 = 7. Explanation: The division", "for the boxes. They happen to be convenient. Note that (for example) if $a$ and $b$ are in box 5, then $a$ divides $b$ or $b$ divides $a$. Same with the other boxes. – André Nicolas Apr 26 '14 at 7:27", "multiplication and division remain, we work left to right, => 3 * 3 = 9 (Microsoft excel and most other calculators will give this answer) • engvarta says: Very useful details provided by you. I hope it would be useful for many of the seekers. Thank you for sharing details here with us. • andrey says: • Mike says:", "- 3 \\right) e 48 \\div \\left( - 6 \\right) + \\left( - 4 \\right) f 14 \\div \\left( - 5 + 3\\right) g 63 \\div \\left( - 9 \\right) - \\left( - 8 \\right) h \\left( - 54 \\right) \\div \\left( - 4 - 2\\right) i \\left( - 36 \\right) \\div \\left( - 6 \\right) \\div \\left( - 3 \\right) j 15 \\div \\left( 40 \\div \\left( - 8 \\right)\\right) 4 Fill in the blank: 2+⬚\\times 5=-18 5 Evaluate: a - 8 + \\left( 24 \\div 4\\right)^{2} b - 5 + 12 \\div 2^{2} c \\left(8 - 10 \\div 2\\right)^{2} d \\left( - 56 \\times 5\\right) \\div \\left( - 7 \\right) e 21 \\div \\left( - 7 \\right) \\times 2 f \\left( - 84 \\right) \\div \\left( 2 \\times 7\\right) g \\left( \\left( - 43 + 8\\right) \\div \\left( - 5 \\right)\\right)^{2} h 16 \\times \\left( 26 \\div \\left(9 + 4\\right)\\right) i \\left(48 - \\left( \\left(- 9 \\right) \\times 6 \\right) \\div 3\\right) + 16 j \\left( \\left(36 - \\left(10 + 10\\right)\\right) \\div 2\\right) + 14 \\times 6 6 Explain why the following statement is incorrect, and find the correct answer: 6-\\left(-60\\right)\\div\\left(-3\\right)=-22 Applications 7 Three people on a farm are given five minutes to pick as many apricots as they can. They get to keep all the", "farmer puts 14 onions into 2 bags, with the same number of onions in each bag. $$\\displaystyle 2 \\times \\boxed{\\phantom{3}} = 14$$ $$14 \\div 2 = \\boxed{\\phantom{3}}$$ “Estas dos ecuaciones tienen las mismas partes: 2, 14 y una cantidad desconocida. ¿Por qué están organizadas de una manera diferente si representan la misma situación?” // “The two equations here have the same parts: 2, 14, and an unknown amount. Why are they arranged differently if they represent the same situation?” (In multiplication, the factors are the number of groups and the size of each group. The number on the other side of the equation is the total amount. In division we start with the total and divide by how many groups we have to find the size of the group or we divide by the size of the group to find the number of groups we have, so that is the answer.) “Llamamos cociente al resultado de una ecuación de división. Por ejemplo, en $$14 \\div 2 = \\boxed{\\phantom{3}}$$, no conocemos el resultado, por lo que vamos a encontrar el valor del cociente. En la ecuación completa $$14 \\div 2 = 7$$, vemos que el valor del cociente es 7” // “We call the result in a division equation the quotient. For example, in $$14 \\div 2 = \\boxed{\\phantom{3}}$$,", "times that a number goes into a division (DIV). Let's take a look at a quick example of 10 divided by 7 (you might want to remind yourself about long division): 1 r 3 7)10 7 3 Hopefully that wasn't too hard. We now need to introduce some terminology, MOD and DIV: • MOD = finds the remainder from long division i.e. 10 MOD 7 = 3 • DIV = finds the number of divides in long division i.e. 10 DIV 7 = 1 Exercise: MOD and DIV Try these examples, working out the MOD and DIV of each: 7 / 2 Answer : MOD = 1 DIV = 3 17 / 5 Answer : MOD = 2 DIV = 3 8 / 2 Answer : MOD = 0 DIV = 4 6 / 9 Answer : MOD = 6 DIV = 0 (9 does not divide into 6 at all!) Now try these explicit calculations: 11 MOD 8 Answer : = 3 8 MOD 4 Answer : = 0 6 DIV 5 Answer : = 1 600 DIV 20 Answer : = 30 Hopefully you are now pretty good with MOD and DIV, but what exactly is the point of all this? A very common example in past exam paper has been using the MOD and DIV" ]

[ "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "statement? No, because $$16\\div (4\\div 2)=16\\div 2=16\\div 2=8$$ and $$(16\\div 4)\\div 2=(4)\\div 2=2$$. Division is not associative, the general property (using $$a$$, $$b$$, and $$c$$) does not exist.", "# How do you evaluate 8.58\\div 2? Dec 16, 2016 $4.29$ #### Explanation: You can also do this mentally by breaking $8.58$ into two parts and using the distributive law: $8.58 \\div 2$ $= \\left(8 + 0.58\\right) \\div 2$ Divide each part by 2 separately... $= 4 + 0.29 \\text{ } \\leftarrow$ now add the answers together: $= 4.29$", "# How do you solve 22y = -88? Apr 3, 2016 $22 y = - 88$ means $22 \\times y = - 88$ $y = - \\frac{88}{22}$, or $y = - 4$ #### Explanation: $22 \\left(- 4\\right) = - 88$ When there is a variable next to a number e.g. \"22y\" that means 22 is being multiplied by a number in order to get the answer. So $22 \\left(- 4\\right) = - 88$", "answer of $224 \\Rightarrow \\boxed{(C)}$.", "The only answer choice that is divisible by $4$ is $\\boxed{\\textbf{(B)}\\ 40}$.", "$m+n=2+89=\\boxed{091}$.", "is \"2nd Y\". Hi, I don't want to pick at you but the conversion tool is under $\\boxed{\\text{2ND}}\\ \\boxed{\\text{MODE}}$ 7. I use a TI-92 and the 'conversion arrow' is under $\\boxed{2nd}, \\;\\ \\boxed{Y}$ on it. It can also be found under MATH as the poster pointed out. I always just typed it in. But, then again, the 89 doesn't have the keyboard the 92 and Voyage 200 has.", "can think of $$4878 \\div 900=5.42$$ as saying, “There are 900 groups of 5.42 in 4878.” How might we show that both statements are true? 1. Explain why $$51.2 \\div 6.4$$ has the same value as $$5.12 \\div 0.64$$. 2. Write a division expression that has the same value as $$51.2 \\div 6.4$$ but is easier to use to find the value. Then, find the value using long division. Student Response Student responses to this activity are available at one of our IM Certified Partners Activity Synthesis The goal of this discussion is for students to contrast the two division methods used in the task. Discuss: • Why is the value of $$48.78 \\div 9$$ the same as the value of $$4,\\!878 \\div 900$$? (The numbers in the second expression are both multiplied by 100, and this does not change the value of the quotient.) • What are some advantages of calculating $$48.78 \\div 9$$ with the decimal intact (the method on the left)? (It is fast, and we don’t need to deal with a bunch of 0’s. Also, if the numbers are from a contextual problem, we could better make meaning of them in their original form.) • What are some advantages of calculating $$4,\\!878 \\div 900$$ with long division? (These are whole numbers, and we are", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "you forgot to divide the right side by 880 ... $440 = 880e^{3.3k}$ divide both sides by 880 ... $\\displaystyle \\frac{1}{2} = e^{3.3k} $ continue ... 7. yeah i did that, then i did ln1/2 = lne^3.3k which is ln1/2 = 3.3k and i was told that ln(1/2) = -ln(2) so i had k = -ln(2)/3.3 is that right so far? 8. Nvm got it. 9. Originally Posted by lilwayne Nvm got it. you're welcome.", "2)\\ =(16\\ \\div \\ 4\\ )\\ \\div\\ (16\\ \\div \\ 2 )$$ (14.2,-0.5)(0,-1)1 (14.2,-1.5)(1,0)8.7 (23.0,-1.5)(0,1)1 (10.9,1.5)(0,1)2 (10.9,2.5)(1,0)2.0 (12.9,2.5)(0,-1)1 (10.9,3.5)(1,0)4.5 (15.3,3.5)(0,-1)2.0 $$16\\div (4\\div 2)=16\\div 2=8$$ and $$(16\\div 4)\\div(16\\div 2)=4\\div 8=\\displaystyle \\frac{1}{2}$$ which is different. The example shows that the distributive property does not apply to division over division. ***************************************************", "of $96$. Hence: $$1 \\equiv -5(19) \\mod 96 \\equiv 1 \\mod 96$$", "Now we know the complement is $274 + 81 + 88 = 443$, and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \\boxed{581}$.", "Question 5 # Solve the following:$$78 \\div [-4 + (-3) of \\left\\{27 \\div (-18 \\div -2)\\right\\}] = ?$$ Solution We know BODMAS rule . So, $$78\\div[-4+(-3)of\\left\\{27\\div(-18\\div-2)\\right\\}]$$ = $$\\\\78\\div[-4+(-3)of\\left\\{27\\div9\\right\\}]$$ =$$\\\\78\\div[-4+(-3)of\\left\\{3\\right\\}]$$ =$$78\\div[-4+\\left(-9\\right)]$$ =$$78\\div[-13]$$ =$$-6$$ . So, Option A is correct answer.", "both sides by 5 $\\Rightarrow 88 + x = 23(5) = 115$ ........now we have 88 being added to our variable, so let's subtract 88 from both sides $\\Rightarrow x = 115 - 88$ ....................now, simplify $\\Rightarrow \\boxed {x = 27}$ therefore, the team scored 27 points in their 5th game Page 2 of 2 First 12", "$4n=96$ and $n=24$. Hence, the only two $n$ that work are $n=24$ and $n=47$, so our answer is $24+47=\\boxed{071}$. -Stormersyle", "# 1975 AHSME Problems/Problem 10 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) We see that the result of this expression will always be in the form $(100\\text{ some number of zeros}001)^2.$ Multiplying these together yields: $$110\\text{ some number of zeros}011$$. This works because of the way they are multiplied. Therefore, the answer is $\\boxed{(A) 4}$.", "1 + 6 = \\boxed{77} \\Rightarrow \\boxed{D}$", "approximately equal to 80 $\\div$ 20 = 8 $\\div$ 2 = 4. Question 24: 75 $\\div$ 23 Solution: The estimated quotient of 75 $\\div$ 23 is approximately equal to 80 $\\div$ 20 = 8 $\\div$ 2 = 4. Question 25: 193 $\\div$ 24 Solution: The estimated quotient of 193 $\\div$ 24 is approximately equal to 200 $\\div$ 20 = 20 $\\div$ 2 = 10. Question 26: 725 $\\div$ 23 Solution: The estimated quotient of 725 $\\div$ 23 is approximately equal to 700 $\\div$ 20 = 70 $\\div$ 2 = 35. Question 27: 275 $\\div$ 25 Solution: The estimated quotient of 275 $\\div$ 25 is approximately equal to 300 $\\div$ 30 = 30 $\\div$ 3 = 10. Question 28: 633 $\\div$ 33 Solution: The estimated quotient of 633 $\\div$ 33 is approximately equal to 600 $\\div$ 30 = 60 $\\div$ 3 = 20. Question 29: 729 $\\div$ 29 Solution: 729 $\\div$ 29 is approximately equal to 700 $\\div$ 30 or 70 $\\div$ 3, which is approximately equal to 23. Question 30: 858 $\\div$ 39 Solution: 858 $\\div$ 39 is approximately equal to 900 $\\div$ 40 or 90 $\\div$ 4, which is approximately equal to 23. Question 31: 868 $\\div$ 38 Solution: 868 $\\div$ 38 is approximately equal to 900 $\\div$ 40 or 90 $\\div$ 4, which is approximately equal" ]

[ "+0 # Hmmmm? Can you teach me how to do this +1 163 5 A five-digit integer will be chosen at random from all possible positive five-digit integers. What is the probability that the number's units digit will be less than 5? Express your answer as a common fraction. Suppose that after I wrote this problem, Kev thought he could be more clever than I could, so he wrote his own problem. Suppose that both of our problems are in the set of 12 problems you are currently working on. If you sat down at your computer this morning and randomly loaded 4 of the 12 problems, what is the probability that both this problem and Kev's problem were among the four you loaded? May 26, 2019 edited by Guest May 26, 2019 edited by Guest May 26, 2019 #1 0 Removed May 26, 2019 edited by Guest May 26, 2019 #2 +106080 +1 A five-digit integer will be chosen at random from all possible positive five-digit integers. What is the probability that the number's units digit will be less than 5? Express your answer as a common fraction. the last digit can be 0,1,2,3 or 4 out of 10 possible digits, so that is 5/10 or 0.5 Suppose that after I wrote this problem, Kev thought he", "# More or Less Probability Level 2 Among the positive integers less than $10,000,000$, do the majority contain the digit 1? ×", "# 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? ×", "# You understood Me Number Theory Level 3 Consider a 5 digit number, $$\\overline{3a25b}$$ for single digits non-negative integers $$a,b$$. If both values of $$a,b$$ are selected at random, what is the probability that the resultant number is divisible by $$33$$? ×", "the rest are of course greater than 5. Therefore the probability should be $\\frac{10!}{10^{10}}.$ The authors solution assumes that there are distinct combinations, and the binomial coefficient (combinations) must be used. In fact his answer is $\\displaystyle \\binom{10}{5}(1/2)^5(1/2)^5$. I do not agree that there are (10 choose 5) choices for the total number of distinct choices. What is the author's question really asking??? The number 10C5 refers to the 5 places where the numbers less than 5 will be put. So these numbers can be in places {1,2,3,4,5}, or in {3,4,6,7,8} , etc. After the place assignment has been made, there are 5^5 choices for each place. Then the remaining numbers in {5,6,7,8,9} can be made in 5^5 ways, but the placements have been fixed. Last edited: Ray Vickson Homework Helper Dearly Missed ## Homework Statement A number is chosen at random between 0 and 1. What is the probability that exactly 5 of its first 10 decimal places consists of digits less than 5? ## Homework Equations Binomial Coefficient = $\\displaystyle \\binom{N}{n1}$ Where n1 denotes \"n1\" objects of an indistinguishable type. ## The Attempt at a Solution Obviously, 5 digits of the first 10 decimal places of this \"number\" is less than 5 and the rest are of course greater than 5. Therefore the probability should be", "# Digit product Probability Level 3 How many positive integers less than $1,000,000,000$ have the product of their digits equal to 49? ×", "## mathmath333 one year ago 6 positive number are taken at random and multiplied together. Then what is the probability that the product ends in an odd digit other than 5. 1. mathmath333 \\large \\color{black}{\\begin{align} & \\normalsize \\text{ 6 positive number are taken at random and multiplied together.}\\hspace{.33em}\\\\~\\\\ & \\normalsize \\text{ Then what is the probability that the product ends in an odd }\\hspace{.33em}\\\\~\\\\ & \\normalsize \\text{ digit other than 5. }\\hspace{.33em}\\\\~\\\\ \\end{align}} 2. mathmate I assume \"random\" means that the probability of the last digit is uniformly distributed over the range 0-9. The resulting last digit depends only on the last digit of the six numbers, so without loss of generality (WLOG), we will consider only the product of six \"random\" digits. The resulting digit will be odd if and only if all six digits are odd. So what is the probability of getting 6 odd digits out of a uniform distribution? Out of the set {0,1,2,3,4,5,6,7,8,9}, there are 5 odd digits. So the probability of randomly selecting an odd digit is 5/10=1/2, resulting in an odd number. However, we know that 5 multiplied by any odd number will have 5 as a terminating digit, so we need to exclude 5 in ALL of the six numbers. That leaves us with 4, A={1,3,7,9} to choose from (out", "Discrete Mathematics Level pending What is the probability that the last digit of a sum of two perfect fourth powers, out of the first 10,000 fourth powers, is 1? If this probability can be expressed as $$\\dfrac mn$$, where $$m$$ and $$n$$ are coprime positive integers, submit your answer as $$m+n$$. ×", "25 Q25. A 5-digit number is formed by the digits 1,2,3,4, and 5 without repetition. What is the probability that the number formed is a multiple of 4? A. 1/2 B. 1/3 C. 1/4 D. 1/5 Hide Number formats Decimals Lofoya.com 2017 You may drag this calculator", "since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955. Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5. N = ab5 --> more than one values of N are possible: 315, 515, 555. Not sufficient. Hope it's clear. As done in option (2), why not choose 551 in option (1), as 1 is multiple of 5? I guess you meant that 1 is a factor of 5. Anyway, since N = a5c, then 5 must a factor of N, which means that the units digit of N (c) must be 5, so it cannot be 1: 5 is not a factor of 551. Hope it's clear. Manager Joined: 22 Aug 2014 Posts: 162 Re: If N is a positive three-digit number that is greater than 2 [#permalink] ### Show Tags 09 Mar 2015, 03:02 (1) The tens digit of N is 5. N = a5c. This implies that 5 is a factor of N, which means", "# numbers + probability • January 18th 2009, 10:57 PM numbers + probability What is the probability that a two digit number chosen at random will be a multiple of 3 and not of 5 . • January 18th 2009, 11:19 PM mr fantastic Quote: What is the probability that a two digit number chosen at random will be a multiple of 3 and not of 5 . One way is just to write out all two digit numbers that are multiples of 3 and cross out those that are also a multiple of 5 .... How many are there ....? Now divide by the total number of two digit numbers. • January 20th 2009, 10:20 AM Soroban Quote: What is the probability that a two digit-number chosen at random will be a multiple of 3 and not of 5? There are 90 two-digit numbers. One-third of them are mutliples of 3: . $\\frac{90}{3} \\:=\\:30$ multiples of 3. But every fifth one of them is a multiple of 5: . $\\frac{30}{5} \\:=\\: 6$ multiples of 15. Hence, there are: . $30 - 6 \\:=\\:{\\color{blue}24}$ multiples of 3 which are not multiples of 5. Therefore: . $P(\\text{multiple of 3, not 5}) \\:=\\:\\frac{24}{90} \\:=\\:\\frac{4}{15}$", "## Probability Let k greater than or equal to 3 be any given integer. What is the probability that a random k-digit number will have at least one 0, at least one 1, and at least one 2? (Note: as usual, every number starts with either a 1, or a 2, .... or a 9 (but not a 0).).", "# Probability An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least one digit? ×", "N itself . Sufficient. (2) The units digit of N is 5. N = ab5 --> more than one values of N are possible: 315, 515, 555. Not sufficient. Hope it's clear. As done in option (2), why not choose 551 in option (1), as 1 is multiple of 5? I guess you meant that 1 is a factor of 5. Anyway, since N = a5c, then 5 must a factor of N, which means that the units digit of N (c) must be 5, so it cannot be 1: 5 is not a factor of 551. Hope it's clear. _________________ Intern Joined: 18 Dec 2011 Posts: 2 Re: If N is a positive three-digit number that is greater than 2 [#permalink] ### Show Tags 25 Sep 2014, 05:10 Bunuel wrote: parry wrote: Bunuel wrote: If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N? Given: N = abc, where a>1 and each of a, b, and c is a factor of N. (1) The tens digit of N is 5. N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0", "• mathmath333 6 positive number are taken at random and multiplied together. Then what is the probability that the product ends in an odd digit other than 5. Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "immediately before you? b. What is the probability that you are not the first performer? Question 56. THOUGHT PROVOKING How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, and 4? Repetition of digits is allowed. Let fixed number 1 on the first place and in the second place can be one of 5 digits 0, 1, 2, 3 or 4. Because repetition of digits is allowed. In third and fourth place it can also be one of five digits. Therefore we see that 5 . 5 . 5 = 125 integers which begin with 1, can be formed. Second, If we fixed 2 on the first place, by the same logic as in the first part, we get that 5³ = 125 integers which begin with 2. Next, if we fixed number 3 on the first place, we obtain 5³ = 125 integers which begin with 2. 5³ + 5³ + 5³ = 375 375 integers greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, 4. Question 57. PROBLEM SOLVING There are 30 students in your class. Your science teacher chooses 5 students at random to complete a group project. Find the probability that you and your", "it never happens that the reverse of a power of two is a power of five. The digits in a big power of two seem to occur in a random way without any regular pattern. If it ever happened that the reverse of a power of two was a power of five, this would be an unlikely accident, and the chance of it happening grows rapidly smaller as the numbers grow bigger. If we assume that the digits occur at random, then the chance of the accident happening for any power of two greater than a billion is less than one in a billion. It is easy to check that it does not happen for powers of two smaller than a billion. So the chance that it ever happens at all is less than one in a billion. That is why I believe the statement is true. But the assumption that digits in a big power of two occur at random also implies that the statement is unprovable. Any proof of the statement would have to be based on some non-random property of the digits. The assumption of randomness means that the statement is true just because the odds are in its favor. It cannot be proved because there is no deep mathematical reason why it has to", "post for a discussion of why this seems unresolvable, and how it may actually put forward a case for time being discrete rather than continuous. Crazy thought. There’s something deeply unsettling about this paradox and also the Unexpected Hanging paradox. Anytime we deal with probabilities and certainty, paradox seems to be lurking nearby. I sometimes ask my students this somewhat related question–perhaps you’ve heard it too: How many positive integers have a 3 in them? (That is, in their decimal representation. 6850104302 has a 3 but 942009947 does not.) If you haven’t ever considered this question, take the time to do it now. Though I actually once worked out the result using limits (like Alexander Bogomolny does marvelously here), it’s easy enough to work out the result in our heads: First ask yourself how many digits a randomly selected integer has. The number of digits is almost certainly greater than 2, right? There are only 90 two-digit positive integers, a finite number, and there are an infinite number of integers with more than two digits. It follows that if you were to pick one at random from among all positive integers*, it would be almost certain to contain more than two digits. The same argument could be applied to a larger number of digits. By the same logic", "is an integer between 1 and 5. Maybe there are other interpretations. So tell me if it makes a logical sense.", "below. The fraction of this region covered by the range from 1.3 to 1.4 is The fraction covered by the other regions (such as 2.3 to 3.3, and so on) can be found similarly, and we can add them together to give the total probability that the first digit following the first non-zero digit will be a 3: In general, the probability of the 2nd digit of d in a base-B number (taken from a logarithmic population) is Extending this analysis to the case of the nth digit following the first non-zero digit, we arrive at the general formula This applies to the case of the leading non-zero digit, with the understanding that with n = 0 the summation reduces to just the single term k = 0. This formula shows that the non-uniformity in the distribution of digits becomes much less as we consider less significant digits. For example, we have P0{1} = 0.301019995..., P1{1} = 0.113890103..., and P2{1} = 0.101375977... Thus the probability of a \"1\" quickly approaches 1/10 as we proceed to less significant digits. 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[ "# Elijah forgot his pass code for a certain website. He knows it is made up of 5 digits and no digit can be repeated. If he randomly chooses 5 different digits, what is the probability that he chooses his pass code? Feb 20, 2017 The chances are $\\frac{1}{30 , 240}$. #### Explanation: Use the permutation formula, as order matters, to determine the number of permutations that are possible. Call the number of permutations $P$. There are 10 possible digits to choose from, and he needs to choose $5$ different digits, so $P$ is given by P = (n!)/((n - r)!) P = (10!)/((10 - 5)!) $P = 30240$ If he randomly chooses one set of $5$ digits, than his chances that he chooses the correct pass code is $\\frac{1}{30 , 240}$ Hopefully this helps!", "\\cdot 4 \\cdot 5=120$. This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from. The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal $(1 \\cdot 2)^2$. Therefore, the probability for each of the orderings of $RRBB$ is $\\frac{4}{120}=\\frac{1}{30}$. There are 6 of these, so the total probability is $\\boxed{\\bf{(B)} \\frac{1}{5}}$. ~IceMatrix", "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 $\\frac{80}{120}=\\frac{2}{3}$.", "the digits are 0,1,4,9 in any order, but that isn't what $$(1/10) ^ 4$$ tells you - to calculate that you'd need: • the probability that the first digit is 0,1,4, or 9 = 4/10 • the probability that the second digit is one of the three not yet selected = 3/10 • the probability that the third digit is one of the two not yet selected = 2/10 • the probability that the final digit is the remaining one from the set of 0,1,4,9 = 1/10 i.e. $$\\prod_{i=1}^{4} \\frac{i}{10}$$ You could then worry about permutations to pick 4109 specifically, but this is needless complication in comparison the way your colleague said. Your colleague is correct. Think of it this way: there are 10,000 ($$10^4$$) 4-digit numbers: 0000-9999. Your target number is one of those 10,000. Thus, if you picked a 4-digit number randomly, you'd have a one in 10,000 chance of picking that number (or any other specific 4-digit number). You're right that what you want to calculate here is a permutation, not a \"simple\" (as you called it) combination. But permutations are actually simpler to calculate than combinations. In a permutation, you're essentially just counting all possible outcomes. In a combination, you additionally have to calculate how many of those outcomes are the same combination of", "1. ## formulating this problem.. a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? 2. Originally Posted by transgalactic a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? If they are both odd, there are 20 numbers to choose from, only one of which is correct, so the probability is .. CB 3. 5 posibilities for first 4 forthe second why 20 numbers to choose from the number are 0..9 The digits are odd so we have {1, 3, 5, 7, 9} to choose them from. The first can be any of these five and the second any of the remaining 4 (since the digits are different) - so total is 5x4=20 CB", "Notice that for binary numbers, i.e., numbers expressed in the base 2, the probability of the leading digit being 1 is 1.000, as it must be, since the leading non-zero digit of a binary number is necessarily 1. The distributions of probabilities of the digits 1 to B-1 for each base B from 2 to 10 are shown below. We can also easily verify that the sum of all the probabilities for digits 1 through b-1 equals 1.0000, as it must, since the leading digit must be one of these. This implies To verify this, recall the fundamental law of logarithms, ln(ab) = ln(a) + ln(b). With this we can re-write this sum of logarithms as the logarithm of a product: which confirms the result. By the same kind of analysis we can determine the probability that the second digit will have a certain value. It's only necessary to consider a single order of magnitude, since the pattern is repeated on each order. For example, in the base 10, the probability of the second digit being \"3\" is equal to the sum of the probabilities of the first two digits being \"1.3\", \"2.3\", \"3.3\", ... or \"9.3\" for numbers in the range from 1 to 10. This is indicated by the shaded regions in the logarithmic scale shown", "focus on the last one. It's similar to how the probability of getting the sum $7$ when throwing two dice can be seen to be $\\frac16$ by noting that no matter what the first die shows, the result on the second die can make the sum $7$ in exactly one way. • I've thought of doing something like 10 x 10 x 10 x 2, but I'm not sure if it's correct. – Broadsword93 Aug 23 '17 at 12:36 • @Broadsword93 Arthur is saying that you choose randomly any digit for first three places. This is done in $10^3$ ways. Now, the sum of these digits can be of form $5k+r$ where $0\\le r < 5$. So make $5$ cases for the sum being $5k+r$. Now take suitable values for units digit. Eg. Say we choose 1,3,5. Now this sum is 9, or $5\\times 1 + 4$. So units digit can be 1 or 6. Similarly there will be 2 choices for units digit in all cases. So answer should be $10^3\\times 2$ – samjoe Aug 23 '17 at 12:45 • @Broadsword93 so yes, it is correct. – Omnomnomnom Aug 23 '17 at 12:45 • @Broadsword93 That's if you want to count all the possible ways and find the probability as $\\frac{2000}{10000}$. I like to think of it", "by taking a sequence of numbers from 0 to 9, always selection a digit with the probability of 1/10. This is abstraction, since we cannot do that to infinity. Still taking it from your perspective, you are saying that for each number in the interval (0, 0.3), we have a match from the interval [0.3, 1). Indeed we have 1:1 mapping. However, the process of abstract selection must be a little bit different: you need to select a number from 0 to 3 for the first digit and from 4 to 9 for another group. If we make this abstract selection in sort of uniform way, we should better say that we need to pick the same number of digits for each interval, one digit in both, then two digits in both and so on. Otherwise, we cannot compare the intervals. You can see immediately that we have the probability of $\\frac{1}{4}$ for the first and $\\frac{1}{6}$ for the second interval. The second digit can be any from 0 to 9 for both intervals so the probabilities are $\\frac{1}{4} \\cdot \\frac{1}{9}$, and $\\frac{1}{6} \\cdot \\frac{1}{9}$ respectively. After n digits we have $\\frac{1}{4} \\cdot (\\frac{1}{9})^{n-1}$, and $\\frac{1}{6} \\cdot (\\frac{1}{9})^{n-1}$. Although both probabilities tend to 0, they are never the same. The ratio for any n, and with that the", "To find the probability of starting with the three digits 247, we integrate f(x) from 2.47 to 2.48. The general equation for two leading digits, D1D2, is: $P(D_1D_2) = \\log_{10}(D_1.D_2 + 0.1) - \\log_{10}(D_1.D_2)$ Which is equivalent to: $P(D_1D_2) = \\log_{10}(D_1D_2 + 1) - \\log_{10}(D_1D_2)$ For example, the probability of a number starting with “2” followed by “4” is log10(25)-log10(24) = 1.77%. Similarly, the equation for three leading digits, D1D2D3, is: $P(D_1D_2D_3) = \\log_{10}(D_1D_2D_3 + 1) - \\log_{10}(D_1D_2D_3)$", "1. ## Probability A four digit number is to be made by choosing four different digits from the set {1,2,3,4,5,6,7}. Find the probability that the no. is even and greater than 6000. 2. Hello, BabyMilo A 4-digit number is to be made by choosing four different digits from the set {1,2,3,4,5,6,7}. Find the probability that the number is even and greater than 6000. There are: . $_7P_4 \\:=\\:\\frac{7!}{3!} \\:=\\:840$ possible 4-digit numbers. How many of them are even and greater than 6000? There are two cases to consider. [1] The number begins with \"6\": .6 _ _ _ We have: .{1, 2, 3, 4, 5, 7} to work with. The last digit must be even: 2 choices. The middle two digits are selected from the remaining 5 digits: . $_5P_2= 20$ choices. Hence, there are: . $2\\cdot20 = 40$ even numbers that begin with \"6\". [2] The number begins with \"7\": .7 _ _ _ We have: .{1, 2, 3, 4, 5, 6} to work with. The last digit must be even: 3 choices. The middle two digits are selected from the remaining 5 digits: . $_5P_2 = 20$ choices. Hence, there are: . $3\\cdot20 = 60$ even numbers that begin with \"7\". Then there are: . $40 + 60 \\:=\\:100$ even numbers greater than 6000. Therefore: .", "\"1\"? Clearly, the sample space is S = \\set{0000, \\ldots, 9999}, with 10000 possible outcomes. Clearly, all possible outcomes are equally likely because the PIN codes are selected randomly. Clearly, if the number is even, then the last digit must be in \\set{0, 2, 4, 6, 8}. Clearly, there are 5 possible last digits, and the other three digits can be any of the ten digits. So there are 10 \\times 10 \\times 10 \\times 5 possible outcomes, or 5000. So the probability of an even number is \\frac{5000}{10000}, or 50%. To find the probability of containing at least one \"1\" digit, we consider the cases where there is one \"1\" digit, two, three, or four. Alternatively, we can consider all the numbers that contain no \"1\" digits - the complement of the number containing at least one \"1\". This is a more efficient way of finding the answer. Clearly, if there are no \"1\" digits, then every digit can be anything but \"1\", so there are 9 possibilities. There are therefore 9 \\times 9 \\times 9 \\times 9 = 6561 possible PIN codes without any \"1\" digits. Clearly, the event of having no \"1\" digits is the complement of the event of interest, so P(1 occurs) = 1 - P(1 does not occur). Since P(1 does not", "the probability that 2 digits are the same is: $\\frac{4}{7}$ The possibility that it's the incorrect number is: $\\frac{6}{7}$ Just like Soroban said. The last question has ofcourse nothing to do with how the numbers look like: • Feb 12th 2010, 05:47 AM FengWei 9331, 8242, 7153, 6064, 8602, 7513, 6424", "in a game where digits accumulate but subsequent digits can never be larger than the previous digit (see the page for a better description, especially the example). Figuring out the expected value for all scores is really hard. But we don't have to figure out all possible scores and the probability of their occurrence. We only need to figure out the expected value for each digit and then sum those expected values. What's the expected value for the most significant digit? What's the expected value for the second-most significant digit? etc. Then sum. Pretend the game ended after just one roll. Then the expected value of our score is: \\sum_{d=0}^{9}\\frac{d}{10}=0.45 But the game continues (the value must go up) so we figure out the expected value of the next digit. Let's consider 9. What's the probability the second digit is a 9? In order for the second digit to be 9, the first must also be a 9. So $P(d_2=9)=0.1*0.1=0.01$. What about 0? If our first digit was 9, then there's still just a $\\frac{1}{10}$ chance the next digit is a 0. But if our first digit was a 1, then the outcomes are either 0 or 1, so the probability is $\\frac{1}{2}$. More generally: P(d_{i+1}=n) = \\sum_{m=0}^{9}P(d_{i+1}=n|d_i=m)*P(d_i=m) where P(d_{i+1}=n|d_i=m) = \\begin{cases} \\frac{1}{m+1},& \\text{if } n\\leq m\\\\ 0,", "between 0 and 1. The probability that you select any particular number is zero. But at the same time, you will end up selecting some number. Whichever number you end up selecting, is a number that you would have said had a 0% chance of being selected! For situations like these, the term “almost never” is used. Rather than saying that any particular number is impossible, you say that it will “almost never” be picked. While this linguistic trick might make you feel less uneasy about the situation, there still seems to be some remaining confusion to be resolved here. So in the case of our thought experiment, no matter how many rounds the game ends up going on, you have a 0% chance of being kidnapped. At the same time, by stipulation you have been kidnapped. Making sense of this is only the first puzzle. The second one is to figure out if it makes sense to talk about some theories making it more likely than others that you’ll be kidnapped (is $\\frac{11}{\\infty}$ smaller than $\\frac{111}{\\infty}$?) An even more trouble infinity is in the expected number of people that are kidnapped. No matter how many rounds end up being played, there are always only a finite number of people that are ever kidnapped. But let’s calculate the", "Question # Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat. Probability and combinatorics Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat. Assume that numbers can start with 0. What is the probability of choosing a random number that starts with 8 from this group? Enter a fraction or round your answer to 4 decimal places, if necessary. 2021-09-05 The first and last digits should be even. Hence, Total number of numbers possible $$\\displaystyle={5}\\cdot{4}\\cdot{8}\\cdot{7}={1120}$$ Total number that starts with $$\\displaystyle{2}={4}\\cdot{8}\\cdot{7}={224}$$ Hence, P(Stars with 2) $$\\displaystyle=\\frac{224}{{1120}}=\\frac{1}{{5}}$$", "five possibilities. Next you choose a second thing and there are four objects remaining. So for each of the five choices of the first thing there are four choices for the second thing. This gives twenty ways of choosing. If you do not account for order then there are one half as many since choosing thing three then thing four is the same as choosing thing four then thing three. So the final result is ten, as expected. There is also a formula for computing this and goes like this: $\\binom{n}{k} = \\frac{n!}{(n-k)!n)}$ and in our case we get the following: $\\binom{5}{2} = \\frac{5!}{(5-2)!2!} =$ $\\frac{120}{(3!)(2!)} =$ $\\frac{120}{(6)(2)} =$ $\\frac{120}{12} = 10$ as expected. (11 O’Clock) $0x0B$ – This is a hexadecimal number (base 16). When you count by a different number base you always reach the number ten when you get to the base. For example in base $10$ the numer that follows nine is ten. And this number is always written as the number $1$ stuck to the number $0$, or $10$ (see how to convert from one base to another and positional value above for why a one and a zero are chosen for the numer ten. So, for example if you count in base three you get the following, $0,1,2,10,11,12,20, ...$ If you count", "+0 0 111 1 Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit? Guest Jun 27, 2018 #1 +964 +2 From 20 to 69, they are a total of 50 integers. To choose 5 of them, we do: $$\\binom{50}{5}$$ Now, we need to find out the number of successful outcomes. We need an integer between 20 and 29 Then another one from 30 to 39, 40 to 49, 50 to 59, and 60 to 69. There are 10 numbers in each category, and therefore $$10^5$$ ways of doing this. Your final answer is: $$\\boxed{\\frac{2500}{52969}}$$ I hope this helped, Gavin GYanggg Jun 27, 2018", "1. ## [SOLVED] probability sampling When I open a bank account i am allocated a 4 digit personal identification number(which may begin with one or more zeros) at random with all possibilites equally likely. Find the probability that: The digits im my number are in strictly increaing order. Thanks 2. If we take any selection of four digits, ex. (4,9,4,8), we can always arrange the four is ascending order: (4,4,8,9). Thus, all we need to finish is to determine how many possible four-selections are there. ${ {4+10-1} \\choose 4}$ and there are $10^4$ possible pin numbers. 3. Originally Posted by Plato If we take any selection of four digits, ex. (4,9,4,8), we can always arrange the four is ascending order: (4,4,8,9). Thus, all we need to finish is to determine how many possible four-selections are there. ${ {4+10-1} \\choose 4}$ and there are $10^4$ possible pin numbers. I understand that there are 10^4 possible pin numbers but i do not understand what means. can you please clarify what you mean by this. I thought the the answers was: 10! / (4!(10-4)!))/10 ^ 4 which is equall to 210/10 ^ 4 thanks 4. That is the selection formula: ${ {K+N-1} \\choose K}$, ${ {M} \\choose J}=\\frac {M!}{(M-J)!(J!)}$ . If we have N difference types of items from which we", "Math Help - probability 1. probability What is the probability that a 6-digit phone number contains at least one 2? (Repetition of numbers and lead zero are allowed). 2. Imagine the situation as being a room with 6 phones. Each phone has 10 digits on it, and you have to pick one digit from each phone! As such there are 10 different ways you can choose 1 from 10. And you have to choose 6 of these. Hence there are $10^6$ combinations! Now lets consider the same situation, only now all the 2s on the phones have been removed. So now there are 6 phones, each with only 9 digits, (0,1,3,4,5,6,7,8,9). So you are choosing 1 button from 9, 6 times! Hence $9^6$. So we have concluded that there are $10^6$ combinations, and $9^6$ of them have no 2! Hence the probability that there is at least one 2 is : $10^6 - 9^6 = 468559$ 3. The response is attached as a gif image, hope this helps you.", "Without repetition of the numbers, Question: Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is A. $\\frac{1}{5}$ B. $\\frac{4}{5}$ C. $\\frac{1}{30}$ D. $\\frac{5}{9}$ Solution: D. 5/9 Explanation: We have digits $0,2,3,5$. We know that, if unit place digit is ‘ 0 ‘ or ‘ 5 ‘ then the number is divisible by 5 If unit place is ‘ 0 ‘ Then first three places can be filled in $3 !$ ways $=3 \\times 2 \\times 1 \\times 1=6$ If unit place is ‘ 5 ‘ Then first place can be filled in two ways and second and third place can be filled in $2 !$ ways $=2 \\times 2 \\times 1 \\times 1=4$ $\\therefore$ Total number of ways $=6+4=10=\\mathrm{n}(\\mathrm{E})$ Total number of ways of arranging the digits $0,2,3,5$ to form 4 – digit numbers without repetition is $3 \\times 3 \\times 2 \\times 1=18$ Probability $=\\frac{\\text { Number of favourable outcomes }}{\\text { Total number of outcomes }}$ $\\therefore$ Required probability $=\\frac{10}{18}=\\frac{5}{9}$ Hence, the correct option is (D)." ]

[ "question. Kudos [?]: [0], given: 3 Math Expert Joined: 02 Sep 2009 Posts: 42606 Kudos [?]: 135613 [0], given: 12705 Re: If n is a positive even integer, what is the probability that the unit [#permalink] ### Show Tags 30 Oct 2017, 22:04 Expert's post 1 This post was BOOKMARKED gotonishu wrote: JeffTargetTestPrep wrote: nkmungila wrote: If n is a positive even integer, what is the probability that the unit’s digit of $$n^4$$ is 6? A. 1 B. $$\\frac{4}{5}$$ C. $$\\frac{3}{5}$$ D. $$\\frac{2}{5}$$ E. $$\\frac{1}{2}$$ Let’s write out all possible units digits of even numbers: 0^4 = 0 2^4 = 6 4^4 = 6 6^4 = 6 8^4 = 6 Thus, the probability is 4/5. But the question says n is positive. So we cant take 0 as a value for n. Also, where does it say that n has to be a single digit number. This seems like a bad question. Jeff there considers the units digits of n, not n itself and not single-digits. So, while n itself cannot be 0, its units digit can, for example n can be 10. _________________ Kudos [?]: 135613 [0], given: 12705 Re: If n is a positive even integer, what is the probability that the unit [#permalink] 30 Oct 2017, 22:04 Display posts from previous: Sort by", "+0 0 111 1 Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit? Guest Jun 27, 2018 #1 +964 +2 From 20 to 69, they are a total of 50 integers. To choose 5 of them, we do: $$\\binom{50}{5}$$ Now, we need to find out the number of successful outcomes. We need an integer between 20 and 29 Then another one from 30 to 39, 40 to 49, 50 to 59, and 60 to 69. There are 10 numbers in each category, and therefore $$10^5$$ ways of doing this. Your final answer is: $$\\boxed{\\frac{2500}{52969}}$$ I hope this helped, Gavin GYanggg Jun 27, 2018", "force search is sequential, then a low number like 0012345 will be found earlier. Do not mix up your intuition about what might be a good lock combination with probability in random events like lotteries. The way password spaces are attacked does not obey a uniform, random distribution, because the choice of combinations in the attack follows some cunning strategy driven by an intellect. The lottery balls neither not prefer nor avoid \"nice\" numbers whose digits follow patterns. To believe that they do is to anthropomorphize the machines: endow them with human qualities, or to endow probability itself with intelligent qualities (like that it is driven by supernatural forces or beings which make choices that guide human fate). There is one more angle to this and it is the mistake of interpreting the probability of a pattern with that of a single instance of a pattern. Suppose we are dealing with seven digit numbers whose digits are 0-9. A number with all digits which are the same might be called \"seven of a kind\". There are ten seven-of-a-kind numbers, which makes them seem rare. On the other hand, say, numbers in which all digits are different are far more numerous: $\\frac{10!}{3!} = 604800$. So in fact it is far less likely that a randomly drawn number will be", "and exactly as often as any other finite string of digits. That’s in the full representation. If you look at all the infinitely many digits the normal number has to offer. If all you have is a slice then some digits are going to be more common and some less common. That’s similar to how if you fairly toss a coin (say) forty times, there’s a good chance you’ll get tails something other than exactly twenty times. Look at the first 35 or so digits of π there’s not a zero to be found. But as you survey more digits you get closer and closer to the expected average frequency. It’s the same way coin flips get closer and closer to 50 percent tails. Zero is a rarity in the first 35 digits. It’s about one-tenth of the first 3500 digits. The digits of a specific number are not random, not if we know what the number is. But we can be presented with a subset of its digits and have no good way of guessing what the next digit might be. That is getting into the same strange territory in which we can speak about the “chance” of a month having a Friday the 13th even though the appearances of Fridays the 13th have absolutely no randomness", "N itself . Sufficient. (2) The units digit of N is 5. N = ab5 --> more than one values of N are possible: 315, 515, 555. Not sufficient. Hope it's clear. As done in option (2), why not choose 551 in option (1), as 1 is multiple of 5? I guess you meant that 1 is a factor of 5. Anyway, since N = a5c, then 5 must a factor of N, which means that the units digit of N (c) must be 5, so it cannot be 1: 5 is not a factor of 551. Hope it's clear. _________________ Intern Joined: 18 Dec 2011 Posts: 2 Re: If N is a positive three-digit number that is greater than 2 [#permalink] ### Show Tags 25 Sep 2014, 05:10 Bunuel wrote: parry wrote: Bunuel wrote: If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N? Given: N = abc, where a>1 and each of a, b, and c is a factor of N. (1) The tens digit of N is 5. N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0", "the rest are of course greater than 5. Therefore the probability should be $\\frac{10!}{10^{10}}.$ The authors solution assumes that there are distinct combinations, and the binomial coefficient (combinations) must be used. In fact his answer is $\\displaystyle \\binom{10}{5}(1/2)^5(1/2)^5$. I do not agree that there are (10 choose 5) choices for the total number of distinct choices. What is the author's question really asking??? The number 10C5 refers to the 5 places where the numbers less than 5 will be put. So these numbers can be in places {1,2,3,4,5}, or in {3,4,6,7,8} , etc. After the place assignment has been made, there are 5^5 choices for each place. Then the remaining numbers in {5,6,7,8,9} can be made in 5^5 ways, but the placements have been fixed. Last edited: Ray Vickson Homework Helper Dearly Missed ## Homework Statement A number is chosen at random between 0 and 1. What is the probability that exactly 5 of its first 10 decimal places consists of digits less than 5? ## Homework Equations Binomial Coefficient = $\\displaystyle \\binom{N}{n1}$ Where n1 denotes \"n1\" objects of an indistinguishable type. ## The Attempt at a Solution Obviously, 5 digits of the first 10 decimal places of this \"number\" is less than 5 and the rest are of course greater than 5. Therefore the probability should be", "+0 # Hmmmm? Can you teach me how to do this +1 163 5 A five-digit integer will be chosen at random from all possible positive five-digit integers. What is the probability that the number's units digit will be less than 5? Express your answer as a common fraction. Suppose that after I wrote this problem, Kev thought he could be more clever than I could, so he wrote his own problem. Suppose that both of our problems are in the set of 12 problems you are currently working on. If you sat down at your computer this morning and randomly loaded 4 of the 12 problems, what is the probability that both this problem and Kev's problem were among the four you loaded? May 26, 2019 edited by Guest May 26, 2019 edited by Guest May 26, 2019 #1 0 Removed May 26, 2019 edited by Guest May 26, 2019 #2 +106080 +1 A five-digit integer will be chosen at random from all possible positive five-digit integers. What is the probability that the number's units digit will be less than 5? Express your answer as a common fraction. the last digit can be 0,1,2,3 or 4 out of 10 possible digits, so that is 5/10 or 0.5 Suppose that after I wrote this problem, Kev thought he", "## mathmath333 one year ago 6 positive number are taken at random and multiplied together. Then what is the probability that the product ends in an odd digit other than 5. 1. mathmath333 \\large \\color{black}{\\begin{align} & \\normalsize \\text{ 6 positive number are taken at random and multiplied together.}\\hspace{.33em}\\\\~\\\\ & \\normalsize \\text{ Then what is the probability that the product ends in an odd }\\hspace{.33em}\\\\~\\\\ & \\normalsize \\text{ digit other than 5. }\\hspace{.33em}\\\\~\\\\ \\end{align}} 2. mathmate I assume \"random\" means that the probability of the last digit is uniformly distributed over the range 0-9. The resulting last digit depends only on the last digit of the six numbers, so without loss of generality (WLOG), we will consider only the product of six \"random\" digits. The resulting digit will be odd if and only if all six digits are odd. So what is the probability of getting 6 odd digits out of a uniform distribution? Out of the set {0,1,2,3,4,5,6,7,8,9}, there are 5 odd digits. So the probability of randomly selecting an odd digit is 5/10=1/2, resulting in an odd number. However, we know that 5 multiplied by any odd number will have 5 as a terminating digit, so we need to exclude 5 in ALL of the six numbers. That leaves us with 4, A={1,3,7,9} to choose from (out", "# numbers + probability • January 18th 2009, 10:57 PM numbers + probability What is the probability that a two digit number chosen at random will be a multiple of 3 and not of 5 . • January 18th 2009, 11:19 PM mr fantastic Quote: What is the probability that a two digit number chosen at random will be a multiple of 3 and not of 5 . One way is just to write out all two digit numbers that are multiples of 3 and cross out those that are also a multiple of 5 .... How many are there ....? Now divide by the total number of two digit numbers. • January 20th 2009, 10:20 AM Soroban Quote: What is the probability that a two digit-number chosen at random will be a multiple of 3 and not of 5? There are 90 two-digit numbers. One-third of them are mutliples of 3: . $\\frac{90}{3} \\:=\\:30$ multiples of 3. But every fifth one of them is a multiple of 5: . $\\frac{30}{5} \\:=\\: 6$ multiples of 15. Hence, there are: . $30 - 6 \\:=\\:{\\color{blue}24}$ multiples of 3 which are not multiples of 5. Therefore: . $P(\\text{multiple of 3, not 5}) \\:=\\:\\frac{24}{90} \\:=\\:\\frac{4}{15}$", "since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be: 355; 555; 755; 955. Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient. (2) The units digit of N is 5. N = ab5 --> more than one values of N are possible: 315, 515, 555. Not sufficient. Hope it's clear. As done in option (2), why not choose 551 in option (1), as 1 is multiple of 5? I guess you meant that 1 is a factor of 5. Anyway, since N = a5c, then 5 must a factor of N, which means that the units digit of N (c) must be 5, so it cannot be 1: 5 is not a factor of 551. Hope it's clear. Manager Joined: 22 Aug 2014 Posts: 162 Re: If N is a positive three-digit number that is greater than 2 [#permalink] ### Show Tags 09 Mar 2015, 03:02 (1) The tens digit of N is 5. N = a5c. This implies that 5 is a factor of N, which means", "82, 91 Probability that the sum of digits of that number will be 10 = $$\\frac { 9 }{ 90 }$$ = $$\\frac { 1 }{ 10 }$$ Short Answer Type Questions (Score 3) Question 28. In selecting a two-digit number up to 50. a. What is the probability of the digit in the ten’s place to be larger than the digit in the one’s place? b. What is the probability of the digit in the tens place to be smaller than the digit in the one’s place? Total numbers of two-digit numbers up to 50 = 41 a. Number of numbers with the digit in the tens place to be larger than the digit in the units place = 11 Probability that the number with tens place digit is larger than the digit in the unit place = 11/41 b. Probability that the number with the tenth place digit is smaller than the digit in the unit place = 26/41 Question 29. A black dice and a white dice are thrown at the same time, a. Write all the possible outcomes. b. What is the probability that the sum of the two numbers is to be 8. c. What is the probability that being the same number to be on both dice? a. Total outcomes (1,1), (1,2),", "even, so N should be even too i.e. it must have a 2 in it. karishma, thanks. So is there a way to make this factoring process generic? It seems like your thought process is to find the lowest factor of the denominator, which is 2, ie not 4 or 8. Then make sure you have that same factor in the numerator. Then the product of all factors in the numerator must be >= than the denominator. Meaning if in numerator there are other products that are not factors you just ignore them. like 5 below (5x2^3)/8 and if you happeneted to choose 3 (5x3^3)/8 = (5x3x3x3)/8 since 5 and 3 are both not factors you still dont have the factor portion of the numerator >=8 Intern Joined: 04 Aug 2012 Posts: 1 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Probability - Integer [#permalink] 10 Jul 2013, 14:10 VeritasPrepKarishma wrote: DeeptiM wrote: If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer? Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers For 5N^3/8 to be an integer, 5N^3 should be completely divisible by 8. This will happen only if N^3 is divisible by 8 i.e. if N has", "the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit? A. 16 B. 15 C. 14 D. 12 E. 10 Are You Up For the Challenge: 700 Level Questions We can rewrite N as 50!(60P10 + 55P5 + 1). We see that the units digit of 60P10 + 55P5 + 1 is 1 since the first two addends have units digits of 0 (notice that 60P10 has 60 as a factor and 55P5 has 55 and 54 as factors). Therefore, the number of trailing zeros of N is the same as the number of trailing zeros of 50!. Recall that the number of trailing zeros in a factorial is determined by the number of 5-and-2 pairs that occur in the factorial. Since the number of 5s is less than the number of 2s in 50!, we see that the number of 5-and-2 pairs is limited by the number of 5s that occur in 50!. From 10, 20, 30, 40, and 50, we obtain 6 factors of 5. From 5, 15, 25, 35, and 45, we obtain 6 additional factors of 5. The total number of factors of 5 in 50! is, therefore, 6 + 6 = 12. Thus, we have twelve 5-and-2", "posted something similar. - Looking at Igors post, it would seem that with 10 digits chosen at random there is a better than 99.89% chance of a perfect partition (the results there are for base 11 not base 10). For under 10 digits a search (enlightened or even just brute force) is very easy. – Aaron Meyerowitz Feb 13 2011 at 20:17", "you draw from the random number. Look at triplets of digits. ‘141’ should turn up about once in every thousand sets of three digits. ‘1415’ should turn up about once in every ten thousand sets of four digits. Any finite string of digits should turn up, and exactly as often as any other finite string of digits. That’s in the full representation. If you look at all the infinitely many digits the normal number has to offer. If all you have is a slice then some digits are going to be more common and some less common. That’s similar to how if you fairly toss a coin (say) forty times, there’s a good chance you’ll get tails something other than exactly twenty times. Look at the first 35 or so digits of π there’s not a zero to be found. But as you survey more digits you get closer and closer to the expected average frequency. It’s the same way coin flips get closer and closer to 50 percent tails. Zero is a rarity in the first 35 digits. It’s about one-tenth of the first 3500 digits. The digits of a specific number are not random, not if we know what the number is. But we can be presented with a subset of its digits and have no", "Browse Questions # If four digit numbers greater than 5000 are randomly formed from the digits 0,1,3,5 and 7,what is the probability of forming number divisible by 5 when repetition of digits is not allowed. $\\begin{array}{1 1}(A)\\;\\large\\frac{3}{8}\\\\(B)\\;\\large\\frac{3}{5}\\\\(C)\\;\\large\\frac{4}{7}\\\\(D)\\;\\large\\frac{7}{8}\\end{array}$ Toolbox: • Required probability=$\\large\\frac{n(E)}{n(S)}$ Step 1: 4 digit number greater than 5000 are to be formed Digits -0,1,3,5,7 Number to be divisible by 5 $\\therefore$ A number can be divisible by 5 only when its units digit is 0,5 Step 2: Since the number have to be greater than 5000.The thousand place will be 5 or 7 and the rest three digits will have options of 4,3,2 outcomes respectively. $\\therefore$ Total number of outcomes n(S)=24+24=48 Step 3: Number of favorable outcomes : For a number to be divisible by 5.The unit digit should be 0 or 5.The unit digit should be 0 or 5 and thousands digits are 5 or 7. $\\therefore$ Options for tens and hundreth digits are 3 and 2. $\\therefore$ Favorable outcomes n(E)=6+6+6=18 Step 4: $\\therefore$ Required probability =$\\large\\frac{n(E)}{n(S)}$ $\\Rightarrow \\large\\frac{18}{48}$ $\\Rightarrow \\large\\frac{3}{8}$ Hence (A) is the correct answer.", "1. ## formulating this problem.. a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? 2. Originally Posted by transgalactic a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? If they are both odd, there are 20 numbers to choose from, only one of which is correct, so the probability is .. CB 3. 5 posibilities for first 4 forthe second why 20 numbers to choose from the number are 0..9 The digits are odd so we have {1, 3, 5, 7, 9} to choose them from. The first can be any of these five and the second any of the remaining 4 (since the digits are different) - so total is 5x4=20 CB", "# Air draft The numbers 1,2,3,4,5 are written on five tickets on the table. Air draft randomly shuffled the tickets and composed a 5-digit number from them. What is the probability that he passed: and, the largest possible number b, the smallest possible number c, a number divisible by five d, even number e, odd number f, a number divisible by three g, a number divisible by nine Correct result: a = 0.0083 b = 0.0083 c = 0.2 d = 0.4 e = 0.6 f = 1 g = 0 #### Solution: $a=\\frac{1}{5!}=\\frac{1}{120}=0.0083$ $b=a=0.0083=\\frac{1}{120}$ 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 Would you like to compute count of combinations? ## Next similar math problems: • Probability How probable is a randomly selected three-digit number divisible by five or seven? • PIN code PIN on Michael credit card is a four-digit number. Michael told this to his friend: • It is a prime number - that is, a number greater than 1, which is only divisible by number one and by itself. • The first digit is larger than the second. • The second d • Three digits number From the numbers 1, 2, 3, 4,", "# Air draft The numbers 1,2,3,4,5 are written on five tickets on the table. Air draft randomly shuffled the tickets and composed a 5-digit number from them. What is the probability that he passed: and, the largest possible number b, the smallest possible number c, a number divisible by five d, even number e, odd number f, a number divisible by three g, a number divisible by nine a = 0.0083 b = 0.0083 c = 0.2 d = 0.4 e = 0.6 f = 1 g = 0 ### Step-by-step explanation: $a=\\frac{1}{5!}=\\frac{1}{120}=0.0083$ $b=a=0.0083=0.0083$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Would you like to compute count of combinations? ## Related math problems and questions: • Five-digit number Anna thinks of a five-digit number that is not divisible by three or four. If he increments each digit by one, it gets a five-digit number that is divisible by three. If he reduces each digit by one, he gets a five-digit number divisible by four. If it sw • Three digits number From the numbers 1, 2, 3, 4, 5 create three-digit numbers that digits not repeat and number is divisible by 2. How many numbers are there? • One dice Calculate the probability of one dice", "# Air draft The numbers 1,2,3,4,5 are written on five tickets on the table. Air draft randomly shuffled the tickets and composed a 5-digit number from them. What is the probability that he passed: and, the largest possible number b, the smallest possible number c, a number divisible by five d, even number e, odd number f, a number divisible by three g, a number divisible by nine a = 0.0083 b = 0.0083 c = 0.2 d = 0.4 e = 0.6 f = 1 g = 0 ### Step-by-step explanation: $a=\\frac{1}{5!}=\\frac{1}{120}=0.0083$ $b=a=0.0083=\\frac{1}{120}$ We will be pleased if You send us any improvements to this math problem. Thank you! Tips to related online calculators Would you like to compute count of combinations? ## Related math problems and questions: • Five-digit number Anna thinks of a five-digit number that is not divisible by three or four. If he increments each digit by one, it gets a five-digit number that is divisible by three. If he reduces each digit by one, he gets a five-digit number divisible by four. If it sw • Three digits number From the numbers 1, 2, 3, 4, 5 create three-digit numbers that digits not repeat and number is divisible by 2. How many numbers are there? • One dice Calculate the probability of" ]

[ "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", "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({...}), }))", "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", "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!", "with x friends, how many cupcakes does each friend get?...", "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.", "to the face during the process 1 guests", "and I've no clue how to snap them out of it. But at least we've defined it. 1 guests", "• Wedding Venues Where", "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", "multiple measures within each level of party.list ? I have also removed the objection to the binomial. – Robert Long Dec 16 '19 at 18:42 • Yes there are up to three measures of party.list in each ibge7 (one per year). Thanks for that response. That is my biggest concern, and your response helped me realize I could confirm it by ensuring (1 | ibge7) + (1 | party.list) is equal to (1 | ibge7) + (1 | ibge7:SIGLA_PARTIDO). – dcoy Dec 19 '19 at 20:53", "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?", "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?", "Dave Horner's Website - Yet another perspective on things... 98 guests Rough Hits : 3310042 how did u find my site? onomastics (seen in code) \"Whenever two men meet there are really six people present. There is each man as he sees himself, each man as the other sees him, and each man as he really is.\" -William James $$e = \\sum_{n=0}^\\infty \\frac{1}{n!}$$", "adjacent seats. To shake things up a bit, they have a rule that nobody is allowed to sit next to their partner. How many seating arrangements are there for this party? I thought of this problem while watching a bad (Hollywood) movie on a bad date ;) × Problem Loading... Note Loading... Set Loading...", "events and parties *", "# 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 $$n$$.", "# 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" ]

[ "not shake hands with himself, nor Mrs. Li with herself. Therefore the number of men should be equal to 3 times the number of women PLUS 1. The number of women therefore must be ONE MORE THAN 1/4 the number of men. Let's set w = no. of Women and m = no. of Men. Then we can set up two equations: m = 3w+1 and w = (m/4)+1 Here we’re assuming that the Li’s did shake hands with each other, taking what it says literally. (Otherwise, they would have shaken hands with exactly the same people, and what is described could not be true.) The first equation could have been written as $$m-1 = 3w$$, meaning that the number of men, minus Mr. Li, is three times the number of women; and the second equation could be $$m = 4(w-1)$$, since the number of men is four times the number of women, excluding Mrs. Li. From the first equation, we can solve for w to get: w = (m - 1)/3 and substitute this into our second equation to get: (m - 1)/3 = (m/4) + 1 Let's simplify the right side by changing the 1 to 4/4 and adding. Now we have: (m - 1)/3 = (m + 4)/4 (you'll want to work out the steps", "of ways to get a $$3$$-cycle from the remaining 3 persons is $$\\frac{3!}{2\\times3}=1$$. So we get $$672$$ ways that the links can form a $$5$$-cycle and a $$3$$-cycle. Adding up the number shown in the fat sentences, we get that there are $$2520+315+672=3507$$ ways to get have 8 people do handshakes with each other if each person shakes hands with exactly 2 different people. Link each person to the two other people that they shake hands with. Then each of the $$8$$ people is part of a cycle of at least $$3$$ people. So either there is one $$8$$-cycle, or two $$4$$-cycles, or a $$3$$ cycle and a $$5$$ cycle. In your count of $$7!=5040$$, every $$8$$-cycle is counted twice (because $$(12345678)$$ is also counted separately as $$(18765432)$$ etc.). So the actual number of $$8$$-cycles is $$5040/2 = 2520$$. There are $$(7 \\times 6 \\times 5) / 2 = 105$$ different $$4$$-cycles, and for each of these the remaining $$4$$ people can be arranged in $$3$$ different $$4$$-cycles (either $$(ABCD)$$ or $$(ACBD)$$ or $$(ACDB)$$). So this adds a further $$105 \\times 3 = 345$$ configurations. There are $$(7 \\times 6 \\times 5 \\times 4) / 2 = 420$$ different $$5$$-cycles and a single $$3$$-cycle for each one. There are $$(7 \\times 6) / 2 = 21$$ different $$3$$-cycles", "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 .", "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", "the index finger • Eleven → middle link of the index finger • Twelve → lower link of the index finger ### Two-handed counting up to 60 After the first dozen has been counted using the thumb as a pointer with the three phalanges of the remaining four fingers of the same hand (4 × 3 = 12), the counting capacity of one hand is initially exhausted. • The other hand is clenched in a fist. In order to remember that a dozen has been counted, one now extends a finger, e.g. B. the thumb. • Now you continue to count by starting again at one with your first hand . At twelve , the second dozen is full. • In order to remember that two dozen have been counted, one now extends the next finger of the other hand, e.g. B. after the thumb out the index finger. • With the five fingers of the first hand you can count five times a dozen, so 5 × 12 = 60. • Now you can count the next dozen again with the first hand, i.e. count up to 72 with two hands (12 on the first plus 60 on the other hand). This finger counting system still exists in parts of Turkey , Iraq , India and Indochina", "## 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.", "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", "can find the value of $$k$$ by equating the power of $$x$$ inside the sigma with $$-2$$. \\begin{align*} &-2=10-2k\\\\ &2k=12\\\\ &k=6\\\\ \\end{align*} Step 3: Write down the coefficient of $$x$$ by substituting the value of $$k$$ $${10 \\choose 6} (2)^{10-6}=3360$$ ## Concept Check Questions 1. At a meeting involving $$5$$ men and $$8$$ women, each person shook hands once with every one person. 1. What was the total number of handshakes that took place? 2. How many handshakes were there between two women? 3. How many handshakes were there between a man and a women? 2. In how many ways can a study answer a four-option multiple choice exam with $$10$$ questions? 3. In a circular boardroom, how many ways can one arrange four female and five male colleagues if the males and females are to alternate? 4. How many different $$5$$ letter words can be made from the alphabet if replacement is not allowed? 5. Expand $$(3x-2y)^5$$ 6. Find the middle term in the expansion of $$(3a+5b)^20$$ leaving your answer in terms of $${20 \\choose k}$$ 7. Find the term independent of $$x$$ in the expansion of $$(5x+ \\frac{3}{x})^6$$ ## Concept Check Solutions 1. (i) $${16 \\choose 2}$$ (ii) $${6 \\choose 2}$$ (iii) $${10 \\choose 1} × {6 \\choose 1}$$ 2. $$4^10$$ 3. $$4!×5!$$ 4. $${26 \\choose", "# Reliance Jio Permutation and combination Quiz 1 Question 1 Time: 00:00:00 In how many different ways can the letters of the word \"MOBILE\" be arranged ? 720 720 360 360 1440 1440 2880 2880 1240 1240 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 2 Time: 00:00:00 A President and Vice - President are to be chosen out of a team having 40 Members .How many ways are there to achieve this ? 1850 1850 1870 1870 1560 1560 1910 1910 1200 1200 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 3 Time: 00:00:00 In a Marriage ceremony there are 50 People present in which 30 Males and 20 females.How many hands shake are possible if every person shakes hands with Others? 1250 1250 1225 1225 1325 1325 1520 1520 1200 1200 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 4 Time: 00:00:00 Find $$^{25}\\textrm{C}_{2}$$ 200 200 300 300 400 400 100 100 150 150 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 5 Time: 00:00:00 Seven people sit around a circular table. Find number of possible arrangements 120 120 360 360 720 720 5050 5050 5040 5040 Once you attempt the question then PrepInsta explanation will", "of three examinations -- one in Algebra, one in Geometry, and one in Calculus. For each examination, there was one mark of $x$, one mark of $y$, and one mark of $z$, where $x$, $y$, and $z$ are distinct positive integers. The total of the marks obtained by each of the students was: Alice (20), Bob (10), and Carol (9). If Bob placed first in the algebra examination, who placed second in the geometry examination? Prove your result. 19. Given any set of ten natural numbers between 1 and 99 inclusive, prove that there are two disjoint nonempty subsets of the set with equal sums of their elements. 20. I invite 10 couples to a party at my house. I ask everyone present, including my wife, how many different people they shook hands with. It turns out that everyone shook hands with a different number of people. If we assume that no one shook hands with his or her partner, how many people did my wife shake hands with? (I did not ask myself any questions.) 21. Suppose you are given 80 seemingly identical gold coins. Indeed, 79 of the coins are identical in every way -- but one of them is counterfeit and slightly lighter than the others. Using a pan balance, how can you locate the", "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", "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.", "• In how many different ways, can the letters of the word 'VENTURE' be arranged? A) 840 B) 5040 C) 1260 D) 2520 D) 2520 The required different ways = $$\\large\\frac{7!}{2!} = \\frac{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2!}{2!} = 2520$$ ##### Similar Questions 1). How many different signals , can be made by 5 flags from 8 flags of different colors? A). 6270 B). 1680 C). 20160 D). 6720 2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets? A). 12 B). 64 C). 256 D). 60 3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together? A). 10! B). 14!/(4!) C). 151200 D). 3628800 4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible? A). 120 B). 100000 C). 6720 D). 30240 5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible? A). 72 B). 144 C). 288 D). 234 6). Find the", "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", "sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 # There are 7 persons in a party. The number of hands shaken all together in the party is: 1. 21 2. 6 3. 10 4. 2 Option 1 : 21 ## Detailed Solution Given: Number of person in a party = 7 $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ Where, n = number of the person r = minimum number of person required for a handshake Calculation: $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ $$\\Rightarrow {7_{{C_2}}} = \\;\\frac{{7!}}{{2! \\times \\left( {7 - 2} \\right)!}}$$ $$\\Rightarrow \\frac{{7 \\times 6 \\times 5!}}{{2! \\times 5!}}$$ ⇒ 42/2 ⇒ 21 ∴ the number of hands shaken is 21. # A boy contains 5 pink , 3 blue & 6 black ball. How many solution of ball can be formed by taking at least one ball form the boy 1. 157 2. 187 3. 167 4. 197 5.", "handshaking into a graph where the people are vertex and handshakes are represented by edges, we will get one or more disjoint cycles. Each cycle must have at least $3$ people so that everyone shakes hand with two other. So we have $4$ cases. 1. We can take all of the people and put them in a cycle of length $9$. 2. We can divide them into $5$-cycle and $4$-cycle. 3. We can divide them into $6$-cycle and $3$-cycle. 4. We can divide them into three $3$-cycles For the first case, this is just cyclic permutation and it can be done in $\\frac{9!}{2\\times9}$ since rotation and reflection results in same handshaking. For the second case, we can divide $9$ people into group of $5$ people and $4$ people in ${9\\choose4}=\\frac{9!}{4!\\cdot5!}$ ways. For each grouping, there are $\\frac{5!}{2\\times5}\\cdot\\frac{4!}{2\\times4}$ ways to put them in the cycles. So, by multiplying this two results, we get total $\\frac{9!}{2\\times4\\times2\\times5}$ ways Similarly, for the third case, there are total $\\frac{9!}{2\\times3\\times2\\times6}$ ways to do that. And for the fourth case, there are just $\\frac{9!}{2\\times3\\times2\\times3\\times2\\times3}$ ways similarly. So, $N=9!\\left(\\frac{1}{2\\times9}+\\frac{1}{2^2\\times4\\times5}+\\frac{1}{2^2\\times3\\times6}+\\frac{1}{2^3\\times3\\times3\\times3}\\right)$ $\\Longrightarrow \\boxed{N=31416}$ In the 4th case, we need to divide by $3!$. And the rest are correct. And don't forget to divide $N$ by 1000 to find the remainder! $N=9!\\left(\\frac{1}{2\\times9}+\\frac{1}{2^2\\times4\\times5}+\\frac{1}{2^2\\times3\\times6}+\\frac{1}{2^3\\times3\\times3\\times3\\times3!}\\right)$ $\\Longrightarrow N=30016$ So, answer should be $16$.", "first person gives $5$ counters to the second. The second person then gives the first $3$ times as many as the first person holds in his hand. How many counters has the second person got in his hand?", "0 or 1 of the second, and so on. If we have a multiset with n_k copies of item k, k=1,2,\\dots,m then we can include from 0 to n_1 copies of the first item, from 0 to n_2 copies of the second, and so on, so the ... 1 We can think like there are two groups such as group A and group B Lets say that they both are in group A ,then the the probability of being A can be calculated \\frac{C(8,3) }{C(10,5)} \\times \\frac {C(5,5) }{C(5,5)}=\\frac{2}{9} .However , they might have been in group B , so we should multiply by 2 .Hence , the result is \\frac{4}{9} 1 Hint: Number of handshakes of american woman with each other would be ^5C_2=10. Absolutely. However second time you chose ^{15}C_2=10 you have again took the number of second american woman so you need to subtract ^{15}C_2-^5C_2 1 The problem is that you have counted handshakes between pairs of American women twice: once in the handshakes not involving men and once in the handshakes not involving Indian women. There are ^5C_2=10 of these, which need to be subtracted. 1 We can even make a stronger statement that$${n \\choose k} = \\Theta\\left (\\left( \\frac{n}{k} \\right)^k \\right) For the proof, see this. Only top voted, non community-wiki answers of", "are wearing gloves, and $$3/4$$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove? $$\\ 3$$ $$\\ 5$$ $$\\ 8$$ $$\\ 15$$ $$\\ 20$$ ###### Solution(s): Since our room has $$\\dfrac 25$$ of the people wearing gloves, the number of people must be a multiple of $$5.$$ Since our room has $$\\dfrac 34$$ of the people wearing hats, the number of people must be a multiple of $$4.$$ Therefore, the people in the room must be a multiple of $$20.$$ Now, we can also use the following formula by the principle of inclusion exclusion: Fraction of people wearing both = Fraction of people wearing gloves + Fraction of people wearing hats - Fraction of people wearing either. This makes our desired fraction equal to $$\\dfrac{2}{5} + \\dfrac 34 -$$ Number of people who wear either. If we wish to minimize the number of who wear both, we maximize the number of people who wear both, to make it $$1.$$ Therefore, the fraction of people that wear both is $$\\dfrac{2}{5} + \\dfrac 34- 1 = \\dfrac 3{20}.$$ Since our number is a multiple of $$20,$$ we have the number of people wearing both as $$3$$ if we choose to have just $$20$$ people. Therefore, E" ]

[ "+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", "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", "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", "attended the wedding, and averaged three drinks per person.", "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 .", "# 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", "sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 # There are 7 persons in a party. The number of hands shaken all together in the party is: 1. 21 2. 6 3. 10 4. 2 Option 1 : 21 ## Detailed Solution Given: Number of person in a party = 7 $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ Where, n = number of the person r = minimum number of person required for a handshake Calculation: $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ $$\\Rightarrow {7_{{C_2}}} = \\;\\frac{{7!}}{{2! \\times \\left( {7 - 2} \\right)!}}$$ $$\\Rightarrow \\frac{{7 \\times 6 \\times 5!}}{{2! \\times 5!}}$$ ⇒ 42/2 ⇒ 21 ∴ the number of hands shaken is 21. # A boy contains 5 pink , 3 blue & 6 black ball. How many solution of ball can be formed by taking at least one ball form the boy 1. 157 2. 187 3. 167 4. 197 5.", "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?", "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,", "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.", "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", "of three examinations -- one in Algebra, one in Geometry, and one in Calculus. For each examination, there was one mark of $x$, one mark of $y$, and one mark of $z$, where $x$, $y$, and $z$ are distinct positive integers. The total of the marks obtained by each of the students was: Alice (20), Bob (10), and Carol (9). If Bob placed first in the algebra examination, who placed second in the geometry examination? Prove your result. 19. Given any set of ten natural numbers between 1 and 99 inclusive, prove that there are two disjoint nonempty subsets of the set with equal sums of their elements. 20. I invite 10 couples to a party at my house. I ask everyone present, including my wife, how many different people they shook hands with. It turns out that everyone shook hands with a different number of people. If we assume that no one shook hands with his or her partner, how many people did my wife shake hands with? (I did not ask myself any questions.) 21. Suppose you are given 80 seemingly identical gold coins. Indeed, 79 of the coins are identical in every way -- but one of them is counterfeit and slightly lighter than the others. Using a pan balance, how can you locate the", "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", "# Moderate Permutation-Combination Solved QuestionAptitude Discussion Q. There are 20 couples in a party. Every person greets every person except his or her spouse. People of the same sex shake hands and those of opposite sex greet each other with a namaskar (It means bringing one's own palms together and raising them to the chest level). What is the total number of handshakes and namaskar's in the party? ✖ A. $760$ ✔ B. $1,140$ ✖ C. $780$ ✖ D. $720$ Solution: Option(B) is correct There are 20 men and 20 women. When a man meets a woman, there are two namaskars, whereas when a man meets a man (or a woman) there is only 1 handshake. Number of handshakes $= 2\\times {^{20}C_2}$ (men and women ) $= 2 \\times \\dfrac{20 \\times 19}{2}=380$ For number of Namskars Every man does 19 namaskars (to the 20 women excluding his wife) and they respond in the same way. $= 2\\times (20)(19)$ $= 2\\times 380$ $=760$ Total, $= 760 + 380$ $= \\textbf{1140}$ Edit: Thank you meghna for pointing our the error in the solution. Solution has been updated. ## (6) Comment(s) Tejas Patil () in ur explanation u havent consider women greeting women (they fall under category of same sex) as stated in question. solution: same sex greetings(HANDSHAKES) case1 : man", "times Note: \"a couple\" doesn't always mean exactly two, although it often does. As mentioned by Mick in the comments, thrice is quite old fashioned and while most people in the UK would understand, it's not commonly used. You might also reference the fact that 12 is also known as a dozen (and therefore 6 is half a dozen): • 6 times = Half a dozen times (or \"a half dozen times\" in the US sometimes) • 12 times = A dozen times There are also some other ways to reference numbers of \"things\" which don't really apply or work in the \"times\" example (some of which are a little archaic and more likely to be seen in literature or poetry than in everyday conversational speech/writing): • 2 ants -> a pair/duo of ants • 3 mice -> a trio/trinity of mice • 4 cats -> a quartet of cats • 5 dogs -> a quintet of dogs • 20 birds -> a score of birds • 144 eggs -> a gross of eggs (more often called 12 dozen eggs) In the UK, we also have some slang for certain amounts of money: • £5 = a fiver • £10 = a tenner • £20 = a score (like above) • £25 = a pony • £500 =", "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.", "# 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," ]

[ "\\cdot |T|$$.", "\\cdots$$", "\\ldots . . = 1 - \\frac{1}{2} = \\frac{1}{2}$", "+ \\frac{\\partial v}{\\partial h}\\cdot dh$", "\\frac{11}{9} \\cdot 3^n \\\\ &= (4 n - 7) \\cdot 3^{n - 2} \\end{align*}", "like: ${\\displaystyle {\\frac {1}{1-2^{-s}}}\\cdot {\\frac {1}{1-3^{-s}}}\\cdot {\\frac {1}{1-5^{-s}}}\\cdot {\\frac {1}{1-7^{-s}}}\\ldots }$", "\\cdot \\frac{20}{3} = \\frac{20}{3}$.", "13}+\\frac{10}{21\\cdot 13}\\approx 0.110$$", "\\cdot \\frac{x - 3}{1}$ this is the same as $\\frac{x - 3}{x + 2}$", "\\cdot x) + C = \\frac{x^2 |x|}{3} + C.$$", "4}{14\\cdot 13}\\\\ \\frac {11}{91}$$", "$\\frac{1\\cdot\\tfrac{1}{3}}{2} = \\boxed{\\textbf{(E)} \\: \\frac{1}{6}}$ ~Lemma proof by sakshamsethi", "\\cdot {10}^{- 14}$", "\\frac{4}{3} \\cdot \\frac{4}{5} \\cdot \\frac{6}{5} \\cdot \\frac{6}{7} \\cdot \\frac{8}{7} \\cdot \\frac{8}{9} \\cdots$", "\\cdot 767.1 \\sqrt{\\frac{1}{14} + \\frac{1}{11}}$ $\\boxed{910 \\pm 639.47} \\text{ or } \\boxed{(270.53, 1549.47)}$", "- \\ \\frac{1}{3} \\cdot [-1] \\ ) \\ = \\ \\frac{2}{3} \\ \\ ,$$ with the area enclosed by the complete curve being four times this. -", "f)\\cdot \\frac{df}{dt} \\\\ = \\frac{f'}{f} \\\\ = \\frac{f'(t)}{f(t)}.$$", "\\cdots) ).$$", "\\cdot \\frac{m_3}{m_2} \\cdot \\frac{m_1}{m_3} = 1$$", "simply $$\\frac{\\partial z}{\\partial x} = f'(x^2 + y^2) \\cdot 2x$$" ]

[ "indicated, we currently are only ready to prove this theorem for the integer powers. We start with positive integers. We have already proved the result for $p=1, 2, 3, 4$ as part of our discovering the pattern. As we developed that pattern, we discovered that we were using a recursive argument each time. The powers $p=3$ and $p=4$ were based on knowing the results for $p=2$ and $p=3\\text{,}$ respectively. Our proof builds on this recursive argument to create a general statement. Suppose that we know for $p=n\\text{,}$ \\begin{equation*} \\frac{d}{dx}[x^n] = n x^{n-1}\\text{.} \\end{equation*} Next consider the power $p=n+1\\text{,}$ and rewrite it $x^{n+1} = x \\cdot x^{n}\\text{.}$ The product rule guarantees \\begin{align*} \\frac{d}{dx}[x^{n+1}] &= \\frac{d}{dx}[x \\cdot x^n]\\\\ &= \\frac{d}{dx}[x] \\cdot x^n + x \\cdot \\frac{d}{dx}[x^n]\\\\ &= 1 \\cdot x^n + x \\cdot nx^{n-1}\\\\ &= x^n + nx^n\\\\ &= (n+1)x^n. \\end{align*} This general recursive argument shows that if the rule is satisfied for an initial value of $p=n\\text{,}$ the statement will be immediately known to be true for the sequence of values $p \\in (n, n+1, n+2, \\ldots)\\text{.}$ Having earlier shown the rule was true for $p=1\\text{,}$ the recursive argument shows it will be true for all positive integers. In addition, because we know the rule is true for $p=\\frac{1}{2}\\text{,}$ the same argument shows that the rule is true for", "motivation, notice that since the divisors of a prime power $p^a$ are given by $\\{1,p,p^2,\\cdots,p^a\\}$, we have (13) \\begin{align} \\begin{split} \\sigma(p^a) &= 1 + p + p^2 + \\cdots + p^a = \\sum_{i=0}^a p^i. \\end{split} \\end{align} This is a geometric series, so we can use the formula for the sum of a geometric series to compute (14) \\begin{align} \\begin{split} \\sigma(p^a) &= \\sum_{i=0}^a p^i = \\frac{p^{a+1}-1}{p-1}. \\end{split} \\end{align} Corollary: For a number $n = p_1^{a_1}\\cdots p_k^{a_k}$, we have $\\displaystyle \\sigma(n) = \\prod_{i=1}^k \\frac{p_i^{a_i+1}-1}{p_i-1}$. #### Example: Summing divisors of some larger numbers Suppose we wanted to compute $\\sigma(28)$. Since $28 = 4 \\cdot 7 = 2^2 \\cdot 7^1$, our formula gives (15) \\begin{align} \\sigma(28) = \\left(\\frac{2^3-1}{2-1}\\right)\\left(\\frac{7^2-1}{7-1}\\right) = 7 \\cdot 8 = 56. \\end{align} Likewise, if we want to compute $\\sigma(120)$, then since $120 = 8 \\cdot 3 \\cdot 5 = 2^3 \\cdot 3^1 \\cdot 5^1$ we have (16) \\begin{align} \\sigma(120) = \\left(\\frac{2^4-1}{2-1}\\right)\\left(\\frac{3^2-1}{3-1}\\right)\\left(\\frac{5^2-1}{5-1}\\right) = 360. \\end{align} $\\square$ Carl pointed out that if a prime appears to the first power in the prime decomposition of n, then the corresponding term in $\\sigma(n)$ is just $p+1$: (17) \\begin{align} \\left(\\frac{p^2-1}{p-1}\\right) = p+1. \\end{align} Note only does this observation match up with our formula, but it also follows from the definition of $\\sigma$: if p is prime then its only divisors are 1 and p,", "25 \\equiv 4 \\mod{11}. \\end{split} Notice that there are 2 residues on this list which are greater than $\\frac{11}{2} = 5.5$, and so Gauss' Lemma says that (6) \\begin{align} \\left(\\frac{5}{11}\\right) = (-1)^2 = 1. \\end{align} $\\square$ Proof of Gauss' Lemma: To start off, we'll split the least non-negative residues from the set (7) \\begin{align} 1\\cdot a, 2\\cdot a, \\cdots, \\frac{p-1}{2} a \\end{align} into two groups: the set $r_1,\\cdots, r_n$ will be the set of those residues that are greater than $\\frac{p}{2}$, and the set $s_1,\\cdots, s_m$ of residues which are less than $\\frac{p}{2}$. Notice that none of these residues actually hit $\\frac{p}{2}$ on the nose, since this number is not an integer. So really we have $\\{r_1,\\cdots, r_n\\} \\subseteq \\{\\frac{p+1}{2},\\cdots,p-1\\}$ and $\\{s_1,\\cdots, s_m\\}\\subseteq \\{1,\\cdots,\\frac{p-1}{2}\\}$. Now we claim that the integers (8) \\begin{align} p-r_1,p-r_2,\\cdots,p-r_n,s_1,s_2,\\cdots,s_m \\end{align} are the same modulo p as the integers (9) \\begin{align} 1,2,3,\\cdots,\\frac{p-1}{2}. \\end{align} To see that this is true, notice that the integers from 8 are all between 1 and $\\frac{p-1}{2}$ — the $s_j$ certainly live in this range, but since we have $\\frac{p+1}{2}\\leq r_i \\leq p-1$ we also have $1 \\leq p-r_i \\leq \\frac{p-1}{2}$. So to show that these collections are the same, we just need to show that the top collection has no repeats. For this, we'll check three things: 1. there are", "integer. Solution Using the Product Rule, we can write \\begin{align} y’ & =(\\underbrace{xx\\cdots x}_{n-1\\text{ times}})\\frac{dx}{dx}+\\cdots+(\\underbrace{xx\\cdots x}_{n-1\\text{ times}})\\frac{dx}{dx}\\\\ & =\\underbrace{x^{n-1}\\cdot1+\\cdots+x^{n-1}\\cdot1}_{n\\text{ times}}\\\\ & =nx^{n-1}.\\end{align} Example 3 Find the derivative of $$y$$ with respect to $$x$$ if $y=4x^{3}-2x^{2}+6x+5.$ Solution \\begin{align} y’ & =4\\times3x^{2}-2\\times2x+6+0.\\\\ & =12x^{2}-4x+6.\\end{align} Example 4 Prove $${\\displaystyle \\frac{d}{dx}x^{n}=nx^{n-1}}$$, where $$n$$ is a negative integer. Solution Let $$n=-m$$, where $$m$$ is a positive integer. Thus $x^{n}=x^{-m}=\\frac{1}{x^{m}}.$ Using the Quotient Rule, we can write \\begin{align} \\frac{d}{dx}\\frac{1}{x^{m}} & =\\frac{0\\times x^{m}-mx^{m-1}\\times1}{\\left(x^{m}\\right)^{2}}\\\\ & =\\underbrace{-m}_{=n}\\frac{x^{m-1}}{x^{2m}}\\\\ & =nx^{m-1-2m}\\\\ & =nx^{-m-1}\\\\ & =nx^{n-1}.\\tag{-m=n}\\end{align} That is, $\\frac{d}{dx}x^{n}=nx^{n-1}$ Example 5 Given $${\\displaystyle y=\\frac{1}{x^{3}}}$$, find $$\\dfrac{dy}{dx}$$. Solution Because $$y=x^{-3}$$, \\begin{align} \\frac{d}{dx}y & =-3x^{-3-1}\\\\ & =-3x^{-4}\\\\ & =-\\frac{3}{x^{4}}.\\end{align} Example 6 Given $$y=\\sqrt{x}$$ ($$x>0$$), find $$\\dfrac{dy}{dx}$$. Solution Let $$f(x)=\\sqrt{x}$$ and $$g(x)=f(x)f(x)=x$$. \\begin{align} g'(x) & =f'(x)f(x)+f(x)f'(x)\\\\ & =2f(x)f'(x)\\end{align} We know $$g'(x)=1$$ (Theorem 1). Thus \\begin{align} 1 & =2f(x)f'(x)\\\\ & =2\\sqrt{x}f'(x)\\end{align} Finally $f'(x)=\\frac{1}{2\\sqrt{x}}.$ Example 7 Evaluate $${\\displaystyle \\frac{d}{dx}\\sqrt{1-3x^{2}}}$$. Solution Let $$u=1-3x^{2}$$ and $$y=\\sqrt{u}$$. Then \\begin{align} \\frac{dy}{dx} & =\\frac{dy}{du}\\frac{du}{dx}\\\\ & =\\left(\\frac{1}{2\\sqrt{u}}\\right)(-6x)\\\\ & =\\frac{-6x}{2\\sqrt{1-3x^{2}}}\\\\ & =\\frac{-3x}{\\sqrt{1-3x^{2}}}.\\end{align} Example 8 Find $$\\dfrac{dy}{dx}$$ if $$y=x^{1/n}$$ where $$n$$ is an integer. Solution Let $$f(x)=x^{1/n}$$ and $$g(x)=[f(x)]^{n}=\\underbrace{f(x)\\cdots f(x)}_{n\\text{ times}}=x^{n/n}=x$$. Using the Product Rule, we have \\begin{align} g'(x) & =\\underbrace{[f(x)]^{n-1}f'(x)+\\cdots+[f(x)]^{n-1}f'(x)}_{n\\text{ times}}\\\\ & =n[f(x)]^{n-1}f'(x)\\\\ 1 & =n\\left[x^{1/n}\\right]^{n-1}f'(x).\\end{align} Thus \\begin{align} f'(x) & =\\frac{1}{nx^{(n-1)/n}}\\\\ & =\\frac{1}{nx^{1-1/n}}\\\\ & =\\frac{1}{n}x^{\\frac{1}{n}-1}.\\end{align} That is $\\frac{d}{dx}x^{\\frac{1}{n}}=\\frac{1}{n}x^{\\frac{1}{n}-1}.$ Example 9 Evaluate $\\frac{d}{dx}\\sqrt[5]{4x^{2}-\\frac{3}{x}}.$ Solution Let $$y=u^{1/5}$$ and $$u=4x^{2}-\\frac{3}{x}$$. Then \\begin{align} \\frac{dy}{du} & =\\frac{1}{5}u^{\\frac{1}{5}-1}\\\\", "proof of the power rule for negative integers. POWER RULE (VERSION 2) Let $f(x)=x^n$ for a negative integer $n$. Then, $f^{\\prime}(x)=n x^{n-1}$. For the proof, let $n$ be a negative integer. Then $-n$ is a positive integer (if $n=-3$, then $-n=3$ ), and the power rule can be applied to $x^{-n}$. Then, \\begin{aligned} \\frac{d}{d x} x^n &=\\frac{d}{d x}\\left(\\frac{1}{x^{-n}}\\right)=\\frac{x^{-n} \\cdot 0-1 \\cdot-n x^{-n-1}}{\\left(x^{-n}\\right)^2} \\ &=\\frac{n x^{-n-1}}{x^{-2 n}}=n x^{-n-1-(-2 n)}=n x^{n-1} . \\end{aligned} Example 13 Find $f^{\\prime \\prime}(x)$ for $f(x)=\\frac{1}{4 x}$. Solution We could use the quotient rule to find the derivative, but we can also use the power rule if we rewrite the function using a negative exponent: \\begin{aligned} f(x) &=\\frac{1}{4} x^{-1} \\ f^{\\prime}(x) &=\\frac{1}{4}\\left(-1 \\cdot x^{-2}\\right)=-\\frac{1}{4} x^{-2}=-\\frac{1}{4 x^2} \\end{aligned} The last step, expressing the derivative in terms of a fraction instead of a negative exponent, is not necessary but can be helpful for interpreting the result. On the other hand, thinking of $f^{\\prime}(x)=-\\frac{1}{4} x^{-2}$ is helpful for taking the second derivative using the power rule: $$f^{\\prime \\prime}(x)=-\\frac{1}{4}\\left(-2 x^{-3}\\right)=\\frac{1}{2} x^{-3}=\\frac{1}{2 x^3} .$$ Notice that as we take derivatives, the exponent in the denominator is getting larger, not smaller. A polynomial’s derivatives eventually reach zero, but this function’s derivatives never reach zero. ## 数学代写|微积分代写Calculus代写|Equations of tangent lines revisited The procedure we used in section $1.8$ to find the equation of", "{2}{7} + \\frac{4}{7^2} + \\frac{7}{7^3} + \\frac{11}{7^4} + \\cdots$ is pretty good too! – user26872 Jul 18 '12 at 7:15 Note that starting with your first element, $a_0=1$; to get to $a_1=2$ we sum $1$, to get to $a_2=4$, we sum $2$, to get to $a_3=7$, we sum $3$. So in general, we can say that $$a_{n}=a_{n-1}+n$$ This is \\begin{align} 2&=1+1 \\\\[8pt] 4& =2+2\\\\[8pt] 7&=4+3\\\\[8pt] 11& =7+4\\\\[8pt] \\cdots&=\\cdots\\\\[8pt] a_n&=a_{n-1}+n \\end{align} We can get several solutions to you problem, I will share $2$: Solution 1 Use the recursion to obtain a closed formula: Since we know that $a_{n}=a_{n-1}+n$ we can write $a_{n-1}=a_{n-2}+n-1$ so we get $$a_{n}=a_{n-2}+(n-1)+n$$ Repeating this process, we get that $$a_{n}=a_{n-3}+(n-2)+(n-1)+n$$ $$a_{n}=a_{n-4}+(n-3)+(n-2)+(n-1)+n$$ $$a_{n}=a_{n-5}+(n-4)+(n-3)+(n-2)+(n-1)+n$$ ...so in general we can say that $$a_n=a_{n-k}+(n-k+1)+\\cdots+n$$ for any $k$ a natural number. (actually we should be proving the above by induction, but let it be) So we can choose $k=n$, which means... $$a_n=a_{n-n}+(n-n+1)+\\cdots+n$$ $$a_n=a_0+(1+\\cdots+n)$$ $$a_n=1+\\frac{n(n+1)}{2}$$ Solution 2 Starting from $$a_{n}=a_{n-1}+n$$ we use generating functions: \\eqalign{ & \\sum\\limits_{n = 1}^\\infty {{a_n}} {x^n} = \\sum\\limits_{n = 1}^\\infty {{a_{n - 1}}} {x^n} + \\sum\\limits_{n = 1}^\\infty n {x^n} \\cr & \\sum\\limits_{n = 1}^\\infty {{a_n}} {x^n} = x\\sum\\limits_{n = 1}^\\infty {{a_{n - 1}}} {x^{n - 1}} + x\\sum\\limits_{n = 1}^\\infty {n{x^{n - 1}}} \\cr & \\sum\\limits_{n = 0}^\\infty {{a_n}} {x^n} - {a_0} = x\\sum\\limits_{n = 0}^\\infty", "\\end{split}$ As $$y''-2xy'+2ny = 0$$ we have $(k+2)\\,(k+1) \\, a_{k+2} + ( - 2k+ 2n) a_k = 0 , \\qquad \\text{or} \\qquad a_{k+2} = \\frac{(2k-2n)}{(k+2)(k+1)} a_k .$ This recurrence relation actually includes $$a_2 = -na_0$$ (which comes about from $$2a_2+2na_0 = 0). Again \\(a_0$$ and $$a_1$$ are arbitrary. \\begin{align*} & a_2 = \\frac{-2n}{(2)(1)}a_0, \\qquad a_3 = \\frac{2(1-n)}{(3)(2)} a_1, \\\\ & a_4 = \\frac{2(2-n)}{(4)(3)} a_2 = \\frac{2^2(2-n)(-n)}{(4)(3)(2)(1)} a_0 , \\\\ & a_5 = \\frac{2(3-n)}{(5)(4)} a_3 = \\frac{2^2(3-n)(1-n)}{(5)(4)(3)(2)} a_1 , \\quad \\ldots \\end{align*} Let us separate the even and odd coefficients. We find that \\begin{align*} a_{2m} &=\\frac{2^m(-n)(2-n)\\cdots(2m-2-n)}{(2m)!} , \\\\ a_{2m+1} &=\\frac{2^m(1-n)(3-n)\\cdots(2m-1-n)}{(2m+1)!} . \\end{align*} Let us write down the two series, one with the even powers and one with the odd. \\begin{align*} y_1(x) & = 1+\\frac{2(-n)}{2!} x^2 + \\frac{2^2(-n)(2-n)}{4!} x^4 + \\frac{2^3(-n)(2-n)(4-n)}{6!} x^6 + \\cdots , \\\\ y_2(x) & = x+\\frac{2(1-n)}{3!} x^3 + \\frac{2^2(1-n)(3-n)}{5!} x^5 + \\frac{2^3(1-n)(3-n)(5-n)}{7!} x^7 + \\cdots . \\end{align*} We then write $y(x) = a_0 y_1(x) + a_1 y_2(x) .$ We also notice that if $$n$$ is a positive even integer, then $$y_1(x)$$ is a polynomial as all the coefficients in the series beyond a certain degree are zero. If $$n$$ is a positive odd integer, then $$y_2(x)$$ is a polynomial. For example, if $$n=4$$, then $y_1(x) = 1 + \\frac{2(-4)}{2!} x^2 + \\frac{2^2(-4)(2-4)}{4!} x^4 = 1 -", "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", "\\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} • \\begin{align*} \\frac{-3x^{5}} {x^{5}} &= -3 \\cdot \\frac{ x^{5}} {x^{5}} \\\\ &= -3 \\cdot x^{5-5} \\\\ &= -3 \\cdot x^{0} \\\\ &= -3 \\cdot 1 \\\\ &= -3 \\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} • \\begin{align*} \\frac{(j^{2}k)^{4}} {(j^{2}k) \\cdot (j^{2}k)^{3}} &= \\frac{(j^{2}k)^{4}} {(j^{2}k)^{1+3}} \\\\ &= \\frac{(j^{2}k)^{4}} {(j^{2}k)^{4}} \\\\ &= (j^{2}k)^{4-4} \\\\ &= (j^{2}k)^{0} \\\\ &= 1 \\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} • \\begin{align*} \\frac {5(rs^{2})^{2}} {(rs^{2})^{2}} &= 5(rs^{2})^{2-2} \\\\ &= 5(rs^{2})^{0} \\\\ &= 5 \\cdot 1 \\\\ &= 5 \\\\ \\ \\ \\ \\ \\\\ \\ \\ \\ \\ \\\\ \\end{align*} ## 1.2.5 Using the Negative Rule of Exponents Another useful result occurs if we relax the condition that $m>n$ in the quotient rule even further. For example, can we simplify $\\frac { h^{3}} {h^{5}}$? When $m<n$—that is, where the difference $m-n$ is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, $\\frac { h^{3}} {h^{5}}$. \\begin {align*} \\frac { h^{3}} {h^{5}} &= \\frac {h \\cdot h \\cdot h} {h \\cdot h \\cdot h \\cdot h \\cdot h} \\\\ &= \\frac { \\cancel{h} \\cdot", "of positive integers \\begin{aligned} A - \\frac{B^2+B}{2} + \\frac{n^2+n}{2} &= [\\cdots (a-2) + (a-1) + a + (a+1) + (a+2) +\\cdots] \\\\&\\hspace{7mm}- [u + (u+1) + (u+2) \\cdots] + [1 + 2 + 3 \\cdots + n] \\\\&= 0 \\end{aligned} we can observe a property as follows $$A = \\frac{(B-n)(B+n+1)}{2}$$ Among the numerous cases of $$A = a \\times b$$, only $$A \\in$$ {prime number} $$\\bigcup$$ {perfect power of 2} are represented as $$A = 1 \\times A$$. The reason for a prime number being represented as $$A = 1 \\times A$$ is obvious. The reason for the case when $$A$$ is a perfect power of 2, $$2^e$$ can be represented only as a multiplication between powers of 2, which are all even, unless $$2^e = 1 \\times 2^e$$. The procedure of algorithm repeats the addition and subtraction of positive integers which let the value of D (the RHS of equation (1)) to oscillate between 0, until it reaches 0. So the longest expected computation time for the algorithm of $$A$$ is the case when $$A$$ is represented as $$A = \\frac{[A-(A-1)][A+(A-1)+1]}{2}$$ or $$A = \\frac{[(\\frac{A-1}{2}+1)-(\\frac{A-1}{2}-1)][(\\frac{A-1}{2}+1)+(\\frac{A-1}{2}-1)+1]}{2}$$ The fact that $$2^e$$ can be only represented as $$2^e = 1 \\times 2^e$$, particularly as $$2^e = \\frac{[2^e - (2^e-1)][2^e + (2^e - 1)+1]}{2}$$ gives us a good tool to assume", "There are four 1-tuples from a four-element set. 50 = │ { () } │ = 1. There is exactly one empty tuple. ### Negative integer exponents By definition, raising a nonzero number to the −1 power produces its reciprocal: $a^{-1} = \\frac{1}{a}.$ One also defines $a^{-n} = \\frac{1}{a^n}$ for any nonzero a and any positive integer n. Raising 0 to a negative power would imply division by 0, so it is left undefined. The definition of an for nonzero a is made so that the identity aman = am+n, initially true only for nonnegative integers m and n, holds for arbitrary integers m and n. In particular, requiring this identity for m = −n is requiring $a^{-n} \\, a^{n} = a^{-n\\,+\\,n} = a^0 = 1,$ where a0 is defined above, and this motivates the definition an = 1/an shown above. 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 The most important identity satisfied by integer exponentiation is $a^{m + n} = a^m \\cdot a^n$ This identity has the consequence $a^{m - n} =\\frac{a^m}{a^n}$ for a ≠ 0, and $(a^m)^n = a^{m\\cdot n}$ Another basic identity is $(a \\cdot b)^n = a^n \\cdot b^n$ While", "b}$$ We just add exponents. Look back at the example and make sure you understand why. Now division, again with powers of two: $$\\frac{2^3}{2^2} = \\frac{2\\cdot 2\\cdot 2}{2\\cdot 2} = 2^1 = 2^{3 - 2}$$ So to divide powers of the same base, we just subtract exponents. $$\\frac{x^a}{x^b} = x^{a - b}$$ And finally, let's raise a power to a power: \\begin{align} (2^3)^2 &= (2 \\cdot 2 \\cdot 2)^2 \\\\ &= (2 \\cdot 2 \\cdot 2)(2 \\cdot 2 \\cdot 2) \\\\ &= 2^3 \\cdot 2^3 = 2^{3 \\cdot 2} \\end{align} To raise a power to a power, simply multiply exponents: $$(x^a)^b = x^{a \\cdot b}$$ Here is a table of the laws of exponents. Tap/click it to download a copy for your notes. Negative exponents Let's begin with the exponent -1. It means \"take the reciprocal.\" So It's just that simple. We define x-1 to mean 1/x. Now we can use the power law of exponents to extend that property to any negative exponent. For example, here are two ways of looking at x-2: \\begin{align} x^{-2} &= (x^{-1})^2 = \\left( \\frac{1}{x} \\right)^2 = \\frac{1}{x^2} \\\\ \\\\ &= (x^2)^{-1} = \\frac{1}{x^2} \\end{align} In the first way we take the square of 1/x, and in the second we take the inverse of x2. Both lead to the same result. You", "\\cdots + 2 +1\\notag\\\\ &= 2^{m} - 1 = n - 1\\notag\\end{align} From now on we keep $n$ always as power of $2$ i.e. $n = 2^{m}$. Then the highest power of $2$ in term $2^{k}\\cdot n!/k!$ is at least $k + n - 1 - k = n - 1$. It follows now that sum $S$ is an integer divisible by $2^{n - 1}$. Similarly $n!a$ is also divisible by $2^{n - 1}$. Thus dividing equation $(1)$ by $2^{n - 1}$ we get \\begin{align}\\frac{n!a}{2^{n - 1}} &= b\\cdot \\frac{S}{2^{n - 1}} + b\\cdot\\frac{R}{2^{n - 1}}\\notag\\\\ A &= b.S' + bR'\\tag{2}\\end{align} where $A$ and $S'$ are integers and $R' = R/(2^{n - 1})$ and $bR'$ is also an integer because of the above equation. We have earlier proved that $$\\frac{2^{n + 1}}{n + 1} < R < \\frac{2^{n + 1}}{n - 1}$$ and therefore \\begin{align} 0 &< \\frac{4b}{n + 1} < b\\frac{R}{2^{n - 1}} < \\frac{4b}{n - 1}\\notag\\\\ \\Rightarrow 0 &< bR' < \\frac{4b}{n - 1} < 1\\notag\\end{align} provided $n > 4b + 1$ and $n$ a power of $2$ say $n = 2^{m}$. It now follows that we can choose a suitable value of $n$ such that $0 < bR' < 1$ and hence $bR'$ is not an integer. This is the contradiction we needed to achieve", "relax the condition that $m>n$ in the quotient rule even further. For example, can we simplify $\\dfrac{{h}^{3}}{{h}^{5}}$? When $m<n$—that is, where the difference $m-n$ is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, $\\dfrac{{h}^{3}}{{h}^{5}}$. \\begin{align} \\frac{{h}^{3}}{{h}^{5}}& = \\frac{h\\cdot h\\cdot h}{h\\cdot h\\cdot h\\cdot h\\cdot h} \\\\ & = \\frac{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}}{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}\\cdot h\\cdot h} \\\\ & = \\frac{1}{h\\cdot h} \\\\ & = \\frac{1}{{h}^{2}} \\end{align} If we were to simplify the original expression using the quotient rule, we would have \\begin{align} \\frac{{h}^{3}}{{h}^{5}}& = {h}^{3 - 5} \\\\ & = {h}^{-2} \\end{align} Putting the answers together, we have ${h}^{-2}=\\dfrac{1}{{h}^{2}}$. This is true for any nonzero real number, or any variable representing a nonzero real number. A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa. ${a}^{-n}=\\dfrac{1}{{a}^{n}} \\text{ and } {a}^{n}=\\dfrac{1}{{a}^{-n}}$ We have shown that the exponential expression ${a}^{n}$ is defined when $n$ is a natural number, 0, or the negative of a natural number. That means that ${a}^{n}$ is defined for any integer $n$. Also, the product and quotient rules and all of the rules we will look at soon hold for", "# binomial proof of positive integer power rule We will use the difference quotient in this proof of the power rule for positive integers. Let $f(x)=x^{n}$ for some integer $n\\geq 0$. Then we have $\\displaystyle f^{\\prime}(x)=\\lim_{h\\rightarrow 0}\\frac{(x+h)^{n}-x^{n}}{h}.$ We can use the binomial theorem to expand the numerator $\\displaystyle f^{\\prime}(x)=\\lim_{h\\rightarrow 0}\\frac{C_{0}^{n}x^{0}h^{n}+C_{% 1}^{n}x^{1}h^{n-1}+\\cdots+C_{n-1}^{n}x^{n-1}h^{1}+C_{n}^{n}x^{n}h^{0}-x^{n}}{h}$ where $C_{k}^{n}=\\frac{n!}{k!(n-k)!}$. We can now simplify the above $\\displaystyle f^{\\prime}(x)$ $\\displaystyle=\\lim_{h\\rightarrow 0}\\frac{h^{n}+nxh^{n-1}+\\cdots+nx^{n-1}h+x^{n% }-x^{n}}{h}$ $\\displaystyle=\\lim_{h\\rightarrow 0}(h^{n-1}+nxh^{n-2}+\\cdots+nx^{n-1})$ $\\displaystyle=nx^{n-1}$ $\\displaystyle=nx^{n-1}.$ Title binomial proof of positive integer power rule BinomialProofOfPositiveIntegerPowerRule 2013-03-22 12:29:43 2013-03-22 12:29:43 mathcam (2727) mathcam (2727) 8 mathcam (2727) Proof msc 26A03", "than $m(m+1)(m+2)(m+3)$. Think symmetry! \\begin{align} (m - \\tfrac 3 2)(m - \\tfrac 1 2)(m + \\tfrac 1 2)(m + \\tfrac 3 2) &= (m^2 - \\tfrac 9 4)(m^2 - \\tfrac 1 4) \\\\ &= m^4 - \\tfrac 5 2 m^2 + \\tfrac{9}{16} \\\\ &= (m^2 - \\tfrac 5 4)^2 - 1 \\end{align} Finally, if $m - \\frac 1 2$ is an integer, $m = k + \\frac 1 2$ where $k$ is an integer. Therefore $m^2 - \\frac 5 4 = k^2 + k - 1$ is also an integer. I will show that $16m(m+1)(m+2)(m+3) = 16(m^2 + 3m + 1)^2 - 16$ It will follow that $$m(m+1)(m+2)(m+3) = (m^2 + 3m + 1)^2 - 1$$ Proof: \\begin{align} 16 m(m+1)(m+2)(m+3) &=(2m)(2m+2)(2m+4)(2m+6)\\\\ & ---\\text{let $2m = n-3$}---\\\\ &=(n-3)(n-1)(n+1)(n+3)\\\\ &=(n-3)(n+3) \\cdot (n-1)(n+1) \\\\ &=(n^2 - 9)(n^2 - 1) \\\\ &=n^4 - 10n^2 + 9 \\\\ &=(n^2 - 5)^2 - 16 \\\\ & ---\\text{Note $n = 2m+3$}---\\\\ &=(4m^2 + 12m + 4)^2 - 16\\\\ &=16(m^2 + 3m + 1)^2 - 16\\\\ \\end{align}", "the regular integer space, we’d carry out arithmetic operations like so: \\begin{aligned} 2 + 3 &= 5 \\\\ 2 \\times 3 &= 6 \\end{aligned} However, in $$\\mathbb{Z}_m$$, this works a bit differently. Let’s take again $$\\mathbb{Z}_4$$ again. How would the above operations look? \\begin{aligned} 2 + 3 &= 5 = 1 \\\\ 2 \\times 3 &= 6 = 2 \\end{aligned} We simplify the results of operations to the base member of their congruence class - in other words, those values between 0 and $$m - 1$$. We’ll use this to solve our next problem. By working modulo 7 ($$\\mathbb{Z}_7$$), we need to prove that $$2^{n + 2} + 3^{2n + 1}$$ is divisible by 7. The way to approach this problem is simply to try and simplify each part of the expression in modulo 7. Starting with $$2^{n + 2}$$, we have: \\begin{aligned} 2^{n + 2} &= 2^2 \\cdot 2^n (\\mathrm{mod} \\textrm{ 7}) \\\\ &= 4 \\cdot 2^n \\end{aligned} This does not simply further in modulo 7. Looking now at $$3^{2n + 1}$$: \\begin{aligned} 3^{2n + 1} &= 3 \\cdot 3^{2n} (\\mathrm{mod} \\textrm{ 7}) \\\\ &= 3 \\cdot 9^n \\\\ &= 3 \\cdot 2^n \\end{aligned} We now add these two simplified results back together to get: \\begin{aligned} 4(2^n) + 3(2^n) = 7(2^n) (\\mathrm{mod} \\textrm{ 7}) \\end{aligned} Therefore in", "is very similar. Although it is often unwise to draw general conclusions from specific examples, we note that when we differentiate $f(x)=x^3$, the power on $x$ becomes the coefficient of $x^2$ in the derivative and the power on $x$ in the derivative decreases by 1. The following theorem states that this power rule holds for all positive integer powers of $x$. We will eventually extend this result to negative integer powers. Later, we will see that this rule may also be extended first to rational powers of $x$ and then to arbitrary powers of $x$. Be aware, however, that this rule does not apply to functions in which a constant is raised to a variable power, such as $f(x)=3^x$. ### The Power Rule Let $n$ be a positive integer. If $f(x)=x^n$, then $f^{\\prime}(x)=nx^{n-1}$. Alternatively, we may express this rule as $\\frac{d}{dx}(x^n)=nx^{n-1}$. ## Proof For $f(x)=x^n$ where $n$ is a positive integer, we have $f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{(x+h)^n-x^n}{h}$. Since $(x+h)^n=x^n+nx^{n-1}h+\\binom{n}{2}x^{n-2}h^2+\\binom{n}{3}x^{n-3}h^3+\\cdots+nxh^{n-1}+h^n$, we see that $(x+h)^n-x^n=nx^{n-1}h+\\binom{n}{2}x^{n-2}h^2+\\binom{n}{3}x^{n-3}h^3+\\cdots+nxh^{n-1}+h^n$. Next, divide both sides by $h$: $\\large \\frac{(x+h)^n-x^n}{h}=\\frac{nx^{n-1}h+\\binom{n}{2}x^{n-2}h^2+\\binom{n}{3}x^{n-3}h^3+\\cdots+nxh^{n-1}+h^n}{h}$. Thus, $\\frac{(x+h)^n-x^n}{h}=nx^{n-1}+\\binom{n}{2}x^{n-2}h+\\binom{n}{3}x^{n-3}h^2+\\cdots+nxh^{n-2}+h^{n-1}$. Finally, $\\begin{array}{ll}f^{\\prime}(x) & =\\underset{h\\to 0}{\\lim}(nx^{n-1}+\\binom{n}{2}x^{n-2}h+\\binom{n}{3}x^{n-3}h^2+\\cdots+nxh^{n-1}+h^n) \\\\ & =nx^{n-1} \\end{array} _\\blacksquare$ ### Applying the Power Rule Find the derivative of the function $f(x)=x^{10}$ by applying the power rule. #### Solution Using the power rule with $n=10$, we obtain $f^{\\prime}(x)=10x^{10-1}=10x^9$. Find the derivative of $f(x)=x^7$. #### Hint Use the", "using the law \\begin{align}\\dfrac{2^5}{2^3}&= 2^{5-3}\\\\&=2^2\\end{align} \\begin{aligned} \\frac{2^{5}}{2^{3}} &=\\frac{2 \\cdot 2 \\cdot 2 \\cdot 2}{2 \\cdot 2 \\cdot 2} \\\\ &=2^{2} \\end{aligned} Law of Zero Exponent: $$\\mathbf{a^0=1}$$ Using the law Without using the law $$2^0 =1$$ \\begin{aligned}2^{0} &=2^{5-5} \\\\&=\\frac{2^{5}}{2^{5}}\\left[\\because \\frac{a^{m}}{a^{n}}=a^{m-n}\\right] \\\\&=\\frac{2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2}{2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2} \\\\&=1\\end{aligned} Law of Negative Exponent: $$\\mathbf{a^{-m} = \\dfrac{1}{a^m}}$$ Using the law Without using the law $$2^{-2} = \\dfrac{1}{2^2}$$ \\begin{aligned} 2^{-2} &=2^{0-2} \\\\ &=\\frac{2^{0}}{2^{2}}\\left[\\because \\frac{a^{m}}{a^{n}}=a^{m-n}\\right] \\\\ &=\\frac{1}{2^{2}}\\left[\\because 2^{0}=1\\right] \\end{aligned} Law of Power of a Power: $$\\mathbf{(a^m)^n=a^{mn}}$$ Using the law Without using the law $$(2^2)^3 = 2^6$$ \\begin{aligned} \\left(2^{2}\\right)^{3} &=\\left(2^{2}\\right)\\left(2^{2}\\right)\\left(2^{2}\\right) \\\\ &=(2 \\cdot 2)(2 \\cdot 2)(2 \\cdot 2) \\\\ &=2^{6} \\end{aligned} Law of Power of a Product: $$\\mathbf{(ab)^m=a^mb^m}$$ Using the law Without using the law $$(xy)^3 = x^3y^3$$ \\begin{aligned} (x y)^{3} &=(x y)(x y)(x y) \\\\ &=(x x x)(y y y) \\\\ &=x^{3} y^{3} \\end{aligned} Law of Power of a Quotient: $$\\mathbf{\\left(\\dfrac{a}{b}\\right)^m= \\dfrac{a^m}{b^m}}$$ Using the law Without using the law $$\\left(\\dfrac{x}{y}\\right)^3= \\dfrac{x^3}{y^3}$$ \\begin{aligned} \\left(\\frac{x}{y}\\right)^{3} &=\\frac{x}{y} \\cdot \\frac{x}{y} \\cdot \\frac{x}{y} \\\\ &=\\frac{x^{3}}{y^{3}} \\end{aligned} From these tables above, you would have got an idea of how and why a particular law works and how these laws of exponents make our work easier. Let's try and solve a few examples using these laws. Important Notes The laws", "we’ll choose to eliminate $$2^{k+1}$$, but we could do it the other way, and it would be fine. \\begin{align} & P(k)=2^{k+1} + 5\\cdot 9^{k} \\\\ & 2^{k+1} = P(k) - 5\\cdot 9^{k} \\end{align} We can now apply this to $$P(k+1)$$, by taking out a factor of $$2$$ from $$2^{k+1+1}$$ and then substituting. \\begin{align} P(k+1) &= 2^{k+1+1} + 5\\cdot 9^{k+1} \\\\ &= 2\\cdot 2^{k+1} + 5\\cdot 9^{k+1} \\\\ &= 2\\left (P(k) - 5\\cdot 9^{k}\\right ) + 5\\cdot 9^{k+1} \\end{align} Now we can factorise the expression, and hopefully find a factor of $$7$$ lurking somewhere! \\begin{align} P(k+1) &= 2\\left (P(k) - 5\\cdot 9^{k}\\right ) + 5\\cdot 9^{k+1} \\\\ &= 2P(k) - 10\\cdot 9^{k} + 5\\cdot 9^{k+1} \\\\ &= 2P(k) - 10\\cdot 9^{k} + 5\\cdot 9\\cdot 9^{k} \\\\ &= 2P(k) + 35\\cdot 9^{k} \\end{align} This means that $$P(k)$$ being divisible by $$7$$ implies that $$P(k+1)$$ is also divisible by $$7$$. Now onto the completion step! This is pretty much the same every time: “ Thus $$P(n)$$ implies $$P(n+1)$$ and as $$P(1)$$ is true, $$P(n)$$ holds for all $$n \\in \\mathbb {N}$$ where $$n \\geqq 1$$. ”" ]

[ "indicated, we currently are only ready to prove this theorem for the integer powers. We start with positive integers. We have already proved the result for $p=1, 2, 3, 4$ as part of our discovering the pattern. As we developed that pattern, we discovered that we were using a recursive argument each time. The powers $p=3$ and $p=4$ were based on knowing the results for $p=2$ and $p=3\\text{,}$ respectively. Our proof builds on this recursive argument to create a general statement. Suppose that we know for $p=n\\text{,}$ \\begin{equation*} \\frac{d}{dx}[x^n] = n x^{n-1}\\text{.} \\end{equation*} Next consider the power $p=n+1\\text{,}$ and rewrite it $x^{n+1} = x \\cdot x^{n}\\text{.}$ The product rule guarantees \\begin{align*} \\frac{d}{dx}[x^{n+1}] &= \\frac{d}{dx}[x \\cdot x^n]\\\\ &= \\frac{d}{dx}[x] \\cdot x^n + x \\cdot \\frac{d}{dx}[x^n]\\\\ &= 1 \\cdot x^n + x \\cdot nx^{n-1}\\\\ &= x^n + nx^n\\\\ &= (n+1)x^n. \\end{align*} This general recursive argument shows that if the rule is satisfied for an initial value of $p=n\\text{,}$ the statement will be immediately known to be true for the sequence of values $p \\in (n, n+1, n+2, \\ldots)\\text{.}$ Having earlier shown the rule was true for $p=1\\text{,}$ the recursive argument shows it will be true for all positive integers. In addition, because we know the rule is true for $p=\\frac{1}{2}\\text{,}$ the same argument shows that the rule is true for", "is very similar. Although it is often unwise to draw general conclusions from specific examples, we note that when we differentiate $f(x)=x^3$, the power on $x$ becomes the coefficient of $x^2$ in the derivative and the power on $x$ in the derivative decreases by 1. The following theorem states that this power rule holds for all positive integer powers of $x$. We will eventually extend this result to negative integer powers. Later, we will see that this rule may also be extended first to rational powers of $x$ and then to arbitrary powers of $x$. Be aware, however, that this rule does not apply to functions in which a constant is raised to a variable power, such as $f(x)=3^x$. ### The Power Rule Let $n$ be a positive integer. If $f(x)=x^n$, then $f^{\\prime}(x)=nx^{n-1}$. Alternatively, we may express this rule as $\\frac{d}{dx}(x^n)=nx^{n-1}$. ## Proof For $f(x)=x^n$ where $n$ is a positive integer, we have $f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{(x+h)^n-x^n}{h}$. Since $(x+h)^n=x^n+nx^{n-1}h+\\binom{n}{2}x^{n-2}h^2+\\binom{n}{3}x^{n-3}h^3+\\cdots+nxh^{n-1}+h^n$, we see that $(x+h)^n-x^n=nx^{n-1}h+\\binom{n}{2}x^{n-2}h^2+\\binom{n}{3}x^{n-3}h^3+\\cdots+nxh^{n-1}+h^n$. Next, divide both sides by $h$: $\\large \\frac{(x+h)^n-x^n}{h}=\\frac{nx^{n-1}h+\\binom{n}{2}x^{n-2}h^2+\\binom{n}{3}x^{n-3}h^3+\\cdots+nxh^{n-1}+h^n}{h}$. Thus, $\\frac{(x+h)^n-x^n}{h}=nx^{n-1}+\\binom{n}{2}x^{n-2}h+\\binom{n}{3}x^{n-3}h^2+\\cdots+nxh^{n-2}+h^{n-1}$. Finally, $\\begin{array}{ll}f^{\\prime}(x) & =\\underset{h\\to 0}{\\lim}(nx^{n-1}+\\binom{n}{2}x^{n-2}h+\\binom{n}{3}x^{n-3}h^2+\\cdots+nxh^{n-1}+h^n) \\\\ & =nx^{n-1} \\end{array} _\\blacksquare$ ### Applying the Power Rule Find the derivative of the function $f(x)=x^{10}$ by applying the power rule. #### Solution Using the power rule with $n=10$, we obtain $f^{\\prime}(x)=10x^{10-1}=10x^9$. Find the derivative of $f(x)=x^7$. #### Hint Use the", "There are four 1-tuples from a four-element set. 50 = │ { () } │ = 1. There is exactly one empty tuple. ### Negative integer exponents By definition, raising a nonzero number to the −1 power produces its reciprocal: $a^{-1} = \\frac{1}{a}.$ One also defines $a^{-n} = \\frac{1}{a^n}$ for any nonzero a and any positive integer n. Raising 0 to a negative power would imply division by 0, so it is left undefined. The definition of an for nonzero a is made so that the identity aman = am+n, initially true only for nonnegative integers m and n, holds for arbitrary integers m and n. In particular, requiring this identity for m = −n is requiring $a^{-n} \\, a^{n} = a^{-n\\,+\\,n} = a^0 = 1,$ where a0 is defined above, and this motivates the definition an = 1/an shown above. 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 The most important identity satisfied by integer exponentiation is $a^{m + n} = a^m \\cdot a^n$ This identity has the consequence $a^{m - n} =\\frac{a^m}{a^n}$ for a ≠ 0, and $(a^m)^n = a^{m\\cdot n}$ Another basic identity is $(a \\cdot b)^n = a^n \\cdot b^n$ While", "$\\ \\ \\ \\ \\ \\ \\$ 3 factors $= p \\cdot p \\cdot p \\cdot q \\cdot q \\cdot q$ $=p^{3} \\cdot q^{3}$ In other words, $(pq)^{3} = p^{3} \\cdot q^{3}$. ### THE POWER OF A PRODUCT RULE OF EXPONENTS For any real numbers $a$ and $b$ and any integer $n$, the power of a product rule of exponents states that $(ab)^{n}=a^{n}b^{n}$ #### Example 7 Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents. • $(ab^{2})^{3}$ • $(2t)^{15}$ • $(-2w^{3})^{3}$ • $\\frac{1} {(-7z)^{4}}$ • $(e^{-2}f^{2})^{7}$ Solution Use the product and quotient rules and the new definitions to simplify each expression. • $(ab^{2})^{3}=(a)^{3} \\cdot (b^{2})^{3} = a^{1 \\cdot 3} \\cdot b^{2 \\cdot 3} = a^{3}b^{6}$ • $(2t)^{15} = (2)^{15} \\cdot (t)^{15} = 2^{15}t^{15} = 32,768t^{15}$ • $(-2w^{3})^{3} = (-2)^{3} \\cdot (w^{3})^{3} = -8 \\cdot w^{3 \\cdot 3} = -8w^{9}$ • $\\frac{1} {(-7z)^{4}} = \\frac{1} {(7)^{4} \\cdot (z)^{4}} = \\frac {1} {2,401z^{4}}$ • $(e^{-2}f^{2})^{7} = (e^{-2})^{7} \\cdot (f^{2})^{7} = e^{-2 \\cdot 7} \\cdot f^{2 \\cdot 7} = e^{-14}f^{14} = \\frac {f^{14}} {e^{14}}$ ## 1.2.6b Finding the Power of a Quotient To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a", "power, we write the result with the common base and the product of the exponents. $(a^{m})^{n} = a^{m \\cdot n}$ Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents. Power Rule Product Rule $5^{3} \\cdot 5^{4}$ $=$ $5^{3+4}$ $=$ $5^{7}$ but $(5^{3})^{4}$ $=$ $5^{3 \\cdot 4}$ $=$ $5^{12}$ $x^{5} \\cdot x^{2}$ $=$ $x^{5+2}$ $=$ $x^{7}$ but $(x^{5})^{2}$ $=$ $x^{5 \\cdot 2}$ $=$ $x^{10}$ $(3a)^{7} \\cdot (3a)^{10}$ $=$ $(3a)^{7+10}$ $=$ $(3a)^{17}$ but $((3a)^{7})^{10}$ $=$ $(3a)^{7 \\cdot 10}$ $=$ $(3a)^{70}$ ### THE POWER RULE OF EXPONENTS For any real number $a$ and positive integers $m$ and $n$, the power rule states that $(a^{m})^{n} = a^{m \\cdot n}$ #### Example 3 Write each of the following products with a single base. Do not simplify further. • $(x^{2})^{7}$ • $((2t)^{5})^{3}$ • $((-3)^{5})^{11}$ Solution Use the power rule to simplify each expression. • $(x^{2})^{7} = x^{2 \\cdot 7} = x^{14}$ • $((2t)^{5})^{3} = (2t)^{5 \\cdot 3} = (2t)^{15}$ • $((-3)^{5})^{11} = (-3)^{5 \\cdot 11} = (-3)^{55}$ ## 1.2.4 Using the Zero Exponent Rule of", "proof of the power rule for negative integers. POWER RULE (VERSION 2) Let $f(x)=x^n$ for a negative integer $n$. Then, $f^{\\prime}(x)=n x^{n-1}$. For the proof, let $n$ be a negative integer. Then $-n$ is a positive integer (if $n=-3$, then $-n=3$ ), and the power rule can be applied to $x^{-n}$. Then, \\begin{aligned} \\frac{d}{d x} x^n &=\\frac{d}{d x}\\left(\\frac{1}{x^{-n}}\\right)=\\frac{x^{-n} \\cdot 0-1 \\cdot-n x^{-n-1}}{\\left(x^{-n}\\right)^2} \\ &=\\frac{n x^{-n-1}}{x^{-2 n}}=n x^{-n-1-(-2 n)}=n x^{n-1} . \\end{aligned} Example 13 Find $f^{\\prime \\prime}(x)$ for $f(x)=\\frac{1}{4 x}$. Solution We could use the quotient rule to find the derivative, but we can also use the power rule if we rewrite the function using a negative exponent: \\begin{aligned} f(x) &=\\frac{1}{4} x^{-1} \\ f^{\\prime}(x) &=\\frac{1}{4}\\left(-1 \\cdot x^{-2}\\right)=-\\frac{1}{4} x^{-2}=-\\frac{1}{4 x^2} \\end{aligned} The last step, expressing the derivative in terms of a fraction instead of a negative exponent, is not necessary but can be helpful for interpreting the result. On the other hand, thinking of $f^{\\prime}(x)=-\\frac{1}{4} x^{-2}$ is helpful for taking the second derivative using the power rule: $$f^{\\prime \\prime}(x)=-\\frac{1}{4}\\left(-2 x^{-3}\\right)=\\frac{1}{2} x^{-3}=\\frac{1}{2 x^3} .$$ Notice that as we take derivatives, the exponent in the denominator is getting larger, not smaller. A polynomial’s derivatives eventually reach zero, but this function’s derivatives never reach zero. ## 数学代写|微积分代写Calculus代写|Equations of tangent lines revisited The procedure we used in section $1.8$ to find the equation of", "\\frac{b^x}{b^y} = b^{x-y}$ • Power of a power: $\\displaystyle (b^x)^y = b^{xy}$ • Product with Common Exponent: $b^x \\, c^x = (bc)^x$ • Quotient with Common Exponent: $\\displaystyle \\frac{b^x}{c^x} = \\left(\\frac{b}{c}\\right)^x$ We illustrate several of the properties in the context of integer powers. For integer exponents, a power means repeated multiplication (similar to how multiplication by an integer means repeated addition). So $b^3 = b \\cdot b \\cdot b\\text{.}$ The product properties are just about counting. ###### Example2.4.1 \\begin{gather*} b^3 \\cdot b^2 = (bbb) \\cdot (bb) = b^5 = b^{3+2}\\\\ (b^2)^3 = (bb)(bb)(bb) = b^{2 \\cdot 3}\\\\ (ab)^3 = (ab)(ab)(ab) = (aaa)(bbb) = a^3 b^3 \\end{gather*} The quotient properties for integer powers result from simplifying fractions, canceling common factors when possible. ###### Example2.4.2 \\begin{gather*} \\frac{b^6}{b^2} = \\frac{(bb)(bbbb)}{bb} = bbbb = b^{6-2}\\\\ \\left(\\frac{b}{c}\\right)^3 = \\frac{b}{c} \\cdot \\frac{b}{c} \\cdot \\frac{b}{c} = \\frac{b^3}{c^3} \\end{gather*} The zero power property is necessary for the power of a sum rule to remain consistent. We know that $b^{x+0}=b^x\\text{.}$ But the properties of powers also mean $b^{x+0} = b^x \\cdot b^0\\text{.}$ For these to both be true requires $b^x = b^x \\cdot b^0$ or that $b^0 = 1\\text{.}$ Addition and powers have no convenient properties. Many mistakes occur when students forget this and imagine that powers distribute like multiplication. (It doesn't!) ###### Example2.4.3 To illustrate that", "That is not the case, as $x = -1$ has power $19$. Hence, $x=-1$ can't be a solution. 3. Power is equal to $0$: $$x^2-7x+11=0$$ $$x_{1/2}=\\frac{7\\pm\\sqrt{7^2-4\\cdot1\\cdot11}}{2}=\\frac{7\\pm\\sqrt{5}}{2}$$ so far, I have calculated that as $1=x^0$ That's not the only way to get to the result of 1. Recall that $$\\overbrace{1\\cdot 1\\cdot 1\\cdot 1 ... \\cdot 1\\cdot 1}^{\\text{n times}}=1$$ and $$\\overbrace{-1\\cdot -1\\cdot -1\\cdot -1 ... \\cdot -1\\cdot -1}^{\\text{2n times}}=1$$ That means that you should also find solutions for $x =-1$ and $x=1$ which create solutions in the form of $$1=(-1)^{2n}$$ and $$1=1^{n}$$ with $x =-1$, that is $$(-1)^2 - 7(-1)+11 = 2n$$ and $x =1$, that is $$1^2 - 7\\cdot 1+11 = n$$", "of positive integers \\begin{aligned} A - \\frac{B^2+B}{2} + \\frac{n^2+n}{2} &= [\\cdots (a-2) + (a-1) + a + (a+1) + (a+2) +\\cdots] \\\\&\\hspace{7mm}- [u + (u+1) + (u+2) \\cdots] + [1 + 2 + 3 \\cdots + n] \\\\&= 0 \\end{aligned} we can observe a property as follows $$A = \\frac{(B-n)(B+n+1)}{2}$$ Among the numerous cases of $$A = a \\times b$$, only $$A \\in$$ {prime number} $$\\bigcup$$ {perfect power of 2} are represented as $$A = 1 \\times A$$. The reason for a prime number being represented as $$A = 1 \\times A$$ is obvious. The reason for the case when $$A$$ is a perfect power of 2, $$2^e$$ can be represented only as a multiplication between powers of 2, which are all even, unless $$2^e = 1 \\times 2^e$$. The procedure of algorithm repeats the addition and subtraction of positive integers which let the value of D (the RHS of equation (1)) to oscillate between 0, until it reaches 0. So the longest expected computation time for the algorithm of $$A$$ is the case when $$A$$ is represented as $$A = \\frac{[A-(A-1)][A+(A-1)+1]}{2}$$ or $$A = \\frac{[(\\frac{A-1}{2}+1)-(\\frac{A-1}{2}-1)][(\\frac{A-1}{2}+1)+(\\frac{A-1}{2}-1)+1]}{2}$$ The fact that $$2^e$$ can be only represented as $$2^e = 1 \\times 2^e$$, particularly as $$2^e = \\frac{[2^e - (2^e-1)][2^e + (2^e - 1)+1]}{2}$$ gives us a good tool to assume", "$$\\frac{4}{5}-\\frac{2}{5} i$$ ## Simplify Powers of $$i$$ The powers of $$i$$ make an interesting pattern that will help us simplify higher powers of $$i$$. Let’s evaluate the powers of $$i$$ to see the pattern. $$\\begin{array}{ccc}{i^{1}} & {i^{2}} & {i^{3}} & {i^{4}} \\\\ {i} & {-1} & {i^{2}\\cdot i} & {i^{2}\\cdot i^{2}}\\\\ {}&{}&{-1\\cdot i}&{(-1)(-1)}\\\\ {}&{}&{-i}&{1}\\end{array}$$ $$\\begin{array}{cccc}{i^{5}} & {i^{6}} & {i^{7}} & {i^{8}} \\\\ {i^{4} \\cdot i} & {i^{4} \\cdot i^{2}} & {i^{4} \\cdot i^{3}} & {i^{4} \\cdot i^{4}} \\\\ {1 \\cdot i} & {1 \\cdot i^{2}} & {1 \\cdot i^{3}} & {1 \\cdot 1} \\\\ {i} & {i^{2}} & {i^{3}} & {1} \\\\ {}&{-1} & {-i}\\end{array}$$ We summarize this now. $$\\begin{array}{ll}{i^{1}=i} & {i^{5}=i} \\\\ {i^{2}=-1} & {i^{6}=-1} \\\\ {i^{3}=-i} & {i^{7}=-i} \\\\ {i^{4}=1} & {i^{8}=1}\\end{array}$$ If we continued, the pattern would keep repeating in blocks of four. We can use this pattern to help us simplify powers of $$i$$. Since $$i^{4}=1$$, we rewrite each power, $$i^{n}$$, as a product using $$i^{4}$$ to a power and another power of $$i$$. We rewrite it in the form $$i^{n}=\\left(i^{4}\\right)^{q} \\cdot i^{r}$$, where the exponent, $$q$$, is the quotient of $$n$$ divided by $$4$$ and the exponent, $$r$$, is the remainder from this division. For example, to simplify $$i^{57}$$, we divide $$57$$ by $$4$$ and we get $$14$$ with a remainder of $$1$$.", "subsets, however, we find that $1$-element subsets give a full power of $a_i$; $2$-element subsets give a $\\frac{1}{2}$ power of $a_i$; $3$-element subsets give a $\\frac{1}{3}$ power of $a_i$; and so on. So when we multiply these all together, we get an exponent of $\\displaystyle 1+(n-1)C1 \\cdot \\frac{1}{2}+(n-1)C2 \\cdot \\frac{1}{3}+ \\cdots +(n-1)C(n-1) \\cdot \\frac{1}{n} = \\sum_{k=1}^n \\frac{1}{k} \\cdot (n-1)C(k-1)$. It remains to evaluate this sum, but we can do that through generating functions. Since $\\displaystyle (1+x)^{n-1} = \\sum_{k=1}^n (n-1)C(k-1) \\cdot x^{k-1}$, we integrate both sides to get $\\displaystyle \\frac{(1+x)^n}{n} = C+\\sum_{k=1}^n \\frac{1}{k}(n-1)C(k-1) \\cdot x^k$. Using $x = 0$, we obtain $C = \\frac{1}{n}$, so $\\displaystyle \\frac{(1+x)^n-1}{n} = \\sum_{k=1}^n \\frac{1}{k}(n-1)C(k-1) \\cdot x^k$. Finally, by the substitution $x = 1$, we get the desired summation: $\\displaystyle \\sum_{k=1}^n \\frac{1}{n} \\cdot (n-1)C(k-1) = \\frac{2^n-1}{n}$. So prior to the last geometric mean, we have $(a_i)^{\\frac{2^n-1}{n}}$, but we know that there are $2^n-1$ non-empty subsets of $S$, which means the last geometric mean is a power of $\\frac{1}{2^n-1}$, and we have $\\left(a_i^{\\frac{2^n-1}{n}}\\right)^{\\frac{1}{2^n-1}} = (a_i)^{\\frac{1}{n}}$, the same power as the original geometric mean we computed. QED. -------------------- Comment: A very nice problem because it branched together several aspects of mathematics all into a single problem, which is always cool. There is probably a non-calculus method of evaluating that sum, but basically any time you see", "than or equal to the largest negative positive power of 3, for the number to be nonnegative. ## Solution 5 (Casework and Recursion) This solution has a similar idea to that of above. We will start off with casework on $a_{7}$: If $a_7 = -1$, we can show that there is no way to assign the values of $-1, 0, 1$ to the other $a_{i}$'s for $0 \\leq i \\leq 6$. This is because even if we make all of the other $a_{i}$ equal to $1$, we will have that our number is $(-1)(3^{7}) + (3^{6}+3^{5}+...+3^{0}) = -3^{7} + \\frac{3^{7}-1}{3-1}$. However, note that since $\\frac{3^{7}-1}{3-1} < 3^{7}$, $-3^{7} + \\frac{3^{7}-1}{3-1} < 0$. If $a_{7}=0$, we can ignore $a_{7}$ and pretend like it never even existed. This then simplifies the problem: the question remains the same, except that instead of having $a_{0}...a_{7}$, we only have $a_{0}...a_{6}$. Now, we introduce some notation: let the number of nonnegative integers that can be written in the form of $a_{n} \\cdot 3^{n} + ... + a_{0} \\cdot 3^{0}$ be denoted as $S_{n}$. Then, the problem is asking us for $S_{7}$, and the total number of nonnegative integers we get from this case is $S_{6}$. Let's keep this in mind and revisit this idea later. If $a_{7}=1$, we can show that no matter what", "# What is the quotient of powers property? ##### 1 Answer Dec 19, 2014 $\\frac{{a}^{m}}{{a}^{n}} = {a}^{m - n}$ This property allows you to simplify problems where you have a fraction of the same numbers ( $a$) raised to different powers ($m \\mathmr{and} n$). For example: $\\frac{{3}^{3}}{{3}^{2}} = \\frac{3 \\cdot 3 \\cdot 3}{3 \\cdot 3} = {3}^{3 - 2} = 3$ You can see how the power of 3, in the numerator, is \"reduced\" by the presence of the power 2 in the denominator. You can also check te result by doing the multiplications: $\\frac{{3}^{3}}{{3}^{2}} = \\frac{3 \\cdot 3 \\cdot 3}{3 \\cdot 3} = \\frac{27}{9} = 3$ As a challenge try to find out what happens when $m = n$ !!!!!", "Rule $\\dfrac{a^m}{a^n}=a^{m-n}$ When powers with the same base are divided, they can be written as one power where the exponent in the denominator is subtracted from the exponent in the numerator. For example, $3^6$ divided by $3^4$ can be rewritten using the rule as follows. $\\dfrac{3^6}{3^4} \\Rightarrow 3^{6-4} = 3^2$ This rule can be explained by writing the powers as a product. $\\dfrac{3^6}{3^4}$ $\\dfrac{3\\cdot 3\\cdot 3\\cdot3\\cdot3\\cdot3}{3\\cdot3\\cdot3\\cdot3}$ $\\dfrac{3\\cdot 3\\cdot \\cancel{3}\\cdot\\cancel{3}\\cdot\\cancel{3}\\cdot\\cancel{3}}{\\cancel{3}\\cdot\\cancel{3}\\cdot\\cancel{3}\\cdot\\cancel{3}}$ $3\\cdot3$ $3^2$ This rule is valid for all values of $m,$ $n$ and when $a\\neq 0.$ Note that the rule might lead to non-real solutions when $a<0$ and at least one of the exponents is not an integer. Rule ## Power of a Power Property ### Rule $\\left(a^m\\right)^n=a^{mn}$ When a power is raised to another exponent, the powers can be combined into one by multiplying them. For example $\\left(5^2\\right)^3$ can be rewritten with this rule as follows. $\\left(5^2\\right)^3 \\Rightarrow 5^{2 \\cdot 3} = 5^6$ This rule can be explained by writing the powers as products. $\\left(5^2\\right)^3$ $a^3=a\\cdot a\\cdot a$ $5^2 \\cdot 5^2 \\cdot 5^2$ $5\\cdot 5\\cdot 5\\cdot 5\\cdot 5\\cdot 5$ $5^{6}$ This rule is valid for all values of $m$ and $n$ when $a\\neq 0.$ For the rule to be true also for $a=0$ it is necessary that both exponents are greater than $0.$ Note that the rule might", "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", "# binomial proof of positive integer power rule We will use the difference quotient in this proof of the power rule for positive integers. Let $f(x)=x^{n}$ for some integer $n\\geq 0$. Then we have $\\displaystyle f^{\\prime}(x)=\\lim_{h\\rightarrow 0}\\frac{(x+h)^{n}-x^{n}}{h}.$ We can use the binomial theorem to expand the numerator $\\displaystyle f^{\\prime}(x)=\\lim_{h\\rightarrow 0}\\frac{C_{0}^{n}x^{0}h^{n}+C_{% 1}^{n}x^{1}h^{n-1}+\\cdots+C_{n-1}^{n}x^{n-1}h^{1}+C_{n}^{n}x^{n}h^{0}-x^{n}}{h}$ where $C_{k}^{n}=\\frac{n!}{k!(n-k)!}$. We can now simplify the above $\\displaystyle f^{\\prime}(x)$ $\\displaystyle=\\lim_{h\\rightarrow 0}\\frac{h^{n}+nxh^{n-1}+\\cdots+nx^{n-1}h+x^{n% }-x^{n}}{h}$ $\\displaystyle=\\lim_{h\\rightarrow 0}(h^{n-1}+nxh^{n-2}+\\cdots+nx^{n-1})$ $\\displaystyle=nx^{n-1}$ $\\displaystyle=nx^{n-1}.$ Title binomial proof of positive integer power rule BinomialProofOfPositiveIntegerPowerRule 2013-03-22 12:29:43 2013-03-22 12:29:43 mathcam (2727) mathcam (2727) 8 mathcam (2727) Proof msc 26A03", "for the Power of Product property is: $$(b \\cdot c)^x=b^x \\cdot c^x$$ ## Power of Quotient The Power of Quotient property is similar in that the exponent of a rational base is multiplied by the exponents in both the numerator and denominator. Let’s look at some examples: $$(\\frac{b^3}{c^2})^2$$ expands to $$(\\frac{b^3}{c^2})(\\frac{b^3}{c^2})$$ The Product of Powers property results in the simplified expression $$\\frac{b^{6}}{c^{4}}$$. However, applying the Power of Quotient property directly is often more efficient: $$(\\frac{b^3}{c^2})^2 = \\frac{b^{3 \\cdot 2}}{c^{2 \\cdot 2}} = \\frac{b^6}{c^4}$$ Here’s another example: $$(\\frac{2b}{c^2d})^3$$ expands to $$(\\frac{2b}{c^2d})(\\frac{2b}{c^2d})(\\frac{2b}{c^2d})$$ Rearranging factors and applying the Product of Powers property results in: $$\\frac{2 \\cdot 2 \\cdot 2 \\cdot b \\cdot b\\cdot b}{c^2 \\cdot c^2 \\cdot c^2 \\cdot d \\cdot d \\cdot d} = \\frac{8b^3}{c^6d^3}$$ If we were to apply the Power of Quotient property, it would look like this: $$(\\frac{2^{1 \\cdot 3}b^{1 \\cdot 3}}{c^{2 \\cdot 3}d^{1 \\cdot 3}}) = \\frac{2^3b^3}{c^6d^3} = \\frac{8b^3}{c^6d^3}$$ The general rule for the Power of Quotient property is: $$(\\frac{b}{c})^x = \\frac{b^x}{c^x}$$ As you can see, these rules are essential in simplifying expressions within our work with various algebraic functions. Thanks for watching this review of exponents! Happy studying! ### What is the formula for product of powers? #### A The product of powers rule says that when multiplying exponents with the same base, you can", "\\frac{1}{2^{k-1}} + \\cdots + \\frac{1}{2^{k - 1}} = 1$ Thus each chunk is at most 1, and taken together with the extra term $\\frac{1}{2^k}$, we have $H_n \\leq \\log_2 n + \\frac{1}{2^k} \\leq \\log_2 n + 1$. On the other hand, if we lower bound the elements by the next power of 2, $1 \\geq 1$ $\\frac{1}{2} + \\frac{1}{3} \\geq \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}$ $\\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} \\geq \\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8} =\\frac{1}{2}$ $\\dots$ $\\frac{1}{2^{k-1}} + \\cdots + \\frac{1}{2^k - 1} \\geq \\frac{1}{2^k} + \\cdots + \\frac{1}{2^k} = \\frac{1}{2}$ Every chunk is at least $\\frac{1}{2}$, hence $H_n \\geq \\frac{1}{2}\\log_2 n$. This essentially is the proof. One technicality: we’ve only shown $H_n = \\Theta (\\log_2 n)$ for $n$ a power of 2. Monotonicity of $H_n$ completes the proof for all $n$: taking the nearest powers of 2 above and below $n$, call them $n_-$ and $n_+$, $n/2 < n_- < n < n_+ < 2n$ Applying our bounds on $H_{n_-}$ and $H_{n_+}$, $H_n \\leq H_{n_+} \\leq \\log_2 n_+ + 1 < \\log_2 (2n) +1= \\log_2 n + 2$ $H_n \\geq H_{n_-} \\geq \\frac{1}{2}\\log_2 n_- > \\frac{1}{2}\\log_2 (n/2) = \\frac{1}{2}\\log_2 n - \\frac{1}{2}$ Thus $H_n = \\Theta(\\log_2 n)$. (In general, for any monotonic linear or sublinear function, it suffices to show $\\Theta(\\cdot)$", "X 7 X 7 X 7 X 7}{7 X 7 X 7 X 7 X 7 X 7}$$ and the quotient as a power we use rule for finding $$\\frac{a^{m}}{a^{n}}$$ a quotient of two powers with the same base as am-n,so $$\\frac{7^{9}}{7^{6}}$$ is 79-6 = 73 Question 7. = $$\\frac{-4.5 X -4.5 X -4.5 X -4.5 X -4.5 X -4.5}{-4.5 X -4.5}$$ = (-4.5)4. Explanation: Given $$\\frac{-4.5^{6}}{-4.5^{2}}$$ the repeated multiplication is $$\\frac{-4.5 X -4.5 X -4.5 X -4.5 X -4.5 X -4.5}{-4.5 X -4.5}$$ and the quotient as a power we use rule for finding $$\\frac{a^{m}}{a^{n}}$$ a quotient of two powers with the same base as am-n ,so $$\\frac{-4.5^{6}}{-4.5^{2}}$$ is (-4.5)6-2 = (-4.5)4. Question 8. = $$\\frac{m X m X m X m X m X m X m X m X m X m}{m X m X m X m X m }$$ = m5 Explanation: Given $$\\frac{m^{10}}{m^{5}}$$ the repeated multiplication is and $$\\frac{m X m X m X m X m X m X m X m X m X m}{m X m X m X m X m }$$ the quotient as a power we use rule for finding $$\\frac{a^{m}}{a^{n}}$$ a quotient of two powers with the same base as am-n so $$\\frac{m^{10}}{m^{5}}$$ is (m)10-5 = m5. DIVIDING POWERS WITH THE SAME BASE Simplify the", "$$f(x) = \\frac{1}{4}\\cdot \\overline{B}_2(2x) - \\overline{B}_2(x) + \\frac{1}{2}\\cdot \\overline{B}_1(x-\\frac{1}{2})$$ (to find such identities systematically, it is convenient to compute the Fourier coefficients of $f$ and remark that the Bernoulli polynomials have nice Fourier coefficients). Then we find : $$\\sum_{k=1}^{\\frac{N-1}{2}} k\\cdot \\log(k) \\equiv \\sum_{k \\in (\\mathbf{Z}/N\\mathbf{Z})^{\\times}} f(\\frac{k}{N})\\cdot \\log(k) \\text{ (mod } 3\\text{)}$$ which you can simplify greatly with the Bernoulli functions (the $\\log(2)$ comes from $2x$ inside $\\overline{B_2}(2x)$). In particular if $N \\equiv 1 \\text{ (mod } 9\\text{)}$ then your $S$ is a $3$-rd power. If $N = 2^p-1 \\geq 31$ is a Mersenne prime , then $\\log(2) \\equiv 0 \\text{ (mod } 3\\text{)}$ since $\\log(2^p) \\equiv p\\log(2) \\equiv0 \\text{ (mod } 3\\text{)}$ and $p>3$. Although I didn't state it, I think I can also prove the following regarding the $3$-th power part: if $N = a^2-ab+b^2$ is a prime $\\equiv 1 \\text{ (mod } 9\\text{)}$ then $b^{-2}\\cdot(ab^2+ba^2)$ is a $3$-rd power modulo $N$. It follows from the same methods as in the preprint." ]

[ "# The Other Angle Geometry Level 1 $$ABCD$$ is a parallelogram. If $$\\angle ABC = 35^\\circ$$, what is the measure (in degrees) of $$\\angle ADC$$? ×", "# IN A PARALLELOGRAM ABCD,IF ANGLE B =65 DEGREE,WHAT IS THE MEASURE OF ANGLE C? Dear Student, Since, in a parallelogram, opposite sides are parallel and equal. And, as opposite sides are parallel, thus, in the figure, angle B and x will be equal as they are corresponding angles. Thus, x=65 Now, as sum of angle C and angle x is 180 (angles on the line)", "$$B$$ and $$D$$ on $$PR$$, and $$C$$ on $$QR$$. Find $$\\angle PQR$$ (in degrees). ×", "# In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D. Question: In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects B as well as ∠D. Then, ∠AMB = ? (a) 45° (b) 60° (c) 90° (d) 30° Solution: (c) ​90° Explanation: ∠B = ∠D $\\Rightarrow \\frac{1}{2} \\angle B=\\frac{1}{2} \\angle D$", "of parallelogram \"ABCD=10 \" sq. units}$", "# The largest angle of a parallelogram measures 120 degrees. If the sides measure 14 inches and 12 inches, what is the exact area of the parallelogram? Dec 24, 2015 $A = 168$ inches #### Explanation: We can get the area of parallelogram even though the angle is not given, since you gave the length of the two sides. Area of parallelogram $= b h$ $b = 14$ $h = 12$ $A = b h$ $A = \\left(14\\right) 12$ $A = 168$", "### A. বৃত্তঃস্থ সামান্তরিক ##### $ABCD$ is a cyclic parallelogram. Bisector of $\\angle BAD$ intersects the circle at point $E$ such that $E$ stays in between $B$ and $C$. If $AC = 2 CD$, then what is the value of $\\angle EAC$ ? Calibration Round #1", "degree trigo - In a right tringle ,angle ABC has a value of 37 degree while its ...", "is the measure of angle D? 25. The diagonals of quadrilateral ABCD intersect at E(-1,4). Two vertices of ABCD are A(2,7) and B(-3,5). What must the coordinates of C and D be in order to ensure that ABCD is a parallelogram? You need to be a HelpTeaching.com member to access free printables. Already a member? Log in for access. | Go Back To Previous Page", "# Vectors – Calculate a vertex in a parallelogram and an angle between diagonals – Exercise 4480 Exercise Given the points $$A(-3,-2,0),B(3,-3,1),C(5,0,2)$$ Calculate the point D and the angle between the diagonals AC and BD in parallelogram ABCD. $$D(-1,1,1)$$ $$\\alpha=60^{\\circ},\\beta=120^{\\circ}$$", "= ____ 56 d. m∠WAX=____ 107° m∠B= ____ 73° m∠YCZ = ___ 107° m∠D=____ 73° e. What kind of quadrilateral is ABCD? Parallelogram ### Eureka Math Geometry Module 1 Lesson 29 Exit Ticket Answer Key Use the properties of midsegments to solve for the unknown value in each question. Question 1. R and S are the midpoints of XW and WV, respectively. What is the perimeter of ∆WXY?", "# A symmetric triangle! Triangle $ABC$ has the property that $\\angle A+\\angle B= \\angle C$. Find the measure of $\\angle C$ in degrees. ×", "W, and Z in parallelogram ABCD several... ) the degree measure of angles x, y, Z in parallelogram ABCD parallelogram by testing if all satisfy...", "Operations, Marketing GPA: 3.5 Re: In the parallelogram above, what is the measure of angle ABC? [#permalink] ### Show Tags 29 Aug 2018, 05:10 Angle ABC = Angle ADC = 55 ( Opposite angle of a parallelogram ) Sufficient (2) The sum of angles ABC and BCD is 180°. No new information Insufficient Hence, A. _________________ The few, the fearless ! Thanks Re: In the parallelogram above, what is the measure of angle ABC? [#permalink] 29 Aug 2018, 05:10 Display posts from previous: Sort by", ", what is the measure of \\angle ABC(in …", "# In a Parallelogram Abcd, If ∠D = 115°, Then Write the Measure of ∠A. - Mathematics In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A. #### Solution In Parallelogram ABCD , ∠A and are ∠D Adjacent angles. We know that in a parallelogram, adjacent angles are supplementary. $\\text { Now}, \\angle A + \\angle D = 180^\\circ$ $\\Rightarrow \\angle A + 115^\\circ = 180^\\circ$ $\\Rightarrow \\angle A = 180^\\circ - 115^\\circ$ $\\Rightarrow \\angle A = 65^\\circ$ $\\text{So, measure of } \\angle\\text { A is } 65^\\circ .$ Is there an error in this question or solution? #### APPEARS IN RD Sharma Mathematics for Class 9", "Day 47 # 47 of 100: A Simple Angle Geometry Level 1 A three-dimensional cube is shown above. What is the measure of $$\\angle ABC$$ in degrees? There are a lot of ways to solve this puzzle; look for a fast one, then try to find a second technique!", "+0 +1 217 1 $ABCD$ is a parallelogram such that $AB=\\sqrt{3}\\cdot BC.$ Diagonals $\\overline{AC}$ and $\\overline{BD}$ intersect at $X$ and $\\angle ABC=\\angle BXC.$ Find the measure of $\\angle ACB$ in degrees. Dec 21, 2019 #1 +111360 +1", "all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 273 [0], given: 0 Re: ABCD is a parallelogram. BD = 2. The angles of triangle BCD are all eq [#permalink] 18 Sep 2017, 12:06 Display posts from previous: Sort by", "? Free Version Moderate # Properties of Parallelograms ACTMAT-QEY1G5 For parallelogram ABCD, if $m{\\angle}A = 4x$ and $m{\\angle}D = 7x-18$, what is the measure of ${\\angle}C$, in degrees? A $6°$ B $18°$ C $24°$ D $72°$ E $108°$" ]

[ "+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 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 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 # 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", "about $\\cos XTY$ to obtain the result $XY^2=\\boxed{717}$. (Notice that $MXTY$ is a parallelogram, which is an important theorem 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. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label(\"\\omega\", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 =", "to get the point $D^\\prime$. $\\angle PCD$ is congruent to $\\angle D^\\prime CD$, since $P$, $D^\\prime$, and $C$ are collinear. Therefore, we can just find $\\cos(\\angle D^\\prime CD)$. Note that $\\triangle CD^\\prime D$ is a right triangle with $\\angle CD^\\prime D$ as a right angle. $[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); R=midpoint(AA--CC); pair X=intersectionpoints(P--CC,BB--R)[0]; draw(AA--CC,dashed); draw(DD--X,dashed); draw(X--CC,dashed); draw(rightanglemark(CC,X,DD)); dot(P); dot(Q); dot(X); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); label(\"D^\\prime\",X,W); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]$ As given by the problem, $CD=1$. Note that $D^\\prime$ is the centroid of equilateral $\\triangle ABC$. Additionally, since $\\triangle ABC$ is equilateral, $D^\\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\\prime$ is $\\frac{2}{3}$ of the altitude of $\\triangle ABC$. Therefore, $CD^\\prime=\\frac{\\sqrt{3}}{3}$. Since $\\cos(\\angle D^\\prime CD)=\\cos(\\angle PCQ)=\\frac{CD^\\prime}{CD}$, $\\cos(\\angle PCQ)=\\frac{\\frac{\\sqrt{3}}{3}}{1}=\\frac{\\sqrt{3}}{3}$ $$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)$$ Simplifying, $PQ^2=\\frac{1}{2}$. Therefore, $PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow$ $\\boxed{\\textbf{C}}$ Solution by treetor10145 ## Solution 2 (less overkill) Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\\triangle ADP$. We find that $\\angle{DAP} = 60 ^{\\circ}$ (since $\\triangle ABD$ is equilateral), and $AD=\\frac{1}{2}$. Use the", "to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so $$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}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. ## Solution 2 (Linearity) Let $O_{1}$,", "+0 # We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB 0 42 1 We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB [asy] size(5cm); pair A,B,C; A=(0,0); B=(5,0); C=(1,1); draw(A--B--C--A); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); [/asy] Nov 6, 2020", "# In the diagram below, point $u$ is on segment $\\overline{by}$ and the measure of $\\angle bug$ is $30$ degrees greater In the diagram below, point $u$ is on segment $\\overline{by}$ and the measure of $\\angle bug$ is $30$ degrees greater than the measure of $\\angle guy$: [asy] unitsize(50); pair y=(cos(45*pi/180),sin(45*pi/180)); pair g=(cos(120*pi/180),sin(120*pi/180)); pair b=(cos(225*pi/180),sin(225*pi/180)); pair u=(0,0); dot(y); dot(g); dot(b); dot(u); label(“$y$”,y,se); label(“$g$”,g,sw); label(“$b$”,b,se); label(“$u$”,u,se); draw(b–y); draw(u–g); [/asy] how many degrees are in the measure of $\\angle guy$?.", "+0 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "= \\angle XBW = 90^{\\circ} = \\angle AMH = \\angle ATH$$ 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)", "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", "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}", "+0 # Pls Help 0 86 1 Point $X$ is on $\\overline{AC}$ such that $AX = 4\\cdot CX = 12$. We know $\\angle ABC = \\angle BXA = 90^\\circ.$ What is $BX$? [asy] size(5cm); pair A,B,C,X; A = (0,0); X = (12,0); C = (16,0); B = (12,sqrt(48)); draw(X--B--A--C--B); draw(rightanglemark(A,B,C,20)); draw(rightanglemark(B,X,A,20)); label(\"$A$\",A,SW); label(\"$B$\",B,N); label(\"$C$\",C,SE); label(\"$X$\",X,S); label(\"$12$\",(A+X)/2,S); [/asy] Nov 1, 2019", "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]$", "+0 # Can I get some help with these? 0 68 1 Hi... I've been strugling with these three problems for a few weeks: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6},$$ where the angle$$\\frac{\\pi}{6}$$ is in radians. (2)Find the length AC, to two decimal places. Image: https://artofproblemsolving.com/texer/sywndnao (3)Find the length BC, to two decimal places. Image: https://artofproblemsolving.com/texer/enqsdotw Nov 15, 2020 #1 0 Quick note: to get the images for (2) and (3), post these codes to get the figures. Make sure you say Render as Image in the TeXeR (2) [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3) [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] Nov 15, 2020", "+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", "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);", "+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", "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" ]

[ "+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", "# 2001 AMC 12 Problems/Problem 24 ## Problem In $\\triangle ABC$, $\\angle ABC=45^\\circ$. Point $D$ is on $\\overline{BC}$ so that $2\\cdot BD=CD$ and $\\angle DAB=15^\\circ$. Find $\\angle ACB$. $\\text{(A) }54^\\circ \\qquad \\text{(B) }60^\\circ \\qquad \\text{(C) }72^\\circ \\qquad \\text{(D) }75^\\circ \\qquad \\text{(E) }90^\\circ$ ## Solution $[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); pair ortho=rotate(-90)*(D-A); pair E=intersectionpoint(A--D, C--(C+10*ortho)); draw(A--B--C--cycle); draw(A--D); draw(C--E, dashed); draw(B--E, dashed); draw( rightanglemark(C,E,D) ); draw( scale(2)*anglemark(B,A,D) ); draw( anglemark(D,B,A) ); label(\"15^\\circ\",(0.6,0),5*ENE); label(\"45^\\circ\",B,5*WNW,UnFill); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",E,S); [/asy]$ We start with the observation that $\\angle ADB = 180^\\circ - 15^\\circ - 45^\\circ = 120^\\circ$, and $\\angle ADC = 15^\\circ + 45^\\circ = 60^\\circ$. We can draw the height $CE$ from $C$ onto $AD$. In the triangle $CED$, we have $\\frac {ED}{CD} = \\cos EDC = \\cos 60^\\circ = \\frac 12$. Hence $ED = CD/2$. By the definition of $D$, we also have $BD=CD/2$, therefore $BD=DE$. This means that the triangle $BDE$ is isosceles, and as $\\angle BDE=120^\\circ$, we must have $\\angle BED = \\angle EBD = 30^\\circ$. Then we compute $\\angle ABE = 45^\\circ - 30^\\circ = 15^\\circ$, thus $\\angle ABE = \\angle BAE$ and the triangle $ABE$ is isosceles as well. Hence $AE=BE$. Now we can note that $\\angle DCE = 180^\\circ - 90^\\circ - 60^\\circ = 30^\\circ$,", "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 $\\angle guy$?.", "+0 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "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}", "about $\\cos XTY$ to obtain the result $XY^2=\\boxed{717}$. (Notice that $MXTY$ is a parallelogram, which is an important theorem 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. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label(\"\\omega\", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 =", "∠A. Label the angles as shown. (∠ACD ≅ ∠DCB since $$\\overline{C D}$$ bisects ∠ACB) Since AB = BC, ∆ABC is isosceless, therefore 2x = a. m∠A + m∠ACE + m∠E = 180° a + (x + 42) + 90 = 180 2x + x + 132 = 180 x = 16 Since a = 2x, m∠A = 32° Question 3. In the figure below, $$\\overline{A D}$$ is the angle bisector of ∠BAC. $$\\overline{B A P}$$ and $$\\overline{B D C}$$ are straight lines, and $$\\overline{A D}$$ || $$\\overline{P C}$$. Prove that AP = AC. Label w, x, y and z as shown. Statements Reasons 1. $$\\overline{A D}$$ is the angle bisector of ∠BAC Given 2. $$\\overline{A D}$$ || $$\\overline{P C}$$ Given 3. z = w Definition of angle bisector 4. z = y If two parallel lines are cut by a transversal, alt. internal angles are equal in measure. 5. w = x If two parallel lines are cut by a tranversal, corr. Angles are equal in measure. 6. x = y Transitive property 7. ∆ACP is isosceless Base angles are congruent 8. AC = AP Definition of isosceles triangle. Question 4. ∆ ABC and ∆ DEF, in the figure below are such that $$\\overline{A B}$$ ≅ $$\\overline{D E}$$, $$\\overline{A C}$$ ≅ $$\\overline{D F}$$ and ∠A ≅∠D. a. Which", "# 2005 CEMC Gauss (Grade 7) Problems/Problem 8 ## Problem In the diagram, what is the measure of $\\angle ACB$ in degrees? $[asy] size(300); draw((-60,0)--(0,0)); draw((0,0)--(64.3,76.6)--(166,0)--cycle); label(\"A\",(64.3,76.6),N); label(\"93^\\circ\",(64.3,73),S); label(\"130^\\circ\",(0,0),NW); label(\"B\",(0,0),S); label(\"D\",(-60,0),S); label(\"C\",(166,0),S); [/asy]$ $\\text{(A)}\\ 57^\\circ \\qquad \\text{(B)}\\ 37^\\circ \\qquad \\text{(C)}\\ 47^\\circ \\qquad \\text{(D)}\\ 60^\\circ \\qquad \\text{(E)}\\ 17^\\circ$ ## Solution Since $\\angle ABC + \\angle ABD = 180^\\circ$ (in other words, $\\angle ABC$ and $\\angle ABD$ are supplementary) and $\\angle ABD = 130^\\circ$, then $\\angle ABC = 50^\\circ$. Since the sum of the angles in triangle $ABC$ is $180^\\circ$ and we know two angles $93^\\circ$ and $50^\\circ$ which add to $143^\\circ$, then $\\angle ACB = 180^\\circ - 143^\\circ = 37^\\circ$. The answer is $B$.", "Let $\\overline{AB}$ and $\\overleftrightarrow{CD}$ be two straight lines intersecting at the point $E .$ According to the figure $\\angle A E D$ and $\\angle C E B$ and $\\angle A E C$ and $\\angle B E D$ are pair wise vertically opposite angles. We have to prove that, $\\angle A E D=\\angle C E B$ and $\\angle A E C=\\angle B E D$ Now, $E$ is the source and $\\overrightarrow{E C}$ is a ray. Then according to the Linear Pair axioms $\\angle A E C+\\angle C E B=180^{\\circ}$. Similar arguments ,assuming $\\overrightarrow{E B}$ as ray, gives $\\angle C E B+\\angle B E D=180^{\\circ} .$ Therefore, we get, $\\angle A E C+\\angle C E B=\\angle C E B+\\angle B E D$ or $, \\angle A E C=\\angle B E D$ Similarly we can show that,$\\angle A E D=\\angle C E B . \\text { (Q.E.D.) }$", "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 # Pls Help 0 86 1 Point $X$ is on $\\overline{AC}$ such that $AX = 4\\cdot CX = 12$. We know $\\angle ABC = \\angle BXA = 90^\\circ.$ What is $BX$? [asy] size(5cm); pair A,B,C,X; A = (0,0); X = (12,0); C = (16,0); B = (12,sqrt(48)); draw(X--B--A--C--B); draw(rightanglemark(A,B,C,20)); draw(rightanglemark(B,X,A,20)); label(\"$A$\",A,SW); label(\"$B$\",B,N); label(\"$C$\",C,SE); label(\"$X$\",X,S); label(\"$12$\",(A+X)/2,S); [/asy] Nov 1, 2019", "+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", "# Difference between revisions of \"2014 AMC 8 Problems/Problem 9\" ## Problem In $\\bigtriangleup ABC$, $D$ is a point on side $\\overline{AC}$ such that $BD=DC$ and $\\angle BCD$ measures $70^\\circ$. What is the degree measure of $\\angle ADB$? $[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot(\"A\", A, SW); dot(\"B\", B, NE); dot(\"C\", C, SE); dot(\"D\", D, S); label(\"70^\\circ\",C,2*dir(180-35));[/asy]$ $\\textbf{(A) }100\\qquad\\textbf{(B) }120\\qquad\\textbf{(C) }135\\qquad\\textbf{(D) }140\\qquad \\textbf{(E) }150$ ## Solution $BD = DC$, so $\\angle DBC = \\angle DCB = 70$. Then $\\angle CDB = 180-(70+70) = 40$. Since $\\angle ADB$ and $\\angle BDC$ are supplementary, $\\angle ADB = 180 - 40 = \\boxed{\\textbf{(D)}~140}$.", "= \\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)", "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", "+0 # Can I get some help with these? 0 68 1 Hi... I've been strugling with these three problems for a few weeks: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6},$$ where the angle$$\\frac{\\pi}{6}$$ is in radians. (2)Find the length AC, to two decimal places. Image: https://artofproblemsolving.com/texer/sywndnao (3)Find the length BC, to two decimal places. Image: https://artofproblemsolving.com/texer/enqsdotw Nov 15, 2020 #1 0 Quick note: to get the images for (2) and (3), post these codes to get the figures. Make sure you say Render as Image in the TeXeR (2) [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3) [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] Nov 15, 2020", "# 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°.", "+0 # We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB 0 42 1 We know $\\angle A=45^\\circ, \\angle B=15^\\circ,$ and $BC=\\sqrt{6}.$ What is AB [asy] size(5cm); pair A,B,C; A=(0,0); B=(5,0); C=(1,1); draw(A--B--C--A); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); [/asy] Nov 6, 2020", "+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" ]

[ "= 660 ‘`USft`’, though the metric equivalents differ slightly between the two cases. The ‘`US`’ prefix or the ‘`survey`’ prefix can also be used to obtain the US survey mile and the value of the US yard prior to 1959, e.g., ‘`USmile`’ or ‘`surveymile`’ (but not`USsurveymile`’). To get the US value of the statute mile, use either ‘`USstatutemile`’ or ‘`USmile`’. Except for distances that extend over hundreds of miles (such as in the US State Plane Coordinate System), the differences in the miles are usually insignificant: ```You have: 100 surveymile - 100 mile You want: inch * 12.672025 / 0.078913984``` The pre-1959 UK values for these units can be obtained with the prefix ‘`UK`’. In the US, the acre is officially defined in terms of the US survey foot, but `units` uses a definition based on the international foot. If you want the official US acre use ‘`USacre`’ and similarly use ‘`USacrefoot`’ for the official US version of that unit. The difference between these units is about 4 parts per million. ## Unit Expressions ### Operators You can enter more complicated units by combining units with operations such as multiplication, division, powers, addition, subtraction, and parentheses for grouping. You can use the customary symbols for these operators when `units` is invoked with its default options. Additionally, `units` supports", "square mile = about 2.6 square km (2.58998811 square kilometers). One square mile = about 2.6 km2 (2.58998811 sq kilometers).", "US yard prior to 1959, e.g., 'USmile' or 'surveymile' (but not 'USsurveymile'). To get the US value of the statute mile, use either 'USstatutemile' or 'USmile'. Except for distances that extend over hundreds of miles (such as in the US State Plane Coordinate System), the differences in the miles are usually insignificant: ```You have: 100 surveymile - 100 mile You want: inch * 12.672025 / 0.078913984 ``` The pre-1959 UK values for these units can be obtained with the prefix 'UK'. In the US, the acre is officially defined in terms of the US survey foot, but 'units' uses a definition based on the international foot. If you want the official US acre use 'USacre' and similarly use 'USacrefoot' for the official US version of that unit. The difference between these units is about 4 parts per million. # UNIT EXPRESSIONS¶ ## Operators¶ You can enter more complicated units by combining units with operations such as multiplication, division, powers, addition, subtraction, and parentheses for grouping. You can use the customary symbols for these operators when 'units' is invoked with its default options. Additionally, 'units' supports some extensions, including high priority multiplication using a space, and a high priority numerical division operator ('|') that can simplify some expressions. You multiply units using a space or an asterisk ('*'). The", "## Thinking Mathematically (6th Edition) We know that 1 mile = 1 km We know that 1 hectare = 0.01 square kilometer We can find how many square kilometers are in 1 square mile. $1~mi^2 = 1~mi\\times 1~mi$ $1~mi^2 = 1.6~km\\times 1.6~km$ $1~mi^2 = 2.56~km^2$ We can find how many square miles are in one hectare. $1 ~ha = \\frac{1~ha}{1}\\times \\frac{0.01~km^2}{1~ha}\\times \\frac{1~mi^2}{2.56~km^2}$ $1~ha = 0.00391~mi^2$ We can express the population density in units of people per square mile. $\\frac{46,690~people}{1000~ha} = \\frac{46,690~people}{1000~ha}\\times \\frac{1~ha}{0.00391~mi^2} = \\frac{11,941~people}{1~mi^2}$ The population density is 11,941 people per square mile.", "Math and Arithmetic Units of Measure Area Length and Distance # One square mile is equal to how many square kilometers? ###### Wiki User 1 square mile=2.58998811 square kilometers 🙏 0 🤨 0 😮 0 😂 0 ## Related Questions There are 2.58998811 square kilometers in a mile. 1 mile = 1.609 kilometers, so 1 square mile = 1.609 x 1.609 km2 = 2.59 km2 One square mile is equal to about 2.6 square kilometers. as a kilometer is smaller there are 0.386 square kilometers in a square mile One square mile is 2.6 (2.58999) square kilometers. A square mile means an area, equal to a square of side 1 mile There are about 2.59 square kilometers in 1 square mile. One square mile equates to about 2.6 square km 27,878,400 square feet equal 1 square mile. 1 square mile = 2.5899881 square kilometres. So, 163,707 square miles = 424,232.28 square kilometres A square mile is equal to 640 acres A square kilometers is smaller as a km = 0.6213711922 miles So a square kilometer would be 398 acres One square mile is equal to 3,097,600 square yards The same amount of that would equal a mile on the road... 1.6km = 1 Mile. 1 (square mile) = 2.58998811 square kilometers1 (square kilometer) = 0.386102159 square mile 1", "prefix can also be used to obtain the US survey mile and the value of the US yard prior to 1959, e.g., 'USmile' or 'surveymile' (but not 'USsurveymile'). To get the US value of the statute mile, use either 'USstatutemile' or 'USmile'. Except for distances that extend over hundreds of miles (such as in the US State Plane Coordinate System), the differences in the miles are usually insignificant: You have: 100 surveymile - 100 mile You want: inch * 12.672025 / 0.078913984 The pre-1959 UK values for these units can be obtained with the prefix 'UK'. In the US, the acre is officially defined in terms of the US survey foot, but 'units' uses a definition based on the international foot. If you want the official US acre use 'USacre' and similarly use 'USacrefoot' for the official US version of that unit. The difference between these units is about 4 parts per million. # UNIT EXPRESSIONS¶ ## Operators¶ You can enter more complicated units by combining units with operations such as multiplication, division, powers, addition, subtraction, and parentheses for grouping. You can use the customary symbols for these operators when 'units' is invoked with its default options. Additionally, 'units' supports some extensions, including high priority multiplication using a space, and a high priority numerical division operator ('|') that", "cm = 500000 x 10 mm = 5000000 mm Question 2: Convert 10 m to km. Solution: 1 m = 10-3 km 10 m = 10 x 10-3 km = $\\frac{10}{1000}$ km = $\\frac{1}{100}$ km = 10 -2 km ## Converting between Metric Units ### Solved Examples Question 1: To convert 1000 acres to Hectares. Solution: Here we have to convert acres to Hectares. 1 acres = 0.404685642 hectares Therefore, 1000 acres = 0.404685642 hectares [xx] 1000 acres = 404.685642 The answers for 1000 acres are equal to 404.685642 Hectares. Question 2: To convert 700 Hectares to Square feet. Solution: Here we have to convert Hectares to Square feet. 1 hectare = 107 639.104 square feet Therefore, 700 Hectares = 107 639.104 square feet [xx] 700 Hectares = 75347372.8 sq.ft The answers for 700 Hectares are equal to 75347372.8 Square feet. Question 3: To convert 100 Kilometers (km) to Miles. Solution: Here we have to convert Kilometers (km) to Miles. 1 kilometer (km) = 0.621371192 miles Therefore, 100 km = 0.621371192 miles [xx] 100 km = 62.1371192 miles The answers for 100 Kilometers (km) are equal to 62.1371192 miles. Question 4: To convert 300 millimeters to inches. Solution: Here we have to convert millimeters (mm) to inches 1 millimeter (mm) = 0.0393701 inches Therefore, 300 mm = 0.0393701", "2 = 72 , 000 , 000 43 , 560 = 1652.892 acres Monday March 29th, 2021 Spelling Distance Learning Assignments Hey all, So today something that has been in the works for a while and I think it's (finally) ready for the public. Now on the Spelling Worksheet Maker you'll see an option for 'Distance Learning'. Just click on that and instead of outputing worksheets it'll generate distance learning assignments. And as a bonus, any previously created spelling list is compatible with the distance learning option. Huzzah! I've been testing and tweaking it for a few months now, so I'm absolutely 100% positive there are no bugs or glitches. That being said. when you find any bug or glitches please let me know in the comments. :P Easter Coloring Sheets Also just in time for easter there are a few coloring sheets available here: Coloring Sheets", "square miles (\"mi\"^2) etc. We have twelve built-in units of area that can be used in vCalc code to make both inputs and outputs appear with their respective units. And this vCalc Area Conversion Factors equation will let you choose any two area units and see the conversion factor used to convert one unit into another unit. The area unit conversion factors available inside vCalc are: 1. meters-squared (\"m\"^2) 2. micrometers-squared (mu\"m\"^2) 3. millimeters-squared (\"mm\"^2) 4. centimeters-squared (\"cm\"^2) 5. kilometers-squared (\"km\"^2) 6. inches-squared (\"in\"^2) 7. feet-squared (\"ft\"^2) 8. yards-squared (\"yd\"^2) 9. miles-squared (\"mi\"^2) 10. acres 11. ares(a) 12. hectares (ha)", "# Square mile The square mile was a European and North American measure of area as widely used as the mile into the 19th century . Their extent varied from country to country due to the different mile lengths. ## Square mile in Germany One unit of the square mile used in German-speaking countries up to the 19th century corresponds to an area of ​​around 55–57 square kilometers (cf. mile , in Germany from 7419 to 7533 m). Physical unit Unit name Square mile Unit symbol ${\\ displaystyle \\ mathrm {sqmi, mi ^ {2}}}$ Physical quantity (s) Area Formula symbol ${\\ displaystyle A}$ dimension ${\\ displaystyle {\\ mathsf {L ^ {2}}}}$ system Anglo-American system of measurement In SI units ${\\ displaystyle \\ mathrm {1 \\, sqmi = 2 \\, 589 \\, 988 {,} 110 \\, 336 \\; m ^ {2}}}$ Derived from Statute mile ## Anglo-American Square Mile As a square mile , it is used in the Anglo-American system of measurement and, in the 21st century, especially in the United States . A square mile is the area of ​​a square with an edge length of 1 mile each (\" statute mile \"; in contrast to the \" nautical mile \" = nautical mile ). This area corresponds to the equivalent of 2.589988110336 km² , i.e. almost", "groups of three digits. Furthermore, in documentation, the comma is used as a separator for groups of three digits when four or more digits appear on either side of the decimal marker. The IEEE standard is to use a space separator. Unit Conversion Factors In table A.1 of the IEEE/ASTM standard, the floating-point approximations for converting degrees, grad, and mil (angle) to radians, kilometers per hour to meters per second and lamberts to candela per square meter are labeled as exact conversions. In the Units package, the exact, rational conversion rates are used. The table indicates that an acre equals 43560 square US survey feet, a chain equals 66 US survey feet, and a rod equals 16.5 US survey feet. Because these units are not intrinsically associated with US survey units, these conversions are used for US survey acres (acre[US_survey]), US survey chains (chain[US_survey]), and US survey rods (rod[US_survey]). In all other cases, the Units package uses either the exact conversion rate from or a floating-point approximation at least as accurate as that listed in table A.1 of the IEEE/ASTM standards.", "answer is given correctly as a repeating decimal .00142857 (Braid uses dots over the digits to indicate the beginning and end of a repeating group. We have used an \"overline\".) 3. Reduce $\\large\\frac{49}{52}\\normalsize$ to a decimal. Solution. Again long division until the remainder repeats when the answer is given correctly as a repeating decimal. .94230769 4. Reduce $\\large\\frac{655}{2149}\\normalsize$ to a decimal. Solution. Long division gives .3047929+. It is not clear how he decides the number of places to calculate. In fact this fraction has a repeating part of length 306. 5. Reduce 157 yds 1 ft 3 in to a mile. Solution. He divides 3 in by 12 to get .25 in, then 1.25 ft by 3 to get .416 yards, then 157.416 by 1760 to get .089441287 mile. 6. Reduce 1 rood 20 poles to an acre. 7. Reduce 5 hrs 48 min 50 sec of a day. 8. Reduce 17' 44'' 48''' to a degree. 9. Value .603125 acres. Solution. He multiplies by 4 to get 2.412500 roods. Then he multiplies .412500 by 40 to get 16.500000 (square) poles, then .500000 by $30\\large\\frac{1}{4}\\normalsize$ to get 15.125000. Finally he converts .125000 to $\\large\\frac{1}{8}\\normalsize$. So the answer is 2 roods 16 (square) poles $15\\large\\frac{1}{8}\\normalsize$ (square) yards. 10. Value .483125 miles. Solution. He converts to furlongs, yds, ins. (There", "Your World, Your Turn (Florida), mathematical transformation of points from the sphere or ellipsoid onto a plane, term used for surfaces in geometric form used in projection process. Highly accurate units that are capable of detecting location differences of less than 1 centimeter of latitude, longtitude, and elevation. }{0^{\\circ} \\text {LONG} } \\ { 79^{\\circ} \\mathrm{W}}\\\\ Ex: there are 63,360 inches in a mile, so on a map with a fractional scale of 1:63,360, we can say 1 inch represents 1 mile, One that has a relatively large representative fraction or the denominator is smaller, One that has a relatively small representative fraction or the denominator is larger, a system in which the spherical surface of Earth is transformed for display on a flat surface. You can control some of these but never all, Cartographers often strive to maintain accuracy either of size or shape using map properties known as, In an equivalent map projection or equal area map projection, the correct size ratio of area on the map to the corresponding actual area on Earth's surface is maintained over the entire map. More recently, scientists have begun inserting Bt genes into other plant types, such as rice, canola, and cotton. Which statement regarding a global environmental indicator is NOT correct? What is the most important active sensing", "what now? The x-axis is tied to the infrastructure number. 7 is less than a typical infrastructure number. In the original post, the bell curve is scaled to 100. For 400 hexes, we have to compress it (assuming 100 is the max). For now, the easiest thing to do is just divide 400 by 100 and say that each point of infrastructure is worth between 1 and 4 extra hexes of not-wilderness. These thoughts are still a bit incoherent, so I apologize. I'm trying to wrap my brain around the rule-defined system. Nobody tell Jeremy Crawford that someone is trying to use rules in D&D. ## June 1, 2018 ### Demoland While I work on the big map, inching towards perfection, I want to create a test case to work through the trade system. This will keep my creative juice flowing. Demoland is such a test case. The names come from a Markov chain generator I'm working on, based here on a selection of Welsh names. Once I like the results more, I'll put up a full post about it. These are 20-mile hexes (nominally a day's travel), with an area of ~346 square miles. The total map has 68 hexes (~23,562 square miles). That's roughly the size of West Virginia: a lot of land for people to", "map makers contented themselves with placing upon their maps a linear scale of miles, deduced from the central meridian or the equator. They now add the proportion which these units of length have to nature, or state how many of these units are contained within some local measure of length. The former method, usually called the “ natural scale,” may be described as “ international,” for it is quite independent of local measures of length, and depends exclusively upon the size and figure of the earth. Thus a scale of 1:1,000,000 signifies that each unit of length on the map represents one million of such units in nature. The second method is still employed in many cases, and we find thus:— 1 in. = 1 statute mile (of 63,366 in.) corresponds to 1 : 63,366 6 in. = 1 ,, ,, ,, ,, 1 : 10,560 1 in. = 5 chains (of 858 in.) . . ,, 1 : 4,890 1 in. = 1 nautical mile (of 73,037 in.) . ,, 1 : 73,037 1 in. = 1 verst (of 42,000 in.) . . ,, 1 : 42,000 2 Vienna in. = 1 Austrian mile (of 288,000 in.) . . . . ,, 1 : 144,000 1 cm. = 500 metres (of 100 cm.) . . ,, 1", "The ‘US’ prefix or the ‘survey’ prefix can also be used to obtain the US survey mile and the value of the US yard prior to 1959, e.g., ‘USmile’ or ‘surveymile’ (but notUSsurveymile’). To get the US value of the statute mile, use either ‘USstatutemile’ or ‘USmile’. Except for distances that extend over hundreds of miles (such as in the US State Plane Coordinate System), the differences in the miles are usually insignificant: You have: 100 surveymile - 100 mile You want: inch * 12.672025 / 0.078913984 The pre-1959 UK values for these units can be obtained with the prefix ‘UK’. In the US, the acre is officially defined in terms of the US survey foot, but units uses a definition based on the international foot. If you want the official US acre use ‘USacre’ and similarly use ‘USacrefoot’ for the official US version of that unit. The difference between these units is about 4 parts per million. Next: , Previous: Unit Definitions, Up: Top ## 5 Unit Expressions ### 5.1 Operators You can enter more complicated units by combining units with operations such as multiplication, division, powers, addition, subtraction, and parentheses for grouping. You can use the customary symbols for these operators when units is invoked with its default options. Additionally, units supports some extensions, including high", "may be used as bases for general maps. The sheets are drawn on the scale of 1:500,000, and photolithographs on this scale are available for use as bases for geologic or other maps. The adaptability of the 1:1,000,000 scale map to use as a base for general geologic maps is shown in the geologic maps of the southern peninsula of Michigan and of Indiana in Monograph 53 (Pls. IV and VII), the map of Florida in Bulletin 60 (Pl. I), and the map of Vermont in Water-Supply Paper 424 (Pl. I). 3. The Survey's two-sheet wall map of the United States, 49 by 76 inches, scale 1:2,500,000 (approximately 40 miles to 1 inch). Parts of this map can be used as bases for general geologic or other maps and as copy for index and other small diagrammatic maps. This map is published both with and without contours. 4. Land Office maps and township plats. These maps are now being published on a scale of 12 miles to 1 inch; they are also photo-lithographed on one-half that scale, or 24 miles to 1 inch. The township [15] plats are printed on a scale of one-half mile to 1 inch. The maps are especially useful in compiling maps in which land lines (townships and sections) are essential, and the township", "land use will include multiple land uses. For these situations we recommend using the recommended value for mixed land uses (0.290 mg/L), adjusting this emc based on local data, or calculating the emc. An emc can be calculated if the total area of interest (Atotal), the area of each land use in the area of interest, and the emc for each land use in the area of interest are known. Site emc = Σ1n ((AArea 1 * emcArea 1)/ (Atotal) + ... ((AArea n * emcArea n) / (Atotal) where A = area in acres. Example calculation • 10 acres of residential; emc = 0.325 mg/L • 10 acres of commercial; emc = 0.200 mg/L • 10 acres of industrial; emc = 0.235 mg/L • 1 acre of transportation; emc = 0.280 mg/L Overall emc = ((0.325 * 10)/31) + ((10 * 0.200)/31) + ((10 * 0.235)/31) + ((1 * 0.280)/31) = 0.254 mg/L ### Agricultural land use emcs We did not conduct a rigorous literature review of emcs for agricultural land uses. We briefly reviewed several papers and compiled the data into ranges. The following papers were reviewed. Note that there is considerable variability in land uses described as agricultural (e.g. row crop, pasture, small grain, etc.) and in the practices employed within these land uses, particularly", "# Jesse's path A 18 mile by 57 mile plot of land is divided into $$1026$$ counties of 1 mile by 1 mile squares. A man starts at the top right corner of the plot and travels a straight line path to the bottom left corner. In each county that he travels through, he spends $$-4$$ times the sum of the amounts he spent in the previous two counties. He spends $1 in the first county and$2 in the second county. If he spends $$a\\times 2^b$$ dollars in the last county, where $$a$$ is odd and $$b \\geq 0$$, find $$a+b$$. This problem is posed by Jesse Z. Details and assumptions Spending a negative amount of money is equivalent to receiving money. As an explicit example, he receives \\$12 in the third county. ×", "multiple land uses. For these situations we recommend using the recommended value for mixed land uses (0.290 mg/L), adjusting this emc based on local data, or calculating the emc. An emc can be calculated if the total area of interest (Atotal), the area of each land use in the area of interest, and the emc for each land use in the area of interest are known. Site emc = Σ1n ((AArea 1 * emcArea 1)/ (Atotal) + ... ((AArea n * emcArea n) / (Atotal) where A = area in acres. Example calculation • 10 acres of residential; emc = 0.325 mg/L • 10 acres of commercial; emc = 0.200 mg/L • 10 acres of industrial; emc = 0.235 mg/L • 1 acre of transportation; emc = 0.280 mg/L Overall emc = ((0.325 * 10)/31) + ((10 * 0.200)/31) + ((10 * 0.235)/31) + ((1 * 0.280)/31) = 0.254 mg/L ### Agricultural land use emcs We did not conduct a rigorous literature review of emcs for agricultural land uses. We briefly reviewed several papers and compiled the data into ranges. The following papers were reviewed. Note that there is considerable variability in land uses described as agricultural (e.g. row crop, pasture, small grain, etc.) and in the practices employed within these land uses, particularly with the management of" ]

[ "base $$b-c$$. This allows us to calculate the area of the whole trapezium:$A=\\left(\\frac{b\\,- c}{2}\\right)h + c\\,h = \\left(\\frac{b+c}{2}\\right)h$The midpoints of the two diagonals are both at a height $$h/2$$ so the signpost is also at this height. It is now easy to see that the areas of the lower and upper fields are:$A_1=\\frac{b h}{4}\\;\\;\\;\\;A_3 = \\frac{c h }{4}$ With the left and right field areas as $$A_2$$ and $$A_4$$, we hence know that:$A=2(A_1+A_3)=2(A_2+A_4)$ where one of the fields has an area of $$3$$ acres and the opposite field has an area of $$3 – x$$ acres with $$x < 1$$. So we have:$A = 12 - 2x \\;\\;\\;\\text{with} \\;\\;\\;0 < x < 1$. Since $$A$$ is an integer in acres, the total area of the four fields can only be $$11$$ acres.", "# 7th Class Mathematics Mensuration Standard Units of Area Standard Units of Area Category : 7th Class ### Standard Units of Area The inter relationship among various units of measurement of area are listed below. $1{{m}^{2}}$ $=$ $100\\times 100c{{m}^{2}}={{10}^{4}}c{{m}^{2}}$ $1{{m}^{2}}$ $=$ $10\\times 10\\,d{{m}^{2}}=100d{{m}^{2}}$ $1\\,d{{m}^{2}}$ $=$ $10\\times 10\\,c{{m}^{2}}=100{{m}^{2}}$ $1\\,da{{m}^{2}}$ $=$ $10\\times 10\\,{{m}^{2}}=100\\,{{m}^{2}}$ $1\\,h{{m}^{2}}$ $=$ $100\\times 100\\,{{m}^{2}}={{10}^{4}}\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $1000\\times 1000\\,{{m}^{2}}={{10}^{6}}\\,{{m}^{2}}$ $1\\,hectare$ $=$ $10000\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $100\\,hectares$ #### Other Topics You need to login to perform this action. You will be redirected in 3 sec", "> > a map has a scale of 1: 50,000 the area of the farm on the map is 6cm^2. # a map has a scale of 1: 50,000 the area of the farm on the map is 6cm^2. ## What is the real area of the farm in hectares? (1 hectare = 10,000... 352 Views ### a map has a scale of 1: 50,000 the area of the farm on the map is 6cm^2. What is the real area of the farm in hectares? (1 hectare = 10,000 m^2= 0.01 km^2) $$a\\ scale\\ s=1:50,000=\\frac{1}{50,000}=\\frac{1}{5\\cdot10^4}\\\\\\\\A_m=6cm^2-the\\ area o\\ the\\ map\\\\\\\\A_r=? -the eal\\ area\\\\\\\\\\frac{A_m}{A_r}=s^2\\\\\\\\\\frac{6}{A_r}=\\left(\\frac{1}{5\\cdot10^4} ight)^2\\\\\\\\\\frac{6}{A_r}=\\frac{1}{(5\\cdot10^5)^2}\\\\\\\\\\frac{6}{A_r}=\\frac{1}{5^2\\cdot(10^4)^2}$$", "$$2b$$ Perimeter of original land ABCD = $$2 (l + b)$$ Acc to ques => $$\\frac{2b}{2 (l + b)} = \\frac{3}{8}$$ => $$8b = 3l + 3b$$ => $$5b = 3l$$ Only one combination is possible, i.e. $$l = 5$$ and $$b = 3$$ ($$\\because$$ It is given that : $$l < 2b$$) => Area of land = $$5 \\times 3 = 15$$ sq. ft => Cost of land = $$15 \\times 1000 = Rs. 15,000$$ For overall profit to be 10 %, S.P. = $$15,000 \\times \\frac{110}{100}$$ = $$Rs. 16,500$$ Side of AEFD = $$b = 3$$ => S.P. of AEFD = $$3^2 \\times 1200 = Rs. 10,800$$ Side of EBHG = $$(l - b) = 5 - 3 = 2$$ S.P. of EBHG = $$2^2 \\times 1150 = Rs. 4,600$$ Length of remaining rectangular land GHCF = $$(l - b) = 5 - 3 = 2$$ Breadth = $$(2b - l) = 6 - 5 = 1$$ Let selling price per sq. ft of GHCF = $$Rs. x$$ S.P. of GHCF = $$2 \\times 1 \\times = Rs. 2x$$ $$\\therefore$$ Total S.P. => $$10,800 + 4,600 + 2x = 16,500$$ => $$2x = 16,500 - 15,400$$ => $$x = \\frac{1100}{2} = Rs. 550$$", "cm = 500000 x 10 mm = 5000000 mm Question 2: Convert 10 m to km. Solution: 1 m = 10-3 km 10 m = 10 x 10-3 km = $\\frac{10}{1000}$ km = $\\frac{1}{100}$ km = 10 -2 km ## Converting between Metric Units ### Solved Examples Question 1: To convert 1000 acres to Hectares. Solution: Here we have to convert acres to Hectares. 1 acres = 0.404685642 hectares Therefore, 1000 acres = 0.404685642 hectares [xx] 1000 acres = 404.685642 The answers for 1000 acres are equal to 404.685642 Hectares. Question 2: To convert 700 Hectares to Square feet. Solution: Here we have to convert Hectares to Square feet. 1 hectare = 107 639.104 square feet Therefore, 700 Hectares = 107 639.104 square feet [xx] 700 Hectares = 75347372.8 sq.ft The answers for 700 Hectares are equal to 75347372.8 Square feet. Question 3: To convert 100 Kilometers (km) to Miles. Solution: Here we have to convert Kilometers (km) to Miles. 1 kilometer (km) = 0.621371192 miles Therefore, 100 km = 0.621371192 miles [xx] 100 km = 62.1371192 miles The answers for 100 Kilometers (km) are equal to 62.1371192 miles. Question 4: To convert 300 millimeters to inches. Solution: Here we have to convert millimeters (mm) to inches 1 millimeter (mm) = 0.0393701 inches Therefore, 300 mm = 0.0393701", "$$\\frac{481}{6} + \\frac{121}{3}+50$$ 5- The distance between two cities is 3,768 feet. What is the distance of the two cities in yards? ☐A. 1,256 yd ☐B. 11,304 yd ☐C. 45,216 yd ☐D. 3,768 yd 6- Mr. Jones saves $3,400 out of his monthly family income of$74,800. What fractional part of his income does Mr. Jones save? ☐A. $$\\frac{1}{22}$$ ☐B. $$\\frac{1}{11}$$ ☐C. $$\\frac{3}{25}$$ ☐D. $$\\frac{2}{15}$$ 7- What is the lowest common multiple of 12 and 20? ☐A. 60 ☐B. 40 ☐C. 20 ☐D. 12 8- Based on the table below, which expression represents any value of f in term of its corresponding value of $$x$$? ☐A. $$f=2x-\\frac{3}{10}$$ ☐B. $$f=x+\\frac{3}{10}$$ ☐C. $$f=2x+2 \\frac{2}{5}$$ ☐D. $$2x+\\frac{3}{10}$$ 9- 96 kg $$=$$… ? ☐A. 96 mg ☐B. 9,600 mg ☐C. 960,000 mg ☐D. 96,000,000 mg 10- Calculate the approximate area of the following circle? (the diameter is 25) ☐A. 78 ☐B. 491 ☐C. 157 ☐D. 1963 ## Best 6th Grade SBAC Math Prep Resource for 2021 1- A $$420=2^2×3^1×5^1×7^1$$ 2- B The area of the trapezoid is: area= $$\\frac{(base 1+base 2)}{2})×height= ((\\frac{10 + 16}{2})x = A$$ $$→13x = A→x = \\frac{A}{13}$$ 3- A $$\\frac{104}{8}=13, \\frac{1352}{104}=13, \\frac{17576}{1352}=13$$ Therefore, the factor is 13 4- C Simplify each option provided. $$A. -20-(3×10)+(6×40)=-20-30+240=190$$ $$B. (\\frac{15}{8})×72 + (\\frac{125}{5}) =135+25=160$$ $$C. ((\\frac{30}{4} + \\frac{15}{2})×8) – \\frac{11}{2} + \\frac{222}{4} =", "7th Class Mathematics Mensuration Standard Units of Area Standard Units of Area Category : 7th Class Standard Units of Area The inter relationship among various units of measurement of area are listed below. $1{{m}^{2}}$ $=$ $100\\times 100c{{m}^{2}}={{10}^{4}}c{{m}^{2}}$ $1{{m}^{2}}$ $=$ $10\\times 10\\,d{{m}^{2}}=100d{{m}^{2}}$ $1\\,d{{m}^{2}}$ $=$ $10\\times 10\\,c{{m}^{2}}=100{{m}^{2}}$ $1\\,da{{m}^{2}}$ $=$ $10\\times 10\\,{{m}^{2}}=100\\,{{m}^{2}}$ $1\\,h{{m}^{2}}$ $=$ $100\\times 100\\,{{m}^{2}}={{10}^{4}}\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $1000\\times 1000\\,{{m}^{2}}={{10}^{6}}\\,{{m}^{2}}$ $1\\,hectare$ $=$ $10000\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $100\\,hectares$ Other Topics You need to login to perform this action. You will be redirected in 3 sec", "# Tag Archives: grade 6 mtap reviewers ## 2011 Grade 6 MTAP Questions – Part 2 This is the 2011 Math Challenge for Grade 6 questions 26-50. Questions 1-25 can be found here. 26.) A pyramid-like circus tent has a 32 m by 28 m rectangular base. If it is 15 m high, what volume of air does it enclose? Solution: v = (lwh)/3 v = (32 × 28 × 15)/3 v = (13440)/3 v = 4480 cu m. 27.) The parallel sides of a trapezoidal farm are 210 m by 150 m. If they are 120 m apart, what is the area of the farm in hectares? Solution: A = ((a+b))/2 h A = ((210+150))/2 120 A = 360/2 (120) A = 21600 sq m Convert meter to hectares = 2.16 hectares ## 2011 Grade 6 MTAP Questions – Part 1 These are the 2011 Math Challenge Questions for Grade 6 questions 1-25. Questions 26-50 can be found here. Solve each item and write the answer in the blank before the number. 1.) What number is 17 more than the difference of 95 and 37? Solution: 95 – 37 = 58 58 + 17 = 75 2.) Use exponents to express 5040 as a product of its prime factors. Answer: $2^4 \\times 3^2 \\times 5 \\times 7$", "can write: Area $\\approx$≈ $\\frac{340}{2}(410+2\\times(310+450)+360)$3402​(410+2×(310+450)+360) Area $\\approx$≈ $389300$389300m2 The trapezoidal rule extended For multiple applications of the trapezoidal rule: $Area\\approx\\frac{h}{2}(d_f+2\\times$Areah2(df+2×(The middle distances)$+d_l)$+dl) Where: $d_f$df​ $=$= first measurement $d_l$dl​ $=$= last measurement The middle distances $=$= all distances excluding the first and the last $h$h $=$= distance between successive measurements Be careful to notice that $d_f$df and $d_l$dl are now the first and last distances of the whole area, not of each trapezium. #### Practice questions ##### Question 1 The following piece of land has straight boundaries on the east, west and south borders and follows a creek at the north. The land has been divided into two sections to allow us to use the trapezoidal rule to approximate the area of the land. Note: the diagram is not to scale. 1. Approximate Area $1$1 using one application of the trapezoidal rule. 2. Approximate Area $2$2 using one application of the trapezoidal rule. 3. Hence, find the approximate area of the piece of land using your answers to parts (a) and (b). 4. Is the actual area of the land greater than or less than this calculated area? Greater than A Less than B Greater than A Less than B ##### Question 2 A surveyor provided the following diagram with measurements for a property she was mapping out. 1.", "## Elementary Technical Mathematics $$12.5 \\ ha$$ The area of the field is $$\\frac{1}{4} \\times \\frac{1}{2} \\ km^2=\\frac{1}{8} \\ km^2=\\frac{10^6}{8} \\ m^2 .$$ Now change it to $ha$ by dividing by $10^4$; that is $$\\frac{10^6}{8} \\ m^2= \\frac{10^6}{8\\times 10^4} \\ ha =\\frac{100}{8} \\ ha=12.5 \\ ha.$$", "in the image. Solution: $$\\text{Area of trapezoid} = \\frac{1}{2}\\times \\text{h}\\left( b_{1}+b_{2} \\right)$$ = $$\\frac{1}{2}\\times 7\\times \\left( 6+9 \\right)$$ [Substitute] = $$\\frac{1}{2}\\times 7\\times \\left( 15 \\right)$$ [Add] = 52.5 [Multiply] The area of the trapezoid is 52.5 square feet Example 2: Find the area of the trapezoid. Solution: $$\\text{Area of trapezoid} = \\frac{1}{2}\\times \\text{h}\\left( b_{1}+b_{2} \\right)$$ = $$\\frac{1}{2}\\times 6\\times \\left( 9+12 \\right)$$ [Substitute] = $$\\frac{1}{2}\\times 6\\times \\left( 21 \\right)$$ [Substitute] = 63 [Multiply] The area of the trapezoid is 63 square inches. Example 3: The shape of a trapezoid can be applied to find the area of a county in Illinois. Given that the population of this county is about 28750 people, what will be the number of people per square mile of the same county? Solution: You are given the population and the dimensions of a country shaped like a trapezoid. You are asked to find the number of people per square mile. Use the formula for the area of a trapezoid to find the area of the county Then divide the population by the area to find the number of people per square mile. $$\\text{Area of trapezoid} = \\frac{1}{2}\\times \\text{h}\\left( b_{1}+b_{2} \\right)$$ = $$\\frac{1}{2}\\times 25\\times \\left( 20+30 \\right)$$ [Substitute] = $$\\frac{1}{2}\\times 25\\times \\left( 50 \\right)$$ [Substitute] = 625 [Multiply] The area of the county is 625 square", "such factors. 3. Jun 27, 2005 ### Enos I think I got the right answer. I just disagreed with the text book because I forgot to round the number to two significant figures. the answer I got was 1070 and the answer in the text book was 1.1x10^3 All I did was 26x1000^2x3.281^2x(1/12)/43560 4. Jun 27, 2005 ### OlderDan Actually, what you did was $$V = 26 km^2 \\cdot 2 in \\cdot \\left[ \\frac{1000\\ \\ m}{km} \\right]^2\\cdot \\left[ \\frac{3.281\\ \\ ft}{m} \\right]^2 \\cdot \\left[ \\frac{1\\ \\ft}{12\\ \\ in} \\right] \\cdot \\left[ \\frac{1\\ \\ acre}{43560\\ \\ ft^2} \\right] = 1071\\ \\ acre-feet$$ Besides leaving out all the units, you left out the 2 in your posted equation. Writing the units is always a good idea and will help you keep track of what you are doing. 5. Jun 28, 2005 ### Enos Whoops, meant to put (2/12) in my posted equation rather than (1/12). I'm starting to get a hang of this stuff.", "# Convert Matric Units of Area to Hectares $\\require{cancel}$ $\\require{bbox}$ Convert metric units of area, such as, $\\text{m}^2, \\text{cm}^2, \\text{mm}^2, \\text{km}^2, ...$ to hectares ( $\\text{ ha}$ ), examples with solutions are presented including more More questions with solutions . A hectare whose abbreviation is written as $\\text{ ha}$ is defined by $1 \\text{ ha} = 10000 \\text{ m}^2$ and since $1 \\text{ hm} = 100 \\text{ m}$ Square both sides $(1 \\text{ hm})(1 \\text{ hm}) = (100 \\text{ m})(100 \\text{ m})$ Simplify the above and rewrite as $1 \\text{ hm}^2 = 10000 \\text{ m}^2$ Hence $1 \\text{ ha} = 1 \\text{ hm}^2$ and therefore conversting to $\\text{ ha}$ is the same as converting to $\\text{ hm}^2$ The table shown below helps in finding factors of conversion between metric units of length which in turn helps converting units of area to $\\text{ hm}^2$ and therefore to $\\text{ ha}$. ## Examples of Conversion to $\\text{ ha}$ with Solutions For a thorough understanding of the conversion, we show all steps with details in examples 1 and 2. Example 1 Convert $5680 \\text{ dam}^2$ to $\\text{ ha}$ Solution to Example 1 We are given $\\text{ dam}^2$, we therefore use Table 1 to find the conversion between $\\text{ dam}$ and $\\text{ hm}$. Using the Table 1 above, we have $1 \\text{ hm}", "= ${\\displaystyle {\\frac {\\text{5 (F° - 32°)}}{\\text{9}}}}$ , F° = ${\\displaystyle {\\frac {\\text{9}}{\\text{5}}}}$C° + 32° , and K° = C° + 273.15° . • Maximum length of sea waves can be found by the formula: W = 1.5${\\displaystyle {\\sqrt {\\text{fetch in nautical miles}}}}$ . • Wave height = 0.026 S2 where S is the wind speed in knots. • Wave speed in knots = 1.34 ${\\displaystyle {\\sqrt {\\text{wavelength in feet}}}}$ , or = 3.03 × wave period in seconds. UNIT CONVERSION Use the conversion tables that appear on the following pages to convert between different systems of units. Conversions followed by an asterisk are exact relationships. MISCELLANEOUS DATA Area 1 square inch _ _ _ _ _ _ _ _ _ _ _ _ _ _ = 6.4516 square centimeters* 1 square foot _ _ _ _ _ _ _ _ _ _ _ _ _ _ = 144 square inches* = 0.09290304 square meter* = 0.000022957 acre 1 square yard _ _ _ _ _ _ _ _ _ _ _ _ _ _ = 9 square feet* = 0.83612736 square meter 1 square (statute) mile_ _ _ _ _ _ _ _ _ _ _ = 27,878,400 square feet* = 640 acres* = 2.589988110336 square kilometers* 1 square centimeter_ _ _ _ _ _ _ _", "# Sunflower Field The trapezoidal sunflower field is located between two parallel paths which are spaced 230 meters apart. The lengths of the parallel sides of the field are 255 m and 274 m. How many tons of sunflower will come from this field if the hectare yield is 2.25 tons? Result m = 13.7 t #### Solution: $h = 230 \\ m \\ \\\\ a = 255 \\ m \\ \\\\ c = 274 \\ m \\ \\\\ \\ \\\\ S = h \\cdot \\ (a+c)/2 = 230 \\cdot \\ (255+274)/2 = 60835 \\ m^2 \\ \\\\ S_{ 2 } = S \\rightarrow ha = S / 10000 \\ ha = 6.0835 \\ ha \\ \\\\ \\ \\\\ m = S_{ 2 } \\cdot \\ 2.25 = 6.0835 \\cdot \\ 2.25 \\doteq 13.6879 = 13.7 \\ \\text { t }$ 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: Looking for help with calculating arithmetic mean? Looking for a statistical calculator? Do you want to convert length units? Do you want to convert mass units? ## Next similar math problems: 1. Area of ditch How great content area will have", "## Elementary Technical Mathematics $\\frac{1}{2}$ acre Let us consider that a homeowner has a lot that measures 1$\\frac{1}{2}$ acres, of that area $\\frac{1}{2}$ acres is wooded , $\\frac{1}{6}$ acres is covered by house , $\\frac{1}{3}$ acres is covered by driveway Area covered by lawn = 1$\\frac{1}{2}$ -$\\frac{1}{2}$ -$\\frac{1}{6}$ - $\\frac{1}{3}$ =$\\frac{3}{2}$ -$\\frac{1}{2}$ -$\\frac{1}{6}$ - $\\frac{1}{3}$ = $\\frac{2}{2}\\times$ $\\frac{3}{3}$ -$\\frac{1}{6}$ - $\\frac{1}{3}\\times$ $\\frac{2}{2}$ =$\\frac{6}{6}$ -$\\frac{1}{6}$ -$\\frac{2}{6}$ =$\\frac{3}{6}$ =$\\frac{1}{2}$ acre", "exercise class in the non-forested area of Stanley Park? ## Solution 1. We need to figure out the number of $$2$$ m by $$2$$ m squares in the field. Since there are $$200\\div 2 =100$$ squares along the long side of the park, and $$100\\div 2 = 50$$ squares along the short side, there are $$100 \\times 50 = 5000$$ squares in total. That means the field could accommodate $$5000$$ people. 2. Since Imaginary Park is $$1$$ km by $$1$$ km (or $$1000$$ m by $$1000$$ m), there could be $$1000\\div 2 = 500$$ people in each row. Since there are $$1000\\div 2 = 500$$ such rows, there could be $$500 \\times 500=250\\,000$$ people in $$100$$ ha of space. This works out to $$250\\,000\\div 100 = 2500$$ people per ha. 3. The non-forested area of Stanley Park is $$\\frac{1}{5}$$ of $$405$$ ha, or $$\\frac{1}{5}\\times 405 = 81$$ ha. This area will accommodate $$2500$$ people per ha. This means a total of $$2500 \\times 81 = 202\\,500$$ people could be present in the non-forested area of Stanley Park at one time.", "of the rainforest. As she reads, Julie finds herself getting more and more irritated. “Are you alright Julie,” Mr. Gibbons asks, as he pauses in his walk around the room checking on students. “No, I’m not,” Julie says, and she proceeds to tell Mr. Gibbons all about what she has learned about the rainforest. “Look here,” she says, pointing to her book. “It says that we lose 112\\begin{align*}1 \\frac{1}{2}\\end{align*} acres of land every second!” Wow! Julie is shocked by that fact. Are you? How much land is lost in one minute given this statistic? How much is lost in three minutes? Working on multiplying mixed numbers is the way to figure out how much acreage is lost. The first question is how much land is lost in one minute. To start, we must convert minutes to seconds since we lose 112\\begin{align*}1 \\frac{1}{2}\\end{align*} acre of land every second. 60 seconds = 1 minute We will be multiplying by 60. Next, we move on to writing an equation. 60×112=\\begin{align*}60 \\times 1\\frac{1}{2} = \\underline{\\;\\;\\;\\;\\;\\;\\;}\\end{align*} To solve this equation, we need to change the whole number to a fraction over one and the mixed number to an improper fraction. \\begin{align*}\\frac{60}{1} \\times \\frac{3}{2} = \\frac{180}{2} = 90\\end{align*} We lose 90 acres of rainforest land every minute. We can figure out how many acres", "\\cdot 6\\right)\\\\ A &= 27+135-9\\\\ A &= 153 \\ units^2\\end{align*} ### Examples #### Example 1 Find the area of the rectangles and triangle. Rectangle #1: \\begin{align*}\\text{Area }= 24(9+12)=504 \\ units^2\\end{align*}. Rectangle #2: \\begin{align*}\\text{Area }=15(9+12)=315 \\ units^2\\end{align*}. Triangle: \\begin{align*}\\text{Area }=\\frac{15(9)}{2}=67.5 \\ units^2\\end{align*}. #### Example 2 Find the area of the whole shape. You need to subtract the area of the triangle from the bottom right corner. \\begin{align*}\\text{Total Area }=504+315+67.5-\\frac{15(12)}{2}=796.5 \\ units^2\\end{align*} ### Review Use the picture below for questions 1-2. Both figures are squares. 1. Find the area of the unshaded region. 2. Find the area of the shaded region. Find the area of the figure below. You may assume all sides are perpendicular. Find the areas of the composite figures. Use the figure to answer the questions. 1. What is the area of the square? 2. What is the area of the triangle on the left? 3. What is the area of the composite figure? Find the area of the following figures. 1. Find the area of the unshaded region. 2. Lin bought a tract of land for a new apartment complex. The drawing below shows the measurements of the sides of the tract. Approximately how many acres of land did Lin buy? You may assume any angles that look like right angles are \\begin{align*}90^\\circ\\end{align*}. (1 acre \\begin{align*}\\approx\\end{align*}", "+0 # Something 0 105 1 A trapezoid has an area of 341 square miles. One base is 8 miles long. The height measures 22 miles. What is the length of the other base Guest May 12, 2017 Sort: #1 +18379 +1 A trapezoid has an area of 341 square miles. One base is 8 miles long. The height measures 22 miles. What is the length of the other base ? $$\\begin{array}{|rcll|} \\hline A &=& \\left(\\frac{x+8}{2} \\right) \\cdot 22 \\\\ 341 &=& \\left(\\frac{x+8}{2} \\right) \\cdot 22 \\\\ 341 &=& (x+8) \\cdot 11 \\quad &| \\quad : 11 \\\\ 31 &=& x+8 \\quad &| \\quad - 8 \\\\ 31-8 &=& x \\\\ 23 &=& x \\\\ \\hline \\end{array}$$ The other base is 23 miles long. heureka May 12, 2017 ### 8 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" ]

[ "> > a map has a scale of 1: 50,000 the area of the farm on the map is 6cm^2. # a map has a scale of 1: 50,000 the area of the farm on the map is 6cm^2. ## What is the real area of the farm in hectares? (1 hectare = 10,000... 352 Views ### a map has a scale of 1: 50,000 the area of the farm on the map is 6cm^2. What is the real area of the farm in hectares? (1 hectare = 10,000 m^2= 0.01 km^2) $$a\\ scale\\ s=1:50,000=\\frac{1}{50,000}=\\frac{1}{5\\cdot10^4}\\\\\\\\A_m=6cm^2-the\\ area o\\ the\\ map\\\\\\\\A_r=? -the eal\\ area\\\\\\\\\\frac{A_m}{A_r}=s^2\\\\\\\\\\frac{6}{A_r}=\\left(\\frac{1}{5\\cdot10^4} ight)^2\\\\\\\\\\frac{6}{A_r}=\\frac{1}{(5\\cdot10^5)^2}\\\\\\\\\\frac{6}{A_r}=\\frac{1}{5^2\\cdot(10^4)^2}$$", "## Thinking Mathematically (6th Edition) We know that 1 mile = 1 km We know that 1 hectare = 0.01 square kilometer We can find how many square kilometers are in 1 square mile. $1~mi^2 = 1~mi\\times 1~mi$ $1~mi^2 = 1.6~km\\times 1.6~km$ $1~mi^2 = 2.56~km^2$ We can find how many square miles are in one hectare. $1 ~ha = \\frac{1~ha}{1}\\times \\frac{0.01~km^2}{1~ha}\\times \\frac{1~mi^2}{2.56~km^2}$ $1~ha = 0.00391~mi^2$ We can express the population density in units of people per square mile. $\\frac{46,690~people}{1000~ha} = \\frac{46,690~people}{1000~ha}\\times \\frac{1~ha}{0.00391~mi^2} = \\frac{11,941~people}{1~mi^2}$ The population density is 11,941 people per square mile.", "# 7th Class Mathematics Mensuration Standard Units of Area Standard Units of Area Category : 7th Class ### Standard Units of Area The inter relationship among various units of measurement of area are listed below. $1{{m}^{2}}$ $=$ $100\\times 100c{{m}^{2}}={{10}^{4}}c{{m}^{2}}$ $1{{m}^{2}}$ $=$ $10\\times 10\\,d{{m}^{2}}=100d{{m}^{2}}$ $1\\,d{{m}^{2}}$ $=$ $10\\times 10\\,c{{m}^{2}}=100{{m}^{2}}$ $1\\,da{{m}^{2}}$ $=$ $10\\times 10\\,{{m}^{2}}=100\\,{{m}^{2}}$ $1\\,h{{m}^{2}}$ $=$ $100\\times 100\\,{{m}^{2}}={{10}^{4}}\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $1000\\times 1000\\,{{m}^{2}}={{10}^{6}}\\,{{m}^{2}}$ $1\\,hectare$ $=$ $10000\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $100\\,hectares$ #### Other Topics You need to login to perform this action. You will be redirected in 3 sec", "question in entirety, here Iis the equation...(will someone else take the time to explain it) $=3.3miles^2\\times\\left(\\frac{1609meters}{1mile} \\right)^2\\times\\left(\\frac{1.093613298yards}{1mete r} \\right)^2$ Quote: Originally Posted by Malay Verify: 1mile =2 yard(suppose) You will find your equation gives wrong answer but mine is correct. I arrived at a different answer from pahsah because the original poster has approximated the value of \\sqrt{3.3} but I haven't approximated anything. Keep Smiling Malay alright lets see...lets try finding yards (m=miles and y=yards) malay's equation: $=1m$ $=1y\\times\\frac{m}{y}$ $=1y\\times\\frac{1}{2}\\frac{m}{y}$ $=\\frac{1}{2}m$ oh look, your equation turned yards into miles quite nicely, except wait! we were trying to turn miles into yards weren't we? my equation: $=1m$ $=1m\\times\\frac{y}{m}$ $=1m\\times2\\frac{y}{m}$ $=2y$ tsk tsk, looks like I win • July 14th 2006, 04:52 AM topsquark Quote: Originally Posted by pashah Hello I was given the following problem and have come up with what i believ to be the correct answer. I would appreciate it if someone could proof read my answer for accuracy. 1 yard = 0.9144 meters 1 mile = 1609 meters if the area of a piece of land is 3.3 square-miles, then how much is it in square-yards? In order to determine the are we must find the length of two sides of the area in question. Therefore miles for one side of our area. So we can", "cm = 500000 x 10 mm = 5000000 mm Question 2: Convert 10 m to km. Solution: 1 m = 10-3 km 10 m = 10 x 10-3 km = $\\frac{10}{1000}$ km = $\\frac{1}{100}$ km = 10 -2 km ## Converting between Metric Units ### Solved Examples Question 1: To convert 1000 acres to Hectares. Solution: Here we have to convert acres to Hectares. 1 acres = 0.404685642 hectares Therefore, 1000 acres = 0.404685642 hectares [xx] 1000 acres = 404.685642 The answers for 1000 acres are equal to 404.685642 Hectares. Question 2: To convert 700 Hectares to Square feet. Solution: Here we have to convert Hectares to Square feet. 1 hectare = 107 639.104 square feet Therefore, 700 Hectares = 107 639.104 square feet [xx] 700 Hectares = 75347372.8 sq.ft The answers for 700 Hectares are equal to 75347372.8 Square feet. Question 3: To convert 100 Kilometers (km) to Miles. Solution: Here we have to convert Kilometers (km) to Miles. 1 kilometer (km) = 0.621371192 miles Therefore, 100 km = 0.621371192 miles [xx] 100 km = 62.1371192 miles The answers for 100 Kilometers (km) are equal to 62.1371192 miles. Question 4: To convert 300 millimeters to inches. Solution: Here we have to convert millimeters (mm) to inches 1 millimeter (mm) = 0.0393701 inches Therefore, 300 mm = 0.0393701", "is 0.1rb by 0.1rb so it's 0.01 barns. - I still do not get it :( – Dukunocil Jul 10 '12 at 16:51 I think I got it :) First part is obvious. Second part: 10fm X 10fm = 100 fm^2 OK. So what is it expressed in barns? I calculate: 100 fm^2 = 10^2 x (10^-15)^2 = 10^2 x 10^-30 = 10^-28 = barn. They should use brackets i.e. 100 [fm]^2 – Dukunocil Jul 10 '12 at 17:02 You wouldn't normally use brackets. After all it's very common to talk of a square centimeter and denote it by $cm^2$, or a square kilometer, $km^2$. You wouldn't write these as $[cm]^2$ or $[km]^2$. – John Rennie Jul 10 '12 at 17:13 ..then if 100[fm]^2 = 1 barn, therefore 1[fm]^2 = 10^-2 barn = 10 X 10^-3 barn = 10 mb :) – Dukunocil Jul 10 '12 at 17:13 I agree, but not 100%. Units should be in brackets to avoid confusion like this one, particularly when dealing with 'exotic' types of units :) Anyway, thanks for allowing me to see this. – Dukunocil Jul 10 '12 at 17:16 An area with a side of 1u has an area of 1 u^2. ▦ An area with a side of 2u has an area of 4 u^2. ▦▦ ▦▦ An", "answer is given correctly as a repeating decimal .00142857 (Braid uses dots over the digits to indicate the beginning and end of a repeating group. We have used an \"overline\".) 3. Reduce $\\large\\frac{49}{52}\\normalsize$ to a decimal. Solution. Again long division until the remainder repeats when the answer is given correctly as a repeating decimal. .94230769 4. Reduce $\\large\\frac{655}{2149}\\normalsize$ to a decimal. Solution. Long division gives .3047929+. It is not clear how he decides the number of places to calculate. In fact this fraction has a repeating part of length 306. 5. Reduce 157 yds 1 ft 3 in to a mile. Solution. He divides 3 in by 12 to get .25 in, then 1.25 ft by 3 to get .416 yards, then 157.416 by 1760 to get .089441287 mile. 6. Reduce 1 rood 20 poles to an acre. 7. Reduce 5 hrs 48 min 50 sec of a day. 8. Reduce 17' 44'' 48''' to a degree. 9. Value .603125 acres. Solution. He multiplies by 4 to get 2.412500 roods. Then he multiplies .412500 by 40 to get 16.500000 (square) poles, then .500000 by $30\\large\\frac{1}{4}\\normalsize$ to get 15.125000. Finally he converts .125000 to $\\large\\frac{1}{8}\\normalsize$. So the answer is 2 roods 16 (square) poles $15\\large\\frac{1}{8}\\normalsize$ (square) yards. 10. Value .483125 miles. Solution. He converts to furlongs, yds, ins. (There", "# Convert Matric Units of Area to Hectares $\\require{cancel}$ $\\require{bbox}$ Convert metric units of area, such as, $\\text{m}^2, \\text{cm}^2, \\text{mm}^2, \\text{km}^2, ...$ to hectares ( $\\text{ ha}$ ), examples with solutions are presented including more More questions with solutions . A hectare whose abbreviation is written as $\\text{ ha}$ is defined by $1 \\text{ ha} = 10000 \\text{ m}^2$ and since $1 \\text{ hm} = 100 \\text{ m}$ Square both sides $(1 \\text{ hm})(1 \\text{ hm}) = (100 \\text{ m})(100 \\text{ m})$ Simplify the above and rewrite as $1 \\text{ hm}^2 = 10000 \\text{ m}^2$ Hence $1 \\text{ ha} = 1 \\text{ hm}^2$ and therefore conversting to $\\text{ ha}$ is the same as converting to $\\text{ hm}^2$ The table shown below helps in finding factors of conversion between metric units of length which in turn helps converting units of area to $\\text{ hm}^2$ and therefore to $\\text{ ha}$. ## Examples of Conversion to $\\text{ ha}$ with Solutions For a thorough understanding of the conversion, we show all steps with details in examples 1 and 2. Example 1 Convert $5680 \\text{ dam}^2$ to $\\text{ ha}$ Solution to Example 1 We are given $\\text{ dam}^2$, we therefore use Table 1 to find the conversion between $\\text{ dam}$ and $\\text{ hm}$. Using the Table 1 above, we have $1 \\text{ hm}", "7th Class Mathematics Mensuration Standard Units of Area Standard Units of Area Category : 7th Class Standard Units of Area The inter relationship among various units of measurement of area are listed below. $1{{m}^{2}}$ $=$ $100\\times 100c{{m}^{2}}={{10}^{4}}c{{m}^{2}}$ $1{{m}^{2}}$ $=$ $10\\times 10\\,d{{m}^{2}}=100d{{m}^{2}}$ $1\\,d{{m}^{2}}$ $=$ $10\\times 10\\,c{{m}^{2}}=100{{m}^{2}}$ $1\\,da{{m}^{2}}$ $=$ $10\\times 10\\,{{m}^{2}}=100\\,{{m}^{2}}$ $1\\,h{{m}^{2}}$ $=$ $100\\times 100\\,{{m}^{2}}={{10}^{4}}\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $1000\\times 1000\\,{{m}^{2}}={{10}^{6}}\\,{{m}^{2}}$ $1\\,hectare$ $=$ $10000\\,{{m}^{2}}$ $1\\,k{{m}^{2}}$ $=$ $100\\,hectares$ Other Topics You need to login to perform this action. You will be redirected in 3 sec", "GMAT 1: 730 Q48 V44 GMAT 2: 730 Q49 V40 GPA: 3.98 WE: Corporate Finance (Health Care) Re: On an aerial photograph, the surface of a pond appears as [#permalink] ### Show Tags 03 Aug 2017, 11:42 1 Why doesn't converting the area to square inches and then converting from inches to miles work? $$(\\frac{7}{16})^2=\\frac{49}{256}$$ $$\\frac{49}{256}*\\frac{22}{7}=\\frac{77}{128}$$ $$\\frac{77}{12}*2\\approx1.2$$ I chose 1.3 because it's the closest answer, but obviously the conversion to miles needs to happen before calculating the area. Why does this matter, and is there a decent framework to do conversions of this sort that involve squaring and cubing? This seems to pop up often, and I find it slightly confusing. Thanks! SR Manager Joined: 09 Oct 2016 Posts: 89 Location: United States GMAT 1: 740 Q49 V42 GPA: 3.49 Re: On an aerial photograph, the surface of a pond appears as [#permalink] ### Show Tags 30 Aug 2017, 04:55 I chose 1.3miles as well but I think they key word here is \"square miles\". That's the indicator that you need to convert before using the area formula Senior SC Moderator Joined: 22 May 2016 Posts: 2204 On an aerial photograph, the surface of a pond appears as [#permalink] ### Show Tags 31 Aug 2017, 12:17 2 spence11 wrote: Why doesn't converting the area to square inches and", "lb / in 2 = 6 . 895 × 10 3 Pa 1 lb / in 2 = 6 . 895 × 10 3 Pa size 12{1 {\"lb\"} slash {\"in\" rSup { size 8{2} } } =6 \".\" \"895\" times \"10\" rSup { size 8{3} } \" Pa\"} {}$ Table C7 Selected British Units Length $1 light year ( ly ) = 9 . 46 × 10 15 m 1 light year ( ly ) = 9 . 46 × 10 15 m size 12{1\" light\"\" year\" $$\"ly\"$$ =9 \".\" \"46\" times \"10\" rSup { size 8{\"15\"} } \" m\"} {}$ $1 astronomical unit ( au ) = 1 . 50 × 10 11 m 1 astronomical unit ( au ) = 1 . 50 × 10 11 m size 12{1\" astronomical\"\" unit\" $$\"au\"$$ =1 \".\" \"50\" times \"10\" rSup { size 8{\"11\"} } \" m\"} {}$ $1 nautical mile = 1 . 852 km 1 nautical mile = 1 . 852 km size 12{1\" nautical\"\" mile\" =1 \".\" \"852\"\" km\"} {}$ $1 angstrom ( Å ) = 10 − 10 m 1 angstrom ( Å ) = 10 − 10 m size 12{1\" angstrom\" $$Å$$ =\"10\" rSup { size 8{ - \"10\"} } \" m\"} {}$ Area $1 acre ( ac ) = 4 . 05 ×", "base $$b-c$$. This allows us to calculate the area of the whole trapezium:$A=\\left(\\frac{b\\,- c}{2}\\right)h + c\\,h = \\left(\\frac{b+c}{2}\\right)h$The midpoints of the two diagonals are both at a height $$h/2$$ so the signpost is also at this height. It is now easy to see that the areas of the lower and upper fields are:$A_1=\\frac{b h}{4}\\;\\;\\;\\;A_3 = \\frac{c h }{4}$ With the left and right field areas as $$A_2$$ and $$A_4$$, we hence know that:$A=2(A_1+A_3)=2(A_2+A_4)$ where one of the fields has an area of $$3$$ acres and the opposite field has an area of $$3 – x$$ acres with $$x < 1$$. So we have:$A = 12 - 2x \\;\\;\\;\\text{with} \\;\\;\\;0 < x < 1$. Since $$A$$ is an integer in acres, the total area of the four fields can only be $$11$$ acres.", "such factors. 3. Jun 27, 2005 ### Enos I think I got the right answer. I just disagreed with the text book because I forgot to round the number to two significant figures. the answer I got was 1070 and the answer in the text book was 1.1x10^3 All I did was 26x1000^2x3.281^2x(1/12)/43560 4. Jun 27, 2005 ### OlderDan Actually, what you did was $$V = 26 km^2 \\cdot 2 in \\cdot \\left[ \\frac{1000\\ \\ m}{km} \\right]^2\\cdot \\left[ \\frac{3.281\\ \\ ft}{m} \\right]^2 \\cdot \\left[ \\frac{1\\ \\ft}{12\\ \\ in} \\right] \\cdot \\left[ \\frac{1\\ \\ acre}{43560\\ \\ ft^2} \\right] = 1071\\ \\ acre-feet$$ Besides leaving out all the units, you left out the 2 in your posted equation. Writing the units is always a good idea and will help you keep track of what you are doing. 5. Jun 28, 2005 ### Enos Whoops, meant to put (2/12) in my posted equation rather than (1/12). I'm starting to get a hang of this stuff.", "of the rainforest. As she reads, Julie finds herself getting more and more irritated. “Are you alright Julie,” Mr. Gibbons asks, as he pauses in his walk around the room checking on students. “No, I’m not,” Julie says, and she proceeds to tell Mr. Gibbons all about what she has learned about the rainforest. “Look here,” she says, pointing to her book. “It says that we lose 112\\begin{align*}1 \\frac{1}{2}\\end{align*} acres of land every second!” Wow! Julie is shocked by that fact. Are you? How much land is lost in one minute given this statistic? How much is lost in three minutes? Working on multiplying mixed numbers is the way to figure out how much acreage is lost. The first question is how much land is lost in one minute. To start, we must convert minutes to seconds since we lose 112\\begin{align*}1 \\frac{1}{2}\\end{align*} acre of land every second. 60 seconds = 1 minute We will be multiplying by 60. Next, we move on to writing an equation. 60×112=\\begin{align*}60 \\times 1\\frac{1}{2} = \\underline{\\;\\;\\;\\;\\;\\;\\;}\\end{align*} To solve this equation, we need to change the whole number to a fraction over one and the mixed number to an improper fraction. \\begin{align*}\\frac{60}{1} \\times \\frac{3}{2} = \\frac{180}{2} = 90\\end{align*} We lose 90 acres of rainforest land every minute. We can figure out how many acres", "Oct 2016 Posts: 85 Location: United States GMAT 1: 740 Q49 V42 GPA: 3.49 Re: On an aerial photograph, the surface of a pond appears as [#permalink] Show Tags 30 Aug 2017, 04:55 I chose 1.3miles as well but I think they key word here is \"square miles\". That's the indicator that you need to convert before using the area formula Senior SC Moderator Joined: 22 May 2016 Posts: 3844 On an aerial photograph, the surface of a pond appears as [#permalink] Show Tags 31 Aug 2017, 12:17 2 1 spence11 wrote: Why doesn't converting the area to square inches and then converting from inches to miles work? $$(\\frac{7}{16})^2=\\frac{49}{256}$$ $$\\frac{49}{256}*\\frac{22}{7}=\\frac{77}{128}$$ $$\\frac{77}{12}*2\\approx1.2$$ I chose 1.3 because it's the closest answer, but obviously the conversion to miles needs to happen before calculating the area. Why does this matter, and is there a decent framework to do conversions of this sort that involve squaring and cubing? This seems to pop up often, and I find it slightly confusing. Thanks! SR spence11 , these problems can be confusing. We have a \"scale factor\" issue, as you note. In one sense, however, when scaling up or down, linear change is one kind of step (ONE change in one length), and area change is a different kind of step (TWO changes, one for each", "to one billionth of a barn. See the barn unit to learn more.$1\\ nb = 10^{-9}\\ b = 10^{-37}\\ m^2$ Show source$...$ picobarn Show source$pb$ pb Show source$\\text{...}$ - Derived unit of cross-section. One picobarn is equal to one trillionth of a barn. See the barn unit to learn more.$1\\ pb = 10^{-12}\\ b = 10^{-40}\\ m^2$ Show source$...$ femtobarn Show source$fb$ fb Show source$\\text{...}$ - Derived unit of cross-section. One femtobarn is equal to one quadrillionth of a barn. See the barn unit to learn more.$1\\ fb = 10^{-15}\\ b = 10^{-43}\\ m^2$ Show source$...$ # Some facts# • The basic unit of measure for the area is the square meter (m2) or derived units created by adding SI prefixes, eg. one square kilometer (km2) or one square centimeter (cm2). • It is still common to use some non-SI units in some areas of daily life. For example land area is still measured in ares (a) or hectares (ha). • In Anglo-Saxon countries, it is common to use units based on traditional imperial length units like miles or inches. In this way, we get new area units such as square mile (mi2), square feet or square inch. # How to convert# • Enter the number to field \"value\" - enter the NUMBER only, no other words, symbols", "Example 4: metric/imperial conversion 1 \\ hectare \\approx 2.5 \\ acres Convert 14.8 acres to hectares. The conversion factor for hectares and acres is 2.5 As we are converting from acres to hectares, we divide by the conversion factor. 14.8 \\div 2.5=5.92 We can write down the approximated conversion. 14.8 \\ hectare\\ \\approx 5.92 \\ acres ### Example 5: miles/kilometres conversion 5 \\ miles \\approx 8 \\ km Convert 30 \\ miles to km . We can find the conversion factor by first dividing by 5 to work out what 1 mile is equivalent to. \\begin{aligned} 5\\ miles &\\approx 8\\ km\\\\\\\\ 1\\ mile &\\approx 1.6\\ km \\end{aligned} The conversion factor from miles to kilometres is 1.6 As we are converting from miles to km , we multiply by the conversion factor. 30 \\times 1.6=48 We can write down the approximated conversion. 30\\ miles \\approx 48 \\ km Alternatively we could have used a fractional conversion factor of \\frac{8}{5}. 30\\times \\frac{8}{5}=48 ### Example 6: miles/kilometres conversion 5 \\ miles \\approx 8 \\ km Convert 120 \\ km to miles . We can find the conversion factor by first dividing by 5 to find what 1 \\ mile is equivalent to. \\begin{aligned} 5\\ miles &\\approx 8\\ km\\\\\\\\ 1\\ mile &\\approx 1.6\\ km \\end{aligned} The conversion factor from miles to kilometres", "exercise class in the non-forested area of Stanley Park? ## Solution 1. We need to figure out the number of $$2$$ m by $$2$$ m squares in the field. Since there are $$200\\div 2 =100$$ squares along the long side of the park, and $$100\\div 2 = 50$$ squares along the short side, there are $$100 \\times 50 = 5000$$ squares in total. That means the field could accommodate $$5000$$ people. 2. Since Imaginary Park is $$1$$ km by $$1$$ km (or $$1000$$ m by $$1000$$ m), there could be $$1000\\div 2 = 500$$ people in each row. Since there are $$1000\\div 2 = 500$$ such rows, there could be $$500 \\times 500=250\\,000$$ people in $$100$$ ha of space. This works out to $$250\\,000\\div 100 = 2500$$ people per ha. 3. The non-forested area of Stanley Park is $$\\frac{1}{5}$$ of $$405$$ ha, or $$\\frac{1}{5}\\times 405 = 81$$ ha. This area will accommodate $$2500$$ people per ha. This means a total of $$2500 \\times 81 = 202\\,500$$ people could be present in the non-forested area of Stanley Park at one time.", "# Formula - Mensuration MENSURATION Mensuration is the science of measurement of the lenghts of lines, areas of surfaces and volumes of solids. Perimeter : Perimeter is sum of all the sides. It is measured in cm,m, etc. Area : The area of any figure is the amount of surface enclosed within its boundary lines. This is measured in square unit like cm2, m2, etc. Volume : If an object is solid, then the space occupied by such an object is called its volume. This is measured in cubic unit like cm3, m3, etc. Basic Conversions : I. 1 km = 10 hm 1 hm = 10 dam 1 dam = 10 m 1 m = 10 dm 1 dm = 10 cm 1 cm = 10 mm 1 m = 100 cm = 1000 mm 1 km = 1000 m II. 1 km = 58 miles 1 mile = 1.6 km 1 inch = 2.54 cm 1 mile = 1760 yd = 5280 ft. 1 nautical mile (knot) = 6080 ft III. 100 kg = 1 quintal 10 quintal = 1 tonne 1 kg = 2.2 pounds (approx.) IV. l litre = 1000 cc 1 acre = 100 m2 1 hectare = 10000 m2(100 acre) ### Part I : Plane Figures TRIANGLE Perimeter (P) = a +", "4.5 kilometers away from the river. b. Apply the distance formula again letting Mr. Farmer at (1, 2) be (x1, y1) and pond (-4, -2) be (x2, y2) and f be the distance of the pond and Mr. Farmer f = $$\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}$$ Formula. = $$\\sqrt{((-4)-1)^{2}+((-2)-2)^{2}}$$ Substitution. = $$\\sqrt{(-5)^{2}+(-4)^{2}}$$ Subtract. = $$\\sqrt{25+16}$$ Evaluate the squares. = $$\\sqrt{41}$$ Add. ≈ ± 6.4 Evaluate. Therefore, the distance of Mr. Farmer to the pond is about 6.4 kilometers. c) Use the distance formula then let Mr. Farmer at (1, 2) be (x1, y1) while the town at (-4, 5) be (x2, y2) and g be the distance of Mr. Framer to the town. g = $$\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}$$ Formula. = $$\\sqrt{((-4)-1)^{2}+(5-2)^{2}}$$ Substitution. = $$\\sqrt{(-5)^{2}+3^{2}}$$ Subtract. = $$\\sqrt{25+9}$$ Evaluate the squares. = $$\\sqrt{34}$$ Add. ≈ ± 5.8 Evaluate. Therefore, Mr Farmer is around 5.8 kilometers away from the town d) Let Mr Farmer (1, 2) be (x1, y1) while Orchard (6, -4) be (x2, y2) and h be the distance of Mr. Farmer and the orchard then apply the distance formula. h = $$\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}$$ Formula. = $$\\sqrt{(6-1)^{2}+((-4)-2)^{2}}$$ Substitution. = $$\\sqrt{5^{2}+(-6)^{2}}$$ Subtract. = $$\\sqrt{25+36}$$ Evaluate the squares. = $$\\sqrt{61}$$ Add. ≈ ± 7.8 Evaluate. Therefore, Mr Farmer is approximately 7.8 kilometers away from the orchard. Question 6. The farmer wants to go fishing in either the" ]