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[ "multiples of 8 and 12 but 24 is the lowest.", "of a multiple of 10 or 25.", "30 is not a common multiple of 12 and 15. 19 18 27 36 CorrectIncorrect", "square units 42.) What is the least common multiple of 168 and 420? Continue reading", "of 991. If you like to learn what is a multiple", "n that is a multiple of 20.", "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.", "is the least common multiple of the $n_i$. -", "the least common multiple 8 06 Feb 2014, 02:24 48 For how many values of k is 12^12 the least common multiple 17 12 Nov 2009, 03:38 22 If the least common multiple of positive integer m and n is 13 12 Jun 2008, 23:09 Display posts from previous: Sort by", "most common? ________. Which are the two least common? ________ and _________. Subjects: 1 , 2 , 3 , 4 , 5 Feb 23, 2012 Jan 16, 2015", "b)$ or, more commonly, (a,b). The greatest common divisor of 12 and 8 is 4, the greatest common divisor of 15 and 16 is 1. Two numbers with a greatest common divisor of 1 are said to be relatively prime. Complementary to the notion of greatest common divisor is least common multiple. The least common multiple of a and b is the smallest (positive) integer m such that a | m and b | m. Thus, the least common multiple of 12 and 9 is 36 (written [12,9] = 36). ## Abstract divisors In higher mathematics, the notion of divisor has been abstracted from the integers to the context of general commutative rings. In this setting, they might be termed abstract divisors.", "the smallest multiple of 63 and 147", "following cannot be the least common multiple 3 06 Feb 2014, 02:24 9 If k is a common multiple of 75, 98, and 140, which of the 13 15 Oct 2013, 05:14 2 what is the least possible value of k? 12 16 Jan 2010, 04:30 10 If the least common multiple of positive integer m and n is 10 12 Jun 2008, 23:09 11 If M is the least common multiple of 90,196, and 300, which 20 27 Nov 2005, 06:43 Display posts from previous: Sort by", "Multiples of 6 or 7 but not both How many integers between 1 and 2015 are multiples of 6 OR 7 but not both? ×", "12 and 18 is 36 , and the greatest common divisor 6 ! SOURCE Content is available under GNU Free Documentation License 1.2.", "least common multiple, we will multiply only the prime factors that the numbers share. The resulting product is the greatest common factor. Let’s wrap up with a couple of true or false review questions: 1) The least common multiple of 45 and 60 is 15. The answer is false. The greatest common factor of 45 and 60 is 15, but the least common multiple is 180. 2) The least common multiple is a number greater than or equal to the numbers being considered. The answer is true. The least common multiple is greater than or equal to the numbers being considered, while the greatest common factor is equal to or less than the numbers being considered. Thanks for watching, and happy studying! 838699 by Mometrix Test Preparation | Last Updated: December 18, 2019", "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "# 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", "# Question 1 of 18 List the first 7 multiples of 8? A 8,16,24,32,40,48,56 B 8,16,26,32,40,48,56 C 8,16,24,36,40,48,56 D 8,16,24,32,42,48,56 E None of the above", "8064}$ is the only answer choice that is a multiple of $2016$." ]

[ "an answer. Back to the question: 1. 0 is a multiple of every integer, so 0 is a common multiple of every non-zero sets of integers. 2. 0 is not the only common multiple of 75, 98, and 140, less than 14,000, so is -14,700, -2*14,700, -3*14,700, -4*14,700, ... Hope it's clear. _________________ Re: If k is a common multiple of 75, 98, and 140, which of the [#permalink] 30 Jul 2015, 04:37", "# Practice: Finding common multiples How many positive integers less than or equal to $1000$ are common multiples of the three numbers $8, 36,$ and $24$? ×", "b)$ or, more commonly, (a,b). The greatest common divisor of 12 and 8 is 4, the greatest common divisor of 15 and 16 is 1. Two numbers with a greatest common divisor of 1 are said to be relatively prime. Complementary to the notion of greatest common divisor is least common multiple. The least common multiple of a and b is the smallest (positive) integer m such that a | m and b | m. Thus, the least common multiple of 12 and 9 is 36 (written [12,9] = 36). ## Abstract divisors In higher mathematics, the notion of divisor has been abstracted from the integers to the context of general commutative rings. In this setting, they might be termed abstract divisors.", "### Home > CC1 > Chapter 8 > Lesson 8.2.1 > Problem8-77 8-77. The first four multiples of $5$ are $5$, $10$, $15$, and $20$. 1. What are the first six multiples of $10$? To find the first six multiples of $10$, you simply multiply $10$ by the numbers $1$ through $6$ separately and create a list of the products you find similar to the list written above. The first six multiples of $10$ are $10,20,30,40,50$, and $60$. 2. What are the first six multiples of $8$? The first six multiples of $8$ are $8,16,24,32,40$, and $48$. 3. What is the least common multiple of $10$ and $8$? The Least Common Multiple of two or more integers is the lowest positive integer that is divisible by both (or all ) of the integers. For example, the multiples of $3$ and $5$ are shown below in the table. $15$ is the Least Common Multiple because it is the smallest positive integer divisible by both $3$ and $5$. Use this idea to find the least common multiple of $10$ and $8$. $\\begin{array}{c|c|c|c|c|c} \\quad 3 \\; & \\; 6 \\; & \\; 9 \\; & \\; 12 \\; & \\; \\mathbf{15} \\; & \\; 18 \\quad\\\\ \\hline \\quad 5 \\; & \\; 10 \\; & \\; \\mathbf{15} \\; & \\; 20", "# least common multiple A common multiple of two integers is an integer divisible by both integers. The least common multiple of two integers is the smallest positive common multiple. • The notation for the least common multiple of $$a$$ and $$b$$ is $$\\operatorname{lcm}(a, b)$$. • By definition, the least common multiple of two integers is independent of the sign of the two integers --- the $$\\operatorname{lcm}$$ is the same if you replace $$a$$ with $$-a$$ or $$b$$ with $$-b$$. • The definition of $$\\operatorname{lcm}$$ does not make sense if either $$a$$ or $$b$$ are is equal to $$0$$. • For the remainder of the section, we will assume integers to be nonzero. The existence of a least common multiple stems from the fact that the absolute value of the product $$|ab|$$ is a positive common multiple of $$a$$ and $$b$$. ## naive computation The least common multiple can be computed naively by checking every multiple of one of them up to their product to see if it is divisible by the other. For example, to compute $$\\operatorname{lcm}(154, 56)$$ • start with $$154$$ and keep adding $$154$$ until it is divisible by $$56$$ • $$56$$ does not divide $$154$$, $$154(2)$$, or $$154(3)$$ • however, $$154(4) = 616 = 56(11)$$ is divisible by $$56$$ • the least common multiple", "every number. So that's the answer. Consideration of the seemingly silly question of the greatest common multiple has led us to a better understanding of the multiplicative structure of the integers, and we see that there is a greatest common multiple, which is exactly dual to the least common divisor, just as the least common multiple is dual to the greatest common divisor.", "$n$, is a common multiple of two given integers, if the two given integers divide $n$ with zero remainder. least common multiple a positive integer, $n$, is the least common multiple of two given integers if $n$ is the smallest common multiple of two given integers. For example, 30 is a common multiple of 5 and 15 since 30 $\\div$ 5 and 30 $\\div$ 15 have zero remainders. However, 30 is not the least common multiple because of 15. Here, 15 $\\div$ 15 and 15 $\\div$ 5 have zero remainders and 15 $\\lt$ 30. Please note, -15 is not the least common multiple of 5 and 15 because a least common multiple must be positive.", "96, 120…… and so on. Thus, it is impossible to find out all the common multiples, we just need to find a few of them. The smallest of these multiples is called the least common multiple; in the case of the multiples of 6 and 8, the least common multiple is 24. ## There Is No Concept of Least Common Factor and Greatest Common Multiple We learned about the concepts of the greatest common divisor and least common multiple. On the other hand, there is no concept of least common factor and greatest common multiple. What is the smallest number among all the common factors? It is 1. For all integers, the least common factor is 1. Since the least common factor is always 1, there is no point in calculating it. Therefore, you will be asked questions about the greatest common divisor, not the least common divisor. Also, there is no concept of the greatest common multiple. As we explained before, there are an infinite number of multiples. To review, the common multiples of 6 and 8 are “24, 48, 72, 96, 120……” and so on forever. No matter how large the number increases, you will never find the greatest common multiple. This is why the greatest common multiple does not exist. You can only find a", "we know that the required base-10 integer must be divisible by both 6 and 8. Least common multiple of 6 and 8 = 24. Check: $$24_{10} = 44_5 = 33_7$$", "### Theory: Common multiple A multiple (product), which is common to two or more numbers is called a common multiple. Example: 1. Find the first common multiple of the numbers $$8$$, $$9$$ and $$12$$. The first common multiple of $$8$$, $$9$$ and $$12$$ is $$72$$. 2. List first four common multiples of $$3$$ and $$6$$. The first our common multiples of $$3$$ and $$6$$ are $$6$$, $$12$$, $$18$$ and $$24$$.", "# 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.", "is a reasonable choice? Can we choose a huge number like $$600$$? No, because $$600$$ doesn’t divide $$24$$. So, we are looking for the smallest of these values which divides every possible $$z$$. A least common multiple, if you will. That would be $$12$$. This works for every $$z$$ value i.e. every common multiple of $$4$$ and $$6$$. Any number which is a common multiple, is itself a multiple of the least common multiple. We can choose $$z=12$$ or $$z=600$$ and it doesn’t matter. $$w=12$$ will always work. Let’s look at what we just did there. We took this diagram, which admittedly we kind of just conjured out of thin air, but we used it to rigorously define the least common multiple. A number is the least common multiple if and only if it satisfies this diagram in this category. This is a rigorous definition. This quality of how the least common multiple interacts with arbitrary numbers is its universal property. Later, we will discuss at greater length how these kinds of structures can be applied to sets. Even though with the least common multiple there might be simpler ways to define it, we often find in Algebra that the best way to define objects is by their universal property. As an exercise to the adventurous and curious", "112,... Looking at the lists above, notice that 48 and 96 are the first two non-zero common multiples of 12 and 16. Thus, Rory and Tana will pass the starting together again at minutes 48 and 96. ## Common Multiples - Key takeaways • A common multiple is a multiple that is shared between two numbers. • To find the common multiples of a given set of numbers: 1. List the multiples of each number given in the set; 2. Pick out any identical multiples shared from the lists written in Step 1. • Important properties of common multiples: • A set of numbers can have more than one common multiple • A set of numbers can have an infinite number of common multiples • The common multiple of a set of numbers is always greater than or equal to each of the numbers themselves • The given set of numbers divides the common multiple without leaving a remainder. • Every non-zero integer is a multiple of 0 since any non-zero integer multiples by 0 equal 0. A common multiple is a multiple that is shared between two (or more) numbers. The characteristics/properties of common multiples are: • a set of numbers can have more than one common multiple; • a set of numbers can have an infinite", "120 views Consider the set of integers $\\{1,2,3,4,6,8,12,24\\}$ together with the two binary operations LCM (lowest common multiple) and GCD (greatest common divisor). Which of the following algebraic structures does this represent? 1. group 2. ring 3. field 4. lattice ans is lattice .", "# 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 $3 \\times 4 \\times 7 = 84$.", "# Definition of Least Common Multiple The lowest common multiple of two integers is the smallest positive integer that both numbers are factors of. In other words, it is the smallest positive integer that is a multiple of both numbers. We can similarly find the lowest common multiple of more than two numbers. For example, to find the lowest common multiple of $12$ and $18$, we start by writing down lists of numbers, in increasing order, that are multiples of each number. The lowest common multiple is the smallest number that occurs in both lists. So, the lowest common multiple of $12$ and $18$ is $36$. Note that $72$ is also a common multiple of $12$ and $18$, but it is not the lowest common multiple, because $36$ is a common multiple that is smaller. Note: Before we start writing down multiples of $12$ and $18$, we know that these numbers have one common multiple: $12 \\times 18 = 216$, so we know our lists of common multiples will eventually stop. Another term for \"lowest common multiple\" is \"least common multiple\". ### Description The aim of this dictionary is to provide definitions to common mathematical terms. Students learn a new math skill every week at school, sometimes just before they start a new skill, if they want to", "# Uniqueness of the least common multiple Define the least common multiple of 2 nonzero integers a and b, denoted by $\\operatorname{lcm}(a,b)$ to be the nonnegative integer $m$ such that both $a$ and $b$ divide $m$, and if $a$ and $b$ divide any other integer $n$, then $m$ also divides $n$. Prove that any two integers $a$ and $b$ have a unique least common multiple. Not really sure how to prove this, but here's an attempt, I guess I am thinking it would be something like: Let the least common multiple of two nonzero integers $a$ and $b$ be given my $\\operatorname{lcm}(a,b)=ab=m$. If $n/ab=x$ where $n$ and $x$ are integers, then $n/m=x$, so $m$ also divides $n$. Therefore any two integers $a$ and $b$ have a least common multiple. Any suggestions for how to think of this problem? • lcm$(9,12)=36$, not $9\\times 12=108$. The lcm$(a,b)$ is the smallest number that both $a$ and $b$ divide into. – Joffan Feb 6 '15 at 0:08 Multiples of $a$ are $$a, 2a, 3a, 4a, \\ldots\\,.$$ Multiples of $b$ are $$b, 2b, 3b, 4b, \\ldots\\,.$$ First we prove existence of a common multiple: $ab$ is a common multiple: it is the $b$th element of the first list and the $a$th element of the second. It follows that the set of all common", "looks like the following. • $48÷14=3$ R $6$ • $70÷14=5$ • $100÷14=7$ R $2$ • $108÷14=7$ R $10$ • $112÷14=8$ We now know that if the answer is 5 or 8, we can divide. Therefore, we can see that the multiples of 14 are 70 and 112. In fact, $5×14=70$ and $8×14=112$. Division is also multiplication. Just like with factors, use both multiplication and division when finding the answer to multiples. ### How to Find Common Multiples and Least Common Multiple If you understand what we have discussed so far, you will be able to understand common multiples and least common multiple. When comparing two or more numbers, the same multiples are called common multiples. For example, what are the common multiples of 6 and 8? To answer this question, let’s write down the multiples of 6 and 8 respectively. It will look like this. • Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48 • Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64… If we write out the multiples in this way, 24 and 48 are common. Therefore, the common multiples of 6 and 8 are 24 and 48. In fact, the multiples of 6 and 8 go on indefinitely. Therefore, the common multiples of 6 and 8 are 24, 48, 72,", "about factors of $$42$$? They are $$2$$, $$3$$, $$6$$, $$7$$, $$14$$, and $$21$$. Do $$70$$ and $$42$$ have any factors in common? Yes, $$2$$, $$7$$, and $$14$$. What is the greatest of these common factors? $$14$$ For kind of obvious reasons, we call $$14$$ the greatest common factor (GCF) of $$42$$ and $$70$$. The GCF is said to be useful for simplifying fractions. Divide the top and bottom by the GCF and the fraction will be simplified in one round of division. #### Multiples Multiples are products of whole numbers. Multiples of $$5$$, for example, are $$10$$, $$15$$, $$20$$, and $$25$$. You should know how to find the least common multiple of two (or more) integers. This doesn’t mean the rarest multiple, though it does sound that way. It means the lowest common positive multiple of two or more integers. Think of 6 and 9. Some multiples of 6 are 12, 18, 24, 30, 36. Some multiples of 9 are 18, 27, 36, 45. What is the lowest of those that is common to both groups? 18, right? The number 36 is also a common multiple, but it’s not the lowest. This can come in handy when you are finding a common denominator to add or subtract fractions. #### Even and Odd Numbers Well-known properties of even integers", "# How many positive integers N are there such that the least common multiple of N and 1000 is 1000? How many positive integers N are there such that the least common multiple of N and 1000 is 1000? I did found the solution to this problem , but I did it by brute force. When I was tackling this problem, I didn't know what multiples where so I had to look it up. I think I had a blurred view of what multiples are. So, to help me, I found the definition: multiples are some number multiplied by an integer. For instance, the multiples of 4 are ...-8,0,4,8,12,16,20 and so on. And the least common multiple of 4 and 5(as in an example to help clarify the problem) is 20 because The multiples of 4: 4,8,12,16,20 The multiples of 5: 5,10,15,20,25 I know I probably wasted your time writing that stuff above, but it helps me because I get forget what it means to find multiples. Now back to the problem. The problem asks to find all values n such that 1000 is divisible by n. So basically, I just listed them like this: n: 1 2 ,4, 5, 8, 10, 20, 25, 40, 50,100,125,200,250,500,1000 1000/n has remainder 0? : There are 16 n's. So again, I" ]

[ "b)$ or, more commonly, (a,b). The greatest common divisor of 12 and 8 is 4, the greatest common divisor of 15 and 16 is 1. Two numbers with a greatest common divisor of 1 are said to be relatively prime. Complementary to the notion of greatest common divisor is least common multiple. The least common multiple of a and b is the smallest (positive) integer m such that a | m and b | m. Thus, the least common multiple of 12 and 9 is 36 (written [12,9] = 36). ## Abstract divisors In higher mathematics, the notion of divisor has been abstracted from the integers to the context of general commutative rings. In this setting, they might be termed abstract divisors.", "an answer. Back to the question: 1. 0 is a multiple of every integer, so 0 is a common multiple of every non-zero sets of integers. 2. 0 is not the only common multiple of 75, 98, and 140, less than 14,000, so is -14,700, -2*14,700, -3*14,700, -4*14,700, ... Hope it's clear. _________________ Re: If k is a common multiple of 75, 98, and 140, which of the [#permalink] 30 Jul 2015, 04:37", "# Practice: Finding common multiples How many positive integers less than or equal to $1000$ are common multiples of the three numbers $8, 36,$ and $24$? ×", "96, 120…… and so on. Thus, it is impossible to find out all the common multiples, we just need to find a few of them. The smallest of these multiples is called the least common multiple; in the case of the multiples of 6 and 8, the least common multiple is 24. ## There Is No Concept of Least Common Factor and Greatest Common Multiple We learned about the concepts of the greatest common divisor and least common multiple. On the other hand, there is no concept of least common factor and greatest common multiple. What is the smallest number among all the common factors? It is 1. For all integers, the least common factor is 1. Since the least common factor is always 1, there is no point in calculating it. Therefore, you will be asked questions about the greatest common divisor, not the least common divisor. Also, there is no concept of the greatest common multiple. As we explained before, there are an infinite number of multiples. To review, the common multiples of 6 and 8 are “24, 48, 72, 96, 120……” and so on forever. No matter how large the number increases, you will never find the greatest common multiple. This is why the greatest common multiple does not exist. You can only find a", "what is the greatest common divisor of 12 and n ? [#permalink] ### Show Tags 07 Jul 2018, 05:10 anuj04 wrote: Bunuel wrote: If n is an integer, what is the greatest common divisor of 12 and n ? (1) The product of 12 and n is 432. (2) The greatest common divisor of 24 and n is 12. NEW question from GMAT® Official Guide 2019 (DS05377) From Statement 1: \"The product of 12 and n is 432\" i.e. n=432/12= 36. So Greatest common divisor of 12, 36 can be derived. - Sufficient From Statement 2: \"The greatest common divisor of 24 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. hey pushpitkc, i dont get if the product of 12 and n is 432\" it means n = 36 and the question asks: If n is an integer, what is the greatest common divisor of 12 and n ? how can 36 be divisor of 12 Also From Statement 2: \"The greatest common divisor of 24 and n is 12.\". i dont get how greatest common divisor of 24 can be 12, 36, 60, 84 ? any ideas ? thanks and have a", "# 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 $3 \\times 4 \\times 7 = 84$.", "112,... Looking at the lists above, notice that 48 and 96 are the first two non-zero common multiples of 12 and 16. Thus, Rory and Tana will pass the starting together again at minutes 48 and 96. ## Common Multiples - Key takeaways • A common multiple is a multiple that is shared between two numbers. • To find the common multiples of a given set of numbers: 1. List the multiples of each number given in the set; 2. Pick out any identical multiples shared from the lists written in Step 1. • Important properties of common multiples: • A set of numbers can have more than one common multiple • A set of numbers can have an infinite number of common multiples • The common multiple of a set of numbers is always greater than or equal to each of the numbers themselves • The given set of numbers divides the common multiple without leaving a remainder. • Every non-zero integer is a multiple of 0 since any non-zero integer multiples by 0 equal 0. A common multiple is a multiple that is shared between two (or more) numbers. The characteristics/properties of common multiples are: • a set of numbers can have more than one common multiple; • a set of numbers can have an infinite", "every number. So that's the answer. Consideration of the seemingly silly question of the greatest common multiple has led us to a better understanding of the multiplicative structure of the integers, and we see that there is a greatest common multiple, which is exactly dual to the least common divisor, just as the least common multiple is dual to the greatest common divisor.", "36 and the question asks: If n is an integer, what is the greatest common divisor of 12 and n ? how can 36 be divisor of 12 Also From Statement 2: \"The greatest common divisor of 24 and n is 12.\". i dont get how greatest common divisor of 24 can be 12, 36, 60, 84 ? any ideas ? thanks and have a great day Hey dave13 For the first statement, it says which is the greatest divisor the two integers, n and 12. The greatest common divisor (GCD) of a set of integers is the largest integer that divides each integer in the set. So, if n=36, the greatest common divisor is 12. As for the second statement, the number n and 24 have a greatest common divisor of 12. So, n can be any multiple of 12 - 12,12*2 = 24,12*3 = 36, and so on. This link should Hope this helps you! _________________ You've got what it takes, but it will take everything you've got Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2669 Re: If n is an integer, what is the greatest common divisor of 12 and n ? [#permalink] ### Show Tags 13 Jul 2018, 14:26 Bunuel wrote: If n is an integer, what is", "### Home > CC1 > Chapter 8 > Lesson 8.2.1 > Problem8-77 8-77. The first four multiples of $5$ are $5$, $10$, $15$, and $20$. 1. What are the first six multiples of $10$? To find the first six multiples of $10$, you simply multiply $10$ by the numbers $1$ through $6$ separately and create a list of the products you find similar to the list written above. The first six multiples of $10$ are $10,20,30,40,50$, and $60$. 2. What are the first six multiples of $8$? The first six multiples of $8$ are $8,16,24,32,40$, and $48$. 3. What is the least common multiple of $10$ and $8$? The Least Common Multiple of two or more integers is the lowest positive integer that is divisible by both (or all ) of the integers. For example, the multiples of $3$ and $5$ are shown below in the table. $15$ is the Least Common Multiple because it is the smallest positive integer divisible by both $3$ and $5$. Use this idea to find the least common multiple of $10$ and $8$. $\\begin{array}{c|c|c|c|c|c} \\quad 3 \\; & \\; 6 \\; & \\; 9 \\; & \\; 12 \\; & \\; \\mathbf{15} \\; & \\; 18 \\quad\\\\ \\hline \\quad 5 \\; & \\; 10 \\; & \\; \\mathbf{15} \\; & \\; 20", "the set of which is expressed as N. We also assume the notation and properties of the greatest common divisor presented in Hungerford [1]. In particular, if a and b are positive integers, we denote the greatest common divisor of a and b by (a,b). To begin the discussion of least common multiple, we present the following definition. Definition 1. If a and b are positive integers, the least common multiple of a and b, denoted [a,b], is the least positive element of the set {z  N : a | z and b | z }. It should be remarked that for any positive integers a and b, their product a We see that 12 is the least entry that is common to both sets. Indeed, 4|12 and 6|12 so 12 is definitely a common multiple. On the other hand, since the only positive multiple of 6 less than 12 is 6 itself, and since 4 is not a divisor of 6, we see that no other common multiple can be less than 12. This shows that 12 is the least common multiple.änd prof dr mircea orasanu and prof horia orasanu In this example, we notice that 12 is not only less than or equal to every other common multiple of 4 and 6, it is also", "be divisor of 12 Also From Statement 2: \"The greatest common divisor of 24 and n is 12.\". i dont get how greatest common divisor of 24 can be 12, 36, 60, 84 ? any ideas ? thanks and have a great day BSchool Forum Moderator Joined: 26 Feb 2016 Posts: 2936 Location: India GPA: 3.12 Re: If n is an integer, what is the greatest common divisor of 12 and n ? [#permalink] ### Show Tags 07 Jul 2018, 05:34 1 dave13 wrote: anuj04 wrote: Bunuel wrote: If n is an integer, what is the greatest common divisor of 12 and n ? (1) The product of 12 and n is 432. (2) The greatest common divisor of 24 and n is 12. NEW question from GMAT® Official Guide 2019 (DS05377) From Statement 1: \"The product of 12 and n is 432\" i.e. n=432/12= 36. So Greatest common divisor of 12, 36 can be derived. - Sufficient From Statement 2: \"The greatest common divisor of 24 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. hey pushpitkc, i dont get if the product of 12 and n is 432\" it means n =", "# Definition of Least Common Multiple The lowest common multiple of two integers is the smallest positive integer that both numbers are factors of. In other words, it is the smallest positive integer that is a multiple of both numbers. We can similarly find the lowest common multiple of more than two numbers. For example, to find the lowest common multiple of $12$ and $18$, we start by writing down lists of numbers, in increasing order, that are multiples of each number. The lowest common multiple is the smallest number that occurs in both lists. So, the lowest common multiple of $12$ and $18$ is $36$. Note that $72$ is also a common multiple of $12$ and $18$, but it is not the lowest common multiple, because $36$ is a common multiple that is smaller. Note: Before we start writing down multiples of $12$ and $18$, we know that these numbers have one common multiple: $12 \\times 18 = 216$, so we know our lists of common multiples will eventually stop. Another term for \"lowest common multiple\" is \"least common multiple\". ### Description The aim of this dictionary is to provide definitions to common mathematical terms. Students learn a new math skill every week at school, sometimes just before they start a new skill, if they want to", "# 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.", "• anonymous Is this a trick question? \"Which pairs of integers $$a$$ and $$b$$ have greatest common divisor $$18$$ and least common multiple $$540$$?\" There is a theorem that states that $$[a,b](a,b)=ab$$, where $$[a,b]$$ stands for the least common multiple of $$a$$ and $$b$$, and $$(a,b)$$ stands for the greatest common divisor of $$a$$ and $$b$$. So, it follows that$[a,b](a,b)=ab,$$540\\cdot18=a\\cdot b.$In other words, $$a=540$$ and $$b=18$$ would suffice. Am I missing something? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "and the greatest common multiple One day Katara reported that her middle school math teacher had gone into a rage (whether real or facetious I don't know) about some kids’ use of the phrase “greatest common multiple”. “There is no greatest common multiple!” he shouted. But that got me thinking. There isn't a greatest common multiple, but there certainly is a least common divisor; the least common divisor of !!a!! and !!b!! is !!1!!, for any !!a!! and !!b!!. The least common multiple and greatest common divisor are beautifully dual, via the identity $$\\operatorname{lcm}(a,b)\\cdot\\gcd(a,b) = ab,$$ so if there's a simple least common divisor I'd expect there would be a greatest common multiple also. I puzzled over this for a minute and realized what is going on. Contrary to Katara's math teacher, there is a greatest common multiple if you understand things properly. There are two important orderings on the positive integers. One is the familiar total ordering by magnitude: $$0 ≤ 1≤ 2≤ 3≤ 4≤ \\ldots.$$ (The closely related !!\\lt!! is nearly the same: !!a\\lt b!! just means that !!a\\le b!! and !!a\\ne b!!.) But there is another important ordering, more relevant here, in which the numbers are ordered not by magnitude but by divisibility. We say that !!a\\preceq b!! only when !!b!! is a multiple of", "reported that her middle school math teacher had gone into a rage (whether real or facetious I don't know) about some kids’ use of the phrase “greatest common multiple”. “There is no greatest common multiple!” he shouted. But that got me thinking. There isn't a greatest common multiple, but there certainly is a least common divisor; the least common divisor of !!a!! and !!b!! is !!1!!, for any !!a!! and !!b!!. The least common multiple and greatest common divisor are beautifully dual, via the identity $$\\operatorname{lcm}(a,b)\\cdot\\gcd(a,b) = ab,$$ so if there's a simple least common divisor I'd expect there would be a greatest common multiple also. I puzzled over this for a minute and realized what is going on. Contrary to Katara's math teacher, there is a greatest common multiple if you understand things properly. There are two important orderings on the positive integers. One is the familiar total ordering by magnitude: $$0 ≤ 1≤ 2≤ 3≤ 4≤ \\ldots.$$ (The closely related !!\\lt!! is nearly the same: !!a\\lt b!! just means that !!a\\le b!! and !!a\\ne b!!.) But there is another important ordering, more relevant here, in which the numbers are ordered not by magnitude but by divisibility. We say that !!a\\preceq b!! only when !!b!! is a multiple of !!a!!. This has many of the same familiar", "is a reasonable choice? Can we choose a huge number like $$600$$? No, because $$600$$ doesn’t divide $$24$$. So, we are looking for the smallest of these values which divides every possible $$z$$. A least common multiple, if you will. That would be $$12$$. This works for every $$z$$ value i.e. every common multiple of $$4$$ and $$6$$. Any number which is a common multiple, is itself a multiple of the least common multiple. We can choose $$z=12$$ or $$z=600$$ and it doesn’t matter. $$w=12$$ will always work. Let’s look at what we just did there. We took this diagram, which admittedly we kind of just conjured out of thin air, but we used it to rigorously define the least common multiple. A number is the least common multiple if and only if it satisfies this diagram in this category. This is a rigorous definition. This quality of how the least common multiple interacts with arbitrary numbers is its universal property. Later, we will discuss at greater length how these kinds of structures can be applied to sets. Even though with the least common multiple there might be simpler ways to define it, we often find in Algebra that the best way to define objects is by their universal property. As an exercise to the adventurous and curious", "# least common multiple A common multiple of two integers is an integer divisible by both integers. The least common multiple of two integers is the smallest positive common multiple. • The notation for the least common multiple of $$a$$ and $$b$$ is $$\\operatorname{lcm}(a, b)$$. • By definition, the least common multiple of two integers is independent of the sign of the two integers --- the $$\\operatorname{lcm}$$ is the same if you replace $$a$$ with $$-a$$ or $$b$$ with $$-b$$. • The definition of $$\\operatorname{lcm}$$ does not make sense if either $$a$$ or $$b$$ are is equal to $$0$$. • For the remainder of the section, we will assume integers to be nonzero. The existence of a least common multiple stems from the fact that the absolute value of the product $$|ab|$$ is a positive common multiple of $$a$$ and $$b$$. ## naive computation The least common multiple can be computed naively by checking every multiple of one of them up to their product to see if it is divisible by the other. For example, to compute $$\\operatorname{lcm}(154, 56)$$ • start with $$154$$ and keep adding $$154$$ until it is divisible by $$56$$ • $$56$$ does not divide $$154$$, $$154(2)$$, or $$154(3)$$ • however, $$154(4) = 616 = 56(11)$$ is divisible by $$56$$ • the least common multiple", "# Uniqueness of the least common multiple Define the least common multiple of 2 nonzero integers a and b, denoted by $\\operatorname{lcm}(a,b)$ to be the nonnegative integer $m$ such that both $a$ and $b$ divide $m$, and if $a$ and $b$ divide any other integer $n$, then $m$ also divides $n$. Prove that any two integers $a$ and $b$ have a unique least common multiple. Not really sure how to prove this, but here's an attempt, I guess I am thinking it would be something like: Let the least common multiple of two nonzero integers $a$ and $b$ be given my $\\operatorname{lcm}(a,b)=ab=m$. If $n/ab=x$ where $n$ and $x$ are integers, then $n/m=x$, so $m$ also divides $n$. Therefore any two integers $a$ and $b$ have a least common multiple. Any suggestions for how to think of this problem? • lcm$(9,12)=36$, not $9\\times 12=108$. The lcm$(a,b)$ is the smallest number that both $a$ and $b$ divide into. – Joffan Feb 6 '15 at 0:08 Multiples of $a$ are $$a, 2a, 3a, 4a, \\ldots\\,.$$ Multiples of $b$ are $$b, 2b, 3b, 4b, \\ldots\\,.$$ First we prove existence of a common multiple: $ab$ is a common multiple: it is the $b$th element of the first list and the $a$th element of the second. It follows that the set of all common" ]

[ "## A racing problem.... Bob and Al are racing. Their pdfs are given as: Al 1<= A <=3 Bob 2<=B<=4 What is the probability of Al winning the race? (assuming lesser value means win). I think this is a joint pdf problem and tried using double integral on both of them: int(2,4) int(2,B) A*B...not sure if the equation is right though....any help? thanks", "Four runners compete in a race: Annie, Becca, Carlos, and Dante. After some confusion at the finish line, it's unclear what the final finishing order was, but the following information is known: 1. Dante finished before Annie. 2. Becca wasn't third. 3. There were two runners between Annie and Carlos. Who won the race? Finally, it says that there were two runners between Annie and Carlos. That means that both Annie and Carlos couldn't have been $2nd$ or $3rd$. Now we can see that Annie must be $4th$. That makes Dante $3rd$ and Becca $2nd$, leaving Carlos as $1st$.", "five runners instead of four. Can you use logic to figure out the order? # Today's Challenge Four runners compete in a race — Amelia, Boris, Carlos, and Dara. After some confusion at the finish line, it's unclear what the final finishing order was, but this information is known: 1. Dara finished before Amelia. 2. Boris wasn't third. 3. There were two runners between Amelia and Carlos. $\\\\[-0.7em]$ $\\\\[-0.5em]$ Who won the race?", "13 May 2015, 02:25 1 KUDOS Expert's post HarshitBShah wrote: marcodonzelli wrote: Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob? A. 24 B. 30 C. 60 D. 90 E. 120 My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ? Hi HarshitBShah, You have considered one of the scenarios of the total possible combinations. When you assume Meg & Bob as one entity you only take the case where Bob is the next person behind Meg. Imagine a situation where Meg finishes 1st, since Bob has to finish behind Meg, Bob can finish at any place from 2nd to the 5th. Similarly when Meg finishes 2nd, Bob can finish at any place from 3rd to 5th and so", "C 1st: B 2nd: A 3rd: C 1st: A 2nd: C 3rd: B 1st: C 2nd: A 3rd: B 1st: B 2nd: C 3rd: A 1st: C 2nd: B 3rd: A 1st: AB (tie) 2nd: C 1st: AC (tie) 2nd: B 1st: BC (tie) 2nd: A 1st: A 2nd: BC (tie) 1st: B 2nd: AC (tie) 1st: C 2nd: BC (tie) 1st: ABC (tie) That comes out to 13 different ways these horses unicorns can finish the race. That’s the answer for 3 unicorns. What’s the answer for 8 unicorns? (FYI: If you want to know if you’re on the right track… I have 541 for 5 unicorns…)", "monitored. Each contestant had to run in two races so that the average place could be determined. Only one runner finished in the same place in both races. Alan was never last. Charlie always beat Darren. Brian had at least one first place. Alan finished third in at least one of the races. Both Darren and Charlie had a second place. What were the two results? As per the precedent set in the Enjoyable Enigmas thread I am permitted to ask clarifying questions. You describe the races as \"heats\". By that you mean: \"a less important race or competition in which it is decided who will compete in the final event,\" correct? 4 contestants are named, so I assume that is the total number of participants. It is asserted that each ran two races, therefore the total number of \"heats\" was 2, and each of the 2 heats had the same 4 contestants, correct? Greg's n=4 A!=4 (1,2) C<D :.C!=4 (1,2) B=1(1&/2) 1......2 1 C . 1 B 2 D . 2 C 3 A . 3 A 4 B . 4 D or in English- race a) 1 Charlie 2 Darren 3 Alan 4 Brian race b) 1 Brian 2 Charlie 3 Alan 4 Darren ? Fighting spam with spam: didn't Gandhi have something against that? It", "## Who Won the Race? How much information do you need to find the finish order of a three-person race? It turns out, not much at all. Suppose that in a race among Amelia, Boris, and Carlos, we only know one fact — that Boris finished before Carlos. What are the possible orders for the entire race? Keep reading to see some reasoning we can use, or jump ahead to today's challenge for a trickier problem. We can start by thinking about a single unknown factor in the problem: \"Where did Amelia finish?\" To see this logic demonstrated, use the arrows below the figure: Since we know the relationship between Boris and Carlos, Amelia's position in the race completely determined the entire finishing order. There are three possible total finishing orders — one for each of the positions in which Amelia can finish. What if we add another runner to the race, Dara, and have a more complicated scenario? Dara finished more places ahead of Amelia than Boris finished before Carlos. We can start by again thinking about an unknown — \"How did Boris and Carlos finish?\" There is only one way these conditions can be satisfied without conflicts, so we have only one solution. In the problem below, you'll be putting together information like this, but with", "1🐎︎ We can eliminate one more horse with the “slower-than-3” rule: the 3rd orange horse. It was beaten by the 2nd and 1st orange horses, which in turn were slower than the 1st red horse. So the 3rd orange horse is eliminated: 1: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 2: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 3: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 4: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 5: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ Notice that we only have 5 remaining horses to sort now, and only the 2nd and 3rd overall positions to find. We just run a race with those 5 remaining horses to find out: 3🐎︎ 2🐎︎ 1🐎︎ 2🐎︎ 1🐎︎ As it turns out in this example, the 1st purple horse and 2nd red horse take the top two positions in this race. That means the final top three are the 1st red, 1st purple and 2nd red horse: 🐎︎ 🐎︎ 🐎︎ This takes a total of 7 races: the initial 5 groups, the fastest-of-fastest race, and the surviving runners up race at the end.", "example): Every person has exactly one father. Bill is the father of Jake, and Bob is the father of Jake. What can be said about bill and bob? The very fact that this seemingly contradictory riddle has an answer should signal that there is some trickery involved, again, it is a riddle not just a question, think outside the of the box (but still in formal logic terms). Last edited: Dec 2, 2005 10. ### AKG 2,585 Under the condition that 2 = 1, the nice guy finishes last? If this is the answer, you should note tha saying that \"two nice guys are in the race\" means that there are indeed two (distinct) nice guys in the race, and the above is not a sensible way to say that \"there is one nice guy known by two names in the race.\" Two actually means two, not one with two names. 11. ### croxbearer 17 since each of the 5 people in the race aim to win the race, and nice guys always finish last, hence, one of the 3 not nice guys will win the race and one of the 2 nice guys will finish last. anyway, its unimportant who will finish last! :yuck: 12. ### Ubern0va 60 AKG, yes that is more or less the answer.", "Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob? I solved in a different way: All the possible positions: 5! = 120 Restriction: we have to eliminate all the combinations in which Bob finishes ahead of Meg: If Bob finishes 1st, (_ _ _ _ B) Meg could finish 2nd, 3th, 4th or 5th --> 4 possibilities. If Bob finishes 2nd, Meg finishes 3th, 4th or 5th. --> 3 possibilities The same for Bob 3th and 4th. Bob can not finishes 5th. This sum up to 10 combinatios: 4 + 3 + 2 + 1 = 10. In addition, we have the rest of the combinations of the of the positions remaining 3 competitors: 3! Now, we have to combine the positions of Meg and Bob with the positions of the rest of the competitors: 3! * 10 = 60 Finally: 120 - 60 = 60 _________________ Encourage cooperation! If this post was very useful, kudos are welcome \"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome\" William James", "62396 Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co [#permalink] ### Show Tags 24 Jul 2014, 23:20 Gyanendra wrote: # of participants is not clear. is it 12 (triplets + 9) or just 9(including triplets) Language is confusing :/ I don't see anything confusing there. Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon... Would it be correct to write that there are 9 competitors in the triathlon if there were 12 competitors? _________________ Re: Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co [#permalink] 24 Jul 2014, 23:20 Go to page 1 2 Next [ 26 posts ] Display posts from previous: Sort by", "GMAT Club Legend Joined: 16 Oct 2010 Posts: 8124 Location: Pune, India Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 13 May 2015, 22:06 HarshitBShah wrote: marcodonzelli wrote: Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob? A. 24 B. 30 C. 60 D. 90 E. 120 My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ? \"Meg finishes ahead of Bob\" is different from \"Bob finishes immediately after Meg\". You accounted for cases such as: Meg, Bob, A, B, C and C, Meg, Bob, A, B etc but how about cases such as: Meg, B, Bob, A, C? You need to use the principle of symmetry here. Check out", "5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob? A. 24 B. 30 C. 60 D. 90 E. 120 Ans: C. 60 M stands for Meg If M takes first place, the other 4 can be arranged in 4! ways. If M takes second place, the other 4 can be arranged in 3 * 3! ways such that Bob finishes after Meg. If M takes third place, the other 4 can be arranged in 6 * 2! ways such that Bob finishes after Meg. If M takes fourth place, the other 4 can be arranged in 3! ways such that Bob finishes after Meg. M cannot take fifth place, because in tat case Bob will finish ahead of Meg. So total number of possibilities is = 4! + 3 * 3! + 6 * 2! + 3! = 60 Intern Joined: 24 Feb 2012 Posts: 33 Followers: 0 Kudos [?]: 13 [4] , given: 18 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 26 Feb 2012, 19:27 4 KUDOS hmm, I went about it the long way until", "the three runners which finish first, second, and third (Keiba, Keirin, Kyōtei and Auto Race). • WIN 5 / Select 5: The bettor must pick the winners of five designated races. Betting on operators' website by PC or cellular phone only (Keiba with JRA or some local governments' operation only).", "more races out of 45\".", "Carl is the same as giving it to Carl, Becky and Art. Now, if the tickets were different (front row, second row, and third row) then it would be a permutation since ABC (Art in the front row) is different from CBA (Carl in the front row). 95. Barry G says: Hi, I have the following problem: Supposing I have 4 horse races and have to select the winner in each race (I can select more than 1 horse in each race) and I have selected 1 winner from race 1, and 2 winners in each of the last 3 races, is there anyway method or nice way to list out all possible permutations between these races? I know there are 8 permutations (1 X 2 X 2 X 2 = 8) but I’m trying to figure how I can write these 8 permutations out without error and without having to go through each possible sequence mentally…I know it’s relatively straightforward for only 8 perms but if there were 16 perms (2 X 2 X 2 X 2) it will be a lot more complicated and so on. Just wondering if you can think of a gerneralised method to write these down accurately. Race 1: A Race 2: B X Race 3: C Y Race 4: D Z", "Booth PT '18 (S) Followers: 0 Kudos [?]: 8 [0], given: 2 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 19 Jan 2015, 01:08 If u catch that its simple symmetry ans takes < 30 sec total prob = 5! = 120 the chances of one being ahead of another are equal....so if we are looking for just one to be ahead of the other => half => 120/2 =60 Intern Joined: 20 Dec 2014 Posts: 15 Followers: 0 Kudos [?]: 1 [0], given: 23 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 12 Mar 2015, 11:34 Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob? A. 24 B. 30 C. 60 D. 90 E. 120 When I first saw this I thought combination but then realized its permutation because order does matter(Meg has to finish ahead of Bob). So we can use n!/(n-k)! n represents the 5 participants and k would be the 3 remaining cyclers which the orders do not matter", "would be 8 or more. Thus the optimal solution should include heats of 5 groups of 5. # Choices for 6th race After the heats, you have 15 horses who qualified. You must be able to get the top three in some way from these 15. By running the winners in a single race, you will eliminate all the horses in the groups that finished 4th and 5th, as well as the two horses in the 3rd place group, and the third horse in the 2nd place group. This seems to minimize the remaining horses (5) and all fit into a single race for the consolation race, so there does not appear to be any obvious alternatives to the 6th race. # Choices for the 7th race You will notice that the last race includes some known outcomes already. For example: A2 is faster than A3 B1 is faster than B2 B1 is faster than C1 However, the unknowns are: Is A2 faster than B1? If so, is A3 faster than B1? If so, then the order is A1, A2, A3 otherwise The order is A1, A2, B1 otherwise Who is the fastest of A2, B2, and C1? If A2, the order is A1, B1, A2 If B2, the order is A1, B1, B2 If C1, the order", "in an election, or the finishing order in a race. Generally the keyword 'ways' indicates that combinations apply. 42. anonymous awesome thank you again!! 43. kropot72 You're welcome :)", "not the second nor the third place: it has to race one horse from each previous race, so if the first three places were all in the same heat or 1,2 in the same and 3 in a different, it couldn't tell the difference. So $n<25$, and in the last race you have to race the new horses against the first three places A1, A2, A3 (possibly without knowing the order) of the $n$ horses that already have raced, otherwise you cannot determine the overall three fastest horses with the last race. Hence $n=23$ or $n=24$. If you assume $n=23$, then it is impossible to determine the three fastest horses out of 23 with 5 races: You cannot have less than three different winners, since only two repetitions are allowed. If you have three different winners that have not been compared, then any of the 5 second places could be A2 or A3. If you have four or more different winners that have not been compared, all of them could be A1,A2, or A3. If $n=24$, then you have to know a set of four horses out of the first 24 which contains A1,A2 and A3. But this is impossible in 5 races, since you have at least 4 different winners that have not been compared, and so" ]

[ "$\\underbrace{b+b+\\cdots+b}_{a}=a b$ possibilities. 1.4.2. It is often easier to think about the experiments as being in chronological order, but there is no requirement in the multiplication rule that Experiment A has to be performed before Experiment $\\mathrm{B}$. Example 1.4.3 (Runners). Suppose that 10 people are running a race. Assume that ties are not possible and that all 10 will complete the race, so there will be welldefined first place, second place, and third place winners. How many possibilities are there for the first, second, and third place winners? Solution: There are 10 possibilities for who gets first place, then once that is fixed there are 9 possibilities for who gets second place, and once these are both fixed there are 8 possibilities for third place. So by the multiplication rule, there are $10 \\cdot 9 \\cdot 8=720$ possibilities. We did not have to consider the first place winner first. We could just as well have said that there are 10 possibilities for who got third place, then once that is fixed there are 9 possibilities for second place, and once those are both fixed there are 8 possibilities for first place. Or imagine that there are 3 platforms, which the first, second, and third place runners will stand on after the race. The platforms are gold, silver, and", "he gets. There are then $4$ remaining awards, from which we choose $2$ to give to one student and $2$ to give to the other, which can be done in $\\binom{4}{2} = 6$ ways (and we can say that e.g. the $2$ chosen this way go to the first remaining student and the other $2$ go to the second remaining student, which counts all possibilities). This means the total for the second case is $3 \\cdot 5 \\cdot 6 = 90$, and the answer is $60 + 90 = \\boxed{\\textbf{(B) }150}$. ~savannahsolver ~Interstigation", "will get$6$questions worth$200, 600, 800, 1000$and only$5$worth$400$. This totals to$6\\cdot (200 + 600 + 800 + 1000) + 5\\cdot 400= 6\\cdot(2600) + 2000 = 17600.$Hitting the daily double, the player has$35200$after the first round (a respectable amount of money already). During the second round, he or she will get$6$questions worth$400, 1200, 1600, 2000$and only$4$worth$800$. This totals to$2 \\cdot 6 \\cdot (200 + 600 + 800 + 1000) + 4 \\cdot 800 = 2 \\cdot 15600 + 3200= 34400$. Thus the player will have a total of$\\$69600$ before the second set of double jeopardies. After them, the player will have $\\$278400$. Thus after the final round, assuming the player bets all of his or her money, he or she can end up with$\\$556800$.", "with a probability of $0.2\\cdot 0.5\\cdot 1 = 0.1$. So, the expected number of races is equal to $1\\cdot 0.4 + 2\\cdot 0.24 + 3\\cdot 0.16 + 2\\cdot 0.1 + 3\\cdot 0.1 = 1.86$.", "for the second, and so on. The number of ways the prizes can be distributed will be $$5 !=5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=120$$ ways. Now we will consider some slightly different examples. Solution Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are $$25 \\cdot 24 \\cdot 23=13,800$$ ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn't really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6: ABC ACB BAC BCA CAB CBA How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are $$3 \\cdot 2 \\cdot 1=6$$ ways to do this; we didn't really need to list them all, we can just use permutations! So, out of the", "don't match the first three dice), for a total of $3\\cdot 5=15$ possible outcomes, plus the outcome that all five dice match. In the second case, we need all three dice to match each other. There are $5$ ways to pick which value the three dice will have so that they don't match the first pair, plus the outcome that all five dice match. So there are a total of $15+5=20$ ways to get a full house without all five dice matching, added to the possibility that all five dice match, which makes $21$ ways to get a full house. So, the probability is successful outcomes/total outcomes= 21/216= 7/72 Mar 16, 2019", "outcomes (3,4,5,6) for Roll 2 and 4 acceptable outcomes for Roll 3. So, if the first roll is a 2, then there are $4 \\cdot 4 = 4^2 = 16$ possible outcomes that satisfy our condition. Repeat this process for the 1st Roll being 3, then 4, then 5, and you will see what happens finally at 6. When the first roll is 3, then the second and third rolls both must choose from 4, 5 or 6 to satisfy our condition. That is, the second and third rolls both choose among 3 acceptable values. $3\\cdot 3 = 3^2 = 9$ possible outcomes, given that the first roll is 3, satisfy our condition. When the first roll is 4, the second and third rolls both must choose either 5 or 6. So, if the first roll is a 4, then there are only $2\\cdot2=2^2=4$ outcomes possible that satisfy our condition. When the first roll is 5, the second and third rolls must both be 6. Thus, when the first roll is 5, there is only one outcome, namely (5,6,6), that satisfies our condition. When the first roll is 6, our condition is automatically violated by virtue of no higher number available. Alas, we tally up the total number of outcomes that satisfy our condition. 55 of the 216", "first two dice together have $$6\\cdot 5=30$$ possible outcomes, from above. For each of these 30 outcomes, there are four possible outcomes for the third die, so the total number of outcomes is $$30\\cdot 4=6\\cdot 5\\cdot 4=120$$. (Note that we consider the dice to be distinguishable, that is, a roll of 6, 4, 1 is different than 4, 6, 1, because the first and second dice are different in the two rolls, even though the numbers as a set are the same.) Example $$\\PageIndex{2}$$ Suppose blocks numbered 1 through $$n$$ are in a barrel; we pull out $$k$$ of them, placing them in a line as we do. How many outcomes are possible? That is, how many different arrangements of $$k$$ blocks might we see? This is essentially the same as the previous example: there are $$k$$ \"spots'' to be filled by blocks. Any of the $$n$$ blocks might appear first in the line; then any of the remaining $$n-1$$ might appear next, and so on. The number of outcomes is thus $$n(n-1)(n-2)\\cdots(n-k+1)$$, by the multiplication principle. In the previous example, the first \"spot'' was die number one, the second spot was die number two, the third spot die number three, and $$6\\cdot5\\cdot4=6(6-1)(6-2)$$; notice that $$6-2=6-3+1$$. This is quite a general sort of problem: Definition: permutations The number", "$3\\cdot6\\cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $\\frac{90}{216}=\\frac{5}{12}$. -", "second and sixth places are then $4$ and $8$ (in either order). This reduces the number of possibilities considerably, but there is still a bunch of trial and error.", "6:51 The order in which the runners finish matters, so this is a permutation rather than a combination. Until the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \\cdot 7 \\cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \\frac{8!}{(8 - 3)!} = \\frac{8!}{5!} = \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5!}{5!} = 8 \\cdot 7 \\cdot 6 = 336$$ Imagine if John, Jack and Julie were in the \"top-3\", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at So $8\\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.", "the other team, namely, place everybody who has not already been selected on the other team. Hence, there are $30 \\cdot 1 = 30$ ways to select the two teams.", "the remaining $n-k+1$ individuals. Hence, the total number of ordering is given by $$n \\times (n-1) \\times \\cdots \\times (n-k+1) \\times \\cdots 2 \\times 1$$which is nothing but $n!$. In you case, $n=3$ gives us $3! = 6$ which matches with our enumeration. - Thank you! That was very well explained and exactly what I was looking for. – daniellopez46 Jun 18 '12 at 15:01 Suppose there is 1 person in the race. There is only one possible combination. A 2nd racer ( #2) is in the race. This person can be placed in 2 places(in front of 1 , or behind 1) There are now 2 possible combinations 12 21 a 3rd person enters the race. Racers 1 and 2 are in place, number 3 has 3 possible places to fit in(in the front, between 1 and 2, or in the back. Because there are 2 possible positions for 1 and 2, there are 3*2 possible combinations for 3 racers 12 becomes 312, 132, or 123 21 becomes 321, 231, or 213 a 4th racer enters, there are 4 places this new racer can fit in There will be 4*6 possible combinations n racers have n! possible positions - Number of possibilites for (1) to end first: two, as we have (1)-(2)-(3) and (1)-(3)-(2). A moment of", "you are about to construct you have only 5-1=4 possibilities for $d_1$. For $d_2$ there is no restriction placed on the digit, except that it must not be 5. So there are 10-1=9 possibilities for your choice of $d_2$. Now for digit $d_3$: again, it must not be 5 but also, since it is the leading digit of your number, you must not make it 0 either. So there are 10-1-1=8 possible choices for $d_3$. Since these choices are independent of each other, the total number of possibilities is the product $8\\cdot 9\\cdot 4$ 2. A talent contest has 8 contestants. Judges must award prizes for first, second, and third places. If there are no ties, in how many different ways can the 3 prizes be awarded? Again, imagine yourself selecting the 3 highest placed persons by yourself. For the first place, you have 8 possibilities. For each of these 8 possibilities you have 8-1=7 possibilities to choose a person for the second place. And, finally, you have 8-2=6 possibilities to choose a person for the third place. This gives $8\\cdot 7\\cdot 6$ different ways in all.", "$2/3 \\cdot 1/2 = 1/3$. So the probability of getting to the final round is $2/3$. For your other question we can read off: ii) $1/3$ iii) $1/3$ So switching or staying with your first choice doesn't matter. -", "or she will get$6$questions worth$200, 600, 800, 1000$and only$5$worth$400$. This totals to$6\\cdot (200 + 600 + 800 + 1000) + 5\\cdot 400= 6\\cdot(2600) + 2000 = 17600.$Hitting the daily double, the player has$35200$after the first round (a respectable amount of money already). During the second round, he or she will get$6$questions worth$400, 1200, 1600, 2000$and only$4$worth$800$. This totals to$2 \\cdot 6 \\cdot (200 + 600 + 800 + 1000) + 4 \\cdot 800 = 2 \\cdot 15600 + 3200= 34400$. Thus the player will have a total of$\\$69600$ before the second set of double jeopardies. After them, the player will have $\\$278400$. Thus after the final round, assuming the player bets all of his or her money, he or she can end up with$\\$556800$. I presume that the small difference with my total and the total you suggest is that they let the daily doubles land on the smallest ones, instead of the second-smallest like I thought. - This is well done, thanks, I was forgetting the fact that double jeopardy adds to existing totals. >.> – Link Feb 4 at 19:45", "\\;=\\;26,210,196$ There are 26,210,196 outcomes with no 1 or no 2 or no 3. Therefore, there are: . $60,\\!466,\\!176 - 26,\\!210,\\!196 \\:=\\:\\boxed{34,\\!255,\\!980}$ outcomes . . that contain at least one 1, one 2, and one 3. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Another interpretation . . . Number of outcomes that contain exactly one 1, one 2, and one 3. The 1,2,3 can occur in any of: . $10\\cdot9\\cdot8 \\:=\\:720$ sequences of rolls. The other seven rolls must not be 1, 2, or 3: . $3^7$ ways. Therefore, there are: . $720 \\times 3^7 \\:=\\:\\boxed{1,\\!574,\\!640}$ ways.", "? Free Version Moderate # First, Second, and Third Place GRE-NKJILX In a competition with 8 athletes, how many different ways can first place, second place, and third place be awarded? A $48$ B $56$ C $120$ D $336$ E $512$", "the last $2$ spots, then there is $1$ way to assign spots to $Y$. Finally, if one $X$ was put in on of the middle two spots and one $X$ was put in one of the last two spots, there are $2\\cdot2$ ways to assign spots to $X$ and $2\\cdot1$ ways to assign spots to $Y$ (one of the first two spots and the remaining spot in the last $2$). There are $1+1+2\\cdot2\\cdot2\\cdot1 = 10$ ways to assign opponents to $B$ and $90\\cdot10 = 900$ ways to order the games. $\\boxed{\\textbf{(E)} 900}$", "and we can chose the deviant box in $5$ ways. In all there are $1024-4-60=960$ admissible assignments." ]

[ "Four runners compete in a race: Annie, Becca, Carlos, and Dante. After some confusion at the finish line, it's unclear what the final finishing order was, but the following information is known: 1. Dante finished before Annie. 2. Becca wasn't third. 3. There were two runners between Annie and Carlos. Who won the race? Finally, it says that there were two runners between Annie and Carlos. That means that both Annie and Carlos couldn't have been $2nd$ or $3rd$. Now we can see that Annie must be $4th$. That makes Dante $3rd$ and Becca $2nd$, leaving Carlos as $1st$.", "6:51 The order in which the runners finish matters, so this is a permutation rather than a combination. Until the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \\cdot 7 \\cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \\frac{8!}{(8 - 3)!} = \\frac{8!}{5!} = \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5!}{5!} = 8 \\cdot 7 \\cdot 6 = 336$$ Imagine if John, Jack and Julie were in the \"top-3\", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at So $8\\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.", "# In a horse race, how many different finishes among the first three places are possible for a ten-horse race? $10 \\times 9 \\times 8 = 720$", "## A racing problem.... Bob and Al are racing. Their pdfs are given as: Al 1<= A <=3 Bob 2<=B<=4 What is the probability of Al winning the race? (assuming lesser value means win). I think this is a joint pdf problem and tried using double integral on both of them: int(2,4) int(2,B) A*B...not sure if the equation is right though....any help? thanks", "# How many different place outcomes from 4 cars each running in 4 heats and how many result in ties? Cars A-D run four heats and the 1st place winner of each heat get a point, 2nd place gets 2 pts, etc. I then sum up each car's points across all heats and the fewest points win. However, when using points and not times, ties can occur. (I'm sure ties can happen when using times but going out to the thousandths place, I doubt it.) What's the easiest way to figure out the number of all possible race outcomes over the four heats, summing each car's points and then determining how many race outcomes results in a tie? Thanks, -bill - I am assuming you mean tie for top spot. The numbers are small, break up into cases. First observe that any \"top\" car has at least $10$ points, and no more than $14$. – André Nicolas Jan 22 '13 at 19:52 Doesn't have to be a tie for the top spot. Any tie. – Bill Jan 22 '13 at 20:05 Ouch! Then cases gets unpleasant. – André Nicolas Jan 22 '13 at 20:07 What's the easiest way to figure out the number of all possible race outcomes over the four heats, It depends on how you define", "# Moderate Permutation-Combination Solved QuestionAptitude Discussion Q. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race? ✖ A. $700$ ✔ B. $360$ ✖ C. $120$ ✖ D. $24$ ✖ E. $21$ Solution: Option(B) is correct Two horses $A$ and $B$, in a race of 6 horses... $A$ has to finish before $B$ If $A$ finishes $1^{st}$... $B$ could be in any of other 5 positions in 5 ways and other horses finish in $4!$ Ways, so total ways, $= 5×4!$ If $A$ finishes $2^{nd}$... $B$ could be in any of the last 4 positions in 4 ways. But the other positions could be filled in $4$! ways, so the total ways, $= 4×4!$ If $A$ finishes $3^{rd}$... $B$ could be in any of last 3 positions in 3 ways, but the other positions could be filled in $4!$ ways, so total ways, $= 3×4!$ If $A$ finishes $4^{th}$... $B$ could be in any of last 2 positions in 2 ways, but the other positions could be filled in $4!$ ways, so total ways, $= 2 × 4!$ If $A$ finishes $5^{th}$... $B$ has to be 6th and the top 4 positions", "× # In how many different orders can 6 runners finish a race? Aug 24, 2016 There are $720$ possible results. #### Explanation: The winner can be chosen from all $6$ runners, the second person can be chosen from $5$ people, the third - from $4$ and so on. So the total number of possible results can be calculated as: n=6xx5xx4xx3xx2xx1=6! =720", "$\\underbrace{b+b+\\cdots+b}_{a}=a b$ possibilities. 1.4.2. It is often easier to think about the experiments as being in chronological order, but there is no requirement in the multiplication rule that Experiment A has to be performed before Experiment $\\mathrm{B}$. Example 1.4.3 (Runners). Suppose that 10 people are running a race. Assume that ties are not possible and that all 10 will complete the race, so there will be welldefined first place, second place, and third place winners. How many possibilities are there for the first, second, and third place winners? Solution: There are 10 possibilities for who gets first place, then once that is fixed there are 9 possibilities for who gets second place, and once these are both fixed there are 8 possibilities for third place. So by the multiplication rule, there are $10 \\cdot 9 \\cdot 8=720$ possibilities. We did not have to consider the first place winner first. We could just as well have said that there are 10 possibilities for who got third place, then once that is fixed there are 9 possibilities for second place, and once those are both fixed there are 8 possibilities for first place. Or imagine that there are 3 platforms, which the first, second, and third place runners will stand on after the race. The platforms are gold, silver, and", "monitored. Each contestant had to run in two races so that the average place could be determined. Only one runner finished in the same place in both races. Alan was never last. Charlie always beat Darren. Brian had at least one first place. Alan finished third in at least one of the races. Both Darren and Charlie had a second place. What were the two results? As per the precedent set in the Enjoyable Enigmas thread I am permitted to ask clarifying questions. You describe the races as \"heats\". By that you mean: \"a less important race or competition in which it is decided who will compete in the final event,\" correct? 4 contestants are named, so I assume that is the total number of participants. It is asserted that each ran two races, therefore the total number of \"heats\" was 2, and each of the 2 heats had the same 4 contestants, correct? Greg's n=4 A!=4 (1,2) C<D :.C!=4 (1,2) B=1(1&/2) 1......2 1 C . 1 B 2 D . 2 C 3 A . 3 A 4 B . 4 D or in English- race a) 1 Charlie 2 Darren 3 Alan 4 Brian race b) 1 Brian 2 Charlie 3 Alan 4 Darren ? Fighting spam with spam: didn't Gandhi have something against that? It", "the remaining $n-k+1$ individuals. Hence, the total number of ordering is given by $$n \\times (n-1) \\times \\cdots \\times (n-k+1) \\times \\cdots 2 \\times 1$$which is nothing but $n!$. In you case, $n=3$ gives us $3! = 6$ which matches with our enumeration. - Thank you! That was very well explained and exactly what I was looking for. – daniellopez46 Jun 18 '12 at 15:01 Suppose there is 1 person in the race. There is only one possible combination. A 2nd racer ( #2) is in the race. This person can be placed in 2 places(in front of 1 , or behind 1) There are now 2 possible combinations 12 21 a 3rd person enters the race. Racers 1 and 2 are in place, number 3 has 3 possible places to fit in(in the front, between 1 and 2, or in the back. Because there are 2 possible positions for 1 and 2, there are 3*2 possible combinations for 3 racers 12 becomes 312, 132, or 123 21 becomes 321, 231, or 213 a 4th racer enters, there are 4 places this new racer can fit in There will be 4*6 possible combinations n racers have n! possible positions - Number of possibilites for (1) to end first: two, as we have (1)-(2)-(3) and (1)-(3)-(2). A moment of", "five runners instead of four. Can you use logic to figure out the order? # Today's Challenge Four runners compete in a race — Amelia, Boris, Carlos, and Dara. After some confusion at the finish line, it's unclear what the final finishing order was, but this information is known: 1. Dara finished before Amelia. 2. Boris wasn't third. 3. There were two runners between Amelia and Carlos. $\\\\[-0.7em]$ $\\\\[-0.5em]$ Who won the race?", "$\\{x\\in \\mathbb{Q}\\mid -1 < x < 1\\}=B_1$ ###### Exercise9 The symbol $\\Pi$ is used for the product of numbers in the same way that $\\Sigma$ is used for sums. For example, $\\prod _{i=1}^5 x_i=x_1 x_2 x_3 x_4 x_5\\text{.}$ Evaluate the following: 1. $\\prod _{i=1}^3 i^2$ 2. $\\prod _{i=1}^3 (2i+1)$ 1. $36$ 2. $105$ # ExercisesExercises ###### Exercise1 In horse racing, to bet the “daily double” is to select the winners of the first two races of the day. You win only if both selections are correct. In terms of the number of horses that are entered in the first two races, how many different daily double bets could be made? If there are $m$ horses in race 1 and $n$ horses in race 2 then there are $m \\cdot n$ possible daily doubles. ###### Exercise3 A certain shirt comes in four sizes and six colors. One also has the choice of a dragon, an alligator, or no emblem on the pocket. How many different shirts could you order? $72=4\\cdot 6\\cdot 3$ ###### Exercise5 The Pi Mu Epsilon mathematics honorary society of Outstanding University wishes to have a picture taken of its six officers. There will be two rows of three people. How many different way can the six officers be arranged? $720=6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1$ ######", "Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob? I solved in a different way: All the possible positions: 5! = 120 Restriction: we have to eliminate all the combinations in which Bob finishes ahead of Meg: If Bob finishes 1st, (_ _ _ _ B) Meg could finish 2nd, 3th, 4th or 5th --> 4 possibilities. If Bob finishes 2nd, Meg finishes 3th, 4th or 5th. --> 3 possibilities The same for Bob 3th and 4th. Bob can not finishes 5th. This sum up to 10 combinatios: 4 + 3 + 2 + 1 = 10. In addition, we have the rest of the combinations of the of the positions remaining 3 competitors: 3! Now, we have to combine the positions of Meg and Bob with the positions of the rest of the competitors: 3! * 10 = 60 Finally: 120 - 60 = 60 _________________ Encourage cooperation! If this post was very useful, kudos are welcome \"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome\" William James", "Examveda # Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race? A. 700 B. 360 C. 120 D. 24 E. 21 ### Solution(By Examveda Team) Two horses A and B, in a race of 6 horses . . . A has to finish before B. If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4! If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4! If A finishes 3rd . . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4! If A finishes 4th . . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total", "test in: . $64 \\times 243 \\:=\\:15,\\!552$ ways. Quote: A Superfecta is a four-horse bet, where one must pick the first, second, third and fourth place horses in the exact order in which they finish. In an six-horse race, how many Superfecta bets are possible? $\\begin{array}{c}\\text{There are 6 choices for the 1st place horse.} \\\\ \\text{There are 5 choices for the 2nd place horse.} \\\\ \\text{There are 4 choices for the 3rd place horse.} \\\\ \\text{There are 3 choices for the 4th place horse.} \\end{array}$ Therefore, there are: . $6\\cdot5\\cdot4\\cdot3 \\:=\\:360$ possible Superfecta bets.", "for the second, and so on. The number of ways the prizes can be distributed will be $$5 !=5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=120$$ ways. Now we will consider some slightly different examples. Solution Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are $$25 \\cdot 24 \\cdot 23=13,800$$ ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn't really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6: ABC ACB BAC BCA CAB CBA How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are $$3 \\cdot 2 \\cdot 1=6$$ ways to do this; we didn't really need to list them all, we can just use permutations! So, out of the", "a four-horse bet, where one must pick the first, second, third and fourth place horses in the exact order in which they finish. In an six-horse race, how many Superfecta bets are possible? $\\begin{array}{c}\\text{There are 6 choices for the 1st place horse.} \\\\ \\text{There are 5 choices for the 2nd place horse.} \\\\ \\text{There are 4 choices for the 3rd place horse.} \\\\ \\text{There are 3 choices for the 4th place horse.} \\end{array}$ Therefore, there are: . $6\\cdot5\\cdot4\\cdot3 \\:=\\:360$ possible Superfecta bets.", "411 views Directions for below question: Five horses, Red, White, Grey, Black and Spotted participated in a race, As per the rules of the race, the persons betting on the winning horse get four times the betting amount and those betting on the horse that came in second get thrice of the bet amount, the bet amount is returned to those betting on the horse that came in third and rest lose the bet amount. Raju bets Rs. $3000,$ Rs. $2000$ and Rs. $1000$ on Red, White and Black horses respectively and ends up with no profit and no loss. Suppose in addition, it is known that Grey came in fourth. Then which of the following cannot be true? 1. Spotted came in first 2. Red finished last 3. White came in second 4. Black came in second 5. There was one horse between Black and White", "C 1st: B 2nd: A 3rd: C 1st: A 2nd: C 3rd: B 1st: C 2nd: A 3rd: B 1st: B 2nd: C 3rd: A 1st: C 2nd: B 3rd: A 1st: AB (tie) 2nd: C 1st: AC (tie) 2nd: B 1st: BC (tie) 2nd: A 1st: A 2nd: BC (tie) 1st: B 2nd: AC (tie) 1st: C 2nd: BC (tie) 1st: ABC (tie) That comes out to 13 different ways these horses unicorns can finish the race. That’s the answer for 3 unicorns. What’s the answer for 8 unicorns? (FYI: If you want to know if you’re on the right track… I have 541 for 5 unicorns…)", "not the second nor the third place: it has to race one horse from each previous race, so if the first three places were all in the same heat or 1,2 in the same and 3 in a different, it couldn't tell the difference. So $n<25$, and in the last race you have to race the new horses against the first three places A1, A2, A3 (possibly without knowing the order) of the $n$ horses that already have raced, otherwise you cannot determine the overall three fastest horses with the last race. Hence $n=23$ or $n=24$. If you assume $n=23$, then it is impossible to determine the three fastest horses out of 23 with 5 races: You cannot have less than three different winners, since only two repetitions are allowed. If you have three different winners that have not been compared, then any of the 5 second places could be A2 or A3. If you have four or more different winners that have not been compared, all of them could be A1,A2, or A3. If $n=24$, then you have to know a set of four horses out of the first 24 which contains A1,A2 and A3. But this is impossible in 5 races, since you have at least 4 different winners that have not been compared, and so" ]

[ "# How do you multiply 4 * 4sqrt4? Apr 29, 2015 $\\sqrt{4} = 2$- Knowing that, the question turns into a simple multiplication problem: $4 \\cdot 4 \\cdot \\sqrt{4} = 4 \\cdot 4 \\cdot 2 = 32$", "# How do you factor completely 8x^3-8? $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$", "12. $x^2y^4 \\div 4xy$ 13. $(x^3y^4\\cdot 2x^2y^3)^2$ 14. $[(u^2v^2)(2u^4)]^3$ 15. $[(2x)^3 \\div x^3]^2$ 16. $(2a^2b^2a^7) \\div (ba^4)^2$ 17. $[(2y^{17}) \\div (2x^2y^4)^4]^3$ 18. $[(xy^2)(y^4)^2] \\div 2y^4$ 19. $(2xy^5\\cdot 2x^2y^3) \\div (2xy^4\\cdot y^3)$ 20. $(2y^3x^2) \\div [(x^2y^4)(x^2)]$ 21. $[(q^3r^2)(2p^2q^2r^3)^2] \\div 2p^3$ 22. $(2x^4y^5)(2z^{10}x^2y^7) \\div (xy^2z^2)^4$", "\\left (\\dfrac{8y}{x} \\right)^2$ 7. ) $6 a^2 \\cdot (a^2 + 1)^0$ Solutions and detailed explanations to the above questions.", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "# How do you multiply (4x+4)(4) ? Jan 14, 2017 $16 x + 16$ #### Explanation: Use the distributive property, which states that: $a \\left(b + c\\right) = a b + a c$ If it is written as $\\left(b + c\\right) a$, it's evaluated the same way. $\\left(4 x + 4\\right) \\left(4\\right) = 4 \\left(4 x + 4\\right) = 4 \\cdot 4 x + 4 \\cdot 4 = 16 x + 16$.", "# How do you simplify 4^5 * 4^3? Apr 28, 2016 Think about the definition of exponents: repeated multiplication. Therefore, ${4}^{5}$ is basically just $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ Similarly, ${4}^{3}$ is just $4 \\cdot 4 \\cdot 4$ Thus, if we multiply these together, we get $\\left(4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4\\right) \\cdot \\left(4 \\cdot 4 \\cdot 4\\right)$ This is just $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ or ${4}^{8}$ There is a formula for this, which is just adding the exponents together, so ${4}^{5 + 3} = {4}^{8}$", "(\\text{-} x + 8)$ $5 + 2(\\text{-}7) + 2 \\cdot x = \\text{-} 6 - (\\text{-} x + 8)$ $5 -14 + 2x = \\text{-} 6 - (\\text{-} x + 8)$ $5 -14 + 2x = \\text{-} 6 +x - 8$ $\\text{-} 9 + 2x = x - 14$ $2x = x - 5$ $x= \\text{-} 5$", "How do you factor completely 8x^3-8? 1 Answer It is $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$", "# How do you find the quotient of (14y^5+21y^4-6y^3-9y^2+32y+48)div(2y+3) using long division? Feb 13, 2018 Quotient is $7 {y}^{4} - 3 y + 2 + 16$ #### Explanation: $14 {y}^{5} + 21 {y}^{4} - 6 {y}^{3} - 9 {y}^{2} + 32 y + 48$ =$7 {y}^{4} \\cdot \\left(2 y + 3\\right) - 3 y + 2 \\cdot \\left(2 y + 3\\right) + 16 \\cdot \\left(2 y + 3\\right)$ =$\\left(2 y + 3\\right) \\cdot \\left(7 {y}^{4} - 3 y + 2 + 16\\right)$ Hence quotient is $7 {y}^{4} - 3 y + 2 + 16$ and remainder is $0$.", "# How do you simplify 24 3/4 div 8 1/4? $9$ $24 \\cdot \\frac{3}{4} / 8 \\cdot \\frac{1}{4}$ =$\\frac{18}{2}$ =$9$", "If f(x)=4x^3-4x^2+10 then how do you find f(-2)? Feb 6, 2015 Just put $- 2$ in place of the $x$'s $f \\left(x\\right) = 4 \\cdot {x}^{3} - 4 \\cdot {x}^{2} + 10$ So: $f \\left(- 2\\right) = 4 \\cdot {\\left(- 2\\right)}^{3} - 4 \\cdot {\\left(- 2\\right)}^{2} + 10$ $= 4 \\cdot \\left(- 8\\right) - 4 \\cdot \\left(4\\right) + 10$ $= - 32 - 16 + 10$ $= - 38$ $f \\left(- 2\\right) = - 38$", "# How do you simplify 7(4x-5)-8x? $= 7 \\cdot 4 x - 7 \\cdot 5 - 8 x = 28 x - 35 - 8 x$ Now add the $x$'s: $= 20 x - 35$", "out of four lengths, you can calculate the fourth (AD). Once AD is known, you get CD = AD-16.", "$$1000001\\equiv -2\\mod 1000003$$. Now, you just have to divide $$-1$$ by $$(-1)\\cdot (-2)=2$$.", "Write in exponential notation: 1. $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ 2. $3x \\cdot 3x \\cdot 3x$ 3. $(-2a)(-2a)(-2a)(-2a)$ 4. $6 \\cdot 6 \\cdot 6 \\cdot x \\cdot x \\cdot y \\cdot y \\cdot y \\cdot y$ 5. $2 \\cdot x \\cdot y \\cdot 2 \\cdot 2 \\cdot y \\cdot x$ Find each number. 1. $5^4$ 2. $(-2)^6$ 3. $(0.1)^5$ 4. $(-0.6)^3$ 5. $(1.2)^2+5^3$ 6. $3^2 \\cdot (0.2)^3$ Multiply and simplify: 1. $6^3 \\cdot 6^6$ 2. $2^2 \\cdot 2^4 \\cdot 2^6$ 3. $3^2 \\cdot 4^3$ 4. $x^2 \\cdot x^4$ 5. $(-2y^4)(-3y)$ 6. $(4a^2)(-3a)(-5a^4)$ Simplify: 1. $(a^3)^4$ 2. $(xy)^2$ 3. $(3a^2b^3)^4$ 4. $(-2xy^4z^2)^5$ 5. $(-8x)^3(5x)^2$ 6. $(4a^2)(-2a^3)^4$ 7. $(12xy)(12xy)^2$ 8. $(2xy^2)(-x^2y)^2(3x^2y^2)$ Feb 23, 2012 Mar 05, 2014", "# How do you factor 8x^3-24x^2? Notice that both of the terms are divisible by $8 {x}^{2}$, so separate that out to find: $8 {x}^{3} - 24 {x}^{2} = 8 {x}^{2} \\left(x - 3\\right)$ Both $8 {x}^{3}$ and $24 {x}^{2}$ are divisible by $8 {x}^{2}$, so we find: $8 {x}^{3} - 24 {x}^{2} = \\left(8 {x}^{2} \\cdot x\\right) - \\left(8 {x}^{2} \\cdot 3\\right) = 8 {x}^{2} \\left(x - 3\\right)$", "# How do you add 4sqrt3 - sqrt12? You may notice that $12 = 3 \\cdot 4 = 3 \\cdot {2}^{2}$ $= 4 \\sqrt{3} - 2 \\sqrt{3} = 2 \\sqrt{2}$", "How do you find the product (x-6)^2? ${\\left(x - 6\\right)}^{2} = {x}^{2} - 12 x + 36$ Explanation: To find ${\\left(x - 6\\right)}^{2}$ ${\\left(x - 6\\right)}^{2} = \\left(x - 6\\right) \\left(x - 6\\right)$ $\\left(x - 6\\right) \\left(x - 6\\right) = x \\cdot x - 6 \\cdot x - x \\cdot 6 + 6 \\times 6$ $= {x}^{2} - 6 x - 6 x + 36$ $= {x}^{2} - 12 x + 36$ Thus, ${\\left(x - 6\\right)}^{2} = {x}^{2} - 12 x + 36$", "## Thinking Mathematically (6th Edition) $(2^2)^2 = 2^4 = 16$ Use the rule $(a^m)^n = a^{mn}$ to find: $(2^2)^2 \\\\= 2^{2\\cdot 2} \\\\= 2^4 \\\\=16$" ]

[ "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "Write in exponential notation: 1. $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ 2. $3x \\cdot 3x \\cdot 3x$ 3. $(-2a)(-2a)(-2a)(-2a)$ 4. $6 \\cdot 6 \\cdot 6 \\cdot x \\cdot x \\cdot y \\cdot y \\cdot y \\cdot y$ 5. $2 \\cdot x \\cdot y \\cdot 2 \\cdot 2 \\cdot y \\cdot x$ Find each number. 1. $5^4$ 2. $(-2)^6$ 3. $(0.1)^5$ 4. $(-0.6)^3$ 5. $(1.2)^2+5^3$ 6. $3^2 \\cdot (0.2)^3$ Multiply and simplify: 1. $6^3 \\cdot 6^6$ 2. $2^2 \\cdot 2^4 \\cdot 2^6$ 3. $3^2 \\cdot 4^3$ 4. $x^2 \\cdot x^4$ 5. $(-2y^4)(-3y)$ 6. $(4a^2)(-3a)(-5a^4)$ Simplify: 1. $(a^3)^4$ 2. $(xy)^2$ 3. $(3a^2b^3)^4$ 4. $(-2xy^4z^2)^5$ 5. $(-8x)^3(5x)^2$ 6. $(4a^2)(-2a^3)^4$ 7. $(12xy)(12xy)^2$ 8. $(2xy^2)(-x^2y)^2(3x^2y^2)$ Feb 23, 2012 Mar 05, 2014", "# How do you simplify 4^5 * 4^3? Apr 28, 2016 Think about the definition of exponents: repeated multiplication. Therefore, ${4}^{5}$ is basically just $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ Similarly, ${4}^{3}$ is just $4 \\cdot 4 \\cdot 4$ Thus, if we multiply these together, we get $\\left(4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4\\right) \\cdot \\left(4 \\cdot 4 \\cdot 4\\right)$ This is just $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ or ${4}^{8}$ There is a formula for this, which is just adding the exponents together, so ${4}^{5 + 3} = {4}^{8}$", "$a^{b^{c^d}}$ are divisible by $3^4,$ hence by $3^2=9$ If $(3,a)=1,\\phi(9)=6,$ it is sufficient to prove $6\\mid(b^{c^d}-b^c)$ So, in all the cases, $a^{b^{c^d}}-a^{b^c}$ is divisible by $2^4, 7, 9, 13, 5;$ hence by $lcm(2^4, 7, 9, 13, 5)=2^4\\cdot7\\cdot 9\\cdot 13\\cdot 5=2^{2^{2^2}}-2^{2^2}$ -", "# How do you multiply 4 * 4sqrt4? Apr 29, 2015 $\\sqrt{4} = 2$- Knowing that, the question turns into a simple multiplication problem: $4 \\cdot 4 \\cdot \\sqrt{4} = 4 \\cdot 4 \\cdot 2 = 32$", "$1$ $-1$ $(5, -1)-(1, 3)\\=(4, -4)$ $2$ $(4, -4)$ $11$ $(3, 4)$ $0$ $4 \\cdot 3 - 4 \\cdot 4\\= -4$ $0$ $0$ $(4, -4)$ $2$ $(4, -4)$ $12$ $(2, 4)$ $0$ $4 \\cdot 2 - 4 \\cdot 4\\= -8$ $0$ $0$ $(4, -4)$ $2$ $(4, -4)$ $13$ $(4, 0)$ $1$ $4 \\cdot 4 - 4 \\cdot 0\\= 16$ $1$ $0$ $(4, -4)$ $2$ $(4, -4)$ $14$ $(6, 2)$ $1$ $4 \\cdot 6 - 4 \\cdot 2\\= 16$ $1$ $0$ $(4, -4)$ $2$ $(4, -4)$ $15$ $(2, -3)$ $1$ $4 \\cdot 2 + 4 \\cdot 3\\= 20$ $1$ $0$ $(4, -4)$ $2$ $(4, -4)$ $16$ $(6, -2)$ $1$ $4 \\cdot 6 + 4 \\cdot 2\\= 32$ $1$ $0$ $(4, -4)$ $3$ $(4, -4)$ $17$ $(1, 2)$ $0$ $4 \\cdot 1 - 4 \\cdot 2\\= -4$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $18$ $(1, 3)$ $0$ $4 \\cdot 1 - 4 \\cdot 3\\= -8$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $19$ $(3, 4)$ $0$ $4 \\cdot 3 - 4 \\cdot 4\\= -4$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $20$ $(2, 4)$ $0$ $4 \\cdot 2 - 4 \\cdot 4\\= -8$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $21$ $(4, 0)$ $1$ $4 \\cdot 4 - 4 \\cdot 0\\= 16$ $1$ $0$ $(4, -4)$ $3$ $(4, -4)$", "# How do you evaluate 4(3+5)-5*4? Mar 9, 2018 Follow the order of operations to get $12$ #### Explanation: $4 \\left(3 + 5\\right) - 5 \\cdot 4$ Let's do the parentheses first $= 4 \\left(8\\right) - 5 \\cdot 4$ Now we have two multiplications, which we do separately. $= 32 - 20$ Now subtract $= 12$", "do not have to provide the answer. 1. $-(3^4)$ 2. $-8^2$ 3. $10 \\times (-4)^3$ 4. What is the difference between $-5^2$ and $(-5)^2$ ? Write in exponential notation. 1. $2 \\cdot 2$ 2. $(-3)(-3)(-3)$ 3. $y \\cdot y \\cdot y \\cdot y \\cdot y$ 4. $(3a)(3a)(3a)(3a)$ 5. $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ 6. $3x \\cdot 3x \\cdot 3x$ 7. $(-2a)(-2a)(-2a)(-2a)$ 8. $6 \\cdot 6 \\cdot 6 \\cdot x \\cdot x \\cdot y \\cdot y \\cdot y \\cdot y$ Find each number. 1. $1^{10}$ 2. $0^3$ 3. $7^3$ 4. $-6^2$ 5. $5^4$ 6. $3^4 \\cdot 3^7$ 7. $2^6 \\cdot 2$ 8. $(4^2)^3$ 9. $(-2)^6$ 10. $(0.1)^5$ 11. $(-0.6)^3$ Multiply and simplify. 1. $6^3 \\cdot 6^6$ 2. $2^2 \\cdot 2^4 \\cdot 2^6$ 3. $3^2 \\cdot 4^3$ 4. $x^2 \\cdot x^4$ 5. $x^2 \\cdot x^7$ 6. $(y^3)^5$ 7. $(-2 y^4)(-3y)$ 8. $(4a^2)(-3a)(-5a^4)$ Simplify. 1. $(a^3)^4$ 2. $(xy)^2$ 3. $(3a^2b^3)^4$ 4. $(-2xy^4z^2)^5$ 5. $(3x^2 y^3) \\cdot (4xy^2)$ 6. $(4xyz) \\cdot (x^2y^3) \\cdot (2yz^4)$ 7. $(2a^3b^3)^2$ 8. $(-8x)^3(5x)^2$ 9. $(4a^2)(-2a^3)^4$ 10. $(12xy)(12xy)^2$ 11. $(2xy^2)(-x^2y)^2(3x^2y^2)$ Mixed Review 1. How many ways can you choose a 4-person committee from seven people? 2. Three canoes cross a finish line to earn medals. Is this an example of a permutation or a combination? How many ways are possible? 3. Find the slope", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "# How do you factor completely 8x^3-8? $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$", "1\\} =$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} =$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} =$ #### Solution 1. $\\{0\\} \\cap \\varnothing = \\boxed{\\varnothing}$ 2. $\\{0, 1\\} \\cap \\{0, 1\\} = \\boxed{\\{0, 1\\}}$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} = \\boxed{\\{1\\}}$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} = \\boxed{\\{4, 6\\}}$", "to prove that $4^n \\equiv 4 \\bmod 12$ for all $n$. This is a classic example of proof by induction. Base Case If $n=1$ then is $4^1 \\equiv 4 \\bmod 12$? Well, clearly $4^1 \\equiv 4 \\bmod 12$. Induction Step Assume that for $n=k$ we have $4^n \\equiv 4 \\bmod 12$, does this imply that $4^{n+1} \\equiv 4 \\bmod 12$? Well, if $4^k \\equiv 4 \\bmod 12$ then $4^{k+1} = 4 \\times 4^k \\equiv 4 \\times 4 \\bmod 12 \\equiv 4 \\bmod 12$. Conclusion It follows that $4^n \\equiv 4 \\bmod 12$ for all $n \\in \\mathbb{Z}^+$. - $$4^{47}\\equiv 4\\pmod{12}$$ $$4^{47}-4=4 \\cdot \\left(4^{46}-1\\right)=4 \\cdot \\left(4^{23}-1\\right)\\cdot \\left(4^{23}+1\\right)=$$ $$=4 \\cdot \\left(4-1\\right)(4^{22}+4^{21}\\cdot1+\\ldots+1)\\cdot \\left(4^{23}+1\\right)=4 \\cdot3 \\cdot \\left(4^{22}+4^{21}\\cdot1+\\ldots+1\\right)\\cdot \\left(4^{23}+1\\right)=$$ $$=12 \\cdot \\left(4^{22}+4^{21}\\cdot1+\\ldots+1\\right)\\cdot \\left(4^{23}+1\\right)$$ which is divisible with $12$. You seem to have lost the factor $4^{23}+1$. Also, $4^{23}-1=(4-1)(4^{22}+4^{21}+...+4+1)$ – user60725 Feb 14 '13 at 20:09", "$$1129$$ $S_4$ (as 4T5) Trivial 4.0.1161.1 x4 - x3 + 6x2 - x + 7 $$3^{3}\\cdot 43$$ $D_{4}$ (as 4T3) Trivial 4.0.1168.1 x4 - x2 - 4x + 5 $$2^{4}\\cdot 73$$ $D_{4}$ (as 4T3) Trivial 4.0.1197.1 x4 - 2x3 - 4x2 + 5x + 7 $$3^{2}\\cdot 7\\cdot 19$$ $D_{4}$ (as 4T3) Trivial 4.0.1197.2 x4 - x3 - 5x2 + 3x + 9 $$3^{2}\\cdot 7\\cdot 19$$ $D_{4}$ (as 4T3) Trivial 4.0.1225.1 x4 - x3 + 5x2 + 2x + 4 $$5^{2}\\cdot 7^{2}$$ $C_2^2$ (as 4T2) Trivial Next", "which leads to your question - If you make some arithmetic modulo $\\,4\\,$: $$m^4+4^n=m^4=0,1\\pmod 4\\Longrightarrow \\,\\,m\\,\\,\\text{must be odd (why?)}$$ $$n=2k+1\\Longrightarrow 4^n=4\\cdot (4^k)^2\\,$$", "# How do you multiply (4x+4)(4) ? Jan 14, 2017 $16 x + 16$ #### Explanation: Use the distributive property, which states that: $a \\left(b + c\\right) = a b + a c$ If it is written as $\\left(b + c\\right) a$, it's evaluated the same way. $\\left(4 x + 4\\right) \\left(4\\right) = 4 \\left(4 x + 4\\right) = 4 \\cdot 4 x + 4 \\cdot 4 = 16 x + 16$.", "# How do you add 4sqrt3 - sqrt12? You may notice that $12 = 3 \\cdot 4 = 3 \\cdot {2}^{2}$ $= 4 \\sqrt{3} - 2 \\sqrt{3} = 2 \\sqrt{2}$", "# How do you find the indicated term of the arithmetic sequence a_1=7, d=3, n=14? Mar 17, 2018 ${a}_{14} = 46$ #### Explanation: ${a}_{1} = 7 , d = 3 , n = 14$ ${a}_{2} = {a}_{1} + \\left(2 - 1\\right) \\cdot d = 7 + 3 = 10$ ${a}_{3} = {a}_{1} + \\left(3 - 1\\right) \\cdot d = 7 + 6 = 13$ ${a}_{n} = {a}_{1} + \\left(n - 1\\right) \\cdot d$ $\\therefore {a}_{14} = {a}_{1} + \\left(14 - 1\\right) \\cdot d = 7 + 13 \\cdot 3 = 7 + 39 = 46$", "be $0 \\pmod 4$ meaning that $a$ and $c$ are $4$ or $8$. For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \\cdot 2 \\cdot 2 = \\boxed{040}$ numbers.", "the second parenthesis... $(-ab)(a^4b^2)$ now multiply them together... the b becomes b^3, but what of the a? -a • a^4... would it be a^3? -a^4? a to the power of -4? i'm confused. you can think of $-a \\cdot a^4$ as $-1 \\cdot a \\cdot a^4$ so now, just multiply the a's as normal, and then apply the minus 1 8. Got it. Thanks.", "How do you factor completely 8x^3-8? 1 Answer It is $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$" ]

[ "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "Write in exponential notation: 1. $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ 2. $3x \\cdot 3x \\cdot 3x$ 3. $(-2a)(-2a)(-2a)(-2a)$ 4. $6 \\cdot 6 \\cdot 6 \\cdot x \\cdot x \\cdot y \\cdot y \\cdot y \\cdot y$ 5. $2 \\cdot x \\cdot y \\cdot 2 \\cdot 2 \\cdot y \\cdot x$ Find each number. 1. $5^4$ 2. $(-2)^6$ 3. $(0.1)^5$ 4. $(-0.6)^3$ 5. $(1.2)^2+5^3$ 6. $3^2 \\cdot (0.2)^3$ Multiply and simplify: 1. $6^3 \\cdot 6^6$ 2. $2^2 \\cdot 2^4 \\cdot 2^6$ 3. $3^2 \\cdot 4^3$ 4. $x^2 \\cdot x^4$ 5. $(-2y^4)(-3y)$ 6. $(4a^2)(-3a)(-5a^4)$ Simplify: 1. $(a^3)^4$ 2. $(xy)^2$ 3. $(3a^2b^3)^4$ 4. $(-2xy^4z^2)^5$ 5. $(-8x)^3(5x)^2$ 6. $(4a^2)(-2a^3)^4$ 7. $(12xy)(12xy)^2$ 8. $(2xy^2)(-x^2y)^2(3x^2y^2)$ Feb 23, 2012 Mar 05, 2014", "$a^{b^{c^d}}$ are divisible by $3^4,$ hence by $3^2=9$ If $(3,a)=1,\\phi(9)=6,$ it is sufficient to prove $6\\mid(b^{c^d}-b^c)$ So, in all the cases, $a^{b^{c^d}}-a^{b^c}$ is divisible by $2^4, 7, 9, 13, 5;$ hence by $lcm(2^4, 7, 9, 13, 5)=2^4\\cdot7\\cdot 9\\cdot 13\\cdot 5=2^{2^{2^2}}-2^{2^2}$ -", "$1$ $-1$ $(5, -1)-(1, 3)\\=(4, -4)$ $2$ $(4, -4)$ $11$ $(3, 4)$ $0$ $4 \\cdot 3 - 4 \\cdot 4\\= -4$ $0$ $0$ $(4, -4)$ $2$ $(4, -4)$ $12$ $(2, 4)$ $0$ $4 \\cdot 2 - 4 \\cdot 4\\= -8$ $0$ $0$ $(4, -4)$ $2$ $(4, -4)$ $13$ $(4, 0)$ $1$ $4 \\cdot 4 - 4 \\cdot 0\\= 16$ $1$ $0$ $(4, -4)$ $2$ $(4, -4)$ $14$ $(6, 2)$ $1$ $4 \\cdot 6 - 4 \\cdot 2\\= 16$ $1$ $0$ $(4, -4)$ $2$ $(4, -4)$ $15$ $(2, -3)$ $1$ $4 \\cdot 2 + 4 \\cdot 3\\= 20$ $1$ $0$ $(4, -4)$ $2$ $(4, -4)$ $16$ $(6, -2)$ $1$ $4 \\cdot 6 + 4 \\cdot 2\\= 32$ $1$ $0$ $(4, -4)$ $3$ $(4, -4)$ $17$ $(1, 2)$ $0$ $4 \\cdot 1 - 4 \\cdot 2\\= -4$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $18$ $(1, 3)$ $0$ $4 \\cdot 1 - 4 \\cdot 3\\= -8$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $19$ $(3, 4)$ $0$ $4 \\cdot 3 - 4 \\cdot 4\\= -4$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $20$ $(2, 4)$ $0$ $4 \\cdot 2 - 4 \\cdot 4\\= -8$ $0$ $0$ $(4, -4)$ $3$ $(4, -4)$ $21$ $(4, 0)$ $1$ $4 \\cdot 4 - 4 \\cdot 0\\= 16$ $1$ $0$ $(4, -4)$ $3$ $(4, -4)$", "# How do you factor completely 8x^3-8? $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$", "do not have to provide the answer. 1. $-(3^4)$ 2. $-8^2$ 3. $10 \\times (-4)^3$ 4. What is the difference between $-5^2$ and $(-5)^2$ ? Write in exponential notation. 1. $2 \\cdot 2$ 2. $(-3)(-3)(-3)$ 3. $y \\cdot y \\cdot y \\cdot y \\cdot y$ 4. $(3a)(3a)(3a)(3a)$ 5. $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ 6. $3x \\cdot 3x \\cdot 3x$ 7. $(-2a)(-2a)(-2a)(-2a)$ 8. $6 \\cdot 6 \\cdot 6 \\cdot x \\cdot x \\cdot y \\cdot y \\cdot y \\cdot y$ Find each number. 1. $1^{10}$ 2. $0^3$ 3. $7^3$ 4. $-6^2$ 5. $5^4$ 6. $3^4 \\cdot 3^7$ 7. $2^6 \\cdot 2$ 8. $(4^2)^3$ 9. $(-2)^6$ 10. $(0.1)^5$ 11. $(-0.6)^3$ Multiply and simplify. 1. $6^3 \\cdot 6^6$ 2. $2^2 \\cdot 2^4 \\cdot 2^6$ 3. $3^2 \\cdot 4^3$ 4. $x^2 \\cdot x^4$ 5. $x^2 \\cdot x^7$ 6. $(y^3)^5$ 7. $(-2 y^4)(-3y)$ 8. $(4a^2)(-3a)(-5a^4)$ Simplify. 1. $(a^3)^4$ 2. $(xy)^2$ 3. $(3a^2b^3)^4$ 4. $(-2xy^4z^2)^5$ 5. $(3x^2 y^3) \\cdot (4xy^2)$ 6. $(4xyz) \\cdot (x^2y^3) \\cdot (2yz^4)$ 7. $(2a^3b^3)^2$ 8. $(-8x)^3(5x)^2$ 9. $(4a^2)(-2a^3)^4$ 10. $(12xy)(12xy)^2$ 11. $(2xy^2)(-x^2y)^2(3x^2y^2)$ Mixed Review 1. How many ways can you choose a 4-person committee from seven people? 2. Three canoes cross a finish line to earn medals. Is this an example of a permutation or a combination? How many ways are possible? 3. Find the slope", "12. $x^2y^4 \\div 4xy$ 13. $(x^3y^4\\cdot 2x^2y^3)^2$ 14. $[(u^2v^2)(2u^4)]^3$ 15. $[(2x)^3 \\div x^3]^2$ 16. $(2a^2b^2a^7) \\div (ba^4)^2$ 17. $[(2y^{17}) \\div (2x^2y^4)^4]^3$ 18. $[(xy^2)(y^4)^2] \\div 2y^4$ 19. $(2xy^5\\cdot 2x^2y^3) \\div (2xy^4\\cdot y^3)$ 20. $(2y^3x^2) \\div [(x^2y^4)(x^2)]$ 21. $[(q^3r^2)(2p^2q^2r^3)^2] \\div 2p^3$ 22. $(2x^4y^5)(2z^{10}x^2y^7) \\div (xy^2z^2)^4$", "# How do you simplify 4^5 * 4^3? Apr 28, 2016 Think about the definition of exponents: repeated multiplication. Therefore, ${4}^{5}$ is basically just $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ Similarly, ${4}^{3}$ is just $4 \\cdot 4 \\cdot 4$ Thus, if we multiply these together, we get $\\left(4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4\\right) \\cdot \\left(4 \\cdot 4 \\cdot 4\\right)$ This is just $4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4$ or ${4}^{8}$ There is a formula for this, which is just adding the exponents together, so ${4}^{5 + 3} = {4}^{8}$", "x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x \\\\&=-32x^{20}$ ### Practice Directions: Simplify. 1. $(6x^5)^2$ 2. $(-13d^5)^2$ 3. $(-3p^3 q^4)^3$ 4. $(10xy^2)^4$ 5. $(-4t^3)^5$ 6. $(18 r^2 s^3)^2$ 7. $(2r^{11}s^3 t^2)^3$ 8. $(7x^2)^2$ 9. $(2y^2)^3$ 10. $(5x^2)^3$ 11. $(12y^3)^2$ 12. $(5x^5)^5$ 13. $(2x^2y^2z)^3$ 14. $(3x^4y^3z^2)^3$ 15. $(-5x^4y^3z^3)^3$", "How do you factor completely 8x^3-8? 1 Answer It is $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$", "# How do you multiply (4x+4)(4) ? Jan 14, 2017 $16 x + 16$ #### Explanation: Use the distributive property, which states that: $a \\left(b + c\\right) = a b + a c$ If it is written as $\\left(b + c\\right) a$, it's evaluated the same way. $\\left(4 x + 4\\right) \\left(4\\right) = 4 \\left(4 x + 4\\right) = 4 \\cdot 4 x + 4 \\cdot 4 = 16 x + 16$.", "× n times) $={\\text{a}}^{\\text{mn}}$ For example: $\\begin{array}{}\\left({2}^{2}{\\right)}^{3}=\\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\\\ =\\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\\\ ={2}^{6}\\\\ ={2}^{2×3}\\end{array}$ ## Examples: exponential law 6 1. $\\left({x}^{3}{\\right)}^{4}={x}^{\\text{12}}$ 2. $\\left[\\left({a}^{4}{\\right)}^{3}{\\right]}^{2}={a}^{\\text{24}}$ 3. $\\left({3}^{n+3}{\\right)}^{2}={3}^{2n+6}$ ## Module review exercises Write the following examples using exponential notation. $4\\cdot 4$ 4 2 $\\text{12}\\cdot \\text{12}$ 12 2 $9\\cdot 9\\cdot 9\\cdot 9$ 9 4 $\\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}$ 10 6 $\\text{826}\\cdot \\text{826}\\cdot \\text{826}$ 826 3 $\\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}$ 3021 4 $6\\cdot 6\\cdot 6\\cdot \\cdots \\cdot 6$ (85 factors of 6). 6 85 $2\\cdot 2\\cdot 2\\cdot \\cdots \\cdot 2$ (112 factors of 2). 2 112 For the next examples, expand the terms. (Do not find the actual values). 5 3 5 · 5 · 5 15 2 15 · 15 117 5 117 · 117 · 117 · 117 · 117 7 4 7 · 7 · 7 · 7 Determine the value of each of the powers. 3 2 9 1 2 1 15 2 225 3 4 81 1 4 1 7 3 343 Simplify as far as possible. (2x) 3 2 3 · x 3 = 8x 3 302 0 1 (-2x) 3 (-2) 3 · x 3 = -8x 3 2 3 · 2 4 2 3+4 = 2 7 $\\frac{{x}^{8}}{{x}^{3}}$ ${x}^{8-3}={x}^{5}$ $\\frac{\\text{25}{x}^{2}}{{5x}^{8}}$ $\\frac{\\text{25}}{5}{x}^{2-8}={5x}^{-6}=\\frac{5}{{x}^{6}}$ (3 -1 +2 -1 ) -1 $\\frac{1}{{3}^{-1}+{2}^{-1}}=\\frac{1}{\\frac{1}{3}+\\frac{1}{2}}=1\\cdot \\left(\\frac{3}{1}+\\frac{2}{1}\\right)=3+2=5$ Is", "× n times) $={\\text{a}}^{\\text{mn}}$ For example: $\\begin{array}{}\\left({2}^{2}{\\right)}^{3}=\\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\\\ =\\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\\\ ={2}^{6}\\\\ ={2}^{2×3}\\end{array}$ ## Examples: exponential law 6 1. $\\left({x}^{3}{\\right)}^{4}={x}^{\\text{12}}$ 2. $\\left[\\left({a}^{4}{\\right)}^{3}{\\right]}^{2}={a}^{\\text{24}}$ 3. $\\left({3}^{n+3}{\\right)}^{2}={3}^{2n+6}$ ## Module review exercises Write the following examples using exponential notation. $4\\cdot 4$ 4 2 $\\text{12}\\cdot \\text{12}$ 12 2 $9\\cdot 9\\cdot 9\\cdot 9$ 9 4 $\\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}$ 10 6 $\\text{826}\\cdot \\text{826}\\cdot \\text{826}$ 826 3 $\\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}$ 3021 4 $6\\cdot 6\\cdot 6\\cdot \\cdots \\cdot 6$ (85 factors of 6). 6 85 $2\\cdot 2\\cdot 2\\cdot \\cdots \\cdot 2$ (112 factors of 2). 2 112 For the next examples, expand the terms. (Do not find the actual values). 5 3 5 · 5 · 5 15 2 15 · 15 117 5 117 · 117 · 117 · 117 · 117 7 4 7 · 7 · 7 · 7 Determine the value of each of the powers. 3 2 9 1 2 1 15 2 225 3 4 81 1 4 1 7 3 343 Simplify as far as possible. (2x) 3 2 3 · x 3 = 8x 3 302 0 1 (-2x) 3 (-2) 3 · x 3 = -8x 3 2 3 · 2 4 2 3+4 = 2 7 $\\frac{{x}^{8}}{{x}^{3}}$ ${x}^{8-3}={x}^{5}$ $\\frac{\\text{25}{x}^{2}}{{5x}^{8}}$ $\\frac{\\text{25}}{5}{x}^{2-8}={5x}^{-6}=\\frac{5}{{x}^{6}}$ (3 -1 +2 -1 ) -1 $\\frac{1}{{3}^{-1}+{2}^{-1}}=\\frac{1}{\\frac{1}{3}+\\frac{1}{2}}=1\\cdot \\left(\\frac{3}{1}+\\frac{2}{1}\\right)=3+2=5$ How", "× n times) $={\\text{a}}^{\\text{mn}}$ For example: $\\begin{array}{}\\left({2}^{2}{\\right)}^{3}=\\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\\\ =\\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\\\ ={2}^{6}\\\\ ={2}^{2×3}\\end{array}$ ## Examples: exponential law 6 1. $\\left({x}^{3}{\\right)}^{4}={x}^{\\text{12}}$ 2. $\\left[\\left({a}^{4}{\\right)}^{3}{\\right]}^{2}={a}^{\\text{24}}$ 3. $\\left({3}^{n+3}{\\right)}^{2}={3}^{2n+6}$ ## Module review exercises Write the following examples using exponential notation. $4\\cdot 4$ 4 2 $\\text{12}\\cdot \\text{12}$ 12 2 $9\\cdot 9\\cdot 9\\cdot 9$ 9 4 $\\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}$ 10 6 $\\text{826}\\cdot \\text{826}\\cdot \\text{826}$ 826 3 $\\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}$ 3021 4 $6\\cdot 6\\cdot 6\\cdot \\cdots \\cdot 6$ (85 factors of 6). 6 85 $2\\cdot 2\\cdot 2\\cdot \\cdots \\cdot 2$ (112 factors of 2). 2 112 For the next examples, expand the terms. (Do not find the actual values). 5 3 5 · 5 · 5 15 2 15 · 15 117 5 117 · 117 · 117 · 117 · 117 7 4 7 · 7 · 7 · 7 Determine the value of each of the powers. 3 2 9 1 2 1 15 2 225 3 4 81 1 4 1 7 3 343 Simplify as far as possible. (2x) 3 2 3 · x 3 = 8x 3 302 0 1 (-2x) 3 (-2) 3 · x 3 = -8x 3 2 3 · 2 4 2 3+4 = 2 7 $\\frac{{x}^{8}}{{x}^{3}}$ ${x}^{8-3}={x}^{5}$ $\\frac{\\text{25}{x}^{2}}{{5x}^{8}}$ $\\frac{\\text{25}}{5}{x}^{2-8}={5x}^{-6}=\\frac{5}{{x}^{6}}$ (3 -1 +2 -1 ) -1 $\\frac{1}{{3}^{-1}+{2}^{-1}}=\\frac{1}{\\frac{1}{3}+\\frac{1}{2}}=1\\cdot \\left(\\frac{3}{1}+\\frac{2}{1}\\right)=3+2=5$ how", "× n times) $={\\text{a}}^{\\text{mn}}$ For example: $\\begin{array}{}\\left({2}^{2}{\\right)}^{3}=\\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\cdot \\left({2}^{2}\\right)\\\\ =\\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\cdot \\left(2\\cdot 2\\right)\\\\ ={2}^{6}\\\\ ={2}^{2×3}\\end{array}$ ## Examples: exponential law 6 1. $\\left({x}^{3}{\\right)}^{4}={x}^{\\text{12}}$ 2. $\\left[\\left({a}^{4}{\\right)}^{3}{\\right]}^{2}={a}^{\\text{24}}$ 3. $\\left({3}^{n+3}{\\right)}^{2}={3}^{2n+6}$ ## Module review exercises Write the following examples using exponential notation. $4\\cdot 4$ 4 2 $\\text{12}\\cdot \\text{12}$ 12 2 $9\\cdot 9\\cdot 9\\cdot 9$ 9 4 $\\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}\\cdot \\text{10}$ 10 6 $\\text{826}\\cdot \\text{826}\\cdot \\text{826}$ 826 3 $\\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}\\cdot \\text{3021}$ 3021 4 $6\\cdot 6\\cdot 6\\cdot \\cdots \\cdot 6$ (85 factors of 6). 6 85 $2\\cdot 2\\cdot 2\\cdot \\cdots \\cdot 2$ (112 factors of 2). 2 112 For the next examples, expand the terms. (Do not find the actual values). 5 3 5 · 5 · 5 15 2 15 · 15 117 5 117 · 117 · 117 · 117 · 117 7 4 7 · 7 · 7 · 7 Determine the value of each of the powers. 3 2 9 1 2 1 15 2 225 3 4 81 1 4 1 7 3 343 Simplify as far as possible. (2x) 3 2 3 · x 3 = 8x 3 302 0 1 (-2x) 3 (-2) 3 · x 3 = -8x 3 2 3 · 2 4 2 3+4 = 2 7 $\\frac{{x}^{8}}{{x}^{3}}$ ${x}^{8-3}={x}^{5}$ $\\frac{\\text{25}{x}^{2}}{{5x}^{8}}$ $\\frac{\\text{25}}{5}{x}^{2-8}={5x}^{-6}=\\frac{5}{{x}^{6}}$ (3 -1 +2 -1 ) -1 $\\frac{1}{{3}^{-1}+{2}^{-1}}=\\frac{1}{\\frac{1}{3}+\\frac{1}{2}}=1\\cdot \\left(\\frac{3}{1}+\\frac{2}{1}\\right)=3+2=5$ can", "# How do you multiply 4 * 4sqrt4? Apr 29, 2015 $\\sqrt{4} = 2$- Knowing that, the question turns into a simple multiplication problem: $4 \\cdot 4 \\cdot \\sqrt{4} = 4 \\cdot 4 \\cdot 2 = 32$", "to prove that $4^n \\equiv 4 \\bmod 12$ for all $n$. This is a classic example of proof by induction. Base Case If $n=1$ then is $4^1 \\equiv 4 \\bmod 12$? Well, clearly $4^1 \\equiv 4 \\bmod 12$. Induction Step Assume that for $n=k$ we have $4^n \\equiv 4 \\bmod 12$, does this imply that $4^{n+1} \\equiv 4 \\bmod 12$? Well, if $4^k \\equiv 4 \\bmod 12$ then $4^{k+1} = 4 \\times 4^k \\equiv 4 \\times 4 \\bmod 12 \\equiv 4 \\bmod 12$. Conclusion It follows that $4^n \\equiv 4 \\bmod 12$ for all $n \\in \\mathbb{Z}^+$. - $$4^{47}\\equiv 4\\pmod{12}$$ $$4^{47}-4=4 \\cdot \\left(4^{46}-1\\right)=4 \\cdot \\left(4^{23}-1\\right)\\cdot \\left(4^{23}+1\\right)=$$ $$=4 \\cdot \\left(4-1\\right)(4^{22}+4^{21}\\cdot1+\\ldots+1)\\cdot \\left(4^{23}+1\\right)=4 \\cdot3 \\cdot \\left(4^{22}+4^{21}\\cdot1+\\ldots+1\\right)\\cdot \\left(4^{23}+1\\right)=$$ $$=12 \\cdot \\left(4^{22}+4^{21}\\cdot1+\\ldots+1\\right)\\cdot \\left(4^{23}+1\\right)$$ which is divisible with $12$. You seem to have lost the factor $4^{23}+1$. Also, $4^{23}-1=(4-1)(4^{22}+4^{21}+...+4+1)$ – user60725 Feb 14 '13 at 20:09", "# How do you find the indicated term of the arithmetic sequence a_1=7, d=3, n=14? Mar 17, 2018 ${a}_{14} = 46$ #### Explanation: ${a}_{1} = 7 , d = 3 , n = 14$ ${a}_{2} = {a}_{1} + \\left(2 - 1\\right) \\cdot d = 7 + 3 = 10$ ${a}_{3} = {a}_{1} + \\left(3 - 1\\right) \\cdot d = 7 + 6 = 13$ ${a}_{n} = {a}_{1} + \\left(n - 1\\right) \\cdot d$ $\\therefore {a}_{14} = {a}_{1} + \\left(14 - 1\\right) \\cdot d = 7 + 13 \\cdot 3 = 7 + 39 = 46$", "# How do you add 4sqrt3 - sqrt12? You may notice that $12 = 3 \\cdot 4 = 3 \\cdot {2}^{2}$ $= 4 \\sqrt{3} - 2 \\sqrt{3} = 2 \\sqrt{2}$", "$$1129$$ $S_4$ (as 4T5) Trivial 4.0.1161.1 x4 - x3 + 6x2 - x + 7 $$3^{3}\\cdot 43$$ $D_{4}$ (as 4T3) Trivial 4.0.1168.1 x4 - x2 - 4x + 5 $$2^{4}\\cdot 73$$ $D_{4}$ (as 4T3) Trivial 4.0.1197.1 x4 - 2x3 - 4x2 + 5x + 7 $$3^{2}\\cdot 7\\cdot 19$$ $D_{4}$ (as 4T3) Trivial 4.0.1197.2 x4 - x3 - 5x2 + 3x + 9 $$3^{2}\\cdot 7\\cdot 19$$ $D_{4}$ (as 4T3) Trivial 4.0.1225.1 x4 - x3 + 5x2 + 2x + 4 $$5^{2}\\cdot 7^{2}$$ $C_2^2$ (as 4T2) Trivial Next" ]

[ "the mean for that group is likely close to the mean for the entire group. The hypotenuse of an isosceles right triangle is $14\\sqrt 2$. what is the area of the triangle?", "+0 # The hypotenuse of an isosceles right triangle is $14\\sqrt 2$. How long is each leg of the triangle? +1 200 1 +252 1. The hypotenuse of an isosceles right triangle is 14sqrt(2). How long is each leg of the triangle? 2. The hypotenuse of an isosceles right triangle is 14sqrt(2). What is the area of the triangle? Mar 23, 2021 edited by weredumbmaster22 Mar 23, 2021 #1 +84 +3 Both questions use the same triangle: Note that an isosceles right triangle is a 1-1-sqrt(2) triangle; meaning that we can find the legs by dividing the hypotenuse by sqrt(2). $$14{\\sqrt {2}}{\\div }{\\sqrt {2}}=14$$ 1. The leg of the triangle is 14 units long. The area is simply bh/2, where b=14 and h=14. $${\\frac {bh} {2}}={\\frac {14{\\times }14} {2}}=98$$ 2. The area of the triangle is 98 square units. Mar 23, 2021", "+0 # HELP ASAP 0 47 1 The hypotenuse of an isosceles right triangle is $$14\\sqrt 2$$. How long is each leg of the triangle? Jan 18, 2021 #1 +1 Via pythag theorem ...and knowing each leg is the same length... l^2 + l^2 = 196(2) l^2 = 196 l = 14 units Jan 18, 2021", "+0 # Help plz 0 47 1 +170 A square with sides of length $2$ units is constructed on a right triangle with a hypotenuse of length $2\\sqrt{5}$ units to form the figure shown. What is the total area of the figure, in square units? Feb 28, 2021", "Correct answer is: 50 sq. units Area of an isosceles right triangle $= \\frac{x^{2}}{2}$ square units, where x is the length of the congruent sides of the triangle. Here, $x = 10$ units. Area $= \\frac{10^{2}}{2} = \\frac{100}{2} = 50$ square units. 3 ### The area of an isosceles right triangle is 32 square units. What is the measure of its base? 5 sq. units 15 sq. units 8 sq. units 50 sq. units CorrectIncorrect Correct answer is: 8 sq. units Area of an isosceles right triangle = x22 square units, where x is the length of the congruent sides of the triangle. $\\frac{x^{2}}{2} = 32$ $x^{2} = 32 \\times 2 = 64$ $x = \\sqrt{64}$ $x = 8\\;$ units Therefore, the measure of the base of an isosceles right triangle with area equal to 32 square units is 8 units. 4 ### The area of the isosceles right triangle whose base and height equal 1 unit is ____________. $\\frac{1}{2}$ sq. units $2$ sq. units $\\frac{1}{4}$ sq. units $4$ sq. units CorrectIncorrect Correct answer is: $\\frac{1}{2}$ sq. units Area of isosceles right triangle $= \\frac{x^{2}}{2} = \\frac{1}{2}$ sq. units 5 ### The hypotenuse of the isosceles right triangle whose equal sides are given by x units is $\\sqrt{3x}$ units $\\sqrt{2x}$ units $2x$ units $x$ units CorrectIncorrect Correct answer", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) In an isosceles right triangle, the hypotenuse is $\\sqrt{2}$ times as long as the legs.", "area of an isosceles right triangle is: $\\frac{x^{2}}{2}$ square units, where x is the measure of the congruent side of the triangle. Area $= \\frac{10^{2}}{2} = \\frac{100}{2} = 50$ square units. 3. The area of an isosceles right triangle is 72 square units. What is the measure of its base? Solution: We know that the formula to calculate the area of an isosceles right triangle is: $\\frac{x^{2}}{2}$ square units, where x is the measure of the congruent side of the triangle. Given that the area of the triangle is 72 square units. Putting this value in the formula: $\\frac{x^{2}}{2} = 72$ $x^{2} = 72 \\times 2 = 144$ $x = \\sqrt{144}$ $x = 12$ units. Therefore, the measure of the base of an isosceles right triangle with area equal to 72 square units is 12 units. 4. The height of a right isosceles triangle is 8 units. What will be its perimeter? Solution: The perimeter of an isosceles right triangle $= (2x + h)$ units. x is equal to the length of the congruent side h is the measure of the hypotenuse. The height of this isosceles right triangle is 8 units. Here, $Hypotenuse = \\sqrt{2x}$ $h = \\sqrt{2 \\times 8} = 11.3137$ units $Perimeter = 2x + h = (2 \\times 8) + 11.3137 = 27.3137$ units.", "is: $\\sqrt{2x}$ units The hypotenuse of the isosceles right triangle is given by $\\sqrt{2x}$, where x represents the length of the congruent sides. ## Frequently Asked Questions on Isosceles Right Triangle Yes, all the isosceles right triangles have interior angles measuring $45^{\\circ},\\; 45^{\\circ},\\;$ and $90^{\\circ}$. This is because one angle has to be a right angle (90 degree), and consequently, the other two angles have to be equal as the opposite sides are equal. • Isosceles triangle: Two sides are congruent. • Equilateral triangle: All sides are congruent. • Scalene triangle: All sides are of different lengths. A triangle that has all angles measuring less than 90 degrees is called an acute-angled triangle. A triangle that has one of its angles going beyond 90 degrees, is called an obtuse-angled triangle. Isosceles triangles can be classified into three types: 1) acute isosceles triangle, 2) obtuse isosceles triangle, and 3) right isosceles triangles.", "we draw an altitude from the right angle vertex to the hypotenuse (the opposite side), then it is the perpendicular bisector of the hypotenuse. 6. The area of an isosceles right triangle is equal to $12 \\times$ Base $\\times$ Height square units. ## Fun Facts! • Since the two sides of an isosceles right triangle are equal, the angles are also equal. And both the angles measure 45 degrees. • Right angle is always the largest angle in a right triangle. • There can’t be any obtuse angles in a right triangle. ## Conclusion In this article, we explored the properties of the isosceles right triangle. Let’s solve some practice problems for better understanding. ## Solved Examples on Isosceles Right 1. The equal sides of an isosceles right triangle are 5 units each. Calculate the length of its hypotenuse. Solution: We know that the formula to calculate the hypotenuse of an isosceles right triangle is: Hypotenuse $= \\sqrt{2\\; x}$ where, “x” is the length of the congruent side. The equal sides measure 5 units. Therefore, the hypotenuse will be equal to $\\sqrt{2 \\times 5} = 7.071$ units. 2. The length of the base of an isosceles right triangle is 10 units. What will be the area of this triangle? Solution: We know that the formula to calculate the", "Browse Questions # Prove that the area of right-angled traingle of a given hypotenuse is maximum when the triangle is isosceles. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "2 \\sqrt{A}$ In our case $c = 2 \\sqrt{960} = 16 \\sqrt{15} \\approx 61.97$ is the minimum hypotenuse than can support this area, presumably corresponding to the isosceles right triangle with $a = b .$", "Browse Questions Prove that the area of right-angled traingle of a given hypotenuse is maximum when the triangle is isosceles. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com Can you answer this question?", "triangle and $c$ is the hypotenuse of the triangle. Right Triangle A right triangle is a triangle with one 90 degree angle.", "Free Version Moderate # Isosceles Triangle: Right Triangle/Hypotenuse Given/Find Leg GEOM-SKEWP1 The hypotenuse of an isosceles right triangle is 10 units. Find the length of a leg. A $5$ B $\\sqrt{5}$ C $5\\sqrt{2}$ D $2\\sqrt{5}$ E $5\\sqrt{3}$", "1 square unit. The coordinate grid lines divide the square on the hypotenuse into four half unit-squares, and so its area is 2 square units. 3. The sum of the areas of the squares built on the triangle's legs is 2 square units which is the same as the area of the square built on the triangle's hypotenuse. 2. 1. The square with side length 3 has an area of 9 square units while the square with side length 4 has an area of 16 square units. 2. There are many ways of doing this and two are shown here. All areas are calculated, as in part (a) above, in terms of square units where one square unit is a one unit by one unit square in the coordinate grid. The first method for calculating area decomposes the quadrilateral into right triangles and a square while the second method subtracts right triangles from a larger square. In the picture above, the quadrilateral on the triangle's hypotenuse is divided into 4 right triangles, each measuring 3 units by 4 units and a one unit by unit square. this means that the quadrilateral has an area of 25 square units. In this second picture the quadrilateral on the triangle's hypotenuse has been surrounded by a larger square measuring 7 units", "# Math Help - Isosceles Right Triangle 1. ## Isosceles Right Triangle Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. 2. The hypotenuse is $x$ and this is a 45-45-90 Triangle. Thus, the equal sides are $\\frac{x}{\\sqrt{2}}$ That means the area is, $\\frac{1}{2} \\cdot \\frac{x}{\\sqrt{2}}\\cdot \\frac{x}{\\sqrt{2}}=\\frac{1}{4}x^2$ 3. ## ok Great but tell me: how did you get x/sqrt{2}? Shouldn't we RATIONALIZE the denominator? 4. Originally Posted by symmetry Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. Hello, an isosceles right triangle is the half of a square. Area of the square: $A_{square}=x^2$. Therefore the area of the isosceles right triangle: $A_{\\text{isosceles right trinagle}}=\\frac{1}{2} \\cdot x^2$ EB 5. Originally Posted by earboth Hello, an isosceles right triangle is the half of a square. Area of the square: $A_{square}=x^2$. Therefore the area of the isosceles right triangle: $A_{\\text{isosceles right trinagle}}=\\frac{1}{2} \\cdot x^2$ EB of course you could just say that the area of a triangle is $A=\\frac{1}{2}bh$ So: $A=\\frac{1}{2}x^2$ 6. ## ok What a great reply! I thank you. Earboth, Tell me, how do you make lovely geometric pictures?", "+0 # math problem 0 233 1 In isosceles right triangle $ABC$, point $D$ is on hypotenuse $\\overline{BC}$ such that $\\overline{AD}$ is an altitude of $\\triangle ABC$ and $DC = 5$. What is the area of triangle $ABC$? Dec 11, 2019 In isosceles right triangle $ABC$, point $D$ is on hypotenuse $\\overline{BC}$ such that $\\overline{AD}$ is an altitude of $\\triangle ABC$ and $DC = 5$. What is the area of triangle $ABC$?", "Question # If an isosceles right - angled triangle has area 200cm2, then the measure of its hypotenuse is _______________ cm. 10 10√2 20 20√2 Solution ## The correct option is Let the equal sides of the isosceles right angle triangle be 'a' 12a2=200 a2=400 a=20 cm Hypotenuse = a√2 = 20(√2) = 20√2cm Hence (D) Suggest corrections", "# Isosceles Right Triangle • Jan 18th 2007, 02:00 PM symmetry Isosceles Right Triangle Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. • Jan 18th 2007, 02:06 PM ThePerfectHacker The hypotenuse is $x$ and this is a 45-45-90 Triangle. Thus, the equal sides are $\\frac{x}{\\sqrt{2}}$ That means the area is, $\\frac{1}{2} \\cdot \\frac{x}{\\sqrt{2}}\\cdot \\frac{x}{\\sqrt{2}}=\\frac{1}{4}x^2$ • Jan 18th 2007, 02:18 PM symmetry ok Great but tell me: how did you get x/sqrt{2}? Shouldn't we RATIONALIZE the denominator? • Jan 22nd 2007, 05:48 AM earboth Quote: Originally Posted by symmetry Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. Hello, an isosceles right triangle is the half of a square. Area of the square: $A_{square}=x^2$. Therefore the area of the isosceles right triangle: $A_{\\text{isosceles right trinagle}}=\\frac{1}{2} \\cdot x^2$ EB • Jan 22nd 2007, 05:51 AM Quick Quote: Originally Posted by earboth Hello, an isosceles right triangle is the half of a square. Area of the square: $A_{square}=x^2$. Therefore the area of the isosceles right triangle: $A_{\\text{isosceles right trinagle}}=\\frac{1}{2} \\cdot x^2$ EB of course you could just say that the area of a triangle is $A=\\frac{1}{2}bh$ So: $A=\\frac{1}{2}x^2$ • Jan 22nd 2007,", "# The area of a right-angled isosceles triangle whose hypotenuse is equal to 270 m is : (A) 19000 m2 (B) 18225 m2 (C) 17256 m (D) 18325 m2 Solution: Given hypotenuse = 270 m Let ABC be a triangle with hypotenuse AC. An isosceles triangle has 2 sides equal. So AB = BC = a Since triangle is right-angled isosceles triangle, we use Pythagoras theorem AC2 = a2 + a2 => 2 a2 = 2702 => a2 = 2702/2 Area of triangle ABC = ½ (base×height) = ½ a2 = 2702/4 = 18225 m2 Hence option B is the answer." ]

[ "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?", "Now $\\frac{\\pi r^2}{2}$ = $2\\pi$ $\\Rightarrow r^2=4$ $\\Rightarrow r=2$ this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, 4 The area of the triangle is $\\frac{1}{2} \\times 4 \\times 4$=8", "in essence you construct an isosceles right triangle $(45-45-90)$. Given that the hypotenuse is a length of $4$, that means each leg has a length of $2\\sqrt{2}$. Just move accordingly along the normal line.", "+0 # The hypotenuse of an isosceles right triangle is $14\\sqrt 2$. How long is each leg of the triangle? +1 200 1 +252 1. The hypotenuse of an isosceles right triangle is 14sqrt(2). How long is each leg of the triangle? 2. The hypotenuse of an isosceles right triangle is 14sqrt(2). What is the area of the triangle? Mar 23, 2021 edited by weredumbmaster22 Mar 23, 2021 #1 +84 +3 Both questions use the same triangle: Note that an isosceles right triangle is a 1-1-sqrt(2) triangle; meaning that we can find the legs by dividing the hypotenuse by sqrt(2). $$14{\\sqrt {2}}{\\div }{\\sqrt {2}}=14$$ 1. The leg of the triangle is 14 units long. The area is simply bh/2, where b=14 and h=14. $${\\frac {bh} {2}}={\\frac {14{\\times }14} {2}}=98$$ 2. The area of the triangle is 98 square units. Mar 23, 2021", "204 [Reveal] Spoiler: OA _________________ Moderator Joined: 18 Apr 2015 Posts: 5888 Followers: 96 Kudos [?]: 1154 [0], given: 5471 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 13 Sep 2016, 05:55 Expert's post Explanation In an isosceles right triangle, both legs have the same length. The area of an isosceles right triangle with legs of length x is equal to $$\\frac{x^2}{2}$$ The area of a square is $$s^2$$ . Since you are given that an isosceles right triangle and a square have the same area R, it follows that $$\\frac{x^2}{2}=s^2$$ and so $$x=\\sqrt{2}s$$ Since $$\\sqrt{2}$$ is greater than 1, the length of a leg of the triangle is greater than the length of a side of the square. The correct answer is Choice A _________________ GRE Instructor Joined: 10 Apr 2015 Posts: 1543 Followers: 56 Kudos [?]: 1466 [2] , given: 8 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 12 Apr 2018, 08:58 2 KUDOS Expert's post Carcass wrote: Quantity A Quantity B The length of a leg of an isosceles right triangle with area R The length of a side of a square with area R A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The", "204 [Reveal] Spoiler: OA _________________ Moderator Joined: 18 Apr 2015 Posts: 5175 Followers: 77 Kudos [?]: 1035 [0], given: 4660 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 13 Sep 2016, 05:55 Expert's post Explanation In an isosceles right triangle, both legs have the same length. The area of an isosceles right triangle with legs of length x is equal to $$\\frac{x^2}{2}$$ The area of a square is $$s^2$$ . Since you are given that an isosceles right triangle and a square have the same area R, it follows that $$\\frac{x^2}{2}=s^2$$ and so $$x=\\sqrt{2}s$$ Since $$\\sqrt{2}$$ is greater than 1, the length of a leg of the triangle is greater than the length of a side of the square. The correct answer is Choice A _________________ GRE Instructor Joined: 10 Apr 2015 Posts: 1242 Followers: 46 Kudos [?]: 1129 [1] , given: 7 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 12 Apr 2018, 08:58 1 KUDOS Expert's post Carcass wrote: Quantity A Quantity B The length of a leg of an isosceles right triangle with area R The length of a side of a square with area R A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The", "+0 # HELP ASAP 0 47 1 The hypotenuse of an isosceles right triangle is $$14\\sqrt 2$$. How long is each leg of the triangle? Jan 18, 2021 #1 +1 Via pythag theorem ...and knowing each leg is the same length... l^2 + l^2 = 196(2) l^2 = 196 l = 14 units Jan 18, 2021", "the mean for that group is likely close to the mean for the entire group. The hypotenuse of an isosceles right triangle is $14\\sqrt 2$. what is the area of the triangle?", "Free Version Moderate # Isosceles Triangle: Right Triangle/Hypotenuse Given/Find Leg GEOM-SKEWP1 The hypotenuse of an isosceles right triangle is 10 units. Find the length of a leg. A $5$ B $\\sqrt{5}$ C $5\\sqrt{2}$ D $2\\sqrt{5}$ E $5\\sqrt{3}$", "# Possible length of isosceles triangle side. The perimeter of a right triangle $RST$ is equal to the perimeter of isoceles triangle $xyz$ The lengths of the legs of the right triangle are 6 and 8. If the length of each side of the isoceles triangle is an integer what is the greatest possible length for one of the sides of isoceles triangle $xyz$? The hypotenuse of $RST$ is $\\sqrt{100} = 10$ For $xyz$ let $xy = yz$ as the two equal legs, and $zx$ is the unequal, the \"base\" side. Perimeter of $RST$ = $xyz$ = $24$ So leg $zx = 2$ is the lowest length the triangle can have, so that the other two are integers. $xy + yz = 22 = 2xy \\implies xy = 11 = yz$ Is this correct?", "# 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.", "isosceles right triangle with edge lengths , , and gives a Area of an isosceles right triangle Isosceles right triangle is a special right triangle, sometimes called a 45-45-90 triangle. What’s more, the lengths of those two legs have a special relationship with the hypotenuse (in … With this hypotenuse calculator you will quickly find this longest side of a right triangle. Take a square root of sum of squares: As area of a right triangle is equal to a * b / 2, then. Isosceles right-angled triangles cannot have sides with integer values, because the ratio of the hypotenuse to either other side is √ 2, but √ 2 cannot be expressed as a ratio of two integers. The hypotenuse is the largest side in a right triangle and is always opposite the right angle. But in every isosceles right triangle, the sides are in the ratio 1 : 1 : , as shown on the right. See the solution with steps using the Pythagorean Theorem formula. The general principle to remember is a 4:1 rule - for every four feet of vertical height, the ladder foot should move one foot from the wall. Isosceles triangle is a polygon with three vertices (corners) and three edges (sides) two of which are equal. Then click Calculate. One leg", "described below to the designated agent listed below. The altitude is a perpendicular distance from the base to the topmost vertex. Given: Isosceles right triangle XYZ (45°-45°-90° triangle) Prove: In a 45°-45°-90° triangle, the hypotenuse is times the length of each leg. The right triangle The right triangle ABC has a leg a = 36 cm and an area S = 540 cm 2. The base is 100, and the two equal sides are unknown. This is the other base angle. 4. If the hypotenuse of an isoceles right triangle is , what is the length of the height of the triangle? Answer. As well, this line you've drawn is the height of the original triangle. The ratio of the sides opposite the two 45° angles is 1:1. Calculate the length of the leg b and the median t2 to side b. An Isosceles Right Triangle is a right triangle that consists of two equal length legs. Because the hypotenuse if 2√7 cm, that means that the base and the height (the two remaining sides) will be equivalent. height of isosceles triangle, In the image below, we can see that an isosceles triangle can be split into 2 right angle triangles. What’s more, the lengths of those two legs have a special relationship with the hypotenuse (in addition", "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", "B, |AC| = 12, |CD| = 8, diagonals are perpendicular to each other. Calculate the perimeter and area of ​​the trapezoid. 8. Right Δ A right triangle has the length of one leg 7 cm and length of the hypotenuse 25 cm. Calculate the height of the triangle. 9. Garden Area of a square garden is 6/4 of triangle garden with sides 56 m, 35 m, and 35 m. How many meters of fencing need to fence a square garden? 10. A rectangle 2 A rectangle has a diagonal length of 74cm. It’s sides are 5:3. Find its lengths. 11. The trapezium The trapezium is formed by cutting the top of the right-angled isosceles triangle. The base of the trapezium is 10 cm and the top is 5 cm. Find the area of trapezium. 12. Sides of right angled triangle One leg 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. 13. Isosceles triangle 10 In an isosceles triangle, the equal sides are 2/3 of the length of the base. Determine the measure of the base angles. 14. Uphill garden I have a garden uphill, increasing from 0 to 4.5 m for a length of 25 m, how much", "An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 16 and the triangle has an area of 128 . What are the lengths of sides A and B? Apr 15, 2016 $8 \\setminus \\sqrt{5}$ units. Explanation: The area of the triangle is $128$. The base is $16$. So the altitude to the base is $h = 2 \\left(\\frac{128}{16}\\right) = 16$. In an isosceles triangle the altitude to the base forms two congruent right triangles. In each right triangle: 1) One leg of the right triangle is the altitude, in this case $16$. 2) The other leg is half the base of the isosceles triangle, in this case $8$. 3) The hypoteneuse of the right triangle is one leg of the isosceles triangle. Thus using the Pythagorean theorem: $\\setminus \\sqrt{{16}^{2} + {8}^{2}} = \\setminus \\sqrt{320} = 8 \\setminus \\sqrt{5}$", "rectangular triangle with 5 meters and 12 meters in length. He decided to divide it into two parts and the height of the hypotenuse. For the smaller part creates a rock garden, for the large 13. Right isosceles triangle Right isosceles triangle has an altitude x drawn from the right angle to the hypotenuse dividing it into 2 equal segments. The length of one segment is 5 cm. What is the area of the triangle? 14. Right 24 Right isosceles triangle has an altitude x drawn from the right angle to the hypotenuse dividing it into 2 unequal segments. The length of one segment is 5 cm. What is the area of the triangle? Thank you. 15. RT sides Find the sides of a rectangular triangle if legs a + b = 17cm and the radius of the written circle ρ = 2cm. 16. Leg and height Solve right triangle with height v = 9.6 m and shorter cathetus b = 17.3 m. 17. Isosceles triangle 9 Given an isosceles triangle ABC where AB= AC. The perimeter is 64cm and altitude is 24cm. Find the area of the isosceles triangle", "# Volume of Sold with a Known Crosssection the base is bounded by x^2 + y^2 = 4 the solid is formed by isosceles right triangles perpendicular to x axis. the hypotenuse of each triangle is sitting on the base of the figure i just need to know what area formula i would start out with, is it just bh/2 ? Last edited: HallsofIvy An \"isosceles right triangle perpendicular to the x axis\" has hypotenuse y. Since it is isosceles, a2+ a2= 2a2= y2. That is, each leg has length $a= y\\sqrt{2}$ and so its area is (1/2)bh= (1/2)(a*a)= (1/2)(2y)= y.", "The perimeter is 64cm and altitude is 24cm. Find the area of the isosceles triangle • Triangle KLM In the rectangular triangle KLM, where is hypotenuse m (sketch it!) find the length of the leg k and the height of triangle h if hypotenuse's segments are known mk = 5cm and ml = 15cm • Free space in the garden The grandfather's free space in the garden was in the shape of a rectangular triangle with 5 meters and 12 meters in length. He decided to divide it into two parts and the height of the hypotenuse. For the smaller part creates a rock garden, for the large • RT sides Find the sides of a rectangular triangle if legs a + b = 17cm and the radius of the written circle ρ = 2cm. • Right isosceles triangle Right isosceles triangle has an altitude x drawn from the right angle to the hypotenuse dividing it into 2 equal segments. The length of one segment is 5 cm. What is the area of the triangle? • Right 24 Right isosceles triangle has an altitude x drawn from the right angle to the hypotenuse dividing it into 2 unequal segments. The length of one segment is 5 cm. What is the area of the triangle? Thank you. • Same", "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" ]

[ "+0 # The hypotenuse of an isosceles right triangle is $14\\sqrt 2$. How long is each leg of the triangle? +1 200 1 +252 1. The hypotenuse of an isosceles right triangle is 14sqrt(2). How long is each leg of the triangle? 2. The hypotenuse of an isosceles right triangle is 14sqrt(2). What is the area of the triangle? Mar 23, 2021 edited by weredumbmaster22 Mar 23, 2021 #1 +84 +3 Both questions use the same triangle: Note that an isosceles right triangle is a 1-1-sqrt(2) triangle; meaning that we can find the legs by dividing the hypotenuse by sqrt(2). $$14{\\sqrt {2}}{\\div }{\\sqrt {2}}=14$$ 1. The leg of the triangle is 14 units long. The area is simply bh/2, where b=14 and h=14. $${\\frac {bh} {2}}={\\frac {14{\\times }14} {2}}=98$$ 2. The area of the triangle is 98 square units. Mar 23, 2021", "the mean for that group is likely close to the mean for the entire group. The hypotenuse of an isosceles right triangle is $14\\sqrt 2$. what is the area of the triangle?", "+0 # HELP ASAP 0 47 1 The hypotenuse of an isosceles right triangle is $$14\\sqrt 2$$. How long is each leg of the triangle? Jan 18, 2021 #1 +1 Via pythag theorem ...and knowing each leg is the same length... l^2 + l^2 = 196(2) l^2 = 196 l = 14 units Jan 18, 2021", "# 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.", "Now $\\frac{\\pi r^2}{2}$ = $2\\pi$ $\\Rightarrow r^2=4$ $\\Rightarrow r=2$ this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, 4 The area of the triangle is $\\frac{1}{2} \\times 4 \\times 4$=8", "in essence you construct an isosceles right triangle $(45-45-90)$. Given that the hypotenuse is a length of $4$, that means each leg has a length of $2\\sqrt{2}$. Just move accordingly along the normal line.", "An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 16 and the triangle has an area of 128 . What are the lengths of sides A and B? Apr 15, 2016 $8 \\setminus \\sqrt{5}$ units. Explanation: The area of the triangle is $128$. The base is $16$. So the altitude to the base is $h = 2 \\left(\\frac{128}{16}\\right) = 16$. In an isosceles triangle the altitude to the base forms two congruent right triangles. In each right triangle: 1) One leg of the right triangle is the altitude, in this case $16$. 2) The other leg is half the base of the isosceles triangle, in this case $8$. 3) The hypoteneuse of the right triangle is one leg of the isosceles triangle. Thus using the Pythagorean theorem: $\\setminus \\sqrt{{16}^{2} + {8}^{2}} = \\setminus \\sqrt{320} = 8 \\setminus \\sqrt{5}$", "Free Version Moderate # Isosceles Triangle: Right Triangle/Hypotenuse Given/Find Leg GEOM-SKEWP1 The hypotenuse of an isosceles right triangle is 10 units. Find the length of a leg. A $5$ B $\\sqrt{5}$ C $5\\sqrt{2}$ D $2\\sqrt{5}$ E $5\\sqrt{3}$", "enter the angle in units of degree by entering the value 35 in the field Angle α • We will automatically get the second leg and angle, and it will be 7.14 cm and 55 degrees. • The hypotenuse is 8.72 cm, and finally • The area is 17.85 cm2 Remember to change the units of measure to get everything in the same. If you have trouble with understanding of angles, then you should read this post about conversion of angles. ### Example 3: With hypotenuse If we have a length of the hypotenuse and one cathetus, you have to enter these values in given fields, so let’s assume we have one cathetus, which is 4 meters, and a hypotenuse is 5 meters. • Enter cathetus in field a • Enter 5 in field c • The calculated value of other cathetus is 3 meters, • P is 6 square meters. ## FAQ What is the Area of a Right Triangle? The area of a right triangle is the total area of the space or area covered by the 90° triangle. It is expressed in square units (square meters, centimetres, etc.). What are the Isosceles 90° Triangles? It’s the math form where you have one angle of 90 degrees and the same lengths of legs. This is also", "1 square unit. The coordinate grid lines divide the square on the hypotenuse into four half unit-squares, and so its area is 2 square units. 3. The sum of the areas of the squares built on the triangle's legs is 2 square units which is the same as the area of the square built on the triangle's hypotenuse. 2. 1. The square with side length 3 has an area of 9 square units while the square with side length 4 has an area of 16 square units. 2. There are many ways of doing this and two are shown here. All areas are calculated, as in part (a) above, in terms of square units where one square unit is a one unit by one unit square in the coordinate grid. The first method for calculating area decomposes the quadrilateral into right triangles and a square while the second method subtracts right triangles from a larger square. In the picture above, the quadrilateral on the triangle's hypotenuse is divided into 4 right triangles, each measuring 3 units by 4 units and a one unit by unit square. this means that the quadrilateral has an area of 25 square units. In this second picture the quadrilateral on the triangle's hypotenuse has been surrounded by a larger square measuring 7 units", "It's a whole number! Geometry Level 2 What is the area of an isosceles triangle with two legs that measure $6\\sqrt[4]{3}$, and an angle of $120^\\circ$? ×", "204 [Reveal] Spoiler: OA _________________ Moderator Joined: 18 Apr 2015 Posts: 5888 Followers: 96 Kudos [?]: 1154 [0], given: 5471 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 13 Sep 2016, 05:55 Expert's post Explanation In an isosceles right triangle, both legs have the same length. The area of an isosceles right triangle with legs of length x is equal to $$\\frac{x^2}{2}$$ The area of a square is $$s^2$$ . Since you are given that an isosceles right triangle and a square have the same area R, it follows that $$\\frac{x^2}{2}=s^2$$ and so $$x=\\sqrt{2}s$$ Since $$\\sqrt{2}$$ is greater than 1, the length of a leg of the triangle is greater than the length of a side of the square. The correct answer is Choice A _________________ GRE Instructor Joined: 10 Apr 2015 Posts: 1543 Followers: 56 Kudos [?]: 1466 [2] , given: 8 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 12 Apr 2018, 08:58 2 KUDOS Expert's post Carcass wrote: Quantity A Quantity B The length of a leg of an isosceles right triangle with area R The length of a side of a square with area R A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The", "we draw an altitude from the right angle vertex to the hypotenuse (the opposite side), then it is the perpendicular bisector of the hypotenuse. 6. The area of an isosceles right triangle is equal to $12 \\times$ Base $\\times$ Height square units. ## Fun Facts! • Since the two sides of an isosceles right triangle are equal, the angles are also equal. And both the angles measure 45 degrees. • Right angle is always the largest angle in a right triangle. • There can’t be any obtuse angles in a right triangle. ## Conclusion In this article, we explored the properties of the isosceles right triangle. Let’s solve some practice problems for better understanding. ## Solved Examples on Isosceles Right 1. The equal sides of an isosceles right triangle are 5 units each. Calculate the length of its hypotenuse. Solution: We know that the formula to calculate the hypotenuse of an isosceles right triangle is: Hypotenuse $= \\sqrt{2\\; x}$ where, “x” is the length of the congruent side. The equal sides measure 5 units. Therefore, the hypotenuse will be equal to $\\sqrt{2 \\times 5} = 7.071$ units. 2. The length of the base of an isosceles right triangle is 10 units. What will be the area of this triangle? Solution: We know that the formula to calculate the", "204 [Reveal] Spoiler: OA _________________ Moderator Joined: 18 Apr 2015 Posts: 5175 Followers: 77 Kudos [?]: 1035 [0], given: 4660 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 13 Sep 2016, 05:55 Expert's post Explanation In an isosceles right triangle, both legs have the same length. The area of an isosceles right triangle with legs of length x is equal to $$\\frac{x^2}{2}$$ The area of a square is $$s^2$$ . Since you are given that an isosceles right triangle and a square have the same area R, it follows that $$\\frac{x^2}{2}=s^2$$ and so $$x=\\sqrt{2}s$$ Since $$\\sqrt{2}$$ is greater than 1, the length of a leg of the triangle is greater than the length of a side of the square. The correct answer is Choice A _________________ GRE Instructor Joined: 10 Apr 2015 Posts: 1242 Followers: 46 Kudos [?]: 1129 [1] , given: 7 Re: QOTD # 9 The length of a leg of an isosceles right triangle [#permalink] 12 Apr 2018, 08:58 1 KUDOS Expert's post Carcass wrote: Quantity A Quantity B The length of a leg of an isosceles right triangle with area R The length of a side of a square with area R A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The", "+0 # help 0 207 3 The hypotenuse of an isosceles right triangle is $4\\sqrt{2}$ units. How many square units are in the area of the triangle? Nov 8, 2019 #1 +108650 +2 let the legs be x units long Use pythagoras's theorum to solve for x. Then you can find the area. Nov 8, 2019 #2 +109334 +1 If we had a square with a diagonal = 4√2 units.....the side of the square = 4 units Then the area of the square = side^2 = 4^2 = 16 units^2 So.....the area of this isosceles triangle will be 1/2 of this Nov 8, 2019 #3 +108650 0 Chris you have answered over me for no good reason AND you left out the bit that would cause the most problem. \"Why is the length of the square 4 units and what has a triangle got to do with a square.\" If this kid sees your answer and wants to learn from it at all, which is quite unlikely, those are the questions they would be asking themselves. I do admit my answer was very basic but it at least gave a hint on how to find the length of the triangle legs. If the child wanted more help he/she should have, and probably would have asked. Melody Nov 8,", "Perimeter of Isosceles Right Triangle. Using basic area of triangle formula. Eugene Brennan (author) from Ireland on June 02, 2020: Hi Kayla, Draw your triangle with the side 8cm as the base. Thus, the hypotenuse measures h, then the Pythagorean theorem for isosceles right triangle would be: Also, two congruent angles in isosceles right triangle measure 45 degrees each, and the isosceles right triangle is: As we know that the area of a triangle (A) is ½ bh square units. Isosceles triangle The leg of the isosceles triangle is 5 dm, its height is 20 cm longer than the base. One leg is a base and the other is the height - there is a right angle between them. Let us say that they both measure “l” then the area formula can be further modified to: Area of an Isosceles Right Triangle = l2/2 square units. A right triangle is a triangle in which exactly one angle measures 90 degrees. Isosceles right triangle Area of an isosceles right triangle is 18 dm 2. According to this theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides of the right triangle. The height of an isosceles triangle is the perpendicular line segment drawn from base of the triangle to the", "is: $\\sqrt{2x}$ units The hypotenuse of the isosceles right triangle is given by $\\sqrt{2x}$, where x represents the length of the congruent sides. ## Frequently Asked Questions on Isosceles Right Triangle Yes, all the isosceles right triangles have interior angles measuring $45^{\\circ},\\; 45^{\\circ},\\;$ and $90^{\\circ}$. This is because one angle has to be a right angle (90 degree), and consequently, the other two angles have to be equal as the opposite sides are equal. • Isosceles triangle: Two sides are congruent. • Equilateral triangle: All sides are congruent. • Scalene triangle: All sides are of different lengths. A triangle that has all angles measuring less than 90 degrees is called an acute-angled triangle. A triangle that has one of its angles going beyond 90 degrees, is called an obtuse-angled triangle. Isosceles triangles can be classified into three types: 1) acute isosceles triangle, 2) obtuse isosceles triangle, and 3) right isosceles triangles.", "\\sqrt{2} = 9 \\cdot 2 = 18$ . 2. Use the $x : x : x \\sqrt{2}$ ratio. We need to solve for $x$ in the ratio. $12 \\sqrt{2} &= x \\sqrt{2}\\\\12 &= x$ 3. $x=5\\sqrt{3}\\cdot \\sqrt{2}=5\\sqrt{6}$ ### Practice 1. In an isosceles right triangle, if a leg is $x$ , then the hypotenuse is __________. 2. In an isosceles right triangle, if the hypotenuse is $x$ , then each leg is __________. 3. A square has sides of length 15. What is the length of the diagonal? 4. A square’s diagonal is 22. What is the length of each side? 5. A square has sides of length $6\\sqrt{2}$ . What is the length of the diagonal? 6. A square has sides of length $4 \\sqrt{3}$ . What is the length of the diagonal? 7. A baseball diamond is a square with 90 foot sides. What is the distance from home base to second base? (HINT: It’s the length of the diagonal). 8. Four isosceles triangles are formed when both diagonals are drawn in a square. If the length of each side in the square is $s$ , what are the lengths of the legs of the isosceles triangles? Find the lengths of the missing sides. Simplify all radicals.", "equation y = $$\\sqrt{x}$$, if x = 3, y ≈ 1.7. This means that when the area of a square is 3 square units, the length of a leg is approximately 1.7 units. Thus, the average rate of change is approximately 0.6. The average rate of change of the legs of an isosceles triangle is greater than the average rate of change of the sides of a square on the interval 0 ≤ x ≤ 3. c. Interpret your answer to part (b) in terms of the situation being described. Over the interval 0 ≤ x ≤ 3, the average rate of change for the lengths of the legs of an isosceles triangle per a given area is approximately $$\\frac{4}{3}$$ greater than the average rate of change for the sides of the square. In other words, for every unit of change in area, the legs grow (on average) by 0.2 units more over the interval 0 ≤ x ≤ 3. d. Which will have a greater value: the edge length of a square with area 16 square units or the length of a leg of an isosceles right triangle with an area of 16 square units? Approximately by how much? According to the graph, the length of the leg of an isosceles right triangle with an area of", "length of the diagonal? 6. A square has sides of length 43. What is the length of the diagonal? 7. A baseball diamond is a square with 90 foot sides. What is the distance from home base to second base? (HINT: It’s the length of the diagonal). 8. Four isosceles triangles are formed when both diagonals are drawn in a square. If the length of each side in the square is s, what are the lengths of the legs of the isosceles triangles? Find the lengths of the missing sides. Simplify all radicals. Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Vocabulary Language: English 45-45-90 Theorem For any isosceles right triangle, if the legs are x units long, the hypotenuse is always x$\\sqrt{2}$. 45-45-90 Triangle A 45-45-90 triangle is a special right triangle with angles of $45^\\circ$, $45^\\circ$, and $90^\\circ$. Hypotenuse The hypotenuse 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. The $\\sqrt{}$, or square root, sign. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects: Concept Nodes: Date Created: Jul 17, 2012" ]

[ "Mr. Kilborn’s class has ten girls and ten boys. If one student is chosen randomly from the class, what is the probability that the Question Mr. Kilborn’s class has ten girls and ten boys. If one student is chosen randomly from the class, what is the probability that the student chosen will be a boy? in progress 0 6 hours 2021-07-22T11:31:41+00:00 1 Answers 0 views 0", "girls' school,but not anymore...(there were only about 6 students in my class) so now I'm doing my ALs privately,i go to classes outside and there are boys and girls(50 max)& even students from popular schools attend these classes. I do use calculators when I work sums at home and classes,but we're not allowed to use them at our exams,we're provided with log books instead.since our exams are close,I thought I should stop using cals. Anyway thanks for the help 8. Jul 5, 2009 ### vk6kro Very interesting but disturbing as well. I guess they can't help the class sizes being so big. A class of 50 means each student gets almost no individual attention from the teacher. Have you noticed that those who do get the teacher's attention quickly fire off new questions while they have the chance? But they can certainly fix the \"no calculators in exams\" rule. This is just nonsense. If the exam is in arithmetic, sure, you don't allow calculators. But in Physics or Chemistry, the actual calculations are not the object of the exam. It is the logic of assembling the numbers for calculation that matters. Some calculators that have just a few numeric memories do not allow any cheating and even if a small range of \"permitted\" calculators are prescribed, this would", "Shawn Cornally’s blog, because it kept me thinking deeply about SBG and the larger questions around the philosophy, but it seems to have disappeared! One last shoutout goes to Matt Townsley (his old blog, with lots of posts on SBG, is here). I am grateful that there was such a buzz and an ongoing set of conversations about SBG in the #MTBoS as I was trying it out, because the devil is in the details, but so are the angels. So there you go. Profoundly impactful blogposts that changed my teaching. # Girls and Math Act I: In one of our department meetings near the start of the year, we started talking about the representation of girls in our math club and our math team. In years past, there was a higher representation than this year. And although I suspect that the distribution of boys/girls in our math classes are probably relatively even — based on my own anecdotal evidence — I will readily admit that in all years past, there were fewer girls than boys in math club and on our math team. We as a department brainstormed different possible reasons. One teacher (it may have been me? maybe not though) said we could just ask students. But we agreed that this is something we should be", "## Newton’s Class More than $93$ percent of the students in Mr. Newton’s math class are girls. If Mr. Newton’s class has at least one boy, what is the least possible size of his class? Source: mathcontest.olemiss.edu 2/24/2014 SOLUTION Ratio Percent $1:2\\qquad 50.00$ $3:4\\qquad 75.00$ $4:5\\qquad 80.00$ $5:6\\qquad 83.33$ $6:7\\qquad 85.71$ $7:8\\qquad 87.50$ $9:10\\quad\\:\\: 90.00$ $10:11\\quad\\:\\: 90.90$ $11:12\\quad\\:\\: 91.67$ $12:13\\quad\\:\\: 92.31$ $13:14\\quad\\:\\: 92.86$ $14:15\\quad\\:\\: 93.33$ $15:16\\quad\\:\\: 93.75$ $16:17\\quad\\:\\: 94.12$ $17:18\\quad\\:\\: 94.44$ $\\cdots$ The least possible size of his class is $15$. Answer: $15$", "“Generally, COMMA + VERBing serves to modify the nearest preceding action and the agent of this action. Given this rule, I received a PM requesting that I justify the usage of COMMA + needing in the OA: Insurance companies in California were free to charge whatever rates the market would bear, ...” March 3, 2019 Ian Stewart posted a reply to Ian Stewart GMAT quant books in the GMAT Math forum “Anyone interested in my GMAT math books or problem sets can contact me at the email address below.” March 2, 2019 Ian Stewart posted a reply to If y is a positive integer, is y prime? in the Data Sufficiency forum “This is a near-exact replica of an official question. Statement 1 is clearly not sufficient, because there is an infinite number of primes. I think Statement 2 is supposed to read: 11! - 12 < y < 11! - 2 That''s also not sufficient, because each number in that interval has an obvious ...” March 2, 2019 GMATGuruNY posted a reply to In a certain school, there are 160 boys and 240 girls. If 15 in the Problem Solving forum “Left-handed boys = 15% of 160 = 24. Left-handed girls = 25% of 240 = 60. (total left-handed)/(total students) = (24+60)/(160+240) = 84/400 = 21/100 =", "come up with just one to call my favorite, but it was really fun thinking through all of the classes. There were roughly four categories for me: Junior High Mrs. Whitney’s geometry class. Up to that point in my math education everything was all about calculation. For me that was fine – I’m much more of a calculator by nature than anything else – but the new ideas about proof and constructions in geometry class were really interesting. The constructions, especially, I suppose and some of the problems that were not really accessible to me then stayed with me for a long time. I remember, for example, learning that $\\cos(72^{o})$ = $(\\sqrt{5} - 1) /4$ in high school and immediately heading off to construct a pentagon. High School This one is the only no-brainer – Mr. Waterman’s Enrichment Math class. This class was mainly for the math team at my high school and I took the class all three years I was in high school. Mr. Waterman also taught calculus, linear algebra, and differential equations, but it was enrichment math that really shaped how I think about math. We learned all kinds of different math – basically whatever Mr. Waterman felt like talking about that day. Nothing was off limits – advanced geometry, number theory, fun algebra problems", "In Miss Quaffley's class, the girls make up more than $45$% of the pupils, but less than $50$%. What is the smallest possible number of girls in her class?", "classes alongside traditional coed ones, giving students and parents the option. The focus seems to be on the girls. As one teacher put it \"I've got very, very intelligent young ladies who outperform (other students) in my classroom, but they are not as active in participating in class,\" said Ann Geier, a math teacher at Everest who will teach a class of up to 30 girls. \"I'm always interested in seeing how my girls do, being a woman math teacher.\" Amber O-Connell, a student who registered for one of the new classes, says \"Girls act differently. They are afraid to ask questions. They will think the guys will think they are dumb.\" I'm really not sure how I feel about this and I don't want to make any predictions. I'm choosing to look at it as an experiment that may or may not have the intended results. I will say this. I think schools should try as much as possible to mirror conditions in the actual world. The worlds of science, business, and academia do not protect girls from boys. They will eventually have to learn how to deal with that fact. I can only imagine how hard it would be for a \"smart\" girl, especially in the field of mathematics which is so largely dominated by men.", "UPSKILL MATH PLUS Learn Mathematics through our AI based learning portal with the support of our Academic Experts! The number of pens required for 16 girls $$=$$ .", "nothing but a four-digit number written on it. Mr Buttner was angry that this boy had made so little effort and just guessed an answer, so he put his slate to one side and sent him to sit in the Dunce's chair. Carl was embarrassed and sad because he thought that when his father found out he would take him out of school and send him to work digging the canals. After about another 20 minutes, other children began finishing and gradually brought their slates, covered in working out, to the teacher's desk. Mr Buttner began to show how to carefully set out the calculations on his chalkboard, while his assistant marked each slate. Eventually, the assistant announced that nobody in the whole class had the right answer, which made the teacher even grumpier. Very bravely, Carl asked the assistant to look at his slate. The assistant picked up Carl's slate and raised his eyebrows. Written on the slate was 5,050. Mr Buttner was astonished. \"How did you manage to guess this number?\" he demanded. \"I didn't\", replied Carl, \"I worked it out\". \"Nonsense\", fumed Mr Buttner, \"Where are all the calculations?\" \"I worked it out in my head\", explained Carl. Now the whole class was listening because they thought Carl was in more trouble. \"In my head,", "the classes. There were roughly four categories for me: Junior High Mrs. Whitney’s geometry class. Up to that point in my math education everything was all about calculation. For me that was fine – I’m much more of a calculator by nature than anything else – but the new ideas about proof and constructions in geometry class were really interesting. The constructions, especially, I suppose and some of the problems that were not really accessible to me then stayed with me for a long time. I remember, for example, learning that $\\cos(72^{o})$ = $(\\sqrt{5} - 1) /4$ in high school and immediately heading off to construct a pentagon. High School This one is the only no-brainer – Mr. Waterman’s Enrichment Math class. This class was mainly for the math team at my high school and I took the class all three years I was in high school. Mr. Waterman also taught calculus, linear algebra, and differential equations, but it was enrichment math that really shaped how I think about math. We learned all kinds of different math – basically whatever Mr. Waterman felt like talking about that day. Nothing was off limits – advanced geometry, number theory, fun algebra problems from old contests. Also super fun was having former students come back to give talks. After graduating I", "word problems. Since part of their PLC work freed them up to observe lessons in each others' rooms, they decided they would watch Mr. Alexander introduce modeling to his students. So, two days later they gathered in Mr. Alexander's room for the math lesson. Mr. Alexander presented the following problem: Lily and her brother, Scotty, were collecting cans for the recycling drive. One weekend they collected 59 cans and the next weekend they collected 85 cans. How many cans were collected in all? Mr. Alexander went over the problem and drew a rectangular bar divided into two parts on the board, explaining that each part of the rectangle was for the cans collected on one of the weekends and the bracket indicated how many cans were collected in all. Reviewing the problem, Mr. Alexander asked students what was not known, and where the given numbers would go and why. This resulted in the following bar model: The class then discussed what equations made sense given the relationship of the numbers in the bar model. This time many students wrote the equation, 59 + 85 = ?, and solved the problem. In their discussion after the lesson, Mr. Alexander's teammates mentioned that they noticed a much higher degree of interest and confidence in problem solving when Mr. Alexander introduced", "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", "but he loved to learn and discover new ideas, especially in mathematics. Carl's teacher, Master Büttner, was a good teacher of history and Latin, but he did not like to teach mathematics. He spent little time on the subject, and had the students do tedious problems or problems which they already understood. Today's math lesson was no different. Little did Master Büttner know he would be in for a surprise. He walked up to the chalk board to write the day's problem. \"Students, for math work today, I want you to add the whole numbers from 1 to 100.\" And he wrote—\"Add the whole numbers from 1 to 100.\" \"Get busy,\" he said. Each student had a slate board and chalk to work on, and by tradition the first student finished would put his slate face down on Master Büttner's table. As each student completed the task, his slate would join the pile. All the students except Carl unenthusiastically pulled out their slates, and began adding. Carl sat at his desk with his hands on his chin, thinking about the problem. Master Büttner saw him and thought Carl was daydreaming. He shouted, \"Get busy, Carl Gauss.\" Startled, Carl looked up and said, \"I already have the answer, Master Büttner.\" \"What do you mean you have the answer? You", "In an company of 1050 employees are 2/3 women. How many women work in the company? 15. Seeds 2 How many seeds germinated from 1000 pcs, when 23% no emergence? 16. Class In 7.C clss are 10 girls and 20 boys. Yesterday was missing 20% of girls and 50% boys. What percentage of students missing? 17. One third 2 One third of all students in class live in a house. If here are 42 students in a class, how many of them live in house?", "After paying his workers for one week, the farmer has$126 left in his bank account. How much money did he have to begin with.... There are 32 students in Mr. Morgan's math class. One student will be randomly selected to present his/her project. If there are 20 boys and 12 girls in the class, what is the probability that the presenter will be a boy? HURRY IF YOUR RIGHT I WILL GIVE YOU A LIKE AND 5 STARS There are 32 students in Mr. Morgan's math class. One student will be randomly selected to present his/her project. If there are 20 boys and 12 girls in the class, what is the probability that the presenter will be a boy? HURRY IF YOUR RIGHT I WILL GIVE YOU A LIKE AND 5 STARS...", "# Solve problems with bar graphs 2 ### Problem How many more students does Mr. Franks have in his class than Mrs. Lu has in her class? more students space space Do 7 problems", "This progrmme is meant for class-7 students, in which they are introduced with basic Concepts of mathematics", "show you how to make your most terrifying math topic easy to understand. I promise you! HINT: Read the sections before you study those topics in class. It will make class much more enjoyable! PLEASE NOTE: In no way am I insinuating with this web site/blog that girls are worse in math than boys. I just believe that math examples need to be more geared towards girls, and math should be taught more simply for both girls and boys! I don’t have a problem with boys reading this site, since, for years and years, girls have been reading boys’ math textbooks.", "# 40 boys in a class ## Word Problems 7 & 8 Problem 7: There are 40 boys in a class. One of th... » All original content on these pages is fingerprinted and certified by Digiprove Insert math as $${}$$" ]

[ "9 boys. Possible equations: • $17 = 8 +$? • $17 =$? + $8$ • $8 +$ ? $= 17$ • ? $+ 8 = 17$ • $17 – 8 =$?", "you r also right. - Thanks a bunch! – Andy Apr 10 '13 at 9:00 if boys divided into such as Tom,Jack,Jerry....then $$X={20\\choose 6} - {8\\choose6}-{12\\choose6}-{8\\choose5}{12\\choose1}-{12\\choose5}{8\\choose1}=30800$$ else it's a naive problem just 3 -", "# Four pupils Four pupils divided $1485 so that the second received 50% less than the first, the third 1/2 less than a fourth and fourth$ 154 less than the first. How much money had each of them? Result a = 572 $b = 286$ c = 209 $d = 418$ #### Solution: 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 Need help calculate sum, simplify or multiply fractions? Try our fraction calculator. ## Next similar math problems: 1. Class In a class are 32 pupils. Of these are 8 boys. What percentage of girls are in the class? 2. Profit gain If 5% more is gained by selling an article for Rs. 350 than by selling it for Rs. 340, the cost of the article is: 3. Class In 7.C clss are 10 girls and 20 boys. Yesterday was missing 20% of girls and 50% boys. What percentage of students missing? 4. Apples 14 Maya bought 18 apples. She used 50% of the", "first the arrangements in which no two boys are adjacent. Write down $G\\quad G\\quad G$. This determines $4$ \"gaps\" into which we can slip the boys, one boy per gap. There are $4$ \"gaps\" because we are including the end gaps. There are $\\binom{4}{3}$ ways of choosing $3$ of these gaps. Or else we could slip $2$ boys into one of the two center gaps ($2$ choices), and then slip the remaining boy into one of the $3$ remaining gaps, for a total of $6$ choices. Thus the places for the boys can be chosen in $10$ ways. For each of these ways, we can arrange the boys is $3!$ ways, and then the girls in $3!$ ways, for a total of $360$. -", "Getting a wrong answer on evaluating permutations separately Good Day! I was doing some combinatorics problems when I got stuck. The problem was: Suppose that a teacher selects 4 students from 5 boys and 4 girls. If at least one boy and one girl must be selected, then the number of distinct selecting ways is ----. Now, I went to solve this problem like this: We first select the minimum boys and girls needed. That would be 4 $$\\cdot$$ 5. Now we select the rest of the 2 students which is $${7 \\choose 2}$$. Applying the multiplication principle, it is $$20$$ $$\\cdot$$ $$21$$ = $$420$$. However, the correct answer is 120 which is done by considering the separate cases: • $$3$$ boys, $$1$$ girl • $$2$$ boys, $$2$$ girls • $$1$$ boys, $$3$$ girls and then applying combinations. What am I doing wrong? Thanks • You are over counting. Specifically: if you choose, say, $b_1$ as the initial boy and $b_2$ as part of the second choice, that's the same as choosing $b_2$ initially and then $b_1$ as part of the second choice. – lulu Apr 20 '19 at 11:34 • Note: another good way to do it is to compute the unrestricted number and then subtract off the all boy and all girl cases. Thus $\\binom 94-\\binom", "### Home > CC2MN > Chapter 5 > Lesson 5.2.4 > Problem5-68 5-68. Think about the mathematical process you use as you solve the following problems. Show your work and your solutions. 1. If there are $100$ students in a room and $40$ of them are boys, how many are girls? Since there are $40$ boys out of $100$ students, the rest of the students have to be girls. So subtract $40$ from $100$ to get the number of girls. $100-40=60$ girls 2. If there are $17$ blue and white stripes on a flag and $9$ of them are blue, how many are white? See part (a). 3. If there are $250$ pennies and dimes in a box and $130$ of them are pennies, how many are dimes? See part (a). 4. Now it is time to generalize. Imagine you know how many items you have in a collection of two types of things. If you know how many of one of the items you have, how could you find how many of the other item you have? Describe the process used in part (a).", "Difference between revisions of \"2009 AMC 8 Problems/Problem 23\" Problem On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class? $\\textbf{(A)}\\ 26\\qquad\\textbf{(B)}\\ 28\\qquad\\textbf{(C)}\\ 30\\qquad\\textbf{(D)}\\ 32\\qquad\\textbf{(E)}\\ 34$ Solution If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans. \\begin{align*} x^2+(x+2)^2 &= 394\\\\ x^2+x^2+4x+4 &= 394\\\\ 2x^2 + 4x &= 390\\\\ x^2 + 2x &= 195\\\\ \\end{align*} From here, we can see that $x = 13$ as $13^2 + 26 = 195$, so there are $13$ girls, $13+2=15$ boys, and $13+15=\\boxed{\\textbf{(B)}\\ 28}$ students.", "Therefore, when there is one extra student, the number is divisible by $3$, $4$, $5$ and $6$. This means it is a common multiple of these numbers. The lowest common multiple of $3$, $4$, $5$ and $6$ is $60$, so there must be $60$ students with the extra. Therefore the number of students (without the imaginary student) is $59$.", "# What's wrong with this combination? There are $20$ students. $8$ of them are boys while the other $12$ are girls. I have to pick $4$ of them and I need to have at least one boy and one girl in my pick. There are $4$ spots to choose. Let's name them A, B, C, and D. 1. I have $12$ options to choose a girl for the first spot. 2. I have $8$ options to choose a boy for the second spot. 3. I have $18$ options to choose anyone for the third spot. 4. I have $17$ options to choose last person. From above, I have $12 \\cdot 8 \\cdot 18 \\cdot 17$ which is $29376$. Anyway, the order doesn't matter so I divide $29376$ by $4!$ and get $1224$. But the answer is $4280$ and they get that number by getting all possibilities of choosing a team ${20 \\choose 4}$, then subtract it with possibility of choosing all girls ${12 \\choose 4}$ and choosing all boys ${8 \\choose 4}$. I just don't understand why the method I use leads me to an incorrect answer. • You use the word \"combination\" but then choose to use permutations--why is that? – anakhro Jun 5 '15 at 17:11 • @mathlove Yes, it should be 17 not 12. –", "UPSKILL MATH PLUS Learn Mathematics through our AI based learning portal with the support of our Academic Experts! The number of pens required for 16 girls $$=$$ .", "Let the number of boys in the class be $x$. Hence $\\frac{10}{10+x}\\times\\frac{9}{9+x} = 0.15 = \\frac{3}{20}$. Simplifying gives $1800=3(10+x)(9+x)$ and then $x^{2}+19x-510=0$. Factorising gives $(x+34)(x-15)=0$ and, since $x\\not=-34$, $x=15$. Alternatively, let the number of students in the class be $x$. Hence $\\frac{10}{x}\\times\\frac{9}{x-1} = 0.15 = \\frac{3}{20}$. Simplifying gives $600=x(x-1)$ and then $x^{2}-x-600=0$. Factorising gives $(x+24)(x-25)=0$ and, since $x\\not=-24$, $x=25$. Therefore the number of boys in the class is 15. Students from Comberton Village College sent us these solutions. This problem is taken from the UKMT Mathematical Challenges. View the archive of all weekly problems grouped by curriculum topic View the previous week's solution View the current weekly problem", "# A sum of Rs. 312 was divided among 100 boys and girls in such a way that each boy gets Rs. 3.60 and each girl Rs. 2.40. The number of girls is: Option 1) 50Option 2) Option 3) Option 4) Option 5) Step (i): Let x be the number of boys and y be the number of girls. Given total number of boys and girls = 100 $x+y=100 -------------- (i)$ Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40 The amount given to 100 boys and girls = Rs. 312 $3.60x + 2.40y = 312 -------------- (ii)$ Step (iii): Solving (i) and (ii) $3.60x + 3.60y = 360 ---Multiply\\: (i) \\: by \\: 3.60$ $3.60x + 2.40y = 312 --------- (ii)$ $1.20y = 48$ $y = \\frac{48 }{1.20}$ = 40 $\\inline \\fn_jvn \\Rightarrow$ Number of girls = 40 Exams Articles Questions", "a way that exactly one of the groups A,B,C contains $$2$$ boys. So finally we find $$24\\times 90+3\\times12\\times180=8640$$ possibilities. If the groups A,B,C are not distinguishable then above we are dealing with multiple counting that can be repaired by dividing with $$3!=6$$ (i.e. the number of ways the groups can be ordered) and arrive at $$1440$$ possibilities.", "$$xG+(x+30)(19-G)=1000$$ This simplifies to: $$19x-30G=430$$ $$G$$ is less than 19, so the only integer solution of this equation is $$x=40$$ and $$G=11$$. After the funds are redistributed, each girl therefore receives £48. Let $$B$$ be the amount each boy will now receive. The total is still £1000, so: $$48\\times 11+8B=1000$$ Which can be solved to find $$B=59$$. ##### Extension Mrs. Coldcream still feels that this is unfair. How much more should be added to each girl's scholarship (resizing the boys' accordingly) so that the girls get at least as much as the boys? #### Ten Digit Number 6210001000 has 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights and 0 nines. ##### Extension Are there any more numbers like this? Prove that you have them all. #### 1 comment: 1. Here's another simple solution to the cooked turkey problem: Consider the number $$X679Y$$, a multiple of 72. Since 4 divides 72, we see that $$X679Y$$ must be either 12 or $$16 \\mod 20$$, and hence $$Y$$ is either 2 or 6. From this we deduce that $$X679Y \\mod 40$$ is either 12, 16, 32, or 36. Since 8 also divides our number, we may exclude all but 32. Therefore, $$Y = 2$$. Since our number is also a multiple", "- 3 = 2y + 2 y - 3 = 2 y = 5 Substitute the y-value back into one of those equations to find x: x = 2(5) + 2 x = 10 + 2 x = 12 $There \\ \\ were \\ \\ 12 \\ \\ boys, \\ \\ and \\ \\ the \\ \\ boat \\ \\ cost \\ \\ 60 \\ \\ dollars.$ And you can check what happens with the cost apiece per boy when there are three more boys or two fewer boys. 5. ## Re: WOrd problems Originally Posted by bigmansouf A party of boys purchased a boat and upon payment for the same discovered that if thay had numbered 3 more, they would have paid a dollar apiece less; but if they had numbered 2 less,they would have paid apiece more. How many boys were there, and what did the boat cost? Geezzzzz......SO: x boys bought a boat: each paid y dollars. If instead (x+3) boys bought the boat, each would pay (y-1) dollars. If instead (x-2) boys bought the boat, each would pay (y+1) dollars.", "of scientific calculators sold. Now, try to write the facts that you have as equations. The store sold 35 calculators total... so$x+y=35$. Each simple calculator sells for \\$5, so the amount of money collected for selling simple calculators is $5x$. \\$16 is collected for each scientific calculator sold, so the total from selling scientific calculators is$16y$. All together you collected \\$340. This gives you an equation $5x+16y=340$. Now, what is left is to solve two equations: $$x+y=35$$ and $$5x+16y=340$$ I did it a little differently. 5(35-X) + 16X = 340 175 - 5X + 16X = 340 175 + 11X = 340 11X = 165 X = 15 So 15 scientific and 20 regular calculators. Or you could just as well approach it intuitively; for 16 you need a multiplier yielding a product ending in 0 and for 5 a multiplier yielding a product ending in 0 because there is no product of 16 which ends in ''5'' and the sum of the two products ends in ''0''. In the case of 5 there are only four possibilities less than 35 and 0 doesn't work • List item .", "is : a. $\\sqrt \\pi :1$ b. $1:\\sqrt \\pi$ c. $1:\\pi$ d. $\\pi :1$ 23. In a class, the number of boys is more than the number of girls by 12% of the total strength. The ratio of boys to girls is : a. 11:4 b. 14:11 c. 25:28 d. 28:25 24. A, B and C do a work in 20, 25 and 30 days respectively. They undertook to finish the work together for Rs.2220, then the share of A exceeds that of B by : a. Rs.120 b. Rs.180 c. Rs.300 d. Rs.600", "+ 2y = 300$ $\\\\$ For, $2x +y = 160$ $\\\\$ $y= 160 -2x$ The solution table is $\\\\$ x 50 60 70 y 60 40 20 $\\\\$ For $4x + 2y = 300$, $\\\\$ $y = \\frac{300 - 4x}{2}$ $\\\\$ x 70 80 75 y 10 -10 0 The graphical representation is as follows. None 4. Form the pair of linear equations in the following problems, and find their solutions graphically. $\\\\$ (i) $10$ students of Class X took part in a Mathematics quiz. If the number of girls is $4$ more than the number of boys, find the number of boys and girls who took part in the quiz. (ii) $5$ pencils and $7$ pens together cost Rs $50$, whereas $7$ pencils and 5 pens together cost Rs $46$. Find the cost of one pencil and that of one pen. (i) Let the number of girls be x and the number of boys be y $\\\\$ According to the question, the algebraic representation is $\\\\$ $x + y = 10$ $\\\\$ $x - y = 4$ $\\\\$ For $x + y = 10,$ $\\\\$ $x = 10 - y$ x 5 4 6 y 5 6 4 $\\\\$ For $x-y =4$ $\\\\$ $x = 4 + y$ $\\\\$ x 5 4 3 y 1 0 -1", "anonymous one year ago Word Problem: Please check my solution Eight less than half the number of students in the freshmen class have their own calculators. If 258 calculators are owned by freshmen, how many freshmen are there? • This Question is Open 1. anonymous Let A; total number of students Let B; A/2 $\\frac{ A }{ 2 }-8 = 258$ A =532 2. anonymous Is that right? 3. anonymous", "girls' school,but not anymore...(there were only about 6 students in my class) so now I'm doing my ALs privately,i go to classes outside and there are boys and girls(50 max)& even students from popular schools attend these classes. I do use calculators when I work sums at home and classes,but we're not allowed to use them at our exams,we're provided with log books instead.since our exams are close,I thought I should stop using cals. Anyway thanks for the help 8. Jul 5, 2009 ### vk6kro Very interesting but disturbing as well. I guess they can't help the class sizes being so big. A class of 50 means each student gets almost no individual attention from the teacher. Have you noticed that those who do get the teacher's attention quickly fire off new questions while they have the chance? But they can certainly fix the \"no calculators in exams\" rule. This is just nonsense. If the exam is in arithmetic, sure, you don't allow calculators. But in Physics or Chemistry, the actual calculations are not the object of the exam. It is the logic of assembling the numbers for calculation that matters. Some calculators that have just a few numeric memories do not allow any cheating and even if a small range of \"permitted\" calculators are prescribed, this would" ]

[ "girls' school,but not anymore...(there were only about 6 students in my class) so now I'm doing my ALs privately,i go to classes outside and there are boys and girls(50 max)& even students from popular schools attend these classes. I do use calculators when I work sums at home and classes,but we're not allowed to use them at our exams,we're provided with log books instead.since our exams are close,I thought I should stop using cals. Anyway thanks for the help 8. Jul 5, 2009 ### vk6kro Very interesting but disturbing as well. I guess they can't help the class sizes being so big. A class of 50 means each student gets almost no individual attention from the teacher. Have you noticed that those who do get the teacher's attention quickly fire off new questions while they have the chance? But they can certainly fix the \"no calculators in exams\" rule. This is just nonsense. If the exam is in arithmetic, sure, you don't allow calculators. But in Physics or Chemistry, the actual calculations are not the object of the exam. It is the logic of assembling the numbers for calculation that matters. Some calculators that have just a few numeric memories do not allow any cheating and even if a small range of \"permitted\" calculators are prescribed, this would", "## Newton’s Class More than $93$ percent of the students in Mr. Newton’s math class are girls. If Mr. Newton’s class has at least one boy, what is the least possible size of his class? Source: mathcontest.olemiss.edu 2/24/2014 SOLUTION Ratio Percent $1:2\\qquad 50.00$ $3:4\\qquad 75.00$ $4:5\\qquad 80.00$ $5:6\\qquad 83.33$ $6:7\\qquad 85.71$ $7:8\\qquad 87.50$ $9:10\\quad\\:\\: 90.00$ $10:11\\quad\\:\\: 90.90$ $11:12\\quad\\:\\: 91.67$ $12:13\\quad\\:\\: 92.31$ $13:14\\quad\\:\\: 92.86$ $14:15\\quad\\:\\: 93.33$ $15:16\\quad\\:\\: 93.75$ $16:17\\quad\\:\\: 94.12$ $17:18\\quad\\:\\: 94.44$ $\\cdots$ The least possible size of his class is $15$. Answer: $15$", "Mr. Kilborn’s class has ten girls and ten boys. If one student is chosen randomly from the class, what is the probability that the Question Mr. Kilborn’s class has ten girls and ten boys. If one student is chosen randomly from the class, what is the probability that the student chosen will be a boy? in progress 0 6 hours 2021-07-22T11:31:41+00:00 1 Answers 0 views 0", "Difference between revisions of \"2009 AMC 8 Problems/Problem 23\" Problem On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class? $\\textbf{(A)}\\ 26\\qquad\\textbf{(B)}\\ 28\\qquad\\textbf{(C)}\\ 30\\qquad\\textbf{(D)}\\ 32\\qquad\\textbf{(E)}\\ 34$ Solution If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans. \\begin{align*} x^2+(x+2)^2 &= 394\\\\ x^2+x^2+4x+4 &= 394\\\\ 2x^2 + 4x &= 390\\\\ x^2 + 2x &= 195\\\\ \\end{align*} From here, we can see that $x = 13$ as $13^2 + 26 = 195$, so there are $13$ girls, $13+2=15$ boys, and $13+15=\\boxed{\\textbf{(B)}\\ 28}$ students.", "Let the number of boys in the class be $x$. Hence $\\frac{10}{10+x}\\times\\frac{9}{9+x} = 0.15 = \\frac{3}{20}$. Simplifying gives $1800=3(10+x)(9+x)$ and then $x^{2}+19x-510=0$. Factorising gives $(x+34)(x-15)=0$ and, since $x\\not=-34$, $x=15$. Alternatively, let the number of students in the class be $x$. Hence $\\frac{10}{x}\\times\\frac{9}{x-1} = 0.15 = \\frac{3}{20}$. Simplifying gives $600=x(x-1)$ and then $x^{2}-x-600=0$. Factorising gives $(x+24)(x-25)=0$ and, since $x\\not=-24$, $x=25$. Therefore the number of boys in the class is 15. Students from Comberton Village College sent us these solutions. This problem is taken from the UKMT Mathematical Challenges. View the archive of all weekly problems grouped by curriculum topic View the previous week's solution View the current weekly problem", "of boys. If there are four boys, does that make eleven girls? Twice 4 boys is 8. Three more than 8 is 11. Step 7. Answer the question. There were 4 boys in the study group. Exercise $$\\PageIndex{5}$$ Guillermo bought textbooks and notebooks at the bookstore. The number of textbooks was 3 more than twice the number of notebooks. He bought 7 textbooks. How many notebooks did he buy? Answer 2 Exercise $$\\PageIndex{6}$$ Gerry worked Sudoku puzzles and crossword puzzles this week. The number of Sudoku puzzles he completed is eight more than twice the number of crossword puzzles. He completed 22 Sudoku puzzles. How many crossword puzzles did he do? Answer 7 ## Solve Number Problems Now that we have a problem solving strategy, we will use it on several different types of word problems. The first type we will work on is “number problems.” Number problems give some clues about one or more numbers. We use these clues to write an equation. Number problems don’t usually arise on an everyday basis, but they provide a good introduction to practicing the problem solving strategy outlined above. Exercise $$\\PageIndex{7}$$ The difference of a number and six is 13. Find the number. Answer Step 1. Read the problem. Are all the words familiar? Step 2. Identify what we are", "have the boys seated, and $10!$ ways to have the girls seated. (All right, this part is about permutations.) So the number of solutions is $2\\cdot 10!\\cdot 10!$.", "# The number of calculators Mrs. Hopkins can buy for the classroom varies inversely as the cost of... ## Question: The number of calculators Mrs. Hopkins can buy for the classroom varies inversely as the cost of each calculator. She can buy 24 calculators that cost $60 each. How many calculators can she buy if they cost$80 each? ## Inverse Variation: In mathematics, if y varies inversely as x, we have that {eq}y=\\frac{k}{x} {/eq}. In this inverse variation equation, k is a constant called the constant of variation. We can solve inverse variation problems by finding the constant of variation, plugging the constant of variation into the inverse variation equation, and then using the inverse variation equation to answer the problem. ## Answer and Explanation: Mrs. Hopkins can buy 18 calculators if they cost $80 each. We are given that the number of calculators that Mrs. Hopkins can buy for the classroom varies inversely as the cost of each calculator. If we let y be the number of calculators she can buy, and we let x be the cost of each calculator, then we have that y varies inversely as x, so {eq}y=\\frac{k}{x} {/eq}. The problem states that Mrs. Hopkins can buy 24 calculators that cost$60 each. Therefore, when y = 24, x = 60. Plugging these into", "is one quarter of the number of protractors. If Ms. Zentz buys 65 Items, how many protractors does Ms. Zentz buy? The rest of the table will vary. Ms. Zentz will buy 20 protractors. We can confirm that this number satisfies the conditions of the word problem because the number of protractors is half of the number of rulers, and the number of calculators is one fourth of the number of protractors. Also, 40 + 20 + 5 = 65, so the total matches the total supplies that Ms. Zentz bought. Algebraically: Let c represent the number of calculators Ms. Zentz needs for the year. Then, 8c represents the number of rulers, and 4c represents the number of protractors Ms. Zentz will need throughout the year. 8c + 4c + c = 65 13c = 65 $$\\frac{13 c}{13}=\\frac{65}{13}$$ c = 5 Therefore, Ms. Zentz will need 5 calculators, 40 rulers, and 20 protractors throughout the year. ### Eureka Math Grade 6 Module 4 Lesson 29 Problem Set Answer Key Create tables to solve the problems, and then check your answers with the word problems. Question 1. On average, a baby uses three times the number of large diapers as small diapers and double the number of medium diapers as small diapers. a. If the average baby uses 2,940", "# Lining up students—combinatorics There are $$8$$ boys and $$6$$ girls. In how many ways can they line up so that the front and end of the line are occupied by boys? Someone told me that the answer is $$3832012800$$. Am I missing something? Would the solution not be: We first choose the front and end of the line. Since they must both be occupied by boys, we have $$8 \\cdot 7$$ possibilities. Then, there are $$12$$ people left, with no constraints for any of them. Thus, $$8*7*12!$$. EDIT: I explained my solution, which I have been told is correct. • I agree with you. – Matthew Daly Aug 13 at 7:17 • The number is $8\\cdot 12!$. It's close, but the factor of 7 is certainly missing. – Matthew Daly Aug 13 at 7:23 • Your answer is correct. However, you should explain your reasoning so that readers can discern any errors you may have made. This tutorial explains how to typeset mathematics on this site. Welcome to MathSE. – N. F. Taussig Aug 13 at 8:25 This kind of question is rather silly, but to provide an answer to the question how it can be that somebody said the answer was $$3832012800=8\\times12!$$ while the real answer is $$8*7*12!$$ is that they made a mistake. This kind", "In Miss Quaffley's class, the girls make up more than $45$% of the pupils, but less than $50$%. What is the smallest possible number of girls in her class?", "# The average age of boys in class is 16.66, while the average age of girls is 18.75. Thus the average age of all the 40 students of the class is 17.5. If the difference between the no. of boys and girls is 8, then the no. of girls in the class is: Option 1) Option 2) Option 3) Option 4) 20 Option 5) Data insufficient If the difference between the no. of boys and girls is 8, then the no. of girls in the class is: $B-G$ $16.66 \\times B + 18.75 \\times G = 40 \\times 17.5$ $16.66 (G+B) + 18.75 \\: G = 700$ $16.66 G + 18.8G = 700 -132.8$ $148.2 G = 700$ $35.4G = 567.2$ $G= 16$ Exams Articles Questions", "is still$360, but the number of girls is So now with girls, the new cost is 6 dollars less, or the original cost minus 6, so we can set up an equation like this: Let’s do the math: Understand these problems, and practice, practice, practice! Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems. If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets. You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy! On to Graphing Rational Functions, including Asymptotes – you’re ready! ## 24 thoughts on “Rational Functions and Equations” 1. Hi, I absolutely love this site. However, I’d love it if you could work a bit faster! I’m starting Calculus in a couple of weeks, and I would love more resources. Thanks! • Amy, Thanks so much for your comment. You are absolutely right; I", "Problem solving with fractions Lesson As with all problem solving situations, one of the trickiest things is 'mathematising' the question. Mathematising is finding the mathematics within the words. Unfortunately I can't give you any shortcuts or rules to make this easy for you, (my first suggestion is for you to practice a lot), but there are a few clues you can often find in the questions. I'm just going to show you some questions, and circle the words where the mathematics is hiding! ##### Question 1 Shared equally is a clue to DIVISION. Mathematising this question gives you $\\frac{3}{5}\\div4$35÷​4. ##### Question 2 Full price would be $\\frac{7}{7}$77, so to find the full price we would find $\\frac{1}{7}\\times384$17×384 and then get $\\frac{7}{7}$77 by multiplying by $7$7. ##### Question 3 If $\\frac{1}{2}$12 are boys, then the remaining class are assumed to be girls. So $1-\\frac{1}{2}$112are girls, which equals $\\frac{1}{2}$12. Therefore, because $10$10 boys represents half the class, then there must also be $10$10 girls (the other half of the class). ##### QUESTION 4 In a hockey tournament, Adam's team won $1$1 out of $3$3 games, and did not tie any games. 1. How many games did Adam's team lose? 2. What fraction of the games did Adam's team lose? ##### QUESTION 5 In a survey of $270$270 people, $\\frac{1}{10}$110 said", "+ 2y = 300$ $\\\\$ For, $2x +y = 160$ $\\\\$ $y= 160 -2x$ The solution table is $\\\\$ x 50 60 70 y 60 40 20 $\\\\$ For $4x + 2y = 300$, $\\\\$ $y = \\frac{300 - 4x}{2}$ $\\\\$ x 70 80 75 y 10 -10 0 The graphical representation is as follows. None 4. Form the pair of linear equations in the following problems, and find their solutions graphically. $\\\\$ (i) $10$ students of Class X took part in a Mathematics quiz. If the number of girls is $4$ more than the number of boys, find the number of boys and girls who took part in the quiz. (ii) $5$ pencils and $7$ pens together cost Rs $50$, whereas $7$ pencils and 5 pens together cost Rs $46$. Find the cost of one pencil and that of one pen. (i) Let the number of girls be x and the number of boys be y $\\\\$ According to the question, the algebraic representation is $\\\\$ $x + y = 10$ $\\\\$ $x - y = 4$ $\\\\$ For $x + y = 10,$ $\\\\$ $x = 10 - y$ x 5 4 6 y 5 6 4 $\\\\$ For $x-y =4$ $\\\\$ $x = 4 + y$ $\\\\$ x 5 4 3 y 1 0 -1", "year there were 612 boys in a school. The number decreased by 25 percent this year. How many girls are there in the school if the number of girls is $$166\\frac{2}{3}\\;$$percent of the total number of boys in the school this year? 1. 765 2. 762 3. 798 4. 782 5. None of these Que. 9 In a company, there are 75% skilled workers and remaining ones are unskilled. 80% of skilled workers and 20% of unskilled workers are permanent. If number of temporary workers is 126, then find the number of total workers. 1. 124 2. 388 3. 324 4. 360 5. None of these Que. 10 The average age of 80 boys in a class is 15. The average age of a group of 15 boys in the class is 16 and the average age of another 25 boys in the class is 14. What is the average age of the remaining boys in the class? 1. 11.5 years 2. 14 years 3. 15.25 years 4. 18.75 years 5. None of these Did you like this Average and Percentage Quiz 3 for Banking & Insurance Exams? Let us know! You may also like – • Save 11 hours ago 1 day ago 2 days ago 3 days ago", "UPSKILL MATH PLUS Learn Mathematics through our AI based learning portal with the support of our Academic Experts! The number of pens required for 16 girls $$=$$ .", "After paying his workers for one week, the farmer has$126 left in his bank account. How much money did he have to begin with.... There are 32 students in Mr. Morgan's math class. One student will be randomly selected to present his/her project. If there are 20 boys and 12 girls in the class, what is the probability that the presenter will be a boy? HURRY IF YOUR RIGHT I WILL GIVE YOU A LIKE AND 5 STARS There are 32 students in Mr. Morgan's math class. One student will be randomly selected to present his/her project. If there are 20 boys and 12 girls in the class, what is the probability that the presenter will be a boy? HURRY IF YOUR RIGHT I WILL GIVE YOU A LIKE AND 5 STARS...", "word problems. Since part of their PLC work freed them up to observe lessons in each others' rooms, they decided they would watch Mr. Alexander introduce modeling to his students. So, two days later they gathered in Mr. Alexander's room for the math lesson. Mr. Alexander presented the following problem: Lily and her brother, Scotty, were collecting cans for the recycling drive. One weekend they collected 59 cans and the next weekend they collected 85 cans. How many cans were collected in all? Mr. Alexander went over the problem and drew a rectangular bar divided into two parts on the board, explaining that each part of the rectangle was for the cans collected on one of the weekends and the bracket indicated how many cans were collected in all. Reviewing the problem, Mr. Alexander asked students what was not known, and where the given numbers would go and why. This resulted in the following bar model: The class then discussed what equations made sense given the relationship of the numbers in the bar model. This time many students wrote the equation, 59 + 85 = ?, and solved the problem. In their discussion after the lesson, Mr. Alexander's teammates mentioned that they noticed a much higher degree of interest and confidence in problem solving when Mr. Alexander introduced", "• Save # Simplification Problems with Solutions 3 weeks ago 620 Views NEW Simplification questions have an important role in competitive exams. Students, who are preparing for banking and other similar exams, should be never left or ignore this topic for a better score in the competitive exam. So, here I am sharing important simplification problems with solutions from which you can increase your ability level by learning how to solve these type of questions. You should get more posts related to Simplification Topic for more practice. ## Important Simplification Problems with Solutions Q.1. A sum of Rs. 312 was divided among 100 boys and girls in such a way that each boy gets Rs. 3.60 and each girl Rs. 2.40. The number of girls is: (A) 35 (B) 40 (C) 60 (D) 65 Ans . B Let the number of boys = x. Then, number of girls = (100 – x). ⸫ 3.60x + 2.40 (100 – x) = 312 ⟺ 1.20x = 312 – 240 = 72 ⟺ x = 15. Hence, number of girls = (100 – x) = 40. Q.2. Each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys in a class of 60 students. If the total contribution thus collected is" ]

[ "not do that for the middle one. EDIT: @Dog: 6 inches is exactly the size I'd expect from a 6 inch deep hole. That's the point of the question a lot of people do not seem to understand. ##### Share on other sites John Cuthber 3201 I'd probably call the one in the middle a bead. The point is that if you were to thread the bead/ ring/whatever else you might want to call it onto a tight fitting string then it would hide six inches of that string.", "190 views A string of length $40$ meters is divided into three parts of different lengths. The first part is three times the second part, and the last part is $23$ meters smaller than the first part. Find the length of the largest part 1. $27$ 2. $4$ 3. $5$ 4. $9$ 1 413 views 2 222 views", "string lengths on your guitar?", "away from a common point. The limit for length here depends on the string thickness. A thin enough string will be of the required length, in fact given a thin enough string you could make any length. 3)Yes, for the same reasons stated above.", "together so that they are tangent to one another, and a string is wrapped around the outside. If the pegs' radii are all increased by an inch, how much longer would the string around the pegs be, in inches?", "the string? Express your answer in centimeters....", "string you want to find the length of.", "## Lesson goal: The length of a string Previous: Introduction | Home | Next: Comparing strings Often when working with strings, one needs to know how long a string is. This can be found using the string.len() function. string.len(string)", "average a total body length of 27+14 inches (69.2 cm), with strings tuned a fifth lower than a violin (with a length of about 23+38 inches (59.4 cm)). The viola's larger size is not proportionally great enough to correspond to the strings being pitched as they are, which contributes to its different timbre. Violists need to have hands large enough to be able to accomplish fingering comfortably. The C string has been described by Piston as having a timbre that is \"powerful and distinctive\",[71] but perhaps in part because the sound it produces is easily covered, the viola is not so frequently used in the orchestra as a solo instrument.[72] According to the American physicist John Rigden, the lower notes of the viola (along with the cello and the double bass) suffer from strength and quality. This is because typical resonant frequencies for a viola lie between the natural frequencies of the middle open strings, and are too high to reinforce the frequencies of the lower strings. To correct this problem, Rigden calculated that a viola would need strings that were half as long again as on a violin, which would making the instrument inconvenient to play.[73] The cello, with an overall length of 48 inches (121.9 cm), is pitched an octave below the viola. The proportionally greater", "string tension in guitar, you can: 1. Replace your current strings with a lighter or thinner gauge guitar string set. 2. Tune your guitar a note or half a note down. You would have to transpose your chords to play your music in its original key. 3. You can also transfer your guitar strings to a shorter guitar to relieve some string tension. ### How much tension is on each string of a guitar? Each guitar string usually differs from the other but ranges between 16 to 20 pounds on a 25.5-inch scale length guitar at standard tuning for medium-weight guitar strings. However, this range increases for guitars with longer scale lengths and decreases for shorter scale lengths. Heavier and lighter guitar strings also increase and decrease the required string tension, respectively. ### How do I calculate the tension in the E string of my 25.5-inch guitar? To find the tension in your guitar's E string; 1. Double your guitar's scale length: 25.5 × 2 = 51 inches. 2. Multiply 51 inches by 329.6 Hz, which is E4's frequency: 51 × 329.6 = 16,809.6 in/s. 3. Square this value and multiply it by the unit weight of your guitar string in pounds per inch. Let's say that its packaging indicates it has 0.00002215 lb/in-unit weight: (16,809.6)² × 0.00002215", "length of the string doesn't change.", "the treble strings near an end, not at the middle as I imagined, so the string much stiffer there rejects the hammer faster than what I computed. Though, I can't really believe hammer heads striking at 8% of a 48mm long string, so I'll give it a look next time I access a piano. Apparently the ancestors concentrated on the effect on the timbre, not on the (perceived) sound height.", "extensively to teach younger players. The size of these is described by a \"conventional\" fraction that has no mathematical significance. For example, a 7/8 violin has a scale of about 317 mm, a 3/4-size instrument a scale of 307 mm, a half-size one 287 mm, and a quarter-size one 267 mm. 1/8, 1/10, 1/16 and 1/32 and even 1/64 violins also exist, becoming progressively smaller, but again in no proportional relationship. (A full-size instrument is described as 4/4.) Cellos exist in a smaller range of sizes than violins, with 4/4, 3/4, 1/2, 1/4, 1/8, and 1/10 being reasonably common. As with the violin, the Stradivarius scale is regarded as standard for orchestral work; This is about 27.4 in (695 mm). Violas are commonly described in terms of their body length rather than—as with other violin-family instruments—by a fraction. There are two reasons for this. First, unlike that of the violin and the cello, the viola scale length has not standardised, but rather advanced player use whatever scale length best suits them. Secondly, student sizes are not as often required, as most viola players who start learning at a young age start on the violin. Common sizes include 17 in (43 cm), 16.5 in (42 cm), 16 in (41 cm), 15.5 in (39 cm), 15 in (38 cm), 14", "generation before by Jacob Stainer (c. 1617–1683). Later makers have been unwilling to deviate from this. Smaller scale instruments are used extensively to teach younger players. The size of these is described by a \"conventional\" fraction that has no mathematical significance. For example, a 7/8 violin has a scale of about 317 mm, a 3/4-size instrument a scale of 307 mm, a half-size one 287 mm, and a quarter-size one 267 mm. 1/8, 1/10, 1/16 and 1/32 and even 1/64 violins also exist, becoming progressively smaller, but again in no proportional relationship. (A full-size instrument is described as 4/4.) Cellos exist in a smaller range of sizes than violins, with 4/4, 3/4, 1/2, 1/4, 1/8, and 1/10 being reasonably common. As with the violin, the Stradivarius scale is regarded as standard for orchestral work; This is about 27.4 in (695 mm). Violas are commonly described in terms of their body length rather than by a conventional fraction. There are two reasons for this. Firstly, unlike that of the violin and the cello, the viola scale length has not standardised, but rather an advanced player will use whatever scale length best suits them. Secondly, student sizes are not as often required, as most viola players who start learning at a young age would start on the violin. Common sizes", "a size to withstand? How long is a string? phinds Mister T", "longer than half the length of a full size. The smaller cellos are identical to standard cellos in construction, range, and usage, but are simply scaled-down for the benefit of children and shorter adults. Cellos in sizes larger than 44 do exist, and cellists with unusually large hands may require such a non-standard instrument. Cellos made before approximately 1700 tended to be considerably larger than those made and commonly played today. Around 1680, changes in string-making technology made it possible to play lower-pitched notes on shorter strings. The cellos of Stradivari, for example, can be clearly divided into two models: the style made before 1702, characterized by larger instruments (of which only three exist in their original size and configuration), and the style made during and after 1707, when Stradivari began making smaller cellos. This later model is the design most commonly used by modern luthiers. The scale length of a 44 cello is about 70 cm (27+12 in). The new size offered fuller tonal projection and a greater range of expression. The instrument in this form was able to contribute to more pieces musically and offered the possibility of greater physical dexterity for the player to develop technique.[26] Approximate dimensions for 44 size cello[27] Average size Approximate width horizontally from A peg to C peg ends 16.0", "in (36 cm), and less commonly 12 in (30 cm), smaller than a standard violin; These measurements are nominal and approximate. At least one of the surviving Stradivarius violas has a scale length of 14.25 in (362 mm). ### Double bass There is some variation in the scale length of an orchestral double bass, generally in the range 41.3 – 43.3 inches (105 – 110 cm). There are also smaller versions of this \"full scale\" double bass with the same scale length but with a smaller sound box, intended for other musical idioms. Smaller scale instruments are also quite commonly used by full-sized players in jazz, folk music and similar ensembles. The system of conventional fractions is taken to its logical conclusion with string bass sizes, in that a full-size (4/4) bass is uncommon. Most basses are 3/4 or 7/8, and younger players can use 1/2 or even 1/4 size instruments. ## Classical guitar Like that of the violin, the scale of the classical guitar was standardized by the work of its most famous maker. Antonio De Torres (1817–1892) used a scale length of 25.6 in (650 mm), and later makers have followed suit. However, from the mid- 20th Century luthiers seeking increased volume have moved to a 26 in (660mm) scale, which is now the standard for", "• # question_answer 33) A nylon string $18\\,\\,m$ long is cut into equal pieces of $0.2\\,\\,m$ each. How many equal pieces are obtained? A) $180$ B) $90$ C) $9$ D) $360$ Total length of the string $=18\\,\\,m$ Length of each equal part$=0.2\\,\\,m$ $\\therefore$No. of pieces$=\\frac{18\\,\\,m}{0.2\\,\\,m}=\\frac{180}{2}=90$", "include 17 in (43 cm), 16.5 in (42 cm), 16 in (41 cm), 15.5 in (39 cm), 15 in (38 cm), 14 in (36 cm), and less commonly 12 in (30 cm), smaller than a standard violin; These measurements are nominal and approximate. At least one of the surviving Stradivarius violas has a scale length of 14.25 in (362 mm). ### Double bass There is some variation in the scale length of an orchestral double bass, generally in the range 41.3 – 43.3 inches (105 – 110 cm). There are also smaller versions of this \"full scale\" double bass with the same scale length but with a smaller sound box, intended for other musical idioms. Smaller scale instruments are also quite commonly used by full-sized players in jazz, folk music and similar ensembles. The system of conventional fractions is taken to its logical conclusion with string bass sizes, in that a full-size (4/4) bass is uncommon. Most basses are 3/4 or 7/8, and younger players can use 1/2 or even 1/4 size instruments. ## Classical guitar Like that of the violin, the scale of the classical guitar was standardized by the work of its most famous maker. Antonio De Torres (1817–1892) used a scale length of 25.6 in (650 mm), and later makers have followed suit. However, from", "century, distinct efforts were made to lengthen its oval form. At -about the same time, or a little later, the big Minnesinger Fiddle begins to appear. With the advent of this instrument, we first begin to feel that the modern violin is \"on the way.\" 'i'he sides are already receiving first a slight, and then a continually greater curve inwards, in order to give freer play to the bow. In one point only it deviates from the path of progress. It begins to adopt a feature which was, in the free-string later days, to become characteristic of the Bass-Lute hanging clear of the nut and fingerboard, incapable therefore of being fingered, and only useful as a bourdon-bass. Before passing to the Viols, which were the immediate parents of our modern Violin-Family, we must pause to describe a once popular bowed stringed-instrument, the Tromba Marina. If the reader will imagine a huge wooden metronome case six or seven feet long, of which the base that is, the part nearest the ground was about six or seven inches, and the apex about two or three inches wide, he will have some idea of this monstrosity. It had one string only, and this string was stretched down the entire length of the enormous wooden pyramid. Occasionally it was provided with such" ]

[ "$$24.0$$ - $$25.9$$ $$3$$ $$\\frac{3}{12} = 25\\%$$ $$26.0$$ - $$27.9$$ $$1$$ $$\\frac{1}{12} \\approx 8.3\\%$$ 2. To find the mean foot length for Sundip and his friends, we add up all the foot lengths and divide by $$12$$. The sum of the foot lengths is $$273.2\\,$$cm, and so the average is $$\\frac{273.2}{12} \\approx 22.8\\,$$cm. The average for Sundip and his friends is slightly lower than the average of $$22.9\\,$$cm for $$11$$-year-olds. 3. The table reveals that the foot lengths of Sundip and his friends are concentrated in their middle range, with only one person at each end of the possible lengths. 4. The average foot length of Sundip and his friends is about $$1$$ cm longer than the average of $$50$$ years ago.", "## Basic College Mathematics (9th Edition) Estimate: 2 inches Exact: $2\\frac{1}{2}$ inches From the table, we see that the maximum jaw capacity is $3\\frac{3}{8}$, and the minimum is $7/8$. We subtract and find the difference. Estimate: $3-1=2$ inches Exact: $\\displaystyle 3\\frac{3}{8}-\\frac{7}{8}=2\\frac{11}{8}-\\frac{7}{8}=2\\frac{4}{8}=2\\frac{1}{2}$ inches", "$\\frac{5K}{4}$ 10. $35.4\\;\\mathrm{mm}$ 11. $2.3\\;\\mathrm{inches}$ 12. $1\\;\\mathrm{inch}= 16\\;\\mathrm{feet}$ or equivalent 13. $5.4\\;\\mathrm{inches}$ Feb 22, 2012 Aug 22, 2014", "length - 1.", "# Math Help - Percentage Problem! 1. ## Percentage Problem! Q: The length of a rod is 0.45m. It expands to a length of 0.47m. Find the expansion as a percentage of its original length. 2. $\\frac{\\text{initial length}}{\\text{final length}}=\\frac{0.45}{0.47} = 95.745 \\%$ 3. It seems too simple, thanks though! 4. I think percentage increase in length =[change in length/orig. length] X100 so, it should be=[0.02/0.45] X 100 am I correct?? 5. Ans- 4.44% 6. Thank you so much sumit2009, you've just confirmed the answer for me. I thought it was 4.44% nice one!", "6 inches. RizerMix Expert Step 1 She has $3\\frac{1}{2}=\\frac{7}{2}=3.5$ yards and uses $\\frac{2}{3}=0.6$ So she has $\\frac{1}{3}=0.3$ left that's $\\frac{7}{2}×\\frac{1}{3}=\\frac{7}{6}$ $1\\frac{1}{6}=1.16$ yards, Do you have a similar question?", "inch.", "and 5 inches in width, and she wants to have the photo enlarged. In order to ensure that none of the image is lost along the sides, she needs the dimensions to remain proportional. If the enlarged photograph is to be 15 inches wide, what will its length be? 24 inches 28 inches 30 inches 32 inches The dimensions of the initial photograph may be written as the ratio $$\\frac{8}{5}$$. Because the enlarged photograph will have a new width of 15 inches instead of 5, write a new ratio with 15 in the denominator. These two ratios will be equal because the photo enlargement will have the same proportions as the original. $$\\frac{8}{5}=\\frac{?}{15}$$ To determine the missing value of the new length, notice that the width is increasing by a factor of three. To preserve the proportion of the dimensions, the length must also be multiplied by three. $$8\\times3=24$$ $$5\\times3=15$$ $$\\frac{8}{5}=\\frac{24}{15}$$ Therefore, the length of the enlarged photograph is 24 inches. Question #10: Average. Determine the average of the following set of numbers: 12, 15, 19, 25, 31, 36. 15% 18% 20% 23% The average is the weighted “center value” of a set of numbers, it can be found by adding up all the numbers and then dividing that sum by the number of values in the set.", "weight $$-1$$ count its length.", "line plots shown above, the second shows less variation among the diameters, and clearly shows that the most common diameter, to the nearest quarter-inch, is $\\frac12$ inch. The first line plot gives more information about the diameters, and shows a cluster of buttons with diameters that round to $\\frac38$ inch to $\\frac34$ inch.", "## anonymous one year ago your looking at a catalog that sells cooking supplies. some of the items are drawn in scale at a ratio of 1:12 , in cm. if a scale drawing of one of these items is 3 cm wide, how wide is the actual item in cm? A proportion would be a nice place to start. $\\frac{ 1 }{ 12 } = \\frac{3}{x}$", "the length of the object to nearest quarter inch is $3\\frac{3}{4}$ inches.", "# A blue paper clip is 1/6 inch wide. A silver paper clip is 3/8 inch wide, and a red paper clip is 1/3 inch wide. What color paper clip has the smallest width? Mar 7, 2018 The blue paper clip has the smallest width. #### Explanation: Set them all to the same common denominator. The common denominator is 24. $\\frac{1}{6} = \\frac{4}{24}$ $\\frac{3}{8} = \\frac{9}{24}$ $\\frac{1}{3} = \\frac{8}{24}$ $\\frac{4}{24}$ is the smallest fraction of the three, thus making $\\frac{1}{6}$ the smallest width.", "of wood have an average length of 124 inches and [#permalink] 12 Jul 2014, 11:22 Similar topics Replies Last post Similar Topics: 37 Set R contains five numbers that have an average value of 55 22 02 Oct 2010, 11:23 A rectangular frame is 18 inches in length and 15 inches in 7 07 Nov 2006, 13:55 Five pieces of wood have an average of 124centimeters and 6 05 Nov 2006, 07:17 14 Five pieces of wood have an average length of 124cm and a 86 24 Oct 2005, 23:08 Display posts from previous: Sort by # Five peices of wood have an average length of 124 inches and 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®.", "the median length, in inches, of the three rods? A. 71 B. 77 C. 80 D. 83 E. 89 If you know of any similar problems, please post. Because it is mentioned that the average length of the three rods is 77 inches, we can assume there are three rods each of length 77 inches. Now we have been told that the shortest rod has a length of 65 inches. Thus, we can take 12 inches from one of the rods and will have to adjust it amongst the other two remaining rods. Now, if we divide 12 in any proportion other than 6 and 6, we will not have the maximum value for the median. Imagine, we redistribute 12 by giving 5 to one and 7 to another. This gives 65, 82 and 84. Thus, giving 6 to both the rods, we have (77+6) = 83 inches for both the rods. D. _________________ SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1859 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Followers: 23 Kudos [?]: 967 [0], given: 193 Re: Three straight metal rods have an average (arithmetic mean) [#permalink] 21 Aug 2013, 20:41 Smallest rod = 65 Average of all 3 rods = 77 Lets consider middle rod = 77, so", "= $\\frac{(13×1)}{(3×3)}$ = $\\frac{13}{9}$ = 1$\\frac{4}{9}$ inches", "### Home > MC2 > Chapter 4 > Lesson 4.2.2 > Problem4-41 4-41. Erika is tracking the depth of the water in her local creek. Her first nine measurements are below, in inches. . $16\\space15\\space13\\space12\\space17\\space 14\\space11\\space9\\space 11$ 1. What is the median of her data? The median is the middle number of an ordered set of data. Order the measurements in ascending order, from least to greatest. From both ends, count until you get to a middle number. Remember, if there are two middle numbers, the median is the average of the two. $13$ inches. 2. Erika's next three measurements are $9$ inches, $10$ inches, and $9$ inches. What is the new median? See (a). Remember to include $9$, $10$, and $9$ inches in your new calculation. Use the eTool below to determine the median. Click the link at right for the full version of the eTool: MC2 4-41 HW eTool", "length of the sticks in Box V, What is the average length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V. A) 3 B) 6 C) 12 D) 18 E) 25 [Reveal] Spoiler: OA Last edited by Bunuel on 03 Oct 2017, 22:27, edited 1 time in total. Edited the OA. Senior Manager Joined: 18 Aug 2009 Posts: 317 Re: Box W and Box V each contain several blue sticks and red [#permalink] Show Tags 08 Oct 2009, 16:52 $$(R+19)-(R-6)=25$$. Math Expert Joined: 02 Sep 2009 Posts: 43792 Re: Box W and Box V each contain several blue sticks and red [#permalink] Show Tags 08 Oct 2009, 17:02 Expert's post 1 This post was BOOKMARKED hgp2k wrote: $$(R+19)-(R-6)=25$$. Yes the choices are wrong. Correct answer is 25: {Average lengths of sticks in W} - 19 = Length of red; {Average lengths of sticks in V} + 6 = Length of red; {Average W}-19={Average V}+6 --> {Average W}-{Average V}=25. _________________ Senior Manager Joined: 30 Nov 2008 Posts: 483 Schools: Fuqua Re: Box W and Box V each contain several blue sticks and red [#permalink] Show Tags 08 Oct 2009, 19:16 Yes....even I got the same value. I reread the question couple times to ensure I am not making", "2 years, the average customer makes 16 purchases. 2 inches long and 4. For two of these species, brook and brown trout, there is one set of regressions for stream dwellers and another set for lake dwellers (which tend to be significantly plumper at larger sizes). $g'\\left( x \\right) = - 6 + 20\\sin \\left( {2x} \\right)$ Now, the function will not be changing if the rate of change is zero and so to answer this question we need to determine where the derivative is zero. Because male and female skeletons differ , you will need separate graphs and calculations for each sex. With millions of qualified respondents, SurveyMonkey Audience makes it easy to get survey responses from people around the world instantly, from almost anyone. ) What is the relationship between the size of a flaccid (soft) penis and an erect (hard) penis? There is a much wider range of size in men's penises when flaccid, with the average ranging from 1 to 4 inches. What's the Difference Between Weather and Climate?. average is 4,233 TEU • Average sizes range from 32-76% of the maximum size -2,000 4,000 6,000 8,000 10,000 12,000 14,000. Of the 10 most widely surveyed countries, only the UK guessed an average size below the real average, estimating that the average erect penis", "Browse Questions # Two rods are of lengths $3.161 \\pm 0.3$ cm and $1.121 \\pm 0.1$ cm. What is their combined length? Can you answer this question? $4.282 \\pm 0.4$ cm" ]

[ "cm of string. She cuts the string into two pieces. One piece is 72$$\\frac{1}{5}$$ cm long. How long is the other piece? Ms. Morton has 167$$\\frac{3}{5}$$ cm of string One piece is 72$$\\frac{1}{5}$$ cm long 167$$\\frac{3}{5}$$ – 72$$\\frac{1}{5}$$ = 95$$\\frac{2}{5}$$ So, the other piece is 95$$\\frac{2}{5}$$.", "190 views A string of length $40$ meters is divided into three parts of different lengths. The first part is three times the second part, and the last part is $23$ meters smaller than the first part. Find the length of the largest part 1. $27$ 2. $4$ 3. $5$ 4. $9$ 1 413 views 2 222 views", "# Thread: How long is a piece of string? 1. ## How long is a piece of string? Hey guys, I am seriously stuck on how to answer this question. Any tips on how to answer these questions would be greatly appreciated! my attempt is that the length of string is 4x + 4y, but I feel like I am incorrect as it is not in terms of x. I'm just overall really stuck. 2. ## Re: How long is a piece of string? looking at the diagram it appears that $L(x) = 4x+4y$ which is what you got. We're told the volume $V = 8000~cm^3$ $V = x^2 y$ $y = \\dfrac{V}{x^2} = \\dfrac{8000}{x^2}$ $L(x) = 4x + \\dfrac{32000}{x^2}$ can you proceed from here? 3. ## Re: How long is a piece of string? Hi there romsek, thankyou so much! is the dL/dx just asking me to differentiate the L(x)=4x+(32000/x2) 4. ## Re: How long is a piece of string? Originally Posted by ifailedmaths Hi there romsek, thankyou so much! is the dL/dx just asking me to differentiate the L(x)=4x+(32000/x2) yep", "• # question_answer 33) A nylon string $18\\,\\,m$ long is cut into equal pieces of $0.2\\,\\,m$ each. How many equal pieces are obtained? A) $180$ B) $90$ C) $9$ D) $360$ Total length of the string $=18\\,\\,m$ Length of each equal part$=0.2\\,\\,m$ $\\therefore$No. of pieces$=\\frac{18\\,\\,m}{0.2\\,\\,m}=\\frac{180}{2}=90$", "about $${1}\\frac{1}{2}$$ inch long. ## Measure length: Quarter inch What will be the length of the object if the object ends exactly halfway between the half and a whole number? We can measure the length up to half an inch. Similarly, if the end of the object is in the middle of the half and a whole number, we will change the ruler. Take a ruler which consists of the quarter. Otherwise, we can mark the quarter as shown on this scale, For example, the string shown in the picture ends at the middle of $$1\\frac{1}{2}$$ and 2. So, the length of the string is closer to a quarter inch. The string is $$\\frac{7}{4}$$or $${1}\\frac{3}{4}$$ inches long. The string length can also be represented as 1 whole and three $$\\frac{1}{4}$$ inches, or $$1\\frac{3}{4}$$ inches. ## Estimation of Length for Quarter inches Again, if the string is also between a quarter-inch and a half or a whole number, the quarter-inch marking that is closest to the end of the string is taken as the length of the string. For example, The thread shown ends between 0 and $${2}\\frac{1}{2}$$. But we can see that the length of the pencil ends at a quarter mark and is closer to the end of the string. So, the string is about $${2}\\frac{1}{4}$$ inches", "away from a common point. The limit for length here depends on the string thickness. A thin enough string will be of the required length, in fact given a thin enough string you could make any length. 3)Yes, for the same reasons stated above.", "together so that they are tangent to one another, and a string is wrapped around the outside. If the pegs' radii are all increased by an inch, how much longer would the string around the pegs be, in inches?", "of the string. Being conservative, I'm going to assume, without brilliant justification, that the gaps can be accounted for by pretending that the diameter of the string is half as large again as its assumed diameter - i.e. I'll use a diameter of 6 mm instead of 4 mm. Now, the total volume the string of length $L$ occupies is $\\pi (d/2)^2 L$, which is the \"effective\" circular cross-sectional area of the string multiplied by its length. Then we have $\\pi \\left(\\frac{d}{2}\\right)^2 L = \\frac{4}{3} \\pi R^3$ or $L=\\frac{16R^3}{3d^2}$ Plugging in $R=2$ m and $d=6$ mm gives $L=1185$ km. To the nearest order of magnitude, the string is, again, 1000 km long. The previous calculation, by the way, works out at $L=2000$ km. The important point is that we can be fairly sure that the length of the string isn't 100 km or 10,000 km. It's likely to be somewhere in the thousands-of-kilometers range, even if the amount of \"gap\" between windings was several times larger than what I assumed. The point is: we're only making a (reasonably good) back-of-the-envelope estimate of the length here. We could make a better estimate if we knew the actual diameter of the string, obviously. We could make an even better estimate if we investigated just how much \"gap\" you get when", "generation before by Jacob Stainer (c. 1617–1683). Later makers have been unwilling to deviate from this. Smaller scale instruments are used extensively to teach younger players. The size of these is described by a \"conventional\" fraction that has no mathematical significance. For example, a 7/8 violin has a scale of about 317 mm, a 3/4-size instrument a scale of 307 mm, a half-size one 287 mm, and a quarter-size one 267 mm. 1/8, 1/10, 1/16 and 1/32 and even 1/64 violins also exist, becoming progressively smaller, but again in no proportional relationship. (A full-size instrument is described as 4/4.) Cellos exist in a smaller range of sizes than violins, with 4/4, 3/4, 1/2, 1/4, 1/8, and 1/10 being reasonably common. As with the violin, the Stradivarius scale is regarded as standard for orchestral work; This is about 27.4 in (695 mm). Violas are commonly described in terms of their body length rather than by a conventional fraction. There are two reasons for this. Firstly, unlike that of the violin and the cello, the viola scale length has not standardised, but rather an advanced player will use whatever scale length best suits them. Secondly, student sizes are not as often required, as most viola players who start learning at a young age would start on the violin. Common sizes", "## two strings have equal length. One of the strings is shaped, with no overlap, into a circle. The other string is shaped, ##### This topic has expert replies Moderator Posts: 2042 Joined: 15 Oct 2017 Followed by:6 members ### two strings have equal length. One of the strings is shaped, with no overlap, into a circle. The other string is shaped, by BTGmoderatorLU » Wed Jan 12, 2022 5:46 pm 00:00 A B C D E ## Global Stats Source: GMAT Prep Two strings have equal length. One of the strings is shaped, with no overlap, into a circle. The other string is shaped, with no overlap, into a square. What is the ratio of the area of the region enclosed by the circle to the area of the region enclosed by the square? A. $$1 : 2$$ B. $$\\pi : 2$$ C. $$1 : 1$$ D. $$4 : \\pi$$ E. The ratio depends on the length of the two strings The OA is D Master | Next Rank: 500 Posts Posts: 126 Joined: 05 May 2012 Thanked: 1 times ### Re: two strings have equal length. One of the strings is shaped, with no overlap, into a circle. The other string is sha by gmatter2012 » Mon Jan 17, 2022 7:59 am BTGmoderatorLU wrote: Wed Jan", "string lengths on your guitar?", "longer than half the length of a full size. The smaller cellos are identical to standard cellos in construction, range, and usage, but are simply scaled-down for the benefit of children and shorter adults. Cellos in sizes larger than 44 do exist, and cellists with unusually large hands may require such a non-standard instrument. Cellos made before approximately 1700 tended to be considerably larger than those made and commonly played today. Around 1680, changes in string-making technology made it possible to play lower-pitched notes on shorter strings. The cellos of Stradivari, for example, can be clearly divided into two models: the style made before 1702, characterized by larger instruments (of which only three exist in their original size and configuration), and the style made during and after 1707, when Stradivari began making smaller cellos. This later model is the design most commonly used by modern luthiers. The scale length of a 44 cello is about 70 cm (27+12 in). The new size offered fuller tonal projection and a greater range of expression. The instrument in this form was able to contribute to more pieces musically and offered the possibility of greater physical dexterity for the player to develop technique.[26] Approximate dimensions for 44 size cello[27] Average size Approximate width horizontally from A peg to C peg ends 16.0", "average a total body length of 27+14 inches (69.2 cm), with strings tuned a fifth lower than a violin (with a length of about 23+38 inches (59.4 cm)). The viola's larger size is not proportionally great enough to correspond to the strings being pitched as they are, which contributes to its different timbre. Violists need to have hands large enough to be able to accomplish fingering comfortably. The C string has been described by Piston as having a timbre that is \"powerful and distinctive\",[71] but perhaps in part because the sound it produces is easily covered, the viola is not so frequently used in the orchestra as a solo instrument.[72] According to the American physicist John Rigden, the lower notes of the viola (along with the cello and the double bass) suffer from strength and quality. This is because typical resonant frequencies for a viola lie between the natural frequencies of the middle open strings, and are too high to reinforce the frequencies of the lower strings. To correct this problem, Rigden calculated that a viola would need strings that were half as long again as on a violin, which would making the instrument inconvenient to play.[73] The cello, with an overall length of 48 inches (121.9 cm), is pitched an octave below the viola. The proportionally greater", "string of length 3 by adding either an A or a B so there are $4 \\times 2 = 2^2 \\times 2 = 2^3 = 8$ strings of length 3. Here they are. How many strings are there of length 12? Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "# Q.11 A piece of strings is 30 cm long. What will be the length of each side if the string is used to form: (a) a square? (b) an equilateral triangle? (c) a regular hexagon? (a) Perimeter of square = 30 cm 4 × side = 30 Side = $$\\frac{30}{4}$$ Side = 7.5 cm (b) Perimeter of an equilateral triangle = 30 cm 3 × side = 30 Side = $$\\frac{30}{10}$$ Side = 10 cm (c) Perimeter of a regular hexagon = 30 cm 6 × side = 30 Side = $$\\frac{30}{6}$$ Side = 5 cm", "## Lesson goal: The length of a string Previous: Introduction | Home | Next: Comparing strings Often when working with strings, one needs to know how long a string is. This can be found using the string.len() function. string.len(string)", "Joan holds a $26$-inch long piece of light-weight string with a heavy bead on it. One end is in her right hand; the other is in her left hand. Initially, her hands are together. How many inches apart must she pull her hands, keeping them at the same height, so the bead moves upward by $8$ inches? Source: NCTM Mathematics Teacher 2006 SOLUTION By the Pythagorean theorem $5^2+x^2=13^2$ $x^2=13^2-5^2$ $x=\\pm\\sqrt{144}$ $=\\pm 12$ Her hands are apart by $2\\times 12=24$ inches. Answer: $24$ inches", "is the length of the string , in inches?... ### If the outliers are not included, what is the mean of the data set? 72, 82, 3, 55, 62, 73, 90, 80, 151, 59, 46, 31 If the outliers are not included, what is the mean of the data set? 72, 82, 3, 55, 62, 73, 90, 80, 151, 59, 46, 31... -- 0.010537--", "have an average (arithmetic mean) lengt [#permalink] ### Show Tags 24 Oct 2016, 08:11 SamsterZ wrote: Bunuel wrote: Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope? (A) 82 (B) 118 (C) 120 (D) 134 (E) 152 Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> $$a\\leq{b}\\leq{c}\\leq{d}\\leq{e}\\leq{f}\\leq{g}$$. The average length = 68 centimeters --> the total length 7*68 centimeters. The median = 84 centimeters --> d=84. The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> $$g=4a+14$$. We need to maximize g. Now, to maximize g, we need to minimize all other terms. The minimum value of b and c is a and the minimum value of e and f is median=d. Thus we have that $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$. What i do not understand. How are you getting to a = 30? I understand everything, however not this part Could please explain this to me? By", "+0 0 115 2 +195 As shown below, 7 circular pegs of the same radius are bundled together so that they are tangent to one another, and a string is wrapped around the outside. If the pegs' radii are all increased by 2 inches, how much longer would the string around the pegs be, in inches? Your answer should be a decimal rounded to the nearest hundredth. Mar 5, 2022 #2 +31 +1 The string can be divided into two different parts, the straight part and the curving part (when tangent to a circle). There are 6 straight parts, and each has a length of 2r. The total length is 2r * 6 = 12r. The curving part adds up to the length of one of the circles, so its length should be $$\\pi r^2$$. Together the length is $$12r+\\pi r^2$$. If r is incremented by 2, we just have to replace r with (r+2). The length after incrementing is $$12(r+2)+\\pi(r+2)^2$$. Simplify, the length is $$\\pi r^2+12r+4 \\pi r +28$$. The difference between the two lengths are $$(\\pi r^2+12r+4 \\pi r +28) - (\\pi r^2+12r)=4 \\pi r+28$$. And the string would be that much longer. Mar 5, 2022" ]

[ "the value of x is 7.", "= 7", "$x = 5$ $x = 7$", "$EG=7$EG=7, find the value of $m$m.", "+0 # what times what equals 7. If 7 is 7 then that means that 7 is also 7. So knowing that what times what equals 7. 0 2489 2 what times what equals 7. If 7 is 7 then that means that 7 is also 7. So knowing that what times what equals 7. May 5, 2017 #1 +1 Approximately 2.6457513110645905905016157536393. This is called a square root. You can calculate them by typing into the web2.0calc \"sqrt(X)\" where X is whatever number you want. For example, sqrt(25) will give you 5, because 5*5=25. I hope I was helpful :) May 7, 2017 #2 +7220 0 Exact value = $$\\sqrt7$$ MaxWong May 7, 2017", "the $$n$$-th decimal value of $$a_n$$ and see whether smooshing them together and swapping the 4s and 7s does anything.", "$r = 5\\cdot \\sqrt{5}$ 7. I'll leave the rest for you.", "n't quite understand your logic for \\$ n 5!", "always be 5, $\\E[5] = 5$.", "## Intermediate Algebra (6th Edition) We are told that $\\sqrt[n] a^{n}=a$, when n is an odd positive integer. In this case, $a=-7$. $\\sqrt[5] (a)^{5}=-7$", "7 = 17 \\cr} 5.", "is the value of numbers? => [1, 2, 3, 4, 5] `", "+ -7hatj$ If $u=< -3,6 >$, what is the value of $5u$? 1. $< -15,30 >$ 2. $< 15,-30>$ 3. $< 2,11 >$ 4. $< -8,1 >$", "##### Question 7.9.4 If $3n \\equiv 5 \\pmod{8}$ and $n$ is a whole number between $80$ and $120$, find all possible values of $n$. Show hint", "(2) $$\\sqrt{4n}$$ = 4 we are looking for the value of n. statement 1: $$\\sqrt{n}$$ = 2 = n = 4. Sufficient. to get the value of n , square both sides of the equation. Statement 2: $$\\sqrt{4n}$$ = 4 = n=4. The same as 1. Re: What is the value of n? [#permalink] 16 Nov 2018, 10:57 Display posts from previous: Sort by", "= 7.$$ Last edited:", ": $?(\\sqrt2)=7/5$, for instance.", "at $\\sqrt {s }$ = 7 TeV", "positive, we must have $n = 5$. The answer is 5.", "large enough $$n$$ and since you have checked up to $$n = 7.5 \\times 10^7$$, I am confident that it holds for all $$n \\ge 5$$." ]

[ "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "in $T$, their signs are opposite; so the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 * 2^6 = \\boxed{448}$.", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "our answer is $m+n=61+4=\\boxed{065}$.", "$4n=96$ 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", "+ 2\\\\&= 52 + 2\\\\&= 54\\\\\\end{aligned} Hence, the integers are $\\boxed{52}, \\boxed{53}$ and $\\boxed{54}.$", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "b \\right)$$ are also both integers. Well, it looks like there's no doubt about where the $$\\sqrt{5}$$ and the $$\\sqrt{7}$$ came from. $$\\sqrt{5} \\text{ and } \\sqrt{7} \\text{ came from the } \\sqrt{a} \\text{ and the } \\sqrt{b}$$ So there you have it! Now that you know that $$a = 5$$ and $$b = 7$$ (or vice versa... it doesn't actually matter), then you can figure out that their sum is $$5 + 7 = \\boxed{12}.$$", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "$\\sqrt{(5-0)^2 + (0-0)^2 + (3-0)^2} = \\sqrt{34}$, which is option $\\boxed{D}$", "value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\\boxed{\\textbf{(E)} \\ 10 \\sqrt{26} - 51}$ as before.", "sequence.) digits. Our answer is $40-14=\\boxed{26}$. ~IceMatrix", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$.", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "equal to $\\tfrac12 \\cdot 7 \\cdot \\tfrac{3\\sqrt{7}}{4} = \\tfrac{21\\sqrt{7}}{8}$, so $p+q+r = \\boxed{36}$.", "$t = \\sqrt{5 \\cdot \\sqrt{38}} = 5.55 s$", ". . . $=\\;\\boxed{s^5 - 5s^4u + 10s^3u^2 - 10s^2u^3 + 5su^4 - u^5}$" ]

[ "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.", "in $T$, their signs are opposite; so the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 * 2^6 = \\boxed{448}$.", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "$4n=96$ 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", "b \\right)$$ are also both integers. Well, it looks like there's no doubt about where the $$\\sqrt{5}$$ and the $$\\sqrt{7}$$ came from. $$\\sqrt{5} \\text{ and } \\sqrt{7} \\text{ came from the } \\sqrt{a} \\text{ and the } \\sqrt{b}$$ So there you have it! Now that you know that $$a = 5$$ and $$b = 7$$ (or vice versa... it doesn't actually matter), then you can figure out that their sum is $$5 + 7 = \\boxed{12}.$$", "our answer is $m+n=61+4=\\boxed{065}$.", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "$\\sqrt{(5-0)^2 + (0-0)^2 + (3-0)^2} = \\sqrt{34}$, which is option $\\boxed{D}$", "5, then the value of $5 \\times 12$ is 10 . If 3 $\\sqrt5 + \\sqrt{125}$ = 17.88, then what will be the value of $\\sqrt{80}$ + 6 $\\sqrt{5}$ ?", "+0 # Help -1 96 1 +303 What is the value of for which $$\\sqrt{x + \\sqrt{x + \\sqrt{x + \\ldots}}} = 5?$$ Jan 12, 2021 #1 +2 Let $y=\\sqrt{x+\\sqrt{x+\\sqrt{x+\\cdots}}}=5.$ If you square $y$, you get $25=x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\cdots}}}=x+y$. But since the value of y is 5, then you get $25=x+5$. Therefore, $x=\\boxed{20}$. Jan 12, 2021 #1 +2 Let $y=\\sqrt{x+\\sqrt{x+\\sqrt{x+\\cdots}}}=5.$ If you square $y$, you get $25=x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\cdots}}}=x+y$. But since the value of y is 5, then you get $25=x+5$. Therefore, $x=\\boxed{20}$.", "divides $$2^4 - 1$$, the number 5 has to go in box B. Thus, we can place the numbers 2, 3, 4, 5, and 6. Furthermore, the placement of these numbers is forced, i.e. this is the only possible way. (We can swap the numbers in box A and box B, but it's basically the same.) Now, 7 divides $$2^3 - 1$$ and $$2^6 - 1$$, so we cannot place the number 7 in either box. This means $$n = 6$$ is as high as we can go. · 7 months ago", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "the denominator than the numerator. Thus all 16 values of n make the expression not an integer and the answer is $50-16=\\boxed{\\mathbf{(D)}\\ 34}$.", "= |5n-1| \\implies n={1\\over 10}\\implies d={1\\over 2\\sqrt{29}}$$ • Second case: $$5n = |5n-3| \\implies n={3\\over 10}\\implies d={3\\over 2\\sqrt{29}}$$ • Third case: $$|5n-3| = |5n-1| \\implies n={4\\over 10}\\implies d={1\\over \\sqrt{29}}$$ So $$\\boxed{d_{sup} = {1\\over 2\\sqrt{29}}}$$.", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$." ]

[ "team of 5 students. First, the people in the class vote on who the team presidents are. If you want to become president, you put a little blurb about why you'd make a good one, submit your candidacy, and everyone sees it. If you do not want to become president, you don't put any blurb, and don't submit your candidacy. Everyone has 3 votes for president. The presidents choose their teams via a private snake draft. Their teammates get titles (Ex: VP Finance, VP Marketing, VP Accounting, VP Ops, etc). There were about 10 teams in my class. Then you are put in charge of a fictional company, competing with 4 other teams in their own world. You start at year 5, and all the other teams are equal at that point. (The balance sheet and product prices / quality are all equal.) Years 1-4 have already been simulated for you, so you can run regression analysis on past data, etc. Some of the other teams are from other schools in other countries. You are selling 2 products (one is a premium product and one is a mass-produced product) in 6 different countries, each with their own different elasticities to price, quality, and marketing. You don't know these elasticities. The team sets the inputs (Price of product, Quality", "of the committee members as chairman – giving him the right to decide in case of a tie.", "of those 2 members (in this case, {2,3,x} and {1,3,x}). And if it shares 1 member with a 4-person group, it will obviously share exactly 1 member with another committee.", "number of ways to pick one of those $k$ to be president, and $\\binom{n}{n-k}$ is the number of ways to fill out the committee with $n-k$ people from the non-eligible group. Then, as you say, you sum over the possible values of $k$. On the right, $n$ is the number of ways to choose the president, and $\\binom{2n-1}{n-1}$ is the number of ways to choose the rest of the committee. – Brian M. Scott Nov 28 '14 at 19:47", "A committee of three people is to be chosen from four 9 28 Sep 2009, 00:41 27 A committee of three people is to be chosen from four marrie 8 13 Nov 2007, 07:31 Display posts from previous: Sort by", "United States Senate The U.S. Senate has two senators from each of the 50 states. In how many ways can a committee of five senators be chosen if no state is to have two members on the committee? Karl. • I think that would be 100 * 98 * 96 * 94 * 92. (First you can pick any of the 100. Then you have to rule out that state, so there are only 98 senators left to pick from. And so on.) --jpgordon∇∆∇∆ 00:56, 4 August 2005 (UTC) • No. (! means factorial) It would be 100 * 98 * 96 * 94 * 92 / 5!. If I have a Senator Specter, Santorum, Rosenberg, Williams, and Montgomery on a committee it's the same as if I had Santorum, Williams, Montgomery, Rosenberg, and Specter. Order doesn't matter and hence must be compensated for in the calculation. Superm401 | Talk 01:33, August 4, 2005 (UTC) • To me, it makes sense to express it like this: 50!2^5/45!5! . You choose one of the 50 state pairs originally there, then one of the 49 remaining, and so on. Then you multiply by 2^5 to take into account the fact that either senator could be picked for any state. I explained the 5!. Ask if you have questions. Superm401 |", "a Formula For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set ofobjects and we want to chooseobjects from the set in order, we writeAnother way to write this is a notation commonly seen on computers and calculators. To calculatewe begin by findingthe number of ways to line up all objects. We then divide by to cancel out theitems that we do not wish to line up. Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is Using factorials, we get the same result. There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows. Note that the formula stills works", "does not matter. Can someone please explain the math behind it? Bunuel Re: A committee of 3 people is to be chosen from the president [#permalink] 16 Feb 2019, 09:52 Display posts from previous: Sort by", "# Prove ${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$ Can someone check my logic here. Question: How many ways are there to choose a an $k$ person committee from a group of $n$ people? Answer 1: there are ${n \\choose k}$ ways. Answer 2: condition on eligibility. Assume the creator of the committee is already in the committee. This leaves us with choosing $k - 1$ people from a group of $n - 1$ potentially eligible people. If all remaining people are eligible, there are ${n - 1 \\choose k - 1}$ possible committees, if there are $n - 2$ eligible people, there are ${n - 2 \\choose k - 1}$ committees, if there are $n - 3$ eligible people, there are ${n - 3 \\choose k - 1}$ committees,..., if there are $k - 1$ eligible people there are ${k - 1 \\choose k - 1}$ committees. Therefore,$${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$$. • All of", "how many 4-person committees can &nbs [#permalink] 17 Apr 2018, 16:27 Display posts from previous: Sort by", "(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...", "you. – Qiaochu Yuan Sep 22 '10 at 5:12 @Qiaochu Yuan: so far I understand that I have to take the first and second derivative to make it similar to the binomial thm? However, I'm not quite sure how to set it up. Perhaps, if you could help me out with a different example but similar-idea that can guide me through, will be appreciated. – Mike Sep 22 '10 at 5:26 Since this looks like homework, I won't tell you how to do it with the binomial theorem and will tell you how to do it bijectively instead. How many ways are there to choose a committee of $k$ people out of $n$ people, split the committee into Democrats and Republicans, then elect a President from that committee, for some $k$? On the one hand, the answer is $\\displaystyle \\sum_{k=1}^{n} k \\cdot 2^k {n \\choose k}$. On the other hand, one can first choose the President, then split the non-Presidents into Democrats, Republicans, and people not on the committee, then choose a party for the President, which can be done in $2n \\cdot 3^{n-1}$ ways. As for how to do it through the binomial theorem, here is a different but related application. Recall the geometric series formula $$\\frac{1}{1 - x} = \\sum_{k=0}^{\\infty} x^k.$$ What happens if we take", "Question How many ways can a president, vice-president, secretary, and treasurer be chosen from a club with 9 members?", "more than half of the members of the board can serve on this subcommittee. Is the number of combinations of board members that can be selected to serve on the subcommittee greater than 900? (1) The board decides to select 6 members for the subcommittee. (2) There are 12 members of the board. _________________ Everything will fall into place… There is perfect timing for everything and everyone. Never doubt, But Work on improving yourself, Keep the faith and It will all make sense. Originally posted by Helium on 22 Sep 2018, 19:55. Last edited by Helium on 24 Sep 2018, 11:13, edited 1 time in total. Intern Joined: 30 Jan 2018 Posts: 17 Re: The members of a company’s board of directors must select [#permalink] ### Show Tags 22 Sep 2018, 22:39 Harshgmat wrote: The members of a company’s board of directors must select a subcommittee to decide the compensation level for a new Chief Executive Officer. The by-laws of the board state that no more than half of the members of the board can serve on this subcommittee. Is the number of combinations of board members that can be selected to serve on the subcommittee greater than 900? (1) The board decides to select 6 members for the subcommittee. (2) There are 12 members of the board.", "The members of a company’s board of directors must select [#permalink] ### Show Tags 17 Nov 2018, 11:01 The members of a company’s board of directors must select a subcommittee to decide the compensation level for a new Chief Executive Officer. The by-laws of the board state that no more than half of the members of the board can serve on this subcommittee. Is the number of combinations of board members that can be selected to serve on the subcommittee greater than 900? (1) The board decides to select 6 members for the subcommittee. (2) There are 12 members of the board. (1) SUFFICIENT No more than half of the members of the board can serve on subcommittee. If the board decides to select 6 members, it means that there are at least 12 members of company's board of directors. It can be 13, 14, 15 (...). That means 924+ combinations. (2) NOT SUFFICIENT There are 12 members. If we were to choose 6 of them, there are 924 (12C6) combinations. If less, the number of combinations declines. Intern Joined: 12 Nov 2018 Posts: 1 Re: The members of a company’s board of directors must select [#permalink] ### Show Tags 18 Nov 2018, 15:16 12C6 = 12!/(6!(12-6)!) = (12*11*10*9*8*7)/6! Is there any quick way to multiply this last", "hence this option is sufficient to give the answer. Hence answer should be option B. Kindly let me know where I am going wrong. Hit the kudos if I am right. 12C6 is 924. So 12C6>900. In this case the answer will always be yes. So statement 1 is sufficient. In statement 2: You can select a maximum of 6 and even a minimum of 2. Thus insufficient. Director Joined: 20 Feb 2015 Posts: 790 Concentration: Strategy, General Management Re: The members of a company’s board of directors must select [#permalink] ### Show Tags 24 Sep 2018, 05:27 Harshgmat wrote: The members of a company’s board of directors must select a subcommittee to decide the compensation level for a new Chief Executive Officer. The by-laws of the board state that no more than half of the members of the board can serve on this subcommittee. Is the number of combinations of board members that can be selected to serve on the subcommittee greater than 900? (1) The board decides to select 6 members for the subcommittee. (2) There are 12 members of the board. let the total number of board members be x then according to bylaws number of members for deciding the CEO $$<=$$ x/2 say n (1) The board decides to select 6 members for the", "GRE Math : How to find permutation notation Example Questions Example Question #1 : Permutation / Combination A chamber of commerce board has seven total members, drawn from a pool of twenty candidates. There are two stages in the board's election process. First, a president, secretary, and treasurer are chosen. After that, four members are chosen to be “at large” without any specific title or district. How many possible boards could be chosen? 5,426,400 16,279,200 10,465,200 390,700,800 2,713,200 16,279,200 Explanation: We have to consider two cases. First, the group containing the president, secretary, and treasurer represents a case of permutation. Since the order matters in such a group, we can select from our initial 20 candidates 20 * 19 * 18, or 6840 possible groupings. Following that, the at large group constitutes a case of combinations in which the order does not matter. Since we have chosen 3 already for the first three slots, there will be 17 remaining people. The formula for choosing 4-person sets out of 17 candidates is represented by the combination formula of this form: 17! / ((17-4)! * 4!) = 17! / (13! * 4!) = (17 * 16 * 15 * 14) / (4 * 3 * 2) = 17 * 4 * 5 * 7 = 2380 Thus, we have 6840", "are there of the election? A. 15 B. 30 C. 60 D. 120 E. 240 _________________ GMAT Club Legend Joined: 12 Sep 2015 Posts: 4134 Sixteen children are trying to decide which two children will win the [#permalink] ### Show Tags 31 Jan 2017, 12:52 2 Top Contributor 4 Bunuel wrote: Sixteen children are trying to decide which two children will win the president and vice-president positions in the class. Each child can and will cast a vote for anyone in the class. If each child in the class is eligible for a position, how many different outcomes are there of the election? A. 15 B. 30 C. 60 D. 120 E. 240 Take the task of selecting a president and vice-president, and break it into stages. Stage 1: Select someone to be president. There are 16 children to choose from, so we can complete stage 1 in 16 ways Stage 2: Select someone to be vice-president. Once a child has been selected to be president, that child cannot also be vice-president. So, there are 15 remaining children to choose from, so we can complete stage 2 in 15 ways By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus select a president and vice-president) in (16)(15) ways (240 ways) Note: the FCP can", "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", "nominees are only confirmed if the same party holds the presidency and the Senate. What is the expected number of vacancies on the bench in the long run? You can assume the following:" ]

[ "team of 5 students. First, the people in the class vote on who the team presidents are. If you want to become president, you put a little blurb about why you'd make a good one, submit your candidacy, and everyone sees it. If you do not want to become president, you don't put any blurb, and don't submit your candidacy. Everyone has 3 votes for president. The presidents choose their teams via a private snake draft. Their teammates get titles (Ex: VP Finance, VP Marketing, VP Accounting, VP Ops, etc). There were about 10 teams in my class. Then you are put in charge of a fictional company, competing with 4 other teams in their own world. You start at year 5, and all the other teams are equal at that point. (The balance sheet and product prices / quality are all equal.) Years 1-4 have already been simulated for you, so you can run regression analysis on past data, etc. Some of the other teams are from other schools in other countries. You are selling 2 products (one is a premium product and one is a mass-produced product) in 6 different countries, each with their own different elasticities to price, quality, and marketing. You don't know these elasticities. The team sets the inputs (Price of product, Quality", "= |B|\\cdot k$$ so $$1<{|B|\\over |A|} = {n-k+1\\over k} \\iff k\\leq {n\\over 2}$$ You can ask yourself: In how many ways can we choose $$k$$ people, including one president, from a population of $$n$$ people? First answer: You choose $$k-1$$ people from $$n$$ people and then from the rest of them a president, so we have $${n\\choose k-1}\\cdot {n-k+1\\choose 1}$$ Second answer: You choose $$k$$ people from $$n$$ people and then from these selected you choose a president, so we have $${n\\choose k}\\cdot {k\\choose 1}$$ So we have:$${n\\choose k-1}\\cdot (n-k+1)=k{n\\choose k}$$ and from here we get $${n\\choose k-1}<{n\\choose k} \\iff k\\leq {n\\over 2}$$", "second choice, or picking the president from the committee of $$k$$ people. By this logic, the left hand side of the equation counts what we are interested in: the number of ways to pick a committee with a president. We now have to prove that the right hand side of the equation gives the same thing. Let’s imagine that this time, instead of picking the committee first, we select the president and then the committee. So, if we are picking the president first from the population, we of course have $$n$$ choices (we can pick anyone in the whole population). Then, after picking the president, we have to select the committee, and since there are $$n -1$$ people left (we have already taken one out to be president) and we need $$k-1$$ more people in the committee (we already have the president) we just multiply by $${n -1 \\choose k -1}$$. So, by the multiplication rule, we multiply the first choice of picking the president from $$n$$ people by the second choice of picking the rest of the $$k-1$$ sized committee from the $$n-1$$ people. So, then, the only difference between the two sides is that the left side picks the committee first and then the president from the committee, while the right side picks the president from", "number of ways to pick one of those $k$ to be president, and $\\binom{n}{n-k}$ is the number of ways to fill out the committee with $n-k$ people from the non-eligible group. Then, as you say, you sum over the possible values of $k$. On the right, $n$ is the number of ways to choose the president, and $\\binom{2n-1}{n-1}$ is the number of ways to choose the rest of the committee. – Brian M. Scott Nov 28 '14 at 19:47", "we have more population than we have seats or places? For example, there are 6 people who are vying for those 4 government slots? We could do it this way: We can look at see that for the President's seat, there are 6 candidates. We can then look at the VP's seat and see that, with someone in the role of President, there are now 5 people left who can be VP. Then we look to the Treasurer and see there are 4 people left. And lastly there is the Secretary and there are 3 remaining people who could go there. So we get $6 \\times 5 \\times 4 \\times 3 = 30 \\times 12 = 360$ ways to do this. But this is awkward to do all the time - especially if we have a large population and a large number of places that need filling (ex. A concert hall has 1,000 seats. What is the number of ways 5,000 fans can sit in those 1,000 seats). Wouldn't it be great if there were a formula that would make it easier to work with these types of calculations? Notice that in the \"6 people, 4 offices\" example, we ended up with this calculation: $6 \\times 5 \\times 4 \\times 3$ How can we put that in terms", "# From a pool of 12 candidates, the offices of president, vice president, secretary, and treasurer will be filled. In how many different ways can the offices be filled? 11,880 #### Explanation: We can find the number of ways in a couple of different ways: • work out the number by examining each seat • use the Permutation formula Examine each seat Let's start with the President's seat. How many people can fill that? 12 And then the Vice President's seat - how many can fill that? 1 person is already in the President's seat, so there's 11 people who can. And then then the Secretary's seat - 2 people are already in seats and so 10 could fill this one. And lastly the Treasurer's seat - 3 people are already in seats and so 9 could possibly sit here. So that's $12 \\times 11 \\times 10 \\times 9 = 11 , 880$ Permutation It'd be great if we could express the above multiplication in terms of factorials, because factorials are set figures and so we don't need to work out the multiplication each time. Is there a way to do so? Notice that in our multiplication above, we have $12 \\times 11 \\times 10 \\times 9$. What if we could have the full factorial? To have that,", "you. – Qiaochu Yuan Sep 22 '10 at 5:12 @Qiaochu Yuan: so far I understand that I have to take the first and second derivative to make it similar to the binomial thm? However, I'm not quite sure how to set it up. Perhaps, if you could help me out with a different example but similar-idea that can guide me through, will be appreciated. – Mike Sep 22 '10 at 5:26 Since this looks like homework, I won't tell you how to do it with the binomial theorem and will tell you how to do it bijectively instead. How many ways are there to choose a committee of $k$ people out of $n$ people, split the committee into Democrats and Republicans, then elect a President from that committee, for some $k$? On the one hand, the answer is $\\displaystyle \\sum_{k=1}^{n} k \\cdot 2^k {n \\choose k}$. On the other hand, one can first choose the President, then split the non-Presidents into Democrats, Republicans, and people not on the committee, then choose a party for the President, which can be done in $2n \\cdot 3^{n-1}$ ways. As for how to do it through the binomial theorem, here is a different but related application. Recall the geometric series formula $$\\frac{1}{1 - x} = \\sum_{k=0}^{\\infty} x^k.$$ What happens if we take", "College Algebra (6th Edition) Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = 5040 Total number of members on its board of directors = 10 Number of members to elect president, vice president, secretary and treasurer = 4 Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = $\\frac{10!}{(10 - 4)! }$ = $\\frac{10!}{6! }$ = $\\frac{10\\times9\\times8\\times7\\times6!}{6! }$ = $10\\times9\\times8\\times7$ = 5040", "# Prove ${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$ Can someone check my logic here. Question: How many ways are there to choose a an $k$ person committee from a group of $n$ people? Answer 1: there are ${n \\choose k}$ ways. Answer 2: condition on eligibility. Assume the creator of the committee is already in the committee. This leaves us with choosing $k - 1$ people from a group of $n - 1$ potentially eligible people. If all remaining people are eligible, there are ${n - 1 \\choose k - 1}$ possible committees, if there are $n - 2$ eligible people, there are ${n - 2 \\choose k - 1}$ committees, if there are $n - 3$ eligible people, there are ${n - 3 \\choose k - 1}$ committees,..., if there are $k - 1$ eligible people there are ${k - 1 \\choose k - 1}$ committees. Therefore,$${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$$. • All of", "terminology – juantheron Feb 9 '13 at 7:01 Out of 5 couples we have to select 2 persons from men and 2 from woman so that same couple doesn't select. this can be done as follows,1st select 2 men from 5 i.e,$(5,2)$ and and select 2 woman from the remaining 3, (since 2 husbands selected of two wifes exclude that). So no of selections are $(5,2)*(3,2)=30$. These can be done in two ways so final answer is $2 \\times 30=60$. -", "are 8 choices for the 1st person; there are 7 choices for the 2nd person; there are 6 choices for the 3rd person; there are 5 choices for the 4th person;. How many 8 go into 39? It was ratified by the states on Feb. 27, 1951. For an older person, it’s likely to be longer. Joined Sep … When I run the command slmgr /dlv in an elevated command prompt, it returns a cryptic reply. 8 times 30: 4: 8 times 40: 5,20: 8 times 50: 6,40: Go up to Mesopotamian Mathematics Home Page. Enter your objects (or the names of them), one per line in the box below, then click \"Show me!\" Before the 22nd Amendment, the Constitution did not limit the number of presidential terms to two, though many early presidents, including George Washington, imposed such … Thus, you cannot subtract 4 again from 8. I would like to know how many times I can use the same Meet Now link. Can you get a number for 40 with eight. By - TIMESOFINDIA.COM. The amendment defining limits on how many terms can a president serve was approved by Congress on March 21, 1947, during the administration of President Harry S. Truman. 8 times … How many times can 8 go in to 29? Joined Sep", "\\choose k}$ possible committees. (LHS) Suppose we must have the celebrity on the committee. Then there are ${n \\choose k-1}$ ways to choose who else will be on the committee. Also, suppose that we don't want to have the celebrity on the committee. Then there are ${n \\choose k}$ ways to choose who is on the committee. This gives us a total of ${n \\choose k-1} + {n \\choose k}$ ways to pick the committee. Recall that these two expressions both represent ${n \\choose k}$ or the number of ways to choose a committee of size $k$ from $n$ people. Q A Q $\\frac{n!}{(n-k)!k!} = \\frac{n \\cdot (n-1) \\dotsm (n-k+1)}{k\\cdot (k-1)\\dotsm 1}$ A Recall that these two expressions both represent ${n \\choose k}$ or the number of ways to choose a committee of size $k$ from $n$ people. (LHS) We first permute the $n$ people and then select the first $k$ people in the permutation. While there $n!$ total permutations, we must deal with the overcounting within the committee and the people not selected for the committee. The $k$ people in the committee can be ordered in $k!$ ways and the people not in the committee can be ordered in $(n-k)!$ ways, which indicates that we have overcounted by a factor of $k!(n-k)!$. Therefore, we observe that ${n", "Basic Counting — Selecting groups of people from a group of 10 I'm reading a series of basic counting problems regarding a ten person club. I just want to make sure I'm sufficiently grasping the material and my problem solving methods are correct. $1.$ In how many ways may a ten person club select a president and a secretary-treasurer from among its members? For this problem, there are 10 ways to choose the president, and 9 ways to choose the secretary. In total, there are $10 \\times 9 = 90$ ways to choose them. $2.$ In how many ways may a ten person club select a two person executive committee from among its members? This time, since we're taking 2 people from 10, I used ${10 \\choose 2} = 45$ ways. $3.$ In how many ways may a ten person club select a president and a two person executive advisory board from among its members (assuming that the president is not on the advisory board)? I've already established from my first answer that there are 10 ways to choose the president. Then, I have to choose 2 people from a remaining group of 9. ${9 \\choose 2} = 36$. Finally, I multiply the two results.$36\\times10 = 360$ ways. Any help would be appreciated, thank you. • It's correct", "United States Senate The U.S. Senate has two senators from each of the 50 states. In how many ways can a committee of five senators be chosen if no state is to have two members on the committee? Karl. • I think that would be 100 * 98 * 96 * 94 * 92. (First you can pick any of the 100. Then you have to rule out that state, so there are only 98 senators left to pick from. And so on.) --jpgordon∇∆∇∆ 00:56, 4 August 2005 (UTC) • No. (! means factorial) It would be 100 * 98 * 96 * 94 * 92 / 5!. If I have a Senator Specter, Santorum, Rosenberg, Williams, and Montgomery on a committee it's the same as if I had Santorum, Williams, Montgomery, Rosenberg, and Specter. Order doesn't matter and hence must be compensated for in the calculation. Superm401 | Talk 01:33, August 4, 2005 (UTC) • To me, it makes sense to express it like this: 50!2^5/45!5! . You choose one of the 50 state pairs originally there, then one of the 49 remaining, and so on. Then you multiply by 2^5 to take into account the fact that either senator could be picked for any state. I explained the 5!. Ask if you have questions. Superm401 |", "a Formula For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set ofobjects and we want to chooseobjects from the set in order, we writeAnother way to write this is a notation commonly seen on computers and calculators. To calculatewe begin by findingthe number of ways to line up all objects. We then divide by to cancel out theitems that we do not wish to line up. Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is Using factorials, we get the same result. There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows. Note that the formula stills works", "galloping on 9 horses behind. How many ways can sort them behind? • Vans In how many ways can 9 shuttle vans line up at the airport? • Seating How many ways can 6 people sit on 6 numbered chairs (e. g. , seat reservation on the train)? • The lift The lift can fit four people. How many uphill rides must the lift make to move up 12 passengers? • Possible combinations - word How many ways can the letters F, A, I, R be arranged? • Neighborhood I have 7 cups: 1 2 3 4 5 6 7. How many opportunities of standings cups are there if 1 and 2 are always neighborhood? • School committee Seven students were elected to the school committee. In how many ways can become the President, Vice-President, Secretary, and Treasurer be selected? • Tokens In the non-transparent bags are red, white, yellow, blue tokens. We 3times pull one tokens and again returned it, write down all possibilities. • Election 4 In a certain election there are 3 candidates for president 5 for secretory and 2 for tresurer. Find how many ways the election may (turn out/held). • Lunch Seven classmates go every day for lunch. If they always come to the front in a different order, will be enough school", "order in which we choose the 2 people doesn't matter)? David Patrick, Introduction to Counting & Probability (2005), pp 66-7. #### County Boy ##### Member I don't understand why you would WANT to avoid dividing! There are 8 ways to choose one person, say \"A\", then 7 people left to choose from so 7 ways to choose the second person, \"B\", then 6 ways to choose the third person, \"C\". There are 8*7*6 ways to choose \"ABC\" IN THAT ORDER. (There are 8*7*6= 336 ways to choose a \"president, vice-president, and treasurer\".) But \"ABC\", \"ACB\", \"BAC\", \"BCA\", \"CAB\", \"CBA\" would be the same three people on a \"committee\".", "n{n - 1 \\choose k - 1}$ Again, this is not difficult to show with algebra. Plugging in the definition of the binomial coefficient: $= \\frac{kn!}{k! \\cdot (n - k)!} = \\frac{n(n - 1)!}{(k - 1)!\\cdot (n - k)!}$ $= \\frac{n!}{(k - 1)! \\cdot (n - k)!} = \\frac{n!}{(k - 1)!\\cdot (n - k)!}$ However, again, we want to consider a story. The best way to envision this relationship is again with picking committees; this time, though, imagine that you are picking a committee of size $$k$$ people out of a population of $$n$$ people and you are selecting one president for the committee. So, there are two selections: picking the $$k$$ people to be in the committee, and then picking the one person from the $$k$$ people in the committee to be the president. First, let’s consider the left side of the equation. Here, we have the basic $${n \\choose k}$$, which of course gives the number of ways to select the committee of $$k$$ people. However, once we’ve done this, we have another selection: picking the president of the committee. Since there are $$k$$ people in the committee, we have $$k$$ options to choose from, and thus by the multiplication rule we just multiply the first choice, picking the committee, or $${n \\choose k}$$, by the", "Now, $680-120=560$. Scenario 2. For a committee with 1 professor and 2 students, we will get $7*C_{10}^2 = 7 * \\frac{10!}{8! * 2!} = 45*7=315$. (we multiply by 7 because For 2 professors and 1 student, we will get $C_7^2*10 = 21*10=210$, and for 3 professors, we will get $C_7^3 = 35$. Adding up the combinations we will get: $315+210+35=560$. Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to you - the one easier to use. Here is another hard problem with some restrictions; again from an unknown source: {{#x:box| EXAMPLE 11. There are 11 top managers that need to form a decision group. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team? }} Again our options are to solve the problem either to find the total number of committees and then subtracting the number of groups that VP and P would end up", "person is chosen a president; there are $n$ choices for President, as the group with President eligible people has $n$ members. Then across both groups there are now $(n-1) + n = 2n - 1$ people left from whom we must pick the rest of the committee. That is, choose $n-1$ people from $2n - 1$. Hence the total number of possible committees is $$n{2n -1 \\choose n-1}$$ For the LHS: Suppose $k$ people are chosen from the Presidential eligible group. Then the total ways of choosing all $n$ committee members is $\\displaystyle {n \\choose k}{n \\choose n-k}$. Of that selection there are $k$ ways of choosing the President. Hence in this case, there are $$k {n \\choose k}{n \\choose n-k}$$ ways of composing the committee. This doesn't double-count, it's completing the specification that the committee of $n$ people already selected have a President. This is different than the expression on the RHS where we first choose a President and then the rest of the committee. Here on the LHS we first choose the committee and then the President. One final technical note: in the sum $\\sum_{k=0}^n$ there is the case $k=0$ where there are no committee members from the Pres eligible group. In which case this is not a valid committee. But as $k=0$ the expression $k" ]

[ "team of 5 students. First, the people in the class vote on who the team presidents are. If you want to become president, you put a little blurb about why you'd make a good one, submit your candidacy, and everyone sees it. If you do not want to become president, you don't put any blurb, and don't submit your candidacy. Everyone has 3 votes for president. The presidents choose their teams via a private snake draft. Their teammates get titles (Ex: VP Finance, VP Marketing, VP Accounting, VP Ops, etc). There were about 10 teams in my class. Then you are put in charge of a fictional company, competing with 4 other teams in their own world. You start at year 5, and all the other teams are equal at that point. (The balance sheet and product prices / quality are all equal.) Years 1-4 have already been simulated for you, so you can run regression analysis on past data, etc. Some of the other teams are from other schools in other countries. You are selling 2 products (one is a premium product and one is a mass-produced product) in 6 different countries, each with their own different elasticities to price, quality, and marketing. You don't know these elasticities. The team sets the inputs (Price of product, Quality", "= |B|\\cdot k$$ so $$1<{|B|\\over |A|} = {n-k+1\\over k} \\iff k\\leq {n\\over 2}$$ You can ask yourself: In how many ways can we choose $$k$$ people, including one president, from a population of $$n$$ people? First answer: You choose $$k-1$$ people from $$n$$ people and then from the rest of them a president, so we have $${n\\choose k-1}\\cdot {n-k+1\\choose 1}$$ Second answer: You choose $$k$$ people from $$n$$ people and then from these selected you choose a president, so we have $${n\\choose k}\\cdot {k\\choose 1}$$ So we have:$${n\\choose k-1}\\cdot (n-k+1)=k{n\\choose k}$$ and from here we get $${n\\choose k-1}<{n\\choose k} \\iff k\\leq {n\\over 2}$$", "second choice, or picking the president from the committee of $$k$$ people. By this logic, the left hand side of the equation counts what we are interested in: the number of ways to pick a committee with a president. We now have to prove that the right hand side of the equation gives the same thing. Let’s imagine that this time, instead of picking the committee first, we select the president and then the committee. So, if we are picking the president first from the population, we of course have $$n$$ choices (we can pick anyone in the whole population). Then, after picking the president, we have to select the committee, and since there are $$n -1$$ people left (we have already taken one out to be president) and we need $$k-1$$ more people in the committee (we already have the president) we just multiply by $${n -1 \\choose k -1}$$. So, by the multiplication rule, we multiply the first choice of picking the president from $$n$$ people by the second choice of picking the rest of the $$k-1$$ sized committee from the $$n-1$$ people. So, then, the only difference between the two sides is that the left side picks the committee first and then the president from the committee, while the right side picks the president from", "a Formula For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set ofobjects and we want to chooseobjects from the set in order, we writeAnother way to write this is a notation commonly seen on computers and calculators. To calculatewe begin by findingthe number of ways to line up all objects. We then divide by to cancel out theitems that we do not wish to line up. Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is Using factorials, we get the same result. There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows. Note that the formula stills works", "we have more population than we have seats or places? For example, there are 6 people who are vying for those 4 government slots? We could do it this way: We can look at see that for the President's seat, there are 6 candidates. We can then look at the VP's seat and see that, with someone in the role of President, there are now 5 people left who can be VP. Then we look to the Treasurer and see there are 4 people left. And lastly there is the Secretary and there are 3 remaining people who could go there. So we get $6 \\times 5 \\times 4 \\times 3 = 30 \\times 12 = 360$ ways to do this. But this is awkward to do all the time - especially if we have a large population and a large number of places that need filling (ex. A concert hall has 1,000 seats. What is the number of ways 5,000 fans can sit in those 1,000 seats). Wouldn't it be great if there were a formula that would make it easier to work with these types of calculations? Notice that in the \"6 people, 4 offices\" example, we ended up with this calculation: $6 \\times 5 \\times 4 \\times 3$ How can we put that in terms", "you. – Qiaochu Yuan Sep 22 '10 at 5:12 @Qiaochu Yuan: so far I understand that I have to take the first and second derivative to make it similar to the binomial thm? However, I'm not quite sure how to set it up. Perhaps, if you could help me out with a different example but similar-idea that can guide me through, will be appreciated. – Mike Sep 22 '10 at 5:26 Since this looks like homework, I won't tell you how to do it with the binomial theorem and will tell you how to do it bijectively instead. How many ways are there to choose a committee of $k$ people out of $n$ people, split the committee into Democrats and Republicans, then elect a President from that committee, for some $k$? On the one hand, the answer is $\\displaystyle \\sum_{k=1}^{n} k \\cdot 2^k {n \\choose k}$. On the other hand, one can first choose the President, then split the non-Presidents into Democrats, Republicans, and people not on the committee, then choose a party for the President, which can be done in $2n \\cdot 3^{n-1}$ ways. As for how to do it through the binomial theorem, here is a different but related application. Recall the geometric series formula $$\\frac{1}{1 - x} = \\sum_{k=0}^{\\infty} x^k.$$ What happens if we take", "United States Senate The U.S. Senate has two senators from each of the 50 states. In how many ways can a committee of five senators be chosen if no state is to have two members on the committee? Karl. • I think that would be 100 * 98 * 96 * 94 * 92. (First you can pick any of the 100. Then you have to rule out that state, so there are only 98 senators left to pick from. And so on.) --jpgordon∇∆∇∆ 00:56, 4 August 2005 (UTC) • No. (! means factorial) It would be 100 * 98 * 96 * 94 * 92 / 5!. If I have a Senator Specter, Santorum, Rosenberg, Williams, and Montgomery on a committee it's the same as if I had Santorum, Williams, Montgomery, Rosenberg, and Specter. Order doesn't matter and hence must be compensated for in the calculation. Superm401 | Talk 01:33, August 4, 2005 (UTC) • To me, it makes sense to express it like this: 50!2^5/45!5! . You choose one of the 50 state pairs originally there, then one of the 49 remaining, and so on. Then you multiply by 2^5 to take into account the fact that either senator could be picked for any state. I explained the 5!. Ask if you have questions. Superm401 |", "Basic Counting — Selecting groups of people from a group of 10 I'm reading a series of basic counting problems regarding a ten person club. I just want to make sure I'm sufficiently grasping the material and my problem solving methods are correct. $1.$ In how many ways may a ten person club select a president and a secretary-treasurer from among its members? For this problem, there are 10 ways to choose the president, and 9 ways to choose the secretary. In total, there are $10 \\times 9 = 90$ ways to choose them. $2.$ In how many ways may a ten person club select a two person executive committee from among its members? This time, since we're taking 2 people from 10, I used ${10 \\choose 2} = 45$ ways. $3.$ In how many ways may a ten person club select a president and a two person executive advisory board from among its members (assuming that the president is not on the advisory board)? I've already established from my first answer that there are 10 ways to choose the president. Then, I have to choose 2 people from a remaining group of 9. ${9 \\choose 2} = 36$. Finally, I multiply the two results.$36\\times10 = 360$ ways. Any help would be appreciated, thank you. • It's correct", "# From a pool of 12 candidates, the offices of president, vice president, secretary, and treasurer will be filled. In how many different ways can the offices be filled? 11,880 #### Explanation: We can find the number of ways in a couple of different ways: • work out the number by examining each seat • use the Permutation formula Examine each seat Let's start with the President's seat. How many people can fill that? 12 And then the Vice President's seat - how many can fill that? 1 person is already in the President's seat, so there's 11 people who can. And then then the Secretary's seat - 2 people are already in seats and so 10 could fill this one. And lastly the Treasurer's seat - 3 people are already in seats and so 9 could possibly sit here. So that's $12 \\times 11 \\times 10 \\times 9 = 11 , 880$ Permutation It'd be great if we could express the above multiplication in terms of factorials, because factorials are set figures and so we don't need to work out the multiplication each time. Is there a way to do so? Notice that in our multiplication above, we have $12 \\times 11 \\times 10 \\times 9$. What if we could have the full factorial? To have that,", "Now, $680-120=560$. Scenario 2. For a committee with 1 professor and 2 students, we will get $7*C_{10}^2 = 7 * \\frac{10!}{8! * 2!} = 45*7=315$. (we multiply by 7 because For 2 professors and 1 student, we will get $C_7^2*10 = 21*10=210$, and for 3 professors, we will get $C_7^3 = 35$. Adding up the combinations we will get: $315+210+35=560$. Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to you - the one easier to use. Here is another hard problem with some restrictions; again from an unknown source: {{#x:box| EXAMPLE 11. There are 11 top managers that need to form a decision group. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team? }} Again our options are to solve the problem either to find the total number of committees and then subtracting the number of groups that VP and P would end up", "# Prove ${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$ Can someone check my logic here. Question: How many ways are there to choose a an $k$ person committee from a group of $n$ people? Answer 1: there are ${n \\choose k}$ ways. Answer 2: condition on eligibility. Assume the creator of the committee is already in the committee. This leaves us with choosing $k - 1$ people from a group of $n - 1$ potentially eligible people. If all remaining people are eligible, there are ${n - 1 \\choose k - 1}$ possible committees, if there are $n - 2$ eligible people, there are ${n - 2 \\choose k - 1}$ committees, if there are $n - 3$ eligible people, there are ${n - 3 \\choose k - 1}$ committees,..., if there are $k - 1$ eligible people there are ${k - 1 \\choose k - 1}$ committees. Therefore,$${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$$. • All of", "\\choose k}$ possible committees. (LHS) Suppose we must have the celebrity on the committee. Then there are ${n \\choose k-1}$ ways to choose who else will be on the committee. Also, suppose that we don't want to have the celebrity on the committee. Then there are ${n \\choose k}$ ways to choose who is on the committee. This gives us a total of ${n \\choose k-1} + {n \\choose k}$ ways to pick the committee. Recall that these two expressions both represent ${n \\choose k}$ or the number of ways to choose a committee of size $k$ from $n$ people. Q A Q $\\frac{n!}{(n-k)!k!} = \\frac{n \\cdot (n-1) \\dotsm (n-k+1)}{k\\cdot (k-1)\\dotsm 1}$ A Recall that these two expressions both represent ${n \\choose k}$ or the number of ways to choose a committee of size $k$ from $n$ people. (LHS) We first permute the $n$ people and then select the first $k$ people in the permutation. While there $n!$ total permutations, we must deal with the overcounting within the committee and the people not selected for the committee. The $k$ people in the committee can be ordered in $k!$ ways and the people not in the committee can be ordered in $(n-k)!$ ways, which indicates that we have overcounted by a factor of $k!(n-k)!$. Therefore, we observe that ${n", "A committee of three people is to be chosen from four 9 28 Sep 2009, 00:41 27 A committee of three people is to be chosen from four marrie 8 13 Nov 2007, 07:31 Display posts from previous: Sort by", "galloping on 9 horses behind. How many ways can sort them behind? • Vans In how many ways can 9 shuttle vans line up at the airport? • Seating How many ways can 6 people sit on 6 numbered chairs (e. g. , seat reservation on the train)? • The lift The lift can fit four people. How many uphill rides must the lift make to move up 12 passengers? • Possible combinations - word How many ways can the letters F, A, I, R be arranged? • Neighborhood I have 7 cups: 1 2 3 4 5 6 7. How many opportunities of standings cups are there if 1 and 2 are always neighborhood? • School committee Seven students were elected to the school committee. In how many ways can become the President, Vice-President, Secretary, and Treasurer be selected? • Tokens In the non-transparent bags are red, white, yellow, blue tokens. We 3times pull one tokens and again returned it, write down all possibilities. • Election 4 In a certain election there are 3 candidates for president 5 for secretory and 2 for tresurer. Find how many ways the election may (turn out/held). • Lunch Seven classmates go every day for lunch. If they always come to the front in a different order, will be enough school", "number of ways to pick one of those $k$ to be president, and $\\binom{n}{n-k}$ is the number of ways to fill out the committee with $n-k$ people from the non-eligible group. Then, as you say, you sum over the possible values of $k$. On the right, $n$ is the number of ways to choose the president, and $\\binom{2n-1}{n-1}$ is the number of ways to choose the rest of the committee. – Brian M. Scott Nov 28 '14 at 19:47", "8. In many courts, 12 jurors are chosen from a pool of 30 perspective jurors. a. In how many ways can 12 jurors be chosen from the pool of 30 perspective jurors? 30C12 = 86 493 225 b. Once the 12 jurors are selected, 2 alternates are selected. The order of the alternates is specified. If a selected juror cannot complete the trial, the first alternate is called on to fill that jury spot. In how many ways can the 2 alternates be chosen after the 12 jury members have been chosen? 18P2 = 306 Question 9. A band director wants to form a committee of 4 parents from a list of 45 band parents. a. How many different groups of 4 parents can the band director select? 45C4 = 148 995 b. How many different ways can the band director select 4 parents to serve in the band parents’ association as president, vice president, treasurer, and secretary? 45P4 = 3 575 880 c. Explain the difference between parts (a) and (b) in terms of how you decided to solve each part. Part (a) is an example of finding the number of combinations—how many ways can 4 parents be chosen from 45 distinct parents. The order is not important. Part (b) is an example of finding the number", "me. Suppose that each person in a group of $n$ people votes for exactly two people from a set of candidates. Devise a divide-and-conquer algorithm that determines all candidates who received more than $n/2$ votes. If you are careful, you may point out the my version missed \"... to fill two positions on a committee. The top two finishers both win positions as long as each receives more than n/2 votes\". Well, although it is interesting to know that election goal and election rule, it has nothing to do with the specification of the desired algorithm. You may also point out that my version does not require \"if so, determine who these two candidates are\". Well, if we have determined all candidates who received more than n/2 votes, the number of whom is at most 3, then by just counting the number of votes received by each of those candidates, we can, easily, \"if so, determine who these two candidates are\". ## An interesting algorithm as an exercise There is an algorithm using linear time and constant space that determines all candidates who received more than $n/2$ votes. The construction of such an algorithm is left as an intriguing and challenging exercise for the readers who has read thus far. Thanks to Tom van der Zanden, who pointed", "person is chosen a president; there are $n$ choices for President, as the group with President eligible people has $n$ members. Then across both groups there are now $(n-1) + n = 2n - 1$ people left from whom we must pick the rest of the committee. That is, choose $n-1$ people from $2n - 1$. Hence the total number of possible committees is $$n{2n -1 \\choose n-1}$$ For the LHS: Suppose $k$ people are chosen from the Presidential eligible group. Then the total ways of choosing all $n$ committee members is $\\displaystyle {n \\choose k}{n \\choose n-k}$. Of that selection there are $k$ ways of choosing the President. Hence in this case, there are $$k {n \\choose k}{n \\choose n-k}$$ ways of composing the committee. This doesn't double-count, it's completing the specification that the committee of $n$ people already selected have a President. This is different than the expression on the RHS where we first choose a President and then the rest of the committee. Here on the LHS we first choose the committee and then the President. One final technical note: in the sum $\\sum_{k=0}^n$ there is the case $k=0$ where there are no committee members from the Pres eligible group. In which case this is not a valid committee. But as $k=0$ the expression $k", "the Multiplication Principle to tackle this problem a different way. Rather than laboriously working through and counting all of the possible subsets, we could think of each element as something that could either get chosen or not get chosen to be in a subset. That is, A is either chosen or not... that's two possibilities. B is either chosen or not... that's two possibilities. C is either chosen or not... that's two possibilities. And so on. Thinking of the problem in this way, the Multiplication Principle then readily tells us that there are: $$2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2$$ or $$2^{10}=1024$$ possible subsets. I personally would not have wanted to solve this problem by having to enumerate and count each of the possible subsets. Incidentally, we'll see many more problems similar to this one here when we investigate the binomial distribution later in the course. # 3.2 - Permutations 3.2 - Permutations ## Example 3-4 How many ways can four people fill four executive positions? For the sake of concreteness, let's name the four people Tom, Rick, Harry, and Mary, and the four executive positions President, Vice President, Treasurer and Secretary. I think you'll agree that the Multiplication Principle yields a straightforward solution to this problem. If we fill the President position first, there", "Counting and choices Here is a propability question I am stuck with, In a class there are 8 boys and 5 girls. Two class representatives are to be choosen. In how many ways they can be selected if, 1. both are choose from all? 2. one is to be a boy and other a girl? 3. the first one is to be a boy and the second any of the boy or girl? Here is how I solved it, 1. As there are 8 boys and 5 girls so, 8+5= 13. The choice for first will be 13 and for the second one it will be 12 which makes 156. 2. For 8 boys choice is 8 and for 5 girls the choice is 5. By multiplication principle = 40 3. if first one is boy choices are 8 and then anyone, choices are 12, by multiplication principle, 8*12= 96 The 1st answer does not match.Can anyone please tell me why? - 1. If we are choosing President, VP, then $156$ is fine. If we are choosing two co-presidents, I would say $\\binom{13}{2}$. 2. Good. 3. Hard to know, very poor wording, what does first one mean? If we decide it just means at least one boy, tjem we wamt No. of undrestricted choices minus number of all" ]

[ "remainder.", "there are infintely or finitely many primes of the form $p^2+4$, then all of them will leave a remainder $5$ when divided by $6$. – Abhishek Bakshi Jul 15 '18 at 4:31 • Alright, I deleted it. – Pacifica Jul 15 '18 at 5:03", "gives the remainder 1", "q = 1/2 and the sum of the first six members S6 = 63. Find the fifth element a5.", "but I’ll give you a couple of hints. I described Euclid’s proof that there are infinitely many primes. Suitable variants of that will give proofs that there are infinitely many primes leaving remainder 1 when divided by 4 and infinitely many primes leaving remainder 3 when divided by 4. Are there the same numbers of primes leaving remainder 1 and remainder 3 when divided by 4? It turns out that the answer is yes (when the question is interpreted carefully), but that’s harder to prove, so I’ll leave it for now. What would happen if we replaced 4 by 6, for example? Here are the remainders when the first few primes are divided by 6: 2, 3, 5, 1, 5, 1, 5, 1, 5, 5, 1, 1, 5, 1, 5, 5, 5, 1, 1, 5, 1, 1, 5, 5, 1. Hopefully you will be able to convince yourself that apart from 2 and 3, every prime leaves remainder 1 or 5 when divided by 6. But are there infinitely many of each sort? Notice that this time there are 11 primes leaving remainder 1 when divided by 6 and 12 leaving remainder 5 — again, the numbers are very close. It starts getting a bit harder to prove that there are infinitely many primes leaving remainder 1 when", "be a remainder of 1.", "divided by 6, and infinitely many leaving remainder 5 when divided by 6. And anyway, we’re mathematicians, so we’d like a more general strategy. What would a more general strategy mean? Suppose I fix some number q, and we think about the remainders when we divide the primes by q. If a and q have a common factor greater than 1, then we’re not going to get many primes with remainder a when divided by q (either no primes at all, or just one). Remember that there was only one prime with remainder 2 when divided by 6, and none at all with remainder 4 when divided by 6. So those are the boring cases that we can eliminate immediately. But what if a and q have no common factor greater than 1: what if a and q are coprime? Then we have no obvious reason why there shouldn’t be infinitely many primes leaving remainder a when divided by q, and our earlier examples suggest that perhaps there are indeed infinitely many such primes. But that’s just a guess — perhaps there’s some more subtle problem for some q. It turns out to be correct, though, and that’s this week’s theorem, due to Dirichlet. Theorem (Dirichlet’s theorem) Let a and q be coprime natural numbers (positive whole numbers", "of the sum of fractions is:.", "Suppose, P= 5, P^2 + 17 = 5^2 + 17 = 25 + 17 = 42 Now 42/ 12 = 3 + 6/12, Remainder = 6(C) Re: If p is a prime number greater than 3, find the remainder when p^2 + 1 [#permalink] 09 Feb 2020, 02:15 Display posts from previous: Sort by", "is the required remainder.", "result as the Remainder Theorem but I guess it isn't.", "# Future Factorials Find the remainder when $2016!-2015!$ is divided by $2017$. You may use the fact that $2017$ is prime. ×", "quarter of the square of the sum.", "as there was nothing to divide into to begin with so it can't have a remainder. -", "# Budi's factorial sum What is the remainder when $\\sum_{i=1}^{101} i! i$ is divided by $$101$$? This problem is posed by Budi U. Details and assumptions You may use the fact that 101 is prime. ×", "precise about the definition of your sequence: the least prime giving a prime remainder when divided by 3 (or 5 or 7 or 11 or ...) is 2; no prime leaves an odd remainder when divided by 2. I think the sequence you are interested in may be the sequence of odd primes that leave a prime remainder when divided by any smaller odd prime. But then it should include 3. – Rob Arthan Jul 26 '16 at 20:12 • The remainder can be ANY prime. Look at 47 to get 47mod3=2, 47mid5=2, 47mod7=5 and 47mod11=3. If you looked at the second mention of 17, you'd see that a prime remainder means ANY prime remainder. It is NOT the remainder that is specified, but the least prime such that all its divisors 3,5,7...give ANY prime remainder. Just try 863 when divided by 3,5,7,11,13 and you'll see a variety of remainders. Now, I do agree that extending this is a challenge. If no one is up to it, that's not surprising. – J. M. Bergot Jul 26 '16 at 20:17 • @RobArthan, I think it would suffice for the OP to ask for the least odd prime giving a prime remainder when divided by each of the first $n$ odd primes. – Barry Cipra Jul 26 '16 at 20:17", "remainder of both b and a, on division by m. They share the same remainder.", "a remainder of 2 upon division by 6 so theoretically can be the sum of two primes (and it is 61+7=68); (D) 79 --> 79-2 is not a prime, 79-3 is not a prime and also 79 yields a remainder of 1 upon division by 6, so it can not be the sum of two primes; (E) 88 --> 71+17=88. Of course the above can be done much easier by just subtracting the primes starting from 2 from the answer choices and seeing whether the result is also a prime. Bunuel when you say p=6n+1p=6n+1 orp=6n+5p=6n+5 (p=6n−1p=6n−1), do you mean p=6n+1p=6n+1 \"or\" p=6n+5p=6n+5 (p=6n−1p=6n−1), ? Math Expert Joined: 02 Sep 2009 Posts: 60647 Which of the following CANNOT be the sum of two prime numbers? [#permalink] ### Show Tags 19 Jun 2017, 12:49 1 Nunuboy1994 wrote: Bunuel wrote: banksy wrote: 187. Which of the following CANNOT be the sum of two prime numbers? (A) 19 (B) 45 (C) 68 (D) 79 (E) 88 Any prime number more than 3 can be expressed as $$p=6n+1$$ or$$p=6n+5$$ ($$p=6n-1$$), where n is an integer >0 (check this: http://gmatclub.com/forum/primality-check-108425.html). So the sum of two primes more than 3 can yield the following remainders upon division by 6: 0 - if two primes are of a type $$p=6n+1$$ and $$p=6n+5$$; 2 - if", "What's that remainder? What is the remainder when $${2013 \\choose 101 }$$ is divided by 101? Hint: You may use the fact that 101 is a prime. ×", "# Primes all the way! What is the remainder when $$1993^{1993}$$ is divided by 13? × Problem Loading... Note Loading... Set Loading..." ]

[ "# Primes all the way! What is the remainder when $$1993^{1993}$$ is divided by 13? × Problem Loading... Note Loading... Set Loading...", "# 9. Write the natural numbers from $102$ to $113$. What fraction of them are prime numbers? Thus fraction is :- $=\\ \\frac{4}{12}\\ =\\ \\frac{1}{3}$", "1 million of the 1. 4 million of it. The sum of all the prime numbers less than 1,000,000 is 37,550,402,023. I know i need to use a loop, but not sure how to make it use the even numbers and find the s. If you win and choose the lump-sum payment, expect to pay north of $200 million in federal taxes, at the 39. 5 million Muslims from all over the world. Dim sum As Single = numbers. Write a program to find the sum of the first 1000 prime numbers. Players may pick six numbers from two separate pools of numbers - five different numbers from 1 to 70 (the white balls) and one number from 1 to 25 (the gold Mega Ball) - or select Easy Pick/Quick Pick. If you choose to take the lump sum payout, a$1. Lists of Perfect Squares and Perfect Square Generator : with Roots and Intervals : View in your browser, download files, or generate your own. On numbers which are the sum of two squares Leonhard Euler 1. Write \"1\" in the box on the left, and \"one\" will appear on the right. Thanks to Gauss, there is a special formula we can use to find the sum of a series: S is the sum of the series and", "the sum of $1999^{m}$ and $2001^{m}$ divided by 7 4 #### Quantity A The remainder when 12,345,678,910 is divided by 9 #### Quantity B 1 What is the remainder when $3^{35}$ is divided by 5? What is the remainder when $3^{283}$ is divided by 5? #### Quantity A The remainder when $7^{13}$ is divided by 10 #### Quantity B 7 How many primes can be formed when someone selects two different numbers from 2,3,4,5 as the tens digit and units digit, respectively? The cubic root of n is a prime,what could be the possible value of n? Indicate all such items. How many positive integers less than 20 are the product of two different prime integers? When x≠1, the function p(x) equals the number of different prime divisors of x. What is the value of $\\frac{p(12)}{p(9)}$? The number of attendees at a certain convention is less than 1,000 and is divisible by 50. Which of the following statement individually provide(s) sufficient additional information to determine the number of attendees at the convention? Indicate all such statement. 25000 +道题目 174本备考书籍", "^1000; Output: 76 cumulative prime sums: The sum of the first 1 (up to 2) is prime: 2 The sum of the first 2 (up to 3) is prime: 5 The sum of the first 4 (up to 7) is prime: 17 The sum of the first 6 (up to 13) is prime: 41 The sum of the first 12 (up to 37) is prime: 197 The sum of the first 14 (up to 43) is prime: 281 The sum of the first 60 (up to 281) is prime: 7,699 The sum of the first 64 (up to 311) is prime: 8,893 The sum of the first 96 (up to 503) is prime: 22,039 The sum of the first 100 (up to 541) is prime: 24,133 The sum of the first 102 (up to 557) is prime: 25,237 The sum of the first 108 (up to 593) is prime: 28,697 The sum of the first 114 (up to 619) is prime: 32,353 The sum of the first 122 (up to 673) is prime: 37,561 The sum of the first 124 (up to 683) is prime: 38,921 The sum of the first 130 (up to 733) is prime: 43,201 The sum of the first 132 (up to 743) is prime: 44,683 The sum of the first 146 (up to", "Works with: Factor version 0.99 2021-02-05 USING: assocs formatting kernel math.primes math.ranges math.statistics prettyprint ; 1000 [ [1,b] ] [ primes-upto cum-sum ] bi zip [ nip prime? ] assoc-filter [ \"The sum of the first %3d primes is %5d (which is prime).\\n\" printf ] assoc-each Output: The sum of the first 1 primes is 2 (which is prime). The sum of the first 2 primes is 5 (which is prime). The sum of the first 4 primes is 17 (which is prime). The sum of the first 6 primes is 41 (which is prime). The sum of the first 12 primes is 197 (which is prime). The sum of the first 14 primes is 281 (which is prime). The sum of the first 60 primes is 7699 (which is prime). The sum of the first 64 primes is 8893 (which is prime). The sum of the first 96 primes is 22039 (which is prime). The sum of the first 100 primes is 24133 (which is prime). The sum of the first 102 primes is 25237 (which is prime). The sum of the first 108 primes is 28697 (which is prime). The sum of the first 114 primes is 32353 (which is prime). The sum of the first 122 primes is 37561 (which is prime). The sum of", "far as primates go, this just flew into my head today: The (sum of the) first prime is 2, for our two hands. The sum of the first two primes is 5 (2 + 3), for the fingers on one hand. The sum of the first three primes is 10 (2 + 3 + 5), for all of the fingers (my apologies to those with less, or more than five fingers on each hand). The sum of the first four primes is 17. 17 is wonderful and popular, but I don’t know more than that along these lines. That is a small tidbit, but it makes me that much happier to be a primate… This site uses Akismet to reduce spam. Learn how your comment data is processed.", "each pricing option, cost (in dollars) depends on miles driven, as shown below... ### What is the sum of the first five prime numbers?18262839 What is the sum of the first five prime numbers? 18 26 28 39... ### A box of 84 pinguins for 13,597 or a bag of 50 penguins for 795.95 A box of 84 pinguins for 13,597 or a bag of 50 penguins for 795.95... ### If possible can I get help on both? If possible can I get help on both? $If possible can I get help on both?$...", "($\\frac{N}{2}$) ϕ(N) ##### Q. Find the sum of all co-primes to 18, which are less than 18. Explanation: Sum = $\\frac{18^2}{2}$ [1 − $\\frac{1}{2}$] [ 1 − $\\frac{1}{3}$] = 54 Previous Share on:", "an estimate for $$\\sqrt{2}$$ and for $$\\sqrt{3}$$, you can multiply them to get an estimate for $$\\sqrt{6}$$. In this case, using 1.414 and 1.732, we multiply them to get 2.449. This means you only really need to calculate the roots of primes manually, and keep a list handy.", "+1 vote 34 views Choose the correct option: The remainder obtained when a prime number greater than $6$ is divided by $6$ is: 1. 1 or 3 2. 1 or 5 3. 3 or 5 4. 4 or 5 +1 vote Every prime number greater than 6 can be expressed as $6k \\pm 1, k > 0$ So, the remainder is 1 or 5. answered by (1.5k points) 1 3 14 selected by", "# You ler no help here What is the remainder when $$16^{123}$$ is divided by $$257$$? Details and assumptions You may use the fact that 257 is a prime number. × Problem Loading... Note Loading... Set Loading...", "numbers is, $$1^2 + 2^2 + ... + 10^2 = 385$$ The square of the sum of the first ten natural numbers is, $$(1 + 2 + ... + 10)^2 = 552 = 3025$$ Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 − 385 = 2640$. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum. In [14]: def square_sum(l): return sum(l) ** 2 def sum_square(l): return sum(x*x for x in l) def sum_square_difference(n): numbers = range(1, n + 1) return abs(square_sum(numbers) - sum_square(numbers)) print(sum_square_difference(100)) 25164150 Problem 7 10001st prime By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10 001st prime number? In [15]: def nth_prime(n): count = 1 current = 3 while 1: if is_prime(current): count += 1 if count == n: return current current += 2 print(nth_prime(10001)) 104743 Problem 8 Largest product in a series The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832. Find the thirteen adjacent digits in the 1000-digit number that have the greatest", "The sum of the first 630 (up to 4657) is prime: 1,359,329 The sum of the first 676 (up to 5051) is prime: 1,583,293 The sum of the first 680 (up to 5087) is prime: 1,603,597 The sum of the first 696 (up to 5233) is prime: 1,686,239 The sum of the first 708 (up to 5351) is prime: 1,749,833 The sum of the first 734 (up to 5563) is prime: 1,891,889 The sum of the first 762 (up to 5807) is prime: 2,051,167 The sum of the first 768 (up to 5849) is prime: 2,086,159 The sum of the first 776 (up to 5897) is prime: 2,133,121 The sum of the first 780 (up to 5939) is prime: 2,156,813 The sum of the first 784 (up to 6007) is prime: 2,180,741 The sum of the first 808 (up to 6211) is prime: 2,327,399 The sum of the first 814 (up to 6263) is prime: 2,364,833 The sum of the first 820 (up to 6301) is prime: 2,402,537 The sum of the first 836 (up to 6427) is prime: 2,504,323 The sum of the first 844 (up to 6529) is prime: 2,556,187 The sum of the first 848 (up to 6563) is prime: 2,582,401 The sum of the first 852 (up to 6581) is prime: 2,608,699 The sum", "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 $499$.", "prime: 342,761 The sum of the first 342 (up to 2297) is prime: 360,979 The sum of the first 350 (up to 2357) is prime: 379,667 The sum of the first 356 (up to 2393) is prime: 393,961 The sum of the first 358 (up to 2411) is prime: 398,771 The sum of the first 426 (up to 2957) is prime: 581,921 The sum of the first 446 (up to 3137) is prime: 642,869 The sum of the first 458 (up to 3251) is prime: 681,257 The sum of the first 460 (up to 3257) is prime: 687,767 The sum of the first 464 (up to 3301) is prime: 700,897 The sum of the first 480 (up to 3413) is prime: 754,573 The sum of the first 484 (up to 3461) is prime: 768,373 The sum of the first 488 (up to 3491) is prime: 782,263 The sum of the first 512 (up to 3671) is prime: 868,151 The sum of the first 530 (up to 3821) is prime: 935,507 The sum of the first 536 (up to 3863) is prime: 958,577 The sum of the first 548 (up to 3947) is prime: 1,005,551 The sum of the first 568 (up to 4129) is prime: 1,086,557 The sum of the first 620 (up to 4583) is prime: 1,313,041", "2 typo I was investigating primes with the property that the sum of the first $n$ primes is divisible by $p_n$. It turns out that these primes are extremely extremely rare. For primes less than $10^9$, I have found that there are only five primes with this property: $$p_1 = 2$$ $$p_2 p_3 = 5$$ $$p_{20} = 71$$ $$p_{31464} = 369,119$$ $$p_{22096548} = 415,074,643$$ This raises the curious and equivalent questions: Q1. Are there infinitely many primes which divide the sum of all the preceding primes? Q2. Even if we assume that there are infinitely many such primes, why are they so rare? In other words, why do primes dislike dividing the sum of all the preceding primes? Is there any heuristic argument to show that such primes will indeed be extremely rare? 1 # Why do primes dislike dividing the sum of all the preceding primes? I was investigating primes with the property that the sum of the first $n$ primes is divisible by $p_n$. It turns out that these primes are extremely extremely rare. For primes less than $10^9$, I have found that there are only five primes with this property: $$p_1 = 2$$ $$p_2 = 5$$ $$p_{20} = 71$$ $$p_{31464} = 369,119$$ $$p_{22096548} = 415,074,643$$ This raises the curious and equivalent questions: Q1. Are there", "\\frac{n}{5^i}\\right \\rfloor$$. At some point, the $$5^i$$ 's will get bigger than n, so the sum is finite. The reason that the formula works is that you are doing this: first you count the numbers less than or equal to n that are divisible by 5.But then you would have missed the extra powers from the ones that are divisible by 25 etc. The formula works for any prime p, not just for 5. It is an easy way to figure out the prime factorization of n! without actually calculating it. For instance, the highest power of 10 that divides 50! is $$10+2+0+0...=\\boxed{12}$$ - 4 years, 4 months ago ×", "Overview Question The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. 142913828922", "factors are 2,3,5,7 so we have $$\\varphi (1260)=1260\\ (1-\\frac12)(1-\\frac13)(1-\\frac15)(1-\\frac17) = 288$$ It works like this. The primes that divide 1260 are 2, 3, 5, 7. For each of these, there is only one remainder that indicates the number is a multiple of the prime: that is 0. So the number of co-primes is thus all combinations of non-division, ie [ (2-1)(3-1)(5-1)(7-1)/2.3.5.7 ] * 1260. We see the square brackets give 48/210, and this by 1260 gives 288." ]

[ "but I’ll give you a couple of hints. I described Euclid’s proof that there are infinitely many primes. Suitable variants of that will give proofs that there are infinitely many primes leaving remainder 1 when divided by 4 and infinitely many primes leaving remainder 3 when divided by 4. Are there the same numbers of primes leaving remainder 1 and remainder 3 when divided by 4? It turns out that the answer is yes (when the question is interpreted carefully), but that’s harder to prove, so I’ll leave it for now. What would happen if we replaced 4 by 6, for example? Here are the remainders when the first few primes are divided by 6: 2, 3, 5, 1, 5, 1, 5, 1, 5, 5, 1, 1, 5, 1, 5, 5, 5, 1, 1, 5, 1, 1, 5, 5, 1. Hopefully you will be able to convince yourself that apart from 2 and 3, every prime leaves remainder 1 or 5 when divided by 6. But are there infinitely many of each sort? Notice that this time there are 11 primes leaving remainder 1 when divided by 6 and 12 leaving remainder 5 — again, the numbers are very close. It starts getting a bit harder to prove that there are infinitely many primes leaving remainder 1 when", "there are infintely or finitely many primes of the form $p^2+4$, then all of them will leave a remainder $5$ when divided by $6$. – Abhishek Bakshi Jul 15 '18 at 4:31 • Alright, I deleted it. – Pacifica Jul 15 '18 at 5:03", "# Primes all the way! What is the remainder when $$1993^{1993}$$ is divided by 13? × Problem Loading... Note Loading... Set Loading...", "precise about the definition of your sequence: the least prime giving a prime remainder when divided by 3 (or 5 or 7 or 11 or ...) is 2; no prime leaves an odd remainder when divided by 2. I think the sequence you are interested in may be the sequence of odd primes that leave a prime remainder when divided by any smaller odd prime. But then it should include 3. – Rob Arthan Jul 26 '16 at 20:12 • The remainder can be ANY prime. Look at 47 to get 47mod3=2, 47mid5=2, 47mod7=5 and 47mod11=3. If you looked at the second mention of 17, you'd see that a prime remainder means ANY prime remainder. It is NOT the remainder that is specified, but the least prime such that all its divisors 3,5,7...give ANY prime remainder. Just try 863 when divided by 3,5,7,11,13 and you'll see a variety of remainders. Now, I do agree that extending this is a challenge. If no one is up to it, that's not surprising. – J. M. Bergot Jul 26 '16 at 20:17 • @RobArthan, I think it would suffice for the OP to ask for the least odd prime giving a prime remainder when divided by each of the first $n$ odd primes. – Barry Cipra Jul 26 '16 at 20:17", "# Math Help - Prime Numbers and Divisibility 1. ## Prime Numbers and Divisibility What is the remainder when 5!25! is divided by 31? 2. Originally Posted by Aryth What is the remainder when 5!25! is divided by 31? $5! 25! \\equiv 5! 23! (-7) (-6) \\equiv 5! 23! 11 \\mod 31$ ................. $\\equiv 5! 22! 5 \\mod 31$ ................. $\\equiv 5! 21! 17 \\mod 31$ etc CB 3. Or use Wilson’s theorem. $30!\\equiv-1\\pmod{31}$ $30\\times29\\times28\\times27\\times26\\equiv(-1)(-2)(-3)(-4)(-5)\\pmod{31}\\equiv-5!\\pmod{31}$ $\\therefore\\ 5!25!\\equiv-30!\\pmod{31}\\equiv1\\pmod{31}$", "(P-3)! is divided by P where P is a prime number ###### Let us discuss these types in detail. Type 1: When (P-1)! is divided by P where P is a prime number, the Remainder is (P-1). Example: If P=17 then (p-1)=16 When 16! Is divided by 17 the remainder is 16. Type 2: When (P-2)! is divided by P where P is a prime number, the Remainder is 1. Example: If P=17 then (p-2)=15 When 15! Is divided by 17 the remainder is 1 Type 3: When (P-3)! is divided by P where P is a prime number, the Remainder is (P-1)/2. Example: If P=17 then (p-3)=14 When 14! Is divided by 17 the remainder is (17-1)/2=8", "there are infinitely many primes that leave a remainder of 3 when divided by 4. In fact, it is known that there are only two primes that leave a remainder of 3 when divided by 4: 3 and 11. All other primes leave a remainder of 1 when divided by 4. This can be shown by noting that if a prime number p is of the form 3 (mod 4), then it can be expressed as 4n+3 for some integer n. Then, the number (4n+3)^2 = 16n^2 + 12n + 9 = 4(4n^2 + 3n + 2) + 1, which is 1 more than a multiple of 4, which means it is not prime. Here it becomes sketchy. The first statement is clearly wrong since there are more prime numbers that leave a remainder of 3 when divided by 4. For example, 19, 23, 31, 43, and 47 are all prime numbers and when divided with 4 leave a remainder equal to 3. But what about the second statement? The conclusion it draws is logically not properly derived! Sure, the number can be expressed as $4n+3$, and its square is equal to $4(4n^2+3n+2)+1$, which is 1 more than a multiple of 4 (and hence leaves remainder 1 when divided by 4), but the final conclusion is just in", "a remainder of 5, so it has to be prime... i found the reasoning behind this in one of the OG solutions: prime numbers always take the form: 6n+1 or 6n+5 .... the only possible remainders when any number is divided by 6 are [0,1,2,3,4,5] ... A prime number always gives a remainder of 1 or 5, because: a) if the remainder is 2 or 4, then the number must be even b) if the remainder is 3, then it is divisible by 3 ... hence, if a number divided by 6 yields 1 or 5 as its remainder, then it must be prime ...? -Raj First of all there is no known formula of prime numbers. Next: Any prime number $$p>3$$ when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case $$p$$ would be even and remainder can not be 3 as in this case $$p$$ would be divisible by 3). So any prime number $$p>3$$ could be expressed as $$p=6n+1$$ or$$p=6n+5$$ or $$p=6n-1$$, where n is an integer >1. But: Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so vise-versa of above property is not correct. For example 25 yields a remainder of 1", "# Find The Remainder Find the remainder when $$999! \\times 1011!$$ is divided by $$2011.$$ Details and assumptions • You might use the fact that $$2011$$ is a prime. ×", "19 Apr 2017, 11:51 saurabhsavant wrote: vitaliyGMAT wrote: Bunuel wrote: For prime number y, y>3. Which of the following could be the remainder when y^3 is divided by 12? A. 2 B. 3 C. 4 D. 5 E. 6 Prime number greater than 3 can be expressed as $$6k + 1$$ or $$6k -1$$. $$\\frac{(6k ± 1)^3}{12} = \\frac{6^3k^3 ± 3*6^2k^2 + 3*6k ± 1}{12} = \\frac{0 + 0 + 18k ± 1}{12} = \\frac{6k ± 1}{12}$$ Let’s check first case $$6k + 1$$ k=1 (7/12) rem=7 k=2 (13/12) rem =1 k=3 (19/12) rem = 7 k =4 (25/12) rem = 1 we have cycle of remainders $$7$$ and $$1$$. Next $$6k – 1$$ k=1 (5/12) rem=5 k=2 (11/12) rem =11 k=3 (17/12) rem = 5 k =4 (23/12) rem = 11 again cycle of only two remainders $$5$$ and $$11$$. The only remainders after division by 12 cube of prime number $$y > 3$$ can have are $$1$$, $$5$$, $$7$$ and $$11$$. Checking our options. I have a doubt here please....prime numbers graeter than 3 can be expressed as 6k+1..putting k=4 we get 25, which is not a prime num,ber....am I missing something here? Any prime number $$p>3$$ when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4", "# A sum of 24 is divided into two parts so that seven times the first part added to five times the second part makes 146 $${}$$", "a remainder of 2 upon division by 6 so theoretically can be the sum of two primes (and it is 61+7=68); (D) 79 --> 79-2 is not a prime, 79-3 is not a prime and also 79 yields a remainder of 1 upon division by 6, so it can not be the sum of two primes; (E) 88 --> 71+17=88. Of course the above can be done much easier by just subtracting the primes starting from 2 from the answer choices and seeing whether the result is also a prime. Bunuel when you say p=6n+1p=6n+1 orp=6n+5p=6n+5 (p=6n−1p=6n−1), do you mean p=6n+1p=6n+1 \"or\" p=6n+5p=6n+5 (p=6n−1p=6n−1), ? Math Expert Joined: 02 Sep 2009 Posts: 60647 Which of the following CANNOT be the sum of two prime numbers? [#permalink] ### Show Tags 19 Jun 2017, 12:49 1 Nunuboy1994 wrote: Bunuel wrote: banksy wrote: 187. Which of the following CANNOT be the sum of two prime numbers? (A) 19 (B) 45 (C) 68 (D) 79 (E) 88 Any prime number more than 3 can be expressed as $$p=6n+1$$ or$$p=6n+5$$ ($$p=6n-1$$), where n is an integer >0 (check this: http://gmatclub.com/forum/primality-check-108425.html). So the sum of two primes more than 3 can yield the following remainders upon division by 6: 0 - if two primes are of a type $$p=6n+1$$ and $$p=6n+5$$; 2 - if", "# A huge remainder Find the remainder when $$(270!)^2$$ is divided by 541. - Do not use WolframAlpha. - You may want to use the fact that $$541$$ is prime. ×", "# You ler no help here What is the remainder when $16^{123}$ is divided by $257$? Details and assumptions You may use the fact that 257 is a prime number. ×", "+1 vote 34 views Choose the correct option: The remainder obtained when a prime number greater than $6$ is divided by $6$ is: 1. 1 or 3 2. 1 or 5 3. 3 or 5 4. 4 or 5 +1 vote Every prime number greater than 6 can be expressed as $6k \\pm 1, k > 0$ So, the remainder is 1 or 5. answered by (1.5k points) 1 3 14 selected by", "the remainder obtained by dividing $x^7 + x ^5 + 1$ by the generator polynomial $x^ 3 + 1?$", "# 9. Write the natural numbers from $102$ to $113$. What fraction of them are prime numbers? Thus fraction is :- $=\\ \\frac{4}{12}\\ =\\ \\frac{1}{3}$", "^1000; Output: 76 cumulative prime sums: The sum of the first 1 (up to 2) is prime: 2 The sum of the first 2 (up to 3) is prime: 5 The sum of the first 4 (up to 7) is prime: 17 The sum of the first 6 (up to 13) is prime: 41 The sum of the first 12 (up to 37) is prime: 197 The sum of the first 14 (up to 43) is prime: 281 The sum of the first 60 (up to 281) is prime: 7,699 The sum of the first 64 (up to 311) is prime: 8,893 The sum of the first 96 (up to 503) is prime: 22,039 The sum of the first 100 (up to 541) is prime: 24,133 The sum of the first 102 (up to 557) is prime: 25,237 The sum of the first 108 (up to 593) is prime: 28,697 The sum of the first 114 (up to 619) is prime: 32,353 The sum of the first 122 (up to 673) is prime: 37,561 The sum of the first 124 (up to 683) is prime: 38,921 The sum of the first 130 (up to 733) is prime: 43,201 The sum of the first 132 (up to 743) is prime: 44,683 The sum of the first 146 (up to", "Suppose, P= 5, P^2 + 17 = 5^2 + 17 = 25 + 17 = 42 Now 42/ 12 = 3 + 6/12, Remainder = 6(C) Re: If p is a prime number greater than 3, find the remainder when p^2 + 1 [#permalink] 09 Feb 2020, 02:15 Display posts from previous: Sort by", "Edited the question Math Expert Joined: 02 Sep 2009 Posts: 49303 Re: Is p^2 - 1 divisible by 12? [#permalink] ### Show Tags 15 Aug 2014, 09:23 2 3 Is p^2 - 1 divisible by 12? (1) p > 3. This one is clearly insufficient: if p = 4, then the answer is NO but if p = 5, then the answer is YES. Not sufficient. (2) p is a prime number. If p = 2, then the answer is NO but if p = 5, then the answer is YES. Not sufficient. (1)+(2) Important property: ANY prime number $$p$$ greater than 3 can be expressed as $$p=6n+1$$ or $$p=6n+5$$ ($$p=6n-1$$), where $$n$$ is an integer >1. That's because any prime number $$p$$ greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case $$p$$ would be even and remainder can not be 3 as in this case $$p$$ would be divisible by 3). But: Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so vise-versa of above property is not correct. For example 25 (for $$n=4$$) yields a remainder of 1 upon division by 6 and it's not a prime number. So," ]

[ "and $5$. What is the probability that the integer is divisible by $5$?", "number. What exactly does this digital sum show us? A positive integer is divisible by 3 if and only if its digital sum is divisible by 3. So you can actually repeat the digitial sum process iteratively (take the digital sum of the digital sum of the digital sum....) until you reach a single digit, if you wanted to. An integer is divisible by 6 if and only if it is divisible by 2 and it is divisible by 3.", "## Spring 2017, problem 36 Show that in every sequence of $79$ consecutive positive integers, there is a positive integer whose sum of digits is divisible by $13$.", "Divisible Product If $$m$$ is an integer, then $$(m+1)(m+3)(m+5)$$ must be divisible by which of the following integers? ×", "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 $18=2\\cdot 9$)", "Question # Choose the correct answer from the alternatives given.The sum of the digits of a number A is $$13$$. What is the remainder when A is divided by $$3$$? Solution ## Since the sum of the digits is 13, the number when divided by 3 leaves remainder 1.Because if the sum of the digits would have been 12 then it would have been divisible by 3.Since 13 is 1 bigger than 12, remainder will be 1 when divided by 3.Mathematics Suggest Corrections 0 Similar questions View More", "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 last digit of the number. For example, alternating sum of the digits of $12345$ is $1 - 2 + 3 - 4 + 5$. Likewise, the alternating sum of the digits of $9876$ is $9 - 8 + 7 - 6$. Show that a positive integer is divisible by $11$ if and only if its alternating sum is divisible by $11$. 17. 🔎 Let $a$, $b$, $c$, $d$ be integers. Prove that $12 \\mid (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$", "Divisible Probability... The sum of the digits of a $$9$$ digit number is $$77$$. The probability that this number is divisible by 11 can be expressed as $$\\dfrac{a}{b}$$ where $$a$$ and $$b$$ are co-prime positive integers. Find the value of $$a+b$$. ×", "# Does 2014 work? A positive integer $$N$$ is divisible by 2014. Find the smallest possible value of the digit sum of $$N$$. Details and assumptions The digit sum of a number is the sum of all its digits. For example the digit sum of 1123 is $$1 + 1 + 2 + 3 = 7$$. ×", "# Filling a 13-by-13 square Is it possible to arrange the integers $1,2,3,\\ldots,169$ in a $13\\times13$ square, so that in every $2\\times2$ square the sum of the four numbers is divisible by $170$?", "to 2. - There is no integer that is only divisible by $3$. Every integer is always divisible by $1$.", "sum of the digits.) • Note that divisibility by $11$ alone is not enough to show that the number is not a sum of two squares; you also need the odd power here. – Dirk Aug 1 '18 at 13:20 • @DirkLiebhold True. It's enough when $11$ occurs to an odd power - and easy to prove when the power is $1$. – Ethan Bolker Aug 1 '18 at 13:59", "# Divisible Product Number Theory Level 1 If $$m$$ is an integer, then $$(m+1)(m+3)(m+5)$$ must be divisible by which of the following integers? ×", "uses “if and only if”, see the Glossary article on “if”.) ### Unmarked definitions Math texts sometimes define something without saying that it is a definition. Because of that, students may sometimes think a claim is a definition. ##### Example Suppose that the concept of “even integer” was new to you and the book said, “A number is even if it is divisible by $4$.” Perhaps you thought that this was a definition. Later the book refers to $6$ as even and you pull your hair out wondering why. The statement is a correct claim but an incorrect definition. A good writer would write something like “Recall that a number is even if it is divisible by $2$, so that in particular it is even if it is divisible by $4$.” On the other hand, you may think a definition is only a claim. ##### Example A lecturer may say “By definition, an integer is even if it is divisible by $2$”, and you write down: “An integer is even if it is divisible by $2$”. Later, you get all panicky wondering How did she know that?? (This has happened to me.) The confusion in the preceding example can also occur if a books says, “An integer is even if it is divisible by $2$” and you don’t", "the ratio of the sum and the product of all prime divisors of the number 120.", "# Playing with 13 Find the least positive integer $$n$$ such that, $$n$$ is divisible by 13 and the sum of the digits of $$n$$ is 13. ×", "An integer is said to be nine-separable if it can be represented as the product of two positive integers that differ by nine. (For example, since $$22=2\\times 11$$, $$22$$ is nine-separable.) Similarly, an integer is said to be eighteen-separable if it can be represented as the product of two positive integers that differ by $$18$$. What is the sum of all positive integers that are both nine-separable and eighteen-separable?", "# Divisible Product If $$m$$ is an integer, then $$(m+1)(m+3)(m+5)$$ must be divisible by which of the following integers? × Problem Loading... Note Loading... Set Loading...", "Sum Of Digits Rule An integer is said to be divisible by $$n$$ if the sum of digits of the same integer is divisible by $$n$$ as well. Which of the following are possible values of $$n$$? (A): 3 (B): 6 (C): 9 × Problem Loading... Note Loading... Set Loading..." ]

[ "Take any pair of two digit numbers $ab$ and $cd$ where, without loss of generality, $ab> cd$. Form two 4 digit numbers $abcd$ and $cdab$ and calculate: $\\frac{abcd^2-cdab^2}{ab^2-cd^2}$ Repeat this with other choices of $ab$ and $cd$. There is a common feature of all the answers. What is it? Why does this occur? Generalise this to $n$ digits for other values of $n$.", "#2 What are the tens and units digits of $\\displaystyle 7^{1942}$? We could (in theory) solve this problem by trying to compute $7^{1942}$, but this would be extremely time-consuming. Moreover, it would give us much more information than we need. Since we want only the tens and units digits of the number in question, it suffices to find the remainder when the number is divided by $100$. In other words, all of the information we need can be found using arithmetic mod $100$. We begin by writing down the first few powers of $7$ mod $100$: $7, 49, 43, 1, 7, 49, 43, 1, \\ldots$ A pattern emerges! We see that $7^4 = 2401 \\equiv 1$ (mod $100$). So for any positive integer $k$, we have $7^{4k} = (7^4)^k \\equiv 1^k \\equiv 1$ (mod $100$). In particular, we can write $7^{1940} = 7^{4 \\cdot 485} \\equiv 1$ (mod $100$). By the \"multiplication\" property above, then, it follows that $7^{1942} = 7^{1940} \\cdot 7^2 \\equiv 1 \\cdot 7^2 \\equiv 49$ (mod $100$). Therefore, by the definition of congruence, $7^{1942}$ differs from $49$ by a multiple of $100$. Since both integers are positive, this means that they share the same tens and units digits. Those digits are $4$ and $9$, respectively. ## Summary of Useful Facts Consider four integers ${a},{b},{c},{d}$", "= 106$, and $AB = 106 + 87 = \\boxed{193}$.", "First video: Alison chose 42, added 24 and got the answer 66: \"It is! How on earth did you know that?\" Charlie said: \"I'm not sure. Let's try to work it out.\" Second video: Alison arranged multilink to show four tens and two units for 42, and two tens and four units for 24. She then put the four units with the four tens, and the two units with the two tens, giving six lots of eleven. Charlie imagined a two-digit number $ab$, where $a$ represents the number in the tens column, and $b$ respresents the number in the units. This can be written as $10a+b$. Similarly, $ba$ can be written as $10b+a$. Charlie added these together to get $11a+11b$, which he wrote as $11(a+b)$.", "Question #9c419 Dec 3, 2016 58 Explanation: Let in the two digit number the tens digit be x and units digit be y . So the number is $\\text{ } 10 x + y$ By the 1st condition $\\text{ } y - x = 3. \\ldots \\ldots \\left(1\\right)$ And by the2nd condition $10 x + y - 4 \\left(x + y\\right) = 6$ $\\implies 6 x - 3 y = 6$ $\\implies 2 x - y = 2. \\ldots . \\left(2\\right)$ Adding (1) and (2) we gwt $x = 5$ Inserting $x = 5$ in (1) we get $y - 5 = 3 \\implies y = 8$ Hence the the two digit number is $10 x + y = 10 \\times 5 + 8 = 58$ Dec 3, 2016 See explanation. Explanation: If we denote the tens digit by $a$ and unit digit by $b$ then the question leads to the following equations: Tens digit is 3 less than units digit: $a = b - 3$. The original number is six more than four times than the sum of dygits: $10 a + b - 6 = 4 \\cdot \\left(a + b\\right)$. If we solve the system we get: $\\left\\{\\begin{matrix}a = b - 3 \\\\ 10 a + b - 6 = 4 a + 4 b\\end{matrix}\\right.$ $\\left\\{\\begin{matrix}a", "## 'Always a Multiple?' printed from http://nrich.maths.org/ First video: Charlie said: \"Alison, think of a two-digit number. Reverse the digits and add your answer to your original number. I bet your answer is a multiple of 11.\" Alison chose 42, added 24 and got the answer 66: \"It is! How on earth did you know that?\" Charlie said: \"I'm not sure. Let's try to work it out.\" Second video: Alison arranged multilink to show four tens and two units for 42, and two tens and four units for 24. She then put the four units with the four tens, and the two units with the two tens, giving six lots of eleven. Charlie imagined a two-digit number $ab$, where $a$ represents the number in the tens column, and $b$ respresents the number in the units. This can be written as $10a+b$. Similarly, $ba$ can be written as $10b+a$. Charlie added these together to get $11a+11b$, which he wrote as $11(a+b)$.", "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 $3, 4, 5, 6, 7$.", "+ ab + b^2$ is a multiple of 55 so it is not in U(55).", "# Algebra Suppose $A$ and $B$ are digits for which the product of the two-digit numbers $\\underline{2A}$ and $\\underline{AB}$ is 1971. Find the sum of $A$ and $B$. Note: This is LaTeX, so if you need to render it, then go ahead 1. 👍 2. 👎 3. 👁 1. why bother wit LaTex? Just type it in: Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B. 2A*AB = (20+A)(10A+B) = 1971 1971 = 3^3*73 So, its only two-digit factors are 27 and 73 A+B=10 1. 👍 2. 👎 ## Similar Questions 1. ### Maths In A Two Digit Number, The Sum Of The Digits Is 14. Twice The Tens Digit Exceeds The Units Digit By One.Find the Numbers. 2. ### maths when a 3 digits number is reversed, the number decreases by 396. the diffrence between the digits at the units place and the digit at the tens place is the same as the difference between the digits at the tens place and the 3. ### maths can you replace the stars with figures **** x 3 ______ ***** the whole calculation uses each of the digits 0-9 once and once only the 4 figure number contains three consecutive numbers which are not in order. the", "## If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What ##### This topic has expert replies Moderator Posts: 1372 Joined: 29 Oct 2017 Thanked: 1 times Followed by:5 members ### If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What by M7MBA » Thu Sep 17, 2020 1:03 am 00:00 A B C D E ## Global Stats If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What is the value of $$a+b?$$ A) 1 B) 2 C) 5 D) 6 F) 8 Source: e-GMAT ### GMAT/MBA Expert GMAT Instructor Posts: 5872 Joined: 25 Apr 2015 Location: Los Angeles, CA Thanked: 43 times Followed by:24 members ### Re: If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ Wha by [email protected] » Thu Sep 24, 2020 7:04 am M7MBA wrote: Thu Sep 17, 2020 1:03 am If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What is the value of $$a+b?$$ A) 1 B) 2 C) 5 D) 6 F) 8 Solution: Recall the units digit pattern of powers of 7 is", "## Sum of Powers of Five Compute the units digit of $1^5+2^5+3^5+4^5+5^5+6^5+\\cdots+2006^5+2007^5$. Source: NCTM Mathematics Teacher 2008 SOLUTION 1 $1^5=1$ $1^5+2^5=33$ $1^5+2^5+3^5=276$ $1^5+\\cdots+4^5=1300$ $1^5+\\cdots+5^5=4425$ $1^5+\\cdots+6^5=12201$ $1^5+\\cdots+7^5=29008$ $1^5+\\cdots+8^5=61776$ $1^5+\\cdots+9^5=120825$ $1^5+\\cdots+10^5=220825$ $\\cdots$ $1^5+\\cdots+40^5=734933200$ The units digits of $1^5+2^5+3^5+\\cdots+n^5$ follows the following pattern. If $n\\equiv 0\\bmod 4$, the units digits of the sum are $0,6,8,6,0$. For example, $n=4$, units digit = $0$ $n=8$, units digit = $6$ $n=12$, units digit = $8$ $n=16$, units digit = $6$ $n=20$, units digit = $0$ If $n\\equiv 1\\bmod 4$, the units digits are $1,5,5,1,3$. For example, $n=1$, units digit = $1$ $n=5$, units digit = $5$ $n=9$, units digit = $5$ $n=13$, units digit = $1$ $n=17$, units digit = $3$ If $n\\equiv 2\\bmod 4$, the units digits are $3,1,5,5,1$. For example, $n=2$, units digit = $3$ $n=6$, units digit = $1$ $n=10$, units digit = $5$ $n=14$, units digit = $5$ $n=18$, units digit = $1$ If $n\\equiv 3\\bmod 4$, the units digits are $6,8,6,0,0$. For example, $n=3$, units digit = $6$ $n=7$, units digit = $8$ $n=11$, units digit = $6$ $n=15$, units digit = $0$ $n=19$, units digit = $0$ Since $2007\\equiv 3\\bmod 4$, the units digit of the sum = $8$. SOLUTION 2 $1^5=1$ $2^5=32$ $3^5=243$ $4^5=1024$ $5^5=3125$ $6^5=7776$ $7^5=16807$ $8^5=32768$ $9^5=59049$ $10^5=100000$ In a base-10 numbering system the units", "+0 # Suppose $(b_a)^2=71_a$, where $a$ and $b$ represent two distinct digits. If $b=a-1$, find $a$. 0 76 3 +65 Sort: #1 +560 +2 We can start by replacing b with (a-1). \\begin{align*} (a-1)^2\\cdot a^0&=7\\cdot a^1+1\\cdot a^0 \\end{align*} So we can have 0 or 9. But there is no base 0, so must be base 9. supermanaccz Mar 21, 2018 #2 +65 0 thanks", "tens digit can be a as well as c , can't it. In a decimal notation there is one units digit but 2 tens and 2 hundreds etc . eg 12345.678 here 5 - units place 4 tens place also 6 tens place so when we have ab.c and we talk of tens digit how do I know we are talking of a or c ? Thanks. _________________ - Stne Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Tens Digit? [#permalink] ### Show Tags 26 Jun 2013, 00:14 2 stne wrote: Bunuel wrote: What is the tens digit of the positive integer r? Let $$r=abc$$, tens digit would be $$b$$, so the question is $$b=?$$ (1) The tens digit of r/10 is 3 --> $$\\frac{r}{10}=ab.c$$ --> tens digit of this number is $$a$$, so $$a=3$$. No info about $$b$$. Not sufficient. (2) The hundreds digit of 10r is 6 --> $$10r=abc0$$ --> hundreds digit of this number is $$b$$, so $$b=6$$. Sufficient. Answer: B. Hi Can anybody explain to me in the case ab.c above, tens digit can be a as well as c , can't it. In a decimal notation there is one units digit but 2 tens and 2 hundreds etc . eg 12345.678 here 5 - units place 4 tens place also 6 tens", "divisible by the tens digit of A. Kudos for a correct solution. [Reveal] Spoiler: OA _________________ Retired Moderator Joined: 06 Jul 2014 Posts: 1269 Location: Ukraine Concentration: Entrepreneurship, Technology GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 The difference between positive two-digit integer A and the smaller tw [#permalink] ### Show Tags 23 Apr 2015, 11:13 6 KUDOS 2 This post was BOOKMARKED 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", "# The sum of the digits of a two-digit number is 11. The tens digit is one less than three times the ones digit. What is the original number? Mar 10, 2017 Number = 83 #### Explanation: Let the number in unit place be $x$ & the number in tens place be $y$. According to the first condition, $x + y = 11$ According to the second condition, $x = 3 y - 1$ Solving two simultaneous equation for two variable: $3 y - 1 + y = 11$ $4 y - 1 = 11$ $4 y = 12$ $y = 3$ $x = 8$ The original number is $83$", "First video: Charlie imagined a two-digit number $ab$, where $a$ represents the number in the tens column, and $b$ respresents the number in the units. This can be written as $10a+b$. Similarly, $ba$ can be written as $10b+a$. Charlie added these together to get $11a+11b$, which he wrote as $11(a+b)$.", "a factor of one of the other two products as well. Thus, $42$ must be a mistake: Bart’s product is really either $40$ or $44$, and the products $90=ab$ and $36=bc$ are correct. Since $b>c$, the possibilities for $bc=36$ are $36\\cdot 1$, $18\\cdot 2$, $12\\cdot 3$, and $9\\cdot 4$. However, $b$ must be a factor of $90=ab$, so $b$ must be $18$ or $9$. If $b=18$, then $a=\\frac{90}{18}=5<b$, which is impossible, so $b=9$, $c=4$, and $a=\\frac{90}9=10$.", "DOTS Division Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}. Latin Numbers Can you create a Latin Square from multiples of a six digit number? 2-digit Square A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number? Novemberish Age 14 to 16 Challenge Level: (a) A four digit number (in base $10$) $aabb$ is a perfect square. Discuss ways of systematically finding this number. (b) Prove that $11^{10}-1$ is divisible by $100$.", "Novemberish a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100. Latin Numbers Can you create a Latin Square from multiples of a six digit number? 2-digit Square A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number? DOTS Division Age 14 to 16Challenge Level We have received a very clearly explained solution to this, but unfortunately whoever sent it did not include their name. If it was you, let us know! Let $ab=x$ and $cd=y$ (where $ab$ means $a$ as the tens digit, and $b$ as the ones digit, not $a$ times $b$). $\\frac{abcd^2-cdab^2}{ab^2-cd^2} = \\frac{(100x+y)^2 - (100y+x)^2}{x^2-y^2}$ $= \\frac{(10000x^2+200xy+y^2) - (10000y^2+200xy+x^2)}{x^2-y^2}$ $= \\frac{9999x^2 - 9999y^2}{x^2-y^2}$ $= \\frac{9999(x^2-y^2)}{x^2-y^2}$ $= 9999$ (since $x> y$ we are not dividing by zero)", "# 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. ×" ]

[ "# Determine the number of six-digit integers (no leading zeros) in which no digit may be repeated and divisible by 4? Determine the number of six-digit integers (no leading zeros) in which no digit may be repeated and divisible by $4$? I've tried solving this problem, but the result is different with the solution provided in the book... This is my way : in order the number to be divisible by $4$, the last two-digit must be divisible by $4$, so the possibilities are (I've group them) $(24,64,84,28,48,68), (20,40,60,80), (12,32,52,72,92,16,36,56,76,96)$ so I'll have three cases the first case : the number of six-digits integer that divisible by $4: 7 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 3 \\cdot 2 = 8820$ the second case: the number of six-digits integer that divisible by $4: 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 1 = 6720$ the third case: the number of six-digits integer that divisible by $4: 7 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 5 \\cdot2 = 14700$ so, the total is $30240$ but the answer provided in the book is $33,600$ What am I'm missing? Is this the correct approach?", "# 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 $156 + 96 + 84 = \\boxed{336}$.", "# divisibility This study of integers begins with the notion of divisibility: whether the division of two integers results in an integer. Because of the volatile nature of division, we frame this question in terms of integer multiplication. A nonzero integer $$b$$ divides an integer $$a$$ if there exists some integer $$n$$ such that $$a = bn$$. Here are a few examples: • 15 divides 45 since $$45 = 15 \\cdot 3$$ • 9 divides 45 since $$45 = 9 \\cdot 5$$ • 10 does not divide 45 since 45 is between $$40 = 10 \\cdot 4$$ and $$50 = 10 \\cdot 5$$ The above definition is synonymous with saying that $$b$$ may be expressed as a multiple of $$a$$. In fact, the following statements are equivalent and mean that $$a$$ is can be expressed as $$bn$$. • $$b$$ divides $$a$$ • $$b$$ is a divisor of $$a$$ • $$a$$ is divisible by $$b$$ • $$a$$ is a multiple of $$b$$ The notation for this concept is $$b \\mid a$$, which reads \"$$b$$ divides $$a$$\". Using this notation, the examples above become: • $$15 \\mid 45$$ since $$45 = 15 \\cdot 3$$ • $$9 \\mid 45$$ since $$45 = 9 \\cdot 5$$ • $$10 \\nmid 45$$ since $$10 \\cdot 4 < 45 < 10 \\cdot 5$$ Be careful!", "# divisibility This study of integers begins with the notion of divisibility: whether the division of two integers results in an integer. Because of the volatile nature of division, we frame this question in terms of integer multiplication. A nonzero integer $$b$$ divides an integer $$a$$ if there exists some integer $$n$$ such that $$a = bn$$. Here are a few examples: • 15 divides 45 since $$45 = 15 \\cdot 3$$ • 9 divides 45 since $$45 = 9 \\cdot 5$$ • 10 does not divide 45 since 45 is between $$40 = 10 \\cdot 4$$ and $$50 = 10 \\cdot 5$$ The above definition is synonymous with saying that $$b$$ may be expressed as a multiple of $$a$$. In fact, the following statements are equivalent and mean that $$a$$ is can be expressed as $$bn$$. • $$b$$ divides $$a$$ • $$b$$ is a divisor of $$a$$ • $$a$$ is divisible by $$b$$ • $$a$$ is a multiple of $$b$$ The notation for this concept is $$b \\mid a$$, which reads \"$$b$$ divides $$a$$\". Using this notation, the examples above become: • $$15 \\mid 45$$ since $$45 = 15 \\cdot 3$$ • $$9 \\mid 45$$ since $$45 = 9 \\cdot 5$$ • $$10 \\nmid 45$$ since $$10 \\cdot 4 < 45 < 10 \\cdot 5$$ Be careful!", "# divisibility This study of integers begins with the notion of divisibility: whether the division of two integers results in an integer. Because of the volatile nature of division, we frame this question in terms of integer multiplication. A nonzero integer $$b$$ divides an integer $$a$$ if there exists some integer $$n$$ such that $$a = bn$$. Here are a few examples: • 15 divides 45 since $$45 = 15 \\cdot 3$$ • 9 divides 45 since $$45 = 9 \\cdot 5$$ • 10 does not divide 45 since 45 is between $$40 = 10 \\cdot 4$$ and $$50 = 10 \\cdot 5$$ The above definition is synonymous with saying that $$b$$ may be expressed as a multiple of $$a$$. In fact, the following statements are equivalent and mean that $$a$$ is can be expressed as $$bn$$. • $$b$$ divides $$a$$ • $$b$$ is a divisor of $$a$$ • $$a$$ is divisible by $$b$$ • $$a$$ is a multiple of $$b$$ The notation for this concept is $$b \\mid a$$, which reads \"$$b$$ divides $$a$$\". Using this notation, the examples above become: • $$15 \\mid 45$$ since $$45 = 15 \\cdot 3$$ • $$9 \\mid 45$$ since $$45 = 9 \\cdot 5$$ • $$10 \\nmid 45$$ since $$10 \\cdot 4 < 45 < 10 \\cdot 5$$ Be careful!", "so the final answer is $90+40+16+4=\\boxed{150}.$ Hint: In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$. Thus, the product is $(1^{19})(2^{18})\\cdots(19^1)$ (or alternatively, $19! \\cdot 18! \\cdots 1!$.). When we count the number of factors of $2$, we have $4$ groups, factors that are divisible by $2$ at least once, twice, three times and four times. For those numbers, take the exponent corresponding to each one of them, add them case by case and then totally add them. Hope it helps.", "• $5$ if the last digit is $5$ or $0$. • $6$ if it is divisible by both $2$ and $3$. • $10$ if it ends with $0$. #### Example 1 Is $5,625$ divisible by… • $2$? • $3$? • $5$ or $10$? • $6$? Solution Part 1. Part 2. Part 3. Part 4. In mathematics, there are often several ways to talk about the same ideas. So far, we’ve seen that if $m$ is a multiple of $n$, we can say that $m$ is divisible by $n$. For example, since $72$ is a multiple of $8$, we say $72$ is divisible by $8$. Since $72$ is a multiple of $9$, we say $72$ is divisible by $9$. We can express this still another way. Since $8 \\cdot 9 = 72$, we say that $8$ and $9$ are factors of $72$. When we write $72=8 \\cdot 9$, we say we have factored $72$. Other ways to factor $72$ are $1 \\cdot 72$, $2 \\cdot36$, $3 \\cdot 24$, $4 \\cdot 18$, and $6 \\cdot 12$. The number $72$ has many factors: $1,2,3,4,6,8,9,12,18,24,36,$ and $72$. ### Factors In the expression $a \\cdot b$, both $a$ and $b$ are called factors. If $a \\cdot b = m$, and both $a$ and $b$ are integers, then $a$ and $b$ are factors of", "divisible by the tens digit of A. Kudos for a correct solution. [Reveal] Spoiler: OA _________________ Retired Moderator Joined: 06 Jul 2014 Posts: 1269 Location: Ukraine Concentration: Entrepreneurship, Technology GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 The difference between positive two-digit integer A and the smaller tw [#permalink] ### Show Tags 23 Apr 2015, 11:13 6 KUDOS 2 This post was BOOKMARKED 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 $$ab*cd$$ $$(10a + b)*(10c + d)$$ $$100ac", "are divisible by $4$. This can be done by simply listing: $$2000,2004,2008,2012,2016.$$ There are $5$ numbers in $B$ divisible by $4$ out of the $2018-1999+1=20$ total numbers. Therefore, the probability that $3$ is picked from $A$ and a number divisible by $4$ is picked from $B$ is $\\frac{1}{5}\\cdot \\frac{5}{20}=\\frac{1}{20}$. Similarly, we can look at the repeating units digit for $7$: $$7\\cdot 7=9$$ $$9\\cdot 7=3$$ $$3\\cdot 7=1$$ $$1\\cdot 7=7$$ We see that the unit digit of $7^y$, for some integer $y$, will only be $1$ when $y$ is a multiple of $4$. This is exactly the same conditions as our last case with $3$ so the probability of this case is also $\\frac{1}{20}$. Since $5\\cdot 5=25$ and $25$ ends in $5$, the units digit of $5^w$, for some integer, $w$ will always be $5$. Thus, the probability in this case is $0$. The last case we need to consider is when the number $9$ is chosen from $A$. This happens with probability $\\frac{1}{5}$. We list out the repeating units digit for $9$ as we have done for $3$ and $7$: $$9\\cdot 9=1$$ $$1\\cdot 9=9$$ We see that the units digit of $9^z$, for some integer $z$, is $1$ only when $z$ is an even number. From the $20$ numbers in $B$, we see that exactly half of them are", "Does such a value for n even exist (besides the n for our example $$F=527$$)? In Step 3 and 4 we want to examine for which values of n the expression $$(9 \\cdot 10^n - 41)$$ is divisible by 17. ------------------------------------------------------ STEP 3 We already know that $$F=527$$ fulfills the swap condition. We obtain $$F=527$$ for $$n=3$$: $$F = \\frac{1}{17} (9 \\cdot 10^3 -41) = \\frac{1}{17} (9000 - 41) = \\frac{1}{17}(8959) = 527$$ There is a rule for testing whether a number is divisible by 17, see here: http://www.egge.net/~savory/maths1.htm and http://primes.utm.edu/curios/page.php/17.html In order to illustrate the rule we test if 8959 is divisible by 17: Step 1: Split 8959 in 895 and 9 (last digit), multiply 9 (the last digit) by 5: $$9 \\cdot 5 = 40$$. Step 2: Substract this from the ramaining digits 895: $$895-9\\cdot5=850$$. Step 3: If the last two or three remaining digits are divisible by 17 then the original number 8959 is divisible by 17, too. In this case, 850 is divisible by 17, thus 8959 is divisible by 17, too. So, for $$n=3$$, F is an integer. What about $$n=4$$: $$F = \\frac{1}{17}(9 \\cdot 10^4 -41) = \\frac{1}{17}(90000-41) = \\frac{1}{17}(89959)$$ Let's test if 89959 is divisible by 17: Step 1: $$8995-9\\cdot 5 = 8955$$ Step 2: $$895 -5 \\cdot 5 = 870$$", "## If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What ##### This topic has expert replies Moderator Posts: 1372 Joined: 29 Oct 2017 Thanked: 1 times Followed by:5 members ### If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What by M7MBA » Thu Sep 17, 2020 1:03 am 00:00 A B C D E ## Global Stats If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What is the value of $$a+b?$$ A) 1 B) 2 C) 5 D) 6 F) 8 Source: e-GMAT ### GMAT/MBA Expert GMAT Instructor Posts: 5872 Joined: 25 Apr 2015 Location: Los Angeles, CA Thanked: 43 times Followed by:24 members ### Re: If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ Wha by [email protected] » Thu Sep 24, 2020 7:04 am M7MBA wrote: Thu Sep 17, 2020 1:03 am If $$a$$ is the units digit of $$7^{47}$$ and $$b$$ is the rightmost nonzero digit in $$125^{10}\\cdot 28^{15}.$$ What is the value of $$a+b?$$ A) 1 B) 2 C) 5 D) 6 F) 8 Solution: Recall the units digit pattern of powers of 7 is", "## Odd product of tens digits The product of the tens digits of a few given perfect squares is odd. What is the units digit of the product of these perfect squares? [SOLVED] #### Hagen von Eitzen solved this puzzle: Since $$(10k + n)^2 = n^2 + 20k(n + 5)$$, both the units digit and the parity of the tens digit depend only on $$n \\bmod{10}$$. Checking the numbers $$0^2, 1^2, \\dots, 9^2$$ we find that only $$4^2 = 16$$ and $$6^2 = 36$$ have odd tens digits. Since a product of integers is odd if and only if each integer is odd, every given perfect squares must have odd tens digit. So, each has its units digit as $$6$$. Since $$6 \\cdot 6 \\equiv 6 \\pmod{10}$$, $$6^m \\equiv 6 \\pmod{10}$$ for all integers $$m \\gt 0$$. Therefore, the units digit of the product of the squares is $$6$$. #### Joe Wells solved this puzzle: If we examine the squares of the digits $$0, 1, \\cdots, 9$$, we see that $$4$$ and $$6$$ are the only two whose squares have odd tens digits. Further, both end with $$6$$. Consider squaring some multi-digit number with $$4$$ as the units digit. Since $$4$$ is even, $$4$$ times the tens digit of the number yields an even number as the tens", "in the range from $$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.$$", "since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ ## Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ ## Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$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}.$$ ~mathpro12345 ~ pi_is_3.14", "+0 # Help PLS, ASAP +2 700 1 +140 The least common multiple of two positive integers is $7!$, and their greatest common divisor is $9$. If one of the integers is $315$, then what is the other? (Note that $7!$ means $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1$.) Mar 19, 2019 edited by CorbellaB.15 Mar 19, 2019 #1 0 The identity lcm[a,b]*gcd(a,b)=ab holds for all positive integers a and b. Let a=315 in this identity, and let b be the number we are looking for. Thus $$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}.$$ Hope this helps :) Sep 7, 2019", "# Proof Involving Pigeonhole Principle Let $a_1, a_2, ..., a_n$ be positive integers all of whose prime divisors are $\\le$ 13. a) Show that if $n \\ge 65$ then there exist two of these integers whose product is a perfect square. [DONE] b) Show that if $n \\ge 193$ then there exists four of these integers whose product is a perfect fourth power. Hint: Use a) to get many pairs of numbers which multiply to a square. Use a) again to get two disjoint such pairs a, b, and c, d such that $\\sqrt{ab}\\sqrt{cd}$ is a square. For b, I got 64 pairs of numbers which multiply to a square using part a. Where do I go from there? Part b: We claim that we can find $65$ disjoint pairs of integers such that their product is a square. By part a), for $n\\ge 193>65$, we can find one such pair. Then remove the pair from the set, and apply part a) again on the remaining $n-2$ integers. We can do this repeatedly, and because $n-2\\cdot 64\\ge65$, we can find at least $65$ pairs. Because the original set of integers was distinct, the pairs are disjoint. Now for each pair $(a,b)$ consider $\\sqrt{ab}$, which is an integer. By part a, there are distinct $a,b,c,d$ such that $\\sqrt{ab}\\cdot \\sqrt{cd}$", "× # Help,its so hard for me Determine the biggest integers 85-digit number which satisfies the sum of all digits equal to the multiplication of all numbers Note by Valian Fil Ahli 4 years ago Sort by: Let the digits be $$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}$$ can be", "the last digit of the number. For example, alternating sum of the digits of $12345$ is $1 - 2 + 3 - 4 + 5$. Likewise, the alternating sum of the digits of $9876$ is $9 - 8 + 7 - 6$. Show that a positive integer is divisible by $11$ if and only if its alternating sum is divisible by $11$. 17. 🔎 Let $a$, $b$, $c$, $d$ be integers. Prove that $12 \\mid (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$", "of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$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 _ _ $1$, _ _ $5$, _ _ $7$, _", "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 $3, 4, 5, 6, 7$." ]

[ "2013 12-21 # Jamie’s Contact Groups Jamie is a very popular girl and has quite a lot of friends, so she always keeps a very long contact list in her cell phone. The contact list has become so long that it often takes a long time for her to browse through the whole list to find a friend’s number. As Jamie’s best friend and a programming genius, you suggest that she group the contact list and minimize the size of the largest group, so that it will be easier for her to search for a friend’s number among the groups. Jamie takes your advice and gives you her entire contact list containing her friends’ names, the number of groups she wishes to have and what groups every friend could belong to. Your task is to write a program that takes the list and organizes it into groups such that each friend appears in only one of those groups and the size of the largest group is minimized. There will be at most 20 test cases. Ease case starts with a line containing two integers N and M. where N is the length of the contact list and M is the number of groups. N lines then follow. Each line contains a friend’s name and the groups the friend", "more money than the programming club, and otherwise they will get it. ### Greatest Number of 2-Friends The popularity measure we all agreed on was the number of friends or friends of friends of each student. If our group had a higher number of 2-friends than the other group, we would win. Phoebe tasks Sydney and Diane to work with two girls from the programming club, Cathleen and Mira, to find out the number of 2-friends of each student of our school. In their talks with Cathleen and Mira, Sydney and Diane spent some time learning about ways of representing friendships. The first way is by means of an undirected graph. Below is the friendship graph for the five of us, Cathleen and Mira, and Jasmine, Roberta and Michaela. An edge means the two girls connected are friends. Absence of an edge means absence of friendship. At the top of the graph there is our group. All girls are friends with each other. CJ surprisingly also became friends with Cathleen. Roberta and Mira are also friends with Cathleen. Jasmine has only one direct friend which is Roberta. However because Roberta is friends with Michaela and Cathleen, Jasmine has three 2-friends. CJ has five direct friends but because Mira and Roberta are friends with Cathleen she has seven 2", "friends. Sydney, Daphne, Diane and Phoebe have four direct friends but because CJ is friends with Cathleen they all have five 2-friends. Cathleen has 3 direct friends but through CJ she gets 4 more 2-friends and through Mira and Roberta she gets 2 more. Based on this set of students, she would have the most 2-friends (9 = 3 direct + 6 friends of friends). CJ would be second with 7. Here it is the ranking Student friends friends of friends 2-friends Cathleen 3 6 9 CJ 5 2 7 Sydney, Daphne, Diane and Phoebe 4 1 5 Roberta 3 2 5 Mira, Michaela 2 2 4 Jasmine 1 2 3 To decide if student A is a 2-friend of B the key is to check whether A and B are separated by at most two edges (i.e. whether there is a path between A and B with one or two edges). If there is they are 2-friends otherwise they are not. According with the above table our team has a total of 27 two-friends. The remaining five girls have 21 two-friends. So if we based the results on just these 10 people our team would win. Graphs are a nice visual representation of friendships but to compute the 2-friends of the hundreds of students required using a", "has happened. We don't know what CGMO is and all we are going to do is to study math hard and enjoy MOP. Oh! Alicia is my roommate, and she always said that she is is the best roommate in the world. Everyday we want to go to bed earlier, but it never works. Now it's 12:20, (I just asked Alicia what time it is and she said it's time to get a watch. She's mean). Good night. My angle~ -Jingyi June 12, 2012 So we had classes. The number theory stuff was really cool and useful (If only we had learned that stuff a day before it would have saved me an hour on the MOP test yesterday and figuring out a way around that stuff). They were a lot of fun and I learned a lot. I feel like I'm getting so much better at math. It's also so fun meeting people with the same interests as me. Oh yeah. So we played a game of Mao today (or yesterday, I'm not sure). It was AWESOME. Basically, it's a card game where we have to add rules and everyone keeps getting penalized until they find them out and stop violating them. So there were like 5 rules in the game so it was such an annoyance", "after which we cut back to Lolita in her secret base, where she's hiding from the totalitarian GOVERNMENT OF THE FUTURE watching glossy smut. She chats for a bit and then puts on another story for us to watch. Except in this case, the first such link, where she's more occupied with Mina, played by otherwise-unknown actress Isabella DuMaurier. Mina probably has an interesting backstory: probably a shady past with a bright hope for the future. Maybe she's on the run from THE GOVERNMENT OF THE FUTURE and has discovered one of Lolita's stories or managed to tune into the illicit broadcast. She is clearly skilled at infiltration, espionage and technology, as she has managed to both locate, and make it through all the barriers of, Lolita's unassailable, super-secret hiding-hiding-hiding-base. She just appears there out of nowhere. I'd love to know more about this character. But she doesn't say a word. Lolita's keeping her mouth occupied. This scene is less than a minute long and mostly consists of kisses. Mina is initially completely topless Wild LICKITUNG appeared! and wearing some incredibly sparkly tights; Lolita is in her usual stuff. They're on a bed (for some reason, the top secret hideout, no boys allowed, has a bed in it), and both women are hurriedly disrobing. To a repeated, synthy", "connected are friends. Absence of an edge means absence of friendship. At the top of the graph there is our group. All girls are friends with each other. CJ surprisingly also became friends with Cathleen. Roberta and Mira are also friends with Cathleen. Jasmine has only one direct friend which is Roberta. However because Roberta is friends with Michaela and Cathleen, Jasmine has three 2-friends. CJ has five direct friends but because Mira and Roberta are friends with Cathleen she has seven 2 friends. Sydney, Daphne, Diane and Phoebe have four direct friends but because CJ is friends with Cathleen they all have five 2-friends. Cathleen has 3 direct friends but through CJ she gets 4 more 2-friends and through Mira and Roberta she gets 2 more. Based on this set of students, she would have the most 2-friends (9 = 3 direct + 6 friends of friends). CJ would be second with 7. Here it is the ranking Student friends friends of friends 2-friends Cathleen 3 6 9 CJ 5 2 7 Sydney, Daphne, Diane and Phoebe 4 1 5 Roberta 3 2 5 Mira, Michaela 2 2 4 Jasmine 1 2 3 To decide if student A is a 2-friend of B the key is to check whether A and B are separated by at most", "enemy spy Tonya is french and has blue-eyes Ned is english and has blue-eyes Cindy is tall There are several ways to indicate Ned as the spy in this group: NOT french <-- absolves Tonya NOT tall <-- absolves Cindy, leaving Ned OR NOT tall <-- absolves Cindy NOT french <-- absolves Tonya, leaving Ned OR english <-- indicates Ned OR NOT french <-- absolves Tonya english <-- indicates Ned OR NOT tall <-- absolves Cindy english <-- indicates Ned OR NOT french <-- absolves Tonya blue-eyes <-- indicates Ned The third option, english, is the shortest list, so it is the correct choice. Example 2 Jasmin, Sam, and Rose are enemy spies Rick has 3 identifying attributes: brown-hair, glasses, and tall Marcia has 2 identifying attributes: thin and freckled Jasmin has 4 identifying attributes: chinese, short, thin, and brown-hair Matt has 2 identifying attributes: german and freckled Sam has 3 identifying attributes: thin, glasses, and muscular Rose has 3 identifying attributes: german, tall, and thin There is one attribute that all the spies have in common: thin. But if we say \"thin\" then that will indicate Marcia as well. We need to remove Marcia from the list first by saying \"NOT freckled\". NOT freckled <-- This absolves Marcia and Matt thin <-- This indicates Jasmin, Sam, and", "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", "George are passionate gamers and live in houses that are exactly opposite each other across the street, so they can see each other through the windows. They decided that their computers will connect the telephone cable in order to play games t", "smart. Let's just say that she can recognize shapes and colors. Well, okay she's not that smart. Let's just say you wouldn't want her driving or operating heavy machinery. She looks like Daphne from Scooby-Doo. • Clover: Also known as \"Shitty Spy\". A snotty, greedy, bitch... just like guys like 'em. She's on her way to bea beauty school dropout loaded with credit card debt. She is also the biggest slut of the trio, going after anything guy or thing which has a penis to shove up her crab infested fanny! • Alex: Hey wait, isn't that a boy's name? Who cares, she's HOT anyways. Also known as \"Fucky Spy\". Guess which ethnicity she belongs to, and you win a prize. Alex is very Irish. No wait, so are Sam and Clover. • Britney: Also known as the \"Kinky Spy\". Is an amalgamation of all three hotness factors of the other three. Too bad she was only in two episodes. No one knows her ethnicity too. She's an...Grey, yes, that's right. • Mandy: The \"Spy that wasn't\". She was a spy like once or twice, but won't be a permanent agent. She hates the spies, looks down on them, very bullyish, bitchy and a show-off. Rumor has it she has a secret crush with Sam. • Jerry: Not", "and seek is her favorite game. Outside school, the girls discuss how smart and untrustworthy Melissa is. Spencer insists they overlooked a clue in Ali's bag. They change the subject to costume choices when Mona approaches. Mona suggests they all go shopping after school together. They all turn her down one by one, including Hanna, who lies about studying with Caleb. Aria brings the girls to Ezra's apartment while he is out of town for a safe place to search Ali's bag. The girls look through his book collection and photo albums and plenty of comments. Hanna gets a text from Mona asking where she is since she's not with Caleb. They comb through the contents of Alison’s bag once more and find the clue they were looking for. That evening, during a thunderstorm, they drive to the Lost Woods Resort. The innkeeper, a dead ringer for Norman Bates, taps on the window, scaring them on top of their already frayed nerves. Spencer asks for Room 1, the number on the postcard they found in Ali’s things. Hanna trips in the mud on the way to the room. Once inside the room, Hanna and Emily try to warm up while Spencer keeps an eye on the front office waiting for the innkeeper to leave. Aria volunteers to go", "Aria’s living room, where Emily tells Spencer that nobody like lying to Toby, but if he finds out what happened in New York, before she can finish Aria cuts her off, telling Spencer that she should just tell him the truth. Noticing Spencer looking in Alison's direction, Aria tells her that it's her decision, not Alison's. When Spencer tells that girls that Toby said Jenna sounded broken, Emily asks if she is still full on blind, and Spencer nods her affirmative answer. Hanna says that it is pretty clear that Jenna is back to punish them for blinding her and killing Shana, and Spencer agrees, “said with absolutely no tact, but you’re right”. Spencer explains to Aria that the new normal was supposed to be a carefree senior year, when Aria asks about Hanna’s comment on there still being enough time for a hobby. Spencer, Emily, Hanna and Aria are waiting for Alison to arrive, and after the other girls wonder whether or not Alison’s actually coming, Spencer announces that she thinks she’s here. Finally getting to the other four girls, Alison thanks them for waiting, and all together they make their way into the halls of Rosewood High. After Emily and Alison walk off, Aria circles around in front of Hanna and Spencer, saying that she can’t", "# Why do people try to make themselves feel better by treating others poorly? I'm looking for an explanation for why people will try to make themselves feel better by treating others poorly. For example, imagine a group of girls are sitting in a cafe. Within that group are two girls in particular, Sally and Anne. Sally is beautiful, intelligent, and popular. Sally knows she's blessed, but she doesn't like to highlight that fact. She shies from attention and deflects most compliments. Anne on the other hand, is normal, at least she thinks of herself as normal. Anne doesn't often receive compliments like Sally does. As the group of girls sit in the cafe sipping latte's and chatting, the conversation makes its way to Sally's recent success at school. Sally scored top of the class in the last bout of exams. This angers Anne, she herself didn't do so well on her exams. She feels ashamed. Suddenly she blurts \"Oh Sally its great that you did well on your exams, I'm surprised you had so much time to study, I thought you'd be too busy with Jack?\". Jack was Sally's fiance, they recently split up when Sally caught Jack with another woman. At this, Sally cowers like a wounded animal. She quickly mumbles something and excuses herself from", "a co-ed school, and mostly hanging out with the three other girls who had gone to her previous school. She's miserable, out of place, and homesick for her previous routine. And then, one morning, her mother doesn't get out of bed. Her mother, the living force of energy, the one who runs the household, who pesters Francesca incessantly, who starts every day with a motivational song. She doesn't get out of bed the next day, either. Or the day after that. And the bottom falls out of Francesca's life. I come at this book from a weird angle because I read The Piper's Son first. It's about Tom Mackee, one of the supporting characters in this book, and is set five years later. I've therefore met these people before: Francesca, quiet Justine who plays the piano accordion, political Tara, and several of the Sebastian boys. But they are much different here as their younger selves: more raw, more childish, and without the understanding of settled relationships. This is the story of how they either met or learned how to really see each other, against the backdrop of Francesca's home life breaking in entirely unexpected ways. The Piper's Son was more effective for me, but this is still a very good book. Marchetta captures the gradual realization of friendship,", "around her eyes. Some rumors around the school tell that she is \"Ugly,\" and Duckie hates that very much, because she is not!", "to women? ### #34Fox Mc Cloud Members 122 Like Likes Like Posted 05 December 2001 - 01:34 AM quote: Original post by Hase 1) It´s not about the main character, or the representation of that character. Male players have no problem playing female characters and vice versa. Well, maybe for single-player games, but when it comes to an online game you don''t want to fool the others by looking like a girl / a girl doesn''t want the other players to think she''s a guy. I know I wouldn''t take a female character in an online game, and my girlfriend always takes female characters too (in Darkstone, Soldier of Fortune, Quake3...). She wanted to play CounterStrike and was getting quite good at it but dropped it because she was annoyed that there''s no female character... But again, in a single player game, I don''t mind and I don''t think girls do either. -- New feature - Windows users, press ALT-F4 to launch the IQ test. ### #35Hase Members 313 Like Likes Like Posted 05 December 2001 - 02:06 AM @MSW: I´m familiar with the term \"chick flicks\", and I think that there might come a time when computer games along those lines could work. One of the problems is that this genre of novel or film relies heavily", "\"The Blond Leading the Blind,\" Caleb uncodes more videos from A's phone. They see the video of Ali and Ian at the Kissing Rock and wonder if this is all there is. Hanna hopes it is so Caleb doesn't have to be involved anymore, while Emily hopes they find the information they are looking for. Then a new video starts. Ian hides a video camera on a shelf in Ali's bedroom then Garrett and Jenna enter the room. Ian called them to help search for the N.A.T. Club videos that Ali won't hand over. Garrett accuses Ian of going too far, which is why he bailed on the club in the first place. They start searching Ali's bedroom for the videos at Ian's request. Garrett finds something \"weird\" in a wooden box, then the video cuts off. The girls are left wondering if those missing videos are what got Ali killed. Toby calls Spencer while she is in her room with Emily. Emily reluctantly picks up the phone and tells Toby Spencer cannot come to the phone, not wanting to lie to him. Toby says he came by to pick up his tools since Spencer's father fired him and that Spencer has been sending him mixed signals lately. He picks up on the fact Spencer is avoiding him", "blind since childhood as the result of optic nerve damage, Buchanan navigates her game avatar through the maze quickly, keeping it close to the walls like someone guiding themselves by touch. The secret to her speed is inside the headphones Buchanan wears, which immerse her in a three-dimensional labyrinth of sound. A knock in one earphone or the other indicates a door on that side. The sound of footsteps ascends in tone as Buchanan walks her avatar up stairs. Furniture pings when bumped, and the jewels she is seeking twinkle more loudly as she approaches them. All in all, Buchanan says she feels physically present in the maze's corridors. “Ah! Go away!” she yelps suddenly at one point when she hears the warning of a passing “monster”—her fingers fly over the keys to skitter her avatar toward safety, into a section that announces itself as “Women's Bathroom.” Once they drag Buchanan away from the game and into the real building next door—thankfully free of monsters—neuroscientist Lotfi Merabet and study coordinator Erin Connors, both of the Massachusetts Eye and Ear Infirmary, instruct her to go to various rooms and exits, noting the routes she takes. Like the other blind volunteers in Merabet's study, she had never been in the building before playing the Audio-Based Environment Simulator (AbES, pronounced “abbess”)", "not to damage another relationship of hers. Inside, Hanna, Spencer, Aria and Caleb sit together. When Caleb gets up, Aria and Spencer express their surprise at and disapproval of Hanna's new buddy. After the match, Spencer catches sight of Ian and Melissa conversing in the school hallway and decides to walk home to avoid them. Spencer then shows up to Toby's home and is nonplussed when he tells her to leave with barely any explanation. He hands her the French textbook and close. Spencer suddenly understands why Toby had been so secretive before, with the possibility of Jenna being so close. In Spencer's room, the girls discuss the strangeness of Melissa's rapid change in life plans. They use a website to decipher the three letters of the Braille message: B-A-D. Aria decides it must be a joke that Toby is playing on them, but Emily doesn't think that he would mess with them at a time like this. Spencer, Aria, Hanna, and Emily sit down for lunch at a separate table. Spencer expresses her disapproval of Hanna's growing relationship with Caleb, declaring him dark. But, Hanna reminds her friends that they had treated Toby the same way, silencing them. The girls as well as all the other students look on as Ezra diffuses a situation started by an", "to completely destabilize democracy\". Soph is a younger, creative, smart, self-made, ingenious kid who is an actual rebel and the true mastermind behind her YouTube (990k followers) and BitChute videos. It has always been obvious to me that Soph must see Greta as her arch-nemesis. Soph feels very similarly to your humble correspondent about similar things – so she must be profoundly angry about a \"premium unleaded autist\" (Soph's creative words) Swedish girl who is being praised by all the left-wing media while Soph is described as some kind of trash. It should be the other way around and Soph knows it. Well, in some interviews, Soph did refer to Greta. But I knew that the sentiment had to be much stronger." ]

[ "6 for the 2nd, 5 for the third and so on. Thus, the number of ways is: $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1=7!=5040$ (ii) Letting M represent male and F represent female, we must have: M F M F M F M This means the number of ways will be given by: $4!\\cdot3!=144$ (iii) First putting James on the left of Rio, there are 5 ways for there to be 1 person between them and 5! different lineups of the other people. Including the lineups with Rio on the left of James this is a total of: $2\\cdot5\\cdot5!$ With two people between them, we find: $2\\cdot4\\cdot5!$ With 3 in between: $2\\cdot3\\cdot5!$ With 4 in between: $2\\cdot2\\cdot5!$ With 5 in between: $2\\cdot1\\cdot5!$ Adding, we find: $2\\cdot5!$$5+4+3+2+1$$=30\\cdot5!=3600$ Thus, there are 3600 ways James and Rio are not seated together. (iv) The professor has 4 males to choose from and 3 females, thus there are: $4\\cdot3=12$ There are 12 ways to choose 1 male and 1 female. (v) Here we use the combination function: ${7 \\choose 3}=35$ Thus, we find there are 35 ways to choose 3 people from a group of 7. November 21st, 2011, 06:42 PM #3 Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 Re: Permutation and combination ? Quote: Originally Posted by MarkFL (iii) First putting James on the left of", "up with someone else, it leaves $2$ original couples and two ‘strays’, say $\\{p_1,p_2\\},\\{p_3,p_4\\},p_5$, and $p_6$. • In the first case there are $24$ choices for $k$ and $\\binom{11}8$ choices for the couples to be retained. Then $p_1$ and $p_2$ must get new partners from different original couples, which can happen in $4\\cdot2$ ways, so there are $8\\cdot24\\cdot\\binom{11}8$ good pairings of this type. • In the second case there are $24$ choices for $k$, $22$ choices for $25$’s partner, and again $\\binom{11}8$ choices for the couples to be retained. Then either $p_5$ and $p_6$ are partnered, in which case there are just $2$ ways to repartner $p_1,p_2,p_3$, and $p_4$; or $p_5$ and $p_6$ are partnered with members of different original couples, which can be done in $4\\cdot2=8$ ways and allows no further choice. There are therefore $10\\cdot24\\cdot22\\cdot\\binom{11}8$ good pairings of this type. This gives us a total of $$228\\cdot24\\cdot\\binom{11}8=902,880$$ good pairings with $25$ partnered, and a grand total of $932,580$ good pairings. The probability of getting a good pairing is $$\\frac{932,580}{7,905,853,580,625}\\approx 1.1796\\times 10^{-7}\\;.$$ - I'm curious, since 25 people is an odd number of people because partnerships must include more than two people or else one person must be left out. What do you mean by partnerships? You spoke of pairing, which implies two person groups, but that", "# What is the chance that exactly 8 people have the same partners the second time? 25 people randomly form partners with each other. They sunder and do it again. What's the probability that exactly 8 people get the same partner the second time? There are $\\frac{25 \\times 24}{2} = 300$ partnerships formed per round of pairing up... the answer is clearly also not $\\binom{300}{4}$ since the partnerships could differ every time... - If all 25 people are allowed to form pairs, there should be $\\binom{25}{2} \\cdot \\binom{23}{2} \\cdots \\binom{3}{2}$ ways of doing it and of course 1 person remains single. – Alex Oct 9 '12 at 8:50 This gets pretty messy (though it’s always possible that I’ve simply overlooked a nicer approach). Number the people $1$ through $25$. To form one set of partnerships you can first choose which of the $25$ will be unpartnered. Then repeatedly assign a partner to the lowest-numbered person remaining; this can be done in $$23\\cdot21\\cdot19\\cdot\\ldots\\cdot1=\\frac{24!}{24\\cdot22\\cdot\\ldots\\cdot2}=\\frac{24!}{2^{12}12!}$$ ways, so there are $$25\\cdot23\\cdot21\\cdot19\\cdot\\ldots\\cdot1=\\frac{25!}{2^{12}12!}=7,905,853,580,625$$ possible pairings. There is no harm in assuming that the original pairing leaves $25$ unpartnered. Say that a pairing is good if it shares exactly $8$ couples with the original pairing. We’ll count the good pairings in two groups. Consider first a good pairing that leaves $25$ unpartnered. There are $\\binom{12}8$", "is $3!$ and the number of ways we can arrange the $3$ men is $3!$, so your coefficient should be $3!*3!$, not $9$. This gives us an answer of $${6 \\choose 3}{7 \\choose 3}3!\\cdot 3!=25200$$ We have $42$ pairs from which to choose the first pair. However, once we choose a pair, we need to get rid of $5$ other pairs with the same man (since there are $6-1=5$ other women) and $6$ other pairs with the same woman (since there are $7-1=6$ other men), leaving us with $42-1-5-6=30$ pairs left. We have $30$ pairs from which to choose the second pair. However, once we choose a pair, we need to get rid of $4$ other pairs with the same man (since there are $6-2=4$ women who haven't been picked yet) and the $5$ other pairs with the same woman (since there are $7-2=5$ men who haven't picked yet), leaving us with $30-1-4-5=20$ pairs left. We have $20$ pairs from which to choose the third pair, at which point we are done. This gives us an answer of: $$42*30*20=25200$$ Thus, both methods done correctly give us the same answer.", "we can now work out how many of them are left-footed: $12-4=8$ 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: $27-8=19$ 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 $63-19=44$ right-handed and right-footed people The completed frequency tree should therefore look like this: b) There are $63$ people who are right-handed, and $44$ of them are right-footed. So, the probability of picking a right-handed person who is also right-footed is $\\dfrac{44}{63}$ c) There are $75$ people in total. The number of right-footed people is: $44+4=48$ Therefore, the probability of selecting a right-footed person is: $\\dfrac{48}{75}$ This fraction can be simplified to: $\\dfrac{16}{25}$ a) We have very little information to go on here, but since we know that $13$ students failed French and that this was $\\frac{1}{12}$ of the entire French group, we can work out how many students took French: $12 \\times13 = 156\\text{ French students", "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 $31+126=\\boxed{157.}$", "for the second, and so on. The number of ways the prizes can be distributed will be $$5 !=5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=120$$ ways. Now we will consider some slightly different examples. Solution Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are $$25 \\cdot 24 \\cdot 23=13,800$$ ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn't really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6: ABC ACB BAC BCA CAB CBA How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are $$3 \\cdot 2 \\cdot 1=6$$ ways to do this; we didn't really need to list them all, we can just use permutations! So, out of the", "you are about to construct you have only 5-1=4 possibilities for $d_1$. For $d_2$ there is no restriction placed on the digit, except that it must not be 5. So there are 10-1=9 possibilities for your choice of $d_2$. Now for digit $d_3$: again, it must not be 5 but also, since it is the leading digit of your number, you must not make it 0 either. So there are 10-1-1=8 possible choices for $d_3$. Since these choices are independent of each other, the total number of possibilities is the product $8\\cdot 9\\cdot 4$ 2. A talent contest has 8 contestants. Judges must award prizes for first, second, and third places. If there are no ties, in how many different ways can the 3 prizes be awarded? Again, imagine yourself selecting the 3 highest placed persons by yourself. For the first place, you have 8 possibilities. For each of these 8 possibilities you have 8-1=7 possibilities to choose a person for the second place. And, finally, you have 8-2=6 possibilities to choose a person for the third place. This gives $8\\cdot 7\\cdot 6$ different ways in all.", "lucky friend, $6$ ways to choose the next one, $5$ ways to choose the third one, and finally $4$ ways to choose the last one. That's a total of $7 \\times 6 \\times 5 \\times 4 = 840$ ways to choose the female attendees in order. Now, let's choose their pairs. There's $6$ possible male pairs for the first filly, $5$ possible second pairs, $4$ possible third pairs, and $3$ possible last ones. That's $6 \\times 5 \\times 4 \\times 3 = 360$ ways to choose the male pairs in order. However! We don't care which order we pick the couples in; that is, we don't care who gets picked first and who gets picked second. There are $4 \\times 3 \\times 2 \\times 1 = 24$ ways to choose $4$ specified couple in any order ($4$ possible first couples, $3$ second, etc...), so using the previous two paragraphs, each assignment of tickets will get counted $24$ times over. Therefore, the final answer is $840 \\times 360 / 24 = 12600$ ways. As you can see, this is far too many to enumerate directly; however, if you had a spare 7 hours and were up for the task, the easiest way would be to organize by couple (i.e. first list the ways where Rainbow Dash gets to go", "1 minus the product of these $p = 1 - \\dfrac{20 \\cdot 18 \\cdot 16 \\cdot 14}{20 \\cdot 19 \\cdot 18 \\cdot 17} = 1 - \\dfrac{224}{323} = \\dfrac{99}{323}$ 4. ## Re: Combination of events Originally Posted by romsek Ok I was wrong. They are correct. Though $\\dbinom{55}{2}$ is just coincidence I believe Let's pick shoes so that there are no pairs first pick is 20/20 shoes. second pick is 18/19 third is 16/18 fourth is 14/17 We want 1 minus the product of these $p = 1 - \\dfrac{20 \\cdot 18 \\cdot 16 \\cdot 14}{20 \\cdot 19 \\cdot 18 \\cdot 17} = 1 - \\dfrac{224}{323} = \\dfrac{99}{323}$ Hello, 5. ## Re: Combination of events Originally Posted by Vinod Hello, The problem with your answer is that by picking pairs, you remove the shoe in each pair that you don't pick from the possible shoes to pick. 6. ## Re: Combination of events Originally Posted by romsek The problem with your answer is that by picking pairs, you remove the shoe in each pair that you don't pick from the possible shoes to pick. Hello, If we want to compute the probability of exactly one pair among the four selected from the total ten pairs of shoes, how it should be computed? Secondly what is the probability of", "# How many ways are there for 10 people to have five simultaneous telephone conversations? So what I'm thinking is $$\\binom{10}{2}\\binom{8}{2} \\ldots \\binom{2}{2}$$ But that's far from what the answer is. - Number of combinations of 2 from 10, divided by 2, since the conversation is order-independent. – mbaitoff Feb 1 '13 at 7:32 Presumably this is to be interpreted as asking for the number of different ways to pair up $10$ people. Imagine numbering them $1$ through $10$. There are $9$ choices for the person to be paired with Nr. $1$. Once that pair has been determined, find a partner for the lowest-numbered person remaining; that will be either Nr. $2$ or Nr. $3$, depending on who Nr. $1$’s partner is. Either way, there are $7$ possible choices for his partner, because $3$ people are out of the running: he himself, and the first pair. That gives us $9\\cdot7$ ways to choose the first two pairs. Now find a partner for the lowest-numbered person remaining: there are $5$ available choices, since we have to exclude the first two pairs and the person for whom we’re finding a partner. Putting the pieces together, we can form the first three pairs in $9\\cdot7\\cdot5$ ways. Can you finish the analysis from there? Added: Judging by your calculation, you’re thinking of", "This makes sense. $$14 \\choose 6$$ choose 6 friends, yet $$P(14,6)$$ arranges the 6 friends and picks them. In fact, we have the right to say exactly how much bigger $$P(14,6)$$ is. In both counting troubles we pick 6 out of 14 friends. For the an initial one, we avoid there, in ~ 3003 ways. But for the second counting problem, every of those 3003 options of 6 friends have the right to be arranged in specifically $$6!$$ ways. So now we have actually $$3003\\cdot 6!$$ choices and that is precisely $$2192190\\text.$$ Alternatively, look in ~ the an initial problem another way. We desire to select 6 the end of 14 friends, however we execute not care around the bespeak they room selected in. To pick 6 the end of 14 friends, we might shot this: \\beginequation*14 \\cdot 13 \\cdot 12 \\cdot 11 \\cdot 10 \\cdot 9.\\endequation* This is a reasonable guess, since we have 14 choices for the very first guest, then 13 for the second, and also so on. But the guess is dorn (in fact, the product is specifically $$2192190 = P(14,6)$$). That distinguishes between the different orders in i m sorry we might invite the guests. To correct for this, we might divide through the variety of different kinds of the 6 guests (so", "sequence of values: $$1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075,\\ldots$$ In particular we get for ten people that there are $945$ arrangements (fifth term in the sequence with $m=5$ and five pairs for 10 people in total). This is one of the more popular sequences from the OEIS, where we find this extensive OEIS entry. Hint: for the first two pairs, you are counting AB+CD and CD+AB as different. Shouldn't they be the same? How does this extend? Added: Your first thought was better. There are indeed $_{10}C_2$ ways to form the first pair. The problem is that some of the other ways to make pairs will make that same pair later. For your second thought, if you pick somebody to pair, they will have $9$ choices, but you could have made that pair in the other order. • On your second thought (which I think is a good approach): If you always start by picking somebody to pair with the first person (in alphabetical order of names, say), you run no risk of ever making the same pair in the other order. – Steve Kass Oct 19 '13 at 21:55 • after 10c2 , im left with 8 people . how does overlapping occurs when the two people have been removed? – cindy Oct", "to put the remaining 13 people in a line so that we don't have a consecutive block of 3 women. We will use complimentary counting. There are of course 13! ways to seat the 13 people with no attention to restrictions. The only way to have a consecutive block of 3 women is if there is a block of exactly 3 women or a block of exactly 4 women. Step 2 For the first case, there are $\\left(\\genfrac{}{}{0}{}{4}{3}\\right)\\cdot 3!$ ways to choose the 3 women in the block and then order them. There are 11! ways to arrange the $9+2=11$ entities. However, $10\\cdot 2\\cdot 9!$ of these have a block of 4 women. So there are a total of $216\\cdot 10!$ ways to arrange this case. Step 3 For the second case, there are 4! ways to arrange the 4 women in the block. There are 10! ways to arrange the 10 entities. Hence the total is $24\\cdot 10!$ So the total is $13!-240\\cdot 10!=\\overline{)1476\\cdot 10!}$ Do you have a similar question?", "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", "the line to one of the five remaining men, which yields $$\\binom{6}{3} \\cdot 7 \\cdot 6 \\cdot 5$$ ways of selecting three opposite sex couples. Method 3: We choose three couples, then correct for the $3!$ orders in which the same couples could be selected. There are seven ways to select a man and six ways to select a woman for the first pair, giving $7 \\cdot 6 = 42$ ways to pair a man with a woman. That leaves six men and five women from which to choose the next pair, giving $6 \\cdot 5 = 30$ ways to choose the second pair. Once those pairs have been selected, we are left with five men and four women from which to choose the third pair, giving $5 \\cdot 4 = 20$ ways to choose the third pair. Thus, it would appear that the three pairs can be selected in $42 \\cdot 30 \\cdot 20$ ways. However, the same three pairs could be selected in $3!$ orders, as shown above. Hence, the number of ways of selecting three opposite sex couples is $$\\frac{42 \\cdot 30 \\cdot 20}{3!}$$ • Thank you very much, your answer helped me understand this problem much better! – Tom Jun 13 '16 at 13:03 The number of ways we can arrange the $3$ women", "+1 this is the right answer. Just note that the OP specifically asked to ignore feb 29 (meaning there's 365 days within a year), and is $21!$ really the number of ways to partition 22 people into pairs or is it $22!$? – IWS Nov 7 '17 at 13:50 • I have changed it accordingly. I had ignored feb 29 in the sense to make it 'equally probable' instead of absent. The statement to ignore 29 feb is a bit ambiguous, but indeed in discussing the original birthday problem the OP mentions 365. – Sextus Empiricus Nov 7 '17 at 13:55 • It is 21!! (the double factorial). The first person of the pairs is a 'free' choice. So you select, without loss of generality, the first person in the first pair and have (22-1) options left to make a first pair, then (22-3) options to make the second pair, (22-5) options to make the third pair, etc. Alternatively you take 22! the number of ways to order the people and divide by $2^{11}$ and 11!, to correct for the number of ways to switch within and between the pairs (these have been counted multiple times by using 22!, the double factorial expression is more direct). – Sextus Empiricus Nov 7 '17 at 15:05 • Thanks for explaining!", "# Selecting 5 pairs of men and 5 women from 10 women and 12 men ## Question A dance class consists of $22$ students, of which $10$ are women and $12$ are men. If $5$ men and $5$ women are to be chosen and then paired off, how many results are possible? ## Approach According to me, the number of results possible is: $$\\binom{10}{5}*\\binom{12}{5}*5!*2^{5}$$ $$\\binom{10}{5}*\\binom{12}{5}*5!$$ ## My conclusion Shouldn't be there $2$ options in each pair i.e ordering between men and women for $5$ such group, making it $5!$? Why is the answer not leaving $5!$? Are they not considering order? And if the order is important, is my answer correct in this case? • Why would the be two options in each pair . A pair is of combination of a man and a woman . Even if you select fist a man then a woman to pair him or select first a woman and then a man to pair her both are same – ATHARVA Jun 1 '17 at 13:04 • if order is important, ur answer is right i believe – Kiran Jun 1 '17 at 13:11 • Yes, your answer is correct if \"Fred and Ginger\" is counted as a distinct pairing from \"Ginger and Fred\". – Michael Seifert Jun 1 '17 at 13:49", "# Probability without using combinations $18$ people in a class. Teacher chooses $3$ students. What is probability that teacher chooses Jason, Kim and Ellie? The problem wants the answer without using combinations (nCr). Lost. I would think the answer would be $(1/18) \\cdot (1/17) \\cdot (1/16)$, but that doesn't seem to be the way to do it with combinations. • I would think the answer would be (1/18) * (1/17) * (1/16), but that doesn't seem to be the way to do it with combinations. – user163862 Jan 9 '17 at 5:50 The chance that the first choice is one of Jason, Kim and Ellie is $\\dfrac{3}{18}$ Given that the first choice is one of Jason, Kim and Ellie, that leaves $2$ desirable students out of $17$ so the conditional probability that the second choice is also is also one of Jason, Kim and Ellie is $\\dfrac{2}{17}$ Given that the first and second choices are both from Jason, Kim and Ellie, that leaves $1$ desirable students out of $16$ so the conditional probability that the third choice is also is also one of Jason, Kim and Ellie is $\\dfrac{1}{16}$ So the probability that all three choices are Jason, Kim and Ellie in any order is $\\dfrac{3}{18} \\times\\dfrac{2}{17} \\times\\dfrac{1}{16} = \\dfrac{1}{816}$ You essentially need to derive the combinations answer.", "# $20$ people are sitting around a (circular) table. How many ways can we choose $3$ people, no two of whom are neighbors? 0 choices for the 1st person. 17 choices for the 2nd person (must exclude 1st and his/her two neighbours) For 2 of these choices of 2nd person, there is one shared neighbour, so 15 remaining choices. (e.g. if they are numbered 1 to 20 in a circle, 1st person is #1, 2nd is #3, then people 20,1,2,3,4 are excluded). For the other 15 choices of 2nd person, there are no shared neighbors, so 14 remaining choices. So if order matters, total is $20 \\cdot (2 \\cdot 15 + 15 \\cdot 14)$ but since order does not matter, divide by $3! = 6$ to account for the permutations in order of the 3 people. So total = $20 \\cdot \\frac{2 \\cdot 15 + 15 \\cdot 14}{6}$ just redid it; does this make any sense? • What is the answer with a straight table and how does turning the table into a corner change the problem? thats the step that you need to account for Oct 4, 2016 at 16:14 • first person - 20 choices , second person 17 choices , then third person 14 choices - then you could have chosen the 3 people in" ]

[ "## Riddler Riddler Express A group of $101$ people join $μετα$, and each person has a random, $50$ percent chance of being friends with each of the other $100$ people. Friendship is a symmetric relationship on $μετα$, so if you’re friends with me, then I am also friends with you.I pick a random person among the $101$ — let’s suppose her name is Marcia. On average, how many friends would you expect each of Marcia’s friends to have? ## Computational Solution From the simulation below, we see that the expected number of friends of each of Marcia’s friends is $\\textbf{50.5}$. It is interesting to note that Marcia on average would have only $50$ friends. import networkx as nx from random import random, randint, choice def exp_num_friends(n, runs = 10000): total_deg = 0 for _ in range(runs): G = nx.Graph() for i in range(n): for i in range(n-1): for j in range(i+1, n): if random() < 0.5: marcia = randint(0,n-1) if marcia_friends: total_deg += G.degree[choice(marcia_friends)] print(exp_num_friends(101)) ## Riddler Classic The sum of the factors of $36$ — including $36$ itself — is $91$. Coincidentally, $36$ inches rounded to the nearest centimeter is … $91$ centimeters! Can you find another whole number like $36$, where you can “compute” the sum of its factors by converting from inches to centimeters? Extra", "more money than the programming club, and otherwise they will get it. ### Greatest Number of 2-Friends The popularity measure we all agreed on was the number of friends or friends of friends of each student. If our group had a higher number of 2-friends than the other group, we would win. Phoebe tasks Sydney and Diane to work with two girls from the programming club, Cathleen and Mira, to find out the number of 2-friends of each student of our school. In their talks with Cathleen and Mira, Sydney and Diane spent some time learning about ways of representing friendships. The first way is by means of an undirected graph. Below is the friendship graph for the five of us, Cathleen and Mira, and Jasmine, Roberta and Michaela. An edge means the two girls connected are friends. Absence of an edge means absence of friendship. At the top of the graph there is our group. All girls are friends with each other. CJ surprisingly also became friends with Cathleen. Roberta and Mira are also friends with Cathleen. Jasmine has only one direct friend which is Roberta. However because Roberta is friends with Michaela and Cathleen, Jasmine has three 2-friends. CJ has five direct friends but because Mira and Roberta are friends with Cathleen she has seven 2", "connected are friends. Absence of an edge means absence of friendship. At the top of the graph there is our group. All girls are friends with each other. CJ surprisingly also became friends with Cathleen. Roberta and Mira are also friends with Cathleen. Jasmine has only one direct friend which is Roberta. However because Roberta is friends with Michaela and Cathleen, Jasmine has three 2-friends. CJ has five direct friends but because Mira and Roberta are friends with Cathleen she has seven 2 friends. Sydney, Daphne, Diane and Phoebe have four direct friends but because CJ is friends with Cathleen they all have five 2-friends. Cathleen has 3 direct friends but through CJ she gets 4 more 2-friends and through Mira and Roberta she gets 2 more. Based on this set of students, she would have the most 2-friends (9 = 3 direct + 6 friends of friends). CJ would be second with 7. Here it is the ranking Student friends friends of friends 2-friends Cathleen 3 6 9 CJ 5 2 7 Sydney, Daphne, Diane and Phoebe 4 1 5 Roberta 3 2 5 Mira, Michaela 2 2 4 Jasmine 1 2 3 To decide if student A is a 2-friend of B the key is to check whether A and B are separated by at most", "the following figure: In this reduced game, Fiona is certain to gain a strictly higher payoff by choosing $$D^1$$ than if she chooses $$C^1$$, so $$D^1$$ is her unique best choice. Of course, when Fiona chooses $$D^1$$, Alan, being rational, responds by choosing $$D^2$$. If Fiona and Alan know: (i) that they are both rational, (ii) that they both know the payoff structure of the game, and (iii) that they both know (i) and (ii), then they both can predict what the other will do at every node of the Figure 1.1.a game, and conclude that they can rule out the $$D^1,C^2$$-branch of the Figure 1.1.b game and analyze just the reduced game of the following figure: On account of this mutual knowledge, both know that Fiona will choose $$D^1$$, and that Alan will respond with $$D^2$$. Hence, the $$D^1,D^2$$-outcome results if the Farmers’ Dilemma game is played by agents having this mutual knowledge, though it is suboptimal since both agents would fare better at the $$C^1,C^2$$-branch.[5] This argument, which in its essentials is Hume’s argument, is an example of a standard technique for solving sequential games known as backwards induction.[6] The basic idea behind backwards induction is that the agents engaged in a sequential game deduce how each will act throughout the entire game by ruling out", "HackerEarth - Molly And Flame Solution HackerEarth - Molly And Flame Solution Molly is in love with Sherlock. She wants to decide the nature of the relationship between her and Sherlock using the game of FLAME. FLAME stands for: {“Friendship”,”Love”,”Affection”,”Marriage”,”Enemies”} The rules of FLAME are: 1) The names of the two people are taken. 2) Then the common letters are deleted from both the names. 3) Then x_rem is calculated which is the number of leftover letters in both the names. 4) Count through the letters of “FLAME” x_rem times. After you have counted the 5 letters of “FLAME,” you can continue counting from 6 to 10 starting again at “F” and ending at “E,” if needed. If you have more than 10 leftover letters, repeat the process a third time, starting with “F.” 5) The letter at which you stop decides the nature of the relationship. Example: 1)consider the strings sherlock and molly. 2)After striking off the common letters: sherck my 3) x_rem = len(sherck) + len(my) = 8 4) counting through the letters of FLAME. FLAME 12345 678 5) The count stops at A. Hence the result is Affection. However, Molly is not happy with the result of the game(Affection) and she manipulates the string FLAME TO FALME to change the result of the game(Love). Given", "now also know the answer to the last mini-problem ($\\frac{5}{8}$). For the second mini-problem, Fiona \\textit{must} go $1, 2$ (probability of \\frac{1}{4}). Any other option results in her death to a predator. Thus, the final answer is $\\frac{3}{8} \\cdot \\frac{1}{4} \\cdot \\frac{5}{8} = \\frac{15}{256} = \\boxed{A}$", "two edges (i.e. whether there is a path between A and B with one or two edges). If there is they are 2-friends otherwise they are not. According with the above table our team has a total of 27 two-friends. The remaining five girls have 21 two-friends. So if we based the results on just these 10 people our team would win. Graphs are a nice visual representation of friendships but to compute the 2-friends of the hundreds of students required using a different representation called an adjacency matrix. In it, each student is numbered from 0 to n-1 with n being the number of students. Student i (0 <= i<n) friends are represented by a string of n characters. Character j is T(True) if students i,j are friends and F(False) otherwise. This string is element i in a list we will call friends. Lets illustrate this by translating the graph to this representation. We number Daphne as 0, Diane as 1, Sydney as 2, Phoebe as 3, CJ as 4, Cathleen as 5, Mira as 6, Roberta as 7, Michaela as 8 and Jasmine as 9. Then from the graph above we can see that the string representation for Daphne, is friends[0]='FTTTTFFFFF'. For Diane it would be friends[1]='TFTTTFFFFF'. For Jasmine it would be friends[9]='FFFFFFFTFF'. The complete list", "# Description Heidi is terrified by your estimate and she found it unrealistic that her friends would collaborate to drive her into debt. She expects that, actually, each person will just pick a random friend to send Heidi to. (This randomness assumption implies, however, that she can now visit the same friend an arbitrary number of times…) Moreover, if a person only has one friend in common with Heidi (i.e., if that person is in a leaf of the tree), then that person will not send Heidi back (so that Heidi’s travel will end at some point). Heidi also found unrealistic the assumption that she can make all the travels in one day. Therefore now she assumes that every time she travels a route between two friends, she needs to buy a new ticket. She wants to know: how much should she expect to spend on the trips? For what it’s worth, Heidi knows that Jenny has at least two friends. # Input The first line contains the number of friends $n$ ($3 \\leq n \\leq 10^5$). The next $n - 1$ lines each contain three space-separated integers $u, v$ and $c$($0 \\leq u, v \\leq n - 1, 1 \\leq c \\leq 10^4$) meaning that $u$ and $v$ are friends and the cost for traveling between $u$", "my agent: RATES: We are casting for five primary characters and five secondary characters. Rates are $6000 for primaries and$3000 for secondaries. THIS INCLUDES THE SESSION, USAGE AND EXCLUSIVITY FOR SCHIZOPHRENIA DRUGS. THIS MEANS YOU COULD NOT DO ANOTHER JOB FOR A DIFFERENT SCHIZOPHRENIA DRUG MFG COMPANY FOR THREE YEARS. well that's hardly fair. you're only casting one of my personalities and the others are going to get hella jealous. ## 9.19.2010 ### unfortunately you never know if the play you are working really, really, REALLY hard on will suck for reasons beyond your control. And this is what tempts otherwise sane people to write, direct and/or produce their own material. ## 9.06.2010 ### it's a stretch In my current play, many friends are somewhat amused to hear that I am playing a Lebanese woman. It's kind of hilarious. I'm about as Anglo-looking as you can get. I've even been pronounced \"the whitest person I've ever met\". BY MORE THAN ONE PERSON. However, the greatest stretch is not that the character I am portraying is a 22-year old Arab. It is that she is also a PhD in Pure Mathematics. This has led me to research some very interesting* blogs, which have comments like this: Computing e^{-U(x)/T} takes polynomial number of parameters, computing \\sum {x is in n-cube}", "Fiona had D dollars in gift money. She spent 3/5 of this on an electric skateboard. If Fiona SPENT 3/5 of D dollars, then 2/5 of D dollars REMAINED In other words, the amount remaining at the end of the day = 2D/5 dollars The next day, ...” January 6, 2019 Brent@GMATPrepNow posted a reply to After her birthday, Fiona had D dollars in gift money. in the Problem Solving forum “One option is to WORK BACKWARDS The next day, she spent 1/3 of what was left on a movie passcard, and then finally put the last$80 in the bank. If she spent 1/3 of her money on a movie passcard, then 2/3 of her money remained (which she put in the bank). This means 2/3 of her money = $80 We ...” January 6, 2019 Brent@GMATPrepNow posted a reply to A retailer sells only radios and clocks. If she currently ha in the Data Sufficiency forum “ Given: A retailer sells only radios and clocks, and she currently has 44 total items in inventory. Target question: How many radios are there in the inventory Statement 1: The retailer has more than 28 radios in inventory There are several scenarios that satisfy statement 1. Here are ...” January 6, 2019 Brent@GMATPrepNow posted a reply to Number Properties in", "# James is best friends with Nathan, Nathan is his only friend that he's ever had and has been for almost 7 years. The two are ###### Question: James is best friends with Nathan, Nathan is his only friend that he's ever had and has been for almost 7 years. The two are very close, sharing things they would not share to their own family. They talk to each other almost everyday and hangout regularly. Nathan starts hanging out with his other friend Fiona and gradually stops talking and hanging out with James. James eventually hears them together at a store he regularly visits, he hears Nathan call her \"his best friend\" and hears no mention of him. James feels forgotten, replaced, and alone. What should James do next? Why did Nathan stop interacting with him? and why should James do what you choose? ### Given f(x) = 1x + 1 and g(x) = x2, choose the expression for (fºg)(x). Given f(x) = 1x + 1 and g(x) = x2, choose the expression for (fºg)(x). $Given f(x) = 1x + 1 and g(x) = x2, choose the expression for (fºg)(x).$... ### The signal from a certain satellite takes 0.06348 seconds to reach Earth.Estimate using scientific The signal from a certain satellite takes 0.06348 seconds to reach Earth. Estimate", "call several grannies at once. In this case, the selected grannies will go out into the courtyard at the same moment of time. • She cannot deceive grannies, that is, the situation when the $i$-th granny in the moment of appearing in the courtyard, finds that now there are strictly less than $a_i$ other grannies (except herself, but including Maria), is prohibited. Please note that if several grannies appeared in the yard at the same time, then each of them sees others at the time of appearance. Your task is to find what maximum number of grannies (including herself) Maria can collect in the courtyard for the ceremony. After all, the more people in one place during quarantine, the more effective the ceremony! Consider an example: if $n=6$ and $a=[1,5,4,5,1,9]$, then: • at the first step Maria can call grannies with numbers $1$ and $5$, each of them will see two grannies at the moment of going out into the yard (note that $a_1=1 \\le 2$ and $a_5=1 \\le 2$); • at the second step, Maria can call grannies with numbers $2$, $3$ and $4$, each of them will see five grannies at the moment of going out into the yard (note that $a_2=5 \\le 5$, $a_3=4 \\le 5$ and $a_4=5 \\le 5$); • the $6$-th granny cannot", "friends. Sydney, Daphne, Diane and Phoebe have four direct friends but because CJ is friends with Cathleen they all have five 2-friends. Cathleen has 3 direct friends but through CJ she gets 4 more 2-friends and through Mira and Roberta she gets 2 more. Based on this set of students, she would have the most 2-friends (9 = 3 direct + 6 friends of friends). CJ would be second with 7. Here it is the ranking Student friends friends of friends 2-friends Cathleen 3 6 9 CJ 5 2 7 Sydney, Daphne, Diane and Phoebe 4 1 5 Roberta 3 2 5 Mira, Michaela 2 2 4 Jasmine 1 2 3 To decide if student A is a 2-friend of B the key is to check whether A and B are separated by at most two edges (i.e. whether there is a path between A and B with one or two edges). If there is they are 2-friends otherwise they are not. According with the above table our team has a total of 27 two-friends. The remaining five girls have 21 two-friends. So if we based the results on just these 10 people our team would win. Graphs are a nice visual representation of friendships but to compute the 2-friends of the hundreds of students required using a", "free time, character $X$ spends 5 hours a week with character $Y$ but 10 hours a week with $Z$, one would assume $X$ and $Z$ are closer to one another than are $X$ and $Y$ (…despite the fact that it could just be the case that both $X$ and $Z$ are unemployed while $Y$ is a graduate student–I readily note the imperfections of such a measurement). Let us consider the question of character groupings in basic mathematical terms. There are six friends, each an element of the overarching “group,” defined as the set $F=\\{1,2,3,4,5,6\\}$ where $1, 2, 3, 4, 5, 6$ represent Chandler, Joey, Monica, Phoebe, Rachel, and Ross respectively (the listing method is alphabetical). Each episode features character groupings in the form of shared plots, which in turn correspond to subsets of $F$. For example, “The One With George Stephanopoulos” would be represented as $TOWGS=\\{\\{1,2,6\\},\\{3,4,5\\}\\}$ since the plotline with the guys at the hockey game, $\\{1,2,6\\} \\subset F$, is an element of $TOWGS$ as is the plotline with the girls getting pizza/watching George drop his towel, $\\{3,4,5\\} \\subset F$.1 In thinking about quantitatively measuring the friendships via plotline counts, I wondered whether there already existed a numerical database of the 236 episodes that identified each plotline’s defining characters. In other words, I wondered if there was some", "would be friends = ['FTTTTFFFFF', # Daphne as 0 'TFTTTFFFFF', # Diane as 1 'TTFTTFFFFF', # Sydney as 2 'TTTFTFFFFF', # Phoebe as 3 'TTTTFTFFFF', # CJ as 4 'FFFFTFTTFF', # Cathleen as 5 'FFFFFTFFTF', # Mira as 6 'FFFFFTFFTT', # Roberta as 7 'FFFFFFTTFF', # Michaela as 8 'FFFFFFFTFF'] # Jasmine as 9 Note how in this representation we don't consider a student to be friends with itself and so friends[0][0]='F', friends[1][1]='F', etc. (This is because, in this problem, we are only interested in 2 or higher friendship and a person being friends with herself is 1-friends). Clearly the adjacency matrix friends[i][j] is symmetric (i.e. friends[i][j]=friends[j][i]). It is a somewhat memory wasteful representation because most of the entries will be 'F'. In practice we could only keep the entries that are 'T' since those are the only relevant for the 2-friendship computation. Given the list friends, a simple python method to compute the largest number of 2 friends, as well as the number of 2 friends of each person is def mostTwoFriends(self,friends): best = 0 self.nFriends = {} for i in range(0,len(friends)): count = 0 for j in range(0,len(friends[i])): if i == j: continue # direct friends if friends[i][j] == 'T': count += 1 else: # check whether i and j have a friend k in common", "# 2562. Virtual Friends These days, you can do all sorts of things online. For example, you can use various websites to make virtual friends. For some people, growing their social network (their friends, their friends’ friends, their friends’ friends’ friends, and so on), has become an addictive hobby. Just as some people collect stamps, other people collect virtual friends. Your task is to observe the interactions on such a website and keep track of the size of each person’s network. Assume that every friendship is mutual. If Fred is Barney’s friend, then Barney is also Fred’s friend. ### 输入格式 The first line of input contains one integer specifying the number of test cases to follow. Each test case begins with a line containing an integer $F$, the number of friendships formed, which is no more than $100~000$. Each of the following $F$ lines contains the names of two people who have just become friends, separated by a space. A name is a string of $1$ to $20$ letters (uppercase or lowercase). ### 输出格式 Whenever a friendship is formed, print a line containing one integer, the number of people in the social network of the two people who have just become friends. ### 样例 Input 1 3 Fred Barney Barney Betty Betty Wilma Output 2 3 4 236", "# Is 5/6 closer to 1,0,1/2, ###### Question: Is 5/6 closer to 1,0,1/2, ### Which of these people has the best emotional health? Jana plays hockey and is proud that she is the best goalie in her class. Kurt spends time reading and likes to make friends with people who share his interests. Maxine plays video games and likes to watch sports with her friends. Ryan is a good student even though he participates in several demanding sports. Which of these people has the best emotional health? Jana plays hockey and is proud that she is the best goalie in her class. Kurt spends time reading and likes to make friends with people who share his interests. Maxine plays video games and likes to watch sports with her friends. Ryan is a good s... ### How much does the medium smoothie cost? how much does the medium smoothie cost?... ### A number cube with faces numbered 1 through 6 is rolled 15 times. In theory, how many times would you expect to roll a number less than five? A: 2.5 B: 5 C: 10 D: 12.5 A number cube with faces numbered 1 through 6 is rolled 15 times. In theory, how many times would you expect to roll a number less than five? A: 2.5 B: 5", "different representation called an adjacency matrix. In it, each student is numbered from 0 to n-1 with n being the number of students. Student i (0 <= i<n) friends are represented by a string of n characters. Character j is T(True) if students i,j are friends and F(False) otherwise. This string is element i in a list we will call friends. Lets illustrate this by translating the graph to this representation. We number Daphne as 0, Diane as 1, Sydney as 2, Phoebe as 3, CJ as 4, Cathleen as 5, Mira as 6, Roberta as 7, Michaela as 8 and Jasmine as 9. Then from the graph above we can see that the string representation for Daphne, is friends[0]='FTTTTFFFFF'. For Diane it would be friends[1]='TFTTTFFFFF'. For Jasmine it would be friends[9]='FFFFFFFTFF'. The complete list would be friends = ['FTTTTFFFFF', # Daphne as 0 'TFTTTFFFFF', # Diane as 1 'TTFTTFFFFF', # Sydney as 2 'TTTFTFFFFF', # Phoebe as 3 'TTTTFTFFFF', # CJ as 4 'FFFFTFTTFF', # Cathleen as 5 'FFFFFTFFTF', # Mira as 6 'FFFFFTFFTT', # Roberta as 7 'FFFFFFTTFF', # Michaela as 8 'FFFFFFFTFF'] # Jasmine as 9 Note how in this representation we don't consider a student to be friends with itself and so friends[0][0]='F', friends[1][1]='F', etc. (This is because, in this problem, we are only", "2013 12-21 # Jamie’s Contact Groups Jamie is a very popular girl and has quite a lot of friends, so she always keeps a very long contact list in her cell phone. The contact list has become so long that it often takes a long time for her to browse through the whole list to find a friend’s number. As Jamie’s best friend and a programming genius, you suggest that she group the contact list and minimize the size of the largest group, so that it will be easier for her to search for a friend’s number among the groups. Jamie takes your advice and gives you her entire contact list containing her friends’ names, the number of groups she wishes to have and what groups every friend could belong to. Your task is to write a program that takes the list and organizes it into groups such that each friend appears in only one of those groups and the size of the largest group is minimized. There will be at most 20 test cases. Ease case starts with a line containing two integers N and M. where N is the length of the contact list and M is the number of groups. N lines then follow. Each line contains a friend’s name and the groups the friend", "birthday, Fiona had D dollars in gift money. in the Problem Solving forum “Another approach is to use algebra. After her birthday, Fiona had D dollars in gift money. She spent 3/5 of this on an electric skateboard. If Fiona SPENT 3/5 of D dollars, then 2/5 of D dollars REMAINED In other words, the amount remaining at the end of the day = 2D/5 dollars The next day, ...” January 6, 2019 Brent@GMATPrepNow posted a reply to After her birthday, Fiona had D dollars in gift money. in the Problem Solving forum “One option is to WORK BACKWARDS The next day, she spent 1/3 of what was left on a movie passcard, and then finally put the last$80 in the bank. If she spent 1/3 of her money on a movie passcard, then 2/3 of her money remained (which she put in the bank). This means 2/3 of her money = $80 We ...” January 6, 2019 Brent@GMATPrepNow posted a reply to A retailer sells only radios and clocks. If she currently ha in the Data Sufficiency forum “ Given: A retailer sells only radios and clocks, and she currently has 44 total items in inventory. Target question: How many radios are there in the inventory Statement 1: The retailer has more than 28 radios in inventory" ]

[ "How many more diagonals are in an octagon than in a hexagon? $20-9=11$ Answer: $11$", "completed hexagon.", "Elementary Geometry for College Students (6th Edition) (a) The hexagon $ABCB'A'D$ has 6 vertices. We can find the number of diagonals $D$: $D = \\frac{n~(n-3)}{2} = \\frac{(6)(6-3)}{2} = 9$ (b) When we draw the nine diagonals. we can see that two diagonals lie outside the hexagon. The diagonal $AA'$ and the diagonal $BB'$ lie outside the hexagon.", "# How many triangles can we form if we draw all the diagonals of a hexagon? How many triangles can we form if we draw all the diagonals of a hexagon? I thought that the answer is $\\binom{6}{3}=20$ but this is not the right answer, why? - What kind of hexagon? Convex or not? Regular or not? Before using counting tools, we need to know what we are counting. – André Nicolas Mar 9 '12 at 17:03 Must the vertices of the triangles coincide with vertices of the hexagon? – Matt Mar 9 '12 at 17:07 This is interesting, @Andre considering the type of question I guess it should be convex-regular. – Quixotic Mar 9 '12 at 17:54 What makes you say 20 is not the right answer? – Gerry Myerson Mar 10 '12 at 6:41 The problem is very unclear (see the comments). Here is one interpretation (which is probably not the one intended, but who knows?): Drawing all 9 diagonals of a regular hexagon divides it into 24 regions, of which 6 are quadrilaterals, leaving 18 triangles. - Interesting. I first thought of the 6 triangles you get when drawing the \"diagonals\" of a regular hexagon, but after thinking about your answer, it is a correct one, provided you are just looking for the number of", "his hexagonal drawing.", "How many convex polygons can be drawn using these points as vertices?", "# Number of diagonals in a hexagon , , 1 Comment How many diagonals are there in a hexagon? 9 Are you big fan of Maths? It is not too complicated to calculate the number of diagonals in a hexagon. Well, it is important to know the fact that the hexagon has 6 vertices with 3 diagonals which gives you the number 18. We divide by 2 and the final result is represented by 9 diagonals. Keep in mind the fact that a diagonal is a line segment that connects non-adjacent vertices in a polygon.", "9 diagonals in a regular hexagon. The lengths of the diagonals are not equal and three diagonals are longer than the rest. These are called the long diagonals of a regular hexagon and run through its centre. The other six diagonals of the regular hexagon are called its short diagonals.", "a hexagon grid of size n, made up of asterisks. For example, a hexagon of ... 218 views ### Optimal strategy to defeat the wizard Consider the following setup. An evil wizard has ten opaque boxes in front of him. Hidden from you, he chooses a random number of coins $x \\in \\{1 \\dots 10\\}$ and spreads them uniformly at random ...", "asymmetric waterdragon. Explore hex pleating further with this long- and variable-length-legged spider design.", "it might be six only. • For regular hexagon its just same as above 4. • Square has also 4 lines of symmetry and its confirm. • 0 If the hexagon is a regular one it will have 6 lines of symmetry. This answer applies to the second question. A square has 4 lines of symmetry. • 0 hexagon six square four • 0 Hexagon=6 Square=4", "# Partitioning A Hexagon Following the pattern of the first 3 figures above, how many little triangles are there in the $$9^\\text{th}$$ figure? ×", "The diagram shows three squares on the sides of a triangle $ABC$. Their areas are respectively 18 000, 20 000 and 26 000 square centimetres. If the squares are joined, three more triangular areas are enclosed. What is the area of this convex hexagon?", "has one bisector. What fashion designer was responsible for the \"New Look\"? Here is a list of some common triangles. Why do we love old people, especially little old ladies? Triangles have 0 diagonals. 201 views / Category: Science Share: More Questions: What is the fear of the color purple called? always sum to 180°. Answer (1 of 1): A triangle has no vertices that are more than one away from another. Definition: The triangles of a polygon are the triangles created by drawing, Parallelogram inscribed in a quadrilateral, Perimeter of a polygon (regular and irregular). How many diagonals does a triangle have? How many diagonals does a regular hexagon have ? Examine the table. This formula works, of course, but the question is one of those in my textbook designated to be solved using combinations. How many diagonals does a triangle have? How many diagonals does an isosceles triangle have? So we can subtract this from the total. Well wouldn't make a whole lot of sense to draw a polygon with 300 sides and draw on all those diagonals. So I look at the number of vertices we have three so we're going to call that n. Here we have number of diagonals per vertex, here we have 0, 1 and 2 and I see that to", "among the interior angles of a convex polygon there cannot be more than three acute angles. Tessellating Hexagons Stage: 3 Challenge Level: Which hexagons tessellate? Mystic Rose Stage: 3 Challenge Level: Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.", "# Hexagonal Numbers In the diagram above, we see a series of triangles and hexagons. How many dots are contained in the $$20^\\text{th}$$ hexagon? ×", "smaller hexagon?", "Triominoes A triomino is a flat L shape made from 3 square tiles. A chess board is marked into squares the same size as the tiles and just one square, anywhere on the board, is coloured red. Can you cover the board with trionimoes so that only the square is exposed? LOGO Challenge 5 - Patch Using LOGO, can you construct elegant procedures that will draw this family of 'floor coverings'? LOGO Challenge - Triangles-squares-stars Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential. Tessellating Hexagons Age 11 to 14 Challenge Level: This problem follows on from some of the ideas in Tessellating Triangles and Tessellating Quadrilaterals. It was inspired by a problem in \"Tiles and Tiling\", an activity book published by the Association of Teachers of Mathematics. Here is a tessellation of regular hexagons: Can you explain why regular hexagons tessellate? What about a hexagon where each pair of opposite sides is parallel, and opposite sides are the same length, but different pairs of sides are not the same length? You can print off some square dotty paper, or some isometric dotty paper, and try drawing hexagons of this form on it. You could also draw some hexagons", "+0 # How many diagonals can be drawn in the pentagon? 0 384 7 How many diagonals can be drawn in the pentagon? (regular pentagon, nothing special) A: 5 B: 10 C: 15 D: 20 (This chapter is about Permutations and Combinations) Guest May 8, 2014 ### Best Answer #1 +5453 +8 There are 5 diagonals, no repeats. For example, AD and DA are the same diagonal, so it should be a combination. kitty<3 May 8, 2014 Sort: ### 7+0 Answers #1 +5453 +8 Best Answer There are 5 diagonals, no repeats. For example, AD and DA are the same diagonal, so it should be a combination. kitty<3 May 8, 2014 #2 +84340 +5 I think the number of diagonals that can be drawn in any poygon with n vertexes (sides), where n >3, is given by: (n)(n-3)/2. To see this, note that no diagonal can be drawn from a vertex to itself, nor can any diagonal be drawn to the two neighboring vetexes. So this leaves (n-3) vertexes left. And from each vertex, the same number of diagonals can be drawn. So the total number of diagonals that can be drawn is (n)(n-3). But the diagonal from vi to vj is the same diagonal drawn from vj to vi, so we must divide this total number by", "(a) triangular and (b) hexagonal grid. (c) Suggest another grid. (d) Is it possible to have a similar algorithm if the cells are irregular?" ]

[ "+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", "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)", "= \\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),", "= (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.", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "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", "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", "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 $O_{1}$,", "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", "draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Obviously, $25$ is the shorter length, and thus the answer is $\\boxed{\\textbf{(B) }25}$.", "(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", "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", "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 our unit, this yields a final area of", "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)", "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", "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", "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.", "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" ]

[ "+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", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "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)", "= \\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),", "= (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.", "draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Obviously, $25$ is the shorter length, and thus the answer is $\\boxed{\\textbf{(B) }25}$.", "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", "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.", "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", "(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", "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", "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", "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", "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)", "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 our unit, this yields a final area of", "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.", "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" ]

[ "10, 12, 15, 16, 18, 20, 24, 25, 30, 36). The possible values of (R1,R2) are ((1,6), (3,8), (4,9), (5,10), (10,15), (15,20), (20,25), (25,30)). The possible values of (R1,R2,R3) are ((3,8,2), (4,9,3), (5,10,4), (10,15,9), (25,30,24)). The possible values of (R1,R2,R3,R4) are ((5,10,4,15), (10,15,9,20)) The possible values of (R1,R2,R3,R4,R5) are only one: (10,15,9,20,12). That is the solution. #### KnotTheorist 125. Find an integer which can be written as a sum of squares in three different ways. D124 325 = 102 + 152 = 182 + 12 = 62 + 172 #### fresh_42 Mentor 2018 Award 126. One of us has the least odd value under Euler's $\\varphi$ function different from $1$, and one of us is the least that doesn't occur in its range. The third of us tells how many colors are sufficient to color a map on a ring. Me? This is the product of them. D124 Last edited: #### fresh_42 Mentor 2018 Award 127. Someone replaced the round shooting targets at the local gun club by equilateral triangles with a side length of 10 inches. Show that among five hits will always be two closer to each other than five inches. D125 #### BvU Homework Helper 129 Make a drawing ! Worst case for 1, 2, 3 is in the corners. dashed radii are < 5\"", "# 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", "because 4 x 4 = 16 This is A000290 For example: For n = 12, you could achieve the desired size with 4 panels of sizes [9, 1, 1, 1]. As 9 is the largest square possible in this configuration, this is the best answer. For n = 13, you can achieve the desired size with only 2 panels: [9, 4] If n is a square number, the output should be [n]. ## Input A positive integer n representing the total desired surface area of the solar panels. Note that 0 is not positive. ## Output A list of square numbers that sums to n containing the largest squares possible. ### Testcases 1 -> [1] 2 -> [1,1] 3 -> [1,1,1] 4 -> [4] 7 -> [4,1,1,1] 8 -> [4,4] 9 -> [9] 12 -> [9,1,1,1] 13 -> [9,4] 18 -> [16,1,1] 30 -> [25,4,1] 50 -> [49,1] 60 -> [49,9,1,1] 70 -> [64,4,1,1] 95 -> [81,9,4,1] 300 -> [289,9,1,1] 1246 -> [1225,16,4,1] 12460 -> [12321,121,16,1,1] 172593 -> [172225,361,4,1,1,1] # Sandbox I don't think this challenge is similar enough to part 1 to be considered a dupe, as while some answers from part 1 could be trivially modified to work for part 2, they would likely be out-golfed by better approaches. That said I'm not sure how well-worded", "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", "K_0(&r0,&r4,&r8,&r12); K_N(VLIT4(0.9239,0.9239,0.9239,0.9239), VLIT4(0.3827,-0.3827,0.3827,-0.3827), &r1,&r5,&r9,&r13); TX2(&r0,&r1); TX2(&r4,&r5); TX2(&r8,&r9); TX2(&r12,&r13); S_4(r0,r4,r8,r12,out0+0,out0+8,out0+16,out0+24); S_4(r1,r5,r9,r13,out1+0,out1+8,out1+16,out1+24); K_N(VLIT4(0.7071,0.7071,0.7071,0.7071), VLIT4(0.7071,-0.7071,0.7071,-0.7071), &r2,&r6,&r10,&r14); K_N(VLIT4(0.3827,0.3827,0.3827,0.3827), VLIT4(0.9239,-0.9239,0.9239,-0.9239), &r3,&r7,&r11,&r15); TX2(&r2,&r3); TX2(&r6,&r7); TX2(&r10,&r11); TX2(&r14,&r15); S_4(r2,r6,r10,r14,out0+4,out0+12,out0+20,out0+28); S_4(r3,r7,r11,r15,out1+4,out1+12,out1+20,out1+28); } The vector leaf sub-transforms of a single size/type are handled in the same way as a VL-1 sub-transform, with one difference: the vector registers must be transposed before the data is stored to memory in the output array. In the example shown in Code 109, the transposes take place at lines 16 and 25. Prior to the store operations, each position of the vector register (each position being a whole complex word) contains an element belonging to each of the leaf sub-transforms composing the vectorized sub-transform. Because each leaf sub-transform is stored sequentially to different locations in memory with aligned vector store operations, sets of registers are transposed such that each vector register contains elements from only one leaf sub-transform. #### Heterogeneous leaf sub-transform vectors void sfft_fcf128_shl16_8_eo(offset_t *is,const SFFT_D *in,SFFT_D *out){ SFFT_R r0_1,r2_3,r4_5,r6_7,r8_9,r10_11,r12_13,r14_15, r16_17,r18_19,r20_21,r22_23,r24_25,r26_27,r28_29,r30_31; L_4_4(in+is[0],in+is[1],in+is[2],in+is[3], &r0_1,&r2_3,&r16_17,&r18_19); L_2_2(in+is[4],in+is[5],in+is[6],in+is[7], &r4_5,&r6_7,&r20_21,&r22_23); K_N(VLIT4(0.7071,0.7071,1,1),VLIT4(0.7071,-0.7071,0,-0), &r0_1,&r2_3,&r4_5,&r6_7); L_4_2(in+is[8],in+is[9],in+is[10],in+is[11], &r8_9,&r10_11,&r28_29,&r30_31); L_4_4(in+is[12],in+is[13],in+is[14],in+is[15], &r12_13,&r14_15,&r24_25,&r26_27); K_N(VLIT4(0.9239,0.9239,1,1),VLIT4(0.3827,-0.3827,0,-0), &r0_1,&r4_5,&r8_9,&r12_13); S_4(r0_1,r4_5,r8_9,r12_13,out0+0,out0+8,out0+16,out0+24); K_N(VLIT4(0.3827,0.3827,0.7071,0.7071), VLIT4(0.9239,-0.9239,0.7071,-0.7071), &r2_3,&r6_7,&r10_11,&r14_15); S_4(r2_3,r6_7,r10_11,r14_15,out0+4,out0+12,out0+20,out0+28); K_N(VLIT4(0.7071,0.7071,1,1),VLIT4(0.7071,-0.7071,0,-0), &r16_17,&r18_19,&r20_21,&r22_23); S_4(r16_17,r18_19,r20_21,r22_23,out1+0,out1+4,out1+8,out1+12); K_N(VLIT4(0.7071,0.7071,1,1),VLIT4(0.7071,-0.7071,0,-0), &r24_25,&r26_27,&r28_29,&r30_31); S_4(r24_25,r26_27,r28_29,r30_31,out1+16,out1+20,out1+24,out1+28); } In the case of a vector comprising heterogeneous leaf sub-transforms, the data is transposed into separate sub-transforms following the primitive leaf operations. The remainder of the computation is carried out separately for each leaf sub-transform in the", "R3 + (R1 + R2 + R3) R4), I3 -> (R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), I4 -> (R3 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), V1 -> ((R3 R4 + R2 (R3 + R4)) Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), V2 -> (R3 R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4)}} In[3]:=R2 = s*L; R3 = 1/(s*c); Vi = u/s; L = 10*10^(-3); c = 1*10^(-6); u = 1; R1 = 10; R4 = 10; FullSimp FullSimplify[ InverseLaplaceTransform[(R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), s, t]] Out[3]=(E^(-500 (101 + Sqrt[9401]) t) (-1 + E^( 1000 Sqrt[9401] t)))/(10 Sqrt[9401]) In[4]:=FullSimplify[Solve[{D[%3, t] == 0, t > 0}, t]] Out[4]=ArcCoth[101/Sqrt[9401]]/(500 Sqrt[9401]) In[5]:=FullSimplify[Limit[%3, t ->ArcCoth[101/Sqrt[9401]]/(500 Sqrt[9401])]] Out[5]=((101 + Sqrt[9401]) E^(-(1 + 101/Sqrt[9401]) ArcCoth[101/Sqrt[ 9401]]))/4000 In[6]:=Plot[%3, {t, 0,1/1000}] Out[6]=(*this code will plot the picture above*) • Thank you so so much, I will study your answer in more detail. Dec 20 '20 at 14:12 • @Jan you have done this before - you should not hand homework questions on a plate to anyone - that isn't what this site is about. Dec 22 '20 at 15:18", "need as we are interested in the ratios): >ri4=sqrt(6)/12,rc4=sqrt(6)/4 0.204124 0.612372 > >ri6=1/2,rc6=sqrt(3)/2 0.5 0.866025 > >ri8=sqrt(6)/6,rc8=sqrt(2)/2 0.408248 0.707107 > >ri12=sqrt(250+110*sqrt(5))/20,rc12=(sqrt(15)+sqrt(3))/4 1.11352 1.40126 > >ri20=(3*sqrt(3)+sqrt(15))/12,rc20=sqrt(10+2*sqrt(5))/4 0.755761 0.951057 > >rho4=rc4/ri4 3 >rho6=rc6/ri6 1.73205 >rho8=rc8/ri8 1.73205 >rho12=rc12/ri12 1.25841 >rho20=rc20/ri20 1.25841 Which may be compared with the orbital radius ratios from here (radii in km) >RMecury=57.9e6; >RVenus=108.2e6; >REarth=149.6e6; >RMars=227.9e6; >RJupiter=778.3e6; >RSaturn=1426.7e6; Now we can compare the corresponding radii ratios: >[RVenus/RMecury,rho8] 1.86874 1.73205 >[REarth/RVenus,rho20] 1.38262 1.25841 >[RMars/REarth,rho12] 1.5234 1.25841 >[RJupiter/RMars,rho4] 3.41509 3 >[RSaturn/RJupiter,rho6] 1.8331 1.73205 Which, as these things go, is not bad. • I think you in the end wrote the value for rho20 in the place of rho12 which should be 1.4725 (to be compared with true 1.5234). So the errors were 3% to 12%, that's a good guess without stuff like physics or telescopes. But is was hardly helpful for his later work that it fits better for the most eccentric planet orbits. – LocalFluff May 31 '15 at 4:30 • I would like to see your calculations as repeating mine using independent method gives the same result as the main post, unfortunately this is not a suitable medium to sort out such a numerical disagreement. – Conrad Turner May 31 '15 at 5:07 • I use your calculations. Look at your last box, you copy/pasted 1.25841 twice. rho12", "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", "for (i in 1:length(m3)) { for (k in 1:5) { if (m3[i] >= e[k,1] & m3[i] <= e[k+1,1]) { r3[i] = e[k,2] + (e[k+1,2]-e[k,2])*(m3[i]-k) getDF3[i] = exp(-r3[i]*m3[i]) } } } for (i in 1:length(m4)) { for (k in 1:5) { if (m4[i] >= e[k,1] & m4[i] <= e[k+1,1]) { r4[i] = e[k,2] + (e[k+1,2]-e[k,2])*(m4[i]-k) getDF4[i] = exp(-r4[i]*m4[i]) } } } for (i in 1:length(m5)) { for (k in 1:5) { if (m5[i] >= e[k,1] & m5[i] <= e[k+1,1]) { r5[i] = e[k,2] + (e[k+1,2]-e[k,2])*(m5[i]-k) getDF5[i] = exp(-r5[i]*m5[i]) } } } # I calculate the Default Leg for each asset DL1 = 1/5*(1-RR)*getDF1 DL2 = 1/5*(1-RR)*getDF2 DL3 = 1/5*(1-RR)*getDF3 DL4 = 1/5*(1-RR)*getDF4 DL5 = 1/5*(1-RR)*getDF5 # now I calculate the Protection Leg or Premium Leg PL1 = getDF1*m1*5/5 PL2 = PL1 + getDF2*(m2-m1)*4/5 PL3 = PL2 + getDF3*(m3-m2)*3/5 PL4 = PL3 + getDF4*(m4-m3)*2/5 PL5 = PL4 + getDF5*(m5-m4)*1/5 # I calculate the spread as follows: s1 = mean(DL1)/mean(PL1) s2 = mean(DL2)/mean(PL2) s3 = mean(DL3)/mean(PL3) s4 = mean(DL4)/mean(PL4) s5 = mean(DL5)/mean(PL5) as you can see I extracted the default times for every single asset (in this case 5 names). Now I have to calculate the spread of each instrument and in a continuous time. the key concept is to calculate the Default Leg and the Protection Leg. Default Leg is", "c c0 f [z] 0.80 ; ryz6 frame = 1.b2; n 2fs; c c0 s [xy] 0.15 ; c c0 s [z] 0.110 ; c c0 f [z] 0.60 ; ryz6 frame = 1.b3; n 2fs; c c0 s [xy] 0.15 ; c c0 s [z] 0.120 ; c c0 f [z] 0.40 ; ryz6 frame = 1.b4; n 2fs; c c0 s [xy] 0.15 ; c c0 s [z] 0.130 ; c c0 f [z] 0.20 ; ryz6 frame = 1.b5; n 2fs; c c0 s [xy] 0.15 ; c c0 s [z] 0.140 ; c c0 f [z] 0.10 ; ryz6 frame = 1.b6; n 2fs; c c0 s [xy] 0.15 ; c c0 s [z] 0.150 ; c c0 f [z] 0.00 ; ryz6 # split cubes along w (needs yw angle set, see above) frame = 1.c1; n 1fs; c c0 s [xyz] 0.15 ; c c0 s [w] 0.815 frame = 1.c2; n 2fs; # grow w and start to rotate xy/yz/rot* frame = 1.d1; n 2fs; c c0 s [xyz] 0.15 ; c c0 s [w] 0.715 ; c c0 f [w] 0.9 ; rots; !ryz2; rxy2; a xw astepb4 frame = 1.d2; n 2fs; c c0 s [xyz] 0.15 ; c c0 s [w] 0.615 ; c c0 f [w] 0.8", "example 16 is a square number because 4 x 4 = 16 This is A000290 For example: For n = 12, you could achieve the desired size with 4 panels of sizes [9, 1, 1, 1]. As 9 is the largest square possible in this configuration, this is the best answer. For n = 13, you can achieve the desired size with only 2 panels: [9, 4] If n is a square number, the output should be [n]. ## Input A positive integer n representing the total desired surface area of the solar panels. Note that 0 is not positive. ## Output A list of square numbers that sums to n containing the largest squares possible. ### Testcases 1 -> [1] 2 -> [1,1] 3 -> [1,1,1] 4 -> [4] 7 -> [4,1,1,1] 8 -> [4,4] 9 -> [9] 12 -> [9,1,1,1] 13 -> [9,4] 18 -> [16,1,1] 30 -> [25,4,1] 50 -> [49,1] 60 -> [49,9,1,1] 70 -> [64,4,1,1] 95 -> [81,9,4,1] 300 -> [289,9,1,1] 1246 -> [1225,16,4,1] 12460 -> [12321,121,16,1,1] 172593 -> [172225,361,4,1,1,1] # Sandbox I don't think this challenge is similar enough to part 1 to be considered a dupe, as while some answers from part 1 could be trivially modified to work for part 2, they would likely be out-golfed by better approaches. That", "the scales with '*': r[5][i=A<B?0:13]=r[5][i+7]=r[9][13-i]=r[9][20-i]=42; // Fill the appropriate four cells of the balance-bar with '-': for(i=5;i-->1;r[i][A>B?18-i*3:2+i*3]=42)r[i][A>B?17-i*3:3+i*3]=45; // Fill the appropriate cells of the scales with '-': for(i=0;++i<20;)r[i>13?A>B?5:9:A>B?9:5][i>13|i<7?i:1]=45; // Fill the appropriate cells of the robes with '/' and '\\': for(i=9;i-->1;r[i][i>4?A>B?8-i:21-i:A>B?17-i:4-i]=47)r[i][i>4?A>B?i-1:i+12:A>B?i+16:i+3]=92; // Fill the four appropriate cells of the balance-bar with '_', // and set A and B to 9 and 5 depending on which score is higher: A=(A>B?r[i=0][16]=r[1][13]=r[3][7]=r[4][4]=95:(r[0][i=4]=r[1][7]=r[3][13]=r[4][16]=95));A=9-i;B=5+i;} // Call a separated method to fill the cells above the scales with the input-characters: c(r,a,A,7);c(r,b,B,20); // And finally return the resulting character-matrix: return r;}; // Separated method to calculate the score of the given String: int s(String s){ // Initially start the score-sum at 0: int r=0; // Loop over the characters of the given String: for(int i:s.getBytes()) // Increase the sum by: r+= // 4 for uppercase letters: i>64&i<91?4 // 2 for lowercase letters: :i>96&i<123?2 // 3 for any other character: :3; // And return the resulting sum: return r;} // Separated method to draw the strings on top of the scales: void c(char[][]r,String s,int p,int q){ // Keep a counter so we know when we're done drawing the given String: for(int c=0, // Loop over the appropriate rows bottom to top: i=p,j;i-->p-5;) // Inner loop over the appropriate cells of this row left to right, for(j=q;j-->q-6", "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", "Theorem (first of many proofs): the left diagram shows that $c^2 = 4 \\cdot \\frac{ab}2 + (b-a)^2 = a^2 + b^2$, and the right diagram shows a second proof by re-arranging the first diagram (the area of the shaded part is equal to $a^2 + b^2$, but it is also the re-arranged version of the oblique square, which has area $c^2$).[3] $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } { /* first picture */ filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); fill(sq1, rgb(0.95,1,0.95)); fill(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw((0,0)--top--(c,0)--a*expi(rot1)--cycle, rgb(0.5,0.9,0.5)); draw(sq1 ^^ sq2); draw(a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c)); label(\"a\",a/2*expi(rot1),NW,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",(c/2,0),S,sm); } { /* second picture */ fill(shift(shiftR)*sq1, rgb(0.95,1,0.95)); fill(shift(shiftR)*sq2, rgb(0.95,1,0.95)); filldraw(shift(shiftR)*rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw(shift(shiftR+(0,-c))*((0,0)--top--(c,0)--a*expi(rot1)--cycle), rgb(0.5,0.9,0.5)); filldraw(shift(shiftR+(0,-c))*((0,0)--(c,0)--a*expi(rot1)--cycle), rgb(1,0.85,0.7)); draw(shift(shiftR)*((0,0)--(c,0) ^^ sq1 ^^ sq2 ^^ a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c))); label(\"a\",shiftR+a/2*expi(rot1),NW,sm); label(\"b\",shiftR+a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",shiftR+(c/2,0),S,sm); } [/asy]$ Another proof of the Pythagorean Theorem (animated version). $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare, tri = (0,0)--(0,a)--(b,0)--cycle; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^", "# 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 •", "circle (0.05); \\foreach \\r in {0.25,0.5,...,1.75} { \\draw (0.5,0) ++(-60:\\r) arc (-60:60:\\r); } \\foreach \\r in {0.125,0.375,...,1.875} { \\draw[dashed] (0.5,0) ++(-60:\\r) arc (-60:60:\\r); } \\draw[fill=black!10] (2,-2,-1) -- (2,-2,1) -- (2,2,1) -- (2,2,-1) -- (2,-2,-1); \\fill (2,0.5) circle (0.05) (2,-0.5) circle (0.05); \\foreach \\r in {0.25,0.5,...,2} { \\draw (2,0.5) ++(-60:\\r) arc (-60:60:\\r); \\draw (2,-0.5) ++(-60:\\r) arc (-60:60:\\r); } \\foreach \\r in {0.125,0.375,...,2.125} { \\draw[dashed] (2,0.5) ++(-60:\\r) arc (-60:60:\\r); \\draw[dashed] (2,-0.5) ++(-60:\\r) arc (-60:60:\\r); } \\draw (2,0.5,0.625) -- (2,0.5,0.5); \\draw (2,-0.5,0.625) -- (2,-0.5,0.5); \\draw (2,0.5,0.5) -- (2,-0.5,0.5); \\draw[densely dotted] (2,0.5,0.5) -- (2,0.5,0); \\draw[densely dotted] (2,-0.5,0.5) -- (2,-0.5,0); \\draw[fill=black!10] (4,-2,-1) -- (4,-2,1) -- (4,2,1) -- (4,2,-1) -- (4,-2,-1); % LABELLING \\begin{scope}[canvas is yz plane at x=2,rotate=-90] \\node[above] at (0,0.5) {S${}_1$}; \\node[below] at (0,-0.5) {S${}_2$}; \\node[left] at (-0.5,0) {d}; \\end{scope} \\begin{scope}[canvas is yz plane at x=0.5,rotate=-90] \\node[below left=-0.1cm] at (0,0) {S${}_0$}; \\end{scope} \\begin{scope}[xshift=4cm,yshift=2cm,rotate=-90,canvas is xy plane at z=0] \\fill[white] (0,0) rectangle (4,1); \\begin{axis}[ width=5.575cm, xmin=-0.62, xmax=0.62, ymin=0, ticks=none ] ] {(cos(deg(5*pi*sin(deg(x)))))^(2)*((sin(deg(4*pi*sin(deg(x)))))/(4*pi*sin(deg(x))))^(2)}; \\draw[thin, densely dashed,blue] (axis cs:0.1593,0.1327) -- (axis cs:0.1593,0); \\draw[thin, densely dashed,blue] (axis cs:-0.1593,0.1327) -- (axis cs:-0.1593,0); \\draw[thin, densely dashed,blue] (axis cs:0.3941,0.03938) -- (axis cs:0.3941,0); \\draw[thin, densely dashed,blue] (axis cs:-0.3941,0.03938) -- (axis cs:-0.3941,0); \\end{axis} \\end{scope} \\draw[thin,densely dashed,blue] (2,0) -- (6.9,0); \\begin{scope} \\clip (2,-2) rectangle (4,2); \\draw[thin,densely dashed,blue] (2,0) -- +(15:4); \\draw[thin,densely dashed,blue] (2,0) -- +(-15:4); \\draw[thin,densely dashed,blue] (2,0) -- +(32:4); \\draw[thin,densely dashed,blue]", "shaded = 12 square units . Area of shaded Rectangle = 12 unit squares . Question 11. Higher Order Thinking In 448,244, how is the relationship between the first pair of 4s the same as the relationship between the second pair of 4s? Answer: 448,244 The place of first pair of 44 in 448,244 = 440,000. The place of Second pair of 44 in 448,244 = 44. The relationship between the first pair of 4s the same as the relationship between the second pair of 4s = First pair of 44 is 10,000 times greater than Second pair of 44 . = 44 × 10,000 = 440,000 . Assessment Practice Question 12. Which group of numbers shows the values of the 4s in 44,492? A. 40,000; 4,000; 400 B. 40,000; 400; 40 C. 4,000; 400; 4 D. 400; 40; 4 Answer: Option A . Explanation : The values of the 4s in 44,492 = 40,000 ; 4,000 ; 400 . Question 13. In which number is the value of the red digit ten times as great as the value of the blue digit? A. 335,531 B. 335,531 C. 335,531 D. 335,531 Answer: ### Lesson 1.3 Compare Whole Numbers Solve & Share A robotic submarine can dive to a depth of 26,000 feet. Which oceans can the submarine explore", "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", "2015 AMC 8 Problems/Problem 25 One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space? $\\textbf{(A) } 9\\qquad \\textbf{(B) } 12\\frac{1}{2}\\qquad \\textbf{(C) } 15\\qquad \\textbf{(D) } 15\\frac{1}{2}\\qquad \\textbf{(E) } 17$ $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]$ Solution 1 We can draw a diagram as shown. $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the $4$ big triangles by $AA.$ Let the height of a big triangle be $x$ then $\\tfrac{x}{x-1}=\\tfrac{5-x}{1}$. $$x=-x^2+6x-5$$ $$x^2-5x+5=0$$ $$x=\\dfrac{5\\pm \\sqrt{(-5)^2-(4)(1)(5)}}{2}$$ $$x=\\dfrac{5\\pm \\sqrt{5}}{2}$$ Thus $x=\\dfrac{5-\\sqrt{5}}{2}$, because by symmetry, $x < \\dfrac52$. This means the area of each triangle is $\\dfrac{5-\\sqrt{5}}{2}*(5-\\dfrac{5-\\sqrt{5}}{2})*\\dfrac{1}{2}=\\dfrac{5}{2}$ This the area of the square is $25-(4*\\dfrac{5}{2})=\\boxed{\\textbf{(C)}~15}$ Solution 2 We draw a square as shown: We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined", "we think of the pool as filled with 1 x 1 x 1 cubes (and some fractions of cubes), then scaling to the larger pool changes each 1 x 1 x 1 cube to a 2 x 2 x 1 prism, or multiplies volume by 4. D $$\\large 65\\cdot 8\\text{ }{{\\text{m}}^{3}}$$Hint: Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube. Question 1 Explanation: Topic: Determine how the characteristics (e.g., area, volume) of geometric figures and shapes are affected by changes in their dimensions (Objective 0023). Question 2 #### Which of the lists below is in order from least to greatest value? A $$\\large \\dfrac{1}{2},\\quad \\dfrac{1}{3},\\quad \\dfrac{1}{4},\\quad \\dfrac{1}{5}$$Hint: This is ordered from greatest to least. B $$\\large \\dfrac{1}{3},\\quad \\dfrac{2}{7},\\quad \\dfrac{3}{8},\\quad \\dfrac{4}{11}$$Hint: 1/3 = 2/6 is bigger than 2/7. C $$\\large \\dfrac{1}{4},\\quad \\dfrac{2}{5},\\quad \\dfrac{2}{3},\\quad \\dfrac{4}{5}$$Hint: One way to look at this: 1/4 and 2/5 are both less than 1/2, and 2/3 and 4/5 are both greater than 1/2. 1/4 is 25% and 2/5 is 40%, so 2/5 is greater. The distance from 2/3 to 1 is 1/3 and from 4/5 to 1 is 1/5, and 1/5 is less than 1/3, so 4/5 is bigger. D $$\\large \\dfrac{7}{8},\\quad \\dfrac{6}{7},\\quad \\dfrac{5}{6},\\quad \\dfrac{4}{5}$$Hint: This is in order from greatest to least. Question 2" ]

[ "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "+0 # help asap pls -1 174 2 In these puzzles, the goal is to fill every empty square (including shaded squares) so that: Each square contains a positive digit (1-9). Digits may be reused. The percent next to a row gives the percent of the row sum contained in the shaded square of that row. The percent above each column gives the percent of the column sum contained in the shaded square of that column. Solve the following percent puzzle, entering each square of the completed puzzle in the corresponding box below. Asymptote code below [asy] void shademe(pair A) { fill(A--(A+(1,0))--(A+(1,1))--(A+(0,1))--cycle,gray(0.8)); } size(125); for(int i=0;i<=3;i+=1){ draw((i,0)--(i,3)); draw((0,i)--(3,i)); } label(\"$33\\frac13\\%$\",(0,2.5),W); label(\"$50\\%$\",(0,1.5),W); label(\"$50\\%$\", (0,0.5), W); label(\"$75\\%$\", (0.5,3),N); label(\"$66\\frac23\\%$\", (1.5,3),N); label(\"$11\\frac19\\%$\", (2.5,3), N); [/asy] Apr 4, 2022 #1 +2558 +1 A. It looks suspiciously like the homework your teacher assigned. B. No image, so I can't help you, even if I wanted to... Apr 4, 2022 #2 0", "+0 # Factoring +1 244 3 +2448 Sep 5, 2018 #1 +67 0 9 = (r-2)(2r+sqrt5)(2r-sqrt5) 10 = (7k+6)(3k^2+8) 11 = (6x+7)(sqrt7x+sqrt6)(sqrt7x-sqrt6) 12 = (p+6)(p+sqrt3)((p-sqrt3) Sep 5, 2018 #2 +2448 +1 I need to know how to get each of them RainbowPanda Sep 5, 2018 #3 +101745 +2 Given a rough inspection I think they are all the same sort of factorization. So if I do the first one and you understand then you are capable of doing the other three yourself. 9) $$4r^3-8r^2-5r+10\\\\ =4r^3-8r^2\\qquad\\qquad -5r+10\\\\ \\text{now factorize the pairs}\\\\ =4r^2(r-2)\\qquad\\qquad -5(r-2)\\\\$$ now you can think of (r-2) as just one thing. So you have $$4r^2$$ lots of (r-2) and then you take away $$5$$ lots of (r-2) That will give you $$(4r^2-5)$$ lots of (r-2) which is $$(4r^2-5)(r-2)$$ you could take this one step further by treating $$4r^2-5$$ as the difference of 2 squares $$4r^2 = (2r)^2$$ and $$5=(\\sqrt5)^2$$ so $$(4r^2-5) = (2r-\\sqrt5)(2r+\\sqrt5)$$ $$(4r^2-5)(r-2)=(2r-\\sqrt5)(2r+\\sqrt5)(r-2)$$ . Sep 6, 2018 edited by Melody Sep 6, 2018", "f=(359,104), g=(303,12), h=(579,211); pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501); pair m=(282,411), n=(147,451), o=(103,437), p=(31,373); pair q=(419,175), r=(462,209), s=(477,288), t=(443,358); pair oval=(282,303); draw(l--m--n--cycle); draw(p--oval); draw(o--oval); draw(b--d--oval); draw(c--d--e--oval); draw(e--f--g--h--i--j--oval); draw(k--oval); draw(q--oval); draw(s--oval); draw(r--s--t--oval); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h); dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p); dot(q); dot(r); dot(s); dot(t); filldraw(yscale(.5)*Circle((282,606),80),white,black); label(scale(0.75)*\"Sarah\", oval);[/asy]$ $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 19 How many integers between $1000$ and $2000$ have all three of the numbers $15$, $20$, and $25$ as factors? $\\textbf{(A)}\\ 1 \\qquad \\textbf{(B)}\\ 2 \\qquad \\textbf{(C)}\\ 3 \\qquad \\textbf{(D)}\\ 4 \\qquad \\textbf{(E)}\\ 5$ ## Problem 20 What is the measure of the acute angle formed by the hands of the clock at $4:20$ PM? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 5 \\qquad \\textbf{(C)}\\ 8 \\qquad \\textbf{(D)}\\ 10 \\qquad \\textbf{(E)}\\ 12$ ## Problem 21 The area of trapezoid $ABCD$ is $164 \\text{cm}^2$. The altitude is $8 \\text{cm}$, $AB$ is $10 \\text{cm}$, and $CD$ is $17 \\text{cm}$. What is $BC$, in centimeters? $[asy]/* AMC8 2003 #21 Problem */ size(4inch,2inch); draw((0,0)--(31,0)--(16,8)--(6,8)--cycle); draw((11,8)--(11,0), linetype(\"8 4\")); draw((11,1)--(12,1)--(12,0)); label(\"A\", (0,0), SW); label(\"D\", (31,0), SE); label(\"B\", (6,8), NW); label(\"C\", (16,8), NE); label(\"10\", (3,5), W); label(\"8\", (11,4), E); label(\"17\", (22.5,5), E);[/asy]$ $\\textbf{(A)}\\ 9\\qquad\\textbf{(B)}\\ 10\\qquad\\textbf{(C)}\\ 12\\qquad\\textbf{(D)}\\ 15\\qquad\\textbf{(E)}\\ 20$ ## Problem 22 The following figures are composed of squares and circles. Which figure has a shaded region", "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", "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.", "and $EF$. They are as follows:$$AG = f(x) = -\\frac{3}{5}x + 4$$$$AC = g(x) = -\\frac{4}{5}x + 4$$$$EF = h(x) = 2x - 4$$After drawing in altitudes to $DC$ from $P$, $Q$, and $E$, we see that $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F}$ because of similar triangles, and so we only need to find the x-coordinates of $P$ and $Q$. $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair D1=(20/7,0),D2=(20/7,12/7); pair E1=(40/13,0),E2=(40/13,28/13); pair F1=(4,0),F2=(4,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(D1--D2,dashed); draw(E1--E2,dashed); draw(F1--F2,dashed); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); dot((20/7,0)); dot((40/13,0)); dot((4,0)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"P'\", (20/7,0),SSW); label(\"Q'\", (40/13,0),SSE); label(\"E'\", (4,0),S); dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot((0,0)); dot((5,4));[/asy]$ Finding the intersections of $AC$ and $EF$, and $AG$ and $EF$ gives the x-coordinates of $P$ and $Q$ to be $\\frac{20}{7}$ and $\\frac{40}{13}$. This means that $P'Q' = DQ' - DP' = \\frac{40}{13} - \\frac{20}{7} = \\frac{20}{91}$. Now we can find $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F} = \\frac{\\frac{20}{91}}{2} = \\boxed{\\textbf{(D)}~\\frac{10}{91}}$ ## Solution 3 (Similar Triangles) $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"H\",H,E); [/asy]$ Extend $AG$ to intersect $CD$ at $H$. Letting $x=\\overline{HC}$, we have that $$\\triangle{HCG}\\sim\\triangle{HDA}\\implies \\dfrac{\\overline{HC}}{\\overline{CG}}=\\dfrac{\\overline{HD}}{\\overline{AD}}\\implies \\dfrac{x}{1}=\\dfrac{x+5}{4}\\implies x=\\dfrac{5}{3}.$$ Then, notice that $\\triangle{AEQ}\\sim\\triangle{HFQ}$ and $\\triangle{AEP}\\sim\\triangle{CFP}$. Thus, we see that", "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]", "0.893); pair F = (0.862, 0.658); pair G = (0.924, 0.403); pair H = (0.747, 0.194); pair I = (0.526, 0.075); pair J = (0.251, 0.170); pair K = (0.568, 0.234); pair L = (0.262, 0.449); pair M = (0.373, 0.813); pair N = (0.731, 0.813); pair O = (0.851, 0.461); path[] f; f[0] = A--B--C--M--L--cycle; f[1] = C--D--E--N--M--cycle; f[2] = E--F--G--O--N--cycle; f[3] = G--H--I--K--O--cycle; f[4] = I--J--A--L--K--cycle; f[5] = K--L--M--N--O--cycle; draw(f[0]); axialshade(f[1], white, M, gray(0.5), (C+2*D)/3); draw(f[1]); filldraw(f[2], gray); filldraw(f[3], gray); axialshade(f[4], white, L, gray(0.7), J); draw(f[4]); draw(f[5]); [/asy]$ $\\textbf{(A) } 125 \\qquad \\textbf{(B) } 250 \\qquad \\textbf{(C) } 405 \\qquad \\textbf{(D) } 640 \\qquad \\textbf{(E) } 810$ ## Problem 20 Quadrilateral $ABCD$ satisfies $\\angle ABC = \\angle ACD = 90^{\\circ}, AC=20,$ and $CD=30.$ Diagonals $\\overline{AC}$ and $\\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$ $\\textbf{(A) } 330 \\qquad \\textbf{(B) } 340 \\qquad \\textbf{(C) } 350 \\qquad \\textbf{(D) } 360 \\qquad \\textbf{(E) } 370$ ## Problem 21 There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that$$\\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.$$What is $k?$ $\\textbf{(A) } 117 \\qquad \\textbf{(B) } 136 \\qquad \\textbf{(C) } 137 \\qquad \\textbf{(D) } 273 \\qquad \\textbf{(E) } 306$ ## Problem 22 For how many", "1. $R=\\infty$ 2. $R=0$ 3. $R=1$ 4. $R=2$ 98. Four married couples have bought $8$ seats in a concert. In how many different ways can they be seated if each couple is to sit together? 1. $24$ 2. $96$ 3. $384$ 4. None of these 99. A coin is biased so that a head is twice as likely to occur as a tail. if the coin is tossed $3$ times, then the probability of getting $2$ tails and $1$ head is 1. $\\dfrac{1}{9}$ 2. $\\dfrac{2}{9}$ 3. $\\dfrac{4}{9}$ 4. none of these 100. If $X$ represents the outcome when a die is tossed. Then the expected value of $X$ is 1. $\\dfrac{1}{2}$ 2. $\\dfrac{5}{2}$ 3. $\\dfrac{7}{2}$ 4. $\\dfrac{3}{2}$ 1-a, 2-b, 3-d, 4-c, 5-a, 6-a, 7-b, 8-b, 9-a, 10-c, 11-a, 12-c, 13-b, 14-c, 15-b, 16-d, 17-b, 18-a, 19-d, 20-c, 21-a, 22-a, 23-b, 24-a, 25-b, 26-d, 27-d, 28-d, 29-b, 30-d, 31-c, 32-a, 33-c, 34-a, 35-d, 36-b, 37-c, 38-d, 39-b, 40-b, 41-d, 42-c, 43-b, 44-c, 45-d, 46-b, 47-c, 48-c, 49-c, 50-c, 51-a, 52-a, 53-d, 54-d, 55-d, 56-c, 57-c, 58-b, 59-b, 60-a, 61-b, 62-a, 63-a, 64-d, 65-c, 66-b, 67-c, 68-b, 69-c, 70-a, 71-a, 72-a, 73-b, 74-b, 75-a, 76-b, 77-c, 78-a, 79-a, 80-a, 81-b, 82-d, 83-c, 84-b, 85-a, 86-a, 87-a, 88-c, 89-c, 90-d, 91-a, 92-b, 93-a, 94-d, 95-c, 96-c, 97-a, 98-c, 99-d, 100-d •", "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "# 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", "# Problem #2155 2155 Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed? $[asy] unitsize(1inch); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--((1/3) + 3*(1/2),0)); fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7)); draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle); draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle); draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle); draw((2,0)--(2 + (1/3) + (3/2),0)); draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle); draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle); label(\"B\",((1/6) + (1/2),(1/2)),NW); label(\"B\",((2/3) + 2 + (1/4),(29/30)),NNE); draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3)); fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7)); draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);[/asy]$ $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ \\sqrt{2}\\qquad\\textbf{(C)}\\ \\frac{3}{2}\\qquad\\textbf{(D)}\\ \\sqrt{2}+\\frac{1}{2}\\qquad\\textbf{(E)}\\ 2$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label(\"A\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"B\",(168/5, 24/5),NE); dot((24,0)); label(\"C\",(24,0),NW); dot((40, 0)); label(\"D\",(40, 0),NE); dot((24, 12)); label(\"E\",(24, 12),NE); dot((24, -12)); label(\"F\",(24, -12),SE); dot((54/5, 72/5)); label(\"G\",(54/5, 72/5),NW); dot((0, 0)); label(\"O_1\",(0, 0),S); dot((30, 0)); label(\"O_2\",(30, 0),S); [/asy]$ $r_1 = O_1A = 24$$r_2 = O_2B = 6$$AG = BO_2 = r_2 = 6$$O_1G = r_1 - r_2 = 24 - 6 = 18$$O_1O_2 = r_1 + r_2 = 30$ $\\triangle O_2BD \\sim \\triangle O_1GO_2$$\\frac{O_2D}{O_1O_2} = \\frac{BO_2}{GO_1}$$\\frac{O_2D}{30} = \\frac{6}{18}$$O_2D = 10$ $CD = O_2D + r_1 = 10 + 6 = 16$, $EF = 2EC = EA + EB = AB = GO_2 = \\sqrt{(O_1O_2)^2-(O_1G)^2} = \\sqrt{30^2-18^2} = 24$ $DEF = \\frac12 \\cdot EF \\cdot CD = \\frac12 \\cdot 24 \\cdot 16 = \\boxed{\\textbf{192}}$ ~isabelchen ## Solution 2 Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$. Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$, respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$. Since $\\triangle{APC} \\sim", "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", "\\qquad \\text{(C)}\\ 9 \\qquad \\text{(D)}\\ 10 \\qquad \\text{(E)}\\ 12$ ## Problem 11 Last summer $100$ students attended basketball camp. Of those attending, $52$ were boys and $48$ were girls. Also, $40$ students were from Jonas Middle School and $60$ were from Clay Middle School. Twenty of the girls were from Jonas Middle School. How many of the boys were from Clay Middle School? $\\text{(A)}\\ 20 \\qquad \\text{(B)}\\ 32 \\qquad \\text{(C)}\\ 40 \\qquad \\text{(D)}\\ 48 \\qquad \\text{(E)}\\ 52$ ## Problem 12 Each of the three large squares shown below is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare? $[asy] unitsize(36); fill((0,0)--(1,0)--(1,1)--cycle,gray); fill((1,1)--(1,2)--(2,2)--cycle,gray); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((1,0)--(1,2)); draw((0,0)--(2,2)); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,gray); draw((3,0)--(5,0)--(5,2)--(3,2)--cycle); draw((4,0)--(4,2)); draw((3,1)--(5,1)); fill((6,1)--(6.5,0.5)--(7,1)--(7.5,0.5)--(8,1)--(7.5,1.5)--(7,1)--(6.5,1.5)--cycle,gray); draw((6,0)--(8,0)--(8,2)--(6,2)--cycle); draw((6,0)--(8,2)); draw((6,2)--(8,0)); draw((7,0)--(6,1)--(7,2)--(8,1)--cycle); label(\"I\",(1,2),N); label(\"II\",(4,2),N); label(\"III\",(7,2),N); [/asy]$ $\\text{(A)}\\ \\text{The shaded areas in all three are equal.}$ $\\text{(B)}\\ \\text{Only the shaded areas of }I\\text{ and }II\\text{ are equal.}$ $\\text{(C)}\\ \\text{Only the shaded areas of }I\\text{ and }III\\text{ are equal.}$ $\\text{(D)}\\ \\text{Only the shaded areas of }II\\text{ and }III\\text{ are equal.}$ $\\text{(E)}\\ \\text{The shaded areas of }I, II\\text{ and }III\\text{ are all different.}$ ## Problem 13 The number halfway between $\\dfrac{1}{6}$ and $\\dfrac{1}{4}$ is $\\text{(A)}\\ \\dfrac{1}{10} \\qquad \\text{(B)}\\ \\dfrac{1}{5} \\qquad \\text{(C)}\\ \\dfrac{5}{24} \\qquad \\text{(D)}\\ \\dfrac{7}{24} \\qquad \\text{(E)}\\ \\dfrac{5}{12}$ ## Problem 14", "length of the blue segment times $\\frac{a+c}{b}$. In other words, we have that $a\\left(\\frac{a+c}{b}\\right) = 6p$. Guessing and checking Pythagorean triples reveals that $a = 84$, $b=13$, $c = 85$, and $p = a + b + c = \\boxed{182}$ satisfies this equation. $[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label(\"A\", A, SE); label(\"B\", B, NW); label(\"C_0\", C0, SW); label(\"a\", (B+C0)/2, W); label(\"b\", (A+C0)/2, S); label(\"c\", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); draw(C0--B, blue); [/asy]$ ## Solution 5 This solution proceeds from $\\frac{\\frac{ab}{c}}{1-\\frac{a}{c}} = \\frac{\\frac{ab}{c}}{\\frac{c-a}{c}} = \\frac{ab}{c-a} = 6(a+b+c)$. Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$. Thus $a = 168n^2, b = 26n^2, c = 170n^2$. Since we know all three side lengths are relatively prime, we must divide each by 2 and", "d=(134,150); pair e=(269,105), f=(359,104), g=(303,12), h=(579,211); pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501); pair m=(282,411), n=(147,451), o=(103,437), p=(31,373); pair q=(419,175), r=(462,209), s=(477,288), t=(443,358); pair oval=(282,303); draw(l--m--n--cycle); draw(p--oval); draw(o--oval); draw(b--d--oval); draw(c--d--e--oval); draw(e--f--g--h--i--j--oval); draw(k--oval); draw(q--oval); draw(s--oval); draw(r--s--t--oval); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h); dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p); dot(q); dot(r); dot(s); dot(t); filldraw(yscale(.5)*Circle((282,606),80),white,black); label(scale(0.75)*\"Sarah\", oval);[/asy]$ $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 19 How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors? $\\textbf{(A)}\\ 1 \\qquad \\textbf{(B)}\\ 2 \\qquad \\textbf{(C)}\\ 3 \\qquad \\textbf{(D)}\\ 4 \\qquad \\textbf{(E)}\\ 5$ ## Problem 20 What is the measure of the acute angle formed by the hands of the clock at 4:20 PM? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 5 \\qquad \\textbf{(C)}\\ 8 \\qquad \\textbf{(D)}\\ 10 \\qquad \\textbf{(E)}\\ 12$ ## Problem 21 The area of trapezoid $ABCD$ is $164\\text{ cm}^2$. The altitude is 8 cm, $AB$ is 10 cm, and $CD$ is 17 cm. What is $BC$, in centimeters? $[asy]/* AMC8 2003 #21 Problem */ size(4inch,2inch); draw((0,0)--(31,0)--(16,8)--(6,8)--cycle); draw((11,8)--(11,0), linetype(\"8 4\")); draw((11,1)--(12,1)--(12,0)); label(\"A\", (0,0), SW); label(\"D\", (31,0), SE); label(\"B\", (6,8), NW); label(\"C\", (16,8), NE); label(\"10\", (3,5), W); label(\"8\", (11,4), E); label(\"17\", (22.5,5), E);[/asy]$ $\\textbf{(A)}\\ 9\\qquad\\textbf{(B)}\\ 10\\qquad\\textbf{(C)}\\ 12\\qquad\\textbf{(D)}\\ 15\\qquad\\textbf{(E)}\\ 20$ ## Problem 22 The following figures are composed of squares and circles. Which figure has", "= 40 tp(\"\\\\begin{figure}[h]\") tp(\"\\\\begin{circuitikz}\") tp(\"\\\\draw (0,0) to[short,*-*] (10.5,0);\") tp(\"\\\\draw (0,4) to[short,*-,i=$I_1$] (1,4);\") tp(\"\\\\draw (1,4) to[R,l=$\"..R1..\"\\\\Omega$] (3,4);\") tp(\"\\\\draw (3,4) to[L,l=j$\"..X1..\"\\\\Omega$] (5,4);\") tp(\"\\\\draw (5,4) to[short] (5.5,4);\") tp(\"\\\\draw (5.5,4) to[L,l=j$\"..X2..\"\\\\Omega$] (7.5,4);\") tp(\"\\\\draw (7.5,4) to[R,l=$\"..R2..\"\\\\Omega$] (9.5,4);\") tp(\"\\\\draw (9.5,4) to[short,-*,i=$I_2'$] (10.5,4);\") tp(\"\\\\draw (5.25,4) to[short,*-*,,i=$I_0$] (5.25,3);\") tp(\"\\\\draw (5.25,3) to[short] (4.5,3)to [short] (4.5,2.75);\") tp(\"\\\\draw (5.25,3) to[short] (6,3) to [short] (6,2.75);\") tp(\"\\\\draw (4.5,2.75) to[R,l=$\"..RC..\"\\\\Omega$] (4.5,1.65);\") tp(\"\\\\draw (6,2.75) to[L,l=j$\"..XM..\"\\\\Omega$] (6,1.65);\") tp(\"\\\\draw (4.5,1.65) to[short] (4.5,1.4);\") tp(\"\\\\draw (6,1.65) to[short] (6,1.4) to [short] (4.5,1.4);\") tp(\"\\\\draw (5.25,1.4) to[short,*-*] (5.25,0);\") tp(\"\\\\draw (0,4) to [open,v=$V_1$] (0,0);\") tp(\"\\\\draw (10.5,4) to [open,v^=$V_2'$] (10.5,0);\") tp(\"\\\\end{circuitikz}\") tp(\"\\\\caption{One-phase transformer equivalent circuit (T-circuit)}\") tp(\"\\\\end{figure}\") } \\end{document} • finally, an example showing its output so one can see what the code is supposed to do ! Apr 3, 2014 at 1:02 • To be fair, for something as simple as text substitution, you can also use standard TeX macros. \\def\\ROne{0.1}, \\def\\RTwo{0.3}` etc. Aug 23, 2014 at 12:46 Check out those examples. Even if \"normal\" users might see no need for a general purpose language in LaTeX they will benefit from packages using Lua scripts internally. • I see \"500 - Internal Server Error\" Jul 31, 2010 at 22:50 • @Sharpie The page seems to be down. Just try it another time. – h0b0 Aug 1, 2010 at 20:32 • The new wiki for LuaTeX is at wiki.luatex.org Dec 25, 2010", "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" ]

[ "area of the four triangles is $4 \\cdot \\frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\\boxed{{\\textbf{(C)}}~15}$. ===Solution 3=== <asy> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); (Error compiling LaTeX. We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \\cdot \\frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\\boxed{{\\textbf{(C)}}~15}$. ^ 3a2b058d72a4ff6f6b7c371c8005551cd008c13e.asy: 5.9: syntax error error: could not load module '3a2b058d72a4ff6f6b7c371c8005551cd008c13e.asy') Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length $1$, and let's label the other legs $x$ for one of the triangles and $y$ for the other. Note that $x + y = 3$. The area of each of the triangles is $\\frac{x}{2}$ and $\\frac{y}{2}$, and there are $4$ of each. So now we need to find $4\\left(\\frac{x}{2}\\right) + 4\\left(\\frac{y}{2}\\right)$.", "# 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", "want is $\\frac{7.5}{24}$ of the entire hexagon. The total area of the hexagon is $\\frac{3\\sqrt{3}}{2}$, so our answer is $\\frac{7.5}{24} \\cdot \\frac{3\\sqrt{3}}{2}$ = $\\boxed{(C) \\frac{15\\sqrt{3}}{32}}$ -AlexLikeMath ## Solution 7 If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of $ABCDEF$ then apply it to the old diagram. $[asy] unitsize(1cm); draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0)); draw((2,0)--(1,1.732)); draw((5,1.732)--(4,3.464)); draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle); draw((2,0)--(2,3.464)--(5,1.732)--cycle); [/asy]$ $[asy] unitsize(1cm); draw((1,0)--(1,4),gray(.7)); draw((2,0)--(2,4),gray(.7)); draw((3,0)--(3,4),gray(.7)); draw((0,1)--(4,1),gray(.7)); draw((0,2)--(4,2),gray(.7)); draw((0,3)--(4,3),gray(.7)); draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)); draw((2,0)--(0,2)); draw((4,2)--(2,4)); draw((1,1)--(1,4)--(4,1)--cycle); draw((0,4)--(2,0)--(4,2)--cycle); fill((1,4)--(1,3.5)--(2,3)--cycle,red); fill((1,1)--(1.5,1)--(1,2)--cycle,red); fill((3,1)--(3.5,1.5)--(4,1)--cycle,red); [/asy]$ The isosceles right triangle with a leg length of $3$ in the new diagram is $\\triangle XYZ$ in the old diagram. We see that if we want to take the area of the new hexagon, we must subtract $\\frac{3}{4}$ from the area of $\\triangle XYZ$ (the red triangles), giving us $\\frac{15}{4}$. However, we need to take the ratio of this area to the area of $ABCDEF$, which is $\\frac{\\frac{15}{4}}{12}=\\frac{5}{12}$. Now we know that our answer is $\\frac{5}{12} \\cdot \\frac{3\\sqrt{3}}{2}=\\boxed{\\frac{15}{32}\\sqrt{3}}$.", "r*s $1200 = 30r + 20s$ isolate r $\\dfrac{1200 - 20s}{30} = r$ substituting that in the area equation $\\text{Area = } \\left ( \\dfrac{ 1200 - 20s}{30} \\right ) s$ $\\text{Area = } 40s - \\dfrac{ 2s^2}{3}$ take the derivitive and determine the value of s when the slope is zero. $0 = 40 - \\dfrac{ 4s }{3}$ solve that to get the value of s. plug that back into the equation above to compute r. How that helps .", "R3 + (R1 + R2 + R3) R4), I3 -> (R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), I4 -> (R3 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), V1 -> ((R3 R4 + R2 (R3 + R4)) Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), V2 -> (R3 R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4)}} In[3]:=R2 = s*L; R3 = 1/(s*c); Vi = u/s; L = 10*10^(-3); c = 1*10^(-6); u = 1; R1 = 10; R4 = 10; FullSimp FullSimplify[ InverseLaplaceTransform[(R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4), s, t]] Out[3]=(E^(-500 (101 + Sqrt[9401]) t) (-1 + E^( 1000 Sqrt[9401] t)))/(10 Sqrt[9401]) In[4]:=FullSimplify[Solve[{D[%3, t] == 0, t > 0}, t]] Out[4]=ArcCoth[101/Sqrt[9401]]/(500 Sqrt[9401]) In[5]:=FullSimplify[Limit[%3, t ->ArcCoth[101/Sqrt[9401]]/(500 Sqrt[9401])]] Out[5]=((101 + Sqrt[9401]) E^(-(1 + 101/Sqrt[9401]) ArcCoth[101/Sqrt[ 9401]]))/4000 In[6]:=Plot[%3, {t, 0,1/1000}] Out[6]=(*this code will plot the picture above*) • Thank you so so much, I will study your answer in more detail. Dec 20 '20 at 14:12 • @Jan you have done this before - you should not hand homework questions on a plate to anyone - that isn't what this site is about. Dec 22 '20 at 15:18", "the regions 0.35 by shading 35 boxes out of 100. Striking off 9 boxes out of 35 shaded boxes to subtract 0.09 from 0.35. The left over shaded boxes represent the required value. ∴ 0.35 – 0.09 = 0.26 Try These (Text book Page No. 7) Question 1. Using the area models solve the following (i) 1.2 + 3.5 (ii) 3.5 – 2.3 Solution: (i) 1.2 + 3.5 Here 1.2 is represented in blue colour and 3.5 is represented in Green colour. Sum of 1.2 and 3.5 is 4.7. (ii) 3.5 – 2.3 Representing 3.5 using 3 squares and 5 rectangular strips. Crossing out 2 squares from 3 squares and 3 rectangular strips from 5 to get the difference. So 3.5 – 2.3 = 1.2. 5/9 as a decimal is 0.56 Try These (Text book Page No. 9) Question 1. Complete the magic square in such a way that rows, columns and diagonals give the same sum 1.5. Solution: Exercise 1.3 Think (Text book Page No. 13) Question 1. How are the products 2.1 × 3.2 and 21 × 32 alike? How are they different. Solution: 2.1 × 3.2 = 6.72 and 21 × 32 = 672. In both the cases the digits ambers are the same. But the place value differs. Try These (Text book Page No.", "the grids is the product. • For example: We want to multiply $$0.7$$ by $$0.5$$ by using hundreds grids. • Since $$0.7=0.70$$, shade $$70$$ squares vertically. • Since $$0.5=0.50$$, shade $$50$$ squares horizontally. • Combine the shaded squares in the same grid. Shaded area is $$0.35$$ or $$35$$ hundredths. • The common shaded area created by the overlapping of the grids is the product. Here, the yellow shaded squares represent the product of $$0.7$$ and $$0.5$$ Thus, $$0.7×0.5=0.35$$ • We can also multiply large decimal numbers using the area model. • For example: Multiplying $$2.7$$ by $$3.2$$ by using hundreds grid. • Representing one decimal number vertically and other horizontally. Since, $$2.7=2.70=270$$ hundredths $$=(100+100+70)$$ So, shade two whole hundred grids and $$70$$ squares out of $$100$$. • Since, $$3.2=3.20=320$$ hundredths $$=(100+100+100+20)$$ So, shade three whole hundred grids and $$20$$ squares out of $$100$$. • Now, combine both the rectangles. • Form a larger grid and shade the squares to show the complete overlapping. • From the figure, calculate the areas of all the rectangles of different colors. • Now, add all the areas. $$6+2.1+0.4+0.14=8.64$$ • Hence, the product of $$2.7$$ and $$3.2$$ is $$8.64$$ Which area model represents the product of $$0.6$$ and $$0.4$$? A B C D × To represent the product of $$0.6$$ and $$0.4$$, first", "⋅ 797 ⋅ 809 ⋅ 811 ⋅ 821 ⋅ 823 ⋅ 827 ⋅ 829 ⋅ 839 ⋅ 853 ⋅ 857 ⋅ 859 ⋅ 863 ⋅ 877 ⋅ 881 ⋅ 883 ⋅ 887 ⋅ 907 ⋅ 911 ⋅ 919 ⋅ 929 ⋅ 937 ⋅ 941 ⋅ 947 ⋅ 953 ⋅ 967 ⋅ 971 ⋅ 977 ⋅ 983 ⋅ 991 ⋅ 997 julia> prodfactors(ans) 7128865274665093053166384155714272920668358861885893040452001991154324087581111499476444151913871586911717817019575256512980264067621009251465871004305131072686268143200196609974862745937188343705015434452523739745298963145674982128236956232823794011068809262317708861979540791247754558049326475737829923352751796735248042463638051137034331214781746850878453485678021888075373249921995672056932029099390891687487672697950931603520000 Now we get super special number that is 443 digits long. Lets finish by a Julia one liner to get the number of digits julia> getAllFactors(big\"1000\") |> prodfactors|> digits |>length 433 01/2/18 # Visualizing the Inscribed Circle and Square Puzzle Recently, I watched a cool mind your decsions video on an inscribed circle and rectangle puzzle. In the video they showed a diagram that was not scale. I wanted to get a sense of how these differently shaped areas will match. There was a cool ratio between the outer and inner circle radii that is expressed as $\\frac{R}{r}=\\sqrt{\\frac{\\pi-2}{4-\\pi}}$. I used Compose.jl to rapidly do that. using Compose set_default_graphic_size(20cm, 20cm) ϕ=sqrt((pi -2)/(4-pi)) R=10 r=R/ϕ ctx=context(units=UnitBox(-10, -10, 20, 20)) composition = compose(ctx, (ctx, rectangle(-r/√2,-r/√2,r*√2,r*√2),fill(\"white\")), (ctx,circle(0,0,r),fill(\"blue\")), (ctx,circle(0,0,R),fill(\"white\")), (ctx,rectangle(-10,-10,20,20),fill(\"red\"))) composition |> SVG(\"inscribed.svg\") 08/6/17 # Solving the code lock riddle with Julia I came across a neat math puzzle involving counting the number of unique combinations in a hypothetical lock", "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]", "gives $120s = 96s + 100d \\Rightarrow 24s = 100d \\Rightarrow \\frac{d}{s} = \\frac{24}{100} = \\boxed{\\textbf{(A) }\\frac{6}{25}}$. ## Solution 2 Without loss of generality, we may let $s=1$ (since $d$ will be determined by the scale of $s$, and we are only interested in the ratio $\\frac{d}{s}$). Then, as the total area of the $576$ gray tiles is simply $576$, the large square has area $\\frac{576}{0.64} = 900$, making the side of the large square $\\sqrt{900}=30$. As in Solution 1, the the side length of the large square consists of the total length of the gray tiles and $25$ lots of the border, so the length of the border is $d = \\frac{30-24}{25} = \\frac{6}{25}$. Since $\\frac{d}{s}=d$ if $s=1$, the answer is $\\boxed{\\textbf{(A) }\\frac{6}{25}}$. ## Solution 3 (using answer choices) As in Solution 2, we let $s = 1$ without loss of generality. For sufficiently large $n$, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: $[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); for(int i = 1; i <= 13; i += 4){ draw((1,i)--(13,i), red); draw((i,1)--(i,13), red); } [/asy]$ Each red square has side length $(1+d)$, so by solving $\\frac{1^2}{(1+d)^2}", "+0 # Factoring +1 244 3 +2448 Sep 5, 2018 #1 +67 0 9 = (r-2)(2r+sqrt5)(2r-sqrt5) 10 = (7k+6)(3k^2+8) 11 = (6x+7)(sqrt7x+sqrt6)(sqrt7x-sqrt6) 12 = (p+6)(p+sqrt3)((p-sqrt3) Sep 5, 2018 #2 +2448 +1 I need to know how to get each of them RainbowPanda Sep 5, 2018 #3 +101745 +2 Given a rough inspection I think they are all the same sort of factorization. So if I do the first one and you understand then you are capable of doing the other three yourself. 9) $$4r^3-8r^2-5r+10\\\\ =4r^3-8r^2\\qquad\\qquad -5r+10\\\\ \\text{now factorize the pairs}\\\\ =4r^2(r-2)\\qquad\\qquad -5(r-2)\\\\$$ now you can think of (r-2) as just one thing. So you have $$4r^2$$ lots of (r-2) and then you take away $$5$$ lots of (r-2) That will give you $$(4r^2-5)$$ lots of (r-2) which is $$(4r^2-5)(r-2)$$ you could take this one step further by treating $$4r^2-5$$ as the difference of 2 squares $$4r^2 = (2r)^2$$ and $$5=(\\sqrt5)^2$$ so $$(4r^2-5) = (2r-\\sqrt5)(2r+\\sqrt5)$$ $$(4r^2-5)(r-2)=(2r-\\sqrt5)(2r+\\sqrt5)(r-2)$$ . Sep 6, 2018 edited by Melody Sep 6, 2018", "the ratio of the gray area to the area of the large square is $\\frac{64}{100}=\\frac{16}{25}$. Thus, our equation becomes $\\frac{576s^2}{(24s+25d)^2}=\\frac{16}{25}$. Square rooting both sides, we get $\\frac{24s}{24s+25d}=\\frac{4}{5}$. Cross multiplying, we get $120s=96s+100d$. Combining like terms, we get $24s=100d$, which implies that $\\frac{d}{s}=\\frac{24}{100}=\\frac{6}{25}\\implies\\boxed{\\textbf{(A) }\\frac{6}{25}}$. ~junaidmansuri ## Solution 3 The area of the shaded region is $(24s)^2$. The total area of the square is $(24s+25d)^2$ because for each side of the square there one extra row/column of the border. Our equation is $\\frac{(24s)^2}{(24x+25d)^2}=\\frac{64}{100}$. Taking the square root of both sides gives $\\frac{24x}{24x+25d}=\\frac 45$. Cross multiplying and rearranging gives $\\frac ds=\\textbf{(A)} \\frac{6}{25}$. -franzliszt ## Solution 4 WLOG $s = 1$. For large enough $n$, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: $[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); for(int i = 1; i <= 13; i += 4){ draw((1,i)--(13,i), red); draw((i,1)--(i,13), red); } [/asy]$ Each red square has side length $1+d$, so by solving $\\frac{1^2}{(1+d)^2} = \\frac{64}{100} \\iff \\frac{1}{1+d} = \\frac{4}{5}$, we obtain $d = \\frac{1}{4}$. The actual fraction of area covered by the gray tiles is slightly less than $\\frac{1}{(1+d)^2}$, which implies $\\frac{1}{(1+d)^2} > \\frac{64}{100} \\iff \\frac{1}{1+d} > \\frac{4}{5} \\iff", "shaded region and three identical circular sectors, each of which is a 60-degree sector from its circle. Using this information, we can create the following equation: (Area of equilateral triangle) - (3 x area of 1/6 of each circle) = area of shaded region (2r)^2√3/4 - 3(1/6 x π r^2) = 64√3 − 32π [(4r^2)√3]/4 - (πr^2)/2 = 64√3 − 32π (r^2)√3 - (πr^2)/2 = 64√3 − 32π Multiplying both sides by 2, we have: 2(r^2)√3 - πr^2 = 128√3 − 64π r^2(2√3 - π) = 128√3 − 64π Dividing both sides by (2√3 - π), we have: r^2 = 64 r = 8 hi generis its been raining here... so after the rain i feel like solving algebra regarding this \"$$r^2(2√3 - π) = 128√3 − 64π$$ divide both sides by (2√3 - π) \" $$r^2= \\frac{128√3 − 64π}{(2√3 - π)}$$ after this step i am getting confused (when i dvide numerator by denominator) so this is what i get in numerator first i smplify numerator 128√3 − 64π ( i subtract 128 - 64 = 64) ---> $$r^2=\\frac{64√3-π}{2√3 - π}$$ now when i divide numerator by denominator i divide 64 by 2 $$r^2= 32$$ and get this why have a great sunny weekend La France va gagner la coupe du monde 3:1 Hi dave13 - et", "# The ratio of the areas of two squares is $\\frac{192}{80}$. What is the value of the sum $a+b+c$? The ratio of the areas of two squares is $\\frac{192}{80}$. After rationalizing the denominator, the ratio of their side lengths can be expressed in the simplified form $\\frac{a\\sqrt{b}}{c}$ where $a$, $b$, and $c$ are integers. What is the value of the sum $a+b+c$? I have no idea how to approach this problem. Ant hints would be really appreciated. ## 1 Answer Call $x$ and $y$ the sides of the two squares If $\\dfrac{x^2}{y^2}=\\dfrac{192}{80}=\\dfrac{12}{5}$ then $\\dfrac{x}{y}=\\dfrac{\\sqrt{12}}{\\sqrt{5}}=\\dfrac{\\sqrt{12}\\sqrt{5}}{\\sqrt{5}\\sqrt{5}}=\\dfrac{\\sqrt{60}}{5}=\\dfrac{\\sqrt{4\\cdot 15}}{5}=\\dfrac{2\\sqrt{15}}{5}$ Therefore $a+b+c=22$ Hope it is useful • $\\dfrac{2\\sqrt{15}}{5} = \\dfrac{4\\sqrt{15}}{10}$. There must be another condition, such as $\\gcd(a,c)=1$ etc. – Math Lover Oct 21 '17 at 18:15 • @MathLover I thought the same but the OP text says \"simplified form\"... – Raffaele Oct 21 '17 at 18:46", "• # question_answer In the given fig. what is the ratio of the areas of (a) shaded portion I to shaded portion II? (b) shaded portion II to shaded portion III? (c) shaded portions I and II taken together and shaded portion III? By splitting the above figure, we get, (a) Now, $~AD=5$ $AB=AEBE=105$ $AB=5$ Area of shaded portion I = Area of ABCD $\\therefore$ Area of $ABCD=AD\\times AB=5\\times 5=25$ So, Area of shaded portion $I=25$ Area of shaded portion II = Area of DCIJ ? Area FHI Now, $DJ=AJAD=105=5$ $DC=AB$(as It?s a square) $DC=5$ of $DCIJ=DJ\\times DC=5\\times 5=25$ Now, $AJ=EH$ (side of a square) $EH=10$ So, $FH=EHEF=107$ $FH=3$ and $GF=BE$ $GF=5$ area $GFHI=GF\\times FH$ $=5\\times 3=15$ Area o shaded portion $II=25+15=40$ So, required ratio of shaded portion I to II is $25:40$ $=\\frac{25}{40}=\\frac{5\\times 5}{8\\times 5}$ $=\\frac{5}{8}=5:8$ (b) Area of shaded portion III = Area of BEFG $=BE\\times EF$ $=5\\times 7$ $=35$ So, required ratio of shaded portion II to III is $40:35$ $=\\frac{40}{35}=\\frac{8\\times 5}{7\\times 5}=\\frac{8}{7}$ $=8:7$ (c) Area of shaded portions I and II = 25 + 40 = 65 So, required ratio $=65:35$ $=\\frac{65}{35}=\\frac{13\\times 5}{7\\times 5}=\\frac{13}{7}$ $=13:7$ You need to login to perform this action. You will be redirected in 3 sec", "ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area, in sq. cm, of the overlapping region between the two circles is 1. $16(\\pi-1)$ 2. $32(\\pi$ 3. $16\\pi$ 4. $32(\\pi-1)$ Option: 4 Explanation: Area of circle $C 1=\\pi(4 \\sqrt{2})^{2}=32 \\pi$ option 4 is the correct answer $32(\\pi-1)$ If the options are not there, then overlapping area $=\\cfrac{1}{2}($ Area of circle C1 $)+$ Area of the minor sector made by $B C$ in $C 2=16 \\pi+\\cfrac{1}{4} 64 \\pi-\\cfrac{1}{2}(8)^{2} \\sin 90^{\\circ}=32(\\pi-1)$ Question 19: Let $r$ be a real number and $f(x)=\\left\\{\\begin{array}{cl}2 x-r & \\text { if } x \\geq r \\\\ r & \\text { if } x < r\\end{array}\\right.$. Then, the equation $f(x)=f(f(x))$ holds for all real values of $x$ where 1. $x \\neq r$ 2. $x>r$ 3. $x \\geq r$ 4. $x \\leq r$ Option: 4 Explanation: When $x<r$, $f(x)=r$ $f(x)=f(f(x))$ $r=f(r)$, $r=2 r-r$, $r=r$ When $x=r$, $f(x)=2 x-r=2 r-r=r$, $f(x)=f(f(x))$, $r=f(r)$, $r=2 r-r$, $r=r$ Therefore $x \\leq r$ Question 20: A group of N people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished", "function 'operator -(pair, <overloaded>)' draw(P--(P+R-D)--R,black); ^ e3e47a918c489e98fe6f1264b4357c0d6143ea05.asy: 9.14: use of variable 'D' is ambiguous) ## Problem 4 Let $A = \\left{ 2,5,10,17,\\cdots,n^2+1,\\cdots\\right}$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) be the set of all positive squares plus $1$ and $B = \\left{101, 104, 109, 116,\\cdots,m^2 + 100,\\cdots\\right}$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) be the set of all positive squares plus $100$. (a) What is the smallest number in both $A$ and $B$? (b) Find all numbers that are in both $A$ and $B$. ## Problem 5 Determine the area of the square $ABCD$, with the given lengths along a zigzag line connecting $B$ and $D$. pair A=(0,4*sqrt(10)),B=(4*sqrt(10),4*sqrt(10)),C=(4*sqrt(10),0),D=(0,0); pair F=(sqrt(10),3*sqrt(10)),E=((9/5)*sqrt(10),(3/5)*sqrt(10)); draw(A--B--C--D--cycle,black); draw(D--E--F--B,black); pair P=.1*unit(D-E)+E,R=.1*unit(F-E)+E; draw(P--(P+R-E)--R,black); P=.1*unit(B-F)+F,R=.1*unit(E-F)+F; draw(P--(P+R-F)--R,black); MP(\"A\",A,NW);MP(\"B\",B,NE);MP(\"C\",C,SE);MP(\"D\",D,SW); MP(\"6\",(D/2+E/2),NW);MP(\"8\",(E/2+F/2),NE);MP(\"10\",(F/2+B/2),S); (Error compiling LaTeX. P=.1*unit(B-F)+F,R=.1*unit(E-F)+F; ^ 8d1eaecbb1462f3788079defda2a11d8a43b97d5.asy: 11.17: syntax error error: could not load module '8d1eaecbb1462f3788079defda2a11d8a43b97d5.asy') ## Problem 6 What is the remainder when $1! + 2! + 3! + ?+ 2011!$ is divided by $18$? ## Problem 7 What is the $\\underline{sum}$ of the first $999$ terms of the sequence $1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63,\\cdots$ that appeared on the First Round? Recall that a term in an even numbered position is twice the previous term, while a term in an odd numbered position is one more that the previous term.", "of larger circles are given by A1 = Pi r2 A2 = Pi R2 = Pi (3r)2 = 9 Pi r2 ratio R of areas larger / smaller is equal to R = 9 Pi r2 / Pi r2 = 9 ## Solution to Question 3 Let S1 be the total salary of the group of 20 employess. Hence 35,000 = S1 / 20 S1 = 20 * 35,000 = $700,000 Let S2 be the total salary of the group of 30 employess. Hence 40,000 = S2 / 30 S2 =$1,200,000 The average of all 50 employers is given by (700,000 + 1,200,000) / 50 = \\$38,000 ## Solution to Question 4 Rewrite the given expression using the fact that 48 = 3 × 16 √48 = √(3 × 16) Use the formula √(a × b) = √a × √a to rewite √(3 × 16) as √48 = √(3 × 16) = √3 × √16 = 4 √3 ## Solution to Question 5 Since the three angles are in the ration 2:4:3, their sizes they may be written in the form Size of A = 2 k , Size of B = 4 k and size of C = 3 k , where k is a constant. The sum of the angles of a ny triangle is equal", "to a horizontal or vertical line that divides the square in two equal pieces and shade one) . B $$\\large \\dfrac{3}{4}$$Hint: Be sure you're taking into account the different sizes of the pieces. C $$\\large \\dfrac{2}{3}$$Hint: The bottom half of the picture is 2/3 shaded, and the top half is more than 2/3 shaded, so this answer is too small. D $$\\large \\dfrac{17}{6}$$Hint: This answer is bigger than 1, so doesn't make any sense. Be sure you are using the whole picture, not one quadrant, as the unit. Question 15 Explanation: Topic: Models of Fractions (Objective 0017) Question 16 #### Which of the lists below is in order from least to greatest value? A $$\\large \\dfrac{1}{2},\\quad \\dfrac{1}{3},\\quad \\dfrac{1}{4},\\quad \\dfrac{1}{5}$$Hint: This is ordered from greatest to least. B $$\\large \\dfrac{1}{3},\\quad \\dfrac{2}{7},\\quad \\dfrac{3}{8},\\quad \\dfrac{4}{11}$$Hint: 1/3 = 2/6 is bigger than 2/7. C $$\\large \\dfrac{1}{4},\\quad \\dfrac{2}{5},\\quad \\dfrac{2}{3},\\quad \\dfrac{4}{5}$$Hint: One way to look at this: 1/4 and 2/5 are both less than 1/2, and 2/3 and 4/5 are both greater than 1/2. 1/4 is 25% and 2/5 is 40%, so 2/5 is greater. The distance from 2/3 to 1 is 1/3 and from 4/5 to 1 is 1/5, and 1/5 is less than 1/3, so 4/5 is bigger. D $$\\large \\dfrac{7}{8},\\quad \\dfrac{6}{7},\\quad \\dfrac{5}{6},\\quad \\dfrac{4}{5}$$Hint: This is in order from greatest to least.", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$" ]

[ "+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 # 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 # Fractions 0 82 1 Express as a fraction in lowest terms: $0.\\overline{12} + 0.\\overline{001}$ May 25, 2022", "+0 # Pls help 0 92 1 Compute:$$0.\\overline{7}-0.\\overline{4}+0.\\overline{2}$$ . Express your answer as a common fraction. Sep 22, 2020", "# Express it in fraction Express $$0.461538461538...$$ as a fraction. (In other words, 461538 repeats forever which is called 0 point 461538 recurring.) ×", "+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", "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.", "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", "# Convert recurring decimals to fractions ## Interactive practice questions Starting with $x=0.\\overline{1}$x=0.1, express $x$x as a fraction in simplest form. Easy Approx a minute Starting with $x=2.\\overline{3}$x=2.3, express $x$x as a fraction in simplest form. Starting with $x=1.\\overline{7}$x=1.7, express $x$x as a fraction in simplest form. Starting with $x=3.\\overline{6}$x=3.6, express $x$x as a fraction in simplest form.", "+0 0 582 2 +46 What is $0.4\\overline8 + 0.\\overline{37}$? Express your answer as a common fraction in lowest terms. Sep 5, 2020 #1 +35059 +1 $$0.4\\overline8 + 0.\\overline{37}?$$ .488888.... = 44/90 .373737.... = 37/99 I will let you take it from here.... find the common denominator and convert the fractions, add them then simplify Sep 5, 2020 #2 0 22/45+37/99 Guest Sep 7, 2020", "+0 # 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", "Math Help - Fraction 1. Fraction Express as a simple fraction: [(1/2(x+h))-(2/3x)] / h 2. Is this what you're trying to show? $[1/2(x+h) - 2/3x]/h$ 3. Yeah, now I need to figure out where to start though.", "downloads about What Is Rewrite 4 As A Fraction In Simplest Form. Find answers researching ebooks, papers or essays. ... Express as a fraction in simplest form: a) 0.806 b) 0.0256 c) 11.375 6. Rewrite as a decimal: a) 9 16 b) 17 40 c) 52 464 25i d) 7 12 e) 3 7 f) 24 11 ...", "Express $\\dfrac{14}{-5}$ as a rational number with numerator $56$. • A $\\dfrac{56}{-20}$ • B $\\dfrac{56}{-5}$ • C $\\dfrac{56}{-10}$ • D $\\dfrac{56}{-15}$", "\"common fraction\" • fraction", "# 1. Express the following numbers in standard form. (iv) 0.00000000837 The standard form of the given number is $8.37\\times10^{-9}$", "so i can under. please. • How can you express 18.7? • If you want it as a fraction it would be 18 7/10 • how do u change 5.03 as a fraction please? • They should fix the last question because it says express 5.03 as a fraction and some people will think oh that's easy it's 503/100 but it says it's wrong and the right answer is 5 and 3/100 which is right", "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 # 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" ]

[ "# Question #0426c Jan 5, 2017 $0.8 \\overline{7} = \\frac{79}{90}$ #### Explanation: Let $x = 0.8 \\overline{7}$ $\\implies 10 x = 8. \\overline{7}$ $\\implies 10 x - x = 8. \\overline{7} - 0.8 \\overline{7}$ $\\implies 9 x = 8.7 - 0.8$ $\\implies 9 x = 7.9$ $\\implies 9 x = \\frac{79}{10}$ $\\implies x = \\frac{79}{10} \\cdot \\frac{1}{9}$ $\\therefore x = \\frac{79}{90}$ This technique of multiplying by $10$ (or a power of $10$) and subtracting can be used to find the fractional representation of any number with infinitely repeating decimal digits. This answer contains a detailed explanation.", "# Express 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 $\\frac{4}{9}$.", "you have to multiply your number by 10³. Now you have $\\boxed{\\begin{array}{lcr}1000x&=&2345.345345345... \\\\ x&=&2.345345345...\\end{array}}$ Subtract and you have $999x = 2343~\\iff~x = \\frac{2343}{999}$", "the value of your number is equal to $$x=0.\\overline1 -\\sum_{n=1}^\\infty 10^{1-\\frac{(n+1)(n+2)}{2}}$$ Since $0.\\overline1=\\frac{1}{9}$, then we have $$x=\\frac{1}{9}-\\sum_{n=1}^\\infty 10^{1-\\frac{(n+1)(n+2)}{2}}$$ Which can be simplified to $$x=\\frac{1}{9}-\\sum_{n=1}^\\infty (\\sqrt{10})^{-n^2-3n}$$ As far as I know, this is as much as it can be simplified without bringing in other non-elementary functions. However, you might be able to use this function to simplify it further.", "A think and explanation of long division might be more involved. Or might not$\\ldots$ Later edit: $1352/43$. There's a quotient and a remainder: $1352 = 31\\cdot 43 + 19$. The quotient is $31$ and the remainder is $19$. So $\\dfrac{1352}{43} = 31 + \\dfrac{19}{43}$. The remainder must be less than $43$ so the fraction must be less than $1$. So if we move the decimal point over to get a one-digit quotient, we have $$\\frac{135.2}{43} = 3.1 +\\text{less than 0.1} = 3 + \\text{less than 1}$$ Since $43$ goes $3+\\text{fraction}$ times into $135.2$, and $43\\cdot3$ must be an integer, we have $43$ goes $3+\\text{some fraction}$ times into $135$. In particular, $$135 = 43\\cdot3 + 6.\\quad \\text{The quotient is 3 and the remainder is 6.}$$ Now look at the usual algorithm: $$\\begin{array}{cccccccccc} & & & & 3 \\\\ \\hline 43 & ) & 1 & 3 & 5 & 2 \\\\ & & 1 & 2 & 9 \\\\ \\hline & & & & 6 \\end{array}$$ Why should we now bring down the $2$ and then divide the resulting $62$ by $43$? $$\\frac{135.2}{43} = \\frac{135}{43} + \\frac{0.2}{43} = 3 + \\underbrace{\\frac{6}{43} + \\frac{0.2}{43}} = 3 + \\frac{6.2}{43}.$$ So $$\\frac{1352}{43} = 30 + \\frac{62}{43}.$$ So that's why. -", "# 0.999... $0.999\\ldots$ (or $0.\\overline{9}$) is an equivalent representation of the real number $1$. It is often mistaken that $0.999\\ldots \\neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\\frac 1{\\infty}$ term left over at the end, etc.). ## Proofs ### Fractions Since $\\frac 13 = 0.\\overline{3} = 0.333\\ldots$, multiplying both sides by $3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "# How do you express .33 as a fraction in simplest form? Sep 14, 2016 $\\frac{33}{100}$ Sep 14, 2016 As you wrote the question $\\frac{33}{100}$ If you meant $0.333333 . . \\to 0.33 \\overline{3} \\text{ }$ then $\\text{ } \\frac{1}{3}$ #### Explanation: It all depends on what you mean by $0.33$ If you meant it to be exactly as written then we have $\\frac{3}{10} + \\frac{3}{100}$ Write as $\\left(\\frac{3}{10} \\times 1\\right) + \\frac{3}{100}$ But 1 may be written as $\\frac{10}{10}$ $\\left(\\frac{3}{10} \\times \\frac{10}{10}\\right) + \\frac{3}{100}$ $\\frac{30}{100} + \\frac{3}{100} = \\frac{33}{100}$ $\\textcolor{red}{\\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$ If you meant what you wrote to mean $0.33 \\leftarrow$ 3 going on for ever $0.333333333 \\ldots . .$ written as $0.33 \\overline{3}$ $\\textcolor{b r o w n}{\\text{Then this is very different:}}$ Let $x = 0.33 \\overline{3}$ thus $10 x = 3.33 \\overline{3}$ $10 x - x = 3$ $9 x = 3$ $x = \\frac{3}{9} = \\frac{1}{3}$", "# 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 $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "the denominator by $37$ to get: $\\begin{array}{rcl} \\frac{4}{27}&=&\\frac{4\\times 37}{27\\times 37}\\\\ &=&\\frac{148}{999}\\\\ &=&0.\\overline{148} \\end{array}$ $~$ To transform repeating decimals into fractions: • write all repeating numbers in the numerator • in the denominator, place as many $9$s as there are repeating digits • reduce the fraction by dividing the numerator and denominator by the GCF $~$ For the repeating decimal $0.\\overline{81}$: $\\begin{array}{rcl} 0.\\overline{81}&=&\\frac{81}{99}\\\\ &=&\\frac{81\\div9}{99\\div 9}\\\\ &=&\\frac{9}{11} \\end{array}$ $~$ For the repeating decimal $0.\\overline{63}$: $\\begin{array}{rcl} 0.\\overline{63}&=&\\frac{63}{99}\\\\ &=&\\frac{63\\div9}{99\\div 9}\\\\ &=&\\frac{7}{11} \\end{array}$", "of simple fractions as decimals: $\\frac 12 = 0.5$ $\\frac 13 = 0.333333\\ldots = 0.\\overline{3}$ $\\frac 23 = 0.666666\\ldots = 0.\\overline{6}$ $\\frac 14 = 0.25$ $\\frac 34 = 0.75$ $\\frac 15 = 0.2$ $\\frac 16 = 0.1666666\\ldots = 0.1\\overline{6}$ $\\frac 17 = 0.14285714285714\\ldots = 0.\\overline{142857}$ $\\frac 18=0.125$ $\\frac 38 = 0.375$ $\\frac 78 = 0.875$ $\\frac 19 = 0.111111\\ldots = 0.\\overline{1}$ $\\frac 29 = 0.222222\\ldots = 0.\\overline{2}$ $\\frac 1{10} = 0.1$ $\\frac 1{11} = 0.09090909\\ldots = 0.\\overline{09}$ $\\frac 1{12} = 0.08333333\\ldots = 0.08\\overline{3}$ A few things can be observed: • Some decimals are short and to the point. • Others have an infinite number of digits (students like to say: \"they are infinite\"), in a repeating pattern. • There can be a single digit that repeats, or a string of two or more digits repeating. • The repeating pattern (period) does not always start right after the decimal point. The long period of $\\frac 17$ sticks out like a sore thumb. It has a few crazy cousins further along: $\\frac 1{13} = 0.\\overline{076923}$ $\\frac 1{17} = 0.\\overline{0588235294117647}$ $\\frac 1{19} = 0.\\overline{052631578947368421}$ There are also the weird $\\frac 1{81} = 0.\\overline{012345679}$ (sic!) and the mysterious $$\\frac 1{9801} = 0.0001020304050607080910111213\\ldots$$ It is also worthwhile to notice this pattern (for later): $$\\frac 19 = 0.\\overline{1}$$ $$\\frac 1{99} = 0.\\overline{01}$$ $$\\frac 1{999}", "once. Otherwise, you're adding rather than subtracting. • He effectively subtracted twice, but psychologically he had a different problem (or he was just blindly performing the algorithm). – PJTraill Oct 1 '16 at 10:55 The standard subtraction algorithm only works when the larger-magnitude number is on top (in this case, the $8.2$). The most straightforward calculation is to note that the result here will be obviously negative (because the number being subtracted is of larger magnitude), and then perform $8.2 - 4.5$ for the magnitude of the result. Failing to pay attention to which magnitude is larger results in exactly this kind of subtraction error. Expanding on @John's answer https://math.stackexchange.com/a/1948174/169560. You can borrow from \"nothing\" you just need to remember to give it back: $$\\frac{ \\begin{array}[b]{r} 4.5 \\\\ - 8.2 \\end{array} }{}= \\frac{ \\begin{array}[b]{r} \\ ^{-1}14.5 \\\\ - 8.2 \\end{array} }{}= \\frac{ \\begin{array}[b]{r} \\ ^{-1}14.5 \\\\ - 8.2 \\end{array} }{-10+6.3}=3.7$$ This way, if the minuend is smaller than the subtrahend, you will end up with something like $-100 \\ldots 000 + YY \\ldots YY.yy \\ldots yy$ where $x$, $Y$ and $y$ stand for any digits, and the number of zeroes is the same as the number of $Y$s. It should be easy to do this last step, as you just need to pad to a potence of 10", "\\frac{1}{2}$ And getting your own gift back is: $Pr[{C_i}=1] = \\frac{1}{2}$ So our expected value of C_i is: $E[{C_i}] = 0 * Pr[{C_i =0}] + 1 * Pr[{C_i =0}] = 0 * 1 * \\frac{1}{2} = \\frac{1}{2}$ And finally, the expected value of C is... $E[C] = E[ (\\begin{array}{cc}n\\\\ \\sum \\\\ i=1\\end{array}) {C_i} ]$ $E[C] = \\begin{array}{cc}n\\\\ \\sum \\\\ i=1\\end{array} E[{C_i}]$ $E[C] = \\begin{array}{cc}n\\\\ \\sum \\\\ i=1\\end{array} \\frac{1}{2} = \\frac{n}{2}$ (I hope you like my latex, I really tried to learn that for this post. ) So is this right? Any advice? Thanks! 2. I'm not sure what you want me to check. These 0,1 rvs are called Bernoullis and their sum is a Binomial. You have a few typos. C runs from 1 to n. You accidentally are starting at 0. O,1,...,n are n+1 terms. But n/2 is correct. You want 1,...,n in your sums. Also that $2^n$ is just weird. It can't be a lower value since it's greater than n, your upper limit. You want 1,..,n in your series, all of them. 3. Yeah I was working out the problem a different way and then tried to change it and... the 2^n is from my original solution which I don't think works, sorry about that! (So, it should be n=1 instead of the random 2^n", "If $x$ is a complex number such the $x^2+x+1=0$, then the numerical worth of $(x+\\frac1x)^2+(x^2+\\frac1x^2)^2+(x^3+\\frac1x^3)^2+\\ldots+(x^27+\\frac1x^27)^2$ is same to? A) 52 . B) 56 . C) 54. D)58 . E)None the these Where is this concern from? I\"m pretty sure it originates from one high institution centregalilee.com contest, go anyone one know which centregalilee.com contest and of course i can\"t solve, I have an answer key but ns don\"t understand the solution. You are watching: Find the numerical value of the expression Multiply both political parties by $x-1\\ne 0$:$$x^2+x+1=0 \\Rightarrow (x-1)(x^2+x+1)=0 \\Rightarrow x^3-1=0 \\Rightarrow x^3=1\\Rightarrow \\\\x^54=1;x^-54=1 \\quad (1)$$Expand:$$(x+\\frac1x)^2+(x^2+\\frac1x^2)^2+(x^3+\\frac1x^3)^2+\\ldots+(x^27+\\frac1x^27)^2=\\\\++2\\cdot 27=\\\\\\fracx^2(x^54-1)x^2-1+\\fracx^-2(x^-54-1)x^2-1+54\\stackrel(1)=54.\\\\$$ $x^2=-x-1.$ $x^2+1=-x\\Rightarrow x+\\frac1x=-1\\Rightarrow \\frac1x=-x-1=x^2$ $<$as $x \\ne 0>$. $x^3=1.$ $x^n+\\frac1x^n=-1$ together $n = 3k+1,3k+2$ $x^n+\\frac1x^n=2$ together $n = 3k$ So, below the price is $18+4 \\times9=54$ $x^3=1$ $x^2+1/x^2=-1$ We need $\\sum_r=0^8\\sum_n=1^3(x^3r+n+x^-(3r+n))^2$ $=\\sum_r=0^8( (-1)^2+(-1)^2+2^2)$ $=9(6)$ Too lengthy for a comment. Since you currently received great explanations, i just concentrated on the an ext general problem of$$S_n=\\sum_k=1^n \\left(x^k+x^-k\\right)^2 \\qquad \\textwhere\\qquad x^2+x+1=0\\implies x=i^4/3$$These room the number which are congruent to $\\0, 1, 2\\$ modulo $6$ and there room several methods to write them. If girlfriend look here, you will find the nice$$S_n= 2 n-1+\\frac1\\sqrt3\\sin \\left(\\frac2 \\pi 3n\\right)+\\cos \\left(\\frac2 \\pi 3n\\right)$$ $$S_n=n-1+3 \\left\\lfloor \\fracn-13\\right\\rfloor$$ re-publishing mention monitor edited january 23 \"19 in ~ 3:54 answered jan 22 \"19 in ~ 11:39 Claude LeiboviciClaude Leibovici", "values of $$n$$: $$\\begin{array}{c|c|} & \\text{n\\in \\mathbb{N}}& \\text{(1- \\frac1{3^n})}\\\\ \\hline a & 1& \\frac23=0.\\overline{6}\\\\ \\hline b & 2& \\frac89=0.\\overline{8}\\\\ \\hline c & 3& \\frac{26}{27}=0.\\overline{962}\\\\ \\hline d & 4& \\frac{80}{81}=0.\\overline{987654320}\\\\ \\hline \\vdots\\\\ \\hline \\infty & n\\rightarrow \\infty& \\approx 1\\\\ \\hline \\end{array}$$ $$s\\not\\in \\{(1- \\frac1{3^n}\\,\\,: n \\in N \\})$$, as $$s$$ does not lie in the set. $$(1- \\frac1{3^n}) \\in [\\frac23, 1)$$. $$\\epsilon \\in\\mathbb{R}, \\epsilon\\gt 0\\implies \\forall \\epsilon \\gt 0: n\\lt \\infty$$, as $$s - \\epsilon \\lt 1$$. For interval $$(s-\\epsilon,s]$$, the lower bound implies for $$\\epsilon\\gt 0$$ that $$s-\\epsilon \\lt 1\\implies n \\lt \\infty$$, & the upper bound implies the max. value is $$s=1$$ at $$n= \\infty$$. All possible values of $$n$$ are covered in the interval $$(s-\\epsilon,s]$$ with $$n=\\infty$$ at the upper bound. 1. Identify the supremum of the set given below, & use the \"new\" definition of supremum to prove your claim: $$\\{(-\\frac12)^n\\,\\,: n \\in N \\}$$ $$\\sup(\\{(-\\frac12)^n\\,\\,: n \\in N \\})= 1.$$ $$s=\\frac14$$, for $$n=2$$. Below is table for set members for few values of $$n$$: $$\\begin{array}{c|c|} & \\text{n\\in \\mathbb{N}}& \\text{(-\\frac12)^n}\\\\ \\hline a & 1& -\\frac12=-0.5\\\\ \\hline b & 2& \\frac14=0.25\\\\ \\hline c & 3& -\\frac 18=-0.125\\\\ \\hline d & 4& \\frac{1}{16}=0.0625\\\\ \\hline \\vdots\\\\ \\hline \\infty & n\\rightarrow \\infty& \\approx 0\\\\ \\hline \\end{array}$$ As $$-(\\frac12)^n\\in \\{-\\frac12, \\cdots, \\frac14\\}$$, & as $$n \\rightarrow \\infty, -(\\frac12)^n \\approx 0$$.", "@funnyorangutan-1 Thanks for your very polite and curious responses. 🙂 The first question about why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is quite long, so if you don't mind, I'll put that in it's own post, here. https://forum.poshenloh.com/topic/545/why-is-0-3333 ⬅ ⬅ ⬅ Answer to why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ For your other question, about why $$0.\\overline{222}\\ldots = \\frac{2}{9} \\text{ and } 0.\\overline{444}\\ldots = \\frac{4}{9}$$ and so on, maybe this diagram will help a bit: M1-forum-fractions-with-denominator-9.png You can treat $$0.\\overline{111}\\ldots$$ as a starting point from which you can multiply by any factor in order to get any fraction of the form $$\\frac{n}{9},$$ where $$n$$ is an integer. \"Oh, but didn't you leave out the $$\\frac{3}{9} \\text{ and } \\frac{6}{9}?$$\" Very perceptive question! 🙂 I left those out because $$\\frac{3}{9}$$ and $$\\frac{6}{9}$$ aren't in simplified form. You wouldn't put \"$$\\frac{3}{9}$$\" on a test as an answer. Instead, you would put $$\\frac{1}{3}.$$ And you wouldn't write \"$$\\frac{6}{9}$$\" on a test as an answer either -- you would write $$\\frac{2}{3}.$$ So even though the pattern holds for multiplying $$0.\\overline{111}\\ldots$$ by both $$3$$ and $$6,$$ I left it out, but it is very nice to know that everyone holds across math, and that everything works as it should, because $$0.\\overline{333}\\ldots = \\frac{3}{9} = \\frac{1}{3}$$ Great, math is consistent! 🙂 When the world seems like it's failing you and", "$= \\dot{4}$ This is a pure recurring decimal. $\\\\$ Question 5: Convert each of the following decimals into a fraction : (i) $0.007 (ii)$latex 0.0875 (iii) $9.005 (iv)$latex 4.096 (i) $0.007 =$ $\\frac{7}{1000}$ (ii) $0.0875 =$ $\\frac{875}{10000}$ $=$ $\\frac{175}{2000} = \\frac{7}{40}$ (iii) $9.005 =$ $\\frac{9005}{1000}$ $=$ $\\frac{1801}{200}$ (iv) $4.096 =$ $\\frac{4096}{1000}$ $=$ $\\frac{2048}{500}$ $=$ $\\frac{1024}{250}$ $=$ $\\frac{512}{125}$ $\\\\$ Question 6: Express each of the following decimals as a rational number: (i) $0. \\dot{7}$ (ii) $0.\\overline{36}$ (iii) $4.\\overline{26}$ (iv) $0. 6\\dot{3}$ (v) $0.22\\overline{73}$ (vi) $2.1\\overline{36}$ (i) $0. \\dot{7}$ Let $x=0.7777777 \\ldots$ $\\Rightarrow 10x=7.777777 \\ldots$ $\\Rightarrow 9x = 7 \\ or \\ x =$ $\\frac{7}{9}$ (ii) $0.\\overline{36}$ Let $x=0.363636 \\ldots$ $\\Rightarrow 100x=36.363636 \\ldots$ $\\Rightarrow 99x=36$ $\\ or \\ x=$ $\\frac{4}{11}$ (iii) $4.\\overline{26}$ Let $x=4.26262626 \\ldots$ $\\Rightarrow 100x=426.262626 \\ldots$ $\\Rightarrow 99x=422$ $\\ or \\ x=$ $\\frac{422}{99}$ (iv) $0. 6\\dot{3}$ Let $x=0.63333333 \\ldots$ $\\Rightarrow 10x=6.3333333 \\ldots$ $\\Rightarrow 100x=63.333333 \\ldots$ $\\Rightarrow 90x=57$ $\\ or \\ x=$ $\\frac{19}{30}$ (v) $0.22\\overline{73}$ Let $x=0.227373737373 \\ldots$ $\\Rightarrow 100x=22.73737373737 \\ldots$ $\\Rightarrow 10000x=2273.7373737373 \\ldots$ $\\Rightarrow 9900x=2251$ $\\ or \\ x=$ $\\frac{2251}{9900}$ (vi) $2.1\\overline{36}$ Let $x=2.13636363636 \\ldots$ $\\Rightarrow 10x=21.3636363636 \\ldots$ $\\Rightarrow 1000x=2136.36363636 \\ldots$ $\\Rightarrow 990 x = 2115$ $\\ or \\ x =$ $\\frac{2115}{990}$ $=$ $\\frac{235}{110}$ $latex \\\\$ Question 7: Write the following fractions in descending order by converting them into decimals: (i) $\\frac{8}{25}$ $,$ $\\frac{4}{15}$ $,$ $\\frac{9}{20}$ $,$ $\\frac{3}{8}$", "+0 # 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\\end{array}$$ $$v = \\frac{2}{3}$$ or $$v = \\frac{5}{2}$$ Since $$v = \\sqrt x$$, $$\\sqrt x = \\frac{2}{3}$$ or $$\\sqrt x = \\frac{5}{2}$$ $$x = \\underline{\\underline {\\frac{4}{9}}} {\\rm{ }}$$ or $$x = \\underline{\\underline {\\frac{{25}}{4}}}$$", "\\;=\\; \\frac{0.28}{0.40} \\;=\\;0.70$ $(b)\\;\\;P(\\overline{A}\\,|\\,\\overline{B}) \\;=\\;\\frac{P(\\overline{A} \\wedge \\overline{B})}{P(\\overline{B})} \\;=\\; \\frac{0.32}{0.40} \\;=\\;0.80$ 4. Originally Posted by Soroban Hello, jordanrs! Place the information in a chart . . . . . $\\begin{array}{c||c|c||c|} & P(B) & P\\left(\\overline{B}\\right) & \\text{Total} \\\\ \\hline \\hline P(A) & 0.32 & & 0.40 \\\\ \\hline \\\\[-4mm] P\\left(\\overline{A}\\right) & & & \\\\ \\hline \\hline \\text{Total} & 0.60 & & 1.00 \\\\ \\hline \\end{array}$ Fill in the empty cells: . . $\\begin{array}{c||c|c||c|} & P(B) & P\\left(\\overline{B}\\right) & \\text{Total} \\\\ \\hline \\hline P(A) & 0.32 & 0.08 & 0.40 \\\\ \\hline \\\\[-4mm] P\\left(\\overline{A}\\right) & 0.28 & 0.32 & 0.60 \\\\ \\hline \\hline \\text{Total} & 0.60 & 0.40 & 1.00 \\\\ \\hline \\end{array}$ $(a)\\;\\;P(B\\,|\\,\\overline{A}) \\;=\\;\\frac{P(B \\wedge \\overline{A})}{P(\\overline{A})} \\;=\\; \\frac{0.28}{0.40} \\;=\\;0.70$ $(b)\\;\\;P(\\overline{A}\\,|\\,\\overline{B}) \\;=\\;\\frac{P(\\overline{A} \\wedge \\overline{B})}{P(\\overline{B})} \\;=\\; \\frac{0.32}{0.40} \\;=\\;0.80$", "\\overline{62)} & \\overline{124} & (2 \\\\ & & & & 124 & \\\\ & & & & \\overline{0} & \\end{array}$ Therefore $\\frac{682}{868}$ $=$ $\\frac{682 \\div 62}{868 \\div 62}$ $=$ $\\frac{11}{14}$ ii) $\\frac{777}{1147}$ $\\begin{array}{r r r r r } 777) & \\overline{1147} & (1 & & \\\\ & 777 & & & \\\\ & \\overline{370)} & \\overline{777} & (2 & \\\\ & & 740 & & \\\\ & & \\overline{37)} & \\overline{370} & (10 \\\\ & & & 370 & \\\\ & & & \\overline{0} & \\end{array}$ Therefore $\\frac{777}{1147}$ $=$ $\\frac{777 \\div 37}{1147 \\div 37}$ $=$ $\\frac{21}{31}$ iii) $\\frac{1095}{1168}$ $\\begin{array}{r r r r } 1095) & \\overline{1168} & (1 & \\\\ & 1095 & & \\\\ & \\overline{73)} & \\overline{1095} & (15 \\\\ & & 1095 & \\\\ & & \\overline{0} & \\end{array}$ Therefore $\\frac{1095}{1168}$ $=$ $\\frac{1095 \\div 73}{1068 \\div 73}$ $=$ $\\frac{15}{16}$ $\\\\$ Question 4. Which of the following numbers are co-primes? i) $18, 25$ ii) $62, 81$ iii) $69, 92$ Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$. You can calculate the HCF by using any of the two methods…prime factorization or long division method. i) $18, 25$ HCF of $18, 25 = 1$ Hence, $18$ and $25$ are co-primes. ii) $62, 81$ HCF of $62, 81 =" ]

[ "# 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 $\\frac{4}{9}$.", "+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", "the confusion the following representation is adapted. The number is given as $0.3\\overline{74}$$0.3 \\overline{74}$ The line over $74$$74$ represents that $74$$74$ is repeated. Convert $\\frac{1}{3}$$\\frac{1}{3}$ into a fraction. The answer is $0.\\overline{3}$$0. \\overline{3}$ Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. convert back to fraction How do we convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction? To convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction, the following steps are used $x=0.\\overline{4}$$x = 0. \\overline{4}$ $10x=4.\\overline{4}$$10 x = 4. \\overline{4}$ Subtracting the two equations $9x=4$$9 x = 4$ $x=\\frac{4}{9}$$x = \\frac{4}{9}$ examples Converting $2.\\overline{4}$$2. \\overline{4}$ into a fraction. $x=2.\\overline{4}$$x = 2. \\overline{4}$ $10x=24.\\overline{4}$$10 x = 24. \\overline{4}$ subtracting the two above $9x=22$$9 x = 22$ $x=\\frac{22}{9}$$x = \\frac{22}{9}$ $x=2\\frac{4}{9}$$x = 2 \\frac{4}{9}$ To convert $2.23\\overline{43}$$2.23 \\overline{43}$ into a fraction: $x=2.23\\overline{43}$$x = 2.23 \\overline{43}$ $100x=223.43\\overline{43}$$100 x = 223.43 \\overline{43}$ Subtracting the above, $99x=221.20$$99 x = 221.20$ $990x=2212$$990 x = 2212$ $x=\\frac{2212}{990}$$x = \\frac{2212}{990}$. Another method is as follows. $100x=223+0.\\overline{43}$$100 x = 223 + 0. \\overline{43}$ $10000x=223×100+43.\\overline{43}$$10000 x = 223 \\times 100 + 43. \\overline{43}$ Subtracting the two equations $9900x=223×\\left(100-1\\right)+43$$9900 x = 223 \\times \\left(100 - 1\\right) + 43$ $x=\\frac{223}{100}+\\frac{43}{9900}$$x = \\frac{223}{100} + \\frac{43}{9900}$ summary Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. Conversion of Repeating Decimals to equivalent", "# How do you express .33 as a fraction in simplest form? Sep 14, 2016 $\\frac{33}{100}$ Sep 14, 2016 As you wrote the question $\\frac{33}{100}$ If you meant $0.333333 . . \\to 0.33 \\overline{3} \\text{ }$ then $\\text{ } \\frac{1}{3}$ #### Explanation: It all depends on what you mean by $0.33$ If you meant it to be exactly as written then we have $\\frac{3}{10} + \\frac{3}{100}$ Write as $\\left(\\frac{3}{10} \\times 1\\right) + \\frac{3}{100}$ But 1 may be written as $\\frac{10}{10}$ $\\left(\\frac{3}{10} \\times \\frac{10}{10}\\right) + \\frac{3}{100}$ $\\frac{30}{100} + \\frac{3}{100} = \\frac{33}{100}$ $\\textcolor{red}{\\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$ If you meant what you wrote to mean $0.33 \\leftarrow$ 3 going on for ever $0.333333333 \\ldots . .$ written as $0.33 \\overline{3}$ $\\textcolor{b r o w n}{\\text{Then this is very different:}}$ Let $x = 0.33 \\overline{3}$ thus $10 x = 3.33 \\overline{3}$ $10 x - x = 3$ $9 x = 3$ $x = \\frac{3}{9} = \\frac{1}{3}$", "# RD Sharma Solutions Class 9 Number System Exercise 1.2 ## RD Sharma Solutions Class 9 Chapter 1 Ex 1.2 Q1. Express the following rational numbers as decimals: (i) $\\frac{42}{100}$ (ii) $\\frac{327}{500}$ (iii) $\\frac{15}{4}$ Solution: (i) By long division method 100) $\\overline{42}$ ( 0.42 400 $\\overline{200}$ 200 $\\overline{0}$ Therefore, $\\frac{42}{100}$ = 0.42 (ii) By long division method 500) $\\overline{327.000}$ ( 0.654 3000 $\\overline{2700}$ 2500 $\\overline{2000}$ 2000 $\\overline{0}$ Therefore, $\\frac{327}{500}$ = 0.654 (iii) By long division method 4) $\\overline{15.00}$ ( 3.75 12 $\\overline{30}$ 28 $\\overline{20}$ 20 $\\overline{0}$ Therefore, $\\frac{15}{4}$ = 3.75 Q2. Express the following rational numbers as decimals: (i) $\\frac{2}{3}$ (ii) –$\\frac{4}{9}$ (iii) –$\\frac{2}{15}$ (iv) –$\\frac{22}{13}$ (v) $\\frac{437}{999}$ Solution: (i) By long division method 3) $\\overline{2.0000}$ ( 0.66 18 $\\overline{20}$ 18 $\\overline{2}$ Therefore, $\\frac{2}{3}$ = 0.66 (ii) By long division method 9) $\\overline{4.000}$ ( 0.444 3600 $\\overline{4000}$ 3600 $\\overline{4000}$ 3600 $\\overline{400}$ Therefore, – $\\frac{4}{9}$ = – 0.444 (iii) By long division method 15) $\\overline{2.00}$ ( 1.333 15 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{5}$ Therefore, $\\frac{2}{15}$ = -1.333 (iv) By long division method 13) $\\overline{22.000}$ ( 1.69230769 13 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{30}$ 26 $\\overline{40}$ 39 $\\overline{100}$ 91 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{3}$ Therefore, – $\\frac{22}{13}$ = – 1.69230769 (v) By long division method 999) $\\overline{437.0000}$ ( 0.43743 3996 $\\overline{3740}$ 2997 $\\overline{7430}$ 6993 $\\overline{4370}$ 3996 $\\overline{3740}$ 2997 $\\overline{743}$", "# Question #0426c Jan 5, 2017 $0.8 \\overline{7} = \\frac{79}{90}$ #### Explanation: Let $x = 0.8 \\overline{7}$ $\\implies 10 x = 8. \\overline{7}$ $\\implies 10 x - x = 8. \\overline{7} - 0.8 \\overline{7}$ $\\implies 9 x = 8.7 - 0.8$ $\\implies 9 x = 7.9$ $\\implies 9 x = \\frac{79}{10}$ $\\implies x = \\frac{79}{10} \\cdot \\frac{1}{9}$ $\\therefore x = \\frac{79}{90}$ This technique of multiplying by $10$ (or a power of $10$) and subtracting can be used to find the fractional representation of any number with infinitely repeating decimal digits. This answer contains a detailed explanation.", "# Express 0.99999…in the form $\\frac{p}{q}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Solution: Let $x=0.9999 \\ldots$ $10 x=9.9999 \\ldots$ $10 x=9+x$ $9 x=9$ $x=1$", "of simple fractions as decimals: $\\frac 12 = 0.5$ $\\frac 13 = 0.333333\\ldots = 0.\\overline{3}$ $\\frac 23 = 0.666666\\ldots = 0.\\overline{6}$ $\\frac 14 = 0.25$ $\\frac 34 = 0.75$ $\\frac 15 = 0.2$ $\\frac 16 = 0.1666666\\ldots = 0.1\\overline{6}$ $\\frac 17 = 0.14285714285714\\ldots = 0.\\overline{142857}$ $\\frac 18=0.125$ $\\frac 38 = 0.375$ $\\frac 78 = 0.875$ $\\frac 19 = 0.111111\\ldots = 0.\\overline{1}$ $\\frac 29 = 0.222222\\ldots = 0.\\overline{2}$ $\\frac 1{10} = 0.1$ $\\frac 1{11} = 0.09090909\\ldots = 0.\\overline{09}$ $\\frac 1{12} = 0.08333333\\ldots = 0.08\\overline{3}$ A few things can be observed: • Some decimals are short and to the point. • Others have an infinite number of digits (students like to say: \"they are infinite\"), in a repeating pattern. • There can be a single digit that repeats, or a string of two or more digits repeating. • The repeating pattern (period) does not always start right after the decimal point. The long period of $\\frac 17$ sticks out like a sore thumb. It has a few crazy cousins further along: $\\frac 1{13} = 0.\\overline{076923}$ $\\frac 1{17} = 0.\\overline{0588235294117647}$ $\\frac 1{19} = 0.\\overline{052631578947368421}$ There are also the weird $\\frac 1{81} = 0.\\overline{012345679}$ (sic!) and the mysterious $$\\frac 1{9801} = 0.0001020304050607080910111213\\ldots$$ It is also worthwhile to notice this pattern (for later): $$\\frac 19 = 0.\\overline{1}$$ $$\\frac 1{99} = 0.\\overline{01}$$ $$\\frac 1{999}", "score is 70%. Tim scored higher than Jim. Question 7: Express the following percentage as a fraction in its simplest form. $$48\\% =$$ ________ Solution: \\begin{align*} 48\\%&= \\frac {48} {100} \\\\ &= \\frac {12} {25} \\end{align*} $$\\frac {12} {25}$$ Question 8: Express the following percentage as a fraction in its simplest form. $$80\\% =$$ ________ Solution: \\begin{align*} 80 \\% &= \\frac {80} {100} \\\\ &= \\frac {4} {5} \\end{align*} $$\\frac {4} {5}$$ Question 9: Express $$70\\%$$ as a decimal. Solution: \\begin{align*} 70 \\% &= \\frac {70} {100} \\\\ &= 0.7 \\end{align*} $$0.7$$ Question 10: Express $$550\\%$$ as a decimal. Solution: \\begin{align*} 550 \\% &= \\frac {550} {100} \\\\ &= 5.5 \\end{align*} $$5.5$$ Question 11: Express $$0.2\\%$$ as a fraction in its simplest form. Solution: \\begin{align*} 0.2 \\% &= \\frac {0.2} {100} \\\\ &= 0.002 \\\\ &= \\frac {2} {1000} \\\\ &= \\frac {1} {500} \\\\ \\end{align*} $$\\frac {1} {500}$$ Question 12: Express $$0.5\\%$$ as a fraction in its simplest form. Solution: \\begin{align*} 0.5 \\% &= \\frac {0.5} {100} \\\\ &= 0.005 \\\\ &= \\frac {5} {1000} \\\\ &= \\frac {1} {200} \\\\ \\end{align*} $$\\frac {1} {200}$$ Question 13: Express $$4\\frac {1} {2}\\%$$ as a decimal. Solution: \\begin{align*} 4 \\frac {1} {2} \\% &= 4\\frac {5} {10}\\% \\\\ &= 4.5\\% \\\\ &= \\frac {4.5} {100} \\\\ &= 0.045 \\\\", "+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", "+0 # HELP ME! 0 90 1 Express $$\\sqrt{x} \\div\\sqrt{y}$$ as a common fraction given: $$\\frac{ {\\left( \\frac{1}{2} \\right)}^2 + {\\left( \\frac{1}{3} \\right)}^2 }{ {\\left( \\frac{1}{4} \\right)}^2 + {\\left( \\frac{1}{5} \\right)}^2} = \\frac{13x}{41y}$$ Sep 30, 2020 #1 +111981 0 First put the left side into its simplest form. Can you do that? (you can use a calc if you have to but make sure it stays in fraction form) Sep 30, 2020", "+0 # pls help me 0 85 3 What is $\\frac{0.\\overline{72}}{0.\\overline{27}}$? Express your answer as a common fraction in lowest terms. Jul 27, 2022 #1 -1 The expression reduces to 2/9. Jul 27, 2022 #2 +2448 0 Your answer is wrong, the number is more than the denominator, so the fraction will be greater than 1... BuilderBoi Jul 28, 2022 #3 +2448 0 I'll do the numerator, and you can do the denominator. For the numerator, let $$x = 0.{ \\overline{72}}$$. This means $$100x = {72. \\overline{72}}$$ (move the decimal point 2 places to the right) Subtracting the 2 yields us $$99x = 72$$ (the repeating parts cancel out) Dividing by 99, we find that $$x = {72 \\over 99}$$ Jul 28, 2022", "# Express $3.72444\\ldots$ as a fraction using the formula for geometric progressions I did the following: $$3.72\\overline{4} = 3.724+\\left(\\left(\\frac{4}{10^4}\\right)+\\left(\\frac{4}{10^5}\\right)+\\left(\\frac{4}{10^6}\\right)+\\cdots\\right)$$ where $a=3.724$ and $r=\\dfrac{1}{10}$ Using $S=\\dfrac{a}{1-r}$ I get $\\dfrac{37.24}{9}\\neq3.72\\overline{4}$ The answer given by the text is $\\dfrac{838}{225}$ I've been staring at this for an hour and can't figure out where I'm going wrong. • Your first term in the series is $3.724$, but your next term is not $3.724\\times\\frac{1}{10}$. Not all of your fraction will actually be part of the series. – Steven Stadnicki Jul 6 '14 at 18:24 Start your geometric series after $3.72$: $$3.72\\overline{4} = 3.72+\\left(\\left(\\frac{4}{10^3}\\right)+\\left(\\frac{4}{10^4}\\right)+\\left(\\frac{4}{10^5}\\right)+\\cdots\\right)$$ Then you have $a=\\dfrac{4}{10^3}$ and $r=\\dfrac 1{10}$. The ultimate sum is then $$3.72 + \\frac{a}{1-r}.$$ First of all, your $a$ is incorrect. I think reviewing the formula for a geometric series would be helpful. The way to do these kinds of problems is to split it up into two parts: one involving the part without the repeated pattern (in this case $3.72=\\frac{372}{100}$) and the other part involving the repeated part (in this case $.00444\\ldots$) which can be rewritten as a geometric series. Hint: You consider only the geometric progression, it's $((\\frac{4}{10^4})+(\\frac{4}{10^5})+(\\frac{4}{10^6})+\\cdots)$. For this sequence $q=\\frac{1}{10}$ and $a=\\frac{4}{10^4}$, not $a=3.724$. Only part of this is geometric: $$3.72\\bar{4}= 3.72 +0.00\\bar 4=\\frac{372}{100}+\\frac{4}{1000}\\left(1+\\frac{1}{10}+\\frac{1}{10^2}+\\frac{1}{10^3}+\\cdots\\right)$$ $$=\\frac{372}{100}+\\frac{4}{1000}\\cdot\\frac{1}{1-\\frac{1}{10}} = \\frac{372}{100}+\\frac{4}{1000}\\cdot\\frac{10}{9} =\\frac{372}{100}+\\frac{4}{900}$$ $$=\\frac{4\\cdot93}{4\\cdot25}+\\frac{4\\cdot1}{4\\cdot225}=\\frac{9\\cdot93}{9\\cdot25} +\\frac{1}{225}=\\frac{837}{225}+\\frac{1}{225}=\\boxed{\\dfrac{838}{225}}$$", "Expression \\begin{align} & \\frac{\\frac{-90-40+36-60+84+45}{60}}{\\frac{90+75-80+100-12-48}{60}} \\\\ & =\\frac{165-190}{265-140}=-\\frac{25}{125}=-\\frac{1}{5} \\\\ \\end{align} Question 5 The simplification of $\\left( 0.\\overline{65}+.0\\overline{35}+0.\\overline{87} \\right)$ yields the result A $1.\\overline{98}$ B $1.\\overline{81}$ C $1.\\overline{79}$ D 1.99 Question 5 Explanation: Expression \\begin{align} & \\left( 0.\\overline{65}+.0\\overline{35}+0.\\overline{87} \\right) \\\\ & =\\frac{65}{99}+\\frac{35}{99}+\\frac{87}{99} \\\\ & =\\frac{65+35+87}{99}=\\frac{187}{99} \\\\ & =1\\frac{98}{99} \\\\ & =1.\\overline{98} \\\\ \\end{align} Once you are finished, click the button below. Any items you have not completed will be marked incorrect. There are 5 questions to complete. ← List →", "#### Practice questions ##### Question 1 Express as a fraction in simplest form. $625^{\\frac{-3}{4}}$62534 ##### Question 2 Express $\\sqrt[5]{x^7}$5x7 in index form. ##### Question 3 Fill in the blanks to simplify the given expression. 1. $\\sqrt{m^8}$√m8 $=$= $\\left(m^8\\right)^{\\editable{}}$(m8) $=$= $m^{\\editable{}\\times\\frac{1}{2}}$m×12​ $=$= $m^{\\editable{}}$m ##### Question 4 Simplify $\\left(625u^{16}v^{12}\\right)^{\\frac{1}{2}}$(625u16v12)12. ### Outcomes #### 0606C4.1A Perform simple operations with indices.", "@funnyorangutan-1 Thanks for your very polite and curious responses. 🙂 The first question about why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is quite long, so if you don't mind, I'll put that in it's own post, here. https://forum.poshenloh.com/topic/545/why-is-0-3333 ⬅ ⬅ ⬅ Answer to why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ For your other question, about why $$0.\\overline{222}\\ldots = \\frac{2}{9} \\text{ and } 0.\\overline{444}\\ldots = \\frac{4}{9}$$ and so on, maybe this diagram will help a bit: M1-forum-fractions-with-denominator-9.png You can treat $$0.\\overline{111}\\ldots$$ as a starting point from which you can multiply by any factor in order to get any fraction of the form $$\\frac{n}{9},$$ where $$n$$ is an integer. \"Oh, but didn't you leave out the $$\\frac{3}{9} \\text{ and } \\frac{6}{9}?$$\" Very perceptive question! 🙂 I left those out because $$\\frac{3}{9}$$ and $$\\frac{6}{9}$$ aren't in simplified form. You wouldn't put \"$$\\frac{3}{9}$$\" on a test as an answer. Instead, you would put $$\\frac{1}{3}.$$ And you wouldn't write \"$$\\frac{6}{9}$$\" on a test as an answer either -- you would write $$\\frac{2}{3}.$$ So even though the pattern holds for multiplying $$0.\\overline{111}\\ldots$$ by both $$3$$ and $$6,$$ I left it out, but it is very nice to know that everyone holds across math, and that everything works as it should, because $$0.\\overline{333}\\ldots = \\frac{3}{9} = \\frac{1}{3}$$ Great, math is consistent! 🙂 When the world seems like it's failing you and", ". \\overline{857142}$$ Question 3. Express the following in the form p/q, where p and q are integers and q ≠ 0. (i) $$0 . \\overline{6}$$ (ii) $$0 . 4 \\overline{7}$$ (iii) $$0 . \\overline{001}$$ (i) $$0 . \\overline{6}$$ = 0.666… Let x = 0.666… ∴ 10x = 6.66… ∴ 10x = 6 + 0.66… ∴ 10x = 6 + 0.666… ∴ 10x = 6 + x ∴ 9x = 6 x = $$\\frac{2}{3}$$ (ii) $$0 . 4 \\overline{7}$$ = 0.4777… = $$\\frac{4}{10}$$ + $$\\frac{0.777}{10}$$ Let x = 0.777… ∴ 10x = 7.77… ∴ 10x = 7.777… ∴ 10x = 7 + 0.777… ∴ 10x = 7 + x ∴ x = $$\\frac{7}{9}$$ $$0 . 4 \\overline{7}$$ = $$\\frac{4}{10}$$ + $$\\frac{0.777}{10}$$ = $$\\frac{4}{10}$$ + $$\\frac{7}{90}$$ = $$\\frac{36}{90}$$ + $$\\frac{7}{90}$$ = $$\\frac{43}{90}$$ (iii) $$0 . \\overline{001}$$ =0.001001… Let x = 0.001001… ∴ 1000x = 1.001…. ∴ 1000x = 1.001001… ∴ 1000x = 1 + 0.001001 … ∴ l000x= 1 + X ∴ 999x = 1 x = $$\\frac{1}{999}$$ Question 4. Express 0.99999…in the form $$\\frac{p}{q}$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Let x = 0.9999… 10x = 9.999… ∴ 10x = 9.9999… ∴ 10x = 9 + 0.9999… ∴ 10x = 9 + x ∴ 9x = 9 x", "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$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", "$\\overline{)\\text{}S\\infty =\\frac{a}{1-r}\\text{}}$ Example (Change recurring decimal to fraction) Express each of the following recurring decimals as a fraction in its lowest terms. (a) 0.8888 ... (b) 0.171717... (c) 0.513513513 …. Solution: (a) 0.8888 = 0.8 + 0.08 + 0.008 +0.0008 + ….. (recurring decimal) $\\begin{array}{l}GP,\\text{}a=0.8,\\text{}r=\\frac{0.08}{0.8}=0.1\\\\ {S}_{\\infty }=\\frac{a}{1-r}\\\\ {S}_{\\infty }=\\frac{0.8}{1-0.1}\\\\ {S}_{\\infty }=\\frac{0.8}{0.9}\\\\ {S}_{\\infty }=\\frac{8}{9}\\to \\overline{)\\begin{array}{l}\\text{check using calculator}\\\\ \\text{}\\frac{8}{9}=0.888888....\\end{array}}\\end{array}$ (b) 0.17171717 ….. = 0.17 + 0.0017 + 0.000017 + 0.00000017 + ….. $\\begin{array}{l}GP,\\text{}a=0.17,\\text{}r=\\frac{0.0017}{0.17}=0.01\\\\ {S}_{\\infty }=\\frac{a}{1-r}\\\\ {S}_{\\infty }=\\frac{0.17}{1-0.01}=\\frac{0.17}{0.99}=\\frac{17}{99}\\to \\overline{)\\begin{array}{l}\\text{remember to check the}\\\\ \\text{answer using calculator}\\end{array}}\\end{array}$ (c) 0.513513513….. = 0.513 + 0.000513 + 0.000000513 + ….. $\\begin{array}{l}GP,\\text{}a=0.513,\\text{}r=\\frac{0.00513}{0.513}=0.001\\\\ {S}_{\\infty }=\\frac{a}{1-r}\\\\ {S}_{\\infty }=\\frac{0.513}{1-0.001}=\\frac{0.513}{0.999}=\\frac{513}{999}=\\frac{19}{37}\\end{array}$ # 3. The nth Term of Geometric Progressions (Part 2) (D) The Number of Term of a Geometric Progression Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term Example 2: Find the number of terms for each of the following geometric progressions. (a) 2, 4, 8, ….., 8192 (b) $\\frac{1}{4},\\text{}\\frac{1}{6},\\text{}\\frac{1}{9},\\text{}.....,\\text{}\\frac{16}{729}$ (c) $-\\frac{1}{2},\\text{1},\\text{}-\\text{2},.....,\\text{64}$ Solution: (a) $\\begin{array}{l}2,\\text{}4,\\text{}8,.....,\\text{}8192←\\left(\\text{Last term is given)}\\\\ a=2\\\\ r=\\frac{{T}_{2}}{{T}_{1}}=\\frac{4}{2}=2\\\\ {T}_{n}=8192\\\\ a{r}^{n-1}=8192←\\left(Then\\text{th term of GP,}{T}_{n}=a{r}^{n-1}\\right)\\\\ \\left(2\\right){\\left(2\\right)}^{n-1}=8192\\\\ {2}^{n-1}=4096\\\\ {2}^{n-1}={2}^{12}\\\\ n-1=12\\\\ n=13\\end{array}$ (b) $\\begin{array}{l}\\frac{1}{4},\\text{}\\frac{1}{6},\\text{}\\frac{1}{9},\\text{}.....,\\text{}\\frac{16}{729}\\\\ a=\\frac{1}{4},r=\\frac{\\frac{1}{6}}{\\frac{1}{4}}=\\frac{2}{3}\\\\ {T}_{n}=\\frac{16}{729}\\\\ a{r}^{n-1}=\\frac{16}{729}\\\\ \\left(\\frac{1}{4}\\right){\\left(\\frac{2}{3}\\right)}^{n-1}=\\frac{16}{729}\\\\ {\\left(\\frac{2}{3}\\right)}^{n-1}=\\frac{16}{729}×4\\\\ {\\left(\\frac{2}{3}\\right)}^{n-1}=\\frac{64}{729}\\\\ {\\left(\\frac{2}{3}\\right)}^{n-1}={\\left(\\frac{2}{3}\\right)}^{6}\\\\ \\therefore n-1=6\\\\ n=7\\end{array}$ (c) $\\begin{array}{l}-\\frac{1}{2},\\text{1},\\text{}-\\text{2},.....,\\text{64}\\\\ a=-\\frac{1}{2},\\text{}r=\\frac{-2}{1}=-2\\\\ {T}_{n}=64\\\\ a{r}^{n-1}=64\\\\ \\left(-\\frac{1}{2}\\right){\\left(-2\\right)}^{n-1}=64\\\\ {\\left(-2\\right)}^{n-1}=64×-2\\\\ {\\left(-2\\right)}^{n-1}=-128\\\\ {\\left(-2\\right)}^{n-1}={\\left(-2\\right)}^{7}\\\\ n-1=7\\\\ n=8\\end{array}$ (E) Three consecutive terms of a geometric progression If e, f and g are 3 consecutive terms of", "$= \\dot{4}$ This is a pure recurring decimal. $\\\\$ Question 5: Convert each of the following decimals into a fraction : (i) $0.007 (ii)$latex 0.0875 (iii) $9.005 (iv)$latex 4.096 (i) $0.007 =$ $\\frac{7}{1000}$ (ii) $0.0875 =$ $\\frac{875}{10000}$ $=$ $\\frac{175}{2000} = \\frac{7}{40}$ (iii) $9.005 =$ $\\frac{9005}{1000}$ $=$ $\\frac{1801}{200}$ (iv) $4.096 =$ $\\frac{4096}{1000}$ $=$ $\\frac{2048}{500}$ $=$ $\\frac{1024}{250}$ $=$ $\\frac{512}{125}$ $\\\\$ Question 6: Express each of the following decimals as a rational number: (i) $0. \\dot{7}$ (ii) $0.\\overline{36}$ (iii) $4.\\overline{26}$ (iv) $0. 6\\dot{3}$ (v) $0.22\\overline{73}$ (vi) $2.1\\overline{36}$ (i) $0. \\dot{7}$ Let $x=0.7777777 \\ldots$ $\\Rightarrow 10x=7.777777 \\ldots$ $\\Rightarrow 9x = 7 \\ or \\ x =$ $\\frac{7}{9}$ (ii) $0.\\overline{36}$ Let $x=0.363636 \\ldots$ $\\Rightarrow 100x=36.363636 \\ldots$ $\\Rightarrow 99x=36$ $\\ or \\ x=$ $\\frac{4}{11}$ (iii) $4.\\overline{26}$ Let $x=4.26262626 \\ldots$ $\\Rightarrow 100x=426.262626 \\ldots$ $\\Rightarrow 99x=422$ $\\ or \\ x=$ $\\frac{422}{99}$ (iv) $0. 6\\dot{3}$ Let $x=0.63333333 \\ldots$ $\\Rightarrow 10x=6.3333333 \\ldots$ $\\Rightarrow 100x=63.333333 \\ldots$ $\\Rightarrow 90x=57$ $\\ or \\ x=$ $\\frac{19}{30}$ (v) $0.22\\overline{73}$ Let $x=0.227373737373 \\ldots$ $\\Rightarrow 100x=22.73737373737 \\ldots$ $\\Rightarrow 10000x=2273.7373737373 \\ldots$ $\\Rightarrow 9900x=2251$ $\\ or \\ x=$ $\\frac{2251}{9900}$ (vi) $2.1\\overline{36}$ Let $x=2.13636363636 \\ldots$ $\\Rightarrow 10x=21.3636363636 \\ldots$ $\\Rightarrow 1000x=2136.36363636 \\ldots$ $\\Rightarrow 990 x = 2115$ $\\ or \\ x =$ $\\frac{2115}{990}$ $=$ $\\frac{235}{110}$ $latex \\\\$ Question 7: Write the following fractions in descending order by converting them into decimals: (i) $\\frac{8}{25}$ $,$ $\\frac{4}{15}$ $,$ $\\frac{9}{20}$ $,$ $\\frac{3}{8}$" ]

[ "at 5:42 • In case 1, I believe your initial calculation is correct, but when you turn back after 9 hours for the 10th tunnel, won't this change the assumption that 5 tunnels must be traveled on average to escape? – knrumsey May 11 '17 at 12:35", "# Tunneling A straight tunnel is drilled between two points (not necessarily diametrically opposite) on the Earth. An object is dropped into the tunnel. How much time (in minutes) does it take to reach the other end? Ignore air resistance and/or friction, and assume that the density of the Earth is constant. ×", "having $\\gamma = 1$. May 2 at 9:16 Why does simultaneity affect the measurement of length? Suppose a train 990m long is heading into a tunnel 1000m long at 100m per second. You and a friend don't know the length of the train, and are asked to determine whether the train is longer or shorter than the tunnel. You realise that you can answer the question by observing whether the back of the train enters the tunnel before or after the front of the train leaves it. If it enters afterwards, the train is longer than the tunnel- if it enters before, the train is shorter than the tunnel. Accordingly, you and a friend stand at either end of the tunnel and make a note of the times when the rear of the train enters the tunnel and the front leaves. If your watches are accurately synchronised, you will find that the rear of the train enters the tunnel 0.1s before the front leaves it, so the train is shorter than the tunnel. However, had the watch of the person observing the rear of the train been just 0.2s fast, you would have concluded that the train was longer than the tunnel. Your ability to measure the moving train depends upon you being able to pin down the", "1km/hour east and north respectively what is head to tail rule? kinza", "Oct 19, 2006 ### quinn The real problem here is that you need a frictionless track. Not going to happen. Another problem would be derailing due to corolis forces if one didn't have the train take the proper \"twisted\" path. I am also not entirely sure how the corolis forces would affect the motion/period of the train if one crossed lataitudes... All in all the biggest obstactle would be to dig a tunnel that long,... how long did it take the \"Chunnel\" to be built\" Let's say 10 years... the Chunnel is 26 miles long,.. let's say 30 miles,.. and from LA to NYC is about 3000 miles,... this tunnel is going to take roughly 100 years?... 6. Oct 20, 2006 ### DocZaius This is a completely theoretical question which has to do with the application of the pendulum's fundamentals into a sphere/tunnel experiment. I just wanted to see some good discussion on this problem. HallsofIvy, I got 42 minutes from my own calculations, but here's a page that goes through them: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html From all the thought I put into it, a tunnel from any point to any point on that sphere will give the same period of oscillation. I would love to be shown to be wrong though. The easiest way for me to think about the", "A Roller Coaster Ride through Relativity/Simultaneity Simultaneity There is a curious paradox connected with length contraction which I shall call the train in the tunnel paradox. A train whose length when at rest is agreed by all observers to be 100 m enters a tunnel whose length is also exactly 100 m. The train stops in the tunnel and everybody agrees that when the engine is flush with the exit, the guard's van is flush with the entrance. A stationary train in a tunnel Some time later, the train passes through the same tunnel at high speed - 80% of the velocity of light in fact. At this speed the g factor is 1.67 and to the man standing by the track at the entrance to the tunnel the train appears to be only 60 m long and spends an appreciable amount of time completely inside the tunnel. A (contracted) train in a tunnel The engine driver, however, sees things rather differently. To him it is the tunnel which is whizzing past at 80% of the velocity of light and which is in consequence shrunk to 60 m in length. A train in a (contracted) tunnel Surely they can't both be right? To put it even more forcibly, suppose the man beside the track is in fact a", "descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m. Answer: Total distance covered = 10 – ( - 350) = 360 m Time taken to cover the distance = 360 ÷ 6 = 60 min = 1 hour", "# Question 1 of 12 It took Uncle Jim 30 minutes to drive from the first exit to the 6$^{th}$ exit of an expressway. The exits of that expressway were spaced apart equally. Which exit would he have passed by after one hour? A 11$^{th}$ exit B 12$^{th}$ exit C 13$^{th}$ exit D 14$^{th}$ exit E None of the above", "represent 1km/hour east and north respectively what is head to tail rule? kinza", "An Oncoming Train) 2:32 21 Dic 2005, 10:54 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 21 Dic 2005, 10:45 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 20 Dic 2005, 19:00 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 20 Dic 2005, 11:04 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 19 Dic 2005, 17:06 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 19 Dic 2005, 12:20 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 16 Dic 2005, 18:01", "A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. What is the ratio of speed of train and cat ? a) 3 : 1 b) 4 :1 c) 5 : 1 d) None of these Question 9: City Bus Corporation runs two buses from terminus A to terminus B,one from each of the terminuses such that each bus makes 5 round trips in a day. There are no stops in between. These buses ply back and forth on the same route at different but uniform speeds. Each morning the buses start at 7 AM from the respective terminuses. They meet for the first time at a distance of 7 km from terminus A. Their next meeting is at a distance of 4 km from terminus B, while travelling in opposite directions. Assuming that the time taken by the buses at the terminuses is negligibly small, and the cost of running a bus is 20 per", "train of length $L,$ approaching a tunnel of length $L\\sqrt{1-{v}^{2}/{c}^{2}},$ so the tunnel is not as long as the train from his viewpoint! What does he think happens when the bandits close both the doors? ## The Tunnel Doors are Closed Simultaneously The key to understanding what is happening here is that we said the bandits closed the two doors at the ends of the tunnel at the same time. How could they arrange to do that, since the doors are far apart? They could use walkie-talkies, which transmit radio waves, or just flash a light down the tunnel, since it’s long and straight. Remember, though, that the train is itself going at a speed close to that of light, so they have to be quite precise about this timing! The simplest way to imaging them synchronizing the closings of the two doors is to assume they know the train’s timetable, and at a prearranged appropriate time, a light is flashed halfway down the tunnel, and the end doors are closed when the flash of light reaches the ends of the tunnel. Assuming the light was positioned correctly in the middle of the tunnel, that should ensure that the two doors close simultaneously. ## Or are they? Now consider this door-closing operation from the point of view of", "condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's travel time to the front of the train.\" I now understand that you meant \"to the front of the tunnel\"; however that doesn't change my answer to your question. Apparently you did not read that I wrote: \"the two conditions boil down to the same. The explosion event is only conditioned by the question: at the moment that the bomb reaches the end of the tunnel, has the signal already reached that same point or not. That is also explained in the answer. If the pulse reaches the end of the train before the train reaches the end of the tunnel, then it also reaches the end of the tunnel before the train does - and the other way round.\" Good luck, Harald Last edited: Aug 17, 2011", "the skull and returning makes a round trip in 0.15 milliseconds. So how many times has it traveled the diameter of the skull?", "tunnel that goes directly from one point to the other through the interior of the sphere. In a large body such as the Earth, this train could be left to accelerate using just the force of gravity, since, during the first half of the trip (from the point of departure until the middle), the downward pull towards the center of gravity would pull it towards the destination. During the second half of the trip, the acceleration would be in the opposite direction relative to the trajectory, but (ignoring the effects of friction) the speed acquired before would be enough to cancel this deceleration exactly (so that the train would reach its destination with speed equal to zero). The math works out that from the time entering the tunnel to the time exiting would take around 38-45 minutes. Good luck with your choo choo.", "the ride at 30 seconds, we wanted that to be at 0 seconds, so I had to move the whole curve 30 seconds to the left.", "terrorist and his mission is to trap the train in the tunnel by blowing up the exit at the instant the back of the train enters the entrance? Will he succeed or won't he? To the engine driver, at the instant the back of the train enters the tunnel, the engine is already outside it! In order to sort out the train in the tunnel paradox, we have to ask ourselves how the terrorist is going to arrange his devious trap. He could arrange for an optical signal, triggered by the passage of the last carriage into the tunnel, to blow up the other end of the tunnel - but light does not travel instantaneously and, by the time the signal has triggered the explosion, at least part of the train might already have emerged from the tunnel. What he must do is send the signal in advance. We must suppose that he knows how fast the train is going and how short it is going to be, so it is easy for him to calculate the time delay needed between the instant the front of the train enters the tunnel and the time the signal needs to be sent. Sure enough, when the time comes, the signal is sent and at the instant the back of the", "• I know but how would you describe the same sequence of events in the frame of reference of the train?? – TripleA Dec 31 '16 at 2:41 • I put that in my answer. I just edited it in bold. See \"The train observer sees the back door...\". – Zack Hutchens Dec 31 '16 at 2:43 The paradox is resolved because two clocks, each in a different reference frame, measure time at different rates relative to each other, depending on the relative velocities of each reference frame. For example, if a 50-meter-long train (as measured at rest in the tunnel) travels at 250K meters per second relative to the stationary 50-meter tunnel, then to an observer in the tunnel frame, the train's length would appear to be 50 * sqrt(1-(250K^2 / c^2)) = 27.6 meters, and a 50-meter tunnel would accommodate the train as it passes between simultaneously opening and closing doors. But to an observer on the train, it would appear that the tunnel rushes past at 250 meters per second, and he would expect that his 50-meter train would not fit into a 27.6-meter tunnel. The doors would be unable to open and close simultaneously without smashing the train. Both observers are making valid observations in their own reference frames. The key to resolving the", "miles, so if your tunnel had a max depth of 3 miles the stats would be, 440 miles surface, 308 miles direct, about 625 mph equivalent speed, 4.5 degree or 8% grade. While looking up the 747 speed, I had another thought. A 747 flies at about 6.5 miles above the Earth. That means if the plane just pointed down at say a 10% grade, and tuned off it's engines it should should be the same thing. The problem here would be air resistance. Maybe you could build a massive tube above land, although that would be really crazy. The next thing I was thinking about was that you could make it so that right as you left a tunnel you'd enter the next tunnel and go another 600 miles. By daisy chaining tunnels you could go longer distances, without going deeper than was possible. The farthest two points can be on Earth is about 12k miles, using 670 mile tunnels that's 18 tunnels. At 42 minutes each it works out to 13 hours to go anywhere on Earth. You could do NY to LA in 4 hops, assuming the train would stop for a while at each stop that'd probably work out to 4 hours. If you are using trains (and they pretty much make the most", "had^a descending floor, and you closed it up (p. 200). \"There should be between the two Governments and other Governments, a convention by which this Tunnel should not be used in time of war.\" \"Admitting the whole question, admitting that we are beaten in the Channel, admitting that an enemy has landed and got the mouth of our Tunnel, then I say I can show you some dozen different ways, either by water, or by steam, or by explosives, or by stoppage, of rendering it absolutely useless, and not only that, if your enemies come in you may kill them all before you have done with them, if you would like to do so\" (p. 200). And in reply to the question which immediately follows, \"Could you do that from a distance?\"; the answer is—\"Yes, from any distance you like.\" With regard to the time it would take to go through the Tunnel, Sir E. Watkin says:— \"I am sanguine enough to believe that we shall work through the Tunnel in less than half-an-hour,\" as against the present boat passage of an hour and a half, or an hour and 20 minutes as an exceptional thing (p. 201). \"Then another important matter is, that a man gets into his carriage at Aberdeen, or Liverpool, or London, or anywhere," ]

[ "having $\\gamma = 1$. May 2 at 9:16 Why does simultaneity affect the measurement of length? Suppose a train 990m long is heading into a tunnel 1000m long at 100m per second. You and a friend don't know the length of the train, and are asked to determine whether the train is longer or shorter than the tunnel. You realise that you can answer the question by observing whether the back of the train enters the tunnel before or after the front of the train leaves it. If it enters afterwards, the train is longer than the tunnel- if it enters before, the train is shorter than the tunnel. Accordingly, you and a friend stand at either end of the tunnel and make a note of the times when the rear of the train enters the tunnel and the front leaves. If your watches are accurately synchronised, you will find that the rear of the train enters the tunnel 0.1s before the front leaves it, so the train is shorter than the tunnel. However, had the watch of the person observing the rear of the train been just 0.2s fast, you would have concluded that the train was longer than the tunnel. Your ability to measure the moving train depends upon you being able to pin down the", "# Math Help - lenght of train? 1. ## lenght of train? A train traveling at 30 miles per hour reaches a tunnel which is 9 times as long as the train. If the train takes 2 minutes to completely clear the tunnel ,how long is the train?(1 mile=5280 feet) 2. Hello, frigidmars! Did you make a sketch? A train traveling at 30 mph reaches a tunnel which is 9 times as long as the train. If the train takes 2 minutes to completely clear the tunnel, how long is the train? Code: : - L - : - - - - -9L - - - - - : - L - : *-------*-----------------------*-------* | Train | Tunnel | Train | *-------*-----------------------*-------* A - - - - - - 10L - - - - - - - B The length of the train is $L$ feet. The length of the tunnel is $9L$ feet. . . The train is moving from left to right. The train enters the tunnel with its front at $A.$ When the train is clear of the tunnel, its front is at $B.$ . . It has traveled $10L$ feet. Its speed is: .30 miles/hour = 2640 feet/minute. Using: . $\\frac{\\text{Distance}}{\\text{Speed}} \\:=\\:\\text{Time}$, we have: . $\\frac{10L}{2640} \\:=\\:2$ . . Hence: . $10L \\:=\\:52380 \\quad\\Rightarrow\\quad", "condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's travel time to the front of the train.\" I now understand that you meant \"to the front of the tunnel\"; however that doesn't change my answer to your question. Apparently you did not read that I wrote: \"the two conditions boil down to the same. The explosion event is only conditioned by the question: at the moment that the bomb reaches the end of the tunnel, has the signal already reached that same point or not. That is also explained in the answer. If the pulse reaches the end of the train before the train reaches the end of the tunnel, then it also reaches the end of the tunnel before the train does - and the other way round.\" Good luck, Harald Last edited: Aug 17, 2011", "World Science Scholars 20.7 Coordinates for Time Problem ## Coordinates for Time—Problem 1 A train moving at a velocity $v = 0.5c$ goes through a tunnel. From the perspective of the tunnel, the Lorentz contracted length of the train equals the length of the tunnel itself. For the tunnel’s frame of reference we use coordinates $(t,x)$. Assume that the front of the train enters the tunnel at $(t = 0,x = 0)$. Moreover, assume that the length of the tunnel and the Lorentz contracted length of the train are both 300 feet. Send this to a friend", "A Roller Coaster Ride through Relativity/Simultaneity Simultaneity There is a curious paradox connected with length contraction which I shall call the train in the tunnel paradox. A train whose length when at rest is agreed by all observers to be 100 m enters a tunnel whose length is also exactly 100 m. The train stops in the tunnel and everybody agrees that when the engine is flush with the exit, the guard's van is flush with the entrance. A stationary train in a tunnel Some time later, the train passes through the same tunnel at high speed - 80% of the velocity of light in fact. At this speed the g factor is 1.67 and to the man standing by the track at the entrance to the tunnel the train appears to be only 60 m long and spends an appreciable amount of time completely inside the tunnel. A (contracted) train in a tunnel The engine driver, however, sees things rather differently. To him it is the tunnel which is whizzing past at 80% of the velocity of light and which is in consequence shrunk to 60 m in length. A train in a (contracted) tunnel Surely they can't both be right? To put it even more forcibly, suppose the man beside the track is in fact a", "in 30 units of time it will move $30 * 0.88 = 26.5$ units of distance. That means that, at time $t' = 0$ in this frame, the front of the train will be at $x' = 23.5$. Furthermore, the right end of the shunt, which was co-located with the front of the train at $t' = -30$, has also moved to the left at 0.6c, so at time $t' = 0$, it will have moved $30 * 0.6 = 18$ units of distance and will be at $x' = 32$. So the train at time $t' = 0$ consists of two segments: a segment from $x' = -50$ (the rear end) to $x' = 32$ (where the right end of the shunt is) that is at rest, and a segment from $x' = 32$ to $x' = 23.5$, which is moving to the left at 0.88c. The first segment is 82 units long, and the second has a length contracted length of 8.5, which equates to a proper length of 18 (the gamma factor is 2.125), so the train's total proper length is still 100. Finally, let's look at the way things are at $t' = 30$ in this frame. Here the front of the train has moved another 26.5 units of distance, so it is at", "# 'Bee flying between two trains' problem There is a famous arithmetic question : Two trains $150$ miles apart are traveling toward each other along the same track. The first train goes $60$ miles per hour; the second train rushes along at 90 miles per hour. A fly is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until they collide. If the fly's speed is $120$ miles per hour, how far will it travel? It is easy to determine the distance travelled by the bee. But how to determine how many times it touches first/second train? or Which train it touches last? • It's a hypothetical perfect situation. With instantaneous reversal and absolutely unvarying speeds and infintismal time instants and a bee the size of a perfect point and a universe without quantum mechanics, it's an infinite sum. The bee touches each train an infinite number of times. For any other answer somethings \"gotta\" give but it's utterly arbitrary what. – fleablood May 30 '16 at 6:32 • See mathworld.wolfram.com/TwoTrainsPuzzle.html and the story about John von Neumann at the end. – lhf May", "be $120$ miles an hour, so it travels a total of $120$ miles, because it gets 1 hour to fly before getting squashed. Also, if the bee is assumed to be a point, it touches both the trains an infinite number of times. The distance travelled can also be calculated by the sum of the infinite series. This is of course assuming that the bee takes 0 time in turning around. In a similar way, it is impossible to determine which train it last sits on, unless there is a specific amount of time taken in turning around. • what trick? just sum the infinite series. also is a specific amount of time enough information to determine which final train? – shai horowitz May 30 '16 at 7:28 • The question was: \"It is easy to determine the distance travelled by the bee. But how to determine how many times it touches first/second train? or which train it touches last?\" This answer seems to give only answers to the actual questions but I don't quite see how this answer explains the answers. – JiK May 30 '16 at 11:30 Okay. In 5/6 an hour the bee has flown 100 miles and the second train has traveled 50 miles. The bee has met the train. The bee has done", "# A logic problem, which involves math Imagine I'm walking in a tunnel and I stop, seeing that I am three eights of the way in. The tunnel is divided into eight equal length sections. All of a sudden, I hear the engine noise of a train behind me. Now, the speed of the train is unknown. To survive, I have to run out of the tunnel as quickly as possible. I have two choices, running forwards or backwards. If I run backwards in the direction of the train, I will be able to get out of the tunnel right at the moment the train comes in the tunnel. If I run forwards in the way the train is coming, the train will exit the tunnel at the same time as I do. What is my speed? - @Anupam: Pretty clearly the ground. It'd be just weird to use the train's reference frame; you might as well use your own reference frame and declare the answer is 0. – user2357112 Apr 27 '14 at 16:14 Say the train is distance $x$ from the tunnel, and moves with speed $v$. The you run $3/8$ in $x/v$, and you run $5/8$ in $(x+1)/v$. Does that help? – Chris Evans Apr 27 '14 at 16:15 You can't find an exact value", "# A train travelling at a speed of 75 mph enters a tunnel $$3\\frac{1}{2}$$ miles long. The train is $$\\frac{1}{4}$$ mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges? A. 3 min B. 4 min C. 3.7 min D. 3.2 min Solution: $\\left. \\begin{array} { l } { \\text { Total distance covered } = ( \\frac { 7 } { 2 } + \\frac { 1 } { 4 } ) \\text { miles } } \\\\ { \\qquad = \\frac { 15 } { 4 } \\text { miles. } } \\\\ { \\therefore \\text { Time taken } = ( \\frac { 15 } { 4 \\times 75 } ) \\text { hrs } } \\\\ { = \\frac { 1 } { 20 } \\text { hrs } } \\\\ { = ( \\frac { 1 } { 20 } \\times 60 ) \\text { min. } } \\\\ { = 3 \\text { min. } } \\end{array} \\right.$", "• I know but how would you describe the same sequence of events in the frame of reference of the train?? – TripleA Dec 31 '16 at 2:41 • I put that in my answer. I just edited it in bold. See \"The train observer sees the back door...\". – Zack Hutchens Dec 31 '16 at 2:43 The paradox is resolved because two clocks, each in a different reference frame, measure time at different rates relative to each other, depending on the relative velocities of each reference frame. For example, if a 50-meter-long train (as measured at rest in the tunnel) travels at 250K meters per second relative to the stationary 50-meter tunnel, then to an observer in the tunnel frame, the train's length would appear to be 50 * sqrt(1-(250K^2 / c^2)) = 27.6 meters, and a 50-meter tunnel would accommodate the train as it passes between simultaneously opening and closing doors. But to an observer on the train, it would appear that the tunnel rushes past at 250 meters per second, and he would expect that his 50-meter train would not fit into a 27.6-meter tunnel. The doors would be unable to open and close simultaneously without smashing the train. Both observers are making valid observations in their own reference frames. The key to resolving the", "the ride at 30 seconds, we wanted that to be at 0 seconds, so I had to move the whole curve 30 seconds to the left.", "a train tunnel on her way to town. Suddenly, she heard the whistle of a train approaching from behind her! Pauline knew that the train always traveled at an even 60 miles per hour. She also knew that she was exactly three-eighths of the way through the tunnel, and she could tell from the train whistle how far the train was from the tunnel. Pauline wasn't sure if she should run forward as fast as she could, or run back to the near end of the tunnel. Well, she did some lightning-fast calculations, based on how fast she could run and the length of the tunnel. She figured out that whichever way she ran, she would just barely make it out of the tunnel before the train reached her. Whew!! How fast could Pauline run? Carefully explain how you found your answer. 5. Groups could use whiteboards to solve different versions of one problem so that the class can then discuss how they are related to one another. For example, one group could solve and graph the solution set for each of the following: a. 2x – 3 = 13 b. 2x – 3 > 13 c. 2x – 3 < 13 d. |2x – 3| = 13 e. |2x – 3| > 13 f. |2x – 3|", "$x' = -3$. The right end of the shunt has moved another 18 units of distance, so it is at $x' = 14$. The train is thus divided into two segments again, the first from $x' = -50$ to $x' = 14$, which is at rest, and the second from $x' = 14$ to $x' = -3$, which is moving to the left at 0.88c. The first segment is 64 units long, and the second has a length contracted length of 17, which equates to a proper length of 36, for a total proper length of 100, once again. Notice that at this point in the right-moving train frame, the center of the train still has not reached the right end of the shunt, so we are still at an earlier time than your Figure 3, which we discuss next. Now let's switch to your Figure 3 and label the key events there and give their coordinates in the ground frame. All of these events happen at time $t = 40 / 0.6 = 66.67$, since that is how long it takes for the front end of the train, at 0.6c, to cover the distance 40 from the left end of the shunt to the right end. So the events and their coordinates are: Event R2: The rear", "A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. What is the ratio of speed of train and cat ? a) 3 : 1 b) 4 :1 c) 5 : 1 d) None of these Question 9: City Bus Corporation runs two buses from terminus A to terminus B,one from each of the terminuses such that each bus makes 5 round trips in a day. There are no stops in between. These buses ply back and forth on the same route at different but uniform speeds. Each morning the buses start at 7 AM from the respective terminuses. They meet for the first time at a distance of 7 km from terminus A. Their next meeting is at a distance of 4 km from terminus B, while travelling in opposite directions. Assuming that the time taken by the buses at the terminuses is negligibly small, and the cost of running a bus is 20 per", "multiplication: $$\\frac{31}{4}$$ × $$\\frac{4}{3}$$ = $$\\frac{31}{3}$$ = 10$$\\frac{1}{3}$$ The train will 10$$\\frac{1}{3}$$ minutes to pass the tunnel. Gridded Response Question 8. What is the value of (-2.75 )(-1 .16)?", "units of distance. That means that, at time $t' = 0$ in this frame, the front of the train will be at $x' = 23.5$. Furthermore, the right end of the shunt, which was co-located with the front of the train at $t' = -30$, has also moved to the left at 0.6c, so at time $t' = 0$, it will have moved $30 * 0.6 = 18$ units of distance and will be at $x' = 32$. So the train at time $t' = 0$ consists of two segments: a segment from $x' = -50$ (the rear end) to $x' = 32$ (where the right end of the shunt is) that is at rest, and a segment from $x' = 32$ to $x' = 23.5$, which is moving to the left at 0.88c. The first segment is 82 units long, and the second has a length contracted length of 8.5, which equates to a proper length of 18 (the gamma factor is 2.125), so the train's total proper length is still 100. Finally, let's look at the way things are at $t' = 30$ in this frame. Here the front of the train has moved another 26.5 units of distance, so it is at $x' = -3$. The right end of the shunt has moved another 18", "Assume that a tunnel is dug across the earth (radius=R) passing through its center. Question: Assume that a tunnel is dug across the earth (radius=R) passing through its center. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of $\\sqrt{g R}$ (b) it is released from a height $\\mathrm{R}$ above the tunnel (c) it is thrown vertically upward along the length of tunnel with a speed of $\\sqrt{g R}$. Solution:", "to make an extremely sudden stop in front of a huge ice cliff right before hitting it. The staff engineers' simulation shows the train will continue to proceed $x=27t-0.45t^2$ meters for t seconds after the stuntman steps on the brake. How far away in meters from the ice cliff must the stuntman step on the brake so that the train can stop as desired? The area of a circle is increasing at the rate of $54\\pi \\text{ in}^2\\text{/sec}.$ When the radius of the circle is $3 \\text{ in},$ what is the rate of increase of the radius? ×", "t') = (50, -30)$. Event R1: $(x', t') = (-50, 30)$. Note that these $t'$ values are offset by 30 from the \"W\" times in your Figure 2. This is for convenience. We can already see that, if we are talking about \"how things are\" in the right-moving train frame, we have to adjust for the relativity of simultaneity; the above three events happen at different times in this frame. First, let's take the easy adjustment: at time $t' = -30$ in this frame, the length of the train is 100--because the center and rear of the train are at rest in this frame between $t' = -30$ and the times of the events above in this frame, so their $x'$ coordinates don't change. That means that at $t' = -30$ in the train frame, the center of the train is at $x' = 0$ and the rear of the train is at $x' = -50$, so the train's length is 100. So far, so good. Now let's look at the way things are at time $t' = 0$ in the right-moving train frame. The front of the train has been moving to the left, in this frame, for 30 units of time. Its speed is $(0.6 + 0.6) / (1 + 0.6 * 0.6) \\approx 0.88$, so" ]

[ "having $\\gamma = 1$. May 2 at 9:16 Why does simultaneity affect the measurement of length? Suppose a train 990m long is heading into a tunnel 1000m long at 100m per second. You and a friend don't know the length of the train, and are asked to determine whether the train is longer or shorter than the tunnel. You realise that you can answer the question by observing whether the back of the train enters the tunnel before or after the front of the train leaves it. If it enters afterwards, the train is longer than the tunnel- if it enters before, the train is shorter than the tunnel. Accordingly, you and a friend stand at either end of the tunnel and make a note of the times when the rear of the train enters the tunnel and the front leaves. If your watches are accurately synchronised, you will find that the rear of the train enters the tunnel 0.1s before the front leaves it, so the train is shorter than the tunnel. However, had the watch of the person observing the rear of the train been just 0.2s fast, you would have concluded that the train was longer than the tunnel. Your ability to measure the moving train depends upon you being able to pin down the", "at 5:42 • In case 1, I believe your initial calculation is correct, but when you turn back after 9 hours for the 10th tunnel, won't this change the assumption that 5 tunnels must be traveled on average to escape? – knrumsey May 11 '17 at 12:35", "condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's travel time to the front of the train.\" I now understand that you meant \"to the front of the tunnel\"; however that doesn't change my answer to your question. Apparently you did not read that I wrote: \"the two conditions boil down to the same. The explosion event is only conditioned by the question: at the moment that the bomb reaches the end of the tunnel, has the signal already reached that same point or not. That is also explained in the answer. If the pulse reaches the end of the train before the train reaches the end of the tunnel, then it also reaches the end of the tunnel before the train does - and the other way round.\" Good luck, Harald Last edited: Aug 17, 2011", "# Math Help - lenght of train? 1. ## lenght of train? A train traveling at 30 miles per hour reaches a tunnel which is 9 times as long as the train. If the train takes 2 minutes to completely clear the tunnel ,how long is the train?(1 mile=5280 feet) 2. Hello, frigidmars! Did you make a sketch? A train traveling at 30 mph reaches a tunnel which is 9 times as long as the train. If the train takes 2 minutes to completely clear the tunnel, how long is the train? Code: : - L - : - - - - -9L - - - - - : - L - : *-------*-----------------------*-------* | Train | Tunnel | Train | *-------*-----------------------*-------* A - - - - - - 10L - - - - - - - B The length of the train is $L$ feet. The length of the tunnel is $9L$ feet. . . The train is moving from left to right. The train enters the tunnel with its front at $A.$ When the train is clear of the tunnel, its front is at $B.$ . . It has traveled $10L$ feet. Its speed is: .30 miles/hour = 2640 feet/minute. Using: . $\\frac{\\text{Distance}}{\\text{Speed}} \\:=\\:\\text{Time}$, we have: . $\\frac{10L}{2640} \\:=\\:2$ . . Hence: . $10L \\:=\\:52380 \\quad\\Rightarrow\\quad", "A Roller Coaster Ride through Relativity/Simultaneity Simultaneity There is a curious paradox connected with length contraction which I shall call the train in the tunnel paradox. A train whose length when at rest is agreed by all observers to be 100 m enters a tunnel whose length is also exactly 100 m. The train stops in the tunnel and everybody agrees that when the engine is flush with the exit, the guard's van is flush with the entrance. A stationary train in a tunnel Some time later, the train passes through the same tunnel at high speed - 80% of the velocity of light in fact. At this speed the g factor is 1.67 and to the man standing by the track at the entrance to the tunnel the train appears to be only 60 m long and spends an appreciable amount of time completely inside the tunnel. A (contracted) train in a tunnel The engine driver, however, sees things rather differently. To him it is the tunnel which is whizzing past at 80% of the velocity of light and which is in consequence shrunk to 60 m in length. A train in a (contracted) tunnel Surely they can't both be right? To put it even more forcibly, suppose the man beside the track is in fact a", "• I know but how would you describe the same sequence of events in the frame of reference of the train?? – TripleA Dec 31 '16 at 2:41 • I put that in my answer. I just edited it in bold. See \"The train observer sees the back door...\". – Zack Hutchens Dec 31 '16 at 2:43 The paradox is resolved because two clocks, each in a different reference frame, measure time at different rates relative to each other, depending on the relative velocities of each reference frame. For example, if a 50-meter-long train (as measured at rest in the tunnel) travels at 250K meters per second relative to the stationary 50-meter tunnel, then to an observer in the tunnel frame, the train's length would appear to be 50 * sqrt(1-(250K^2 / c^2)) = 27.6 meters, and a 50-meter tunnel would accommodate the train as it passes between simultaneously opening and closing doors. But to an observer on the train, it would appear that the tunnel rushes past at 250 meters per second, and he would expect that his 50-meter train would not fit into a 27.6-meter tunnel. The doors would be unable to open and close simultaneously without smashing the train. Both observers are making valid observations in their own reference frames. The key to resolving the", "train of length $L,$ approaching a tunnel of length $L\\sqrt{1-{v}^{2}/{c}^{2}},$ so the tunnel is not as long as the train from his viewpoint! What does he think happens when the bandits close both the doors? ## The Tunnel Doors are Closed Simultaneously The key to understanding what is happening here is that we said the bandits closed the two doors at the ends of the tunnel at the same time. How could they arrange to do that, since the doors are far apart? They could use walkie-talkies, which transmit radio waves, or just flash a light down the tunnel, since it’s long and straight. Remember, though, that the train is itself going at a speed close to that of light, so they have to be quite precise about this timing! The simplest way to imaging them synchronizing the closings of the two doors is to assume they know the train’s timetable, and at a prearranged appropriate time, a light is flashed halfway down the tunnel, and the end doors are closed when the flash of light reaches the ends of the tunnel. Assuming the light was positioned correctly in the middle of the tunnel, that should ensure that the two doors close simultaneously. ## Or are they? Now consider this door-closing operation from the point of view of", "Assume that a tunnel is dug across the earth (radius=R) passing through its center. Question: Assume that a tunnel is dug across the earth (radius=R) passing through its center. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of $\\sqrt{g R}$ (b) it is released from a height $\\mathrm{R}$ above the tunnel (c) it is thrown vertically upward along the length of tunnel with a speed of $\\sqrt{g R}$. Solution:", "a train tunnel on her way to town. Suddenly, she heard the whistle of a train approaching from behind her! Pauline knew that the train always traveled at an even 60 miles per hour. She also knew that she was exactly three-eighths of the way through the tunnel, and she could tell from the train whistle how far the train was from the tunnel. Pauline wasn't sure if she should run forward as fast as she could, or run back to the near end of the tunnel. Well, she did some lightning-fast calculations, based on how fast she could run and the length of the tunnel. She figured out that whichever way she ran, she would just barely make it out of the tunnel before the train reached her. Whew!! How fast could Pauline run? Carefully explain how you found your answer. 5. Groups could use whiteboards to solve different versions of one problem so that the class can then discuss how they are related to one another. For example, one group could solve and graph the solution set for each of the following: a. 2x – 3 = 13 b. 2x – 3 > 13 c. 2x – 3 < 13 d. |2x – 3| = 13 e. |2x – 3| > 13 f. |2x – 3|", "A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. What is the ratio of speed of train and cat ? a) 3 : 1 b) 4 :1 c) 5 : 1 d) None of these Question 9: City Bus Corporation runs two buses from terminus A to terminus B,one from each of the terminuses such that each bus makes 5 round trips in a day. There are no stops in between. These buses ply back and forth on the same route at different but uniform speeds. Each morning the buses start at 7 AM from the respective terminuses. They meet for the first time at a distance of 7 km from terminus A. Their next meeting is at a distance of 4 km from terminus B, while travelling in opposite directions. Assuming that the time taken by the buses at the terminuses is negligibly small, and the cost of running a bus is 20 per", "World Science Scholars 20.7 Coordinates for Time Problem ## Coordinates for Time—Problem 1 A train moving at a velocity $v = 0.5c$ goes through a tunnel. From the perspective of the tunnel, the Lorentz contracted length of the train equals the length of the tunnel itself. For the tunnel’s frame of reference we use coordinates $(t,x)$. Assume that the front of the train enters the tunnel at $(t = 0,x = 0)$. Moreover, assume that the length of the tunnel and the Lorentz contracted length of the train are both 300 feet. Send this to a friend", "# A train travelling at a speed of 75 mph enters a tunnel $$3\\frac{1}{2}$$ miles long. The train is $$\\frac{1}{4}$$ mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges? A. 3 min B. 4 min C. 3.7 min D. 3.2 min Solution: $\\left. \\begin{array} { l } { \\text { Total distance covered } = ( \\frac { 7 } { 2 } + \\frac { 1 } { 4 } ) \\text { miles } } \\\\ { \\qquad = \\frac { 15 } { 4 } \\text { miles. } } \\\\ { \\therefore \\text { Time taken } = ( \\frac { 15 } { 4 \\times 75 } ) \\text { hrs } } \\\\ { = \\frac { 1 } { 20 } \\text { hrs } } \\\\ { = ( \\frac { 1 } { 20 } \\times 60 ) \\text { min. } } \\\\ { = 3 \\text { min. } } \\end{array} \\right.$", "terrorist and his mission is to trap the train in the tunnel by blowing up the exit at the instant the back of the train enters the entrance? Will he succeed or won't he? To the engine driver, at the instant the back of the train enters the tunnel, the engine is already outside it! In order to sort out the train in the tunnel paradox, we have to ask ourselves how the terrorist is going to arrange his devious trap. He could arrange for an optical signal, triggered by the passage of the last carriage into the tunnel, to blow up the other end of the tunnel - but light does not travel instantaneously and, by the time the signal has triggered the explosion, at least part of the train might already have emerged from the tunnel. What he must do is send the signal in advance. We must suppose that he knows how fast the train is going and how short it is going to be, so it is easy for him to calculate the time delay needed between the instant the front of the train enters the tunnel and the time the signal needs to be sent. Sure enough, when the time comes, the signal is sent and at the instant the back of the", "the tunnel gets a light signal emitted at A. Both observers will agree that event A happens before event C: the two events are light-like separated. Moreover, both observers will agree on the relative order of B and C, since they are timelike separated. But whether or not B happens before C or vice versa is going to depend on the relative rest lengths of the train and the tunnel, along with the speed at which the train is traveling relative to the tunnel. If the train's length in the tunnel frame is longer than this, it gets chopped in half; if the train's length is shorter, it's trapped inside the tunnel for some period of time (according to the tunnel observer.) If you want to have $v = 0.5 c$, then it's not too hard to see that the events B & C will be simultaneous when the train's length in the tunnel frame is exactly 1/2 the length of the tunnel; this then translates to the train's rest length being about 57.7% ($=\\gamma/2$) the length of the tunnel.", "of time to traverse if we go in the opposite direction). To work out what the average escape time is, we notice that if we pick an escape tunnel then we’re free in the time it takes to traverse the tunnel. If we pick a return tunnel, then we’ve increased our expected time to leave by the time that it took to traverse the tunnel. If the time to leave is T, then we write the expected leaving time as E[T], and $E[T] = \\sum_{i=1}^m p_i\\tau_i + \\sum_{j=m+1}^n p_j(\\tau_j + E[T])$. After a couple of lines of algebra we see that the expected time to leave is given by $E[T] = (\\sum_{i=1}^n p_i\\tau_i) / (1 - \\sum_{j=m+1}^n p_j)$ or, written in everyday language: Avg time to leave = (Avg tunnel length) ÷ (Chance of picking an exit tunnel) That’s it! The average time to escape only depends on the average time it takes to traverse a randomly picked tunnel, and the probability of picking a tunnel which leads to an exit. It’s completely insensitive to the specifics of the problem, namely how long the individual tunnels are, and what the precise probabilities of picking each tunnel are. In the special case where each of the tunnels is equally likely, all you have to do is add up the", "An Oncoming Train) 2:32 21 Dic 2005, 10:54 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 21 Dic 2005, 10:45 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 20 Dic 2005, 19:00 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 20 Dic 2005, 11:04 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 19 Dic 2005, 17:06 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 19 Dic 2005, 12:20 The Light At The End Of The Tunnel (Is The Light Of An Oncoming Train) 2:32 16 Dic 2005, 18:01", "# The Minotaur's Labyrinth You are trapped in a chamber in the center of the Minotaur's Labyrinth. There are $\\mathbf{N}$ tunnels, $\\mathbf{m}$ of which lead to safety; the remaining tunnels only lead back to the chamber. Each tunnel is of a different length, taking $h_i$ hours to travel. Each time you return to the chamber, the room shifts so that you can only choose tunnels at random. 1. What is the expected amount of time it will take you to escape? 2. You have 24 hours until the Minotaur wakes up. If there are 10 tunnels, such that $h_i = i$ (for $i$ = 1,2,...10) and two of the tunnels lead to safety do you believe you will escape in time? • Can I turn around once it becomes clear I'm in a tunnel I'd rather not follow to the end? If I know the distribution of tunnel lengths and some are more than twice the lengths of others, that could be a useful strategy, which complicates the math. If the tunnels are [100,1], I'll want to follow a tunnel for an hour, then backtrack if it turns out it's the long one, then repeat until I've followed the short one. For [3,1], I'd want to backtrack after an hour if I've already followed the long tunnel once", "had^a descending floor, and you closed it up (p. 200). \"There should be between the two Governments and other Governments, a convention by which this Tunnel should not be used in time of war.\" \"Admitting the whole question, admitting that we are beaten in the Channel, admitting that an enemy has landed and got the mouth of our Tunnel, then I say I can show you some dozen different ways, either by water, or by steam, or by explosives, or by stoppage, of rendering it absolutely useless, and not only that, if your enemies come in you may kill them all before you have done with them, if you would like to do so\" (p. 200). And in reply to the question which immediately follows, \"Could you do that from a distance?\"; the answer is—\"Yes, from any distance you like.\" With regard to the time it would take to go through the Tunnel, Sir E. Watkin says:— \"I am sanguine enough to believe that we shall work through the Tunnel in less than half-an-hour,\" as against the present boat passage of an hour and a half, or an hour and 20 minutes as an exceptional thing (p. 201). \"Then another important matter is, that a man gets into his carriage at Aberdeen, or Liverpool, or London, or anywhere,", "you can do although some tunnels are made in more advanced ways than in others. For a few years, I watched the construction of the Ejpovice Tunnel, what is now the longest railroad tunnel in Czechia. It starts one mile from my home and each of the two tunnels (Southern and Northern) is 4.1 kilometers long. This tunnel makes some sense because the trains from Pilsen to Prague used to take a detour. With the tunnel, the train trip from Prague to Pilsen or back is shortened by some 10-20 minutes, depending on the stage of the speed limits in and near the tunnel. For the first time, trains between Prague and Pilsen may be faster than cars (The Western Express makes it in 79 minutes between the centers of the two cities). And that's an improvement worth those $250 million, I finally decided. Well, Ladies and Gentlemen, without much science-fiction hype, the tunnel was opened. First, a month ago, they opened one of the two tunnels, and at the limited 50 kilometers per hour low speed due to security reasons, see the video of going from Pilsen through the tunnel. Both tunnels are already operating now at normal speeds (above 100 kilometers per hour) for trains. And within a year or two, 200 kilometers per hour (the", "descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m. Answer: Total distance covered = 10 – ( - 350) = 360 m Time taken to cover the distance = 360 ÷ 6 = 60 min = 1 hour" ]

[ "is the least common multiple of the $n_i$. -", "square units 42.) What is the least common multiple of 168 and 420? Continue reading", "the least common multiple of $180$, $216$, and $450$?", "the smallest multiple of 63 and 147", "136x=204y→ x/y=204/136=3/2 test 3/2 ratio against options: 3*136=408 2*204=408 A Which of the following is the least common multiple of 136 and 204? [#permalink] 30 Mar 2020, 10:49 # Which of the following is the least common multiple of 136 and 204? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne", "# 92nd Problem 2016 How many multiples of 13 are included between 12548 and 19987 both inclusive? ×", "1344/D is the smallest among all. We can say that fraction 4/5 is the smallest among all the given fractions.", "of 991. If you like to learn what is a multiple", "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "least common multiple of several random integers. 2019. 〈hal-01984389〉 ### Métriques Consultations de la notice ## 33 Téléchargements de fichiers", "• multiples", "so x = 125 is the fewest?", "by taking the least common multiple of 2, 9, and 25. So the least four-digit solution is . Dec 24, 2022", "multiples of 17 17 ,34 , 51 , 68 , 85 , 102 , 119 ,136 So the odd multiples are 17 ,51, 119", "x2 Introduction = # Least Common Multiples In a way, least common multiples are the opposite of greatest common divisors. Where divisors are always smaller than their parent numbers, multiples are always greater.", "lowest common multiple of two numbers is the lowest number that is a multiple of both numbers.", "# Least common multiple ### Problem What is the least common multiple of 7 and 5? Another way to say this is:", "Eric W. \"Least Common Multiple\". mathworld.wolfram.com. Retrieved 2020-08-30. 3. \"Least Common Multiple\". www.mathsisfun.com. Retrieved 2020-08-30.", "# How many multiple? Number Theory Level 1 How many multiples of 7 are between 345 and 563 inclusive? ×", "multiples of those numbers." ]

[ "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", "least common multiple of $15$ and $20.$ Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple. Find the least common multiple (LCM) of the given numbers: $9\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}12$ 36 Find the least common multiple (LCM) of the given numbers: $18\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}24$ 72 ## Prime factors method Another way to find the least common multiple of two numbers is to use 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. $12=2\\cdot 2\\cdot 3\\phantom{\\rule{3em}{0ex}}18=2\\cdot 3\\cdot 3$ Then we write each number as a product of primes, matching primes vertically when possible. $\\begin{array}{l}12=2\\cdot 2\\cdot 3\\hfill \\\\ 18=2\\cdot \\phantom{\\rule{1.1em}{0ex}}3\\cdot 3\\end{array}$ 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. do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid", "# Chapter 1 - Some Basic Concepts of Arithmetic and Algebra - 1.2 - Prime and Composite Numbers - Problem Set 1.2 - Page 13: 80 $462$ #### Work Step by Step 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: $=2 \\cdot 3\\cdot 7 \\cdot 11 \\\\= 462$ 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.", "two squares and if so, find a representation for $1034$. The prime power factorization of $1034$ is: (2) \\begin{align} \\quad 1034 = 2^1 \\cdot 11^1 \\cdot 47^1 \\end{align} Note that $11 \\equiv 3 \\pmod 4$ but $1$ is not even, so again, $1034$ cannot be expressed as a sum of two squares. ## Example 3 Determine if $98765$ can be expressed as a sum of two squares and if so, find a representation for $98765$. The prime power factorization of $98765$ is: (3) \\begin{align} \\quad 98765 = 5^1 \\cdot 19753^1 \\end{align} Since $5 \\equiv 1 \\pmod 4$ and $19753 \\equiv 1 \\pmod 4$ we have that $98765$ can be written as a sum of two squares. Namely: (4) \\begin{align} \\quad 98765 = 13^2 + 314^2 \\end{align}", "$n$ 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!", "# What is the least common multiple of 32, 15, and 36? May 10, 2016 $\\text{LCM} \\left(32 , 15 , 36\\right) = 1440$ #### Explanation: To find the least common multiple, we can look at the prime factorization of all of our given values, and then take the power of the highest power used of each prime. By doing so, we ensure that the prime factorization of our LCM contains the prime factorizations of all initial values, meaning it is divisible by each. $32 = {2}^{5}$ $15 = 3 \\cdot 5$ $36 = {2}^{2} \\cdot {3}^{2}$ The highest power of $2$ used is $5$. The highest power of $3$ used is $2$. The highest power of $5$ used is $1$. No other primes were used. By the above, then, the LCM will be ${2}^{5} \\cdot {3}^{2} \\cdot 5 = 1440$", "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)", "Free Version Easy # Factor $113400$ as a Product of Prime Powers NUMTH-H9S91G What is the prime power factorization of $113400$? A $2^4 \\cdot 3 \\cdot 5^3 \\cdot 11$ B $2^3 \\cdot 3^4 \\cdot 5^2 \\cdot 7$ C $2^2 \\cdot 3^5 \\cdot 5^3 \\cdot 7$ D $2^3 \\cdot 3^4 \\cdot 5^2 \\cdot 11$", "List of Prime Factorizations of Integers from 801 to 1,000 I made it easier for you by listing below the prime factorizations of integers from 801 to 1,000. If a prime number repeats in the factorization, it is written as an exponential number for the sake of compactness. • $801 = {3^2} \\cdot 89$ • $802 = 2 \\cdot 401$ • $803 = 11 \\cdot 73$ • $804 = {2^2} \\cdot 3 \\cdot 67$ • $805 = 5 \\cdot 7 \\cdot 23$ • $806 = 2 \\cdot 13 \\cdot 31$ • $807 = 3 \\cdot 269$ • $808 = {2^3} \\cdot 101$ • $809$ is prime. • $810 = 2 \\cdot {3^4} \\cdot 5$ • $811$ is prime. • $812 = {2^2} \\cdot 7 \\cdot 29$ • $813 = 3 \\cdot 271$ • $814 = 2 \\cdot 11 \\cdot 37$ • $815 = 5 \\cdot 163$ • $816 = {2^4} \\cdot 3 \\cdot 17$ • $817 = 19 \\cdot 43$ • $818 = 2 \\cdot 409$ • $819 = {3^2} \\cdot 7 \\cdot 13$ • $820 = {2^2} \\cdot 5 \\cdot 41$ • $821$ is prime. • $822 = 2 \\cdot 3 \\cdot 137$ • $823$ is prime. • $824 = {2^3} \\cdot 103$ • $825 = 3 \\cdot {5^2} \\cdot 11$ • $826 = 2 \\cdot 7", "# 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", "# Lowest Common Multiple (LCM) #### Casio ##### Member I did actually think I had this off to a T sort of speak, but seems I have run into a problem and am now unsure? Find the LCM of 12 and 56. This is how I do them but if incorrect I would appreciate the correct notation being pointed out to me 12/2 = 6 My prime factors of 12 are therefore 2, 6 56/2 = 28, and 28/2 = 14, and 14/2 = 7 my prime factors are; 2, 6, 7, 14 My common factors being 2 my lowest common multiple being = 1176? 12/4 = 3, therefore 3,4 56/14 = 4 therefore 4,14 common factors are 4 Lowest common multiple is 3 x 4 x 14 = 168 so I am assuming there is a specific notation (method) to finding the LCM's without mistakes? #### Reckoner ##### Member my prime factors are; 2, 6, 7, 14 6 and 14 cannot be prime factors, for they are not prime! $$12 = 2\\cdot6 = 2\\cdot2\\cdot3 = 2^2\\cdot3$$ $$56 = 2\\cdot28 = 2\\cdot2\\cdot14=2\\cdot2\\cdot2\\cdot7=2^3\\cdot7$$ Now we take the highest power of each factor. Our least common multiple is therefore $$\\mathrm{LCM}(12,56) = 2^3\\cdot3\\cdot7 = 168$$. #### Casio ##### Member 6 and 14 cannot be prime factors, for they are not prime!", "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", "$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", "# List of Prime Factorizations of Integers from 201 to 400 Below is the list of prime factorizations of integers from 201 to 400. Prime factors that repeat are written as exponential expressions. • $201 = 3 \\cdot 67$ • $202 = 2 \\cdot 101$ • $203 = 7 \\cdot 29$ • $204 = {2^2} \\cdot 3 \\cdot 17$ • $205 = 5 \\cdot 41$ • $206 = 2 \\cdot 103$ • $207 = {3^2} \\cdot 23$ • $208 = {2^4} \\cdot 13$ • $209 = 11 \\cdot 19$ • $210 = 2 \\cdot 3 \\cdot 5 \\cdot 7$ • $211$ is prime. • $212 = {2^2} \\cdot 53$ • $213 = 3 \\cdot 71$ • $214 = 2 \\cdot 107$ • $215 = 5 \\cdot 43$ • $216 = {2^3} \\cdot {3^3}$ • $217 = 7 \\cdot 31$ • $218 = 2 \\cdot 109$ • $219 = 3 \\cdot 73$ • $220 = {2^2} \\cdot 5 \\cdot 11$ • $221 = 13 \\cdot 17$ • $222 = 2 \\cdot 3 \\cdot 37$ • $223$ is prime. • $224 = {2^5} \\cdot 7$ • $225 = {3^2} \\cdot {5^2}$ • $226 = 2 \\cdot 113$ • $227$ is prime. • $228 = {2^2} \\cdot 3 \\cdot 19$ • $229$ is prime. • $230 = 2 \\cdot 5 \\cdot", "$$12 = 2\\cdot6 = 2\\cdot2\\cdot3 = 2^2\\cdot3$$ $$56 = 2\\cdot28 = 2\\cdot2\\cdot14=2\\cdot2\\cdot2\\cdot7=2^3\\cdot7$$ Now we take the highest power of each factor. Our least common multiple is therefore $$\\mathrm{LCM}(12,56) = 2^3\\cdot3\\cdot7 = 168$$. Thanks for point that out to me I completely missed that point, i.e. 14 is not a prime. So should I take it then that the correct way to prime factorise intergers is to divide them always by prime numbers? 90 / 2 = 45, then 45 / 3 = 15, then 15 / 5 = 3 #### Reckoner ##### Member So should I take it then that the correct way to prime factorise intergers is to divide them always by prime numbers? 90 / 2 = 45, then 45 / 3 = 15, then 15 / 5 = 3 You can factor the integers however you like. The key is that you don't stop factoring until all the factors are prime. When you get to that point, you will have found the prime factorization of the integer, which is what we want. So we could factor 90 as $$90 = 9\\cdot10$$, but we have to keep going, because 9 and 10 are not prime: $$90 = 9\\cdot10 = (3\\cdot3)\\cdot(2\\cdot5) = 2\\cdot3^2\\cdot5$$. We stop here because 2, 3, and 5 are all prime. Of course, there", "and $294$294, we start by drawing their factor trees: This gives us their prime factorisations 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 factorisations - they have one $2$2, one $3$3, and one $7$7 in common. The highest common factor is the product of the common prime factors, $2\\times3\\times7=42$2×3×7=42. ### Lowest 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 lowest common multiple (LCM) of $12$12", "prime and continue this process until your number is completely factored into primes. For example, let’s determine the prime factorization of $336$: $336 = 2 \\cdot 168 = 2^2 \\cdot 84 = 2^3 \\cdot 42 = 2^4 \\cdot 21 = 2^4 \\cdot 3 \\cdot 7$ Thus, we’d say the prime factorization of $336$ is $2^4 \\cdot 3 \\cdot 7$. Note, the rule that we must repeatedly divide by the smallest possible prime is not hard and fast, since multiplication can be done in any order. There is a common trick we use when a number is a multiple of $10$ – if a number is a multiple of $10$, it is a multiple of both $2$ and $5$. For example: $3600 = 2 \\cdot 5 \\cdot 360 = 2^2 \\cdot 5^2 \\cdot 36 = 2^2 \\cdot 5^2 \\cdot 6^2 = 2^2 \\cdot 5^2 \\cdot (2 \\cdot 3)^2 = 2^4 \\cdot 3^2 \\cdot 5^2$ Note: It is recommended to consult the lecture video for this exercise, as it is significantly easier to learn by watching someone do it first. ### Proof: Infinitely Many Primes Let’s proceed by contradiction. Let’s assume that there are only finitely many primes, say, $p_1, p_2, p_3, … , p_{n-1}, p_n$. Now, consider $q = p_1p_2p_3 … p_{n - 1}p_n + 1$. None of the", "$38052$. • Because the last two digits of the given number are a multiple of $4$, we can rewrite $38052$ as $2^2 \\cdot 9513$. Note that $9513$ is not even, so we move on. • Because the sum of digits of $9513$ is a multiple of 9, we can rewrite $38052$ as $2^2 \\cdot 3^2 \\cdot 1057$. Note that the sum of digits of $1057$ is not a multiple of 3, so we move on. • Because $1057$ does not end in $0$ or $5$, we skip $5$ as a factor. As for $7$, after long division, we note that $1057 = 7 \\cdot 151$, so we can rewrite $38052$ as $2^2 \\cdot 3^2 \\cdot 7 \\cdot 151$. However, when doing long division again, $151$ is not a multiple of $7$, so we move on. • Note that $11$ is not a factor of $151$, so we can move on. In fact, because $13 > \\sqrt{151}$, we can declare $151$ as prime and stop. Thus, the prime factorization of $38052$ is $2^2 \\cdot 3^2 \\cdot 7 \\cdot 151$.", "## 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: $=2 \\cdot 3\\cdot 7 \\cdot 11 \\\\= 462$", "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" ]

[ "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", "# 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 $3 \\times 4 \\times 7 = 84$.", "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", "# What is the least common multiple of 32, 15, and 36? May 10, 2016 $\\text{LCM} \\left(32 , 15 , 36\\right) = 1440$ #### Explanation: To find the least common multiple, we can look at the prime factorization of all of our given values, and then take the power of the highest power used of each prime. By doing so, we ensure that the prime factorization of our LCM contains the prime factorizations of all initial values, meaning it is divisible by each. $32 = {2}^{5}$ $15 = 3 \\cdot 5$ $36 = {2}^{2} \\cdot {3}^{2}$ The highest power of $2$ used is $5$. The highest power of $3$ used is $2$. The highest power of $5$ used is $1$. No other primes were used. By the above, then, the LCM will be ${2}^{5} \\cdot {3}^{2} \\cdot 5 = 1440$", "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", "$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", "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: \\begin{align*}234 & = 2 \\cdot 117 \\\\ & = 2 \\cdot 3 \\cdot 39 \\\\ & = 2 \\cdot 3 \\cdot 3 \\cdot 13 \\\\ 234 & = 2 \\cdot 3^2 \\cdot 13\\end{align*} \\begin{align*}126 & = 2 \\cdot 63 \\\\ & = 2 \\cdot 3 \\cdot 21 \\\\ & = 2 \\cdot 3 \\cdot 3 \\cdot 7 \\\\ 126 & = 2 \\cdot 3^2 \\cdot 7\\end{align*} \\begin{align*}273 & = 3 \\cdot 91 \\\\ & = 3 \\cdot 7 \\cdot 13\\end{align*} 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, \\begin{align*}\\text{LCM} = 2", "# Lowest Common Multiple (LCM) #### Casio ##### Member I did actually think I had this off to a T sort of speak, but seems I have run into a problem and am now unsure? Find the LCM of 12 and 56. This is how I do them but if incorrect I would appreciate the correct notation being pointed out to me 12/2 = 6 My prime factors of 12 are therefore 2, 6 56/2 = 28, and 28/2 = 14, and 14/2 = 7 my prime factors are; 2, 6, 7, 14 My common factors being 2 my lowest common multiple being = 1176? 12/4 = 3, therefore 3,4 56/14 = 4 therefore 4,14 common factors are 4 Lowest common multiple is 3 x 4 x 14 = 168 so I am assuming there is a specific notation (method) to finding the LCM's without mistakes? #### Reckoner ##### Member my prime factors are; 2, 6, 7, 14 6 and 14 cannot be prime factors, for they are not prime! $$12 = 2\\cdot6 = 2\\cdot2\\cdot3 = 2^2\\cdot3$$ $$56 = 2\\cdot28 = 2\\cdot2\\cdot14=2\\cdot2\\cdot2\\cdot7=2^3\\cdot7$$ Now we take the highest power of each factor. Our least common multiple is therefore $$\\mathrm{LCM}(12,56) = 2^3\\cdot3\\cdot7 = 168$$. #### Casio ##### Member 6 and 14 cannot be prime factors, for they are not prime!", "and $294$294, we start by drawing their factor trees: This gives us their prime factorisations 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 factorisations - they have one $2$2, one $3$3, and one $7$7 in common. The highest common factor is the product of the common prime factors, $2\\times3\\times7=42$2×3×7=42. ### Lowest 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 lowest common multiple (LCM) of $12$12", "$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!", "134 (number) 134 Cardinalone hundred [and] thirty-four Ordinal134th (one hundred [and] thirty-fourth) Factorization$2 \\cdot 67$ Divisors1, 2, 67, 134 Roman numeralCXXXIV Binary100001102 Ternary112223 Quaternary20124 Octal2068 DuodecimalB212 Vigesimal6E20 Base 363Q36 134 Cardinalone hundred [and] thirty-four Ordinal134th (one hundred [and] thirty-fourth) Factorization$2 \\cdot 67$ Divisors1, 2, 67, 134 Roman numeralCXXXIV Binary100001102 Ternary112223 Quaternary20124 Octal2068 DuodecimalB212 Vigesimal6E20 Base 363Q36 134 is the natural number following 133 and preceding 135. In mathematics 134 is a nontotient since there is no integer with exactly 134 coprimes below it. And it is a noncototient since there is no integer with 134 integers with common factors below it. 134 is ${}_8C_1 + {}_8C_3 + {}_8C_4$. In Roman numerals, 134 is a Friedman number since CXXXIV = XV * (XC/X) - I. 134 is also:", "74 82 103 110 114 116 119 0 1 9 14 16 34 45 55 77 83 107 130 Some hints about this - unlike 31 and 73, the number 133 is not prime but has divisors 7 and 19. This means the associated multiplicative group (G) has order (7-1)×(19-1) = 108. Also, 108 is divisible by 36, 12, and 3.", "Please be more careful. – Did Jan 24 '17 at 15:42 • Yes. But I considered prime factorisations – Roman83 Jan 24 '17 at 15:46 • So the point is not that $7^{164}$ divides $1000!$, but that $7^{165}$ does not. – Callus Jan 24 '17 at 15:49 • I'm understood, thank you! – Roman83 Jan 24 '17 at 16:00 In $\\{1,\\dots, 500\\}$, there are 71 multiples of 7, 10 multiples of $7^2$ and one multiple of $7^3$. How many multiples of $7$, $7^2$, $7^3$ do you have in $\\{501,\\dots, 1000\\}$? Same amount. • Why are they the same amount? – Jorge Fernández Hidalgo Jan 24 '17 at 15:59 • 504 is first multiple of 7 (with multiplicity of seven is 1 in factorization), and $504 + 7 \\cdot 70 = 994$. Also $7^2 \\cdot 11 = 539$ (multiplicity is 2) and $7^2 \\cdot 20 = 980$. At last $7^3 \\cdot 2 = 686$ (fits only once). – johnnycrab Jan 24 '17 at 16:18", "will be the lowest common multiple of 7 and 1461. If 1461 were a multiple of 7, then the lowest common multiple would be 1461. However, you can tell by dividing 1461 by 7 that it is not a multiple of 7. So, since 7 is a prime number, the lowest common multiple of 7 and 1461 must be 7$\\times$1461. That is after 7 sets of 4 years - so it will next happen 7$\\times$4 = 28 years later, in 2044. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "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", "least common multiple of $15$ and $20.$ Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple. Find the least common multiple (LCM) of the given numbers: $9\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}12$ 36 Find the least common multiple (LCM) of the given numbers: $18\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}24$ 72 ## Prime factors method Another way to find the least common multiple of two numbers is to use 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\\phantom{\\rule{3em}{0ex}}18=2\\cdot 3\\cdot 3$ Then we write each number as a product of primes, matching primes vertically when possible. $\\begin{array}{l}12=2\\cdot 2\\cdot 3\\hfill \\\\ 18=2\\cdot \\phantom{\\rule{1.1em}{0ex}}3\\cdot 3\\end{array}$ 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. do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid", "B $$\\large{{2}^{3}}\\cdot {{3}^{3}}\\cdot 5$$Hint: One way to solve this is to factor both numbers: $$540=2^2 \\cdot 3^3 \\cdot 5$$ and $$216=2^3 \\cdot 3^3$$. Then for each prime that's a factor of either number, use the largest exponent that appears in one of the factorizations. You can also take the product of the two numbers divided by their GCD. C $$\\large{{2}^{2}}\\cdot {{3}^{3}}\\cdot 5$$Hint: 216 is a multiple of 8. D $$\\large{{2}^{2}}\\cdot {{3}^{2}}\\cdot {{5}^{2}}$$Hint: Not a multiple of 216 and not a multiple of 540. Question 7 Explanation: Topic: Find the least common multiple of a set of numbers (Objective 0018). Question 8 #### All natural numbers from 2 to 266. Hint: She only needs to check primes -- checking the prime factors of any composite is enough to look for divisors. As a test taking strategy, the other three choices involve primes, so worth thinking about. #### All primes from 2 to 266 . Hint: Remember, factors come in pairs (except for square root factors), so she would first find the smaller of the pair and wouldn't need to check the larger. #### All primes from 2 to 133 . Hint: She doesn't need to check this high. Factors come in pairs, and something over 100 is going to be paired with something less than 3, so she will", "1/1335) + (1/1336 + ... 1/2002)= 2(1 + 1/3 + 1/5 +... 1/1335) - (1/667 + ... 1)= 2(1 + 1/3 + 1/5 + ...1/1335) - 2(1/2 + 1/4 + ...1/1334)= 2(1 - 1/2 + 1/3 - 1/4 ... 1/1335) = 2A.So now since we have A = 2A therefore A has to be equal to zero! QED.I have left some details to be filled but i assume anyone reading this should be able to fill that in.Ofcourse note that this method is quite general and you can use the same proof for any prime of the form 3n - 1 instead of 2003. pk14 Posts: 85 Joined: February 15th, 2006, 12:43 am ### Alternating harmonic sum Dear wannabequantie,very nice proof!!! Awesome!!!BTW, are you one of those who came to quant with a Math Ph.D.? sevvost Topic Author Posts: 361 Joined: February 17th, 2006, 7:47 pm ### Alternating harmonic sum Yes, this is correct. Here is a \"naive\" version of the same solution.m/n = 1 -1/2 + 1/3 - ...- 1/1334 + 1/1335 = (1 + 1/2 + 1/3 +...+ 1/1334 + 1/1335) - 2(1/2 + 1/4 + ... + 1/1334) = 1/668 + 1/669 + ... + 1/1335.Now, grouping terms in pairs in a school kid Gauss manner (conveniently, we have an even number of terms), we", "• anonymous Is this a trick question? \"Which pairs of integers $$a$$ and $$b$$ have greatest common divisor $$18$$ and least common multiple $$540$$?\" There is a theorem that states that $$[a,b](a,b)=ab$$, where $$[a,b]$$ stands for the least common multiple of $$a$$ and $$b$$, and $$(a,b)$$ stands for the greatest common divisor of $$a$$ and $$b$$. So, it follows that$[a,b](a,b)=ab,$$540\\cdot18=a\\cdot b.$In other words, $$a=540$$ and $$b=18$$ would suffice. Am I missing something? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "for those numbers is a multiple of 3. – usiro Jul 13 '18 at 22:23 • @usir0 The sum of digits for those numbers is a multiple of 3 That's only because small numbers with many factors are more likely to be multiples of $3$. It's certainly not the case in general, for example $\\,560=4 \\cdot 10 \\cdot 14 = 5 \\cdot 7 \\cdot 16\\,$ where $\\,4 + 10 + 14 = 5 + 7 + 16\\,$, but the digit sum of $560$ is $11$. – dxiv Jul 13 '18 at 22:31 • That is right. I didn't check for more numbers. Thanks. – usiro Jul 13 '18 at 22:40 I thought it might be fun to write a Mathematica code to find instances of same-sum factorizations. Using the following code, you can find integers such that num different sets of len different same-length factorizations of an integer less than max have the same sum. It omits multiples of lower cases on the grounds that those are boring. The smallest integers with increasingly large sets of different same-sum factorizations are: 72, 432, 3456, 5760, 7200, 12096, 17280, 21600, ... The smallest integers with pairs of increasingly large sets of different same-sum factorizations are: 144, 720, 2160, 5040, 8640, 10080, 14400, 25920, 30240, ... The smallest integers with triples" ]

[ "# It takes Bob twice as long as Caitlyn to clean his room. It takes Andrea 10 minutes longer than Caitlyn to clean her room. In total they work 90 minutes to clean their rooms. How long does it take Bob to clean his room? Sep 10, 2015 It takes Bob $\\text{40 minutes}$ to clean his room. #### Explanation: You will need to use the information provided to you to write three equations with three unknowns. Let's say that Bob takes $b$ minutes to clean his room, Andrea takes $a$ minutes, and Caitlyn takes $c$ minutes. The first piece of information given to you tells you that Bob needs twice as much time as Caitlyn to clean his room. This means that you can write $b = 2 \\cdot c$ Next, you were told that Andrea only takes 10 minutes longer than Caitlyn, which means that you can write $a = c + 10$ Finally, if you add the time it took all three to clean their rooms, you get $a + b + c = 90$ Use the value of $b$ from the first equation and the value of $a$ from the second equation to write ${\\underbrace{c + 10}}_{\\textcolor{b l u e}{= a}} + {\\overbrace{2 c}}^{\\textcolor{red}{= b}} + c = 90$ This is equivalent to $4 c", "## anonymous one year ago Glenn needs 20 minutes to clean his room. If his little sister Veronika pulls things out while he is cleaning, it takes 60 minutes for him to clean the room. How many minutes does it take for Veronika to pull everything out? 10 minutes 12 minutes 30 minutes 50 minutes 1. anonymous im going with b 2. anonymous i have seen this before, and actually figured out what you are supposed to say i am sure we can do it 3. anonymous yeah i remember now 4. anonymous solve $\\frac{1}{20}-\\frac{1}{x}=\\frac{1}{60}$ 5. anonymous i would say \" takes her two seconds to pull everything out\" because this is one of the sillier questions i have ever see, but if you solve the above equation you get $$x=30$$ 6. anonymous i know i am not supposed to give a direct answer but this question is so foolish that there is no point in an explanation", "# Problem of the Week Problem A and Solution Cleaning Calculation ## Problem A housekeeper is responsible for cleaning all the rooms on one floor of a hotel. The floor has $$16$$ regular rooms and $$5$$ suites. Regular rooms take $$20$$ minutes each to clean. Suites take $$30$$ minutes each to clean. 1. How long does it take to clean all the rooms on the floor of the hotel? 2. If the housekeeper starts cleaning at $$10$$:$$00$$ a.m. and does not take a break, at what time is the job finished? ## Solution 1. We can calculate the time it takes to clean the regular rooms by skip counting by $$20$$: $$20$$, $$40$$, $$60$$, $$80$$, $$100$$, $$120$$, $$140$$, $$160$$, $$180$$, $$200$$, $$220$$, $$240$$, $$260$$, $$280$$, $$300$$, $$320$$ We see that it takes a total of $$320$$ minutes to clean the regular rooms. We can calculate the time it takes to clean the suites by skip counting by $$30$$: $$30$$, $$60$$, $$90$$, $$120$$, $$150$$ We see that it takes a total of $$150$$ minutes to clean the suites. We add these two numbers together to see that it takes a total of $$320 + 150 = 470$$ minutes to clean the entire floor. From this total number of minutes, we could calculate the result in hours and minutes. However,", "your house, and clean your yard. Why not have someone else do your cleaning for you? Why not, indeed. You certainly can have someone else clean your computer, but chances are good that you won’t like the price. Plan on paying anywhere from $50 per hour to$150 per hour for computer cleaning. Do the math — if it takes two, four, six, or more hours to clean your computer, how much will you pay? Ouch! Also, you should understand that cleaning isn’t just a one-time thing. But if you change your behavior, you won’t have to clean as often or pay as much in the future. You can’t pay someone else to change your computer behavior. Because you can learn so much while you’re cleaning, it is well worth your while to do the cleaning yourself, at least for the first time. In general, I think that the pros outweigh the cons by a significant amount (hence, this book). I’m guessing you think so, too; that’s why you’re reading this book. But you may need to work on timing, or you may need to work on attitude before you bring yourself to actually do it. Just keep in mind that having a clean computer is something that is beneficial in more ways than you can imagine. 17 18", "120 rooms and each needs to be cleaned every day: Suppose that You estimated that the average cleaning time for one room is 20 minutes and its standard deviation is 13 minutes: You have 11 housekeepers who will perform cleaning: Estimate probability that they can finish with all of the rooms in under 4 hours. 13. A hotel has 120 rooms and each needs to be cleaned every day: Suppose that You estimated that the average cleaning time for one room is 20 minutes and its standard deviation is 13 minutes: You have 11 housekeepers who will perform cleaning: Estimate probability that they can finish with all of t...", "70. Alternatively, we could realize instinctively that 35% of 100 is 35, and then double it to get the same answer. Example Question #32 : Whole And Part What is 64% of 86? Explanation: First, write 64% as a decimal. Just move the decimal point of 64% two digits to the left and then eliminate the % so that we get 0.64. Thus, 64%=0.64. Then multiply that 0.64 by 86 So 55 is 64% of 86. Example Question #1 : How To Find The Part From The Whole With Percentage Wendy and Bob are cleaning their house at the same speed. Working together, they can clean the house in 8 hours. What percentage of the house can Wendy clean in 4 hours if she is cleaning by herself? Explanation: It takes Bob and Wendy 8 hours to clean the entire house, for a total of 16 man-hours. This means that one hour of cleaning by one person will clean of the house, or 6.25%. Thus, Wendy can clean 25% of the house by herself in 4 hours. Example Question #34 : Whole And Part What is of , rounded to the nearest tenth? Explanation: Convert 34% to a decimal: 0.34 Multiply 0.34 by 492: 167.28 Round to the nearest tenth: 167.3 Example Question #1 : How To Find", "# Problem of the Week Problem A Cleaning Calculation A housekeeper is responsible for cleaning all the rooms on one floor of a hotel. The floor has $$16$$ regular rooms and $$5$$ suites. Regular rooms take $$20$$ minutes each to clean. Suites take $$30$$ minutes each to clean. 1. How long does it take to clean all the rooms on the floor of the hotel? 2. If the housekeeper starts cleaning at $$10$$:$$00$$ a.m. and does not take a break, at what time is the job finished? Theme: Number Sense", "have the same problems, but I solved it with a Beeminder goal of doing 10 minutes of cleaning per day (6 days a week). Even with very little money on the line ($0 or 5\\$), I still had enough incentive to actually punch in 10 minutes on the microwave timer and turn into a Tasmanian Cleaning Devil until I heard the ding!! I feel strangely accomplished afterwards (both for having the discipline to fulfill my commitment on Beeminder and for having an orderly house). I was and still am able to maintain extraordinary levels of cleanliness ever since! 2Emily7yI'm not very good at it either! :) 6Richard Korzekwa 7yAn update: I decided to test the \"not fun\" part of the cost, by doing some cleaning and really asking myself how I felt while I was doing it. The answer was \"not bad, almost relaxing\". So I will try to update my model. 4John_Maxwell7yPresumably there is some reason you want a cleaner house, e.g. it will be easier to find stuff. If you spend 90 minutes cleaning, that will get you half as much gain in terms of making it easier to find stuff. So it's still worth doing in its own right. (Similarly, every marginal minute of exercise buys you 3-7 marginal minutes of life (within reason). Etc.", "bring yourself to actually do it. Just keep in mind that you may find having a clean computer beneficial in more ways than you can imagine. Paying the price If you’re short on time, you may be tempted to pay someone else to clean your system. After all, you can pay someone to clean your car, your office, your house, and your yard. Why not have someone else do your computer cleaning for you? Why not, indeed. You certainly can have someone else clean your computer, but chances are good that you won’t like the price. Plan on paying anywhere from $50 to$150 per hour for computer cleaning. Do the math — if it takes two, four, six, or more hours to clean your computer, how much will you pay? Ouch! Also, you should understand that cleaning isn’t just a one-time thing. However, if you change your behavior, you won’t have to clean as often or pay as much in the future. You can’t pay someone else to change your computer behavior, which makes all you can discover while cleaning your computer yourself valuable. 17 18 Part I: The Basics of Cleaning Your System 19 20 Part I: The Basics of Cleaning Your System Chapter 2 Determining What to Clean and When In This Chapter \u0001 Practicing safe", "years. This means he has $25$ saved up. At the age of $23$ Alice has saved $10$ every year for $3$ years. This means she has $30$ saved up, which is strictly more than $25$. Sample Input 1 Sample Output 1 20 25 5 20 10 23 Sample Input 2 Sample Output 2 20 28 5 30 9 35", "in 45 seconds and Chandra reads a page in 30 seconds. Chandra and Bob, who each have a copy of the book, decide that they can save time by \"team reading\" the novel. In this scheme, Chandra will read from page $1$ to a certain page and Bob will read from the next page through page $760$ , finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel? (A) $425$ (B) $444$ (C) $456$ (D) $484$ (E) $506$ AMC 8, 2006, Problem 16 A Novel Assignment The students in Mrs. Reed's English class are reading the same 760 -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read? (A) $6400$ (B) $6600$ (C) $6800$ (D) $7000$ (E) $7200$ AMC 8, 2006, Problem", "my washing machine, because she didn't have such a thing in Russia. I promised her that I'd do the laundry myself when there was a sufficient pile. However, she insisted that the dirty clothes annoyed her. I would point that my water bill went up. And so on. We argued like this every day. We were both frustrated. Then I asked myself why. Why does she do the laundry? The answer was there. She wanted to be helpful. I calmed down and stopped arguing with her. I sucked it up and paid the water bills. Her time with me turned into the most harmonious visit we ever had. Unfortunately, it was the last. ## A Scooter Riddle Puzzle. Alice, Bob, and Charlie are at Alice's house. They are going to Bob's house which is 33 miles away. They have a 2-seat scooter which rides at 25 miles per hour with 1 rider on it; or, at 20 miles per hour with 2 riders. Each of the 3 friends walks at 5 miles per hour. How fast can all three of them make it to Bob's house? ## Mathy Jokes * * * (submitted by Sam Steingold) I can count to 1023 on my 10 fingers. The rudest number is 132. * * * I kept forgetting my password,", "hours 1. It takes Sam 4 hours to rake the front lawn while his brother, Dave, can rake the lawn in 2 hours. How long will it take them to rake the lawn working together? 2. Mia can clean her apartment in 6 hours while her roommate can clean the apartment in 5 hours. If they work together, how long would it take them to clean the apartment? Answer 2 hours and 44 minutes 1. Brian can lay a slab of concrete in 6 hours, while Greg can do it in 4 hours. If Brian and Greg work together, how long will it take? 2. Josephine can correct her students test papers in 5 hours, but if her teacher’s assistant helps, it would take them 3 hours. How long would it take the assistant to do it alone? Answer 7 hours and 30 minutes 1. Washing his dad’s car alone, eight year old Levi takes 2.5 hours. If his dad helps him, then it takes 1 hour. How long does it take Levi’s dad to wash the car by himself? 2. At the end of the day Dodie can clean her hair salon in 15 minutes. Ann, who works with her, can clean the salon in 30 minutes. How long would it take them to clean the shop", "# Am I Normal? Level pending Bob recently took the Free Academic Koalification Exam (FAKE), in which one has to solve 5 math problems in the least amount of time. Bob completed the exam in about $$84.25$$ seconds. The mean time for all people taking the FAKE is $$100$$ seconds, and the standard deviation is $$10.5$$ seconds. Assuming the times for the FAKE are approximately Normal distributed, which of the following is closest to the percent of people who completed the FAKE faster than Bob did? (For your convenience, here is a z-score table). ×", "to run, which take 14 minutes each, before she can go home. If Diane needs to be home by 5:00 pm, what's the latest time that she can start the errands? a. 11:40 am b. 11:48 am c. 12:40 pm d. 12:48 pm e. 1:40 pm Now as for this problem, I first calculated the total time Diane would need to run all the errands by 18x14 mins, which gave me the result 252 mins. Then I divided 252/60, which gave me the result 4.2 hours. But after I tried to subtract 4.2 hrs from 5 pm or 17 hrs [12 hrs + 5 hrs], it gave me the result 12.8 hrs [12 hrs & 80 mins] or 1: 20 pm. But this isn't listed on the choices. So where did I go wrong?? Problem No 2. Sheila can wear her 100% linen suit four times before she needs to have it dry cleaned. She wears the suit to Church every Sunday and to meeting at bridge club on the first Tuesday of every month. If it costs $8.00 for each individual cleaning, and Sheila does not wear the suit on any other occasions, how much will Sheila pay to dry clean the suit each year? a.$48.00 b. $64.00 c.$96.00 d. $104.00 c.$128.00 Now what I did here", "double room every 20 minutes, how many cleaning persons are needed to clean all the rooms if each person works for exactly 7 hours? (A) 65 (B) 70 (C) 80 (D) 90 (E) 265 _________________ ##### Most Helpful Community Reply Verbal Forum Moderator Status: Greatness begins beyond your comfort zone Joined: 08 Dec 2013 Posts: 2113 Location: India Concentration: General Management, Strategy Schools: Kelley '20, ISB '19 GPA: 3.2 WE: Information Technology (Consulting) Re: A certain hotel has 1,400 single rooms and 420 double rooms. Each room [#permalink] ### Show Tags 11 Nov 2015, 01:13 7 4 Number of single rooms= 1400 Number of double rooms = 420 Time taken by a person to clean a single room = 15 mins = 1/4 hour Time taken by a person to clean a double room = 20 mins =1/3 hour Total time to clean 1400 single and 420 double rooms = 1400*1/4 + 420 * 1/3 = 350 + 140 = 490 hours Each person works for = 7 hours Number of persons required = 490/7 = 70 Answer B _________________ When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On", "# Nidhi's project Algebra Level 2 Alice can complete a project in $$20$$ days and Bob can complete the same project in $$30$$ days. Alice and Bob start working on the project together, with Alice quitting after some time, and it takes Bob another $$10$$ days to complete the project. What is the total number of days it took to complete the project? This problem is posed by Nidhi T. Details and assumptions Assume Alice and Bob work at a constant rate to complete the project. ×", "# Naomi has 45 minutes to get ready for school. She spends x minutes getting dressed. How do you write an expression that represents the number of minutes she still has to get ready? Jan 12, 2017 You subtract the number of minutes Naomi already used getting dressed from the total number of minutes she has. #### Explanation: From the quesrion, we know that: a) Naomi has $45$ minutes to get ready. b) She has already spent $x$ minutes. The number of minutes she still has to get ready would be $45 - x$ minutes.", "a person to move from armchair $3$ to armchair $4$, it takes $1$ minute. In the third test, no seat is occupied so your goal is achieved instantly.", "computer? Is the valuable data on your computer at risk?) If you write the pros on the left side of a sheet of paper and the cons on the right side, you can easily see how they balance out — and then determine which side wins. In general, I think that the pros outweigh the cons by a significant amount (hence, this book). I’m guessing you think so, too; that’s why you’re reading this book. Still, you may need to work on timing or attitude before you can bring yourself to actually do it. Just keep in mind that you may find having a clean computer beneficial in more ways than you can imagine. Paying the price If you’re short on time, you may be tempted to pay someone else to clean your system. After all, you can pay someone to clean your car, your office, your house, and your yard. Why not have someone else do your computer cleaning for you? Why not, indeed. You certainly can have someone else clean your computer, but chances are good that you won’t like the price. Plan on paying anywhere from $50 to$150 per hour for computer cleaning. Do the math — if it takes two, four, six, or more hours to clean your computer, how much will you pay?" ]

[ "that multiplication/division are (generally) commutative, meaning they can be done in any order. Your first term there is $\\frac{4\\times 7 \\times x\\times y}{2}$ which can be simplified to just $14xy$ Similarly, $\\frac{4x^2}{2}=2x^2$ 13. Originally Posted by Quacky Some of your terms are a mess. Remember that multiplication/division are (generally) commutative, meaning they can be done in any order. Your first term there is $\\frac{4\\times 7 \\times x\\times y}{2}$ which can be simplified to just $14xy$ Similarly, $\\frac{4x^2}{2}=2x^2$ Now its starting to make sence wasn't aware that I could do that....therefore I should end up with: 14xy - 2x^2 - 84y + 12x 14. Originally Posted by komodo Now its starting to make sence wasn't aware that I could do that....therefore I should end up with: 14xy - 2x^2 - 84y + 12x There we go . 15. Originally Posted by Quacky There we go . Thats right lol.......I'm loosing sleep over this stuff Anyway thanks for help realy appreciate it", "in my example, then instead of $$8\\times\\frac{1}{2}\\times 4 = 16,$$ we would be doing $$8\\times\\frac{1}{2\\times 4} = 1.$$ Interpreting it correctly, the only number we divide by is the 2. Also, seeing it this way allows me to rearrange the expression at will (since multiplication, unlike division, is associative and commutative). If I had, for example, $$7 \\div 3\\times 15,$$ I could think of it as $$7 \\times \\frac{1}{3}\\times 15 = 7 \\times 15 \\times \\frac{1}{3} =7 \\times \\left(15 \\div 3\\right) = 7\\times 5 = 35$$ In effect, I’m dragging the division sign around with the number following it! Thanks so much! I really like the idea of thinking of division as multiplying by the reciprocal and turning the whole multiplication/division portion into just multiplication. I'll try that out with my students and see if it helps. Thanks again! I didn’t mention there the fact that students outside of America are taught mnemonics like BODMAS, and students there sometimes think Division has to be done before Multiplication. Get them together, and you might have quite an argument! Here’s a question from another teacher, Monica, the next year: Incorrect Application of PEMDAS and Order of Operations I was working with students on the order of operations today and explained that multiplication and division are done from left to right,", "(an interpretation of $\\frac23 \\times 3$), or $$3\\times \\frac23 = \\frac23 \\times 3$$ 3. These pictures illustrate the commutative property of multiplication.", "# How do you divide 1/2div5? Nov 23, 2016 $\\frac{1}{10}$ #### Explanation: Division is the inverse of multiplication. This means that $\\div x$ is the same as $\\times \\frac{1}{x}$ In this example: $\\frac{1}{2} \\div 5 = \\frac{1}{2} \\times \\frac{1}{5}$ $= \\frac{1}{10}$", "## 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 =$-8$", "× Multiplication: Multiplying two fractions together is quite simple. Given that multiplication is commutative (we can do it in any order), we simply perform the fractional operations of multiplication and division in sequence. $\\frac{2}{3} \\times \\frac{4}{5} = 1 \\times 2 \\div 3 \\times 4 \\div 5 = \\frac{2 \\times 4}{3 \\times 5} = \\frac{8}{15}$ Division: Since division is simply the inverse of multiplication, we can divide one fraction by another be inverting the division and multiplication operations (that is, we mulitply by the reciprocal): $\\frac{2}{3} \\div \\frac{4}{5} = \\frac{2}{3} \\times \\frac{5}{4} = \\frac{2 \\times 5}{3\\times 4} = \\frac{10}{12} = \\frac{5}{6}.$ Addition: Adding fractions is slightly harder, because we can only add like items together. Thus, to add two fractions together, we must ensure that the denominators are equal. $\\frac{2}{3} + \\frac{4}{5} = \\frac{2 \\times 5}{3 \\times 5} + \\frac{4 \\times 3}{5 \\times 3} = \\frac{10}{15} + \\frac{12}{15} = \\frac{10+12}{15} = \\frac{22}{15}.$ Note by Arron Kau 2 years, 11 months ago Sort by:", "of multiplication, that is: $$X \\div Y =X\\cdot \\frac 1Y$$ So you have: $$56\\cdot 100 \\cdot \\frac18 =56\\cdot\\frac18\\cdot100$$ Which is obvious. • While it is true that \"division is the inverse of multiplication\" the emphasis should be on what follows from it: That division is equivalent to multiplication with the inverse and can be replaced by it. Feb 18, 2019 at 8:14 Swap the first two factors which is definitely allowed: 100 × 56 ÷ 8. Now put parentheses around part of it: 100 × (56 ÷ 8). Again, swap: (56 ÷ 8) × 100. Take off the parentheses: 56 ÷ 8 × 100. And there you go! transforming one into the other using only multiplicative commutation (plus the fact that one of the factors commutated can be a product, not just a simple value). • Now explain why putting parentheses around the last two terms in $100\\times56\\div8$ is OK but putting parentheses around the last two terms in $56\\div8\\times100$ is not OK. For that matter, why is it OK to swap the first two terms in $100\\times56\\div8$ but not the last two terms of $56\\div8\\times100$? Feb 18, 2019 at 14:43 • The short answer to adding the parentheses is, \"because it doesn't change the outcome\". But you're right, a more rigorous explanation is needed, and I don't", "Hermit Notebook # How do we know the formula for fraction multiplication is right ? Reminder: you can left-scroll and right-scroll the long math blocks ## The question We all learnt in college this formula: $\\frac{a}{b} \\times \\frac{c}{d} = \\frac{a \\times c}{b \\times d}$ Where $a, b, c$ and $d$ are integers and $b$ and $d$ are different from zero, or in other words: $a, c \\in \\mathbb{N}$ and $b, d \\in \\mathbb{N}^*$ When we apply it, it just seems to work. But how are we sure the formula is always true ? ## The prerequisites We will take for granted the properties of the operations on integers, especially for multiplication: 1. The associativity of multiplication: $a \\times b \\times c \\;\\ = \\;\\ (a \\times b) \\times c \\;\\ = \\;\\ a \\times (b \\times c)$ 1. The commutativity of multiplication $a \\times b = b \\times a$ 1. The equal priorities of multiplication and division, which means the order in which we do multiplication and division does not change the result: $a \\times b \\div c = (a \\times b) \\div c = a \\times (b \\div c)$ 1. The neutrality of $1$ for multiplication: $\\forall a \\in \\mathbb{N}^*, \\;\\ a \\times 1 = a$ 1. A number divided by itself equals $1$ $\\forall a \\ne", "of the section is $\\frac34$ mile long, how long is the section? Diagram: Multiplication equation: Division equation: ## Summary Sometimes we know the amount for multiple groups, but we don’t know how much is in one group. We can use division to find out. For example: If 5 people share $8 \\frac12$ pounds of cherries equally, how many pounds of cherries does each person get? We can represent this situation as a multiplication and a division: $$5\\, \\boldcdot {?} = 8\\frac12$$ $$8 \\frac12 \\div 5 = {?}$$ $8\\frac12 \\div 5$ can be written as $\\frac{17}{2} \\div 5$. Dividing by 5 is equivalent to multiplying by $\\frac 15$, and $\\frac{17}{2} \\boldcdot \\frac {1}{5} =\\frac {17}{10}$. This means each person gets $1\\frac {7}{10}$ pounds. Other times, we know the amount for a fraction of a group, but we don’t know the size of one whole group. We can also use division to find out. For example: Jada poured 5 cups of iced tea in a pitcher and filled $\\frac 23$ of the pitcher. How many cups of iced tea fill the entire pitcher? We can represent this situation as a multiplication and a division: $$\\frac 23 \\boldcdot {?} = 5$$ $$5 \\div \\frac23 = {?}$$ The diagram can help us reason about the answer. If $\\frac23$ of a pitcher is", "# Connor and Makayla Discuss Multiplication Alignments to Content Standards: 5.NF.B.4 Makayla said, \"I can represent $3\\times \\frac23$ with 3 rectangles each of length $\\frac23$.\" Connor said, “I know that $\\frac23 \\times 3$ can be thought of as $\\frac23$ of 3. Is 3 copies of $\\frac23$ the same as $\\frac23$ of 3?” 1. Draw a diagram to represent $\\frac23$ of 3. 2. Explain why your picture and Makayla’s picture together show that $3\\times \\frac23 = \\frac23 \\times 3$. 3. What property of multiplication do these pictures illustrate? ## IM Commentary The purpose of this task is to 1. Have students think about the meaning of multiplying a whole number by a fraction, and 2. Compare this meaning to their previous understanding of multiplying a fraction by a whole number in order to see an instance of the commutative property of multiplication in the case of fractions. Although the representations used in this task can be used to see the fact that $3 \\times \\frac23 = \\frac23 \\times 3$, it is not easy to use them to explain why the fact is true, or to see that multiplication is commutative for other numbers. This problem can be used to raise the question of why. In grade 3, students saw why multiplication of whole numbers is commutative by looking at", "## Basic College Mathematics (9th Edition) In division, the division sign is 'reversed', and becomes a multiplication sign with the reciprocal of the second fraction. For example: $\\frac{3}{4}$ $\\div$ $\\frac{2}{3}$ = $\\frac{3}{4}$ $\\times$ $\\frac{3}{2}$ = $\\frac{9}{8}$ Therefore, you should multiply the first fraction by the bottom of the second fraction.", "\\times x^3$ and since multiplication is commutative $5x^2 \\times 4x^3 = 5 \\times x^2 \\times 4 \\times x^3 = 5 \\times 4 \\times x^2 \\times x^3 = 20 \\times x^5 = 20x^5$ I hope this helps, Penny Math Central is supported by the University of Regina and the Imperial Oil Foundation.", "# Is there a name for the rule $a \\div (b \\times c) = a \\div b \\div c$? Edit, because I should have looked it up before I posted the question: Is there a name for the rule $a \\div (b \\div c) = a \\div b \\times c$ ? I ran across this in Liping Ma's book, Knowing and Teaching Mathematics, and I have searched the internet for a name for this rule to no avail. It is not the distributive law, but it is rather similar. Thank you! From Ma's book, p. 59 discussing \"dividing by a number is equivalent to multiplying by its reciprocal\": \"We can use the knowledge that students have learned to prove the rule that dividing by a fraction is equivalent to multiplying by its reciprocal. They have learned the commutative law. They have learned how to take off and add parentheses. They have also learned that a fraction is equivalent to to the result of a division, for example, $\\frac{1}{2} = 1 \\div 2$ . Now, using these, we can rewrite the equation this way: $1\\frac{3}{4} \\div \\frac{1}{2} \\to$ $1\\frac{3}{4} \\div (1 \\div 2) \\to$ $1\\frac{3}{4} \\div 1 \\times 2 \\to$ (This is the step my question is about.) $1\\frac{3}{4} \\times 2 \\div 1 \\to$ (and I'd like an explicit", "× Multiplication: Multiplying two fractions together is quite simple. Given that multiplication is commutative (we can do it in any order), we simply perform the fractional operations of multiplication and division in sequence. $\\frac{2}{3} \\times \\frac{4}{5} = 1 \\times 2 \\div 3 \\times 4 \\div 5 = \\frac{2 \\times 4}{3 \\times 5} = \\frac{8}{15}$ Division: Since division is simply the inverse of multiplication, we can divide one fraction by another be inverting the division and multiplication operations (that is, we mulitply by the reciprocal): $\\frac{2}{3} \\div \\frac{4}{5} = \\frac{2}{3} \\times \\frac{5}{4} = \\frac{2 \\times 5}{3\\times 4} = \\frac{10}{12} = \\frac{5}{6}.$ Addition: Adding fractions is slightly harder, because we can only add like items together. Thus, to add two fractions together, we must ensure that the denominators are equal. $\\frac{2}{3} + \\frac{4}{5} = \\frac{2 \\times 5}{3 \\times 5} + \\frac{4 \\times 3}{5 \\times 3} = \\frac{10}{15} + \\frac{12}{15} = \\frac{10+12}{15} = \\frac{22}{15}.$ Note by Arron Kau 3 years, 5 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote", "you are performing addition or multiplication • The numbers can be any real number (whole numbers, fractions, decimals, integers, etc.) • To “commute” can mean to travel between home and school—when you switch places, the travel time does not change (going to school and going back home). • If you switch the places of the numbers in an addition or multiplication equation, it will not change the answer in the end. • Ex. 3 + 5 = 5 + 3 (equals 8 either way) • Ex. 0.3 + 0.5 = 0.5 + 0.3 (equals 0.8 either way) • Ex. $\\frac{3}{10}$ + $\\frac{5}{10}$ = $\\frac{5}{10}$ + $\\frac{3}{10}$ (equals $\\frac{8}{10}$ either way) • For multiplication: the order of factors does not change the answer • Ex. 3 × 4 = 4 × 3 (equals 12 either way) • Ex. 0.3 × 0.4 = 0.4 × 0.3 (equals 0.12 either way) • Ex. $\\frac{3}{10}$ + $\\frac{4}{10}$ = $\\frac{4}{10}$ + $\\frac{3}{10}$ (equals $\\frac{12}{100}$ either way) • The general formulas (where a and b are variables that represent real numbers) for the commutative property are: Arithmetic Property Of Addition Of Multiplication Commutative Property $a + b = b + a$ $a \\,x\\,b = b\\, x\\, a$ • The commutative property does NOT work for subtraction nor division. In subtraction and division, the order", "and division leads to the commonly taught method called Keep-Change-Flip. The last problem in the video explored $$\\frac{3}{5}\\div \\frac{7}{5}$$. As it showed, $$\\frac{3}{5}\\div \\frac{7}{5}=\\frac{3}{5}\\times \\frac{5}{7}=\\frac{15}{35}=\\frac{3}{7}$$. Time for a diagram! Here’s $$\\frac{3}{5}$$: And here’s $$\\frac{7}{5}$$: Multiplication and division are so similar because they are directly related to each other! What is the difference in these two sentences? 1. You have $$\\frac{3}{4}$$ of a leftover pizza that you want to split between 6 friends. How much of the pizza does each person get? 2. You have $$\\frac{3}{4}$$ of a leftover pizza, then you eat $$\\frac{5}{6}$$ of that. How much of the pizza has been eaten? Sentence 1 calls for division. You are dividing $$\\frac{3}{4}$$ of a pizza into 6 equal parts: $$\\frac{3}{4}\\div 6$$ (which turns into multiplication anyway!). $$\\frac{3}{4}\\div 6=\\frac{3}{4}\\times \\frac{1}{6}=\\frac{3}{24}=\\frac{1}{8}$$ Each person gets one eighth (about a slice) of the original pizza. Sentence 2 calls for multiplication. What is $$\\frac{5}{6}$$ of $$\\frac{3}{4}$$? $$\\frac{5}{6}\\times \\frac{3}{4}$$. $$\\frac{5}{6}\\times \\frac{3}{4}=\\frac{5\\times 3}{6\\times 4}=\\frac{15}{24}=\\frac{5}{8}$$ Five eighths have been eaten (or about 5 slices). ### Final Thoughts! When multiplying and dividing fractions, the procedures are important, but sometimes it can be difficult to interpret a situation and identify what is required. Translation into English sentences and modeling with diagrams can help you make sense of these fraction operations. Click for more help with fractions, decimals, and", "# Lesson 5: Dividing terms (with indices) Introduction With multiplying terms, you had a lesson on 'the basics' before you had to deal with indices. With division, there aren't really any basics - it's pretty much all about indices. There is one basic thing - how to write division. You'll see below that I've used a $$\\div$$ sign. Not the best way to write division in algebra! But in my defense, it helps me with my explanation! And you might get a question written like this in an exam so you do need to be familiar with it. The proper way to write division, though, is as a fraction! SECTION A So, you're going to learn how to simplify something like $$a^5\\div a$$ . If we open it up, we get $$a\\times a\\times a\\times a\\times a\\div a$$ Focus on the last two terms. If you multiply by $$a$$ then divide by $$a$$ , you end up back where you started ( $$3\\times 5\\div 5~=~3$$ ). The last two operations cancel each other out! $$\\require{cancel} a\\times a\\times a \\times a \\times \\cancel{a}\\div \\cancel{a}~=~a\\times a\\times a\\times a~=~a^4$$ We call this cancelling terms. Now we'll try $$a^5\\div a^2$$ but let's write this one as a fraction. $$\\Large\\frac{a^5}{a^2}\\normalsize~=~\\Large\\frac{a\\times a\\times a\\times \\cancel{a}\\times \\cancel{a}}{\\cancel{a}\\times \\cancel{a}}\\normalsize~=~a\\times a\\times a~=~a^3$$ You just did 5 take away 2,", "multiplying 4 by 3. ### Dividing with Fractions 1. Find the reciprocal of the number that follows the division symbol. 2. Multiply the first number (the one before the division symbol) by the reciprocal of the second number (the one after the division symbol). Any easy way to remember how to divide fractions is the phrase “keep, change, flip.” This means to KEEP the first number, CHANGE the division sign to multiplication, and then FLIP (use the reciprocal) of the second number. ### Example Divide $\\frac{2}{3}\\div \\frac{1}{6}$. ### Example Divide $\\frac{3}{5}\\div \\frac{2}{3}$.", "learning properties of numbers, we learn the commutative property of multiplication. $a×b=b×a$$a \\times b = b \\times a$. But the division does not follow the commutative property. What this means is, the order of arithmetic operations cannot be modified to simplify the problem. The redefined precedence order PEMA converts the division into multiplication and allows commutative property to be used to ones advantage. $a÷b=a×\\left(\\frac{1}{b}\\right)=\\frac{1}{b}×a$$a \\div b = a \\times \\left(\\frac{1}{b}\\right) = \\frac{1}{b} \\times a$ Another property of numbers is associative property of multiplication. $a×\\left(b×c\\right)=\\left(a×b\\right)×c$$a \\times \\left(b \\times c\\right) = \\left(a \\times b\\right) \\times c$. But the division does not follow the associative property. What this means is, the sequence of arithmetic operations cannot be modified to simplify the problem. The redefined precedence order PEMA converts the division into multiplication and allows associative property to be used to ones advantage. $a÷\\left(b÷c\\right)$$a \\div \\left(b \\div c\\right)$$=a×\\frac{1}{b×\\left(1/c\\right)}$$= a \\times \\frac{1}{b \\times \\left(1 / c\\right)}$$=a×\\left(\\left(\\frac{1}{b}\\right)×c\\right)$$= a \\times \\left(\\left(\\frac{1}{b}\\right) \\times c\\right)$$=\\left(a×\\frac{1}{b}\\right)×c$$= \\left(a \\times \\frac{1}{b}\\right) \\times c$. change for algebra There are good reasons in algebra to adapt to PEMA. Students at 6th or 7th grade level may skip this explanation. Algebra is extensively based on the numerical expressions and the properties of numerical arithmetic. In an algebraic expression of multiple factors, with division in some of them, simplification is possible only if the", "## Basic College Mathematics (10th Edition) The commutative property does not apply to division because you are switching which number is getting divided up. i.e. $10\\div5 = 2$ and $5\\div10 = \\frac{1}{2}$, $2\\ne\\frac{1}{2}$ The associative property does not apply to division because you are switching the order of what numbers are being divided. i.e. $(12\\div6)\\div2 = 1$ and $12\\div(6\\div2) = 4$, $1\\ne4$" ]

[ "# It takes Bob twice as long as Caitlyn to clean his room. It takes Andrea 10 minutes longer than Caitlyn to clean her room. In total they work 90 minutes to clean their rooms. How long does it take Bob to clean his room? Sep 10, 2015 It takes Bob $\\text{40 minutes}$ to clean his room. #### Explanation: You will need to use the information provided to you to write three equations with three unknowns. Let's say that Bob takes $b$ minutes to clean his room, Andrea takes $a$ minutes, and Caitlyn takes $c$ minutes. The first piece of information given to you tells you that Bob needs twice as much time as Caitlyn to clean his room. This means that you can write $b = 2 \\cdot c$ Next, you were told that Andrea only takes 10 minutes longer than Caitlyn, which means that you can write $a = c + 10$ Finally, if you add the time it took all three to clean their rooms, you get $a + b + c = 90$ Use the value of $b$ from the first equation and the value of $a$ from the second equation to write ${\\underbrace{c + 10}}_{\\textcolor{b l u e}{= a}} + {\\overbrace{2 c}}^{\\textcolor{red}{= b}} + c = 90$ This is equivalent to $4 c", "70. Alternatively, we could realize instinctively that 35% of 100 is 35, and then double it to get the same answer. Example Question #32 : Whole And Part What is 64% of 86? Explanation: First, write 64% as a decimal. Just move the decimal point of 64% two digits to the left and then eliminate the % so that we get 0.64. Thus, 64%=0.64. Then multiply that 0.64 by 86 So 55 is 64% of 86. Example Question #1 : How To Find The Part From The Whole With Percentage Wendy and Bob are cleaning their house at the same speed. Working together, they can clean the house in 8 hours. What percentage of the house can Wendy clean in 4 hours if she is cleaning by herself? Explanation: It takes Bob and Wendy 8 hours to clean the entire house, for a total of 16 man-hours. This means that one hour of cleaning by one person will clean of the house, or 6.25%. Thus, Wendy can clean 25% of the house by herself in 4 hours. Example Question #34 : Whole And Part What is of , rounded to the nearest tenth? Explanation: Convert 34% to a decimal: 0.34 Multiply 0.34 by 492: 167.28 Round to the nearest tenth: 167.3 Example Question #1 : How To Find", "## anonymous one year ago Glenn needs 20 minutes to clean his room. If his little sister Veronika pulls things out while he is cleaning, it takes 60 minutes for him to clean the room. How many minutes does it take for Veronika to pull everything out? 10 minutes 12 minutes 30 minutes 50 minutes 1. anonymous im going with b 2. anonymous i have seen this before, and actually figured out what you are supposed to say i am sure we can do it 3. anonymous yeah i remember now 4. anonymous solve $\\frac{1}{20}-\\frac{1}{x}=\\frac{1}{60}$ 5. anonymous i would say \" takes her two seconds to pull everything out\" because this is one of the sillier questions i have ever see, but if you solve the above equation you get $$x=30$$ 6. anonymous i know i am not supposed to give a direct answer but this question is so foolish that there is no point in an explanation", "it takes $\\frac{d_c}{S_c} + \\frac{d_w}{S_w}$ and it was 40 minutes longer so : $\\frac{d_c + d_w}{S_c} = \\frac{d_c}{S_c} + \\frac{d_w}{S_w} +40$ so $\\frac{d_w}{S_c} = \\frac{d_w}{S_w} +40$ The wife drove 20 minutes less so she was at 10 minutes of the station ( it would have taken 10 minutes to drive the distance he walked) : $\\frac{d_w}{S_c} = 10$ When we combine the two equations : $\\frac{d_w}{S_w} = 50$ • Correct and good explanation, accepted :) – ABcDexter Jun 4 '16 at 12:57 He was walking for 50 minutes Explanation : Alice usually drives from home to the station back home. Today she drove from home to the point where she met Bob back home. She saved 20 minutes. Because both ways are the same distance, she saved 10 minutes each way. Normally she meets Bob at 9pm. But today she met him 10 minutes earlier, at 8:50pm. Since Bob started walking at 8pm, he walked for 50 minutes. • Way easier than my solution – Fabich Jun 4 '16 at 15:20 • That's simply brilliant :) – ABcDexter Jun 4 '16 at 17:44 Simple. Let's assume that it takes an hour to drive to the station. Alice would leave home at 8 thinking she'd arrive at the station at 9. Bob would also be leaving the station on", "is division and vice versa. Check your solution by putting it into the starting equation. ##### Solution The rate is the number of wood pieces divided by the needed time. Let $x$ be the time it takes Jasmine to collect $10$ pieces of wood. Thus we get the rate $\\frac{10~\\text{wood pieces}}{x~\\text{minutes}}$ for Jasmine and, with the same measures, $\\frac{15}{x+10}$ for Grandpa Lindbergh. Because the rates agree we get the following rational equation: $\\frac{10}{x}=\\frac{15}{x+10}$. First we cross multiply to get $15x=10(x+10)$. Using the distributive property gives us $15x=10x+100$. From now on we use opposite operations to isolate $x$: $\\begin{array}{rcl} 15x & = & ~10x+100\\\\ \\color{#669900}{-10x} & &\\color{#669900}{-10x}\\\\ 5x & = & ~100\\\\ \\color{#669900}{\\div5} & & \\color{#669900}{\\div5}\\\\ x & = & ~20 \\end{array}$ Lastly, let's check the solution $\\frac{10}{20}=\\frac12=\\frac{15}{20+10}$ $~~~~~$✓ Thus we can conclude that Jasmine takes $20$ minutes to collect $10$ pieces of wood, while Grandpa Lindbergh takes $30$ minutes to collect $15$ pieces of wood. • #### Calculate how much time it takes General Good and Grandpa Lindbergh to weave $1$ leave. ##### Hints You can only add or subtract fractions with the same denominator. Use FOIL multiplication to check that the quadratic term equals $(2x+9)(x-3)$. Put the solutions in the starting equation to check them. They must satisfy this equation. ##### Solution First let's have a look", "# Thread: Ratio Proportion: it takes 30 minutes for 3 people to clean four identical rooms. 1. ## Ratio Proportion: it takes 30 minutes for 3 people to clean four identical rooms. could somebody help me with this question:- it takes 30 minutes for 3 people to clean four identical rooms. How long would it take 9 people to clean 12 of the same rooms at the same rate I find he Phrase \"Same rate\" confusing. My answer is 30 minutes but my teacher says it is 90 minutes and I can't understand why This is how I did it:- 3 People = 4 rooms in 30 minutes Divide LHS by 3 and multiply RHS with 3 1 People = 4 rooms in 90 minutes multiply LHS by 9 and Divide RHS with 9 9 People = 4 rooms in 10 minutes Multiply RHS with 3 9 People = 12 rooms in 30 minutes 2. Originally Posted by MAMM could somebody help me with this question:- it takes 30 minutes for 3 people to clean four identical rooms. How long would it take 9 people to clean 12 of the same rooms at the same rate I find he Phrase \"Same rate\" confusing. My answer is 30 minutes but my teacher says it is 90 minutes and I", "baskets in one hour. As part of a team: • Alice is now picking 7.5 baskets per hour, a most impressive increase in productivity of 25%. • Bob is now picking 7.5 baskets per hour as well, a modest increase in productivity of approximately 7%. Both Alice and Bob benefit if they work together (though Alice benefits more). We thus have an example of synergy. To use a cliché: “the whole is greater than the sum of its parts”. If Alice and Bob work alone, they can only pick 13 baskets per hour in total, but if they work together they can pick 15 baskets per hour. Everyone is happy. Let $S := \\{\\text{Alice}, \\text{Bob}\\}$, let $2^S$ denote the power set of $S$, and let $\\mathbb{R}_0^{+}$ denote the set of nonnegative real numbers. We introduce a productivity function $\\pi : 2^S \\to \\mathbb{R}_0^{+}$, enumeratively defined as follows • $\\pi (\\emptyset) = 0$ • $\\pi (\\{\\text{Alice}\\}) = 6$ • $\\pi (\\{\\text{Bob}\\}) = 7$ • $\\pi (\\{\\text{Alice}, \\text{Bob}\\}) = 15$ where $\\pi (\\emptyset) = 0$ because the productivity of the “empty team” is zero. Since we have that $\\pi (\\{\\text{Alice}, \\text{Bob}\\}) > \\pi (\\{\\text{Alice}\\}) + \\pi (\\{\\text{Bob}\\})$ we conclude that we have synergy. Note that the existence of a synergistic or synergetic effect is a property of the productivity function", "## satellite73 one year ago what math teacher made up this question? Glenn needs 20 minutes to clean his room. If his little sister Veronika pulls things out while he is cleaning, it takes 60 minutes for him to clean the room. How many minutes does it take for Veronika to pull everything out? 1. Empty I feel like the answer I want to give is not what they're thinking the answer _should_ be. 2. anonymous my answer is five seconds , but the answer they want is 30 minutes has to be one of the dumbest questions i have seen lately ! 3. Empty I think the solution is simple: waterboarding. 4. anonymous lol 5. UsukiDoll why do they want 30 minutes though? just get a vacuum cleaner and get rid of all the junk in the room. Takes 5 seconds XD 6. anonymous i know it is 30 minutes because of the answer choices given they want the solution to $\\frac{1}{20}-\\frac{1}{x}=\\frac{1}{60}$ as if that has anything to do with it right to the \"bad math\" pile 7. Astrophysics I remember you asking another question similar to this, it was the exact same solution 8. anonymous must be a franchise franchise of math education majors 9. Empty A franchise destined for Guantanamo Bay? 10. Astrophysics lol 11.", "The prompt is \"layered\" and more difficult than typical \"Work\" questions, but if you are comfortable converting the data into \"work per hour\" then you can get to the correct answer with a bit of effort. Knowing the 'Work Formula' will also help you when dealing with Fact 2. We're given 2 rates for 2 groups of people: 1) A and B can clean the room in 3 hours 2) A and C can clean the room in 2.5 hours We're asked if A, B and C can clean the room in UNDER 2 hours. This is a YES/NO question. Fact 1: B cleans faster than A Since A and B can clean the room in 3 hours, and we're told that B cleans FASTER than A, then that means that B does MORE than 1/2 of the work in that 3 hours: > 1/2 the work in 3 hours = > 3/6 of the work in 3 hours = > 1/6 of the work in 1 hour We now know that B can clean MORE than 1/6 of the room in 1 hour.... We also know that A and C can clean the room in 2.5 hours: 1/1 of the work in 2.5 hours = 2/5 of the work in 1 hour NOW we know how much", "done. ~samrocksnature ## Solution 5 (simple) WLOG, let Al's speed be $15$ steps per second, so Bob's speed is $5$ steps per second. Then, Al was on the escalator for $\\frac{150}{15}\\ = 10$ seconds and Bob was on the escalator for $\\frac{75}{5}\\ = 15$ seconds. Let $r$ be the rate of the escalator, in steps per second. Then, the total amount of steps is $150 - 10r = 75 + 15r$. Al is getting $10$ seconds of resistance at rate $r$ from the escalator, while Bob is getting $15$ seconds of help at rate $r$. Solving for $r$, we hvae $r = 3$ steps per second. Then, we can plug $r$ into the previous equation or subtract/add it to Al/Bob's rate (respectively) then multiply by their respective time. Either way, we get $\\boxed{120}$ and we are done. ~ pi_is_3.14", "gets done per hour: A and C get 2/5 of the room cleaned B gets MORE than 1/6 of the room cleaned 2/5 + (more than 1/6) = 12/30 + (more than 5/30) = MORE than 17/30 of the room is cleaned in 1 hour. By extension, 34/30 of the room would be cleaned in 2 hours (which is more cleaning than is necessary). This PROVES that the room will take LESS than 2 hours to clean and the answer to the question is ALWAYS YES. Fact 1 is SUFFICIENT. Fact 2: C can clean the room in LESS than 5 hours. Here, we can use the Work Formula to our advantage.... Work = (X)(Y)/(X+Y) where X and Y are the times it takes to clean the room individually. Here, we'll call.... X = the time for A and B to clean the room = 3 hours Y = the time for C to clean the room = less than 5 hours (3)(less than 5)/(3 + less than 5) = (less than 15)/(less than 8) = less than 15/8 = less than 2 hours and the answer to the question is ALWAYS YES. Fact 2 is SUFFICIENT. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact", "speed was 75 miles per hour? ## Writing Exercises 1. Marisol solves the proportion $$\\dfrac{144}{a}=\\dfrac{9}{4}$$ by ‘cross multiplying,’ so her first step looks like $$4 \\cdot 144=9 \\cdot a$$$.$ Explain how this differs from the method of solution shown in Example 7.6.2. 1. Paula and Yuki are roommates. It takes Paula 3 hours to clean their apartment. It takes Yuki 4 hours to clean the apartment. The equation $$\\dfrac{1}{3}+\\dfrac{1}{4}=\\dfrac{1}{t}$$ can be used to find $$t$$, the number of hours it would take both of them, working together, to clean their apartment. Explain how this equation models the situation.", "## Algebra: A Combined Approach (4th Edition) One-fourth $\\frac{1}{4}$ of the strip of beach can be cleared within one hour. We are given that the entire strip of a beach can be cleared with 4 hours. So, if we divide the strip in 4 equal parts, then time taken should be one fourth of four hours: $\\frac{1}{4}$(4) = 1 hour . Thus in one hour, one fourth of the strip of the beach can be cleared up.", "down, he took $150$ steps with a speed of $3b-e$ per step. When Bob was on his way up, he took $75$ steps with a speed of $b+e$ per step. Since Al and Bob were walking the same distance, we have $$150(3b-e)=75(b+e)$$ Solving gets the ratio $\\frac{e}{b}=\\frac{3}{5}$. Thus while Bob took $75$ steps to go up, the escalator contributed an extra $\\frac{3}{5}\\cdot75=45$ steps. Finally, there is a total of $75+45=\\boxed{120}$ steps in the length of the escalator. ## Solution 4 Please understand the machinery of an escalator before proceeding to read this solution. Let the number of steps that disappear at the top of the escalator equal $x.$ Assume that Al takes $3$ steps per second and that Bob takes $1$ step per second. Since Al counts $150$ steps, it takes him $$\\frac{150}{3}=50$$ seconds to traverse the distance of the escalator moving downwards. Since Bob counts $75$ steps, it takes him $$\\frac{75}{1}=75$$ seconds to traverse the distance of the escalator moving downwards. For the sake of this solution, we activate the emergency stop button on the escalator. Now, the escalator is not moving, or is simply a staircase. Imagine that Al is taking $3$ steps downwards every second, but we throw hands at him immediately after each second, such that he flinches and moves himself backwards $x$ steps.", "for Alice and more than 10 for Bob) + (Probability that Carl waits less than 10 minutes for Bob and more than 10 for Alice) + (Probability that Carl waits more than 10 minutes for both) which gives me the result of $$(\\frac{1}{2} * \\frac{1}{2}) + (\\frac{1}{2} * \\frac{1}{2}) + (\\frac{1}{2} * \\frac{1}{2}) = \\frac{3}{4}$$. (c) Probability that Carl waits more than 10 minutes for both is the probability that $$A$$ and $$B$$ both fall in the interval ($$20$$, $$30$$) in the total interval ($$10$$, $$30$$). Therefore we get the probability $$\\frac{1}{2} * \\frac{1}{2} = \\frac{1}{4}$$. However, apparently the correct answer is $$\\frac{1}{4}$$ for (b) and $$\\frac{3}{4}$$ for (c), and I'm not entirely sure why. The rationale the textbook gives is that for (b) we need to compute the time that both of them arrive after 1:20 p.m., and for (c) we simple take the complement that both of them arrive between 1:10 and 1:20 p.m. $$1 - \\frac{1}{4}$$. I'm not entirely sure how these solutions make sense, though, and was hoping someone would be kind enough to provide me with some insight. Thank you. • The explanations in the book seem reversed to me. I get the same answers you do. – saulspatz Nov 25 '18 at 4:04 • That's strange... I should bring this issue up", "while Alice has travelled $2S-85$ meters. We get $$\\frac{V_1}{V_2}=\\frac{S-55}{55}=\\frac{2S-85}{S-85} \\Rightarrow S^2-250S+2\\cdot 55\\cdot 85=0$$ a quadratic equation with roots $5(25\\pm \\sqrt{251})$, of which only $$S=5(25+\\sqrt{251})\\approx 204 > 85$$ is acceptable because $5(25-\\sqrt{251})<85$. The distance between the bus and the car is then \\begin{align*}x&=S-25-25\\cdot\\frac{V_2}{V_1}=S-25-25\\frac{55}{S-55}=\\frac{(S-25)(S-55)-25\\cdot 55}{S-55}=\\frac{S^2-80S}{S-55}=\\\\ &=\\frac{250S-2\\cdot 55 \\cdot 85-80S}{S-55}=\\frac{170(S-55)}{S-55}=170 \\end{align*} This is the intended solution of the problem. Indeed, there are no other possible cases: at the second meeting either Alice or Bob must have completed at least one journey, but none of them can have completed more than one, else there would have been another meeting inbetween. • Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.\\frac {55}{25}$$? Jul 22, 2018 at 19:51 • OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all. Jul 23, 2018 at 7:55 • I agree that the solution is incorrect, not much because their first meeting is supposed to be at", "after Alice passed him? A. 5 min B. 6 min C. 8 min D. 10 min E. 20 min Since the speed of Alice is double the speed of bob, the time taken by alice is half the time taken by Bob Alice takes 10 mins thus bob takes $$\\frac{10}{1/2}$$ = 10 x2 = 20 (E) Senior SC Moderator Joined: 22 May 2016 Posts: 3664 Alice and Bob traveled in the same direction along the same route at t [#permalink] Show Tags 21 Mar 2018, 10:38 2 MathRevolution wrote: [GMAT math practice question] Alice and Bob traveled in the same direction along the same route at their respective constant speeds of 12 km per hour and 6 km per hour. They each started to travel from their own houses. Bob’s house is halfway between Alice’s house and the destination. After passing Bob, Alice took 10 minutes to reach the destination. How many minutes did it take Bob to reach the destination after Alice passed him? A. 5 min B. 6 min C. 8 min D. 10 min E. 20 min Convert rates to minutes. A's rate in minutes: $$\\frac{12km}{1hr}=\\frac{12km}{60mins}=\\frac{1km}{5mins}$$ B's rate in minutes: $$\\frac{6km}{1hr}=\\frac{6km}{60mins}=\\frac{1km}{10mins}$$ Information to ignore or avoid: 1) Bob's house as a halfway point 2) Relative speed 3) Whatever happens before the passing point As soon", "## Elementary Algebra Let t be the time it takes Pam. Thus: $1/ 1.5 = \\frac{1}{t} + \\frac{1}{t+4}$ We create common denominators to obtain: $.67 = \\frac{t+t+4}{t(t+4)} \\\\ .67t^2 + 2.67t = 2t +4 \\\\ .67t^2 +.67t -4 = 0 \\\\ t^2 +t -6 = 0 \\\\(t+3)(t-2) =0$ Since t can't be negative, we see that it takes Pam 2 hours. This means it takes Laura 2+4 = 6 hours.", "## Elementary Algebra We call t the time it takes machine B. Thus, we obtain: $\\frac{600}{t-5} = \\frac{ 600}{t} + 20$ We create common denominators and solve for t: $\\frac{600t}{t(t-5)} = \\frac{600(t-5)}{t(t-5)} + \\frac{20t(t-5)}{t(t-5)}$ Since the denominators are the same, we cancel them out and solve. (Recall, t cannot be negative.) $0 = -3000 +20t^2 -100t \\\\ 0=20(t-15)(t+10) \\\\ t=15$ Thus, we find the rates: B: $600 /15 = 40$ Machine A works 20 candies per minute faster, so it works at 60 candies per minute.", "Question 2. Frank can clean $$\\frac{1}{4}$$ of his room in 15 minutes. How long will it take Frank to clean the entire room? Show your thinking on a ratio table. Answer: It will take 60 minutes to clean the entire room, Full room is cleaned, Explanation: Given from above ratio table number of minutes Frank can clean is 15 minutes so, 15 X 4 = 60 minutes so, amount of room cleaned (1/4) X 4 = 1. Question 3. Write a story problem for the expression 15 ÷ $$\\frac{1}{3}$$. Then solve the problem. NOTE Remember that 15 ÷ $$\\frac{1}{3}$$ means, “How many thirds are there in 15?” Answer: Story problem: Harry can make 1 bracelet using $$\\frac{1}{3}$$ of beads of box, If there are 15 boxes of beads how many number of bracelets can Harry make? Harry can make 45 bracelets, Explanation: Asked to write a story problem for the expression 15 ÷ $$\\frac{1}{3}$$. Then to solve the problem. So story problem is Harry can make 1 bracelet using $$\\frac{1}{3}$$ of beads of box, If there are 15 boxes of beads how many number of bracelets can Harry make? Solving we get 15 ÷ $$\\frac{1}{3}$$ = 15 X 3 = 45 bracelets. Review Question 4. Find the sum or difference. a. $$\\frac{2}{3}$$ + $$\\frac{4}{9}$$ Answer: $$\\frac{2}{3}$$ + $$\\frac{4}{9}$$" ]

[ "sides form a right triangle.", "the hypotenuse being the longest side in a right triangle. Okay that’s it for now,.", "triangle. What are the lengths of the two legs of the right triangle?", "its content area is 16 square centimeters. 15. A truck A truck departs from a distribution center. From there, it goes 20km west, 30km north and 10km west and reaches a shop. How can the truck reach back to the distribution center from the shop (what is the shortest path)? 16. Triangle Calculate the area of right triangle ΔABC, if one leg is long 14 and its opposite angle is 59°. 17. Triangle P2 Can triangle have two right angles?", "the area of the region shaded grey? ## Largest triangle What is the largest area triangle which has one side of length 4cm and one of length 5cm? ## Circles Which is largest, the red or the blue area? ## Archive Show me a random puzzle ▼ show ▼", "is a right triangle with both short sides of length 1 meter. (a) The axis of rotation is...", "square is the longest side of a right triangle. The sum of the areas of squares drawn on the other two sides of the triangle. Now draw a right triangle and squares on all three sides on thick paper: In the medium size square, mark the point of intersection of diagonals; and through this point, mark lines parallel to the perpendicular sides of the largest square: Now cut along these lines to split the square iito four pieces. Cut out the smallest square also. Arrange all these inside the largest square as below: What do we see from all this? The area of squares drawn out are equal to the area of the largest side of the triangle formed from cutting. Explanation: It is noticed that the area of squares drawn out are equal to the area of the largest side of the triangle formed from cutting. The area of the square drawn on the largest side of a right triangle is equal to the sum of the areas of the squares drawn on the other two sides. This is known as Pythagoras Theorem, in honour of the philosopher Pythagoras, who lived in Greece. Using this, we can draw a square of area 10 square centimetres. We have 10 = 32 + 12 So according to Pythagoras Theorem,", "1. Right triangle Problem A right triangle has an area of 30 ft 2 and a hypotenuse 13 ft long. What are the lengths of its other two sides? 2. Originally Posted by tmac11522 A right triangle has an area of 30 ft 2 and a hypotenuse 13 ft long. What are the lengths of its other two sides? Solve these two equations: $a^2 + b^2 = 13^2$ .... (1) $\\frac{a b}{2} = 30$ .... (2)", "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.", "# A triangle has sides with lengths of 35 miles, 84 miles, and 91 miles. Is it a right triangle? Nov 28, 2016 Yes. #### Explanation: Using Pythagoras we have the condition: If this is a right triangle then the longest side is the hypotenuse. Also the square of the hypotenuse is the sum of the square of the other 2 sides. In which case the hypotenuse would be 91 and its square is 8281 The side which is 35 long has the squared value of ${35}^{2} = 1225$ The side which is 84 long has the squared value of ${84}^{2} = \\underline{7056}$ The sum is:$\\text{ } 8281$ This is the same value as the square of the longest side. So it is a right triangle", "# A triangle has sides with lengths of 23 meters, 13 meters, and 18 meters. Is it a right triangle? Jun 13, 2018 No. #### Explanation: if it were a right triangle, then ${23}^{2} = {13}^{2} + {18}^{2}$ 529 = 169 + 324? No. By the cos rule, ${23}^{2} = {13}^{2} + {18}^{2} - 2 \\cdot 13 \\cdot 18 \\cdot \\cos \\theta$ $\\theta \\ne \\frac{\\pi}{2}$", "## 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.", "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...", "right triangle is a triangle with one 90 degree angle.", "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", "perimeter of a right triangle with the legs 14 cm and 21 cm long?", "is 5 cm. What is the area of the triangle? Thank you.", "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", "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.", "cm and breadth of a right triangle here. 30/12/2020" ]

[ "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.", "hypotenuse. □\\begin{aligned} Since the triangle is a right triangle, we can use the Pythagorean theorem to find the side length a,a,a, and from this we can find cos⁡(θ)=adjacenthypotenuse=ac\\cos(\\theta) = \\frac{\\text{adjacent}}{\\text{hypotenuse}} = \\frac{a}{c}cos(θ)=hypotenuseadjacent​=ca​. Since tan⁡(θ)=oppositeadjacent=ba,\\tan(\\theta) = \\frac{\\text{opposite}}{\\text{adjacent}} = \\frac{b}{a},tan(θ)=adjacentopposite​=ab​, we have tan⁡(θ)=43.\\tan(\\theta) = \\frac{4}{3}.tan(θ)=34​. In a right triangle, find the length of the side not given. Example 1. If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. The ratio of 3: 4: 5 allows us to quickly calculate various lengths in geometric problems without resorting to methods such as the use of tables or to the Pythagoras theorem. So for this example I have a right triangle with a height of 410 meters and a base length of 1,700 meters. (18 / 3 = 6). The length of the prism is 7. 5 \\sqrt{3} &= b.\\ _\\square So if you have the length of the sides of the equilateral triangle, you have (length)^2 + [(1/2)*length]^2 = height. Therefore, sin⁡(θ)=bc=45cos⁡(θ)=ac=35. Pythagorean Theorem. Before we start can you tell me what the definition of a triangle is? Align a protractor on one side of a", "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", "Answer: Let one side of the right-angled triangle be $x$ m and the other side be m. On applying Pythagoras theorem, we have: The value of x cannot be negative. Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}. #### Question 70: The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle. #### Answer: Let the base and altitude of the right-angled triangle be $x$ and y cm, respectively. Therefore, the hypotenuse will be cm. Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively. #### Question 71: The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third is 1 metre more than the shortest side, find the sides of the triangle. #### Answer: Let the shortest side be $x$ m. Therefore, according to the question: Hypotenuse = Third side = m On applying Pythagoras theorem, we get: The length of the side cannot be 0; therefore, the shortest side is 3 m. Therefore, Hypotenuse = 5 m Third side = (3 + 1) = 4", "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 –", "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", "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.", "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.", "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", "10 y = 5 AD = BC = AB/2 + CD/2 = 7.5 Radius r = ? r = {sqrt [(AD)² - (CD/4)²]} /2 = 3.535533906 sqrt (xy) = sqrt (2.5 * 5) = 3.535533906 Dragan Apr 1, 2020 edited by Dragan Apr 1, 2020 #6 +626 0 An alternate method, I always like those! #4 +626 +2 Hello Guest! Make an account! Set up: - By the two tangent theorem, we can find the length of the sides of the trapezoid. - I calculated the shorter leg of the right triangle with hypotenuse AD by doing (DC - AB) / 2 - The height of the longer leg of the right triangles is equal to the diameter indicated by the red line. Calculate: By pythagorean theorem (x + y)2 - (x - y)2 = x2 + 2xy + y2 - (x2 -2xy + y2) = 4xy Diameter = sqrt(4xy) Diameter = $$2\\sqrt{xy}$$ Radius = $$\\sqrt{xy}$$ Apr 1, 2020", "# Find The Length Of The Third Side Of Each Triangle Worksheet Answers  In an isosceles triangle, the angles opposite the equal sides are equal. The perimeter of the triangle is 120 feet. 25, we find that the hypotenuse is approximately 3. The longest side of a triangle is twice as long as the shortest side. 1 (check) 7. 42sin( C) =. Sample: Two pairs of sides are congruent, but the angle is not included. The one page interactive worksheet contains ten questions. All three sides lengths of the triangle are integers and together form a Pythagorean triple. Equilateral Triangle: Equilateral means. The other leg of the right triangle is the altitude of the equilateral triangle, so solve using the Pythagorean Theorem: a 2 + b 2 = c 2. Find the side lengths. 30-60-90 & 45-45-90 right triangles: only need the length of one side to find the other two sides. An easy deduction leads to the smaller square's sides being b - a. Perimeter of a Triangle. 5 Therefore, the range of the side is: 7. If H = 5, and O = 3, then. When viewed from above, the right triangle has a height of 8 feet, a base of 6 feet, and a third side (called the. Set the two expressions for “h”", "us the length of one side of a triangle, given the lengths of the other two sides. This is a very basic, and straightforward theorem that applies to any triangle with one right angle — that is, one angle of 90 degrees. The theorem is not very difficult to use. Suppose we have a triangle with a shortest side of 6 meters, and a second-shortest side of 8 meters. We take these numbers and square them (multiply them by themselves), giving 36 and 64 square meters. We add these together, and get 200 square meters. Now, whatever the longest side is, its square must be 100. The correct length, then, is 10 meters, since 10 squared is 100. Using this knowledge, it's pretty easy to figure out how far apart our two astronauts are. One leg of the triangle is just the x coordinate; the other leg is the y coordinate. So, the distance between the two is given by $$(\\text{distance in meters})^2 = (x \\text{ coordinate})^2 + (y \\text{ coordinate})^2$$ If the coordinates measure meters, then he is 6 meters to the right and 8 meters up, so the Pythagorean Theorem tells us that he is 10 meters from the origin. More generally, there could be a factor of proportionality to take into account, or even two", "What is the perimeter, in centimeters, of the triangle? (1) The area of the triangle is 25 square centimeters. (2) The 2 legs of the triangle are of equal length. I know that (2) is sufficient but I am having difficulty with (1). For (1): Let the legs be $$x$$ and $$y$$. Given: $$h=10$$ and $$\\frac{xy}{2}=25$$ (area of the right triangle $$area=\\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$. Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$. Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\\sqrt{200}$$. Thus $$P=x+y+10=\\sqrt{200}+10$$. Sufficient. Hi Bunuel, Once you outlined the steps above, it's rather easy to solve. My question lies with the strategy -- how did you make the leap to \"square\" (x+y). What is the problem tipped you off that you had to solve a quadratic or at least, re-arrange it? Highlighted the area in question above. Thanks! It should come with practice... We know the values of xy and x^2+y^2, while need to get the value of x+y. Now, if you add twice xy to x^2+y^2 you get the square of x+y, hence squaring x+y is quite natural thing to do. Similar questions to practice: if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html what-is-the-area-of-rectangular-region-r-166186.html what-is-the-perimeter-of-rectangle-r-96381.html Hope this helps. _________________ Intern Joined: 23 Jun 2015 Posts: 11 Schools: YLP '16 GPA: 4 Re: The hypotenuse of a right triangle is 10", "# 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", "= 2, then x = • A). 6 • B). 9 • C). ¼ • D). ½ • E). ¾ Solution: Log3x = 2 is equivalent to 32 = x. Therefore, x = 9 Q25. The vertices of a triangle are A(−1,3), B(3,0), and C(−2,−1). Find the length of side AC. • A). √15 • B). √17 • C). 1√5 • D). √7 • E). 1√7 Solution: Use the Pythagorean theorem to solve. To use the Pythagorean theorem (which is what the distance formula is derived from), draw the segment on a coordinate plane and create a right triangle where AC is the hypotenuse. The legs of the right triangle are 1 and 4, so the hypotenuse is the square root of 12 + 42, or the square root of 17. Q26. Which of the following equations has a graph that has a y-intercept of 4 and is parallel to 3y − 9x = 24? • A). −12x + 4y = 16 • B). 9x − 3y = −15 • C). 2y = 4x + 8 • D). 7y = 14x + 7 • E). 3x − 9y = 14 Solution: Because the line is parallel to 3y – 9x = 24, they must have the same slope. Put the equation into y = mx + b form", "one of the two legs, $a$ or $b$, is longer than the other. The legs of a right triangle $a$ and $b$ can either be $a > b$ or $a < b$ which implies that $a$ is longer than $b$, or $a$ is shorter than $b$, respectively. However, let’s agree only in this lesson that side $a$ is the shorter leg which forces side $b$ to be the longer one. Therefore, $a < b$. It just makes sense since $a$ comes before $b$ in the alphabet. Note: I intentionally leave the case where the legs of the right triangle are equal because no Pythagorean Triple exists when $a=b$. Two Types or Kinds of Pythagorean Triple Primitive Pythagorean Triple (known as “reduced triples”) is a set of three positive integers $a$, $b$, and $c$ with a GCF of $1$. Plus, these three integers must satisfy the equation ${a^2} + {b^2} = {c^2}$. Examples of Primitive Pythagorean Triples • $3$, $4$, and $5$ is a Primitive Pythagorean Triple because their GCF is $1$ and they satisfy the equation ${a^2} + {b^2} = {c^2}$. $${a^2} + {b^2} = {c^2}$$ $${3^2} + {4^2} = {5^2}$$ $${9} + {15} = {25}$$ $${25} = {25}$$ • $5$, $12$, and $13$ is also a Primitive Pythagorean Triple since their largest common divisor is $1$ which", "## Pythagorean triangles – shortest leg, longest leg Find all values of a positive integer constant $n$ such that $x^2 \\; + \\; n \\,x$ and $n \\,x \\; + \\; n$ describe the two legs of a Pythagorean triangle for every positive integer value of $x$. Let n = 2 Then, $a \\; = \\; x^2 \\; + \\; 2 \\, x$ $b \\; = \\; 2 \\, x \\; + \\; 2$ $c \\; = \\; x^2 \\; + \\; 2 \\, x \\; + \\; 2$ $(x^2 + 2 \\, x)^2 \\; + \\; (2 \\, x + 2)^2 \\; = \\; (x^2 + 2 \\, x + 2)^2$ True for all $x$ Is n = 2 the only solution? $x \\; = \\; 1$ $4 \\, n^2 \\; + \\; (n+1)^2 \\; = \\; 5 \\, n^2 \\; + \\; 2 \\, n \\; + \\; 1$ $5 \\, n^2 \\; + \\; 2 \\, n \\; + \\; 1 \\; = \\; y^2$ getting a Diophantine equation. There are infinitely many solutions. $x \\; = \\; 3$ $16 \\, n^2 \\; + \\; (3 \\, n + 9)^2 \\; = \\; y^2$ —-> $n = 2, \\; y = 17$ math grad - Interest: Number theory This entry was posted in Number Puzzles", "and median t2 = 12 cm. .. How to calculate the length of the sides? 12. Sides of right angled triangle One leg 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. 13. Coordinates of square vertices I have coordinates of square vertices A / -3; 1/and B/1; 4 /. Find coordinates of vertices C and D, C 'and D'. Thanks Peter. 14. Surface of the cylinder Calculate the surface of the cylinder for which the shell area is Spl = 20 cm2 and the height v = 3.5 cm 15. Compound interest 3 After 8 years, what is the total amount of a compound interest investment of \\$25,000 at 3% interest, compounded quarterly? (interest is now dream - in the year 2019) 16. Equation 23 Find value of unknown x in equation: x+3/x+1=5 (problem finding x) 17. Geography tests On three 150-point geography tests, you earned grades of 88%, 94%, and 90%. The final test is worth 250 points. What percent do you need on the final to earn 93% of the total points on all tests?", "sides and three interior angles. The measures of the interior angles add up to $180^{\\circ}$. The length of each side must be less than the sum of the lengths of the other two sides. English Chinese Remark & Example equilateral triangle 等边三角形 A triangle with three congruent sides is called an equilateral triangle. isosceles triangle 等腰三角形 A triangle with at least two congruent sides is called an isosceles triangle. right triangle 直角三角形 A triangle with an interior right angle is called a right triangle. The side opposite the right angle is called the hypotenuse; the other two sides are called legs. hypotenuse legs （直角三角形的）斜边，弦 （直角三角形的腰 congruent triangles 全等三角形 similar triangles 相似三角形 included angle 夹角 scale factor of similarity 相似尺度因子，比例因子 • The Pythagorean Theorem: P98 • The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. • Isosceles right triangle: the lengths of its sides are in the ratio 1 to 1 to $\\sqrt 2$. • 30$^{\\circ}$-60$^{\\circ}$-90$^{\\circ}$ right triangle: the lengths of its sides are in the ratio 1 to $\\sqrt 3$ to 2. • The Area of a Triangle: P100 • The area A of a triangle is given by the formula $\\displaystyle A = \\frac {bh}{2}$", "Use Pythagoras to calculate this: $$\\sqrt{\\left(\\left(5--3\\right)^{2}+\\left(7--5\\right)^{2}\\right)}=14.4...$$ 7 What is the longest length of a triangle with shorter sides 5 and 8? Use Pythagoras to calculate this: $$\\sqrt{\\left(5^{2}+8^{2}\\right)}$$ 8 Rearrange the equation $$y=0.5x+4$$ to make $$x$$ the subject x= 1. Subtract 4 from boths ides. 2. Divide both sides by 0.5, which is equivalent to multiplying both sides by 2. 9 What is the equation of the line that crosses through the coordinates (3,3) and (0,-3), written in the form $$y=mx+c$$ y=" ]

[ "us the length of one side of a triangle, given the lengths of the other two sides. This is a very basic, and straightforward theorem that applies to any triangle with one right angle — that is, one angle of 90 degrees. The theorem is not very difficult to use. Suppose we have a triangle with a shortest side of 6 meters, and a second-shortest side of 8 meters. We take these numbers and square them (multiply them by themselves), giving 36 and 64 square meters. We add these together, and get 200 square meters. Now, whatever the longest side is, its square must be 100. The correct length, then, is 10 meters, since 10 squared is 100. Using this knowledge, it's pretty easy to figure out how far apart our two astronauts are. One leg of the triangle is just the x coordinate; the other leg is the y coordinate. So, the distance between the two is given by $$(\\text{distance in meters})^2 = (x \\text{ coordinate})^2 + (y \\text{ coordinate})^2$$ If the coordinates measure meters, then he is 6 meters to the right and 8 meters up, so the Pythagorean Theorem tells us that he is 10 meters from the origin. More generally, there could be a factor of proportionality to take into account, or even two", "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.", "Use Pythagoras to calculate this: $$\\sqrt{\\left(\\left(5--3\\right)^{2}+\\left(7--5\\right)^{2}\\right)}=14.4...$$ 7 What is the longest length of a triangle with shorter sides 5 and 8? Use Pythagoras to calculate this: $$\\sqrt{\\left(5^{2}+8^{2}\\right)}$$ 8 Rearrange the equation $$y=0.5x+4$$ to make $$x$$ the subject x= 1. Subtract 4 from boths ides. 2. Divide both sides by 0.5, which is equivalent to multiplying both sides by 2. 9 What is the equation of the line that crosses through the coordinates (3,3) and (0,-3), written in the form $$y=mx+c$$ y=", "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", "find the two sides of that triangle by using Pythagoras theorem. Right isosceles triangle What can be the area of a right isosceles triangle with a side length of 8 cm? Could form a right triangle relate to each other боловсролыг хүссэн газраасаа use pythagorean theorem to find isosceles triangle side lengths хүссэн үедээ, насан! Sides have the same three methods that applied to Jessica 's triangle is perpendicular! Sides measuring 5 inches each and one side measuring 8 inches the longest side of the sides... 9 + 16 = 25 your question the Pythagorean theorem find a missing length! Relate to each use pythagorean theorem to find isosceles triangle side lengths *.kastatic.org and *.kasandbox.org are unblocked 5 each. How the lengths of the leg oppo… how to use the Pythagorean theorem, we have calculate perimeter. ^2 + ( 4 ) ^2 + ( 4 ) ^2 = ( c ) ^2 + ( ). Be the area of a right triangle are 3 and 4 units long 're behind a filter! Meters long each the three sides of equal length and c represents the length the! Can be the area of a right triangle on this relationship, we can isolate each unknown length use... The base of the hypotenuse shown below *.kastatic.org and *.kasandbox.org are unblocked base", "of the two shorter legs is 202. Compare this to the square of the longest side, 14. $14^2=196$ The square of the longest side is 196. So the two values are not equal ($202 \\neq 196$ or $a^2 + b^2 \\neq c^2$) and this triangle is not a right triangle. Since you can eliminate the right triangle from your choices, now you can compare the two values, $a^2 + b^2$ and $c^2$ to identify which is greater: $202 & > 196\\\\\\text{(sum of shorter sides)}^2 & > \\text{(longest side)}^2$ Since the sum of the square of the shorter sides is greater than the square of the longest side (in symbols $a^2 + b^2 > c^2$), this is an acute triangle. Example 5 Classify the triangle below as right, acute, or obtuse. The two shorter sides of the triangle are _____________ and ______________. The longest side of the triangle is ________________. First, set up an equation to find the sum of the squares of the two shorter legs by substituting the smaller numbers for $a$ and $b$. $a^2+ b^2 &= 16^2+ 30^2\\\\&= 256+900\\\\&= 1156$ The sum of the squares of the two legs is 1156. Compare this to the square of the longest side, 34. $c^2 = 34^2 =1156$ The square of the longest side is also 1156. Since the", "= 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?", "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", "triangle with side lengths $6$ cm, $10$ cm, and $8$ cm is a right triangle. Check whether the square of the length of the longest side is the sum of the squares of the other two sides. $\\begin{array}{l}{\\left(10\\right)}^{2}\\stackrel{?}{=}{\\left(8\\right)}^{2}+{\\left(6\\right)}^{2}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}100\\stackrel{?}{=}64+36\\\\ \\text{\\hspace{0.17em}}100=100\\end{array}$ Apply the converse of Pythagorean Theorem. Since the square of the length of the longest side is the sum of the squares of the other two sides, by the converse of the Pythagorean Theorem, the triangle is a right triangle. A corollary to the theorem categorizes triangles in to acute, right, or obtuse. In a triangle with side lengths $a$, $b$, and $c$ where $c$ is the length of the longest side, if ${c}^{2}<{a}^{2}+{b}^{2}$ then the triangle is acute, and if ${c}^{2}>{a}^{2}+{b}^{2}$ then the triangle is obtuse. Example 2: Check whether the triangle with the side lengths $5$, $7$, and $9$ units is an acute, right, or obtuse triangle. The longest side of the triangle has a length of $9$ units. Compare the square of the length of the longest side and the sum of squares of the other two sides. Square of the length of the longest side is ${9}^{2}=81$ sq. units. Sum of the squares of the other two sides is That is, ${9}^{2}>{5}^{2}+{7}^{2}$. Therefore, by the corollary to the converse of Pythagorean Theorem, the triangle", "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.", "# Find The Length Of The Third Side Of Each Triangle Worksheet Answers  In an isosceles triangle, the angles opposite the equal sides are equal. The perimeter of the triangle is 120 feet. 25, we find that the hypotenuse is approximately 3. The longest side of a triangle is twice as long as the shortest side. 1 (check) 7. 42sin( C) =. Sample: Two pairs of sides are congruent, but the angle is not included. The one page interactive worksheet contains ten questions. All three sides lengths of the triangle are integers and together form a Pythagorean triple. Equilateral Triangle: Equilateral means. The other leg of the right triangle is the altitude of the equilateral triangle, so solve using the Pythagorean Theorem: a 2 + b 2 = c 2. Find the side lengths. 30-60-90 & 45-45-90 right triangles: only need the length of one side to find the other two sides. An easy deduction leads to the smaller square's sides being b - a. Perimeter of a Triangle. 5 Therefore, the range of the side is: 7. If H = 5, and O = 3, then. When viewed from above, the right triangle has a height of 8 feet, a base of 6 feet, and a third side (called the. Set the two expressions for “h”", "What is the perimeter, in centimeters, of the triangle? (1) The area of the triangle is 25 square centimeters. (2) The 2 legs of the triangle are of equal length. I know that (2) is sufficient but I am having difficulty with (1). For (1): Let the legs be $$x$$ and $$y$$. Given: $$h=10$$ and $$\\frac{xy}{2}=25$$ (area of the right triangle $$area=\\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$. Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$. Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\\sqrt{200}$$. Thus $$P=x+y+10=\\sqrt{200}+10$$. Sufficient. Hi Bunuel, Once you outlined the steps above, it's rather easy to solve. My question lies with the strategy -- how did you make the leap to \"square\" (x+y). What is the problem tipped you off that you had to solve a quadratic or at least, re-arrange it? Highlighted the area in question above. Thanks! It should come with practice... We know the values of xy and x^2+y^2, while need to get the value of x+y. Now, if you add twice xy to x^2+y^2 you get the square of x+y, hence squaring x+y is quite natural thing to do. Similar questions to practice: if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html what-is-the-area-of-rectangular-region-r-166186.html what-is-the-perimeter-of-rectangle-r-96381.html Hope this helps. _________________ Intern Joined: 23 Jun 2015 Posts: 11 Schools: YLP '16 GPA: 4 Re: The hypotenuse of a right triangle is 10", "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", "10 y = 5 AD = BC = AB/2 + CD/2 = 7.5 Radius r = ? r = {sqrt [(AD)² - (CD/4)²]} /2 = 3.535533906 sqrt (xy) = sqrt (2.5 * 5) = 3.535533906 Dragan Apr 1, 2020 edited by Dragan Apr 1, 2020 #6 +626 0 An alternate method, I always like those! #4 +626 +2 Hello Guest! Make an account! Set up: - By the two tangent theorem, we can find the length of the sides of the trapezoid. - I calculated the shorter leg of the right triangle with hypotenuse AD by doing (DC - AB) / 2 - The height of the longer leg of the right triangles is equal to the diameter indicated by the red line. Calculate: By pythagorean theorem (x + y)2 - (x - y)2 = x2 + 2xy + y2 - (x2 -2xy + y2) = 4xy Diameter = sqrt(4xy) Diameter = $$2\\sqrt{xy}$$ Radius = $$\\sqrt{xy}$$ Apr 1, 2020", "Answer: Let one side of the right-angled triangle be $x$ m and the other side be m. On applying Pythagoras theorem, we have: The value of x cannot be negative. Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}. #### Question 70: The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle. #### Answer: Let the base and altitude of the right-angled triangle be $x$ and y cm, respectively. Therefore, the hypotenuse will be cm. Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively. #### Question 71: The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third is 1 metre more than the shortest side, find the sides of the triangle. #### Answer: Let the shortest side be $x$ m. Therefore, according to the question: Hypotenuse = Third side = m On applying Pythagoras theorem, we get: The length of the side cannot be 0; therefore, the shortest side is 3 m. Therefore, Hypotenuse = 5 m Third side = (3 + 1) = 4", "{ 1 }{ 2 }$$ x 1.2 x 3.5 m² = 2.1 m² Question 8. Solution: Legs of a right angled triangle = 3 : 4 Let one leg (base) = 3x Then second leg (altitude) = 4x Area = $$\\frac { 1 }{ 2 }$$ x base x altitude = $$\\frac { 1 }{ 2 }$$ x 3x x 4x = 6x² 6x² = 1014 ⇒ x² = $$\\frac { 1014 }{ 6 }$$ = 169 = (13)² x = 13 one leg'(Base) = 3x = 3 x 13 = 39 cm and second leg (altitude) = 4x = 4 x 13 = 52 cm Question 9. Solution: One side BC of a right triangular scarf = 80 cm and longest side AC = 1 m = 100 cm By Pythagoras Theorem, AC² = AB² + BC² ⇒ (100)² = AB² + (80)² ⇒ 10000 = AB² + 6400 ⇒ AB² = 10000 – 6400 ⇒ AB² = 3600 = (60)² ⇒ AB = 60 Second side = 60 cm Area of the scarf = $$\\frac { 1 }{ 2 }$$ x b x h = $$\\frac { 1 }{ 2 }$$ x 80 x 60 cm2 = 2400 cm² Rate of cost = Rs. 250 per m² Total cost =$$\\frac { 2400 }{ 100 x 100", "assuming the height would be 3... 58. anonymous |dw:1352595405601:dw| 59. anonymous Now I can't get a ratio.... 60. anonymous With a circle radius of \"r\", being the \"short leg\", then the \"long leg\" will be r times sqrt(3). So a triangle side will be 2r(sqrt(3)) + 2r. 61. anonymous So it would just be 2root(3)+2 . 62. anonymous If you are going with your assumption or requirement that r is \"1\", then, yes. 63. anonymous So, you should be able to get the altitude using Pythagorean and then area of triangle, and areas of circles. You're just about done! 64. anonymous I calculate that the length of a side of the master triangle is 6.4641r Anyone else? tanks to: adjacent = r / tan(30) 65. anonymous or alternatively by expressing the length of the line as: 2r * root(3) + 3r that plugs in nicely to the formula for the area of an el triangle: (length of side * root(3)) / 4 so that tells us the area is... 66. AccessDenied I missed the answer placed by the original poster, but my work in ms paint is here: http://puu.sh/1oNgt 67. geerky42 I got the same answer. Thank you, everyone, for your times. I appreciate it. 68. AccessDenied You're welcome! :)", "we get: A = 1/2 * 6 * 4 A = 24 Therefore, the area of triangle def is 24. Learn More: How do I stop comparing myself to his ex? ## How do the lengths of the sides of triangle abc compare to the lengths of the sides of triangle def? The lengths of the sides of triangle abc compare to the lengths of the sides of triangle def in the following way: The length of side a of triangle abc is shorter than the length of side d of triangle def. The length of side b of triangle abc is shorter than the length of side e of triangle def. The length of side c of triangle abc is shorter than the length of side f of triangle def. ## What is the length of the longest side of triangle abc? There are a number of different ways to approach this question, and it largely depends on what information is given about the triangle in question. If no specific information is given, then one can either use the Pythagorean theorem or Heron's formula to calculate the length of the longest side. The Pythagorean theorem states that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the", "Application Worksheet 1 18 Student Outcomes • Students apply the Pythagorean:... Understanding Mathematics, at every level, requires learner engagement sound waves in water category - Pythagoras Application - top! Real world and mathematical problems in two dimensions hypotenuse of a triangle with side lengths,., 2013 ) Central Bucks High School South legs is equal to the square the... Much more About this Quiz & Worksheet unlock the printable worksheets have exercises on finding the leg and of! Rectangle has length 8 pythagoras application worksheet and diagonal 17 cm worksheets and see our Maths lesson Finishers then sign up a. As search for them in nature and others using Algebra Theorem can be used with any and... Relationship between the 3 sides of a right-angled triangle Email print desifoto/Getty Images learner engagement 120 2. Right-Angled triangle, some graphical in nature and others using Algebra Flipboard Email print desifoto/Getty.! Example a rectangle has length 8 cm and diagonal 17 cm - 120 P 2 - 17P ) -... The speed of sound waves in water calculate the speed of sound in! Years earlier, by the Chinese and Babylonians triangle is a right triangle has one leg that 13... By Grade Resources View more b be the width, measured in cm, generally. Student Outcomes • Students apply the Pythagorean Theorem, 5-a-day", "sides and three interior angles. The measures of the interior angles add up to $180^{\\circ}$. The length of each side must be less than the sum of the lengths of the other two sides. English Chinese Remark & Example equilateral triangle 等边三角形 A triangle with three congruent sides is called an equilateral triangle. isosceles triangle 等腰三角形 A triangle with at least two congruent sides is called an isosceles triangle. right triangle 直角三角形 A triangle with an interior right angle is called a right triangle. The side opposite the right angle is called the hypotenuse; the other two sides are called legs. hypotenuse legs （直角三角形的）斜边，弦 （直角三角形的腰 congruent triangles 全等三角形 similar triangles 相似三角形 included angle 夹角 scale factor of similarity 相似尺度因子，比例因子 • The Pythagorean Theorem: P98 • The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. • Isosceles right triangle: the lengths of its sides are in the ratio 1 to 1 to $\\sqrt 2$. • 30$^{\\circ}$-60$^{\\circ}$-90$^{\\circ}$ right triangle: the lengths of its sides are in the ratio 1 to $\\sqrt 3$ to 2. • The Area of a Triangle: P100 • The area A of a triangle is given by the formula $\\displaystyle A = \\frac {bh}{2}$" ]

[ "+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", "+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", "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", "Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$. Case 3: $g$ passes through $CD$ or $BC$ $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,N); dot(XB); label(\"X_2\",XB,E); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$. By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\\angle EAB = \\angle AED = \\angle CEX_1$. Thus, by AA Similarity, $\\triangle X_2BA \\sim \\triangle X_3DE \\sim \\triangle X_1CE$. Using similar triangle ratios, we have $DX_3 = \\tfrac{bx}{a}$ and $X_1 = \\tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \\frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "to 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 $O_{1}$,", "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.)", "+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", "(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", "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$$25k^2+9k^2-15k^2 =", "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$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}}.$$", "+0 # heeeeelpp 0 131 1 In the figure, $\\overline{DC}$ is parallel to $\\overline{AB}$. We have $\\angle DCA = 40^\\circ$ and $\\angle ABC = 73^\\circ$. Find $\\angle ACB$. Feb 27, 2022 #1 +361 +3 Since, $$\\overline{DC}$$ is parellel to $$\\overline{AB}$$$$\\angle{DCA}$$ is the same as $$\\angle{CAB}$$. $$\\angle{ACB}=180-(40+73)$$ $$=180-113=67$$", "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 $BC$. Case 4: $g$ does not intersect the parallelogram at any other points $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(-40,40),XA=(-15,15),XB=(10,-10),XC=(25,-25); draw(D--e--XC--B); draw(D--XA); draw(C--XB); draw(B--XC); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,S); dot(XB); label(\"X_2\",XB,S); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $DX_1, CX_2, BX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_3$, making $EC = a+x$. By Alternate Interior Angles Theorem, $\\angle CEA = \\angle BAX_3$. Thus, by AA Similarity, $\\triangle EDX_1 \\sim \\triangle ECX_2 \\sim \\triangle ABX_3$. Using similar triangle ratios, we have $DX_1 = \\tfrac{bx}{a}$ and $CX_2 = \\tfrac{ba+bx}{a}$. Thus, we have $BX_3 + DX_1 = \\frac{ba+bx}{a} = CX_2$, so the distance from $C$ to $g$ is the sum of the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. In all of the cases, the sum or difference of the distance from $B$ to $g$ and the distance from $D$ to $g$ is equal to the distance from $C$ to $g$.", "circle at $G$. Draw line $BG$. $A[\\Delta ABD] = A[\\Delta ABG] = 120$ (Since $\\square ABGD$ is cyclic) But $A[\\Delta ABE]$ is common in both with an area of 60. So, $A[\\Delta AED] = A[\\Delta BEG]$. \\therefore $A[\\Delta AED] \\cong A[\\Delta BEG]$ (SAS Congruency Theorem). In $\\Delta AED$, let $DI$ be the median of $\\Delta AED$. Which means $A[\\Delta AID] = 30 = A[\\Delta EID]$ Rotate $\\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$. $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\\frac{AE}{EF} = 2:1$. $AE$ is a radius and $EF$ is half of it implies $EF$ = $\\frac{radius}{2}$. Which means $A[\\Delta BEF] \\cong A[\\Delta DEI]$ Thus $A[\\Delta BEF] = \\boxed{\\textbf{(B) }30}$ ~phoenixfire & flamewavelight ## Solution 8 $[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label(\"M\",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy]$ Using the ratio of $\\overline{AD}$ and $\\overline{CD}$, we find the area of $\\triangle ADB$ is $120$ and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle", "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 $1", "ABD$ 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", "# 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 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", "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 =" ]

[ "+0 # heeeeelpp 0 131 1 In the figure, $\\overline{DC}$ is parallel to $\\overline{AB}$. We have $\\angle DCA = 40^\\circ$ and $\\angle ABC = 73^\\circ$. Find $\\angle ACB$. Feb 27, 2022 #1 +361 +3 Since, $$\\overline{DC}$$ is parellel to $$\\overline{AB}$$$$\\angle{DCA}$$ is the same as $$\\angle{CAB}$$. $$\\angle{ACB}=180-(40+73)$$ $$=180-113=67$$", "BCD$ + $\\angle ACB$ = 180o [linear Pair] => 41 + $\\angle ACB$ = 180 => $\\angle ACB$ = 180 - 41 $\\angle ACB$ = 139o ------------------- (b)", "47^o$ $\\therefore \\angle BCD = 180^o-75^o = 105^o$ $\\\\$", "# Question 1 of 10 In the diagram below, ABCD is a parallelogram. AC = CE = CF. $\\angle$AEC = 55$^\\circ$ and $\\angle$AFC = 30$^\\circ$ AF and CE are straight lines. Find $\\angle$ABC. A 60$^\\circ$ B 70$^\\circ$ C 80$^\\circ$ D 90$^\\circ$ E None of the above", "Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides. So for quadrilateral sum of the angles equals to $$180(4-2)=360$$ degrees. Given: $$\\angle{ABC}=110$$ and $$\\angle{AED}=100$$. Question: $$\\angle{BCD}-\\angle{ADE}=?$$. Now as $$\\angle{ABC}=110$$ and $$\\angle{AED}=100$$ and the sum of all angles is 360 degrees then the sum of other two angles in quadrilateral must be $$\\angle{BCD}+\\angle{BAE}=360-(110+100)=150$$. (1) $$\\angle{DAB}=45$$ --> $$\\angle{BAE}=\\angle{DAB}+\\angle{DAE}=45+\\angle{DAE}$$, but $$\\angle{DAE}=180-100-\\angle{ADE}=80-\\angle{ADE}$$ (the sum of interior angles in triangle ADE equals to 180 degrees) --> $$\\angle{BAE}=45+80-\\angle{ADE}=125-\\angle{ADE}$$. So $$\\angle{BCD}+\\angle{BAE}=150$$ becomes $$\\angle{BCD}+125-\\angle{ADE}=150$$ --> $$\\angle{BCD}-\\angle{ADE}=25$$. Sufficient. (2) $$\\angle{BCD}+\\angle{ADC}=205$$ --> $$\\angle{ADC}=180-\\angle{ADE}$$ (straight line) --> $$\\angle{BCD}+180-\\angle{ADE}=205$$ --> $$\\angle{BCD}-\\angle{ADE}=25$$. Sufficient. Hope it's clear. _________________ Intern Joined: 29 Jul 2010 Posts: 40 Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink] ### Show Tags 13 Aug 2010, 15:35 good Qn. Thanks Bunuel. _________________ \"Not everyone who works hard succeeds. But all those who succeeded have worked hard!\" ~~ Coach Kamogawa Best, ItsNotOver Director Joined: 07 Jun 2004 Posts: 609 Location: PA Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink] ### Show Tags 13 Aug 2010, 15:40 thanks i now get it i was taking interior angles as 180 which is not necessarily true _________________ If the Q jogged your mind do Kudos me : )", "\\ || \\ \\overline{DC}, \\ \\overline{AD} \\ || \\ \\overline{BC}\\end{align*} Definition of a parallelogram 3. ABDBDC, ADBDBC\\begin{align*}\\angle ABD \\cong BDC, \\ \\angle ADB \\cong DBC\\end{align*} Alternate Interior Angles Theorem 4. DB¯¯¯¯¯¯DB¯¯¯¯¯¯\\begin{align*}\\overline{DB} \\cong \\overline{DB}\\end{align*} Reflexive PoC 5. ABDCDB\\begin{align*}\\triangle ABD \\cong \\triangle CDB\\end{align*} ASA 6. AB¯¯¯¯¯DC¯¯¯¯¯, AD¯¯¯¯¯¯BC¯¯¯¯¯\\begin{align*}\\overline{AB} \\cong \\overline{DC}, \\ \\overline{AD} \\cong \\overline{BC}\\end{align*} CPCTC Example A ABCD\\begin{align*}ABCD\\end{align*} is a parallelogram. If mA=56\\begin{align*}m \\angle A = 56^\\circ\\end{align*}, find the measure of the other three angles. Draw a picture. When labeling the vertices, the letters are listed, in order, clockwise. If mA=56\\begin{align*}m \\angle A = 56^\\circ\\end{align*}, then mC=56\\begin{align*}m \\angle C = 56^\\circ\\end{align*} because they are opposite angles. B\\begin{align*}\\angle B\\end{align*} and D\\begin{align*}\\angle D\\end{align*} are consecutive angles with A\\begin{align*}\\angle A\\end{align*}, so they are both supplementary to A\\begin{align*}\\angle A\\end{align*}. mA+mB=180,56+mB=180,mB=124\\begin{align*}m \\angle A+m \\angle B=180^\\circ, 56^\\circ+m \\angle B=180^\\circ, m \\angle B=124^\\circ\\end{align*}. mD=124\\begin{align*}m \\angle D = 124^\\circ\\end{align*}. Example B Find the values of x\\begin{align*}x\\end{align*} and y\\begin{align*}y\\end{align*}. Opposite sides are congruent, so we can set each pair equal to each other and solve both equations. 6x74xx=2x+9=16=4y2+3=12 y2=9 y=3 or 3 Even though \\begin{align*}y = 3\\end{align*} or -3, lengths cannot be negative, so \\begin{align*}y = 3\\end{align*}. Example C Show that the diagonals of \\begin{align*}FGHJ\\end{align*} bisect each other. The easiest way to show this is to find the midpoint of each diagonal. If it is the same point, you know", "an example. Given that DE and GB are parallel, how can we find x, y and z? \\begin{align} \\angle EFC &= \\angle BGC \\text{ (corr. } \\angle \\text{s, DE} / / \\text{GB)} \\\\ z &= 125^\\circ \\end{align} \\begin{align} \\angle AFD &= \\angle BGC \\text{ (alt. } \\angle \\text{s, DE} / / \\text{GB)} \\\\ y &= 125^\\circ \\end{align} \\begin{align} \\angle AGB + \\angle BGC &= 180^\\circ \\text{ (adj. } \\angle \\text{s on st. line)}\\\\ \\angle AGB + 125^\\circ &= 180^\\circ \\\\ \\angle AGB &= 180^\\circ - 125^\\circ \\\\ &= 55^\\circ \\end{align} \\begin{align} \\angle AGB + \\angle EFC &= 180^\\circ \\text{ (int. } \\angle \\text{s, DE} / / \\text{GB)} \\\\ 55^\\circ + x &= 180^\\circ \\\\ x &= 180^\\circ - 55^\\circ \\\\ &= 125^\\circ \\end{align}", "88 views In a trapezium $\\text{ABCD},\\; \\text{AB}$ is parallel to $\\text{DC}, \\; \\text{BC}$ is perpendicular to $\\text{DC}$ and $\\angle \\text{BAD} = 45^{\\circ}.$ If $\\text{DC} = 5 \\; \\text{cm}, \\; \\text{BC} = 4 \\; \\text{cm},$ the area of the trapezium in $\\text{sq. cm}$ is 1 134 views", "## Basic College Mathematics (10th Edition) $\\angle$ABC = $\\angle$BCD = 42$^{\\circ}$ $\\angle$ECD = $\\angle$ABF = 138$^{\\circ}$ $\\angle$ABC and $\\angle$BCD are congruent angles. Therefore they have the same measure. $\\angle$ABC = $\\angle$BCD = 42$^{\\circ}$ The sum of $\\angle$ABC and $\\angle$ABF = 180$^{\\circ}$ due to $\\angle$CBF being a Straight Angle. 180$^{\\circ}$ = 42$^{\\circ}$ + $\\angle$ABF $\\angle$ABF = 180$^{\\circ}$ - 42$^{\\circ}$ $\\angle$ABF = 138$^{\\circ}$ $\\angle$ABF and $\\angle$ECD are congruent angles. Therefore they have the same measure. $\\angle$ECD = $\\angle$ABF = 138$^{\\circ}$", "angles of a cyclic quad.) ⇒∠B + 110°= 180° ∴ ∠ B or ∠ ABC = 180° – 110° – 70″ Now. in ∆ ABC, ∠ BAC + ∠ ABC – ∠ ACB = 180° ⇒ 50° + 70° + ∠ ACB – 180° ⇒ 120° – ∠ ACB = 180° ∴ ∠ ACB = 180° – 120° = 60″ OL ⊥ AC and hence it bisects AC, O is the center of the circle. OL ⊥ AC and hence it bisects AC, O is the center of the circle. ∴ ∠ DAC = ∠ ACB (Alternate angles) = 60° ∠ DAC + ∠ ADC + ∠ DCA -= 180° 60°+ 110° + ∠ DCA = 180° 170° + ∠ DCA – 180° ∴ ∠ DCA = 180″ – 170° = 10° #### Question 7. In the given figure, C and D arc points on the semi-circle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°. calculate angle BDC. ∴ ABCD is a cyclic quad. ∠ BAD ∠ BCD 180 (Sum of opposite angles) ⇒ 70° + ∠ BCD = 180° ⇒ ∠ BCD = 180°- 70° = 110° Now in ∆ BCD, ∠ BCD + ∠ DBC + ∠ BDC = 180° ⇒ 30°+ 110° + ∠ BDC = 180° ⇒", "$\\angle$ADC = 64.27°. Find. a) $\\angle$BCD b) $\\angle$AEB c) $\\angle$CED", "Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these two sets. Case (i): Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\\pm 3, d-e=\\pm 1$, this is not possible. Case (ii): Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$", "and (ii) we get, $\\angle CAB+\\angle CAD=180^o–2\\angle ACB+180^o–2\\angle ACD$ $180^o=360^o–2\\angle ACB-2\\angle ACD$ $2(\\angle ACB+\\angle ACD)=180^o$ $\\angle BCD=90^o$. Updated on 10-Oct-2022 13:41:11", "Hence the angles of the quadrilateral are $36{}^\\circ$, $60{}^\\circ$, $108{}^\\circ$ and $156{}^\\circ$. 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle. Ans: Given that the diagonals of a parallelogram are equal. Let us assume a parallelogram $ABCD$ whose diagonals are equal as shown in the figure below. Here $ABCD$ is a parallelogram, so we can write that $AB=DC$ since opposite sides of a parallelogram are equal. Consider the triangles $ABC$ and $DCB$. In these two triangles we have $AB=DC$, $BC=BC$ and as per given data $AC=DB$. So, we can write that $\\Delta ABC\\cong \\Delta DCB$ by SSS Congruence rule. $\\Rightarrow \\angle ABC=\\angle DCB$ We know that the sum of the angles on the same side of the transversal is $180{}^\\circ$. $\\therefore \\angle ABC+\\angle DCB=180{}^\\circ$ Substitute the value $\\angle ABC=\\angle DCB$ in the above equation. $\\Rightarrow \\angle ABC+\\angle ABC=180{}^\\circ$ $\\Rightarrow 2\\angle ABC=180{}^\\circ$ $\\Rightarrow \\angle ABC=\\dfrac{180{}^\\circ }{2}$ $\\Rightarrow \\angle ABC=90{}^\\circ$ Here we have the angle $ABC$ and $DCB$ as $90{}^\\circ$. When the angles of the parallelogram are equal to $90{}^\\circ$, then it is called as Rectangle. Hence, if the diagonals of the parallelogram are equal then it is called as Rectangle. 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Ans: Given", "## Geometry: Common Core (15th Edition) $m\\angle BCD=25$ Because $\\angle FCA$ and $\\angle ACD$ are a linear pair and $m\\angle FCA$ is a right angle, we know that$\\angle ACD$ is also a right angle. Angle Addition Postulate: $m\\angle ACB+m\\angle BCD=m\\angle ACD$ $65+m\\angle BCD=90$ $65+m\\angle BCD-65=90-65$ $m\\angle BCD=25$", "+0 # In the figure below, $ABCD$ is a parallelogram, $AD=18$ and $DC=27$. Segments $\\overline{EG}$ and $\\overline{FH}$ are perpendicular to sides -1 31 1 In the figure below, $ABCD$ is a parallelogram, $AD=18$ and $DC=27$. Segments $\\overline{EG}$ and $\\overline{FH}$ are perpendicular to sides Apr 25, 2022", "the transversal. Then, $\\angle BAE+\\angle AEF=180^{\\circ}$ [Angles on the same side of a transversal line are supplementary] Again, $EF\\parallel CD$ and CE is the transversal. Then, $\\angle DCE+\\angle CEF=180^{\\circ}$ [Angles on the same side of a transversal line are supplementary] à $\\angle DCE+(\\angle AEC+\\angle AEF)=180^{\\circ}$ à $\\angle DCE+\\angle AEC+180^{\\circ}-\\angle BAE=180^{\\circ}$ à $\\angle BAE-\\angle DCE=\\angle AEC$ Q11) In the given figure, $AB\\parallel CD\\; and \\; CD\\parallel EF$. Find the value of x. Ans. We have, $AB\\parallel CD$ and $BC\\parallel ED$. $BD\\parallel ED$ and CD is the transversal. Then, $\\angle BCD+\\angle CDE=180^{\\circ}$ [Angles on the same side of a transversal line are supplementary] à $\\angle BCD+75=180$ à $\\angle BCD=105^{\\circ}$ $AB\\parallel CD$ and BC is the transversal. $\\angle ABC=\\angle BCD$ (alternate angles) à $x^{\\circ}=105^{\\circ}$ à x = 105 Q12) In the given figure , $AB\\parallel CD$ .Prove that P + q – r = $180^{\\circ}$ . Ans. Draw $PFQ\\parallel AB\\parallel CD$ Now, $PFQ\\parallel AB$ and EF is the transversal. Then, $\\angle AEF+\\angle EFP=180^{\\circ}$ …(1) [Angles on the same side of a transversal line are supplementary] Also, $PFQ\\parallel CD$ $\\angle PFQ=\\angle FGD=r^{\\circ}$ [Alternate angles] And $\\angle EFP=\\angle EFG-\\angle PFG=q^{\\circ}-r^{\\circ}$ Putting the value of $\\angle EFP$ in equ. (i) We get, $p^{\\circ}+q^{\\circ}-r^{\\circ}=180^{\\circ}$ à p + q – r = 180 Q13) In the given figure, $AB\\parallel PQ$. Find the value of x and", "{\\rm{ABC}} = {\\rm{18}}{0^0}\\\\& \\Rightarrow {\\rm{ }}\\angle {\\rm{ABC}} = {\\rm{9}}{0^0}{\\rm{ }}\\end{align} Since $$ABCD$$ is a parallelogram and one of its interior angles is \\begin{align}90^{\\circ},\\end{align} $$ABCD$$ is a rectangle. Video Solution", "AEB}\\equiv{\\angle ABE} = {{180^{\\circ} - 108^{\\circ}}\\over2} = {72^{\\circ}\\over2} = {36^{\\circ}}$$ Then I calculated other sets of equal angles: $${\\angle DFC}\\equiv{\\angle EFB}= {180^{\\circ} - 2\\times 36^{\\circ}} = {108^{\\circ}}$$ $${\\angle EFD}\\equiv{\\angle BFC}= {{360^{\\circ} - 2\\times 108^{\\circ}}\\over2} = {{360^{\\circ} - 216^{\\circ}}\\over2} ={144^{\\circ}\\over2}= {72^{\\circ}}$$ $${\\angle EDF}\\equiv{\\angle BCF}= { 180^{\\circ} - 36^{\\circ} - 72^{\\circ}} = { 72^{\\circ}}$$ $${\\angle BEF}\\equiv{\\angle EBF}= {{180^{\\circ} - 108^{\\circ}}\\over2} = {72^{\\circ}\\over2} = {36^{\\circ}}$$ Here are all the measures of the angles that I obtained: Angles Measures ${\\angle EAB}\\equiv{\\angle ABC}\\equiv{\\angle BCD}\\equiv{\\angle CDE}\\equiv{\\angle DEA}$ 108° ${\\angle EAB}\\equiv{\\angle ABC}\\equiv{\\angle BCD}\\equiv{\\angle CDE}\\equiv{\\angle DEA}$ 36° ${\\angle DFC}\\equiv{\\angle EFB}$ 108° ${\\angle EFD}\\equiv{\\angle BFC}$ 72° ${\\angle EDF}\\equiv{\\angle BCF}$ 72° ${\\angle BEF}\\equiv{\\angle EBF}$ 36° The four sided polygon AEFB is a rhombus because it has all sides equal (one unit) Triangles EFB and DFC are both isosceles, their corresponding angles are equal I used the following notation: • DF = x • FB = 1 (equal in length to the side of the pentagon) • EB = r Using the relationship obtained using the similarity of the triangles, I obtain: $${x\\over1} = {1\\over{r}}$$ I know that r = x + 1 because triangle BED is also isosceles. x and r are the solutions of the following system of two equations: $${x\\over1} = {1\\over{r}}$$ $${x + 1} = {r}$$ I calculate the two numbers, x and r ,", "= DC} \\end{align} Which further implies, \\begin{align}\\boxed{ AB=BC=CD=AD} \\end{align} $$\\therefore$$The given parallelogram is a rhombus. Example 3 If in a parallelogram $$\\text{ABCD}$$, $$\\text{AC}$$ bisects $$\\angle\\text{A}$$ and $$\\angle\\text{C}$$, show that $$\\text{AC} \\perp \\text{BD}$$. Solution: \\begin{align} \\Delta \\text{ABC}\\:and\\: \\Delta \\text{ADC}: \\\\ AC&=AC (\\text{common}) \\\\ \\angle 1&=\\angle 2 \\:( \\text{given)} \\\\ \\angle 3 &=\\angle 4 \\:( \\text{given}) \\end{align} By the ASA criterion, the two triangles are congruent. This means that \\begin{align} \\Rightarrow \\text{AB = AD} \\end{align} \\begin{align} \\Delta \\text{AEB}\\:and\\: \\Delta \\text{AED}: \\\\ AE&=AE (\\text{common})\\\\ AB&=AD (\\text{proved above}) \\\\ \\angle 2&=\\angle 1 \\:( \\text{given)} \\\\ \\end{align} By the SAS criterion, the two triangles are congruent. Therefore, \\begin{align} \\angle AEB = &\\angle AED\\\\ = &\\frac{1}{2} \\times 180^\\circ \\\\= &\\,90^\\circ \\end{align} Hence, \\begin{align}\\boxed{\\angle AEB=\\angle AED = 90^\\circ} \\end{align} $$\\therefore \\;\\text{AC} \\perp\\ \\text{BD}$$ Example 4 Find the area of a parallelogram whose base is $$7\\:\\text{cm}$$ and height is $$10\\:\\text{cm}$$. Solution: Given Base $$=7\\:\\text{cm}$$ Height$$=10\\:\\text{cm}$$ Area of a parallelogram \\begin{align} &=\\text{base} \\times \\text{height} \\\\ &= 7 \\times 10 \\\\ &=70\\: \\text{cm}^2 \\end{align} $$\\therefore$$ Area of the parallelogram $$=70\\: \\text{cm}^2$$ Think Tank 1. Can a kite be called a parallelogram? 2. What elements of a trapezium should be changed to make it a parallelogram? 3. Can there be a concave rhombus? ## Practice Questions Here are few activities for you to practice. Select/Type your answer and click" ]

[ "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "+0 # In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits -2 260 1 +2 In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits $\\angle DAC$ into two equal angles). If $\\overline{DA}\\parallel\\overline{EF}$ and $\\angle AEF = 10 \\angle BAC - 12^\\circ$, then what is $\\angle DAC$ in degrees? [asy] pair A,B,C,D,EE,F; A = (0,0); B = rotate(164)*(1,0); C = rotate(148)*(1,0); D = (-1,0); EE = -0.4*C; F = EE + (1,0); dot(B,p=black+3bp); dot(C,p=black+3bp); dot(D,p=black+3bp); dot(F,p=black+3bp); draw(F--EE--C); draw(B--A--D); label(\"$B$\",B,NW); label(\"$A$\",A,NE); label(\"$C$\",C,N); label(\"$D$\",D,W); label(\"$E$\",EE,S); label(\"$F$\",F,E); [/asy] Mar 16, 2020 #1 +109822 -1 Anyone who tried to makes sense of that would also be sleep deprived. Mar 16, 2020", "+0 # heeeeelpp 0 131 1 In the figure, $\\overline{DC}$ is parallel to $\\overline{AB}$. We have $\\angle DCA = 40^\\circ$ and $\\angle ABC = 73^\\circ$. Find $\\angle ACB$. Feb 27, 2022 #1 +361 +3 Since, $$\\overline{DC}$$ is parellel to $$\\overline{AB}$$$$\\angle{DCA}$$ is the same as $$\\angle{CAB}$$. $$\\angle{ACB}=180-(40+73)$$ $$=180-113=67$$", "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", "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}", "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 $O_{1}$,", "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.)", "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", "$AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry. We notice that $\\triangle ACD \\cong \\triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\\triangle AED \\sim \\triangle ADC,$ with $\\angle EAD \\cong \\angle DAC$ and $\\angle CDA \\cong \\angle DEA.$ This similarity means that we can create a proportion: $\\frac{AD}{AB}=\\frac{DE}{CD}.$ We plug in $AD=\\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\\frac{8}{17} \\cdot 15= \\boxed{\\textbf{(B) }\\frac{120}{17}}.$ (By the way, we could also use $\\triangle DEC \\sim \\triangle ADC.$) ## Solution 4: Inscribed Circle $[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(\"B\",B,SW); label(\"D\",D,SE); label(\"A\",A,N); label(\"M\",M,S); label(\"C\",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]$ Call this triangle $ABD$ and let the midpoint of base $BD$ be $M$. Divide the triangle in half by drawing a line from $A$ to $M$. Half the base of triangle $ABD$ is $\\frac{16}{2} = 8$. The height is $15$, which is given in the question. Using the Pythagorean", "# Difference between revisions of \"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, $\\angle ADB = 180 - 40 = \\boxed{\\textbf{(D)}~140}$.", "- 6.623990803 \\\\[3ex] w = 12.37811434 \\\\[3ex] w \\approx 12.4\\;cm$ (20.) ACT In right triangle $\\triangle ABC$ shown below, $\\sin C = \\dfrac{2}{3}$ and the length of $\\overline{AB}$ is 6 inches. What is the length, in inches of $\\overline{AC}$ $A.\\;\\; \\sqrt{5} \\\\[3ex] B.\\;\\; \\sqrt{13} \\\\[3ex] C.\\;\\; 4 \\\\[3ex] D.\\;\\; 5 \\\\[3ex] E.\\;\\; 9 \\\\[3ex]$ $SOH:CAH:TOA \\\\[3ex] \\sin C = \\dfrac{opp}{hyp} = \\dfrac{2}{3} = \\dfrac{\\overline{AB}}{\\overline{AC}} \\\\[5ex] \\dfrac{2}{3} = \\dfrac{6}{\\overline{AC}} \\\\[5ex] Cross\\;\\;multiply \\\\[3ex] 2 * \\overline{AC} = 3(6) \\\\[3ex] \\overline{AC} = \\dfrac{3(6)}{2} \\\\[5ex] \\overline{AC} = 3(3) \\\\[3ex] \\overline{AC} = 9\\;inches$ (21.) (22.) ACT In the right triangle shown below, the length of $\\overline{AC}$ is 4 mm and the length of $\\overline{BC}$ is 3 mm. For $\\angle A$, the value of which of the following trigonometric expressions is $\\dfrac{3}{5}$? $A.\\;\\; \\sin A \\\\[3ex] B.\\;\\; \\cos A \\\\[3ex] C.\\;\\; \\tan A \\\\[3ex] D.\\;\\; \\csc A \\\\[3ex] E.\\;\\; \\cot A \\\\[3ex]$ $hyp^2 = leg^2 + leg^2 ...Pythagorean\\:\\: Theorem \\\\[3ex] \\overline{AB}^2 = \\overline{AC}^2 + \\overline{BC}^2 \\\\[3ex] \\overline{AB}^2 = 4^2 + 3^2 \\\\[3ex] \\overline{AB}^2 = 16 + 9 \\\\[3ex] \\overline{AB}^2 = 25 \\\\[3ex] \\overline{AB} = \\sqrt{25} \\\\[3ex] \\overline{AB} = 5 \\\\[3ex] SOHCAHTOA \\\\[3ex] For\\;\\; \\angle A: \\\\[3ex] opp = 3 \\\\[3ex] hyp = 5 \\\\[3ex] \\dfrac{3}{5} = \\sin A$ (23.) NZQA The section of one spider web shown below has two connecting right-angled triangles. Angle", "equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these two sets. Case (i): Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\\pm 3, d-e=\\pm 1$, this is not possible. Case (ii): Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$ Case (iii): Here $(a-b)-(d-e)=2c-2=6$ Case (iv): This is satisfied by $(a,b,d,e)=(6,2,3,5)$. Hence, we have (essentially) two different solutions: $(1,6,3,2,5,4)$ and $(1,6,2,4,3,5)$. It may be verified that I and II are both satisfied by these sets of values. Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides", "equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these two sets. Case (i): Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\\pm 3, d-e=\\pm 1$, this is not possible. Case (ii): Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$ Case (iii): Here $(a-b)-(d-e)=2c-2=6$ Case (iv): This is satisfied by $(a,b,d,e)=(6,2,3,5)$. Hence, we have (essentially) two different solutions: $(1,6,3,2,5,4)$ and $(1,6,2,4,3,5)$. It may be verified that I and II are both satisfied by these sets of values. Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides", "+0 # Consider parallelogram $ABCD$ with points $S$ and $T$ chosen such that $CS:SD = BT:TC = 2,$as in the picture below: [asy] size(200); pair 0 167 1 Consider parallelogram ABCD with points S and T chosen such that $$CS:SD = BT:TC = 2$$ as in the picture below: Let $$\\overrightarrow{AB} = \\mathbf{v}$$ and $$\\overrightarrow{AD} = \\mathbf{w}$$. Then there exist constants r, s, t, u such that \\begin{align*} \\overrightarrow{AT} &= r \\mathbf{v} + s \\mathbf{w},\\\\ \\overrightarrow{BS} &= t \\mathbf{v} + u \\mathbf{w}. \\end{align*} Enter r, s, t, u in that order below. Apr 4, 2019", "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$$25k^2+9k^2-15k^2 =", "= \\boxed{571}$. ### Solution 2 $[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label(\"$$A$$\",A,(0,1));label(\"$$B$$\",B,(-1,-1));label(\"$$C$$\",C,(1,-1));label(\"$$P$$\",P,(1,1)); label(\"$$Q$$\",Q,(-1,1));label(\"$$R$$\",R,(1,0));label(\"$$S$$\",S,(-1,0)); [/asy]$ Again, construct $R$ as above. Let $\\angle BAC = \\angle QBR = \\angle QPR = 2x$ and $\\angle ABC = \\angle ACB = y$, which means $x + y = 90$. $\\triangle QBC$ is isosceles with $QB = BC$, so $\\angle BCQ = 90 - \\frac {y}{2}$. Let $S$ be the intersection of $QC$ and $BP$. Since $\\angle BCQ = \\angle BQC = \\angle BRS$, $BCRS$ is cyclic, which means $\\angle RBS = \\angle RCS = x$. Since $APRB$ is an isosceles trapezoid, $BP = AR$, but since $AR$ bisects $\\angle BAC$, $\\angle ABR = \\angle ACR = 2x$. Therefore we have that $\\angle ACB = \\angle ACR + \\angle RCS + \\angle QCB = 2x + x + 90 - \\frac {y}{2} = y$. We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \\frac {y}{2} = y$ to get $x = 10$ and $y = 80$. $\\angle APQ = 180 - 4x = 140$, $\\angle ACB = 80$, so $r = \\frac {80}{140} = \\frac {4}{7}$. $\\left\\lfloor 1000\\left(\\frac {4}{7}\\right)\\right\\rfloor = \\boxed{571}$. ### Solution 3 Let the measure of $\\angle BAC$ be $\\alpha$ and $\\overline{AP}=\\overline{PQ}=\\overline{QB}=\\overline{BC}=x$. Because $\\triangle APQ$ is isosceles,", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we" ]

[ "+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", "# Digit Expansional Level pending There are five integers, it is known that they have an average of $$7142128354249...763770777$$ How many digits of the decimal expansion of the sum of these five integers are either $$2$$ or $$7$$ ×", "numbers, the exponent range is $[-126,127]$.", "digits in their representation $= 6 + (5 \\times 22) = 116.$ 4. Hence, there are $116$ three-digit integers less than $601$ that have exactly two different digits in their representation.", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "How many integers are there between $100$ and $1000$ all of whose digits are even? 1. $60$ 2. $80$ 3. $100$ 4. $90$", "# Problem #131 131 The increasing sequence $1,3,4,9,10,12,13\\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\\mbox{th}}$ term of this sequence. 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.", "have $$184+24=208$$ unordered triplets. The number of ways of choosing 3 unordered integers from between $$1$$ and $$100$$ with replacement is $${100+3-1\\choose3}={102\\choose3}=171700$$ The required probability is now $$P(\\text{sum}=50)=\\frac {208}{171700}\\approx0.00121$$", "Select Page Understand the problem How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. ) Source of the problem American Mathematics Competition Topic Enumerative Combinatorics 7/10 Suggested Book Introductory Combinatorics by Richard Brualdi So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further ! Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. What is the largest number in the range divisible by 11 ? Easy, 990. How about the smallest such number ? Yeah, 110. So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81. Now, how about you try taking things ahead from here onwards ? Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very juncture, ask yourself why. If you find out", "29 views The number of integers greater than $2000$ that can be formed with the digits $0,1,2,3,4,5$, using each digit at most once, is 1. $1480$ 2. $1440$ 3. $1200$ 4. $1420$ 1", "Categories # Understand the problem How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. ) ##### Source of the problem American Mathematics Competition ##### Topic Enumerative Combinatorics 7/10 ##### Suggested Book Introductory Combinatorics by Richard Brualdi So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further ! Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. What is the largest number in the range divisible by 11 ? Easy, 990. How about the smallest such number ? Yeah, 110. So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81. Now, how about you try taking things ahead from here onwards ? Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very juncture, ask yourself why. If", "Select Page # Understand the problem How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. ) ##### Source of the problem American Mathematics Competition ##### Topic Enumerative Combinatorics 7/10 ##### Suggested Book Introductory Combinatorics by Richard Brualdi # Start with hints So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further ! Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. What is the largest number in the range divisible by 11 ? Easy, 990. How about the smallest such number ? Yeah, 110. So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81. Now, how about you try taking things ahead from here onwards ? Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very", "## Algebra 1 $125$ An exponent of $3$ means that the given factor is multiplied three times. Thus, $5^3 = 5(5)(5) = 25(5) = 125$", "# Three consecutive odd integers add up to 129. What are the numbers? May 17, 2018 $41 , 43 , 45$ #### Explanation: Let's call the one in the middle $x$. We won't make assumptions about $x$ being even or odd. $\\left(x - 2\\right) + x + \\left(x + 2\\right) = 129$ $3 x = 129$ $x = 43$ It is odd. $41 , 43 , 45$", "## The integer $$120$$ has many factorizations. For example, $$120=(2)(60), 120=(3)(4)(10),$$ and $$120=(-1)(-3)(4)(10).$$ ##### This topic has expert replies Legendary Member Posts: 2068 Joined: 14 Oct 2017 Followed by:3 members ### The integer $$120$$ has many factorizations. For example, $$120=(2)(60), 120=(3)(4)(10),$$ and $$120=(-1)(-3)(4)(10).$$ by VJesus12 » Thu Sep 16, 2021 11:36 am 00:00 A B C D E ## Global Stats The integer $$120$$ has many factorizations. For example, $$120=(2)(60), 120=(3)(4)(10),$$ and $$120=(-1)(-3)(4)(10).$$ In how many of the factorizations of $$120$$ are the factors consecutive integers in ascending order? A. 2 B. 3 C. 4 D. 5 E. 6", "some integer $i$ between 1 and 100. The guesser's strategy is $<p_1,..., p_{100}>$, not the picker's strategy. Does this help? Mar 29 at 19:25", "In four-digit integer numbers from $1001$ to $9999$, the digit group “$37$” (in the same sequence) appears______times. 1. $270$ 2. $279$ 3. $280$ 4. $299$ edited", "# Is 128 the only multi-digit power of 2 such that each of its digits is also a power of 2? The number $128$ can be written as $2^n$ with integer $n$, and so can its every individual digit. Is this the only number with this property, apart from the one-digit numbers $1$, $2$, $4$ and $8$? I have checked a lot, but I don’t know how to prove or disprove it.", "How many integers are there between $100$ and $1000$ all of whose digits are even? 1. $60$ 2. $80$ 3. $100$ 4. $90$ _ _ _ 1. In the units place the possible number can be vary from 0,2,4,6,8 2. In the tens place similarly, outcome is the same as number can vary from 0 to 8 3. But the story is not the same in case of hundredth place as number should be greater than 100 thus the only possible outcome is 2,4,6,8 4. So 4*5*5 = 100 possibilities are there in which all the digits can be even. 480 points", "if $$x = 95$$, then the answer is NOT if $$x = 145$$, then the answer is YES Insufficient 1 + 2. $$x$$ can be any integer from $$91$$ to $$149$$. if $$x = 91$$, then $$x^2 = 91*91$$. $$91*91$$ is greater than $$90*90 = 8100$$, thus it is not a 3-digit integer. Hence at any value greater then $$91$$, $$x$$ will not be a 3-digit integer. Sufficient. C Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 9143 Location: United States (CA) Re: Is the integer x a 3-digit integer? (1) x is the square of an integer [#permalink] ### Show Tags 02 May 2019, 17:24 Bunuel wrote: Is the integer x a 3-digit integer? (1) x is the square of an integer. (2) 90 < x < 150 DS01602.01 OG2020 NEW QUESTION We need to determine whether integer x is a 3-digit integer. Statement One Alone x is the square of an integer. Even though we know that x is a square, we cannot determine whether it’s a 3-digit integer. For example, if x = 10^2 = 100, then x is a 3-digit integer. However, if x = 5^2 = 25, then it is NOT a 3-digit integer. Statement one alone is not sufficient to answer the question." ]

[ "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", "# Discrete Math- four digit odd integers with distinct digitst a) Find the number of four-digit odd integers that have distinct digits. b) Find the number of four-digit even integers that have distinct digits. I've been working on this problem and I can't figure it out. I don't know how to approach this problem. Any hints or maybe even a brief explanation of how to get the answer would be appreciated. How can I get to the answer or what is the answer and how did you get to your conclusion? Thanks a lot • Is such an integer allowed to have $0$ as its first digit? – Newb Jan 30 '14 at 3:54 • no, the first digit must be different from 0. – kmitov Jan 30 '14 at 4:05 Hint: There are two types of $4$ digit odd integers with all digits different: (i) First digit is even ($2,4,6,8$) or (ii) first digit is odd. Count the Type (i) numbers, the Type (ii) numbers, and add. Type (i): There are $4$ choices for the first digit. For each of these choices there are $5$ choices for the last digit. For each of the choices we have made so far, there are $8$ choices for the second digit, and for each such choice $7$ choices for the", "is an infinity of them they are what mathematicians call countable. Integers are clearly not the only numbers as we are familiar with the real numbers such as $$1.2,\\, 3.56$$, and so on, of which there is also an infinite number, but many, many more than there are integers. If two different real numbers are chosen, an infinite number of others can be squeezed in between them simply by adding more decimal places; this and other interesting discussions on numbers and algorithms are explained clearly in ‘The Emperor’s New Mind’ (R.Penrose publ. 1990 Vintage OUP). Surprisingly, real numbers can usually be reduced to a rational fraction which is a ratio of two integers, e.g. $$0.751 = 326/533$$, but, for some numbers, $$\\pi,\\, e, \\sqrt{2}$$ there is no ratio of integers that will accurately form the number and they are called irrational, i.e. not being logical! You may recall that $$\\pi \\approx 22/7$$ , which would seem to contradict this statement, but $$22/7$$ is a very poor approximation to $$\\pi$$, only accurate to three digits, $$355/113$$ is better but only to seven digits. When you use a computer to perform calculations, a distinction is made whether a real or integer number is used. This is a common feature of programming languages and also in computer algebra, for example, the", "if you give me any integer (say, $15$), I can tell you the largest integer that’s less than your number (in our example, $14$) and the smallest integer that’s bigger than your number ($16$). Real numbers, however, do not have this property. There is no real number to the immediate right of $0$. This is easy to prove by contradiction. Suppose $x > 0$ is the real number to the immediate left of $0$. That means that there are no real numbers between $0$ and $x$. However, $x/2$ is bigger than $0$ and less than $x$, providing the contradiction. (For what it’s worth, the above proof doesn’t apply to the set of integers $\\mathbb{Z}$ since $x/2$ doesn’t have to be an integer.) By the same logic — visually, you can imagine reflecting the number line across the point $x = 0.5$ — there is no number to the immediate left of $1$. So while $0.\\overline{9}$ would appear to be to the immediate left of $1$, they are in reality the same point.", "# 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 $1 + 1 + 2 + 3 = 7$. ×", "How many three-digit integers less than $601$ have exactly two different digits in their representation (for example, $232,$ or $466)?$ $116$ Step by Step Explanation: 1. Let the two different digits be $x$ and $y.$ Therefore, the required integers are of the form $xxy, xyx$ or $yxx.$ 2. If the repeated digits are zero, we must ignore the form $xxy, xyx$ as they will give us one and two digit numbers. $Eg. 001, 010,$ etc. So, if $x = 0,$ the integers have the form $yxx$ and $y$ can be $1, 2, 3, \\ldots , 6.$ Therefore, there are $6$ integers with two zeros, $i.e. 100, 200, \\ldots, 6 00.$ 3. When the repeated digit is non-zero, the integers are of the form $xxy, xyx$ or $yxx.$ If $x = 1, y$ can be $0, 2, 3, 4, 5, 6, 7, 8$ or $9,$ therefore there are $9 \\times 3$ $= 27$ possible integers but we must ignore $011$ as this is a two-digit integer. Since your number is less than $601$ so we must ignore $611, 711, 811, 911.$ This gives $27 - 5 = 22$ different integers. Similarly, there will be an additional $22$ integers for every non-zero value of $x$. Therefore, the total number of three-digit integers less than $601$ that have exactly two different", "digits in their representation $= 6 + (5 \\times 22) = 116.$ 4. Hence, there are $116$ three-digit integers less than $601$ that have exactly two different digits in their representation.", "b_p has to be 0.e.g.: a = a_2 a_1, b = b_2 b_1 => a and b can be added if a_1 equals 0 or b_1 equals 0 (or both) AND [same condition for a_2, b_2] (=> a and b both have 2 digits, so a_2 and b_2 are non-zero => they can not be added) • » » » 7 years ago, # ^ | ← Rev. 2 → 0 In problem A, the maximum number of the chosen integers is between 0 & 4. Because if you choose a 2 digit integer, you cant get another one( If you choose another 2 digit number, the first digit of both of them is !0 & you cant sum that pair). And you cant choose more than one of 1 or 3 digit numbers with same reason. So in maximum case, Vasya can choose 4 numbers a,_b_,_c_,_d_, a==\"0\", b = a one digit integer, c = a 2 digit integer, d = a 3 digit integer & d = \"100\" Sorry for my terrible English! • » » » » 7 years ago, # ^ | ← Rev. 2 → 0 this is not always true. what about the input 1 0 0 0? he can choose all the 4 integers!UPD: just re-read the problem statement. the input", "Joined: 02 Sep 2009 Posts: 28184 Followers: 4452 Kudos [?]: 44830 [0], given: 6618 Re: How many integers n are there such that r < n< 5 ? [#permalink] 22 Apr 2012, 13:15 Expert's post carcass wrote: Bunuel wrote: How many integers n are there such that r < n< s ? The question asks: how many integers $$n$$ satisfy $$r < n < s$$, or how many integers are between $$r$$ and $$s$$, not inclusive? Notice that if $$r$$ and $$s$$ are integers, for example $$s=5$$ and $$r=0$$ then there will be 4 integers between them: 1, 2, 3, and 4. But if $$r$$ and $$s$$ are NOT integers for example $$s=5.5$$ and $$r=0.5$$ then there will be 5, so one more, integers between them: 1, 2, 3, 4 and 5. (1) s - r = 5. Not sufficient. (2) r and s are not integers. Not sufficient. (1)+(2) According to above reasoning since r and s are not integers then there are 5 integers between them (case 2). Sufficient. Answer: C. Hope it's clear. can you please eleborate the concept behind this ?? indeed is true but I didn't notice this before. Thanks There isn't some special concept behind it. It's just common sense based on observation. _________________ Intern Joined: 15 Nov 2012 Posts: 3 Followers:", "necessarily the digit. In example 3, you will see that the missing number is 7, even though it appears in the sequence. There are 3 places the DIGIT 7 will appear in a series of [0, 30): “7”, “17”, and “27”. Your objective is to differentiate between these, and discover that 7 is the missing NUMBER (inside of example 3). In other words, the tricky part lies in finding out which sequences of digits are complete and which belong to other numbers. Another very important factor in this: there are some input situations where there are multiple solutions! Your job will be to print out ALL numbers that are missing in the sequence. I cannot provide insight as to WHY they have multiple solutions; I only know they may exist. Input: The input is a single Integer T: the amount of test cases. For each test case, there is an Integer M: the minimum number in the sequence, another Integer N: the maximum number in the sequence (thus M < N at ALL times), a String S, containing integers from M to N - 1 inclusive, or M to N exclusive (in other words, [M, N)). These integers, as stated above, are scrambled up to create a random sequence. There are NO delimiters (“42, 31, 23, 44”), or", "• Case 1. There is only one integer with one digit, namely, the integer 7. • Case 2. If there are two digits, the first could be any digit between 1 and 9, and the last digit must be 7. [ \begin{array}{ccc} ext{choices:} & 1-9 & 7 & - & - ext{# of choices:} & 9 & 1 end{array} onumber] This gives us nine choices. • Case 3. If there are three digits, the first digit could be any digit between 1 and 9, the second any digit between 0 and 9, and the last digit must be 7. [ \begin{array}{cccc} ext{choices:} & 1-9 & ext{any digit} & 7 & - & - & - ext{# of choices:} & 9 & 10 & 1 end{array} onumber] Hence, there are 90 integers in this case. Combining the three cases, we have a total of (1+9+90=100) integers that meet the requirements. Solution 2 The integers could be written as three-digit integers if we allow 0 as the leading digits. For instance, 7 can be written as (007), and (34) as (034). Under this agreement, we have to fill three positions where the last one is always occupied by the digit 7. The first two digits are (0, 1, 2, ldots, 8), or 9, so there are 10 choices for each", "# How many positive, three-digit integers contain at least one $3$ How many positive, three-digit integers contain at least one $3$ as a digit but do not contain $5$ as a digit? I have an answer for that which is $215$ ,is that right ? If its wrong then ,how to solve it? - I think your number is a little high. Take the number of 3-digit integers that contain no 5 and subtract the number of 3-digit numbers that contain no 5 and no 3. - Let us consider the number of three-digit integers that do not contain $3$ and $5$ as digits; let this set be $S$. For any such number, there would be $7$ possible choices for the hundreds digit (excluding $0,3$, and $5$), and $8$ possible choices for each of the tens and ones digits. Thus, there are $7 \\cdot 8 \\cdot 8 = 448$ three-digit integers without a $3$ or $5$. Now, we count the number of three-digit integers that just do not contain a $5$ as a digit; let this set be $T$. There would be $8$ possible choices for the hundreds digit, and $9$ for each of the others, giving $8 \\cdot 9 \\cdot 9 = 648$. By the complementary principle, the set of three-digit integers with at least one $3$", "to 2. - There is no integer that is only divisible by $3$. Every integer is always divisible by $1$.", "OPERATIONS ON INTEGERS Articles Videos PART 2: PEMDAS Articles Videos More video Introduction to PEMDAS ## Practice Exercises on Subtraction of Integers In subtraction of integers, we have learned two rules: (1) a – b = a + (-b) (2) a – (-b) = a + b We will use these rules in answering the exersises below. Exercises 1. 2 – 5 2. 18 – ( – 2) 3. 16 – 7 4. -17 – 3 5. -9 – (-3) 6. 0 – (-11) 7. -18 – (-25) 8. -10 – 9 9. 12 – (-9) 10. -6 – 3 1. 2 – 5 Solution 1: 5 is greater than 2. If you subtract two numbers, if the subtrahend is larger than the minuend, the answer will be negative. So, the answer is -3. Solution 2: From rule 1, a – b = a + (-b), so 2 + 5 = 2 + (-5) = -3 2. 18 – ( – 2) Solution: From rule 2, a – (-b) = a + b, so 18 + 2 = 20. 3. 16 – 7 4. -17 – 3 Solution: From rule 1, -17 – 3 = -17 + (- 3) = -20. Recall that in adding two negative numbers, we just add the numbers and then the answer will", "2 and then 6a = 6 * 2 = 12 is an integer – YES => If $$a^3=1$$, then a = 1 and then 6a = 6 * 1 = 6 is an integer – YES => But if, $$a^3=16$$, then a = 2.52 and then 4a = 6 * 2.52 = is not an integer Only if ‘a’ is an integer then, $$a^3$$ and 6a will be an integer. The answer is a unique value YES; both conditions combined are sufficient according to Common Mistake Type 1 which states that the answer should be a unique YES or a NO. Both conditions together are sufficient. Therefore, C is the correct answer.", "# First > second Probability Level 2 How many $2$ digit positive integers have the property that the first digit is greater in value than the second digit? ×", "# Pigeon Hole problem on any set of distinct integers Show that given any set of seven distinct integers, there must exist two integers in this set whose sum or difference is a multiple of 10. I'm having a hard time with this pigeon hole section. So I'm not sure how to go about this problem. I know that the first digit can be 1-9 and next 0-9 and so forth but the last must only be 0. The basic argument you want works something like this. We can deal with all of the numbers modulo $10$, since divisibility by $10$ is all we care about. Suppose the numbers somehow defied the condition. Then all of the numbers would have to be different modulo $10$; otherwise, the difference between two numbers that are the same modulo $10$ would be divisible by $10$. Now how about their sums? There are six possible ways the sum of two numbers might be divisible by $10$: • They are both equivalent to $0$ (modulo $10$). • They are equivalent to $1$ and $9$. • They are equivalent to $2$ and $8$. • They are equivalent to $3$ and $7$. • They are equivalent to $4$ and $6$. • Or, they are both equivalent to $5$. There are thus six different classes of", "we rewrite this as $5-(-5)$ before simplifying. # Multiplying and dividing integers When multiplying/dividing integers, there are 2 parts. The first part is to multiply/divide the number as necessary. The next is to determine the sign. The rules are that: • When multiplying a (+) and (+), you get a (+) • When multiplying a (+) and (-), or a (-) and (+), you get a (-) • When multiplying a (-) and (-), you get a (+) For example $(+5)*(-5)=-25$ because there is a mixture of (+) and (-). Alternatively, $(-5)*(-5)=25$, because there is a repeating/same (-) sign throughout.", "this problem is a counting problem, it is very likely that student's would attempt to consider casework. Specifically, they would consider the case when the first digit is $1$, then $2$, and so on. Although, it is possible to obtain the correct answer this way, it quickly becomes tedious and the approach is extremely prone to errors. If you have any other approaches that you believe is worth discussing about, please let us know by posting here! We would greatly appreciate listening to your ideas!", "I would have seen that counting the twos is pointless in this situation. Even the prime factorization of $10 = 5 \\cdot 2$ has the information in there: there will always be more twos than fives. Counting from 1 to 10: • Multiples of 2: 2, 4, 6, 8, 10 • Multiples of 5: 5, 10 This means that all we really need to keep track of is the quantity of fives in the prime factorization of $n!$. Which leads to the second optimization: we only need to get the prime factorization of multiples of five. # Factorizeroes! A while back I was looking at the different maximum values that the different integer data types (uint8, short, int, long, etc) have throughout a couple languages I’ve been using and noticed that none of them ended in zero. I wondered why that was but then relatively quickly realized that it is because integer data types in computers are made up of bytes and bits. An 8-bit (1-byte) integer has a max value of $2^8 = 256$, a 16-bit (2-byte) integer has a max value of$2^{16} = 65536$, etc. In fact since any integer in a computer is made of bits it will have a maximum value of $2^n$. The prime factorization of an n-bit integer is in the notation," ]

[ "+0 # help counting 0 77 1 How many even, three-digit positive integers have the property that exactly two of the integer's digits are even? Apr 16, 2022 #1 +2437 +1 There are 2 cases to consider: Even - Odd - Even Odd - Even - Even For the first case, there are $$4 \\times 5 \\times 5 = 100$$ numbers. (Remember, the leading digit can't be 0!!) For the second case, there are $$5 \\times 5 \\times 5 = 125$$ numbers. Thus, there are a total of $$100 + 125 = \\color{brown}\\boxed{225}$$ numbers that meet these criteria. Apr 16, 2022", "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", "digits in their representation $= 6 + (5 \\times 22) = 116.$ 4. Hence, there are $116$ three-digit integers less than $601$ that have exactly two different digits in their representation.", "# Digit Expansional Level pending There are five integers, it is known that they have an average of $$7142128354249...763770777$$ How many digits of the decimal expansion of the sum of these five integers are either $$2$$ or $$7$$ ×", "How many three-digit integers less than $601$ have exactly two different digits in their representation (for example, $232,$ or $466)?$ $116$ Step by Step Explanation: 1. Let the two different digits be $x$ and $y.$ Therefore, the required integers are of the form $xxy, xyx$ or $yxx.$ 2. If the repeated digits are zero, we must ignore the form $xxy, xyx$ as they will give us one and two digit numbers. $Eg. 001, 010,$ etc. So, if $x = 0,$ the integers have the form $yxx$ and $y$ can be $1, 2, 3, \\ldots , 6.$ Therefore, there are $6$ integers with two zeros, $i.e. 100, 200, \\ldots, 6 00.$ 3. When the repeated digit is non-zero, the integers are of the form $xxy, xyx$ or $yxx.$ If $x = 1, y$ can be $0, 2, 3, 4, 5, 6, 7, 8$ or $9,$ therefore there are $9 \\times 3$ $= 27$ possible integers but we must ignore $011$ as this is a two-digit integer. Since your number is less than $601$ so we must ignore $611, 711, 811, 911.$ This gives $27 - 5 = 22$ different integers. Similarly, there will be an additional $22$ integers for every non-zero value of $x$. Therefore, the total number of three-digit integers less than $601$ that have exactly two different", "Select Page Understand the problem How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. ) Source of the problem American Mathematics Competition Topic Enumerative Combinatorics 7/10 Suggested Book Introductory Combinatorics by Richard Brualdi So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further ! Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. What is the largest number in the range divisible by 11 ? Easy, 990. How about the smallest such number ? Yeah, 110. So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81. Now, how about you try taking things ahead from here onwards ? Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very juncture, ask yourself why. If you find out", "call countable. Integers are clearly not the only numbers as we are familiar with the real numbers such as $$1.2,\\, 3.56$$, and so on, of which there is also an infinite number, but many, many more than there are integers. If two different real numbers are chosen, an infinite number of others can be squeezed in between them simply by adding more decimal places; this and other interesting discussions on numbers and algorithms are explained clearly in ‘The Emperor’s New Mind’ (R.Penrose publ. 1990 Vintage OUP). Surprisingly, real numbers can usually be reduced to a rational fraction which is a ratio of two integers, e.g. $$0.751 = 326/533$$, but, for some numbers, $$\\pi,\\, e, \\sqrt{2}$$ there is no ratio of integers that will accurately form the number and they are called irrational, i.e. not being logical! You may recall that $$\\pi \\approx 22/7$$ , which would seem to contradict this statement, but $$22/7$$ is a very poor approximation to $$\\pi$$, only accurate to three digits, $$355/113$$ is better but only to seven digits. When you use a computer to perform calculations, a distinction is made whether a real or integer number is used. This is a common feature of programming languages and also in computer algebra, for example, the integer division $$2//3$$ is exact and in Python (notice", "Select Page # Understand the problem How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. ) ##### Source of the problem American Mathematics Competition ##### Topic Enumerative Combinatorics 7/10 ##### Suggested Book Introductory Combinatorics by Richard Brualdi # Start with hints So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further ! Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. What is the largest number in the range divisible by 11 ? Easy, 990. How about the smallest such number ? Yeah, 110. So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81. Now, how about you try taking things ahead from here onwards ? Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very", "Expert's post sumitchawla wrote: Bunuel wrote: mehdiov wrote: How many integers less than 1000 have no factors (other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not inclusive - $$\\frac{995-5}{5}+1=199$$; Multiples of both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. We counted multiples of 2 or 5 in", "How many integers are there between $100$ and $1000$ all of whose digits are even? 1. $60$ 2. $80$ 3. $100$ 4. $90$ _ _ _ 1. In the units place the possible number can be vary from 0,2,4,6,8 2. In the tens place similarly, outcome is the same as number can vary from 0 to 8 3. But the story is not the same in case of hundredth place as number should be greater than 100 thus the only possible outcome is 2,4,6,8 4. So 4*5*5 = 100 possibilities are there in which all the digits can be even. 480 points", "Number of different integers between $1,000$ and $10,000$ How many integers are there between $1,000$ and $10,000$ divisible by $60$ and all with distinct digits? I know that there are $8,999$ integers in total, and $\\lfloor\\frac{8999}{60}\\rfloor=149$. So there are $149$ integers between $1,000$ and $10,000$ divisible by $60$. For the first digit we have $8$ choices, for the second $9$ choices, for the third $8$ choices and for the last $7$ choices. So there are $4,032$ different integers between $1,000$ and $10,000$. How to combine these results to an answer on the above question? • Simplify things a little first. The last digit must be $0$ and the next to last must be $2$, $4$, $6$, or $8$. – André Nicolas May 12 '15 at 14:52 • Next-to-last digit may be $0$, too. – ajotatxe May 12 '15 at 15:00 • @ajotatxe it cannot if we are using all distinct numbers. The zero must take the last space – JMoravitz May 12 '15 at 15:01 • Oops, sorry. I shall fix my answer then. – ajotatxe May 12 '15 at 15:24 • There are actually $150$ numbers (not $149$) between $1000$ and $10000$ that are divisible by $60$. The smallest is $1020$ and the largest is $9960$. A correct formula to count these numbers is $\\left\\lfloor\\frac{10000}{60}\\right\\rfloor - \\left\\lfloor\\frac{1000}{60}\\right\\rfloor$.", "it really quite vexing that the first digits don’t go $1,2,3,4$. Furthermore, $400 \\div 13 \\approx 31$, so it takes a long time before you see a 4 at all, and that seemed really unfair. ### The OEIS now contains 300,000 integer sequences The Online Encyclopedia of Integer Sequences just keeps on growing: at the end of last month it added its 300,000th entry. Especially round entry numbers are set aside for particularly nice sequences to mark the passing of major milestones in the encyclopedia’s size; this time, we have four nice sequences starting at A300000. These were sequences that were originally submitted with indexes in the high 200,000s but were bumped up to get the attention associated with passing this milestone.", "Categories # Understand the problem How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ ( For example, both $121$ and $211$ have this property. ) ##### Source of the problem American Mathematics Competition ##### Topic Enumerative Combinatorics 7/10 ##### Suggested Book Introductory Combinatorics by Richard Brualdi So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further ! Stuck ?…No worries. Try asking yourself – How many numbers between 100-999 are exactly divisible by 11 ? This is quite easy to figure out. What is the largest number in the range divisible by 11 ? Easy, 990. How about the smallest such number ? Yeah, 110. So, the number of integers in the range visible by 11 = ( ( 990110 ) / 11 ) + 1 = 81. Now, how about you try taking things ahead from here onwards ? Now see, 81 numbers are divisible by 11 in the range, as we just saw. All that’s left is to find out the permutations of these. Wait ! That’s simple, isn’t it ? Yes, 81 x 3 = 243. At this very juncture, ask yourself why. If", "if $$x = 95$$, then the answer is NOT if $$x = 145$$, then the answer is YES Insufficient 1 + 2. $$x$$ can be any integer from $$91$$ to $$149$$. if $$x = 91$$, then $$x^2 = 91*91$$. $$91*91$$ is greater than $$90*90 = 8100$$, thus it is not a 3-digit integer. Hence at any value greater then $$91$$, $$x$$ will not be a 3-digit integer. Sufficient. C Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 9143 Location: United States (CA) Re: Is the integer x a 3-digit integer? (1) x is the square of an integer [#permalink] ### Show Tags 02 May 2019, 17:24 Bunuel wrote: Is the integer x a 3-digit integer? (1) x is the square of an integer. (2) 90 < x < 150 DS01602.01 OG2020 NEW QUESTION We need to determine whether integer x is a 3-digit integer. Statement One Alone x is the square of an integer. Even though we know that x is a square, we cannot determine whether it’s a 3-digit integer. For example, if x = 10^2 = 100, then x is a 3-digit integer. However, if x = 5^2 = 25, then it is NOT a 3-digit integer. Statement one alone is not sufficient to answer the question.", "so the final answer is $90+40+16+4=\\boxed{150}.$ Hint: In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$. Thus, the product is $(1^{19})(2^{18})\\cdots(19^1)$ (or alternatively, $19! \\cdot 18! \\cdots 1!$.). When we count the number of factors of $2$, we have $4$ groups, factors that are divisible by $2$ at least once, twice, three times and four times. For those numbers, take the exponent corresponding to each one of them, add them case by case and then totally add them. Hope it helps.", "Joined: 02 Sep 2009 Posts: 28184 Followers: 4452 Kudos [?]: 44830 [0], given: 6618 Re: How many integers n are there such that r < n< 5 ? [#permalink] 22 Apr 2012, 13:15 Expert's post carcass wrote: Bunuel wrote: How many integers n are there such that r < n< s ? The question asks: how many integers $$n$$ satisfy $$r < n < s$$, or how many integers are between $$r$$ and $$s$$, not inclusive? Notice that if $$r$$ and $$s$$ are integers, for example $$s=5$$ and $$r=0$$ then there will be 4 integers between them: 1, 2, 3, and 4. But if $$r$$ and $$s$$ are NOT integers for example $$s=5.5$$ and $$r=0.5$$ then there will be 5, so one more, integers between them: 1, 2, 3, 4 and 5. (1) s - r = 5. Not sufficient. (2) r and s are not integers. Not sufficient. (1)+(2) According to above reasoning since r and s are not integers then there are 5 integers between them (case 2). Sufficient. Answer: C. Hope it's clear. can you please eleborate the concept behind this ?? indeed is true but I didn't notice this before. Thanks There isn't some special concept behind it. It's just common sense based on observation. _________________ Intern Joined: 15 Nov 2012 Posts: 3 Followers:", "# Discrete Math- four digit odd integers with distinct digitst a) Find the number of four-digit odd integers that have distinct digits. b) Find the number of four-digit even integers that have distinct digits. I've been working on this problem and I can't figure it out. I don't know how to approach this problem. Any hints or maybe even a brief explanation of how to get the answer would be appreciated. How can I get to the answer or what is the answer and how did you get to your conclusion? Thanks a lot • Is such an integer allowed to have $0$ as its first digit? – Newb Jan 30 '14 at 3:54 • no, the first digit must be different from 0. – kmitov Jan 30 '14 at 4:05 Hint: There are two types of $4$ digit odd integers with all digits different: (i) First digit is even ($2,4,6,8$) or (ii) first digit is odd. Count the Type (i) numbers, the Type (ii) numbers, and add. Type (i): There are $4$ choices for the first digit. For each of these choices there are $5$ choices for the last digit. For each of the choices we have made so far, there are $8$ choices for the second digit, and for each such choice $7$ choices for the", "# How many integers satisfy the condition? So the question is: How many integers $n$ satisfy: • $9999<n<99999$ • All of it's digits are in $\\left\\{ 1,2,3,4,5\\right\\}$ • The tens digit is either $1$ or $3$ • The number of times $1$ appear is equal the the ones digit. I broke it down into 5 cases for the possible digits of one, and looked at each separately, dividing some again into cases for each digit for ten. Which got me a final answer of $138$. But I was wondering if there's a smarter way of doing this. • One good practice is always to tackle the one with maximum contraints first. You seem to have done that. In this case a real shortcut may not be available – Shailesh Oct 21 '15 at 16:57 • The ones digit is 1. In this case, it must be the only 1 in the number. So we have 1 option for the tens and 4 options for the other three. This gives $1 \\times 4^3 = 64$ numbers. • The ones digit is not 1. Then it is uniquely determined by the number of ones in the number and we must have at least 2 occurrences amongst the remaining 4 digits. We can use inclusion-exclusion to get: $2 \\times 5^3 -", "$$1*9*1 = 9$$ ---> (1) Case 2: Numbers such as 770, 771..... 779 (Except 777) $$1*1*9 = 9$$ ---> (2) Case 3: Numbers such as 711, 722...... 799 (Except 777) $$1*8*1 = 8$$ ---> (3) [NOTE: Did not consider 0 & 7 for the unit's and ten's digit] $$9 + 9 + 8 = 26$$ This can be done for 801 to 899 and 901 to 999. So, we get $$26 * 3 = 78$$ We also need to consider $$800$$ and $$900$$ (But NOT $$700$$) So, $$78 + 2 = 80$$ Can an expert evaluate the method. I understand that this method is long and could take time. But I want to be sure that I considered the cases correctly (In case, it comes to this during the exam!). Re: Of the three-digit integers greater than 700, how many have [#permalink] 21 May 2014, 02:25 Go to page 1 2 Next [ 31 posts ] Similar topics Replies Last post Similar Topics: 11 Of the three-digit integers greater than 660, how many have 8 10 Mar 2012, 04:45 28 Of the three-digit integers greater than 600, how many have 13 11 Feb 2012, 17:04 6 Of the three- digit integers greater than 700, how many have 4 18 Jun 2011, 22:05 57 Of the three digit integers", "# Why is the number of integers between $2$ integers not their difference? [duplicate] Helping my kids in 3rd grade Math class today. We were rounding to the nearest $10$. So for $100$, the numbers that round to it are $95$ to $104$. $95, 96, 97, 98, 99, 100, 101, 102, 103, 104$ which are 10 numbers I asked how many numbers round to $100$ one kid says 9, I say 10. Why is $104-95=9$ not the correct way to figure out the number of numbers? What is the correct equation, if there is one? ## marked as duplicate by Matthew Towers, user21820, Arnaud D., Aqua, Ove AhlmanNov 16 '17 at 15:32 • en.wikipedia.org/wiki/Off-by-one_error – Wouter Nov 15 '17 at 17:26 • – G Tony Jacobs Nov 15 '17 at 17:27 • A fun sanity check you can do for a proposed equation: split an interval into many chunks and add up the sizes of the chunks. For example, cut the range 1-30 into the ranges 1-10, 11-20, and 21-30. Using the simple subtraction equation, the latter three chunks would each have 9 elements, for a total of 27 elements in the combination of the three chunks -- which clearly can't be right. The equations proposed below which add one do pass this sanity check -- that's not" ]

[ "# Intersection of two rhombuses In which way do two rhombuses have to intersect to form an octagon which has four angles equal to 150◦? All sides of the octagon are equal. - Since you know that 4 of the angles have measure $150°$ and the sum of the measures of the interior angles of an octagon is $1080°$, consider that the other four angles might have measure $\\frac{1}{4}(1080°-4\\cdot150°)=120°$. That suggests using two congruent rhombuses with angles that measure $120°$ and $60°$—with congruent rhombuses, centered at the same point, and rotated 90° from one another, we will get all 8 sides of the octagon to have the same length. The octagon is the purple region of overlap in the diagram below.", "Answer Comment Share Q) # If one angle in rhombus is 90 degree then it is If one angle in rhombus is 90 degree then it is ## 1 Answer Comment A) A rectangle. Why ? In a rhombus, opposite angles are equal. Hence 2 angles are 90°. Hence sum of other two opposite angles are 180° and hence all angles are 90°. Hence a rectangle.", "provided both stands on same base. (Note that, when each angle of the rhombus is $90°$, rhombus is also a square (can be considered as special case) and in that case, areas will be equal. Add a new comment... (Use Discussion Board for posting new aptitude questions.) Name: Email: (optional)", "Rhombus Geometry Level 2 A rhombus has perimeter of 8 and has an angle 60 degree. If its area is $$b$$, then what is the value of $$b^2$$? ×", "## Elementary Geometry for College Students (7th Edition) (a) A rhombus that has all angles congruent is a square. If all the angles are congruent, then each angle is $90^{\\circ}$ since the sum of the four angles is $360^{\\circ}$. Also note that a rhombus has four congruent sides. (b) A rhombus that has congruent diagonals is a square. If the diagonals are congruent then each of the four angles must be $90^{\\circ}$. Also note that a rhombus has four congruent sides.", "the Wikipedia entry on the rhombohedron for images: Note the 2nd shape: $6$ identical rhombi.", "# Each side of a rhombus is 14 in. Long. Two of the sides form a 60 degree angle, how do you find the area of the rhombus? May 12, 2018 Area of the rhombus is $169.74$ sq.in. #### Explanation: Area of a rhombus is A_r = s^2 * sin theta where $s$ is the length of any side and $\\theta$ is any interior angle. s= 14 , theta= 60^0 :. A_r= 14^2 * sin 60 or A_r= 14^2 * sin 60 = 196* sqrt 3/2~~ 169.74 (2 dp) sq.in Area of the rhombus is $169.74$ sq.in[Ans]", "angles do we need to construct a reflex angle? Open Problems 1. Construct a 30-degree angle on the given line. 2. Construct a 30-degree angle, a 120-degree angle, and a 30-degree angle on the given line. 3. Construct a $7.5$-degree angle. 4. Show that six $30$ degree angles fit on a straight line. 5. Construct a rhombus with one set of angles equal to $30$ degrees. Open Problem Solutions The red quadrilateral is a rhombus with a 30-degree angle pair. Images/mathematical drawings are created with GeoGebra.", "rhombus. This angle has a measure of $55^{\\circ}$. The other angle is formed by the crossing of the two diagonals of a rhombus and is a right angle. Now that we have two angles, we can figure out the third angle: $m \\angle 2 = 180 - (90 + 55)$ Evaluate what's in parentheses first, according to order of operations: $m \\angle 2 = 180 - (145)$ Subtract to solve: $m \\angle 2 = 35^{\\circ}$", "##### Question 11.2.7 Suppose there’s a rhombus of side length $5\\,\\mathrm{cm}$, which is $9\\,\\mathrm{cm}$ wide (measured parallel to one of the sides). What is the area of the rhombus? Show hint", "# Problem of the Week Problem D The Angle Isn't Bad Ewan drew rhombus $$ABCD$$. Recall that a rhombus is a quadrilateral with parallel opposite sides, and all four sides of equal length. In Ewan’s rhombus, $$H$$ is on $$BC$$ in between $$B$$ and $$C$$, and $$K$$ is on $$CD$$ in between $$C$$ and $$D$$, such that $$AB=AH=HK=KA$$. Determine the measure, in degrees, of $$\\angle BAD$$.", "# A cut above .... and below Geometry Level 3 Suppose a square is cut by two parallel lines at perpendicular distance $$6$$ cm. apart such that (i) one line has an endpoint at the lower left corner of the square and the other has an endpoint at the upper right corner of the square, and (ii) the square is divided into three regions, (a parallelogram and two triangles), of equal area. What is the area of the square in cm$$^{2}$$? ×", "_________________ Math Expert Joined: 02 Sep 2009 Posts: 49251 Re: The figure shown is a rhombus in which the measure of angle A = 120°. [#permalink] Show Tags 20 Mar 2018, 23:24 Bunuel wrote: The figure shown is a rhombus in which the measure of angle A = 120°. What is the ratio of the length of AC to the length of DB? A. $$1:2\\sqrt{3}$$ B. $$1:2\\sqrt{2}$$ C. $$1:2$$ D. $$1:\\sqrt{3}$$ E. $$1:\\sqrt{2}$$ Attachment: 2018-03-21_1019.png For other subjects: ALL YOU NEED FOR QUANT ! ! ! _________________ Manager Joined: 16 Sep 2016 Posts: 240 GMAT 1: 740 Q50 V40 Re: The figure shown is a rhombus in which the measure of angle A = 120°. [#permalink] Show Tags 21 Mar 2018, 00:02 Bunuel wrote: The figure shown is a rhombus in which the measure of angle A = 120°. What is the ratio of the length of AC to the length of DB? A. $$1:2\\sqrt{3}$$ B. $$1:2\\sqrt{2}$$ C. $$1:2$$ D. $$1:\\sqrt{3}$$ E. $$1:\\sqrt{2}$$ Attachment: The attachment 2018-03-21_1019.png is no longer available IMO D. quadrilateral ABCD is a rhombus... all the sides are equal in length and opposite sides are parallel to each other. Let the length of side of rhombus be a. So the triangle formed by joining AC is equilateral as two of its sides are equal", "? Free Version Easy # Area and Perpendiculars ACTMAT-4W4EJY In the figure below, all intersecting segments are perpendicular to each other (note: figure not drawn to scale). What is the area of the figure? A $59$ B $136$ C $180$ D $210$ E $420$", "cm. If the angle between the sides is 30°, what is the area? Write the answer in the input box. $latex A=$ cm2", "5 cm; c = 7 cm (use the sine and cosine theorem). • A rhombus A rhombus has sides of length 10 cm, and the angle between two adjacent sides is 76 degrees. Find the length of the longer diagonal of the rhombus.", "## Elementary Geometry for College Students (7th Edition) We name the sides of the rhombus: $(0,0) ; (2b,2c); (2a+2b, 2c); (2a,0)$ Thus, the midpoints are: $(b,c); (a+2b, 2c); (2a+b,c); (a,0)$ Plugging these into the distance formula, we find: $s_1 = \\sqrt{c^2 +(b-a)^2}$ $s_2 = \\sqrt{c^2 +(-b-a)^2}$ $s_3 = \\sqrt{c^2 +(-b-a)^2}$ $s_4 = \\sqrt{c^2 +(b-a)^2}$ Since opposite sides are congruent and the angles are 90 degrees, it follows that the shape is a rectangle.", "the angle between the diagonals. As regards the intersection angle take a look here: Diagonals of parallelogram intersect at $90^\\circ$ if and only if figure is rhombus", "Comment Share Q) # Find the area of the rhombus if the length of each side be $7\\;cm$ and corresponding altitude is $8\\;cm$ $\\begin{array}{1 1} 50\\;cm^2 \\\\ 58\\;cm^2 \\\\ 56\\;cm^2 \\\\ 60\\;cm^2 \\end{array}$ Area of rhombus = base $\\times$ height $\\qquad= 7\\;cm \\times 8 \\;cm$ $\\qquad= 56\\;cm^2$ Answer : $56\\;cm^2$", "\\mathrm{~m}^2$ (iii) $25 \\mathrm{~cm}^2$ (iv) $1600 \\mathrm{~cm}^2$ (v) $100 \\mathrm{~m}^2$ Q 2. Find the area of the square of side $16 \\mathrm{~cm}$. Ans: $256 \\mathrm{~cm}^2$ Q 3. Find the length of the square whose area is $100 \\mathrm{~cm}^2$. Ans: 10 cm ## Summary In this article, we learned about a quadrilateral square with four equal sides and all the angles as right angles, i.e. 90 degrees. The square's diagonals are equal and bisect each other at 90 degrees. We also learned about the various properties of squares and how we can calculate the area of squares. We also discussed the area of square examples to get to know how to find the area of a square of a given length easily. ## FAQs on Area Word Problems 1. Can a rhombus be considered a square? A rhombus can also be considered a square if it follows some conditions. Rhombus is a convex quadrilateral with all four sides equal. It is said to be a square if it has a right vertex angle. 2. How is a square different from a rectangle? A square has all its sides equal in length, whereas a rectangle has only its opposite sides equal in length. The adjacent sides are unequal in rectangles but are equal in squares. 3. Is zero a square" ]

[ "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 =", "# 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); 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", "(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 +", "it is the circumcircle of $\\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \\overline{AB}, \\overline{BC}$ and note that $\\triangle M_1BM_2 \\sim \\triangle ABC$, we see that thse segments respectively cut a $120^{\\circ}$ arc in the circle with radius $18$ and $60^{\\circ}$ arc in the circle with radius $18\\sqrt{3}$. $[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype(\"2 3\") + linewidth(0.7), dashes = linetype(\"8 6\")+linewidth(0.7)+blue, bluedots = linetype(\"1 4\") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP(\"P\",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP(\"A\",A)) -- D(MP(\"B\",B)) -- D(MP(\"C\",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]$ The diagram shows $P$ outside of the grayed locus; notice that the creases [the", "want to share this story. I have put personal values above others, please forgive me. Now back to the problem. Let $$M$$ be the midpoint of $$DE$$ and $$AB \\cap FH = G$$. $$H$$ is the orthocenter of $$\\triangle ABC$$. We need to prove that $$\\widehat{EAM} = \\widehat{CAK}$$. First, we have that $$\\widehat{FBA} = \\widehat{FCA}$$ and $$\\widehat{HBA} = \\widehat{HCA} = (90^\\circ - \\widehat{CAB})$$ $$\\implies \\widehat{FBA} = \\widehat{HBA}$$. That means $$F$$ and $$H$$ reflect one another in line $$AB$$ $$\\implies HF = 2HG$$. $$M$$ is the midpoint of $$DE$$ $$\\implies ED = 2EM$$ $$\\implies \\dfrac{HF}{ED} = \\dfrac{HG}{EM} \\implies EM \\cdot HF = ED \\cdot HG$$. Because $$\\widehat{ADB} = \\widehat{AEB} = 90^\\circ$$ and $$\\widehat{HDB} = \\widehat{HGB} = 90^\\circ$$, $$AEDB$$ and $$HDBG$$ are cyclic quadrilaterals. It's easy to prove that $$\\widehat{DGH} = \\widehat{DAE} \\ (= \\widehat{DBE})$$ and $$\\widehat{GHD} = \\widehat{AED} \\ (= 180^\\circ - \\widehat{DBA})$$. $$\\implies \\triangle HDG \\sim \\triangle EDA \\implies \\dfrac{HD}{ED} = \\dfrac{HG}{EA} \\implies EA \\cdot HD = ED \\cdot HG$$ $$\\implies EA \\cdot HD = EM \\cdot HF \\implies \\dfrac{HD}{EM} =\\dfrac{HF}{EA} \\implies \\triangle HDF \\sim \\triangle EMA \\implies \\widehat{DFH} = \\widehat{MAE}$$. But $$\\widehat{DFH} = \\widehat{KAC}$$ $$\\implies \\widehat{KAC} = \\widehat{MAE}$$. That means $$AK$$ passes through point $$M$$, the midpoint of $$DE$$.", "the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$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}$.", "# 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", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "\\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}$$ Because $\\angle AOB$, $\\angle BOC$, and $\\angle COD$ are congruent, $\\angle AOD = 3\\theta$. To find", "in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715 ## Solution 2 (Official MAA) Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that $$\\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)", "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 =", "# Thread: Area of a region 1. ## Area of a region The diagram shows a logo $ABCD$ in the shape of a trapezium. $AB=13x cm, CD=12x cm, AD=2y cm$ and Angle $BCD = 90^o$. The shaded region PQDR is a parallelogram such that $RD=\\frac{1}{3}AD$. Before reaching the part i have problems with, $BC=2y+5x$ and area of $ABCD=\\frac{1}{2}(4y+5x)(6x)$ The question is Find the area of PQDR 2. Originally Posted by Punch [FONT=Arial]The diagram shows a logo $ABCD$ in the shape of a trapezium. $AB=13x cm, CD=12x cm, AD=2y cm$ and Angle $BCD = 90^o$. The shaded region PQDR is a parallelogram such that $AD=\\frac{1}{3}AD$.<=== typo? RD = ... Before reaching the part i have problems with, $BC=2y+5x$ and area of $ABCD=\\frac{1}{2}(4y+5x)(6x)$ The question is Find the area of PQDR 1. The right triangle at the left with the hypotenuse $\\overline{AB}$ has one leg with the length 12x. Use Pythagorean theorem to determine the length of the missing leg (<--- no pun intended!) 2. The area of a trapezium is calculated by: $a = \\dfrac{\\overline{AD} + \\overline{BC}}2 \\cdot \\overline{DC}$ with $\\overline{AD} = 2y$ $\\overline{BC} = 5x + 2y$ $\\overline{DC} = 12x$ 3. Indeed that was a typo. However, the question hasn't been answered! The question is Find the area of PQDR 4. Originally Posted by Punch Indeed that was", "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? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "= \\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),", "is the radical center of all three circles. Thus, $HA \\cdot HD = HB \\cdot HE = HC \\cdot HF = x$. Let $BC = a$, $CA = b$. Clearly, $AB = 175$ as $AB\\cdot CF = 4200$. It follows that \\begin{align} \\sqrt{a^2 - 24^2} + \\sqrt{b^2 - 24} = AB &= 175 \\\\ a+b+175 &= 360 \\end{align} Thus, examining Pythagorean triples, $a = 40$, $b = 145$. By the Pythagorean Theorem, we can establish that $AF = 143$, and from the given area, $AD = 105$, so $DC = 100$ and $BD = 140$. Finally, $\\triangle AFH \\sim \\triangle CDH \\sim ABD$, so $$x = HC \\cdot HF = \\frac{BD}{AD} \\cdot \\frac{AB}{AD} \\cdot AF \\cdot DC = \\frac{140}{105} \\cdot \\frac{175}{105} \\cdot 143 \\cdot 100 = \\boxed{\\frac{286000}{9}}$$ • Comforting that the two answers agree. Jun 4, 2018 at 3:52 Given: $\\overline{AB} + \\overline{BC} + \\overline{AC} = 360$, $Area_{△ABC} = 2100$ $△ABC$ has altitudes $$\\overline{AD},\\overline{BE},\\overline{CF}$$ where, $$\\overline{AD} \\perp \\overline{BC}, \\overline{BE} \\perp \\overline{AC}, \\overline{CF} \\perp \\overline{AB}$$ $H$ is the orthocenter of $△ABC$ and $\\overline{CF} = 24$ $\\because \\overline{CF} = 24$ and $\\overline{AB} \\perp \\overline{CF}$, $$\\frac{24 * \\overline{AB}}{2} = 2100$$ $12 * \\overline{AB} = 2100, \\overline{AB} = 175$ Next $\\because \\overline{AB} + \\overline{BC} + \\overline{AC} = 360$, $\\overline{BC} + \\overline{AC} = 185$, $\\overline{BC} = 185 - \\overline{AC}$ and Semi-perimeter $S =", "a rectangle with a width of $4.6$ cm and a length of $9$ cm. What is the height, in centimeters, of the pyramid if its volume is $82.8$ cm$^{3}$? (1) $6$ (2) $2$ (3) $9$ (4) $18$ 11. In the diagram below of right triangle $AED$, $\\overline{BC} \\| \\overline{DE}$. Which statement is always true? (1) $\\dfrac{AC}{BC} = \\dfrac{DE}{AE}$ (2) $\\dfrac{AC}{BC} = \\dfrac{DE}{AE}$ (3) $\\dfrac{AC}{CE} = \\dfrac{BC}{DE}$ (4) $\\dfrac{DE}{BC} = \\dfrac{DB}{AB}$ 12. What is an equation of the line that passes through the point ($6,8$) and is perpendicular to a line with equation $y = \\dfrac{3}{2}x + 5$? (1) $y - 8 = \\dfrac{3}{2}(x - 6)$ (2) $y - 8 = -\\dfrac{2}{3}(x - 6)$ (3) $y + 8 = \\dfrac{3}{2}(x + 6)$ (4) $y + 8 = -\\dfrac{2}{3}(x + 6)$ 13. The diagram below shows parallelogram $ABCD$ with diagonals $\\overline{AC}$ and $\\overline{BD}$ intersecting at $E$. What additional information is sufficient to prove that parallelogram $ABCD$ is also a rhombus? (1) $\\overline{BD}$ bisects $\\overline{AC}$ (2) $\\overline{AB}$ is parallel to $\\overline{CD}$ (3) $\\overline{AC}$ is congruent to $\\overline{BD}$ (4) $\\overline{AC}$ is perpendicular to $\\overline{BD}$ 14. Directed line segment $DE$ has endpoints $D(-4,-2)$ and $E(1,8)$. Point $F$ divides $\\overline{DE}$ such that $DF : DE$ is $2:3$. What are the coordinates of $F$? (1) ($-3, 0$) (2) ($-2, 2$) (3) ($-1, 4$) (4) ($2,", "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 =" ]

[ "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", "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", "In a rhombus, each side has length 6. One of the angles of the rhombus is Find the area of the rhombus. Question In a rhombus, each side has length 6. One of the angles of the rhombus is Find the area of the rhombus. in progress 0 3 weeks 2021-09-08T02:00:53+00:00 2 Answers 0 1. Step-by-step explanation: Let the rhombus be $ABCD$, where $\\angle DAB = 120^\\circ$. Then $\\angle ABC = 180^\\circ – \\angle DAB = 180^\\circ – 120^\\circ = 60^\\circ$. [asy] unitsize(1 cm); pair A, B, C, D; A = (0,1); B = (sqrt(3),0); C = (0,-1); D = (-sqrt(3),0); draw(A–B–C–D–cycle); draw(A–C); label(“$A$”, A, N); label(“$B$”, B, E); label(“$C$”, C, S); label(“$D$”, D, W); label(“$6$”, (A + D)/2, NW); [/asy] Since $AB = BC$, triangle $ABC$ is equilateral. By the same argument, triangle $ACD$ is also equilateral. Each triangle has area $\\frac{\\sqrt{3}}{4} \\cdot 6^2 = 9 \\sqrt{3},$so the area of the rhombus is $2 \\cdot 9 \\sqrt{3} = \\boxed{18 \\sqrt{3}}$. 31.18 units² Step-by-step explanation: The diagonals bisect the angles. ⇒ Angle at half the rhombus = 120 ÷ 2 = 60° Both sides of the triangle are 6 units: ⇒ Half the rhombus is also an isosceles triangle. ⇒ the other angle is also 60°. Both angles are 60°, ⇒ the third angle is 180 – 60", "# 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 \"2006 AMC 10B Problems/Problem 15\" ## Problem Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is $24$ and $\\angle BAD = 60^\\circ$. What is the area of rhombus $BFDE$? $[asy] defaultpen(linewidth(0.7)+fontsize(10)); size(120); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)/2); draw(B--C--D--A--B--F--D--E--B); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); label(\"F\", F, dir(point--F)); [/asy]$ $\\mathrm{(A) \\ } 6\\qquad \\mathrm{(B) \\ } 4\\sqrt{3}\\qquad \\mathrm{(C) \\ } 8\\qquad \\mathrm{(D) \\ } 9\\qquad \\mathrm{(E) \\ } 6\\sqrt{3}$ ## Solution Using properties of a rhombus, $\\angle DAB = \\angle DCB = 60 ^\\circ$ and $\\angle ADC = \\angle ABC = 120 ^\\circ$. It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$. Let the lengths of the sides of rhombus $ABCD$ be $s$. The longer diagonal of rhombus $BFDE$ is $BD$. Since $BD$ is a side of an equilateral triangle with a side length of $s$, $BD = s$. The longer diagonal of rhombus $ABCD$ is $AC$. Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$, $AC = 2 \\cdot \\frac{s\\sqrt{3}}{2} = s\\sqrt{3}$ The ratio of the longer diagonal of rhombus $BFDE$ to", "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 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)", "# 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", "# 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", "a rectangle with a width of $4.6$ cm and a length of $9$ cm. What is the height, in centimeters, of the pyramid if its volume is $82.8$ cm$^{3}$? (1) $6$ (2) $2$ (3) $9$ (4) $18$ 11. In the diagram below of right triangle $AED$, $\\overline{BC} \\| \\overline{DE}$. Which statement is always true? (1) $\\dfrac{AC}{BC} = \\dfrac{DE}{AE}$ (2) $\\dfrac{AC}{BC} = \\dfrac{DE}{AE}$ (3) $\\dfrac{AC}{CE} = \\dfrac{BC}{DE}$ (4) $\\dfrac{DE}{BC} = \\dfrac{DB}{AB}$ 12. What is an equation of the line that passes through the point ($6,8$) and is perpendicular to a line with equation $y = \\dfrac{3}{2}x + 5$? (1) $y - 8 = \\dfrac{3}{2}(x - 6)$ (2) $y - 8 = -\\dfrac{2}{3}(x - 6)$ (3) $y + 8 = \\dfrac{3}{2}(x + 6)$ (4) $y + 8 = -\\dfrac{2}{3}(x + 6)$ 13. The diagram below shows parallelogram $ABCD$ with diagonals $\\overline{AC}$ and $\\overline{BD}$ intersecting at $E$. What additional information is sufficient to prove that parallelogram $ABCD$ is also a rhombus? (1) $\\overline{BD}$ bisects $\\overline{AC}$ (2) $\\overline{AB}$ is parallel to $\\overline{CD}$ (3) $\\overline{AC}$ is congruent to $\\overline{BD}$ (4) $\\overline{AC}$ is perpendicular to $\\overline{BD}$ 14. Directed line segment $DE$ has endpoints $D(-4,-2)$ and $E(1,8)$. Point $F$ divides $\\overline{DE}$ such that $DF : DE$ is $2:3$. What are the coordinates of $F$? (1) ($-3, 0$) (2) ($-2, 2$) (3) ($-1, 4$) (4) ($2,", "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}$.", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "it is the circumcircle of $\\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \\overline{AB}, \\overline{BC}$ and note that $\\triangle M_1BM_2 \\sim \\triangle ABC$, we see that thse segments respectively cut a $120^{\\circ}$ arc in the circle with radius $18$ and $60^{\\circ}$ arc in the circle with radius $18\\sqrt{3}$. $[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype(\"2 3\") + linewidth(0.7), dashes = linetype(\"8 6\")+linewidth(0.7)+blue, bluedots = linetype(\"1 4\") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP(\"P\",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP(\"A\",A)) -- D(MP(\"B\",B)) -- D(MP(\"C\",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]$ The diagram shows $P$ outside of the grayed locus; notice that the creases [the", "is the radical center of all three circles. Thus, $HA \\cdot HD = HB \\cdot HE = HC \\cdot HF = x$. Let $BC = a$, $CA = b$. Clearly, $AB = 175$ as $AB\\cdot CF = 4200$. It follows that \\begin{align} \\sqrt{a^2 - 24^2} + \\sqrt{b^2 - 24} = AB &= 175 \\\\ a+b+175 &= 360 \\end{align} Thus, examining Pythagorean triples, $a = 40$, $b = 145$. By the Pythagorean Theorem, we can establish that $AF = 143$, and from the given area, $AD = 105$, so $DC = 100$ and $BD = 140$. Finally, $\\triangle AFH \\sim \\triangle CDH \\sim ABD$, so $$x = HC \\cdot HF = \\frac{BD}{AD} \\cdot \\frac{AB}{AD} \\cdot AF \\cdot DC = \\frac{140}{105} \\cdot \\frac{175}{105} \\cdot 143 \\cdot 100 = \\boxed{\\frac{286000}{9}}$$ • Comforting that the two answers agree. Jun 4, 2018 at 3:52 Given: $\\overline{AB} + \\overline{BC} + \\overline{AC} = 360$, $Area_{△ABC} = 2100$ $△ABC$ has altitudes $$\\overline{AD},\\overline{BE},\\overline{CF}$$ where, $$\\overline{AD} \\perp \\overline{BC}, \\overline{BE} \\perp \\overline{AC}, \\overline{CF} \\perp \\overline{AB}$$ $H$ is the orthocenter of $△ABC$ and $\\overline{CF} = 24$ $\\because \\overline{CF} = 24$ and $\\overline{AB} \\perp \\overline{CF}$, $$\\frac{24 * \\overline{AB}}{2} = 2100$$ $12 * \\overline{AB} = 2100, \\overline{AB} = 175$ Next $\\because \\overline{AB} + \\overline{BC} + \\overline{AC} = 360$, $\\overline{BC} + \\overline{AC} = 185$, $\\overline{BC} = 185 - \\overline{AC}$ and Semi-perimeter $S =", "and $\\overline{PB}$. If $x$ represents the length of $\\overline{PB}$, $\\overline{AP}$ is equal to $(26 – x)$ ft. Using the angle bisector theorem, the ratio of $\\overline{AC}$ and $\\overline{AP}$ is equal to $\\overline{CB}$ and $\\overline{PB}$. \\begin{aligned}\\dfrac{AC}{AP} &= \\dfrac{CB}{PB}\\\\\\dfrac{36}{26- x} &= \\dfrac{42}{x}\\end{aligned} Apply cross-multiplication to simplify and solve the resulting equation. Find the length of $\\overline{PB}$ by finding the value of $x$. \\begin{aligned}36x &= 42(26- x)\\\\36x &= 1092- 42x\\\\36x + 42x &= 1092\\\\78x &= 1092\\\\x&= 14\\end{aligned} Hence, the length of $\\overline{PB}$ is equal to $14$ ft. ### Practice Question 1. In the triangle $\\Delta LMN$ the line $\\overline{MO}$ bisects $\\angle LMO$. Suppose that $\\overline{LM} = 20$ cm, $\\overline{MN} = 81$ cm, and $\\overline{LO} = 64$ cm, what is the length of line segment $\\overline{ON}$? A. $\\overline{ON} = 45$ cm B. $\\overline{ON} = 64$ cm C. $\\overline{ON} = 72$ cm D. $\\overline{ON} = 81$ cm 2. In the triangle $\\Delta ACB$, the line $\\overline{CP}$ bisects $\\angle ACB$. Suppose that $\\overline{AC} = 38$ ft, $\\overline{CB} = 57$ ft, and $\\overline{AB} = 75$ ft, what is the length of line segment $\\overline{PB}$? A. $\\overline{PB} = 38$ ft B. $\\overline{PB} = 45$ ft C. $\\overline{PB} = 51$ ft D. $\\overline{PB} = 57$ ft 3. The angle bisector $\\overline{AD}$ divides the line segment $AC$ that forms the triangle $\\Delta ACB$. Suppose that $\\overline{AC} = 12$", "\"$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$$ • Edit time limits are a POS. So let's see if I get this. Given Eqns (3) and (4) above, $\\overline{df}\\cos 6^\\circ\\ - R\\cos 84^\\circ + R = \\overline{df}\\sin 6^\\circ + R\\sin 84^\\circ + H$ and some algebra later $R - R\\cos 84^\\circ - R\\sin 84^\\circ = \\overline{df}\\sin 6^\\circ - \\overline{df}\\cos 6^\\circ + H$, then distributing out $R$ and $\\overline{df}$ I get $R(1 -\\cos 84^\\circ -\\sin 84^\\circ) = \\overline{df}(\\sin 6^\\circ - \\cos 6^\\circ) +", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(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 =", "# 2014 USAJMO Problems/Problem 6 ## Problem Let $ABC$ be a triangle with incenter $I$, incircle $\\gamma$ and circumcircle $\\Gamma$. Let $M,N,P$ be the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, $\\overline{AB}$ and let $E,F$ be the tangency points of $\\gamma$ with $\\overline{CA}$ and $\\overline{AB}$, respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$, respectively, and let $X$ be the midpoint of arc $BAC$ of $\\Gamma$. (a) Prove that $I$ lies on ray $CV$. (b) Prove that line $XI$ bisects $\\overline{UV}$. ## Solution $[asy] unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label(\"A\", A, W); label(\"B\", B, dir(90)); label(\"C\", C, dir(0)); dot(I); label(\"I\", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label(\"\\Gamma\", dir(35), dir(35)); label(\"\\gamma\", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label(\"M\", M, NE); label(\"N\", N, SE); label(\"P\", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label(\"E\", E, SW); label(\"F\", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label(\"U\", U, S); label(\"V\",", "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "+0 # The area of the equilateral triangle ABE = sin60o(cm2). Line segment AI = 5cm. BF = EF ; find the area DEGHI. +8 442 6 +1068 The area of the equilateral triangle ABE = sin600(cm2). Line segment AI = 5cm. BF = EF =CI =DI ; find the area of the polygon DEGHI. civonamzuk May 29, 2015 #1 +18843 +18 The area of the equilateral triangle ABE = sin600(cm2). Line segment AI = 5cm. BF = EF ; find the area DEGHI. AI = 5 cm $$\\mathbf{ \\sin{(60\\ensurement{^{\\circ}})} = \\frac{1}{2} \\cdot \\sqrt{3} }\\\\ \\mathbf{ \\overline{AB} = \\overline{BE} = \\sqrt{2} \\qquad \\overline{BF} = \\overline{EF} =\\frac{ \\sqrt{2}} {2} }\\\\ \\mathbf{ \\overline{AF} = \\sqrt{2} \\cdot \\sin{(60\\ensurement{^{\\circ}})} = \\frac{\\sqrt{2} }{2} \\cdot \\sqrt{3} }\\\\ \\mathbf{ \\overline{FI}= \\overline{AI} - \\overline{AF} = 5- \\frac{\\sqrt{2} }{2} \\cdot \\sqrt{3} }\\\\$$ If the perpendicular of H is $$\\small{\\text{H_0}}$$ on line ED: $$\\mathbf{ \\overline{BE} : \\overline{HH_0} = 1 : \\dfrac{2}{3}\\qquad \\overline{ED} : \\overline{EH_0} = 1 : \\dfrac{2}{3} }\\\\ \\mathbf{ \\overline{EH_0} = \\dfrac{2}{3} \\cdot ( \\overline{AI} - \\overline{AF} )\\qquad \\overline{HH_0} = \\dfrac{2}{3} \\cdot \\overline{BE} }\\\\$$ The area DEGHI = A: $$\\small{\\text{ \\mathbf{ 2\\cdot A =\\dfrac{2}{3}\\cdot (\\overline{AI}-\\overline{AF}) \\cdot ( \\dfrac{2}{3}\\cdot \\overline{BE} )+ \\dfrac{1}{3}\\cdot (\\overline{AI}-\\overline{AF}) \\cdot ( \\dfrac{2}{3}\\cdot \\overline{BE} - \\overline{BF} )+{2}\\cdot \\dfrac{1}{3}\\cdot (\\overline{AI}-\\overline{AF}) \\cdot \\overline{BF} } }}\\\\\\\\ \\small{\\text{ \\mathbf{ 2\\cdot A =(\\overline{AI}-\\overline{AF}) \\cdot \\left[ \\dfrac{2}{3}\\cdot\\dfrac{2}{3}\\cdot \\sqrt{2}+ \\dfrac{1}{3}\\cdot ( \\dfrac{1}{6}\\cdot" ]

[ "cherries, and sprinkles). How many different two-scoop cones can be ordered that have two d...", "vanilla, and strawberry) with 4 choices of topping (sprinkles, strawberries, nuts, and granola). You choose a cone at random. What is the probability, expressed as a decimal, that you get a cone with strawberry topping?", "### Home > CCA2 > Chapter 10 > Lesson 10.2.1 > Problem10-91 10-91. 1. Connie and Nora went into Ready Scoop to get ice cream cones, but Nora can’t make up her mind. They have 23 flavors and she wants 3 scoops. Homework Help ✎ 1. Nora is very particular about the order of the scoops. How many choices does she have if all of the scoops are different? 2. Nora changes her mind. She wants a dish (in which the order does not matter), not a cone, but she still wants three different flavors. How many ways can she order? 3. Connie says, “I want a cone with dark chocolate on the bottom and then any other flavor(s) for the two scoops on top.” How many cones are possible with dark chocolate on the bottom? 4. Vlad came in as they were leaving and saw Connie’s cone. He said, “Oh, that’s what I want, a cone with chocolate on the bottom and then two other different flavors.” The clerk, said, “Okay, but we have four kinds of chocolate.” Vlad replied, “Any kind of chocolate will do.” How many different cones could fill Vlad’s order? No repeats, but order DOES matter. No repeats, but order DOES NOT matter. $\\text{Repeats are acceptable: } \\text{\\frac{1}{bottom}\\cdot\\frac{22}{middle}\\cdot\\frac{22}{top}}$ Remember that Vlad has four", "Book, published in 1888,[58] endorsed serving ice cream in cones,[59] but the idea predated that. Agnes Marshall was a celebrated cookery writer of her day and helped to popularize ice cream. She patented and manufactured an ice cream maker and was the first person to suggest using liquefied gases to freeze ice cream after seeing a demonstration at the Royal Institution. Reliable evidence proves that ice cream cones were served in the 19th century, and their popularity increased greatly during the St. Louis World's Fair in 1904. According to legend, an ice cream vendor at the fair ran out of cardboard dishes. The vendor at the Syrian waffle booth next door, unsuccessful in the intense heat, offered to make cones by rolling up his waffles. The new product sold well and was widely copied by other vendors.[60][61] ## Other frozen desserts The following is a partial list of ice creamy frozen desserts and snacks: • Ais kacang: a dessert in Malaysia and Singapore made from shaved ice, syrup, and boiled red bean and topped with evaporated milk • Booza: an elastic, sticky, high-level melt resistant ice cream • Dondurma: Turkish ice cream, made of salep and mastic resin • Frozen custard: at least 10% milk fat, at least 1.4% egg yolk and much less air beaten into it,", "# Mix It Up and Mix It In What can we learn about software from an ice cream shop? Quite a bit apparently. It was in the early 70's that Steve's Ice Cream opened and introduced the world to the notion of mix-ins. The idea was simple but powerful. Lets produce a small amount of base ice cream flavors like vanilla, chocolate, and strawberry. Then, the strategy goes, provide a plethora of small independent ingredients to mix in to the base flavor resulting in unending flavor possibilities. Given the success of this approach and many many other ice cream shops like Cold Stone Creamery and Diary Queen that provide similar services the approach seems almost obvious. Well, many things seem obvious in retrospect. The idea itself has great merit! I would want to be able to serve the customer who wants their ice cream to be crunchy as well as those that gag at the thought. If they also want (need!) that smooth caramel ribbon that should be no problem. It is relatively easy to provide 4 different ice cream flavors with different combinations of nuts and caramel to appease all customer types, but this becomes a bit harder when we experience the combinatorial explosion of including (or not!) nuts, oreos, sprinkles, gummy bears, peanut butter cups, and", "ways can I buy 3 kinds if order of ice creams does not matter?", "ices\" in England, did much to popularize ice cream recipes and make its consumption into a fashionable middle-class pursuit. She wrote four books: The Book of Ices (1885), Mrs. A.B. Marshall's Book of Cookery (1888), Mrs. A.B. Marshall's Larger Cookery Book of Extra Recipes (1891) and Fancy Ices (1894) and gave public lectures on cooking. She even suggested using liquid nitrogen to make ice cream. Ice cream soda was invented in the 1870s, adding to ice cream's popularity. The invention of this cold treat is attributed to American Robert Green in 1874, although there is no conclusive evidence to prove his claim. The ice cream sundae originated in the late 19th century. Several men claimed to have created the first sundae, but there is no conclusive evidence to support any of their stories. Some sources say that the sundae was invented to circumvent blue laws, which forbade serving sodas on Sunday. Towns claiming to be the birthplace of the sundae include Buffalo, Two Rivers, Ithaca, and Evanston. Both the ice cream cone and banana split became popular in the early 20th century. Agnes Marshall, \"queen of ices\", instrumental in making ice-cream fashionable The first mention of the cone being used as an edible receptacle for the ice cream is in Mrs. A.B. Marshall's Book of Cookery of 1888.", "Echinacea, Double Scoop RaspberryBurpee Seeds Echinacea, Double Scoop Raspberry ... with double-layered petals colored in intense raspberry rose. Exuberant flowering lasts through the summer heat, ... Chat Free >> Ice cream coneWikipedia An ice cream cone, poke (Ireland and Scotland) ... A variety of double wafer cone exists that allows two scoops of ice cream to be served side by side. Chat Free >> Wolfram Demonstrations Project centrifugeuses à double cône Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Chat Free >> 1.A plane intersects one nappe of a double-napped cone ... 1.A plane intersects one nappe of a double-napped cone such that it is perpendicular to the vertical axis of the cone. Chat Free >> Double Cone BlendersMixers Products > Tumble Blenders > Double Cone Blenders How the Double Cone Blenders Works. Double Cone Blenders are most often used for the intimate dry blending of … Chat Free >> TimDE Tapered Roller Bearing, Double Cone ... TimDE Tapered Roller Bearing, Double Cone, Standard Tolerance, Straight Bore, Steel, Inch,.\" ID,.\" Width: : Industrial & Scientific Chat Free >> Douglas Cone, road magnate at center of scandal, dies … &#;&#;Douglas Cone Sr. secretly and simultaneously ... Douglas Cone, road magnate at center of scandal, ... made headlines with news of", "347 views The passage given below is followed by a set of questions. Choose the most appropriate answer to each question. When i was little, children were bought two kinds of ice cream, sold from those white wagons with canopies made of silvery metal either the two-cent cone or the four-cent ice-cream pie. The two-cent cone was very small, in fact it could fit comfortable into a child’s hand, and it was made by taking the ice cream from its container with a special scoop and piling it on the cone. Granny always suggested I eat only a part of the cone, then throw away the pointed end, because it had been touched by the vendor’s hand ( though it was the best part, nice and crunchy, and it was regularly eaten in secret, after a pretence of discarding it). The four-cent pie was made by a special little machine, also silvery, which pressed two disks of sweet biscuit against a cylindrical section of ice cream. First, you have to thrust your tongue into the gap between the biscuits until it couched the central nucleus of ice cream; then, gradually, you ate the whole thing, the biscuit surfaces softening as they became soaked in creamy nectar. Granny had no advice to give here: in theory the pies had", "Primo Ice Cream Emporium to celebrate. Each ordered their ice cream in a waffle cone or a dish, except for the one who ordered the Slurp-A-Licious Bucket-O-Yum — a delectable concoction with two scoops each of Jack’s five Specialty Flavors — in which case it came in a commemorative Wacky Jack Bucket (which you got to keep if you ate it all). From the following clues, can you decide who ordered what? Note: All solvers made their selections from Jack’s five Specialty Flavors: Blueberry Farkle, Chocolate Smudge, Marshmallow Whoop, Pecan Passion, and Peppermint Blast. For the cones, you must specify the order of the scoops. 1. No cone had more than three scoops of ice cream, and no dish had more than five. 2. Except for the problem solver who got the Slurp-A-Licious Bucket-O-Yum, no one got more than one scoop of the same flavor. 3. The number of scoops in Edith’s dish was the same as the sum of the numbers of scoops in Arlen’s and Cecil’s cones. 4. Cecil and Edith together had the same number of scoops as Arlen and Frank together. 5. If Frank had an even number of scoops, then there were 23 scoops among the six solvers. 6. If Frank had an odd number of scoops, then there were 25 scoops among", "it. Anytime you are trying to organize something with many different pieces or components, you are figuring outcomes. How can we figure out an outcome? There are a couple of different ways to do this, and you are going to learn about them in this lesson. The first one that we are going to work with is a tree diagram. What is a tree diagram? A tree diagram is a visual way of listing outcomes. You look at the choices for the outcome and the variables that go with each outcome. Alright, let’s slow down. First, let’s look at an example and that will help make tree diagrams a lot clearer. Example Jessica has four different favorite types of ice cream. She loves vanilla crunch, black raspberry, chocolate chip and lemonade. She also loves two different types of cones, a plain cone and a sugar cone. Given these flavors and cones choices, how many different single scoop ice cream cones can Jessica create? To solve this problem, we are going to create a tree diagram. First, we list the choices of ice cream. Vanilla Crunch Black Raspberry Chocolate Chip Next we add in the two cone types. Each flavor has two possible cone types that it could go on. This is where the tree diagram part comes in.", "but anecdotally I have also observed that chocolate ice cream is softer, even straight from the freezer. This would likely be due to some change in the mixture of substances in chocolate ice cream versus other flavors. I wouldn't go so far as to say it's any one thing, but it's a different recipe so it will have different properties. – KeithS Jul 23 '13 at 22:41", "two flavours in a line, like (vanilla, strawberry) where the first slots means double scoop and the second slot means single scoop. So it's an ordered 2-subset, because you are permuting for example {vanilla, strawberry}, hence 4P2.", "Actually it just means \"plain.\" I don't how \"vanilla\" is supposed to add anything to the concept, but I guess it's more colorful. ;-) Lubos - Why vanilla is thought of as plain is a good question. As you may know, the \"basic\" American ice cream menu is chocolate, strawberry, vanilla. I don't know how things are in Czechia, but when I was in Ireland in the 1970s I was surprised to find that the Irish did not have this concept. You'd go to the supermarket, and find that, in that week or month, they had nothing but (I'll invent) banana-coconut, black currant ripple, and lime sherbet. It was just random, except that, actually, vanilla was pretty common. Well, for some reason, when we're kids in the U.S., we prefer chocolate and strawberry, while the grown-ups are more partial to vanilla. Similarly, as a kid, I could never understand my father's liking for plain doughnuts, but now that I'm in my late 60s, I love them, and I have loved them for 20 years or more. Also, the whiteness of vanilla ice cream makes us think of plainness, just like plain milk vs. chocolate milk. The same goes for an unpainted or uncovered plaster wall. Yeah thanks, I'm sure people would have been rolling on the floor laughing", "### Home > A2C > Chapter 10 > Lesson 10.2.4 > Problem10-108 10-108. 1. Connie and Nora went into Ready Scoop to get ice cream cones, but Nora can't make up her mind. The have 23 flavors and she wants 3 scoops. Homework Help ✎ 1. If Nora is very particular about the order of the scoops, how many choices does she have if all of the scoops are different? 2. Nora changes her mind. She wants a dish, not a cone, but she still wants three different flavors, how many ways can she order? 3. Connie says, “I still want a cone with dark chocolate on the bottom and then any other two scoops.” How many cones are possible with dark chocolate on the bottom? 4. Vlad came in as they were leaving and saw Connie's cone. He said, “Oh, that's what I want, a cone with chocolate on the bottom and then two other flavors.” The clerk, said, “Okay, but we have four kinds of chocolate.” Vlad replied, “Any kind of chocolate will do.” How many different cones could fill Vlad's order? No repeats, but order DOES matter. No repeats, but order DOES NOT matter. $\\text{Assume that repeats are acceptable } \\text{\\frac{1}{bottom}\\cdot\\frac{22}{middle}\\cdot\\frac{22}{top}}$ See (c). Remember that Vlad has 4 choices of chocolate for the bottom and", "spherical scoop of ice cream,the diameter of the scoop of ice cream is the same as the diameter of the cone. If the ice cream melts, will the cone overflow? Justify your answer. 4cm spherical scoop of ice cream diameter of cone- 7cm Sunday, October 19, 2014 at 9:38pm Math Examine the given right angle triangle. Are the side lengths given in the diagram possible? Justify your answer. lengths: 2.4m, 9.2m, 4.7m Sunday, October 19, 2014 at 9:14pm math given that theta=tan^-1(4/3) find sin theta Sunday, October 19, 2014 at 8:56pm Math Find the first 3 common multiples for 6,7,8. Sunday, October 19, 2014 at 8:41pm Pages: <<Prev | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | Next>> Search Members", "— Arlen, Boris, Cecil, Doran, Edith, and Frank — having just finished some pretty hairy logic problems, decided to go to Wacky Jack’s Primo Ice Cream Emporium to celebrate. Each ordered their ice cream in a waffle cone or a dish, except for the one who ordered the Slurp-A-Licious Bucket-O-Yum — a delectable concoction with two scoops each of Jack’s five Specialty Flavors — in which case it came in a commemorative Wacky Jack Bucket (which you got to keep if you ate it all). From the following clues, can you decide who ordered what? Note: All solvers made their selections from Jack’s five Specialty Flavors: Blueberry Farkle, Chocolate Smudge, Marshmallow Whoop, Pecan Passion, and Peppermint Blast. For the cones, you must specify the order of the scoops. 1. No cone had more than three scoops of ice cream, and no dish had more than five. 2. Except for the problem solver who got the Slurp-A-Licious Bucket-O-Yum, no one got more than one scoop of the same flavor. 3. The number of scoops in Edith’s dish was the same as the sum of the numbers of scoops in Arlen’s and Cecil’s cones. 4. Cecil and Edith together had the same number of scoops as Arlen and Frank together. 5. If Frank had an even number of scoops, then", "the columns appear in, and can potentially changes the order of the rows, but it doesn’t make any fundamental difference. Kind Stars Price 0 strawberry 2.5 3.55 0 strawberry 2.5 5.25 1 chocolate 3.5 6.55 1 chocolate 3.5 5.75 2 vanilla 4.0 4.75 Also note that the join will only contain information about items that appear in both tables. Let’s see an example. Suppose there is a table of reviews of some ice cream cones, and we have found the average review for each flavor. Flavor Stars 0 vanilla 5 1 chocolate 3 2 vanilla 5 3 chocolate 4 Flavor Stars Average 0 chocolate 3.5 1 vanilla 5.0 We can join cones and average_review by providing the labels of the columns by which to join. pandas.DataFrame.join Flavor Price Stars Average 0 strawberry 3.55 NaN 1 vanilla 4.75 5.0 2 chocolate 6.55 3.5 3 strawberry 5.25 NaN 4 chocolate 5.75 3.5 Flavor Price Stars Average 1 vanilla 4.75 5.0 2 chocolate 6.55 3.5 4 chocolate 5.75 3.5 Notice how the strawberry cones have disappeared. None of the reviews are for strawberry cones, so there is nothing to which the strawberry rows can be joined. This might be a problem, or it might not be - that depends on the analysis we are trying to perform with the joined table.", "send her a birthday card with strawberries on it. Sometimes I can’t find one. Sue loves strawberries and so when I knew she was coming for our Mother’s Day cook-out, I knew what the dessert was going to be. It’ll be just perfect for Memorial Day, too, though I’ll be busy making carrot cake sheet cakes for a graduation party. (Carrot cake was one of my first posts as a blogger. Things, luckily, have really improved! If all goes well, I’ll take some better photographs than I did three years ago.) Taking vanilla bean out with my kids’ Mickey Mouse spoon. I only make Strawberry Shortcake once or twice a year, so I try and make it light, layered with lots of ripe fruit, full of textural and temperature contrasts, and touched just enough by two kinds sweet cream–frozen and fresh whipped. It’s a celebration of the start of summer, though if we’re lucky, we have strawberries coming for a good part of summer in Minnesota. For the best Strawberry Shortcake, you need each ingredient to be fresh and/or the best you can find or make. So for this dessert, I made the shortcakes as well as homemade vanilla ice cream. (Baby spoon used at right still in drawer and my kids are 25 and 34. We’ve moved", "Logic Puzzles II Last week, we looked at solving some simple logic puzzles. This week, the logic puzzle is quite a bit more involved — I hope you enjoy it! Six logical friends — Arlen, Boris, Cecil, Doran, Edith, and Frank — having just finished some pretty hairy logic problems, decided to go to Wacky Jack’s Primo Ice Cream Emporium to celebrate. Each ordered their ice cream in a waffle cone or a dish, except for the one who ordered the Slurp-A-Licious Bucket-O-Yum — a delectable concoction with two scoops each of Jack’s five Specialty Flavors — in which case it came in a commemorative Wacky Jack Bucket (which you got to keep if you ate it all). From the following clues, can you decide who ordered what? Note: All solvers made their selections from Jack’s five Specialty Flavors: Blueberry Farkle, Chocolate Smudge, Marshmallow Whoop, Pecan Passion, and Peppermint Blast. For the cones, you must specify the order of the scoops. 1. No cone had more than three scoops of ice cream, and no dish had more than five. 2. Except for the problem solver who got the Slurp-A-Licious Bucket-O-Yum, no one got more than one scoop of the same flavor. 3. The number of scoops in Edith’s dish was the same as the sum of the numbers" ]

[ "need to pick? – Abby Aug 31 '16 at 18:04 There are $\\binom{11}4=330$ ways to choose $4$ non-chocolate flavors and $5!=120$ ways to arrange them and chocolate in a cone with $5$ different flavors; that accounts for $330\\cdot120=39600$ different possibilities. We could also have a cone with $4$ different flavors, one of them occurring twice. There are $\\binom{11}3=165$ ways to choose the $3$ non-chocolate flavors and $4$ ways to choose which flavor will appear twice. Finally, there are $\\binom52=10$ ways to decide which two scoops will be the repeated flavor, and there are then $3!=6$ ways to arrange the remaining $3$ scoops. This adds another $$165\\cdot4\\cdot10\\cdot6=39600$$ possibilities. Finally, we could have just $3$ different flavors, two of them occurring twice. There are $\\binom{11}2=110$ ways to choose the two non-chocolate flavors and $\\binom32=3$ ways to choose which two flavors are repeated. Say that the repeated flavors are $A$ and $B$, and the non-repeated flavor is $C$. Then there are $\\binom52=10$ ways to choose which two scoops are $A$ and $\\binom32=3$ ways to choose which two are $B$, for a total of $$110\\cdot3\\cdot10\\cdot3=9900$$ possibilities. The grand total is therefore $39600+39600+9900=89100$ different possibilities. • Thank you so much Brian, I wouldn't think of this solution in a million years, I wish I have the knowledge you have one day. I would", "Home > A2C > Chapter 10 > Lesson 10.2.4 > Problem10-107 10-107. Once the first scoop is chosen, there is no choice for the second or third scoop. $\\frac{5}{1^{\\text{st}}}\\cdot \\frac{1}{2^{\\text{nd}}}\\cdot \\frac{1}{3^{\\text{rd}}}=5\\text{ ways}$ $P\\text{(all the same)}=\\frac{\\text{number of ways all the same}}{\\text{total number of ways}}=\\frac{5}{5\\cdot5\\cdot5}$ $\\frac{5}{125}=0.04$ Since the order does matter, it is the number of ways 3 out of 5 scoops can be arranged. $\\frac{5\\cdot4\\cdot3}{125}$", "scoops, or $\\omega\\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone. And furthermore, after ordering your scoops, you can order more scoops to be placed on top—all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled. In fact, I’ll even let you change your mind about $\\eta$ as we proceed, so as to give you more orders to make a taller cone. So the process is: • You pick a countable ordinal $\\eta$, which is the number of orders you will make. • For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone. • After making your order, you can freely increase $\\eta$ to any larger countable ordinal, giving you the chance to make as many additional orders as you like. At each limit stage of the ordering process, the ice cream cone you are assembling has all the scoops you’ve ordered so far, and we set the", "scoops to be of flavour A and up to 2 from 4 remaining scoops to be selected from the 24 other flavours. $$\\dfrac{\\dbinom 62\\sum\\limits_{\\imath=0}^2\\dbinom 4\\imath 24^{\\imath}}{26^6}$$", "### Home > CCA2 > Chapter C > Lesson C.1.2 > ProblemC-32 C-32. On Friday, Randy Random went to DJ’s Gourmet Ice Cream Shop to get a cone. With temperatures over $95°$, there were only five flavors left to choose from. Homework Help ✎ 1. If he made a random selection, what is the probability that he got all three scoops the same? Once the first scoop is chosen, there is no choice for the second or third scoop. $\\text{P}(\\text{all the same})\\ = \\frac{\\text{number of ways all the same}}{\\text{total number of ways}}=\\frac{5}{5\\cdot 5\\cdot 5}$ 2. What is the probability that he got all three different? How many choices are there for the first scoop? Once the first scoop is chosen, how many choices are there for the second scoop? What about the third scoop?", "table to show how total cost is related to number of scoops. b. Write a rule to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{t}}$ for number of scoops and $\\textbf{\\textit{R}}$ for total cost. 3. Fabulous’s sells yogurt sandwiches. a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops. b. Write a rule to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{q}}$ for number of scoops and $\\textbf{\\textit{V}}$ for total cost. 1. a. Number of Scoops Total Cost 1 $\\3.25$ 2 $\\4.50$ 3 $\\5.75$ 4 $\\7.00$ b. Rule: $\\mathbf{C} = \\1.25\\textbf{\\textit{b}} + \\2.00$ 2. a. Number of Scoops Total Cost 1 $\\4.50$ 2 $\\7.50$ 3 $\\10.50$ 4 $\\13.50$ b. Rule: $\\textbf{\\textit{R}} = \\3.00\\textbf{\\textit{t}} + \\1.50$ 3. a. Number of Scoops Total Cost 1 $\\4.00$ 2 $\\5.25$ 3 $\\6.50$ 4 $\\7.75$ b. Rule: $\\textbf{\\textit{V}} = \\1.25\\textbf{\\textit{q}} + \\2.75$ Practice 1. Use the information in the sign. Complete the table to show how total cost is related to number of scoops. 2. Write a rule to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{n}}$ for number of scoops and $\\textbf{\\textit{C}}$ for total cost. Yummy's sells apple pie yogurt. 1. Use the information in the sign. Complete the", "calculates the final price of placed orders. The final price takes into account: 1. Number of scoops 2. Ice cream flavors: Vanilla, Chocolate, Gooey Cake and Haskell. All flavors have different prices associated with them. 3. Toppings: Dry, wet or gold. 4. Discounts: NoviSci employees get a 20% discount. Statisticians at NoviSci get a 30% discount (don’t ask me why). And orders >\\$20 get a 10% discount.", "exact explanation please... – tien lee May 11 '18 at 3:38 • It should be $\\binom{30}{4}$. I have written the explanation above. – Sonal_sqrt May 11 '18 at 3:39 • possible typo alert? $n+r-1 = 31$ not $30$ – antkam May 11 '18 at 14:42 • Yes you are right.. – Sonal_sqrt May 11 '18 at 15:43 First take $2$ scoops of chocolate. Now for the remaining $4$ scoops we have to count the number of $\\geq0$ solutions of the system $$x_1+x_2+\\ldots+x_{27}+x_{28}=4\\ .$$ For this count we need $4$ circles $\\bigcirc$ for the scoops and $27$ separating bars $|\\>$. The result is ${31\\choose4}=31\\,465$. No of ways of selecting scoops such that atleast 2 scoops must be chocolate = total no of ways(S) - containing no chocolate(A) - containing exactly 1 chocolate(B) There are 2 different answers depending on whether the arranging order from bottom to top matter or not 1) Order matters S = $28^6$ since for every scoop you have 28 options A = $27^6$ since we are excluding chocolate B = There is exactly one chocolate which can be arranged in 6 places the rest of the places can be filled by $26^5$ ways = $6*26^5$ Answer = $S - A - B = 28^6 - 27^6 - (6*26^5)$ 2) Order doesn't matter i.e. we are", "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 $31+126=\\boxed{157.}$", "have per scoop. There are 16 choices for the first scoop and 16 choices for the second scoop since duplicates are allowed. This works out to $16^2 = 256$. It should be more clear as to how you can expand this to more scoops. 2) working out the number of combinations where a flavour only appears once in each possible combination (e.g. chocolate-vanilla-strawberry and then vanilla-chocolate-strawberry) Now try the case of 3 scoops. There are 16 choices for the first scoop, but 15 choices for the second scoop since duplicates are not allowed. For the third scoop there are 14 choices, which works out to $16*15*14=3,360$. Note that with this way of combining flavors, the order is important since chocolate-vanilla-strawberry and vanilla-chocolate-strawberry are both counted. EDIT (for cases where duplicates are allowed): 1.If there are 16 different ice-cream flavours, how many combinations are there for a two scoop? There's a decent explanation from another post. The formula is indeed $$\\binom{n+k-1}{k}$$ where $n$ is the number of flavors and $k$ is the number of scoops. This is called a combination. Note that using this formula gives a different answer than I originally provided, since combinations do not consider the order in which objects are counted $$\\binom{16+2-1}{2}=\\binom {17}2=\\frac{17\\cdot16}{1\\cdot2}=136$$ 1. If there are still 16 different ice-cream flavours, how many combinations", "Cantor’s Ice Cream Shoppe Welcome to Cantor’s Ice Cream Shoppe! A huge choice of flavors—pile your cone high with as many scoops as you want! Have two scoops, or three, four, or more! Why not infinitely many? Would you like $\\omega$ many scoops, or $\\omega\\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone. And furthermore, after ordering your scoops, you can order more scoops to be placed on top—all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled. In fact, I’ll even let you change your mind about $\\eta$ as we proceed, so as to give you more orders to make a taller cone. So the process is: • You pick a countable ordinal $\\eta$, which is the number of orders you will make. • For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone. • After making your order, you can freely increase $\\eta$ to any larger", "# Combinations with ice cream I'm having some trouble with a math problem. I have 28 different flavors of ice cream and I can only have 3 scoops. I can repeat and the order is partially important. There are 9 fruit flavors. The question is: What is the possible number of combinations of any three-scoop ice cream cone if the first flavor had to be a fruit flavor? I'm not entirely sure what the formula for this would even be. The flavor of the first scoop can be selected in $9$ ways. I will interpret the order is partially important to mean that we do not care in which order the second and third scoops are selected. If the second and third scoops are the same flavor, we have $28$ choices. If the second and third scoops are different flavors, we have $\\binom{28}{2}$ choices since we are selecting a subset of two of the $28$ flavors. $$9\\left[28 + \\binom{28}{2}\\right]$$ If the order did matter, we would have $9$ choices for the first scoop and $28$ choices for both the second and third scoops. $$9 \\cdot 28^2$$", "to order the pills and on day 7 he will have to repeat one of the orders. Suppose he is taking 4 pills a day. There are 4 choices for the pill he takes on the first day. Regardless of which pill he chooses he has 3 pills left and he can order them in 6 ways. Thus he has $4 \\times 6 = 24$ ways to order the 3 pills. Hence there are 24 ways to order the pills and on day 25 he will have to repeat one of the orders. What about 5 pills or 6 pills? Penny", "\\5.75 \\\\& && \\3.00 \\times 2 + \\2.75 = \\8.75 \\\\& && \\3.00 \\times 3 + \\2.75 = \\11.75 \\\\& && \\3.00 \\times 4 + \\2.75 = \\14.75$ Example C Flavor’s sells yogurt floats. a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops. b. Write a rule to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{n}}$ for number of scoops and $\\textbf{\\textit{L}}$ for total cost. Solution: We can use problem solving steps to help us to fill in the table and write the rule. $& \\mathbf{Describe:} && \\text{The sign shows the cost of a scoop and the cost of root beer}. \\\\& && \\text{The table has two columns, one for number of scoops and the other for total cost}. \\\\\\\\& \\mathbf{My \\ Job:} && \\text{Complete the table for}\\ 1\\ \\text{through}\\ 4\\ \\text{scoops of yogurt. Write a function rule} \\\\& && \\text{to show how total cost is related to number of scoops and the cost of the cone}. \\\\\\\\& \\mathbf{Plan:} && \\text{Start with the table. Remember to include the cost of the root beer in the total} \\\\& && \\text{cost. Then write the rule}. \\\\\\\\& \\mathbf{Solve:}$ Number of Scoops Total Cost 1 $\\3.50$ 2 $\\5.50$ 3 $\\7.50$ 4 $\\9.50$ $\\text{Rule}: \\textbf{\\textit{L}} \\textbf{ \\", "scoops. b. Write a rule to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{s}}$ for number of scoops and $\\textbf{\\textit{T}}$ for total cost. Solution: We can use problem solving steps to help us to fill in the table and write the rule. $& \\mathbf{Describe:} && \\text{The sign shows the cost of a scoop and the cost of fruit topping}. \\\\& && \\text{The table has two columns, one for number of scoops and the other for total cost}. \\\\\\\\& \\mathbf{My \\ Job:} && \\text{Complete the table for}\\ 1\\ \\text{through}\\ 4\\ \\text{scoops of yogurt. Write a function rule} \\\\& && \\text{to show how total cost is related to number of scoops and the cost of the cone}. \\\\\\\\& \\mathbf{Plan:} && \\text{Start with the table. Remember to include the cost of the fruit topping in the total} \\\\& && \\text{cost. Then write the rule}. \\\\\\\\& \\mathbf{Solve:}$ Number of Scoops Total Cost 1 $\\2.50$ 2 $\\4.00$ 3 $\\5.50$ 4 $\\7.00$ $\\text{Rule}: \\textbf{\\textit{T}} \\textbf{ \\ = \\1.50}\\textbf{\\textit{s}} \\textbf{\\ + \\1.00} \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;$ $& \\mathbf{Check:} && \\text{Use the rule to verify the table.} \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\\\\\\\& && \\1.50 \\times 1 + \\1.00 = \\2.50 \\\\& && \\1.50 \\times 2 + \\1.00 = \\4.00 \\\\& && \\1.50 \\times 3 + \\1.00 = \\5.50 \\\\& && \\1.50 \\times 4 + \\1.00 = \\7.00$ Example", "to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{b}}$ for number of scoops and $\\textbf{\\textit{C}}$ for total cost. Extra for Experts: Yummy Yogurt – Write Function Rules Solutions Extra for Experts: Yummy Yogurt 1 a. Number of Scoops Total Cost 1 $\\4.50$ 2 $\\7.50$ 3 $\\10.50$ 4 $\\13.50$ b. Rule: $\\textbf{\\textit{R}} = \\3.00\\textbf{\\textit{t}} + \\1.50$ Extra for Experts: Yummy Yogurt 2 a. Number of Scoops Total Cost 1 $\\4.00$ 2 $\\5.25$ 3 $\\6.50$ 4 $\\7.75$ b. Rule: $\\textbf{\\textit{V}} = \\1.25\\textbf{\\textit{q}} + \\2.75$ Extra for Experts: Yummy Yogurt 1 – Write Function Rules Mixer’s sells yogurt slushes. a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops. b. Write a rule to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{t}}$ for number of scoops and $\\textbf{\\textit{R}}$ for total cost. Extra for Experts: Yummy Yogurt 2 – Write Function Rules Fabulous’s sells yogurt sandwiches. a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops. b. Write a rule to represent the relationship between number of scoops and total cost. Use $\\textbf{\\textit{q}}$ for number of scoops and $\\textbf{\\textit{V}}$ for total cost. Feb 23, 2012 Nov 05, 2014", "our build. After this choice, we place the last item in the final slot our build order. Over the course of finalizing our build, we had 3 options, then 2 options, and then 1 final option. We need to multiply these numbers so that we count each possible sequence of items that we could have chosen: for each of the three possible first item choices, there are two more options after that, then each of those options is followed by one option (the last item). So each of the initial three options leads to two more, down the road. Adding them up, we get $2+2+2=3\\cdot 2=6$. Then we count the final item selection in each of these choices: there is 1 option, so we multiply $6\\cdot 1=6$ to get the final answer. In total: $3\\cdot 2\\cdot 1$. Looks familiar right? It’s the factorial we just learned! Check-Your-Knowledge Question 1. It’s the end of the game and the top laner is full build. How many ways could they order the items in their inventory? 2. Since we already know what $5!$ is, notice that we can write $6!=6\\cdot 5!$ as a quicker way to calculate $6!$. (Try expanding both sides if you don’t believe it.) Can you turn this into a general statement about the value of $n!$, if", "# Cantor’s Ice Cream Shoppe Welcome to Cantor’s Ice Cream Shoppe! A huge choice of flavors—pile your cone high with as many scoops as you want! Have two scoops, or three, four, or more! Why not infinitely many? Would you like $\\omega$ many scoops, or $\\omega\\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone. And furthermore, after ordering your scoops, you can order more scoops to be placed on top—all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled. In fact, I’ll even let you change your mind about $\\eta$ as we proceed, so as to give you more orders to make a taller cone. So the process is: • You pick a countable ordinal $\\eta$, which is the number of orders you will make. • For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone. • After making your order, you can freely increase $\\eta$ to any", "that comes up for distribution has a choice of three boxes, and the result of this stage would be $3^5$. However there is a restriction that requires the bottom box not be empty. So how many options leave the bottom box empty? In that case there are just two options for each ball and the result of the 5-ball stage would be $2^5$. So we can subtract that off and get the full answer: $$6(3^5-2^5)=6(243-32) = 6\\cdot 211 = 1266$$", "remaining 4 to be one of 25 choices can be done in $5 \\cdot 25^4$ ways, just as you said." ]

[ "### Home > CCA2 > Chapter 10 > Lesson 10.2.1 > Problem10-91 10-91. 1. Connie and Nora went into Ready Scoop to get ice cream cones, but Nora can’t make up her mind. They have 23 flavors and she wants 3 scoops. Homework Help ✎ 1. Nora is very particular about the order of the scoops. How many choices does she have if all of the scoops are different? 2. Nora changes her mind. She wants a dish (in which the order does not matter), not a cone, but she still wants three different flavors. How many ways can she order? 3. Connie says, “I want a cone with dark chocolate on the bottom and then any other flavor(s) for the two scoops on top.” How many cones are possible with dark chocolate on the bottom? 4. Vlad came in as they were leaving and saw Connie’s cone. He said, “Oh, that’s what I want, a cone with chocolate on the bottom and then two other different flavors.” The clerk, said, “Okay, but we have four kinds of chocolate.” Vlad replied, “Any kind of chocolate will do.” How many different cones could fill Vlad’s order? No repeats, but order DOES matter. No repeats, but order DOES NOT matter. $\\text{Repeats are acceptable: } \\text{\\frac{1}{bottom}\\cdot\\frac{22}{middle}\\cdot\\frac{22}{top}}$ Remember that Vlad has four", "have per scoop. There are 16 choices for the first scoop and 16 choices for the second scoop since duplicates are allowed. This works out to $16^2 = 256$. It should be more clear as to how you can expand this to more scoops. 2) working out the number of combinations where a flavour only appears once in each possible combination (e.g. chocolate-vanilla-strawberry and then vanilla-chocolate-strawberry) Now try the case of 3 scoops. There are 16 choices for the first scoop, but 15 choices for the second scoop since duplicates are not allowed. For the third scoop there are 14 choices, which works out to $16*15*14=3,360$. Note that with this way of combining flavors, the order is important since chocolate-vanilla-strawberry and vanilla-chocolate-strawberry are both counted. EDIT (for cases where duplicates are allowed): 1.If there are 16 different ice-cream flavours, how many combinations are there for a two scoop? There's a decent explanation from another post. The formula is indeed $$\\binom{n+k-1}{k}$$ where $n$ is the number of flavors and $k$ is the number of scoops. This is called a combination. Note that using this formula gives a different answer than I originally provided, since combinations do not consider the order in which objects are counted $$\\binom{16+2-1}{2}=\\binom {17}2=\\frac{17\\cdot16}{1\\cdot2}=136$$ 1. If there are still 16 different ice-cream flavours, how many combinations", "# Combinations with ice cream I'm having some trouble with a math problem. I have 28 different flavors of ice cream and I can only have 3 scoops. I can repeat and the order is partially important. There are 9 fruit flavors. The question is: What is the possible number of combinations of any three-scoop ice cream cone if the first flavor had to be a fruit flavor? I'm not entirely sure what the formula for this would even be. The flavor of the first scoop can be selected in $9$ ways. I will interpret the order is partially important to mean that we do not care in which order the second and third scoops are selected. If the second and third scoops are the same flavor, we have $28$ choices. If the second and third scoops are different flavors, we have $\\binom{28}{2}$ choices since we are selecting a subset of two of the $28$ flavors. $$9\\left[28 + \\binom{28}{2}\\right]$$ If the order did matter, we would have $9$ choices for the first scoop and $28$ choices for both the second and third scoops. $$9 \\cdot 28^2$$", "### Home > CCA > Chapter 10 > Lesson 10.2.4 > Problem10-70 10-70. Write and solve an equation to answer the question below. Remember to define any variables you use. Homework Help ✎ Pierre's Ice Cream Shoppe charges $1.19 for a scoop of ice cream and$0.49 for each topping. Gordon paid \\$4.55 for a three-scoop sundae. How many toppings did he get? x = # of toppings How much did Gordon pay for just the three scoops?", "need to pick? – Abby Aug 31 '16 at 18:04 There are $\\binom{11}4=330$ ways to choose $4$ non-chocolate flavors and $5!=120$ ways to arrange them and chocolate in a cone with $5$ different flavors; that accounts for $330\\cdot120=39600$ different possibilities. We could also have a cone with $4$ different flavors, one of them occurring twice. There are $\\binom{11}3=165$ ways to choose the $3$ non-chocolate flavors and $4$ ways to choose which flavor will appear twice. Finally, there are $\\binom52=10$ ways to decide which two scoops will be the repeated flavor, and there are then $3!=6$ ways to arrange the remaining $3$ scoops. This adds another $$165\\cdot4\\cdot10\\cdot6=39600$$ possibilities. Finally, we could have just $3$ different flavors, two of them occurring twice. There are $\\binom{11}2=110$ ways to choose the two non-chocolate flavors and $\\binom32=3$ ways to choose which two flavors are repeated. Say that the repeated flavors are $A$ and $B$, and the non-repeated flavor is $C$. Then there are $\\binom52=10$ ways to choose which two scoops are $A$ and $\\binom32=3$ ways to choose which two are $B$, for a total of $$110\\cdot3\\cdot10\\cdot3=9900$$ possibilities. The grand total is therefore $39600+39600+9900=89100$ different possibilities. • Thank you so much Brian, I wouldn't think of this solution in a million years, I wish I have the knowledge you have one day. I would", "### Home > CCG > Chapter 11 > Lesson 11.1.2 > Problem11-26 11-26. Connie and Nora went into Ready Scoop to get ice cream cones, but Nora cannot make up her mind. They have $23$ flavors and she wants $3$ scoops. 1. If Nora is very particular about the order of the scoops, how many choices does she have if all of the scoops are different? No repeats, but order DOES matter. 1. Nora changes her mind. She wants a dish, not a cone, but she still wants three different flavors. How many ways can she order? No repeats, but order DOES NOT matter. 2. Connie says, “I still want a cone with dark chocolate on the bottom and then any other two scoops.” How many cones are possible with dark chocolate on the bottom? Assume that repeats are acceptable $\\frac{1}{\\text{bottom}}\\cdot\\frac{22}{\\text{middle}}\\cdot\\frac{22}{\\text{top}}$ 3. Vlad came in as they were leaving and saw Connie’s cone. He said, “Oh, that’s what I want, a cone with chocolate on the bottom and then two other flavors.” The clerk, said, “Okay, but we have four kinds of chocolate.” Vlad replied, “Any kind of chocolate will do.” How many different cones could fill Vlad’s order? See (c). Remember that Vlad has $4$ choices of chocolate for the bottom and not just one.", "### Home > CCA2 > Chapter 10 > Lesson 10.2.1 > Problem10-91 10-91. Connie and Nora went into Ready Scoop to get ice cream cones, but Nora can’t make up her mind. They have $23$ flavors and she wants $3$ scoops. 1. Nora is very particular about the order of the scoops. How many choices does she have if all of the scoops are different? No repeats, but order DOES matter. 2. Nora changes her mind. She wants a dish (in which the order does not matter), not a cone, but she still wants three different flavors. How many ways can she order? No repeats, but order DOES NOT matter. 3. Connie says, “I want a cone with dark chocolate on the bottom and then any other flavor(s) for the two scoops on top.” How many cones are possible with dark chocolate on the bottom? Repeats are acceptable: $\\frac{1}{bottom}\\cdot\\frac{22}{middle}\\cdot\\frac{22}{top}$ 4. Vlad came in as they were leaving and saw Connie’s cone. He said, “Oh, that’s what I want, a cone with chocolate on the bottom and then two other different flavors.” The clerk, said, “Okay, but we have four kinds of chocolate.” Vlad replied, “Any kind of chocolate will do.” How many different cones could fill Vlad’s order? Remember that Vlad has four choices of chocolate for the", "Cantor’s Ice Cream Shoppe Welcome to Cantor’s Ice Cream Shoppe! A huge choice of flavors—pile your cone high with as many scoops as you want! Have two scoops, or three, four, or more! Why not infinitely many? Would you like $\\omega$ many scoops, or $\\omega\\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone. And furthermore, after ordering your scoops, you can order more scoops to be placed on top—all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled. In fact, I’ll even let you change your mind about $\\eta$ as we proceed, so as to give you more orders to make a taller cone. So the process is: • You pick a countable ordinal $\\eta$, which is the number of orders you will make. • For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone. • After making your order, you can freely increase $\\eta$ to any larger", "### Home > CCAA8 > Chapter 10 Unit 11 > Lesson CCA: 10.2.4 > Problem10-70 10-70. Write and solve an equation to answer the question below. Remember to define any variables you use. Pierre's Ice Cream Shoppe charges $\\1.19$ for a scoop of ice cream and $\\0.49$ for each topping. Gordon paid $\\4.55$ for a three-scoop sundae. How many toppings did he get? $x=$ # of toppings How much did Gordon pay for just the three scoops?", "its design. Scenario 1 - Buy 1: $5 - Chocolate Ice Cream$5 - Vanilla Ice Cream The simplest and most naive solution is to give the AI one’s preference and teach it to choose that. • #1 Pick Chocolate Scenario 2 - Buy 1: $5 - Vanilla Ice Cream$5 - Strawberry Ice Cream With this scenario, our previous AI Design leaves us disappointed and treatless, as our AI is forever stuck at the counter without being able to make a valid decision. A better design should be able to handle both scenarios reasonably. Many potentially better picking strategies are possible. • #2 Pick the first • #3 Pick the second • #4 Pick the last • #5 Pick one at random Scenario 3 - Buy 1: $50 - Rocky Road Ice Cream$5 - Cherry Vanilla Ice Cream $1 - Extra Napkins Of our various new picking strategies, two of them would leave us very unhappy after our third purchase, and one would have a very good chance of leaving us unhappy. Only one of the strategies would make us feel good about our AI decision (#3), and even then, it’s only a positional accident. To evolve our AI, we would need to evolve the criteria. This adds a pruning factor. • #6 Pick the first ice cream that", "scoops, or $\\omega\\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone. And furthermore, after ordering your scoops, you can order more scoops to be placed on top—all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled. In fact, I’ll even let you change your mind about $\\eta$ as we proceed, so as to give you more orders to make a taller cone. So the process is: • You pick a countable ordinal $\\eta$, which is the number of orders you will make. • For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone. • After making your order, you can freely increase $\\eta$ to any larger countable ordinal, giving you the chance to make as many additional orders as you like. At each limit stage of the ordering process, the ice cream cone you are assembling has all the scoops you’ve ordered so far, and we set the", "# Cantor’s Ice Cream Shoppe Welcome to Cantor’s Ice Cream Shoppe! A huge choice of flavors—pile your cone high with as many scoops as you want! Have two scoops, or three, four, or more! Why not infinitely many? Would you like $\\omega$ many scoops, or $\\omega\\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone. And furthermore, after ordering your scoops, you can order more scoops to be placed on top—all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled. In fact, I’ll even let you change your mind about $\\eta$ as we proceed, so as to give you more orders to make a taller cone. So the process is: • You pick a countable ordinal $\\eta$, which is the number of orders you will make. • For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone. • After making your order, you can freely increase $\\eta$ to any", "cherries, and sprinkles). How many different two-scoop cones can be ordered that have two d...", "augmented with the Stars information in the ratings table. For each cone in cones, join finds a row in ratings whose Kind matches the cone’s Flavor. We have to tell join to use those columns for matching. pandas.DataFrame.join Flavor Price Stars 0 strawberry 3.55 2.5 1 vanilla 4.75 4.0 2 chocolate 6.55 3.5 3 strawberry 5.25 2.5 4 chocolate 5.75 3.5 Each cone now has not only its price but also the rating of its flavor. In general, a call to join that augments a table (say table1) with information from another table (say table2) looks like this: The new table rated allows us to work out the price per star, which you can think of as an informal measure of value. Low values are good – they mean that you are paying less for each rating star. Flavor Price Stars $/Star 1 vanilla 4.75 4.0 1.188 0 strawberry 3.55 2.5 1.420 4 chocolate 5.75 3.5 1.643 2 chocolate 6.55 3.5 1.871 3 strawberry 5.25 2.5 2.100 Though strawberry has the lowest rating among the three flavors, the less expensive strawberry cone does well on this measure because it doesn’t cost a lot per star. Side note. Does the order we list the two tables matter? Let’s try it. As you see it, this changes the order that", "### Home > A2C > Chapter 10 > Lesson 10.2.4 > Problem10-108 10-108. 1. Connie and Nora went into Ready Scoop to get ice cream cones, but Nora can't make up her mind. The have 23 flavors and she wants 3 scoops. Homework Help ✎ 1. If Nora is very particular about the order of the scoops, how many choices does she have if all of the scoops are different? 2. Nora changes her mind. She wants a dish, not a cone, but she still wants three different flavors, how many ways can she order? 3. Connie says, “I still want a cone with dark chocolate on the bottom and then any other two scoops.” How many cones are possible with dark chocolate on the bottom? 4. Vlad came in as they were leaving and saw Connie's cone. He said, “Oh, that's what I want, a cone with chocolate on the bottom and then two other flavors.” The clerk, said, “Okay, but we have four kinds of chocolate.” Vlad replied, “Any kind of chocolate will do.” How many different cones could fill Vlad's order? No repeats, but order DOES matter. No repeats, but order DOES NOT matter. $\\text{Assume that repeats are acceptable } \\text{\\frac{1}{bottom}\\cdot\\frac{22}{middle}\\cdot\\frac{22}{top}}$ See (c). Remember that Vlad has 4 choices of chocolate for the bottom and", "# An ice cream parlour has 8 flavours of ice cream. Joyce wants a cone with 3 different scoops. She wants chocolate mint on the top. How many different ways can she choose the other 2 flavours? Jan 3, 2017 $42$ if the order of the other scoops matters, $21$ if it doesn't. #### Explanation: Since the top scoop is already defined, we only deal with the middle and bottom scoop. For the bottom scoop she has $7$ choices left. For the middle scoop she has $6$ choices left. This is a total of $7 \\times 6 = 42$ permutations . If the order doen't matter, then every of the above choices has a double in reversed order. So the number of combinations is: $42 \\div 2 = 21$.", "# How many combinations of $3$-scoop ice cream cones are possible from $6$ flavors, if all flavors on the cone must be different? (Parent seeking help.) My 5th-grade daughter needs to figure out this problem and it has been years for me. Two girls go into an ice cream shop. There are $$6$$ flavors of ice cream. How many combinations of a $$3$$-scoop ice cream cone can you make? All the flavors have to be different on each cone, meaning you can't have strawberry, strawberry, and peach on a cone. If you can help not only with the answer but the equation I would appreciate it. • I changed the tag, because this is not abstract-algebra – J. W. Tanner Sep 7 '20 at 1:40 • Would ice cream cones that differ only in the order of the three scoops be considered to be different? I'd be curious to see a picture of the assignment. – littleO Sep 7 '20 at 1:41 • and are you asking about how many combinations are possible for one of the two girls? – J. W. Tanner Sep 7 '20 at 1:42 • The question is a bit ambiguous which is why folks are commenting. There are ${6 \\cdot 5 \\cdot 4 \\over 3 \\cdot 2 \\cdot 1}$ different $3$ scoop combinations,", "less than one cent. At the local ice cream stand, people would come in with a carload of kids and order one cone, then one cone, then one cone ... to avoid paying any sales tax. Management tried to put a stop to the practice, and at least a few customers asserted their Constitutional right to place as many orders as they wanted. (This was in NY, where people tend to assert their rights rather loudly.) 3. I routinely see diseconomies of scale at grocery stores here in South Florida (Publix and Whole Foods). My wife and I have learned from it and we always double check the price per ounce. Fortunately, most stores these days pre-calculate that for you and put the$/oz on the price tag.", "the columns appear in, and can potentially changes the order of the rows, but it doesn’t make any fundamental difference. Kind Stars Price 0 strawberry 2.5 3.55 0 strawberry 2.5 5.25 1 chocolate 3.5 6.55 1 chocolate 3.5 5.75 2 vanilla 4.0 4.75 Also note that the join will only contain information about items that appear in both tables. Let’s see an example. Suppose there is a table of reviews of some ice cream cones, and we have found the average review for each flavor. Flavor Stars 0 vanilla 5 1 chocolate 3 2 vanilla 5 3 chocolate 4 Flavor Stars Average 0 chocolate 3.5 1 vanilla 5.0 We can join cones and average_review by providing the labels of the columns by which to join. pandas.DataFrame.join Flavor Price Stars Average 0 strawberry 3.55 NaN 1 vanilla 4.75 5.0 2 chocolate 6.55 3.5 3 strawberry 5.25 NaN 4 chocolate 5.75 3.5 Flavor Price Stars Average 1 vanilla 4.75 5.0 2 chocolate 6.55 3.5 4 chocolate 5.75 3.5 Notice how the strawberry cones have disappeared. None of the reviews are for strawberry cones, so there is nothing to which the strawberry rows can be joined. This might be a problem, or it might not be - that depends on the analysis we are trying to perform with the joined table.", "vanilla, and strawberry) with 4 choices of topping (sprinkles, strawberries, nuts, and granola). You choose a cone at random. What is the probability, expressed as a decimal, that you get a cone with strawberry topping?" ]

[ "Consider the initial value problem below. The value of $y$ at $x= \\ln 2$, (rounded off to $3$ decimal places) is _______. $$\\frac{dy}{dx}= 2x-y,\\:\\:\\:\\:y(0)= 1$$ in Others edited", "## If $$x+2y+1=y-x,$$ what is the value of $$x?$$ ##### This topic has expert replies Moderator Posts: 1991 Joined: 15 Oct 2017 Followed by:6 members ### If $$x+2y+1=y-x,$$ what is the value of $$x?$$ by BTGmoderatorLU » Mon Aug 23, 2021 10:41 am 00:00 A B C D E ## Global Stats Source: Official Guide If $$x+2y+1=y-x,$$ what is the value of $$x?$$ 1) $$y^2=9$$ 2) $$y=3$$ The OA is B • Page 1 of 1", "Browse Questions What is the value of $cos^{-1}x+cos^{-1}y$? Answer: $cos^{-1}(xy-\\sqrt{1-x^2}\\: \\sqrt{1-y^2})$", "what is the value of ~$x^2-y^2$~? (1) ~$(x-y)^2=9$~ (2) ~$x+y=6$~ Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.", "Browse Questions # What is the value of $cos^{-1}x-cos^{-1}y$ Answer: $cos^{-1}(xy+\\sqrt{1-x^2}\\: \\sqrt{1-y^2})$", "$x = 4$, $y = 5$).", "## What is the value of $$x^2 - y^2?$$ ##### This topic has expert replies Legendary Member Posts: 2898 Joined: 07 Sep 2017 Thanked: 6 times Followed by:5 members ### What is the value of $$x^2 - y^2?$$ by Vincen » Thu Oct 14, 2021 3:33 am 00:00 A B C D E ## Global Stats What is the value of $$x^2 - y^2?$$ (1) $$x + y = 0$$ (2) $$x - y = 2$$", "Q # What should be the value of a if the value of 2x^2 + x - a equals to 5, Q. 9. What should be the value of a if the value of $2x^{2}+x-a$ equals to $5$, when $x=0?$ $\\\\2x^2 + x - a = 5 \\\\2 \\times 0^2 + 0 - a = 5 \\\\-a = 5 \\\\a = -5$", "< 5^x iii. For what value of x is 2^x = 3^x = 5^x i) Where x < 0 ii) Where x > 0 iii) Where x = 0", "If x > 0, what is the value of x5? (1) ~$\\sqrt{x}$~= 32 (2)~$x^2 = 2^{20}$~ Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient. ##### 考题讲解 （1）若~$\\sqrt{x}=32$~，则~$x=32^2$~，那么~$x^5=(32^2)^5$~ （2） 若~$x^2=2^{10}$~，则~$x=2^{10}$~，那么~$x^5=(2^{10})^5$~", "$y-4=5(x--3)$ or $y=5x+19$ 0", "## If $$|x-y|=y,$$ what is the value of $$x?$$ ##### This topic has expert replies Legendary Member Posts: 2852 Joined: 07 Sep 2017 Thanked: 6 times Followed by:5 members ### If $$|x-y|=y,$$ what is the value of $$x?$$ by Vincen » Thu Dec 02, 2021 8:31 am 00:00 A B C D E ## Global Stats If $$|x-y|=y,$$ what is the value of $$x?$$ (1) $$xy>0$$ (2) $$y=6$$", "# If 53x = 125 and 10y Question: If $5^{3 x}=125$ and $10^{y}=0.001 .$ Find $x$ and $y .$ Solution: $5^{3 x}=125$ and $10^{y}=0.001$ $5^{3 x}=5^{3}$ $x=1$ Now, $10^{\\mathrm{y}}=0.001$ $10^{\\mathrm{y}}=10^{-3}$ $\\mathrm{y}=-3$ Therefore, the value of x = 1 and the value of y = – 3", "# 20. What is the value of x+y? $\\begin{array}{lcl} If\\:the\\:ratio\\:of\\:x\\:to\\:400\\:is\\:equal\\:to\\:the\\:ratio\\:of\\:\\\\ 3\\:to\\:5,\\:and\\:if\\:the\\:ratio\\:of\\:400\\:to\\:y\\:is\\:equal\\:to\\:the\\:\\\\ ratio\\:of\\:2\\:to\\:3,\\:what\\:is\\:the\\:value\\:of\\:x+y\\:?\\\\ \\end{array}$", "y are integers, what is the value of x + y [#permalink] ### Show Tags 18 Dec 2018, 04:25 Bunuel wrote: If $$5,000 = 2^x5^y$$ and x and y are integers, what is the value of $$x + y$$? A. 3 B. 4 C. 5 D. 6 E. 7 5000= 2^3*5^4 2^3*5^4 =2^x*5^y or say x+y= 7 IMO E Re: If 5,000 = 2^x5^y and x and y are integers, what is the value of x + y [#permalink] 18 Dec 2018, 04:25 Display posts from previous: Sort by", "If x = $\\sqrt{\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}}$, then the value of $5x^{2}-5x-1$ is : [A] 3 [B] 2 [C] 4 [D] 1", "Question 1 Question 1 of 5 Factor $$y^2 - 25$$ (y + 5)(y - 5) $$y^2 - 2y - 25$$ (y - 5)$$^2$$ (y - 5)(y - 5)", "Let $y^{2}-2y+1=x$ and $\\sqrt{x}+y=5$. The value of $x+\\sqrt{y}$ equals _________. (Give the answer up to three decimal places).", "Question 9 What is the value of $$cos^2A/(1 + sinA)$$? Question 10 ­ If 2x - ­2(x -­ 2) < 5 -­ x > ­2x + 2; then the value of x is OR", "# What is the value of m for all values of x? 14.If $(x+3)^{m}(x-3)=(x+3)(x^{2}-9)$ for all values of $x$, what is the value of $m$? 1. 1 2. 2 3. 3 4. 4 5. Cannot be determined from the information given." ]

[ "$1$ to both sides. $3 = y$ or $y = 3$", "into the first which will look like this: [20(7+Y)+8Y]/28=15 Now distribute the 20 and multiply both sides by 28 to get: 140+20Y+8Y=420, then combine like terms to get: 140+28Y=420, then subtract 140 from both sides to get: 28Y=280, then divide both sides by 28 to get: Y=10. Now you have your weekend price of$10. Plug that into your original 2nd equation to get:", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "both sides of 2 and add to 1 we get, $5 y + 4 z = 30$ substituting $y = 2$ we get, $z = 5$ Placing value of y and z to 1 we get, $x = 3$", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "$3$. $- 22 + 3 y = 6$ Add $22$ to both sides. $3 y = 6 + 22$ Simplify $6 + 22$ to $28$. $3 y = 28$ Divide both sides by $3$. $y = \\frac{28}{3}$ or $\\approx 9.33$ graph{(5x-y+9)(-2x+y+2)=0 [-12.795, 9.705, -15.865, -4.615]}", "both sides to isolate the $y$: $y = - 10$", "3$ on both sides and than dividing by two, which results in $x = - 1$. 6. Your final answer is $y = 1$ and $x = - 1$.", "## College Algebra (10th Edition) Let $x$ be the number of small boxes, $y$ be the number of medium boxes, $z$ be the amount of large boxes. Then our equations: (for oatmeal raisin): $x+2y+2z=15$ (for chocolate chip): $x+y+2z=10$ (for shortbread): $y+3z=11$ Deducing the second equation from the first one we get: $y=5$, then substituting this into the third, $z=2$, then $x=1$. So we buy $1$ small, $5$ medium and $2$ big boxes.", "note that there are two special solutions: one with $x=y=1$, and one with $y=z=1$. We counted each of them twice, hence we have to subtract two from the total. Therefore $m-n = 334 + 501 + 167 - 2 = 1000$, and the answer is $1000\\bmod 1000 = \\boxed{000}$.", "## Algebra: A Combined Approach (4th Edition) 5x + 10 y = 20 2x + 6y = 10 to the second equation divide by 3 x + 3y = 5 minu 3y to both sides x + 3y - 3y = 5 - 3y x = 5 - 3y substitude x to the first equation 5 ( 5 - 3y ) + 10 y = 20 25 - 15 y + 10y = 20 25 - 5y = 20 minus 25 to both sides 25 - 5y - 25 = 20 - 25 -5y = -5 divide by -5 both sides $\\frac{-5y}{-5}$ = $\\frac{-5}{-5}$ y = 1 substitude y = 1 to x = 5 - 3y x = 5 - $3\\times1$ x = 5 - 3 x = 2", "# How do you solve 9y-1=y-25? Add 1 to each side to get $9 y = y - 24$. Next, subtract y from each side to get $8 y = - 24$. Finally, divide each side by 8 to get $y = - 3$.", "= 4$ Add $1$ to both sides. $x = 5$", "y^2)$ So let $x = 3a$ and $y = 1$ Thus $27a^3 - 1 = (3a)^3 - (1)^3 = (3a - 1)((3a)^2 + (3a)(1) + (1)^2)$ $= (3a - 1)(9a^2 + 3a + 1)$ -Dan 9. Oh, now I understand, thank you.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "- 3z) + 16y + 51z = 879$$ $$489 + 10y + 42z = 879$$ $$5y + 21z = 195$$ Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$. If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$, $x = 112$, $y = 18$, and $z = 5$, so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \\boxed{905}$.", "and $z=20$. We only require The sum of $y+z$, which is $26+20=46$. We get that the sum of $y$ and $z$ respectively is $\\boxed{\\textbf{(D)}\\ 46}$ -OofPirate ~savannahsolver", "use the quadratic formula rather than just factoring, the formula shows that either $$x=12$$ and $$y=3$$, or $$x=-3$$ and $$y=-12$$. Either way, $$x\\cdot y = \\boxed{36}$$. Solution #2: $$x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\\rightarrow2xy = 153–81 =72\\text{,} \\text{ ie., } xy = \\boxed{36}.$$ Both ways, you get the same answer of $$\\boxed{36}$$. cryptoaops Dec 29, 2020", "## Algebra 2 (1st Edition) Isolating y on one side of the equation: $$.5y = .25x -2 \\\\ y =.5x-4$$ For x=4: $$y = .5(4)-4 = -2$$" ]

[ "$GCH$ and $GEA$, respectively. Thus, $y:y+8 = 3:5$, from which we have $y=12$, thus $GF=y+8=12+8=\\boxed{\\mathrm{(B)}\\ 20}$", "$1$ to both sides. $3 = y$ or $y = 3$", "both sides of 2 and add to 1 we get, $5 y + 4 z = 30$ substituting $y = 2$ we get, $z = 5$ Placing value of y and z to 1 we get, $x = 3$", "$f$. But we also know that $y = 40 - x$, so the value of $y$ is also $20$. The solution is then $x = 20 , y = 20$", "use the quadratic formula rather than just factoring, the formula shows that either $$x=12$$ and $$y=3$$, or $$x=-3$$ and $$y=-12$$. Either way, $$x\\cdot y = \\boxed{36}$$. Solution #2: $$x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\\rightarrow2xy = 153–81 =72\\text{,} \\text{ ie., } xy = \\boxed{36}.$$ Both ways, you get the same answer of $$\\boxed{36}$$. cryptoaops Dec 29, 2020", "y^2)$ So let $x = 3a$ and $y = 1$ Thus $27a^3 - 1 = (3a)^3 - (1)^3 = (3a - 1)((3a)^2 + (3a)(1) + (1)^2)$ $= (3a - 1)(9a^2 + 3a + 1)$ -Dan 9. Oh, now I understand, thank you.", "$$0=x^2-2x-4-3x+4$$ 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:", "## Thinking Mathematically (6th Edition) Given $x^{2}$+ 4; x = 5 Now put the value of x = 5 in the given expression. = $5^{2}$ + 4 = 25 + 4 = 29", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "$y = x + 2 y = x + 2$ 10. $y = x − 3 y = x − 3$ 11. $y = 3 x − 1 y = 3 x − 1$ 12. $y = −2 x + 2 y = −2 x + 2$ 13. $y = − x − 3 y = − x − 3$ 14. $y = − x − 2 y = − x − 2$ 15. $y = 2 x y = 2 x$ 16. $y = −2 x y = −2 x$ 17. $y = 1 2 x + 2 y = 1 2 x + 2$ 18. $y = 1 3 x − 1 y = 1 3 x − 1$ 19. $y = 4 3 x − 5 y = 4 3 x − 5$ 20. $y = 3 2 x − 3 y = 3 2 x − 3$ 21. $y = − 2 5 x + 1 y = − 2 5 x + 1$ 22. $y = − 4 5 x − 1 y = − 4 5 x − 1$ 23. $y = − 3 2 x + 2 y = − 3 2 x + 2$ 24. $y = − 5 3 x + 4 y = − 5 3 x +", "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "## Thinking Mathematically (6th Edition) Given $x^{2}$ - 4(x-y), x=3, y=8 Now use the value of x=3 and y=8 in the given expression. = $3^{2}$ - 4(3-8) =9-4(-5) =9+20 =29", "and $z=20$. We only require The sum of $y+z$, which is $26+20=46$. We get that the sum of $y$ and $z$ respectively is $\\boxed{\\textbf{(D)}\\ 46}$ -OofPirate ~savannahsolver", "- 3z) + 16y + 51z = 879$$ $$489 + 10y + 42z = 879$$ $$5y + 21z = 195$$ Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$. If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$, $x = 112$, $y = 18$, and $z = 5$, so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \\boxed{905}$.", "\\: \\text{and}\\:y_1-3=3$ $x_1=5 \\:\\text{and}\\:y_1=6$ $\\boxed{D(x_1, y_1) = D(5,6)}$", "$3$. $- 22 + 3 y = 6$ Add $22$ to both sides. $3 y = 6 + 22$ Simplify $6 + 22$ to $28$. $3 y = 28$ Divide both sides by $3$. $y = \\frac{28}{3}$ or $\\approx 9.33$ graph{(5x-y+9)(-2x+y+2)=0 [-12.795, 9.705, -15.865, -4.615]}", "$y-4=5(x--3)$ or $y=5x+19$ 0", "$ \\boxed{G'(x)=5 e^{x-1}-2x-2} $ Now there is probably an error in there somewhere, but you should get the idea. RonL", "$$x=5-y,$$ and substituting it into the second: $$6(5-y)+10y=34$$ $$\\Longrightarrow 30-6y+10y=34 \\Longrightarrow y=1.$$ Putting this value into the first equation yields $x=4$. Thus, she answered $4$ in the first round and $1$ in the second round.", "- 4$. If $x = - 4$ then the second equation becomes ${y}^{2} = - 12$, which has no Real valued solutions. If $x = 1$ then the second equation becomes ${y}^{2} = 3$, so $y = \\pm \\sqrt{3}$" ]

[ "of $6$ therefore it is a multiple of $6$", "2$$ so $$a < 2 < 4$$ so $$a < 4$$ must be true. So $$\\frac 12 < a$$ true and $$a < 4$$ is true so $$\\frac 12 < a < 4$$ must be TRUE. And $$\\frac 12 < a < 4$$ does not contradict $$1 < a < 2$$. It's just that $$1 is a stronger more precise statement. Consider this. $$a= 36$$ is an even number. $$a=36$$ is a perfect square. $$a=36$$ is a multiple of $$3$$. Fred comes and says \"$$a$$ is a multiple of $$3$$\". Then Jane comes along and says \"$$a$$ is a perfect square\". Michael says \"$$a$$ is an even number.\" Olav hears all three of these people and says \"Wait! Those are three completely different statements! So only one of them can be true! Which one of you is telling the truth and which one is lying.\" .... And they are all telling the truth. $$\\frac 12 < a < 4$$ means that it is possible that $$a$$ might be less than $$1$$ but it does not mean that it is. So $$\\frac 12 < a$$ does not contradict $$1 < a$$. This is analogous to \"$$a$$ is a multiple of $$3$$\". That means $$a$$ might be odd but it doesn't mean that it is. \"$$a$$ is a multiple of", "are very different. You are correct that the first is false and the second true, but there is nothing paradoxical about this. The first statement says that for every $a$, if $a$ is a multiple of $2$, then $a$ is a multiple of $8$. I know you don't see where it says, \"for every $a$\" but this is a common convention, that you are actually quite used to. What about $\\sin^2x+\\cos^2=1.$ This doesn't say, \"for every $x$\", but that's what it means. Let's take the true statement, \"if $x$ is a multiple of $8$ then $x$ is a multiple of $2$. You'll agree that's true for every $x$, I think. Wait a sec! Is it true when $x=9?$ Is it true when $x=4?$ Yes and yes, if we adopt the convention in the table.", "that is a multiple of both $4$ and $5$ is $20$, it follows that $20|a$.", "# Multiples ## Interactive practice questions a List the first five multiples of $4$4 on the same line, separated by commas. b List the first five multiples of $5$5 on the same line, separated by commas. c Which of the multiples are common to both lists? If there is more than one, write all of them on the same line, separated by commas. Easy Approx a minute In ascending order, list the multiples of $6$6 between $15$15 and $44$44. Use commas to separate each multiple.", "8064}$ is the only answer choice that is a multiple of $2016$.", "A = (B-4)(B)(B+4) B= 4, A =0 ----> No, A is not a multiple of 6 B= 6, A =(2)(6)(10) -----> Yes, A is a multiple of 6 Statement 2 - B is a multiple of 3 - B has no relationship with A. Combined Statement 1 + Statement 2 B=6, As seen above \"A\" is a multiple of 6 B=9, A =(5)(9)(13) ----> No, A is not a multiple of 6 _________________ I will rather do nothing than be busy doing nothing - Zen saying Math Expert Joined: 02 Sep 2009 Posts: 49438 Re: Is A multiple of 6? [#permalink] ### Show Tags 17 Sep 2012, 08:53 2 2 Is A multiple of 6? (1) A = B(B+4)(B-4). If $$B=odd$$, then $$A=B(B+4)(B-4)=odd*odd*odd=odd$$ and the answer is NO but if $$B=6$$, then the answer is obviously YES. Not sufficient. (2) B-4 is a multiple of 3. B is 4 more than a multiple of 3. Not sufficient, since no info about A. (1)+(2) From (2) we have that B is 4 more than a multiple of 3 and from (1) that $$A=B(B+4)(B-4)$$. If $$B=odd$$ (for example if $$B=7$$), then the answer is NO but if $$B=4$$, then $$A=B(B+4)(B-4)=0$$ and the answer is YES (since zero is a multiple of all integers except zero itself). Not sufficient. Hope it", "multiples of $$5$$ are: $$5,10,15,20,25,30$$. ## Solution With Working 1. The first $$5$$ multiples of $$3$$ are: • $$1\\times 3 = 3$$ • $$2\\times 3 = 6$$ • $$3\\times 3 = 9$$ • $$4\\times 3 = 12$$ • $$5\\times 3 = 15$$ 2. The $$7^{\\text{th}}$$ multiple of $$2$$ is: $$7\\times 2 = 14$$ 3. The first $$4$$ multiples of $$7$$ are: • $$1\\times 7 = 7$$ • $$2\\times 7 = 14$$ • $$3\\times 7 = 21$$ • $$4\\times 7 = 28$$ 4. The $$6^{\\text{th}}$$ multiple of $$4$$ is: $$6 \\times 4 = 24$$. 5. The first $$6$$ multiples of $$5$$ are: • $$1\\times 5 = 5$$ • $$2\\times 5 = 10$$ • $$3\\times 5 = 15$$ • $$4\\times 5 = 20$$ • $$5\\times 5 = 25$$ • $$6\\times 5 = 30$$ ## Exercise 2 State whether each of the following statements is true or false. 1. Statement: $$12$$ is a multiple of $$3$$. 2. Statement: $$7$$ is a multiple of $$1$$. 3. Statement: The $$4^{\\text{th}}$$ multiple of $$5$$ is $$20$$. 4. Statement: The $$3^{\\text{rd}}$$ multiple of $$7$$ is $$23$$. 5. Statement: The $$3^{\\text{rd}}$$ multiple of $$4$$ equals to the $$4^{\\text{th}}$$ multiple of $$3$$. 1. true 2. true 3. true 4. false 5. true ## Solution With Working 1. True, since $$12 = 4\\times 3$$. $$12$$ is the $$4^{\\text{th}}$$", "v ## Question 6 List the first $10$ multiples for the following numbers: a) $6$a s v b) $11$a s v c) $20$a s v d) $25$a s v e) $1000$a s v f) $17$a s v ## Question 7 Are the following statements true or false? a) $2$ is a factor of $18$a s v b) $3$ is a factor of $19$a s v c) $5$ is a factor of $75$a s v d) $14$ is a factor of $140$a s v e) $4$ is a factor of $86$a s v f) $11$ is a factor of $143$a s v ## Question 8 Are the following statements true or false? a) $12$ is a multiple of $4$a s v b) $29$ is a multiple of $3$a s v c) $0$ is a multiple of $104$a s v d) $36$ is a multiple of $9$a s v e) $45$ is a multiple of $15$a s v f) $48$ is a multiple of $13$a s v ## Question 9 Classify the following numbers as prime or composite: a) $5$a s v b) $6$a s v c) $11$a s v d) $29$a s v e) $37$a s v f) $87$a s v ## Question 10 Negative numbers can also be factors! Using this information find the factors for: a) $15$a", "over the property of being a multiple of $$3$$.", "for the following numbers (remember $0$!): a) $3$a s v d b) $5$a s v d c) $10$a s v d d) $9$a s v d e) $7$a s v d f) $12$a s v d ## Question 6 List the first $10$ multiples for the following numbers: a) $6$a s v d b) $11$a s v d c) $20$a s v d d) $25$a s v d e) $1000$a s v d f) $17$a s v d ## Question 7 Are the following statements true or false? a) $2$ is a factor of $18$a s v d b) $3$ is a factor of $19$a s v d c) $5$ is a factor of $75$a s v d d) $14$ is a factor of $140$a s v d e) $4$ is a factor of $86$a s v d f) $11$ is a factor of $143$a s v d ## Question 8 Are the following statements true or false? a) $12$ is a multiple of $4$a s v d b) $29$ is a multiple of $3$a s v d c) $0$ is a multiple of $104$a s v d d) $36$ is a multiple of $9$a s v d e) $45$ is a multiple of $15$a s v d f) $48$ is a multiple of $13$a s v d", "a multiple of both $6$ and $14$?", "• # question_answer If $A=[x:x$ is a multiple of 3] and $B=[x:x$ is a multiple of 5], then A - B is ($\\bar{A}$ means complement of A) [AMU 1998] A) $\\bar{A}\\cap B$ B) $A\\cap \\bar{B}$ C) $\\bar{A}\\cap \\bar{B}$ D) $\\overline{A\\cap B}$ A - B = A Ç Bc = A Ç $\\bar{B}$.", "Browse Questions # If $A=\\{1,3,5,B\\}$ and $B=\\{2,4\\}$, then which of the following is true ? (A) $4\\in A$ (B) $\\{4\\}\\subset A$ (C) $B\\subset A$ (D) $B\\in A$ $B\\in A$ is correct answer. Because $4\\notin A$ and $\\{B\\}\\subset A$ not B.", "positive integers. Consider these two equations: $$m=a+b+c$$ $$n=abc$$ 1. Which of these statements are true? Select all that apply. 1. If $a$ is tripled, $m$ is tripled. 2. If $a$, $b$, and $c$ are all tripled, then $m$ is tripled. 3. If $a$ is tripled, $n$ is tripled. 4. If $a$, $b$, and $c$ are all tripled, then $n$ is tripled. 2. Create a true statement of your own about one of the equations. ### Student Response Details about student responses to this activity are available at IM Certified Partner LearnZillion (requires a district subscription) or without a subscription here (requires registration). ### Activity Synthesis Ask previously identified students to share their reasoning about which statements are true (or not true). Display any examples (or counterexamples) for all to see and have students refer to them while sharing. If using the algebraic structure is not brought up in students’ explanations, display for all to see: • If $a$, $b$, and $c$ are all tripled, the expression becomes $3a +3b+3c$, which can be written as $3(a+b+c)$ by using the distributive property to factor out the 3. So if all the addends are tripled, their sum, $m$, is also tripled. • Looking at the third statement, if $a$ is tripled, the expression becomes $(3a)bc$, which, by using the associative property,", "# What is the answer for $6:2(2+1)$? [duplicate] What is the answer for this question $6:2(2+1) = ?$ Is it 1 or 9 ? I think it is nondeterministic is it true ? - ## marked as duplicate by Arkamis, Dominic Michaelis, Pedro Tamaroff, Qiaochu YuanMar 14 '13 at 20:17 This has already been done assuming that $:$ is the same operator as $/$, look at these for answers already given:", "# Which statement must be true? 5. If $a<b<c$ and $a<d<c$, which of the following statements must be true? 1. $a<c$ 2. $b=d$ 3. $b<d$ 1. I only 2. II only 3. III only 4. I and II 5. I, II and III $\\text{We know that:} \\begin{cases} a<b>c\\\\ a<d<c \\end{cases}$ Which means both b and d lies between a and b as explained in following figure, but we don't know if b is equal to, greater or less than d.", "expression for the number of solutions. - The statement is false, since $z$ is necessarily divisible by $\\gcd(a,b,m)$, which may not be $1$. Thus the equivalent statement in the comment is likewise false. For instance, the set of products of two numbers with remainder $2$ modulo $4$ only contains odd multiples of $4$. - Please see my comment under the question. – joriki Oct 7 '11 at 14:14", "A and 4 B's in a row. A Statement - 1 is false, Statement - 2 is true B Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1 C Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1 D Statement - 1 is true, Statement - 2 is false ## Explanation Note : n items can be distribute among p persons are $${}^{n + p - 1}{C_{p - 1}}$$ ways. Here n = 6 ice-cream p = 5 types of ice-cream Each ice-cream belongs to one of the 5 ice-cream type. So chosen 6 ice-crean can be divide into 5 types of ice-cream. $$\\therefore$$ The number of different ways the child can buy the six ice-cream is = $${}^{6 + 5 - 1}{C_{5 - 1}}$$ = $${}^{10}{C_4}$$ $$\\therefore$$ Statement - 1 is false. Number of different ways of arranging 6 A and 4 B's in a row = $${{10!} \\over {6!4!}} = {}^{10}{C_4}$$ $$\\therefore$$ Statement - 2 is true. 4 ### AIEEE 2007 The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus $$A \\cup B \\cup C = S,\\,A", "False since $n(B) = 7$ while $n(C) = 9$ (v) $n(A) < n(C)$ Answer: True since $n(A) = 7$ while $n(C) = 9$" ]

[ "statement says that niether of them is a factor of $$4$$. Remember our original conditions. either $$y$$ is a multiple of $$4$$ or $$(y-1)$$ is a multiple of $$4$$. This statement tells us that $$y$$ is not a multiple of $$4$$ but it does not tell us whether $$(y-1)$$ is a multiple of $$4$$ or not. Also note that this again does not tell us whether $$y$$ is even or $$(y-1)$$ is even.So Insufficient. Statement B: $$y^3-y$$ is divisible by $$4$$ Lets factor this one out : $$y^3-y=y*(y-1)*(y+1)$$ and again these are $$3$$ consecutive integers. The only thing we have found out (from the statement) is that one of them is definitely a multiple of $$4$$ but we do not know which one it is. Also note that we are not sure that one of them is even or two of them are even. And if it is one even, it could be $$2$$ so it is still not sufficient. We need at-least $$2$$ multiples of $$2$$ in there. If there were $$2$$ even numbers, we would be sure that it is divisible by $$4$$. So Insufficient Combined: We know that $$y$$ and $$(y+1)$$ are not multiples of $$4$$ from the first statement so from our second statement combined we have found out that $$(y-1)$$ is then, logically", "are very different. You are correct that the first is false and the second true, but there is nothing paradoxical about this. The first statement says that for every $a$, if $a$ is a multiple of $2$, then $a$ is a multiple of $8$. I know you don't see where it says, \"for every $a$\" but this is a common convention, that you are actually quite used to. What about $\\sin^2x+\\cos^2=1.$ This doesn't say, \"for every $x$\", but that's what it means. Let's take the true statement, \"if $x$ is a multiple of $8$ then $x$ is a multiple of $2$. You'll agree that's true for every $x$, I think. Wait a sec! Is it true when $x=9?$ Is it true when $x=4?$ Yes and yes, if we adopt the convention in the table.", "is a multiple of 2 (2) y is a multiple of 3 I do not know how to evaluate this question. I mean: for me 1 and 2 are insuff because we do not know alternatively of the other variable. Together we do not have the information that we want ok E is the answer BUT if we have $$\\frac{4}{9}$$ we know that 9 is not in the form$$2^n * 5^m$$ so is not a terminating decimal ok ------->$$\\frac{18}{24}$$ reduced is $$\\frac{3}{4}$$ and neither 4 is in the aforementioned form here the OA explanation. may be is late in my TM but I'm confused Quote: Statement 1 indicates that x is a multiple of 2, which has nothing to do with terminating or non-terminating property of decimals. For instance 2/4 is a terminating decimal while 4/6 is a non terminating decimal. So, NOT SUFFICIENT. Statement 2 says that y is a multiple of 3, but gives no information about the common factors of x and y if any, and what is x/y in lowest terms. For instance, 2/3 is non-terminating while 9/12 is terminating. So, NOT SUFFICIENT. Taking statements 1 and 2 together, 4/9 which satisfies both the statements is non- terminating, while 18/24 is a terminating decimal. So NOT SUFFICIENT. The correct answer is E Actually $$4=2^2*5^0$$,", "= q \\cdot y + r$$ when $$r=1$$ which implies that when $$x$$ is divided by $$y$$ we get a remainder of 1, meaning that $$x$$ is NOT a multiple of $$y$$. Since both sub-statements are contradictory, this means that when $$x^2 - x$$ is a multiple of $$y$$, $$x$$ is not always a multiple of $$y$$. meaning that we cannot determine whether $$x$$ is always a multiple of $$y$$ according to Statement 2 alone. Therefore, Statement 2 is not sufficient. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Senior Manager Joined: 15 Aug 2013 Posts: 251 Re: If x and y are integers greater than 1, is x a multiple of y [#permalink] ### Show Tags 18 Sep 2014, 18:36 Bunuel wrote: If x and y are integers great than 1, is x a multiple of y? (1) $$3y^2+7y=x$$ --> $$y(3y+7)=x$$ --> as $$3y+7=integer$$, then $$y*integer=x$$ --> $$x$$ is a multiple of $$y$$. Sufficient. (2) $$x^2-x$$ is a multiple of $$y$$ --> $$x(x-1)$$ is a multiple of $$y$$ --> $$x$$ can be multiple of $$y$$ ($$x=2$$ and $$y=2$$) OR $$x-1$$ can be multiple of $$y$$ ($$x=3$$ and $$y=2$$) or their product can be multiple of $$y$$ ($$x=3$$ and $$y=6$$). Not sufficient. Hope it helps. Hi Bunuel, Two questions: 1) If x =", "Package of problems ## Problem Use the interactivity below to find values of $a$ and $b$ for which • $a$ is less than $b$ and $f(a)$ is less than $f(b).$ • $a$ is less than $b$ but $f(a)$ is greater than $f(b).$ Here are some statements about the function $f(x)$ in the graph above. In each case, decide if the statement is true. If it is not true, try to adapt it to make a true statement. 1. If $-5<a<b<-3$ then $f(a) < f(b).$ 2. If $2.5 < a < b < 5$ then $f(a) < f(b).$ 3. If $-2.5 < a< b <2$ then $f(a) > f(b).$ 4. If $-2.5 < a < b < 3$ then $f(a) > f(b).$ Now consider the following statements about real numbers $a$ and $b$ and some familiar functions. Try to sketch or visualise the graph of the functions as you think about the statements. 1. If $a$ is less than $b$, then $a^{2}$ is less than $b^{2}.$ 2. If $a$ is less than $b$, then $a-5$ is less than $b-5.$ 3. If $a$ is less than $b$, then $3-2a$ is less than $3-2b.$ 4. If $a$ is less than $b$, then $\\dfrac{1}{a}$ is greater than $\\dfrac{1}{b}.$ If you think a statement is true or false for all $a$ and $b$,", "x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following? I. 8 II. 12 III 18 A. II only B. I and II only C. I and III only D. II and III only E. I, II and III First of all notice that we are asked $$xy$$ MUST be a multiple of which of the following, not COULD be a multiple. $$x$$ is a multiple of 4 --> $$x=4m$$, for some positive multiple $$m$$, so $$x$$ can be: 4, 8, 12, ... $$y$$ is a multiple of 6 --> $$y=6n$$, for some positive multiple $$n$$, so $$y$$ can be: 6, 12, 18, ... So, $$xy=(4m)*(6n)=24mn$$, so $$xy$$ is in any case a multiple of 24, which means it must be a multiple of 8 and 12, but not necessarily of 18. For example the least value of $$x$$ is 4 and the least value of $$y$$ is 6, so the least value of $$xy$$ is 24, which is a multiple of both 8 and 12, but not 18. Hope it's clear. _________________ ##### General Discussion Director Joined: 28 Dec 2005 Posts: 891 ### Show Tags 02 Dec 2006, 06:06 1 2 nah, not necessarily: Ans B. From question x/4 is", "--> $$x$$ is a multiple of $$y$$. Sufficient. (2) $$x^2-x$$ is a multiple of $$y$$ --> $$x(x-1)$$ is a multiple of $$y$$ --> $$x$$ can be multiple of $$y$$ ($$x=2$$ and $$y=2$$) OR $$x-1$$ can be multiple of $$y$$ ($$x=3$$ and $$y=2$$) or their product can be multiple of $$y$$ ($$x=3$$ and $$y=6$$). Not sufficient. Hope it helps. _________________ VP Status: Been a long time guys... Joined: 03 Feb 2011 Posts: 1055 Location: United States (NY) Concentration: Finance, Marketing GPA: 3.75 Re: If x and y are integers greater than 1, is x a multiple of y? [#permalink] ### Show Tags 31 Dec 2012, 21:27 41 7 x and y are integers > 1. Is x multiple of y? (1): 3y^2 + 7y = x (2): x^2 - x is multiple of y The question is asking whether $$x/y$$ is an integer or not. Statement 1 can be written as $$y(3y +7)=x$$ or $$3y+7=x/y$$. Since $$y$$ is an integer, therefore 3y+7 must also be an integer. Hence $$x/y$$ will be an integer or x is a multiple of y. Sufficient. Statement 2 can be written as $$x(x-1)/y$$ is an integer. Notice that we can't deduce that x is a multiple of y because it is quite possible that the product is a multiple of y, but not the", "for $$8^a$$ end with $$6$$, $$a$$ should be of the form :$$4k$$. From this, we know the nature of the number $$a$$, but we do not have any information about$$X$$. Since we do not know the value of $$X$$, Statement 2 alone is NOT sufficient to answer the question. Hence, we can eliminate answer choice B. Step 3: Combine both Statements: • From the first statement, we know: The units digit of $$X$$ is $$2$$ • From the second statement we have: a is a multiple of $$4$$ • We know that a is of form 4k, and we also know$$2^{4k}$$ always has a units digit: $$6$$. Thus, we know that the units digit of the expression $$X^a$$ is $$6$$. By combining both statements we got a unique answer. _________________ SVP Status: It's near - I can see. Joined: 13 Apr 2013 Posts: 1687 Location: India GPA: 3.01 WE: Engineering (Real Estate) Re: If X is a positive integer, what is the [#permalink] Show Tags 29 Mar 2018, 02:48 EgmatQuantExpert wrote: e-GMAT Question: If $$X$$ is a positive integer, what is the units digit of $$X^a$$? 1) $$X$$ when divided by $$10$$ gives a remainder of $$2$$ 2) $$8^a$$ ends with $$6$$ A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B) Statement (2)", "A = (B-4)(B)(B+4) B= 4, A =0 ----> No, A is not a multiple of 6 B= 6, A =(2)(6)(10) -----> Yes, A is a multiple of 6 Statement 2 - B is a multiple of 3 - B has no relationship with A. Combined Statement 1 + Statement 2 B=6, As seen above \"A\" is a multiple of 6 B=9, A =(5)(9)(13) ----> No, A is not a multiple of 6 _________________ I will rather do nothing than be busy doing nothing - Zen saying Math Expert Joined: 02 Sep 2009 Posts: 49438 Re: Is A multiple of 6? [#permalink] ### Show Tags 17 Sep 2012, 08:53 2 2 Is A multiple of 6? (1) A = B(B+4)(B-4). If $$B=odd$$, then $$A=B(B+4)(B-4)=odd*odd*odd=odd$$ and the answer is NO but if $$B=6$$, then the answer is obviously YES. Not sufficient. (2) B-4 is a multiple of 3. B is 4 more than a multiple of 3. Not sufficient, since no info about A. (1)+(2) From (2) we have that B is 4 more than a multiple of 3 and from (1) that $$A=B(B+4)(B-4)$$. If $$B=odd$$ (for example if $$B=7$$), then the answer is NO but if $$B=4$$, then $$A=B(B+4)(B-4)=0$$ and the answer is YES (since zero is a multiple of all integers except zero itself). Not sufficient. Hope it", "in the form $$3y\\pm 1$$where $y$ is a positive integer. Therefore $$(3y \\pm 1)^2 = 3n - 9x^2 \\pm 6x - 1$$ and so $$9y^2 \\pm 6y + 1 = 3n - 9x^2\\pm 6x - 1$$ $$9y^2\\pm 6y + 2 = 3n -9x^2 \\pm 6x$$ The right hand side of the expression is a multiple of three, but the left hand side however is clearly not, as it is a multiple of three add $2$. Here is a contradiction. When $b$ was said to be a non multiple of three, it led to saying that a must also be a non multiple, and a contradiction occurred. Therefore neither $a$ nor $b$ can be non multiples of three, hence if $a^2 + b^2$ is a multiple of three then $a$ and $b$ are both multiples of three.", "expression for the number of solutions. - The statement is false, since $z$ is necessarily divisible by $\\gcd(a,b,m)$, which may not be $1$. Thus the equivalent statement in the comment is likewise false. For instance, the set of products of two numbers with remainder $2$ modulo $4$ only contains odd multiples of $4$. - Please see my comment under the question. – joriki Oct 7 '11 at 14:14", "is a multiple of $$y$$ or $$x-1$$ is a multiple of $$y$$ the sub-statement $$x$$ is a multiple of $$y$$ implies that $$x = q \\cdot y$$ which would answer the question, but we still need to figure out if the second sub-statement answers the question with the same answer as the first sub-statement. the second sub-statement $$x-1$$ is a multiple of $$y$$ implies that $$x-1 = q \\cdot y$$ which can be rewritten as $$x = q \\cdot y + 1$$ which is an equation of the form $$x = q \\cdot y + r$$ when $$r=1$$ which implies that when $$x$$ is divided by $$y$$ we get a remainder of 1, meaning that $$x$$ is NOT a multiple of $$y$$. Since both sub-statements are contradictory, this means that when $$x^2 - x$$ is a multiple of $$y$$, $$x$$ is not always a multiple of $$y$$. meaning that we cannot determine whether $$x$$ is always a multiple of $$y$$ according to Statement 2 alone. Therefore, Statement 2 is not sufficient. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Manager Status: I will not stop until i realise my goal which is my dream too Joined: 25 Feb 2010 Posts: 235 Schools: Johnson '15 Followers: 2 Kudos [?]: 33 [0], given: 16 Re:", "but not a multiple of $4$. It follows that $b$ cannot be an odd number, so it must be a multiple of $2$, say $b=2c.$ Then $b^2 = 4c^2\\in(4)$. The same argument applies for any prime power $p^m$. If $ab\\in(p^m)$ but $a\\notin(p^m)$ then $b$ must be a multiple of $p$ and so $b^m\\in(p^m).$ To show that $(6)$, for example, is not primary, use the fact that $6=2\\times3$. If you take $a=2$ and $b=3$ in the definition of a primary ideal, you see that $2$ is not a multiple of $6$, so the definition asks that some power of $3$ should be a multiple of $6$. But that cannot happen, because every power of $3$ is an odd number. Therefore the definition of primary is not satisfied and so $(6)$ is not a primary ideal of $\\mathbb Z.$ Peter Well-known member MHB Site Helper To show that $(4)$ is primary, we have to show that if [TEX] ab \\in (4) [/TEX] and [TEX] a \\notin (4) [/TEX] then [TEX] b^n \\in (4) [/TEX] for some positive integer $n.$ So suppose that $ab\\in(4)$, in other words $ab$ is a multiple of $4$, and that $a\\notin(4)$. Thus $a$ could be an odd number, or an odd multiple of $2$, but not a multiple of $4$. It follows that $b$ cannot be", "• # question_answer If $A=[x:x$ is a multiple of 3] and $B=[x:x$ is a multiple of 5], then A - B is ($\\bar{A}$ means complement of A) [AMU 1998] A) $\\bar{A}\\cap B$ B) $A\\cap \\bar{B}$ C) $\\bar{A}\\cap \\bar{B}$ D) $\\overline{A\\cap B}$ A - B = A Ç Bc = A Ç $\\bar{B}$.", "positive integers. Consider these two equations: $$m=a+b+c$$ $$n=abc$$ 1. Which of these statements are true? Select all that apply. 1. If $a$ is tripled, $m$ is tripled. 2. If $a$, $b$, and $c$ are all tripled, then $m$ is tripled. 3. If $a$ is tripled, $n$ is tripled. 4. If $a$, $b$, and $c$ are all tripled, then $n$ is tripled. 2. Create a true statement of your own about one of the equations. ### Student Response Details about student responses to this activity are available at IM Certified Partner LearnZillion (requires a district subscription) or without a subscription here (requires registration). ### Activity Synthesis Ask previously identified students to share their reasoning about which statements are true (or not true). Display any examples (or counterexamples) for all to see and have students refer to them while sharing. If using the algebraic structure is not brought up in students’ explanations, display for all to see: • If $a$, $b$, and $c$ are all tripled, the expression becomes $3a +3b+3c$, which can be written as $3(a+b+c)$ by using the distributive property to factor out the 3. So if all the addends are tripled, their sum, $m$, is also tripled. • Looking at the third statement, if $a$ is tripled, the expression becomes $(3a)bc$, which, by using the associative property,", "Summary Combining like terms allows us to write expressions more simply with fewer terms. But it can sometimes be tricky with long expressions, parentheses, and negatives. It is helpful to think about some common errors that we can be aware of and try to avoid: • $$6x-x$$ is not equivalent to 6. While it might be tempting to think that subtracting $$x$$ makes the $$x$$ disappear, the expression is really saying take 1 $$x$$ away from 6 $$x$$'s, and the distributive property tells us that $$6x-x$$ is equivalent to $$(6-1)x$$ • $$7-2x$$ is not equivalent to $$5x$$. The expression $$7-2x$$ tells us to double an unknown amount and subtract it from 7. This is not always the same as taking 5 copies of the unknown. • $$7-4(x+2)$$ is not equivalent to $$3(x+2)$$. The expression tells us to subtract 4 copies of an amount from 7, not to take $$(7-4)$$ copies of the amount. If we think about the meaning and properties of operations when we take steps to rewrite expressions, we can be sure we are getting equivalent expressions and are not changing their value in the process. ### Glossary Entries • expand To expand an expression, we use the distributive property to rewrite a product as a sum. The new expression is equivalent to the original expression.", "b+1$ and $a$ is a multiple of $b$ iff $a$ belongs to every set that contains $a$ and is closed under addition by $a$. – hot_queen Jun 16 '16 at 4:39 Addition is automatic. If $x$, $y$ and $z$ are three words of numerals of the same length over base $[b] = \\{0,1,\\dots,b-1\\}$, then the predicate $x+y=z$, seen as a language of words over the triple alphabet $[b]^3$, is recognized by a finite automaton. Hence it is definable in $MSO[<]$ using this particular kind of interpretation, and so is all of Presburger arithmetic $FO[+]$. A 1994 paper by Bruy`ere, Hansel, Michaux and Villemaire studies this question fairly comprehensively.", "# Data Sufficiency: Yes/No Basic Technique If $$t$$ is a positive integer, is $$t$$ a multiple of $$3$$? >(1) $$3$$ is a factor of $$\\frac{t}{4}$$. >(2) $$4t+3$$ is a multiple of $$3$$. Incorrect. [[snippet]] In this question, either one of the statements is __Sufficient__. Correct. [[snippet]] Stat. (1): If 3 is a factor of $$\\frac{t}{4}$$, then $$\\frac{t}{4}$$ is a multiple of 3, so >$$\\frac{t}{4} = 3 \\times \\text{Some integer}$$. Multiply this equation by 4 to get >$$t = 12 \\times \\text{Some integer}$$. Thus, It follows that $$t$$ is a multiple of 12 and is therefore also a multiple of 3. This is a definite \"Yes,\" so **Stat.(1) → S → AD**. Stat. (2): First isolate the equation: >$$4t+3 = \\text{Multiple of 3}$$ >$$4t = (\\text{Multiple of 3}) - 3$$. Now recall the rules of adding and subtracting multiples: >$$\\text{Multiple of } n \\pm \\text{Multiple of } n = \\text{Multiple of } n$$. Since $$4t$$ is the difference between multiples of 3, it _must_ be a multiple of 3. Since 4 is not a multiple of 3, $$t$$ itself must be a multiple of 3 for the product to be divisible by 3. This is a definite \"Yes,\" so **Stat.(2) → S → D**. Incorrect. [[snippet]] In this question, either one of the statements is __Sufficient__. Incorrect. [[snippet]] Yes,", "sense. – Hagen von Eitzen Mar 8 '13 at 16:33 • @HagenvonEitzen: Right. – random maths student Mar 8 '13 at 16:40 Your main problem, I think, is not understanding exactly what is meant by the statement that $3$ is a common divisor of $6,9$, and $15$. This simply means that $3$ is a divisor of $6$, $3$ is a divisor of $9$, and $3$ is a divisor of $15$. And when you paraphrase it that way, its structure becomes very evident: it’s just $S\\land N\\land F$, where $S,N$, and $F$ are the statements ‘$3$ is a divisor of $6$’, ‘$3$ is a divisor of $9$’, and ‘$3$ is a divisor of $15$’, respectively. You could go a little further and notice that these three statements themselves all have the same basic structure. If $T(n)$ is the ‘open’ statement ‘$3$ is a divisor of $n$’, then your original sentence has the logical form $T(6)\\land T(9)\\land T(15)$. $a \\mod b = c$ means exist $p\\in \\mathbb{N}$ such that $a = p*b +c$. So $6 \\mod 3 =0$ exactly means $3$ is a divisor of $6$. For your answer, your $T$ as no meaning \"3 is a common divisor\" of what? But $S,N,F$ or ok. Remark that $S\\equiv 6 \\mod 3 =0$, $N\\equiv 9 \\mod 3 =0$ and $F\\equiv 15", "statement 1 is NOT SUFFICIENT Statement 2: n³ + 2n² + n is a multiple of 3 More FACTORING n³ + 2n² + n = n(n² + 2n + 1) = n(n + 1)(n + 1) So, statement 2 tells us that n(n + 1)(n + 1) is a multiple of 3 This means that EITHER n is a multiple of 3 OR (n + 1) is a multiple of 3 Let's examine both cases: Case a: n is a multiple of 3 If n is a multiple of 3, then n - 1 CANNOT be a multiple of 3. Why not? Well, we already know that 1 out of every 3 consecutive integers is a multiple of 3 So, if n is a multiple of 3, the n+3 is also a multiple of 3 AND n+6 is a multiple of 3, AND n+9 is a multiple of 3, etc Likewise, n-3 is a multiple of 3 AND n-6 is a multiple of 3, AND n-9 is a multiple of 3, etc. Notice that n-1 is NOT among the possible multiples of 3 So, in this case, the answer to the target question is NO, n - 1 is NOT a multiple of 3 Case b: (n + 1) is a multiple of 3 If n+1 is a" ]

[ "are very different. You are correct that the first is false and the second true, but there is nothing paradoxical about this. The first statement says that for every $a$, if $a$ is a multiple of $2$, then $a$ is a multiple of $8$. I know you don't see where it says, \"for every $a$\" but this is a common convention, that you are actually quite used to. What about $\\sin^2x+\\cos^2=1.$ This doesn't say, \"for every $x$\", but that's what it means. Let's take the true statement, \"if $x$ is a multiple of $8$ then $x$ is a multiple of $2$. You'll agree that's true for every $x$, I think. Wait a sec! Is it true when $x=9?$ Is it true when $x=4?$ Yes and yes, if we adopt the convention in the table.", "statement says that niether of them is a factor of $$4$$. Remember our original conditions. either $$y$$ is a multiple of $$4$$ or $$(y-1)$$ is a multiple of $$4$$. This statement tells us that $$y$$ is not a multiple of $$4$$ but it does not tell us whether $$(y-1)$$ is a multiple of $$4$$ or not. Also note that this again does not tell us whether $$y$$ is even or $$(y-1)$$ is even.So Insufficient. Statement B: $$y^3-y$$ is divisible by $$4$$ Lets factor this one out : $$y^3-y=y*(y-1)*(y+1)$$ and again these are $$3$$ consecutive integers. The only thing we have found out (from the statement) is that one of them is definitely a multiple of $$4$$ but we do not know which one it is. Also note that we are not sure that one of them is even or two of them are even. And if it is one even, it could be $$2$$ so it is still not sufficient. We need at-least $$2$$ multiples of $$2$$ in there. If there were $$2$$ even numbers, we would be sure that it is divisible by $$4$$. So Insufficient Combined: We know that $$y$$ and $$(y+1)$$ are not multiples of $$4$$ from the first statement so from our second statement combined we have found out that $$(y-1)$$ is then, logically", "positive integers. Consider these two equations: $$m=a+b+c$$ $$n=abc$$ 1. Which of these statements are true? Select all that apply. 1. If $a$ is tripled, $m$ is tripled. 2. If $a$, $b$, and $c$ are all tripled, then $m$ is tripled. 3. If $a$ is tripled, $n$ is tripled. 4. If $a$, $b$, and $c$ are all tripled, then $n$ is tripled. 2. Create a true statement of your own about one of the equations. ### Student Response Details about student responses to this activity are available at IM Certified Partner LearnZillion (requires a district subscription) or without a subscription here (requires registration). ### Activity Synthesis Ask previously identified students to share their reasoning about which statements are true (or not true). Display any examples (or counterexamples) for all to see and have students refer to them while sharing. If using the algebraic structure is not brought up in students’ explanations, display for all to see: • If $a$, $b$, and $c$ are all tripled, the expression becomes $3a +3b+3c$, which can be written as $3(a+b+c)$ by using the distributive property to factor out the 3. So if all the addends are tripled, their sum, $m$, is also tripled. • Looking at the third statement, if $a$ is tripled, the expression becomes $(3a)bc$, which, by using the associative property,", "A = (B-4)(B)(B+4) B= 4, A =0 ----> No, A is not a multiple of 6 B= 6, A =(2)(6)(10) -----> Yes, A is a multiple of 6 Statement 2 - B is a multiple of 3 - B has no relationship with A. Combined Statement 1 + Statement 2 B=6, As seen above \"A\" is a multiple of 6 B=9, A =(5)(9)(13) ----> No, A is not a multiple of 6 _________________ I will rather do nothing than be busy doing nothing - Zen saying Math Expert Joined: 02 Sep 2009 Posts: 49438 Re: Is A multiple of 6? [#permalink] ### Show Tags 17 Sep 2012, 08:53 2 2 Is A multiple of 6? (1) A = B(B+4)(B-4). If $$B=odd$$, then $$A=B(B+4)(B-4)=odd*odd*odd=odd$$ and the answer is NO but if $$B=6$$, then the answer is obviously YES. Not sufficient. (2) B-4 is a multiple of 3. B is 4 more than a multiple of 3. Not sufficient, since no info about A. (1)+(2) From (2) we have that B is 4 more than a multiple of 3 and from (1) that $$A=B(B+4)(B-4)$$. If $$B=odd$$ (for example if $$B=7$$), then the answer is NO but if $$B=4$$, then $$A=B(B+4)(B-4)=0$$ and the answer is YES (since zero is a multiple of all integers except zero itself). Not sufficient. Hope it", "x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following? I. 8 II. 12 III 18 A. II only B. I and II only C. I and III only D. II and III only E. I, II and III First of all notice that we are asked $$xy$$ MUST be a multiple of which of the following, not COULD be a multiple. $$x$$ is a multiple of 4 --> $$x=4m$$, for some positive multiple $$m$$, so $$x$$ can be: 4, 8, 12, ... $$y$$ is a multiple of 6 --> $$y=6n$$, for some positive multiple $$n$$, so $$y$$ can be: 6, 12, 18, ... So, $$xy=(4m)*(6n)=24mn$$, so $$xy$$ is in any case a multiple of 24, which means it must be a multiple of 8 and 12, but not necessarily of 18. For example the least value of $$x$$ is 4 and the least value of $$y$$ is 6, so the least value of $$xy$$ is 24, which is a multiple of both 8 and 12, but not 18. Hope it's clear. _________________ ##### General Discussion Director Joined: 28 Dec 2005 Posts: 891 ### Show Tags 02 Dec 2006, 06:06 1 2 nah, not necessarily: Ans B. From question x/4 is", "• # question_answer If $A=[x:x$ is a multiple of 3] and $B=[x:x$ is a multiple of 5], then A - B is ($\\bar{A}$ means complement of A) [AMU 1998] A) $\\bar{A}\\cap B$ B) $A\\cap \\bar{B}$ C) $\\bar{A}\\cap \\bar{B}$ D) $\\overline{A\\cap B}$ A - B = A Ç Bc = A Ç $\\bar{B}$.", "in the form $$3y\\pm 1$$where $y$ is a positive integer. Therefore $$(3y \\pm 1)^2 = 3n - 9x^2 \\pm 6x - 1$$ and so $$9y^2 \\pm 6y + 1 = 3n - 9x^2\\pm 6x - 1$$ $$9y^2\\pm 6y + 2 = 3n -9x^2 \\pm 6x$$ The right hand side of the expression is a multiple of three, but the left hand side however is clearly not, as it is a multiple of three add $2$. Here is a contradiction. When $b$ was said to be a non multiple of three, it led to saying that a must also be a non multiple, and a contradiction occurred. Therefore neither $a$ nor $b$ can be non multiples of three, hence if $a^2 + b^2$ is a multiple of three then $a$ and $b$ are both multiples of three.", "# Data Sufficiency: Yes/No Basic Technique If $$t$$ is a positive integer, is $$t$$ a multiple of $$3$$? >(1) $$3$$ is a factor of $$\\frac{t}{4}$$. >(2) $$4t+3$$ is a multiple of $$3$$. Incorrect. [[snippet]] In this question, either one of the statements is __Sufficient__. Correct. [[snippet]] Stat. (1): If 3 is a factor of $$\\frac{t}{4}$$, then $$\\frac{t}{4}$$ is a multiple of 3, so >$$\\frac{t}{4} = 3 \\times \\text{Some integer}$$. Multiply this equation by 4 to get >$$t = 12 \\times \\text{Some integer}$$. Thus, It follows that $$t$$ is a multiple of 12 and is therefore also a multiple of 3. This is a definite \"Yes,\" so **Stat.(1) → S → AD**. Stat. (2): First isolate the equation: >$$4t+3 = \\text{Multiple of 3}$$ >$$4t = (\\text{Multiple of 3}) - 3$$. Now recall the rules of adding and subtracting multiples: >$$\\text{Multiple of } n \\pm \\text{Multiple of } n = \\text{Multiple of } n$$. Since $$4t$$ is the difference between multiples of 3, it _must_ be a multiple of 3. Since 4 is not a multiple of 3, $$t$$ itself must be a multiple of 3 for the product to be divisible by 3. This is a definite \"Yes,\" so **Stat.(2) → S → D**. Incorrect. [[snippet]] In this question, either one of the statements is __Sufficient__. Incorrect. [[snippet]] Yes,", "but not a multiple of $4$. It follows that $b$ cannot be an odd number, so it must be a multiple of $2$, say $b=2c.$ Then $b^2 = 4c^2\\in(4)$. The same argument applies for any prime power $p^m$. If $ab\\in(p^m)$ but $a\\notin(p^m)$ then $b$ must be a multiple of $p$ and so $b^m\\in(p^m).$ To show that $(6)$, for example, is not primary, use the fact that $6=2\\times3$. If you take $a=2$ and $b=3$ in the definition of a primary ideal, you see that $2$ is not a multiple of $6$, so the definition asks that some power of $3$ should be a multiple of $6$. But that cannot happen, because every power of $3$ is an odd number. Therefore the definition of primary is not satisfied and so $(6)$ is not a primary ideal of $\\mathbb Z.$ Peter Well-known member MHB Site Helper To show that $(4)$ is primary, we have to show that if [TEX] ab \\in (4) [/TEX] and [TEX] a \\notin (4) [/TEX] then [TEX] b^n \\in (4) [/TEX] for some positive integer $n.$ So suppose that $ab\\in(4)$, in other words $ab$ is a multiple of $4$, and that $a\\notin(4)$. Thus $a$ could be an odd number, or an odd multiple of $2$, but not a multiple of $4$. It follows that $b$ cannot be", "1 tells us that the LCM of (a,b) is 42. Therefore, we have that b=7,a=6 remainder is 1, but if b=14 and a=3 remainder is 2, therefore two different answers–> Insufficient. Statement 2, says that the GCF of (a,b) is 2. Therefore since both are even and b is not a multiple of a the remainder will always be 2, therefore Sufficient Cheers J e-GMAT Representative Joined: 04 Jan 2015 Posts: 1897 Re: We are given two integers a and b, such that, 2 < a < b and b is not a [#permalink] ### Show Tags 21 Mar 2018, 01:19 Solution: We are given: • ‘$$b$$’ is not a multiple of ‘$$a$$’. • Both, ‘$$a$$’ and ‘$$b$$’, are greater than $$2$$ and ‘$$b$$’ is greater than ‘$$a$$’. Statement-1 is “The least common multiple of ‘a’ and ‘b’ is 42”. $$42= 2*3*7$$ ‘$$a$$’ and ‘$$b$$’ can have multiple values based on prime factorization of $$42$$. Since the value of remainder may or may not greater than 1. Hence, Statement 1 alone is not sufficient to answer the question. Statement-2 is “The greatest common factor of ‘a’ and ‘b’ is 2.” Thus, • $$a=2x$$ and $$b=2y$$, where $$x$$ and $$y$$ are co-prime numbers. Since ‘$$b$$’ is greater than ‘$$a$$’, ‘$$b$$’ on dividing by ‘$$a$$’ can be written as: •", "even numbers. The reason for this pattern is that a product $a \\times b$ is even when at least one of its factors, $a$ or $b$, is even. 2. The multiples of three are shaded blue in the table below while the other numbers in the table are left white: Every third row is filled with multiples of three and similarly every third column is filled with multiples of three. The reason for this pattern is that a product $a \\times b$ is divisible by three when at least one of $a$ or $b$ is divisible by three. 3. The multiples of four are shaded blue in the table below while the other numbers in the table are left white: There is a clear difference between the pattern of multiples of $4$ and those of $2$ and $3$ seen above in parts (a) and (b). As in parts (a) and (b), every fourth column and every fourth row is shaded as these are multiples of $4$. There are, however, additional isolated shaded boxes. One of these is $2 \\times 2 = 4$. The reason why these isolated boxes occur for multiples of $4$ while they did not occur for multiples of $2$ or $3$ is that $4$ can be written as a product of smaller positive whole numbers", "proof is not correct. The statement that no element of $$B$$ can be used in more than one linear combination giving a vector of $$G$$ is not true. Suppose $$B$$ and $$G$$ are two different bases of $$V$$. If we accept your statement, we must conclude each element of $$G$$ is a scalar multiple of some element of $$B$$, which is false. • @s If $B$ and $G$ are both bases, they have the same number of elements. If you can use each element of $B$ only once, they you can't use more than one. So each element of $G$ is a scalar multiple of an element of $B$. (I said the bases are the same; that was a mistake, which I'll correct.) – saulspatz Oct 12 '19 at 21:31 • Ignore that last comment; what I meant to ask was: if the statement that \"no element of $B$ can be used in more than one linear combination giving a vector of $G$\" is false, then what is the true condition? I don't know. – sandbag66 Oct 12 '19 at 22:45", "Now, $$x!$$ is >$$1\\times 2\\times 3\\times$$ … $$(x-1)\\times x$$, so $$(x-1)$$ and 2 are included in $$x!$$ (that is, provided that $$x$$ is greater than 2). Hence, this is the correct answer. Incorrect. [[snippet]] Yes, statements II and III are true. But what about statement I? Can you find an example eliminating this statement? Incorrect. [[snippet]] Yes, statement I and III are true. But what about statement II? Can you find an example eliminating this statement? Incorrect. [[snippet]] Yes, statement I is true. But what about statements II and III? Can you find an example eliminating these statements? Incorrect. [[snippet]] Yes, statement II is true. But what about statements I and III? Can you find an example eliminating these statements? Correct. [[snippet]] __Plug In__ $$x = 3$$. In this case, the product could be $$3 \\times 4 \\times 5 = 60$$. >I. 60 is divisible by $$x - 1 = 3 - 1 = 2$$. >II. 60 is divisible by $$2x = 2 (3) = 6$$. >III. 60 is divisible by $$x! = 3! = 3 \\times 2 \\times 1 = 6$$. __Plug In__ another value of $$x$$ to be sure because the question says *must be*. This time try an even integer (such as $$x= 4$$). In this case, the product could be $$2 \\times 3 \\times", "for $$8^a$$ end with $$6$$, $$a$$ should be of the form :$$4k$$. From this, we know the nature of the number $$a$$, but we do not have any information about$$X$$. Since we do not know the value of $$X$$, Statement 2 alone is NOT sufficient to answer the question. Hence, we can eliminate answer choice B. Step 3: Combine both Statements: • From the first statement, we know: The units digit of $$X$$ is $$2$$ • From the second statement we have: a is a multiple of $$4$$ • We know that a is of form 4k, and we also know$$2^{4k}$$ always has a units digit: $$6$$. Thus, we know that the units digit of the expression $$X^a$$ is $$6$$. By combining both statements we got a unique answer. _________________ SVP Status: It's near - I can see. Joined: 13 Apr 2013 Posts: 1687 Location: India GPA: 3.01 WE: Engineering (Real Estate) Re: If X is a positive integer, what is the [#permalink] Show Tags 29 Mar 2018, 02:48 EgmatQuantExpert wrote: e-GMAT Question: If $$X$$ is a positive integer, what is the units digit of $$X^a$$? 1) $$X$$ when divided by $$10$$ gives a remainder of $$2$$ 2) $$8^a$$ ends with $$6$$ A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B) Statement (2)", "statement 1 is NOT SUFFICIENT Statement 2: n³ + 2n² + n is a multiple of 3 More FACTORING n³ + 2n² + n = n(n² + 2n + 1) = n(n + 1)(n + 1) So, statement 2 tells us that n(n + 1)(n + 1) is a multiple of 3 This means that EITHER n is a multiple of 3 OR (n + 1) is a multiple of 3 Let's examine both cases: Case a: n is a multiple of 3 If n is a multiple of 3, then n - 1 CANNOT be a multiple of 3. Why not? Well, we already know that 1 out of every 3 consecutive integers is a multiple of 3 So, if n is a multiple of 3, the n+3 is also a multiple of 3 AND n+6 is a multiple of 3, AND n+9 is a multiple of 3, etc Likewise, n-3 is a multiple of 3 AND n-6 is a multiple of 3, AND n-9 is a multiple of 3, etc. Notice that n-1 is NOT among the possible multiples of 3 So, in this case, the answer to the target question is NO, n - 1 is NOT a multiple of 3 Case b: (n + 1) is a multiple of 3 If n+1 is a", "is only true if it always is true. It is false if it is false in even one case. For this reason, the statement “all prime numbers are odd” is false. There is only one even prime number, 2, but that is enough to make the entire statement false. To prove a statement is false, it is necessary to find just one counterexample. In this case, it is required to find a three numbers $a, b,$ and $c$ such that $a=c$ but $a\\neq b$ and $c\\neq b$. One possible counter example is if $a=1$, $b=0$, and $c=1$. In this case, the transitive property of equality states that since $a=1$ and $c=1$, $a=c$. But, $a\\neq b$ and $c\\neq b$. Therefore, the inverse of the transitive property of equality is not true. ### Example 5 Let $w, x, y$ and $z$ be real numbers such that: $3y-2w+2z=7z+2y$ and $-4x+4w-3z=2z+6w-5x$ Use the transitive property to show that $x=y$. ### Solution This problem requires first solving for $x$ and $y$ using the addition and subtraction properties of equality. If $3y-2w+2z=7z+2y$, the subtraction property of equality states it is possible to subtract $2y$ from both sides. $3y-2y-2w+2z=7z+2y-2y$ This simplifies to: $y-2w+2z=7z$ Then, add $2w-2z$ to both sides. The addition property of equality says it is possible to do this and maintain the equality.", "for the following numbers (remember $0$!): a) $3$a s v d b) $5$a s v d c) $10$a s v d d) $9$a s v d e) $7$a s v d f) $12$a s v d ## Question 6 List the first $10$ multiples for the following numbers: a) $6$a s v d b) $11$a s v d c) $20$a s v d d) $25$a s v d e) $1000$a s v d f) $17$a s v d ## Question 7 Are the following statements true or false? a) $2$ is a factor of $18$a s v d b) $3$ is a factor of $19$a s v d c) $5$ is a factor of $75$a s v d d) $14$ is a factor of $140$a s v d e) $4$ is a factor of $86$a s v d f) $11$ is a factor of $143$a s v d ## Question 8 Are the following statements true or false? a) $12$ is a multiple of $4$a s v d b) $29$ is a multiple of $3$a s v d c) $0$ is a multiple of $104$a s v d d) $36$ is a multiple of $9$a s v d e) $45$ is a multiple of $15$a s v d f) $48$ is a multiple of $13$a s v d", "b+1$ and $a$ is a multiple of $b$ iff $a$ belongs to every set that contains $a$ and is closed under addition by $a$. – hot_queen Jun 16 '16 at 4:39 Addition is automatic. If $x$, $y$ and $z$ are three words of numerals of the same length over base $[b] = \\{0,1,\\dots,b-1\\}$, then the predicate $x+y=z$, seen as a language of words over the triple alphabet $[b]^3$, is recognized by a finite automaton. Hence it is definable in $MSO[<]$ using this particular kind of interpretation, and so is all of Presburger arithmetic $FO[+]$. A 1994 paper by Bruy`ere, Hansel, Michaux and Villemaire studies this question fairly comprehensively.", "is a power of two $\\le X$ you can subtract that results in a number without the property. It can be seen that the following property satisfies both requirements: $X$ is not a multiple of $3$ Obviously $X = 0$ doesn't have this property (because it is a multiple of $3$). Now, if $X$ is a multiple of $3$, then subtracting a power of two from it results in a number that is not a multiple by $3$, because no power of two is a multiple of $3$. But if $X$ is not a multiple of $3$, then there are two cases: • If $X \\bmod 3 = 1$, then we can subtract $2^0 = 1$ from $X$. • If $X \\bmod 3 = 2$, then we can subtract $2^1 = 2$ from $X$. In both cases, the resulting number is a multiple of $3$. Thus, the property satisfies all requirements! So the solution is simply: If $X$ is divisible by $3$, then the second player wins. Otherwise, the first player wins. Finally, checking whether $X$ is a multiple of $3$ can be done with the Horner method modulo $3$: (here we write X[i] as the $i$th bit of $X$ from the left): def is_multiple_of_3(X): r = 0 for i in 1..n: r = (2*r + X[i]) %", "question, but we still need to figure out if the second sub-statement answers the question with the same answer as the first sub-statement. the second sub-statement $$x-1$$ is a multiple of $$y$$ implies that $$x-1 = q \\cdot y$$ which can be rewritten as $$x = q \\cdot y + 1$$ which is an equation of the form $$x = q \\cdot y + r$$ when $$r=1$$ which implies that when $$x$$ is divided by $$y$$ we get a remainder of 1, meaning that $$x$$ is NOT a multiple of $$y$$. Since both sub-statements are contradictory, this means that when $$x^2 - x$$ is a multiple of $$y$$, $$x$$ is not always a multiple of $$y$$. meaning that we cannot determine whether $$x$$ is always a multiple of $$y$$ according to Statement 2 alone. Therefore, Statement 2 is not sufficient. (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Manager Joined: 15 Aug 2013 Posts: 239 Re: If x and y are integers greater than 1, is x a multiple of y? [#permalink] ### Show Tags 18 Sep 2014, 18:36 Bunuel wrote: If x and y are integers great than 1, is x a multiple of y? (1) $$3y^2+7y=x$$ --> $$y(3y+7)=x$$ --> as $$3y+7=integer$$, then $$y*integer=x$$ --> $$x$$ is a multiple of $$y$$. Sufficient." ]

[ "with. For a polygon, the diameter, what is the radius from 1 to 25 sure they fit diameter. Π. π is roughly equal to 14 cm Circles in this lesson, you can work the! Perimeter or circumference of Circles which has radius 2 cm, 10 m, circumference and area of circles mm and 78.!", "Whose Diagonals is 24 Cm. Concept: Area of a Polygon. S", "48 cm and the perimeter is 38 cm then what does the shape look like?", "# A strange side length Geometry Level 4 A regular polygon inscribed in a unit circle has a side length of $\\large \\sqrt{2-\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{3}}}}} .$ Find the number of sides of the polygon. ×", "what is the length The perimeter of quadrilateral is 58 centimeters. If the lengths of three sides are 14 cm, 15 cm, and 17 cm, what is the length of the missing side?...", "the above information, answer the following questions. Q1) The perimeter of the given protractor (in cm) is a) 22 b) 44 c) 36 d) 58 Q2) The area of the protractor given in the question in sq. cm is a) 77 b) 154 c) 308 d) 616 Q3) If the diameter of the protractor is increased by 2.1 cm, then the perimeter will increased by a) 3.5 cm b) 6.6 cm c) 5.4 cm d) 7.3 cm Q4) If the original protractor is divided into six equal sectors, what will be the area of each sector (in sq. cm) ? a) 77/6 b) 77/3 c) 154/3 d) 308/3 Q5) The perimeter of each of the above sectors (in cm) will be a) 53/3 b) 64/3 c) 95/3 d) 106/3 Question Answer 1 c 2 a 3 c 4 a 5 a AREA RELATED TO CIRCLES A regular polygon of side 14 cm is drawn. Further circles of equal radii are constructed inside the polygon as shown below. Based on the above information, answer the following questions. Q1) What is the angle at each vertex of the polygon ? a) 108 b) 120 c) 135 d) 150 Q2) What fraction of a circle is drawn at each vertex ? a) 3/10 b) 1/3 c) 3/8 d) 5/12 Q3) What", "10 Cm, Am = 20 Cm, An = 50 Cm, Ao = 60 Cm And Ad = 90 Cm. Concept: Area of a Polygon. S", "Rectangle A rectangle has four sides, whose two sides are equal. Then, its perimeter is; P = x + x + y +y = 2 (x+y) ### Square It has four equal sides. In fact, it is one of the regular polygons whose n = 4. Thus, P = n\\cdot S = 4\\cdot S", "• # question_answer 8) Find the perimeter of a regular hexagon with each side measuring 8 m. A regular hexagon has 6 sides and each side is of 8 cm. $\\therefore$ Perimeter of a regular hexagon $=6\\times$Length of one side $=6\\times 8m=48m$ Hence, the perimeter of a regular hexagon is 48 cm.", "# arithmetic sequences • May 24th 2009, 04:52 AM Tweety arithmetic sequences A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675 cm and the length of the longest side is twice that of the shortest side. Find, for this series: (a) The common difference. (b) The first term. • May 24th 2009, 05:55 AM Jester Quote: Originally Posted by Tweety A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675 cm and the length of the longest side is twice that of the shortest side. Find, for this series: (a) The common difference. (b) The first term. Let $a$ be the shortest side and $d$ the difference so the sides are $a, a+d, a+2d,\\; \\cdots \\; a + 9d$ Adding gives $10a + 45d = 675\\, (*)$, the condition that the longest is twice the shortes gives $2a = a + 9d$ or $a = 9d$. Use this in (*).", "is 4 Cm. If One of Its Diagonals is 8 Cm Long, Find the Length of the Other Diagonal. Concept: Area of a Polygon. S", "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? Concept: Area of a Polygon. S", "# Polygon MCQ Quiz - Objective Question with Answer for Polygon - Download Free PDF Last updated on Mar 15, 2023 Exams like SSC CGL, GD, MTS etc commonly feature Mensuration in their syllabus. One of the most important parts of Mensuration is Polygon. Candidates must prepare Polygon MCQs Quiz effectively as they can be quite confusing at times. Testbook has curated these questions for candidates to prepare Polygon objective questions well. This article also lists down a few tips and shortcuts to solve Polynomial questions quickly with accuracy. Check out these Polynomial question answers and practice these problems with Testbook now! ## Latest Polygon MCQ Objective Questions #### Polygon Question 1: If the perimeter of a regular septagon is 490 cm, then the length of one side is: 1. 70cm 2. 30cm 3. 50cm 4. 35cm Option 1 : 70cm #### Polygon Question 1 Detailed Solution Given: The perimeter of a regular septagon = 490 cm Concept used: The perimeter of a regular septagon = 7 × side Calculation: According to the question, ⇒ 7 × side = 490 cm ⇒ side = $$\\dfrac{490}{7}$$ = 70 cm ∴ The length of one side is 70 cm. #### Polygon Question 2: If the perimeter of a regular pentagon is 320 cm, then the length of one side is:", "many diagonals does a 63-sided convex polygon have? [#permalink] 08 Aug 2017, 13:58 Display posts from previous: Sort by", "length of side when the perimeter of the hexagon is 120 cm. This implies that; P = Perimeter of the hexagon = 120 cm a = P / 6 a = 120 / 6 a = 20 Therefore, the length of side (a) is 20 cm.", "interior angle of a regular octagon is 135°. 4 ### Find the perimeter of an irregular octagon whose side lengths are 3 cm, 7 cm, 8 cm, 5 cm, 4 cm, 6 cm, 10 cm, and 9 cm. 50 cm 72 cm 52 cm 34 cm CorrectIncorrect Perimeter $= (3 + 7 + 8 + 5 + 4 + 6 + 10 + 9)$ cm Perimeter $= 52$ cm.", "have perimeter of 12 square sides, or 24 cm. • I'm sorry for the late reply, but I'm accepting your answer. It is very nice for you to add colors that made me understand it. Great! – Chris Steinbeck Bell Mar 21 at 2:26", "# Perimeter of regular hexagon Geometry Level 1 A regular hexagon has side length $$8$$ cm. What is the perimeter (in cm) of the hexagon? ×", "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", "but with side-lengths that have variables and numbers it can be tricky—you need to know how to identify like-terms and how to sum them correctly. Therefore, the perimeter of the polygon is 13x-34 units. But getting an answer such as 13x-34 is not necessarily what students have in mind as an answer. Often, they expect and, perhaps, want a number. Continue reading “Algebraic Expressions vs Algebraic Equations”" ]

[ "# Perimeter Geometry Level 1 In the polygon to the right, all adjacent edges are at right angles so that all its interior angles are either $90^\\circ$ or $270^\\circ.$ The four longer edges are equal in length, and all of the shorter edges are also equal in length. The area of the shape is 528. What is the perimeter? ×", "## Thinking Mathematically (6th Edition) perimeter = $44 \\text{ m}$ The height of the polygon is 10 meters. This means that the unknown vertical side length is: $=10-2 = 8$ m The width of the polygon is 12 meters. This means that the unknown horizontal side length is: $=12 - 7 = 5$ m The perimeter of a polygon is equal to the sum of all the lengths of its sides. Thus, the perimeter of the polygon is: $=(12 + 10 + 7 + 8 + 5 + 2) \\text{ m} \\\\=44 \\text{ m}$", "# 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", "=24 / 12 =2 units. Nov 14, 2018 #3 +46 +1 The inradius of a polygon can be found by $$A=rs$$, where $$A$$ is the area, $$r$$ is the inradius, and $$s$$ is the semi-perimeter. We know that the area is $$\\frac{1}{2}*6*8=24$$, based on the information given, and $$\\frac{6+8+10}{2}=12$$ which is considered the semi-perimeter. Now, we have to solve for the radius, and plugging our values in, we reach $$24=12r, 12r=24, \\boxed{r=2}$$ Nov 14, 2018", "If a polygon has its sum of interior angles smaller than $4190^0$, what is the maximum number of sides of the polygon? A. 28 B. 22 C. 25 D. 27", "# Polygon MCQ Quiz - Objective Question with Answer for Polygon - Download Free PDF Last updated on Mar 15, 2023 Exams like SSC CGL, GD, MTS etc commonly feature Mensuration in their syllabus. One of the most important parts of Mensuration is Polygon. Candidates must prepare Polygon MCQs Quiz effectively as they can be quite confusing at times. Testbook has curated these questions for candidates to prepare Polygon objective questions well. This article also lists down a few tips and shortcuts to solve Polynomial questions quickly with accuracy. Check out these Polynomial question answers and practice these problems with Testbook now! ## Latest Polygon MCQ Objective Questions #### Polygon Question 1: If the perimeter of a regular septagon is 490 cm, then the length of one side is: 1. 70cm 2. 30cm 3. 50cm 4. 35cm Option 1 : 70cm #### Polygon Question 1 Detailed Solution Given: The perimeter of a regular septagon = 490 cm Concept used: The perimeter of a regular septagon = 7 × side Calculation: According to the question, ⇒ 7 × side = 490 cm ⇒ side = $$\\dfrac{490}{7}$$ = 70 cm ∴ The length of one side is 70 cm. #### Polygon Question 2: If the perimeter of a regular pentagon is 320 cm, then the length of one side is:", "# Geometry Geometry Level 1 If a polygon has $27$ diagonals, then the number of sides of the polygon is ×", "## Elementary Geometry for College Students (6th Edition) As the number of sides gets larger, the polygon becomes more like a circle. Thus, we treat it like a circle with radius 7: $A = \\pi r^2 \\\\ A = \\pi (7^2) \\approx 154$", "## Thinking Mathematically (6th Edition) RECALL: The perimeter of a polygon is equal to the sum of the lengths of all its sides. Add the lengths of all the sides to find the perimeter: perimeter = $3+12+3+12= 30$ cm", "= 279.90 cm2. Question 2: Find the area of a 12 sided polygon with a circumscribed circle radius of 4 cm. Solution: Area of the 12 sided polygon in a circumscribed circle of radius R is, Area = 3R2. Area = 3 * 4 = 3 * 16 = 48 cm2. ## Practice Problems on 12 Sided Polygon Given below are some of the practice problems on regular 12 sided polygon. ### Practice Problems Question 1: Calculate the area of the dodecagon with a side length of 8 cm. Question 2: Calculate the area of the dodecagon with a circumscribed circle radius of 6.5 cm. Related Topics Math Help Online Online Math Tutor *AP and SAT are registered trademarks of the College Board.", "radius $r$. To find $r$, we put, $πr² = 314$ cm$^²$ $\\Rightarrow r ≈ 10$ cm Now, vertices $1$ and $501$ of our $1000$ sided polygon will correspond to the opposite ends of the diameter of the circum-circle of this polygon. ∴ The distance between them will be approximately = $2 × r = 20$ cm Hence, option $3$. -", "# Perimeter of regular hexagon Geometry Level 1 A regular hexagon has side length $$8$$ cm. What is the perimeter (in cm) of the hexagon? ×", "# arithmetic sequences • May 24th 2009, 04:52 AM Tweety arithmetic sequences A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675 cm and the length of the longest side is twice that of the shortest side. Find, for this series: (a) The common difference. (b) The first term. • May 24th 2009, 05:55 AM Jester Quote: Originally Posted by Tweety A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675 cm and the length of the longest side is twice that of the shortest side. Find, for this series: (a) The common difference. (b) The first term. Let $a$ be the shortest side and $d$ the difference so the sides are $a, a+d, a+2d,\\; \\cdots \\; a + 9d$ Adding gives $10a + 45d = 675\\, (*)$, the condition that the longest is twice the shortes gives $2a = a + 9d$ or $a = 9d$. Use this in (*).", "## Thinking Mathematically (6th Edition) perimeter = $1000$ inches RECALL: (1) The perimeter of a polygon is equal to the sum of the lengths of all its sides. (2) A square has four equal sides. This means that the three unknown side lengths are all 250 inches. Add the lengths of all the sides to find the perimeter: perimeter = $250+250+250+250= 1000$ inches", "($-10^5 \\le x_i, y_i \\le 10^5$) — coordinates of polygon's vertices in counterclockwise order. All internal angles of the polygon are strictly less than $\\pi$. The length $l$ exceeds the perimeter of the polygon by at least $10^{-3}$. Output Output a single floating-point number — the area of the fiber shape $F(S, l)$. Your answer will be considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Examples Input 3 4 0 0 1 0 0 1 Output 3.012712585980357 Input 4 5 0 0 1 0 1 1 0 1 Output 5.682061989789656 Input 5 17 0 0 2 -1 3 0 4 3 -1 4 Output 37.719371276930820 Note The following pictures illustrate the example tests.", "## Prealgebra (7th Edition) The perimeter of the polygon is the sum of the lengths of the sides. $2+3+1+3+4=5+1+3+4=6+3+4=9+4=13$", "# Can you determine the sides? Geometry Level 1 Find the number of sides of the polygon if the sum of it's interior angles is equal to $$1260^\\circ$$. ×", "$V = \\dfrac{4}{3} \\pi r^3$ = 2720.7 ................... If the sphere is INSIDE the cube which has side lengths of 10, then the diameter of the cube will equal the side length of the cube. $V = \\dfrac{4}{3} \\pi r^3$ = 523.6", "Solution Solving each the equations, we find that either $$x = 108$$, $$x = 60$$, $$x = 45$$, or $$x = 72$$. The exterior angle of a regular polygon must be factor of 360, so this rules out $$x = 108$$. For the remaining three cases, $$60$$º is the exterior angle for regular hexagons, $$45$$º is for regular octagons, and $$72$$º is for regular pentagons. Only the pentagon has a prime number of sides. Therefore, the polygon has 5 sides.", "a 12-sided polygon, to a 24-sided polygon, to a 48-sided polygon, and ended up with a 96-sided polygon. This final estimate gave a range for π between 3.1408 and 3.1428, which is accurate to two places." ]

[ "# arithmetic sequences • May 24th 2009, 04:52 AM Tweety arithmetic sequences A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675 cm and the length of the longest side is twice that of the shortest side. Find, for this series: (a) The common difference. (b) The first term. • May 24th 2009, 05:55 AM Jester Quote: Originally Posted by Tweety A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675 cm and the length of the longest side is twice that of the shortest side. Find, for this series: (a) The common difference. (b) The first term. Let $a$ be the shortest side and $d$ the difference so the sides are $a, a+d, a+2d,\\; \\cdots \\; a + 9d$ Adding gives $10a + 45d = 675\\, (*)$, the condition that the longest is twice the shortes gives $2a = a + 9d$ or $a = 9d$. Use this in (*).", "# Perimeter of regular hexagon Geometry Level 1 A regular hexagon has side length $$8$$ cm. What is the perimeter (in cm) of the hexagon? ×", "# Polygon MCQ Quiz - Objective Question with Answer for Polygon - Download Free PDF Last updated on Mar 15, 2023 Exams like SSC CGL, GD, MTS etc commonly feature Mensuration in their syllabus. One of the most important parts of Mensuration is Polygon. Candidates must prepare Polygon MCQs Quiz effectively as they can be quite confusing at times. Testbook has curated these questions for candidates to prepare Polygon objective questions well. This article also lists down a few tips and shortcuts to solve Polynomial questions quickly with accuracy. Check out these Polynomial question answers and practice these problems with Testbook now! ## Latest Polygon MCQ Objective Questions #### Polygon Question 1: If the perimeter of a regular septagon is 490 cm, then the length of one side is: 1. 70cm 2. 30cm 3. 50cm 4. 35cm Option 1 : 70cm #### Polygon Question 1 Detailed Solution Given: The perimeter of a regular septagon = 490 cm Concept used: The perimeter of a regular septagon = 7 × side Calculation: According to the question, ⇒ 7 × side = 490 cm ⇒ side = $$\\dfrac{490}{7}$$ = 70 cm ∴ The length of one side is 70 cm. #### Polygon Question 2: If the perimeter of a regular pentagon is 320 cm, then the length of one side is:", "# A strange side length Geometry Level 4 A regular polygon inscribed in a unit circle has a side length of $\\large \\sqrt{2-\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{3}}}}} .$ Find the number of sides of the polygon. ×", "with. For a polygon, the diameter, what is the radius from 1 to 25 sure they fit diameter. Π. π is roughly equal to 14 cm Circles in this lesson, you can work the! Perimeter or circumference of Circles which has radius 2 cm, 10 m, circumference and area of circles mm and 78.!", "# Perimeter Geometry Level 1 In the polygon to the right, all adjacent edges are at right angles so that all its interior angles are either $90^\\circ$ or $270^\\circ.$ The four longer edges are equal in length, and all of the shorter edges are also equal in length. The area of the shape is 528. What is the perimeter? ×", "sides of len [#permalink] ### Show Tags 19 May 2017, 23:37 For any polygon, sum of its interior and exterior angle (at that particular vertex) = 180 If interior angle = 144, exterior angle = 180-144 = 36 For a regular polygon of n sides, each exterior angle = 360/n (sum of all exterior angles of a polygon is 360) So 360/n = 36 or n = 10. Perimeter of regular polygon = n* length of each side = 10*12 = 120 Hence C Director Joined: 04 Dec 2015 Posts: 739 Location: India Concentration: Technology, Strategy Schools: ISB '19, IIMA , IIMB, XLRI WE: Information Technology (Consulting) Re: What is the perimeter of a regular polygon with sides of len [#permalink] ### Show Tags 22 May 2017, 15:40 phlaym39 wrote: What is the perimeter of a regular polygon with sides of length 12 cm and internal angles measuring 144° each? A. 96 cm B. 108 cm C. 120 cm D. 132 cm E. 144 cm Let number of sides of the polygon be n. (Given its a regular polygon. Regular polygon has all sides and interior angles equal.) Internal Angle = $$\\frac{180 (n-2)}{n}$$ Given internal Angle is $$144^{\\circ}$$ Therefore; $$\\frac{180 (n-2)}{n}$$ = $$144^{\\circ}$$ 180 (n-2) = 144 x n 180n - 180 x 2 = 144n 180n -", "## Thinking Mathematically (6th Edition) perimeter = $44 \\text{ m}$ The height of the polygon is 10 meters. This means that the unknown vertical side length is: $=10-2 = 8$ m The width of the polygon is 12 meters. This means that the unknown horizontal side length is: $=12 - 7 = 5$ m The perimeter of a polygon is equal to the sum of all the lengths of its sides. Thus, the perimeter of the polygon is: $=(12 + 10 + 7 + 8 + 5 + 2) \\text{ m} \\\\=44 \\text{ m}$", "# Problem #125 125 Three 12 cm $\\times$12 cm squares are each cut into two pieces $A$ and $B$, as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon, as shown in the second figure, so as to fold into a polyhedron. What is the volume (in $\\mathrm{cm}^3$) of this polyhedron? 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.", "10 Cm, Am = 20 Cm, An = 50 Cm, Ao = 60 Cm And Ad = 90 Cm. Concept: Area of a Polygon. S", "OpenStudy (anonymous): The perimeter of a octagon is 124 inches. Find the length of each side 4 years ago OpenStudy (anonymous): We start with the definition of a regular octagon. Any regular polygon is equilateral. That is, it's sides are of equal length. Also by definition an octagon is an eight sided polygon. Thus a regular octagon has 8 sides of equal lengths. The perimeter of a polygon is the sum of the lengths of each side. This leads to the following equation: 8*(length of a side) = 124 Now we divide both sides by 8 to get length of a side = 124/8 = 15.5 inches 4 years ago OpenStudy (austinl): $8x = 124$ $x = \\frac{124}{8}$ 4 years ago OpenStudy (austinl): This is of course assuming that it is a Regular Octagon. 4 years ago", "### Home > INT2 > Chapter 11 > Lesson 11.2.1 > Problem11-54 11-54. What is the area of a regular polygon with $100$ sides and a perimeter of $100$ units? See the math notes box in lesson 7.4.1. $A \\approx 795.51\\text{ u}^2$", "## Thinking Mathematically (6th Edition) RECALL: The perimeter of a polygon is equal to the sum of the lengths of all its sides. Add the lengths of all the sides to find the perimeter: perimeter = $3+12+3+12= 30$ cm", "= 279.90 cm2. Question 2: Find the area of a 12 sided polygon with a circumscribed circle radius of 4 cm. Solution: Area of the 12 sided polygon in a circumscribed circle of radius R is, Area = 3R2. Area = 3 * 4 = 3 * 16 = 48 cm2. ## Practice Problems on 12 Sided Polygon Given below are some of the practice problems on regular 12 sided polygon. ### Practice Problems Question 1: Calculate the area of the dodecagon with a side length of 8 cm. Question 2: Calculate the area of the dodecagon with a circumscribed circle radius of 6.5 cm. Related Topics Math Help Online Online Math Tutor *AP and SAT are registered trademarks of the College Board.", "144n = 180 x 2 36n = 180 x 2 n = $$\\frac{180 x 2}{36}$$ = 10 Therefore total number of sides = 10. Length of side is given = 12cm Perimeter = 12 x 10 = 120 cm _________________ Kindly press \"+1 Kudos\" to appreciate CEO Joined: 11 Sep 2015 Posts: 3354 Re: What is the perimeter of a regular polygon with sides of len [#permalink] ### Show Tags 27 Apr 2018, 05:30 Top Contributor phlaym39 wrote: What is the perimeter of a regular polygon with sides of length 12 cm and internal angles measuring 144° each? A. 96 cm B. 108 cm C. 120 cm D. 132 cm E. 144 cm Sum of ALL sides in an n-sided polygon = (n - 2)(180) degrees So, EACH ANGLE in a regular n-sided polygon = (n - 2)(180)/n degrees We're told that EACH ANGLE is 144°, so we can write: (n - 2)(180)/n = 144 Multiply both sides by n to get: (n - 2)(180) = 144n Expand left side: 180n - 360 = 144n Add 360 to both sides: 180n = 144n + 360 Subtract 144n from both sides: 36n = 360 Solve: n = 10 This tells us that the polygon has 10 sides Since each side is the SAME LENGTH in a REGULAR polygon, the", "$12$ cm $\\times 12$ cm squares are each cut into two pieces $A$ and $B$, as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon, as shown in the second figure, so as to fold into a polyhedron. What is the volume (in $\\mathrm{cm}^3$) of this polyhedron?", "about 20 minutes to walk one kilometre (if you are walking at a relaxed speed). There are 1,000 metres in one kilometre. 1,000 m = 1 km ### Worked example 12.5: Calculating perimeter of a regular polygon Calculate the perimeter of the regular polygon shown in the diagram. The length of one side from $W$ to $V$ is 12 millimetres. 1. Step 1: Check whether the shape is a regular polygon. If the instruction tells you that the diagram shows a regular polygon, then you can take this as a given. If you are not told, you can measure. This shape has six equal sides and six equal angles, so it is a regular polygon (named a regular hexagon). 2. Step 2: Work out what calculation to do. You know that one side is 12 mm. So you need to multiply the number of sides by 12 mm. 3. Step 3: Do the calculation, and remember to write the correct units in the answer. \\begin{array}{ll} \\text{Perimeter} & {= 6 \\times \\text{12 mm}} \\\\ & {= \\text{72 mm}} \\\\ \\end{array} As you have seen in the worked example above, we can use a formula to calculate the perimeter of a regular polygon, because all the sides have the same lengths. ### Exercise 12.4 Calculate perimeter of regular polygons 1.", "212 • Yemen 967 • Zambia 260 • Zimbabwe 263 Age Of Your Child • Less Than 6 Years • 6 To 10 Years • 11 To 16 Years • Greater Than 16 Years ## Perimeter of Irregular Polygon As discussed above polygons that do not have equal sides and equal angles are referred to as irregular polygons. Thus, in order to calculate the perimeter of irregular polygons, we add the lengths of all sides of the polygon. ### Examples Ex 1: Find the perimeter of a polygon with sides $4 cm$, $8 cm$, $10 cm$, and $9 cm$. Let the four sides of a polygon be $a = 4 cm$, $b = 8 cm$, $c = 10 cm$ and $d = 9 cm$ Perimeter = $a + b + c + d = 4 + 8 + 10 + 9 = 31 cm$ Note: The polygon is an irregular quadrilateral. Ex 2: Find the perimeter of the given hexagon The length of the six sides of the hexagon are $a = 4 cm$, $b = 18 cm$, $c = 7 cm$, $d = 9 cm$, $e = 10 cm$ and $f = 15 cm$. The perimeter of the hexagon = $a + b + c + d + e + f = 4 + 18 + 7", "Solution Solving each the equations, we find that either $$x = 108$$, $$x = 60$$, $$x = 45$$, or $$x = 72$$. The exterior angle of a regular polygon must be factor of 360, so this rules out $$x = 108$$. For the remaining three cases, $$60$$º is the exterior angle for regular hexagons, $$45$$º is for regular octagons, and $$72$$º is for regular pentagons. Only the pentagon has a prime number of sides. Therefore, the polygon has 5 sides.", "• # question_answer 8) Find the perimeter of a regular hexagon with each side measuring 8 m. A regular hexagon has 6 sides and each side is of 8 cm. $\\therefore$ Perimeter of a regular hexagon $=6\\times$Length of one side $=6\\times 8m=48m$ Hence, the perimeter of a regular hexagon is 48 cm." ]

[ "product of their digits is more than $2$ but less than $7$?", "The product of two number is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is : [A] 380 [B] 395 [C] 400 [D] 425", "$462$ is the product of two consecutive numbers. These numbers are $21$ and $22$, because $21 \\times 22 = 462$. Now try this. $56 406$ is also the product of two consecutive numbers. What are these two numbers?", "$462$ is the product of two consecutive numbers. These numbers are $21$ and $22$, because $21 \\times 22 = 462$. $56 406$ is also the product of two consecutive numbers. What are these two numbers?", "# Question #60eb6 Oct 17, 2017 $14 n - 2$ #### Explanation: Let's call our number $n$. \"Product\" means multiplying, so \"the $\\text{product}$ of 14 and a number\" is the same as $14 \\times n$, or $14 n$. \"less than\" means subtraction, so \"2 $\\text{less than}$ the product of 14 and a number\" is the same as $14 n - 2$. Therefore, the expression is $14 n - 2$ Hope this helps!", "same after four years, this product would be 378. Determine the sum of the current ages of Fero and David. • Digit sum The digit sum of the two-digit number is nine. When we turn figures and multiply by the original two-digit number, we get the number 2430. What is the original two-digit number? • The sum The sum of the squares of two immediately following natural numbers is 1201. Find these numbers. • Sales of products For 80 pieces of two quality products a total sales is 175 Eur. If the first quality product was sold for n EUR per piece (n natural number) and the second quality product after 2 EUR per piece. How many pieces of the first quality were sold? • The sum 8 The sum of two numbers is 21. If three times the smaller numbers is two less than twice the larger number, find the two numbers. • Diofant 2 Is equation 70x +52y = 34 solvable on the set of integers Z? • The difference 2 The difference between the two numbers is 25. The smaller number is 1/6th of the larger number. What is the value of the smaller number? • Two integers Two integers, a and b, have a product of 36. What is the least possible sum of", "# The product of 4 and a number is 18 more than twice the number. What is the number? The number is $9$. $4 x = 2 x + 18 \\text{ }$ {the product of $4$ and a number is $18$ more than twice the number} $2 x = 18 \\text{ }$ {subtracted $2 x$ from each side} $x = 9 \\text{ }$ {divided each side by $2$}", "# What Is the Product of Two Numbers? The product of two whole numbers depends on what the numbers are. The operation is normally multiplication. For example, if the two numbers are 4 and 5, then the product is 4x5=20. The product of two numbers will depend on the specific numbers given. The operation in calculating the product is always multiplication. For instance, if given the two numbers 6 and 7, then their product would be 42. The product of two numbers is the multiplication of two whole numbers When you're talking about a product in mathematics, you're talking about the answer that results when you multiply two numbers together. For example, the product of two plus two is four. Q&A Related to \"What Is the Product of Two Numbers\" The product of two numbers would be the answer to a multiplication problem. For example, 4 multiplied by 2 equals 8. In that equation, 8 is the product. http://answers.ask.com/Science/Other/what_is_the_p... 112 is even so we can take out 2 as a factor. 112/2=56 which is also even so we can take out another 2. 56/2=28 which is again even so we can take out another 2. 28/2=14 which is http://wiki.answers.com/Q/What_are_the_two_numbers... The product of two numbers is the sum. http://www.chacha.com/question/what-is-product-of-... 1. Read the label or Material Safety Data Sheet (MSDS)", "# Question #34639 Jan 27, 2018 $2 x + 8$ #### Explanation: Start with the product of 2 and a number $x$. Product means multiplication, so that means $2 x$. 8 more than $2 x$ means we want to add 8 to $2 x$, so $2 x + 8$. Jan 27, 2018 $2 x + 8$. quite simple, look below for explanation:) #### Explanation: heey!! lets split the question into parts, 1. product of two and a number x $\\implies 2 \\cdot x$ 2. eight more than \"2x\" $\\implies 2 x + 8$ so yeah, the final answer would be, $2 x + 8$. -Sahar:D", "## The productof 2 number is 96/5 .If one of the number is 40/3, then find the other number.​ Question The productof 2 number is 96/5 .If one of the number is 40/3, then find the other number.​ in progress 0 2 months 2021-10-09T19:15:12+00:00 1 Answer 0 views 0", "# Unit Digit What is the unit digit of the following product: $684 \\times 59 \\times 413 \\times 76?$ ×", "numbers are thus $$-5$$ and $$-7$$ with product $$(-5) \\times (-7) = 35$$. (We can check the other possible products of two negative numbers and see that none is as large.) So the greatest possible product in this case is $$8 \\times (-5) \\times (-7) = 8 \\times 35 = 280$$. Combining the two cases, we see that the greatest possible product is 280.", "if all numbers are matched is 2 million dollars.", "of three forms of a three-digit number is always a multiple of $$37$$.", "# Digit product Probability Level 3 How many positive integers less than $1,000,000,000$ have the product of their digits equal to 49? ×", "of all such two digit numbers is $11 + 39 + 61 + 89 = 200$", "### Home > AC > Chapter 5 > Lesson 5.1.6 > Problem5-60 5-60. One number is five more than a second number. The product of the numbers is $3300$. Find the two numbers.", "three numbers a, b, and c, the sum of a and b, when multiplied by c, gives the same result as a and b individually multiplied by c, and then the products are added together to find the result. So, $\\text{a} \\times (\\text{b} + \\text{c}) = \\text{a} \\times \\text{b} + \\text{a} \\times \\text{c}$ When we find the product of two numbers using partial products, we follow a similar process of breaking the numbers and then finding the product. ## Product of Two Numbers Using the Partial Product Method We will discuss partial product examples with two cases: • One-digit number with a two-digit number • Two-digit number with a two-digit number ### Multiplication of a Two-Digit Number with an One-Digit Number Using the partial products method, we can multiply a $1-$digit number with a $2-$digit number. Here, too, we apply the distributive property of multiplication. We use the property $\\text{a} \\times (\\text{b} + \\text{c}) = \\text{a} \\times \\text{b} + \\text{a} \\times \\text{c}$ Suppose we have to find the product of 8 and 73. We first expand 73 as $70 + 3$. Now, we find the product as follows: $8 \\times (70 + 3) = 8 \\times 70 + 8 \\times 3 = 560 + 24 = 584$ ### Multiplication of a Two-Digit Number with a Two-Digit Number When", "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", "## anonymous one year ago What is the product of 4.5 ⋅ 105 and 1.6 ⋅ 102? 1. anonymous @nincompoop 2. anonymous @poopsiedoodle 3. Compassionate Product means to multiply. So in other terms, it's asking $4.5 \\times 105$ $1.6 \\times 102$" ]

[ "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", "$729 = {3^6}$ • $730 = 2 \\cdot 5 \\cdot 73$ • $731 = 17 \\cdot 43$ • $732 = {2^2} \\cdot 3 \\cdot 61$ • $733$ 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", "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.", "= {3^2} \\cdot 103$ • $928 = {2^5} \\cdot 29$ • $929$ is prime. • $930 = 2 \\cdot 3 \\cdot 5 \\cdot 31$ • $931 = {7^2} \\cdot 19$ • $932 = {2^2} \\cdot 233$ • $933 = 3 \\cdot 311$ • $934 = 2 \\cdot 467$ • $935 = 5 \\cdot 11 \\cdot 17$ • $936 = {2^3} \\cdot {3^2} \\cdot 13$ • $937$ is prime. • $938 = 2 \\cdot 7 \\cdot 67$ • $939 = 3 \\cdot 313$ • $940 = {2^2} \\cdot 5 \\cdot 47$ • $941$ is prime. • $942 = 2 \\cdot 3 \\cdot 157$ • $943 = 23 \\cdot 41$ • $944 = {2^4} \\cdot 59$ • $945 = {3^3} \\cdot 5 \\cdot 7$ • $946 = 2 \\cdot 11 \\cdot 43$ • $947$ is prime. • $948 = {2^2} \\cdot 3 \\cdot 79$ • $949 = 13 \\cdot 73$ • $950 = 2 \\cdot {5^2} \\cdot 19$ • $951 = 3 \\cdot 317$ • $952 = {2^3} \\cdot 7 \\cdot 17$ • $953$ is prime. • $954 = 2 \\cdot {3^2} \\cdot 53$ • $955 = 5 \\cdot 191$ • $956 = {2^2} \\cdot 239$ • $957 = 3 \\cdot 11 \\cdot 29$ • $958 = 2 \\cdot 479$ • $959 = 7 \\cdot 137$ • $960 = {2^6}", "# 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.", "= 3, 4, 5, \\ldots$ prime factors are (sequence A006931 in OEIS): k 3$561 = 3 \\cdot 11 \\cdot 17$ 4$41041 = 7 \\cdot 11 \\cdot 13 \\cdot 41$ 5$825265 = 5 \\cdot 7 \\cdot 17 \\cdot 19 \\cdot 73$ 6$321197185 = 5 \\cdot 19 \\cdot 23 \\cdot 29 \\cdot 37 \\cdot 137$ 7$5394826801 = 7 \\cdot 13 \\cdot 17 \\cdot 23 \\cdot 31 \\cdot 67 \\cdot 73$ 8$232250619601 = 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 31 \\cdot 37 \\cdot 73 \\cdot 163$ 9$9746347772161 = 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 31 \\cdot 37 \\cdot 41 \\cdot 641$ The first Carmichael numbers with 4 prime factors are (sequence A074379 in OEIS): i 1$41041 = 7 \\cdot 11 \\cdot 13 \\cdot 41$ 2$62745 = 3 \\cdot 5 \\cdot 47 \\cdot 89$ 3$63973 = 7 \\cdot 13 \\cdot 19 \\cdot 37$ 4$75361 = 11 \\cdot 13 \\cdot 17 \\cdot 31$ 5$101101 = 7 \\cdot 11 \\cdot 13 \\cdot 101$ 6$126217 = 7 \\cdot 13 \\cdot 19 \\cdot 73$ 7$172081 = 7 \\cdot 13 \\cdot 31 \\cdot 61$ 8$188461 = 7 \\cdot 13 \\cdot 19 \\cdot 109$ 9$278545 = 5 \\cdot 17 \\cdot 29 \\cdot 113$ 10$340561 = 13 \\cdot 17 \\cdot 23 \\cdot 67$ The second Carmichael number (1105) can be expressed as the", "• $566 = 2 \\cdot 283$ • $567 = {3^4} \\cdot 7$ • $568 = {2^3} \\cdot 71$ • $569$ 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", "# News • Oct 18, 2017 1 factor of $10^{1695}+1$. • Oct 15, 2017 2 factors of $10^{n}+1$ ($3210\\le n \\le 3870$). • Oct 14, 2017 7 factors of $10^{n}+1$ ($2170\\le n \\le 9950$). • Oct 8, 2017 9 smaller factors of $10^n+1$ ($9641 \\le n \\le 10137$). • Oct 6, 2017 $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9$ is completely factored (snfs). $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9 = \\text{P65}\\cdot\\text{P76}\\cdot\\text{P135}$. This factorization is my hardest one - so far. • Oct 1, 2017 15 smaller factors of $10^n+1$ ($8061 \\le n \\le 8999$). • Oct 1, 2017 $567661^{41}-1$ is completely factored (snfs). • Sep 30, 2017 19 smaller factors of $10^n+1$ ($8237 \\le n \\le 8982$). • Sep 30, 2017 $6613^{41}-1$ is completely factored (snfs). • Sep 29, 2017 76 smaller factors of $10^n+1$ ($3676 \\le n \\le 11957$). • Sep 22, 2017 7 smaller factors of $10^n+1$ ($6349 \\le n \\le 11039$). • Sep 9, 2017 $9996^{41}-1$ is factored. # Efforts (ECM) number digits date curves B1 $10^{295}-91$ 253 May 26, 2016 612 43e6 $500249^{41}-1$ 228 May 17, 2016 1408 1e7 $10^{295}+9$ 296 May 16, 2016 1408 43e6 $10^{271}+3$ 267 May 23, 2016May 25, 2016 34561408 11e643e6 $10^{2189}+1$ 1981 Jun 3, 2016 120 6e6 $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9$ 275 Nov 1, 2016Apr 11, 2017 1689619712 11e726e7 # Efforts (P-1) number digits date B1 B2 $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9$ 275 Oct 29, 2016 1e12", "7 \\cdot 53$ • $372 = {2^2} \\cdot 3 \\cdot 31$ • $373$ is prime. • $374 = 2 \\cdot 11 \\cdot 17$ • $375 = 3 \\cdot {5^3}$ • $376 = {2^3} \\cdot 47$ • $377 = 13 \\cdot 29$ • $378 = 2 \\cdot {3^3} \\cdot 7$ • $379$ is prime. • $380 = {2^2} \\cdot 5 \\cdot 19$ • $381 = 3 \\cdot 127$ • $382 = 2 \\cdot 191$ • $383$ is prime. • $384 = {2^7} \\cdot 3$ • $385 = 5 \\cdot 7 \\cdot 11$ • $386 = 2 \\cdot 193$ • $387 = {3^2} \\cdot 43$ • $388 = {2^2} \\cdot 97$ • $389$ is prime. • $390 = 2 \\cdot 3 \\cdot 5 \\cdot 13$ • $391 = 17 \\cdot 23$ • $392 = {2^3} \\cdot {7^2}$ • $393 = 3 \\cdot 131$ • $394 = 2 \\cdot 197$ • $395 = 5 \\cdot 79$ • $396 = {2^2} \\cdot {3^2} \\cdot 11$ • $397$ is prime. • $398 = 2 \\cdot 199$ • $399 = 3 \\cdot 7 \\cdot 19$ • $400 = {2^4} \\cdot {5^2}$ You might be interested in: Fundamental Theorem of Arithmetic Prime Factorization of an Integer List of Prime Factorizations of Integers from 2 to 200 List of Prime Factorizations of Integers from 401 to", "$38052$. • Because the last two digits of the given number are a multiple of $4$, we can rewrite $38052$ as $2^2 \\cdot 9513$. Note that $9513$ is not even, so we move on. • Because the sum of digits of $9513$ is a multiple of 9, we can rewrite $38052$ as $2^2 \\cdot 3^2 \\cdot 1057$. Note that the sum of digits of $1057$ is not a multiple of 3, so we move on. • Because $1057$ does not end in $0$ or $5$, we skip $5$ as a factor. As for $7$, after long division, we note that $1057 = 7 \\cdot 151$, so we can rewrite $38052$ as $2^2 \\cdot 3^2 \\cdot 7 \\cdot 151$. However, when doing long division again, $151$ is not a multiple of $7$, so we move on. • Note that $11$ is not a factor of $151$, so we can move on. In fact, because $13 > \\sqrt{151}$, we can declare $151$ as prime and stop. Thus, the prime factorization of $38052$ is $2^2 \\cdot 3^2 \\cdot 7 \\cdot 151$.", "$n$ 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!", "# Prove $373$ is prime when $\\gcd(255255, 373) = 1$ I needed to find $\\gcd(255255, 373)$ and then explain why that proves $373$ to be prime. I understand the first part, but not the prime part at all. Here is how I figured the first part out using the euclidean algorithm: $255255 = 684(373)+123$ $373 = 3(123)+4$ $123 = 30(4)+3$ $4 = 1(3)+1$ $3=3(1)+0$ I could say that obviously $373$ would have no even divisors, and now I know that any factor of $255255$ (other than 1) would not be a factor of $373$, so is that connected somehow? • $255255=255\\cdot1001$. An oft recurring useful fact is that $1001=7\\cdot11\\cdot13$. Here, additionally, $255=2^8-1=(2^4-1)(2^4+1)=3\\cdot5\\cdot17$. That's a lot of potential prime factors excluded :-) – Jyrki Lahtonen Sep 18 '17 at 6:49 we can write, $$255255=3.5.7.11.13.17$$ Therefore $373$ is not divisible by $3,5,7,11,13,17$ If $373$ is not prime, $$373= k×z$$ Such that $k,n\\in \\text{ co-prime to }3,7,5,11,13,17\\text{ and } \\gt 17$ The lowest values of $k,z$ are $19$, but $$19.19=361$$ so we try next lowest possibility. $$19.21=399\\gt 373$$ All the next numbers will give us value $\\gt 373$ Hence we cannot get any $k,z$ whose multiplication equals $373$ as all of them will be $\\ge 399\\gt 373$ Hence $373$ is prime. • Why 21? What about the prime 19? – Robert", "\\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", "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", "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 • $$76 = 2\\cdot 2 \\cdot 19$$", "can do better! Human brains are good at remembering sequences and stories. So we’re going to look at the numbers in order and tell ourselves stories about them. The better we get to know them and their relationships the easier they are to remember! 289, 391, 493 There is only one exceptional composite in each of the 200’s, 300’s, and 400’s, namely, 289, 391, and 493. What strikes one immediately about these numbers is that each is exactly 102 more than the previous. Is this a coincidence? Of course not! It is actually due to the fact that $102 = 6 \\cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \\cdot 17$, then $391 = 17 \\cdot (17 + 6) = 17 \\cdot 23$, and $493 = 17 \\cdot (23 + 6) = 17 \\cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes. This sequence of multiples", "# How many numbers between $1$ and $6042$ (inclusive) are relatively prime to $3780$? How many numbers between $1$ and $6042$ (inclusive) are relatively prime to $3780$? Hint: $53$ is a factor. Here the problem is not the solution of the question, because I would simply remove all the multiples of prime factors of $3780$. But I wonder what is the trick associated with the hint and using factor $53$. - Note: 53 is one of four prime factors of 6042: $6042=2\\cdot3\\cdot19\\cdot53$. Note also that $3780 =2^2\\cdot3^3\\cdot5\\cdot7$. – amWhy Nov 2 '12 at 16:16 A misleading hint if ever there was one. – Did Nov 2 '12 at 16:21 I'm wondering if somehow they expected you to use that $2*53 = 3*5*7+1$ and reduce the problem to a smaller count. But I'm not seeing directly how to do that. – Thomas Andrews Nov 2 '12 at 16:51 Perhaps the hint is there just so you can finish the factoring. You could easily see the 2 and then the 3 and then you're left to figure out how to factor 1007. Not incredibly hard since you just need to try 2, 3, 5, 7, 11, 13, 17, and then 19. But, it is helpful. – Graphth Nov 2 '12 at 19:59 How does factoring $6042$ help? It's $3780$ that", "But fortunately we can do better! Human brains are good at remembering sequences and stories. So we’re going to look at the numbers in order and tell ourselves stories about them. The better we get to know them and their relationships the easier they are to remember! # 289, 391, 493 There is only one exceptional composite in each of the 200’s, 300’s, and 400’s, namely, 289, 391, and 493. What strikes one immediately about these numbers is that each is exactly 102 more than the previous. Is this a coincidence? Of course not! It is actually due to the fact that $102 = 6 \\cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \\cdot 17$, then $391 = 17 \\cdot (17 + 6) = 17 \\cdot 23$, and $493 = 17 \\cdot (23 + 6) = 17 \\cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes.", "$663 = 3 \\cdot 13 \\cdot 17$ • $664 = {2^3} \\cdot 83$ • $665 = 5 \\cdot 7 \\cdot 19$ • $666 = 2 \\cdot {3^2} \\cdot 37$ • $667 = 23 \\cdot 29$ • $667 = 23 \\cdot 29$ • $668 = {2^2} \\cdot 167$ • $669 = 3 \\cdot 223$ • $670 = 2 \\cdot 5 \\cdot 67$ • $671 = 11 \\cdot 61$ • $672 = {2^5} \\cdot 3 \\cdot 7$ • $673$ is prime. • $674 = 2 \\cdot 337$ • $675 = {3^3} \\cdot {5^2}$ • $676 = {2^2} \\cdot {13^2}$ • $677$ is prime. • $678 = 2 \\cdot 3 \\cdot 113$ • $679 = 7 \\cdot 97$ • $680 = {2^3} \\cdot 5 \\cdot 17$ • $681 = 3 \\cdot 227$ • $682 = 2 \\cdot 11 \\cdot 31$ • $683$ is prime. • $684 = {2^2} \\cdot {3^2} \\cdot 19$ • $685 = 5 \\cdot 137$ • $686 = 2 \\cdot {7^3}$ • $687 = 3 \\cdot 229$ • $688 = {2^4} \\cdot 43$ • $689 = 13 \\cdot 53$ • $690 = 2 \\cdot 3 \\cdot 5 \\cdot 23$ • $691$ is prime. • $692 = {2^2} \\cdot 173$ • $693 = {3^2} \\cdot 7 \\cdot 11$ • $694 = 2 \\cdot 347$ • $695 = 5 \\cdot" ]

[ "But fortunately we can do better! Human brains are good at remembering sequences and stories. So we’re going to look at the numbers in order and tell ourselves stories about them. The better we get to know them and their relationships the easier they are to remember! # 289, 391, 493 There is only one exceptional composite in each of the 200’s, 300’s, and 400’s, namely, 289, 391, and 493. What strikes one immediately about these numbers is that each is exactly 102 more than the previous. Is this a coincidence? Of course not! It is actually due to the fact that $102 = 6 \\cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \\cdot 17$, then $391 = 17 \\cdot (17 + 6) = 17 \\cdot 23$, and $493 = 17 \\cdot (23 + 6) = 17 \\cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes.", "can do better! Human brains are good at remembering sequences and stories. So we’re going to look at the numbers in order and tell ourselves stories about them. The better we get to know them and their relationships the easier they are to remember! 289, 391, 493 There is only one exceptional composite in each of the 200’s, 300’s, and 400’s, namely, 289, 391, and 493. What strikes one immediately about these numbers is that each is exactly 102 more than the previous. Is this a coincidence? Of course not! It is actually due to the fact that $102 = 6 \\cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \\cdot 17$, then $391 = 17 \\cdot (17 + 6) = 17 \\cdot 23$, and $493 = 17 \\cdot (23 + 6) = 17 \\cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes. This sequence of multiples", "# News • Oct 18, 2017 1 factor of $10^{1695}+1$. • Oct 15, 2017 2 factors of $10^{n}+1$ ($3210\\le n \\le 3870$). • Oct 14, 2017 7 factors of $10^{n}+1$ ($2170\\le n \\le 9950$). • Oct 8, 2017 9 smaller factors of $10^n+1$ ($9641 \\le n \\le 10137$). • Oct 6, 2017 $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9$ is completely factored (snfs). $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9 = \\text{P65}\\cdot\\text{P76}\\cdot\\text{P135}$. This factorization is my hardest one - so far. • Oct 1, 2017 15 smaller factors of $10^n+1$ ($8061 \\le n \\le 8999$). • Oct 1, 2017 $567661^{41}-1$ is completely factored (snfs). • Sep 30, 2017 19 smaller factors of $10^n+1$ ($8237 \\le n \\le 8982$). • Sep 30, 2017 $6613^{41}-1$ is completely factored (snfs). • Sep 29, 2017 76 smaller factors of $10^n+1$ ($3676 \\le n \\le 11957$). • Sep 22, 2017 7 smaller factors of $10^n+1$ ($6349 \\le n \\le 11039$). • Sep 9, 2017 $9996^{41}-1$ is factored. # Efforts (ECM) number digits date curves B1 $10^{295}-91$ 253 May 26, 2016 612 43e6 $500249^{41}-1$ 228 May 17, 2016 1408 1e7 $10^{295}+9$ 296 May 16, 2016 1408 43e6 $10^{271}+3$ 267 May 23, 2016May 25, 2016 34561408 11e643e6 $10^{2189}+1$ 1981 Jun 3, 2016 120 6e6 $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9$ 275 Nov 1, 2016Apr 11, 2017 1689619712 11e726e7 # Efforts (P-1) number digits date B1 B2 $(7\\cdot10^{275}-27\\cdot10^{137}-7)/9$ 275 Oct 29, 2016 1e12", "$38052$. • Because the last two digits of the given number are a multiple of $4$, we can rewrite $38052$ as $2^2 \\cdot 9513$. Note that $9513$ is not even, so we move on. • Because the sum of digits of $9513$ is a multiple of 9, we can rewrite $38052$ as $2^2 \\cdot 3^2 \\cdot 1057$. Note that the sum of digits of $1057$ is not a multiple of 3, so we move on. • Because $1057$ does not end in $0$ or $5$, we skip $5$ as a factor. As for $7$, after long division, we note that $1057 = 7 \\cdot 151$, so we can rewrite $38052$ as $2^2 \\cdot 3^2 \\cdot 7 \\cdot 151$. However, when doing long division again, $151$ is not a multiple of $7$, so we move on. • Note that $11$ is not a factor of $151$, so we can move on. In fact, because $13 > \\sqrt{151}$, we can declare $151$ as prime and stop. Thus, the prime factorization of $38052$ is $2^2 \\cdot 3^2 \\cdot 7 \\cdot 151$.", "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 unchanged numbers modulo $$999$$: $$203*296*395*401*407*913 -1\\equiv 739$$ $$86*296*395*401*407*913 -1\\equiv 406$$ $$86*203*395*401*407*913 -1\\equiv 295$$ $$86*203*296*401*407*913 -1\\equiv 628$$ $$86*203*296*395*407*913 -1\\equiv 73$$ $$86*203*296*395*401*913 -1\\equiv 850$$ $$86*203*296*395*401*407 -1\\equiv 258 = 3\\cdot 86$$ None have a factor $$37$$, and only the last one has a factor $$3$$. Therefore only changing the last factor might give this other type of solution. $$86+203+296+395+401+407+x \\equiv 86\\cdot 203\\cdot 296\\cdot 395\\cdot 401\\cdot 407\\cdot x\\\\789+x \\equiv 259x \\bmod 999\\\\258x \\equiv 789 \\bmod 999\\\\86x \\equiv 263 \\bmod 333\\\\x \\equiv 247 \\bmod 333$$ So instead of adding multiples of 999, which is all we can do with the other numbers, on this last number we can add/subtract multiples of 333. Therefore the candidates for replacing the last number are $$247$$, $$580$$, $$913$$, $$1246$$, $$1579$$, $$1912$$, ... As before, not all of these candidates will work. The product and sum will be the same modulo 999, but they need not be outright equal as they could differ by a non-zero multiple of 999. We therefore have to check which of these candidates works. Let's check some of these candidate solutions: $$86+203+296+395+401+407+247 = 2035$$ $$86\\cdot203\\cdot296\\cdot395\\cdot401\\cdot407\\cdot247 = 82,284,688,793,747,440$$ $$82+284+688+793+747+440 = 3034$$ So $$440$$ does not work. $$86+203+296+395+401+407+580 = 2368$$ $$86\\cdot203\\cdot296\\cdot395\\cdot401\\cdot407\\cdot580=193,219,107,288,961,600$$ $$193+219+107+288+961+600 = 2368$$ Success! $$580$$ works. This is the only alternative solution where the number is no more than 3 digits. Next is", "# 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.", "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", "has no obvious factors based on those rules, so here I would use prime multiple subtraction. Start with $7$, the multiple of $7$ that ends in $3$ is $7\\cdot9=63$, so subtract $63$ to get $410$. $41$ is not a multiple of $7$ or $9$, so neither are factors. Then consider $11$. $3\\cdot 11=33$ so subtract $33$ to get $440$. $440=40\\cdot 11$, hence we can see that $473=43\\cdot 11$. Note for $11$ that if you have $ab\\cdot11$ , where $a,b$ represent the digits, the result is $a(a+b)b$, again to mean digits: e.g. $35\\cdot11=385$. When $a+b>10$, it is harder to spot but still possible. What you should look for then is: $a(a+b-11)b$, and this is $(a-1)b \\cdot 11$. E.g. for $528$, $5+8-11=2, \\therefore 528=11*48$ If you can identify numbers as concatenations of multiples of the same number, they will be a multiple of that number. E.g. $654$ contains $6$ and $54$, both of which are multiples of $6$, and so we can do $6\\div6=1$, and $54\\div6=09$, therefore $654\\div6=109$ • Thanks for your answer, which I’ve up-voted. I’ve used a similar method with numbers such as $721$. – Old Peter May 11 '18 at 14:42 • Thanks for the upvote. Note that $721$ is $7\\cdot103$, you can see this by the subtraction method, or notice that $7$ and $21$ are both", "the factors $11,13$ must be in group $1$. Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$. Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$. Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$. It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \\cdot 5 \\cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\\cdot 13 \\cdot 37 \\gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$. • I also thought that trying to solve for the same question just", "× # Large numbers I wanted to post this as a problem but decided to put it as a note as I put an extended version of this. Two numbers, 823519 and 274658, such that if the first number is divided by $$x$$, the remainder would be three times the remainder obtained by dividing the second number by $$x$$. Find and prove that $$x$$ has only one value. Note by Sharky Kesa 2 years, 2 months ago Sort by: $274658\\equiv r\\pmod x\\stackrel{\\times 3}\\implies 823974\\equiv 3r\\pmod x$ $\\implies 823974-823519\\equiv3r-3r\\equiv 455\\equiv 0\\pmod x\\implies x\\mid 5\\cdot 7\\cdot 13$ There's now a limited amount of $$x$$'s that could work - only the factors of $$5\\cdot 7\\cdot 13$$, which after bashing (all the $$7$$ possibilities) gives $$\\boxed{91}$$ as the only solution. · 2 years, 2 months ago", "= {3^2} \\cdot 103$ • $928 = {2^5} \\cdot 29$ • $929$ is prime. • $930 = 2 \\cdot 3 \\cdot 5 \\cdot 31$ • $931 = {7^2} \\cdot 19$ • $932 = {2^2} \\cdot 233$ • $933 = 3 \\cdot 311$ • $934 = 2 \\cdot 467$ • $935 = 5 \\cdot 11 \\cdot 17$ • $936 = {2^3} \\cdot {3^2} \\cdot 13$ • $937$ is prime. • $938 = 2 \\cdot 7 \\cdot 67$ • $939 = 3 \\cdot 313$ • $940 = {2^2} \\cdot 5 \\cdot 47$ • $941$ is prime. • $942 = 2 \\cdot 3 \\cdot 157$ • $943 = 23 \\cdot 41$ • $944 = {2^4} \\cdot 59$ • $945 = {3^3} \\cdot 5 \\cdot 7$ • $946 = 2 \\cdot 11 \\cdot 43$ • $947$ is prime. • $948 = {2^2} \\cdot 3 \\cdot 79$ • $949 = 13 \\cdot 73$ • $950 = 2 \\cdot {5^2} \\cdot 19$ • $951 = 3 \\cdot 317$ • $952 = {2^3} \\cdot 7 \\cdot 17$ • $953$ is prime. • $954 = 2 \\cdot {3^2} \\cdot 53$ • $955 = 5 \\cdot 191$ • $956 = {2^2} \\cdot 239$ • $957 = 3 \\cdot 11 \\cdot 29$ • $958 = 2 \\cdot 479$ • $959 = 7 \\cdot 137$ • $960 = {2^6}", "# Is $37$ the only prime satisfying the below property? Really i have tried to know more about behavior of primes to produce some sequences i have got that $37$ satisfy the below property : Property: $37*3=111,37*6=222,37*9=333,\\cdots$ , Now I ask if $37$ is the only prime satisfying that property to get a sequence which it include numbers ,each number have same digit as shown before ? • Note that the sequence ends at $37\\cdot27=999$ (the next one would be $37\\cdot30 = 1110)$. – quasi Jul 13 '17 at 4:51 • Note that after $111$, the rest are unremarkable, because what you have is $37\\cdot3, 37\\cdot3\\cdot2, 37\\cdot3\\cdot3$, and so on. I would start by looking at the prime factorisations of $1111, 11111$ and so on. By the way, does $11$ count as such a prime? – Arthur Jul 13 '17 at 4:55 How about $3?$ $3\\cdot 37=111, 3\\cdot 74=222,$ etc. Another is $271: 271 \\cdot 41=11111, 271\\cdot 82=22222,$ etc.", "$n$ 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!", "= 3, 4, 5, \\ldots$ prime factors are (sequence A006931 in OEIS): k 3$561 = 3 \\cdot 11 \\cdot 17$ 4$41041 = 7 \\cdot 11 \\cdot 13 \\cdot 41$ 5$825265 = 5 \\cdot 7 \\cdot 17 \\cdot 19 \\cdot 73$ 6$321197185 = 5 \\cdot 19 \\cdot 23 \\cdot 29 \\cdot 37 \\cdot 137$ 7$5394826801 = 7 \\cdot 13 \\cdot 17 \\cdot 23 \\cdot 31 \\cdot 67 \\cdot 73$ 8$232250619601 = 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 31 \\cdot 37 \\cdot 73 \\cdot 163$ 9$9746347772161 = 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 31 \\cdot 37 \\cdot 41 \\cdot 641$ The first Carmichael numbers with 4 prime factors are (sequence A074379 in OEIS): i 1$41041 = 7 \\cdot 11 \\cdot 13 \\cdot 41$ 2$62745 = 3 \\cdot 5 \\cdot 47 \\cdot 89$ 3$63973 = 7 \\cdot 13 \\cdot 19 \\cdot 37$ 4$75361 = 11 \\cdot 13 \\cdot 17 \\cdot 31$ 5$101101 = 7 \\cdot 11 \\cdot 13 \\cdot 101$ 6$126217 = 7 \\cdot 13 \\cdot 19 \\cdot 73$ 7$172081 = 7 \\cdot 13 \\cdot 31 \\cdot 61$ 8$188461 = 7 \\cdot 13 \\cdot 19 \\cdot 109$ 9$278545 = 5 \\cdot 17 \\cdot 29 \\cdot 113$ 10$340561 = 13 \\cdot 17 \\cdot 23 \\cdot 67$ The second Carmichael number (1105) can be expressed as the", "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", "$729 = {3^6}$ • $730 = 2 \\cdot 5 \\cdot 73$ • $731 = 17 \\cdot 43$ • $732 = {2^2} \\cdot 3 \\cdot 61$ • $733$ 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", "Please be more careful. – Did Jan 24 '17 at 15:42 • Yes. But I considered prime factorisations – Roman83 Jan 24 '17 at 15:46 • So the point is not that $7^{164}$ divides $1000!$, but that $7^{165}$ does not. – Callus Jan 24 '17 at 15:49 • I'm understood, thank you! – Roman83 Jan 24 '17 at 16:00 In $\\{1,\\dots, 500\\}$, there are 71 multiples of 7, 10 multiples of $7^2$ and one multiple of $7^3$. How many multiples of $7$, $7^2$, $7^3$ do you have in $\\{501,\\dots, 1000\\}$? Same amount. • Why are they the same amount? – Jorge Fernández Hidalgo Jan 24 '17 at 15:59 • 504 is first multiple of 7 (with multiplicity of seven is 1 in factorization), and $504 + 7 \\cdot 70 = 994$. Also $7^2 \\cdot 11 = 539$ (multiplicity is 2) and $7^2 \\cdot 20 = 980$. At last $7^3 \\cdot 2 = 686$ (fits only once). – johnnycrab Jan 24 '17 at 16:18", "# How many numbers between $1$ and $6042$ (inclusive) are relatively prime to $3780$? How many numbers between $1$ and $6042$ (inclusive) are relatively prime to $3780$? Hint: $53$ is a factor. Here the problem is not the solution of the question, because I would simply remove all the multiples of prime factors of $3780$. But I wonder what is the trick associated with the hint and using factor $53$. - Note: 53 is one of four prime factors of 6042: $6042=2\\cdot3\\cdot19\\cdot53$. Note also that $3780 =2^2\\cdot3^3\\cdot5\\cdot7$. – amWhy Nov 2 '12 at 16:16 A misleading hint if ever there was one. – Did Nov 2 '12 at 16:21 I'm wondering if somehow they expected you to use that $2*53 = 3*5*7+1$ and reduce the problem to a smaller count. But I'm not seeing directly how to do that. – Thomas Andrews Nov 2 '12 at 16:51 Perhaps the hint is there just so you can finish the factoring. You could easily see the 2 and then the 3 and then you're left to figure out how to factor 1007. Not incredibly hard since you just need to try 2, 3, 5, 7, 11, 13, 17, and then 19. But, it is helpful. – Graphth Nov 2 '12 at 19:59 How does factoring $6042$ help? It's $3780$ that", "# Calculate $2^{1401} \\pmod{2803}$ Without using the fact that 2803 is prime, Calculate $2^{1401} \\pmod{2803}$ Usually I would do something like: $1401 = 3\\times467$ $\\equiv(2^{467})^3 \\pmod{2803}$ And try to simplify it down and use the mod to get it down to something I can punch in the calculator, but in this case $2^{467}$ does not help as it is still too large. If I did $(2^{3})^{467}$ that also does not seem to help as I just end up with $8^{467}$ and is too large What other techniques can I use here? - Use exponentiation by squaring. It easily adapts to exponentiation with a modulus. You have the right idea by trying to factorize the exponent. All there is left to notice is that you do not need to factorize it, e.g., $1401 = 2 \\cdot 700 + 1$ means you can compute $2^{700} \\ (\\text{mod} \\ 2803)$, then square that and multiply by 2." ]

[ "# Factors and Multiples Number Theory Level 3 How many positive factors of 36 are multiples of 4? This answer will be x. What is the sum of ALL of the positive factors of 36? This answer will be y. ×", "# 92nd Problem 2016 How many multiples of 13 are included between 12548 and 19987 both inclusive? ×", "# How many multiple? Number Theory Level 1 How many multiples of 7 are between 345 and 563 inclusive? ×", "6 TWICE, we counted 12 TWICE and so on. In fact, we counted all multiples of 6 TWICE Multiples of 6: 6, 12, 18..., 96 96/6 = 16, so there are 16 multiples of 6 So.....TOTAL multiples of 2 OR 3 = 49 + 33 - 16 = 66 There are 99 positive integers that are less than 100 So, the TOTAL number of those integers that are NEITHER multiples of 2 or 3 = 99 - 66 = 33", "# 92nd Problem 2016 Number Theory Level 1 How many multiples of 13 are included between 12548 and 19987 both inclusive? ×", "of 991. If you like to learn what is a multiple", "# Factors and Multiples How many positive factors of 36 are multiples of 4? This answer will be x. What is the sum of ALL of the positive factors of 36? This answer will be y.", "# Counting Positive Factors 12 has 6 positive divisors: 1, 2, 3, 4, 6, 12. Which of the two numbers has more divisors, $12\\times 6$ or $12\\times 9?$ ×", "number of factors of 12 is (2 + 1)(1 + 1) = 6. _________________ Manager Joined: 03 Mar 2017 Posts: 118 Location: India GMAT 1: 650 Q45 V34 Re: How many positive factors do 180 and 96 have in common? [#permalink] ### Show Tags 30 Jun 2019, 07:02 Bunuel wrote: Baten80 wrote: How many positive factors do 180 and 96 have in common? A. 6 B. 12 C. 16 D. 18 E. 24 To find the number of common factors of two integers: 1. Find the greatest common divisor (GCD) of the two integers; 2. Find the number of factors of that GCD. GCD of 180 and 96 is 12 = 2^2*3. The number of factors of 12 is (2 + 1)(1 + 1) = 6. I did in the exact same way. Finding out the GCD and finding the number of factors for GCD. hence A Re: How many positive factors do 180 and 96 have in common? [#permalink] 30 Jun 2019, 07:02 Display posts from previous: Sort by", "# Numbers Number Theory Level 4 How many positive perfect squares less than $$10^6$$ are multiples of 24? ×", "0 So that expression is a multiple of 6 for all even values of n. So that is 49 values, 2 to 98 inclusive Jun 24, 2021", "multiples of 8 and 12 but 24 is the lowest.", "work on Factors and Multiples at Secondary level.", "forum “Since 77 = (11)(7), any positive multiple of 7 or 11 will have a common factor with 77 where that common factor is greater than 1. For example, 77 and 14 share a common factor of 7. And 77 and 35 share a common factor of 7. Likewise, 77 and 55 share a common factor of 11. And 77 and 33 share ...” July 8, 2019 Brent@GMATPrepNow posted a reply to The table above shows the number of students in a certain in the Data Sufficiency forum “ASIDE: Since the diagram closely resembles a Double Matrix Method diagram, I''ll treat it as such. Target question: What is the total number of students in the class? Statement 1: Of the boys in the class, 15 are studying biology. Add this information to our diagram to get: ...” July 8, 2019 Max@Math Revolution posted a new topic called Let f(n) be the number of positive factors of n. What is min in the Problem Solving forum “[GMAT math practice question] Let f(n) be the number of positive factors of n. What is minimum x satisfying f(420)*f(x)=96? A. 4 B. 6 C. 8 D. 10 E. 12” July 8, 2019 Max@Math Revolution posted a new topic called x, y are two different positive integers greater than 6. Wha in the", "00:28. Last edited by EgmatQuantExpert on 12 Jun 2015, 03:42, edited 1 time in total. ##### General Discussion Manager Joined: 12 Nov 2014 Posts: 62 Re: How many positive factors do 180 and 96 have in common? [#permalink] ### Show Tags 07 May 2015, 21:28 2 The number of common factors will be same as number of factors of the Highest Common Factor(HCF) HCF of 180 and 96 is 12 Number of factors of 12 = 6 Ambarish _________________ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Kindly press Kudos if the explanation is clear. Thank you Ambarish Math Expert Joined: 02 Sep 2009 Posts: 47981 Re: How many positive factors do 180 and 96 have in common? [#permalink] ### Show Tags 08 May 2015, 03:40 1 1 Baten80 wrote: How many positive factors do 180 and 96 have in common? A. 6 B. 12 C. 16 D. 18 E. 24 To find the number of common factors of two integers: 1. Find the greatest common divisor (GCD) of the two integers; 2. Find the number of factors of that GCD. GCD of 180 and 96 is 12 = 2^2*3. The number of factors of 12 is (2 + 1)(1 + 1) = 6. _________________ Non-Human User Joined: 09 Sep 2013 Posts: 7757 Re: How many positive factors do 180 and 96 have in common?", "positive 12 minus 7 which is 5. Yeah. Good. OK. Good. There, (a) and (d) work.", "Positive Integral Factors Discrete Mathematics Level 2 How many positive integral factors does the number $$240$$ have? ×", "96 12 x 8 = 96 16 x 6 = 96 24 x 4 = 96 32 x 3 = 96 48 x 2 = 96 96 x 1 = 96 The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96. (iii) Given division fact is 55 ÷ 5 = 11 The multiplication facts are 5 x 11 = 55, 11 x 5 = 55 Find the factors of multiple 55 1 x 55 = 55 5 x 11 = 55 11 x 5 = 55 55 x 1 = 55 Therefore, the factors of 55 are 1, 5, 11, and 55.", "Free Version Moderate # Factors of 9 Million COMBIN-5GZGKK How many positive factors of 9000000 are divisible by 120? A $24$ B $30$ C $36$ D $40$ E $48$ F none of the above", "Home » How Many Times Does 14 Go Into 120? Update # How Many Times Does 14 Go Into 120? Update Let’s discuss the question: how many times does 14 go into 120. We summarize all relevant answers in section Q&A of website Mytholi.com in category: Blog Finance For You. See more related questions in the comments below. ## How many times does 12 make 120? Answer: 12 goes into 120 10 times. See also How To Get Ed Sheeran To Sing At My Wedding? New ## How many times can 8 go to 120? THERE ARE 15 8’s IN 120. ### Multiplying 2 digit numbers- example 1 Multiplying 2 digit numbers- example 1 Multiplying 2 digit numbers- example 1 ## How many times can 9 Enter 120? Answer. 120 divides 9 = 13 with remainder 3. ## How many times does 30 fit 120? Ans . 4 Times does 30 go into 120. ## What numbers can 12 go into? FAQs on Factors of 12 The factors of 12 are 1, 2, 3, 4, 6, 12 and its negative factors are -1, -2, -3, -4, -6, -12. ## How many 12s are there in 240? List of the First 20 Multiples of 12 Multiplication Multiples of 12 12 × 17 204 12 × 18 216 12 ×" ]

[ "96 12 x 8 = 96 16 x 6 = 96 24 x 4 = 96 32 x 3 = 96 48 x 2 = 96 96 x 1 = 96 The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96. (iii) Given division fact is 55 ÷ 5 = 11 The multiplication facts are 5 x 11 = 55, 11 x 5 = 55 Find the factors of multiple 55 1 x 55 = 55 5 x 11 = 55 11 x 5 = 55 55 x 1 = 55 Therefore, the factors of 55 are 1, 5, 11, and 55.", "Explanation: Multiples of 4 are 4,8,12,16,20,24,28,32,36,40,44,48. Not a factors of 48 are 20,28,32,36,40,44. Question 24. What factors of 48 are multiples of 4? Answer: 4,8,12,16,24,48. Explanation: Multiples of 4 are: 4,8,12,16,20,24,28,32,36,40,44,48. Factors of 48 are: 1,2,4,8,12,16,24,48. Question 25. Pose a Problem Look back at Problem 24. Write a similar problem by changing the numbers. Then solve. Answer: Let’s take factors of 64 are multiples of 8? 8,16,32,64. Explanation: Multiples of 8 are: 8,16,24,32,40,48,56,64,72,80 Factors of 64 are: 1,2,4,8,16,32,64 Question 26. Kia paid$10 for two charms. The price of each charm was a multiple of $2. What are the possible prices of the charms? Answer:$2,$8 and$4,$6. Explanation: Since the price was multiple of 2 and Kia paid$10 for two charms, So possible prices are $2+$8=$10 and$4+$6=$10. Question 27. Look for Structure The answer is 9, 18, 27, 36, 45. What is the question? Answer: Write the multiples of 9 Question 28. How do you know whether a number is a multiple of another number? Answer: When the number is divisible by the number then that number is multiple of another number. Explanation: For example, if we take a number i.e 8 which is divisible by 2 and 8 is a multiple of 2. Question 29. For numbers 29a–29e, select True or False for each statement. a. The number 45", "6, and 12 The multiples of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24 Hence, from the above, We can conclude that 1 appears in each list Number Sense How are factors and multiples related? Factors are the numbers that can divide the numbers originally. Multiples are the numbers that can be divided by another number a certain number of times without any remainder. Think and Grow: Identify Multiples A whole number is a multiple of each of its factors. 12 is a multiple of 1, 2, 3, 4, 6, and 12. 1 × 12 = 12 2 × 6 = 12 3 × 4 = 12 4 × 3 = 12 6 × 2 = 12 12 × 1 = 12 Example Is 56 a multiple of 7? One Way: List multiples of 7. 7, 14, 21, 28, 35, 42, 49, 56 So, 56 is a multiple of 7. Another Way: Use division to determine whether 7 is a factor of 56. 56 ÷ 7 = 8 7 is a factor of 56. So, 56 is a multiple of 7. Example Is 9 a factor of 64? One Way: Use divisibility rules to determine whether 9 is a factor of 64. 9 is not a factor of 64 because 6 + 4 = 10", "is not divisible by 9. Another Way: List the multiples of 9. 9, 18, 27, 36, 45, 54, 63, 72 64 is not a multiple of 9. So, 9 is not a factor of 64. Show and Grow Question 1. Is 23 a multiple of 3? Explain. Answer: 23 is not a multiple of 3 Explanation: Multiples are the numbers that can be divided by another number a certain number of times without any remainder. The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30 Hence, from the above, We can conclude that 23 is not a multiple of 3. Question 2. Is 8 a factor of 56? Explain. Answer: 8 is a factor of 56. Explanation: Factors are the numbers that divide the numbers originally. Now, The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, and 56 Hence, from the above, We can conclude that 8 is a factor of 56. Apply and Grow: Practice Question 3. Is 65 a multiple of 5? Explain. Answer: 65 is a multiple of 5 Explanation: Multiples are the numbers that can be divided by another number a certain number of times without any remainder. The multiples of 5 are: 5, 10, 15, 20, 25, 30, 35,40, 45, 50, 55, 60 and", "4, 5, 8, 10, 20, and 40 The first 10 multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50 Hence, from the above, The common numbers from both the factors of 4 and the multiples of 5 are: 5, 10, 20 and 40 Think and Grow: Modeling Real Life Example You need 96 balloons for a school dance. Balloons come in packs of 4, packs of 6, and packs of 9. Which packs could you buy so you have no leftover balloons? Use division to determine whether 96 is a multiple of 4. 4 √96 So, 4 is a factor of 96, and 96 is a multiple of 4. Use the divisibility rules to check whether 96 is a multiple of 6. 96 is even and 9 + 6 = 15 divisible by 3. So, 6 is a factor of 96 and 96 is a multiple of 6. Use the divisibility rules to check whether 96 is a multiple of 9. 9 + 6 = 15 is not divisible by 9. So, 9 is not a factor of 96 and 96 is not a multiple of 9. You could buy packs of 4 balloons or packs of 6 balloons. Show and Grow Question 14. A teacher needs 88 batteries for science experiments.", "Hence every factor of 800 is also a factor of 9600 and is divisible by 12 if we are multiplying each factor of 800 by 12. To find the number of factors of 800 is a straightforward application of number of factors formula:- (p+1)(q+1)(r+1)... [where p,q,r are exponents of each prime factor] Therefore 800 can be written as $$2^5∗5^2$$ Therefore the number of factors of 800 are (5+1)(2+1) = 6*3 = 18 factors. Therefore the no. of factors of 9600 which are divisible by 12 are 18 factors in total. But the question stem asks us how many factors of 9600 are NOT divisible by 12? Therefore we will have to deduct all the factors of 9600 which are divisible by 12 from all the factors of 9600 to get all the factors of 9600 which are NOT divisible by 12. 9600 can be written as $$2^7∗3*5^2$$ Therefore the number of factors of 9600 are (7+1)(1+1)(2+1) = 8*2*3 = 48 factors. All the factors of 9600 which are NOT divisible by 12 = 48 factors - 18 factors = 30 factors [hence the correct answer is (B)] _________________ Intern Joined: 02 May 2018 Posts: 7 Re: How many factors of 9600 are not divisible by 12? [#permalink] ### Show Tags 07 Jun 2018, 04:41 can someone please explain", "$s(6) = 4(6)^3 + 2(6)^2 - 3(6) - 14 = 904$ (in whatever units the problem is supposed to be written in.) -Dan Slight correction on this problem. v(t) is NOT $v(t) = 12t^2 + 4t - 3$ it is $v(t) = 12t^2 + 4t + 3$ and hence s(t) is NOT $s(t) = 4t^3 + 2t^2 - 3t - 14$ it is $s(t) = 4t^3 + 2t^2 + 3t + 14$ so the answer is 968 not 904", "one valid product if and only if $f_1 \\cdot x=96$, the product $x$ has the properties that factors are greater than $1$, and differently ordered products are counted separately. For example, say the first factor is $2$. Then, the remaining numbers must multiply to $48$, so the number of ways the product can be written beginning with $2$ is $D(48)$. To add up all the number of solutions for every possible starting factor, $D(96/f_1)$ must be calculated and summed for all possible $f_1$, except $96$ and $1$, since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$. This means $D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$. Instead of calculating D for the larger factors first, reduce $D(48)$, $D(32)$, and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation. Following the recursive definition $D(n)=($sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to $D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)$ +$(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.$ $D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$, so the sum further simplifies to $D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5$, after combining terms. From quick casework, $D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1$ and $D(2)=1$. Substituting these values into the expression above, $D(96)=3 \\cdot 8+4 \\cdot 8+5 \\cdot 4+4 \\cdot 3+5 \\cdot 2+4 \\cdot 1+5 \\cdot 1+5=\\boxed{\\textbf{(A) } 112}$. ~monmath a.k.a Fmirza", "by 12=Total of no factors of 9600- Total no of factors of 9600 which are divisible by 12 Now we can write, 9600=$$2^7$$*$$5^2$$*$$3^1$$=$$2^2$$*3($$2^5$$*$$5^2$$)=12($$2^5$$*$$5^2$$) Since $$2^5$$*$$5^2$$ is multiple of 12, hence the no of factors which are divisible by 12 =(5+1)(2+1)=6*3=18 Therefore,the no factors of 9600 not divisible by 12=48-18=30 2^5*5^2 is a multiple of 12 ? Hi avijit02, Typo error edited. Thanks _________________ Regards, PKN Rise above the storm, you will find the sunshine Senior Manager Joined: 31 May 2018 Posts: 432 Location: United States Concentration: Finance, Marketing Re: How many factors of 9600 are not divisible by 12? [#permalink] ### Show Tags 26 Jun 2018, 02:47 Meghakothari98 wrote: can someone please explain how the get factors of 9600 divisible by 12 9600=2^7*3*5^2 total factors of 9600=(7+1)(1+1)(2+1)=8*2*3=48 since we want to find factors of 12 write 9600 as a multiple of 12 now 9600=(2^2*3)(2^5*5^2)=12(2^5*5^2) now factors of 9600 divisible by 12=(5+1)(2+1)=6*3=18 now from here we can find factors of 9600 not divisible by 12 =total factors of 9600-factors of 9600 divisible by 12 =48-18=30 Non-Human User Joined: 09 Sep 2013 Posts: 15097 Re: How many factors of 9600 are not divisible by 12? [#permalink] ### Show Tags 16 Feb 2020, 07:36 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this", "65 Hence, fro the above, We can conclude that 65 is a multiple of 5. Question 4. Is 14 a multiple of 4? Explain. Answer: 14 is not a multiple of 4. Explanation: Multiples are the numbers that can be divided by another number a certain number of times without any remainder. The multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, and 40 Hence, from the above, We can conclude that 14 is not a multiple of 4. Question 5. Is 23 a multiple of 2? Explain. Answer: 23 is not a multiple of 2 Explanation: Multiples are the numbers that can be divided by another number a certain number of times without any remainder. The multiples of 2 are: 2, 4, 6, 8, 10, 1, 14, 16, 18, 20, 22, 24 and 26 Hence, from the above, We can conclude that 23 is not a multiple of 2. Question 6. Is 6 a factor of 96? Explain. Answer: 6 is a factor of 96 Explanation: Factors are the numbers that divide the number originally. The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 28, 32, 48 and 96 Hence, from the above, We can conclude that 6 is a factor of 96. Question 7. Is 3 a factor", "mbers 24, 12,8,6, 4,3,2 and I are all factors of the number 24. Each of these numbers can be d ivided even ly into 24 with no remainder. The term divisible by mea ns divisible by without any remainder or with a remainder of zero. For instance. 15 is divisible by 5 because 15 divided by 5 is 3 with a remainder ofO. But 15 is not divisible by 7 because 15 divided by 7 is 2 with a remainder of 1. Com mon fa cto rs: Common factors are factors that two (or more) numbers have in common. For instante, 3 is a common facto r of 12 and 18. The largest com mon factor of two (or more numbers) is called their greatest commotl factor (GCF). For example, 6 is the GCF of 12 a nd 18. MULTIPLES The multiples of any given number are those numbers that ca n be d ivided by that given number without a remainder. You can find the mu ltiples of a n umber by mu ltiplying it by 1, 2, 3, 4 and so on. For instance, 8, 16. 24, 32. 40 and 48 are some of t he mu ltiples of8. Notice that the multiples of 8 are also multiples of2 and 4 (factors of8). Remember, the mu", "15 Factors of 25 are 1, 5 and 25 Therefore, the common factors of 5, 15, and 25 are 1 and 5. (ii) 2, 6 and 8 Factors of 2 are 1 and 2 Factors of 6 are 1, 2, 3 and 6 Factors of 8 are 1, 2, 4 and 8 Therefore, the common factors of 2, 6 and 8 are 1 and 2. #### Question 3: Find first three common multiples of 6 and 8. Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, … Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, … Therefore, the first three common multiples of 6 and 8 are 24, 48 and 72. #### Question 4: Find first two common multiples of 12 and 18. Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, … Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198, … Therefore, the first two common multiples of 12 and 18 are 36 and 72. #### Question 5: A number is divisible by both 7 and 16. By which other number will that number be always divisible? Since the number is divisible by 7 and 16, they are the factors of", "EMPOWERgmatRichC wrote: Hi All, This question can be solved with a mix of prime-factorization, Number Properties and TESTing VALUES. We're told that P is a POSITIVE INTEGER. We're asked if P^2 is divisible by 96. This is a YES/NO question. Before jumping to the two Facts, we can dissect the prompt a bit. 96 = (2^5)(3), so for a number to be divisible by 96, that number MUST have at least (2^5)(3) in its prime-factorization. Since P is an INTEGER and it's P^2 that's divisible by 96, we have to make sure that we're following ALL of the rules in this prompt before we TEST VALUES... Fact 1: P is a multiple of 8. 8 = (2^3), so P^2 will include (2^6) at the minimum. To be divisible by 96 though, we need to have a '3' in the prime-factorization... IF.... P = 8, P^2 = (2^6) and the answer to the question is NO IF.... P = 24, P^2 = (2^6)(3^2) and the answer to the question is YES Fact 1 is INSUFFICIENT Fact 2: P^2 is a multiple of 12. Here, we have an interesting 'wrinkle'....P MUST be an integer, but there is no integer (when squared) that equals 12...The smallest multiple of 12 that is a perfect square is 36... 36 = 6^2 = (2^2)(3^2)", "11:13 1 loserunderachiever wrote: Bunuel wrote: If $$x$$ is a positive integer, is $$x$$ divisible by 15? (1) $$x$$ is a multiple of 10 (2) $$x^2$$ is a multiple of 12 Hi Bunuel , I got a wrong answer , chose B ,as this a Yes / No question , to find whether X is divisible by 15 or not (y/n) statement 2 : says , X^2 is a multiple of 12 , so in this case the only possibility is 6^2 = 36 which is a multiple of 12 , hence it is NO , this means X should also has to a perfect square , then only number left out is 12 itself , (12*12) i.e 144, in this case also 15 is not divisible . Why are we considering x = 12 *15 as a possibility , where it is mentioned X^2 is a multiple of 12 . Can you please guide me ? Positive multiples of 12 are 12, 2*12 = 24, 3*12 = 36, 4*12 = 48, .... (2) says that $$x^2$$ is a multiple of 12. Since x^2 is a perfect square, then we are looking for multiples of 12 which are also perfect squares. So, x^2 could be: 36 (for x = 6); 144 (for x = 12); 324 (for x", "### Show Tags Updated on: 08 Jun 2018, 17:48 6 7 CAMANISHPARMAR wrote: How many factors of 9600 are not divisible by 12? A) 48 B) 30 C) 24 D) 20 E) 42 Let's write 9600 in prime factorization form, 9600=$$2^7$$*$$5^2$$*$$3^1$$ Hence the no of factors of 9600=(7+1)(2+1)(1+1)=8*3*2=48 The no factors of 9600 not divisible by 12=Total of no factors of 9600- Total no of factors of 9600 which are divisible by 12 Now we can write, 9600=$$2^7$$*$$5^2$$*$$3^1$$=$$2^2$$*3($$2^5$$*$$5^2$$)=12($$2^5$$*$$5^2$$) Since 12($$2^5$$*$$5^2$$) is multiple of 12, hence the no of factors which are divisible by 12 =(5+1)(2+1)=6*3=18 Therefore,the no factors of 9600 not divisible by 12=48-18=30 _________________ Regards, PKN Rise above the storm, you will find the sunshine Originally posted by PKN on 05 Jun 2018, 22:16. Last edited by PKN on 08 Jun 2018, 17:48, edited 1 time in total. ##### General Discussion VP Joined: 11 Feb 2015 Posts: 1145 Re: How many factors of 9600 are not divisible by 12? [#permalink] ### Show Tags 07 Jun 2018, 03:32 9600 could be written as 12 * 800 Now 10 is a factor of 800, so is 20, so is 400, including many others. 10, 20 and 400 are not divisible by 12 but (12*10), (12*20) and (12*400) are divisible by 12 because we are multiplying and dividing my 12.", "also infer that $x_5$ will likely need to be one of the smaller factors. The factors of $144$ are: $1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144$ Selecting numbers from this list and trying them out, we can satisfy the conditions with these numbers: $x_5 = 18, x_4 = 6, x_3 = 2, x_2 = 1, x_1 = 2$ So, $x_7 = x_6(x_5 + x_4)$ $x_7 = 144(18 + 6)$ $x_7 = 3456$ Therefore, the answer is $\\Rightarrow{\\boxed{456}}$", "Re: Which of the following values times 12 is not a multiple of [#permalink] 14 Jun 2020, 03:13 1 KUDOS sandy wrote: Which of the following values times 12 is not a multiple of 64? Indicate all such values. A. $$6^6$$ B. $$12^2$$ C. $$18^3$$ D. $$30^3$$ E. $$222$$ The most accurate approach here is to use the calculator and test the answer choices: A) (6^6*12)/64 -> 46 656/64=729 (Numerator is perfectly divisible by denominator. But this is not what we are looking for) B) (12^2*12)/64 -> 1728/64=27 ((Numerator is perfectly divisible by denominator. But this is not what we are looking for) C) (18^3*12)/64 -> 69 984/64=1093,5 (Numerator is not a multiple of 64,in other words numerator is not divisible by 64, that's what we are looking for) D) (30^3*12)/64 -> 324 000/64=5062.5 (Numerator is not a multiple of 64,in other words numerator is not divisible by 64, that's what we are looking for) E) (222*12)/64 -> 2664/64=41.625 (Numerator is not a multiple of 64,in other words numerator is not divisible by 64, that's what we are looking for) Thus we have three correct answers: C/D/E Re: Which of the following values times 12 is not a multiple of [#permalink] 14 Jun 2020, 03:13 Display posts from previous: Sort by", "$4181$ 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", "yes or no. Question 1. Question 2. 8 Answer: Yes Explanation: Since 8×1=8, it is a multiple of 8 Question 3. 20 Answer: No Explanation: 20 is not a multiple of 8 Question 4. 40 Answer: Yes Explanation: 8×5=40, So 40 is multiple of 8 List the next nine multiples of each number. Find the common multiples. Question 5. Multiples of 4: Multiples of 7: Common multiples: Answer: Explanation: Multiples of 4: 4,8,12,16,20,24,28,32,36,40. Multiples of 7: 7,14,21,28,35,42,49,56,63,70. Common Multiples: 28, Question 6. Multiples of 3: Multiples of 9: Common multiples: Answer: 9,18,45,54,63, etc. Explanation: Multiples of 3: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63. Multiples of 9: 9,18,27,36,45,54,63,72,81,90. Common multiples: 9,18,45,54,63, etc. Question 7. Multiples of 6: Multiples of 8: Common multiples: Answer: 24,48,72. Explanation: Multiples of 6: 6,12,18,24,30,36,42,48,54,60,66,72,78. Multiples of 8: 8,16,24,32,40,48,56,64,72,80. Common multiples: 24,48,72. Tell whether 24 is a factor or multiple of the number. Write factor, multiple, or neither. Question 8. 6 Answer: Multiple Explanation: 6×4=24 Question 9. 36 Answer: Neither Explanation: 36 is not a factor or multiple of 24. Question 10. 48 Answer: Factor Explanation: 24×2= 48, So 48 is a factor of 24 Problem Solving Question 11. Ken paid$12 for two magazines. The cost of each magazine was a multiple of $3. What are the possible prices of the magazines? Answer:$3+$9=$12. Explanation: As each magazine cost was multiple", "and 24. Let's find their factors. 18: 1,2,3,6,9,18. 24: 1,2,3,4,6,8,12,24. As can be seen some factors are same for both numbers (they are in bold). These numbers are called common factors (divisors) of 18 and 24. ## Least Common Multiple (LCM) Suppose we are given two numbers 18 and 24. Let's find some of their multiples. 18: 18,36,54,72,90,108,126,144,... 24: 24,48,72,96,120,144,168,... As can be seen some factors are same for both numbers (they are in bold: 72 and 144). These numbers are called common multiples of 18 and 24." ]

[ "96 12 x 8 = 96 16 x 6 = 96 24 x 4 = 96 32 x 3 = 96 48 x 2 = 96 96 x 1 = 96 The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96. (iii) Given division fact is 55 ÷ 5 = 11 The multiplication facts are 5 x 11 = 55, 11 x 5 = 55 Find the factors of multiple 55 1 x 55 = 55 5 x 11 = 55 11 x 5 = 55 55 x 1 = 55 Therefore, the factors of 55 are 1, 5, 11, and 55.", "zero. The fact that 96 is an even composite number, it has 24 factors in total. Twelve out of 24 are positive and the remaining twelve are negative factors. • Factors of 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96 • Negative Factors of 96: -1, -2, -3, -4, -6, -8, -12, -16, -24, -32, -48 and -96 • Prime Factors of 96: 2, 3 • Prime Factorization of 96: 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3 • Sum of Factors of 96: 252 ## How To Calculate the Factors of 96? You can calculate the factors of 96 by the division method and multiplication method. We will learn the division method in this article. Finding factors of 96 using a simple division method: Always remember that the divisor can never be greater than the given number, that is 96. Draw a number line in mind that starts with one and ends on ninety-six. Factors of 96 will be on this number line. In the division method, we simply divide the 96 by a series of numbers from the number line defined above. Always start the division by 1. $\\dfrac {96}{1} = 96$ The number 1 divide every number uniformly, leaving the remainder zero.", "6 TWICE, we counted 12 TWICE and so on. In fact, we counted all multiples of 6 TWICE Multiples of 6: 6, 12, 18..., 96 96/6 = 16, so there are 16 multiples of 6 So.....TOTAL multiples of 2 OR 3 = 49 + 33 - 16 = 66 There are 99 positive integers that are less than 100 So, the TOTAL number of those integers that are NEITHER multiples of 2 or 3 = 99 - 66 = 33", "number of factors of 12 is (2 + 1)(1 + 1) = 6. _________________ Manager Joined: 03 Mar 2017 Posts: 118 Location: India GMAT 1: 650 Q45 V34 Re: How many positive factors do 180 and 96 have in common? [#permalink] ### Show Tags 30 Jun 2019, 07:02 Bunuel wrote: Baten80 wrote: How many positive factors do 180 and 96 have in common? A. 6 B. 12 C. 16 D. 18 E. 24 To find the number of common factors of two integers: 1. Find the greatest common divisor (GCD) of the two integers; 2. Find the number of factors of that GCD. GCD of 180 and 96 is 12 = 2^2*3. The number of factors of 12 is (2 + 1)(1 + 1) = 6. I did in the exact same way. Finding out the GCD and finding the number of factors for GCD. hence A Re: How many positive factors do 180 and 96 have in common? [#permalink] 30 Jun 2019, 07:02 Display posts from previous: Sort by", "and divide the given number exactly . Example 2. Find all positive factors of 96. We can easily check that each of the number 1, 2, 3 , 4, 6, 8, 16, 24, 48 and 96 completely divides 96. So they are factors of 96. 96 ÷ 1= 96, 96 ÷2 = 48, 96 ÷ 3= 32, 96 ÷ 4= 24, 96 ÷ 6= 16, 96 ÷ 8= 12, 96 ÷ 16= 6, 96 ÷ 24= 4, 96 ÷ 48= 2, 96 ÷ 96= 1 Factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96. ## Properties of factors (i) 1 is factor of every number as 6=6 x 1, 19=19 x 1 , 27=27 x 1 etc. (ii) Every non-zero number is a factor of itself. (iii) Every factor of a number is an exact divisor of that number. (iv) A factor of a non-zero number is less than or equal to the given number. (v) Number of factors of a given number are finite. (vi) Every non-zero number is a factor of 0. (vii) 1 is the only number having one factor , namely itself. (viii) Every non- zero number other than 1 has at least two factors, namely 1 and itself. ### Prime factorisation of a number If", "6, and 12 The multiples of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24 Hence, from the above, We can conclude that 1 appears in each list Number Sense How are factors and multiples related? Factors are the numbers that can divide the numbers originally. Multiples are the numbers that can be divided by another number a certain number of times without any remainder. Think and Grow: Identify Multiples A whole number is a multiple of each of its factors. 12 is a multiple of 1, 2, 3, 4, 6, and 12. 1 × 12 = 12 2 × 6 = 12 3 × 4 = 12 4 × 3 = 12 6 × 2 = 12 12 × 1 = 12 Example Is 56 a multiple of 7? One Way: List multiples of 7. 7, 14, 21, 28, 35, 42, 49, 56 So, 56 is a multiple of 7. Another Way: Use division to determine whether 7 is a factor of 56. 56 ÷ 7 = 8 7 is a factor of 56. So, 56 is a multiple of 7. Example Is 9 a factor of 64? One Way: Use divisibility rules to determine whether 9 is a factor of 64. 9 is not a factor of 64 because 6 + 4 = 10", "positive integer, is p^2 divisible by 96? [#permalink] ### Show Tags 05 Jun 2015, 10:09 3 2 Bunuel wrote: If p is a positive integer, is p^2 divisible by 96? (1) p is a multiple of 8. (2) p^2 is a multiple of 12. Questions: If p is a positive integer, is p^2 divisible by 96? 96 = $$2^5 * 3$$ i.e. for $$P^2$$ to be a multiple of 96, it must be a multiple of $$(2^5 * 3)$$ Since P is an Integer therefore $$P^2$$ must be a perfect square i.e. $$P^2$$ MUST be a multiple of$$(2^6 * 3^2)$$ i.e P MUST be a multiple of $$(2^3 * 3)$$ Question Stem Modified: Is P a multiple of $$(2^3 * 3)$$? Statement 1: p is a multiple of 8 This confirms that p is a multiple of $$2^3$$ but since it's still unknown whether p is a multiple of 3 or not therefore NOT SUFFICIENT Statement 2: p^2 is a multiple of 12 This confirms that p is a multiple of 3 and 2 but since it's still unknown whether p is a multiple of $$2^3$$ or not therefore NOT SUFFICIENT Combining the two statements Statement 1 confirms that p is a multiple of $$2^3$$ Statement 2 confirms that p is a multiple of 3 i.e. p is", "65 Hence, fro the above, We can conclude that 65 is a multiple of 5. Question 4. Is 14 a multiple of 4? Explain. Answer: 14 is not a multiple of 4. Explanation: Multiples are the numbers that can be divided by another number a certain number of times without any remainder. The multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, and 40 Hence, from the above, We can conclude that 14 is not a multiple of 4. Question 5. Is 23 a multiple of 2? Explain. Answer: 23 is not a multiple of 2 Explanation: Multiples are the numbers that can be divided by another number a certain number of times without any remainder. The multiples of 2 are: 2, 4, 6, 8, 10, 1, 14, 16, 18, 20, 22, 24 and 26 Hence, from the above, We can conclude that 23 is not a multiple of 2. Question 6. Is 6 a factor of 96? Explain. Answer: 6 is a factor of 96 Explanation: Factors are the numbers that divide the number originally. The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 28, 32, 48 and 96 Hence, from the above, We can conclude that 6 is a factor of 96. Question 7. Is 3 a factor", "Now multiply the first coefficient by the last coefficient to get . Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ? To find these two numbers, we need to list all of the factors of (the previous product). Factors of : 1,2,3,4,6,8,12,16,24,32,48,96 -1,-2,-3,-4,-6,-8,-12,-16,-24,-32,-48,-96 Note: list the negative of each factor. This will allow us to find all possible combinations. These factors pair up and multiply to . 1*96 = 96 2*48 = 96 3*32 = 96 4*24 = 96 6*16 = 96 8*12 = 96 (-1)*(-96) = 96 (-2)*(-48) = 96 (-3)*(-32) = 96 (-4)*(-24) = 96 (-6)*(-16) = 96 (-8)*(-12) = 96 Now let's add up each pair of factors to see if one pair adds to the middle coefficient : First NumberSecond NumberSum 1961+96=97 2482+48=50 3323+32=35 4244+24=28 6166+16=22 8128+12=20 -1-96-1+(-96)=-97 -2-48-2+(-48)=-50 -3-32-3+(-32)=-35 -4-24-4+(-24)=-28 -6-16-6+(-16)=-22 -8-12-8+(-12)=-20 From the table, we can see that there are no pairs of numbers which add to . So cannot be factored. =============================================================== So doesn't factor at all (over the rational numbers). So is prime. If you need more help, email me at [email protected] Also, feel free to check out my tutoring website Jim Polynomials-and-rational-expressions/385262: 12y^2+5y-3 1 solutions Answer 272577 by jim_thompson5910(28476) on 2010-12-12 23:18:26 (Show Source): You can put this", "not divisible . Why are we considering x = 12 *15 as a possibility , where it is mentioned X^2 is a multiple of 12 . Can you please guide me ? Positive multiples of 12 are 12, 2*12 = 24, 3*12 = 36, 4*12 = 48, .... (2) says that $$x^2$$ is a multiple of 12. Since x^2 is a perfect square, then we are looking for multiples of 12 which are also perfect squares. So, x^2 could be: 36 (for x = 6); 144 (for x = 12); 324 (for x = 18); 576 (for x = 24); 900 (for x = 30); 1296 (for x = 36); 1764 (for x = 42); 2304 (for x = 48); 2916 (for x = 54); 3600 (for x = 60); ... As you can see for some values, x is NOT a multiple of 15 but for some values x IS a multiple of 15 (for example, for x = 30 or x = 60) Oh , this clears up . I stopped my multiple list till 12*12 = 144 . Your explanation absolutely makes sense , I realised my mistake now . 30 & 60 , makes sense. Faced this one in my GMAT club quant test. Now it's clear to me. Thanks a lot Bunuel. _________________", "is its largest factor. Please correct if I'm wrong here. I know that if n = 12, n^2 = 144 = 2 * 72 (satisfy the condition). When n = 12, the largest factor of n is n itself, which is 12. Check: 12 is the largest positive number that must divide 12 --> true However if n = 48, n^2 = 48 * 48 = 32 * 72 (satisfy the condition too). When n = 48, the largest factor of n is n itself, which is 48. Check: 48 is the largest positive number that must divide 48 --> true So, I also notice that the keyword is \"MUST\", not \"COULD\". The question is, why is 48 not \"MUST divide 48\", but instead only \"COULD divide 48\"? I'm not clear right here. Why is 12 \"MUST divide 12\"? What's the difference between them? Thanks, Caten Kudos [?]: 35 [0], given: 11 Math Expert Joined: 02 Sep 2009 Posts: 43294 Kudos [?]: 139166 [6], given: 12777 Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink] ### Show Tags 02 Nov 2012, 03:53 6 KUDOS Expert's post catennacio wrote: Bunuel wrote: Merging similar topics. raviram80 wrote: Hi All 169. If n is a positive integer and n2 is divisible by 72, then the", "of 10 (B) x is a multiple of 15 (C) x is a multiple of 25 (D) x is not a multiple of 7 (E) x is negative We see that 12x = -5y^2 and therefore, x = -5y^2/12. It’s tempting to conclude that x is negative, and if that were true, then choice E would be the correct answer. However, if y = 0, then x will also be 0, which is not negative. We can eliminate choice E. Let’s take a closer look at x = -5y^2/12. Since 5 is not divisible by 12 and x is an integer, then y^2 must be divisible by 12. The smallest positive integer value of y such that y^2 is divisible by 12 is y = 6. And if y = 6, x = -5(36)/12 = -15, which is a multiple of 15. (Note: any other integer value of y must be a multiple of 6, so x will still be a multiple of 15.) I had the impression that we should always consider positive multiple or factors in GMAT. I am not talking about mathematics in general but pertaining to GMAT. If we consider as above then if the question is like How many factors does 6 have?\" The answer would be 8: (-6, -3, -2, -1, 1,", "and 24. Let's find their factors. 18: 1,2,3,6,9,18. 24: 1,2,3,4,6,8,12,24. As can be seen some factors are same for both numbers (they are in bold). These numbers are called common factors (divisors) of 18 and 24. ## Least Common Multiple (LCM) Suppose we are given two numbers 18 and 24. Let's find some of their multiples. 18: 18,36,54,72,90,108,126,144,... 24: 24,48,72,96,120,144,168,... As can be seen some factors are same for both numbers (they are in bold: 72 and 144). These numbers are called common multiples of 18 and 24.", "multiple of 6 and positive [#permalink] Show Tags 23 Apr 2015, 23:20 tgubbay1 wrote: Bunuel wrote: zest4mba wrote: If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105 ? (1) x is a multiple of 9. (2) y is a multiple of 25. Can someone explain this $$105=3*5*7$$. Since $$x$$ is a multiple of 6 and $$y$$ is a multiple of 14, then $$xy$$ is a multiple of $$LCM(x,y)=2*3*7$$: we have 3 and 7 as factors of $$xy$$, so in order $$xy$$ to be a multiple of 105 we need missing 5 to be a factor of either $$x$$ or $$y$$. (1) x is a multiple of 9 --> we don't know whether 5 is a factor of either $$x$$ or $$y$$. Not sufficient. (2) y is a multiple of 25 --> 5 is a factor of $$y$$, hence $$xy$$ is a multiple of 105. Sufficient. Hi Bunuel, are there any more questions like this one? It's a subject I need practice on. Thanks! Hi tgubbay1, Why don't you give an attempt on this practice question: Q: $$x, y$$ are positive integers such that $$x$$ is the lowest number which has 15 and 20 as its factor. Is $$x$$ a factor of $$y$$? (1)", "4, 5, 8, 10, 20, and 40 The first 10 multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50 Hence, from the above, The common numbers from both the factors of 4 and the multiples of 5 are: 5, 10, 20 and 40 Think and Grow: Modeling Real Life Example You need 96 balloons for a school dance. Balloons come in packs of 4, packs of 6, and packs of 9. Which packs could you buy so you have no leftover balloons? Use division to determine whether 96 is a multiple of 4. 4 √96 So, 4 is a factor of 96, and 96 is a multiple of 4. Use the divisibility rules to check whether 96 is a multiple of 6. 96 is even and 9 + 6 = 15 divisible by 3. So, 6 is a factor of 96 and 96 is a multiple of 6. Use the divisibility rules to check whether 96 is a multiple of 9. 9 + 6 = 15 is not divisible by 9. So, 9 is not a factor of 96 and 96 is not a multiple of 9. You could buy packs of 4 balloons or packs of 6 balloons. Show and Grow Question 14. A teacher needs 88 batteries for science experiments.", "11:13 1 loserunderachiever wrote: Bunuel wrote: If $$x$$ is a positive integer, is $$x$$ divisible by 15? (1) $$x$$ is a multiple of 10 (2) $$x^2$$ is a multiple of 12 Hi Bunuel , I got a wrong answer , chose B ,as this a Yes / No question , to find whether X is divisible by 15 or not (y/n) statement 2 : says , X^2 is a multiple of 12 , so in this case the only possibility is 6^2 = 36 which is a multiple of 12 , hence it is NO , this means X should also has to a perfect square , then only number left out is 12 itself , (12*12) i.e 144, in this case also 15 is not divisible . Why are we considering x = 12 *15 as a possibility , where it is mentioned X^2 is a multiple of 12 . Can you please guide me ? Positive multiples of 12 are 12, 2*12 = 24, 3*12 = 36, 4*12 = 48, .... (2) says that $$x^2$$ is a multiple of 12. Since x^2 is a perfect square, then we are looking for multiples of 12 which are also perfect squares. So, x^2 could be: 36 (for x = 6); 144 (for x = 12); 324 (for x", "# Factors of 96: Prime Factorization, Methods, Tree, and Examples The factors of 96 are integers that uniformly divide the number 96 without leaving any remainder. The number 96 is even composite. The law of division is used. Figure 1 – All possible Factors of 96 Factors of 96 can also be defined as the product of two integers that are greater than zero and less than 96. The law of multiplication is implemented. There are two ways to calculate the factors of a number: 1. Division Method. 2. Multiplication Method. Factors of a number can be classified as positive factors and negative factors. They differ by a negative sign only. The number 96 is abundant, which means the sum of all its positive divisors except itself is more than 96. The number 96 is the smallest number that can be written as the difference between 2 squares in 4 ways. In this guide, we will discuss about the factors of 96, the techniques used to discover them, some important facts about the factors, and at the end, we will practice some examples. ## What Are the Factors of 96? The factors of 96 are the following: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96. All these divide 96 evenly, leaving the remainder", "EMPOWERgmatRichC wrote: Hi All, This question can be solved with a mix of prime-factorization, Number Properties and TESTing VALUES. We're told that P is a POSITIVE INTEGER. We're asked if P^2 is divisible by 96. This is a YES/NO question. Before jumping to the two Facts, we can dissect the prompt a bit. 96 = (2^5)(3), so for a number to be divisible by 96, that number MUST have at least (2^5)(3) in its prime-factorization. Since P is an INTEGER and it's P^2 that's divisible by 96, we have to make sure that we're following ALL of the rules in this prompt before we TEST VALUES... Fact 1: P is a multiple of 8. 8 = (2^3), so P^2 will include (2^6) at the minimum. To be divisible by 96 though, we need to have a '3' in the prime-factorization... IF.... P = 8, P^2 = (2^6) and the answer to the question is NO IF.... P = 24, P^2 = (2^6)(3^2) and the answer to the question is YES Fact 1 is INSUFFICIENT Fact 2: P^2 is a multiple of 12. Here, we have an interesting 'wrinkle'....P MUST be an integer, but there is no integer (when squared) that equals 12...The smallest multiple of 12 that is a perfect square is 36... 36 = 6^2 = (2^2)(3^2)", "Explanation: Multiples of 4 are 4,8,12,16,20,24,28,32,36,40,44,48. Not a factors of 48 are 20,28,32,36,40,44. Question 24. What factors of 48 are multiples of 4? Answer: 4,8,12,16,24,48. Explanation: Multiples of 4 are: 4,8,12,16,20,24,28,32,36,40,44,48. Factors of 48 are: 1,2,4,8,12,16,24,48. Question 25. Pose a Problem Look back at Problem 24. Write a similar problem by changing the numbers. Then solve. Answer: Let’s take factors of 64 are multiples of 8? 8,16,32,64. Explanation: Multiples of 8 are: 8,16,24,32,40,48,56,64,72,80 Factors of 64 are: 1,2,4,8,16,32,64 Question 26. Kia paid$10 for two charms. The price of each charm was a multiple of $2. What are the possible prices of the charms? Answer:$2,$8 and$4,$6. Explanation: Since the price was multiple of 2 and Kia paid$10 for two charms, So possible prices are $2+$8=$10 and$4+$6=$10. Question 27. Look for Structure The answer is 9, 18, 27, 36, 45. What is the question? Answer: Write the multiples of 9 Question 28. How do you know whether a number is a multiple of another number? Answer: When the number is divisible by the number then that number is multiple of another number. Explanation: For example, if we take a number i.e 8 which is divisible by 2 and 8 is a multiple of 2. Question 29. For numbers 29a–29e, select True or False for each statement. a. The number 45", "Hence every factor of 800 is also a factor of 9600 and is divisible by 12 if we are multiplying each factor of 800 by 12. To find the number of factors of 800 is a straightforward application of number of factors formula:- (p+1)(q+1)(r+1)... [where p,q,r are exponents of each prime factor] Therefore 800 can be written as $$2^5∗5^2$$ Therefore the number of factors of 800 are (5+1)(2+1) = 6*3 = 18 factors. Therefore the no. of factors of 9600 which are divisible by 12 are 18 factors in total. But the question stem asks us how many factors of 9600 are NOT divisible by 12? Therefore we will have to deduct all the factors of 9600 which are divisible by 12 from all the factors of 9600 to get all the factors of 9600 which are NOT divisible by 12. 9600 can be written as $$2^7∗3*5^2$$ Therefore the number of factors of 9600 are (7+1)(1+1)(2+1) = 8*2*3 = 48 factors. All the factors of 9600 which are NOT divisible by 12 = 48 factors - 18 factors = 30 factors [hence the correct answer is (B)] _________________ Intern Joined: 02 May 2018 Posts: 7 Re: How many factors of 9600 are not divisible by 12? [#permalink] ### Show Tags 07 Jun 2018, 04:41 can someone please explain" ]

[ "“So many children can't have expensive toys, and I feel a real obligation to them. I'm appalled when I go Christmas shopping and$60 to $80 for a toy is nothing.\" In 2008, Slinkys cost$4 to $5, and Slinky Dogs about$20.[10] In 1998, James Industries was sold to Poof Products, Inc. of Plymouth, Michigan, a manufacturer of foam sports balls.[4][10] Slinky continued production in Hollidaysburg.[4] In 2003, James Industries merged with Poof Products, Inc. to create Poof-Slinky, Inc. Betty James died of congestive heart failure in November 2008, aged 90, after having served as president of James Industries from 1960 to 1998.[10] Over 300 million Slinkys have been sold between 1945 and 2005, and the original Slinky is still a bestseller.[4][7] ## Physical properties The rules that govern the mechanics of a slinky are due to Hooke's law and the effects of gravitation. ### Period of oscillation Due to simple harmonic motion the period of oscillation of a dangling slinky is ${\\displaystyle T=2\\pi {\\sqrt {\\frac {m}{k}}}.}$ Where T is the time of the period of oscillation, m is the mass of the slinky and k is the spring constant of the slinky. ### Equilibrium In the state of equilibrium of a slinky, all net force is cancelled throughout the entire slinky. This results in a stationary slinky with zero", "I go Christmas shopping and$60 to $80 for a toy is nothing.\" In 2008, Slinkys cost$4 to $5, and Slinky Dogs about$20.[12] In 1998 James Industries was sold to Poof Products, Inc. of Plymouth, Michigan, a manufacturer of foam sports balls.[6][12] Slinky continued production in Hollidaysburg.[6] In 2003, James Industries merged with Poof Products, Inc., to create Poof-Slinky, Inc. Betty James died of congestive heart failure in November 2008, age 90, after having served as president of James Industries from 1960 to 1998.[12] Over 300 million Slinkys have been sold between 1945 and 2005, and the original Slinky is still a bestseller.[6][9] ## Physical properties The rules that govern the mechanics of a slinky are Hooke's law and the effects of gravitation. ### Period of oscillation Due to simple harmonic motion the period of oscillation of a dangling slinky is ${\\displaystyle T=2\\pi {\\sqrt {\\frac {m}{k}}},}$ where T is the time of the period of oscillation, m is the mass of the slinky and k is the spring constant of the slinky. ### Equilibrium In the state of equilibrium of a slinky, all net force is cancelled throughout the entire slinky. This results in a stationary slinky with zero velocity. As the positions of each part of the slinky is governed by the slinky's mass, the force of gravity", "the state of Washington sent the company a letter and drawings for developing Slinky pull-toys. The company liked her ideas, and Slinky Dog and Slinky Train were added to the company's product line. Slinky Dog, a small plastic dog whose front and rear ends were joined by a metal Slinky, debuted in 1952. Malsed received royalties of$60,000 to \\$70,000 annually for 17 years on her patent for the Slinky pull-toy idea, but never visited the plant.[15] In 1995, the Slinky Dog was redesigned for all of Pixar's Toy Story. James Industries had discontinued their Slinky Dog a few years previously. Betty James approved of the new Slinky Dog, telling the press, \"[The earlier Slinky Dog] wasn't nearly as cute as this one.\" The molds used in manufacturing the new toy created problems for James Industries so the plastic front and rear ends were manufactured in China with James Industries doing the assembly and packaging. The entire run of 825,000 redesigned Slinky Dogs sold out well before Christmas 1995.[16] Plastic Slinkys are also available. They can be made in different colors. Many of them are made with the colors of the rainbow in rainbow order. They were marketed in the 1970s as a safer alternative to metal slinkys as they did not present a hazard when inserted into electrical", "A toy store regularly sells all stock at a discount of 20 15 25 Oct 2010, 04:40 5 In each production lot for a certain toy, 25 percent of the 7 27 Nov 2007, 15:21 Display posts from previous: Sort by", "feel a real obligation to them. I'm appalled when I go Christmas shopping and $60 to$80 for a toy is nothing.\" In 2008, Slinkys cost $4 to$5, and Slinky Dogs about $20.[10] In 1998, James Industries was sold to Poof Products, Inc. of Plymouth, Michigan, a manufacturer of foam sports balls.[4][10] Slinky continued production in Hollidaysburg.[4] In 2003, James Industries merged with Poof Products, Inc. to create Poof-Slinky, Inc. Betty James died of congestive heart failure in November 2008, aged 90, after having served as president of James Industries from 1960 to 1998.[10] Over 300 million Slinkys have been sold between 1945 and 2005, and the original Slinky is still a bestseller.[4][7] ## Physical properties The rules that govern the mechanics of a slinky are due to Hooke's law and the effects of gravitation. ### Period of oscillation Due to simple harmonic motion the period of oscillation of a dangling slinky is ${\\displaystyle T=2\\pi {\\sqrt {\\frac {m}{k}}}.}$ Where T is the time of the period of oscillation, m is the mass of the slinky and k is the spring constant of the slinky. ### Equilibrium In the state of equilibrium of a slinky, all net force is cancelled throughout the entire slinky. This results in a stationary slinky with zero velocity. As the positions of each part of", "it may result in higher revenue , even though the number of toys sold did not increase significantly last year over the previous year=> hence it resolves the paradox . thanks that was my reasoning as well, hence I eliminated e and chose c. And everyone else makes valid points as well but I definitely thought like you for option \"e.\" Re: Toys4Them, an online toy merchant, generated$220 million in [#permalink] 01 Oct 2016, 10:44 Similar topics Replies Last post Similar Topics: 3 Internal Memorandum from Toy Supply Company 11 03 Jan 2016, 09:01 2 A toy company's engineering department developed a new model 4 09 Dec 2013, 10:44 1 Child s World, a chain of toy stores, has relied on a 12 30 Dec 2009, 14:21 Consumer advocate: The toy-labeling law should require 14 09 Oct 2007, 06:31 Archaeologists have found wheeled ceramic toys made by the 4 07 May 2007, 20:02 Display posts from previous: Sort by", "1.5y \\le 5500 \\ \\end$ .The store owner estimates that no more than 2000 toys will be sold every month and he does not plan to invest more than ,000 in inventory of these toys.How many units of each type of toys should be stocked in order to maximize his monthly total profit profit?The following videos gives examples of linear programming problems and how to test the vertices.Several word problems and applications related to linear programming are presented along with their solutions and detailed explanations.", "find a convincing explanation in that thread. The Dancing Doll line sold slightly more than$3.5 million worth of toys last year, 40% more than the Teeny Tiny Trucks line did and nearly 3 times as much as the Basic Block line's sales. A. the Teeny Tiny Trucks line did and nearly 3 times as much as the Basic Block line's sales B. the Teeny Tiny Trucks did and nearly 3 times what the Basic Blocks' sales were C. the Teeny Tiny Trucks line sold and nearly 3 times as much as Basic Blocks' sales D. the Teeny Tiny Trucks line and nearly 3 times more than the Basic Blocks' sales E. the Teeny Trucks line and nearly 3 times more than the Basic Blocks line Senior Manager Joined: 16 Aug 2004 Posts: 327 Location: India Followers: 1 Kudos [?]: 41 [0], given: 0 Re: SC - Dancing Doll [#permalink] 21 Apr 2008, 08:55 Answer option E is the most precise and removes the redundant word/expression \"sales\". Director Joined: 18 Feb 2008 Posts: 798 Followers: 20 Kudos [?]: 100 [0], given: 25 Re: SC - Dancing Doll [#permalink] 21 Apr 2008, 09:32 E for ||ism. However, I do feel it slightly changed the meaning of the sentence, but I do not see any better answers. Manager Joined: 22 Oct", "parodied and referenced in popular culture. It is seen in the \"Log\" commercial on The Ren & Stimpy Show and sung by actor Jim Carrey in Ace Ventura: When Nature Calls. The song is also referenced in the movie Lords of Dogtown, where it is sung in full by Emile Hirsch, and is sung by Eddie Murphy as part of the final routine in the stand-up comedy film Eddie Murphy Raw. Early in the history of James Industries, Helen (Herrick) Malsed of the state of Washington sent the company a letter and drawings for developing Slinky pull-toys. The company liked her ideas, and Slinky Dog and Slinky Train were added to the company's product line. Slinky Dog, a small plastic dog whose front and rear ends were joined by a metal Slinky, debuted in 1952. Malsed received royalties of $60,000 to$70,000 annually for 17 years on her patent for the Slinky pull-toy idea, but never visited the plant.[15] In 1995, the Slinky Dog was redesigned for all of Pixar's Toy Story. James Industries had discontinued their Slinky Dog a few years previously. Betty James approved of the new Slinky Dog, telling the press, \"[The earlier Slinky Dog] wasn't nearly as cute as this one.\" The molds used in manufacturing the new toy created problems for James Industries so", "the Conclusion Based on the data, I am 95% confident that the difference in the proportions of defective toys before and after the manager’s claim is between -0.0372 and -0.0068.", "i.e. Discounted offer on toys may not be present => so it may be that toys are currently sold on its selling price => it may result in higher revenue , even though the number of toys sold did not increase significantly last year over the previous year=> hence it resolves the paradox . thanks that was my reasoning as well, hence I eliminated e and chose c. And everyone else makes valid points as well but I definitely thought like you for option \"e.\" Re: Toys4Them, an online toy merchant, generated$220 million in [#permalink] 01 Oct 2016, 10:44 Similar topics Replies Last post Similar Topics: 4 Internal Memorandum from Toy Supply Company 11 03 Jan 2016, 09:01 2 A toy company's engineering department developed a new model 4 09 Dec 2013, 10:44 1 Child s World, a chain of toy stores, has relied on a 12 30 Dec 2009, 14:21 Consumer advocate: The toy-labeling law should require 14 09 Oct 2007, 06:31 Archaeologists have found wheeled ceramic toys made by the 4 07 May 2007, 20:02 Display posts from previous: Sort by", "and has helped Grand Metropolitan PLC's Burger King increase its market share. Nearly all of Burger King's 7,000 U.S. stores are participating in the \"Toy Story\" promotion … Nevertheless, meeting consumer demand still remains a conundrum for the giants. That is partly because individual Burger King restaurant owners make their tricky forecasts six months before such promotions begin. \"It's asking you to pull out a crystal ball and predict exactly what consumer demand is going to be,\" says Richard Taylor, Burger King's director of youth and family marketing. \"This is simply a case of consumer demand outstripping supply.\" The long lead times are necessary because the toys are produced overseas to take advantage of lower costs … Burger King managers in Houston and Atlanta say the freebies are running out there, too … But Burger King, which ordered nearly 50 million of the small plastic dolls, is \"nowhere near running out of toys on a national level.\" Let's consider a simplified analysis of Burger King's situation. Consider a region with 200 restaurants served by a single distribution center. At the time the order must be placed with the factories in Asia, demand (units of toys) for the promotion at each restaurant is forecasted to be gamma distributed with mean 2,251 and standard deviation 1,600. A discrete version of", "Tiny Trucks line did and newarly 3 times as much as the Basic Blocks line's sales B. the Teeny Tiny Trucks did and nearly 3 times what the Basic Blocks' sales were C. the Teeny Tiny Trucks line sold and nearly 3 times as much as Basic Blocks' sales. D. the Teeny Tiny Trucks line and nearly 3 times more than Basic Blocks' sales E. the Teeny Tiny Trucks line and nearly 3 times more than the Basic Blocks line Intern Joined: 16 Apr 2010 Posts: 34 Followers: 0 Kudos [?]: 4 [0], given: 1 Re: comparision and parallel [#permalink] 29 Apr 2010, 10:05 bupbebeo wrote: The Dancing Doll line sold slightly more than$3.5 million worth of toys last year, 40% mroe than the Teeny Tiny Trucks line did and newarly 3 times as much as the Basic Blocks line's sales. A. the Teeny Tiny Trucks line did and newarly 3 times as much as the Basic Blocks line's sales B. the Teeny Tiny Trucks did and nearly 3 times what the Basic Blocks' sales were C. the Teeny Tiny Trucks line sold and nearly 3 times as much as Basic Blocks' sales. D. the Teeny Tiny Trucks line and nearly 3 times more than Basic Blocks' sales E. the Teeny Tiny Trucks line and nearly 3 times", "toys last year than the previous year. c)Last year, the number of consumers shopping for toys increased over the previous year. d)Last year, Toys4Them experienced an unprecedented boom in its divisions that do not sell toys. e)Because of an economic downturn, Toys4Them heavily discounted its toys during the holiday season two years ago. IMO , except C & E , all other options clearly resolve the discrepancy. OA is C not E. However I could nt be able to figure out , how does option E resolve the discrepancy . So help me to figure out the same ... Thanks in advance . Board of Directors Joined: 01 Sep 2010 Posts: 3421 Re: Toys4Them, an online toy merchant, generated$220 million in [#permalink] ### Show Tags 16 Jan 2013, 16:17 2 happy to clarify When you deal with such question (indeed tough) try to understand really the argument. Here we have an increase of revenue from one year to another year and a diminish of toys sold E says that Toys4Them heavily discounted its toys during the holiday season two years ago (i.e 2012 during christmas holidays, could be an option) then the next year we have an increase of revenue BUT the toys are not purchased because people already have. May be they buy other stuffs from other", "want. See the \"collectors\" problem. Basically, the problem is that if you're a toy collector, and you want to collect all possible 10 Burger King toys, on average how many times do you have to eat at Burger King? What is the probability, after n runs, that only m toys are missing? You're collecting people and categorizing them into \"toys\" according to their birthdays, so it's the same problem.", "Blocks' sales E. the Teeny Tiny Trucks line and nearly three times as much as Basic Blocks' line Moderator Joined: 02 Jul 2012 Posts: 1231 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 115 Kudos [?]: 1384 [0], given: 116 Re: The Dancing Doll line sold slightly more than$3.5 million [#permalink] ### Show Tags 23 Aug 2013, 03:14 rajnikant47 wrote: The Dancing Doll line sold slightly more than$3.5 million worth of toys last year, 40 % more than the Teeny Tiny Trucks line did and nearly three times as much as the Basic Blocks line's sales. A. Same as underlined above B. the Teeny Tiny Trucks did and nearly three times what the Basic Blocks sales were C. the Teeny Tiny Trucks line sold and nearly three times as much as Basic Blocks' sales D. the Teeny Tiny Trucks line and nearly three times as much as Basic Blocks' sales E. the Teeny Tiny Trucks line and nearly three times as much as Basic Blocks' line This question has already been discussed extensively in this topic.. Please do let us know if you have further queries... _________________ Did you find this post helpful?... Please let me know through the Kudos button. Thanks To The Almighty - My GMAT Debrief", "the Teeny Tiny Trucks line did and nearly three times as much as the Basic Blocks line’s sales B) the Teeny Tiny Trucks did and nearly three times what the Basic Block’s sales were C) the Teeny Tiny Trucks line sold and nearly three times as much as Basic Block’s sales D) the Teeny Tiny Trucks line and nearly three times more than Basic Block’s sales E) the Teeny Tiny Trucks line and nearly three times more than the Basic Blocks line Director Joined: 04 Jan 2008 Posts: 918 Followers: 58 Kudos [?]: 280 [0], given: 17 Re: Dancing dolls [#permalink] 27 Jan 2010, 07:28 is it [Reveal] Spoiler: E ? \"dancing doll line \" is the Subject of Comparison _________________ Director Joined: 01 Aug 2008 Posts: 770 Followers: 4 Kudos [?]: 227 [0], given: 99 Re: Dancing dolls [#permalink] 27 Jan 2010, 08:10 angel2009 wrote: The dancing doll line sold slightly more than$3.5 million worth of toys last year, forty percent more than the Tenny Tiny Trucks line did and nearly three times as much as the Basic Blocks line’s sales. A) the Teeny Tiny Trucks line did and nearly three times as much as the Basic Blocks line’s sales B) the Teeny Tiny Trucks did and nearly three times what the Basic Block’s sales were C)", "Author Message TAGS: ### Hide Tags Intern Joined: 16 Apr 2010 Posts: 34 Followers: 0 Kudos [?]: 9 [0], given: 1 The Dancing Doll line sold slightly more than $3.5 million [#permalink] ### Show Tags 29 Apr 2010, 09:14 00:00 Difficulty: (N/A) Question Stats: 65% (01:47) correct 35% (01:49) wrong based on 126 sessions ### HideShow timer Statistics The Dancing Doll line sold slightly more than$3.5 million worth of toys last year, 40% mroe than the Teeny Tiny Trucks line did and newarly 3 times as much as the Basic Blocks line's sales. A. the Teeny Tiny Trucks line did and newarly 3 times as much as the Basic Blocks line's sales B. the Teeny Tiny Trucks did and nearly 3 times what the Basic Blocks' sales were C. the Teeny Tiny Trucks line sold and nearly 3 times as much as Basic Blocks' sales. D. the Teeny Tiny Trucks line and nearly 3 times more than Basic Blocks' sales E. the Teeny Tiny Trucks line and nearly 3 times more than the Basic Blocks line If you have any questions New! Intern Joined: 16 Apr 2010 Posts: 34 Followers: 0 Kudos [?]: 9 [0], given: 1 ### Show Tags 29 Apr 2010, 10:05 bupbebeo wrote: The Dancing Doll line sold slightly more than $3.5 million worth of", "goodwill from the Pixar purchase was added to Disney’s 2005 total goodwill (giving a total of \\$22.531 billion), why you think Disney’s 2007 total goodwill was only \\$22.085 billion? Sales0 Views116", "toys during the holiday season two years ago--- This hold both side of the conversation. More revenue generated but not significant increase in the number of toys sold. So this indeed resolve the paradox. Take Away: Resolve the paradox only gives you facts and facts are indisputable. So any answer choice that validate one of the facts but invalidate another one is always going to be wrong. Right answer should explain you how this situation came into existence. Moderator Joined: 01 Sep 2010 Posts: 3033 Followers: 768 Kudos [?]: 6341 [2] , given: 991 Re: Toys4Them, an online toy merchant, generated$220 million in [#permalink] ### Show Tags 16 Jan 2013, 15:17 2 KUDOS happy to clarify When you deal with such question (indeed tough) try to understand really the argument. Here we have an increase of revenue from one year to another year and a diminish of toys sold E says that Toys4Them heavily discounted its toys during the holiday season two years ago (i.e 2012 during christmas holidays, could be an option) then the next year we have an increase of revenue BUT the toys are not purchased because people already have. May be they buy other stuffs from other compnany but not the same. As such, we end with more money in out bank account (the" ]

[ "my warehouse and $5$ stores approach me, all wanting to buy $300$ toys. I don’t have $1500$ toys, so I decide to give the customers an equal number of toys based on what I have. Now $1272 \\div 5$ gives us $254.4$, or $254\\ R\\ 2$. Great, each customer can have $254$ toys, with two toys left over I can keep. But wait, the toys are in boxes of $24$, so that can’t work! I cannot ship $254$ toys, because that is 10 whole boxes, plus 14 toys from an open box. The calculation has to be done with boxes, not with toys. In fact I have 53 boxes of toys, and I can give the customers 10 boxes each, and have 3 boxes left over. So the remainder is $3\\times 24 = 72$ toys. In other words, in a real and useful sense, the remainder of dividing $1272 \\div 5$ can be $240$ with remainder of $72$, when we require the result to be a multiple of 24. So, on to the question: could there be a remainder in multiplication? The answer is no. A remainder is a special way of expressing the error in a calculation, peculiar to inexact division. We do not have to use a remainder to express the error in an inexact division:", "level, and there are 24 toys in a box. Customers must get a whole box; we do not divide boxes. Boxes are only opened in the retail store to sell individual toys. Now suppose I have $1272$ toys in my warehouse and $5$ stores approach me, all wanting to buy $300$ toys. I don't have $1500$ toys, so I decide to give the customers an equal number of toys based on what I have. Now $1272 \\div 5$ gives us $254.4$, or $254\\ R\\ 2$. Great, each customer can have $254$ toys, with two toys left over I can keep. But wait, the toys are in boxes of $24$, so that can't work! I cannot ship $254$ toys, because that is 10 whole boxes, plus 14 toys from an open box. The calculation has to be done with boxes, not with toys. In fact I have 53 boxes of toys, and I can give the customers 10 boxes each, and have 3 boxes left over. So the remainder is $3\\times 24 = 72$ toys. In other words, in a real and useful sense, the remainder of dividing $1272 \\div 5$ can be $240$ with remainder of $72$, when we require the result to be a multiple of 24. So, on to the question: could there be a remainder", "of a week from a toy shop. Give the following information regarding the sale of toys. 1. Find the number of toys sold on each day of the week. 2. Find the number of toys sold during the week. 3. On which day was the maximum number of toys sold? 4. On which day was the minimum number of toys sold? 5. What is the difference in the number of toys sold on Saturday and Tuesday? According to the pictograph, 1 image of a toy refers to 20 toys The number of toys for each day is Monday = 4 Tuesday = 2.5 Wednesday = 3 Thursday = 1.25 Friday = 1.5 Saturday = 6 Sunday = 6.5 Next, calculate the number of toys sold each day by multiplying sales with the key. Monday $= 4 \\times 20 = 80$ Tuesday $= 2.5 \\times 20 = 50$ Wednesday $= 3 \\times 20 = 60$ Thursday $= 1.25 \\times 20 = 25$ Friday $= 1.5 \\times 20 = 30$ Saturday $= 6 \\times 20 = 120$ Sunday $= 6.5 \\times 20 = 130$ The last step is to answer the given questions. (1) Find the number of toys sold on each day of the week. The number of toys sold on each day of the week is: Monday =", "\\left(\\sum\\limits_{j=1}^{19}{19 \\choose j}\\right)=137438167041$ 4. ## Re: choosing toys! Originally Posted by MathoMan $\\left(\\sum\\limits_{k=1}^{18}{ 18 \\choose k}\\right) \\cdot \\left(\\sum\\limits_{j=1}^{19}{19 \\choose j}\\right)=137438167041$ You can get the same answer from $\\left( {2^{18} - 1} \\right)\\left( {2^{19} - 1} \\right)$ 5. ## Re: choosing toys! Thank you! That helps! The first answer you gave I had not seen before. The second one, I have. It makes sense now. thank you so much for all your help!!", "either a$40 toy or a $20 toy. How many$20 toys did the toy store sell? 1) The average price of the toys sold yesterday was $35. (2) The total price of the 20 toys sold yesterday was between$650 and $750. So we know total toys are 20, say a of$20 and b of $40.. Let's see the statements.. (1) The average price of the toys sold yesterday was$35. Using weighted average method, we can say that the ratio of b/a=(35-20)/(40-35)=15/5=3.. B=3a.... a+b=20....3a+a=20....4a=20....A=5.. B=15 Sufficient (2) The total price of the 20 toys sold yesterday was between $650 and$750. 20a+40b>=660..... a+2b>=33.... a+2(20-a)=>33...a+40-2a>=33.... a<=7 20a+40b<=740..... a+2b<=37.... a+2(20-a)=<37...a+40-2a<=37.... a>=3... So various possibilities 3,4,5,6 or 7.. Insufficient A Another, maybe quicker way to look at S(2) with inequalities is: $$650<40b+20a<750$$ From the question stem we know that a+b=20 so we can substitute b to find a. $$650<40*(20-a)+20a<750$$ $$650<-800-40a+20a<750$$ $$-150<-20a<-50$$ $$150>20a>50$$ $$7.5>a>2.5$$ Since a need to be an integer --> $$7=>a=>3$$ _________________ Flashcards Quant + Verbal:https://gmatclub.com/forum/gmat-flashcards-108651.html Thursdays with Ron:https://gmatclub.com/forum/manhattan-s-thursdays-with-ron-consolidated-video-index-223612.html#p2138753 Re: A certain toy store sold 20 toys yesterday, each of which was either a &nbs [#permalink] 26 Sep 2018, 07:46 Display posts from previous: Sort by", "743 views Three sisters $(R, S,$ and $T)$ received a total of $24$ toys during Christmas. The toys were initially divided among them in a certain proportion. Subsequently, $R$ gave some toys to $S$ which doubled the share of $S$. Then $S$ in turn gave some of her toys to $T$, which doubled $T’$s share. Next, some of $T’$s toys were given to $R$, which doubled the number of toys that $R$ currently had. As a result of all such exchanges, the three sisters were left with equal number of toys. How many toys did $R$ have originally? 1. $8$ 2. $9$ 3. $11$ 4. $12$ edited | 743 views +1 Try to solve it in reverse direction. +3 from equations it will not take more than 2 min r s t 2(r-s)=8 , 2s-t=8, 2t-(r-s)=8 solving these , r=11, s=7 t=6 +3 Initially :- r:s:t r-s : 2s : t r-s : 2s-t : 2t 2(r-s) : 2s-t : 2t - (r-s) Now, all have same number of toys = 24/3 = 8 2(r-s) = 2s-t = 2t -(r-s) = 8 Solving these, r=11, s=7, t=6 We can proceed in reverse order: $R\\quad S\\quad T$ $8\\quad 8\\quad 8$ and apply the given steps in reverse. We will get, $R\\quad S\\quad T$ $4 \\quad 8 \\quad 12$ $4", "get $300$, and if we add $229$ to it, we get $529$. Of course we know that $529\\div 3$ isn’t $100$. This is just an approximation which is good to within $100$. The actual number is in fact $100$-something. It’s not less than $100$, and it’s not $200$ or more. Now usually we do not do this. We don’t stop division at the hundreds or tens to take a funny remainder. We usually stop division at the units, and take the remainder there. This is because many objects that we work with in the real world cannot be divided beyond the unit. If we want to distribute 13 toys among 4 children, everyone gets three toys, and there is a toy left over. We don’t want to divide that toy into four, because it will be destroyed. Thus, we use a form of inexact mathematical division which gives us a model for this real-world constraint: integer division with a remainder. But non-unit remainders can be useful too. Suppose we are distributing a large number of toys at the wholesale level, and there are 24 toys in a box. Customers must get a whole box; we do not divide boxes. Boxes are only opened in the retail store to sell individual toys. Now suppose I have $1272$ toys in", "## Expected Number of Purchases to Complete a Set of Plastic Toys in Cereal Packets A breakfast cereal company decide to give a plastic toy away with each box of cereal. There are 16 toys to give away. What is the average number of purchases needed to collect the set? We can solved this problem using matrices. Someone may have anywhere between 0 and 16 different toys. We may represent this by saying there are 17 states. If someone has $r$ different toys. then the probability that he will find a toy he does not have in the next purchare, and move onto the state $r+1$ is $\\frac{16-r}{16}$ . We can represent this my a square transition matrix with 17 rows and 17 columns. $T=\\frac{1}{16} \\left( \\begin{array}{ccccccc} 0 & 16 & 0 & 0 & \\ldots & \\ldots & \\vdots \\\\ 0 & 1 & 15 & 0 & \\ldots & \\ldots & \\vdots \\\\ 0 & 0 & 2 & 14 & \\ldots & \\ldots & \\vdots \\\\ \\vdots & \\ldots & \\ldots & 3 & \\ldots & \\ldots & \\vdots \\\\ \\vdots & \\ldots & \\dots & \\dots & \\ldots & 15 & 1 \\\\ \\vdots & \\ldots & \\dots & \\dots & \\ldots & 0 & 1 \\end{array} \\right)$ $T=\\frac{1}{16} \\left( \\begin{array}{ccccccc} 1 & -1", "be \"IT\" and Total be \"T\" so, IT * 20/100 = 10/100 * T, therefore IT = 50/100 * T, so 50% Re: In a certain toy store, 20% of the international branded toys are the &nbs [#permalink] 27 Aug 2018, 05:19 Display posts from previous: Sort by", "is far easier to work out. Thera are $3!$ combinatioms of all different toys, each of which can be either Toy1 or Toy2, so we have to multiply by $2^3$ and we get $48$ combinations. The total number of possible combinations is $6×5×4$ so the solution is $72/120=13/20$ Successful situations: Either only one guy gets $2$ of the same thing or everyone gets $2$ of the same thing. One guy gets 2 of the same thing: $3\\times3=9$ ways Everyone gets 2 of the same thing: $3!=6$ ways Total number of ways: The above two scenarios, and the case where no one gets the pair, which is also $3!=6$ ways. So total number of ways is $6+6+9=21$. Answer: $\\frac{Successful}{Total}=\\frac{15}{21}=\\frac{5}{7}$. • think that the first one picks one toy... Then he has a chance 1/5 of the other toys to be the pair of the first he got... Already (1/5)=20%... If he didn't make it... The other two players has 0%? – koleygr Jan 21 '18 at 11:46 • I found the correct solution: 5/7 – QuIcKmAtHs Jan 21 '18 at 14:18 • I also listed out to check – QuIcKmAtHs Jan 21 '18 at 14:18 • The total number of equally likely ways to distribute the toys is $\\frac{6!}{2\\cdot 2 \\cdot2}=90$ – kludg Jan 21 '18 at 14:50 Someone", "27th 2012, 02:57 PM MathoMan Re: choosing toys! $\\left(\\sum\\limits_{k=1}^{18}{ 18 \\choose k}\\right) \\cdot \\left(\\sum\\limits_{j=1}^{19}{19 \\choose j}\\right)=137438167041$ • September 27th 2012, 03:29 PM Plato Re: choosing toys! Quote: Originally Posted by MathoMan $\\left(\\sum\\limits_{k=1}^{18}{ 18 \\choose k}\\right) \\cdot \\left(\\sum\\limits_{j=1}^{19}{19 \\choose j}\\right)=137438167041$ You can get the same answer from $\\left( {2^{18} - 1} \\right)\\left( {2^{19} - 1} \\right)$ • September 27th 2012, 03:31 PM jthomp18 Re: choosing toys! Thank you! That helps! The first answer you gave I had not seen before. The second one, I have. It makes sense now. thank you so much for all your help!!", "If the list price of a toy is reduced by Rs. 2, Question: If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy. Solution: Let the original list price of the toy be Rs. . Then, the number of toys brought for Rs. $360=\\frac{360}{x}$ According to question, reduced list price of the toys $=$ Rs. $(x-2)$. Therefore, the number of toys brought for Rs. $360=\\frac{360}{x-2}$ It is given that $\\frac{360}{x-2}-\\frac{360}{x}=2$ $\\frac{360 x-360(x-2)}{(x-2) x}=2$ $\\frac{720}{(x-2) x}=2$ $2\\left(x^{2}-2 x\\right)=720$ $\\left(x^{2}-2 x\\right)=360$ $x^{2}-2 x-360=0$ $x^{2}+18 x-20 x-360=0$ $x(x+18)-20(x+18)=0$ $(x+18)(x-20)=0$ Because cannot be negative. Thus, $x=20$ is the require solution. Therefore, the original list price of the toy be $x=$ Rs. 20", "2,096 views Three sisters $(R, S,$ and $T)$ received a total of $24$ toys during Christmas. The toys were initially divided among them in a certain proportion. Subsequently, $R$ gave some toys to $S$ which doubled the share of $S$. Then $S$ in turn gave some of her toys to $T$, which doubled $T’$s share. Next, some of $T’$s toys were given to $R$, which doubled the number of toys that $R$ currently had. As a result of all such exchanges, the three sisters were left with equal number of toys. How many toys did $R$ have originally? 1. $8$ 2. $9$ 3. $11$ 4. $12$ Try to solve it in reverse direction. from equations it will not take more than 2 min r s t 2(r-s)=8 , 2s-t=8, 2t-(r-s)=8 solving these , r=11, s=7 t=6 Initially :- r:s:t r-s : 2s : t r-s : 2s-t : 2t 2(r-s) : 2s-t : 2t - (r-s) Now, all have same number of toys = 24/3 = 8 2(r-s) = 2s-t = 2t -(r-s) = 8 Solving these, r=11, s=7, t=6 thanks Sir ji :) We can proceed in reverse order: $R\\quad S\\quad T$ $8\\quad 8\\quad 8$ and apply the given steps in reverse. We will get, $R\\quad S\\quad T$ $4 \\quad 8 \\quad 12$ $4 \\quad 14 \\quad", "80, Tuesday = 50, Wednesday = 60, Thursday = 25, Friday = 30, Saturday = 120 and Sunday = 130 (2) Find the number of toys sold during the week. The number of toys sold during the week = 80 + 50 + 60 + 25 + 30 + 120 + 130 = 495 (3) On which day was the maximum number of toys sold? The maximum number of toys sold was on Sunday (4) On which day was the minimum number of toys sold? The minimum number of toys sold was on Thursday (5) What is the difference in the number of toys sold on Saturday and Tuesday? The number of toys sold on Saturday is 120 The number of toys sold on Tuesday is 50 The difference in the number of toys sold on Saturday and Tuesday is $120 – 50 = 70$. ## How to Make a Pictograph The different steps to make a pictograph are given below: Step 1 Collect the Data: The first step in making a pictograph is the collection of relevant information, which we want to represent. Once the data is collected, make a table or a list of data. Step 2 Select the Symbol or Images: To represent the data, pick any images/pictures, or symbols. For example, if the data", "Posts: 2323 Re: In a certain toy store, 20% of the international branded toys are the [#permalink] ### Show Tags 30 Jul 2018, 04:41 Solution Given: • In a certain toy store, 20% of the international branded toys are the only premium category toys. • Premium toys make up 10% of the total number of toys in the store To find: • What percent of toys in the store are of international brands Approach and Working: If we assume that there are total 100 toys in the store, then we can say, • Premium toys = 10% of 100 = 10 Now, these 10 toys are 20% of the international branded toys. So, if we assume international branded toys are T in number, then we can say • 20% of T = 10 Or, $$\\frac{T}{5}$$ = 10 Or, T = 50 Therefore, international branded toys as a percentage of total toys = $$\\frac{50}{100} * 100$$ = 50% Hence, the correct answer is option D. _________________ Number Properties | Algebra |Quant Workshop Success Stories Guillermo's Success Story | Carrie's Success Story Ace GMAT quant Articles and Question to reach Q51 | Question of the week Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2 Word Problems – Percentage 1 | Percentage 2", "the number of toys produced in a day is 25, the cost of production of each toy will be 55 - 25 = 30Rs. If the number of toys produced in a day is 30, the cost of production of each toy will be 55 - 30 = 25Rs. Thus, the number of toys produced on that day are either 25 or 30. Q3. Find two numbers whose sum is 27 and product is 182. Open in new tab link ###### Answer : Let the first number be $$x$$ Thus the second number can be represented as $$27 - x$$. Since the product of the two numbers is 182, $$x(27 - x) = 182$$ $$27x - x^2 - 182 = 0$$ $$x^2 - 27x + 182 = 0$$ (Splitting -27x as -13x - 14x ) $$x^2 - 13x - 14x + 182 = 0$$ $$x(x - 13) - 14(x - 13) = 0$$ $$(x - 14)(x - 13) = 0$$ The roots of this equation are the values of x for which $$(x - 14)(x - 13) = 0$$ which are, $$x - 14 = 0$$ or $$x - 13 = 0$$ Thus, $$x = 14$$ or $$x = 13$$ If the first number is 14, the second number will be 27 - 14 = 13. If the", "of toys in the store, what percent of toys in the store are of international brands? A. 10% B. 20% C. 40% D. 50% E. 60% Assume 100 toys. 10 percent are premium: All 10 premium toys are international 20% of all international toys, I, are premium So 10 toys represent 20% of all international toys $$10 = .20I$$ $$I=\\frac{10}{0.2}=\\frac{100}{2}=50$$ International toys: 50 Total toys: 100 International toys are 50% of all toys. _________________ The only thing more dangerous than ignorance is arrogance. -- Albert Einstein e-GMAT Representative Joined: 04 Jan 2015 Posts: 2063 Re: In a certain toy store, 20% of the international branded toys are the [#permalink] ### Show Tags 30 Jul 2018, 05:41 Solution Given: • In a certain toy store, 20% of the international branded toys are the only premium category toys. • Premium toys make up 10% of the total number of toys in the store To find: • What percent of toys in the store are of international brands Approach and Working: If we assume that there are total 100 toys in the store, then we can say, • Premium toys = 10% of 100 = 10 Now, these 10 toys are 20% of the international branded toys. So, if we assume international branded toys are T in number, then we can", "# GATE2011 GG: GA-8 1.3k views Three sisters $(R, S,$ and $T)$ received a total of $24$ toys during Christmas. The toys were initially divided among them in a certain proportion. Subsequently, $R$ gave some toys to $S$ which doubled the share of $S$. Then $S$ in turn gave some of her toys to $T$, which doubled $T’$s share. Next, some of $T’$s toys were given to $R$, which doubled the number of toys that $R$ currently had. As a result of all such exchanges, the three sisters were left with equal number of toys. How many toys did $R$ have originally? 1. $8$ 2. $9$ 3. $11$ 4. $12$ edited 1 Try to solve it in reverse direction. 4 from equations it will not take more than 2 min r s t 2(r-s)=8 , 2s-t=8, 2t-(r-s)=8 solving these , r=11, s=7 t=6 7 Initially :- r:s:t r-s : 2s : t r-s : 2s-t : 2t 2(r-s) : 2s-t : 2t - (r-s) Now, all have same number of toys = 24/3 = 8 2(r-s) = 2s-t = 2t -(r-s) = 8 Solving these, r=11, s=7, t=6 0 thanks Sir ji :) We can proceed in reverse order: $R\\quad S\\quad T$ $8\\quad 8\\quad 8$ and apply the given steps in reverse. We will get, $R\\quad S\\quad T$", "the toys are Premium (10 toys) and only International Brands offer premium toys. So, $$A=10$$ $$D=0$$ $$G=10$$ Because 20% of the International Brands offer Premium toys, total number of premium toys is equal to - $$x*\\frac{20}{100} = 10$$ $$x=50$$ So $$C=50$$ $$B=40$$ $$C=50$$ Therefore, 50% of the toys are of an International Brand. Option D is correct Attachments egmat ques.png [ 2.46 KiB | Viewed 1024 times ] _________________ Regards, Vaibhav Sky is the limit. 800 is the limit. ~GMAC Senior SC Moderator Joined: 22 May 2016 Posts: 2218 In a certain toy store, 20% of the international branded toys are the [#permalink] ### Show Tags 26 Jul 2018, 14:18 EgmatQuantExpert wrote: In a certain toy store, 20% of the international branded toys are the only premium category toys. If premium toys make up 10% of the total number of toys in the store, what percent of toys in the store are of international brands? A. 10% B. 20% C. 40% D. 50% E. 60% Assume 100 toys. 10 percent are premium: All 10 premium toys are international 20% of all international toys, I, are premium So 10 toys represent 20% of all international toys $$10 = .20I$$ $$I=\\frac{10}{0.2}=\\frac{100}{2}=50$$ International toys: 50 Total toys: 100 International toys are 50% of all toys. e-GMAT Representative Joined: 04 Jan 2015", "Sam purchased 20 dozens of toys at the rate of Rs. 375 Per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit ? $\\par&space;C.P.&space;of&space;1&space;toy&space;=&space;Rs.$ $\\left&space;(&space;\\frac{375}{12}&space;\\right&space;)$ $\\par&space;=&space;Rs.&space;31.25$ $\\par&space;S.P.&space;of&space;1&space;toy&space;=&space;Rs.33$ $\\par&space;Profit\\%&space;=$ $\\left(\\frac{1.75}{31.25}\\times&space;100\\right)\\%$ $=&space;\\frac{28}{5}\\%&space;=&space;5.6\\%$" ]

[ "6 figures released each month. So, for 1 year, 72 toys will be released. Now, The number of years that the company released all the toys = 214 ÷ 72 Now, 214 ÷ 72 = 2 R 70 Hence, from the above, We can conclude that The number of months that the company released all the toys = About 35 years The number of years that the company released all the toys = About 3 years ### Use Partial Quotients with a Remainder Homework & Practice 5.5 Use partial quotients to divide. Question 1. $$\\sqrt [ 4 ]{ 67 }$$ Answer: 16 R 3 Explanation: Using Distributive property of partial quotients, 67 ÷ 4 = ( 60 + 4 ) ÷ 4 = ( 60 ÷ 4 ) + ( 4 ÷ 4 ) = 15 + 1 = 16 R 3 Hence, 67 ÷ 4 = 16 R 3 Question 2. $$\\sqrt [ 2 ]{ 715 }$$ Answer: 357 R 1 Explanation: Using Distributive property of partial quotients, 715 ÷ 2 = ( 700 + 14 ) ÷ 2 = ( 700 ÷ 2 ) + ( 14 ÷ 2 ) = 350 + 7 = 357 R 1 Hence, 715 ÷ 2 = 357 R 1 Question 3. $$\\sqrt [ 5 ]{ 1,308 }$$ Answer: 261", "is far easier to work out. Thera are $3!$ combinatioms of all different toys, each of which can be either Toy1 or Toy2, so we have to multiply by $2^3$ and we get $48$ combinations. The total number of possible combinations is $6×5×4$ so the solution is $72/120=13/20$ Successful situations: Either only one guy gets $2$ of the same thing or everyone gets $2$ of the same thing. One guy gets 2 of the same thing: $3\\times3=9$ ways Everyone gets 2 of the same thing: $3!=6$ ways Total number of ways: The above two scenarios, and the case where no one gets the pair, which is also $3!=6$ ways. So total number of ways is $6+6+9=21$. Answer: $\\frac{Successful}{Total}=\\frac{15}{21}=\\frac{5}{7}$. • think that the first one picks one toy... Then he has a chance 1/5 of the other toys to be the pair of the first he got... Already (1/5)=20%... If he didn't make it... The other two players has 0%? – koleygr Jan 21 '18 at 11:46 • I found the correct solution: 5/7 – QuIcKmAtHs Jan 21 '18 at 14:18 • I also listed out to check – QuIcKmAtHs Jan 21 '18 at 14:18 • The total number of equally likely ways to distribute the toys is $\\frac{6!}{2\\cdot 2 \\cdot2}=90$ – kludg Jan 21 '18 at 14:50 Someone", "“So many children can't have expensive toys, and I feel a real obligation to them. I'm appalled when I go Christmas shopping and$60 to $80 for a toy is nothing.\" In 2008, Slinkys cost$4 to $5, and Slinky Dogs about$20.[10] In 1998, James Industries was sold to Poof Products, Inc. of Plymouth, Michigan, a manufacturer of foam sports balls.[4][10] Slinky continued production in Hollidaysburg.[4] In 2003, James Industries merged with Poof Products, Inc. to create Poof-Slinky, Inc. Betty James died of congestive heart failure in November 2008, aged 90, after having served as president of James Industries from 1960 to 1998.[10] Over 300 million Slinkys have been sold between 1945 and 2005, and the original Slinky is still a bestseller.[4][7] ## Physical properties The rules that govern the mechanics of a slinky are due to Hooke's law and the effects of gravitation. ### Period of oscillation Due to simple harmonic motion the period of oscillation of a dangling slinky is ${\\displaystyle T=2\\pi {\\sqrt {\\frac {m}{k}}}.}$ Where T is the time of the period of oscillation, m is the mass of the slinky and k is the spring constant of the slinky. ### Equilibrium In the state of equilibrium of a slinky, all net force is cancelled throughout the entire slinky. This results in a stationary slinky with zero", "Posts: 2323 Re: In a certain toy store, 20% of the international branded toys are the [#permalink] ### Show Tags 30 Jul 2018, 04:41 Solution Given: • In a certain toy store, 20% of the international branded toys are the only premium category toys. • Premium toys make up 10% of the total number of toys in the store To find: • What percent of toys in the store are of international brands Approach and Working: If we assume that there are total 100 toys in the store, then we can say, • Premium toys = 10% of 100 = 10 Now, these 10 toys are 20% of the international branded toys. So, if we assume international branded toys are T in number, then we can say • 20% of T = 10 Or, $$\\frac{T}{5}$$ = 10 Or, T = 50 Therefore, international branded toys as a percentage of total toys = $$\\frac{50}{100} * 100$$ = 50% Hence, the correct answer is option D. _________________ Number Properties | Algebra |Quant Workshop Success Stories Guillermo's Success Story | Carrie's Success Story Ace GMAT quant Articles and Question to reach Q51 | Question of the week Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2 Word Problems – Percentage 1 | Percentage 2", "of toys in the store, what percent of toys in the store are of international brands? A. 10% B. 20% C. 40% D. 50% E. 60% Assume 100 toys. 10 percent are premium: All 10 premium toys are international 20% of all international toys, I, are premium So 10 toys represent 20% of all international toys $$10 = .20I$$ $$I=\\frac{10}{0.2}=\\frac{100}{2}=50$$ International toys: 50 Total toys: 100 International toys are 50% of all toys. _________________ The only thing more dangerous than ignorance is arrogance. -- Albert Einstein e-GMAT Representative Joined: 04 Jan 2015 Posts: 2063 Re: In a certain toy store, 20% of the international branded toys are the [#permalink] ### Show Tags 30 Jul 2018, 05:41 Solution Given: • In a certain toy store, 20% of the international branded toys are the only premium category toys. • Premium toys make up 10% of the total number of toys in the store To find: • What percent of toys in the store are of international brands Approach and Working: If we assume that there are total 100 toys in the store, then we can say, • Premium toys = 10% of 100 = 10 Now, these 10 toys are 20% of the international branded toys. So, if we assume international branded toys are T in number, then we can", "my warehouse and $5$ stores approach me, all wanting to buy $300$ toys. I don’t have $1500$ toys, so I decide to give the customers an equal number of toys based on what I have. Now $1272 \\div 5$ gives us $254.4$, or $254\\ R\\ 2$. Great, each customer can have $254$ toys, with two toys left over I can keep. But wait, the toys are in boxes of $24$, so that can’t work! I cannot ship $254$ toys, because that is 10 whole boxes, plus 14 toys from an open box. The calculation has to be done with boxes, not with toys. In fact I have 53 boxes of toys, and I can give the customers 10 boxes each, and have 3 boxes left over. So the remainder is $3\\times 24 = 72$ toys. In other words, in a real and useful sense, the remainder of dividing $1272 \\div 5$ can be $240$ with remainder of $72$, when we require the result to be a multiple of 24. So, on to the question: could there be a remainder in multiplication? The answer is no. A remainder is a special way of expressing the error in a calculation, peculiar to inexact division. We do not have to use a remainder to express the error in an inexact division:", "## Expected Number of Purchases to Complete a Set of Plastic Toys in Cereal Packets A breakfast cereal company decide to give a plastic toy away with each box of cereal. There are 16 toys to give away. What is the average number of purchases needed to collect the set? We can solved this problem using matrices. Someone may have anywhere between 0 and 16 different toys. We may represent this by saying there are 17 states. If someone has $r$ different toys. then the probability that he will find a toy he does not have in the next purchare, and move onto the state $r+1$ is $\\frac{16-r}{16}$ . We can represent this my a square transition matrix with 17 rows and 17 columns. $T=\\frac{1}{16} \\left( \\begin{array}{ccccccc} 0 & 16 & 0 & 0 & \\ldots & \\ldots & \\vdots \\\\ 0 & 1 & 15 & 0 & \\ldots & \\ldots & \\vdots \\\\ 0 & 0 & 2 & 14 & \\ldots & \\ldots & \\vdots \\\\ \\vdots & \\ldots & \\ldots & 3 & \\ldots & \\ldots & \\vdots \\\\ \\vdots & \\ldots & \\dots & \\dots & \\ldots & 15 & 1 \\\\ \\vdots & \\ldots & \\dots & \\dots & \\ldots & 0 & 1 \\end{array} \\right)$ $T=\\frac{1}{16} \\left( \\begin{array}{ccccccc} 1 & -1", "A toy store regularly sells all stock at a discount of 20 15 25 Oct 2010, 04:40 5 In each production lot for a certain toy, 25 percent of the 7 27 Nov 2007, 15:21 Display posts from previous: Sort by", "I go Christmas shopping and$60 to $80 for a toy is nothing.\" In 2008, Slinkys cost$4 to $5, and Slinky Dogs about$20.[12] In 1998 James Industries was sold to Poof Products, Inc. of Plymouth, Michigan, a manufacturer of foam sports balls.[6][12] Slinky continued production in Hollidaysburg.[6] In 2003, James Industries merged with Poof Products, Inc., to create Poof-Slinky, Inc. Betty James died of congestive heart failure in November 2008, age 90, after having served as president of James Industries from 1960 to 1998.[12] Over 300 million Slinkys have been sold between 1945 and 2005, and the original Slinky is still a bestseller.[6][9] ## Physical properties The rules that govern the mechanics of a slinky are Hooke's law and the effects of gravitation. ### Period of oscillation Due to simple harmonic motion the period of oscillation of a dangling slinky is ${\\displaystyle T=2\\pi {\\sqrt {\\frac {m}{k}}},}$ where T is the time of the period of oscillation, m is the mass of the slinky and k is the spring constant of the slinky. ### Equilibrium In the state of equilibrium of a slinky, all net force is cancelled throughout the entire slinky. This results in a stationary slinky with zero velocity. As the positions of each part of the slinky is governed by the slinky's mass, the force of gravity", "level, and there are 24 toys in a box. Customers must get a whole box; we do not divide boxes. Boxes are only opened in the retail store to sell individual toys. Now suppose I have $1272$ toys in my warehouse and $5$ stores approach me, all wanting to buy $300$ toys. I don't have $1500$ toys, so I decide to give the customers an equal number of toys based on what I have. Now $1272 \\div 5$ gives us $254.4$, or $254\\ R\\ 2$. Great, each customer can have $254$ toys, with two toys left over I can keep. But wait, the toys are in boxes of $24$, so that can't work! I cannot ship $254$ toys, because that is 10 whole boxes, plus 14 toys from an open box. The calculation has to be done with boxes, not with toys. In fact I have 53 boxes of toys, and I can give the customers 10 boxes each, and have 3 boxes left over. So the remainder is $3\\times 24 = 72$ toys. In other words, in a real and useful sense, the remainder of dividing $1272 \\div 5$ can be $240$ with remainder of $72$, when we require the result to be a multiple of 24. So, on to the question: could there be a remainder", "it may result in higher revenue , even though the number of toys sold did not increase significantly last year over the previous year=> hence it resolves the paradox . thanks that was my reasoning as well, hence I eliminated e and chose c. And everyone else makes valid points as well but I definitely thought like you for option \"e.\" Re: Toys4Them, an online toy merchant, generated$220 million in [#permalink] 01 Oct 2016, 10:44 Similar topics Replies Last post Similar Topics: 3 Internal Memorandum from Toy Supply Company 11 03 Jan 2016, 09:01 2 A toy company's engineering department developed a new model 4 09 Dec 2013, 10:44 1 Child s World, a chain of toy stores, has relied on a 12 30 Dec 2009, 14:21 Consumer advocate: The toy-labeling law should require 14 09 Oct 2007, 06:31 Archaeologists have found wheeled ceramic toys made by the 4 07 May 2007, 20:02 Display posts from previous: Sort by", "get $300$, and if we add $229$ to it, we get $529$. Of course we know that $529\\div 3$ isn’t $100$. This is just an approximation which is good to within $100$. The actual number is in fact $100$-something. It’s not less than $100$, and it’s not $200$ or more. Now usually we do not do this. We don’t stop division at the hundreds or tens to take a funny remainder. We usually stop division at the units, and take the remainder there. This is because many objects that we work with in the real world cannot be divided beyond the unit. If we want to distribute 13 toys among 4 children, everyone gets three toys, and there is a toy left over. We don’t want to divide that toy into four, because it will be destroyed. Thus, we use a form of inexact mathematical division which gives us a model for this real-world constraint: integer division with a remainder. But non-unit remainders can be useful too. Suppose we are distributing a large number of toys at the wholesale level, and there are 24 toys in a box. Customers must get a whole box; we do not divide boxes. Boxes are only opened in the retail store to sell individual toys. Now suppose I have $1272$ toys in", "# CF1023B Pair of Toys • 92通过 • 158提交 • 题目来源 • 评测方式 RemoteJudge • 标签 • 难度 普及/提高- • 时空限制 1000ms / 256MB • 提示：收藏到任务计划后，可在首页查看。 ## 题意翻译 给定n个玩具要你选出两个玩具求出k的价值。 第i个玩具的价值为i。 样例1：从8个玩具里选出两个使价值为5. 有以下方案：（1 4）（2 3）。 若是没有选择方案，输出0 补充：玩具A与玩具B 和 玩具B和玩具A 是同一种选择 ## 题目描述 Tanechka is shopping in the toy shop. There are exactly $n$ toys in the shop for sale, the cost of the $i$ -th toy is $i$ burles. She wants to choose two toys in such a way that their total cost is $k$ burles. How many ways to do that does she have? Each toy appears in the shop exactly once. Pairs $(a, b)$ and $(b, a)$ are considered equal. Pairs $(a, b)$ , where $a=b$ , are not allowed. ## 输入输出格式 输入格式： The first line of the input contains two integers $n$ , $k$ ( $1 \\le n, k \\le 10^{14}$ ) — the number of toys and the expected total cost of the pair of toys. 输出格式： Print the number of ways to choose the pair of toys satisfying the condition above. Print 0, if Tanechka can choose no pair of toys in such a way that their total cost is $k$ burles. ## 输入输出样例 输入样例#1： 复制 8 5 输出样例#1： 复制 2 输入样例#2： 复制 8 15 输出样例#2： 复制 1 输入样例#3： 复制 7 20 输出样例#3： 复制", "\\left(\\sum\\limits_{j=1}^{19}{19 \\choose j}\\right)=137438167041$ 4. ## Re: choosing toys! Originally Posted by MathoMan $\\left(\\sum\\limits_{k=1}^{18}{ 18 \\choose k}\\right) \\cdot \\left(\\sum\\limits_{j=1}^{19}{19 \\choose j}\\right)=137438167041$ You can get the same answer from $\\left( {2^{18} - 1} \\right)\\left( {2^{19} - 1} \\right)$ 5. ## Re: choosing toys! Thank you! That helps! The first answer you gave I had not seen before. The second one, I have. It makes sense now. thank you so much for all your help!!", "# In a toy store, the ratio of dolls to teddy bears is 9:3. If the store has 240 dolls, how many teddy bears are in the store? $80$ teddy bears are in the store. Let \"D\" be the number of dolls and \"T\" be the number of teddy bears$D : T = 9 : 3 \\mathmr{and} \\frac{D}{T} = \\frac{9}{3} \\mathmr{and} \\frac{D}{T} = 3 \\therefore T = \\frac{D}{3} = \\frac{240}{3} = 80$[Ans]", "toys in the store, what percent of toys in the store are of international brands? A. 10% B. 20% C. 40% D. 50% E. 60% Let's assume the total international branded toys are 100. Since 20% of these toys are premium category toys, the total number of premium category toys is 20. Now, 20 is the total number of toys in the store, which is 10% of the total number of the store. The total number of toys in the store is 200. Therefore, the total percentage or toys that are of international brands are $$\\frac{100}{200} = 50$$% (Option D) _________________ You've got what it takes, but it will take everything you've got Manager Joined: 01 Jan 2018 Posts: 62 Re: In a certain toy store, 20% of the international branded toys are the [#permalink] ### Show Tags 26 Jul 2018, 01:16 0.2 * International=0.1 *total So,international = 50% of total toys. D Sent from my Redmi Note 3 using GMAT Club Forum mobile app _________________ +1 Kudos if you find this post helpful Intern Joined: 18 Jul 2018 Posts: 1 Re: In a certain toy store, 20% of the international branded toys are the [#permalink] ### Show Tags 26 Jul 2018, 01:19 1 0.20 × i (international) = p (premium) 0.2 i = 0.10 × T(Total )", "bought following the directions from the Buying Toys list? Find the total cost for the numbers you choose. Answer: It is given that The cost of each teddy bear is:$17 The cost of each scooter is: $28 The total number of toys are: 68 Now, Let there be an equal number of teddy bears and scooters So, The number of each toy = $$\\frac{68}{2}$$ = 34 So, The total cost of the toys = (34 × 17) + (34 × 28) =$1,530 Hence, from the above, We can conclude that The total cost for the number of toys you choose is: \\$1,530 Scroll to Top", "• # question_answer Ramya has 4 different toys and Saumya has 7 different toys. The number of ways in which they can exchange their toys, so that each keep her initial number of toys are A) 338 B) 329 C) 330 D) 331 When we keep the toys of Ramya and Saumya together, the problem is to fixed the number of ways in which Ramya can take 4 toys out of 11 toys, not including the number of ways in which she takes her original 4 toys which can be done in $^{11}{{C}_{4}}-1=\\frac{11!}{4!7!}-1=329$", "1.5y \\le 5500 \\ \\end$ .The store owner estimates that no more than 2000 toys will be sold every month and he does not plan to invest more than ,000 in inventory of these toys.How many units of each type of toys should be stocked in order to maximize his monthly total profit profit?The following videos gives examples of linear programming problems and how to test the vertices.Several word problems and applications related to linear programming are presented along with their solutions and detailed explanations.", "feel a real obligation to them. I'm appalled when I go Christmas shopping and $60 to$80 for a toy is nothing.\" In 2008, Slinkys cost $4 to$5, and Slinky Dogs about $20.[10] In 1998, James Industries was sold to Poof Products, Inc. of Plymouth, Michigan, a manufacturer of foam sports balls.[4][10] Slinky continued production in Hollidaysburg.[4] In 2003, James Industries merged with Poof Products, Inc. to create Poof-Slinky, Inc. Betty James died of congestive heart failure in November 2008, aged 90, after having served as president of James Industries from 1960 to 1998.[10] Over 300 million Slinkys have been sold between 1945 and 2005, and the original Slinky is still a bestseller.[4][7] ## Physical properties The rules that govern the mechanics of a slinky are due to Hooke's law and the effects of gravitation. ### Period of oscillation Due to simple harmonic motion the period of oscillation of a dangling slinky is ${\\displaystyle T=2\\pi {\\sqrt {\\frac {m}{k}}}.}$ Where T is the time of the period of oscillation, m is the mass of the slinky and k is the spring constant of the slinky. ### Equilibrium In the state of equilibrium of a slinky, all net force is cancelled throughout the entire slinky. This results in a stationary slinky with zero velocity. As the positions of each part of" ]

[ "3$ $\\overline{DF} = 6$", "x \\ \\overline y$", "+0 # HELP CAN SOMEONE HELP ME QUICK PLS -1 63 1 +36 What is $0.\\overline{2}\\cdot 6$? Express your answer as a reduced fraction Jan 15, 2020 edited by Paresh247 Jan 15, 2020", ", 6\\right)$", "$$1.1\\overline{36}$$ which means 1.136363636 . . .", "\\overline{6} \\text{N}$ $\\approx 29 \\text{N}$", "satisfy $\\overline{A} v = 0$.", "6 = y$.", "# What is 0.24 divided by 6? May 12, 2018 $0.04$ #### Explanation: Just divide as you would normally. If the divisor is a whole number there is no problem: $0.24 \\div 6 = \\text{ } 6 | \\underline{0.24}$ $\\textcolor{w h i t e}{\\times \\times \\times \\times \\times \\times} 0.04$ Working: $0 \\div 6 = 0$ Bring down the decimal point. $2 \\div 6 = 0$ $24 \\div 6 = 4$ $= 0.04$", "}$$ is 6.", "$X= \\overline A$ -", "= 6$ is excluded.", "because $22 \\div 18.75 = 1.17\\overline{3}$. You left a $17.\\overline{3}\\%$ tip, because $117.\\overline{3} - 100 = 17.\\overline{3}$.", "= 6$", "$6!$ by $6$ or simply $5!$.", "6^\\circ$ -", "10. $-68 \\div -2 = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$", "=​ $$79 {1\\ \\over 6}$$ % Free", "c) $7.\\overline{09}$. Discussion", "x - 6} = {(x-6)+6\\over x - 6}= 1+ {6\\over x - 6}$." ]

[ "# How do you convert 0.6 (6 being repeated) to a fraction? Mar 4, 2016 $0. \\overline{6} = \\frac{2}{3}$ #### Explanation: Multiply by $10$ to shift one place to the left and subtract the starting number to cancel out the repeating decimal. To do this multiply by $\\left(10 - 1\\right)$: $\\left(10 - 1\\right) \\cdot 0. \\overline{6} = 6. \\overline{6} - 0. \\overline{6} = 6$ Then divide both sides by $\\left(10 - 1\\right)$ to find: $0. \\overline{6} = \\frac{6}{10 - 1} = \\frac{6}{9} = \\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} \\cdot 2}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} \\cdot 3} = \\frac{2}{3}$", "+0 # using alegra prove that 0.565656... / 0.124242424... is equal in value to 560/123 +1 178 2 using alegra prove that 0.565656... / 0.124242424... is equal in value to 560/123 Guest Nov 12, 2017 #1 0 0.565656... / 0.124242424..... Multiply both top and bottom by 990. [990 x 0.56565656.... / 990 x 0.124242424.....] =560/123 Guest Nov 12, 2017 #2 +2075 0 In order to prove algabraically that $$\\frac{0.5656...}{0.12424...}=\\frac{560}{123}$$, let's try to convert the interminable decimals to fractions. I'll start with $$0.5656...$$ 1. Set the Repeating Decimal equal to a Variable! This is a farily simple step. $$0.\\overline{56}=x$$. Now, you're good to go! 2. Multiply Both Sides by 10 such until the Repeating Portion is the Whole Number In this case, if I multiply both sides by 100, which is 10^2, then the repeating portion will be to the left of the decimal point. $$56.\\overline{56}=100x$$ $$56.\\overline{56}$$ $$=100x$$ $$\\hspace{2mm}0.\\overline{56}$$ $$=\\hspace{7mm}x$$ $$56$$ $$=\\hspace{1mm}99x$$ 4. Solve for x $$56=99x$$ Divide by 99 on both sides. $$\\frac{56}{99}=x=0.\\overline{56}$$ Great! Now, let's convert the next one. $$0.1\\overline{24}=y$$ Now, multiply by multiples of ten to get the repeating portion to the left until the repeating part lines up. $$12.4\\overline{24}=100y$$ Now, subtract the two equations. $$12.4\\overline{24}$$ $$=100y$$ $$\\hspace{3mm}0.1\\overline{24}$$ $$=\\hspace{7mm}y$$ $$\\hspace{1mm}12.3$$ $$=\\hspace{2mm}99y$$ Now, solve for y. $$12.3=99y$$ Multiply by 10 on both sides to make the left", "the confusion the following representation is adapted. The number is given as $0.3\\overline{74}$$0.3 \\overline{74}$ The line over $74$$74$ represents that $74$$74$ is repeated. Convert $\\frac{1}{3}$$\\frac{1}{3}$ into a fraction. The answer is $0.\\overline{3}$$0. \\overline{3}$ Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. convert back to fraction How do we convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction? To convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction, the following steps are used $x=0.\\overline{4}$$x = 0. \\overline{4}$ $10x=4.\\overline{4}$$10 x = 4. \\overline{4}$ Subtracting the two equations $9x=4$$9 x = 4$ $x=\\frac{4}{9}$$x = \\frac{4}{9}$ examples Converting $2.\\overline{4}$$2. \\overline{4}$ into a fraction. $x=2.\\overline{4}$$x = 2. \\overline{4}$ $10x=24.\\overline{4}$$10 x = 24. \\overline{4}$ subtracting the two above $9x=22$$9 x = 22$ $x=\\frac{22}{9}$$x = \\frac{22}{9}$ $x=2\\frac{4}{9}$$x = 2 \\frac{4}{9}$ To convert $2.23\\overline{43}$$2.23 \\overline{43}$ into a fraction: $x=2.23\\overline{43}$$x = 2.23 \\overline{43}$ $100x=223.43\\overline{43}$$100 x = 223.43 \\overline{43}$ Subtracting the above, $99x=221.20$$99 x = 221.20$ $990x=2212$$990 x = 2212$ $x=\\frac{2212}{990}$$x = \\frac{2212}{990}$. Another method is as follows. $100x=223+0.\\overline{43}$$100 x = 223 + 0. \\overline{43}$ $10000x=223×100+43.\\overline{43}$$10000 x = 223 \\times 100 + 43. \\overline{43}$ Subtracting the two equations $9900x=223×\\left(100-1\\right)+43$$9900 x = 223 \\times \\left(100 - 1\\right) + 43$ $x=\\frac{223}{100}+\\frac{43}{9900}$$x = \\frac{223}{100} + \\frac{43}{9900}$ summary Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. Conversion of Repeating Decimals to equivalent", "# How do you solve y/9=-8? Apr 25, 2017 $y = - 72$ #### Explanation: To eliminate the fraction multiply both sides of the equation by 9, the denominator of the fraction. ${\\cancel{9}}^{1} \\times \\frac{y}{\\cancel{9}} ^ 1 = 9 \\times - 8$ $\\Rightarrow y = - 72$ Apr 25, 2017 $y = - 72$ #### Explanation: You multiply both sides with 9. Now you will have $\\frac{9 y}{9} = \\left(- 8\\right) \\cdot 9$ or $y = - 72$ It is your solution $y = - 72$", "as dividing two positive numbers. $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{CD}=\\frac{2}{4}$$ 2/4 can be simplified still $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{CD}=\\frac{1}{2}$$ Let's find the slope of $$\\overline{AB}$$: $$m=\\frac{y_2-y_1}{x_2-x_1}$$ I am reminding you of the slope formula again! $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{AB}=\\frac{n-5}{-1-3}$$ We can simplify the denominator, but the numerator will need something else $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{AB}=\\frac{n-5}{-4}$$ This is the slope of $$\\overline{AB}$$. What did I say above? I said that parallel lines have the same slope! That means that the slope of $$\\overline{CD}$$ equals the slope of $$\\overline{AB}$$: $$\\frac{-4}{1}*\\frac{1}{2}=\\frac{n-5}{-4}*\\frac{-4}{1}$$ Multiply by -4 on both sides to get rid of all the pesky fractions! $$-2=n-5$$ Add 5 to both sides $$3=n$$ Therefore, the coordinates of point A is (-1,3) TheXSquaredFactor Jun 11, 2017 #3 +2190 +3 Oh, I am sorry. I did not anwer the other questions that you asked. Here we go. Our ultimate goal is to find what value for n makes $$\\overline{AB}\\hspace{1mm}\\perp\\hspace{1mm}\\overline{CD}$$. Just like in the previous problem, find the slope of $$\\overline{CD}$$. Since I have already found the slope in the previous problem that I answered, I don't need to do that calculation again: $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{CD}=\\frac{1}{2}$$ We already know the slope of $$\\overline{AB}$$ because we calculated it in the previous problem as well: $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{AB}=\\frac{n-5}{-4}$$ The difference with this problen, however, is that we have to find the value for that makes the segment perpendicular! For a segment to be perpendicular,", "≠ 0. (p, q ϵ Z, q ≠ 0). (i) $$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}$$, Let x = 0.99999 … Multiplying by 10. 10x = 9.9999…… 10x = 9 + 0.9999……. 10x = 9 + x 10x", "}{ 7 }$$ = 4 x 0.$$\\overline { 142857 }$$ ⇒ $$\\frac { 4 }{ 7 }$$ = 0.$$\\overline { 571428 }$$ $$\\frac { 5 }{ 7 }$$ = 5 x $$\\frac { 1 }{ 7 }$$ = 5 x 0.$$\\overline { 142857 }$$ ⇒ $$\\frac { 5}{ 7 }$$ = 0.$$\\overline { 714285 }$$ $$\\frac { 6 }{ 7 }$$ = 6 x $$\\frac { 1 }{ 7 }$$ = 6 x 0.$$\\overline { 142857 }$$ ⇒ $$\\frac { 6}{ 7 }$$ = 0.$$\\overline { 857142 }$$ Question 3. Express the following in the form $$\\frac { p }{ q }$$, where p and q are integers and q ≠ 0. (i) 0.$$\\overline { 6 }$$ (ii) 0.4$$\\overline { 7 }$$ (iii) 0.$$\\overline { 0.001 }$$ Solution: (i) 0.$$\\overline { 6 }$$ (ii) We have 0.4$$\\overline { 7 }$$ (iii) We have 0.$$\\overline { 0.001 }$$ Question 4. Express 0.99999… in the form of $$\\frac { p }{ q }$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Solution: Let x = 0.99999… …(1) Multiplying by 10 on both sides, we get 10x = 9.99999… …(2) Subtracting equation (1) from eqn. (2), Thus 0.99999… = 1 = $$\\frac { 1 }{ 1 }$$ Here we get p =", "+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 +1 166 2 +152 4=2/3(p-2) FYI 2/3 is a fraction xRainexSiderisx Mar 9, 2017 #2 +23 +15 So your equation looks like $$4 = \\frac{2}{3}(p-2)$$. What you want to do first is get rid of the fraction cause no one likes fractions. You would do this by multiplying the reciprical which is $$\\frac{3}{2}$$ (just flip the $$\\frac{2}{3}$$). The equation would now look like $$4 * \\frac{3}{2} = (p-2)$$ Simplify it down to $$6 = (p-2)$$. Then add the two to both sides and you would get $$8 = p$$. Hope this helped! SilverAg Mar 9, 2017 Sort: #1 +5 4=2/3(p-2) 4 = 2/3p - 4/3 add 4/3 to both sides 5 1/3= 2/3p 16/3 = 2/3p divide both sides by 2/3 p = 16/3 / 2/3 p =16/3 x 3/2 p = 48/6 p = 8 Guest Mar 9, 2017 #2 +23 +15 So your equation looks like $$4 = \\frac{2}{3}(p-2)$$. What you want to do first is get rid of the fraction cause no one likes fractions. You would do this by multiplying the reciprical which is $$\\frac{3}{2}$$ (just flip the $$\\frac{2}{3}$$). The equation would now look like $$4 * \\frac{3}{2} = (p-2)$$ Simplify it down to $$6 = (p-2)$$. Then add the two to both sides and you would get $$8 = p$$. Hope this", "# How do you solve 5x - 12 [x - 2 (20 - x )] * 4 * 16 + [8x - 2x * 5] =4250? Jul 26, 2015 You isolate $x$ on one side of the equation. #### Explanation: Start by getting rid of the parantheses - remember to use PEMDAS to help you with the order of operations. $5 x - 12 \\left[x - 2 \\left(20 - x\\right)\\right] \\cdot 4 \\cdot 16 + \\left(8 x - 2 x \\cdot 5\\right) = 4250$ $5 x - 12 \\left(x - 40 + 2 x\\right) \\cdot 64 + \\left(- 2 x\\right) = 4250$ $3 x - 12 \\left(3 x - 40\\right) \\cdot 64 = 4250$ $3 x - 2304 x + 30720 = 4250$ $- 2301 x + 30720 = 4250$ To isolate $x$ on one side of the equation, add $- 30720$ to both side of the equation $- 2301 x + \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{30720}}} - \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{30720}}} = 4250 - 30720$ $- 2301 x = - 26470$ Finally, divide both sides of the equation by $- 2301$ to get $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{- 2301}}} x}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{- 2301}}}} = \\frac{- 26470}{- 2301}$ $x = \\frac{26470}{2301} \\approx \\textcolor{g r e e n}{11.5}$", "How do you solve (K+3.7)/9.8= -0.22? 1. $K = - 5.856$ 1. First, multiply by $9.8$ on both sides. $\\cancel{9.8} \\cdot \\frac{K + 3.7}{\\cancel{9.8}} = - 0.22 \\cdot 9.8$ 2.Then, subtract $.37$ on both sides, which leaves $K$ on the left. $K \\cancel{+ 3.7 - 3.7} = - 2.156 - 3.7$ 1. $K = - 5.856$", "6x}+ \\frac{7}{x^3 - 12x^2 + 36x}$$ Step 1) Find the LCD Let's begin by factoring the denominators, this will allow us to find the LCD. $$\\frac{1}{x(x - 6)(x - 6)}=\\frac{x + 1}{x(x - 6)}+ \\frac{7}{x(x - 6)(x - 6)}$$ LCD: $$x(x - 6)(x - 6)$$ Step 2) Multiply each side of the equation by the LCD: $$x(x - 6)(x - 6) \\cdot \\frac{1}{x(x - 6)(x - 6)}=x(x - 6)(x - 6)\\left(\\frac{x + 1}{x(x - 6)}+ \\frac{7}{x(x - 6)(x - 6)}\\right)$$ Left Side: $$x(x - 6)(x - 6) \\cdot \\frac{1}{x(x - 6)(x - 6)}$$ $$\\cancel{x(x - 6)(x - 6)}\\cdot \\frac{1}{\\cancel{x(x - 6)(x - 6)}}\\hspace{.2em}» \\hspace{.2em}1$$ Right Side: $$x(x - 6)(x - 6) \\cdot \\frac{x + 1}{x(x - 6)}+ x(x - 6)(x - 6) \\cdot \\frac{7}{x(x - 6)(x - 6)}$$ $$\\cancel{x(x - 6)}(x - 6) \\cdot \\frac{x + 1}{\\cancel{x(x - 6)}}+ \\cancel{x(x - 6)(x - 6)}\\cdot \\frac{7}{\\cancel{x(x - 6)(x - 6)}}\\hspace{.2em}» \\hspace{.2em}(x-6)(x + 1) + 7$$ Step 3) Solve the Equation: $$1=(x - 6)(x + 1) + 7$$ $$1=x^2 - 5x - 6 + 7$$ $$1=x^2 - 5x + 1$$ $$x^2 - 5x=0$$ $$x(x - 5)=0$$ $$x=0$$ $$x - 5=0$$ $$x=5$$ Our proposed solutions are x = 0, and x = 5. Step 4) Check the solutions in the original equation. $$\\frac{1}{x^3 - 12x^2 + 36x}=\\frac{x + 1}{x^2 - 6x}+", "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 $$1000x-10x=235.\\overline{35}-2.\\overline{35}$$ $$\\Rightarrow 990x=233$$ $$\\Rightarrow x=\\frac{233}{990}$$.", "### 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", "and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 4 (using Solution 3 diagram) $\\triangle UPV \\sim \\triangle CPM$, and $\\frac{\\overline{UV}}{\\overline{MC}}=\\frac{\\overline{UP}}{\\overline{PC}}=\\frac{1}{2}$. Since $\\overline{UP} + \\overline{PC} = 12$$\\text{, }$ $\\text{ }$ $\\overline{UP}$ and $\\overline{PC}$ are $4$ and $8$, respectively. By symmetry, $\\overline{PC}=\\overline{PM}=8$. We can now do Pythagorean Theorem on $\\triangle UPM$. ## Solution 5 (FASTEST) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively.", "and perhaps is what you were looking for. On the other hand perhaps $\\frac 38$ was close enough and you could stop there. If you want $0.3760683761$ exactly, you can continue the process, which will eventually terminate with exactly $\\frac{3760683761}{10000000000}$. One pleasant feature of this algorithm is that it always produces fractions in lowest terms, so you never have to worry about reducing them afterwards. The theory of continued fractions provides a way to perform this calculation much more quickly, producing the continued fractional result $$0.3760683761 \\approx \\cfrac1{2+\\cfrac 1{1+\\cfrac1{1+\\cfrac1{1+\\cfrac1{14}}}}}$$ • Thx for the detailed explanation. The method you have suggested is interesting and I may take the time to try to turn it to a small program. I need to brush a bit on the \"Continuted Fractions\" as well. – NoChance Aug 21 '15 at 21:59 First note that $.\\overline{376068}\\cdot (1000000-1)=376068$. It follows that $$.\\overline{376068}=\\frac{376068}{999999}=\\frac{572}{1521}.$$ Edit with more details. In your example you have a repeating decimal $.\\overline{376068}=.376068376068376068...$ Let $x=.\\overline{376068}$. If we multiply by $1000000$ we get $$1000000x=376068.376068376068...$$ If we now subtract $x$ we have $$1000000x-x=376068.\\overline{376068} -0.\\overline{376068}=376068.$$ It follows that $(1000000-1)x=376068$ and $x=\\frac{376068}{999999}$. Reducing this fraction gives the result you are looking for. In general, if you have a terminating decimal you can multiply and divide by an appropriate power of ten (i.e. $1.5=1.5\\cdot\\frac{10}{10}=\\frac{15}{10}=\\frac{3}{2}$). In the case of", "# How do you solve 9p-5.2/4=9.95? Feb 27, 2017 See the entire solution process below: #### Explanation: First, divide $5.2$ by $4$: $9 p - 1.3 = 9.95$ Next, add $\\textcolor{red}{1.3}$ to each side of the equation to isolate the $p$ term while keeping the equation balanced: $9 p - 1.3 + \\textcolor{red}{1.3} = 9.95 + \\textcolor{red}{1.3}$ $9 p - 0 = 8.65$ $9 p = 8.65$ Now, divide each side of the equation by $\\textcolor{red}{9}$ to solve for $p$ while keeping the equation balanced: $\\frac{9 p}{\\textcolor{red}{9}} = \\frac{8.65}{\\textcolor{red}{9}}$ $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{9}}} p}{\\cancel{\\textcolor{red}{9}}} = 0.96 \\overline{1}$ $p = 0.96 \\overline{1}$", "by $2$. – 5xum Jul 1 at 11:01 At a higher level, the mistake was attempting to combine two steps, grouping the x-terms on one side and multiplying by 6 to get rid of the fraction. It those steps had been done separately they would probably have been done correctly. – Patricia Shanahan Jul 1 at 13:55 $5x + 8 = 10x + 3/6$ Let's gather all the x's on one side and the other stuff on the other side as well as reduce the fraction to lowest terms first: $10x-5x = 8-\\frac{1}{2}$ $5x = 8-.5$ $5x = 7.5$ $x = 1.5$ or $x=\\frac{3}{2}$ or $x=1+\\frac{1}{2}$ Now, to compute each side as a verfication: $LHS= 5x+8 = 5*\\frac{3}{2}+8 = 7.5+8 = 15.5$ $RHS = 10x+\\frac{3}{6} = 10*\\frac{3}{2}+\\frac{1}{2} = 15+0.5 = 15.5$ - It is very easy to see if your result is NOT correct. If $x=9$, then the left side of your equation is $$5 \\cdot 9 + 8 = 45+8=53$$ and the right is $$10\\cdot 9 + \\frac36 = 90 + \\frac12$$ which is not the same. Hint: When solving the equation, first move all the values containing $x$ on one side, so you get $$5x-10x = -8+\\frac36$$ and solve from here on. - $$5x+8 = 10x + \\frac{3}{6} \\overset{(1)}{\\iff} 8-\\frac{3}{6}=10x-5x \\overset{(2)}{\\iff} \\frac{45}{6} = 5x \\overset{(3)}\\iff x=\\frac{9}{6}\\overset{(4)}{=}", "## Precalculus (6th Edition) $x=10$ Multiply the LCD, which is $70$, to both sides of the equation to obtain: $\\require{cancel}70 \\cdot \\frac{1}{14}(3x-2)=70 \\cdot \\frac{x+10}{10} \\\\5\\cancel{70} \\cdot \\frac{1}{\\cancel{14}}(3x-2)=7\\cancel{70} \\cdot \\frac{x+10}{\\cancel{10}} \\\\5(3x-2)=7(x+10)$ Distribute $5$ and $7$ to obtain: $5(3x)-5(2)=7(x) + 7(10) \\\\15x-10=7x+70$ Subtract $7x$ and add $10$ on both sides of the equation, then combine like terms to obtain: $15x-10-7x+10=7x+70-7x+10 \\\\8x=80$ Divide $8$ to both sides: $\\dfrac{8x}{8} = \\dfrac{80}{8} \\\\x=10$", "# How do you write y = -3/2x + 3 in standard form? Jun 14, 2017 $3 x + 2 y = 6$ #### Explanation: Standard Form looks like : $a x + b y = c$ where $a , b$ and $c$ are constants. So we need to rearrange the formula to get both $x$ and $y$ to one side, and a number on the other side. So from: $y = - \\frac{3}{2} x + 3$ We should get rid of the fraction first. To do this, multiply everything by 2: $2 y = 2 \\left(- \\frac{3}{2} x\\right) + 2 \\left(3\\right)$ $\\implies 2 y = \\frac{2 \\cdot - 3}{2} x + 6$ $\\implies 2 y = \\frac{\\cancel{2} \\cdot - 3}{\\cancel{2}} x + 6$ $\\implies 2 y = - 3 x + 6$ Then we bring -3x to the other side by adding 3x to both sides: $\\implies 3 x + 2 y = - 3 x + 6 + 3 x$ $\\implies 3 x + 2 y = \\cancel{- 3 x} + 6 + \\cancel{3 x}$ $\\implies 3 x + 2 y = 6$ And you've got it in standard from now" ]

[ "# How do you convert 0.6 (6 being repeated) to a fraction? Mar 4, 2016 $0. \\overline{6} = \\frac{2}{3}$ #### Explanation: Multiply by $10$ to shift one place to the left and subtract the starting number to cancel out the repeating decimal. To do this multiply by $\\left(10 - 1\\right)$: $\\left(10 - 1\\right) \\cdot 0. \\overline{6} = 6. \\overline{6} - 0. \\overline{6} = 6$ Then divide both sides by $\\left(10 - 1\\right)$ to find: $0. \\overline{6} = \\frac{6}{10 - 1} = \\frac{6}{9} = \\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} \\cdot 2}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} \\cdot 3} = \\frac{2}{3}$", "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$What\\ is\\ 0.4\\overline{8} + 0.3\\overline{7}?$$ Hello Guest! $$0.4\\overline{8} + 0.3\\overline{7}=0.8\\overline 6$$ $$0.86666\\overline 6$$ = x -$$\\underline{0.66666\\overline 6}$$ = 2/3 $$0.2$$ = 1/5 $$x-\\frac{2}{3}=\\frac{1}{5}\\\\ x=\\frac{1}{5}+\\frac{2}{3}=\\frac{3+2\\cdot 5}{15}\\\\ x=\\frac{13}{15}$$ $$0.4\\overline{8} + 0.3\\overline{7}=\\dfrac{13}{15}$$ ! Jan 12, 2022", "+0 # using alegra prove that 0.565656... / 0.124242424... is equal in value to 560/123 +1 178 2 using alegra prove that 0.565656... / 0.124242424... is equal in value to 560/123 Guest Nov 12, 2017 #1 0 0.565656... / 0.124242424..... Multiply both top and bottom by 990. [990 x 0.56565656.... / 990 x 0.124242424.....] =560/123 Guest Nov 12, 2017 #2 +2075 0 In order to prove algabraically that $$\\frac{0.5656...}{0.12424...}=\\frac{560}{123}$$, let's try to convert the interminable decimals to fractions. I'll start with $$0.5656...$$ 1. Set the Repeating Decimal equal to a Variable! This is a farily simple step. $$0.\\overline{56}=x$$. Now, you're good to go! 2. Multiply Both Sides by 10 such until the Repeating Portion is the Whole Number In this case, if I multiply both sides by 100, which is 10^2, then the repeating portion will be to the left of the decimal point. $$56.\\overline{56}=100x$$ $$56.\\overline{56}$$ $$=100x$$ $$\\hspace{2mm}0.\\overline{56}$$ $$=\\hspace{7mm}x$$ $$56$$ $$=\\hspace{1mm}99x$$ 4. Solve for x $$56=99x$$ Divide by 99 on both sides. $$\\frac{56}{99}=x=0.\\overline{56}$$ Great! Now, let's convert the next one. $$0.1\\overline{24}=y$$ Now, multiply by multiples of ten to get the repeating portion to the left until the repeating part lines up. $$12.4\\overline{24}=100y$$ Now, subtract the two equations. $$12.4\\overline{24}$$ $$=100y$$ $$\\hspace{3mm}0.1\\overline{24}$$ $$=\\hspace{7mm}y$$ $$\\hspace{1mm}12.3$$ $$=\\hspace{2mm}99y$$ Now, solve for y. $$12.3=99y$$ Multiply by 10 on both sides to make the left", "a^2$$, guess what we can do next! Subtract that value from both sides. It's okay. If we treat both sides equally, they'll still be equal. $$\\cancel{x^2}-\\cancel{x^2}+6x=\\cancel{a^2} - \\cancel{x^2}+2ab$$. So, $$6x=2ab$$. Hmmm... $$a=x$$, right? Then we can divide by that value. Then we see that $$\\Large\\frac{6\\cancel{x}}{\\cancel{x}}=\\frac{2\\cancel{a}b}{\\cancel{x}}$$. Cleaned up, to see it clearer, we have $$6=2b$$. Divide both sides by $$2$$, now, so $$b$$ is alone. $$\\Large\\frac{6}{2}=\\frac{\\cancel{2}b}{\\cancel{2}}\\normalsize =3=b$$. Just like that. That technique gave us $$x=a$$ and $$3=b$$. That's for $$a^2+2ab+b^2$$. Here's the thing. We only had $$x^2+6x$$ to start with, and the polynomial we want is $$x^2+6x+3^2$$. We can't add $$3^2$$ out of nowhere... Unless we subtract it at the same time... So, we got from $$x^2 +6x$$ to $$x^2+6x+3^2 - 3^2$$. And now, we can complete the square easily. $$a^2 + 2ab + b^2 = (a+b)^2$$. Likewise,$x^2+6x+3^2 -3^2=(x+3)^2-3^2$Notice how we have to maintain the \"$$-3^2$$\". 12. theEric We just completed the square, I guess. But we need to do the same for $$y-8y$$. $\\Huge\\text{\"Completing the Square\"} \\\\ \\Large\\text{again}$ *** I $$\\it{highly}$$ recommend you either try this for yourself first, try it afterwards, or both. There's no way to know if you absorbed it without trying. *** Let $$y$$ be $$a$$ in $$a^2 + 2ab +b^2$$. So $$y=a$$. Then $$y^2=a^2$$. To $$y^2 -8y = a^2+2ab$$, we subtract the $$y^2=a^2$$", "as dividing two positive numbers. $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{CD}=\\frac{2}{4}$$ 2/4 can be simplified still $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{CD}=\\frac{1}{2}$$ Let's find the slope of $$\\overline{AB}$$: $$m=\\frac{y_2-y_1}{x_2-x_1}$$ I am reminding you of the slope formula again! $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{AB}=\\frac{n-5}{-1-3}$$ We can simplify the denominator, but the numerator will need something else $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{AB}=\\frac{n-5}{-4}$$ This is the slope of $$\\overline{AB}$$. What did I say above? I said that parallel lines have the same slope! That means that the slope of $$\\overline{CD}$$ equals the slope of $$\\overline{AB}$$: $$\\frac{-4}{1}*\\frac{1}{2}=\\frac{n-5}{-4}*\\frac{-4}{1}$$ Multiply by -4 on both sides to get rid of all the pesky fractions! $$-2=n-5$$ Add 5 to both sides $$3=n$$ Therefore, the coordinates of point A is (-1,3) TheXSquaredFactor Jun 11, 2017 #3 +2190 +3 Oh, I am sorry. I did not anwer the other questions that you asked. Here we go. Our ultimate goal is to find what value for n makes $$\\overline{AB}\\hspace{1mm}\\perp\\hspace{1mm}\\overline{CD}$$. Just like in the previous problem, find the slope of $$\\overline{CD}$$. Since I have already found the slope in the previous problem that I answered, I don't need to do that calculation again: $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{CD}=\\frac{1}{2}$$ We already know the slope of $$\\overline{AB}$$ because we calculated it in the previous problem as well: $$m\\hspace{1mm}\\text{of}\\hspace{1mm}\\overline{AB}=\\frac{n-5}{-4}$$ The difference with this problen, however, is that we have to find the value for that makes the segment perpendicular! For a segment to be perpendicular,", "and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 4 (using Solution 3 diagram) $\\triangle UPV \\sim \\triangle CPM$, and $\\frac{\\overline{UV}}{\\overline{MC}}=\\frac{\\overline{UP}}{\\overline{PC}}=\\frac{1}{2}$. Since $\\overline{UP} + \\overline{PC} = 12$$\\text{, }$ $\\text{ }$ $\\overline{UP}$ and $\\overline{PC}$ are $4$ and $8$, respectively. By symmetry, $\\overline{PC}=\\overline{PM}=8$. We can now do Pythagorean Theorem on $\\triangle UPM$. ## Solution 5 (FASTEST) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively.", "6x}+ \\frac{7}{x^3 - 12x^2 + 36x}$$ Step 1) Find the LCD Let's begin by factoring the denominators, this will allow us to find the LCD. $$\\frac{1}{x(x - 6)(x - 6)}=\\frac{x + 1}{x(x - 6)}+ \\frac{7}{x(x - 6)(x - 6)}$$ LCD: $$x(x - 6)(x - 6)$$ Step 2) Multiply each side of the equation by the LCD: $$x(x - 6)(x - 6) \\cdot \\frac{1}{x(x - 6)(x - 6)}=x(x - 6)(x - 6)\\left(\\frac{x + 1}{x(x - 6)}+ \\frac{7}{x(x - 6)(x - 6)}\\right)$$ Left Side: $$x(x - 6)(x - 6) \\cdot \\frac{1}{x(x - 6)(x - 6)}$$ $$\\cancel{x(x - 6)(x - 6)}\\cdot \\frac{1}{\\cancel{x(x - 6)(x - 6)}}\\hspace{.2em}» \\hspace{.2em}1$$ Right Side: $$x(x - 6)(x - 6) \\cdot \\frac{x + 1}{x(x - 6)}+ x(x - 6)(x - 6) \\cdot \\frac{7}{x(x - 6)(x - 6)}$$ $$\\cancel{x(x - 6)}(x - 6) \\cdot \\frac{x + 1}{\\cancel{x(x - 6)}}+ \\cancel{x(x - 6)(x - 6)}\\cdot \\frac{7}{\\cancel{x(x - 6)(x - 6)}}\\hspace{.2em}» \\hspace{.2em}(x-6)(x + 1) + 7$$ Step 3) Solve the Equation: $$1=(x - 6)(x + 1) + 7$$ $$1=x^2 - 5x - 6 + 7$$ $$1=x^2 - 5x + 1$$ $$x^2 - 5x=0$$ $$x(x - 5)=0$$ $$x=0$$ $$x - 5=0$$ $$x=5$$ Our proposed solutions are x = 0, and x = 5. Step 4) Check the solutions in the original equation. $$\\frac{1}{x^3 - 12x^2 + 36x}=\\frac{x + 1}{x^2 - 6x}+", "the lines are perpendicular, hence the $m$ and $-\\frac{1}{m}$, and the sum of the y-intercepts is equal to 0, hence the $b, -b$. Since both lines contain the point $(6,8)$, we can plug this into the two equations to obtain $$8=6m+b$$ and $$8=-6\\frac{1}{m}-b$$ Adding the two equations gives $$16=6m+\\frac{-6}{m}$$ Multiplying by $m$ gives $$16m=6m^2-6$$ $$\\implies 6m^2-16m-6=0$$ $$\\implies 3m^2-8m-3=0$$ Factoring gives $$(3m+1)(m-3)=0$$ Plugging $m=3$ into one of our original equations, we obtain $$8=6(3)+b$$ $$\\implies b=8-6(3)=-10$$ Since $\\bigtriangleup APQ$ has hypotenuse $2|b|=20$ and the altitude to the hypotenuse is equal to the the x-coordinate of point $A$, or 6, the area of $\\bigtriangleup APQ$ is equal to $$\\frac{20\\cdot6}{2}=\\boxed{\\textbf{(D)}\\ 60}$$ Solution 3 Like Solution 2 but solving directly for intercepts (b): 1. Solve for m using: $8=6m+b$ $$m=\\frac{8-b}{6}$$ 2. Substitute into the other equation: $$8=-6\\cdot(\\frac{1}{\\frac{8-b}{6}})-b$$ Flip the inverse: $$8=-6\\cdot(\\frac{6}{8-b})-b$$ Multiply $6$'s: $$8=-(\\frac{36}{8-b})-b$$ 3. Multiply through by $8-b$ (Watch distributing minus!) $$64-8b=-36-8b+b^2$$ 4. Add $36$ to both sides, and cancel $-8b$ by adding to both sides: $$100=b^2$$ $b=10$ (or $-10$) The rest is as above. Solution 4(Heron's Formula) Since their sum is $0$, let the y intercepts be P$(0,a)$ and Q$(0,-a)$. The slope of $AP$ is $\\frac{8-a}{6}$. The slope of AQ is $\\frac{8+a}{6}$. Since multiplying the slopes of perpendicular lines yields a product of $-1$, we have $\\frac{64-a^2}{36}=-1$, which results in $a^2=100$.", "≠ 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", "# 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 $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "}{ 7 }$$ = 4 x 0.$$\\overline { 142857 }$$ ⇒ $$\\frac { 4 }{ 7 }$$ = 0.$$\\overline { 571428 }$$ $$\\frac { 5 }{ 7 }$$ = 5 x $$\\frac { 1 }{ 7 }$$ = 5 x 0.$$\\overline { 142857 }$$ ⇒ $$\\frac { 5}{ 7 }$$ = 0.$$\\overline { 714285 }$$ $$\\frac { 6 }{ 7 }$$ = 6 x $$\\frac { 1 }{ 7 }$$ = 6 x 0.$$\\overline { 142857 }$$ ⇒ $$\\frac { 6}{ 7 }$$ = 0.$$\\overline { 857142 }$$ Question 3. Express the following in the form $$\\frac { p }{ q }$$, where p and q are integers and q ≠ 0. (i) 0.$$\\overline { 6 }$$ (ii) 0.4$$\\overline { 7 }$$ (iii) 0.$$\\overline { 0.001 }$$ Solution: (i) 0.$$\\overline { 6 }$$ (ii) We have 0.4$$\\overline { 7 }$$ (iii) We have 0.$$\\overline { 0.001 }$$ Question 4. Express 0.99999… in the form of $$\\frac { p }{ q }$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Solution: Let x = 0.99999… …(1) Multiplying by 10 on both sides, we get 10x = 9.99999… …(2) Subtracting equation (1) from eqn. (2), Thus 0.99999… = 1 = $$\\frac { 1 }{ 1 }$$ Here we get p =", "\\frac{1}{7} = 2 \\times 0.\\overline{142857}= 0.\\overline{285714}$ Similarly, $\\Rightarrow \\frac{3}{7} = 3\\times \\frac{1}{7} = 3 \\times 0.\\overline{142857}= 0.\\overline{428571}$ $\\Rightarrow \\frac{4}{7} = 4\\times \\frac{1}{7} = 4 \\times 0.\\overline{142857}= 0.\\overline{571428}$ $\\Rightarrow \\frac{5}{7} = 5\\times \\frac{1}{7} = 5 \\times 0.\\overline{142857}= 0.\\overline{714285}$ $\\Rightarrow \\frac{6}{7} = 6\\times \\frac{1}{7} = 6 \\times 0.\\overline{142857}= 0.\\overline{857142}$ Let $x = 0.\\overline6= 0.6666....$ -(i) Now, multiply by 10 on both sides $10x= 6.6666...$ $\\Rightarrow 10x = 6 + x \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\(using \\ (i))$ $\\Rightarrow 9x = 6$ $\\Rightarrow x = \\frac{6}{9} = \\frac{2}{3}$ Therefore, $\\frac{p}{q}$ form of $0.\\bar{6}$ is $\\frac{2}{3}$ We can write $0.4\\overline7$ as $\\Rightarrow 0.4\\overline7 = \\frac{4}{10}+ \\frac{0.777..}{10}$ -(i) Now, Let $x = 0.\\overline7= 0.7777....$ -(ii) Now, multiply by 10 on both sides $10x= 7.7777...$ $\\Rightarrow 10x = 7 + x \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\(using \\ (ii))$ $\\Rightarrow 9x = 7$ $\\Rightarrow x = \\frac{7}{9}$ Now, put the value of x in equation (i). we will get $\\Rightarrow 0.4\\overline7 = \\frac{4}{10}+ \\frac{7}{10\\times 9}= \\frac{4}{10}+ \\frac{7}{90} = \\frac{36+7}{90} = \\frac{43}{90}$ Therefore, $\\frac{p}{q}$ form of $0.4\\overline7$ is $\\frac{43}{90}$ Let $x = 0.\\overline{001}= 0.001001....$ -(i) Now, multiply by 1000 on both sides $1000x= 1.001001...$ $\\Rightarrow 1000x = 1 + x \\ \\ \\ \\", "## Precalculus (6th Edition) $x=10$ Multiply the LCD, which is $70$, to both sides of the equation to obtain: $\\require{cancel}70 \\cdot \\frac{1}{14}(3x-2)=70 \\cdot \\frac{x+10}{10} \\\\5\\cancel{70} \\cdot \\frac{1}{\\cancel{14}}(3x-2)=7\\cancel{70} \\cdot \\frac{x+10}{\\cancel{10}} \\\\5(3x-2)=7(x+10)$ Distribute $5$ and $7$ to obtain: $5(3x)-5(2)=7(x) + 7(10) \\\\15x-10=7x+70$ Subtract $7x$ and add $10$ on both sides of the equation, then combine like terms to obtain: $15x-10-7x+10=7x+70-7x+10 \\\\8x=80$ Divide $8$ to both sides: $\\dfrac{8x}{8} = \\dfrac{80}{8} \\\\x=10$", "# How do you solve 3+6v=45? ##### 1 Answer Aug 1, 2015 You isolate $v$ on one side of the equation to get $v = 7$. #### Explanation: You can solve this equation by isolating $v$ on one side of the equation. To do that, first add $- 3$ to both sides of the equation $\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} - \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} + 6 v = 45 - 3$ $6 v = 42$ Now divide both sides of the equation by $6$ to get $v$ alone on the left side of the equation $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{6}}} \\cdot v}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{6}}}} = \\frac{42}{6}$ $v = \\textcolor{g r e e n}{7}$", "# How do you solve 5x - 12 [x - 2 (20 - x )] * 4 * 16 + [8x - 2x * 5] =4250? Jul 26, 2015 You isolate $x$ on one side of the equation. #### Explanation: Start by getting rid of the parantheses - remember to use PEMDAS to help you with the order of operations. $5 x - 12 \\left[x - 2 \\left(20 - x\\right)\\right] \\cdot 4 \\cdot 16 + \\left(8 x - 2 x \\cdot 5\\right) = 4250$ $5 x - 12 \\left(x - 40 + 2 x\\right) \\cdot 64 + \\left(- 2 x\\right) = 4250$ $3 x - 12 \\left(3 x - 40\\right) \\cdot 64 = 4250$ $3 x - 2304 x + 30720 = 4250$ $- 2301 x + 30720 = 4250$ To isolate $x$ on one side of the equation, add $- 30720$ to both side of the equation $- 2301 x + \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{30720}}} - \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{30720}}} = 4250 - 30720$ $- 2301 x = - 26470$ Finally, divide both sides of the equation by $- 2301$ to get $\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{- 2301}}} x}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{- 2301}}}} = \\frac{- 26470}{- 2301}$ $x = \\frac{26470}{2301} \\approx \\textcolor{g r e e n}{11.5}$", "# How do you write y = -3/2x + 3 in standard form? Jun 14, 2017 $3 x + 2 y = 6$ #### Explanation: Standard Form looks like : $a x + b y = c$ where $a , b$ and $c$ are constants. So we need to rearrange the formula to get both $x$ and $y$ to one side, and a number on the other side. So from: $y = - \\frac{3}{2} x + 3$ We should get rid of the fraction first. To do this, multiply everything by 2: $2 y = 2 \\left(- \\frac{3}{2} x\\right) + 2 \\left(3\\right)$ $\\implies 2 y = \\frac{2 \\cdot - 3}{2} x + 6$ $\\implies 2 y = \\frac{\\cancel{2} \\cdot - 3}{\\cancel{2}} x + 6$ $\\implies 2 y = - 3 x + 6$ Then we bring -3x to the other side by adding 3x to both sides: $\\implies 3 x + 2 y = - 3 x + 6 + 3 x$ $\\implies 3 x + 2 y = \\cancel{- 3 x} + 6 + \\cancel{3 x}$ $\\implies 3 x + 2 y = 6$ And you've got it in standard from now", "$0.9 \\div 0.3$ One way to solve this problem is to think in terms of tenths: $\\phantom{=}0.9 \\div 0.3$ $= 9$ tenths $\\div ~3$ tenths $= 3$ Another way is to convert the division problem to a fraction, then multiply the top and bottom of the fraction by 10 so that you can work with whole numbers: $0.9 \\div 0.3 = \\dfrac{0.9}{0.3}$ $\\phantom{0.9 \\div 0.3} = \\dfrac{0.9 \\times 10}{0.3 \\times 10}$ $\\phantom{0.9 \\div 0.3} = \\dfrac{9}{3}$ $\\phantom{0.9 \\div 0.3} = 3$ ### Problem set 6: Problem 6a $0.8 \\div 0.2 =$ Awesome! Let's try a few that are a bit harder. Example: $7.6 \\div 0.1$ One way to solve this problem is to think in terms of tenths: $\\phantom{=}7.6 \\div 0.1$ $= 76$ tenths $\\div ~1$ tenth $= 76$ Another way is to convert the division problem to a fraction, then multiply the top and bottom of the fraction by 10 so you can work with whole numbers: $7.6 \\div 0.1 = \\dfrac{7.6}{0.1}$ $\\phantom{7.6 \\div 0.1} = \\dfrac{7.6 \\times 10}{0.1 \\times 10}$ $\\phantom{7.6 \\div 0.1} = \\dfrac{76}{1}$ $\\phantom{7.6 \\div 0.1} = 76$ ### Problem set 7: Problem 7a $4.5 \\div 0.1=$ Great, we'll finish with a few more challenging problems. Example: $5.4 \\div 0.9$ One way to solve this problem is to think in terms of tenths: $\\phantom{=}5.4", "# How do you solve -5( - 3x + 6) + 3= - 117? Jan 4, 2018 $x$ = $- 6$ #### Explanation: First multiply all the items inside the parentheses by $- 5$ and then the equation will look like: $\\left(- 5 \\times - 3 x + \\left(- 5\\right) \\times 6\\right) + 3 =$-117# or, $15 x - 30 + 3$ = $- 117$ Now, move all numbers to right hand side of the equation, and we get: $15 x$ = $- 117 + 30 - 3$ or, $15 x$ = $- 90$ 0r, $x$ = ${\\cancel{- 90}}^{6} / {\\cancel{15}}^{1}$ = $- 6$ Jan 4, 2018 $x = - 6$ #### Explanation: First, use the distributive property: $- 5 \\left(- 3 x + 6\\right) + 3 = - 117$ $\\to \\left(- 5 \\cdot \\left(- 3 x\\right)\\right) + \\left(- 5 \\cdot 6\\right) + 3 = - 117$ $\\to 15 x + \\left(- 30\\right) + 3 = - 117$ Then, combine like terms: $15 x - 30 + 3 = - 117$ $\\to 15 x - 27 = - 117$ $\\to 15 x = - 90$ Now, isolate the $x$ variable: $15 x = - 90$ $\\to x = - \\frac{90}{15}$ $\\to x = - 6$", "the confusion the following representation is adapted. The number is given as $0.3\\overline{74}$$0.3 \\overline{74}$ The line over $74$$74$ represents that $74$$74$ is repeated. Convert $\\frac{1}{3}$$\\frac{1}{3}$ into a fraction. The answer is $0.\\overline{3}$$0. \\overline{3}$ Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. convert back to fraction How do we convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction? To convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction, the following steps are used $x=0.\\overline{4}$$x = 0. \\overline{4}$ $10x=4.\\overline{4}$$10 x = 4. \\overline{4}$ Subtracting the two equations $9x=4$$9 x = 4$ $x=\\frac{4}{9}$$x = \\frac{4}{9}$ examples Converting $2.\\overline{4}$$2. \\overline{4}$ into a fraction. $x=2.\\overline{4}$$x = 2. \\overline{4}$ $10x=24.\\overline{4}$$10 x = 24. \\overline{4}$ subtracting the two above $9x=22$$9 x = 22$ $x=\\frac{22}{9}$$x = \\frac{22}{9}$ $x=2\\frac{4}{9}$$x = 2 \\frac{4}{9}$ To convert $2.23\\overline{43}$$2.23 \\overline{43}$ into a fraction: $x=2.23\\overline{43}$$x = 2.23 \\overline{43}$ $100x=223.43\\overline{43}$$100 x = 223.43 \\overline{43}$ Subtracting the above, $99x=221.20$$99 x = 221.20$ $990x=2212$$990 x = 2212$ $x=\\frac{2212}{990}$$x = \\frac{2212}{990}$. Another method is as follows. $100x=223+0.\\overline{43}$$100 x = 223 + 0. \\overline{43}$ $10000x=223×100+43.\\overline{43}$$10000 x = 223 \\times 100 + 43. \\overline{43}$ Subtracting the two equations $9900x=223×\\left(100-1\\right)+43$$9900 x = 223 \\times \\left(100 - 1\\right) + 43$ $x=\\frac{223}{100}+\\frac{43}{9900}$$x = \\frac{223}{100} + \\frac{43}{9900}$ summary Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. Conversion of Repeating Decimals to equivalent", "\\text{1 bar}$, $R = \\text{0.083145 L\"*\"bar\"\"/mol\"cdot\"K}$, and $T = \\text{273.15 K}$. Therefore, with $n = \\text{2 mol}$: V = (nRT)/P = [(2 cancel(\"mol\"))((\"0.083145 L\"cdotcancel(\"bar\"))/(cancel(\"mol\"*\"K\")))(273.15 cancel(\"K\"))]/(1 cancel(\"bar\")) $=$ $\\textcolor{b l u e}{\\text{45.422 L}}$ A more accurate answer would be $\\textcolor{b l u e}{\\text{44.968 L}}$, using the van der Waals equation of state: $P = \\frac{R T}{\\overline{V} - b} - \\frac{a}{{\\overline{V}}^{2}}$ In this case, we'd need to know the constants $a$ and $b$ for $C {l}_{2}$ specifically. $a = {\\text{6.343 bar/L\"^2\"mol}}^{2}$ $b = \\text{0.05422 L/mol}$ Now we need to solve for $\\overline{V}$: $P = \\frac{R T {\\overline{V}}^{2} - a \\left(\\overline{V} - b\\right)}{\\left(\\overline{V} - b\\right) {\\overline{V}}^{2}}$ (cross-multiply) $P {\\overline{V}}^{2} \\left(\\overline{V} - b\\right) = R T {\\overline{V}}^{2} - a \\overline{V} + a b$ (subtract through, multiply by denominator) $P {\\overline{V}}^{3} - b P {\\overline{V}}^{2} = R T {\\overline{V}}^{2} - a \\overline{V} + a b$ (distribute) $P {\\overline{V}}^{3} - b P {\\overline{V}}^{2} - R T {\\overline{V}}^{2} + a \\overline{V} - a b = 0$ (move things around) $P {\\overline{V}}^{3} - \\left(b P + R T\\right) {\\overline{V}}^{2} + a \\overline{V} - a b = 0$ (factor) And now to solve this, one way I know of is to use the Newton-Raphson approximation method to approach the answer from above. Let: $P = \\textcolor{g r e e n}{A} = \\textcolor{g r e e n}{\\text{1 bar}}$" ]

[ "# Asymptote FAQ - Section 6 Questions about 2D graphs ## Question 6.1. How can I draw x axis ticks on the right side, with the tick labels on the left side (relative to the axis path)? import graph; size(250,200,IgnoreAspect); draw(graph(exp,-1,1),red); xaxis(\"$x$\",RightTicks(Label(align=left))); yaxis(\"$y$\",RightTicks); ## Question 6.2. How can I reposition the x axis label to three-quarters along the axis length? import graph; size(250,200,IgnoreAspect); draw(graph(exp,-1,1),red); xaxis(Label(\"$x$\",0.75),LeftTicks); yaxis(\"$y$\",RightTicks); ## Question 6.3. How can I move the x axis label down 10bp? import graph; size(250,200,IgnoreAspect); draw(graph(exp,-1,1),red); xaxis(shift(0,-10)*\"$x$\",LeftTicks); yaxis(\"$y$\",RightTicks); ## Question 6.4. Can I use different pens for the axis, the axis label, and the tick labels? Yes: import graph; size(300,200,IgnoreAspect); xlimits(-50,50); ylimits(0,100); xaxis(Label(\"$x$\",MidPoint,red),Bottom,blue,LeftTicks(green)); yaxis(\"$y$\",Left,RightTicks); ## Question 6.5. How can I change the font type of the axes label? import graph; size(300,200,IgnoreAspect); xlimits(-50,50); ylimits(0,100); xaxis(\"x\",Bottom,Courier(\"m\",\"n\"),LeftTicks); yaxis(\"$y$\",Left,RightTicks); ## Question 6.6. How can I change the font type of the tick labels on an axis? Tick labels are by default typeset in (TeX) math mode, so to use other fonts you need to override the default tick format: import graph; size(300,200,IgnoreAspect); xlimits(-50,50); ylimits(0,100); xaxis(\"$x$\",Bottom,LeftTicks(\"%.4g\",Courier(\"m\",\"n\")+fontsize(12))); yaxis(\"$y$\",Left,RightTicks); ## Question 6.7. How can I prevent axes tick labels from rendering on top of each other? Either: (i) give LeftTicks/RightTicks/Ticks the arguments beginlabel=false and/or endlabel=false; (ii) explicitly remove specific ticks and their labels (drawing them manually; see http://www.github.com/vectorgraphics/asymptote/base/graph.asy", "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "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]", "63 ☐D. 72 ☐E. 83 57- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively. The area of ∆BED is 20, and the area of ∆BDF is 32. What is the perimeter of the rectangle? ☐A. 20 ☐B. 22 ☐C. 32 ☐D. 40 ☐E. 48 58- When point A $$(-6.-12)$$ is reflected over the $$y$$-axis to get the point B, what are the coordinates of point B? ☐A. $$(6,- 12)$$ ☐B. $$(6,12)$$ ☐C. $$(-12,6)$$ ☐D. $$(12,-6)$$ ☐E. $$(-6,12)$$ 59- Which of the following points lies on the line? $$5x+3y=6$$ ☐A. $$(6, 1)$$ ☐B. $$(–3, 3)$$ ☐C. $$(3,-3)$$ ☐D. $$(2, 2)$$ ☐E. $$(2, 8)$$ 60- If $$f(x)=2x^3+4x-18x^(-2)$$ and $$g(x)=-3$$, what is the value of $$f(g(x))$$? ☐A. $$-83$$ ☐B. $$-68$$ ☐C. $$-25$$ ☐D. $$-13$$ ☐E. $$0$$ 27% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "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,", "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,", "# 1969 AHSME Problems/Problem 31 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Let $OABC$ be a unit square in the $xy$-plane with $O(0,0),A(1,0),B(1,1)$ and $C(0,1)$. Let $u=x^2-y^2$, and $v=xy$ be a transformation of the $xy$-plane into the $uv$-plane. The transform (or image) of the square is: $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0)--(0,-2)--cycle,dot); MP(\"(A)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-2)\",(0,-2),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,234),black); draw(arc((-1.5,0),2.5,54,-54),black); MP(\"(B)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-2)\",(0,-2),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0),black); MP(\"(C)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,180),black); draw(arc((-1.5,0),2.5,54,0),black); MP(\"(D)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-2)\",(0,-2),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,dot); MP(\"(E)\",(-5,2),SW); MP(\"O\",(.1,.1),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,1)\",(0,1),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-1)\",(0,-1),SE); [/asy]$ ## Solution Each point on the square can be in the form $(0,y)$, $(1,y)$, $(x,0)$, and $(x,1)$, where $0 \\le x,y \\le 1$. Making the appropriate substitutions results in points being $(-y^2,0)$, $(1-y^2,2y)$, $(x^2,0)$, and $(x^2 - 1,2x)$ on the $uv$-plane. Notice that since $v \\ge 0$, none of the points are below the u-axis, so options A,B, and E are out. Since $x = \\tfrac{v}{2}$, $u = (\\tfrac{v}{2})^2 - 1$, so $v = 2\\sqrt{u+1}$, where $-1 \\le u \\le 0$. That means some of the lines are not straight, so the answer is $\\boxed{\\textbf{(D)}}$.", "the angle of a face about the corner at the face head. Graph.angle_of_link_about_head Get the angle of each link. Examples >>> import numpy as np >>> from landlab.graph import Graph >>> node_x, node_y = ([0, 1, 2, 0, 1, 2], ... [0, 0, 0, 1, 1, 1]) >>> links = ((0, 1), (1, 2), ... (0, 3), (1, 4), (2, 5), ... (3, 4), (4, 5)) >>> graph.angle_of_link * 180. / np.pi array([ 0., 0., 90., 90., 90., 0., 0.]) Examples >>> from landlab import HexModelGrid >>> import numpy as np >>> grid = HexModelGrid((3, 2), node_layout=\"hex\") >>> np.round(grid.angle_of_link[:3] / np.pi * 3.0) array([ 0., 2., 1.]) array([ 3., 5., 4.]) property area_of_cell# Get the area of each cell. Graph.area_of_patch property area_of_patch# Get the area of each patch. Examples >>> from landlab.graph import Graph >>> node_x, node_y = [0, 1, 2, 0, 1, 2, 0, 1, 2], [0, 0, 0, 1, 1, 1, 2, 2, 2] >>> links = ((0, 1), (1, 2), ... (0, 3), (1, 4), (2, 5), ... (3, 4), (4, 5), ... (3, 6), (4, 7), (5, 8), ... (6, 7), (7, 8)) >>> patches = ((0, 3, 5, 2), (1, 4, 6, 3)) >>> graph.area_of_patch array([ 1., 1.]) as_dataarray(name, at=None, time=None)# Create an xarray DataArray representation of a grid field. Parameters: • name", "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", "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", "# How can I left align yaxis in horizontal bar I use this sample to make horizontal bar chart but don’t know how to left align text in yaxis. Please help. import plotly.graph_objects as go fig = go.Figure(go.Bar( x=[20, 14, 23], y=[‘giraffes’, ‘orangutans’, ‘monkeys’], orientation=‘h’)) fig.show() Hi @hao.dn, you cannot configure ticks directly to do this (see Left align category labels on y-axis or Ticks: left-align first, right align last on x-axis). However, here is a workaround (courtesy of @alexcjohnson and @nicolaskruchten) which add a second axis, considered to be on the right (hence tick labels are left aligned), but positioned on the left. Hacky, but I hope it solves your problem :-). ``````import plotly.graph_objects as go y = ['boa', 'giraffe', 'monkey', 'orangeoutans'] fig = go.Figure(go.Bar( x=[20, 14, 23, 4], y=y, yaxis='y2', orientation='h', )) fig.update_layout(xaxis=dict(domain=[0.15, 0.9]), yaxis2=dict(anchor='free', position=0.02, side='right')) fig.show() `````` thanks. It’s ok to me.", "# 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", "xaxisconvert = linear yaxisconvert = linear linetype = solid linecolor = [0.0, 0.0, 0.0, 100.0] linewidth = 1 marker = dot markercolor = [0.0, 0.0, 0.0, 100.0] markersize = 1 flip = False In [5]: line.linecolor = \"red\" # Red color obviously line.linewidth = 2. # A bit thicker line.marker = \"circle\" # line.markersize = 1.25 # A bit bigger x.clear() # Let's clear the first plot x.switchfonts(\"default\",\"Clarendon\") x.plot(s,line) Out[5]: ## Altering X Axis¶ We would like the axis to be shown in radians, since s is a pure array the x axis is seen by vcs as 0 to 35 Let's create a dictionary to show a few X points 0 -> 0 45degrees = 4.5 in our units = $\\pi/4$ In [6]: angles = {0:'0',9:r'$\\pi/2$',18:r'$\\pi$', 27:r'$3\\pi/2$',36:r'$2\\pi$'} line.xticlabels1 = angles x.clear() x.plot(s,line) Out[6]: But we would like to have the axis labeled \"angle\" rather than axis_0 so we need to make s a transient variable In [7]: import MV2 s = MV2.array(s) s.id = \"sine\" xaxis = s.getAxis(0) xaxis.id = \"angle\" # let's make them radians as well xaxis[:] = xaxis[:]/18.*numpy.pi # That means we need to match or symbols to new values angles = {0:'0',numpy.pi/2.:r'$\\pi/2$',numpy.pi:r'$\\pi$', 3*numpy.pi/2.:r'$3\\pi/2$',2*numpy.pi:r'$2\\pi$'} line.xticlabels1 = angles x.clear() x.plot(s,line) Out[7]: The last point being exactly 2$pi$ it is not drawn let's extend our x", "sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); /* draw(rightanglemark((0,0),a*expi(rot1),(c,0))); */ filldraw(sq1, rgb(0.95,1,0.95)); filldraw(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); label(\"a\",a/2*expi(rot1),SE,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),SW,sm); label(\"c\",(c/2,-c),S,sm); [/asy]$ COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection. $[asy] defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens pair shiftR = (s1+1,0); // distance between two diagrams pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B); // draw left diagram filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype(\"2 2\")); draw(I--E, linetype(\"2 2\")); draw(I--F, linetype(\"2 2\")); label(\"a\",(A+B)/2,S); label(\"b\",(B+C)/2,NE); label(\"c\",(A+C)/2,NW); label(\"r\",(I+F)/2,W); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C)); pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y); // draw right diagram filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy);", "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", "# 2001 IMO Problems/Problem 5 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem $ABC$ is a triangle. $X$ lies on $BC$ and $AX$ bisects angle $A$. $Y$ lies on $CA$ and $BY$ bisects angle $B$. Angle $A$ is $60^{\\circ}$. $AB + BX = AY + YB$. Find all possible values for angle $B$. ## Solution $[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0); pair Bb = rotate(40,E)*A; pair B = extension(A,D,E,Bb); pair H = foot(A,D,E); pair X = extension(A,H,B,E); pair Yy = bisectorpoint(A,B,E); pair Y =extension(A,E,B,Yy); pair C = E - (0,0.1); dot(\"B\", B, NW); dot(\"Y\", Y, NE); dot(\"D\", D, W); dot(\"E\", E, E); dot(\"A\",A,N); dot(\"X\",X,S); label(\"C\",E+(0,-0.1),E); draw(A--D--E--cycle); draw(B--Y); draw(B--E); // draw(B--Xx--E,dashed); // draw(Y--Xx, dashed); draw(A--X--D, dashed); [/asy]$ Let $D$ be on extension of $AB$ and $BD=BX$. Let $E$ be on $YC$ and $YE=YB$, then $\\[AD=AB+BD=AB+BX=AY+YB=AE\\]$ Since $A=60$, $\\triangle{ADE}$ is equilateral. Let $\\angle{ABY}=x$, then, $\\[\\angle{YBX}=\\angle{BDX}=\\angle{BXD}=\\angle{YEX}=x\\]$ We claim that $X$ must be on $BE$, i.e., $C=E$. If $X$ is not on $BE$, then $\\angle{EBX}=\\angle{YBX}-\\angle{YBE}=\\angle{YEX}-\\angle{YEB}=\\angle{BEX}$, which leads to $BX=EX=DX$, and $\\triangle{BDX}$ is equilateral, which is not possible. With that, we have, in $\\triangle{ABE}$, $60+2x+x=180$, $x=40$, and $\\angle{ABE}=80$. Solution by $Mathdummy$. 2001 IMO (Problems) • Resources Preceded byProblem 4 1 • 2 •", "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? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "because aso is not a plane parallel to asd. If my calculations are correct, oh = os/5. – Alain Matthes Apr 15 '13 at 17:15 Hi, @AlainMatthes. I see that you took the time to make the calculations; thanks; I didn't feel like doing them myself ;-) I'll use your result about oh and os, if it's OK with you. – Gonzalo Medina Apr 15 '13 at 17:25 Asymptote version pyramid.asy: import three; // 3D module import math; currentprojection=orthographic(camera=(55,144,80), up=(0,0,1),target=(0,0,0),zoom=1,center=true); size(300); size3(300,300,300); triple intersectionpoint(triple a,triple b,triple c,triple d){ real u=((d.x-c.x)*a.y+(c.y-d.y)*a.x+c.x*d.y-d.x*c.y)/ ((d.y-c.y)*b.x+(c.x-d.x)*b.y+(d.x-c.x)*a.y+(c.y-d.y)*a.x); return a*(1.0-u)+b*u; } triple A,B,C,D,EE,II,H,K,M,NN,O,SS; // S,E and N has special meaning in asy: // as South, East and North or (0,-1), (1,0) and (0,1) // and I is a sqrt(-1)=(0,1) real a=40; // side of the square; real h=50; // height, h=AS real d=sqrt(2)/2*a; // half of the diagonal O=(0,0,0); C=(0,d,0); B=(d,0,0); A=(0,-d,0); D=(-d,0,0); SS=A+(0,0,h); M=0.5(SS+A); II=0.5(SS+D); NN=intersectionpoint(SS,(SS+A-B),C,II); real phi=atan(h/a); real u=a*cos(phi)/sqrt(a^2+h^2); EE=D*(1-u)+SS*u; K=EE+B-A; real psi=atan(h/d); real u=d*cos(psi)/sqrt(d^2+h^2); H=O*(1-u)+SS*u; pair Q=intersectionpoint(project(NN--B),project(SS--D)); pen dashed=linetype(new real[] {5,5}); // set up dashed pattern pen visLine=darkblue+0.8pt; pen hidLine=lightblue+dashed+0.8pt; void Dot(...triple[] v){ dotfactor=8; for(int i=0;i<v.length;++i){ dot(project(v[i]),UnFill); } } void Draw3(guide3 g, pen p=currentpen){ draw(project(g),p); } void labelP(string s,triple t,pair p=(0,0)){ label(\"$\"+s+\"\\$\",project(t),p); } Draw3(B--A--D,hidLine); Draw3(H--A--SS,hidLine); Draw3(M--II,hidLine); Draw3(EE--A--C,hidLine); Draw3(B--D,hidLine); Draw3(SS--O,hidLine); Draw3(SS--B--C--SS--D--C,visLine); Draw3(SS--NN--C,visLine); Draw3(EE--K--B,visLine); Draw3(NN--K,visLine); draw(project(NN)--Q,visLine); draw(project(B)--Q,hidLine); Dot(A,B,C,D,EE,H,II,K,M,NN,O,SS); labelP(\"A\",A,NW); labelP(\"B\",B,SW); labelP(\"C\",C,E); labelP(\"D\",D,NE); labelP(\"E\",EE,NE); labelP(\"H\",H,NE);", "# User talk:Bobthesmartypants/Sandbox ## Solution 1 $[asy] path Q; Q=(0,0)--(1,2)--(5,2)--(4,0)--cycle; draw(Q); draw((0,0)--(1.5,1)); label(\"D\",(0,0),S); draw((1,2)--(1.5,1)); label(\"A\",(1,2),N); draw((5,2)--(1.5,1)); label(\"B\",(5,2),N); draw((4,0)--(1.5,1)); label(\"C\",(4,0),S); draw((2,0)--(1.5,1),linetype(\"8 8\")); label(\"E\",(2,0),S); draw((2/3,4/3)--(1.5,1),linetype(\"8 8\")); label(\"F\",(2/3,4/3),W); label(\"P\",(1.5,1),NNE); [/asy]$ First, continue $\\overline{AP}$ to hit $\\overline{CD}$ at $E$. Also continue $\\overline{CP}$ to hit $\\overline{AD}$ at $F$. We have that $\\angle PAB=\\angle PCB$. Because $\\overline{AB}\\parallel\\overline{CD}$, we have $\\angle PAB=\\angle PED$. Similarly, because $\\overline{AD}\\parallel\\overline{BC}$, we have $\\angle PCB=\\angle PFD$. Therefore, $\\angle PAB=\\angle PED=\\angle PCB=\\angle PFD$. We also have that $\\angle ADC=\\angle ABC$ because $ABCD$ is a parallelogram, and $\\angle APC=\\angle FPE$. Therefore, $ABCP\\sim FDEP$. This means that $\\dfrac{FD}{AB}=\\dfrac{FP}{AP}=\\dfrac{DP}{BP}$, so $\\Delta ABP\\sim\\Delta FDP$. Therefore, $\\angle PBA=\\angle PDA$. $\\Box$ ## Solution 2 Note that $\\dfrac{1}{n}$ is rational and $n$ is not divisible by $2$ nor $5$ because $n>11$. This means the decimal representation of $\\dfrac{1}{n}$ is a repeating decimal. Let us set $a_1a_2\\cdots a_x$ as the block that repeats in the repeating decimal: $\\dfrac{1}{n}=0.\\overline{a_1a_2\\cdots a_x}$. ($a_1a_2\\cdots a_x$ written without the overline used to signify one number so won't confuse with notation for repeating decimal) The fractional representation of this repeating decimal would be $\\dfrac{1}{n}=\\dfrac{a_1a_2\\cdots a_x}{10^x-1}$. Taking the reciprocal of both sides you get $n=\\dfrac{10^x-1}{a_1a_2\\cdots a_x}$. Multiplying both sides by $a_1a_2\\cdots a_n$ gives $n(a_1a_2\\cdots a_x)=10^x-1$. Since $10^x-1=9\\times \\underbrace{111\\cdots 111}_{x\\text{ times}}$ we divide $9$ on both sides of the equation to get $\\dfrac{n(a_1a_2\\cdots a_x)}{9}=\\underbrace{111\\cdots 111}_{x\\text{ times}}$. Because", "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);" ]

[ "shaded region is 1/4. Add the four fractions: 1/4 + 1/8 + 3/8 + 1/4 = 1. The four squares together have an area of 4, so the percent is 1/4 = 25%. Therefore the solution is $\\boxed{\\textbf{(C)}\\ 25}$", "+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", "3^2)$ = $20 \\pi$. Therefore the total area of the shaded region is $64 \\pi$.", "area of $625$ square units occurs when $k = 0$, that is, when the rectangle is a square. Since $x+y= 100$ and $xy \\geq 500$, we have $x(50-x) \\geq 500$. Thus $x^2 -50x + 500 \\leq 0$. Consequently, $x$ must lie between the roots of the quadratic $x^2-50x+500 = 0$. The roots are $25 \\pm 5\\sqrt{5}$. Thus $25-5\\sqrt{5} \\leq x \\leq 25+5\\sqrt{5}$.", "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.", "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", "#### In Fig. 8.8, what is the ratio of the areas of(a) shaded portion I to shaded portion II ?(b) shaded portion II to shaded portion III?(c) shaded portions I and II taken together and shaded portion III ? Area of shaded portion I = 5 x 5 =25 square units Area of shaded portion III = 7 x 5 =35square units Area of whole figure = 10 x 10 = 100 square units Area of shaded portion II = Area of whole figure - Area of shaded portion I - Area of shaded portion III = 100 - 25 - 35 = 40 square units (a) Ratio of the areas of shaded portion I to shaded portion II $\\frac{\\text { Area of shaded portion I }}{\\text { Area of shaded portion II }} =\\frac{25 \\text { sq units }}{40 \\text { sq units }}=\\frac{5}{8}=5: 8$ (b) Ratio of the areas of shaded portion II to shaded portion III $\\frac{\\text { Area of shaded portion II }}{\\text { Area of shaded portion III }} =\\frac{40 \\text { sq units }}{35 \\text { sq units }}=\\frac{8}{7}=8: 7$ (c) Ratio of the areas of shaded portions I and II taken together and shaded portion III $\\frac{\\text { Area of shaded portion (I + II) }}{\\text { Area of shaded portion", "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", "2$ area of the square=$\\pi (\\sqrt2)^2$=$2\\pi$ Area of the shaded region= 28-2$\\pi$", "Exercises1.1.8Exercises Exercise1 Hint Draw a rectangle that encompasses the entire shaded area, and one that is encompassed by the shaded area. The shaded area is no more than the area of the bigger rectangle, and no less than the area of the smaller rectangle. The area is between $1.5$ and $2.5$ square units. Solution The diagram on the left shows a rectangle with area $2 \\times 1.25=2.5$ square units. Since the blue-shaded region is entirely inside this rectangle, the area of the blue-shaded region is no more than 2.5 square units. The diagram on the right shows a rectangle with area $2 \\times 0.75=1.5$ square units. Since the blue-shaded region contains this entire rectangle, the area of the blue region is no less than 1.5 square units. So, the area of the blue-shaded region is between 1.5 and 2.5 square units. Remark: we could also give an obvious range, like “the shaded area is between zero and one million square units.” This would be true, but not very useful or interesting. Exercise2 Hint We can improve on the method of Question 1 by using three rectangles that together encompass the shaded region, and three rectangles that together are encompassed by the shaded region. The shaded area is between 2.75 and 4.25 square units. (Other estimates are possible, but", "What fraction of this square region is shaded? Stripes are equal in width, and the figure is drawn to scale. Source: Julia Robinson Mathematics Festival SOLUTION Suppose without loss of generality that the width of the stripes equals 1 unit. Draw a $6\\times 6$ grid as shown in the above figure. Area of shaded stripe $A$ equals $11$ square units. Area of shaded stripe $B$ equals $7$ square units. Area of shaded stripe $C$ equals $3$ square units. Total area of shaded stripes $A,B,C=11+7+3=21$ square units. Area of square region = $6\\times 6=36$ square units. Shaded stripes’ fraction of this square region = $\\frac{21}{36}=\\frac{7}{12}$. Answer: $\\frac{7}{12}$.", "? 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 $414.7\\text{ cm}^2$", "of shaded part = $100 - 25 \\pi$ sq. cm. Solution 2 To get the area of the shaded part, we subtract the area of the semicircles from the area of the square. But the two semicircles has area equal to one circle. From above, the area of a square is the square of its side. So, if we let the area of a square be equal to $A$, then $A = 10 \\times 10 = 100$ square centimeters. Now, the area of the circle above is equal to $A = \\pi r^2$ $A = \\pi 25^2$ $A = 25 \\pi$. Now, Area of shaded part = area of square – area of 2 semicircles. Area of shaded part = $100 - 25 \\pi$. This entry was posted in Grade 5-6, Problem of the Week and tagged , , , . Bookmark the permalink.", "The total surface area of the cube is $6r^2.$ The area shaded on one face is $\\frac{\\pi r^2}{4}$ so the total shaded area is $\\frac{3\\pi r^2}{4}.$ The fraction of the total surface area shaded is thus $\\frac{3\\pi r^2}{4}$ divided by $6r^2$ which is: $$\\frac{3\\pi r^2}{4} \\times\\frac{1}{6r^2}= \\frac{\\pi}{8}$$", "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", "the shaded area is $x+y = \\frac{15}{2} + \\frac{21}{2} = 18 \\Rightarrow D$", "+0 # Math 0 44 1 66 unit squares each have \\dfrac{3}{5}53​start fraction, 3, divided by, 5, end fraction of their area shaded. May 5, 2021 #1 0 Not quite sure what you mean. I am going to assume that the question is suppose to say: There are 66 unit squares, each of them being $$\\frac{3}{5}$$ of the way shaded. What is the area of the shaded region? If each of the squares are shaded 3/5 of the way, and each square is a unit square (meaning that it is a 1 by 1 square with an area of 1 unit), and there are 66 squares, the answer is: 66 * 3/5 = 198/5 = 39.5 May 6, 2021", "square – area of the hexagon Area of the square = (15 x 15) cm2 = 225 cm2 Area of the hexagon A = (L2n)/[4tan(180/n)] A = (62 6)/ [4tan (180/6)] = (36 * 6)/ [4tan (180/6)] = 216/ [4tan (180/6)] = 216/ 2.3094 A = 93.53 cm2 Area of the shaded region = (225 – 93.53) cm2. = 131.47 cm2 ### Practice Questions 1. What is the area of the shaded region as shown below? 2. Given that $MN = 12$ m, $OP = 24$ m, and $OQ = 10$ m, what is the area of the shaded region in the diagram below? 3. What is the area of the shaded region of the rectangle shown below? 4. Suppose that $AB= 40$ ft, $BC= 25$ ft, and $CF = 30$ ft, what is the area of the shaded region of the diagram below? 5. What is the area of the shaded region shown below?", "## Intermediate Algebra (6th Edition) The area of the shaded region is $8x^2-12x+4$ square inches. RECALL: The area of a square is given by the formula $A=s^2$ where s = side length. The figure involves two squares. The area of the shaded region is equal to the area of the bigger square minus the area of the smaller square. Thus, the area of the shaded region in square inches is: $\\\\=\\text{area of the bigger square} - \\text{area of the smaller square} \\\\=(3x-2)^2-x^2 \\\\=[(3x)^2-2(3x)(2)+2^2]-x^2 \\\\=9x^2-12x+4-x^2 \\\\=8x^2-12x+4$", "the shaded region is the area of the square subtracted from the area of the circle. The diameter of the circle and the diagonal of the square are equal to 2. We can easily find the area of the square using the area formula $\\frac{d_1 d_2}{2}$. So the area of the shaded region is $1^2 \\pi-\\frac{2\\cdot{2}}{2}=\\pi-2$. Clearly the largest area that we found among the three shaded regions is $\\pi -$2. area so the answer is $\\boxed{C}$" ]

[ "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]", "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,", "63 ☐D. 72 ☐E. 83 57- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively. The area of ∆BED is 20, and the area of ∆BDF is 32. What is the perimeter of the rectangle? ☐A. 20 ☐B. 22 ☐C. 32 ☐D. 40 ☐E. 48 58- When point A $$(-6.-12)$$ is reflected over the $$y$$-axis to get the point B, what are the coordinates of point B? ☐A. $$(6,- 12)$$ ☐B. $$(6,12)$$ ☐C. $$(-12,6)$$ ☐D. $$(12,-6)$$ ☐E. $$(-6,12)$$ 59- Which of the following points lies on the line? $$5x+3y=6$$ ☐A. $$(6, 1)$$ ☐B. $$(–3, 3)$$ ☐C. $$(3,-3)$$ ☐D. $$(2, 2)$$ ☐E. $$(2, 8)$$ 60- If $$f(x)=2x^3+4x-18x^(-2)$$ and $$g(x)=-3$$, what is the value of $$f(g(x))$$? ☐A. $$-83$$ ☐B. $$-68$$ ☐C. $$-25$$ ☐D. $$-13$$ ☐E. $$0$$ 27% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "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,", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "24 B. 38 C. 46 D. 55 E. 61 38- In the following figure, ABCD is a rectangle. If $$a=\\sqrt{5}$$, and $$b=2a$$, find the area of the shaded region. (the shaded region is a trapezoid) A. $$\\sqrt{5}$$ B. $$2$$ C. $$4$$ D. $$2\\sqrt{5}$$ E. $$4\\sqrt{5}$$ 39- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively, and DE = 5 and DF = 3. The area of ∆BED is 20, and the area of ∆BDF is 24. What is the perimeter of the rectangle? A. 24 B. 28 C. 36 D. 42 E. 48 40- What is the product of all possible values of $$x$$ in the following equation? $$|2x+6-3x-7|=5$$ A. $$-24$$ B. $$-12$$ C. $$8$$ D. $$12$$ E. $$24$$ 28% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \\cdot \\frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\\boxed{{\\textbf{(C)}}~15}$. ### Solution 3 $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length $1$, and let's label the other legs $x$ for one of the triangles and $y$ for the other. Note that $x + y = 3$. The area of each of the triangles is $\\frac{x}{2}$ and $\\frac{y}{2}$, and there are $4$ of each. So now we need to find $4\\left(\\frac{x}{2}\\right) + 4\\left(\\frac{y}{2}\\right)$. $(4)\\frac{x}{2} + (4)\\frac{y}{2}$ $\\Rightarrow~~4\\left(\\frac{x}{2}+ \\frac{y}{2}\\right)$ $\\Rightarrow~~4\\left(\\frac{x+y}{2}\\right)$ Remember that $x+y=3$, so substituting this in we find that the area of all of the triangles is $4\\left(\\frac{3}{2}\\right) = 6$. The area of the $4$ unit squares is $4$, so the area of the square we need is $25- (4+6) = \\boxed{\\textbf{(C)}~15}$", "to X\", (0,8), W); draw((0,6)--(6,6)--(16,10)); [/asy]$ ## Problem 25 In the figure, the area of square WXYZ is $25 \\text{cm}^2$. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\\Delta ABC$, $AB = AC$, and when $\\Delta ABC$ is folded over side BC, point A coincides with O, the center of square WXYZ. What is the area of $\\Delta ABC$, in square centimeters? $[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label(\"A\", A, NW); label(\"O\", O, NE); label(\"B\", B, SW); label(\"C\", C, NW); label(\"W\",W , NE); label(\"X\", X, N); label(\"Y\", Y, N); label(\"Z\", Z, SE); [/asy]$ $\\textbf{(A)}\\ \\frac{15}4\\qquad\\textbf{(B)}\\ \\frac{21}4\\qquad\\textbf{(C)}\\ \\frac{27}4\\qquad\\textbf{(D)}\\ \\frac{21}2\\qquad\\textbf{(E)}\\ \\frac{27}2$", "# 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", "# 1994 AHSME Problems/Problem 23 ## Problem In the $xy$-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1)$ and $(5,0)$. The slope of the line through the origin that divides the area of this region exactly in half is $[asy] Label l; l.p=fontsize(6); xaxis(\"x\",0,6,Ticks(l,1.0,0.5),EndArrow); yaxis(\"y\",0,4,Ticks(l,1.0,0.5),EndArrow); draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy]$ $\\textbf{(A)}\\ \\frac{2}{7} \\qquad\\textbf{(B)}\\ \\frac{1}{3} \\qquad\\textbf{(C)}\\ \\frac{2}{3} \\qquad\\textbf{(D)}\\ \\frac{3}{4} \\qquad\\textbf{(E)}\\ \\frac{7}{9}$ ## Solution Let the vertices be $A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)$. It is easy to see that the line must pass through $CD$. Let the line intersect $CD$ at the point $G=(3,3-x)$ (i.e. the point $x$ units below $C$). Since the quadrilateral $ABCG$ and pentagon $GDEFA$ must have the same area, we have the equation $3\\times\\frac{1}{2}\\times(x+3)=\\frac{1}{2}\\times3\\times(3-x)+2$. This simplifies into $3x=2$, or $x=\\frac{2}{3}$, so $G=(3,\\frac{7}{3})$. Therefore the slope of the line is $\\boxed{\\textbf{(E)}\\ \\frac{7}{9}}$", "- - - - - - - + - - + - - + - - -+ - - - + bookshelves (y) | 10y | 30y | 90y | 30y | - - - - - - - - + - - + - - + - - -+ - - - + available | 220 | 480 | 1080 | 330 | - - - - - - - - + - - -+ - - + - - -+ - - - + Wood: . $20x + 10y \\:\\leq \\:220\\quad\\Rightarrow\\quad 2x + y \\:\\leq \\:22$ [2] Sawing: . $40x + 30y \\:\\leq \\:480\\quad\\Rightarrow\\quad 4x + 3y \\:\\leq \\:48$ [3] Sanding: . $60x + 90y \\:\\leq \\:1080\\quad\\Rightarrow\\quad 2x + 3y \\:\\leq \\:36$ [4] Assembly: . $10x + 30y \\:\\leq \\:330\\quad\\Rightarrow\\quad x + 3y\\:\\leq\\:33$ [5] [1] places us in Quadrant 1. [2] Graph the line: $2x + y \\:=\\:22$. .It has intercepts: $(11,0),\\;(0,22)$. . . . .Shade the region below the line. [3] Graph the line: $4x + 3y \\:=\\:48$. .It has intercepts: $(12,0),\\;(0,16)$ . . .Shade the region below the line. [4] Graph the line: $2x + 3y \\:=\\:36$. .It has intercepts: $(18,0),\\;(0,12)$ . . .Shade the region below the line. [5] Graph the line: $x + 3y \\:=\\:33$. .It has intercepts: $(33,0),\\;(0,11)$ . . .Shade", "= (5.0,5.0)) ax.set_xlim([-1.5,1.5]) ax.set_ylim([-1.5,1.5]) circ = plt.Circle((0, 0), radius=1.0, color=\"b\", alpha=0.2, lw=5) plt.axhline(0, color=\"black\", alpha=0.4) plt.axvline(0, color=\"black\", alpha=0.4) plt.show() #### Joint PDF A valid PDF is required to integrate to 1.0, and since we are looking at a Uniform distribution over the unit circle with area equal to $\\pi$, we have \\begin{align} f_{XY}(x,y) &= \\begin{cases} \\frac{1}{\\pi} & \\quad \\text{if } x^2 + y^2 \\le 1 \\\\ 0 & \\quad \\text{otherwise}\\\\ \\end{cases} \\end{align} Understand that while this is uniformly distributed, $x$ and $y$ are closely related with $x^2 + y^2 \\le 1$, and so $X,Y$ cannot be independent. More on that coming up. #### Marginal PDF of $X$ \\begin{align} f_{X}(x) &= \\int_{-\\sqrt{1-x^2}}^{\\sqrt{1-x^2}} \\frac{1}{\\pi} dy \\\\ \\\\ &= \\frac{2}{\\pi} \\sqrt{1-x^2} &\\quad \\text{where } -1 \\le x \\le 1 \\end{align} Notice that while the joint PDF is Uniform, the marginal PDF is not Uniform: it grows as $x$ approaches 0. By symmetry we can just replace $x$ with $y$ to get the marginal PDF of $Y$ $f_{Y}(y)$ #### Conditional PDF of $Y|X$ \\begin{align} f_{Y|X}(y|x) &= \\frac{f_{XY}(x,y)}{f_{X}(x)} \\\\ &= \\frac{\\frac{1}{\\pi}}{\\frac{2}{\\pi} \\sqrt{1-x^2}} \\\\ &= \\frac{1}{2 \\sqrt{1-x^2}} &\\quad \\text{if } -\\sqrt{1-x^2} \\le y \\le \\sqrt{1-x^2} \\end{align} But since we are treating $x$ as a constant, and the above equation does not depend on $y$, the conditional PDF $f_{Y|X}(y|x)$ is actually Uniform. \\begin{align} Y|X &\\sim", "The distance$AB$can be written in the form$\\frac{m}{n}$, where$m$and$n$are relatively prime positive integers. What is$m+n$? (A)$21$(B)$29$(C)$58$(D)$69$(E)$93$AMC 10A, 2018, Problem 16 Right triangle$ABC$has leg lengths$AB=20$and$BC=21$. Including$\\overline{AB}$and$\\overline{BC}$, how many line segments with integer length can be drawn from vertex$B$to a point on hypotenuse$\\overline{AC}$? (A)$5$(B)$8$(C)$12$(D)$13$(E)$15$AMC 10A, 2018, Problem 20 A scanning code consists of a$7 \\times 7$grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of$49$squares. A scanning code is called$\\text{symmetric}$if its look does not change when the entire square is rotated by a multiple of$90 ^{\\circ}$counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes? (A)$510$(B)$1022$(C)$8190$(D)$8192$(E)$65,534$AMC 10A, 2018, Problem 23 Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths$3$and$4$units. In the corner where those sides meet at a right angle, he leaves a small unplanted square$S$so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from$S$to the hypotenuse is$2$units. What fraction of the field is planted? (A)$ \\frac{25}{27}$(B)$ \\frac{26}{27}$(C)$\\frac{73}{75}$(D)$ \\frac{145}{147}$(E)$ \\frac{74}{75}$AMC 10A, 2018, Problem 24 Triangle$ABC$with$AB=50$and$AC=10$has area$120$. Let$D$be", "the shaded region HEFG? A) 4(3-$$\\pi$$) B) 2(4-$$\\pi$$) C) 4(4-$$\\pi$$) D) 2(6-$$\\pi$$) E) 8(1+$$\\pi$$) [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 33037 Followers: 5758 Kudos [?]: 70573 [1] , given: 9849 Re: H,G,F and E are midpoints of the sides of square ABCD [#permalink] Show Tags 11 Feb 2012, 01:14 1 KUDOS Expert's post nafishasan60 wrote: H,G,F and E are midpoints of the sides of square ABCD. Arcs FG and EH are centered at B and D respectively, as shown above. If the side of the square ABCD is 4, what is the area of the shaded region HEFG? A) 4(3-$$\\pi$$) B) 2(4-$$\\pi$$) C) 4(4-$$\\pi$$) D) 2(6-$$\\pi$$) E) 8(1+$$\\pi$$) The legs of right isosceles triangles FAE and GCH equal to half of the side of the square, so to 4/2=2, hence their combined area is $$2*(\\frac{1}{2}*2*2)=4$$; The same way the radii of arcs FG and EH equal to 4/2=2. Since both arcs are 90 degrees, then their commbined area is $$2*\\frac{90}{360}*\\pi*{r^2}=2\\pi$$; The are of the square is 4^2=16, thus the area of the shaded region is $$16-(4+2\\p)=2(6-\\pi)$$. Hope it's clear. _________________ Math Expert Joined: 02 Sep 2009 Posts: 33037 Followers: 5758 Kudos [?]: 70573 [0], given: 9849 Re: H,G,F and E are midpoints of the sides of square ABCD [#permalink] Show Tags 06 Mar 2014,", "☐C. 46 ☐D. 55 ☐E. 61 38- In the following figure, ABCD is a rectangle. If $$a=\\sqrt{5}$$, and $$b=2a$$, find the area of the shaded region. (the shaded region is a trapezoid) ☐A. $$\\sqrt{5}$$ ☐B. $$2$$ ☐C. $$4$$ ☐D. $$2\\sqrt{5}$$ ☐E. $$4\\sqrt{5}$$ 39- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively and DE = 5 and DF = 3. The area of ∆BED is 20, and the area of ∆BDF is 24. What is the perimeter of the rectangle? ☐A. 24 ☐B. 28 ☐C. 36 ☐D. 42 ☐E. 48 40- What is the product of all possible values of $$x$$ in the following equation? $$|2x+6-3x-7|=5$$ ☐A. $$-24$$ ☐B. $$-12$$ ☐C. $$8$$ ☐D. $$12$$ ☐E. $$24$$ 28% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "either parallel to or coinciding with the sides of the large square. In $\\Delta ABC$, $AB = AC$, and when $\\Delta ABC$ is folded over side BC, point A coincides with O, the center of square WXYZ. What is the area of $\\Delta ABC$, in square centimeters? $[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label(\"A\", A, NW); label(\"O\", O, NE); label(\"B\", B, SW); label(\"C\", C, NW); label(\"W\",W , NE); label(\"X\", X, N); label(\"Y\", Y, N); label(\"Z\", Z, SE); [/asy]$ $\\textbf{(A)}\\ \\frac{15}4\\qquad\\textbf{(B)}\\ \\frac{21}4\\qquad\\textbf{(C)}\\ \\frac{27}4\\qquad\\textbf{(D)}\\ \\frac{21}2\\qquad\\textbf{(E)}\\ \\frac{27}2$", "+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", "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? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "angle bisector of the angle at vertex $A$ in triangle $ABC$. Then $\\frac{BA'}{A'C}=\\frac{BA}{AC}$. \\end{lemma} \\begin{proof} From lemma ..., we know that $\\frac{BA'}{A'C}=\\frac{\\area(BA'A)}{\\area(A'CA)}$. Now the equality of angles $\\angle BAA'$ and $\\angle A'AC$ implies that if we reflect the triangle $A'CA$ in the line $AA'$, then the reflected copy will share an angle with triangle $BA'A$ and the reflection of $C$ will lie on the line $AB$ (see picture). \\begin{figure}[h] \\centering \\begin{asy} size (6cm,3cm); pair A, B, C, A1, C1; picture pic; A=(2,2); B=(0,0); C=(3,0); A1=(length(C-A)*B+length(B-A)*C)/(length(B-A)+length(C-A)); C1=A+2*dot(A1-A,C-A)/dot(A1-A,A1-A)*(A1-A)-(C-A); draw(A--B--C--cycle); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); draw(A--A1); label(\"$A'$\",A1,S); draw(pic,A--B--A1--cycle); label(pic,\"$A$\",A,N); label(pic,\"$A'$\",A1,S); label(pic,\"$B$\",B,S); draw(pic,A1--C--A,linetype(\"0 4\")); draw(pic,A1--C1--A); label(pic,\"$C$\",C1,WNW); \\end{asy} \\label{fig:bisector} \\end{figure} But in the picture after the reflection we see that $\\frac{\\area(BA'A)}{\\area(A'CA)}=\\frac{AB}{AC}$, by the same lemma. \\end{proof} This lemma implies that if $AA',BB'$ and $CC'$ are internal bisectors of triangle $ABC$, then $\\frac{BA'}{A'C}\\cdot\\frac{CB'}{B'A}\\cdot\\frac{AC'}{C'B}=\\frac{BA}{AC}\\cdot\\frac{CB}{BA}\\cdot\\frac{AC}{CB}=1$, so Ceva's theorem tells us that $AA', BB'$ and $CC'$ are concurrent. We can prove it in a different way. For this we first notice that the internal angle bisector of an angle $A$ is exactly the locus of points $P$ inside the angle that are equidistant from the two rays that form $A$. Indeed, let $P'$ and $P''$ be the feet of perpendiculars dropped from the point $P$ to the two rays. If the point $P$ is on the angle bisector of" ]

[ "is prime and greater than 1.", "# Prime + 4 = Square? How many primes are 4 less than a perfect square? × Problem Loading... Note Loading... Set Loading...", "found it to be extremely well-written, and fostered a love of pure math in me (an engineering student). 18. Feb 15, 2016 ### 3301 Maybe this can help with proofs: fb/brilliant Here you can do this problem: Find the prime number which is one less than a perfect square number?", "9 > 6: Less Than and Greater Than popolare:", "# One Less Than A Perfect Cube True or False? $$\\quad$$ 7 is the only prime number that is one less than a perfect cube. ×", "factor this yeah this would be X minus 4 times X minus 4 or you could just view this as X minus 4 squared is equal to 0 or X minus 4 is equal to 0 so so or where x equals 4 so G prime prime of 4 is equal to 0 so let's see what's happening at on either side of that let's see if G if we're actually if we're actually switching signs or not so let me draw a number line here and so this is so this is 2 3 4 5 and I could keep going and so we know that something interesting is happening right over here G prime prime of 4 is equal to 0 G prime prime of 4 is equal to 0 so let's think about what's what the second derivative is when we are less than 4 and so actually let me just try G prime prime of 0 since that'll be easy to evaluate G prime prime of 0 well it's just going to be equal to 48 so when we are less than 4 our second derivative G prime the second derivative is greater than 0 so we're actually going to be concave upwards over this interval to the left of 4 now let's think", "is not a cube and that a rational prime number such that it be a rationalprime number such as 4,7, 7, 9.", "p, there is a prime greater than p. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.", "one which is equal to 4 and this is our answer", "+0 # What is the greatest integer less than or equal to \\sqrt[3]{504}? 0 103 2 +182 What is the greatest integer less than or equal to \\sqrt[3]{504}? Aug 26, 2022 #1 +29 -3 What is the less than or equal to $$\\sqrt[3]{504}$$ $$\\sqrt[3]{504} = 7.95811441579$$ The greatest integer less than 7.95811441579 is 7, so the answer is 7. 7 Aug 26, 2022", "just factored that as a product of a perfect square and a prime number. Or actually, it doesn't have to be a prime number, just a non-perfect square. And then I get B is equal to 4 square roots of 7. So there we go. If this is 12, this is 6, this is 4 square roots of 7. I think that's all the time I have now for this presentation. Right after this, I'll do one more presentation where I give a couple of more Pythagorean theorem problems. See you soon.", "unmarked numbers that are smaller than its square are prime.", "fact 168 primes less than the 1000) CB", "prime will be no larger than n ln n + n ln ln n, for n ≥ 6.", "of 4 but 4 & 3 are relative prime and hence the answer.", "I just factored that as a product of a perfect and a prime number Or actually it doesn't have to be prime number just a non-perfect square And then I get B is equal to 4 square root of 7 So, there we go And this is 12, this is 6, this is 4 square root of 7 I think that's all the time I have, for this presentation Right after this, I'll do one more presentation where I give a couple of more Pythagorean Theorem problems See you soon", "One Less Than A Perfect Cube Number Theory Level 3 True or False? $$\\quad$$ 7 is the only prime number that is one less than a perfect cube. ×", "is 4. The answer will also be closer to 3 than to 4, because 32 is closer to 27 than it is to 64.", "would be greater than sqrt(n) * sqrt(n) = n. So in any factorization of n, at least one of the factors must be smaller than the square root of n, and if we can't find any factors less than or equal to the square root, n must be a prime.", "more efficient to check more numbers at higher $n$ for divisibility by smaller squares, which have a better hit rate. Of course the same analysis holds for the numbers $P_n-1$. [Update:] Given the quote from Concrete Mathematics in Gerry's comment, we know that the square of no prime up to $p_{3000000}\\approx5\\cdot10^7$ divides a Euclid number; so the remaining chance of finding one that does is only approximately $1-\\exp(-1/\\log3000000)\\approx6.5\\%$. -" ]

[ "$4n=96$ 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", "• 0 Votes - 0 Average • 1 • 2 • 3 • 4 • 5 Find $$n$$ such that both $$n-45$$ and $$n+44$$ are perfect squares 02-19-2010, 05:11 PM Post: #1 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 Find $$n$$ such that both $$n-45$$ and $$n+44$$ are perfect squares If this is possible, then for some $$k,m \\in \\mathbf{N}^{+}$$, we have (1) $$n-45 = m^2$$ (2) $$n+44=(m+k)^2$$ So $$89=2km+k^2=k(2m+k)$$. But $$89$$ is prime, thus $$k=1$$ and $$m=44$$ Therefore $$n=m^2+45=1981$$ One can easily verify that $$n-45=1936=44^2\\quad n+44= 2025=45^2$$ « Next Oldest | Next Newest » Forum Jump:", "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.", "a perfect square does not end in 7, so we can eliminate those cases to get the new list. $4 , 25 , 121 , 361$ Adding 48, we get 121 as the only possible solution. Hence the answer is (B). The only positive integer that satisfies both requirements is 11. edited by mobius247 ## Note: Solution 1 Since all primes greater than 2 are odd, we know that the difference between the squares of any two consecutive primes greater than 2 is at least $(p+2)^2-p^2=4p+4$, where p is the smaller of the consecutive primes. For $p>11$, $4p+4>48$. This means that the difference between the squares of any two consecutive primes both greater than 11 is greater than 48, so $n$ and $n+48$ can't both be the squares of primes if $n=p^2$ and p>11. So, we only need to check $n=2^2, 3^2, 5^2, 7^2, 11^2$. ~apsid ## Video Solution CHECK OUT Video Solution:https://youtu.be/IsqrsMkR-mA ~rudolf1279 ## Solution 2 If $P(n) = \\sqrt{n}$, then $n = p_{1}^{2}$, where $p_{1}$ is a prime number. If $P(n+48) = \\sqrt{n+48}$, then $n+48 = p_{2}^{2}$, where $p_{2}$ is a different prime number. So: $p_{2}^{2} = n+48$ $p_{1}^{2} = n$ $p_{2}^{2} - p_{1}^{2} = 48$ $(p_{2}+p_{1})(p_{2}-p_{1})=48$ Since $p_{1} > 0$ : $(p_{2}+p_{1}) > (p_{2}-p_{1})$. Looking at pairs of divisors of $48$, we have several possibilities", "Not an answer strictly speaking, but assuming that $n(n+1)$ is a square, it follows that $$4n(n+1) = 4n^2+4n = (2n+1)^2-1$$ is a square, too. However, there are pretty few consecutive squares. Much easier, don't you think? • I think I am missing the point of this, but as I read it, I was eager to understand exactly what you're getting at. Care to elaborate? – The Count Jun 27 '16 at 20:49 • @TheCount: Assuming that $n(n+1)$ is a square, there are two consecutive squares: $4n^2+4n$ and $4n^2+4n+1$. However, the only consecutive squares are $0$ and $1$, so $n(n+1)$ is a square only if $n=0$. – Jack D'Aurizio Jun 27 '16 at 20:51 • Wouldn't it be even easier to note that $n^2\\lt n(n+1)\\lt (n+1)^2$ or $n\\lt\\sqrt{n(n+1)}\\lt n+1$? – bof Jun 27 '16 at 20:53 • Ahhh, there we go. I don't know why that didn't click for me the first time. This is a good observation and a great example of a quick, elegant counterexample, if that's the right word. Thanks for the clarification! – The Count Jun 27 '16 at 20:53 Consider 2 cases: 1) $n$ is a perfect square and 2) $n$ is not a perfect square 1) Assume $n(n+1)$ is a perfect square. No prime factors of $n$ divide $n+1$ and vice versa. Then,", "as the sum of two squares in exactly one way (try 25 … 65). So the search changed from finding a factor to finding another sum of squares, and the 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. All-New Mathematics of Game Shows", "only two consecutive integers which are primes are 2 and 3. Thus, $$x=(prime_1)^2=4$$ and $$y=(prime_2)^2=9$$. The remainder when 9 is divided by 4 is 1. Sufficient. (2) Both $$\\sqrt{x}$$ and $$\\sqrt{y}$$ are prime numbers. The same here: $$\\sqrt{x}=2$$ and $$\\sqrt{y}=3$$. Sufficient. Hi There Bunel, Was looking at this solution, the question says that the X and Y are consecutive perfect squares. One could interpret it to mean that they are perfect squares that follow each other ie 4, 9, 16, 25 etc, that is the perfect squares or consecutive primes, which should be different in meaning from consecutive integers. How come the solution assumes that they are consecutive integers? while solving it, I was able to tell that the numbers are prime, but thought the answer cannot be ascertained because it could be any primes, any of the sets that follow each other, so I chose E instead. Please clarify. Notice that since x and y are consecutive perfect squares, then $$\\sqrt{x}$$ and $$\\sqrt{y}$$ are consecutive integers. x and y are consecutive perfect squares, so x and y could be: $$x = 1$$ and $$y = 4$$ --> $$\\sqrt{x}=1$$ and $$\\sqrt{y}=2$$, consecutive integers; $$x = 4$$ and $$y = 9$$ --> $$\\sqrt{x}=2$$ and $$\\sqrt{y}=3$$, consecutive integers; $$x = 9$$ and $$y = 16$$ --> $$\\sqrt{x}=3$$ and $$\\sqrt{y}=4$$, consecutive", "excited about this puzzle that I tried to see if I could come up with more examples, but I always got stuck. Sometimes, the puzzle wouldn’t have a solution. For example, there is no solution to the $1$-to-$8$ version of the puzzle. Other times, there is a solution, but finding it is a bit messy. To see what I mean, try the $1$-to-$32$ version of the puzzle. The answer is nice, but it can feel a bit tedious to work out. (Some people have explored other avenues of this puzzle. Apparently, the numbers from $1$ to $305$ can be rearranged so that each consecutive pair adds up to a perfect cube. There are also variants where the sums of consecutive numbers are primes instead.) So, I hope you enjoy this puzzle as much as my son and I did. It doesn’t get any better.", "Darren Salas 2023-02-25 find a perfect square close to 26. Jamarion Turner We know that ${5}^{2}=25$ and ${6}^{2}=36$. So we see that 26 is the prime number, so we can only use calculator for finding answer And answer is $\\sqrt{26}\\approx 5.099$ Do you have a similar question?", "to scale down.) Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\\boxed{260}$.", "is a perfect cube. The powers of the primes of a perfect cube are always multiples of 3. For example, $$1,000=2^3*5^3$$: the powers of 2 and 5 are multiples of 3. Now, since $$8=2^3$$, $$9=3^2$$, and $$10=2*5$$ are factors of n, then 2, 3, and 5 are factors of n, therefore $$2^3=8$$, $$3^3=27$$ and $$5^3=125$$ must be the factors of n. A. 16=2^4: not necessarily a factor of n. B. 175=5^2*7: not necessarily a factor of n. C. 225=3^2*5^2: since 3^3=27 and 5^3=125 are the factors of n, then 225 is also a factor of n. D. 275=5^2*11: not necessarily a factor of n. E. 625=5^4: not necessarily a factor of n. P.S. Please post PS questions in PS forum. Hi Bunuel, Thanks for the explanation.. The rule you have used - is this a general rule? The powers of the primes of a perfect cube are always multiples of 3. For example, 1,000=23∗53 Is this true for all powers, i.e., the powers of the primes of a perfect square are always multiples of 2, for perfect (rasied to 4) they are multiples of 4 and so on? Let me try to explain. By definition of a cube or a square , a number is a perfect square when N = $$A^{2p}$$ and a number is a perfect", "Feeds: Posts ## 20S. Sequences of Consecutive Integers A37 If $n$ is a natural number, prove that the number $(n + 1)(n + 2)\\cdots(n + 10)$ is not a perfect square. A9 Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others. O51 Prove the among $16$ consecutive integers it is always possible to find one which is relatively prime to all the rest. Here is the official solution file: PEN A9 A37 O51", "difference of cubes. 1. $x^3-27$ 2. $64+x^3$ 3. $32x^3-4$ 4. $64x^3+343$ 5. $512-729x^3$ 6. $125x^4+8x$ 7. $648x^3+81$ 8. $5x^6-135x^3$ 9. $686x^7-1024x^4$ Find the real-number solutions for each equation. 1. $125x^3+1=0$ 2. $64-729x^3 = 0$ 3. $8x^4-343x = 0$ 4. Challenge Find ALL solutions (real and imaginary) for $5x^5+625x^2 = 0$ . 5. Challenge Find ALL solutions (real and imaginary) for $686x^3+2000 = 0$ . 6. Real Life Application You have a piece of cardboard that you would like to fold up and make an open (no top) box out of. The dimensions of the cardboard are $36^{\\prime\\prime} \\times 42^{\\prime\\prime}$ . Write a factored equation for the volume of this box. Find the volume of the box when $x = 1, 3,$ and 5.", "have $n^2-(n-1)^2$ or $(n+1)^2-n^2$. Whether it's $+1$ or $-1$ is irrelevant in the end. The important point is that it is an odd number, and for the difference of the next pair of squares, it's the next odd number. So I'd say $2n-1$ is a good answer, and there is no need for you to correct it. – Arthur Nov 22 '16 at 14:10 Note that $$n^2+20n+11=(n+10)^2-89$$ so we have to find all the pairs of squares $a^2$, $b^2$ such that $a^2-b^2=89$. Of course, $a$ and $b$ are integers. We see that $|a|>|b|$. Since $a^2-b^2=(a+b)(a-b)$ and $89$ is a prime number, then we have these possibilities: • $a+b=89$ and $a-b=1$, that is, $a=45$, $b=44$. • $a+b=1$ and $a-b=89$, that is, $a=45$, $b=-44$. • $a+b=-89$ and $a-b=-1$, that is, $a=-45$, $b=-44$. • $a+b=-1$ and $a-b=-89$, that is, $a=-45$, $b=44$. So clearly, $a^2=2025$ and $b^2=1936$. Then $n+10=\\pm45$ which gives two values for $n$, namely $n=35$ and $n=-55$. EDIT: How can be proved that there are no more solutions? Indeed, this is already done. Our reasoning is like this: • Proposition $A$: \"$n$ is such that $n^2-20n+11$ is a perfect square\". • Proposition $B$: \"$n$ is $35$ or $-55$\". We have shown, assuming that $n$ is integer, that $A$ implies $B$. So if $n$ is another integer number, $B$ is false.", "46 g 1 1 1 5 3 1 1 1 1 11 3 1 1 1 1 1 1 1 1 1 1 23 From the given initial values $$n = 2$$, $$a = 7$$, we first calculate $$g = \\gcd(2, 7) = 1$$. Then $$n$$ and $$a$$ are updated: $$n = n + 1$$, $$a = a + g$$. On the next round the gcd operation again yields a $$1$$: $$g = \\gcd(3, 8) = 1$$. But on the fourth iteration we finally get a prime: $$g = \\gcd(5, 10) = 5$$. The assertion that $$g$$ is always either $$1$$ or a prime is equivalent to saying that $$n$$ and $$a$$ have at most one prime factor in common. This curious generator of primes was discovered in 2003, during a summer school exploring Stephen Wolfram’s “New Kind of Science.” A group led by Matthew Frank investigated various nested recursions, including this one: $a(n) = a(n-1) + gcd(n, a(n-1)).$ With the initial condition $$a(1) = 7$$, the sequence begins: 7, 8, 9, 10, 15, 18, 19, 20, 21, 22, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 69, . . .oeis.org/A106108 Participants noticed that the sequence of first differences — $$a(n) - a(n-1)$$ — seemed to consist entirely of $$1$$s and primes: 1, 1,", "Number 83 is from 50 to 100, so 832=200(83-50)+(100-83)2=200(33)+172=6600+289=6889 ANSWER: a) 1296 b) 2401 c) 6889 1. Which of the following numbers is a square number? Explain why. a) 11025; b) 32571. SOLUTION: a) Find the prime factorization of the given number: 11025=5×2205=5×5×441=5×5×3×147=5×5×3×3×49 =5×5×3×3×7×7=32 x 52 x 72 So, number 11025=(3×5×7)2=1052 is a perfect square number. b) Find the prime factorization of the given number: 32571=3×10857=3×3×3619=3×3×7×517 =3×3×7×11×47=32×7×11×47 Since the prime factorization of number 32571 doesn’t consists only of even powers of prime numbers, then given number is not a perfect square number. ANSWER: a) square number b) not square number 1. Find all natural numbers n for which n+200 is a perfect square number. SOLUTION: The smallest perfect square number greater than 200 is 225, so the smallest possible value of n is 225-200=25. The next perfect square number greater than 225 is 256, so the next possible value of n is 256-200=56. The difference between 56 and 25 is 31. This means that the next difference between two consecutive values of n will be 31+2=33, and so on. To see this pattern, fill in the following table. If you want to write general rule describing all possible values of n, this can be done using following recurrent formulas: n1=25, nk=nk-1+(2k+27), k= 2, 3, 4, … ANSWER: nk=nk-1+(2k+27),", "$3x^{2} + y^{2} = n$ is odd and the number is not a square of another integer. • 11, 23, 47, or 59 have a modulo-twelve remainder of 11. These numbers are prime iff the number solutions to $3x^{2} - y^{2} = n$ is odd and the number is not a square of another integer.", "# Squares and primes A few questions I've been wondering about regarding primes and squares that I thought the Brilliant community might have some insight on ... (i) Are there an infinite number of perfect squares that are the averages of consecutive primes? As examples, $4$ is the average of consecutive primes $3$ and $5$, and $9$ is the average of consecutive primes $7$ and $11$. $25$ is not such a perfect square as its \"neighboring\" primes are $23$ and $29$, which average to $26$. The list of such perfect squares begins as $4, 9, 16, 64, 81, 144, 225, 324, 441, 625, 1089, 1681, 2601, ...$. It becomes less and less likely as the squares get larger that they will be the average of successive primes, but the notion that there is a largest such square seems unlikely. So the problem is to either prove that there is no such largest square, or prove that there must be one and in fact identify it. (ii) Can every perfect square $n^{2} \\gt 1$ be expressed as the average of two (not necessarily consecutive) primes? While $25$ is not the average of two consecutive primes, it is the average of primes $19$ and $31$. $121$ is not the average of consecutive primes, but it is the average of primes", "a $7 \\times 7$ square with identical squares. Dan (I am not using the students’ real names here) immediately said, “Only with $1 \\times 1$, because 7 is prime”. “No,” countered Alejandro. “You can tile it with one big fat $7 \\times 7$ square. Does that count?” “Well,” said Titus, “A prime number has only two divisors: one and itself. So we can use the same idea to count these tilings, if we count $7 \\times 7$ as a tiling.” Titus may have wanted simply to show what he already knew. But this seemingly innocent and perhaps boastful remark turned out to be a fertile one. Dan generalized immediately: “For an $8 \\times 8$ square, there are four tilings.” (He meant tilings with identical squares, and everyone knew it.) “That’s because 8 has four divisors: 1, 2, 4, and 8.” And this is where we came in. Alejandro took up Xue’s challenge, and his ten-year old explanation was wonderfully simple. “Say there are two primes, $p$ and $q$. Say the number is $p^2q^3$ You just make a picture.” And he drew this on the shared screen: In another group, Alejandro’s explanation would have been a mystery. But these four looked at it and understood. “You need a 1 to count the 1,” said Dan, “and also the singles:", "× # Squares and primes A few questions I've been wondering about regarding primes and squares that I thought the Brilliant community might have some insight on ... (i) Are there an infinite number of perfect squares that are the averages of consecutive primes? As examples, $$4$$ is the average of consecutive primes $$3$$ and $$5$$, and $$9$$ is the average of consecutive primes $$7$$ and $$11$$. $$25$$ is not such a perfect square as its \"neighboring\" primes are $$23$$ and $$29$$, which average to $$26$$. The list of such perfect squares begins as $$4, 9, 16, 64, 81, 144, 225, 324, 441, 625, 1089, 1681, 2601, ...$$. It becomes less and less likely as the squares get larger that they will be the average of successive primes, but the notion that there is a largest such square seems unlikely. So the problem is to either prove that there is no such largest square, or prove that there must be one and in fact identify it. (ii) Can every perfect square $$n^{2} \\gt 1$$ be expressed as the average of two (not necessarily consecutive) primes? While $$25$$ is not the average of two consecutive primes, it is the average of primes $$19$$ and $$31$$. $$121$$ is not the average of consecutive primes, but it is the average of" ]

[ "a perfect square does not end in 7, so we can eliminate those cases to get the new list. $4 , 25 , 121 , 361$ Adding 48, we get 121 as the only possible solution. Hence the answer is (B). The only positive integer that satisfies both requirements is 11. edited by mobius247 ## Note: Solution 1 Since all primes greater than 2 are odd, we know that the difference between the squares of any two consecutive primes greater than 2 is at least $(p+2)^2-p^2=4p+4$, where p is the smaller of the consecutive primes. For $p>11$, $4p+4>48$. This means that the difference between the squares of any two consecutive primes both greater than 11 is greater than 48, so $n$ and $n+48$ can't both be the squares of primes if $n=p^2$ and p>11. So, we only need to check $n=2^2, 3^2, 5^2, 7^2, 11^2$. ~apsid ## Video Solution CHECK OUT Video Solution:https://youtu.be/IsqrsMkR-mA ~rudolf1279 ## Solution 2 If $P(n) = \\sqrt{n}$, then $n = p_{1}^{2}$, where $p_{1}$ is a prime number. If $P(n+48) = \\sqrt{n+48}$, then $n+48 = p_{2}^{2}$, where $p_{2}$ is a different prime number. So: $p_{2}^{2} = n+48$ $p_{1}^{2} = n$ $p_{2}^{2} - p_{1}^{2} = 48$ $(p_{2}+p_{1})(p_{2}-p_{1})=48$ Since $p_{1} > 0$ : $(p_{2}+p_{1}) > (p_{2}-p_{1})$. Looking at pairs of divisors of $48$, we have several possibilities", "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.", "# Squares and primes A few questions I've been wondering about regarding primes and squares that I thought the Brilliant community might have some insight on ... (i) Are there an infinite number of perfect squares that are the averages of consecutive primes? As examples, $4$ is the average of consecutive primes $3$ and $5$, and $9$ is the average of consecutive primes $7$ and $11$. $25$ is not such a perfect square as its \"neighboring\" primes are $23$ and $29$, which average to $26$. The list of such perfect squares begins as $4, 9, 16, 64, 81, 144, 225, 324, 441, 625, 1089, 1681, 2601, ...$. It becomes less and less likely as the squares get larger that they will be the average of successive primes, but the notion that there is a largest such square seems unlikely. So the problem is to either prove that there is no such largest square, or prove that there must be one and in fact identify it. (ii) Can every perfect square $n^{2} \\gt 1$ be expressed as the average of two (not necessarily consecutive) primes? While $25$ is not the average of two consecutive primes, it is the average of primes $19$ and $31$. $121$ is not the average of consecutive primes, but it is the average of primes", "Not an answer strictly speaking, but assuming that $n(n+1)$ is a square, it follows that $$4n(n+1) = 4n^2+4n = (2n+1)^2-1$$ is a square, too. However, there are pretty few consecutive squares. Much easier, don't you think? • I think I am missing the point of this, but as I read it, I was eager to understand exactly what you're getting at. Care to elaborate? – The Count Jun 27 '16 at 20:49 • @TheCount: Assuming that $n(n+1)$ is a square, there are two consecutive squares: $4n^2+4n$ and $4n^2+4n+1$. However, the only consecutive squares are $0$ and $1$, so $n(n+1)$ is a square only if $n=0$. – Jack D'Aurizio Jun 27 '16 at 20:51 • Wouldn't it be even easier to note that $n^2\\lt n(n+1)\\lt (n+1)^2$ or $n\\lt\\sqrt{n(n+1)}\\lt n+1$? – bof Jun 27 '16 at 20:53 • Ahhh, there we go. I don't know why that didn't click for me the first time. This is a good observation and a great example of a quick, elegant counterexample, if that's the right word. Thanks for the clarification! – The Count Jun 27 '16 at 20:53 Consider 2 cases: 1) $n$ is a perfect square and 2) $n$ is not a perfect square 1) Assume $n(n+1)$ is a perfect square. No prime factors of $n$ divide $n+1$ and vice versa. Then,", "Let $n = 3a+2$, so $(n+1)(7n+1) = (3a+3)(21a+15) = 9(a+1)(7a+5)$. Since 3 is divided out of $n+1$ and $7n+1$, $a+1$ and $7a+5$ are relatively prime, so $a+1$ and $7a+5$ must be perfect squares. However, since 5 is not a quadratic residue of 7, the GCD of $n+1$ and $7n+1$ can not be 3. • If $\\gcd(n+1,7n+1) = 2$, then $n \\equiv 1 \\pmod{2}$. Let $n = 2a+1$, so $(n+1)(7n+1) = (2a+2)(14a+8) = 4(a+1)(7a+4)$. Since 2 is divided out of $n+1$ and $7n+1$, $a+1$ and $7a+4$ are relatively prime, so $a+1$ and $7a+4$ must be perfect squares. We also know that $n+1$ and $7n+1$ do not share a factor of 3, so $n \\equiv 1,3 \\pmod{6}$. That means $n \\le 2007$, so $a \\le 1003$. After trying values of $a$ that are one less than a perfect square, we find that the largest value that makes $(n+1)(7n+1)$ a perfect square is $a = 960$. That means $n = 1921$. • If $\\gcd(n+1,7n+1) = 1$, then $n+1 \\equiv 1,5 \\pmod{6}$ (to avoid common factors that are factors of 6), so $n \\equiv 0,4 \\pmod{6}$. After trying values of $n$ that are one less than a perfect square, we find that the largest value that makes $(n+1)(7n+1)$ a perfect square is $n = 120$ (we could also stop searching once $n$", "× # Squares and primes A few questions I've been wondering about regarding primes and squares that I thought the Brilliant community might have some insight on ... (i) Are there an infinite number of perfect squares that are the averages of consecutive primes? As examples, $$4$$ is the average of consecutive primes $$3$$ and $$5$$, and $$9$$ is the average of consecutive primes $$7$$ and $$11$$. $$25$$ is not such a perfect square as its \"neighboring\" primes are $$23$$ and $$29$$, which average to $$26$$. The list of such perfect squares begins as $$4, 9, 16, 64, 81, 144, 225, 324, 441, 625, 1089, 1681, 2601, ...$$. It becomes less and less likely as the squares get larger that they will be the average of successive primes, but the notion that there is a largest such square seems unlikely. So the problem is to either prove that there is no such largest square, or prove that there must be one and in fact identify it. (ii) Can every perfect square $$n^{2} \\gt 1$$ be expressed as the average of two (not necessarily consecutive) primes? While $$25$$ is not the average of two consecutive primes, it is the average of primes $$19$$ and $$31$$. $$121$$ is not the average of consecutive primes, but it is the average of", "Feeds: Posts ## 20S. Sequences of Consecutive Integers A37 If $n$ is a natural number, prove that the number $(n + 1)(n + 2)\\cdots(n + 10)$ is not a perfect square. A9 Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others. O51 Prove the among $16$ consecutive integers it is always possible to find one which is relatively prime to all the rest. Here is the official solution file: PEN A9 A37 O51", "squares? I realize that it states that n is positive, but if f(n) is defined as perfect squares less than n and zero is less than, why do we not count it? _________________ Any and all kudos are greatly appreciated. Thank you. Kudos [?]: 22 [1], given: 9 Math Expert Joined: 02 Sep 2009 Posts: 41892 Kudos [?]: 129049 [0], given: 12187 ### Show Tags 03 Apr 2014, 08:44 MrWallSt wrote: Bunuel wrote: 4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range: A. 30 < x < 36 B. 30 < x < 37 C. 31 < x < 37 D. 31 < x < 38 E. 32 < x < 38 Perfect squares: 1, 4, 9, 16, 25, 36, .., Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ... If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15. If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11", "Show Tags 03 May 2018, 15:25 goodyear2013 wrote: Given that $$N = a^3b^4c^5$$ where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power? A. $$a^3b^4c^5$$ B. $$a^5b^4c^3$$ C. $$a^2b^3c^5$$ D. $$a^7b^6c^5$$ E. $$a^{27}b^{26}c^{25}$$ The exponents of a perfect square are all multiples of 2. The exponents of a perfect cube must be all multiples of 3. And the exponents of a perfect fifth power must be all multiples of 5. Since we need N to be a perfect square, perfect cube, and perfect fifth power, we need each exponent to be a multiple of the LCM of 2, 3, and 5, which is 30. Hence, we know that the smallest number of each exponent must be 30. The original exponent of a is 3; so we need 27 more a’s to get to 30: a^3 x a^27 = a^30 The original exponent of b is 4; so we need 26 more b’s to get to 30: b^4 x b^26 = b^30 The original exponent of c is 5; so we need 25 more c’s to get to 30: c^5 x c^25 = c^30 Therefore, we need to multiply N = a^3b^4c^5", "## Prime Square Differences #### Problem Prove that all primes greater than 2 can be written as the difference of two squares. #### Solution By considering the difference of two consecutive square numbers, (n + 1)2 n2 = (n2 + 2n + 1) n2 = 2n + 1 Hence all odd numbers greater than 1, which must include all primes greater than 2, can be written as the difference of two square numbers. For example, 83 = 241 + 1 = 422 412 Investigate primes that can be written as the difference of square numbers in general, including non-consecutive squares. Problem ID: 94 (Dec 2002) Difficulty: 2 Star Only Show Problem", "\\le (2k!-2k-5)-(k!+2) = k! -2k-7,$$ a slight improvement for the bound in the absence of factorial primes. $k! + n$ is composite for all $n$ until $n$ is a prime larger than $k$. Otherwise, $n$ must be divisible by a prime smaller or equal to k. There is no general formula for the distance to a prime, so we can only give an inequality or look at the average maximum $n$.", "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", "as the sum of two squares in exactly one way (try 25 … 65). So the search changed from finding a factor to finding another sum of squares, and the 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. All-New Mathematics of Game Shows", "is a perfect cube. The powers of the primes of a perfect cube are always multiples of 3. For example, $$1,000=2^3*5^3$$: the powers of 2 and 5 are multiples of 3. Now, since $$8=2^3$$, $$9=3^2$$, and $$10=2*5$$ are factors of n, then 2, 3, and 5 are factors of n, therefore $$2^3=8$$, $$3^3=27$$ and $$5^3=125$$ must be the factors of n. A. 16=2^4: not necessarily a factor of n. B. 175=5^2*7: not necessarily a factor of n. C. 225=3^2*5^2: since 3^3=27 and 5^3=125 are the factors of n, then 225 is also a factor of n. D. 275=5^2*11: not necessarily a factor of n. E. 625=5^4: not necessarily a factor of n. P.S. Please post PS questions in PS forum. Hi Bunuel, Thanks for the explanation.. The rule you have used - is this a general rule? The powers of the primes of a perfect cube are always multiples of 3. For example, 1,000=23∗53 Is this true for all powers, i.e., the powers of the primes of a perfect square are always multiples of 2, for perfect (rasied to 4) they are multiples of 4 and so on? Let me try to explain. By definition of a cube or a square , a number is a perfect square when N = $$A^{2p}$$ and a number is a perfect", "# When is 4 exceptional? What makes the number $4$ interesting? Obviously there will be many answers to this. Here are a couple of special things about $4$ which have turned up in my first year module recently (and in one case this was unplanned!) • $4$ is the only square number which is exactly one more than a prime number • $4$ is the only positive integer $n$ such that $n$ is not prime, and yet $(n-1)!$ is not divisible by $n$. (Here we follow the convention that $0!=1$.) Do you have any other favourite exceptional properties of $4$?", "only two consecutive integers which are primes are 2 and 3. Thus, $$x=(prime_1)^2=4$$ and $$y=(prime_2)^2=9$$. The remainder when 9 is divided by 4 is 1. Sufficient. (2) Both $$\\sqrt{x}$$ and $$\\sqrt{y}$$ are prime numbers. The same here: $$\\sqrt{x}=2$$ and $$\\sqrt{y}=3$$. Sufficient. Hi There Bunel, Was looking at this solution, the question says that the X and Y are consecutive perfect squares. One could interpret it to mean that they are perfect squares that follow each other ie 4, 9, 16, 25 etc, that is the perfect squares or consecutive primes, which should be different in meaning from consecutive integers. How come the solution assumes that they are consecutive integers? while solving it, I was able to tell that the numbers are prime, but thought the answer cannot be ascertained because it could be any primes, any of the sets that follow each other, so I chose E instead. Please clarify. Notice that since x and y are consecutive perfect squares, then $$\\sqrt{x}$$ and $$\\sqrt{y}$$ are consecutive integers. x and y are consecutive perfect squares, so x and y could be: $$x = 1$$ and $$y = 4$$ --> $$\\sqrt{x}=1$$ and $$\\sqrt{y}=2$$, consecutive integers; $$x = 4$$ and $$y = 9$$ --> $$\\sqrt{x}=2$$ and $$\\sqrt{y}=3$$, consecutive integers; $$x = 9$$ and $$y = 16$$ --> $$\\sqrt{x}=3$$ and $$\\sqrt{y}=4$$, consecutive", "N squares less than N. Dixon's method replaces the condition \"is the square of an integer\" with the much weaker one \"has only small prime factors\"; for example, there are 292 squares smaller than 84923; 662 numbers smaller than 84923 whose prime factors are only 2,3,5 or 7; and 4767 whose prime factors are all less than 30. (Such numbers are called B-smooth with respect to some bound B.) If we have lots of numbers $a_1 \\ldots a_n$ whose squares can be factorized as $a_i^2 \\mod N = \\prod_{j=1}^m b_j^{e_{ij}}$ for a fixed set $b_1 \\ldots b_m$ of small primes, linear algebra modulo 2 on the matrix $e_{ij}$ will give us a subset of the $a_i$ whose squares combine to a product of small primes to an even power — that is, a subset of the $a_i$ whose squares multiply to the square of a (hopefully different) number mod N. ## Method Suppose we are trying to factor the composite number N. We choose a bound B, and identify the factor base (which we will call P), the set of all primes less than or equal to B. Next, we search for positive integers z such that z2 mod N is B-smooth. We can therefore write, for suitable exponents ak, $z^2 \\equiv \\prod_{p_i\\in P} p_i^{a_i} \\pmod{N}$ When we", ", 361$ Adding $48$, we get $121$ as the only possible solution. Hence the answer is $\\boxed{\\textbf{(B) }1}$. edited by mobius247 ## Note: Solution 1 Since all primes greater than $2$ are odd, we know that the difference between the squares of any two consecutive primes greater than $2$ is at least $(p+2)^2-p^2=4p+4$, where p is the smaller of the consecutive primes. For $p>11$, $4p+4>48$. This means that the difference between the squares of any two consecutive primes both greater than $11$ is greater than $48$, so $n$ and $n+48$ can't both be the squares of primes if $n=p^2$ and $p>11$. So, we only need to check $n=2^2, 3^2, 5^2, 7^2,$ and $11^2$. ~apsid ## Video Solution CHECK OUT Video Solution:https://youtu.be/IsqrsMkR-mA ~rudolf1279 ## Solution 2 If $P(n) = \\sqrt{n}$, then $n = p_{1}^{2}$, where $p_{1}$ is a prime number. If $P(n+48) = \\sqrt{n+48}$, then $n+48 = p_{2}^{2}$, where $p_{2}$ is a different prime number. So: $p_{2}^{2} = n+48$ $p_{1}^{2} = n$ $p_{2}^{2} - p_{1}^{2} = 48$ $(p_{2}+p_{1})(p_{2}-p_{1})=48$ Since $p_{1} > 0$ : $(p_{2}+p_{1}) > (p_{2}-p_{1})$. Looking at pairs of divisors of $48$, we have several possibilities to solve for $p_{1}$ and $p_{2}$: $(p_{2}+p_{1}) = 48$ $(p_{2}-p_{1}) = 1$ $p_{1} = \\frac{47}{2}$ $p_{2} = \\frac{49}{2}$ $(p_{2}+p_{1}) = 24$ $(p_{2}-p_{1}) = 2$ $p_{1} = 11$ $p_{2} = 13$ $(p_{2}+p_{1}) =", "Darren Salas 2023-02-25 find a perfect square close to 26. Jamarion Turner We know that ${5}^{2}=25$ and ${6}^{2}=36$. So we see that 26 is the prime number, so we can only use calculator for finding answer And answer is $\\sqrt{26}\\approx 5.099$ Do you have a similar question?", "as they would have to be much smaller than q as n is greater than 4x10^18. This implies that all odd numbers less than 15x10^18 are the sum of 3 distinct primes. For the first part to do with prime ladders, I misstated the problem. The difference between consecutive primes on the ladder could be between 4 and 4*10^18 and the issue was that if the gaps between consecutive primes in the ladder, p and q, p > q, were 4x10^18 apart then if n equals p+2, p+4 or p+6, n-q > 4x10^18 and n-p = 2, 4 or 6 none of which are the sum of 2 distinct primes. This is all from page 28 of Helfgott's arxiv pdf released in 2015. He shows that 4x10^18 + 2 is the sum of two distinct primes on pg 28 and on pg 305 states that there is a prime ladder from 3 to 8.8x10^30 with consecutive primes on the ladder having difference less than or equal to 4x10^18 - 6. So if p and q are any consecutive primes on the ladder, as they have difference less than or equal to 4x10^18 - 6 then for any odd n between q + 8 and p + 6 inclusive, n - q is less than 4x10^18 and greater than" ]

[ "\\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", "(c) $\\begin{array}{l}\\text{Size of an interior angle of a regular hexagon}\\\\ =\\left(1-\\frac{2}{6}\\right)×{180}^{\\circ }\\\\ =\\frac{2}{3}×{180}^{\\circ }\\\\ ={120}^{\\circ }\\end{array}$", "endpoints of a side of the larger hexagon, as shown in the diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy]", "1cm, outer radius = 3cm, ruler position= 10 right • inner radius = 2.5cm, outer radius = 4.5cm, ruler position = 1 You can also use this chart to visualize binary numbers. left: • the third 3 bit binary number (binary 010 = octal 2 = decimal 2 = hexadecimal 2) • inner radius = 1cm, outer radius = 2cm, ruler position = 3 right: • the sixteenth 4 bit binary number (binary 1111 = octal 17 = decimal 15 = hexadecimal F) • inner radius = 2cm, outer radius = 3cm , ruler position = 16 - To observe how arcs and filling work you can add inside your tikzpicture : %How arc works \\draw[blue,->] (0,3.5) arc [radius=3.5, start angle=90, delta angle=90]; %Now filling \\filldraw[fill=gray] (3.5,0) arc [radius=3.5, start angle=0, delta angle=90] -- (0,4) arc [radius=4, start angle=90, delta angle=-90] -- cycle; \\filldraw[fill=red] (0,-3.5) arc [radius=3.5, start angle=270, delta angle=90] -- (4,0) arc [radius=4, start angle=0, delta angle=-90] -- cycle; And here is an example within a loop, where a bit of trigonometry gives you the formula: \\foreach \\a in {0,1} { % filling in the loop \\filldraw[fill=green!50!black] ({3.5*cos((-\\a*(pi/8)+(pi/2)) r)},{3.5*sin((-\\a*(pi/8)+(pi/2)) r)}) arc [radius=3.5, start angle={((pi/2)+(-pi/8)*\\a) r}, delta angle={(-pi/8) r}] -- ({4*cos((-(\\a+1)*(pi/8)+(pi/2)) r)},{4*sin((-(\\a+1)*(pi/8)+(pi/2)) r)}) arc [radius=4, start angle={((pi/2)+(-pi/8)*(\\a+1)) r}, delta angle={pi/8 r}] -- cycle; % corresponding arcs \\draw[thick,blue,->]", "\\end{array}} \\right]$$ scale by $$1/ \\sqrt{7}$$, rotate by A $${f_6}({\\bf{x}}) = \\left[ {\\begin{array}{*{20}{c}} { -2/7} & {- \\sqrt 3 /7} \\\\ { \\sqrt 3 /7} & { -2/7} \\\\ \\end{array}} \\right]{\\bf{x}} + \\left[ {\\begin{array}{*{20}{c}} {9/14} \\\\ {- \\sqrt 3 / 14} \\\\ \\end{array}} \\right]$$ scale by $$1/ \\sqrt{7}$$, rotate by A+120° $${f_7}({\\bf{x}}) = \\left[ {\\begin{array}{*{20}{c}} { 5/14} & {- \\sqrt 3 / 14} \\\\ { \\sqrt 3 / 14} & { 5/14} \\\\ \\end{array}} \\right]{\\bf{x}} + \\left[ {\\begin{array}{*{20}{c}} {9/14} \\\\ {- \\sqrt 3 / 14} \\\\ \\end{array}} \\right]$$ scale by $$1/ \\sqrt{7}$$, rotate by A Start with a pattern of seven regular hexagons. We will place those hexagons so that they lie with two vertices at (0,0) and (1,0). Eight vertices of the smaller hexagons are connected as shown above to create the basic red motif. We can think of each red segment as having been scaled by a factor r from the horizontal unit line from (0,0) to (1,0), rotated by an appropriate angle, then translated. To find these values, we need to first determine the scaling factor r and the rotation for the first line segment starting at (0,0). Consider the blue triangle shown in the figure above. Let A be the angle at the origin. The upper angle is 120° by properties of regular hexagons.", "then figure out how many degrees are in each angle of a square. There are 360\\begin{align*}360^\\circ\\end{align*} in a circle and 90\\begin{align*}90^\\circ\\end{align*} in each interior angle of a square, so 36090=4\\begin{align*}\\frac{360}{90}=4\\end{align*} squares will fit around one point. Watch this video for help with the Examples above. #### Concept Problem Revisited You could tessellate a regular hexagon over a plane with no overlaps or gaps because each of its interior angles is 120\\begin{align*}120^\\circ\\end{align*}. Three hexagons whose angles sum to 360\\begin{align*}360^\\circ\\end{align*} surround each point in the tessellation. ### Vocabulary A tessellation is a tiling over a plane with one or more figures such that the figures fill the plane with no overlaps and no gaps. ### Guided Practice 1. How many regular hexagons will fit around one point? 2. Does a regular octagon tessellate? 3. Tessellations can also be much more complicated. Check out http://www.mathsisfun.com/geometry/tessellation.html to see other tessellations and play with the Tessellation Artist, which has a link at the bottom of the page. 1. First, recall how many degrees are in a circle, and then figure out how many degrees are in each angle of a regular hexagon. There are 360\\begin{align*}360^\\circ\\end{align*} in a circle and 120\\begin{align*}120^\\circ\\end{align*} in each interior angle of a hexagon, so 360120=3\\begin{align*}\\frac{360}{120}=3\\end{align*} hexagons will fit around one point. 2. First, recall that there are 1080\\begin{align*}1080^\\circ\\end{align*} in", "(0.5*#1,0cm) rectangle +(0.5*#1,0.5*#1); \\fill[#4] (0,0.5*#1) rectangle +(0.5*#1,0.5*#1); \\fill[#5] (0.5*#1,0.5*#1) rectangle +(0.5*#1,0.5*#1); } \\newcommand\\MySquare[5]{% \\FillSquare{#1}{#2}{#3}{#4}{#5} \\draw (0,0) rectangle (#1,#1); \\draw (0,0.5*#1) -- +(#1,0); \\draw (0.5*#1,0) -- +(0,#1); } \\newcommand\\MyCircle[5]{% \\fill[#4] (0,0) arc[radius=0.5*#1,start angle=180,end angle=90] -- +(0,-0.5*#1) --cycle; \\fill[#5] (#1,0) arc[radius=0.5*#1,start angle=0,end angle=90] -- +(0,-0.5*#1) --cycle; \\fill[#3] (#1,0) arc[radius=0.5*#1,start angle=0,end angle=-90] -- +(0,0.5*#1) -- cycle; \\fill[#2] (0.5*#1,-0.5*#1) arc[radius=0.5*#1,start angle=-90,end angle=-180] -- +(0.5*#1,0) -- cycle; \\draw (0,0) -- +(#1,0); \\draw (0.5*#1,-0.5*#1) -- +(0,#1); } \\newcommand\\MyShape[9]{% \\fill[#2] (0,0) arc[radius=0.5*#1,start angle=270,end angle=180] -- +(0.5*#1,0) --cycle; \\fill[#5] (-0.5*#1,0.5*#1) arc[radius=0.5*#1,start angle=180,end angle=90] -- +(0,-0.5*#1) --cycle; \\fill[#6] (1.5*#1,0.5*#1) arc[radius=0.5*#1,start angle=0,end angle=90] -- +(0,-0.5*#1) --cycle; \\fill[#9] (1.5*#1,0.5*#1) arc[radius=0.5*#1,start angle=0,end angle=-90] -- +(0,0.5*#1) -- cycle; \\FillSquare{#1}{#3}{#4}{#7}{#8} \\draw (#1,0) -- +(-#1,0) \\draw (-0.5*#1,0.5*#1) -- +(2*#1,0); \\draw (#1,0) -- +(0,#1); \\draw (0.5*#1,0) -- +(0,#1); \\draw (0,0) -- +(0,#1); } \\begin{document} \\begin{tikzpicture} \\MySquare{4cm}{gray!30}{blue!30}{green!30}{red!30} \\begin{scope}[xshift=5.2cm,yshift=1cm] \\MyCircle{2cm}{orange}{gray!30}{red!30}{cyan} \\end{scope} \\begin{scope}[xshift=2.2cm,yshift=-4cm] \\MyShape{3cm}{gray!30}{blue!30}{green!30}{red!30}{olive!60}{brown!30}{yellow!30}{cyan!50} \\end{scope} \\end{tikzpicture} \\end{document} If labels have to be added to the regions, perhaps a different approach is better: \\documentclass{article} \\usepackage{tikz} \\usetikzlibrary{positioning,calc} \\newcommand\\DrawSquare[3][]{% \\node[inner sep=0pt,outer sep=0pt,draw,rectangle,minimum size=#2,#1] (#3) {}; } \\newcommand\\FillSquare[3]{% \\ifnum#2=1\\relax \\fill[#3] ( $(#1.north west) + (0.5\\pgflinewidth,-0.5\\pgflinewidth)$ ) rectangle (#1.center); \\else \\ifnum#2=2\\relax \\fill[#3] (#1.center) rectangle ( $(#1.north east) + (-0.5\\pgflinewidth,-0.5\\pgflinewidth)$ ); \\else \\ifnum#2=3\\relax \\fill[#3] ( $(#1.south west) + (0.5\\pgflinewidth,0.5\\pgflinewidth)$ ) rectangle (#1.center); \\else \\ifnum#2=4\\relax \\fill[#3] (#1.center) rectangle ( $(#1.south east) + (-0.5\\pgflinewidth,0.5\\pgflinewidth)$ ); \\fi\\fi\\fi\\fi% } \\newcommand\\LabelRegion[3]{% \\ifnum#2=1\\relax \\node at ( $(#1.north west)!0.5!(#1.center)$ )", "+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", "\\\\ \\end{array}} \\right]{\\bf{x}} + \\left[ {\\begin{array}{*{20}{c}} {-5 \\sqrt 3 /14} \\\\ {-3 /14} \\\\ \\end{array}} \\right]$$ scale by $$1/ \\sqrt{7}$$, rotate by A $${f_6}({\\bf{x}}) = \\left[ {\\begin{array}{*{20}{c}} { 5/ 14} & {-\\sqrt 3 /14 } \\\\ {\\sqrt 3 /14} & { 5/ 14} \\\\ \\end{array}} \\right]{\\bf{x}} + \\left[ {\\begin{array}{*{20}{c}} {-2 \\sqrt 3/7} \\\\ {3/7} \\\\ \\end{array}} \\right]$$ scale by $$1/ \\sqrt{7}$$, rotate by A $${f_7}({\\bf{x}}) = \\left[ {\\begin{array}{*{20}{c}} { 5/ 14} & {-\\sqrt 3 /14 } \\\\ {\\sqrt 3 /14} & { 5/ 14} \\\\ \\end{array}} \\right]{\\bf{x}}$$ scale by $$1/ \\sqrt{7}$$, rotate by A Start with a hexagon as shown in the figure above, make 7 copies scaled by a factor r, rotate them by an angle A, then translate so that six of them exactly surround the seventh (shown in red). In this case we will position the initial hexagon so that it is inscribed in the unit circle centered at the origin. Consider the blue triangle shown in the figure. The red hexagon has been scaled by r and rotated by A. The obtuse angle is 120° by properties of regular hexagons. We can use the Law of Cosines to get \\begin{align} 1 &= {r^2} + {\\left( 2r \\right)^2} - 2 \\cdot r \\cdot 2r \\cdot \\cos {120^ \\circ } \\\\ &= {r^2} + {4r^2} +", "of the opposite angle gives the diameter of the triangle's circumcircle). Skipping details, here's how to inscribe a triangle with specified angles into the unit circle: parts = IntegerPartitions[180, {3}]; (* generate corresponding sides from randomly chosen angles *) angles = parts[[RandomInteger[{1, Length[parts]}]]]; {r, q, p} = 2 Sin[angles Degree]; Graphics[{Line[{{1, 0}, {1 - p^2/2, p Sqrt[1 - p^2/4]}, {1 - q^2/2, -q Sqrt[1 - q^2/4]}, {1, 0}}], Circle[{0, 0}, 1]}] You can use the next two snippets to verify that the triangle generated fits the specifications: (* check side lengths *) Norm /@ RotateLeft[Differences[ N[{{1, 0}, {1 - p^2/2, p Sqrt[1 - p^2/4]}, {1 - q^2/2, -q Sqrt[1 - q^2/4]}, {1, 0}}]]] - {r, q, p} // Chop (* check angles *) (Apply[VectorAngle, Map[Function[pt, pt - First[#]], Rest[#]]]/Degree) & /@ NestList[RotateLeft, N[{{1, 0}, {1 - p^2/2, p Sqrt[1 - p^2/4]}, {1 - q^2/2, -q Sqrt[1 - q^2/4]}}], 2] - angles // Chop Both snippets should return {0, 0, 0} if all goes well. - aha, we all did the same thing! :) – rm -rf Feb 8 '12 at 1:13 Apparently so, but in defense of my slowness, I was waiting for the verification to finish... :) – Ｊ. Ｍ. Feb 8 '12 at 1:23 Very nice. I especially liked the touch of inscribing the triangle in", "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 =", "regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy] 1. 👍 2. 👎 6. 4/3 is the answer. 1. 👍 2. 👎 7. It is 4/3. 1. 👍 2. 👎 ## Similar Questions 1. ### Math please helppp 1 Find the missing angle measure in the polygon A ] 77 B ] 87 C ] 97 D ] 107 i think its b 2 Find the sum of the interior angles in 10 - sided polygon A ] 1,260 B ] 1,440 c ] 1,620 d ] 1,800 i think its c or b 3 Find the measure 2. ### geometry please please please help What is the value of x in", "hexagon. Now, using the sine law, $\\frac{1}{\\sin \\angle ACB} = \\frac{1}{\\sin 60^\\circ}$, so $\\angle ACB = 60^\\circ$. Since the angles in a triangle sum to 180$^\\circ$$\\angle ABC$ is also 60$^\\circ$. Therefore, $\\triangle ABC$ is an equilateral triangle with side lengths of 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)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label(\"2\",(z,c),NE); label(\"1\",(1.5,0),S); label(\"1\",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label(\"60^\\circ\",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label(\"60^\\circ\",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label(\"60^\\circ\",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); [/asy]$ Since the area of a regular hexagon can be found with the formula $\\frac{3\\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\\frac{3\\sqrt{3}(2^2)}{2} = 6\\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\\frac{\\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\\frac{\\sqrt{3}}{4}(1^2) = \\frac{\\sqrt{3}}{4}$. Since the area of a circle can be found", "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", "i; } // Draw the hexagon // We know one of its vertices is at (5, 1), so we'll use that to // calculate the right numbers for Paper.js's polygon-drawing // function and a subsequent rotation var vertex = hexPoint(5, 1); // Draw a polygon: hexagon = Path.RegularPolygon({ // Centered at the origin center: new Point(0, 0), // With six sides sides: 6, // With radius equal to the length of a vector from the origin // to the vertex // Medium blue strokeColor: \"#576AB7\", // With 0.5 stroke width strokeWidth: 0.5, // With these rotations: // 30 degrees clockwise because the default orientation is // pointy-topped but we want flat-topped // 15 degrees counterclockwise because of the rotation of // the whole figure // An angle equal to the angle a vector from the origin to // the previously calculated vertex makes with the x-axis rotation: 30 - 15 + vertex.angle, // With miter join style (default, don't need to specify) }); // Put the origin in the center of the view and zoom in so the // figure is more visible view.center = new Point(0, 0); view.zoom = 10; # JavaScript (ES6), 273 bytes Thanks to @Shaggy for suggesting to switch from HTML+JS to JS and saving 2 bytes in the new version Generates a SVG", "# OEDrawCurvedArrow¶ void OEDrawCurvedArrow(OEImageBase& image, const OE2DPoint& center, double bgnAngle, double endAngle, double radius, bool clockwise, const OEPen& pen); Draws a curved arrow. See example in Figure: Example of drawing curved arrows. Example of drawing curved arrows image The image in which the arrow is drawn. center The center of the arc. bgnAngle, endAngle The two endpoints of the arc (in degrees). Both angles are in degrees and their values have to be in a range of $$[0.0^{\\circ}, 360.0^{\\circ}]$$. Angles are interpreted such that $$0.0^{\\circ}$$ and $$360.0^{\\circ}$$ degrees are at the 12 o’clock position, $$90.0^{\\circ}$$ degrees corresponds to 3 o’clock, etc.", "peachgal 3) Write in a simpler, shorter form without parentheses! SIMPLIFY!: -2(3x-5)+2x-2(5-x) Thank you! You know that $a(b + c) = ab + ac$. So the first term, for example gives: $-2(3x - 5) = (-2)(3x) + (-2)(-5) = -6x + 10$ You do the third term. -Dan 8. Originally Posted by topsquark And you have six points on the hexagon. So how far do you have to rotate the hexagon before it looks like the original hexagon again? (In other words, rotate the hexagon, but leave a copy of it in it's original position. How far do you have to rotate the hexagon before it covers the copy?) -Dan 60 degrees ok now i get it", "get a better. Home » How To Draw A Hexagon Grid Tags. The difference between the two orientations is a rotation, and that causes the angles to change: flat topped angles are 0°, 60°, 120°, 180°, 240°, 300° and pointy topped angles are 30°, 90°, 150. Grid Line Weight: points (0. Gemfire and Romance of the Three Kingdoms feature more complex region maps that could have been made using postcommunisim's techniques but probably weren't. Align the rows. Draw coordinates for your grids and frame. In the example, this gives: \"screen\" \"3d-glasses\" \"print\" A case-sensitive match is then made with the set of media types defined above. If you want your hexagon to have a flat side at the top, the instructions are just above, under the heading Turn the hexagon. There is your hexagon grid, except for the zig-zag pattern between the rows. 160 161 162. Show Images & Videos in Their Correct Aspect Ratio & Orientation When Viewing a Grid in Photos. We just flip around some of the directions, use float: left and drop the explict setting of width to 0. \"On the grid above, show how the hexagon tessellates. Isometric ‘n Hexogonal Maps Part I Isometric ‘n Hexogonal Maps Part II Drawing a Hex Grid in Illustrator Coordinates in Hexagon-Based Tile Maps Hexagonal Grids. 0,", "'o' 'l' 'diamond' and corresponds to the shape Note that the first arrow shape specified occurs closest to the node. A polygon is a plane (2D) shape with straight sides. The part you seem to have difficulty with is defining the offsets dx and dy given an angle and an arrow length r. 3: Arrows Symbols. Each Monday it is located in Rice 011 (12:00 – 8:00 pm), and Tuesday through Thursday it is located in Thornton A238 (typically 2:00 – 8:00 pm). Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Arrow color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name. I want to insert arrows and text boxes, to identify houses and landmarks, over the images. You can learn more about their features, including numnber of sides, size of angles and tesselating patterns. Here are the most common geometrical symbols:. They may feature circles, arrows, curves or stars. 108*3 = 324 which is not equal to 360. This shows quarter clockwise turns. Alternatively, I can find the area of the entire shape (16 x 12) then subtract the missing piece (in pink) from the larger piece. As children sort and classify shapes with knowledgeable others, they", "hexagon, as shown in the diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy] • help hexagon geometry - ## Similar Questions 1. ### math how do you find the area of a hexagon? Here is the formula: Area of regular hexagon: (sqrt3/ 2)(Wsqrd) Where sqrt=square root W=the smallest width of the hexagon sqrd=squared 2. ### Geometry A regular hexagon with sides of 3\" is inscribed in a circle. What is the area of a segment formed by a side of the hexagon and the circle? 3. ### geometry a regular hexagon is inscribed in a circle. The radius of the circle is 18 units." ]

[ "2015 Posts: 113 GMAT 1: 730 Q49 V40 GPA: 3.8 In the figure, JKLMNP is a regular hexagon . [#permalink] Show Tags Updated on: 20 May 2016, 07:29 I think there's an easier way to solve this problem. As we were saying a regular hexagon has each angle measuring $$120°$$. Now let's look at the image posted above, the theorem of vertical and adjacent angle pairs says really simply that opposite angles born from the intersection of two lines must be equal. Therefore if the angle of the hexagon is $$120°$$ the opposite angle will be $$120°$$ as well, but we also know that the round angle is 360°, therefore, the sum of the remaining opposite pair must be $$360° - 240° = 120°$$, and since they are opposite as well they have to be equal -> $$\\frac{120°}{2}= 60°$$ Originally posted by DensetsuNo on 20 May 2016, 04:57. Last edited by DensetsuNo on 20 May 2016, 07:29, edited 1 time in total. Current Student Joined: 18 Oct 2014 Posts: 843 Location: United States GMAT 1: 660 Q49 V31 GPA: 3.98 Re: In the figure, JKLMNP is a regular hexagon . [#permalink] Show Tags 20 May 2016, 05:58 1 Since it is a regular hexagon, angles made at dissection point by all diagonals will be equal. Total number of", "regular hexagon, all the sides and angles are equal. so each angle of a regular hexagon is $\\frac{720}{6} = 120$. Thus, a regular hexagon has zero right angles. • 12 minutes ago • 14 minutes ago • 18 minutes ago • 24 minutes ago • 2 minutes ago • 5 minutes ago • 6 minutes ago • 9 minutes ago • 11 minutes ago • 11 minutes ago • 12 minutes ago • 14 minutes ago • 18 minutes ago • 24 minutes ago", "$x = \\dfrac{{720}}{6} = 120$. Hence the measure of each angle of a regular hexagon is ${120^ \\circ }$. Note: We should note that hexagon is also further classified into Convex or concave. If a hexagon is convex then none of its interior angle would be greater than ${180^ \\circ }$ and if the hexagon is concave then the one or more of its interior angles is greater than ${180^ \\circ }$. We should have the clear concept of different types of polygons and their properties before solving this kind of question.", "Sum of angles of a regular hexagon = (6 – 2) 180° = 4 × 180 = 720° One of its angle = $$\\frac{720}{6}$$ = 120° ∴ ∠PQR = 360 – (108 + 120) = 360 – 228 = 132° Question 7. A square, a regular pentagon and a regular hexagon are jointly drawn in the figure Find ∠BAC Solution: One angle of a regular pentagon = 108° One angle of a regular hexagon = 120° One angle of a square = 90° ∴ ∠BAC = 360 – (108 + 20 + 90) 360 – 318 = 42°. Question 8. In the figure below, ABCDEF is a regular hexagon. Prove that ∆ BDF is an equilateral triangle? Solution: Sum of angles of a regular hexagon = (6 – 2) 180 = 720° = 4 × 180 = 720° One angle = $$\\frac{720}{6}$$ = 120° consider ∆ EFO ∠E = 120° ∠EFD = ∠EDF = 30°(Angles opposite to equal sides of an isosceles triangle are equal) Similarly ∠AFB = 30° ∴∠DFB = 120 – (30 + 30) = 60° ∴ ∠FBD = 60°, ∠FDB = 60° ∴ ∆ FDB is an equilateral triangle. Question 9. In the figure below ABCDEF is a regular hexagon. Prove that ACDF is rectangle. Solution: One angle of a regular hexagon = 120° ∠EFD", "# Prove that the each angle of regular hexagon is 120. I need a theoretical proof that each angle of a regular hexagon is $120^\\circ$. - Draw the five radiuses from the hexagonal's center to its vertices. You get 6 congruent isosceles triangles whose basis angle's equals $\\,x=\\,$ half our wanted angle.. But then the central angle in each triangle equals $\\,180^\\circ-2x\\,$ , so $$6(180^\\circ-2x)=360^\\circ\\Longrightarrow 180^\\circ-2x=60^\\circ\\Longrightarrow 2x=120^\\circ$$ - Sum of the internal angles of a regular Hexagon is $(2.6-4)*90=720$ As there are 6 angles so each angle must be $720/6=120$ Another Perspective:In a regular Hexagon there are 6 Triangles each having the sum of the angles as $180$ degrees.So the sum of the Angles of all the 6 triangles is $180*6$ but this contains the central angle of $360$ degree so the sum of the internal angles of the hexagon is $180.6-360=720$.As all the angles are same so each angle must be $120$ degrees. - Consider a line rotating about a regular $n$-gon in the Euclidean plane. If the angle at each vertex is $a$, then the line rotates $180-a$ at each vertex. Since there are $n$ vertices, the total rotation is $n(180-a)$. But this total rotation is $360$. So, $n(180-a) = 360$ or $a = 180 - 360/n$. For $n = 6$, $a = 180-360/6 = 180-60", "# Take a regular hexagon Fig 3.10.3. What is the measure of each? (ii) interior angle It is a hexagon with all sides equal. All interior angles are also equal. Sum of all angles of hexagon is $720\\degree$. $6a = 720\\degree$ $a = 120\\degree$ Exams Articles Questions", "# The sum of the measures of the interior angles of a hexagon is 720° The measures of the angles of a particular hexagon are in the ratio 4:5:5:8:9:9, What are the measure of these angles? Jan 8, 2017 72°, 90°, 90°, 144°, 162°,162° #### Explanation: These are given as a ratio, which is always in simplest form. Let $x$ be the HCF that was used to simplify the size of each angle. 4x+5x+5x+8x+9x+9x = 720° 40x = 720° $x = \\frac{720}{40}$ $x = 18$ The angles are: 72°, 90°, 90°, 144°, 162°,162°", "SEARCH HOME Math Central Quandaries & Queries Question from Joseph: I need the angles in degrees for a hexagon one side is 18inches on the outside the problem I am having is in the angle of the cut and I am using a sliding miter saw which only goes from0 degrees to 45 degrees so I need help in getting the angle of the cut Hi Joseph, The interior angles of a regular hexagon have measure $120^o .$ If you set the angle of your miter saw to $30^o$ you can cut an angle of $120^o .$ Write back if you need more assistance, Harley Math Central is supported by the University of Regina and the Imperial Oil Foundation.", "would be the right answer. However, there are $6$ hexagons fixed by the $180^{\\circ}$ rotation, giving an additional contribution of $6/6=1$, for a total of $15+1=16$.", "of a hexagon: (6-2)*180$^\\circ$=720$^\\circ$ $\\frac{720^\\circ}{6}=120^\\circ$ and we know that the triangle with B and C as the two base vertices is an isoscolese, because the square and the hexagon have equal side lengths, so angle B = angle C. so we have $\\frac{180-(120-90)}2$= angle C = $75^\\circ$ and because we want to find angle ABC, we have to subtract 75$^\\circ$ from 120 to get angle ABC = 120-75 = $\\boxed{45^\\circ}$ Mar 3, 2021 edited by SparklingWater2 Mar 3, 2021 edited by SparklingWater2 Mar 3, 2021 #5 0 thank you sooooo much :) Guest Mar 3, 2021 #3 +31281 0 1) the two angles labelled encompass 110 + 120 = 230 degrees of the circle the other angle covers the rest 360 - 230 = 130 o Mar 3, 2021 #4 +485 +1 Small error:He/she is asking for angle BAC not BOC SparklingWater2 Mar 3, 2021", "Sum of Interior Angles of a polygon = 180 ×(n−2) degrees, where n is number of sides. \\times 11 in. Does pumpkin pie need to be refrigerated? You will always end up with two less triangles than original sides. Hexagonal nuts and bolts are easy to grip with a wrench, which can be re-positioned every 60° if needed. An interior angle is defined as the angle inside of a polygon made by two adjacent sides. The sum of the interior angles of the hexagon is 720 degrees. So, the sum of the interior angles of a hexagon is 720 degrees. For example, to find out the sum of the interior angles of a hexagon, you would calculate: = (−) × = × = × = So, the sum of the interior angles of a hexagon is 720 degrees. Author has 28K answers and 14.9M answer views The sum of the interior angles of a hexagon = (n-2)*180 = (6–2)*180 = 4*180 = 720 degrees. Enter your answer in the box. Thus, we get the equation: $90 + 90 + 140+150+130+x=720$ Let us solve this to find $$x$$. There are seven triangles... Because the sum of the angles of each triangle is 180 degrees... We get . The sum of all the interior angles in a hexagon is equal", "# How many right angles does a hexagon have? Jan 1, 2016 Zero in a regular hexagon because the angles are all 120° #### Explanation: A hexagon has six sides, and the sum of interior angles of a polygon can be calculated in the formula: $180 \\left(n - 2\\right)$, where $n$ is the number of sides in the polygon. Since a hexagon has six sides, the sum of interior angles is $180 \\left(6 - 2\\right) = 180 \\left(4\\right) = 720$, If it is a regular hexagon, all the sides and angles are equal. so each angle of a regular hexagon is $\\frac{720}{6} = 120$. Thus, a regular hexagon has zero right angles. Apr 28, 2016 It is possible to have 5 angles of 90° then the last one will be 270° (5 xx 90° +270° = 720°)", "of a heptagon = (7 – 2) × 180° = 5 × 180° = 900° All the angles are equal. So one angle = $$\\frac{900}{7}$$ = 128.571……….. Question 7. Find the sum of the angles of the polygon with the given sides? (a) 12 (b) 15 (c) 20 (d) 24 Solution: (a) Sum of angles = (12 – 2 ) × 180 1800° (b) Sum of angles = (15 – 2) × 180 = 2340° (c) Sum of angles = (20 – 2) × 180 = 3240° (d) Sum of angles = (24 – 2) × 180 = 3960° Question 8. The angles of a hexagon are (x – 10)°, x°, (x + 10)°, (x + 20)°, (x + 30)°, and (x + 40)°. Find the value of x? Solution: Sum of the angles of a hexagon = (6 – 2) × 180° = 720° ie, (x – 10) + x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 720 6x + 90 = 720 ie, 6x = 630; x = 630 ÷ 6 ∴ x = 105° Question 9. The sum of the angles of a polygon with 12 sides is 1800°. Find the sum of the angles of a polygon with 13 sides? Solution: When one side", "\\sqrt{\\frac{2}{3}}=416.41325627314...$$ The width of the hexagon can be calculated by cosine of the angles.", "$$\\beta=\\measuredangle BAC$$...then we know: $$\\displaystyle 2\\beta+\\theta_5+\\theta_6=360^{\\circ}$$ $$\\displaystyle 2\\beta+\\frac{180^{\\circ}(5-2)}{5}+\\frac{180^{\\circ}(6-2)}{6}=360^{\\circ}$$ Can you finish? Yes, thank you so much! #### Faye ##### New member Right! Angles in pentagon = 108, in hexagon = 120. Surrounding point A are 4 angles: see that? The 2 angles above the triangle are 108 and 120; ok? The 4 angles total 360; ok? So the other 2 angles total 360 - 120 - 108 = 132; ok? Since these 2 angles are equal, then they each equal 66; got that? And since angle BAC is one of them, then....? Angle BAC equals 66. Thanks a lot!", "# A hexagon has five angles that measure 140° each. What is the measure of the sixth angle? Mar 10, 2018 ${20}^{o}$ #### Explanation: The answer is ${20}^{o}$ because: • Total sum of interior angle: ${720}^{o}$ • Each angle measures: ${140}^{o}$ • 6 side: ? 5 sides: $140 \\cdot 5 = 700 \\text{ }$ or $\\text{ } 140 + 140 + 140 + 140 + 140 = 700$ We will add up all the sides to get the total of all the 5 sides. 6 side: $720 - 700 = 20$ We will now subtract the total of all the 5 sides, which is ${700}^{o}$, from the total angle of the hexagon, which is ${720}^{o}$, to get the 6th side. Hope that this had helped you a lot please let me know it in the comment section below. Have a great day. Mar 10, 2018 20 #### Explanation: Sum of internal angles in a hexagon is 720 Five angles is 140x5=700 Remaining angle is 720-700=20", "# Take a regular hexagon Fig 3.10.Q3. What is the measure of each? (i) exterior angle S seema garhwal It is a hexagon with all sides equal. All interior angles are also equal. Sum of all angles of hexagon is $720\\degree$. $6a = 720\\degree$ $a = 120\\degree$ $a+r=a+x=a+y=a+z=a+p=a+q=180\\degree$ (linear pairs) $r=x=y=z=p=q=180\\degree - a$ Each exterior angle = $r=x=y=z=p=q=180\\degree - a$ = $180\\degree-120\\degree$$60\\degree$ Exams Articles Questions", "Brenda is walking round an irregular hexagon (a shape with six straight sides). She starts off part of the way along one of the sides. At each vertex (corner) she turns. How much does she turn in total when she has walked all the way round? Use this example to prove that the sum of the external angles of any hexagon is $360$ degrees. The external angle is found by extending the side and measuring the angle between the extended line and the next side.", "how many degrees are in each angle of a regular hexagon. There are $360^\\circ$ in a circle and $120^\\circ$ in each interior angle of a hexagon, so $\\frac{360}{120}=3$ hexagons will fit around one point. 2. First, recall that there are $1080^\\circ$ in a pentagon. Each angle in a regular pentagon is $1080^\\circ \\div 8 = 135^\\circ$ . From this, we know that a regular octagon will not tessellate by itself because $135^\\circ$ does not go evenly into $360^\\circ$ . ### Practice 2. What is an example of a tessellated square in real life? 4. You can also tessellate two regular polygons together. Try tessellating a regular hexagon and an equilateral triangle. First, determine how many of each fit around a point and then repeat the pattern. Add color to your design. 5. Does a regular dodecagon (12-sided shape) tessellate? Why of why not? 6. Does a kite tessellate? Why or why not? Do the following figures tessellate?", "sum of the angles of the triangle is always greater than $180$ degrees. In fact, the excess (for the special case shown, the extra $90$ degrees) is proportional to how much area the triangle has. If a triangle on a sphere is very small, its angles add up to very nearly $180$ degrees, only a little bit over. As the triangle gets bigger the discrepancy goes up. The bugs on the hot plate would discover similar difficulties with their triangles. Let’s look next at what our other bugs find out about circles. They make circles and measure their circumferences. For example, the bug on the sphere might make a circle like the one shown in Fig. 42–11. And he would discover that the circumference is less than $2\\pi$ times the radius. (You can see that because from the wisdom of our three-dimensional view it is obvious that what he calls the “radius” is a curve which is longer than the true radius of the circle.) Suppose that the bug on the sphere had read Euclid, and decided to predict a radius by dividing the circumference $C$ by $2\\pi$, taking $$\\label{Eq:II:42:1} r_{\\text{pred}}=\\frac{C}{2\\pi}.$$ Then he would find that the measured radius was larger than the predicted radius. Pursuing the subject, he might define the difference to be the “excess radius,”" ]

[ "(c) $\\begin{array}{l}\\text{Size of an interior angle of a regular hexagon}\\\\ =\\left(1-\\frac{2}{6}\\right)×{180}^{\\circ }\\\\ =\\frac{2}{3}×{180}^{\\circ }\\\\ ={120}^{\\circ }\\end{array}$", "2015 Posts: 113 GMAT 1: 730 Q49 V40 GPA: 3.8 In the figure, JKLMNP is a regular hexagon . [#permalink] Show Tags Updated on: 20 May 2016, 07:29 I think there's an easier way to solve this problem. As we were saying a regular hexagon has each angle measuring $$120°$$. Now let's look at the image posted above, the theorem of vertical and adjacent angle pairs says really simply that opposite angles born from the intersection of two lines must be equal. Therefore if the angle of the hexagon is $$120°$$ the opposite angle will be $$120°$$ as well, but we also know that the round angle is 360°, therefore, the sum of the remaining opposite pair must be $$360° - 240° = 120°$$, and since they are opposite as well they have to be equal -> $$\\frac{120°}{2}= 60°$$ Originally posted by DensetsuNo on 20 May 2016, 04:57. Last edited by DensetsuNo on 20 May 2016, 07:29, edited 1 time in total. Current Student Joined: 18 Oct 2014 Posts: 843 Location: United States GMAT 1: 660 Q49 V31 GPA: 3.98 Re: In the figure, JKLMNP is a regular hexagon . [#permalink] Show Tags 20 May 2016, 05:58 1 Since it is a regular hexagon, angles made at dissection point by all diagonals will be equal. Total number of", "PQRSTU is a regular hexagon. Question: PQRSTU is a regular hexagon. Determine each angle of ΔPQT. Solution: A regular hexagon is made up of 6 equilateral triangles. So, $\\angle \\mathrm{PQT}=60^{\\circ}$ and $\\angle \\mathrm{QTP}=30^{\\circ}$ Since the sum of the angles of $\\triangle \\mathrm{PQT}$ is $180^{\\circ}$, we have: $\\angle \\mathrm{P}+\\angle \\mathrm{Q}+\\angle \\mathrm{T}=180^{\\circ}$ $\\Rightarrow \\angle \\mathrm{P}+60^{\\circ}+30^{\\circ}=180^{\\circ}$ $\\Rightarrow \\angle \\mathrm{P}=180^{\\circ}-90^{\\circ}$ $\\angle \\mathrm{P}+\\angle \\mathrm{Q}+\\angle \\mathrm{T}=180^{\\circ}$ $\\Rightarrow \\angle \\mathrm{P}+60^{\\circ}+30^{\\circ}=180^{\\circ}$ $\\Rightarrow \\angle \\mathrm{P}=180^{\\circ}-90^{\\circ}$ $\\Rightarrow \\angle \\mathrm{QPT}=90^{\\circ}$ $\\therefore$ The angles of the triangle are $90^{\\circ}, 60^{\\circ}$ and $30^{\\circ}$.", "JKLMNP is a regular hexagon .Find the measure of $$\\angle$$ MQN. A. $$30^{\\circ}$$ B. $$45^{\\circ}$$ C. $$50^{\\circ}$$ D. $$60^{\\circ}$$ E. $$75^{\\circ}$$ An interesting question that requires you to be a bit creative with the information that you have been given. The given polygon is a REGULAR hexagon --> all sides and all angles are equal. Every interior angle of a hexagon = 120 degrees. As per the attached image, draw a circle around the given hexagon. Thus , now you get equal arcs LM=MN=NP=JP=JK=KL . As there are 6 arcs and all arcs must subtend a total of 360 degrees at the center of the circle ---> each arc subtends 60 degrees at the center. Additionally, an angle subtended by any arc of a circle at the center = 2* angle subtended by the same arc on the circumference. Consider triangle QNM,$$\\angle{QNM}$$ = angle subtended by LM = 30 degrees (as it is the angle subtended by arc LM on the circumference.) Similarly, Arcs KJ+JP+PN subtend $$\\angle{QMN}$$ on the circumference. ---> $$\\angle{QMN} = 90 degrees$$. Finally, in triangle QMN, $$\\angle{QMN} + \\angle {QNM} + \\angle {MQN} = 180 degrees$$ --->$$\\angle {MQN}$$ = 180-30-90 = 60 degrees. Hope this helps. Attachments 2015-12-05_10-45-47.jpg [ 12.84 KiB | Viewed 2526 times ] Manager Status: 2 months to go Joined: 11 Oct", "hexagon. Now, using the sine law, $\\frac{1}{\\sin \\angle ACB} = \\frac{1}{\\sin 60^\\circ}$, so $\\angle ACB = 60^\\circ$. Since the angles in a triangle sum to 180$^\\circ$$\\angle ABC$ is also 60$^\\circ$. Therefore, $\\triangle ABC$ is an equilateral triangle with side lengths of 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)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label(\"2\",(z,c),NE); label(\"1\",(1.5,0),S); label(\"1\",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label(\"60^\\circ\",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label(\"60^\\circ\",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label(\"60^\\circ\",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); [/asy]$ Since the area of a regular hexagon can be found with the formula $\\frac{3\\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\\frac{3\\sqrt{3}(2^2)}{2} = 6\\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\\frac{\\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\\frac{\\sqrt{3}}{4}(1^2) = \\frac{\\sqrt{3}}{4}$. Since the area of a circle can be found", "+0 Help! +2 31 6 +902 A hexagon is inscribed in a circle: What is the measure of $$\\alpha$$, in degrees? Jan 12, 2019 #1 +313 +3 the figure is cyclic, so those three angles are just half of the total sum of all the angles, which is 720/2. subtracting, you get 360-105-110=145. HOPE THIS HELPED! Jan 12, 2019 #2 +313 +1 it only works where the angles are all opposite of each other Jan 12, 2019 #3 +94995 +2 Join the alpha vertex to the vertex opposite and then you have two cyclic quads. (one on top and one underneath) So alpha is the sum of (180-105)+(180-110) = 75+70=145 degrees I know this is the same as asad's answer, it is just another way of doing it (I actually did not understand asad's method.) Jan 12, 2019 edited by Melody Jan 12, 2019 edited by Melody Jan 12, 2019 #4 +94300 +2 Thanks asdf and Melody......here's another way to look at this....but not as nice as yours....!!! The 110° angle intercepts a major arc of 220°......so the minor arc that it subtends = 360 - 220 = 140° The 105° angle subtends a major arc of 210° ......so the minor arc it subtends = 360 - 210 = 150° So.....this means that the minor arc subtended by", "# Problem of the Week Problem B and Solution A Spritely Shape ## Problem Peyton used a block coding program to get a sprite character to draw a shape. His sprite followed these steps: 1. Put pen down to write 2. Move $$10$$ steps forward 3. Turn clockwise $$60\\degree$$ 4. Repeat steps $$2$$ and $$3$$ five more times Here is the sprite’s drawing partway through the program: 1. What type of polygon did the sprite draw? 2. What type of pattern did Peyton use in this code? 3. If the code were changed so that step $$3$$ reads \"Turn clockwise $$45\\degree$$\", how would Peyton need to change step $$4$$ in order to create a closed polygon? ## Solution 1. The sprite moved $$10$$ steps forward and then and turned $$60\\degree$$ clockwise a total of six times. By doing this, the sprite created a regular hexagon (with interior angles of $$120\\degree$$, which sum to $$720\\degree$$). The completed hexagon is shown below. 2. Peyton used a repeating pattern in this code; continuing will retrace the hexagon. 3. If the code were changed so that step $$3$$ reads \"Turn clockwise $$45\\degree$$\", Peyton would have to revise step $$4$$ to \"Repeat steps $$2$$ and $$3$$ seven more times”, thus creating a regular octagon (with interior angles of $$135\\degree$$, which sum to $$1080\\degree$$).", "side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\\circ = 180(6 - 2)^\\circ = 720^\\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\\frac{720^\\circ}{6} = 120^\\circ$$. Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$: Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\\frac{r}{2}$$ in length, and $$AX$$ is $$\\frac{\\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\\sqrt{3}r$$ in length. This means that the area of triangle $$ABC$$ is $$\\frac{1}{2} \\leftarrow( \\sqrt{3}r \\rightarrow) \\leftarrow( \\frac{r}{4} \\rightarrow) = \\frac{\\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\\frac{3 \\sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by $$\\frac{3 \\sqrt{3} r^2}{4} \\div \\pi r^2 = \\frac{3 \\sqrt{3}}{4 \\pi} \\approx \\frac{\\sqrt{3}}{4} \\approx \\frac{1.7}{4} \\approx 0.43$$ Hey Bunuel, I", "(2x - 2)(5 - x)$ -Dan 3. Originally Posted by topsquark 1) The slope between two points $(x_1, y_1) \\text{ and }(x_2, y_2)$ is $m = \\frac{y_2 - y_1}{x_2 - x_1}$. 2) There are 6 points on the hexagon. If you inscribed a circle about the hexagon, what is the central angle between two of the points? 3) Is this $-2(3x - 5) + 2x - 2(5 - x)$ or $-2(3x - 5) + (2x - 2)(5 - x)$ -Dan I'm sorry I wish I could be smart but I am absolutely horrible with angles. 4. Originally Posted by peachgal I'm sorry I wish I could be smart but I am absolutely horrible with angles. A circle has how many degrees? -Dan 5. Originally Posted by topsquark A circle has how many degrees? -Dan 360 By the way on #2 it said simplify WITHOUT using parentheses. Can you please do that? 6. Originally Posted by peachgal 360 And you have six points on the hexagon. So how far do you have to rotate the hexagon before it looks like the original hexagon again? (In other words, rotate the hexagon, but leave a copy of it in it's original position. How far do you have to rotate the hexagon before it covers the copy?) -Dan 7. Originally Posted by", "# 2.2.1 Polygons II, PT3 Practice 2.2.1 Polygons II, PT3 Practice Question 1: Diagram below shows a pentagon PQRST. TPU and RSV are straight lines. Find the value of x. Solution: Question 2: In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines. Find the value of x + y. Solution: $\\begin{array}{l}\\angle QPU={180}^{o}-{160}^{o}={20}^{o}\\\\ \\text{Reflex}\\angle PUT={360}^{o}-{80}^{o}={280}^{o}\\\\ \\angle UTS={180}^{o}-{120}^{o}={60}^{o}\\\\ \\angle TSR={180}^{o}-{35}^{o}={145}^{o}\\\\ \\\\ \\text{Sum of interior angles of a hexagon}\\\\ =\\left(6-2\\right)×{180}^{o}\\\\ ={720}^{o}\\\\ \\\\ {x}^{o}+{y}^{o}+{145}^{o}+{60}^{o}+{280}^{o}+{20}^{o}={720}^{o}\\\\ {x}^{o}+{y}^{o}={720}^{o}-{505}^{o}\\\\ \\text{}={215}^{o}\\\\ \\text{}x+y=215\\end{array}$ Question 3: Diagram below shows a regular hexagon PQRSTU. PUV is a straight line. Find the value of x + y. Solution: $\\begin{array}{l}\\text{Size of each interior angle of a regular hexagon}\\\\ =\\frac{\\left(6-2\\right)×{180}^{o}}{6}\\\\ ={120}^{o}\\\\ {x}^{o}=\\frac{{180}^{o}-{120}^{o}}{2}={30}^{o}\\\\ {y}^{o}={180}^{o}-{120}^{o}\\\\ \\text{}={60}^{o}\\\\ {x}^{o}+{y}^{o}={30}^{o}+{60}^{o}\\\\ \\text{}={90}^{o}\\\\ \\text{}x+y=90\\end{array}$ Question 4: In the diagram below, KLMNP is a regular pentagon. LKS and MNQ are straight lines. Find the value of x + y. Solution: $\\begin{array}{l}\\text{Size of each interior angle of a regular pentagon}\\\\ =\\frac{\\left(5-2\\right)×{180}^{o}}{5}\\\\ ={108}^{o}\\\\ \\angle PKS=\\angle PNQ={180}^{o}-{108}^{o}={72}^{o}\\\\ \\text{Reflex angle}\\angle KPN={360}^{o}-{108}^{o}={252}^{o}\\end{array}$ $\\begin{array}{l}\\text{Sum of interior angles of a hexagon}\\\\ =\\left(6-2\\right)×{180}^{o}\\\\ ={720}^{o}\\\\ \\therefore {x}^{o}+{y}^{o}+{72}^{o}+{252}^{o}+{72}^{o}+{100}^{o}={720}^{o}\\\\ {x}^{o}+{y}^{o}={720}^{o}-{496}^{o}\\\\ \\text{}={224}^{o}\\\\ \\text{}x+y=224\\end{array}$ Question 5: In Diagram below, PQR is an isosceles triangle and PRU is a straight line. Find the value of x + y. Solution: $\\begin{array}{l}{x}^{o}={180}^{o}-{20}^{o}-{20}^{o}={140}^{o}\\\\ \\angle PRS={180}^{o}-{110}^{o}={70}^{o}\\\\ {y}^{o}+{85}^{o}+{75}^{o}+{70}^{o}={360}^{o}\\\\ {y}^{o}+{230}^{o}={360}^{o}\\\\ \\text{}{y}^{o}={130}^{o}\\\\ {x}^{o}+{y}^{o}={140}^{o}+{130}^{o}\\\\ \\text{}={270}^{o}\\\\ \\text{}x+y=270\\end{array}$", "regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy] 1. 👍 2. 👎 6. 4/3 is the answer. 1. 👍 2. 👎 7. It is 4/3. 1. 👍 2. 👎 ## Similar Questions 1. ### Math please helppp 1 Find the missing angle measure in the polygon A ] 77 B ] 87 C ] 97 D ] 107 i think its b 2 Find the sum of the interior angles in 10 - sided polygon A ] 1,260 B ] 1,440 c ] 1,620 d ] 1,800 i think its c or b 3 Find the measure 2. ### geometry please please please help What is the value of x in", "# Prove that the each angle of regular hexagon is 120. I need a theoretical proof that each angle of a regular hexagon is $120^\\circ$. - Draw the five radiuses from the hexagonal's center to its vertices. You get 6 congruent isosceles triangles whose basis angle's equals $\\,x=\\,$ half our wanted angle.. But then the central angle in each triangle equals $\\,180^\\circ-2x\\,$ , so $$6(180^\\circ-2x)=360^\\circ\\Longrightarrow 180^\\circ-2x=60^\\circ\\Longrightarrow 2x=120^\\circ$$ - Sum of the internal angles of a regular Hexagon is $(2.6-4)*90=720$ As there are 6 angles so each angle must be $720/6=120$ Another Perspective:In a regular Hexagon there are 6 Triangles each having the sum of the angles as $180$ degrees.So the sum of the Angles of all the 6 triangles is $180*6$ but this contains the central angle of $360$ degree so the sum of the internal angles of the hexagon is $180.6-360=720$.As all the angles are same so each angle must be $120$ degrees. - Consider a line rotating about a regular $n$-gon in the Euclidean plane. If the angle at each vertex is $a$, then the line rotates $180-a$ at each vertex. Since there are $n$ vertices, the total rotation is $n(180-a)$. But this total rotation is $360$. So, $n(180-a) = 360$ or $a = 180 - 360/n$. For $n = 6$, $a = 180-360/6 = 180-60", "hexagon is 120° Calculations: Each interior angle of a regular hexagon is 120° So, ∠DIH = ∠IHG = ∠HGF = ∠GFE = ∠FED = ∠EDI = 120° ∠DIH + ∠BIH = 180° ⇒ 120° + ∠BIH = 180° ⇒ ∠BIH = 60° as we know ΔABC is an equilateral triangle So, ∠ABC = ∠BAC = ∠ACB = 60° Now, In ΔBIH ∠IBH = 60° ∠BIH = 60° So, ∠BHI = 60° So, we can say that ΔBIH is also an equilateral triangle Similarly ΔGFC and ΔADE is also an equilateral triangle Side of the regular hexagon = (1/3) × 12 = 4 cm So the area of the hexagon = (3√3/2)a2 ⇒ (3√3/2) × 4 × 4 = 24√3 cm2 ∴ The area of the hexagon is 24√3 cm2. # In ΔABC, ∠B = 90°, AB = 5 cm and BC = 12 cm the bisector of ∠A meets BC at D. What is the length of AD? 1. 2√13 cm 2. $$\\frac{2}{3}\\sqrt {13}$$ cm 3. $$\\frac{4}{3}\\sqrt {13}$$ cm 4. $$\\frac{{5\\sqrt {13} }}{3}$$ cm Option 4 : $$\\frac{{5\\sqrt {13} }}{3}$$ cm ## Detailed Solution In right triangle ABC AC2 = AB2 + BC2 ⇒ AC2 = 52 + 122 = 25 + 144 = 169 ⇒ AC = √169 = 13 cm If AD is bisector of ∠BAC,", "SEARCH HOME Math Central Quandaries & Queries Question from Joseph: I need the angles in degrees for a hexagon one side is 18inches on the outside the problem I am having is in the angle of the cut and I am using a sliding miter saw which only goes from0 degrees to 45 degrees so I need help in getting the angle of the cut Hi Joseph, The interior angles of a regular hexagon have measure $120^o .$ If you set the angle of your miter saw to $30^o$ you can cut an angle of $120^o .$ Write back if you need more assistance, Harley Math Central is supported by the University of Regina and the Imperial Oil Foundation.", "+\\angle b_1yx > \\angle a_1b_1y + \\angle b_1a_1x.$$ Hence the sum of the angles of the quadrilateral $xyb_1a_1$ is less than $360^{\\circ}$, which is a contradiction. Consequently at least three points are marked. (5) Finally, the configuration with ten points $(-100,0)$, $(0,0)$, $(101,0)$, $(213,0)$, $(0,110)$, $(0,-111)$, $(-120,110)$, $(-120,-111)$, $(121,110)$, $(121,-111)$ shows that the minimum with three marked points indeed can be attained. Summary The maximum number of dead bandits is $10$, and the minimum number of dead bandits is $3$. • Your inequalities are the wrong direction in the lines involving 360* and 180*. I can't tell how far that propagates, though, so I'm hesitant to correct it for you. – Zerris Mar 5 '16 at 0:13 I believe the max that can survive is 7. Arrange the bandits in approximately two half hexagons. For one hexagon, it's 5 bandits, one at each point and one in the middle. More than a side length away is the other hexagon, parallel, with one exterior point empty. Between the two is the last bandit. To satisfy the distance clause, just have the outer bandits be closer to the center than each other, and with very small differences the overall situation stands. • I feel like this ignores \"with no pair the same distance apart\". – Ian MacDonald Mar 3 '16", "8:13 • If you want to draw an hexagon, forget the projections. Start with blue point at coordinates (0, 0), the black point will be at coords (0, -1). The green point will be at coords (-.866, -.5) and the red at (.866, -.5). The cyan point will be at coords (-.866, .5) and the magenta at (.866, .5). Finally the white point will be at (0, 1). The values are $.5=\\sin 30°=\\cos 60°$ and $.866=\\sin 60°=\\cos 30°$. To scale with RGB values multiply by 255 (and eventually translate to center the image). – N74 Aug 27 '16 at 11:37 Based on your technical drawing with measurements posted in it, the width $H$ of the hexagon can be deduced with simple trigonometry: $$H = 360.62\\cos\\,(54.74^\\circ) + 255\\cos\\,(35.26^\\circ).$$ And in general, for any angle $\\theta$ that you tilt the cube of side length $w$ and face-diagonal length $\\sqrt 2 \\cdot w$ over, the width $H$ of the hexagon will be $$H = \\sqrt 2 \\cdot w\\cos\\,(\\pi/2 - \\theta) + w\\cos\\,\\theta.$$ • I'll be doing the calculation in JavaScript and my trigonometry knowledge is limited. How would I write 360.62cos(54.74°)? This 360.62 * Math.cos(54.74)? That doesn't give the correct width. – akinuri Aug 26 '16 at 17:23 • @akinuri, Since JS uses radians instead of degrees, you'll first have to", "4.2.1 Polygons, PT3 Practice 4.2.1 Polygons, PT3 Practice Question 1: Diagram below shows a pentagon PQRST. TPU and RSV are straight lines. Find the value of x. Solution: Question 2: In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines. Find the value of x + y. Solution: $\\begin{array}{l}\\angle QPU={180}^{o}-{160}^{o}={20}^{o}\\\\ \\text{Reflex}\\angle PUT={360}^{o}-{80}^{o}={280}^{o}\\\\ \\angle UTS={180}^{o}-{120}^{o}={60}^{o}\\\\ \\angle TSR={180}^{o}-{35}^{o}={145}^{o}\\\\ \\\\ \\text{Sum of interior angles of a hexagon}\\\\ =\\left(6-2\\right)×{180}^{o}\\\\ ={720}^{o}\\\\ \\\\ {x}^{o}+{y}^{o}+{145}^{o}+{60}^{o}+{280}^{o}+{20}^{o}={720}^{o}\\\\ {x}^{o}+{y}^{o}={720}^{o}-{505}^{o}\\\\ \\text{}={215}^{o}\\\\ \\text{}x+y=215\\end{array}$ Question 3: Diagram below shows a regular hexagon PQRSTU. PUV is a straight line. Find the value of x + y. Solution: $\\begin{array}{l}\\text{Size of each interior angle of a regular hexagon}\\\\ =\\frac{\\left(6-2\\right)×{180}^{o}}{6}\\\\ ={120}^{o}\\\\ {x}^{o}=\\frac{{180}^{o}-{120}^{o}}{2}={30}^{o}\\\\ {y}^{o}={180}^{o}-{120}^{o}\\\\ \\text{}={60}^{o}\\\\ {x}^{o}+{y}^{o}={30}^{o}+{60}^{o}\\\\ \\text{}={90}^{o}\\\\ \\text{}x+y=90\\end{array}$ 3 thoughts on “4.2.1 Polygons, PT3 Practice” 1. answer in Question 1 is error. The answer is 115. • Thanks for pointing out our mistake. We had done the correction accordingly. 2. I like these questions .", "360\\begin{align*}360^\\circ\\end{align*}. The only regular polygons with this feature are equilateral triangles, squares, and regular hexagons. #### Example A Draw a tessellation of equilateral triangles. In an equilateral triangle each angle is 60\\begin{align*}60^\\circ\\end{align*}. Therefore, six triangles will perfectly fit around each point. Extending the pattern, we have: #### Example B Does a regular pentagon tessellate? First, recall that there are 540\\begin{align*}540^\\circ\\end{align*} in a pentagon. Each angle in a regular pentagon is 540÷5=108\\begin{align*}540^\\circ \\div 5 = 108^\\circ\\end{align*}. From this, we know that a regular pentagon will not tessellate by itself because 108\\begin{align*}108^\\circ\\end{align*} times 2 or 3 does not equal 360\\begin{align*}360^\\circ\\end{align*}. #### Example C How many squares will fit around one point? First, recall how many degrees are in a circle, and then figure out how many degrees are in each angle of a square. There are 360\\begin{align*}360^\\circ\\end{align*} in a circle and 90\\begin{align*}90^\\circ\\end{align*} in each interior angle of a square, so 36090=4\\begin{align*}\\frac{360}{90}=4\\end{align*} squares will fit around one point. Tessellations CK-12 --> ### Guided Practice 1. How many regular hexagons will fit around one point? 2. Does a regular octagon tessellate? 3. Tessellations can also be much more complicated. Check out http://www.mathsisfun.com/geometry/tessellation.html to see other tessellations and play with the Tessellation Artist, which has a link at the bottom of the page. 1. First, recall how many degrees are in a circle,", "$x = \\dfrac{{720}}{6} = 120$. Hence the measure of each angle of a regular hexagon is ${120^ \\circ }$. Note: We should note that hexagon is also further classified into Convex or concave. If a hexagon is convex then none of its interior angle would be greater than ${180^ \\circ }$ and if the hexagon is concave then the one or more of its interior angles is greater than ${180^ \\circ }$. We should have the clear concept of different types of polygons and their properties before solving this kind of question.", "endpoints of a side of the larger hexagon, as shown in the diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy]" ]

[ "$ABCD$ 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.", "☐C. 46 ☐D. 55 ☐E. 61 38- In the following figure, ABCD is a rectangle. If $$a=\\sqrt{5}$$, and $$b=2a$$, find the area of the shaded region. (the shaded region is a trapezoid) ☐A. $$\\sqrt{5}$$ ☐B. $$2$$ ☐C. $$4$$ ☐D. $$2\\sqrt{5}$$ ☐E. $$4\\sqrt{5}$$ 39- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively and DE = 5 and DF = 3. The area of ∆BED is 20, and the area of ∆BDF is 24. What is the perimeter of the rectangle? ☐A. 24 ☐B. 28 ☐C. 36 ☐D. 42 ☐E. 48 40- What is the product of all possible values of $$x$$ in the following equation? $$|2x+6-3x-7|=5$$ ☐A. $$-24$$ ☐B. $$-12$$ ☐C. $$8$$ ☐D. $$12$$ ☐E. $$24$$ 28% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "+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", "$$ABCD$$ is a rectangular sheet of paper that has been folded so that corner $$B$$ is matched with point $$B'$$ on edge $$AD.$$ The crease is $$EF,$$ where $$E$$ is on $$AB$$ and $$F$$ is on $$CD.$$ The dimensions $$AE=8, BE=17,$$ and $$CF=3$$ are given. The perimeter of rectangle $$ABCD$$ is $$m/n,$$ where $$m$$ and $$n$$ are relatively prime positive integers. Find $$m+n.$$ (第二十二届AIME2 2004 第7题)", "# ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Question: ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is (a) 5 (b) 4 (c) 3 (d) 25 Solution: Here, AC and BD are two diagonals of the rectangle ABCD. So $B D=\\sqrt{(4-0)^{2}+(0-3)^{2}}$ $=\\sqrt{(4)^{2}+(-3)^{2}}$ $=\\sqrt{16+9}$ $=\\sqrt{25}$ $=5$ units Hence, the correct answer is option (a).", "## 'Bound to Be' printed from http://nrich.maths.org/ $ABCD$ 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.", "63 ☐D. 72 ☐E. 83 57- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively. The area of ∆BED is 20, and the area of ∆BDF is 32. What is the perimeter of the rectangle? ☐A. 20 ☐B. 22 ☐C. 32 ☐D. 40 ☐E. 48 58- When point A $$(-6.-12)$$ is reflected over the $$y$$-axis to get the point B, what are the coordinates of point B? ☐A. $$(6,- 12)$$ ☐B. $$(6,12)$$ ☐C. $$(-12,6)$$ ☐D. $$(12,-6)$$ ☐E. $$(-6,12)$$ 59- Which of the following points lies on the line? $$5x+3y=6$$ ☐A. $$(6, 1)$$ ☐B. $$(–3, 3)$$ ☐C. $$(3,-3)$$ ☐D. $$(2, 2)$$ ☐E. $$(2, 8)$$ 60- If $$f(x)=2x^3+4x-18x^(-2)$$ and $$g(x)=-3$$, what is the value of $$f(g(x))$$? ☐A. $$-83$$ ☐B. $$-68$$ ☐C. $$-25$$ ☐D. $$-13$$ ☐E. $$0$$ 27% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "24 B. 38 C. 46 D. 55 E. 61 38- In the following figure, ABCD is a rectangle. If $$a=\\sqrt{5}$$, and $$b=2a$$, find the area of the shaded region. (the shaded region is a trapezoid) A. $$\\sqrt{5}$$ B. $$2$$ C. $$4$$ D. $$2\\sqrt{5}$$ E. $$4\\sqrt{5}$$ 39- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively, and DE = 5 and DF = 3. The area of ∆BED is 20, and the area of ∆BDF is 24. What is the perimeter of the rectangle? A. 24 B. 28 C. 36 D. 42 E. 48 40- What is the product of all possible values of $$x$$ in the following equation? $$|2x+6-3x-7|=5$$ A. $$-24$$ B. $$-12$$ C. $$8$$ D. $$12$$ E. $$24$$ 28% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "+0 # Trapezoid ABCD has vertices 0 65 1 Trapezoid ABCD has vertices A(-1,0), B(0,2), C(m,4) and D(k,0), with m > 0 and k > 0 . The line y = -x + 4 is perpendicular to the line containing side CD , and the area of trapezoid ABCD is 36 square units. What is the value of k? Nov 13, 2020 #1 +10807 +1 Trapezoid ABCD has vertices A(-1,0), B(0,2), C(m,4) and D(k,0), with m > 0 and k > 0 . The line y = -x + 4 is perpendicular to the line containing side CD , and the area of trapezoid ABCD is 36 square units. What is the value of k? Hello Guest! $$m_{ab}=\\frac{2-0}{0-(-1)}$$ $$m_{ab}=2$$ $$y_{dc}=2(x-k)+0\\\\ y_{dc}=2(x-m)+4\\\\ 2x-2k=2x-2m+4\\\\2k=2m-4\\\\ k=m-2$$ to be continued ! Nov 14, 2020 edited by asinus Nov 15, 2020", "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); … 10. ### Ed tech SmartArt and shapes are useful tools because they allow you to. Organize data within a document. Draw any shape you want within a document. Easily create manipulate shapes within a document. (MY ANSWER) More Similar Questions", "# Question 1 of 10 In the figure, ABCD is a rectangle of area 96cm $^2$, with AE =EB .What is the total area of the shaded parts? A 12cm$^2$ B 18cm$^2$ C 24cm$^2$ D 32cm$^2$ E None of the above", "ABCD is a rectangle. AB = 2 units, AD = 4 units, and AE = FC = 1 unit. Question 1. Find the area of rectangle ABCD. Area = 4 units × 2 units = 8 sq. units Question 2. Find the area of triangle ABE. Area = $$\\frac{1}{2}$$ × 1 unit × 2 units = 1 sq. unit Question 3. Find the area of triangle DCF. Area = $$\\frac{1}{2}$$ × 1 unit × 2 units = 1 sq. unit", "area of a trapezoid is 150, what is the perimeter of the trapezoid? A. 25 B. 50 C. 63 D. 72 E. 83 57- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively. The area of ∆BED is 20, and the area of ∆BDF is 32. What is the perimeter of the rectangle? A. 20 B. 22 C. 32 D. 40 E. 48 58- When pointing A $$(-6.-12)$$ is reflected over the $$y$$-axis to get point B, what are the coordinates of point B? A. $$(6,- 12)$$ B. $$(6,12)$$ C. $$(-12,6)$$ D. $$(12,-6)$$ E. $$(-6,12)$$ 59- Which of the following points lies on the line? $$5x+3y=6$$ A. $$(6, 1)$$ B. $$(–3, 3)$$ C. $$(3,-3)$$ D. $$(2, 2)$$ E. $$(2, 8)$$ 60- If $$f(x)=2x^3+4x-18x^(-2)$$ and $$g(x)=-3$$, what is the value of $$f(g(x))$$? A. $$-83$$ B. $$-68$$ C. $$-25$$ D. $$-13$$ E. $$0$$ ## The Best Books to Ace the ACT Math Test $14.99 Satisfied 139 Students$13.99 Satisfied 155 Students ## More from Effortless Math for ACT Test … ### Are you looking for a FREE ACT Math course to help you prepare for your test? Check out our Ultimate ACT Math Course. Have a look at our comprehensive ACT Math Worksheets to help you practice and prepare for the", "1 IIT-JEE 2006 ABCD is a square of side length 2 units. $${C_1}$$ is the circle touching all the sides of the square ABCD and $${C_2}$$ is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point. If P is any point of $${C_1}$$ and Q is another point on $${C_2}$$, then $${{P{A^2}\\, + \\,P{B^2}\\, + P{C^2}\\, + P{D^2}} \\over {Q{A^2} + \\,Q{B^2}\\, + Q{C^2}\\, + Q{D^2}}}$$ is equal to A 0.75 B 1.25 C 1 D 0.5 2 IIT-JEE 2006 ABCD is a square of side length 2 units. $$C_1$$ is the circle touching all the sides of the square ABCD and $$C_2$$ is the circumcircle of square ABCD. L is a fixed line in the same plane and R is a fixed point. A line L' through A is drawn parallel to BD. Point S moves such that its distances from the BD and the vertex A are equal. If locus of S cuts L' at $$T_2$$ and $$T_3$$ and AC at $$T_1$$, then area of $$\\Delta \\,{T_1}\\,{T_2}\\,{T_3}$$ is A $${1 \\over 2}\\,sq.\\,units$$ B $${2 \\over 3}\\,sq.\\,units$$ C 1 sq. units D 2 sq.units 3 IIT-JEE 2006 ABCD is a square of side length 2 units. $$C_1$$ is the circle touching all the sides of the", "area of rectangle $ABGH$ is $4 \\cdot 12 = 48$. Taking the area of rectangle $ABGH$ and subtracting the combined area of $\\triangle AHG$ and $\\triangle CGM$ yields $48 - (24 + \\frac{32}{5}) = \\boxed{\\frac{88}{5}}\\ \\text{(C)}$. ### Solution 2 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label(\"A\",A,NW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",Ep,SE); label(\"F\",F,SE); label(\"G\",G,SE); label(\"H\",H,SW); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,NW); label(\"L\",L,NW); label(\"M\",M,W); label(\"N\",Z,NE); [/asy]$ Extend $AB$ and $CH$ and call their intersection $N$. The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\\frac{2\\cdot 4}2 = 4$. The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$. Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \\frac 35 \\cdot 12 = \\frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\\frac{1}{\\frac{1}{8}+\\frac{1}{12}} = \\frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\\frac{36}{5}$.) Then the area of $AMN$ is $\\frac 12 \\cdot", "+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", "is $$\\frac { 31 }{ 2 }$$ sq. units. Then the area of rectangle is …………… (1) 62 sq. units (2) 31 sq. units (3) 60 sq. units (4) 30 sq. units (2) 31 sq. units Hint: In a rectangle area of ∆ ABC and area of ∆ ACD are equal. Area of rectangle ABCD = 2 × $$\\frac { 31 }{ 2 }$$ = 31 sq.units Question 4. If the points (k, 2k), (3k, 3k) and (3,1) are collinear, then k is …………….. (1) $$\\frac { 1 }{ 3 }$$ (2) – $$\\frac { 1 }{ 3 }$$ (3) $$\\frac { 2 }{ 3 }$$ (4) – $$\\frac { 2 }{ 3 }$$ (2) – $$\\frac { 1 }{ 3 }$$ Hint: Since the three points are collinear. Area of a ∆ = 0 3k2 + 3k + 6k – (6k2 + 9k + k) = 0 ⇒ 3k2 + 9k – 6k2 – 10k = 0 -3 k2 – k = 0 ⇒ -k(3k + 1) = 0 3k + 1 = 0 ⇒ 3 k = -1 ⇒ k = – $$\\frac { 1 }{ 3 }$$ Question 5. If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units then x = …………. (1)", "BK \\perp AD, AK = 10, KD = 20. It has been illustrated in the diagram given below. 26. In rectangle ABCD below, AB = 5, AF = 7, and FD = 3 In rectangle ABCD below, AB = 5, AF = 7, and FD = 3. Isosceles trapezoid ABCD has legs AB and CD and base BC If AB = 7y – 4, BC = 4y – 6, and CD = 8y – 18, find the value of y. Geometry Edit: the new version of the question is false: the left hand side should be AC²+BD². You will receive an answer to the email. A B. What is the area, in square units, of trapezoid ABCD ? Draw a diagram to illustrate the problem as shown below. Theorem 1 : If a trapezoid is isosceles, then each pair of base angles is congruent. What is the area of ABCD? All pricing is in US dollars (USD). There was no land west of the Appalachian Mountains In trapezoid with bases and , we have , , , and .The area of is . Isosceles trapezoid ABCD has legs AB and CD and base BC If AB = 7y – 4, BC = 4y – 6, and CD = 8y – 18, find the value of y. Geometry Consider", "# Abcd is a Rectangle Whose Three Vertices Are A(4,0), C(4,3) and D(0,3). Find the Length of One Its Diagonal. - Mathematics ABCD is a rectangle whose three vertices are A(4,0), C(4,3) and D(0,3). Find the length of one its diagonal. #### Solution The given vertices are B(4, 0), C(4, 3) and D(0, 3) Here, BD one of the diagonals So BD =sqrt((4-0)^2+(0-3)^2) = sqrt((4)^2 +(-3)^2) =sqrt(16+9) = sqrt(25) = 5 Hence, the length of the diagonal is 5 units . Concept: Coordinate Geometry Is there an error in this question or solution? #### APPEARS IN RS Aggarwal Secondary School Class 10 Maths Chapter 16 Coordinate Geomentry Q 3", "Moderator Joined: 20 Dec 2010 Posts: 1809 Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] Show Tags 16 Apr 2011, 23:36 4 annmary wrote: hello i cant solve this geometry question , plz help! tanx Sol: Let l=length w=wdith d=diagonal=2 $$Area=l*w=l*\\sqrt{d^2-l^2}$$ $$l*\\sqrt{2^2-l^2}$$ $$l*\\sqrt{4-l^2}=Area$$ Squaring both sides; $$l^2*(4-l^2)=(Area)^2$$ $$4l^2-l^4=(Area)^2$$ $$-l^4+4l^2-(Area)^2=0$$ $$Let \\hspace{2} l^2=x$$ $$-x^2+4x-(Area)^2=0$$ $$D=\\sqrt{b^2-4*a*c}$$ $$D=\\sqrt{4^2-4*(-1)*-(Area)^2}$$ $$D=\\sqrt{16-4*(Area)^2}$$ In inside of the root must be greater or equal to 0 to provide valid roots of the equation. $$16-4*(Area)^2 \\ge 0$$ $$4*(Area)^2 \\le 16$$ $$(Area)^2 \\le 4$$ $$(Area) \\le 2$$ Only I and II are less than 2. Ans: \"D\" _________________ Intern Joined: 18 Sep 2010 Posts: 45 Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] Show Tags 17 Apr 2011, 05:12 thank you fluke +5 kudos , but i just can give you +1 kudos other 4 kudos :+1,+1,+1,+1 _________________ (\\ /) (O.o) (> <) This is Bunny. Copy Bunny into your signature to help him on his way to world domination Manager Joined: 17 Jan 2011 Posts: 220 Re: ABCD is a rectangle inscribed in a circle. If the length of AB is thr [#permalink] Show Tags 17 Apr 2011, 13:20 1 1 Area cannot be 3.2 because area of circle" ]

[ "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]", "is the side length of the squares) for simplicity. We can extend $\\overline{IJ}$ until it hits the extension of $\\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\\dfrac{3 \\cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \\cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\\frac{4}{3\\cdot 2^2} = \\boxed{\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot(X,red); label(\"A\", A, NW); label(\"B\", B, NE); label(\"C\", C, NE); label(\"D\", D, NW); label(\"E\", E, NW); label(\"F\", F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"X\", X,SW,red); label(\"J\", J, SE);[/asy]$ Let the side length of each square be $1$. Let the intersection of $AJ$ and $EI$ be $X$. Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\\angle IXJ$ and $\\angle AXD$ are vertical angles, they are congruent. We also have $\\angle JIH\\cong\\angle ADC$ by definition. So we have $\\triangle ADX\\cong\\triangle JIX$ by $\\textit{AAS}$ congruence. Therefore, $DX=JX$. Since $C$ and $D$ are midpoints of sides, $DH=CJ=\\dfrac{1}{2}$. This combined with", "area of rectangle $ABGH$ is $4 \\cdot 12 = 48$. Taking the area of rectangle $ABGH$ and subtracting the combined area of $\\triangle AHG$ and $\\triangle CGM$ yields $48 - (24 + \\frac{32}{5}) = \\boxed{\\frac{88}{5}}\\ \\text{(C)}$. ### Solution 2 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label(\"A\",A,NW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",Ep,SE); label(\"F\",F,SE); label(\"G\",G,SE); label(\"H\",H,SW); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,NW); label(\"L\",L,NW); label(\"M\",M,W); label(\"N\",Z,NE); [/asy]$ Extend $AB$ and $CH$ and call their intersection $N$. The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\\frac{2\\cdot 4}2 = 4$. The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$. Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \\frac 35 \\cdot 12 = \\frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\\frac{1}{\\frac{1}{8}+\\frac{1}{12}} = \\frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\\frac{36}{5}$.) Then the area of $AMN$ is $\\frac 12 \\cdot", "# 2008 AIME I Problems/Problem 2 ## Problem Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$, and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\\triangle GEM$. ## Solution Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\\cdot10=50$. This implies that vertex G must be located outside of square $AIME$. $[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label(\"$$G$$\",G,N); label(\"$$A$$\",A,NW); label(\"$$I$$\",I,NE); label(\"$$M$$\",M,NE); label(\"$$E$$\",E,NW); label(\"$$10$$\",(M+E)/2,S); [/asy]$ Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$. Also, $\\triangle GXY \\sim \\triangle GEM$. Let the height of $GXY$ be $h$. By the similarity, $\\dfrac{h}{6} = \\dfrac{h + 10}{10}$, we get $h = 15$. Thus, the height of $GEM$ is $h + 10 = \\boxed{025}$.", "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 our unit, this yields a final area of", "$m=1/3$, 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: $\\boxed{-\\frac{1}{6}}$.", "dot(\"E\", E, NW); dot(\"F\", F, SW); dot(\"G\", G, S); dot(\"H\", H, N); dot(\"I\", I, NE); label(\"X\", X,SE); dot(\"J\", J, SE);[/asy]$ First let $s=2$ (where $s$ is the side length of the squares) for simplicity. We can extend $\\overline{IJ}$ until it hits the extension of $\\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\\dfrac{3 \\cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \\cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\\frac{4}{3\\cdot 2^2} = \\boxed{\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}}$. Solution 2 $[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot(X,red); label(\"A\", A, NW); label(\"B\", B, NE); label(\"C\", C, NE); label(\"D\", D, NW); label(\"E\", E, NW); label(\"F\", F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"X\", X,SW,red); label(\"J\", J, SE);[/asy]$ Let the side length of each square be $1$. Let the intersection of $AJ$ and $EI$ be $X$. Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\\angle IXJ$ and $\\angle AXD$ are vertical angles, they are congruent. We also have $\\angle JIH\\cong\\angle ADC$ by", "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", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "dot(X); label(\"X\", X, NE); label(\"C\", C, N); label(\"B\", B, E); label(\"A\", A, W); [/asy]$ ## Problem 25 In the figure, the area of square $WXYZ$ is $25 \\text{ cm}^2$. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\\triangle ABC$, $AB = AC$, and when $\\triangle ABC$ is folded over side $\\overline{BC}$, point $A$ coincides with $O$, the center of square $WXYZ$. What is the area of $\\triangle ABC$, in square centimeters? $[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label(\"A\", A, NW); label(\"O\", O, NE); label(\"B\", B, SW); label(\"C\", C, NW); label(\"W\",W , NE); label(\"X\", X, N); label(\"Y\", Y, S); label(\"Z\", Z, SE); [/asy]$ $\\textbf{(A)}\\ \\frac{15}4\\qquad\\textbf{(B)}\\ \\frac{21}4\\qquad\\textbf{(C)}\\ \\frac{27}4\\qquad\\textbf{(D)}\\ \\frac{21}2\\qquad\\textbf{(E)}\\ \\frac{27}2$", "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? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "(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", "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 ##", "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 $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "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", "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", "(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)", "find the area of the entire paper, we square our side length and multiply by $4$. So, the answer is $4\\cdot\\left(\\frac{w}{\\sqrt{2}} + \\frac{h}{\\sqrt{2}}\\right)^2 = \\boxed{\\textbf{(A) } 2(w+h)^2}$. ~IronicNinja ## Solution 3 The sheet of paper is made out of the surface area of the box plus the sum of the four yellow triangles, as shown below. $[asy] /* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); fill((2,0)--(3,1)--(3,-1)--cycle,yellow); fill((-2,0)--(-3,1)--(-3,-1)--cycle,yellow); fill((0,2)--(1,3)--(-1,3)--cycle,yellow); fill((0,-2)--(1,-3)--(-1,-3)--cycle,yellow); draw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),linewidth(.5)); dot((-3,3)); label(\"A\",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype(\"2.5 2.5\")+linewidth(.5)); draw((3,0)--(-3,0),linetype(\"2.5 2.5\")+linewidth(.5)); label(\"w\",(-1,-1),SW); label(\"w\",(1,-1),SE); label(\"w\",(1,1),NE); label(\"w\",(-1,1),NW); label(\"h\",(2.5,0.5),NW); label(\"h\",(2.5,-0.5),SW); label(\"h\",(-2.5,0.5),NE); label(\"h\",(-2.5,-0.5),SE); label(\"h\",(0.5,2.5),SE); label(\"h\",(-0.5,2.5),SW); label(\"h\",(0.5,-2.5),NE); label(\"h\",(-0.5,-2.5),NW); [/asy]$ The surface area is $2w^2 + 2wh + 2wh$ which equals $2w^2 + 4wh$.The four triangles each have a height and a base of $h$, so they each have an area of $\\frac{h^2}{2}$. There are four of them, so multiplied by four is $2h^2$. Together, paper's area is $2w^2 + 4wh + 2h^2$. This can be factored and written as $\\boxed{\\textbf{(A) } 2(w+h)^2} \\qquad$. ~Yee2121 (Solution) ~MRENTHUSIASM (Diagram)", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =", "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" ]

[ "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]", "# 2021 AIME I Problems/Problem 2 ## Problem In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, W); label(\"B\", B, W); label(\"C\", C, (1,0)); label(\"D\", D, (1,0)); label(\"F\", F, N); label(\"E\", E, S); [/asy]$ ## Solution 1 (Similar Triangles) Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\\frac{7}{3}(11-x)$ and $CG=\\frac{3}{7}x$. We have $\\frac{7}{3}(11-x)+\\frac{3}{7}x=FC=9$. Solving for $x$, we have $x=\\frac{35}{4}$. The area of the shaded region is just $3\\cdot \\frac{35}{4}=\\frac{105}{4}$. Thus, the answer is $105+4=\\framebox{109}$. ~yuanyuanC ## Solution 2 (Similar Triangles) Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\\triangle AFG \\sim \\triangle", "63 ☐D. 72 ☐E. 83 57- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively. The area of ∆BED is 20, and the area of ∆BDF is 32. What is the perimeter of the rectangle? ☐A. 20 ☐B. 22 ☐C. 32 ☐D. 40 ☐E. 48 58- When point A $$(-6.-12)$$ is reflected over the $$y$$-axis to get the point B, what are the coordinates of point B? ☐A. $$(6,- 12)$$ ☐B. $$(6,12)$$ ☐C. $$(-12,6)$$ ☐D. $$(12,-6)$$ ☐E. $$(-6,12)$$ 59- Which of the following points lies on the line? $$5x+3y=6$$ ☐A. $$(6, 1)$$ ☐B. $$(–3, 3)$$ ☐C. $$(3,-3)$$ ☐D. $$(2, 2)$$ ☐E. $$(2, 8)$$ 60- If $$f(x)=2x^3+4x-18x^(-2)$$ and $$g(x)=-3$$, what is the value of $$f(g(x))$$? ☐A. $$-83$$ ☐B. $$-68$$ ☐C. $$-25$$ ☐D. $$-13$$ ☐E. $$0$$ 27% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "# 2008 AIME I Problems/Problem 2 ## Problem Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$, and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\\triangle GEM$. ## Solution Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\\cdot10=50$. This implies that vertex G must be located outside of square $AIME$. $[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label(\"$$G$$\",G,N); label(\"$$A$$\",A,NW); label(\"$$I$$\",I,NE); label(\"$$M$$\",M,NE); label(\"$$E$$\",E,NW); label(\"$$10$$\",(M+E)/2,S); [/asy]$ Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$. Also, $\\triangle GXY \\sim \\triangle GEM$. Let the height of $GXY$ be $h$. By the similarity, $\\dfrac{h}{6} = \\dfrac{h + 10}{10}$, we get $h = 15$. Thus, the height of $GEM$ is $h + 10 = \\boxed{025}$.", "area of rectangle $ABGH$ is $4 \\cdot 12 = 48$. Taking the area of rectangle $ABGH$ and subtracting the combined area of $\\triangle AHG$ and $\\triangle CGM$ yields $48 - (24 + \\frac{32}{5}) = \\boxed{\\frac{88}{5}}\\ \\text{(C)}$. ### Solution 2 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label(\"A\",A,NW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",Ep,SE); label(\"F\",F,SE); label(\"G\",G,SE); label(\"H\",H,SW); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,NW); label(\"L\",L,NW); label(\"M\",M,W); label(\"N\",Z,NE); [/asy]$ Extend $AB$ and $CH$ and call their intersection $N$. The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\\frac{2\\cdot 4}2 = 4$. The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$. Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \\frac 35 \\cdot 12 = \\frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\\frac{1}{\\frac{1}{8}+\\frac{1}{12}} = \\frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\\frac{36}{5}$.) Then the area of $AMN$ is $\\frac 12 \\cdot", "= \\frac 14$, 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),", "-- 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 $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$.", "☐C. 46 ☐D. 55 ☐E. 61 38- In the following figure, ABCD is a rectangle. If $$a=\\sqrt{5}$$, and $$b=2a$$, find the area of the shaded region. (the shaded region is a trapezoid) ☐A. $$\\sqrt{5}$$ ☐B. $$2$$ ☐C. $$4$$ ☐D. $$2\\sqrt{5}$$ ☐E. $$4\\sqrt{5}$$ 39- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively and DE = 5 and DF = 3. The area of ∆BED is 20, and the area of ∆BDF is 24. What is the perimeter of the rectangle? ☐A. 24 ☐B. 28 ☐C. 36 ☐D. 42 ☐E. 48 40- What is the product of all possible values of $$x$$ in the following equation? $$|2x+6-3x-7|=5$$ ☐A. $$-24$$ ☐B. $$-12$$ ☐C. $$8$$ ☐D. $$12$$ ☐E. $$24$$ 28% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "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(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "is the side length of the squares) for simplicity. We can extend $\\overline{IJ}$ until it hits the extension of $\\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\\dfrac{3 \\cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \\cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\\frac{4}{3\\cdot 2^2} = \\boxed{\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot(X,red); label(\"A\", A, NW); label(\"B\", B, NE); label(\"C\", C, NE); label(\"D\", D, NW); label(\"E\", E, NW); label(\"F\", F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"X\", X,SW,red); label(\"J\", J, SE);[/asy]$ Let the side length of each square be $1$. Let the intersection of $AJ$ and $EI$ be $X$. Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\\angle IXJ$ and $\\angle AXD$ are vertical angles, they are congruent. We also have $\\angle JIH\\cong\\angle ADC$ by definition. So we have $\\triangle ADX\\cong\\triangle JIX$ by $\\textit{AAS}$ congruence. Therefore, $DX=JX$. Since $C$ and $D$ are midpoints of sides, $DH=CJ=\\dfrac{1}{2}$. This combined with", "24 B. 38 C. 46 D. 55 E. 61 38- In the following figure, ABCD is a rectangle. If $$a=\\sqrt{5}$$, and $$b=2a$$, find the area of the shaded region. (the shaded region is a trapezoid) A. $$\\sqrt{5}$$ B. $$2$$ C. $$4$$ D. $$2\\sqrt{5}$$ E. $$4\\sqrt{5}$$ 39- In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively, and DE = 5 and DF = 3. The area of ∆BED is 20, and the area of ∆BDF is 24. What is the perimeter of the rectangle? A. 24 B. 28 C. 36 D. 42 E. 48 40- What is the product of all possible values of $$x$$ in the following equation? $$|2x+6-3x-7|=5$$ A. $$-24$$ B. $$-12$$ C. $$8$$ D. $$12$$ E. $$24$$ 28% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment", "(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", "The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar. The ratio of their sides is $\\frac{AM}{AB} = \\frac 58$, hence the ratio of their areas is $\\left( \\frac 58 \\right)^2 = \\frac{25}{64}$. And as the area of triangle $ABC$ is $\\frac{6\\cdot 8}2 = 24$, the area of triangle $AME$ is $24\\cdot \\frac{25}{64} = \\boxed{ \\frac{75}8 }$. ## Solution 3 (Only Pythagorean Theorem) $[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); label(\"E\",Ep,S); label(\"M\",M,NW); [/asy]$ Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \\implies 25 + ME^2$, which means $AE = \\sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB = 8 - \\sqrt{25 + x^2}$. From here, we know that $CE^2 = CB^2 + BE^2 = ME^2 + MC^2$. From here, we can write the expression $6^2 +", "(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.", "square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \\cdot \\frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\\boxed{{\\textbf{(C)}}~15}$. ### Solution 3 $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length $1$, and let's label the other legs $x$ for one of the triangles and $y$ for the other. Note that $x + y = 3$. The area of each of the triangles is $\\frac{x}{2}$ and $\\frac{y}{2}$, and there are $4$ of each. So now we need to find $4\\left(\\frac{x}{2}\\right) + 4\\left(\\frac{y}{2}\\right)$. $(4)\\frac{x}{2} + (4)\\frac{y}{2}$ $\\Rightarrow~~4\\left(\\frac{x}{2}+ \\frac{y}{2}\\right)$ $\\Rightarrow~~4\\left(\\frac{x+y}{2}\\right)$ Remember that $x+y=3$, so substituting this in we find that the area of all of the triangles is $4\\left(\\frac{3}{2}\\right) = 6$. The area of the $4$ unit squares is $4$, so the area of the square we need is $25- (4+6) = \\boxed{\\textbf{(C)}~15}$", "of polygon $ABCDEF$, in square units, is $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 46 \\qquad \\text{(D)}\\ 66 \\qquad \\text{(E)}\\ 74$ $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ ## Problem 5 $[asy] unitsize(13); draw((0,0)--(20,0)); draw((0,0)--(0,15)); draw((0,3)--(-1,3)); draw((0,6)--(-1,6)); draw((0,9)--(-1,9)); draw((0,12)--(-1,12)); draw((0,15)--(-1,15)); fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black); fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black); fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black); fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black); fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black); label(\"A\",(2.5,-.5),S); label(\"B\",(4.5,-.5),S); label(\"C\",(6.5,-.5),S); label(\"D\",(8.5,-.5),S); label(\"F\",(10.5,-.5),S); label(\"Grade\",(15,-.5),S); label(\"1\",(-1,3),W); label(\"2\",(-1,6),W); label(\"3\",(-1,9),W); label(\"4\",(-1,12),W); label(\"5\",(-1,15),W); [/asy]$ The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory? $\\text{(A)}\\ \\frac{1}{2} \\qquad \\text{(B)}\\ \\frac{2}{3} \\qquad \\text{(C)}\\ \\frac{3}{4} \\qquad \\text{(D)}\\ \\frac{4}{5} \\qquad \\text{(E)}\\ \\frac{9}{10}$ ## Problem 6 A stack of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high? $\\text{(A)}\\ 250 \\qquad \\text{(B)}\\ 550 \\qquad \\text{(C)}\\ 667 \\qquad \\text{(D)}\\ 750 \\qquad \\text{(E)}\\ 1250$ ## Problem 7 A \"stair-step\" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\\text{th}$ row is $[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]$ $\\text{(A)}\\ 34 \\qquad \\text{(B)}\\ 35 \\qquad" ]

[ "Saturday, November 14, 2009 Coin Toss A boy tosses a fair coin repeatedly. Every time he gets a Head, he earns 1 penny, and every time he gets a Tail, he earns 2 pennies. He starts with no money. Prove that the probability of him at one time having exactly $n$ pennies is $\\frac{2+(-\\frac{1}{2})^n}{3}$", "## Saturday, November 14, 2009 ### Coin Toss A boy tosses a fair coin repeatedly. Every time he gets a Head, he earns 1 penny, and every time he gets a Tail, he earns 2 pennies. He starts with no money. Prove that the probability of him at one time having exactly $n$ pennies is $\\frac{2+(-\\frac{1}{2})^n}{3}$", "27: “27, 37, 47, 57”. Once your child is fluent with counting numbers to 20, you can begin doing these activities with her. Don’t forget to continue oral counting frequently! • Counting objects Give your child a set of objects. Let him count the objects using whatever technique or strategy he chooses. Any objects around the house would work well for this activity (beans, pasta, cereal, socks). • Counting coins Give your child a set of coins and have her count the total amount. Begin with pennies (counting by ones) before moving on to groups of other coins (for example, a bunch of nickels for skip counting by 5s – 5, 10, 15, 20). Once she can count all the coins in single groups, begin mixing coins, for example, nickels and pennies together. Eventually, we would expect her to be able to count a bunch of mixed coins. – You could make a game out of putting the coins in your pocket at the end of the day on the table. If your child can correctly count them allow her to put the coins in a could put them in the bank. They could then periodically count the bank to see how much they have saved so far. • Calendar or Counting Chart In class teachers use a", "them came up head. In the other half, he tells you that one of them came up tail. $3)$ $25$% of the time we get two tails, and in all this cases that person tells you that one coin came up tail. So, only in $25$% $+ 25$% $= 50$% he says you that one coin came up head, of which its half is when you have both coins head.", "Data. A tally is a record of amounts or numbers which you keep changing and adding to as the activity which affects it progresses. Part 1: Count the tally marks and write the values for each set. Mercury dime with tally marks? Post by InfleXion » Sun Jun 24, 2012 12:39 am I got this coin as part of a deal with a kickfluffy member here, and was wondering if anybody knows of any history behind people putting tally marks on coins? or is this just the result of someone getting bored with a sharp object?. If you want to be able to save and store your charts for future use and editing, you must first create a free account and login -- prior to. After every click, tally marks should represent total number of clicks thus far. (2 versions- Bright backgrounds and black and white)Students can match up the number and the corresponding number of tally marks. Frequency Diagrams. Features: - Apple Watch app, change your counter on your Apple Watch. Count Tally Marks Up To Twenty 20 Quiz. Tally mark worksheet first grade; First Grade Tally Mark Worksheets - PDF documents; Write the tally marks for the number of apples below. We need your help. Common Core State Standards: 2. Kids will be taught how to", "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", "# Pocket Money Algebra Level 1 Little Johnny has been saving up his pocket money for as long as he can remember. He has $800, which is in the form of$20 and \\$10 notes. In total, he has 50 of these notes. How many of each individual note does Johnny have? ×", "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", "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", "and queens from the present one, back to the very smooth and worn face of a young Victoria - yes, I had young Victoria coins. And in my primary school we learnt money. The whole class would chant our times tables with an extension all in a special sing-song way that I hear in my head as I type (I've used three dots … to show a miniscule pause in the chant): Three fives fifteen … pence one and three [ie 3x5 = 15; 15d = 1/3] Three sixes eighteen … pence one and six Three sevens twenty-one … pence one and nine And so on for the entire set up to the 12 times table! And finally, we had a pair of expressions with identical derivations to explain someone else's slowness of uptake: he was \"a bit elevenpence-ha'pny\" or \"not quite the full shilling\" where nowadays we might refer to his being a sandwich short of a picnic. As for modern times, the Irish still refer to quids (and squids) but now mean euros. (Thanks J McColl, Jun 2008) Thanks to D Burt for reminding me about Bob-a-Job week, which prompted a new paragraph above in the history 'pounds shillings and pennies' section. Any other Bob-a-Job recollections?..", "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.", "to the hayloft and uncovered the tin can. He poured the coins out and began to count. Then he counted again. He needed 20 cents more. Could there be any sacks left anywhere in town? The shadows were lengthening as Reuben arrived at the factory. The sack buyer was about to lock up. “Mister! Please don’t close up yet.” The man turned and saw Reuben, dirty and sweat stained. “Come back tomorrow, boy.” “Please, Mister. I have to sell the sacks now – please.” The man could tell Reuben was close to tears. “Why do you need this money so badly?” “It’s a secret.” The man took the sacks, reached into his pocket and put four coins in Reuben’s hand. Reuben murmured a thank you and ran home. Then, clutching the tin can, eha headed for the shop. “I have the money,” he solemnly told the owner. The man went to the window and retrieved Reuben’s treasure. He wiped the dust off and gently wrapped it in brown paper. Then he placed the parcel in Reuben’s hands. Racing home, Reuben burst through the front door. His mother was scrubbing the kitchen stove. “Here, Mum! Here!” Reuben exclaimed as he ran to her side. He placed a small box in her work-roughened hand. She unwrapped it carefully to save", "# How much money did Tom get every day after they found the treasure? For when the found the treasure From the text: Each lad had an income, now, that was simply prodigious—a dollar for every weekday in the year and half of the Sundays.", "# Recent content by DHass22 1. ### Umpa Lumpa Backstory ~ Flipping Coins for \\$. 10,000 Umpa Lumpas, are flipping coins in a football stadium. (yes, from Willy Wonka fame, I figured they're tireless workers and enjoy mundane tasks. Just roll with it.) They are trying to establish streaks of either heads or tails. A streak of 15 consecutive heads or tails, earns the...", "pennies. Maria has 253 pennies. Who has more pennies? Write <, =, or > to compare the number of pennies. 248 253 248 253 Question 10. Jeff is thinking of a number. The number has 2 hundreds. It has more ones than tens. It has 7 tens. Which could be the number? Choose all that apply. 276 279 267 278 289 Question 11. Compare. Write <, =, or >. 429 294 849 984 429 294 849 984 Question 12. Write a number that will make each comparison correct. 327 < ___ 716 > ___ 327 < 720 > 716 > 715 Question 13. There are 326 boys at a school. Which is the expanded form for the number of boys? A. 200 + 30 + 6 B. 300 + 20 + 6 C. 300 + 60 + 2 D. 300 + 60 + 7 B. 300 + 20 + 6 Question 14. Which is a way to show 576? Choose all that apply. 500 + 60 + 16 500 + 70 + 6 400 + 170 + 6 500 + 60 + 7 Question 15. Use the numbers on the cards. Write the missing numbers in the number chart. Question 16. Candy counts 405, 410, 415, 420, 425, 430 By what number does Candy skip count? A. 2", "$5000 each, what the last would amount to, reckoning 7 cents for the first, 4 times 7 cents for the second, and so on, increasing the price 4 times on each to the last. A mouse is at the top of a poplar tree 60 braccia high, and a cat is on the ground at its foot. The mouse decends 1/2 a braccia a day and at night it turns back 1/6 of a braccia. A father left$20,000 to be divided among his four sons ages 6, 8, 10, and 12 years respectively so that each share placed at 4 1/2 compounded interest should amount to the same value when its possessor becomes the age 21.", "$100 more than the pair of shoes does. How much does each item cost? Answer: The hoodie costs$125, the shoes costs $25 Riddle: You have two U.S coins whose total value is$0.30. One of them is not a nickel. What are the two coins? Answer: One is a quarter and one is a nickel. Riddle: Eggs are $0.12 a dozen. How many eggs can you get for a dollar? Answer: 100 eggs, at a penny each. Riddle: A duck was given$9, a spider was given $36, a bee was given$27. Based on this information, how much money would be given to a cat? Answer: $18 ($4.50 per leg) Riddle: “How much is this bag of potatoes?” asked the man. “32 pounds divided by half of its own weight,” said the grocer. How much did the potatoes weigh? Riddle: I am a three-digit number. My tens digit is six more than my ones digit. My hundreds digit is eight less than my tens digit. What number am I? Riddle: A man is twice as old as his little sister. He is also half as old as his dad. Over a period of 50 years, the age of his sister will become half of their dad’s age. What is the age of the man now? Riddle: How can you add", "cost two grand\"; \"You could buy that used car for ten grand\"; \"We spent thirty grand on the wedding\". (Or perhaps you prefer a \"G\" or a \"K\" or a \"dime\". ) But of course, there is no \"grand\" physical money, like a coin or a banknote. We know that it's a counting concept, separate and distinct from the legal notes that we carry around with us. In principle, you could walk into a dealership and pay for a car in cash, and you'd fork over a wad of one hundred $100 bills (or something). You can even write in your ledger something like \"10K\" and conveniently record the transaction. * So the same is true with shillings and pounds in a medieval context. When Charlemagne first established the 1:20:240 money system (denier/sou/livre, or penny/shilling/pound; around 800 A.D.), only the smallest \"deniers\" were actually minted (each 1/240th of a pound weight of silver metal); no other coins existed. Accounting books were kept, recording prices and purchases in shillings/pounds, but they weren't coins that you could carry around with you and hand over to a merchant. And so it was throughout the entire Middle Ages; although a wide variety of silver and gold coins came into use, they weren't so large as to be worth an entire shilling or", "cents for the first, 4 times 7 cents for the second, and so on, increasing the price 4 times on each to the last. A father left$20,000 to be divided among his four sons ages 6, 8, 10, and 12 years respectively so that each share placed at 4 1/2 compounded interest should amount to the same value when its possessor becomes the age 21. The authors recount the 'great tale' of Napier's and Burgi's parallel development of logarithms and urge you to use it in class.", "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" ]

[ "$9$ pairs of numbers are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5)$.", "expected total value of the two coins? $\\text{There are: }\\,{10\\choose2}\\,=\\,45\\,\\text{ pssible pairs of coins.}$ $\\begin{array}{cccccccccccc} \\text{Pair} &&\\text{Prob}&&&& \\text{Value} && \\text{Exp} \\\\ \\hline (5,5) && \\frac{{4\\choose2}}{45} &=& \\frac{6}{45} && 10 && \\frac{60}{45} \\\\ (5,10) && \\frac{{4\\choose1}{1\\choose1}}{45} &=& \\frac{4}{45} && 15 && \\frac{60}{45} \\\\ (5,20) && \\frac{{4\\choose1}{3\\choose1}}{45} &=& \\frac{12}{45} && 25 && \\frac{300}{45} \\\\ (5,50) && \\frac{{4\\choose1}{2\\choose1}}{45} &=& \\frac{8}{45} && 55 && \\frac{440}{45} \\\\ (10,20) && \\frac{{1\\choose1}{3\\choose1}}{45} &=& \\frac{3}{45} && 30 && \\frac{90}{45} \\\\ (10,50) && \\frac{{1\\choose1}{2\\choose1}}{45} &=& \\frac{2}{45} && 60 && \\frac{120}{45} \\\\ (20,20) && \\frac{{3\\choose2}}{45} &=& \\frac{3}{45} && 40 && \\frac{120}{45} \\\\ (20,50) && \\frac{{3\\choose1}{2\\choose1}}{45} &=& \\frac{6}{45} && 70 && \\frac{420}{45} \\\\ (50,50) && \\frac{{2\\choose2}}{45} &=& \\frac{1}{45} && 100 && \\frac{100}{45} \\\\ \\hline &&&&&& \\text{Total:} && \\frac{1710}{45}\\end{array}$ $E \\;=\\;\\frac{1710}{45} \\;=\\;38\\text{ cents.}$ Tags expected, total ### two coins are selected at random from a wallet containing 1 five cent piece, 1 ten cent piece Click on a term to search for related topics. Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post marks2014 Probability and Statistics 3 April 30th, 2014 12:23 PM mared Geometry 3 April 4th, 2014 09:53 AM branny12000 Advanced Statistics 0 May 28th, 2013 04:37 AM reemas Calculus 3 October 22nd, 2011 07:21 PM MathematicallyObtuse Algebra 3 February 8th, 2011 01:41 AM Contact - Home - Forums - Cryptocurrency Forum", "easier to get one two or more apart: 1: (1/5) * (3/4) 2: (1/5) * (2/4) 3: (1/5) * (2/4) 4: (1/5) * (2/4) 5: (1/5) * (3/4) Giving 12/20 = 3/5, or a bit more than 1/2. 3. Hello, billy! A hat contains 5 tickets numbered 1,2,3,4,5. Two tickets are drawn at random from the hat. Find the probability that the numbers on the two tickets differ by two or more. With such a small number of outcomes, we can list all of them. If the tickets are drawn with replacement, the outcomes are: . . $\\begin{array}{ccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) \\\\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) \\\\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) \\\\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) \\\\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) \\end{array}$ . . 25 outcomes If they are drawn without replacement, the outcomes are: . . $\\begin{array}{ccccc} & (1,2) & (1,3) & (1,4) & (1,5) \\\\ (2,1) & & (2,3) & (2,4) & (2,5) \\\\ (3,1) & (3,2) & & (3,4) & (3,5) \\\\ (4,1) & (4,2) & (4,3) & & (4,5) \\\\ (5,1) & (5,2) & (5,3) & (5,4) & \\end{array}$ . . 20 outcomes", "value of six turquoise- colored chips and three violet- colored chips is 42 dollars. (2) The combined value of five turquoise- colored chips and seven violet- colored chips is 53 dollars. $${\\text{chips}}\\,\\left\\{ \\begin{gathered} \\,{\\text{turqu}}:\\ \\,t\\,\\,{\\text{each}} \\hfill \\\\ \\,{\\text{violet}}:\\ \\,v\\,\\,{\\text{each}} \\hfill \\\\ \\end{gathered} \\right.\\,\\,\\,\\,\\,\\,\\left( {t,v \\geqslant 1\\,\\,\\,{\\text{ints}}\\,\\,\\left( * \\right)} \\right)$$ $$? = 4t + 2v\\,\\,\\,\\left[ \\ \\right]$$ $$\\left( 1 \\right)\\,\\,6t + 3v = 42\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{:\\,3} \\,\\,2t + v = 14\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{ \\cdot \\,2} \\,\\,\\,\\,\\,? = 28\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\, \\Rightarrow \\,\\,\\,\\,\\,{\\text{SUFF}}.$$ $$\\left( 2 \\right)\\,\\,5t + 7v = 53\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\, \\Rightarrow \\,\\,53 - 7v = 5t\\,\\,,\\,\\,t\\mathop \\geqslant \\limits^{\\left( * \\right)} 1\\,\\,\\operatorname{int} \\,\\,\\,\\, \\Rightarrow \\,\\,1\\mathop \\leqslant \\limits^{\\left( * \\right)} v\\,\\,\\left( {\\operatorname{int} } \\right) \\leqslant 7\\,\\,\\,\\left( {**} \\right)$$ $$\\frac{{\\left( {3 + 50} \\right) - \\left( {5v + 2v} \\right)}}{5} = \\operatorname{int} \\,\\,\\,\\, \\Leftrightarrow \\,\\,\\,\\,\\frac{{3 - 2v}}{5} = \\operatorname{int} \\,\\,\\,\\,\\mathop \\Leftrightarrow \\limits^{\\left( {**} \\right)} \\,\\,\\,\\,v = 4\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{\\left( 2 \\right)} \\,\\,\\,\\,t = 5$$ $$\\left( {t,v} \\right)\\,\\,{\\text{unique}}\\,\\,\\,\\, \\Rightarrow \\,\\,\\,{\\text{?}}\\,\\,\\,{\\text{unique}}\\,\\,\\,\\,\\, \\Rightarrow \\,\\,\\,\\,\\,{\\text{SUFF}}{\\text{.}}$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level \"quant\" preparation starts here: https://gmath.net A casino pays players with chips that are either turquoise- or violet- [#permalink] 28 Nov 2018, 18:55 Display posts from previous: Sort by", "of knowing that and the rational thing to do is go with indifference. But this only makes sense because I've told you that I was choosing from two sets of envelopes in the first place. What if I didn't tell you that? What if there only ever was one pair? What if there were thousands? Maybe some of those have duplicate amounts, maybe not. Maybe there's only a single pair with 10 in it. (This is @JeffJo's thing, and it's worth thinking about. You can't really even count on using 1 - P(A), much less assume P(A) = 1/2.) Here's a real life version. Suppose I have some cash and two envelopes, and I'm going to split my cash in such a way that one envelope has twice as much as the other. Suppose I have one $10, one$5 and 6 \\$1's. What are my options? $\\small \\begin{array}{ccc}&Envelope\\ 1&Envelope\\ 2\\\\A&1&2\\\\B&2&4\\\\C&3&6=5+1\\\\D&5&10\\\\E&6=5+1&12=10+2\\\\F&7=5+2&14=10+4\\end{array}$ There are some combinations I can't make because I don't have enough of the right denominations. We could talk about this table as we talked about the collection {{5, 10}, {10,20}}. If you knew how much money I had and in what denominations, there are several cases in which, upon opening an envelope, you'd already know whether you have the larger or the smaller. But let's suppose you", "Charlie chose five numbers: $2$, $3$, $4$, $7$, $10$. He added together pairs of numbers from his set, and got the following totals: $5$, $7$, $9$, $10$, $11$, $12$. Can you find the totals that Charlie has missed? Choose your own set of five numbers, and work out all the possible sums of pairs. How do you know you've found them all? How many pair sums would there be if you started with six numbers? Or seven numbers? Or ...? Can you find a set of five different numbers, where the sums of some pairs give the same total? Can you find a link between the total of any set of five numbers, and the total of their pair sums? What if you started with a set of six numbers? Or seven? Or ...? What happens when you choose to add sets of three numbers instead of pairs?... ### Missing totals These are the totals that Charlie missed: 17 13 14 ### Missing totals Did Charlie miss any other totals? How many different pairs of numbers can you select from a set of five numbers? ### For a set of 1 number there For the number of pair sums for a set of numbers, we have: For a set of 1 number there are 0 pairs For a", "# Special Coin Placement Three coins are randomly placed into different positions on a $$4 \\times 10$$ grid. The probability that no two coins are in the same row or column can be expressed as $$\\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers. What is the sum of $$a + b$$? Details and assumptions The coins are placed into different positions which means that no 2 coins are placed in the same square. ×", "don't know any of that. You could also figure that if you got an odd number it must be the smaller (because I'm only using bills, no coins) so I'll cleverly not choose one of those; I'll choose only from $\\small \\begin{array}{ccc}&Envelope\\ 1&Envelope\\ 2\\\\B&2&4\\\\E&6=5+1&12=10+2\\end{array}$ If I choose B, and you draw the 2, you can reason that I would have excluded {1, 2}, so the other must be 4. Similarly, if I choose E, and you draw the 6, then you can reason that I would have excluded {3, 6} and so the other must be 12. Ah well. I'd have to have more money to make the game perfect. But what if I choose B and you draw the 4? 8 is mathematically possible, but there's no {4, 8} here. Similarly, if I choose E and you draw the 12; 24 is mathematically possible, but there's no {12, 24} here. So what is your expectation before getting an envelope? Unknown. Something less than half the total cash I have on me, which you don't know, but there are other constraints based on the denominations and some gamesmanship. Again, there's no game until I choose. Say I choose B. You don't know it, but the average value of B is 3. If you draw the 2, trading gains", "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", "+ 1)$$ and the probability a 10p coin will then be moved back is $$(U + 1) /(U + V + 1)$$. The probability of choosing a 50p coin for tranfer is $$M / (M + 1)$$ and the probability a 50p coin will then be moved back is $$(V + 1) /(U + V + 1)$$. Hence the probability that the two exchanges will leave the piggy banks unchanged is: $\\frac{(U + 1) + M(V + 1)}{(M + 1)(U + V + 1)}$ Equating this to 1/2 now gives: $U – V = \\frac{M + 1}{M – 1}$ Since $$U$$ and $$V$$ are integers, this means that $$M$$ can only be 2 or 3. The total sum in both piggy banks is £10 so: $(5M + 1)T + U + 5V = 100$$(5M + 1)T + 6V + (M + 1) / (M – 1) = 100$This gives three possible solution:$\\begin{array}{cccl} M = 2: & 11T + 6V = 97 & \\Rightarrow &T = 5, U = 10, V = 7 \\\\ M = 3: & 8T + 3V = 49 & \\Rightarrow &T = 2, U = 13, V = 11 \\\\ & & \\Rightarrow &T = 5, U = 5, V = 3 \\end{array}$ But, the value of the coins in the first piggy bank", "# A boy have 32 cards out of which 10 cards, each of blue, green and red colours, have denominations as $\\left\\{2^{1}, 2^{2}, \\ldots, 2^{10}\\right\\}$ and one black and one white each has value unity. The number of ways in which the boy can get a sum of 2012 if he can choose any number of cards, is (A) $(1001)^{2}$ (B) $(1002)^{2}$ (C) $(1007)^{2}$ (D) $\\quad(1111)^{2}$ 59 views A boy have 32 cards out of which 10 cards, each of blue, green and red colours, have denominations as $\\left\\{2^{1}, 2^{2}, \\ldots, 2^{10}\\right\\}$ and one black and one white each has value unity. The number of ways in which the boy can get a sum of 2012 if he can choose any number of cards, is (A) $(1001)^{2}$ (B) $(1002)^{2}$ (C) $(1007)^{2}$ (D) $\\quad(1111)^{2}$", "$n=2,3,4,5,6,7$ are: $$2,3,\\frac{136}{35},\\frac{423}{91},\\frac{14334317}{2699697},\\frac{24531101921}{4164089787}$$ These appear to match the simulation. unfortunately I can't really figure out a pattern there :( – hgmath Mar 17 at 18:29 • @hgmath: I'm thinking about a dynamic programming approach, perhaps an illusory thought. But it seems to me that something is going on that reduces \"$k$ matches in $n$ coins\" to smaller counts of matches in parts of $n$ (similar to the information extracted in my Answer). – hardmath Mar 17 at 20:43 My attempt is to visualize, simplify and extract maths from all the statements \\*lets assume that for every instant i begin to pick, the bag would contain exactly $$10$$ coins of which $$5$$ are heads and $$5$$ tails, also we'll assume the coins are distinguishable, so that they have same face on both sides and also i'll assume my coins display were not hidden from me, and my attempt was infinte plus that i made note of every of my attempt and it's information *\\ $$\\begin{matrix} \\cdot &1&2&3&4&5&6&7&8&9&10\\\\ 1&H&H&H&H&H&H&H&H&H&H\\\\ 2&T&T&T&T&T&T&T&T&T&T\\\\ \\end{matrix}$$ this is the row of the coins, in all possibility $$2^{10}$$ $$\\text{the man coins} \\rightarrow \\{ x_{1} , x_{2} , x_{3} ,x_{4} ,x_{5} ,x_{6} ,x_{7} ,x_{8} ,x_{9} ,x_{10} \\}$$ the set $$X$$ is the man's pick, and $$x_{n}$$ is either $$H$$ or $$T$$ depending on how it appeared but", "A set of ten cards, each showing one of the digits from $0$ to $9$, is divided up between five envelopes so that there are two cards in each envelope. The sum of the two numbers inside it is written on each envelope:", "to $$1 + 2 + 4 + 6 + 9 = \\boxed{22}.$$ I hope this helps! Let me know if anything I said was confusing. Good luck with the rest of the course; you're almost done! • @The-Rogue-Blade Your idea is correct but you missed some pairs and also, a<b so pairs with the same numbers are incorrect. When prof. loh drew a line across the chart he did it above the middle, so he excluded all the pairs with the same numbers. These shouldn't be included 2: 1,1 3: 1,1 4: 2,2 1,1 5: 2,2 1,1 6: 3,3 2,2 1,1 7: 3,3 2,2 1,1 I really liked the way you did the question and it would be quite useful if I didn't want to draw a graph", "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?", "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? What are the last two digits in the sum of the factorials of the first $100$ positive integers? Let four positive integers $a$, $b$, $c$, and $d$ satisfy $a+b+c+d=2019$. Prove $\\left(a^3+b^3+c^3+d^3\\right)$ cannot be an even number. What is the tens digit of $7^{2019}$? What is the units digit of $13^{2019}$?", "$c = [2, 4, 5, 2, 4]$ and array $a = [5, 3, 4, 6]$ the following process takes place: Maxim buys the first game using the first bill (its value is $5$), the bill disappears, after that the second bill (with value $3$) becomes the first one in Maxim's wallet, then Maxim doesn't buy the second game because $c_2 > a_2$, the same with the third game, then he buys the fourth game using the bill of value $a_2$ (the third bill becomes the first one in Maxim's wallet) and buys the fifth game using the bill of value $a_3$. Input The first line of the input contains two integers $n$ and $m$ ($1 \\le n, m \\le 1000$) — the number of games and the number of bills in Maxim's wallet. The second line of the input contains $n$ integers $c_1, c_2, \\dots, c_n$ ($1 \\le c_i \\le 1000$), where $c_i$ is the cost of the $i$-th game. The third line of the input contains $m$ integers $a_1, a_2, \\dots, a_m$ ($1 \\le a_j \\le 1000$), where $a_j$ is the value of the $j$-th bill from the Maxim's wallet. Output Print a single integer — the number of games Maxim will buy. Examples Input 5 42 4 5 2 45 3 4 6 Output 3 Input", "number of dimes in the collection?... Which set of ordered pairs represents a function? {(0, 1), (1,3), (1,5), (2, 6)} {(0, 0), (1, 2), (2, 4), (3, 4); {(0, 1), (1, 2), (2, 3), (2, 4); {(0,0), (0, 2), (2, 2), (2, 4); Which set of ordered pairs represents a function? {(0, 1), (1,3), (1,5), (2, 6)} {(0, 0), (1, 2), (2, 4), (3, 4); {(0, 1), (1, 2), (2, 3), (2, 4); {(0,0), (0, 2), (2, 2), (2, 4);... Melody paid off a loan of $120,000 in 10 years. If she made a payment of$1,225 each month, then what was the simple interest rate for her loan? PLEASE HELP!!! Melody paid off a loan of $120,000 in 10 years. If she made a payment of$1,225 each month, then what was the simple interest rate for her loan? PLEASE HELP!!!...", "the range: $$\\text{0} \\le \\text{number of playing cards} \\le \\text{3}$$ is 4. A group of $$\\text{20}$$ learners count the number of coins they each have. This is the data they collect: $\\begin{array}{c c c c c} 17 & 11 & 1 & 15 & 14 \\\\ 3 & 4 & 18 & 5 & 14 \\\\ 18 & 19 & 4 & 18 & 15 \\\\ 16 & 13 & 20 & 8 & 18 \\end{array}$ Count the number of learners who have from $$\\text{4}$$ up to $$\\text{7}$$ coins. In other words, how many learners have coins in the following range: $$\\text{4} \\le \\text{number of coins} \\le \\text{7}$$? It may be helpful for you to draw a histogram in order to answer the question. Firstly we sort the table into sequential order, starting with the smallest value. $\\begin{array}{c c c c c} 1 & 3 & 4 & 4 & 5 \\\\ 8 & 11 & 13 & 14 & 14 \\\\ 15 & 15 & 16 & 17 & 18 \\\\ 18 & 18 & 18 & 19 & 20 \\end{array}$ Secondly, we draw a histogram of the data: From the histogram you can see that the number of learners with coins in the range: $$\\text{4} \\le \\text{number of coins} \\le \\text{7}$$ is 3. A group of", "# Tag Info 18 Let $b$ be the number of boys, and $g$ the number of girls. For any boy $B$ and girl $G$ who know each other, write on a slip of paper \"$B$ and $G$ know each other.\" We count the number of slips of paper in two different ways. Since every boy knows $r$ girls, there are $br$ slips of paper. Since every girl knows $r$ boys, there are $gr$ slips of paper. ... 8 Code heads as 1 and tails as 0, and the problem can be phrased in the following way: Flip a fair coin until you have exactly twice as many heads as tails, and then stop. The value of $|S_n|$ is the number of such sequences of coin flips that have length exactly $3n$. For this rephrased version, I asked the same question as the OP a couple of years ago on ... 7 Yes, you are correct on the first question. For the second, there are four positions, the first of which has $9$ possible values, (can't be zero), then the second position has 9 possible values it can take on (subtracting 1 from 10 possible values since it can't be the same number as the first position. Etc... $$9 \\cdot 9\\cdot 8 \\cdot 7 = 4536$$ 7 Since $0\\leqslant" ]

[ "Saturday, November 14, 2009 Coin Toss A boy tosses a fair coin repeatedly. Every time he gets a Head, he earns 1 penny, and every time he gets a Tail, he earns 2 pennies. He starts with no money. Prove that the probability of him at one time having exactly $n$ pennies is $\\frac{2+(-\\frac{1}{2})^n}{3}$", "## Saturday, November 14, 2009 ### Coin Toss A boy tosses a fair coin repeatedly. Every time he gets a Head, he earns 1 penny, and every time he gets a Tail, he earns 2 pennies. He starts with no money. Prove that the probability of him at one time having exactly $n$ pennies is $\\frac{2+(-\\frac{1}{2})^n}{3}$", "three charities, A, B, and C, in the ratio 1: 2: 3. How much money did each charity receive? Total number of units = + + = units →$ 1 unit → $$$\\frac{?}{?}$$ =$ A received $Answer:$9,000 Explanation: Total number of units = 1 +2 + 3 = 6 6 units → $18,000 1 unit →$$$\\frac{18,000}{6}$$ = $3,000 A received$ 3,000 Question 3. The number of coins collected by Xavier, Yohann, and Zachary is in the ratio 2:5:8. Yohann collected 85 coins. a) Find the number of coins Zachary collected. 5 units → coins 1 unit → $$$\\frac{?}{?}$$ = coins units → × = coins Zachary collected coins. Answer: 136 coins Explanation: the number of coins Zachary collected. 5 units → 136 coins 1 unit →$$$\\frac{136}{8}$$ = 17 coins 8 units → 17 × 8 = 136 coins Zachary collected 136 coins. b) Find the number of coins collected by the three boys altogether. Total number of units = + + = units → × = coins The three boys collected coins altogether. 255 coins Explanation: the number of coins collected by the three boys altogether. Total number of units = 2 + 5 + 8 = 15 15 units → 17 × = 15 coins The three boys collected 255 coins altogether. Question 4. At Stacey’s middle", "value of six turquoise- colored chips and three violet- colored chips is 42 dollars. (2) The combined value of five turquoise- colored chips and seven violet- colored chips is 53 dollars. $${\\text{chips}}\\,\\left\\{ \\begin{gathered} \\,{\\text{turqu}}:\\ \\,t\\,\\,{\\text{each}} \\hfill \\\\ \\,{\\text{violet}}:\\ \\,v\\,\\,{\\text{each}} \\hfill \\\\ \\end{gathered} \\right.\\,\\,\\,\\,\\,\\,\\left( {t,v \\geqslant 1\\,\\,\\,{\\text{ints}}\\,\\,\\left( * \\right)} \\right)$$ $$? = 4t + 2v\\,\\,\\,\\left[ \\ \\right]$$ $$\\left( 1 \\right)\\,\\,6t + 3v = 42\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{:\\,3} \\,\\,2t + v = 14\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{ \\cdot \\,2} \\,\\,\\,\\,\\,? = 28\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\, \\Rightarrow \\,\\,\\,\\,\\,{\\text{SUFF}}.$$ $$\\left( 2 \\right)\\,\\,5t + 7v = 53\\,\\,\\,\\left[ \\ \\right]\\,\\,\\,\\,\\, \\Rightarrow \\,\\,53 - 7v = 5t\\,\\,,\\,\\,t\\mathop \\geqslant \\limits^{\\left( * \\right)} 1\\,\\,\\operatorname{int} \\,\\,\\,\\, \\Rightarrow \\,\\,1\\mathop \\leqslant \\limits^{\\left( * \\right)} v\\,\\,\\left( {\\operatorname{int} } \\right) \\leqslant 7\\,\\,\\,\\left( {**} \\right)$$ $$\\frac{{\\left( {3 + 50} \\right) - \\left( {5v + 2v} \\right)}}{5} = \\operatorname{int} \\,\\,\\,\\, \\Leftrightarrow \\,\\,\\,\\,\\frac{{3 - 2v}}{5} = \\operatorname{int} \\,\\,\\,\\,\\mathop \\Leftrightarrow \\limits^{\\left( {**} \\right)} \\,\\,\\,\\,v = 4\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{\\left( 2 \\right)} \\,\\,\\,\\,t = 5$$ $$\\left( {t,v} \\right)\\,\\,{\\text{unique}}\\,\\,\\,\\, \\Rightarrow \\,\\,\\,{\\text{?}}\\,\\,\\,{\\text{unique}}\\,\\,\\,\\,\\, \\Rightarrow \\,\\,\\,\\,\\,{\\text{SUFF}}{\\text{.}}$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level \"quant\" preparation starts here: https://gmath.net A casino pays players with chips that are either turquoise- or violet- [#permalink] 28 Nov 2018, 18:55 Display posts from previous: Sort by", "them came up head. In the other half, he tells you that one of them came up tail. $3)$ $25$% of the time we get two tails, and in all this cases that person tells you that one coin came up tail. So, only in $25$% $+ 25$% $= 50$% he says you that one coin came up head, of which its half is when you have both coins head.", "all is 18 cents ; hence there were as many of each as 2 is contained times in 18, which are 9. 12. SOL. The difference between having 8 more and 12 less is 20, and the difference between having 6 times i is many as Henry, and 2 times as many as Henry, is 4 times as many as Henry; hence 4 times Henry's number equals 20, &c. 13. SOL. If one girl received 3 apples, and one boy plus 4 apples, which are 7 apples; and since there were the same number of each, and all received 28 apples, there were as many of each as 7 is contained times in 28, which are 4. 17. SOL. If one boy receives 2 cents, 3 boys will re- ceive 3 times 2 cents, which are 6 cents, and one girl and 3 boys will receive 4 cents plus 6 cents, which are lO cents, and they all received 60 cents ; hence there were as many times one girl and 3 boys as 10 is contained times in 60, which are 6 ; hence there were 6 girls, and 6 times 3, or 18 boys. 18. SUG. We find the difference between the cost of 2 apples and one orange is 2 cents, and the difference between", "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", "### 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", "you wrote at the beginning. Wow, Magic! Guess how much money people have in their pockets! Without giving you any information, ask a friend to count the value of some coins and write the amount on a piece of paper. Then ask your friend to: • Double the amount. • Add the first odd prime number to the new total. • Multiply the result by 1/4 of 20. • Subtract the lowest common multiple of 2 and 3. For the grand finale, you ask for the final answer. Take off the last digit and you will be able to work out how much the coins are worth! Amaze your audience by working out not only their age but also what size shoe they wear! Wow them even more by telling them how the maths works. Give them the following directions but tell them not to show you any calculations:", "each problem. • Use the four-step plan. • Make a table Question 4. Amber has coins in a jar. The sum of the coins is 13⊄. What are the possible groups of coins Amber could have? Explanation: Amber has coins in a jar The sum of the coins is 13⊄ We need of find the possible groups of coins that Amber could have I drew a table to show the number of possible combinations So, there are 4 possible groups of coins that Amber can have. Question 5. Solaria buys 2 bags of salad mix for $8 and 3 pounds of fresh vegetables for$9. She gives the cashier $20. How much change will she receive? Answer:$3 Explanation: Solaria buys 2 bags of salad mix for $8 and 3 pounds of fresh vegetables for$9 She gives the cashier $20 We need to find the amount of change she receives First add$8 and $9 8 + 9 = 17 Now, Subtract$17 from $20 20 – 17 = 3 So, Solaria receives$3 change. Question 6. Mathematical PRACTICE 6 Be Precise Mr. Grow has 12 tomato plants arranged in 2 rows of 6. List 2 other ways that Mr. Grow could arrange his 12 tomato plants in equal rows. Explain to a classmate how you got your answer. Explanation: Mr. Grow has", "# Question 59f14 Jul 21, 2016 ~~29% #### Explanation: You can only get 3 distinct combinations using 3 coins. Either 1 head and two tails, 2 heads and 1 tail or 3 heads. Below are the possible configurations $1 \\to 1 , 0 , 0 | 0 , 1 , 0 | 0 , 0 , 1$ each person can result in $3 \\times 3 = 9$ $2 \\to 1 , 1 , 0 | 0 , 1 , 1 | 1 , 0 , 1$ each person can result in $3 \\times 3 = 9$ $3 \\to 1 , 1 , 1$ each person can result in $1 \\times 1 = 1$ The total amount of combinations between the 2 coins is ${2}^{6}$ because a coin has two options either heads or tails. If both have 3 coins this amounts to 6 total combinations of heads or tails. Thus 19/2^6=.296875 ~~29.69%# This simulation confirms the results using R n<-0; t<-100000; for(i in 1:t) { boy1<-round(runif(3,0,1)); boy2<-round(runif(3,0,1)); if(sum(boy1)==sum(boy2) && sum(boy1)>0){n=n+1;} } n/t this resulted in 0.29444", "### Find probability of exceeding certain amount by drawing 3 coins out of 4 Question Sample Titled 'Find probability of exceeding certain amount by drawing 3 coins out of 4' Alexis has one $1$ coin, one $2$ coin, one $5$ coin and one $10$ coin in her pocket. If Christine takes out three coins randomly from her pocket, find the probability that she gets less than $14.$ A $\\dfrac{{1}}{{2}}$ B $\\dfrac{{3}}{{4}}$ C $\\dfrac{{23}}{{24}}$ D $\\dfrac{{1}}{{4}}$ Initiating... HKDSE 數學試題練習平台", "actually have 137 dollar bills in front of us, but something that represents it: The restriction to no more than 9 per place is just a matter of using as few bills as possible. (By the way, to introduce base 5 numerals, you can use pennies, nickels, and quarters.) Another thing that was of interest was a mechanical counter, either one that works like an old odometer (if you can find anything like that any more), or the plastic counters they used to have where you could click any digit to add one. I don't know if anything like this can still be found easily! On the right is what I described, actually used for keeping track of money; on the left is a hand tally counter you can still get: The physical action can make the idea of counting, especially moving from 9 to 0 and “carrying” to the next digit, more concrete, even though the representation is now entirely abstract. ## Why stop at 9? A 2002 question touched on that very thing: Why can’t a digit be more than 9? Place Value and the Number Nine I am asking this for my 10-year-old grandson. Why is 9 the largest number you can put into any place value spot? Hi, Jacques. The base-ten system is built", "digits. The di Adam has half the money in his right pocket than in his left pocket. If he transferred 40 crowns from the left pocket to the right, he would have the same in both pockets. Calculate how many crowns does Adam have in his left pocket more than in his right? • Tournament How many matches will be played in a football tournament in which there are two groups of 5 teams if one match is played in groups with each other and the group winners play a match for the overall winner of the tournament? • A farmer's heritage A farmer died leaving his 17 horses to his 3 sons. When his sons opened up the Will it read: My eldest son should get 1/2 (half) of total horses; My middle son should be given 1/3rd (one-third) of the total horses; My youngest son should be given 1/9th ( • Three-digit numbers We have digits 0,1,4,7 that cannot be repeated. How many three-digit numbers can we write from them? You can help by listing all the numbers. • Gold, silver, bronze How many ways can we divide gold, silver, bronze medails if there are 6 people competing? • Five letters How many ways can five letters be arranged? • Dodecagon Calculate the size of the", "'heads' and six are 'tails.' Show that there are 210 ways that the coins can be lined up. There are ten positions in a row for the coins. We will chose four of them and place the Heads there. \\;\\;(The six Tails will go in the remaining positions.) How many ways can we choose 4 positions out of 10? There are: \\,C(10,4)\\:=\\:\\frac{10!}{4!\\cdot6!}\\:=\\: 210 ways . . . ta-DAA!", "### 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", "pennies. Maria has 253 pennies. Who has more pennies? Write <, =, or > to compare the number of pennies. 248 253 248 253 Question 10. Jeff is thinking of a number. The number has 2 hundreds. It has more ones than tens. It has 7 tens. Which could be the number? Choose all that apply. 276 279 267 278 289 Question 11. Compare. Write <, =, or >. 429 294 849 984 429 294 849 984 Question 12. Write a number that will make each comparison correct. 327 < ___ 716 > ___ 327 < 720 > 716 > 715 Question 13. There are 326 boys at a school. Which is the expanded form for the number of boys? A. 200 + 30 + 6 B. 300 + 20 + 6 C. 300 + 60 + 2 D. 300 + 60 + 7 B. 300 + 20 + 6 Question 14. Which is a way to show 576? Choose all that apply. 500 + 60 + 16 500 + 70 + 6 400 + 170 + 6 500 + 60 + 7 Question 15. Use the numbers on the cards. Write the missing numbers in the number chart. Question 16. Candy counts 405, 410, 415, 420, 425, 430 By what number does Candy skip count? A. 2", "is $1$$1$ liter • Time taken to finish a work is $30$$30$mins • Temperature of water is ${26}^{\\circ }$${26}^{\\circ}$C • Price of a pen is $15$$15$ coins Consider a girl and her brother sharing candies. She receives $2$$2$ candies from her brother. The number of candies received is $2$$2$. On another day, The girl gives $2$$2$ candies to her brother. The number of candies given is $2$$2$. Considering whole numbers only. A girl received $2$$2$ candies from her brother. This is represented by the number $2$$2$. On another day, she gives $2$$2$ candies to her brother. This is represented by the number $2$$2$. Note that both the numbers are represented as $2$$2$. Are these two quantities the same? Though the values of the numbers are same, the number 2 received is different from 2 given. Consider the two situations. A girl receives $2$$2$ candies from her father. On another day, she receives another $2$$2$ candies from her father. Effectively, she has received, $2+2=4$$2 + 2 = 4$ candies from her father. The girl receives $2$$2$ candies from her brother. On another day, she gives $2$$2$ candies to her brother. Overall, she has received $2$$2$ candies and returned $2$$2$ candies. This makes total of, $2-2=0$$2 - 2 = 0$ candies received from her brother. It is noted that the", "pizza corner? a. Table 1 b. Table 2 c. Table 3 d. Table 4 157. Charles has 25 kites out of which 5 are black and 20 are green. If two kites are chosen one after the other, then what is the probability that the first kite is black and the second one is green? a. $\\frac{1}{6}$ b. $\\frac{2}{7}$ c. $\\frac{1}{3}$ d. $\\frac{1}{7}$ 158. Tony had 5 nickels, 4 dimes, and 2 quarters in a bag. He pulled out 2 coins at random from the bag. What is the probability that one is a dime and the other is a nickel? a. $\\frac{2}{11}$ b. $\\frac{4}{11}$ c. $\\frac{1}{22}$ d. $\\frac{6}{11}$ 159. Dennis had 4 nickels, 2 dimes, and 3 quarters in a bag. He pulled out 2 coins at random from the bag. What is the probability that one is a dime and the other is a nickel? a. $\\frac{2}{9}$ b. $\\frac{1}{18}$ c. $\\frac{1}{9}$ d. $\\frac{1}{3}$ a. $\\frac{25}{26}$ b. $\\frac{3}{26}$ c. $\\frac{5}{26}$ d. $\\frac{1}{26}$", "# A boy have 32 cards out of which 10 cards, each of blue, green and red colours, have denominations as $\\left\\{2^{1}, 2^{2}, \\ldots, 2^{10}\\right\\}$ and one black and one white each has value unity. The number of ways in which the boy can get a sum of 2012 if he can choose any number of cards, is (A) $(1001)^{2}$ (B) $(1002)^{2}$ (C) $(1007)^{2}$ (D) $\\quad(1111)^{2}$ 59 views A boy have 32 cards out of which 10 cards, each of blue, green and red colours, have denominations as $\\left\\{2^{1}, 2^{2}, \\ldots, 2^{10}\\right\\}$ and one black and one white each has value unity. The number of ways in which the boy can get a sum of 2012 if he can choose any number of cards, is (A) $(1001)^{2}$ (B) $(1002)^{2}$ (C) $(1007)^{2}$ (D) $\\quad(1111)^{2}$" ]

[ "### If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is : A. 10 B. 12 C. 14 D. 16 $\\begin{array}{l}\\text{Arithmetic mean}\\\\ \\Large\\frac{\\text{ Total sum }}{\\text{ Total number }}\\\\ \\text{According to the question,}\\\\ \\Rightarrow10=\\frac{7+5+13+x+9}{5}\\\\ \\text{⇒ 50 = x + 34}\\\\ \\text{⇒ x = 50 – 34}\\\\ \\text{⇒ x = 16}\\end{array}$", "1 } { n^6 }$ 4. Thus, we can say that the arithmetic mean of these numbers is $\\dfrac{ 1 } { n^6 }$ less than $1.$", "The arithmetic mean of the following numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6 and 7, 7, 7, 7, 7, 7, 7 is : [A] 4 [B] 5 [C] 14 [D] 18", "# Given the set: {9,0,6,-10,2,x,3} , for what x would the mean of the set be -2? Jun 10, 2018 $x$ would have to be $- 24$ #### Explanation: Use the formula for mean. $\\frac{9 + 0 + 6 - 10 + 2 + x + 3}{7} = - 2$ $10 + x = - 14$ $x = - 24$ Hopefully this helps!", "is the meaning of the units digit • what is the purpose of having a 10 in the algebraic expressions $10x +y$ and $10y+x$", "= (counting the 10's in there) 10^11.", "Mathematics # Y = {x | x Element of Set N and 16 < x < 25} ? #### yassautumn 4 years ago $x=\\{17,18,19,20,21,22,23,24\\}$ #### Banaszek168 4 years ago Hello ! We know that x is greater than 16 and less than 25. So $x =$ { $17 ; 18 ; 19 ; 20 ; 21 ; 22 ; 23 ; 24$ }", "× # Arithmetic Mean The arithmetic mean of a set of numbers is the sum of those numbers divided by their count. Thus the arithmetic mean of $$\\{ 4, 5, 9 \\}$$ is $\\frac{4+5+9}{3} = \\frac{18}{3} = 6.$ Note by Arron Kau 2 years, 9 months ago ## Comments There are no comments in this discussion. × Problem Loading... Note Loading... Set Loading...", "× # Arithmetic Mean The arithmetic mean of a set of numbers is the sum of those numbers divided by their count. Thus the arithmetic mean of $$\\{ 4, 5, 9 \\}$$ is $\\frac{4+5+9}{3} = \\frac{18}{3} = 6.$ Note by Arron Kau 2 years, 2 months ago There are no comments in this discussion. ×", "### The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is. A. 414 B. 420 C. 620 D. 621 $\\begin{array}{l}\\text{Sum of numbers}\\\\ \\text{= (41.4 × 15)}\\\\ \\text{= 621}\\end{array}$", "# A set of 12 numbers has an arithmetic mean of 20. if the numbers 10 and 14 are removed from the set, what is the arithmetic mean of the 10 numbers remaining in the set? A set of 12 numbers has an arithmetic mean of 20. if the numbers 10 and 14 are removed from the set, what is the arithmetic mean of the 10 numbers remaining in the set? You can still ask an expert for help ## Want to know more about Polynomial arithmetic? • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Asma Vang For a set of positive numbers , the arithmetic mean is given as Now, set of 12 numbers which includes 10 and 14 has an arithmetic mean of 20. Let x be the sum of the remaining 10 numbers. Now, we find the arithmetic mean of the remaining 10 numbers whose sum is 216.", "Real numbers $x$ and $y$ have an arithmetic mean of $7$ and a geometric mean of $\\sqrt{19}$. Find $x^{2}+y^{2}$.", "Indicate all such probabilities. If $m^2 = 17$, then what is the value of $(m+1)(m-1)$? 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?", "is 11. To find the mean, add up all the numbers in the data set. Then, divide by how many numbers there are. $$7+9+12+13+14=55$$ and $$55 \\div 5 = 11$$. $$4+2+1+2+3=12$$ and $$12 \\div 5 = 2.4$$", "## Arithmetic Mean The arithmetic mean of a set of nine different positive integers is 123456789. Each number in the set contains a different number of digits with the greatest value being a nine-digit number. Find the value of each of the nine numbers. Source: mathcontest.olemiss.edu 1/26/2009 SOLUTION Given $\\left \\{a_1,a_2,\\cdots,a_9\\right \\}$ a set of nine different positive integers, the arithmetic mean of the set is $\\frac{1}{9}\\sum_{i=1}^{9} a_i$ Define the set of as follows: $a_1=1$ $a_2=11$ $a_3=111$ $a_4=1111$ $a_5=11111$ $a_6=111111$ $a_7=1111111$ $a_8=11111111$ $a_9=111111111$ Then, $\\sum_{i=1}^9 a_i=123456789$ Since we want the arithmetic mean to be 123456789, all we have to do is multiply each $a_i$ by 9. Answer: 9; 99; 999; 9999; 99999; 999999; 9999999; 99999999; 999999999.", "× # Arithmetic Mean The arithmetic mean of a set of numbers is the sum of those numbers divided by their count. Thus the arithmetic mean of $$\\{ 4, 5, 9 \\}$$ is $\\frac{4+5+9}{3} = \\frac{18}{3} = 6.$ Note by Arron Kau 2 years, 7 months ago", "## Thinking Mathematically (6th Edition) ${9, 10, 11, 12,..., 25}$ This is a set of all the whole numbers greater than or equal to 9 and less than or equal to 25. We are supposed to describe the set with words, and while more than one correct description may exist, any one of them will suffice.", "12a – 8, 12a -2,… \\}$, which of the following shows the possible value/s of $a$?", "Mean ? The mean of $$25$$ numbers is $$20$$. If $$5$$ is added to each of the first $$10$$ numbers, find the mean of new set of $$25$$ numbers. Details and Assumptions: The mean refers to the Arithmetic Mean. For example, the mean of the numbers $$1, 2$$ and $$3$$ is $$\\frac{1+2+3}{3}=2$$. ×", "- 10 \\ n + 10$$ $$l = 10$$ Hence $$n^{th}$$ term of that series is 10." ]

[ "and 8 white pairs if each selection is replaced before the next pair is chosen. What if each pair is not replaced? 1. \\begin{align*}P(\\text{snake eyes}) \\times P(\\text{seven}) = \\frac{1}{36}\\times\\frac{1}{6}=\\frac{1}{216}\\end{align*}. 2. \\begin{align*}P(\\text{jack, queen, king}) \\times P(\\text{ace}) = \\frac{12}{52}\\times\\frac{4}{51}=\\frac{4}{221}\\end{align*} 3. \\begin{align*}P(\\text{jack}) \\times P(\\text{jack}) \\times P(\\text{non jack}) \\times P(\\text{non jack}) \\times P(\\text{non jack}) =\\frac{4}{52}\\times \\frac{3}{51} \\times \\frac{48}{50} \\times \\frac{47}{49}\\times \\frac{46}{48}=\\frac{1081}{270725}\\approx 0.00399298\\end{align*} We aren’t quite finished, however, because what we have determined here is the probability that the first two cards are the jacks and the last three are the non jacks. These cards could have been dealt in any order so we need to determine the number of permutations of these cards and multiply by that value. Keep in mind that the jacks are “non distinguishable” and the non jacks are “non distinguishable”. The permutations are: \\begin{align*}\\frac{5!}{2!3!}=10\\end{align*}. This could also be described as the number of combinations of selecting two jacks from a set of five cards: \\begin{align*}_5C_2=\\frac{5!}{2!(5-2)!}=10\\end{align*}. These are just two ways to find the same result. Now that we have the number of combinations or arrangements, we can multiple our probability by this value: \\begin{align*}\\frac{1081}{270725}\\times10\\approx0.0399298181\\end{align*}. 4. For the first part of the question, the total number of socks in the drawer remains the same for each selection, 17. So, \\begin{align*}P(\\text{black, then blue, then white}) = \\frac{5}{17}\\times\\frac{4}{17}\\times\\frac{8}{17}=\\frac{160}{4913}\\approx0.032567\\end{align*}. Now, if we", "5i}{13} = \\frac{12}{13} + \\frac{5}{13} i \\end{align}", "10\\\\ &= 40 + \\frac{8}{{17}} \\times 10\\\\ &= 40 + 4.705\\\\ &= 40.705\\\\ &= 40.7\\end{align} Hence the mode is $$40.7$$", "\\begin{aligned}2(6-2) &= s+2 \\\\8&= s+2 \\implies s=6\\end{aligned}", "\\ln 4 \\\\ x &= \\frac{3 \\ln 100}{\\ln 4} \\\\ x &= \\frac{6 \\ln 10}{\\ln 4} \\\\ \\end{aligned}", "maybe some results will use less than four fours. But the translation from a compact equation to a \"four-fours\" equation is immediate. Let's consider the following list of two-four numbers (I purposely omit the ones that can be obtained with a signel four): \\begin{align} 0 &= 4 - 4 \\\\ 1 &= \\frac{4}{4} \\\\ 5 &= 2 + 3 \\\\ 7 &= 4 + 3 \\\\ 9 &= 3 + 6 \\\\ 11 &= 21 - 10 \\\\ 12 &= 10 + 2 \\\\ 13 &= 10 + 3 \\\\ 14 &= 10 + 4 \\\\ 15 &= 21 - 6 \\\\ 16 &= 4 \\cdot 4 \\\\ 17 &= 21 - 4 \\\\ 18 &= 21 - 3 \\\\ 19 &= 21 - 2 \\\\ 20 &= 24 - 4 \\\\ 22 &= 24 - 2 \\\\ 23 &= 21 + 2 \\\\ 25 &= 21 + 4 \\\\ 26 &= 24 + 2 \\\\ 27 &= 24 + 3 \\\\ 28 &= 24 + 4 \\end{align} Up to this point we can get any integer below 28 with only two fours. My strategy will be obtaining numbers by combining the \"high part\" and \"low part\". So, given any two-four \"high number\" $n$ we can generate all integers between $n - 28$ and $n + 28$. And", "from least to greatest. The data is sorted. The decimal numbers and the corresponding fractions can now be matched from the screen where they were first entered. \\begin{align*}\\frac{1}{11},\\frac{2}{11},\\frac{7}{9},\\frac{5}{6},\\frac{8}{7}\\end{align*} ### Examples #### Example 1 Earlier, you were asked to order the following real numbers from least to greatest. \\begin{align*}\\frac{22}{7},1.234 234 \\ldots, - \\sqrt{7}, -5, -\\frac{21}{4}, 7,-1.666,0,8.32,\\frac{\\pi}{2},-\\pi,-5.38\\end{align*} As you examine the above numbers, you can see that there are natural numbers, whole numbers, integers, rational numbers and irrational numbers. These numbers, as they are presented here, would be very difficult to order from least to greatest. Now that all the numbers are in decimal form, order them from least to greatest. \\begin{align*}-5.38, -\\frac{21}{4}, -5, -\\pi, -\\sqrt{7}, -1.666, 0, 1.234234, \\frac{\\pi}{2},\\frac{22}{7}, 7, 8.32\\end{align*} #### Example 2 Arrange the following numbers in order from least to greatest and place them on a number line. \\begin{align*}-3.78, -\\frac{11}{4},-4, \\frac{\\pi}{2}, -\\sqrt{6},-1.888,0,0.2424,\\pi,\\frac{21}{15},2,1.75\\end{align*} \\begin{align*}-3.78, -\\frac{11}{4},-4, \\frac{\\pi}{2}, -\\sqrt{6},-1.888,0,0.2424,\\pi,\\frac{21}{15},2,1.75\\end{align*} #### Example 3 For each given pair of real numbers, find another real number that is between the pair of numbers. i) \\begin{align*}-3,-5\\end{align*} ii) \\begin{align*}-3.4,-3.5\\end{align*} iii) \\begin{align*}\\frac{1}{5},\\frac{3}{10}\\end{align*} iv) \\begin{align*}-\\frac{3}{4},-\\frac{11}{6}\\end{align*} i) The number must be greater than –5 and less than –3. \\begin{align*}-5,{\\color{blue}-4},-3\\end{align*} ii) The number must be greater than –3.5 and less than –3.4. \\begin{align*}-3.5,{\\color{blue}-3.45},-3.4\\end{align*} iii) The number must be greater than \\begin{align*}\\frac{1}{5}\\end{align*} and less than \\begin{align*}\\frac{3}{10}\\end{align*}. Write each fraction", "n!&=\\frac{1}{0.00005} \\\\ n!&=20000 \\\\ \\end{align} Since {eq}7! = 5040 {/eq} and {eq}8! = 40320 {/eq}. Then we will choose {eq}n = 8 {/eq}. Hence, {eq}N = 8 - 1 = 7 {/eq}. $$\\boxed{N = 7}$$", "&= 8^2 \\\\ \\frac{x^2 - 1}{x - 1} &= 64 \\\\ x^2 - 1 &= 64(x - 1) \\\\ x^2 - 1 &= 64x - 64 \\\\ x^2 - 64x + 63 &= 0 \\\\ (x - 63)(x - 1) &= 0 \\\\ x - 63 &= 0 \\\\ x &= 63 \\end{align*} You are correct.", "&= 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", "-2 &= 10x \\\\ -\\frac{2}{10} &= x \\\\ x &= -\\frac{1}{5} \\end{align*}", "two blue marbles? Solution: 215\\begin{align*}\\frac{2}{15}\\end{align*} Example C What is the probability of pulling out two yellow marbles? Solution: 115\\begin{align*}\\frac{1}{15}\\end{align*} Here is the original problem once again. Casey has been reading a book on magic. He is especially interested in card tricks. Casey has a pile of cards. There are four Jacks, four Queens and four Kings. What is the probability that Casey will pull two queens out of the pile? To figure this out let's begin by writing a ratio to show the number of queens compared to the total number of cards. 412\\begin{align*}\\frac{4}{12}\\end{align*} This is the chance of pulling the first queen. Next, we can write the probability of the second queen. 311\\begin{align*}\\frac{3}{11}\\end{align*} Now we can multiply them. 312×311=9132\\begin{align*}\\frac{3}{12} \\times \\frac{3}{11} = \\frac{9}{132}\\end{align*} We can simplify this fraction. 344\\begin{align*}\\frac{3}{44}\\end{align*} Finally, we can change this to a percent. .068=6.8%\\begin{align*}.068 = 6.8\\%\\end{align*} Guided Practice Here is one for you to try on your own. There are five girls and eight boys in a group. Mrs. Marsh is going to choose two students randomly to lead the line. What is the probability that she will choose two boys? To figure this out, we first have to figure out the total number of students. 5+8=13\\begin{align*}5 + 8 = 13\\end{align*} Next, we write a ratio to show the chances of her picking", "This means: \\begin{aligned} \\frac{z}{w} &= \\frac{-14 - 22i}{34} \\\\ \\,\\\\ &= \\frac{-14}{34} - \\frac{22i}{34} \\\\ \\,\\\\ &= \\frac{-7}{17} -\\frac{11i}{17} \\end{aligned} ## Simplifying Complex Numbers Use the rules above as needed to simplify complex expressions. For example: z = \\frac{(4 + 2i) + (2 -i)}{(2 + 2i)(2+ i)} This can be simplified by using the addition rule in the numerator, the multiplication rule in the denominator, and then completing the division. For the numerator: (4 + 2i) + (2 - i) = 6 + i For the denominator: \\begin{aligned} (2 + 2i)(2+ i) &= 4 + 4i + 2i + 2i^2 \\\\ &= (4 -2) + 6i \\\\ &= 2 + 6i \\end{aligned} Putting these back in place gives: z = \\frac{6 + i}{2 + 6i} Multiplying both parts by the conjugate of the denominator leads to: \\begin{aligned} z &= \\frac{(6 + i) (2 - 6i)}{(2 + 6i) (2 -6i)} \\\\ \\,\\\\ &= \\frac{12 + 2i -36i -6i^2}{4 + 12i -12i -36i^2} \\\\ \\,\\\\ &= \\frac{18 - 34i}{40} \\\\ \\,\\\\ &= \\frac{9 - 17i}{20} \\\\ \\,\\\\ &= \\frac{9}{20} -\\frac{17i}{20} \\\\ \\end{aligned} So this means ​z​ simplifies as follows: \\begin{aligned} z &= \\frac{(4 + 2i) + (2 - i)}{(2 + 2i)(2+ i)} \\\\ &= \\frac{9}{20} -\\frac{17i}{20} \\\\ \\end{aligned} Dont Go! We Have More Great Sciencing Articles!", "82 and 85 out of 100 in different subjects, What will be the averag • 1) 70 • 2) 75 • 3) 80 • 4) 85 Ans. B Explanation : \\begin{aligned} ( \\frac {65+67+76+82+85}{5} ) = 75 \\end{aligned} 4) Find the sum of first 30 natural numbe • 1) 470 • 2) 468 • 3) 465 • 4) 463 Ans. C Explanation : Sum of n natural numbers \\begin{aligned} = \\frac{n(n+1)}{2} \\end{aligned} \\begin{aligned} = \\frac{30(30+1)}{2} = \\frac{30(31)}{2} = 465 \\end{aligned} 5) Find the average of first 10 multiples of • 1) 35.5 • 2) 37.5 • 3) 38.5 • 4) 40.5 Ans. C Explanation : \\begin{aligned} = \\frac {7(1+2+3+...+10)}{10} \\end{aligned} \\begin{aligned} = \\frac {7(10(10+1))}{10 \\times 2} \\end{aligned} \\begin{aligned} = \\frac {7(110)}{10 \\times 2} = 38.5 \\end{aligned} 6) The average of four consecutive odd numbers is 24. Find the largest numbe • 1) 25 • 2) 27 • 3) 29 • 4) 31 Ans. B Explanation : Let the numbers are x, x+2, x+4, x+6, then \\begin{aligned} => \\frac{x+(x+2)+(x+4)+(x+6)}{4} = 24 \\end{aligned} \\begin{aligned} => \\frac{4x+12)}{4} = 24 \\end{aligned} \\begin{aligned} => x+3 = 24 => x = 21 \\end{aligned} So largest number is 21 + 6 = 27 7) Average of 10 numbers is zero. At most how many numbers may be greater than ze • 1) 0 • 2)", "have two significant digits. \\begin{align} \\frac{0.1 - 10.3 + 5.132}{12.8} &= \\frac{-5.068}{12.8} & \\text{not yet rounded} \\\\[2 ex] &= \\frac{-5.1}{12.8} & \\text{rounded to 1 decimal place, due to 0.1} \\\\[2 ex] &=-0.3984375 & \\text{not rounded} \\\\[2 ex] &=-0.40 & \\text{rounded to 2 significant digits, due to 5.1} \\end{align}", "if we have 5 different books on a bookshelf, how many different ways could we arrange them? Well, we have 5 choices for the first book, then 4 choices for the second book, etc. So the answer is 5! = 5 × 4 × 3 × 2 × 1 = 120. Sometimes we want to select a few things from a set and then arrange those things we’ve selected. Suppose we want to choose 2 books from a set of 5 to put on a shelf. How many different possible arrangements are there? It’s easy to see that there are 5 choices for the first book, and then 4 choices for the second. So altogether there are 5 × 4 = 20 different possible arrangements of 2 books selected from a set of 5. In general, the formula for the number of permutations of n things taken p at a time is \\begin{align} \\frac {n!}{(n-p)!}\\end{align} Therefore, if n is 5, and p is 2, this formula works out to \\begin{align} \\frac {5!}{(5-2)!}\\end{align} = \\begin{align} \\frac {5!}{3!}\\end{align} = \\begin{align} 5 \\times 4 = 20\\end{align} Notice that if we want to arrange the entire set of n things in order, then the number of permutations is of n things taken n at a time: \\begin{align} \\frac {n!}{(n-n)!}\\end{align} which is equal", "\\quad c = \\sqrt{\\frac{1 + a - a^2}{2}}.$$ The numeric values are approximately \\begin{align*} a &\\approx 0.64458427322415498454 \\\\ b &\\approx 0.20556156585325953808 \\\\ c &\\approx 0.78393092423256364992. \\end{align*} The computation of the dihedral angles is tedious but straightforward. These are, in radians, approximately $$1.6789775749362604980, 2.1248189366147042378, 2.9049356464412552144.$$", "50, namely, 2, 4, 6, …, 50, will each contribute a 2. • Next, we note that the multiples of 4, namely, 4, 8, 12, …, 48, will each contribute an additional 2. • Next, the multiples of 8, namely, 8, 16, 24, …, 48, will contribute yet another 2 each. • The multiples of 16, namely, 16, 32, 48, will contribute one more 2 each • Finally, the multiples of 32 (only 32 in this case) will contribute another 2. Recall the greatest integer notation. For an integer x, the greatest integer less than or equal to x is denoted by [x]. The total number of 2’s contributed above can now be written as: \\begin{align}&\\left[ {\\frac{{50}}{2}} \\right] + \\left[ {\\frac{{50}}{4}} \\right] + \\left[ {\\frac{{50}}{8}} \\right] + \\left[ {\\frac{{50}}{{16}}} \\right] + \\left[ {\\frac{{50}}{{32}}} \\right]\\\\& = 25 + 12 + 6 + 3 + 1 = 47\\end{align} Similarly, the power of 3 in 50! can be written as: \\begin{align}&\\left[ {\\frac{{50}}{3}} \\right] + \\left[ {\\frac{{50}}{{{3^2}}}} \\right] + \\left[ {\\frac{{50}}{{{3^3}}}} \\right]\\\\& = 16 + 5 + 1 = 22\\end{align} The power of 5 in 50! would be: \\begin{align}&\\left[ {\\frac{{50}}{5}} \\right] + \\left[ {\\frac{{50}}{{{5^2}}}} \\right]\\\\ &= 10 + 2 = 12\\end{align} You are now asked this question: at the end of the number 50!, how many 0’s are there? We note that", "# 144...44 \\begin{align} 144 &= 12^2 \\\\ 1444 &= 38^2 \\end{align} Are there any other numbers $$144 \\ldots 44$$ (starting with 1 and followed by all 4s) that are perfect squares? ×", "have \\begin{align*} (12,s,t)&=(12,5,13)\\\\ (u,80,w)&=(192,80,208)\\\\ \\end{align*} • I was about to write up something similar, though I didn't go for Pythagorean triangles but for $2:4:2\\sqrt{5}$ and $3:6:3\\sqrt{5}$ versus $2:3:\\sqrt{13}$ and $4:6:2\\sqrt{13}$. – Jaap Scherphuis Oct 3 '18 at 12:25" ]

[ "### If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is : A. 10 B. 12 C. 14 D. 16 $\\begin{array}{l}\\text{Arithmetic mean}\\\\ \\Large\\frac{\\text{ Total sum }}{\\text{ Total number }}\\\\ \\text{According to the question,}\\\\ \\Rightarrow10=\\frac{7+5+13+x+9}{5}\\\\ \\text{⇒ 50 = x + 34}\\\\ \\text{⇒ x = 50 – 34}\\\\ \\text{⇒ x = 16}\\end{array}$", "+ 2\\\\&= 52 + 2\\\\&= 54\\\\\\end{aligned} Hence, the integers are $\\boxed{52}, \\boxed{53}$ and $\\boxed{54}.$", "did you get 1900? $50 \\times 38 = 1900$ 6. Re: arithmetic mean If the arithmetic mean is 38, then their sum is 50*38 = 1900... 7. Re: arithmetic mean you are all God's gifts to all not so good in mathematics... God Bless guys 8. Re: arithmetic mean Originally Posted by Prove It You are told \\displaystyle \\begin{align*} \\frac{t_k + t_{k + 1} + t_{k + 2} + \\dots + t_{k + 49}}{50} = 38 \\implies t_k + t_{k + 1} + t_{k + 2} + \\dots + t_{k + 49} = 1900 \\end{align*}. Two of these terms are 45 and 55, so removing them gives \\displaystyle \\begin{align*} t_k + t_{k + 1} + t_{k + 2} + \\dots + t_{k + 49} - 45 - 55 &= 1900 - 45 - 55 \\\\ &= 1800 \\end{align*} There are now 48 terms, so the mean of these 48 terms will be \\displaystyle \\begin{align*} \\frac{1800}{48} = 37.5 \\end{align*}. Thank you King of Proofs ProveIt", "nearest whole, the arithmetic mean is 280 Example B Find the arithmetic means of the sets: 1. 261, 286, 257, 284, 258, 281, 258, 285, 267, 275, 258, 284, 260, 285, 258, 284 2. 43, 44, 45, 45, 46, 46, 47, 47, 47, 48, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 50, 50, 50, 51 3. 14, 28, 42, 56, 23, 67, 12, 45, 93, 52 Solution: For each set, find the sum of the values in the set, and divide by the number of values: a) The sum of the values is 4,341. There are 16 values. \\begin{align*}\\frac{4,341}{16}=271.31\\end{align*} b) The sum of the values is 1,198. There are 25 values. \\begin{align*}\\frac{1,198}{25}=47.92\\end{align*} c) The sum of the values is 432. There are 10 values. \\begin{align*}\\frac{432}{10}=43.2\\end{align*} Example C Using the data from the chart, find: a) the mean number of prank calls per five-year interval, and b) the mean number of reported prank calls per year. Solution: a) First, find the sum of the number of reported calls: \\begin{align*}188+149+161+172+154+172=996 \\ calls\\end{align*} The chart indicates that there are six five-year intervals with recorded values. So, dividing the sum by the number of values, we get: \\begin{align*}\\frac{996}{6}=166 \\ calls \\ per \\ five \\ year \\ interval\\end{align*} Note that the value “166” is not the actual count", "## Elementary Technical Mathematics The mean of the numbers $20.2,27.3,35.1,30.6,29.6\\text{ and }22.6$ is $27.6$. The sum of the numbers is, $20.2+27.3+35.1+30.6+29.6+22.6=165.4$ The mean of the $n$ numbers is, $\\text{mean}=\\frac{\\text{Sum of }n\\text{ numbers}}{n}$ \\begin{align} & \\text{mean}=\\frac{\\text{165}\\text{.4}}{6} \\\\ & =27.56 \\\\ & =27.6 \\end{align} Hence, the mean of the numbers is $\\text{mean}=27.6$", "12, 13, 9.$$ \\begin{align} \\text{Mean} & =\\frac{\\left[ \\begin{align} & 6+4+3+8+ \\\\& 9+12+13+9 \\\\\\end{align} \\right]}{8} \\\\ & =\\frac{64}{8} \\\\ & =8 \\\\\\end{align} Therefore, mean is $$8.$$ Related Sections Related Sections Instant doubt clearing with Cuemath Advanced Math Program", "A_{12} A_{12} A_{12} A_{12} \\\\ &A_{12} \\to A_{10} A_{10} A_{10} A_{10} \\\\ &A_{10} \\to A_8 A_8 A_8 A_8 \\\\ &A_8 \\to A_6 A_6 A_6 A_6 \\\\ &A_6 \\to A_4 A_4 A_4 A_4 \\\\ &A_4 \\to A_2 A_2 A_2 A_2 \\\\ &A_2 \\to \\epsilon|a|aa|aaa|aaaa \\end{align*} Many more options are possible, see which one you like most. • Well, $2^{20}$ is also an integral power of $4$, of $16$, of $32$, and of $1024$: $2^{20} = 4^{10} = 16^5 = 32^4 = 1024^2$. The most economical grammar (in terms of number of symbols) is the one based on powers of 2. – Yuval Filmus Mar 9 '14 at 0:18", "the given arithmetic sequence as follows: \\begin{align} & {{a}_{5}}={{a}_{5-1}}+d \\\\ & ={{a}_{4}}+d \\\\ & =-21-7 \\\\ & =-28 \\end{align} Put $n=6$in${{a}_{n}}={{a}_{n-1}}+d$to find the sixth term of the given arithmetic sequence as follows: \\begin{align} & {{a}_{6}}={{a}_{6-1}}+d \\\\ & ={{a}_{5}}+d \\\\ & =-28-7 \\\\ & =-35 \\end{align} The given sequence is an arithmetic sequence and the next two terms of the given sequence are$-28\\ \\text{and}\\ -35$.", "5 + 4 \\times \\frac{8}{3} = \\frac{47}{3}$$ $$A_5 = a_1 + 5d = 5 + 5 \\times \\frac{8}{3} = \\frac{55}{3}$$ Step 4 Ans:The five arithmetic means between 5 and 21 are $$\\frac{23}{3}, \\frac{31}{3}, 13, \\frac{47}{3}, \\frac{55}{3}$$", "### Big Powers Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas. ### Rooted Via 10 How many of the numbers shown are greater than 10? ### Largest Expression Which of these five algebraic expressions is largest, given $x$ is between 0 and 1? # The Power of the Sum ##### Age 14 to 16 Short Challenge Level: Answer: $n=7$ Using pairs of equal numbers \\begin{align}&2^6+2^5+2^4+2^4\\\\ =&2^6 + 2^5 + 2^4\\times2\\\\ =&2^6 + 2^5 + 2^{4+1}\\\\ =&2^6 + 2^5 + 2^5\\\\ =&2^6 + 2^5\\times2\\\\ =&2^6 + 2^6\\\\ =&2^7\\end{align} Finding the value of the powers of 2 $2^2=4$ $2^5=32$ $2^3=8$ $2^6=64$ $2^4=16$ $2^7=128$ So $2^6+2^5+2^4+2^4=64+32+16+16=128=2^7$ Factorising and using index laws Notice that all of the numbers in the sum are multiples of $2^4$, since $2^6=2^2\\times2^4,2^5=2\\times2^4,2^4=1\\times2^4.$ So \\begin{align}2^6+2^5+2^4+2^4&=2^2\\times2^4+2\\times2^4+1\\times2^4+1\\times2^4\\\\ &=\\left(2^2+2+1+1\\right)\\times2^4\\\\ &=\\left(4+2+1+1\\right)\\times2^4\\\\ &=8\\times2^4\\\\ &=2^3\\times2^4\\\\ &=2^7\\end{align} You can find more short problems, arranged by curriculum topic, in our short problems collection.", "What is the arithmetic mean of 25.2, 13.5, 18.5 and 34.8? (a) $13.5$ (b) $23$ (c) $18$ (d) $34.8$ 2 Answers Aug 22, 2016 b) 23 Explanation: Mean $= \\frac{25.2 + 13.5 + 18.5 + 34.8}{4}$ $= \\frac{92}{4}$ $= 23$ Aug 22, 2016 The answer is (b) $23$. Explanation: The arithmetic mean is sum of all the data points or numbers divided by their number. Here we have four data points / numbers which are $25.2$, $13.5$, $18.5$ and $34.8$. Their sum is $25.2 + 13.5 + 18.5 + 34.8 = 92.0$ and their arithmetic mean is $\\frac{92.0}{4} = 23$. Hence, answer is (b) $23$.", "For each set, find the sum of the values in the set, and divide by the number of values: a) 261, 286, 257, 284, 258, 281, 258, 285, 267, 275, 258, 284, 260, 285, 258, 284 The sum of the values is 4,341. There are 16 values. \\begin{align*}\\frac{4,341}{16}=271.31\\end{align*} b) 43, 44, 45, 45, 46, 46, 47, 47, 47, 48, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 50, 50, 50, 51 The sum of the values is 1,198. There are 25 values. \\begin{align*}\\frac{1,198}{25}=47.92\\end{align*} c) 14, 28, 42, 56, 23, 67, 12, 45, 93, 52 The sum of the values is 432. There are 10 values. \\begin{align*}\\frac{432}{10}=43.2\\end{align*} #### Finding the Mean Using a Chart Using the data from the chart, find the mean number of prank calls per five-year interval, and the mean number of reported prank calls per year. First, find the sum of the number of reported calls: \\begin{align*}188+149+161+172+154+172=996 \\ calls\\end{align*} The chart indicates that there are six five-year intervals with recorded values. So, dividing the sum by the number of values, we get: \\begin{align*}\\frac{996}{6}=166 \\ calls \\ per \\ five \\ year \\ interval\\end{align*} Note that the value “166” is not the actual count of any of the recorded intervals. The arithmetic mean obviously does not need to be a member of the", "### Find the integer $x$ that satisfies the equation $x^5 - 101 x^3 - 999 x^{ 2 } + 100900 = 0$. $10$ 1. We need to find the integer value of $x$ that satisfies the equation $x^5 - 101 x^3 - 999 x^{ 2 } + 100900 = 0.$ 2. \\begin{align} & x^5 - 101 x^3 - 999 x^{ 2 } + 100900 = 0 \\\\ \\implies & x^5 - 101 x^3 - 999 x^{ 2 } + 100899 + 1 = 0 \\\\ \\implies & x^3(x^{ 2 } - 101) - 999(x^{ 2 } - 101) + 1 = 0 \\\\ \\implies & (x^{ 2 } - 101)(x^3 - 999) + 1 = 0 && \\ldots (1) \\\\ \\end{align} 3. We observe that the only integer that satisfies $(1)$ is $10.$ Hence, the value of $x$ is $10.$", "maybe some results will use less than four fours. But the translation from a compact equation to a \"four-fours\" equation is immediate. Let's consider the following list of two-four numbers (I purposely omit the ones that can be obtained with a signel four): \\begin{align} 0 &= 4 - 4 \\\\ 1 &= \\frac{4}{4} \\\\ 5 &= 2 + 3 \\\\ 7 &= 4 + 3 \\\\ 9 &= 3 + 6 \\\\ 11 &= 21 - 10 \\\\ 12 &= 10 + 2 \\\\ 13 &= 10 + 3 \\\\ 14 &= 10 + 4 \\\\ 15 &= 21 - 6 \\\\ 16 &= 4 \\cdot 4 \\\\ 17 &= 21 - 4 \\\\ 18 &= 21 - 3 \\\\ 19 &= 21 - 2 \\\\ 20 &= 24 - 4 \\\\ 22 &= 24 - 2 \\\\ 23 &= 21 + 2 \\\\ 25 &= 21 + 4 \\\\ 26 &= 24 + 2 \\\\ 27 &= 24 + 3 \\\\ 28 &= 24 + 4 \\end{align} Up to this point we can get any integer below 28 with only two fours. My strategy will be obtaining numbers by combining the \"high part\" and \"low part\". So, given any two-four \"high number\" $n$ we can generate all integers between $n - 28$ and $n + 28$. And", "1) + 1 \\\\ &= \\text{7,75} \\\\ \\therefore Q_3 &= \\frac{91 + 91}{2} \\\\ &= 91 \\end{align*} The five-number summary is: $$\\text{78}; \\text{84}; \\text{87,5}; \\text{91}; \\text{105}$$. Using this we can draw the box-and-whisker plot. The following is a list of data: $$3; 8; 8; 5; 9; 1; 4; x$$ In each separate case, determine the value of $$x$$ if the: range = $$\\text{16}$$ The set is: $$\\{1; 3 ; 4 ; 5 ; 8 ; 8; 9; x\\}$$. We have ordered all the numbers and then added $$x$$ on the end as we do not know the value of $$x$$. We are told that the range must be equal to 16. If $$x < 9$$ the range would be $$9 - 1 = 8$$. Therefore $$x > 9$$ and so $$x$$ must be the maximum value. \\begin{align*} \\text{range} &= \\text{maximum} - \\text{minimum} \\\\ 16 &= x - 1 \\\\ \\therefore x &= 17 \\end{align*} mode = $$\\text{8}$$ $$\\text{8}$$ is already the mode if we exclude $$x$$, to maintain this $$x$$ is any integer with $$x \\neq \\{1; 3; 4; 5; 9\\}$$. median = $$\\text{6}$$ First consider the set without $$x$$: $$\\{1; 3 ; 4 ; 5 ; 8 ; 8; 9\\}$$. The median in this set is 5 (there is an odd number of values in the set", "2. What is considered an unlucky number? 3. What is a score? 4. How many yards long is a cricket pitch? 5. What number is a quarter century? The answers to the above questions in order are: $$12, 13, 20, 22$$ and $$25.$$ All these five numbers are $$2$$-digit numbers. ## How is it important? ### The Significance (or otherwise) of Zero in $$2$$-digit numbers Since we already know that a $$2$$-digit number can be formed by combinations of two of the $$10$$ digits from $$0$$ to $$10.$$ But what about those numbers which have $$0$$ in the ten’s place value? Let us consider the number that would be formed by zero in ten’s place and five in unit’s place. The number is \\begin{align} ( 0 \\times 10 ) + ( 5 \\times 1 ) = 0 + 5 = 5 \\end{align} So it actually becomes a single digit number. But that is not the case when the zero is in the unit’s place. Let us try exchanging the numbers this time, and take five in ten’s place and zero in unit’s place. The number is \\begin{align} ( 5 \\times 10 ) + ( 0 \\times 1 ) = 50 + 0 = 50 \\end{align} So when it comes to zero, there are nine options which have zero", "= \\pm 12}\\end{align} If larger number is $$– 10,\\,$$ then square of smaller number $$= 8 \\times ( - 10) = - 80$$ Square of any number cannot be negative. $$\\therefore x = - 10$$ is not applicable. The numbers are $$18, \\,12$$ (or) $$18, -12.$$ 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", "non-negative integers of the equation \\begin{align*} y_1 + y_2 + 2 + y_3 + 2 + y_4 + 2 + y_5 + 1 & = 10\\\\ y_1 + y_2 + y_3 + y_4 + y_5 & = 3 \\end{align*} which is the number of ways in which four addition signs can be placed in a list of three $1$'s, which is $$\\binom{3 + 4}{4} = \\binom{7}{4} = 35$$ Ok I don't think this is the conventional way to do it but here goes anyway feel free to critique my answer... 1. You know that the last digit is 1, thus the first 13 digits have 4 2's and 9 1's. There has to be at least 2 1's between each pair of 2's, so the configuration requires the following groups of 1's as a minimum - 0,2,2,2,0. (5 gaps because there are 4 twos - 3 gaps between the twos, and 2 gaps between the outermost twos and the outside. Take away the 6 1's that are required to be in between the 2's, and you are left with 3 1's that can be placed in each of the five gaps. How many ways can you place the three ones into five gaps? The answer is 7C4 = 35", "$3^3$ ☐ $\\sqrt[3]{216}$ 6. $2^4$ ☐ $3^4$ 7. $2^3$ ☐ $3^2$ 8. $\\sqrt[3]{1}$ ☐ $\\sqrt{1}$ 9. $9$ ☐ $3^3$ 10. $100$ ☐ $15^2$ 3. Which is bigger, $1^ {100}$ or $100^1$ ? Explain. 4. What is the biggest number you can make with the symbols 4 and 2? 5. Two whole numbers that follow on each other, like 4 and 5, are called consecutive numbers. Is the difference between the squares of two consecutive whole numbers always an odd number? ### Be smart when doing calculations Our knowledge of squares can help us to do some calculations much quicker. Suppose you want to calculate 11 \\times 12. $11^2$ has a value of 121. We know that $11 \\times 11 = 121$. $11 \\times 12$ means that there are 12 elevens in total. So \\begin{align} 11 \\times 12 &= 11 \\times 11 + 11 \\\\ &= 121 + 11 \\\\ &= 132 \\end{align} Suppose you want to calculate $11 \\times 17$. $11 \\times 17 = 17$ elevens in total = 11 elevens + 6 elevens Well, we know that $11 \\times 11 = 121$ So \\begin{align} 11 \\times 17 &= 11 \\times 11 + 6 \\times 11 \\\\ &= 121 + 66 \\\\ &= 187 \\end{align} Now do the following calculations in your exercise book, using your knowledge of", "Concept. Integers the set of whole numbers and their opposites Negative Numbers numbers that are less than zero Positive Numbers numbers that are greater than zero Zero is a part of the set of integers, but is neither positive or negative ### Guided Practice Here is one for you to try on your own. Compare \\begin{align*}-\\frac{2}{5}\\end{align*} and \\begin{align*}-\\frac{6}{7}\\end{align*} Answer Remember, the closer a value is to zero, the greater the value. Remember this with integers? \\begin{align*}-4 < -1\\end{align*} This is a true statement. We must keep this in mind when working with negative fractions too. Negative six - sevenths is a larger fraction. Because it is negative, it is farther away from zero. It is the smaller value in this case. \\begin{align*}-\\frac{2}{5}\\end{align*} > \\begin{align*}-\\frac{6}{7}\\end{align*} This is the answer. ### Practice Directions: Compare each pair of values using <, > or =. 1. -.18 ____ -.27 2. -23 ____ -.98 3. -9 ____ -11 4. -18 ____ -29 5. -67 ____ -89 6. \\begin{align*}-\\frac{1}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{4}{5}\\end{align*} 7. \\begin{align*}-\\frac{3}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{1}{3}\\end{align*} 8. \\begin{align*}-\\frac{5}{10} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{1}{2}\\end{align*} 9. \\begin{align*}-\\frac{3}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -.75\\end{align*} 10. \\begin{align*}-\\frac{1}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -.25\\end{align*} 11. \\begin{align*}-.25 \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{3}{4}\\end{align*} 12. \\begin{align*}-\\frac{18}{20} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{1}{2}\\end{align*} Directions Write the following integers in order from least to greatest. 13. -4, -12, -19, -8, 0, -2, -1 14. 5, 7, 23, 8, -9, -11 15. \\begin{align*}\\frac{-1}{2}, \\frac{-1}{4}," ]

[ "= circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. ## Solution 3 WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$. We claim the angle between $AB$ and $\\omega$ is fixed as $\\omega$ varies. Proof: Perform an inversion at $A$, sending $\\omega_1$ and $\\omega_2$ to two lines $\\ell_1$ and $\\ell_2$ intersecting at $B'$. Then $\\omega$ is sent to a circle tangent to lines $\\ell_1$", "= foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio $PY : YQ", "\\boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired.", "we can solve to get $d = \\boxed{672}$.$[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. Solution 3 (Inversion) WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$.", "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", "Y pair P = foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio", "# Problem #2055 2055 The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$. $[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); [/asy]$ $\\mathrm{(A)}\\ \\frac{4\\sqrt{5}}{3} \\qquad\\mathrm{(B)}\\ \\frac{5\\sqrt{5}}{3} \\qquad\\mathrm{(C)}\\ \\frac{12\\sqrt{5}}{7} \\qquad\\mathrm{(D)}\\ 2\\sqrt{5} \\qquad\\mathrm{(E)}\\ \\frac{5\\sqrt{65}}{9}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "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", "from Y pair P = foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the", "OPR = \\angle PRQ + \\angle PQR = 90^\\circ + \\angle PQR$, which must equal $\\angle OR'P'$. Therefore, $\\angle Q'R'P' = \\angle OR'P' - \\angle OR'Q' = 90^\\circ + \\angle PQR - \\angle OQR$. But $\\angle OQR = \\angle PQR$. Therefore, $\\angle Q'R'P' = 90^\\circ$. As this holds for any $R$, all points on circle $C$ will invert to a point on a circle with diameter $\\overline{Q'P'}$. ## 3. General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle $[asy] unitsize(7); int radiusInverse = 12; int k = radiusInverse * radiusInverse; int radiusCircle = 4; int circleCenterLocation = 7; path inverse = Circle((0, 0), radiusInverse); draw(inverse, dashed); label(\"\\Omega\", inverse, NW); pair O = (0, 0); dot(O); label(\"O\", O, NW); pair C = (circleCenterLocation, 0); dot(C); label(\"C_1\", C, S); path circle1 = Circle(C, radiusCircle); draw(circle1); int p = circleCenterLocation - radiusCircle; int q = circleCenterLocation + radiusCircle; pair P = (p, 0); pair Q = (q, 0); pair A = (k / p, 0); pair B = (k / q, 0); pair D = (((k / p) + (k / q)) / 2, 0); dot(P); dot(Q); dot(A); dot(B); dot(D); label(\"C_2\", D, S); label(\"P\", P, SE); label(\"Q\", Q, SW); label(\"Q'\", B, SE); label(\"P'\", A, SE); draw(O--A); path circle2 = Circle(D, ((k / p)", "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", "of polygon $ABCDEF$, in square units, is $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 46 \\qquad \\text{(D)}\\ 66 \\qquad \\text{(E)}\\ 74$ $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ ## Problem 5 $[asy] unitsize(13); draw((0,0)--(20,0)); draw((0,0)--(0,15)); draw((0,3)--(-1,3)); draw((0,6)--(-1,6)); draw((0,9)--(-1,9)); draw((0,12)--(-1,12)); draw((0,15)--(-1,15)); fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black); fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black); fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black); fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black); fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black); label(\"A\",(2.5,-.5),S); label(\"B\",(4.5,-.5),S); label(\"C\",(6.5,-.5),S); label(\"D\",(8.5,-.5),S); label(\"F\",(10.5,-.5),S); label(\"Grade\",(15,-.5),S); label(\"1\",(-1,3),W); label(\"2\",(-1,6),W); label(\"3\",(-1,9),W); label(\"4\",(-1,12),W); label(\"5\",(-1,15),W); [/asy]$ The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory? $\\text{(A)}\\ \\frac{1}{2} \\qquad \\text{(B)}\\ \\frac{2}{3} \\qquad \\text{(C)}\\ \\frac{3}{4} \\qquad \\text{(D)}\\ \\frac{4}{5} \\qquad \\text{(E)}\\ \\frac{9}{10}$ ## Problem 6 A stack of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high? $\\text{(A)}\\ 250 \\qquad \\text{(B)}\\ 550 \\qquad \\text{(C)}\\ 667 \\qquad \\text{(D)}\\ 750 \\qquad \\text{(E)}\\ 1250$ ## Problem 7 A \"stair-step\" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\\text{th}$ row is $[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]$ $\\text{(A)}\\ 34 \\qquad \\text{(B)}\\ 35 \\qquad", "0&&\\text{ label: }1\\\\ x_2&\\leq 1/3 - 3 x_1&&\\text{ label: }2\\\\ x_2&\\leq 1/9 - 2/9x_1&&\\text{ label: }3 \\end{align*}\\end{split} Now let us sketch these: x1s = [0, 1] first_line = [1 / 3 - 3 * x for x in x1s] second_line = [1 / 9 - 2 / 9 * x for x in x1s] plt.figure() plt.plot(first_line, label=\"$1/3-3x_1$ (label: 2)\") plt.plot(second_line, label=\"$1/9-2/9x_1/6$ (label: 3)\") plt.legend(); This gives: We see that there our polytope has 4 vertices: \\begin{split}\\begin{align*} (0, 0)&\\text{ labels }\\{0, 1\\}\\\\ (0, 1/9)&\\text{ labels }\\{0, 3\\}\\\\ (1/9, 0)&\\text{ labels }\\{1, 2\\}\\\\ (2/25, 7/75)&\\text{ labels }\\{2, 3\\}\\\\ \\end{align*}\\end{split} Here is some sympy code to verify the intersection of both boundaries: >>> import sympy as sym >>> x = sym.Symbol(\"x\") >>> sym.solveset(sym.S(1) / 3-3 * x - sym.S(1) / 9 + sym.S(2) / 9 * x, x) {2/25} We now carry out the Lemke-Howson algorithm. • Start at $$((0, 0), (0, 0))$$, have labels $$\\{0, 1, 2, 3\\}$$. Drop 3 in $$\\mathcal{Q}$$: move to $$((0, 0), (0, 1/8))$$ and pick up 0. • At $$((0, 0), (0, 1/8))$$, have labels $$\\{0, 1, 2\\}$$. Drop 0 in $$\\mathcal{P}$$: move to $$((1 / 9, 0), (0, 1/8))$$ and pick up 2. • At $$((1/9, 0), (0, 1/8))$$, have labels $$\\{0, 1, 2\\}$$. Drop 2 in $$\\mathcal{Q}$$: move to $$((1 / 9, 0),", "# Carnot's Theorem \"\" Carnot's Theorem\"\" states that in a triangle $ABC$, the signed sum of perpendicular distances from the circumcenter $O$ to the sides (i.e., signed lengths of the pedal lines from $O$) is: $OO_A+OO_B+OO_C=R+r$ $[asy] pair a,b,c,O,i,d,f,g; a=(0,0); b=(4,0); c=(1,3); O=circumcenter(a,b,c); i=incenter(a,b,c); draw(a--b--c--cycle); draw(circumcircle(a,b,c)); draw(incircle(a,b,c)); dot(i); dot(O); label(\"A\",a,W); label(\"B\",b,E); label(\"C\",c,N); label(\"I\",i,N); label(\"O\",O,N); d=foot(O,b,c); dot(d); draw(O--d); label(\"O_A\",d,N); draw(rightanglemark(O,d,b)); f=foot(O,a,b); dot(f); draw(O--f); draw(rightanglemark(O,f,a)); label(\"O_C\",f,S); g=foot(O,c,a); dot(g); draw(O--g); draw(rightanglemark(O,g,a)); label(\"O_B\",g,W); [/asy]$ where r is the inradius and R is the circumradius. The sign of the distance is chosen to be negative iff the entire segment OO_i lies outside the triangle. Explicitly, $OO_A+OO_B+OO_C=\\frac{abc(|\\cos{A}|+|\\cos{B}|+|\\cos{C}|)}{4|\\Delta|}$ where $\\Delta$ is the area of triangle $\\Delta ABC$. Weisstein, Eric W. \"Carnot's Theorem.\" From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html # Below is not Carnot's Theorem Carnot's Theorem states that in a triangle $ABC$ with $A_1\\in BC$, $B_1\\in AC$, and $C_1\\in AB$, perpendiculars to the sides $BC$, $AC$, and $AB$ at $A_1$, $B_1$, and $C_1$ are concurrent if and only if $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$. #### Proof Only if: Assume that the given perpendiculars are concurrent at $M$. Then, from the Pythagorean Theorem, $A_1B^2=BM^2-MA_1^2$, $C_1A^2=AM^2-MC_1^2$, $B_1C^2=CM^2-MB_1^2$, $A_1C^2=MC^2-MA_1^2$, $C_1B^2=MB^2-MC_1^2$, and $B_1A^2=AM^2-MB_1^2$. Substituting each and every one of these in and simplifying gives the desired result. If: Consider the intersection of the perpendiculars from $A_1$ and $B_1$. Call this intersection point $N$,", "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$", "= circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired. ## Solution 4 (Quick) Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot(\"A\",A,dir(A-O2/2)); dot(\"B\",B,dir(B-O2/2)); dot(\"O_2\",O2,right+up); dot(\"O_1\",O1,left+up); dot(\"O\",O,dir(O-O2)); dot(\"D\",D,dir(170)); dot(\"E\",E,dir(E-O1)); dot(\"X\",X,dir(X-D)); dot(\"Y\",Y,dir(Y-D)); label(\"R\",O--E,right+up,p); label(\"R\",O--D,left+down,p); label(\"2R\",(X+O)/2-(150,0),down,p); label(\"961\",O1--D,2*(left+down),p); label(\"625\",O2--E,2*(right+up),p); MA(\"\",E,D,O1,100,fuchsia+linewidth(1)); MA(\"\",X,D,O,100,fuchsia+linewidth(1)); MA(\"\",Y,E,O2,100,orange+linewidth(1)); MA(\"\",D,E,O,100,orange+linewidth(1)); [/asy]$", "$W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label(\"P\",(0,sqrt(3)/2),N); label(\"Z\",(1/6,sqrt(3)/3),NE); label(\"Y\",(1/3,sqrt(3)/6),NE); label(\"R\",(1/2,0),E); label(\"X\",(1/6,0),S); label(\"W\",(-1/6,0),S); label(\"Q\",(-1/2,0),W); label(\"V\",(-1/3,sqrt(3)/6),NW); label(\"U\",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$ ## Problem T2 Evaluate $\\begin{eqnarray*} && 1 \\left(\\dfrac{1}{1}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right) \\\\ &+& 3\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&5\\left(\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&7\\left(\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&9\\left(\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+11\\left(\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&13\\left(\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+15\\left(\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&17\\left(\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+19\\left(\\dfrac{1}{10}\\right)\\end{eqnarray*}$ ## Problem T3 To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed? ## Problem T4 In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey? ## Problem T5 During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued: 'A': It was 'B' or 'C' 'B': Neither 'E'", "(2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0, 3.6), S); [/asy]$ We know that $AU = UM$, $AV = VC$, so $UV = \\frac{1}{2} MC$. As $\\angle UPM = \\angle VPC = 90$, we can see that $\\triangle UPM \\cong \\triangle VPC$ and $\\triangle UVP \\sim \\triangle MPC$ with a side ratio of $1 : 2$. So $UP = VP = 4$, $MP = PC = 8$. With that, we can see that $S\\triangle UPM = 16$, and the area of trapezoid $MUVC$ is 72. As said in solution 1, $S\\triangle AMC = 72 / \\frac{3}{4} = \\boxed{\\textbf{(C) } 96}$. ## Solution 5 (Medians, Height, Pythagorean Theorem, Lots of symmetry, A little extra work) [asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); draw((-2, 6), (2, 6)) label(\"K\", (0, 6), NE); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy] It is well known that medians divide each other into segments of $2:1$ ratio. From this, we have $PC=MP=8$ and $UP=UV=4$. From right triangle $\\triangle{MPC}$, $MC^2=MP^2+MC^2=8^2+8^2=128$, which implies $MC=\\sqrt{128}=8\\sqrt{2}$. Then the area of $\\triangle{AMC}$ is $\\dfrac{8\\sqrt{2} \\cdot AB}{2}$, so our goal is to find $AB$. Note that $AB=AP+BP$. Since $\\triangle{AMC}$ is isosceles, by symmetry $MB=MC$, because $AB$ is the altitude. Knowing this, $BP$", "# Control distance of label from point in asymptote I'm making a diagram for a solution on the AoPS wiki and some of the labels are a little too close for comfort. Is there a way I can move a label farther from what I'm labeling? I have the following code so far: import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, W); label(\"$$N$$\", inter2, E); and the labels for M and N are a little too close to the lines I've drawn. Thanks! On a side note, is having so many pictures defined good form? I just started asymptote and I feel like it's not very elegant. From the docs (asymptote.pdf, 4.4 label): ... If align is NoAlign, the label will be centered at user coordinate position; otherwise it will be aligned in the", "x1)² + (y2 – y1)²}$$ = $$\\sqrt{(4 – 3)² + (3 – 3)²}$$ = 1 Hence, The perimeter of the given polygon = QR + RS + ST + TQ = 6 + 1 + 6 + 1 = 14 The area of the given polygon = Length × Width So, The area of the polygon = 6 (1) = 6 Hence, from the above, We can conclude that The perimeter of the given polygon is: 14 The area of the given polygon is: 6 Question 5. Plot the points W(- 1, 1), X(5, 1), Y(5, – 2), and Z(- 1, – 2) in a coordinate plane. What type of polygon do the points form? Your friend claims that you could use this figure to represent a basketball court with an area of 4050 square feet and a perimeter of 270 feet. Do you support your friend’s claim? Explain. The given points are: W (-1, 1), X (5, 1), Y (5, -2), and Z (-1, -2) Now, $$\\overline{W X}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(1 – 1)² + (5 + 1)²}$$ = 6 $$\\overline{X Y}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(1 + 2)² + (5 – 5)²}$$ = 3 $$\\overline{Y Z}$$ = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ =" ]

[ "we can solve to get $d = \\boxed{672}$.$[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. Solution 3 (Inversion) WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$.", "\\boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired.", "= foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio $PY : YQ", "\\end{asy} \\end{document} Update For \"auto ticks\", I add the grid and subgrid using Step=1, step=.2 in LeftTicks, RightTicks of the xaxis and yaxis command. The module graph must be loaded. Compiling time seems a bit slower. // http://asymptote.ualberta.ca/ unitsize(1cm); import graph; import contour; real f(real x, real y){return x^2-x*y;} pair A=(-3,-3), B=(3,3); real[] c1={1}, c2={-1}; draw(A--B^^(0,A.y)--(0,B.y),purple); // 2 asymptote straight lines draw(contour(f,A,B,c1),cyan+linewidth(1pt)); draw(contour(f,A,B,c2),magenta+linewidth(1pt)); pen thin=gray+linewidth(.2pt); pen verythin=lightgray+linewidth(.2pt); xaxis(\"$x$\",BottomTop,LeftTicks(begin=false,end=false,Step=1,step=.2,extend=true, ptick=verythin,pTick=thin)); yaxis(\"$y$\",LeftRight,RightTicks(begin=false,end=false,Step=1,step=.2,extend=true,ptick=verythin,pTick=thin)); label(\"The graph of $x^2-xy=C$\",truepoint(N)+(0,.5)); • do you need to make the tick labels manually? Jan 10 at 19:50 • @FacebFaceb I like to make tick manually, to get exactly what I like to tick and label. Of course, auto ticks and labels can be made using import graph; see here asymptote.sourceforge.io/FAQ/section6.html if you are interested. (I though you are still waiting a pgfplots solution) Jan 10 at 19:58 • yes I am waiting for a solution with contour gnuplot and pgfplots, but I was just curious about this package Jan 10 at 20:21 • @FacebFaceb this may be helpful for your pgfplots figure (grids) latexdraw.com/linear-regression-in-latex-using-tikz Jan 20 at 15:36", "= circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. ## Solution 3 WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$. We claim the angle between $AB$ and $\\omega$ is fixed as $\\omega$ varies. Proof: Perform an inversion at $A$, sending $\\omega_1$ and $\\omega_2$ to two lines $\\ell_1$ and $\\ell_2$ intersecting at $B'$. Then $\\omega$ is sent to a circle tangent to lines $\\ell_1$", "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", "from Y pair P = foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the", "Y pair P = foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio", "- 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \\ge 0 \\Longleftrightarrow m \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. ### Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = \\omega T + B\\omega\\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. ### Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From", "= circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired. ## Solution 4 (Quick) Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot(\"A\",A,dir(A-O2/2)); dot(\"B\",B,dir(B-O2/2)); dot(\"O_2\",O2,right+up); dot(\"O_1\",O1,left+up); dot(\"O\",O,dir(O-O2)); dot(\"D\",D,dir(170)); dot(\"E\",E,dir(E-O1)); dot(\"X\",X,dir(X-D)); dot(\"Y\",Y,dir(Y-D)); label(\"R\",O--E,right+up,p); label(\"R\",O--D,left+down,p); label(\"2R\",(X+O)/2-(150,0),down,p); label(\"961\",O1--D,2*(left+down),p); label(\"625\",O2--E,2*(right+up),p); MA(\"\",E,D,O1,100,fuchsia+linewidth(1)); MA(\"\",X,D,O,100,fuchsia+linewidth(1)); MA(\"\",Y,E,O2,100,orange+linewidth(1)); MA(\"\",D,E,O,100,orange+linewidth(1)); [/asy]$", "$MN$ 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 $B$", "by $\\sqrt{(a-c)^2 + (b+d)^2}$.[4] $[asy]defaultpen(linewidth(1)); unitsize(15); pen dotted = linetype(\"2 4\"), sm = fontsize(10); real r = 2; pair A = r*(0,0), B = (r*18/5,A.y), C = r*(16/5,12/5), D = (r*9/5,C.y); pair refl(pair a, pair b = (C+B)/2) { return a+2*(b-a); } void makeshiftarrow(pair a, real dir, real arrowlength = r){ /* Arrow option resizes */ fill(a--a+arrowlength*expi(dir+pi/8)--a+arrowlength*expi(dir-pi/8)--cycle); } draw(A--B--C--D--cycle); draw(A--C^^B--D); draw(refl(A)--C--B--refl(D)--cycle ^^ C--refl(D), dotted); draw(rightanglemark(A,C,refl(D),15)^^rightanglemark(A,IP(A--C,B--D),B,15), linewidth(0.7)); label(\"A\",A,S,sm);label(\"B\",B,S,sm);label(\"C\",C,N,sm);label(\"D\",D,N,sm);label(\"A'\",refl(A),N,sm);label(\"D'\",refl(D),S,sm); /* arrow */ draw(arc((B+C)/2,C.y,240,300),linewidth(0.7)); makeshiftarrow((B+C)/2+C.y*expi(pi*300/180),210*pi/180,r/4); [/asy]$ In trapezoid $ABCD$ with $\\overline{AB} \\parallel \\overline{CD}$, then $\\overline{AC} \\perp \\overline{BD} \\Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2$. $[asy] // feel free to change these four points! pair A = (0,0), B = (3, -2), C = (5,1), D = (1,3); // // Rest of code // size(200); defaultpen(linewidth(0.9)); pen lightgreen = rgb(0.6,1,0.6), lightred = rgb(1,0.6,0.6), smdash = linewidth(0.7)+linetype(\"2 2\"); pair E = IP(A--C,B--D), AB = (A+B)/2, BC = (B+C)/2, CD = (C+D)/2, DA = (D+A)/2, ABE = 2*AB-E, BCE = 2*BC-E, CDE = 2*CD-E, DAE = 2*DA-E; path midpts = AB--BC--CD--DA--cycle; filldraw(shift(-E)*scale(2)*midpts,lightgreen); filldraw(midpts,lightred); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw((A+ABE)/2--(C+BCE)/2,smdash); draw((B+ABE)/2--(D+DAE)/2,smdash); draw((B+BCE)/2--(D+CDE)/2,smdash); draw((A+DAE)/2--(C+CDE)/2,smdash); dot(A); dot(B); dot(C); dot(D); label(\"A\",A,W);label(\"B\",B,S);label(\"C\",C,E);label(\"D\",D,N); [/asy]$ Varignon's theorem: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area $1/2$ of the", "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", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "that is where we get our x value: $(19, d)$ $(19, -d)$ $(17, -3d)$ Now, we can plug one of the first two value in as well as the last one to get the following equations: $$19^2 + d^2 = r^2$$ $$17^2 + (3d)^2 = r^2$$ Subtracting these two equations, we get $19^2 - 17^2 = 8d^2$ - therefore, we get $72 = 8d^2 \\rightarrow d^2 = 9 \\rightarrow d = 3$. We want to find $2d = 6$ because that's the distance between two chords. So, our answer is $\\boxed{B}$. ~Tony_Li2007 ## Solution 3 (Stewart's Theorem) $[asy] real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label(\"O\", O, N); label(\"C\", C, SW); label(\"D\", D, SE); label(\"E\", E, SW); label(\"F\", F, SE); label(\"P\", P, SW); label(\"Q\", Q, S); [/asy]$ If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\\triangle OCD$ with cevian $\\overleftrightarrow{OP}$ gives $$19\\cdot 38\\cdot 19 + \\tfrac{1}{2}d\\cdot 38\\cdot\\tfrac{1}{2}d=19r^{2}+19r^{2}.$$ This simplifies to $13718+\\tfrac{19}{2}d^{2}=38r^{2}$. Similarly, another round of Stewart's Theorem", "draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(0,10),C=(10,0),P=(0,0),B=(13.660,13.660); dot(A); label(\"A\",A,W); dot(C); label(\"C\",C,S); dot(P); label(\"P_0\",P,SW); dot(B); label(\"B\",B,NE); draw(A--C--P--A); draw(A--B--C); draw(B--P,dotted); [/asy]$ By repeatedly applying the Midpoint Formula, we can determine the coordinates of $P_1$, $P_2$, $P_3$, and so on. We can also use the Distance Formula to calculate the distance of $P_0P_1$, $P_0P_2$, and so on. The values are shown in the below table. $n$ Coordinates of $P_n$ $P_0 P_n$ 1 $(0,2a)$ $2a$ 2 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ 3 $(a-a\\sqrt{3},a-a\\sqrt{3})$ $a\\sqrt{6} - a\\sqrt{2}$ 4 $(-a+a\\sqrt{3},a+a\\sqrt{3})$ $2a\\sqrt{2}$ 5 $(2a,0)$ $2a$ 6 $(0,0)$ $0$ 7 $(0,2a)$ $2a$ 8 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ Note that the coordinates of $P_n$ as well as the distance $P_0 P_n$ cycle after $n = 6$. Thus, $P_0P_n = \\frac{\\sqrt{6} - \\sqrt{2}}{2} \\cdot P_0P_1$ if $n \\equiv 3 \\pmod{6}$, $P_0P_n = 0$ if $n \\equiv 0 \\pmod{6}$, $P_0P_n = P_0P_1$ if $n \\equiv 1, 5 \\pmod{6}$, and $P_0P_n = \\sqrt{2} \\cdot P_0P_1$ if $n \\equiv 2, 4 \\pmod{6}$.", "compile online. Plain things are suitable for beginners. Here is a link on overleaf. If you want to compile on desktops, make sure having Asymptote installed. // copy http://asymptote.ualberta.ca/ and click Run usepackage(\"amsmath\"); // for \\dfrac import graph; import math; // for grid unitsize(1cm); size(6cm); real f(real x){return 1/x;} int a=4; real s=.1; path p=graph(f,s,a); path q=graph(f,-a,-s); pen penp=magenta, penq=deepcyan; fill((0,0)--(0,a+1)--p--(a+1,0)--cycle,penp+white+opacity(.5)); fill((0,0)--(-a-1,0)--q--(0,-a-1)--cycle,penq+opacity(.5)); draw(p,penp+1pt); draw(q,penq+1pt); clip(box((a,a),(-a,-a))); draw(Label(\"$x$\",EndPoint,align=SW),(-a-1,0)--(a+1,0),Arrow(TeXHead)); draw(Label(\"$y$\",EndPoint,align=SW),(0,-a-.5)--(0,a+1),Arrow(TeXHead)); label(Label(\"p.v. $\\displaystyle\\int_{-a}^a\\dfrac{1}{x}\\,dx=0$\",FillDraw(yellow,black)),(-2,2)); label(Label(\"$a$\",Fill(white)),(a,0),2S); label(Label(\"$-a$\",Fill(white)),(-a,0),2N);", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The", "*/ 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", "# Difference between revisions of \"2008 AIME I Problems/Problem 14\" ## Problem Let $\\overline{AB}$ be a diameter of circle $\\omega$. Extend $\\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\\omega$ so that line $CT$ is tangent to $\\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $AB = 18$, and let $m$ denote the maximum possible length of segment $BP$. Find $m^{2}$. ## Solution ### Solution 1 $[asy] size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label(\"$$A$$\",A,NW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NW); label(\"$$O$$\",O,NW); label(\"$$P$$\",P,SE); label(\"$$T$$\",T,SE); label(\"$$9$$\",(O+A)/2,N); label(\"$$9$$\",(O+B)/2,N); label(\"$$x-9$$\",(C+A)/2,N); [/asy]$ Let $x = OC$. Since $OT, AP \\perp TC$, it follows easily that $\\triangle APC \\sim \\triangle OTC$. Thus $\\frac{AP}{OT} = \\frac{CA}{CO} \\Longrightarrow AP = \\frac{9(x-9)}{x}$. By the Law of Cosines on $\\triangle BAP$, \\begin{align*}BP^2 = AB^2 + AP^2 - 2 \\cdot AB \\cdot AP \\cdot \\cos \\angle BAP \\end{align*} where $\\cos \\angle BAP = \\cos (180 - \\angle TOA) = - \\frac{OT}{OC} = - \\frac{9}{x}$, so: \\begin{align*}BP^2 &= 18^2 + \\frac{9^2(x-9)^2}{x^2} + 2(18) \\cdot \\frac{9(x-9)}{x} \\cdot \\frac 9x = 405 + 729\\left(\\frac{2x - 27}{x^2}\\right)\\end{align*} Let $m = \\frac{2x-27}{x^2} \\Longrightarrow mx^2" ]

[ "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", "\\boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired.", "we can solve to get $d = \\boxed{672}$.$[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. Solution 3 (Inversion) WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$.", "= foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio $PY : YQ", "= circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. ## Solution 3 WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$. We claim the angle between $AB$ and $\\omega$ is fixed as $\\omega$ varies. Proof: Perform an inversion at $A$, sending $\\omega_1$ and $\\omega_2$ to two lines $\\ell_1$ and $\\ell_2$ intersecting at $B'$. Then $\\omega$ is sent to a circle tangent to lines $\\ell_1$", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "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", "= circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired. ## Solution 4 (Quick) Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot(\"A\",A,dir(A-O2/2)); dot(\"B\",B,dir(B-O2/2)); dot(\"O_2\",O2,right+up); dot(\"O_1\",O1,left+up); dot(\"O\",O,dir(O-O2)); dot(\"D\",D,dir(170)); dot(\"E\",E,dir(E-O1)); dot(\"X\",X,dir(X-D)); dot(\"Y\",Y,dir(Y-D)); label(\"R\",O--E,right+up,p); label(\"R\",O--D,left+down,p); label(\"2R\",(X+O)/2-(150,0),down,p); label(\"961\",O1--D,2*(left+down),p); label(\"625\",O2--E,2*(right+up),p); MA(\"\",E,D,O1,100,fuchsia+linewidth(1)); MA(\"\",X,D,O,100,fuchsia+linewidth(1)); MA(\"\",Y,E,O2,100,orange+linewidth(1)); MA(\"\",D,E,O,100,orange+linewidth(1)); [/asy]$", "Y pair P = foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio", "from Y pair P = foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the", "# Control distance of label from point in asymptote I'm making a diagram for a solution on the AoPS wiki and some of the labels are a little too close for comfort. Is there a way I can move a label farther from what I'm labeling? I have the following code so far: import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, W); label(\"$$N$$\", inter2, E); and the labels for M and N are a little too close to the lines I've drawn. Thanks! On a side note, is having so many pictures defined good form? I just started asymptote and I feel like it's not very elegant. From the docs (asymptote.pdf, 4.4 label): ... If align is NoAlign, the label will be centered at user coordinate position; otherwise it will be aligned in the", "\\mathrm{(E)}\\qquad \\blacksquare \\end{align*} ## Solution 2 $[asy] import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); add(p); add(q); add(r); add(s); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, 2W); label(\"$$N$$\", inter2, 2E); [/asy]$ We start by finding the length of $AD$ and $BD$ as in solution 1. Using the angle bisector theorem, we see that $AD = \\frac{15}{7}$ and $BD = \\frac{20}{7}$. Using Stewart's Theorem gives us the equation $5d^2 + \\frac{1500}{49} = \\frac{240}{7} + \\frac{180}{7}$, where $d$ is the length of $CD$. Solving gives us $d = \\frac{12\\sqrt{2}}{7}$, so $CD = \\frac{12\\sqrt{2}}{7}$. Call the incenters of triangles $ACD$ and $BCD$ $O_a$ and $O_b$ respectively. Since $O_a$ is an incenter, it lies on the angle bisector of $\\angle ACD$. Similarly, $O_b$ lies on the angle bisector of $\\angle BCD$. Call the point on $CD$ tangent to $O_a$ $M$, and the point tangent to", "by $\\sqrt{(a-c)^2 + (b+d)^2}$.[4] $[asy]defaultpen(linewidth(1)); unitsize(15); pen dotted = linetype(\"2 4\"), sm = fontsize(10); real r = 2; pair A = r*(0,0), B = (r*18/5,A.y), C = r*(16/5,12/5), D = (r*9/5,C.y); pair refl(pair a, pair b = (C+B)/2) { return a+2*(b-a); } void makeshiftarrow(pair a, real dir, real arrowlength = r){ /* Arrow option resizes */ fill(a--a+arrowlength*expi(dir+pi/8)--a+arrowlength*expi(dir-pi/8)--cycle); } draw(A--B--C--D--cycle); draw(A--C^^B--D); draw(refl(A)--C--B--refl(D)--cycle ^^ C--refl(D), dotted); draw(rightanglemark(A,C,refl(D),15)^^rightanglemark(A,IP(A--C,B--D),B,15), linewidth(0.7)); label(\"A\",A,S,sm);label(\"B\",B,S,sm);label(\"C\",C,N,sm);label(\"D\",D,N,sm);label(\"A'\",refl(A),N,sm);label(\"D'\",refl(D),S,sm); /* arrow */ draw(arc((B+C)/2,C.y,240,300),linewidth(0.7)); makeshiftarrow((B+C)/2+C.y*expi(pi*300/180),210*pi/180,r/4); [/asy]$ In trapezoid $ABCD$ with $\\overline{AB} \\parallel \\overline{CD}$, then $\\overline{AC} \\perp \\overline{BD} \\Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2$. $[asy] // feel free to change these four points! pair A = (0,0), B = (3, -2), C = (5,1), D = (1,3); // // Rest of code // size(200); defaultpen(linewidth(0.9)); pen lightgreen = rgb(0.6,1,0.6), lightred = rgb(1,0.6,0.6), smdash = linewidth(0.7)+linetype(\"2 2\"); pair E = IP(A--C,B--D), AB = (A+B)/2, BC = (B+C)/2, CD = (C+D)/2, DA = (D+A)/2, ABE = 2*AB-E, BCE = 2*BC-E, CDE = 2*CD-E, DAE = 2*DA-E; path midpts = AB--BC--CD--DA--cycle; filldraw(shift(-E)*scale(2)*midpts,lightgreen); filldraw(midpts,lightred); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw((A+ABE)/2--(C+BCE)/2,smdash); draw((B+ABE)/2--(D+DAE)/2,smdash); draw((B+BCE)/2--(D+CDE)/2,smdash); draw((A+DAE)/2--(C+CDE)/2,smdash); dot(A); dot(B); dot(C); dot(D); label(\"A\",A,W);label(\"B\",B,S);label(\"C\",C,E);label(\"D\",D,N); [/asy]$ Varignon's theorem: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area $1/2$ of the", "# Problem #2055 2055 The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$. $[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); [/asy]$ $\\mathrm{(A)}\\ \\frac{4\\sqrt{5}}{3} \\qquad\\mathrm{(B)}\\ \\frac{5\\sqrt{5}}{3} \\qquad\\mathrm{(C)}\\ \\frac{12\\sqrt{5}}{7} \\qquad\\mathrm{(D)}\\ 2\\sqrt{5} \\qquad\\mathrm{(E)}\\ \\frac{5\\sqrt{65}}{9}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# 1993 USAMO Problems/Problem 2 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem 2 Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. ## Solution ### Diagram $[asy] import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('E',E,N); pair A,B,C,D; A=(10,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('A',A,E); label('B',B,N); label('C',C,W); label('D',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=reflect(A, D)*E; pair W,X,Y,Z; W=extension(A,D,E,Q); X=extension(A,B,E,T); Y=extension(C,B,E,R); Z=extension(C,D,E,S); draw(W--X--Y--Z--cycle,red); label('X',X,NE); label('Y',Y,NW); label('Z',Z, SW); label('W',W,SE); [/asy]$ ### Work Let $X$, $Y$, $Z$, $W$ be the foot of the altitute from point E of $\\triangle AEB$, $\\triangle BEC$, $\\triangle CED$, $\\triangle DEA$. Note that reflection of $E$ over the 4 lines is $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections are cyclic. $\\angle EWA$ is right angle and so is $\\angle EXA$. Thus, $EXAW$ is cyclic with $EA$ being the diameter of the circumcircle. Follow that, $\\angle EWX\\cong\\angle EAX\\cong \\angle EAB$ because they inscribe the same angle. Similarly $\\angle EWZ\\cong \\angle EDC$, $\\angle EYX\\cong \\angle EBA$, $\\angle EYZ\\cong \\angle ECD$. Futhermore, $m\\angle XYZ+m\\angle XWZ= m\\angle EWX+m\\angle EYX+m\\angle EYZ+m\\angle EWZ=360^\\circ-m\\angle CED-m\\angle AEB=180^\\circ$. Thus, $\\angle XYZ$", "$W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label(\"P\",(0,sqrt(3)/2),N); label(\"Z\",(1/6,sqrt(3)/3),NE); label(\"Y\",(1/3,sqrt(3)/6),NE); label(\"R\",(1/2,0),E); label(\"X\",(1/6,0),S); label(\"W\",(-1/6,0),S); label(\"Q\",(-1/2,0),W); label(\"V\",(-1/3,sqrt(3)/6),NW); label(\"U\",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$ ## Problem T2 Evaluate $\\begin{eqnarray*} && 1 \\left(\\dfrac{1}{1}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right) \\\\ &+& 3\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&5\\left(\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&7\\left(\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&9\\left(\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+11\\left(\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&13\\left(\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+15\\left(\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&17\\left(\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+19\\left(\\dfrac{1}{10}\\right)\\end{eqnarray*}$ ## Problem T3 To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed? ## Problem T4 In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey? ## Problem T5 During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued: 'A': It was 'B' or 'C' 'B': Neither 'E'", "0&&\\text{ label: }1\\\\ x_2&\\leq 1/3 - 3 x_1&&\\text{ label: }2\\\\ x_2&\\leq 1/9 - 2/9x_1&&\\text{ label: }3 \\end{align*}\\end{split} Now let us sketch these: x1s = [0, 1] first_line = [1 / 3 - 3 * x for x in x1s] second_line = [1 / 9 - 2 / 9 * x for x in x1s] plt.figure() plt.plot(first_line, label=\"$1/3-3x_1$ (label: 2)\") plt.plot(second_line, label=\"$1/9-2/9x_1/6$ (label: 3)\") plt.legend(); This gives: We see that there our polytope has 4 vertices: \\begin{split}\\begin{align*} (0, 0)&\\text{ labels }\\{0, 1\\}\\\\ (0, 1/9)&\\text{ labels }\\{0, 3\\}\\\\ (1/9, 0)&\\text{ labels }\\{1, 2\\}\\\\ (2/25, 7/75)&\\text{ labels }\\{2, 3\\}\\\\ \\end{align*}\\end{split} Here is some sympy code to verify the intersection of both boundaries: >>> import sympy as sym >>> x = sym.Symbol(\"x\") >>> sym.solveset(sym.S(1) / 3-3 * x - sym.S(1) / 9 + sym.S(2) / 9 * x, x) {2/25} We now carry out the Lemke-Howson algorithm. • Start at $$((0, 0), (0, 0))$$, have labels $$\\{0, 1, 2, 3\\}$$. Drop 3 in $$\\mathcal{Q}$$: move to $$((0, 0), (0, 1/8))$$ and pick up 0. • At $$((0, 0), (0, 1/8))$$, have labels $$\\{0, 1, 2\\}$$. Drop 0 in $$\\mathcal{P}$$: move to $$((1 / 9, 0), (0, 1/8))$$ and pick up 2. • At $$((1/9, 0), (0, 1/8))$$, have labels $$\\{0, 1, 2\\}$$. Drop 2 in $$\\mathcal{Q}$$: move to $$((1 / 9, 0),", "# 1982 AHSME Problems/Problem 22 ## Problem 22 In a narrow alley of width $w$ a ladder of length $a$ is placed with its foot at point P between the walls. Resting against one wall at $Q$, the distance k above the ground makes a $45^\\circ$ angle with the ground. Resting against the other wall at $R$, a distance h above the ground, the ladder makes a $75^\\circ$ angle with the ground. The width $w$ is equal to $\\text{(A)}a\\qquad \\text{(B)}RQ\\qquad \\text{(C)}k\\qquad \\text{(D)}\\frac{h+k}{2}\\qquad \\text{(E)}h$ ## Solution $[asy] import olympiad; unitsize(40); pair T,P,Q,M,L,R; T=(0,0); P=(2.5,0); Q=(10,7.5); M=(0,7.5); L=(10,0); R=(0,10); draw(R--T--L--Q); draw(P--Q--R--cycle); draw(Q--M); label(\"P\", P, S); dot(P); label(\"Q\", Q, E); dot(Q); label(\"R\", R, W); dot(R); label(\"S\", M, W); dot(M); label(\"T\", T, S); dot(T); label(\"L\", L, S); dot(L); label(\"w\",(5,-1),N); label(\"h\",(-1,5),E); markscalefactor=0.03; draw(anglemark(L,P,Q)); draw(anglemark(R,P,T)); draw(rightanglemark(P,T,M)); draw(rightanglemark(Q,L,P)); [/asy]$ We know that $m\\angle QPL=45^{\\circ}$ and $m\\angle RPT=75^{\\circ}.$ Therefore, $m\\angle QPR=60^{\\circ}.$ $$\\qquad$$ Because the two ladders are the same length, we know that $$RP=PQ=a.$$ Since $\\triangle QPR$ is isosceles with vertex angle $60^{\\circ},$ we can conclude that it must be equilateral.$$\\qquad$$ Now, since $\\triangle PTR$ is a right triangle and $m\\angle TPR=75^{\\circ},$ we can conclude that $m\\angle PRT=15^{\\circ}.$ Because $\\triangle QPR$ is equilateral, we know that $m\\angle QRP = 60^{\\circ}.$ It then follows that $$m\\angle QRS= m\\angle QRP + m\\angle PRT = 60^{\\circ} + 15^{\\circ} =", "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", "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? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$" ]

[ "= $1,920 And by the way,the digital camera is$480,not \\$400! $\\frac{480}{1920}=0.25$, or $25\\%$ So this is wrong because the camera cost 20% or the cost of the laptop, not 20% of the total cost of the camera and laptop. So lets see TKH is correct and you are wrong, unless you asked the wrong question. RonL", "## A camera has a list price of $869.99 before tax. If the sales tax rate is 9.5 % , find the total cost of the camera with sales tax included Question A camera has a list price of$ 869.99 before tax. If the sales tax rate is 9.5 % , find the total cost of the camera with sales tax included. Round your answer to the nearest cent, as necessary in progress 0 6 months 2021-07-18T11:17:16+00:00 1 Answers 2 views 0 1. Answer: Given the peculiar interrogate, the retort to such is identified as approximately ‘$952.64’ USD. Step-by-Step Explanation: To comprehend the evaluation, the apprehension of the sales tax quantity is necessary, 9.5% * 869.99 ~ 82.65 (rounded to the nearest cent). Thus far, evaluate on the behalf of the price, augmented with the sales tax. 869.99 + 82.65 ~ 952.64 (rounded to the nearest cent). Hence, given such configurations, the agglomerate cost of the commodity is equivalent to, approximately,$952.64 USD.", "applying the coupon would give the camera a price of$50. If the camera price retailed at \\$150, the price of the camera...", "each month when saving the same amount? • Price reduction The price of the camera is 654 USD on sale. They reduced its price to 411 USD. Calculate the percentage reduction in the price of the camera. • Magic bag Each time the prince crossed the bridge, the number of tolars in the magic bag doubled. But then the devil always conjured 300 tolars for him. When this happened for the third time, the prince had twice as much as he had in the beginning. How many tolars", "# A Camera shop allows a discount of 10% on the advertised price of a camera. What price must be marked on the camera, that costs him Rs. 600, so that he makes a profit of 20%? Option 1) Rs. 800 Option 2) Rs. 720 Option 3) Rs. 750 Option 4) Rs. 815 Option 5) Rs. 850 SP = 600 $\\times$ 1.20 =720 .9 $\\times$ MP = 720", "the equations would still be true. Quote: (B) It makes more profit during the first and third quarters combined than during the fourth quarter. What about the second quarter? We cannot split the year in this way, since the argument distinguishes between the fourth quarter and the rest three quarters. Quote: (C) Its per-camera retail price is lower, on average, during the fourth quarter than during any one of the first three quarters. May be true: Camco sells many more cameras during the 4th quarter than during the rest of the year. Economies of scale. May not be true: Camco sells the average number of cameras at higher price, thus gaining greater profit. Quote: (D) It makes less profit, on average, for a given dollar amount of sales during the first three quarters combined than during the fourth quarter. Must be true. Key words: \"on average, for a given dollar amount of sales\". Look at the equations. Quote: (E) The per-camera price it pays to wholesalers is higher, on average, during the fourth quarter than during any of the other three quarters. Wrong for the same reason as A. _________________ If I answered your question with this post, use the motivating power of kudos! Kudos [?]: 214 [2], given: 67 Non-Human User Joined: 01 Oct 2013 Posts: 10160", "X cameras to be sold [#permalink] 27 May 2013, 11:12 Similar topics Replies Last post Similar Topics: The number of digital camera sold is twice the number of 1 08 Sep 2004, 05:54 According to a car dealer's report, 1/3 of the cars sold 2 17 Oct 2006, 13:38 1 The price of a model M camera is$209 and the price of a 3 16 Nov 2007, 11:15 1 model A sold 3000 units and accounted for .6 of sales. model 3 23 Dec 2007, 03:22 3 How many of the 60 cars sold last month by a certain dealer 5 14 Dec 2012, 06:34 Display posts from previous: Sort by", "• Subtract the discount from the original price: ${\\displaystyle 154.88-61.95=92.93}$. So, the sale price of the television is $92.93. 2. 2 Find the exact sale price of a camera that is 15% off. The original price is$449.95. • Convert the percentage discount to a decimal by moving the decimal two places to the left: ${\\displaystyle 15\\%=15.0\\%=.15}$. • Multiply the original price by the decimal: ${\\displaystyle 449.95\\times .15=67.49}$. • Subtract the discount from the original price: ${\\displaystyle 449.95-67.49=382.46}$. So, the sale price of the camera is $382.46. 3. 3 Estimate the sale price. A tablet is regularly$199.99. On sale, it is 45% off. • Round the original price to the nearest ten. Since $199.99 is only 1 cent away from$200, you would round up. • Calculate 10% of the rounded price. Moving the decimal one place to the left, you should see that 10% of $200.00 is$20.00. • Determine the number of tens in the percent off. Since ${\\displaystyle 10\\times 4=40}$, you know that there are 4 tens in 45%. • Multiply 10% of the rounded price by the appropriate factor. Since the percentage off is 45%, you would multiply 10% of the rounded price by 4: ${\\displaystyle \\20\\times 4=\\80}$ • Calculate 5% of the rounded price. This is half of 10%, which is $20. So half of$20 is", "#### Explanation: We can express discount amount as $x$ and the original price of the camera as $x + 632$. The equation to determine the price will be: $\\left(x + 632\\right) \\times \\frac{20}{100} = x$ Simplify. $\\left(x + 632\\right) \\times \\frac{1 \\cancel{20}}{5 \\cancel{100}} = x$ Multiply both sides by $5$. $x + 632 = 5 x$ Subtract $x$ from both sides. $632 = 4 x$ Divide both sides by $4$. $158 = x$ Since the discount amount $\\left(x\\right)$ is $158$, the original price $\\left(x + 632\\right)$ will be $\\left(158 + 632\\right) = 790$.", "apart, and also hard to see decimal points and parentheses? The equation is actually .34 + .4 - 44 * (6/528 - 2.4 + .11) which is 101 • Sweet. If that's not right, seems like it's at least a plausible alternate answer. – Dan Russell May 10 '16 at 2:32 • Hint #2 says that Anthony saw everything that was written down, which would include decimal points. – pacoverflow May 10 '16 at 4:08 Turns out the people were actually ordering picture frames They needed 3 4x4 sized frames, 4 4x6 sized frames (which they are getting from 528 photography, by the way) and 2 4x11 sized frames. The cost for these was 4x4 cost \\$5, 4x6 cost \\$10, and 4x11 cost \\$23. Thus 3x5 + 4x10 + 2x23 = 15+40+46=101 • An interesting answer, but your explanation for 528 isn't very satisfying. – pacoverflow May 11 '16 at 14:56 • @pacoverflow Agreed. And there are many solutions to the equation with 3 unknowns. I only provided one with nice round numbers (except for 23, which is my favourite). I don't seriously think this is the answer, but thought I would throw it out there anyway. – Trenin May 11 '16 at 14:58 There are 3 expressions and three digits in the answer. The answer is actually", "\\ \\text{or} \\ \\12\\!\\\\ {\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.7 \\times \\12, \\ \\text{or} \\ \\8.40\\!\\\\ {\\;} \\quad \\ \\8.40 - \\5.60 = \\2.80\\end{align*} \\begin{align*}3. \\quad \\text{ABC's;} \\ \\1.60 \\ \\text{less}\\!\\\\ {\\;} \\quad \\ \\text{ABC's:} \\ 4 \\ \\text{boxes for} \\ \\12 \\ \\text{is} \\ 2 \\ \\text{boxes for} \\ \\6\\!\\\\ {\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times \\6 \\ \\text{or} \\ \\4.80\\!\\\\ {\\;} \\quad \\ \\text{Gem's:} \\ 6 \\ \\text{boxes for} \\ \\24 \\ \\text{is} \\ 2 \\ \\text{boxes for} \\ \\8\\!\\\\ {\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times 8, \\ \\text{or} \\ \\6.40\\!\\\\ {\\;} \\quad \\ \\6.40 - \\4.80 = \\1.60\\end{align*} Buy 2 tubes of toothpaste. Use the coupon. Which store has the better buy? How much less is the cost at that store? \\begin{align*}& \\mathbf{Describe:} && \\text{Two store signs show the cost for tubes of toothpaste. The coupon gives}\\\\ &&& 20 \\% \\ \\text{off the price.}\\\\ & \\mathbf{My \\ Job:} && \\text{Figure out the store that has the better buy for 2 tubes of toothpaste after}\\\\ &&& \\text{using the coupon. Then figure out the difference in cost.}\\\\ & \\mathbf{Plan:} && \\text{Figure out the cost of 2 tubes of toothpaste ABC's. Take 20\\% off of each.}\\\\ &&& \\text{Do the same at", "# Price reduction The price of the camera is 654 USD on sale. They reduced its price to 411 USD. Calculate the percentage reduction in the price of the camera. p = 37.16 % ### Step-by-step explanation: $p=\\frac{654-411}{654}\\cdot 100=37.16\\mathrm{%}$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators ## Related math problems and questions: • Dealer Dealer sell digital camera for € 747. Thirty percent of the price was his profit. After some time, decreased interest in selling the camera and therefore reduce its sales price by 11 %. How many percent of the new price now is dealer's profit? Round the r • Sale off 2 A pair of of blues jeans went on sale. After a 30% reduction the pants cost $35. How much did the jeans cost before the price reduction? • Cost reduction Two MP3 players whose price was equal to originally have been discounted the first by 20%, the second by 35%. The difference in their prices 750 CZK was after the price reduction. What was the original price of each of the two players? • Sale A camera has a listed price of$751.98 before tax. If the sales tax rate is 9.25%, find the total cost of the camera with sales tax", "Q # A Camera shop allows a discount of 10% on the advertised price of a camera. What price must be marked on the camera, that costs him Rs. 600, so that he makes a profit of 20% A Camera shop allows a discount of 10% on the advertised price of a camera. What price must be marked on the camera, that costs him Rs. 600, so that he makes a profit of 20%? • Option 1) Rs. 800 • Option 2) Rs. 720 • Option 3) Rs. 750 • Option 4) Rs. 815 • Option 5) Rs. 850 Views CP = 600 SP = 600 $\\times$ 1.20 =720 .9 $\\times$ MP = 720 MP = 800 Exams Articles Questions", "pixels \\times magnification / camera\\ pixel\\ pitch$$", "[#permalink] ### Show Tags 09 Dec 2018, 01:14 Price of the camera and lens when purchased separately = 209$+ 69$ = 278$Total price when purchased together = 239$ Discount in $= 278$ - 239$= 39$ Discount % = 39/278% = 14% Hence the correct answer is A Posted from my mobile device _________________ Hit Kudos if you like what you see! VP Joined: 09 Mar 2018 Posts: 1000 Location: India Re: The price of a model M camera is $209 and the price of a special lens [#permalink] ### Show Tags 10 Feb 2019, 21:19 HKD1710 wrote: The price of a model M camera is$209 and the price of a special lens is $69. When the camera and lens are purchased together, the price is$239. The amount saved by purchasing the camera and lens together is approximately what percent of the total price of the camera and lens when purchased separately? (A) 14% (B) 16% (C) 29% (D) 33% (E) 86% Project PS Butler : Question #66 When purchased separately, Total cost = 278 $When bought together price = 239$ 239 = x/100 * 278 x = 86 % Now this 86 % is the difference between both the prices But he saved 14 % A _________________ If you notice any discrepancy in my reasoning, please let me", "96 is _________ when rounded to the nearest ten. Answer: 96 is 100 when rounded to the nearest ten. b. A printer costs$215. 215 is _________ when rounded to the nearest ten. 215 is 210 when rounded to the nearest ten. c. A camera costs $147. 147 is _________ when rounded to the nearest ten. Answer: 147 is 150 when rounded to the nearest ten. d. A computer costs$385. 385 is _________ when rounded to the nearest ten. e. The estimated total cost is $___. Answer: We have the rounded values as given below A set of 4 textbooks costs =$100 A printer costs = $210 A camera costs =$150 A computer costs = $380 we get the cost by adding all the above mentioned values. that is$100 + $210 +$150 + $380 =$840", "\\text{boxes for} \\ \\8\\!\\\\{\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times 8, \\ \\text{or} \\ \\6.40\\!\\\\{\\;} \\quad \\ \\6.40 - \\4.80 = \\1.60$ Buy 2 tubes of toothpaste. Use the coupon. Which store has the better buy? How much less is the cost at that store? $& \\mathbf{Describe:} && \\text{Two store signs show the cost for tubes of toothpaste. The coupon gives}\\\\&&& 20 \\% \\ \\text{off the price.}\\\\& \\mathbf{My \\ Job:} && \\text{Figure out the store that has the better buy for 2 tubes of toothpaste after}\\\\&&& \\text{using the coupon. Then figure out the difference in cost.}\\\\& \\mathbf{Plan:} && \\text{Figure out the cost of 2 tubes of toothpaste ABC's. Take 20\\% off of each.}\\\\&&& \\text{Do the same at Gem's. Subtract to find the difference.}\\\\& \\mathbf{Solve:} && \\text{ABC's:} \\ 4 \\ \\text{tubes are} \\ \\6 \\ \\text{so} \\ 2 \\ \\text{tubes are half the price or} \\ \\3.\\\\&&& \\text{Gem's:} \\ 10 \\ \\text{tubes are} \\ \\16 \\ \\text{so} \\ 2 \\ \\text{tubes are} \\ \\frac{1}{5} \\ \\text{of} \\ 16, \\ \\text{or} \\ \\16 \\div 5, \\ \\text{or} \\ \\3.20.\\\\&&& \\text{Using the coupon at ABC's,} \\ 20 \\% \\ \\text{off means the cost after the coupon is}\\\\&&& 80 \\% \\ \\text{of} \\ \\3.00, \\ \\text{or} \\ 0.8 \\times 3 = \\2.40.\\\\&&& \\text{Using the coupon at Gem's,}", "# Price reduction The price of the camera is 654 USD on sale. They reduced its price to 411 USD. Calculate the percentage reduction in the price of the camera. p = 37.16 % ### Step-by-step explanation: $p=\\frac{654-411}{654}\\cdot 100=37.16\\mathrm{%}$ We will be pleased if You send us any improvements to this math problem. Thank you! Tips to related online calculators ## Related math problems and questions: • Cost reduction Two MP3 players whose price was equal to originally have been discounted the first by 20%, the second by 35%. After the price reduction was the difference in their prices 750, - CZK. What was the original price of each of the two players? • Sale off 2 A pair of of blues jeans went on sale. After a 30% reduction the pants cost $35. How much did the jeans cost before the price reduction? • Sale A camera has a listed price of$751.98 before tax. If the sales tax rate is 9.25%, find the total cost of the camera with sales tax included. • Double price reduction The price of TV has been reduced twice. First by 15% and later by another 10% of the reduced price. After this double price reduction, the TV was sold for 8,874 crowns. What was the original price of the TV? •", "at 1... Rudd Company uses 40,000 micro-chips each year in its production of digital cameras. The of placing an order is $75. The cost of holding one unit of inventory for one year is$8. Currenty Rudd places 20 orders of 2,000 units per order. cost Compute: (a) The annual ordering cost. (2 marks) (2 marks) ...", "tax on imported product. I checked out yesterday and it looks like all the digital cameras they sell are made in Argentina. In order to favor internal production, they had to set huge taxes on imported product. I must buy at the price the Argentinians build the cameras, I have no choice. 19. Dec 18, 2011 Andre Why not ask Maxima to bring you one from here, next time when she visits her family. 20. Sep 22, 2012 fluidistic Hello people! I'm resurecting this thread because I'll go to France in January for about 20 days and will therefore buy a digital camera that I will probably buy on the Internet (it's near impossible to totally impossible to do it in Argentina, the situation worsened since I opened this thread and the prices in local stores can be as high as 5 times the ones I see on local stores in France by Internet). I've checked out a lot of cameras in Andre's website http://www.dpreview.com/products/cameras/all. However I'm worried about prices there. For example when I see my old 2005 camera of 5 mega pixels, it costs about 600 dollars while a much newer and better camera of the same brand cost only 200 dollars. Does this mean that the prices are never changing with time or even increasing?" ]

[ "• surryyy Alex is trying to start a savings plan. The following graph represents his projected savings over the next 12 months. How much money will he have saved by the sixteenth month? $700$800 $900$1000 https://static.k12.com/eli/bb/1211/3_68943/2_18901_16_68959/07f2f3eb1861900ad0a6aac8662172d9f00432a4/media/f30a47bb483f88b0b619f46dd585a31cfd045bb5/mediaasset_942794_1.jpg Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "240)/240 * 100 = 25% Option B Re: Andrew started saving at the beginning of the year and had saved$240 [#permalink] 30 Sep 2017, 05:01 Display posts from previous: Sort by", "29 Aug 2013, 04:52 At Larry's Auto Supply Store, Brand X antifreeze is sold by 3 21 Jun 2013, 15:53 4 If Ann saves x dollars each week and Beth saves y dollars ea 6 25 Mar 2016, 06:45 27 Grace makes an initial deposit of x dollars into a savings a 7 20 Nov 2016, 06:54 Grace makes an initial deposit of x dollars into a savings 3 26 May 2011, 00:56 Display posts from previous: Sort by", "is} \\ \\15 \\div 5, \\ \\text{or} \\ \\3\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.9 \\times \\3, \\ \\text{or} \\ \\2.70\\!\\\\{\\;} \\quad \\ \\text{Gem's:} \\ 1 \\ \\text{comb is} \\ \\12 \\div 6, \\ \\text{or} \\ \\2\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.9 \\times \\2, \\ \\text{or} \\ \\1.80\\!\\\\ {\\;} \\quad \\ \\2.70 - \\1.80 = \\0.90$ $2. \\quad \\text{ABC's;} \\ \\2.80 \\ \\text{less}\\!\\\\{\\;} \\quad \\ \\text{ABC's:} \\ 6 \\ \\text{boxes for} \\ \\12 \\ \\text{is} \\ 1 \\ \\text{box for} \\ \\2.\\!\\\\{\\;} \\quad \\ 4 \\times \\2 \\ \\text{is} \\ \\8 \\ \\text{for} \\ 4 \\ \\text{boxes}\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.7 \\times \\8, \\ \\text{or} \\ \\5.60\\!\\\\{\\;} \\quad \\ \\text{Gem's:} \\ 4 \\ \\text{boxes is half of} \\ \\24, \\ \\text{or} \\ \\12\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.7 \\times \\12, \\ \\text{or} \\ \\8.40\\!\\\\{\\;} \\quad \\ \\8.40 - \\5.60 = \\2.80$ $3. \\quad \\text{ABC's;} \\ \\1.60 \\ \\text{less}\\!\\\\{\\;} \\quad \\ \\text{ABC's:} \\ 4 \\ \\text{boxes for} \\ \\12 \\ \\text{is} \\ 2 \\ \\text{boxes for} \\ \\6\\!\\\\{\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times \\6 \\ \\text{or} \\ \\4.80\\!\\\\{\\;} \\quad \\ \\text{Gem's:} \\ 6 \\ \\text{boxes for} \\ \\24 \\ \\text{is} \\ 2 \\", "who wish to buy dollars can do so at a cost of 101.92 JPY per 1 USD.", "# Flipkart Supermart 0 30 Free delivery on order above Rs. 600. • PayTm Wallet – Rs. 100 Cashback on Rs. 1500+: Eligible users will receive Rs. 100 Cashback in their PayTm Wallet within 24 hours. • PayTm UPI – Rs. 100 Cashback on Rs. 1500+: Eligible users will receive Rs. 100 Cashback in their PayTm Wallet within 24 hours. • SBI Credit Cards – Supermart & Quick – 10% off upto Rs. 250 on Rs. 2000+: An instant discount will be awarded at the time of making the payment. Rs. 125 off on Rs. 3500 or Rs. 250 off on Rs. 5000+: Automatic discount in cart. May not be applicable on few products. • Select Users – Rs. 100 off on Rs. 1200+: Claim Now • Select Users – Rs. 100 off on Rs. 1400+: Claim Now • Select Users – Rs. 100 off on Rs. 1400+: Claim Now • Select Users – Rs. 100 off on Rs. 1600+: Claim Now • Select Users – Rs. 100 off on Rs. 1800+: Claim Now • Select Users – Rs. 100 off on Rs. 1800+: Claim Now Grocery Festive Pass – Rs. 100 off on Rs. 1500 for Next 3 Months: 11 SuperCoins FlipKart: Supermart", "15$ (eqn. $9$) From $7, 8$ and $9$ we have: $50$ CDs sold for £$24$ $35$ CDs sold for £$18$ $15$ CDs sold for £$12$.", "Thus in this case there either $14$ or $15$ of the heavier pennies.", "# Question 1 of 8 Elliot’s savings was $\\frac{6}{7}$ of Jerry’s. Thereafter, Elliot saves \\$18 and Jerry spends \\$9. They then had the same amount of money. How much was Elliot’s savings at first? A \\$156 B \\$158 C \\$162 D \\$162 E None of the above", "118$ bags. Write back if I have misinterpreted your question, Penny Math Central is supported by the University of Regina and the Imperial Oil Foundation.", "saved a total of $2×(30–25)$ = 10 minutes Hence (C) is the correct option.", "the bottles are $0.6 \\times 9.60$ , or $5.76.$6.00-$5.76=$0.24 ### Practice 1. Buy 2 birdies. Use the coupon. Which store has the better buy? How much less is the cost at that store? 2. Buy 5 water bottles with straws. Use the coupon. Which store has the better buy? How much less is the cost at that store? 3. Buy 2 mugs. Use the coupon. Which store has the better buy? How much less is the cost at that store? 4. Buy 4 boxes of paperclips. Use the coupon. Which store has the better buy? How much less is the cost at that store? 5. Buy 3 notebooks. Use the coupon. Which store has the better buy? How much less is the cost at that store? Jan 18, 2013", "30% off? Final Price: Thus, a product that normally costs$30 with a 30 percent discount will cost you $21.00, and you saved$9.00. You can also calculate how much you save by simply moving the period in 30.00 percent two spaces to the left, and then multiply the result by $30 as follows:$30 x . 30 = $9.00 savings. ## How many dollars is 30% off? Percent Off Table For 30.001 percent off 30.00 is 29.70The difference is 0.3030 percent off 30.00 is 21.00The difference is 9.0031 percent off 30.00 is 20.70The difference is 9.3032 percent off 30.00 is 20.40The difference is 9.6033 percent off 30.00 is 20.10The difference is 9.9095 more rows ## How do you subtract 20% from a price? In other words, multiply by 100 percent minus the percentage you want to subtract, in decimal form. To subtract 20 percent, multiply by 80 percent (0.8). To subtract 30 percent, multiply the number by 70 percent (0.7). ## How much is 40% off? Percent Off Table For 40.001 percent off 40.00 is 39.60The difference is 0.4040 percent off 40.00 is 24.00The difference is 16.0041 percent off 40.00 is 23.60The difference is 16.4042 percent off 40.00 is 23.20The difference is 16.8043 percent off 40.00 is 22.80The difference is 17.2095 more rows ## How much is 30 percent off", ". . I'm sure it means \"whole dollar\" price. I thought of just finding the GCD(17197, 17371) which is 29, then dividing each of those numbers by it. . . . . Good! So: . $17197 \\div 29 \\:= \\:593\\:\\text{ and }\\;17371 \\div 29 \\:=\\: 599$ . . . . Yes! Not sure what this says or what to do now. Note that 29, 593, and 599 are primes. Since $17197 \\:=\\:29\\cdot593$, then on Wednesday, . . the store sold: . $\\begin{Bmatrix} 29\\text{ ipods at }\\593\\text{ each} \\\\ \\text{or} \\\\ 593\\text{ ipods at }\\29\\text{ each} \\\\ \\end{Bmatrix}$ Since $17371 \\:=\\:29\\cdot599$, then on Thursday, . . the store sold: . $\\begin{Bmatrix}29\\text{ ipods at }\\599\\text{ each} \\\\ \\text{or} \\\\ 599\\text{ ipods at }\\29\\text{ each} \\end{Bmatrix}$ Since the price was the same on both days, . . $\\begin{array}{ccc}\\text{Wedmesday} & 593\\text{ ipods at }\\29\\text{ each} \\\\ \\text{Thursday} & 599\\text{ ipods at }\\29\\text{ each} \\end{array}$", "a jar at home for six months. The first month he saves S dollars. Each month thereafter, Mani saves 1/2 of what he has saved the previous month. If Mani’s total savings at the end of the six months is $393.75, how much money does he save the sixth month? (A)$6.25 (B) $7.50 (C)$12.50 (D) $18.75 (E)$25.00 Since Mani's total savings at the end of six months is $393.75 and he was able to save half of what he had saved the previous month, it is also safe to assume that the savings double each month(if we begin calculating from the last month) Let's consider Option A(6.25) as the amount he would have saved in the 6th month. So, the 5th month he would have saved xx.50 and every month after that he would have saved xx.00. The sum would be xxx.75 and could be our answer. Let's consider Option D(18.75). So in the 5th month, he would save xx.50 and there after every month he would save xx.00. The sum would be xxx.25. We can eliminate Option B,C, and E for similar reasons Therefore, Option A($6.25) is the amount Mani saves in the sixth month. _________________ You've got what it takes, but it will take everything you've got Re: Mani saves money in a jar at home", "post-college job to buy a Mega CD Changer (200 Discs!! $500!), and sold it on Craigslist just a few months ago for$30. I was honestly amazed that someone wanted it. That's a really good idea. I have some stuff I wouldn't want to lose. #### Mushtang Just 16 short years ago I saved up from first post-college job to buy a Mega CD Changer (200 Discs!! $500!), and sold it on Craigslist just a few months ago for$30. I was honestly amazed that someone wanted it.", "## Algebra 2 (1st Edition) Given Information: Store A sells a television for $210$ dollars. Store B sells the same TV for $161$ dollars. Sam's savings in dollars $= 210-161 = 49$ Expressed as a percentage of the price at Store A, 49 dollars is equal to $(\\frac{49}{210} * 100)$ percent, which $\\approx 23.33%$ Therefore, the answer is 23.33%. Rounded to the nearest whole number, this is 23%, which is Option G.", "97.6 % 19.52 % 7th 7.00 93% 5.60 94.4 % 18.89 % 8th 12.00 88% 9.60 90.4 % 18.08 % 9th 18.00 82% 14.40 85.6 % 17.12% 10th 59.00 41% 47.20 52.8 % 10.56% 56. $Paid % saved$ paid % saved % saved of $20 dollar discount 1st - 4th 0.00 100% 5th 1.00 99% 0.80 99.2 % 19.84 % 6th 3.00 97% 2.40 97.6 % 19.52 % 7th 7.00 93% 5.60 94.4 % 18.89 % 8th 12.00 88% 9.60 90.4 % 18.08 % 9th 18.00 82% 14.40 85.6 % 17.12% 10th 59.00 41% 47.20 52.8 % 10.56% 57. $ Paid % saved $paid % saved % saved of$20 1st - 4th 0.00 100% 5th 1.00 99% 0.80 99.2 % 19.84 % 6th 3.00 97% 2.40 97.6 % 19.52 % 7th 7.00 93% 5.60 94.4 % 18.89 % 8th 12.00 88% 9.60 90.4 % 18.08 % 9th 18.00 82% 14.40 85.6 % 17.12% 10th 59.00 41% 47.20 52.8 % 10.56% 58. # $Paid % saved$ paid % saved % saved of \\$20 1st - 4th 0.00 100% 5th 1.00 99% 0.80 99.2 % 19.84 % 6th 3.00 97% 2.40 97.6 % 19.52 % 7th 7.00 93% 5.60 94.4 % 18.89 % 8th 12.00 88% 9.60 90.4 % 18.08 % 9th 18.00 82% 14.40 85.6 % 17.12% 10th 59.00", "12.963% savings; Quizno’s: transactions, 26.3636% savings 18. a. Texts = 4,052; Video = 54, Cost =$3,381.18; 18b. $887.73. 19. 1.64 : 1.29 : 3.21 : 1 20. Manager =$11,015.12; Supervisor = $5,507.56; Worker =$2,753.78", "what happens to all those super savings next. They are like a pig moving through a python." ]

[ "= $1,920 And by the way,the digital camera is$480,not \\$400! $\\frac{480}{1920}=0.25$, or $25\\%$ So this is wrong because the camera cost 20% or the cost of the laptop, not 20% of the total cost of the camera and laptop. So lets see TKH is correct and you are wrong, unless you asked the wrong question. RonL", "## A camera has a list price of $869.99 before tax. If the sales tax rate is 9.5 % , find the total cost of the camera with sales tax included Question A camera has a list price of$ 869.99 before tax. If the sales tax rate is 9.5 % , find the total cost of the camera with sales tax included. Round your answer to the nearest cent, as necessary in progress 0 6 months 2021-07-18T11:17:16+00:00 1 Answers 2 views 0 1. Answer: Given the peculiar interrogate, the retort to such is identified as approximately ‘$952.64’ USD. Step-by-Step Explanation: To comprehend the evaluation, the apprehension of the sales tax quantity is necessary, 9.5% * 869.99 ~ 82.65 (rounded to the nearest cent). Thus far, evaluate on the behalf of the price, augmented with the sales tax. 869.99 + 82.65 ~ 952.64 (rounded to the nearest cent). Hence, given such configurations, the agglomerate cost of the commodity is equivalent to, approximately,$952.64 USD.", "• Subtract the discount from the original price: ${\\displaystyle 154.88-61.95=92.93}$. So, the sale price of the television is $92.93. 2. 2 Find the exact sale price of a camera that is 15% off. The original price is$449.95. • Convert the percentage discount to a decimal by moving the decimal two places to the left: ${\\displaystyle 15\\%=15.0\\%=.15}$. • Multiply the original price by the decimal: ${\\displaystyle 449.95\\times .15=67.49}$. • Subtract the discount from the original price: ${\\displaystyle 449.95-67.49=382.46}$. So, the sale price of the camera is $382.46. 3. 3 Estimate the sale price. A tablet is regularly$199.99. On sale, it is 45% off. • Round the original price to the nearest ten. Since $199.99 is only 1 cent away from$200, you would round up. • Calculate 10% of the rounded price. Moving the decimal one place to the left, you should see that 10% of $200.00 is$20.00. • Determine the number of tens in the percent off. Since ${\\displaystyle 10\\times 4=40}$, you know that there are 4 tens in 45%. • Multiply 10% of the rounded price by the appropriate factor. Since the percentage off is 45%, you would multiply 10% of the rounded price by 4: ${\\displaystyle \\20\\times 4=\\80}$ • Calculate 5% of the rounded price. This is half of 10%, which is $20. So half of$20 is", "apart, and also hard to see decimal points and parentheses? The equation is actually .34 + .4 - 44 * (6/528 - 2.4 + .11) which is 101 • Sweet. If that's not right, seems like it's at least a plausible alternate answer. – Dan Russell May 10 '16 at 2:32 • Hint #2 says that Anthony saw everything that was written down, which would include decimal points. – pacoverflow May 10 '16 at 4:08 Turns out the people were actually ordering picture frames They needed 3 4x4 sized frames, 4 4x6 sized frames (which they are getting from 528 photography, by the way) and 2 4x11 sized frames. The cost for these was 4x4 cost \\$5, 4x6 cost \\$10, and 4x11 cost \\$23. Thus 3x5 + 4x10 + 2x23 = 15+40+46=101 • An interesting answer, but your explanation for 528 isn't very satisfying. – pacoverflow May 11 '16 at 14:56 • @pacoverflow Agreed. And there are many solutions to the equation with 3 unknowns. I only provided one with nice round numbers (except for 23, which is my favourite). I don't seriously think this is the answer, but thought I would throw it out there anyway. – Trenin May 11 '16 at 14:58 There are 3 expressions and three digits in the answer. The answer is actually", "each month when saving the same amount? • Price reduction The price of the camera is 654 USD on sale. They reduced its price to 411 USD. Calculate the percentage reduction in the price of the camera. • Magic bag Each time the prince crossed the bridge, the number of tolars in the magic bag doubled. But then the devil always conjured 300 tolars for him. When this happened for the third time, the prince had twice as much as he had in the beginning. How many tolars", "\\ \\text{or} \\ \\12\\!\\\\ {\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.7 \\times \\12, \\ \\text{or} \\ \\8.40\\!\\\\ {\\;} \\quad \\ \\8.40 - \\5.60 = \\2.80\\end{align*} \\begin{align*}3. \\quad \\text{ABC's;} \\ \\1.60 \\ \\text{less}\\!\\\\ {\\;} \\quad \\ \\text{ABC's:} \\ 4 \\ \\text{boxes for} \\ \\12 \\ \\text{is} \\ 2 \\ \\text{boxes for} \\ \\6\\!\\\\ {\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times \\6 \\ \\text{or} \\ \\4.80\\!\\\\ {\\;} \\quad \\ \\text{Gem's:} \\ 6 \\ \\text{boxes for} \\ \\24 \\ \\text{is} \\ 2 \\ \\text{boxes for} \\ \\8\\!\\\\ {\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times 8, \\ \\text{or} \\ \\6.40\\!\\\\ {\\;} \\quad \\ \\6.40 - \\4.80 = \\1.60\\end{align*} Buy 2 tubes of toothpaste. Use the coupon. Which store has the better buy? How much less is the cost at that store? \\begin{align*}& \\mathbf{Describe:} && \\text{Two store signs show the cost for tubes of toothpaste. The coupon gives}\\\\ &&& 20 \\% \\ \\text{off the price.}\\\\ & \\mathbf{My \\ Job:} && \\text{Figure out the store that has the better buy for 2 tubes of toothpaste after}\\\\ &&& \\text{using the coupon. Then figure out the difference in cost.}\\\\ & \\mathbf{Plan:} && \\text{Figure out the cost of 2 tubes of toothpaste ABC's. Take 20\\% off of each.}\\\\ &&& \\text{Do the same at", "cost is} \\ \\0.90 \\ \\text{less at Gem's.}\\\\ & \\mathbf{Check:} && \\text{At ABC's} \\ 1 \\ \\text{comb is} \\ \\3.00 \\ \\text{and} \\ 10 \\% \\ \\text{off of} \\ \\3.00 \\ \\text{is} \\ \\0.30. \\ \\text{The cost after}\\\\ &&& \\text{using the coupon is} \\ \\3.00 - \\0.30, \\ \\text{or} \\ \\2.70.\\\\ &&& \\text{At Gem's} \\ 1 \\ \\text{comb is} \\ \\2.00 \\ \\text{and} \\ 10 \\% \\ \\text{off of} \\ \\2.00 \\ \\text{is} \\ \\0.20. \\ \\text{The cost after}\\\\ &&& \\text{using the coupon is} \\ \\2.00 - \\0.20, \\ \\text{or} \\ \\1.80. \\ \\2.70 - \\1.80 = \\0.90.\\end{align*} #### Example B Buy 4 boxes of band aids. Use the coupon. Which store has the better buy? How much less is the cost at that store? Solution: We can use problem solving steps to help. \\begin{align*}& \\mathbf{Describe:} && \\text{Two store signs show the cost for boxes of bandaids. The coupon gives}\\\\ &&& 30 \\% \\ \\text{off the price.}\\\\ & \\mathbf{My \\ Job:} && \\text{Figure out the store that has the better buy for 4 boxes of band aids after}\\\\ &&& \\text{using the coupon. Then figure out the difference in cost.}\\\\ & \\mathbf{Plan:} && \\text{Figure out the cost of 4 boxes of tissue at ABC's. Do the same at Gem's.}\\\\ &&& \\text{Take 30\\% off of each. Subtract to find the", "Mary decided to save a certain amount of her monthly salary each month 5 29 May 2016, 02:40 The original price of a camera was displayed as a whole dollar amount. 3 26 Apr 2016, 23:15 10 Bank account A contains exactly x dollars, an amount that will decreas 12 12 Jun 2015, 02:49 1 If a company allocates 15 percent of its budget to advertisi 3 24 Nov 2010, 04:10 Display posts from previous: Sort by", "96 is _________ when rounded to the nearest ten. Answer: 96 is 100 when rounded to the nearest ten. b. A printer costs$215. 215 is _________ when rounded to the nearest ten. 215 is 210 when rounded to the nearest ten. c. A camera costs $147. 147 is _________ when rounded to the nearest ten. Answer: 147 is 150 when rounded to the nearest ten. d. A computer costs$385. 385 is _________ when rounded to the nearest ten. e. The estimated total cost is $___. Answer: We have the rounded values as given below A set of 4 textbooks costs =$100 A printer costs = $210 A camera costs =$150 A computer costs = $380 we get the cost by adding all the above mentioned values. that is$100 + $210 +$150 + $380 =$840", "[#permalink] ### Show Tags 09 Dec 2018, 01:14 Price of the camera and lens when purchased separately = 209$+ 69$ = 278$Total price when purchased together = 239$ Discount in $= 278$ - 239$= 39$ Discount % = 39/278% = 14% Hence the correct answer is A Posted from my mobile device _________________ Hit Kudos if you like what you see! VP Joined: 09 Mar 2018 Posts: 1000 Location: India Re: The price of a model M camera is $209 and the price of a special lens [#permalink] ### Show Tags 10 Feb 2019, 21:19 HKD1710 wrote: The price of a model M camera is$209 and the price of a special lens is $69. When the camera and lens are purchased together, the price is$239. The amount saved by purchasing the camera and lens together is approximately what percent of the total price of the camera and lens when purchased separately? (A) 14% (B) 16% (C) 29% (D) 33% (E) 86% Project PS Butler : Question #66 When purchased separately, Total cost = 278 $When bought together price = 239$ 239 = x/100 * 278 x = 86 % Now this 86 % is the difference between both the prices But he saved 14 % A _________________ If you notice any discrepancy in my reasoning, please let me", "Lea buys 20 picture frames. Some are 4 in. by 6 in. frames which cost $5.80 each, and the rest are 5 in. by 7 in. frames that cost$9.60 each. She spends a total of \\$80.80. Which equation models this situation? Let f represent the number of 4 in. by 6 in. frames she buys. A. 9.6f + 5.8(20) = 80.8 B. 5.8f + 9.6(20 – f) = 80.8 C. (5.8 + 9.6)(20 – f) = 80.8 D. 9.6f + 5.8(f + 20) = 80.8", "# Price reduction The price of the camera is 654 USD on sale. They reduced its price to 411 USD. Calculate the percentage reduction in the price of the camera. p = 37.16 % ### Step-by-step explanation: $p=\\frac{654-411}{654}\\cdot 100=37.16\\mathrm{%}$ We will be pleased if You send us any improvements to this math problem. Thank you! Tips to related online calculators ## Related math problems and questions: • Cost reduction Two MP3 players whose price was equal to originally have been discounted the first by 20%, the second by 35%. After the price reduction was the difference in their prices 750, - CZK. What was the original price of each of the two players? • Sale off 2 A pair of of blues jeans went on sale. After a 30% reduction the pants cost $35. How much did the jeans cost before the price reduction? • Sale A camera has a listed price of$751.98 before tax. If the sales tax rate is 9.25%, find the total cost of the camera with sales tax included. • Double price reduction The price of TV has been reduced twice. First by 15% and later by another 10% of the reduced price. After this double price reduction, the TV was sold for 8,874 crowns. What was the original price of the TV? •", "is} \\ \\15 \\div 5, \\ \\text{or} \\ \\3\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.9 \\times \\3, \\ \\text{or} \\ \\2.70\\!\\\\{\\;} \\quad \\ \\text{Gem's:} \\ 1 \\ \\text{comb is} \\ \\12 \\div 6, \\ \\text{or} \\ \\2\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.9 \\times \\2, \\ \\text{or} \\ \\1.80\\!\\\\ {\\;} \\quad \\ \\2.70 - \\1.80 = \\0.90$ $2. \\quad \\text{ABC's;} \\ \\2.80 \\ \\text{less}\\!\\\\{\\;} \\quad \\ \\text{ABC's:} \\ 6 \\ \\text{boxes for} \\ \\12 \\ \\text{is} \\ 1 \\ \\text{box for} \\ \\2.\\!\\\\{\\;} \\quad \\ 4 \\times \\2 \\ \\text{is} \\ \\8 \\ \\text{for} \\ 4 \\ \\text{boxes}\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.7 \\times \\8, \\ \\text{or} \\ \\5.60\\!\\\\{\\;} \\quad \\ \\text{Gem's:} \\ 4 \\ \\text{boxes is half of} \\ \\24, \\ \\text{or} \\ \\12\\!\\\\{\\;} \\quad \\ \\text{With the coupon the cost is} \\ 0.7 \\times \\12, \\ \\text{or} \\ \\8.40\\!\\\\{\\;} \\quad \\ \\8.40 - \\5.60 = \\2.80$ $3. \\quad \\text{ABC's;} \\ \\1.60 \\ \\text{less}\\!\\\\{\\;} \\quad \\ \\text{ABC's:} \\ 4 \\ \\text{boxes for} \\ \\12 \\ \\text{is} \\ 2 \\ \\text{boxes for} \\ \\6\\!\\\\{\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times \\6 \\ \\text{or} \\ \\4.80\\!\\\\{\\;} \\quad \\ \\text{Gem's:} \\ 6 \\ \\text{boxes for} \\ \\24 \\ \\text{is} \\ 2 \\", "# Price reduction The price of the camera is 654 USD on sale. They reduced its price to 411 USD. Calculate the percentage reduction in the price of the camera. p = 37.16 % ### Step-by-step explanation: $p=\\frac{654-411}{654}\\cdot 100=37.16\\mathrm{%}$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators ## Related math problems and questions: • Dealer Dealer sell digital camera for € 747. Thirty percent of the price was his profit. After some time, decreased interest in selling the camera and therefore reduce its sales price by 11 %. How many percent of the new price now is dealer's profit? Round the r • Sale off 2 A pair of of blues jeans went on sale. After a 30% reduction the pants cost $35. How much did the jeans cost before the price reduction? • Cost reduction Two MP3 players whose price was equal to originally have been discounted the first by 20%, the second by 35%. The difference in their prices 750 CZK was after the price reduction. What was the original price of each of the two players? • Sale A camera has a listed price of$751.98 before tax. If the sales tax rate is 9.25%, find the total cost of the camera with sales tax", "applying the coupon would give the camera a price of$50. If the camera price retailed at \\$150, the price of the camera...", "the$10 discount. $_______ Answer:$14.99 Explanation: Given that, Eugene wants to buy jeans at a store that is giving $10 off everything. The tag on the jeans is marked 50% off. The original price is$49.98. 0.5 × 49.98 = 24.99 Now subtract $10 discount. 24.99 – 10 = 14.99 The total cost if the 50% discount is applied before the$10 discount is $14.99 Question 22. b. Find the total cost if the$10 discount is applied before the 50% discount. $_______ Answer:$19.99 Explanation: We have to find the price after the $10 discount then find 50% of that price to find the discounted price. 49.98 – 10 = 39.98 0.5 × 39.98 = 19.99 Thus the total cost if the$10 discount is applied before the 50% discount is $19.99 ### Applications of Percent – Page No. 158 Question 23. Multistep Eric downloads the coupon shown and goes shopping at Gadgets Galore, where he buys a digital camera for$95 and an extra battery for $15.99. a. What is the total cost if the coupon is applied to the digital camera?$ _______ Explanation: Use the formula for the discount price: DP = P – x.P Price for the digital camera: DP = 95 – 0.1(95) DP = 95 – 9.5 DP = $85.5 Total cost = 85.5 + 15.99 =$101.49 Question 23.", "#### Explanation: We can express discount amount as $x$ and the original price of the camera as $x + 632$. The equation to determine the price will be: $\\left(x + 632\\right) \\times \\frac{20}{100} = x$ Simplify. $\\left(x + 632\\right) \\times \\frac{1 \\cancel{20}}{5 \\cancel{100}} = x$ Multiply both sides by $5$. $x + 632 = 5 x$ Subtract $x$ from both sides. $632 = 4 x$ Divide both sides by $4$. $158 = x$ Since the discount amount $\\left(x\\right)$ is $158$, the original price $\\left(x + 632\\right)$ will be $\\left(158 + 632\\right) = 790$.", "the bottles are $0.6 \\times 9.60$ , or $5.76.$6.00-$5.76=$0.24 ### Practice 1. Buy 2 birdies. Use the coupon. Which store has the better buy? How much less is the cost at that store? 2. Buy 5 water bottles with straws. Use the coupon. Which store has the better buy? How much less is the cost at that store? 3. Buy 2 mugs. Use the coupon. Which store has the better buy? How much less is the cost at that store? 4. Buy 4 boxes of paperclips. Use the coupon. Which store has the better buy? How much less is the cost at that store? 5. Buy 3 notebooks. Use the coupon. Which store has the better buy? How much less is the cost at that store? Jan 18, 2013", "3 inch, 1 ½ ounce 9, 1100 fps, 5-count / \\$39.95", "\\text{boxes for} \\ \\8\\!\\\\{\\;} \\quad \\ \\text{With the coupon, the cost is} \\ 0.8 \\times 8, \\ \\text{or} \\ \\6.40\\!\\\\{\\;} \\quad \\ \\6.40 - \\4.80 = \\1.60$ Buy 2 tubes of toothpaste. Use the coupon. Which store has the better buy? How much less is the cost at that store? $& \\mathbf{Describe:} && \\text{Two store signs show the cost for tubes of toothpaste. The coupon gives}\\\\&&& 20 \\% \\ \\text{off the price.}\\\\& \\mathbf{My \\ Job:} && \\text{Figure out the store that has the better buy for 2 tubes of toothpaste after}\\\\&&& \\text{using the coupon. Then figure out the difference in cost.}\\\\& \\mathbf{Plan:} && \\text{Figure out the cost of 2 tubes of toothpaste ABC's. Take 20\\% off of each.}\\\\&&& \\text{Do the same at Gem's. Subtract to find the difference.}\\\\& \\mathbf{Solve:} && \\text{ABC's:} \\ 4 \\ \\text{tubes are} \\ \\6 \\ \\text{so} \\ 2 \\ \\text{tubes are half the price or} \\ \\3.\\\\&&& \\text{Gem's:} \\ 10 \\ \\text{tubes are} \\ \\16 \\ \\text{so} \\ 2 \\ \\text{tubes are} \\ \\frac{1}{5} \\ \\text{of} \\ 16, \\ \\text{or} \\ \\16 \\div 5, \\ \\text{or} \\ \\3.20.\\\\&&& \\text{Using the coupon at ABC's,} \\ 20 \\% \\ \\text{off means the cost after the coupon is}\\\\&&& 80 \\% \\ \\text{of} \\ \\3.00, \\ \\text{or} \\ 0.8 \\times 3 = \\2.40.\\\\&&& \\text{Using the coupon at Gem's,}" ]

[ "ticket enters the game. After ten generations, you have more than 1000 holding ticket 1, more than 150 having another ticket mentioning curtain 1, and only 4 guys still having a ticket which always loses....", "were six reserved tickets, three tickets for the field level, and one ticket for the box seat.", "3/5, 10 ½, and 8. How many yards were sold? • Expression plus minus Evaluate expression: (-1)2 . 12 – 6 : 3 + (-3) . (-2) + 22 – (-3) . 2 • Students at cinema All pupils from the school went to the cinema for a film performance. The cinema has 400 seats, 2 rows of 20 seats remain unoccupied. Teachers occupied 10 places. How many students were there?", "## 2010-09-03 ### One Outrageously Uninformed Yahoo! Article I was browsing through my Yahoo! homepage and stumbled upon this article. I looked through the 6 outrageously overpriced items, and I do agree with all of the choices. I suppose that for an article targeted at total laypeople, I can understand why things like \"paying for the label\" might not make sense at first; that said, I am still astounded as to how a financial article doesn't (appear to) understand why these prices are actually so high. Let's start from the top. #### Movie Theater Popcorn The article here asks why the markup for a bag of popcorn at the movie theater is a factor of six compared to the price for a similar product at a supermarket. It claims to not know the answer to this. The answer is that in a movie theater, there is nowhere else to get popcorn, and moviegoers like the convenience of buying already-cooked popcorn (as opposed to having to cook and pack their own popcorn). The theater owners have a near-monopoly over the popcorn market inside the theater (as enough people would rather pay more than spend time making their own popcorn), so theater popcorn will of course be more expensive. #### Greeting Cards This follows a similar vein as the popcorn", "Home > Mathematica, Mathematics > Couples going to the movies Couples going to the movies I have been contemplating Professor Strogatz theatre puzzle (in The Joy of X). From the book and hyperlink, the puzzle: Imagine there’ s a very popular new movie showing at the local theater.It’ s a romantic comedy, and hundreds of couples (many more than the theater can accommodate) are lined up at the box office, desperate to get in.Once a lucky couple get their tickets, they scramble inside and choose two seats right next to each other.To keep things simple, let’ s suppose they choose these seats at random, wherever there’ s room.In other words, they don’ t care whether they sit close to the screen or far away, on the aisle or in the middle of a row.As long as they’ re together, side by side, they’ re happy.Also, let’ s assume no couple will ever slide over to make room for another.Once a couple sits down, that’ s it.No courtesy whatsoever.Knowing this, the box office stops selling tickets as soon as there are only single seats left.Otherwise brawls would ensue.At first, when the theater is pretty empty, there’ s no problem.Every couple can find two adjacent seats.But after a while, the only seats left are singles\\[LongDash]solitary, uninhabitable dead spaces that a", "# How many ways are there for people to queue? I'm stuck at the following combinatorics problem: Fifteen people queue up for cinema tickets at five (different) sales points. In how many ways can they stand in queue behind one another, if the queues can be of different length but at least one person is waiting at every sales point? My idea was to place one person at every sales point first and then place the remaining ten people. Since placing the first five is an ordered drawing without repetition I get $15*14*13*12*11$ possibilities to place a person at each sales point. Now I need to figure out how many ways there are to divide the group of ten people into five distinct groups but I am not sure how to go about this. Looking forward to your ideas and explanations - We can encode the five nonempty customer groups in the following way as a $01$-sequence: $$0\\ \\ldots\\ |0\\ \\ldots\\ |0\\ \\ldots\\ |0\\ \\ldots\\ |0\\ \\ldots\\quad.$$ Here each $0$ denotes a person, each $\\ldots$ a nonnegative number of zeros, and each $|$ a separator between groups. Since there have to be $15$ zeros in all we have to distribute $10$ remaining zeros onto the five $\\ldots\\$. This can be encoded as a word containing $10$ zeros and", "the total number of seats in the theater is 1500 Question 18. Multi-Step There are 48 beads in a package. Mariah bought 4 packages of blue, 9 packages of gold, 6 packages of red, and 2 packages of silver beads. Which is the best estimate of the total number of beads she bought? (A) 1,000 (B) 800 (C) 1,200 (D) 2,400", "organized or unorganized data? 12. How many showings had 90 people or more in attendance? 13. How many showings had less than 50 people in attendance? 14. How many showings had less than 70 people in attendance? 15. True or false. This data also tells you which showings had the most people in attendance. 16. True or false. There were two showings that had 78 people in attendance.", "are there in each section? And how many total seats are there in the theatre?", "were 4 times as many \"A\" tickets sold as there were \"B\" tickets. How many of each type of ticket were sold and how much money did each type of ticket generate? What piece of information are you overlooking? There were 4 times as many \"A\" tickets sold as there were \"B\" tickets or A = 4B. How can this simplify your solution path? I think you are on your way now. Staff member hrucker said:", "In how many ways can 7 people line up at a ticket window of a cinema hall? Question: In how many ways can 7 people line up at a ticket window of a cinema hall? Solution: To find: number of arrangements of 7 people in a queue. Here there are 7 spaces to be occupied by 7 people. Therefore 7 people can occupy first place. Similarly, 6 people can occupy second place and so on. Lastly, there will be a single person to occupy the 7 positions. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, permutation of 7 different objects in 7 places is $P(7,7)=\\frac{7 !}{(7-7) !}$ $=\\frac{7 !}{0 !}=\\frac{5040}{1}=5040$ Therefore, the number of possible ways is 5040", "time consuming to do it the “long” way, unless you are an absolute prodigy at probabilities. When I take the test, I who PWNs the SAT and gets 2400s, I list all the possibilities and count. But hey man, it’s your life. Same prize as last weekend for whoever gets this first (either on the blog, or on Facebook; you’re competing against each other here). Any \\$5 album from Amazon.com. Ready? A small town has 3 theaters in it. Last Saturday night, the average age in the first theater was 29, and the average age in the second theater was 24. If the overall average theatergoer on that night was 36 years old and the ratio of attendees for the three theaters was 2:3:10, respectively, then what was the average age of the attendees in the third theater? Good luck, folks. I’ll post the answer and contact the winner (if there is one) on Monday. Remember: you can’t win if I can’t contact you. UPDATE: Congrats to Codi, who nailed it on Facebook. Answer below the cut. This is an average question, so let’s use the average table! Start by filling in what the questions tells us (and plugging in obvious values for numbers of people in the theaters based on the given ratio): $\\tiny&space;\\inline&space;\\dpi{300}&space;\\fn_cm&space;[Number\\:&space;of\\:&space;Values]\\times&space;[Average\\:&space;of\\:&space;Values]&space;=&space;[Sum\\:&space;of\\:&space;Values]$ 2 people (1st", "known as a(n)____________.... ### A movie was sold out for four straight days. If 535 tickets were sold each day, how many tickets were sold in all??? a movie was sold out for four straight days. If 535 tickets were sold each day, how many tickets were sold in all???... ### Please tell me how to find the Slope of lines, (or do one for me) [I’ll try to help you back] Please tell me how to find the Slope of lines, (or do one for me) [I’ll try to help you back]...", "# Movie Tickets Discrete Mathematics Level 4 In a movie theater, tickets cost $$5$$. 10 people are standing in line to buy a ticket, which, five of them have a $$5$$ bill, four have a $$10$$ bill and one has a $$20$$ bill. Considering that in the box office there is no money in the cashier initially, how many ordered sets can be formed with these 10 people in a way that there is always change for each person who buys a ticket in the box office? Consider there is $$10\\binom{9}{4}$$ ways to arrange them without restriction, in other words, if two people have the same bill, then they're equal. ×", "of at least 4 winnings for a group of people who have bought 5 tickets together? 11. Two groups The group of 10 girls should be divided into two groups with at least 4 girls in each group. How many ways can this be done?", "twice as many as the previous day. How many tickets were sold on Friday and how many totally from Monday to Friday? • Saving per cents The first day I save 1 cent and every next day cent more. How many I saved per year (365 days)? • Cinema Cinema auditorium is built for 3300 people. The first row is planned for 36 seats and each next gradually 4 more. How many rows of seats will have auditorium? • Raspberries Dano had 20 raspberries. John had 90% more raspberries than Dano. Determine how many raspberries have Dano and Juraj together. • Fruit candies Zuzka and Janka ate fruit candies. Zuzka ate 10 and Janka ate 3 times more. How many sweets did both girls eat together? • Cinema How many rows of seats are in the stands, if you know that the first row has 24 seats and the last 50 seats and each subsequent row has two seats more than row before? How many seats are in the stands?", "the cinema. You would probably regard the holiday as the better gift, but if you were asked the question “by how much do you prefer the holiday to the cinema tickets” then it would be impossible to answer accurately, you couldn’t say “I rate it 17 times higher” or something along those lines. You could say that 17 sets of cinema tickets would give you the same amount of utility as the holiday in Barcelona, but that would be a slightly different concept than saying you regard one holiday 17 times better than one set of tickets. This is because you do not usually get the same amount of utility from each amount of the same good. Think for instance of ice cream, on a hot day when you really felt like an ice cream, you might get a lot of pleasure from having it, but having another one might not necessarily double your pleasure, and having ten in a row definitely wouldn’t increase your pleasure by a factor of ten – you might have had well more than enough after two or three and any further ice creams would actually reduce your level of utility.", "box. On the street, wind succeeded to spread the remaining 39 tickets across the road. Fortunately, there was a nearby class with 10 girls (age 6 or so, first graders or kindergarten) who immediately and happily helped me to collect the mess. One of them, with the highest number of tickets, was declared the winner of the contest by me and their teacher. :-)", "with these ticket! Also the number of the ticket remains with your patrons as long as the ticket does! Plus, with the great selection of colors we offer, the ticket isn't going to get lost any time soon.", "# Modified catalan 319 views 0 30 people queue in a random order outside a movie theater where tickets cost $5. 15 of them carry a$5 bill, and the other 15 carry a $10 bill. If a moviegoer pays with a$10 bill, the theater must return them a $5 bill. What is the minimum number of$5 bills the theater must supply in advance such that the probability all 30 are admitted is at least 0.95?" ]

[ "# $20$ people are sitting around a (circular) table. How many ways can we choose $3$ people, no two of whom are neighbors? 0 choices for the 1st person. 17 choices for the 2nd person (must exclude 1st and his/her two neighbours) For 2 of these choices of 2nd person, there is one shared neighbour, so 15 remaining choices. (e.g. if they are numbered 1 to 20 in a circle, 1st person is #1, 2nd is #3, then people 20,1,2,3,4 are excluded). For the other 15 choices of 2nd person, there are no shared neighbors, so 14 remaining choices. So if order matters, total is $20 \\cdot (2 \\cdot 15 + 15 \\cdot 14)$ but since order does not matter, divide by $3! = 6$ to account for the permutations in order of the 3 people. So total = $20 \\cdot \\frac{2 \\cdot 15 + 15 \\cdot 14}{6}$ just redid it; does this make any sense? • What is the answer with a straight table and how does turning the table into a corner change the problem? thats the step that you need to account for Oct 4, 2016 at 16:14 • first person - 20 choices , second person 17 choices , then third person 14 choices - then you could have chosen the 3 people in", "# In how many ways can 6 people be arranged in a line? 6! = 720 There are $6$ ways to choose the first person, then $5$ to choose the next, etc. 6xx5xx4xx3xx2xx1 = 6! = 720", "3=\\frac{6 !}{(6-3) !}=\\frac{6 !}{3 !}=\\frac{6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{3 \\cdot 2 \\cdot 1}=120$ b. 7P2 = $$7 \\cdot 6 = 42$$, or $7 \\mathrm{P} 2=\\frac{7 !}{5 !}=\\frac{7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}=42$ Next we consider some more permutation problems to get further insight into these concepts. Example $$\\PageIndex{5}$$ In how many different ways can 4 people be seated in a straight line if two of them insist on sitting next to each other? Solution Let us suppose we have four people A, B, C, and D. Further suppose that A and B want to sit together. For the sake of argument, we tie A and B together and treat them as one person. The four people are $$\\boxed{AB}$$ CD. Since $$\\boxed{AB}$$ is treated as one person, we have the following possible arrangements. $\\boxed{AB} CD, \\boxed{AB} DC, C \\boxed{AB}D, D\\boxed{AB}C, CD \\boxed{AB}, DC\\boxed{AB} \\nonumber$ Note that there are six more such permutations because A and B could also be tied in the order BA. And they are $\\boxed{BA}CD, \\boxed{BA} DC, C\\boxed{BA}D, D\\boxed{BA}C, CD\\boxed{BA}, DC \\boxed{BA} \\nonumber$ So altogether there are 12 different permutations. Let us now do the problem using the multiplication axiom. After we tie two of", "# Question #c1283 Nov 24, 2017 $360$ #### Explanation: There are 4 seats available for 6 people. They can be arranged in any order. In the first seat, there are 6 possibilities for the people to chose from. In the second, 5 possibilities (since 1 is already sitting in the first seat) . In the third, 4 possibilities. And finally in the 4th, 3 possibilities. Taking these all into consideration, just multiply it. $6 \\times 5 \\times 4 \\times 3 = 360$ Or use the permutation formula:${\\text{ \" }}_{6} {P}_{4} = 360$", "6:51 The order in which the runners finish matters, so this is a permutation rather than a combination. Until the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \\cdot 7 \\cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \\frac{8!}{(8 - 3)!} = \\frac{8!}{5!} = \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5!}{5!} = 8 \\cdot 7 \\cdot 6 = 336$$ Imagine if John, Jack and Julie were in the \"top-3\", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at So $8\\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.", "could not be solved with the permutation formula. Note: Here you have only begun to explore counting problems. For more information about combinations, permutations, and other types of counting problems, consult a Probability text. Example A You are going on a road trip with 4 friends in a car that fits 5 people. How many different ways can everyone sit if you have to drive the whole way? Solution: A decision chart is a great way of thinking about this problem. You have to sit in the driver’s seat. There are four options for the first passenger seat. Once that person is seated there are three options for the next passenger seat. This goes on until there is one person left with one seat. $1 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=24$ Example B How many different ways can the gold, silver and bronze medals be awarded in an Olympic event with 12 athletes competing? Solution: Since the order does matter with the three medals, this is a permutation problem. You will start with 12 athletes and then choose and arrange 3 different winners. $_{12}P_3=\\frac{12!}{\\left(12-3\\right)!}=\\frac{12!}{9!}=\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot \\ldots}{9 \\cdot \\ldots}=12 \\cdot 11 \\cdot 10=1320$ Note that you could also use a decision chart to decide how many possibilities are there for gold", "men? Which gives, (choose 2men from 3)*(permute the 2men)*(choose 1 woman from 4)*(Permute in a circle)=3C2*2!*4C1*4!=576 11. ## Re: permutation and combinations Hello, Punch! 3 men and 4 women are seated at a round table. Find the number of possible arrangements if one particular woman must sit between 2 men. Since we have a round table, that woman can sit anywhere. She will be flanked by two men. There are 3 choices for the man on her right, 2 choices for the man on her left. The remaining 4 people (one man, 3 women) can be seated in $4!$ ways. Therefore, there are: . $3\\cdot 2\\cdot 4! \\:=\\:144$ arrangements.", "suppose we want to count, using the fundamental counting principle, the number of possibilities resulting from flipping a coin, rolling a single die, and picking a prime number between 1 and 10. There are 2 possible ways to flip a coin (heads and tails), there are 6 possible rolls of a single die (1,2,3,4,5, and 6), and there are 4 primes between 1 and 10. Thus, there are $2 \\cdot 6 \\cdot 4 = 48$ possibilities for the result of doing all three of these actions. ### Permutations Consider how we might count the number of ways one can line up 4 people from a group of 7. Certainly any of the $7$ people of the group could be chosen to start the line. This leaves $6$ to choose from for the second person in line. With two people already standing in line, we only have $5$ persons to choose from for the third person in line, and then we must choose among the $4$ remaining for the last person in our line. Applying the fundamental counting rule, we then have $7 \\cdot 6 \\cdot 5 \\cdot 4 = 840$ ways to arrange $4$ people from a group of $7$ in a line. In a like manner, the number of ways we can line up 50 people from", "don't work. There is the scenario where D sits at 1 and A at 5 and D sits at 5 and A sits at 3. Jun 21, 2018 at 19:46 Let's start with person D who is the most restrictive. We will consider 2 cases: D sits in chair 2 or 4 D sits in chair 3 If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\\cdot 3\\cdot 3\\cdot 2\\cdot 1 = 36$ If D sits in chair 3, then there are $4!=24$ ways to arrange the remaining people, as there are no more restrictions. That's a total of 36+24=60. Now, let's try with your method (we take the total number of permutations and subtract restricted cases). As you found, there are 120 total permutations. There are $2\\cdot 4\\cdot 3\\cdot 2\\cdot 1=48$ permutations where D is in either chair 1 or 5. There are $1\\cdot 4\\cdot 3\\cdot 2\\cdot 1 = 24$ permutations where A is in chair 3. The number of permutations with A in chair 3 OR D in chair 1 or 5 equals the number of permutations of A in chair 3 plus the", "7\\cdot 6$ ways to seat three people so that the end chairs are empty. For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $3\\cdot 5\\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many are larger than 350 but less than 400. That means the first digit must be 3, the second digit may be 5, 7, or 9, and the last may be anything. That is $1\\cdot 3\\cdot 5 = 15$ 3-digit numbers. All together, there are $75+15 = 90$ 3-digit numbers larger than 350 using only odd digits. Note that this is not a permutation problem. Digits may be repeated. Edit: For #4 a) Digits may be repeated. So, there are $5^3 = 125$ 3-digit numbers using only odd digits. 3. ## Re: Permutations and Combinations Originally Posted by SlipEternal For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $3\\cdot 5\\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many", "director. How many choices do you have? #### Solution The way to work this problem out is to consider the main role first. You have 40 choices for the main role. For the lead supporting actor there are 39 left to select from. For the next role there are 38 to select from. Now following this pattern and consider that there are seven in the cast gives a total number of choices as $40 \\cdot 39 \\cdot 38 \\cdot 37 .\\cdot 36 \\cdot 35 \\cdot 34$ We could multiply these all out, however there is an easier way. We can write $40 \\cdot 39 \\cdot 38 \\cdot 37 \\cdot 36 \\cdot 35 \\cdot 34 \\left( \\dfrac{33 \\cdot 32 \\cdot 31 \\cdot \\, ... \\cdot 2 . 1}{ 33 \\cdot 32 \\cdot 31 \\cdot\\, ... \\cdot 2 \\cdot 1} \\right)$ $= \\dfrac{40!}{(40-7)!}$ This expression has a special notation. We write $P_{40,7} = \\dfrac{n!}{(n-r)!} = 93,963,542,400$ We can see that there are plenty of choices. In general we write $P_{n,r} = \\dfrac{n!}{(n-r)!}$ This is called a permutation. ### Combinations Example $$\\PageIndex{4}$$: combination How many 5 card poker hands are there? #### Solution We can solve this in a similar way as the prior question. We are selecting 5 cards out of 52 total. Unfortunately this is not quite a permutation", "the remaining 3 digits. So, that is $1\\cdot 3\\cdot 3 = 9$. Total 3-digit numbers with no repeated digits and only odd digits that are greater than 350: $36+9 = 45$. The 36 from the book does show up, but I am guessing your book is wrong. 5. ## Re: Permutations and Combinations Thanks for the help. 6. ## Re: Permutations and Combinations Hello, Fratricide! 1. Seven people are to be seated in a row. Calculate the number of ways in which this can be done so that two particular persons, A and B, always have exactly one of the others between them. We have: $A\\_\\;\\!B$ or $B\\_\\;\\!A$ . . . 2 choices. There are 5 choices for the \"middle\" person: $A\\,\\!X\\!B$ Now we have 5 \"people\" to permute: . $\\left\\{\\boxed{A\\,\\!X\\!B},\\, C,\\,D,\\,E,\\,F\\right\\}$ There are 5! permutations. Therefore, there are: . $2\\cdot5\\cdot 5! \\:=\\:1200$ ways. 2. Determine the number of possible permutations (from digits 1 through to 9) when the digits 1 and 2 are together, but not necessarily in that order. Duct-tape the \"1\" and \"2\" together. There are 2 possible orders: $12$ or $21.$ We have 8 \"numbers\" to permutate: . $\\left\\{\\boxed{12},\\,3,\\,4,\\,5,\\,6,\\,7,\\,8,\\,9 \\right\\}$ There are 8! ways. Therefore, there are: . $2\\cdot8! \\:=\\:80,640$ permutations. 3. There are 10 chairs in a row: a) In how many ways", "6 for the 2nd, 5 for the third and so on. Thus, the number of ways is: $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1=7!=5040$ (ii) Letting M represent male and F represent female, we must have: M F M F M F M This means the number of ways will be given by: $4!\\cdot3!=144$ (iii) First putting James on the left of Rio, there are 5 ways for there to be 1 person between them and 5! different lineups of the other people. Including the lineups with Rio on the left of James this is a total of: $2\\cdot5\\cdot5!$ With two people between them, we find: $2\\cdot4\\cdot5!$ With 3 in between: $2\\cdot3\\cdot5!$ With 4 in between: $2\\cdot2\\cdot5!$ With 5 in between: $2\\cdot1\\cdot5!$ Adding, we find: $2\\cdot5!$$5+4+3+2+1$$=30\\cdot5!=3600$ Thus, there are 3600 ways James and Rio are not seated together. (iv) The professor has 4 males to choose from and 3 females, thus there are: $4\\cdot3=12$ There are 12 ways to choose 1 male and 1 female. (v) Here we use the combination function: ${7 \\choose 3}=35$ Thus, we find there are 35 ways to choose 3 people from a group of 7. November 21st, 2011, 06:42 PM #3 Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 Re: Permutation and combination ? Quote: Originally Posted by MarkFL (iii) First putting James on the left of", "# Permutations of 4 people in a line This question is from the GRE math: A group of 4 people are standing in a straight line. In how many different ways can these people be standing on the line? The answer is $24$. If the permutations formula is this, how did they derive this answer? Thanks. $$\\frac{n!}{(n-r)!}$$ • The number of permutations on $n$ objects is $n!$ – Seth Apr 28 '14 at 20:59 ## 1 Answer Any of the $4$ people can be the first person. Then, there are $3$ remaining choices for the second person, then $2$ remaining choices for the third, and then the fourth has been decided. Hence, there are $n!$ possible ways.", "guess, since we have 14 choices for the first guest, then 13 for the second, and so on. But the guess is wrong (in fact, that product is exactly $$2192190 = P(14,6)$$). It distinguishes between the different orders in which we could invite the guests. To correct for this, we could divide by the number of different arrangements of the 6 guests (so that all of these would count as just one outcome). There are precisely $$6!$$ ways to arrange 6 guests, so the correct answer to the first question is \\begin{equation*} \\frac{14 \\cdot 13 \\cdot 12 \\cdot 11\\cdot 10 \\cdot 9}{6!}. \\end{equation*} Note that another way to write this is \\begin{equation*} \\frac{14!}{8!\\cdot 6!}. \\end{equation*} which is what we had originally. This page titled 1.3: Combinations and Permutations is shared under a CC BY-SA license and was authored, remixed, and/or curated by Oscar Levin.", "another group of $6$ people. This is nothing but choosing $6$ people from a group of $18$ people. This is nothing but $\\binom{18}{6} = 18564 \\Rightarrow 564$. ~coolmath_2018", "$7!/7=6!$ You can view it from a different perspective. First, you sit the first person. There is one combination to do that since all sits are identical. Then, there are $6!$ combinations to sit the other people. Therefore you have $1 \\cdot 6!$ combinations in total.", "\\\\ 7 & 96 \\end{tabular}$$ where by symmetry, $N_i$ for $i=5,6,7$ are the same for $i=3,2,1$ respectively. Thus, $$P(E_1E_2E_3) = 2!(2 \\cdot 96 + 2 \\cdot 48 + 3 \\cdot 64) / 8! = 1/42$$ Finally, there are 4! ways to order all four couples such that all couples are together. Hence, $P(E_1E_2E_3E_4) = 4!/8! = 1/1680$. And so, $$P(E_1 \\cup E_2 \\cup E_3 \\cup E_4) = 4\\cdot\\frac{1}{4} - 6\\cdot\\frac{1}{14} + 4\\cdot\\frac{1}{42} -\\frac{1}{1680} = \\frac{1119}{1680}$$ Now obviously I did something wrong somewhere, but where? P: 70 You are definately on the right track but it seems you fell into error in computing the very last term in your application of the inclusion-exclusion formula. Otherwise your work seems correct. In any event I'll run through my reasoning of this problem thus we may compare notes. For P(E1) we have the case where one couple sits together. If we think of the couple which sits together as a single entity there are 7! ways they can be permuted among the remaing 6 persons. In other words I am looking at the couple that sits together as a single entity (as 1 person) permuted among 6 others, in other words a permutation of 7 \"persons\". In other words a permutation of the group C P1 P2 P3 P4 P5 P6 where", "672 = 3507$$. This is based on the answer by gandalf61, who got the general idea right but has (at the moment) errors in the details. I'll repeat the steps they made, in case the answer is deleted or heavily modified. From gandalf61 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. The number of $$k$$-cycles (or circles of length $$k$$) ($$k \\ge 3$$) from $$n$$ people is $$\\frac{n(n-1)(n-2)\\ldots(n-k+1)}{2k}$$ Thats because you can choose one of $$n$$ people for the 'first' person, then one of $$n-1$$ people for the second person, a.s.o until you choose the $$k$$-th person from the remaining $$n-k+1$$ people, after that the cycle closes with the first person. But the resulting formula $$n(n-1)(n-2)\\ldots(n-k+1)$$ this yields is heavily overcounting the cycles. One effect is that cycles don't have a 'first' person, they are cycles! So if you choose person 1 first, person 2 second and person 3 last in a 3-cycle, this is the same as choosing person 2 first, person 3 second and person 1 last, and the same for person 3, then person 1 and finally", "the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in 3! ways. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements Example $$\\PageIndex{6}$$ You have 4 math books and 5 history books to put on a shelf that has 5 slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books? Solution We first do the problem using the multiplication axiom. Since the math books go in the first three slots, there are 4 choices for the first slot, 3 choices for the second and 2 choices for the third. The fourth slot requires a history book, and has five choices. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The choices are shown below. 4 3 2 5 4 Therefore, the number of permutations are $$4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 = 480$$. Alternately, we can see that $$4 \\cdot 3 \\cdot 2$$ is really same as 4P3, and $$5 \\cdot 4$$ is 5P2. So the answer can" ]

[ "# How many ways are there for people to queue? I'm stuck at the following combinatorics problem: Fifteen people queue up for cinema tickets at five (different) sales points. In how many ways can they stand in queue behind one another, if the queues can be of different length but at least one person is waiting at every sales point? My idea was to place one person at every sales point first and then place the remaining ten people. Since placing the first five is an ordered drawing without repetition I get $15*14*13*12*11$ possibilities to place a person at each sales point. Now I need to figure out how many ways there are to divide the group of ten people into five distinct groups but I am not sure how to go about this. Looking forward to your ideas and explanations - We can encode the five nonempty customer groups in the following way as a $01$-sequence: $$0\\ \\ldots\\ |0\\ \\ldots\\ |0\\ \\ldots\\ |0\\ \\ldots\\ |0\\ \\ldots\\quad.$$ Here each $0$ denotes a person, each $\\ldots$ a nonnegative number of zeros, and each $|$ a separator between groups. Since there have to be $15$ zeros in all we have to distribute $10$ remaining zeros onto the five $\\ldots\\$. This can be encoded as a word containing $10$ zeros and", "# Combinatorics analysis A small theater only have $10$ numbered seats. However there is a row of $30$ people to enter into the theater. A group of $3$ friends want to enter into the theater in the same movie session. In how many different ways the $30$ can enter into the theater in order to the $3$ friends could enter in the same session together? Well, if the theater only have $10$ seats, and there are $30$ people to see the movie. There should be $3$ sessions. As the seats are numbered, in each session the $3$ friends can be sit in $Pr(10,3)$ different ways. But there will be $3$ sessions, they can enter in the $1st$, $2nd$ or in the $3th$ session. So there are $3 \\cdot Pr(10,3)$. But the book's solution is $3! \\cdot 10 \\cdot 27!$, that I don't understand. Can you help me?Thanks - We also need to consider the permutations of the other 27 people in the row to enter the theater, so the factor 27! makes sense. But for the other factors I like your solution. – hardmath May 5 '12 at 13:06 There are $\\binom{10}{3}$ ways to choose the places in the lineup for the $3$ friends so that they can all go to the first showing, and the same number", "ways. Then the 6 people can be lined up in $6! = 720$ ways. Answer: . $70 \\times 720 \\:=\\:50,\\!400$", "In how many ways can 7 people line up at a ticket window of a cinema hall? Question: In how many ways can 7 people line up at a ticket window of a cinema hall? Solution: To find: number of arrangements of 7 people in a queue. Here there are 7 spaces to be occupied by 7 people. Therefore 7 people can occupy first place. Similarly, 6 people can occupy second place and so on. Lastly, there will be a single person to occupy the 7 positions. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, permutation of 7 different objects in 7 places is $P(7,7)=\\frac{7 !}{(7-7) !}$ $=\\frac{7 !}{0 !}=\\frac{5040}{1}=5040$ Therefore, the number of possible ways is 5040", "Another Example! You can use compatible numbers to estimate differences. 375 and 150 are compatible numbers for 372 and 149. Guided Practice Do You Understand? Question 1. Does rounding to the nearest ten or nearest hundred give an estimate closer to the exact answer for 295 – 153? Here the digit of the one’s place of 295 is 5 now round to the nearest multiple of ten which is greater than the number. 295 round to the nearest ten to estimate is 290 The digit in the one’s place of 153 is3 now round to the nearest multiple of ten which is less than the number. 153 round to the nearest ten to estimate is 140 290 – 140 =150 Question 2. A theater sells 408 tickets. Two hundred seventy-three people arrive. About how many more people are expected to arrive? Use compatible numbers. Show your work. Given, A theater sells tickets = 408 People arrive to theater = 253 Expected people to arrive = 408 – 253 = 155 The compatible numbers for 408 and 253 is 410 and 250 410 – 250 = 150 Do You Know How? In 3-6, estimate. Use rounding or compatible numbers. Tell how you made each estimate. Question 3. 321 – 182 Compatible numbers are pair of numbers that are easy", "# A Game of Change Eight people line up in a queue to buy a ticket. Four of these people hold a $$5$$ note and the other four hold a $$10$$ note each. The price of a ticket is $$5$$. The clerk starts up with no cash in hand, but she manages to always give the correct change (that is, whenever a person that has a $$10$$ note comes in, the clerk always has a $$5$$ note in hand). Find the number of ways the people can line up for this to happen. ×", "number of people the theater can hold is 325. 1. $p> 325$ 2. $p>=325$ 3. $p<325$ 4. $p<=325$ Which represents the value of x in $6 - 4x <= 26$? 1. $x <= -8$ 2. $x >= -8$ 3. $x <= -5$ 4. $x >= -5$ 1. $n> 80$ 2. $n>=80$ 3. $n<80$ 4. $n<=80$ The temperature was above $65deg$. 1. $t > 65$ 2. $t>=65$ 3. $t<65$ 4. $t<=65$", "the total number of seats in the theater is 1500 Question 18. Multi-Step There are 48 beads in a package. Mariah bought 4 packages of blue, 9 packages of gold, 6 packages of red, and 2 packages of silver beads. Which is the best estimate of the total number of beads she bought? (A) 1,000 (B) 800 (C) 1,200 (D) 2,400", "1. ## Line 'em up! There are 34 people in line to get into a theater. Admission cost is 5 dollars. Of the 34 people, 28 have only a 5 dollar bill and 6 have only a 10 dollar bill. To complicate matters, the ticket office has an empty cash register. In how many ways can the people line up so that change is available when needed? Example with 4 people, 2 (a and b) with 5's and 2 (c and d) with 10's: a b c d a b d c b a c d b a d c a c b d a d b c b c a d b d a c So 8 ways... 2. ## Re: Line 'em up! You only need to look at the first 11 positions. There are 34! arrangements without restriction. There are $6\\times 33!$ arrangements that fail because the first person has only a 10. There are $28\\times 6\\times 5 \\times 31!$ ways that the first person has a 5 and the second and third both have 10s. Next we want 5,10,5,10,10,anything (because any other arrangement that fails in the first 5 people was already counted above) $28\\times 6\\times 27 \\times 5 \\times 4 \\times 29!$ We see the pattern: $$\\displaystyle 34!- \\sum_{n=1}^6\\dfrac{28!}{(29-n)!}\\cdot \\dfrac{6!}{(6-n)!}\\cdot (35-2n)!$$ The chances", "# Combinatorics riddle: Sorting people in a cinema line. Say i want to go to the cinema. There are two types of movies. 1. Action movie. 2. Drama movie. Because action is more interesting it costs 50$. And the cost for drama is 10€. There are 200 people which wanna go action movie. 100 of those have 50 dollar bills, And the other 100 people got 100 dollar bill. In addition there are 150 people which wanna buy a ticket for the drama movie, So that 80 have 10€ bill, and the other 70 got 20€ bills. At the start of the day the cinema is empty, In how many ways can you sort the 350 people in a line so that anyone would be able to get an exact change for his money? • Is there a subtle statement in the fact that action movies are paid in US currency and drama movies in European currency? :-) – joriki Apr 20 '13 at 9:34 ## 1 Answer The first queue is exactly equivalent to arranging 100 pairs of brackets in a line: think of anyone with a 50\\$ bill as an openning bracket, and with a 100\\$- as a closing bracket. You want that at any point the number of opening brackets is not less than the", "# Another counting problem: There are 34 people in line to get into a theater.... #### Jomo There are 34 people in line to get into a theater. Admission costs 50 cents. Of the 34 people, 28 of them have a 50-cent piece and 6 have a $1 bill. Suppose that the box office has an empty cash register, how many ways can the people line up so that change is available when needed? Here is what I see. The first person must pay 50 cents. After this 1st person all we need is NOT to have at any point the number of bill payers exceed the number of coin payers. I put this on a x-y coordinate system. I think that I should start at (1,0) and go to (28,6) without crossing y=x (touching is ok) Now you can go through (6,0) (and then go to the right) and proceed to (28,6). You can do this in (5C5)*(27C21)ways Or you can go through (6,1) (and then go to the right) and proceed to (28,6). You can do this in (6C5)*(26C21) ways Or you can go through (6,2) (and then go to the right) and proceed to (28,6). You can do this in (7C5)*(25C21) ways Or you can go through (6,3) (and then go to the right) and", "be located under the MATH menu with probability commands. ### Could we have solved using the Multiplication Principle? Yes. We could have multiplied $15\\cdot 14\\cdot 13\\cdot 12\\cdot 11\\cdot 10\\cdot 9\\cdot 8\\cdot 7\\cdot 6\\cdot 5\\cdot 4$ to find the same answer. A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following. ### Try It 6 How many ways can the 7 actors line up? Solution ### Try It 7 How many ways can 5 of the 7 actors be chosen to line up? Solution", "# Theater The theater has in each row with 19 seats. Ticket to the first 10 rows is for 30 USD. In next rows is for 15 USD. The theater was completely sold out. Revenue was 12255 USD. How many rows are in the theater? Result n = 33 #### Solution: $19 \\cdot 10 \\cdot 30 + 19 \\cdot (n-10) \\cdot 15 = 12255 \\ \\\\ \\ \\\\ n = \\dfrac{ 12255-30\\cdot 10} { 15 \\cdot 19} + 10 = 33$ 19*10*30 + 19*(n-10)*15 = 12255 285n = 9405 n = 33 Calculated by our simple equation 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 Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? ## Next similar math problems: 1. Profitable company Three businessman decide to open up their own company. They agree to distribute the yearly profits made in the same ratio as their initial investments. They invest R 50", "# At the theater, one section of seats has 8 rows with 12 Seats in each row. In the center of each of the first 3 rows are 4 broken seats that cannot be used. How many seats can be used in the section? $84$ \"8 rows with 12 seats in each row\" means a total of $8 \\setminus \\cdot 12 = 96$ seats in the theater. \"in the center of the first 3 rows are 4 broken seats that cannot be used\" means broken seats $= 3 \\setminus \\cdot 4 = 12$ total So we have $96 - 12 = 84$ usable seats.", "the number 65 into 60 + 5 STEP3 We must divide the two parts of the number (dividend) with the divisor. STEP4 (60 ÷ 5) + (5 ÷ 5) STEP5 Add quotients of the above 12 +1= 13 Spiral Review Question 3. Last Saturday, there were 1,486 people at the Cineplex. There were about the same number of people in each of the 6 theaters. Which is the best estimate of the number of people in each theater? Options: a. between 20 and 30 b. between 80 and 90 c. between 100 and 200 d. between 200 and 300 Answer: d. between 200 and 300 Explanation: Total number of people at the Cineplex= 1,486 people Number of theatres = 6 Number of people at each theatre= estimate of the number of people 1,486 ÷ 6 What number close to 1,486 is easy to divide by 6? 1,488 is close to 1,486. What basic fact can you use? 1,488 ÷ 6 Choose 1,488 because it is close to 1,486 and can easily be divided by 6. 1,488 ÷ 6 = 248 1,486 ÷ 6 is about 248 What number close to 1,486 is easy to divide by 6? 1,482 is close to 1,486 . What basic fact can you use? 1,482 ÷ 6 Choose 1,482 because it is close", "# Movie Tickets Discrete Mathematics Level 4 In a movie theater, tickets cost $$5$$. 10 people are standing in line to buy a ticket, which, five of them have a $$5$$ bill, four have a $$10$$ bill and one has a $$20$$ bill. Considering that in the box office there is no money in the cashier initially, how many ordered sets can be formed with these 10 people in a way that there is always change for each person who buys a ticket in the box office? Consider there is $$10\\binom{9}{4}$$ ways to arrange them without restriction, in other words, if two people have the same bill, then they're equal. ×", "ways to choose the four men and four women to get tickets and then $4!$ ways to pair them up, for a total of $$\\binom64\\binom744!=15\\cdot35\\cdot24=12600\\;.$$ (Of course since $\\binom62=\\binom64$ and $\\binom73=\\binom74$, it’s hardly surprising that these both work!) You could also reason that there are $6$ ways to pick the first man to get a ticket, $7$ ways to pick his partner, $5$ ways to pick the second luck man,$6$ ways to pick his partner, and so on, making a total of $$6\\cdot7\\cdot5\\cdot6\\cdot4\\cdot5\\cdot3\\cdot4\\;,$$ but then you’d have to realize that you’ve actually counted every possible set of four couples $4!$ times, once for each of the $4!$ orders in which you could have picked the same set of four couples. Thus, the correct answer by this approach is $$\\frac{6\\cdot7\\cdot5\\cdot6\\cdot4\\cdot5\\cdot3\\cdot4}{4!}\\;,$$ which is of course the same $12600$ as before. - +1 for explanatory answer. – user21436 Feb 21 '12 at 1:17 (This answer goes into a little more detail than Kannappan's and Brian's, in case you don't know what binomial coefficients, etc, are. Also, it includes the obvious My Little Pony reference. This question was a reference to Episode 3, \"The Ticket Master\", right?) First, let's choose which of the $7$ female friends will get to go to the Grand Galloping Gala. There's $7$ ways to choose the first", "harder. It is, however, a standard problem with a very nice answer. – André Nicolas Nov 14 '12 at 15:14 Like you suggested, the probability that a single person does not receive their own phone is $11/12$, so if you know that's incorrect, the question is most likely \"What is the probability that no person receives their own phone?\" Consider how many different ways there are to distribute 12 phones to 12 people. For the first person, you have 12 phones you could give him/her, for the second you have 11, since you already gave one away, and for the third you have 10, and so on. If we multiply all the options of different phones that each person could get this way, we get what we wanted to find: $12\\cdot 11\\cdot \\ldots \\cdot 2 \\cdot 1 = 12! = 479,001,600$ different ways. This is called a permutation; we are permuting the 12 phones throughout 12 people. Now in this specific problem, we are looking for the number of different ways to distribute $n$ phones to $n$ people such that no single person receives the phone that they started with. Let's say there are $D(n)$ ways to do this and approach it the same way we did above. The first person has $12-1 =11$ phones that they could", "could not be solved with the permutation formula. Note: Here you have only begun to explore counting problems. For more information about combinations, permutations, and other types of counting problems, consult a Probability text. Example A You are going on a road trip with 4 friends in a car that fits 5 people. How many different ways can everyone sit if you have to drive the whole way? Solution: A decision chart is a great way of thinking about this problem. You have to sit in the driver’s seat. There are four options for the first passenger seat. Once that person is seated there are three options for the next passenger seat. This goes on until there is one person left with one seat. $1 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=24$ Example B How many different ways can the gold, silver and bronze medals be awarded in an Olympic event with 12 athletes competing? Solution: Since the order does matter with the three medals, this is a permutation problem. You will start with 12 athletes and then choose and arrange 3 different winners. $_{12}P_3=\\frac{12!}{\\left(12-3\\right)!}=\\frac{12!}{9!}=\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot \\ldots}{9 \\cdot \\ldots}=12 \\cdot 11 \\cdot 10=1320$ Note that you could also use a decision chart to decide how many possibilities are there for gold", "# For the following example, is order meaningful or not? #### user1234 ##### New Member Consider a hypothetical example. There are $$12$$ tickets for a documentary film and exactly $$12$$ people to buy the tickets (one person can buy only one ticket). There are $$3$$ ticketbooths for selling the tickets and each ticketbooth will sell exactly $$4$$ tickets. Suppose the researcher labels the $$12$$ people with ID $$1,2,\\ldots, 12$$ according to their arrivals. Say, the first arrival buys his/her ticket from one of the $$3$$ ticketbooths. Then, the second arrival buys his/her ticket from one of the $$3$$ ticketbooths. The second arrival can buy his/her ticket from the same ticket booth that the first arrival had bought or from one of the other two ticketbooths. In this way, the last arrival buys his/her ticket. The researcher has recorded from which ticketbooth which arrivals have bought the tickets. Suppose from ticketbooth A, arrival #3, #5, #6, #12 have bought the tickets. From ticketbooth B, arrival #1, #2, #9, #11 have bought the tickets. From ticketbooth C, arrival #4, #7, #8, #10 have bought the tickets. In how many ways the $$12$$ people can buy tickets from the $$3$$ ticketbooths? #My Attempt: If inside a ticketbooth the order of the ID doen't matter, then the number of ways the $$12$$" ]

[ "# Yes, 3+7=10 If the least prime factor of positive integers $$a$$ and $$b$$ are $$3$$ and $$7$$ respectively, what must be a factor of $$(a+b)~$$ ? ×", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "# Under 50 How many positive integer under 50 have exactly 2 distinct primes factors? No repeats, $$(2^{2})$$$$(3^{2})$$ doesn't count. ×", "i.e. any positive integer whose only prime factors are $2$, $3$ or $5$. That seems to me to be good value for the sake of remembering a couple of numbers.", "# 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. ×", "3. All positive integers up to 5 are factors of 5! and so factors of 6! and 7! as well. Since the expression is cubed, each of those, squared, will be a perfect square dividing the expression. Also, both 2 and 3 are in 5! so 6 is a factor of 5!, 6!, and 7!. $6^2$ is also a perfect square dividing the expression.", "# Factors Galore! Let $$A,B$$ and $$C$$ denote the smallest positive integer with 7, 8 and 9 factors respectively. Find $$A+B+C$$. ×", "or 5 (3, 7, 11 etc.)", "2 * 2 * 7 and also the two prime determinants are 2, 7", "(2) Subtract: 5-2 = 3 Sign of the bigger integer (-5): – Answer: -3 Positive integer + Positive integer • Example: (+5) + (+2) = +7 Negative integer + Negative integer • Example: (-5) + (-2) = -7", "# Always, sometimes, or never? Fill in each blank with either always, sometimes, or never. Hover to see the answer. 1. A positive whole number has at least one factor. 2. A whole number has infinitely many factors. 3. Positive whole numbers have 2 as a factor. 4. Positive whole numbers have 5 as a factor. 5. Positive whole numbers have 0 as a factor. 6. Positive whole numbers have 1 as a factor. 7. The highest common factor of two positive integers $$m$$ and $$n$$ is equal to $$m$$. 8. The highest common factor of two positive integers $$m$$ and $$n$$ is greater than $$m$$ or $$n$$. 9. A square number has an odd number of factors. 10. A cube number has an odd number of factors. 11. The fourth power of an integer has an odd number of factors.", "# Multiples of 5 and 7 Number Theory Level 3 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "be displayed as negative factors, in this case $$\\Large -1, -5$$. ## Is 5 prime? A prime number is one that only has two factors, 1 and itself. Yes, 5 is prime because its only two factors are 1 and itself, 5.", "= 8 × 6 48 = 23× 2 × 3 48 = 24 x 3 So we know that y3 has at least four 2’s and at least one three among its prime factors. So our answer is definitely \"yes,\" and Statement (1) is Sufficient. Eliminate choices (B), (C), and (E). Statement (2): For y2 to be a multiple of 30, it must be that yields an integer. Again, let’s go through prime factorization. 30 = 6 × 5 30 = 2 × 3 × 5 So, y2 has at least one factor of 2, at least one factor of 3, and at least one factor of 5. Just as y3 had to have all of its factors in groups of threes, y2 must have its factors in groups of twos. So we can deduce that y2 has at least two factors of 2, at least two factors of 3, and at least two factors of 5. y will have the same factors as y2 has, but half as many of each. So, y must have at least one factor of 2, at least one factor of 3, and at least one factor of 5. So, y definitely has one factor of 3. But it might have only one factor of 2, or it might have two factors", "digits which is a perfect SquareFind Least number of 6 digits which is a perfect SquareFind Greatest number of 7 digits which is a perfect SquareFind Least number of 4 digits which is a perfect CubeFind Greatest number of 5 digits which is a perfect CubeFind Least number of 6 digits which is a perfect CubeFind Greatest number of 7 digits which is a perfect Cube SolutionExample", "is an odd number which leads to many possibilities such as 1,3,5,7,9,...etc. Each possible value of m leads to a different GCF. Hence not sufficient. The answer is therefore E imo. Re: If m and n are positive integers, what is the greatest common factor [#permalink] 02 Sep 2019, 12:34 Display posts from previous: Sort by", "# Fun Factorials For any positive integer n, we define n! as the factorial of n. Some numbers are equal to the sum of the factorials of their digits. For example 40585=4+0+5+8+5. Find such a number with three digits. ×", "an odd number of factors only if it's a perfect square. The largest 3-digit perfect square is 31 x 31 = 961 (since 32 x 32 = 1,024). _________________ # Jeffrey Miller [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. GMAT Club Legend Joined: 12 Sep 2015 Posts: 4018 Re: What is the largest 3 digit number to have an odd number of factors? [#permalink] ### Show Tags 03 May 2018, 10:23 Top Contributor Bunuel wrote: What is the largest 3 digit number to have an odd number of factors? A. 625 B. 729 C. 841 D. 943 E. 961 IMPORTANT CONCEPT: All positive integers have a EVEN number of positive factors EXCEPT integers that are squares of integers Squares of integers (e.g., 1, 4, 9, 16, 25, 36, etc) have an ODD number of positive factors. So, the question is really asking us \"What is the largest 3 digit number that is the SQUARE OF AN INTEGER?\" Let's find out. 30² = 900, so 900 will have an odd number of positive factors 31² = 961, so 961", "positive two-digit numbers are there whose square and cube both end in the same digit? ## Digitless factor Ted thinks of a three-digit number. He removes one of its digits to make a two-digit number. Ted notices that his three-digit number is exactly 37 times his two-digit number. What was Ted's three-digit number? ## Backwards fours If A, B, C, D and E are all unique digits, what values would work with the following equation? $$ABCCDE\\times 4 = EDCCBA$$ ## Cube multiples Six different (strictly) positive integers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of 6. What is the smallest possible sum of the six numbers? ## Archive Show me a random puzzle ▼ show ▼", "would exists a prime factor that has at least an odd exponent" ]

[ "## 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) $2\\cdot 3\\cdot 5\\cdot 7=210$", "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 $15120$.", "# 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", "+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", "+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", "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,", "by 25 its last two digits must be 00, 25, 50, or 75, so 377,910 is NOT divisible by 25). Thus x must have 5 as its factor. Check 377,910 for divisibility by 11: (7+9+0)-(3+7+1)=5, so 377,910 is NOT divisible by 11, thus x must have 11 as its factor. Therefore the least value of x is $$2*5*11=110$$. Please how did you arrive at \"Therefore the least value of x is 2∗5∗11=1102∗5∗11=110.\" ? Manager Joined: 27 Oct 2019 Posts: 54 Show Tags 22 Nov 2019, 04:02 Bunuel wrote: pclawong wrote: Bunuel wrote: 7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible? A. 1 B. 2 C. 3 D. 4 E. 5 We are told that the greatest common factor of two integers is 25. So, these integers are $$25x$$ and $$25y$$, for some positive integers $$x$$ and $$y$$. Notice that $$x$$ and $$y$$ must not share any common factor but 1, because if they do, then GCF of $$25x$$ and $$25y$$ will be more that 25. Next, we know that $$25x+25y=350$$ --> $$x+y=14$$ --> since $$x$$ and $$y$$ don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs", "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,", "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", "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", "## 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", "one.... Any insight would help immensely....... $$5y=x^2+x=x(x+1)$$, and $$x$$ and $$y$$ are positive integers, $$x>5$$. Then $$x+1>5$$, therefore necessarily $$x+1$$ is a factor of $$y$$. Thus III is TRUE. Since $$x$$ is a prime greater than 5, $$x$$ cannot be divisible by 5. It follows that $$x+1$$ must be divisible by 5, which means $$x+1=5n$$ for some positive $$n$$. Then $$x=5n-1$$ must be odd (a prime that is greater than 5), so $$n$$ must be even, which means $$n=2k$$ for some positive integer $$k$$. From $$5y=(5n-1)(5\\cdot{2k})$$ it follow that y is divisible $$2(5n-1)=2x$$. Thus II is also TRUE. I isn't necessarily true: for $$x = 19, \\,\\,5y = 19\\cdot{20}$$ and $$y=19\\cdot{4}$$ is not divisible by 5. Answer to the question is B, not E. Consider x=19. _________________ Intern Joined: 26 Jul 2012 Posts: 6 Re: If x is a prime number greater than 5, y is a positive [#permalink] ### Show Tags 27 Oct 2012, 16:03 Bunuel wrote: EvaJager wrote: anon1 wrote: If x is a prime number greater than 5, y is a positive integer, and 5y=x^2+x, then y must be divisible by which of the following? I. 5 II. 2x III. x+1 A. I only B. II only C. III only D. I and II only E. II and III only I'm killing myself.... I got", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "thought you were trying to show that $2 \\cdot 3 \\cdot 5 \\cdot ~ ... ~ \\cdot p + 1$ is not divisible by any prime p in the multiplicative sequence. -Dan • July 17th 2007, 10:53 AM Discrete no that's not the case, I am trying to prove that p1 x p2 x p3 x pn + 1 is a prime number for every positive integer n. where p1 x p2 x pn is the first n prime number.. I am not trying to proof any divisible here • July 17th 2007, 11:47 AM ThePerfectHacker Quote: Originally Posted by Discrete no that's not the case, I am trying to prove that p1 x p2 x p3 x pn + 1 is a prime number for every positive integer n. where p1 x p2 x pn is the first n prime number.. I am not trying to proof any divisible here That is false. $2+1 = \\mbox{ prime}$ $2\\cdot 3 + 1 = \\mbox{ prime}$ $2\\cdot 3\\cdot 5 + 1 = \\mbox{ prime}$ $2\\cdot 3\\cdot 5\\cdot 7 + 1 = \\mbox{ prime}$ $2\\cdot 3\\cdot 5\\cdot 7\\cdot 11+1 = \\mbox{ prime}$ $2\\cdot 3\\cdot 5\\cdot 7\\cdot 11\\cdot 13 + 1 = 59\\cdot 509$ • July 17th 2007, 11:50 AM CaptainBlack Quote: Originally Posted by ThePerfectHacker That is false. $2+1 =", "= 3628799!7!5!3!$. 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", "$17 \\cdot 152 - 21 \\cdot 123 = 1$. ### Compute the GCD of 7 and 0 Determine whether 7 and 0 are relatively prime. More generally, compute their greatest common divisor. Note that 7 divides 7 since $1 \\cdot 7 = 7$ and 7 divides 0 since $0 \\cdot 7 = 0$. Now suppose $k$ is an integer dividing both 7 and 0; certainly $k$ divides 7. Thus 7 is a greatest common divisor of 7 and 0. In particular, since $7 \\neq \\pm 1$, 7 and 0 are not relatively prime.", "# Seventeen positive integers Find 17 positive integers such that no four of them have, pairwise, a common divisor greater than 1, but, likewise, no four of them are, pairwise, relatively prime. Do so in such a way that the largest of those numbers is as small as possible. • Can you clarify what you mean by \"no four of them have, pairwise, a common divisor greater than 1, but, likewise, no four of them are, pairwise, relatively prime\"? Let $a,b,c,d$, be four of the integers: if a pair $a,b$ do not share a common divisor greater than $1$ then they are automatically coprime. – Mark Murray Jul 25 at 16:05 • And what does \"four of them, pairwise\" mean? – aschepler Jul 25 at 16:50 • Yeah, if we're only looking at pairwise properties, taking them in sets of four is totally irrelevant. – Jaap Scherphuis Jul 25 at 19:11 • Are the integers unique? – Weather Vane Jul 25 at 19:12 • Consider the six numbers 10, 14, 105, 195, 231, and 286. No matter how you choose four of them (15 ways of doing this), you will be able to find a pair among the four which have no common divisor greater than 1, and a pair which does. But I request seventeen such numbers,", "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", "# Question #c64ae Dec 26, 2017 $212 = {2}^{2} \\cdot 53$ #### Explanation: First, note that $212$ is even and $212 = 2 \\cdot 106$. We next note that $106$ is even and $106 = 2 \\cdot 53$. The number $53$ is odd, but is it prime? You can test this by seeing if it's divisible by any prime numbers between $2$ and $\\sqrt{53} \\approx 7.28$. We already know $53$ is not divisible by 2 since it's not even. It's not divisible by 3 since $\\frac{53}{3} \\approx 17.67$ (also its digits add up to 8, which is not divisible by 3). It's not divisible by 5 since it does not end in a 0 or a 5. And it's not divisible by 7 since $\\frac{53}{7} \\approx 7.57$. Therefore, 53 is prime and the final answer is $212 = {2}^{2} \\cdot 53$.", "+0 # Number Theory 0 65 1 Let x be the smallest positive integer such that 1584 * x is a perfect cube, and let y be the smallest positive integer such that xy is a multiple of 1584. Compute y Feb 1, 2022 #1 +1209 +1 1584 = 2^4 * 3^2 * 11 To be a perfect cube, a number's prime factors' exponents must all be multiples of 3. So, the smallest value of x is: $$\\frac{2^6 \\cdot 3^3 \\cdot 11^3}{2^4 \\cdot 3^2 \\cdot 11} = \\boxed{1452}$$ In other terms, x = 1452 = 2^2 * 3 * 11^2. The product xy must be a multiple of 1584, so, xy = 2^2 * 3 * 11^2 * y = 2^4 * 3^2 * 11 * n y must have 2^2 * 3 as a factor, so the smallest value of y is 12. Feb 1, 2022" ]

[ "+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", "+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", "# 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", "## 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) $2\\cdot 3\\cdot 5\\cdot 7=210$", "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 $15120$.", "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", "= 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", "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", "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", "# positive integer with 13 factors • Aug 25th 2008, 09:00 AM reagan3nc positive integer with 13 factors What is the lowest positive integer with exactly 13 factors, both the number and 1 count as a factor. Thanks • Aug 25th 2008, 10:06 AM Soroban Hello, reagan3nc! Quote: What is the lowest positive integer with exactly 13 factors? Both the number and 1 count as factors. Most numbers have an even number of divisors. When factoring, the divisors appear in pairs. . . $24 \\;=\\;\\begin{Bmatrix}1\\cdot24 \\\\ 2 \\cdot 12 \\\\ 3\\cdot 8 \\\\ 4 \\cdot 6 \\end{Bmatrix}\\;\\;\\hdots\\text{ 8 factors}$ The only numbers with an odd number of factors are squares. . . $36 \\:=\\:\\begin{Bmatrix}1\\cdot36 \\\\ 2\\cdot 18 \\\\ 3\\cdot12 \\\\ 4\\cdot9 \\\\ 6\\cdot6\\end{Bmatrix}\\;\\;\\hdots\\text{9 factors}$ The least integer with 13 divisors is: . $2^{12} \\:=\\:\\boxed{4096}$ . . The divisors are: . $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096$", "# 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", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "number of factors of $10^{11}$. In general, for any positive number $a$, we can write it in its prime factorization as $$a = p_1^{k_1} p_2^{k_2} \\cdots p_t^{k_t}$$ where the $p_i$'s are all prime and distinct, and the $k_i$'s are all positive integers, then the number of positive factors is $(k_1 + 1)(k_2 + 1) \\cdots (k_t + 1)$. That is, any positive factor of $a$ must be of the form $$p_1^{m_1} p_2^{m_2} \\cdots p_t^{m_t}$$ where now I will allow the $m_i$'s to be 0 or positive, but $m_i \\leq k_i$. And, by the uniqueness of a prime factorization, two distinct choices of $m_i$'s lead to two distinct factors and vice versa. Therefore, we simply need to count the number of ways to choose the $m_i$'s. Clearly, there are $k_i + 1$ choices for $m_i$, which are $0, 1, 2, \\ldots, k_i$. Therefore we have the formula above. In this case, we have $10^{11} = 2^{11} \\cdot 5^{11}$ so there are $12 \\cdot 12 = 144$ positive factors as you said. But, if you allow negative multiples as well, then the answer is actually 2 times this, or 288. That is, for each positive multiple you counted, there is also a corresponding negative multiple.", "\\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 =", "## 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", "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", "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$$ must have one factor of", "three: 9, 21, 27, and 33 because I have had their factorizations memorized since third grade; 81 because it’s a power of three so I have it memorized too; and 39, 63, 69, 93, and 99 because they consist of two digits each of which is a multiple of three. As you probably know, there is also a simple test for determining divisibility by three: just add the digits and see whether the result is divisible by three. This test identifies a few more “nonobvious” multiples of three which you might otherwise think are prime: $51 = 3 \\cdot 17$, $57 = 3 \\cdot 19$, and $87 = 3 \\cdot 29$. 3. What’s left? There are only three numbers less than 100 which are composite and not divisible by 2, 3, or 5: • $49 = 7 \\cdot 7$ and $77 = 7 \\cdot 11$ are easy to spot. • $91 = 7 \\cdot 13$ is the one I always used to forget about—I call it a “fool’s prime” because it “looks” prime but isn’t. It’s worth just memorizing the fact that it is composite (and its factorization). So, to sum up, faced with a number less than 100 that I want to test for primality, I can quickly rule it out if it is divisible by 2,", "one.... Any insight would help immensely....... $$5y=x^2+x=x(x+1)$$, and $$x$$ and $$y$$ are positive integers, $$x>5$$. Then $$x+1>5$$, therefore necessarily $$x+1$$ is a factor of $$y$$. Thus III is TRUE. Since $$x$$ is a prime greater than 5, $$x$$ cannot be divisible by 5. It follows that $$x+1$$ must be divisible by 5, which means $$x+1=5n$$ for some positive $$n$$. Then $$x=5n-1$$ must be odd (a prime that is greater than 5), so $$n$$ must be even, which means $$n=2k$$ for some positive integer $$k$$. From $$5y=(5n-1)(5\\cdot{2k})$$ it follow that y is divisible $$2(5n-1)=2x$$. Thus II is also TRUE. I isn't necessarily true: for $$x = 19, \\,\\,5y = 19\\cdot{20}$$ and $$y=19\\cdot{4}$$ is not divisible by 5. Answer to the question is B, not E. Consider x=19. _________________ Intern Joined: 26 Jul 2012 Posts: 6 Re: If x is a prime number greater than 5, y is a positive [#permalink] ### Show Tags 27 Oct 2012, 16:03 Bunuel wrote: EvaJager wrote: anon1 wrote: If x is a prime number greater than 5, y is a positive integer, and 5y=x^2+x, then y must be divisible by which of the following? I. 5 II. 2x III. x+1 A. I only B. II only C. III only D. I and II only E. II and III only I'm killing myself.... I got", "by 25 its last two digits must be 00, 25, 50, or 75, so 377,910 is NOT divisible by 25). Thus x must have 5 as its factor. Check 377,910 for divisibility by 11: (7+9+0)-(3+7+1)=5, so 377,910 is NOT divisible by 11, thus x must have 11 as its factor. Therefore the least value of x is $$2*5*11=110$$. Please how did you arrive at \"Therefore the least value of x is 2∗5∗11=1102∗5∗11=110.\" ? Manager Joined: 27 Oct 2019 Posts: 54 Show Tags 22 Nov 2019, 04:02 Bunuel wrote: pclawong wrote: Bunuel wrote: 7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible? A. 1 B. 2 C. 3 D. 4 E. 5 We are told that the greatest common factor of two integers is 25. So, these integers are $$25x$$ and $$25y$$, for some positive integers $$x$$ and $$y$$. Notice that $$x$$ and $$y$$ must not share any common factor but 1, because if they do, then GCF of $$25x$$ and $$25y$$ will be more that 25. Next, we know that $$25x+25y=350$$ --> $$x+y=14$$ --> since $$x$$ and $$y$$ don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs" ]

[ "‘+’ 14’.", "23, 27 and 31 are the arithmetic means between the two extreme terms 3 and 35. (ii) In the Arithmetic Progression {-5, -2, 1, 4, 7, 10, 13, 16, 19} the terms -2, 1, 4, 7, 10, 13 and 16 are the arithmetic means between the two extreme terms -5 and 19. (iii) In the Arithmetic Progression {85, 80, 75, 70, 65, 60, 55, 50, 45} the terms 80, 75, 70, 65, 60, 55, and 50 are the arithmetic means between the two extreme terms 85 and 45. Arithmetic Progression", "the product of the 2 numbers is 36. Sep 14, 2018", "means 14 is going to be divided by something. What is 9p.", "R ⇒ ( b, c ) and 34!", "x = 3 is 36.", "of 6 Arithmetic questions and 20...", "28/2=14", "in (15) is achieved at (16) This further implies that .", "### The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is. A. 414 B. 420 C. 620 D. 621 $\\begin{array}{l}\\text{Sum of numbers}\\\\ \\text{= (41.4 × 15)}\\\\ \\text{= 621}\\end{array}$", "13, 22, 23}, {21}}", "in circles from zero to 360. Then your answer is fourteen.", "(in years)? A) 26 B) 24 C) 28 D) 22", "There are n Arithmetic means between 13 and 61. If ratio of the fi | Filo Class 11 Math Algebra Sequences and Series 578 150 There are Arithmetic means between and If ratio of the first . and the may be then find Solution: Let be the arithmetic means between and . Then, the common difference is given by: Now, And Since,", "*} 33 (*<*) 34 end 35 (*>*)", "hence, the original number is 14/19.", "What is the arithmetic mean of 25.2, 13.5, 18.5 and 34.8? (a) $13.5$ (b) $23$ (c) $18$ (d) $34.8$ 2 Answers Aug 22, 2016 b) 23 Explanation: Mean $= \\frac{25.2 + 13.5 + 18.5 + 34.8}{4}$ $= \\frac{92}{4}$ $= 23$ Aug 22, 2016 The answer is (b) $23$. Explanation: The arithmetic mean is sum of all the data points or numbers divided by their number. Here we have four data points / numbers which are $25.2$, $13.5$, $18.5$ and $34.8$. Their sum is $25.2 + 13.5 + 18.5 + 34.8 = 92.0$ and their arithmetic mean is $\\frac{92.0}{4} = 23$. Hence, answer is (b) $23$.", "The arithmetic mean of the following numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6 and 7, 7, 7, 7, 7, 7, 7 is : [A] 4 [B] 5 [C] 14 [D] 18", "### A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found mean to be 12. What should be the number in place of x? A. 3 B. 7 C. 17 D. 31 Answer: Option B ### Solution(By Apex Team) \\begin{aligned}&\\text{ Clearly, we have }\\\\ &\\Rightarrow\\frac{3+11+7+9+15+13+8+19+17+21+14+x}{12}=12\\\\ &\\Rightarrow137+x=144\\\\ &\\Rightarrow x=144-137\\\\ &\\Rightarrow x=7\\end{aligned} 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", "Question # Find the arithmetic mean between $$15$$ and $$-7$$. Solution ## As we know, arithmetic mean between $$a$$ & $$b$$ is $$=\\dfrac{a+b}{2}$$$$mean=\\dfrac{15-7}{2}=4$$Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More" ]

[ "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "add all three cases: $1+17+243=\\boxed{261}$", "our answer is $m+n=61+4=\\boxed{065}$.", "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 $35 * 4 + 29 = \\boxed{169}$.", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$.", "3 + 4 = \\boxed{21}$", "48, can be divided by 3.", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "sequence.) digits. Our answer is $40-14=\\boxed{26}$. ~IceMatrix", "48 is 24. =24", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "$4n=96$ and $n=24$. Hence, the only two $n$ that work are $n=24$ and $n=47$, so our answer is $24+47=\\boxed{071}$. -Stormersyle", "(16 - 1)^2 + 1 = \\boxed{406}$.", "$14$ of these extra pairs. The answer is $28+21+14 = \\boxed{063}$", "left. $(27*8)/{9 \\choose 3}{6 \\choose3}$ simplifies down to our fraction m/n, which is $9/70$. Adding them up gives $9 + 70 = \\boxed{79}$.", "filled in the boxes above? A) 28，40 B)27，40 C)30，35 D)26，37 E)25，39 B 10. $$11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 =$$ x Which of the following numbers are suitable to put in the boxes? A) 16,10 B) 15,9 C) 14,8 D) 135,0 E) 20,21 C Pages ( 1 of 2 ): 1 2Next »", "+ 31 = \\boxed{058}$.", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "31+x = 42\\\\ x = \\boxed{{\\textbf{(D)}\\ 11}}$", "the answer is $3 + 11 = \\boxed{\\mathrm{(A)}\\ 14}$" ]

[ "### The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is. A. 414 B. 420 C. 620 D. 621 $\\begin{array}{l}\\text{Sum of numbers}\\\\ \\text{= (41.4 × 15)}\\\\ \\text{= 621}\\end{array}$", "Math Help - Arithmetic Mean 1. Arithmetic Mean The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? 2. Originally Posted by Dragon The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? Hello, Dragon, the sum of the first nine numbers is: 9 * 54 = 486. So your problem becomes: $\\frac{9*54+u+v}{11}=66$ That means: u + v = 240 or: u = 240 - v So you can construct a pair of numbers with this property. EB", "# What ia the arithmetric means between 68 and 3? Jun 17, 2018 The arithmetic mean of 68 and 3 is $35 \\frac{1}{2}$ #### Explanation: To find the answer we need to know what the arithmetic mean is. I can point you to for instance http://www.statisticshowto.com/arithmetic-mean/ If we have n numbers ${a}_{1} , {a}_{2} , \\ldots , {a}_{n}$, the arithmetic mean is simply $\\frac{{a}_{1} , {a}_{2} , \\ldots , {a}_{n}}{n}$ In our instance we have ${a}_{1} = 68 , {a}_{2} = 3$ Therefore, the arithmetic mean is $\\frac{68 + 3}{2} = 35 \\frac{1}{2}$", "What is the arithmetic mean of 25.2, 13.5, 18.5 and 34.8? (a) $13.5$ (b) $23$ (c) $18$ (d) $34.8$ 2 Answers Aug 22, 2016 b) 23 Explanation: Mean $= \\frac{25.2 + 13.5 + 18.5 + 34.8}{4}$ $= \\frac{92}{4}$ $= 23$ Aug 22, 2016 The answer is (b) $23$. Explanation: The arithmetic mean is sum of all the data points or numbers divided by their number. Here we have four data points / numbers which are $25.2$, $13.5$, $18.5$ and $34.8$. Their sum is $25.2 + 13.5 + 18.5 + 34.8 = 92.0$ and their arithmetic mean is $\\frac{92.0}{4} = 23$. Hence, answer is (b) $23$.", "# Arithmetic mean There are n arithmetic means between 3 and 17 . The ratio of the first mean to the last mean is 1:3 . Find n .", "48 is 24. =24", "# Arithmetic Mean • November 8th 2006, 06:37 AM Dragon Arithmetic Mean The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? • November 8th 2006, 07:01 AM earboth Quote: Originally Posted by Dragon The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? Hello, Dragon, the sum of the first nine numbers is: 9 * 54 = 486. $\\frac{9*54+u+v}{11}=66$", "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 $35 * 4 + 29 = \\boxed{169}$.", "Question # Find the arithmetic mean between $$15$$ and $$-7$$. Solution ## As we know, arithmetic mean between $$a$$ & $$b$$ is $$=\\dfrac{a+b}{2}$$$$mean=\\dfrac{15-7}{2}=4$$Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "Math Help - arithmetic mean 1. arithmetic mean the arithmetic mean of a set of 50 numbers is 38. if two numbers 45 and 55 are discarded the mean of the remaining set of numbers is ____ . need little help on this guys. thanks 3. Re: arithmetic mean Originally Posted by rcs the arithmetic mean of a set of 50 numbers is 38. if two numbers 45 and 55 are discarded the mean of the remaining set of numbers is ____ . need little help on this guys. thanks You are told \\displaystyle \\begin{align*} \\frac{t_k + t_{k + 1} + t_{k + 2} + \\dots + t_{k + 49}}{50} = 38 \\implies t_k + t_{k + 1} + t_{k + 2} + \\dots + t_{k + 49} = 1900 \\end{align*}. Two of these terms are 45 and 55, so removing them gives \\displaystyle \\begin{align*} t_k + t_{k + 1} + t_{k + 2} + \\dots + t_{k + 49} - 45 - 55 &= 1900 - 45 - 55 \\\\ &= 1800 \\end{align*} There are now 48 terms, so the mean of these 48 terms will be \\displaystyle \\begin{align*} \\frac{1800}{48} = 37.5 \\end{align*}. 4. Re: arithmetic mean thanks sir... but how did you get 1900? 5. Re: arithmetic mean Originally Posted by rcs thanks sir... but how", "# What is an Arithmetic means What is an Arithmetic means ? When three numbers a, A and b are in A.P., then A is called the arithmetic mean of numbers 'a' and 'b'. Given that, a, A , b is in A.P. Then As they are in A.P. so their common difference will be constant. A - a = d and b - A = d ∴ A - a = b - A 2A = a + b ∴ A = $\\frac{a + b}{2}$ Thus the required arithmetic mean (A.M) of two numbers 'a' and 'b' is $\\frac{a + b}{2}$ Example 1 : Consider the following A.P. 3,8,13,18,23,28,33 Here the first term is 3 and the last term is 33 and there are 5 numbers between the two . So all these five numbers are called 'arithmetic means' between 3 and 33 Example 2 : Consider the A.P. as 4,8,12,16,20 Here the first term is 4 and the last term is 20. There are three terms ( 8,12,and 16) between the two terms. So these three terms are 'arithmetic means' between 4 and 20. ## What is an arithmetic means between the following ? 1) 7,13,19 Solution:Arithmetic mean between 7 and 19 is 13. 2) 6,9,12,15,18 Solution:Arithmetic means between 6 and 18 is 9,12 and 15.", "Find three arithmetic means between 6 and – 6. Question: Find three arithmetic means between 6 and – 6. Solution: let the three AM be $\\mathrm{X}_{1}, \\mathrm{X}_{2}, \\mathrm{X}_{3}$. So new AP will be $6, x_{1}, x_{2}, x_{3},-6$ Also – 6 = 6 + 4d $d=-3$ $x_{1}=3$ $x_{2}=0$ $x_{3}=-3$", "3) 'a' is an arithmetic between 6 and 18 .Find 'a' Solution: 'a' is an A.M between 6 and 18 ∴ a = $\\frac{6 + 18}{2}$ a = $\\frac{24}{2}$ ∴ a = 12 In the given two numbers, any number of arithmetic means can be inserted between them. Let the two numbers be 'a' and 'b' and the arithmetic means inserted will be $A_{1},A_{2},A_{3},...A_{n}$. That means 'n' number of arithmetic means can be inserted between the two numbers 'a' and 'b'. Let a, $A_{1},A_{2},A_{3},...A_{n}$, b is in A.P. Here b is the (n + 2)th term So, b = a + [(n + 2) - 1]d b = a + (n + 1)d b - a = (n + 1)d ∴ d = $\\frac{b - a }{n + 1}$ Thus 'n' arithmetic means between 'a' and 'b' are as follows $A_{1}$ = a + d = a + $\\frac{b - a }{n + 1}$ $A_{2}$ = a + 2d = a + $\\frac{2(b - a) }{n + 1}$ $A_{3}$ = a + 3d = a + $\\frac{3(b - a) }{n + 1}$ - - - - $A_{n}$ = a + nd = a + $\\frac{n(b - a) }{n + 1}$ ## Examples on inserting arithmetic means between the numbers 1) Insert three arithmetic means between 8 and 26.", "The arithmetic mean of the following numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6 and 7, 7, 7, 7, 7, 7, 7 is : [A] 4 [B] 5 [C] 14 [D] 18", ", 03.08.2020 14:01 lolololol21 What is the arithmetic mean of the following numbers? 6 , 4 , 1 , 9 , 3 , 8 , 3 , 5 , 10 Another question on Mathematics Mathematics, 21.06.2019 20:00 Find the slope of the line passing through a pair of points", "+ 2\\\\&= 52 + 2\\\\&= 54\\\\\\end{aligned} Hence, the integers are $\\boxed{52}, \\boxed{53}$ and $\\boxed{54}.$", "### If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is : A. 10 B. 12 C. 14 D. 16 $\\begin{array}{l}\\text{Arithmetic mean}\\\\ \\Large\\frac{\\text{ Total sum }}{\\text{ Total number }}\\\\ \\text{According to the question,}\\\\ \\Rightarrow10=\\frac{7+5+13+x+9}{5}\\\\ \\text{⇒ 50 = x + 34}\\\\ \\text{⇒ x = 50 – 34}\\\\ \\text{⇒ x = 16}\\end{array}$", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$.", "# Arithmetic mean The number $a$ obtained by dividing the sum of several numbers $a_1,\\ldots,a_n$ by their number $n$: $$a=\\frac{a_1+\\ldots+a_n}{n}.$$", "Mathematics # How can you write 72+ (8+19) to make easier to add? $72+(8+19)= \\\\\\\\ =70+2+8+10+9= \\\\\\\\ =80+10+9= \\\\\\\\ =90+9= \\\\\\\\=\\boxed{99}$" ]

[ "many numbers are less than 222 with a digit sum is 8?", "Solution: Since 2 is a prime number, the prime factorization of 8 = 2 × 2 × 2.", "of the sum of these two numbers anyone...", "Fermat's Primes and Bases (This problem was inspired by this problem) How many digits does the number $$2^{2^{8}}$$ contain in base 8 ? ×", "21 such numbers. So there are 45 - 21 = 24 two-digit numbers that cannot be expressed as the sum of two primes. Denis needs to go sit in the corner for misreading the question AGAIN.", "Both 4 and 8 can be written as the sum of two prime numbers ($4=2+2$, $8=5+3$). How many numbers less than 20 cannot be written as the sum of two prime numbers?", "• Two-digit number Digit sum of thinking two-digit natural number is 11. When it exchanging a sequence of digits, given a number which is 27 less than the thinking number. Find out which number I think.", "the product of the 2 numbers is 36. Sep 14, 2018", "# Sum of Primes Number Theory Level 2 How many ways can the number 100,001 be written as the sum of two primes? ×", "# Fermat's Primes and Bases (This problem was inspired by this problem) How many digits does the number $$2^{2^{8}}$$ contain in base 8 ? × Problem Loading... Note Loading... Set Loading...", "to get sum 8 are = 4 Published on 21-Jan-2020 11:11:47 Advertisements", "# Quite Binary What is the sum (in base 10) of all the natural numbers less than 64 which have exactly two ones in their base 2 representation? ×", "# 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. ×", "less than 1,001 has digit sum digit of the digit sum number 1. 16. Six-digit primes Find all six-digit prime numbers that contain each one of digits 1,2,4,5,7 and 8 just once. How many are they? 17. Primes 2 Which prime numbers is number 2025 divisible?", "of two integers.", "# How many ways we can represent 50! as the sum of two primes? in how many ways we can represent 50! as the sum of two primes?", "# Express 36 as the sum of two primes numbers? 291 views Express 36 as the sum of two primes numbers? posted Oct 7, 2018 29 + 7 = 36 31 + 5 = 36 23 + 13 = 36 19 + 17 = 36 Similar Puzzles ``````_ _ (6) _ _ _ _ _ _ (-4) _", "Sums V What is the sum of all numbers multiple by 6 that are also multiples by 8 between 100 and 1000? ×", "What is the sum of the digits of two-digit positive integer M? (1) If 5 22 Apr 2015, 03:35 9 The number A is a two-digit positive integer; the number B i 8 22 Jul 2011, 09:18 X is a two-digit number. If the ratio between the units 4 18 May 2011, 02:00 Display posts from previous: Sort by", "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" ]

[ "3 are the same). Bunuel wrote: scaredshikless wrote: 25. John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? A. 15/16 B. 11/16 C. 11/12 D. 1/2 E. 5/8 I don't understand the explanation to this one (so I obviously got it wrong...). Also, at first, I read it as 1 only appears once in the last three places. If that were the case, how would you solve this? Thanks, folks. The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}. {X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \\frac{4}{8}=\\frac{1}{2} and probability of {X} will not be a prime is again \\frac{1}{2}. We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes. Let's count the opposite probability and subtract it from 1. Opposite probability of at least 2 primes is 0 or 1 prime: So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}. Scenario 1 prime - {P}{NP}{NP}{NP}: \\frac{4!}{3!}*\\frac{1}{2}*(\\frac{1}{2})^3=\\frac{4}{16}. We are", "last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? A. 15/16 B. 11/16 C. 11/12 D. 1/2 E. 5/8 I don't understand the explanation to this one (so I obviously got it wrong...). Also, at first, I read it as 1 only appears once in the last three places. If that were the case, how would you solve this? Thanks, folks. The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}. {X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \\frac{4}{8}=\\frac{1}{2} and probability of {X} will not be a prime is again \\frac{1}{2}. We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes. Let's count the opposite probability and subtract it from 1. Opposite probability of at least 2 primes is 0 or 1 prime: So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}. Scenario 1 prime - {P}{NP}{NP}{NP}: \\frac{4!}{3!}*\\frac{1}{2}*(\\frac{1}{2})^3=\\frac{4}{16}. We are multiplying by \\frac{4!}{3!} as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P} - 4 ways (basically the # of permutations of 4 objects out ow which 3 are the same). Scenario 0", "$$h$$ is 36. No two digits of $$h$$ are identical. How many different numbers are possible values of $$h$$? A. 6 B. 12 C. 24 D. 36 E. 48 We start with the prime factorization of $$h$$, which is $$2^23^2$$. In other words, $$h$$ is the product of two 2's and two 3's. All the factors of 36, except for 1, can be constructed as products of some or all of these 2's and 3's: 2, 3, 4, 6, 9, 12, 18, and 36. At this stage, we can see that the possible values of any digit of h only include 1, 2, 3, 4, 6, and 9. Any other digit would contain the wrong primes (5, 7) or too many 2's (8), or it would turn the product to zero (0). We now test subgroups of four digits, imposing the conditions that the product of all four digits is 36 and that no two digits are the same. Start with one digit at a time; do not worry about the order of the digits yet. 9: If one of the digits is 9, then the product of the other three digits must be 4. The only possible sets of positive digits that have 3 members and multiply together to 4 are {2, 2, 1} and {4, 1,", "$$h$$ is 36. No two digits of $$h$$ are identical. How many different numbers are possible values of $$h$$? A. 6 B. 12 C. 24 D. 36 E. 48 We start with the prime factorization of $$h$$, which is $$2^23^2$$. In other words, $$h$$ is the product of two 2's and two 3's. All the factors of 36, except for 1, can be constructed as products of some or all of these 2's and 3's: 2, 3, 4, 6, 9, 12, 18, and 36. At this stage, we can see that the possible values of any digit of h only include 1, 2, 3, 4, 6, and 9. Any other digit would contain the wrong primes (5, 7) or too many 2's (8), or it would turn the product to zero (0). We now test subgroups of four digits, imposing the conditions that the product of all four digits is 36 and that no two digits are the same. Start with one digit at a time; do not worry about the order of the digits yet. 9: If one of the digits is 9, then the product of the other three digits must be 4. The only possible sets of positive digits that have 3 members and multiply together to 4 are {2, 2, 1} and {4, 1,", "there are two such numbers 42 and 24. Verification : Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.", "contest, translated to english), I figured out a case where it fails: 2 524287 524287 (actually, it could be any prime numbers greater than 40,050.) In this case the solution is -1 (because the gcd of two equal numbers is the number itself), however, my code prints 1 as the answer. Is there anything that can be done about this? • • +19 • By lixolas, history, 18 months ago, , Hello everyone!", "digits is 2 mod 3 then, then none of the digits can be 2 mod 3. So, every such number is comprised of at most 0, 3, 6, 9, and 1, 4, 7 OR 0, 3, 6, 9, and 2, 5, 8. 3. These should be much more rare if you generalize to more than 1 delete. If there were 2 deletes, the mod 3 argument would rule out most cases. I believe the only case possible would be the digits add to 1 mod 3 and all the digits are 1 mod 3, or the digits add to 2 mod 3 and all the digits are 2 mod 3. So, they could only contain 1, 4, and 7 OR 2, 5, 8. In the second case, ending with any of those numbers makes the original not prime. So, we only have the first case. They must have the digits 1, 4, 7 only. And, they would have to contain 4, 7, 10, 13, 16, ... digits to ensure the digit totals worked out. - Here's my C code for generating these numbers. It should be very fast since it uses a sieve, but also consumes a lot of memory for the sieve[] array (we could compress by a factor of 8 by using a bitfield). // find", "did not appear at all. What is the probability that the phone number contains at least two prime digits? Here's my issue with this problem and similar ones. Hope somebody can help me grasp this. The posted solution is this: Since 1 appears exactly three times, we can solve for the other four digits only. For every digit we can choose out of 8 digits only (without 1 and 0). Since we have 4 prime digits (2, 3, 5, 7) and 4 non-prime digits (4, 6, 8, 9), the probability of choosing a prime digit is ½. We need at least two prime digits: One minus (the probability of having no prime digits + having one prime digit): There are 4 options of one prime digit, each with a probability of (1/2)4. There is only one option of no prime digit with a probability of (1/2)4. So: [1- ((1/2)4+(1/2)4*4)] = 11/16. I understand how they calculated this, but here's what I don't get: say for one prime digit scenario, the solution calculates different arrangements, as in prob of 4*(1/2)^4. I get the 1/2^4 as being the probability of getting one prime number and three non-prime numbers. But why take into account the arrangements by multiplying by 4? Whether the prime number comes in position 1, 2, 3 or", "1, 2, 3 and 6 So, the prime numbers we get in range from 1 to 6 are 2, 3 and 5 The count of prime number is 3 $\\therefore {\\text{The probability of getting any prime number is }}$ $\\Rightarrow {\\text{(Count of Prime Number)/(Total Count) = (3)/(6)}}$ $= 1/2$ $\\therefore {\\text{The value of m is 2}}$ $\\therefore {\\text{B is the correct option}}$ Note-In these types of questions, we should know how to calculate probability. The general form to calculate the probability of anything is to divide the count of favourable things by the total number of things.", "84 (E) 252 21x=12y which is 7x=4y after dividing by 3 on both sides. 7 and 4 are co-primes that is neither of them is factor of each other. So the first common multiple will be 7*4 = 28 hence x=4 and y=7 and hence x*y=28 Re: If x and y are positive integers and 21x=12y [#permalink] 11 Mar 2017, 08:45", "has its own article. 497 = 7 × 71, sum of five consecutive primes (89 + 97 + 101 + 103 + 107) 498 = 2 × 3 × 83, sphenic number, untouchable number 499 prime number 03-10-2013 05:06:04", "# Easy Probability Solved QuestionAptitude Discussion Q. What is the probability that a number selected from numbers 1,2,3,...,30, is prime number, when each of the given numbers is equally likely to be selected? ✖ A. 9/30 ✖ B. 8/30 ✔ C. 10/30 ✖ D. 11/30 Solution: Option(C) is correct $X = \\{2,3,5,7,11,13,17,19,23,29\\}$ $n(X)= 10, n(S)=30$ Hence required probability, $= \\dfrac{n(X)}{n(S)}$ $= \\dfrac{\\textbf{10}}{\\textbf{30}}$ ## (1) Comment(s) Puskar () Prime numbers from 1 to 30 = ( 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ) P( a prime number) = 10C1/30C1", "see statement two and we see the P-Q is 36. Let's see which odd number combinations could result in a difference where the ones digit is 6. They are 9 and 3, 7 and 1, 3 and 7, and 5 and 9. We can get rid of 9 and 3 and 5 and 9 because 9 isn't prime. That leaves 7 and 1, 5 and 9, and 3 and 7. When you combine the digits of 5 with any number for a two digit number in this equation, at some point, 5 is in the ones digit and it ain't prime so get rid of 5 and 9. Next there is 7 and 1 and 3 and 7. combine 7 and 1 first and you see that you get 71 and 17 and see if we can get a result with the ones digit at 6. It's impossible, so 3 and 7 is left so we're back to 37 and 73. We find out that 73-37 is 36 and 73=P and 37=Q and we can tell which value is bigger so that proves sufficiency, and Statement 2 is sufficient. Since Statements 1 and 2 are sufficient, D is the answer. In your mind, this shouldn't take longer than 2 minutes, though this response took me like five minutes", "in: 26, 6, 32 or Due to the number of X's in line one(19) and line two(16), add those together and subtract them from 62. 62 - (19 + 16) = 62 - 35 = 27 Resulting in: 26, 6, 27 or Just subtract the previous numbers from 62. 62 - (26 + 6) = 62 - 32 = 30 Resulting in: 26, 6, 30 The combination is: 26, 6, 31 (do this last one twice...) First: 26 as in I: THE NUMBER OF LETTERS IN THE ALPHABET. (LATIN, NOT GREEK). Second: 6 as in II: THE NUMBER OF WHOLE TONES IN AN OCTAVE. And finally: 31 (performed twice) This one is a bit of a stretch. 62 is the number of X's on the wall. Obviously, the input can't be greater than 50.... but 62 is a semiprime number. It is formed as the product of two prime numbers in this case 31 x 2 . So my vote is the third combination is 31, performed twice.", "random number then 3 has chance 2/3 of dividing a or a + 2, but only chance 1/3 of dividing a and a + 6, so the latter pair is conjectured twice as likely to both be prime.", "A number is selected from numbers 1 to 25. The probability that it is prime is Question: A number is selected from numbers 1 to 25. The probability that it is prime is (a) $\\frac{2}{3}$ (b) $\\frac{1}{6}$ (c) $\\frac{1}{3}$ (d) $\\frac{5}{6}$ Solution: GIVEN: A number is selected from 1 to 25 TO FIND: Probability that, the number is a prime Total number is 25 Numbers from 1 to 25 that are primes are 2, 3, 5, 7, 11, 13, 17, 19 and 23 Total numbers that are primes from 1 to 25 is 9 We know that PROBABILITY = Hence probability of getting a prime number from 1 to 25 is equal to Hence no option is correct", "< 32336, then there are infinitely many primes of the form (k*24^n-1)/gcd(k-1,24-1) with integer n>=1 * if k < 105, then there are infinitely many primes of the form (k*25^n-1)/gcd(k-1,25-1) with integer n>=1 * if k < 149, then there are infinitely many primes of the form (k*26^n-1)/gcd(k-1,26-1) with integer n>=1 * if k < 13, then there are infinitely many primes of the form (k*27^n-1)/gcd(k-1,27-1) with integer n>=1 * if k < 3769, then there are infinitely many primes of the form (k*28^n-1)/gcd(k-1,28-1) with integer n>=1 * if k < 4, then there are infinitely many primes of the form (k*29^n-1)/gcd(k-1,29-1) with integer n>=1 * if k < 4928, then there are infinitely many primes of the form (k*30^n-1)/gcd(k-1,30-1) with integer n>=1 * if k < 145, then there are infinitely many primes of the form (k*31^n-1)/gcd(k-1,31-1) with integer n>=1 * if k < 10, then there are infinitely many primes of the form (k*32^n-1)/gcd(k-1,32-1) with integer n>=1 * if k < 545, then there are infinitely many primes of the form (k*33^n-1)/gcd(k-1,33-1) with integer n>=1 * if k < 6, then there are infinitely many primes of the form (k*34^n-1)/gcd(k-1,34-1) with integer n>=1 * if k < 5, then there are infinitely many primes of the form (k*35^n-1)/gcd(k-1,35-1) with integer n>=1 * if k < 33791,", "increasing in this interval. Therefore, the largest possible value of X(99 − X) is 49(99 − 49) = 2450. Hence, XY ≤ 2450 Let me write down all the constraints we have found so far neatly. 1. 2 ≤ X ≤ 49 2. 3 ≤ Y ≤ 97 3. 5 ≤ X + Y ≤ 99 4. 6 ≤ XY ≤ 2450 Now, let us analyze what the two mathematicians said. P says \"I cannot find these numbers.\" If X and Y are two prime numbers, P can easily factor the product to find the numbers X and Y. For example, if XY = 35, he can easily determine that the two numbers are 5 and 7. The only other way of factoring it is 1 × 35 but X = 1 is not allowed in this puzzle. If XY is a cube of a prime number, he can find the two numbers as X = ∛(XY) and Y = [∛(XY)]2. For example, if XY = 125, he can say that the two numbers are 5 and 25. If XY is a fourth power of a prime number, he can find the two numbers as X = ∜(XY) and [∜(XY)]3. For example, if XY = 81, he can say that the two numbers are 3 and 27. Factoring", "are less than 41, and the last two are greater than 41, then 41 doesn’t have the form of a primoral prime. • Notice that: $$2^5\\;-\\;1=32-1=31$$ $$2^6\\;-\\;1=64-1=63$$ Therefore, 41 doesn’t have the form of a Mersenne prime. • Let’s consider the values that takes 2p + 1, for different primes p: Since 41 is not in the right column, it doesn’t have the form of a safe prime. We invite you to read other articles on prime numbers, in our blog, to find out which other prime numbers belong to these classes. Do you know that 43 is also a prime number?", "only the 32 has to be changed to make it work for, e.g., 64-bit integers. • Hardcoding a list of the prime numbers below 128 is not allowed in a challenge that involves primality testing, even if the language of choice only supports signed 8-bit integers. Dennis • 204.6k • 5 • 90 • 165" ]

[ "specifically, but I hope you’ll take my advice and check your work on the problems at the end of this post. ## And finally… Practice! 1. Mrs. Anderson has to choose a group of two boys and two girls to represent her class. If there are six boys and eight girls in the class, how many groups are possible? 2. In the imaginary country of Mathematica, a ZIP code is five digits long. The first digit must be a prime number, the third digit must not be 0, and the last digit must be even or zero. How many ZIP codes are possible for this country? 3. A hand of four cards is dealt from a standard 52-card deck. What is the probability that it contains at least one ace? (Hint: Find the probability of it containing no aces first.) 4. How many positive integers less than 700 can be made using only digits that are prime numbers? (Hint: Make a case for each possible number of digits.) 5. How many even positive integers can be made using the digits 1, 2, and 4, using each digit only once? 6. There are 7 red marbles, 2 blue marbles, and 1 green marble in a jar. In how many ways can three marbles be selected from the jar, if", "specifically, but I hope you’ll take my advice and check your work on the problems at the end of this post. ## And finally… Practice! 1. Mrs. Anderson has to choose a group of two boys and two girls to represent her class. If there are six boys and eight girls in the class, how many groups are possible? 2. In the imaginary country of Mathematica, a ZIP code is five digits long. The first digit must be a prime number, the third digit must not be 0, and the last digit must be even or zero. How many ZIP codes are possible for this country? 3. A hand of four cards is dealt from a standard 52-card deck. What is the probability that it contains at least one ace? (Hint: Find the probability of it containing no aces first.) 4. How many positive integers less than 700 can be made using only digits that are prime numbers? (Hint: Make a case for each possible number of digits.) 5. How many even positive integers can be made using the digits 1, 2, and 4, using each digit only once? 6. There are 7 red marbles, 2 blue marbles, and 1 green marble in a jar. In how many ways can three marbles be selected from the jar, if", "(D) $$\\frac{5}{6}$$ (E) 1 Diagnostic # 11 -------CONCEPT------------ Single digit prime numbers are 2,3,5,7 Now, except 2 all are odd so we can only get an odd sum if one of our prime number is even while other is odd (Only an odd number can be prime). 2+3 = 5 2+5 = 7 are only 2 cases to satisfy the condition. while 2 + 7 = 9, it is not prime & Now the total cases to get a sum by selecting any 2 prime numbers out of these 4 are 4C2 = 6 So, probability for sum NOT to be prime number = 4/6 = 2/3 Re: What is the probability that the sum of two different [#permalink] 12 Jun 2019, 02:16 Display posts from previous: Sort by", "# Math Help - The sum on two numbers in 28... 1. ## The sum on two numbers in 28... The sum of two numbers is 28 and their product is seven. Find the sum of the reciprocals of the numbers. This one has me confused. How can their product equal 7 if 7 is prime? 1+27 027 14+14 196 2+26 052 13+15 195 3+25 075 2. This is just a very basic algebra problem. x+y=28 xy=7 (1/x)+(1/y)=? 3. Hello, ceasar! The sum of two numbers is 28 and their product is seven. Find the sum of the reciprocals of the numbers. This one has me confused. How can their product equal 7 if 7 is prime? WHO said these are whole numbers ?? There is a marvelously sneaky trick for this one . . . We have: .x + y .= .28 . [1] . . . . . . . . .xy . = . .7 . .[2] Divide [1] by [2]: . x + y . . .28 . . . . . . . . . . . . . ------- .= .--- . . . . . . . . . . . . . . .xy . . . . 7 . . . . . . . . . . .", "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", "number, not a product). • Six-digit primes Find all six-digit prime numbers that contain each one of digits 1,2,4,5,7 and 8 just once. How many are they? • MO Z8-I-1 2018 Fero and David meet daily in the elevator. One morning they found that if they multiply their current age, they get 238. If they did the same after four years, this product would be 378. Determine the sum of the current ages of Fero and David. • Z9–I–4 MO 2017 Numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 were prepared for a train journey with three wagons. They wanted to sit out so that three numbers were seated in each carriage and the largest of each of the three was equal to the sum of the remaining two. The conduct • Digit sum How many are three-digit numbers that have a digit sum of 6? • Z9–I–1 In all nine fields of given shape to be filled natural numbers so that: • each of the numbers 2, 4, 6, and 8 is used at least once, • four of the inner square boxes containing the products of the numbers of adjacent cells of the outer square, • in the cir", "21 such numbers. So there are 45 - 21 = 24 two-digit numbers that cannot be expressed as the sum of two primes. Denis needs to go sit in the corner for misreading the question AGAIN.", "Both 4 and 8 can be written as the sum of two prime numbers ($4=2+2$, $8=5+3$). How many numbers less than 20 cannot be written as the sum of two prime numbers?", "a prime units digit? Director Joined: 20 Apr 2016 Posts: 761 Followers: 6 Kudos [?]: 517 [0], given: 94 Re: Similar problems [#permalink] 15 Oct 2018, 19:44 ruposh6240 wrote: 1. How many integers between 2,000 and 3,999 have a ones digit?? 2. How many integers between 0 and 1570 have a prime tens digit and a prime units digit? The first ques- is not complete 2. there are 252 integers. Next time would request you to read the rules before posting, as this type of post are not encouraged https://greprepclub.com/forum/rules-for ... -1083.html _________________ If you found this post useful, please let me know by pressing the Kudos Button Supreme Moderator Joined: 01 Nov 2017 Posts: 213 Followers: 3 Kudos [?]: 70 [0], given: 1 Re: How many integers between 1 and 10^21 are such that the sum [#permalink] 15 Oct 2018, 22:15 Expert's post Bunuel wrote: How many integers between 1 and $$10^{21}$$ are such that the sum of their digits is 2? A. 190 B. 210 C. 211 D. 230 E. 231 Kudos for correct solution. How many digits will 10^n have ;- n+1, so 10^21 will have 22 digits - 1 of one and 21 of zeroes but we are looking for < 10^21, so we will have 22-1=21 digits to fill up.. we can get", "# Let's re-share it! Two numbers $$x,y$$ with $$x>y$$ are chosen at random from the set$$\\left\\{1,2,3,.....,299,300\\right\\}.$$ If the probability that $$x^{2} - y^{2}$$ is divisible by 3 is $$\\frac{a}{b},$$ then what is the digit sum of $$a \\times b$$? Details and Assumptions: 1. Digit sum refers to the sum of all the digits of the number. For example, the digit sum of 123 is $$1+2+3=6.$$ 2. $$a$$ and $$b$$ are co-prime numbers. ×", "+= 2) { if (num % i == 0) return false; } return true; } - Instead of checking numbers, check the sets of digits in sorted order. Let's say you are looking for six digit solutions. Sum of squares of digits is at most 6 x 81, so you first create an array bool prime [6*81 + 1] and fill it. Then for (a = 1; a <= 9; ++a) for (b = a; b <= 9; ++b) for (c = b; c <= 9; ++c) for (d = c; d <= 9; ++d) for (e = d; e <= 9; ++e) for (f = e; f <= 9; ++f) if (prime [a + b + c + d + e + f]) if (prime [a*a + b*b + c*c + d*d + e*e + f*f) print all numbers made from the digits a,b,c,d,e,f. Even if you go up to 19 digit numbers, there are only about 1.5 million combinations of digits to check which should take only a few dozen milliseconds. Finding the numbers is then easy. Even for 63 digit numbers, there are less than 10 billion combinations of digits to check. -", "1, 2, 3 and 6 So, the prime numbers we get in range from 1 to 6 are 2, 3 and 5 The count of prime number is 3 $\\therefore {\\text{The probability of getting any prime number is }}$ $\\Rightarrow {\\text{(Count of Prime Number)/(Total Count) = (3)/(6)}}$ $= 1/2$ $\\therefore {\\text{The value of m is 2}}$ $\\therefore {\\text{B is the correct option}}$ Note-In these types of questions, we should know how to calculate probability. The general form to calculate the probability of anything is to divide the count of favourable things by the total number of things.", "has its own article. 497 = 7 × 71, sum of five consecutive primes (89 + 97 + 101 + 103 + 107) 498 = 2 × 3 × 83, sphenic number, untouchable number 499 prime number 03-10-2013 05:06:04", "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", "last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? A. 15/16 B. 11/16 C. 11/12 D. 1/2 E. 5/8 I don't understand the explanation to this one (so I obviously got it wrong...). Also, at first, I read it as 1 only appears once in the last three places. If that were the case, how would you solve this? Thanks, folks. The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}. {X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \\frac{4}{8}=\\frac{1}{2} and probability of {X} will not be a prime is again \\frac{1}{2}. We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes. Let's count the opposite probability and subtract it from 1. Opposite probability of at least 2 primes is 0 or 1 prime: So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}. Scenario 1 prime - {P}{NP}{NP}{NP}: \\frac{4!}{3!}*\\frac{1}{2}*(\\frac{1}{2})^3=\\frac{4}{16}. We are multiplying by \\frac{4!}{3!} as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P} - 4 ways (basically the # of permutations of 4 objects out ow which 3 are the same). Scenario 0", "test is simple. To make the sum of two numbers odd, one of the numbers must be odd and the other even. There is only one even prime, so that limits you to sums of the form $2+p$. Thus the odd numbers that are the sum of two primes are exactly the ones that are two more than a prime. The first few are $$5, 7, 9, 13, 15, 19, 21, 25, 31, 33, 39, 43\\ldots$$ Also note that it is not known whether every even number is the sum of two primes. Every single even number that has been checked has been verified to be the sum of two primes, but we don't know whether it is always true.", "3 are the same). Bunuel wrote: scaredshikless wrote: 25. John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? A. 15/16 B. 11/16 C. 11/12 D. 1/2 E. 5/8 I don't understand the explanation to this one (so I obviously got it wrong...). Also, at first, I read it as 1 only appears once in the last three places. If that were the case, how would you solve this? Thanks, folks. The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}. {X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \\frac{4}{8}=\\frac{1}{2} and probability of {X} will not be a prime is again \\frac{1}{2}. We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes. Let's count the opposite probability and subtract it from 1. Opposite probability of at least 2 primes is 0 or 1 prime: So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}. Scenario 1 prime - {P}{NP}{NP}{NP}: \\frac{4!}{3!}*\\frac{1}{2}*(\\frac{1}{2})^3=\\frac{4}{16}. We are", "6 ⋅ 6 = 36 Prime numbers are 2,3,5,7 and 11 Favorable outcome is F = 15(11,12,14,16,21,23,25,32,34,41,43,52,56,61,65) Probability of getting sum as a prime number is P (Sp) = F T = 15 36 = 5 12 [Ans] Answer link, 3 and 5. (Often probability is expressed as a percentage in which case 1 3 = 331 3 %) Answer link. 9K answer views Updated 4 y There is a 3 in 6 chance to not get a prime number on a single roll (1,4,6). 25. (iv) A multiple of 2 on on. So, y). Odd numbers on a dice are 1, 5, well thought and well explained computer science and programming articles, and the probability of getting an odd number (1, the ones that contain both prime are (2, well thought and well explained computer science and programming articles, the probability of getting the two numbers add up to a prime When one dice is rolled, 6 }, 3, 4), the odds of either even occuring should be: $\\frac{1}{36} + \\frac{1}{2} = \\frac{19}{36}$ A Computer Science portal for geeks. This gives us a total of 1 + 2 + 4 + 6 + 2 = 15 possible ways to get a prime number. The profitability of getting an even prime number = 61 Was this answer helpful?", "# A 2n digit number starts with 2 and all its digits are prime, then the probability that the sum of all 2 consecutive digits of the number is prime is Question : A 2n digit number starts with 2 and all its digits are prime, then the probability that the sum of all 2 consecutive digits of the number is prime is (A) $4\\times 2^{3n}$ (B) $4\\times 2^{-3n}$ (C) $2^{3n}$ (D) none of these Answer : option (B) Solution : The prime digits are 2, 3, 5, 7. If we fix 2 at first place, then other 2n-1 places are filled by all four digits. So the total number of cases is $4^{2n-1}$ Now, sum of 2 consecutive digits is prime when consecutive digits are (2,3) or (2, 5) Then 2 will be fixed at all alternative places So favourable number of cases is $2^{n}$. Therefore, probability is $\\dfrac{2^{n}}{4^{2n-1}}=2^{n}2^{-4n+2}=2^{2}2^{-3n}=\\dfrac{4}{2^{3n}}$ Tags: No tags", "rhyming primes (e.g., 3 and 7) is twice the central number ($$2 \\times 5 = 10$$). Sums are quick operations, so it might be good to rephrase our question in terms of taking sums of two primes. We are asked which integer between 1 and 20, inclusive, has the greatest number of rhyming primes around it. So we should list out the primes up to 40, since the larger number in any pair of rhyming primes that average to 20 would have to be below 40 (primes are restricted to positive integers). Here are the primes less than 40: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37. Rephrasing the question in terms of sums, we can ask: what number between 1 and 20, when multiplied by 2, can be expressed as a sum of two different primes from this list in the greatest number of different ways? We should now start from the answer choices, rather than test all 20 theoretical possibilities. Unfortunately, there is no shortcut; you actually have to check the possibilities. Primes are unevenly distributed, so there’s no way to intuit the answer. We should start by checking the highest number, because we will probably be able to construct more valid pairs around larger numbers than around smaller numbers. Construct the" ]

[ "1/2, 1/3 and 1/6. Each man takes some money from the pile until nothing is left. Of the two water reeds, the one grows 3 feet and the other 1 foot on the first day. The growth of the first becomes every day half of that of the preceding day, while the other grows twice as much as on the day before. A person has a circular yard that is 150 ft. in diameter, and wishes a walk of equal width made round it within the fence...", "their mature size. We use cookies to give you the best experience on our website. Square foot gardening is the practice of dividing the growing area into small square sections, typically 1 foot (30 cm) on a side, hence the name. This method relies on structure to create small yet concentrated gardens that typically have high yields. I would never go back to the traditional way of gardening. A well protected box for a small garden: Boxes in which the plants are grown are 3 feet wide and 4 … This system is especially good for beginners as it makes plant spacing much easier and also makes more sense for growing in a small space as most beginners do. Square Foot Gardening is an easy to understand method of planting vegetables which splits a gardeners plot into square foot areas rather than into rows. Next, plants are then sown or transplanted into … There are several resources out there to help you with the square foot gardening methods. I especially loved the trellises for my tomatoes. amzn_assoc_marketplace = \"amazon\"; 573 Now, if you don’t know what square foot gardening is, it’s simply a garden segmented into 1 foot by 1 foot squares (I mean this method isn’t called square foot gardening for nothing, right?) Mel Bartholomew’s book Square", "wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 10 feet by 1,000 feet. it will have 2,020 feet of wall for 10,000 square feet, or 4.950 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 20 feet by 500 feet. it will have 1,040 feet of wall for 10,000 square feet, or 9.615 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 40 feet by 250 feet. it will have 580 feet of wall for 10,000 square feet, or 17.241 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 80 feet by 125 feet. it will have 410 feet of wall for 10,000 square feet, or 24.390 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 100 feet by 100 feet. it will have 400 feet of wall for 10,000 square feet, or 25 square feet for every foot of wall. So, out of all possible rectangles, a square shape gives the least amount of wall to be defended for the same square footage inside. But a square", "of logarithms and urge you to use it in class. An oblong garden is a half yard longer than it is wide and consists entirely of a gravel walk...", "has been working on her tomato plants, Alex has been working on designing the garden plot. He knows that he wants two plots, one to be in the shape of a square and one to be the shape of a rectangle. His square plot has a length and width of 9 feet. His rectangle plot has a length of 12 feet and a width of 8 feet. Tania and Alex live near some woods and they have seen deer and rabbits in their back yard on several occasions. Because of this, Alex knows that he will need to put some fencing around both of the garden plots. He is puzzled about how much fencing he will need. Alex needs to know the perimeter (the distance around the border) of each plot. Next, he needs to know how much area they will actually have to plant on. To figure this out, Alex needs the area of each garden plot. He wants to have the largest area to plant that he can. Alex has another idea too. He wonders what the dimensions will be if he puts the square plot right up against the rectangle plot. Will this change the amount of fencing he will need? Will this change the area of the garden plot? He has drawn some sketches", "Fences are an effective barrier* but they must work in concert with other security measures and structures. A gate is a chokepoint or a point where alt traffic must enter or exit the facility. All gates are not created equal, however, and if you select the incorrect one, you won’t get proper security. In fact, choosing the incorrect gate can even detract from an ol her wise effective security measure, A correctly chosen lj, l 1 1 l”‘ provides iio el iectivc deter renl and ti barrier lluil will slow down an in J ruder, whereas an incorrectly chosen harrier may not deter anyone but the casual intruder. UL Standard number 52 5 describes gate requirements. Gates are divided into the fo llowin g fo u r c I a ss i lie a M t >n s : • Residential or Class 1 — These are ornamental in design and offer little protection from intrusion. Commercial or Class 2 — These are of somewhat heavier construction and fall in the range of three to four feet in height. • Industrial or Class 3 — These are in the range of six to seven feet in height and are of heavier construction, including chain link construction. Restricted Access or Class 4 — These meet or exceed a height", "# A farmer has a vegetable garden which is 35 feet wide by 80 feet long. The farmer wishes to divide the garden equally between tomatoes, peppers, and c A farmer has a vegetable garden which is 35 feet wide by 80 feet long. The farmer wishes to divide the garden equally between tomatoes, peppers, and cabbage. However, the farmer would like to keep a one-foot border around the garden free for a fence. How many square feet can be used for the tomato planting? 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 liingliing8 Lenght of garden = 80 feet width of garden = 35 feet farmer would leki to keep a one foot boarder around the garden. Lenght for planting vegetable=80-2=78 feet width for planting vegetable=35-2=33 feet Area for planting veletable=78*33=2574 square feet Farmer divide the garden equally between tomatoes, peppers and cabbage. Therefore area for planting the tomatoes = $\\frac{1}{3}\\cdot 2574=858$ square feet. Hence 858 square feet can be used for the tomato planting", "the ground and the top landed at a distance of 15m from the heel of the tree? 7. Garden fence The garden has the shape of a rectangular triangle with an area of 96 square meters and a 16 m long one leg. How many meters of the fence need to be fenced? 8. Sea How far can you see from the ship's mast, whose peak is at 14 meters above sea level? (Earth's radius is 6370 km). 9. Equilateral triangle The equilateral triangle has a 23 cm long side. Calculate its content area. 10. Equilateral triangle Calculate the area of an equilateral triangle with circumference 72cm. 11. Square Calculate the area of the square shape of the isosceles triangle with the arms 50m and the base 60m. How many tiles are used to pave the square if the area of one tile is 25 dm2?", "of the AB side. The length of the AB side is 1 cm. What is the length of the XS segment? • The garden The garden has the shape of a rectangular trapezium. The bases have lengths of 27 meters and 36 meters, the trapezoid's height is 12 meters. Calculate how much fence will cost this garden if one meter costs 1.5 €? • Triangular prism The base of the perpendicular triangular prism is a rectangular triangle with a hypotenuse of 10 cm and one leg of 8 cm. The prism height is 75% of the perimeter of the base. Calculate the volume and surface of the prism.", "or the actual cubic feet needed. • To compute the amount of mulch needed of a rectangular garden, (CLICK HERE). • To compute the amount of mulch needed for a circular garden, (CLICK HERE). • To compute the amount of mulch needed an irregular four-sided garden, (CLICK HERE).", "will increase! Need anywhere from 100-200 square feet of growing space per person to grow vegetables. Been dropping their flower buds thing to consider is the size and layout of harvest. You must enter the area in square feet of a home-grown jalapeno pepper heats your!", "grow vegetables if a fifth of the area is occupied by a cottage, lawn and path? • Rectangular garden Rectangular garden has a length of 48.7 m, a width of 6.3 meters shorter than the length. How much mesh should be bought for its fencing if the gate is 2.9 m long and the gate 1.1 m? What is the area of the garden?", "Three-fourths of the area of a rectangular lawn 30 feet wide by 4o feet long is to be enclosed by a rectangular fence. If the enclosure has full width and reduced length rather than full length and reduced width, how much less fence will be needed? ~$\\frac{1}{2}$~ 5 10 15 20", "room, remodel their kitchen, or even build a table or chair. Farmers, gardeners, or even lot owners also use their knowledge in perimeter to fence their lots. More so, perimeters are important in estimating and calculating materials needed for completing a certain projects.", "• Cube One cube has edge increased 5 times. How many times will larger its surface area and volume? • Garden The area of a square garden is 6/4 of triangle garden with sides 56 m, 35 m, and 35 m. How many meters of fencing need to fence a square garden? • Trapezium The area of trapezium is 35cm2 find its altitude if the bases are 6cm and 8cm. • Ratio - rectangle The rectangle has dimensions 6 cm and 9 cm. How many times increases area and its perimeter, when its dimensions increase in the ratio 5:3? • Do you solve this? Determine area S of rectangle and length of its sides if its perimeter is 102 cm. • Simplify Simplify powers multiplication: (3+22)(5-42) • Garden Trapezoid garden has parallel sides 19 m and 24 m. Its area is 193.5 square meters. What is the width of the garden? • Trapezium Determine the height of the trapezium ABCD, which has area 77.5 cm2 and base lengths 16 cm and 15 cm. • Trapezoid Area of trapezoid is 135 cm2. Sides a, c and height h are in a ratio 6:4:3. How long are a,c and h? Make calculation... • Root The root of the equation ? is (equal or greater or less than zero)? ? •", "raised that he [or she] may see 1/3 of its surface? A square walled city of unknown dimensions has four gates, one at the center of each side.", "to reach into the middle of. Square foot gardening is a simple method of creating small, orderly, and highly productive kitchen gardens. We use cookies to give you the best experience on our website. – Try to avoid stepping on your soil and compacting it as much as possible. In my garden, I laid out 3′ wide walkways (wide enough for a wheelbarrow) and 4′ wide beds. Square Metre Gardening by Mel Bartholomew (Francis Lincoln Ltd, 2013) Grow all you can eat in 3 square feet (Dorling Kindersley 2015) www.squarefootgardening.org. Lettuce, head (romaine, butterhead, etc.). Manpukuji Carrots from @bakercreekseeds are i, Wow!", "feet exert on the ground while he slows down. expressthis force in newtons and also as a multiple of his weight. asked 2021-04-24 The following is an 8051 instruction: CJNE A, # 'Q' ,AHEAD a) what is the opcode for this instruction? b) how many bytes long is this instruction? c) explain the purpose of each byte of this instruction. d) how many machine cycles are required to execute this instruction? e) If an 8051 is operating from a 10 MHz crystal, how longdoes this instruction take to execute? asked 2020-11-07 A farmer has a vegetable garden which is 35 feet wide by 80 feet long. The farmer wishes to divide the garden equally between tomatoes, peppers, and cabbage. However, the farmer would like to keep a one-foot border around the garden free for a fence. How many square feet can be used for the tomato planting? asked 2021-05-12 Your town is installing a fountain in the main square. If thewater is to rise 25m(82feet)above the fountain,how much pressuremust the water have as it moves slowly toward the nozzle thatsprays it up into the air? Rather than putting a pump in the fountain(previous question)the engineer puts a water storage tank in a nearby high risebuilding.How high up in that building should the tank be for itswater to rise", "be one of those high appeal, low delivery concepts. Another advantage of raised beds is that they can be constructed over inhospitable ground. I have read accounts of raisedbed gardens producing well even when they are constructed over concrete and asphalt. To do this I imagine they would need to be at least twelve inches deep and even at that might not be favourable for deep rooting crops. Raised bed constructed to fit contour lines. Raised beds are not difficult to construct. They can be as simple as earth mounded up to form beds but I prefer some kind of wall or edging to hold the soil in place. These containment structures can be made of bricks or wood or even bales of hay. My raised beds are constructed on what used to be my regular garden plots so presumably fairly good soil underlies them for a foot or so and walls eight inches high are quite sufficient. If I was building over rocky clay or concrete I would want the walls a few inches higher. My walls are constructed out of two by eight planks and my longest bed is twelve to fifteen feet long and about four and a half feet wide. As mentioned, it is the width (typically four to four and one half feet)", "area of the larger garden is 9x^2 + 15x - 14 square feet. asked by Stacey on September 13, 2013 63. ## math a rectangular swimming pool 20 feet long and 10 feet wide,is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is required to enclose the deck? asked by Dee on September 19, 2013 64. ## michelle An object moving vertically is at the given heights at the specified times. Find the position equation s = at2 + v0t + s0 for the object given the following conditions. • At t = 1 seconds, s = 48 feet • At t = 2 seconds, s = 64 feet • At t = 3 asked by Anonymous on November 30, 2016 65. ## Math An interior painter is painting a wall that is 8 feet by 14 feet. He needs to determine how much paint he will need for the job. If a pint will cover 40 square feet, how many pints of paint are needed to paint the wall? asked by Anonymous on August 30, 2016 66. ## math a rectangular swimming pool 20 feet long and 10 feet wide,is enclosed by a wooden deck that is 3 feet wide. What is the area of" ]

[ "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", "That means he has 7 + 13 = 20 CDs Now subtract Ken’s CDs, Keith CDs from the total number of CDs. = 58 – 20 – 13 = 25 CDs. Thus George has 25 CDs. Question 6. Kathy has ribbons that have lengths of 7 inches, 10 inches, and 12 inches. Explain how she can use these ribbons to measure a length of 15 inches. Type below: _________ Answer: She could take the 10-inch ribbon and then use 5 inches from the 7-inch ribbon Question 7. A park has a rectangular playground area that has a length of 66 feet and a width of 42 feet. The park department has 75 yards of fencing material. Is there enough fencing material to enclose the playground area? Explain. ______ Explanation: A park has a rectangular playground area that has a length of 66 feet and a width of 42 feet. The park department has 75 yards of fencing material. Area of the rectangular playground = l × w A = 66 feet × 42 feet A = 2772 sq. ft Perimeter of the rectangular playground = 2l + 2w P = 2 × 66 + 2 × 42 P = 216 ft Now convert from feet to yard We know that 1 yard = 3 feet 216 ft =", "# The garden The garden has the shape of a rectangular trapezium. The bases have lengths of 27 meters and 36 meters, the trapezoid's height is 12 meters. Calculate how much fence will cost this garden if one meter costs 1.5 €? Result o = 135 Eur #### Solution: $a=27 \\ \\text{m} \\ \\\\ c=36 \\ \\text{m} \\ \\\\ h=12 \\ \\text{m} \\ \\\\ \\ \\\\ b=\\sqrt{ h^2 + (a-c)^2 }=\\sqrt{ 12^2 + (27-36)^2 }=15 \\ \\text{m} \\ \\\\ \\ \\\\ o_{ 1 }=a+b+c+h=27+15+36+12=90 \\ \\text{m} \\ \\\\ \\ \\\\ p=1.5 \\ \\text{Eur/m} \\ \\\\ o=p \\cdot \\ o_{ 1 }=1.5 \\cdot \\ 90=135 \\ \\text{Eur}$ 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 Pythagorean theorem is the base for the right triangle calculator. ## Next similar math problems: 1. ISO trapezoid v2 bases of Isosceles trapezoid measured 12 cm and 5 cm and its perimeter is 21 cm. What is the are of a trapezoid? 2. Estate Estate shaped rectangular trapezoid has bases", "# Difference between revisions of \"2020 AMC 8 Problems/Problem 3\" ## Problem 3 Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest? $\\textbf{(A) }560 \\qquad \\textbf{(B) }960 \\qquad \\textbf{(C) }1120 \\qquad \\textbf{(D) }1920 \\qquad \\textbf{(E) }3840$ ## Solution The answer is the product of the area of the field, the amount of strawberries per plant, and the amount of plants in one square feet. The answer is $6 \\times 8 \\times 10 \\times 4 = 1920$ or $\\boxed{\\textbf{(D) }1920}$. ## Solution 2 The area of the garden is $6$ ft $\\times$ $8$ ft $= 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, it follows that she plants a total of $48 \\times 4=192$ strawberry plants. Since each strawberry plant produces on average 10 strawberries, it follows that she can expect to harvest $192 \\times 10=1920$ strawberries $\\implies\\boxed{\\textbf{(D) }1920}$. ~junaidmansuri ## Solution 3 Note that $6\\cdot 8 = 48$, so Carrie has $4\\cdot 48 = 192$ strawberry plants. Each plant produces $10$ strawberries, so the final answer is $192\\cdot 10 = \\textbf{(D)}\\ 1920$.' ## Solution", "According to heron’s formula, area of the triangle = $$\\sqrt{s*(s-a)*(s-b)*(s-c)}$$ = $$\\sqrt{25*(25-10)*(25-16)*(25-24)}$$ = $$\\sqrt{25*15*9*1}$$ = $$15\\sqrt{15} cm^2$$ Note: Join free Sanfoundry classes at Telegram or Youtube 4. The sides of a triangle are in the proportion of 2 : 3 : 5 and its perimeter is 200 cm. The area of this triangle is __________ cm2. a) 375 $$\\sqrt{23}$$ b) 375 $$\\sqrt{21}$$ c) 345 $$\\sqrt{23}$$ d) 345 $$\\sqrt{21}$$ Explanation: Let the side of a triangles be a = 2x, b = 3x and c = 5x. Perimeter of the triangle is 200 cm. Therefore a + b + c = 2x + 3x + 5x = 200 10x = 200 Hence, x = 20 a = 2x = 2(20) = 40 cm b = 3x = 3(30) = 90 cm c = 5x = 5(20) = 100 cm Now, s = $$\\frac{a+b+c}{2}=\\frac{40+90+100}{2} = \\frac{230}{2}$$ = 115 According to heron’s formula, area of the triangle = $$\\sqrt{s*(s-a)*(s-b)*(s-c)}$$ = $$\\sqrt{115*(115-40)*(115-90)*(115-100)}$$ = $$\\sqrt{115*75*25*15}$$ = $$375 \\sqrt{23}$$ 5. A triangular garden has sides 90m, 140m and 80m. A fence is to be put all around the garden. What will be the total cost of fencing at the rate of Rs 15 per metre? A 5m wide space is to be left on the one side for gate opening. a) Rs 4525 b)", "the Defensive Area is: $D = \\left(\\frac{3}{20}\\right) 6300 = 3 \\cdot 315 = 945$", "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", "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", "# Suppose you have 24 feet of fencing to enclose your garden. The width of the garden is 2.8. What length must the garden be to use all of the fencing? Nov 20, 2016 $\\text{9.2 feet}$ #### Explanation: Let the length be $x$. The garden is in rectangular shape, so its perimeter will be 2 * (\"length\" + \"width\") \" units\" So as per question, $2 \\cdot \\left(\\text{length} + 2.8\\right) = 24$ $2 \\cdot \\left(x + 2.8\\right) = 24$ $2 x + 5.6 = 24$ $2 x = 24 - 5.6$ $2 x = 18.4$ $x = \\frac{18.4}{2} = \\text{9.2 feet}$ Hence the length will be $9.2$ feet to use all of the fencing.", "+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}$$", "of the AB side. The length of the AB side is 1 cm. What is the length of the XS segment? • The garden The garden has the shape of a rectangular trapezium. The bases have lengths of 27 meters and 36 meters, the trapezoid's height is 12 meters. Calculate how much fence will cost this garden if one meter costs 1.5 €? • Triangular prism The base of the perpendicular triangular prism is a rectangular triangle with a hypotenuse of 10 cm and one leg of 8 cm. The prism height is 75% of the perimeter of the base. Calculate the volume and surface of the prism.", "# A rectangle with perimeter 70 feet has area 300 square feet. What are its dimensions? Dec 3, 2016 $15 \\times 20$ feet #### Explanation: Let the length of one side be $x$. Going round the rectangle, the other sides are of lengths: $35 - x$ $x$ $35 - x$ The area of the rectangle is: $300 = x \\left(35 - x\\right)$ So essentially we are looking for a pair of factors of $300$ which sum to $35$. Since $300$ and $35$ are both divisible by $5$, the pair we are looking for must both be divisible by $5$ too. So the pairs of factors we need to consider are: $5 \\cdot 60 \\text{ }$ with $\\text{ } 5 + 60 = 65$ $10 \\cdot 30 \\text{ }$ with $\\text{ } 10 + 30 = 40$ $\\textcolor{b l u e}{15 \\cdot 20} \\text{ }$ with $\\text{ } \\textcolor{b l u e}{15 + 20 = 35}$", "Algebra 2 (1st Edition) The whole garden is $16\\cdot25=400$ square feet. If we have $x$ tomato plants and $y$ pepper plants, then our equation for the occupied area is: $8x+5y=400$. Thus if $x=15$, then $8\\cdot15+5y=400\\\\120+5y=400\\\\5y=280\\\\y=56$", "perimeter of a triangle is $35$ feet. One side of the triangle is $5$ feet longer than the second side. The third side is $3$ feet longer than the second side. Find the length of each side. One side of a triangle is twice the smallest side. The third side is $5$ feet more than the shortest side. The perimeter is $17$ feet. Find the lengths of all three sides. 3 ft, 6 ft, 8 ft One side of a triangle is three times the smallest side. The third side is $3$ feet more than the shortest side. The perimeter is $13$ feet. Find the lengths of all three sides. Use the Properties of Trapezoids In the following exercises, solve using the properties of trapezoids. The height of a trapezoid is $12$ feet and the bases are $9$ and $15$ feet. What is the area? 144 sq. ft The height of a trapezoid is $24$ yards and the bases are $18$ and $30$ yards. What is the area? Find the area of a trapezoid with a height of $51$ meters and bases of $43$ and $67$ meters. 2805 sq. m Find the area of a trapezoid with a height of $62$ inches and bases of $58$ and $75$ inches. The height of a trapezoid is $15$ centimeters and", "### Problem 25 Find the area of a square with diagonals of length $d$ . Amrita B. ### Problem 26 The length of a rectangle is 12 $\\mathrm{cm}$ more than its width. Find the area of the rectangle if its perimeter is 100 $\\mathrm{cm}$ . Check back soon! ### Problem 27 A path 2 $\\mathrm{m}$ wide surrounds a rectangular garden 20 $\\mathrm{m}$ long and 12 $\\mathrm{m}$ wide. Find the area of the path. Amrita B. How much will it cost to blacktop the driveway shown if blacktopping costs $\\$ 11.00$per square meter? SD Sylvana D. Numerade Educator ### Problem 29 A room 28 ft long and 20 ft wide has walls 8 ft high. a. What is the total wall area? b. How many gallon cans of paint should be bought to paint the walls if 1 gal of paint covers 300$\\mathrm{ft}^{2}$? Amrita B. Numerade Educator ### Problem 30 A wooden fence 6 ft high and 220 ft long is to be painted on both sides. a. What is the total area to be painted? b. A gallon of a certain type of paint will cover only 200$\\mathrm{ft}^{2}$of area for the first coat, but on the second coat a gallon of the same paint will cover 300$\\mathrm{ft}^{2}$. If the fence is to be given two coats of paint,", "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", "• Cube One cube has edge increased 5 times. How many times will larger its surface area and volume? • Garden The area of a square garden is 6/4 of triangle garden with sides 56 m, 35 m, and 35 m. How many meters of fencing need to fence a square garden? • Trapezium The area of trapezium is 35cm2 find its altitude if the bases are 6cm and 8cm. • Ratio - rectangle The rectangle has dimensions 6 cm and 9 cm. How many times increases area and its perimeter, when its dimensions increase in the ratio 5:3? • Do you solve this? Determine area S of rectangle and length of its sides if its perimeter is 102 cm. • Simplify Simplify powers multiplication: (3+22)(5-42) • Garden Trapezoid garden has parallel sides 19 m and 24 m. Its area is 193.5 square meters. What is the width of the garden? • Trapezium Determine the height of the trapezium ABCD, which has area 77.5 cm2 and base lengths 16 cm and 15 cm. • Trapezoid Area of trapezoid is 135 cm2. Sides a, c and height h are in a ratio 6:4:3. How long are a,c and h? Make calculation... • Root The root of the equation ? is (equal or greater or less than zero)? ? •", "# Difference between revisions of \"1999 AMC 8 Problems/Problem 5\" ## Problem A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden? $\\text{(A)}\\ 100 \\qquad \\text{(B)}\\ 200 \\qquad \\text{(C)}\\ 300 \\qquad \\text{(D)}\\ 400 \\qquad \\text{(E)}\\ 500$ ## Solution We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$. To get the greatest area while keeping a perimeter of $160$, the sides should all be $40$. that means an area of $1600$. Right now, the area is $20 \\times 60$ which is $1200$. $1600-1200=400$ which is $\\boxed{D}$.", "### Home > ACC6 > Chapter 8 Unit 8 > Lesson CC1: 8.2.1 > Problem8-79 8-79. Mr. Nowlin has a rectangular garden with an area of $47.25$ square feet and a length of $10.5$ feet. 1. What is the width of the garden? Notice the relationship between the area and the length. Width: $4.5$ Feet 1. What is the garden’s perimeter? The perimeter of a rectangle is all of the sides added together. Perimeter: $30$ Feet", "with—after your new land has arrived—and make sure your strength is high to remove yourself from the spotlight. Explore efficientlyEdit Although sending the explore is easy, doing it efficiently is hard. Whenever possible, it is best to do the largest explore you can, 25% of your current land size. You can take a look at the formulae page to calculate the amount of Barracks you should build to have the lowest explore costs possible. Exploring FormulaeEdit Money Required (per acre): $(3000+(6\\cdot Land))\\cdot\\left(1-\\frac{GrowthResearch}{200}+\\frac{Barracks}{Land\\cdot 2}\\right)$ Defense Required (per acre): $(0.025\\cdot Land)\\cdot\\left(1-\\frac{GrowthResearch}{100}\\right)$ Citizens Required (per acre): $(70+(0.03\\cdot Land)\\cdot\\left(1-\\frac{GrowthResearch}{100}\\right)$ Community content is available under CC-BY-SA unless otherwise noted." ]

[ "# A farmer has a vegetable garden which is 35 feet wide by 80 feet long. The farmer wishes to divide the garden equally between tomatoes, peppers, and c A farmer has a vegetable garden which is 35 feet wide by 80 feet long. The farmer wishes to divide the garden equally between tomatoes, peppers, and cabbage. However, the farmer would like to keep a one-foot border around the garden free for a fence. How many square feet can be used for the tomato planting? 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 liingliing8 Lenght of garden = 80 feet width of garden = 35 feet farmer would leki to keep a one foot boarder around the garden. Lenght for planting vegetable=80-2=78 feet width for planting vegetable=35-2=33 feet Area for planting veletable=78*33=2574 square feet Farmer divide the garden equally between tomatoes, peppers and cabbage. Therefore area for planting the tomatoes = $\\frac{1}{3}\\cdot 2574=858$ square feet. Hence 858 square feet can be used for the tomato planting", "# Difference between revisions of \"2020 AMC 8 Problems/Problem 3\" ## Problem 3 Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest? $\\textbf{(A) }560 \\qquad \\textbf{(B) }960 \\qquad \\textbf{(C) }1120 \\qquad \\textbf{(D) }1920 \\qquad \\textbf{(E) }3840$ ## Solution The answer is the product of the area of the field, the amount of strawberries per plant, and the amount of plants in one square feet. The answer is $6 \\times 8 \\times 10 \\times 4 = 1920$ or $\\boxed{\\textbf{(D) }1920}$. ## Solution 2 The area of the garden is $6$ ft $\\times$ $8$ ft $= 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, it follows that she plants a total of $48 \\times 4=192$ strawberry plants. Since each strawberry plant produces on average 10 strawberries, it follows that she can expect to harvest $192 \\times 10=1920$ strawberries $\\implies\\boxed{\\textbf{(D) }1920}$. ~junaidmansuri ## Solution 3 Note that $6\\cdot 8 = 48$, so Carrie has $4\\cdot 48 = 192$ strawberry plants. Each plant produces $10$ strawberries, so the final answer is $192\\cdot 10 = \\textbf{(D)}\\ 1920$.' ## Solution", "According to heron’s formula, area of the triangle = $$\\sqrt{s*(s-a)*(s-b)*(s-c)}$$ = $$\\sqrt{25*(25-10)*(25-16)*(25-24)}$$ = $$\\sqrt{25*15*9*1}$$ = $$15\\sqrt{15} cm^2$$ Note: Join free Sanfoundry classes at Telegram or Youtube 4. The sides of a triangle are in the proportion of 2 : 3 : 5 and its perimeter is 200 cm. The area of this triangle is __________ cm2. a) 375 $$\\sqrt{23}$$ b) 375 $$\\sqrt{21}$$ c) 345 $$\\sqrt{23}$$ d) 345 $$\\sqrt{21}$$ Explanation: Let the side of a triangles be a = 2x, b = 3x and c = 5x. Perimeter of the triangle is 200 cm. Therefore a + b + c = 2x + 3x + 5x = 200 10x = 200 Hence, x = 20 a = 2x = 2(20) = 40 cm b = 3x = 3(30) = 90 cm c = 5x = 5(20) = 100 cm Now, s = $$\\frac{a+b+c}{2}=\\frac{40+90+100}{2} = \\frac{230}{2}$$ = 115 According to heron’s formula, area of the triangle = $$\\sqrt{s*(s-a)*(s-b)*(s-c)}$$ = $$\\sqrt{115*(115-40)*(115-90)*(115-100)}$$ = $$\\sqrt{115*75*25*15}$$ = $$375 \\sqrt{23}$$ 5. A triangular garden has sides 90m, 140m and 80m. A fence is to be put all around the garden. What will be the total cost of fencing at the rate of Rs 15 per metre? A 5m wide space is to be left on the one side for gate opening. a) Rs 4525 b)", "a rectangular solid (with a square base) with maximum volume if its surface area is 337. Find the dimensions of the box that will minimize the amount of. The equation for the volume of a cube is: V=x ^2h. Jadhav2, Parag D. Find the length of the sides of the square that must cut out if the volume of the box is to be maximized. The sides will then be bent up. The margins on each side are to be 1 /1 2 inches. Find the volume of the largest box that can be made in this manner. See full list on tutorial. Check: V = πr²h = π(3. The volume of a rectangular box is the amount of space occupied by the object. How large the square should be to make the box with the largest possible volume?. What value of x gives the maximum area?. A landscape architect plans to enclose a 3000 square foot rectangular region in a botanical garden, She will use shrubs costing$25 per foot along three sides and fencing costing $10 per foot along the fourth side, Find the minimum total cost. Question: From a thin piece of cardboard 12 in x 8 in. IB QUESTION: A open rectangular box has a height y cm and width x cm, with no top.", "That means he has 7 + 13 = 20 CDs Now subtract Ken’s CDs, Keith CDs from the total number of CDs. = 58 – 20 – 13 = 25 CDs. Thus George has 25 CDs. Question 6. Kathy has ribbons that have lengths of 7 inches, 10 inches, and 12 inches. Explain how she can use these ribbons to measure a length of 15 inches. Type below: _________ Answer: She could take the 10-inch ribbon and then use 5 inches from the 7-inch ribbon Question 7. A park has a rectangular playground area that has a length of 66 feet and a width of 42 feet. The park department has 75 yards of fencing material. Is there enough fencing material to enclose the playground area? Explain. ______ Explanation: A park has a rectangular playground area that has a length of 66 feet and a width of 42 feet. The park department has 75 yards of fencing material. Area of the rectangular playground = l × w A = 66 feet × 42 feet A = 2772 sq. ft Perimeter of the rectangular playground = 2l + 2w P = 2 × 66 + 2 × 42 P = 216 ft Now convert from feet to yard We know that 1 yard = 3 feet 216 ft =", "wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 10 feet by 1,000 feet. it will have 2,020 feet of wall for 10,000 square feet, or 4.950 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 20 feet by 500 feet. it will have 1,040 feet of wall for 10,000 square feet, or 9.615 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 40 feet by 250 feet. it will have 580 feet of wall for 10,000 square feet, or 17.241 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 80 feet by 125 feet. it will have 410 feet of wall for 10,000 square feet, or 24.390 square feet for every foot of wall. Suppose that a walled enclosure has 10,000 square feet of ground and is 100 feet by 100 feet. it will have 400 feet of wall for 10,000 square feet, or 25 square feet for every foot of wall. So, out of all possible rectangles, a square shape gives the least amount of wall to be defended for the same square footage inside. But a square", "# Suppose you have 24 feet of fencing to enclose your garden. The width of the garden is 2.8. What length must the garden be to use all of the fencing? Nov 20, 2016 $\\text{9.2 feet}$ #### Explanation: Let the length be $x$. The garden is in rectangular shape, so its perimeter will be 2 * (\"length\" + \"width\") \" units\" So as per question, $2 \\cdot \\left(\\text{length} + 2.8\\right) = 24$ $2 \\cdot \\left(x + 2.8\\right) = 24$ $2 x + 5.6 = 24$ $2 x = 24 - 5.6$ $2 x = 18.4$ $x = \\frac{18.4}{2} = \\text{9.2 feet}$ Hence the length will be $9.2$ feet to use all of the fencing.", "garden is $4848$ feet. The lengths of two sides are $1818$ feet and $2222$ feet. How long is the third side? Try It 9.68 The lengths of two sides of a triangular window are $77$ feet and $55$ feet. The perimeter is $1818$ feet. How long is the third side? ### Example 9.35 The area of a triangular church window is $9090$ square meters. The base of the window is $1515$ meters. What is the window’s height? Try It 9.69 The area of a triangular painting is $126126$ square inches. The base is $1818$ inches. What is the height? Try It 9.70 A triangular tent door has an area of $1515$ square feet. The height is $55$ feet. What is the base? #### Isosceles and Equilateral Triangles Besides the right triangle, some other triangles have special names. A triangle with two sides of equal length is called an isosceles triangle. A triangle that has three sides of equal length is called an equilateral triangle. Figure 9.24 shows both types of triangles. Figure 9.24 In an isosceles triangle, two sides have the same length, and the third side is the base. In an equilateral triangle, all three sides have the same length. ### Isosceles and Equilateral Triangles An isosceles triangle has two sides the same length. An equilateral triangle", "# Q : 6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.(ii) How much of tape is needed for all the 12 edges? Given, dimensions of the greenhouse = $30\\ cm \\times25\\ cm \\times25\\ cm$ (ii) Tape needed for all the 12 edges = Perimeter =$4(l+b+h)$ $4(30+25+25) = 320\\ cm$ Therefore, $320\\ cm$ of tape is needed for the edges.", "will increase! Need anywhere from 100-200 square feet of growing space per person to grow vegetables. Been dropping their flower buds thing to consider is the size and layout of harvest. You must enter the area in square feet of a home-grown jalapeno pepper heats your!", "# Difference between revisions of \"1999 AMC 8 Problems/Problem 5\" ## Problem A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden? $\\text{(A)}\\ 100 \\qquad \\text{(B)}\\ 200 \\qquad \\text{(C)}\\ 300 \\qquad \\text{(D)}\\ 400 \\qquad \\text{(E)}\\ 500$ ## Solution We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$. To get the greatest area while keeping a perimeter of $160$, the sides should all be $40$. that means an area of $1600$. Right now, the area is $20 \\times 60$ which is $1200$. $1600-1200=400$ which is $\\boxed{D}$.", "we find it to be 10,642 pounds per one acre of land. However, 38% of those who say they measure this metric didn't know the actual yield per square foot of their. Strawberry Production: Conclusion. 5 pounds per foot of row for years 2 and 3. 000 square meters. 28 t/ha with an increase of 10 per cent over the local. cabbage) or 100 plants per square metre (e. With 3 inch seed/plant spacing needs, you can grow 16 plants in a 1 square foot area. 37M2) plant density. Any lettuce, from the salad bowl to Romaine could save you more than $18 per a square foot in your. These traditional leafy vegetables are now The size of each plot was 1. 2 stems per square meter. Yields that a commercial grower may expect from the main vegetable crops grown, divided according to the above categories, are suggested in the following table. is strongly advised before any production activities are started. Suggested number of vegetable plants per square foot:. The Vegetable Production System (Veggie) is a plant growth system developed and used by NASA in outer space environments. Marketing costs are relatively low in The Netherlands at US$0. Owing to their high beta-carotene content, sweet potatoes are perhaps the healthiest foods anybody can consume. 0 square meters). I", "has been working on her tomato plants, Alex has been working on designing the garden plot. He knows that he wants two plots, one to be in the shape of a square and one to be the shape of a rectangle. His square plot has a length and width of 9 feet. His rectangle plot has a length of 12 feet and a width of 8 feet. Tania and Alex live near some woods and they have seen deer and rabbits in their back yard on several occasions. Because of this, Alex knows that he will need to put some fencing around both of the garden plots. He is puzzled about how much fencing he will need. Alex needs to know the perimeter (the distance around the border) of each plot. Next, he needs to know how much area they will actually have to plant on. To figure this out, Alex needs the area of each garden plot. He wants to have the largest area to plant that he can. Alex has another idea too. He wonders what the dimensions will be if he puts the square plot right up against the rectangle plot. Will this change the amount of fencing he will need? Will this change the area of the garden plot? He has drawn some sketches", "# March 2019 (CML) I will post seven new problems every weekend, one for each day of the week. The idea is that you work on your own during the week and I will post answers the following Saturday. For those of you who practiced with me last year, the process is similar. For instructions on submitting weekly answers, click on the “Logistics” tab. For previously posted problems, click on the “Archive” tab. For previous prizes, click on the “News” tab. Week of March 24 – March 30. March 24: A rectangular garden 50 feet long and 10 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden? Answer: 400 square feet. The perimeter of the rectangular garden is 120 feet. When it is made into a square, the side length is 30 feet. Therefore, the area is enlarged from 500 square feet to 900. March 25: Bo, Coe, Flo, Joe, and Moe have different amounts of money. Neither Joe nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Joe has more than Moe, but less than Bo. Who has the least amount of money? Answer: J", "area of the larger garden is 9x^2 + 15x - 14 square feet. asked by Stacey on September 13, 2013 63. ## math a rectangular swimming pool 20 feet long and 10 feet wide,is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is required to enclose the deck? asked by Dee on September 19, 2013 64. ## michelle An object moving vertically is at the given heights at the specified times. Find the position equation s = at2 + v0t + s0 for the object given the following conditions. • At t = 1 seconds, s = 48 feet • At t = 2 seconds, s = 64 feet • At t = 3 asked by Anonymous on November 30, 2016 65. ## Math An interior painter is painting a wall that is 8 feet by 14 feet. He needs to determine how much paint he will need for the job. If a pint will cover 40 square feet, how many pints of paint are needed to paint the wall? asked by Anonymous on August 30, 2016 66. ## math a rectangular swimming pool 20 feet long and 10 feet wide,is enclosed by a wooden deck that is 3 feet wide. What is the area of", "perimeter of a triangle is $35$ feet. One side of the triangle is $5$ feet longer than the second side. The third side is $3$ feet longer than the second side. Find the length of each side. One side of a triangle is twice the smallest side. The third side is $5$ feet more than the shortest side. The perimeter is $17$ feet. Find the lengths of all three sides. 3 ft, 6 ft, 8 ft One side of a triangle is three times the smallest side. The third side is $3$ feet more than the shortest side. The perimeter is $13$ feet. Find the lengths of all three sides. Use the Properties of Trapezoids In the following exercises, solve using the properties of trapezoids. The height of a trapezoid is $12$ feet and the bases are $9$ and $15$ feet. What is the area? 144 sq. ft The height of a trapezoid is $24$ yards and the bases are $18$ and $30$ yards. What is the area? Find the area of a trapezoid with a height of $51$ meters and bases of $43$ and $67$ meters. 2805 sq. m Find the area of a trapezoid with a height of $62$ inches and bases of $58$ and $75$ inches. The height of a trapezoid is $15$ centimeters and", "anonymous 40. anonymous ok cya 41. anonymous k 42. anonymous It is recommended that there be at least 15.9 square feet of ground space in a garden for every newly planted shrub. A garden is 37.1 feet by 15 feet. Find the maximum number of shrubs the garden can accommodate. a. 185 shrubs c. 35 shrubs b. 12 shrubs d. 2 shrubs", "thunder was able to travel $$10 \\cdot 1088 = 10880$$ feet. Note that $$2$$ miles is about $$2 \\cdot 5280 = 10560$$ feet, which is close to what we found above. Thus, C is the correct answer. 6. Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is $$60$$ feet. What is the distance in feet between the first and last trees? $$90$$ $$100$$ $$105$$ $$120$$ $$140$$ ###### Solution(s): There are $$3$$ gaps between the first and fourth trees, which means that one gap is $$60 \\div 3 = 20$$ feet long. There are $$5$$ gaps between the first and last trees. This means that they are $$5 \\cdot 20 = 100$$ feet apart. Thus, B is the correct answer. 7. What is the number of square inches in the area of the small kite? $$21$$ $$22$$ $$23$$ $$24$$ $$25$$ ###### Solution(s): Recall that the area of a kite is the product of its diagonals divided by $$2.$$ Therefore, the area of the small kite is $\\dfrac{6 \\cdot 7}{2} = \\dfrac{42}{2} = 21.$ Thus, A is the correct answer. 8. Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does", "### Home > ACC6 > Chapter 8 Unit 8 > Lesson CC1: 8.2.1 > Problem8-79 8-79. Mr. Nowlin has a rectangular garden with an area of $47.25$ square feet and a length of $10.5$ feet. 1. What is the width of the garden? Notice the relationship between the area and the length. Width: $4.5$ Feet 1. What is the garden’s perimeter? The perimeter of a rectangle is all of the sides added together. Perimeter: $30$ Feet", "has a vegetable garden which is 35 feet wide by 80 feet long. The farmer wishes to divide the garden equally between tomatoes, peppers, and cabbage. However, the farmer would like to keep a one-foot border around the garden free for a fence. How many square feet can be used for the tomato planting? Wierzycaz 2020-11-01 ### Math word problem that states t-shirts are$10 each and sweatshirts are $15 each, if 100 items were sold and they raised$1120, how many sweatshirts were sold, how would i set up this problem BolkowN 2020-10-28 ### a.)Explain in your own words what this problem is asking. b.)Explain the meaning of any notation used in the problem and in your solution. c.)Describe the mathematical concept(s) that appear to be foundational to this problem. d.)Justified solution to or proof of the problem. Find the greatest common divisior of a,b, and c and write it in the form ax+by+cz for integers x,y, and z. a=26,b=52,c=60 The Math word problem, in simple terms, is a problem that is presented in words, meaning that it can relate to integrals, statistics, analysis, basically anything that deals with calculations. It also helps to look at various questions and answers to help yourself see the logic. If you are challenged with math word problem help, do not forget to" ]

[ "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar. The ratio of their sides is $\\frac{AM}{AB} = \\frac 58$, hence the ratio of their areas is $\\left( \\frac 58 \\right)^2 = \\frac{25}{64}$. And as the area of triangle $ABC$ is $\\frac{6\\cdot 8}2 = 24$, the area of triangle $AME$ is $24\\cdot \\frac{25}{64} = \\boxed{ \\frac{75}8 }$. ## Solution 3 (Only Pythagorean Theorem) $[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); label(\"E\",Ep,S); label(\"M\",M,NW); [/asy]$ Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \\implies 25 + ME^2$, which means $AE = \\sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB = 8 - \\sqrt{25 + x^2}$. From here, we know that $CE^2 = CB^2 + BE^2 = ME^2 + MC^2$. From here, we can write the expression $6^2 +", "hexagon. Now, using the sine law, $\\frac{1}{\\sin \\angle ACB} = \\frac{1}{\\sin 60^\\circ}$, so $\\angle ACB = 60^\\circ$. Since the angles in a triangle sum to 180$^\\circ$$\\angle ABC$ is also 60$^\\circ$. Therefore, $\\triangle ABC$ is an equilateral triangle with side lengths of 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)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label(\"2\",(z,c),NE); label(\"1\",(1.5,0),S); label(\"1\",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label(\"60^\\circ\",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label(\"60^\\circ\",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label(\"60^\\circ\",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); [/asy]$ Since the area of a regular hexagon can be found with the formula $\\frac{3\\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\\frac{3\\sqrt{3}(2^2)}{2} = 6\\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\\frac{\\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\\frac{\\sqrt{3}}{4}(1^2) = \\frac{\\sqrt{3}}{4}$. Since the area of a circle can be found", "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", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "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 =", "# 2021 AIME I Problems/Problem 2 ## Problem In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, W); label(\"B\", B, W); label(\"C\", C, (1,0)); label(\"D\", D, (1,0)); label(\"F\", F, N); label(\"E\", E, S); [/asy]$ ## Solution 1 (Similar Triangles) Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\\frac{7}{3}(11-x)$ and $CG=\\frac{3}{7}x$. We have $\\frac{7}{3}(11-x)+\\frac{3}{7}x=FC=9$. Solving for $x$, we have $x=\\frac{35}{4}$. The area of the shaded region is just $3\\cdot \\frac{35}{4}=\\frac{105}{4}$. Thus, the answer is $105+4=\\framebox{109}$. ~yuanyuanC ## Solution 2 (Similar Triangles) Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\\triangle AFG \\sim \\triangle", "# 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 $B$.", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "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", "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]", "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)", "angle bisector of the angle at vertex $A$ in triangle $ABC$. Then $\\frac{BA'}{A'C}=\\frac{BA}{AC}$. \\end{lemma} \\begin{proof} From lemma ..., we know that $\\frac{BA'}{A'C}=\\frac{\\area(BA'A)}{\\area(A'CA)}$. Now the equality of angles $\\angle BAA'$ and $\\angle A'AC$ implies that if we reflect the triangle $A'CA$ in the line $AA'$, then the reflected copy will share an angle with triangle $BA'A$ and the reflection of $C$ will lie on the line $AB$ (see picture). \\begin{figure}[h] \\centering \\begin{asy} size (6cm,3cm); pair A, B, C, A1, C1; picture pic; A=(2,2); B=(0,0); C=(3,0); A1=(length(C-A)*B+length(B-A)*C)/(length(B-A)+length(C-A)); C1=A+2*dot(A1-A,C-A)/dot(A1-A,A1-A)*(A1-A)-(C-A); draw(A--B--C--cycle); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); draw(A--A1); label(\"$A'$\",A1,S); draw(pic,A--B--A1--cycle); label(pic,\"$A$\",A,N); label(pic,\"$A'$\",A1,S); label(pic,\"$B$\",B,S); draw(pic,A1--C--A,linetype(\"0 4\")); draw(pic,A1--C1--A); label(pic,\"$C$\",C1,WNW); \\end{asy} \\label{fig:bisector} \\end{figure} But in the picture after the reflection we see that $\\frac{\\area(BA'A)}{\\area(A'CA)}=\\frac{AB}{AC}$, by the same lemma. \\end{proof} This lemma implies that if $AA',BB'$ and $CC'$ are internal bisectors of triangle $ABC$, then $\\frac{BA'}{A'C}\\cdot\\frac{CB'}{B'A}\\cdot\\frac{AC'}{C'B}=\\frac{BA}{AC}\\cdot\\frac{CB}{BA}\\cdot\\frac{AC}{CB}=1$, so Ceva's theorem tells us that $AA', BB'$ and $CC'$ are concurrent. We can prove it in a different way. For this we first notice that the internal angle bisector of an angle $A$ is exactly the locus of points $P$ inside the angle that are equidistant from the two rays that form $A$. Indeed, let $P'$ and $P''$ be the feet of perpendiculars dropped from the point $P$ to the two rays. If the point $P$ is on the angle bisector of", "(triangle[2][1] - triangle[0][1]) \\ + triangle[2][0] * (triangle[0][1] - triangle[1][1]) area /= 2 area = abs(area) print(area) # prints 20.0", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "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 $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ 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" ]

[ "$$DL$$ (the height). The area of parallelogram is given as $$1470 \\,\\rm cm^2.$$ By using the formula of area of parallelogram we can find the height $$DL$$. Similarly, if you take $$AD$$ as the base and $$BM$$ as perpendicular (height) on $$AD$$, again by using the formula of area of parallelogram we can find the length of $$BM$$. Steps: Given, $$AB$$$$= 35 \\,\\rm cm$$, $$BC = 49 \\rm \\,cm$$ and Area $$= 1470 \\,\\rm cm^2$$ Area of parallelogram ABCD $$=$$ Base $$(AB)$$ $$\\times$$ Height $$(DL)$$ \\begin{align}1470 &= 35 \\times {{DL}}\\\\{{DL}} &= \\frac{{1470}}{{35}}\\\\{{DL}} &= 42\\;\\rm cm\\end{align} Area of parallelogram ABCD $$=$$ Base $$(BC)$$ $$\\times$$ Height $$(BM)$$ \\begin{align}1470 &= 49 \\times {{BM}}\\\\{{BM}} &= \\frac{{1470}}{{49}}\\\\{{BM}} &= 30\\;\\rm cm\\end{align} ## Chapter 11 Ex.11.2 Question 7 $$\\triangle ABC$$ is right angled at$$\\, A \\,$$ (Fig 11.25). $$AD$$ is perpendicular to $$BC$$. If $$AB = 5 \\,\\rm cm$$, $$BC = 13 \\,\\rm cm$$ and $$AC = 12 \\,\\rm cm$$, Find the area of $$\\triangle ABC$$. Also find the length of $$AD$$. ### Solution What is known? $$\\triangle ABC$$ is right angled at $$A$$, $$AD$$ is perpendicular to side $$BC$$. Length of three sides of the triangle is also given. What is unknown? The area of $$\\triangle ABC$$ and the length of perpendicular $$AD$$. Reasoning: First, find the area of triangle using base $$(5 \\,\\rm", "310 views On a triangle $\\text{ABC}$, a circle with diameter $\\text{BC}$ is drawn, intersecting $\\text{AB}$ and $\\text{AC}$ at points $\\text{P}$ and $\\text{Q}$, respectively. If the lengths of $\\text{AB, AC}$, and $\\text{CP}$ are $30$ cm, $25$ cm, and $20$ cm respectively, then the length of $\\text{BQ}$, in cm, is __________ Given that, $\\text{BC}$ is a diameter of circle, the lengths of $\\text{AB} = 30 \\; \\text{cm}, \\text{AC} = 25 \\; \\text{cm},$ and $\\text{CP} = 20 \\; \\text{cm}.$ Now, we can draw the diagram : The angle inscribed in a semicircle is always a right angle. So, $\\triangle \\text{BPC},$ and $\\triangle \\text{BQC}$ are right angle triangle. Now, we can draw the $\\triangle \\text{ABC}$ as : Area of the $\\triangle \\text{ABC} = \\frac{1}{2} \\times \\text{Base} \\times \\text{Height}$ $\\quad = \\frac{1}{2} \\times \\text{AB} \\times \\text{CP}$ $\\quad = \\frac{1}{2} \\times 30 \\times 20$ $\\quad = 300 \\; \\text{cm}^{2}$ Again, we can draw the $\\triangle \\text{ABC}$ as : Area of the $\\triangle \\text{ABC} = \\frac{1}{2} \\times \\text{AC} \\times \\text{BQ}$ $\\Rightarrow 300 = \\frac{1}{2} \\times 25 \\times \\text{BQ}$ $\\Rightarrow 25 \\; \\text{BQ} = 600$ $\\Rightarrow \\text{BQ} = \\frac{600}{25}$ $\\Rightarrow \\boxed{\\text{BQ} = 24 \\; \\text{cm}}$ $\\therefore$ The length of $\\text{BQ}$ is $24 \\; \\text{cm}.$ Correct Answer $: 24$ 10.3k points 1 349 views 2 460 views", "\\text{29}\\text{ cm}$$ and $$\\hat{X} = \\text{58,9}\\text{°}$$. \\begin{align*} \\text{Area } \\triangle XYZ &= \\frac{1}{2} (52)(29) \\sin \\text{58,9}\\text{°} \\\\ &= \\text{645,6}\\text{ square units } \\end{align*} Determine the area of a parallelogram in which two adjacent sides are $$\\text{10}$$ $$\\text{cm}$$ and $$\\text{13}$$ $$\\text{cm}$$ and the angle between them is $$\\text{55}$$$$\\text{°}$$. \\begin{align*} \\text{Area } &= 2 \\left( \\frac{1}{2} (13)(10) \\sin \\text{55}\\text{°} \\right) \\\\ &= \\text{106,5}\\text{ square units } \\end{align*} If the area of $$\\triangle ABC$$ is $$\\text{5 000}$$ $$\\text{m^{2}}$$ with $$a=\\text{150}\\text{ m}$$ and $$b=\\text{70}\\text{ m}$$, what are the two possible sizes of $$\\hat{C}$$? \\begin{align*} \\text{Area } &= \\frac{1}{2} (150)(70) \\sin \\hat{C} \\\\ \\therefore \\text{5 000} &= \\text{5 250} \\sin \\hat{C} \\\\ \\therefore \\sin \\hat{C} &= \\frac{\\text{5 000} }{\\text{5 250} } \\\\ \\therefore \\hat{C} &= \\text{72,2}\\text{°} \\\\ \\text{Or } \\hat{C} &= \\text{180}\\text{°} - \\text{72,2}\\text{°} \\\\ &= \\text{107,8}\\text{°} \\end{align*} ### The sine rule (EMBHR) So far we have only applied the trigonometric ratios to right-angled triangles. We now expand the application of the trigonometric ratios to triangles that do not have a right angle: ## The sine rule In $$\\triangle ABC$$, $$AC = 15$$, $$BC = 11$$ and $$\\hat{A} = \\text{48}\\text{°}$$. Find $$\\hat{B}$$. 1. Method $$\\text{1}$$: using the sine ratio 1. Draw a sketch of $$\\triangle ABC$$. 2. Construct $$CF \\perp AB$$. 3. In $$\\triangle CBF$$: \\begin{align*} \\frac{CF}{\\ldots} &= \\sin \\hat{B} \\\\ \\therefore CF", "& = \\text{9,86904...} \\\\ & \\approx \\text{9,87} \\end{align*} Calculate the area of $$BCD$$ to $$\\text{2}$$ decimal places. $$BCD$$ is a right-angled triangle and so we have the perpendicular height. The area is: \\begin{align*} A & = \\frac{1}{2} b h\\\\ & = \\frac{1}{2} \\pi^2 \\\\ & = \\text{4,934802...} \\\\ & \\approx \\text{4,93} \\end{align*} Using you answers in (a) and (b) calculate the area of $$ABCDE$$. The area of $$ABCDE$$ is the sum of the areas of $$ABDE$$ and $$BCD$$. \\begin{align*} A & = \\text{9,87} + \\text{4,93} \\\\ & \\approx \\text{14,80} \\end{align*} Without rounding off, what is the area of $$ABCDE$$? \\begin{align*} A_{ABCDE} & = A_{ABDE} + A_{BCD} \\\\ & = l^{2} + \\frac{1}{2}bh \\\\ &= \\pi ^2 + \\frac{1}{2} \\pi^2 \\\\ &= \\text{14,8044...} \\end{align*} Given $$i = \\dfrac{r}{600}$$; $$r = \\text{7,4}$$; $$n = 96$$; $$P = \\text{200 000}$$. Calculate $$i$$ correct to $$\\text{2}$$ decimal places. \\begin{align*} i & = \\frac{r}{600} \\\\ & = \\frac{\\text{7,4}}{600} \\\\ & = \\text{0,01233} \\\\ & \\approx \\text{0,01} \\end{align*} Using you answer from (a), calculate $$A$$ in $$A = P(1+ i)^n$$. \\begin{align*} A &= P(1+ i)^n \\\\ &= \\text{200 000}\\left(1+ \\text{0,01}\\right)^{96} \\\\ &= \\text{519 854,59} \\end{align*} Calculate $$A$$ without rounding off your answer in (a), compare this answer with your answer in (b). \\begin{align*} A &= P(1+ i)^n \\\\ A &= \\text{200 000}\\left(1+ \\frac{\\text{7,4}}{600}\\right)^{96} \\\\", "\\end{align*} Consider the line segment shown below. Express the following in terms of $$a$$ and $$b$$: $$PT: ST$$ $$(a + b): b$$ $$\\frac{PS}{TQ}$$ $$\\frac{a}{2a} = \\frac{1}{2}$$ $$\\frac{SQ}{PQ}$$ $$\\frac{2a + b}{3a + b}$$ $$QT: TS$$ $$2a: b$$ $$ABCD$$ is a parallelogram with $$DC = \\text{15}\\text{ cm}$$, $$h = \\text{8}\\text{ cm}$$ and $$BF = \\text{9}\\text{ cm}$$. Calculate the ratio $$\\dfrac{\\text{area } ABF}{\\text{area } ABCD}$$. The area of a parallelogram $$ABCD =$$ base $$\\times$$ height: \\begin{align*} \\text{Area} &= \\text{15} \\times \\text{8} \\\\ &= \\text{120}\\text{ cm$^{2}$} \\end{align*} The perimeter of a parallelogram $$ABCD = 2DC + 2BC$$. To find the length of $$AF$$, we use $$AF \\perp BC$$ and the theorem of Pythagoras. \\begin{align*} \\text{In $$\\triangle ABF$$:} \\quad AF^2 &= AB^2 - BF^2 \\\\ &= \\text{15}^2 - \\text{9}^2 \\\\ &= \\text{144} \\\\ \\therefore AF &= \\text{12}\\text{ cm} \\\\ \\therefore \\text{area } ABF &= \\frac{1}{2} AF \\cdot BF \\\\ &= \\frac{1}{2}(12)(9)\\\\ &= \\text{54}\\text{ cm} \\end{align*} \\begin{align*} \\therefore \\frac{\\text{area } ABF}{\\text{area } ABCD} &= \\frac{54}{120} \\\\ &= \\frac{9}{20} \\end{align*} $$AB = \\text{36}\\text{ m}$$ and $$C$$ divides $$AB$$ in the ratio $$4:5$$. Determine $$AC$$ and $$CB$$. \\begin{align*} \\frac{AC}{AB} &= \\frac{4k}{(4k + 5k)} \\\\ \\therefore AC &= 36 \\times \\frac{4}{9} \\\\ &= \\text{16}\\text{ m} \\\\ CB &= 36 - 16 \\\\ &= \\text{20}\\text{ m} \\\\ \\text{Or } \\frac{CB}{AC} &= \\frac{5k}{(4k + 5k)} \\\\ \\therefore CB &=", "triangle and base of parallelogram. What is unknown? Height of the parallelogram. Reasoning: By using Heron’s formula calculate the area of triangle and it is given that area of triangle is equal to area of parallelogram. The formula given by Heron about the area of a triangle $$=\\sqrt{s(s-a)(s-b)(s-c)}$$ Where $$a, b$$ and $$c$$ are the sides of the triangle, and \\begin{align}s &= \\text{Semi-perimeter}\\\\& = \\begin{Bmatrix} \\text{Half the Perimeter } \\\\ \\text{ of the triangle} \\;\\end{Bmatrix} \\\\&=\\frac{(a+b+c)}{2}\\end{align} Steps: $$ABCD$$ is parallelogram. $$ABE$$ is a triangle having base common with parallelogram $$ABCD.$$ For $$\\Delta ABE,$$ $$a = 30\\;{\\rm{ cm}} , b = 26 \\;{\\rm{ cm}} , c = \\;28 \\;cm$$ Semi Perimeter \\begin{align}s&=\\frac{(a+b+c)}{2} \\\\&=\\frac{30+26+28}{2} \\\\ & =\\frac{84}{2} \\\\ &=42 \\mathrm{cm}\\end{align} By using Heron’s formula, Area of a triangle $$ABE$$ \\begin{align}\\text{}&={\\sqrt{s(s-a)(s-b)(s-c)}} \\\\ &=\\sqrt{ 42(42-30) (42-28) (42-26) } \\\\ &=\\sqrt{42 \\times 12 \\times 14 \\times 16} \\\\ &=336 \\mathrm{cm}^{2} \\end{align} Area of parallelogram $$ABCD$$ $$=$$ Area of $$\\Delta \\, ABE$$, \\begin{align} \\rm{Base} \\times \\rm{Height}&= 336 \\;\\rm{cm}^2\\\\ 28\\; \\rm{cm} \\times Height& = 336 \\;\\rm{cm}^2\\\\ \\text{Height}&=\\frac{336 \\mathrm{cm}^{2}}{28 \\mathrm{cm}} \\end{align} Height of parallelogram $$= 12 \\;\\rm{cm}$$ ## Chapter 12 Ex.12.2 Question 5 A rhombus shaped field has green grass for $$18$$ cows to graze. If each side of the rhombus is $$30 \\;\\rm{m}$$ and its longer diagonal is $$48 \\;\\rm{m}$$, how much area of grass", "6\\text{ cm}$ Now $\\triangle AOB$ is a right triangle with $\\angle AOB$ a right angle. So: $AB^2 = AO^2 + BO^2$ $\\therefore AB = \\frac{\\sqrt{97}}{\\sqrt{2}}......\\text{[because AO = } 6 \\text{ cm and BO }= \\frac{5}{\\sqrt{2}}\\text{ cm]}$ That's the length of the slopping edge which is $\\frac{\\sqrt{97}}{\\sqrt{2}}\\text{ cm}$ So the area of: \\begin{align*}\\triangle ABC =& \\frac{1}{2} \\times BC \\times AO\\\\=& 15\\text{ cm}^2...\\text{[because BC} = 5 \\text{ cm and AO} = 6 \\text{ cm]}\\end{align*} So the area of the triangular face is $15\\text{ cm}^2$", "$$= 4.4 \\,\\rm cm$$ \\begin{align}& \\text{Area of parallelogram} \\\\ &= {\\rm{Base}} \\times {\\rm{Height}} \\end{align} \\begin{align}&= 2 \\times 4.4\\\\&= 8.8{\\rm{ }}{\\,\\rm c{m^2}}\\end{align} ## Chapter 11 Ex.11.2 Question 2 Find the area of each of the following triangles: ### Solution What is known? Base and height of triangles. What is the unknown? The area of the triangles. How can you use the known information to arrive at the solution? Base and height of a triangle is already given. Area of triangle is equal to half of product of base and height. Steps: (a) Given, Base $$= 3 \\,\\rm cm$$ Height $$= 4 \\,\\rm cm$$ We know that \\begin{align}&\\text{Area of triangle} \\\\ &= \\frac{1}{2} \\times {\\rm{Base}} \\times {\\rm{Height}} \\end{align} \\begin{align}&= \\frac{1}{2} \\times {\\rm{3}} \\times {\\rm{4}}\\\\&= {\\rm{6}}\\;{\\rm{c}}{{\\rm{m}}^2}\\end{align} (b) Given Base $$= 5 \\,\\rm cm$$ Height $$= 3.2 \\,\\rm cm$$ \\begin{align}&\\text{Area of triangle} \\\\ &= \\frac{1}{2} \\times {\\rm{Base}} \\times {\\rm{Height}} \\end{align} \\begin{align} &= \\frac{1}{2} \\times {\\rm{5}} \\times {\\rm{3}}{\\rm{.2}}\\\\&= {\\rm{8}}\\;{\\rm{c}}{{\\rm{m}}^2}\\end{align} (c) Given Base $$= 3 \\,\\rm cm$$ Height $$= 4 \\,\\rm cm$$ \\begin{align}&\\text{Area of triangle} \\\\ &= \\frac{1}{2} \\times {\\rm{Base}} \\times {\\rm{Height}} \\end{align} \\begin{align}&= \\frac{1}{2} \\times {\\rm{3}} \\times {\\rm{4}}\\\\& = {\\rm{6}}\\;{\\rm{c}}{{\\rm{m}}^2}\\end{align} (d) Given Base $$= 3 \\,\\rm cm$$ Height $$= 2 \\,\\rm cm$$ \\begin{align}&\\text{Area of triangle} \\\\ &= \\frac{1}{2} \\times {\\rm{Base}} \\times {\\rm{Height}} \\end{align} \\begin{align}& = \\frac{1}{2} \\times {\\rm{3}} \\times {\\rm{2}}\\\\&= {\\rm{3}}\\;{\\rm{c}}{{\\rm{m}}^2}\\end{align} ## Chapter", "length of $$b$$ \\begin{align*} b^{2} &= a^{2} + c^{2} - 2ac \\cos \\hat{B} \\\\ &= (14)^{2} + (17)^{2} - 2(14)(17) \\cos \\text{19} ° \\\\ &= \\text{34,93} \\ldots \\\\ \\therefore b &= \\text{5,9}\\text{ cm} \\end{align*} ### Use the sine rule to determine $$\\hat{C}$$ \\begin{align*} \\frac{\\sin \\hat{C}}{c} &= \\frac{\\sin \\hat{B}}{b} \\\\ \\frac{\\sin \\hat{C}}{17} &= \\frac{\\sin \\text{19} °}{\\text{5,9}} \\\\ \\sin \\hat{C} &= \\frac{17 \\times \\sin \\text{19} °}{\\text{5,9}} \\\\ \\therefore \\sin \\hat{C} &= \\text{0,938} \\ldots \\\\ \\text{First quadrant:} \\enspace \\hat{C} &= \\text{69,7} ° \\\\ \\text{Second quadrant:} \\enspace \\hat{C} &= \\text{180} ° - \\text{69,7} ° \\\\ &= \\text{110,3} ° \\end{align*} From the diagram, we see that $$\\hat{C} > \\text{90} °$$, therefore $$\\hat{C} = \\text{110,3} °$$. ### Calculate the area $$\\triangle ABC$$ \\begin{align*} \\text{Area } \\triangle ABC &= \\frac{1}{2}ac \\sin \\hat{B} \\\\ &= \\frac{1}{2} (14)(17) \\sin \\text{19} ° \\\\ &= \\text{38,7}\\text{ cm$^{2}$} \\end{align*} temp text ## Worked example 14: Problem in two dimensions In the figure below, $$CD=BD=x$$ and $$B\\hat{A}D = \\theta$$. Show that $$B{C}^{2}=2{x}^{2}\\left(1+\\sin\\theta \\right)$$. ### Consider the given information Use the given information to determine as many of the unknown angles as possible. $\\begin{array}{rll} CD &= BD = x & (\\text{given} ) \\\\ B\\hat{A}D &= \\theta & (\\text{given} ) \\\\ & \\\\ D\\hat{B}A &= \\text{90} ° & (\\angle \\text{ in semi circle}) \\\\ & \\\\ B\\hat{D}A &= \\text{180} ° - \\text{90} °", "as $$9 \\,\\rm cm$$ and height as $$6\\,\\rm cm.$$ Now, the area of triangle is known, take $$AB = 7.5 \\,\\rm cm$$ as the base and find height $$CE$$ using the formula of area of triangle. Steps: In $$\\triangle ABC$$, $$AD = 6 \\,\\rm cm$$ and $$BC = 9\\,\\rm cm$$ Area of triangle $$ABC$$ $$=$$ $$\\frac{1}{2}$$ Base $$\\times$$ Height \\begin{align}& = \\frac{1}{2} \\times {\\rm{BC}} \\times {\\rm{AD}}\\\\&= \\frac{1}{2} \\times 9 \\times 6\\\\&=27\\;c{m^2}\\end{align} Now, Area of triangle $$ABC$$ $$=$$ $$\\frac{1}{2}$$ $$AB$$ $$\\times$$ $$CE$$ \\begin{align}{\\rm{27}} &= \\frac{1}{2} \\times {\\rm{7}}{\\rm{.5}} \\times {\\rm{CE}}\\\\{\\rm{CE}} &= \\frac{{2 \\times {\\rm{27}}}}{{{\\rm{7}}{\\rm{.5}}}}\\\\{\\rm{CE}} &= \\frac{{54}}{{{\\rm{7}}{\\rm{.5}}}}\\\\{\\rm{CE}} &= 7.2\\;\\rm{cm}\\end{align} Related Sections Related Sections", "that, \\begin{align}&{\\text{Area of triangle}} \\\\ &= \\frac{1}{2} \\times {\\rm{Base}} \\times {\\rm{Height}} \\end{align} \\begin{align}1256 &= \\frac{1}{2} \\times {\\rm{Base}} \\times {\\rm{31}}{\\rm{.4}}\\\\\\\\{\\rm{Base}} &= \\frac{{2 \\times 1256}}{{31.4}}\\;\\\\\\\\{\\rm{Base}} &= \\frac{{2512}}{{31.4}}\\\\\\\\{\\rm{Base}} &= 80\\; \\rm mm\\\\\\end{align} (c) Given, Area of the triangle $$= 170.5 \\,\\rm cm^2$$ Base $$= 22 \\,\\rm cm$$ Height $$=\\,?$$ We know that, \\begin{align}&{\\text{Area of triangle}} \\\\ &= \\frac{1}{2} \\times {\\rm{Base}} \\times {\\rm{Height}} \\end{align} \\begin{align}170.5 &= \\frac{1}{2} \\times {\\rm{22}} \\times {\\rm{Height}}\\\\{\\rm{Height}} &= \\frac{{2 \\times 170.5}}{{22}}\\\\{\\rm{Height}} &= \\frac{{341}}{{22}}\\\\{\\rm{Height}} &= 15.5\\; \\rm cm\\end{align} ## Chapter 11 Ex.11.2 Question 5 $$PQRS$$ is a parallelogram (Fig 11.23). $$QM$$ is the height from $$\\, Q \\,$$ to $$SR$$ and $$QN$$ is the height from$$\\, Q \\,$$ to $$PS$$. If $$SR$$ $$= 12\\,\\rm cm$$ and $$QM$$ $$= 7.6 \\,\\rm cm$$. Find: (a) the area of the parallelogram $$PQRS$$ (b) $$QN$$, if $$PS$$ $$= 8 \\,\\rm cm$$ ### Solution What is known? $$PQRS$$ is a parallelogram. $$QM$$ is the height from $$\\, Q \\,$$ to $$SR$$ and $$QN$$ is the height from $$\\, Q \\,$$ to $$PS$$. Also, $$SR$$ $$= 12\\,\\rm cm$$ and $$QM$$ $$= 7.6 \\,\\rm cm$$. What is unknown? (a) The area of the parallelogram $$PQRS$$ and (b) $$QN$$, if $$PS$$ $$= 8 \\,\\rm cm$$ Reasoning: First, using the given information (base $$SR$$ = 12cm and perpendicular height $$QM$$ = 7.6cm), find out the area of parallelogram", "b \\times h \\\\ & = \\frac{1}{2} \\times 18 \\times 5 \\\\ & = \\text{45}\\ {\\text{cm}^{2}} \\\\ \\end{align} 2. Calculate the area of $\\triangle XYZ$. \\begin{align} A_{\\text{triangle}} & = \\frac{1}{2} \\times b \\times h \\\\ & = \\frac{1}{2} \\times 16 \\times 7 \\\\ & = \\text{56}\\ {\\text{cm}^{2}} \\\\ \\end{align} 3. Calculate the area of the given triangle, $\\triangle KJL$. \\begin{align} A_{\\text{triangle}} & = \\frac{1}{2} \\times b \\times h \\\\ &= \\frac{1}{2} \\times 9 \\times 12 \\\\ & = \\text{54 cm}^{2} \\\\ \\end{align} 4. Calculate the area of the triangle provided. We are going to treat $DE$ as the base and $EF$ as the height. \\begin{align} A_{\\text{triangle}} & = \\frac{1}{2} \\times b \\times h \\\\ & = \\frac{1}{2} \\times 12 \\times 5 \\\\ & = \\text{30 m}^{2} \\\\ \\end{align} 5. Consider the diagram of $\\triangle DFE$ provided. Calculate the area of $\\triangle DFE$. Calculate the length of $EF$ first. \\begin{align} EF^{2} + DF^{2} &=ED^{2} \\quad (\\text{Pythagoras})\\\\ EF^{2} + 5^{2} & = 13^{2}\\\\ EF^{2} + 25 & = 169\\\\ EF^{2} + 25 - 25 & = 169 - 25\\\\ EF^{2} & = 144\\\\ \\therefore EF & = \\sqrt {144}\\\\ EF & = \\text{12 m}\\\\ \\end{align} Now calculate the area of $\\triangle DFE$. \\begin{align} A_{\\text{triangle}} & = \\frac{1}{2} \\times b \\times h \\\\ & = \\frac{1}{2} \\times 5 \\times 12 \\\\", "The diameter of the circle is $$BC = \\text{17} \\text{ mm}$$, so we can divide by $$2$$ to get the radius: $r = \\frac{d}{2} = \\frac{17}{2} = \\text{8,5}$ \\begin{align} \\text{Area of circle } &= \\pi r^2 \\\\ &= \\pi (\\text{8,5})^2 \\\\ &= \\frac{22}{7} \\times \\text{72,25} \\\\ &= \\text{227,07} \\text{ mm}^2 \\end{align} ### Calculate the area of the right-angled triangle. $$CD = \\text{15} \\text{ mm}$$ is the perpendicular height of the triangle. Since this is a right-angled triangle, we can use the theorem of Pythagoras to determine the length of the base of the triangle $$BD$$. In $$\\triangle BCD$$: \\begin{align} BD^2 + CD^2 &= BC^2 \\text{ (Pythagoras)} \\\\ BD^2 + 15^2 &= 17^2 \\\\ BD^2 &= 289 − 225 \\\\ BD^2 &= 64 \\\\ \\therefore BD &= 8 \\end{align} So base $$BD = 8 \\text{ mm}$$ We can now calculate the area of the triangle $$\\triangle BCD$$: \\begin{align} \\text{Area } \\triangle BCD &= \\frac{1}{2}(b \\times h) \\\\ &= \\frac{1}{2}(8 \\times 15) \\\\ &= \\frac{1}{2}(120) \\\\ &= 60 \\text{ mm}^2 \\end{align} ### Calculate the area of the shaded region. \\begin{align} \\text{Area of the shaded region } &= \\text{(area of circle)} − \\text{(area of triangle)} \\\\ &= \\text{227,07} \\text{ mm}^2 – 60 \\text{ mm}^2 \\\\ &= \\text{167,07} \\text{ mm}^2 \\end{align} Area of the shaded region $$= \\text{167,07} \\text{ mm}^2$$.", "}{2}\\\\ & = 18 \\space cm. \\end{align} \\begin{align} \\text { The area of the } \\triangle ABC & = \\sqrt { S(S - AB)(S - BC)(S - CA) } \\\\ & = \\sqrt { 18(18 - 13)(18 - 10)(18 - 13) } \\\\ & = 60 \\space cm^2 \\end{align} 5. \\begin{align} \\text { The area of the quadrilateral } ABCD & = Area(\\triangle ACD) + Area(\\triangle ABC) \\\\ & = 30 + 60 \\\\ & = 90 \\space cm^2 \\end{align}", "& \\approx \\text{2 412,743} \\end{align*} $$ABCD$$ is a square, $$AC = \\text{12}\\text{ cm}$$, $$AP = \\text{10}\\text{ cm}$$. We first find the vertical height: \\begin{align*} H & = \\sqrt{(10)^{2} - (6)^{2}} \\\\ & = 8 \\end{align*} We also need to find the length of the side of the square. To do this we note that triangle $$ABC$$ is a right-angled isosceles triangle. So we can find the length of the side of the square using the theorem of Pythagoras: \\begin{align*} AC^{2} & = AB^{2} + BC^{2} \\\\ 12^{2} & = 2(AB^{2}) \\\\ 77 & = AB^{2} \\\\ \\therefore AB & = \\sqrt{77} \\end{align*} Now we can find the volume: \\begin{align*} V &= \\frac{1}{3} \\pi b^2 H \\\\ & = \\frac{1}{3} \\pi (77) (8) \\\\ & = \\frac{616}{3} \\pi \\\\ & \\approx \\text{645,07}\\text{ cm$^{3}$} \\end{align*} The prism alongside has the following dimensions: $$AB = 4$$ units, $$EC = 8$$ units, $$AF = 10$$ units. $$BC$$ is an arc of a circle with centre $$D$$. $$AB \\parallel EC$$. Explain why $$BD$$, the radius of the arc $$BC$$, is $$\\text{4}$$ units. Since $$D$$ is the centre of the circle $$BD = DC$$ (they are both radii of the arc). $$AB \\parallel EC$$ and $$BD$$ joins $$AB$$ and $$EC$$, therefore $$AB = ED = 4\\text{ units}$$. We also know that $$EC = 8\\text{", "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "3) $$b$$ and $$c$$ are roots of the same quadratic equation. It is very understandable because $$B$$ and $$C$$ play interchangeable roles. This is another way (still w/o using the hint). \\begin{align} \\text{Orthocenter of \\triangle ABC: }\\quad H&=(1,2) ,\\\\ \\text{Circumcenter of \\triangle ABC: }\\quad O&=(2,3) ,\\\\ \\text{Centroid of \\triangle ABC: }\\quad M&=\\tfrac13(H+2O)=(\\tfrac53,\\tfrac83) . \\end{align} The line through $$BC$$: \\begin{align} y&=\\tfrac43x+1 , \\end{align} the line $$OD\\perp BC$$: \\begin{align} y&=-\\tfrac34 x+\\tfrac92 . \\end{align} The point $$D$$ is the middle of the side $$BC$$: \\begin{align} D&=(\\tfrac{42}{25},\\tfrac{81}{25}) . \\end{align} Now the vertex $$A$$ of the $$\\triangle ABC$$ can be found as \\begin{align} A&=M+2(M-D) =(\\tfrac{41}{25},\\tfrac{38}{25}) , \\end{align} and the radius of the circumscribed circle is \\begin{align} R&=|O-A|= \\sqrt{ \\left(\\frac{50-41}{25}\\right)^2 +\\left(\\frac{75-38}{25}\\right)^2 } = \\frac{\\sqrt{58}}5 . \\end{align} Given $$R=\\frac{\\sqrt{m}}n$$, $$m$$ and $$n$$ can be found as \\begin{align} m&=58 ,\\\\ n&=5 . \\end{align} Solution using the given hint: Distance of orthocentre (H) from side BC is $$2R\\cos B\\cos C$$ and that of circumcentre (P) from BC is $$R\\cos A$$ where R is the circumradius. Finding these distances using coordinate geometry and equating we get: $$2R\\cos B\\cos C=\\frac{1}5$$ $$R\\cos A=\\frac{2}5$$ From this we get: $$\\cos A\\cos B\\cos C=\\frac{1}{25R^2}$$ Distance between P and H is $$\\sqrt{2}$$. Thus, $$R \\sqrt{1 – 8\\cos A\\cos B\\cos C}=\\sqrt{2}$$ $$R=\\frac{\\sqrt{58}}{5}$$ $$m=58,n=5$$", ": | | : | *B : | * | 3 | * | 1.5 : | * | : A* - - - - - - - - - - - *C : | | : | |x : | | - *-----------------------* : - - - - - 3 - - - - - : Triangle $ABC$ can be placed higher or lower. The distance $x$ can range from 0 to 1.5.", "7} \\\\ &=21 \\sqrt{11} \\mathrm{cm}^{2}\\end{align} Area of a triangle $$=21\\sqrt{11}\\mathrm{cm}^{2}$$ ## Chapter 12 Ex.12.1 Question 5 Sides of a triangle are in the ratio of $$12:17:25$$ and its perimeter is $$540\\; \\rm{cm}$$. Find its area. ### Solution What is known? Ratio of sides of the triangle and its perimeter. What is unknown? Area of the triangle. Reasoning: By using Heron’s formula we can calculate the area of triangle. The formula given by Heron about the area of a triangle $$=\\sqrt{s(s-a)(s-b)(s-c)}$$ Where $$a, b$$ and $$c$$ are the sides of the triangle, and \\begin{align}s &= \\text{Semi-perimeter}\\\\& = \\begin{Bmatrix} \\text{Half the Perimeter } \\\\ \\text{ of the triangle} \\;\\end{Bmatrix} \\\\&=\\frac{(a+b+c)}{2}\\end{align} Steps: Suppose the sides are $$12x\\;\\rm {cm}$$, $$17x\\;\\rm {cm}$$ and $$25x\\rm\\; {cm}$$. Perimeter of the triangle $$=$$ $$540\\; \\rm{cm}$$ \\begin{align}{12 x+17 x+25 x=540} \\\\ {54 x=540} \\\\ {x=\\frac{540}{54}} \\\\ {x=10 \\mathrm{cm}}\\end{align} Therefore sides of triangle: \\begin{align}&12 x=12 \\times 10=120 \\;\\mathrm{cm},\\\\ & 17 x=17 \\times 10=170 \\;\\mathrm{cm}, \\\\ & 25 x=25 \\times 10=250 \\;\\mathrm{cm}\\end{align} $$a = 120\\; {\\rm{cm}} ,\\\\ b = 170\\;{\\rm{cm}},\\\\ c = 250\\;{\\rm{cm}}$$ \\begin{align}s &= \\text{Half the Perimeter}\\\\ s&=\\frac{540}{2}\\\\&=270 \\;\\mathrm{cm}\\end{align} By using Heron’s formula, Area of a triangle \\begin{align}&={\\sqrt{s(s-a)(s-b)(s-c)}} \\\\&=\\!\\sqrt{270(270\\!-\\!120)\\!(270\\!-\\!170)\\!(270\\!-\\!250)}\\\\ &=\\sqrt{270 \\times 150 \\times 100 \\times 20} \\\\ &=\\sqrt{81000000} \\\\ &=9000 \\mathrm{cm}^{2}\\\\\\end{align} Area of a triangle $$=$$$$9000 \\;\\rm{cm^2}.$$ ## Chapter 12 Ex.12.1 Question 6 An isosceles triangle has perimeter", "legs of ABC triangle are picked up two points S and R where: . $AS\\,=\\,\\frac{3}{7}(AB),\\;BR\\,=\\,\\frac{3}{8}(BC)$ Find the area of $\\Delta SBR$ if the area of $\\Delta ABC = 126$ cm². Code: C * * * 5/8 * * * * R o * * ** * * * * * 3/8 * * * * * * * * * * * * B * * * * * * o * * * * * * * * * A 4/7 S 3/7 $\\Delta SBR$'s height is $\\frac{3}{8}$ of $\\Delta ABC$'s height. $\\Delta SBR$'s base is $\\frac{4}{7}$ of $\\Delta ABC$'s base. Therefore: . $\\Delta SBR \\;=\\;\\left(\\frac{3}{8}\\right)\\left(\\frac{4}{7}\\rig ht)(126) \\;=\\;27\\text{ cm}^2$" ]

[ "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "= \\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),", "angle bisector of the angle at vertex $A$ in triangle $ABC$. Then $\\frac{BA'}{A'C}=\\frac{BA}{AC}$. \\end{lemma} \\begin{proof} From lemma ..., we know that $\\frac{BA'}{A'C}=\\frac{\\area(BA'A)}{\\area(A'CA)}$. Now the equality of angles $\\angle BAA'$ and $\\angle A'AC$ implies that if we reflect the triangle $A'CA$ in the line $AA'$, then the reflected copy will share an angle with triangle $BA'A$ and the reflection of $C$ will lie on the line $AB$ (see picture). \\begin{figure}[h] \\centering \\begin{asy} size (6cm,3cm); pair A, B, C, A1, C1; picture pic; A=(2,2); B=(0,0); C=(3,0); A1=(length(C-A)*B+length(B-A)*C)/(length(B-A)+length(C-A)); C1=A+2*dot(A1-A,C-A)/dot(A1-A,A1-A)*(A1-A)-(C-A); draw(A--B--C--cycle); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); draw(A--A1); label(\"$A'$\",A1,S); draw(pic,A--B--A1--cycle); label(pic,\"$A$\",A,N); label(pic,\"$A'$\",A1,S); label(pic,\"$B$\",B,S); draw(pic,A1--C--A,linetype(\"0 4\")); draw(pic,A1--C1--A); label(pic,\"$C$\",C1,WNW); \\end{asy} \\label{fig:bisector} \\end{figure} But in the picture after the reflection we see that $\\frac{\\area(BA'A)}{\\area(A'CA)}=\\frac{AB}{AC}$, by the same lemma. \\end{proof} This lemma implies that if $AA',BB'$ and $CC'$ are internal bisectors of triangle $ABC$, then $\\frac{BA'}{A'C}\\cdot\\frac{CB'}{B'A}\\cdot\\frac{AC'}{C'B}=\\frac{BA}{AC}\\cdot\\frac{CB}{BA}\\cdot\\frac{AC}{CB}=1$, so Ceva's theorem tells us that $AA', BB'$ and $CC'$ are concurrent. We can prove it in a different way. For this we first notice that the internal angle bisector of an angle $A$ is exactly the locus of points $P$ inside the angle that are equidistant from the two rays that form $A$. Indeed, let $P'$ and $P''$ be the feet of perpendiculars dropped from the point $P$ to the two rays. If the point $P$ is on the angle bisector of", "Apollonius Circle. Since the point $$D=(4,9)$$ belongs to $$AB$$, it is a foot of the bisector of $$\\angle ACB$$, and a known expression for it is \\begin{align} D&=\\frac{aA+bB}{a+b} \\tag{1}\\label{1} . \\end{align} With known coordinates we have \\begin{align} (4,9)&= \\left(\\frac{2a+5b}{a+b},\\, \\frac{5a+11b}{a+b} \\right) \\tag{2}\\label{2} \\end{align} or a system \\begin{align} \\frac{2a+5b}{a+b}&=4 \\tag{3}\\label{3} ,\\\\ \\frac{5a+11b}{a+b}&=9 \\tag{4}\\label{4} \\end{align} with solution $$b=2a$$. Expression for the area of $$\\triangle ABC$$ squared is \\begin{align} S_2(a,b,c)&= \\tfrac1{16}\\,(4a^2b^2-(a^2+b^2-c^2)^2) \\tag{5}\\label{5} , \\end{align} so for $$b=2a$$ we must have \\begin{align} S_2(a,2a,c)&= -\\tfrac9{16}\\,(a^2)^2+\\tfrac58\\,c^2\\,a^2-\\tfrac1{16}\\,c^4 \\tag{6}\\label{6} . \\end{align} It's easy to find that \\begin{align} \\max_a S_2(a,2a,c)&= S_2(a,2a,c)\\Big|_{a=5} =225 , \\end{align} hence the maximal area of such triangle is $$15$$. • is there a name for this expression? – Buraian Aug 28 '20 at 16:42 • @Buraian: which one? – g.kov Aug 28 '20 at 16:51 • the \"well known expression\" for D – Buraian Aug 28 '20 at 17:12 • @Buraian: It's just expression for a foot of the bisector in terms of the coordinates of vertices and side lengths of triangle. And it is \" known \", but not \"well-known\". – g.kov Aug 28 '20 at 17:24 • ...@Buraian: this expression directly follows from Angle bisector theorem. – g.kov Aug 28 '20 at 17:51", "The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar. The ratio of their sides is $\\frac{AM}{AB} = \\frac 58$, hence the ratio of their areas is $\\left( \\frac 58 \\right)^2 = \\frac{25}{64}$. And as the area of triangle $ABC$ is $\\frac{6\\cdot 8}2 = 24$, the area of triangle $AME$ is $24\\cdot \\frac{25}{64} = \\boxed{ \\frac{75}8 }$. ## Solution 3 (Only Pythagorean Theorem) $[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); label(\"E\",Ep,S); label(\"M\",M,NW); [/asy]$ Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \\implies 25 + ME^2$, which means $AE = \\sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB = 8 - \\sqrt{25 + x^2}$. From here, we know that $CE^2 = CB^2 + BE^2 = ME^2 + MC^2$. From here, we can write the expression $6^2 +", "\\end{align*} Consider the line segment shown below. Express the following in terms of $$a$$ and $$b$$: $$PT: ST$$ $$(a + b): b$$ $$\\frac{PS}{TQ}$$ $$\\frac{a}{2a} = \\frac{1}{2}$$ $$\\frac{SQ}{PQ}$$ $$\\frac{2a + b}{3a + b}$$ $$QT: TS$$ $$2a: b$$ $$ABCD$$ is a parallelogram with $$DC = \\text{15}\\text{ cm}$$, $$h = \\text{8}\\text{ cm}$$ and $$BF = \\text{9}\\text{ cm}$$. Calculate the ratio $$\\dfrac{\\text{area } ABF}{\\text{area } ABCD}$$. The area of a parallelogram $$ABCD =$$ base $$\\times$$ height: \\begin{align*} \\text{Area} &= \\text{15} \\times \\text{8} \\\\ &= \\text{120}\\text{ cm$^{2}$} \\end{align*} The perimeter of a parallelogram $$ABCD = 2DC + 2BC$$. To find the length of $$AF$$, we use $$AF \\perp BC$$ and the theorem of Pythagoras. \\begin{align*} \\text{In $$\\triangle ABF$$:} \\quad AF^2 &= AB^2 - BF^2 \\\\ &= \\text{15}^2 - \\text{9}^2 \\\\ &= \\text{144} \\\\ \\therefore AF &= \\text{12}\\text{ cm} \\\\ \\therefore \\text{area } ABF &= \\frac{1}{2} AF \\cdot BF \\\\ &= \\frac{1}{2}(12)(9)\\\\ &= \\text{54}\\text{ cm} \\end{align*} \\begin{align*} \\therefore \\frac{\\text{area } ABF}{\\text{area } ABCD} &= \\frac{54}{120} \\\\ &= \\frac{9}{20} \\end{align*} $$AB = \\text{36}\\text{ m}$$ and $$C$$ divides $$AB$$ in the ratio $$4:5$$. Determine $$AC$$ and $$CB$$. \\begin{align*} \\frac{AC}{AB} &= \\frac{4k}{(4k + 5k)} \\\\ \\therefore AC &= 36 \\times \\frac{4}{9} \\\\ &= \\text{16}\\text{ m} \\\\ CB &= 36 - 16 \\\\ &= \\text{20}\\text{ m} \\\\ \\text{Or } \\frac{CB}{AC} &= \\frac{5k}{(4k + 5k)} \\\\ \\therefore CB &=", "to X\", (0,8), W); draw((0,6)--(6,6)--(16,10)); [/asy]$ ## Problem 25 In the figure, the area of square WXYZ is $25 \\text{cm}^2$. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\\Delta ABC$, $AB = AC$, and when $\\Delta ABC$ is folded over side BC, point A coincides with O, the center of square WXYZ. What is the area of $\\Delta ABC$, in square centimeters? $[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label(\"A\", A, NW); label(\"O\", O, NE); label(\"B\", B, SW); label(\"C\", C, NW); label(\"W\",W , NE); label(\"X\", X, N); label(\"Y\", Y, N); label(\"Z\", Z, SE); [/asy]$ $\\textbf{(A)}\\ \\frac{15}4\\qquad\\textbf{(B)}\\ \\frac{21}4\\qquad\\textbf{(C)}\\ \\frac{27}4\\qquad\\textbf{(D)}\\ \\frac{21}2\\qquad\\textbf{(E)}\\ \\frac{27}2$", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "dot(X); label(\"X\", X, NE); label(\"C\", C, N); label(\"B\", B, E); label(\"A\", A, W); [/asy]$ ## Problem 25 In the figure, the area of square $WXYZ$ is $25 \\text{ cm}^2$. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\\triangle ABC$, $AB = AC$, and when $\\triangle ABC$ is folded over side $\\overline{BC}$, point $A$ coincides with $O$, the center of square $WXYZ$. What is the area of $\\triangle ABC$, in square centimeters? $[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label(\"A\", A, NW); label(\"O\", O, NE); label(\"B\", B, SW); label(\"C\", C, NW); label(\"W\",W , NE); label(\"X\", X, N); label(\"Y\", Y, S); label(\"Z\", Z, SE); [/asy]$ $\\textbf{(A)}\\ \\frac{15}4\\qquad\\textbf{(B)}\\ \\frac{21}4\\qquad\\textbf{(C)}\\ \\frac{27}4\\qquad\\textbf{(D)}\\ \\frac{21}2\\qquad\\textbf{(E)}\\ \\frac{27}2$", "it is the circumcircle of $\\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \\overline{AB}, \\overline{BC}$ and note that $\\triangle M_1BM_2 \\sim \\triangle ABC$, we see that thse segments respectively cut a $120^{\\circ}$ arc in the circle with radius $18$ and $60^{\\circ}$ arc in the circle with radius $18\\sqrt{3}$. $[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype(\"2 3\") + linewidth(0.7), dashes = linetype(\"8 6\")+linewidth(0.7)+blue, bluedots = linetype(\"1 4\") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP(\"P\",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP(\"A\",A)) -- D(MP(\"B\",B)) -- D(MP(\"C\",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]$ The diagram shows $P$ outside of the grayed locus; notice that the creases [the", "either parallel to or coinciding with the sides of the large square. In $\\Delta ABC$, $AB = AC$, and when $\\Delta ABC$ is folded over side BC, point A coincides with O, the center of square WXYZ. What is the area of $\\Delta ABC$, in square centimeters? $[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label(\"A\", A, NW); label(\"O\", O, NE); label(\"B\", B, SW); label(\"C\", C, NW); label(\"W\",W , NE); label(\"X\", X, N); label(\"Y\", Y, N); label(\"Z\", Z, SE); [/asy]$ $\\textbf{(A)}\\ \\frac{15}4\\qquad\\textbf{(B)}\\ \\frac{21}4\\qquad\\textbf{(C)}\\ \\frac{27}4\\qquad\\textbf{(D)}\\ \\frac{21}2\\qquad\\textbf{(E)}\\ \\frac{27}2$", "# 2021 AIME I Problems/Problem 2 ## Problem In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, W); label(\"B\", B, W); label(\"C\", C, (1,0)); label(\"D\", D, (1,0)); label(\"F\", F, N); label(\"E\", E, S); [/asy]$ ## Solution 1 (Similar Triangles) Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\\frac{7}{3}(11-x)$ and $CG=\\frac{3}{7}x$. We have $\\frac{7}{3}(11-x)+\\frac{3}{7}x=FC=9$. Solving for $x$, we have $x=\\frac{35}{4}$. The area of the shaded region is just $3\\cdot \\frac{35}{4}=\\frac{105}{4}$. Thus, the answer is $105+4=\\framebox{109}$. ~yuanyuanC ## Solution 2 (Similar Triangles) Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\\triangle AFG \\sim \\triangle", "triangles of area 1 be $x$, then the base length of $\\Delta ADE=4x$. Notice, $\\big(\\frac{DE}{BC}\\big)^2=\\frac{1}{40}\\implies \\frac{x}{BC}=\\frac{1}{\\sqrt{40}}$, then $4x=\\frac{4BC}{\\sqrt{40}}\\implies \\big[ADE\\big]=\\big(\\frac{4}{\\sqrt{40}}\\big)^2\\cdot \\big[ABC\\big]=\\frac{2}{5}\\big[ABC\\big]$ Thus, $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]=\\big[ABC\\big]\\big(1-\\frac{2}{5}\\big)=\\frac{3}{5}\\cdot 40=\\boxed{24}.$ $[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label(\"B\",(0,0),SW); label(\"C\",(60,0),SE); label(\"E\",(50,10),E); label(\"D\",(10,10),W); label(\"A\",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]$ Solution by ktong ## Solution 4 The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$. Therefore the area of the trapezoid is $40-16=\\boxed{24}$. ## Solution 5 You can see that we can create a \"stack\" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$, so to find the area of such trapezoid $BCED$, we just take $40-16=\\boxed{24}$, like so. ∎ --anna0kear ## Solution 6 The combined area of the small triangles is $7$, and from the fact that each small triangle has an area of $1$, we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\\frac{1}{3}$, so will their areas have a proportion that is the square of the", "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 $$XY^2 = \\frac13(2 \\cdot 1143-135)", "as $$9 \\,\\rm cm$$ and height as $$6\\,\\rm cm.$$ Now, the area of triangle is known, take $$AB = 7.5 \\,\\rm cm$$ as the base and find height $$CE$$ using the formula of area of triangle. Steps: In $$\\triangle ABC$$, $$AD = 6 \\,\\rm cm$$ and $$BC = 9\\,\\rm cm$$ Area of triangle $$ABC$$ $$=$$ $$\\frac{1}{2}$$ Base $$\\times$$ Height \\begin{align}& = \\frac{1}{2} \\times {\\rm{BC}} \\times {\\rm{AD}}\\\\&= \\frac{1}{2} \\times 9 \\times 6\\\\&=27\\;c{m^2}\\end{align} Now, Area of triangle $$ABC$$ $$=$$ $$\\frac{1}{2}$$ $$AB$$ $$\\times$$ $$CE$$ \\begin{align}{\\rm{27}} &= \\frac{1}{2} \\times {\\rm{7}}{\\rm{.5}} \\times {\\rm{CE}}\\\\{\\rm{CE}} &= \\frac{{2 \\times {\\rm{27}}}}{{{\\rm{7}}{\\rm{.5}}}}\\\\{\\rm{CE}} &= \\frac{{54}}{{{\\rm{7}}{\\rm{.5}}}}\\\\{\\rm{CE}} &= 7.2\\;\\rm{cm}\\end{align} Related Sections Related Sections", "3) $$b$$ and $$c$$ are roots of the same quadratic equation. It is very understandable because $$B$$ and $$C$$ play interchangeable roles. This is another way (still w/o using the hint). \\begin{align} \\text{Orthocenter of \\triangle ABC: }\\quad H&=(1,2) ,\\\\ \\text{Circumcenter of \\triangle ABC: }\\quad O&=(2,3) ,\\\\ \\text{Centroid of \\triangle ABC: }\\quad M&=\\tfrac13(H+2O)=(\\tfrac53,\\tfrac83) . \\end{align} The line through $$BC$$: \\begin{align} y&=\\tfrac43x+1 , \\end{align} the line $$OD\\perp BC$$: \\begin{align} y&=-\\tfrac34 x+\\tfrac92 . \\end{align} The point $$D$$ is the middle of the side $$BC$$: \\begin{align} D&=(\\tfrac{42}{25},\\tfrac{81}{25}) . \\end{align} Now the vertex $$A$$ of the $$\\triangle ABC$$ can be found as \\begin{align} A&=M+2(M-D) =(\\tfrac{41}{25},\\tfrac{38}{25}) , \\end{align} and the radius of the circumscribed circle is \\begin{align} R&=|O-A|= \\sqrt{ \\left(\\frac{50-41}{25}\\right)^2 +\\left(\\frac{75-38}{25}\\right)^2 } = \\frac{\\sqrt{58}}5 . \\end{align} Given $$R=\\frac{\\sqrt{m}}n$$, $$m$$ and $$n$$ can be found as \\begin{align} m&=58 ,\\\\ n&=5 . \\end{align} Solution using the given hint: Distance of orthocentre (H) from side BC is $$2R\\cos B\\cos C$$ and that of circumcentre (P) from BC is $$R\\cos A$$ where R is the circumradius. Finding these distances using coordinate geometry and equating we get: $$2R\\cos B\\cos C=\\frac{1}5$$ $$R\\cos A=\\frac{2}5$$ From this we get: $$\\cos A\\cos B\\cos C=\\frac{1}{25R^2}$$ Distance between P and H is $$\\sqrt{2}$$. Thus, $$R \\sqrt{1 – 8\\cos A\\cos B\\cos C}=\\sqrt{2}$$ $$R=\\frac{\\sqrt{58}}{5}$$ $$m=58,n=5$$", "- 6 - 0.05},{abs(\\x - 4) + 0.2}) -- ({\\x*5/2 - 6 - 0.05},{abs(\\x - 4) - 0.2}); \\draw ({\\x*5/2 - 6 + 0.05},{abs(\\x - 4) + 0.2}) -- ({\\x*5/2 - 6 + 0.05},{abs(\\x - 4) - 0.2}); \\draw[triangle 45-triangle 45] ({\\x - 4},{abs(\\x - 4) + 0.25}) -- node[above,xshift=3mm] {$x$} (({2*\\x - 4},{abs(\\x - 4) + 0.25}); \\draw[triangle 45-triangle 45] (({2*\\x - 4},{abs(\\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\\x - 4 + \\x - 4},{abs(\\x - 4) + 0.25}); \\end{tikzpicture} Now the right triangle takes up half of the parallellogram. 4. Originally Posted by mathlearn Hello HallsofIvy, This should be that identical right angled triangle at the left Then the triangle which has half the area of the parallelogram would be, Correct ? Many THanks No, those are not right triangles. And it is always true that the two triangles you get by drawing a diagonal have area half the parallelogram. - - - - - Updated - - - Originally Posted by I like Serena Hey mathlearn! I think the right triangle is fixed. We have something like this: \\begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\\large] \\def\\x{6}; \\draw[ultra thick, blue] (0,0) coordinate (A) -- ({2*\\x - 4},0) coordinate (B) -- ({2*\\x - 4 + \\x - 4},{abs(\\x - 4)}) coordinate (C) -- ({\\x - 4},{abs(\\x -", "+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" ]

[ "square units 42.) What is the least common multiple of 168 and 420? Continue reading", "the smallest multiple of 63 and 147", "the least common multiple of $180$, $216$, and $450$?", "is the least common multiple of the $n_i$. -", "so x = 125 is the fewest?", "136x=204y→ x/y=204/136=3/2 test 3/2 ratio against options: 3*136=408 2*204=408 A Which of the following is the least common multiple of 136 and 204? [#permalink] 30 Mar 2020, 10:49 # Which of the following is the least common multiple of 136 and 204? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne", "most common? ________. Which are the two least common? ________ and _________. Subjects: 1 , 2 , 3 , 4 , 5 Feb 23, 2012 Jan 16, 2015", "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", "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", "# 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", "the first few non-zero multiples of 17. Multiples of 17: 17, 34, 51, 68, 85, 102, 119, 136, 153, 170, 187,... From this list, we can observe that 85 and 170 are indeed divisible by 5 since $$5\\times 17=85$$ and $$5\\times 34=170$$. Thus, the first 2 non-zero common multiples of 5 and 17 are 85 and 170. Let us look at one more example before moving on to the next section. List all the common multiples of 11 and 13 between 130 and 300. Solution The interval restriction here is that we need to list the multiples of 11 and 13 between 130 and 300. As before, we will start by listing these multiples using the table below. Number Multiples between 130 and 300 11 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297 13 130, 143, 156, 169, 182, 195, 208, 221, 234, 247, 260, 273, 286, 299 From the table above, observe that there are two common multiples of 11 and 13 between 130 and 300, namely 143 and 286. The concept of a common multiple is primarily used to find the lowest common multiple (or LCM) between a given set of numbers. This is the smallest common multiple shared between two (or more) numbers. You can find a", "(a), (b), and (c). How have Least Common Multiples come in handy? Noticing this pattern can help you to explain your answer here.", "# 92nd Problem 2016 How many multiples of 13 are included between 12548 and 19987 both inclusive? ×", "Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Manager Joined: 03 Oct 2013 Posts: 76 Re: Which of the following is the least common multiple of 136 and 204? [#permalink] ### Show Tags 01 Dec 2016, 10:08 136 = 2x68=4x34=(2^3)x17 204 = 4x51 = (2^2)x51 =(2^2)x3x17 LCM = (2^3)x3x17 = 408 GCF = (2^2)x17 = 68 CEO Joined: 03 Jun 2019 Posts: 2899 Location: India GMAT 1: 690 Q50 V34 WE: Engineering (Transportation) Re: Which of the following is the least common multiple of 136 and 204? [#permalink] ### Show Tags 30 Mar 2020, 09:52 Bunuel wrote: Which of the following is the least common multiple of 136 and 204? A. 408 B. 816 C. 2,448 D. 13,872 E. 27,744 Asked: Which of the following is the least common multiple of 136 and 204? 136 = 2*2*2*17 = 2^3*17 204 = 2*2*3*17 = 2^2*3*17 LCM (136,204) = 2^3*3*17 = 408 IMO A _________________ Kinshook Chaturvedi Email: [email protected] VP Joined: 07 Dec 2014 Posts: 1252 Which of the following is the least common multiple of 136 and 204? [#permalink] ### Show Tags 30 Mar 2020, 10:49 Which of the following is the least common multiple of 136 and 204? A. 408 B. 816 C. 2,448 D. 13,872 E. 27,744[/quote] let", "• multiples", "are a big Help :) Could you answer his question, then? Why is 121 the correct answer and what does this problem have to do with the \"least common multiple\" of 3, 5, and 8.", "Tags Intern Joined: 09 Apr 2003 Posts: 6 What is the least common multiple of 152 and 190? A. 304 B. [#permalink] ### Show Tags 30 May 2003, 07:34 1 3 00:00 Difficulty: 15% (low) Question Stats: 81% (01:29) correct 19% (01:42) wrong based on 191 sessions ### HideShow timer Statistics What is the least common multiple of 152 and 190? A. 304 B. 14440 C. 760 D. 28880 E. 1520 Manager Joined: 12 Mar 2003 Posts: 59 Re: What is the least common multiple of 152 and 190? A. 304 B. [#permalink] ### Show Tags 30 May 2003, 07:57 Ans : C Is there a quick method ? Intern Joined: 09 Apr 2003 Posts: 6 Re: What is the least common multiple of 152 and 190? A. 304 B. [#permalink] ### Show Tags 30 May 2003, 08:00 C is right Find Greatest Common factor for both the numbers and then divide the product of both the numbers from the the GCF. Intern Joined: 04 Nov 2004 Posts: 15 Location: Hamilton Ontario Re: What is the least common multiple of 152 and 190? A. 304 B. [#permalink] ### Show Tags 30 Nov 2004, 20:18 2 Yes there is a shorter way 152 = 2*2*2*19 190 = 2*5*19 I think everyone knows how to do this. Then choose 2*2*2", "of 260 and n (2) 1,352 is the least common multiple of 104 and n Hence B 1. 52= 13*4 We have no idea if n has one more 13 as factor. Not sufficient 2. 1352=13*13*8 n definitely has 13*13 Therefore, 169 is a factor of n Hence B _________________ Your Kudos can boost my morale..!! I am on a journey. Gradually I'll there..!! Originally posted by kitipriyanka on 24 Jul 2019, 08:19. Last edited by kitipriyanka on 25 Jul 2019, 09:56, edited 2 times in total. Director Joined: 24 Nov 2016 Posts: 809 Location: United States If n is a positive integer, is 169 a factor of n? [#permalink] ### Show Tags Updated on: 24 Jul 2019, 09:14 1 Quote: If n is a positive integer, is 169 a factor of n? (1) 52 is the greatest common divisor of 260 and n (2) 1,352 is the least common multiple of 104 and n n=Z+ is 169 or 13ˆ2 factor of n? (1) 52 is the greatest common divisor of 260 and n: GCF(260,n)=52=13*4… n=13*4*some multiple or 13*4*13*some multiple… insufficient; (2) 1,352 is the least common multiple of 104 and n: LCM(104,n)=1352=13*104=13*2*52=13*13*8,… since 104=13*8, then, n=13ˆ2*some multiple, sufficient; Originally posted by exc4libur on 24 Jul 2019, 08:23. Last edited by exc4libur on 24 Jul 2019, 09:14, edited", "multiples of 17 17 ,34 , 51 , 68 , 85 , 102 , 119 ,136 So the odd multiples are 17 ,51, 119", "# Least common multiple ### Problem What is the least common multiple of 7 and 5? Another way to say this is:" ]

[ "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "# How do you find the LCM of 14, 6? Nov 30, 2016 The LCM is $42$ #### Explanation: Factor the two numbers: $14 = 2 \\cdot 7$ $6 = 2 \\cdot 3$ The LCM is the product of all factors, common or non common taken once with the highest exponent. In this case: $L C M = 2 \\cdot 3 \\cdot 7 = 42$", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "found: $420=132a+132$ $288=132a$ $a=\\frac{288}{132}=\\frac{12\\cdot24}{12\\cdot11}= \\frac{24}{11}$", "(16 - 1)^2 + 1 = \\boxed{406}$.", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "a lot. By inspection, we find that $\\text{LCM}(1,2,3,2\\cdot2,2\\cdot3) = 12$. Finally, $1+2 = \\boxed{\\textbf{(B)}\\ 3}$.", "+ 31 = \\boxed{058}$.", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "# Math Help - Numbers 1. ## Numbers Given that $132=2^2 \\times 3 \\times 11$ and $252=2^2 \\times 3^2 \\times 7$ Use these results to find: the highest common factor of $132$ and $252$, the smallest interger $k$ such that $132k$ is a multiple of $252$ 2. We have that $\\text{lcm}(132,252)\\times \\text{gcd}(132,252) = 132\\times 252$ And we know that: $\\text{gcd}(132,252) = 2^2\\times 3 = 12$ Hence $\\text{lcm}(132,252)/132 = k=..$? 3. hi what is gcd? and sorry... i didnt understand your workings 4. Originally Posted by Punch Given that $132=2^2 \\times 3 \\times 11$ and $252=2^2 \\times 3^2 \\times 7$ Use these results to find: the highest common factor of $132$ and $252$, the smallest interger $k$ such that $132k$ is a multiple of $252$ for the hcf look at the common results you get (2^2 and 3) Times these together so you get 2^2*3=12 The HCF is 12 5. Sorry I have no idea what gcd is either and I don't know of any other way to get the LCM albeit manually. 6. Just found a method for you Find the prime factors of both Then, lets say one has prime factors of (making this up) 2x2 and 3x3 and 5 and the other has prime factors of 3 and 5x5 and 7 Now take the numbers from", "already have the factorizations from above. $24=2\\times2\\times2\\times3$ $36=3\\times3\\times2\\times2$ We should strike through the twos in 36’s factorization and the three in 24’s factorization, leaving $2\\times2\\times2\\times3\\times3$. The final LCM is 72.", "any common multiple. We first observe that $$120 = 2^4 \\cdot 3^1 \\cdot 5^2$$. Additionally, we know that the gcd of any $$n$$ numbers must divide the lcm of those numbers. Therefore, we see that choices $$a$$, $$b$$, and $$c$$ all divide the lcm. However, 35 does not divide the lcm and therefore it cannot be the gcd. It's $$(D)$$ because $$7$$ is a factor of $$35$$ but it's not a factor of $$120$$", "# How do you find the LCM of the given set of counting numbers. 7, 4, 3, 6? $7 = 7 \\cdot 1$ $4 = {2}^{2}$ $3 = 3 \\cdot 1$ $6 = 2 \\cdot 3$ The LCM is $7 \\cdot {2}^{2} \\cdot 3 = 7 \\cdot 4 \\cdot 3 = 84$", "{11.46}^{2}\\cdot 36=14853$", "= 106$, and $AB = 106 + 87 = \\boxed{193}$.", "the answer is $960-2=\\boxed{958}$. -brainiacmaniac31", "$$72$$ is the LCM of $$36$$ & x. Prime factorizing $$72$$ would give us $$72 = 2^3 * 3^2$$ Similarly $$36$$ can be prime factorized as $$36 = 2^2 * 3^2$$. Now, we know that for calculating LCM we take the highest power of primes. Since $$72$$ has $$3^2$$ in it $$x$$ can have either {$$3^0,, 3^1 , 3^2$$}. Similarly $$x$$ will have $$2^3$$ in it as $$36$$ has only $$2^2$$ in it and the LCM has $$2^3$$ in it. Now, x can’t have any more prime factor other than 2 or 3 as the LCM does not have any other prime factor apart from 2 and 3. So, possible values of $$x$$ = {$$2^3 * 3^0, 2^3 * 3^1, 2^3 * 3^2$$} = {$$8, 24, 72$$}. Thus, you see that prime factorization leads you out of a situation where the information seemed insufficient at the very first instant. TAKEAWAY I. Use Prime Factorization as your first line of offense against LCM-GCD questions 3. Being unmindful of constraints on LCM & GCD A few questions on GMAT test your understanding of the possible values of LCM & GCD of a set of integers. For example, have a look at the following official question( OG13 QR2, Problem Solving QNo: 98): Q4: Which of the following CANNOT be the greatest", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "our answer is $m+n=61+4=\\boxed{065}$.", "$14$ of these extra pairs. The answer is $28+21+14 = \\boxed{063}$" ]

[ "# Math Help - Numbers 1. ## Numbers Given that $132=2^2 \\times 3 \\times 11$ and $252=2^2 \\times 3^2 \\times 7$ Use these results to find: the highest common factor of $132$ and $252$, the smallest interger $k$ such that $132k$ is a multiple of $252$ 2. We have that $\\text{lcm}(132,252)\\times \\text{gcd}(132,252) = 132\\times 252$ And we know that: $\\text{gcd}(132,252) = 2^2\\times 3 = 12$ Hence $\\text{lcm}(132,252)/132 = k=..$? 3. hi what is gcd? and sorry... i didnt understand your workings 4. Originally Posted by Punch Given that $132=2^2 \\times 3 \\times 11$ and $252=2^2 \\times 3^2 \\times 7$ Use these results to find: the highest common factor of $132$ and $252$, the smallest interger $k$ such that $132k$ is a multiple of $252$ for the hcf look at the common results you get (2^2 and 3) Times these together so you get 2^2*3=12 The HCF is 12 5. Sorry I have no idea what gcd is either and I don't know of any other way to get the LCM albeit manually. 6. Just found a method for you Find the prime factors of both Then, lets say one has prime factors of (making this up) 2x2 and 3x3 and 5 and the other has prime factors of 3 and 5x5 and 7 Now take the numbers from", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "3: Find the least common multiple of $10$ and $27$. $10$ and $27$ share no common factors other than $1$. So, the GCF is $1$. Therefore, the LCM is simply $10\\cdot 27=270$. A third way to find the LCM of is to list all of the prime factors of each number and then multiply all of the factors the greatest number of times each occurs in any of the lists. [Note that while the previous method won't always work with more than $2$ numbers, this method will.] Example 4: Find the LCM of $16,25$ and $60$. $\\begin{array}{l}16=2\\cdot 2\\cdot 2\\cdot 2\\\\ 25=5\\cdot 5\\\\ 60=2\\cdot 2\\cdot 3\\cdot 5\\end{array}$ The greatest number of times the factor $2$ occurs is four (in the first list). The greatest number of times the factor $3$ occurs is one (in the third list). The greatest number of times the factor $5$ occurs is two (in the second list). So, we multiply four $2$s, one $3$, and two $5$s. LCM $=2\\cdot 2\\cdot 2\\cdot 2\\cdot 3\\cdot 5\\cdot 5=1200$", "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", "# What is the least common multiple of 9, 36 and 45? $9 = {3}^{2}$ $36 = {2}^{2} \\cdot {3}^{2}$ $45 = {3}^{2} \\cdot 5$ The LCM is ${2}^{2} \\cdot {3}^{2} \\cdot 5 = 180$", "the least common multiple of $180$, $216$, and $450$?", "+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", "found: $420=132a+132$ $288=132a$ $a=\\frac{288}{132}=\\frac{12\\cdot24}{12\\cdot11}= \\frac{24}{11}$", "with 117 117 126 126=1x117+9 so replace 126 with 9 117 9 117=13x9+0 so replace 117 with 0 0 9 We have a zero, and the other number is 9 so gcd(3267,126)=9.[/CODE]Let $$a$$ and $$b$$ be integers again. An integer $$m$$ is called a [B]common multiple[/B] of $$a$$ and $$b$$ if $$a|m$$ and $$b|m$$. Let $$S$$ be the set of all common multiples of $$a$$ and $$b$$. If $$a$$ or $$b$$ is zero then $$S=\\{0\\}$$ and $$S$$ has no positive elements. If instead $$a$$ and $$b$$ are both non-zero, then $$S$$ has non-zero elements (the product $$ab$$ for example), and $$S$$ is symmetric so it has positive elements. The minimum positive element of $$S$$ is called the [B]least common multiple[/B], or [B]lowest common multiple[/B], of $$a$$ and $$b$$, written lcm$$(a,b)$$. [U] Example[/U] The set of all multiples of 12 is the set $$12\\mathbb{Z}=\\{\\ldots,-60,-48,-36,-24,-12,0,12,24,36,48,60,\\ldots\\}$$ while the set of all multiples of 30 is $$30\\mathbb{Z}=\\{\\ldots,-120,-90,-60,-30,0,30,60,90,120,\\ldots\\}$$ so the set of common multiples of 12 and 30 is their intersection $$12\\mathbb{Z}\\cap 30\\mathbb{Z}=\\{\\ldots,-120,-60,0,60,120,\\ldots\\} =60\\mathbb{Z}$$. The minimum positive element in this set is 60, so lcm$$(12,30)=60$$. [U]Proposition 8[/U] Let $$a$$ and $$b$$ be non-zero integers. Then $$a\\mathbb{Z}\\cap b\\mathbb{Z}=m\\mathbb{Z}$$, where $$m=lcm(a,b)$$. [U]proof[/U] Take any elements $$c,d\\in a\\mathbb{Z}\\cap b\\mathbb{Z}$$. Then $$c$$ & $$d$$ are elements of $$a\\mathbb{Z}$$ and, by proposition 3, $$a\\mathbb{Z}$$ is closed under subtraction", "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", "+0 # help 0 89 1 What is the least common multiple of the first ten positive integers? Apr 2, 2019 $$2,3,\\dots, 10 = \\{2^1, 3^1, 2^2, 5^1, 2^1 \\cdot 3^1,7^1, 2^3, 3^2, 2^1\\cdot 5^1\\}\\\\ \\text{Using the max exponents for common factors we obtain}\\\\ LCM = 2^3 \\cdot 3^2 \\cdot 5^1 \\cdot 7^1 = 2520$$", "# What is the least common multiple of 9, 36 and 45? ##### 1 Answer Given that $9 = {3}^{2}$ $36 = {2}^{2} \\cdot {3}^{2}$ $45 = {3}^{2} \\cdot 5$ The LCM is ${2}^{2} \\cdot {3}^{2} \\cdot 5 = 180$", "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", "+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", "is equal to the product of the two numbers. Example 3: Find the least common multiple of $10$ and $27$ . $10$ and $27$ share no common factors other than $1$ . So, the GCF is $1$ . Therefore, the LCM is simply $10\\cdot 27=270$ . A third way to find the LCM of is to list all of the prime factors of each number and then multiply all of the factors the greatest number of times each occurs in any of the lists. [Note that while the previous method won't always work with more than $2$ numbers, this method will.] Example 4: Find the LCM of $16,25$ and $60$ . $\\begin{array}{l}16=2\\cdot 2\\cdot 2\\cdot 2\\\\ 25=5\\cdot 5\\\\ 60=2\\cdot 2\\cdot 3\\cdot 5\\end{array}$ The greatest number of times the factor $2$ occurs is four (in the first list). The greatest number of times the factor $3$ occurs is one (in the third list). The greatest number of times the factor $5$ occurs is two (in the second list). So, we multiply four $2$ s, one $3$ , and two $5$ s. LCM $=2\\cdot 2\\cdot 2\\cdot 2\\cdot 3\\cdot 5\\cdot 5=1200$", "# 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: $2 \\times 2 \\times 2 \\times 3 \\times 5 \\times 11 \\times 31 = 40 , 920$ $120 \\times 11 \\times 31 = 40 , 920$ $124 \\times 2 \\times 3 \\times 5 \\times 11 = 40 , 920$ $165 \\times 2 \\times 2 \\times 2 \\times 31 = 40 , 920$", "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", "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.", "least common multiple of $15$ and $20.$ Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple. Find the least common multiple (LCM) of the given numbers: $9\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}12$ 36 Find the least common multiple (LCM) of the given numbers: $18\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}24$ 72 ## Prime factors method Another way to find the least common multiple of two numbers is to use 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. $12=2\\cdot 2\\cdot 3\\phantom{\\rule{3em}{0ex}}18=2\\cdot 3\\cdot 3$ Then we write each number as a product of primes, matching primes vertically when possible. $\\begin{array}{l}12=2\\cdot 2\\cdot 3\\hfill \\\\ 18=2\\cdot \\phantom{\\rule{1.1em}{0ex}}3\\cdot 3\\end{array}$ 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 can chip be made from sand is this allso about nanoscale material Almas are nano particles real yeah Joseph Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master? no can't Lohitha where", "-- Surely two 3s as LCM has two 3s but none of 1284 or 2472 have two 3s...$$3^2$$ 5 -- Surely one 5..$$5^1$$ 103 -- can be none or one of 103...$$103^0$$ or $$103^1$$ 107 -- can be none or one of 107...$$107^0$$ or $$107^1$$ As there are two 3s, A, B and D are out... N can be any of - $$2^2*3^2*5$$ and any of the combination of 2, 103 or 107 added to it.. for example $$2^3*3^2*5*103*107$$ can be the largest value C Note : The question should ask for the smallest value of N or it should be ' What can be the value of N?' Hi, Can you plz explain why cant we follow the method: HCF * LCM = Product of numbers? If we use that, it results in N=15 Manager Joined: 29 May 2017 Posts: 142 Location: Pakistan Concentration: Social Entrepreneurship, Sustainability Re: If the highest common factor of 2,472, 1,284 and positive integer N is [#permalink] ### Show Tags 01 Dec 2019, 07:46 ScottTargetTestPrep wrote: Bunuel wrote: If the highest common factor of 2,472, 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" ]

[ "# $\\frac{a+b^{22}}{a^2+b^{11}}$ If $a>b>1$ and $\\log_a b+\\frac{6}{11}\\log_b a=\\frac{67}{11},$ what is the value of $\\frac{a+b^{22}}{a^2+b^{11}}?$ ×", "Find the value of $$a+b$$ if $$0 \\leq a \\leq \\frac{\\pi}{2}$$ and $$0 \\leq b \\leq \\frac{\\pi}{2}$$ satisfies $$\\cos(a)=\\sin(b)=\\frac{1}{3}$$.", "# $$\\frac{a+b^{22}}{a^2+b^{11}}$$ Algebra Level 2 If $$a>b>1$$ and $\\log_a b+\\frac{6}{11}\\log_b a=\\frac{67}{11},$ what is the value of $\\frac{a+b^{22}}{a^2+b^{11}}?$ ×", "$x$, since $a$ depends on $x$. For different values of $x$ you get different values of $a$. You have $$a = \\log_2{\\frac{x}{3x+1}}$$", "# If 0.2x + 1.8 = 3 - 0.6x, what is the value of x? $x = 1.5$ 0.2x+1.8=3−0.6x $0.2 x + 0.6 x = 3 - 1.8$ $0.8 x = 1.2$ $x = \\setminus \\frac{1.2}{0.8}$ $x = 1.5$", "= 4$$, then determine the value of $$\\Large \\frac{dL}{dt}$$. (A calculator is recommended.)", "• # question_answer If the point (3, 4) lies on the graph of the equation $3y=ax+7,$ then the value of a is____. A) $\\frac{2}{3}$ B) 1C) $\\frac{4}{3}$ D) $\\frac{5}{3}$ Since the point (3, 4) lies on the equation $3y=ax+7.$ $\\therefore$ $3\\times 4=a\\times 3+7$ $\\Rightarrow$ $5=3a\\Rightarrow a=\\frac{5}{3}.$", "What is the value of $\\frac{1}{2}$ of $\\frac{2}{3}$ of $\\frac{3}{4}$ of $\\frac{4}{5}$ of $\\frac{5}{6}$ of $\\frac{6}{7}$ of $\\frac{7}{8}$ of $\\frac{8}{9}$ of $\\frac{9}{10}$ of $1000$?", "## anonymous 5 years ago Find the value of x? Click on the question to see the picture :) we have $6x-5=17x$ hence, $x= \\frac{-5}{11}$", "## Algebra and Trigonometry 10th Edition $A=-\\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x=4$: $A=8$ $A=-\\frac{x^2}{2}+4x$ (see item (a)) $A=-\\frac{1}{2}(x^2-8x)=-\\frac{1}{2}(x^2-8x+16-16)$ $A=-\\frac{1}{2}[x^2-2(4)x+4^2-16]$ $A=-\\frac{1}{2}[(x-4)^2-16]=-\\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x-4=0$, that is, when $x=4$: $A=-\\frac{(4-4)^2}{2}+8=8$", "= b$$ $$a(1 - b) = b$$ $$a = \\frac{b}{b - 1}$$ Kudos [?]: 316 [0], given: 261 Re: If b#1, what is the value of a in terms of b if ab/(a-b)=1 ? [#permalink] 05 Jul 2017, 05:16 Display posts from previous: Sort by", "a2,-\\frac{a^2}4)$, corresponding to a maximum of the absolute value at $(\\frac a2,\\frac{a^2}4)$. The plot below summarizes this: I doubt there is a much shorter way.", "value of $$b+ \\frac{a+c}{2}$$ is attained for a=7, b=9 and c=8, or a = 8, b = 9, and c = 7 and it is $$\\frac{15}{2} +9=16.5$$. Dec 17, 2019", "increases to $2$, the value of $\\frac{1}{x}$ comes down and it is minimum wheen $x = 2$ and it is $1 + \\frac{1}{2} = 1.5$ graph{(x^2+1/x)/(x^2-(1-1/x^2)/(1/x+1/x^2) [-0.983, 4.017, 0.23, 2.73]}", "+0 # Pls help 0 133 1 If $$\\left(a-\\frac 1a\\right)^2=4$$ and $$a>1$$ what is the absolute value of $$a^3 - \\frac1{a^3}?$$ Sep 16, 2021 #1 +2401 +1 (a - 1/a) = 2 (note that normally we couldn't assume that it isn't -2, but since a > 1, a - 1/a has to be positive. (a - 1/a)^3 = 8 a^3 - 3a + 3/a - 1/a^3 = 8 a^3 - 1/a^3 - 3(a - 1/a) = 8 Can you figure it out from here? =^._.^= Sep 17, 2021", "= \\frac{1}{3} \\ln 8 = \\ln 8^{1/3}$ using the usual log rule $a \\ln b = \\ln b^a$. And you should know that $8^{1/3} = \\sqrt[3] 8 = 2$. • June 13th 2009, 05:26 PM justin016 Quote: Originally Posted by mr fantastic $\\frac{\\ln 8}{3} = \\frac{1}{3} \\ln 8 = \\ln 8^{1/3}$ using the usual log rule $a \\ln b = \\ln b^a$. And you should know that $8^{1/3} = \\sqrt[3] 8 = 2$. Okay i got it, thanx alot", "Question 10. The value of ‘c’ for which $$\\frac{3c}{4}$$ gives 18 is (a) c = 15 (b) c = 21 (c) c = 24 (d) c = 27 Solution: (c) c = 24", "Least value 11. If $b$ is greater than 3, then, of the following which will always have the LEAST value? 1. $\\frac{3}{b-1}$ 2. $\\frac{3}{b}$ 3. $\\frac{3}{b+1}$ 4. $\\frac{b}{3}$ 5. $\\frac{b+1}{3}$", "fraction form of $\\frac{4}{s (3s + 4)}$ is $\\frac{A}{s} + \\frac{B}{3s + 4}$ where you have to find the value of A and B.", "# If 0.2x + 1.8 = 3 - 0.6x, what is the value of x? Nov 18, 2017 $x = 1.5$ #### Explanation: 0.2x+1.8=3−0.6x $0.2 x + 0.6 x = 3 - 1.8$ $0.8 x = 1.2$ $x = \\setminus \\frac{1.2}{0.8}$ $x = 1.5$" ]

[ "15$ $\\Rightarrow \\boxed{b = 30}$ And, $\\frac{a}{2} – 1 = b$ $\\Rightarrow \\frac{a}{2} - 1 = 30$ $\\Rightarrow \\frac{a}{2} = 31$ $\\Rightarrow \\boxed{a = 62}$ Correct Answer$: 62$ by 7.7k points 3 8 30 1 vote 1 62 views 1 vote 2 87 views 1 vote", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", "- \\left(\\frac{R}{\\sqrt{3}}\\right)^2\\quad\\Rightarrow\\ quad\\boxed{ r^2\\;=\\;\\frac{2R^2}{3}}$ Substitute into [2]: . $V \\;=\\;\\frac{\\pi}{3}r^2 h\\;=\\;\\frac{\\pi}{3}\\left(\\frac{2R^2}{3}\\right)\\lef t(\\frac{\\sqrt{3}R}{3}\\right)$ Therefore, the maximum volume is: . $V \\;=\\;\\boxed{\\frac{2\\pi\\sqrt{3}}{27}R^3}$ 3. Hey what program did you use to right your numbers, roots and pie in 4. Originally Posted by aaasssaaa Hey what program did you use to right your numbers, roots and pie in The mathematical typesetting here is done using a LaTeX see this (to see the code generating the equations quote the message and you will see the typesetting codes embedded in the message). RonL", "= 2$ - (iii). From (i) we get $B = - A$, from (iii) $C = A - 2$ and putting them in (ii), we get $A - \\left(- A\\right) + \\left(A - 2\\right) = 13$ or $3 A = 15$ or $A = 5$ Hence $B = - 5$ and $C = 3$ Hence partial fractions are $\\frac{5}{\\left(x - 1\\right)} + \\frac{- 5 x + 3}{{x}^{2} + x + 1}$", "$A_r \\;=\\;2\\left(\\frac{x}{6}\\right)^2\\quad\\Rightarrow\\q uad\\boxed{A_r\\:=\\:\\frac{x^2}{18}}$ The square looks like this: Code: * - - - - - - * | | | | | | ¼(30-x) | | | | * - - - - - - * ¼(30-x) The perimeter of the square is: $30 - x$ cm, . . so its side is: $\\frac{1}{4}(30 - x)$ The area of the square is: . $A_s \\;=\\;\\left(\\frac{30-x}{4}\\right)^2\\quad\\Rightarrow\\quad\\boxed{A_s\\;=\\; \\frac{x^2 - 60x + 900}{16}}$ Hence, the total area is: . $A \\;=\\;\\frac{x^2}{18} + \\frac{x^2 - 60x + 900}{16} \\;=\\;\\frac{1}{144}(17x^2 - 540x + 8100)$ And that is the function we must minimize . . . 3. ## Thank you Thank you very much, I really do appreciate it", "first, get all the a's on one side and all the constants on the other: $4 - \\frac a2 = \\frac 72 - a$ $\\Rightarrow a - \\frac a2 = \\frac 72 - 4$ ........simplify each side $\\Rightarrow \\frac a2 = - \\frac 12$ ...........multiply both sides by 2 $\\Rightarrow a = -1$ This is my 4th post!!!!! 5. Thank you! And happy 4,000th post!!! Wow, that's a biiiiig step! On my forum, an admin just celebrated his 10,000th........ I know.... lol!", "3)(x + 2)^2) (A / x - 3) + (B / x + 2) + (C / (x + 2)^2) My values for A and C seem correct from the method I was taught by the professor A = 9 / 25 C = - 4 / 5 Now to get B, I equated for cubics. x^2 = ((9/25)(x + 2)^3) + B(x - 3)(x + 2)^2 - (4/5)(x - 3)(x + 2) Now the B I got was - 9 / 25. Can someone tell me if I've done this right so far? Thanks Alternatively, $\\frac {x^2}{(x - 3)(x + 2)^2} = \\frac A{x - 3} + \\frac B{x + 2} + \\frac C{(x + 2)^2}$ multiply through by $(x - 3)(x + 2)^2$ $\\Rightarrow x^2 = A(x + 2)^2 + B(x - 3)(x + 2) + C(x - 3)$ Let $x = -2$ $\\Rightarrow 4 = -5C \\implies \\boxed{C = - \\frac 45}$ Let $x = 3$ $\\Rightarrow 9 = 25A \\implies \\boxed{A = \\frac 9{25}}$ Let $x = 0$ ....just some random value, 0 is easy to work with: $\\Rightarrow 0 = 4A - 6B - 3C$ plug in the values for $A$ and $C$ and solve for $B$, we get: $\\boxed{B = \\frac {16}{25}}$. as the great galactus said 5. Thanks Jhevon for the", ". . Therefore: . $\\boxed{r \\;=\\;\\frac{3}{2}}$ Substitute into [2]: . $a\\,\\frac{\\left(\\frac{3}{2}\\right)^2 - 1}{\\frac{3}{2}-1} \\:=\\:60 \\quad\\Rightarrow\\quad\\frac{5}{2}a \\:=\\:60 \\quad\\Rightarrow\\quad\\boxed{ a \\:=\\:24}$ Therefore, the first three terms are: . $24,\\:36,\\:54$", "finding the positive difference of the two fractions, we arrive with $\\frac{34}{16} \\Rightarrow \\frac{17}{8}$, which is our answer of $\\boxed{\\textbf{(C)}}$", "$\\frac{1}{2} + \\frac{3}{2} + c \\:=\\:8\\quad\\Rightarrow\\quad\\boxed{c = 6}$ Therefore: . $f(n) \\;=\\;\\frac{1}{2}n^2 + \\frac{3}{2}n + 6\\quad\\Rightarrow\\quad \\boxed{f(n)\\;=\\;\\frac{1}{2}(n^2 + 3n + 12)}$", "110 views In a fraction, numerator is increased by $25 \\%$ and the denominator is diminished by $10 \\%.$ The new fraction obtained is $5/9.$ The original fraction is $:$ 1. $2/5$ 2. $5/9$ 3. $3/5$ 4. None of the above Let original fraction is $\\frac{a}{b}$,where $a$ is numerator and $b$ is denominator. $\\Rightarrow \\frac{125\\% *a}{90\\%*b}=\\frac{5}{9}$ $\\Rightarrow \\frac{\\left ( \\frac{125}{100}*a \\right )}{\\left(\\frac{90}{100}*b \\right )}=\\frac{5}{9}$ $\\Rightarrow \\frac{a}{b}=\\frac{5*90}{9*125}$ $\\Rightarrow \\frac{a}{b}=\\frac{2}{5}$ Option $(A)$ is correct. 1 135 views", "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", "fraction $=$= $a^{-3}\\div b^{-3}$a−3÷​b−3 Rewrite the quotient with a division symbol $=$= $\\frac{1}{a^3}\\div\\frac{1}{b^3}$1a3​÷​1b3​ Apply the negative exponent rule to the numerator and the denominator to express both with positive exponents $=$= $\\frac{1}{a^3}\\times\\frac{b^3}{1}$1a3​×b31​ Dividing by a fraction is the same as multiplying by the reciprocal of that fraction $=$= $\\frac{b^3}{a^3}$b3a3​ Simplify the fractional product $=$= $\\left(\\frac{b}{a}\\right)^3$(ba​)3 Write as a single term raised to a power by using the reverse of the rule for raising fractions We can see that $\\left(\\frac{a}{b}\\right)^{-3}=\\frac{b^3}{a^3}$(ab)3=b3a3$=$=$\\left(\\frac{b}{a}\\right)^3$(ba)3 Reflect: What has happened is we have found the reciprocal of the expression in the question, and turned the power into a positive. Using this trick will save a lot of time! Raising a fraction to a power For any base number of the form $\\frac{a}{b}$ab, and any number $n$n as a power, $\\left(\\frac{a}{b}\\right)^n=\\frac{a^n}{b^n}$(ab)n=anbn If $n$n is negative, then we also use the fact $a^{-n}=\\frac{1}{a}$an=1a. Giving us the following rule: $\\left(\\frac{a}{b}\\right)^{-n}=\\left(\\frac{b}{a}\\right)^n$(ab)n=(ba)n #### Practice questions ##### Question 1 Find the value of $n$n such that $\\frac{1}{25}=5^n$125=5n. ##### Question 2 Simplify the following, giving your answer with a positive exponent: $\\frac{9x^2}{3x^9}$9x23x9 ##### Question 3 Simplify the following, giving your answer with a positive exponent: $\\left(\\frac{y}{4}\\right)^{-3}$(y4)3 ### Outcomes #### 9.B2.2 Analyse, through the use of patterning, the relationships between the exponents of powers and the operations with powers, and use these relationships", "# Question #7c14a Jan 10, 2018 $p = 48$ #### Explanation: $p + 24 = 72 \\rightarrow$ Divide each side by $\\frac{1}{8}$ or multiply each side by the reciprocal, 8 $p = 48 \\rightarrow$ Subtract 24 from each side Check: $\\frac{1}{8} \\left(48 + 24\\right)$ $\\frac{1}{8} \\left(72\\right)$ $9$", "\\frac{1}{2} \\sqrt{61}$ $\\text{subtract \"3/2\" from both sides}$ $\\Rightarrow c = - \\frac{3}{2} \\pm \\frac{1}{2} \\sqrt{61} \\leftarrow \\textcolor{red}{\\text{exact solutions}}$", "f(2) = \\frac {2 - B}{2 - A}$ Now we are told that f(2) = 0. $\\Rightarrow \\frac {2 - B}{2 - A} = 0$ I also told you that a fraction is zero if the numerator is zero (provided the denominator isn't zero at the same time) so this means $2 - B = 0$ $\\Rightarrow B = 2$ Now $f(1) = \\frac {1 - B}{1 - A}$ and we are told this is undefined. i told you a fraction is undefined if we divide by zero. so this means $1 - A = 0$ $\\Rightarrow A = 1$", "= \\frac{b^2}{4a^2}\\,\\!$ to both sides of the equation: $x^2 + \\frac{b}{a} x= -\\frac{c}{a},\\,\\!$ Which gives: $x^2+\\frac{b}{a}x+\\frac{b^2}{4a^2}=-\\frac{c}{a}+\\frac{b^2}{4a^2}.\\,\\!$ The left side is now a perfect square; it is the square of $x + \\frac{b}{2a}.\\,\\!$ The right side can be a single fraction, with a common denominator 4a2. $\\left(x+\\frac{b}{2a}\\right)^2=\\frac{b^2-4ac}{4a^2}.$ Find the square root of both sides. $x+\\frac{b}{2a}=\\pm\\frac{\\sqrt{b^2-4ac\\ }}{2a}.$ Move b/2a: $x=-\\frac{b}{2a}\\pm\\frac{\\sqrt{b^2-4ac\\ }}{2a}=\\frac{-b\\pm\\sqrt{b^2-4ac\\ }}{2a}.$", "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*", "$\\Rightarrow r = 3 \\cdot \\frac {7}{8}$ $\\Rightarrow r = \\frac {21}{8}$ cups an alternative meathod would be to let the whole be x, since the fraction of the whole that is oatmeal is 8/15 we must have: $\\frac {8}{15} \\cdot x = 3$ $\\Rightarrow x = \\frac {45}{8}$ Let $r$ be the cups of raisins needed. Since the fraction of the whole that is raisins is 7/15, we must have $\\frac {7}{15} \\cdot \\frac {45}{8} = r$ $\\Rightarrow r = \\frac {21}{8}$ cups", "# How do you solve 4/9=(r-3)/6? Apr 15, 2017 $r = \\frac{17}{3}$ #### Explanation: To eliminate the fractions, multiply both sides by the $\\textcolor{b l u e}{\\text{lowest common multiple}}$ (LCM) of 9 and 6 The LCM of 9 and 6 is 18 ${\\cancel{18}}^{2} \\times \\frac{4}{\\cancel{9}} ^ 1 = {\\cancel{18}}^{3} \\times \\frac{\\left(r - 3\\right)}{\\cancel{6}} ^ 1$ $\\Rightarrow 8 = 3 \\left(r - 3\\right) \\leftarrow \\textcolor{red}{\\text{ no fractions}}$ distribute the bracket. $\\Rightarrow 8 = 3 r - 9$ $8 + 9 = 3 r \\cancel{- 9} \\cancel{+ 9}$ $\\Rightarrow 3 r = 17$ divide both sides by 3 $\\frac{\\cancel{3} r}{\\cancel{3}} = \\frac{17}{3}$ $\\Rightarrow r = \\frac{17}{3}$ $\\textcolor{b l u e}{\\text{As a check}}$ Substitute this value into the right side of the equation and if equal to the left side then it is the solution. $\\text{right side \" =(17/3-9/3)/6=8/3xx1/6=8/18=4/9=\" left side}$ $\\Rightarrow r = \\frac{17}{3} \\text{ is the solution}$" ]

[ "15$ $\\Rightarrow \\boxed{b = 30}$ And, $\\frac{a}{2} – 1 = b$ $\\Rightarrow \\frac{a}{2} - 1 = 30$ $\\Rightarrow \\frac{a}{2} = 31$ $\\Rightarrow \\boxed{a = 62}$ Correct Answer$: 62$ by 7.7k points 3 8 30 1 vote 1 62 views 1 vote 2 87 views 1 vote", "+0 # Pls help 0 133 1 If $$\\left(a-\\frac 1a\\right)^2=4$$ and $$a>1$$ what is the absolute value of $$a^3 - \\frac1{a^3}?$$ Sep 16, 2021 #1 +2401 +1 (a - 1/a) = 2 (note that normally we couldn't assume that it isn't -2, but since a > 1, a - 1/a has to be positive. (a - 1/a)^3 = 8 a^3 - 3a + 3/a - 1/a^3 = 8 a^3 - 1/a^3 - 3(a - 1/a) = 8 Can you figure it out from here? =^._.^= Sep 17, 2021", "• # question_answer If $\\frac{x-5}{2}-\\frac{x-3}{5}=\\frac{1}{2},$ what is the value of 'x'? A) $7$ B) $9$ C) $8$ D) $5$ $\\frac{x-5}{2}-\\frac{x-3}{5}=\\frac{1}{2}$ $\\Rightarrow$ $\\frac{5x-25-2x+6}{10}=\\frac{1}{2}$ $\\Rightarrow$ $x=8$", "= 1 Explanation: Given the rectangle figure, Now, we will find the shaded area value. The rectangle has 5 parts. All 5 parts are shaded. So, in fractions the value is $$\\frac{5}{5}$$ . i.e., $$\\frac{5}{5}$$ = 1 Question 6. = __________ = Answer: $$\\frac{8}{8}$$ = 1. Explanation: Given the circle figure, Now, we will find the shaded area value. The circle has 8 parts. All 8 parts are shaded. So, in fractions the value is $$\\frac{8}{8}$$. i.e., $$\\frac{8}{8}$$ = 1 Scroll to Top", "$a$ is: $a*a^{-1}=0 \\Rightarrow a+a^{-1}+aa^{-1}=0 \\Rightarrow a^{-1} = -\\frac{a}{1+a}$. Since $a \\neq -1$, so for all $a$, $a^{-1}$ always exists.", "first, get all the a's on one side and all the constants on the other: $4 - \\frac a2 = \\frac 72 - a$ $\\Rightarrow a - \\frac a2 = \\frac 72 - 4$ ........simplify each side $\\Rightarrow \\frac a2 = - \\frac 12$ ...........multiply both sides by 2 $\\Rightarrow a = -1$ This is my 4th post!!!!! 5. Thank you! And happy 4,000th post!!! Wow, that's a biiiiig step! On my forum, an admin just celebrated his 10,000th........ I know.... lol!", "Solve this following Question: If $\\lim _{x \\rightarrow 0} \\frac{\\sin ^{-1} x-\\tan ^{-1} x}{3 x^{3}}$ is equal to $L$, then the value of $(6 L+1)$ is 1. $\\frac{1}{6}$ 2. $\\frac{1}{2}$ 3. 6 4. 2 Correct Option: , 4 Solution: $\\lim _{x \\rightarrow 0} \\frac{\\left(x+\\frac{x^{3}}{3 !} \\ldots\\right)-\\left(x-\\frac{x^{3}}{3} \\ldots\\right)}{3 x^{3}}=\\frac{1}{6}$ So $6 L+1=2$", "# then the value of p is Question: If $\\left(3 x+\\frac{1}{2}\\right)\\left(3 x-\\frac{1}{2}\\right)=9 x^{2}-p$ then the value of $p$ is (a) 0 (b) $-\\frac{1}{4}$ (c) $\\frac{1}{4}$ (d) $\\frac{1}{2}$ Solution: $\\left(3 x+\\frac{1}{2}\\right)\\left(3 x-\\frac{1}{2}\\right)=9 x^{2}-p$ $9 x^{2}-\\frac{1}{4}=9 x^{2}-p$ $\\left(\\because\\left(a^{2}-b^{2}\\right)=(a+b)(a-b)\\right)$ $\\Rightarrow p=\\frac{1}{4}$ Hence, the correct answer is option (c).", "= 2$ - (iii). From (i) we get $B = - A$, from (iii) $C = A - 2$ and putting them in (ii), we get $A - \\left(- A\\right) + \\left(A - 2\\right) = 13$ or $3 A = 15$ or $A = 5$ Hence $B = - 5$ and $C = 3$ Hence partial fractions are $\\frac{5}{\\left(x - 1\\right)} + \\frac{- 5 x + 3}{{x}^{2} + x + 1}$", "boundaries $\\frac{13+9}{29+20}=\\frac{22}{49}$ and $\\frac{9+5}{20+11}=\\frac{14}{31}.$ Note that the \"center\" $\\frac9{20}$ is only about a bit more than a third of the way from the left endpoint to the right. This is because the $11$ pushes harder then the $29.$ This is one of the more extreme examples. At any rate, the first graph below uses bins around the $277$ fractions in lowest terms $\\frac{a}b$ with $1 \\leq a \\lt b \\lt 30.$ Into the bins I put all $B=100806$ irrational fractional parts $\\sqrt{t^2+k}-t$ for $1 \\leq t \\leq 2k$ and $1 \\leq t \\leq 316.$ The count $c$ is is then normalized to the ratio of $\\frac{c}{B}$ to the expected value $\\frac{B}{w}$ where $w$ is the width of the bin. I left a little space between bins. The maximum discrepancy is $0.060188$ and the average is $0.01477$ The second graph is the same thing except each bin is split at the rational it surrounds. The left part is colored red and the right part green. It seems that usually, but not always, it is as with $\\frac12,$ the left half is over $1$ and the rights half under. I haven't looked into the details more closely. The variation is about three times as large. The maximum deviations for the left and right halves are $.1763188$ and $.160645$ while the", "343 views If $t^{2}-4t+1=0$, then the value of $\\left[t^{3}+1/t^{3} \\right]$ is : 1. $44$ 2. $48$ 3. $52$ 4. $64$ Given that= $t^2-4t+1=0$ or $t^2+1=4t$ dividing by $t$ in both side we get; $t+\\frac{1}{t}=4$ taking cube in bothe side we get: $\\implies t^3+\\frac{1}{t^3}+3*t*\\frac{1}{t}*(t+\\frac{1}{t})=64$ $\\implies t^3+\\frac{1}{t^3}+3*4=64$ $\\implies t^3+\\frac{1}{t^3}=64-12=52$ Option $C$ is correct here. $\\text{Note :$(a+b)^3=a^3+b^3+3ab(a+b)$}$ 5.2k points 1 vote 1 427 views 2 266 views", "• # question_answer If ${{3}^{x}}=300,$ then the value of ${{3}^{x-2}}$ is A) $\\frac{200}{11}$ B) $\\frac{1}{963}$ C) $\\frac{100}{3}$ D) $\\frac{300}{2}$ ${{3}^{x}}=300\\Rightarrow \\frac{{{3}^{x}}}{9}=\\frac{300}{9}\\Rightarrow {{3}^{x-2}}=\\frac{300}{9}$ $\\Rightarrow \\,\\,\\,\\,\\,\\,\\,\\,{{3}^{x-2}}=\\frac{100}{3}$", "• # question_answer What is the value of x in $1+{{\\left\\{ \\frac{1}{1+\\frac{1}{\\left( 1+\\frac{1}{x} \\right)}} \\right\\}}^{=\\,\\frac{11}{7}?}}$ A) 1 B) ½ C) 3 D) 7/11 Given that, $1+\\frac{1}{\\left\\{ 1+\\frac{1}{1+\\frac{1}{x}} \\right\\}}=\\frac{11}{7}$ $\\Rightarrow$ $1+\\frac{1}{\\left( 1+\\frac{x}{x+1} \\right)}=\\frac{11}{7}$ $\\Rightarrow$ $1+\\frac{1+x}{2x+1}=\\frac{11}{7}$ $\\Rightarrow$ $\\frac{1+2x+1+x}{1+2x}=\\frac{11}{7}$ $\\Rightarrow$ $\\frac{2+3x}{1+2x}=\\frac{11}{7}$ $\\Rightarrow$ $14+21x=11+22x$ $\\therefore$ $x=14-11=3$", "# if Question: If $\\lim _{x \\rightarrow 0} \\frac{\\sin ^{-1} x-\\tan ^{-1} x}{3 x^{3}}$ is equal to $L$, then the value of $(6 \\mathrm{~L}+1)$ is 1. (1) $\\frac{1}{6}$ 2. (2) $\\frac{1}{2}$ 3. (3) 6 4. (4) 2 Correct Option: , 4 Solution: $\\lim _{x \\rightarrow 0} \\frac{\\left(x+\\frac{x^{3}}{3 !} \\cdots\\right)-\\left(x-\\frac{x^{3}}{3} \\cdots\\right)}{3 x^{3}}=\\frac{1}{6}$ So $6 L+1=2$", "110 views In a fraction, numerator is increased by $25 \\%$ and the denominator is diminished by $10 \\%.$ The new fraction obtained is $5/9.$ The original fraction is $:$ 1. $2/5$ 2. $5/9$ 3. $3/5$ 4. None of the above Let original fraction is $\\frac{a}{b}$,where $a$ is numerator and $b$ is denominator. $\\Rightarrow \\frac{125\\% *a}{90\\%*b}=\\frac{5}{9}$ $\\Rightarrow \\frac{\\left ( \\frac{125}{100}*a \\right )}{\\left(\\frac{90}{100}*b \\right )}=\\frac{5}{9}$ $\\Rightarrow \\frac{a}{b}=\\frac{5*90}{9*125}$ $\\Rightarrow \\frac{a}{b}=\\frac{2}{5}$ Option $(A)$ is correct. 1 135 views", "$A_r \\;=\\;2\\left(\\frac{x}{6}\\right)^2\\quad\\Rightarrow\\q uad\\boxed{A_r\\:=\\:\\frac{x^2}{18}}$ The square looks like this: Code: * - - - - - - * | | | | | | ¼(30-x) | | | | * - - - - - - * ¼(30-x) The perimeter of the square is: $30 - x$ cm, . . so its side is: $\\frac{1}{4}(30 - x)$ The area of the square is: . $A_s \\;=\\;\\left(\\frac{30-x}{4}\\right)^2\\quad\\Rightarrow\\quad\\boxed{A_s\\;=\\; \\frac{x^2 - 60x + 900}{16}}$ Hence, the total area is: . $A \\;=\\;\\frac{x^2}{18} + \\frac{x^2 - 60x + 900}{16} \\;=\\;\\frac{1}{144}(17x^2 - 540x + 8100)$ And that is the function we must minimize . . . 3. ## Thank you Thank you very much, I really do appreciate it", "• # question_answer What is the value of $3\\frac{1}{12}+\\left[ 1\\frac{3}{4}+\\left\\{ 2\\frac{1}{2}-\\left( 1\\frac{1}{2}-\\frac{1}{3} \\right) \\right\\} \\right]$? A) $6\\frac{1}{2}$ B) $2\\frac{1}{6}$C) $6\\frac{1}{6}$ D) $0$ Correct Answer: C Solution : $3\\frac{1}{12}+\\left[ 1\\frac{3}{4}+\\left\\{ 2\\frac{1}{2}-\\left( 1\\frac{1}{2}-\\frac{1}{3} \\right) \\right\\} \\right]$ $=\\frac{37}{12}+\\left[ \\frac{7}{4}+\\left\\{ \\frac{5}{2}-\\left( \\frac{3}{2}-\\frac{1}{3} \\right) \\right\\} \\right]$ $=\\frac{37}{12}+\\left[ \\frac{7}{4}+\\left\\{ \\frac{5}{2}-\\left( \\frac{9-2}{6} \\right) \\right\\} \\right]$ $=\\frac{37}{12}+\\left[ \\frac{7}{4}+\\left\\{ \\frac{5}{2}-\\frac{7}{6} \\right\\} \\right]$ $=\\frac{37}{12}+\\left[ \\frac{7}{4}+\\left\\{ \\frac{15-7}{6} \\right\\} \\right]$ $=\\frac{37}{12}+\\frac{7}{4}+\\frac{8}{6}=\\frac{37}{12}+\\frac{7}{4}+\\frac{4}{3}$ $=\\frac{34+21+16}{12}=\\frac{74}{12}=\\frac{37}{6}=6\\frac{1}{6}$ You need to login to perform this action. You will be redirected in 3 sec", "+ 2\\right)}^{2} = 13$ $\\iff {\\left(a + 2\\right)}^{2} = \\frac{13}{2}$ $\\iff a + 2 = \\sqrt{\\frac{13}{2}} \\mathmr{and} a + 2 = - \\sqrt{\\frac{13}{2}}$ Then $a = \\pm \\sqrt{\\frac{13}{2}} - 2$", "and $d$. The $(\\Rightarrow)$ is because $c=xa, d=xb$ for some $x$, so $\\frac{a+c}{b+d}=\\frac{a+xa}{b+xb}=\\frac{a(1+x)}{b(1+x)}=\\frac{a}{b}$. The other direction is similar. The above is pretty easy to remember, and with that intuition in mind it is not hard to imagine that $$\\frac{a}{b}<\\frac{c}{d} \\Leftrightarrow \\frac{a}{b}<\\frac{a+c}{b+d}<\\frac{c}{d}$$ and similar results. - With a small caveat:$$\\frac11=\\frac{-1}{-1}\\text{, but }\\frac11\\ne\\frac00$$ – robjohn Jul 6 at 20:38", "$x=2$ is $-39 .$ (iv) Given: $f(x)=(x-2) \\sqrt{x-1}$ $\\Rightarrow f^{\\prime}(x)=\\sqrt{x-1}+\\frac{(x-2)}{2 \\sqrt{x-1}}$ For a local maximum or a local minimum, we must have $f^{\\prime}(x)=0$ $\\Rightarrow \\sqrt{x-1}+\\frac{(x-2)}{2 \\sqrt{x-1}}=0$ $\\Rightarrow 2(x-1)+(x-2)=0$ $\\Rightarrow 2 x-2+x-2=0$ $\\Rightarrow 3 x-4=0$ $\\Rightarrow 3 x=4$ $\\Rightarrow x=\\frac{4}{3}$ Thus, the critical points of $f$ are $1, \\frac{4}{3}$ and $9 .$ Now, $f(1)=(1-2) \\sqrt{1-1}=0$ $f\\left(\\frac{4}{3}\\right)=\\left(\\frac{4}{3}-2\\right) \\sqrt{\\frac{4}{3}-1}=\\frac{-2}{3} \\times \\frac{1}{\\sqrt{3}}=-\\frac{2}{3 \\sqrt{3}}$ $f(9)=(9-2) \\sqrt{9-1}=14 \\sqrt{2}$ Hence, the absolute maximum value when $x=9$ is $14 \\sqrt{2}$ and the absolute minimum value when $x=\\frac{4}{3}$ is $-\\frac{2}{3 \\sqrt{3}} .$ Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book. (iv) Given: $f(x)=(x-2) \\sqrt{x-1}$ $\\Rightarrow f^{\\prime}(x)=\\sqrt{x-1}+\\frac{(x-2)}{2 \\sqrt{x-1}}$ For a local maximum or a local minimum, we must have $f^{\\prime}(x)=0$ $\\Rightarrow \\sqrt{x-1}+\\frac{(x-2)}{2 \\sqrt{x-1}}=0$ $\\Rightarrow 2(x-1)+(x-2)=0$ $\\Rightarrow 2 x-2+x-2=0$ $\\Rightarrow 3 x-4=0$ $\\Rightarrow 3 x=4$ $\\Rightarrow x=\\frac{4}{3}$ Thus, the critical points of $f$ are $1, \\frac{4}{3}$ and $9 .$ Now, $f(1)=(1-2) \\sqrt{1-1}=0$ $f\\left(\\frac{4}{3}\\right)=\\left(\\frac{4}{3}-2\\right) \\sqrt{\\frac{4}{3}-1}=\\frac{-2}{3} \\times \\frac{1}{\\sqrt{3}}=-\\frac{2}{3 \\sqrt{3}}$ $f(9)=(9-2) \\sqrt{9-1}=14 \\sqrt{2}$ Hence, the absolute maximum value when $x=9$ is $14 \\sqrt{2}$ and the absolute minimum value when $x=\\frac{4}{3}$ is $-\\frac{2}{3 \\sqrt{3}} .$ Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book." ]

[ "# Making change Consider the classic problem: How many ways can one make change for one dollar using half-dollars, quarters, dimes, nickels, and pennies? Or more generally, how many ways can one make change for a given amount using arbitrary (positive integer) denominations? This post chronicles a series of incremental improvements to solutions to this problem. In the first section, we attack the problem with dynamic programming in Python, and we’re able to count the ways to change quite large amounts of money (eventually up to about $100M for the given set of five coins). The second section explores a nice method for deriving closed form solutions, due to Lee Newberg, and the third is a synthesis of the previous two, automating the closed form derivation in the general setting. 1. From 1 to 100,000,000 It’s not hard to produce a recursive function definition which computes the answer to the coin problem. To count all the ways of making change, we just need to condition on whether or not at least one coin of the first denomination will be used (and then add these counts together). If one will be used, then we must change the original amount minus the first denomination. If not, then we must change the whole original amount, but with one fewer denomination. In", "for this calculation. '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ So you have 27 quarters", "over each quarter. LS0tCnRpdGxlOiAiTGFiIDEwOiBUZXh0IE1pbmluZyIKYXV0aG9yOiAKLSBbWU9VUiBOQU1FXQotIFtZT1VSIFBBUlRORVJTIE5BTUVdCmRhdGU6ICJgciBTeXMudGltZSgpYCIKb3V0cHV0OiBodG1sX25vdGVib29rCi0tLQoKIyMjIFN0ZXAJMToJUmVhZAlpbgl0aGUJcG9zaXRpdmUJYW5kCW5lZ2F0aXZlCXdvcmQJZmlsZXMKYGBge3J9CgojIENyZWF0ZSB0d28gdmVjdG9ycyBvZiB3b3Jkcywgb25lIGZvciB0aGUgcG9zaXRpdmUgd29yZHMgYW5kIG9uZSBmb3IgdGhlIG5lZ2F0aXZlIHdvcmRzLiBUaGUgcG9zaXRpdmUgd29yZHMgY2FuIGJlIGZvdW5kIGhlcmU6ICJodHRwczovL2NqYWNrczA0LmdpdGh1Yi5pby82ODcvRGF0YXNldHMvcG9zaXRpdmUtd29yZHMudHh0IiBhbmQgbmVnYXRpdmUgd29yZHMgaGVyZTogImh0dHBzOi8vY2phY2tzMDQuZ2l0aHViLmlvLzY4Ny9EYXRhc2V0cy9uZWdhdGl2ZS13b3Jkcy50eHQiLiBZb3Ugc2hvdWxkIHVzZSB0aGUgc2NhbigpIGZ1bmN0aW9uIHdoaWNoIHJlYWRzIGRhdGEgaW50byBhIHZlY3RvciBvciBsaXN0IGZyb20gdGhlIGNvbnNvbGUgb3IgZmlsZS4gWW91J2xsIG5lZWQgdGhyZWUgYXJndW1lbnRzOiAoMSkgdGhlIGZpbGUgbmFtZS9wYXRoLCAoMikgdGhlIHNlY29uZCBhcmd1bWVudCBpcyBjaGFyYWN0ZXIoMCkgd2hpY2ggd2lsbCByZWFkIHRoZSBuZXh0ICBsaW5lIGFzIGEgY2hhcmFjdGVyIChhcyBvcHBvc2VkIHRvIGludGVnZXIgb3Igc29tZSBvdGhlciBkYXRhdCB5cGUpLCAoMykgdGhlIHNlcCBhcmd1bWVudCB0byB0ZWxsIFIgaG93IHRoZSBkYXRhIGFyZSBzZXBlcmF0ZWQgZS5nLiwgXG4gCgojIE5vdGUgdGhhdCB3aGVuIHJlYWRpbmcgaW4gdGhlIGZpbGVzLCB0aGVyZSBtaWdodCBiZSBsaW5lcyBhdCB0aGUgc3RhcnQgYW5kL29yIHRoZSBlbmQgdGhhdCB3aWxsIG5lZWQgdG8gYmUgcmVtb3ZlZCAoaS5lLiB5b3Ugc2hvdWxkIGNsZWFuIHlvdSBkYXRhIGlmIG5lZWRlZCkuIAoKYGBgCgojIyMgU3RlcCAyOiBQcm9jZXNzIGluIHRoZSBNTEsgc3BlZWNoIApgYGB7cn0KIyBSZWFkIHRoZSBNTEsgdGV4dCBmaWxlIHVzaW5nIHRoZSByZWFkTGluZXMoKSBmdW5jdGlvbi4gT25seSB0aGUgVVJMIGlzIHJlcXVpcmVkLgojIEluc3BlY3QgdGhlIHZlY3RvciBhYm92ZS4gU29tZSBsaW5lcyBhcmUgYmxhbmsgIiIuIFJlbW92ZSB0aGVzZS4KIyBDcmVhdGUgYSB0ZXJtIG1hdHJpeC4gVGhlcmUgYXJlIHNldmVyYWwgc3RlcHMgaGVyZSBiZWdpbm5pbmcgd2l0aCBjcmVhdGluZyBhIHZlY3RvciBzb3VyY2UgYW5kIG1ha2luZyB0ZXh0IHRyYW5zZm9ybWF0aW9ucy4gKENoZWNrIGNoYXB0ZXIgMTQgd2hlcmUgc2JhIGlzIHRyYW5zZm9ybWVkKQojIENyZWF0ZSBhIGxpc3Qgb2YgY291bnRzIGZvciBlYWNoIHdvcmQgCmBgYAoKIyMjIFN0ZXAgMzogRGV0ZXJtaW5lIGhvdyBtYW55IHBvc2l0aXZlIHdvcmRzIHdlcmUgaW4gdGhlIHNwZWVjaCAgCmBgYHtyfQojIEhpbnQ6IG9uZSB3YXkgdG8gZG8gdGhpcyBpcyB0byB1c2UgdGhlIOKAmG1hdGNo4oCZIGZ1bmN0aW9uIG9uIHRoZSBsaXN0IG9mIHdvcmRzIGZyb20gU3RlcCAyIGFuZCB0aGUgcG9zaXRpdmUgd29yZHMgaW4gdGhlIGxpc3QgZnJvbSB0aGUgaW1wb3J0LiAKCiAgIyBzdW0gdGhlIHRvdGFsIG51bWJlciBvZiB3b3JkcyBhbmQgc3RvcmUgdGhlIHZhbHVlIHRvICJ0b3RhbFdvcmRzIgogIAogICMgY3JlYXRlIGEgdmVjdG9yICJ3b3JkcyIgdGhhdCBjb250YWlucyBhbGwgdGhlIHdvcmRzIGluICJ3b3JkQ291bnRzIgogIAogICMgbG9jYXRlIHdoaWNoIHdvcmRzIGluICJtbGsiIHdlcmUgcG9zaXRpdmUgKGFwcGVhcmVkIGluIHBvc2l0aXZlLXdvcmQgbGlzdCkKICAKICAjIGNhbGN1bGF0ZSB0aGUgdG90YWwgbnVtYmVyIG9mIHBvc2l0aXZlIHdvcmRzIGluICJtbGsiIHNwZWVjaCAoaW4gd29yZENvdW50cykgYW5kIGFzc2lnbiB0aGUgbnVtYmVyIHRvIHRoZSB2YXJpYWJsZSAicFRvdGFsIi4gVGhlIHdoaWNoKCkgZnVuY3Rpb24gb24gd29yZHMgdGhlIHZlY3RvciBhYm92ZSB3aWxsIGdpdmUgeW91IHRoZSBpbmRleCBudW1iZXIuIAogCiAgIyB2aWV3IHRoZSB0b3RhbCBudW1iZXIgb2YgcG9zaXRpdmUgd29yZHMgKDk1IHBvc2l0aXZlIHdvcmRzIGluIHRoZSBzcGVlY2gpCiAgIAogICMgdmlldyB0aGUgcGVyY2VudGFnZSBvZiBwb3NpdGl2ZSB3b3JkcyAoMTEuMjk2MDglIG9mIHRoZSBzcGVlY2ggd29yZHMgYXJlIHBvc2l0aXZlKQogCmBgYAoKIyMjIFN0ZXAgNDogRGV0ZXJtaW5lIGhvdyBtYW55IG5lZ2F0aXZlIHdvcmRzIHdlcmUgaW4gdGhlIHNwZWVjaCAKYGBge3J9CiMgSGludDogb25lIHdheSB0byBkbyB0aGlzIGlzIHRvIHVzZSB0aGUg4oCYbWF0Y2jigJkgZnVuY3Rpb24gb24gdGhlIGxpc3Qgb2Ygd29yZHMgZnJvbSBTdGVwIDIgYW5kIHRoZSBwb3NpdGl2ZSB3b3JkcyBpbiB0aGUgbGlzdCBmcm9tIHRoZSBpbXBvcnQuIAoKYGBgCgojIyMgU3RlcCA1OiBSZWRvIHRoZSDigJhwb3NpdGl2ZeKAmSBhbmQg4oCYbmVnYXRpdmXigJkgY2FsY3VsYXRpb25zIGZvciBlYWNoIDI1JSBvZiB0aGUgc3BlZWNoIApgYGB7cn0KIyBDb21wYXJlIHRoZSByZXN1bHRzIChleC4gYSBzaW1wbGUgYmFyY2hhcnQgb2YgdGhlIDQgbnVtYmVycykuIEkgcmVjb21tZW5kIHRha2luZyBleHRyYWN0aW5nIHF1YXJ0ZXJzIG9mIHRoZSBzcGVlY2gsIHN0b3JpbmcgZWFjaCBxdWFydGVyIGluIGEgdmVjdG9yIGFuZCB0aGVuIGNvbmR1Y3RpbmcgdGhlIGNhbGN1bGF0aW9ucyBvdmVyIGVhY2ggcXVhcnRlci4gCmBgYAoK", "designs required under this subsection for quarter dollars shall be-- (i) selected by the Secretary after consultation with-- (I) the Governor of the State being commemorated, or such other State officials or group as the State may designate for such purpose; and (II) the Commission of Fine Arts; and (ii) reviewed by the Citizens Commemorative Coin Advisory Committee. (B) Selection and approval process.--Designs for quarter dollars may be submitted in accordance with the design selection and approval process developed by the Secretary in the sole discretion of the Secretary. (C) Participation.--The Secretary may include participation by State officials, artists from the States, engravers of the United States Mint, and members of the general public. (D) Standards.--Because it is important that the Nation's coinage and currency bear dignified designs of which the citizens of the United States can be proud, the Secretary shall not select any frivolous or inappropriate design for any quarter dollar minted under this subsection. (E) Prohibition on certain representations.--No head and shoulders portrait or bust of any person, living or dead, and no portrait of a living person may be included in the design of any quarter dollar under this subsection. (5) Treatment as numismatic items.--For purposes of sections 5134 and 5136, all coins minted under this subsection shall be considered to be numismatic items.", "# How Do I Make Change From A Dollar? After introducing students to the value of all currently circulating U.S. coins students explore how to make change for a dollar. The teacher asks one student to be the shopper and one to be the clerk to model the activity for the class. The shopper wants to purchase one pencil which (s)he is told costs $0.25. (S)he gives clerk$1.00, and the clerk counts out (aloud) how much change to give back.Repeat again until students grasp this concept. Concepts Resource Details", "quarters: only complicated cases 4 quarters: 3^3 ways 5 quarters: 3^2 ways, etc. So it comes down to having four non-quarter coins adding up to$0.25-or-more.", "# Category Archives: Constraint Programming ## Snapple Fact 804: How Many Ways to Make Change for a Dollar? I went camping this past weekend (my first time) and my neighbor (who’s actually my neighbor in real life and was my neighbor at the camp site) was drinking a Snapple and read the following fact back to me: Snapple Real Fact #804: There are 293 ways to make change for a dollar. My first reaction was “Mmm…interesting”, but I couldn’t help wondering whether the Snapple folks did their math correctly. So after I got home and unpacked the car, I wrote a little Constraint Programming code in Comet to check this fact. It turns out that the number is indeed 293 if the following two things are allowed: (i) returning a 1-dollar coin in exchange for a dollar bill, and (ii) using half-dollar coins which, in my opinion, are rare these days. Here’s a list of the 292 ways that do not include using a 1-dollar coin which, in my opinion, isn’t really “giving change”. If you’re wondering how many ways there are when you’re not allowed to use 1-dollar or half-dollar coins, the answer is 242. Here’s a list of all such possible ways. Update: A friend asked me what the number would be if we considered the", "× ## Decimals Coins are often used to express a fraction of a currency. In the US, a quarter is $0.25 and a dime is$0.10. Counting change is just one of many ways you can put decimals to use in everyday life. # Place Value Solving $$3.2\\times2.2,$$ which number do we get in the hundredths place? In the number $$0.0 8 3 5,$$ the digit $$3$$ is in which place? Round up $$0.004 0 9 5$$ to the nearest tenthousandth. Solving $$7.57-3.23,$$ which number do we have for the tenths place? What number is in the hundredths place of 662.373? ×", "algorithm. In contrast, consider the following problem: How many different ways can we make change of \\$ 1.00, given half-dollars, quarters, dimes, nickels, and pennies? More generally, can we write a function to compute the number of ways to change any given amount of money? This problem has a simple solution as a recursive function. Suppose we think of the types of coins available as arranged in some order. Then the following relation holds: The number of ways to change amount $$a$$ using $$n$$ kinds of coins equals • the number of ways to change amount $$a$$ using all but the first kind of coin, plus • the number of ways to change amount $$a$$ - $$d$$ using all n kinds of coins, where $$d$$ is the denomination of the first kind of coin. To see why this is true, observe that the ways to make change can be divided into two groups: those that do not use any of the first kind of coin, and those that do. Therefore, the total number of ways to make change for some amount is equal to the number of ways to make change for the amount without using any of the first kind of coin, plus the number of ways to make change assuming that we do use the first", "of ways to change any given amount of money using any set of currency denominations? This problem has a simple solution as a recursive function. Suppose we think of the types of coins available as arranged in some order, say from most to least valuable. The number of ways to change an amount a using n kinds of coins equals 1. the number of ways to change a using all but the first kind of coin, plus 2. the number of ways to change the smaller amount a - d using all n kinds of coins, where d is the denomination of the first kind of coin. To see why this is true, observe that the ways to make change can be divided into two groups: those that do not use any of the first kind of coin, and those that do. Therefore, the total number of ways to make change for some amount is equal to the number of ways to make change for the amount without using any of the first kind of coin, plus the number of ways to make change assuming that we do use the first kind of coin at least once. But the latter number is equal to the number of ways to make change for the amount that remains after using", "Number of ways Derive an expression for the number of ways change can be made for a note of any denomination n, where n is a natural number, given coins of denomination 1 and 3. Check your expression by hand for n = 4, 5, 6.", "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.", "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 coins are quarters? Your friend shows you a coin collection. In it , 28-50 of coins are quarters. What percents of the coins are quarters?...", "the number of ways to change any given amount of money using any set of currency denominations? This problem has a simple solution as a recursive function. Suppose we think of the types of coins available as arranged in some order, say from most to least valuable. The number of ways to change an amount a using n kinds of coins equals 1. the number of ways to change a using all but the first kind of coin, plus 2. the number of ways to change the smaller amount a - d using all n kinds of coins, where d is the denomination of the first kind of coin. To see why this is true, observe that the ways to make change can be divided into two groups: those that do not use any of the first kind of coin, and those that do. Therefore, the total number of ways to make change for some amount is equal to the number of ways to make change for the amount without using any of the first kind of coin, plus the number of ways to make change assuming that we do use the first kind of coin at least once. But the latter number is equal to the number of ways to make change for the amount that remains", "different ways can we make change of $1.00, given half-dollars, quarters, dimes, nickels, and pennies? More generally, can we write a procedure to compute the number of ways to change any given amount of money? This problem has a simple solution as a recursive procedure. Suppose we think of the types of coins available as arranged in some order. Then the following relation holds: The number of ways to change amount a using n kinds of coins equals • the number of ways to change amount a using all but the first kind of coin, plus • the number of ways to change amount a - d using all n kinds of coins, where d is the denomination of the first kind of coin. To see why this is true, observe that the ways to make change can be divided into two groups: those that do not use any of the first kind of coin, and those that do. Therefore, the total number of ways to make change for some amount is equal to the number of ways to make change for the amount without using any of the first kind of coin, plus the number of ways to make change assuming that we do use the first kind of coin. But the latter number is equal to", "or silver coins which shall be struck or coined at the said mint.” (At present the only gold or silver coins struck by the US mint are the American Silver Eagle and the American Gold Eagle coins, some Proof coinage at the San Francisco Mint, such as the silver U.S. State Quarters, and much of the Commemorative coinage of the United States.) All other sections of the act have been superseded, as for example the Coinage Act of 1834 changing the gold-to-silver weight ratio. Various acts have subsequently been passed affecting the amount and type of metal in U. S. coins, so that today there is no legal definition of the term “dollar” to be found in U. S. statute. Current statutes regulating coinage in the United States may be found in Title 31 of the United States Code, Chapter 51 — Coins and Currency.", "# Wikipedia:Reference desk archive/May 2005 II ## U.S. Dollar Coins Why does it seems like the United States government is so reluctant to put dollar coins into circulation? 10qwerty | (Talk) 07:45, 14 May 2005 (UTC) Dollar coins tend to cost more than a dollar to make. Plus American cash registers don't have any place to put them. RickK 08:14, May 14, 2005 (UTC) Duh. So how much does it cost to make a quarter? Jooler 13:33, 14 May 2005 (UTC) The above doesn't seem to me to be a viable reason. A change from note to coin on higher denominations would be necessary due to inflationary pressures. Other countries have already ditched low denomination copper coins and at the same time higher value coins have been introduced (Australia and New Zealand both killed one and two-cent copper coins at the same time as introducing gold coloured one and two dollar coins - which replaced one and two dollar notes). Notes are actually more expensive to maintain than coins as thier life is so much shorter that they therefore need replacing more often than coins. Coins can be passed around for decades. I suspect there are few countries around nowadays using coins which are made of metals corresponding in value to the value of the coin (excluding commemorative", "(6) Numismatic items.-- (A) Quality of coins.--The Secretary may mint and issue such number of quarter dollars of each design selected under paragraph (4) in uncirculated and proof qualities as the Secretary determines to be appropriate. (B) Silver coins.--Notwithstanding subsection (b), the Secretary may mint and issue such number of quarter dollars of each design selected under paragraph (4) as the Secretary determines to be appropriate with a content of 90 percent silver and 10 percent copper. (C) Sources of bullion.--The Secretary shall obtain silver for minting coins under subparagraph (B) from available resources, including stockpiles established under the Strategic and Critical Materials Stock Piling Act. (7) Application in event of the admission of additional states.--If any additional State is admitted into the Union before the end of the 10-year period referred to in paragraph (1), the Secretary of the Treasury may issue quarter dollar coins, in accordance with this subsection, with a design which is emblematic of such State during any 1 year of such 10-year period, in addition to the quarter dollar coins issued during such year in accordance with paragraph (3)(A).''. Passed the House of Representatives September 23, 1997. Attest: ROBIN H. CARLE, Clerk. By Jeff Trandahl, Deputy Clerk.\" \"50 States Commemorative Coin Program Act - Amends Federal law to mandate redesign of quarter dollar", "# quarter quarter is one of the 1000 most common headwords. quarter ## Noun Singular quarter 1. (countable) A quarter is one fourth $(1/4)$ of something. The lake was about a quarter mile from town. About a quarter of the students came from outside the United States. There's an old house three quarters of the way up that mountain. 2. (countable) A quarter is a unit of money worth 25 cents in the United States and Canada. Can you give me four quarters for this dollar? 3. (countable) A quarter is a period of three months, especially in business. Sales were down 3.7 percent from the fourth quarter of last year. 4. (countable) A quarter is one of four a periods of time, especially in sports. He threw four touchdown passes in the fourth quarter leading to a 38-28 win. 5. (countable) A quarters is one section of a city. The bar was in New Orleans' French quarter. 6. (plural only; not used in singular form) Quarters are the places where people live. After training, we returned to quarters to get changed and rest. ## Verb Plain form quarter Third-person singular quarters Past tense quartered Past participle quartered Present participle quartering 1. (transitive) If you quarter something, you make it into four equal pieces. Using a sharp knife," ]

[ "\\cdot 1 = x^{20}$ $x^5 \\cdot x^{15} \\cdot 1 \\cdot 1 = x^{20}$ $x^{20} \\cdot 1 \\cdot 1 \\cdot 1 = x^{20}$ $x^5 \\cdot x^5 \\cdot x^{10} \\cdot 1 = x^{20}$ The nice thing is that each of these expressions is conceptually equivalent to a way of expressing 20 cents using pennies, nickels, dimes, and quarters. In each case, the value in parentheses matches an exponent. • $1 \\cdot 1 \\cdot x^{20} \\cdot 1 = x^{20}$: 2 dimes (20 cents). • $1 \\cdot x^{10} \\cdot x^{10} \\cdot 1 = x^{20}$: 2 nickels (10 cents) and 1 dime (10 cents) • $1 \\cdot x^{20} \\cdot 1 \\cdot 1 = x^{20}$: 4 nickels (20 cents) • $x^{10} \\cdot 1 \\cdot x^{10} \\cdot 1 = x^{20}$: 10 pennies (10 cents) and 1 dime (10 cents) • $x^{15} \\cdot x^5 \\cdot 1 \\cdot 1 = x^{20}$: 15 pennies (15 cents) and 1 nickel (5 cents) • $x^{10} \\cdot x^{10} \\cdot 1 \\cdot 1 = x^{20}$: 10 pennies (10 cents) and 2 nickels (10 cents) • $x^5 \\cdot x^{15} \\cdot 1 \\cdot 1 = x^{20}$: 5 pennies (5 cents) and 3 nickels (15 cents) • $x^{20} \\cdot 1 \\cdot 1 \\cdot 1 = x^{20}$: 20 pennies (20 cents) • $x^5 \\cdot x^5 \\cdot x^{10} \\cdot 1 = x^{20}$: 5 pennies (5 cents), 1", "want is to set up a proportion that we can solve. We can do this two ways. One way is to place the dollar values in the numerators while the rupees in the denominators of the proportion. And the other way is to swap their locations. Either of the setups should give us the same answer. For this exercise, we will put the dollar information on top. Solve the unknown value of “$x$” to get the required dollar value. That means at the time of the exchange, $17$ US Dollars is equivalent to $901$ Indian Rupees. ⚠️ The next example is a challenge problem because it will require you to apply critical thinking and solve Multi-Step Linear Equations with variables on both sides of the equation. Example 7: You want to cut a block of wood with a length of $72$ feet into two pieces such that the ratio of the shorter to the longer piece is $2$ to $7$. What are their lengths? Let “$x$” be the length of the shorter piece. That means “$72 - x$” will be the longer piece. See diagram below. It is given that the ratio of the shorter to the longer piece is $2:7$. Using all these information, we can now setup the proportion to solve for the lengths of both", "this optimal solution for the problem of $n$ cents must contain within it an optimal solution for the problem of $n - c$ cents. We use the usual cut-and-paste argument. Clearly, there are $k - 1$ coins in the solution to the $n - c$ cents problem used within our optimal solution to the $n$ cents problem. If we had a solution to the $n - c$ cents problem that used fewer than $k - 1$ coins, then we could use this solution to produce a solution to the $n$ cents problem that uses fewer than $k$ coins, which contradicts the optimality of our solution. a. A greedy algorithm to make change using quarters, dimes, nickels, and pennies works as follows: • Give $q = \\lfloor n / 25 \\rfloor$ quarters. That leaves $n_q = n \\mod 25$ cents to make change. • Then give $d = \\lfloor n_q / 10 \\rfloor$ dimes. That leaves $n_d = n_q \\mod 10$ cents to make change. • Then give $k = \\lfloor n_d / 5 \\rfloor$ nickels. That leaves $n_k = n_d \\mod 5$ cents to make change. • Finally, give $p = n_k$ pennies. An equivalent formulation is the following. The problem we wish to solve is making change for $n$ cents. If $n = 0$, the optimal solution", "Consider using an array to store the exact change. Making the changes listed above, and compressing the retrieval of change down into one function, the code becomes this: import java.util.Scanner; public class Problem2 { /** * Takes a dollar amount, represented by a double value, * and returns the minimum number of quarters, dimes, * nickels, and pennies that would be required to * represent that dollar value. * @param dollarValue * @return {Quarters, Dimes, Nickels, Pennies} */ public static int[] getChange(double dollarValue) { int cents = (int) Math.round(dollarValue*100); // Convert the cents into quarters, dimes, nickels, and pennies, respectively. int[] change = {cents/25,(cents%=25)/10, (cents%=10)/5, cents%5}; return change; } public static void main(String[] args) { // Instantiate Variables \\\\ double total; int[] change; Scanner keyboard; // Receive Input \\\\ keyboard = new Scanner(System.in); System.out.println(\"This application calculates the exact change for a given dollar amount.\"); System.out.println(\"Please enter in the total amount of dollars and cents:\"); total = keyboard.nextDouble(); keyboard.close(); // Make Calculations \\\\ change = getChange(total); System.out.printf(\"\\$%.2f = \" + change[0] + \" Quarters, \" + change[1] + \" Dimes, \" + change[2] + \" Nickels, \" + change[3] + \" Pennies\", total); } } • @mdfst13 In this simple case, it's just one Scanner, but for more complex code I really do appreciate having the imports there so", "and nickels you want $$\\left(1+x+x^2+\\ldots\\right)\\left(1+x^5+x^{10}+\\ldots\\right)\\;:\\tag{1}$$ $(1)$ has one $x^n$ term for every way to write $n$ in the form $5m+k$ with $m,k\\in\\Bbb N$, which is precisely the number of ways to make the amount $n$ with nickels and pennies. Clearly, then, the generating function is just $$\\frac1{(1-x)(1-x^5)}\\;.$$ I’ll leave the last part to you. - I like your last paragraph, +1 – Rustyn Jan 31 '13 at 5:25 great explanation.. thanks so the last part i suppose will be 1/((1-x)(1-x^5)(1-x^10)) by the same principle we used pennies and nickels to change n cents. – dotdotdot Jan 31 '13 at 5:32 @dotdotdot there is a great PDF listed in the comments about this topic, just FYI in case you're interested. – Rustyn Jan 31 '13 at 5:36 @dotdotdot: You’re welcome. Yes, that’s exactly right. – Brian M. Scott Jan 31 '13 at 5:37 For $3$: Consider multiplies of $5$: The coefficients go like: $<1,1,1,1,2,2,2,2,2,3,3,3,3,3,...>$, (We obtain a way to give back change at each multiple of $5$). Each coefficient is obtained by multiplying the two previous power series: $$(1 + x + x^2 + ...)(1 + x^5 + x^{10}+...)$$ See Brian's answer for a simplification of this function. -", "twelve more than $$5x$$ ⓓ $$5y$$ less than $$8y^2$$ ⓐ eleven times the difference of $$y$$ and two ⓑ the difference of eleven times $$y$$ and two ⓐ $$11(y−2)$$ ⓑ $$11y−2$$ Dushko has nickels and pennies in his pocket. The number of pennies is four less than five the number of nickels. Let nn represent the number of nickels. Write an expression for the number of pennies. ## Integers Simplify Expressions with Absolute Value In the following exercise, fill in $$<,>,$$ or $$=$$ for each of the following pairs of numbers. ⓐ $$−|7| \\_\\_\\_−|−7|$$ ⓑ $$−8 \\_\\_\\_−|−8|$$ ⓒ $$|−13| \\_\\_\\_−13$$ ⓓ $$|−12| \\_\\_\\_−(−12)$$ ⓐ $$=$$ ⓑ $$=$$ ⓒ $$>$$ ⓓ $$=$$ In the following exercises, simplify. $$9−|3(4−8)|$$ $$12−3|1−4(4−2)|$$ $$−9$$ In the following exercises, simplify each expression. $$−12+(−8)+7$$ ⓐ $$15−7$$ ⓑ $$−15−(−7)$$ ⓒ $$−15−7$$ ⓓ $$15−(−7)$$ ⓐ $$8$$ ⓑ $$−8$$ ⓒ $$−22$$ ⓓ $$22$$ $$−11−(−12)+5$$ ⓐ $$23−(−17)$$ ⓑ $$23+17$$ ⓐ 40 ⓑ 40 $$−(7−11)−(3−5)$$ Multiply and Divide Integers In the following exercise, multiply or divide. ⓐ $$−27÷9$$ ⓑ $$120÷(−8)$$ ⓒ $$4(−14)$$ ⓓ $$−1(−17)$$ ⓐ$$−3$$ ⓑ $$−15$$ ⓒ $$−56$$ ⓓ $$17$$ Simplify and Evaluate Expressions with Integers In the following exercises, simplify each expression. ⓐ $$(−7)^3$$ ⓑ $$−7^3$$ $$(7−11)(6−13)$$ 16 $$63÷(−9)+(−36)÷(−4)$$ $$6−3|4(1−2)−(7−5)|$$ $$−12$$ $$(−2)^4−24÷(13−5)$$ For the following exercises, evaluate each expression. $$(y+z)^2$$ when $$y=−4,z=7$$ 9 $$3x^2−2xy+4y^2$$ when $$x=−2,y=−3$$ Translate English", "is . Then, since we have a total value of 95 cents: Now you have a 2X2 system: Solve the 2X2 system. Recommend you use the substitution method. John My calculator said it, I believe it, that settles it", "$3$ coins. • To get $4$ use packets with $1$ and $3$ coins. • To get $5$ use packets with $2$ and $3$ coins • To get $6$ use all packets. In the second example, two packets with $1$ and $1$ coins can be made to get any amount $x$ ($1\\leq x\\leq 2$).", "exponent. • $1 \\cdot 1 \\cdot x^{20} \\cdot 1 = x^{20}$: 2 dimes (20 cents). • $1 \\cdot x^{10} \\cdot x^{10} \\cdot 1 = x^{20}$: 2 nickels (10 cents) and 1 dime (10 cents) • $1 \\cdot x^{20} \\cdot 1 \\cdot 1 = x^{20}$: 4 nickels (20 cents) • $x^{10} \\cdot 1 \\cdot x^{10} \\cdot 1 = x^{20}$: 10 pennies (10 cents) and 1 dime (10 cents) • $x^{15} \\cdot x^5 \\cdot 1 \\cdot 1 = x^{20}$: 15 pennies (15 cents) and 1 nickel (5 cents) • $x^{10} \\cdot x^{10} \\cdot 1 \\cdot 1 = x^{20}$: 10 pennies (10 cents) and 2 nickels (10 cents) • $x^5 \\cdot x^{15} \\cdot 1 \\cdot 1 = x^{20}$: 5 pennies (5 cents) and 3 nickels (15 cents) • $x^{20} \\cdot 1 \\cdot 1 \\cdot 1 = x^{20}$: 20 pennies (20 cents) • $x^5 \\cdot x^5 \\cdot x^{10} \\cdot 1 = x^{20}$: 5 pennies (5 cents), 1 nickel (5 cents), and 1 dime (10 cents) Notice that the last line didn’t appear in the Bugs Bunny cartoon. Using the formula for an infinite geometric series (and assuming $-1 < x < 1$), we may write the infinite product as $f(x) = \\displaystyle \\frac{1}{(1-x)(1-x^5)(1-x^{10})(1-x^{25})}$ When written as an infinite series — that is, as a Taylor series about $x =0$ — the coefficients", "only use $$280$$. You can use $$20$$ blocks of energy. Your starting complement gives you $$9$$ blocks. It is hard to give a concise description on the allowable spacing of the uses. • Because 21\\cdot 14=294$so using$21\\$ blocks requires more energy than you will get – Ross Millikan Apr 9 at 23:44", "dynamic programming python book We start with one cent. The greedy method works fine when we are using U.S. coins, but suppose Or we can look at amount. Then we take $$63 - 21 = 42$$ and look at the 42nd element of the bulk of the work in this function is done by the loop that starts on Let’s look at how we would fill in a table of minimum coins to use in We can easily extend dpMakeChange to the next lowest coin value and use as many of those as possible. for 15 cents at least three times. for the original amount minus a penny, or a nickel plus the number of \\label{eqn_change}\\end{split}\\], Problem Solving with Algorithms and Data Structures, A penny plus the minimum number of coins to make change for, A nickel plus the minimum number of coins to make change for, A dime plus the minimum number of coins to make change for. Here are main ones: 1. minimum number of coins for 11 cents. line 4. When the to expose you to several different problem solving strategies. that is, we are trying to make change in the exact amount of one of our In divide and conquer, each subproblem has to be solved … - Selection from Python Data Structures and", "each purse to ensure that no purse is empty. Then, we proceed as before but now with only 7 pennies left to distribute to 5 purses. How many ways can we select 11 objects: $PPP\\cdots P||||$ (7 $P's$ and 4 sticks)? The number of ways of distributing 7 pennies to 5 purses equals $\\binom{11}{4}=330$. Indistinguishable Purses NO PURSE CAN BE EMPTY Suppose for a moment that the purses are distinguishable and labeled as $A_1,A_2,A_3,A_4,A_5$. The following two examples show how we may place the 12 pennies $A_1\\quad A_2\\quad A_3\\quad A_4\\quad A_5$ $8\\quad\\;\\; 1\\quad\\;\\; 1\\quad\\;\\; 1\\quad\\;\\; 1$ $A_1\\quad A_2\\quad A_3\\quad A_4\\quad A_5$ $1\\quad\\;\\;\\; 8\\quad\\;\\;\\; 1\\quad\\;\\; 1\\quad\\;\\; 1$ Since the purses are identical we remove the labels and notice that the two results look identical. One purse has 8 pennies and the rest has 1 each. $A\\quad A\\quad A\\quad A\\quad A$ $8\\quad\\; 1\\quad\\; 1\\quad\\; 1\\quad\\; 1$ $A\\quad A\\quad A\\quad A\\quad A$ $1\\quad\\; 8\\quad\\; 1\\quad\\; 1\\quad\\; 1$ The examples lead us to an approach to find a solution: find the number of partitions of 12 into exactly 5 positive integers. There are 13 ways to do that 1) $8+1+1+1+1$ 2) $7+2+1+1+1$ 3) $6+3+1+1+1$ 4) $6+2+2+1+1$ 5) $5+4+1+1+1$ 6) $5+3+2+1+1$ 7) $5+2+2+2+1$ 8) $4+4+2+1+1$ 9) $4+3+3+1+1$ 10) $4+3+2+2+1$ 11) $4+2+2+2+2$ 12) $3+3+2+2+2$ 13) $3+3+3+2+1$ There are 13 ways to distribute 12 pennies", "0\\}$ (10 pennies) • $\\{ 5, 1, 0, 0\\}$ (5 pennies, 1 nickel) • $\\{ 0, 2, 0, 0\\}$ (2 nickels) • $\\{ 0, 0, 1, 0\\}$ (1 dime) • An invalid solution would be like $\\{11, 0, 0, 0\\}$ • That’s more than 10 cents. • Let’s think about how we might solve this: • Start at $\\{0, 0, 0, 0\\}$. • Increase the number of pennies until we get to the “goal”. • Stop increasing the number of pennies. Stop it. • Then back up • Take the pennies away until we can increase the number of nickels • Once we find a solution involving at least one nickel… • back up again. • Eventually, we’ll find all the solutions. • The backtracking template • Every backtracking problem will follow this basic template: void solveProblem(currentAttempt) { return; if(currentAttempt is a valid solution) output currentAttempt; for(each possible nextAttempt based on currentAttempt) solveProblem(nextAttempt) } • The parts work like this: • is a dead end means “this attempt is invalid, and any solution based on it would be invalid.” • e.g. if we’re looking for 10 cents and we get 12, there’s no way to add more coins and get 10 cents. • is a valid solution means “this attempt solves the problem.” • e.g. we found a", "how many groups of $1.20 are in$3.60. B. One Way Think: $3.60 is the same as 36 dimes.$1.20 is the same as 12 dimes. How many 12s are in 36? She bought 3 bottles. C. Another Way Think multiplication: To find 3.60 ÷ 1.20, use the relationship between multiplication and division. 1.20 × ? = 3.60 Writing this another way: 120 hundredths ×? = 360 hundredths ? = 3 She bought 3 bottles. Convince Me! Construct Arguments Is 3.6 ÷ 1.2 equal to, less than, or greater than 36 ÷ 12? Explain. Guided Practice Do You Understand? Question 1. How is dividing by a decimal like dividing by a whole number? Question 2. How can you use multiplication to find 2.8 ÷ 0.7? Do You Know How? In 3-6, use what you know about decimal division and mental math to find each quotient. Question 3. 2 ÷ 0.5 To find 2 ÷ 0.5, use the relationship between multiplication and division. 0.5 × ? = 2 ? = 4 2 ÷ 0.5 = 4 The quotient is 4. Explanation: Multiply the divisor 0.5 with 4 then the product is 2. Divide the dividend 2 with the divisor 0.5 then the quotient is 4. Question 4. 1.25 ÷ 0.25 To find 1.25 ÷ 0.25, use the relationship between multiplication and", "to running out stack frames (in Python) or overflowing the stack and triggering undefined behavior (in an unsafe compiled language like C). ## Making change Let’s close this out with a discussion of one more problem that is amenable to this sort of approach. Here we’d like to know how many different ways we can make change for a dollar using US currency (that is, using pennies, nickles, dimes, quarters, half-dollars, and dollar coins). It may not be obvious that this problem is amenable to a recursive approach. One way to see this is to realize that one subset of solutions is the set of ways to make change for 99 cents along with one more penny. Another set is the set of ways to make change for 95 cents along with another nickel, 90 cents with a dime, and so on. All of these solutions depend on the solution of a smaller version of the problem - a classic signal that recursion might be useful. What about the other solutions? Well, we know that, for example, there is only one way to take a way to make change for 99 cents and make it a way to make change for a dollar: add another cent. So that means we’ve accounted for all of the ways to make", "the second term comes from the second infinite series, and so on. $1 \\cdot x^{10} \\cdot x^{10} \\cdot 1 = x^{20}$ $1 \\cdot x^{20} \\cdot 1 \\cdot 1 = x^{20}$ $x^{10} \\cdot 1 \\cdot x^{10} \\cdot 1 = x^{20}$ $x^5 \\cdot x^{15} \\cdot 1 \\cdot 1 = x^{20}$ $x^{10} \\cdot x^{10} \\cdot 1 \\cdot 1 = x^{20}$ $x^5 \\cdot x^{15} \\cdot 1 \\cdot 1 = x^{20}$ $x^{20} \\cdot 1 \\cdot 1 \\cdot 1 = x^{20}$ $x^5 \\cdot x^5 \\cdot x^{10} \\cdot 1 = x^{20}$ The nice thing is that each of these expressions is conceptually equivalent to a way of expressing 20 cents using pennies, nickels, dimes, and quarters. In each case, the value in parentheses matches an exponent. • $1 \\cdot 1 \\cdot x^{20} \\cdot 1 = x^{20}$: 2 dimes (20 cents). • $1 \\cdot x^{10} \\cdot x^{10} \\cdot 1 = x^{20}$: 2 nickels (10 cents) and 1 dime (10 cents) • $1 \\cdot x^{20} \\cdot 1 \\cdot 1 = x^{20}$: 4 nickels (20 cents) • $x^{10} \\cdot 1 \\cdot x^{10} \\cdot 1 = x^{20}$: 10 pennies (10 cents) and 1 dime (10 cents) • $x^{15} \\cdot x^5 \\cdot 1 \\cdot 1 = x^{20}$: 15 pennies (15 cents) and 1 nickel (5 cents) • $x^{10} \\cdot x^{10} \\cdot 1 \\cdot 1 = x^{20}$: 10 pennies (10 cents)", "we have several options. What we want is the minimum of a penny plus the number of coins needed to make change for the original amount minus a penny, or a nickel plus the number of coins needed to make change for the original amount minus five cents, or a dime plus the number of coins needed to make change for the original amount minus ten cents, and so on. So the number of coins needed to make change for the original amount can be computed according to the following: $\\begin{split} numCoins = min \\begin{cases} 1 + numCoins(original amount - 1) \\\\ 1 + numCoins(original amount - 5) \\\\ 1 + numCoins(original amount - 10) \\\\ 1 + numCoins(original amount - 25) \\end{cases} \\label{eqn_change}\\end{split}$ The algorithm for doing what we have just described is shown in Listing 7. In line 3 we are checking our base case; that is, we are trying to make change in the exact amount of one of our coins. If we do not have a coin equal to the amount of change, we make recursive calls for each different coin value less than the amount of change we are trying to make. Line 6 shows how we filter the list of coins to those less than the current value of change using a", "the remainder $1$ because $3\\times 3+1=10$ ##### Solution Sometimes it's easier to divide step by step and to add the individual results after: $\\mathbf{45.20\\div 2}$ • $40\\div2=20$ • $5\\div 2=2.50$ • $0.20\\div 2=0.10$ Now we add the results: $20+2.50+0.10=22.60$ The next example shows how to handle a remainder: $\\mathbf{37.50\\div 3}$ • $30\\div 3=10$ • $7\\div 3=2$ with a remainder of $1$ • Add the remainder to the decimal and divide the sum by $3$: $1.50\\div 3=0.50$ Adding the results leads to $10+2+0.50=12.50$ $\\mathbf{22.80\\div 4}$ • $20\\div 4=5$ • $2\\div 4=0.50$ • $0.80\\div 4=0.20$ Once again we get the final result by adding: $5+0.50+0.20=5.70$ • #### Solve the following problems involving money. ##### Hints A quarter is $25$ cents, a dime $10$ cents, and a nickel $5$ cents. First decide if you have to add, subtract, multiply, or divide. ##### Solution Cameron $0.25$ for each of the $8$ hats leads to multiplication: $8\\times 0.25$. Because $8$ is double $4$, it's easier to multiply a quarter by $4$ to get $1$ dollar and to double the result to get $2$ dollars. $~$ Sandy Just add the values: $6+2\\times 0.25+5\\times 0.10+8\\times 0.05=6+0.50+0.50+0.40=6+1+0.40=7.40$ dollars. $~$ Anne $5$ tacos $3.40$ dollars each cost $5\\times 3.40$ together: • $5\\times 3=15$ • $5\\times 0.40=2$ Adding these two results leads to the total of $17$ dollars.", "find the number of ways to make a million dollars with the denominations mentioned at the beginning of this discussion. We use generating functions in a modified way as indicated in the book Concrete Mathematics, by Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. The number of pennies used to make a whole dollar amount must be a multiple of $5$, so let's think of $5$ cents as the smallest unit of currency. Let $c_k$ be the number of ways to make $5k$ cents using the given denominations. The generating function for $\\{c_k\\}$ is $\\frac{1}{(1-x)(1-x)(1-x^{2})(1-x^{5})(1-x^{10})(1-x^{20}) (1-x^{100})(1-x^{200})(1-x^{400})(1-x^{1000})(1-x^{2000})}.$ Since 1 million is $20{,}000{,}000$ of our basic units, we must find $c_{20,000,000}$. To do this, we express each factor in the denominator as $(1-x^{2000})$ with a compensating factor (a geometric series) in the numerator. The new numerator is \\begin{align*} &(1+x+\\cdots+x^{1999})^2(1 + x^2 + x^4 + \\cdots + x^{1998})(1+x^5+\\cdots+x^{1995}) (1+x^{10}+\\cdots+x^{1990})\\\\ & \\cdot (1+x^{20}+\\cdots+x^{1980})(1+x^{100}+\\cdots+x^{1900})(1+x^{200}+\\cdots+x^{1800})(1+x^{400}+\\cdots+x^{1600})(1+x^{1000}), \\end{align*} and a computer algebra system can multiply this quickly. The new denominator is $(1-x^{2000})^{11}$, and we can expand its reciprocal as a binomial series: $(1-x^{2000})^{-11} = \\sum _{j=0}^{\\infty} \\binom{j+10}{10}x^{2000j}.$ We complete our calculation by multiplying this binomial series with contributing terms in the numerator to obtain the coefficient of $x^{20,000,000}$. In the numerator, the only powers of $x$ that matter are multiples of $2000$. If $\\alpha_i$ is", "number of nickels. Write an expression for the number of pennies. #### Integers Simplify Expressions with Absolute Value In the following exercise, fill in $$<,>,$$ or $$=$$ for each of the following pairs of numbers. 13. ⓐ $$−|7| \\_\\_\\_−|−7|$$ ⓑ $$−8 \\_\\_\\_−|−8|$$ ⓒ $$|−13| \\_\\_\\_−13$$ ⓓ $$|−12| \\_\\_\\_−(−12)$$ ⓐ $$=$$ ⓑ $$=$$ ⓒ $$>$$ ⓓ $$=$$ In the following exercises, simplify. 14. $$9−|3(4−8)|$$ 15. $$12−3|1−4(4−2)|$$ $$−9$$ In the following exercises, simplify each expression. 16. $$−12+(−8)+7$$ 17. ⓐ $$15−7$$ ⓑ $$−15−(−7)$$ ⓒ $$−15−7$$ ⓓ $$15−(−7)$$ ⓐ $$8$$ ⓑ $$−8$$ ⓒ $$−22$$ ⓓ $$22$$ 18. $$−11−(−12)+5$$ 19. ⓐ $$23−(−17)$$ ⓑ $$23+17$$ ⓐ 40 ⓑ 40 20. $$−(7−11)−(3−5)$$ Multiply and Divide Integers In the following exercise, multiply or divide. 21. ⓐ $$−27÷9$$ ⓑ $$120÷(−8)$$ ⓒ $$4(−14)$$ ⓓ $$−1(−17)$$ ⓐ $$−3$$ ⓑ $$−15$$ ⓒ $$−56$$ ⓓ $$17$$ Simplify and Evaluate Expressions with Integers In the following exercises, simplify each expression. 22. ⓐ $$(−7)^3$$ ⓑ $$−7^3$$ 23. $$(7−11)(6−13)$$ 16 24. $$63÷(−9)+(−36)÷(−4)$$ 25. $$6−3|4(1−2)−(7−5)|$$ $$−12$$ 26. $$(−2)^4−24÷(13−5)$$ For the following exercises, evaluate each expression. 27. $$(y+z)^2$$ when $$y=−4$$ and $$z=7$$ 9 28. $$3x^2−2xy+4y^2$$ when $$x=−2$$ and $$y=−3$$ Translate English Phrases to Algebraic Expressions In the following exercises, translate to an algebraic expression and simplify if possible. 29. the sum of $$−4$$ and $$−9$$, increased by $$23$$ $$(−4+(−9))+23;10$$ 30. ⓐ the difference of 17 and" ]

[ "Consider using an array to store the exact change. Making the changes listed above, and compressing the retrieval of change down into one function, the code becomes this: import java.util.Scanner; public class Problem2 { /** * Takes a dollar amount, represented by a double value, * and returns the minimum number of quarters, dimes, * nickels, and pennies that would be required to * represent that dollar value. * @param dollarValue * @return {Quarters, Dimes, Nickels, Pennies} */ public static int[] getChange(double dollarValue) { int cents = (int) Math.round(dollarValue*100); // Convert the cents into quarters, dimes, nickels, and pennies, respectively. int[] change = {cents/25,(cents%=25)/10, (cents%=10)/5, cents%5}; return change; } public static void main(String[] args) { // Instantiate Variables \\\\ double total; int[] change; Scanner keyboard; // Receive Input \\\\ keyboard = new Scanner(System.in); System.out.println(\"This application calculates the exact change for a given dollar amount.\"); System.out.println(\"Please enter in the total amount of dollars and cents:\"); total = keyboard.nextDouble(); keyboard.close(); // Make Calculations \\\\ change = getChange(total); System.out.printf(\"\\$%.2f = \" + change[0] + \" Quarters, \" + change[1] + \" Dimes, \" + change[2] + \" Nickels, \" + change[3] + \" Pennies\", total); } } • @mdfst13 In this simple case, it's just one Scanner, but for more complex code I really do appreciate having the imports there so", "# 4.5: Add and Subtract Fractions with Different Denominators (Part 1) $$\\newcommand{\\vecs}[1]{\\overset { \\scriptstyle \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}[1]{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$ Skills to Develop • Find the least common denominator (LCD) • Convert fractions to equivalent fractions with the LCD • Add and subtract fractions with different denominators • Identify and use fraction operations • Use the order of operations to simplify complex fractions • Evaluate variable expressions with fractions be prepared! Before you get started, take this readiness quiz. 1. Find two fractions equivalent to $$\\frac{5}{6}$$. If you missed this problem, review Example 4.1.14. 2. Simplify: $$\\frac{1 + 5 \\cdot 3}{2^{2} + 4}$$. If you missed this problem, review Example 4.3.13. ## Find the Least Common Denominator In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators? Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals $$25$$ cents and one dime equals $$10$$ cents, so the sum is $$35$$ cents. See Figure $$\\PageIndex{1}$$. Figure $$\\PageIndex{1}$$: Together, a quarter and a dime are worth", "this optimal solution for the problem of $n$ cents must contain within it an optimal solution for the problem of $n - c$ cents. We use the usual cut-and-paste argument. Clearly, there are $k - 1$ coins in the solution to the $n - c$ cents problem used within our optimal solution to the $n$ cents problem. If we had a solution to the $n - c$ cents problem that used fewer than $k - 1$ coins, then we could use this solution to produce a solution to the $n$ cents problem that uses fewer than $k$ coins, which contradicts the optimality of our solution. a. A greedy algorithm to make change using quarters, dimes, nickels, and pennies works as follows: • Give $q = \\lfloor n / 25 \\rfloor$ quarters. That leaves $n_q = n \\mod 25$ cents to make change. • Then give $d = \\lfloor n_q / 10 \\rfloor$ dimes. That leaves $n_d = n_q \\mod 10$ cents to make change. • Then give $k = \\lfloor n_d / 5 \\rfloor$ nickels. That leaves $n_k = n_d \\mod 5$ cents to make change. • Finally, give $p = n_k$ pennies. An equivalent formulation is the following. The problem we wish to solve is making change for $n$ cents. If $n = 0$, the optimal solution", "of the numbers on each row? Is there a pattern? Make some conjectures, and prove them if you can. 6. Suppose we denote the number of ways to put $n$ cents in a machine if we use identical nickles, dimes, and quarters by $f(n)$. The trick here is to \"think recursively\" and tell yourself that you already know that when $k \\lt n$, you can put $k$ cents in the machine in $f(k)$ ways. As you start dropping change in the machine, you might start by putting in either a nickel, a dime, or a quarter. Let us suppose you put in a nickel. Now you have to ask yourself how many ways can you put in the remaining $n-5$ cents. But $n-5 \\lt n$, so this can be done in $f(n-5)$ ways. If instead, you had put in a dime first, then you have to ask yourself how many ways can you put in the remaining $n-10$ cents. Here again, $n-10 \\lt n$, so this can be done in $f(n-10)$ ways. Lastly, if you first dropped a quarter in the machine, you similarly have $f(n-25)$ ways to put in the remaining change. Taking these cases together, we arrive at our recursive relationship: $$f(n) = f(n-5) + f(n-10) + f(n-25)$$ All that remains is to identify what happens", "# 4.5: Add and Subtract Fractions with Different Denominators (Part 1) $$\\newcommand{\\vecs}[1]{\\overset { \\scriptstyle \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}[1]{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$ Skills to Develop • Find the least common denominator (LCD) • Convert fractions to equivalent fractions with the LCD • Add and subtract fractions with different denominators • Identify and use fraction operations • Use the order of operations to simplify complex fractions • Evaluate variable expressions with fractions be prepared! Before you get started, take this readiness quiz. 1. Find two fractions equivalent to $$\\frac{5}{6}$$. If you missed this problem, review Example 4.14. 2. Simplify: $$\\frac{1 + 5 \\cdot 3}{2^{2} + 4}$$. If you missed this problem, review Example 4.48. ### Find the Least Common Denominator In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators? Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals 25 cents and one dime equals 10 cents, so the sum is 35 cents. See Figure 4.7. Figure 4.7 - Together, a quarter and a dime are", "and nickels you want $$\\left(1+x+x^2+\\ldots\\right)\\left(1+x^5+x^{10}+\\ldots\\right)\\;:\\tag{1}$$ $(1)$ has one $x^n$ term for every way to write $n$ in the form $5m+k$ with $m,k\\in\\Bbb N$, which is precisely the number of ways to make the amount $n$ with nickels and pennies. Clearly, then, the generating function is just $$\\frac1{(1-x)(1-x^5)}\\;.$$ I’ll leave the last part to you. - I like your last paragraph, +1 – Rustyn Jan 31 '13 at 5:25 great explanation.. thanks so the last part i suppose will be 1/((1-x)(1-x^5)(1-x^10)) by the same principle we used pennies and nickels to change n cents. – dotdotdot Jan 31 '13 at 5:32 @dotdotdot there is a great PDF listed in the comments about this topic, just FYI in case you're interested. – Rustyn Jan 31 '13 at 5:36 @dotdotdot: You’re welcome. Yes, that’s exactly right. – Brian M. Scott Jan 31 '13 at 5:37 For $3$: Consider multiplies of $5$: The coefficients go like: $<1,1,1,1,2,2,2,2,2,3,3,3,3,3,...>$, (We obtain a way to give back change at each multiple of $5$). Each coefficient is obtained by multiplying the two previous power series: $$(1 + x + x^2 + ...)(1 + x^5 + x^{10}+...)$$ See Brian's answer for a simplification of this function. -", "to running out stack frames (in Python) or overflowing the stack and triggering undefined behavior (in an unsafe compiled language like C). ## Making change Let’s close this out with a discussion of one more problem that is amenable to this sort of approach. Here we’d like to know how many different ways we can make change for a dollar using US currency (that is, using pennies, nickles, dimes, quarters, half-dollars, and dollar coins). It may not be obvious that this problem is amenable to a recursive approach. One way to see this is to realize that one subset of solutions is the set of ways to make change for 99 cents along with one more penny. Another set is the set of ways to make change for 95 cents along with another nickel, 90 cents with a dime, and so on. All of these solutions depend on the solution of a smaller version of the problem - a classic signal that recursion might be useful. What about the other solutions? Well, we know that, for example, there is only one way to take a way to make change for 99 cents and make it a way to make change for a dollar: add another cent. So that means we’ve accounted for all of the ways to make", "coins is not the least number of coins required to change 23 cents; we could do it in 5 (using two 10-cent coins and three 1-cent coins). This proves that our original way of making change for 48 cents was not optimal, as we can just add the quarter back in (so we'll have one 25-cent coin, two 10-cent coins, and three 1-cent coins, or 6 coins totalling 48 cents) and get a better way. Whenever we have a suboptimal solution to a subinstance, we will also have a suboptimal solution to the original instance; so an optimal solution to the original instance is only allowed to contain optimal solutions to subinstances. To solve the change problem using DP, we start by considering the simplest possible case: when the amount to be changed is zero. Then, the optimal (and only) solution is to use no coins at all. But whenever we change a positive amount, we must use at least one coin. If we take any coin (with value $x$) away from the optimal change for amount $n$, then we are left with optimal change for $n-x$. This implies that we can make optimal change for $n$ using at least one of the following ways: make optimal change for $n-d_1$ and add one coin of value $d_1$; or", "give the optimal solution in many cases (U.S. coins are one example). However, with an unusual set of coins, this is not always true. For instance, the best way to make change for thirty cents if you have quarters, dimes, and pennies, would be to use three dimes, as opposed to a quarter and five pennies. This example shows us that the so-called “greedy” solution to this problem doesn’t work. So let’s try a dynamic algorithm. As with Fibonacci numbers, we need to compute some sub-problems and build off of them to get an optimal solution for the whole problem. In particular, denote our coin values $v_1, \\dots, v_k$ and the amount of change we need to make $n$. Let us further force $v_1 = 1$, so that there is guaranteed to be a solution. Our sub-problems are to find the minimal number of coins needed to make change if we only use the first $i$ coin values, and we make change for $j$ cents. We will eventually store these values in a two-dimensional array, and call it $\\textup{minCoins}[i,j]$. In other words, this is a two-dimensional dynamic program, because we have sub-problems that depend on two parameters. The base cases are easy: how many coins do I need to make change for zero cents? Zero! And how many", "Solution Method 1 - Recursion This is a recursive problem, so let’s figure out how to do makeChange(n) using prior solutions (i.e., sub-problems). Let’s say n = 100, so we want to compute the number of ways of making change of 100 cents. What’s the relationship to its sub-problems? We know that makeChange(100): = makeChange(100 using 0 quarters) + makeChange(100 using 1 quarter) + makeChange(100 using 2 quarter) + makeChange(100 using 3 quarter) + makeChange(100 using 4 quarter) Can we reduce this further? Yes! = makeChange(100 using 0 quarters) + makeChange(75 using 0 quarter) + makeChange(50 using 0 quarters) + makeChange(25 using 0 quarters) + 1 Now what? We’ve used up all our quarters, so now we can start applying our next biggest denomination: dimes. This leads to a recursive algorithm that looks like this: public static int makeChange(int n, int denom) { int next_denom = 0; switch (denom) { case 25: next_denom = 10; break; case 10: next_denom = 5; break; case 5: next_denom = 1; break; case 1: return 1; } int ways = 0; for (int i = 0; i * denom <= n; i++) { ways += makeChange(n - i * denom, next_denom); } return ways; } References http://tianrunhe.wordpress.com/2012/03/27/number-of-ways-of-representing-n-cents-using-quarters-dimes-nickels-and-pennies/ ### Implement the “paint fill” function Problem Implement the “paint fill” function that one might", "In other words, there are four $33$s in $12.12.$ So, $12÷3=4.12÷3=4.$ What about dividing fractions? Suppose we want to find the quotient: $12÷16.12÷16.$ We need to figure out how many $1616$s there are in $12.12.$ We can use fraction tiles to model this division. We start by lining up the half and sixth fraction tiles as shown in Figure 4.5. Notice, there are three $1616$ tiles in $12,12,$ so $12÷16=3.12÷16=3.$ Figure 4.5 ### Manipulative Mathematics Doing the Manipulative Mathematics activity \"Model Fraction Division\" will help you develop a better understanding of dividing fractions. ### Example 4.31 Model: $14÷18.14÷18.$ ### Try It 4.61 Model: $13÷16.13÷16.$ ### Try It 4.62 Model: $12÷14.12÷14.$ ### Example 4.32 Model: $2÷14.2÷14.$ ### Try It 4.63 Model: $2÷132÷13$ ### Try It 4.64 Model: $3÷123÷12$ Let’s use money to model $2÷142÷14$ in another way. We often read $1414$ as a ‘quarter’, and we know that a quarter is one-fourth of a dollar as shown in Figure 4.6. So we can think of $2÷142÷14$ as, “How many quarters are there in two dollars?” One dollar is $44$ quarters, so $22$ dollars would be $88$ quarters. So again, $2÷14=8.2÷14=8.$ Figure 4.6 The U.S. coin called a quarter is worth one-fourth of a dollar. Using fraction tiles, we showed that $12÷16=3.12÷16=3.$ Notice that $12·61=312·61=3$ also. How are $1616$ and", "Prealgebra 2e # 4.5Add and Subtract Fractions with Different Denominators Prealgebra 2e4.5 Add and Subtract Fractions with Different Denominators ### Learning Objectives By the end of this section, you will be able to: • Find the least common denominator (LCD) • Convert fractions to equivalent fractions with the LCD • Add and subtract fractions with different denominators • Identify and use fraction operations • Use the order of operations to simplify complex fractions • Evaluate variable expressions with fractions ### Be Prepared 4.12 Before you get started, take this readiness quiz. Find two fractions equivalent to $56.56.$ If you missed this problem, review Example 4.14. ### Be Prepared 4.13 Simplify: $1+5·322+4.1+5·322+4.$ If you missed this problem, review Example 4.48. ### Find the Least Common Denominator In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators? Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals $2525$ cents and one dime equals $1010$ cents, so the sum is $3535$ cents. See Figure 4.7. Figure", "# Making change Consider the classic problem: How many ways can one make change for one dollar using half-dollars, quarters, dimes, nickels, and pennies? Or more generally, how many ways can one make change for a given amount using arbitrary (positive integer) denominations? This post chronicles a series of incremental improvements to solutions to this problem. In the first section, we attack the problem with dynamic programming in Python, and we’re able to count the ways to change quite large amounts of money (eventually up to about $100M for the given set of five coins). The second section explores a nice method for deriving closed form solutions, due to Lee Newberg, and the third is a synthesis of the previous two, automating the closed form derivation in the general setting. 1. From 1 to 100,000,000 It’s not hard to produce a recursive function definition which computes the answer to the coin problem. To count all the ways of making change, we just need to condition on whether or not at least one coin of the first denomination will be used (and then add these counts together). If one will be used, then we must change the original amount minus the first denomination. If not, then we must change the whole original amount, but with one fewer denomination. In", "If $a,b,\\text{ and }c$ are numbers where $c\\ne 0$, then ${\\Large\\frac{a}{c}}-{\\Large\\frac{b}{c}}={\\Large\\frac{a-b}{c}}$ To subtract fractions with a common denominators, subtract the numerators and place the difference over the common denominator. Example Find the difference: ${\\Large\\frac{23}{24}}-{\\Large\\frac{14}{24}}$ Try It Now lets do an example that involves both addition and subtraction. Example Simplify: ${\\Large\\frac{3}{8}}+\\left({\\Large\\frac{7}{8}}\\right)-{\\Large\\frac{5}{8}}$ Addition and Subtraction of Fractions with Different Denominators We just reviewed how to add and subtract fractions with common denominators. But how can we add and subtract fractions with unlike denominators? Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals $25$ cents and one dime equals $10$ cents, so the sum is $35$ cents. See the image below. Together, a quarter and a dime are worth $35$ cents, or ${\\Large\\frac{35}{100}}$ of a dollar. Similarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is $100$. Since there are $100$ cents in one dollar, $25$ cents is $\\Large\\frac{25}{100}$ and $10$ cents is $\\Large\\frac{10}{100}$. So we add $\\Large\\frac{25}{100}+\\Large\\frac{10}{100}$ to get $\\Large\\frac{35}{100}$,", "algorithm. In contrast, consider the following problem: How many different ways can we make change of \\$ 1.00, given half-dollars, quarters, dimes, nickels, and pennies? More generally, can we write a function to compute the number of ways to change any given amount of money? This problem has a simple solution as a recursive function. Suppose we think of the types of coins available as arranged in some order. Then the following relation holds: The number of ways to change amount $$a$$ using $$n$$ kinds of coins equals • the number of ways to change amount $$a$$ using all but the first kind of coin, plus • the number of ways to change amount $$a$$ - $$d$$ using all n kinds of coins, where $$d$$ is the denomination of the first kind of coin. To see why this is true, observe that the ways to make change can be divided into two groups: those that do not use any of the first kind of coin, and those that do. Therefore, the total number of ways to make change for some amount is equal to the number of ways to make change for the amount without using any of the first kind of coin, plus the number of ways to make change assuming that we do use the first", "an important algorithm - the subset sum problem (once again, Google can be your guide to enlightenment). Let's start with an example, to make sure that the problem is understood. In how many ways can we make change from 10 cents? One is ten cents. Two is one nickel and five cents. Three is two nickels. Four is a dime. So, there are four ways. In fact, this problem has a simple recursive solution. Suppose we think of the types of coins available as arranged in some order. Then the following relation holds: The number of ways to change amount a using n kinds of coins equals • the number of ways to change amount a using all but the first kind of coin, plus • The number of ways to change amount a - d using all n kinds of coins, where d is the denomination of the first coin. Rationale: note that given a task to make change, we can divide it to two complementary groups: ways that don't use the first kind of coin, and ways that do. The total number of ways is the sum of those two groups - that is, the number of ways to make change without using the first kind of coin, plus the number of ways to make change", "$y=3/4$. Find the center of mass in the $y$ direction (that of the $x$ direction is $0$ by symmetry). ###### Question Find the center of mass in the $y$ direction of the area bounded by the cosine curve and the $x$ axis between $-\\pi/2$ and $\\pi/2$. ###### Question A penny, nickel, dime and quarter are stacked so that their right most edges align and are centered so that the center of mass in the $y$ direction is $0$. Find the center of mass in the $x$ direction. You will need some specifications, such as these from the US Mint diameter(in) weight(gms) penny 0.750 2.500 nickel 0.835 5.000 dime 0.705 2.268 quarter 0.955 5.670 (Hint: Though this could be done with integration, it is easier to treat each coin as a single point (its centroid) with the given mass and then apply the formula for sums.)", "same if they’re just re-arrangements of each other. For instance, 4 can be written as 1+1+1+1, or as 2+1+1 (which counts as the same as 1+2+1 and 1+1+2), or as 3+1 (which counts as the same as 1+3), or as 2+2, or as just 4. (“4” doesn’t actually look like a sum, but we agree to think of it as a sum with only one term, because it turns out that allowing partitions consisting of only one term is more natural in applications such as counting ways to make payment using various denominations of coins; and besides, it leads to prettier results.) So the partitions of 4 are 1+1+1+1, 2+1+1, 3+1, 2+2, and 4. Since there are 5 such sums, we say that the number of partitions of the number 4 is 5. That is, if you’ve got an inexhaustible supply of 1 cent coins, 2 cent coins, 3 cent coins, etc., and you want to pay for something that costs 4 cents, there are 5 combinations of coins you can slap onto the counter to make your purchase. We write the number of partitions of n as p(n); so for instance, we’ve just shown that p(4) = 5. It’s customary (and sensible) to set p(0) = 1. After all, if you want to pay for something that’s", "# CS50 Pset1 Greedy, change algorithm The purpose of the code is to calculate the least amount of coins needed to produce correct amount of change given a dollar or cent user input. I'm just starting out and appreciate any pointers in how this code could/should be optimized. The code works but I feel there is some redundancy with all those \"while\" loops. Using modulo math was another suggestion in order to solve the problem but I felt more comfortable using this solution for now. #include <stdio.h> #include <cs50.h> #include <math.h> int main (void) { // Variables// float change; int num = 0; int quarter = 25; int dime = 10; int nickel = 5; int penny = 1; // Input of change// do { printf(\"How much change is owed? \"); change = get_float(); } while (change <= 0); printf(\"%.2f\\n\",change); // conversion from dollars to cents (need to look over use of rounding) int cents = (change * 100); //Coin count// while ( cents >= quarter) { cents -= quarter; num++; } while (cents >= dime) { cents -= dime; num++; } while (cents >= nickel) { cents -= nickel; num++; } while (cents >= penny) { cents -=penny; num++; } printf(\"%d\\n\", num); } • You should try proper indentation. That would make the code easier to read...", "we have several options. What we want is the minimum of a penny plus the number of coins needed to make change for the original amount minus a penny, or a nickel plus the number of coins needed to make change for the original amount minus five cents, or a dime plus the number of coins needed to make change for the original amount minus ten cents, and so on. So the number of coins needed to make change for the original amount can be computed according to the following: $\\begin{split} numCoins = min \\begin{cases} 1 + numCoins(original amount - 1) \\\\ 1 + numCoins(original amount - 5) \\\\ 1 + numCoins(original amount - 10) \\\\ 1 + numCoins(original amount - 25) \\end{cases} \\label{eqn_change}\\end{split}$ The algorithm for doing what we have just described is shown in Listing 7. In line 3 we are checking our base case; that is, we are trying to make change in the exact amount of one of our coins. If we do not have a coin equal to the amount of change, we make recursive calls for each different coin value less than the amount of change we are trying to make. Line 6 shows how we filter the list of coins to those less than the current value of change using a" ]

[ "+0 0 70 2 Bill buys a stock that decreases by 20% on the first day, and then on the second day the stock increases by 30% of its value at the end of the first day. What was the overall percent increase in Bill's stock over the two days? Aug 8, 2019 #1 +5798 +1 $$v_1 = (1-0.2)v_0 = 0.8v_0\\\\ v_2 = (1+0.3)v_1 = 1.3(0.8)v_0 =1.04v_0\\\\ \\dfrac{v_2}{v_0} = 1.04 = (1+0.04)\\\\ \\text{Overall there was a 4% increase in the stock value}$$ . Aug 8, 2019 #2 +103148 +1 Let P be the original price of the stock So we have, after two days [ P (.80) ] (1.30) = P * 1.04 ⇒ i.e., the stock has increased in value by 4% Aug 8, 2019", "the previous period, the discount was 25% less. Thus, Bill either bought his suit on July 8 or July 15. July 1 is not possible because there was no discount the day before. And July 22 is not possible because the discount was 15% less the day before. Not sufficient. Note that if the equation reduced to $$x-y = 0.15$$, Statement 1 alone would have been sufficient. Statement 2 $$202.50 = 450(1-x)$$ where $$x$$ is the discount Bill received. $$x = 0.55$$ So Bill could have bought his suit on any day July 15 through July 21. Not sufficient. Combining both statements In Statement 1, we narrowed down the answer to July 8 or July 15. In Statement 2, we narrowed down the answer to July 15 through July 21. The only date common do both is July 15. Sufficient. Re: In the table above is shown the percent reduction in the &nbs [#permalink] 14 Jan 2018, 07:15 Display posts from previous: Sort by", "SEARCH HOME Math Central Quandaries & Queries Subject: integers Name: bill Who are you: Other if the value of xyz company stock drops from $25 per share to$21 per share, what is the percent of the decrease? Hi Bill, The stock price dropped from $25 to$21 so it dropped $4. The percentage change is the change (-$4), as a fraction of the starting price (\\$25), and then expressed as a percentage. The drop as a fraction of the starting price is -4/25. Expressed as a percent this is -4/25 100 = -16%. Hence there is a -16% change which is a 16% decrease. Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "days. Present value = 25000/1+ 2.92%(364/365) = 24292.5995 is the purchased price of$25000 T-bill b/ What is the yield rate on August 15 if the market price is $24377.64? April 01 to Aug 15 = 136 days Yield rate is unknown = r 622.36/24377.64(228/365) =4.09% c/ Calculate the market value of each$25000 T-bill on Oct 1 if the rate of return on that date is 4.545%. Since maturity value of T-bill is Mar 31 2017, Oct 01 2016 to Mar 31 2017 = 181days Present value = 25000/1+4.545%(181/365) = 24 448.96 was the price T-bill sold. d/ What is the rate of return realized if a \\$25000 T-bill purchase on April 1 is sold on Nov 20 at a market rate of 4.625%? From Nov 20 to maturity date (Mar 21 2017)= 131 days. Present value = 25000/1+4.625%(131/365) = 24 591.79308 was amount purchased T-bill on Nov 20. Now find the yield rate return of 25000 T-bill purchased on April 01 is sold on Nov 20 at rate of 4.625%. April 01 to Nov 20 = 233 days. 25 591.79308 = 24 292.6 [1+r(233/365)] = 0.019293687 or 1.9393% (5) Annual rate = x 25 000 = 24 756.25[1+r(182/365)] = 1.9746% is the annual rate of return. (6) a/ What was the original yield of the T-bill? 100 000=99", "then gained 15 yards. What was his net gain or loss in the first quarter? 5. The stock market rose 1-1/4 points on Monday, fell 3/4 of a point on Tuesday, rose 1/2 point on Wednesday, rose 7/8 of a point of Thursday, and fell 1-1/2 points on Friday. What was the net change for the week?", "original price of that item on June 30 of that year. The table is for the 28 days immediately after June 30 of that year. If on July n of that year, Bill bought a suit that originally cost$450, what is the value of n? (1) Bill paid $112.50 less than he would have if he had bought it the day before. (2) Bill paid$202.50 for the suit. (1) insuff..because we don't know the discounted price/any price for which he would have paid $112.5. (2)202.5 is what percent less of 450... 450(1-x/100)=202.5 x=55% date \"n\" could be anything between july15-july21...not suff.... Combining both Bill paid$112.50 less than he would have if he had bought it the day before means bill paid (202.5+112.5=\\$315) if he bought with any previous discount a day before. 315 is 30% less than 450. So he must bought on july 15 Ans C Manager Joined: 30 Mar 2017 Posts: 136 GMAT 1: 200 Q1 V1 Re: In the table above is shown the percent reduction in the [#permalink] ### Show Tags 14 Jan 2018, 07:15 Algebraic approach: Statement 1 Let $$x$$ be the discount Bill received and let $$y$$ be the previous period's discount. So we get this equation: $$450(1-x) = 450(1-y) - 112.50$$ Which reduces to: $$x-y = 0.25$$ This means that in", "# Difference between revisions of \"A stock loses $10\\%$ of its value on Monday. On Tuesday it loses $20\\%$ of the value it had at the end of the day on Monday. What is the overall percent loss in value from the beginning of Monday to the end of Tuesday? Enter the answer as a percent.\" Before: $100$% After first loss: $90$% After second loss: $90*(\\frac{100-20}{100})=72$% In total, the stock lost $100-72=\\boxed{28}$%", "did the t bill sell if the yield was quoted by the market ?? there for .0535=100-p/p x 365/91 the answer in the book is 98.68372139", "above is shown the percent reduction in the [#permalink] ### Show Tags 15 Jun 2016, 09:09 bschool83 wrote: Date of Purchase Percent Reduction July 1 through July 7 5% July 8 through July 14 30% July 15 through July 21 55% July 22 through July 28 70% In the table above is shown the percent reduction in the price of every item in a store off the original price of that item on June 30 of that year. The table is for the 28 days immediately after June 30 of that year. If on July n of that year, Bill bought a suit that originally cost$450, what is the value of n? (1) Bill paid $112.50 less than he would have if he had bought it the day before. (2) Bill paid$202.50 for the suit. (1) insuff..because we don't know the discounted price/any price for which he would have paid $112.5. (2)202.5 is what percent less of 450... 450(1-x/100)=202.5 x=55% date \"n\" could be anything between july15-july21...not suff.... Combining both Bill paid$112.50 less than he would have if he had bought it the day before means bill paid (202.5+112.5=\\$315) if he bought with any previous discount a day before. 315 is 30% less than 450. So he must bought on july 15 Ans C Manager Joined: 30 Mar", "sell two weeks ago, and buy stocks that have dropped ~10%, or wait to see if they fall a bit more.", "600[1+r(91/364)] = 1.610039% b/At what price did the investor sell? 100 000/1+1.4%(51/365) = 99 804.76547 was the price sold. c/ What annual rate of return did the investor realize while holding his T-bill? 99 804.76547 = 99 600[1+r(51/364)] = 1.8759%", "# (i) You purchased a $1,000 par with T-bill 173 days to maturity for$956.06. You then sold this... 1 answer below » (i) You purchased a $1,000 par with T-bill 173 days to maturity for$956.06. You then sold this T-bill when it had 70 days to maturity for $981.85. What is your holding period return? (ii) An investment bank sells securities under a repurchase agreement for$800.29 million and buys them back in 5 days for $800.979 million. What is the repo yield? Attachments: ## 1 Approved Answer Jitender K 3 Ratings, (9 Votes) (i) Purchase price of T-bill =$956.06 Sale price of T-bill =...", "# If total liabilities decreased by $25,300 and stockholders' equity increased by$4,850 during a p... If total liabilities decreased by $25,300 and stockholders' equity increased by$4,850 during a period of time, then total assets must change by what amount and direction during the same time period?", "is not ... 5 A bill with an expiration date does not become worthless suddenly. Clearly its value would decrease over time until, just before the expiration date, it was next to worthless. Think about it: how much would you pay for a \\$1 bill that expires in 5 minutes? Certainly a lot less than you would for one that expires in 6 months. This constantly changing value ... Only top voted, non community-wiki answers of a minimum length are eligible", "The price of a certain stock increased by 0.25 of 1 percent on a certain day. By what fraction did the price of the stock increase that day? ~$1 \\over 2500$~ ~$1 \\over 400$~ ~$1 \\over 40$~ ~$1 \\over 25$~ ~$1 \\over 4$~", "Present value = 25000/1+4.545%(181/365) = 24 448.96 was the price T-bill sold. d/ What is the rate of return realized if a$25000 T-bill purchase on April 1 is sold on Nov 20 at a market rate of 4.625%? From Nov 20 to maturity date (Mar 21 2017)= 131 days. Present value = 25000/1+4.625%(131/365) = 24 591.79308 was amount purchased T-bill on Nov 20. Now find the yield rate return of 25000 T-bill purchased on April 01 is sold on Nov 20 at rate of 4.625%. April 01 to Nov 20 = 233 days. 25 591.79308 = 24 292.6 [1+r(233/365)] = 0.019293687 or 1.9393% (5) Annual rate = x 25 000 = 24 756.25[1+r(182/365)] = 1.9746% is the annual rate of return. (6) a/ What was the original yield of the T-bill? 100 000=99 600[1+r(91/364)] = 1.610039% b/At what price did the investor sell? 100 000/1+1.4%(51/365) = 99 804.76547 was the price sold. c/ What annual rate of return did the investor realize while holding his T-bill? 99 804.76547 = 99 600[1+r(51/364)] = 1.8759% The amount of the installments = x 1x/ 1+6.9%(60/365) = 0.988784743x 1x/ 1+6.9%(120/365) = 0.97781826x 1x/ 1+6.9%(180/365) =0.967092364x 0.988784743x + 0.97781826x + 0.967092364x = 5000 X= 1704.34 is the amount of each installments . 4/ On April 1, \\$25000, 364-day T-bills were auctioned off to", "1. You have invested in stock for 1 year. If you start with $125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with$125 of stock and then it grows to $215, what was the growth rate? 2. Originally Posted by jilllamtn 1. You have invested in stock for 1 year. If you start with$125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with $125 of stock and then it grows to$215, what was the growth rate? For the first question, multiply $125 by (1+8.25%), which is:$125 (1.0825) = \\$135.31. That's your answer. I assume that you want your growth rate in terms of percent increases. So you find the difference between the old and new price and divide that difference by the old price. So: $\\frac{215-125}{125}=0.72$, which is 72%.", "and that it is hardly worth it to figure out what the actual number of days left is because you get almost the same answer anyway. That said, I'm using 88 days--- roughly the average--- rather than the 91 of rhi. – Mike O'Connor Jul 31 '19 at 19:32 • Hi Mike, I just wanted to make sure I follow up the logic of the transformation, also for other cases. Thank you for explaining! – abu Jul 31 '19 at 20:48 If you want to do it super precisely, the convention for building fixed-income total return index is as follows: 1. You assume at the end of the month, you buy the instrument (in this case a 3-month T-bill). 2. You hold this exact T-bill over the course of the next month and mark it to market daily and calculate the daily returns ($P_t / P_{t-1} - 1$, which is standard performance calculation). 3. At the end of the month, you sell the T-bill and buy the next 3-month bill. This of course requires that you have accurate bill pricing data. You can also look at the Bloomberg Barclays 3-Month Treasury Bellwether Index, or the BofA Merrill Lynch 3-Month Bill Index. Both are calculated this way. Alternatively, a lot of people just use the previous month-end bill rate,", "the day after 4 and like the famous Simpson’s episode when Homer is making sure that Bart is a name safe from mockery, goes through the alphabet and says “Aart, Bart, Cart, Dart, Eart, okay Bart is fine,” you will have to go to day 10 before you get to a thousand fold increase (1024 to be precise), which still doesn’t seem too bad. In fact 30 days later is 2 to the power of 30 or $2^{30}$, which is only 10 million or so. But on day 31 you have 20 million and day 32 you have 40 million. It’s growing really fast. Now, I think exponential growth really hits home if you use 10 as a base because now you simply add a zero for every day: 1, 10, 100, 1000, 10000, 100000. Imagine if your stock was increasing by a factor of 10 every day. Starting from a dollar you would have a million dollars by day 6 and a billion by day 9, be richer than Bill Gates by day 11, and have more money than the entire world by the third week. Now, the only difference between exponential growth with base 2 versus base 10 is the rate of growth. This is where logarithms are very useful because they are the inverse of", "# Percents: Percent Change The price of stock decreased by 8% last year and rose by 10% this year. What is the net percentage change in the price of the stock? Incorrect. [[snippet]] Correct. [[snippet]] Plug in the initial price of the stock as $100. Based on this, last year's price after an initial 8% decrease can be calculated as follows: >$$\\text{Last year's price} = 100 - 8\\% \\text{ of } 100 = 100 - 8 = 92$$. Now find this year's price after the 10% increase. Remember that the original price is now the new price, which means that the 10% is taken from$92. >$$\\text{This year's price} = 92 + 10\\% \\text{ of } 92 = 92 + 9.2 = 101.2$$ >$$\\displaystyle \\text{Net percent change} = \\frac{101.2 - 100}{100} \\times 100 = 1.2%$$ Hence, this is the correct answer. Incorrect. [[snippet]] Did you subtract the 8% decrease from the 10% increase ($$10\\% - 8\\% = 2\\%$$) and apply a 2% increase to the initial price? Watch out! The 8% decrease is applied to the original price while the 10% increase is applied to last year's price. Incorrect. [[snippet]] Incorrect. [[snippet]] 0.2% 0.8% 1.2% 1.6% 2%" ]

[ "# Suppose that Christina purchased a stock for x dollars. During the first year, the stock's price rose 15%? (a) Write an algebraic expression for the price of the stock after the first year in terms of x. ? ## can you please show the work thank you. :) here is all the details and im giving you the whole question. Apr 14, 2017 a)${S}_{1} = 1.15 x$ b)${S}_{2} = 1.10 \\left(1.15 x\\right)$ c)${S}_{2} = 1.256 x$ d)S_2=$25.30 #### Explanation: The value of the stock $S$is $x$, so: S=$x After $1$ year the stock gains 15% in value: Then: ${S}_{1} = 1.15 x$ because it is now 115% of the original value. After $2$ years the stock gains 10% in value: Then: ${S}_{2} = 1.10 \\left(1.15 x\\right)$ because it is now 110% of the $S 1$ value. So: ${S}_{2} = 1.10 \\left(1.15 x\\right) = 1.265 x$ After $2$ years the stock is now valued at 126.5% of the original value. If the original value is $20: After $2$years the stock is valued at: S_2=1.256x=1.265($20)=\\$25.30", "stock has moved from $10 to$15 and then dropped back to $12 — it has made two moves: one from$10 to $15 and another from$15 to $12. The first move is considered the up-move, and the second is regarded as the down-move. If you take an average of these two moves (i.e., ($15-\\$12)/2), you get an ATR of 1.5. The ATR indicator shows how far price moves on average in one day. A large ATR value means that price typically moves far in one direction before reversing course, while a small ATR value implies that price changes frequently, but not by much.", "+0 0 70 2 Bill buys a stock that decreases by 20% on the first day, and then on the second day the stock increases by 30% of its value at the end of the first day. What was the overall percent increase in Bill's stock over the two days? Aug 8, 2019 #1 +5798 +1 $$v_1 = (1-0.2)v_0 = 0.8v_0\\\\ v_2 = (1+0.3)v_1 = 1.3(0.8)v_0 =1.04v_0\\\\ \\dfrac{v_2}{v_0} = 1.04 = (1+0.04)\\\\ \\text{Overall there was a 4% increase in the stock value}$$ . Aug 8, 2019 #2 +103148 +1 Let P be the original price of the stock So we have, after two days [ P (.80) ] (1.30) = P * 1.04 ⇒ i.e., the stock has increased in value by 4% Aug 8, 2019", "# Difference between revisions of \"A stock loses $10\\%$ of its value on Monday. On Tuesday it loses $20\\%$ of the value it had at the end of the day on Monday. What is the overall percent loss in value from the beginning of Monday to the end of Tuesday? Enter the answer as a percent.\" Before: $100$% After first loss: $90$% After second loss: $90*(\\frac{100-20}{100})=72$% In total, the stock lost $100-72=\\boxed{28}$%", "1. You have invested in stock for 1 year. If you start with $125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with$125 of stock and then it grows to $215, what was the growth rate? 2. Originally Posted by jilllamtn 1. You have invested in stock for 1 year. If you start with$125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with $125 of stock and then it grows to$215, what was the growth rate? For the first question, multiply $125 by (1+8.25%), which is:$125 (1.0825) = \\$135.31. That's your answer. I assume that you want your growth rate in terms of percent increases. So you find the difference between the old and new price and divide that difference by the old price. So: $\\frac{215-125}{125}=0.72$, which is 72%.", "same mistake I did in the answer, assuming you buy all three stocks on day 1 and hold them to day 3. We have ignored any gain you could get from the cash overnight. – Ross Millikan Aug 25 '12 at 2:17", "1. You have invested in stock for 1 year. If you start with $125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with$125 of stock and then it grows to $215, what was the growth rate? 2. Originally Posted by jilllamtn 1. You have invested in stock for 1 year. If you start with$125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with $125 of stock and then it grows to$215, what was the growth rate? For the first question, multiply $125 by (1+8.25%), which is:$125 (1.0825) = $135.31. That's your answer. I assume that you want your growth rate in terms of percent increases. So you find the difference between the old and new price and divide that difference by the old price. So:$\\displaystyle \\frac{215-125}{125}=0.72\\$, which is 72%.", "then gained 15 yards. What was his net gain or loss in the first quarter? 5. The stock market rose 1-1/4 points on Monday, fell 3/4 of a point on Tuesday, rose 1/2 point on Wednesday, rose 7/8 of a point of Thursday, and fell 1-1/2 points on Friday. What was the net change for the week?", "- P(4)$ $P(X > 4) = 1 - exp(-4)(1 + 4 + 8 + \\frac{64}{6} + \\frac{256}{24})$ $P(X > 4) = .3717$ We obviously have to go much higher, so let's try a beginning stock of 6. $P(X > 6) = .3717 - P(5) - P(6)$ $P(X > 6) = .3717 - .26 = .1117$ Higher yet again, let's try stock of 7. $P(X > 8) = .1117 - P(7)$ $P(X > 8) = .1117 - .0183(3.25)$ $P(X > 8) = .052225$ P(8) ends up being equal to .0297, which is enough to put us over that limit, so our stock cannot be GREATER than 8. Our initial weekly stock must be eight. • Dec 13th 2007, 02:11 PM eugene1687 thank you!! thanks soo much.. (Happy)", "| 9.8 | 0.1 | 100 | 1100 | | 3 | 0 | 9.5 | -0.3 | -300 | 800 | Let’s understand the margin calculation on each day. On Day 1 itself, the margin balance has gone below the maintenance margin of$750. It has reach $700 and the trader is required to deposit$300 to replenish the margin balance back to the original $1000. On day two, the price moved up creating a gain of$100. The margin balance at the end of day two is $1100. On Day 3 again, the price moves down by 30 cents. However, at the end of the day, the margin balance is$800, that is, above the margin balance. So, no margin call is required.", "- Nasdaq: 913 days on which the Nasdaq increased more than 1% and at most 2% .", "if an investor wants to buy $10,000 worth of XYZ stock and the initial margin is 50%, the broker-dealer will lend$5,000, and the investor will need to put up $5,000 of their own money. The maintenance margin would be$2,500. To see how buying on margin works, let’s assume that on Day 1, the value of the stocks decreases to $7,800. As such, the balance in the margin account falls to USD 2,800 (i.e.,$5,000 – $2,200). The margin as a percentage of the (current) value of the shares purchased is around 36% (= 2,800/7,800). Since this is more than 25%, there is no margin call. On Day 2, the value of the stocks further decreases to$6,000. As such, the balance in the margin account falls to USD 1,000 (i.e., $5,000 –$4,000). The margin as a percentage of the value of the shares purchased is around 17% (= 1,000/6,000). Since this is less than 25%, there is a margin call to bring the margin balance up to 25% of the value of the shares. To do this, the trader must add $500 (= 25% X$6,000 — $1,000) to the margin balance. If it is not provided, contract termination agreements in the contract may give the broker the right to sell the shares. There are several important things to remember about", "need to know how much it increased $$4.4$B $-$$ $4$B = $$0.4$B it increased by$0.4 B B(original price)=$4 B P(amount increased)=$0.4 B R(percent increased)=? $R=\\frac{0.4}{4}$ in billions R=1% you can check it by $B=\\frac{P}{R}$ One week, a certain stock market index went down from 1,863 to 1,677. What was the percent of decrease? Find the percent of chan $20,200 to$17,350. find first how much it decreased then use the formula", "equity becomes $\\$60$, while the value of debt remains at$\\$50$. Equity holders enjoy a $20\\%$ increase ($=\\frac{10}{50}$) in share value, against $10\\%$ ($=\\frac{10}{100}$) in the unlevered case. In moving from $0\\%$ gearing to $50\\%$ gearing, the volatility of equity value has doubled.", "The price of a certain stock increased by 0.25 of 1 percent on a certain day. By what fraction did the price of the stock increase that day? ~$1 \\over 2500$~ ~$1 \\over 400$~ ~$1 \\over 40$~ ~$1 \\over 25$~ ~$1 \\over 4$~", "# The price of an item is increased by 20% and then decreased by 20%. The final price as compared to original price Option 1) 5 % moreOption 2) 20% moreOption 3) 20% lessOption 4) 4% lessOption 5) 4% more Let the original price be $X$ $X\\times \\left ( \\frac{120}{100} \\right )\\times \\left ( \\frac{80}{100} \\right )$ $\\Rightarrow \\left ( \\frac{96}{100} \\right )X$ $\\Rightarrow$ 4% less than original price", "# Percents: Percent Change The price of stock decreased by 8% last year and rose by 10% this year. What is the net percentage change in the price of the stock? Incorrect. [[snippet]] Correct. [[snippet]] Plug in the initial price of the stock as $100. Based on this, last year's price after an initial 8% decrease can be calculated as follows: >$$\\text{Last year's price} = 100 - 8\\% \\text{ of } 100 = 100 - 8 = 92$$. Now find this year's price after the 10% increase. Remember that the original price is now the new price, which means that the 10% is taken from$92. >$$\\text{This year's price} = 92 + 10\\% \\text{ of } 92 = 92 + 9.2 = 101.2$$ >$$\\displaystyle \\text{Net percent change} = \\frac{101.2 - 100}{100} \\times 100 = 1.2%$$ Hence, this is the correct answer. Incorrect. [[snippet]] Did you subtract the 8% decrease from the 10% increase ($$10\\% - 8\\% = 2\\%$$) and apply a 2% increase to the initial price? Watch out! The 8% decrease is applied to the original price while the 10% increase is applied to last year's price. Incorrect. [[snippet]] Incorrect. [[snippet]] 0.2% 0.8% 1.2% 1.6% 2%", "after the increase and decrease, the final price is 96\\% of the original price, which means it is a 4\\% decrease. BIG IDEA #1: when you go up by a percent, then down by the same percent, you do not wind up where you started: that’s the trap. BIG IDEA #2: in this situation, as in any situation in which you have a series of percentage changes, simply multiply the respective multipliers.", "# Consider an increase in the volatility of the stock in the previous problem. Suppose that if the 1. Consider an increase in the volatility of the stock in the previous problem. Suppose that if the stock increases in price, it will increase to $130, and that if it falls, it will fall to$70. Show that the value of the call option is higher than the value derived using the original assumptions. (LO 16-2)", "# Say a stock has increased from $4.34 a share to$6.55 a share. How much would that be percentage-wise? Aug 9, 2016 50.92% increase. #### Explanation: Calculate the percentage increase or percentage decrease in the same way. Find the difference - whether increased or decreased. Percent change = \"change\"/\"original value\" xx 100% Change = $6.55-$4.34 =\\$2.21 2.21/4.34 xx 100% = 50.92% increase" ]

[ "+0 0 70 2 Bill buys a stock that decreases by 20% on the first day, and then on the second day the stock increases by 30% of its value at the end of the first day. What was the overall percent increase in Bill's stock over the two days? Aug 8, 2019 #1 +5798 +1 $$v_1 = (1-0.2)v_0 = 0.8v_0\\\\ v_2 = (1+0.3)v_1 = 1.3(0.8)v_0 =1.04v_0\\\\ \\dfrac{v_2}{v_0} = 1.04 = (1+0.04)\\\\ \\text{Overall there was a 4% increase in the stock value}$$ . Aug 8, 2019 #2 +103148 +1 Let P be the original price of the stock So we have, after two days [ P (.80) ] (1.30) = P * 1.04 ⇒ i.e., the stock has increased in value by 4% Aug 8, 2019", "SEARCH HOME Math Central Quandaries & Queries Subject: integers Name: bill Who are you: Other if the value of xyz company stock drops from $25 per share to$21 per share, what is the percent of the decrease? Hi Bill, The stock price dropped from $25 to$21 so it dropped $4. The percentage change is the change (-$4), as a fraction of the starting price (\\$25), and then expressed as a percentage. The drop as a fraction of the starting price is -4/25. Expressed as a percent this is -4/25 100 = -16%. Hence there is a -16% change which is a 16% decrease. Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "original price of that item on June 30 of that year. The table is for the 28 days immediately after June 30 of that year. If on July n of that year, Bill bought a suit that originally cost$450, what is the value of n? (1) Bill paid $112.50 less than he would have if he had bought it the day before. (2) Bill paid$202.50 for the suit. (1) insuff..because we don't know the discounted price/any price for which he would have paid $112.5. (2)202.5 is what percent less of 450... 450(1-x/100)=202.5 x=55% date \"n\" could be anything between july15-july21...not suff.... Combining both Bill paid$112.50 less than he would have if he had bought it the day before means bill paid (202.5+112.5=\\$315) if he bought with any previous discount a day before. 315 is 30% less than 450. So he must bought on july 15 Ans C Manager Joined: 30 Mar 2017 Posts: 136 GMAT 1: 200 Q1 V1 Re: In the table above is shown the percent reduction in the [#permalink] ### Show Tags 14 Jan 2018, 07:15 Algebraic approach: Statement 1 Let $$x$$ be the discount Bill received and let $$y$$ be the previous period's discount. So we get this equation: $$450(1-x) = 450(1-y) - 112.50$$ Which reduces to: $$x-y = 0.25$$ This means that in", "days. Present value = 25000/1+ 2.92%(364/365) = 24292.5995 is the purchased price of$25000 T-bill b/ What is the yield rate on August 15 if the market price is $24377.64? April 01 to Aug 15 = 136 days Yield rate is unknown = r 622.36/24377.64(228/365) =4.09% c/ Calculate the market value of each$25000 T-bill on Oct 1 if the rate of return on that date is 4.545%. Since maturity value of T-bill is Mar 31 2017, Oct 01 2016 to Mar 31 2017 = 181days Present value = 25000/1+4.545%(181/365) = 24 448.96 was the price T-bill sold. d/ What is the rate of return realized if a \\$25000 T-bill purchase on April 1 is sold on Nov 20 at a market rate of 4.625%? From Nov 20 to maturity date (Mar 21 2017)= 131 days. Present value = 25000/1+4.625%(131/365) = 24 591.79308 was amount purchased T-bill on Nov 20. Now find the yield rate return of 25000 T-bill purchased on April 01 is sold on Nov 20 at rate of 4.625%. April 01 to Nov 20 = 233 days. 25 591.79308 = 24 292.6 [1+r(233/365)] = 0.019293687 or 1.9393% (5) Annual rate = x 25 000 = 24 756.25[1+r(182/365)] = 1.9746% is the annual rate of return. (6) a/ What was the original yield of the T-bill? 100 000=99", "the previous period, the discount was 25% less. Thus, Bill either bought his suit on July 8 or July 15. July 1 is not possible because there was no discount the day before. And July 22 is not possible because the discount was 15% less the day before. Not sufficient. Note that if the equation reduced to $$x-y = 0.15$$, Statement 1 alone would have been sufficient. Statement 2 $$202.50 = 450(1-x)$$ where $$x$$ is the discount Bill received. $$x = 0.55$$ So Bill could have bought his suit on any day July 15 through July 21. Not sufficient. Combining both statements In Statement 1, we narrowed down the answer to July 8 or July 15. In Statement 2, we narrowed down the answer to July 15 through July 21. The only date common do both is July 15. Sufficient. Re: In the table above is shown the percent reduction in the &nbs [#permalink] 14 Jan 2018, 07:15 Display posts from previous: Sort by", "# MandysNotes Tuesday, 22 April 2014 00:00 ## Changes in Percent By Gideon Rate this item Changes in percent can be counter-intuitive. For example, if the value of a stock falls by fifty percent on one day, and then rises by fifty percent on the next, it will not have recovered all of its value. It will in fact be seventy-five percent of its original value. It would need to increase by one hundred percent to recover all its value. Say that the value of a certain stock is $$X$$ dollars on Monday morning when the market opens. By the time the market closes it has lost half its value--it is down fifty percent--and so is $$\\frac{X}{2}$$ dollars. On Tuesday morning the stock starts out at $$Y = \\frac{X}{2}$$ dollars. When the market closes on Tuesday the stock is up fifty precent (from where it was in the morning) and so is worth $\\frac{100 + 50}{100}Y = \\frac{150}{100}Y = 1.5 Y \\text{ dollars.}$ But $$Y = \\frac{X}{2},$$ and so $$1.5 Y = \\frac{1.5}{2} X = .75 X$$ The reason for the confusion is that when we say on Monday that a stock is \"down fifty-percent,\" we mean down fifty-percent from its value that morning, that is it is down fifty-percent relative to $$X.$$ When, on Tuesday afternoon, we", "# Difference between revisions of \"A stock loses $10\\%$ of its value on Monday. On Tuesday it loses $20\\%$ of the value it had at the end of the day on Monday. What is the overall percent loss in value from the beginning of Monday to the end of Tuesday? Enter the answer as a percent.\" Before: $100$% After first loss: $90$% After second loss: $90*(\\frac{100-20}{100})=72$% In total, the stock lost $100-72=\\boxed{28}$%", "certain day. By what fraction did the price of the stock increase that day? (A) 1/2500 (B) 1/400 (C) 1/40 (D) 1/25 (E) 1/4 [Reveal] Spoiler: OA Senior Manager Joined: 18 Jan 2010 Posts: 256 Re: The price of a certain stock increased by 0.25 of 1 percent on a...... [#permalink] ### Show Tags 27 May 2016, 03:21 1 KUDOS 0.25 of 1 percent 0.25 is $$\\frac{1}{4}$$. 1 percent is $$\\frac{1}{100}$$ Of means multiplication. 0.25 of 1 percent = $$\\frac{1}{400}$$ Intern Joined: 18 Sep 2016 Posts: 49 Re: The price of a certain stock increased by 0.25 of 1 percent on a...... [#permalink] ### Show Tags 28 Sep 2016, 03:18 nalinnair wrote: The price of a certain stock increased by 0.25 of 1 percent on a certain day. By what fraction did the price of the stock increase that day? (A) 1/2500 (B) 1/400 (C) 1/40 (D) 1/25 (E) 1/4 0.25 of 1% is equal to say $$0.25*\\frac{1}{100}$$ = $$\\frac{0.25}{100}$$ = $$\\frac{25}{(100*100)}$$ = $$\\frac{1}{400}$$ _________________ Please press Kudos if you like my post Intern Joined: 04 Jun 2017 Posts: 4 Re: The price of a certain stock increased by 0.25 of 1 percent on a...... [#permalink] ### Show Tags 16 Dec 2017, 01:47 I am not a native English speaker, and is a bit confused towards the meaning", "working days later. E. Six working days later. $$r = 100*[\\sqrt{\\frac{(v+q)}{p}} - 1]$$ $$=>\\frac{r}{100} =\\sqrt{\\frac{(v+q)}{p}} - 1$$ $$=>\\frac{r}{100} +1=\\sqrt{\\frac{(v+q)}{p}}$$, Now square both sides to get $$=>(\\frac{r}{100}+1)^2 =\\frac{(v+q)}{p}$$ $$=>p*(\\frac{r}{100}+1)^2 =v+q$$ $$=>p*(\\frac{r}{100}+1)^2-q=v$$ Now as he sells at price $$v$$, hence he sells the bond after $$2$$ days of increase and the third day of decrease Option B ##### General Discussion VP Joined: 11 Feb 2015 Posts: 1146 Re: An investor purchased a bond for p dollars on Monday. For a certain nu [#permalink] ### Show Tags 25 May 2018, 18:15 2 If the brackets are not there then how did you take the 100 on left hand side in step number two in your solution? niks18 wrote: CAMANISHPARMAR wrote: Did you mean $$r = 100*[\\sqrt{[fraction](v+q)/p[/fraction]} - 1]$$? 101mba101 wrote: An investor purchased a bond for p dollars on Monday. For a certain number of days, the value of the bond increased by r percent per day. After this period of constant increase, the bond decreased the next day by q dollars and the investor decided to sell the bond that day for v dollars. When did the investor sell the bond if $$r = 100*\\sqrt{\\frac{(v+q)}{p}} - 1$$? A. Two working days later. B. Three working days later. C. Four working days later. D. Five working days later. E. Six working", "Present value = 25000/1+4.545%(181/365) = 24 448.96 was the price T-bill sold. d/ What is the rate of return realized if a$25000 T-bill purchase on April 1 is sold on Nov 20 at a market rate of 4.625%? From Nov 20 to maturity date (Mar 21 2017)= 131 days. Present value = 25000/1+4.625%(131/365) = 24 591.79308 was amount purchased T-bill on Nov 20. Now find the yield rate return of 25000 T-bill purchased on April 01 is sold on Nov 20 at rate of 4.625%. April 01 to Nov 20 = 233 days. 25 591.79308 = 24 292.6 [1+r(233/365)] = 0.019293687 or 1.9393% (5) Annual rate = x 25 000 = 24 756.25[1+r(182/365)] = 1.9746% is the annual rate of return. (6) a/ What was the original yield of the T-bill? 100 000=99 600[1+r(91/364)] = 1.610039% b/At what price did the investor sell? 100 000/1+1.4%(51/365) = 99 804.76547 was the price sold. c/ What annual rate of return did the investor realize while holding his T-bill? 99 804.76547 = 99 600[1+r(51/364)] = 1.8759% The amount of the installments = x 1x/ 1+6.9%(60/365) = 0.988784743x 1x/ 1+6.9%(120/365) = 0.97781826x 1x/ 1+6.9%(180/365) =0.967092364x 0.988784743x + 0.97781826x + 0.967092364x = 5000 X= 1704.34 is the amount of each installments . 4/ On April 1, \\$25000, 364-day T-bills were auctioned off to", "then gained 15 yards. What was his net gain or loss in the first quarter? 5. The stock market rose 1-1/4 points on Monday, fell 3/4 of a point on Tuesday, rose 1/2 point on Wednesday, rose 7/8 of a point of Thursday, and fell 1-1/2 points on Friday. What was the net change for the week?", "1. You have invested in stock for 1 year. If you start with $125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with$125 of stock and then it grows to $215, what was the growth rate? 2. Originally Posted by jilllamtn 1. You have invested in stock for 1 year. If you start with$125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with $125 of stock and then it grows to$215, what was the growth rate? For the first question, multiply $125 by (1+8.25%), which is:$125 (1.0825) = \\$135.31. That's your answer. I assume that you want your growth rate in terms of percent increases. So you find the difference between the old and new price and divide that difference by the old price. So: $\\frac{215-125}{125}=0.72$, which is 72%.", "1$$ $$=>\\frac{r}{100} +1=\\sqrt{\\frac{(v+q)}{p}}$$, Now square both sides to get $$=>(\\frac{r}{100}+1)^2 =\\frac{(v+q)}{p}$$ $$=>p*(\\frac{r}{100}+1)^2 =v+q$$ $$=>p*(\\frac{r}{100}+1)^2-q=v$$ Now as he sells at price $$v$$, hence he sells the bond after $$2$$ days of increase and the third day of decrease Option B ##### General Discussion Director Joined: 12 Feb 2015 Posts: 899 Re: An investor purchased a bond for p dollars on Monday. For a certain nu [#permalink] ### Show Tags 25 May 2018, 19:15 1 If the brackets are not there then how did you take the 100 on left hand side in step number two in your solution? niks18 wrote: CAMANISHPARMAR wrote: Did you mean $$r = 100*[\\sqrt{[fraction](v+q)/p[/fraction]} - 1]$$? 101mba101 wrote: An investor purchased a bond for p dollars on Monday. For a certain number of days, the value of the bond increased by r percent per day. After this period of constant increase, the bond decreased the next day by q dollars and the investor decided to sell the bond that day for v dollars. When did the investor sell the bond if $$r = 100*\\sqrt{\\frac{(v+q)}{p}} - 1$$? A. Two working days later. B. Three working days later. C. Four working days later. D. Five working days later. E. Six working days later. Hi CAMANISHPARMAR the brackets are not there. bracket before 100 would imply that -1", "1+6.9%(60/365) = 0.988784743x 1x/ 1+6.9%(120/365) = 0.97781826x 1x/ 1+6.9%(180/365) =0.967092364x 0.988784743x + 0.97781826x + 0.967092364x = 5000 X= 1704.34 is the amount of each installments . 4/ On April 1, \\$25000, 364-day T-bills were auctioned off to yield 2.92% a/ What is the price of each \\$25000 T-bill on April? Purchased on April 01. 364 T-bill. April 01(assuming 2016) to Mar 31 2017 = 364 days. Present value = 25000/1+ 2.92%(364/365) = 24292.5995 is the purchased price of \\$25000 T-bill b/ What is the yield rate on August 15 if the market price is \\$24377.64? April 01 to Aug 15 = 136 days Yield rate is unknown = r 24377.64 = 24292.5995 [1+r(136/365)]= 0.009395139 or 0.9395% 1.Problem number 12 on page 301 of the text. 2.Problem number 14 on page 301 of the text. c/ Calculate the market value of each \\$25000 T-bill on Oct 1 if the rate of return on that date is 4.545%. Since maturity value of T-bill is Mar 31 2017, Oct 01 2016 to Mar 31 2017 = 181days Present value = 25000/1+4.545%(181/365) = 24 448.96389 was the price T-bill sold. d/ What is the rate of return realized if a \\$25000 T-bill purchase on April 1 is sold on Nov 20 at a market rate of 4.625%? From Nov 20 to maturity", "1. You have invested in stock for 1 year. If you start with $125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with$125 of stock and then it grows to $215, what was the growth rate? 2. Originally Posted by jilllamtn 1. You have invested in stock for 1 year. If you start with$125 worth, a 8.25% increase in value will cause it to be worth how much? If you start with $125 of stock and then it grows to$215, what was the growth rate? For the first question, multiply $125 by (1+8.25%), which is:$125 (1.0825) = $135.31. That's your answer. I assume that you want your growth rate in terms of percent increases. So you find the difference between the old and new price and divide that difference by the old price. So:$\\displaystyle \\frac{215-125}{125}=0.72\\$, which is 72%.", "The price of a certain stock increased by 0.25 of 1 percent on a certain day. By what fraction did the price of the stock increase that day? ~$1 \\over 2500$~ ~$1 \\over 400$~ ~$1 \\over 40$~ ~$1 \\over 25$~ ~$1 \\over 4$~", "installments. The amount of the installments = x 1x/ 1+6.9%(60/365) = 0.988784743x 1x/ 1+6.9%(120/365) = 0.97781826x 1x/ 1+6.9%(180/365) =0.967092364x 0.988784743x + 0.97781826x + 0.967092364x = 5000 X= 1704.34 is the amount of each installments . 4/ On April 1, \\$25000, 364-day T-bills were auctioned off to yield 2.92% a/ What is the price of each \\$25000 T-bill on April? Purchased on April 01. 364 T-bill. April 01(assuming 2016) to Mar 31 2017 = 364 days. Present value = 25000/1+ 2.92%(364/365) = 24292.5995 is the purchased price of \\$25000 T-bill b/ What is the yield rate on August 15 if the market price is \\$24377.64? April 01 to Aug 15 = 136 days Yield rate is unknown = r 24377.64 = 24292.5995 [1+r(136/365)]= 0.009395139 or 0.9395% 1.Problem number 12 on page 301 of the text. 2.Problem number 14 on page 301 of the text. c/ Calculate the market value of each \\$25000 T-bill on Oct 1 if the rate of return on that date is 4.545%. Since maturity value of T-bill is Mar 31 2017, Oct 01 2016 to Mar 31 2017 = 181days Present value = 25000/1+4.545%(181/365) = 24 448.96389 was the price T-bill sold. d/ What is the rate of return realized if a \\$25000 T-bill purchase on April 1 is sold on Nov 20 at a", "# Percents: Percent Change The price of stock decreased by 8% last year and rose by 10% this year. What is the net percentage change in the price of the stock? Incorrect. [[snippet]] Correct. [[snippet]] Plug in the initial price of the stock as $100. Based on this, last year's price after an initial 8% decrease can be calculated as follows: >$$\\text{Last year's price} = 100 - 8\\% \\text{ of } 100 = 100 - 8 = 92$$. Now find this year's price after the 10% increase. Remember that the original price is now the new price, which means that the 10% is taken from$92. >$$\\text{This year's price} = 92 + 10\\% \\text{ of } 92 = 92 + 9.2 = 101.2$$ >$$\\displaystyle \\text{Net percent change} = \\frac{101.2 - 100}{100} \\times 100 = 1.2%$$ Hence, this is the correct answer. Incorrect. [[snippet]] Did you subtract the 8% decrease from the 10% increase ($$10\\% - 8\\% = 2\\%$$) and apply a 2% increase to the initial price? Watch out! The 8% decrease is applied to the original price while the 10% increase is applied to last year's price. Incorrect. [[snippet]] Incorrect. [[snippet]] 0.2% 0.8% 1.2% 1.6% 2%", "# Suppose that Christina purchased a stock for x dollars. During the first year, the stock's price rose 15%? (a) Write an algebraic expression for the price of the stock after the first year in terms of x. ? ## can you please show the work thank you. :) here is all the details and im giving you the whole question. Apr 14, 2017 a)${S}_{1} = 1.15 x$ b)${S}_{2} = 1.10 \\left(1.15 x\\right)$ c)${S}_{2} = 1.256 x$ d)S_2=$25.30 #### Explanation: The value of the stock $S$is $x$, so: S=$x After $1$ year the stock gains 15% in value: Then: ${S}_{1} = 1.15 x$ because it is now 115% of the original value. After $2$ years the stock gains 10% in value: Then: ${S}_{2} = 1.10 \\left(1.15 x\\right)$ because it is now 110% of the $S 1$ value. So: ${S}_{2} = 1.10 \\left(1.15 x\\right) = 1.265 x$ After $2$ years the stock is now valued at 126.5% of the original value. If the original value is $20: After $2$years the stock is valued at: S_2=1.256x=1.265($20)=\\$25.30", "told all of us that they thought if everything really went well the shares of stock would become worth $10. The original option price was$1. Those shares today after splits are worth, I think, somewhere around \\$1,000 or more. Nobody had any idea where the company was going, and for that matter even as the company grew and got its own buildings, moved to its own campus, a lot of us still thought that it was a little over-hyped, things couldn’t keep going as well as they did. But, in fact, they kept going way better. HOW BILL GATES BECAME A MORE DISTANT FIGURE AS THE COMPANY GREW Instead of the casual everyone has contact with Bill, he’s just another person, it got to the stage where there were the meetings to prepare for the meetings with Bill, or the meetings to prepare for the meetings to prepare for the meetings with Bill, and especially new hires were just terrified to talk to him. Some of us old timers never really got it, it was it’s just Bill and you can still yell at him and he respects that. But for those who weren’t there in the early days, it’s a lot harder to come into a company and say, “Oh yes, I’m going to yell at the" ]

[ "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", "shaded crescents to the area of the square. ## Dartboard Concentric circles with radii 1, $$\\frac{1}{2}$$, $$\\frac{1}{3}$$, $$\\frac{1}{4}$$, ... are drawn. Alternate donut-shaped regions are shaded. What is the total shaded area? ## Circles Which is largest, the red or the blue area? ## Archive Show me a random puzzle ▼ show ▼", "One hundred concentric circles with radii $$1, 2, 3, \\dots, 100$$ are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius 100 can be expressed as $$m/n,$$ where $$m$$ and $$n$$ are relatively prime positive integers. Find $$m + n.$$ (第二十一届AIME1 2003 第2题)", "# 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", "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", "shaded gray? Explanation: For the sake of simplicity, we will assume that ; this reasoning is independent of the actual length. The four concentric circles have radii 1, 2, 3, and 4, respectively, and their areas can be found by substituting each radius for in the formula : The outer gray ring is the region between the largest and second-largest circles, and has area The inner gray ring is the region between the second-smallest and smallest circles, and has area The total area of the gray regions is Since out of total area is gray, the percent of the figure that is gray is . ### Example Question #42 : Circles In the above figure, . Give the ratio of the area of the outer ring to that of the inner circle. 12 to 1 7 to 1 16 to 1 9 to 1 7 to 1 Explanation: For the sake of simplicity, we will assume that ; this reasoning is independent of the actual length. The four concentric circles have radii 1, 2, 3, and 4, respectively, and their areas can be found by substituting each radius for in the formula The areas of the largest circle and the second-largest circle are, respectively, The difference of their areas, which is the area of the outer ring, is", "# Area of Infinite Pointed Star Geometry Level pending Suppose we have 2 concentric circles with radii $$r$$ and $$\\frac{r}{n}$$ for some integer $$n \\geq 2$$. Place $$n$$ equally spaced points on the circumference of the larger circle. For each such point, draw two tangent line segments to the smaller circle and shade in the area enclosed by the new line segments and the smaller circle. As $$n$$ tends to infinity, what portion of the larger circle do we shade with the above construction? CLARIFICATION: once a region is shaded, shading it again has no effect. note: sorry there's no picture... message me if you know of a good site to draw geometry and export images! ×", "# Let's color circles. Geometry Level 4 150 concentric circles with radii 1, 2, 3, . . . , 150 are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius 150 can be expressed as m/n, where m and n are relatively prime positive integers. Find m + n. ×", "# Problem #2069 2069 Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region. $[asy] import graph; unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$ $\\text{(A)}\\ \\pi \\qquad \\text{(B)}\\ 1.5\\pi \\qquad \\text{(C)}\\ 2\\pi \\qquad \\text{(D)}\\ 3\\pi \\qquad \\text{(E)}\\ 3.5\\pi$ 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.", "\\mathrm{(E) \\ } \\frac{7}{8}$ ## Problem 12 An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$, with $b>c$. Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$, and let $OY$ be the radius of the larger circle that contains $Z$. Let $a=XZ$, $d=YZ$, and $e=XY$. What is the area of the annulus? $[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label(\"O\",O,SW); label(\"X\",X,E); label(\"Y\",Y,N); label(\"Z\",Z,SW); label(\"a\",X--Z,N); label(\"b\",0.25*X,SE); label(\"c\",O--Z,E); label(\"d\",Y--Z,W); label(\"e\",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]$ $\\mathrm{(A) \\ } \\pi a^2 \\qquad \\mathrm{(B) \\ } \\pi b^2 \\qquad \\mathrm{(C) \\ } \\pi c^2 \\qquad \\mathrm{(D) \\ } \\pi d^2 \\qquad \\mathrm{(E) \\ } \\pi e^2$ ## Problem 13 In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack? $\\mathrm{(A) \\ } 7 \\qquad \\mathrm{(B) \\ } 8 \\qquad \\mathrm{(C) \\ } 9 \\qquad \\mathrm{(D) \\ } 10 \\qquad \\mathrm{(E) \\ } 11$ ## Problem 14 A bag initially contains red marbles and blue", "Archived This topic is now archived and is closed to further replies. OpenGL How to draw a circle or a arc filled with color in OpenGL? This topic is 6151 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. Recommended Posts How to draw a circle or a arc filled with color in OpenGL? Share on other sites I guess you can use a triangle fan. Try something like this: glBegin(GL_TRIANGLE_FAN);glVertex3f(0.0f, 0.0f, 0.0f)float i;for (i = 0; i <= 360.0f; i += 360/num_steps) glVertex3f(cos(DEGTORAD * i) * radius, sin(DEGTORAD * i) * radius, 0.0f);glEnd(); Assume that num_steps is sorta the level of detail of the circle. Basically it''s how circular your circle is going to be. In actuality, this really just creates a regular polygon with num_steps sides. If num_steps is around 30 maybe, it will look close to a circle. radius is the radius of the circle, and DEGTORAD is pi/180. 1. 1 Rutin 37 2. 2 3. 3 4. 4 5. 5 • 11 • 10 • 13 • 103 • 11 • Forum Statistics • Total Topics 632976 • Total Posts 3009672 • Who's Online (See full list) There are no registered users currently online ×", "\\path[fill=gray!30] (-2,0) circle (1); \\path[fill=gray!30] (2,0) circle (1); \\end{tikzpicture} \\end{document} • I want the gray region to be uniform. – Adelyn May 13 '15 at 16:26 • All the shaded region. I guess I don't need inner shading and outer shading. – Adelyn May 13 '15 at 16:28 • Also \\draw[fill] can be \\filldraw – user11232 May 13 '15 at 16:31 • @Adelyn (-3,0) arc (180:0:2) means, start from (-3,0), draw an arc with starting angle 180 with a radius 2 units and end with an angle 0. Both 180 and 0 are measured from origin. You can also use (-3,0) arc [start angle=180,end angle=0,radius=2] See section 14.7 on page 153 of pgfmanual. – user11232 May 13 '15 at 16:39 • \\draw[fill=white] (-3,1) arc (180:0:2) -- (-3,1) -- cycle;. – user11232 May 13 '15 at 16:45", "neighbors. Find the area of the shaded region. $[asy] import graph; unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$ $\\text{(A)}\\ \\pi \\qquad \\text{(B)}\\ 1.5\\pi \\qquad \\text{(C)}\\ 2\\pi \\qquad \\text{(D)}\\ 3\\pi \\qquad \\text{(E)}\\ 3.5\\pi$ ## Problem 6 For how many positive integers $m$ does there exist at least one positive integer $n$ such that $m \\cdot n \\le m + n$? $\\mathrm{(A) \\ } 4\\qquad \\mathrm{(B) \\ } 6\\qquad \\mathrm{(C) \\ } 9\\qquad \\mathrm{(D) \\ } 12\\qquad \\mathrm{(E) \\ }$ infinitely many ## Problem 7 A $45^\\circ$ arc of circle A is equal in length to a $30^\\circ$ arc of circle B. What is the ratio of circle A's area and circle B's area? $\\text{(A)}\\ 4/9 \\qquad \\text{(B)}\\ 2/3 \\qquad \\text{(C)}\\ 5/6 \\qquad \\text{(D)}\\ 3/2 \\qquad \\text{(E)}\\ 9/4$ ## Problem 8 Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct? $[asy] unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); [/asy]$ $\\text{(A)}\\ B = W \\qquad \\text{(B)}\\ W = R \\qquad \\text{(C)}\\ B = R \\qquad \\text{(D)}\\ 3B = 2R \\qquad \\text{(E)}\\ 2R", "circle = new fabric.Circle({ left: 215, top: 100, fill: \"white\", stroke: \"#c154c1\", strokeWidth: 5, hasControls: false }); // Adding it to the canvas </html>", "\\dots, 21$. 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", "to radian measure. Leave the answer in terms of $\\pi$. $$30^{\\circ}$$ Convert to radian measure. Leave the answer in terms of $\\pi$. $$30^{\\circ}$$... Find the area of region $A C D B . \\widehat{A B}$ and $\\widehat{C D}$ are arcs of concentric circles with center $O .|\\overline{O A}|=8$ and $|\\overline{A C}|=2$ (Angle can't copy) Find the area of region $A C D B . \\widehat{A B}$ and $\\widehat{C D}$ are arcs of concentric circles with center $O .|\\overline{O A}|=8$ and $|\\overline{A C}|=2$ (Angle can't copy)... In & public park, triangular region of grass bounded shown in the figure Each sprinkler head can be hir its BnalesofSprinkler Feads are Placed at the vertices of the triangle as set limit its angle of rotation_ angle of rotation be set so as to keep the wal For each sprinkler head, Ikways dry and to conserve water? Round how large - should the the nearest degree_91 f135,7 ft127The angles of rotation are LALB~ Id In & public park, triangular region of grass bounded shown in the figure Each sprinkler head can be hir its BnalesofSprinkler Feads are Placed at the vertices of the triangle as set limit its angle of rotation_ angle of rotation be set so as to keep the wal For each sprinkler head, Ikways dry an... Find € correct", "Sketch region $2 < |z - 3 - 2i| < 3$. I believe it would be a circle with radius $2$ and $3$ about the coordinates $(3, 2i)$ and shaded between the two radius. Am I correct?", "$ABC$ and $CDE$, divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is $[asy] pair A,B,C,D,EE; A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0); fill(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)--arc((5,0),EE,A,CCW)--cycle,gray); draw(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)); draw(circle((5,0),5)); dot(A); dot(B); dot(C); dot(D); dot(EE); label($ $\\text{(A)}\\ 2:3 \\qquad \\text{(B)}\\ 1:1 \\qquad \\text{(C)}\\ 3:2 \\qquad \\text{(D)}\\ 9:4 \\qquad \\text{(E)}\\ 5:2$ Jan 28: Two $4 \\times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares? $[asy] unitsize(6mm); draw(unitcircle); filldraw((0,1)--(1,2)--(3,0)--(1,-2)--(0,-1)--(-1,-2)--(-3,0)--(-1,2)--cycle,lightgray,black); filldraw(unitcircle,white,black); [/asy]$ $\\textbf{(A)}\\ 16-4\\pi\\qquad \\textbf{(B)}\\ 16-2\\pi \\qquad \\textbf{(C)}\\ 28-4\\pi \\qquad \\textbf{(D)}\\ 28-2\\pi \\qquad \\textbf{(E)}\\ 32-2\\pi$ Week of Jan 15 – Jan 21: Click GoogleForm to submit answers. Version $\\alpha$: Thank you to all who gave me feedback. Just like I’ve found with my own personal experience, counting and probability are the hardest types of problems. We will continue practicing counting problems this week. Jan 15: How many ways are there to put 5 balls into 2 boxes if the balls are distinguishable and boxes are distinguishable? [Answer:] $2^5=32$ Jan 16: How many ways are there to", "radius of circle in ex's \\def\\dotsize{1.4}% size of dots relative to \\mygraphic \\def\\charwidth{4.5}% overall characterwidth in circle radii \\def\\mygraphic{\\smiley}% \\dotcircle \\end{document} • Thank you very much for your detailed solution. It is delightful to see a non-TikZ approach. While quite detailed, it is illuminating. – d3pd Sep 21 '15 at 9:13", "The total area of the $13$ smaller circles is $13\\pi$, so the shaded area is $\\left(13\\pi + 4\\pi\\sqrt{3}\\right) - 13\\pi = \\boxed{\\textbf{(A) } 4 \\pi \\sqrt{3}}$. ## Solution 4 $[asy] unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair A,B,C; A=(0,sqrt(3)*2); B=(-2,0); C=(0,0); draw(A--B); draw(A--C); draw(B--C); dot(A); dot(B); dot(C); label(\"A\",A, N); label(\"B\",B, W); label(\"C\",C, S); [/asy]$ In the diagram above, $AB=4$ and $BC=2$, so $AC=\\sqrt{4^2-2^2}=2\\sqrt{3}$. The larger circle's radius is $AC+1=2\\sqrt{3}+1$, so the larger circle's area is $\\pi\\left(2\\sqrt{3}+1\\right)^2=\\pi\\left(13+4\\sqrt{3}\\right)=13\\pi+4\\pi\\sqrt{3}$. Now, subtracting the combined area of the smaller circles gives $13\\pi+4\\pi\\sqrt{3}-13\\pi=\\boxed{\\textbf{(A) } 4 \\pi \\sqrt{3}}$. ~savannahsolver" ]

[ "3^2)$ = $20 \\pi$. Therefore the total area of the shaded region is $64 \\pi$.", "2$ area of the square=$\\pi (\\sqrt2)^2$=$2\\pi$ Area of the shaded region= 28-2$\\pi$", "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", "+0 # circles -4 77 2 +-291 In the diagram, each circle is divided into two equal areas and $O$ is the center of the larger circle. The area of the larger circle is $64\\pi.$ What is the total area of the shaded regions? Apr 5, 2020", "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", "the bigger circle is two times that of the smaller circles(the diameter), the radius of the larger circle in terms of $r$ would be $2r$. The area of the larger circle would come to $(2r)^2{\\pi} = 4{\\pi}r^2$. Subtracting the area of the smaller circles from that of the larger circle(since that would be the shaded region), we have $$4{\\pi}r^2 - 2{\\pi}r^2 = 2{\\pi}r^2 = 1.$$ Therefore, the area of the shaded region is $1$ or $\\boxed{D}.$", "Now the area of the circle is $\\pi$ cm$^2$, and hence the shaded area is 2 cm$^2$. This problem is taken from the UKMT Mathematical Challenges. View the archive of all weekly problems grouped by curriculum topic View the previous week's solution View the current weekly problem", "third that of the large circle. The area of each small circle is $\\pi(r/3\\: \\times\\: r/3)$. The total shaded area is $9({\\pi}{r^2}/9)$. This can be simplified to ${\\pi}{r^2}$. We can go one step further by saying that with $n$ circles the area is $n({\\pi}{r^2}/n)$ - which can again be simplified to ${\\pi}{r^2}$. Therefore the answer is that the shaded area is the same in each picture.", "the bigger circle B = $\\pi R^{2}$ Area of the shaded circular region = Area of circle A – Area of circle B Area of the shaded circular region = $\\pi R^{2} – \\pi r^{2}$ = $\\pi ( r^{2}- R^{2})$ Lets say if $R = 2r$, then the area of the shaded region would be: Area of shaded region = Area of circle A – Area of circle B = $\\pi (2r)^{2} – \\pi r^{2}$ Area of shaded region = $4\\pi r^{2} – \\pi r^{2} = 3 \\pi r^{2}$ The area of the circular shaded region can also be determined if we are only given the diameter of the circle by replacing “$r$” with “$2r$”. ### Example 7: Find the area of the shaded region in terms of pi for the figure given below. ### Solution: The radius of the smaller circle is = $5$ cm The radius of the larger/bigger circle is = $8$ cm Area of the shaded circular region = Area of circle A – Area of circle B Area of the shaded circular region = $\\pi R^{2} – \\pi r^{2}$ Area of the shaded circular region = $\\pi 8^{2} – \\pi 5^{2}$ Area of the shaded circular region = $\\pi (64 – 25) = 39\\pi$. Hopefully, this guide helped you develop the concept of", "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", "# How do i calculate the area of shaded region? I wouldn like to find the area of shaded region which it's circulated by a triangle as show in the below picture ? Note: I tried to draw other circle arround triangle ,but it's seems hard to me to deduce the shaded area Thank you for any help Let the area of the triangle be $\\Delta$, and the inradius be $r$ Then dividing the triangle into three by joining each vertex to the incentre we get $r(a+b+c)=2\\Delta$. The area of the circle is $\\pi r^2$. So all we need is the formula for $\\Delta$ which is $\\Delta=\\sqrt{s(s-a)(s-b)(s-c)}$, where $s=(a+b+c)/2$. With $a=15,b=14,c=13$, we have $s=21$, so $\\Delta=84$ and hence $r=168/42=4$. So the shaded area is $84-16\\pi\\approx33.73$.", "-2a) (E) (-a, -3a) 5) In the figure below, the circles are all congruent and tangent. All lines that contain the centers of two of the circles would be either horizontal or vertical. The total area of the circles is 100 pi. What is the area of the shaded region? (A) $25 - 10\\pi$ (B) $50 - 10\\pi$ (C) $100 - 10\\pi$ (D) $100 - 25\\pi$ (E) $100 - 50\\pi$ If you have questions about these problems or anything else to do with the SAT, leave a comment below or send me an email at [email protected] .", "we got is 4pir square = pi r square. Now area is four times instead of 3 times. So please clarify this The stem says that the area of the shaded region is 3 times the area of the smaller circular region, not that the area of the larger circular region is 3 times the area of the smaller circular region. Does this make sense? _________________ Kudos [?]: 139370 [0], given: 12787 VP Joined: 22 May 2016 Posts: 1245 Kudos [?]: 457 [0], given: 679 Re: In the figure shown, if the area of the shaded region is 3 t [#permalink] Show Tags 06 Dec 2017, 11:09 monirjewel wrote: Attachment: Untitled2.png In the figure shown, if the area of the shaded region is 3 times the area of the smaller circular region, then the circumference of the larger circle is how many times the circumference of the smaller circle? A. $$4$$ B. $$3$$ C. $$2$$ D. $$\\sqrt{3}$$ E. $$\\sqrt{2}$$ [Reveal] Spoiler: Attachment: 1111220830.jpg Algebraically Let a = Small circle's area Let s = Shaded region's area Let A = Large circle's area $$s = 3a$$ $$A = s + a$$ $$A = 3a + a$$ $$A = 4a$$ $$a = \\pi r^2$$ $$A = \\pi R^2$$ From above: $$A = 4a$$ $$\\pi R^2 = 4\\pi r^2$$ $$R^2 =", "+0 # geo problem 0 170 2 Find the area and the perimeter of the shaded regions below. Aug 4, 2020 #1 +32709 +1 TWO triangle hypotenuses find via pythag theorem h^2 = 4^2 + 10^2 solve for h rememeber: there are 2 semicircle perimeter = 1/2 pi d = 1/2 pi * 8 add them together Aug 4, 2020 #2 +1164 +1 AREA: 10*8/2 for the area of the triangle and π4^2 so we add them up in a equation to get that the area 80/2+π4^2=40+16π so the area of the shaded region is 40+16π or $$\\boxed{40+16π}$$ Circumfrence=8π/2 for the rounded part and 2*(sqrt(10^2+(8/2)^2)) again we make them into an equation 4π+2*(sqrt(10^2+4^2))=4π+2*(sqrt(116)) then we simplify it to get 4π+4*(sqrt(29)) or $$4π+4\\sqrt{29}$$ Aug 4, 2020", "## terricherri 3 years ago Find the area of the shaded portion in the square. (assuming the central point of each arc is the corresponding corner) 1. terricherri 2. TfGuy44 $( 8 - 2 \\pi )$ About 1.71681... Find more explanations on OpenStudy", "# 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.", "the shaded region is the area of the square subtracted from the area of the circle. The diameter of the circle and the diagonal of the square are equal to 2. We can easily find the area of the square using the area formula $\\frac{d_1 d_2}{2}$. So the area of the shaded region is $1^2 \\pi-\\frac{2\\cdot{2}}{2}=\\pi-2$. Clearly the largest area that we found among the three shaded regions is $\\pi -$2. area so the answer is $\\boxed{C}$", "## yociyoci 4 years ago Find the shaded region, which UV=VW=WX=XY=XZ=10 cm 1. vishal_kothari diagram? 2. yociyoci |dw:1327117815897:dw| 3. GT Assuming they are ALL semi-circles.......you can compute areas of those semi-circles and add/subtract the right amounts. 4. MrYantho Well done. I didn't even think they were semi-circles. Area = BigSemi + 3(small-semis) $A=((\\pi)r ^{2})/2 + 3(((\\pi)r ^{2})/2)$ $A=((\\pi)(50) ^{2})/2 + 3(((\\pi)(10) ^{2})/2)$ 5. yociyoci 312.5pi+12.5*3-12.5*2 =312.5pi+12.5 Is it like this or is it wrong? 6. MrYantho Whoops. My earlier one didn't subtract the 2 small semi-circle cutouts. So my answer above, you only add one semi-circle. 7. yociyoci ok, thank you!", "area is $$3 \\pi r^2 + 2 \\pi rh .$$ Therefore anything that comes after that is naturally wrong. Please check your answer before posting. – Leo Granger Jun 30 '19 at 9:24", "\\pi (6)^2=12\\pi$So the area of the region is 12pi-9root(3). Putting everything together you get:$6\\pi+6\\pi+12\\pi-9\\sqrt{3}=24\\pi-9\\sqrt{3}$A fair warning, arithmetic may be off, check the math to make sure its correct. 13. grayp 24 pi - 9 root 3 is the full answer?" ]

[ "The total area of the $13$ smaller circles is $13\\pi$, so the shaded area is $\\left(13\\pi + 4\\pi\\sqrt{3}\\right) - 13\\pi = \\boxed{\\textbf{(A) } 4 \\pi \\sqrt{3}}$. ## Solution 4 $[asy] unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair A,B,C; A=(0,sqrt(3)*2); B=(-2,0); C=(0,0); draw(A--B); draw(A--C); draw(B--C); dot(A); dot(B); dot(C); label(\"A\",A, N); label(\"B\",B, W); label(\"C\",C, S); [/asy]$ In the diagram above, $AB=4$ and $BC=2$, so $AC=\\sqrt{4^2-2^2}=2\\sqrt{3}$. The larger circle's radius is $AC+1=2\\sqrt{3}+1$, so the larger circle's area is $\\pi\\left(2\\sqrt{3}+1\\right)^2=\\pi\\left(13+4\\sqrt{3}\\right)=13\\pi+4\\pi\\sqrt{3}$. Now, subtracting the combined area of the smaller circles gives $13\\pi+4\\pi\\sqrt{3}-13\\pi=\\boxed{\\textbf{(A) } 4 \\pi \\sqrt{3}}$. ~savannahsolver", "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", "Sketch region $2 < |z - 3 - 2i| < 3$. I believe it would be a circle with radius $2$ and $3$ about the coordinates $(3, 2i)$ and shaded between the two radius. Am I correct?", "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", "## yociyoci 4 years ago Find the shaded region, which UV=VW=WX=XY=XZ=10 cm 1. vishal_kothari diagram? 2. yociyoci |dw:1327117815897:dw| 3. GT Assuming they are ALL semi-circles.......you can compute areas of those semi-circles and add/subtract the right amounts. 4. MrYantho Well done. I didn't even think they were semi-circles. Area = BigSemi + 3(small-semis) $A=((\\pi)r ^{2})/2 + 3(((\\pi)r ^{2})/2)$ $A=((\\pi)(50) ^{2})/2 + 3(((\\pi)(10) ^{2})/2)$ 5. yociyoci 312.5pi+12.5*3-12.5*2 =312.5pi+12.5 Is it like this or is it wrong? 6. MrYantho Whoops. My earlier one didn't subtract the 2 small semi-circle cutouts. So my answer above, you only add one semi-circle. 7. yociyoci ok, thank you!", "neighbors. Find the area of the shaded region. $[asy] import graph; unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$ $\\text{(A)}\\ \\pi \\qquad \\text{(B)}\\ 1.5\\pi \\qquad \\text{(C)}\\ 2\\pi \\qquad \\text{(D)}\\ 3\\pi \\qquad \\text{(E)}\\ 3.5\\pi$ ## Problem 6 For how many positive integers $m$ does there exist at least one positive integer $n$ such that $m \\cdot n \\le m + n$? $\\mathrm{(A) \\ } 4\\qquad \\mathrm{(B) \\ } 6\\qquad \\mathrm{(C) \\ } 9\\qquad \\mathrm{(D) \\ } 12\\qquad \\mathrm{(E) \\ }$ infinitely many ## Problem 7 A $45^\\circ$ arc of circle A is equal in length to a $30^\\circ$ arc of circle B. What is the ratio of circle A's area and circle B's area? $\\text{(A)}\\ 4/9 \\qquad \\text{(B)}\\ 2/3 \\qquad \\text{(C)}\\ 5/6 \\qquad \\text{(D)}\\ 3/2 \\qquad \\text{(E)}\\ 9/4$ ## Problem 8 Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct? $[asy] unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); [/asy]$ $\\text{(A)}\\ B = W \\qquad \\text{(B)}\\ W = R \\qquad \\text{(C)}\\ B = R \\qquad \\text{(D)}\\ 3B = 2R \\qquad \\text{(E)}\\ 2R", "# Problem #2069 2069 Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region. $[asy] import graph; unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$ $\\text{(A)}\\ \\pi \\qquad \\text{(B)}\\ 1.5\\pi \\qquad \\text{(C)}\\ 2\\pi \\qquad \\text{(D)}\\ 3\\pi \\qquad \\text{(E)}\\ 3.5\\pi$ 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.", "\\mathrm{(E) \\ } \\frac{7}{8}$ ## Problem 12 An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$, with $b>c$. Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$, and let $OY$ be the radius of the larger circle that contains $Z$. Let $a=XZ$, $d=YZ$, and $e=XY$. What is the area of the annulus? $[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label(\"O\",O,SW); label(\"X\",X,E); label(\"Y\",Y,N); label(\"Z\",Z,SW); label(\"a\",X--Z,N); label(\"b\",0.25*X,SE); label(\"c\",O--Z,E); label(\"d\",Y--Z,W); label(\"e\",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]$ $\\mathrm{(A) \\ } \\pi a^2 \\qquad \\mathrm{(B) \\ } \\pi b^2 \\qquad \\mathrm{(C) \\ } \\pi c^2 \\qquad \\mathrm{(D) \\ } \\pi d^2 \\qquad \\mathrm{(E) \\ } \\pi e^2$ ## Problem 13 In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack? $\\mathrm{(A) \\ } 7 \\qquad \\mathrm{(B) \\ } 8 \\qquad \\mathrm{(C) \\ } 9 \\qquad \\mathrm{(D) \\ } 10 \\qquad \\mathrm{(E) \\ } 11$ ## Problem 14 A bag initially contains red marbles and blue", "Advertisement Remove all ads # In Figure 3, two concentric circles with centre 0, have radii 21cm and 42 em. If ∠ AOB = 60°, find the area of the shaded region. - Mathematics In Figure , two concentric circles with centre O, have radii 21cm and 42 cm. If ∠ AOB = 60°, find the area of the shaded region. [use π=22/7] Advertisement Remove all ads #### Solution 1 Solution: Given : Radii of inner circle = 21 cm = r Radii of outer circle = 42 cm = R ∠AOB = θ = 60° Also, Area of ring = π(R2-r2) Area of a sector = theta/360 pir^2 The area of shaded region = Area of ring – Area of ABCD = Area of ring – Area of sector of Outer Circle - Area of sector of Inner Circle =pi(R^2-r^2)-(pi(R^2-r^2))/1 xx theta/360 =pi(R^2-r^2)[1-theta/360] =22/7(42^2-21^2)(1-60/360) =3465 cm^2 #### Solution 2 Solution: Given : Radii of inner circle = 21 cm = r Radii of outer circle = 42 cm = R ∠AOB = θ = 60° Also, Area of ring = π(R2-r2) Area of a sector = theta/360 pir^2 The area of shaded region = Area of ring – Area of ABCD = Area of ring – Area of sector of Outer Circle - Area of sector of", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of \"2002 AMC 12A Problems/Problem 5\" The following problem is from both the 2002 AMC 12A #5 and 2002 AMC 10A #5, so both problems redirect to this page. ## Problem Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region. $[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$ $\\textbf{(A)}\\ \\pi \\qquad \\textbf{(B)}\\ 1.5\\pi \\qquad \\textbf{(C)}\\ 2\\pi \\qquad \\textbf{(D)}\\ 3\\pi \\qquad \\textbf{(E)}\\ 3.5\\pi$ ## Solution The outer circle has radius $1+1+1=3$, and thus area $9\\pi$. The little circles have area $\\pi$ each; since there are 7, their total area is $7\\pi$. Thus, our answer is $9\\pi-7\\pi=\\boxed{2\\pi\\Rightarrow \\textbf{(C)}}$.", "Let the answer be $$f_R(d)$$. Position the two planes symmetrically about the equator of the sphere. Then surface area between the planes and above (or below) the equator is half the total area between the planes, so $$f_R(d)=2f_R(d/2)$$, which implies that $$f_R(d)$$ is linear in $$d$$. Since $$f_R(2R)=4\\pi R^2$$ (the surface area of the whole sphere), we must have $$f_R(d)=2\\pi R d$$. From Catriona Agg: Find the area of the shaded region: There are many different configurations of quarter circles which satisfy the constraints shown. In particular, you can make the radius of one of the shaded quarter-circles take any value in $$[0,8\\sqrt{2}]$$, and there exist radii for the other two circles which make the diagram true. However, miraculously, the shaded area does not depend on the exact configuration. Knowing this, you can easily find the shaded area by: • Taking the two shaded quarter circles to have equal radii. Then the shaded area is a semicircle of radius 8, so the area is $$\\frac{\\pi}{2}8^2 = 32 \\pi$$. • Taking one of the two shaded quarter circles to have radius zero. Then the shaded area becomes a single quarter circle of radius $$8\\sqrt{2}$$, so its area is $$\\frac{\\pi}{4}(8\\sqrt{2})^2 = 32 \\pi$$. There's the old one that goes something like: \"I'm thinking of a number. I double it, add", "# 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", "you need to use radians for the start and end locations of the arc. Lets inspect Figure 3 a bit. You will need several arc Method commands. • Each of those commands will need the center coordinates, which were (300, 300), a radius, which was 100 for the circle and 3 to the center, and the direction, which was clockwise. • To go clockwise around the circle you start at $$0*\\pi$$, then go to $$0.5*\\pi$$, then go to $$1*\\pi$$, then go to $$1.5*\\pi$$, then end at $$2*\\pi$$ (alias at $$0*\\pi$$). • In order to graph the circle you would start at $$0*\\pi$$ and end at $$2*\\pi$$. A separate arc Method command is needed. • In order to graph the center you would need a separate arc Method command to start at $$0*\\pi$$ and end at $$2*\\pi$$ and a way to fill the tiny circle so it looks like dot. • In order to get the tan arc you would need a separate arc Method command that starts at $$1.2*\\pi$$ and ends at $$1.7*\\pi$$. In addition we needed a line width ,here 10, and a way to fill the width of the arc with the color tan. • The radii were drawn using another command. • All the labels were put on after the circular features were drawn using", "shaded gray? Explanation: For the sake of simplicity, we will assume that ; this reasoning is independent of the actual length. The four concentric circles have radii 1, 2, 3, and 4, respectively, and their areas can be found by substituting each radius for in the formula : The outer gray ring is the region between the largest and second-largest circles, and has area The inner gray ring is the region between the second-smallest and smallest circles, and has area The total area of the gray regions is Since out of total area is gray, the percent of the figure that is gray is . ### Example Question #42 : Circles In the above figure, . Give the ratio of the area of the outer ring to that of the inner circle. 12 to 1 7 to 1 16 to 1 9 to 1 7 to 1 Explanation: For the sake of simplicity, we will assume that ; this reasoning is independent of the actual length. The four concentric circles have radii 1, 2, 3, and 4, respectively, and their areas can be found by substituting each radius for in the formula The areas of the largest circle and the second-largest circle are, respectively, The difference of their areas, which is the area of the outer ring, is", "# In Figure 2, Two Concentric Circles with Centre O, Have Radii 21 Cm and 42 Cm. If ∠Aob = 60°, Find the Area of the Shaded Region. - Mathematics Sum In Figure 2, two concentric circles with centre O, have radii 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. #### Solution Radius of inner circle, OC = 21 cm Radius of outer circle, OA = 42 cm Area of a circle with radius R = piR^2 = pi(42)^2 Area of a circle with radius r = pir^2 = pi(21)^2 Area of sector AOB = theta/360 xx piR^2 = 60/360 xx pi(42)^2 = (pi(42)^2)/6 Area of sector COD = theta/360 xx pir^2 = 60/360 xx pi(21)^2 = (pi(21)^2)/6 Area of shaded portion = Area of circle with radius R - Area of circle with radius r - [Area of sector AOB - Area of sector COD] = pi(42)^2 - pi(21)^2 - [(pi(42)^2)/6 - (pi(21)^2)/6] = pi[(42)^2 - (21)^2 - 1/6[(42)^2 - (21)^2]] = pi[((42)^2 - (21)^2)(1 - 1/6)] = pi[(42 - 21)(42 + 21)5/6] = 22/7 xx 5/6 xx 21 xx 63 = 3465 cm2 Concept: Circumference of a Circle Is there an error in this question or solution?", "# The area of circles tangent inside another circle Consider the following shape: Three identical circles are tanged inside a circle. Which is greater: 1. The area of the shaded regions. 2. Twice the area of the unshaded regions. My attempt: I tried to solve it approximately. Let $R=2$ be the radius of the big circle, and its area is $4\\pi$. The diameter of the smaller circles is about $R$ and their radius is $R/2=1$, the area of each small circle is $\\pi$. So, the area of the shaded region is $3\\pi$. The area of the unshaded region is about $4\\pi - 3\\pi =\\pi$. Twice of it is about $2\\pi$ which is smaller than $3\\pi$, the area of the shaded region. However, the correct answer is 2! • – user Feb 15 '18 at 8:55 • How can the radius of the big circle be $\\;R=2\\;$ and the diameter of the smaller circles be \"about\" (?) $\\;R(=2)\\;$ ? That'd mean the three circles would intersect at the big circle's center...but they don't. Feb 15 '18 at 9:04 • Take a peek at math.stackexchange.com/questions/666491/… . It is even poetic... Feb 15 '18 at 9:08 • @DonAntonio As I said, I tried to make an approximation! Feb 15 '18 at 9:46 • @Ahmad Well, that seems like a crude approximation...why", "r } 2=\\pi r$ centers each... Is 9, times pi -- 9 pi Ind the perimeter of shaded region design. Taking anything from my office be considered as a theft how much did J. Robert Oppenheimer get paid overseeing! Squared pi geomagnetic field because of the semicircle diameter that things that look like circles are,! Is $\\color { red } { 20\\pi }$ to find the of... Now the perimeter of the shaded region = Circumference of the shaded region = Circumference of semicircle P r... – area of the square the best way to play a chord larger than hand! Combustion chamber per nozzle, things that look like squares are squares, that! Let the radius of the shaded fighure is 14 π + 2 ) or 9, pi! 2 + 3 + 4 ) = 10 hence 10pi is perimeter do we not observe a greater Casimir than., if radii of 1,2,3,4 respectively in term of π ) rectangle a. Semicircles have radii of 1,2,3,4 respectively with you? is 14 π 2... Shape below is made from 7 hexagon tiles it by two d of... To 3 significant figures. 0 are 7 cm = 44 + 28 = 72 cm AQO. By two the 4 small semi circles + 3 + 4 ) = 7 cm =", "One hundred concentric circles with radii $$1, 2, 3, \\dots, 100$$ are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius 100 can be expressed as $$m/n,$$ where $$m$$ and $$n$$ are relatively prime positive integers. Find $$m + n.$$ (第二十一届AIME1 2003 第2题)", "# Area of Infinite Pointed Star Geometry Level pending Suppose we have 2 concentric circles with radii $$r$$ and $$\\frac{r}{n}$$ for some integer $$n \\geq 2$$. Place $$n$$ equally spaced points on the circumference of the larger circle. For each such point, draw two tangent line segments to the smaller circle and shade in the area enclosed by the new line segments and the smaller circle. As $$n$$ tends to infinity, what portion of the larger circle do we shade with the above construction? CLARIFICATION: once a region is shaded, shading it again has no effect. note: sorry there's no picture... message me if you know of a good site to draw geometry and export images! ×", "area of the shaded area if a song is tuned in half-step down, against... 11 $\\pi cm^2$ D. 60 $\\pi cm^2$ E. None of the shaded is... That of the semi-circles given figure = Length of RS is the distance from the centre to the outside the... Any level and professionals in related fields 134188: find the area of ARPQ I defeat a Minecraft zombie picked! \\Color { red } { 20\\pi } \\$ of sector OBD one side of the shaded =! This becomes six pie, which is about 44 0.85 enslave ability 4... Overseeing the Manhattan Project comment dit-on what 's the area and perimeter – HW # 72 find the of... Diameter ( d ) of the shaded region in semicircle if four small semicircles have radii 1,2,3,4! Πr + 2r, where r is the distance from the centre to the outside of the semi-circle AQO ! The outside of the 4 semicircle the formula for area of circle: 49 the radius the! Figure, and then we can divide that by two following figures. = 40° respectively." ]

[ "clans. People would eat when hungry, sleep when tired, and everyone would have their own schedule. In this case timekeeping would actually be quite important to help people with different schedules co-ordinate. Perhaps bells would be sounded every hour or two (say) as a means of marking time. But how long is \"an hour or two\"? Well, the time does not need to be exact. One possibility is a traditional unbroken song is sung by the religious order (with individuals joining and leaving the song to eat, rest etc.) and at particular points in the (long) song, the bell is struck. Alternatively the constant migration might lead to a religious practice where a group is constantly marching, perhaps in a large circle, at a slow and very steady pace. Every thousand (say) paces the bell is struck. That said, it is quite complex for everyone to be working to their own rhythm, especially in simple societies. To create a more orderly approach, there could be a bell sounded every 12 hours, with people \"swapping over\" from work to leisure and rest and vice versa. Or there could be 3 \"phases\" in everyone's day (basically work, leisure/socialising, sleep), each of around 8 hours, with 3 different cohorts doing different things at different times. How would such long time periods", "tower before or after a service or wedding. While on these everyday occasions the ringers must usually content themselves with shorter \"touches,\" each lasting a few minutes, for special occasions they often attempt a quarter-peal or peal, lasting approximately 45 minutes or three hours respectively. If a peal attempt succeeds, towers sometimes mark the occasion with a peal board mounted on the wall of the ringing chamber; at St Peter Mancroft in Norwich there is one documenting what is generally considered to be the first true peal: 5040 changes of Plain Bob Triples (a method still popular today), rung 2 May 1715. Today over 4000 peals are rung each year. Dorothy L. Sayers's mystery novel The Nine Tailors is famous for the central part played by change ringing. Much of the action centres on a bell tower and the peals rung in it, and to draw the reader in Sayers takes care to explain change ringing and analyze its improbable popularity; quotations from the book are popular with ringers. Moreover, the entire book is infused with an air of change ringing to the extent that her chapter titles all employ campanological terminology; and indeed, one of the book's conceits is that it is a sort of multi-part peal. ### Organization and extent The Central Council of Church Bell", "Related Searches Definitions # change ringing Traditional English art of ringing tower or hand bells in a succession of different orders. Groups of swinging bells in English church towers date from the 10th century. Control of bell ringing was much enhanced with the invention of the bell wheel in the 14th century; further improvements in bearings and fittings allowed the bells to swing at the same speed and led to recognizably modern change ringing by the 17th century. A “ring” (set) of 6 bells can be tolled in 720 different “changes” (orders); 8 bells allow 40,320 different changes, and 10 bells allow more than 3,000,000. In practice, only a selection of the possible changes is rung, derived by a “method” (a shuffling algorithm), usually involving switching pairs of bells in a certain order (e.g., 123456 becomes 214365, which can then become 241635, etc.). “Rounds”—all the bells in order from highest to lowest—is rung before and after a method is completed. Learn more about change ringing with a free trial on Britannica.com. Campanology (from late Latin campana, \"bell\"; and Greek -λογία, -logia) is the study of bells. It encompasses the physical realities of bells — how they are cast, tuned and sounded — as well as the various methods devised to perform bell-ringing. In particular, it is common to", "just under 18 hours.[10]) An extent on 12 bells would take over thirty years. Since extents are obviously not always practicable, ringers more often undertake shorter performances. Such ringing starts and ends with rounds, having meanwhile visited only a subset of the available permutations; but truth is still considered essential — no row can ever be repeated; to do so would make the ringing false. A peal is an extended performance; it must last at least 5000 changes on eight or more bells and at least 5040 on seven or fewer bells (5040 being 7!, the length of a full extent on seven). A performance of 1250 (on 8 or more) or 1260 (on 7 or fewer) changes likewise makes a quarter peal (quarter for short); a peal or a quarter tends to last about three hours or 45 minutes, respectively. ## Changes on handbells Change ringing can also be performed on handbells, and is quite popular in its own right. Many record-length peals, including the longest peal ever rung, are by handbell ringers. Normally each ringer has a bell in each hand and sit or stand in a circle (like tower ringers). The tower bell terms of handstroke and backstroke are retained, referring to an upwards and downwards ring of the bell respectively; and as in towers,", "So, you've got groups of people who need to meet up for some collective activity at a certain time. But you don't want any mechanical timekeeping device. Here's a solution. You've got a guy with a big bell who decides what time of day it is. He doesn't have a clock and doesn't know the time precisely. But he can tell by the light, or just his gut sense, when it's about mid-morning or noon or mid-afternoon and so forth. So, perhaps six times a day he rings his bell at approximately the right time, and everyone hears it. So, you can schedule a meeting: \"meet in the west field after third bell.\" When people in town hear the third bell, they head to the west field, and arrive at roughly the same time for the meeting, perhaps within about fifteen minutes of the bell ringing, depending on the size of the town. If you want, there's a way to be a bit more accurate about the time than the bell-ringer's \"gut feeling,\" still without using any timekeeping device. The bell-ringer can be a guy who spends his day doing some repetitive job, and he rings the bell based on how much of his job he got done. Perhaps he's the baker. Perhaps he's plowing the fields or", "Wikipedia Pens Pencils Six bells, commencing tolling together, toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? ×", "# URGENT HELP PLEASE at 10am three bells ring together. The large bell rings every four minutes, the old bell rings every 7 minutes and the small bell rings every three minutes. When will be the next time that all the three bells ring together? ###### Question: at 10am three bells ring together. The large bell rings every four minutes, the old bell rings every 7 minutes and the small bell rings every three minutes. When will be the next time that all the three bells ring together? ### His awkwardness at digging the hole with a shovel whose handle was twice as long as he was made us laugh, and we covered our mouths with our hands so he wouldn't hear.\" The reader can infer that Doodle's family______________________________. His awkwardness at digging the hole with a shovel whose handle was twice as long as he was made us laugh, and we covered our mouths with our hands so he wouldn't hear.\" The reader can infer that Doodle's family______________________________.... ### Balance Ca(OH)2 + H3PO4 yields Ca3(PO4)2 + H2O Balance Ca(OH)2 + H3PO4 yields Ca3(PO4)2 + H2O... ### Read the following sentence. More students voted for healthy lunch choices at school, making it clear that they were in the majority. Which of the following phrases provides a context clue", "Since the 20th century, in Belgium and The Netherlands, clock chime bells have inner tuning and produce complex fully harmonized music. The here described chimes, often singular chime, should neither be confounded with another musical instrument called chimes or tubular bells, nor with a wind chime. ## Russian Orthodox bells The bells in Russian tradition are sounded by their clappers, attached to ropes; a special system of ropes is developed individually for every belltower. All the ropes are gathered in one place, where the bell-ringer stands. The ropes (usually - all ropes) are not pulled, but rather pressed with hands or legs. Since one end of every rope is fixed, and the ropes are kept in tension, a press or even a punch on a rope makes a clapper move. The Russian Tsar Bell is the largest extant bell in the world. ## Change ringing In England the bells in church towers are generally hung for full circle ringing: every bell swings through a complete circle (actually a little more than 360 degrees) each time it sounds. Between strokes, it sits poised 'upside-down', with the mouth pointed upwards; pulling on a rope connected to the bell swings it down and its own momentum swings it back up again on the other side. These rings of bells have relatively", "an electronic clock. Every quarter hour, the clock triggers the strikers in a melodic sequence, and the bells sound out over the campus. Depending on wind and ambient noise conditions, they can be heard up to half a mile from the tower. Several thousand people are within earshot of the clock chimes.(2) [4] Since the original installation of bells in the library tower in 1930, the clock has chimed the Westminster Quarters, so named because of its adoption on the tower clock at Westminster Palace, where the British Houses of Parliament meet. (The lowest bell there, which tolls the hours and weighs thirteen and a half tons, is the famous “Big Ben.”) Having been reproduced in clocks both large and small all over the world, the Westminster Quarters is a well-known chime tune—perhaps the only tune people think of when the subject of clock chimes is broached. Its popularity certainly owes much to the attractive melodic sequence, but perhaps more to sentimental associations of British imperial pomp and circumstance, reinforced by the use of the Westminster Quarters by the BBC World Service for many years to preface the top-of-the-hour world news report on short-wave radio. [5] As University Carilloneur, I am steward of the bells and the clock mechanism. Over the years, I had become curious about the", "that while an extent on 6 bells can be accomplished in half an hour, a full peal on 8 bells should take nearly twenty-two and a half hours and one on 12 bells would take over thirty years! Naturally, then, except in towers with only a few bells, ringers typically can only ring a subset of the available permutations. But the key stricture of an extent, uniqueness (any row may only be rung once), is considered essential. This is called truth; to repeat any row would make the performance false. Another key limitation keeps a given bell from moving up or back more than a single place from row to row; if it rings (for instance) fourth in one row, in the next row it can only ring third, fourth, or fifth. Thus from row to row each bell either keeps its place or swaps places with one of its neighbours.[1] This rule has its origins in the physical reality of tower bells: a bell, swinging through a complete revolution with every row, has considerable inertia and the ringer has only a limited ability to accelerate or retard its cycle. A third key rule mandates rounds as the start and end of all ringing. So to summarize: any performance must start out from rounds, visit a number of", "mins = 31 mins Electronic devices will beep after every 31 minutes. Hence, both devices will beep together again at 10:31 a.m. again. Question 26: The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m., then at what time will they again change simultaneously? Solution: The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. Find the LCM of 48, 72 and 108: 48 = 2 × 2 × 2 × 2 × 3 = 24 × 3 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 108 = 2 × 2 × 3 × 3 × 3= 22 × 3 3 LCM = 432 or 432 sec Convert into mins: 432 sec = 7 mins 12 sec So, at 8:7:12 hrs traffic lights will again change. Question 27: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together? Solution: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. Find LCM using prime factorization: 2 = 2 (prime number)", "and the nunnery’s ding occur at the same moment, and that is exactly what is due to happen at noon today. Between consecutive Feasts the dongs from the monastery and the dings from the nunnery occur alternately and, although the two noises only coincide on Feast days, they do occur a minute apart at some other times. When the bells coincided last time (at noon, a prime number of days ago) this whole island indulged in its usual orgy of eating and drinking. How many days ago was that? This puzzle was included in the book The Sunday Times Book of Brain-Teasers: Book 1 (1980, edited by Victor Bryant and Ronald Postill). The puzzle text above is taken from the book. After this puzzle was published The Sunday Times was hit by industrial action, and the next issue was not published until 18th November 1979. [teaser903] • #### Jim Randell 9:47 am on 11 January 2022 Permalink | Reply If the shorter interval is i minutes, then, working forwards from the previous feast (= 0) the bell rings at (minutes): i, 2i, 3i, 4i, … ni And if the shorter interval is (i + x) minutes, then that bell rings at: (i + x), 2(i + x), 3(i + x), … (n − 1)(i + x) And the", "together and toll respectively at intervals of 6, 7, 8, 9 and 12 seconds. How many times they will toll together in one hour? A. 5 B. 14 C. 6 D. 7 E. 9 [Reveal] Spoiler: OA Senior Manager Joined: 30 Aug 2009 Posts: 286 Location: India Concentration: General Management Followers: 3 Kudos [?]: 149 [0], given: 5 Re: Five bells begin to toll together and toll respectively at i [#permalink] ### Show Tags 02 Dec 2009, 01:26 1 This post was BOOKMARKED kirankp wrote: Five bells begin to toll together and toll respectively at intervals of 6,7,8,9 and 12 seconds.How many times they will toll together in one hour? A)5 B)14 C)6 D)7 E)9 D-7 all bells will toll together once every 504seconds Manager Joined: 22 Jul 2008 Posts: 94 Location: Bangalore,Karnataka Followers: 3 Kudos [?]: 159 [0], given: 11 Re: Five bells begin to toll together and toll respectively at i [#permalink] ### Show Tags 02 Dec 2009, 09:19 7 is right. LCM of 6,7,8,9 and 12 = 504 so the bells will toll together after 504 sec i.e 3600/504 =7 Intern Joined: 31 Oct 2009 Posts: 38 Followers: 1 Kudos [?]: 48 [0], given: 0 Re: Five bells begin to toll together and toll respectively at i [#permalink] ### Show Tags 02 Dec 2009, 13:47", "changes without memorising huge quantities of data, without any written prompts. Ringers can also ring different methods, with different \"works\" on different numbers of bells - so there is a huge variety of ways of ringing changes in method ringing. ## Peals and quarter peals A peal board recording the details of a notable peal. Thousands of these boards exist in change ringing belfries For some people, the ultimate goal of this system is to ring all the permutations, to ring a tower's bells in every possible order without repeating – what is called an extent (or sometimes, formerly, a full peal). The feasibility of this depends on how many bells are involved: if a tower has ${\\displaystyle n}$ bells, they have ${\\displaystyle n!}$ (read factorial) possible permutations, a number that becomes quite large as ${\\displaystyle n}$ grows. For example, while six bells have 720 permutations, eight bells have 40,320; furthermore, 10! = 3,628,800, and 12! = 479,001,600. Estimating two seconds for each change (a reasonable pace), one finds that while an extent on six bells can be accomplished in half an hour, an extent on eight bells should take nearly twenty-two and a half hours. (When in 1963 ringers in Loughborough became the only band in history to achieve this feat on tower bells, it took them", "following a method, a relatively simple pattern they have learned to direct them from row to row. Where the treble bell follows exactly the same pattern as the other bells it is called a principle. For some people, the ultimate goal of this system is to ring all the permutations, to ring a tower's bells in every possible order without repeating – what is called an extent (or sometimes, formerly, a full peal). The feasibility of this depends on how many bells are involved: if a tower has ${\\displaystyle n}$ bells, they have ${\\displaystyle n!}$ (read factorial) possible permutations, a number that becomes quite large as ${\\displaystyle n}$ grows. For example, while six bells have 720 permutations, eight bells have 40,320; furthermore, 10! = 3,628,800, and 12! = 479,001,600. Estimating two seconds for each change (a reasonable pace), one finds that while an extent on six bells can be accomplished in half an hour, a full peal on eight bells should take nearly twenty-two and a half hours. (When in 1963 ringers in Loughborough became the only band in history to achieve this feat on tower bells, it took them just under 18 hours.[6]) An extent on 12 bells would take over thirty years. Since extents are obviously not always practicable, ringers more often undertake shorter performances. Such", "were spread over the north tower, south tower and the two pagan towers. The bells 5-9 were hanging in the northern pagan tower. In the southern Heidenturm the bells 3-4, in the north tower the half-pummerin (St. John's bell) and in the south tower the pummerin (Josefinic bell) with a few single bells. The disposition of the old bell was: H °°, e °, g sharp °, c ', dis', dis', f sharp', g sharp ', c' ', g' '. #### Clock bells Two bells for the clock are hung in the belfry of the tower; the quarter hours are struck on the prim bell , the full hours on the clock bell : No. Surname Strike tone (a 1 = 435 Hz) Weight (kg, approx.) Diameter (mm, approx.) Casting year Caster 1 Clock sounds of 1 + 4 / 16 1500 1490 1449 J. Straiffing & P. ​​Obrecht 2 Primbelllein g 2 - 4 / 16 140 600 1771 Franz Josef Sheichel #### More bells Two small bells that escaped being melted down as part of the metal collection in 1942 are placed in the Pummerin's bell room . According to their names, they were mainly rung individually. The Speisglöckl (dining bell) was rung for communion , the Zügenglöckl (train bell) exhorted the people to pray for", "Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively. How many times they ring together in 2 hours? 1. 120 2. 60 3. 121 4. 112 Option 3 : 121 ## Detailed Solution GIVEN: Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively. CONCEPT: LCM: It is a number which is a multiple of two or more numbers. CALCULATION: LCM of (6, 12, 15, 20) = 60 All 4 bells ring together again after every 60 seconds Now, In 2 Hours, they ring together = [(2 × 60 × 60)/60] times + 1 (at the starting) = 121 times ∴ In 2 hours they ring together for 121 times Mistake Points In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1. # A tree breaks due to storm and the broken part bends, so that the top of the tree touches the ground making an angle of 30° with the ground. The distance between the foot of the tree to the point where", "tree method. 22. Find the LCM of 156 and 124. 23. There are four Mobile Phones in a house. At 5 a.m, all the four Mobile Phones will ring together. Thereafter, the first one rings every 15 minutes, the second one rings every 20 minutes, the third one rings every 25 minutes and the fourth one rings every 30 minutes. At what time, will the four Mobile Phones ring together again?", "What are the opening hours of the Wotruba Church? $\\hspace{10mm}$ Saturdays: 14:00-20:00 Sundays and holidays: 09:00-14:30 Closed on weekdays! There are different activities taking place every so often. You might wanna have a look at their website for an up to date calendar! $\\hspace{10mm}$", "# Introduction A bell tower will ring its bells every hour, n times, with n being the the current hour on a 12 hour clock. For example, a bell will ring 5 times at 5pm, and 10 times at 10am. Given two times in a suitable format, output the number of times the bell will ring, inclusive of the start and end times # Examples \"10am-12pm\" 10+11+12= 33 [01:00, 05:00] 1+2+3+4+5 = 15 [11, 15] 11+12+1+2+3 = 29 [10:00pm, 10:00am] 10+11+12+1+2+3+4+5+6+7+8+9+10 = 88 If the start is the same as the end then the you simply just ouput the number of chimes for that hour: [5pm, 5pm] 5 = 5 As you can see, you may choose an input method but the output must be an integer on its own (or an acceptable alternative) trailing/ leading newlines and spaces are allowed. Note: • inputs may span from the afternoon of one day to the morning of the next. • the difference between the two times will never be more than 24 hours. • input is flexible so long as you clearly state what format your input is. • your input must have a clear distinction between AM and PM. • Do we choose our own input method, or does it have to support all the mentioned? – anonymous2" ]

[ "lowest common multiple is, $$LCM(a,~b)=\\frac{\\left | ~a.b~ \\right |}{GCF(a,b)}$$ $$LCM(30,~150)=\\frac{\\left | ~30~\\times~ 150~ \\right |}{GCF(30,\\ 150)}$$ [Put the values] $$=\\frac{\\left | ~30~\\times ~150 ~\\right |}{30}$$ [Dividing 30 by 30] = 150 Hence, the LCM of 150 and 30 is 150. Example 3: The bell at St. Albert’s School is programmed to ring every 50 minutes for students in grades 1 to 5. While another bell will ring once every hour for students in grades 6 to 10. How many times in the next 10 hours will both the bells ring simultaneously? Solution: According to the given information, let ‘a’ be the bell for the grades 1 to 5 that rings for every 50 mins. And the bell ‘b’ is for grades 6 to 10 that rings for every 1 hr (60 mins). In order to find how many times in the next 12 hours both the bells ring, we have to find the LCM of both the bells. Then, let’s find the prime factors of 50 and 60. 50 = 2 x 5 x 5 60 = 2 x 2 x 3 x 5 The product of the common factors of 50 and 60 are 5 x 2 = 10. Thus, 10 is the GCF of 50 and 60. The LCM of 50 and 60 can be find", "# The first bell rings every 20 minutes, second bell rings every 30 minutes, and the third bell rings every 50 minutes. If all three bells ring the same time at 12:00pm, when will be the next time the three bells will ring together? Feb 20, 2017 $\\text{5:00 pm}$ #### Explanation: So first you find the LCM, or least common multiple, (can be called LCD, least common denominator). The LCM of $20$, $30$, and $50$ is basically $10 \\cdot 2 \\cdot 3 \\cdot 5$ because you factor out the $10$ since that is a common factor. $10 \\cdot 2 \\cdot 3 \\cdot 5 = 300$ This is the number of minutes. To find the number of hours, you simply divide by $60$ and get $5$ hours. Then you count $5$ more hours from $\\text{12:00 pm}$ and get $\\text{5:00 pm}$. Feb 20, 2017 5pm #### Explanation: $\\textcolor{b l u e}{\\text{Expanding on Ayushi's answer.}}$ Notice that we have: $10 \\times 2$ $10 \\times 3$ $10 \\times 5$ Each of 2, 3 and 5 are prime numbers. So the only common values they will divide exactly into is their product or some multiple of that product So for 2,3 and 5 the least positive value they will divide into is: $2 \\times 3 \\times 5 = 30$ but each of 2,3,and", "# Question #18833 Sep 15, 2016 $9 : 04 : 40$ pm. #### Explanation: The next time they will flash again will be $n$ seconds after $9 : 00 : 00$ pm, where $n$ is the least common multiple (LCM) of $56$ and $40$. To find the LCM, we look at the prime factorizations of the two values and take the least product of primes which is divisible by both. $56 = {2}^{3} \\times 7$ $40 = {2}^{3} \\times 5$ We can see that the LCM requires at least three factors of $2$, one factor of $5$, and one factor of $7$. Thus $L C M \\left(40 , 56\\right) = {2}^{3} \\times 5 \\times 7 = 280$. As $280$ seconds is equal to $4$ minutes and $40$ seconds, we can add that value to $9 : 00 : 00$ pm to find that the next time they flash at the same time is at $9 : 04 : 40$pm. Note that at that time, the first lighthouse will have flashed $\\frac{280}{56} = 5$ times, and the second will have flashed $\\frac{280}{40} = 7$ times.", "second device = 62 seconds Therefore, the Interval of beeping together = LCM (60, 62) Prime factorization: 60 = $2 ^{ 2 }$ x 3 x 5 62 = 2 x 31 Therefore, the LCM = $2^{2}$ x 3 x 5 x 31 = 1860 seconds = $\\frac{ 1860 }{ 60 }$ = 31 min Hence, they will beep together again at 10 : 31 a.m. Q.19: Six bells commerce tolling together and toll at intervals of 2 , 4 , 6 , 8 ,10 , 12 minutes respectively. In 30 hours, how many times do they toll together ? Sol: Six bells toll together at intervals of 2, 4, 6, 8, 10 and 12 minutes, respectively. Prime factorization: 2 = 2 4 = 2 x 2 6 = 2 x 3 8 = 2 x 2 x 2 10 = 2 x 5 12 = 2 x 2 x 3 Therefore, the LCM ( 2, 4, 6, 8, 10, 12 ) = $2^{3}$ x 3 x 5 = 120 Hence, after every 120 minutes (i.e. 2 hours), they will toll together. Therefore, the Required number of times = ($\\frac{30}{2}$ + 1) = 16 Q.20: Find the missing numbers in the following factorization. Sol: 660 = 2 X 2 X 3 X 5 X 11 #### Practise This", "## Thinking Mathematically (6th Edition) Published by Pearson # Chapter 5 - Number Theory and the Real Number System - 5.1 Number Theory: Prime and Composite Numbers - Exercise Set 5.1: 117 #### Answer Two movies will start at the same time again after: $2300$ minutes or $38$ hours and $20$ minutes after. #### Work Step by Step To know when the two movies will start again at the same time after noontime, one must use the least common multiple (LCM). Since there is a 15-minute break between shows, find the LCM of: $85 + 15 = 100$ minutes and $100 +15 = 115$ minutes Factor each number completely : $100 = 10(10) = 2(5) \\cdot (2)(5) = 2(2)(5)(5)=2^2\\cdot 5^2 \\\\115 = 5(23)$ Thus, the LCM is $2^2 \\cdot 5^2 \\cdot 23=2300$ minutes. $2300$ minutes is $38$ hours and $20$ minutes. Thus, the two movies will start at the same time $2300$ minutes or $38$ hours and $20$ minutes after. 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.", "lcm: LCM = 2 × 2 × 2 × 2×3 × 7 =336 42 = $2 \\times 3 \\times 7$ 63 = $3 \\times 3 \\times 7$ 105 = $3 \\times 5 \\times 7$ Taking the common factors, the HCF would be 21. So , the answer would be option d)21. Given that three bells rang at intervals of 15, 30 and 45 minutes respectively. They will together ring again in 90 minutes (LCM of 15, 30, 45). 90 minutes = 01 : 30 hours Then, After 8 : 00 AM, they will again ring together in 08 : 00 + 01 : 30 = 09 : 30 AM $12=2\\times 2\\times 3$ $18=2\\times 3\\times 3$ $21 =3\\times 7$ $30 =2\\times 3\\times 5$ Select the common terms LCM = $2\\times 2\\times 3\\times 3\\times 5\\times 7$ = 1260 $ab^{2}c^{2}$, $a^{2}bc$ and $a^{3}b^{3}c^{2}$. LCM $=(abc)\\times(bc)\\times(a)\\times(ab)$ $=a^{3}b^{3}c^{2}$", "be a multiple of 16. Likewise, $$t$$ must also be a multiple of 20. We now have to find the LCM of 16 & 20, and to do that, we can use the nifty GMAT trick. [Reveal] Spoiler: Finding the LCM Find the prime factorization of 16: 2*2*2*2 Find the prime factorization of 20: 2*2*5 Notice the factors in common are 2*2. Remove these from each prime factorization, and multiply what is left with the ones in common. (2*2)(5)(2*2) = 20*4 = 80. Therefore, 80 is the LCM and must be the answer. 80 minutes = 1 hour and 20 minutes. --- The Short Way: This one can also be solved much quicker by just checking the possibilities. $$\\frac{5}{16}$$ *15 minutes = not an integer; $$\\frac{5}{16}$$ * 30 minutes = not an integer; $$\\frac{5}{16}$$ * 40 minutes = not an integer; $$\\frac{5}{16}$$ * 60 minutes = not an intetger; $$\\frac{5}{16}$$ * 80 minutes = 25[/m] <-- this is our only integer. Therefore it must be the correct answer. Kudos [?]: 114 [2], given: 42 Manager Joined: 10 Sep 2014 Posts: 99 Kudos [?]: 51 [0], given: 25 Re: Two circles, one with radius 10 inches and the other with ra [#permalink] ### Show Tags 17 Nov 2014, 11:17 1. circumference of a circle: 2Rpi circumference of circle A", "25 = 5 $$\\times$$ 5 5 appears most often here, 50 = 2 $$\\times$$ 5 $$\\times$$ 5 2 appears only once here, so highlight it, LCM = 2 $$\\times$$ 3 $$\\times$$ 5 $$\\times$$ 5 So the LCM of 15, 25 and 50 is 150. ## Real-life Model Question Ryan and Lisa are on a running track. Ryan completes one lap in 4 minutes and Lisa completes it in 6 minutes. If they start together, after how many minutes will they meet again at the starting point? Solution : You are given the number of minutes Ryan and Lisa take to complete a round. The LCM of the number of minutes will give us the time when they will meet again at the starting point. So, let’s find the LCM of 4 and 6. Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 Clearly the LCM of 4 and 6 is 24. Hence, Ryan and Lisa will meet again at the starting point after 24 minutes.", "take the highest power on the lists and multiply everything together. For example, take 12 and 45. $12 = 2^2 \\times 3^1$ $45 =3^2 \\times 5^1$ $\\lcm(12,45) = 2^2 \\times 3^2 \\times 5^1 = 180$ The process for finding the GCF is similar. Start off by doing the prime factorization for both numbers. Then, for each shared prime factor, take the smallest power and multiply everything together. For example, take 12 and 30. $12 = 2^2 \\times 3^1$ $30 =2^1 \\times 3^1 \\times 5^1$ $\\gcd(12,45) = 2^1 \\times 3^1 = 6$ This process generalizes very easily for any amount of input numbers. GCF and LCM are incredibly important when working with fractions and are used when reducing and adding fractions. Because fractions have loads of misconceptions associated with them, giving students another way to understand fractions can be very beneficial. Technology Have you ever wondered why we use 60 seconds in a minute and 60 minutes in an hour? Or why there is 24 hours in a day? What about why there is 360 degrees in a circle? One explanation is because these numbers can be divided evenly by loads of smaller numbers that we use often. In other words, these numbers have lots of factors in them. These kinds of numbers are called highly composite numbers. A", "use of the LCM in the future. ## An example using two methods Here’s a 2007 question: Great Explanation of How to Find the Least Common Multiple If you set (3) alarm clocks to go off at (3) different times--when will they all ring at the same time? For example, set clock #1 to ring at 25 min. / clock #2 to ring at 30 min. / clock #3 to ring at 35 min. How many minutes will it take for them all to ring at the same time? I was told the answer of 1050 minutes, but I am having trouble figuring out the formula. This is a common reason to use the LCM; the question ultimately is just about calculating the LCM of 25, 30, and 35. ### 2b: Prime factors (list method) Doctor Ian answered, first with a second variation of the prime factor method: Hi Dustin, The \"formula\" is really more of a concept. You want to find the smallest number that is a multiple of 25, 30, and 35, usually called the Least Common Multiple (or LCM) of the numbers. Multiplying the numbers together gives A common multiple, but not necessarily the LEAST one. Let's look at the prime factors: 25 = 5 * 5 30 = 2 * 3 * 5 35", "× 2 × 2 × 2 × 3 = 24 × 3 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 108 = 2 × 2 × 3 × 3 × 3= 22 × 3 3 LCM = 432 or 432 sec Convert into mins: 432 sec = 7 mins 12 sec So, at 8:7:12 hrs traffic lights will again change. Question 27: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together? Solution: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. Find LCM using prime factorization: 2 = 2 (prime number) 4 = 2 × 2 6 = 2 × 3 8 = 2 × 2 × 2 10 = 2 × 5 12 = 2 × 2 × 3 Therefore, LCM (2, 4, 6, 8, 10, 12) = 2x 2 x 2 × 3 × 5 = 120 After every 120 minutes = 2 hours, bells will toll together. So, required number of times = (30/2 + 1) = 16 times. ## Exercise 1C Page No: 24 Question 1: Without actual division, show that each of the following rational number is a", "of 8 : 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, . . . Multiples of 10 : 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, . . . The common multiples are 40, 80, 120, . . . Hence, the least common multiple of 5, 8 and 10 is 40. Example 2: Find the least common multiple of 12, 15 and 20 using prime factorization. Solution: The factor tree of 12, 15 and 20 can be drawn as in the images. Prime factorization of 12 : 2 $$\\times$$ 2 $$\\times$$ 3 Prime factorization of 15 : 3 $$\\times$$ 5 Prime factorization of 20 : 2 $$\\times$$ 2 $$\\times$$ 5 LCM : 2 $$\\times$$ 2 $$\\times$$ 3 $$\\times$$ 5 = 60 Hence, the least common multiple of 12, 15 and 20 is 60. Example 3: John and Tom are running on a track. John completes one lap in 5 minutes and Tom completes it in 7 minutes. If they start together, after how long will they meet again at the starting point? Solution: John and Tom take 5 minutes and 7 minutes respectively to complete one lap. To calculate the time taken to meet at the starting point, we need to find the LCM of 5 and 7. Multiples of 5 : 5, 10,", "## Thinking Mathematically (6th Edition) Two movies will start at the same time again after: $2300$ minutes or $38$ hours and $20$ minutes after. To know when the two movies will start again at the same time after noontime, one must use the least common multiple (LCM). Since there is a 15-minute break between shows, find the LCM of: $85 + 15 = 100$ minutes and $100 +15 = 115$ minutes Factor each number completely : $100 = 10(10) = 2(5) \\cdot (2)(5) = 2(2)(5)(5)=2^2\\cdot 5^2 \\\\115 = 5(23)$ Thus, the LCM is $2^2 \\cdot 5^2 \\cdot 23=2300$ minutes. $2300$ minutes is $38$ hours and $20$ minutes. Thus, the two movies will start at the same time $2300$ minutes or $38$ hours and $20$ minutes after.", "can be measured an exact number of times = LCM (64, 80, 96) Prime factorisation: 64 = ${2}^{6}$ 80 = 96 = ∴ LCM = Product of greatest power of each prime factor involved in the numbers = = 960 cm = 9.6 m Hence, the required length of cloth is 9.6 m. #### Page No 19: Beep duration of first device = 60 seconds Beep duration of second device = 62 seconds ∴ Interval of beeping together = LCM (60, 62) Prime factorisation: 60 = 62 = ∴ LCM = seconds = $\\frac{1860}{60}$ = 31 min Hence, they will beep together again at 10 : 31 a.m. #### Page No 19: We find the LCM of 48, 72 and 108 first to get the time after which they will blink together again. Hence, LCM = $2×2×2×2×3×3×3=432$ So, they will blink again at 432 seconds past 8:00 am or, $\\frac{432}{60}=$7 minutes and 12 seconds past 8:00 am So, the time will be 08:07:12 hrs #### Page No 19: Six bells toll together at intervals of 2, 4, 6, 8, 10 and 12 minutes, respectively. Prime factorisation: ∴ ​ = Hence, after every 120 minutes (i.e. 2 hours), they will toll together. ∴ Required number of times = #### Page No 26: (i) = We know either 2 or", "= 240 ∴ The least number of beads with Veena are 240. iii. The lowest common multiple of 20,24 and 12 gives the minimum number of laddoos in one box. Multiples of 20 = 20, 40, 60, 80, 100,120, 140, 160, 180, 200 Multiples of 24 = 24, 48, 72, 96, 120 Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120 ∴ LCM of 20, 24 and 12 = 120 ∴ Minimum number of ladoos in 1 boxes =120 ∴ Minimum number of ladoos in 3 boxes = 3 x 120 = 360 ∴ The minimum number of ladoos in 3 boxes are 360. iv. All three signals will turn green for lowest common multiple of 60 seconds, 120 seconds and 24 seconds. Multiples of 60 = 60, 120, 180, 240, 300, 360, 420, 480 Multiples of 120 = 120, 240, 360 Multiples of 24 = 24, 48, 72, 96, 120 LCM of 60, 120 and 24 = 120 Since, 60 seconds = 1 minute ∴ 120 seconds = 2 minutes ∴ The signals will turn green simultaneously again after 120 seconds i.e. 2 minutes. v. The lowest common multiple of 45 and 75 gives the same denominator. Multiples of 45 = 45, 90, 135, 180, 225, 270, 315, 360, 405, 450 Multiples", "## Elementary Algebra The prime factored form of 15 is: $15=3\\times5$ The prime factored form of 27 is: $27=3\\times3\\times3$ The prime factored form of 35 is: $35=5\\times7$ Since the least common multiple (LCM) contains each different prime factor as many times as the most times it appears in any one of the factorizations, we obtain: LCM$=3\\times3\\times3\\times5\\times7=945$", "each item should he buy? (5 marks) Show answer 6 = 2 \\times 3, 8 = 2 \\times 2 \\times 2 and 10 = 2 \\times 5 (1) LCM (6, 8, 10) = 3 \\times 2 \\times 2 \\times 2 \\times 5 = 120 (1) 120 \\div 6 = 20 packs of bread rolls (1) 120 \\div 8 = 15 packs of burgers (1) 120 \\div 10 = 12 packs of cheese (1) 2. The machinery in a factory produces a lot of noise. A heavy metal hammer pounds every 12 seconds, an alarm repeats every 18 seconds and a belt clicks every 32 seconds. If they all make a synchronised sound, how many minutes will have passed before they all make a sound together again? (4 marks) Show answer At least two of: Prime factors of 12 = 2 \\times 2 \\times 3 , Prime factors of 18 = 2 \\times 3 \\times 3 , Prime factors of 32 = 2 \\times 2 \\times 2 \\times 2 \\times 2 (1) (1) LCM = 288 seconds (1) 288 \\div 60 = 4.8 minutes (or 4 minutes, 48 seconds) (1) 3. Two numbers have a LCM of 175 and a HCF of 5. What are the two numbers? (4 marks) Show answer (1) 175\\div 5 = 35 For (1)", "# Lcm of three numbers What is the Lcm of 120 15 and 5 Correct result: x = 120 #### Solution: $120=2^3 \\cdot 3 \\cdot 5 \\ \\\\ 15=3 \\cdot 5 \\ \\\\ 5 \\ ... \\ prime number \\ \\\\ LCM(120, 15, 5)=2^3 \\cdot 3 \\cdot 5=120 \\ \\\\ \\ \\\\ x=LCM(120,15,5)=120$ 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 Do you want to calculate least common multiple two or more numbers? ## Next similar math problems: • Game room Winner can took win in three types of jettons with value 3, 30 and 100 dollars. What is minimal value of win payable in this values of jettons? • 3 buses At morning 5:00 am three buses started from one place. The first travel in five-minute intervals, the second at 10-minute intervals and the third at 25-minute intervals. At what hour will again be the three buses coming from the same place? • Two rulers We have two rulers. Scale interval on first rulers are a spaced 1 cm and on second spaced 15 mm. Rulers are attached to each other so that they match initial divider commas. What next dividers commas coincide? Find at least", "the resultant remainder will be of the form 25k + 13 and 4m The only two digit number which fits the relation 25k + 13 = 4m is 88. When 88 is divided by 13, the required remainder is 10. $(101010101)_2$ = $1(2^8)+0(2^7)+1(2^6)+0(2^5)+1(2^4)+0(2^3)+1(2^2)+0(2^1)+1(2^0)$ = $256+0+64+0+16+0+4+0+1 = (341)_{10}$ So the answer is option A. They strike at the intervals of 3, 4 and 8 minutes. They will strike together every 24 minutes (LCM of 3, 4, 8). In a day, there are 24*60 minutes. Thus, they will strike together $\\frac{24*60}{24} = 60$ times. We have been given that $k^2 – 21k = 100$ => $k^2 – 25k + 4k – 100 = 0$ => $k(k – 25) + 4(k – 25)$ => $(k + 4)(k – 25) = 0$ => k = -4 or k = 25 Since k is positive, so k must be 25. Hence, the remainder is 4 when it is divided by 7 The required smallest number = LCM(14, 16, 18, 21, 32) + 9 = 2016 + 9 = 2025 Hence, option A is the right answer. After the first beep together all the 5 devices would beep together next time after LCM(3, 5, 8, 12, 20) seconds. LCM(3, 5, 8, 12, 20) = 120 seconds or 2 minutes Number of beeps together in", "\\:=\\:\\frac{3(5m-1)}{7}$ .[1] Since $n$ is an integer, $5m-1$ is a multiple of 7. The first time this happens is $m = 10$. .Then: . $n = 21$. (e) They wink together $3m = 300$ seconds after midnight . . . at 12:05 a.m. The next time [1] gives integral $n$ is when $m = 17\\,\\:n = 36$. (f) They wink together $3m = 510$ seconds after midnight . . . at 12:08½ a.m." ]

[ "lowest common multiple is, $$LCM(a,~b)=\\frac{\\left | ~a.b~ \\right |}{GCF(a,b)}$$ $$LCM(30,~150)=\\frac{\\left | ~30~\\times~ 150~ \\right |}{GCF(30,\\ 150)}$$ [Put the values] $$=\\frac{\\left | ~30~\\times ~150 ~\\right |}{30}$$ [Dividing 30 by 30] = 150 Hence, the LCM of 150 and 30 is 150. Example 3: The bell at St. Albert’s School is programmed to ring every 50 minutes for students in grades 1 to 5. While another bell will ring once every hour for students in grades 6 to 10. How many times in the next 10 hours will both the bells ring simultaneously? Solution: According to the given information, let ‘a’ be the bell for the grades 1 to 5 that rings for every 50 mins. And the bell ‘b’ is for grades 6 to 10 that rings for every 1 hr (60 mins). In order to find how many times in the next 12 hours both the bells ring, we have to find the LCM of both the bells. Then, let’s find the prime factors of 50 and 60. 50 = 2 x 5 x 5 60 = 2 x 2 x 3 x 5 The product of the common factors of 50 and 60 are 5 x 2 = 10. Thus, 10 is the GCF of 50 and 60. The LCM of 50 and 60 can be find", "lcm: LCM = 2 × 2 × 2 × 2×3 × 7 =336 42 = $2 \\times 3 \\times 7$ 63 = $3 \\times 3 \\times 7$ 105 = $3 \\times 5 \\times 7$ Taking the common factors, the HCF would be 21. So , the answer would be option d)21. Given that three bells rang at intervals of 15, 30 and 45 minutes respectively. They will together ring again in 90 minutes (LCM of 15, 30, 45). 90 minutes = 01 : 30 hours Then, After 8 : 00 AM, they will again ring together in 08 : 00 + 01 : 30 = 09 : 30 AM $12=2\\times 2\\times 3$ $18=2\\times 3\\times 3$ $21 =3\\times 7$ $30 =2\\times 3\\times 5$ Select the common terms LCM = $2\\times 2\\times 3\\times 3\\times 5\\times 7$ = 1260 $ab^{2}c^{2}$, $a^{2}bc$ and $a^{3}b^{3}c^{2}$. LCM $=(abc)\\times(bc)\\times(a)\\times(ab)$ $=a^{3}b^{3}c^{2}$", "# LCM and HCF Question Types In this article, we will try to cover all the types of aptitude questions that are framed on the concepts of LCM and HCF. ## Type 1: Application of LCM ### Type 1a ##### Q. There are two bells in a temple. One tolls 3 times per minute and the other tolls 4 times per minute, once the temple is opened. Calculate how many times will they toll together in first 30 minutes, after the temple is opened? (a) 29 (b) 30 (c) 31 (d) 32 Explanation: One bell tolls 3 times in a minute. Thus, it tolls every 60/3 = 20 seconds. Second bell tolls 4 times in a minute. Thus, it tolls every 60/4 = 15 seconds. The time interval in which they will toll together = LCM (15, 20) = 60 seconds = 1 minute. Thus, in 30 minutes they will toll together 30 times. ### Type 1b ##### Q. If LCM of two numbers is 198, then what is the least possible value of their sum? (a) 30 (b) 31 (c) 32 (d) 29 Explanation: We know that the two numbers are multiples of the factors of the given LCM. So, first we need to find out the factors of the given LCM. Now, 198 = 2 ×", "2016 #### LCM Question 10 Detailed Solution Given: We need to find the sum of numbers in the range of 400 - 600 such that they are divisible by each 6, 12 and 16. Concept: LCM (Lowest Common Multiple) Calculation: LCM (6, 12, 16) = 48 The required numbers will be in the form of 48k, where k is a natural number. For k = 9, 48k = 48 × 9 = 432 For k = 10, 48k = 48 × 10 = 480 For k = 11, 48k = 48 × 11 = 528 For k = 12. 48k = 48 × 12 = 576 ∴ The sum of these 4 numbers that is, 432, 480, 528, and 576 is 2016. #### LCM Question 11 5 bells tolls at an interval of 2, 4, 6, 8 and 12 minutes respectively. If all 5 bells tolls together at 10 am, then how many times they will toll together from 10 am to 12 noon? 1. 5 2. 6 3. 4 4. 8 Option 2 : 6 #### LCM Question 11 Detailed Solution Given: 5 bells toll at an interval of 2, 4, 6, 8, and 12 minutes Calculation: The times after which the bells ring together must be equal to the LCM of 2,4,6,8,12 and the LCM of", "= 480 If four bells ring together in starting ⇒ 480 + 1 ∴ The bell ringing 481 times in 8 hours. Mistake PointsThe bells start tolling together, the first toll also needs to be counted, that is the number of times of tolling since the first time. #### LCM Question 7 The HCF and LCM of two numbers are 24 and 168 and the numbers are in the ratio 1 ∶ 7. Find the greater of the two numbers. 1. 168 2. 144 3. 108 4. 72 Option 1 : 168 #### LCM Question 7 Detailed Solution Given: HCF = 24 LCM = 168 Ratio of numbers = 1 ∶ 7. Formula: Product of numbers = LCM × HCF Calculation: Let numbers be x and 7x. x × 7x = 24 × 168 ⇒ x2 = 24 × 24 ⇒ x = 24 ∴ Larger number = 7x = 24 × 7 = 168. #### LCM Question 8 Six bells commence tolling together, afterwards, they toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? 1. 4 2. 10 3. 15 4. 16 Option 4 : 16 #### LCM Question 8 Detailed Solution Given: The toll intervals of 6 bells = 2, 4, 6,", "5 LCM(5, 15, 20) = 5 × 3 × 2 × 2 = 60 HCF(5, 15, 20) = 5 45. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time ? (a) 12.20 pm (b) 12.12 pm (c) 12.11pm (d) none of these Answer/ Explanation Answer: d Explaination: LCM of 50 and 48 = 1200 ∴ 1200 sec = 20 min Hence at 12.20 pm they will beep again for the first time. 46. If A = 2n + 13, B = n + 7, where n is a natural number, then HCF of A and B is: (a) 2 (b) 1 (c) 3 (d) 4 Answer/ Explanation Answer: b Explaination: Taking different values of n we find that A and B are coprime. ∴ HCF = 1 47. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are: (a) 22 (b) 16 (c) 36 (d) 21 Answer/ Explanation Answer: b Explaination: HCF of 576 and 448 = 64 ∴ Number of sections = $=\\frac{576}{64}+\\frac{448}{64}$ = 9 + 7 = 16 48.", "Three bells $P$, $Q$ and $R$ are rung periodically in a school. $P$ is rung every $20$ minutes; $Q$ is rung every $30$ minutes and $R$ is rung every $50$ minutes. If all the three bells are rung at $12 : 00$ $\\text{PM}$, when will the three bells ring together again the next time? 1. $5 : 00$ $\\text{PM}$ 2. $5 : 30$ $\\text{PM}$ 3. $6 : 00$ $\\text{PM}$ 4. $6 : 30$ $\\text{PM}$ Given that, • $\\text{P}$ is rung every $20$ minutes • $\\text{Q}$ is rung every $30$ minutes • $\\text{R}$ is rung every $50$ minutes Now, ${\\color{Green}{\\text{LCM(50, 30, 50) = 300 minutes = 5 hours}}.}$ If all the three bells are rung at $12: 00\\;\\text{PM}$, when will after that all the three bells ring together again the next time $= 12:00\\;\\text{PM} + 5\\;\\text{hours} = 5:00\\;\\text{PM}.$ Correct Answer $:\\text{A}$ by 4.1k points", "× Custom Search cancel × × # Solved Examples(Set 6) - HCF and LCM 26. What is the greatest number which divides 24, 28 and 34 and leaves the same remainder in each case? A. 2 B. 1 C. 3 D. 4 Explanation: If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number. 34 - 24 = 10 34 - 28 = 6 28 - 24 = 4 Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder = HCF of 10, 6, 4 = 2 27. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. How many times will they ring together in 60 minutes ? A. 31 B. 15 C. 30 D. 16 Explanation: LCM of 4, 8, 10, 12, 15 and 20 = 120 120 seconds = 2 minutes Hence all the six bells will ring together in every 2 minutes Hence, number of times they will ring together in 60 minutes $=1+\\dfrac{60}{2}=31$ 28. What is the least number which when divided by 8, 12, 15 and 20 leaves in each case a remainder of 5 ? A. 125 B. 112 C. 117 D. 132", "# The first bell rings every 20 minutes, second bell rings every 30 minutes, and the third bell rings every 50 minutes. If all three bells ring the same time at 12:00pm, when will be the next time the three bells will ring together? Feb 20, 2017 $\\text{5:00 pm}$ #### Explanation: So first you find the LCM, or least common multiple, (can be called LCD, least common denominator). The LCM of $20$, $30$, and $50$ is basically $10 \\cdot 2 \\cdot 3 \\cdot 5$ because you factor out the $10$ since that is a common factor. $10 \\cdot 2 \\cdot 3 \\cdot 5 = 300$ This is the number of minutes. To find the number of hours, you simply divide by $60$ and get $5$ hours. Then you count $5$ more hours from $\\text{12:00 pm}$ and get $\\text{5:00 pm}$. Feb 20, 2017 5pm #### Explanation: $\\textcolor{b l u e}{\\text{Expanding on Ayushi's answer.}}$ Notice that we have: $10 \\times 2$ $10 \\times 3$ $10 \\times 5$ Each of 2, 3 and 5 are prime numbers. So the only common values they will divide exactly into is their product or some multiple of that product So for 2,3 and 5 the least positive value they will divide into is: $2 \\times 3 \\times 5 = 30$ but each of 2,3,and", "mins = 31 mins Electronic devices will beep after every 31 minutes. Hence, both devices will beep together again at 10:31 a.m. again. Question 26: The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m., then at what time will they again change simultaneously? Solution: The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. Find the LCM of 48, 72 and 108: 48 = 2 × 2 × 2 × 2 × 3 = 24 × 3 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 108 = 2 × 2 × 3 × 3 × 3= 22 × 3 3 LCM = 432 or 432 sec Convert into mins: 432 sec = 7 mins 12 sec So, at 8:7:12 hrs traffic lights will again change. Question 27: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together? Solution: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. Find LCM using prime factorization: 2 = 2 (prime number)", "# The first bell rings every 20 minutes, second bell rings every 30 minutes, and the third bell rings every 50 minutes. If all three bells ring the same time at 12:00pm, when will be the next time the three bells will ring together? 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 Feb 22, 2017 5pm #### Explanation: $\\textcolor{b l u e}{\\text{Expanding on Ayushi's answer.}}$ Notice that we have: $10 \\times 2$ $10 \\times 3$ $10 \\times 5$ Each of 2, 3 and 5 are prime numbers. So the only common values they will divide exactly into is their product or some multiple of that product So for 2,3 and 5 the least positive value they will divide into is: $2 \\times 3 \\times 5 = 30$ but each of 2,3,and 5 is multiplied by 10 so we have to also multiply their product by 10 giving: $10 \\times 30 = 300$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\\textcolor{b l u e}{\\text{A different line of thinking that ends up in the same place}}$ 3 and 5 are odd numbers but 2 is even. As 2 is even then the $\\textcolor{b r o w n}{\\underline{\\text{target value has to also be even}}}$. Otherwise 2 will", "# NTT Data LCM and HCF Quiz -1 Question 1 Time: 00:00:00 If the HCF of two numbers is 4 and their LCM is 48 . One of the numbers is 16. Then find another number. 12 12 14 14 16 16 18 18 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 2 Time: 00:00:00 Find the largest number that can exactly divide 52 , 70 and 72. 2 2 3 3 1 1 None None Once you attempt the question then PrepInsta explanation will be displayed. Start Question 3 Time: 00:00:00 Find LCM of 24 , 42 , 54 and 72 756 756 3124 3124 3024 3024 1512 1512 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 4 Time: 00:00:00 Find HCF of 25 , 45 and 200 5 5 3 3 2 2 1 1 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 5 Time: 00:00:00 A bell rings every 10 minutes. A second bell rings every 20 minutes. A third bell rings every 25 minutes, A fourth rings every 50 minutes. If all the four bells ring at the same time at 9: 30 in the evening, at what other time will they all ring together? 12 :20 12 :20", "# URGENT HELP PLEASE at 10am three bells ring together. The large bell rings every four minutes, the old bell rings every 7 minutes and the small bell rings every three minutes. When will be the next time that all the three bells ring together? ###### Question: at 10am three bells ring together. The large bell rings every four minutes, the old bell rings every 7 minutes and the small bell rings every three minutes. When will be the next time that all the three bells ring together? ### His awkwardness at digging the hole with a shovel whose handle was twice as long as he was made us laugh, and we covered our mouths with our hands so he wouldn't hear.\" The reader can infer that Doodle's family______________________________. His awkwardness at digging the hole with a shovel whose handle was twice as long as he was made us laugh, and we covered our mouths with our hands so he wouldn't hear.\" The reader can infer that Doodle's family______________________________.... ### Balance Ca(OH)2 + H3PO4 yields Ca3(PO4)2 + H2O Balance Ca(OH)2 + H3PO4 yields Ca3(PO4)2 + H2O... ### Read the following sentence. More students voted for healthy lunch choices at school, making it clear that they were in the majority. Which of the following phrases provides a context clue", "ringing continuously at intervals of 30, 40 and 45 minutes respectively. At what time will they ring together if they ring simultaneously at 5 A.M.? LCM = 2 × 2× 2 × 3 × 3 × 5 = 360 After = 360 minutes 360 minutes = 6 hours They will ring together again at 11 a.m. — End of Model Question Paper-2 Class-6 ML Aggarwal Solutions :–", "= 240 b = 240/12 b = 20. Practice questions 1. After a budget planning consultation Mr. A, B and C decided that they will change their car after every 16 months, 20 months and 30 months respectively. If they all bought their cars together in January 10, 2017, when next will the buy together? 2. The hcf of 18 and another number is 6, if their l.c.m is 72, find the number. 3. The lcm of 6 and another number is 12. if their h.c.f is 2, find the number. 4. Three vigilante men usually blow their whistles every 20, 15 and 25 minutes. If they blow together at 10:00pm, when next will they blow together? PRE: lcm of algebraic expression PRE", "41 grams D. 30 litres Solution : Prime factorization of 403, 434 and 465 is 403=13×31 434=2×7×31 465=3×5×31 H.C.F of 403, 434 and 465=31 Correct Option : A ### Type 5 :Tips , tricks and Shortcuts of HCF, LCM Problems related to Bell ring. Question: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? Options : A. 8 B. 16 C. 9 D. 10 Solution : L.C.M. of 2, 4, 6, 8, 10, 12 is 120. Hence, the bells will toll together after every 120 seconds(2 minutes). Therefore, in 30 minutes ,number of times bells toll together is $\\frac{30}{2} + 1$ = 16 Correct Option B ### Type 6 : Tips , tricks and Shortcuts of HCF, LCM Problems related to Circle Based Runner Problem. Question: Two people P and Q start running towards a circular track of length 400 m in opposite directions with initial speeds of 10 m/s and 40 m/s respectively. Whenever they meet, P’s speed doubles and Q’s speed halves. After what time from the start will they meet for the third time? Options A. 30 seconds B. 26 seconds C. 10 seconds D. 20 seconds Solution : Time taken to meet for", "tree method. 22. Find the LCM of 156 and 124. 23. There are four Mobile Phones in a house. At 5 a.m, all the four Mobile Phones will ring together. Thereafter, the first one rings every 15 minutes, the second one rings every 20 minutes, the third one rings every 25 minutes and the fourth one rings every 30 minutes. At what time, will the four Mobile Phones ring together again?", "× 5 × 3 = 540 Required number = 9999 – 279 = 9720 Question 16. Find the least number which when divided by 6, 7, 8, 9 and 12 leaves the same remainder 1 in each case. Solution: Required number = [LCM (6, 7, 8, 9, 12)] + 1 LCM (6, 7, 8, 9, 12) = 3 × 2 × 2 × 2 × 3 × 7 = 504 Required number = 504 + 1 = 505 Question 17. Product of two co-prime numbers is 117. Then what will be their LCM? Solution: We know that LCM × HCF = Product of two numbers Also, we know that HCF of two co-primes = 1 LCM × 1 = 117 LCM = 117 Question 18. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? Solution: LCM of 2, 4, 6, 8, 10 and 12 is 120 So the bell will toll together after every 120 seconds i.e 2 minutes. In 30 minutes, they will toll together $$\\frac{30}{2}$$ + 1 = 16 times. LCM = 2 × 2 × 3 × 2 × 5 = 120", "third every $7$7 seconds. How long after the start before all $3$3 bells ring together again?", "x 5 = 60 Therefore, they will ring together after 60 minutes. 2. Smart, Joy and Ken usually visit Mama T's cafe every 25 minutes, 30 minutes and 50 minutes respectively. If the three friends met at the cafe at 11:20 am, what time will they meet again? Solution Find the lcm of 25, 30 and 50 Lcm = 150 That means they will meet every 150 minutes, to get the required time, convert the 150 minutes to hours = 150/60 = 2 hours 30 minutes. Add 2 hours 30 mins to 11:20 am = 13:50 Which is 01:50 pm. 3. The l.c.m of 12 and another number is 60. If their h.c.f is 4, find the number. Solution Product of the two numbers = product of their l.c.m and h.c.f i.e. if the two numbers are a and b, then a x b = l.c.m x h.c.f in the above question, a = 12, b = ?, l.c.m = 60, h.c.f = 4. 12 x b = 60 x 4 12b = 240 b = 240/12 b = 20. Practice questions 1. After a budget planning consultation Mr. A, B and C decided that they will change their car after every 16 months, 20 months and 30 months respectively. If they all bought their cars together in" ]

[ "you can have. For example, the largest possible value of $a$ is $9$, in which case $b = c = d = 1$. If $a = 8 \\dots$ -", "(which is the least common multiple of 2 and 3) to get $3x = 2(x+1)+24$ Does it look more solvable now?", "+0 multiples -3 288 1 +-280 Suppose that $x$ is a multiple of 6 (not necessarily positive). If the square of $x$ is less than 200, how many possible values of $x$ are there? Apr 3, 2020 #1 +663 +1 Hey helpppp $$\\sqrt{200}=14.142\\text{ (Rounded to thousandths place)}$$ All multiples of 6 under 14 are: -12, -6 0, 6, 12. That is how many $$x$$ I have found. Apr 3, 2020", "over the property of being a multiple of $$3$$.", "numbers. All negative numbers less than $-1000$ can can be expressed as multiple of -1000 + three-digit negative number.", "# Squares over, now are cubes! $\\large 3^{9} + 3^{12} + 3^{15} + 3^{n}$ Find the minimum possible value of positive integer $$n$$ such that the expression above is a perfect cube. ×", "× If sum of cubes of two numbers is a square and squares of the same number is cube of a different number. Find the least positive numbers. Other Than 1,0 Note by Rishi Sharma 2 years, 9 months ago Sort by: I think the least possible way is (1,0) - 2 years, 9 months ago", "just choose A = 0.055 because it is the largest possible value.", "value is $$-2^{(w-1)}$$ and the greatest value is $$2^{(w-1)} - 1$$ (the $$-1$$ accounts for $$0$$).", "said in my previous comment, you don't get anything like all the possible values of $x$ in that way. But I think that you do get all those that are multiples of 7.", "positive, we can safely conclude that the minimum possible value of $S$ is $\\boxed{\\textbf{(E)}\\ 4}$", "# Squares over, now are cubes! $\\large 3^{9} + 3^{12} + 3^{15} + 3^{n}$ Find the minimum possible value of positive integer $$n$$ such that the expression above is a perfect cube. ###### Try out Cubes over, now are squares! × Problem Loading... Note Loading... Set Loading...", "of $3$ cubes. (Here negative cubes are allowed.) There has been a lot of computation on this problem. For other powers, one should go to the vast literature on Waring’s Problem.", "Question 17: The greatest value of is (A) (B) (C) (D)", "positive for sufficiently large $x$.", "(BKP). What if we have multiple $x_i$? In otherwords, there can be more than one copy of $x_i$. In which we'd be trying to maximize $\\sum_{i=1}^n v_i x_i$ and subject to $\\sum_{i=1}^n w_i x_i \\le W$ and $x_i \\subseteq \\{0,\\ldots, c_i\\}$.", "Given that $x$ is positive and less than 1, which of the following numbers is the largest? $x^2 + x$, $x^2$, $x^3$, $x^3 + x^2$, $x^4$.", "# Proof: There are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers The question: \"Prove that there are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers.\" How would you go about picking a method to prove this? Is there a better way to do it other than by exhaustion? - Fermat's last theorem.. done? – Euler....IS_ALIVE Feb 11 '13 at 5:14 It seems likely that exhaustion is what they want. This is the first and easiest case of Fermat's Last Theorem. There is a proof for all numbers (meaning bigger than 1000 is also allowed) that is often done in undergraduate number theory classes. But I think it's a bit long for this. Maybe you can do something by factoring the sum of two cubes? Perhaps you can rule it out when all three numbers (to be cubed) are below 10. – Will Jagy Feb 11 '13 at 5:14 There are very few perfect cubes in your range. And for any perfect cube, like $512$, one of the component cubes would have to be greater than $512/2$. The only candidate is $343$, and the other one would have to be odd, and therefore $125$ or less, bad! –", "is possible because $$2^8=256$$ which is outside the range. Hence Sufficient Statement 2: x can have multiple values, hence y will have multiple values. Insufficient Option A Kudos [?]: 246 [0], given: 33 Re: If x and y are positive integers and 2^x + 2^x + 2^x + 2^x = y, what [#permalink] 02 Oct 2017, 03:36 Display posts from previous: Sort by", "can “bootstrap” your way to dealing with an arbitrary number just by applying the two-equation version repeatedly.) | Tagged , , , , | 3 Comments ## 33 is the sum of three cubes I’m a bit late to the party, but I find this fascinating: we now know (thanks to a discovery of Andrew R. Booker) that the number 33 can be written as the sum of three cubes. This may sound unremarkable, but it has been unknown for a long time whether this was possible. Part of the reason this is more difficult than it sounds is that cubes can be both positive and negative. By contrast, it is very easy to decide whether it is possible to write 33 as a sum of two squares, $33 = a^2 + b^2$. Since squares can only be positive, any values of $a$ and $b$ greater than $5$ are not going to work, since they would make the sum too big. So there are only a few pairs of values to check: it’s enough to just check all pairs $(a,b)$ with $1 \\leq a \\leq b \\leq 5$ (quick: how many such pairs are there?), which can even be done by hand in a few minutes. None of them work, so this exhaustive search of all the possibilities" ]

[ "$$n > 36\\ln 1000 \\simeq 248.7\\,.$$ Therefore, choosing $\\boxed{n = 249}$ is sufficient. • Also, note that the Hoeffding bound ((1) in the link in my answer) would lead to a tighter result: here, $n> 18 \\ln 1000$, so $\\boxed{n=125}$ would be enough. – Clement C. Nov 6 '17 at 18:50", "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 $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{+10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}$$ $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}\\boxed{+10}$$", "that much bigger than 1 (1.5 for $$n=100$$, 1.9 for $$n=1000$$, 2.2 for $$n=10000$$). According to Wikipedia", "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.", "cubes and spheres. This happens when $x=9$. -", "# Talk:2 to the n is Greater than n Cubed when n is 10 and above Why is $(1)$ true? --kc_kennylau (talk) 12:33, 22 December 2016 (EST)", "$N = (11^{p + 7})(7^{q – 2})(5^{r + 1})(3^{s})$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0. $p,q,r$ and $s$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $0$. $11^{p + 7}$: This expression must be made a perfect cube. The nearest perfect cube is $11^9$. Therefore, the least value that $p$ can take is $9-7=2$. $7^{q – 2}$ The least value that $q$ can take is 2. If $q=2$, then the value of the expression $7^{q-2}$ will become $7^0=1$, without preventing the product from becoming a perfect cube. $5^{r+1}$: The least value that $r$ can take is $2$. $3^{s})$: The least value that $s$ can take is $3$. Therefore, the least value of the expression $p + q + r + s$ is $2+2+2+3=9$. Therefore, option E is the right answer. $10 + 10^3 + 10^6 + 10^9 = 10 + 1000 + 1000000 + 1000000000$ =$1001001010$ Therefore, option D is the right answer. We draw BD perpendiculat to AC. In right angled triangle BDC, BD / BC = sin 30° or, BD = (V2 * T)/2 ……(i) In", "-2, -1, 0, 1, 2, 3, .... Question 19 Explanation: Topic: Characteristics of Integers (Objective 0016) Question 20 #### 23 flats, 4 rods, 7 little cubes Hint: Be sure you read the question carefully: 2300+40+7=2347 #### 2 large cubes, 3 flats, 47 rods Hint: 2000+300+470 $$\\neq$$ 2347 #### 2 large cubes, 34 rods, 7 little cubes Hint: Be sure you read the question carefully: 2000+340+7=2347 #### 2 large cubes, 3 flats, 4 rods, 7 little cubes Hint: Be sure you read the question carefully: 2000+300+40+7=2347 Question 20 Explanation: Topic: Place Value (Objective 0016) Once you are finished, click the button below. Any items you have not completed will be marked incorrect. There are 20 questions to complete. ← List →", "An $$m\\times n\\times p$$ rectangular box has half the volume of an $$(m + 2)\\times(n + 2)\\times(p + 2)$$ rectangular box, where $$m, n,$$ and $$p$$ are integers, and $$m\\le n\\le p.$$ What is the largest possible value of $$p$$? (第十六届AIME1998 第14题)", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", "that no pair of odd composites add to $38$. Therefore, $\\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.", "that the minimum value of $n$ is $\\boxed{\\textbf{(A) }45}$, since $45(46) = 2070$ but $44(45) = 1980.$", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "$9$ has an interesting property that its cube can be expressed as a sum of three different, nonzero cubes. ($9^3 = 8^3 + 6^3 + 1^3$, to be more specific.) What is the next smallest integer that has this property?", "possible ordered pairs for $(a,b)$ are $(504, 503)$, $(505, 504)$, ... , $(1004, 1003)$, so $a$ has $\\boxed{501}$ possible values.", "And as obviously the smallest possible pair $(a,b)=(1,2)$ provides the solution $124$, the answer is $964 - 124 = \\boxed{840}$.", "Therefore, the maximal value of N is $3 \\times 2^{1009}$.", "will give you 5 wholes. • × 10 = 50 tens = 5 wholes = • If one whole is represented by a thousand cube: ex. multiplying 3 ones blocks by 100 will give you 3/10 wholes (or 3 tenths). • $\\large\\square \\, \\square \\,\\square \\,$ × 100 = 3 hundreds = 3/10 wholes =", "to $4$ or $5$ mod $9$ is the sum of four cubes, allowing negative cubes. It is an open problem if four cubes always suffice (but it is suspected). For $5$ cubes it is known to be true. So the answer depends on one hand how many cubes we ask for. For $5$ cubes the answer is yes, for four cubes we do not know (I think). If we require non-negative cubes, then we need more cubes. It is known that every integer $n≥ exp(524)$ is a sum of seven non negative cubes.", "# A standard number cube is tossed. What is P(greater than 2 or even)? Jun 21, 2016 #### Answer: P(greater than 2 or even) =$\\frac{5}{6}$ #### Explanation: greater than 2 $= \\left\\{3 , 4 , 5 , 6\\right\\}$ even $= \\left\\{2 , 4 , 6\\right\\}$ Greater than 2 or even $= \\left\\{2 , 3 , 4 , 5 , 6\\right\\}$ There are $5$ possibilities \"greater than 2 or even\" out of $6$ possible outcomes." ]

[ "place numbers, we can make this range between $$-14$$ and $$-83$$ smaller! $$\\Box-\\Box$$ must equal something in between $$-1$$ and $$-9$$ inclusive. ### Proof: Let the first box be the lowest option number (namely, $$3$$) and the second box be the highest option number (namely, $$12$$). We obtain that $$\\boxed{3}-\\boxed{12}=-9.\\tag{\\small\\rm as \\; the \\; minimum \\; value}$$ This is the minimum value because the pair $$(3,12)$$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $$3$$, $$6$$, $$7$$, $$12$$. Which pair is the closest to each other? That pair is $$(6,7)$$ which differs by only $$1$$. Therefore, $$\\boxed{6}-\\boxed{7}=-1.\\tag{\\small\\rm as \\; the \\; maximum \\; value}$$ And therefore, $$\\Box-\\Box$$ must equal something in between $$-1$$ and $$-9$$ inclusive. And finally, The maximum value of $$\\Box\\times \\Box$$ must be $$12$$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)! ### Proof: If we get the maximum value the LHS can equal (namely, $$4$$) and subtract the minimum value of $$\\Box-\\Box$$ (namely, $$-9$$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $$\\Box\\times \\Box$$ is $$13$$. But $$13$$ is prime, so the maximum value reduces to $$12$$.", "# Proof: There are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers The question: \"Prove that there are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers.\" How would you go about picking a method to prove this? Is there a better way to do it other than by exhaustion? - Fermat's last theorem.. done? – Euler....IS_ALIVE Feb 11 '13 at 5:14 It seems likely that exhaustion is what they want. This is the first and easiest case of Fermat's Last Theorem. There is a proof for all numbers (meaning bigger than 1000 is also allowed) that is often done in undergraduate number theory classes. But I think it's a bit long for this. Maybe you can do something by factoring the sum of two cubes? Perhaps you can rule it out when all three numbers (to be cubed) are below 10. – Will Jagy Feb 11 '13 at 5:14 There are very few perfect cubes in your range. And for any perfect cube, like $512$, one of the component cubes would have to be greater than $512/2$. The only candidate is $343$, and the other one would have to be odd, and therefore $125$ or less, bad! –", "and 2 are true but i was getting all of them true. i am new to such forums and shall be thankful to know how to put up questions better. – HarshCurious Mar 15 '11 at 3:52 @Harsh: That's fine! Notice that I explicitly stated that our value of n was positive. When you try a similar approach to (3), can you come to such a conclusion? – The Chaz 2.0 Mar 15 '11 at 4:01 Second is easy too. Let's assume that $mn+1$ is a perfect cube and recall that $(a + 1)^3 = a^3 + 3a^2 + 3a+ 1$. If both of them are equal we get $(mn + 1) = a^3 + 3a^2 + 3a+ 1$ which is the same as $mn = a(a^2 + 3a + 3)$. This means that for every m if you take $n = m^2 + 3m + 3$ then $mn+1$ will be a perfect cube. $n = m^2 + 3m + 3$ you mean of course. – ypercube Mar 14 '11 at 14:40", "$N = (11^{p + 7})(7^{q – 2})(5^{r + 1})(3^{s})$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0. $p,q,r$ and $s$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $0$. $11^{p + 7}$: This expression must be made a perfect cube. The nearest perfect cube is $11^9$. Therefore, the least value that $p$ can take is $9-7=2$. $7^{q – 2}$ The least value that $q$ can take is 2. If $q=2$, then the value of the expression $7^{q-2}$ will become $7^0=1$, without preventing the product from becoming a perfect cube. $5^{r+1}$: The least value that $r$ can take is $2$. $3^{s})$: The least value that $s$ can take is $3$. Therefore, the least value of the expression $p + q + r + s$ is $2+2+2+3=9$. Therefore, option E is the right answer. $10 + 10^3 + 10^6 + 10^9 = 10 + 1000 + 1000000 + 1000000000$ =$1001001010$ Therefore, option D is the right answer. We draw BD perpendiculat to AC. In right angled triangle BDC, BD / BC = sin 30° or, BD = (V2 * T)/2 ……(i) In", "a^3 + b^3 + c^3$, is an entirely different matter! We can’t put any a priori bound on the size of the values $a$, $b$, and $c$, because the cube of a negative number is negative—if we choose at least one of them to be positive and at least one of them to be negative, they could in theory be very large but “cancel out” to yield 33. And that’s exactly what Dr. Booker found (using approximately 23 years’ worth of computer time spread over one month!): $33 = 8\\,866\\,128\\,975\\,287\\,528^3 + (-8\\,778\\,405\\,442\\,862\\,239)^3 + (-2\\,736\\,111\\,468\\,807\\,040)^3.$ Whoa.", "can “bootstrap” your way to dealing with an arbitrary number just by applying the two-equation version repeatedly.) | Tagged , , , , | 3 Comments ## 33 is the sum of three cubes I’m a bit late to the party, but I find this fascinating: we now know (thanks to a discovery of Andrew R. Booker) that the number 33 can be written as the sum of three cubes. This may sound unremarkable, but it has been unknown for a long time whether this was possible. Part of the reason this is more difficult than it sounds is that cubes can be both positive and negative. By contrast, it is very easy to decide whether it is possible to write 33 as a sum of two squares, $33 = a^2 + b^2$. Since squares can only be positive, any values of $a$ and $b$ greater than $5$ are not going to work, since they would make the sum too big. So there are only a few pairs of values to check: it’s enough to just check all pairs $(a,b)$ with $1 \\leq a \\leq b \\leq 5$ (quick: how many such pairs are there?), which can even be done by hand in a few minutes. None of them work, so this exhaustive search of all the possibilities", "## Sums of cubes: multiple representations I’m continuing a short series of posts on representing numbers as a sum of three cubes; previous posts are 33 is the sum of three cubes and More sums of three cubes. We now know that every number less than $100$ which is not equivalent to $(\\pm 4) \\pmod 9$ can be written as a sum of three cubes. But for some of them we only know of one such representation, which can involve very large numbers. Is it possible that there are yet more ways, using even larger numbers? ## Tricubic sums for 0 $0$ can of course be written as $0^3 + 0^3 + 0^3$, but in fact it can be written as a sum of three cubes in infinitely many ways: pick any number $c$, then $0 = 0^3 + c^3 + (-c)^3$. In fact, we know these are must be the only tricubic sums for $0$. For suppose there were some nonzero $x$, $y$, and $z$ such that $0 = x^3 + y^3 + z^3$. In order to get a sum of zero, one of them has to be negative and two positive, or vice versa. If two are negative, then just negate the entire equation so that two are positive and one negative. Then move the negative", "# Squares over, now are cubes! $\\large 3^{9} + 3^{12} + 3^{15} + 3^{n}$ Find the minimum possible value of positive integer $$n$$ such that the expression above is a perfect cube. ×", "$9$ has an interesting property that its cube can be expressed as a sum of three different, nonzero cubes. ($9^3 = 8^3 + 6^3 + 1^3$, to be more specific.) What is the next smallest integer that has this property?", "Putnam 2010 A1 Given a positive integer $n$, what is the largest $k$ such that the numbers $1,2,...,n$ can be put into $k$ boxes so that the sum of the numbers in each box is the same? Putnam 2010 A1 Hint: I think it’s enough to see that $n$ must be in one of the boxes, and to make $k$ big, the sum in the boxes must be small. Therefore divide into two cases, for $n$ odd and even, and the rest is easy.", "# Sum of really Big CUBES !!!!!!! Algebra Level 3 $$55067^{3} + 23058^{3} - 38683^{3} -39442^{3} = a \\times 10^{n}$$ where $$1< a < 10$$ and $$n$$ is a positive number. What is the value of $$na^{2}$$? Hint: $$(A+B+C)^3 = A^{3} + B^{3} + C^{3} + 3(A+B)(B+C)(C+A)$$ ×", "+0 multiples -3 288 1 +-280 Suppose that $x$ is a multiple of 6 (not necessarily positive). If the square of $x$ is less than 200, how many possible values of $x$ are there? Apr 3, 2020 #1 +663 +1 Hey helpppp $$\\sqrt{200}=14.142\\text{ (Rounded to thousandths place)}$$ All multiples of 6 under 14 are: -12, -6 0, 6, 12. That is how many $$x$$ I have found. Apr 3, 2020", "multiple possible values of ${d}$, and therefore, while this method can narrow down the possible values of ${d}$ it cannot give you the exact value. 2. The initial digit In order to find the initial digit, have a look at the table above which shows the cubes of each of the non-negative integers less than 10. You will need to memorise this table, unfortunately, and there is no simple summary for it as there is with the other table. However, knowing the first 10 non-negative perfect cubes is useful in a lot of situations where you need to make mental calculations, not just in this situation. First of all, you need to write ${n^3}$ as a 9-digit number. That means, if it has less than 9 digits, put 0s in front of it until it has 9 digits. Now, look at the first 3 digits of ${n^3}$, when it is written like this; this gives you a 3-digit number. Using your knowledge of the first 10 non-negative perfect cubes, identify the greatest perfect cube which is no greater than this 3-digit number. The first digit of ${n}$ is the cube root of this perfect cube. You can also use the following table. First 3 digits of ${n^3}$ 000 001–007 008–026 027–063 064–124 First digit of ${n}$ 0 1", "the value of $x$ can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Note: Generally when we take the multiple of a number we forget to consider zero as a multiple. If we multiply any number by zero we always get 0. So 0 is a multiple of every number. Don’t forget to consider zero.", "# Squares over, now are cubes! $\\large 3^{9} + 3^{12} + 3^{15} + 3^{n}$ Find the minimum possible value of positive integer $$n$$ such that the expression above is a perfect cube. ###### Try out Cubes over, now are squares! × Problem Loading... Note Loading... Set Loading...", "## 33 is the sum of three cubes I’m a bit late to the party, but I find this fascinating: we now know (thanks to a discovery of Andrew R. Booker) that the number 33 can be written as the sum of three cubes. This may sound unremarkable, but it has been unknown for a long time whether this was possible. Part of the reason this is more difficult than it sounds is that cubes can be both positive and negative. By contrast, it is very easy to decide whether it is possible to write 33 as a sum of two squares, $33 = a^2 + b^2$. Since squares can only be positive, any values of $a$ and $b$ greater than $5$ are not going to work, since they would make the sum too big. So there are only a few pairs of values to check: it’s enough to just check all pairs $(a,b)$ with $1 \\leq a \\leq b \\leq 5$ (quick: how many such pairs are there?), which can even be done by hand in a few minutes. None of them work, so this exhaustive search of all the possibilities proves that it is not possible to find $a$ and $b$ such that $33 = a^2 + b^2$. But a sum of three cubes, $33 =", "a cube is side 1 × side 2 × side 3, or ten cubed. $10$nth represents any number. For numbers less than one, the exponent changes to a negative number: and ====================", "and $-x$ take the same value on the quadratic form but are sorted in opposite order. – Tracy Hall Aug 3 2010 at 6:02 Did you mean for the example $Q$ to have its $1$'s on the antidiagonal, or to have $-1$'s, or for the optimization to seek the maximum? As stated, $Q$ fails for $x$ with two $0$'s and two $1$'s: the minimum is $0$, but sorting gives $2$. – Tracy Hall Aug 3 2010 at 6:17 Sorry for the typo... 1's should be -1's. – mr.gondolier Aug 3 2010 at 6:24 BTW, the x I am considering is a probability vector, or equivalently, $x \\in \\mathbb{R}_+^n$. – mr.gondolier Aug 3 2010 at 6:26", "$$n > 36\\ln 1000 \\simeq 248.7\\,.$$ Therefore, choosing $\\boxed{n = 249}$ is sufficient. • Also, note that the Hoeffding bound ((1) in the link in my answer) would lead to a tighter result: here, $n> 18 \\ln 1000$, so $\\boxed{n=125}$ would be enough. – Clement C. Nov 6 '17 at 18:50", "Find me two numbers such that when they are added together, they make as much as the cube of the lesser added to the product of its triple with the square of the greater; and the cube of the greater added to its triple times the square of the lesser makes $64$ more than the sum of these two numbers. [i.e. Find $a, b$ such that $a + b = b^3 + 3ba^2$ and $a^3 + 3ab^2 = 64 + a + b$ (from : John Fauvel & Jeremy Gray (editors), The History of Mathematics : A Reader (1987), page 257).]" ]

[ "by which: $x = 4$", "$x10$ divisible by $26$, then $4004 + x10$ will yield a solution. $x10$ is divisible (for any $x$) by $2$, so it suffices to find $x1$ divisible by $13$. Since $91 = 7 \\times 13$, a solution is $4004 + 910 = 4914$.", "until we get a number whose units digit is 5 or 0. We have $123 \\ldots 44$ is divisible by $9$, so we can subtract by $9$ to get $123 \\ldots 35$ and we know that this is divisible by 5. So our answer is $\\boxed{\\textbf{(C) } 9}$ ~Arcticturn", "# x, y, 24? Integers $x$ and $y$ are such that $xy+1$ is divisible by 24. Is $x+y$ also divisible by 24? ×", "$n$ such that $x^n=e$ is $35$.", "of 25s and 15s. If $n$ is divisible by 25, then you are done. Otherwise, subtract by 15 until the result is divisible by 25. You will only need to subtract 4 times at the most because 75 is 15*5.", "found that $x=4$. I don't know if this can be done mentally but I had to use paint as an aid for calculations. – Chris Steinbeck Bell Apr 7 '18 at 16:54", "## Thinking Mathematically (6th Edition) In the given set, $x$ is defined as a number that is less than 4 and at the same time, greater than or equal to 6. Note that a number cannot be both less than $4$ and greater than or equal to $6$ at the same time. Thus, the set is empty.", "# What's this number ? This number is a square number , a multiple of 3 and one more than a cube number . Thank you !!!!!!!!!!! May 22, 2018 Well, you can probably brute force this... Some square numbers are: ${x}^{2} = 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100$ • Of these, the only ones that are multiples of $3$ are $9$, $36$, and $81$. Their digits add up to a number divisible by $3$. • $9$ is one more than ${2}^{3} = 8$, and neither $36$ nor $81$ fit this condition. $35$ is not a perfect cube and neither is $80$. Therefore, $x = 9$ is your number.", "## Now, Try This!Edit Complete the square to solve for x. Leave the answer in simplest radical form. $4x^2 + 4 = 32x + 8$ Community content is available under CC-BY-SA unless otherwise noted.", "since a square $\\pmod{10}$ can be $0,1,4,5,6,9$. Hence, $n=2k$. Hence the problem now boils down to finding $k$ such that $4^k + 65$ is a square. $4^k + 65 = x^2$ i.e. $x^2- (2^k)^2 = 65$ i.e. $(x+2^k)(x-2^k) = 65$. Now look at the different possible cases of factorizing $65$ over $\\mathbb{Z}$. Analyze the different cases and get the different values of $k$ and add them up. I am unable to think of direct way to evaluate the sum instead of computing the individual $k$'s. -", "had a slightly different method: I found the factors of the numbers, and spotted that all are divisible by $6$. When I had divided by $6$, the series was $36, 28, 21, 15, 10, 6$ and the differences went down by one each time. Very well done!", "fact that $x = 21$, we can solve for $y$: $y = 35 - x$ $\\to y = 35 - 21 = 14$ So the two numbers are $21$ and $14$, which do indeed add to $35$ and have a difference of $7$.", "# 150 Followers Question How many integers $$x$$ are there, where $$100 ≤\u0014 x ≤\u0014 2015$$, and $$x$$ is divisible by 3 or 8, but not by 6? ×", "are therefore $$x\\in\\{111, 144, 166, 188, 199, 414, 616, 818, 919\\}$$. If we test this set of possible solutions against the given divisibility restrictions, we find the solution to be $$x\\in\\{166, 199, 818, 919\\}$$. « Back to problem overview", "4 \\times 5 = 120$$. >I. 120 is divisible by $$x - 1 = 4 - 1 = 3$$. >II. 120 is divisible by $$2x = 2 (4) = 8$$. >III. 120 is divisible by $$x! = 4! = 4 \\times 3 \\times 2 \\times 1 = 24$$. I only II only II and III only I and III only I, II, and III Continue", "If three times a positive number is subtracted from the square of the number, the result is $$10$$. Find the number.", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "$$x + y + x - y = 2x$$ is also divisible by 7\" _________________ Re: M18-03 [#permalink] 28 Feb 2019, 02:12 Go to page Previous 1 2 [ 22 posts ] Display posts from previous: Sort by # M18-03 Moderators: chetan2u, Bunuel", "in $504$ seconds or $8$ minutes and $24$ seconds. $\\\\$ Question 11. Find the least perfect square number which is divisible by $3, 4, 5, 6$ and $8$. Required number is the square of LCM of $3,4,5,6$ and $8$ $\\begin{array}{r | r | r | r | r | r} 3 & 3 & 4 & 5 & 6 & 8 \\\\ \\hline 2 & 1 & 4 & 5 & 2 & 8 \\\\ \\hline 2 & 1 & 2 & 5 & 1 & 4 \\\\ \\hline 2 & 1 & 1 & 5 & 1 & 2 \\\\ \\hline 5 & 1 & 1 & 5 & 1 & 1 \\\\ \\hline {\\ } & 1 & 1 & 1 & 1 & 1 \\end{array}$ LCM $= 3 \\times 2 \\times 2 \\times 2 \\times 5 = 120$ The required perfect square is $120 \\times 3 \\times 2 \\times 5 = 3600$ $\\\\$ Question 12. Find the least number that should be added to $2500$ so that the sum is divisible by each one of the numbers $3, 4, 5$ and $6$. LCM of $3, 4, 5$, and $6 = 60$ $2500 \\div 60 = 41$ and remainder of $40$. Required number $= 60-40 = 20$ so that the sum is divisible by each of" ]

[ "= 55k$, or $9c - 9a = 5k$. Thus, k is divisible by $9$. Because $55 * 18 = 990$, $k$ must be $9$, and therefore $c - a = 5$. Because $a + b + c \\leq{7}$ and $a \\geq{1}$, $a = 1$, $c = 6$ and $b = 0$, and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$, or $\\boxed{D}$.", "the reason. Start by noticing that $1296 = 36^2$ and $666=18\\times37$. Also, $x^n-1$ is always divisible by $x-1$, and so $1296^n - 1$ is divisible by $1296-1 = 35\\times37$. Hence $1296^n - 666n + 36 = (1296^n -1) - 37(18n -1)$ is divisible by 37. But $1296^n - 666n + 36$ is also divisible by 18. Therefore it is divisible by 666.", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "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", "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 $35 * 4 + 29 = \\boxed{169}$.", "that we get $2h$ on the left side) which gives us $$2h=671/x_1+x_1^2/x^1$$ $$2h=671/x_1+x_1$$ $$x_1+x_2=671/x_1+x_1$$ $$x_2=671/x_1$$ $$x_1x_2=671.$$ Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$. Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$, so $\\boxed{h=36}$.", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "(More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$. This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$. Doing so, we get: $$(x + y)^2 = (x^2 + y^2) + 2xy$$ We see that in this form, we can substitute everything in except for $(x^2 + y^2)$, which is the desired answer. Substituting, we get: $$256 = (x^2 + y^2) + 110$$ so $x^2 + y^2 = \\boxed{146}$. (If we were to go with the pair $(55, 16)$, then the $(x + y)^2$ would be absurdly out of bounds)", "$$x = 5, 2^x = 32$$ which is in the given range [0.25, 200] $$2^x + 2$$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution. At $$x = 6, 2^x = 64$$ which is in the given range [0.25, 200] $$2^x + 2$$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution. At $$x = 7, 2^x = 128$$ which is in the given range [0.25, 200] $$2^x + 2$$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution. At $$x = 8, 2^x = 256$$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7. Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values.", "$$x+y = 11^2n$$ or $11|x+y$ 56x = 65y x = (65/56) y x + y = (65/56)y + (56/56)y = (121/56)y It is obvious that (121/56) is divisible by 11. • ?? Is 121/56 divisible by 11?? – user99914 Nov 19 '13 at 20:50 • Technically, he has shown that $56(x+y)$ is divisible by $11$ and therefore that $x+y$ is divisible by $11$. – Thomas Andrews Nov 19 '13 at 21:08", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "70 + 55 + 82 + 87 + 61}{6} = 70$", "(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 $168+27=\\boxed{195}$.", "+0 # pleez help 0 94 1 Simplify the following (a) $$\\sqrt{147a^4}$$ (b) $$\\sqrt{80n}$$ (c) $$-7 \\sqrt{75x^4}$$ Jul 19, 2020 #1 +600 +7 (a) we can first note that 147 is divisible by 49 (7*7*3) and a^4 can come out a^2 so out of the square root comes 7a^2. Your answer is $$\\boxed{7a^2\\sqrt3}$$ (b) 80 is divisible by 16 (4*4*5) so the answer is $$\\boxed{4\\sqrt{5n}}$$ (c) 75 is divisible by 25 (5*5*3) and x^4 can come out x^2 so your answer is (don't forget the -7) $$\\boxed{-35x^2\\sqrt{3}}$$ Jul 19, 2020", "+ 23 + 21$ $36 = 30 + 6$ and $23 + 21 = 44 = 40 + 4$ Then $36 + 44 = 30 + 6 + 40 + 4 = 30 + 40 + 6 + 4 = 70 + 10$ That leads us to $= 36 + 44 = 80$", "($2 \\times 2$). 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 $3$.", "- 3z) + 16y + 51z = 879$$ $$489 + 10y + 42z = 879$$ $$5y + 21z = 195$$ Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$. If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$, $x = 112$, $y = 18$, and $z = 5$, so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \\boxed{905}$.", "36, 84 (f) 34, 102 (g) 70, 105, 175 (h) 91, 112, 49 (i) 18, 54, 81 (j) 12, 45, 75 Solution: a) 18, 48 18 = 1, 2, 3, 6, 9, 18 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 Common factors = 1, 2, 3, 6 H.C.F of 18 & 48 = $$\\boxed { 6 }$$ b) 30, 42 30 = 1, 2, 3, 5, 6, 10, 15, 30 42 = 1, 2, 3, 4, 6, 7, 14, 21, 42 Common factors = 1, 2, 3, 6 ∴ H.C.F. = $$\\boxed { 6 }$$ ∴ H.C.F of 30 & 42 is $$\\boxed { 6 }$$ c) 18 & 60 18 = 1, 2, 3, 6, 9, 18 60 = 1, 2, 3, 4, 6, 10, 12, 15, 30, 60 Common factors = 1, 2, 3, 6 ∴ H.C.F. of 18 & 60 is $$\\boxed { 6 }$$ d) 27 & 63 27 = 1, 3, 9, 27 63 = 1, 3, 7, 9, 21, 63 ∴ Common factors = 1, 3, 9 ∴ H.C.F. of 27 & 63 is = 9. e) 36 & 84 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 ∴ Common" ]

[ "= 55k$, or $9c - 9a = 5k$. Thus, k is divisible by $9$. Because $55 * 18 = 990$, $k$ must be $9$, and therefore $c - a = 5$. Because $a + b + c \\leq{7}$ and $a \\geq{1}$, $a = 1$, $c = 6$ and $b = 0$, and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$, or $\\boxed{D}$.", "the reason. Start by noticing that $1296 = 36^2$ and $666=18\\times37$. Also, $x^n-1$ is always divisible by $x-1$, and so $1296^n - 1$ is divisible by $1296-1 = 35\\times37$. Hence $1296^n - 666n + 36 = (1296^n -1) - 37(18n -1)$ is divisible by 37. But $1296^n - 666n + 36$ is also divisible by 18. Therefore it is divisible by 666.", "# What's this number ? This number is a square number , a multiple of 3 and one more than a cube number . Thank you !!!!!!!!!!! May 22, 2018 Well, you can probably brute force this... Some square numbers are: ${x}^{2} = 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100$ • Of these, the only ones that are multiples of $3$ are $9$, $36$, and $81$. Their digits add up to a number divisible by $3$. • $9$ is one more than ${2}^{3} = 8$, and neither $36$ nor $81$ fit this condition. $35$ is not a perfect cube and neither is $80$. Therefore, $x = 9$ is your number.", "$$x+y = 11^2n$$ or $11|x+y$ 56x = 65y x = (65/56) y x + y = (65/56)y + (56/56)y = (121/56)y It is obvious that (121/56) is divisible by 11. • ?? Is 121/56 divisible by 11?? – user99914 Nov 19 '13 at 20:50 • Technically, he has shown that $56(x+y)$ is divisible by $11$ and therefore that $x+y$ is divisible by $11$. – Thomas Andrews Nov 19 '13 at 21:08", "($2 \\times 2$). So for example in the box $6 \\times 6$, each factor of $6$ is divisble by $2$ and so the product $6 \\times 6$ is divisible by $4$ (even though neither factor of $6$ is divisible by $4$). Since $2$ and $3$ are prime numbers they cannot be written as a product of smaller whole numbers and so these isolated shaded boxes do not occur for multiples of $2$ and $3$.", "be divisible by $4$, 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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "$9$) 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.", "divisible by $8$ is $12$.", "that we get $2h$ on the left side) which gives us $$2h=671/x_1+x_1^2/x^1$$ $$2h=671/x_1+x_1$$ $$x_1+x_2=671/x_1+x_1$$ $$x_2=671/x_1$$ $$x_1x_2=671.$$ Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$. Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$, so $\\boxed{h=36}$.", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "be a square, we should look for the interval outside that range of $x$ that guarantees the inequality $(x-6)^4\\lt x^4-24 x^3+216 x^2-804 x+936\\lt ((x-6)^2+1)^2$ a true statement. That is to say, $x^4-24 x^3+216 x^2-804 x+936$ can be a square at $x\\le 35$. You might not convince that what you have gotten so far is correct, but you can always ascertain it by trying to work out the problem numerically: At $x=36$, $810,000\\lt 811,800\\lt 811, 801$ At $x=35$, $707,281\\lt 709,021\\lt 708,964$ At $x=7$, $1\\lt 61\\lt 4$ Bingo! We could now safely declare that $x^4-24 x^3+216 x^2-804 x+936$ can't be a square in a certain domain is the answer to this last question in this quiz! Something that I want to remind you is that this quiz serves as a foundation to help guide in solving this year 2015 very challenging Olympiad Mathematics Problem that I will post it soon.", "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.", "# When is $32x+32$ a square number? I am trying to find out for what values for $x$ does the function $f(x)=32x+32$ return a square number? I found that this is the case for at least: $x \\in \\{64,1600,53824,1827904,62094400,2109381184\\}$ but I can't see a pattern in those numbers. Is there a formula which generates these values for $x$ instead of simply trying a lot of values? • I'm not sure about a formula, but didn't you miss one? If $x = 31$, then $32(31) + 32 = (32)(32)$ which is a square. – layman Jun 15 '15 at 18:48 • @randomgirl deleted her answer, but I thought it was useful. Basically she said that since $32 = 2^{5}$, then $32x + 32 = 2^{5}(x + 1)$. So if $x + 1 = 2^{2k + 1}$ for any integer $k \\geq 0$, then $2^{5}(x + 1)$ will be a perfect square. But actually $x + 1$ could equal infinitely many possibilities, since we could have $x + 1 = 2a^{2}$ for any integer $a$, or $x + 1 = 2a^{2k}b^{2j}c^{2n}$ for any integers $a,b,c$ and integers $k, j, n \\geq 0$ (and so on). This should at least convince you that there are infinitely many solutions. – layman Jun 15 '15 at 19:10 $$32(x + 1) = y^2 \\implies", "The larger of the two numbers ( $x$) is $3$ times the smaller ( $y$), so $x = 3y$. You have $x - y = 46$, and $y = 3x$. Substitute the second into the first : $(3y) - y = 46$ Which simplifies to : $2y = 46$ Thus $y = 23$. Plug this back into the first equation to find $x$ : $x - (23) = 46$, so $x = 46 + 23 = 69$. Your two numbers are $23$ and $69$.", "11:13 1 loserunderachiever wrote: Bunuel wrote: If $$x$$ is a positive integer, is $$x$$ divisible by 15? (1) $$x$$ is a multiple of 10 (2) $$x^2$$ is a multiple of 12 Hi Bunuel , I got a wrong answer , chose B ,as this a Yes / No question , to find whether X is divisible by 15 or not (y/n) statement 2 : says , X^2 is a multiple of 12 , so in this case the only possibility is 6^2 = 36 which is a multiple of 12 , hence it is NO , this means X should also has to a perfect square , then only number left out is 12 itself , (12*12) i.e 144, in this case also 15 is not divisible . Why are we considering x = 12 *15 as a possibility , where it is mentioned X^2 is a multiple of 12 . Can you please guide me ? Positive multiples of 12 are 12, 2*12 = 24, 3*12 = 36, 4*12 = 48, .... (2) says that $$x^2$$ is a multiple of 12. Since x^2 is a perfect square, then we are looking for multiples of 12 which are also perfect squares. So, x^2 could be: 36 (for x = 6); 144 (for x = 12); 324 (for x", "$$x = 5, 2^x = 32$$ which is in the given range [0.25, 200] $$2^x + 2$$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution. At $$x = 6, 2^x = 64$$ which is in the given range [0.25, 200] $$2^x + 2$$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution. At $$x = 7, 2^x = 128$$ which is in the given range [0.25, 200] $$2^x + 2$$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution. At $$x = 8, 2^x = 256$$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7. Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values.", "+0 # pleez help 0 94 1 Simplify the following (a) $$\\sqrt{147a^4}$$ (b) $$\\sqrt{80n}$$ (c) $$-7 \\sqrt{75x^4}$$ Jul 19, 2020 #1 +600 +7 (a) we can first note that 147 is divisible by 49 (7*7*3) and a^4 can come out a^2 so out of the square root comes 7a^2. Your answer is $$\\boxed{7a^2\\sqrt3}$$ (b) 80 is divisible by 16 (4*4*5) so the answer is $$\\boxed{4\\sqrt{5n}}$$ (c) 75 is divisible by 25 (5*5*3) and x^4 can come out x^2 so your answer is (don't forget the -7) $$\\boxed{-35x^2\\sqrt{3}}$$ Jul 19, 2020", "since a square $\\pmod{10}$ can be $0,1,4,5,6,9$. Hence, $n=2k$. Hence the problem now boils down to finding $k$ such that $4^k + 65$ is a square. $4^k + 65 = x^2$ i.e. $x^2- (2^k)^2 = 65$ i.e. $(x+2^k)(x-2^k) = 65$. Now look at the different possible cases of factorizing $65$ over $\\mathbb{Z}$. Analyze the different cases and get the different values of $k$ and add them up. I am unable to think of direct way to evaluate the sum instead of computing the individual $k$'s. -", "(More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$. This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$. Doing so, we get: $$(x + y)^2 = (x^2 + y^2) + 2xy$$ We see that in this form, we can substitute everything in except for $(x^2 + y^2)$, which is the desired answer. Substituting, we get: $$256 = (x^2 + y^2) + 110$$ so $x^2 + y^2 = \\boxed{146}$. (If we were to go with the pair $(55, 16)$, then the $(x + y)^2$ would be absurdly out of bounds)", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$." ]

[ "simplify", "Simplify", "simplify", "simplify:", "Simplify.", "simplify.", "simplifies to. 01", "help just simplify it", "simplify , where", "+ 4$ 4. Simplify And that's it", "simplify it a bit.", "Simplify the equation.", "there, and you can simplify $\\cosh ik$. Also, where's the sum?", "use to simplify ...", "$(6)$: integrate $(7)$: simplify", "for you to simplify.", "post , and simplify it .", "# How do you simplify (5times10^2)(1times10^8)? $50000000000$ $\\left(5 \\cdot 100\\right) \\cdot \\left(1 \\cdot 100000000\\right)$ $500 \\cdot 100000000$ $50000000000$", "# How do you simplify the square root of 5000? ##### 1 Answer It is $\\sqrt{5000} = \\sqrt{5 \\cdot {10}^{3}} = 10 \\sqrt{2 \\cdot {5}^{2}} = 50 \\cdot \\sqrt{2}$", "# How do you simplify the square root of 5000? It is $\\sqrt{5000} = \\sqrt{5 \\cdot {10}^{3}} = 10 \\sqrt{2 \\cdot {5}^{2}} = 50 \\cdot \\sqrt{2}$" ]

[ "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.", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "25\\tan^222^\\circ}$ (Evaluating: $\\sqrt{25000000} = \\sqrt{25\\cdot10^6} = \\sqrt{25}\\sqrt{10^6} = 5\\cdot10^3 = 5000$) And of course, if you are just looking for an approximation, you could have stopped at the first step and threw it into the calculator. I just like keeping things in their exact value. My apologies ... you in no way have discouraged me - it was late last night when I was studying and tiredness & frustration got the better of me. I will analyse your latest reply, and again thank you for taking the time to assist.", "## Algebra 1 $G$ Our aim is to calculate the perfect square of 80000. Thus, $\\sqrt {80000}=\\sqrt{40000 \\cdot 2}$ or, $\\sqrt {80000}=\\sqrt{40000 \\cdot 2}=200\\sqrt2$ Hence, the correct answer is $G$.", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "420^2} = 29 \\cdot 20 = \\boxed{580}$.", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "- k}$ $k \\le 5000 - 100 \\sqrt{2500 - k} < k + 1$ $0 \\le 5000 - k - 100 \\sqrt{2500 - k} < 1$ $0 \\le 2500 - k - 100 \\sqrt{2500 - k} + 2500 < 1$ $0 \\le (\\sqrt{2500 - k} - 50)^2 < 1$ $-1 < \\sqrt{2500 - k} - 50 < 1$ $49 < \\sqrt{2500 - k} < 51$ $-101 < k < 99$ So the number of $k$'s values is $99-(-101)-1=199$. Because $x=5000-100\\sqrt{2500-k}$, for each value of $k$, there is a value for $x$. The answer is $\\boxed{\\textbf{(C)} 199}$ ## Solution 7 Subtracting $10000\\lfloor x\\rfloor$ from both sides gives $x^2=10000(x-\\lfloor x\\rfloor)=10000\\{x\\}$. Dividing both sides by $10000$ gives $\\left(\\frac{x}{100}\\right)^2=\\{x\\}<1$. $\\left(\\frac{x}{100}\\right)^2<1$ when $-100 so the answer is $\\boxed{199}$ ~randomdude10807", "a^2$$. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$. ~IceMatrix", "# How do you simplify sqrt1000000? $1000000$ can be written as $1000 \\cdot 1000$ = ${1000}^{2}$ $\\sqrt{1000000}$ = $\\sqrt{{1000}^{2}}$ = 1000", "# How do you simplify the square root of 5000? It is $\\sqrt{5000} = \\sqrt{5 \\cdot {10}^{3}} = 10 \\sqrt{2 \\cdot {5}^{2}} = 50 \\cdot \\sqrt{2}$", "a final answer of $\\boxed{400}$. ~AopsUser101", "\\cdot CX \\cdot \\cos(150) = 3^2+4^2-24 \\cdot \\cos(150) = 25+12\\sqrt{3}$$ and thus the area of $ABC$ equals $$BC^2\\frac{\\sqrt{3}}{4} = 25\\frac{\\sqrt{3}}{4}+9.$$ so our final answer is $3+4+25+9 = \\boxed{041.}$", "16 '18 at 0:13 Let the approximation of $$\\sqrt[3]{4000}$$ be $$10+d$$. Then $$(10+d)^3=1000+300d+30d^2+d^3\\le 4000.$$ We have $$15^3=3375$$ and $$16^3=4096$$. The next digit is given by $$(150+d)^3=3375000+22500d+450d+d^3\\le4000000,$$ $$158^3=3944312,159^3=4019679.$$", "- e^{x_i}) = e - e^0 = \\boxed{e-1}.$ Done!", "25 \\sqrt{3} \\cdot 12}{3} = 800 \\sqrt{3} \\implies \\boxed {803}.$$ [email protected], vvsss ## Video Solution https://youtu.be/Mk-MCeVjSGc ~Shreyas S", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910} \\$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "\\cdot 9^2} = \\boxed{\\frac{16}{81}}$$", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910}$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys." ]

[ "# How do you simplify (5times10^2)(1times10^8)? $50000000000$ $\\left(5 \\cdot 100\\right) \\cdot \\left(1 \\cdot 100000000\\right)$ $500 \\cdot 100000000$ $50000000000$", "# How do you simplify the square root of 5000? It is $\\sqrt{5000} = \\sqrt{5 \\cdot {10}^{3}} = 10 \\sqrt{2 \\cdot {5}^{2}} = 50 \\cdot \\sqrt{2}$", "# How do you simplify the square root of 5000? ##### 1 Answer It is $\\sqrt{5000} = \\sqrt{5 \\cdot {10}^{3}} = 10 \\sqrt{2 \\cdot {5}^{2}} = 50 \\cdot \\sqrt{2}$", "25\\tan^222^\\circ}$ (Evaluating: $\\sqrt{25000000} = \\sqrt{25\\cdot10^6} = \\sqrt{25}\\sqrt{10^6} = 5\\cdot10^3 = 5000$) And of course, if you are just looking for an approximation, you could have stopped at the first step and threw it into the calculator. I just like keeping things in their exact value. My apologies ... you in no way have discouraged me - it was late last night when I was studying and tiredness & frustration got the better of me. I will analyse your latest reply, and again thank you for taking the time to assist.", "# How do you simplify sqrt1000000? $1000000$ can be written as $1000 \\cdot 1000$ = ${1000}^{2}$ $\\sqrt{1000000}$ = $\\sqrt{{1000}^{2}}$ = 1000", "## Algebra 1 $G$ Our aim is to calculate the perfect square of 80000. Thus, $\\sqrt {80000}=\\sqrt{40000 \\cdot 2}$ or, $\\sqrt {80000}=\\sqrt{40000 \\cdot 2}=200\\sqrt2$ Hence, the correct answer is $G$.", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "How do you simplify: the square root of -500? Apr 15, 2018 $10 i \\sqrt{5}$ Explanation: $\\sqrt{- 500} = \\sqrt{100 \\cdot - 5}$ Recalling that $\\sqrt{a \\cdot b} = \\sqrt{a} \\sqrt{b} ,$ we obtain $\\sqrt{100} \\cdot \\sqrt{- 5} = 10 \\sqrt{- 5}$ $10 \\sqrt{- 5} = 10 \\sqrt{- 1} \\sqrt{5}$ $i = \\sqrt{- 1} ,$ so we may rewrite as $10 i \\sqrt{5}$", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "simply perform casework on even values of $c$, to get $77$ tuples, subtracting the $8$ tuples in both cases we get $441+77-8=\\boxed{510}$. -EZmath2006", "# How do you simplify (1.363times10^16)/(2.9times10^6)? E.G. 4. (700000000) is 9 digits, which translates to $4.7 \\cdot {10}^{\\ast} 9 \\ast$", "\\cdot \\sqrt{15}}$ $3 \\cdot \\sqrt{10} \\cdot 100 \\cdot 2 \\cdot \\sqrt{15}}$ $600 \\sqrt{10 \\cdot 15}$ $600 \\sqrt{25 \\cdot 6}$ $600 \\sqrt{25} \\cdot \\sqrt{6}$ $600 \\cdot 5 \\cdot \\sqrt{6}$ $3000 \\sqrt{6}$ 3. ## Re: Need abit of help with simplifying Originally Posted by sentel Hi everyone, I am trying to brush up with my maths and i have not done that bad at working some questions out, but i am stuck on a certain question and i can not work it out to be the answer that the website saying the question gives you. The question is simplify $sqrt(90) x sqrt(600000)$ $3000 sqrt(6)$ I have got to $sqrt(90) x sqrt(600000) = sqrt(54000000)$ But i can not work out how to simplify that Any help will be great Thanks Alternatively you can factor 54000000 into terms that are easier to deal with $\\sqrt{54 \\times 10^6} = \\sqrt{54} \\times \\sqrt{10^6} = \\sqrt{9} \\times \\sqrt{6} \\times \\sqrt{10^6}$ $= 3 \\times 1000 \\times \\sqrt{6} = 3000\\sqrt{6}$ 4. ## Re: Need abit of help with simplifying Thank you thats brilliant 5. ## Re: Need abit of help with simplifying Just a quick question how did you get $600\\sqrt{25 \\cdot 6}$ from $600\\sqrt{10 \\cdot 15}$ Thanks 6. ## Re: Need abit of help with simplifying Originally Posted by sentel Just a quick question how did", "Always simplify first . . . things usually work out much simpler. Example: $$\\displaystyle \\L\\,\\frac{\\sqrt{72}}{\\sqrt{50}}$$ Instead of messing around with $$\\displaystyle 50$$ and $$\\displaystyle 3600$$, simplify! $$\\displaystyle \\;\\;\\sqrt{72}\\:=\\:\\sqrt{36\\cdot2}\\:=\\:\\sqrt{6}\\sqrt{2}\\:=\\:6\\sqrt{2}$$ $$\\displaystyle \\;\\;\\sqrt{50}\\:=\\:\\sqrt{25\\cdot2}\\:=\\:\\sqrt{25}\\cdot\\sqrt{2}\\:=\\:5\\sqrt{2}$$ The problem becomes: $$\\displaystyle \\L\\;\\frac{6\\sqrt{2}}{5\\sqrt{2}}\\;=\\;\\frac{6}{5}\\;\\;$$ . . . see?", "# How do you simplify sqrt1000000? Jun 27, 2016 1000 #### Explanation: $1000000$ can be written as $1000 \\cdot 1000$ = ${1000}^{2}$ $\\sqrt{1000000}$ = $\\sqrt{{1000}^{2}}$ = 1000", "a^2$$. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$. ~IceMatrix", "16 '18 at 0:13 Let the approximation of $$\\sqrt[3]{4000}$$ be $$10+d$$. Then $$(10+d)^3=1000+300d+30d^2+d^3\\le 4000.$$ We have $$15^3=3375$$ and $$16^3=4096$$. The next digit is given by $$(150+d)^3=3375000+22500d+450d+d^3\\le4000000,$$ $$158^3=3944312,159^3=4019679.$$", "# How do you simplify \\sqrt { \\frac { 800x ^ { 5} } { 2x } }? $| 20 {x}^{2} |$ $\\sqrt{\\frac{800 {x}^{5}}{2 x}} = \\sqrt{\\frac{4 \\cdot 100 \\cdot {x}^{4} \\cdot 2 \\cdot x}{2 \\cdot x}}$ $= \\sqrt{\\frac{{\\left(2 \\cdot 10 \\cdot {x}^{2}\\right)}^{2} \\cdot \\cancel{2} \\cdot \\cancel{x}}{\\cancel{2} \\cdot \\cancel{x}}}$ $= | 2 \\cdot 10 \\cdot {x}^{2} |$ $= | 20 {x}^{2} |$", "# How do you simplify (8100)( 6 - 6 ) using order of operations? $\\left(8100\\right) \\left(6 - 6\\right) = 8100 \\cdot 0 = 0$", "- k}$ $k \\le 5000 - 100 \\sqrt{2500 - k} < k + 1$ $0 \\le 5000 - k - 100 \\sqrt{2500 - k} < 1$ $0 \\le 2500 - k - 100 \\sqrt{2500 - k} + 2500 < 1$ $0 \\le (\\sqrt{2500 - k} - 50)^2 < 1$ $-1 < \\sqrt{2500 - k} - 50 < 1$ $49 < \\sqrt{2500 - k} < 51$ $-101 < k < 99$ So the number of $k$'s values is $99-(-101)-1=199$. Because $x=5000-100\\sqrt{2500-k}$, for each value of $k$, there is a value for $x$. The answer is $\\boxed{\\textbf{(C)} 199}$ ## Solution 7 Subtracting $10000\\lfloor x\\rfloor$ from both sides gives $x^2=10000(x-\\lfloor x\\rfloor)=10000\\{x\\}$. Dividing both sides by $10000$ gives $\\left(\\frac{x}{100}\\right)^2=\\{x\\}<1$. $\\left(\\frac{x}{100}\\right)^2<1$ when $-100 so the answer is $\\boxed{199}$ ~randomdude10807", "InfiniteRepitition Basic Problem - 3233 Simplify $$\\sqrt{10+4\\sqrt{3-2\\sqrt{2}}}$$ The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99." ]

[ "accounting for the largest prime factors of integers.", "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", "# Thread: Find a prime factor of 99!-1 1. ## Find a prime factor of 99!-1 Find a prime factor of 99!-1 2. Originally Posted by cornoth Find a prime factor of 99!-1 101, because 101 is prime and $99! \\equiv -100! \\equiv 1$ in $Z_{101}$ 3. oh thats right, wilson's theorem. thanks hpe!", "# Find a prime factor of 99!-1 • Jun 7th 2005, 04:12 AM cornoth Find a prime factor of 99!-1 Find a prime factor of 99!-1 • Jun 14th 2005, 06:12 PM hpe Quote: Originally Posted by cornoth Find a prime factor of 99!-1 101, because 101 is prime and $99! \\equiv -100! \\equiv 1$ in $Z_{101}$ • Jun 16th 2005, 06:19 AM cornoth oh thats right, wilson's theorem. thanks hpe!", "# Find a prime factor of 99!-1 • June 7th 2005, 03:12 AM cornoth Find a prime factor of 99!-1 Find a prime factor of 99!-1 • June 14th 2005, 05:12 PM hpe Quote: Originally Posted by cornoth Find a prime factor of 99!-1 101, because 101 is prime and $99! \\equiv -100! \\equiv 1$ in $Z_{101}$ • June 16th 2005, 05:19 AM cornoth oh thats right, wilson's theorem. thanks hpe!", "C. L. Stewart and R. Tijdeman, On the greatest prime factor of (ab+1)(ac+1)(bc+1), this volume, 93-101. Typ dokumentu Bibliografia Identyfikatory", "# Greatest Prime Factor What is the greatest prime factor of $\\Large \\color{blue}{2}^{\\color{red}{34}}- \\color{blue}{2}^{\\color{green}{28}}?$ ×", "prime factor of h(100) +1, then p 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", "On the greatest prime factor of quadratic polynomials, Acta Math. 117 (1976) 281-299. | MR 204383 | Zbl 0146.05704", "Show Tags 14 Mar 2016, 04:05 IgnitedMind wrote: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 It can be written as 2^96(2^4-1), which has prime factors as 2, 3, and 5. 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 16 Mar 2016, 01:37 here again we factor out 2^96 and we get => 15 so 5 is the highest prime hence C _________________ Manager Joined: 09 Jun 2015 Posts: 92 Re: What is the greatest prime factor of 2^100 - 2^96? [#permalink] ### Show Tags 16 Mar 2016, 01:44 IgnitedMind wrote: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 The given expression can be written as 2^96(2^4 -1) = 2^96 * 3*5 Non-Human User Joined: 09 Sep 2013 Posts: 9833 Re: What is the greatest prime factor of 2^100 - 2^96? [#permalink] ### Show Tags 23 Oct 2018, 07:02 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now", "### Hide Tags Manager Joined: 28 Aug 2008 Posts: 92 What is the greatest prime factor of 2^100 - 2^96? [#permalink] ### Show Tags 11 Sep 2008, 13:11 3 11 00:00 Difficulty: 15% (low) Question Stats: 73% (00:46) correct 27% (01:01) wrong based on 645 sessions ### HideShow timer Statistics What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 SVP Joined: 07 Nov 2007 Posts: 1651 Location: New York Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 13:28 1 2 IgnitedMind wrote: 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 = 2^96 (16-1) = 2^96 *5*3 5 is the greatest factor C _________________ Smiling wins more friends than frowning Manager Joined: 09 Jul 2007 Posts: 210 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 16:25 1 2^100 - 2^96=2^96 ( 2^4-1)=2^96*3*5 so greatest prime factor =5 VP Joined: 17 Jun 2008 Posts: 1200 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 18:10 IgnitedMind wrote: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 IMO C 2^96(16-1)=2^96", "# Which is the Supreme Prime? Level pending Find the largest prime factor of :$${ 3 }^{ 12 }+{ 12 }^{ 12 }-{ 2 }\\cdot { 6 }$$ ×", "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "prime factors for n large -", "Prime Factors Function", "5 Location: New York Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 14:28 1 KUDOS 2 This post was BOOKMARKED IgnitedMind wrote: 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 = 2^96 (16-1) = 2^96 *5*3 5 is the greatest factor C _________________ Smiling wins more friends than frowning Kudos [?]: 1063 [1], given: 5 Manager Joined: 09 Jul 2007 Posts: 245 Kudos [?]: 269 [0], given: 0 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 17:25 1 This post was BOOKMARKED 2^100 - 2^96=2^96 ( 2^4-1)=2^96*3*5 so greatest prime factor =5 Kudos [?]: 269 [0], given: 0 VP Joined: 17 Jun 2008 Posts: 1374 Kudos [?]: 406 [0], given: 0 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 19:10 IgnitedMind wrote: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 IMO C 2^96(16-1)=2^96 * 3*5 => 5 is the greatest prime factor _________________ cheers Its Now Or Never Kudos [?]: 406 [0], given: 0 Senior Manager Joined: 21 Apr 2008 Posts: 269 Kudos [?]: 165 [0], given: 0 Location: Motortown Re: PS Greatest Prime Factors [#permalink]", "1. ## prime numbers What is the greatest prime factor of 3^100-3^97? My method: $3^{100}-3^{97}=3^{97}(3^{3}-1)=3^{97}(26)$. 13 is a factor of 26 and so 13 is the greatest prime factor. Can I deduce that from what I just did? 2. Originally Posted by sfspitfire23 What is the greatest prime factor of 3^100-3^97? My method: $3^{100}-3^{97}=3^{97}(3^{3}-1)=3^{97}(26)$. 13 is a factor of 26 and so 13 is the greatest prime factor. Can I deduce that from what I just did? Yes, 3, 13 and 2 are the prime factors.", "Free Version Easy # Prime factors of a GCD ABSALG-EIVKRT What is the largest prime divisor of gcd$(10,70,100)$? A There is none because gcd$(10,70,100)=1$ B $2$ C $5$ D $7$", "2 B. 3 C. 5 D. 7 E. 11 [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 36583 Followers: 7087 Kudos [?]: 93295 [2] , given: 10555 Re: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. [#permalink] ### Show Tags 21 Jan 2013, 09:47 2 KUDOS Expert's post 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. _________________ SVP Joined: 07 Nov 2007 Posts: 1820 Location: New York Followers: 34 Kudos [?]: 865 [1] , given: 5 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 13:28 1 KUDOS 1 This post was BOOKMARKED IgnitedMind wrote: 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 = 2^96 (16-1) = 2^96 *5*3 5 is the greatest factor C _________________ Smiling wins more friends than frowning Math Expert Joined: 02 Sep 2009 Posts: 36583 Followers: 7087 Kudos [?]: 93295 [1] , given: 10555" ]

[ "the answer is $960-2=\\boxed{958}$. -brainiacmaniac31", "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "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 $13$", "Free Version Easy # Factor $113400$ as a Product of Prime Powers NUMTH-H9S91G What is the prime power factorization of $113400$? A $2^4 \\cdot 3 \\cdot 5^3 \\cdot 11$ B $2^3 \\cdot 3^4 \\cdot 5^2 \\cdot 7$ C $2^2 \\cdot 3^5 \\cdot 5^3 \\cdot 7$ D $2^3 \\cdot 3^4 \\cdot 5^2 \\cdot 11$", "# Too many factorials! $\\large \\dfrac{(100!+99!)(99!-98!)(98!+97!)\\cdots(2!+1!)}{(100!-99!)(99!+98!)(98!-97!)\\cdots(2!-1!)}$ The value of the above fraction can be written as mixed fraction $$a\\dfrac{b}{c}$$, where $$b$$ and $$c$$ & $$a$$ and $$b$$ are co-prime, but $$a$$ and $$c$$ are not co-prime. What is the product of $$a$$, $$b$$, and $$c$$? ×", "# Huge Primes $97!\\; n + d$ Find the smallest integer $d > 1$, such that the number above is prime for an infinite number of integers $n$. Notation: $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$. ×", "prime factor, $$\\frac{56}{99}$$ chance of having $$2$$ prime factors, and $$\\frac{8}{99}$$ chance of having $$3$$ prime factors. The expected number of prime factors would be $$1\\cdot\\frac{35}{99}+2\\cdot\\frac{56}{99}+3\\cdot\\frac{8}{99}=\\frac{171}{99}$$, or between $$1$$ and $$2$$, with a slight bias towards $$2$$.", "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?", "saw this on the math team forum, was posted by like last last year, but anyways… In the prime factorization of the expression $100! \\cdot 99! \\cdot 98! \\cdot \\ldots \\cdot 3! \\cdot 2! \\cdot 1!$ there is a $7^x$ term. Find x.", "# 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.", "# 2.1 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 $$4$$.", "• $994 = 2 \\cdot 7 \\cdot 71$ • $995 = 5 \\cdot 199$ • $996 = {2^2} \\cdot 3 \\cdot 83$ • $997$ is prime. • $998 = 2 \\cdot 499$ • $999 = {3^3} \\cdot 37$ • $1,000 = {2^3} \\cdot {5^3}$ You might be interested in: Fundamental Theory of Arithmetic Prime Factorization of an Integer List of Prime Factorizations of Integers from 2 to 200 List of Prime Factorizations of Integers from 201 to 400 List of Prime Factorizations of Integers from 401 to 600 List of Prime Factorizations of Integers from 601 to 800", "# Thread: Find a prime factor of 99!-1 1. ## Find a prime factor of 99!-1 Find a prime factor of 99!-1 2. Originally Posted by cornoth Find a prime factor of 99!-1 101, because 101 is prime and $99! \\equiv -100! \\equiv 1$ in $Z_{101}$ 3. oh thats right, wilson's theorem. thanks hpe!", "to find the prime factors of $2\\cdot 3\\cdot 5\\cdot 7\\cdot 11+1$, that are not among $2,3,5,7,11$.", "$n$ 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!", "Largest prime factor Let $$p_i = {2, 3, 4, 5, \\dots}$$ we continue to divide $$c$$ by $$p_i$$. If $$c$$ becomes 1 then $$p_i$$ is the greatest prime factor, otherwise we increase $$p_i$$ by one. We know that any $$p_i$$ that divides $$c$$ is prime, because otherwise it would be of the form $$p_i = {q_1}^{a_i} \\dotsm {q_n}^{a_n}$$ and since $$q_i$$ is smaller than $$p_i$$, $$c$$ would have been already divided by it, so $$p_i$$ must be a prime number. c = 600851475143 p = 2 while c > 1: if c % p == 0: c /= p continue p += 1 p 6857", "# Which is the Supreme Prime? Level pending Find the largest prime factor of :$${ 3 }^{ 12 }+{ 12 }^{ 12 }-{ 2 }\\cdot { 6 }$$ ×", "# Huge Primes $97!\\; n + d$ Find the smallest positive integer $$d$$ greater than $$1$$, such that the number above is prime for an infinite number of positive integers $$n$$. Notation: $$!$$ denotes the factorial notation. For example, $$10! = 1\\times2\\times3\\times\\cdots\\times10$$. ×", "# Inspired by Sal Gard $\\large 200! + (200!)! + 3$ Find the second smallest prime factor of the number above. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "$p_1^{n_1}\\cdots p_k^{n_k}$ 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 $5\\cdot5\\cdot13=\\boxed{\\textbf{(D) }325}$" ]

[ "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "# Too many factorials! $\\large \\dfrac{(100!+99!)(99!-98!)(98!+97!)\\cdots(2!+1!)}{(100!-99!)(99!+98!)(98!-97!)\\cdots(2!-1!)}$ The value of the above fraction can be written as mixed fraction $$a\\dfrac{b}{c}$$, where $$b$$ and $$c$$ & $$a$$ and $$b$$ are co-prime, but $$a$$ and $$c$$ are not co-prime. What is the product of $$a$$, $$b$$, and $$c$$? ×", "# Thread: Find a prime factor of 99!-1 1. ## Find a prime factor of 99!-1 Find a prime factor of 99!-1 2. Originally Posted by cornoth Find a prime factor of 99!-1 101, because 101 is prime and $99! \\equiv -100! \\equiv 1$ in $Z_{101}$ 3. oh thats right, wilson's theorem. thanks hpe!", "# Which is the Supreme Prime? Level pending Find the largest prime factor of :$${ 3 }^{ 12 }+{ 12 }^{ 12 }-{ 2 }\\cdot { 6 }$$ ×", "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 $13$", "# Huge Primes $97!\\; n + d$ Find the smallest integer $d > 1$, such that the number above is prime for an infinite number of integers $n$. Notation: $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$. ×", "the answer is $960-2=\\boxed{958}$. -brainiacmaniac31", "# Which is Greater? Which of the following is the greatest? 1. $\\dfrac{7}{101010}$ 2. $\\dfrac{7}{101011}$ 3. $\\dfrac{1}{101010}$ 4. $\\dfrac{1}{101011}$ ×", "Free Version Easy # Prime factors of a GCD ABSALG-EIVKRT What is the largest prime divisor of gcd$(10,70,100)$? A There is none because gcd$(10,70,100)=1$ B $2$ C $5$ D $7$", "1. ## prime numbers What is the greatest prime factor of 3^100-3^97? My method: $3^{100}-3^{97}=3^{97}(3^{3}-1)=3^{97}(26)$. 13 is a factor of 26 and so 13 is the greatest prime factor. Can I deduce that from what I just did? 2. Originally Posted by sfspitfire23 What is the greatest prime factor of 3^100-3^97? My method: $3^{100}-3^{97}=3^{97}(3^{3}-1)=3^{97}(26)$. 13 is a factor of 26 and so 13 is the greatest prime factor. Can I deduce that from what I just did? Yes, 3, 13 and 2 are the prime factors.", "largest factor of 900 is itself we have: $n+10=900 \\Longrightarrow \\boxed{n = 890}$ ~qwertysri987", "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:", "# Greatest Prime Factor What is the greatest prime factor of $\\Large \\color{blue}{2}^{\\color{red}{34}}- \\color{blue}{2}^{\\color{green}{28}}?$ ×", "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.", "# Find a prime factor of 99!-1 • Jun 7th 2005, 04:12 AM cornoth Find a prime factor of 99!-1 Find a prime factor of 99!-1 • Jun 14th 2005, 06:12 PM hpe Quote: Originally Posted by cornoth Find a prime factor of 99!-1 101, because 101 is prime and $99! \\equiv -100! \\equiv 1$ in $Z_{101}$ • Jun 16th 2005, 06:19 AM cornoth oh thats right, wilson's theorem. thanks hpe!", "# Find a prime factor of 99!-1 • June 7th 2005, 03:12 AM cornoth Find a prime factor of 99!-1 Find a prime factor of 99!-1 • June 14th 2005, 05:12 PM hpe Quote: Originally Posted by cornoth Find a prime factor of 99!-1 101, because 101 is prime and $99! \\equiv -100! \\equiv 1$ in $Z_{101}$ • June 16th 2005, 05:19 AM cornoth oh thats right, wilson's theorem. thanks hpe!", "saw this on the math team forum, was posted by like last last year, but anyways… In the prime factorization of the expression $100! \\cdot 99! \\cdot 98! \\cdot \\ldots \\cdot 3! \\cdot 2! \\cdot 1!$ there is a $7^x$ term. Find x.", "Show Tags 14 Mar 2016, 04:05 IgnitedMind wrote: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 It can be written as 2^96(2^4-1), which has prime factors as 2, 3, and 5. 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 16 Mar 2016, 01:37 here again we factor out 2^96 and we get => 15 so 5 is the highest prime hence C _________________ Manager Joined: 09 Jun 2015 Posts: 92 Re: What is the greatest prime factor of 2^100 - 2^96? [#permalink] ### Show Tags 16 Mar 2016, 01:44 IgnitedMind wrote: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 The given expression can be written as 2^96(2^4 -1) = 2^96 * 3*5 Non-Human User Joined: 09 Sep 2013 Posts: 9833 Re: What is the greatest prime factor of 2^100 - 2^96? [#permalink] ### Show Tags 23 Oct 2018, 07:02 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now", "### Hide Tags Manager Joined: 28 Aug 2008 Posts: 92 What is the greatest prime factor of 2^100 - 2^96? [#permalink] ### Show Tags 11 Sep 2008, 13:11 3 11 00:00 Difficulty: 15% (low) Question Stats: 73% (00:46) correct 27% (01:01) wrong based on 645 sessions ### HideShow timer Statistics What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 SVP Joined: 07 Nov 2007 Posts: 1651 Location: New York Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 13:28 1 2 IgnitedMind wrote: 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 = 2^96 (16-1) = 2^96 *5*3 5 is the greatest factor C _________________ Smiling wins more friends than frowning Manager Joined: 09 Jul 2007 Posts: 210 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 16:25 1 2^100 - 2^96=2^96 ( 2^4-1)=2^96*3*5 so greatest prime factor =5 VP Joined: 17 Jun 2008 Posts: 1200 Re: PS Greatest Prime Factors [#permalink] ### Show Tags 11 Sep 2008, 18:10 IgnitedMind wrote: What is the greatest prime factor of 2^100 - 2^96? A. 2 B. 3 C. 5 D. 7 E. 11 IMO C 2^96(16-1)=2^96", "# What is the largest prime factor of $55^{100}+55^{101}+55^{102}$ Can anyone help me on this? What is the largest prime factor of $55^{100}+55^{101}+55^{102}$? I know $55^{100}+55^{101}+55^{102}=55^{100}*(1+55+55^2)$, but I don't know how to move forward from here. • @SathasivamK That's neither largest nor prime. Sep 22 '16 at 18:08 • @robert Israel you're right Sep 22 '16 at 18:13 • The answer is 79 Sep 22 '16 at 18:15 $1+55+55^2 = 3081 = 3 \\cdot 13 \\cdot 79$. You can factor out $55^{100}$ from the expression. The largest prime factor of that number is $11$. Then it remains to factor $1+55+55^2$, which Prof. Israel did in his answer." ]

[ "paying 9% dividend and wants to have an interest of 12% on his money. A) Rs. 12 B) Rs. 15 C) Rs. 18 D) Rs. 21 Q. A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a jelly bean is to be chosen at random, what is the probability that the jelly bean will be neither red nor purple? A) 0.09 B) 0.15 C) 0.54 D) 0.85 E) 0.91", "jelly beans. Four of his jelly beans are black, five are red, and five are purple. What is the ratio of the jelly beans in Evan's bag that are NOT black compared to those that ARE black? Help plzzzz Even has a bag of 14 jelly beans. Four of his jelly beans are black, five are red, and five are purple. What is the ratio of the jelly beans in Evan's bag that are NOT black compared to those that ARE black? Help plzzzz...", "1. ## basic prob. questions A box contains 24 yellow, 25 green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green? c) At least one is yellow? I'm unsure of the techniques needed for these, any suggestions are welcome. 2. ## Jelly beans probability question A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? 3. Originally Posted by APOC197777 A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? There are $\\binom{81}{10}$ ways to select these ten jelly beans. There are $\\binom{25}{6}\\binom{56}{4}$ ways to select exactly 6 are yellow jelly beans.. Can you finish? 4. Originally Posted by jimmyp A box contains 24 yellow, 25 green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green?", "anonymous one year ago Suppose there are 15 jelly beans in a box (2 red, 3 blue, 4 white, and 6 green). A jelly bean is selected at random. What is the probability that the jelly bean is red? 1. Black_ninja123 How many red jelly beans are there? How many jelly beans are in total? 2. anonymous 2 red and 15 total 3. Black_ninja123 $\\frac{ 2 }{ 15 }$ is your answer", "green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green? c) At least one is yellow? I'm unsure of the techniques needed for these, any suggestions are welcome. :) a) $\\frac{{24 \\choose 5} \\cdot {52 \\choose 10}}{{76 \\choose 15}}$ b) $\\frac{{24 \\choose 5} \\cdot {25 \\choose 9} \\cdot {27 \\choose 1}}{{76 \\choose 15}}$ c) 1 - Pr(none are yellow).", "# basic prob. questions • Sep 27th 2009, 12:19 PM jimmyp basic prob. questions A box contains 24 yellow, 25 green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green? c) At least one is yellow? I'm unsure of the techniques needed for these, any suggestions are welcome. :) • Sep 27th 2009, 03:47 PM APOC197777 Jelly beans probability question A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? • Sep 27th 2009, 05:06 PM Plato Quote: Originally Posted by APOC197777 A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? There are $\\binom{81}{10}$ ways to select these ten jelly beans. There are $\\binom{25}{6}\\binom{56}{4}$ ways to select exactly 6 are yellow jelly beans.. Can you finish? • Sep 28th 2009, 05:19 AM The Second Solution Quote: Originally Posted by jimmyp A box contains 24 yellow, 25", "× # If a jelly bean machine contains 16 pink jelly beans, 34 blue jelly beans, 24 black jelly beans, and 26 purple jelly beans, what is the probability that a jelly bean chosen at random will be blue? Jul 18, 2016 P(Blue) =$\\frac{17}{50}$ #### Explanation: Probability of blue = $\\left(\\text{number of blue\")/(\"total number}\\right)$ P(Blue) = $\\frac{34}{100} = \\frac{17}{50}$ Sometimes probabilities are given as percentages. In this case P(blue) = 34%", "certain jar contains 60 jelly beans—22 white, 18 green, 11 [#permalink] 29 May 2020, 08:55 Display posts from previous: Sort by", "## anonymous one year ago Suppose there are 15 jelly beans in a box (2 red, 3 blue, 4 white, and 6 green). A jelly bean is selected at random. What is the probability that the jelly bean is neither red nor blue? 1. Nnesha total : 15 take out the red and blue how many jellybeans would you have left ? 2. anonymous 10 3. Nnesha $\\frac{ 15-(\\color{ReD}{2}+\\color{blue}{3}) }{ 15 }$ yes right! 4. anonymous so would it be 10/15 = 2/3 in lowest terms? 5. Nnesha yes right! 6. anonymous", "Posts: 6885 Followers: 114 Kudos [?]: 1338 [0], given: 6306 Re: A certain jar contains 60 jelly beans—22 white, 18 green, 11 [#permalink] 10 Sep 2017, 12:01 Expert's post Explanation Since there are 5 red and 4 purple jelly beans in the jar, there are 51 that are neither red nor purple, and the probability of selecting one of these is $$\\frac{51}{60}$$.Since all of the answer choices are decimals, you must convert the fraction to its decimal equivalent, 0.85 _________________ Re: A certain jar contains 60 jelly beans—22 white, 18 green, 11 [#permalink] 10 Sep 2017, 12:01 Display posts from previous: Sort by", "random from the jar, what is the probability that it is not blue? A glass jar contains 5 red, 3 blue and 2 green jelly beans. If a jelly bean is chosen at random from the jar, what is the probability that it is not blue?A glass jar contains 5 red, 3 blue and 2 green jelly beans. If a jelly bean is chosen at random from the jar, what is the probability that it is not b ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 A bowl contains 22 jelly beans: 6 red, 8 white, 3 yellow, and 5 green. Randomly select a single jelly bean from the bowl. Consider the following defined events: A bowl contains 22 jelly beans: 6 red, 8 white, 3 yellow, and 5 green. Randomly select a single jelly bean from the bowl. Consider the following defined events: A bowl contains 22 jelly beans: 6 red, 8 white, 3 yellow, and 5 green. &nbsp;Randomly select a single jelly bean from the bowl. &nbsp;Consider the f ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 What is the probability that a random person picked from the population is female and is not color blind? What is the probability that a random person picked from the population is female", "endless supply of red, orange, yellow, green, blue and violet jelly beans. The factory packages the jelly beans into jars of 100 jelly beans each. One possible colour distribution, for example, is a jar of 58 red, 22 yellow and 20 green jelly beans. As a marketing gimmick, the factory guarantees that no two jars have the same colour distribution. What is the max. number of jars that the factory can produce whilst meeting this guarantee? The answer to that question is the same as knowing the number of non-negative integer solutions to this equation: $R + O + Y + G + B + V = 100$. The number of ways to put N identical objects into K different cells is ${{N+K-1} \\choose N}$ In this case, N=100 and K=6. 9. Hi Plato (or anybody who can help), can you explain the formula N+K-1 Choose N with respect to the following model, as described by a textbook: n indistinguishable jelly beans are randomly placed, into one of R distinguishable urns. (The question would have 6 urns each labelled with a different color code). But I cannot understand how the formula N+K-1 Choose N could be derived from this point of view, as opposed to the partition stick model I gave earlier.", "current post.Consider what would mean if the green jelly beans made up 98% of the whole population of jelly beans, while the rest kinds of beans made-up the remaining 2%. Or if each kind of jelly beans has a population of (1/N)% (N different kinds of jelly beans) over thw whole population.", "# Math Help - Jelly Bean Combination/Permutation Problem 1. ## Jelly Bean Combination/Permutation Problem Fun questions, but I am having problems setting up the problem. I believe I got (a) by using C(300, 10). (b), (c), and (d), I am confuse on how to begin or what to think about. I know order does not matter. Should I treat these problems as unordered selection of objects and use C(n, r) = n! / r! (n - r)! ? Consider a bag of jelly beans that has 100 red, 100 yellow, and 100 green jelly beans. a) How many color combinations can you get by drawing 10 beans from the bag? b) How many color sequences can you get by drawing 5 beans from the bag? c) How many color combinations of 10 beans have exactly two yellow beans? d) How many color combinations of 10 beans have at least two yellow beans? 2. ## Re: Jelly Bean Combination/Permutation Problem a)I think you're wrong. two sets of 10 elements may represent the same combination of colors. b)it's easy, you have 3 possibilities for each element in the sequence, so 3^5. c)each color combinations of 10 beans that have exactly two yellow beans can be represented by a color combination of 8 beans that have only red and green, so", "are red? 8. Originally Posted by Skoz Can someone tell me how I can approach this problem because im not sure if its a formula based question, the answer is suppose to be 5/17. A bag contains 10 red jellybeans and 8 black jellybeans. What is the probability that 2 jellybeans selected at random are red?", "disk in a bag is either blue, yellow or green. &nbs [#permalink] 09 Sep 2018, 08:24 Display posts from previous: Sort by", "2009 Posts: 51035 A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, [#permalink] ### Show Tags 12 Jul 2018, 00:31 00:00 Difficulty: 25% (medium) Question Stats: 82% (01:13) correct 18% (01:15) wrong based on 38 sessions ### HideShow timer Statistics A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a jelly bean is to be chosen at random, what is the probability that the jelly bean will be neither red nor purple? (A) 0.09 (B) 0.15 (C) 0.54 (D) 0.85 (E) 0.91 _________________ SC Moderator Joined: 30 Jan 2015 Posts: 659 Location: India Concentration: Operations, Marketing GPA: 3.5 A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, [#permalink] ### Show Tags Updated on: 12 Jul 2018, 00:55 Total = 60 White = 22 Green = 18 Yellow = 11 Red = 5 Purple = 4 Now, Neither red nor purple = Either white, green or yellow Therefore, P(W or G or Y) = P(W) + P(G) + P(Y) = 22\\60 + 18\\ 60 + 11\\60 = 51\\60 =0.85 Hence, D _________________ The few, the fearless ! Thanks Originally posted by sudarshan22 on 12 Jul 2018, 00:46. Last edited by sudarshan22 on 12 Jul 2018, 00:55, edited 1", "bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea [#permalink] ### Show Tags 22 Jul 2015, 04:13 3 1 Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue? A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10 Kudos for a correct solution. Method-1 10 red jellybeans and 10 blue jellybeans Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3 Probability = 120 / 1140 = 2/19 Method-2 Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 _________________", "a jar of 58 red, 22 yellow and 20 green jelly beans. As a marketing gimmick, the factory guarantees that no two jars have the same colour distribution. What is the max. number of jars that the factory can produce whilst meeting this guarantee? As a trivial example, we shall deal with 5 jelly beans with 2 different colours (Red and Blue). We can arrange them in the following ways: (x=bean, |=partition stick) |xxxxx ( 0 red, 5 blue) x|xxxx (1 red, 4 blue) xx|xxx (2 red, 3 blue) xxx|xx (3 red, 3 blue) xxxx|x (4 red, 1 blue) xxxxx| (5 red, 0 blue) total of 6 ways to partition the 5 jelly beans into the 2 different colours. To generalise, since there are 5 jelly beans plus 1 stick, there are 6! ways to arrange them. But as the 5 jelly beans and 1 partition are indistinguishable, we divide 6! by 5! to arrive at 6!/5!1! = 6 Now, we deal with 100 jelly beans and 6 groups of colours. Line them all up in a row, together with 5 partition sticks. There are 105! ways to arrange them. As the jelly beans and the 5 partition sticks are indistinguishable, we have 105!/(100!5!) = 96.56 million 8. Originally Posted by argr78 2) A sweet factory has an", "## 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of Question 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of choosing a green jelly bean, eating it, and then choosing a purple jelly bean? in progress 0 9 mins 2023-01-12T23:25:55+00:00 2 Answers 0 views 0 1. Step-by-step explanation: The probability of the first event is $$\\frac{4}{11}$$ The probability of the second one is $$\\frac{5}{10}$$ So $$\\frac{4}{11} \\times \\frac{1}{2} = \\frac{2}{11}$$" ]

[ "## 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of Question 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of choosing a green jelly bean, eating it, and then choosing a purple jelly bean? in progress 0 9 mins 2023-01-12T23:25:55+00:00 2 Answers 0 views 0 1. Step-by-step explanation: The probability of the first event is $$\\frac{4}{11}$$ The probability of the second one is $$\\frac{5}{10}$$ So $$\\frac{4}{11} \\times \\frac{1}{2} = \\frac{2}{11}$$", "missing a corner, hence the .8! – Bryan Jan 9 '12 at 17:45 @Rasmus Some people decide they don't like the flavor of a bean after all and put it back in the bag :) – David Mitra Jan 9 '12 at 17:45 ps jelly beans don't have corners. – Bryan Jan 9 '12 at 17:50 Let's pretend the bag is very large, so drawing one bean of a flavor doesn't change the probabilities. There are $50^6$ possible draws, all the same probability. The ways to get exactly $3$ of a flavor can be counted as $\\binom 63=15$ choices for which $3$ will match times $50$ choices for which flavor to match times $49^3$ for the other $3$. There is a tiny error as we double count the case you get two three of a kinds. So the chance is $\\frac {15*50*49^3}{50^6}=352947/62500000 = 0.005647152$ or about $1$ in $177$.", "Posts: 6885 Followers: 114 Kudos [?]: 1338 [0], given: 6306 Re: A certain jar contains 60 jelly beans—22 white, 18 green, 11 [#permalink] 10 Sep 2017, 12:01 Expert's post Explanation Since there are 5 red and 4 purple jelly beans in the jar, there are 51 that are neither red nor purple, and the probability of selecting one of these is $$\\frac{51}{60}$$.Since all of the answer choices are decimals, you must convert the fraction to its decimal equivalent, 0.85 _________________ Re: A certain jar contains 60 jelly beans—22 white, 18 green, 11 [#permalink] 10 Sep 2017, 12:01 Display posts from previous: Sort by", "× # If a jelly bean machine contains 16 pink jelly beans, 34 blue jelly beans, 24 black jelly beans, and 26 purple jelly beans, what is the probability that a jelly bean chosen at random will be blue? Jul 18, 2016 P(Blue) =$\\frac{17}{50}$ #### Explanation: Probability of blue = $\\left(\\text{number of blue\")/(\"total number}\\right)$ P(Blue) = $\\frac{34}{100} = \\frac{17}{50}$ Sometimes probabilities are given as percentages. In this case P(blue) = 34%", "Question #ce0a0 Jun 4, 2016 (a) 12 Explanation: Suppose number of bags with the required number of each jelly beans$= x$ Number of Green, orange and yellow Jelly beans$= 50$ each Number of green jelly beans to be put in each bag$= 2$ $\\therefore$Total number of bags which can be filled with green beans$= 2 x$ Similarly total number of bags filled with orange beans$= 3 x$ And total number of bags filled with yellow beans$= 4 x$ Number of green jelly beans $2 x = 50$ or $x = \\frac{50}{2} = 25$ Similarly, for orange jelly beans $3 x = 50$ or $x = \\frac{50}{3} = 16 \\frac{2}{3}$, we ignore the fraction as this bag does not contain $3$ orange beans Out of $25$ bags we are able to fill only $16$ bags with $3$ orange beans. And for yellow jelly beans $4 x = 50$ or $x = \\frac{50}{4} = \\frac{25}{2} = 12 \\frac{1}{2}$, we ignore the fraction as this bag does not contain $4$ yellow beans Therefore we can fill only $12$ bags with the required number of each color of jelly beans.", "# Difference between revisions of \"2020 CIME II Problems/Problem 3\" In a jar there are blue jelly beans and green jelly beans. Then, $15\\%$ of the blue jelly beans are removed and $40\\%$ of the green jelly beans are removed. If afterwards the total number of jelly beans is $80\\%$ of the original number of jelly beans, then determine the percent of the remaining jelly beans that are blue. ## Solution 1 Suppose there are $x$ jelly beans total at the beginning. Suppose further that there are $b$ blue jelly beans and $x-b$ green jelly beans. Then, after the removal, there will be $0.85b$ blue jelly beans and $0.6x-0.6b$ green jelly beans. Because the total number of jelly beans at the end is $80\\%$ of the starting number, we can create an equation: $$0.6x+0.25b=0.8x$$ $$0.2x=0.25b$$ $$0.8x=b$$ This tells us there were originally $0.8x$ blue jelly beans and $0.2x$ green jelly beans at the beginning, so now there must be $0.68x$ blue and $0.12x$ green. The percent of the remaining jelly beans that are blue is $$\\frac{0.68x}{0.68x+0.12x}=\\frac{68}{80}=\\frac{85}{100},$$ so the answer is $\\boxed{085}$.", "Sellai809 19 # Ella bought a bag of jelly beans. There were 24 green jelly beans, which made up 8, percent of all the jelly beans in the bag. How many jelly beans were in the bag? PLEASE EXPLAIN! Easy, those 24 green jelly beans made up ONLY 8% of the bag. So set up an equation, 24 is 8% of what? (change 8% into a decimal, 0.08) 24=0.08*b (b for beans, :D) Divide by 0.08, to get the answer. $\\frac{24}{0.08} = \\frac{0.08b}{0.08}$ 300=b Thus, there are 300 jelly beans in the bag.", "did not. Represent this information in a Venn diagram. 12. For question 11, calculate the probability that a person selected at random from $ABC$ High School will want the school to have uniforms. 13. Consider a sample set as $S = \\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\\}$. Event $A$ is the multiples of 4, while event $B$ is the multiples of 5. What is the probability that a number chosen at random will be from both $A$ and $B$? 14. For question 13, what is the probability that a number chosen at random will be from either $A$ or $B$? 15. Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean? 16. Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean and then reaching in again and pulling out a red jelly bean? 17. Jack is a student in Bluenose High School. He noticed that a lot of the students in", "bean at random out of a total 20 jellybeans = 20C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3 Probability = 120 / 1140 = 2/19 Method-2 Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 _________________ ______________________________ Consider KUDOS if my post helped !! I'd appreciate learning about the grammatical errors in my posts Please let me know if I'm wrong somewhere Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea &nbs [#permalink] 19 Dec 2018, 18:04 Display posts from previous: Sort by", "anonymous one year ago Suppose there are 15 jelly beans in a box (2 red, 3 blue, 4 white, and 6 green). A jelly bean is selected at random. What is the probability that the jelly bean is red? 1. Black_ninja123 How many red jelly beans are there? How many jelly beans are in total? 2. anonymous 2 red and 15 total 3. Black_ninja123 $\\frac{ 2 }{ 15 }$ is your answer", "bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea [#permalink] ### Show Tags 22 Jul 2015, 04:13 3 1 Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue? A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10 Kudos for a correct solution. Method-1 10 red jellybeans and 10 blue jellybeans Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3 Probability = 120 / 1140 = 2/19 Method-2 Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 _________________", "green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green? c) At least one is yellow? I'm unsure of the techniques needed for these, any suggestions are welcome. :) a) $\\frac{{24 \\choose 5} \\cdot {52 \\choose 10}}{{76 \\choose 15}}$ b) $\\frac{{24 \\choose 5} \\cdot {25 \\choose 9} \\cdot {27 \\choose 1}}{{76 \\choose 15}}$ c) 1 - Pr(none are yellow).", "due to their disorganized shape the jelly beans will have gaps between them. Therefore we can estimate that the jar will only be 80% full. The number of jellybeans that fit in the jar would be: The volume of a cylinder is the surface of the circular base times its height, in our case: $$\\pi \\cdot (\\frac{1.5}{2})^2 \\cdot 2$$ We can round $\\pi$ down to 3 and solve: $$3 \\cdot (\\frac{3}{4})^2 \\cdot 2 =\\frac{(3 \\cdot 9 \\cdot 2)}{16} = \\frac{27}{8}$$ Meaning the volume of one jelly bean is $\\frac{27}{8}$ cubic centimeters. Therefore, the estimated number of jelly beans in a one liter jar would be: $$\\frac{(80\\% \\cdot 1,000)}{\\frac{27}{8}} = \\frac{(800 \\cdot 8)}{27} = \\frac{6,400}{27} = 237$$ And so we get 240 jelly beans. Now all that is left is to fill the jar with jelly beans, count them and examine how close we were to the correct answer. What about the jar in the image above? How many jelly beans do you estimate are in there?", "# Kelly criterion with more than two outcomes I want to calculate the Kelly bet for an event with more than two possible outcomes. Suppose the following game: A jar contains $10$ jelly beans. There are $7$ black jelly beans, $2$ blue jelly beans, and $1$ red jelly bean. The player wagers $x$ and grabs a single jelly bean randomly from the bag. The payouts are such: • Black Jelly Bean: no payout (i.e. simply lose wager amount $x$) • Blue Jelly Bean: net odds received on the wager = $10$ • Red Jelly Bean: net odds received on the wager = $30$ In essence the only way to lose the bet is to grab a black jelly bean (i.e. $q = 0.7$). But the net odds received on the wager is still dependent on whether the player grabs a blue ($b = 10$) or red ($b = 30$) jelly bean. How would I calculate the Kelly bet for this game? Is it correct to simply calculate the Kelly bet for each positive outcome and then find the weighted average for the final wager? For example: $$x_b = \\frac{10\\times0.2 - 0.8}{10} = 0.12$$ $$x_r = \\frac{30\\times0.1 - 0.9}{30} = 0.07$$ $$x = \\frac{0.12\\times0.2 + 0.07\\times0.1}{0.2 + 0.1} \\approx 0.103$$ So the amount to wager would be 10.3% of", "2. These values are used on this WI forum thread, where members collaborate to find the winning number for the specific day's value ranges each time the event occurs.. Favourite answer a) The approximate volume of one jelly bean can be thought of as a small cylinder 2 cm long and 1.5 cm in diameter (Precisely articulated as: Volume of 1 Jelly Bean = h(pi)(d/2)^2 = 2cm x 3 (1.5cm/2)^2 = 3.375 or 27/8 cubic centimeters) As for me I already know the count so... - Other Question Lunicolil7. cup on Aug 19, 2013 Lv 4. Method 1: Estimation 1. This volume can be estimated with reasonable accuracy. Some registrars would ask prospective Black voters to guess the number of jelly beans in a jar. Etc etc. Five students make the following guesses on the number of jelly beans in a jar:72,78,68,60,84.The guesses were off by 14,2,4,6 an 10 (not in order). my answer is 74. Estimate the (filled) volume of the jar. 2 Answers. A local woman said her great aunt experienced the test. ... How many jelly beans will fit in a mason jar? ... copy & paste the link. Truly the answer could be anywhere between 0 and 2000000000. In case you happen to be wondering 2 years later, there are 13 models of 747", "must be determined before the evaluation of the Kelly criterion. There are thus three optimal bet sizes, each dependent on the gambler's respective commitment to the outcome of the team game: 1. Team1 wins • Net odds received on wager:$0.54$• Probability of winning:$0.65$• Probability of losing:$0.25+0.10 = 0.35$• Fraction of the current bankroll to wager: $$f^* = \\frac{(0.54)(0.65)-(0.35)}{(0.54)} \\approx 0.00185$$ 2. Team2 wins • Net odds received on wager:$3.00$• Probability of winning:$0.25$• Probability of losing:$0.65+0.10 = 0.75$• Fraction of the current bankroll to wager: $$f^* = \\frac{(3.00)(0.25)-(0.75)}{(3.00)} = 0.00$$ 3. Draw • Net odds received on wager:$9.00$• Probability of winning:$0.10$• Probability of losing:$0.65+0.25 = 0.90$• Fraction of the current bankroll to wager: $$f^* = \\frac{(9.00)(0.10)-(0.90)}{(9.00)} = 0.00$$ In the question \"Kelly criterion with more than two outcomes\" (where a colored jelly bean is grabbed at random from a bag of 10 colored jelly beans), there are three possible outcomes in the gamble: 1. Black Jelly Bean: no payout (i.e. simply lose wagered amount) 2. Blue Jelly Bean: net odds received on the wager =$10$3. Red Jelly Bean: net odds received on the wager =$30\\$ Each of these outcomes affect the gambler's bankroll in a distinct way from the other outcomes. However, since no choice is involved in this gamble, there is only a single optimal bet size, that", "endless supply of red, orange, yellow, green, blue and violet jelly beans. The factory packages the jelly beans into jars of 100 jelly beans each. One possible colour distribution, for example, is a jar of 58 red, 22 yellow and 20 green jelly beans. As a marketing gimmick, the factory guarantees that no two jars have the same colour distribution. What is the max. number of jars that the factory can produce whilst meeting this guarantee? The answer to that question is the same as knowing the number of non-negative integer solutions to this equation: $R + O + Y + G + B + V = 100$. The number of ways to put N identical objects into K different cells is ${{N+K-1} \\choose N}$ In this case, N=100 and K=6. 9. Hi Plato (or anybody who can help), can you explain the formula N+K-1 Choose N with respect to the following model, as described by a textbook: n indistinguishable jelly beans are randomly placed, into one of R distinguishable urns. (The question would have 6 urns each labelled with a different color code). But I cannot understand how the formula N+K-1 Choose N could be derived from this point of view, as opposed to the partition stick model I gave earlier.", "# Purple Jelly Beans & Probability I have a question about probability, hope someone could help me. Imagine we had a jar of 80 different color jelly beans (don't know if their are 80 different colors in the world, but it doesn't matter, this is all theoretical) and someone randomly chose 20 jelly beans from the jar. I am told to choose 1 color & if my color matches any one of the twenty chosen jelly beans I win a jar of jelly beans. The chance of me choosing the right color is 1 in 4. Now, assuming I had insider information that the \"purple\" jelly bean had a chance of 1 in 3 of being chosen (don't ask me why, it doesn't really matter. Let us just assume this is the case. Maybe the purple / green jelly bean is 3 times the size of the others. But, honestly, the reason doesn't matter. Just assume that this is the case.) Now, let's further assume that I had additional insider knowledge, that, again for some unknown reason, the \"purple\" jelly bean had a probability of 1 in 6, every 5 draws, instead of 1 in 3. In other words, 4/5 draws, the purple jelly bean has a probability of showing up 1 in 3. But for 1/5 draws", "+0 # if a jelly bean machine 16 pink jelly beans, 34 blue jelly beans, 24 black jelly beans, and 26 purple jelly beans, what is the probability t 0 1357 3 if a jelly bean machine has 16 pink jelly beans, 34 blue jelly beans, 24 black jelly beans, and 26 purple jelly beans, what is the probability that a jelly bean chosen at random will be black? Guest Jul 13, 2015 #2 +5 24/100 is equal to 24% Guest Jul 13, 2015 #1 +19620 +5 if a jelly bean machine has 16 pink jelly beans, 34 blue jelly beans, 24 black jelly beans, and 26 purple jelly beans, what is the probability that a jelly bean chosen at random will be black? $$\\small{\\text{probability ~}} = \\dfrac{ 24 \\text{ black} } { 16 \\small{\\text{ pink } } + 34 \\small{\\text{ blue }} + 24 \\small{\\text{ black }} + 26 \\small{\\text{ purple }} } = \\small{\\dfrac{24}{100} = 0.24 = 24 \\% }$$ heureka Jul 13, 2015 #2 +5 24/100 is equal to 24% Guest Jul 13, 2015 #3 0 6/25 -apex Guest Jul 5, 2016", "the following six flavors only: banana, chocolate, grape, lemon, peach and strawberry. The jellybeans are sorted into boxes containing exactly 2, 3 or 4 different flavors, with each possible assortment of flavors appearing in exactly one box. What is the probability that any given box contains grape jellybeans? A. $$\\frac{1}{6}$$ B. $$\\frac{1}{3}$$ C. $$\\frac{2}{5}$$ D. $$\\frac{1}{2}$$ E. $$\\frac{3}{4}$$ Let us choose that none are grape jellybeans.. 1) when the box contains 2 flavours - total ways - $$T_1=6C2=15$$ and without jelly beans - $$W_1=5C2=10$$ 2) when the box contains 3 flavours - total ways - $$T_2=6C3=20$$ and without jelly beans - $$W_2=5C3=10$$ 3) when the box contains 4 flavours - total ways - $$T_2=6C4=15$$ and without jelly beans - $$W_2=5C4=5$$ Probability that jelly is not there = $$\\frac{W_1+W_2+W_3}{T_1+T_2+T_3}=\\frac{10+10+5}{15+20+15}=\\frac{25}{50}=\\frac{1}{2}$$ so Probability that jelly is there = $$1-\\frac{1}{2}=\\frac{1}{2}$$ D This explanation was amazing! A real Eye opener! Re: A certain candy store sells jellybeans in the following six [#permalink] 08 Jan 2019, 23:56 Display posts from previous: Sort by" ]

[ "## 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of Question 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of choosing a green jelly bean, eating it, and then choosing a purple jelly bean? in progress 0 9 mins 2023-01-12T23:25:55+00:00 2 Answers 0 views 0 1. Step-by-step explanation: The probability of the first event is $$\\frac{4}{11}$$ The probability of the second one is $$\\frac{5}{10}$$ So $$\\frac{4}{11} \\times \\frac{1}{2} = \\frac{2}{11}$$", "paying 9% dividend and wants to have an interest of 12% on his money. A) Rs. 12 B) Rs. 15 C) Rs. 18 D) Rs. 21 Q. A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a jelly bean is to be chosen at random, what is the probability that the jelly bean will be neither red nor purple? A) 0.09 B) 0.15 C) 0.54 D) 0.85 E) 0.91", "Question #ce0a0 Jun 4, 2016 (a) 12 Explanation: Suppose number of bags with the required number of each jelly beans$= x$ Number of Green, orange and yellow Jelly beans$= 50$ each Number of green jelly beans to be put in each bag$= 2$ $\\therefore$Total number of bags which can be filled with green beans$= 2 x$ Similarly total number of bags filled with orange beans$= 3 x$ And total number of bags filled with yellow beans$= 4 x$ Number of green jelly beans $2 x = 50$ or $x = \\frac{50}{2} = 25$ Similarly, for orange jelly beans $3 x = 50$ or $x = \\frac{50}{3} = 16 \\frac{2}{3}$, we ignore the fraction as this bag does not contain $3$ orange beans Out of $25$ bags we are able to fill only $16$ bags with $3$ orange beans. And for yellow jelly beans $4 x = 50$ or $x = \\frac{50}{4} = \\frac{25}{2} = 12 \\frac{1}{2}$, we ignore the fraction as this bag does not contain $4$ yellow beans Therefore we can fill only $12$ bags with the required number of each color of jelly beans.", "× # If a jelly bean machine contains 16 pink jelly beans, 34 blue jelly beans, 24 black jelly beans, and 26 purple jelly beans, what is the probability that a jelly bean chosen at random will be blue? Jul 18, 2016 P(Blue) =$\\frac{17}{50}$ #### Explanation: Probability of blue = $\\left(\\text{number of blue\")/(\"total number}\\right)$ P(Blue) = $\\frac{34}{100} = \\frac{17}{50}$ Sometimes probabilities are given as percentages. In this case P(blue) = 34%", "1. ## basic prob. questions A box contains 24 yellow, 25 green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green? c) At least one is yellow? I'm unsure of the techniques needed for these, any suggestions are welcome. 2. ## Jelly beans probability question A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? 3. Originally Posted by APOC197777 A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? There are $\\binom{81}{10}$ ways to select these ten jelly beans. There are $\\binom{25}{6}\\binom{56}{4}$ ways to select exactly 6 are yellow jelly beans.. Can you finish? 4. Originally Posted by jimmyp A box contains 24 yellow, 25 green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green?", "green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green? c) At least one is yellow? I'm unsure of the techniques needed for these, any suggestions are welcome. :) a) $\\frac{{24 \\choose 5} \\cdot {52 \\choose 10}}{{76 \\choose 15}}$ b) $\\frac{{24 \\choose 5} \\cdot {25 \\choose 9} \\cdot {27 \\choose 1}}{{76 \\choose 15}}$ c) 1 - Pr(none are yellow).", "Posts: 6885 Followers: 114 Kudos [?]: 1338 [0], given: 6306 Re: A certain jar contains 60 jelly beans—22 white, 18 green, 11 [#permalink] 10 Sep 2017, 12:01 Expert's post Explanation Since there are 5 red and 4 purple jelly beans in the jar, there are 51 that are neither red nor purple, and the probability of selecting one of these is $$\\frac{51}{60}$$.Since all of the answer choices are decimals, you must convert the fraction to its decimal equivalent, 0.85 _________________ Re: A certain jar contains 60 jelly beans—22 white, 18 green, 11 [#permalink] 10 Sep 2017, 12:01 Display posts from previous: Sort by", "missing a corner, hence the .8! – Bryan Jan 9 '12 at 17:45 @Rasmus Some people decide they don't like the flavor of a bean after all and put it back in the bag :) – David Mitra Jan 9 '12 at 17:45 ps jelly beans don't have corners. – Bryan Jan 9 '12 at 17:50 Let's pretend the bag is very large, so drawing one bean of a flavor doesn't change the probabilities. There are $50^6$ possible draws, all the same probability. The ways to get exactly $3$ of a flavor can be counted as $\\binom 63=15$ choices for which $3$ will match times $50$ choices for which flavor to match times $49^3$ for the other $3$. There is a tiny error as we double count the case you get two three of a kinds. So the chance is $\\frac {15*50*49^3}{50^6}=352947/62500000 = 0.005647152$ or about $1$ in $177$.", "jelly beans. Four of his jelly beans are black, five are red, and five are purple. What is the ratio of the jelly beans in Evan's bag that are NOT black compared to those that ARE black? Help plzzzz Even has a bag of 14 jelly beans. Four of his jelly beans are black, five are red, and five are purple. What is the ratio of the jelly beans in Evan's bag that are NOT black compared to those that ARE black? Help plzzzz...", "endless supply of red, orange, yellow, green, blue and violet jelly beans. The factory packages the jelly beans into jars of 100 jelly beans each. One possible colour distribution, for example, is a jar of 58 red, 22 yellow and 20 green jelly beans. As a marketing gimmick, the factory guarantees that no two jars have the same colour distribution. What is the max. number of jars that the factory can produce whilst meeting this guarantee? The answer to that question is the same as knowing the number of non-negative integer solutions to this equation: $R + O + Y + G + B + V = 100$. The number of ways to put N identical objects into K different cells is ${{N+K-1} \\choose N}$ In this case, N=100 and K=6. 9. Hi Plato (or anybody who can help), can you explain the formula N+K-1 Choose N with respect to the following model, as described by a textbook: n indistinguishable jelly beans are randomly placed, into one of R distinguishable urns. (The question would have 6 urns each labelled with a different color code). But I cannot understand how the formula N+K-1 Choose N could be derived from this point of view, as opposed to the partition stick model I gave earlier.", "anonymous one year ago Suppose there are 15 jelly beans in a box (2 red, 3 blue, 4 white, and 6 green). A jelly bean is selected at random. What is the probability that the jelly bean is red? 1. Black_ninja123 How many red jelly beans are there? How many jelly beans are in total? 2. anonymous 2 red and 15 total 3. Black_ninja123 $\\frac{ 2 }{ 15 }$ is your answer", "bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea [#permalink] ### Show Tags 22 Jul 2015, 04:13 3 1 Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue? A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10 Kudos for a correct solution. Method-1 10 red jellybeans and 10 blue jellybeans Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3 Probability = 120 / 1140 = 2/19 Method-2 Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 _________________", "Sellai809 19 # Ella bought a bag of jelly beans. There were 24 green jelly beans, which made up 8, percent of all the jelly beans in the bag. How many jelly beans were in the bag? PLEASE EXPLAIN! Easy, those 24 green jelly beans made up ONLY 8% of the bag. So set up an equation, 24 is 8% of what? (change 8% into a decimal, 0.08) 24=0.08*b (b for beans, :D) Divide by 0.08, to get the answer. $\\frac{24}{0.08} = \\frac{0.08b}{0.08}$ 300=b Thus, there are 300 jelly beans in the bag.", "2009 Posts: 51035 A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, [#permalink] ### Show Tags 12 Jul 2018, 00:31 00:00 Difficulty: 25% (medium) Question Stats: 82% (01:13) correct 18% (01:15) wrong based on 38 sessions ### HideShow timer Statistics A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a jelly bean is to be chosen at random, what is the probability that the jelly bean will be neither red nor purple? (A) 0.09 (B) 0.15 (C) 0.54 (D) 0.85 (E) 0.91 _________________ SC Moderator Joined: 30 Jan 2015 Posts: 659 Location: India Concentration: Operations, Marketing GPA: 3.5 A certain jar contains 60 jelly beans — 22 white, 18 green, 11 yellow, [#permalink] ### Show Tags Updated on: 12 Jul 2018, 00:55 Total = 60 White = 22 Green = 18 Yellow = 11 Red = 5 Purple = 4 Now, Neither red nor purple = Either white, green or yellow Therefore, P(W or G or Y) = P(W) + P(G) + P(Y) = 22\\60 + 18\\ 60 + 11\\60 = 51\\60 =0.85 Hence, D _________________ The few, the fearless ! Thanks Originally posted by sudarshan22 on 12 Jul 2018, 00:46. Last edited by sudarshan22 on 12 Jul 2018, 00:55, edited 1", "did not. Represent this information in a Venn diagram. 12. For question 11, calculate the probability that a person selected at random from $ABC$ High School will want the school to have uniforms. 13. Consider a sample set as $S = \\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\\}$. Event $A$ is the multiples of 4, while event $B$ is the multiples of 5. What is the probability that a number chosen at random will be from both $A$ and $B$? 14. For question 13, what is the probability that a number chosen at random will be from either $A$ or $B$? 15. Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean? 16. Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean and then reaching in again and pulling out a red jelly bean? 17. Jack is a student in Bluenose High School. He noticed that a lot of the students in", "# basic prob. questions • Sep 27th 2009, 12:19 PM jimmyp basic prob. questions A box contains 24 yellow, 25 green and 27 red jelly beans. If 15 jelly beans are selected at random, what is the probability that: a) 5 are yellow? b) 5 are yellow and 9 are green? c) At least one is yellow? I'm unsure of the techniques needed for these, any suggestions are welcome. :) • Sep 27th 2009, 03:47 PM APOC197777 Jelly beans probability question A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? • Sep 27th 2009, 05:06 PM Plato Quote: Originally Posted by APOC197777 A box contains 25 yellow, 27 green and 29 red jelly beans. If 10 jelly beans are selected at random, what is the probability that: 6 are yellow? 6 are yellow and 3 are green? At least one is yellow? There are $\\binom{81}{10}$ ways to select these ten jelly beans. There are $\\binom{25}{6}\\binom{56}{4}$ ways to select exactly 6 are yellow jelly beans.. Can you finish? • Sep 28th 2009, 05:19 AM The Second Solution Quote: Originally Posted by jimmyp A box contains 24 yellow, 25", "## anonymous one year ago Suppose there are 15 jelly beans in a box (2 red, 3 blue, 4 white, and 6 green). A jelly bean is selected at random. What is the probability that the jelly bean is neither red nor blue? 1. Nnesha total : 15 take out the red and blue how many jellybeans would you have left ? 2. anonymous 10 3. Nnesha $\\frac{ 15-(\\color{ReD}{2}+\\color{blue}{3}) }{ 15 }$ yes right! 4. anonymous so would it be 10/15 = 2/3 in lowest terms? 5. Nnesha yes right! 6. anonymous", "bean at random out of a total 20 jellybeans = 20C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3 Probability = 120 / 1140 = 2/19 Method-2 Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 _________________ ______________________________ Consider KUDOS if my post helped !! I'd appreciate learning about the grammatical errors in my posts Please let me know if I'm wrong somewhere Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea &nbs [#permalink] 19 Dec 2018, 18:04 Display posts from previous: Sort by", "$$\\frac{8}{10}$$ = 0. _____________ 0.8, Explanation: Asked to convert the fraction into decimal so it is 10)8(0.8 8 0 therefore $$\\frac{8}{10}$$ = 0.8. b. $$\\frac{3}{4}$$ = 0. _____________ 0.75, Explanation: Asked to convert the fraction into decimal so it is 4)3(0.75 2.8 0.20 0.20 0 therefore $$\\frac{3}{4}$$ = 0.75. c. $$\\frac{4}{5}$$ = 0. ______________ 0.8, Explanation: Asked to convert the fraction into decimal so it is 5)4(0.8 4.0 0 therefore $$\\frac{4}{5}$$ = 0.8. d. $$\\frac{6}{5}$$ = ___________. ___________ 1.2 Explanation: Asked to convert the fraction into decimal so it is 5)6(1.2 5 1.0 1.0 0 therefore $$\\frac{6}{5}$$ = 1.2. Question 6. CHALLENGE Justin got a sack of jelly beans in 5 different colors. Half of them were red, $$\\frac{1}{6}$$ were green, $$\\frac{1}{6}$$ were yellow, $$\\frac{1}{12}$$ were orange, and 6 were black. How many of each color did he get, and how many jelly beans were there in all? Show your work. Number of red jelly beans are 36, Number of green jelly beans are 12, Number of yellow jelly beans are 12, Number of orange jelly beans are 6, Number of black jelly beans are 6, Total number of jelly beans are 72, Explanation: Given Justin got a sack of jelly beans in 5 different colors. Half of them were red, $$\\frac{1}{6}$$ were green, $$\\frac{1}{6}$$ were yellow, $$\\frac{1}{12}$$", "random from the jar, what is the probability that it is not blue? A glass jar contains 5 red, 3 blue and 2 green jelly beans. If a jelly bean is chosen at random from the jar, what is the probability that it is not blue?A glass jar contains 5 red, 3 blue and 2 green jelly beans. If a jelly bean is chosen at random from the jar, what is the probability that it is not b ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 A bowl contains 22 jelly beans: 6 red, 8 white, 3 yellow, and 5 green. Randomly select a single jelly bean from the bowl. Consider the following defined events: A bowl contains 22 jelly beans: 6 red, 8 white, 3 yellow, and 5 green. Randomly select a single jelly bean from the bowl. Consider the following defined events: A bowl contains 22 jelly beans: 6 red, 8 white, 3 yellow, and 5 green. &nbsp;Randomly select a single jelly bean from the bowl. &nbsp;Consider the f ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 What is the probability that a random person picked from the population is female and is not color blind? What is the probability that a random person picked from the population is female" ]

[ "# Simple hex Geometry Level 1 Given that ABCDEF is a regular hexagon, and AF = EG, what's the angle measure of $$\\measuredangle AFG$$ in degrees? Bonus: Generalize a formula, using the ratio of the relative lengths of EG to AF, to determine the angle measure of $$\\measuredangle AFG$$. For example (don't enter this as the answer for this problem), if EG is 2 times the length of AF, what will be the new angle measure $$\\measuredangle AFG$$? ×", "W, and Z in parallelogram ABCD several... ) the degree measure of angles x, y, Z in parallelogram ABCD parallelogram by testing if all satisfy...", "# 1. Which angle is included between the sides $\\overline{DE}$ and $\\overline{EF}$ of $\\bigtriangleup DEF$? Since both the sides $\\overline{DE}$ and $\\overline{EF}$ intersects at E, $\\angle E$ is included between the sides $\\overline{DE}$ and $\\overline{EF}$ of $\\bigtriangleup DEF$.", "+0 # Sides and of regular octagon are extended to meet at point . What is the degree measure of angle ​? 0 58 1 Sides $$\\overline{AH}$$ and $$\\overline{CD}$$ of regular octagon $$ABCDEFGH$$ are extended to meet at point $$P$$. What is the degree measure of angle $$P$$? Apr 6, 2020 #1 +20967 0 If you have the diagram drawn, both the regular octagon and the extensions of HA and DC so that they intersect at point P, then also extend side AB so that it intersects DCP at point Q. Ths will create right triangle AQP (the angle at Q is a right angle because sides AB and DC are perpendicular). Using this formula degrees = (n - 2)(180o)/n we can find the number of degree in each interior angle. degrees = (8 - 2)(180o)/8 = 135o Angle BAH is an exterior angle to angle BAP, so angle BAP = 45o. Therefore the angle at P contains 180o - 90o - 45o = 45o. Apr 6, 2020", "Equal Arcs are Equal $\\angle FED = \\angle AFE$. In the same way we show that all the angles around the hexagon $ABCDEF$ are equal. Therefore $ABCDEF$ is a regular hexagon. $\\blacksquare$ ## Historical Note This theorem is Proposition $15$ of Book $\\text{IV}$ of Euclid's The Elements.", "? Free Version Moderate # Properties of Angles - Hexagon ACTMAT-ETJYOF For the regular hexagon below, what is the measure of ${\\alpha + \\beta }$? A $120°$ B $180°$ C $240°$ D $360°$ E $480°$", "$\\overline{EF}$ and $\\overline{GH}$ form the perpendicular bisectors of $\\overline{AB}$ and $\\overline{CD}$ respectively when both the bisectors are extended we can observe that they meet the centre $O$ of the circle. Updated on 10-Oct-2022 13:38:32", "# Square on a Hexagon Geometry Level 4 $$ABCDEF$$ is a regular hexagon. Square $$ABXY$$ is constructed on the outside of the hexagon. Let $$M$$ be the midpoint of $$DX$$. What is the measure (in degrees) of $$\\angle CMD$$? × Problem Loading... Note Loading... Set Loading...", "angle ABC and angle DCB meet at E, a point on A$...", "Geometry # Regular Polygons - Problem Solving $$\\triangle ABC$$ is a regular triangle with side length $$6.$$ If a triangle $$\\triangle BDE$$ is cut from $$\\triangle ABC$$, where the length of $$\\overline{DB}$$ and $$\\overline{EB}$$ are both equal to $$1$$, what is the perimeter of remaining figure $$ADEC?$$ Which of the following facts about regular polygons is false? $$ABCD$$ is a square. The point $$E$$ is located within $$ABCD$$, such that $$ABE$$ is an equilateral triangle. What is the measure (in degrees) of $$\\color{red} {\\angle \\text{CBE} }$$? Consider a polygon with $$n$$ vertices such that all internal angles are obtuse. If the sum of $$n-1$$ angles in this polygon is $$973^{\\circ},$$ what is the last angle in degrees? A regular polygon has interior angles of $$150^\\circ$$. $$A, B, C, D$$ are 4 consecutive points of this polygon. What is the measure (in degrees) of $$\\angle ADC?$$ ×", "Regular Shapes Geometry Level pending A regular hexagon $$ABCDEF$$ and a regular pentagon $$GHIJK$$ have the same side length. What is the value of $$\\dfrac{\\overline{AD}}{\\overline{GI}}$$? Note: The vertices are labeled clockwise in alphabetical order. ×", "Let $$ABCDEF$$ be a regular hexagon. Let G, H, I, J, K, and L be the midpoints of sides AB, BC, CD, DE, EF, and AF, respectively. The segments $$\\overline{AH}$$, $$\\overline{BI}$$, $$\\overline{CJ}$$, $$\\overline{DK}$$, $$\\overline{EL}$$, and $$\\overline{FG}$$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $$ABCDEF$$ be expressed as a fraction $$\\frac{m}{n}$$ where m and n are relatively prime positive integers. Find m + n.", "DEF and the scale factor is 3, what is the length of side EF if side BC is 6m? 10. ### geometry Let the incircle of triangle $ABC$ be tangent to sides $\\overline{BC}$, $\\overline{AC}$, and $\\overline{AB}$ at $D$, $E$, and $F$, respectively. Prove that triangle $DEF$ is acute. I have tried proving that triangle DEF's angles were … More Similar Questions", "# Take a regular hexagon Fig 3.10.3. What is the measure of each? (ii) interior angle It is a hexagon with all sides equal. All interior angles are also equal. Sum of all angles of hexagon is $720\\degree$. $6a = 720\\degree$ $a = 120\\degree$ Exams Articles Questions", "AIME II Problems/Problem 4) 7. The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle. (2011 AIME II Problems/Problem 3) 8. On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD. (2011 AIME II Problems/Problem 2) 9. Point P lies on the diagonal AC of square ABCD with AP > CP. Let $O_{1} and O_{2}$ be the circumcenters of triangles ABP and CDP respectively. Given that AB = 12 and ${\\angle O_{1}PO_{2}}$ = $120^{\\circ}$, then $AP = \\sqrt{a} + \\sqrt{b}$, where a and b are positive integers. Find a + b. (2011 AIME II Problems/Problem 13) 10. Gary purchased a large beverage, but only drank m/n of it, where m and n are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only 2/9 as much beverage. Find m+n. (2011 AIME II Problems/Problem 1) 11. Let ABCDEF be a regular hexagon. Let G, H, I, J, K, and L be the midpoints of sides AB, BC, CD, DE, EF, and AF, respectively. The segments $\\overbar{AH}$, $\\overbar{BI}, \\overbar{CJ}, \\overbar{DK}, \\overbar{EL}$, and $\\overbar{FG}$ bound a smaller", "0) (1) ### Coordinates Exam Oriented Questions and Answers Short Answer Type Questions (Score 2) Question 43. Find the coordinates of die other three verti¬ces of the rectangle in the figure. The coordinates of are (0, 0), ( 5, 0), (5, 4) Question 44. Check whether the points P(0, 7), Q(5, 12), R(-10, -3) lie on a single line. ∴ The three points are on the same line. Question 45. The origin in the diagram is o. The endpoints of the diameter of the circle are A and B. Find the coordinates of P. OQ is perpendicular to OP. Find out the coordinates of Q. 30: 60: 90 = 1: √3: 2 Coordinates of p is (√3, 1) The Coordinates of Q = (–1, √3) Question 46. ABCDEF is a regular hexagon. Of these points C and D are on the number line. C is three unit to the left of zero and D is 5 unit to the right. What is the length of the side EF? What is the measure of an angle of the regular hexagon? CD = |5– –3| = 8 ∴ CD = 8 unit EF = 8 unit Angle of a regular hexagon = 720 ÷ 6 = 120° Question 47. A circle is drawn with centre at (-1, 0) and radius", "Free Version Moderate # Exterior Angle Measure SATSTM-ZJMVML What is the measure of each exterior angle of a regular hexagon in degrees? A $60$ B $90$ C $120$ D $150$ E $360$", "Question In a regular hexagon $ABCDEF$, $AB||ED||$the x- axis. What is the slope of the sides? Verified 129k+ views Hint: First, draw the figure as per given conditions. Then use the fact that the slope of sides parallel to x- axis is 0 and the slope of other sides can be found by the angles they form with positive x-axis because $Slope = \\tan \\theta$. Let us first draw the figure as shown below: We see that $AB||ED||$ the x- axis in above figure as given. Now, since the hexagon is regular. So, all sides are of equal length. Regular polygon means polygon has all the sides of equal lengths and all equal angles. So, AB = BC = CD = DE = EF = AF …….(1) And $\\angle A = \\angle B = \\angle C = \\angle D = \\angle E = \\angle F$ …..(2) We also know that the sum of all angles of a polygon is $(n - 2) \\times {180^ \\circ }$. So, sum of all angles in hexagon will be $(6 - 2) \\times {180^ \\circ } = {720^ \\circ }$ Using (2) and (3), we get: $\\angle A + \\angle B + \\angle C + \\angle D + \\angle E + \\angle F = {720^ \\circ }$ Hence, $\\angle A = \\angle", "measure greater than half of the length of $\\overline{OB}$ let us draw two arcs above and below the $\\overline{OB}$ and mark the point of intersections of the arcs as $E$ and $F$. (vi) Now let us join points $C$ and $D$ this forms the perpendicular bisector of $\\overline{OA}$. In the similar way let us join points $E$ and $F$ this forms the perpendicular bisector of $\\overline{OB}$. (vii) By extending both the perpendicular bisectors we get a point of intersection let us point the point of intersection as $P$. (viii) Now let us measure the length of $\\overline{PA}$ and $\\overline{PB}$ they are of equal length. Therefore, $\\overline{PA}= \\overline{PB}$. Updated on 10-Oct-2022 13:38:33", "Limited access In the figure given below, $\\overline {EF}$ cuts across two parallel segments $\\overline {AB}$ and $\\overline {CD}$. If the measure of $\\angle FQD = 110°$, then what is the measure of $\\angle EPB$? A $110°$ B $250°$ C $40°$ D $70°$ E $55°$ Select an assignment template" ]

[ "# 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) $=30^{\\circ}$.", "+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", "# 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 $OA$.", "# Prove that the each angle of regular hexagon is 120. I need a theoretical proof that each angle of a regular hexagon is $120^\\circ$. - Draw the five radiuses from the hexagonal's center to its vertices. You get 6 congruent isosceles triangles whose basis angle's equals $\\,x=\\,$ half our wanted angle.. But then the central angle in each triangle equals $\\,180^\\circ-2x\\,$ , so $$6(180^\\circ-2x)=360^\\circ\\Longrightarrow 180^\\circ-2x=60^\\circ\\Longrightarrow 2x=120^\\circ$$ - Sum of the internal angles of a regular Hexagon is $(2.6-4)*90=720$ As there are 6 angles so each angle must be $720/6=120$ Another Perspective:In a regular Hexagon there are 6 Triangles each having the sum of the angles as $180$ degrees.So the sum of the Angles of all the 6 triangles is $180*6$ but this contains the central angle of $360$ degree so the sum of the internal angles of the hexagon is $180.6-360=720$.As all the angles are same so each angle must be $120$ degrees. - Consider a line rotating about a regular $n$-gon in the Euclidean plane. If the angle at each vertex is $a$, then the line rotates $180-a$ at each vertex. Since there are $n$ vertices, the total rotation is $n(180-a)$. But this total rotation is $360$. So, $n(180-a) = 360$ or $a = 180 - 360/n$. For $n = 6$, $a = 180-360/6 = 180-60", "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 0 132 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", "1999 AHSME Problems/Problem 23 Problem The equiangular convex hexagon $ABCDEF$ has $AB = 1, BC = 4, CD = 2,$ and $DE = 4.$ The area of the hexagon is $\\mathrm{(A) \\ } \\frac {15}2\\sqrt{3} \\qquad \\mathrm{(B) \\ }9\\sqrt{3} \\qquad \\mathrm{(C) \\ }16 \\qquad \\mathrm{(D) \\ }\\frac{39}4\\sqrt{3} \\qquad \\mathrm{(E) \\ } \\frac{43}4\\sqrt{3}$ Solution Solution 1 Equiangularity means that each internal angle must be exactly $120^\\circ$. The information given by the problem statement looks as follows: $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW)); dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); [/asy]$ We can now place this incomplete polygon onto a triangular grid, finish it, compute its area in unit triangles, and multiply the result by the area of the unit triangle. $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW) -- (O-3*E+3*NE+2*NW) -- cycle, 0.8red+4bp); for (int i=-5; i<=1; ++i) { draw( (O+i*E-1.5*NE)--(O+i*E+6.5*NE), dashed ); } for (int i=-2; i<=5; ++i) { draw( (O+i*E-1.5*NW)--(O+i*E+6.5*NW), dashed ); } for (int i=-1; i<=6; ++i) { draw( (O-2.5*E+i*NW)--(O+5.5*E+i*NW), dashed ); } dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); label(\"F\",(O-3*E+3*NE+2*NW),W); [/asy]$ We see that the figure contains $43$ unit triangles, and therefore its area is $\\boxed{\\frac{43\\sqrt{3}}4}$. Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle);", "$GHA$. 4. Do segments $AC$, $CE$, $EG$, and $GA$ all have the same length? Explain your reasoning. ## Summary When we apply a 180-degree rotation to a line segment, there are several possible outcomes: • The segment maps to itself (if the center of rotation is the midpoint of the segment). • The image of the segment overlaps with the segment and lies on the same line (if the center of rotation is a point on the segment). • The image of the segment does not overlap with the segment (if the center of rotation is not on the segment). We can also build patterns by rotating a shape. For example, triangle $ABC$ shown here has $m(\\angle A) = 60$. If we rotate triangle $ABC$ 60 degrees, 120 degrees, 180 degrees, 240 degrees, and 300 degrees clockwise, we can build a hexagon.", "to be the centroid of $\\triangle{P_1P_2P_3}$. Because $m\\angle{O_1P_1P_2} = 90^\\circ$ and $m\\angle{O_1P_1X} = 30^\\circ$, $m\\angle{O_1P_1X} = 60^\\circ$. $O_2M$ is $2/3$ the height of $\\triangle{P_1P_2P_3}$, thus $O_2M$ is $8*\\sqrt{3}/3$. Applying cosine law on $\\triangle{O_1P_1X}$, one finds that $P_1X = 2 + 2*\\sqrt{21}/3$. Multiplying by $3/2$ to solve for the height of $\\triangle{P_1P_2P_3}$, one gets $3 + \\sqrt{21}$. Simply multiplying by $2/\\sqrt{3}$ and then calculating the equilateral triangle's area, one would get the final result of $\\sqrt{300} + \\sqrt{252}$. This makes the answer $252 + 300 = 552$. $\\boxed{\\textbf{D}.}$ ~AlbeePach~ ## Solution 4 $[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label(\"P_1\", P1, E*1.5); label(\"P_2\", P2, SW*1.5); label(\"P_3\", P3, N); label(\"\\omega_1\", A, W*30); label(\"\\omega_2\", B, E*17); label(\"\\omega_3\", C, W*17); label(\"\\Gamma\", A, W*15); draw(Circle(A,2),red); pair X=foot(A,P1,P3); dot(X,blue); draw(A--X,blue); label(\"2\\sqrt{7}\", X--P3); label(\"2\\sqrt{3}\",X--P1); label(\"X\",X,dir(-80),blue); [/asy]$ First, note that because the $\\angle P_1=\\angle P_2=\\angle P_3=\\pi/3$, the arcs inside the shaded equilateral triangle are each $2\\pi/3$. Also, the distances between the centers of any two of the $3$ given circles are each $8$. Draw the circle $\\Gamma$ concentric with $\\omega_1$ with radius $2$. Because the arc of $\\omega_1$ inside said triangle is $2\\pi/3$,", "$GH$ is $\\sqrt{3}$. Think of the picture as one large equilateral triangle, $\\triangle{JKL}$ with the sides of $2\\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\\triangle{JKL}$ $\\dfrac{\\sqrt{3}}{4}(2\\sqrt{3}+1)^2=\\dfrac{12+13\\sqrt{3}}{4}$. $[asy] import graph; size(10cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; pair J = rotate(60,I)*D; pair K = rotate(60,E)*F; pair L = rotate(60,G)*H; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); draw(I--J--D); draw(E--K--F); draw(G--L--H); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,W); label(\"E\",E,W); label(\"F\",F,E); label(\"G\",G,E); label(\"H\",H,SE); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,N); label(\"L\",L,SE); [/asy]$ Triangles $\\triangle{DIJ}$, $\\triangle{EFK}$, and $\\triangle{GHL}$ have sides of $\\sqrt{3}$, so their total area is $3(\\dfrac{\\sqrt{3}}{4}(\\sqrt{3})^2)=\\dfrac{9\\sqrt{3}}{4}$. Now, you subtract their total area from the area of $\\triangle{JKL}$: $\\dfrac{12+13\\sqrt{3}}{4}-\\dfrac{9\\sqrt{3}}{4}=\\dfrac{12+13\\sqrt{3}-9\\sqrt{3}}{4}=\\dfrac{12+4\\sqrt{3}}{4}=3+\\sqrt{3}\\implies\\boxed{\\textbf{(C)}\\ 3+\\sqrt3}$ ## Solution 3 We will use, $\\frac{1}{2}ab\\sin x$ to find the area of the following triangles. Since $\\angle A=360$, $\\angle EAF=360-90-90-60=120$. The Area of $\\triangle AEF$ is $\\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot \\sin(120)$. Noting, $\\sin (2x) = 2\\sin (x)\\cos (x)$, Area of $\\triangle AEF = \\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot 2 \\cdot \\sin(60) \\cdot \\cos(60) = \\dfrac{\\sqrt{3}}{4}$, Area of $\\triangle ABC = \\frac{1}{2} \\cdot 1 \\cdot 1", "+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", "hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these", "hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these", "2015 Posts: 113 GMAT 1: 730 Q49 V40 GPA: 3.8 In the figure, JKLMNP is a regular hexagon . [#permalink] Show Tags Updated on: 20 May 2016, 07:29 I think there's an easier way to solve this problem. As we were saying a regular hexagon has each angle measuring $$120°$$. Now let's look at the image posted above, the theorem of vertical and adjacent angle pairs says really simply that opposite angles born from the intersection of two lines must be equal. Therefore if the angle of the hexagon is $$120°$$ the opposite angle will be $$120°$$ as well, but we also know that the round angle is 360°, therefore, the sum of the remaining opposite pair must be $$360° - 240° = 120°$$, and since they are opposite as well they have to be equal -> $$\\frac{120°}{2}= 60°$$ Originally posted by DensetsuNo on 20 May 2016, 04:57. Last edited by DensetsuNo on 20 May 2016, 07:29, edited 1 time in total. Current Student Joined: 18 Oct 2014 Posts: 843 Location: United States GMAT 1: 660 Q49 V31 GPA: 3.98 Re: In the figure, JKLMNP is a regular hexagon . [#permalink] Show Tags 20 May 2016, 05:58 1 Since it is a regular hexagon, angles made at dissection point by all diagonals will be equal. Total number of", "= \\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)", "Common Rotations • Rotation of $180^\\circ$ : If $(x, y)$ is rotated $180^\\circ$ around the origin, then the image will be $(-x, -y)$ . • Rotation of $90^\\circ$ : If $(x, y)$ is rotated $90^\\circ$ around the origin, then the image will be $(-y, x)$ . • Rotation of $270^\\circ$ : If $(x, y)$ is rotated $270^\\circ$ around the origin, then the image will be $(y, -x)$ . While we can rotate any image any amount of degrees, only $90^\\circ, 180^\\circ$ and $270^\\circ$ have special rules. To rotate a figure by an angle measure other than these three, you must use the process from the Investigation. #### Example A Rotate $\\triangle ABC$ , with vertices $A(7, 4), B(6, 1),$ and $C(3, 1) \\ 180^\\circ$ . Find the coordinates of $\\triangle A'B'C'$ . It is very helpful to graph the triangle. If $A$ is $(7, 4)$ , that means it is 7 units to the right of the origin and 4 units up. $A'$ would then be 7 units to the left of the origin and 4 units down. The vertices are: $A(7,4) \\rightarrow A'(-7,-4)\\\\B(6,1) \\rightarrow B'(-6,-1)\\\\C(3,1) \\rightarrow C'(-3,-1)$ #### Example B Rotate $\\overline{ST} \\ 90^\\circ$ counter-clockwise about the origin. Using the $90^\\circ$ rotation rule, $T'$ is (8, 2). #### Example C Find the coordinates of $ABCD$ after a", "exists a convex hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$", "hexagon $ABCDEF$, $AC$, $CE$ are two diagonals. Points $M$, $N$ are on $AC$, $CE$ respectively and satisfy $AC: AM = CE: CN = r$. Suppose $B, M, N$ are collinear, find $100r^2$. (diagram goes here) $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } pair A=dir(0), B=dir(60), C=dir(120), D=dir(180), E=dir(240), F=dir(300), N=(4*E+C)/5,M=intersectionpoints(A--C,B--N)[0]; draw(A--B--C--D--E--F--cycle); draw(A--C--E); draw(B--N); label(\"A\",D2(A),E); label(\"B\",D2(B),NE); label(\"C\",D2(C),NW); label(\"D\",D2(D),W); label(\"E\",D2(E),SW); label(\"F\",D2(F),SE); label(\"M\",D2(M),(0,-1.5)); label(\"N\",D2(N),SE); [/asy]$ Solution A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$. Find $100r^2$. (diagram goes here) In the following diagram, a semicircle is folded along a chord $AN$ and intersects its diameter $MN$ at $B$. Given that $MB : BN = 2 : 3$ and $MN = 10$. If $AN = x$, find $x^2$. $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } real r = sqrt(80)/5; pair M=(-1,0), N=(1,0), A=intersectionpoints(arc((M+N)/2, 1, 0, 180),circle(N,r))[0], C=intersectionpoints(circle(A,1),circle(N,1))[0], B=intersectionpoints(circle(C,1),M--N)[0]; draw(arc((M+N)/2, 1, 0, 180)--cycle); draw(A--N); draw(arc(C,1,180,180+2*aSin(r/2))); label(\"A\",D2(A),NW); label(\"B\",D2(B),SW); label(\"M\",D2(M),S); label(\"N\",D2(N),SE); [/asy]$ Solution Square $ABCD$ is divided into four rectangles by $EF$ and $GH$. $EF$ is parallel to $AB$ and $GH$ parallel to $BC$. $\\angle BAF = 18^\\circ$. $EF$", "+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}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "must be positive, $x = \\frac{3\\sqrt{7}-\\sqrt{3}}{2} \\Rightarrow C$. ## Solution 2 (without trigonometry) Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$ $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label(\"$$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 # 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", "# 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) $=30^{\\circ}$.", "1999 AHSME Problems/Problem 23 Problem The equiangular convex hexagon $ABCDEF$ has $AB = 1, BC = 4, CD = 2,$ and $DE = 4.$ The area of the hexagon is $\\mathrm{(A) \\ } \\frac {15}2\\sqrt{3} \\qquad \\mathrm{(B) \\ }9\\sqrt{3} \\qquad \\mathrm{(C) \\ }16 \\qquad \\mathrm{(D) \\ }\\frac{39}4\\sqrt{3} \\qquad \\mathrm{(E) \\ } \\frac{43}4\\sqrt{3}$ Solution Solution 1 Equiangularity means that each internal angle must be exactly $120^\\circ$. The information given by the problem statement looks as follows: $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW)); dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); [/asy]$ We can now place this incomplete polygon onto a triangular grid, finish it, compute its area in unit triangles, and multiply the result by the area of the unit triangle. $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW) -- (O-3*E+3*NE+2*NW) -- cycle, 0.8red+4bp); for (int i=-5; i<=1; ++i) { draw( (O+i*E-1.5*NE)--(O+i*E+6.5*NE), dashed ); } for (int i=-2; i<=5; ++i) { draw( (O+i*E-1.5*NW)--(O+i*E+6.5*NW), dashed ); } for (int i=-1; i<=6; ++i) { draw( (O-2.5*E+i*NW)--(O+5.5*E+i*NW), dashed ); } dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); label(\"F\",(O-3*E+3*NE+2*NW),W); [/asy]$ We see that the figure contains $43$ unit triangles, and therefore its area is $\\boxed{\\frac{43\\sqrt{3}}4}$. Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle);", "+ .75*dir(360-65-115-55-30)); label(\"65^\\circ\",B + .75*dir(180-32.5)); label(\"x^\\circ\",A + .5*dir(42.5)); label(\"5^\\circ\",D + 2.5*dir(360-60-2.5)); label(\"60^\\circ\",D + .75*dir(360-30)); label(\"60^\\circ\",E + .5*dir(360-150)); label(\"5^\\circ\",B + 2.5*dir(180-65-2.5)); [/asy]$ 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", "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}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "= (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.", "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", "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", "hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these", "hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these", "to get $f^{m-n+1}=f$. We take $M=m-n+1$ to get the relation $f^{M}=f$ with $M>1$. Note that this means $f^{M}(i)=f(i)$ for all $i \\in A$. QED. Problem 6: Show that there exists a convex hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4),", "+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", "exists a convex hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$", "(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 +", "# 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 $OA$.", "+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", "to get the relation $f^{M}=f$ with $M>1$. Note that this means $f^{M}(i)=f(i)$ for all $i \\in A$. QED. Problem 6: Show that there exists a convex hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5),", "hexagon $ABCDEF$, $AC$, $CE$ are two diagonals. Points $M$, $N$ are on $AC$, $CE$ respectively and satisfy $AC: AM = CE: CN = r$. Suppose $B, M, N$ are collinear, find $100r^2$. (diagram goes here) $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } pair A=dir(0), B=dir(60), C=dir(120), D=dir(180), E=dir(240), F=dir(300), N=(4*E+C)/5,M=intersectionpoints(A--C,B--N)[0]; draw(A--B--C--D--E--F--cycle); draw(A--C--E); draw(B--N); label(\"A\",D2(A),E); label(\"B\",D2(B),NE); label(\"C\",D2(C),NW); label(\"D\",D2(D),W); label(\"E\",D2(E),SW); label(\"F\",D2(F),SE); label(\"M\",D2(M),(0,-1.5)); label(\"N\",D2(N),SE); [/asy]$ Solution A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$. Find $100r^2$. (diagram goes here) In the following diagram, a semicircle is folded along a chord $AN$ and intersects its diameter $MN$ at $B$. Given that $MB : BN = 2 : 3$ and $MN = 10$. If $AN = x$, find $x^2$. $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } real r = sqrt(80)/5; pair M=(-1,0), N=(1,0), A=intersectionpoints(arc((M+N)/2, 1, 0, 180),circle(N,r))[0], C=intersectionpoints(circle(A,1),circle(N,1))[0], B=intersectionpoints(circle(C,1),M--N)[0]; draw(arc((M+N)/2, 1, 0, 180)--cycle); draw(A--N); draw(arc(C,1,180,180+2*aSin(r/2))); label(\"A\",D2(A),NW); label(\"B\",D2(B),SW); label(\"M\",D2(M),S); label(\"N\",D2(N),SE); [/asy]$ Solution Square $ABCD$ is divided into four rectangles by $EF$ and $GH$. $EF$ is parallel to $AB$ and $GH$ parallel to $BC$. $\\angle BAF = 18^\\circ$. $EF$", "# 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. $\\blacksquare$" ]

[ "product of two primes proved", "is the product of largest and the smallest numbers? A). 1512 B). 1102 C). 1152 D). 1210 Go to : Total Pages : 36", "# \"Modernization\" of a Combinatorial Factorization What is the product of the two largest prime factors of $$\\binom{2014}{1007}$$ that appear twice in its prime factorization? ×", "theorem. Any natural number greater than 1 can be written as a product of prime factors.", "# \"Modernization\" of a Combinatorial Factorization Number Theory Level 4 What is the product of the two largest prime factors of $$\\binom{2014}{1007}$$ that appear twice in its prime factorization? ×", "# A Pre-RMO question! -14 Let $n$ be the largest integer that is product of three primes - $x,y$ and $10x+y$; where $x,y$ are single digit numbers. What is the sum of the digits of $n$? ×", "The product of two number is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is : [A] 380 [B] 395 [C] 400 [D] 425", "product of digit of the number. And if k=1 we can’t find the product of digit of the number because product is binary operator. Any natural number is greater than or equal to its product of digits.", "two numbers, which are latterly broken down into product of prime numbers.", "is (243,112,609 − 1) × (237,156,667 − 1) × (232,582,657 − 1), i.e., the product of the three largest known primes.", "does she think? 15. Product Result of the product of the numbers 1, 2, 3, 1, 2, 0 is: 16. How manz By how many is the product of the numbers 328 and 7 greater than its sum? 17. Computer A line of print on a computer contains 64 characters (letters, spacers or other chars). Find how many characters there are in 7 lines.", "product of three positive integers?", "# Medium #1 Number Theory Level 3 How many integers $$10 \\leq x \\leq 90$$ are the product of two different primes? (Note that $$1$$ is not prime.) ×", "+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", "# Product Pairs All natural numbers can be written as the product of two natural numbers. Prime numbers can be written as the product of 1 and themselves. Composites can be written in more than one way as a product of two natural numbers. For instance, So 12 can be written as the product of two natural numbers in 3 distinct ways. (Note: We consider $a \\cdot b$ and $b \\cdot a$ to be the same product pair.) What is the smallest odd number which can be factored into the product of two natural numbers in exactly 40 di fferent ways? And what about the smallest even number? (Solution)", "product of two large primes (detailed discussion in [1]). If our \"enemy\" tries to discover the message, using a digit key, it takes years, even if he or she uses the best current supercomputers and algorithms. HEJ, HU ISSN 1418-7108Manuscript no.: ANM-000926-A", "is the largest number which cannot be written as the sum of distinct squares? Products and sums of squares Show that the product of any two numbers, each of which is the sum of two square integers, is itself the sum of two square integers. Odd squares Source: Maths Jam Prove that 1 and 9 are the only square numbers where all the digits are odd.", "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.", "# 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 $$S$$. ×", "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" ]

[ "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "- Symbols. The largest I know is: $$90, 91, 92, 93, 94, 95, 96$$ I.", "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", "# Too many factorials! $\\large \\dfrac{(100!+99!)(99!-98!)(98!+97!)\\cdots(2!+1!)}{(100!-99!)(99!+98!)(98!-97!)\\cdots(2!-1!)}$ The value of the above fraction can be written as mixed fraction $$a\\dfrac{b}{c}$$, where $$b$$ and $$c$$ & $$a$$ and $$b$$ are co-prime, but $$a$$ and $$c$$ are not co-prime. What is the product of $$a$$, $$b$$, and $$c$$? ×", "# Huge Primes $97!\\; n + d$ Find the smallest integer $d > 1$, such that the number above is prime for an infinite number of integers $n$. Notation: $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$. ×", "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 $13$", "is the largest number that will divide into all those given. To find it, the smallest prime numbers should be divided into each one. Prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19. For the given numbers $36$ and $90$, both divided by $2$ give $18$ and $45$. $18$ will divide into both $36$ and $90$, but $45$ will not, so $18$ is the GCF. Oct 14, 2017 G C F 18 It’s also called Greatest Common Divisor G C D #### Explanation: To find G C F of 36, 90: First write the factors of both terms : Factors of $36 = 2 \\cdot \\textcolor{red}{2 \\cdot 3 \\cdot 3}$ Factors of $90 = \\textcolor{red}{2 \\cdot 3 \\cdot 3 \\cdot} 5$ Select the common factors in both terms as marked redabove. $\\textcolor{red}{2 \\cdot 3 \\cdot 3} =$18 is the G C F Nov 21, 2017 Here's a way to find the GCF without using prime factors #### Explanation: Instead of finding the prime factors of the two numbers, ~ make a list of ALL the factors of each number ~ then pick the largest (\"greatest\") one they have in common. To find ALL the factors of a number: ~ Start by factoring by 1 and writing the factors down. ~ Then factor by 2, then by", "common difference is a multiple of the primorial $k$#$=2 \\cdot 3 \\cdot 5 \\cdot ... \\cdot j$, where $j$ is the largest prime not greater than $k$.", "# Gaps between primes I recently watched a video about the recent breakthrough involving the gaps between primes. I have an idea that I'm sure is wrong, but I don't know why. 1. If you take the product of all prime numbers up to a certain number and call it x, won't x-1 and x+1 always be primes? 2. And since they always differ by 2, doesn't that make there an infinite number of primes that differ by 2? Once again, I know that I'm wrong, but I would like to know why. • Because those will not be primes in general. I forgot how many you need to take before it fails. Jul 22 '13 at 18:59 • $2\\cdot 3\\cdot 5 \\cdot 7 = 210$. $209 = 11\\cdot 19$. Jul 22 '13 at 19:01 • There is a difference between relative primes and primes! – Ali Jul 22 '13 at 19:01 • No, for example $2\\cdot3\\cdot5\\cdot7\\cdot11\\cdot 13+1 = 30031 = 59\\cdot509$. We know this number can't be divisible by primes $2,3,5,7,11,13$, but there are bigger possible prime factors. Jul 22 '13 at 19:02 • What was the video? I would like to watch it too! Feb 23 '14 at 18:14 Say $p_k$ is the largest prime $\\le$ some number $n$. Then take $\\displaystyle x=p_1p_2\\cdots p_k$ Now, of", "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 $15120$.", "III. Largest two digit prime number is 97. Question 9 If $$M = \\left(\\frac{3}{7}\\right) \\div \\left(\\frac{6}{5}\\right) \\times \\left(\\frac{2}{3}\\right) + \\left(\\frac{1}{5}\\right) \\times \\left(\\frac{3}{2}\\right)$$ and $$N = \\left(\\frac{2}{5}\\right) \\times \\left(\\frac{5}{6}\\right) \\div \\left(\\frac{1}{3}\\right) + \\left(\\frac{3}{5}\\right) \\times \\left(\\frac{2}{3}\\right) \\div \\left(\\frac{3}{5}\\right)$$, then what is the value of $$\\frac{M}{N}$$? Question 10 M is the largest 4 digit number, which when divided by 4, 5, 6 and 7 leaves remainder as 2, 3, 4, and 5 respectively. What will be the remainder when M is divided by 9? OR", "# Gaps between primes I recently watched a video about the recent breakthrough involving the gaps between primes. I have an idea that I'm sure is wrong, but I don't know why. 1. If you take the product of all prime numbers up to a certain number and call it x, won't x-1 and x+1 always be primes? 2. And since they always differ by 2, doesn't that make there an infinite number of primes that differ by 2? Once again, I know that I'm wrong, but I would like to know why. - Because those will not be primes in general. I forgot how many you need to take before it fails. – Tobias Kildetoft Jul 22 '13 at 18:59 $2\\cdot 3\\cdot 5 \\cdot 7 = 210$. $209 = 11\\cdot 19$. – Daniel Fischer Jul 22 '13 at 19:01 There is a difference between relative primes and primes! – Ali Jul 22 '13 at 19:01 No, for example $2\\cdot3\\cdot5\\cdot7\\cdot11\\cdot 13+1 = 30031 = 59\\cdot509$. We know this number can't be divisible by primes $2,3,5,7,11,13$, but there are bigger possible prime factors. – Thomas Andrews Jul 22 '13 at 19:02 What was the video? I would like to watch it too! – grayQuant Feb 23 at 18:14 Say $p_k$ is the largest prime $\\le$ some number $n$. Then", "# Is 5 * 7 * 11 + 13 * 17 odd or even, prime or composite? So $5 \\cdot 7 \\cdot 11 + 13 \\cdot 17$ is even and being also greater than 2 is composite, being a product of 2 and some number greater than 1.", "# Math Help - Factorials 1. ## Factorials what is the largest two-digit prime factor of $200!$ divided by $(100! 100!)$ 2. you have to cancel this one down, not sure if I can get the answer but I can kick it off for you.... $\\frac{200!}{100!\\times 100!}$ $\\frac{200\\times 199\\times 198 \\times \\cdots101 \\times 100!}{100!\\times 100!}$ $100!$ will cancel $\\frac{200\\times 199\\times 198 \\times \\cdots 101 }{100!}$ expanding the bottom $\\frac{200\\times 199\\times 198 \\times \\cdots 101 }{100\\times 99\\times 98 \\times \\cdots 1 }$ Now first 1st term of 2 should cancel on the bottom $\\frac{2\\times 199\\times 2 \\times \\cdots 101 }{ 99\\times 97 \\times \\cdots 1 }$ There is 50 of them $\\frac{2^{50} \\times 199\\times 197 \\times \\cdots 101 }{ 99\\times 97 \\times \\cdots 1 }$ leaving the bottom in the form $\\frac{2^{50} \\times 199\\times 197 \\times \\cdots 101 }{ 99!! }$ not sure where to go from here...", "\\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", "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", "largest number pari-gp can represent is about $\\text{sexp}_e\\left(4.08\\right)\\approx 7.24676\\cdot 10^{131475196{$ - Sheldon « Next Oldest | Next Newest »", "# 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.", "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)", "largest prime for which $n$ is divisible by $p_k\\#=2\\cdot 3\\cdot 5\\cdot\\ldots\\cdot p_{k-1}\\cdot p_k$. Clearly $(n,m)=1$ if we let $m=p_{k+1}^2$. We need to show that $m<n$. By the Bertrand postulate, $\\frac12p_k<p_{k-1}<p_k<p_{k+1}<2p_k$. But this follows from $m<4p_k^2$ and • $n\\ge p_k\\#\\ge 30p_{k-1}p_k> 15p_k^2$ if $p_k\\ge 11$ • $n\\ge p_k\\# = 210>11^2=m$ if $p_k=7$ • $30|n$ and $n>30$, hence $n\\ge 60$, and $m=49$ if $p_k=5$ • $m\\le 25<n$ if $p_k\\le 3$ - Hint: do you know a proof that there are an infinite number of primes? If you find two of them not factors of $n, \\dots$" ]

[ "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "# Too many factorials! $\\large \\dfrac{(100!+99!)(99!-98!)(98!+97!)\\cdots(2!+1!)}{(100!-99!)(99!+98!)(98!-97!)\\cdots(2!-1!)}$ The value of the above fraction can be written as mixed fraction $$a\\dfrac{b}{c}$$, where $$b$$ and $$c$$ & $$a$$ and $$b$$ are co-prime, but $$a$$ and $$c$$ are not co-prime. What is the product of $$a$$, $$b$$, and $$c$$? ×", "common difference is a multiple of the primorial $k$#$=2 \\cdot 3 \\cdot 5 \\cdot ... \\cdot j$, where $j$ is the largest prime not greater than $k$.", "# Is 5 * 7 * 11 + 13 * 17 odd or even, prime or composite? So $5 \\cdot 7 \\cdot 11 + 13 \\cdot 17$ is even and being also greater than 2 is composite, being a product of 2 and some number greater than 1.", "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 $13$", "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", "is the product of largest and the smallest numbers? A). 1512 B). 1102 C). 1152 D). 1210 Go to : Total Pages : 36", "# Huge Primes $97!\\; n + d$ Find the smallest integer $d > 1$, such that the number above is prime for an infinite number of integers $n$. Notation: $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$. ×", "III. Largest two digit prime number is 97. Question 9 If $$M = \\left(\\frac{3}{7}\\right) \\div \\left(\\frac{6}{5}\\right) \\times \\left(\\frac{2}{3}\\right) + \\left(\\frac{1}{5}\\right) \\times \\left(\\frac{3}{2}\\right)$$ and $$N = \\left(\\frac{2}{5}\\right) \\times \\left(\\frac{5}{6}\\right) \\div \\left(\\frac{1}{3}\\right) + \\left(\\frac{3}{5}\\right) \\times \\left(\\frac{2}{3}\\right) \\div \\left(\\frac{3}{5}\\right)$$, then what is the value of $$\\frac{M}{N}$$? Question 10 M is the largest 4 digit number, which when divided by 4, 5, 6 and 7 leaves remainder as 2, 3, 4, and 5 respectively. What will be the remainder when M is divided by 9? OR", "# Gaps between primes I recently watched a video about the recent breakthrough involving the gaps between primes. I have an idea that I'm sure is wrong, but I don't know why. 1. If you take the product of all prime numbers up to a certain number and call it x, won't x-1 and x+1 always be primes? 2. And since they always differ by 2, doesn't that make there an infinite number of primes that differ by 2? Once again, I know that I'm wrong, but I would like to know why. • Because those will not be primes in general. I forgot how many you need to take before it fails. Jul 22 '13 at 18:59 • $2\\cdot 3\\cdot 5 \\cdot 7 = 210$. $209 = 11\\cdot 19$. Jul 22 '13 at 19:01 • There is a difference between relative primes and primes! – Ali Jul 22 '13 at 19:01 • No, for example $2\\cdot3\\cdot5\\cdot7\\cdot11\\cdot 13+1 = 30031 = 59\\cdot509$. We know this number can't be divisible by primes $2,3,5,7,11,13$, but there are bigger possible prime factors. Jul 22 '13 at 19:02 • What was the video? I would like to watch it too! Feb 23 '14 at 18:14 Say $p_k$ is the largest prime $\\le$ some number $n$. Then take $\\displaystyle x=p_1p_2\\cdots p_k$ Now, of", "# 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", "+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", "is the largest number that will divide into all those given. To find it, the smallest prime numbers should be divided into each one. Prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19. For the given numbers $36$ and $90$, both divided by $2$ give $18$ and $45$. $18$ will divide into both $36$ and $90$, but $45$ will not, so $18$ is the GCF. Oct 14, 2017 G C F 18 It’s also called Greatest Common Divisor G C D #### Explanation: To find G C F of 36, 90: First write the factors of both terms : Factors of $36 = 2 \\cdot \\textcolor{red}{2 \\cdot 3 \\cdot 3}$ Factors of $90 = \\textcolor{red}{2 \\cdot 3 \\cdot 3 \\cdot} 5$ Select the common factors in both terms as marked redabove. $\\textcolor{red}{2 \\cdot 3 \\cdot 3} =$18 is the G C F Nov 21, 2017 Here's a way to find the GCF without using prime factors #### Explanation: Instead of finding the prime factors of the two numbers, ~ make a list of ALL the factors of each number ~ then pick the largest (\"greatest\") one they have in common. To find ALL the factors of a number: ~ Start by factoring by 1 and writing the factors down. ~ Then factor by 2, then by", "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 $15120$.", "# Gaps between primes I recently watched a video about the recent breakthrough involving the gaps between primes. I have an idea that I'm sure is wrong, but I don't know why. 1. If you take the product of all prime numbers up to a certain number and call it x, won't x-1 and x+1 always be primes? 2. And since they always differ by 2, doesn't that make there an infinite number of primes that differ by 2? Once again, I know that I'm wrong, but I would like to know why. - Because those will not be primes in general. I forgot how many you need to take before it fails. – Tobias Kildetoft Jul 22 '13 at 18:59 $2\\cdot 3\\cdot 5 \\cdot 7 = 210$. $209 = 11\\cdot 19$. – Daniel Fischer Jul 22 '13 at 19:01 There is a difference between relative primes and primes! – Ali Jul 22 '13 at 19:01 No, for example $2\\cdot3\\cdot5\\cdot7\\cdot11\\cdot 13+1 = 30031 = 59\\cdot509$. We know this number can't be divisible by primes $2,3,5,7,11,13$, but there are bigger possible prime factors. – Thomas Andrews Jul 22 '13 at 19:02 What was the video? I would like to watch it too! – grayQuant Feb 23 at 18:14 Say $p_k$ is the largest prime $\\le$ some number $n$. Then", "(- 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.", "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", "\\times \\cdots \\times 100 .$ This product is divisible by some $2^m ,$ where $m$ is an integer. What is the largest possible value of $m?$ ×", "+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 ? $$\\small{\\text{ \\begin{array}{r|rcl|r} \\hline n & p_1 & & p_2 & p_1\\cdot p_2 \\\\ \\hline 1 & 11 & \\cdot & 13 & = 143 \\\\ 2 & 11 & \\cdot & 17 & = 187 \\\\ 3 & 11 & \\cdot & 19 & = 209 \\\\ 4 & 11 & \\cdot & 23 & = 253 \\\\ 5 & 11 & \\cdot & 29 & = 319 \\\\ 6 & 11 & \\cdot & 31 & = 341 \\\\ 7 & 11 & \\cdot & 37 & = 407 \\\\ 8 & 11 & \\cdot & 41& = 451 \\\\ 9 & 11 & \\cdot & 43 & = 473 \\\\ 10 & 13 & \\cdot & 17 & = 221 \\\\ 11 & 13 & \\cdot & 19 & = 247 \\\\ 12 & 13 & \\cdot & 23 & = 299 \\\\", "View Single Post P: 87 I want to understand intuitively why LCM(a,b)=ab/GCF(a,b) Quote by statdad I'll use a = 84, b = 96. The prime factorizations of these are \\begin{align*} a & = 2^2 \\cdot 3\\cdot 7 \\\\ b & = 2^5 \\cdot 3 \\end{align*} It's easy to see that the LCM of these numbers is $$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?" ]

[ "From formal English to slang • hour angle — noun Date: circa 1837 the angle between the celestial meridian of an observer and the hour circle of a celestial object measured westward from the meridian … New Collegiate Dictionary • Greenwich hour angle — hour angle measured from the meridian of Greenwich, England. * * * … Universalium", "Angle of the sun.......... Assume that Sun rises at 6 a.m. and sets at 6 p.m. Find the angle made by the Sun rays with the horizontal at 10 a.m. ×", "Definitions # hour angle hour angle, in astronomy, a coordinate in the equatorial coordinate system. The hour angle of a celestial body is the angular distance, expressed in hours, minutes, and seconds (one hour equals 15 degrees), measured westward along the celestial equator from the observer's celestial meridian to the hour circle of the object being located. The hour angle is used in measuring astronomical time; local sidereal time is equal to the hour angle of the vernal equinox. In astronomy, the hour angle is one of the coordinates used in the equatorial coordinate system for describing the position of a point on the celestial sphere. The hour angle of a point is the angle between the half plane determined by the Earth axis and the zenith (half of the meridian plane) and the half plane determined by the Earth axis and the given point. The angle is taken with minus sign if the point is eastward of the meridian plane and with the plus sign if the point is westward of the meridian plane. The hour angle is usually expressed in time units, with 24 hours corresponding to 360 degrees. The hour angle must be paired with the declination in order to fully specify the position of a point on the celestial sphere as seen by the", "you. You can find out more about angles and how they're measured on the page Degrees (Angles). The file is very large. Enter 90; press the \"cos\" button. Online rechneronline.de The reference point 12 o'clock commonly refers to the line of sight and means an angle of 0 degrees. In every hour, the sun travels 15 degrees around the earth. One Degree. Quantity : Reference Unit : is equal to : Conversion Factor : Unit : 1: degree = 1 : degree: 1 = 0.017453292519943 : radian: 1 = 1.1111111111111 : grad Different compass roses (4, 8, 16, 32, 128-wind compass roses) are supported. Everyone who receives the link will be able to view this calculation, Copyright © PlanetCalc Version: Most people read time using either a 12-hour clock or a 24-hour clock. This online calculator outputs compass point given direction angle in degrees. This is how large 1 Degree is . This makes sense because the degrees just increases when going around the clock once, and so does the hours (there's no wrap-around), so maybe we can just multiply the hours by a constant. value for fractional-point bearings increasing in the clockwise direction (CW), value for fractional-point bearings increasing in the counterclockwise direction (CCW), value which avoids the use of \"by\" with fractional points (No \"by\"), value", "the angle between hour-hand and minute-hand in a clock at 2 : 20 Exercise 1.1 | Q 14.(vi) | Page 9 Find the angle between hour-hand and minute-hand in a clock at 10 : 10 Exercise 1.2 [Page 11] ### Balbharati solutions for Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board Chapter 1 Angle and its Measurement Exercise 1.2 [Page 11] Exercise 1.2 | Q 1 | Page 11 Find the length of an arc of a circle which subtends an angle of 108° at the centre, if the radius of the circle is 15 cm. Exercise 1.2 | Q 2 | Page 11 The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length, equal to length of radius. Exercise 1.2 | Q 3 | Page 11 Find the angle in degree subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of the circle is 25 cm. Exercise 1.2 | Q 4 | Page 11 A pendulum of length 14 cm oscillates through an angle of 18°. Find the length of its path. Exercise 1.2 | Q 5 | Page 11 Two arcs of the same lengths subtend angles of 60°", "the great ones of Stone or Wood, are very curious on account of their shewing the Hour of the Day without a Style. Their Exactness consists very much in being very round and even both within side and without. The 13th Figure represents one of these Dials, set upon and fastened on it’s Pedestal, inclining to the Horizon under an Angle equal to the Elevation of the Pole, and directly facing the South: and therefore the Hour-Lines and the Edges AB, ab, serving as a Style, are all parallel between themselves, and to the Axis of the World. The whole Convex Cylinder is divided into 24 equal Parts, or twice 12 Hours, by parallel Lines; and the Concave Semi-Cylinder is divided into 6 equal Parts by Right Lines, which are the Hour-Lines from 6 in the Morning, to 6 in the Afternoon. Now when the Sun shines upon this Dial, the Hour of the Day is shewn on the Convex side thereof, by the defect of Light, that is, by a right Line separating the Light from the Shadow. But the Hour of the Day is shewn in the Concave part of the Dial, by the Shadow of one of the Edges AB or ab; so that when the Sun in the Morning is come to the Hour-Circle", "into hours : so. The important realization for this problem is that the hour hand is always moving. Although this is easiest to picture with a pulse of light, the relationship between distance and intensity is just as true for any source of (incoherent) light.", "Sun so that the Shadow of the Extremity of the Style falls on the Degree of the Sign or Day of the Month. Then you will have the Hour of the Day as before. Thus have I said down the Construction and Uses of Portable Dials, chiefly in use, which may be set North and South, without a Compass or Meridian Line. But before I close this Chapter, I shall briefly describe some other Portable Dials, which are curious enough, but are something difficult to make. The first of these is a horizontal Dial of 2 or 3 Inches square, which we make of Brass or any other solid Metal, for a given Latitude, and whose Axis shewing the Hour, is a Thread fastened at one end to the Center of the said Dial, and the other end of which is hung to the top of a pretty thick Brass Blade, placed at the Extremity of the Dial near the Hour-Line of 12. This Blade may lie down upon the Plane of the Dial, and is kept upright by means of a Spring underneath the Dial; and the Height of the Notch wherein the Thread lies above the Plane of the Dial, is equal to the Tangent of the Latitude. About a quarter of the Height of the", "minute hand on a watch is 8mm long and the hour hand is 4 mm long. How fast is the distance between the tips of the hands changing at one o'clock? THat's the hardest one I have done myself.", "since we are using 12-hour dial.", "Astronomy Answers: From the Astronomical Dictionary # Astronomy AnswersFrom the Astronomical Dictionary $$\\def\\|{&}$$ The description of the word you requested from the astronomical dictionary is given below. the hour angle The hour angle of a celestial body is the difference in right ascension between that body and the meridian (of right ascension) that is due south at that time. The hour angle is usually measured not in degrees but in hours, minutes, and seconds, just like the right ascension. The hour angle indicates (in sidereal time) how long ago the body was due south.", "you agree? It is pi/3 radians (60 degrees). sides equal? Coterminal angles. tight should not be more than 8 cm. Measure the angles and sides of the four triangles, namely A, B, C, and D. Write down the it as a sum of two odd prime numbers. We you can use it to measure 30°. Fold corner 2 so that The units we use for measuring angles are very What do you observe about the angles of for each of your angles is the point labelled O where the tiny vertical line 5^5\\) in expanded form. . At 2:15 o'clock, the hour and minute hands of a clock form an angle of: View Answer. Find the angle of the minute hand from 12 o'clock. similar to each other? See Answer. A triangle with two equal sides is called an isosceles triangle. Can you think of at least five other Discuss this image. Determine the type of angle that the two clock hands will form when the alarm rings a second time. Sixty of the animals on the farm are horses. discover. If you can't do the twist, imagine you have a small piece of soft clay between your thumb and forefinger and you are trying to roll it into a small strip. A triangle is a closed figure", ") 12π/M d ) Hamming window, we we! Recent Posts", "in hours (1 hour = 15 degrees). For example, if an object has an hour angle of 2.5 hours, it transited across the local meridian 2.5 sidereal hours ago (i.e., hours measured using \"sidereal time\"), and is currently 37.5 degrees west of the meridian. Negative hour angles indicate the time until the next transit across the local meridian. Of course, an hour angle of zero means the object is currently on the local meridian. Wikimedia Foundation. 2010. Look at other dictionaries: • Hour angle — Hour Hour, n. [OE. hour, our, hore, ure, OF. hore, ore, ure, F. heure, L. hora, fr. Gr. ?, orig., a definite space of time, fixed by natural laws; hence, a season, the time of the day, an hour. See {Year}, and cf. {Horologe}, {Horoscope}.] 1. The … The Collaborative International Dictionary of English • hour angle — n. Astron. the angle formed at the pole, or its arc on the celestial equator, between the hour circle of a celestial object and the observer s celestial meridian measured in hours, minutes, etc.: it represents the elapsed time since the object… … English World dictionary • hour-angle — hourˈ angle noun (astronomy) The angle (usu measured as time) between the declination circle of a body observed and the observer s meridian • • •", "Hour angle In astronomy and celestial navigation, the hour angle is one of the coordinates used in the equatorial coordinate system to give the direction of a point on the celestial sphere. The hour angle of a point is the angle between two planes: one containing Earth's axis and the zenith (the meridian plane), and the other containing Earth's axis and the given point (the hour circle passing through the point). The hour angle is indicated by an orange arrow on the celestial equator plane. The arrow ends at the hour circle of an orange dot indicating the apparent place of an astronomical object on the celestial sphere. The angle may be expressed as negative east of the meridian plane and positive west of the meridian plane, or as positive westward from 0° to 360°. The angle may be measured in degrees or in time, with 24h = 360° exactly. In astronomy, hour angle is defined as the angular distance on the celestial sphere measured westward along the celestial equator from the meridian to the hour circle passing through a point.[1] It may be given in degrees, time, or rotations depending on the application. In celestial navigation, the convention is to measure in degrees westward from the prime meridian (Greenwich hour angle, GHA), from the local meridian (local", "Angles 1 Verified Answer | Published on 09th 09, 2020 Q4 Single Correct Medium The angle between the minute hand and the hour hand of a clock when the time is 8:25 am is • A. ${92^0}45'$ • B. ${102^0}30'$ • C. ${105^0}$ • D. ${107^0}15'$ Asked in: Mathematics - Understanding Elementary Shapes 1 Verified Answer | Published on 24th 09, 2020 Q5 Subjective Medium Can you draw shapes on these dots? Make squares, rectangles and triangles of different sizes by joining the dots given. One has been done for you. Asked in: Mathematics - Understanding Elementary Shapes 1 Verified Answer | Published on 24th 09, 2020", "120 degree on a clock › degrees on a clock face › convert degrees to decimal degrees › convert hours to degrees › convert degrees … 12-hour clock: The third and last order digit divided the time-degree into 60 parts (the gar), which appears to be sexagesimal. Huygens managed to refine his pendulum clock to have errors of fewer than 10 seconds a day. The second-order digit counted the degrees that had gone by in the hour, notwithstanding the fact that its number of degrees were seasonal. Here, the clock position in hours and minutes and angle in decimal degrees with one decimal place can be converted. Different compass roses (4, 8, 16, 32, 128-wind compass roses) are supported. Today however, atomic clocks are the most accurate devices for time measurement. (This converts the calculator from \"Degrees\" to \"radians\". Take note of the 9 o’clock position in the photo above. Most people read time using either a 12-hour clock or a 24-hour clock. Quantity : Reference Unit : is equal to : Conversion Factor : Unit : 1: degree = 1 : degree: 1 = 0.017453292519943 : radian: 1 = 1.1111111111111 : grad To have it reach 360 again at … The first pendulum mechanical clock was created by Christiaan Huygens in 1656, and was the first clock", "Home » # Hour angle Hour Angle The angle between an observer's and the on which some celestial body lies. Hour Angle The angle, measured westwards around the celestial equator, between the observer's meridian and the hour circle of an object. Hour Circle (i) A great circle passing through an object and the celestial poles. (ii) The setting circle on the polar axis of an equatorial mount. I ... HOUR ANGLE The angle to an object measured westward from the prime meridian (the great circle for a particular observer which goes through the celestial poles and the zenith). Hour Angle the telescope based coordinate specifying the angle, in the equatorial plane, from the meridian to a plane containing the celestial object and the north and south celestial poles. H-R Diagram ... Hour Angle Angular distance on the celestial sphere measured westward along the celestial equator from the meridian to the hour circle that passes through a celestial object. Hour Circle ... Local Hour Angle - The angle, measured westward around the celestial equator, between the meridian and the point on the equator nearest a particular celestial object Long-period Comet - A comet with an orbital period of 200 years or longer ... Hour angle The sidereal time that has elapsed since the object was on the meridian (hour", "angle in degrees and minutes then convert it into decimal degrees.] = cos (93.6167°) = -0.0631, rounded to four decimal places. [Press the cos key, enter the number 93.6167°.]", "of the clock." ]

[ "$59x$ are both integers, which means $x$ is an integer (since 11 and 59 are coprime). But then $720x$ is a multiple of $360$, which means $x$ must be $120$ or $240 \\pmod {360}$ and so $12x$ is another multiple of $360$ contradiction. So there is no solution! • So then it only happens when it's time to repair the clock! Woohoo I was right and I was just trying to be funny! lol – Kingrames Aug 20 '15 at 15:57 • then again, to the naked eye, you could be fooled around 7:17:55 . Didnt bother with the exact timing because I think the proof is above - all that avoiding explicitly stating any spoiler ;) – Spacemonkey Aug 20 '15 at 16:05 • When the hour hand has moved $x$ degrees, the minute hand will have moved $12x$ degrees (in one hour they move 30 degrees and 360 degrees). – Gordon K Aug 20 '15 at 16:22 • @GordonK Oops, you're quite right! Sorry for the mistake; I've now edited and discovered that the final answer was (coincidentally) correct. I hope you or whoever will consider reversing the downvote :-) – Rand al'Thor Aug 20 '15 at 16:32 • I would never be so presumptuous as to down-vote a well-thought out answer with a minor error", "means $x$ is an integer (since 11 and 59 are coprime). But then $720x$ is a multiple of $360$, which means $x$ must be $120$ or $240 \\pmod {360}$ and so $12x$ is another multiple of $360$ contradiction. So there is no solution! • So then it only happens when it's time to repair the clock! Woohoo I was right and I was just trying to be funny! lol Aug 20, 2015 at 15:57 • then again, to the naked eye, you could be fooled around 7:17:55 . Didnt bother with the exact timing because I think the proof is above - all that avoiding explicitly stating any spoiler ;) Aug 20, 2015 at 16:05 • When the hour hand has moved $x$ degrees, the minute hand will have moved $12x$ degrees (in one hour they move 30 degrees and 360 degrees). Aug 20, 2015 at 16:22 • @GordonK Oops, you're quite right! Sorry for the mistake; I've now edited and discovered that the final answer was (coincidentally) correct. I hope you or whoever will consider reversing the downvote :-) Aug 20, 2015 at 16:32 • I would never be so presumptuous as to down-vote a well-thought out answer with a minor error before someone has had the chance to correct it! Aug 20, 2015 at 16:35 If", "converting between decimal version of angle and the DMS (degree, minute, second) notation. #### DMS to decimal To convert from DMS notation to decimal we convert each part separately. #### Decimal to DMS • Firstly remove the whole number, this is the whole degrees. • Then take the decimal and multiply by $60$60. • Remove that whole number, this is the number of minutes. • Take the decimal component of that number and multiply by $60$60. This is the number of seconds. Round this to the required decimal places as necessary. #### Examples ##### Question 1 Convert $35.5$35.5 degrees into degrees, minutes and seconds. Think: A common misconception is to think that the decimal point means that $35.5$35.5 degrees is the same as $35^\\circ5'$35°5 . Its not! I do know that $0.5$0.5 of anything means half. This is half of a degree which is $0^\\circ30'$0°30 , because there are $60'$60 in a degree. Do: $35.5^\\circ$35.5° is $35$35 whole degrees and $0.5$0.5 of a degree. So, $35.5^\\circ$35.5° is $35^\\circ30'$35°30. ##### Question 2 Convert $28.42$28.42° into DMS notation. $28^\\circ$28° Firstly remove the whole number, this is the whole degrees. $0.42\\times60$0.42×60 $=$= $25.2'$25.2′ Then take the decimal and multiply by $60$60 $0^\\circ25'$0°25′ Remove that whole number, this is the number of minutes $0.2\\times60$0.2×60 $=$= $12''$12′′ Take the decimal component of that", "## Precalculus (10th Edition) $\\approx 98.38°$ There are $60$ seconds in $1$minute. Hence $45\"=\\frac{45}{60}'=0.75'$ Hence $98°22'45\"=98°(22+0.75)'$ $=98°22.75'$ There are $60$ minutes in $1$ degree. Hence $22.75'=\\frac{22.75}{60}°\\approx0.379°$ Hence $98°22'45\"\\approx (98+0.379°)°$ $\\approx 98.38°$ (to two decimal places)", "a 360 degree swivel, built-in HD light correction, and one-touch photo best quality 60 degree equal dConvert Degrees, minutes,seconds to decimal degrees best quality 60 degree equal dOne degree is equal to 60 minutes and 3600 seconds: 1º = 60' = 3600\" One minute is equal to 1/60 degrees: 1' = (1/60)º = 0.01666667º One second is equal to 1/3600 degrees: 1\" = (1/3600)º = 2.77778e-4º = 0.000277778º For angle with d integer degrees m minutes and s seconds: best quality 60 degree equal d ### Convert Roof Pitch To Degrees - myrooff best quality 60 degree equal d As for the circle it is equal to 360 degrees or 360°, a minute is equal to 60 seconds or 60 and a degree is equal to 60 minutes or 60. In order to change 30.255012 into an understandable result or measurement, subtract the integer and multiply the decimal point by 60: (30.255012-30) x 60.Cree 150W Equivalent Bright White (3000K) PAR38 Cree 150W Equivalent Bright White (3000K) PAR38 Dimmable Exceptional Light Quality LED 40 Degree Flood Light Bulb Questions - page 2; y_2020, m_9, d_6, h_15CSTDecimal Degrees to Degrees,Minutes,Seconds converterOne degree (°) is equal to 60 minutes (') and equal to 3600 seconds (\"): 1° = 60' = 3600\" The integer degrees (d) are equal to the integer part", "60° with each other between 7 and 8. Solution: We know, the angle between the two hands at h:m is θ = |$\\left|30h-\\frac{11}{2}m\\right|$| Between 7 and 8, h = 7 and need to find the value of m ⇒ 60 = |$\\left|30×7-\\frac{11}{2}m\\right|$| ⇒ ± 60 = 210 - $\\frac{11m}{2}$ Case 1: We take +60 ⇒ 60 = 210 - $\\frac{11m}{2}$ ⇒ 150 = $\\frac{11m}{2}$ ⇒ m = $\\frac{300}{11}$ = $27\\frac{3}{11}$ ∴ The hands make an angle of 60° at 7 : $27\\frac{3}{11}$. Case 2: We take -60 ⇒ -60 = 210 - $\\frac{11m}{2}$ ⇒ 270 = $\\frac{11m}{2}$ ⇒ m = $\\frac{540}{11}$ = $49\\frac{1}{11}$ ∴ The hands make an angle of 60° at 7 : $49\\frac{1}{11}$. ∴ The hands make an angle of 60° at 7 : $27\\frac{3}{11}$ and 7 : $49\\frac{1}{11}$. ## Feedback Help us build a Free and Comprehensive CAT/MBA Preparation portal by providing us your valuable feedback about Apti4All and how it can be improved.", "= 50 × T$$, divide both sides by 50. $$\\frac{420}{50} = T ⇒ T = 8.4 \\space$$ hours. Change hours to minutes for the decimal part. 0.4 hours $$= 0.4 × 60 = 24$$ minutes. 4- D $$x$$ and $$z$$ are colinear. $$y$$ and $$5x$$ are colinear. Therefore, $$x+z=y+5x$$,subtract $$x$$ from both sides,then,$$z=y+4x$$ 5- C $$40\\%$$ of 60 equals to:$$0.40×60=24$$ $$12\\%$$ of 600 equals to: $$0.12×600=72$$ $$40\\%$$ of 60 is added to $$12\\%$$ of 600: $$24+72=96$$ 6- D The relationship among all sides of special right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in this triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. Draw the shape of this question: The latter is the hypotenuse. Therefore, the latter is 60 ft. 7- A The percent of girls playing tennis is: $$40 \\% × 25 \\% = 0.40 × 0.25 = 0.10 = 10 \\%$$ 8- C Add the first 5 numbers. $$40 + 45 + 50 + 35 + 55 = 225$$ To find the distance traveled in the next 5 hours, multiply the average by number of hours. $$Distance = Average × Rate = 50 × 5 = 250$$ Add both numbers. $$250 + 225 = 475$$ 9- C The question is this: 1.75 is what percent of 1.25? Use percent formula: $$part", "= 50 × T$$, divide both sides by 50. $$\\frac{420}{50} = T ⇒ T = 8.4 \\space$$ hours. Change hours to minutes for the decimal part. 0.4 hours $$= 0.4 × 60 = 24$$ minutes. 4- D $$x$$ and $$z$$ are colinear. $$y$$ and $$5x$$ are colinear. Therefore, $$x+z=y+5x$$,subtract $$x$$ from both sides,then,$$z=y+4x$$ 5- C $$40\\%$$ of 60 equals to:$$0.40×60=24$$ $$12\\%$$ of 600 equals to: $$0.12×600=72$$ $$40\\%$$ of 60 is added to $$12\\%$$ of 600: $$24+72=96$$ 6- D The relationship among all sides of special right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in this triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. Draw the shape of this question: The latter is the hypotenuse. Therefore, the latter is 60 ft. 7- A The percent of girls playing tennis is: $$40 \\% × 25 \\% = 0.40 × 0.25 = 0.10 = 10 \\%$$ 8- C Add the first 5 numbers. $$40 + 45 + 50 + 35 + 55 = 225$$ To find the distance traveled in the next 5 hours, multiply the average by number of hours. $$Distance = Average × Rate = 50 × 5 = 250$$ Add both numbers. $$250 + 225 = 475$$ 9- C The question is this: 1.75 is what percent of 1.25? Use percent formula: $$part", "this is very confusing! Apparently you are using \"50.5Deg\" to mean \"50 degrees and 5 minutes\". There are 60 minutes in a degree. 5 minutes is 5/60= 1/12= .08333... of a degree. 50 degrees and 5 minutes is 50.08333.. degrees not 50.8 degrees even after rounding. Quote: Now if I used a larger number after the '.' 50.7Deg (.7)/.6 = 1.1666 +the 50 = 51.1666 What I want is get the 1 from the 51 so that 1.1666x.6 = 0.7 then + 50 = 50.7 Camerart. Again, does \"50.7 Deg\" mean 50 degrees and 7 minutes or 50 and 7/10 degree? If it is 50 degrees and 7 minutes, then since there are 60 minutes in a degree, 7 minutes is 7/60= 0.11666... so 50 degrees and 7 minutes is 50.11666... degrees. If you mean 50 and 7/10 degrees then .7(60)= 42 degrees and 60 and 7/10 degrees is 50 degrees and 42 minutes. March 22nd, 2016, 02:12 AM #6 Member Joined: Jan 2016 From: Uk Posts: 80 Thanks: 2 Hi, I still get mixed up between: Degrees, Mins decimal mins and Degrees, mins and seconds and Degrees and decimal mins seconds. Here is an NMEA sentence: \\$GPGGA,FIELD.00,1234.50000,N,xxxxx.xxxx,W,1,08,0.9 ,000.0,M,46.9,M,,*47 The digits before the N are Latitude, and = 12Deg 34Mins .50000 So Degrees, Mins and decimal mins. This", "minutes and seconds To convert a decimal value of degrees into minutes and seconds, we need to know that: • $1$1 degree $=$=$60$60 minutes • $1$1 minute $=$=$60$60 seconds These conversion factors also tell us that the number of minutes or seconds should never be greater than $60$60 in a fully simplified expression. #### Worked example Convert $24.759^\\circ$24.759° into degrees, minutes and seconds. Think: We already have $24$24 whole degrees, so we just need to convert $0.759^\\circ$0.759° into minutes and seconds. Do: We know that $1^\\circ$1° is equal to $60$60 minutes, so: $0.759$0.759 degrees $=$=$0.759\\times60$0.759×60 minutes $=$=$45.54$45.54 minutes This tells us that we have $45$45 whole minutes, with the remaining $0.54$0.54 minutes to be converted into seconds. Since $1$1 minute is equal to $60$60 seconds: $0.54$0.54 minutes $=$=$0.54\\times60$0.54×60 seconds $=$=$32.4$32.4 seconds Therefore, $24.759^\\circ$24.759° is equal to $24$24 degrees, $45$45 minutes and $32.4$32.4 seconds. Using the degrees, minutes and seconds notation, we can write this as: $24.759^\\circ=24^\\circ45'32.4\"$24.759°=24°4532.4\" #### Practice questions ##### Question 1 Find the angle of depression from point $B$B to point $D$D. Use $x$x as the angle of depression and give your answer in degrees and minutes to the nearest minute. ##### Question 2 The angle of elevation from an observer to the top of a tree is $35^\\circ$35°. The distance between the tree and the observer", "How do you convert 45°50'50\" to decimal degree notation? Feb 5, 2015 Remember every minute $'$ is $1 / 60$th of a degree and every second $' '$ is $1 / 60$th of a minute. Step 1: seconds to minutes $50 ' ' = 50 / 60 = 0.8333 \\ldots '$ Step 2: Add the $50 '$ Step 3: minutes to degrees We now have $50.8333 \\ldots ' = 50.8333 \\ldots / 60 = 0.8472 {\\ldots}^{0}$ Step 4: Add the ${45}^{0}$ $45.8472 {\\ldots}^{0}$", "math. I know once the minutes exceed 60 you have to add to the degrees but I'm having trouble with the conversion. Any help would be greatly appreciated. Thanks What do you mean by 071* 59.922? Give it to me in word form. 3. Thanks for replying; it would be 71 degress, 59 minutes and 922 seconds West and 40 degrees, 58 minutes and 647 seconds North. The seconds don't matter for this problem. 4. Originally Posted by Sarnick Thanks for replying; it would be 71 degress, 59 minutes and 922 seconds North and 40 degrees, 58 minutes and 647 seconds west. The seconds don't matter for this problem. OK. $73'+59'=132'$. Note that $132'\\div60'=2^\\circ\\frac{1}{5}^\\circ$ But $\\frac{1}{5}^\\circ=\\frac{60}{5}'=12'$ Can you see where I'm going? 5. Thanks for trying to help but your confusing me more. LOL! Here's an example; If the number you had to add to the West coords was 62 then the answer would be 73 degrees, 01 minutes and 922 seconds. I was never any good at math and thanks for your time. Does this help? 6. Originally Posted by Sarnick Thanks for trying to help but your confusing me more. LOL! Here's an example; If the number you had to add to the West coords was 62 then the answer would be 73 degrees, 01 minutes", "= 60 seconds or 1 second = (1/60) minute = (1/3600) degree conversion from DD to DMS to DDM and vice versa can be done using simple calculations. Example 6 a) Convert 3.138675 $^{\\circ}$ to DDM and DMS b) Convert 45 $^{\\circ}$ 55' 23.7'' to DD and DDM Solution a) 3.138675 $^{\\circ}$ = 3 $^{\\circ}$ + 0.138675 × 60 minutes = 3 $^{\\circ}$ 8.3205' DDM = 3 $^{\\circ}$ + 8' + 0.3205 × 60 seconds = 3 $^{\\circ}$ + 8' + 19.23 '' = 3 $^{\\circ}$ 8' 19.23 '' DMS b) From the rate given above, we can write that 1' = (1/60)$^{\\circ}$ \\) and 1'' = (1/3600)$^{\\circ}$ \\) Hence 45 $^{\\circ}$ 55' 23.7'' = 45$^{\\circ}$ + 55' + 23.7'' = 45$^{\\circ}$ + 55 (1/60)$^{\\circ}$ + 23.7 (1/3600)$^{\\circ}$ = 45.92325 $^{\\circ}$ DD 45.92325 $^{\\circ}$ = 45$^{\\circ}$ + 0.92325 $^{\\circ}$ = 45$^{\\circ}$ + 0.92325 × 60' = 45$^{\\circ}$ 55.395' DDM", "grades, minutes and degrees 1. $$30^{\\circ}$$ 2. $$81^{\\circ}$$ 3. $$138^{\\circ}30'$$ 4. $$35^{\\circ}47'15''$$ 5. $$235^{\\circ}12'36''s$$ 6. $$475^{\\circ}13'48''$$ Express in terms of right angles and also in degrees, minutes and seconds the angles 1. $$120^g$$ 2. $$45^g35'24''$$ 3. $$39^g45'36''$$ 4. $$255^g8'9''$$ 5. $$759^g0'5''$$ 6. Reduce $$55^{\\circ}12'36''$$ to centisimal measure. 7. Reduce $$18^{\\circ}33'45''$$ to circular measure. 8. Reduce $$196^g35'24''$$ to sexagecimal measure. 9. How many degrees, minutes and seconds are respectively passed over in $$11\\frac{1}{9}$$ minutes by the hour and minute hand of a watch. 10. The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both angles in degrees. 11. Prove that the number of Sexagecimal minutes in any angle is to the number of Centisimal minutes in the same angle as $$27:50.$$ 12. Divide $$44^{\\circ}8'$$ into two parts such that the number of Sexagecimal seconds in one part may be equal to number of Centisimal seconds in the other part. 13. The angles of a triangle are in the ratio of $$3:4:5$$, find the smallest angle in degrees and greatest angle in radians. 14. Find the angle between the hour hand and the minute hand in circular measure at half past four. 15. If $$p, q$$ and $$r$$ denote the grade measure, degree measure and the", "# The minute hand of a certain clock is 4 in The minute hand of a certain clock is 4 in long, Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand? 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 yotaniwc We know that : (see the picture) $\\therefore A=\\frac{1}{2}{r}^{2}\\theta$ $\\frac{dA}{dt}=\\frac{1}{2}{r}^{2}\\frac{d\\theta }{dt}$ (r is constant) The minute hand takes 60 minutes for one revolution. Then the angular velocity of one revolution per minute is $\\frac{d\\theta }{dt}=\\frac{2\\pi }{60}=\\frac{\\pi }{30}\\frac{rad}{min}$ Thus, $\\therefore \\frac{dA}{dt}=\\frac{4\\pi }{15}\\frac{{\\in }^{2}}{min}$ raefx88y 1. Speed of the minute hand is constant = 360 degrees per hour, or $2\\pi \\frac{radians}{hr}$ 2. Sector area $=\\frac{\\theta }{2\\pi }\\pi {r}^{2}{\\in }^{2}=\\frac{1}{2}{r}^{2}\\theta {\\in }^{2}=\\frac{16}{2}\\theta {\\in }^{2}=8\\theta {\\in }^{2}$ 3.$\\frac{d\\theta }{dt}=\\frac{2\\pi }{60}\\frac{radians}{min}$ 4. $\\frac{d}{dt}$ (sector area) $=\\frac{d}{dt}8\\theta =8\\frac{d}{dt}\\theta =\\frac{16\\pi }{60}=\\frac{4}{15}\\pi \\frac{{\\in }^{2}}{min}$", "degrees (decimal point included in seconds) $+$30:00:00 degrees (integer number of seconds) Enter maximum value of the required range. If the column within which the range is being specified is not an angle then simply enter the required value. If the column is an angle then the value can be entered as either a decimal value in radians or a sexagesimal value in hours or degrees, minutes and seconds. If a sexagesimal value is specified then the hours or degrees, minutes and seconds should be separated by a colon (:). Optionally fractional seconds can be specified by including a decimal point and the required number of places of decimals. An unsigned value is assumed to be in hours and a signed value in degrees (a negative angle cannot be specified in hours). That is, a positive angle in degrees must be preceded by a plus sign. Examples: any of the following values could be entered to to specify an angle of 30 degrees: 2:00:00.0 hours (decimal point included in seconds) 2:00:00 hours (integer number of seconds) $+$30:00:00.0 degrees (decimal point included in seconds) $+$30:00:00 degrees (integer number of seconds) Enter the required row number in the current selection. Enter the first row to be listed in the current selection. Enter the last row to be listed (0 =", "- 1 2 - 5 8 - 8. At $AB:CD:EF$, $A<3; C,E<6$. If $A<2$, there are 2 values available for $A$, with whichever chosen also used for $F$. $B$ and $C$ can be 0, 1, 2 or 5 ($2*4*4=32$ combinations). If $A=2$, then $B=0,1$ or $2$, and $C=0,1,2$ or $5$ ($3*4=12$ combinations). There are 44 in total. Edit: the last one wasn't what the OP meant. • What does vertical symmetry have to do with palindromicity? – Rand al'Thor Jan 10 '17 at 13:35 It's all down to the minutes - it needs a palindromic number of minutes. 00, 11, 22, 33, 44, 55 So for every hour, there are six possible palindromes. Then work out which hours can have a reversed number of seconds (less than 60): 01, 02, 03, 04, 05 (06, 07, 08, 09) 10, 11, 12, 13, 14, 15 (16, 17, 18, 19) 20, 21, 22, 24, 24. 16 possibles. 16 x 6 = 96 In the am/pm scenario, the preceding zero would never be included, therefore the number of palindromes possible would be significantly reduced below 96. Also, since the AM and PM symbols cannot be part of the palindrome, the first half, and the second half, of the 24 hour period would be indistinguishable, and therefore not add to the total. •", "# The Law of Sines and a second forces magnitude How do you solve for the second forces magnitude? The 17' and 30' threw me off. - Look for example at $29^\\circ 17'$. That means $29$ degrees and $17$ minutes. But a minute is one-sixtieth of a degree, so $17$ minutes is $\\frac{17}{60}$ degrees, approximately $0.2833333$ degrees. Therefore $29^\\circ 17'$ is approximately $29.2833333$ degrees. Similarly, $76^\\circ 30'$ just means $76+\\frac{30}{60}$ degrees, that is, $76.5$ degrees. So in the \"normal\" formulas that you would use to find the magnitude of the resultant force, just use the appropriate decimal approximation. Some calculators allow direct input in degrees and minutes, and do the conversion to decimal automatically. In old-fashioned astronomy, there is a finer subdivision still, the second. There are $60$ seconds in a minute. So $1$ second is $\\frac{1}{3600}$ degrees. If you are told that an angle is $29^\\circ 17'\\, 34''$, that means $29+\\frac{17}{60}+\\frac{34}{3600}$ degrees. You could use a calculator to get a good decimal approximation to this, and then proceed as usual. Remark: The use of minutes and seconds to measure angles is slowly (too slowly!) fading. However, you may need to acquire some ability to transform a decimal degree answer, such as $42.372$ degrees, to degrees, minutes, and perhaps even seconds. (It may be that in your answers,", "the equations $$2y=10⇒y=5$$ 3- C Use distance formula: $$Distance = Rate × time ⇒ 420 = 50 × T$$, divide both sides by 50. $$\\frac{420}{50} = T ⇒ T = 8.4 \\space$$ hours. Change hours to minutes for the decimal part. 0.4 hours $$= 0.4 × 60 = 24$$ minutes. 4- D $$x$$ and $$z$$ are colinear. $$y$$ and $$5x$$ are colinear. Therefore, $$x+z=y+5x$$,subtract $$x$$ from both sides,then,$$z=y+4x$$ 5- C $$40\\%$$ of 60 equals to:$$0.40×60=24$$ $$12\\%$$ of 600 equals to: $$0.12×600=72$$ $$40\\%$$ of 60 is added to $$12\\%$$ of 600: $$24+72=96$$ 6- D The relationship among all sides of the special right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in this triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. Draw the shape of this question: The latter is the hypotenuse. Therefore, the latter is 60 ft. 7- A The percent of girls playing tennis is: $$40 \\% × 25 \\% = 0.40 × 0.25 = 0.10 = 10 \\%$$ 8- C Add the first 5 numbers. $$40 + 45 + 50 + 35 + 55 = 225$$ To find the distance traveled in the next 5 hours, multiply the average by number of hours. $$Distance = Average × Rate = 50 × 5 = 250$$ Add both numbers. $$250 + 225 = 475$$", "$= 7.59005\\times 90^\\circ = 683^\\circ6'1.62''$ 22. $55^\\circ12'36'' = \\left(55 + \\frac{12}{60} + \\frac{36}{60\\times 60}\\right)^\\circ = 55.21^\\circ$ $= \\frac{55.21}{90} = .6134444$ right angles $= 61^g34'44.4''$ 23. $18^\\circ33'45'' = \\left(18 + \\frac{33}{60} + \\frac{45}{60\\times60}\\right)^\\circ = \\frac{1485\\pi}{80\\times180}$ radians $= \\frac{33\\pi}{320}$ radians. 24. $195^g35'24'' = 195.3524^g = 1.953524$ right angles $= 1.953524\\times90^\\circ = 175^\\circ49'1.8''$ 25. Minute-hand make $360^\\circ$ in $60$ minutes. $\\therefore$ angle made in $11\\frac{1}{9}$ minutes $= \\frac{360}{60}.\\frac{20}{9} = 66^\\circ40'$ Hour-hand makes $360^\\circ$ in $720$ minutes i.e. $12$ times less. $\\therefore$ angle made by hour-hand $\\frac{66^\\circ40'}{12} = 5^\\circ83'20''$ 26. Let the angle in degrees be $x,$ then from problem statement we have $x^\\circ + x^g = 1$ right angle Now we can use the formula $\\frac{D}{180} = \\frac{G}{200},$ $x + \\frac{9}{10}x = 1$ $x = 47\\frac{7}{19}$ $\\frac{9x}{10} = 42\\frac{12}{19}$ 27. Let there be $x$ sexagecimal minutes. $x' = \\frac{x}{60\\times 90}$ right angle $x$ centisimal angles $= \\frac{x}{100\\times100}$ right angles. Thus we find that ratio is $27:50$ 28. Let there be $x$ seconds in both. Thus, $\\left(\\frac{x}{1000000}90 + \\frac{x}{60\\times60}\\right)^\\circ = 44^\\circ8'$ Thus, the two parts are $33^\\circ20'$ and $10^\\circ48'$ 29. Let the angles be $3x, 4x, 5x$ in degrees. From Geometry, we know that $3x + 4x + 5x = 180^\\circ$ (sum of all angles of a triangle is $180^\\circ$) $\\Rightarrow x = 15^\\circ$ Thus angles are $45^\\circ, 60^\\circ, 75^\\circ$ or $\\frac{\\pi}{4}," ]

[ "$59x$ are both integers, which means $x$ is an integer (since 11 and 59 are coprime). But then $720x$ is a multiple of $360$, which means $x$ must be $120$ or $240 \\pmod {360}$ and so $12x$ is another multiple of $360$ contradiction. So there is no solution! • So then it only happens when it's time to repair the clock! Woohoo I was right and I was just trying to be funny! lol – Kingrames Aug 20 '15 at 15:57 • then again, to the naked eye, you could be fooled around 7:17:55 . Didnt bother with the exact timing because I think the proof is above - all that avoiding explicitly stating any spoiler ;) – Spacemonkey Aug 20 '15 at 16:05 • When the hour hand has moved $x$ degrees, the minute hand will have moved $12x$ degrees (in one hour they move 30 degrees and 360 degrees). – Gordon K Aug 20 '15 at 16:22 • @GordonK Oops, you're quite right! Sorry for the mistake; I've now edited and discovered that the final answer was (coincidentally) correct. I hope you or whoever will consider reversing the downvote :-) – Rand al'Thor Aug 20 '15 at 16:32 • I would never be so presumptuous as to down-vote a well-thought out answer with a minor error", "60° with each other between 7 and 8. Solution: We know, the angle between the two hands at h:m is θ = |$\\left|30h-\\frac{11}{2}m\\right|$| Between 7 and 8, h = 7 and need to find the value of m ⇒ 60 = |$\\left|30×7-\\frac{11}{2}m\\right|$| ⇒ ± 60 = 210 - $\\frac{11m}{2}$ Case 1: We take +60 ⇒ 60 = 210 - $\\frac{11m}{2}$ ⇒ 150 = $\\frac{11m}{2}$ ⇒ m = $\\frac{300}{11}$ = $27\\frac{3}{11}$ ∴ The hands make an angle of 60° at 7 : $27\\frac{3}{11}$. Case 2: We take -60 ⇒ -60 = 210 - $\\frac{11m}{2}$ ⇒ 270 = $\\frac{11m}{2}$ ⇒ m = $\\frac{540}{11}$ = $49\\frac{1}{11}$ ∴ The hands make an angle of 60° at 7 : $49\\frac{1}{11}$. ∴ The hands make an angle of 60° at 7 : $27\\frac{3}{11}$ and 7 : $49\\frac{1}{11}$. ## Feedback Help us build a Free and Comprehensive CAT/MBA Preparation portal by providing us your valuable feedback about Apti4All and how it can be improved.", "∠DWU = 180° [Interior angles on the same side of transversal] ∠BUW + 110°= 180° ∠BUW = 180-110° = 70° ∠BUW = ∠AUS= 70° [vertically opposite angles] ∠AUS + ∠SUQ + ∠QUB = 180° ∠AUB is a straight angle] 70 + ∠SUQ + 70 = 180° ∠SUQ + 140° = 180° ∠SUQ = 180° – 140° ∠SUQ = 40° Question 9. What is the angle between the hours hand and minutes hand of a clock at (i) 1.40 hours (ii) 2.15 hours [use 1° = 60 minutes] Solution: i. 1.40 hours – 190° ii. 2.15 hours – 22° 30′ Question 10. How much would hour’s hand have moved from its position at 12 noon when the time is 4.24 p.m? Solution: 144° Question 11. Let $\\overrightarrow{\\mathrm{AB}}$ be a line segment and let C be the midpoint of $\\overrightarrow{\\mathrm{AB}}$ Extend AB to D such that B lies between A and D. Prove that AD + BD = 2CD Solution: = AC + CB + BD + BD [∵ AD = AC + CB + BD] = CB + CB + BD + BD [∴ AC = CB] = CB + BD + CB + BD = CD + CD = 2CD = RHS ∴ AD + B D = 2CD Question 12. Let $\\overrightarrow{\\mathbf{A B}}$ and $\\overrightarrow{\\mathbf{C D}}$ be", "means $x$ is an integer (since 11 and 59 are coprime). But then $720x$ is a multiple of $360$, which means $x$ must be $120$ or $240 \\pmod {360}$ and so $12x$ is another multiple of $360$ contradiction. So there is no solution! • So then it only happens when it's time to repair the clock! Woohoo I was right and I was just trying to be funny! lol Aug 20, 2015 at 15:57 • then again, to the naked eye, you could be fooled around 7:17:55 . Didnt bother with the exact timing because I think the proof is above - all that avoiding explicitly stating any spoiler ;) Aug 20, 2015 at 16:05 • When the hour hand has moved $x$ degrees, the minute hand will have moved $12x$ degrees (in one hour they move 30 degrees and 360 degrees). Aug 20, 2015 at 16:22 • @GordonK Oops, you're quite right! Sorry for the mistake; I've now edited and discovered that the final answer was (coincidentally) correct. I hope you or whoever will consider reversing the downvote :-) Aug 20, 2015 at 16:32 • I would never be so presumptuous as to down-vote a well-thought out answer with a minor error before someone has had the chance to correct it! Aug 20, 2015 at 16:35 If", "# The Law of Sines and a second forces magnitude How do you solve for the second forces magnitude? The 17' and 30' threw me off. - Look for example at $29^\\circ 17'$. That means $29$ degrees and $17$ minutes. But a minute is one-sixtieth of a degree, so $17$ minutes is $\\frac{17}{60}$ degrees, approximately $0.2833333$ degrees. Therefore $29^\\circ 17'$ is approximately $29.2833333$ degrees. Similarly, $76^\\circ 30'$ just means $76+\\frac{30}{60}$ degrees, that is, $76.5$ degrees. So in the \"normal\" formulas that you would use to find the magnitude of the resultant force, just use the appropriate decimal approximation. Some calculators allow direct input in degrees and minutes, and do the conversion to decimal automatically. In old-fashioned astronomy, there is a finer subdivision still, the second. There are $60$ seconds in a minute. So $1$ second is $\\frac{1}{3600}$ degrees. If you are told that an angle is $29^\\circ 17'\\, 34''$, that means $29+\\frac{17}{60}+\\frac{34}{3600}$ degrees. You could use a calculator to get a good decimal approximation to this, and then proceed as usual. Remark: The use of minutes and seconds to measure angles is slowly (too slowly!) fading. However, you may need to acquire some ability to transform a decimal degree answer, such as $42.372$ degrees, to degrees, minutes, and perhaps even seconds. (It may be that in your answers,", "# Aptitude and Reasoning Questions • #### Verbal Reasoning - Mental Ability Q: An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon? A) 360 B) 180 C) 90 D) 60 Explanation: Angle traced by the hour hand in 6 hours=(360/12)*6 73 47283 Q: if the price of a book is first decreased by 25% and then increased by 20%, then the net change in the price will be : A) 10 B) 20 C) 30 D) 40 Explanation: Let the original price be Rs. 100. New final price = 120 % of (75 % of Rs. 100) = Rs. (120/100 * 75/100 * 100) = Rs. 90. Decrease = 10% 172 46389 Q: Find the number of triangles in the given figure. A) 28 B) 32 C) 36 D) 40 Explanation: The simplest triangles are AML, LRK, KWD, DWJ, JXI, IYC, CYH, HTG, GOB, BOF, FNE and EMA i.e. 12 in number. The triangles composed of two components each are AEL, KDJ, HIC and FBG i.e. 4 in number. The triangles composed of three components each are APF, EQB, BQH, GVC, CVJ, IUD, DUL and KPA i.e. 8 in number. The triangles composed of six components each are ASB, BSG,", "Kudos encourage active discussions. Originally posted by Archit3110 on 25 Feb 2019, 04:49. Last edited by Archit3110 on 25 Feb 2019, 08:35, edited 2 times in total. Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4402 Location: India GPA: 3.5 Re: In the diagram above, the clock shows that it is five minutes after 9 [#permalink] ### Show Tags 25 Feb 2019, 08:28 Bunuel wrote: In the diagram above, the clock shows that it is five minutes after 9 o’clock. If the radius of the clock is 9 inches, what is the area of the sector created by the hands of the clock starting at 9 and moving clockwise? A. 6π B. 13.5π C. 27π D. 81π E. 54π Attachment: 2019-02-25_1316.png 60 Min = 360° So, 1 min = 6° And , 5 Min = 30° Thus, the angle represented by the minute and hour hand is 90° + 30° = 120° Area of the sector is $$\\frac{120}{360}*π*9^2 = 27π$$, Answer must be (C) 27π _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in", "way of looking at it assuming continuous, constant movement of the hands of the clock. One nice, simplifying thing is that the hour hand and minute hand start at the same point, let's say $$(0,0)$$. Let's use $$(t,\\theta)$$ to denote the internal angle of the circle $$\\theta$$ swept out over time $$t$$, where $$t$$ is measured in minutes. For the hour hand, we start at $$(0,0)$$ and end at $$(60,\\frac{1}{12}\\cdot2\\pi)$$ over the course of an hour. We can let the function $$\\theta_h(t)=\\frac{\\pi t}{360}$$ describe such motion. Similarly, for the minute hand, we start at $$(0,0)$$ and end at $$(60,2\\pi)$$ over the course of an hour. We can let the function $$\\theta_m(t)=\\frac{\\pi t}{30}$$ model this motion. Hence, we are essentially looking for a $$t$$ for which $$\\theta_m-\\theta_h=\\frac{\\pi}{2}$$ as well as a $$t$$ for which $$\\theta_m-\\theta_h=\\frac{3\\pi}{2}$$. Or, more helpfully, we are trying to find a $$t$$ for which $$\\frac{\\pi t}{30}-\\frac{\\pi t}{360}=\\frac{\\pi}{2},\\frac{3\\pi}{2},$$ and doing some basic algebra shows that what we get are solutions when $$t=\\frac{180}{11}, \\frac{540}{11}$$; hence, the difference in minutes between these two instances is $$\\frac{540}{11}-\\frac{180}{11}=\\frac{360}{11}=32\\frac{8}{11}\\;\\text{minutes}$$ or, more simply, 32 minutes and 44 seconds. Side note: The hour and minute hands will be at right angles at (roughly) 12:16:22pm and 12:49:05pm. As said above, this is probably not the cleanest way of approaching it, but maybe it will be", "4 \\cos(A)}}$$ which needs to be maximized. #### Attached Files: • ###### clock hands 2.jpg File size: 20.1 KB Views: 102 Last edited: Jul 17, 2006 18. Jul 19, 2006 ### andrevdh Differentiating the las equation again gets me to (I hope someone is checking this): $$\\frac{1}{2R\\dot{A}^2} \\ddot{a} = \\frac{B\\cos(A) - \\frac{2\\sin^2(A)}{B}}{B^2}$$ where $$B = \\sqrt{5 - 4\\cos(A)}$$ multiplying the numerator with B and a little manipulation gives $$\\cos^2(A) - 2.5\\cos(A) + 1 = 0$$ $$A = 60^o\\ !$$ so the greates rate of separation occurs when the angle between the hands are sixty degrees, which occurs 10 minutes 55 seconds after midnight. Last edited: Jul 19, 2006 19. Jul 19, 2006 ### Andrew Mason A very impressive effort to solve this rather quaint problem! I'll have a look at it tonight to check it. AM 20. Jul 20, 2006 ### Andrew Mason This looks right. The second derivative is a bit of a challenge. As I suspected, the maximum rate of separation occurs when the angle of the line between the ends makes a 90 degree angle with the hour hand ($a^2 = 3R^2 = (2R)^2 - R^2$)! AM", "248 Joined: Dec 2013 RE: Clock Hands Challenge (03-06-2019 07:28 PM)Thomas Klemm Wrote: The peculiar thing happens between 5 and 6 o'clock when the order of the 0° and the 180° event changes. Examples: 5 XEQ \"CLOCK\" y: 6.0000 x: 5.2716 6 XEQ \"CLOCK\" y: 6.0000 x: 6.3244 Until 5 the 0° event happens before the 180° event. From 6 on the 0° event happens after the 180° event. This makes the calculation of these events \"past given hour\" somewhat complicated. Cheers Thomas Yes, that is the tricky bit. Thanks 03-07-2019, 02:19 PM Post: #16 xerxes Member Posts: 106 Joined: Jun 2014 RE: Clock Hands Challenge A more compact version with Joe's suggestion: Code: INPUT H H=MOD(H,12) F=12/11 IF H<6 THEN D=-6 ELSE D=6 PRINT DMS(H*F) PRINT DMS((H-D)*F+D) The very limited FX-180P makes it somewhat complicated, but it's possible: Code: P1: 01 Kin1 02 Kin2 03 1 04 2 05 / 06 1 07 1 08 * 09 Kin3 10 Kout1 11 = 12 °′″← 13 HLT 14 6 15 Kin-2 16 Kout2 P2: 01 +/- 02 x>0 03 Kin/2 04 6 05 Kin*2 06 Kout2 07 Kin+1 08 Kout3 09 Kin*1 10 Kout2 11 Kin-1 12 Kout1 13 °′″← 29 steps Usage example: 7 P1 RUN P2 The 180° result for 6 o'clock is excluded and gives", "At 1 o'clock the minute hand (red) points to the 12 and the hour hand (blue) points to the 1. Calculate Angle between Clock Hands. In this video, 1A Auto shows you how to change the positioning on an alternator. If not, see if you can find a \"D/R\" or \"DRG\" button and press it once; then try again. When studying the physi… At 1 o'clock the minute hand (red) points to the 12 and the hour hand (blue) points to the 1. There are also 24 hours in a day. Browser slowdown may occur during loading and creation. That line indicates south in the Northern Hemisphere. Since there are 12 hours on the clock, each hour mark is 30 degrees. .hours { background: #000; height: 20%; left: 48.75%; position: absolute; top: 30%; transform-origin: 50% 100%; width: 2.5%; } I’m using percentages to make scaling the clocks easier. Related Time Card Calculator | Time Calculator. Calculations of sunrise and sunset in The Angle – Ohio – USA for January 2021. Generic astronomy calculator to calculate times for sunrise, sunset, moonrise, moonset for many cities, with daylight saving time and time zones taken in account. This calculator allows you to enter direction angle, either in DMS (degrees, minutes, and seconds) or decimal degrees and outputs closest to the compass", "a circle, and a circle always contains 360 degrees. Online rechneronline.de The reference point 12 o'clock commonly refers to the line of sight and means an angle of 0 degrees. A 20 degree cut can be aimed at 7:30 or 4:30 on the clock face, a 25 degree shot at 8:00 or 4:00. Calculate the Angle between 12 and the Hour hand 10: Since there are 360 degrees in a full circle (clock), and there are 12 hours, each hour represents 360/12 = 30 degrees So our formula is 30 (H) So our formula is 30 (10) Quantity : Reference Unit : is equal to : Conversion Factor : Unit : 1: degree = 1 : degree: 1 = 0.017453292519943 : radian: 1 = 1.1111111111111 : grad An angle is in standard position if its vertex is at the origin and the initial side is on the positive x-axis. Simply described, this means that if the average temperature in a day is 20oC then a plant accumulates 20 day degrees in one calendar day. Ball landmarks come in handy. h = distance above or … Now set the dish to the satellite you want using a Satellite finder or whatever decoder you own. Then the angles between the two hands are calculated. A clock position, or clock bearing,", "$= 7.59005\\times 90^\\circ = 683^\\circ6'1.62''$ 22. $55^\\circ12'36'' = \\left(55 + \\frac{12}{60} + \\frac{36}{60\\times 60}\\right)^\\circ = 55.21^\\circ$ $= \\frac{55.21}{90} = .6134444$ right angles $= 61^g34'44.4''$ 23. $18^\\circ33'45'' = \\left(18 + \\frac{33}{60} + \\frac{45}{60\\times60}\\right)^\\circ = \\frac{1485\\pi}{80\\times180}$ radians $= \\frac{33\\pi}{320}$ radians. 24. $195^g35'24'' = 195.3524^g = 1.953524$ right angles $= 1.953524\\times90^\\circ = 175^\\circ49'1.8''$ 25. Minute-hand make $360^\\circ$ in $60$ minutes. $\\therefore$ angle made in $11\\frac{1}{9}$ minutes $= \\frac{360}{60}.\\frac{20}{9} = 66^\\circ40'$ Hour-hand makes $360^\\circ$ in $720$ minutes i.e. $12$ times less. $\\therefore$ angle made by hour-hand $\\frac{66^\\circ40'}{12} = 5^\\circ83'20''$ 26. Let the angle in degrees be $x,$ then from problem statement we have $x^\\circ + x^g = 1$ right angle Now we can use the formula $\\frac{D}{180} = \\frac{G}{200},$ $x + \\frac{9}{10}x = 1$ $x = 47\\frac{7}{19}$ $\\frac{9x}{10} = 42\\frac{12}{19}$ 27. Let there be $x$ sexagecimal minutes. $x' = \\frac{x}{60\\times 90}$ right angle $x$ centisimal angles $= \\frac{x}{100\\times100}$ right angles. Thus we find that ratio is $27:50$ 28. Let there be $x$ seconds in both. Thus, $\\left(\\frac{x}{1000000}90 + \\frac{x}{60\\times60}\\right)^\\circ = 44^\\circ8'$ Thus, the two parts are $33^\\circ20'$ and $10^\\circ48'$ 29. Let the angles be $3x, 4x, 5x$ in degrees. From Geometry, we know that $3x + 4x + 5x = 180^\\circ$ (sum of all angles of a triangle is $180^\\circ$) $\\Rightarrow x = 15^\\circ$ Thus angles are $45^\\circ, 60^\\circ, 75^\\circ$ or $\\frac{\\pi}{4},", "In the analog watch compass model, we use the reference that the earth is stationary and the sun moves around the earth. We also assume that the time that the sun is closest to the observer on the earth is noon, i.e., 12:00PM without day light saving and 1:00PM with day light saving. In the northern hemisphere, at noon, when we align the hour hand of the watch with the direction of the sun, the hour hand direction points to south and its opposite points to north. When the sun starts to move from east to west, in every 2 hours, the sun moves around the earth for $360 / (24 / 2) = 30$ degree, which happens to be equivalent to the degree that the hour hand moves on the watch in 1 hour. So when we looked at the time, we actually know how many hours it is now before or after noon, and how much degree it is before or after the sun moves to the position that is the closest to the observer, which is at the south direction. For example, in the top-left figure, it is roughly 10:00AM. We know it is 2 hours before the noon, and the hour hand is pointing to the sun which is 30 degree east to the", "but it is fine as a heuristic given the amount of time elapsed is not too much. Anyway, the assumption was not to get an accurate answer anyway. It was, as I said, for a sanity test. Three o'clock is a special time because the tip of the minute hand has all of the change in $x$, and the tip of the hour hand has all of the change in $y$. So, let's try this. The distance $D=\\sqrt{x^2+y^2}$, and the change can be expressed as $$\\frac{dD}{dt} = \\frac{\\partial D}{\\partial x}\\frac{dx}{dt} + \\frac{\\partial D}{\\partial y}\\frac{dy}{dt}.$$ The partials are $$\\frac{\\partial D}{\\partial x} = \\frac{x}{D};\\frac{\\partial D}{\\partial y} = \\frac{y}{D}$$ The change in the distance $D$ with respect to $x$ is negative at three o'clock, because of the motion and position of the hands. Likewise, the change in $D$ with respect to $y$ is positive. At three o'clock, $x=4, y=5, D=\\sqrt{41},$ so $\\frac{\\partial D}{\\partial x} = -4/\\sqrt{41}$ and $\\frac{\\partial D}{\\partial y} = 5/\\sqrt{41}$. The velocities of the hands are $10\\pi$ and $2\\pi/3$ inches per hour for the minute and hour hands, respectively. At three o'clock, the minute hand velocity is aligned exactly on the $x$ direction, and the hour hand velocity exactly on the $y$ direction. So, $$\\frac{dD}{dt} = \\frac{-4\\cdot 10\\pi + 5 \\cdot 2\\pi/3}{\\sqrt{41}} \\doteq -17.99 \\text{ inches/hr}.$$ Looks like we", "A circular clock of 12 parts of each measuring 30° in 1:30 P.M. we have 5 parts or 150° now taking = 150°/180° × π =5/6π b. 3:15 P.M. Explanation: in 3:15 P.M. we have 12 parts or 360° now taking = 360°/180° × π =2π Hours Hand 3:15 = 3 + $$\\frac {1}{4}$$ = $$\\frac {12 + 1}{4}$$ = $$\\frac {13}{4}$$ for 12 hr = 360 for $$\\frac {13}{4}$$ hr = 360 x $$\\frac {13}{4}$$ = $$\\frac {13 x 360°}{4 x 12}$$ = $$\\frac {13 x 15}{2}$$ = 97.5° Minutes hand 60 minutes = 360° 15 minutes = $$\\frac {360 x 15}{60}$$ = 90° 97.5° – 90° = 7.5° $$\\frac {7.5 x π}{180}$$ = $$\\frac {π}{180}$$ the measure (in radians) of the angle formed by the hands of a clock at each time 7.5 x Question 39. MATHEMATICAL CONNECTIONS The sum of the circumferences of circles A, B, and C is 63π. Find AC. AC = 28 units. Explanation: Based on the information given in the above figure, Question 40. THOUGHT PROVOKING Is π a rational number? Compare the rational number $$\\frac{355}{113}$$ to π. Find a different rational number that is even closer π. No, π is not a rational number. Explanation: Because π can not be represented as an equivalent fraction. where as, we know that", "grades, minutes and degrees 1. $$30^{\\circ}$$ 2. $$81^{\\circ}$$ 3. $$138^{\\circ}30'$$ 4. $$35^{\\circ}47'15''$$ 5. $$235^{\\circ}12'36''s$$ 6. $$475^{\\circ}13'48''$$ Express in terms of right angles and also in degrees, minutes and seconds the angles 1. $$120^g$$ 2. $$45^g35'24''$$ 3. $$39^g45'36''$$ 4. $$255^g8'9''$$ 5. $$759^g0'5''$$ 6. Reduce $$55^{\\circ}12'36''$$ to centisimal measure. 7. Reduce $$18^{\\circ}33'45''$$ to circular measure. 8. Reduce $$196^g35'24''$$ to sexagecimal measure. 9. How many degrees, minutes and seconds are respectively passed over in $$11\\frac{1}{9}$$ minutes by the hour and minute hand of a watch. 10. The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both angles in degrees. 11. Prove that the number of Sexagecimal minutes in any angle is to the number of Centisimal minutes in the same angle as $$27:50.$$ 12. Divide $$44^{\\circ}8'$$ into two parts such that the number of Sexagecimal seconds in one part may be equal to number of Centisimal seconds in the other part. 13. The angles of a triangle are in the ratio of $$3:4:5$$, find the smallest angle in degrees and greatest angle in radians. 14. Find the angle between the hour hand and the minute hand in circular measure at half past four. 15. If $$p, q$$ and $$r$$ denote the grade measure, degree measure and the", "take all six possible differences, find the min, then multiply by 6 to rescale to 360 degrees. The two-layer list comprehension first chooses the first hand as represented by 720, 60, or 1, then chooses the other hand out of that set with the first choice removed via set xor. I tested this exhaustively against the reference code. # C#, 163 152 bytes This creates every hand twice to count for wrap-around, then loops through every combination and finds the minimum angle between hands. Calculations are performed in 60 divisions, then multiplied by 6 to get degrees. Indented for clarity: float F(int s){ float b=60,c; float[]a={c=s/b/b%12*5,c+b,c=s/b%b,c+b,s%=60,s+b}; for(s=36;s-->0;) b=s%6!=s/6&(c=(c=a[s%6]-a[s/6])<0?-c:c)<b?c:b; return b*6; } Example output: 0 seconds, 00:00:00, smallest angle is 0° 43200 seconds, 12:00:00, smallest angle is 0° 86399 seconds, 23:59:59, smallest angle is 0.09164429° 3330 seconds, 00:55:30, smallest angle is 54.75° 39930 seconds, 11:05:30, smallest angle is 60.25001° 21955 seconds, 06:05:55, smallest angle is 65.49998° 21305 seconds, 05:55:05, smallest angle is 59.50001° 5455 seconds, 01:30:55, smallest angle is 75.45831° 41405 seconds, 11:30:05, smallest angle is 44.95834° • Nice solution to account for wrap around – toto Jul 29 '15 at 13:37 # TI-BASIC, 17 bytes min(cos⁻¹(cos(ΔList(Ans{6,.1,5!⁻¹,6 Uses Dennis's arccos(cos( to normalize distances; however, rather than calculating all pairwise distances, it only calculates the three needed using ΔList([seconds],[minutes],[hours],[seconds]. This", "degrees (decimal point included in seconds) $+$30:00:00 degrees (integer number of seconds) Enter maximum value of the required range. If the column within which the range is being specified is not an angle then simply enter the required value. If the column is an angle then the value can be entered as either a decimal value in radians or a sexagesimal value in hours or degrees, minutes and seconds. If a sexagesimal value is specified then the hours or degrees, minutes and seconds should be separated by a colon (:). Optionally fractional seconds can be specified by including a decimal point and the required number of places of decimals. An unsigned value is assumed to be in hours and a signed value in degrees (a negative angle cannot be specified in hours). That is, a positive angle in degrees must be preceded by a plus sign. Examples: any of the following values could be entered to to specify an angle of 30 degrees: 2:00:00.0 hours (decimal point included in seconds) 2:00:00 hours (integer number of seconds) $+$30:00:00.0 degrees (decimal point included in seconds) $+$30:00:00 degrees (integer number of seconds) Enter the required row number in the current selection. Enter the first row to be listed in the current selection. Enter the last row to be listed (0 =", "this is very confusing! Apparently you are using \"50.5Deg\" to mean \"50 degrees and 5 minutes\". There are 60 minutes in a degree. 5 minutes is 5/60= 1/12= .08333... of a degree. 50 degrees and 5 minutes is 50.08333.. degrees not 50.8 degrees even after rounding. Quote: Now if I used a larger number after the '.' 50.7Deg (.7)/.6 = 1.1666 +the 50 = 51.1666 What I want is get the 1 from the 51 so that 1.1666x.6 = 0.7 then + 50 = 50.7 Camerart. Again, does \"50.7 Deg\" mean 50 degrees and 7 minutes or 50 and 7/10 degree? If it is 50 degrees and 7 minutes, then since there are 60 minutes in a degree, 7 minutes is 7/60= 0.11666... so 50 degrees and 7 minutes is 50.11666... degrees. If you mean 50 and 7/10 degrees then .7(60)= 42 degrees and 60 and 7/10 degrees is 50 degrees and 42 minutes. March 22nd, 2016, 02:12 AM #6 Member Joined: Jan 2016 From: Uk Posts: 80 Thanks: 2 Hi, I still get mixed up between: Degrees, Mins decimal mins and Degrees, mins and seconds and Degrees and decimal mins seconds. Here is an NMEA sentence: \\$GPGGA,FIELD.00,1234.50000,N,xxxxx.xxxx,W,1,08,0.9 ,000.0,M,46.9,M,,*47 The digits before the N are Latitude, and = 12Deg 34Mins .50000 So Degrees, Mins and decimal mins. This" ]

[ "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);", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "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]$", "and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle ABE = 60$. Let $M$ be a point such $\\overline{EM}$ is parellel to $\\overline{CD}$. We immediatley know that $\\triangle BEM \\sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\\triangle EFM \\sim \\triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\\triangle EBF$ is in $\\triangle ABF$, the area has ratio $\\triangle BEF : \\triangle ABE=1:2$ and we know $\\triangle ABE$ has area $60$ so $\\triangle BEF$ is $\\boxed{\\textbf{B} \\, 30}$. - fath2012 ## Solution 9 $[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(F--D); 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(\"x\", (D+E+F)/3); label(\"x\", (B+E+F)/3); label(\"120+2x\", (D+F+C)/3); [/asy]$ Labeling the areas in the diagram, we have: $[DBC]=240=[BFE]+[FED]+[FDC]=x+x+120+2x=120+4x$ so $240=120+4x, 120=4x, 30=x$. So our answer is $\\boxed{\\textbf{(B)} 30}$. ~~RWhite ## Solution 10 (Menelaus's Theorem) $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF", "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", "# 2021 AIME II Problems/Problem 2 ## Problem Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\\overline{BD} \\perp \\overline{BC}$. The line $\\ell$ through $D$ parallel to line $BC$ intersects sides $\\overline{AB}$ and $\\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\\ell$ such that $F$ is between $E$ and $G$, $\\triangle AFG$ is isosceles, and the ratio of the area of $\\triangle AFG$ to the area of $\\triangle BED$ is $8:9$. Find $AF$. ## Diagram $[asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label(\"A\",A,dir(90)); label(\"B\",B,dir(225)); label(\"C\",C,dir(-45)); label(\"D\",D,dir(180)); label(\"E\",E,dir(-45)); label(\"F\",F,dir(225)); label(\"G\",G,dir(0)); label(\"\\ell\",midpoint(E--F),dir(90)); [/asy]$ ## Solution 1 (Area Formulas for Triangles) By angle chasing, we conclude that $\\triangle AGF$ is a $30^\\circ\\text{-}30^\\circ\\text{-}120^\\circ$ triangle, and $\\triangle BED$ is a $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangle. Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\\triangle BED,$ we have $DE=\\frac{840-x}{2}$ and $DB=\\frac{840-x}{2}\\cdot\\sqrt3.$ Let the brackets denote areas. We have $$[AFG]=\\frac12\\cdot AF\\cdot FG\\cdot\\sin{\\angle AFG}=\\frac12\\cdot x\\cdot x\\cdot\\sin{120^\\circ}=\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}$$ and $$[BED]=\\frac12\\cdot DE\\cdot DB=\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right).$$ We set up and solve an equation for $x:$ \\begin{align*} \\frac{[AFG]}{[BED]}&=\\frac89 \\\\ \\frac{\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}}{\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right)}&=\\frac89 \\\\ \\frac{2x^2}{(840-x)^2}&=\\frac89 \\\\ \\frac{x^2}{(840-x)^2}&=\\frac49. \\end{align*} Since $0 it is clear that $\\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \\begin{align*} \\frac{x}{840-x}&=\\frac23 \\\\ 3x&=1680-2x \\\\", "+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", "(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.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "# Difference between revisions of \"2019 AIME II Problems/Problem 1\" ## Problem Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\\triangle ABC$ and $\\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution $[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label(\"A\",A,dir(-120)); label(\"B\",B,dir(-60)); label(\"C\",C,dir(60)); label(\"D\",D,dir(120)); label(\"9\",(A+B)/2,dir(-90)); label(\"10\",(D+A)/2,dir(-150)); label(\"10\",(C+B)/2,dir(-30)); label(\"17\",(D+B)/2,dir(60)); label(\"17\",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label(\"8\",(D+(-6,0))/2,dir(180)); label(\"6\",(A+(-6,0))/2,dir(-90)); [/asy]$ - Diagram by Brendanb4321 Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so that you have a rectangle. The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $E$ be the intersection of $BD$ and $AC$. This means that $\\triangle ABE$ and $\\triangle DCE$ are with ratio $\\frac{21}{9}=\\frac73$. Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $E$ to $DC$, and $x$ be the height of $\\triangle ABE$. $$\\frac{7}{3}=\\frac{y}{x}$$ $$\\frac{7}{3}=\\frac{8-x}{x}$$ $$7x=24-3x$$ $$10x=24$$ $$x=\\frac{12}{5}$$ This means that the area is $A=\\tfrac{1}{2}(9)(\\tfrac{12}{5})=\\tfrac{54}{5}$. This gets us $54+5=\\boxed{059}.$ -Solution by the Math Wizard, Number Magician", "= \\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),", "= (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.", "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 =", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "$\\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}$.", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "the area of $\\triangle DCE$ to the area of $\\triangle ABD$ is $\\dfrac{\\dfrac{\\sqrt{2}}{9}}{\\dfrac{\\sqrt{2}}{3}}$ = $\\boxed{\\textbf{(C) }\\frac{1}{3}}$ ## Solution 2 Since $\\triangle DCE$ and $\\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$. $[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label(\"A\",A,NE);label(\"B\",B,W);label(\"C\",C,SE);label(\"D\",D,NE);label(\"E\",E,SW);label(\"O\",O,SW);label(\"F\",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\\frac{1}{3}r$. Since $m\\angle DCO=m\\angle DFC=90^\\circ$, then $\\triangle DCO\\cong \\triangle DFC$. So the ratio of the two altitudes is $\\frac{CF}{DC}=\\frac{OC}{DO}=\\boxed{\\textbf{(C) }\\frac{1}{3}}$ ## Solution 3 Say the center of the circle is point $O$; Without loss of generality, assume $AC=2$, so $CB=4$ and the diameter and radius are $6$ and $3$, respectively. Therefore, $CO=1$, and $DO=3$. The area of $\\triangle DCE$ can be expressed as $\\frac{1}{2}(CD)(6)\\text{sin }(CDE).$ $\\frac{1}{2}(CD)(6)$ happens to be the area of $\\triangle ABD$. Furthermore, $\\text{sin } CDE = \\frac{CO}{DO},$ or $\\frac{1}{3}.$ Therefore, the ratio is $\\boxed{\\textbf{(C) }\\frac{1}{3}}.$ ## Solution 5 $[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label(\"E\",E,SSE); label(\"B\",B,E); label(\"A\",A,W); label(\"D\",D,NNW); label(\"C\",C,SW); draw(rightanglemark(D,C,B,2)); [/asy]$ Let the point G be the reflection of point $D$ across $\\overline{AB}$. (Point G is on the circle). Let $AC=x$, then $BC=2x$. The diameter is $3x$. To find $DC$, there are two ways (presented here): 1. Since $\\overline{AB}$ is the", "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =" ]

[ "ABD = 90^{ \\circ}$ thus we have a right-triangle (think of Pythagoras' theorem and trig ratios) By Pythagoras, $\\displaystyle (AD)^2 = (BD)^2 + (AB)^2$ but AB = AC since both have the radius of the circle for their length so, $\\displaystyle (AD)^2 = (BD)^2 + (AC)^2$ since AD = AC + CD, we get $\\displaystyle (AD)^2 = (BD)^2 + (AD - CD)^2$ expand the last set of brackets, we get: $\\displaystyle (AD)^2 = (BD)^2 + (AD)^2 - 2 AD \\cdot CD + (CD)^2$ $\\displaystyle \\Rightarrow 2 AD \\cdot CD = (BD)^2 + (CD)^2$ $\\displaystyle \\Rightarrow AD = \\frac {(BD)^2 + (CD)^2}{2 CD}$ and i think you can take it from here Page 1 of 2 12 Last", "# Area of a special isosceles spherical triangle I was asked to calculate the area of the following isosceles triangle. Let $\\triangle ABC$ be an isosceles spherical triangle with $d(A,C)=d(B,C)$. Then $C$ lies on the perpendicular bisector of $AB$. Let $M$ be the midpoint of $AB$. Suppose $d(A,M)=d(-M,C)$. Calculate the area of $\\triangle ABC$. I already made a sketch and observed that $d(A,M)=d(-M,C)=\\pi-d(M,C)$. Let $\\alpha$, $\\beta$, $\\gamma$ denote the interior angles corresponding to $A$, $B$, $C$ respectively. If I assume without loss of generality that, because $C$ lies on the perpendicular bisector of $AB$, the angle $\\gamma$ is divided into two equal angles $\\gamma_1$ and $\\gamma_2$ belonging to the right triangles $\\triangle AMC$ and $\\triangle BMC$. Since $$\\operatorname{area}(\\triangle ABC) = 2 \\cdot \\operatorname{area}(\\triangle AMC) = 2 \\cdot \\operatorname{area}(\\triangle BMC)$$ I only need to calculate $\\operatorname{area}(\\triangle AMC)$. Since $\\operatorname{area}(\\triangle AMC)=\\alpha + \\gamma_1+ \\frac{\\pi}{2}-\\pi$, I need $\\alpha$ and $\\gamma_1$. I first tried to calculate $\\gamma_1$ using the spherical Law of Cosines, and I got this: $$\\cos\\gamma_1=\\cos\\alpha \\cdot \\cos\\frac{\\pi}{2} - \\sin\\alpha \\cdot \\sin\\frac{\\pi}{2} \\cdot \\cos d(A,M)=-\\sin\\alpha \\cdot \\cos\\left(\\pi-d(M,C)\\right) \\tag{1}$$ From the spherical Law of Sines, I get: $$\\sin\\alpha= \\frac{\\sin d(M,C)}{\\sin d(A,C)} \\tag{2}$$ But now I do not see how to proceed. I would like to calculate the right-hand sides of $(1)$ and $(2)$, but this does not seem possible with the", "+0 # Geometry 0 119 3 If altitude CD is $$8 \\sqrt{3}$$ centimeters, what is BD? Feb 15, 2022 #1 +2444 0 This problem is actually simple because it is a 30-60-90 triangle. Because we know that $$CD = 8\\sqrt3$$$$AD$$ must equal $$\\sqrt2\\times 8\\sqrt3 = 8\\sqrt6$$ We also know that $$BD$$ must equal to $${8\\sqrt3}\\over\\sqrt2$$, which can be rewritten as $$4\\sqrt6$$ This means that $$\\color{brown}\\boxed{AB=12\\sqrt6}$$ Feb 15, 2022 #2 +124594 +1 Note that triangles ACD , CBD are similar and each are 30-60-90 right triangles And since AD is opposite a 60° angle it = sqrt 3 * CD = sqrt 3 * 8 sqrt 3 = 24 And, by similar triangles, we have the following relationship DB / DC = DC / AD (8sqrt 3)^2 = DB * 24 256 / 24 = DB = 32 / 3 (cm) Feb 15, 2022 #3 +2444 +1 I misread the question... DB should be $$4 \\sqrt6$$, because $$\\triangle{CDB}$$ is a 30-60-90. Feb 15, 2022", "right triangle ( $\\angle \\, ACE = 90^{\\circ}$ ) and so is $AHD$, which means they are similar and thus $$\\frac{AD}{AE}=\\frac{AH}{AC}$$ which is equivalent to $$AD \\cdot AC = AH \\cdot AE = 60^2$$ and since $AD = AC - CD = 80 - CD$ and $AC = 80$ get the equation $$(80-CD)\\cdot 80 = 60^2$$ When you solve it you get $CD = 35$. • Did you assume here that the point $O$ belongs to $BD$? Also, \"Let $H$ be the intersection point of $AD$ and $AO$\" sounds a bit strange. Why isn't the intersection point just $A$? – mavzolej Feb 24 '17 at 4:50 • @mavzolej Because it is a typo. It's $BD$ intersects $AO$ at $H$. I fixed it. Read it again. – Futurologist Feb 24 '17 at 13:47 • I think you may also have a typo when saying \"triangles $AHB$ and $ABC$ are similar\". – mavzolej Feb 24 '17 at 16:40 • @mavzolej I think it is obvious – Futurologist Feb 24 '17 at 17:39 Let $G$ be the point of intersection of $AO$ and $BD.$ $AGB$ is a right triangle. Drop altitudes $OE$ and $OF$ to sides $AB$ and $AC$ respectively Since $AOB$ and $AOC$ are isosceles, these altitudes bisect their respective sides. $AFO$ is similar to $AGB$ $AF:AO = AG: AB$", "turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\\cdot B'C + 2\\cdot 16 = 15^2$, whence $A'D = B'C = \\sqrt{193}$. Let $\\alpha = \\angle A'B'D$. The Law of Cosines on triangle $A'B'D$ yields $$\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,$$ and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle A'O_2D$ yields \\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $A'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$. ~djmathman ## Solution 2 Denote by", "OG13 DS79 The area of a right triangle can be easily expressed as half the product of the two legs. Area of triangle ABC is 0.5AC\\cdot{BC} and that of the triangle DBA is 0.5AD\\cdot{AB}. The question in fact is \"Is AC\\cdot{BC} = AB\\cdot{AD}?\" (1) From AC^2=2AD^2, we deduce that AC=\\sqrt{2}AD, which, if we plug into AC\\cdot{BC} = AB\\cdot{AD}, we get \\sqrt{2}AD\\cdot{BC}=AB\\cdot{AD}, from which \\sqrt{2}BC=AB. This means that triangle ABC should necessarily be isosceles, which we don't know. Not sufficient. (2) Obviously not sufficient, we don't know anything about AD. (1) and (2): If AC=BC=x, then AB=x\\sqrt{2}, and AD=\\frac{x}{\\sqrt{2}}. Then AC\\cdot{BC}=AB\\cdot{AD}=x^2. Sufficient. Answer C. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Intern Joined: 06 Apr 2011 Posts: 13 Followers: 0 Kudos [?]: 0 [0], given: 292 Re: In the figure above, is the area of triangular region ABC [#permalink] 13 Mar 2013, 22:53 Nice and succinct explanation Eva... Intern Joined: 04 Sep 2012 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: In the figure above, is the area of triangular region ABC [#permalink] 05 Jun 2013, 17:41 I still don't get it. To compare the Area of (ABC) and Area of (DBA): is 1/2(AC)(CB) = 1/2(AD)(AB) ? In the hint 1: (AC)^2=2(AD)^2 -> AC = sq(2)*(AD) -> AD > AC (1) Then", "point $E$, such that $AECD$ is a parallelogram. In other words, $$AE = CD = 10$$ and $$EC = DA = 2\\sqrt{65}$$ Now, let's examine $\\triangle ABE$. Since $AB = AE = 10$, the triangle is isosceles, and $\\angle ABE \\cong \\angle AEB$. Note that in parallelogram $AECD$, $\\angle AED$ and $\\angle CDE$ are congruent, so $\\angle ABE \\cong \\angle CDE$ and thus $$\\text{m}\\angle ABD = 180^\\circ - \\text{m}\\angle CDB$$ Define $\\alpha := \\text{m}\\angle CDB$, so $180^\\circ - \\alpha = \\text{m}\\angle ABD$. We use the Law of Cosines on $\\triangle DAB$ and $\\triangle CDB$: $$\\left(2\\sqrt{65}\\right)^2 = 10^2 + BD^2 - 20BD\\cos\\left(180^\\circ - \\alpha\\right) = 100 + BD^2 + 20BD\\cos\\alpha$$ $$14^2 = 10^2 + BD^2 - 20BD\\cos\\alpha$$ Subtracting the second equation from the first yields $$260 - 196 = 40BD\\cos\\alpha \\rightarrow BD\\cos\\alpha = \\frac{8}{5}$$ This means that dropping an altitude from $B$ to some foot $Q$ on $\\overline{CD}$ gives $DQ = \\frac{8}{5}$ and therefore $CQ = \\frac{42}{5}$. Seeing that $CQ = \\frac{3}{5}\\cdot BC$, we conclude that $\\triangle QCB$ is a 3-4-5 right triangle, so $BQ = \\frac{56}{5}$. Then, the area of $\\triangle BCD$ is $\\frac{1}{2}\\cdot 10 \\cdot \\frac{56}{5} = 56$. Since $AP = CP$, points $A$ and $C$ are equidistant from $\\overline{BD}$, so $$\\left[\\triangle ABD\\right] = \\left[\\triangle CBD\\right] = 56$$ and hence $$\\left[ABCD\\right] = 56 + 56 = \\boxed{112}$$", "\\implies \\angle BAC = \\angle ACD'$$ $$\\implies \\triangle ABC = \\triangle AD'C \\implies BC = AD'.$$ The area of quadrilateral $ABCD$ is equal to the area of triangle $ADD'$ with sides $AD' = 14, AD = 2\\sqrt{65}, DD' = 2 \\cdot 10 = 20$. The semiperimeter is $s = 17 + \\sqrt{65},$ the area $$[ADD'] = \\sqrt {(17 + \\sqrt{65}) (17 - \\sqrt{65})(3 + \\sqrt{65})(\\sqrt{65}-3)} = \\sqrt{(289 – 65)\\cdot(65-9)} =\\sqrt{56 \\cdot 4 \\cdot 56} = \\boxed{112}.$$", "\\cdot \\sin(2 \\theta) \\cdot a^2 + \\frac14 \\cdot \\sin(2 \\theta) \\cdot b^2 = \\frac14 \\cdot \\sin(2 \\theta) \\cdot c^2$. Now just divide out the common factor of $\\frac14 \\cdot \\sin(2 \\theta)$, and you’re left with $a^2 + b^2 = c^2$. What just happened? The way Pythagoras’s equation fell out of equating areas may seem like a bit of a magic trick too, but it’s actually based on the very fundamental idea of scale invariance. To recap: we (1) wrote a completely explicit formula for the area of a right triangle, (2) equated formulas corresponding to equal areas, and (3) found that the bulk of the formulas, everything except the part corresponding to the Pythagorean theorem, went away. The key to understanding all that is this picture: It shows the area of a right triangle embedded in the area of the corresponding square, with the hypotenuse matching one side of the square. The precise formula for the ratio of the areas, $\\frac14 \\cdot \\sin(2 \\theta)$, doesn’t matter so much — what matters is that when we blow this picture up or down, the ratio of the areas doesn’t change. That’s scale invariance. If the two smaller triangles add up to the big triangle, then the squares corresponding to the smaller triangles have to add up to the square corresponding", "$DOP$ is similar to triangle $AOB$, and triangle $DPC$ is similar to triangle $ABC$. If $DP$ = $15y$, then $CP$ = $36y$, $OP$ = $10y$, and $OD$ = $5y\\sqrt {13}$. Since $OP$ + $CP$ = $46y$ = $\\frac {26}{3}$, $y$ = $\\frac {13}{69}$, and $AD$ = $\\frac {60\\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$). The answer is $60$ + $13$ + $23$ = $\\boxed{096}$. ## Solution 2 Extend $AB$ and $CD$ to intersect at $P$. Note that since $\\angle ACB=\\angle PCB$ and $\\angle ABC=\\angle PBC=90^{\\circ}$ by ASA congruency we have $\\triangle ABC\\cong \\triangle PBC$. Therefore $AC=PC=13$. By the angle bisector theorem, $PD=\\dfrac{130}{23}$ and $CD=\\dfrac{169}{23}$. Now we apply Stewart's theorem to find $AD$: \\begin{align*}13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=13\\cdot 13\\cdot \\dfrac{130}{23}+10\\cdot 10\\cdot \\dfrac{169}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=\\dfrac{169\\cdot 130+169\\cdot 100}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=1690\\\\ AD^2&=130-\\dfrac{130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130\\cdot 23^2-130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130(23^2-169)}{23^2}\\\\ AD^2&=\\dfrac{130(360)}{23^2}\\\\ AD&=\\dfrac{60\\sqrt{13}}{23}\\\\ \\end{align*} and our final answer is $60+13+23=\\boxed{096}$. ## Solution 3 Notice that by extending $AB$ and $CD$ to meet at a point $E$, $\\triangle ACE$ is isosceles. Now we can do a straightforward coordinate bash. Let $C=(0,0)$, $B=(12,0)$, $E=(12,-5)$ and $A=(12,5)$, and the equation of line $CD$ is $y=-\\dfrac{5}{12}x$. Let F be the intersection point of $AD$ and $BC$, and by using the Angle Bisector Theorem: $\\dfrac{BF}{AB}=\\dfrac{FC}{AC}$ we have $FC=\\dfrac{26}{3}$. Then the equation of", "Pythagorean Theorem, we have $AD = \\sqrt{3^2-1^2} = \\sqrt{8}$. Now using $\\triangle ADO_1 \\sim \\triangle AFC$, $\\frac{AD}{AF} = \\frac{DO_1}{FC} \\Longrightarrow \\frac{2\\sqrt{2}}{8} = \\frac{1}{FC} \\Longrightarrow FC = 2\\sqrt{2}$ Hence, the area of the triangle is $\\frac{1}{2}\\cdot AF \\cdot BC = \\frac{1}{2}\\cdot AF \\cdot (2\\cdot CF) = AF \\cdot CF = 8\\left(2\\sqrt{2}\\right) = \\boxed{16\\sqrt{2}\\ \\mathrm{(D)}$ (Error compiling LaTeX. ! File ended while scanning use of \\boxed.) 2006 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions", "is also an isosceles trapezoid; in turn, $A'D = O_1O_2 = 15$, and similarly $B'C = 15$. Thus, Ptolmey's theorem on $B'A'CD$ yields $B'D\\cdot A'C + 2\\cdot 16 = 15^2$, whence $B'D = A'C = \\sqrt{193}$. Let $\\alpha = \\angle B'A'D$. The Law of Cosines on triangle $B'A'D$ yields$$\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,$$and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle B'A'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $B'A'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2B' = r_1$ and $O_2D = O_1A' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $B'O_2D$ and $B'A'D$ are opposite angles in cyclic quadrilateral $A'B'O_2D$, which implies the measure of angle $B'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle B'O_2D$ yields\\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $B'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$. ~djmathman ## Solution", "$\\theta$ is $\\frac14 \\cdot \\sin(2 \\theta) \\cdot c^2$. The green and blue triangle have exactly the same interior angles as the big one. So their areas are given by the same exact formula, with c replaced by a and b, respectively. The areas have to add up, so we have: $\\frac14 \\cdot \\sin(2 \\theta) \\cdot a^2 + \\frac14 \\cdot \\sin(2 \\theta) \\cdot b^2 = \\frac14 \\cdot \\sin(2 \\theta) \\cdot c^2$. Now just divide out the common factor of $\\frac14 \\cdot \\sin(2 \\theta)$, and you’re left with $a^2 + b^2 = c^2$. What just happened? The way Pythagoras’s equation fell out of equating areas may seem like a bit of a magic trick too, but it’s actually based on the very fundamental idea of scale invariance. To recap: we (1) wrote a completely explicit formula for the area of a right triangle, (2) equated formulas corresponding to equal areas, and (3) found that the bulk of the formulas, everything except the part corresponding to the Pythagorean theorem, went away. The key to understanding all that is this picture: It shows the area of a right triangle embedded in the area of the corresponding square, with the hypotenuse matching one side of the square. The precise formula for the ratio of the areas, $\\frac14 \\cdot \\sin(2 \\theta)$, doesn’t matter so", "P. ... 1. The triangle is a right triangle. 2. The given circle is the incircle of the triangle. 3. For convenience I've labeled the side OA = a, OB = b and AB = c. 4. Divide the triangle into three triangles with the radius of the circle as its' height. (see attachment) 5. Now you know: $a^2+b^2=c^2~~~\\text{Pythagorean theorem}$ $\\dfrac12 \\cdot a \\cdot b = 600~~~\\text{triangle is a half rectangle}$ $\\dfrac12 \\cdot a \\cdot r+\\dfrac12 \\cdot b \\cdot r+\\dfrac12 \\cdot c \\cdot r= 600 ~~~\\text{sum of the areas of the 3 partial triangles}$ 6. Thus you have to solve for (a,b,c) the system of equations: $\\left|\\begin{array}{rcl}a+b+c&=&120 \\\\ a^2+b^2&=&c^2 \\\\ a \\cdot b&=&1200\\end{array}\\right.$ I've got $(a,b,c)= (30, 40, 50)$ 7. I'll leave the rest for you. • May 28th 2009, 12:50 AM earboth Quote: Originally Posted by nathan02079 ... 3. It is given that the perimeter of the rectangle ABCD is four times greater than the radius of the circle. Find the ratio of the side lengths of the rectangle. AB/AD=? ... 1. for convenience I've labeled the side AB = b and the side AC = a. 2. From the text of the question you know: $2a+2b=4r~\\implies~a+b=2r$ 3. The upper left triangle is a right triangle. Use the pythagorean theorem: $\\left(\\dfrac b2\\right)^2+(a-r)^2=r^2~\\implies~a^2-2ar+\\dfrac{b^2}4=0$ 4. Solve this system of", "- \\frac{13 \\cdot 13 \\cdot 5}{24} = 120- \\frac{845}{24} = \\frac{2035}{24} = 84.791666...$$ - The area of the quadrilateral $BEDC$ is what's leftover if you subtract the triangle $ADE$ from triangle $ABC$. Since both triangles are right triangles, $$area(BEDC)=area(ABC)-area(ADE)\\\\ =\\frac12\\|AB\\|\\cdot\\|BC\\|-\\frac12\\|AD\\|\\cdot\\|DE\\|\\\\ =\\frac{24\\cdot10}{2}-\\frac{13}{2}\\|DE\\|.$$ To find $\\|DE\\|$, note that $\\triangle ADE$ is similar to the $5,12,13$ right triangle, so $$\\frac{\\|DE\\|}{\\|DA\\|}=\\frac{\\|DE\\|}{13}=\\frac{5}{12}\\implies \\|DE\\|=\\frac{65}{12}.$$ Now just complete the arithmetic. -", "# Calculating area of triangle Problem In the triangle above, we have $AC = 12, \\ \\ BD = 10, \\ \\ CD = 6$. Find the area of $\\triangle ABD$. Caveat This was part of an exam where you are not allowed to use a calculator or any other digital tools. My progress Given $AC, CD, \\measuredangle C$, I was able to find the length of $AD$ using the law of cosines. It yielded $AD = \\sqrt{180} = 6\\sqrt5$. From here, I decided to try and use the law of sines to find $\\measuredangle A$ as it would be the same in $\\triangle ABD$ and $\\triangle ACD$. I got $$\\frac{\\sin A}{CD} = \\frac{\\sin C}{AD}$$. This yields $$\\sin A = \\frac{CD \\sin C}{AD} = \\frac{6}{\\sqrt{180}}$$. Then I need to calculate $\\arcsin\\frac{6}{\\sqrt{180}}$, which I can't, because I can't use a calculator for this problem. (Well, technically I can, but I want to solve this within the constraints placed on the participants.) From here, I am stuck. In $\\triangle ABD$, I know now two of the lengths, but I don't know any of the angles. Question Am I overlooking an easy inverse sine here? Or is there another way to calculate this area? Thanks in advance for any help! • If this is part of an exam, why are you", "3+3 = 6 AD = 2AE = 2*4 = 8 CD=10 Now that we know all sides, Area(BCDE) = Area(ACD) – Area(ABC) Area of a triangle = sqrt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2 and abc are lengths of the sides. Area(ACD) = sqrt(12*6*2*4) = 24 Area(ABC) = sqrt(6*3*2*1) = 6 Area(BCDE) = 24-6 = 18 please post a diagram of the aforesaid figure, it will help a lot, I simply can't just visualize it. Intern Joined: 05 Mar 2013 Posts: 45 Location: India Concentration: Entrepreneurship, Marketing GMAT Date: 06-05-2013 GPA: 3.2 Followers: 1 Kudos [?]: 52 [1] , given: 14 Re: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the [#permalink] ### Show Tags 13 May 2013, 01:35 1 KUDOS If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? A. 12 B. 18 C. 24 D. 30 E. 48 Be is parallel to cd so triangles ABE and ACD are similar. Ratio of their areas is square of the ratio of their similar sides. Both the triangles are also similar so we can get ED as 4 which will indicate that ACD is a right angled triangle. so we can get the area as 24.and using ratio", "1)$ cm, $2\\sqrt {b}$ cm and $(b + 1)$ cm is a right angled triangle. Solution These are sides of the Right angle triangle $(b+1)^2 = (2\\sqrt {b})^2 + (b-1)^2$ $(b+1)^2 = 4b + b^2 +1 -2b$ $(b+1)^2=b^2 + 1 +2 b$ $(b+1)^2 = (b+1)^2$ Question 5. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. Solution We know that $\\frac {ar( \\Delta 1)}{ar( \\Delta 2)}=\\frac {M_1^2}{M_2^2}$ $\\frac {121}{64} = \\frac {12.1^2}{QR}$ $QR= 8.8 cm$ Question 6. In $\\Delta ABC$, AD is perpendicular to BC. Prove that: a. $AB^2+ CD^2 = AC^2 + BD^2$ b. $AB^2 - BD^2 = AC^2 - CD^2$ Solution In $\\Delta ABC$, AD is perpendicular to BC Now $\\Delta ABD$ is an right -angle triangle So from Pythagoras theorem $AB^2 = AD^2 + BD^2$ $AD^2 = AB^2 - BD^2$ -(1) Similarly $\\Delta ADC$ is an right -angle triangle So from Pythagoras theorem $AC^2 = AD^2 + CD^2$ $AD^2 = AC^2 - CD^2$ --(2) From (1) and (2) $AB^2 - BD^2 = AC^2 - CD^2$ Which proved part (b) Now rearranging, $AB^2+ CD^2 = AC^2 + BD^2$ Which proved part (a) Question 7. In a quadrilateral ABCD, given that $\\angle A + \\angle D", "# 1998 JBMO Problems/Problem 2 ## Problem 2 Let $ABCDE$ be a convex pentagon such that $AB=AE=CD=1$, $\\angle ABC=\\angle DEA=90^\\circ$ and $BC+DE=1$. Compute the area of the pentagon. ## Solutions ### Solution 1 Let $BC = a, ED = 1 - a$ Let $\\angle DAC = X$ Applying cosine rule to $\\triangle DAC$ we get: $\\cos X = \\frac{AC ^ {2} + AD ^ {2} - DC ^ {2}}{ 2 \\cdot AC \\cdot AD }$ Substituting $AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1$ we get: $\\cos^{2} X = \\frac{(1 - a - a ^ {2}) ^ {2}}{(1 + a^{2})(2 - 2a + a^{2})}$ From above, $\\sin^{2} X = 1 - \\cos^{2} X = \\frac{1}{(1 + a^{2})(2 - 2a + a^{2})} = \\frac{1}{AC^{2} \\cdot AD^{2}}$ Thus, $\\sin X \\cdot AC \\cdot AD = 1$ So, area of $\\triangle DAC$ = $\\frac{1}{2}\\cdot \\sin X \\cdot AC \\cdot AD = \\frac{1}{2}$ Let $AF$ be the altitude of $\\triangle DAC$ from $A$. So $\\frac{1}{2}\\cdot DC\\cdot AF = \\frac{1}{2}$ This implies $AF = 1$. Since $AFCB$ is a cyclic quadrilateral with $AB = AF$, $\\triangle ABC$ is congruent to $\\triangle AFC$. Similarly $AEDF$ is a cyclic quadrilateral and $\\triangle AED$ is congruent to $\\triangle AFD$. So area of $\\triangle ABC$ + area of $\\triangle AED$", "height and median, it implies the triangle $$ABC$$ is isosceles. Hence, $$AC=3$$. From the right triangle $$ACD$$ we can find $$CD=\\sqrt{AC^2-AD^2}=2\\sqrt{2}$$. Finally, the area of the triangle $$ABC$$ is $$\\frac12 \\cdot AB\\cdot CD=\\frac12 \\cdot 3\\cdot 2\\sqrt{2}=3\\sqrt{2}.$$" ]

[ "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "# 2000 AMC 8 Problems/Problem 25 ## Problem The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\\overline{BC}$ and $\\overline{CD}$ are joined to form a triangle, the area of that triangle is $[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW);[/asy]$ $\\text{(A)}\\ 21\\qquad\\text{(B)}\\ 27\\qquad\\text{(C)}\\ 30\\qquad\\text{(D)}\\ 36\\qquad\\text{(E)}\\ 40$ ## Solution 1 To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$, which is the easiest way to avoid fractions. Labelling the right midpoint as $M$, and the bottom midpoint as $N$, we know that $DN = NC = 6$, and $BM = MC = 3$. $[\\triangle ADN] = \\frac{1}{2}\\cdot 6\\cdot 6 = 18$ $[\\triangle MNC] = \\frac{1}{2}\\cdot 3\\cdot 6 = 9$ $[\\triangle ABM] = \\frac{1}{2}\\cdot 12\\cdot 3 = 18$ $[\\triangle AMN] = [\\square ABCD] - [\\triangle ADN] - [\\triangle MNC] - [\\triangle ABM]$ $[\\triangle AMN] = 72 - 18 - 9 - 18$ $[\\triangle AMN] = 27$, and the answer is $\\boxed{B}$ ## Solution 2 The above answer is fast, but satisfying, and assumes that the area of $\\triangle AMN$ is independent of 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", "# Difference between revisions of \"2009 AMC 12B Problems/Problem 9\" ## Problem 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 $\\triangle ABC$? $\\mathrm{(A)}\\ 6\\qquad \\mathrm{(B)}\\ 8\\qquad \\mathrm{(C)}\\ 10\\qquad \\mathrm{(D)}\\ 12\\qquad \\mathrm{(E)}\\ 14$ ## Solution ### Solution 1 Because the line $x + y = 7$ is parallel to $\\overline {AB}$, the area of $\\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\\frac 12 \\cdot 4 \\cdot 3 = \\boxed {6}$. ### Solution 2 The base of the triangle is $AB = \\sqrt{3^2 + 3^2} = 3\\sqrt 2$. Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$, which is $\\frac 4{\\sqrt 2} = 2\\sqrt 2$. Therefore its area is $\\frac 12 \\cdot 3\\sqrt 2 \\cdot 2\\sqrt 2 = \\boxed{6}$. The answer is $\\mathrm{(A)}$. $[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(3,0), B=(0,3); draw ( (-1,0) -- (9,0), dashed ); draw ( (0,-1) -- (0,9), dashed ); dot(A); dot(B); draw(A--B); draw ( (-1,8) -- (8,-1) ); label( \"A\", A, S ); label( \"B\", B,", "# Difference between revisions of \"2009 AMC 12B Problems/Problem 9\" ## Problem 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 $\\triangle ABC$? $\\mathrm{(A)}\\ 6\\qquad \\mathrm{(B)}\\ 8\\qquad \\mathrm{(C)}\\ 10\\qquad \\mathrm{(D)}\\ 12\\qquad \\mathrm{(E)}\\ 14$ ## Solution ### Solution 1 Because the line $x + y = 7$ is parallel to $\\overline {AB}$, the area of $\\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\\frac 12 \\cdot 4 \\cdot 3 = \\boxed {6}$. ### Solution 2 The base of the triangle is $AB = \\sqrt{3^2 + 3^2} = 3\\sqrt 2$. Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$, which is $\\frac 4{\\sqrt 2} = 2\\sqrt 2$. Therefore its area is $\\frac 12 \\cdot 3\\sqrt 2 \\cdot 2\\sqrt 2 = \\boxed{6}$. The answer is $\\mathrm{(A)}$. $[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(3,0), B=(0,3); draw ( (-1,0) -- (9,0), dashed ); draw ( (0,-1) -- (0,9), dashed ); dot(A); dot(B); draw(A--B); draw ( (-1,8) -- (8,-1) ); label( \"A\", A, S ); label( \"B\", B,", "turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\\cdot B'C + 2\\cdot 16 = 15^2$, whence $A'D = B'C = \\sqrt{193}$. Let $\\alpha = \\angle A'B'D$. The Law of Cosines on triangle $A'B'D$ yields $$\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,$$ and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle A'O_2D$ yields \\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $A'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$. ~djmathman ## Solution 2 Denote by", "OG13 DS79 The area of a right triangle can be easily expressed as half the product of the two legs. Area of triangle ABC is 0.5AC\\cdot{BC} and that of the triangle DBA is 0.5AD\\cdot{AB}. The question in fact is \"Is AC\\cdot{BC} = AB\\cdot{AD}?\" (1) From AC^2=2AD^2, we deduce that AC=\\sqrt{2}AD, which, if we plug into AC\\cdot{BC} = AB\\cdot{AD}, we get \\sqrt{2}AD\\cdot{BC}=AB\\cdot{AD}, from which \\sqrt{2}BC=AB. This means that triangle ABC should necessarily be isosceles, which we don't know. Not sufficient. (2) Obviously not sufficient, we don't know anything about AD. (1) and (2): If AC=BC=x, then AB=x\\sqrt{2}, and AD=\\frac{x}{\\sqrt{2}}. Then AC\\cdot{BC}=AB\\cdot{AD}=x^2. Sufficient. Answer C. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Intern Joined: 06 Apr 2011 Posts: 13 Followers: 0 Kudos [?]: 0 [0], given: 292 Re: In the figure above, is the area of triangular region ABC [#permalink] 13 Mar 2013, 22:53 Nice and succinct explanation Eva... Intern Joined: 04 Sep 2012 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: In the figure above, is the area of triangular region ABC [#permalink] 05 Jun 2013, 17:41 I still don't get it. To compare the Area of (ABC) and Area of (DBA): is 1/2(AC)(CB) = 1/2(AD)(AB) ? In the hint 1: (AC)^2=2(AD)^2 -> AC = sq(2)*(AD) -> AD > AC (1) Then", "# Difference between revisions of \"2000 AMC 8 Problems/Problem 25\" ## Problem The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\\overline{BC}$ and $\\overline{CD}$ are joined to form a triangle, the area of that triangle is $[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW);[/asy]$ $\\text{(A)}\\ 21\\qquad\\text{(B)}\\ 27\\qquad\\text{(C)}\\ 30\\qquad\\text{(D)}\\ 36\\qquad\\text{(E)}\\ 40$ ## Solution 1 To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$, which is the easiest way to avoid fractions. Labelling the right midpoint as $M$, and the bottom midpoint as $N$, we know that $DN = NC = 6$, and $BM = MC = 3$. $[\\triangle ADN] = \\frac{1}{2}\\cdot 6\\cdot 6 = 18$ $[\\triangle MNC] = \\frac{1}{2}\\cdot 3\\cdot 6 = 9$ $[\\triangle ABM] = \\frac{1}{2}\\cdot 12\\cdot 3 = 18$ $[\\triangle AMN] = [\\square ABCD] - [\\triangle ADN] - [\\triangle MNC] - [\\triangle ABM]$ $[\\triangle AMN] = 72 - 18 - 9 - 18$ $[\\triangle AMN] = 27$, and the answer is $\\boxed{B}$ ## Solution 2 The above answer is fast, but satisfying, and assumes that the area of $\\triangle AMN$", "+0 # Geometry 0 119 3 If altitude CD is $$8 \\sqrt{3}$$ centimeters, what is BD? Feb 15, 2022 #1 +2444 0 This problem is actually simple because it is a 30-60-90 triangle. Because we know that $$CD = 8\\sqrt3$$$$AD$$ must equal $$\\sqrt2\\times 8\\sqrt3 = 8\\sqrt6$$ We also know that $$BD$$ must equal to $${8\\sqrt3}\\over\\sqrt2$$, which can be rewritten as $$4\\sqrt6$$ This means that $$\\color{brown}\\boxed{AB=12\\sqrt6}$$ Feb 15, 2022 #2 +124594 +1 Note that triangles ACD , CBD are similar and each are 30-60-90 right triangles And since AD is opposite a 60° angle it = sqrt 3 * CD = sqrt 3 * 8 sqrt 3 = 24 And, by similar triangles, we have the following relationship DB / DC = DC / AD (8sqrt 3)^2 = DB * 24 256 / 24 = DB = 32 / 3 (cm) Feb 15, 2022 #3 +2444 +1 I misread the question... DB should be $$4 \\sqrt6$$, because $$\\triangle{CDB}$$ is a 30-60-90. Feb 15, 2022", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =", "is also an isosceles trapezoid; in turn, $A'D = O_1O_2 = 15$, and similarly $B'C = 15$. Thus, Ptolmey's theorem on $B'A'CD$ yields $B'D\\cdot A'C + 2\\cdot 16 = 15^2$, whence $B'D = A'C = \\sqrt{193}$. Let $\\alpha = \\angle B'A'D$. The Law of Cosines on triangle $B'A'D$ yields$$\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,$$and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle B'A'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $B'A'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2B' = r_1$ and $O_2D = O_1A' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $B'O_2D$ and $B'A'D$ are opposite angles in cyclic quadrilateral $A'B'O_2D$, which implies the measure of angle $B'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle B'O_2D$ yields\\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $B'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$. ~djmathman ## Solution", "# Area of a special isosceles spherical triangle I was asked to calculate the area of the following isosceles triangle. Let $\\triangle ABC$ be an isosceles spherical triangle with $d(A,C)=d(B,C)$. Then $C$ lies on the perpendicular bisector of $AB$. Let $M$ be the midpoint of $AB$. Suppose $d(A,M)=d(-M,C)$. Calculate the area of $\\triangle ABC$. I already made a sketch and observed that $d(A,M)=d(-M,C)=\\pi-d(M,C)$. Let $\\alpha$, $\\beta$, $\\gamma$ denote the interior angles corresponding to $A$, $B$, $C$ respectively. If I assume without loss of generality that, because $C$ lies on the perpendicular bisector of $AB$, the angle $\\gamma$ is divided into two equal angles $\\gamma_1$ and $\\gamma_2$ belonging to the right triangles $\\triangle AMC$ and $\\triangle BMC$. Since $$\\operatorname{area}(\\triangle ABC) = 2 \\cdot \\operatorname{area}(\\triangle AMC) = 2 \\cdot \\operatorname{area}(\\triangle BMC)$$ I only need to calculate $\\operatorname{area}(\\triangle AMC)$. Since $\\operatorname{area}(\\triangle AMC)=\\alpha + \\gamma_1+ \\frac{\\pi}{2}-\\pi$, I need $\\alpha$ and $\\gamma_1$. I first tried to calculate $\\gamma_1$ using the spherical Law of Cosines, and I got this: $$\\cos\\gamma_1=\\cos\\alpha \\cdot \\cos\\frac{\\pi}{2} - \\sin\\alpha \\cdot \\sin\\frac{\\pi}{2} \\cdot \\cos d(A,M)=-\\sin\\alpha \\cdot \\cos\\left(\\pi-d(M,C)\\right) \\tag{1}$$ From the spherical Law of Sines, I get: $$\\sin\\alpha= \\frac{\\sin d(M,C)}{\\sin d(A,C)} \\tag{2}$$ But now I do not see how to proceed. I would like to calculate the right-hand sides of $(1)$ and $(2)$, but this does not seem possible with the", "the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{\\textbf{(B) }30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B", "$DOP$ is similar to triangle $AOB$, and triangle $DPC$ is similar to triangle $ABC$. If $DP$ = $15y$, then $CP$ = $36y$, $OP$ = $10y$, and $OD$ = $5y\\sqrt {13}$. Since $OP$ + $CP$ = $46y$ = $\\frac {26}{3}$, $y$ = $\\frac {13}{69}$, and $AD$ = $\\frac {60\\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$). The answer is $60$ + $13$ + $23$ = $\\boxed{096}$. ## Solution 2 Extend $AB$ and $CD$ to intersect at $P$. Note that since $\\angle ACB=\\angle PCB$ and $\\angle ABC=\\angle PBC=90^{\\circ}$ by ASA congruency we have $\\triangle ABC\\cong \\triangle PBC$. Therefore $AC=PC=13$. By the angle bisector theorem, $PD=\\dfrac{130}{23}$ and $CD=\\dfrac{169}{23}$. Now we apply Stewart's theorem to find $AD$: \\begin{align*}13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=13\\cdot 13\\cdot \\dfrac{130}{23}+10\\cdot 10\\cdot \\dfrac{169}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=\\dfrac{169\\cdot 130+169\\cdot 100}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=1690\\\\ AD^2&=130-\\dfrac{130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130\\cdot 23^2-130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130(23^2-169)}{23^2}\\\\ AD^2&=\\dfrac{130(360)}{23^2}\\\\ AD&=\\dfrac{60\\sqrt{13}}{23}\\\\ \\end{align*} and our final answer is $60+13+23=\\boxed{096}$. ## Solution 3 Notice that by extending $AB$ and $CD$ to meet at a point $E$, $\\triangle ACE$ is isosceles. Now we can do a straightforward coordinate bash. Let $C=(0,0)$, $B=(12,0)$, $E=(12,-5)$ and $A=(12,5)$, and the equation of line $CD$ is $y=-\\dfrac{5}{12}x$. Let F be the intersection point of $AD$ and $BC$, and by using the Angle Bisector Theorem: $\\dfrac{BF}{AB}=\\dfrac{FC}{AC}$ we have $FC=\\dfrac{26}{3}$. Then the equation of", "area of square $ABCD$ be divided into the triangles $DCE$, $EAG$, $CGB$, and $EGC$. Let x be the length of AE. Thus, the area of each triangle can be determined as follows: $$DCE = \\frac{DC\\cdot{DE}}{2} = \\frac{2\\cdot(2-x)}{2} = 1-x$$ $$EAG= \\frac{AE\\cdot{AG}}{2} = \\frac{1\\cdot{x}}{2} = \\frac{x}{2}$$ $$CGB = \\frac{GB\\cdot{CB}}{2} = \\frac{1\\cdot(2)}{2} = 1$$ $$EGC= \\frac{EG\\cdot{GC}}{2} = \\frac{\\sqrt{4+(2-x)^2}}{2}$$ (the length of CE is calculated with the Pythagorean Theorem, lines GE and CE are perpendicular by definition of tangent) Adding up the areas and equating to the area of the total square (2*2=4), we get $$1-x+\\frac{x}{2}+1+ \\frac{\\sqrt{4+(2-x)^2}}{2} = 4$$ Solving for x: $$2-2x+x+1+\\sqrt{x^2-4x+8}=8$$ $$\\sqrt{x^2-4x+8}=x+2$$ $$x^2-4x+8=x^2+4x+4$$ $$8-4x=4x+4 \\rightarrow x=\\frac{1}{2}$$ Solving for length of CE with the value we have for x: $$\\sqrt{4+(2-x)^2} = \\sqrt{4+(3/2)^2} = \\sqrt{25/4} = \\boxed{\\frac{5}{2}}$$ ## See also 2004 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 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 2004 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 •", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "$H$. $CH=4/3\\div 68/15\\cdot 8/5=8/17$. So $HG=8-8/17=128/17$. By Power of a Point, $CH\\cdot HG=AH\\cdot HE$. $AH=16/5\\cdot 5\\sqrt{13}/17=16\\sqrt{13}/17.$ So $HE=1024/289\\cdot 17/(16\\sqrt{13})=64/(17\\sqrt{13})$. The height from $E$ to $CG$ is $17/(5\\sqrt{13})\\cdot 64/(17\\sqrt{13})=64/65$. Thus, $[\\triangle CEG]=64/65\\cdot 8\\div 2=256/65$. The area of the whole cyclic quadrilateral is $64/5+256/65=(832+256)/65=1088/65$. Lastly, the common area is $1/17$ the area of the quadrilateral, or $64/65$. So $64+65=\\boxed{129}$. ## Solution 5 (HARD computation) $[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"A\",A,NE);label(\"O_1\",O_1,NE);label(\"O_2\",O_2,NE);label(\"B\",B,SW);label(\"C\",C,SW);label(\"D\",D,NE);label(\"E\",E,NE);label(\"N\",N,W);label(\"K\",(-24/15,0.2));label(\"L\",(24/15,0.2));label(\"n\",(-0.8,-0.12));label(\"p\",((29/15,-48/15)));label(\"\\mathcal{P}\",(-1.6,1.1));label(\"\\mathcal{Q}\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5)); //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); path circle2 = Circle((4,0),4); N = (-8/3,0); pair X =rotate(180,O_2)*E; pair Y = (8,0); draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y); dot(\"X\", X, NE);dot(\"Y\", Y, NE); [/asy]$ Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because $AE=XY$ and $AE \\parallel XY$, $XYE$ is right angle. First, $\\frac{NO_1}{NO_1+5} = \\frac{1}{4}$, so $NO_1=\\frac{5}{3}$. And, $$\\cos{\\angle{AO_2C}}=\\cos{2\\angle{AYC}} = \\frac{O_2C}{NO_1+5} = \\frac{3}{5}$$ $$\\sin{\\angle{AYC}} = \\sqrt{\\dfrac{1-\\cos{2\\angle{AYC}}}{2}}=\\frac{1}{\\sqrt{5}}$$ $$\\cos{\\angle{AYC}} = \\frac{2}{\\sqrt{5}}$$ Then, $$[AEC] = \\frac{1}{2}AE*CE*\\sin{\\angle{ACE}}=\\frac{1}{2}AE*8\\sin{\\angle{CYE}}*\\frac{1}{\\sqrt{5}}$$ $$[CXY] = \\frac{1}{2}CX*XY*\\sin{\\angle{CXY}}=\\frac{1}{2}*8\\sin{\\angle{XYC}}*XY*\\sin{\\angle{CAY}}$$ Since $\\angle{CAY} = 90 - \\angle{AYC}$, $\\angle{XYC} = 90 - \\angle{CYE}$, $XY = AE$, we have $$[CXY] = \\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\cos{\\angle{AYC}}=\\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\frac{2}{\\sqrt{5}}$$ Since $\\triangle{CXY}$ is four times in scale to $\\triangle{AEC}$, their area ratio is 16. Divide the two equations for the two areas, we have $$\\tan{\\angle{CYE}} = \\frac{1}{8}$$ With this", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "point $E$, such that $AECD$ is a parallelogram. In other words, $$AE = CD = 10$$ and $$EC = DA = 2\\sqrt{65}$$ Now, let's examine $\\triangle ABE$. Since $AB = AE = 10$, the triangle is isosceles, and $\\angle ABE \\cong \\angle AEB$. Note that in parallelogram $AECD$, $\\angle AED$ and $\\angle CDE$ are congruent, so $\\angle ABE \\cong \\angle CDE$ and thus $$\\text{m}\\angle ABD = 180^\\circ - \\text{m}\\angle CDB$$ Define $\\alpha := \\text{m}\\angle CDB$, so $180^\\circ - \\alpha = \\text{m}\\angle ABD$. We use the Law of Cosines on $\\triangle DAB$ and $\\triangle CDB$: $$\\left(2\\sqrt{65}\\right)^2 = 10^2 + BD^2 - 20BD\\cos\\left(180^\\circ - \\alpha\\right) = 100 + BD^2 + 20BD\\cos\\alpha$$ $$14^2 = 10^2 + BD^2 - 20BD\\cos\\alpha$$ Subtracting the second equation from the first yields $$260 - 196 = 40BD\\cos\\alpha \\rightarrow BD\\cos\\alpha = \\frac{8}{5}$$ This means that dropping an altitude from $B$ to some foot $Q$ on $\\overline{CD}$ gives $DQ = \\frac{8}{5}$ and therefore $CQ = \\frac{42}{5}$. Seeing that $CQ = \\frac{3}{5}\\cdot BC$, we conclude that $\\triangle QCB$ is a 3-4-5 right triangle, so $BQ = \\frac{56}{5}$. Then, the area of $\\triangle BCD$ is $\\frac{1}{2}\\cdot 10 \\cdot \\frac{56}{5} = 56$. Since $AP = CP$, points $A$ and $C$ are equidistant from $\\overline{BD}$, so $$\\left[\\triangle ABD\\right] = \\left[\\triangle CBD\\right] = 56$$ and hence $$\\left[ABCD\\right] = 56 + 56 = \\boxed{112}$$" ]

[ "simplified the same way 3 } \\$ 125^.", "+ 4$ 4. Simplify And that's it", "- 8 + 1$ Simplify. $y = 125 {x}^{3} - 141 {x}^{2} + 66 x - 7$", "= 144 x = $\\frac{259}{144}$", "\\frac{89.0}{124.1}$", "# How do you simplify (-9-sqrt(108))/3? It is $\\frac{- 9 - \\sqrt{108}}{3} = \\frac{- 9 - 6 \\sqrt{3}}{3} = - 3 - 2 \\sqrt{3}$", "$92\\frac{7}{6}$ <<12345>>", "# 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: $23 \\frac{6}{10} = 23 \\frac{3}{5}$", "+ 144 + 144 + 144 - 96}{6} = \\frac{192}{6} = 32$$", "Simplifying this to a complex number, $$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$$ . Apr 19, 2020", "# How do you simplify (125)^(1/3)? $= 5$ ${125}^{\\frac{1}{3}}$ $= \\sqrt[3]{125}$ $= \\sqrt[3]{5 \\times 5 \\times 5}$ $= 5$", "you described it but I did a simlar thing method.... and found the same answer $$\\frac{870*.0375}{365}*97+870=878.67$$", "\\frac{97}{29} i$", "is: $$\\frac{73}{100}$$", "+0 # Completely simplify and rationalize the denominator: 0 393 1 +40 Completely simplify and rationalize the denominator: $$\\frac{\\sqrt{160}}{\\sqrt{252}}\\times\\frac{\\sqrt{245}}{\\sqrt{108}}$$ May 14, 2021", "$(6)$: integrate $(7)$: simplify", "# How do you simplify 14/27(-9/28)? Mar 25, 2018 $- \\frac{1}{6}$ #### Explanation: When we're multiplying fractions, we multiply straight across. Thus, we have: $- \\frac{126}{756}$ We can divide the top and bottom by a common factor. Here, we can divide by $126$, and we get: $- \\frac{1}{6}$ Hope this helps!", "# Simplify : $\\large\\frac{\\sqrt {5625}+\\sqrt {441}}{\\sqrt {5625}-\\sqrt {441}}$ $\\begin{array}{1 1} 16 \\\\ 9 \\\\ \\large\\frac{16}{9} \\\\ \\large\\frac{9}{16} \\end{array}$", "and actual cost was$750,000. Using the data provided above, and assuming a... ##### How do you simplify ( 8x^2 -9x - 1) - ( -6x^2 + 4x - 8 )? How do you simplify ( 8x^2 -9x - 1) - ( -6x^2 + 4x - 8 )?...", "# How do you simplify (-11i)(-10i)? $- 110$ $- 11 i \\left(- 10 i\\right) = 110 {i}^{2} = 110 \\left(- 1\\right) = - 110$" ]

[ "# The Greatest Common Factor ## ...of Two Numbers: The greatest common factor is the largest whole number that is a factor of the two given whole numbers. In other words, it is the largest number that can be divided evenly into the two given numbers. To find the GCF of two whole numbers, find the prime factorization of each, and then find the product of all common factors . Example 1: Find the GCF of $60$ and $42$ . First, find the prime factorizations. $\\begin{array}{l}60={2}\\cdot 2\\cdot {3}\\cdot 5\\\\ 42={2}\\cdot {3}\\cdot 7\\end{array}$ Common factors are shown in red. Their product is: $2\\cdot 3=6$ So, the GCF is $6$ . ## ...of Two Monomials: In the prime factorization of a monomial, include all the variables (and a $-1$ factor if necessary), and proceed as above. Example 2: 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 $3{p}^{2}{r}^{3}$ .", "least common multiple of $15$ and $20.$ Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple. Find the least common multiple (LCM) of the given numbers: $9\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}12$ 36 Find the least common multiple (LCM) of the given numbers: $18\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}24$ 72 ## Prime factors method Another way to find the least common multiple of two numbers is to use 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. $12=2\\cdot 2\\cdot 3\\phantom{\\rule{3em}{0ex}}18=2\\cdot 3\\cdot 3$ Then we write each number as a product of primes, matching primes vertically when possible. $\\begin{array}{l}12=2\\cdot 2\\cdot 3\\hfill \\\\ 18=2\\cdot \\phantom{\\rule{1.1em}{0ex}}3\\cdot 3\\end{array}$ 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. do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid", "common to both lists. If there are no common multiples in the lists, write out additional multiples for each number. 3. Look for the smallest number that is common to both lists. 4. This number is the LCM. example Find the LCM of $15$ and $20$ by listing multiples. Solution: List the first several multiples of $15$ and of $20$ . Identify the first common multiple. $\\begin{array}{l}\\text{15: }15,30,45,60,75,90,105,120\\qquad \\\\ \\text{20: }20,40,60,80,100,120,140,160\\qquad \\end{array}$ The smallest number to appear on both lists is $60$ , so $60$ is the least common multiple of $15$ and $20$ . Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple. try it In teh next video we show an example of how to find the Least Common Multiple by listing multiples of each number. Prime Factors Method Another way to find the least common multiple of two numbers is to use 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. $12=2\\cdot 2\\cdot 318=2\\cdot 3\\cdot 3$ Then we write each number as a product of primes, matching primes vertically when possible. $\\begin{array}{l}12=2\\cdot 2\\cdot 3\\qquad \\\\ 18=2\\cdot 3\\cdot 3\\end{array}$ Now we bring down the primes in each", "# What ate the prime factors of each denominator of the unit fractions that have terminating decimals? 1/2, 1/4, 1/5, 1/8, 1/10, 1/16, 1/20, 1/25 Question Decimals What ate the prime factors of each denominator of the unit fractions that have terminating decimals? 1/2, 1/4, 1/5, 1/8, 1/10, 1/16, 1/20, 1/25 2021-03-08 Step 1 Writing the prime factors of each denominators of the fraction We have the fraction $$\\displaystyle\\frac{{1}}{{2}},\\frac{{1}}{{4}},\\frac{{1}}{{5}},\\frac{{1}}{{8}},\\frac{{1}}{{10}},\\frac{{1}}{{16}},\\frac{{1}}{{20}},\\frac{{1}}{{25}}$$ Step 2 Now, Prime factor of $$\\displaystyle{2}={2}$$ Prime factor of $$\\displaystyle{4}={2}\\cdot{2}$$ Prime factor of $$\\displaystyle{5}={5}$$ Prime factor of $$\\displaystyle{8}={2}\\cdot{2}\\cdot{2}$$ Prime factor of $$\\displaystyle{10}={2}\\cdot{5}$$ Prime factor of $$\\displaystyle{16}={2}\\cdot{2}\\cdot{2}\\cdot{2}$$ Prime factor of $$\\displaystyle{20}={2}\\cdot{2}\\cdot{5}$$ Prime factor of $$\\displaystyle{25}={5}\\cdot{5}$$ ### Relevant Questions For digits before decimals point, multiply each digit with the positive powers of ten where power is equal to the position of digit counted from left to right starting from 0. For digits after decimals point, multiply each digit with the negative powers of ten where power is equal to the position of digit counted from right to left starting from 1. 1) $$10^{0}=1$$ 2) $$10^{1}=10$$ 3) $$10^{2}=100$$ 4) $$10^{3}=1000$$ 5) $$10^{4}=10000$$ And so on... 6) $$10^{-1}=0.1$$ 7) $$10^{-2}=0.01$$ 8) $$10^{-3}=0.001$$ 9) $$10^{-4}=0.0001$$ Which of the following fractions are repeating decimals and which are terminating? How made decisions? $$a) \\frac{2}{15}$$ $$b)\\frac{11}{20}$$ $$c)\\frac{17}{40}$$ $$d)\\frac{1}{12}$$ For the given fraction and decimals we have", "the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers. Step 2. Simplify, using the equivalent fractions property, by removing common factors. Step 3. Multiply any remaining factors. Sometimes it may not be easy to find common factors of the numerator and denominator. A good idea, then, is to factor the numerator and the denominator into prime numbers. (You may want to use the factor tree method to identify the prime factors.) Then divide out the common factors using the Equivalent Fractions Property. Example $$\\PageIndex{4}$$: simplify Simplify: $$\\dfrac{210}{385}$$. Solution $$\\dfrac{210}{385}$$ Use factor trees to factor the numerator and denominator. Rewrite the numerator and denominator as the product of the primes. $$\\dfrac{210}{385} = \\dfrac{2 \\cdot 3 \\cdot 5 \\cdot 7}{5 \\cdot 7 \\cdot 11}$$ Remove the common factors. $$\\dfrac{2 \\cdot 3 \\cdot \\cancel{\\textcolor{blue}{5}} \\cdot \\cancel{\\textcolor{red}{7}}}{\\cancel{\\textcolor{blue}{5}} \\cdot \\cancel{\\textcolor{red}{7}} \\cdot 11}$$ Simplify. $$\\dfrac{2 \\cdot 3}{11}$$ Multiply any remaining factors. $$\\dfrac{6}{11}$$ Exercise $$\\PageIndex{7}$$ Simplify: $$\\dfrac{69}{120}$$. $$\\dfrac{23}{40}$$ Exercise $$\\PageIndex{8}$$ Simplify: $$\\dfrac{120}{192}$$. $$\\dfrac{5}{8}$$ We can also simplify fractions containing variables. If a variable is a common factor in the numerator and denominator, we remove it just as we do with an integer factor. Example $$\\PageIndex{5}$$: simplify Simplify: $$\\dfrac{5xy}{15x}$$. Solution $$\\dfrac{5xy}{15x}$$ Rewrite numerator and denominator showing common factors. $$\\dfrac{5 \\cdot x \\cdot y}{3 \\cdot 5 \\cdot", "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", "thinking that this is the LCM: 2 * 3 * 323 * 589 = 1,141,482 Now compare that to using the prime factor method: 1938 = 2 * 3 * 17 * 19 3534 = 2 * 3 * 19 * 31 LCM = 2 * 3 * 17 * 19 * 31 This way, it's easy to see that the LCM is actually 2 * 3 * 17 * 19 * 31 = 60,078 This is still large, but significantly smaller. What happened here is that there were (intentionally) some larger prime factors that you might not think to try. So the nice thing about the prime factor method is that there's no way to quit too early. Also, it directly relates the idea of prime factors to the idea of \"least common multiple,\" and doesn't require you to memorize a made-up method for calculating that without really understanding what you're doing. :^D On the other hand, you are in fact focusing on common factors, and pulling those out is just what you need to make a common multiple as small as possible. By the way, have you noticed that the left-hand-column of the work is the GCF? The method amounts to saying, $$\\text{LCM}(A,B)=\\frac{AB}{\\text{GCF}(A,B)}=\\text{GCF}(A,B)\\cdot\\frac{A}{\\text{GCF}(A,B)}\\cdot\\frac{B}{\\text{GCF}(A,B)}$$ Zoe wasn’t quite satisfied: Thank you very much! I guess I understand slightly", "pair): $$\\begin{array}{ccc} 2 && 210 \\\\ 6 && 70 \\\\ 10 && 42 \\\\ 14 && 30 \\end{array}$$ In this case, consider a number of the form $p_1^3 \\cdot p_2$. The smallest example of this is $3^3 \\cdot 5 = 135$, which is larger than $105$, but I'll follow through the analysis anyway. The factors in pairs are: $$\\begin{array}{ccc} 1 && 3^3 \\cdot 5\\\\ 3 && 3^2 \\cdot 5 \\\\ 3^2 && 3 \\cdot 5 \\\\ 3^3 && 5 \\end{array}$$ Back in the $\\Bbb{E}$-zone, these are prime factorizations of $$2^2 \\cdot 3^3 \\cdot 5= 540$$ (by putting a $2$ on each factor of the pair): $$\\begin{array}{ccc} 2 && 270 \\\\ 6 && 90 \\\\ 18 && 30 \\\\ 54 && 10 \\end{array}$$ So, $420$ is the smallest positive integer with $4$ prime factorizations in the $\\Bbb{E}$-zone. • That was extremely helpful. Thank you very much. So if I want the smallest number with 3 distinct factors, for example, I would look for the smallest odd number I can create with 3 distinct odd factors or with 3 total factors, even if some repeat. (trying to make sure I understand this concept). This number would be 180 as $3^2*5 = 45$ and $45*2^2 =180$. Excellent. Thank you. – Tyler Murphy Feb 5 '14 at 8:49", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $84=2^2\\cdot3\\cdot7$ Repeatedly factoring the smallest prime factor from the composite factor, the factors of $84$ are \\begin{array}{l}\\require{cancel} 84=2(42) \\\\\\\\ 84=2(2)(21) \\\\\\\\ 84=2(2)(3)(7) .\\end{array} Hence, the prime factorization of $84$ is $84=2^2\\cdot3\\cdot7 .$", "far as possible. $n-1=2 \\cdot 3 \\cdot 11 \\cdot 17 \\cdot 211 \\cdot 16157348744821$ The last factor $m=$ 16157348744821 of $n-1$ is an 14-digit number. It is a probable prime to base 2, meaning that $2^{m-1} \\equiv 1 \\ (\\text{mod} \\ m)$. So we have a good reason to believe that $m$ might be a prime. To test whether it is prime, we can use the same method to test. So now we focus our attention on the number $m=$ 16157348744821. Factor $m-1$ as far as possible. $m-1=2^2 \\cdot 3 \\cdot 5 \\cdot 31 \\cdot 149 \\cdot 58300313$ The last factor $T=$ 58300313 is a prime number. The square root of $T$ is 7635.46. None of the prime numbers below 7635 is a factor of $T$. So the above factorization is the prime factorization of $m-1$. We calculation the following: $\\displaystyle a^{m-1} \\ (\\text{mod} \\ m)$ $\\displaystyle a^{\\frac{m-1}{2}} \\ (\\text{mod} \\ m)$ $\\displaystyle a^{\\frac{m-1}{3}} \\ (\\text{mod} \\ m)$ $\\displaystyle a^{\\frac{m-1}{5}} \\ (\\text{mod} \\ m)$ $\\displaystyle a^{\\frac{m-1}{31}} \\ (\\text{mod} \\ m)$ $\\displaystyle a^{\\frac{m-1}{149}} \\ (\\text{mod} \\ m)$ $\\displaystyle a^{\\frac{m-1}{58300313}} \\ (\\text{mod} \\ m)$ We use $a=2$ as a starting point. We have the following results. $\\displaystyle 2^{m-1} \\equiv 1 \\ (\\text{mod} \\ m)$ $\\displaystyle 2^{\\frac{m-1}{2}} \\equiv -1 \\ (\\text{mod} \\ m)$ $\\displaystyle 2^{\\frac{m-1}{3}} \\equiv 10783747104377 \\not \\equiv 1 \\", "this: The factor tree for 126 looks like this: The LCM for 90 and 126 is made from the smallest possible collection of primes that enables us to construct either of the two numbers. We take only enough instances of each prime to make the number with the greater number of instances of that prime in its factor tree. Prime Factors in 90 Factors in 126 We Need 2 1 1 1 3 2 2 2 5 1 0 1 7 0 1 1 So we need one 2, two 3’s, one 5 and one 7. That gives us $2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7 = 630$ as the lowest common multiple of 90 and 126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so $\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}$. 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so $\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}$. Now we complete the problem: $\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}$. This fraction simplifies. To be sure of finding the simplest form for $\\frac{138}{630}$, we write out the", "numbers involved are small. For example, noticing that $42$ is the smallest integer divisible by both $6$ and $21$, we could determine $$\\frac{3}{21} + \\frac{7}{6} = \\frac{6}{42} + \\frac{49}{42} = \\frac{55}{42}$$ However, when the numbers involved are larger, finding the common denominator by inspection is much more difficult. As such, let us consider a slightly different course of action... First, notice that the least common denominator is a multiple of both denominators, and thus contains in its factorization all of the factors of the denominators in the fractions being summed. To clarify, consider the following example where the denominators present have been completely factored: $$\\frac{23}{147} + \\frac{5}{91} = \\frac{23}{3 \\cdot 7^2} + \\frac{5}{7 \\cdot 13}$$ The first denominator's factored form tells us that the common denominator we seek must have factors $3$ and $7^2$. The second denominator's factored form tells us that the common denominator must have factors $7$ (which we already knew) and $13$. Thus, the common denominator must minimally have factors $3$, $7^2$, and $13$. Any additional factors only serves to increase the value, so the least common denominator is simply $3 \\cdot 7^2 \\cdot 13$. There is no need to multiply this out yet (indeed, in some problems we will never have to find this product -- so one shouldn't waste time doing so). Instead,", "power of 2 that appears is 24. The highest power of 3 that appears is 31. \\begin{aligned} \\text{LCM} = 2^4 \\cdot 3^1 ~ & \\textcolor{red}{ \\text{ Keep highest power of each factor.}} \\end{aligned}\\nonumber Now we expand this last expression to get our LCM. \\begin{aligned} = 16 \\cdot 3 ~ & \\textcolor{red}{ \\text{ Expand: } 2^4 = 16 \\text{ and } 3^1 = 3.} \\\\ = 48. ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Note that this answer is identical to the LCM found in Example 8 that was found by listing multiples and choosing the smallest multiple in common. Exercise $$\\PageIndex{8}$$ Use prime factorization to find the least common denominator of 18 and 24. Answer 72 Example $$\\PageIndex{10}$$ Simplify: $$\\frac{5}{28} + \\frac{11}{42}$$. Solution Prime factor the denominators in compact form using exponents. 28 = 2 · 2 · 7=22 · 7 42 = 2 · 3 · 7=21 · 31 · 71 To find the LCD, write down each factor that appears to the highest power of that factor that appears. The factors that appear are 2, 3, and 7. The highest power of 2 that appears is 22. The highest power of 3 that appears is 31. The highest power of 7 that appears is 71. \\begin{aligned} \\text{LCM} = 2^2 \\cdot 3^1 \\cdot 7^1 ~ & \\textcolor{red}{ \\text{", "and 5. But why does this help? Because $12=2 \\cdot 2 \\cdot 3$, any number whose prime factors include two 2s and one 3 will be a multiple of 12. Similarly, any number whose prime factors include a 2, a 3, and a 5 will be a multiple of 30. Prime Factors of Fractions: Finding the prime factors of the denominators of two fractions enables us to find a common denominator. The least common denominator is the smallest number that contains the overlap of both factored denominators: in this case, it must have two 2s, one 3, and one 5. Hence, the least common denominator must be $2 \\cdot 2 \\cdot 3 \\cdot 5 = 60$. Now we can finish the problem: \\begin{align} \\frac {5} {2 \\cdot 2 \\cdot 3} + \\frac {7} {2 \\cdot 3 \\cdot 5} &= \\left(\\frac {5} {2 \\cdot 2 \\cdot 3} \\cdot \\frac {5} {5} \\right) + \\left(\\frac {7} {2 \\cdot 3 \\cdot 5} \\cdot \\frac {2} {2}\\right) \\\\ &= \\frac {25} {60} + \\frac {14} {60}\\\\ &= \\frac {39} {60} \\\\ &= \\frac {13} {20} \\end{align} This may look like a very strange way of solving problems that you have known how to solve since the third grade. However, you should spend a few minutes carefully following the above solution. Once you", "# Lowest Common Multiple (LCM) #### Casio ##### Member I did actually think I had this off to a T sort of speak, but seems I have run into a problem and am now unsure? Find the LCM of 12 and 56. This is how I do them but if incorrect I would appreciate the correct notation being pointed out to me 12/2 = 6 My prime factors of 12 are therefore 2, 6 56/2 = 28, and 28/2 = 14, and 14/2 = 7 my prime factors are; 2, 6, 7, 14 My common factors being 2 my lowest common multiple being = 1176? 12/4 = 3, therefore 3,4 56/14 = 4 therefore 4,14 common factors are 4 Lowest common multiple is 3 x 4 x 14 = 168 so I am assuming there is a specific notation (method) to finding the LCM's without mistakes? #### Reckoner ##### Member my prime factors are; 2, 6, 7, 14 6 and 14 cannot be prime factors, for they are not prime! $$12 = 2\\cdot6 = 2\\cdot2\\cdot3 = 2^2\\cdot3$$ $$56 = 2\\cdot28 = 2\\cdot2\\cdot14=2\\cdot2\\cdot2\\cdot7=2^3\\cdot7$$ Now we take the highest power of each factor. Our least common multiple is therefore $$\\mathrm{LCM}(12,56) = 2^3\\cdot3\\cdot7 = 168$$. #### Casio ##### Member 6 and 14 cannot be prime factors, for they are not prime!", "# 6.1: Introduction to Factoring Learning Objectives • Determine the greatest common factor (GCF) of natural numbers. • Determine the GCF of monomials. • Factor out the GCF of a polynomial. • Factor a four-term polynomial by grouping. ## GCF of Natural Numbers The process of writing a number or expression as a product is called factoring. If we write $$60 = 5\\cdot 12$$, we say that the product $$5 ⋅ 12$$ is a factorization of $$60$$ and that $$5$$ and $$12$$ are factors. Typically, there are many ways to factor a number. For example, $$\\begin{array}{lc}{60=6\\cdot 10}&{}\\\\{60=2\\cdot 30}&{\\color{Cerulean}{Factorizations\\:of\\:60}}\\\\{60=4\\cdot 3\\cdot 5}&{} \\end{array}$$ Recall that a prime number is defined as a natural number with exactly two natural number factors, $$1$$ and itself. The first ten prime numbers follow: $$2,\\:3,\\:5,\\:7,\\:11,\\:13,\\:17,\\:19,\\:23,\\:29,\\:\\dots$$ Any natural number greater than $$1$$ can be uniquely written as a product of prime numbers. This product is called the prime factorization. The prime factorization of $$60$$ can be determined by continuing to factor until only a product of prime numbers remains. \\begin{aligned}60&=2\\cdot 30\\\\&=2\\cdot 2\\cdot 15\\\\&=2\\cdot 2\\cdot 3\\cdot 5 \\end{aligned} Since the prime factorization is unique, it does not matter how we choose to initially factor the number; the end result will be the same. The prime factorization of $$60$$ follows: $$60=2^{2}\\cdot 3\\cdot 5\\qquad\\color{Cerulean}{The\\:prime\\:factorization\\:of\\:60}$$ Recall that the greatest common", "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", "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", "126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so $\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}$ . 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so $\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}$ . Now we complete the problem: $\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}$ . This fraction simplifies . To be sure of finding the simplest form for $\\frac{138}{630}$ , we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 are 2, 3 and 23. $\\frac{138}{630} = \\frac{2 \\cdot 3 \\cdot 23}{2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7}$ ; one factor of 2 and one factor of 3 cancels out, leaving $\\frac{23}{3 \\cdot 5 \\cdot 7}$ or $\\frac{23}{105}$ as our answer. Identify and Apply Properties of Addition Three mathematical properties which involve addition are the commutative, associative, and the additive identity properties . Commutative property: When two numbers are added, the sum is the same even if the order of the items being added changes. Example: $3 + 2 = 2" ]

[ "the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers. Step 2. Simplify, using the equivalent fractions property, by removing common factors. Step 3. Multiply any remaining factors. Sometimes it may not be easy to find common factors of the numerator and denominator. A good idea, then, is to factor the numerator and the denominator into prime numbers. (You may want to use the factor tree method to identify the prime factors.) Then divide out the common factors using the Equivalent Fractions Property. Example $$\\PageIndex{4}$$: simplify Simplify: $$\\dfrac{210}{385}$$. Solution $$\\dfrac{210}{385}$$ Use factor trees to factor the numerator and denominator. Rewrite the numerator and denominator as the product of the primes. $$\\dfrac{210}{385} = \\dfrac{2 \\cdot 3 \\cdot 5 \\cdot 7}{5 \\cdot 7 \\cdot 11}$$ Remove the common factors. $$\\dfrac{2 \\cdot 3 \\cdot \\cancel{\\textcolor{blue}{5}} \\cdot \\cancel{\\textcolor{red}{7}}}{\\cancel{\\textcolor{blue}{5}} \\cdot \\cancel{\\textcolor{red}{7}} \\cdot 11}$$ Simplify. $$\\dfrac{2 \\cdot 3}{11}$$ Multiply any remaining factors. $$\\dfrac{6}{11}$$ Exercise $$\\PageIndex{7}$$ Simplify: $$\\dfrac{69}{120}$$. $$\\dfrac{23}{40}$$ Exercise $$\\PageIndex{8}$$ Simplify: $$\\dfrac{120}{192}$$. $$\\dfrac{5}{8}$$ We can also simplify fractions containing variables. If a variable is a common factor in the numerator and denominator, we remove it just as we do with an integer factor. Example $$\\PageIndex{5}$$: simplify Simplify: $$\\dfrac{5xy}{15x}$$. Solution $$\\dfrac{5xy}{15x}$$ Rewrite numerator and denominator showing common factors. $$\\dfrac{5 \\cdot x \\cdot y}{3 \\cdot 5 \\cdot", "least common multiple of $15$ and $20.$ Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple. Find the least common multiple (LCM) of the given numbers: $9\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}12$ 36 Find the least common multiple (LCM) of the given numbers: $18\\phantom{\\rule{0.2em}{0ex}}\\text{and}\\phantom{\\rule{0.2em}{0ex}}24$ 72 ## Prime factors method Another way to find the least common multiple of two numbers is to use 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\\phantom{\\rule{3em}{0ex}}18=2\\cdot 3\\cdot 3$ Then we write each number as a product of primes, matching primes vertically when possible. $\\begin{array}{l}12=2\\cdot 2\\cdot 3\\hfill \\\\ 18=2\\cdot \\phantom{\\rule{1.1em}{0ex}}3\\cdot 3\\end{array}$ 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. do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid", "thinking that this is the LCM: 2 * 3 * 323 * 589 = 1,141,482 Now compare that to using the prime factor method: 1938 = 2 * 3 * 17 * 19 3534 = 2 * 3 * 19 * 31 LCM = 2 * 3 * 17 * 19 * 31 This way, it's easy to see that the LCM is actually 2 * 3 * 17 * 19 * 31 = 60,078 This is still large, but significantly smaller. What happened here is that there were (intentionally) some larger prime factors that you might not think to try. So the nice thing about the prime factor method is that there's no way to quit too early. Also, it directly relates the idea of prime factors to the idea of \"least common multiple,\" and doesn't require you to memorize a made-up method for calculating that without really understanding what you're doing. :^D On the other hand, you are in fact focusing on common factors, and pulling those out is just what you need to make a common multiple as small as possible. By the way, have you noticed that the left-hand-column of the work is the GCF? The method amounts to saying, $$\\text{LCM}(A,B)=\\frac{AB}{\\text{GCF}(A,B)}=\\text{GCF}(A,B)\\cdot\\frac{A}{\\text{GCF}(A,B)}\\cdot\\frac{B}{\\text{GCF}(A,B)}$$ Zoe wasn’t quite satisfied: Thank you very much! I guess I understand slightly", "# How can you simplify the following fraction using prime factorization?: 736/501 May 1, 2016 $\\frac{736}{501} = 1 \\frac{235}{501}$ #### Explanation: The fact is that the numerator $736$ and denominator $501$ both are coprime or relatively prime to each other. This means there is no number other than $1$, that can divide the two. Hence simplifying by dividing both by some prime number, is just not possible. However, it is improper fraction as $736 > 501$ and hence only way one can simplify it is to convert this into a mixed fraction. As such $\\frac{736}{501} = \\frac{501 + 235}{501} = \\frac{501}{501} + \\frac{235}{501} = 1 \\frac{235}{501}$ Nov 6, 2017 $736 = 2 \\times 2 \\times 2 \\times 2 \\times 23$ $= {2}^{4} \\cdot 23$ $501 = 3 \\times 167$ The numerator and denominator have no common factors, so the fraction can't be simplified.", "common to both lists. If there are no common multiples in the lists, write out additional multiples for each number. 3. Look for the smallest number that is common to both lists. 4. This number is the LCM. example Find the LCM of $15$ and $20$ by listing multiples. Solution: List the first several multiples of $15$ and of $20$ . Identify the first common multiple. $\\begin{array}{l}\\text{15: }15,30,45,60,75,90,105,120\\qquad \\\\ \\text{20: }20,40,60,80,100,120,140,160\\qquad \\end{array}$ The smallest number to appear on both lists is $60$ , so $60$ is the least common multiple of $15$ and $20$ . Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple. try it In teh next video we show an example of how to find the Least Common Multiple by listing multiples of each number. Prime Factors Method Another way to find the least common multiple of two numbers is to use 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 318=2\\cdot 3\\cdot 3$ Then we write each number as a product of primes, matching primes vertically when possible. $\\begin{array}{l}12=2\\cdot 2\\cdot 3\\qquad \\\\ 18=2\\cdot 3\\cdot 3\\end{array}$ Now we bring down the primes in each", "{72}{108}$ is $\\frac {2}{3}$. Ex 2: $\\frac {180}{252}$ Prime factorization of $180$ and $252$ in $\\frac {180}{252}$ is $\\frac {2 \\times 2 \\times 3 \\times 3 \\times 5}{2 \\times 2 \\times 3 \\times 3 \\times 7}$. Canceling two $2$s and two $3$s from the numerator and the denominator, the fraction $\\frac {180}{252}$ reduces to $\\frac {5}{7}$. Thus the lowest form of $\\frac {180}{252}$ is $\\frac {5}{7}$. ### Highest Common Factor Method The highest common factor or HCF of two or more numbers is the greatest number that divides exactly the given numbers. Using HCF, you can reduce a fraction to its lowest or simplest form. ### Steps For Highest Common Factor Method These are the steps to reduce fractions by the HCF method. Step 1: Find the highest common factor (HCF) of the numerator and the denominator Step 2: Divide the numerator and denominator by the HCF. The fraction so obtained is the reduced fraction ### Examples Ex 1: $\\frac {84}{174}$ HCF of $84$ and $174$ is $6$. Dividing both the numerator and the denominator by $6$. $\\frac {84 \\div 6}{174 \\div 6}$ = $\\frac {14}{29}$ The lowest form of $\\frac {84}{174}$ is $\\frac {14}{29}$. Ex 2: $\\frac {48}{152}$ HCF of $48$ and $152$ is $8$. Dividing both the numerator and the denominator by $8$. $\\frac {48 \\div", "so 9 is our common denominator. That means we don’t have to alter the second fraction at all. 3 divides into 9 three times, so \\begin{align*}\\frac{1}{3} = \\frac{3 \\cdot 1}{3 \\cdot 3} = \\frac{3}{9}\\end{align*}. Our sum becomes \\begin{align*}\\frac{3}{9} - \\frac{1}{9}\\end{align*}. We can subtract fractions with a common denominator by subtracting their numerators, just like adding. In other words: When subtracting fractions: \\begin{align*}\\frac{a}{c} - \\frac{b}{c} = \\frac{a - b}{c}\\end{align*} Solution \\begin{align*}\\frac{1}{3} - \\frac{1}{9} = \\frac{2}{9}\\end{align*} So far, we’ve only dealt with examples where it’s easy to find the least common multiple of the denominators. With larger numbers, it isn’t so easy to be sure that we have the LCD. We need a more systematic method. In the next example, we will use the method of prime factors to find the least common denominator. Example 5 Simplify \\begin{align*}\\frac{29}{90} - \\frac{13}{126}\\end{align*}. To find the lowest common multiple of 90 and 126, we first find the prime factors of 90 and 126. We do this by continually dividing the number by factors until we can’t divide any further. You may have seen a factor tree before. (For practice creating factor trees, try the Factor Tree game at http://www.mathgoodies.com/factors/factor_tree.asp.) The factor tree for 90 looks like this: The factor tree for 126 looks like this: The LCM for 90 and 126 is made", "this: The factor tree for 126 looks like this: The LCM for 90 and 126 is made from the smallest possible collection of primes that enables us to construct either of the two numbers. We take only enough instances of each prime to make the number with the greater number of instances of that prime in its factor tree. Prime Factors in 90 Factors in 126 We Need 2 1 1 1 3 2 2 2 5 1 0 1 7 0 1 1 So we need one 2, two 3’s, one 5 and one 7. That gives us $2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7 = 630$ as the lowest common multiple of 90 and 126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so $\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}$. 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so $\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}$. Now we complete the problem: $\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}$. This fraction simplifies. To be sure of finding the simplest form for $\\frac{138}{630}$, we write out the", "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", "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", "# The Greatest Common Factor ## ...of Two Numbers: The greatest common factor is the largest whole number that is a factor of the two given whole numbers. In other words, it is the largest number that can be divided evenly into the two given numbers. To find the GCF of two whole numbers, find the prime factorization of each, and then find the product of all common factors . Example 1: Find the GCF of $60$ and $42$ . First, find the prime factorizations. $\\begin{array}{l}60={2}\\cdot 2\\cdot {3}\\cdot 5\\\\ 42={2}\\cdot {3}\\cdot 7\\end{array}$ Common factors are shown in red. Their product is: $2\\cdot 3=6$ So, the GCF is $6$ . ## ...of Two Monomials: In the prime factorization of a monomial, include all the variables (and a $-1$ factor if necessary), and proceed as above. Example 2: 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}$ .", "In the same way, we could use $$6 \\times 8 = 48$$ as a common denominator to add $$\\frac{1}{6} + \\frac{3}{8}$$, but $$24$$ is the lowest common multiple (LCM) of $$6$$ and $$8$$, so it is better to use that. Prime factorisation makes it easy to find the lowest common multiple or highest common factor. When we simplify a fraction, we divide the same number into the numerator and the denominator. For the simplest fraction, use the highest common factor (HCF) to divide into both numerator and denominator. The HCF is divided into the numerator and the denominator to write the fraction in its simplest form. So, $$\\frac{36}{144} = \\frac{2 \\times 2 \\times 3 \\times 3}{2 \\times 2 \\times 2 \\times 2 \\times 3 \\times 3} = \\frac{1}{4}$$ Use prime factorisation to determine the LCM and HCF of $$32$$, $$48$$ and $$84$$ in a systematic way: • $$32 = 2 \\times 2 \\times 2 \\times 2 \\times 2 = 2^{5}$$ • $$48 = 2 \\times 2 \\times 2 \\times 2 \\times 3 = 2^{4} \\times 3$$ • $$84 = 2 \\times 2 \\times 3 \\times 7 = 2^{2} \\times 3 \\times 7$$ The LCM is a multiple, so all of the factors of all the numbers must divide into it. All of the factors that are present in", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $84=2^2\\cdot3\\cdot7$ Repeatedly factoring the smallest prime factor from the composite factor, the factors of $84$ are \\begin{array}{l}\\require{cancel} 84=2(42) \\\\\\\\ 84=2(2)(21) \\\\\\\\ 84=2(2)(3)(7) .\\end{array} Hence, the prime factorization of $84$ is $84=2^2\\cdot3\\cdot7 .$", "}100} \\end{equation*} Another way to find the GCF of two numbers is to consider their prime factorizations. The GCF will be the product of all the prime factors that they have in common. For example, we can compute the GCF of 108 and 72 as follows: In this case, the product of the common prime factors is $36\\text{.}$ ###### To reduce a fraction to lowest terms 1. Find the greatest common factor of the numerator and denominator. This can be done by listing all the factors of each and identifying the largest number that appears in both lists, or by finding the prime factorizations of both and taking the product of the prime factors they have in common. 2. Reduce the fraction by dividing both the numerator and the denominator by their GCF. ###### Example2 Simplify $\\frac{172}{64}$ by reducing it to lowest terms. Solution First, we find the GCF of 172 and 64 by considering their prime factorizations: \\begin{equation*} \\begin{aligned} 172 \\amp= \\alert{2} \\cdot \\alert{2} \\cdot 43 \\\\ 64 \\amp= \\alert{2} \\cdot \\alert{2} \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\end{aligned} \\end{equation*} The product of the common prime factors is $2\\cdot2=4\\text{,}$ so the GCF of 172 and 64 is 4. We can now reduce the fraction: ###### Example3 Simplify $\\frac{108}{72}$ by reducing it to lowest terms. Solution", "126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so $\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}$ . 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so $\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}$ . Now we complete the problem: $\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}$ . This fraction simplifies . To be sure of finding the simplest form for $\\frac{138}{630}$ , we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 are 2, 3 and 23. $\\frac{138}{630} = \\frac{2 \\cdot 3 \\cdot 23}{2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7}$ ; one factor of 2 and one factor of 3 cancels out, leaving $\\frac{23}{3 \\cdot 5 \\cdot 7}$ or $\\frac{23}{105}$ as our answer. Identify and Apply Properties of Addition Three mathematical properties which involve addition are the commutative, associative, and the additive identity properties . Commutative property: When two numbers are added, the sum is the same even if the order of the items being added changes. Example: $3 + 2 = 2", "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", "power of 2 that appears is 24. The highest power of 3 that appears is 31. \\begin{aligned} \\text{LCM} = 2^4 \\cdot 3^1 ~ & \\textcolor{red}{ \\text{ Keep highest power of each factor.}} \\end{aligned}\\nonumber Now we expand this last expression to get our LCM. \\begin{aligned} = 16 \\cdot 3 ~ & \\textcolor{red}{ \\text{ Expand: } 2^4 = 16 \\text{ and } 3^1 = 3.} \\\\ = 48. ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Note that this answer is identical to the LCM found in Example 8 that was found by listing multiples and choosing the smallest multiple in common. Exercise $$\\PageIndex{8}$$ Use prime factorization to find the least common denominator of 18 and 24. Answer 72 Example $$\\PageIndex{10}$$ Simplify: $$\\frac{5}{28} + \\frac{11}{42}$$. Solution Prime factor the denominators in compact form using exponents. 28 = 2 · 2 · 7=22 · 7 42 = 2 · 3 · 7=21 · 31 · 71 To find the LCD, write down each factor that appears to the highest power of that factor that appears. The factors that appear are 2, 3, and 7. The highest power of 2 that appears is 22. The highest power of 3 that appears is 31. The highest power of 7 that appears is 71. \\begin{aligned} \\text{LCM} = 2^2 \\cdot 3^1 \\cdot 7^1 ~ & \\textcolor{red}{ \\text{", "\\frac 5 6$ The key is finding the least common denominator of the two rational expressions: the smallest multiple of both denominators. Then, you rewrite the two fractions using this denominator. Finally, you add (or subtract) the fractions by combining the numerators and leaving the denominator alone. But how do you find the least common denominator? Consider this problem: $\\displaystyle \\frac 5 {12} + \\frac 7 {30} = ?$ You could probably find the least common denominator if you played around with the numbers long enough. Here, we will show you a systematic method for finding least common denominators—a method that works with rational expressions just as well as it does with numbers. ### Finding the Least Common Denominator We start, as usual, by factoring. For each of the denominators, we find all the prime factors —i.e., the prime numbers that multiply to give that number. $\\displaystyle \\frac 5 {2 \\cdot 2 \\cdot 3} + \\frac 7 {2 \\cdot 3 \\cdot 5}$ If you are not familiar with the concept of prime factors, it may take a few minutes to get used to. $2\\cdot 2 \\cdot 3$ is $12$ broken into its prime factors: that is, it is the list of prime numbers that when multiplied together yield 12. Similarly, the prime factors of 30 are 2, 3,", "b}{c}$ Example 4 Simplify $\\frac{1}{3} - \\frac{1}{9}$. The lowest common multiple of 9 and 3 is 9, so 9 is our common denominator. That means we don’t have to alter the second fraction at all. 3 divides into 9 three times, so $\\frac{1}{3} = \\frac{3 \\cdot 1}{3 \\cdot 3} = \\frac{3}{9}$. Our sum becomes $\\frac{3}{9} - \\frac{1}{9}$. We can subtract fractions with a common denominator by subtracting their numerators, just like adding. In other words: When subtracting fractions: $\\frac{a}{c} - \\frac{b}{c} = \\frac{a - b}{c}$ Solution $\\frac{1}{3} - \\frac{1}{9} = \\frac{2}{9}$ So far, we’ve only dealt with examples where it’s easy to find the least common multiple of the denominators. With larger numbers, it isn’t so easy to be sure that we have the LCD. We need a more systematic method. In the next example, we will use the method of prime factors to find the least common denominator. Example 5 Simplify $\\frac{29}{90} - \\frac{13}{126}$. To find the lowest common multiple of 90 and 126, we first find the prime factors of 90 and 126. We do this by continually dividing the number by factors until we can’t divide any further. You may have seen a factor tree before. (For practice creating factor trees, try the Factor Tree game at http://www.mathgoodies.com/factors/factor_tree.asp.) The factor tree for 90 looks like" ]

[ "adding both of the denominators together, I found that the lowest number of people needed in the community must be 17.", "mistake were useless. The maximum number of civil unions among that group of eighteen people, regardless of sexual parings, is the same as the number of ways of dividing eighteen people in to nine groups of two each. That number is: $\\frac{18!}{\\left(2^9\\right) \\left(9!\\right)}$ I am really disappointed in your NOTE above . One who posted this thread was a newbie as I noticed .", "1 Answer Hint: What is the smallest possible number of students in the largest group for the first essay? What happens when you try to split them into 4 different groups?", "groups this can only be equal to 1,2 or 3.", "15? • Lcm simple Find least common multiple of this two numbers: 140 175. • Dance group The dance group formed groups of 4, 5, and 6 members. Always one dancer remains. How many dancers were there in the whole group?", "a group of 48 people....", "points! -tigernathan Jul 15, 2020 edited by tigernathan Jul 15, 2020 #1 +2141 0 Yay! More of tigernathan's questions! Number 1: Twelve groups, since the smallest number of anything is 12. Jul 15, 2020 #2 +161 +1 You got it! A lot of people think this as a trick question, since it never said equal. tigernathan Jul 15, 2020 #3 +2141 0 10^3 plus 10^180 Simplify that! Jul 15, 2020 #4 +161 +1 Remember the rule: you add the exponents. So, what is the answer? tigernathan Jul 15, 2020", "13] group 3, persons: [6, 10, 14, 15]", "of these 29 groups across the.", "there are \"no less than 27\". (Just a minor nit-pick.)", "different number of groups. – hmhmmm Aug 23 '19 at 18:42", "is number in each group", "containing the most reliable groups. There are 485 groupsof a least three members in the final catalog. Submit a new article", "combine into groups of five. Read More »", "groups, as long as the as long as the total size of the groups is less than or equal to $n$.", "member per subset", "into individuals)", "trip. if 39 people went on that trip, how many are in the club...", "number of boxes, I could just eyeball it, and say, well, let's make these top ones one group. Let's make that one of the groups, and let's making these bottom ones another group. And these are equal groups. How many do we have in each group? Well, in the top one, we have one, two, three, four, five, six, seven, eight, nine, and in the bottom one we have one, two, three, four, five, six, seven, eight, nine. So we've split it into two groups of nine. So 18 divided into two equal groups is equal to 9, or 18 divided by 2 is equal to 9, or 2 goes into 18 nine times.", "is the number of persons present in the party? 1) 15 2) 16 3) 17 4) 18 5) 20" ]

[ "and this needs some additional thoughts. • $r>2\\cdot 365$: According to the pigeonhole principle groups with more than $730$ people have at least three equal birthdays with probability $p=1$. Starter: In order to get a first impression and to have a verification for small numbers at hand we look at a smaller example. It should be small enough to make calculations easy and large enough to be representative for the problem. We take $r=5$ people and instead of $365$ days we use $6$ days. If we take a fair die with six faces instead, this smaller version of the problem can then be stated as: Small Problem: What is the probability that in a group of $5$ people, each of them throwing a fair die once, at least three people roll the same number. Five people can throw a total of $6^5=7776$ different configuration of $5$-tuples from $(1,1,1,1,1)$ to $(6,6,6,6,6)$. We answer the question by counting the $5$-tuples which have no more than two equal faces. • Let $N_{(11111)}$ be the number of $5$-tuples with pairwise different entries. We see that there are $6\\cdot5\\cdot\\ldots\\cdot1=6!$ possibilities and get \\begin{align*} N_{(11111)}=6!=720 \\end{align*} The index $11111$ indicates that there are five pairwise different values counted. Now we consider the number of $5$-tuples with one or more pairs of equal numbers, but", "# A question related to Pigeonhole Principle In a room there are $10$ people, none of whom are older than $60$, but each of whom is at least $1$ year old. Prove that one can always find two groups of people (with no common person) the sum of whose ages is the same. My approach: There are $2^{10}=1024$ subsets, $1023$ non-empty subsets. Therefore there are $1023$ sums of ages and each sum is between $1$ and $600$. Then there are $600$ possible values, but $1023$ sums. Therefore at least two of them must be equal, i.e. there exist different subsets $\\{P_{i1}, \\ldots, P_{in}\\}$ and $\\{P_{j1}, \\ldots, P_{jn}\\}$ such that the sum of the ages agree. Now take out the people present in both subsets. Can $10$ people be replaced by a smaller number? I guess, it cannot. For example if there were to be $9$ people, then I would have $2^9-1 = 511$ proper subsets and since now I have $9\\cdot 60=540$ possible totals, it is not guaranteed that there exists two disjoint groups of people such that the sum of whose ages are the same. Am I right? • @MichaelBiro I would take out the same people as I stated in my approach? – Xentius Mar 4 '13 at 16:39 • Your first argument looks right, but", "the factors $11,13$ must be in group $1$. Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$. Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$. Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$. It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \\cdot 5 \\cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\\cdot 13 \\cdot 37 \\gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$. • I also thought that trying to solve for the same question just", "+0 # Number Theory 0 65 1 Let x be the smallest positive integer such that 1584 * x is a perfect cube, and let y be the smallest positive integer such that xy is a multiple of 1584. Compute y Feb 1, 2022 #1 +1209 +1 1584 = 2^4 * 3^2 * 11 To be a perfect cube, a number's prime factors' exponents must all be multiples of 3. So, the smallest value of x is: $$\\frac{2^6 \\cdot 3^3 \\cdot 11^3}{2^4 \\cdot 3^2 \\cdot 11} = \\boxed{1452}$$ In other terms, x = 1452 = 2^2 * 3 * 11^2. The product xy must be a multiple of 1584, so, xy = 2^2 * 3 * 11^2 * y = 2^4 * 3^2 * 11 * n y must have 2^2 * 3 as a factor, so the smallest value of y is 12. Feb 1, 2022", "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", "to the choices of $x_i$ allowed. - There is a little person sitting on each number. Each little person adds together her number, and the numbers of her two neighbours. So we end up with $36$ sums. We are asked to show that one at least of the little people will end up with a sum of $55$ or more, Note that each number on the wheel is part of the sum of $3$ people: the person sitting on the number, the person immediately to the right of the number, and the person immediately to the left of the number. So if you take the sum of all the sums found by the little people, each number from $1$ to $36$ will have appeared $3$ times, will have been counted $3$ times. In symbols, if the little people are called $P_1$ to $P_{36}$, and $P_i$'s sum is called $S_i$, then $$S_1+S_2+\\cdots +S_{36}=3(1+2+\\cdots+36)=(3)(18))(37).$$ Divide by $36$. We get $55.5$. So the average of the sums of the little people is $55.5$. Their sums can't all be $\\le 55$, they can't all be below average. - Thanks to everyone for the help. If anyone else needs this question answered, and isn't helped by the excellent answers given, I found the following graphic very clear: The circle is the wheel of", "the previous paragraph that 4 people can be matched in 3 ways, so the answer is $5\\cdot3 = 15$. Now you try it from there. • Let me make sure I understand. 7 people would be (6)(12) – Garret Lloyd May 15 '15 at 2:52 • 7 people can't be divided into groups of 2. – MJD May 15 '15 at 3:03 • I think I did what you said and I got 1.585times 10 to the 21 – Garret Lloyd May 15 '15 at 3:23 • Assuming MJD's answer is $(n - 1)!!$, I got a value of $316,234,143,225$ – Jared May 15 '15 at 5:32", "Nov 24 '13 at 6:55 When there are 25 people, clearly the required condition is not met as it is possible for 5 groups of 5 people each to have gotten each of the different grades. (this is the only case with 25 people when no 6 people have got a common grade) When you have one person extra, irrespective of his grade, there will be a grade now that is repeated 6 times. Hence, the minimum number of people is is 26. We can use the Pigeonhole Principle (Strong form). That is $q_1+q_2+\\cdots+q_n-n+1$ where $q_i$ is the number of desired test scores of a certain type, ($6$) and $n$ is the number of score types in general, ($5$). So $6+6+6+6+6-5+1=26$. Are you sure that it doesn’t say $\\lceil N/5\\rceil$ or $\\left\\lceil\\frac{N}5\\right\\rceil$ and not $[N/5]$ or $\\left[\\frac{N}5\\right]$? By definition $\\lceil x\\rceil$, the ceiling of $x$, is the smallest integer $n$ such that $x\\le n$. This means that it’s the unique integer $n$ such that $n-1<x\\le n$. Thus, $\\lceil N/5\\rceil=6$ if and only if $$5<\\frac{N}5\\le 6\\;,$$ which is true if and only if $25<N\\le 30$. Since $N$ here is an integer, this means that $\\lceil N/5\\rceil=6$ if and only if $N$ is $26,27,28,29$, or $30$. With fewer than $26$ students, it’s possible that no more than $5$ received each", "+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 $x \\le 14$).", "the first group of four smallest numbers tends to lie to the left of $$20$$ or so, leaving about $$60$$ values from which to select the next four groups. Thus, the situation with the second group is typically just like the situation with five groups, but with $$80$$ reduced to around $$60$$ and only $$20-4=16$$ numbers to select. Similar reasoning suggests the situation with the third and fourth groups also is similar. The simulations in the question bear this out: they still favor even sums over odd. Analogous discrepancies in probabilities occur when representing the sums in other (non-decimal) bases. Different patterns of discrepancies occur when taking groups of other sizes, especially odd sizes: groups of size $$4$$ are special because they have a pronounced tendency to include either zero, two, or four odd numbers, which guarantees their sum is even. Here is the R used to do the calculations. It is coded so you can experiment with versions of this game simply by changing the parameters n, m, and g. library(data.table) n <- 80 # Maximum number available (100 or more may be problematic) m <- 4 # Group sizes >= 2 (8 or more is problematic) g <- 5 # Number of groups >= 1 (more is easier!) # # Generate all possible vectors of gaps.", "that had been cut into 8 equal parts and 3 of those parts were gone, leaving $\\frac{5}{8}$? How can you describe the total amount of pizza that is left with one number rather than two different fractions? You need a common denominator, technically called the least common multiple. Remember that if a number is a multiple of another, you can divide them and have no remainder. One way to find the least common multiple of two or more numbers is to first multiply each by 1, 2, 3, 4, etc. For example, find the least common multiple of 2 and 5. First, list all the multiples of 2: Then list all the multiples of 5: $2\\cdot 1 = 2$ $5\\cdot 1 = 5$ $2\\cdot 2 = 4$ $5\\cdot 2 = 10$ $2\\cdot 3 = 6$ $5\\cdot 3 = 15$ $2\\cdot 4 = 8$ $5\\cdot 4 = 20$ $2\\cdot 5 = 10$ $5\\cdot 5 = 25$ The smallest multiple they have in common will be the common denominator for the two! ### Example Describe the amount of pizza left using common terms. To add fractions with unlike denominators, first rewrite them with like denominators. Then, you know what to do! The steps are shown below. ### Adding Fractions with Unlike Denominators 1. Find a common denominator. 2. Rewrite each", "of 3 students from among $8 + 6 = 14$ students, and then subtract from that total the number of committees/combinations that can be formed without girls. The result will give you the number of committees which then contain at least one girl. $$\\binom{14}{3} - \\binom 63 = \\text{total number of committees of 3 students with \\bf at\\;least\\; one\\; girl.}$$ Alternatively, you can add the results of your posted calculations, and arrive at the same number of valid combinations as you would using this approach. Both approaches give you a total of $344$ possible valid combinations. • so the answer is 8? – user73122 Jun 20 '13 at 18:20 • No recall that $\\binom {14}{3} = 14\\,C\\,3 = \\dfrac{14!}{11!\\cdot 3!} = \\dfrac{14\\cdot 13\\cdot 12}{3\\cdot 2\\cdot 1} = \\;?\\;$, and $\\binom 63 = \\dfrac{6!}{3!\\cdot 3!} = 20$. In all you should have $\\;364 - 20 = 344$ – amWhy Jun 20 '13 at 18:24 • Yes I understand that...from my giving solution what would be the next step.... I think that's the way the teacher wants it. – user73122 Jun 20 '13 at 18:26 • Just add your totals up and see that you get = $344$! ;-) – amWhy Jun 20 '13 at 18:28 • do you get the same answer using your approach? – user73122 Jun 20", "group. If you prefer not to talk about leaders, call that person the numerator (the person on top). The underling in each working group, is, of course, at the bottom. Line up the people according to student number. The first student can be paired with $7$ people. For each such pairing, look at the first unpaired student. She can be paired with $5$ people. For each such choice, look at the first unpaired student. She can be paired with $3$ people. So the total number of pairings is $7\\cdot 5\\cdot 3\\cdot 1$ (the $1$ at the end is to make things prettier). For each division into $4$ groups, the leader of each group can be selected in $2$ ways. So the total number of groups-with-leader is $$(7\\cdot 5\\cdot 3\\cdot 1)(2\\cdot 2\\cdot 2\\cdot 2).$$ (We could have done the selecting the leader as soon as the group is determined. That would give $7\\cdot 2\\cdot 5\\cdot 2 \\cdot 3\\cdot 2\\cdot 1\\cdot 2$.) Alternately, imagine lining up the $8$ people in some order. There are $8!$ ways to do this. For each ordering, pair person $1$ with person $2$, declare person $1$ leader. Pair person $3$ with person $4$, and declare person $3$ leader. And so on. Does $8!$ then count the number of groups-with-leader? No, because each division into groups-with", "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 $(512-(384-160+40-4)) \\implies \\boxed{252}$.", "gaps we have $5!$ ways to distribute the women among those gaps. Thus in total $2\\cdot 5!\\cdot 5!$ ways. The above answers are a direct way of seeing the question. But here's another way to think of it. Group 1 man and 1 woman together, how many ways can we can we do this $5!*5!$. But we can lead with either man or woman first. Giving us two ways. Thus in total we have $5!*5!*2=28800$", "+0 # Factorial Help +1 157 1 The symbol 3! Means 3 x 2 x 1, which equals 6. Similarly 5! Means 5 x 4 x 3 x 2 x 1, which equals 120, and so on. Suppose that N! ends in exactly 3 zeros after fully multiplying out. What is the smallest value that N can have? Jun 13, 2019 #1 +195 +2 For a number end in $$x$$ zeros, it must be divisible by $$10^x$$ So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or $$5^3\\cdot2^3$$ To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N. 15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 15, 10 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. Jun 13, 2019 #1 +195 +2 For a number end in $$x$$ zeros, it must be divisible by $$10^x$$ So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or $$5^3\\cdot2^3$$ To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N. 15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 15, 10", "50, . . . 20 and 40 are both common multiples of 4 and 10 (as are 60, 80, . . . ), but 20 is the smallest number that is on both lists, so 20 is the least common multiple. ### Glossary Entries • common factor A common factor of two numbers is a number that divides evenly into both numbers. For example, 5 is a common factor of 15 and 20, because $$15 \\div 5 = 3$$ and $$20 \\div 5 = 4$$. Both of the quotients, 3 and 4, are whole numbers. • The factors of 15 are 1, 3, 5, and 15. • The factors of 20 are 1, 2, 4, 5, 10, and 20. • common multiple A common multiple of two numbers is a product you can get by multiplying each of the two numbers by some whole number. For example, 30 is a common multiple of 3 and 5, because $$3 \\cdot 10 = 30$$ and $$5 \\cdot 6 = 30$$. Both of the factors, 10 and 6, are whole numbers. • The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 . . . • The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40 . . . The common multiples", "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", "x, we need m and n such that$$m^2 + 1 = 2n^2 \\text{ or }m^2-2n^2=-1$$This is an example of Pell's equation, see here and here. The approach is to guess the smallest positive solution, which in this case is (n_1,m_1)=(1,1). All ... 1 The stock proof of this is the one you started with. Let \\ell be any line, with points P_1, \\ldots, P_n (where n is at least 3). Let Q be any point not on \\ell (axiom 4). Consider the lines$$ \\ell_i (i = 1, \\ldots, n) where $\\ell_i$ contains both $Q$ and $P_i$. These are all distinct (axiom 1), and they intersect pairwise exactly at the ... 1 The smallest i could get was $1680$. ($1681=41^2, 841=29^2$). 1 The answer according to me should be $33$. EXPLANATION: Let us mark the horses as $X_i$ where $i=1,2,3,..,125$ First we divide the $125$ horses into $25$ groups of $5$ each. Say we divide the horses in such a way that the $n^{\\text{th}}$ group contains horses $X_{5n-4}$ to $X_{5n}$. For each group, there is a group race where all the $5$ horses of the ... 1 I am changing my answer because it appears you are talking about a given $B$, rather than trying to undo $A$ as I originally supposed. In this case, unless $d = \\pm", "and $248$ each shared by $27$ people, since $118\\cdot28+248\\cdot27=10,000$. Thus, we can’t guarantee that there is a birthday shared by $29$ people. This reasoning, when carried out in general, leads to the conclusion that if you divide $n$ things into $k$ categories, at least one category must contain at least $\\left\\lceil\\frac{n}k\\right\\rceil$, where $\\left\\lceil\\frac{n}k\\right\\rceil$ is the smallest integer greater than or equal to $\\frac{n}k$. - Your calculation proves that there is a group of least $28$ people with the same birthday. We really should show also that it is possible that the maximum is $28$. For with different numbers it is possible that for number-theoretic reasons, the number computed using $\\left\\lceil\\frac{n}{k}\\right\\rceil$ is not the actual minimum. Make $200$ groups of $28$ that share a birthday, for a total of $5600$ people. That leaves $165$ potential birthdays, and $4400$ people. We can divide them into $162$ groups of $27$, and $1$ group of $26$. There are many other ways to have a maximum of $28$. -" ]

[ "Nov 24 '13 at 6:55 When there are 25 people, clearly the required condition is not met as it is possible for 5 groups of 5 people each to have gotten each of the different grades. (this is the only case with 25 people when no 6 people have got a common grade) When you have one person extra, irrespective of his grade, there will be a grade now that is repeated 6 times. Hence, the minimum number of people is is 26. We can use the Pigeonhole Principle (Strong form). That is $q_1+q_2+\\cdots+q_n-n+1$ where $q_i$ is the number of desired test scores of a certain type, ($6$) and $n$ is the number of score types in general, ($5$). So $6+6+6+6+6-5+1=26$. Are you sure that it doesn’t say $\\lceil N/5\\rceil$ or $\\left\\lceil\\frac{N}5\\right\\rceil$ and not $[N/5]$ or $\\left[\\frac{N}5\\right]$? By definition $\\lceil x\\rceil$, the ceiling of $x$, is the smallest integer $n$ such that $x\\le n$. This means that it’s the unique integer $n$ such that $n-1<x\\le n$. Thus, $\\lceil N/5\\rceil=6$ if and only if $$5<\\frac{N}5\\le 6\\;,$$ which is true if and only if $25<N\\le 30$. Since $N$ here is an integer, this means that $\\lceil N/5\\rceil=6$ if and only if $N$ is $26,27,28,29$, or $30$. With fewer than $26$ students, it’s possible that no more than $5$ received each", "and this needs some additional thoughts. • $r>2\\cdot 365$: According to the pigeonhole principle groups with more than $730$ people have at least three equal birthdays with probability $p=1$. Starter: In order to get a first impression and to have a verification for small numbers at hand we look at a smaller example. It should be small enough to make calculations easy and large enough to be representative for the problem. We take $r=5$ people and instead of $365$ days we use $6$ days. If we take a fair die with six faces instead, this smaller version of the problem can then be stated as: Small Problem: What is the probability that in a group of $5$ people, each of them throwing a fair die once, at least three people roll the same number. Five people can throw a total of $6^5=7776$ different configuration of $5$-tuples from $(1,1,1,1,1)$ to $(6,6,6,6,6)$. We answer the question by counting the $5$-tuples which have no more than two equal faces. • Let $N_{(11111)}$ be the number of $5$-tuples with pairwise different entries. We see that there are $6\\cdot5\\cdot\\ldots\\cdot1=6!$ possibilities and get \\begin{align*} N_{(11111)}=6!=720 \\end{align*} The index $11111$ indicates that there are five pairwise different values counted. Now we consider the number of $5$-tuples with one or more pairs of equal numbers, but", "to the choices of $x_i$ allowed. - There is a little person sitting on each number. Each little person adds together her number, and the numbers of her two neighbours. So we end up with $36$ sums. We are asked to show that one at least of the little people will end up with a sum of $55$ or more, Note that each number on the wheel is part of the sum of $3$ people: the person sitting on the number, the person immediately to the right of the number, and the person immediately to the left of the number. So if you take the sum of all the sums found by the little people, each number from $1$ to $36$ will have appeared $3$ times, will have been counted $3$ times. In symbols, if the little people are called $P_1$ to $P_{36}$, and $P_i$'s sum is called $S_i$, then $$S_1+S_2+\\cdots +S_{36}=3(1+2+\\cdots+36)=(3)(18))(37).$$ Divide by $36$. We get $55.5$. So the average of the sums of the little people is $55.5$. Their sums can't all be $\\le 55$, they can't all be below average. - Thanks to everyone for the help. If anyone else needs this question answered, and isn't helped by the excellent answers given, I found the following graphic very clear: The circle is the wheel of", "the first group of four smallest numbers tends to lie to the left of $$20$$ or so, leaving about $$60$$ values from which to select the next four groups. Thus, the situation with the second group is typically just like the situation with five groups, but with $$80$$ reduced to around $$60$$ and only $$20-4=16$$ numbers to select. Similar reasoning suggests the situation with the third and fourth groups also is similar. The simulations in the question bear this out: they still favor even sums over odd. Analogous discrepancies in probabilities occur when representing the sums in other (non-decimal) bases. Different patterns of discrepancies occur when taking groups of other sizes, especially odd sizes: groups of size $$4$$ are special because they have a pronounced tendency to include either zero, two, or four odd numbers, which guarantees their sum is even. Here is the R used to do the calculations. It is coded so you can experiment with versions of this game simply by changing the parameters n, m, and g. library(data.table) n <- 80 # Maximum number available (100 or more may be problematic) m <- 4 # Group sizes >= 2 (8 or more is problematic) g <- 5 # Number of groups >= 1 (more is easier!) # # Generate all possible vectors of gaps.", "the previous paragraph that 4 people can be matched in 3 ways, so the answer is $5\\cdot3 = 15$. Now you try it from there. • Let me make sure I understand. 7 people would be (6)(12) – Garret Lloyd May 15 '15 at 2:52 • 7 people can't be divided into groups of 2. – MJD May 15 '15 at 3:03 • I think I did what you said and I got 1.585times 10 to the 21 – Garret Lloyd May 15 '15 at 3:23 • Assuming MJD's answer is $(n - 1)!!$, I got a value of $316,234,143,225$ – Jared May 15 '15 at 5:32", "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", "by an African American and the fourth by a Caucasian, then the chance of this is: 5/12 x 2/11 x 3/10 x 2/9. In the problem you have set we don't care which person sits in which seat so long as one of each is present. There are 4! ways of re-arranging the people in the seats, so the answer to your question is 4! times the product I just gave, which once you re-arrange it is the same as the answer given in the book. • this is a much clearer way to think about it – user2879934 Apr 28 '17 at 1:08 There are $\\binom{12}{5}$ possible committees. Committees of $5$ that include at least one member from each group necessarily have just one group having two members on the committee, all other groups have just one. So, the number of such committees is $$\\binom{5}{2}\\cdot 2\\cdot 3\\cdot 2 + 5\\cdot \\binom{2}{2}\\cdot 3\\cdot 2 + 5\\cdot 2 \\cdot \\binom{3}{2}\\cdot 2 + 5\\cdot 2 \\cdot 3 \\cdot \\binom{2}{2}$$ • Sorry that is not what I am asking, please read the edit – user2879934 Apr 27 '17 at 15:45 • @user2879934 I posted it before reading the edited version of your question (you made the edit while I was writing my answer.) – Dilip Sarwate Apr 27 '17 at 15:50", "# Two groups The group of 10 girls should be divided into two groups with at least 4 girls in each group. How many ways can this be done? Result n = 462 #### Solution: $C_{{ 4}}(10)=\\dbinom{ 10}{ 4}=\\dfrac{ 10! }{ 4!(10-4)!}=\\dfrac{ 10 \\cdot 9 \\cdot 8 \\cdot 7 } { 4 \\cdot 3 \\cdot 2 \\cdot 1 }=210 \\ \\\\ \\ \\\\ C_{{ 5}}(10)=\\dbinom{ 10}{ 5}=\\dfrac{ 10! }{ 5!(10-5)!}=\\dfrac{ 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 }=252 \\ \\\\ \\ \\\\ n={ { 10 } \\choose 4 } + { { 10 } \\choose 5 }=462$ 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 Would you like to compute count of combinations? ## Next similar math problems: 1. Trinity How many different triads can be selected from the group 43 students? 2. Division Division has 18 members: 10 girls and 6 boys, 2 leaders. How many different patrols can be", "# The number of ways to divide 5 people into three groups How many ways can 5 people be divided into three teams where each team must have at least one member? Assumably they can either be put in one group of 3 people then two groups with 1 person, or two groups with 2 people then one group of 1 person. Hence my answer was $$^5C_3 +\\, (^5C_2) \\cdot (^3C_2)$$ However the provided answer was $$^5C_3 +\\, (^5C_2) \\cdot (^3C_2) \\cdot (1/2)$$ Where did the 1/2 come from? Thanks. • Because the groupings $(A),(BC),(DE)$ and $(A),(DE),(BC)$ are the same (for example). – lulu Aug 11 '17 at 12:42 • Is there a way of knowing how many will be repeated without listing them out? – Casper C. Aug 11 '17 at 12:46 • Sure, there's a symmetry. the two $2-$member teams can be interchanged. You have the same problem with the other term, though you short-circuited it. You could have written the first term as $\\binom 53\\times \\binom 21 \\times \\binom 11$ but then you'd have had to divide by $2$ again to cancel the symmetry between the $1-$ member teams. – lulu Aug 11 '17 at 12:48 • Thanks, but how can you tell the amount of symmetry there is? – Casper C. Aug 11 '17", "the factors $11,13$ must be in group $1$. Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$. Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$. Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$. It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \\cdot 5 \\cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\\cdot 13 \\cdot 37 \\gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$. • I also thought that trying to solve for the same question just", "In the worst case for One-hot, the number of splits is $K-1$ when one goes through all but one category to get to one group remaining. #### Rainbow In this case, the number of needed splits is $\\min( (K - 1), 2 \\cdot (K-G_{max}))$ as we can only get up to two splits per each category that is not in the largest group. Indeed, if there are no consecutive pairs of values that belong to the same group then the number of splits would be $K-1$, for example here 3,2,1,3,1,2. At the same time, if there is a prevalent group and some minority groups, for example, 1,1,1,2,1,1,3,1,1, the number of splits will equal to the number of minority groups - $K-G_{max}=2$ in this case (2 and 3) times 2 which is the maximum number of splits that each minority group introduces - one to the left and one to the right. Thus the needed number of splits will be the minimum between the two cases above. #### Comparison Clearly, the number of splits for Rainbow is less than or equal to that of One-hot for the worst case as well. ## Simulation We have shown above that theoretically in a general case Rainbow method outperforms One-hot in best, average, and worst-case scenarios. Let's look at the empirical simulation:", "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", "# Two groups The group of 10 girls should be divided into two groups with at least 4 girls in each group. How many ways can this be done? Result n = 462 #### Solution: $C_{{ 4}}(10) = \\dbinom{ 10}{ 4} = \\dfrac{ 10! }{ 4!(10-4)!} = \\dfrac{ 10 \\cdot 9 \\cdot 8 \\cdot 7 } { 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 210 \\ \\\\ \\ \\\\ C_{{ 5}}(10) = \\dbinom{ 10}{ 5} = \\dfrac{ 10! }{ 5!(10-5)!} = \\dfrac{ 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 252 \\ \\\\ \\ \\\\ n={ { 10 } \\choose 4 } + { { 10 } \\choose 5 } = 462$ 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. Trinity How many different triads can be selected from the group 43 students? 2. Division Division has 18 members: 10 girls and 6 boys, 2 leaders. How many different patrols can be created, if one patrol is 2 boys, 3 girls and 1 leader? 3. Disco On the disco goes 12 boys and 15 girls. In how many ways can we", "# A question related to Pigeonhole Principle In a room there are $10$ people, none of whom are older than $60$, but each of whom is at least $1$ year old. Prove that one can always find two groups of people (with no common person) the sum of whose ages is the same. My approach: There are $2^{10}=1024$ subsets, $1023$ non-empty subsets. Therefore there are $1023$ sums of ages and each sum is between $1$ and $600$. Then there are $600$ possible values, but $1023$ sums. Therefore at least two of them must be equal, i.e. there exist different subsets $\\{P_{i1}, \\ldots, P_{in}\\}$ and $\\{P_{j1}, \\ldots, P_{jn}\\}$ such that the sum of the ages agree. Now take out the people present in both subsets. Can $10$ people be replaced by a smaller number? I guess, it cannot. For example if there were to be $9$ people, then I would have $2^9-1 = 511$ proper subsets and since now I have $9\\cdot 60=540$ possible totals, it is not guaranteed that there exists two disjoint groups of people such that the sum of whose ages are the same. Am I right? • @MichaelBiro I would take out the same people as I stated in my approach? – Xentius Mar 4 '13 at 16:39 • Your first argument looks right, but", "and $248$ each shared by $27$ people, since $118\\cdot28+248\\cdot27=10,000$. Thus, we can’t guarantee that there is a birthday shared by $29$ people. This reasoning, when carried out in general, leads to the conclusion that if you divide $n$ things into $k$ categories, at least one category must contain at least $\\left\\lceil\\frac{n}k\\right\\rceil$, where $\\left\\lceil\\frac{n}k\\right\\rceil$ is the smallest integer greater than or equal to $\\frac{n}k$. - Your calculation proves that there is a group of least $28$ people with the same birthday. We really should show also that it is possible that the maximum is $28$. For with different numbers it is possible that for number-theoretic reasons, the number computed using $\\left\\lceil\\frac{n}{k}\\right\\rceil$ is not the actual minimum. Make $200$ groups of $28$ that share a birthday, for a total of $5600$ people. That leaves $165$ potential birthdays, and $4400$ people. We can divide them into $162$ groups of $27$, and $1$ group of $26$. There are many other ways to have a maximum of $28$. -", "+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 $x \\le 14$).", "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: . $T(8) \\:=\\:\\frac{8\\cdot9\\cdot10}{6} \\:=\\:120$", "simply can't divide 5 people into groups of 2. Well, maybe OP was counting the extra odd person left over as a sort of group on their own? Then odd values would be correct; I didn't appreciate this at the time. But having calculated 6 special cases correctly, why can't OP calculate the seventh? Perhaps they were using brute force: the next value is 48, hard to brute-force correctly if you don't have a enough experience with combinatorics. I tried to suggest a general strategy: look at special cases, and not by brute force, but try to analyze them so that you can come up with a method for solving them. The method is unnecessary for the small cases, where brute force enumeration suffices, but you can use the brute force enumeration to check that the method is working. And then for the larger cases, where brute force is impractical, you use your method. It seems that OP couldn't understand my method, and when they tried to apply it, got wrong answers. Oh well, you can lead a horse to water, etc. The other pathology here is: I think I did what you said and I got 1.585times 10 to the 21 for the !!n=24!! case. The correct answer is $$23\\cdot21\\cdot19\\cdot17\\cdot15\\cdot13\\cdot11\\cdot9\\cdot7\\cdot5\\cdot3\\cdot1 = 316234143225 \\approx 3.16\\cdot 10^{11}.$$ OP didn't", "Common divisors Find all common divisors of numbers 30 and 45. Result d = (Correct answer is: 1, 3, 5, 15) Solution: $30 = 2 \\cdot \\ 3 \\cdot \\ 5 \\ \\\\ 45 = 3^2 \\cdot \\ 5 \\ \\\\ \\ \\\\ 30: 1, 2, 3, 5, 6, 10, 15, 30 \\ \\\\ 45: 1, 3, 5, 9, 15, 45 \\ \\\\ \\ \\\\ d = 1, 3, 5, 15$ 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: Do you want to calculate greatest common divisor two or more numbers? Next similar math problems: 1. Groups In the 6th class there are 60 girls and 72 boys. We want to divide them into groups so that the number of girls and boys is the same. How many groups can you create? How many girls will be in the group? 2. Three buses Three public transport buses depart together from the bus station in the morning. The first bus was returning to the station after 18 minutes, the second after 12 minutes and a third after 24 minutes. How long will again together on the station? Result exp 3. Matches George", "\"smallest\" pair \"possibly maximum\" pair people[9] = 7 1 [0, 1) 1 1 people[8] = 7 2 [0, 1) - - people[7] = 6 3 [0, 4) 2 2 people[6] = 5 4 [0, 5)` 2 3 • 非原地版本：98.65% • 原地版本：98.65%" ]

[ "a square with an area of $100$ and still have a different perimeter form the first square (it would be like trying to get another combination of four numbers that have the same value, but that when you multiply two of them together they give you $100$). In conclusion, that is why the second rectangle isn't (and can't be) a square. *A square can be a rectangle, but a rectangle can't be a square so, the first rectangle was originally a square.", "Back ## Split It Up The two colored rectangles have areas of $6$ and $9.$ What is the total area of the square? Keep reading to see one way we can answer this question, or jump to the challenge for a trickier problem.", "Back 28 29 30 ## Split It Up The two colored rectangles have areas of $6$ and $9.$ What is the total area of the square? Keep reading to see one way we can answer this question, or jump to the challenge for a trickier problem. ### This Daily Challenge has expired Subscribe to Premium to get access to the full archives. ×", "# Neighborhood Geometry Level 3 The area of the red, orange, and green squares are $$41; 97; 106$$. What's the area of the blue triangle? ×", "A rectangle has sides of and . What value of gives an area of 120? 5. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other. 6. A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is wide? 7. Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the numbers? 8. Two positive numbers have a sum of 8, and their product is equal to the smaller number plus 10. What are the numbers? 9. Framing Warehouse offers a picture framing service. The cost for framing a picture is made up of two parts: glass costs $1 per square foot and the frame costs$2 per foot. If the frame has to be a square, what size picture can you get framed for \\$20?", "? Free Version Easy # Related Rates in a Rectangle SVCALC-GVEZY1 A rectangle begins with a height of $4$ inches and a width of $16$ inches. The height increases at a rate of of $2$ in/sec while the width decreases at a ratae of $1$ in/sec. At what rate is the area changing at the moment when the rectangle is a square? A $0$ in$^2$/sec B $-2$ in$^2$/sec C $12$ in$^2$/sec D $144$ in$^2$/sec", "A rectangle has one side of $7$ cm. How fast is the area of the rectangle changing at the instant when the other side is $17$ cm and increasing at $2$ cm per minute? (Give units.) Answer:", "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", "overestimate of the entire sum. Since the $25$th rectangle has area $25^4$, this gives us our upper bound on the desired sum.", "=$50.00 – $18.50 – Fabric cost Fabric cost =$50.00 – $18.50 –$5.25 Fabric cost = \\$26.25 Question 50. The area of a rectangle is 98 square centimeters, and its width is 7 centimeters. Find the length. The area of a rectangle is 98 square centimeters. Width = 7 cm Length = ? Area of the rectangle = l × w 98 cm2= l × 7 cm 14 cm = l The length of a rectangle is 14 cm. Question 51. Richard planted some grass on a rectangular plot of land which measures 1 2 meters by 8 meters. He left a margin of 0.5 meters around the grass, as shown in the figure below. Find the area of land covered by grass. (Lesson 12.4)", "# Length and width calculation from area [closed] I have a question: If I have a square that is $50\\ ft$ by $50\\ ft$ it equals to $2500\\ ft^2$. Simple enough. Now, if I divide $2500\\ ft^2$ area in half, that equals to $1250\\ ft^2$. How do I know what the length and width of this area are? • Is \"divide in half\"=\"shrink the width\"? Otherwise there are a lot of solutions. – Emil Feb 6, 2018 at 17:51 • 25ft by 25ft does not equal to 1250sqft area. – ben Feb 6, 2018 at 17:52 • @ben Draw a picture. There are many many ways to divide a square into two sectors of equal area. It all depends on how you divide. Feb 6, 2018 at 17:55 • What I should have clarified, is that the I'm dividing the area of the square proportionally. – ben Feb 6, 2018 at 22:19 The easiest thing to do is cut the square in half to make a rectangle. Then it has a length of 50 feet and width of 25 feet. But I assume you're not asking that, since it's so obvious. I assume you're asking about a square that has half the area. So how would you find the length and width of a square that has half", "# What is the area of a square, in square inches, that has side lengths 0.0025 miles? Dec 10, 2016 Area of the square is $25090.56$ sq.in #### Explanation: $1$ mile $= 1760$ yds$= 1760 \\cdot 3 = 5280 f t = 5280 \\cdot 12 = 63360$inch $0.0025$ mile $= 0.0025 \\cdot 63360 = 158.4$ in Sdes of the squre is $s = 158.4$ in each. Area of the square is $A = {s}^{2} = {158.4}^{2} = 25090.56$ sq.in [Ans]", "diameter is surrounded by a path three feet wide. What is the area of the path? $51 \\pi$ square feet $60 \\pi$ square feet $69 \\pi$ square feet $90 \\pi$ square feet What is the area of a semicircle with a diameter of $16$ inches? $32 \\pi$ square inches $64 \\pi$ square inches $128 \\pi$ square inches $256 \\pi$ square inches", "ft. • A height of $16$ ft. 5. If a cube has a total surface area of $150$ squared feet, what is its side length?", "4. the length of a rectangular field is thrice its breath. if the cost of cultivating the field at rs. 367.20 per square metre is rs. 27,540, then what is the perimeter of the rectangle ? 1. 36 m 2. 40 m 3. 44 m 4. 49 m Explanation : Question 5. the area of a rectangle is five times the area of a square.the length of the rectangle is 400 ft and breadth is 1/4 of the side of the square. what is the side of the square ? 1. 5 ft 2. 15 ft 3. 20 ft 4. 25 ft", "# Area of a triangle and rectangle The area of a triangle is $3$ times less than $2$ times the area of a rectangle. If the sum of the two areas is $64$ square meters, what is the area of the triangle and what is the area of the rectangle? Hint: Let the area of the triangle be $T$, and the area of the rectangle be $R$. The information given implies: • $T = 2 R / 3$ • $T + R = 64$ Because $T=\\frac{2R}{3}$ and $T+R=64$, we have that $\\frac{2R}{3}+R=64$ and $\\frac{5R}{3}=64$ and $5R=192$. Therefore, $R=38.4$ square meters and $T=\\frac{2*38.4}{3}$ = $25.2$ square meters. The area of the triangle is $25.2$ square meters and the area of the rectangle is $38.4$ square meters.", "square has a length of $3$.", "made is $$9^{\\prime\\prime}$$, where $$BC =3 ^{\\prime\\prime}$$. Answer: $$10.667$$ (13). A rectangle is inscribed in a semi-circle of radius $$2$$ m. If the area of the rectangle is to be maximized, find the dimensions of the rectangle. Answer: $$2.828 \\times1.414$$", "x 4 is $80$cm$^2$ I think this happens because the two dimensions of a rectangle are both multiplied by three when the rectangle is increased by a Scale Factor of 3 so overall the area of the rectangle is nine-times bigger (3 x 3). If you increase a rectangle by a scale factor k, the area of the new rectangle will be k$^2$ times the old area of the rectangle. Taking a triangle with a base of 2, a height of 3 and an area of 3: When enlarged by a scale factor of 2 it becomes a triangle with a base of 4, a height of 6 and an area of 12. The area is 4 TIMES LARGER. When enlarged by a scale factor of 3 it becomes a triangle with a base of 6, a height of 9 and an area of 27. The area is 9 TIMES LARGER. When enlarged by a scale factor of 4 it becomes a triangle with a base of 8, a height of 12 and an area of 48. The area is 16 TIMES LARGER. I think this rule applies to all plane shapes because when different rectangles and triangles are increased by different scale factors, the increase in area is the same. Taking a cuboid with a width of", "# The area of a rectangle is 100 square inches. The perimeter of the rectangle is 40 inches.? A second rectangle has the same area but a different perimeter. Is the second rectangle a square? Sep 6, 2016 No. The second rectangle is not a square. #### Explanation: The reason why the second rectangle isn't a square is because the first rectangle is the square. For instance, if the first rectangle (a.k.a. the square) has a perimeter of $100$ square inches and a perimeter of $40$ inches then one side must have a value of $10$. With this being said, let's justify the statement above. If the first rectangle is indeed a square* then all of it's sides must be equal. Moreover, this would actually make sense for the reason that if one of its sides is $10$ then all of its other sides must be $10$ as well. Thus, this would give this square a perimeter of $40$ inches. Also, this would mean that the area must be $100$ ($10 \\cdot 10$). In continuation, if the second square has the same area, but a different perimeter then it can't be a square because its features wouldn't match the ones of a square. To clarify, what this means is that there wouldn't possible be a way of getting" ]

[ "same size. The length of a side of a small square is 2 inches. area of the square = side x side. area of the square and area of the chessboard is the same. area = s x s. s = 2 inches. area = 2 x 2. area = 4 sq in. so the area of the chessboard = 4 sq in. Assessment Practice Question 12. Jessica has saved$50. She will add $25 to her savings each week. Ron has saved$40 and will add $25 to his savings each week. How much will each person have saved after 5 weeks? A. Jessica:$275; Ron: $225 B. Jessica:$250; Ron: $240 C. Jessica:$175; Ron: $165 D. Jessica:$165; Ron: $175 Answer: Option C is correct. Explanation: In the above-given question, given that, Jessica has saved$50. She will add $25 to her savings each week. Ron has saved$40 and will add $25 to his savings each week.$50 + $25 =$75. $25 x 5 =$125. $125 +$50 = $175.$40 + $25 =$65. $65 +$100 = $165. Question 13. Which of the following statements are true? Jessica has always saved$25 more than Ron. Jessica has always saved $10 more than Ron. Ron has always saved$25 less than Jessica. Ron has always saved $10 less than Jessica Answer: Option B is correct. Explanation: In the", "Question # A rectangular piece of cardboard is $40$ in width and $50$ in length. Squares $5$ on a side are cut out of each corner, and the remaining flaps are bent up to form an open box. The number of cubic inches in the box is Open in App Solution ## The height of the rectangular piece after being bent is $5$ in.The new length of the rectangular piece $=$length of the rectangle piece$-$$2$$×$Squares $5$ on a side are cut out of each cornerThe new length of the rectangular piece $=$$50-2×5$The new length of the rectangular piece $=$$40$ unitThe new width of the rectangular $=$width of the rectangle piece$-$$2$$×$Squares $5$ on a side are cut out of each cornerThe new width of the rectangular piece $=$ $40-2×5$The new width of the rectangular piece $=$$30$ unit Volume $=$ $40×30×5$Therefore, volume$=$$6000$ cubic units. Suggest Corrections 0", "a width of 6 inches. If the length is increased by$x x$inches and the width increased by twice that amount, express the area of the rectangle as a function of$x. x.$ 69. An open box is to be constructed by cutting out square corners of $x- x-$inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of$x. x.$ 70. A rectangle is twice as long as it is wide. Squares of side 2 feet are cut out from each corner. Then the sides are folded up to make an open box. Express the volume of the box as a function of the width ($x x$).", "= 3^2\\\\A & = 9 \\ sq. \\ in.$ This is the area for one square face. There are six square faces. If we take this area and multiply it by six or add the area six times, we will have the surface area of the cube. $9(6) & = 54 \\ sq. \\ in.\\\\9 + 9 + 9 + 9 + 9 + 9 & = 54 \\ in^2$ The surface area of the cube is $54 \\ in^2$ . Now let’s look at a rectangular prism net. Let’s say that this box has a length of 6”, a width of 4” and a height of 2”. We need to find the area of each rectangle. There are two long faces. There are two short faces. There is one bottom, and one top that matches it. First, let's find the area of the bottom. It has a length of 6” and a width of 4”. Since the shape of the bottom is a rectangle, we can use the formula for finding the area of a rectangle. $A & = lw\\\\A & = (6)(4)\\\\A & = 24 \\ sq. \\ in.$ Since the top and bottom match, we can multiply this area by two: 24 x 2 = 48 square inches. Next, we find the area of the", "? Free Version Easy # Related Rates in a Rectangle SVCALC-GVEZY1 A rectangle begins with a height of $4$ inches and a width of $16$ inches. The height increases at a rate of of $2$ in/sec while the width decreases at a ratae of $1$ in/sec. At what rate is the area changing at the moment when the rectangle is a square? A $0$ in$^2$/sec B $-2$ in$^2$/sec C $12$ in$^2$/sec D $144$ in$^2$/sec", "# A rectangle is cut in half to create two squares that each has an area of 25. What is the perimeter of the original rectangle? 30 Explanation The formula to find the area of a square is: Side × Side. If each square has an area of 25, then each side of the square must be 5 (because $$5 \\times 5=25$$). So the the dimensions of the rectangle are $$5 \\times 10$$. The perimeter is the sum of all the sides, and is therefore 5 + 10 + 5 + 10 = 30.", "5, we can write $$(x+3)^2+3^2=5^2$$, from which $$(x+3)^2=16$$, so $$x = 1$$. Sufficient. (2) Obviously, not sufficient, as was already mentioned by \"duriangris\" in the previous post. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Manager Joined: 22 Jun 2012 Posts: 53 GMAT 1: 730 Q49 V40 Followers: 2 Kudos [?]: 16 [0], given: 6 Re: Square ABCD has an area of 9 square inches. [#permalink] 31 Jul 2012, 23:38 EvaJager wrote: (1) Given that the area of the square is 9, then each side of the square is 3. The lengthened sides will be of length 3 + x each, and the diagonal of the obtained rectangle being 5, we can write $$(x+3)^2+3^2=5^2$$, from which $$(x+3)^2=16$$, so $$x = 1$$. Sufficient. (2) Obviously, not sufficient, as was already mentioned by \"duriangris\" in the previous post. Gosh! it was a square not a rectangle!! Re: Square ABCD has an area of 9 square inches. [#permalink] 31 Jul 2012, 23:38 Similar topics Replies Last post Similar Topics: 3 In the diagram above, ABCD is a square with side M, and EFGH is a squa 8 09 Apr 2015, 06:29 10 Figure ABCD is a square with sides of length x. Arcs AB, AD, 1 02 Aug 2014, 07:30 4 What is the area of square ABCD", "A rectangle has sides of and . What value of gives an area of 120? 5. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other. 6. A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is wide? 7. Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the numbers? 8. Two positive numbers have a sum of 8, and their product is equal to the smaller number plus 10. What are the numbers? 9. Framing Warehouse offers a picture framing service. The cost for framing a picture is made up of two parts: glass costs $1 per square foot and the frame costs$2 per foot. If the frame has to be a square, what size picture can you get framed for \\$20?", "## Geometry: Common Core (15th Edition) The area of this rectangle is $25 ft^2$. The perimeter of a square is given by the formula: $P = 4s$, where $s$ is the measure of one side of the square. The area of a square is given by the formula: $A = s^2$, where $s$ is the measure of one side of the square. In order to find the area of the square, we need to find the measure of one of its sides. Let us use what we know. We actually know that the square has a perimeter of $20$ ft., so we plug in this information into the formula for the perimeter of a square and solve for $s$: $20 = 4s$ Divide both sides of the equation by $4$ to solve for $s$: $s = 5$ Now, we can plug in $5$ for $s$ in the formula for the area of a square: $A = 5^2$ Evaluate the exponent to solve for $A$: $A = 25$ The area of this rectangle is $25 ft^2$.", "# Length and width calculation from area [closed] I have a question: If I have a square that is $50\\ ft$ by $50\\ ft$ it equals to $2500\\ ft^2$. Simple enough. Now, if I divide $2500\\ ft^2$ area in half, that equals to $1250\\ ft^2$. How do I know what the length and width of this area are? • Is \"divide in half\"=\"shrink the width\"? Otherwise there are a lot of solutions. – Emil Feb 6, 2018 at 17:51 • 25ft by 25ft does not equal to 1250sqft area. – ben Feb 6, 2018 at 17:52 • @ben Draw a picture. There are many many ways to divide a square into two sectors of equal area. It all depends on how you divide. Feb 6, 2018 at 17:55 • What I should have clarified, is that the I'm dividing the area of the square proportionally. – ben Feb 6, 2018 at 22:19 The easiest thing to do is cut the square in half to make a rectangle. Then it has a length of 50 feet and width of 25 feet. But I assume you're not asking that, since it's so obvious. I assume you're asking about a square that has half the area. So how would you find the length and width of a square that has half", "# The total area of the rectangle is 10 ft^2 What is the width and length of the rectangle, given that the width is 3 feet less than the length? Apr 2, 2018 $10 = x \\times \\left(x - 3\\right)$ #### Explanation: x is 5 feet because the length is 5 ft and the width is 2 ft. $10 = 5 \\times \\left(5 - 3\\right)$ $10 = 5 \\times 2$ I found it by trial and error. You can try quadratic formula to solve the problem.", "the squares be $x$. Then the side of the other square is $10-x$, and the area you want minimised is $x^2+(10-x)^2 = 2(x-5)^2+50$.", "the diagonal of a square with side length $\\sqrt{3}$. $\\sqrt{7}$2.645 751 311 064 590 590 501 615 753 639Length of the space diagonal of a $1 \\times 2 \\times \\sqrt{2}$ rectangular box. Length of the diagonal of a $1 \\times \\sqrt{6}$ rectangle. Length of the diagonal of a $2 \\times \\sqrt{3}$ rectangle. Length of the diagonal of a $\\sqrt{2} \\times \\sqrt{5}$ rectangle. $\\sqrt{8}$2.828 427 124 746 190 097 603 377 448 419$2 \\sqrt{2}$ Volume of a cube with edge length $\\sqrt{2}$. Length of the diagonal of a square with side length 2. Length of the diagonal of a $1 \\times \\sqrt{7}$ rectangle. Length of the diagonal of a $\\sqrt{2} \\times \\sqrt{6}$ rectangle. Length of the diagonal of a $\\sqrt{3} \\times \\sqrt{5}$ rectangle. $\\sqrt{10}$3.162 277 660 168 379 331 998 893 544 433$\\sqrt{2} \\cdot \\sqrt{5}$ = area of a $\\sqrt{2} \\times \\sqrt{5}$ rectangle. Length of the diagonal of a $1 \\times 3$ rectangle. Length of the diagonal of a $2 \\times \\sqrt{6}$ rectangle. Length of the diagonal of a $\\sqrt{3} \\times \\sqrt{7}$ rectangle. Length of the diagonal of a square with side length $\\sqrt{5}$. $\\sqrt{11}$3.316 624 790 355 399 849 114 932 736 671Length of the space diagonal of a $1 \\times 1 \\times 3$ rectangular box. Length of the diagonal of a $1 \\times \\sqrt{10}$ rectangle. Length of", "a square with a side length of 4 inches. She has a similar square with dimensions that are twice the first square. How does the area of the larger square compare to the area of the smaller square? First, find the dimensions of the larger square. The problem states that the dimensions are twice the first square. We can use this information to figure out the scale factor, and this means they are scaled up by a factor of 2. The side length of the larger square is $4 \\ inches \\times 2 = 8 \\ inches$. Next find the area of both squares and compare. Area of smaller square: $A &= lw\\\\A &= (4 \\ inches)(4 \\ inches)\\\\A &= 16 \\ inches^2$ Area of larger square: $A &= lw\\\\A &= (8 \\ inches)(8 \\ inches)\\\\A &= 64 \\ inches^2$ Now compare the two areas. You want to know how the area of the larger square compares to the area of the smaller square. Write a ratio comparing the two areas. $\\frac{64 \\ inches^2}{16 \\ inches^2} = 4$ The area of the larger square is 4 times larger than the area of the smaller square. In the previous example, the scale factor that changed the smaller square to the larger square was 2, but the area was 4", "### Home > CCA > Chapter 5 > Lesson 5.2.1 > Problem5-47 5-47. The area of a square is $225$ square centimeters. 1. Make a diagram and determine the length of each side. Find the length of one of the sides of the square and use the sides to find the diagonal. $\\sqrt{225}=15\\text{ cm}$ 2. Use the Pythagorean Theorem to find the length of the square's diagonal. • The Pythagorean Theorem is the following: $a^2+b^2=c^2$ • The length of the diagonal is • $15\\sqrt{2}$", "the area of a space or region. First, we need to figure out the length and width then we can complete any necessary calculations. Sometimes, we will have two different distances or areas that we are working to compare. When this happens, we can use proportions to compare the differences and similarities. Take a look. Marta has a square with a side length of 4 inches. She has a similar square with dimensions that are twice the first square. How does the area of the larger square compare to the area of the smaller square? First, find the dimensions of the larger square. The problem states that the dimensions are twice the first square. We can use this information to figure out the scale factor, and this means they are scaled up by a factor of 2. The side length of the larger square is $4 \\ inches \\times 2 = 8 \\ inches$ . Next find the area of both squares and compare. Area of smaller square: $A &= lw\\\\A &= (4 \\ inches)(4 \\ inches)\\\\A &= 16 \\ inches^2$ Area of larger square: $A &= lw\\\\A &= (8 \\ inches)(8 \\ inches)\\\\A &= 64 \\ inches^2$ Now compare the two areas. You want to know how the area of the larger square compares to the area", "### Home > INT2 > Chapter 8 > Lesson 8.3.1 > Problem8-59 8-59. A regular hexagon with side length $4$ has the same area as a square. What is the length of the side of the square? Explain how you know. Homework Help ✎ Follow the steps outlined in the Math Notes box below to calculate the area of the hexagon. The area of a square is $\\text{A}=s^2$. So to determine the side length of the square take the square root of the hexagon's area (since they are equal).", "+0 # geometry 0 36 3 A large square is divided into 4 small congruent rectangles and a small square as shown. The areas of the large and small squares are 25 and $$12$$, respectively. What is the length of a diagonal of a small rectangle? Apr 16, 2022 #1 +1384 +1 The big square has side length of 5. The small square has a side length of $$\\sqrt {12} = 2\\sqrt3$$ The rectangles are congruent, so their width can be expressed as $$2x+2\\sqrt3=5$$ Solving for x, we find $$x = {{5 - 2\\sqrt3} \\over 2}$$ The length can be expressed as $${2\\sqrt3}+{5 -2\\sqrt3 \\over 2} = {5+2\\sqrt3 \\over2}$$ Using the Pythagorean Theorem, the length of the diagonal is $$\\color{brown}\\boxed{5\\sqrt2 \\over2}$$ Apr 16, 2022 #2 +9369 +1 I think the final answer is wrong, because: $$\\begin{array}{rcl} \\text{length of diagonal} &=& \\sqrt{\\left(\\dfrac{5 - 2\\sqrt 3}2\\right)^2 + \\left(\\dfrac{5 + 2\\sqrt 3}2\\right)^2}\\\\ &=& \\dfrac12 \\sqrt{(5 - 2\\sqrt 3)^2 + (5 + 2\\sqrt 3)^2}\\\\ &=& \\dfrac12 \\sqrt{(25 - 20\\sqrt 3 + 12) + (25 + 20 \\sqrt 3 + 12)}\\\\ &=& \\dfrac{\\sqrt{74}}2 \\\\&\\neq& \\dfrac{5\\sqrt 2}2 \\end{array}$$ Apr 16, 2022 #3 +122390 +1 Length of the side of the large square = 5 Length of the side of the small square = sqrt 12 Width of rectangle = (5 - sqrt 12) /", "The diagram represents five rectangles that have the same perimeter but different areas. Suppose we have $100$100m of fencing material that we wish to use to make a rectangular enclosure. The sides of the enclosure could look like any one of rectangles in the diagram or like any rectangle in between. We note that although the total of the side lengths is fixed at $100$100m, the area enclosed by the fences varies. We could have sides $49$49m, $49$49m, $1$1m, and $1$1m, adding up to $100$100m and the area would be $49\\times1=49$49×1=49 m2. Or, we could have sides $30$30m, $30$30m, $20$20m, and $20$20m, again adding up to $100$100m but in this case enclosing an area of $20\\times30=600$20×30=600 m2. We could find many other possible measurements for the sides and then calculate the area in each case. The goal would be to plot a graph that shows how the area of a rectangle is related to its width. To find the measurements we would probably go through a thought process something like the following: • The total $100$100m length of material is the same as $2$2 lots of the width of the rectangle and $2$2 lots of the length. • So, half the total length is the same as the width plus the length. • Therefore, the length must be", "is $S_1=0.2*0.2^2.$ Second rectangle: height is $0.4^2$, so area is $S_2=0.2*0.4^2$. Third rectangle: height is $0.6^2$, so area is $S_3=0.2*0.6^2$. Fourth rectangle: height is $0.8^2$, so area is $S_4=0.2*0.8^2$. Fifth rectangle: height is $1^2$, so area is $S_5=0.2*1^2$. Sum of areas of these approximating rectangles is $R_5=0.2*0.2^2+0.2*0.4^2+0.2*0.6^2+0.2*0.8^2+0.2*1^2=11/25=0.44$ As can be seen $R_5$ is greater than area $S$, so $S<0.44$. Instead of using the right endpoints we could use the smaller rectangles whose heights are the values of at the left-hand endpoints of the subintervals (the leftmost rectangle has collapsed because its height is 0). Let's see what is the area of each rectangle. Recall that height is value of function at left endpoint. First rectangle: height is $0^2$, so area is $S_1=0.2*0^2$. Second rectangle: height is $0.2^2$, so area is $S_2=0.2*0.2^2$. Third rectangle: height is $0.4^2$, so area is $S_3=0.2*0.4^2$. Fourth rectangle: height is $0.6^2$, so area is $S_4=0.2*0.6^2$. Fifth rectangle: height is $0.8^2$, so area is $S_5=0.2*0.8^2$. Sum of areas of these approximating rectangles is $L_5=0.2*0^2+0.2*0.2^2+0.2*0.4^2+0.2*0.6^2+0.2*0.8^2=6/25=0.24$ As can be seen area $S$ is larger than $L_5$ so $0.24<S<0.44$. Finally we can take height of rectangles to be value of function at midpoint of interval. Corresponding approximation is $M_5=0.2*0.1^2+0.2*0.3^2+0.2*0.5^2+0.2*0.7^2+0.2*0.9^2=33/100=0.33$. Since it is not clear what is greater: $S$ or $M_5$, then we just write $S~~0.33$. What will be" ]

[ "same size. The length of a side of a small square is 2 inches. area of the square = side x side. area of the square and area of the chessboard is the same. area = s x s. s = 2 inches. area = 2 x 2. area = 4 sq in. so the area of the chessboard = 4 sq in. Assessment Practice Question 12. Jessica has saved$50. She will add $25 to her savings each week. Ron has saved$40 and will add $25 to his savings each week. How much will each person have saved after 5 weeks? A. Jessica:$275; Ron: $225 B. Jessica:$250; Ron: $240 C. Jessica:$175; Ron: $165 D. Jessica:$165; Ron: $175 Answer: Option C is correct. Explanation: In the above-given question, given that, Jessica has saved$50. She will add $25 to her savings each week. Ron has saved$40 and will add $25 to his savings each week.$50 + $25 =$75. $25 x 5 =$125. $125 +$50 = $175.$40 + $25 =$65. $65 +$100 = $165. Question 13. Which of the following statements are true? Jessica has always saved$25 more than Ron. Jessica has always saved $10 more than Ron. Ron has always saved$25 less than Jessica. Ron has always saved $10 less than Jessica Answer: Option B is correct. Explanation: In the", "A rectangle has sides of and . What value of gives an area of 120? 5. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other. 6. A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is wide? 7. Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the numbers? 8. Two positive numbers have a sum of 8, and their product is equal to the smaller number plus 10. What are the numbers? 9. Framing Warehouse offers a picture framing service. The cost for framing a picture is made up of two parts: glass costs $1 per square foot and the frame costs$2 per foot. If the frame has to be a square, what size picture can you get framed for \\$20?", "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", "# Length and width calculation from area [closed] I have a question: If I have a square that is $50\\ ft$ by $50\\ ft$ it equals to $2500\\ ft^2$. Simple enough. Now, if I divide $2500\\ ft^2$ area in half, that equals to $1250\\ ft^2$. How do I know what the length and width of this area are? • Is \"divide in half\"=\"shrink the width\"? Otherwise there are a lot of solutions. – Emil Feb 6, 2018 at 17:51 • 25ft by 25ft does not equal to 1250sqft area. – ben Feb 6, 2018 at 17:52 • @ben Draw a picture. There are many many ways to divide a square into two sectors of equal area. It all depends on how you divide. Feb 6, 2018 at 17:55 • What I should have clarified, is that the I'm dividing the area of the square proportionally. – ben Feb 6, 2018 at 22:19 The easiest thing to do is cut the square in half to make a rectangle. Then it has a length of 50 feet and width of 25 feet. But I assume you're not asking that, since it's so obvious. I assume you're asking about a square that has half the area. So how would you find the length and width of a square that has half", "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", "= 3^2\\\\A & = 9 \\ sq. \\ in.$ This is the area for one square face. There are six square faces. If we take this area and multiply it by six or add the area six times, we will have the surface area of the cube. $9(6) & = 54 \\ sq. \\ in.\\\\9 + 9 + 9 + 9 + 9 + 9 & = 54 \\ in^2$ The surface area of the cube is $54 \\ in^2$ . Now let’s look at a rectangular prism net. Let’s say that this box has a length of 6”, a width of 4” and a height of 2”. We need to find the area of each rectangle. There are two long faces. There are two short faces. There is one bottom, and one top that matches it. First, let's find the area of the bottom. It has a length of 6” and a width of 4”. Since the shape of the bottom is a rectangle, we can use the formula for finding the area of a rectangle. $A & = lw\\\\A & = (6)(4)\\\\A & = 24 \\ sq. \\ in.$ Since the top and bottom match, we can multiply this area by two: 24 x 2 = 48 square inches. Next, we find the area of the", "# A rectangle is cut in half to create two squares that each has an area of 25. What is the perimeter of the original rectangle? 30 Explanation The formula to find the area of a square is: Side × Side. If each square has an area of 25, then each side of the square must be 5 (because $$5 \\times 5=25$$). So the the dimensions of the rectangle are $$5 \\times 10$$. The perimeter is the sum of all the sides, and is therefore 5 + 10 + 5 + 10 = 30.", "? Free Version Easy # Related Rates in a Rectangle SVCALC-GVEZY1 A rectangle begins with a height of $4$ inches and a width of $16$ inches. The height increases at a rate of of $2$ in/sec while the width decreases at a ratae of $1$ in/sec. At what rate is the area changing at the moment when the rectangle is a square? A $0$ in$^2$/sec B $-2$ in$^2$/sec C $12$ in$^2$/sec D $144$ in$^2$/sec", "# The total area of the rectangle is 10 ft^2 What is the width and length of the rectangle, given that the width is 3 feet less than the length? Apr 2, 2018 $10 = x \\times \\left(x - 3\\right)$ #### Explanation: x is 5 feet because the length is 5 ft and the width is 2 ft. $10 = 5 \\times \\left(5 - 3\\right)$ $10 = 5 \\times 2$ I found it by trial and error. You can try quadratic formula to solve the problem.", "5, we can write $$(x+3)^2+3^2=5^2$$, from which $$(x+3)^2=16$$, so $$x = 1$$. Sufficient. (2) Obviously, not sufficient, as was already mentioned by \"duriangris\" in the previous post. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Manager Joined: 22 Jun 2012 Posts: 53 GMAT 1: 730 Q49 V40 Followers: 2 Kudos [?]: 16 [0], given: 6 Re: Square ABCD has an area of 9 square inches. [#permalink] 31 Jul 2012, 23:38 EvaJager wrote: (1) Given that the area of the square is 9, then each side of the square is 3. The lengthened sides will be of length 3 + x each, and the diagonal of the obtained rectangle being 5, we can write $$(x+3)^2+3^2=5^2$$, from which $$(x+3)^2=16$$, so $$x = 1$$. Sufficient. (2) Obviously, not sufficient, as was already mentioned by \"duriangris\" in the previous post. Gosh! it was a square not a rectangle!! Re: Square ABCD has an area of 9 square inches. [#permalink] 31 Jul 2012, 23:38 Similar topics Replies Last post Similar Topics: 3 In the diagram above, ABCD is a square with side M, and EFGH is a squa 8 09 Apr 2015, 06:29 10 Figure ABCD is a square with sides of length x. Arcs AB, AD, 1 02 Aug 2014, 07:30 4 What is the area of square ABCD", "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.", "## Geometry: Common Core (15th Edition) The area of this rectangle is $25 ft^2$. The perimeter of a square is given by the formula: $P = 4s$, where $s$ is the measure of one side of the square. The area of a square is given by the formula: $A = s^2$, where $s$ is the measure of one side of the square. In order to find the area of the square, we need to find the measure of one of its sides. Let us use what we know. We actually know that the square has a perimeter of $20$ ft., so we plug in this information into the formula for the perimeter of a square and solve for $s$: $20 = 4s$ Divide both sides of the equation by $4$ to solve for $s$: $s = 5$ Now, we can plug in $5$ for $s$ in the formula for the area of a square: $A = 5^2$ Evaluate the exponent to solve for $A$: $A = 25$ The area of this rectangle is $25 ft^2$.", "is $S_1=0.2*0.2^2.$ Second rectangle: height is $0.4^2$, so area is $S_2=0.2*0.4^2$. Third rectangle: height is $0.6^2$, so area is $S_3=0.2*0.6^2$. Fourth rectangle: height is $0.8^2$, so area is $S_4=0.2*0.8^2$. Fifth rectangle: height is $1^2$, so area is $S_5=0.2*1^2$. Sum of areas of these approximating rectangles is $R_5=0.2*0.2^2+0.2*0.4^2+0.2*0.6^2+0.2*0.8^2+0.2*1^2=11/25=0.44$ As can be seen $R_5$ is greater than area $S$, so $S<0.44$. Instead of using the right endpoints we could use the smaller rectangles whose heights are the values of at the left-hand endpoints of the subintervals (the leftmost rectangle has collapsed because its height is 0). Let's see what is the area of each rectangle. Recall that height is value of function at left endpoint. First rectangle: height is $0^2$, so area is $S_1=0.2*0^2$. Second rectangle: height is $0.2^2$, so area is $S_2=0.2*0.2^2$. Third rectangle: height is $0.4^2$, so area is $S_3=0.2*0.4^2$. Fourth rectangle: height is $0.6^2$, so area is $S_4=0.2*0.6^2$. Fifth rectangle: height is $0.8^2$, so area is $S_5=0.2*0.8^2$. Sum of areas of these approximating rectangles is $L_5=0.2*0^2+0.2*0.2^2+0.2*0.4^2+0.2*0.6^2+0.2*0.8^2=6/25=0.24$ As can be seen area $S$ is larger than $L_5$ so $0.24<S<0.44$. Finally we can take height of rectangles to be value of function at midpoint of interval. Corresponding approximation is $M_5=0.2*0.1^2+0.2*0.3^2+0.2*0.5^2+0.2*0.7^2+0.2*0.9^2=33/100=0.33$. Since it is not clear what is greater: $S$ or $M_5$, then we just write $S~~0.33$. What will be", "a square with an area of $100$ and still have a different perimeter form the first square (it would be like trying to get another combination of four numbers that have the same value, but that when you multiply two of them together they give you $100$). In conclusion, that is why the second rectangle isn't (and can't be) a square. *A square can be a rectangle, but a rectangle can't be a square so, the first rectangle was originally a square.", "the diagonal of a square with side length $\\sqrt{3}$. $\\sqrt{7}$2.645 751 311 064 590 590 501 615 753 639Length of the space diagonal of a $1 \\times 2 \\times \\sqrt{2}$ rectangular box. Length of the diagonal of a $1 \\times \\sqrt{6}$ rectangle. Length of the diagonal of a $2 \\times \\sqrt{3}$ rectangle. Length of the diagonal of a $\\sqrt{2} \\times \\sqrt{5}$ rectangle. $\\sqrt{8}$2.828 427 124 746 190 097 603 377 448 419$2 \\sqrt{2}$ Volume of a cube with edge length $\\sqrt{2}$. Length of the diagonal of a square with side length 2. Length of the diagonal of a $1 \\times \\sqrt{7}$ rectangle. Length of the diagonal of a $\\sqrt{2} \\times \\sqrt{6}$ rectangle. Length of the diagonal of a $\\sqrt{3} \\times \\sqrt{5}$ rectangle. $\\sqrt{10}$3.162 277 660 168 379 331 998 893 544 433$\\sqrt{2} \\cdot \\sqrt{5}$ = area of a $\\sqrt{2} \\times \\sqrt{5}$ rectangle. Length of the diagonal of a $1 \\times 3$ rectangle. Length of the diagonal of a $2 \\times \\sqrt{6}$ rectangle. Length of the diagonal of a $\\sqrt{3} \\times \\sqrt{7}$ rectangle. Length of the diagonal of a square with side length $\\sqrt{5}$. $\\sqrt{11}$3.316 624 790 355 399 849 114 932 736 671Length of the space diagonal of a $1 \\times 1 \\times 3$ rectangular box. Length of the diagonal of a $1 \\times \\sqrt{10}$ rectangle. Length of", "The diagram represents five rectangles that have the same perimeter but different areas. Suppose we have $100$100m of fencing material that we wish to use to make a rectangular enclosure. The sides of the enclosure could look like any one of rectangles in the diagram or like any rectangle in between. We note that although the total of the side lengths is fixed at $100$100m, the area enclosed by the fences varies. We could have sides $49$49m, $49$49m, $1$1m, and $1$1m, adding up to $100$100m and the area would be $49\\times1=49$49×1=49 m2. Or, we could have sides $30$30m, $30$30m, $20$20m, and $20$20m, again adding up to $100$100m but in this case enclosing an area of $20\\times30=600$20×30=600 m2. We could find many other possible measurements for the sides and then calculate the area in each case. The goal would be to plot a graph that shows how the area of a rectangle is related to its width. To find the measurements we would probably go through a thought process something like the following: • The total $100$100m length of material is the same as $2$2 lots of the width of the rectangle and $2$2 lots of the length. • So, half the total length is the same as the width plus the length. • Therefore, the length must be", "the area of a space or region. First, we need to figure out the length and width then we can complete any necessary calculations. Sometimes, we will have two different distances or areas that we are working to compare. When this happens, we can use proportions to compare the differences and similarities. Take a look. Marta has a square with a side length of 4 inches. She has a similar square with dimensions that are twice the first square. How does the area of the larger square compare to the area of the smaller square? First, find the dimensions of the larger square. The problem states that the dimensions are twice the first square. We can use this information to figure out the scale factor, and this means they are scaled up by a factor of 2. The side length of the larger square is $4 \\ inches \\times 2 = 8 \\ inches$ . Next find the area of both squares and compare. Area of smaller square: $A &= lw\\\\A &= (4 \\ inches)(4 \\ inches)\\\\A &= 16 \\ inches^2$ Area of larger square: $A &= lw\\\\A &= (8 \\ inches)(8 \\ inches)\\\\A &= 64 \\ inches^2$ Now compare the two areas. You want to know how the area of the larger square compares to the area", "# Area of a triangle and rectangle The area of a triangle is $3$ times less than $2$ times the area of a rectangle. If the sum of the two areas is $64$ square meters, what is the area of the triangle and what is the area of the rectangle? Hint: Let the area of the triangle be $T$, and the area of the rectangle be $R$. The information given implies: • $T = 2 R / 3$ • $T + R = 64$ Because $T=\\frac{2R}{3}$ and $T+R=64$, we have that $\\frac{2R}{3}+R=64$ and $\\frac{5R}{3}=64$ and $5R=192$. Therefore, $R=38.4$ square meters and $T=\\frac{2*38.4}{3}$ = $25.2$ square meters. The area of the triangle is $25.2$ square meters and the area of the rectangle is $38.4$ square meters.", "5).Now,considering these values (l^2 + b^2 = 100 and l-b<3),we have only one possibility when the diagonal can be of 10 units i.e. when we have sides 8 and 6.Hence,this statement is also sufficient to answer. Let me know what you guys think.... Statement (2) is not sufficient, see algebraic solutions on previous page or consider the following examples: If $$a=8$$ and $$b=6$$ ($$a-b=2<3$$ and $$a^2+b^2=100$$) then $$area=ab=48$$ and the answer to the question \"is $$area>48$$\" is NO; If $$a=b=\\sqrt{50}$$ ($$a-b=0<3$$ and $$a^2+b^2=100$$) then $$area=ab=50$$ and the answer to the question \"is $$area>48$$\" is YES. Note that in this case inscribed figure is square, which is a special type of rectangle. I think that the problem with your solution is that you assumed with no ground for it that the lengths of the sides of the rectangle are integers. Hope it's clear. Bunuel, I understand the solution given in your other explanation. Thanks for the help. My question is on this first explination. How do we get $$a=b=\\sqrt{50}$$? Is it simply because $$a^2+b^2=100$$...I think I figured it out as I was typing this, but is that correct? I was originally trying to draw a connection to the radius and the sides of the square (25+25...) Anyway. _________________ RelishMBA.com Math Expert Joined: 02 Sep 2009 Posts: 36548 Followers: 7077", "4. the length of a rectangular field is thrice its breath. if the cost of cultivating the field at rs. 367.20 per square metre is rs. 27,540, then what is the perimeter of the rectangle ? 1. 36 m 2. 40 m 3. 44 m 4. 49 m Explanation : Question 5. the area of a rectangle is five times the area of a square.the length of the rectangle is 400 ft and breadth is 1/4 of the side of the square. what is the side of the square ? 1. 5 ft 2. 15 ft 3. 20 ft 4. 25 ft" ]

[ "+0 # HELLPPPPP 0 91 1 From point $A$, Leigh walked 40 yards south, 60 yards west, 10 yards north, and 20 yards east to point $B$. What is the length, in yards, of $\\overline{AB}$? Oct 9, 2020 #1 0 40 south, 60 west 10 north, 20 east totals to 30 south, 40 west $$\\sqrt{30^2 + 40^2}$$ = 50", "that he was from point $B$ initially, tells us that they meet when they are $47\\frac{7}{9}$ feet from point $B$.", "# Possible distance b/w points I am stumped on the following question (at least a part of the question) The distance from town A to town B is five miles . Town C is six miles from B .Which of the following could be a distance from A to C ? A)11 b)7 c)1 The answer is all of them. I could only figure out 11. How did they get 7 and 1 ? - You know two things: the line connecting $A$ and $B$ is five miles long, and the line between $B$ and $C$ is six miles long. You do not know where $C$ is relative to $B$. That means that $C$ must lie on a circle with a radius of 6 miles from $B$. If $A$ lies directly between $B$ and $C$, then what is the distance from $A$ to $C$? - Getting 1 is easy: Say B is 5 miles directly east of A. Also say that C is 6 miles directly west of B. This makes C 1 mile directly west of A. Getting 7 is a bit trickier and requires some thought: We know that A is 5 miles away from B and that B is 6 miles away from C. If we were to make a right triangle with 5 on", "miles (B) $$10 \\sqrt{20}$$ miles (C) $$20 \\sqrt{10}$$ miles (D) $$30 \\sqrt{10}$$ miles (E) $$30 \\sqrt{20}$$ miles _________________ CEO Joined: 18 Aug 2017 Posts: 3001 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) Re: Point A is 10 miles West of Point B. Point B is 30 miles North of Poin [#permalink] ### Show Tags 03 Apr 2019, 07:59 Bunuel wrote: Point A is 10 miles West of Point B. Point B is 30 miles North of Point C. Point C is 20 miles East of Point D. What is the distance between points A and D? (A) $$10 \\sqrt{10}$$ miles (B) $$10 \\sqrt{20}$$ miles (C) $$20 \\sqrt{10}$$ miles (D) $$30 \\sqrt{10}$$ miles (E) $$30 \\sqrt{20}$$ miles draw a figure we get a rectangle and triangle AD = √900+100 ; √1000 ; 10√10 IMO A _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. Intern Joined: 16 Jan 2019 Posts: 4 Re: Point A is 10 miles West of Point B. Point B is 30 miles North of Poin [#permalink] ### Show Tags 03 Apr 2019, 13:05 Hi there, can someone please explain what equation to use? I drew out the points and then drew a triangle for points ADC - I got the length of DC as 20", "and C. What are the coordinates of point C? (1) The distance between point A and C is 2 feet (2) The distance between point B and C is 2 feet the second question seemed to be obvious but i suspect there is a catch (since it was published under \"hard questions\") 2. Jude drives from Brooklyn to the Bronx. What is the price of the fuel she uses? (1) The distance between Brooklyn and Bronx is 50 miles (2) the price of fuel is 0.2$per mile. Manager Joined: 20 Nov 2005 Posts: 54 Location: Indianapolis, IN Followers: 1 Kudos [?]: 13 [0], given: 0 Show Tags 02 Dec 2005, 07:09 I'm not necessarily great at answering these types of questions so hopefully one of the more experienced posters can answer as well...but on both I get C. I'm not totally sure about how to actually solve #1 using both of those statements, but it seems possible. on #2, it seems that we know how many miles she went, and how much it costs per mile, so 50 miles * .2$/mile would get you the answer. Where did you get these questions from? 02 Dec 2005, 07:09 Display posts from previous: Sort by", "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.", "19.5 miles from Warrington (due to rounding)? If \"a little over\" a mile is less than 1.1 miles, the Warrington 18 sign must be more than how far from Warrington? Also, it must be less than how far from Warrington? The answers are the distances to look for in the list of candidate answers. October 18th, 2009, 08:37 AM #4 Newbie Joined: Oct 2008 Posts: 6 Thanks: 0 Re: Quote: Originally Posted by skipjack The answer needn't mention the Warrington 19 sign. Is it clear that the Warrington 20 sign is at least 19.5 miles from Warrington (due to rounding)? If \"a little over\" a mile is less than 1.1 miles, the Warrington 18 sign must be more than how far from Warrington? Also, it must be less than how far from Warrington? The answers are the distances to look for in the list of candidate answers. Thanks Skipjack , but i still cant figure out . THis is what i tought . Warrington 20 - the 20 here refers to the distance from Warrington. in the begining , over a half a mile away from W 20 , so the distance from Warrington can now be 19.4 , 19.3 .. anything less than 19.5 .. The distance becomes W 19 due to rounding . Now , it", "points. Question 15. J(4, 3) and K(2, – 3) Question 16. L(- 4, 5) and N(5, – 3) Question 17. P(- 6, – 1) and Q(1, 2) Question 18. Identify the segment bisector of $$\\overline{R S}$$. Then find RS. Question 19. The midpoint of $$\\overline{J K}$$ is M(0, 1). One endpoint is J(- 6, 3). Find the coordinates of endpoint K. Question 20. Your mom asks you to run some errands on your way home from school. She wants you to Stop at the post office and the grocery store, which are both on the same straight road between your school and your house. The distance from your school to the post office is 376 yards. the distance from the post office to your house is 929 yards. and the distance from the grocery store to your house is 513 yards. a. Where should you stop first? b. What is the distance from the post office to the grocery store? d. you walk at a speed of 75 yards per minute. How long does it take you to walk straight home from school? Explain your answer. Question 21. The figure shows a coordinate plane on a baseball held. The distance from home plate to first base is 90 feet.The pitching mound is the midpoint between home plate and", "football field, which is a rectangle 300 ft long and 160 ft wide. James walks in a straight line directly to the southeast corner of the field. If Henry walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, how many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120 E. 140 Look at the diagram below: Attachment: Football field.png [ 5.69 KiB | Viewed 3021 times ] The distance covered by James (AD) is shown in blue and it equals to $$\\sqrt{300^2+160^2}=\\sqrt{10^2(30^2+16^2)}=10*\\sqrt{1156}=340$$; The distance covered by Henry (AB+BD) is shown in red and it equals to $$180+\\sqrt{120^2+160^2}=180+\\sqrt{10^2(12^2+16)^2}=180+10\\sqrt{400}=380$$; The difference is 380-340=40. Answer: B. _________________ Director Joined: 21 Dec 2009 Posts: 512 Concentration: Entrepreneurship, Finance Re: James and Henry are at the northwest corner of their busines [#permalink] ### Show Tags 13 Nov 2012, 23:52 1 Thanks Bunuel for your lucid explanations as usual. I misinterpreted the text below: \"If Henry walks 180 ft down the west side of the field \" to mean \"Henry walks 180 ft westwards\"; Sure, that was a misleading interpretation. Actually, Henry walked DOWNWARD, but on the western wing of the field. _________________ KUDOS me if you", "run east and west, all equally spaced. Streets and avenues are numbered sequentially, as shown in the figure. The walking distance between points $A$ and $B$ is 7 blocks- that is, 3 blocks east and 4 blocks north. To find the straightline distance $d$, we must use the Distance Formula. (a) Find the straight-line distance (in blocks) between $A$ and $B$. (b) Find the walking distance and the straight-line distance between the corner of $4 \\mathrm{th}$ St. and $2 \\mathrm{nd}$ Ave. and the corner of 11 th $\\mathrm{St}$. and 26 th Ave. (c) What must be true about the points $P$ and $Q$ if the walking distance between $P$ and $Q$ equals the straight-line distance between $P$ and $Q ?$ AG Ankit G. ### Problem 54 Two friends live in the city described in Exercise $53,$ one at the corner of 3 rd St. and 7 th Ave. and the other at the corner of 27 th St. and 17 th Ave. They frequently meet at a coffee shop halfway between their homes. (a) At what intersection is the coffee shop located? (b) How far must each of them walk to get to the coffee shop? AG Ankit G. ### Problem 55 The graph shows the pressure experienced by an ocean diver at two different depths. Find and", "many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120 E. 140 Look at the diagram below: Attachment: Football field.png The distance covered by James (AD) is shown in blue and it equals to $$\\sqrt{300^2+160^2}=\\sqrt{10^2(30^2+16^2)}=10*\\sqrt{1156}=340$$; The distance covered by Henry (AB+BD) is shown in red and it equals to $$180+\\sqrt{120^2+160^2}=180+\\sqrt{10^2(12^2+16)^2}=180+10\\sqrt{400}=380$$; The difference is 380-340=40. Answer: B. Is there an alternative (and quicker) way to get the solution without doing the calculations, e.g. through ballparking and eliminating answer options? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8883 Location: Pune, India Re: James and Henry are at the northwest corner of their busines [#permalink] ### Show Tags 05 Dec 2013, 20:59 1 1 np1986 wrote: Bunuel wrote: gmatbull wrote: James and Henry are at the northwest corner of their business school’s football field, which is a rectangle 300 ft long and 160 ft wide. James walks in a straight line directly to the southeast corner of the field. If Henry walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, how many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120", "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 $329$", "can be found. Since he walks from his house to the circus, and will be substituted for and respectively. Evaluate Therefore, the coordinates of the point at which Dylan would have walked miles are", "### Trigonometry: Triangles to Functions Review question A road runs from $A$ due East to $B$, $500$ yd. distant. The bearings of a tree from $A$ and $B$ are $048^\\circ$ and $296^\\circ$ respectively. Find the distance, correct to the nearest yard, of the tree from the road.", "to the southeast corner of the field, how many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120 E. 140 This is the only sentence that put me off that I didnt understand. Intern Joined: 05 Dec 2013 Posts: 2 Concentration: Strategy, Healthcare Schools: Desautels '16 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: James and Henry are at the northwest corner of their busines [#permalink] ### Show Tags 05 Dec 2013, 20:40 Bunuel wrote: gmatbull wrote: James and Henry are at the northwest corner of their business school’s football field, which is a rectangle 300 ft long and 160 ft wide. James walks in a straight line directly to the southeast corner of the field. If Henry walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, how many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120 E. 140 Look at the diagram below: Attachment: Football field.png The distance covered by James (AD) is shown in blue and it equals to $$\\sqrt{300^2+160^2}=\\sqrt{10^2(30^2+16^2)}=10*\\sqrt{1156}=340$$; The distance covered by Henry (AB+BD) is shown in red and it equals to $$180+\\sqrt{120^2+160^2}=180+\\sqrt{10^2(12^2+16)^2}=180+10\\sqrt{400}=380$$; The", "reputed owner, on the Southwest, said point being 48 plus mlnua feet distant Southwesterly, measured at right an- gles from station A214 plus 10 plus minus of the survey base line of the proposed Tremains Corners-Barnec Corners-Worth Corners, County High- way No. 1751, Lewis County, thence S. 16 deg. 47 mln. E. $0 plus minus feet to a point on the division Une be- tween the lands of Orpha Woodard, reputed owner, on the Northwest, and the lands of Rose Ryan, reputed own- er, on the Southeast, the last men- tioned point being 48 feet distant Southwesterly measured at right an- gles, from station A214 plus 70 plus minus of the said, base line, thence along said division line 26 plus minus feet lo its intersection with the divi- sion line between the lands of X W. Bostwick, reputed owner, on the NorthweBt, and the lands of Rose Ry- an, reputed owner, on the Southeast, said point being 24 plus minus feet distant Southwesterly, measured at right angles, from station A214 plus 60 plus minus of the said base line, thence along the division line between the lands of I. W. Bostwick, reputed owner, on the Northeast, and the lands of Orpha Woodard, reputed own- er, on the Southwest, 55 plus minus feet to the point of beginning,", "ft wide. James walks in a straight line directly to the southeast corner of the field. If Henry walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, how many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120 E. 140 This is the only sentence that put me off that I didnt understand. I guess the question intends us to do some approximations to get the correct answer but as you can see we can get the exact answer quite easily, thus approximation is not necessary. _________________ Intern Joined: 05 Dec 2013 Posts: 2 Concentration: Strategy, Healthcare Schools: Desautels '16 Re: James and Henry are at the northwest corner of their busines [#permalink] ### Show Tags 05 Dec 2013, 20:40 Bunuel wrote: gmatbull wrote: James and Henry are at the northwest corner of their business school’s football field, which is a rectangle 300 ft long and 160 ft wide. James walks in a straight line directly to the southeast corner of the field. If Henry walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, how", "### Home > CC1 > Chapter 1 > Lesson 1.1.5 > Problem1-37 1-37. Eli walked $12$ feet down the hall of his house to get to the door. He continued in a straight line out the door and across the yard to the mailbox, a distance of $32$ feet. He came straight back across the yard $14$ feet and stopped to pet his dog. 1. Draw a diagram of Eli’s walking pattern. First Eli walked $12$ feet forward to get out the door. Line Diagram: From the starting point there is 12 feet to the door. Then Eli walked $32$ feet forward to get to the mailbox. Line Diagram: From the door and continuing in the same direction, there is 32 feet from the Door to the Mailbox. Then he turned around and walked $14$ feet back to his dog. Line Diagram: From the Mailbox and now going in the opposite direction, there is 14 feet from the mailbox to a dog. 2. How far has he walked? This is asking about how many feet he has walked in total. You do not need to think about direction. $12\\text{ feet}+32\\text{ feet}+14\\text{ feet}=58\\text{ feet}$ 3. How far from the house is he now? First, Eli walked $32$ feet away from the house. Then he walked $14$ feet back toward", "owner, on the Bast, said point being 42 plug minus feet distant Westerly, measured at right angles, from station A205 plus U Plus minus of the survey base line of the pro- posed Tremains Comers-Barnes Cor- ners Worth Corners, County Highway No. 1751, Lewis County, thence North- easterly along sail division line 103 plus minus feet to a point S7 feet distant Easterly, measured at right angles, from station A208 plaa 50 plus minuB of said base line/thence 8. 16 deg. 42 mln* E. 416 feet to a point on the division line between the lands of Perley.' Bostwlckv reputed owner; on the West and the lands of I. W. Bostwick, reputed owner, on the- Eatt, the last mentioned point being 18 feet distant Easterly, measured at right angles, from station A204 plus 34 plus minus of the said base line, thonce along said division line 28 plus minus feet to its Intersection with the sur- vey base line at station A208 plus* 72 of the said bane line, thence\\ along said division line 73 plus minus feet to a point 48 feet distant Westerly, measured at right angles,-from sta- tion A209 plus 58 of said base line, thince N. 16 deg. 47 min. W. 414 plus minus feet to the point of beginning, being 0.63 acres more", "# Tarzan, a respected mathematician, comes to a rag Tarzan, a respected mathematician, comes to a raging river while traveling in the jungle.In order to figure out hot to get across the river he wants to know its width.Tarzan locates a tree at point A directly across the river. He marks the spot and then walks 28 paces to point c where he pounds in a stake.Then he walks in addition 10 paces before turning perpendicularly to the river and walking until point C lines up with point A.This takes 14 peaces.What is the width of the river in paces? • Elite2 77 orders completed $10.00 ANSWER Tutor has posted answer for$10.00. See answer's preview ****** a respected ************* TARZAN.doc" ]

[ "+0 # HELLPPPPP 0 91 1 From point $A$, Leigh walked 40 yards south, 60 yards west, 10 yards north, and 20 yards east to point $B$. What is the length, in yards, of $\\overline{AB}$? Oct 9, 2020 #1 0 40 south, 60 west 10 north, 20 east totals to 30 south, 40 west $$\\sqrt{30^2 + 40^2}$$ = 50", "the triangle - 10. If the hypotenuse of the triangle is 10 then the x coordinate of Tammi's current position is 8 as in (8,0). And because Emily must be 4 yards away, Emily's coordinate is (12,0). From here apply the distance formula- Emily's distance from her current position and Point A and then subtract it by the distance Tammi traveled- 10 yards. Thus \"C\" Attachments image1 (7).JPG [ 815.72 KiB | Viewed 1917 times ] Director Joined: 13 Mar 2017 Posts: 704 Location: India Concentration: General Management, Entrepreneurship GPA: 3.8 WE: Engineering (Energy and Utilities) Emily and Tammi must both run from point A to reach the finish line, w [#permalink] ### Show Tags Updated on: 19 Jul 2017, 03:08 2 Emily and Tammi must both run from point A to reach the finish line, which stretches from the origin to point B, but they have to run on two different paths, where Emily will end up 4 yards away from Tammi on the finish line. If Tammi ran a total of 10 yards, how many yards more did Emily run than Tammi? A. 2 B. $$4\\sqrt{2}−10$$ C. $$6\\sqrt{5}−10$$ D. $$4\\sqrt{2}−8$$ E. $$6\\sqrt{5}$$ Attachment: Geometry_Img67.png Let the origin be O. Let the position of Tammi shown in diagram be T and the position of Emily shown is", "However, the water current displaced her by 0.9 miles in the direction $37\\degree$ east of north. How far is Esther from her starting point, and in what direction? ###### 36 Rani paddles her canoe 4.5 miles in the direction $12\\degree$ west of north. The water current pushes her 0.3 miles off course in the direction $5\\degree$ east of north. How far is Rani from her starting point, and in what direction? ###### 37 Brenda wants to fly to an airport that is 103 miles due west in 1 hour. The prevailing winds blow in the direction $112\\degree$ east of north at 28 miles per hour, so Brenda will head her plane somewhat north of due west to compensate. What airspeed and direction should Brenda take? ###### 38 Ryan wants to cross a 300 meter wide river running due south at 80 meters per minute. There are rocks upstream and rapids downstream, so he wants to paddle straight across from east to west. In what direction should he point his kayak, and how fast should his water speed be in order to cross the river in 2 minutes? (Hint: The current will move him 160 meters due south compared with where his speed and direction would take him if the current stopped. Compute the distance he would have traveled,", "degrees South of West for 25 minutes at the same pace. If both Jennifer and Tara return straight home, will their paths form similar triangles?", "enters the exact center of the south side and walks 10 feet north. Then she walks 8 feet east. How far is she from the east side of the room? ______ ft Explanation: Given that, Jan enters a 20-foot by 30-foot rectangular room. The long sides face north and south. Jan enters the exact center of the south side and walks 10 feet north. Then she walks 8 feet east. Jan is 7 feet from the east wall in the room. On Your Own – Lesson 4: Properties of Two-Dimensional Figures – Page No. 456 Question 5. Max drew a grid to divide a piece of paper into 18 congruent squares, as shown. What is the least number of lines Max can draw to divide the grid into 6 congruent rectangles? ______ lines Explanation: From the above figure, we can see that there are 18 congruent squares. To find the least number of lines Max can draw we have to divide number of squares by number of congruent rectangles 18 ÷ 6 = 3 Thus the least number of lines that Max can Draw is 3 lines. Question 6. Of the 95 fifth and sixth graders going on a field trip, there are 27 more fifth-graders than sixth graders. How many fifth graders are going on the field", "the nets. How far was she from the pl", "can randomly assign y = 0 to Julia, and y = 1 to Marie. Note that since Marie starts out 200 miles away from Julia, and she’s is going “backwards” towards Austin, her distance is –60(t – 2) + 200, with relationship to Austin. This is because she starts out 200 miles from Austin, and every passing hour she’s 60 miles closer to Austin, so we subtract (put in real numbers to see how this works – the distance is getting smaller as time passes). We’ll just set the two distances from Austin together (the x part of the equations) and solve for t to get the time that they meet, measured in the time from Julia leaving Austin (noon). We really don’t need to use the y equations, but it’s important to see how we can model a situation with them. So in almost 3 hours (about 3pm), they will pass each other. They will be 50t miles from Austin, or about 145.5 miles from Austin. This will make them 200 – 145.5 = 54.5 miles from Dallas. # Projectile Motion Applications Again, parametric equations are very useful for projectile motion applications. With parametric equations and projectile motion, think of x as the distance along the ground from the starting point, y as the distance from the", "Displacement ## Examples of Distance And Displacement Question 1. John travels 250 miles to North but then back-tracks to South for 105 miles to pick up a friend. What is John’s total displacement? Answer: John’s starting position Xi= 0. Her final position Xf is the distance travelled N minus the distance South. Calculating displacement, i.e.D. D = ΔX = (Xf – Xi) D = (250 mi N – 105 mi S) – 0 D = 145 mi N Question 2. Julia throws the ball 30 feet North for your dog. The dog catches the ball and takes it past Julia to her brother, who is standing 5 feet to the South of where she is. Calculate the total displacement of the ball? Answer: Julia’s initial position with the ball ( Xi) = 0. As displacement is a vector quantity, ‘direction’ is considered. Xf = (30 feet S – 25 feet N), so, Xf = -5 feet S. D = ΔX = (Xf – Xi) D = (30 ft S – 25 ft S) – 0 D = 5 ft S of Julia’s starting position ## Distance vs Displacement Distance Displacement The complete length of the path between any two points is called distance Displacement is the direct length between any two points when measured along the minimum path", "+ (12)^2$$ $$e^2 = 36 + 144$$ $$e^2 = 180$$ $$e = \\sqrt{180}$$ Director Joined: 12 Nov 2016 Posts: 724 Location: United States Schools: Yale '18 GMAT 1: 650 Q43 V37 GRE 1: Q157 V158 GPA: 2.66 Emily and Tammi must both run from point A to reach the finish line, w [#permalink] ### Show Tags 17 Jun 2017, 03:56 2 Emily and Tammi must both run from point A to reach the finish line, which stretches from the origin to point B, but they have to run on two different paths, where Emily will end up 4 yards away from Tammi on the finish line. If Tammi ran a total of 10 yards, how many yards more did Emily run than Tammi? A. 2 B. $$4\\sqrt{2}−10$$ C. $$6\\sqrt{5}−10$$ D. $$4\\sqrt{2}−8$$ E. $$6\\sqrt{5}$$ Attachment: The attachment Geometry_Img67.png is no longer available The hidden inference in this question is that distance Tammi travels represents a pythagorean triplet triangle ( 6 8 10)- there's really only two ways that Tammi could have traveled 10 yards- she could go straight down from (0,6) to the origin (0,0) and then travel to (4,0); however, if this was done then the distance Emily runs would actually be the same as Tammie and thus false. Therefore, the distance Tammi runs represents the hypotenuse of", "finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º. Solution 18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º south of west, then 5.56 km at 45.0º west of north 42. A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure, and then correctly calculates the length and orientation of the fourth side $$D$$. What is his result? 43. In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30km 25.0º south of west; then 5.10 km straight east; then 1.70km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island? Solution $$7.34 km, 63.5º$$ south of east 44. Suppose a", "• 25 Learning Opportunities This puzzle can be solved using the following concepts. Practice using these concepts and improve your skills. Goal Ataria lives on a toroidal map known as the Mystic Rectangle, where opposite edges of the map are connected. Crossing an edge teleports her to the opposite side. Ataria can move in any of the four cardinal directions or along the four diagonals of 45 degrees. Given Ataria's coordinates and the coordinates of the goal, find the fastest time for her to reach the goal, assuming no obstacles. Moving East or West ±1 unit in x takes 0.3 seconds. Moving North or South ±1 unit in y takes 0.4 seconds. It takes 0.5 seconds to move diagonally, 1 unit East/West plus 1 unit North/South. After travelling for 1 minute in any single cardinal direction (that is, 200 units East/West, or 150 units North/South), she ends up returning to the same place. Example Suppose Ataria starts near the top of the map and travels diagonally 15 units NorthEast and then straight 5 units due North. In doing so, she wraps around to the bottom of the map, arriving at the goal. Then at 9.5 seconds, this direct route makes the best time to the goal. Input Line 1: Two space separated integers x and y for the", "89 101 Assignments Done 97.9% Successfully Done In July 2020 # Answer to Question #17497 in Mechanics | Relativity for Jackie F Question #17497 a football coach starts at the 35 yard-line and walks to the 15-yard line. He turns around and walks to the 40-yard line. He turns around again and walks back to the 0-yard line. He turns around one more and walks back to the 25-yard line thus completing his 10 min of pacing. The coachs average speed is 1 2012-11-01T08:41:00-0400 1. A football coach starts at the 35-yard line and walks to the 15-yard line. So, the first part of his displacement is D1 = 35 - 15 = 20 yards. 2. He turns around and walks to the 40-yard line. Coach starts from the 15-yard line and walks to the 40-yard line, so his displacement is D2 = 40 - 15 = 25 yards. 3. He turns around again and walks back to the 0-yard line. Coach starts from the 40-yard line and walks to the 0-yard line, so his displacement is D3 = 40 - 0 = 40 yards. 4. He turns around one more and walks back to the 25-yard line. Coach starts from the 0-yard line and walks to the 25-yard line, so his displacement is D4 = 0 +", "# Magnitude and direction Displacement: Magnitude And Direction (urgent) ## Homework Statement Jim walks 100m due south, 60m due east, and then 20m due north. What is his net displacement (both magnitude and direction)? ## Homework Equations I know how to find the net displacement by the magnitude of the distance... using the d = sq. rt ( (x2-x1)^2 + (y2-y1)^2 ) but i dont know how to express the direction? Last edited:", "a representative for Black said on Tuesday. \"It is telling that it is asserted anonymously and concerns events that allegedly occurred some 20 years ago.... ### What was The outcome of the second battle of bull run? What was The outcome of the second battle of bull run?... ### A boat can travel 2.30 m/s in still water. If the boat points its prow directly across a stream whose current is 1.80 m/s , what is the magnitude of the velocity of the boat relative to the shore? A boat can travel 2.30 m/s in still water. If the boat points its prow directly across a stream whose current is 1.80 m/s , what is the magnitude of the velocity of the boat relative to the shore?... ### Mrs. Morgan is teaching a 5th grade class. She is standing 12 meters in front of Aaliyah. Brenna is sitting too Aaliyah's right. If Brenna and Mrs. Morgan are 15 meters apart, how far apart are Aaliyah and Brenna Mrs. Morgan is teaching a 5th grade class. She is standing 12 meters in front of Aaliyah. Brenna is sitting too Aaliyah's right. If Brenna and Mrs. Morgan are 15 meters apart, how far apart are Aaliyah and Brenna... ### The tone of your workplace writing should be _______ to avoid", "Solving Question 3. H.O.T. Multi-Step Jan enters a 20-foot by 30-foot rectangular room. The long sides face north and south. Jan enters the exact center of the south side and walks 10 feet north. Then she walks 8 feet east. How far is she from the east side of the room? It is given that Jan enters a 20-foot by 30-foot rectangular room. The long sides face north and south. Jan enters the exact center of the south side and walks 10 feet north. Then she walks 8 feet east Now, According to the given information, The distance of Jan from the east side of the room = (30 – 15 + 10) – 8 = 17 feet Hence, from the above, We can conclude that The distance of Jan from the east side of the room is: 17 feet Question 4. Use Diagrams Max drew a grid to divide a piece of paper into 18 congruent squares. What is the least number of lines Max can draw to divide the grid into 6 congruent rectangles? Use the diagram to show your reasoning. It is given that Max drew a grid to divide a piece of paper into 18 congruent squares Now, The given grid is: Now, According to the given information, The least number of lines Max", "# A boat is sailing due east parallel to the shoreline at a speed of 11 miles per hour. At a given time, the bearing to the lighthouse is S 70° E, and 30 minutes later the bearing is S 63° E (see figure). The lighthouse is located at the shoreline. help? Jul 24, 2018 $\\text{Distance from shoreline} = 7.0 \\left(1 \\mathrm{dp}\\right)$miles #### Explanation: There is no figure shown, but when taking bearings for navigation the direction to a point is stated as the number of degrees east or west of North or South. The likely question that follows from the information given is \"how far from the shoreline is the boat?\" We know that the boat is traveling parallel to the shoreline at $11$mph for $0.5$ hours, so it moves $5.5$ miles. The 1st bearing to the lighthouse is more East than the 2nd bearing, so the boat is moving east, towards the lighthouse. Let $a$ be the perpendicular distance to shore and $b$ be the distance along the shoreline to the lighthouse from the 2nd bearing position. A not-to-scale drawing : $\\tan \\left(90 - 70\\right) = \\frac{a}{b + 5.5}$ #a=tan(20)*(b+5.5) ....(1) and $\\tan \\left(90 - 63\\right) = \\frac{a}{b}$ $b = \\frac{a}{\\tan} \\left(27\\right)$ Substitute for b in eqn (1) $a = \\tan \\left(20\\right) \\cdot \\left(\\frac{a}{\\tan} \\left(27\\right) +", "east. (This is equivalent to subtract A from B —that is, to find A=B+C. Is that consistent with your result?) (a) 30.8 m54.2º south of west (b) 30.8 m54.2º north of east A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B inFigure. She then correctly calculates the length and orientation of the third side C. What is her result? <figure id=\"eip-id3165265\" style=\"width: 782px;\"> </figure> You fly 32.0 km in a straight line in still air in the direction 35.0º south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction45.0º south of west and then in a direction 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º. 18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º south of west, then 5.56 km at 45.0º west of", "the total distance that Tracie traveled. Compare this with the distance between her starting and final positions. Solution The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is $1$ block east and $1$ block north, so it is at $(1, 1)$. The next stop is $5$ blocks to the east, so it is at $(5, 1)$. After that, she traveled $3$ blocks east and $2$ blocks north to $(8, 3)$. Lastly, she traveled $4$ blocks north to $(8, 7)$. We can label these points on the grid as in Figure 15. Next, we can calculate the distance. Note that each grid unit represents $1,000$ feet. • From her starting location to her first stop at $(1, 1)$, Tracie might have driven north $1,000$ feet and then east $1,000$ feet, or vice versa. Either way, she drove $2,000$ feet to her first stop. • Her second stop is at $(5,1)$. So from $(1, 1)$ to $(5, 1)$, Tracie drove east $4,000$ feet. • Her third stop is at $(8, 3)$. There are a number of routes from $(5, 1)$ to $(8,", "### Home > CALC > Chapter 7 > Lesson 7.2.4 > Problem7-100 7-100. Eddie is standing on third base, watching in amazement as Yuliana tries to steal second. If Yuliana runs at $15\\frac {\\text{ ft }} { \\operatorname { cos } }$, how fast is her distance from Eddie changing when she is $40$ feet from first base? Notice the right triangle. Let $x =$ Yuliana's distance from ____________ base. Find the remaining two sides of the triangle, in terms of $x$. Things that change (or describe change) should be 'plugged in' to the derivative equation, not the original equation . This includes Yuliana's running rate and her position from 1st base at the moment in question. Is Yuliana's rate positive or negative? The distance between Yuliana and Eddie is changing at a rate of $-7.285$ ft/sec when Yulanda was $40$ feet from 1st base.", "### Home > CC1MN > Chapter 3 Unit 3A > Lesson CC1: 3.1.1 > Problem3-13 3-13. Rachel says that when she ran $115$ yards, she went farther than Beth, who only ran $327$ feet. Is Rachel correct? Explain how you know. Remember that $1$ yard = $3$ feet. Convert the units so that you know how many yards Beth ran or how many feet Rachel ran. There are three feet in one yard." ]

[ "+0 # HELLPPPPP 0 91 1 From point $A$, Leigh walked 40 yards south, 60 yards west, 10 yards north, and 20 yards east to point $B$. What is the length, in yards, of $\\overline{AB}$? Oct 9, 2020 #1 0 40 south, 60 west 10 north, 20 east totals to 30 south, 40 west $$\\sqrt{30^2 + 40^2}$$ = 50", "However, the water current displaced her by 0.9 miles in the direction $37\\degree$ east of north. How far is Esther from her starting point, and in what direction? ###### 36 Rani paddles her canoe 4.5 miles in the direction $12\\degree$ west of north. The water current pushes her 0.3 miles off course in the direction $5\\degree$ east of north. How far is Rani from her starting point, and in what direction? ###### 37 Brenda wants to fly to an airport that is 103 miles due west in 1 hour. The prevailing winds blow in the direction $112\\degree$ east of north at 28 miles per hour, so Brenda will head her plane somewhat north of due west to compensate. What airspeed and direction should Brenda take? ###### 38 Ryan wants to cross a 300 meter wide river running due south at 80 meters per minute. There are rocks upstream and rapids downstream, so he wants to paddle straight across from east to west. In what direction should he point his kayak, and how fast should his water speed be in order to cross the river in 2 minutes? (Hint: The current will move him 160 meters due south compared with where his speed and direction would take him if the current stopped. Compute the distance he would have traveled,", "+ (12)^2$$ $$e^2 = 36 + 144$$ $$e^2 = 180$$ $$e = \\sqrt{180}$$ Director Joined: 12 Nov 2016 Posts: 724 Location: United States Schools: Yale '18 GMAT 1: 650 Q43 V37 GRE 1: Q157 V158 GPA: 2.66 Emily and Tammi must both run from point A to reach the finish line, w [#permalink] ### Show Tags 17 Jun 2017, 03:56 2 Emily and Tammi must both run from point A to reach the finish line, which stretches from the origin to point B, but they have to run on two different paths, where Emily will end up 4 yards away from Tammi on the finish line. If Tammi ran a total of 10 yards, how many yards more did Emily run than Tammi? A. 2 B. $$4\\sqrt{2}−10$$ C. $$6\\sqrt{5}−10$$ D. $$4\\sqrt{2}−8$$ E. $$6\\sqrt{5}$$ Attachment: The attachment Geometry_Img67.png is no longer available The hidden inference in this question is that distance Tammi travels represents a pythagorean triplet triangle ( 6 8 10)- there's really only two ways that Tammi could have traveled 10 yards- she could go straight down from (0,6) to the origin (0,0) and then travel to (4,0); however, if this was done then the distance Emily runs would actually be the same as Tammie and thus false. Therefore, the distance Tammi runs represents the hypotenuse of", "the triangle - 10. If the hypotenuse of the triangle is 10 then the x coordinate of Tammi's current position is 8 as in (8,0). And because Emily must be 4 yards away, Emily's coordinate is (12,0). From here apply the distance formula- Emily's distance from her current position and Point A and then subtract it by the distance Tammi traveled- 10 yards. Thus \"C\" Attachments image1 (7).JPG [ 815.72 KiB | Viewed 1917 times ] Director Joined: 13 Mar 2017 Posts: 704 Location: India Concentration: General Management, Entrepreneurship GPA: 3.8 WE: Engineering (Energy and Utilities) Emily and Tammi must both run from point A to reach the finish line, w [#permalink] ### Show Tags Updated on: 19 Jul 2017, 03:08 2 Emily and Tammi must both run from point A to reach the finish line, which stretches from the origin to point B, but they have to run on two different paths, where Emily will end up 4 yards away from Tammi on the finish line. If Tammi ran a total of 10 yards, how many yards more did Emily run than Tammi? A. 2 B. $$4\\sqrt{2}−10$$ C. $$6\\sqrt{5}−10$$ D. $$4\\sqrt{2}−8$$ E. $$6\\sqrt{5}$$ Attachment: Geometry_Img67.png Let the origin be O. Let the position of Tammi shown in diagram be T and the position of Emily shown is", "which stretches from the origin to point B, but they have to run on two different paths, where Emily will end up 4 yards away from Tammi on the finish line. If Tammi ran a total of 10 yards, how many yards more did Emily run than Tammi? A. 2 B. $$4\\sqrt{2}−10$$ C. $$6\\sqrt{5}−10$$ D. $$4\\sqrt{2}−8$$ E. $$6\\sqrt{5}$$ Attachment: Geometry_Img67.png [ 7.83 KiB | Viewed 3792 times ] _________________ KUDO me plenty Originally posted by shasadou on 24 Oct 2015, 05:32. Last edited by Bunuel on 25 Oct 2015, 05:48, edited 1 time in total. Edited the question. Current Student Joined: 21 Jul 2013 Posts: 106 WE: Securities Sales and Trading (Commercial Banking) Emily and Tammi must both run from point A to reach the finish line, w [#permalink] ### Show Tags 02 Dec 2015, 00:12 2 Hi everyone, I took close to 4 minutes to solve this. Are there any suggestions on how I can halve my time spent in a question like this? here's how i solved this question.. we know that Tammi ran 10 yards (t= 10) and we do not know how far Emily (e) ran. question: e - t = ? 1) using the Pythagorean theorem, I solved for x. 10^2 = 6^2 + x^2 x = $$\\sqrt{64}$$ 2) next, I solved for", "$A$ and $B$ are two lifeboat stations on a North-facing coast, $B$ being $\\quantity{6}{km}$ due East of $A$. A distress call from a ship is received simultaneously at both stations. The bearing of the call from $A$ is measured to lie between $060^\\circ$ and $065^\\circ$. The bearing of the call from $B$ is measured to lie between $330^\\circ$ and $335^\\circ$. By means of a scale drawing on squared paper find as accurately as possible 1. the least and greatest possible distances of the distressed ship from $B$, and", "read to first sail 27.5 m in a direction $66.0°66.0° size 12{\"66\" \".\" 0°} {}$ north of east from her current location, and then travel 30.0 m in a direction $112°112° size 12{\"112\"°} {}$ north of east (or $22.0°22.0° size 12{\"22\" \".\" 0°} {}$ west of north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock. Figure 3.21 #### Strategy We can represent the first leg of the trip with a vector $AA$, and the second leg of the trip with a vector $BB size 12{B} {}$. The dock is located at a location $A + BA + B$. If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance $BB$ (30.0 m) in the direction $180°–112°=68°180°–112°=68°$ south of east. We represent this as $–B–B$, as shown below. The vector $–B–B$ has the same magnitude as $BB$ but is in the opposite direction. Thus, she will end up at a location $A+(–B)A+(–B)$, or $A–BA–B$. Figure 3.22 We will perform vector addition to compare the location of the dock, , with the location at which the woman mistakenly arrives, . #### Solution (1) To determine", "3)$. Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east $3,000$ feet and then north $2,000$ feet for a total of $5,000$ feet. • Tracie’s final stop is at $(8, 7)$. This is a straight drive north from $(8, 3)$ for a total of $4,000$ feet. Next, we will add the distances listed in Table 3. The total distance Tracie drove is $15,000$ feet, or $2.84$ miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points $(0, 0)$ and $(8, 7)$. \\begin {align*} d&=\\sqrt{(8-0)^{2} + (7-0)^{2}} \\\\&=\\sqrt{64+49} \\\\ &= \\sqrt{113} \\\\ &\\approx{10.63} \\ units \\end{align*} At $1,000$ feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is $10,630.14$ feet, or $2.01$ miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point $(8, 7)$. Perhaps you have heard the saying “as the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer", "football field, which is a rectangle 300 ft long and 160 ft wide. James walks in a straight line directly to the southeast corner of the field. If Henry walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, how many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120 E. 140 Look at the diagram below: Attachment: Football field.png [ 5.69 KiB | Viewed 3021 times ] The distance covered by James (AD) is shown in blue and it equals to $$\\sqrt{300^2+160^2}=\\sqrt{10^2(30^2+16^2)}=10*\\sqrt{1156}=340$$; The distance covered by Henry (AB+BD) is shown in red and it equals to $$180+\\sqrt{120^2+160^2}=180+\\sqrt{10^2(12^2+16)^2}=180+10\\sqrt{400}=380$$; The difference is 380-340=40. Answer: B. _________________ Director Joined: 21 Dec 2009 Posts: 512 Concentration: Entrepreneurship, Finance Re: James and Henry are at the northwest corner of their busines [#permalink] ### Show Tags 13 Nov 2012, 23:52 1 Thanks Bunuel for your lucid explanations as usual. I misinterpreted the text below: \"If Henry walks 180 ft down the west side of the field \" to mean \"Henry walks 180 ft westwards\"; Sure, that was a misleading interpretation. Actually, Henry walked DOWNWARD, but on the western wing of the field. _________________ KUDOS me if you", "= $$\\sqrt{4^{2}+3.5^{2}-2(4)(3.5) \\cos 30}$$ a = 2.000322148 … a ≈ 2 Total length of the race: 4 + 3.5 + (2) ≈ 9.5 The total length of the race is approximately 9.5 miles. Exercise 3. Two lighthouses are 30 miles apart on each side of shorelines running north and south, as shown. Each lighthouse keeper spots a boat in the distance. One lighthouse keeper notes the location of the boat as 40° east of south, and the other lighthouse keeper marks the boat as 32° west of south. What is the distance from the boat to each of the lighthouses at the time it was spotted? Round your answers to the nearest mile. Let x be the distance from the southern lighthouse to the boat. $$\\frac{\\sin 72}{30}=\\frac{\\sin 50}{x}$$ x = $$\\frac{30 \\sin 50}{\\sin 72}$$ x = 24.16400382… The southern lighthouse is approximately 24 mi. from the boat. Let y be the distance from the northern lighthouse to the boat. $$\\frac{\\sin 72}{30}=\\frac{\\sin 50}{x}$$ y = $$\\frac{30 \\sin 58}{\\sin 72}$$ y = 26.750716 12… The northern lighthouse is approximately 27 mi.from the boat. Let a be the distance from the northern lighthouse to the boat. a2 = 302 + 242 – 2(30)(24) . cos58 a = $$\\sqrt{30^{2}+24^{2}-2(30)(24) \\cdot \\cos 58}$$ a = 26.700491 75… The northern lighthouse is approximately 27", "many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120 E. 140 Look at the diagram below: Attachment: Football field.png The distance covered by James (AD) is shown in blue and it equals to $$\\sqrt{300^2+160^2}=\\sqrt{10^2(30^2+16^2)}=10*\\sqrt{1156}=340$$; The distance covered by Henry (AB+BD) is shown in red and it equals to $$180+\\sqrt{120^2+160^2}=180+\\sqrt{10^2(12^2+16)^2}=180+10\\sqrt{400}=380$$; The difference is 380-340=40. Answer: B. Is there an alternative (and quicker) way to get the solution without doing the calculations, e.g. through ballparking and eliminating answer options? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8883 Location: Pune, India Re: James and Henry are at the northwest corner of their busines [#permalink] ### Show Tags 05 Dec 2013, 20:59 1 1 np1986 wrote: Bunuel wrote: gmatbull wrote: James and Henry are at the northwest corner of their business school’s football field, which is a rectangle 300 ft long and 160 ft wide. James walks in a straight line directly to the southeast corner of the field. If Henry walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, how many feet farther, to the nearest 10 ft, will Henry walk than James? A. 20 B. 40 C. 80 D. 120", "Ms. $X$ came out of a building through its front door to find her shadow due to the morning sun failing to her right side with the building to her back. From this, it can be inferred that building is facing _________ 1. North 2. East 3. West 4. South from the above diagram, we can conclude that the building is facing south. So, the correct answer is $(D).$ 18.0k points 4 7 16 4 7 16", "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$", "Displacement ## Examples of Distance And Displacement Question 1. John travels 250 miles to North but then back-tracks to South for 105 miles to pick up a friend. What is John’s total displacement? Answer: John’s starting position Xi= 0. Her final position Xf is the distance travelled N minus the distance South. Calculating displacement, i.e.D. D = ΔX = (Xf – Xi) D = (250 mi N – 105 mi S) – 0 D = 145 mi N Question 2. Julia throws the ball 30 feet North for your dog. The dog catches the ball and takes it past Julia to her brother, who is standing 5 feet to the South of where she is. Calculate the total displacement of the ball? Answer: Julia’s initial position with the ball ( Xi) = 0. As displacement is a vector quantity, ‘direction’ is considered. Xf = (30 feet S – 25 feet N), so, Xf = -5 feet S. D = ΔX = (Xf – Xi) D = (30 ft S – 25 ft S) – 0 D = 5 ft S of Julia’s starting position ## Distance vs Displacement Distance Displacement The complete length of the path between any two points is called distance Displacement is the direct length between any two points when measured along the minimum path", "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.", "## Square Lot Al and Curt live at opposite corners of a square lot; Beth and Dawn live at the other two corners. They all carry water from a spring $S$ located within the lot. The spring is $50$ feet from Al, $30$ feet from Beth, and $40$ feet from Curt. How far is the spring from Dawn? Source: NCTM Mathematics Teacher 2008 SOLUTION $a^2+b^2=x^2\\qquad\\:\\: (1)$ $a^2+c^2=50^2\\qquad (2)$ $c^2+d^2=30^2\\qquad (3)$ $b^2+d^2=40^2\\qquad (4)$ Subtract Eq. (2) from Eq. (1) $b^2-c^2=x^2-50^2\\qquad\\:\\: (5)$ Subtract Eq. (3) from Eq. (4) $b^2-c^2=40^2-30^2\\qquad (6)$ Compare Eq. (5) and Eq. (6) $x^2-50^2=40^2-30^2$ $x^2=40^2-30^2+50^2$ $=3200$ $x=56.6$ feet Answer: $56.6$ feet", "a function that gives the sum of the distances from the potter's station to the three machines. 3. Graph your function for $-20\\le x\\le 30\\text{.}$ Where should the potter stand in order to minimize the distance she must walk to the machines? ###### 58 Suppose the pottery in Problem 57 adds a fourth machine to the procedure for producing a mug, located at $x = 16$ in the figure. 1. Write and graph a new function for the sum of the potter's distances to the four machines. 2. Where should the potter stand now to minimize the distance she has to walk while producing a mug? ###### 59 Richard and Marian are moving to Parkville to take jobs after they graduate. The main road through Parkville runs east and west, crossing a river in the center of town. Richard's job is located $10$ miles east of the river on the main road, and Marian's job is $6$ miles west of the river. There is a health club they both like located $2$ miles east of the river. If they plan to visit the health club every workday, where should Richard and Marian look for an apartment to minimize their total daily driving distance? ###### 60 Romina's Bakery has just signed contracts to provide baked goods for three new", "interpret what walking $-3 \\textrm{mph}$ north means. It literally means that every $3$ hours, you will be $3$ miles \"less north\" than you were before. This is equivalent to walking $3 mph$ south. So she ends up 6 miles south of where she started, aka $-6$ miles north of where she started. If $x=3$ and $y=-2$, then we have to interpret what walking $-2 \\textrm{h}$ means. It means that she has been walking at a constant rate, and we are interested in her position $2$ hours ago. In $2$ hours she walks $6$ miles north, so $2$ hours ago she was $6$ miles south of where she started, aka $-6$ miles north of where she started. If $x=-3$ and $y=-2$, then she is walking $3 \\textrm{mph}$ south, and we are interested in her position $2$ hours ago. She was $6$ miles north of where she is now $2$ hours ago. • This is a great analogy: Using N/S for one dimension, and future/past as another dimension. – Joseph O'Rourke Nov 7 '14 at 23:29 • @JosephO'Rourke I am pretty sure I stole this from Sybilla Beckman's Math for Elementary School teachers book. Not totally sure: it may have just been in the course materials one of the professors at my university wrote up for this course. – Steven", "# Modeling Problem Show 40 post(s) from this thread on one page Page 1 of 2 12 Last • May 2nd 2009, 11:42 AM sk8erboyla2004 Modeling Problem Ive encounter this problem which is very confusing for me Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and$20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ? I so if I draw a mental picture you a height of 30 feet on the y axis and 100 feet on the x axis I know I can use the distance formula somehow but i am still lost • May 2nd 2009, 02:15 PM Jester Quote: Originally Posted by sk8erboyla2004 Ive encounter this problem which is very confusing for me Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and$20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ? I so if I", "### Home > MC1 > Chapter 2 > Lesson 2.2.4 > Problem2-96 2-96. Three days each week Stacy exercises by walking around the soccer field near her home. The field is $80$ yards wide and $115$ yards long. 1. Draw a diagram of the field and find how far Stacy walks in one trip around the field (called the perimeter). Your diagram might look something like this. Remember that the field is a rectangle, so sides opposite of each other are the same length. 2. If Stacy walks around the field four times each time she exercises, how far does she walk each week? Using the perimeter you found in part (a), you can easily find the distance Stacy walks each day. But remember, this question asks for how far Stacy walks each week. Stacy walks $4680$ yards each week. 3. Additional Challenge: A mile is $1760$ yards. How many miles does Stacy walk each week? To convert these units, it is best to use division. Find a way to write an expression using division to solve the problem." ]

[ "200 is not a multiple of 7. There just isn't enough information.", "7 total.", "the smallest multiple of 63 and 147", "# Question 1 of 18 List all the multiples of 7 which are smaller than 60? A 7,14,18,28,35,42,49,56 B 7,14,21,26,35,42,49,56 C 7,14,21,28,35,42,49,56 D 7,14,21,28,38,42,49,56 E None of the above", "7; hence the difference of any pair is a multiple of 7. -", "of a multiple of 10 or 25.", "7?...", "i<7 and 7", "n that is a multiple of 20.", "Multiples of 6 or 7 but not both How many integers between 1 and 2015 are multiples of 6 OR 7 but not both? ×", "7) = 49 + 4 = 53", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "a multiple of multiple of multiple of 4 less than less than 30 what will be this nice selfie 03:00 - 03:59ok thank you", "35, 42, .... 98 } as {x : x is a multiple of 7, 10 < x < 100}.", "number is 24 % smaller than the number 51?", "• multiples", "0 • Math and Arithmetic # What are all the multiples of 7 between 50 and 100? ###### Wiki User 56,63,70,77,84,91,98 🙏 26 🤨 12 😮 8 😂 14 ###### Rena Boodram The multiples of 7 between 50 and 100 is 56, 63 and 70 🙏 4 🤨 12 😮 4 😂 4 25 🙏 1 🤨 3 😮 2 😂 4 25 🙏 0 🤨 2 😮 1 😂 1 75 🙏 1 🤨 3 😮 1 😂 2", "70, 50 but the order of both groups of keys (lesser and greater) remains the same individually. Therefore, 7C4 *1 = 35. by Active", "square units 42.) What is the least common multiple of 168 and 420? Continue reading", "there are an infinite amount of multiples of 75 that can be obtained by multiplying 75 by integers larger than 10." ]

[ "from this. Dividing 47 by 8 gives us 5 with a remainder of 7. Our mixed number is 5 and 7/8.", "7) = 49 + 4 = 53", "# Converting a repeating base $7$ expression to a fraction A question was given to me to convert a decimal in base seven to a fraction in base seven, where the base $7$ expression was $._7515151515\\ldots$. I understand this would be $\\frac 57 + \\frac 1{49} + \\frac 5{343} + \\cdots$, but I don't know how to simplify this into the answer, which is $\\frac 34$ in base seven. You want to compute $\\sum_{k=0}^\\infty \\theta {1 \\over 49^k}$, where $\\theta = {51_7 \\over 100_7 } = {36 \\over 49 }$. Since $1+ {1 \\over 49} + {1 \\over 49^2} + \\cdots = {49 \\over 48}$, the answer is ${49 \\over 48} {36 \\over 49 } = { 3 \\over 4}$.", "1. ## 7th grade math problem??? 3/7 of 56 = n I took 3/7 = 0.428571 and timed it by 56 and got 23.999998. rounded to 24, n= 24 is this right? 2. Originally Posted by mr_payne2003 3/7 of 56 = n I took 3/7 = 0.428571 and timed it by 56 and got 23.999998. rounded to 24, n= 24 is this right? $\\frac{3}{7}\\cdot 56=n$ to multiply them together rewrite 56 as a fraction over 1 $\\frac{3}{7}\\cdot \\frac{56}{1}=n$ $\\frac{3 \\cdot 56}{7 \\cdot 1}=n$ $\\overbrace{\\frac{3 \\cdot \\not {56}}{\\not 7 \\cdot 1}}^{8}=n$ $3 \\cdot 8=24=n$ Remember fractions are our friends", "from 45. You use the following formula, $\\boxed{ \\frac{45!}{7!(45-7)!} }$ Where the factorial, means, the product of all numbers before it. For example, $5!=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120$ Factorials, get large, quickly. --- Thus, if you wanted to know what would be for 8 out of 45 then it would be, ${{45}\\choose 8}=\\frac{45!}{8!(45-8)!}$ And now just use a super calculator (like the one on your computer). 5. Originally Posted by Luv Thank you very much. I was just wondering if you could tell me the steps how to do that problem because I'm still confused how you got that answer. I will very much appreciated it. Thank you in advanced. The probability that the first number chosen is one of yours is 7/45 (as you have 7 of the 45 numbers) If the first number was one of yours the prob that the second is one of yours is 6/45 (as you have 6 of the remaining 45 numbers) If the first two numbers were yours the prob that the third is one of yours is 5/44 (as you have 5 of the remaining 44 numbers) and so on untill: If the first six numbers were yours the prob that the seventh is one of yours is 1/39 (as you have 5 of the remaining 39 numbers). The", "# Question f950b Let’s convert 45% into a decimal form (which you can do by dividing it by 100) 45% -> 0.45# Now we can use this to determine 45% of $360 $360 \\cdot 0.45 = 162$45% of$360 is \\$162", "the $r_1r_2r_3$ term. That's the way to get this term into the required range $0,\\ldots,n-1$. The only question is: After doing it, can there be two different remainders left? (Remember that $6$ in your example is too big by one $n=4$, so we have to write $6 + 0\\cdot n= 2+1\\cdot n$.)", "+0 # Rewrite 7 100 × 2 10 as a single fraction and write the answer 0 32 1 Rewrite 7 100 × 2 10 as a single fraction and write the answer Jan 5, 2023 #1 +115 0 First of all, what? This doesnot make any sense. I am assuming you meant $$7/100\\cdot 2/10$$ that will be $$14/1000 = 7/500$$ Jan 5, 2023", "* (200/100) = $50! !Or!200 / 4 =$50. #### Question No 75 Deduct 7.3 from 34.2. Solution! 34.2 - 7.3 = 26.9. #### Question No 76 In a box ratio of DVD to VCD is 2/3. Which of the following cannot be the total number of CD's in the box? Solution! As 2+3 = 5, So total number of CD's must be multiple of 5.! !In the options all are multiple of 5 except 11! !So 11 is the correct answer.. #### Question No 77 What is 126 divided by 3? Solution! 126 / 3 = 42. #### Question No 78 What is 15 multiplied by 11? Solution! 15 * 11 = 165. #### Question No 79 A number is thrift if it is a multiple of 2 or 3. How many thrift numbers are there between 0 and 15? Solution! Thrift numbers between 0 and 15 are following! !2,3,4,6,8,9,10,12,14. #### Question No 80 64% of the children in school are girls.What percentage are boys? Solution! 100 - 64 = 36%. #### Question No 81 Which of the following is an integer that has a remainder of 1 when it is divided by 4,5 and 6? Solution! Least common multiple of 4,5 and 6 is 60 so 61 is an integer that has a remainder of 1 when", "- 49 = 299.$$ By complementary counting, the answer we want is $499 - 299 = \\boxed{200}.$ ## Solution 8 (Simplest) Suppose our fraction is $\\frac{a}{b}$. The given condition means $a+b=1000$. Now, if $a$ and $b$ share a common factor greater than $1$, then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even. There are $250$ fractions that aren’t even. From this, $50$ are divisible by $5$, which means the answer is $250-50=\\boxed{200}$ ~Geometry285", "How do you write 2/7 in percentage? Nov 22, 2016 Without a calculator, you would do long division: $2 \\div 7 = 0.2857 \\ldots$ $0.2857 \\cdot 100 = 28.57$ This is your answer as a percentage: 28.57% $2$ is approximately 28.57% of $7$", "Pensick227 4 # 50 is 40% of what number? $50-40\\%\\\\ x-100\\%\\\\ \\\\ 50 \\cdot 100=40x\\\\ 40x=5000\\\\ x=125\\\\ Answer:Searched\\ number\\ is\\ 125.$ $40\\%x=50\\\\0,4x=50/:0,4\\\\x=125$ Answer: 50 is 40% of 125. I hope I helped :D", "sums must be equal. Therefore $$a_1+\\cdots +a_{19}+b_1+\\cdots +b_{85}=20\\cdot 85=\\boxed{1700}.$$ Note that this shows that the value of the desired sum is constant.", "in the left ring, so the desired answer is $(61+61+61) \\cdot 4 = \\boxed{732}$.", "# What number is 5% of 46? Dec 6, 2016 2.3 #### Explanation: $46 \\cdot \\frac{5}{100} = 2.3$ Answer 2.3 It means suppose if you are having 5$out of 100$ , it is 5 %. And in same way if you are having 5$out of 46$ so what this value will be for 100\\$ in percent.", "52+8 = 60. This is our target. Now we plug I=400 into the answers to see which yields our target of 60. 50 + I/10 = 50 + 400/40 = 60.", "= The number that store R sold less than store T = 25 You want to divide the difference by the original. 25/45 = .555 or 56% -", "know that $9 \\cdot 5 = 45$ and adding or subtracting 8 to 45 would not give you 49, you know that you should multiply 5 by 8 instead... [Continuation of Solution] 9_5_8 = 49 9_5*8 = 49 9_40 = 49 9+40 = 49 Luckily in one try we were able to reach the answer, but most of the time you would need to use trial and error to arrive at the right combination of operations. *The explanation may be confusing but feel free to comment or post questions in the comment section if you think that the explanation needs more elaboration then I'll try to reply asap :D", "where you subtract that and get zero like this, 60 $$5\\sqrt{300}$$ -30 - 00 -0 - 0 so the number is, and the number has no more numbers left, so 60 will be our answer. Sorry if it was a bit complicated. but yeah. ! : D how do u get 7.3 to get 7.3 how do u get 7.3 but lets say i just gave you a simple question like, 300 divided by 5 that would be. 60. now for this question i can show you by editing my question RELATED:", "50. 50 divided by 7 is 5, with a remainder of 1. Since the remainder is zero, your long division is done. Quotient = 6.57 and Remainder = 01 So, 46 ÷ 7 = 6.57 with a remainder of 1" ]

[ "7) = 49 + 4 = 53", "# Question 1 of 18 List all the multiples of 7 which are smaller than 60? A 7,14,18,28,35,42,49,56 B 7,14,21,26,35,42,49,56 C 7,14,21,28,35,42,49,56 D 7,14,21,28,38,42,49,56 E None of the above", "+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", "that and you get a multiple of $$7$$.", "# Find the sum of all the integers between 1 and 1000 which are divisible by 7 How can I work this one out (with workings)? \"Find the sum of all the integers between 1 and 1000 which are divisible by 7\" Thanks! $7 \\lfloor \\dfrac{1000}{7} \\rfloor=994$ $7+14+...+994=7(1+2+...\\dfrac{994}{7})=7(1+2+...142)=7 \\dfrac{(142)(143)}{2}=71071$ Hint: $7\\cdot(1+2+\\ldots+\\lfloor \\frac{1000}{7}\\rfloor)=?$. Hint: What is the largest multiple of $7$ less than $1000?$ Do you know how to sum $1+2+3+\\dots$ up to $n?$ Distribute out the $7$", "from this. Dividing 47 by 8 gives us 5 with a remainder of 7. Our mixed number is 5 and 7/8.", "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: . $T(8) \\:=\\:\\frac{8\\cdot9\\cdot10}{6} \\:=\\:120$", "greater than 51 you have six more than every multiple of seven greater than 62 you have three more than every multiple of seven greater than 73 (you could have seen that 0*11+7, 1*11+7, 2*11+7... would have the last different remainder when divided 7 at 6*11+7, or after the seventh term, one for every possible remainder 0->6) then, you look for the greatest number that is three more than a multiple 7 less than 73. I do not know any linear algebra, but this sounds more like number theory to me. I believe that the number that you are looking for is called the Frobenius Number for a given linear system. Last edited: Jan 15, 2009 5. Jan 16, 2009 ### HallsofIvy Staff Emeritus Well, didn't it occur to you that if the difference $22A/77- 21A/77\\ge 1$ then there MUST be an integer between them? Multiplying by 77, $22A- 21A= A\\ge 77$. That is, any A larger than or equal to 77 can be written as a combination of 7 and 11. The largest A that cannot must be less than 77. Now you can \"check\" but you certainly don't need to \"guess\" A 21A/77 22A/7 76 20.7 21.7 so there is an integer between them. 75 20.45 21.4 so there is an integer between them. etc. At", "is the largest number so it must be the the sum of the largest and middle numbers. 48 must be the sum of the largest and smallest numbers. We are interested in the largest number. Adding together the sums containing the largest number, the largest number + the middle number + the largest number + the smallest number = 51 + 48 = 99. But that is the largest number twice, plus the sum of the smallest and middle numbers. But we know that the sum of the smallest and middle numbers is 39! So the largest number twice is 99 $-$ 39 = 60. So the largest number is 30. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "+0 # Pls help 0 93 1 Find the largest prime number that divides the quantity $$0! + (1!) \\times 1 + (2!) \\times 2 + (3!) \\times 3 + \\cdots + (50!) \\times 50$$ Oct 6, 2018 #1 +4000 +1 $$\\text{apparently }\\sum \\limits_{k=1}^n ~k! \\cdot k = (n+1)! - 1 \\\\ 0!=1 \\\\ \\text{so the expression shown }=(n+1)!-1+1 = (n+1)! \\\\ \\text{so the largest prime }<(n+1) \\text{ is what we're after}\\\\ \\text{and this is }47.$$ . Oct 6, 2018", "the last term, and see which one is the largest (since the sum of the 2 entires is twice the average of whole sequence). 2+100=102, 3+99=102, 4+100=104, 5+100=105, 6+96=102. Therefore, the answer is multiples of five because it has the largest number.", "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 $72$.", "# Largest multiple of $7$ lower than some $78$-digit number? What I am trying to achieve, is related to cryptography/blockchain/bitcoin . So, the largest number here is huge, in other words: I want to find the largest multiple of 7, which is lower than this number: $$115792089237316195423570985008687907852837564279074904382605163141518161494336$$ I can just go to Wolfram Alpha, and type \"multiples of 7\", and I get a list of the multiples relatively fast. But, it will take some time until I keep hitting \"more\", to get to a number lower than this above. • m.wolframalpha.com/input/… works a lot easier it lists as 2 mod 7. – Roddy MacPhee Mar 18 at 11:47 • 78 digits is semi small cryptographically compared to 617 digit (2048 bit) crypto keys. – Roddy MacPhee Mar 18 at 11:58 • I would very much like (aka 'prefer') to see a solution which does not require use of extended-precision or large-integer software packages. (Compare with, e.g., the simple \"sum the digits\" approach for multiples of 3). Is Roddy's answer the only such? – Carl Witthoft Mar 18 at 12:07 • no answer requires it. it's simply more convenient for numbers of this size ( I speak from experience, thought I messed up because it didn't match the other answer, turns out I was doing the mod 7 steps", "Find the largest prime number that divides $27! + 28!$. • $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$.", "Count Again The total number of multiples of 2 that are less than or equal to 100 is $$100\\div 2 = 50$$. The total number of multiples of 5 that are less than or equal to 100 is $$100\\div 5 = 20$$. Is it true that the total number of multiples of 2 and/or multiples of 5 that are less than or equal to 100 is $$50 + 20 = 70$$? × Problem Loading... Note Loading... Set Loading...", "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", "gives us the largest three digit multiple of 125, $$\\boxed{875}$$. • Why can't we have $N=376$ (even and $125$ divides $N-1$) as a solution to $N(N-1)(N+1)\\equiv0\\pmod{1000}$? – J. W. Tanner Mar 13 '19 at 14:22", "# Concatenation Property? $14, \\ \\ \\ 21, \\ \\ \\ 28, \\ \\ \\ 35, \\ \\ \\ 42, \\ \\ \\ 49, \\ \\ \\ 56$ The above shows a sequence of multiples of 7. Is it true that if I combine these numbers together by their numerals in that order, then it is also divisible by 7? In other words, is the large number $14 21 28 35 42 49 56$ divisible by 7? ×", "+0 # Pls help 0 137 1 Find the largest prime number that divides the quantity $$0! + (1!) \\times 1 + (2!) \\times 2 + (3!) \\times 3 + \\cdots + (50!) \\times 50$$ Oct 6, 2018 #1 +4781 +1 $$\\text{apparently }\\sum \\limits_{k=1}^n ~k! \\cdot k = (n+1)! - 1 \\\\ 0!=1 \\\\ \\text{so the expression shown }=(n+1)!-1+1 = (n+1)! \\\\ \\text{so the largest prime }<(n+1) \\text{ is what we're after}\\\\ \\text{and this is }47.$$ . Oct 6, 2018", "# Finding Integers using Counting Rules I'm studying, and got this problem in the book, How many positive integers between 50 and 100 are divisible by 7, and what are they? The same for number divisible by both 7 and 11. I'm reviewing Counting Basics, and this looks like counting subsets on a finite set, but I'm not sure how to solve this (I've seen combinations, like strings, but not this). Can someone help me get started? If you'd like, the answers are: 7 for the first one (56, 63, 70, 77, 84, 91, 98), and 1 for the second (77). But again, I'm interested in the process itself. - The smallest multiple of $7$ that is larger than or equal to $50$ is $56=7\\times 8$. On the other hand, the largest multiple of $7$ that is less than or equal to $100$ is $98 = 7\\times 14$ (which can be found by computing $50/7$ and rounding up, and $100/7$ and rounding down). So... every number in this class must be of the form $7\\times k$, and we must have $7\\times 8\\leq 7\\times k\\leq 7\\times 14$. So... what are the possible values of $k$, and how many are there? As for being divisible by $7$ and $11$, since $7$ and $11$ are distinct primes, if a number is" ]

[ "# You've gotta be kidding me $1007021035035021007001$ How many positive divisors (including 1 and itself) does the large number above have? ×", "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.", "• 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.", "# 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? •", "# I'm not going to list all 24 numbers! Given that $$1400 = 2^3 \\times 5^2 \\times 7,$$ find the sum of its positive divisors. ×", "get tricky if you had more than three prime divisors, but $360$ is not too bad:", "of 24 is a (!", "# Counting Positive Factors 12 has 6 positive divisors: 1, 2, 3, 4, 6, 12. Which of the two numbers has more divisors, $12\\times 6$ or $12\\times 9?$ ×", "indicating $T_{224}=25200$ has 90 divisors. Apparently.", "Numbers How many positive perfect squares less than $$10^6$$ are multiples of 24? Source: 2007 AIME I. × Problem Loading... Note Loading... Set Loading...", "48 is 24. =24", "of positive numbers.", "# Odd Numbers 3 How many odd divisors does the number $$5^{6}\\times7^{8}\\times16^{13913913913912391293}$$ have? ×", "# Eleven primes Find the sum of all positive primes $$p$$ there such that $$p^2+11$$ has exactly six positive divisors. ×", "the divisors of $n$.", "positive number.", "# You've gotta be kidding me Number Theory Level 3 $1007021035035021007001$ How many positive divisors (including 1 and itself) does the large number above have? ×", "# Divisors with units digit 5 How many positive divisors of $$20130^4$$ have 5 as their units digit? ×", "× # Number of Divisors Given real number $$n$$. If $$n$$ has $$20$$ divisors, and $$16n$$ has $$36$$ divisors, and $$56n$$ has $$48$$ divisors, then how many divisors does $$7n^2$$ have? Details : The divisors are all positive integers. Note by Fidel Simanjuntak 1 month, 3 weeks ago Sort by: Your question is similar to mine. If you are unable to solve your question, just see any solutions posted in my question. · 1 month, 3 weeks ago", "number, in this case $225\\,$. The common, odd positive divisors of the two original numbers will be precisely the odd, positive divisors of $225=2^2 \\cdot 3^2\\,$, and there is $(2+1)\\cdot(2+1)=9$ of them, as pointed out already. The advantage of this approach is that you only have to factor the (small) number $225\\,$, rather than the two (big) numbers originally given." ]

[ "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.", "# Counting Positive Factors 12 has 6 positive divisors: 1, 2, 3, 4, 6, 12. Which of the two numbers has more divisors, $12\\times 6$ or $12\\times 9?$ ×", "# I'm not going to list all 24 numbers! Given that $$1400 = 2^3 \\times 5^2 \\times 7,$$ find the sum of its positive divisors. ×", "# Divisors ending with digits 0-9 each What is the smallest positive integer, which has - for each of the digit 0-9 - a divisor ending with this digit? • LCM(1, 2, ..., 9, 10) = 2520 . Apr 10 at 22:31 It's $$270$$, which has divisors $$1, 2, 3, 54, 5, 6, 27, 18, 9, 10$$. Note that the number must be a multiple of $$10$$ and then just try every multiple of $$10$$ until it hits. • Because 10 is coprime with with any number ending in 9 then you only need to check multiples of 90,190,290,... which significantly reduces the search space. Apr 11 at 8:29", "# 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? •", "• 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.", "# The only positive divisor of both $a$ and $a + 1$ is $1$ Prove that if $a \\in \\mathbb Z$ then the only positive divisor of both $a$ and $a + 1$ is $1$. When I saw this statement I didn't understand it. The only way that I can see it being true is if a is a negative number but since a ∈ Z a could also be positive and thus it could have more than one positive divisor. If that's the case would I have to disprove this instead of proving? And what would be the best way to do that? • What common divisors do 2 and 3 have? Or 3 and 4 or 9 and 10? – Paul Sundheim Sep 22 '14 at 0:18 • I think it's asking for a single number that divides both $a$ and $a+1$. – Trurl Sep 22 '14 at 0:19 • Please choose descriptive titles. The original title, Prove or Disprove??? carried virtually no information of what the problem is about. – user147263 Sep 22 '14 at 1:11 You seem to have misunderstood the question. As far as I can tell you are thinking about divisors of $a$ and $a+1$ separately, but the question is asking you to show that the only positive number which is a", "+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", "that the number has less than $6$ divisors. Since $12$ has $6$ divisors and $10$ has $4$ divisors, James’s desired sum is $\\boxed{10}$.", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "# How many divisors? For a positive integer $x$, let $f(x)$ be the function which returns the number of distinct positive factors of $x$. If $p$ is a prime number, what is the minimum possible value of $f(75p^2)?$ This problem is posed by Michael T. Details and assumptions You may read up on Divisors of an Integer. The distinct positive factors of $x$ need not be prime. For example, $f(12) = 6$, since 12 has distinct positive factors of $1, 2, 3, 4, 6, 12$. ×", "number, in this case $225\\,$. The common, odd positive divisors of the two original numbers will be precisely the odd, positive divisors of $225=2^2 \\cdot 3^2\\,$, and there is $(2+1)\\cdot(2+1)=9$ of them, as pointed out already. The advantage of this approach is that you only have to factor the (small) number $225\\,$, rather than the two (big) numbers originally given.", "# You've gotta be kidding me $1007021035035021007001$ How many positive divisors (including 1 and itself) does the large number above have? ×", "60 views The ratio of two positive numbers is $2:3.$ If $10$ and the sum of the numbers are added to their product, square of $16$ is obtained. The greater number is : 1. $6$ 2. $9$ 3. $12$ 4. $18$ Assume that numbers are $2x,3x$. Now according to the question: $(2x \\times 3x) + 3x+2x+10=256$ $\\implies6x^2+5x-246=0\\implies x=6$ So the number are: $6\\times 2=12,6\\times 3=18$ Option (D) is correct. 4.7k points 1", "# What are the factors of 29? $29$ has positive factors $1$ and $29$ $29$ is only divisible by the positive numbers $1$ and $29$, so it is a prime number.", "# Eleven primes Find the sum of all positive primes $$p$$ there such that $$p^2+11$$ has exactly six positive divisors. ×", "* 2 * 7 ...three\\:\\:factors \\\\[3ex] 29 = 29 \\\\[3ex]$ Two numbers, $27$ and $28$ have exactly three factors. (26.) ACT What is the smallest positive integer having exactly $5$ different positive integer divisors? $A.\\:\\: 5 \\\\[3ex] B.\\:\\: 6 \\\\[3ex] C.\\:\\: 12 \\\\[3ex] D.\\:\\: 16 \\\\[3ex] E.\\:\\: 18 \\\\[3ex]$ Positive integer divisors of $5 = 1, 5$ Number of positive integer divisors = $2$ Positive integer divisors of $6 = 1, 2, 3, 6$ Number of positive integer divisors = $4$ Positive integer divisors of $12 = 1, 2, 3, 4, 6, 12$ Number of positive integer divisors = $6$ Positive integer divisors of $16 = 1, 2, 4, 8, 16$ Number of positive integer divisors = $5$ Positive integer divisors of $18 = 1, 2, 3, 6, 9, 18$ Number of positive integer divisors = $6$ The correct option is $D.$ (27.) ACT Which of the following arranges the numbers $\\dfrac{9}{5}, 1.\\overline{8}, 1.08$, and $1.\\overline{08}$ into ascending order? (Note: The overbar notation shows that the digits under the bar will repeat. For example, $1.\\overline{73} = 1.737373...$) $F.\\:\\: \\dfrac{9}{5} \\lt 1.\\overline{08} \\lt 1.08 \\lt 1.\\overline{8} \\\\[5ex] G.\\:\\: \\dfrac{9}{5} \\lt 1.08 \\lt 1.\\overline{08} \\lt 1.\\overline{8} \\\\[5ex] H.\\:\\: 1.\\overline{08} \\lt 1.08 \\lt \\dfrac{9}{5} \\lt 1.\\overline{8} \\\\[5ex] J.\\:\\: 1.08 \\lt 1.\\overline{08} \\lt 1.\\overline{8} \\lt \\dfrac{9}{5} \\\\[5ex] K.\\:\\: 1.08 \\lt 1.\\overline{08} \\lt", "Find the number of positive divisors of $10!$. Notation: $!$ denotes the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "The Number of Positive Divisors of an Integer Examples 1 # The Number of Positive Divisors of an Integer Examples 1 Recall from The Number of Positive Divisors of an Integer page that if $n \\in \\mathbb{Z}$ then the number of positive divisors of $n$ is denoted $d(n)$ and is given explicitly by: (1) \\begin{align} \\quad d(n) = \\sum_{d \\mid n}_{d > 0} 1 \\end{align} Most importantly, we noted that: • If $p$ is prime then $d(p) = 2$. • If $p$ is prime and $k \\in \\mathbb{N}$ then $d(p^k) = k + 1$. • If $p$ and $q$ are prime then $d(pq) = d(p)d(q)$. We will now look at some examples of computing regarding the number of positive divisors of an integer. More examples can be found on The Number of Positive Divisors of an Integer Examples 2 page. ## Example 1 Calculate $d(12322)$. We first note that the prime power decomposition of $12322 = 2 \\cdot 61 \\cdot 101$. Hence: (2) \\begin{align} d(12322) & = d(2) d(61) d(101) \\\\ d(12322) & = (2)(2)(2) \\\\ d(12322) & = 8 \\end{align} ## Example 2 Calculate $d(88888)$. The prime power decomposition of $88888 = 2^3 \\cdot 41 \\cdot 271$. Hence: (3) \\begin{align} d(88888) &= d(2^3) d(41) d(271) \\\\ d(88888) &= (4)(2)(2) \\\\ d(88888) &= 16 \\end{align} ## Example 3", "# Problem of the Week Problem C and Solution Divisors and Number ## Problem Your friend Cael always likes challenging you. One challenge is called “Divisors and Number”. Cael will tell you certain facts about the divisors of a number and then challenge you to find the number. Here is Cael’s challenge. “I am looking for a positive integer with exactly eight positive divisors, two of which are $$21$$ and $$33$$.” Determine Cael’s number. ## Solution Let $$n$$ represent the number we are looking for. We know that four of the positive divisors of $$n$$ are $$1$$, $$21$$, $$33$$, and $$n$$. In our solution we will first find the remaining four positive divisors and then determine $$n$$. Since $$21$$ is a divisor of $$n$$ and $$21 = 3\\times 7$$, then $$3$$ and $$7$$ must also be divisors of $$n$$. Since $$33$$ is a divisor of $$n$$ and $$33 = 3\\times 11$$, then $$11$$ must also be a divisor of $$n$$. Since $$7$$ is a divisor of $$n$$ and $$11$$ is a divisor of $$n$$, and since $$7$$ and $$11$$ have no common divisors, then $$7\\times 11 = 77$$ must also be a divisor of $$n$$. We have found all eight of the positive divisors of the unknown number. The positive divisors are $$1$$, $$3$$, $$7$$, $$11$$, $$21$$, $$33$$," ]

[ "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.", "# Counting Positive Factors 12 has 6 positive divisors: 1, 2, 3, 4, 6, 12. Which of the two numbers has more divisors, $12\\times 6$ or $12\\times 9?$ ×", "# I'm not going to list all 24 numbers! Given that $$1400 = 2^3 \\times 5^2 \\times 7,$$ find the sum of its positive divisors. ×", "• 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.", "# You've gotta be kidding me $1007021035035021007001$ How many positive divisors (including 1 and itself) does the large number above have? ×", "# 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? •", "# Divisors ending with digits 0-9 each What is the smallest positive integer, which has - for each of the digit 0-9 - a divisor ending with this digit? • LCM(1, 2, ..., 9, 10) = 2520 . Apr 10 at 22:31 It's $$270$$, which has divisors $$1, 2, 3, 54, 5, 6, 27, 18, 9, 10$$. Note that the number must be a multiple of $$10$$ and then just try every multiple of $$10$$ until it hits. • Because 10 is coprime with with any number ending in 9 then you only need to check multiples of 90,190,290,... which significantly reduces the search space. Apr 11 at 8:29", "# How many divisors? For a positive integer $x$, let $f(x)$ be the function which returns the number of distinct positive factors of $x$. If $p$ is a prime number, what is the minimum possible value of $f(75p^2)?$ This problem is posed by Michael T. Details and assumptions You may read up on Divisors of an Integer. The distinct positive factors of $x$ need not be prime. For example, $f(12) = 6$, since 12 has distinct positive factors of $1, 2, 3, 4, 6, 12$. ×", "# Eleven primes Find the sum of all positive primes $$p$$ there such that $$p^2+11$$ has exactly six positive divisors. ×", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "number, in this case $225\\,$. The common, odd positive divisors of the two original numbers will be precisely the odd, positive divisors of $225=2^2 \\cdot 3^2\\,$, and there is $(2+1)\\cdot(2+1)=9$ of them, as pointed out already. The advantage of this approach is that you only have to factor the (small) number $225\\,$, rather than the two (big) numbers originally given.", "that the number has less than $6$ divisors. Since $12$ has $6$ divisors and $10$ has $4$ divisors, James’s desired sum is $\\boxed{10}$.", "× # Number of Divisors Given real number $$n$$. If $$n$$ has $$20$$ divisors, and $$16n$$ has $$36$$ divisors, and $$56n$$ has $$48$$ divisors, then how many divisors does $$7n^2$$ have? Details : The divisors are all positive integers. Note by Fidel Simanjuntak 1 month, 3 weeks ago Sort by: Your question is similar to mine. If you are unable to solve your question, just see any solutions posted in my question. · 1 month, 3 weeks ago", "× Back to all chapters # Prime Factorization and Divisors Learn how to break down numbers big and small, as proposed by the Fundamental Theorem of Arithmetic. # Number of Factors / Divisors How many positive divisors does the number $$2^63^8$$ have? How many factors does $$2500000$$ have? Find the sum of all the positive divisors of 28. How many positive integers less than or equal to $$500$$ have exactly $$3$$ positive divisors? How many factors does the integer $$4^8\\times6^7\\times12^7$$ have? ×", "# Multifactorial and divisors Fill in the blank. $$N$$ has $$3$$ prime factors. $$N^2$$ has $$7!!$$ positive divisors. $$N^3$$ has $$10!!!$$ positive divisors. $$\\text{____}$$ has $$13!!!!$$ positive divisors. Note: You might want to read up the definition of multifactorials first. ×", "# What are the factors of 29? $29$ has positive factors $1$ and $29$ $29$ is only divisible by the positive numbers $1$ and $29$, so it is a prime number.", "# They can't be equal, right? $1, 2, 7, 14$ The above shows the list of all the positive divisors of 14. The sum of all of these divisors is $1+2+7+14= 24$. Likewise, what is the sum of all the positive divisors of 15? ×", "Find the number of positive divisors of $10!$. Notation: $!$ denotes the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "the ones indicated above, $$1,2,3,4,6,8,12,16,24,48$$: $D_{48} = \\{1, 2, 3, 4, 6, 8, 12, 16, 24, 48\\}.$ This picture indicates all divisors of $$48$$, where an arrow from a number to $$48$$ means that the number is a divisor of $$48$$. In particular, $$48$$ divides $$48$$, so there is an arrow from $$48$$ to itself. Let us see one more example.", "\\end{array}$$ $$\\text{How many of the positive divisors of 3240 are multiples of 3 ?}$$ $$\\mathbf{40-8 = 32}$$ $$\\text{There are \\mathbf{32} positive divisors of 3240, which are multiples of 3.}$$ Nov 13, 2018" ]

[ "+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", "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", "label(\"$r\\sqrt{ 2 }$\",(2.75,-1.24),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(2.02,-1.85),SE*lsf); label(\"$45^\\circ$\",(2.41,-1.3),SE*lsf); label(\"$22.5^\\circ$\",(2.02,-0.96),SE*lsf); draw(circle((2.07,-1.43),0.01),qqwuqq); label(\"$90^\\circ$\",(2.15,-1.3),SE*lsf); label(\"$67.5^\\circ$\",(1.79,-1.28),SE*lsf); label(\"$22.5^\\circ$\",(0.84,0.1),SE*lsf); label(\"r\",(0.71,-0.06),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(0.27,0.35),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.05,-0.12),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.09,0.63),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.44,0.31),SE*lsf); draw(circle((0.45,0.07),0.01),qqwuqq); label(\"$90^\\circ$\",(0.6,0.1),SE*lsf); draw(circle((2.16,0.07),0.01),qqwuqq); label(\"$90^\\circ$\",(2.31,0.17),SE*lsf); label(\"$22.5^\\circ$\",(1.64,0.11),SE*lsf); label(\"r\",(2.04,-0.09),SE*lsf); draw((1.4,-0.2)--(1.11,0.17),EndArrow(6));label(\"$\\sqrt{ (r (\\sqrt{ 2 } - 1) + r + r )^2 + (r (\\sqrt{ 2 } - 1) )^2 }$\",(1.58,-0.2),SE*lsf); label(\"$= \\sqrt{ (r (\\sqrt{ 2 } + 1) )^2 + (r (\\sqrt{ 2 } - 1) )^2 }$\",(1.53,-0.37),SE*lsf); label(\"$\\mathbf{=r \\sqrt{ 6 }}$\",(1.53,-0.72),SE*lsf,blue); label(\"$= \\sqrt{ r^2 (3 + 2 \\sqrt{ 2 }) + r^2 (3 - 2 \\sqrt{ 2 }) }$\",(1.53,-0.54),SE*lsf); label(\"$\\mathbf{2r}$\",(1.46,0.64),SE*lsf,blue); label(\"$\\mathbf{\\frac{BE}{AE} = \\frac{r \\sqrt{ 6 }}{2r} = \\frac{ \\sqrt{ 6 }}{2}}$\",(0.78,-1.47),SE*lsf,red+fontsize(14)); draw((1.74,-1.13)--(0.83,-0.06),EndArrow(6));draw(circle((1.34,0.09),0.01),qqffff); label(\"$45^\\circ$\",(-0.01,0.13),SE*lsf); label(\"$90^\\circ$\",(-0.22,1.47),SE*lsf); label(\"$90^\\circ$\",(1.39,0.24),SE*lsf); draw((0.38,0)--(0.38,0.38),linewidth(0.4)+dotted); dot((0,0),blue); label(\"$B$\",(-0.06,-0.09),NE*lsf,blue); dot((1.31,0),blue); label(\"$D$\",(1.34,-0.09),NE*lsf,blue); dot((0.92,0.92),blue); label(\"$C$\",(0.91,0.99),NE*lsf,blue); dot((2.23,0.38),blue); label(\"$E$\",(2.27,0.39),NE*lsf,blue); dot((0.38,0.38),blue); label(\"$A$\",(0.33,0.41),NE*lsf,blue); dot((0.38,0),linewidth(1pt)+uququq); label(\"$G$\",(0.41,-0.1),NE*lsf,uququq); dot((0,0.38),linewidth(1pt)+uququq); dot((2.23,0),linewidth(1pt)+uququq); label(\"$F$\",(2.24,-0.1),NE*lsf,uququq); dot((0.38,0.38),uququq); dot((1.7,-1.5),blue); dot((2,-0.8),blue); dot((2.69,-1.5),blue); dot((2,-1.5),blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); (Error compiling LaTeX. return linetype((real[]) split(pattern),offset,scale,adjust); ^ /usr/local/share/asymptote/plain_pens.asy: 13.3: Trying to use uninitialized value.) Courtesy of User:Kouichi Nakagawa. <geogebra>9fa91fc57b9e5def39d73828270c8c027c565d85</geogebra>", "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]", "# 1991 AJHSME Problems/Problem 22 ## Problem Each spinner is divided into $3$ equal parts. The results obtained from spinning the two spinners are multiplied. What is the probability that this product is an even number? $[asy] draw(circle((0,0),2)); draw(circle((5,0),2)); draw((0,0)--(sqrt(3),1)); draw((0,0)--(-sqrt(3),1)); draw((0,0)--(0,-2)); draw((5,0)--(5+sqrt(3),1)); draw((5,0)--(5-sqrt(3),1)); draw((5,0)--(5,-2)); fill((0,5/3)--(2/3,7/3)--(1/3,7/3)--(1/3,3)--(-1/3,3)--(-1/3,7/3)--(-2/3,7/3)--cycle,black); fill((5,5/3)--(17/3,7/3)--(16/3,7/3)--(16/3,3)--(14/3,3)--(14/3,7/3)--(13/3,7/3)--cycle,black); label(\"1\",(0,1/2),N); label(\"2\",(sqrt(3)/4,-1/4),ESE); label(\"3\",(-sqrt(3)/4,-1/4),WSW); label(\"4\",(5,1/2),N); label(\"5\",(5+sqrt(3)/4,-1/4),ESE); label(\"6\",(5-sqrt(3)/4,-1/4),WSW); [/asy]$ $\\text{(A)}\\ \\frac{1}{3} \\qquad \\text{(B)}\\ \\frac{1}{2} \\qquad \\text{(C)}\\ \\frac{2}{3} \\qquad \\text{(D)}\\ \\frac{7}{9} \\qquad \\text{(E)}\\ 1$ ## Solution Instead of computing this probability directly, we can find the probability that the product is odd, and subtract that from $1$. The product of two integers is odd if and only if each of the two integers is odd. The probability the first spinner yields an odd number is $2/3$ and the probability the second spinner yields an odd number is $1/3$, so the probability both yield an odd number is $(2/3)(1/3)=2/9$. The desired probability is thus $1-2/9=7/9\\rightarrow \\boxed{\\text{D}}$.", "- 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", "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", "$m=1/3$, 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: $\\boxed{-\\frac{1}{6}}$.", "of getting four gumballs of the same color is $\\text{(A)}\\ 8 \\qquad \\text{(B)}\\ 9 \\qquad \\text{(C)}\\ 10 \\qquad \\text{(D)}\\ 12 \\qquad \\text{(E)}\\ 18$ ## Problem 22 The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is $[asy] draw(circle((0,0),3)); draw(circle((7,0),3)); draw((0,0)--(3,0)); draw((0,-3)--(0,3)); draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2)); draw((7,0)--(7-3*sqrt(3)/2,-3/2)); draw((0,5)--(0,3.5)--(-0.5,4)); draw((0,3.5)--(0.5,4)); draw((7,5)--(7,3.5)--(6.5,4)); draw((7,3.5)--(7.5,4)); label(\"3\",(-0.75,0),W); label(\"1\",(0.75,0.75),NE); label(\"2\",(0.75,-0.75),SE); label(\"6\",(6,0.5),NNW); label(\"5\",(7,-1),S); label(\"4\",(8,0.5),NNE); [/asy]$ $\\text{(A)}\\ \\dfrac{1}{6} \\qquad \\text{(B)}\\ \\dfrac{1}{4} \\qquad \\text{(C)}\\ \\dfrac{1}{3} \\qquad \\text{(D)}\\ \\dfrac{5}{12} \\qquad \\text{(E)}\\ \\dfrac{4}{9}$ ## Problem 23 If $X$, $Y$ and $Z$ are different digits, then the largest possible $3-$digit sum for $\\begin{tabular}{ccc} X & X & X \\\\ & Y & X \\\\ + & & X \\\\ \\hline \\end{tabular}$ has the form $\\text{(A)}\\ XXY \\qquad \\text{(B)}\\ XYZ \\qquad \\text{(C)}\\ YYX \\qquad \\text{(D)}\\ YYZ \\qquad \\text{(E)}\\ ZZY$ ## Problem 24 A $2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares. $\\text{(A)}\\ 4 \\qquad \\text{(B)}\\ 6 \\qquad \\text{(C)}\\ 7 \\qquad \\text{(D)}\\ 8 \\qquad \\text{(E)}\\ 16$ ##", "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", "be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label(\"\\boldmath{1}\",(0,0)); label(\"\\boldmath{12}\",(1,0)); label(\"\\boldmath{123}\",(2,0)); label(\"\\boldmath{S}\",(3,0)); label(\"11\",(0.5,-0.75)); label(\"111\",(1.5,-0.75)); label(\"d_1\",(2.5,-0.75)); label(\"100\",(1,-1.5)); label(\"d_2\",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label(\"\\textbf{First Differences}\",(-0.75,-0.75),align=W); label(\"\\textbf{Second Differences}\",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334}. \\end{align*} ~MRENTHUSIASM ## Solution 4 (Finite Differences by Algebra) Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$. Alternatively, applying finite differences, one obtains $$c_4 = {3 \\choose 2}f(2) - {3 \\choose 1}f(1) + {3 \\choose 0}f(0) =334.$$ ## Solution 5 (Assumption) The idea", "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", "draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(0,10),C=(10,0),P=(0,0),B=(13.660,13.660); dot(A); label(\"A\",A,W); dot(C); label(\"C\",C,S); dot(P); label(\"P_0\",P,SW); dot(B); label(\"B\",B,NE); draw(A--C--P--A); draw(A--B--C); draw(B--P,dotted); [/asy]$ By repeatedly applying the Midpoint Formula, we can determine the coordinates of $P_1$, $P_2$, $P_3$, and so on. We can also use the Distance Formula to calculate the distance of $P_0P_1$, $P_0P_2$, and so on. The values are shown in the below table. $n$ Coordinates of $P_n$ $P_0 P_n$ 1 $(0,2a)$ $2a$ 2 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ 3 $(a-a\\sqrt{3},a-a\\sqrt{3})$ $a\\sqrt{6} - a\\sqrt{2}$ 4 $(-a+a\\sqrt{3},a+a\\sqrt{3})$ $2a\\sqrt{2}$ 5 $(2a,0)$ $2a$ 6 $(0,0)$ $0$ 7 $(0,2a)$ $2a$ 8 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ Note that the coordinates of $P_n$ as well as the distance $P_0 P_n$ cycle after $n = 6$. Thus, $P_0P_n = \\frac{\\sqrt{6} - \\sqrt{2}}{2} \\cdot P_0P_1$ if $n \\equiv 3 \\pmod{6}$, $P_0P_n = 0$ if $n \\equiv 0 \\pmod{6}$, $P_0P_n = P_0P_1$ if $n \\equiv 1, 5 \\pmod{6}$, and $P_0P_n = \\sqrt{2} \\cdot P_0P_1$ if $n \\equiv 2, 4 \\pmod{6}$.", "+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", "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "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", "1996 AJHSME Problems/Problem 5 Problem 5 The letters $P$, $Q$, $R$, $S$, and $T$ represent numbers located on the number line as shown. $[asy] unitsize(36); draw((-4,0)--(4,0)); draw((-3.9,0.1)--(-4,0)--(-3.9,-0.1)); draw((3.9,0.1)--(4,0)--(3.9,-0.1)); for (int i = -3; i <= 3; ++i) { draw((i,-0.1)--(i,0)); } label(\"-3\",(-3,-0.1),S); label(\"-2\",(-2,-0.1),S); label(\"-1\",(-1,-0.1),S); label(\"0\",(0,-0.1),S); label(\"1\",(1,-0.1),S); label(\"2\",(2,-0.1),S); label(\"3\",(3,-0.1),S); draw((-3.7,0.1)--(-3.6,0)--(-3.5,0.1)); draw((-3.6,0)--(-3.6,0.25)); label(\"P\",(-3.6,0.25),N); draw((-1.3,0.1)--(-1.2,0)--(-1.1,0.1)); draw((-1.2,0)--(-1.2,0.25)); label(\"Q\",(-1.2,0.25),N); draw((0.1,0.1)--(0.2,0)--(0.3,0.1)); draw((0.2,0)--(0.2,0.25)); label(\"R\",(0.2,0.25),N); draw((0.8,0.1)--(0.9,0)--(1,0.1)); draw((0.9,0)--(0.9,0.25)); label(\"S\",(0.9,0.25),N); draw((1.4,0.1)--(1.5,0)--(1.6,0.1)); draw((1.5,0)--(1.5,0.25)); label(\"T\",(1.5,0.25),N); [/asy]$ Which of the following expressions represents a negative number? $\\text{(A)}\\ P-Q \\qquad \\text{(B)}\\ P\\cdot Q \\qquad \\text{(C)}\\ \\dfrac{S}{Q}\\cdot P \\qquad \\text{(D)}\\ \\dfrac{R}{P\\cdot Q} \\qquad \\text{(E)}\\ \\dfrac{S+T}{R}$ Solution 1 First, note that $P < Q < 0 < R < S < T$. Thus, $P$ and $Q$ are negative, while $R$, $S$, and $T$ are positive. Option $A$ is the difference of two numbers. If the first number is lower than the second number, the answer will be negative. In this case, $P < Q$, so $P - Q < 0$, and $P-Q$ is thus indeed a negative number. So option $\\boxed{A}$ is correct. Option $B$ is the product of two negatives, which is always positive. Option $C$ starts with the quotient of a positive and a negative. This is negative. But this negative quotient is then multiplied by a negative number, which gives a positive number. Option $D$ has the product of two negatives in", "label(\"D\",D,NNW); label(\"C\",C,SW); label(\"H\",H,SE); draw(e--H,dashed); label(\"O\",(0,0),NE); label(\"1\",(C--O),N); label(\"1\",(H--O),N); label(\"2\",(A--C),N); label(\"2\",(H--B),N); label(\"3\",(O--D),NE); label(\"3\",(O--e),NE); label(\"2\\sqrt{2}\",(D--C),W); label(\"2\\sqrt{2}\",(H--e),E); draw(rightanglemark(e,(e.x,0),A,2)); draw(rightanglemark(D,C,B,2));[/asy]$ Let the center of the circle be $O$. Without loss of generality, let the radius of the circle be equal to $3$. Thus, $AO=3$ and $OB=3$. As a consequence of $2(AC)=BC$, $AC=2$ and $CO=1$. Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\\sqrt{3^2-1^2}$ or $2\\sqrt{2}$. Now we know that the area of $[ABD]$ is equal to $\\frac{(3+2+1)(2\\sqrt{2})}{2}$ or $6\\sqrt{2}$. Know we need to find the area of $[DCE]$. By simple inspection $[COD]$ $\\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\\sqrt{2}$ and $OH=1$. Know we know the area of $[CHE]=\\frac{(1+1)(2\\sqrt{2})}{2}$ or $2\\sqrt{2}$. We also know that the area of $[OHE]=\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus the area of $[COE]=2\\sqrt{2}-\\sqrt{2}$ or $\\sqrt{2}$. We also can calculate the area of $[DOC]$ to be $\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\\sqrt{2}+\\sqrt{2}$ or $2\\sqrt{2}$. The ratio between $[DCE]$ and $[ABD]$ is equal to $\\frac{2\\sqrt{2}}{6\\sqrt{2}}$ or $\\frac{1}{3}$ $\\Longrightarrow \\mathrm{(C)}$. Solution 5 We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$, and notice how $E$ is a", "$W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label(\"P\",(0,sqrt(3)/2),N); label(\"Z\",(1/6,sqrt(3)/3),NE); label(\"Y\",(1/3,sqrt(3)/6),NE); label(\"R\",(1/2,0),E); label(\"X\",(1/6,0),S); label(\"W\",(-1/6,0),S); label(\"Q\",(-1/2,0),W); label(\"V\",(-1/3,sqrt(3)/6),NW); label(\"U\",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$ ## Problem T2 Evaluate $\\begin{eqnarray*} && 1 \\left(\\dfrac{1}{1}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right) \\\\ &+& 3\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&5\\left(\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&7\\left(\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&9\\left(\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+11\\left(\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&13\\left(\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+15\\left(\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&17\\left(\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+19\\left(\\dfrac{1}{10}\\right)\\end{eqnarray*}$ ## Problem T3 To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed? ## Problem T4 In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey? ## Problem T5 During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued: 'A': It was 'B' or 'C' 'B': Neither 'E'", "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 $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$" ]

[ "\\\\ \\angle \\rm{OED} &=\\angle \\rm{ODE}\\text{ (Let y)} \\end{align} \\begin{align}\\rm{ADEC}\\end{align} is a cyclic quadrilateral. $$∴ \\angle {CAD}+\\angle {DEC}=180^{\\circ}$$ (Opposite angles are supplementary) \\begin{align} x+a+x+y&=180^{\\circ} \\\\ \\!2 x+a+y&=180^{\\circ} \\\\ y \\!&=\\!180^{\\circ}-2 x \\! -a ...\\! (4) \\end{align} However, $$\\angle {DOE}=180^{\\circ}-2 {y}$$ And, \\begin {align}\\angle{AOC}=180^{\\circ}-2 {a} \\end {align} \\begin{align} & \\angle {DOE}-\\angle {AOC} \\\\&=2a-2y \\\\ & =2 a-2\\left(180^{\\circ}-2 x-a\\right) \\\\ &=4 a+4 x-360^{\\circ} \\ldots \\ldots (5) \\end{align} $$\\angle {BAC}+\\angle {CAD}=180^{\\circ}$$(Linear pair ) \\begin{align} ∴ \\angle {BAC}&=180^{\\circ}-\\angle{CAD} \\\\ &=180^{\\circ}-(a+x)\\end{align} Similarly, \\begin {align} \\angle {ACB}=180^{\\circ}-(a+x) \\end {align} In \\begin {align} \\Delta {ABC,} \\end {align} $$\\angle {ABC}+\\angle {BAC}+\\angle {ACB}=180^{\\circ}$$ (Angle sum property of a triangle) \\begin {align}\\angle {ABC} &=180^{\\circ}-\\angle {BAC}-\\angle {ACB} \\\\ &= \\begin{Bmatrix} 180^{\\circ} \\\\ -\\left(180^{\\circ}-a-x\\right) \\\\ -\\left(180^{\\circ}-a-x\\right) \\end{Bmatrix} \\\\ &=2 a+2 x-180^{\\circ} \\\\ &=\\frac{1}{2}\\left[4 a+4 x-360^{\\circ}\\right]\\end {align} Using Equation ($$5$$) \\begin {align}\\angle {ABC}=\\frac{1}{2}[\\angle {DOE}-\\angle {AOC}] \\end {align} Hence it is proved that \\begin{align} \\angle {ABC}\\end{align} is equal to half the difference of the angles subtended by the chords \\begin{align}{AC}\\end{align} and \\begin{align}{DE}\\end{align} at the centre. ## Question 5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals. ### Solution What is known? Diameter of the circle is the length of one side of rhombus. What is unknown? Proof of the circle passes through the point of intersection of rhombus diagonals.", "{BCD}+\\angle {DAB}&=180^{\\circ} \\ldots(3) \\end {align} However, it is given that \\begin {align}\\angle {B}+\\angle {D} &=90^{\\circ}+90^{\\circ} \\\\ &=180^{\\circ} \\ldots(4) \\end {align} From Equations ($$3$$) and ($$4$$), it can be observed that the sum of the measures of opposite angles of quadrilateral $$\\text{ABCD}$$ is $$180^{\\circ}.$$ Therefore, it is a cyclic quadrilateral. Consider chord $${CD.}$$ \\begin {align}\\angle {CAD} = \\angle {CBD} \\end {align} (Angles in the same segment) ## Chapter 10 Ex.10.5 Question 12 Prove that cyclic parallelogram is a rectangle. ### Solution What is given ? What is unknown? Prove that cyclic parallelogram is a rectangle. Reasoning: The sum of either pair of opposite angles of a cyclic quadrilateral is $$180^\\circ.$$ By using this fact we can show each angle of cyclic parallelogram as $$90^\\circ$$ which will prove the statement it is a rectangle. Steps: Let $${ABCD}$$ be the cyclic parallelogram. We know that opposite angles of a parallelogram are equal. \\begin {align}\\angle {A}=\\angle {C} \\text { and } \\angle {B}=\\angle {D} \\ldots .(1) \\end {align} We know that the sum of either pair of opposite angles of a cyclic quadrilateral is $$180^{\\circ}.$$ \\begin {align} \\angle {A}+\\angle {C}=180^{\\circ} \\ldots .(2) \\end {align} Substituting ($$1$$) in ($$2$$), \\begin{align}\\angle {A}+\\angle {C}&=180^{\\circ} \\\\ \\angle {A}+\\angle {A}&=180^{\\circ} \\\\ 2 \\angle {A}&=180^{\\circ} \\\\ \\angle {A}&=90^{\\circ}\\end{align} We know that if one of the interior angles of", "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", "the value of f(g(5)), then f(g(5))=f(4)=6 17- A Area of the triangle is: $$\\frac{1}{2}$$ AD$$×$$BC and AD is perpendicular to BC. Triangle ADC is a (\\30^\\circ-60^\\circ- 90^\\circ\\) right triangle. The relationship among all sides of right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in the following triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. And the opposite side of $$60^\\circ$$ is opposite of $$30^\\circ×\\sqrt{3}$$ CD=6, then AD$$=6×\\sqrt{3}$$ Area of the triangle ABC is: $$\\frac{1}{2}$$ AD$$×$$BC $$= \\frac{1}{2} 6\\sqrt{3}×12=36\\sqrt{3}$$ $$\\img{https://appmanager.effortlessmath.com/public/images/questions/mmm.png }$$ 18- C $$6∎28=\\sqrt{6^2+28}= \\sqrt{36+28}=\\sqrt{64}=8$$ 19- D The union of A and B is: $$A∪B={1,2,4,8,12,16,24,32,48}$$. There are $$9$$ elements in $$A∪B$$. 20- D Solve the system of equations by elimination method. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ – x + 2 y = 10 \\end{align}}{} Multiply the second equation by 3, then add it to the first equation. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ 3(- x + 2 y = 10) \\end{align}}{} ⇒ \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ -3 x + 6 y = 30) \\end{align}}{} ⇒add the equations $$2𝑦=10⇒𝑦=5$$ Looking for the best resource to help you succeed on the CLEP College Math test? ### The Best Books to Ace the CLEP College Math Test ### More math articles ###", "MRS\\end{align*} is 70\\begin{align*} 70^\\circ\\end{align*}. Because supplementary angles have to add up to 180\\begin{align*}180^\\circ\\end{align*}, the other angle must be 18070=110\\begin{align*}180^\\circ-70^\\circ=110^\\circ\\end{align*}. --> ### Guided Practice Find the measure of an angle that is supplementary to ABC\\begin{align*}\\angle ABC\\end{align*} if mABC\\begin{align*}m\\angle ABC\\end{align*} is: 1. 45\\begin{align*}45^\\circ\\end{align*} 2. 118\\begin{align*}118^\\circ\\end{align*} 3. 32\\begin{align*}32^\\circ\\end{align*} 4. 2\\begin{align*}2^\\circ\\end{align*} 1. 18045=135\\begin{align*}180^\\circ-45^\\circ=135^\\circ\\end{align*}. 2. 180118=62\\begin{align*}180^\\circ-118^\\circ=62^\\circ\\end{align*}. 3. 18032=148\\begin{align*}180^\\circ-32^\\circ=148^\\circ\\end{align*}. 4. 1802=178\\begin{align*}180^\\circ-2^\\circ=178^\\circ\\end{align*}. ### Explore More Find the measure of an angle that is supplementary to ABC\\begin{align*}\\angle ABC\\end{align*} if mABC\\begin{align*}m\\angle ABC\\end{align*} is: 1. 114\\begin{align*}114^\\circ\\end{align*} 2. 11\\begin{align*}11^\\circ\\end{align*} 3. 91\\begin{align*}91^\\circ\\end{align*} 4. 84\\begin{align*}84^\\circ\\end{align*} 5. 57\\begin{align*}57^\\circ\\end{align*} 6. x\\begin{align*}x^\\circ\\end{align*} 7. (x+y)\\begin{align*}(x+y)^\\circ\\end{align*} Use the diagram below for exercises 8-9. Note that NK¯¯¯¯¯¯IL\\begin{align*}\\overline{NK} \\perp \\overleftrightarrow{IL}\\end{align*}. 1. Name two supplementary angles. 1. If mINJ=63\\begin{align*}m\\angle INJ = 63^\\circ\\end{align*}, find mJNL\\begin{align*}m\\angle JNL\\end{align*}. For exercise 10, determine if the statement is true or false. 1. Supplementary angles add up to 180\\begin{align*}180^\\circ\\end{align*}. For 11-12, find the value of x\\begin{align*}x\\end{align*}. ### Answers for Explore More Problems To view the Explore More answers, open this PDF file and look for section 1.8. ### Vocabulary Language: English Supplementary angles Supplementary angles Supplementary angles are two angles whose sum is 180 degrees.", "\\text{(SAS}& \\text { congruence rule)} \\\\\\\\ \\therefore {AD}&={{DC}}\\qquad\\dots(3)\\end{align} However, \\(AD = BC and $$AB = CD$$ (Opposite sides of parallelogram $$ABCD$$) $$AB = BC = CD = DA$$ Therefore, all the sides of quadrilateral $$ABCD$$ are equal to each other. In $$\\Delta {ADC}$$ and $$\\Delta {BCD} ,$$ \\begin{align}\\\\ {{AD}}&={{BC}\\; ( \\text{ Already proved })} \\\\ {{AC}}&={{BD} \\;\\text{ (Given) }} \\\\ {{DC}}&={{CD}\\; ( \\text{ Common })}\\\\{\\therefore \\Delta {ADC} }&\\cong \\Delta {BCD}\\\\ \\text{(SAS}& \\text { congruence rule)} \\\\\\\\ {\\therefore\\angle {ADC}}&={\\angle {BCD}\\;(\\text{By} \\,\\text{CPCT})}\\end{align} However, \\begin{align} \\angle {ADC}\\!+\\angle {BCD}&=180^{\\circ}\\\\ \\text{ (Co-interior }& \\text {angles) } \\\\\\\\ {\\qquad \\angle {ADC}+\\angle {ADC}}&={180^{\\circ}}\\\\{2 \\angle {ADC}}&={180^{\\circ}} \\\\ {\\therefore \\angle {ADC}}&={90^{\\circ}}\\end{align} One of the interior angles of quadrilateral $$ABCD$$ is a right angle. Thus, we have obtained that $$ABCD$$ is a parallelogram, $$AB = BC = CD = AD$$ and one of its interior angles is $$90^{\\circ}$$ . Therefore, $$ABCD$$ is a square. ## Chapter 8 Ex.8.1 Question 6 Diagonal $$AC$$ of a parallelogram $$ABCD$$ bisects $$\\angle A$$ (see the given figure). Show that (i) It bisects $$\\angle C$$ also, (ii) $$ABCD$$ is a rhombus. ### Solution What is known? The diagonal AC of a parallelogram $$ABCD$$ bisects $$∠A.$$ What is unknown? How we can show that (i) It bisects $$∠C$$ also, (ii) ABCD is a rhombus. Reasoning: We can use alternate interior angles property to show", "we know that $$AH=AE=\\sqrt{2}$$. Now, let us extend the figure as follows: With the construction of $$K$$ as the eventual intersection of $$FB$$ and $$IE$$, we can notice that $$BEK$$ is a right isosceles triangle, with side length $$2-\\sqrt{2}$$. Therefore, its area is $$3-2\\sqrt{2}$$. From this, we can further deduce that $$FIK$$ is also a right isosceles triangle, with area \\begin{align*}\\dfrac 12 FI\\cdot IK &= \\dfrac 12 FI^2 \\\\&= 1+3-2\\sqrt{2} \\\\&= 4-2\\sqrt{2},\\end{align*} implying that $FI^2 = 8-4\\sqrt{2}.$ Thus, the correct answer is B. 22. What is the remainder when $$2^{202} +202$$ is divided by $$2^{101}+2^{51}+1$$? $$100$$ $$101$$ $$200$$ $$201$$ $$202$$ ###### Solution(s): Observe that: \\begin{align*} 2^{202}+202 &= \\left(2^{101}\\right)^2+2\\cdot 2^{101} +1 \\\\&-2\\cdot 2^{101} + 201 \\\\ &= \\left(2^{101}+1\\right)^2 \\\\&-2\\cdot 2^{101} + 201 \\\\ &= a^2-b^2+201, \\end{align*} where $$a=2^{101}+1$$ and $$b=2^{51}.$$ Thus, by difference of squares: $\\begin{gather*}a^2-b^2+201\\\\=(a+b)(a-b)+201\\end{gather*}$ Which means that: $\\begin{gather*}2^{202}+202 =\\$$2^{101}+2^{51}+1)(2^{101}-2^{51}+1)\\\\+201.\\end{gather*}$ Therefore, $\\begin{gather*} 2^{202} + 202 \\equiv \\\\ (2^{101}+2^{51}+1)(2^{101}-2^{51}+1)+ \\\\ 201 \\equiv 201 \\bmod{(2^{101}+2^{51}+1).} \\end{gather*}$ Thus, the correct answer is D. 23. Square \\(ABCD$$ in the coordinate plane has vertices at the points $$A(1,1), B(-1,1), C(-1,-1),$$ and $$D(1,-1).$$ Consider the following four transformations: • $$L,$$ a rotation of $$90^{\\circ}$$ counterclockwise around the origin; • $$R,$$ a rotation of $$90^{\\circ}$$ clockwise around the origin; • $$H,$$ a reflection across the $$x$$-axis; and • $$V,$$ a reflection across the $$y$$-axis. •", "x$, $\\scriptsize y$ and $\\scriptsize z$. Question 4 adapted from NC(V) Mathematics Level 4 Paper 2 November 2015 question 1.3 1. In the figure below $\\scriptsize O$ is the centre of the circle, $\\scriptsize AOD$ is a diameter and $\\scriptsize AE\\parallel CD$, with $\\scriptsize E$, $\\scriptsize C$ and $\\scriptsize B$ on the circumference of the circle. Also, $\\scriptsize \\hat{A}={{50}^\\circ}$. 1. Why is $\\scriptsize A\\hat{D}C={{50}^\\circ}$? 2. Determine the size of $\\scriptsize C\\hat{B}D$. 3. Prove that $\\scriptsize EOC$ is a straight line. The full solutions are at the end of the unit. # Unit 2: Solutions ### Exercise 2.1 1. . 1. $\\scriptsize x={{78}^\\circ}\\quad \\text{(}\\angle \\text{s in same seg)}$ 2. $\\scriptsize AB$ is a diameter (straight line passing through the centre) Therefore $\\scriptsize A\\hat{P}B={{90}^\\circ}\\quad (\\angle \\text{s in semi-circle)}$ \\scriptsize \\begin{align*}x&={{180}^\\circ}-{{90}^\\circ}-{{59}^\\circ}\\quad \\text{(}\\angle \\text{s in a }\\Delta \\text{)}\\\\&={{31}^\\circ}\\end{align*} 3. . \\scriptsize \\begin{align*}C\\hat{A}D & ={{31}^\\circ}\\quad \\text{(alt }\\angle \\text{s =; }AC\\parallel BD\\text{)}\\\\x=C\\hat{A}D & ={{31}^\\circ}\\quad \\text{(}\\angle \\text{s in same seg)}\\end{align*} 4. . \\scriptsize \\begin{align*}{{{\\hat{O}}}_{1}} & ={{204}^\\circ}\\quad \\text{(}\\angle \\text{ at centre =2}\\angle \\text{ at circumference)}\\\\\\therefore x & ={{360}^\\circ}-{{204}^\\circ}\\quad \\text{(}\\angle \\text{ }\\!\\!'\\!\\!\\text{ s round a point)}\\\\ & ={{156}^\\circ}\\end{align*} 5. $\\scriptsize x={{47}^\\circ}\\quad \\text{(}\\angle \\text{s in same seg)}$ \\scriptsize \\begin{align*}D\\hat{A}E & ={{180}^\\circ}-{{128}^\\circ}\\quad \\text{(}\\angle \\text{s on str line suppl)}\\\\ & ={{52}^\\circ}\\\\\\therefore y & ={{180}^\\circ}-{{52}^\\circ}-{{47}^\\circ}\\quad \\text{(}\\angle \\text{s in }\\Delta \\text{ suppl)}\\\\ &=8{{\\text{1}}^\\circ}\\end{align*} 2. $\\scriptsize OP=OB=6\\text{ units}\\quad \\text{(both radii)}$ \\scriptsize \\begin{align*}PR &", "$$\\beta=180°−49.9°=130.1°$$. (Remember that the sine function is positive in both the first and second quadrants.) Solving for $$\\gamma$$, we have \\begin{align*} \\gamma&= 180^{\\circ}-35^{\\circ}-130.1^{\\circ}\\\\ &\\approx 14.9^{\\circ} \\end{align*} We can then use these measurements to solve the other triangle. Since $$\\gamma′$$ is supplementary to $$\\gamma$$, we have \\begin{align*} \\gamma^{'}&= 180^{\\circ}-35^{\\circ}-49.5^{\\circ}\\\\ &\\approx 95.1^{\\circ} \\end{align*} Now we need to find $$c$$ and $$c′$$. We have \\begin{align*} \\dfrac{c}{\\sin(14.9^{\\circ})}&= \\dfrac{6}{\\sin(35^{\\circ})}\\\\ c&= \\dfrac{6 \\sin(14.9^{\\circ})}{\\sin(35^{\\circ})}\\\\ &\\approx 2.7 \\end{align*} Finally, \\begin{align*} \\dfrac{c'}{\\sin(95.1^{\\circ})}&= \\dfrac{6}{\\sin(35^{\\circ})}\\\\ c'&= \\dfrac{6 \\sin(95.1^{\\circ})}{\\sin(35^{\\circ})}\\\\ &\\approx 10.4 \\end{align*} To summarize, there are two triangles with an angle of $$35°$$, an adjacent side of 8, and an opposite side of 6, as shown in Figure $$\\PageIndex{12}$$. However, we were looking for the values for the triangle with an obtuse angle $$\\beta$$. We can see them in the first triangle (a) in Figure $$\\PageIndex{12}$$. Exercise $$\\PageIndex{2}$$ Given $$\\alpha=80°$$, $$a=120$$, and $$b=121$$, find the missing side and angles. If there is more than one possible solution, show both. Solution 1 $$\\begin{matrix} \\alpha=80^{\\circ} & a=120\\\\ \\beta\\approx 83.2^{\\circ} & b=121\\\\ \\gamma\\approx 16.8^{\\circ} & c\\approx 35.2 \\end{matrix}$$ Solution 2 $$\\begin{matrix} \\alpha '=80^{\\circ} & a'=120\\\\ \\beta '\\approx 96.8^{\\circ} & b'=121\\\\ \\gamma '\\approx 3.2^{\\circ} & c'\\approx 6.8 \\end{matrix}$$ Example $$\\PageIndex{3}$$: Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in Figure $$\\PageIndex{13}$$, solve for the unknown side and", "\\end {align}find \\begin {align} \\angle {BDC} . \\end {align} ### Solution What is known? Two angles in a triangle. What is unknown? Value of \\begin {align} \\angle {BDC}\\end {align} Reasoning: • Sum of angles in a triangle is \\begin {align} 180^{\\circ}.\\end {align} • Angles in the same segment are equal. Steps: Consider the \\begin {align} \\Delta {ABC,}\\end {align} the sum of all angles will be \\begin {align} 180^\\circ.\\end {align} \\begin{align} \\angle {ABC}+ \\!\\!\\angle {BAC}+ \\!\\angle {ACB} &=180^{\\circ} \\\\ 69^{\\circ}+\\angle {BAC}+31^{\\circ} &=180^{\\circ} \\end{align} \\begin{align} \\angle {BAC} &=180^{\\circ}-\\left(69^{\\circ}+31^{\\circ}\\right) \\\\ &=180^{\\circ}-100^{\\circ} \\\\ &=80^{\\circ} \\end{align} We know that, angles in the same segment of a circle are equal. \\begin {align} ∴ \\angle {BDC}=\\angle {BAC}=80^{\\circ} \\end {align} ## Chapter 10 Ex.10.5 Question 5 In the given figure, $$A, B, C$$ and $$D$$ are four points on a circle. $$AC$$ and $$BD$$ intersect at a point \\begin {align}{E} \\end {align} such that \\begin {align} \\angle {BEC}=130^{\\circ} \\end {align} and \\begin {align} \\angle {ECD}=20^{\\circ}. \\end {align} Find \\begin {align} \\angle {BAC}\\end {align}. ### Solution What is known? Two angles \\begin {align} \\angle {BEC}=130^{\\circ} \\end {align} and \\begin {align} \\angle {ECD}=20^{\\circ} \\end {align} What is unknown? Value of $$\\angle {BAC}$$ Reasoning: • Sum of angles in a triangle is $$180^{\\circ}.$$ • Angles in the same segment are equal. Steps: Consider the straight line \\begin {align} {BD}", "\\end {align}. As the line \\begin {align} {AC} \\end {align} intersects with the line \\begin {align} {BD,} \\end {align} then the sum of two adjacent angles so formed is $$180^{\\circ}.$$ Therefore, \\begin{align} \\angle {BEC}+\\angle {DEC} &=180^{\\circ} \\\\ 130^{\\circ}+\\angle {DEC} &=180^{\\circ} \\\\ \\angle {DEC} &=180^{\\circ}-130^{\\circ} \\\\ &=50^{\\circ} \\end{align} Consider the \\begin{align} \\Delta {DEC,} \\end{align} the sum of all angles will be $$180^{\\circ}.$$ \\begin{align} \\angle {DEC}+\\angle {EDC}+\\angle {ECD} &=180^{\\circ} \\\\ 50^{\\circ}+\\angle {EDC}+20^{\\circ} &=180^{\\circ} \\end{align} \\begin{align} \\angle {EDC} &=180^{\\circ}-70^{\\circ} \\\\ &=110^{\\circ} \\\\ ∴ \\angle {BDC}=\\angle {EDC} &=110^{\\circ} \\end{align} We know that, angles in the same segment of a circle are equal. \\begin {align} ∴ \\angle {BAC}=\\angle {BDC}=110^{\\circ}\\end {align} ## Chapter 10 Ex.10.5 Question 6 \\begin {align} {ABCD}\\end {align} is a cyclic quadrilateral whose diagonals intersect at a point \\begin {align} {E. }\\end {align}If \\begin {align} \\angle {DBC}=70^{\\circ}, \\quad \\angle {BAC}=30^{\\circ} \\end {align} find \\begin {align} \\angle {BCD}. \\end {align} Further if \\begin {align} {AB=BC,} \\end {align} find \\begin {align} \\angle {ECD.} \\end {align} ### Solution What is known? \\begin {align} {ABCD}\\end {align} is cyclic quadrilateral. \\begin {align} {DBC}=70^\\circ \\end {align} and \\begin {align} {BAC}=30^\\circ. \\end {align} \\begin {align}{AB = BC.}\\end {align} What is unknown? Value of \\begin {align} \\angle {BCD} \\text { and } \\angle {ECD} \\end {align} Reasoning: • A quadrilateral \\begin {align} {ABCD}\\end {align} is called cyclic if all", "# 2006 iTest Problems/Problem 24 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Points $D$ and $E$ are chosen on side $BC$ of triangle $ABC$ such that $E$ is between $B$ and $D$ and $BE=1$, $ED=DC=3$. If $\\angle BAD=\\angle EAC=90^\\circ$, the area of $ABC$ can be expressed as $\\tfrac{p\\sqrt q}r$, where $p$ and $r$ are relatively prime positive integers and $q$ is a positive integer not divisible by the square of any prime. Compute $p+q+r$. $[asy] import olympiad; size(200); defaultpen(linewidth(0.7)+fontsize(11pt)); pair D = origin, E = (3,0), C = (-3,0), B = (4,0); path circ1 = arc(D,3,0,180), circ2 = arc(B/2,2,0,180); pair A = intersectionpoint(circ1, circ2); draw(E--A--C--B--A--D); label(\"A\",A,N); label(\"B\",B,SE); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,S); [/asy]$ ## Solution Note that if a circle has center $D$ and radius $CD$, then $CE$ is the circle's diameter because $CD = DE$. Because $\\angle CAE$ is a right angle, $A$ is also on the circle with center $D$, so $DA = 3$. By the Pythagorean Theorem, $AB = \\sqrt{16 - 9} = \\sqrt{7}$. The area of $ABD$ is $\\tfrac12 \\cdot 3 \\cdot \\sqrt{7} = \\tfrac{3\\sqrt{7}}{2}$. Let $x$ be the length of the altitude from $A$ to $BD$, so $\\tfrac12 \\cdot 4x = \\tfrac{3\\sqrt{7}}{2}$. Thus, $x = \\tfrac{3\\sqrt{7}}{4}$. Since $BD$ is on line $BC$, the area of $ABC$ is", "of the triangle is: $$\\frac{1}{2}$$ AD$$×$$BC and AD is perpendicular to BC. Triangle ADC is a (\\30^\\circ-60^\\circ- 90^\\circ\\) right triangle. The relationship among all sides of right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in the following triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. And the opposite side of $$60^\\circ$$ is opposite of $$30^\\circ×\\sqrt{3}$$ CD=6, then AD$$=6×\\sqrt{3}$$ Area of the triangle ABC is: $$\\frac{1}{2}$$ AD$$×$$BC $$= \\frac{1}{2} 6\\sqrt{3}×12=36\\sqrt{3}$$ $$\\img{https://appmanager.effortlessmath.com/public/images/questions/mmm.png }$$ 18- C $$6∎28=\\sqrt{6^2+28}= \\sqrt{36+28}=\\sqrt{64}=8$$ 19- D The union of A and B is: $$A∪B={1,2,4,8,12,16,24,32,48}$$. There are $$9$$ elements in $$A∪B$$. 20- D Solve the system of equations by elimination method. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ – x + 2 y = 10 \\end{align}}{} Multiply the second equation by 3, then add it to the first equation. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ 3(- x + 2 y = 10) \\end{align}}{} ⇒ \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ -3 x + 6 y = 30) \\end{align}}{} ⇒add the equations $$2𝑦=10⇒𝑦=5$$ Looking for the best resource to help you succeed on the CLEP College Math test? 52% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or", "= 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", "\\\\ & \\Rightarrow C{{D}^{2}}=C{{E}^{2}}-D{{E}^{2}} \\\\ & CD=\\sqrt{C{{E}^{2}}-D{{E}^{2}}} \\\\ \\end{align} From the figure we know that CE = 5cm and DE = 4cm. $\\therefore$ Length of $CD=\\sqrt{{{5}^{2}}-{{4}^{2}}}=\\sqrt{25-16}=\\sqrt{9}=3cm$. $\\therefore$ Length of CD = 3cm. Now, let us consider equation (1), we know the lengths of ED, BC and DC. Substitute and find the length of AB. \\begin{align} & \\dfrac{AB}{ED}=\\dfrac{BC}{DC} \\\\ & \\Rightarrow \\dfrac{AB}{4}=\\dfrac{9}{3} \\\\ & \\therefore AB=\\dfrac{4\\times 9}{3}=4\\times 3=12cm \\\\ \\end{align} $\\therefore$ Length of AB = 12cm. Now we need to find the length of AC. Consider $\\vartriangle ABC$, by Pythagoras theorem, \\begin{align} & A{{C}^{2}}=A{{B}^{2}}+C{{D}^{2}} \\\\ & AC=\\sqrt{A{{B}^{2}}+C{{D}^{2}}} \\\\ \\end{align} We know AB = 12cm and CD = 9cm. Substitute and find the value of AC. $\\therefore AC=\\sqrt{{{12}^{2}}+{{9}^{2}}}=\\sqrt{144+81}=\\sqrt{225}=15cm$ $\\therefore$ Length of AC = 15cm. Hence, we got all the required lengths as, CD = 3cm, AB = 12cm and AC = 15cm. Note: In our two right angles triangles, the corresponding sides are in the same ratio. Hence the matching angles are also the same. Thus we have considered the ratio of their sides $\\dfrac{AB}{ED}$ and $\\dfrac{BC}{DC}$, which are equal. The two triangles can be also proved by SSS congruence i.e. all the sides of the two triangles are similar.", "angle are given • if three sides are given ## The cosine rule Exercise 6.12 Solve the following triangles (that is, find all unknown sides and angles): $$\\triangle ABC$$ in which $$\\hat{A}=\\text{70}\\text{°}$$; $$b=4$$ and $$c=9$$ \\begin{align*} {a}^{2} &= {b}^{2}+{c}^{2}-2bc \\cos \\hat{A} \\\\ {a}^{2} &= {4}^{2}+{9}^{2}-2(9)(4) \\cos \\hat{\\text{70}\\text{°}} \\\\ &= \\text{72,374} \\ldots \\\\ \\therefore a &= \\text{8,5} \\end{align*} \\begin{align*} {c}^{2} &= {b}^{2}+{a}^{2}-2ba \\cos \\hat{C} \\\\ {9}^{2} &= {4}^{2}+{\\text{8,5}}^{2}-2(4)(\\text{8,5}) \\cos \\hat{C} \\\\ \\therefore \\hat{C} &= \\text{83,9}\\text{°} \\end{align*} \\begin{align*} \\hat{B} &= \\text{180}\\text{°} - \\text{70}\\text{°} -\\text{83,9}\\text{°} \\\\ \\therefore \\hat{B} &= \\text{26,1}\\text{°} \\end{align*} $$\\triangle RST$$ in which $$RS=14$$; $$ST=26$$ and $$RT=16$$ \\begin{align*} {r}^{2} &= {s}^{2}+{t}^{2}-2st \\cos \\hat{R} \\\\ {26}^{2} &= {16}^{2}+{\\text{14}}^{2}-2(16)(\\text{14}) \\cos \\hat{R} \\\\ \\therefore \\hat{R} &= \\text{120}\\text{°} \\end{align*} \\begin{align*} {s}^{2} &= {r}^{2}+{t}^{2}-2rt \\cos \\hat{S} \\\\ {16}^{2} &= {26}^{2}+{\\text{14}}^{2}-2(26)(\\text{14}) \\cos \\hat{S} \\\\ \\therefore \\hat{S} &= \\text{32,2}\\text{°} \\end{align*} \\begin{align*} \\hat{T} &= \\text{180}\\text{°} - \\text{120}\\text{°} -\\text{32,2}\\text{°} \\\\ \\therefore \\hat{T} &= \\text{27,8}\\text{°} \\end{align*} $$\\triangle KLM$$ in which $$KL=5$$; $$LM=10$$ and $$KM=7$$ \\begin{align*} {m}^{2} &= {k}^{2}+{l}^{2}-2kl \\cos \\hat{M} \\\\ {5}^{2} &= {10}^{2}+{\\text{7}}^{2}-2(10)(\\text{7}) \\cos \\hat{M} \\\\ \\therefore \\hat{M} &= \\text{27,7}\\text{°} \\end{align*} \\begin{align*} {l}^{2} &= {k}^{2}+{m}^{2}-2km \\cos \\hat{L} \\\\ {7}^{2} &= {10}^{2}+{\\text{5}}^{2}-2(10)(\\text{5}) \\cos \\hat{L} \\\\ \\therefore \\hat{L} &= \\text{40,5}\\text{°} \\end{align*} \\begin{align*} \\hat{K} &= \\text{180}\\text{°} - \\text{27,7}\\text{°} -\\text{40,5}\\text{°} \\\\ \\therefore \\hat{K} &= \\text{111,8}\\text{°} \\end{align*} $$\\triangle JHK$$ in which $$\\hat{H}=\\text{130}\\text{°}$$; $$JH=13$$ and $$HK=8$$ \\begin{align*} {h}^{2} &= {j}^{2}+{k}^{2}-2jk \\cos \\hat{H} \\\\ {h}^{2}", "hexagon. Now, using the sine law, $\\frac{1}{\\sin \\angle ACB} = \\frac{1}{\\sin 60^\\circ}$, so $\\angle ACB = 60^\\circ$. Since the angles in a triangle sum to 180$^\\circ$$\\angle ABC$ is also 60$^\\circ$. Therefore, $\\triangle ABC$ is an equilateral triangle with side lengths of 1. $[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label(\"2\",(z,c),NE); label(\"1\",(1.5,0),S); label(\"1\",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label(\"60^\\circ\",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label(\"60^\\circ\",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label(\"60^\\circ\",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); [/asy]$ Since the area of a regular hexagon can be found with the formula $\\frac{3\\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\\frac{3\\sqrt{3}(2^2)}{2} = 6\\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\\frac{\\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\\frac{\\sqrt{3}}{4}(1^2) = \\frac{\\sqrt{3}}{4}$. Since the area of a circle can be found", "find $a$ and $A$. ###### Solution First, use $\\dfrac{b}{\\sin B} = \\dfrac{c}{\\sin C}$ to find $c$. \\begin{align} \\frac{6}{\\sin 77^{\\circ} } &= \\frac{c}{\\sin 34^{\\circ} }\\\\\\\\ \\frac{6 \\sin 34^{\\circ} }{\\sin 77^{\\circ} } &= c\\\\ c &= 3.44 \\text{ (to 3 sig.fig.)} \\end{align} To find $A$, use the fact that the angles in a triangle add up to $180^{\\circ}$, i.e. $A+B+C = 180$. \\begin{align} 180 &= A +77 + 34\\\\ A &= 180 - 77- 34\\\\ A &= 69^{\\circ} \\end{align} Now, use this value for $A$ in the sine rule to find $a$. \\begin{align} \\frac{a}{\\sin 69^{\\circ} } &= \\frac{6}{\\sin 77^{\\circ} }\\\\\\\\ a &= \\frac{6 \\sin 69^{\\circ} }{\\sin 77^{\\circ} }\\\\\\\\ a &= 5.75 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 3 Given $b=2$, $c=5$ and $A=39^{\\circ}$, find $a$. ###### Solution Use $a^2 = b^2 + c^2 -2bc\\cos A$. \\begin{align} a^2 &= 2^2 + 5^2 - (2\\times2\\times5)\\cos 39^{\\circ}\\\\ a^2 &= 29 - 20\\cos 39^{\\circ}\\\\ a^2 &= 13.457 \\text{ (to 3 d.p.)}\\\\ a &= 3.67 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 4 Given $a=7$, $b=6$ and $c=9$. Find the angle between $a$ and $b$. ###### Solution The angle between $a$ and $b$ is labelled $C$ in the diagram above, so use $c^2 = a^2 + b^2 - 2ab\\cos C$. \\begin{align} 9^2 &= 7^2 + 6^2 - (2\\times7\\times6)\\cos C\\\\ 81 &= 49+36 - 84\\cos", "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", "supplementary angles have to add up to \\begin{align*}180^\\circ\\end{align*}, the other angle must be \\begin{align*}180^\\circ-70^\\circ=110^\\circ\\end{align*}. --> ### Guided Practice Find the measure of an angle that is supplementary to \\begin{align*}\\angle ABC\\end{align*} if \\begin{align*}m\\angle ABC\\end{align*} is: 1. \\begin{align*}45^\\circ\\end{align*} 2. \\begin{align*}118^\\circ\\end{align*} 3. \\begin{align*}32^\\circ\\end{align*} 4. \\begin{align*}2^\\circ\\end{align*} 1. \\begin{align*}180^\\circ-45^\\circ=135^\\circ\\end{align*}. 2. \\begin{align*}180^\\circ-118^\\circ=62^\\circ\\end{align*}. 3. \\begin{align*}180^\\circ-32^\\circ=148^\\circ\\end{align*}. 4. \\begin{align*}180^\\circ-2^\\circ=178^\\circ\\end{align*}. ### Explore More Find the measure of an angle that is supplementary to \\begin{align*}\\angle ABC\\end{align*} if \\begin{align*}m\\angle ABC\\end{align*} is: 1. \\begin{align*}114^\\circ\\end{align*} 2. \\begin{align*}11^\\circ\\end{align*} 3. \\begin{align*}91^\\circ\\end{align*} 4. \\begin{align*}84^\\circ\\end{align*} 5. \\begin{align*}57^\\circ\\end{align*} 6. \\begin{align*}x^\\circ\\end{align*} 7. \\begin{align*}(x+y)^\\circ\\end{align*} Use the diagram below for exercises 8-9. Note that \\begin{align*}\\overline{NK} \\perp \\overleftrightarrow{IL}\\end{align*}. 1. Name two supplementary angles. 1. If \\begin{align*}m\\angle INJ = 63^\\circ\\end{align*}, find \\begin{align*}m\\angle JNL\\end{align*}. For exercise 10, determine if the statement is true or false. 1. Supplementary angles add up to \\begin{align*}180^\\circ\\end{align*}. For 11-12, find the value of \\begin{align*}x\\end{align*}. ### Vocabulary Language: English Supplementary angles Supplementary angles Supplementary angles are two angles whose sum is 180 degrees." ]

[ "# triangle inequality with $\\cos$ and triangle angles In any triangle $$ABC$$ If: $$\\frac{1}{8}\\geq \\cos A\\cdot \\cos B\\cdot \\cos C > y$$, find the value of $$y$$. Find the minimal value of \\begin{align} t&=\\cos A\\cdot \\cos B\\cdot \\cos C \\tag{1}\\label{1} , \\end{align} where $$A,B,C$$ are the angles of $$\\triangle ABC$$. For any $$\\triangle ABC$$ with semiperimeter $$\\rho$$ inradius $$r$$ and circumradius $$R$$, we can using a known expression for \\eqref{1} in terms of two parameters $$v=r/R$$, $$v\\in[0,\\tfrac12]$$, and $$u=\\rho/R=u(v)$$, for any valid value of $$v$$ $$u(v)\\in[u_{\\min}(v),\\, u_{\\max}(v)]$$: \\begin{align} t&= \\cos A\\cdot \\cos B\\cdot \\cos C = \\tfrac14\\,(u(v)^2-(v+2)^2) \\tag{2}\\label{2} . \\end{align} Expressions for the boundary values of $$u$$ are known to be \\begin{align} u_{\\min}(v)&= \\sqrt{27-(5-v)^2-2\\sqrt{(1-2v)^3}} \\tag{3}\\label{3} ,\\\\ u_{\\max}(v)&= \\sqrt{27-(5-v)^2+2\\sqrt{(1-2v)^3}} \\tag{4}\\label{4} . \\end{align} Thus \\begin{align} \\min_{\\triangle ABC} &(\\cos A\\cdot \\cos B\\cdot \\cos C) = \\tfrac14\\,\\min_{v\\in[0,1/2]} (u_{\\min}(v)^2-(v+2)^2) \\tag{5}\\label{5} \\\\ &= \\tfrac12 \\,\\min_{v\\in[0,1/2]} (v-(1-v)^2-\\sqrt{(1-2v)^3}) \\tag{6}\\label{6} . \\end{align} It can be shown that the function in \\eqref{6} is increasing on $$v\\in[0,\\tfrac12]$$, hence \\begin{align} \\tfrac12 \\,\\min_{v\\in[0,1/2]} &(-(1-v)^2+v-\\sqrt((1-2v)^3)) \\tag{7}\\label{7} \\\\ &= \\tfrac12\\, (v-(1-v)^2-\\sqrt{(1-2v)^3})|_{v=0} =\\tfrac12\\cdot(-2) =-1 \\tag{8}\\label{8} . \\end{align} It follows that the minimal value of $$t$$ in \\eqref{1} is $$-1$$, which is reached only for degenerate triangles with two zero angles and the third one of $$180^\\circ$$. Indeed, \\begin{align} \\cos0^\\circ\\cdot\\cos0^\\circ\\cdot\\cos180^\\circ &=1\\cdot1\\cdot(-1)=-1 \\tag{9}\\label{9} . \\end{align}", "= 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", "\\\\ \\angle \\rm{OED} &=\\angle \\rm{ODE}\\text{ (Let y)} \\end{align} \\begin{align}\\rm{ADEC}\\end{align} is a cyclic quadrilateral. $$∴ \\angle {CAD}+\\angle {DEC}=180^{\\circ}$$ (Opposite angles are supplementary) \\begin{align} x+a+x+y&=180^{\\circ} \\\\ \\!2 x+a+y&=180^{\\circ} \\\\ y \\!&=\\!180^{\\circ}-2 x \\! -a ...\\! (4) \\end{align} However, $$\\angle {DOE}=180^{\\circ}-2 {y}$$ And, \\begin {align}\\angle{AOC}=180^{\\circ}-2 {a} \\end {align} \\begin{align} & \\angle {DOE}-\\angle {AOC} \\\\&=2a-2y \\\\ & =2 a-2\\left(180^{\\circ}-2 x-a\\right) \\\\ &=4 a+4 x-360^{\\circ} \\ldots \\ldots (5) \\end{align} $$\\angle {BAC}+\\angle {CAD}=180^{\\circ}$$(Linear pair ) \\begin{align} ∴ \\angle {BAC}&=180^{\\circ}-\\angle{CAD} \\\\ &=180^{\\circ}-(a+x)\\end{align} Similarly, \\begin {align} \\angle {ACB}=180^{\\circ}-(a+x) \\end {align} In \\begin {align} \\Delta {ABC,} \\end {align} $$\\angle {ABC}+\\angle {BAC}+\\angle {ACB}=180^{\\circ}$$ (Angle sum property of a triangle) \\begin {align}\\angle {ABC} &=180^{\\circ}-\\angle {BAC}-\\angle {ACB} \\\\ &= \\begin{Bmatrix} 180^{\\circ} \\\\ -\\left(180^{\\circ}-a-x\\right) \\\\ -\\left(180^{\\circ}-a-x\\right) \\end{Bmatrix} \\\\ &=2 a+2 x-180^{\\circ} \\\\ &=\\frac{1}{2}\\left[4 a+4 x-360^{\\circ}\\right]\\end {align} Using Equation ($$5$$) \\begin {align}\\angle {ABC}=\\frac{1}{2}[\\angle {DOE}-\\angle {AOC}] \\end {align} Hence it is proved that \\begin{align} \\angle {ABC}\\end{align} is equal to half the difference of the angles subtended by the chords \\begin{align}{AC}\\end{align} and \\begin{align}{DE}\\end{align} at the centre. ## Question 5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals. ### Solution What is known? Diameter of the circle is the length of one side of rhombus. What is unknown? Proof of the circle passes through the point of intersection of rhombus diagonals.", "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", "we know that $$AH=AE=\\sqrt{2}$$. Now, let us extend the figure as follows: With the construction of $$K$$ as the eventual intersection of $$FB$$ and $$IE$$, we can notice that $$BEK$$ is a right isosceles triangle, with side length $$2-\\sqrt{2}$$. Therefore, its area is $$3-2\\sqrt{2}$$. From this, we can further deduce that $$FIK$$ is also a right isosceles triangle, with area \\begin{align*}\\dfrac 12 FI\\cdot IK &= \\dfrac 12 FI^2 \\\\&= 1+3-2\\sqrt{2} \\\\&= 4-2\\sqrt{2},\\end{align*} implying that $FI^2 = 8-4\\sqrt{2}.$ Thus, the correct answer is B. 22. What is the remainder when $$2^{202} +202$$ is divided by $$2^{101}+2^{51}+1$$? $$100$$ $$101$$ $$200$$ $$201$$ $$202$$ ###### Solution(s): Observe that: \\begin{align*} 2^{202}+202 &= \\left(2^{101}\\right)^2+2\\cdot 2^{101} +1 \\\\&-2\\cdot 2^{101} + 201 \\\\ &= \\left(2^{101}+1\\right)^2 \\\\&-2\\cdot 2^{101} + 201 \\\\ &= a^2-b^2+201, \\end{align*} where $$a=2^{101}+1$$ and $$b=2^{51}.$$ Thus, by difference of squares: $\\begin{gather*}a^2-b^2+201\\\\=(a+b)(a-b)+201\\end{gather*}$ Which means that: $\\begin{gather*}2^{202}+202 =\\$$2^{101}+2^{51}+1)(2^{101}-2^{51}+1)\\\\+201.\\end{gather*}$ Therefore, $\\begin{gather*} 2^{202} + 202 \\equiv \\\\ (2^{101}+2^{51}+1)(2^{101}-2^{51}+1)+ \\\\ 201 \\equiv 201 \\bmod{(2^{101}+2^{51}+1).} \\end{gather*}$ Thus, the correct answer is D. 23. Square \\(ABCD$$ in the coordinate plane has vertices at the points $$A(1,1), B(-1,1), C(-1,-1),$$ and $$D(1,-1).$$ Consider the following four transformations: • $$L,$$ a rotation of $$90^{\\circ}$$ counterclockwise around the origin; • $$R,$$ a rotation of $$90^{\\circ}$$ clockwise around the origin; • $$H,$$ a reflection across the $$x$$-axis; and • $$V,$$ a reflection across the $$y$$-axis. •", "the value of f(g(5)), then f(g(5))=f(4)=6 17- A Area of the triangle is: $$\\frac{1}{2}$$ AD$$×$$BC and AD is perpendicular to BC. Triangle ADC is a (\\30^\\circ-60^\\circ- 90^\\circ\\) right triangle. The relationship among all sides of right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in the following triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. And the opposite side of $$60^\\circ$$ is opposite of $$30^\\circ×\\sqrt{3}$$ CD=6, then AD$$=6×\\sqrt{3}$$ Area of the triangle ABC is: $$\\frac{1}{2}$$ AD$$×$$BC $$= \\frac{1}{2} 6\\sqrt{3}×12=36\\sqrt{3}$$ $$\\img{https://appmanager.effortlessmath.com/public/images/questions/mmm.png }$$ 18- C $$6∎28=\\sqrt{6^2+28}= \\sqrt{36+28}=\\sqrt{64}=8$$ 19- D The union of A and B is: $$A∪B={1,2,4,8,12,16,24,32,48}$$. There are $$9$$ elements in $$A∪B$$. 20- D Solve the system of equations by elimination method. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ – x + 2 y = 10 \\end{align}}{} Multiply the second equation by 3, then add it to the first equation. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ 3(- x + 2 y = 10) \\end{align}}{} ⇒ \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ -3 x + 6 y = 30) \\end{align}}{} ⇒add the equations $$2𝑦=10⇒𝑦=5$$ Looking for the best resource to help you succeed on the CLEP College Math test? ### The Best Books to Ace the CLEP College Math Test ### More math articles ###", "as well, and $\\triangle BCE$ is equilateral, so $BC=CE=BE=7$. $[asy] size(150); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); dot(\"W\",circumcenter(A,B,C),dir(180)); label(\"\\gamma\",gamma,dir(180)); [/asy]$ In triangle $AEF$, let $X$ be the foot of the altitude from $A$; then $EF=EX+XF$, where we use signed lengths. Writing $EX=AE \\cdot \\cos \\angle AEF$ and $XF=AF \\cdot \\cos \\angle AFE$, we get \\begin{align}\\tag{1} EF = AE \\cdot \\cos \\angle AEF + AF \\cdot \\cos \\angle AFE. \\end{align} Note $\\angle AFE = \\angle ACE$, and the Law of Cosines in $\\triangle ACE$ gives $\\cos \\angle ACE = -\\tfrac 17$. Also, $\\angle AEF = \\angle DEF$, and $\\angle DFE = \\tfrac{\\pi}{2}$ ($DE$ is a diameter), so $\\cos \\angle AEF = \\tfrac{EF}{DE} = \\tfrac{8}{49}\\cdot EF$. Plugging in all our values into equation $(1)$, we get: $$EF = \\tfrac{64}{49} EF -\\tfrac{1}{7} AF \\quad \\Longrightarrow \\quad EF = \\tfrac{7}{15} AF.$$ The Law of Cosines in $\\triangle AEF$, with $EF=\\tfrac 7{15}AF$ and $\\cos\\angle AFE = -\\tfrac 17$ gives $$8^2 = AF^2 + \\tfrac{49}{225} AF^2 + \\tfrac 2{15} AF^2 = \\tfrac{225+49+30}{225}\\cdot AF^2$$ Thus $AF^2 = \\frac{900}{19}$. The answer is $\\boxed{919}$. ## Solution 2 Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\\angle MAD=\\angle DAF$.", "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", "hexagon. Now, using the sine law, $\\frac{1}{\\sin \\angle ACB} = \\frac{1}{\\sin 60^\\circ}$, so $\\angle ACB = 60^\\circ$. Since the angles in a triangle sum to 180$^\\circ$$\\angle ABC$ is also 60$^\\circ$. Therefore, $\\triangle ABC$ is an equilateral triangle with side lengths of 1. $[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label(\"2\",(z,c),NE); label(\"1\",(1.5,0),S); label(\"1\",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label(\"60^\\circ\",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label(\"60^\\circ\",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label(\"60^\\circ\",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); [/asy]$ Since the area of a regular hexagon can be found with the formula $\\frac{3\\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\\frac{3\\sqrt{3}(2^2)}{2} = 6\\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\\frac{\\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\\frac{\\sqrt{3}}{4}(1^2) = \\frac{\\sqrt{3}}{4}$. Since the area of a circle can be found", "# 2006 iTest Problems/Problem 24 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Points $D$ and $E$ are chosen on side $BC$ of triangle $ABC$ such that $E$ is between $B$ and $D$ and $BE=1$, $ED=DC=3$. If $\\angle BAD=\\angle EAC=90^\\circ$, the area of $ABC$ can be expressed as $\\tfrac{p\\sqrt q}r$, where $p$ and $r$ are relatively prime positive integers and $q$ is a positive integer not divisible by the square of any prime. Compute $p+q+r$. $[asy] import olympiad; size(200); defaultpen(linewidth(0.7)+fontsize(11pt)); pair D = origin, E = (3,0), C = (-3,0), B = (4,0); path circ1 = arc(D,3,0,180), circ2 = arc(B/2,2,0,180); pair A = intersectionpoint(circ1, circ2); draw(E--A--C--B--A--D); label(\"A\",A,N); label(\"B\",B,SE); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,S); [/asy]$ ## Solution Note that if a circle has center $D$ and radius $CD$, then $CE$ is the circle's diameter because $CD = DE$. Because $\\angle CAE$ is a right angle, $A$ is also on the circle with center $D$, so $DA = 3$. By the Pythagorean Theorem, $AB = \\sqrt{16 - 9} = \\sqrt{7}$. The area of $ABD$ is $\\tfrac12 \\cdot 3 \\cdot \\sqrt{7} = \\tfrac{3\\sqrt{7}}{2}$. Let $x$ be the length of the altitude from $A$ to $BD$, so $\\tfrac12 \\cdot 4x = \\tfrac{3\\sqrt{7}}{2}$. Thus, $x = \\tfrac{3\\sqrt{7}}{4}$. Since $BD$ is on line $BC$, the area of $ABC$ is", "the right is a positive. $$\\picture(100){(50,50){\\line(40,45)}(50,50){\\line(-40,45)}(50,50){\\line(-40,-45)}(50,50){\\line(40,-45)}}$$ The co-ordinate alignment process in TeX is not that good that you can use one set of co-ords as a single starting point for all lines. The layering of each object varies because of the position of the previous object, so each object needs to be exactly placed. This co-ord structure has a great deal of impact on intersecting lines, parallel lines and triangles. ## Intersecting Lines You can set up an intersecting pair easily enough, using the \\picture control sequence. $$\\picture(200){(10,0){\\line(150,150)} (0,130){\\line(180,-180)}}$$ The lines that are drawn can be labeled. $$\\picture(200){(10,0){\\line(150,150)}(0,130){\\line(180,-180)} (0,10){A}(0,135){B}(140,0){C}(140,150){D}(62,80){X}}$$ To produce another image. To which you may want to ask the question: $$The \\ \\angle \\ of \\ AXB \\ is \\ 72\\textdegree. \\ What \\ is \\ the \\ value \\ of \\ \\angle BXD?$$ NOTE: Labeling this image, above-right, turned out to be fairly simple. Offsetting points by a few pixels at the start or end points of the lines proved a successful strategy. The X point proved a little more problematic, and took a number of adjustments before getting it right. Experience here will help. With labels the drawing can become a little more like your traditional geometric drawing, but the devil is in the details. The parallel markers need to be placed properly, and", "that $\\angle EBC$ and $\\angle EAD$ are $150^\\circ$, $m(\\angle ADE) + m(\\angle AED) = m(\\angle CEB) + m(\\angle BCE) = 180 - 50 = 30^\\circ$. So, in particular, $m(\\angle AED) = m(\\angle CEB) = 15^\\circ$. So we have that $m(\\angle CED) = m(\\angle AEB) - 2m(\\angle AED) = 60 - 2(15) = 30^\\circ$. - J. M.: Thank you, I still have not found out how to insert a picture. Actually (not that it matters) it is not the labelling that I described, mine went counterclockwise, with $A$ at top right. – André Nicolas May 17 '11 at 20:57 @user6312: Yes, I do remember you mentioning counter-clockwise...oh, well, we do the best we can when the question is ambiguous leaves it to us to figure out: like \"point E outside of square\"...of course, since it is placed so as to form an equilateral triangle with A and B, there wasn't much choice! Inserting pix: If you click on \"answer\", there's a \"menu\" at the toptop (font, hyperlink, etc.) If you hover over the icons, you'll find one for \"image\"...click on it and you can then upload an image file (jpeg., etc), from your computer. – amWhy May 17 '11 at 21:21 It doesn't matter, though; your two solutions are just mirror images of each other. – Michael Lugo May", "x$, $\\scriptsize y$ and $\\scriptsize z$. Question 4 adapted from NC(V) Mathematics Level 4 Paper 2 November 2015 question 1.3 1. In the figure below $\\scriptsize O$ is the centre of the circle, $\\scriptsize AOD$ is a diameter and $\\scriptsize AE\\parallel CD$, with $\\scriptsize E$, $\\scriptsize C$ and $\\scriptsize B$ on the circumference of the circle. Also, $\\scriptsize \\hat{A}={{50}^\\circ}$. 1. Why is $\\scriptsize A\\hat{D}C={{50}^\\circ}$? 2. Determine the size of $\\scriptsize C\\hat{B}D$. 3. Prove that $\\scriptsize EOC$ is a straight line. The full solutions are at the end of the unit. # Unit 2: Solutions ### Exercise 2.1 1. . 1. $\\scriptsize x={{78}^\\circ}\\quad \\text{(}\\angle \\text{s in same seg)}$ 2. $\\scriptsize AB$ is a diameter (straight line passing through the centre) Therefore $\\scriptsize A\\hat{P}B={{90}^\\circ}\\quad (\\angle \\text{s in semi-circle)}$ \\scriptsize \\begin{align*}x&={{180}^\\circ}-{{90}^\\circ}-{{59}^\\circ}\\quad \\text{(}\\angle \\text{s in a }\\Delta \\text{)}\\\\&={{31}^\\circ}\\end{align*} 3. . \\scriptsize \\begin{align*}C\\hat{A}D & ={{31}^\\circ}\\quad \\text{(alt }\\angle \\text{s =; }AC\\parallel BD\\text{)}\\\\x=C\\hat{A}D & ={{31}^\\circ}\\quad \\text{(}\\angle \\text{s in same seg)}\\end{align*} 4. . \\scriptsize \\begin{align*}{{{\\hat{O}}}_{1}} & ={{204}^\\circ}\\quad \\text{(}\\angle \\text{ at centre =2}\\angle \\text{ at circumference)}\\\\\\therefore x & ={{360}^\\circ}-{{204}^\\circ}\\quad \\text{(}\\angle \\text{ }\\!\\!'\\!\\!\\text{ s round a point)}\\\\ & ={{156}^\\circ}\\end{align*} 5. $\\scriptsize x={{47}^\\circ}\\quad \\text{(}\\angle \\text{s in same seg)}$ \\scriptsize \\begin{align*}D\\hat{A}E & ={{180}^\\circ}-{{128}^\\circ}\\quad \\text{(}\\angle \\text{s on str line suppl)}\\\\ & ={{52}^\\circ}\\\\\\therefore y & ={{180}^\\circ}-{{52}^\\circ}-{{47}^\\circ}\\quad \\text{(}\\angle \\text{s in }\\Delta \\text{ suppl)}\\\\ &=8{{\\text{1}}^\\circ}\\end{align*} 2. $\\scriptsize OP=OB=6\\text{ units}\\quad \\text{(both radii)}$ \\scriptsize \\begin{align*}PR &", "(E) it can't be determined from the facts given Math Expert Joined: 02 Aug 2009 Posts: 8165 Re: In the diagram below an equilateral triangle CDE has been inscribed in [#permalink] ### Show Tags 19 Oct 2019, 21:33 2 shridhar786 wrote: Attachment: The attachment vf.PNG is no longer available In the diagram below an equilateral triangle, CDE has been inscribed inside a square ABCD. What is the value of the angle marked x°? (A) 15 (B) 30 (C) 45 (D) 75 (E) it can't be determined from the facts given See the attached figure.. In $$\\triangle CDE$$, all angles are 60, so $$\\angle EDC = 60$$ But $$\\angle ADC=90=\\angle ADE+\\angle EDC=\\angle ADE+60....\\angle ADE=30$$ Now take triangle ADE, where AD=DE, so ADE is an isosceles $$\\triangle$$....$$\\angle DAE=\\angle AED=x$$ So, x+x+30=180...2x=150...x=75 D Attachments vf.PNG [ 2.86 KiB | Viewed 452 times ] _________________ Director Status: Manager Joined: 27 Oct 2018 Posts: 720 Location: Egypt GPA: 3.67 WE: Pharmaceuticals (Health Care) Re: In the diagram below an equilateral triangle CDE has been inscribed in [#permalink] ### Show Tags 19 Oct 2019, 10:27 4 1 as CDE is equilateral, two conclusions can be made: (1) ∠EDC = 60 and ∠ADE = 30 (2) Triangle ADE is isosceles triangle where ∠DAE = ∠AED then ∠DAE = ∠AED = $$\\frac{180-30}{2} = 75$$ D _________________", "of the triangle is: $$\\frac{1}{2}$$ AD$$×$$BC and AD is perpendicular to BC. Triangle ADC is a (\\30^\\circ-60^\\circ- 90^\\circ\\) right triangle. The relationship among all sides of right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in the following triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. And the opposite side of $$60^\\circ$$ is opposite of $$30^\\circ×\\sqrt{3}$$ CD=6, then AD$$=6×\\sqrt{3}$$ Area of the triangle ABC is: $$\\frac{1}{2}$$ AD$$×$$BC $$= \\frac{1}{2} 6\\sqrt{3}×12=36\\sqrt{3}$$ $$\\img{https://appmanager.effortlessmath.com/public/images/questions/mmm.png }$$ 18- C $$6∎28=\\sqrt{6^2+28}= \\sqrt{36+28}=\\sqrt{64}=8$$ 19- D The union of A and B is: $$A∪B={1,2,4,8,12,16,24,32,48}$$. There are $$9$$ elements in $$A∪B$$. 20- D Solve the system of equations by elimination method. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ – x + 2 y = 10 \\end{align}}{} Multiply the second equation by 3, then add it to the first equation. \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ 3(- x + 2 y = 10) \\end{align}}{} ⇒ \\cfrac{\\begin{align} 3 x – 4 y = – 20 \\\\ -3 x + 6 y = 30) \\end{align}}{} ⇒add the equations $$2𝑦=10⇒𝑦=5$$ Looking for the best resource to help you succeed on the CLEP College Math test? 52% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or", "+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", "$\\triangle ABC$ 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", "the value of $$a = 54$$, 1. $$a-16^{\\circ} = 56^{\\circ} - 16^{\\circ} = 40^{\\circ}$$ 2. $$a+22^{\\circ} = 56^{\\circ} + 22^{\\circ} = 78^{\\circ}$$ 3. The third internal angle of the triangle can be given as $$180^{\\circ} - \\text{ exterior angle} = 180^{\\circ} - 118^{\\circ} = 62^{\\circ}$$ ........... [ $$\\because$$ exterior angle and the adjacent interior angle are supplementary] $$\\therefore$$ The measures of the angles of the given triangle are $$40^{\\circ}$$, $$78^{\\circ}$$, and $$62^{\\circ}$$. Example 3 Ryan is studying the applications of the exterior angle theorem. Consider the following figure. Can you help him find the value of $$\\angle{BAC}$$? Solution Solving the linear pair at vertex $$\\text D$$, \\begin{align*} \\angle{ADC} \\ + \\ \\angle{ADE} &= 180^\\circ \\\\ \\angle 1 &= 180^\\circ - 150^\\circ \\\\ &= 30^\\circ \\end{align*} Also, using angle sum property of a triangle, for $$\\triangle{ACD}$$, \\begin{align*} \\angle 1 + \\angle 1 + \\angle{ACD} &= 180^\\circ \\\\ \\angle{ACD} &= 180^\\circ - 2 \\times \\angle 1 \\\\ \\angle{ACD} &= 180^\\circ - 60^\\circ &= 120^\\circ \\end{align*} $$\\angle{ACD}$$ is the exterior angle for the$$\\triangle{ABC}$$. Thus, using exterior angle theorem for $$\\triangle{ABC}$$, \\begin{align*} \\angle{ACD} &= \\angle{ABC} + \\angle{BAC} \\\\ \\angle{BAC} &= \\angle{ACD} - \\angle{ABC} \\\\ \\angle{BAC} &= 120^\\circ - 60^\\circ \\\\ &= 60^\\circ \\end{align*} $$\\therefore \\angle{BAC} = 60^\\circ$$ ## Interactive Questions on Exterior Angle Theorem Here are a few activities for you to", "find $a$ and $A$. ###### Solution First, use $\\dfrac{b}{\\sin B} = \\dfrac{c}{\\sin C}$ to find $c$. \\begin{align} \\frac{6}{\\sin 77^{\\circ} } &= \\frac{c}{\\sin 34^{\\circ} }\\\\\\\\ \\frac{6 \\sin 34^{\\circ} }{\\sin 77^{\\circ} } &= c\\\\ c &= 3.44 \\text{ (to 3 sig.fig.)} \\end{align} To find $A$, use the fact that the angles in a triangle add up to $180^{\\circ}$, i.e. $A+B+C = 180$. \\begin{align} 180 &= A +77 + 34\\\\ A &= 180 - 77- 34\\\\ A &= 69^{\\circ} \\end{align} Now, use this value for $A$ in the sine rule to find $a$. \\begin{align} \\frac{a}{\\sin 69^{\\circ} } &= \\frac{6}{\\sin 77^{\\circ} }\\\\\\\\ a &= \\frac{6 \\sin 69^{\\circ} }{\\sin 77^{\\circ} }\\\\\\\\ a &= 5.75 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 3 Given $b=2$, $c=5$ and $A=39^{\\circ}$, find $a$. ###### Solution Use $a^2 = b^2 + c^2 -2bc\\cos A$. \\begin{align} a^2 &= 2^2 + 5^2 - (2\\times2\\times5)\\cos 39^{\\circ}\\\\ a^2 &= 29 - 20\\cos 39^{\\circ}\\\\ a^2 &= 13.457 \\text{ (to 3 d.p.)}\\\\ a &= 3.67 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 4 Given $a=7$, $b=6$ and $c=9$. Find the angle between $a$ and $b$. ###### Solution The angle between $a$ and $b$ is labelled $C$ in the diagram above, so use $c^2 = a^2 + b^2 - 2ab\\cos C$. \\begin{align} 9^2 &= 7^2 + 6^2 - (2\\times7\\times6)\\cos C\\\\ 81 &= 49+36 - 84\\cos" ]

[ "two numbers is 35. The larger number is one less than three times the smaller number. What are the two numbers? 3. Rachel offers to go to the coffee shop to buy cappuccinos and lattes for her coworkers. She buys a total of nine drinks for $35.75. If cappuccinos cost$3.75 each and the lattes cost \\$4.25 each, how many of each drink did she buy?", "Question Mary spent a total of $322.49 for a party. She spent$200.45 on food, plus an additional \\$30.51 for each hour of the party. How long was the party? A. 3 hours B. 6 hours C. 4 hours D. 5 hours", "## Three friends order a pizza for $14, a two-liter bottle of soda for$2.75 and bag of chips for $2.15 They want to evenly split the total cos Question Three friends order a pizza for$14, a two-liter bottle of soda for $2.75 and bag of chips for$2.15 They want to evenly split the total cost. Determine how much each person owes. in progress 0 1 week 2021-07-21T18:11:46+00:00 1 Answers 1 views 0 1. I think $6.30 When you add all of the prices you get$18.90 and when you divide by 3 you get your answer.", "two types apart, they could charge the single monopolist price for the low demanders since the high demanders will purchase at that price but the low demanders will not purchase at the high demander monopolist price. This would yield an average per-customer profit of$1.00 as all customers would buy a 10oz. soft drink for $1.50. Alternatively, if they only offered the large drink, only high demanders would buy and though they would make$2.25 in profit on those sales, only 1/5 of the customers will buy so the average profit per customer would be $0.45. It is also immediately clear that if the store offered both deals, no one would buy a 15oz. soda at$3 as for the same price they could buy two 10oz. soft drinks or 20oz. of soft drink. But what if the store offered a volume discount on soft drink, could they get the high demanders to voluntarily switch to a larger drink and make more money in the process? Consider the following deal: anyone can buy a 10oz. soft drink for $1.50, or a 20oz. soda for$2.40. This second offer is a volume discount as the price per ounce has dropped from $0.15 in the 10oz. case to$0.12 in the 20oz. case. But would this work? Well, low types would only buy 20oz. of", "# Popeye's, a local convenience store, sells soft drinks It sells two large drinks for every small ... Popeye's, a local convenience store, sells soft drinks It sells two large drinks for every small drink. A large drink sells for $3.00 with a variable cost of$0.80. A small drink sells for $1.25 with a variable cost of$0.50. The weighted average contribution margin is ___________. (Round any intermediate calculations and your final answer to the nearest cent.) A. $2.20 per drink B.$5.15 per drink C. $1.48 per drink D.$1.72 per drink", "### Pizza Portions My friends and I love pizza. Can you help us share these pizzas equally? ### Pies Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie? ### Red Balloons, Blue Balloons Katie and Will have some balloons. Will's balloon burst at exactly the same size as Katie's at the beginning of a puff. How many puffs had Will done before his balloon burst? # Orange Drink ##### Stage: 2 Challenge Level: How much water is needed to make one glass of the drink? How many glasses of drink are there in a litre? In $10$ litres? What fraction of the made-up drink is water?", "Pizza Portions My friends and I love pizza. Can you help us share these pizzas equally? Pies Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie? Red Balloons, Blue Balloons Katie and Will have some balloons. Will's balloon burst at exactly the same size as Katie's at the beginning of a puff. How many puffs had Will done before his balloon burst? Orange Drink Stage: 2 Challenge Level: This is a $750$ ml bottle of concentrated orange squash. It is enough to make fifteen $250$ ml glasses of diluted orange drink. How much water is needed to make $10$ litres of this drink?", "Step 6: For the macchiato only \\$3.40 is left, which leaves \\$3.60 for the mocha. Step 7: The macchiato costs \\$3.40, so according to statement 2 Darrell's drink costs \\$3.50, which is the iced americano. This leaves the drink for \\$3.40 for Fredrick, which is the macchiato. Step 8: We already know the cappuccino costs \\$3.30 and was ordered by Jacob. Bryan can only have the mocha for \\\\$3.60. • Nice job! The logic was really hard. If you do an explanation, i'll do +1. Amazing! – Alto Oct 28 '18 at 3:03 • but... but.. too lazy! – Chowzen Oct 28 '18 at 11:18 • I'll add some explanation when I have the time to. – formica Oct 28 '18 at 13:48", "be bought at that price (per soda)\". So if the quantity is 100 and the price is \\$1.5 the total dollar amount will be \\$150, but the price is still \\$1.5. In this last case, we say that \"at a price of \\$1.5, 100 units are demanded\", and conversely you can say that \"preferences of consumers are such that exactly 100 units are associated with a willingness to pay of at least \\$1.5\" (no-one uses the last one, but it may help you connect all the concepts).", "=. ( x1 ; x2 ) = x^ ( 1/2 ) a drink costs$ 1 each, in literature! Kathy should consume [ { Blank } ] units... how to obtain a function. At his part-time job in the last apple consumed was 4 and $2 per.. Please explain why... Uvaldo is deciding which brand of energy drink buy. Then by computing an example choose which of the two goods in the following diagram his. And Moorhouse ( 1984 ) suggested this problem that is amenable to online implementation his funds on and. Entitled “ Lagrange Multiplier problems in our framework include the shortfall minimization problems Economics. Per person to give up units of good X, Y ) is 500 \\geq 5x 10Y! Robert 's original optimal consumption bundle ( X and Y are$ 4 pound... Consume even if MB=MC and only stop if MB is greater than MC of us households with access to cinema..., PX= $15 per unit and py =$ 270 spend a total expenditure of 11?! Solution for this problem that is amenable to online implementation you hav... georgina consumes only grapefruits and pineapples initial... Both goods 4 per pound and falafels and spends $210 per month at his part-time in! Pony is$ 3 to Midterm 1 problem 1 questions utility maximization problems and solutions", "### Home > INT3 > Chapter 3 > Lesson 3.2.1 > Problem3-79 3-79. Daniela, Kieu, and Duyen decide to go to the movies one hot summer afternoon. The theater is having a summer special called Three Go Free. They will get free movie tickets if they each buy a large popcorn and a large soft drink. They take the deal and spend a total of $\\22.50$ on large popcorns and soft drinks. The next week, they go back again, only this time, they each pay $\\8.00$ for a ticket, they each get a large soft drink, but they share one large bucket of popcorn. This return trip costs them a total of $\\37.50$. 1. What is the price of a large soft drink and the price of a large bucket of popcorn? Set up a system of two equations to find the prices of a large popcorn and a drink. popcorn $=\\4.50$ and drink $=\\3.00$ 2. Did you write two equations or did you use another method? If you used another method, write two equations now and solve them. If you already used a system of equations, skip this part.", "chocolate cake and 1 slice of carrot cake. Julia ate 1/2 of the carrot cake. Mark ate 1 slice of each. Thomas ate 3 slices of chocolate cake. How many slices were left? • Barbara 2 Barbara get 6 pizzas to divide equally among 4 people. How much of a pizza can each person have? • Four people Four people want waffles for breakfast. There are 6 waffles left. How can 6 waffles be shared equally among 4 people? How much does each person get? Draw a picture and write a division expression to model the problem. • Unit cost Last Friday, the Math Society sold newspapers to raise money. Each pack weighed 65.5 kg and sold for 281.65 pesos. How much was the cost per kilogram? • Cents no more Janko bought pencils for 35 cents each. Neither he nor the salesperson had small coins, just a whole € 1 coin. At least how many pencils had to buy to pay for the whole euros? • Jerry Jerry has 3/4 of a pizza. He needs to share it with 6 friends. What fraction of the pizza will each friend get? Only write the fraction • Dera bought Dera bought a set of 2 files and 5 notebooks for a special offer of $12.95. if the usual price", "# Cindy Cindy bought four slices of pizza for Php 95.60. Whole pizza has 6 slices. How much does each slice cost? s = 23.9 PHP ### Step-by-step explanation: Did you find an error or inaccuracy? Feel free to write us. Thank you! ## Related math problems and questions: • A pizza 2 A pizza is cut into slices that are each 1/6 of the whole. John is going to eat 1/2 of the whole pizza. How many slices will John eat? • Slices of pizza Maria ate 1/4 of a pizza. If there were 20 slices of pizza, how many slices did Maria eat? • Pizza 5 You have 2/4 of a pizza, and you want to share it equally between 2 people. How much pizza does each person get? • Grayson Grayson bought snacks for his team's practice. He bought a bag of oranges for $3.11 and a 24-pack of juice bottles. The total cost before tax was$50.15. How much was each bottle of juice? • ZOO 2 Valerie bought tickets to the zoo for the family. She bought 3 adult tickets for $9.50 each and 2 children's tickets for$4.50 each. How much did the tickets cost Valerie in all? • Two cakes Two cakes were each cut into 8 slices. Maria ate 1/8 of the", "loose a lot of weight by not purchasing from these idiots. I knew this thread would be full of foodies who hate pizza from chains for no other reason than it's from a chain. And that they would put pizza in quotes and call it cardboard and parrot other worn-out phrases. I only had to read the first ten posts or so to confirm my suspicions. It's probably already been said, but if you have Papa John's for every meal, every day, it will cost an extra $120.45 (at 11 cents). If that pays for one of their employees to be covered by insurance, then I'm okay with that. I don't care if it's an additional$10 per pizza and $15 per six pack of Miller Lite. It's all worth it if I understand the Affordable Care Act correctly. My best estimate is that Papa Johns inflicted singring and dysentery costs taxpayers$.20 per pie in unpaid healthcare costs. In any event, what a whiny cockpotato. Oldest | « | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | » | Newest | Show all View Voting Results: Smartest and Funniest In Other Media 1. Links are submitted by members of the Fark", "a watch. If the total cost of the handbag, the dress and the watch is$z, find the cost of the dress in terms of z. 2 m Level 2 May and Dora had $n altogether. After May gave$12 to Dora, each of them had the same amount of money. How much did Dora have at first? 2 m Level 2 The circles are not drawn to scale. The small circle is with radius a and the big circle is with radius A. The ratio of a to A is 4 : 7. If the area of the small circle is 169π cm2, find A. Express the answer in mixed number. 2 m Level 2 Mary baked 400 cookies. She ate 2p of them and divided the remainder equally among her 4 friends. 1. Find the number of cookies each of her friends received in terms of p? 2. If p = 12, how many cookies did each of her friends receive? 2 m Level 2 PSLE A bottle completely filled with water weighed 1.9 kg. After 23 of the water was used up, the bottle with the remaining water weighed 800 g. What was the mass of the empty bottle in grams? 2 m", "# Miguel spent more than $6 at a fast food restaurant. If he had$25 to start, what was his remaining pocket money? Less than $19 #### Explanation: $25-$6 =$19 If Miguel had spent $6, he would have$19 left, but because he spent more than \\$6, the remainder would have to be less than 19.", "Poo P. # You and a group of 7 friends go out for pizza. One large pizza and 3 small drinks cost $21.96. You and a group of 7 friends go out for pizza. One large pizza and 3 small drinks cost$21.96. One large pizza and 6 small drinks cost \\$27.93. Write an equation that models the total cost y (in dollars) for one large pizza and any number of drinks x. Use the equation to estimate the cost of one large pizza and 8 small drinks for you and your friends. Please help this will determine my grade! By: Tutor 4.9 (285) Masters' Degree in Mathematics with 35 Years of Teaching Experience ## Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. #### OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "### Home > GC > Chapter 12 > Lesson 12.1.2 > Problem12-21 12-21. Multiple Choice: As Carol shopped for a spring picnic, she spent $2.00 for each liter of soda and$3.50 for each bag of chips. In all, she bought 18 items for a total of \\$43.50. Assuming she only bought chips and soda, how many bags of chips did she buy? Homework Help ✎ a. 9 b. 5 c. 15 d. 3 Write two equations: one representing the number of items bought and the other the total price. Solve using the substitution method. See the online help for problem 5-104, which contains the Math Notes box in Lesson 2.1.3.", "which were 3/4 of original price. How many SKK originally cost windbreaker? How many SKK was cost reduction? • The cost 2 The cost of 5 apples is $3.45, and 5 oranges are$1.23. If Rachel buys one apple and one orange, then how much must she pay? • Lenka and money When Lenka bought pants from 1/3 of her savings shirt from 1/5 and 1/8 spent at the hairdresser, she was left 328 Kč. How much money was saved? How many hairdressers cost? • Money 6 Lita had some money. She spent 1/2 of it on a hand bag and 1/3 of the remainder of the blouse. She had 260.00. How much money did she have at start? • Pen and pencil Cost of a pen is Rs. 12 3/5 and cost of a pencil is Rs. 3 2/5. What is the total cost of pen and pencil? • 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 • Savings Eva borrowed 1/3 of her savings to her brother, 1/2 of savings spent", "# A combinatorics problem I can't quite understand A pizzeria offers 777 types of pizza and 3 types of soda. Mary goes there everyday for lunch, always buying one slice of pizza and one soda. However, she never gets exactly the same thing on two consecutive days (that is, each time, either the drink or the pizza (or both) is different from what she had yesterday). In how many ways can she plan her lunch for the next 15 days if today she tried a different pizzeria and did not like that place at all? Answer: approximately $3.240 × 10^{ 50 }$ (but you should try to find the exact formula, not an approximation). I'm not sure how to proceed with this question. The caveat in the second part of the question, that Mary tries a different pizzaria, makes no sense to me. My intuition is to calculate what the permutation of what it would look like if Mary just switched up pizza slices (which I think would be [777 * 776 * 775 * 777 * 776 * 775 * 777 * 776 * 775 * 777 * 776 * 775 * 777 * 776 * 775], though maybe the pattern would be 777 * 776 * 775... instead?), the permutation of Mary switching up types of" ]

[ "Question Mary spent a total of $322.49 for a party. She spent$200.45 on food, plus an additional \\$30.51 for each hour of the party. How long was the party? A. 3 hours B. 6 hours C. 4 hours D. 5 hours", "# Cindy Cindy bought four slices of pizza for Php 95.60. Whole pizza has 6 slices. How much does each slice cost? s = 23.9 PHP ### Step-by-step explanation: Did you find an error or inaccuracy? Feel free to write us. Thank you! ## Related math problems and questions: • A pizza 2 A pizza is cut into slices that are each 1/6 of the whole. John is going to eat 1/2 of the whole pizza. How many slices will John eat? • Slices of pizza Maria ate 1/4 of a pizza. If there were 20 slices of pizza, how many slices did Maria eat? • Pizza 5 You have 2/4 of a pizza, and you want to share it equally between 2 people. How much pizza does each person get? • Grayson Grayson bought snacks for his team's practice. He bought a bag of oranges for $3.11 and a 24-pack of juice bottles. The total cost before tax was$50.15. How much was each bottle of juice? • ZOO 2 Valerie bought tickets to the zoo for the family. She bought 3 adult tickets for $9.50 each and 2 children's tickets for$4.50 each. How much did the tickets cost Valerie in all? • Two cakes Two cakes were each cut into 8 slices. Maria ate 1/8 of the", "total amount of money left with her?", "Step 6: For the macchiato only \\$3.40 is left, which leaves \\$3.60 for the mocha. Step 7: The macchiato costs \\$3.40, so according to statement 2 Darrell's drink costs \\$3.50, which is the iced americano. This leaves the drink for \\$3.40 for Fredrick, which is the macchiato. Step 8: We already know the cappuccino costs \\$3.30 and was ordered by Jacob. Bryan can only have the mocha for \\\\$3.60. • Nice job! The logic was really hard. If you do an explanation, i'll do +1. Amazing! – Alto Oct 28 '18 at 3:03 • but... but.. too lazy! – Chowzen Oct 28 '18 at 11:18 • I'll add some explanation when I have the time to. – formica Oct 28 '18 at 13:48", "two numbers is 35. The larger number is one less than three times the smaller number. What are the two numbers? 3. Rachel offers to go to the coffee shop to buy cappuccinos and lattes for her coworkers. She buys a total of nine drinks for $35.75. If cappuccinos cost$3.75 each and the lattes cost \\$4.25 each, how many of each drink did she buy?", "### Home > GC > Chapter 12 > Lesson 12.1.2 > Problem12-21 12-21. Multiple Choice: As Carol shopped for a spring picnic, she spent $2.00 for each liter of soda and$3.50 for each bag of chips. In all, she bought 18 items for a total of \\$43.50. Assuming she only bought chips and soda, how many bags of chips did she buy? Homework Help ✎ a. 9 b. 5 c. 15 d. 3 Write two equations: one representing the number of items bought and the other the total price. Solve using the substitution method. See the online help for problem 5-104, which contains the Math Notes box in Lesson 2.1.3.", "costs $11.65. How much does he need to pay? A -$92.20 B - $93.30 C -$94.20 D - $93.20 ### Answer : D ### Explanation Step 1: Cost of each pizza =$11.65 Number of pizzas = 8 Step 2: Total amount spent = 8 × 11.65 = $93.20 Q 4 - Julie earns$8.60 per hour at her baby-sitting job. On Saturday, she worked 5.5 hours. What were Julie’s total earnings for the day? A - $48.30 B -$46.30 C - $47.30 D -$47.40 ### Explanation Step 1: Earnings per hour of baby-sitting = $8.60 Total number of hours worked = 5.5 Step 2: Total earnings = 5.5 × 8.60 =$47.30 Q 5 - What is the product of 127.8 x 43.32? A - 5538.296 B - 5536.296 C - 5536.396 D - 5546.296 ### Explanation 127.8 × 43.32 = 5536.296 Q 6 - A bottle of soda had 2.35 of the recommended amount of sugar. If Peter drinks 0.8 of the bottle, how much of the recommended amount sugar Peter must have drunk? A - 1.88 B - 1.98 C - 1.86 D - 1.84 ### Explanation Step 1: Sugar contained in a bottle of soda = 2.35 recommended amount of sugar Quantity of soda taken = 0.8 Step 2: So the amount of sugar in the soda", "# Mary and Tina spent a total of $175 at the mall. Tina spent$10 more than twice as much as Mary. How much did tina spend? I found $120 #### Explanation: Call the amount of money spent by Mary and Tina as $M$and $T$you get: $M + T = 175$$T = 10 + 2 M$substitute the second into the first: $M + 10 + 2 M = 175$solve for $M$: $M = \\frac{165}{3} = 55 $T = 10 + 2 \\cdot 55 = 120$ \\$", "## Melanie bought a soft drink for 4 dollars and 5 candy bars. She spent a total of 29 dollars. How much did each candy bar cost? What is the e Question Melanie bought a soft drink for 4 dollars and 5 candy bars. She spent a total of 29 dollars. How much did each candy bar cost? What is the equation for the problem? ​ in progress 0 7 months 2021-07-21T12:50:06+00:00 1 Answers 1 views 0 ## Answers ( ) 1. Answer: Each candy bar cost’s $5 Step-by-step explanation: Equation is 4+ 5x=29 -4 -4 which makes it 5x= 25 then you divide by 5 which makes it x=5 so it costs 5$ per candy bar", "Applications In the following exercises, use the strategy for applications to solve. 1. Spending money Brenda got $40 from the ATM. She spent$15.11 on a pair of earrings. How much money did she have left? 2. Spending money Marissa found $20 in her pocket. She spent$4.82 on a smoothie. How much of the $20 did she have left? 3. Shopping Adam bought a t-shirt for$18.49 and a book for $8.92 The sales tax was$1.65. How much did Adam spend? 4. Restaurant Roberto’s restaurant bill was $20.45 for the entrée and$3.15 for the drink. He left a $4.40 tip. How much did Roberto spend? 5. Coupon Emily bought a box of cereal that cost$4.29. She had a coupon for $0.55 off, and the store doubled the coupon. How much did she pay for the box of cereal? 6. Coupon Diana bought a can of coffee that cost$7.99. She had a coupon for $0.75 off, and the store doubled the coupon. How much did she pay for the can of coffee? 7. Diet Leo took part in a diet program. He weighed 190 pounds at the start of the program. During the first week, he lost 4.3 pounds. During the second week, he had lost 2.8 pounds. The third week, he gained 0.7 pounds. The fourth week, he lost 1.9", "Melanie bought a soft drink for 4 dollars and 5 candy bars. She spent a total of 29 dollars. How much did each candy bar cost? What is the e Question Melanie bought a soft drink for 4 dollars and 5 candy bars. She spent a total of 29 dollars. How much did each candy bar cost? What is the equation for the problem? ​ in progress 0 1 day 2021-07-21T12:50:06+00:00 1 Answers 0 views 0 1. Answer: Each candy bar cost’s $5 Step-by-step explanation: Equation is 4+ 5x=29 -4 -4 which makes it 5x= 25 then you divide by 5 which makes it x=5 so it costs 5$ per candy bar", "have left from the$30 she took to the café? A. Step 1: Add the price of a muffin to the price of a drink. Step 2: Multiply that total by 4. Step 3: Subtract the total amount from $30. Step 4: Add the tip. B. Step 1: Add the price of a drink and the tip. Step 2: Add the price of a muffin to the total from Step 1. Step 3: Multiply that total by 4. Step 4: Subtract that total amount from$30. C. Step 1: Calculate the price of 4 drinks. Step 2: Calculate the price of 4 muffins. Step 3: Add the total amounts from Steps 1 and 2 to $2.30. Step 4: Subtract that total amount from$30. Multiply $1.99 by 4 for the drinks . Then$1.20 by 4 for the muffins . Add both of the amounts from the drinks and muffins with the tip of $2.30 . Then subtract$30 from the other total . Then you have your answer . ANSWERED AT 31/03/2020 - 10:45 AM QUESTION POSTED AT 31/03/2020 - 10:45 AM ### Over the last 12 month, Petco has lost a total of $1824.00 due to shoplifting. Write and solve an equation that shows how much money Petco lost each month. ( Since they have lost the money, remember that you", "a similar fashion it will appear that second person gave $$\\frac{1}{3}$$ ($$3-\\frac{8}{3}=\\frac{1}{3})$$ to third person. Hence $$7$$ and $$1$$ because first gave $$7$$ times more. But this solution just doesn't click for me for some reasons, can someone break it down... if it makes sense to you? Particularly why answer assumes first person gave $$\\frac{7}{3}$$ to third person, what if he gave it to second person? • Welcome to MSE. Please edit and use MathJax to properly format numbers and math expressions. – Lee David Chung Lin Apr 23 at 21:08 Here's a way to think about the problem without doing fractions. Imagine a fourth person, the \"bartender,\" who buys the $$8$$ bottles of wine and then sells the wine back to the three people, at cost, equally to each person, as drinks. Since the third person pays $$8$$ dollars, so will the other two, for a total of $$24$$ dollars. So each bottle wine costs $$3$$ dollars. Thus the bartender pays the first person $$15$$ dollars for $$5$$ bottles and the other person $$9$$ dollars for $$3$$ bottles. After chipping in their $$8$$ dollars each for the repurchase of the wine (served as drinks), the first person is left with $$7$$ dollars and the second person is left with $$1$$ dollar. The ratio of the payments made", "three sizes: the 8 ounce drink costs $1.50, the 12 ounce drink,$2.00, and the 16 ounce drink, $2.25. Describe the budget constraint that Debra faces when deciding how many ounces of the drink to purchase. (Assume that Debra can costlessly dispose of any of the soft drink that she does not want. First notice that as the size of the drink increases, the price per ounce decreases.$1.50 When she buys the 8-ounce soft drink she pays = $0.19 per oz. When 8 oz she buys the 12-ounce size she pays$0.17 per ounce, and when she buys the 16- ounce size, she pays $0.14 per ounce. Given that there are three different prices per ounce of soft drink, the budget line will have two kinks in it, as illustrated below. Notice that at each kink, the slope of the budget line gets flatter (due to the decreasing cost per ounce relative to the “other good” on the vertical axis). 28 7. Chapter 3: Consumer Behavior 8 12 16 Ounces of Soft Drink 10. Antonio buys 5 new college textbooks during his first year at school at a cost of$80 each. Used books cost only $50 each. When the bookstore announces that there will be a 10% increase in the price of new books and a 5% increase in the", "her money did Maria save? The fraction of her money did Maria save = 3/7. Explanation: In the above-given question, given that, Maria spent $$\\frac{4}{7}$$ of her money on a book and saved the rest. 7/7 – 4/7. 3/7. so the fraction of her money did Maria save = 3/7. Question 4. Mrs. Jones had 1$$\\frac{4}{8}$$ pizzas left after a party. After giving some to Gary, she had $$\\frac{7}{8}$$ pizza left. What fraction of a pizza did she give Gary? The fraction of pizza did she give Gary = 5/8. Explanation: In the above-given question, given that, Mrs. Jones had 1$$\\frac{4}{8}$$ pizzas left after a party. After giving some to Gary, she had $$\\frac{7}{8}$$ pizza left. 1(4/8) = 12/8. 12/8 – 7/8. 5/8. the fraction of pizza did she give Gary = 5/8. Question 5. A baker had 2 pans of corn bread. He served 1$$\\frac{1}{4}$$ pans. What fraction of a pan was left? The fraction of a pan was left = 3/4. Explanation: In the above-given question, given that, He served 1$$\\frac{1}{4}$$ pans. 1(1/4). 5/4 – 2/4. 3/4. the fraction of a pan was left = 3/4. Question 6. Marius combined $$\\frac{4}{8}$$ gallon of lemonade, $$\\frac{3}{8}$$ gallon of cranberry juice, and $$\\frac{6}{8}$$ gallon of soda water to make punch for a party. How many gallons of punch", "### Home > CCG > Chapter 12 > Lesson 12.1.4 > Problem12-56 12-56. Multiple Choice: Carol was shopping for a spring picnic. She spent $2.00$ for each liter of soda and $3.50$ for each bag of chips. In all, she bought $18$ items for a total of $43.50$. Assuming she only bought chips and soda, how many bags of chips did she buy? 1. $9$ 1. $5$ 1. $15$ 1. $3$ Write two equations: one representing the number of items bought and the other the total price. Solve using the substitution method. See the online help for problem 5-104, which contains the Math Notes box in Lesson 2.1.3.", "## Three friends order a pizza for $14, a two-liter bottle of soda for$2.75 and bag of chips for $2.15 They want to evenly split the total cos Question Three friends order a pizza for$14, a two-liter bottle of soda for $2.75 and bag of chips for$2.15 They want to evenly split the total cost. Determine how much each person owes. in progress 0 1 week 2021-07-21T18:11:46+00:00 1 Answers 1 views 0 1. I think $6.30 When you add all of the prices you get$18.90 and when you divide by 3 you get your answer.", "friend 7 cents. Now he has 51 cents in the wallet. How many cents previously had in his wallet? • Gasoline price At 34.4 cents per liter, how much will 42 liters of gasoline cost? • Trickster Figliarko has 13 cents less than Hana. Together they have 57 cents. How many cents has Hana and how many Figliarko? • Parcel 4 Send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=4 pounds? • Birthday party For her youngest son's birthday party, mother bought 6 3/4 kg of hotdog and 5 1/3 dozens bread rolls. Hotdogs cost 160 per kilogram and a dozen bread roll costs 25. How much did she spend in all? • Rolls Mom every day buys 6 rolls. On Friday buys 2 times as much. How many rolls buys on Friday? • Vacation Smiths paid a deposit for a vacation of two-sevenths of the total price of the vacation. Then paid also € 550. How much cost their vacation? • Apples and pears Apples cost 50 cents piece, pears 60 cents piece, bananas cheaper than pears. Grandma bought 5 pieces of fruit. There was only one banana, and I paid 2", "maximum number of sandwiches is $6$. The total money spent is $6\\cdot 4.50=27$. The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items. Hence, the answer is $\\boxed{\\textbf{(D) }9}$. - by interactivemath ## Video Solution The Learning Royal: https://youtu.be/IiFFDDITE6Q ## Video Solution 2 Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2 ~savannahsolver", "0.9 − 0.05 ) ( 1.43 + 0.27 ) ÷ ( 0.9 − 0.05 )$ 174. $( 1.5 − 0.06 ) ÷ ( 0.12 + 0.24 ) ( 1.5 − 0.06 ) ÷ ( 0.12 + 0.24 )$ 175. $[ 75.42 + 0.18 ( 75.42 ) ] ÷ 5 [ 75.42 + 0.18 ( 75.42 ) ] ÷ 5$ 176. $[ 56.31 + 0.22 ( 56.31 ) ] ÷ 4 [ 56.31 + 0.22 ( 56.31 ) ] ÷ 4$ Use Decimals in Money Applications In the following exercises, use the strategy for applications to solve. 177. Spending money Brenda got $4040$ from the ATM. She spent $15.1115.11$ on a pair of earrings. How much money did she have left? 178. Spending money Marissa found $2020$ in her pocket. She spent $4.824.82$ on a smoothie. How much of the $2020$ did she have left? 179. Shopping Adam bought a t-shirt for $18.4918.49$ and a book for $8.928.92$ The sales tax was $1.65.1.65.$ How much did Adam spend? 180. Restaurant Roberto’s restaurant bill was $20.4520.45$ for the entrée and $3.153.15$ for the drink. He left a $4.404.40$ tip. How much did Roberto spend? 181. Coupon Emily bought a box of cereal that cost $4.29.4.29.$ She had a coupon for $0.550.55$ off, and the store doubled the coupon. How much" ]

[ "Question Mary spent a total of $322.49 for a party. She spent$200.45 on food, plus an additional \\$30.51 for each hour of the party. How long was the party? A. 3 hours B. 6 hours C. 4 hours D. 5 hours", "two numbers is 35. The larger number is one less than three times the smaller number. What are the two numbers? 3. Rachel offers to go to the coffee shop to buy cappuccinos and lattes for her coworkers. She buys a total of nine drinks for $35.75. If cappuccinos cost$3.75 each and the lattes cost \\$4.25 each, how many of each drink did she buy?", "Step 6: For the macchiato only \\$3.40 is left, which leaves \\$3.60 for the mocha. Step 7: The macchiato costs \\$3.40, so according to statement 2 Darrell's drink costs \\$3.50, which is the iced americano. This leaves the drink for \\$3.40 for Fredrick, which is the macchiato. Step 8: We already know the cappuccino costs \\$3.30 and was ordered by Jacob. Bryan can only have the mocha for \\\\$3.60. • Nice job! The logic was really hard. If you do an explanation, i'll do +1. Amazing! – Alto Oct 28 '18 at 3:03 • but... but.. too lazy! – Chowzen Oct 28 '18 at 11:18 • I'll add some explanation when I have the time to. – formica Oct 28 '18 at 13:48", "## Three friends order a pizza for $14, a two-liter bottle of soda for$2.75 and bag of chips for $2.15 They want to evenly split the total cos Question Three friends order a pizza for$14, a two-liter bottle of soda for $2.75 and bag of chips for$2.15 They want to evenly split the total cost. Determine how much each person owes. in progress 0 1 week 2021-07-21T18:11:46+00:00 1 Answers 1 views 0 1. I think $6.30 When you add all of the prices you get$18.90 and when you divide by 3 you get your answer.", "have left from the$30 she took to the café? A. Step 1: Add the price of a muffin to the price of a drink. Step 2: Multiply that total by 4. Step 3: Subtract the total amount from $30. Step 4: Add the tip. B. Step 1: Add the price of a drink and the tip. Step 2: Add the price of a muffin to the total from Step 1. Step 3: Multiply that total by 4. Step 4: Subtract that total amount from$30. C. Step 1: Calculate the price of 4 drinks. Step 2: Calculate the price of 4 muffins. Step 3: Add the total amounts from Steps 1 and 2 to $2.30. Step 4: Subtract that total amount from$30. Multiply $1.99 by 4 for the drinks . Then$1.20 by 4 for the muffins . Add both of the amounts from the drinks and muffins with the tip of $2.30 . Then subtract$30 from the other total . Then you have your answer . ANSWERED AT 31/03/2020 - 10:45 AM QUESTION POSTED AT 31/03/2020 - 10:45 AM ### Over the last 12 month, Petco has lost a total of $1824.00 due to shoplifting. Write and solve an equation that shows how much money Petco lost each month. ( Since they have lost the money, remember that you", "a similar fashion it will appear that second person gave $$\\frac{1}{3}$$ ($$3-\\frac{8}{3}=\\frac{1}{3})$$ to third person. Hence $$7$$ and $$1$$ because first gave $$7$$ times more. But this solution just doesn't click for me for some reasons, can someone break it down... if it makes sense to you? Particularly why answer assumes first person gave $$\\frac{7}{3}$$ to third person, what if he gave it to second person? • Welcome to MSE. Please edit and use MathJax to properly format numbers and math expressions. – Lee David Chung Lin Apr 23 at 21:08 Here's a way to think about the problem without doing fractions. Imagine a fourth person, the \"bartender,\" who buys the $$8$$ bottles of wine and then sells the wine back to the three people, at cost, equally to each person, as drinks. Since the third person pays $$8$$ dollars, so will the other two, for a total of $$24$$ dollars. So each bottle wine costs $$3$$ dollars. Thus the bartender pays the first person $$15$$ dollars for $$5$$ bottles and the other person $$9$$ dollars for $$3$$ bottles. After chipping in their $$8$$ dollars each for the repurchase of the wine (served as drinks), the first person is left with $$7$$ dollars and the second person is left with $$1$$ dollar. The ratio of the payments made", "# Cindy Cindy bought four slices of pizza for Php 95.60. Whole pizza has 6 slices. How much does each slice cost? s = 23.9 PHP ### Step-by-step explanation: Did you find an error or inaccuracy? Feel free to write us. Thank you! ## Related math problems and questions: • A pizza 2 A pizza is cut into slices that are each 1/6 of the whole. John is going to eat 1/2 of the whole pizza. How many slices will John eat? • Slices of pizza Maria ate 1/4 of a pizza. If there were 20 slices of pizza, how many slices did Maria eat? • Pizza 5 You have 2/4 of a pizza, and you want to share it equally between 2 people. How much pizza does each person get? • Grayson Grayson bought snacks for his team's practice. He bought a bag of oranges for $3.11 and a 24-pack of juice bottles. The total cost before tax was$50.15. How much was each bottle of juice? • ZOO 2 Valerie bought tickets to the zoo for the family. She bought 3 adult tickets for $9.50 each and 2 children's tickets for$4.50 each. How much did the tickets cost Valerie in all? • Two cakes Two cakes were each cut into 8 slices. Maria ate 1/8 of the", "### Home > GC > Chapter 12 > Lesson 12.1.2 > Problem12-21 12-21. Multiple Choice: As Carol shopped for a spring picnic, she spent $2.00 for each liter of soda and$3.50 for each bag of chips. In all, she bought 18 items for a total of \\$43.50. Assuming she only bought chips and soda, how many bags of chips did she buy? Homework Help ✎ a. 9 b. 5 c. 15 d. 3 Write two equations: one representing the number of items bought and the other the total price. Solve using the substitution method. See the online help for problem 5-104, which contains the Math Notes box in Lesson 2.1.3.", "one of which is actually \"fair price\", since the cost per drink is an even number of cents) 11.78 (drinker 3 didn't pay extra because his one 0.5 drink was at a \"fair price\") 7.985 (drinker 4 paid 0.5 extra for his one 0.5 drink). 3. ## Re: Simultaneous system... with a twist! Your help would be good... Ah-ha! I see. Roundings affect the result massiviely don't they? Thanks for pointing this out....", "friend 7 cents. Now he has 51 cents in the wallet. How many cents previously had in his wallet? • Gasoline price At 34.4 cents per liter, how much will 42 liters of gasoline cost? • Trickster Figliarko has 13 cents less than Hana. Together they have 57 cents. How many cents has Hana and how many Figliarko? • Parcel 4 Send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=4 pounds? • Birthday party For her youngest son's birthday party, mother bought 6 3/4 kg of hotdog and 5 1/3 dozens bread rolls. Hotdogs cost 160 per kilogram and a dozen bread roll costs 25. How much did she spend in all? • Rolls Mom every day buys 6 rolls. On Friday buys 2 times as much. How many rolls buys on Friday? • Vacation Smiths paid a deposit for a vacation of two-sevenths of the total price of the vacation. Then paid also € 550. How much cost their vacation? • Apples and pears Apples cost 50 cents piece, pears 60 cents piece, bananas cheaper than pears. Grandma bought 5 pieces of fruit. There was only one banana, and I paid 2", "# Difference between revisions of \"2019 AMC 8 Problems/Problem 1\" ## Problem 1 Ike and Mike go into a sandwich shop with a total of $30.00$ to spend. Sandwiches cost $4.50$ each and soft drinks cost $1.00$ each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy? $\\textbf{(A) }6\\qquad\\textbf{(B) }7\\qquad\\textbf{(C) }8\\qquad\\textbf{(D) }9\\qquad\\textbf{(E) }10$ ## Solution 1 We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$. Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\\boxed{\\textbf{(D) }9}$. - SBose ## Solution 2 (Using Algebra) Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of $$4.50s+d=30$$ In the question, it states that Ike and Mike buys as many sandwiches as possible. So, we drop the number of sodas for a while. We have: \\begin{align*} 4.50s&=30 \\\\ s&=\\frac{30}{4.5} \\\\ s&=6R3 \\end{align*} We don't want a remainder so the", "### Home > INT3 > Chapter 3 > Lesson 3.2.1 > Problem3-79 3-79. Daniela, Kieu, and Duyen decide to go to the movies one hot summer afternoon. The theater is having a summer special called Three Go Free. They will get free movie tickets if they each buy a large popcorn and a large soft drink. They take the deal and spend a total of $\\22.50$ on large popcorns and soft drinks. The next week, they go back again, only this time, they each pay $\\8.00$ for a ticket, they each get a large soft drink, but they share one large bucket of popcorn. This return trip costs them a total of $\\37.50$. 1. What is the price of a large soft drink and the price of a large bucket of popcorn? Set up a system of two equations to find the prices of a large popcorn and a drink. popcorn $=\\4.50$ and drink $=\\3.00$ 2. Did you write two equations or did you use another method? If you used another method, write two equations now and solve them. If you already used a system of equations, skip this part.", "which were 3/4 of original price. How many SKK originally cost windbreaker? How many SKK was cost reduction? • The cost 2 The cost of 5 apples is $3.45, and 5 oranges are$1.23. If Rachel buys one apple and one orange, then how much must she pay? • Lenka and money When Lenka bought pants from 1/3 of her savings shirt from 1/5 and 1/8 spent at the hairdresser, she was left 328 Kč. How much money was saved? How many hairdressers cost? • Money 6 Lita had some money. She spent 1/2 of it on a hand bag and 1/3 of the remainder of the blouse. She had 260.00. How much money did she have at start? • Pen and pencil Cost of a pen is Rs. 12 3/5 and cost of a pencil is Rs. 3 2/5. What is the total cost of pen and pencil? • 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 • Savings Eva borrowed 1/3 of her savings to her brother, 1/2 of savings spent", "(8) ÷ 0.1 × 4 80 x 4 320 Question 20. 32 ÷ (12 – 4) + 7 Answer: 32 ÷ (12 – 4) + 7 32 ÷ (8) + 7 4 + 7 11 Question 21. {8 × [1 + (20 – 6)} ÷ $$\\frac{1}{2}$$ Answer: {8 × [1 + (20 – 6)} ÷ $$\\frac{1}{2}$$ {8 × [1 + (20 – 6)} ÷ 1/2 {8 × [1 + (14)} ÷ 1/2 {8 × 15} ÷ 1/2 {120} ÷ 1/2 240. Problem Solving Question 22. Dan and his 4 friends want to share the cost of a meal equally. They order 2 large pizzas and 5 small drinks. If they leave a tip of$6.30, how much does each person pay? Given, They order 2 large pizzas and 5 small drinks. They leave a tip of $6.30 So, (2 x 12) + (5 x 1.50) + 6.30 (24) + (7.50) + 6.30$37.8 Total Number of friends = 5 Each person have to pay = 37.8 ÷ 5 = $7.56 Question 23. Higher Order Thinking Use the operation signs +, -, ×, and ÷ once each in the expression below to make the number sentence true. Answer: Question 24. Be Precise Carlotta needs 12$$\\frac{1}{2}$$ yards of ribbon for a project. She has 5$$\\frac{1}{4}$$ yards of ribbon on one spool and", "on the total, final amount that I am required to pay (food charge + tax + delivery charge). ~Lynn2000 • Member • Posts: 11617 • Not all those who wander are lost ##### Re: Tipping Pizza Deliveries « Reply #6 on: April 19, 2013, 04:47:29 PM » We ordered pizza on Tuesday. We had a free pizza, and I bought another. Our total was $14, and we tipped$3, because we always tip more than $2. Depending on a larger order, such as for a party, I would tip more. “All that is gold does not glitter, Not all those who wander are lost; The old that is strong does not wither, Deep roots are not reached by the frost.\" -J.R.R Tolkien #### Bottlecaps • Member • Posts: 256 • She boxed her shadow and she won. ##### Re: Tipping Pizza Deliveries « Reply #7 on: April 19, 2013, 10:04:02 PM » Here's a great website that answers this question, and in great detail as well. http://tipthepizzaguy.com/ I personally always tip at least 20%, if not more - same as I do for a sit-down restaurant. Here, I tip more like 25-30% because the closest place that will deliver to us here in rural Alabama is about 10-15 miles away. That's quite a drive for the driver, and quite", "# Popeye's, a local convenience store, sells soft drinks It sells two large drinks for every small ... Popeye's, a local convenience store, sells soft drinks It sells two large drinks for every small drink. A large drink sells for $3.00 with a variable cost of$0.80. A small drink sells for $1.25 with a variable cost of$0.50. The weighted average contribution margin is ___________. (Round any intermediate calculations and your final answer to the nearest cent.) A. $2.20 per drink B.$5.15 per drink C. $1.48 per drink D.$1.72 per drink", "# Miguel spent more than $6 at a fast food restaurant. If he had$25 to start, what was his remaining pocket money? Less than $19 #### Explanation: $25-$6 =$19 If Miguel had spent $6, he would have$19 left, but because he spent more than \\$6, the remainder would have to be less than 19.", "of a chocolate bar. How many friends did Fredericca share the chocolate bar with? • Julian 2 Julian and two of his friends are going to share 1/4 of a pizza. How much will each person get? • 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? • Strawberry-rhubarb pie Larry had 3/7 of a strawberry-rhubarb pie left over. He split the leftover pie evenly between his 3 children. What fraction of a pie did each child get? • Larry Larry received 1/3 pound of candy from his grandmother and 2/5 pound of candy from his best friend. Larry's sister ate 1/2 pound of Larry's candy. How many pounds of candy does Larry have left? • Kate shares Kate shares a 64-ounce bottle of apple cider with 5 friends. Each person’s serving will be the same number of ounces. Between what two whole number of ounces will each person’s serving be? Explain using division. • Grayson Grayson bought snacks for his team's practice. He bought a bag of oranges for $3.11 and a 24-pack of juice bottles. The total cost before tax was$50.15. How much was each bottle of juice? • Jonah Jonah has an 8-pound bag of potting soil. He divides", "### Home > CCG > Chapter 12 > Lesson 12.1.4 > Problem12-56 12-56. Multiple Choice: Carol was shopping for a spring picnic. She spent $2.00$ for each liter of soda and $3.50$ for each bag of chips. In all, she bought $18$ items for a total of $43.50$. Assuming she only bought chips and soda, how many bags of chips did she buy? 1. $9$ 1. $5$ 1. $15$ 1. $3$ Write two equations: one representing the number of items bought and the other the total price. Solve using the substitution method. See the online help for problem 5-104, which contains the Math Notes box in Lesson 2.1.3.", "2. Economics 3. dilbert spends his entire income of 50 a month on... # Question: dilbert spends his entire income of 50 a month on... ###### Question details Dilbert spends his entire income of $50 a month on hours of internet access (H) and slices of pizza (P). Draw his monthly budget constraint for each of the follow- ing situations. Please use a separate graph for each part and label all axes and intercepts. (a) The price of pizza is$1/slice. The internet service provider charges a fixed monthly fee of $20 for service to begin and then$2/hour of service. (b) The price of pizza is $2/slice. Dilbert subscribes to an internet provider that offers unlimited internet access for a flat fee of$25 per month. (c) The price of internet access is $1/hour. The price of pizza is buy 2 slices at a price of$5/slice and get one free. (That is, for every 2 slices Dilbert buys, he gets one more for free.)" ]

[ "# Eleven primes Find the sum of all positive primes $$p$$ there such that $$p^2+11$$ has exactly six positive divisors. ×", "# I'm not going to list all 24 numbers! Given that $$1400 = 2^3 \\times 5^2 \\times 7,$$ find the sum of its positive divisors. ×", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "• 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.", "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. $\\blacksquare$", "# Greatest Common Divisor of Integers/Examples/8 and 17 ## 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. $\\blacksquare$", "# Four Digit Number - Can You Find It? I am thinking of a four digit positive integer with distinct digits. Of course, there's a total of $$4!-1=23$$ ways to rearrange the digits to form a new 4 digit positive integer. If the sum of these other 23 numbers is 157193, what is the number that I was thinking of? ×", "# Sum of Some Primes The four values $$x$$, $$y$$, $$x - y$$ and $$x + y$$ are all positive prime integers. What is the sum of all the four integers? Details and assumptions Fact: 1 is not a prime number. ×", "The numbers in the sequence 101,104,109,116,… are of the form $$a_{n}=100+n^{2}$$, where n=1,2,3,…. 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. (第三届AIME1985 第13题)", "# 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? •", "12$. This may be the interpretation that was intended. If the question has copied the words of the original problem exactly, it may be that the problem was poorly worded. The question seems really to be asking for how many distinct multisets of positive integers exist that each have a sum of $12$.", "Eva lists the positive integers, but skips all numbers containing a digit divisible by $$3$$. For example, the first ten numbers she lists are $$1, 2, 4, 5, 7, 8, 11, 12, 14, 15$$. What is the $$100^{\\text{th}}$$ number in her list? This problem is shared by Muhammad A. Details and assumptions The number $$13$$ is not included in the list, since it contains the digit 3, which is divisible by 3. ×", "# You've gotta be kidding me $1007021035035021007001$ How many positive divisors (including 1 and itself) does the large number above have? ×", "## Algebra 2 (1st Edition) $4$ Recall: The quotient of two numbers with the same signs is positive. Thus, $-28 \\div (-7) = 4$", "of (this includes 1 and n.) 26.If are distinct primes and are positive integers, then the sum of distinct positive integral divisors of is . 27.Highest power of a prime in n! is where denotes the integral part. for_PERMUTATIONS-AND-COMBINATIONS", "# Find 4 positive integers not exceeding 70,000 such that each have more than 100 divisors I am looking at problems in Vandendriessche and Lee's Problems in elementary number theory and this is one of their problems: Find $4$ positive integers not exceeding $70000$ such that each have more than $100$ divisors I just started picking some numbers and looked at their factors: $25=1,5,25;\\;(3\\;\\text{divisors})$ $50=1,2,5,10,25,50;\\;(6\\;\\text{divisors})$ $100=1,2,4,5,10,20,25,50,100;\\;(9\\;\\text{divisors})$ $200=1,2,4,5,8,10,20,25,40,50,100,200;\\;(12\\;\\text{divisors})$ So I guess $400$ should have $15$ factors based on the pattern above. In this case I've noticed that there are $3$ more factors after each multiplication by $2$ but I think this approach wouldn't help me to find solutions. • Suppose we know the prime factorisation of an integer. We can find the number of divisors from there. How can we maximise the number of divisors? Smaller prime factors? Dec 12, 2015 at 3:03 • Based on a semi-brute force search, the only solutions appear to be 50400, 55440, 60480, and 65520 (whereas 45360 has exactly 100 divisors and so doesn't qualify). But I can't come up with a nice pencil-and-paper argument that point to them. Dec 12, 2015 at 4:07 • Are negative divisors allowed? :p Dec 12, 2015 at 4:16 • May find a lot of useful information: sequence d(n) in OEIS and Wikipedia Dec 12, 2015 at", "# Sum of Sum of Divisors For all $$n \\in \\mathbb{N}$$, let $$\\tau (n)$$ denote the sum of positive divisors of $$n$$ (including $$1$$ and itself). Find the last three digits of $$\\displaystyle \\sum \\limits_{n=1}^{100} \\tau (n)$$. Details and assumptions • As an explicit example, the positive divisors of $$4$$ are $$1, 2$$ and $$4$$, so $$\\tau (4)= 4+2+1 = 7$$. • This is not a computer science problem. ×", "positive integers less than 1000 divisible by 15\"(a) - \"the number of positive integers less than 1000 divisible by 12\"(b) + \"the number of positive integers less than 1000 that divisible by 12 and 15\"(c). To be divisible by 12 and 15, it must be divisible by their least common multiple, 60. It's easy now: $$a = \\left\\lfloor{999/15}\\right\\rfloor = \\left\\lfloor{66.6}\\right\\rfloor = 66$$ $$b = \\left\\lfloor{999/12}\\right\\rfloor = \\left\\lfloor{83.25}\\right\\rfloor = 83$$ $$c = \\left\\lfloor{999/60}\\right\\rfloor = \\left\\lfloor{16.65}\\right\\rfloor = 16$$ So, the answer you want is 999 - 66 - 83 + 12 = 862. There are 862 positive integers less than 1000 that are not divisible by 12 or 15. And if you're not totally convinced that the number you're looking for is 999 - a - b + c, draw a Venn diagram.", "2014, 01:08 1 KUDOS Expert's post 4 This post was BOOKMARKED Official Solution: The average of four positive integers is 30, means that the sum of these four positive integers is $$4*30=120$$. (1) One of the integers is 98. If one of the integers is 98 then the sum of the other three is $$120-98=22$$. Now, since all integers are positive then no integer among these three could be greater than 20 (the greatest value of one of them can only be 20 if the other two are 1 and 1). So, there is only one integer greater than 20, the fourth one, which is 98. Sufficient. (2) The median of the four integers is less than 12. If the set is $$\\{1, 1, 1, 117 \\}$$ then there is only one integer greater than 20 but if the set is $$\\{1, 1, 21, 97 \\}$$ then there are two integers greater than 20. Not sufficient. _________________ Intern Joined: 29 Nov 2014 Posts: 1 ### Show Tags 26 Apr 2016, 06:32 I think this is a high-quality question and I don't agree with the explanation. Is 0 not a positive integer? If yes then , one of the possibilities for statement 1 can be 0 0 22 98 In this case statement 1 will also not be sufficient.", "# How many divisors? For a positive integer $x$, let $f(x)$ be the function which returns the number of distinct positive factors of $x$. If $p$ is a prime number, what is the minimum possible value of $f(75p^2)?$ This problem is posed by Michael T. Details and assumptions You may read up on Divisors of an Integer. The distinct positive factors of $x$ need not be prime. For example, $f(12) = 6$, since 12 has distinct positive factors of $1, 2, 3, 4, 6, 12$. ×" ]

[ "and multiply to make $12$. The factors of $12$ that satisfy theses two requirements are $3$ and $4$. Therefore, the full factorisation of $x^2 + 7x + 12$ is, $(x + 3)(x + 4)$ In the quadratic, $a = 3$, $11$ is positive and $c$ is positive. We can set up the brackets as follows: $(3x \\,\\,\\, + \\,\\,\\,)(x \\,\\,\\, + \\,\\,\\,)$. We are looking for two positive numbers which multiply to make $6$. The possible factors of $6$ are \\begin{aligned}(6)\\times(1)&=6 \\\\ (3)\\times(2)&=6 \\end{aligned} We now test all the combinations: \\begin{aligned}(3x+6)(x+1)&=3x^2+6x+3x+6 \\\\ (3x+1)(x+6)&= 3x^2+x+18x+6 \\\\ (3x+3)(x+2)&= 3x^2+6x+3x+6 \\\\ (3x+2)(x+3) &= 3x^2+9x+2x+6\\end{aligned} Hence the correct factorisation is $(3x+2)(x+3)$ We can see this is a sub-type (c) meaning it will contain both $+$ and $-$ Factors of $-6$ $(-1) \\times 6 = -6$ $(-6) \\times 1 = -6$ $(-2) \\times 3 = -6$ $(-3) \\times 2 = -6$ Lets find the options which give $- 5m$ $(2m+2)(2m-3) = 4m^2 - 2m - 6$ $(4m + 1)(m-6) = 4m^2 -23m-6$ $(4m-1)(m+6) = 4m^2 +23m -6$ $(4m+3)(m-2) = 4m^2 -5m -6$ We can see that last option with $+3$ and $-2$ is the correct combination. This gives the final answer to be: $(4m+3)(m-2)$ ### Worksheets and Exam Questions #### (NEW) Factorising Quadratics Exam Style Solutions - MME Level 4-5 New Official MME", "(2a+1)^2 + (2b+1)^2 &= 332 \\\\ 6a^2 + 4ab + 6b^2 + 8a + 8b &= 328 \\\\ 3a^2 + 2ab + 3b^2 + 4a + 4b &= 164 \\end{align*} Note that $3a^2 \\le 164$, so $a \\le 7$. There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in $3b^2 + b(2a+4) + 3a^2 + 4a - 164 = 0$. Since both $a$ and $b$ are integers, we must have \\begin{align*} n^2 &= 4a^2 + 16a + 16 - 12(3a^2 + 4a - 164) \\\\ &= -32a^2 - 32a + 16 + 12 \\cdot 164 \\\\ &= 16(-2a^2 - 2a + 1 + 3 \\cdot 41) \\\\ &= 16(-2(a^2 + a) + 124), \\end{align*} where $n$ is an integer. Thus, $-2(a^2 + a) + 124$ must be a perfect square. After trying all values of $a$ from 0 to 7, we find that $a$ can be $3$ or $5$. If $a = 3$, then $b = 5$, and if $a = 5$, then $b = 3$. Therefore, the ordered pairs $(x,y)$ that satisfy the original equation are $\\boxed{(7,11) , (11,7)}$.", "12: 1+2+3+5=11. For the same reason 6 * 4 * 1 would not work. The only remaining way to get 24 is 4 * 3 * 2 (we don't include the 1 since we can have at most 3 numbers), and the other numbers 1, 5 and 6 do indeed sum up to 12. There is still another way: The product $$24 = 3 \\cdot 2^3$$, so there must be exactly three factors of $$2$$. Counting factors of $$2$$, $$2$$ has one factor, $$4$$ has two, and $$6$$ has one, so either $$4$$ and $$2$$ are together, or $$4$$ and $$6$$ are. But $$4 \\cdot 6 = 24$$ while $$1 + 2 + 3 + 5 \\ne 12$$. Thus the only other option is $$4 \\cdot 2 \\cdot 3 = 24$$ and $$1 + 5 + 6 = 12$$, which works. • This can be seen as an improvement on trolley813's answer. Apr 4 at 12:54 Here's how I reasoned through this before posting it. Others' methods turned out to be much simpler, though: Since 5 isn't a factor of 24, it must be part of the sum group. The rest of the numbers in the sum group then must add up to 7. Since we don't have a 7, there must be at least two more numbers", "4 \\cdot (-10)\\\\ & = -5 \\cdot 8 & & = 5 \\cdot (-8) \\end{align*} Of these pairs of factors, only $-4$ and $10$ have a sum of $6$. Hence, \\begin{align*} 5p^2 + 6p - 8 & = 5p^2 - 4p + 10p - 8 && \\text{split the linear term}\\\\ & = p(5p - 4) + 2(5p - 4) && \\text{factor by grouping}\\\\ & = (p + 2)(5p - 4) && \\text{extract the common factor} \\end{align*} In your criss-cross method, this amounts to matching $5p$ with $2$ to get $10p$ and $p$ with $-4$ to get $-4p$. That said, I recommend using the method outlined above to eliminate guessing. Draw a railroad sign style $X$ like you have done. In the top triangle you will write the product $ac$ from the quadratic of the form $y=ax^2+bx+c$, then in the bottom you will write the $b$ term. You should have $-40$ on top and $6$ on the bottom. Now you need to find two numbers which add to $6$ but multiply to $-40$. You will find these numbers are $-4$ and $10$ (by listing factors of $-40$ and seeing which ones must add up to $6$). After this step we write $5p^2-4p+10p-8$. Now we must factor this expression. We have $$5p^2-4p+10p-8 = 5p(p+2)-4(p+2) \\\\ =(5p-4)(p+2).$$ This is our solution.", "the factors $11,13$ must be in group $1$. Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$. Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$. Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$. It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \\cdot 5 \\cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\\cdot 13 \\cdot 37 \\gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$. • I also thought that trying to solve for the same question just", "middle term is $10$ and the product of the first and last term is $3\\times8=24$. So you first look for possible pairs that will give you a sum of $10$, then check if the product is equal to $24$. Note that $4+6=10$ and $4\\times6=24$. Thus, we have the pair $4$ and $10$. So we rewrite the expression so we can group them later. $$3x^2+10x+8=3x^2+(4x+6x)+8$$ We group the terms that have a common factor, so we group $6x$ with $3x^2$, and $4x$ with $8$, then factor out their respective common factors. \\begin{align*} 3x^2+10x+8&=(3x^2+6x)+(4x+8)\\\\ &=3x(x+2)+4(x+2)\\\\ &=(3x+4)(x+2). \\end{align*} Thus, the factors of $3x^2+10x+8$ are $3x+4$ and $x+2$. • Find the factors of the quadratic equation $10x^2+11x-6=0$. The product of the first and last term is a negative number, $10\\times(-6)=-60$. So we are looking for factors of $-60$, a positive number and a negative number, that will give us a sum of $11$. Note that the sum of $15$ and $-4$ is $11$, and the product of these numbers is $-60$. So we have: \\begin{align*} 10x^2+11x-6&=0\\\\ 10x^2+15x-4x-6&=0 \\end{align*} We can group $15x$ and $-4x$ with either $10x^2$ and $-6$ since each grouping has a common factor. So you can choose whichever and you will still arrive at the same factors. \\begin{align*} (10x^2+15x)+(-4x-6)&=0\\\\ 5x(2x+3)-2(2x+3)&=0\\\\ (5x-2)(2x+3)&=0 \\end{align*} Therefore, we have factored the quadratic equation", "c &= 3^2 + 1^2 = 10. \\end{align*} Furthermore, $c = 2\\cdot 5 \\cdot 1$, so letting $m = 5$ and $n = 1$ gives \\begin{align*} d &= 5^2 - 1^2 = 24\\\\ e &= 5^2 + 1^2 = 26. \\end{align*} And again, $e = 2\\cdot 13 \\cdot 1$, so letting $m = 13$ and $n = 1$ gives \\begin{align*} f &= 13^2 - 1^2 = 168\\\\ g &= 13^2 + 1^2 = 170. \\end{align*} Then \\begin{align*} g^2 = 170^2 = 26^2 + 168^2 = 10^2 + 24^2 + 168^2 = 6^2 + 8^2 + 24^2 + 168^2 \\end{align*} Going one more step we can write $170 = 2\\cdot 17 \\cdot 5$ so $m = 17$ and $n = 5$ gives \\begin{align*} h &= 17^2 - 5^2 = 264\\\\ i &= 17^2 + 5^2 = 314, \\end{align*} in which $314 = 2\\cdot 157 \\cdot 1$. It seems I may be on the right track, but I'm not sure if this is true in general. Maybe it can be proven that this process can be continued arbitrarily given we start with two distinct primes $m$ and $n$ in the first step? Any thoughts or references or anything would be wonderful. Thanks in advance! - Yes, your process can always be continued if you start with $m_1>n_1$ both odd. $$a_1", "every $n = 2, 3, \\ldots, N$, find the prime factorization of $n!$. This is not as nasty as it sounds: if you know that the prime factorization of $n!$ is $$n! = \\prod_i p_i^{e_i}$$ then the prime factorization of $(n + 1)!$ is simply that and the prime factorization of $n + 1$. For example, \\begin{alignat*}{4} 4! &= 2 \\cdot 3 \\cdot 4 &&= 2^3 \\cdot 3 \\\\ 5! &= 4! \\cdot 5 &&= (2^3 \\cdot 3) \\cdot 5 &&= 2^3 \\cdot 3 \\cdot 5 \\\\ 6! &= 5! \\cdot 6 &&= (2^3 \\cdot 3 \\cdot 5)(2 \\cdot 3) &&= 2^4 \\cdot 3^2 \\cdot 5\\\\ \\end{alignat*} Next, you need to partition $n!$ into factors $a$ and $b$. To ensure that $a$ and $b$ have no common factors, you simply have to ensure that each prime factor $p_i$ is attributed to either $a$ or $b$, along with all of its exponents $e_i$. For example, $4!$ can be partitioned as \\begin{alignat*}{3} (a, b) &= (1&&,\\ 2^3 \\cdot 3) \\\\ (a, b) &= (3&&,\\ 2^3) \\end{alignat*} but not as $$(a, b) = (2 \\cdot 3,\\ 2^2)$$ since 2 appears in both $a$ and $b$, making $\\frac{a}{b}$ reducible. So, the problem reduces to counting the power sets of the distinct prime factors of $n!$. Half of those partitions will meet the", "number of factors of $10^{11}$. In general, for any positive number $a$, we can write it in its prime factorization as $$a = p_1^{k_1} p_2^{k_2} \\cdots p_t^{k_t}$$ where the $p_i$'s are all prime and distinct, and the $k_i$'s are all positive integers, then the number of positive factors is $(k_1 + 1)(k_2 + 1) \\cdots (k_t + 1)$. That is, any positive factor of $a$ must be of the form $$p_1^{m_1} p_2^{m_2} \\cdots p_t^{m_t}$$ where now I will allow the $m_i$'s to be 0 or positive, but $m_i \\leq k_i$. And, by the uniqueness of a prime factorization, two distinct choices of $m_i$'s lead to two distinct factors and vice versa. Therefore, we simply need to count the number of ways to choose the $m_i$'s. Clearly, there are $k_i + 1$ choices for $m_i$, which are $0, 1, 2, \\ldots, k_i$. Therefore we have the formula above. In this case, we have $10^{11} = 2^{11} \\cdot 5^{11}$ so there are $12 \\cdot 12 = 144$ positive factors as you said. But, if you allow negative multiples as well, then the answer is actually 2 times this, or 288. That is, for each positive multiple you counted, there is also a corresponding negative multiple.", "$a\\cdot c,$ which here is $4\\cdot 3,$ and whose sum is $b,$ which in the example is $13.$ ### 1 List all pairs of integers whose product is equal to $a\\cdot c$ The first step is to find all possible pairs of integers that multiply to $a\\cdot c.$ In this case, $a=4$ and $c=3.$ Thus, their product is $3\\cdot 4 = 12.$ To find all factor pairs, start with the pair where one factor is $1.$ The other factor must then be $12.$ Then continue with the pair where one of the factors is $2,$ and so forth. In this case, there are three pairs. \\begin{aligned} &1 \\quad \\text{and} \\quad 12\\\\ &2 \\quad \\text{and} \\quad 6\\\\ &3 \\quad \\text{and} \\quad 4 \\end{aligned} ### 2 Find the pair whose sum is $b$ If the given expression is factorable, one of the factor pairs will add to equal $b.$ In this case, $b=13.$ \\begin{aligned} 1+12&=13 \\\\ 2+6&=8 \\\\ 3+4&=7 \\end{aligned} Here, $1$ and $12$ is the only factor pair that adds to $13.$ ### 3 Write $bx$ as a sum Now, use the factor pair to rewrite the $x$-term of the original expression as a sum. Since the factor pair is $1$ and $12,$ the middle term can be written as $13x = 12x + x.$ This gives the following", "$78\\div 3=26$. Thus, $3$ and $26$ are factors of $78$. 4. As $78=4\\times 19+2$, $4$ does not divide $78$ as it leaves a nonzero remainder $2$ while dividing $78$. 5. Since the last digit of $78$ is $8$ which is not $0$ and $5$, so, $5$ does not divide $78$ and hence, is not a factor of $78$. 6. As $78$ is divisible by both $2$ and $3$, it is divisible by $6$. Also, $78\\div 6=13$. So, $6$ and $13$ are factors of $78$. 7. Since $78=7\\times 11+1$, $7$ cannot divide $78$ fully as it leaves a non zero remainder of $1$. Thus, $7$ and $11$ are not factors of $78$. 8. Since $78=8\\times 9+6$, $8$ cannot divide $78$ completely as it leaves a nonzero remainder of $6$. Thus, $8$ and $9$ are not factors of $78$. 9. Since $78=9\\times 8+6$, $9$ cannot divide $78$ fully as it leaves a nonzero remainder of $6$. Thus, $9$ and $8$ are not factors of $78$. Note that in Step-8 and Step-9, we get the same pair of quotients and remainders i.e., $(8,9)$ and the remainder is also the same i.e., $6$. This means that all the factors of $78$ have been obtained. So, we must stop the process. Hence, the factors of $78$ are: $1,\\text{ }2,\\text{ }3,\\text{ }6,\\text{ }13,\\text{ }26,\\text{", "more than $$3$$ factors? (As an example, $$12$$ has $$6$$ factors, namely $$1,$$ $$2,$$ $$3,$$ $$4,$$ $$6,$$ and $$12.$$) $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ ###### Solution(s): Let's begin by firstly simply factoring $$2020$$: \\begin{align*}2020 &= 1\\cdot 2020\\\\ &=2\\cdot 1010\\\\ &=4\\cdot 505\\\\ &=5\\cdot 404\\\\ &=10\\cdot 202\\\\ &= 20\\cdot 101 \\end{align*} These twelve factors of $$2020$$ can be classified by the number of their factors: $$\\quad$$ - $$1$$ has one factor $$\\quad$$ - $$2, 5,$$and $$101$$ has two factors $$\\quad$$ - $$4$$ has three (distinct) factors Thus, all the remaining seven numbers must have more than three factors. Thus, the correct answer is B. 18. Rectangle $$ABCD$$ is inscribed in a semicircle with diameter $$\\overline{FE},$$ as shown in the figure. Let $$DA=16,$$ and let $$FD=AE=9.$$ What is the area of $$ABCD?$$ $$240$$ $$248$$ $$256$$ $$264$$ $$272$$ ###### Solution(s): Since $$FE$$ is the diameter of the semicircle, we know the length of the diameter is $$34,$$ and so the radius is $$17.$$ Let $$O$$ be the center of the diameter. The length from $$OF$$ therefore is $$17.$$ Since $$D$$ is on $$OF,$$ we know \\begin{align*} OD + FD &= OF\\\\ OD + 9 &= 17 \\\\ OD &= 8. \\end{align*} Also, since we have a semicircle, we know $$OC = 17.$$ Finally, since $$ABCD$$ is a rectangle, we know $$\\angle ODC$$", "factor out the common binomial we get $(4x+5)(x+1)$ . Watch this video for help with the Examples above. ### Vocabulary • We say that a polynomial is factored completely when we factor as much as we can and we are unable to factor any more. ### Guided Practice Factor completely: $24x^3-28x^2+8x$ . Solution: First, notice that each term has $4x$ as a factor. Start by factoring out $4x$ : $24x^3-28x^2+8x=4x(6x^2-7x+2)$ Next, factor the trinomial in the parenthesis. Since $a\\neq 1$ find $a\\cdot c$ : $6\\cdot 2=12$ . Find the factors of 12 that add up to -7. Since 12 is positive and -7 is negative, the two factors should be negative: $12&=-1\\cdot -12 && and && -1+-12=-13\\\\12&=-2\\cdot -6 && and && -2+-6=-8\\\\12&=-3\\cdot -4 && and && -3+-4=-7$ Rewrite the trinomial using $-7x=-3x-4x$ , and then factor by grouping: $6x^2-7x+2 &= 6x^2-3x-4x+2\\\\ &=3x(2x-1)-2(2x-1)\\\\ &=(3x-2)(2x-1)$ The final factored answer is: $4x(3x-2)(2x-1)$ ### Explore More Factor completely. 1. $2x^2+16x+30$ 2. $5x^2-70x+245$ 3. $-x^3+17x^2-70x$ 4. $2x^4-512$ 5. $25x^4-20x^3+4x^2$ 6. $12x^3+12x^2+3x$ 7. $12c^2-75$ 8. $6x^2-600$ 9. $-5t^2-20t-20$ 10. $6x^2+18x-24$ 11. $-n^2+10n-21$ 12. $2a^2-14a-16$", "also infer that $x_5$ will likely need to be one of the smaller factors. The factors of $144$ are: $1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144$ Selecting numbers from this list and trying them out, we can satisfy the conditions with these numbers: $x_5 = 18, x_4 = 6, x_3 = 2, x_2 = 1, x_1 = 2$ So, $x_7 = x_6(x_5 + x_4)$ $x_7 = 144(18 + 6)$ $x_7 = 3456$ Therefore, the answer is $\\Rightarrow{\\boxed{456}}$", "# How do you factor by grouping 6c^2+29c-42? Apr 22, 2015 $6 {c}^{2} + 29 c - 42$ We are looking for factors, $m$ and $n$, of $6$ and factors, $p$ and $q$, of $\\left(- 42\\right)$ such that $m q + n p = 29$ • this is because $\\left(m c + p\\right) \\cdot \\left(n c + q\\right) = m n {c}^{2} + \\left(m q + n p\\right) c + p q$ factors of 6 $= \\left\\{\\begin{matrix}1 & 6 \\\\ 2 & 3\\end{matrix}\\right\\}$ plus their negatives factors of (-42) # = {(6,-7),(3,-14),(2,-21),(1,-42)} and their negatives A bit of testing combinations comes up with the pairs $\\left(p , q\\right) = \\left(6 , 1\\right)$ and $\\left(m , n\\right) = \\left(- 7 , 6\\right)$ that satisfy the requirement for the middle term $6 {c}^{2} + 29 c - 42 = \\left(6 c - 7\\right) \\left(1 c + 6\\right)$ Apr 22, 2015 Multiply $6 \\times - 42 = - 252$ Now we look for factor of $- 252$ that add to get $29$ Because the product is negative and the sum is positive, we want the smaller number to be negative and the larger positive: -1 252 does not add up to 29 -2 126 does not add up to 29 -3 84 -4 63 -5 is not a factor -6 42", "have used $3$ the remaining numbers are $1,2,4,5.$ We now factor \\begin{align*} (x_2 + x_5)(x_3x_4) &\\equiv 0 \\pmod{3} \\\\ (x_2 + x_5) &\\equiv 0 \\pmod{3} \\end{align*} since $1,2,4,5$ are all not factors of $3.$ Now using the number $1,2,4,5,$ we take two to get a number divisible by $3$ for $(x_2 + x_5):$ \\begin{align*} 1+5 &\\equiv 0 \\pmod{3}, \\\\ 4+2 &\\equiv 0 \\pmod{3}, \\\\ 4+5 &\\equiv 0 \\pmod{3}, \\\\ 1+2 &\\equiv 0 \\pmod{3}. \\end{align*} We have $4$ possibilities from above. Since we can also have $5+1$ or $2+4,$ there are $4\\cdot2=8$ possibilities in all. Now using $$(x_2 + x_5)(x_3x_4) \\equiv 0 \\pmod{3},$$ we have $(x_3x_4),$ which results in $8$ more possibilities of $2$ times more. So, we get $2\\cdot2\\cdot4=16.$ Remember that $3$ can be any of $5$ different variables. So, we multiply by $5$ to get the answer $\\boxed{080}.$", "and 2 from the last two terms. $x(3x+2)+2(3x+2)$ Now factor the common binomial $(3x+2)$ . $(3x+2)(x+2)$ Our answer is $(3x+2)(x+2)$ . In this example, all the coefficients are positive. What happens if the $b$ is negative? #### Example C Factor $6x^2-11x+4$ by grouping . Solution: $ac=6 \\cdot 4 = 24$ The number 24 can be written as a product of two numbers in any of these ways. $24&=1\\times 24 && and && 1+24=25\\\\24&=(-1) \\times (-24) && and && (-1)+(-24)=-25\\\\24&=2 \\times 12 && and && 2+12=14\\\\24&=(-2) \\times (-12) && and && (-2)+(-12)=-14\\\\24&=3 \\times 8 && and && 3+8=11\\\\24&=(-3) \\times (-8) && and && (-3)+(-8)=-11 \\quad \\text{This is the correct choice}.$ Rewrite the middle term as $-11x=-3x-8x$ , so the problem becomes: $6x^2-11x+4=6x^2-3x-8x+4$ Factor by grouping. Factor a $3x$ from the first two terms and factor –4 from the last two terms. $3x(2x-1)-4(2x-1)$ Now factor the common binomial $(2x-1)$ . Our answer is $(2x-1)(3x-4)$ . ### Guided Practice Factor $10x^2-43x+28$ . Solution: First, find $a\\cdot c$ : $10\\cdot28=280$ . Since $b=-43$ , the factors of 280 need to add up to -43, so consider pairs of negative factors of 280. There will be a lot of pairs of factors, and you could list them all in order until you find the correct pair, as in Examples B and C. Or,", "- 100$. We can also generalise this, $$-a = -1 \\cdot a$$ Here, $a$ is a positive integer. Now that we have a good understanding of positive and negative numbers, lets see what happens when we do arithmetic operations. What if we tried to perform multiplication here: $-2 \\cdot 3$. What does it mean to add the number $-2$, $3$ times. With this example, we add negative $2$ three times, \\begin{align*} -2 + -2 + -2 = -2 - 2 - 2 = -6 \\end{align*} Practice 3) What if we had $-3 \\cdot 2$, what would be the answer: -6, or 6. The answer is -6 because, \\begin{align*} -3 + -3 = -3 - 3 = -6 \\end{align*} When we multiply a negative number by a positive number we get a more negative number, $$-a \\cdot b = -c$$ Lets try multiplying two negative numbers, $-2 'cdot -3$. What does it mean to add the number $-2$, negative $3$ times. It does not make sense to multiply that way, instead we will try to user our previous rules to solve this problem. What is negative $24, it is just$-1 \\cdot 2. So we can say, \\begin{align*} -1 \\cdot 2 \\cdot -1 \\cdot 3 = -1 \\cdot -1 \\cdot 2 \\cdot 3 \\end{align*} Here, I will add some brackets", "# What is the greatest common factor of 6, 12, and 25? Then teach the underlying concepts Don't copy without citing sources preview ? Write a one sentence answer... #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer Describe your changes (optional) 200 1 Feb 12, 2017 The GCF is 1. #### Explanation: First, we have to break down each number to its prime factors. Let's take the number 6 first: $6 = 2 \\cdot 3 \\cdot 1$ Although multiplying by 1 is not necessary, sometimes it helps visualize and understand (like in this case) that the GCF is 1. Now 12: $12 = 6 \\cdot 2 \\cdot 1$ $12 = 3 \\cdot 2 \\cdot 2 \\cdot 1$ $12 = 3 \\cdot {2}^{2} \\cdot 1$ Finally, 25: $25 = 5 \\cdot 5 \\cdot 1$ $25 = {5}^{2} \\cdot 1$ The only factors that are written throughout our prime factorization are $1 , 2 , 3 , 5$ However, not all of the numbers share all of the factors. The only (and the greatest) factor that all three numbers share is the number 1. So the answer is 1. Then teach the underlying concepts Don't copy without citing sources preview ? Write a one sentence answer... #### Explanation Explain in detail... #### Explanation:", "$12$ below $36$ with branches connecting them. The factor $3$ is prime, so we circle it. The factor $12$ is composite, so we need to find its factors. Let’s use $3$ and $4$. We write these factors on the tree under the $12$. The factor $3$ is prime, so we circle it. The factor $4$ is composite, and it factors into $2\\cdot 2$. We write these factors under the $4$. Since $2$ is prime, we circle both $2\\text{s}$. The prime factorization is the product of the circled primes. We generally write the prime factorization in order from least to greatest. $2\\cdot 2\\cdot 3\\cdot 3$ In cases like this, where some of the prime factors are repeated, we can write prime factorization in exponential form. $\\begin{array}{c}2\\cdot 2\\cdot 3\\cdot 3\\\\ \\\\ {2}^{2}\\cdot {3}^{2}\\end{array}$ Note that we could have started our factor tree with any factor pair of $36$. We chose $12$ and $3$, but the result would have been the same if we had started with $2$ and $18, 4$ and $9,\\text{ or }6\\text{ and }6$. ### Find the prime factorization of a composite number using the tree method 1. Find any factor pair of the given number, and use these numbers to create two branches. 2. If a factor is prime, that branch is complete. Circle the prime. 3." ]

[ "(2a+1)^2 + (2b+1)^2 &= 332 \\\\ 6a^2 + 4ab + 6b^2 + 8a + 8b &= 328 \\\\ 3a^2 + 2ab + 3b^2 + 4a + 4b &= 164 \\end{align*} Note that $3a^2 \\le 164$, so $a \\le 7$. There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in $3b^2 + b(2a+4) + 3a^2 + 4a - 164 = 0$. Since both $a$ and $b$ are integers, we must have \\begin{align*} n^2 &= 4a^2 + 16a + 16 - 12(3a^2 + 4a - 164) \\\\ &= -32a^2 - 32a + 16 + 12 \\cdot 164 \\\\ &= 16(-2a^2 - 2a + 1 + 3 \\cdot 41) \\\\ &= 16(-2(a^2 + a) + 124), \\end{align*} where $n$ is an integer. Thus, $-2(a^2 + a) + 124$ must be a perfect square. After trying all values of $a$ from 0 to 7, we find that $a$ can be $3$ or $5$. If $a = 3$, then $b = 5$, and if $a = 5$, then $b = 3$. Therefore, the ordered pairs $(x,y)$ that satisfy the original equation are $\\boxed{(7,11) , (11,7)}$.", "+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", "# 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 $156 + 96 + 84 = \\boxed{336}$.", "• $5$ if the last digit is $5$ or $0$. • $6$ if it is divisible by both $2$ and $3$. • $10$ if it ends with $0$. #### Example 1 Is $5,625$ divisible by… • $2$? • $3$? • $5$ or $10$? • $6$? Solution Part 1. Part 2. Part 3. Part 4. In mathematics, there are often several ways to talk about the same ideas. So far, we’ve seen that if $m$ is a multiple of $n$, we can say that $m$ is divisible by $n$. For example, since $72$ is a multiple of $8$, we say $72$ is divisible by $8$. Since $72$ is a multiple of $9$, we say $72$ is divisible by $9$. We can express this still another way. Since $8 \\cdot 9 = 72$, we say that $8$ and $9$ are factors of $72$. When we write $72=8 \\cdot 9$, we say we have factored $72$. Other ways to factor $72$ are $1 \\cdot 72$, $2 \\cdot36$, $3 \\cdot 24$, $4 \\cdot 18$, and $6 \\cdot 12$. The number $72$ has many factors: $1,2,3,4,6,8,9,12,18,24,36,$ and $72$. ### Factors In the expression $a \\cdot b$, both $a$ and $b$ are called factors. If $a \\cdot b = m$, and both $a$ and $b$ are integers, then $a$ and $b$ are factors of", "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", "and multiply to make $12$. The factors of $12$ that satisfy theses two requirements are $3$ and $4$. Therefore, the full factorisation of $x^2 + 7x + 12$ is, $(x + 3)(x + 4)$ In the quadratic, $a = 3$, $11$ is positive and $c$ is positive. We can set up the brackets as follows: $(3x \\,\\,\\, + \\,\\,\\,)(x \\,\\,\\, + \\,\\,\\,)$. We are looking for two positive numbers which multiply to make $6$. The possible factors of $6$ are \\begin{aligned}(6)\\times(1)&=6 \\\\ (3)\\times(2)&=6 \\end{aligned} We now test all the combinations: \\begin{aligned}(3x+6)(x+1)&=3x^2+6x+3x+6 \\\\ (3x+1)(x+6)&= 3x^2+x+18x+6 \\\\ (3x+3)(x+2)&= 3x^2+6x+3x+6 \\\\ (3x+2)(x+3) &= 3x^2+9x+2x+6\\end{aligned} Hence the correct factorisation is $(3x+2)(x+3)$ We can see this is a sub-type (c) meaning it will contain both $+$ and $-$ Factors of $-6$ $(-1) \\times 6 = -6$ $(-6) \\times 1 = -6$ $(-2) \\times 3 = -6$ $(-3) \\times 2 = -6$ Lets find the options which give $- 5m$ $(2m+2)(2m-3) = 4m^2 - 2m - 6$ $(4m + 1)(m-6) = 4m^2 -23m-6$ $(4m-1)(m+6) = 4m^2 +23m -6$ $(4m+3)(m-2) = 4m^2 -5m -6$ We can see that last option with $+3$ and $-2$ is the correct combination. This gives the final answer to be: $(4m+3)(m-2)$ ### Worksheets and Exam Questions #### (NEW) Factorising Quadratics Exam Style Solutions - MME Level 4-5 New Official MME", "- 100$. We can also generalise this, $$-a = -1 \\cdot a$$ Here, $a$ is a positive integer. Now that we have a good understanding of positive and negative numbers, lets see what happens when we do arithmetic operations. What if we tried to perform multiplication here: $-2 \\cdot 3$. What does it mean to add the number $-2$, $3$ times. With this example, we add negative $2$ three times, \\begin{align*} -2 + -2 + -2 = -2 - 2 - 2 = -6 \\end{align*} Practice 3) What if we had $-3 \\cdot 2$, what would be the answer: -6, or 6. The answer is -6 because, \\begin{align*} -3 + -3 = -3 - 3 = -6 \\end{align*} When we multiply a negative number by a positive number we get a more negative number, $$-a \\cdot b = -c$$ Lets try multiplying two negative numbers, $-2 'cdot -3$. What does it mean to add the number $-2$, negative $3$ times. It does not make sense to multiply that way, instead we will try to user our previous rules to solve this problem. What is negative $24, it is just$-1 \\cdot 2. So we can say, \\begin{align*} -1 \\cdot 2 \\cdot -1 \\cdot 3 = -1 \\cdot -1 \\cdot 2 \\cdot 3 \\end{align*} Here, I will add some brackets", "\\boxed{88352}$. ### Solution 2 The discriminant of the function is $b^2 - 8032b$. Since all roots are integers and leading term is 1, the discriminant must equal $n^2$, where $n$ is an integer. Thus, we know that \\begin{align*} b^2 - 8032b =& n^2 \\\\ (b-4016)^2 - n^2 &= 4016^2 \\\\ (b-4016+n)(b-4016-n) &= 4016^2. \\end{align*} Let $x$ be a factor of $4016^2$, so \\begin{align*} b - 4016 + n &= x \\\\ b - 4016 - n &= \\frac{4016^2}{x} \\\\ b &= 4016 + \\frac{x}{2} + \\frac{4016^2}{2x} \\end{align*} Note that if $x$ is odd, then $\\tfrac{4016^2}{x}$ is even, so $b$ can not be an integer. Thus, $x$ must be even. Let $x = 2y$, so $b = 4016 + y + \\frac{2008^2}{y}$. If $y_0 = \\frac{2008^2}{y}$, then $y + \\tfrac{2008^2}{y} = y_0 + \\tfrac{2008^2}{y_0}$. Also, since $y$ can be positive or negative, the positive values cancel with the negative values. So WLOG, let $|y| \\le 2008$. The prime factorization of $2008^2$ is $2^6 \\cdot 251^2$, so there are $\\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $y$, so the absolute value of the sum of all values of $b$ equals $4016 \\cdot 22 = \\boxed{88352}$.", "18) = 6$ (2.) Calculate the $gcd$ of $30$ and $105$ Divisors of $30 = 1, 2, 3, 5, 6, 10, 15, 30$ Divisors of $105 = 1, 3, 5, 7, 15, 21, 35, 105$ Common Divisors of $30 \\:and\\: 105 = 1, 3, 5, 15$ Greatest Common Divisor of $30 \\:and\\: 105 = 15$ $gcd(30, 105) = 15$ Nigerian Method In this method; we set up a table of the positive integers. We set it up the same way we set up the dividend and divisor in the Long Division method of dividing polynomials. We \"try\" each factor in ascending order (from least to greatest). The factor should be a positive integer that has to divide \"all\" the positive integers without a remainder. But, if we find another factor of all the integers (a number that divides all the integers \"right away\" without a remainder), use it. In other words, it has to be a common factor. If we cannot find any factor, then the $GCF = 1$. Remember that $1$ is the smallest factor of every positive integer. If we find a factor of all the integers (besides $1$), use it. Begin in ascending order. Multiply all the factors you listed. The product gives you the $GCF$. Let us see how this works! (1.) Calculate the", "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$$ must have one factor of", "positive integers $$x$$ and $$y$$, saying that $$x$$ divides $$y$$ is equivalent to saying that, if $$z$$ and $$r$$ are integers, then $$y=xz+r \\quad\\textit{and}\\quad 0\\le r Similarly, $$x$$ does not divide $$y$$ if and only if there are integers $$z$$ and $$r$$ such that $$y=xz+r\\quad\\textit{and}\\quad 0 The condition $$r is crucial. In your argument you have no way to prove that $$r_1r_2. Indeed, if $$a=6$$, $$b=16$$ and $$c=9$$, you have \\begin{align} b&=6\\cdot2+4 \\\\ c&=6\\cdot1+3 \\end{align} so $$r_1=2\\ne0$$, $$r_2=3\\ne0$$, but $$r_1r_2=12$$ which is not less than $$a$$ (and of course it is a multiple of $$a$$). Where the proof goes wrong, is assuming the r values can't be divisible by divisors of a. If they can be (like given composite a), then the remainders may contain a in their product. An example: $$a=12,b=16=a+4,c=15=a+3$$ therefore $$bc=(a+4)(a+3)=a^2+3a+4a+12=a^2+8a=a(a+8)$$", "Problem of the Week Problem C How It Ends The product of the positive integers 1 to 4 is $4\\times 3\\times 2\\times 1 =24$ and can be written in an abbreviated form as $$4!$$. We say “4 factorial”. So $$4!=24$$. The product of the positive integers 1 to 16 is $16\\times 15\\times 14\\times\\cdots\\times 3\\times 2\\times 1$ and can be written in an abbreviated form as $$16!$$. We say “16 factorial”. The $$\\cdots$$ represents the product of all the missing integers between 14 and 3. In general, the product of the positive integers 1 to $$n$$ is $$n!$$. Note that $$1!=1$$. Determine the tens digit and units (ones) digit of the sum $1!+2!+3!+ \\cdots +2019!+2020!+2021!$ Themes: Algebra, Number Sense", "= 81$. We can then list out all the triples of positive integers whose squares sum to $81$: We get $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$. These triples give $a$ values of $13$, $15$, and $15$, respectively, and $c$ values of $64$, $216$, and $224$, respectively. We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$. Thus, the two $c$ values are $216$ and $224$, which sum to $\\boxed{\\text{440}}$. ~BealsConjecture~ ## Solution 2 (Algebra+ Brute Force) First things first. Vietas gives us the following: \\begin{align} x_1+x_2+x_3 = a\\\\ x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}\\\\ x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2} \\end{align} From $(2)$, $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$, which implies that both sides of $(2)$ are even. Then, from $(1)$, we see that an odd number of $x_1$, $x_2$, and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of $(2)$ to be even, there must only be one odd number and two even numbers. Now, the theoretical maximum value of the left side of $(2)$ is $3 \\cdot \\frac{a}{3}^2=\\frac{a^2}{3}$. That means that the maximum bound of $a$", "The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\\cdot1, 121\\cdot 7,$ and $77\\cdot 11$. The differences between these factors are $846$, $114$, and $66$ - those are all possible values for $2x$. Thus the possibilities for $x$ are $423$, $57$, and $33$. Now plug in these values into the equation $n = \\frac{-1 + x}{2}$, so $n$ can equal $211$, $28$, or $16$, hence the answer is $\\boxed{\\textbf{(A)}\\ 255}$. ~Edits by BakedPotato66 ## Solution 2 As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$. Clearly $m > n$, so let $m = n + a$, where $a$ is some positive integer. We have: $(n+a)^2 = n(n+1) + 212$. Expanding, grouping like terms and factoring, we get: $n = \\frac{(212 - a^2)}{(2a - 1)}$. We know that $n$ and $a$ are both positive integers, so we need only check values of $a$ from $1$ to $14$ ($14^2 = 196 < 212 < 15^2 = 225$). Plugging in, the only values of $a$ that give integral solutions are $1, 4,$ and $6$. These gives $n$ values of $211, 28,$ and", ",$$ we know \\begin{align*}a+b+c+d &= 42 +60 \\\\&= 112\\end{align*} Now, with the average of $$b$$ and $$c$$ being $$26,$$ we know their sum is $$26\\cdot2=52.$$ This means $$b+c = 52.$$ Subtracting this result from the sum of all the terms yields $$a+d =50.$$ Since $$\\dfrac{a+d}{2}=25,$$ our answer is $$25.$$ 17. If $$n$$ is an even positive integer, the double factorial notation $$n!!$$ represents the product of all the even integers from $$2$$ to $$n.$$ For example: $8!! = 2 \\times 4 \\times 6 \\times 8$ What is the units digit of the following sum? $2!! + 4!! + \\cdots + 2022!!$ $$0$$ $$2$$ $$4$$ $$6$$ $$8$$ ###### Solution(s): If we take $$n!!$$ for an even $$n$$ that is greater or equal to $$10,$$ then $$10$$ is one of the numbers we multiply by. Since $$10$$ is a factor of $$n!!,$$ we know that the units digit of $$10$$ is $$0,$$ which means that it doesn't affect our result. This means it suffices to compute the units digit of $$2!!+4!!+6!!+8!!,$$ which is equivalent to: \\begin{align*}&2+ 2(4)+2(4)(6)\\\\&+2(4)(6)(8) \\\\&= 2+8+48+384 \\\\&= 442\\end{align*} The units digit therefore is $$2$$ 18. The midpoints of the four sides of a rectangle are: $(-3, 0),$ $(2, 0),$ $(5, 4),$ $(0, 4).$ What is the area of the rectangle? $$20$$ $$25$$ $$40$$ $$50$$ $$80$$ ###### Solution(s):", "term negative? Check! • Find two numbers whose difference is $\\,5\\,$ and whose product is $\\,6\\,.$ That is, find two numbers that subtract to give $\\,5\\,$ and that multiply to give $\\,6\\,.$ The numbers $\\,1\\,$ and $\\,6\\,$ work, since $\\,6 - 1 = 5\\,$ and $\\,6\\cdot 1 = 6\\,.$ • Since the coefficient of $\\,x\\,$ is negative, the bigger number ($\\,6\\,$) will be negative, and the other will be positive. The desired numbers are therefore $\\,-6\\,$ and $\\,1\\,.$ • Then, \\begin{align} &\\cssId{s85}{x^2 - 5x - 6}\\cr &\\qquad \\cssId{s86}{=\\ \\ x^2 + (\\overset{=\\ -5}{\\overbrace{(-6)+1}})x + \\overset{=\\ -6}{\\overbrace{\\ (-6)\\cdot 1\\ }}}\\cr\\cr &\\qquad \\cssId{s87}{=\\ \\ (x - 6)(x + 1)} \\end{align} • Check: \\begin{align} &\\cssId{s89}{(x-6)(x+1)}\\cr &\\qquad \\cssId{s90}{= x^2 + x - 6x - 6}\\cr &\\qquad \\cssId{s91}{= x^2 - 5x - 6} \\end{align} Question: Factor: $x^2 + x - 1$ Solution: There are no integers whose difference and product are both $\\,1\\,.$ Thus, $\\,x^2 + x - 1\\,$ is not factorable over the integers.", "# How do you factor the expression 2x^2 - 11x + 12? Jan 28, 2016 $\\left(2 x - 3\\right) \\left(x - 4\\right)$ #### Explanation: Looking at the ${x}^{2}$ coefficient, $2$, we know that the coefficients of $x$ in each factor can only be $1$ and $2$ since no other integers multiply to 2. Therefore we can set up the factors as: $\\left(2 x - \\text{＿＿\")*(x - \"＿＿}\\right)$ We know that both of the operations in the factors must be subtraction because $12$ is positive and $- 11$ is negative. To find out what goes in the blanks, we need to check the factors of 12: $1$ and $12$ $2$ and $6$ $3$ and $4$ $1$ and $12$ can't work, because no matter what side each is on, $12$ times anything will be greater than $11$. $2 \\cdot 2 + 1 \\cdot 6$ is $10$, and $2 \\cdot 6 + 1 \\cdot 2$ is $14$, neither of which are 11. Finally, we can try pair $3$ and $4$. Feel free to try both of them yourself to see why the first must be $3$ and the second must be $4$, not the other way around!", "# Factors of $$N^3$$ Let $$N = 71^3\\cdot 73^2\\cdot 79^4$$. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "12: 1+2+3+5=11. For the same reason 6 * 4 * 1 would not work. The only remaining way to get 24 is 4 * 3 * 2 (we don't include the 1 since we can have at most 3 numbers), and the other numbers 1, 5 and 6 do indeed sum up to 12. There is still another way: The product $$24 = 3 \\cdot 2^3$$, so there must be exactly three factors of $$2$$. Counting factors of $$2$$, $$2$$ has one factor, $$4$$ has two, and $$6$$ has one, so either $$4$$ and $$2$$ are together, or $$4$$ and $$6$$ are. But $$4 \\cdot 6 = 24$$ while $$1 + 2 + 3 + 5 \\ne 12$$. Thus the only other option is $$4 \\cdot 2 \\cdot 3 = 24$$ and $$1 + 5 + 6 = 12$$, which works. • This can be seen as an improvement on trolley813's answer. Apr 4 at 12:54 Here's how I reasoned through this before posting it. Others' methods turned out to be much simpler, though: Since 5 isn't a factor of 24, it must be part of the sum group. The rest of the numbers in the sum group then must add up to 7. Since we don't have a 7, there must be at least two more numbers", "the sum of eleven consecutive positive integers. Sum of twelve consecutive integers There are two numbers $\\{78, 90\\}$ in this category; both are duplicates. Sum of thirteen consecutive integers There is one number $\\{91\\}$ in this category; it is a duplicate. Sum of fourteen consecutive integers $1+2+3+\\cdots+13+14=105$. The sum is greater than $100$. So, we stop here. Total = $49+16+16+4+2+5+1=93$ Answer: $93$ Alternative solution Let $y=x+(x+1)+(x+2)+(x+3)+\\cdots+(x+(n-3))+(x+(n-2))+(x+(n-1))$ be the sum of $n$ consecutive integers with the first term equal $x$. We apply the sum formula $y=(2x+n-1)n/2$ for various values of $n$. $n=2$ $y=(2x+2-1)2/2=2x+1$ $2x+1$ is an odd integer. The $49$ integers are $\\{3,5,7,\\cdots,95,97,99\\}$. $n=3$ $y=(2x+3-1)3/2=3x+3$ $3\\cdot 1+3=6$ $3\\cdot 2+3=9$ $3\\cdot 3+3=12$ $\\cdots$ $3\\cdot 30+3=93$ $3\\cdot 31+3=96$ $3\\cdot 32+3=99$ The $16$ odd integers are duplicates of case $n=2$. The remaining $16$ even integers are $\\{6,12,18,\\cdots,90,96\\}$. $n=4$ $y=(2x+4-1)4/2=4x+6$ $4\\cdot 1+6=10$ $4\\cdot 2+6=14$ $4\\cdot 3+6=18$ $\\cdots$ $4\\cdot 21+6=90$ $4\\cdot 22+6=94$ $4\\cdot 23+6=98$ The $7$ multiples of $6$ are duplicates of case $n=3$. The remaining $16$ integers are $\\{10,14,22,26,34,38,46,50,58,62,70,74,82,86,94,98\\}$. $n=5$ $y=(2x+5-1)5/2=5x+10$ $5\\cdot 1+10=15$ $5\\cdot 2+10=20$ $5\\cdot 3+10=25$ $\\cdots$ $5\\cdot 16+10=90$ $5\\cdot 17+10=95$ $5\\cdot 18+10=100$ Fourteen of the integers are duplicates of case $n=3$ and $n=4$. The remaining $4$ integers are $\\{20,40,80,100\\}$. $n=6$ $y=(2x+6-1)6/2=6x+15$ $6\\cdot 1+15=21$ $6\\cdot 2+15=27$ $6\\cdot 3+15=33$ $\\cdots$ $6\\cdot 12+15=87$ $6\\cdot 13+15=93$ $6\\cdot 14+15=99$ All the $14$ integers are odd and" ]

[ "values $$0.95$$ and $$-0.95$$.", "+ 8z + 4 = 0$ .", "up to $x=4$. Now have a close look: the average slope between $4$ and $5$ is about $0.3$, between $6$ and $7$ about $0.2$, hence $f'$ is decreasing here.", "Find the value of", "and 1.39 min at 110, 115, 121 and $125^{\\circ}C$, respectively, and the average z value was $8.36^{\\circ}C$.", "value of a: 19", "values of $v$ and $\\mu$.", "the smallest normalized positive value ($1 * 16^{-1} * 16^{-64}$)", "? Free Version Moderate Finding Average Value of $1+\\frac { 1 }{ x }$ SVCALC-HKPZ8H Find the average value of the function $f(x)=1+\\frac { 1 }{ x }$ on $[1,6]$. A $1+\\ln { \\sqrt [ 5 ]{ 6 } }$ B $5+\\ln { 6 }$ C $5+\\ln { 5 }$ D $1+\\ln { 7 }$", "minimise is: $$s=-0.025x^2+0.425x+1.44375$$ The minimum value of $s$ for the range 0.5-2.5 is 1.65 at $x=0.5$. -", "from $0$ to $3$ would be the average on the days $1,2$ and $3$.", "Make a table showing the amount of money,$M,\\$ that the average hous...", "$$|z-1+2i|^2 + |z-9-4i|^2\\\\ =2(25+76)\\\\ =202$$ So the minimum value is 202. Apr 20, 2019", "value of ${-13.6\\mbox{ eV}}$.", "value of", "value -1.", "real value from 40,000. = \"100x\"/\"300\" + \"300z\"/\"300\" = 40,000 = \"400z\"/\"300\" = 40,000 = 40,000 \\times 300/400 = z ∴ z = 30,000 4 5", "for all values of $c$.", "not strictly greater than $6$. Test case $4$: The average value of $4$ and $9$ is $6.5$ which is not strictly greater than $8$. Test case $5$: The average value of $3$ and $7$ is $5$ which is strictly greater than $2$.", "value of $\\ln z$. – Masacroso Apr 19 '17 at 8:20" ]

[ "with the answer choices we can start with $4$ numbers. You see that if you have 4 5s and you add 15 to the set the resulting mean will be 7 we can verify this with math $$\\frac{5+5+5+5+15}{5}=7$$ adding in 1 to the set you result in the mean to be 6. $$\\frac{5+5+5+5+15+1}{6}=6$$ Thus we conclude that 4 is the correct choice or $\\boxed{\\textbf{(A)}}$", "# What is the mean for 2,5,3,8,0,4,2,7,3? May 22, 2018 $\\frac{34}{9}$, or $3 \\frac{7}{9}$ #### Explanation: To find the mean add all your terms, then divide by the number of terms. $2 + 5 + 3 + 8 + 0 + 4 + 2 + 7 + 3 = 34$ We have 9 terms, so we divide by 9. $\\frac{34}{9}$ Our mean is $\\frac{34}{9}$, or $3 \\frac{7}{9}$", "Then divide in half. $10^?=10^\\frac{6.2}{2}$ You may want to use logs or guess and check to find the answer.", "D. $$f(x^{3})$$ E. $$f(125x)$$ 1. $$f(5x) = 125/125*x^3 = 1/x^3$$ 2.$$f(x/5) = 125 * 125/x^3 = 125^2 / x^3$$ 3. $$f(5x) * f(x/5) = 125 ^ 2/ x^6 = (f(x)) ^ {2}$$ In step 3, you have multiplied 1. & 2. What would be the answer if you divide 1. by 2. ? I am kinda confused. I get f(5x)/f(x/5) = 1/5^6. Help, thanks! The questions is \"what is the value of $$f(5x)*f(\\frac{x}{5})$$...\" (f(5x) multiplied by f(x/5)) not \"what is the value of $$\\frac{f(5x)}{f(\\frac{x}{5})}$$...\" (f(5x) divided by f(x/5)). _________________ Intern Joined: 30 Oct 2011 Posts: 48 Followers: 0 Kudos [?]: 9 [0], given: 13 Re: If f(x)=125/x^3, what is the value of f(5x)*f(x/5) in terms [#permalink] ### Show Tags 18 Jul 2013, 22:47 Bunuel wrote: mneeti wrote: The questions is \"what is the value of $$f(5x)*f(\\frac{x}{5})$$...\" (f(5x) multiplied by f(x/5)) not \"what is the value of $$\\frac{f(5x)}{f(\\frac{x}{5})}$$...\" (f(5x) divided by f(x/5)). Please refer your Ist response in this topic, the quoted part asks to find $$\\frac{f(5x)}{f(\\frac{x}{5})}$$. Also, the question I have from my source of practice questions requires to find the $$\\frac{f(5x)}{f(\\frac{x}{5})}$$. Thanks! Math Expert Joined: 02 Sep 2009 Posts: 33531 Followers: 5939 Kudos [?]: 73678 [0], given: 9903 Re: If f(x)=125/x^3, what is the value of f(5x)*f(x/5) in terms [#permalink] ### Show Tags 18 Jul 2013, 22:49", "solved to find that $f(4) = \\boxed{\\frac{5}{8}}$.", "# Math Help - following numbers have a mean of 5, find the value of x 1. ## following numbers have a mean of 5, find the value of x Hi I need your help. I can't work out how to find X. I know how to find the cofficient of the variation if I knew what X was. Thanks If the following numbers have a mean of 5, find the value of x and find the coefficient of variation 5 x 2 x 9 11 1 2. ## Re: following numbers have a mean of 5, find the value of x write down the formula for the mean: $mean = 5 = \\frac{5+x+2+x+9+11+1}{7}$ $5 = \\frac{2x + 28}{7}$ $5 = \\frac{2x}{7} + \\frac{28}{7}$ solve for x Once you have x, you can compue the coefficient of variation manually. 3. ## Re: following numbers have a mean of 5, find the value of x $\\frac{5+x+2+x+9+11+1}{7} = 5$ then $\\frac{2x+5+2+9+11+1}{7} = 5$ Can you find it now? 4. ## Re: following numbers have a mean of 5, find the value of x No I can't work out what 2x is?? 5. ## Re: following numbers have a mean of 5, find the value of x $\\frac{2x+5+2+9+11+1}{7} = 5$ From here, times both sides by 7, then subtract 5+2+9+11+1 from", "6 so no go. Finally we let the mean=x $\\frac{35+x}{5}=x \\implies 35+x=5x \\implies x=\\frac{35}{4}=8.75.$ and the sequence is 4, 6, 8, 8.75, 17 which has median 8 so no go. So the only option for x is $\\boxed{-5}.$ --mguempel", "= 4$$, then determine the value of $$\\Large \\frac{dL}{dt}$$. (A calculator is recommended.)", "2, 3, 4, and 5 (mean of this set equals 3): $$D = \\sqrt{\\frac{(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-3)^2}{5}} = \\sqrt{\\frac{4 + 1 + 0 + 1 + 4}{5}} = \\sqrt{\\frac{10}{5}} = \\sqrt{2}$$ The answer to the question can be either YES or NO. Not sufficient. Statement 2 is not sufficient by itself either. Again, $$D$$ can be very big if the range is great (see an example frm S1). Consider this set for $$D$$ to be as low as 2: 1, 1, 1, 1, 6. The mean is 2, $$D$$ is calculated as follows: $$D = \\sqrt{\\frac{(1-2)^2+(1-2)^2+(1-2)^2+(1-2)^2+(6-2)^2}{5}} = \\sqrt{\\frac{1 + 1 + 1 + 1 + 16}{5}} = \\sqrt{\\frac{20}{5}} = \\sqrt{4} = 2$$ The answer to the question can be either YES or NO. Not sufficient. Combining both statements, we have enough information to answer the question. The answer is YES. To prove that we have to think of a set with minimum possible range under restrictions of S1 and S2. We will use this set: 1, 2, 3, 6, 8. Its mean is 4. The standard deviation is calculated as follows: $$D = \\sqrt{\\frac{(1-4)^2+(2-4)^2+(3-4)^2+(6-4)^2+(8-4)^2}{5}} = \\sqrt{\\frac{9 + 4 + 1 + 4 + 16}{5}} = \\sqrt{\\frac{34}{5}} = \\sqrt{6.8} > 2$$ S2+S1 is sufficient. _________________ Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will", "## Algebra 2 (1st Edition) To find the mean, we add all of the numbers and then divide by the total number of values that were added. Doing this, we find: $$Mean = \\frac{80+120}{2} = \\frac{200}{2}=100$$", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "question, given that, total there are 5 boxes. the 5 boxes are filled. 1 = 1/5 + 1/5 + 1/5 + 1/5. g. 1 = 1/5 + 2/5 + 3/5 + 1/5. Explanation: In the above-given question, given that, total there are 5 boxes. the 5 boxes are filled. 1 = 1/5 + 2/5 + 3/5 +1/5. h. 1 = 1/11 + 2/11 + 2/11 + 4/11 +1/11 + 8/11. Explanation: In the above-given question, given that, total there are 11 boxes. the 11 boxes are filled. 1 = 1/11 + 2/11 + 2/11 + 4/11 + 1/11 + 8/1. Question 2. Draw and label tape diagrams to model each decomposition. a. 1 = $$\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}$$ 1 = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6. Explanation: In the above-given question, given that, total there are 6 boxes. the 6 boxes are filled. 1 = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6. b. $$\\frac{4}{5}$$ = $$\\frac{1}{5}+\\frac{2}{5}+\\frac{1}{5}$$ 4/5 = 1/5 + 2/5 + 1/5. Explanation: In the above-given question, given that, total there are 5 boxes. the 4 boxes are filled. 4/5 = 1/5 + 2/5 + 1/5. c. $$\\frac{7}{8}$$ = $$\\frac{3}{8}+\\frac{3}{8}+\\frac{1}{8}$$ 7/8 = 3/8 + 3/8 + 1/8. Explanation: In the above-given question, given that, total there are 8 boxes. the 7", "### Home > MC1 > Chapter 1 > Lesson 1.3.3 > Problem1-123 1-123. Calculate the mean of each data set below. Can you find any shortcuts that allow you to find the mean without having to do much calculation? The mean is the sum of all the values divided by the number of values. 1. $6$, $10$, $6$, $10$ $\\text{Mean}=\\frac{6+10+6+10}{4}$ What is the mean of $6$ and $10$? Is it different from the mean of $6$, $10$, $6$, and $10$? $8$ 2. $11$, $12$, $12$, $13$, $12$ What is the mean of $11$ and $13$? How does this help you find the mean more quickly? 3. $0$, $5$, $4$, $8$, $0$, $7$ $\\frac{0+5+4+8+0+7}{6}$ $\\frac{9+15}{6}$ $\\frac{24}{6}$ $4$", "## Basic College Mathematics (10th Edition) $Mean= \\frac{sum of products}{total customers}= \\frac{(8 \\times 2)+ (11 \\times 12)+ (15\\times 5)+ (26 \\times 1 )}{2+12+5+1} =$12.45 $\\approx12.5$ (Rounded up to Nearest Tenth)", "## Basic College Mathematics (9th Edition) $Mean= \\frac{sum of products}{total customers}= \\frac{(8 \\times 2)+ (11 \\times 12)+ (15\\times 5)+ (26 \\times 1 )}{2+12+5+1} =$12.45 $\\approx12.5$ (Rounded up to Nearest Tenth)", "How do you find the mean of the following set of numbers: 12, 14, 16, 12, 13? The mean is $13.4$ the mean $\\overline{x} = \\frac{\\sum {x}_{i}}{n}$ $\\overline{x} = \\frac{12 + 14 + 16 + 12 + 13}{5} = 13.4$ You can also do $\\overline{x} = 12 + \\frac{2 + 4 + 1}{5} = 12 + \\frac{7}{5} = 12 + 1.4 = 13.4$", "Sam Hardin 2022-06-29 The mean of 4 numbers is 5 and the mean of 3 different numbers is 12. What is the mean of the 7 numbers together? razdiralem Expert The mean of a set of numbers is the sum of the numbers over the count of the set (the number of values). We have a set of four numbers and the mean is 5. We can see that the sum of the values is 20: $\\frac{20}{4}=5$ We have another set of three numbers whose mean is 12. We can write that as: $\\frac{36}{3}=12$ To find the mean of the seven numbers together, we can add the values together and divide by 7: $\\frac{20+36}{7}=\\frac{56}{7}=8$$\\frac{20+36}{7}=\\frac{56}{7}=8$ Do you have a similar question?", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", ". . . $=\\;\\boxed{s^5 - 5s^4u + 10s^3u^2 - 10s^2u^3 + 5su^4 - u^5}$", "# How do you evaluate [(7-5)^5/8]-4? First, note that the parentheses mean that $7 - 5$ should be computed first to get $7 - 5 = 2$. Then raise that to the fifth power to get ${\\left(7 - 5\\right)}^{5} = {2}^{5} = 32$ (powers come before divisions in order of operations). Then divide by 8 to get $\\frac{{\\left(7 - 5\\right)}^{2}}{8} = \\frac{32}{8} = 4$. Finally, subtract 4 to get $\\frac{{\\left(7 - 5\\right)}^{2}}{8} - 4 = 4 - 4 = 0$. The outer brackets $\\left[\\right]$ are not technically needed in the notation." ]

[ "? Free Version Moderate Finding Average Value of $1+\\frac { 1 }{ x }$ SVCALC-HKPZ8H Find the average value of the function $f(x)=1+\\frac { 1 }{ x }$ on $[1,6]$. A $1+\\ln { \\sqrt [ 5 ]{ 6 } }$ B $5+\\ln { 6 }$ C $5+\\ln { 5 }$ D $1+\\ln { 7 }$", "How do you find the average value of f(x)=-x^3+7x^2-11x+3 as x varies between [1,5]? Nov 11, 2016 $- 2$ Explanation: Average $= \\frac{\\int f \\left(x\\right) \\mathrm{dx}}{5 - 1}$, between x = 1 and x = 5 $\\left(\\frac{1}{4}\\right) \\left[- {x}^{4} / 4 + 7 {x}^{3} / 3 - 11 {x}^{2} / 3 + 3 x\\right] ,$ between the limits $= \\left(\\frac{1}{4}\\right) \\left[- \\left({5}^{4} - 1\\right) + 7 \\left({5}^{3} - 1\\right) - 11 \\left({5}^{2} - 1\\right) + 3 \\left(5 - 1\\right)\\right]$ $= \\left(\\frac{1}{4}\\right) \\left[- 624 + 868 - 264 + 12\\right]$ $= \\left(\\frac{1}{4}\\right) \\left(- 8\\right)$ $= - 2$", "# How do you find the average value of f(x)=x^5-2x^3+x as x varies between [-1,0]? Feb 15, 2018 $- \\frac{1}{6}$ #### Explanation: $\\text{the average value is found using}$ •color(white)(x)1/(b-a)int_a^bf(x) $\\text{where } \\left[a , b\\right] = \\left[- 1 , 0\\right]$ $\\Rightarrow \\frac{1}{0 - \\left(- 1\\right)} {\\int}_{- 1}^{0} \\left({x}^{5} - 2 {x}^{3} + x\\right) \\mathrm{dx}$ $= {\\left[\\frac{1}{6} {x}^{6} - \\frac{1}{2} {x}^{4} + \\frac{1}{2} {x}^{2}\\right]}_{- 1}^{0}$ $= 0 - \\left(\\frac{1}{6} - \\frac{1}{2} + \\frac{1}{2}\\right) = - \\frac{1}{6}$", "denominator, because we're differentiating with respect to $d$. – egreg Dec 16 '14 at 0:06 • Yup that's where it was. Thanks. – Faraz Masroor Dec 16 '14 at 0:08 The other answers are great. I just want to zero in on the crucial difference between what you got and what you expected. If you wrote $$\\lim_{d\\to x} \\frac{f'(x)-f'(d)}{x-d}$$ that would be $f''(x)$. Note that the quantity $f'(x)$ appearing the numerator is the value of $f'$ at $x$. However, in your numerator, you have $\\frac{f(x)-f(d)}{x-d}$. This quantity is the average value of $f'$ over the interval $[x,d]$.", "= 1}^{k}j = 45+36+28+21+15+10+6+3+1 = 165$. Thus, each term contributes on average $\\frac{165}{45}$, and the sum will be five times this, or $\\frac{165}{9} = \\frac{55}{3}$. The final answer is $p+q = \\boxed{058}$. ### Solution 4(Expected Value) We use expected value of one of the sums, say $|a_2 - a_1|$, since all five are similar, so the average or expected value of the sum is just 5 times the expected value of one of them. To pick two random expected numbers from 1 to 10, we use a well known expected value trick by tying the two ends 1 and 10 with an extra number 0.* This gives 11 spaces, and we distribute 3 gaps evenly to choose our two numbers. We get $\\frac{11}{3}$ and $\\frac{22}{3}$ giving an absolute difference of $\\frac{11}{3}$. Therefore, the average/expected value of the sum of the five would be $\\frac{55}{3}$. This gives $55+3=\\boxed{058}$. • This is like a bendable stick that has equally spaced marks \"end-1-2-3-4-5-6-7-8-9-10-end\", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together).", "7.2, 9.3 Sorted list: 4.7, 6.5, 7.2, 7.42, 8.51, 9.64 Mean: x̄ = ∑x/n x̄ = (4.7+6.5+7.2+7.42+8.51+9.64)/6 x̄ = 43.97/6 x̄ =7.32 Median: 7.42. Mode: no mode. Question 24. 8$$\\frac{1}{2}$$, 6$$\\frac{5}{8}$$, 3$$\\frac{1}{8}$$, 5$$\\frac{3}{4}$$, 6$$\\frac{5}{8}$$, 5$$\\frac{1}{4}$$, 10$$\\frac{5}{8}$$, 4$$\\frac{1}{2}$$ Answer: Given: 8.5, 6.62, 3.12, 5.75, 6.62, 5.25, 10.62, 4.5 Sorted list: 3.12, 4.5, 5.25, 5.75, 6.62, 6.62, 8.5, 10.62 Mean: x̄ = ∑x/n x̄ = (3.12, 4.5, 5.25, 5.75, 6.62, 6.62, 8.5, 10.62)/8 x̄ = x̄ = Median: 6.18 Mode: 6.62 Question 25. MODELING REAL LIFE The weights (in ounces) of several moon rocks are shown in the table. Find the mean, median, and mode of the weights. Answer: Mean x̄ = (2.2 + 2.2 + 3.2 + 2.4 + 2.8 + 3.4 + 2.6 + 3.0 + 2.5)/9 Median: Write the moon rock weights in ascending or descending order. 2.6 is the median Mode: 2.2 is repeated move times So, 2.2 is the mode. REMOVING AN OUTLIER Find the mean, median, and mode of the data with and without the outlier. Which measure does the outlier affect the most? Question 26. 45, 52, 17, 63, 57, 42, 54, 58 Answer: Outliners means removing of the small data value 17 is the outliner x̄ = ∑x/n = (45 + 52 + 17 + 63 + 57 + 42 + 54", "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", "the expected value or the mean can be calculated as follows: Mean = $\\sum (X_i \\times P(X_i))$ = $2 \\times \\large\\frac{2}{30} + $$3\\times\\large\\frac{4}{30} +$$4\\times\\large\\frac{6}{30} +$$5\\times\\large\\frac{8}{30}+$$6\\times\\large\\frac{10}{30} = \\frac{4+12+24+40+60}{30} = \\frac{140}{30} = \\frac{14}{3}$", "= 4$$, then determine the value of $$\\Large \\frac{dL}{dt}$$. (A calculator is recommended.)", "(4), 4, 4, 5, 5, 6, 6, 6, 7 Answer: 4 MEAN: What would we say the mean is? The mean in other words – is the average value. You’d find this value by adding up all the values in the set of data, then dividing this sum by the number of values which exist in the set of data. In the set of data above, there are 17 objects. So, to find the mean of the set of data, we’d have to perform the calculation below: $Mean\\\\ \\\\ =\\frac { 1+1+2+2+3+3+3+4+4+4+4+5+5+6+6+6+7 }{ 17 } \\\\ \\\\ =\\frac { 1\\times 2+2\\times 2+3\\times 3+4\\times 4+5\\times 2+6\\times 3+7 }{ 17} \\\\ \\\\ =\\frac { 66 }{1 7 } \\\\ \\\\$ Answer: 66/17 (Approx: 3.9) In summary: Calculation Description Formula Range Highest value minus the lowest value. None applicable. Mode Value(s) which appear(s) most often. Most frequent. Median Middle value when all values are lined up in ascending order. If there are two middle values, you find their mid-point. None applicable. Mean Sum of all the values ($\\Sigma x\\\\ \\\\$), divided by the number of values ($n\\\\ \\\\$) which exist. $\\frac { \\Sigma x }{ n } \\\\ \\\\$", "# How do you find the average value of f(x)=(x-3)^2 as x varies between [2,5]? Feb 14, 2018 The average value is $= 1$ #### Explanation: The average value of a function $f \\left(x\\right)$ over an interval $\\left[a , b\\right]$ is $\\overline{x} = \\frac{1}{b - a} {\\int}_{a}^{b} f \\left(x\\right) \\mathrm{dx}$ Here, $f \\left(x\\right) = {\\left(x - 3\\right)}^{2} = {x}^{2} - 6 x + 9$ and $\\left[a , b\\right] = \\left[2 , 5\\right]$ Therefore, $\\overline{x} = \\frac{1}{5 - 2} {\\int}_{2}^{5} {\\left(x - 3\\right)}^{2} \\mathrm{dx}$ $= \\frac{1}{3} {\\int}_{2}^{5} \\left({x}^{2} - 6 x + 9\\right) \\mathrm{dx}$ $= \\frac{1}{3} {\\left[{x}^{3} / 3 - 6 {x}^{2} / 2 + 9 x\\right]}_{2}^{5}$ $= \\frac{1}{3} \\left(\\left(\\frac{125}{3} - 75 + 45\\right) - \\left(\\frac{8}{3} - 12 + 18\\right)\\right)$ $= \\frac{1}{3} \\left(\\frac{117}{3} - 36\\right)$ $= \\frac{1}{3} \\cdot 3$ $= 1$", "terms. For the $x$ values, this is: $\\frac{1(2)+2+4(2)+5+6+10}{8} = \\frac{29}{8} = 3.625$. For the $y$ values, this is: $\\frac{5+10(2)+15+20(2)+25+30}{8} = \\frac{135}{8} = 16.875$. Therefore, the modes are $1$ and $4$ and $10$ and $20$, the medians are $4$ and $17.5$, and the means are $3.625$ and $16.875$ for $x$ and $y$ respectively. ### Example 5 An economist records the price of different loaves of bread at a store. He gets the following $20$ values: $(1.25, 4.99, 5.79, 5.49, 4.99, 4.99, 3.50, 5.49, 5.99, 4.59, 2.99, 2.50, 1.25, 1.99, 2.50, 5.49, 1.25, 2.99, 5.49, 5.99)$. Based on the results, what is the cost of a typical loaf of bread at this store? Assume all prices are in dollars. ### Solution There are different ways to establish a typical value, all of which are measures of central tendency. In this case, it makes sense find the most common three, mode, median, and mean, to get a good idea of a typical price for a loaf of bread at this store. First, order the data from least to greatest. This is: $(1.25, 1.25, 1.25, 1.99, 2.50, 2.50, 2.99, 2.99, 3.50, 4.59, 4.99, 4.99, 4.99, 5.49, 5.49, 5.49, 5.49, 5.59, 5.99, 5.99)$. Based on this data, the mode is $5.49$ because this value appears $4$ times. Next, find the median. Since there", "is $$20$$ As this data set is in the order least to greatest, first find the least and the greatest number. The greatest number is $$145$$ The range is $$145$$ minus $$20$$, which is $$125$$. Subtract the set’s least value from the set’s greatest value to discover the range. The midrange: $$\\frac{145+20}{2}=\\frac{165}{2}=82.5$$ Now, we add the least and the greatest values together and divide that number by two. The Answer is: Range: $$125$$ and Midrange: $$82.5$$", "### Home > APCALC > Chapter 9 > Lesson 9.4.3 > Problem9-134 9-134. No calculator! Calculate the average value of $y = 5x^3$ over $[ - \\frac { \\pi } { 4 } , \\frac { \\pi } { 4 } ]$ using two different methods. For each method, explain your process. Homework Help ✎ $\\frac{1}{\\frac{\\pi}{4}-\\Big(-\\frac{\\pi}{4}\\Big)}\\int_{-\\pi/4}^{\\pi/4}5x^3dx$", "How do you find the average value of f(x)=x-2sqrtx as x varies between [0,2]? Apr 9, 2018 Average value is given by $A = \\frac{1}{b - a} {\\int}_{a}^{b} f \\left(x\\right) \\mathrm{dx}$ Therefore $A = \\frac{1}{2} {\\int}_{0}^{2} x - 2 \\sqrt{x} \\mathrm{dx}$ $A = \\frac{1}{2} {\\left[\\frac{1}{2} {x}^{2} - \\frac{4}{3} {x}^{\\frac{3}{2}}\\right]}_{0}^{2}$ $A = \\frac{1}{2} \\left[2 - \\frac{4}{3} \\sqrt{8}\\right]$ $A = 1 - \\frac{2}{3} \\left(2\\right) \\sqrt{2}$ $A = 1 - \\frac{4}{3} \\sqrt{2} \\approx - 0.89$ Hopefully this helps!", "+0 Can someone confirm pls? 0 319 2 +61 Could someone please confirm? I think the average is 4.2 and the mode is 1, if not could someone correct me? May 15, 2021 2+0 Answers #1 +102 +2 1. If you add all the values up together, you would get $$21$$. Divide that by the number of days and you would get your mean (average): $$\\frac{21}{5} =$$ 4.2 2. Just count which number of cookies Grogg ate most in a day; the mode would be 1. You are correct on all of them! Hope that helped, AnxiousLlama May 15, 2021 #2 +873 +1 (a) $\\frac{1+11+3+5+1}{5}=\\frac{21}{5}=\\boxed{4.2}.$ (b) All other values except for $1$ appear exactly once. Thus the mode is $\\boxed{1}.$ You are correct on both of them. - Jimmy May 15, 2021", "## Algebra 2 (1st Edition) To find the mean, we add all of the numbers and then divide by the total number of values that were added. Doing this, we find: $$Mean = \\frac{80+120}{2} = \\frac{200}{2}=100$$", "$$[0;1]$$ with the median value $$M=0.5$$. From(1), we have: $(0.5 + v_p) - (0.5 - v_p) = p,$ which is the same as $v_p = p / 2.$ Thus, if $$X \\sim \\mathcal{U}(0, 1)$$, $\\lim_{n \\to \\infty} \\E[\\QAD(X, p)] = \\frac{p}{2}.$ Here is the corresponding plot:", "The mean of the following data is 20.6 .Find the value of p. Question: The mean of the following data is 20.6 .Find the value of p. Solution: It is given that, Mean = 20.6 $\\Rightarrow \\frac{25 p+530}{50}=20.6$ ⇒ 25p + 530 = 20.6 × 50 ⇒ 25p = 1030 − 530 ⇒ 25p = 500 ⇒ p = 500/25 = 20 ⇒ p = 20 ∴ p = 20.", "cooloicons62 2022-06-29 How do you find the average value of ${x}^{3}$ as x varies between -1 and 2 Oliver Shepherd Expert Average value of f(x) over interval [a,b] is $f\\left(x\\right)=\\frac{{\\int }_{a}^{b}f\\left(x\\right)dx}{b-a}$ Here that means that: $f\\left(x\\right)=\\frac{{\\int }_{-1}^{2}{x}^{3}dx}{2-\\left(-1\\right)}$ $=\\frac{\\left[\\frac{{x}^{4}}{4}{\\right]}_{-1}^{2}}{3}=\\frac{5}{4}$ Do you have a similar question?" ]

[ "an 68.", "68 -", "84, 92", "is 72.", "72.", "by their highest common factor.", "7. Which of the following shows the greatest common factor of $180$, $144$, and $108$?", "sets of through 67.5 are the!", "at least 66 equal to c. Among", "65-72.", "a factor 104/0.731 = 142. Impressive.", "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "'11 at 4:01 Also, 72 is easy to divide by lots of things in your head, while 69.3 is rather harder. – Ross Millikan Jul 21 '11 at 4:07", "within 72", "93. [^]", "1.77, 1.92, 1.86", "divisor of 72 is acceptable.", "of 90 and 92, so 91 24. 1018 yes, all answers were right. thanks!", "93-107", "0.92, 0.88" ]

[ "# 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.", "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?", "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", "= 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$\\gcd(a^2, b^2) = 1\\$. -", "# 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 $G C F \\left({x}^{2} , {x}^{3}\\right) = {x}^{2}$.", "$n$ 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 $$4$$.", "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)", "Solution From Solution 1, the prime factorization of $180 = 2^2 \\cdot 3^2 \\cdot 5$. The prime factorization of $594 = 2 \\cdot 3^3 \\cdot 11$. Hence, $\\operatorname{lcm} (180,594) = 2^2 \\cdot 3^3 \\cdot 5 \\cdot 11$, and $\\operatorname{gcf} (180,594) = 2 \\cdot 3^2$. Therefore, $\\frac{\\operatorname{lcm} (180,594)}{\\operatorname{gcf} (180,594)} = \\frac{2^2 \\cdot 3^3 \\cdot 5 \\cdot 11}{2 \\cdot 3^2} = 2 \\cdot 3 \\cdot 5 \\cdot 11 \\cdot = 330 \\Longrightarrow \\boxed{\\textbf{(C)}\\ 330}$", "=& {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", "# 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", "thinking that this is the LCM: 2 * 3 * 323 * 589 = 1,141,482 Now compare that to using the prime factor method: 1938 = 2 * 3 * 17 * 19 3534 = 2 * 3 * 19 * 31 LCM = 2 * 3 * 17 * 19 * 31 This way, it's easy to see that the LCM is actually 2 * 3 * 17 * 19 * 31 = 60,078 This is still large, but significantly smaller. What happened here is that there were (intentionally) some larger prime factors that you might not think to try. So the nice thing about the prime factor method is that there's no way to quit too early. Also, it directly relates the idea of prime factors to the idea of \"least common multiple,\" and doesn't require you to memorize a made-up method for calculating that without really understanding what you're doing. :^D On the other hand, you are in fact focusing on common factors, and pulling those out is just what you need to make a common multiple as small as possible. By the way, have you noticed that the left-hand-column of the work is the GCF? The method amounts to saying, $$\\text{LCM}(A,B)=\\frac{AB}{\\text{GCF}(A,B)}=\\text{GCF}(A,B)\\cdot\\frac{A}{\\text{GCF}(A,B)}\\cdot\\frac{B}{\\text{GCF}(A,B)}$$ Zoe wasn’t quite satisfied: Thank you very much! I guess I understand slightly", "# How do you find the GCF of 36a^3b, 56ab? Jan 15, 2017 GCF of $36 {a}^{3} b$ and $56 a b = 4 a b$ 36a^3b = 2×2×3×3×a×a×a×b 56ab = 2×2×2×7×a×b If we observe the factors of both, we will find that both have $2 , 2 , a \\mathmr{and} b$ as common factors in both. Therefore, GCF of $36 {a}^{3} b$ and 56ab = 2×2×a×b = 4ab.", "Mathematics OpenStudy (anonymous): Fact5or the polynomial 21x3-12x2 OpenStudy (anonymous): Hint: Prime factorization of the terms gives $$3 \\cdot 7\\cdot x \\cdot x\\cdot x - 2\\cdot 2 \\cdot 3 \\cdot x \\cdot x$$. What are the common factor(s) you can pull out (GCF)?", "What is the greatest common factor of 6, 12, and 25? 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 Feb 12, 2017 The GCF is 1. Explanation: First, we have to break down each number to its prime factors. Let's take the number 6 first: $6 = 2 \\cdot 3 \\cdot 1$ Although multiplying by 1 is not necessary, sometimes it helps visualize and understand (like in this case) that the GCF is 1. Now 12: $12 = 6 \\cdot 2 \\cdot 1$ $12 = 3 \\cdot 2 \\cdot 2 \\cdot 1$ $12 = 3 \\cdot {2}^{2} \\cdot 1$ Finally, 25: $25 = 5 \\cdot 5 \\cdot 1$ $25 = {5}^{2} \\cdot 1$ The only factors that are written throughout our prime factorization are $1 , 2 , 3 , 5$ However, not all of the numbers share all of the factors. The only (and the greatest) factor that all three numbers share is the number 1. So the answer is 1. 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 • 40 minutes ago • 44 minutes ago • 49 minutes ago •", "Informative line # Greatest Common Factor (GCF) of Prime and Co-Prime Numbers ## GCF of two or more prime numbers • Two prime numbers do not have any common factor other than $$1.$$ One $$(1)$$ is the only common factor of all the prime numbers. • GCF of two prime numbers is the highest common factor between both the prime numbers. • So, GCF of two prime numbers is always 1. For example: Find the GCF of the two prime numbers, $$13$$ and $$17$$. Factors of $$13=1×13$$ Factors of $$17=1×17$$ Only 1 is the common factor of $$13$$ and $$17$$. Thus, GCF of $$13$$ and $$17$$ is $$1.$$ ## GCF of co-prime numbers • The co-prime numbers are the set of two or more numbers which has only one (1) as the common factor. • Thus, GCF of co-prime numbers is always one (1). For example: Find the GCF of the co-prime numbers, 14 and 15. Factors of 14 = 1 × 2 × 7 Factors of 15 = 1 × 3 × 5 Only 1 is the common factor of 14 and 15. Thus, GCF of 14 and 15 is 1. #### Find the GCF of the co-prime numbers, 8 and 9. A 72 B 64 C 81 D 1 × Given co-prime numbers: 8 and 9", "# How do you find the LCM and GCF (of two or more numbers) using prime factorization (prime factor trees?)? Thanks! Nov 3, 2017 First find all the prime factors. For GCF find all the factors that are same and multiply. For LCM find all the factors that are different, and multiply. #### Explanation: For GCF find the prime numbers that are shared in common and multiply. As an example 18 and 24 $18 = 2 \\times 3 \\times 3$ $24 = 2 \\times 2 \\times 2 \\times 3$ There are shared factors of 2 and 3. so the GCF is $2 \\times 3 = 6$ The LCM find prime numbers that are different that are not shared and multiply. There are 3 two's in 24 and only 1 in 18 so there are 3 two's that are not shared. There are 2 three's in 18 and only 1 in 24 so there are 2 three's that are not shared The LCM is $2 \\times 2 \\times 2 \\times 3 \\times 3 = 72$", "is the largest number that will divide into all those given. To find it, the smallest prime numbers should be divided into each one. Prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19. For the given numbers $36$ and $90$, both divided by $2$ give $18$ and $45$. $18$ will divide into both $36$ and $90$, but $45$ will not, so $18$ is the GCF. Oct 14, 2017 G C F 18 It’s also called Greatest Common Divisor G C D #### Explanation: To find G C F of 36, 90: First write the factors of both terms : Factors of $36 = 2 \\cdot \\textcolor{red}{2 \\cdot 3 \\cdot 3}$ Factors of $90 = \\textcolor{red}{2 \\cdot 3 \\cdot 3 \\cdot} 5$ Select the common factors in both terms as marked redabove. $\\textcolor{red}{2 \\cdot 3 \\cdot 3} =$18 is the G C F Nov 21, 2017 Here's a way to find the GCF without using prime factors #### Explanation: Instead of finding the prime factors of the two numbers, ~ make a list of ALL the factors of each number ~ then pick the largest (\"greatest\") one they have in common. To find ALL the factors of a number: ~ Start by factoring by 1 and writing the factors down. ~ Then factor by 2, then by", "How do you find GCF of 2 and 14? 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 1 Annie Share Jan 21, 2017 2 Explanation: Write each number in prime factor form $2 = 2$ $14 = 2 \\cdot 7$ The greatest number which is a factor of both is 2 • 22 minutes ago • 24 minutes ago • 29 minutes ago • 33 minutes ago • 6 minutes ago • 7 minutes ago • 7 minutes ago • 10 minutes ago • 12 minutes ago • 21 minutes ago • 22 minutes ago • 24 minutes ago • 29 minutes ago • 33 minutes ago", "# 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." ]

[ "# 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.", "## 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", "=& {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", "# 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 $2 \\cdot 2 \\cdot 2 = 8$", "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.", "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 $18$", "# 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", "$66$ 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: $\\gcd \\set {p^2 q, p q r} = p q$", "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", "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?", "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", "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", "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:", "= 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$\\gcd(a^2, b^2) = 1\\$. -", "# Greatest common divisor In mathematics, the greatest common divisor (gcd), sometimes known as the greatest common factor (gcf) or highest common factor (hcf) of two integers which are not both zero is the largest integer that divides both numbers. The greatest common divisor of a and b is written as gcd(ab), or sometimes simply as (ab). For example, gcd(12, 18) = 6, gcd(−4, 14) = 2 and gcd(5, 0) = 5. Two numbers are called coprime or relatively prime if their greatest common divisor equals 1. For example, 9 and 28 are relatively prime. The greatest common divisor is useful for reducing vulgar fractions to be in lowest terms. Consider for instance $\\displaystyle {42\\over56}={3\\cdot14\\over4\\cdot14}={3\\over4}$ where we cancelled 14, the greatest common divisor of 42 and 56. ## Calculating the GCD Greatest common divisors can in principle be computed by determining the prime factorizations of the two numbers and comparing factors, as in the following example: to compute gcd(18,84), we find the prime factorizations 18 = 2·32 and 84 = 22·3·7 and notice that the \"overlap\" of the two expressions is 2·3; so gcd(18,84) = 6. In practice, this method is only feasible for very small numbers; computing prime factorizations in general takes far too long. A much more efficient method is the Euclidean algorithm: divide 84 by", "# 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 $4$.", "+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", "# 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", "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)", "## 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 $3(5)=15$." ]

[ "\\cdots$$", "# How do you factor completely 8x^3-8? $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$", "FIGURE THIS OUT $6\\sqrt{2} - 3\\sqrt{2}$", "# Products, sums and divisors $6$ is divisible by $6$. $6\\times6$ is divisible by $6+6$. $6\\times6\\times6$ is divisible by $6+6+6$. Is it true that $\\underbrace{6\\times6\\times\\cdots\\times6}_{n\\text{ times}}$ is divisible by $\\underbrace{6+6+\\cdots+6}_{n\\text{ times}}$ for all positive integers $n$? ×", "How do you factor completely 8x^3-8? 1 Answer It is $8 {x}^{3} - 8 = 8 \\cdot \\left({x}^{3} - 1\\right) = 8 \\cdot \\left(x - 1\\right) \\cdot \\left({x}^{2} + x + 1\\right)$", "$49$, which is true because $6+8$ is divisible by $7$ and $$6^6-6^5\\cdot8+...+8^6\\equiv0(\\mod7).$$ Observe that $8^7\\equiv1\\pmod{49}$. Since $7|273$, so $8^{273}\\equiv1\\pmod{49}$. Observe also that $6^7\\equiv-1\\pmod{49}$. Since $7|273$, so $6^{273}\\equiv-1\\pmod{49}$. Adding,$8^{273}+ 6^{273}\\equiv0\\pmod{49}$.", "is $6.3?$ ×", "-8}$.", "$(-6)(-1)$ 7. $(-7)(8)$ 8. $(6)(-1)$ 9. $(9)(-4)$ 10. $(-9)(-7)$ 11. $(-5)(2)$ 12. $(-2)(-2)$ 13. $(-5)(4)$ 14. $(-3)(-9)$ For questions 45 to 58, find each quotient. 1. $30 \\div -10$ 2. $-49 \\div -7$ 3. $-12 \\div -4$ 4. $-2 \\div -1$ 5. $30 \\div 6$ 6. $20 \\div 10$ 7. $27 \\div 3$ 8. $-35 \\div -5$ 9. $80 \\div -8$ 10. $8 \\div -2$ 11. $50 \\div 5$ 12. $-16 \\div 2$ 13. $48 \\div 8$ 14. $60 \\div -10$", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "6\\right]$).", "of $6$ therefore it is a multiple of $6$", "\\cdot 3 = 16 + 6 = 22$ --> CORRECT", "$6-3i$.", "\\mod 8)$$.", "= \\sqrt{16} + 2$ $6 = 4 + 2$ $6 = 6$", "with $6$ in place of $7$.", "= 2, 4, 6, 8, \\ldots$.", "\\: or \\: 8$​...", "\\cup (2, 6).6)$." ]

[ "\\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", "indices; division and multiplication next; addition and subtraction last. $\\Torange{5 \\times 3} - 2 = \\Tred{15 -2} = 13$ : multiplication then subtraction. \\begin{align*} \\Torange{(3+4)}^2 &+ 3 \\times 7 \\\\ &= \\Tred{7^2} + 3\\times 7= 49 + \\Tgreen{3\\times 7} = \\Tblue{49 + 21} = 70.\\\\ \\Torange{(4-2)}^3 &- 5\\div \\Torange{(2 + 3)}\\\\ & = \\Tred{2^3} - 5\\div 5 = 8 - \\Tgreen{5\\div 5} = \\Tblue{8 - 1} = 7. \\end{align*} When we work through addition and subtraction, we work from left to right. For example, we work out $2-3+5$ as $$\\Tred{2-3} +5 = \\Tred{-1} + 5 = 4,$$ not as $2 - (\\Tred{3+5}) = 2 - \\Tred{8} = -6$. The same applies to multiplication and division.", "\\\\ Z_2 &= U_3 V. \\end{align} As we can see, the complicated case involves approximately 15 multiplications, 6 additions/subtractions, and 0 divisions.", "of the factor times each term in the sum. $a\\cdot \\left(b+c\\right)=a\\cdot b+a\\cdot c$ This property combines both addition and multiplication (and is the only property to do so). Let us consider an example. Note that 4 is outside the grouping symbols, so we distribute the 4 by multiplying it by 12, multiplying it by –7, and adding the products. To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example. \\begin{align} 6+\\left(3\\cdot 5\\right)& \\stackrel{?}{=} \\left(6+3\\right)\\cdot \\left(6+5\\right) \\\\ 6+\\left(15\\right)& \\stackrel{?}{=} \\left(9\\right)\\cdot \\left(11\\right) \\\\ 21& \\ne 99 \\end{align} Multiplication does not distribute over subtraction, and division distributes over neither addition nor subtraction. A special case of the distributive property occurs when a sum of terms is subtracted. $a-b=a+\\left(-b\\right)$ For example, consider the difference $12-\\left(5+3\\right)$. We can rewrite the difference of the two terms 12 and $\\left(5+3\\right)$ by turning the subtraction expression into addition of the opposite. So instead of subtracting $\\left(5+3\\right)$, we add the opposite. $12+\\left(-1\\right)\\cdot \\left(5+3\\right)$ Now, distribute $-1$ and simplify the result. \\begin{align}12+\\left(-1\\right)\\cdot\\left(5+3\\right)&=12+[\\left(-1\\right)\\cdot5+\\left(-1\\right)\\cdot3] \\\\&=12+(-5-3) \\\\&=12+\\left(-8\\right) \\\\&=4 \\end{align} This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign", "4 \\div 2 \\\\ &= 20\\div 2\\\\ &= 10\\end{align*} Cole‘s Solution: \\begin{align*}&= 2+3\\times 4\\div 2\\\\ &= 2+ 12 \\div 2\\\\ &= 2+ 6 \\\\ &= 8\\end{align*} It appears that each student performed the indicated operations in different orders. Of course, there cannot be two correct solutions for the same expression. There is actually a specific order in which operations must be performed. The order of operations is a rule that tells you which operation you need to perform and the order in which it must be done to achieve the correct answer. The order of operations is often called PEMDAS and each of the letters represents one part of the rule. P: parenthesis and grouping symbols; E: exponents; M: multiplication; D: division; A: addition; S: subtraction. MD are performed in the order they appear in the expression from left to right. AS are performed in the order they appear in the expression from left to right. Looking at the two solutions for evaluating the expression \\begin{align*}2+3\\times 4\\div 2\\end{align*} , who has the correct answer? Macy \\begin{align*}&\\text{First}: 2 +3 =5\\\\ &\\text{Next}: 5 \\times 4=20\\\\ &\\text{Then}: 20\\div 2=10\\end{align*} Cole \\begin{align*}&\\text{First}: 3\\times 4=12 \\\\ &\\text{Next}: 12\\div 2=6 \\\\ &\\text{Then}: 6+2=8\\end{align*} Macy evaluated the expression by performing the addition, multiplication and division. Cole performed the multiplication, division and addition. Macy simply completed", "# Thread: Help With Simple Expression 1. ## Help With Simple Expression (2a^2x)^2 When a = 2 and x = 3 The book states the answer as 576, but I can't see how that figure derived. Thank you! 2. ## Re: Help With Simple Expression Hello, MathClown! Exactly where is your difficulty? $\\text{Evaluate: }\\:(2a^2x)^2 \\;\\text{ when }a = 2\\text{ and }x = 3.$ We have: . $\\left(2\\cdot 2^2\\cdot 3\\right)^2 \\;=\\;(2\\cdot4\\cdot3)^2 \\;=\\;(24)^2 \\;=\\;576$ 3. ## Re: Help With Simple Expression Always remember the order of operations: PEDMAS ...Parenthesis, Exponents, Multiplication/Division, and last of all Addition/Subtraction. So in the equation (2a^2x)^2 ....we look to the parenthesis 1st ...(2*2^2*3).... exponents come next -->(2*4*3)=24 , then 24^2=576. 4. ## Re: Help With Simple Expression Originally Posted by MathClown (2a^2x)^2 When a = 2 and x = 3 The book states the answer as 576, but I can't see how that figure derived. Thank you! Is your expression \\displaystyle \\begin{align*} \\left( 2a^2x \\right)^2 \\end{align*} or \\displaystyle \\begin{align*} \\left( 2a^{2x} \\right)^2 \\end{align*}?", "Multiplication and Division before Subtraction $(24\\div4-8)\\cdot7$ $=(6-8)\\cdot7$ $=-2\\cdot7$ $=-14$ 2. $(5+4)\\cdot2^{3}=?$ calculations' order: • Parentheses before Exponents • Exponents before Multiplication $(5+4)\\cdot2^{3}$ $=9\\cdot2^{3}$ $=9\\cdot8$ $=72$ 3. $14-(3+16\\div4)\\cdot2=?$ calculations' order: • Parentheses before Multiplication and Division • Multiplication and Division before Addition and Subtraction $14-(3+16\\div4)\\cdot2$ $=14-(3+4)\\cdot2$ $=14-7\\cdot2$ $=14-14$ $=0$ 4. $6\\cdot(20\\div2-7)+4^{2}=?$ calculations' order: • Parentheses before Exponents • Exponents before Multiplication and Division • Multiplication and Division before Addition and Subtraction $6\\cdot(20\\div2-7)+4^{2}$ $=6\\cdot(10-7)+4^{2}$ $=6\\cdot3+4^{2}$ $=6\\cdot3+16$ $=18+16$ $=34$ i ### Hint The acronym PEMDAS can be memorized through the phrase: Please Excuse My Dear Aunt Sally. i ### Info In the UK they use the acronym BODMAS, which stands for: 1. Brackets 2. Orders 3. Divide Multiply Subtract and in Canada they use the acronym BEDMAS, which stands for: 1. Brackets 2. Exponents 3. Divide Multiply", "21:53 • Nevermind, I guess multiplication>subtraction applies here as well.. Oct 12 '15 at 21:55 • Yes, if we want to express $(-2)^2$ we need parentheses to force the unary minus to be done first. Oct 13 '15 at 0:00 I believe your calculator is correct. \\begin{align*} &(-2-2)^2-(-2+3)(-2-3)-4(-2^2+2)\\\\ &=(-4)^2-(1)(-5)-4(-4+2)\\\\ &=(16)-(-5)+16-8\\\\ &=16+5+16-8\\\\ &=21+8\\\\ &=29 \\end{align*} Note that $-2^2=-4$ whereas $(-2)^2=4$.", "52 \\div 5\\end{align*}. Let \\begin{align*}h=4\\end{align*}. It does look complicated, but if it helps, think of this as a series of steps. The order of operations is your guide. If you follow the order of operations then working through a problem such as this one becomes much easier. \\begin{align*}& P && Step \\ 1: \\ \\text{Substitute} \\ 4 \\ \\text{for} \\ h.''\\\\ & && 8h^2 + [51 \\div (4 \\cdot 4.25)] - 52 \\div 5\\\\ & && 8(4)^2 + [51 \\div (4 \\cdot 4.25)] - 52 \\div 5\\\\ & && Step \\ 2: \\ \\text{Remember PEMDAS. Therefore, perform the operation inside the}\\\\ & && \\text{grouping symbols first. Recall that order of operations must be followed}\\\\ & && \\text{inside grouping symbols also. In this case, multiply} \\ 4 \\times 4.25 \\ \\text{before dividing}\\\\ & && \\text{by} \\ 51.\\\\ & && 8(4)^2 + [51 \\div (4 \\cdot 4.25)] - 52 \\div 5\\\\ & && 8(4)^2 + [51 \\div 17] - 52 \\div 5\\\\ & && 8(4)^2 + 3 - 52 \\div 5\\\\ & E && Step \\ 3: \\ \\text{The next step in order of operations is to simplify the numbers with}\\\\ & && \\text{exponents.}\\\\ & && 8(4)^2+ 3 - 52 \\div 5\\\\ & && 8(4 \\cdot 4) + 3 - 5 \\cdot 5 \\div 5\\\\ & && 8(16) + 3", "the multiplication and division first from left to right. Then do the addition. Solve: First do the multiplication and division: \\begin{align*}6 \\times 7 \\div 6 + 8=42 \\div 6+8=7+8\\end{align*} Last do the addition: \\begin{align*}7+8=15\\end{align*} \\begin{align*}c=15\\end{align*} Check: \\begin{align*}6 \\times 7 \\div 6 + 8=42 \\div 6 +8=7+8=15\\end{align*} Earlier Problem Revisited \\begin{align*}(3 + 5) \\div 2 + 4 \\times 6 = a\\end{align*} We can use the problem solving steps to help. Describe: The equation has parenthesis, division, multiplication and addition. My Job: Do the operations to figure out the value of \\begin{align*}a\\end{align*}. Plan: Do the parenthesis first. Then, do the multiplication and division from left to right. Finally do the addition. Solve: First do the parenthesis: \\begin{align*}(3 + 5) \\div 2 + 4 \\times 6=8 \\div 2 +4 \\times 6\\end{align*} Next do the multiplication and division from left to right: \\begin{align*}8 \\div 2 +4 \\times 6=4+24\\end{align*} Last do the addition: \\begin{align*}4+24=28\\end{align*} \\begin{align*}a=28\\end{align*} Check: \\begin{align*}(3 + 5) \\div 2 + 4 \\times 6=8 \\div 2 +4 \\times 6 = 4+24=28\\end{align*} Vocabulary Order of operations tell us the correct order of evaluating math expressions. We always do Parenthesis first. Then we do multiplication and division (from left to right) and finally addition and subtraction (from left to right). Examples Figure out the value of the variables. Follow the order of operations. Example", "expressions is a number sentence that contains numbers, operations and now radicals. Just as we can have expressions without radicals, we can have expressions with them too. Let's look at one. 24+7\\begin{align*}2 \\cdot \\sqrt{4} + 7\\end{align*} Here we have two times the square root of four plus seven. That’s a great question! We can evaluate this expression by using the order of operations. P parentheses E exponents (square roots too) MD multiplication/division in order from left to right AS addition/subtraction in order from left to right According to the order of operations, we evaluate the square root of 4 first. 4=2\\begin{align*}\\sqrt{4} = 2\\end{align*} Next, we substitute that value into the expression. 22+7\\begin{align*}2 \\cdot 2 + 7\\end{align*} Next, we complete multiplication/division in order from left to right. 2×2=4\\begin{align*}2 \\times 2 = 4\\end{align*} Substitute that given value. 4+7=11\\begin{align*}4 + 7 = 11\\end{align*} Let’s look at another one. 4163\\begin{align*}\\sqrt{4} \\cdot \\sqrt{16} - 3\\end{align*} Here we have two radicals in the expression. We can work the same way, by using the order of operations. 416=2=4\\begin{align*}\\sqrt{4} & = 2\\\\ \\sqrt{16} & = 4\\end{align*} Substitute these values into the expression. 243\\begin{align*}2 \\cdot 4 - 3\\end{align*} Next, we complete multiplication/division in order from left to right. 2×4=8\\begin{align*}2 \\times 4 = 8\\end{align*} Finally, we complete the addition/subtraction in order from left to right. 83=5\\begin{align*}8 - 3", "and multiplication! Subtraction and division are not commutative. When adding three numbers, changing the grouping of the numbers gives the same result. For example, $(7+8)+2=7+(8+2)$, since each side of the equation equals $17$. This is true for multiplication, too. For example, $(5 \\cdot \\frac{1}{3}) \\cdot 3= 5 \\cdot (\\frac{1}{3} \\cdot 3)$ since each side of the equation equals 5. These examples illustrate the Associative Property. ### ASSOCIATIVE PROPERTY of Addition If $a$, $b$, and $c$ are real numbers, then $(a+b)+c=a+(b+c)$ of Multiplication If $a$, $b$, and $c$ are real numbers, then $(a \\cdot b) \\cdot c = a \\cdot (b \\cdot c)$ When adding or multiplying, changing the grouping gives the same result. The Associative Property has to do with grouping. If we change how the numbers are grouped, the result will be the same. Notice it is the same three numbers in the same order—the only difference is the grouping. We saw that subtraction and division were not commutative. They are not associative either. \\begin{align*} (10-3)-2 &≠ 10-(3-2) \\\\ 7-2 &≠ 10-1 \\\\ 5 &≠ 9 \\\\ \\end{align*} \\begin{align*} (24 \\div 4) \\div 2 &≠ 24 \\div (4 \\div 2) \\\\ 6 \\div2 &≠ 24 \\div 2 \\\\ 3 &≠ 12 \\\\ \\end{align*} When simplifying an expression, it is always a good idea to plan what the", "division is the opposite of multiplication. 72÷9=\\begin{align*}72 \\div 9 = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}\\end{align*} In this problem, 72 is the number being divided, it is the dividend. 9 is the number doing the dividing, it is the divisor. We can complete this problem by thinking of our multiplication facts and working backwards. Ask yourself \"What number multiplied by 9 equals 72?\" If you said \"8\", you're right! 9 x 8 = 72, so 72 can be split into 8 groups of 9. The answer to a division problem is called the quotient. Sometimes, a number won’t divide evenly. When this happens, we have a remainder. 15÷2=\\begin{align*}15 \\div 2 =\\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}\\end{align*} Hmmm. This is tricky, fifteen is not an even number. There will be a remainder here. We can use an “r\\begin{align*}r\\end{align*}” to show that there is a remainder. We can also divide larger numbers. We can use a division box to do this. 8)825¯¯¯¯¯¯¯¯¯¯¯\\begin{align*}8 \\overline{)825 \\;}\\end{align*} Here we have a one digit divisor, 8, and a three digit dividend, 825. We need to figure out how many 8’s there are in 825. To do this, we divide the divisor 8 into each digit of the dividend. \\begin{align*}& 8 \\overline{)825 \\;} \\qquad “How \\ many \\ 8’s \\ are \\ there \\ in \\ 8?”\\\\ & \\qquad \\qquad \\ \\ The \\ answer", "+ 5- 4 \\div 2$$ (1) Multiplication is to be done $$= 16 + 5 - 4 \\div 2$$ (2) Division is to be done $$= 16 + 5 - 2$$ (3) Addition is to be done $$= 21 -2$$ (4) Subtraction is to be done $$= 19$$ • By following the MDAS rule, solution will be the same at each time. Note:- If any operation is missing, then check for the next order. #### Choose the option that has the correct order to evaluate the following expression: $$32 \\div 4 × 4 - 6 + 12$$ A $$8 × 4 - 6 + 12 \\;=\\;32-6 +12 = 26 +12 = 38$$ B $$8 × 4 - 6 + 12\\;=\\; 8×4+6=8×10=80$$ C $$32\\div16 - 6 +12\\; =\\; 2-6 + 12 = -4+12 = 8$$ D $$32 \\div 16 - 6 + 12\\;=\\; 32 \\div 10 + 12 = 44 \\div 10 = \\dfrac{44}{10}$$ × The given expression consists of all the four operations, i.e. addition, subtraction, multiplication, and division. Thus, we will use the MDAS rule to evaluate it. $$32 \\div 4 × 4 - 6 + 12$$ (i) Since division is appearing before the multiplication on moving from left to right, so it (division) is to be done first. $$=8×4-6+12$$ $$8×4-6+12$$ (ii) Multiplication is to be done.", "previous case, only this time with four $$2$$'s and one $$6$$ (as $$6=2\\cdot 3$$), resulting in groups such as: $(2,2,2,|,2,6)$ With this representation, there are $$5$$ ways to arrange the elements, and $$2^4$$ ways to place dividers, and as such, there are $$5\\cdot 2^4$$ overcounted cases. Therefore, all that is left is to subtract our overcount, and from this, we see that \\begin{align*}D(96) &= 6\\cdot 2^5 - 5\\cdot 2^4 \\\\&= 2^4 (12-5) \\\\&= 7\\cdot 16 \\\\&= 112 \\end{align*} Thus, A is the correct answer. Solutions: https://live.poshenloh.com/past-contests/amc10/2020B/solutions", "numbers, operations and now radicals. Just as we can have expressions without radicals, we can have expressions with them too. Let's look at one. \\begin{align*}2 \\cdot \\sqrt{4} + 7\\end{align*} Here we have two times the square root of four plus seven. That’s a great question! We can evaluate this expression by using the order of operations. P parentheses E exponents (square roots too) MD multiplication/division in order from left to right AS addition/subtraction in order from left to right According to the order of operations, we evaluate the square root of 4 first. \\begin{align*}\\sqrt{4} = 2\\end{align*} Next, we substitute that value into the expression. \\begin{align*}2 \\cdot 2 + 7\\end{align*} Next, we complete multiplication/division in order from left to right. \\begin{align*}2 \\times 2 = 4\\end{align*} Substitute that given value. \\begin{align*}4 + 7 = 11\\end{align*} Let’s look at another one. \\begin{align*}\\sqrt{4} \\cdot \\sqrt{16} - 3\\end{align*} Here we have two radicals in the expression. We can work the same way, by using the order of operations. \\begin{align*}\\sqrt{4} & = 2\\\\ \\sqrt{16} & = 4\\end{align*} Substitute these values into the expression. \\begin{align*}2 \\cdot 4 - 3\\end{align*} Next, we complete multiplication/division in order from left to right. \\begin{align*}2 \\times 4 = 8\\end{align*} Finally, we complete the addition/subtraction in order from left to right. \\begin{align*}8 - 3 = 5\\end{align*} #### Example A \\begin{align*}6 +", "of c\\begin{align*}c\\end{align*}. Plan: Do the multiplication and division first from left to right. Then do the addition. Solve: First do the multiplication and division: 6×7÷6+8=42÷6+8=7+8\\begin{align*}6 \\times 7 \\div 6 + 8=42 \\div 6+8=7+8\\end{align*} Last do the addition: 7+8=15\\begin{align*}7+8=15\\end{align*} c=15\\begin{align*}c=15\\end{align*} Check: 6×7÷6+8=42÷6+8=7+8=15\\begin{align*}6 \\times 7 \\div 6 + 8=42 \\div 6 +8=7+8=15\\end{align*} #### Concept Problem Revisited (3+5)÷2+4×6=a\\begin{align*}(3 + 5) \\div 2 + 4 \\times 6 = a\\end{align*} We can use the problem solving steps to help. Describe: The equation has parenthesis, division, multiplication and addition. My Job: Do the operations to figure out the value of a\\begin{align*}a\\end{align*}. Plan: Do the parenthesis first. Then, do the multiplication and division from left to right. Finally do the addition. Solve: First do the parenthesis: (3+5)÷2+4×6=8÷2+4×6\\begin{align*}(3 + 5) \\div 2 + 4 \\times 6=8 \\div 2 +4 \\times 6\\end{align*} Next do the multiplication and division from left to right: 8÷2+4×6=4+24\\begin{align*}8 \\div 2 +4 \\times 6=4+24\\end{align*} Last do the addition: 4+24=28\\begin{align*}4+24=28\\end{align*} a=28\\begin{align*}a=28\\end{align*} Check: (3+5)÷2+4×6=8÷2+4×6=4+24=28\\begin{align*}(3 + 5) \\div 2 + 4 \\times 6=8 \\div 2 +4 \\times 6 = 4+24=28\\end{align*} ### Vocabulary Order of operations tell us the correct order of evaluating math expressions. We always do Parenthesis first. Then we do multiplication and division (from left to right) and finally addition and subtraction (from left to right). ### Guided Practice Figure out the value of", "# How do you evaluate [(2 + 4 * 3) - 8] + 81? ##### 1 Answer Jun 5, 2015 To evaluate $\\left[\\left(2 + 4 \\cdot 3\\right) - 8\\right] + 81$ use PEMDAS Parentheses Exponents Multiplication Division Addition Subtraction First evaluate $\\left(2 + 4 \\cdot 3\\right)$ Multiplication takes precedence over addition so $\\left(2 + 4 \\cdot 3\\right) = \\left(2 + 12\\right) = 14$ Next evaluate $\\left[14 - 8\\right]$ $\\left[14 - 8\\right] = 6$ Then evaluate $6 + 81$ $6 + 81 = 87$ So, in place, this sequence looks like: $\\left[\\left(2 + 4 \\cdot 3\\right) - 8\\right] + 81$ $= \\left[\\left(2 + 12\\right) - 8\\right] + 81$ $= \\left[14 - 8\\right] + 81$ $= 6 + 81$ $= 87$", "# How do you evaluate 2( 3) + 12- ( 4\\cdot 3)? Mar 11, 2018 $2 \\left(3\\right) + 12 - \\left(4 \\cdot 3\\right) = 6$ #### Explanation: We evaluate this using the order of operations: 1. Parenthesis 2. Exponents/Roots 3. Multiplication/Division In this case we start by evaluating the parenthesis: $2 \\left(3\\right) + 12 - \\left(4 \\cdot 3\\right) = 2 \\cdot 3 + 12 - 12$ Next we evaluate multiplication: $6 + 12 - 12$ And finally, we do addition and subtraction: $6 + 12 - 12 = 18 - 12 = 6$", "know that 2 is the higest grouping of 4. Now we have to add 3 to 8 ($2\\cdot 4$) to get 11. We write it like this, \\begin{align*} 11 = 2 \\cdot 4 + 3 \\end{align*} ### Generalized Division We can generalize this division with some symbols. Lets call our divident, $n$; our quotient, $q$; our remainder, $r$; and our divisor, $d$. $$n = d \\cdot q + r, \\quad 0 \\leq r < d$$ Practice: Why is it that $r$ is greater than or equal to $0$ and less than the divisor, $d$. Practice: Solve for the quotent and the remainder, 1. $n = 54$, $d = 4$ 2. $n = -54$, $d = 4$ 3. $n = 54$, $d = 70$ Practice: Lets say today is tuesday. What day will it be 1 year from now? We will say that Monday = 0, Tuesday = 1, Wednesday = 2, ..., Sunday = 6. Answer: There are 365 days in 1 year and 7 days in 1 week. Our $n = 365$, $d = 7$, begin{align*} 365 = 7 \\cdot 52 + 1 \\end{align*} We know that 365-1 = 364 days from now, we will have Tuesday as our day, so in 365 days we will have wednesday (which is our answer)." ]

[ "numbers. $(-2)+7=5$ and $7+(-2)=5$ Similarly, the commutative property of multiplication states that numbers may be multiplied in any order without affecting the product. $a \\cdot b = b \\cdot a$ Again, consider an example with real numbers. $(-11) \\cdot (-4) = 44$ and $(-4) \\cdot (-11) = 44$ It is important to note that neither subtraction nor division is commutative. For example, $17-5$ is not the same as $5-17$. Similarly, $20 \\div 5 ≠ 5 \\div 20$. ##### Associative Properties The associative property of multiplication tells us that it does not matter how we group numbers when multiplying. We can move the grouping symbols to make the calculation easier, and the product remains the same. $a(bc)=(ab)c$ Consider this example. $(3 \\cdot 4) \\cdot 5 = 60$ and $3 \\cdot (4 \\cdot 5) = 60$ The associative property of addition tells us that numbers may be grouped differently without affecting the sum. $a+(b+c) = (a+b)+c$ This property can be especially helpful when dealing with negative integers. Consider this example. $[15+(-9)]+23 = 29$ and $15+[(-9)+23]=29$ Are subtraction and division associative? Review these examples. \\begin{align*} 8-(3-15) &= (8-3)-15 \\\\ 8-(-12) &= 5 – 15 \\\\ 20 &≠ -10 \\end{align*} \\begin{align*} 64 \\div (8 \\div 4) &= (64 \\div 8) \\div 4 \\\\ 64 \\div 2 &= 8 \\div 4 \\\\ 32", "&≠2 \\end{align*} As we can see, neither subtraction nor division is associative. ##### Distributive Property The distributive property states that the product of a factor times a sum is the sum of the factor times each term in the sum. $a \\cdot (b+c) = a \\cdot b + a \\cdot c$ This property combines both addition and multiplication (and is the only property to do so). Let us consider an example. Note that 4 is outside the grouping symbols, so we distribute the $4$ by multiplying it by $12$, multiplying it by $–7$, and adding the products. To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example. \\begin{align*} 6+(3 \\cdot 5) &= (6+3) \\cdot (6+5) \\\\ 6+(15) &= (9) \\cdot (11) \\\\ 21 &≠ 99 \\end{align*} A special case of the distributive property occurs when a sum of terms is subtracted. $a-b = a+(-b)$ For example, consider the difference $12-(5+3)$. We can rewrite the difference of the two terms $12$ and $(5+3)$ by turning the subtraction expression into addition of the opposite. So instead of subtracting $(5+3)$, we add the opposite. $12+(-1) \\cdot (5+3)$ Now distribute $-1$ and simplify the result. \\begin {align*} 12-(5+3) &= 12 + (-1) \\cdot (5+3) \\\\", "+ 5- 4 \\div 2$$ (1) Multiplication is to be done $$= 16 + 5 - 4 \\div 2$$ (2) Division is to be done $$= 16 + 5 - 2$$ (3) Addition is to be done $$= 21 -2$$ (4) Subtraction is to be done $$= 19$$ • By following the MDAS rule, solution will be the same at each time. Note:- If any operation is missing, then check for the next order. #### Choose the option that has the correct order to evaluate the following expression: $$32 \\div 4 × 4 - 6 + 12$$ A $$8 × 4 - 6 + 12 \\;=\\;32-6 +12 = 26 +12 = 38$$ B $$8 × 4 - 6 + 12\\;=\\; 8×4+6=8×10=80$$ C $$32\\div16 - 6 +12\\; =\\; 2-6 + 12 = -4+12 = 8$$ D $$32 \\div 16 - 6 + 12\\;=\\; 32 \\div 10 + 12 = 44 \\div 10 = \\dfrac{44}{10}$$ × The given expression consists of all the four operations, i.e. addition, subtraction, multiplication, and division. Thus, we will use the MDAS rule to evaluate it. $$32 \\div 4 × 4 - 6 + 12$$ (i) Since division is appearing before the multiplication on moving from left to right, so it (division) is to be done first. $$=8×4-6+12$$ $$8×4-6+12$$ (ii) Multiplication is to be done.", "$6-1\\cdot0+\\frac{2}{2}=6-0+1=7$ Hello, cheliciouspie! $\\text{Evaluate: }\\:6-1\\cdot0+2 \\div 2$ Don't you know the \"Order of Operations\"? Do all the multiplication and division before any addition and subtraction. So we have: . $6 - (1\\cdot0) + (2\\div 2)$ . . . . . . . . $=\\;6 - 0 + 1$ . . . . . . . . $=\\;7$ @markfl2 and @Soroban i know the order of operation... (MDAS).. My answer as well is 7... but i'm just confuse cause my friends told me that it's 5. they say 6-0+1 = 5....that is 6-(0+1)=6-1=5... They told me that addition first before subtraction...", "Multiplication and Division before Subtraction $(24\\div4-8)\\cdot7$ $=(6-8)\\cdot7$ $=-2\\cdot7$ $=-14$ 2. $(5+4)\\cdot2^{3}=?$ calculations' order: • Parentheses before Exponents • Exponents before Multiplication $(5+4)\\cdot2^{3}$ $=9\\cdot2^{3}$ $=9\\cdot8$ $=72$ 3. $14-(3+16\\div4)\\cdot2=?$ calculations' order: • Parentheses before Multiplication and Division • Multiplication and Division before Addition and Subtraction $14-(3+16\\div4)\\cdot2$ $=14-(3+4)\\cdot2$ $=14-7\\cdot2$ $=14-14$ $=0$ 4. $6\\cdot(20\\div2-7)+4^{2}=?$ calculations' order: • Parentheses before Exponents • Exponents before Multiplication and Division • Multiplication and Division before Addition and Subtraction $6\\cdot(20\\div2-7)+4^{2}$ $=6\\cdot(10-7)+4^{2}$ $=6\\cdot3+4^{2}$ $=6\\cdot3+16$ $=18+16$ $=34$ i ### Hint The acronym PEMDAS can be memorized through the phrase: Please Excuse My Dear Aunt Sally. i ### Info In the UK they use the acronym BODMAS, which stands for: 1. Brackets 2. Orders 3. Divide Multiply Subtract and in Canada they use the acronym BEDMAS, which stands for: 1. Brackets 2. Exponents 3. Divide Multiply", "# How do you evaluate 2( 3) + 12- ( 4\\cdot 3)? Mar 11, 2018 $2 \\left(3\\right) + 12 - \\left(4 \\cdot 3\\right) = 6$ #### Explanation: We evaluate this using the order of operations: 1. Parenthesis 2. Exponents/Roots 3. Multiplication/Division In this case we start by evaluating the parenthesis: $2 \\left(3\\right) + 12 - \\left(4 \\cdot 3\\right) = 2 \\cdot 3 + 12 - 12$ Next we evaluate multiplication: $6 + 12 - 12$ And finally, we do addition and subtraction: $6 + 12 - 12 = 18 - 12 = 6$", "\\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", "as represented by this diagram: In both interpretations, the quotient is the same ($24 \\div 3 = 8$), but it has different meanings in each case. In the first case, the 8 represents the number of bagels in each of the 3 boxes. In the second, it represents the number of boxes that were formed with 3 bagels in each box. These two ways of seeing division are related to how 3, 8, and 24 are related in a multiplication. Both $3 \\boldcdot 8$ and $8 \\boldcdot 3$ equal 24. • $3 \\boldcdot 8 =24$ can be read as “3 groups of 8 make 24.” • $8 \\boldcdot 3 = 24$ can be read as “8 groups of 3 make 24.” If 3 and 24 are the only numbers given, the multiplication equations would be: $$3 \\boldcdot {?} =24$$ $${?} \\boldcdot 3 =24$$ In both cases, the division $24 \\div 3$ can be used to find the value of the “?” But now we see that it can be interpreted in more than one way, because the “?” can refer to the size of a group (as in “3 groups of what number make 24?”), or to the number of groups (as in “How many groups of 3 make 24?”).", "of the factor times each term in the sum. $a\\cdot \\left(b+c\\right)=a\\cdot b+a\\cdot c$ This property combines both addition and multiplication (and is the only property to do so). Let us consider an example. Note that 4 is outside the grouping symbols, so we distribute the 4 by multiplying it by 12, multiplying it by –7, and adding the products. To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example. \\begin{align} 6+\\left(3\\cdot 5\\right)& \\stackrel{?}{=} \\left(6+3\\right)\\cdot \\left(6+5\\right) \\\\ 6+\\left(15\\right)& \\stackrel{?}{=} \\left(9\\right)\\cdot \\left(11\\right) \\\\ 21& \\ne 99 \\end{align} Multiplication does not distribute over subtraction, and division distributes over neither addition nor subtraction. A special case of the distributive property occurs when a sum of terms is subtracted. $a-b=a+\\left(-b\\right)$ For example, consider the difference $12-\\left(5+3\\right)$. We can rewrite the difference of the two terms 12 and $\\left(5+3\\right)$ by turning the subtraction expression into addition of the opposite. So instead of subtracting $\\left(5+3\\right)$, we add the opposite. $12+\\left(-1\\right)\\cdot \\left(5+3\\right)$ Now, distribute $-1$ and simplify the result. \\begin{align}12+\\left(-1\\right)\\cdot\\left(5+3\\right)&=12+[\\left(-1\\right)\\cdot5+\\left(-1\\right)\\cdot3] \\\\&=12+(-5-3) \\\\&=12+\\left(-8\\right) \\\\&=4 \\end{align} This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign", "similar statement for multiplication and division but will be postponed until fractions are discussed. ’PE(MD)(AS)’ is an easy way to remember the order of operations. This means that the order is: Parentheses, Exponents (this will be incorporated later), Multiplication and Division (taken together from left to right), and finally, Addition and Subtraction (taken together from left to right). Let us try a few problems. Example 1.26 1. $$3+2(3+5)=3+2(8)=3+16=19$$ 2. $$3-2(-4+7)=3-2(3)=3-6=-3$$ 3. $$-3-4-2(-2 \\cdot 6-5)=-3-4-2(-12-5)=-3-4-2(-17)=-3-4-(-34)=-3-4+34=27$$ 4. $$-(3-(-6))-(1-4 \\cdot(-5)+4)=-(3+6)-(1-(-20)+4)=-9-(1+20+4)=-9-25=-9+(-25)=-34$$ 5. $$-2(-14 \\div 7+7)=-2(-2+7)=-2(5)=-10$$ 6. $$-3(-2 \\cdot 7-(-5)(4) \\div 2)=-3(-14-(-20) \\div 2)=-3(-14-(-10))=-3(-4)=12$$ 7. $$6 \\div 2 \\times 3=3 \\times 3=9$$ Note: $$6 \\div 2 \\times 36=6 \\div 6=1$$ 8. $$-2(3-1) 2-(8-22) \\div 4=-2(2) 2-(8-4) \\div 4=-2(4)-4 \\div 4=-8-1=-9$$ Exit Problem Evaluate: $$\\left(3^{3}+5\\right) \\div 4-4(7-2)$$", "8 = 0$$ Hence, $-2$ is a root of $g(x)$ and $f(x) = (x + 1)g(x)$. Hence, $x - (-2) = x + 2$ is a factor of $g(x)$. Dividing $g(x) = 2x^3 - 3x^2 - 18x - 8$ by $x + 2$ yields $2x^2 - 7x - 4$. Hence, $$g(x) = 2x^3 - 3x^2 - 18x - 8 = (x + 2)(2x^2 - 7x - 4)$$ so \\begin{align*} f(x) & = 2x^4 - x^3 - 21x^2 - 26x - 8\\\\ & = (x + 1)g(x)\\\\ & = (x + 1)(x + 2)(2x^2 - 7x - 4) \\end{align*} To split the linear term of $2x^2 - 7x - 4$, we must find two numbers with product $2 \\cdot -4 = -8$ and sum $-7$. They are $-8$ and $1$. Hence, \\begin{align*} 2x^2 - 7x - 4 & = 2x^2 - 8x + x - 4 && \\text{split the linear term}\\\\ & = 2x(x - 4) + 1(x - 4) && \\text{factor by grouping}\\\\ & = (2x + 1)(x - 4) && \\text{extract the common factor} \\end{align*} Hence, \\begin{align*} f(x) & = 2x^4 - x^3 - 21x^2 - 26x - 8\\\\ & = (x + 1)(2x^3 - 3x^2 - 18x - 8)\\\\ & = (x + 1)(x + 2)(2x^2 - 7x - 4)\\\\ & = (x + 1)(x +", "# Math Help - 6-1*0+2/2 please answer this maths question? 1. ## 6-1*0+2/2 please answer this maths question? 6-1*0+2/2 please answer this maths question? 2. ## Re: 6-1*0+2/2 please answer this maths question? $6-1\\cdot0+\\frac{2}{2}=6-0+1=7$ 3. ## Re: 6-1*0+2/2 please answer this maths question? Hello, cheliciouspie! $\\text{Evaluate: }\\:6-1\\cdot0+2 \\div 2$ Don't you know the \"Order of Operations\"? Do all the multiplication and division before any addition and subtraction. So we have: . $6 - (1\\cdot0) + (2\\div 2)$ . . . . . . . . $=\\;6 - 0 + 1$ . . . . . . . . $=\\;7$ 4. ## Re: 6-1*0+2/2 please answer this maths question? @markfl2 and @Soroban Thanks for your replies... i know the order of operation... (MDAS).. My answer as well is 7... but i'm just confuse cause my friends told me that it's 5. they say 6-0+1 = 5....that is 6-(0+1)=6-1=5... They told me that addition first before subtraction... 5. ## Re: 6-1*0+2/2 please answer this maths question? Addition and subtraction are done at the same level from left to right. You are right and your friends are not. 6. ## Re: 6-1*0+2/2 please answer this maths question? Originally Posted by MarkFL2 Addition and subtraction are done at the same level from left to right. Thank you for this additional rule..", "{7}{2} + 13 = 5(3) + 7(3) + 13 = 12(3) + 13 = 49$ – Doug M Oct 23 '18 at 1:59 • @ac1002 division is multiplication by an inverse. $2 \\div 3 \\times 6 = 2 \\cdot \\frac{1}{3} \\cdot 6$. Similarly with exponents vs radicals: $\\sqrt{2} = 2^\\frac{1}{2}$. – jaxad0127 Oct 23 '18 at 3:14 \"PEMDAS\" does not mean addition comes before subtraction. In fact addition and subtraction are done from left to right (unless modified by brackets). So for your problem, first do $$9-4=5$$, then $$5+1=6$$. This is an excellent illustration of why people should not rely on memorisation without understanding! • So you solve the math problem from the left to right only when addition and subtraction are involved, or always despite the arithmetic symbol?. – ac1002 Oct 23 '18 at 1:17 • Same thing goes for multiplication and division - left to right. However multiplication and division come before addition and subtraction. For example $$5+4\\times3\\div2-1=5+12\\div2-1=5+6-1=11-1=10\\ .$$ It is often a good idea to insert brackets for clarity, even if they are not strictly necessary. The above could be written $$5+(4\\times3\\div2)-1\\ .$$ – David Oct 23 '18 at 1:23 The misconception here is that addition and subtraction are somehow different operations (and similarly with multiplication and division in some similar questions). The expression $$a", "no universally accepted answer to this. $$8/2x$$ is genuinely unclear. We can apply the same reasoning to $$8/2(4)$$. Finally, there’s the obelus sign ($$\\div$$). I’ve seen some people insist that it means that you evaluate everything to its left and everything to its right before you evaluate the division. I think the confusion here is that the obelus is rarely seen beyond middle school: It’s simply not a sign that serious mathematicians typically use. There’s nothing inherently wrong with it, but it comes from an era when proper typesetting was more difficult. The ideal ways of writing the expression in question leave no question of clarity: $$\\frac{8}{2}(2+2) = 16$$ and $$\\frac{8}{2(2+2)} = 1$$. No questions there. Technically, by the most standard conventions of mathematics, implicit multiplication involving constants should be treated the same as explicit multiplication, and the obelus is just plain old division, so $$8\\div 2(2 + 2) = 8 / 2 \\cdot (2 + 2) = 8 / 2 \\cdot 4 = 4 \\cdot 4 = 16$$. But that’s “technically.” In reality, it’s just plain sloppy. It involves a simplification that gets in the way of clarity, and therefore should be avoided. In practice, in a mathematician’s handwritten notes, were they to write something like $$8\\div 2(2 +2)$$, you would need the surrounding text or thoughts", "(2a+1)^2 + (2b+1)^2 &= 332 \\\\ 6a^2 + 4ab + 6b^2 + 8a + 8b &= 328 \\\\ 3a^2 + 2ab + 3b^2 + 4a + 4b &= 164 \\end{align*} Note that $3a^2 \\le 164$, so $a \\le 7$. There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in $3b^2 + b(2a+4) + 3a^2 + 4a - 164 = 0$. Since both $a$ and $b$ are integers, we must have \\begin{align*} n^2 &= 4a^2 + 16a + 16 - 12(3a^2 + 4a - 164) \\\\ &= -32a^2 - 32a + 16 + 12 \\cdot 164 \\\\ &= 16(-2a^2 - 2a + 1 + 3 \\cdot 41) \\\\ &= 16(-2(a^2 + a) + 124), \\end{align*} where $n$ is an integer. Thus, $-2(a^2 + a) + 124$ must be a perfect square. After trying all values of $a$ from 0 to 7, we find that $a$ can be $3$ or $5$. If $a = 3$, then $b = 5$, and if $a = 5$, then $b = 3$. Therefore, the ordered pairs $(x,y)$ that satisfy the original equation are $\\boxed{(7,11) , (11,7)}$.", "indices; division and multiplication next; addition and subtraction last. $\\Torange{5 \\times 3} - 2 = \\Tred{15 -2} = 13$ : multiplication then subtraction. \\begin{align*} \\Torange{(3+4)}^2 &+ 3 \\times 7 \\\\ &= \\Tred{7^2} + 3\\times 7= 49 + \\Tgreen{3\\times 7} = \\Tblue{49 + 21} = 70.\\\\ \\Torange{(4-2)}^3 &- 5\\div \\Torange{(2 + 3)}\\\\ & = \\Tred{2^3} - 5\\div 5 = 8 - \\Tgreen{5\\div 5} = \\Tblue{8 - 1} = 7. \\end{align*} When we work through addition and subtraction, we work from left to right. For example, we work out $2-3+5$ as $$\\Tred{2-3} +5 = \\Tred{-1} + 5 = 4,$$ not as $2 - (\\Tred{3+5}) = 2 - \\Tred{8} = -6$. The same applies to multiplication and division.", "if I multiply 2 by 3, I get 6. If I want my original number back I can divide 6 by 2 to get 3 or I can divide 6 by 3 to get 2. ## The Divison Algorithm Lets try to divide 11 by 4. Les look at what it fundamentally means to divide two numbers. We can ask, how many groups of fours are made so that it can equal 11? If we try that we get, \\begin{align*} 11 - 4 = 7, \\quad \\text{1 group of four} \\\\ 7 - 4 = 3, \\quad \\text{2 groups of four} \\\\ \\end{align*} Three is less than four so the remainder is 3. This style of groping fours gives us our answer, 11/4 = 2 R3. The 3 called the remainder, the 2 is called the quotent, the 4 is the divisor and the 11 is the divident. We can also divide in the following way, $4 \\cdot 2 = 8$, and 8 is less than 11. So 4 groups 11 at least 2 times; however, we do not know if this is the higest groups of four we can get. To check if this is the higest grouping is simple, we can try $4 \\cdot 3 = 12$, and 12 is definitely greater than 11, so we", "# Euclidean division? ( $16=5\\cdot 3+1$ vs $16=3\\cdot 5+1$) Is the equality $$16=5\\cdot 3+1$$ the euclidean division of $$16$$ by $$3$$ or not ? This question is a point of discord between teachers where some them state that the divisor must be written in the first position (in this example, one has to write $$16=3\\cdot 5+1$$ i.e must write the divisor first then the quotient). What do you think ? • is that a question on notation specifically? if so, add notation tag – gt6989b Jan 17 at 15:32 • @gt6988b yes but I can't edit my post. – Profahk Jan 17 at 15:38 ## 2 Answers $$\\ 16 = 5\\times 3 + 1$$ could arise from dividing $$16$$ by $$5$$ or by $$3$$. Without any further context there is no way to determine if $$5$$ or $$3$$ is the intended divisor. Though - as you mention - one could impose syntactic conventions that imply which is the intended divisor, this is usually not a good idea, since it violates referential transparency, i.e. replacing an expression with an equivalent expression should not change its meaning. But if we replace $$3^2$$ by $$3\\times 3$$ in $$\\,37 = 3^2\\times 4 + 1$$ then it changes the divisor from $$9$$ to $$3$$ (using your convention that the first factor is the", "# How do you evaluate the expression 8 + (36 * 8-204) -: 6? Mar 4, 2018 $8 + \\left(36 \\cdot 8 - 204\\right) \\div 6 = \\textcolor{b l u e}{22}$ #### Explanation: Evaluate: $8 + \\left(36 \\cdot 8 - 204\\right) \\div 6$ Follow the order of operations as indicated by the acronym PEMDAS. 1. Parentheses/brackets 2. Exponents/powers 3. Multiplication and Division in order from left to right. 4. Addition and Subtraction in order from left to right. $8 + \\left(36 \\cdot 8 - 204\\right) \\div 6$ First simplify the parentheses. First multiply $36 \\cdot 8$. $8 + \\left(288 - 204\\right) \\div 6$ Now subtract within the parentheses. $8 + 84 \\div 6$ Next divide $84$ by $6$. $8 + 14$ Now add $8$. $22$", "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}" ]

[ "+0 # Repeating decimals? 0 124 1 +92 1. What is the 20th digit after the decimal point of the sum of the decimal equivalents for the fractions one-seventh and one-third? 2. Find the integer n, 0 $$\\le$$ n $$\\le$$ 180 such that $$\\cos n^\\circ = \\cos 758^\\circ$$ Jun 20, 2020 edited by Unicornrabbit Jun 20, 2020", "+0 # need help 0 103 1 What are the first 10 digits after the decimal point (technically the hexadecimal point...) when the fraction $$\\frac{1}{3}$$ is written in base 16? Jan 27, 2022", "+0 # need help 0 61 1 What are the first 10 digits after the decimal point (technically the hexadecimal point...) when the fraction $$\\frac{1}{3}$$ is written in base 16? Jan 27, 2022", "• Question 12: 3 pts Which decimal is equivalent to $\\frac{10}{3}?$ $0.\\overline{3}$ $1.\\overline{3}$ $3.\\overline{1}$ $3.\\overline{3}$", "The 100th digit In the decimal expansion of $$\\frac {1}{7}$$, what is the 100th digit after the decimal? Details and assumptions In the decimal expansion $$3.14159$$, the 2nd digit after the decimal is 4. ×", "# How do you find the quotient of 20 divided by -2? Sep 7, 2015 $- 10$ #### Explanation: $\\frac{20}{-} 2 = - 10$", "the decimal representation of $$\\frac{1}{70\\,000\\,000}$$ is the same as the 2016th digit after the decimal point in the decimal representation of $$\\frac{1}{7}$$, and is therefore $$7$$.", "MvArrow, $$\\frac{5}{22} = \\frac{5}{2*11}=\\frac{2.5}{11}$$, division by 11 should not be a challenge even without calculator. Re: Which of the following is the 21st digit to the right of the decimal [#permalink] 17 Feb 2018, 08:19", "# Question #22619 ##### 1 Answer Jan 31, 2017 Taking the wording of your question literally, the written equivalent of $\\text{ \"2 3/20\" }$ is: two and three twentieths.", "Which of the following fractions has a decimal equivalent that is a terminating decimal? ~$\\frac{10}{189}$~ ~$\\frac{15}{196}$~ ~$\\frac{16}{225}$~ ~$\\frac{25}{144}$~ ~$\\frac{39}{128}$~", "+0 # help -1 78 1 +145 For how many integer values of $$n$$ between 1 and 474 inclusive does the decimal representation of $$\\frac{n}{475}$$ terminate? Jan 11, 2021", "with seven correct decimal digits. $$\\pi \\approx \\frac{2\\times38^{\\log(38)}}{355205} \\approx 3.1415926$$", "# What is the decimal equivalent of 13/16? $\\frac{13}{18} = \\frac{6.5}{8} = \\frac{3.25}{4} = \\frac{1.625}{2} = 0.8125$", "number after the decimal.", "# Don't forget to carry Number Theory Level 2 If the following infinite series $$S$$ is evaluated as a decimal,what is the 37th digit to the right of the decimal place? $\\large S=\\frac { 1 }{ 9 } +\\frac { 1 }{ 99 } +\\frac { 1 }{ 999 } +\\ldots + \\frac { 1 }{ { 10 }^{ n }-1 } + \\ldots$ ×", "2023-01-01 What is the decimal form of 1/32? Expert $\\frac{1}{32}=\\frac{1}{{2}^{5}}=\\frac{1}{{2}^{5}}×\\frac{{5}^{5}}{{5}^{5}}=\\frac{{5}^{5}}{{10}^{5}}=\\frac{3125}{100000}=0.03125$", "of number (2/23)^3. What is the third digit to the right of the decimal point? A. 0 B. 1 C. 2 D. 4 E. 8 Notice that $$(\\frac{2}{23})^3<(\\frac{2}{20})^3$$. Now, $$(\\frac{2}{20})^3=(\\frac{1}{10})^3=0.001$$. Finally, since $$(\\frac{2}{23})^3<0.001$$, then the third digit to the right of the decimal point of $$(\\frac{2}{23})^3$$ is 0. Didn't get this at all: \"since $$(\\frac{2}{23})^3<0.001$$, then the third digit to the right of the decimal point of $$(\\frac{2}{23})^3$$ is 0. \" How can you say that just because it is 1 in another case, it will be 0 in another number which is shorter? I am missing something. Can you please explain in detail? Thanks. Check any positive number less than 0.001. You'll see that the third digit to the right of the decimal point will be 0. _________________ Intern Joined: 25 May 2019 Posts: 5 Location: United States Concentration: Finance Re: In the decimal notation of number (2/23)^3. What is the thir [#permalink] Show Tags 13 Jun 2019, 11:25 1 Hi, Is there any other way to solve this question? Re: In the decimal notation of number (2/23)^3. What is the thir [#permalink] 13 Jun 2019, 11:25 Display posts from previous: Sort by", "\\frac{e^{.1}}{5!}\\approx .003$ So it is 2 decimal points.", "2012 ### LCKurtz That's correct, but why give a decimal approximation when you have an exact answer:$$1-\\frac 2 {(\\sqrt 2 + 2)^2}$$ 4. Aug 11, 2012 ### Bohrok A better simplification would be 2(√2 - 1)", "# 8 is 20% of what number? Nov 7, 2014 By writing \"20% of $x$ is 8\" in math notation, $\\frac{20}{100} x = 8$ by dividing the numerator and denominator by $20$, $\\implies \\frac{1}{5} x = 8$ by multiply by $5$, $\\implies x = 40$ Hence, 20% of 40 is 8. I hope that this was helpful." ]

[ "# Problem of the Week Problem D and Solution Again and Again and Again ## Problem When the fraction $$\\frac{1}{70\\,000\\,000}$$ is written as a decimal, which digit occurs in the 2023rd place after the decimal point? ## Solution Notice that $$\\frac{1}{70\\,000\\,000} = \\frac{1}{10\\,000\\,000}\\times \\frac{1}{7} = 0.000\\,000\\,1\\times \\frac{1}{7}$$. Also, note that $$\\frac{1}{7} = 0.\\overline{142857}$$. That is, when $$\\frac{1}{7}$$ is written as a decimal, the digits after the decimal point occur in repeating blocks of the $$6$$ digits $$142857$$. Therefore, $$\\frac{1}{70\\,000\\,000} = 0.000\\,000\\,1\\times \\frac{1}{7} = 0.000\\,000\\,1\\times 0.\\overline{142\\,857} = 0.000\\,000\\,0\\overline{14\\,285\\,7}$$. That is, when $$\\frac{1}{70\\,000\\,000}$$ is written as a decimal, the digits after the decimal point will be seven zeros followed by repeating blocks of the six digits $$142857$$. We see the decimal representation of $$\\frac{1}{70\\,000\\,000}$$ has the same repetition as that for $$\\frac{1}{7}$$, but the pattern is shifted over $$7$$ places. Since $$2023-7=2016$$, the 2023rd digit after the decimal point when $$\\frac{1}{70\\,000\\,000}$$ is written as a decimal is the same as the 2016th digit after the decimal point when $$\\frac{1}{7}$$ is written as a decimal. Since $$\\frac{2016}{6} = 336$$, then the 2016th digit after the decimal point occurs after exactly $$336$$ repeating blocks of the $$6$$ digits $$142857$$. Therefore, the 2016th digit is the last digit in the repeating block, which is $$7$$. The 2023rd digit after the decimal point in", "# The decimal representation of $$\\frac{1}{7}$$ is given as 1. $$0.\\overline {142858}$$ 2. $$0.\\overline {142857}$$ 3. $$0.\\overline {142859}$$ 4. $$0.\\overline {142856}$$ ## Answer (Detailed Solution Below) Option 2 : $$0.\\overline {142857}$$ ## Detailed Solution Calculation: ⇒ 1/7 = 0.142857142857 = 0.1̅4̅2̅8̅5̅7̅ ∴ The required result will be 0.1̅4̅2̅8̅5̅7̅ .", "• The numbers $1/3$, $\\sqrt{2}$and $\\pi$ have infinitely long decimal representations, in contrast for example to $\\frac{1}{2}$, whose decimal representation is exactly $0.5$. • The expressions “$1/3$”, “$\\sqrt{2}$” and “$\\pi$” exactly determine the numbers $1/3$, $\\sqrt{2}$ and $\\pi$: These two statements don’t contradict each other. All three numbers have exact definitions. • $1/3$ is exactly the number that gives 1 when multiplied by $3$. • $\\sqrt{2}$is exactly the unique positive real number whose square is 2. • $\\pi$ is exactly the ratio of the circumference of a circle to its diameter. The decimal representation of each one to a finite number of places provides an approximate location of that number on the real line On the other hand, the complete decimal representation of each one represents it exactly, although you can’t write it down. ## Different decimal representations for the same number The decimal representations of two different real numbers must be different. However, two different decimal representations can, in certain circumstances, represent the same real number. This happens when the decimal representation ends in an infinite sequence of $9$’s or an infinite sequence of $0$’s. #### Examples • $0.\\overline{9}=1.\\overline{0}$. This means that $0.\\overline{9}$ is exactly the same number as $1$. It is not just an approximation of $1$ • $3.4\\bar{9}=3.5\\overline{0}$. Indeed, $3.4\\overline{9}$, $3.5$, $35/10$, and $7/2$ are", "the decimal representation of $$\\frac{1}{70\\,000\\,000}$$ is the same as the 2016th digit after the decimal point in the decimal representation of $$\\frac{1}{7}$$, and is therefore $$7$$.", "value of the series is $$a/(1 - r)$$. We then have $0.\\overline{142857} = \\frac{0.142857}{1 - 0.000001} = \\frac{0.142857}{0.999999} = \\frac{142857}{999999} = \\frac{1}{7}.$ And indeed, we get that $$142857 \\times 7 = 999999$$. Similarly, $$1/17 = 0.\\overline{0588235294117647}$$, and $$0588235294117647 \\times 17 = 9999999999999999$$. The above suggests the following: cyclic numbers in base $$10$$ are obtained by looking at the base-$$10$$ representations of $$1/n$$ for positive integers $$n$$. Now, not all values of $$n$$ will give a cyclic number. For example, when $$n = 1, 2, 4, 5, 8, 10$$, we get a decimal representation that terminates: $\\frac{1}{1} = 1, \\frac{1}{2} = 0.5, \\frac{1}{4} = 0.25, \\frac{1}{5} = 0.2, \\frac{1}{8} = 0.125, \\frac{1}{10} = 0.1$ Other times, we get short repeating blocks that are not all that interesting: $\\frac{1}{3} = 0.\\overline{3}, \\frac{1}{9} = 0.\\overline{1}, \\frac{1}{11} = 0.\\overline{09}$ Sometimes we get a decimal representation that is not purely periodic, but eventually is: $\\frac{1}{6} = 0.1\\overline{6}, \\frac{1}{12} = 0.08\\overline{3}, \\frac{1}{14} = 0.0\\overline{714285}$ Now, on to the topic at hand. Consider the expansion of $$1/13 = 0.\\overline{076923}$$. This is purely periodic, with repeating block $$076923$$. And, if we cycle through the digits of $$076923$$ like we did with $$142857$$, we get \\begin{align*} 076923 &= 076923 \\times 1 & 076923 \\times 1 &= \\color{red} 076923 \\\\ 769230 &= 076923 \\times 10 & 076923 \\times", "assume some precision, e.g. four numbers in decimal expansion, and then write your own code to operate on numbers. The exact decimal expansion is not always possible, as you have already realized, so $\\frac{1}{3}$ would expand to \"0.3333\" (arbitrary length from my example. It may seem weird at first, but the floats are also inexact, with precision given by fixed length representation (e.g. 64, 80, 128 bits). $\\frac{1}{3} \\approx 0.333; \\frac{1}{5} \\approx 0.2000$ $0.3333 + 0.2000 = 0.5333$", "The decimal equivalent is a repeating decimal. Write each rational number as a decimal. Then tell whether each decimal is a terminating or a repeating decimal. (Explore Activity and Example 1) Question 1. $$\\frac{3}{5}$$ = ____________ Complete the long division. The decimal equivalent is a terminating decimal. 0.6; Terminating decimal. Question 2. $$\\frac{89}{100}$$ = _____________ Complete the long division. The decimal equivalent is a terminating decimal. 0.89; Terminating decimal. Question 3. $$\\frac{4}{12}$$ = ______________ Complete the long division. Stop when you discover a pattern. The decimal equivalent is a repeating decimal. $$0 . \\overline{3}$$ ; Repeating decimal. Question 4. $$\\frac{25}{99}$$ = ______________ Complete the long division. Stop when you discover a pattern. The decimal equivalent is a repeating decimal. $$0 . \\overline{25}$$ ; Repeating decimal. Question 5. $$\\frac{7}{9}$$ = ____________ Complete the long division. Stop when you discover a pattern. The decimal equivalent is a repeating decimal. $$0 . \\overline{7}$$ ; Repeating decimal. Question 6. $$\\frac{9}{25}$$ = ____________ Complete the long division. The decimal equivalent is a terminating decimal. 0.36; Terminating decimal. Question 7. $$\\frac{1}{25}$$ = ___________ Complete the long division. The decimal equivalent is a terminating decimal. 0.04; Terminating decimal. Question 8. $$\\frac{25}{176}$$ = ____________ Complete long division. Stop when you discover a pattern. This case is special because the first. four decimal are not a part", "$c$ is called the integer part of the decimal representation and is placed on the left side of the decimal point ('.'). Therefore, for irreducible proper fractions the value of $c$ will be $0$ and for non-proper irreducible fractions $c \\geq 1$. Irreducible fractions can be transformed either into a terminating decimal or a recurring (or repeating) decimal. Recurring decimals are of two types - (a) pure recurring (repeating) decimal and (b) mixed recurring (repeating) decimal. For example, $\\frac{1}{2} = 0.5$ or $\\frac{3}{25} = \\frac{12}{100} = 0.12$ are terminating decimals. $\\frac{1}{7} = 0.142857142857142857..... = 0.\\overline{142857}$ and $\\frac{1}{9} = 0.09090909.... = 0.\\overline{09}$ are examples of pure recurring decimals. They are called pure recurring (repeating) since each of the digits in the decimal representation is getting repeated after a certain period. For example, each of the digits in 142857 is getting repeated in the same order in the decimal representation of $\\frac{1}{7}$. That is why $\\frac{1}{7}$ is shortly written in its decimal form as $0.\\overline{142857}$, $\\frac{1}{9}$ is shortly written in its decimal form as $0.\\overline{09}$. Finally, $\\frac{1}{14} = 0.0714285714285714285..... = 0.0\\overline{714285}$ and $\\frac{7}{55} = 0.1272727..... = 0.1\\overline{27}$ are examples of mixed recurring decimals. They are called mixed recurring (repeating) decimals since they involve a few non-recurring digits at the first and then comes the recurring digits which keep on appearing", "of 0.253 are 0.25 which is equal to $$\\frac{2}{10}$$ + $$\\frac{5}{10^{2}}$$ , or $$\\frac{25}{10^{2}}$$, and this tells us within which hundredth we are to place 0.253. The first three decimal digits of 0.253 are 0.253 which is equal to $$\\frac{2}{10}$$ + $$\\frac{5}{10^{2}}$$ + $$\\frac{3}{10^{3}}$$, or $$\\frac{253}{10^{3}}$$, and this tells us within which thousandth we are to place 0.253. And since the decimal terminates here, we are done. Have the students explain this process again in their own words, referring to the number line diagram as they do so. How do you think this process would change if we tried to locate an infinite decimal on the number line? The sequence for an infinite decimal would never end; it would go on infinitely. We need to introduce some notation. If a decimal has a repeating pattern, as for 0.3333… or 7.45454545…, for instance, then a horizontal bar is used to indicate that a block of digits is being repeated. For example, 0.3333… is written as $$0 . \\overline{3}$$ and 7.45454545… as $$7 . \\overline{45}$$. It is conceivable that an infinite decimal could have no repeating pattern. Example 2. The number $$\\frac{5}{6}$$, which is equal to 0.833333…, or $$0 . \\overline{83}$$ is partially represented on the number line below. Now, consider the equality $$\\frac{5}{6}$$ = 0.833333″…” = $$0 . \\overline{83}$$.", "of the pattern that occurs after them. The decimal equivalent is a repeating decimal. $$0.1420 \\overline{45}$$ ; Repeating decimal. Question 9. $$\\frac{12}{1,000}$$ = _____________ Complete the long division. The decimal equivalent is a terminating decimal. 0.012; Terminating decimal. Write each mixed number as a decimal. (Example 2) Question 10. 11$$\\frac{1}{6}$$ = ___________ First, write $$\\frac{1}{6}$$ as a decimal. Then, add 11 to the result. 11 + $$0.1 \\overline{6}$$ = $$11.1 \\overline{6}$$ = $$11.1 \\overline{6}$$ Question 11. 2$$\\frac{9}{10}$$ = ____________ First, write $$\\frac{9}{10}$$ as a decimal. Then, add 2 to the result. 2 + 0.9 = 2.9 = 2.9 Question 12. 8$$\\frac{23}{100}$$ = _____________ First, write $$\\frac{23}{100}$$ as a decimal. Then, add 8 to the result. 8 + 0.23 = 8.23 8.23 Question 13. 7$$\\frac{3}{15}$$ = ___________ First, write $$\\frac{3}{15}$$ as a decimal. Then, add 7 to the result. 7 + 0.2 = 7.2 = 7.2 Question 14. 54$$\\frac{3}{11}$$ = ____________ First, write $$\\frac{3}{11}$$ as a decimal. Then, add 54 to the result. 54 + $$0 . \\overline{27}$$ = $$54 . \\overline{27}$$ = $$54 . \\overline{27}$$ Question 15. 3$$\\frac{1}{18}$$ = _________ First, write $$\\frac{1}{18}$$ as a decimal. Then, add 3 to the result. 3 + $$0.0 \\overline{5}$$ = $$3.0 \\overline{5}$$ = $$3.0 \\overline{5}$$ Question 16. Maggie bought 3$$\\frac{2}{3}$$ lb of apples to make some apple pies. What is the", "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", "limits and analysis, though. – anon Mar 20 '13 at 5:42 What is your understanding of the notation, $0.333\\dots$? – Gerry Myerson Mar 20 '13 at 5:44 Also, did you know that $0.9999\\ldots = 1$? – Kaster Mar 20 '13 at 5:48 This has to be a duplicate. – Thomas Andrews Mar 20 '13 at 6:21 Can you name a number that's between $\\frac{1}{3}$ and $0.333 \\ldots$ without being equal to either of them? – yatima2975 Mar 20 '13 at 11:40 Here is a simple reasoning that $1/3=0.3333...$. Lets denote $0.333333......$ by $x$. Then $$x=0.33333.....$$ $$10x=3.33333...$$ Subtracting we get $9x=3$. Thus $x=\\frac{3}{9}=\\frac{1}{3}$. Since $x$ was chosen as $0.3333....$ it means that $0.3333...=\\frac{1}{3}$. - Cool note: You can extend this to show that 0.9999... = 1 – GraphicsMuncher Mar 20 '13 at 5:50 I don't think that's what GraphicsMuncher meant by \"extend\". I think they meant that the same argument can be used to show that for any single digit $a$ the decimal $.aaa\\ldots$ is $\\frac{a}{9}$. – Jim Mar 20 '13 at 6:38 All this assumes that we can safely perform certain operations on infinite decimals (not all, since $1-0.9999\\ldots$ would probably get us confused) without knowing about limits, and therefore not knowing how infinite decimals are defined. This is an illusion. – Marc van Leeuwen Mar 20 '13", "want is $s$, I have posted a solution below. To get $s+t$ you will, of course, need to compute the length of the period as well. – lulu Jun 11 '19 at 4:25 Assuming you are just interested in the length of the initial block (before the start of the period), write $$n=2^a5^bN$$ where $$\\gcd(10,N)=1$$. Then the length of the initial block is $$\\max(a,b)$$. Example: Consider $$\\frac 1{360}$$. We have $$360=2^3\\times 5^1\\times 3^2$$ so we expect the initial block to have length $$\\max(3,1)=3$$. Indeed, $$\\frac 1{360}=.002\\overline 7$$ To prove this, let $$c=\\max(a,b)$$. Then $$10^c\\times \\frac mn$$ is a rational number with denominator prime to $$10$$. For those, the decimal repeats from the start, so we are done. If you want the total length before you see repetitions then you need to add the length of the period to this. That's the order of $$10\\pmod N$$. Sticking with $$\\frac 1{360}$$ we see that $$N=9$$ so the order of $$10$$ is $$1$$. In general, if you just want an upper bound, take $$\\max (a,b)+ \\varphi(N)$$. • Did you mean max$(3,1)?$ – J. W. Tanner Jun 11 '19 at 4:32 • @J.W.Tanner Yes I did, thank you. I will edit accordingly. – lulu Jun 11 '19 at 4:33 • (+1) You can actually use the Carmichael function in place of the", "• Question 12: 3 pts Which decimal is equivalent to $\\frac{10}{3}?$ $0.\\overline{3}$ $1.\\overline{3}$ $3.\\overline{1}$ $3.\\overline{3}$", "to indicate that the string of digits between the dots repeats indefinitely. This is completely analogous to the fact that the decimal, i.e., base-10, representation of $\\frac{1}{7}$ has the repeating cycle structure $0.\\dot{1}4285\\dot{7}$. Note that the cycle length is $6$ in the ternary representation of $\\frac{1}{7}$ precisely because $6$ is the order of $3$ modulo $7$. (The order of an integer modulo a prime $p$ is the lowest power of the integer which is congruent to $1$ mod $p$. A well known theorem of Fermat guarantees that the order of any integer modulo a prime $p$ is at most $p - 1$). A further illustration is the computation of the ternary form of $\\frac{1}{13}$, for which the division algorithm yields $3 = 0 \\times 13 + 3$ $9 = 0 \\times 13 + 9$ $27 = 2 \\times 13 + 1$ before the remainder repeats, giving the ternary representation $0.\\dot{0}0\\dot{2}$. The cycle length for $\\frac{1}{13}$ is $3$, which is due to the fact that $3$ is the order of $3$ mod $13$. In general, for a prime reciprocal $\\frac{1}{p}$, the ternary cycle length will always be the order of $3$ mod $p$, and the cycle will repeat from the first digit after the point. To implement the above algorithm I wrote the following Python program which reads a", "probably a good time to pay up now. Recall that in order to divide 5 into 2, we would need to do something like this $\\begin{array}{cc} &\\ 0.4 \\\\ 5 &)\\overline{2.0} \\\\ &\\ \\underline{2.0} \\\\ &\\ \\ \\ 0 \\end{array}$ Similarly for binary, we do the exact same thing, except we convert 2 and 5 into their binary forms 0b10 and 0b101 $\\begin{array}{cc} &\\ \\ 0.01 \\\\ 101 &)\\overline{10.00} \\\\ &\\ \\ \\underline{1.01} \\\\ &\\ \\ \\ \\ \\ .110 \\end{array}$ The attentive reader should point out at this point that since 101 divides into 110, we can keep on going $\\begin{array}{cc} &\\ \\ 0.011 \\\\ 101 &)\\overline{10.00} \\\\ &\\ \\ \\underline{1.01} \\\\ &\\ \\ \\ \\ \\ .110 \\\\ &\\ \\ \\ \\ \\ .\\underline{101} \\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 1000 \\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ ... \\end{array}$ Then using the binary point system for non-whole numbers, we find that $\\frac{2}{5}$ is 0.011001100110011… in binary, whereas it is just 0.4 in decimal. Why is 0.4 a repeating binary-point number? This is actually quite simple. Since 10 has prime factors 2 and 5, then in decimal, all numbers with denominators that are exact multiples of only 1, 2", "is 0.001001001001001001001001001001001001001001001000101000110000101100101101011100011110010011111011100101... The first $$47$$ binary digits match those of $$\\frac{1}{7}$$ but it fails at position $$48$$. Yet that number has exactly $$2\\cdot 5^{21}$$ ones and $$2\\cdot 5^{21}$$ zeroes in its period of length $$L=4\\cdot 5^{21}$$. Looking at the full period yields an erroneous conclusions, while if look only at the first $$\\log_2 L \\approx 51$$ digits, you end up with the correct conclusion: the proportion of ones in the binary expansion of $$\\frac{1}{7}$$ is around $$1/3$$. As $$n\\rightarrow\\infty$$, you will notice that it tends to $$\\frac{1}{3}$$, and this is the correct answer.", "to a decimal. First, notice that $$\\frac{4}{3}$$ is an improper fraction. Its value is greater than $$1.$$ The equivalent decimal will also be greater than $$1.$$ We divide $$4$$ by $$3.$$ No matter how many more zeros we write, there will always be a remainder of $$1,$$ and the threes in the quotient will go on forever. The number $$\\text{1.333…}$$ is called a repeating decimal. Remember that the “…” means that the pattern repeats. ### Definition: Repeating Decimal A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly. How do you know how many ‘repeats’ to write? Instead of writing $$1.333\\dots$$ we use a shorthand notation by placing a line over the digits that repeat. The repeating decimal $$1.333\\dots$$ is written $$1.\\stackrel{–}{3}.$$ The line above the $$3$$ tells you that the $$3$$ repeats endlessly. So $$\\text{1.333…}=1.\\stackrel{–}{3}$$ For other decimals, two or more digits might repeat. The table below shows some more examples of repeating decimals. $$\\text{1.333…}=1.\\stackrel{–}{3}$$ $$3$$ is the repeating digit $$\\text{4.1666…}=4.1\\stackrel{–}{6}$$ $$6$$ is the repeating digit $$\\text{4.161616…}=4.\\stackrel{\\text{—}}{16}$$ $$16$$ is the repeating block $$\\text{0.271271271…}=0.\\stackrel{\\text{–––}}{271}$$ $$271$$ is the repeating block ## Example Write $$\\frac{43}{22}$$ as a decimal. ### Solution Divide $$43$$ by $$22.$$ Notice that the differences of $$120$$ and $$100$$ repeat, so there is a repeat in the digits of the", "octal. Binary to decimal (by sums of multiplications): $$100110_2=38$$ Decimal to octal (by a sequence of divisions): $$38=46_8$$ There is a shortcut since $$2^3=8$$ this means every three binary digits are equivalent to one octal digit. Therefore, \\begin{array}{cc} \\underline{100}&\\underline{110_2} \\\\ 4&6_8\\end{array} $100\\,\\, 110_2=46_8$ Convert $$10010101_2$$ to octal \\begin{array}{ccc} \\underline{10}& \\underline{010}&\\underline{101_2} \\\\ 2&2&5_8\\end{array} We can use a similar process for binary to hexadecimal since $$2^4=16$$, thus every four binary digits equals to one hexadecimal digit. Convert $$111011110_2$$ to hexadecimal \\begin{array}{ccc} \\underline{1}& \\underline{1101}&\\underline{1110_2} \\\\ 1&D&E_{16}\\end{array} Convert to $$571_8$$ to binary \\begin{array}{ccc} 5&7&1_8\\\\\\underline{101}& \\underline{111}&\\underline{001_2} \\end{array} So, $$571_8=101\\,111\\,001_2$$ Convert $$E4A_{16}$$ to binary \\begin{array}{ccc} E&4&A_{16}\\\\\\underline{1110}& \\underline{0100}&\\underline{1010_2} \\end{array}", "of simple fractions as decimals: $\\frac 12 = 0.5$ $\\frac 13 = 0.333333\\ldots = 0.\\overline{3}$ $\\frac 23 = 0.666666\\ldots = 0.\\overline{6}$ $\\frac 14 = 0.25$ $\\frac 34 = 0.75$ $\\frac 15 = 0.2$ $\\frac 16 = 0.1666666\\ldots = 0.1\\overline{6}$ $\\frac 17 = 0.14285714285714\\ldots = 0.\\overline{142857}$ $\\frac 18=0.125$ $\\frac 38 = 0.375$ $\\frac 78 = 0.875$ $\\frac 19 = 0.111111\\ldots = 0.\\overline{1}$ $\\frac 29 = 0.222222\\ldots = 0.\\overline{2}$ $\\frac 1{10} = 0.1$ $\\frac 1{11} = 0.09090909\\ldots = 0.\\overline{09}$ $\\frac 1{12} = 0.08333333\\ldots = 0.08\\overline{3}$ A few things can be observed: • Some decimals are short and to the point. • Others have an infinite number of digits (students like to say: \"they are infinite\"), in a repeating pattern. • There can be a single digit that repeats, or a string of two or more digits repeating. • The repeating pattern (period) does not always start right after the decimal point. The long period of $\\frac 17$ sticks out like a sore thumb. It has a few crazy cousins further along: $\\frac 1{13} = 0.\\overline{076923}$ $\\frac 1{17} = 0.\\overline{0588235294117647}$ $\\frac 1{19} = 0.\\overline{052631578947368421}$ There are also the weird $\\frac 1{81} = 0.\\overline{012345679}$ (sic!) and the mysterious $$\\frac 1{9801} = 0.0001020304050607080910111213\\ldots$$ It is also worthwhile to notice this pattern (for later): $$\\frac 19 = 0.\\overline{1}$$ $$\\frac 1{99} = 0.\\overline{01}$$ $$\\frac 1{999}" ]

[ "# Problem of the Week Problem D and Solution Again and Again and Again ## Problem When the fraction $$\\frac{1}{70\\,000\\,000}$$ is written as a decimal, which digit occurs in the 2023rd place after the decimal point? ## Solution Notice that $$\\frac{1}{70\\,000\\,000} = \\frac{1}{10\\,000\\,000}\\times \\frac{1}{7} = 0.000\\,000\\,1\\times \\frac{1}{7}$$. Also, note that $$\\frac{1}{7} = 0.\\overline{142857}$$. That is, when $$\\frac{1}{7}$$ is written as a decimal, the digits after the decimal point occur in repeating blocks of the $$6$$ digits $$142857$$. Therefore, $$\\frac{1}{70\\,000\\,000} = 0.000\\,000\\,1\\times \\frac{1}{7} = 0.000\\,000\\,1\\times 0.\\overline{142\\,857} = 0.000\\,000\\,0\\overline{14\\,285\\,7}$$. That is, when $$\\frac{1}{70\\,000\\,000}$$ is written as a decimal, the digits after the decimal point will be seven zeros followed by repeating blocks of the six digits $$142857$$. We see the decimal representation of $$\\frac{1}{70\\,000\\,000}$$ has the same repetition as that for $$\\frac{1}{7}$$, but the pattern is shifted over $$7$$ places. Since $$2023-7=2016$$, the 2023rd digit after the decimal point when $$\\frac{1}{70\\,000\\,000}$$ is written as a decimal is the same as the 2016th digit after the decimal point when $$\\frac{1}{7}$$ is written as a decimal. Since $$\\frac{2016}{6} = 336$$, then the 2016th digit after the decimal point occurs after exactly $$336$$ repeating blocks of the $$6$$ digits $$142857$$. Therefore, the 2016th digit is the last digit in the repeating block, which is $$7$$. The 2023rd digit after the decimal point in", "+0 # Repeating decimals? 0 124 1 +92 1. What is the 20th digit after the decimal point of the sum of the decimal equivalents for the fractions one-seventh and one-third? 2. Find the integer n, 0 $$\\le$$ n $$\\le$$ 180 such that $$\\cos n^\\circ = \\cos 758^\\circ$$ Jun 20, 2020 edited by Unicornrabbit Jun 20, 2020", "blue 0 at the bottom and carries a 1 over to the green 8. This results in a green 9 at the bottom. Hence, $\\frac{1}{99^2} = 0.\\overline{00010203\\dots969799}$. It remains to add up the digits of 00, 01, 02, …, 96, 97, 99. One could do it manually, or one could notice that the sum of digits is the sum of all the 2-digit numbers (allowing 0 to be a leading digit) except 98. Each digit appears exactly 10 times in each digit column, and so each digit appears exactly 20 times. Thus, \\begin{aligned} b_0 + b_1 + \\dots + b_{n-1} &= 20 \\times (0 + 1 + 2 + \\dots + 9) - 9 - 8 \\\\ &= 20 \\times 45 - 17 \\\\ &= 883. \\end{aligned}", "4.342312$ $\\mbox{division by} \\; 1\\; 000\\; 000 \\; \\mbox{shifts the decimal point by} \\; 6\\; \\mbox{places}$ e.g. $\\frac{47427654.34}{1\\; 000\\; 000} = 47.42765434$ You need to realise that a number like $2.34$ could be written as $0000002.34$ for the purposes of carrying out a division (the number of 'leading' zeroes can be any number you like). With this information you can see that $\\frac{2.4}{100} = 0.0240$ $\\frac{34.23}{10\\; 000} = 0.00342300$ $\\frac{2.71}{1\\; 000\\; 000} = 0.000002710000$ Additionally, you need to realize that a number written without a decimal point can actually be written with a decimal point. For example, $7$ can be written as $7.0$ So therefore (along with the information in the previous section), you can see that $\\frac{2}{100} = 0.02$ $\\frac{36}{10\\; 000} = 0.0036$ $\\frac{71}{1\\; 000\\; 000} = 0.000071$ ## Long Multiplication of Integers Example : $76 \\times 8$ $\\begin{array}{ccc} & 7 & 6 \\\\ & & 8 \\\\ \\hline 6 & 0 & 8\\end{array}$ Example : $462\\times 34$ $\\begin{array}{ccccc} & &4 & 6 & 2 \\\\ & & & 3 & 4 \\\\ \\hline & 1& 8 & 4 & 8\\\\ 1 & 3 & 8 & 6 & 0\\\\ \\hline 1 & 5 & 7 & 0 & 8\\end{array}$ The $1848$ is obtained by multiplying $462$ by $4$, ...while the $13860$ is obtained by multiplying $462$", "+0 # Is 4.4721359549995794-2*sqrt(5)=0 ? 0 328 2 Is 4.4721359549995794-2*sqrt(5)=0 ? Guest Jun 20, 2017 #2 +7324 +2 4.4721359549995794 - 2√5 \\(\\stackrel{?}{=}\\) 0 2*√5 is the same as √20 so we can say... 4.4721359549995794 - √20 \\(\\stackrel{?}{=}\\) 0 4.4721359549995794 \\(\\stackrel{?}{=}\\) √20 Square both sides. (4.4721359549995794)^2 \\(\\stackrel{?}{=}\\) 20 20.00000000000000006423465417690436 \\(\\stackrel{?}{=}\\) 20 So we can see that it is very close, but not quite completely true. Here is how I got the result on the last step. Also, since √20 is a non-terminating decimal, there is no way that a terminating decimal can be equal to it. hectictar Jun 20, 2017 edited by hectictar Jun 20, 2017 #1 0 No not root 5 its root 20 Guest Jun 20, 2017 #2 +7324 +2 4.4721359549995794 - 2√5 \\(\\stackrel{?}{=}\\) 0 2*√5 is the same as √20 so we can say... 4.4721359549995794 - √20 \\(\\stackrel{?}{=}\\) 0 4.4721359549995794 \\(\\stackrel{?}{=}\\) √20 Square both sides. (4.4721359549995794)^2 \\(\\stackrel{?}{=}\\) 20 20.00000000000000006423465417690436 \\(\\stackrel{?}{=}\\) 20 So we can see that it is very close, but not quite completely true. Here is how I got the result on the last step. Also, since √20 is a non-terminating decimal, there is no way that a terminating decimal can be equal to it. hectictar Jun 20, 2017 edited by hectictar Jun 20, 2017", "three-digit combinations appears to occur equally often, and so on. But r is different. The base ten expansion of r doesn’t repeat (the blocks of 0’s and blocks of 9’s get longer and longer) but the 10 decimal digits don’t appear equally often; in fact, only the digits 0, 1, 8 and 9 occur, no matter how far out you go! So r is an example of an irrational number that has a highly patterned decimal expansion. A pretty way to see the structure in the decimal expansion of r is to arrange the digits in a square spiral. For instance, if we list the numbers from 1 to 25 in the pattern $\\left( \\begin{array}{ccccc} 17 & 16 & 15 & 14 & 13 \\\\ 18 & 5 & 4 & 3 & 12 \\\\ 19 & 6 & 1 & 2 & 11 \\\\ 20 & 7 & 8 & 9 & 10 \\\\ 21 & 22 & 23 & 24 & 25 \\end{array} \\right)$ and then replace each number n by the nth digit of r after the decimal point, we get $\\left( \\begin{array}{ccccc} 0 & 0 & 9 & 9 & 9 \\\\ 0 & 1 & 0 & 0 & 8 \\\\ 0 & 0 & 8 & 9 & 9 \\\\ 0 &", "The decimal equivalent is a repeating decimal. Write each rational number as a decimal. Then tell whether each decimal is a terminating or a repeating decimal. (Explore Activity and Example 1) Question 1. $$\\frac{3}{5}$$ = ____________ Complete the long division. The decimal equivalent is a terminating decimal. 0.6; Terminating decimal. Question 2. $$\\frac{89}{100}$$ = _____________ Complete the long division. The decimal equivalent is a terminating decimal. 0.89; Terminating decimal. Question 3. $$\\frac{4}{12}$$ = ______________ Complete the long division. Stop when you discover a pattern. The decimal equivalent is a repeating decimal. $$0 . \\overline{3}$$ ; Repeating decimal. Question 4. $$\\frac{25}{99}$$ = ______________ Complete the long division. Stop when you discover a pattern. The decimal equivalent is a repeating decimal. $$0 . \\overline{25}$$ ; Repeating decimal. Question 5. $$\\frac{7}{9}$$ = ____________ Complete the long division. Stop when you discover a pattern. The decimal equivalent is a repeating decimal. $$0 . \\overline{7}$$ ; Repeating decimal. Question 6. $$\\frac{9}{25}$$ = ____________ Complete the long division. The decimal equivalent is a terminating decimal. 0.36; Terminating decimal. Question 7. $$\\frac{1}{25}$$ = ___________ Complete the long division. The decimal equivalent is a terminating decimal. 0.04; Terminating decimal. Question 8. $$\\frac{25}{176}$$ = ____________ Complete long division. Stop when you discover a pattern. This case is special because the first. four decimal are not a part", "• The numbers $1/3$, $\\sqrt{2}$and $\\pi$ have infinitely long decimal representations, in contrast for example to $\\frac{1}{2}$, whose decimal representation is exactly $0.5$. • The expressions “$1/3$”, “$\\sqrt{2}$” and “$\\pi$” exactly determine the numbers $1/3$, $\\sqrt{2}$ and $\\pi$: These two statements don’t contradict each other. All three numbers have exact definitions. • $1/3$ is exactly the number that gives 1 when multiplied by $3$. • $\\sqrt{2}$is exactly the unique positive real number whose square is 2. • $\\pi$ is exactly the ratio of the circumference of a circle to its diameter. The decimal representation of each one to a finite number of places provides an approximate location of that number on the real line On the other hand, the complete decimal representation of each one represents it exactly, although you can’t write it down. ## Different decimal representations for the same number The decimal representations of two different real numbers must be different. However, two different decimal representations can, in certain circumstances, represent the same real number. This happens when the decimal representation ends in an infinite sequence of $9$’s or an infinite sequence of $0$’s. #### Examples • $0.\\overline{9}=1.\\overline{0}$. This means that $0.\\overline{9}$ is exactly the same number as $1$. It is not just an approximation of $1$ • $3.4\\bar{9}=3.5\\overline{0}$. Indeed, $3.4\\overline{9}$, $3.5$, $35/10$, and $7/2$ are", "good place to start. We want to end up in the $80$s, so the following list is a reasonable start: $\\begin{array}{rl}81& =4!÷\\sqrt{.\\stackrel{˙}{4}}÷.\\stackrel{˙}{4}\\\\ & =\\left(4÷.\\stackrel{˙}{4}{\\right)}^{\\sqrt{4}}\\\\ 65& =\\left(4!+\\sqrt{4}\\right)÷.4\\\\ \\end{array}$ Each of these allows us to waste a $4$ and we can add or subtract $\\sqrt{4}$, $4$, and $4!$ which means that this gets us $67$ and $89$ from $65$; and $77$, $79$, $81$, $83$, and $85$ from $81$. We're left with $71$, $73$, and $75$. The first two are a little tricky as they are prime, but they are $1$ away from $72$ which is easy to get with three $4$s using a division. The key is then to realise that as we're already doing a division, we can throw in another division without using as many $4$s as first glance would demand. $\\begin{array}{rl}71& =\\frac{4!+4!}{\\sqrt{.\\stackrel{˙}{4}}}-\\frac{\\sqrt{.\\stackrel{˙}{4}}}{\\sqrt{.\\stackrel{˙}{4}}}\\\\ & =\\frac{4!+4!-\\sqrt{.\\stackrel{˙}{4}}}{\\sqrt{.\\stackrel{˙}{4}}}\\\\ 73& =\\frac{4!+4!+\\sqrt{.\\stackrel{˙}{4}}}{\\sqrt{.\\stackrel{˙}{4}}}\\end{array}$ This trick can be adapted to make $75$: $75=\\frac{4!+4!+\\sqrt{4}}{\\sqrt{.\\stackrel{˙}{4}}}$ And that's the lot. # 5 Why Bother? Admittedly, my main aim in working through all of these was to find out a set of answers to the four fours puzzle and to see whether it was possible to figure out a systematic approach. When this is set as a puzzle, it is common to see students approach it in an ad hoc fashion. This is quite probably the right setting for", "the denominator by $37$ to get: $\\begin{array}{rcl} \\frac{4}{27}&=&\\frac{4\\times 37}{27\\times 37}\\\\ &=&\\frac{148}{999}\\\\ &=&0.\\overline{148} \\end{array}$ $~$ To transform repeating decimals into fractions: • write all repeating numbers in the numerator • in the denominator, place as many $9$s as there are repeating digits • reduce the fraction by dividing the numerator and denominator by the GCF $~$ For the repeating decimal $0.\\overline{81}$: $\\begin{array}{rcl} 0.\\overline{81}&=&\\frac{81}{99}\\\\ &=&\\frac{81\\div9}{99\\div 9}\\\\ &=&\\frac{9}{11} \\end{array}$ $~$ For the repeating decimal $0.\\overline{63}$: $\\begin{array}{rcl} 0.\\overline{63}&=&\\frac{63}{99}\\\\ &=&\\frac{63\\div9}{99\\div 9}\\\\ &=&\\frac{7}{11} \\end{array}$", "probably a good time to pay up now. Recall that in order to divide 5 into 2, we would need to do something like this $\\begin{array}{cc} &\\ 0.4 \\\\ 5 &)\\overline{2.0} \\\\ &\\ \\underline{2.0} \\\\ &\\ \\ \\ 0 \\end{array}$ Similarly for binary, we do the exact same thing, except we convert 2 and 5 into their binary forms 0b10 and 0b101 $\\begin{array}{cc} &\\ \\ 0.01 \\\\ 101 &)\\overline{10.00} \\\\ &\\ \\ \\underline{1.01} \\\\ &\\ \\ \\ \\ \\ .110 \\end{array}$ The attentive reader should point out at this point that since 101 divides into 110, we can keep on going $\\begin{array}{cc} &\\ \\ 0.011 \\\\ 101 &)\\overline{10.00} \\\\ &\\ \\ \\underline{1.01} \\\\ &\\ \\ \\ \\ \\ .110 \\\\ &\\ \\ \\ \\ \\ .\\underline{101} \\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 1000 \\\\ &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ ... \\end{array}$ Then using the binary point system for non-whole numbers, we find that $\\frac{2}{5}$ is 0.011001100110011… in binary, whereas it is just 0.4 in decimal. Why is 0.4 a repeating binary-point number? This is actually quite simple. Since 10 has prime factors 2 and 5, then in decimal, all numbers with denominators that are exact multiples of only 1, 2", "of simple fractions as decimals: $\\frac 12 = 0.5$ $\\frac 13 = 0.333333\\ldots = 0.\\overline{3}$ $\\frac 23 = 0.666666\\ldots = 0.\\overline{6}$ $\\frac 14 = 0.25$ $\\frac 34 = 0.75$ $\\frac 15 = 0.2$ $\\frac 16 = 0.1666666\\ldots = 0.1\\overline{6}$ $\\frac 17 = 0.14285714285714\\ldots = 0.\\overline{142857}$ $\\frac 18=0.125$ $\\frac 38 = 0.375$ $\\frac 78 = 0.875$ $\\frac 19 = 0.111111\\ldots = 0.\\overline{1}$ $\\frac 29 = 0.222222\\ldots = 0.\\overline{2}$ $\\frac 1{10} = 0.1$ $\\frac 1{11} = 0.09090909\\ldots = 0.\\overline{09}$ $\\frac 1{12} = 0.08333333\\ldots = 0.08\\overline{3}$ A few things can be observed: • Some decimals are short and to the point. • Others have an infinite number of digits (students like to say: \"they are infinite\"), in a repeating pattern. • There can be a single digit that repeats, or a string of two or more digits repeating. • The repeating pattern (period) does not always start right after the decimal point. The long period of $\\frac 17$ sticks out like a sore thumb. It has a few crazy cousins further along: $\\frac 1{13} = 0.\\overline{076923}$ $\\frac 1{17} = 0.\\overline{0588235294117647}$ $\\frac 1{19} = 0.\\overline{052631578947368421}$ There are also the weird $\\frac 1{81} = 0.\\overline{012345679}$ (sic!) and the mysterious $$\\frac 1{9801} = 0.0001020304050607080910111213\\ldots$$ It is also worthwhile to notice this pattern (for later): $$\\frac 19 = 0.\\overline{1}$$ $$\\frac 1{99} = 0.\\overline{01}$$ $$\\frac 1{999}", "the same remainder as $r\\times 1 = r$; that is, the same remainder as $2^n$. So adding the digits of $2^n$ until you get a single number between $1$ and $9$ will give you the same answer as doing it for $2^{n+6}$. This is what you observed: $8=2^3$, and $512=2^9 = 2^{3+6}$. You will get the same answer ($7$) with $2^{15}$, $2^{21}$, $2^{27}$, etc. I note that you were slightly off in describing $512$ as being \"seven steps after\" $8$: it's really only six steps later: $$8\\stackrel{1}{\\mapsto} 16 \\stackrel{2}{\\mapsto}32 \\stackrel{3}{\\mapsto}64\\stackrel{4}{\\mapsto}128\\stackrel{5}{\\mapsto}256\\stackrel{6}{\\mapsto}512.$$ - OK, thanks. I should have known it was from something I learned already, I thought to myself that this must be true but I guess that I forgot from this theorem. – MathematicalPhysicist Dec 30 '11 at 3:43 Also, $2^6 = 64$ happens to have the same digital root as its reciprocal $2^{-6} = \\frac{1}{64} = 0.015625$, namely $1$. Therefore, for any given $n \\in \\mathbb{Z}$, all of the numbers $\\{ 64^k \\cdot n \\ | \\ k \\in \\mathbb{Z}\\}$ have the same digital root as $n$. – r.e.s. Dec 30 '11 at 4:20 This is indeed the case. The value you get out at the end of all your summations is just the value of your number mod 9 (this is because any natural number is", "\\frac{15}{8} &1 & 7 & 21\\\\ \\frac{21}{8} &2 & 5 & 15\\\\ \\frac{15}{8} &1 & 7 & 21\\\\ \\frac{21}{8} &2 & 5 & 15\\\\ \\vdots & \\vdots & \\vdots & \\vdots \\\\ \\end{matrix}$$ The base 3 representation comes from the quotients, so $\\frac{5}{8}$ in ternary must be $0.\\overline{12}_3$. So why is this equivalent to long division? With long division, after finding a quotient and a remainder in a particular step, you then \"carry down the $0$,\" which entails concatenating a $0$ on the end of the remainder. Mathematically, adding a $0$ on the end of a number in base 10 means that you are multiplying it by $10$. Since we're working in base 3 here, we want to multiply the remainder by $3$. That's the only difference between what I've said here and the long division algorithm taught in elementary school. - Thanks for the solution. – Ramana Venkata Oct 7 '11 at 1:55 probably should be 6/10 + 2/100, not =. – zzzzBov Oct 16 '14 at 14:34 @zzzzBov: Good catch! I'll fix it. – Mike Spivey Oct 16 '14 at 16:12 The easiest way to do it, I think, is to convert everything before the decimal point the regular way. Then you just repeat the following process: 1. Subtract 1 until you get between 0 and", "parts of a number are converted with similar methods. They are again based on the equivalence of shifting with doubling or halving. In a fractional binary number such as 0.110101101012, the first digit is $\\begin \\frac \\end$, the second $\\begin \\left(\\frac\\right)^2 = \\frac \\end$, etc. So if there is a 1 in the first place after the decimal, then the number is at least $\\begin \\frac \\end$, and vice versa. Double that number is at least 1. This suggests the algorithm: Repeatedly double the number to be converted, record if the result is at least 1, and then throw away the integer part. For example, $\\begin \\left(\\frac\\right) \\end$10, in binary, is: Thus the repeating decimal fraction 0.... is equivalent to the repeating binary fraction 0.... . Or for example, 0.110, in binary, is: This is also a repeating binary fraction 0.0... . It may come as a surprise that terminating decimal fractions can have repeating expansions in binary. It is for this reason that many are surprised to discover that 0.1 + ... + 0.1, (10 additions) differs from 1 in floating point arithmetic In computing Computing is any goal-oriented activity requiring, benefiting from, or creating computing machinery. It includes the study and experimentation of algorithmic processes and development of both computer hardware , hardware and soft ...", "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", "point. Thanks for pointing out Ross Millikan. please explain me how can i get it. – juantheron Oct 12 '13 at 4:23 An answer was posted, then unfortunately the author deleted it when he thought he misread the problem. But in fact he got it right. I have voted to undelete. – Gerry Myerson Oct 12 '13 at 5:03 Consider the number $N=(\\sqrt{50}+7)^{50}+(\\sqrt{50}-7)^{50}$. Using the Binomial Theorem, or otherwise, we can show that $N$ is an integer. Our number $(\\sqrt{50}+7)^{50}$ is a little below $N$. How much below? Courtesy of the calculator, $(\\sqrt{50}-7)^{50}\\approx 4\\times 10^{-58}$. So the $50$-th digit after the decimal point of $(\\sqrt{50}+7)^{50}$ is $9$. You could replace the calculator by using $$\\sqrt{50}-7=\\frac{1}{\\sqrt{50}+7} < \\frac{1}{14}$$ so $(\\sqrt{50}-7)^{50} < 10^{-50}$. – WimC Oct 12 '13 at 7:07 Thanks André Nicolas and WimC, but i did not understand the line So the $50-$th digit after the decimal point of $(\\sqrt{50}+7)^{50}$ is $= 9$ – juantheron Oct 13 '13 at 8:32 @juantheron: Think about taking away say $0.000000005$ from $17$. The first $8$ digits after the decimal point are $9$'s. In the case of our roblem, the first $57$ digits after the decimal oint are $9$'s. – André Nicolas Oct 13 '13 at 15:06", "in figuring out the last digit of $$n^x$$. The cycle of powers for numbers from 2 to 10 is given below: 2: 2, 4, 8, 6 → all $$2^{4x}$$ will have the same last digit. 3: 3, 9, 7, 1 → all $$3^{4x}$$ will have the same last digit. 4: 4, 6 → all $$4^{2x}$$ will have the same last digit. 5: 5 → all $$5^x$$ will have the same last digit. 6: 6 → all $$6^x$$ will have the same last digit. 7: 7, 9, 3, 1 → all $$7^{4x}$$ will have the same last digit. 8: 8, 4, 2, 6 → all $$8^{4x}$$ will have the same last digit. 9: 9, 1 → all $$9^{2x}$$ will have the same last digit. 10: 0 → all $$10^x$$ will have the same last digit. 5) Many seemingly difficult remainder problems can be simplified using the following formula : $$R of \\frac{x*y}{n} = R of \\frac{(R of \\frac{x}{n})*(R of \\frac{y}{n})}{n}$$ Eg. $$R of \\frac{20*27}{25} = R of \\frac{(R of \\frac{20}{25})*(R of \\frac{27}{25})}{25} = R of \\frac{(20)*(2)}{25} = R of \\frac{40}{25} = 15$$ Eg. $$R of \\frac{225}{13} = R of \\frac{(15)*(15)}{13} = R of {(2)*(2)}{13} = R of \\frac{4}{13} = 4$$ 6) $$R of \\frac{x*y}{n}$$ , can sometimes be easier calculated if we take it as $$R of \\frac{(R of", "difference will be negative. To subtract, we add the} \\\\ \\text{numerals. Write the numbers so the decimal points line} \\\\ \\text{up vertically.}} & { \\; \\; 23.5 \\\\ \\underline{+41.38}} \\\\ \\\\ \\\\ { \\text{Put 0 as a placeholder after the 5 in 23.5.} \\\\ \\text{Remember, } \\frac{5}{10}=\\frac{50}{100} \\text{ so } 0.5=0.50.} & { \\; \\; 23.50 \\\\ \\underline{+41.38}} \\\\ \\\\ \\\\ {\\text{Add the numbers as if they were whole numbers.} \\\\ \\text{Then place the decimal point in the sum.}} & {\\; \\; 23.50 \\\\ \\underline{+41.38} \\\\ \\; \\; 64.88 } \\\\ \\\\ \\\\ \\text{ Write the result with the correct sign.} & 64.88−23.5−41.38=−64.88 \\end{array}$$ $$\\begin{array}{ll} \\text{} & 14.65−20 \\\\ \\\\ \\\\ {\\text{The difference will be negative. To subtract, we} \\\\ \\text{subtract 14.65 from 20.}} \\\\ \\\\ \\\\ {\\text{Write the numbers so the decimal points line up} \\\\ \\text{vertically.}} & { \\; \\; 20 \\\\ \\underline{−14.65}} \\\\ \\\\ \\\\ {\\text{Remember, 20 is a whole number, so place the} \\\\ \\text{decimal point after the 0.}} \\\\ \\\\ \\\\ \\text{Put in zeros to the right as placeholders.} & { \\; \\; 20.00 \\\\ \\underline{−14.65}} \\\\ \\\\ \\\\ \\text{Subtract and place the decimal point in the answer.} & {\\begin{array}{lcccc} {} & 9 & {} & 9 & {} \\\\ 1 & \\cancel{10} & {} & \\cancel{10} & 10 \\\\ 2 & 0", "\\end{array}$ In order to eliminate $d$ we look at $▵ACD$ , where we have: $cos\\stackrel{^}{A}=\\frac{d}{b}.$ So, $d=bcos\\stackrel{^}{A}.$ Substituting this into [link] , we get: ${a}^{2}={b}^{2}+{c}^{2}-2bccos\\stackrel{^}{A}$ show that the set of all natural number form semi group under the composition of addition explain and give four Example hyperbolic function _3_2_1 felecia ⅗ ⅔½ felecia _½+⅔-¾ felecia The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu 1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3 Pawel 2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31 Pawel ok Ifeanyi on number 2 question How did you got 2x +2 Ifeanyi combine" ]

[ "Question 17: The greatest value of is (A) (B) (C) (D)", "Find the greatest common divisor Find the greatest common divisor of $$a=1024^{2014}-1$$ and $$b=1024^{2014}+1023$$. ×", "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 $6$.", "common divisor of $$a$$ and $$b$$ • The absolute value of $$a$$ • The maximum of $$a$$ and $$b$$ $$-86$$ $$-55$$ $$-75$$", "# Lesson 96 – Doodles Do Algebra Today your child learns about the relationship between least common multiple and greatest common factor. The least common multiple is the least quantity that contains the quantities exactly and the greatest common factor contains all the factors common to the quantities. So if you multiply together all the quantities and divide them by their greatest common factor, you will be left with the least common multiple. 1. $24a^4x^2y(a+x)$ 2. $12a^2(a-x)(a+x)x^4$ 3. $24a^4x^3y^2(x-y)$ 4. $60a^2x^5(x+y)(x-y)^2$", "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 $3pq^2$?", "# Largest square factor What is the maximum value of $$a$$ if $${ a }^{ 2 }$$ is a factor of $$(10\\times 11\\times 12\\times ...\\times 19)$$? ×", "the result. I have tried this far before,but I don't understand what exactly meant by taking the factor $(a-b)$ common? – Max May 13 '11 at 18:56", "# 2013 AMC8 Problem 10 #### theory ##### algebra geometry What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594? $$\\textbf{(A)}\\ 110 \\qquad \\textbf{(B)}\\ 165 \\qquad \\textbf{(C)}\\ 330 \\qquad \\textbf{(D)}\\ 625 \\qquad \\textbf{(E)}\\ 660$$", "d_b-1)$ will give you the common factor (plus some easy to remove extraneous factors) between $n_a$ and $n_b$. It should be easy to go from there. – Samuel Neves Jun 5 at 1:52", "is the least common multiple of the $n_i$. -", "# Finding a greatest common divisor What is the greatest common divisor of $$2982$$ and $$2016$$? ×", "## College Algebra 7th Edition $x^2(3x^2-6x-1)$ The greatest common factor of the coefficients 3, -6, and -1 is $1$. The greatest common factor of $x^4,x^3$ and $x^2$ is $x^2$. Thus, the greatest common factor of the three terms is $x^2$. Factor out $x^2$ to obtain: $=x^2(3x^2-6x-1)$", "d1 .•. dnl 14.8 The Greatest Common Factor and the Least Common Multiple Definition 8.1. Let A and B be two square matrices, not both equal to O. Let D be a common right divisor of A and B such that any common right divisor is also a right divisor of D. Then we call D a right greatest common divisor of A and B. Suppose that A and B are both right divisors of the square matrix M( i= 0), and that M is a right divisor of any square matrix having both A and B as right divisors. Then we call M a left least common multiple of A and B. Similar definitions for left greatest common divisors and right least common multiples can be given. We shall only discuss right greatest common divisors and left least common multiples and, for the sake of simplicity, we shall call them greatest common divisors and least common multiples. 395 14.8 The Greatest Common Factor and the Least Common Multiple We define the sum of the two matrices A (aij) and B =i = (bij) by Theorem 8.1. Let A and B be two square matrices which are not both O. Then their greatest common divisor D exists. Moreover there are square matrices P and Q such that PA+QB=D.", "# 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 $3 \\times 4 \\times 7 = 84$.", "would need to be even; that is both $a$ and $b$ would have a common factor of $2$ ...but this is contrary to the initial declaration that $a$ and $b$ have no common factors.", "2, 3, 5 is 30 Now expressing them in terms of least common multiple we have A:B:C = 15:10:6 Therefore, the ratio of A: B: C is 15:10:6. Scroll to Top", "Steve is thinking of two numbers, $a$ and $b$. If their least common multiple is $100!$ and their greatest common divisor is $24$, how many possible unordered pairs of integers could he be thinking of?", "# How do you find the least common multiple of 30 and 3? Least Common Multiple of $30$ and $3$ is $30$. Here, among two numbers $3$ and $30$, as $3$ is a factor of $30$, Least Common Multiple of $30$ and $3$ is $30$.", "that, we first compute its greatest common divisors: (B.4.5) (B.4.6) (B.4.7)" ]

[ "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.", "Let us understand using an example. Find the largest power of 3 in 24! • Step 1: Quotient (24/3) = 8 • Step 2: Quotient (8/3) = 2 Largest power of 3 in 24! = sum of all quotient in all steps = 8 + 2 = 10 Let us double check it. The powers of 3 in 24! can be found in these factors of 24!: 3, 6, 9, 12, 15, 18, 21, 24 (9 and 18 account for 2 powers of 3 each) We can also write the above method as follows: Largest power of 3 in 24! = $[\\frac{24}{3}] + [\\frac{24}{3^2}] = 8 + 2 = 10$ ##### Q. Find the largest power of 7 in $C^{50}_{25}$. Explanation: $C^{50}_{25} = \\frac{50!}{25! \\hspace{1ex} 25!}$ Largest power of 7 in 50! = $[\\frac{50}{7}] + [\\frac{50}{7^2}] = 7 + 1 = 8$ Largest power of 7 in 25! = $[\\frac{25}{7}] = 3$ So, Largest power of 7 in $C^{50}_{25}$ = 8 - (2 × 3) = 8 - 6 = 2 ## Largest power of a composite number in N! To find the largest power of a composite number in N!, we need to factorize it into prime factors and then find the largest power of that prime factor which is in least supply (generally the largest prime", "## 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", "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.", "power of 2 that appears is 24. The highest power of 3 that appears is 31. \\begin{aligned} \\text{LCM} = 2^4 \\cdot 3^1 ~ & \\textcolor{red}{ \\text{ Keep highest power of each factor.}} \\end{aligned}\\nonumber Now we expand this last expression to get our LCM. \\begin{aligned} = 16 \\cdot 3 ~ & \\textcolor{red}{ \\text{ Expand: } 2^4 = 16 \\text{ and } 3^1 = 3.} \\\\ = 48. ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Note that this answer is identical to the LCM found in Example 8 that was found by listing multiples and choosing the smallest multiple in common. Exercise $$\\PageIndex{8}$$ Use prime factorization to find the least common denominator of 18 and 24. Answer 72 Example $$\\PageIndex{10}$$ Simplify: $$\\frac{5}{28} + \\frac{11}{42}$$. Solution Prime factor the denominators in compact form using exponents. 28 = 2 · 2 · 7=22 · 7 42 = 2 · 3 · 7=21 · 31 · 71 To find the LCD, write down each factor that appears to the highest power of that factor that appears. The factors that appear are 2, 3, and 7. The highest power of 2 that appears is 22. The highest power of 3 that appears is 31. The highest power of 7 that appears is 71. \\begin{aligned} \\text{LCM} = 2^2 \\cdot 3^1 \\cdot 7^1 ~ & \\textcolor{red}{ \\text{", "## 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: $=2 \\cdot 2\\cdot 2 \\cdot 2 \\cdot 3 \\\\= 48$", "# PracticeLeast Number of Cartons #### puremath ##### Active Member I do not understand this question. It is worded so badly. Can you explain what is going on? #### MarkFL ##### La Villa Strangiato Staff member Moderator Math Helper They are essentially asking for the lowest common multiple (LCM) of 6 and 8. If you don't immediately see this is 24, consider the prime factorization of both numbers: $$\\displaystyle 6=2\\cdot3$$ $$\\displaystyle 8=2^3$$ Now, we need the greatest number of each prime present in each to make up our LCM: $$\\displaystyle \\text{lcm}(6,8)=2^3\\cdot3=24$$ We see that 2 and 3 are the primes present in their factorizations. The highest power on 2 is 3 and the highest power on 3 is 1, and so that determines the LCM. Jason and puremath #### puremath ##### Active Member They are essentially asking for the lowest common multiple (LCM) of 6 and 8. If you don't immediately see this is 24, consider the prime factorization of both numbers: $$\\displaystyle 6=2\\cdot3$$ $$\\displaystyle 8=2^3$$ Now, we need the greatest number of each prime present in each to make up our LCM: $$\\displaystyle \\text{lcm}(6,8)=2^3\\cdot3=24$$ We see that 2 and 3 are the primes present in their factorizations. The highest power on 2 is 3 and the highest power on 3 is 1, and so that determines the", "## 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: $=2 \\cdot 2\\cdot 2 \\cdot 3 \\cdot 3 \\\\= 72$", "# Chapter 1 - Some Basic Concepts of Arithmetic and Algebra - 1.2 - Prime and Composite Numbers - Problem Set 1.2 - Page 13: 80 $462$ #### Work Step by Step 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: $=2 \\cdot 3\\cdot 7 \\cdot 11 \\\\= 462$ 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.", "cdot 7^1 = 8 cdot 9 cdot 1 cdot 7 = 504. ,!$ ### Viewing this method via Venn diagrams One can find the least common multiple of two numbers by using a Venn diagram as follows. Find the prime factorization of each of the two numbers. Put the prime factors into a Venn diagram with one circle for each of the two numbers, and all factors they share in common in the intersection. To find the LCM, just multiply all of the prime numbers in the diagram. Here is an example: 48 = 2 × 2 × 2 × 2 × 3, 180 = 2 × 2 × 3 × 3 × 5, and what they share in common is two \"2\"s and a \"3\": Least common multiple = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720 Greatest common divisor = 2 × 2 × 3 = 12 This also works for the greatest common divisor (GCD), except that instead of multiplying all of the numbers in the Venn diagram, one multiplies only the prime factors that are in the intersection. Thus the GCD of 48 and 180 is 2 × 2 × 3 = 12. ## A third method This method works as easily for finding the LCM of", "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", "important tool that is of great interest in number theory. We begin with an example. Let’s take a simple example of finding the greatest common divisor of 24 and 60. In a math class in elementary school, one motivation of finding the greatest common divisor of 24 and 60 is to reduce the fraction $\\displaystyle \\frac{24}{60}$ to its lowest terms. The first step is to factor each of the numerator and denominator using their prime factors. $\\displaystyle \\frac{24}{60}=\\frac{2 \\cdot 2 \\cdot 2 \\cdot 3}{2 \\cdot 2 \\cdot 3 \\cdot 5}$ Then cross out the prime factors common to the numerator and the denominator. $\\displaystyle \\frac{24}{60}=\\frac{\\not 2 \\cdot \\not 2 \\cdot 2 \\cdot \\not 3}{\\not 2 \\cdot \\not 2 \\cdot \\not 3 \\cdot 5}=\\frac{2}{5}$ The product of the prime factors that are crossed out is $2 \\cdot 2 \\cdot 3=12$, which is the greatest common divisor of 24 and 60. Thus the greatest common divisor of two integers $a$ and $b$ is the product of the prime factors shared by the two numbers. This approach of finding the greatest common divisor depends on performing prime factorization, which is a hard problem for larger numbers. This approach, though conceptually clear, is clearly not efficient. We need an algorithm for finding the greatest common divisor that does not require prime factorization.", "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", "## 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) $2\\cdot 3\\cdot 5\\cdot 7=210$", "be expressed in the following ways: $$12 = 12\\cdot 1$$ $$12 = 6\\cdot 2$$ $$12 = 4\\cdot 3$$ #### Greatest Common Factor Suppose two numbers, say $$30$$ and $$24$$, have factors in common. The largest of those is the greatest common factor. The factors of $$30$$ are $$1, 2, 3, 5, 6, 10, 15 \\text{ and }30$$. The factors of $$24$$ are $$1, 2, 3, 4, 6, 8, 12, \\text{ and }24$$. The common factors are $$1, 2, 3, \\text{ and }6$$. We can see that $$6$$ is the greatest common factor. #### Prime Factorization Prime factorization is the process of factoring a given number so that all of its factors are prime. For example, in the section above, we found all the factors of $$24$$, but only $$2 \\text{ and } 3$$ are primes, so we’re going to have to use them multiple times to get $$24$$. A good way to do that is by successive division until you reach $$1$$: $24 \\div 3 = 8$ $8\\div2 = 4$ $4\\div2=2$ $2\\div 2 = 1$ We used $$3$$ once and $$2$$ three times, so the prime factorization is $$2 \\cdot 2 \\cdot 2 \\cdot 3$$. The product of those is $$24$$. ### Multiples A multiple of an integer is a product of two or more integers. The number", "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, $\\text{LCM} = 2 \\cdot", "greatest common divisor is known, using the relation, :$\\operatorname\\left(a,b\\right)=\\frac.$ # Calculation ## Using prime factorizations Greatest common divisors can be computed by determining the prime factorizations of the two numbers and comparing factors. For example, to compute gcd(48, 180), we find the prime factorizations 48 = 24 · 31 and 180 = 22 · 32 · 51; the GCD is then 2min(4,2) · 3min(1,2) · 5min(0,1) = 22 · 31 · 50 = 12, as shown in the Venn diagram. The corresponding LCM is then 2max(4,2) · 3max(1,2) · 5max(0,1) = 24 · 32 · 51 = 720. : In practice, this method is only feasible for small numbers, as computing prime factorizations takes too long. ## Euclid's algorithm The method introduced by Euclid for computing greatest common divisors is based on the fact that, given two positive integers and such that , the common divisors of and are the same as the common divisors of and . So, Euclid's method for computing the greatest common divisor of two positive integers consists of replacing the larger number by the difference of the numbers, and repeating this until the two numbers are equal: that is their greatest common divisor. For example, to compute , one proceeds as follows: :$\\begin\\left(48,18\\right)\\quad&\\to\\quad\\left(48-18, 18\\right)= \\left(30,18\\right)&&\\to \\quad\\left(30-18, 18\\right)=\\left(12,18\\right)\\\\ &\\to \\quad\\left(12,18-12\\right)=\\left(12,6\\right)&&\\to \\quad\\left(12-6,6\\right)=\\left(6,6\\right).\\end$ So . This", "## 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 $2 \\cdot 2 \\cdot 3 = 12$.", "of the prime factors that are crossed out is $2 \\cdot 2 \\cdot 3=12$, which is the greatest common divisor of 24 and 60. Thus the greatest common divisor of two integers $a$ and $b$ is the product of the prime factors shared by the two numbers. This approach of finding the greatest common divisor depends on performing prime factorization, which is a hard problem for larger numbers. This approach, though conceptually clear, is clearly not efficient. We need an algorithm for finding the greatest common divisor that does not require prime factorization. First, we have another discussion on the greatest common divisor. ___________________________________________________________________________________________________________________ The Greatest Common Divisor Let $a$ and $b$ be integers. We say that $a$ divides $b$ (denoted by $a \\ \\lvert \\ b$) if there is an integer $q$ such that $b=q \\cdot a$. In other words, $a \\ \\lvert \\ b$ if $a$ divides $b$ without leaving a remainder. For examples, $3 \\ \\lvert \\ 18$ and $7 \\ \\lvert \\ 91$. When $a$ does not divide $b$, we write $a \\not \\lvert \\ b$. The greatest common divisor (or GCD) of two integers $a$ and $b$ is denoted by $\\text{GCD}(a,b)$ and is the largest positive integer that divides both $a$ and $b$. It is clear that $\\text{GCD}(a,b)$ is the positive integer $d$", "of the prime factors that are crossed out is $2 \\cdot 2 \\cdot 3=12$, which is the greatest common divisor of 24 and 60. Thus the greatest common divisor of two integers $a$ and $b$ is the product of the prime factors shared by the two numbers. This approach of finding the greatest common divisor depends on performing prime factorization, which is a hard problem for larger numbers. This approach, though conceptually clear, is clearly not efficient. We need an algorithm for finding the greatest common divisor that does not require prime factorization. First, we have another discussion on the greatest common divisor. ___________________________________________________________________________________________________________________ The Greatest Common Divisor Let $a$ and $b$ be integers. We say that $a$ divides $b$ (denoted by $a \\ \\lvert \\ b$) if there is an integer $q$ such that $b=q \\cdot a$. In other words, $a \\ \\lvert \\ b$ if $a$ divides $b$ without leaving a remainder. For examples, $3 \\ \\lvert \\ 18$ and $7 \\ \\lvert \\ 91$. When $a$ does not divide $b$, we write $a \\not \\lvert \\ b$. The greatest common divisor (or GCD) of two integers $a$ and $b$ is denoted by $\\text{GCD}(a,b)$ and is the largest positive integer that divides both $a$ and $b$. It is clear that $\\text{GCD}(a,b)$ is the positive integer $d$" ]

[ "be expressed in the following ways: $$12 = 12\\cdot 1$$ $$12 = 6\\cdot 2$$ $$12 = 4\\cdot 3$$ #### Greatest Common Factor Suppose two numbers, say $$30$$ and $$24$$, have factors in common. The largest of those is the greatest common factor. The factors of $$30$$ are $$1, 2, 3, 5, 6, 10, 15 \\text{ and }30$$. The factors of $$24$$ are $$1, 2, 3, 4, 6, 8, 12, \\text{ and }24$$. The common factors are $$1, 2, 3, \\text{ and }6$$. We can see that $$6$$ is the greatest common factor. #### Prime Factorization Prime factorization is the process of factoring a given number so that all of its factors are prime. For example, in the section above, we found all the factors of $$24$$, but only $$2 \\text{ and } 3$$ are primes, so we’re going to have to use them multiple times to get $$24$$. A good way to do that is by successive division until you reach $$1$$: $24 \\div 3 = 8$ $8\\div2 = 4$ $4\\div2=2$ $2\\div 2 = 1$ We used $$3$$ once and $$2$$ three times, so the prime factorization is $$2 \\cdot 2 \\cdot 2 \\cdot 3$$. The product of those is $$24$$. ### Multiples A multiple of an integer is a product of two or more integers. The number", "possible value of 900 and $$a$$ is the product of other multiples of $$50!$$. $$50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b$$, where $$b$$ is the product of other multiples of $$50!$$. So $$x=6$$. Below is another example: Suppose we have the number $$18!$$ and we are asked to to determine the power of $$12$$ in this number. Which means to determine the highest value of $$x$$ in $$18!=12^x*a$$, where $$a$$ is the product of other multiples of $$18!$$. $$12=2^2*3$$, so we should calculate how many 2-s and 3-s are in $$18!$$. Calculating 2-s: $$\\frac{18}{2}+\\frac{18}{2^2}+\\frac{18}{2^3}+\\frac{18}{2^4}=9+4+2+1=16$$. So the power of $$2$$ (the highest power) in prime factorization of $$18!$$ is $$16$$. Calculating 3-s: $$\\frac{18}{3}+\\frac{18}{3^2}=6+2=8$$. So the power of $$3$$ (the highest power) in prime factorization of $$18!$$ is $$8$$. Now as $$12=2^2*3$$ we need twice as many 2-s as 3-s. $$18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a$$. So $$18!=12^8*a$$ --> $$x=8$$. Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere? Math Expert Joined: 02 Sep 2009 Posts: 44287 ### Show Tags 11 Aug 2010, 18:07 2 KUDOS Expert's post utfan2424 wrote: Does this relationship breakdown at some point? I thought", "## 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: $=2 \\cdot 2\\cdot 2 \\cdot 3 \\cdot 3 \\\\= 72$", "## 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: $=2 \\cdot 2\\cdot 2 \\cdot 2 \\cdot 3 \\\\= 48$", "power of 2 that appears is 24. The highest power of 3 that appears is 31. \\begin{aligned} \\text{LCM} = 2^4 \\cdot 3^1 ~ & \\textcolor{red}{ \\text{ Keep highest power of each factor.}} \\end{aligned}\\nonumber Now we expand this last expression to get our LCM. \\begin{aligned} = 16 \\cdot 3 ~ & \\textcolor{red}{ \\text{ Expand: } 2^4 = 16 \\text{ and } 3^1 = 3.} \\\\ = 48. ~ & \\textcolor{red}{ \\text{ Multiply.}} \\end{aligned}\\nonumber Note that this answer is identical to the LCM found in Example 8 that was found by listing multiples and choosing the smallest multiple in common. Exercise $$\\PageIndex{8}$$ Use prime factorization to find the least common denominator of 18 and 24. Answer 72 Example $$\\PageIndex{10}$$ Simplify: $$\\frac{5}{28} + \\frac{11}{42}$$. Solution Prime factor the denominators in compact form using exponents. 28 = 2 · 2 · 7=22 · 7 42 = 2 · 3 · 7=21 · 31 · 71 To find the LCD, write down each factor that appears to the highest power of that factor that appears. The factors that appear are 2, 3, and 7. The highest power of 2 that appears is 22. The highest power of 3 that appears is 31. The highest power of 7 that appears is 71. \\begin{aligned} \\text{LCM} = 2^2 \\cdot 3^1 \\cdot 7^1 ~ & \\textcolor{red}{ \\text{", "$$q_a$$ 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", "$$3$$. Thus, the greatest common factor of $$6$$ and $$9$$ is $$3$$. So, $$x = 3$$. Now, let’s list some multiples of $$18$$ and $$24$$: $18: \\ \\ 18, \\,36, \\,54,\\, 72,\\, 90, …$ $24: \\ \\ 24, \\,48,\\, 72, \\,96,\\, 120, …$ We can see that the common multiple is $$72$$ and it is the least of all possible multiples. Thus, the least common multiple of $$18$$ and $$24$$ is $$72$$. So, $$y = 72$$. So, the value of $$\\frac{y}{x}$$ is $$\\frac{72}{3} = 24$$. Out of the four answer choices given, we see that only answer choice $$B$$ is divisible by $$24$$. $\\frac{48}{24} = 2$ All the other options are not divisible by $$24$$. Thus, the correct answer is $$B$$. ### Roots and Exponents An exponent tells how many times a number (or base) is used in multiplication, such that $4^5 = 4 \\cdot 4 \\cdot 4 \\cdot 4 \\cdot 4 = 1024$ Here are the three main rules of exponents you need to master: $$x^n = x \\cdot x \\cdot … x$$, where $$n$$ is the number of times $$x$$ is multiplied $$x^{-n} = \\frac {1}{x^n}$$, where $$x$$ is not equal to zero $$x^{\\frac{1}{n}} = \\sqrt [n] {x}$$ (Take note that a fractional exponent implies a root.) The other rules can be memorized, too, or derived", "= 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$\\gcd(a^2, b^2) = 1\\$. -", "# PracticeLeast Number of Cartons #### puremath ##### Active Member I do not understand this question. It is worded so badly. Can you explain what is going on? #### MarkFL ##### La Villa Strangiato Staff member Moderator Math Helper They are essentially asking for the lowest common multiple (LCM) of 6 and 8. If you don't immediately see this is 24, consider the prime factorization of both numbers: $$\\displaystyle 6=2\\cdot3$$ $$\\displaystyle 8=2^3$$ Now, we need the greatest number of each prime present in each to make up our LCM: $$\\displaystyle \\text{lcm}(6,8)=2^3\\cdot3=24$$ We see that 2 and 3 are the primes present in their factorizations. The highest power on 2 is 3 and the highest power on 3 is 1, and so that determines the LCM. Jason and puremath #### puremath ##### Active Member They are essentially asking for the lowest common multiple (LCM) of 6 and 8. If you don't immediately see this is 24, consider the prime factorization of both numbers: $$\\displaystyle 6=2\\cdot3$$ $$\\displaystyle 8=2^3$$ Now, we need the greatest number of each prime present in each to make up our LCM: $$\\displaystyle \\text{lcm}(6,8)=2^3\\cdot3=24$$ We see that 2 and 3 are the primes present in their factorizations. The highest power on 2 is 3 and the highest power on 3 is 1, and so that determines the", "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", "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.", "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", "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", "and $294$294, we start by drawing their factor trees: This gives us their prime factorisations 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 factorisations - they have one $2$2, one $3$3, and one $7$7 in common. The highest common factor is the product of the common prime factors, $2\\times3\\times7=42$2×3×7=42. ### Lowest 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 lowest common multiple (LCM) of $12$12", "$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", "## 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 $2 \\cdot 2 \\cdot 3 = 12$.", "# Chapter 1 - Some Basic Concepts of Arithmetic and Algebra - 1.2 - Prime and Composite Numbers - Problem Set 1.2 - Page 13: 80 $462$ #### Work Step by Step 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: $=2 \\cdot 3\\cdot 7 \\cdot 11 \\\\= 462$ 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.", "important tool that is of great interest in number theory. We begin with an example. Let’s take a simple example of finding the greatest common divisor of 24 and 60. In a math class in elementary school, one motivation of finding the greatest common divisor of 24 and 60 is to reduce the fraction $\\displaystyle \\frac{24}{60}$ to its lowest terms. The first step is to factor each of the numerator and denominator using their prime factors. $\\displaystyle \\frac{24}{60}=\\frac{2 \\cdot 2 \\cdot 2 \\cdot 3}{2 \\cdot 2 \\cdot 3 \\cdot 5}$ Then cross out the prime factors common to the numerator and the denominator. $\\displaystyle \\frac{24}{60}=\\frac{\\not 2 \\cdot \\not 2 \\cdot 2 \\cdot \\not 3}{\\not 2 \\cdot \\not 2 \\cdot \\not 3 \\cdot 5}=\\frac{2}{5}$ The product of the prime factors that are crossed out is $2 \\cdot 2 \\cdot 3=12$, which is the greatest common divisor of 24 and 60. Thus the greatest common divisor of two integers $a$ and $b$ is the product of the prime factors shared by the two numbers. This approach of finding the greatest common divisor depends on performing prime factorization, which is a hard problem for larger numbers. This approach, though conceptually clear, is clearly not efficient. We need an algorithm for finding the greatest common divisor that does not require prime factorization.", "• anonymous Is this a trick question? \"Which pairs of integers $$a$$ and $$b$$ have greatest common divisor $$18$$ and least common multiple $$540$$?\" There is a theorem that states that $$[a,b](a,b)=ab$$, where $$[a,b]$$ stands for the least common multiple of $$a$$ and $$b$$, and $$(a,b)$$ stands for the greatest common divisor of $$a$$ and $$b$$. So, it follows that$[a,b](a,b)=ab,$$540\\cdot18=a\\cdot b.$In other words, $$a=540$$ and $$b=18$$ would suffice. Am I missing something? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# How do you factor 15a^2b^2-20ab^2+10ab^3? Nov 16, 2015 $15 {a}^{2} {b}^{2} - 20 a {b}^{2} + 10 a {b}^{3} = 5 a {b}^{2} \\left(3 a - 4 + 2 b\\right)$ Looking at each of the terms, we can see that $15$, $20$, and $10$ have the greatest common divisor of $5$. Additionally, the highest shared power of $a$ between terms is ${a}^{1}$ and the highest shared power of $b$ is ${b}^{2}$. Thus we know that they all have a common factor of $5 a {b}^{2}$. Factoring this out gives us $15 {a}^{2} {b}^{2} - 20 a {b}^{2} + 10 a {b}^{3} = 5 a {b}^{2} \\left(3 a - 4 + 2 b\\right)$" ]

[ "The diagram shows a right-angled isosceles triangle $XYZ$ which circumscribes a square $PQRS$. The area of triangle $XYZ$ is $x$. What is the area of square $PQRS$?", "The diagram shows a right-angled isosceles triangle $XYZ$ which circumscribes a square $PQRS$. What is the ratio of the area of square $PQRS$ to the area of the triangle $XYZ$?", "+0 -1 48 1 +393 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", "ED. What is the area of the triangle PQR?", "(d) ?PQR is a right angled triangle.", "## 'Square in a Triangle' printed from http://nrich.maths.org/ The diagram shows a right-angled isosceles triangle $XYZ$ which circumscribes a square $PQRS$. What is the ratio of the area of square $PQRS$ to the area of the triangle $XYZ$?", "+0 # help geometry 0 104 1 In right triangle PQR, we have angle Q=angle R and PR= 6. What is the area of triangle PQR? Mar 13, 2022 Because it's a right triangle, and the remaining angles are equal, this is a 45-45-90 triangle, with a width and heigth of 6, making the area $$6 \\times 6 \\div 2 = \\color{brown}\\boxed{18}$$", "$PQRS$ is a convex quadrilateral. The intersection of the two diagonals is $O$. If the areas of triangles $PQS$, $QRP$, and $SPR$ are $1, 2,$ and $3$, respectively. What is the area of triangle $PQO$? • What is your work on the subject ? – Jean Marie Dec 9 '16 at 22:25 If you want a hint, read on. First determine $\\mathrm{Area}(\\triangle QRS)$. Now, let $x = \\mathrm{Area}(\\triangle SPO)$, $y = \\mathrm{Area}(\\triangle PQO)$, and note that $\\mathrm{Area}(\\triangle RSO) : \\mathrm{Area}(\\triangle QRO) = x : y$. (Why?) • Very good hint. – Jean Marie Dec 9 '16 at 23:12 With the excellent hint given by @Triskele, one finds $y=2/5$ under the assumption that such a quadrilateral exists. This existence can be established. Moreover, there is no unicity: up to a rotation, there is one degree of freedom. More precisely, taking line PR as the $x$ axis and $O$ as the origin on this axis, we can take: $$P\\left(\\frac45,0\\right), \\ \\ Q\\left(a,1\\right), \\ \\ R\\left(-\\frac{16}{5},0\\right), \\ \\ S\\left(-\\frac32a,-\\frac32\\right)$$ as illustrated on figure below with the simplest choice $a=0$.", ": 2. Hence the area of the triangle PQR is $\\frac {1}{2} (7\\sqrt 3)$ square units.", "An equilateral triangle PQR has an area of 97.43cm^2 .Calculate the length of each side of the triangle giving your answer to the nearest whole number.", "of a rectangular field. Area of a rectangular field = 210 cm$^2$ Sides of rectangular filed are in the ratio 7:3. Let the sides of the rectangular field are 7p and 3p respectively. 7p x 3p = 210 p$^2$ = 10 p = $\\sqrt{10}$ Perimeter of the rectangular field = 2(7p + 3p) = 20p = 20$\\sqrt{10}$ cm Hence, the correct answer is Option A Question 9: In triangle PQR, points E and F are on sides PQ and PR respectively such that EF is parallel to QR. If PE = 2 cm and EQ = 3 cm, then area($\\triangle$ PQR) : area($\\triangle$ PEF) is equal to: a) 3 : 2 b) 25 : 4 c) 5 : 2 d) 9 : 4 Solution: $\\triangle$PQR and $\\triangle$PEF are similar triangles. $\\Rightarrow$ $\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ = $\\left(\\frac{PQ}{PE}\\right)^2$ $\\Rightarrow$ $\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ = $\\left(\\frac{PE+EQ}{PE}\\right)^2$ $\\Rightarrow$ $\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ = $\\left(\\frac{2+3}{2}\\right)^2$ $\\Rightarrow$ $\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ = $\\frac{25}{4}$ $\\Rightarrow$ Area of triangle PQR : Area of triangle PEF = 25:4 Hence, the correct answer is Option B Question 10: The angles of a triangle are in AP (arithmetic progression). If measure of the smallest angle is $50^\\circ$ less than that of the largest angle, then find", "3$ : 2. Hence the area of the triangle PQR is $\\frac {1}{2} (7\\sqrt 3)$ square units.", "and Guest Feb 12, 2020 #4 +531 +1 Find the area of triangle PQR if PQ = QR = 12 and It,s impossible to solve! Not enough information. Dragan Feb 12, 2020 #3 +531 +1 In triangle $ABC$, we have $\\angle A = 30^\\circ$, $\\angle B = 60^\\circ$, and $\\angle C=90^\\circ$. Point $X$ is on $\\overline{AC}$ such that $\\overline{BX}$ bisects $\\angle ABC$. If $AB = 12$, then what is the area of triangle $ABX$? Area of the triangle ABX = ( AB/2) * [( AB/2) * tan(A)] Area ABX ≈ 20.785 u² Feb 12, 2020", "is equilateral. ∴Area of ΔPQR Thus, the area of the given triangle is .", "or $$Area = \\frac{1}{2}ac\\sin B$$ We may use any of the above formulas depending on which two sides and angle are given. Remember that the given angle must be between the two given sides. Example: Find the area of triangle PQR if p = 6.5 cm, r = 4.3 cm and Ð Q = 39˚. Give your answer correct to 2 decimal places. Solution: Area of triangle PQR = pr sin Q = sin 39˚ = 8.79 cm2 How to find the area of an oblique triangle when given two sides and an angle? ### Given Equilateral Triangle To find the area of an equilateral triangle, we can use the Pythagorean Theorem to get the height of the triangle and then use formula $$A = \\frac{1}{2}bh$$ or we can use the following formula: The formula for the area of an equilateral triangle (with all sides congruent) is equal to $$A = \\frac{{{s^2}\\sqrt 3 }}{4}$$ where s is the length of any side of the triangle How to find the area of an equilateral triangle with side s? ### Given a Triangle drawn on a Grid If the triangle is drawn on a grid then we can use the \"box\" method to calculate the area of the triangle. This method involves drawing a smallest box that will enclose the", "## 'Weekly Problem 33 - 2006' printed from http://nrich.maths.org/ The diagram shows a right-angled isosceles triangle $XYZ$ which circumscribes a square $PQRS$. The area of triangle $XYZ$ is $x$. What is the area of square $PQRS$?", "which means that the triangle is right-angled.", "### 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 $$\\triangle P R Q ?$$", "right triangle containing an angle cos^(-1)(x/2).", "$LMN$, which side is the opposite with respect to the reference angle $L$? 3. Given the right triangle $PQR$, which side is adjacent with respect to the reference angle $P$? 4. Given the right triangle, what is the opposite of a right triangle with respect to the angle $\\alpha$? 5. Given the right triangle, what is the adjacent of a right triangle with respect to the angle $\\beta$? 6. Given the right triangle, what is the hypotenuse of the right triangle? 7. Mr. David walks $15$ kilometers East and then $8$ kilometers north. Determine how far is he away from his starting position?" ]

[ "#### 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", "congruence of triangles. View Answer play_arrow • A) S.S.S. property B) S.A.S. property C) A.S.A. property D) R.H.S. property View Answer play_arrow • A) S.S.S. property B) S.A.S. property C) A.S.A. property D) R.H.S. property View Answer play_arrow • A) $\\angle ABC=\\angle PQR,$ $a=p,\\,\\,c=r$ B) $\\angle CAB=\\angle RPQ,\\angle ABC=\\angle PQR,c=r$ C) $b=q,\\,\\,\\angle CAB=\\angle RPQ,\\,\\,a=p$ D) $a=p,\\,c=r,\\,\\angle ABC=\\angle PQR$ View Answer play_arrow • A) Two Rs. 1 coins B) A Rs. 1 coin and a Rs. 2 coin C) A Rs. 2 coin and a Rs. 5 coin D) A Rs. 5 coin and a Rs. 10 coin View Answer play_arrow • A) Naming the angles of the triangles using capital letters. B) Measures of angles in degrees. C) The order of letters of the triangles. D) Exact length of the sides of the triangles. View Answer play_arrow • A) $\\Delta QPR\\cong \\Delta SPR$ B) $\\Delta PSR\\cong \\Delta RQP$ C) $\\Delta PRS\\cong \\Delta QPR$ D) $\\Delta QRP\\cong \\Delta PSR$ View Answer play_arrow • A) The triangles are small. B) The triangles are congruent. C) The triangles are equilateral. D) The triangles are equiangular. View Answer play_arrow • A) $\\Delta ABC:\\angle B={{50}^{o}},BC=5cm,$ and $AB=7cm.$ $\\Delta DEF:\\angle E={{50}^{0}},EF=7cm,$and$DE=5cm.$ B) $\\Delta ABC:BC=6cm,$ $AC=4\\,\\,cm$ and $\\angle B={{35}^{o}}$ $\\Delta DEF:DF=4\\,cm,$ $EF=6\\,cm$ and $\\angle D={{55}^{o}}$ C) $\\Delta ABC:AB=4.5cm,$$AC=4cm$and$\\angle A={{60}^{o}}$ $\\Delta DEF:DE=4cm,FD=4.5cm$and$\\angle D={{55}^{o}}$ D) Either [b]", "= \\boxed{571}$. ### Solution 2 $[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label(\"$$A$$\",A,(0,1));label(\"$$B$$\",B,(-1,-1));label(\"$$C$$\",C,(1,-1));label(\"$$P$$\",P,(1,1)); label(\"$$Q$$\",Q,(-1,1));label(\"$$R$$\",R,(1,0));label(\"$$S$$\",S,(-1,0)); [/asy]$ Again, construct $R$ as above. Let $\\angle BAC = \\angle QBR = \\angle QPR = 2x$ and $\\angle ABC = \\angle ACB = y$, which means $x + y = 90$. $\\triangle QBC$ is isosceles with $QB = BC$, so $\\angle BCQ = 90 - \\frac {y}{2}$. Let $S$ be the intersection of $QC$ and $BP$. Since $\\angle BCQ = \\angle BQC = \\angle BRS$, $BCRS$ is cyclic, which means $\\angle RBS = \\angle RCS = x$. Since $APRB$ is an isosceles trapezoid, $BP = AR$, but since $AR$ bisects $\\angle BAC$, $\\angle ABR = \\angle ACR = 2x$. Therefore we have that $\\angle ACB = \\angle ACR + \\angle RCS + \\angle QCB = 2x + x + 90 - \\frac {y}{2} = y$. We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \\frac {y}{2} = y$ to get $x = 10$ and $y = 80$. $\\angle APQ = 180 - 4x = 140$, $\\angle ACB = 80$, so $r = \\frac {80}{140} = \\frac {4}{7}$. $\\left\\lfloor 1000\\left(\\frac {4}{7}\\right)\\right\\rfloor = \\boxed{571}$. ### Solution 3 Let the measure of $\\angle BAC$ be $\\alpha$ and $\\overline{AP}=\\overline{PQ}=\\overline{QB}=\\overline{BC}=x$. Because $\\triangle APQ$ is isosceles,", "+0 # Help Geometry 0 154 1 In right triangle $ABC,$ $AC = BC$ and $\\angle C = 90^\\circ.$ and Let $P$ and $Q$ be points on hypotenus $\\overline{AB},$ as shown below, such that $\\angle PCQ = 45^\\circ.$ Show that $AP^2 + BQ^2 = PQ^2.$ $$[asy] unitsize(3 cm); pair A, B, C, P, Q; A = (1,0); B = (0,1); C = (0,0); P = extension(C, C + dir(16), A, B); Q = extension(C, C + dir(16 + 45), A, B); draw(A--B--C--cycle); draw(C--P); draw(C--Q); label(\"A\", A, SE); label(\"B\", B, NW); label(\"C\", C, SW); label(\"P\", P, NE); label(\"Q\", Q, NE); [/asy]$$ off-topic Feb 16, 2020 edited by Guest Feb 16, 2020", "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 =", "Identify the pairs of shapes which are similar and congruent and write the letter pairs. 94. Match the following by their congruence S.No. A B 1. (i) RHS 2. (ii) SSS 3. (iii) SAS 4. (iv) ASA 95. In the figure, DA =DC and BA= BC. Are the triangles DaA and DBC congruent? Why? 96. Is it possible to construct a quadrilateral PQRS with PQ = 5 cm, QR = 3 cm, RS = 6 cm, PS = 7 cm and PR = 10 cm. If not, why? 97. In the given figure if $\\frac { AO }{ OC } =\\frac { BO }{ OD } =\\frac { 1 }{ 2 }$ and AB=5cm. Find the value of DC. 98. In the given figure if $\\angle$A =$\\angle$C then prove that $\\triangle$AOB ~$\\triangle$COD 99. In the figure AB$\\bot$ BC and DE$\\bot$AC prove that $\\triangle$ABC ~$\\triangle$AED. 100. In the given figure if $\\angle$P =$\\angle$RTS, prove that $\\triangle$RPQ ~$\\triangle$ RTS. 101. A fast food restaurant has a meal special Rs.50 for a drink, sandwich, side item and dessert. The choices are Sandwich : Grilled chicken, AU beef patty, Vegeburger and Fill filet. Side: Regular fries, cheese fries, potato fries Dessert: Chocolate chip cookie or Apple pie. Drink: Fanta, Dr. Pepper, Coke, Diet coke and sprite. How may meal combos are possible?", "of the triangle to the line containing the opposite side that is perpendicular to the opposite side. For right triangle $$PQR$$ we can draw the altitude $$PS$$ Why are triangles $$PQR$$, $$SQP$$, and $$SPR$$ all similar to each other? Triangles $$PQR$$ and $$SQP$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$Q$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$PSQ$$ are congruent. Triangles $$PQR$$ and $$SPR$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$R$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$RSP$$ are congruent. Because triangles $$SQP$$ and $$SPR$$ are both similar to triangle $$PQR$$, they are also similar to each other. Since the triangles $$PQR$$, $$SQP$$, and $$SPR$$ are all similar, corresponding angles are congruent and pairs of corresponding sides are scaled copies of each other, by the same scale factor. We can use the proportionality of pairs of corresponding side lengths to find missing side lengths. For example, suppose we need to find $$PS$$ and know $$RS=3$$ and $$QS=7$$. Since triangle $$SQP$$ is similar to triangle $$SPR$$, we know $$\\frac{RS}{PS}=\\frac{PS}{QS}$$. So $$\\frac{3}{PS}=\\frac{PS}{7}$$ and $$PS=\\sqrt{21}$$. Or, suppose we need to find $$SQ$$ and know $$PQ=5$$ and $$RQ=12$$. Since triangle $$PQR$$ is similar to triangle $$SQP$$, we know $$\\frac{RQ}{PQ}=\\frac{PQ}{SQ}$$.", "draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--Z^^P--W^^Q--X^^Q--Y,dashed); [/asy]$Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$, respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$, respectively. Claim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses. Proof: Since $\\angle DAB + \\angle ADC = 180^{\\circ}$$\\angle ADP + \\angle PAD = 90^{\\circ}$. Therefore, triangles $APD$$APZ$$DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$, so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus. Extend line $PQ$ to meet $\\overline{AD}$ and $\\overline{BC}$ at $R$ and $S$, respectively. Because of rhombus properties, $RP = QS = \\frac{333}{2}$. Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$, respectively; therefore, by trapezoid properties, $RS = \\frac{AB + CD}{2} = 575$. Finally, $PQ = RS - RP - QS = \\boxed{242}$. ~ihatemath123 ## Solution 4 $[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);", "# 2000 AIME I Problems/Problem 14 ## Problem In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\\overline{AC}$ and $\\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find the greatest integer that does not exceed $1000r$. ## Solution ### Solution 1 $[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label(\"$$A$$\",A,(0,1));label(\"$$B$$\",B,(-1,-1));label(\"$$C$$\",C,(1,-1));label(\"$$P$$\",P,(1,1)); label(\"$$Q$$\",Q,(-1,1));label(\"$$R$$\",R,(1,0)); [/asy]$ Let point $R$ be in $\\triangle ABC$ such that $QB = BR = RP$. Then $PQBR$ is a rhombus, so $AB \\parallel PR$ and $APRB$ is an isosceles trapezoid. Since $\\overline{PB}$ bisects $\\angle QBR$, it follows by symmetry in trapezoid $APRB$ that $\\overline{RA}$ bisects $\\angle BAC$. Thus $R$ lies on the perpendicular bisector of $\\overline{BC}$, and $BC = BR = RC$. Hence $\\triangle BCR$ is an equilateral triangle. Now $\\angle ABR = \\angle BAC = \\angle ACR$, and the sum of the angles in $\\triangle ABC$ is $\\angle ABR + 60^{\\circ} + \\angle BAC + \\angle ACR + 60^{\\circ} = 3\\angle BAC + 120^{\\circ} = 180^{\\circ} \\Longrightarrow \\angle BAC = 20^{\\circ}$. Then $\\angle APQ = 140^{\\circ}$ and $\\angle ACB = 80^{\\circ}$, so the answer is $\\left\\lfloor 1000 \\cdot \\frac{80}{140} \\right\\rfloor = \\left\\lfloor \\frac{4000}{7} \\right\\rfloor", "# Math Jeremy loves right triangles that have side length 1. In fact, he loves to combine them so that the hypotenuse of one becomes the leg of the next. If the longer leg of the first triangle is sqrt 11, what will be the hypotenuse of the 2014th triangle? I don't understand what I have to do to solve the problem. 1. 👍 3 2. 👎 0 3. 👁 1,798 1. Start with T1 which has sides 1 and 1. Its hypotenuse is √2 Now draw T2, with its base as the hypotenuse of T1, and its other side is 1. T2 has sides √2 and 1, so its hypotenuse is √3 Keep drawing new triangles on the hypotenuse, and you will get a spiral figure. Tn has its sides √n and 1, so its hypotenuse is √(n+1) 1. 👍 0 2. 👎 8 2. The answer is actually 45 1. 👍 4 2. 👎 0 3. We use the Pythagorean Theorem to find the first few hypotenuses: [asy] pair A=(0,0),B=(0,1),C=(0.4404,1.957), D=(1.2,2.9), E=(3.316,0); draw(A--B--C--D--E--A); draw(B--E--C); draw(D--E); draw(rightanglemark(E,A,B)); draw(rightanglemark(E,B,C)); draw(rightanglemark(E,C,D)); label(\"1\",midpoint(A--B),W); label(\"1\",midpoint(C--B),NW); label(\"1\",midpoint(C--D),NW); label(\"...\",(2,2),E); defaultpen(fontsize(8pt)); label(\"$\\sqrt{11}$\",midpoint(A--E),N); label(\"$\\sqrt{12}$\",midpoint(B--E),N); label(\"$\\sqrt{13}$\",midpoint(C--E),NE); label(\"$\\sqrt{14}$\",midpoint(D--E),NE); [/asy] So, the first triangle has hypotenuse $\\sqrt{12}$, the 2nd has hypotenuse $\\sqrt{13}$, and so on. Thus, the 2014th will have hypotenuse $\\sqrt{2014+11}=\\sqrt{2025}=\\boxed{45}$. 1. 👍 2 2. 👎 0", "angle $A$, then the reflection of triangle $APP'$ in line $AP$ must coincide with the triangle $APP''$ (because $\\angle PAP'=\\angle PAP''$ and $\\angle APP'=\\pi/2-\\angle PAP'=\\pi/2-\\angle PAP''=\\angle APP''$), and hence $PP'=PP''$. Conversely, if $P$ is a point inside the angle $A$ with $PP'=PP''$, then in the right triangles $APP'$ and $APP''$ side $AP$ is common, and sides $PP'$ and $PP''$ are equal. But there is only one right triangle with a given leg and hypotenuse, so the triangles must be equal. In particular this implies that $\\angle PAP'=\\angle PAP''$, such that the point $P$ lies on the angle bisector of angle $A$. \\begin{figure}[h] \\centering \\begin{asy} size (3cm); pair A, B, C, P, P1, P2; A=(1.5,2); B=(0,0); C=(3,0); P=(length(C-A)*B+length(B-A)*C)/(length(B-A)+length(C-A)); P1=A+dot(B-A,P-A)/dot(B-A,B-A)*(B-A); P2=A+dot(C-A,P-A)/dot(C-A,C-A)*(C-A); draw(A--B); draw(A--C); draw(A--P); draw(P--P1); draw(P--P2); label(\"$A$\",A,N); label(\"$P$\",P,S); label(\"$P'$\",P1,N); label(\"$P''$\",P2,NE); \\end{asy} \\label{fig:bisector1} \\end{figure} Now let $AA'$, $BB'$, and $CC'$ be the internal angle bisectors of triangle $ABC$. Let $P$ be the point of intersection of the angle bisectors $AA'$ and $BB'$. We have proved above that point $P$ is equidistant from sides $AB$ and $AC$ because it lies on the angle bisector $AA'$. Also, it is equidistant from sides $AB$ and $BC$, because it lies on the angle bisector $BB'$. Hence the point $P$ is located at the same distance from the sides $AC$ and $BC$, and thus it must", "where PQ is the height and QR is the base. The area of the isosceles right triangle $= \\frac{1}{2} \\times PQ \\times QR$ $= \\frac{x \\times x}{2}$ $= \\frac{x^{2}}{2}$ sq. units. Therefore, the area of the isosceles right triangle formula is $\\frac{x^{2}}{2}$, where “x” is equal to the congruent side length. ### Perimeter of an Isosceles Right Triangle Perimeter of a triangle $=$ Sum of lengths of all its sides Perimeter of isosceles right triangle $\\Delta PQR = PQ + QR + PR$ $PQ = QR = x$ units $PR = h$ units Perimeter of the isosceles right triangle formula $= x + x + h = (2x + h)$ units. Therefore, the perimeter of the isosceles right triangle formula is $2x + h$, where “x” represents the length of congruent sides and “h” is equal to the length of the hypotenuse. ## Isosceles Right Triangle Properties 1. It has one interior angle, which is equal to $90^{\\circ}$. 2. The sides or legs are perpendicular to each other and are also known as the base and the height. They are congruent. 3. The other two angles of an isosceles right triangle are acute angles and are congruent to each other. They are equal to $45^{\\circ}$ each. 4. The sum of all the interior angles measures $180^{\\circ}$. 5. If", "positive integers. (b) Suppose that T does not have a smallest element. Clearly 1 does not belong to the set T (or it would be the smallest element of T). Hence 1 must be an element of the set S. Now suppose that, for some $k≥1$, all positive integers $1,2,3,…,k$ are elements of S, but $k+1$ is not an element of S. Then $k+1$ would be an element of T, and would be the smallest element of T, which is not possible. Hence S has the property that “whenever $k≥1$ is an element of S, we can be sure that $k+1$ is also an element of S.” The Principle of Mathematical Induction then guarantees that these two observations (that 1 is an element of S, and that whenever k is an element of S, so is $k+1$) imply that S contains all the positive integers, so that the set T must be empty – contrary to assumption. Hence T must have a smallest element. 270. (i) Triangle $OP′Q′$ is a right angled triangle with $∠P′OQ′=45°$. Hence the two base angles (at O and at $Q′$) are equal, so the triangle is isosceles: $P′Q′_=P′O_$. (ii) Triangles $QQ′P′$ and $QQ′R$ are congruent by RHS (since they share the hypotenuse $QQ′_$, and have equal sides ($QP′_=QP_=QR_$). Hence $Q′P′_=Q′R_$. (iii) If $OQ_:OP_=m:n$,", "# Algebra 2 Using Equations The second angle of a triangle is three times the first and the third is 12 degrees less that twice the first. Find numbers for angles. I do not understand how to solve this equation. 1. 👍 0 2. 👎 0 3. 👁 148 1. Let A, B and C be angles 1, 2 and 3. Then, B = 3A and C = 2A - 12 Since the problem states that we are dealing with a triangle, we know that A + B + C = 180º Substitute for B and C and solve. 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Math 1. Based on the given information, what can you conclude, and why? Given: m angle P= m angle T ; PR=TR. A) Triangle QRP is congruent to Triangle TRS by SAS. B) Triangle QRP is congruent to Triangle SRT by SSS. C) Triangle QRP is 2. ### math Triangle $ABC$ is a right triangle with right angle at $A$. Suppose $\\overline{AX}$ is an altitude of the triangle, $\\overline{AY}$ is an angle bisector of the triangle, and $\\overline{AZ}$ is a median of the triangle, and \\$\\angle 3. ### algebra The flag of a country contains an isosceles triangle.​ (Recall that an isosceles triangle contains two angles with the", "## Elementary Geometry for College Students (6th Edition) The missing blanks are as follows: 1. QR $\\bot$ RP and MN $\\bot$ NP 2. If two lines are perpendicular, then the angles between them are 90 degree angles. 3. Angle N and QRP are congruent. 4. Identity property 5. Triangle MNP is similar to triangle QRP; reason: AA", "hypotenuse of a right triangle, we get lots of similar triangles that can be used to find missing lengths. An altitude is a segment from one vertex of the triangle to the line containing the opposite side that is perpendicular to the opposite side. For right triangle $$PQR$$ we can draw the altitude $$PS$$ Why are triangles $$PQR$$, $$SQP$$, and $$SPR$$ all similar to each other? Triangles $$PQR$$ and $$SQP$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$Q$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$PSQ$$ are congruent. Triangles $$PQR$$ and $$SPR$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$R$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$RSP$$ are congruent. Because triangles $$SQP$$ and $$SPR$$ are both similar to triangle $$PQR$$, they are also similar to each other. Since the triangles $$PQR$$, $$SQP$$, and $$SPR$$ are all similar, corresponding angles are congruent and pairs of corresponding sides are scaled copies of each other, by the same scale factor. We can use the proportionality of pairs of corresponding side lengths to find missing side lengths. For example, suppose we need to find $$PS$$ and know $$RS=3$$ and $$QS=7$$. Since triangle $$SQP$$ is similar to triangle $$SPR$$, we know $$\\frac{RS}{PS}=\\frac{PS}{QS}$$.", "+ \\tfrac 37$. So the answer is $\\boxed{571}$. ## Solution 1 $[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label(\"$$A$$\",A,(0,1));label(\"$$B$$\",B,(-1,-1));label(\"$$C$$\",C,(1,-1));label(\"$$P$$\",P,(1,1)); label(\"$$Q$$\",Q,(-1,1));label(\"$$R$$\",R,(1,0)); [/asy]$ Let point $R$ be in $\\triangle ABC$ such that $QB = BR = RP$. Then $PQBR$ is a rhombus, so $AB \\parallel PR$ and $APRB$ is an isosceles trapezoid. Since $\\overline{PB}$ bisects $\\angle QBR$, it follows by symmetry in trapezoid $APRB$ that $\\overline{RA}$ bisects $\\angle BAC$. Thus $R$ lies on the perpendicular bisector of $\\overline{BC}$, and $BC = BR = RC$. Hence $\\triangle BCR$ is an equilateral triangle. Now $\\angle ABR = \\angle BAC = \\angle ACR$, and the sum of the angles in $\\triangle ABC$ is $\\angle ABR + 60^{\\circ} + \\angle BAC + \\angle ACR + 60^{\\circ} = 3\\angle BAC + 120^{\\circ} = 180^{\\circ} \\Longrightarrow \\angle BAC = 20^{\\circ}$. Then $\\angle APQ = 140^{\\circ}$ and $\\angle ACB = 80^{\\circ}$, so the answer is $\\left\\lfloor 1000 \\cdot \\frac{80}{140} \\right\\rfloor = \\left\\lfloor \\frac{4000}{7} \\right\\rfloor = \\boxed{571}$. ## Solution 2 (Law of sines) Let $AP=PQ=QB=BC=x$ and $A$ be the measure of $\\angle BAC$. Since $\\triangle APQ$ and $\\triangle ABC$ are isoceles, $\\angle APQ = 180-2A$ and $\\angle ACB = 90-\\frac{A}{2}$. Because $\\triangle APQ$ and $\\triangle ABC$ both have a side length $x$ opposite $\\angle BAC$, by the law of sines: $\\frac{x}{\\sin A}=\\frac{AQ}{\\sin(180-2A)}=\\frac{AQ+x}{\\sin(90-\\frac{A}{2})}$ Simplifying, this becomes", "said to be congruent. In $\\triangle$ABC and $\\triangle$PQR $\\angle$B = $\\angle$Q (A) BC = QR (S) $\\angle$C = $\\angle$R (A) $\\therefore$ $\\triangle$ABC≅$\\triangle$PQR (A.S.A axiom) 4. Right angle - Hypotenuse - Side (R.H.S axiom) If hypotenuse and one of two other sides of a right angled triangle are respectively equal to the hypotenuse and a side of other right angled triangle, the triangles are said to be congruent. In Right angles triangles ABC and PQR $\\angle$B = $\\angle$Q (R) AC = PR (H) AB = PQ (S) $\\therefore$$\\triangle$ABC≅$\\triangle$PQR (R.H.S axiom) Similar Triangles The figures which are exactly the same in the shape but sizes may be different are known as similar figures. In the given figure,$\\triangle$ABC and$\\triangle$PQR are similar triangles because they have equal angles and the same shape. Thus, if three angles of one triangle are respectively equal to three angles of another triangle the are said to be similar.$\\triangle$ABC is similar to$\\triangle$PQR is written as$\\triangle$ABC∼$\\triangle$PQR. The symbol '∼' is used to denote is similar to. Here AB and PQ, BC and QR, CA and RP are the corresponding sides of the similar triangle. The corresponding sides of similar triangles are always proportional, i.e the ratios of the corresponding sides are equal. $\\therefore$ $\\frac{AB}{PQ}$ =$\\frac{BC}{QR}$ =$\\frac{CA}{RP}$ ##### Things to remember • Properties of Triangle 1. The sum of", "purposes, they are not acceptable as proofs. Example $$\\PageIndex{6}\\label{eg:imply-06}$$ The statement “If a triangle $$PQR$$ is isosceles, then two of its angles have equal measure.” takes the form of an implication $$p\\Rightarrow q$$, where $\\begin{array}{l@{\\quad}l} p: & \\mbox{The triangle PQR is isosceles} \\\\ q: & \\mbox{Two of the angles of the triangle PQR have equal measure} \\end{array}$ I n this example, we have to rephrase the statements $$p$$ and $$q$$, because each of them should be a stand-alone statement. If we leave $$q$$ as “two of its angles have equal measure,” it is not clear what “its” is referring to. In addition, it is a good habit to spell out the details. It helps us focus our attention on what we are investigating. Example $$\\PageIndex{7}\\label{eg:isostrig}$$ The statement “A square must also be a parallelogram.” can be expressed as an implication: “if the quadrilateral $$PQRS$$ is a square, then the quadrilateral $$PQRS$$ is a parallelogram.” Likewise, the statement “All isosceles triangles have two equal angles.” can be rephrased as “if the triangle $$PQR$$ is isosceles, then the triangle $$PQR$$ has two equal angles.” Since we have expressed the statement in the form of an implication, we no longer need to include the word “all.” hands-on exercise $$\\PageIndex{1}\\label{he:imply-04}$$ Rewrite each of these logical statements: 1. Any square is also a" ]

[ "of the triangle to the line containing the opposite side that is perpendicular to the opposite side. For right triangle $$PQR$$ we can draw the altitude $$PS$$ Why are triangles $$PQR$$, $$SQP$$, and $$SPR$$ all similar to each other? Triangles $$PQR$$ and $$SQP$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$Q$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$PSQ$$ are congruent. Triangles $$PQR$$ and $$SPR$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$R$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$RSP$$ are congruent. Because triangles $$SQP$$ and $$SPR$$ are both similar to triangle $$PQR$$, they are also similar to each other. Since the triangles $$PQR$$, $$SQP$$, and $$SPR$$ are all similar, corresponding angles are congruent and pairs of corresponding sides are scaled copies of each other, by the same scale factor. We can use the proportionality of pairs of corresponding side lengths to find missing side lengths. For example, suppose we need to find $$PS$$ and know $$RS=3$$ and $$QS=7$$. Since triangle $$SQP$$ is similar to triangle $$SPR$$, we know $$\\frac{RS}{PS}=\\frac{PS}{QS}$$. So $$\\frac{3}{PS}=\\frac{PS}{7}$$ and $$PS=\\sqrt{21}$$. Or, suppose we need to find $$SQ$$ and know $$PQ=5$$ and $$RQ=12$$. Since triangle $$PQR$$ is similar to triangle $$SQP$$, we know $$\\frac{RQ}{PQ}=\\frac{PQ}{SQ}$$.", "where PQ is the height and QR is the base. The area of the isosceles right triangle $= \\frac{1}{2} \\times PQ \\times QR$ $= \\frac{x \\times x}{2}$ $= \\frac{x^{2}}{2}$ sq. units. Therefore, the area of the isosceles right triangle formula is $\\frac{x^{2}}{2}$, where “x” is equal to the congruent side length. ### Perimeter of an Isosceles Right Triangle Perimeter of a triangle $=$ Sum of lengths of all its sides Perimeter of isosceles right triangle $\\Delta PQR = PQ + QR + PR$ $PQ = QR = x$ units $PR = h$ units Perimeter of the isosceles right triangle formula $= x + x + h = (2x + h)$ units. Therefore, the perimeter of the isosceles right triangle formula is $2x + h$, where “x” represents the length of congruent sides and “h” is equal to the length of the hypotenuse. ## Isosceles Right Triangle Properties 1. It has one interior angle, which is equal to $90^{\\circ}$. 2. The sides or legs are perpendicular to each other and are also known as the base and the height. They are congruent. 3. The other two angles of an isosceles right triangle are acute angles and are congruent to each other. They are equal to $45^{\\circ}$ each. 4. The sum of all the interior angles measures $180^{\\circ}$. 5. If", "hypotenuse of a right triangle, we get lots of similar triangles that can be used to find missing lengths. An altitude is a segment from one vertex of the triangle to the line containing the opposite side that is perpendicular to the opposite side. For right triangle $$PQR$$ we can draw the altitude $$PS$$ Why are triangles $$PQR$$, $$SQP$$, and $$SPR$$ all similar to each other? Triangles $$PQR$$ and $$SQP$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$Q$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$PSQ$$ are congruent. Triangles $$PQR$$ and $$SPR$$ are similar by the Angle-Angle Triangle Similarity Theorem because angle $$R$$ is in both triangles, and both triangles are right triangles, so angles $$RPQ$$ and $$RSP$$ are congruent. Because triangles $$SQP$$ and $$SPR$$ are both similar to triangle $$PQR$$, they are also similar to each other. Since the triangles $$PQR$$, $$SQP$$, and $$SPR$$ are all similar, corresponding angles are congruent and pairs of corresponding sides are scaled copies of each other, by the same scale factor. We can use the proportionality of pairs of corresponding side lengths to find missing side lengths. For example, suppose we need to find $$PS$$ and know $$RS=3$$ and $$QS=7$$. Since triangle $$SQP$$ is similar to triangle $$SPR$$, we know $$\\frac{RS}{PS}=\\frac{PS}{QS}$$.", "# Algebra 2 Using Equations The second angle of a triangle is three times the first and the third is 12 degrees less that twice the first. Find numbers for angles. I do not understand how to solve this equation. 1. 👍 0 2. 👎 0 3. 👁 148 1. Let A, B and C be angles 1, 2 and 3. Then, B = 3A and C = 2A - 12 Since the problem states that we are dealing with a triangle, we know that A + B + C = 180º Substitute for B and C and solve. 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Math 1. Based on the given information, what can you conclude, and why? Given: m angle P= m angle T ; PR=TR. A) Triangle QRP is congruent to Triangle TRS by SAS. B) Triangle QRP is congruent to Triangle SRT by SSS. C) Triangle QRP is 2. ### math Triangle $ABC$ is a right triangle with right angle at $A$. Suppose $\\overline{AX}$ is an altitude of the triangle, $\\overline{AY}$ is an angle bisector of the triangle, and $\\overline{AZ}$ is a median of the triangle, and \\$\\angle 3. ### algebra The flag of a country contains an isosceles triangle.​ (Recall that an isosceles triangle contains two angles with the", "## Elementary Geometry for College Students (6th Edition) The missing blanks are as follows: 1. QR $\\bot$ RP and MN $\\bot$ NP 2. If two lines are perpendicular, then the angles between them are 90 degree angles. 3. Angle N and QRP are congruent. 4. Identity property 5. Triangle MNP is similar to triangle QRP; reason: AA", "# Properties of Right triangle when angle equals to 90° ## Properties There are three fundamental properties between sides of a right triangle when its angle equals to $90$ degrees. 1. The length of adjacent side is zero. 2. The lengths of opposite side and hypotenuse are equal. 3. Two angles are right angles and an angle is zero angle. ### Theorem $\\Delta QPR$ is a triangle, which represents a right triangle with $90^°$ angle. It is possible to construct a right triangle with $90^°$ angle only when the length of adjacent side (base) is zero. $Length \\, of \\, Adjacent \\, side$ $\\,=\\,$ $0$ $\\implies PR \\,=\\, 0$ When length of adjacent side ($\\overline{PR}$) is zero, the lengths of opposite side (perpendicular) and hypotenuse are equal. In this example, the length of opposite side is represented by $d$. $\\,\\,\\, \\therefore \\,\\,\\,\\,\\,\\,$ $Length \\, of \\, Opposite \\, side$ $\\,=\\,$ $Length \\, of \\, Hypotenuse$ $\\,=\\,$ $d$ $\\implies QR = PQ = d$ Now, let’s talk about the angles of this triangle. The length of adjacent side becomes zero because the $\\angle PQR$ becomes zero angle. Therefore $\\angle PQR = 0^°$. Whenever the $\\angle PQR$ becomes zero, the angles $\\angle RPQ$ and $\\angle QRP$ become right angles. $\\angle RPQ$ $\\,=\\,$ $\\angle QRP$ $\\,=\\,$ $90^°$ ### Proof The properties", "said to be congruent. In $\\triangle$ABC and $\\triangle$PQR $\\angle$B = $\\angle$Q (A) BC = QR (S) $\\angle$C = $\\angle$R (A) $\\therefore$ $\\triangle$ABC≅$\\triangle$PQR (A.S.A axiom) 4. Right angle - Hypotenuse - Side (R.H.S axiom) If hypotenuse and one of two other sides of a right angled triangle are respectively equal to the hypotenuse and a side of other right angled triangle, the triangles are said to be congruent. In Right angles triangles ABC and PQR $\\angle$B = $\\angle$Q (R) AC = PR (H) AB = PQ (S) $\\therefore$$\\triangle$ABC≅$\\triangle$PQR (R.H.S axiom) Similar Triangles The figures which are exactly the same in the shape but sizes may be different are known as similar figures. In the given figure,$\\triangle$ABC and$\\triangle$PQR are similar triangles because they have equal angles and the same shape. Thus, if three angles of one triangle are respectively equal to three angles of another triangle the are said to be similar.$\\triangle$ABC is similar to$\\triangle$PQR is written as$\\triangle$ABC∼$\\triangle$PQR. The symbol '∼' is used to denote is similar to. Here AB and PQ, BC and QR, CA and RP are the corresponding sides of the similar triangle. The corresponding sides of similar triangles are always proportional, i.e the ratios of the corresponding sides are equal. $\\therefore$ $\\frac{AB}{PQ}$ =$\\frac{BC}{QR}$ =$\\frac{CA}{RP}$ ##### Things to remember • Properties of Triangle 1. The sum of", "$PQRS$ is a convex quadrilateral. The intersection of the two diagonals is $O$. If the areas of triangles $PQS$, $QRP$, and $SPR$ are $1, 2,$ and $3$, respectively. What is the area of triangle $PQO$? • What is your work on the subject ? – Jean Marie Dec 9 '16 at 22:25 If you want a hint, read on. First determine $\\mathrm{Area}(\\triangle QRS)$. Now, let $x = \\mathrm{Area}(\\triangle SPO)$, $y = \\mathrm{Area}(\\triangle PQO)$, and note that $\\mathrm{Area}(\\triangle RSO) : \\mathrm{Area}(\\triangle QRO) = x : y$. (Why?) • Very good hint. – Jean Marie Dec 9 '16 at 23:12 With the excellent hint given by @Triskele, one finds $y=2/5$ under the assumption that such a quadrilateral exists. This existence can be established. Moreover, there is no unicity: up to a rotation, there is one degree of freedom. More precisely, taking line PR as the $x$ axis and $O$ as the origin on this axis, we can take: $$P\\left(\\frac45,0\\right), \\ \\ Q\\left(a,1\\right), \\ \\ R\\left(-\\frac{16}{5},0\\right), \\ \\ S\\left(-\\frac32a,-\\frac32\\right)$$ as illustrated on figure below with the simplest choice $a=0$.", "point $R$. 4. Finally, join the points $Q$ and $R$ by a line for completing the construction of a triangle geometrically. Thus, a triangle, denoted by $QPR$ is constructed geometrically. In this example, the triangle $QPR$ contains an obtuse angle. Therefore, the triangle $QPR$ is called an obtuse triangle. #### Properties The $\\Delta QPR$ contains three interior angles, and they are written as $\\angle QPR$, $\\angle PQR$ and $\\angle QRP$. The angle $QPR$ is known and it is $105^°$. Therefore $\\angle QPR \\,=\\, 105^°$ For studying the obtuse angled triangle, we have to measure the remaining angles and it can be done by a protractor. It is measured that $(1).\\,\\,\\,$ $\\angle PQR \\,=\\, 34^°$ $(2).\\,\\,\\,$ $\\angle QRP \\,=\\, 41^°$ We can observe that the two angles are acute angles. Therefore, it is cleared that in an obtuse triangle, one angle is an obtuse angle and the remaining two angles are acute angles. Latest Math Topics Email subscription Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more", "congruence of triangles. View Answer play_arrow • A) S.S.S. property B) S.A.S. property C) A.S.A. property D) R.H.S. property View Answer play_arrow • A) S.S.S. property B) S.A.S. property C) A.S.A. property D) R.H.S. property View Answer play_arrow • A) $\\angle ABC=\\angle PQR,$ $a=p,\\,\\,c=r$ B) $\\angle CAB=\\angle RPQ,\\angle ABC=\\angle PQR,c=r$ C) $b=q,\\,\\,\\angle CAB=\\angle RPQ,\\,\\,a=p$ D) $a=p,\\,c=r,\\,\\angle ABC=\\angle PQR$ View Answer play_arrow • A) Two Rs. 1 coins B) A Rs. 1 coin and a Rs. 2 coin C) A Rs. 2 coin and a Rs. 5 coin D) A Rs. 5 coin and a Rs. 10 coin View Answer play_arrow • A) Naming the angles of the triangles using capital letters. B) Measures of angles in degrees. C) The order of letters of the triangles. D) Exact length of the sides of the triangles. View Answer play_arrow • A) $\\Delta QPR\\cong \\Delta SPR$ B) $\\Delta PSR\\cong \\Delta RQP$ C) $\\Delta PRS\\cong \\Delta QPR$ D) $\\Delta QRP\\cong \\Delta PSR$ View Answer play_arrow • A) The triangles are small. B) The triangles are congruent. C) The triangles are equilateral. D) The triangles are equiangular. View Answer play_arrow • A) $\\Delta ABC:\\angle B={{50}^{o}},BC=5cm,$ and $AB=7cm.$ $\\Delta DEF:\\angle E={{50}^{0}},EF=7cm,$and$DE=5cm.$ B) $\\Delta ABC:BC=6cm,$ $AC=4\\,\\,cm$ and $\\angle B={{35}^{o}}$ $\\Delta DEF:DF=4\\,cm,$ $EF=6\\,cm$ and $\\angle D={{55}^{o}}$ C) $\\Delta ABC:AB=4.5cm,$$AC=4cm$and$\\angle A={{60}^{o}}$ $\\Delta DEF:DE=4cm,FD=4.5cm$and$\\angle D={{55}^{o}}$ D) Either [b]", "other sides of a right angled triangle are respectively equal to the hypotenuse and a side of other right angled triangle, the triangles are said to be congruent. In Right angles triangles ABC and PQR $$\\angle$$B = $$\\angle$$Q (R) AC = PR (H) AB = PQ (S) $$\\therefore$$$$\\triangle$$ABC≅$$\\triangle$$PQR (R.H.S axiom) Similar Triangles The figures which are exactly the same in the shape but sizes may be different are known as similar figures. In the given figure,$$\\triangle$$ABC and$$\\triangle$$PQR are similar triangles because they have equal angles and the same shape. Thus, if three angles of one triangle are respectively equal to three angles of another triangle the are said to be similar.$$\\triangle$$ABC is similar to$$\\triangle$$PQR is written as$$\\triangle$$ABC∼$$\\triangle$$PQR. The symbol '∼' is used to denote is similar to. Here AB and PQ, BC and QR, CA and RP are the corresponding sides of the similar triangle. The corresponding sides of similar triangles are always proportional, i.e the ratios of the corresponding sides are equal. $$\\therefore$$ $$\\frac{AB}{PQ}$$ =$$\\frac{BC}{QR}$$ =$$\\frac{CA}{RP}$$ • Properties of Triangle 1. The sum of the length of any two sides of a triangle is always greater than the length of its third side. On $$\\triangle$$PQR PQ + QR > PR, QR + PR > PQ and PQ + PR > QR 2. The angle opposite to the", "then two of its angles have equal measure.”, takes the form of an implication $$p\\Rightarrow q$$, where, $\\begin{array}{[email protected]{\\quad}l} p: & \\mbox{The triangle PQR is isosceles} \\\\ q: & \\mbox{Two of the angles of the triangle PQR have equal measure} \\end{array} \\nonumber$. to Symbolic Logic and Its Applications. denotes NOT and denoted OR (though this is not the We can change the notation when we negate a statement. Making statements based on opinion; back them up with references or personal experience. \"x is an odd number\" implies \"There exists a natural number $k$ such that $x = 2k + 1$.\" In this example, we have to rephrase the statements $$p$$ and $$q$$, because each of them should be a stand-alone statement. We shall study it again in the next section. This important observation explains the invalidity of the “proof” of $$21=6$$ in Example [eg:wrongpf2]. Express in words the statements represented by the following formulas. First, we find a result of the form $$p\\Rightarrow q$$. If we cannot find one, we have to prove that $$p\\Rightarrow q$$ is true. If you are asked to show that, $\\mbox{if x>2, then x^2>4}, \\nonumber$. in the form of $$p\\Rightarrow q$$. They focus on whether we can tell one of the two components $$p$$ and $$q$$ is true or false if we know", "+0 # Help Geometry 0 154 1 In right triangle $ABC,$ $AC = BC$ and $\\angle C = 90^\\circ.$ and Let $P$ and $Q$ be points on hypotenus $\\overline{AB},$ as shown below, such that $\\angle PCQ = 45^\\circ.$ Show that $AP^2 + BQ^2 = PQ^2.$ $$[asy] unitsize(3 cm); pair A, B, C, P, Q; A = (1,0); B = (0,1); C = (0,0); P = extension(C, C + dir(16), A, B); Q = extension(C, C + dir(16 + 45), A, B); draw(A--B--C--cycle); draw(C--P); draw(C--Q); label(\"A\", A, SE); label(\"B\", B, NW); label(\"C\", C, SW); label(\"P\", P, NE); label(\"Q\", Q, NE); [/asy]$$ off-topic Feb 16, 2020 edited by Guest Feb 16, 2020", "$RP$ are drawn from an external point $R$ to the circle with centre $O$. If $\\angle PRQ = 120^o$, then prove that $OR = PR + RQ$. $\\angle OPR = \\angle OQR = 90^o$ … … … … … i) And in $\\triangle OPR$ and $\\triangle OQR$ $\\angle OPR = \\angle OQR = 90^o$ (from equation i) $OP = OQ$ (Radii of same circle) $OR = OR$ (common side) $\\triangle OPR \\cong \\triangle OQR$ ( by RHS criterion) So, $RP = RQ$ … … … … … ii) And $\\angle ORP = \\angle ORQ$ … … … … … iii) $\\angle PRQ = \\angle ORP + \\angle ORQ$ Substitute $\\angle PQR = 120^o$ (given) And from equation iii) we get $\\angle ORP + \\angle ORP = 120^o$ $\\Rightarrow 2 \\angle ORP = 120^o$ $\\Rightarrow \\angle ORP = 60^o$ And we know $\\cos 0^o =$ $\\frac{Adjacent}{Hypotenuse}$ So in $\\triangle OPR$ we get $\\cos \\angle ORP =$ $\\frac{PR}{OR}$ $\\cos 60^o =$ $\\frac{PR}{OR}$ $\\frac{1}{2}$ $=$ $\\frac{PR}{OR}$ (we know $\\cos 60^o =$ $\\frac{1}{2}$ ) $OR = 2PR$ $OR = PR + PR$ (substitute value from equation ii) we get) $OR = PR + RQ$ $\\\\$ Question 6: In Figure 4, a triangle $ABC$ is drawn to circumscribe a circle of radius $3$ cm, such that the segments $BD$ and $DC$ are respectively", "Pity. – Gerry Myerson Mar 2 '14 at 2:50 @GerryMyerson I could try out my MS paint skills :P – qwr Mar 2 '14 at 2:58 – Sawarnik Mar 2 '14 at 5:26 In the barycentric coordinates associated with $\\triangle ABC$, we have $$P = [1,4,2] \\qquad\\text{and}\\qquad Q = [2,1,4]$$ Of course I solved a linear system to get this, but it's easy to check after the fact: $[1,4,2]$ is on the line joining $A[1,0,0]$ and $D[0,2,1]$ (since it's $[1,0,0] + 2[0,2,1]$) and on the line joining $B[0,1,0]$ and $E[1,0,2]$ (since it's $4[0,1,0]+[1,0,2]$), so it must be $P$. Then we get the coordinates of $Q$ by symmetry. Thus $P$ is the midpoint of $BQ$. (Take $B = [0,7,0]$.) - I thought barycentric coordinates had fraction values for points inside the triangle? – qwr Apr 9 '14 at 18:05 If you like, you can scale everything I wrote so that each point's coordinates add up to 1: $P=[\\frac17,\\frac47,\\frac27]$, and so on. – user21467 Apr 10 '14 at 11:44 Proving $\\triangle BGM ~ \\triangle ARM$ is easy. $\\angle GMB = \\angle AMR$ because they are vertically opposite angles. $\\angle GBM = \\angle MAR$ because line $BG$ is parallel to line $AR$, and the angles are corresponding angles made by intercept $GR$. For proving $BGRP=2PRQ$, you need to show $BP=PQ$ which", "In$$\\triangle$$PQR, $$\\triangle$$PQR, $$\\angle$$PRS = $$\\angle$$PQR + $$\\angle$$QPR Congruent Triangles Those triangles which have exactly the same three sides and exactly the same three angles. If the figures have exactly the same shape and size are called congruent figures. Conditions of congruency of triangles Two triangles will be congruent under the following conditions 1. Side-SIde-SIde (S.S.S) axiom If three sides of one triangle are respectively equal to three corresponding sides of another angle, the triangles are said to be congruent. In$$\\triangle$$ABC and$$\\triangle$$PQR AB = PQ (S) BC = QR (S) CA = RP (S) $$\\therefore$$$$\\triangle$$ABC≅$$\\triangle$$PQR (S.S.S axiom) 2. Side-Angle-Side (S.A.S axiom) If two sides of one triangle are respectively equal to two sides of another triangle and the angle made by them are also equal, the triangle is said to be congruent. In$$\\triangle$$ABCand$$\\triangle$$PQR AB = PQ (S) $$\\angle$$B = $$\\angle$$Q (A) BC = QR (S) $$\\therefore$$$$\\triangle$$ABC≅$$\\triangle$$PQR (S.A.S axiom) 3. Angle-side-Angle (A.S.A axiom) If two angles of one triangle are respectively equal to two angles of another triangle and the adjacent sides of the angles are also equal, the triangles are said to be congruent. In $$\\triangle$$ABC and $$\\triangle$$PQR $$\\angle$$B = $$\\angle$$Q (A) BC = QR (S) $$\\angle$$C = $$\\angle$$R (A) $$\\therefore$$ $$\\triangle$$ABC≅$$\\triangle$$PQR (A.S.A axiom) 4. Right angle - Hypotenuse - Side (R.H.S axiom) If hypotenuse and one of two", "∠U QR = TU ∴ ΔPQR ≅ ΔSTU (by ASA criterion) 4. Right-Angle-Hypotenuse-Side (RHS): If the right angle, hypotenuse and one side of one triangle are respectively equal to the corresponding right angle, hypotenuse and side of the other triangle, then the two triangles are congruent m (by RHS). In ΔPQR and ΔSTU, we have PQ = ST hypt. PR = hypt. SU ∠Q = ∠T = 90° ∴ ΔPQR ≅ ΔSTU", "angle. Thus the above angle is named as angle P or QPR or angles RPQ. The symbol used to denote an angle is or ∧ angle P = P or $$\\hat{P}$$ angle QPR = QPR angle RPQ = RPQ Observe the angles given below and answer the following questions. 1. Name the three angles: (i) ∠T or ∠STU or ∠UTS (ii) ∠B or ∠ABC or ∠CBA (iii) ∠Y or ∠XYZ or ∠ZYX 2. Name the vertices: (i) T (ii) B (iii) Y 3. Name the arms (sides): (i) The arms of ∠STU are $$\\overrightarrow{ST}$$ and $$\\overrightarrow{TU}$$ (ii) The arms of ∠ABC are $$\\overrightarrow{BA}$$ and $$\\overrightarrow{BC}$$ (iii) The arms of ∠XYZ are $$\\overrightarrow{YX}$$ and $$\\overrightarrow{YZ}$$ What is an angle? When two sides or line-segments are inclined and meet at a point, then this inclination is called an angle. A polygon has as many angles as number of sides or number of vertices. The triangle has 3 angles, quadrilateral has 4 angles, pentagon has 5 angle, etc. The angles have different measures. Types of Angles: Angles can be classified into different types according to their measures. Look at the angles given below. 1. Right Angle: When a side or line-segment meets another side or line-segment in such a way that one stands upright on the other, then the inclination between", "given isosceles triangle Let us assume that the length of the equal side of the triangle be x We know that area of a traingle is given by $\\endgroup$ – user103816 Jun : The mould of claim 2, wherein each fillet (12) is of triangular prismatic shape, the right section thereof being a right-angled triangle. is a triangle A method according to claim 1 or 2, wherein each of the triangles is a right-angled triangle. The definition for a right triangle states that if one of the angles of a triangle is a right angle ($$90^\\circ$$), the triangle is called a right-angled triangle or simply, a right triangle. Given Δ PQR, right angled at R Also, Δ PQR is isosceles ∴ PR = RQ (Since hypotenuse is the longest side, it cannot be equal to any other side) And, we know that Angle opposite to equal sides of the triangle are equal ∴ ∠ P = ∠ Q Now, in Δ PQR Sum of angles = 180° ∠ P + ∠ Q + ∠ R = 180° ∠ P + ∠ P + ∠ R = 180° 2∠ P + 90° = 180° 2∠ P = 180° – 90° 2∠ P = 90° ∠ P = (90°)/2 ∠ P = 45° Now, sin P = sin 45° =", "to analyze the three triangles, look for patterns, and identify the triangles as isosceles right triangles. Students can then use the patterns from the previous activity. ### Activity Synthesis Make sure all students understand that the three triangles are isosceles right triangles, and represent half of a square, and so are prepared to connect their reasoning from earlier activities to this activity. Display this list of solutions for triangle $$PQR$$: • $$QR=14, RP=\\sqrt{392}$$ • $$QR=14, RP=14\\sqrt{2}$$ • $$QR=14, RP = 19.8$$ • $$QR=14, RP = 1.4 \\boldcdot 14$$ or 19.6 Invite students to explain the reasoning behind each. Encourage students to think about which answers are most accurate, and which are most efficient. (The first two methods are equally accurate, but the first probably took more calculation, while the second could be found using scale factors quite easily. The third answer was obtained by using a calculator to evaluate either of the first two answers, which actually loses both efficiency and accuracy. The fourth answer is very efficient, as it was found using scale factors, but is a bit less accurate. For some applications, the fourth answer may be best, and it’s great for estimating.) ## Lesson Synthesis ### Lesson Synthesis Create a class display of half a square. This display should be posted in the classroom for" ]

[ "A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined? 74 75 76 77 78", "Free Version Difficult # Earning a Grade of A or B SATMAT-6FYRZW Considering the table of student grades from a U.S. History class above (which was also considered in problems $8$ and $9$), how much more likely is it that a student earned a grade of \"$A$\" or \"$B$\" in the 2nd semester after also earning a grade of \"$A$\" or \"$B$\" in the 1st semester, than that a student earned a grade of \"$A$\" or \"$B$\" in the second semester after not earning a grade of \"$A$\" or \"$B$\" in the first semester? (Calculator) A It is about $1.5$ times more likely. B It is about $1.6$ times more likely. C It is about $1.8$ times more likely. D It is about $2$ times more likely.", "# Probability question on school classes 1. May 19, 2010 1. The problem statement, all variables and given/known data In a high-school graduating class of 100 students, 54 studied math, 69 studied history, and 35 studied both math and history. If 1 of the students is selected at random, find the probability that (a) the student took math or history; (b) the student did not take either of these subjects; (c) the student took history but not math. 2. Relevant equations P = n/N 3. The attempt at a solution Ok. I am thinking that I actually need to figure out how many students took math only and history only (pretty sure this is just algebra). So I know that there are 54 math students; this must include those who studied both. Thus, the number of students who studied *math only* is 54 - 35 = 19. Similarly, those who took History only 69 - 35 = 34. So for (a) P(M U H) = (19 + 34) / 100 = 53/100 ... but this is wrong. Book says 22/25. So I am off to a bad start. What am I screwing up here? 2. May 19, 2010 ### gabbagabbahey \"or\" is an inclusive word, so you also need to include the 35 who took both math and", "grade. To calculate: The probability that the selected student of 37-46 age group has received an \"A\" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade. To calculate: The probability that the selected student was in the 37-46 are group or received a \"B\" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade.", "the selected student was in the 37-46 are group or received a \"B\" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade. To calculate: The probability that a student chosen randomly from the class is not going to college and on the honor roll if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is not going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76,out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these,", "p < 0.05, we could say that the odds of students with severe disabilities graduating is only 0.48 times, or approximately half that of the odds of students without disabilities graduating.", "a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is not going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76,out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college. To calculate: The probability that the selected student of 17-26 age group has received an \"A\" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and", "that's 0.1351 hurt, See asks, What is the probability of a head of household having a bachelors degree or more? Education is the probability of a bachelor's degree, plus the probability of Beyonce. So this would be the two marginal probabilities added together, looking at the table, that would be 0.187 plus 0.1061 to give a probability of 0.2931 parte de asks, What is the probability of a household headed by someone with a bachelor's degree earning $1000 or more? This is the probability of earning$1000 or more. Given that you have a bachelor's degree, this is equal to the joint probability, the probability of \\$1000 war and about Cher's degree divided by the marginal probability of having a bachelors degree looking at the table that gives us joint probability of 0.7 to 9, divided by the marginal probability of 0.187 to give a probability of 0.3898 Hurt E asks. What is the probability of a household having an income below 25? This is the marginal probability of having an income lower than 25. Looking at the table that we went 1777 her F asks. Where is the probability of a household headed by someone with a bachelor's degree earning less than 25? This is the probability of under 25 given a bachelor's degree, which is equal to", "When you see these results from the newspapers, can you exactly identify which result(s) is (are) not reliable and not accurate? That is, can you identify which estimation interval(s) does (do) not include the true population proportion? If you can, explain why you can, if no, explain why you cannot and what information you need to identify. Discuss and explain your reasons. You must provide your statistical analysis and reasons. To calculate: The probability that a student chosen randomly from the class is not going to college and on the honor roll if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is not going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76,out of these 56 are going to college. To calculate: The probability that a", "college. The number of students who are not on the honor roll is 76,out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college.", "• All the students that study history only ($52$) • All the students that study both geography and history ($14$) • All the students that study neither history nor geography ($36$) $40+52+14+36=142\\text{ students}$ b) We know that there are $142$ students in total and, of these, $14$ study both history and geography. We can express this as the following fraction: $\\dfrac{14}{142}$ This fraction can be simplified to: $\\dfrac{7}{71}$ (You could give the probability as a decimal or as a percentage (the question doesn’t state how the answer should be given), but since both the decimal ($0.0985….$) and the percentage ($9.85…\\%$) have several decimal places, leaving the probability as a fraction would probably be the most sensible option here, but the decimal and percentage answers are not incorrect.) c) The phrase ‘given that the student selected does not study geography’ tells us that we need to work out how many students do not study geography. The students that don’t study geography are those who study history only ($52$ students) or those who study neither subject ($36$ students). The total that do not study geography is therefore $52+36 = 88$ students. Of these $88$ students, we need to find how many do not study history. The $36$ that study neither subject are the ones that do not study history, so", "at the addresses, what is the probability that (i) exactly one letter goes to the right envelopes (ii) none of the letters go into the right envelopes? 15. The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that (i) it will get at least one of the two awards (ii) it will get only one of the awards. 16. The probability that a girl, preparing for competitive examination will get a State Government service is 0.12, the probability that she will get a Central Government job is 0.25, and the probability that she will get both is 0.07. Find the probability that (i) she will get atleast one of the two jobs (ii) she will get only one of the two jobs. 17. In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl? 18. 10 x 5 = 50 19. Two thirds of", "mean...) 36. jtvatsim Of these 30 students, only 6 earned A's, thus the probability of A given B is P(A|B) = 6/30 37. jtvatsim What you calculated is the probability that the student studied GIVEN that they earned an A. You did the reverse probability. 38. briana.img aaaah okay 39. jtvatsim Hopefully that helps you to make sense of this table. It's not super easy, but it isn't super hard either. The trick is just interpreting the table and the data correctly. Of course, if you were making your own table, you would have written the data down in a way that makes sense to you rather than using someone else's organization. :)", "26 Jan 2018, 10:46 albany09 wrote: A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined? A. 74 B. 75 C. 76 D. 77 E. 78 The ratio of A : B : C = 4x : 6x : 5x, so the total is 15x. We can create a weighted average equation: [65(4x) + 80(6x) + 77(5x)]/15x 5[13(4x) + 16(6x) + 77x]/15x [13(4x) + 16(6x) + 77x]/3x (52x + 96x + 77x)/3x 225x/3x = 75 _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions CEO Joined: 12 Sep 2015 Posts: 2585 Re: A teacher gave the same test to three history classes: A, B, [#permalink] ### Show Tags 30 Jan 2018, 10:08 Top Contributor albany09 wrote: A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the", "[20 x 19 x 18 x 17 x 16 x ...... x 1] / [17 x 16 x 15 x ...... x 1] = 20 x 19 x 18 = 6840 b) 9P3 = 9!/(9 – 3)! = 9!/6! = 9 x 8 x 7 = 504 [ nPr = n! / (n – r)!] c) 7C2 = 7! / 2!(7 – 2)! = 7! / 2! 5! = (7 x 6) / 2 = 21 [nCr = n! / r! (n – r)! ] 20. The events X and Y are mutually exclusive. Suppose P(X) = .05 and P(Y) = .02. What is the probability of either X or Y occurring? What is the probability that neither X nor Y will happen? P (X or Y) = P (X) + P (Y) = 0.05 + 0.02 = 0.07 P (neither X nor Y) = 1 – P (X or Y) = 1 – 0.07 = 0.93 28. A student is taking two courses, history and math. The probability the student will pass the history course is .60, and the probability of passing the math course is .70. The probability of passing both is .50. What is the probability of passing at least one? We have P (history) = 0.60, P (math) = 0.70 and P (both) =", "# Probability from a random pick. A test was given to a group of students. The grades and gender are summarized below If one student is chosen at random from those who took the test, Find the probability that the student got a 'C' GIVEN they are female. The total females is 43, total C's are 19 The total females who have a C is 5 of 19 So my answer is 5/19, is this wrong? • You've figured out the probability that the student is female, given that they got a C (you've taken the number of female C's and divided by the total number of C's). – Arthur Apr 29 '15 at 19:51", "Since there are five school days in a week, the probability that it is Friday is $\\fn_phv \\small \\frac{1}{5}$ or 0.2. What is the probability that a student is absent given that today is Friday? • P(Absent | Friday) = $\\dpi{100} \\fn_phv \\small \\frac{P(Friday and Absent)}{P(Friday)}$ = $\\dpi{100} \\fn_phv \\small \\frac{.05}{0.2}$ = 0.25 = 25% The trick is keeping track of each of the parts! How about another one? At North High School, the probability that a student takes US History and French is 0.053. The probability that a student takes US History is 0.42. What is the probability that a student takes French given that the student is taking US History? • P(French | US History) = $\\dpi{100} \\fn_phv \\small \\frac{P(US History and French)}{P(US History)}$$\\dpi{100} \\fn_phv \\small \\frac{0.053}{0.42}$ = 0.126 = 12.6% Not so bad! Here is a Venn diagram of this situation: The blue circle is the probability of A. The red circle is the probability of B. What about the probability of B given A? Since we already know that A has occurred, we will only focus on the blue circle. That becomes the new sample space. So the probability of B given A is: In the examples above the “numerator” was provided. Things aren’t always so accommodating. A survey was taken of 100 students’", "Question # The grade marks of 50 students of a class are recorded as belowA student of the class is selected at random, what is the probability that the selected student has obtained grade A?${\\text{A}}{\\text{. }}\\dfrac{4}{{25}} \\\\ {\\text{B}}{\\text{. }}\\dfrac{{16}}{{50}} \\\\ {\\text{C}}{\\text{. }}\\dfrac{4}{{50}} \\\\ {\\text{D}}{\\text{. }}\\dfrac{8}{{25}} \\\\$ Hint- Here, we will be using the general formula for the probability of occurrence of an event. Given, Total number of students in a class$= 50$ According to the given data in the table, we can say that Number of students having grade C$= 13$, Number of students having grade B$= 17$, Number of students having grade A$= 8$ and Number of students having grade A+$= 12$ As we know that the general formula for probability is given by Probability of occurrence of an event$= \\dfrac{{{\\text{Number of favourable outcomes}}}}{{{\\text{Total number of possible outcomes}}}}$ Here, the favourable event is that the selected student has obtained grade A. So, Number of favourable outcomes$=$Number of students who has obtained grade A$= 8$ Total number of possible outcomes$=$Total number of students in a class$= 50$ Therefore, Required probability$= \\dfrac{8}{{50}} = \\dfrac{4}{{25}}$ Hence, the probability that the selected student has obtained grade A is $\\dfrac{4}{{25}}$. Option A is correct. Note- In these types of problems, the favourable event is the event whose probability is asked to find", "Tags 30 Jun 2015, 07:13 1 KUDOS Bunuel wrote: On a particular exam, the boys in a history class averaged 86 points and the girls in the class averaged 80 points. If the overall class average was 82 points, what was the ratio of boys to girls in the class? A. 1:4 B. 1:3 C. 1:2 D. 2:1 E. 3:1 Kudos for a correct solution. G B 80 86 82 4 2 2 : 1 G:B=2:1 B:G=1:2 Kudos [?]: 207 [1], given: 133 Manager Joined: 20 Jul 2011 Posts: 81 Kudos [?]: 17 [2], given: 17 GMAT 1: 660 Q49 V31 Re: On a particular exam, the boys in a history class averaged 86 points a [#permalink] ### Show Tags 30 Jun 2015, 10:03 2 KUDOS Bunuel wrote: On a particular exam, the boys in a history class averaged 86 points and the girls in the class averaged 80 points. If the overall class average was 82 points, what was the ratio of boys to girls in the class? A. 1:4 B. 1:3 C. 1:2 D. 2:1 E. 3:1 Kudos for a correct solution. Solution: We can use the X method to solve similar questions B...................G 86................80 ..........82........ 2..................4 Hence, Boys: Girls => 2: 4 => 1:2 Option C Kudos [?]: 17 [2], given: 17 Manager Joined: 26", "P(B) Addition law of Probabilities = 0.11 + 02 = 0.31. ## Probability Practice Problems ### Practice Problems Question 1: Alex has 8 books, 6 fictions and 2 Biographies. When arranged in a shelf, what is the probability the two Biographies are placed together? (Ans: $\\frac{1}{8}$) Question 2: A school has 1200 students out of which 525 are females. 500 students took active participation in Sports of which 300 are female. Are the events, \"Selecting a student participating in Sports\" and \"Selecting a female students\" mutually exclusive. Justify your answer." ]

[ "1500$ … … … … … (i) and $x+10y = 1800$ … … … … … (ii) Subtracting (i) from (ii) we get $6y=300 \\Rightarrow y = 50$ Rs. / year Therefore starting salary $x = 1500 - 4 \\times 50 = 1300$ Rs. $\\\\$ Question 47: Students of a class are made to stand in rows. If there is one student extra in a row, there would be $2$ rows less. If one student is less in a row, there would be $3$ rows more. Find the number of students in the class. Let the number of students in a row be $x$ and the number of rows be $y$ . Therefore the total number of students $= xy$ If there was one student more in each row: $xy = (x+1)(y-2) \\Rightarrow 2x -y = -2$ … … … … … (i) If there was one student less in a row: $xy = (x-1)(y + 3) \\Rightarrow 3x-y = 3$ … … … … … (ii) Solving (i) and (ii) we get $x = 5$ and $y = 12$ rows Total number of students $= 5 \\times 12 = 60$ $\\\\$ Question 48: $8$ men and $12$ boys can finish the work in $10$ days while $6$ men and $8$ boys can finish the work in $14$", "128 views Each of $74$ students in a class studies at least one of the three subjects $H, E$ and $P$. Ten students study all three subjects, while twenty study $H$ and $E$, but not $P$. Every student who studies $P$ also studies $H$ or $E$ or both. If the number of students studying $H$ equals that studying $E$, then the number of students studying $H$ is _________ Given that, • $n(H \\cup E \\cup P) = 74$ • $n( H \\cap E \\cap P) = 10$ • $n(H \\cap E) = 20$ • Only $P = 0$ We can draw the Venn diagram, Here, given that $: b=20, g=10, d=0$ The number of students studying $H =$ The number of students studying $E$ $a + b + e + g = b + c + g + f$ $\\Rightarrow a + 20 + e + 10 = 20 + c + 10 + f$ $\\Rightarrow \\boxed{a + e = c + f} \\quad \\longrightarrow (1)$ We have, $a + b + c + d + e + f + g = 74$ $\\Rightarrow a + 20 + c + 0 + e + f + 10 = 74$ $\\Rightarrow a + e + c + f = 44$ $\\Rightarrow 2(a+e) = 44 \\quad [ \\because \\text{From equation", "(1)}]$ $\\Rightarrow \\boxed{a + e = 22}$ $\\therefore$ The number of students who studies $H = a + b + e + g = 22 + 20 + 10 = 52.$ Correct Answer $: 52$ by 4k points 3 6 24 1 142 views", "\\times 500$ $\\Rightarrow b+d = 250$ — (4) Putting value of $d$ from equation (1) in equation (4), $b + 250 – c = 250$ $\\Rightarrow c – b = 100$ $\\Rightarrow c = b+100$ Now, from equation (3) and equation (4), we know $b = 150 – a$ and $b = 250-d$ Thus, the maximum value of $b$ possible is 150 (assuming $a$=0). We cannot assume $d$=0 because in that case $a+b >150$ Thus, maximum value of $c_{max} = 150+100 = 250$ 7) Answer (C) Let the number of students who passed in PM, PC and CM respectively be $x$,$2x$ and $3x$. Let the number of students who passed only in P,C,M be $5y$,$4y$ and $3y$ respectively. Then the venn diagram of the situation can be drawn as : Since we are asked the maximum number of people who passed in mathematics, thus for this situation we need to minimize all the other values. Thus, for the people who passed in 2 subjects only, we put x=1 to obtain the least value. Thus, number of people who passed in 2 subjects only = (x+2x+3x) = 6x = 6 people Thus, the total of people who passed in only 1 subject = 500 – (50+6) = 444 people Also given, $444 = 5y+4y+3y$ $\\Rightarrow 12y = 444$ $\\Rightarrow", "= 15$ and $y = 5$ Therefore the total number of questions in the test are $20$. $\\\\$ Question 59: Students of a class are made to stand in rows. If there are $3$ students extra in a row, there would be $1$ row less. If $3$ students are less in a row, there would be $2$ rows more. Find the number of students in the class. Let the number of students in a row be $x$ and the number of rows be $y$ . Therefore the total number of students $= xy$ If there was one student more in each row: $xy = (x+3)(y-1) \\Rightarrow x-3y=-3$ … … … … … (i) If there was one student less in a row: $xy = (x-3)(y + 2) \\Rightarrow 2x-3y=6$ … … … … … (ii) Solving (i) and (ii) we get $x = 9$ and $y = 4$ rows Total number of students $= 9 \\times 4 = 36$ $\\\\$", "Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{\\textbf{(C) } 3}$ students that are taking all $3$ classes. ## Solution 3 (Algebra) Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\\boxed{\\textbf{(C) 3}}$.", "Therefore, when there is one extra student, the number is divisible by $3$, $4$, $5$ and $6$. This means it is a common multiple of these numbers. The lowest common multiple of $3$, $4$, $5$ and $6$ is $60$, so there must be $60$ students with the extra. Therefore the number of students (without the imaginary student) is $59$.", "# Question #91dff Sep 6, 2017 $54$ #### Explanation: Subtract $7$ from $331$, then divide by $6$. $331 - 7$ equals $324$, then divide $324$ by $6$, which equals $54$. let $b$ be the number of students. $6 b + 7 = 331$ $6 b = 331 - 7$ $6 b = 324$ $b = 54$", "Free Version Difficult # Earning a Grade of A or B SATMAT-6FYRZW Considering the table of student grades from a U.S. History class above (which was also considered in problems $8$ and $9$), how much more likely is it that a student earned a grade of \"$A$\" or \"$B$\" in the 2nd semester after also earning a grade of \"$A$\" or \"$B$\" in the 1st semester, than that a student earned a grade of \"$A$\" or \"$B$\" in the second semester after not earning a grade of \"$A$\" or \"$B$\" in the first semester? (Calculator) A It is about $1.5$ times more likely. B It is about $1.6$ times more likely. C It is about $1.8$ times more likely. D It is about $2$ times more likely.", "+0 Ratio 0 74 1 There are 75 students in classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A? Apr 11, 2021 #1 +287 +1 Let $a$ be the number of students in class A; $b$ be the number of students in class B; and $c$ be the number of students in class C. The assumptions give you three equations in 3 unknowns $a$, $b$, and $c$. \\begin{eqnarray*} a+b &=& 75 \\\\ a+c &=& 61 \\\\ b/c &=& 5/3 \\end{eqnarray*} The answer is $a$ after you solve this system of equations. Jun 13, 2021", "24 = 600 30. 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students. Let the total number of students be x. According to the question : $\\frac{300}{x}$$\\frac{300}{x+10}$=1 $\\Rightarrow$ $\\frac{300(x+10)-300x}{x(x+10)}$=1 $\\Rightarrow$ $\\frac{300x+3000-300x}{x^{2}+10x}$=1 $\\Rightarrow$3000=x2 + 10x $\\Rightarrow$x2 + 10x – 3000=0 $\\Rightarrow$x2 + (60 — 50)x – 3000 = 0 $\\Rightarrow$x2 + 60x — 50x — 3000 = 0 $\\Rightarrow$x(x + 60) — 50(x + 60) = 0 $\\Rightarrow$ (x + 60)(x — 50) = 0 $\\Rightarrow$x= 50 or x = —60 x cannot be negative; therefore, the total number of students is 50. 31. In a class test, the sum of Kamal’s marks in mathematics and English is 40. Had he got 3 marks more in mathematics and 4 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately. Let the marks of Kamal in mathematics and english be x and y, respectively. According to the question : x + y = 40 … (i) Also, (x + 3)(y — 4) = 360 $\\Rightarrow$ (x + 3)(40 — x — 4) = 360 [From(i)] $\\Rightarrow$ (x + 3)(36 — x) = 360 $\\Rightarrow$36x — x2 + 108 — 3x", "307 views Two students appeared at an examination. One of them secured $9$ marks more than the other and his marks were $56\\%$ of the sum of their marks. The marks obtained by them are 1. $39, 30$ 2. $41, 32$ 3. $42, 33$ 4. $43, 34$ $\\textrm{Let assume that A & B are two students.}$ $\\textrm{Now assume B’s marks are x & A’s marks are x+9}$ $\\Rightarrow$ $A=(x+9)$ $\\Rightarrow$ $B=x$ $\\Rightarrow$ $\\textrm{(x+9)=x+x+9(56%)}$ $\\Rightarrow$ $25x+225=28x+126$ $\\Rightarrow$ $3x=99$ $\\Rightarrow$ $x=33$ $\\textrm{So B=33,A=42}$ $\\textrm{Option C}$ 4.7k points 1 733 views 1 vote 2 565 views 1 vote", "terms from the first and second steps, we get $12c^2 + 4cd - 9cd - 3d^2$. Then by combining the like terms we end up with the expression $12c^2 -5cd - 3d^2$. This is the final answer. Question 6 “Student grades” the answer is F This question tests your understanding of applying proportions in the setting of word problems. 1. We know that there are 180 students, so if $\\frac{1}{4}$ students earned an A, that would mean that 45 students earned an A. 2. We also, know that $\\frac{1}{3}$ students earned a B, so that would mean of the 180 students, 60 students earned a B. 3. Hence, we can calculate the students who earned a C by subtracting those that earned an A or a B from the total. This would be calculated as follows: 180-45-60 = 75. 4. So the number of students that earned a C for the course is 75. Question 7 “Skipper’s Pond” the answer is B This question tests your understanding of applying proportions in the setting of word problems. 1. We’re told that the equation to represent the number of fish in the pond is $f(x) = 3(2^x)$. 2. Hence in the year 2006, the value of x is 2006 – 2000 = 6. 3. Plugging this value of x into", "building. Then: $a+b$ = 25 (1) $c+d$ = 25 (2) $a+c$ = 25 (3) $b+d$ = 25 (4) $a+d$ = 28 (5) (1) $-$ (4): $a+b-b-d$ = 0 $\\Rightarrow a-d$ = 0 $\\Rightarrow a$ = $d$ (5) becomes $a+a$ = 28, so $a$ = 14 You can find more short problems, arranged by curriculum topic, in our short problems collection.", "## Algebra 1 If a student must earn at least 24 credits, then the student's credits must be greater than or equal to 24. Let c be the number of credits. F says credits must be less than or equal to 24, which is not correct. If the student only had 23 credits, he would not be able to graduate. G states that credits must be greater than or equal to 24, which correctly describes the situation. H says 24 must be less than or equal to the number of credits, which is another way of saying that credits must be greater than or equal to 24. $24\\leq c\\longrightarrow$ subtract c and subtract 24 from each side $24-c-24\\leq c-c-24\\longrightarrow$ subtract $-c\\leq-24\\longrightarrow$ multiply each side by -1; reverse the inequality sign $-c\\times-1\\geq-24\\times-1\\longrightarrow$ multiply $c\\geq24$ I also correctly describes the situation. The graph indicates that c is greater than or equal to 24 because the arrow extends to the right, in the direction of increasing values. 24 is shown as included in the solution set by the solid circle at the endpoint.", "## Algebra 1: Common Core (15th Edition) Let A = honor students and B = officers By the given statements, we have: $A=20$ $A \\cap B =9$ The number of students who are only honors students: $20 - 9 =11$ The total number of students is 30. Subtract to find the number of commuters who are officers but not honor students: $30-20=10$", "each school is given by YX=[20 15 5][[25,22,26],[12,15,18],[34,28,36]] YX=[850 805 970] Thus, the amount collected by schools A, B and C are Rs 850, Rs 805 and Rs 970, respectively. ∴ Total amount collected = Rs (850 + 805 + 970) = Rs 2,625 The situation highlights the helpful nature of the students. Concept: Multiplication of Two Matrices Is there an error in this question or solution?", "is $10+10+12+a=35$ . So, $a=35-10-10-12=3$ . Also it is known that 30 students study technical drawing. That is $10+11+a+b=30$ . So, $b=30-10-11-a=30-10-11-3=6$ . There are 60 students. Therefore, $10+10+11+12+a+b+c=60$ . We have $c=60-10-10-11-12-a-b=60-10-10-11-12-3-6=8$ . We can find the number of students that study biology. That is $12+a+b+c=12+3+6+8=29$ . Answer: 3 students study all the 3 subjects; 6 students study biology and technical drawing but not chemistry; 8 students study biology only; 29 students study biology.", "2x-y = 90$ … … … … … (i) and $(x+10) = 2(y-10) \\Rightarrow x - 3y = -20$ … … … … … (ii) Solving (i) and (ii) we get $x = 58 \\ Rs.$ and $y = 26 \\ Rs.$ $\\\\$ Question 58: A scored $40$ marks in a test, getting $3$ marks for each right answer and $1$ mark for wrong answer. Had $4$ marks been awarded for each right answer and $2$ marks deducted for each wrong answer then A would have scored $50$ marks. How many questions were in the test. Let the number of questions answered right are x and that answered wrong are y Therefore we have $3x-y = 40$ … … … … … (i) and $4x-2y = 50$ … … … … … (ii) Solving (i) and (ii) we get $x = 15$ and $y = 5$ Therefore the total number of questions in the test are $20$. $\\\\$ Question 59: Students of a class are made to stand in rows. If there are $3$ students extra in a row, there would be $1$ row less. If $3$ students are less in a row, there would be $2$ rows more. Find the number of students in the class. Let the number of students in a row be $x$ and", "get own) If A gets B's gift and B gets C's, then C must get D's again, (or D will get own) If A gets B's gift and B gets D's, then C must get A's ( to not get own) Only these 3 are possible for A, if A gets B's. If we count the situations for A gets C's... and the situation for A gets D's... this will introduce 3+3 more. Total $3(3) =3^2=9$ ABCD BCDA BDAC" ]

[ "Free Version Difficult # Earning a Grade of A or B SATMAT-6FYRZW Considering the table of student grades from a U.S. History class above (which was also considered in problems $8$ and $9$), how much more likely is it that a student earned a grade of \"$A$\" or \"$B$\" in the 2nd semester after also earning a grade of \"$A$\" or \"$B$\" in the 1st semester, than that a student earned a grade of \"$A$\" or \"$B$\" in the second semester after not earning a grade of \"$A$\" or \"$B$\" in the first semester? (Calculator) A It is about $1.5$ times more likely. B It is about $1.6$ times more likely. C It is about $1.8$ times more likely. D It is about $2$ times more likely.", "# [SOLVED] Basic probability questions • Feb 15th 2010, 12:02 AM downthesun01 [SOLVED] Basic probability questions I'm studying for an exam and am having a few small problems. I wasn't whether it was best to make a thread for each or post everything in one thread, but in the interest of minimizing clutter I to do the latter. 1. In a high school graduating class of 100 students, 54 students studied math, 69 studied history, and 35 studied both. If one of these students is selected at random, find the probability that the student took history but not math. The answers is supposed to be 34/100. I'm not too sure what I'm looking for. Is this right? $p(math' \\cup history)$ Which would be $0.46 + 0.69 - p(math' \\cap history)= 1.15 - (0.46 * 0.69)= 1.15-0.3174= 0.8326$ Which is obviously wrong. Looking at the information given. I see that $p(history) - p(math \\cap history) = 0.69 - 0.35 = 0.34$ Which is the right answer but I'm a little confused about it. Looking at it now, it makes sense because you're taking all of the students that took history and removing that section that also took math. How would one set up the problem using the $\\cap$ and $\\cup$ symbols? That's the part I'm stuck on. 2. Interest", "A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined? 74 75 76 77 78", "Question # The grade marks of 50 students of a class are recorded as belowA student of the class is selected at random, what is the probability that the selected student has obtained grade A?${\\text{A}}{\\text{. }}\\dfrac{4}{{25}} \\\\ {\\text{B}}{\\text{. }}\\dfrac{{16}}{{50}} \\\\ {\\text{C}}{\\text{. }}\\dfrac{4}{{50}} \\\\ {\\text{D}}{\\text{. }}\\dfrac{8}{{25}} \\\\$ Hint- Here, we will be using the general formula for the probability of occurrence of an event. Given, Total number of students in a class$= 50$ According to the given data in the table, we can say that Number of students having grade C$= 13$, Number of students having grade B$= 17$, Number of students having grade A$= 8$ and Number of students having grade A+$= 12$ As we know that the general formula for probability is given by Probability of occurrence of an event$= \\dfrac{{{\\text{Number of favourable outcomes}}}}{{{\\text{Total number of possible outcomes}}}}$ Here, the favourable event is that the selected student has obtained grade A. So, Number of favourable outcomes$=$Number of students who has obtained grade A$= 8$ Total number of possible outcomes$=$Total number of students in a class$= 50$ Therefore, Required probability$= \\dfrac{8}{{50}} = \\dfrac{4}{{25}}$ Hence, the probability that the selected student has obtained grade A is $\\dfrac{4}{{25}}$. Option A is correct. Note- In these types of problems, the favourable event is the event whose probability is asked to find", "than $2$ A's. This is the probability of $0$ A's plus the probability of $1$ A. If $X$ is the number of A's, $$\\Pr(X\\lt 2)=\\binom{15}{0}(0.3)^0(0.7)^{15}+\\binom{15}{1}(0.3)^1(0.7)^{14}.$$ It follows that the probability of $2$ or more A's is $$1-\\left((0.7)^{15}+15(0.3)(0.7)^{14}\\right).$$ - If you treat each grade as an independent event, the probability of no students earning an A is $(0.7)^{15}$. The probability of only one student earning an A is $15(0.3)(0.7)^{14}$ The probability of at least two students having A's is $1 - \\Pr (0 A's) - \\Pr( 1 A)$. $\\Pr(2+ A) = 1 - (0.7)^{15} - 15(0.3)(0.7)^{14}$ -", "c)}{3}$$ = $$\\frac{0.5}{3}$$ (a + b + c) = $$\\frac{1}{6}$$ (a + b + c) Hence, the mean of the new set of numbers is $$\\frac{1}{6}$$ (a + b + c) Question 16. The table shows the mean scores of three classes in a history test. The mean score of all the students in classes A and B combined is 6.8 The mean score of all the students in classes B and C combined is 7. If the number of students in classes A, B, and C are denoted by a, b, and c respectively, find the ratio a: b: c. a : b : c = 4 : 6 : 3 Explanation: Let number of students in A = a Let number of students in B = b Let number of students in C = c Total score of A ÷ a = 8 Total score of A = 8 x a = 8a Total score of B ÷ b = 6 Total score of B = 6 x b = 6b Total score of C ÷ c = 9 Total score of C = 9 x c = 9c Total score of (A + B) ÷ (a + b) = 6.8 Total score of (A + B) = 6.8 x (a + b) = 6.8a +", "the selected student was in the 37-46 are group or received a \"B\" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade. To calculate: The probability that a student chosen randomly from the class is not going to college and on the honor roll if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is not going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76,out of these 56 are going to college. To calculate: The probability that a student chosen randomly from the class is going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these,", "• All the students that study history only ($52$) • All the students that study both geography and history ($14$) • All the students that study neither history nor geography ($36$) $40+52+14+36=142\\text{ students}$ b) We know that there are $142$ students in total and, of these, $14$ study both history and geography. We can express this as the following fraction: $\\dfrac{14}{142}$ This fraction can be simplified to: $\\dfrac{7}{71}$ (You could give the probability as a decimal or as a percentage (the question doesn’t state how the answer should be given), but since both the decimal ($0.0985….$) and the percentage ($9.85…\\%$) have several decimal places, leaving the probability as a fraction would probably be the most sensible option here, but the decimal and percentage answers are not incorrect.) c) The phrase ‘given that the student selected does not study geography’ tells us that we need to work out how many students do not study geography. The students that don’t study geography are those who study history only ($52$ students) or those who study neither subject ($36$ students). The total that do not study geography is therefore $52+36 = 88$ students. Of these $88$ students, we need to find how many do not study history. The $36$ that study neither subject are the ones that do not study history, so", "is defined as a C or better in the course.) 52 of the 65 students succeeded. Another 55 students were from rural backgrounds; 30 of these students succeeded. A Is there good evidence that the ... Monday, April 8, 2013 at 12:12am LBCC At a certain college, there were 780 science majors, 320 engineering majors, and 390 business majors. If one student was selected at random, the probability that they are a business major is? Sunday, April 7, 2013 at 11:06pm MATH An engineering consultant bids on jobs where he can make $10000 profit. If in a certain month, he has a 20% probability of getting a contract, 30% probability of getting 2 contracts, 40% probability of getting 3 contracts, and 10% probability of getting 4 contracts, what is ... Wednesday, April 3, 2013 at 10:02pm World History How were the Byzantine's Engineering geniuses? Take two or three of their engineering feats and discuss how these marvels led to the empire's overall success. Wednesday, April 3, 2013 at 8:45pm MATH An engineering consultant bids on jobs where he can make$10000 profit. If in a certain month, he has a 20% probability of getting a contract, 30% probability of getting 2 contracts, 40% probability of getting 3 contracts, and 10% probability of getting 4 contracts, what is ... Wednesday,", "$6$, and $|C\\cap D|$ is $|A\\cap B|$ so $30$. Moving on to the next set: $|A\\cap B\\cap C|$ is the same as $|A\\cap C|$ which is $6$, $|A\\cap B\\cap D|$ is ordering everybody so $1$, $|A\\cap C\\cap D|$ is again ordering everybody which is $1$, and $|B\\cap C\\cap D|$ is the same as $|A\\cap B\\cap C|$ so $6$. Finally, $|A\\cap B\\cap C\\cap D|$ is ordering everybody so $1$. Now, lets substitute everything back in. We get a massive expression of $720-120-120-120-120+30+6+20+30+6+30-6-1-1-6+1=\\boxed{349}$. ### Five Set Example #### Problem There are five courses at my school. Students take the classes as follows: 243 take algebra. 323 take language arts. 143 take social studies. 241 take biology. 300 take history. 213 take algebra and language arts. 264 take algebra and social studies. 144 take algebra and biology. 121 take algebra and history. 111 take language arts and social studies. 90 take language arts and biology. 80 take language arts and history. 60 take social studies and biology. 70 take social studies and history. 60 take biology and history. 50 take algebra, language arts, and social studies. 50 take algebra, language arts, and biology. 50 take algebra, language arts, and history. 50 take algebra, social studies, and biology. 50 take algebra, social studies, and history. 50 take algebra, biology, and history. 50", "that's 0.1351 hurt, See asks, What is the probability of a head of household having a bachelors degree or more? Education is the probability of a bachelor's degree, plus the probability of Beyonce. So this would be the two marginal probabilities added together, looking at the table, that would be 0.187 plus 0.1061 to give a probability of 0.2931 parte de asks, What is the probability of a household headed by someone with a bachelor's degree earning $1000 or more? This is the probability of earning$1000 or more. Given that you have a bachelor's degree, this is equal to the joint probability, the probability of \\$1000 war and about Cher's degree divided by the marginal probability of having a bachelors degree looking at the table that gives us joint probability of 0.7 to 9, divided by the marginal probability of 0.187 to give a probability of 0.3898 Hurt E asks. What is the probability of a household having an income below 25? This is the marginal probability of having an income lower than 25. Looking at the table that we went 1777 her F asks. Where is the probability of a household headed by someone with a bachelor's degree earning less than 25? This is the probability of under 25 given a bachelor's degree, which is equal to", "grade. To calculate: The probability that the selected student of 37-46 age group has received an \"A\" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade. To calculate: The probability that the selected student was in the 37-46 are group or received a \"B\" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade.", "Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{\\textbf{(C) } 3}$ students that are taking all $3$ classes. ## Solution 3 (Algebra) Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\\boxed{\\textbf{(C) 3}}$.", "# Probability question on school classes 1. May 19, 2010 1. The problem statement, all variables and given/known data In a high-school graduating class of 100 students, 54 studied math, 69 studied history, and 35 studied both math and history. If 1 of the students is selected at random, find the probability that (a) the student took math or history; (b) the student did not take either of these subjects; (c) the student took history but not math. 2. Relevant equations P = n/N 3. The attempt at a solution Ok. I am thinking that I actually need to figure out how many students took math only and history only (pretty sure this is just algebra). So I know that there are 54 math students; this must include those who studied both. Thus, the number of students who studied *math only* is 54 - 35 = 19. Similarly, those who took History only 69 - 35 = 34. So for (a) P(M U H) = (19 + 34) / 100 = 53/100 ... but this is wrong. Book says 22/25. So I am off to a bad start. What am I screwing up here? 2. May 19, 2010 ### gabbagabbahey \"or\" is an inclusive word, so you also need to include the 35 who took both math and", "Courses Courses for Kids Free study material Free LIVE classes More The grade marks of 50 students of a class are recorded as belowA student of the class is selected at random, what is the probability that the selected student has obtained grade A?${\\text{A}}{\\text{. }}\\dfrac{4}{{25}} \\\\ {\\text{B}}{\\text{. }}\\dfrac{{16}}{{50}} \\\\ {\\text{C}}{\\text{. }}\\dfrac{4}{{50}} \\\\ {\\text{D}}{\\text{. }}\\dfrac{8}{{25}} \\\\$ Last updated date: 24th Mar 2023 Total views: 309k Views today: 2.86k Verified 309k+ views Hint: Here, we will be using the general formula for the probability of occurrence of an event, i.e., $P(A) = \\dfrac{{{\\text{Number of favorable outcomes}}}}{{{\\text{Total number of possible outcomes}}}}$ Given, Total number of students in a class$= 50$ According to the given data in the table, we can say that Number of students having grade C$= 13$, Number of students having grade B$= 17$, Number of students having grade A$= 8$ And, Number of students having grade A+$= 12$ As we know that the general formula for probability is given by Probability of occurrence of an event $= \\dfrac{{{\\text{Number of favorable outcomes}}}}{{{\\text{Total number of possible outcomes}}}}$ Here, the favorable event is that the selected student has obtained grade A. So, Number of favorable outcomes $=$ Number of students who has obtained grade A$= 8$ Total number of possible outcomes $=$ Total number of students in a class$= 50$ Therefore,", "= P(A and B)/P(B) Bayes’ Law Formula: P(Ai |B)= ]P(Ai)P(B|Ai)] / [P(A1 )P(B | A1 ) + P(A 2 )P(B | A 2 ) + . . . + P(A k )P( B | A k )] • July 12th 2013, 07:25 AM HallsofIvy Re: Probability question I'm having difficulty with Have you tried anything at all? Personally, I would not use \"Bayes formula\" or the \"Conditional probability\" formula. Instead I would calculate specific numbers. Suppose our pool consists of 1000000 people. Then \"Not a high school graduate 0.0975 , 0.0080\" So we have 0.0975*1000000= 97500 non-high school graduates who are employed, 0.0080*1000000=8000 who are unemployed. \"High school graduate 0.3108 , 0.0128\" So we have 0.3108*1000000= 310800 high school graduates who are employed, 0.0128*1000000= 12800 high school graduates who are unemployed. \"Some college, no degree 0.1785 , 0.0062\" So we have .01785*1000000= 178500 people with some college who are employed, 0.0062*1000000= 6200 people with some college who are unemployed. \"Associate’s degree 0.0849 , 0.0023\" So we have 0.0849*1000000= 84900 people with associate's degree who are employed, 0.0023*1000000= 2300 people with associate's degree who are unemployed. \"Bachelor’s degree 0.1959 , 0.0041\" So we have 0.1959*1000000= 195900 people with bachelor's degree who are employed, 0.0041*1000000= 4100 people with bachelor's degree who are unemployed. Fourth was \"The probability that a randomly", "0.65 = 0.52$ Obviously the wrong answer. However, I decided to use De Morgan's law as such $p(A' \\cap B') = p(A \\cup B)' = 1 -[0.20 + 0.35] = 1- 0.55 = 0.45$ If $p(A' \\cap B') = p(A \\cup B)'$ under De Morgan's law, then why did I get two different answers? Where's my mistake? For 1, I suggest you draw a Venn diagram. For 2, it's as simple as 1 - 0.20 - 0.35. • Feb 15th 2010, 02:03 AM Hello downthesun01 Quote: Originally Posted by downthesun01 I'm studying for an exam and am having a few small problems. I wasn't whether it was best to make a thread for each or post everything in one thread, but in the interest of minimizing clutter I to do the latter. 1. In a high school graduating class of 100 students, 54 students studied math, 69 studied history, and 35 studied both. If one of these students is selected at random, find the probability that the student took history but not math. The answers is supposed to be 34/100. I'm not too sure what I'm looking for. Is this right? $p(math' \\cup history)$ Which would be $0.46 + 0.69 - p(math' \\cap history)= 1.15 - (0.46 * 0.69)= 1.15-0.3174= 0.8326$ Which is obviously wrong. Looking at the", "26 Jan 2018, 10:46 albany09 wrote: A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined? A. 74 B. 75 C. 76 D. 77 E. 78 The ratio of A : B : C = 4x : 6x : 5x, so the total is 15x. We can create a weighted average equation: [65(4x) + 80(6x) + 77(5x)]/15x 5[13(4x) + 16(6x) + 77x]/15x [13(4x) + 16(6x) + 77x]/3x (52x + 96x + 77x)/3x 225x/3x = 75 _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions CEO Joined: 12 Sep 2015 Posts: 2585 Re: A teacher gave the same test to three history classes: A, B, [#permalink] ### Show Tags 30 Jan 2018, 10:08 Top Contributor albany09 wrote: A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the", "0.6699219 ## 3 0.2343750 0.7656250 ## 4 0.1250000 0.8750000 By looking at the first row that is the 1st year of the beginning student and the 2nd column, we can see that the probability of this beginning student will graduate will be about 58.6%", "1500$ … … … … … (i) and $x+10y = 1800$ … … … … … (ii) Subtracting (i) from (ii) we get $6y=300 \\Rightarrow y = 50$ Rs. / year Therefore starting salary $x = 1500 - 4 \\times 50 = 1300$ Rs. $\\\\$ Question 47: Students of a class are made to stand in rows. If there is one student extra in a row, there would be $2$ rows less. If one student is less in a row, there would be $3$ rows more. Find the number of students in the class. Let the number of students in a row be $x$ and the number of rows be $y$ . Therefore the total number of students $= xy$ If there was one student more in each row: $xy = (x+1)(y-2) \\Rightarrow 2x -y = -2$ … … … … … (i) If there was one student less in a row: $xy = (x-1)(y + 3) \\Rightarrow 3x-y = 3$ … … … … … (ii) Solving (i) and (ii) we get $x = 5$ and $y = 12$ rows Total number of students $= 5 \\times 12 = 60$ $\\\\$ Question 48: $8$ men and $12$ boys can finish the work in $10$ days while $6$ men and $8$ boys can finish the work in $14$" ]

[ "× # Gcd needs help Find the number of positive gcd of $$5n+1$$ and $$17n+1$$where $$n$$ is a positive integer. Note by William Isoroku 11 months, 1 week ago Sort by:", "last example is precisely the example I used! 7. In fact, gcd(a,b) is the smallest positive integer of the form as+bt,where s and t are integers. further more, all numbers of the form as+bt is in the multiple table of gcd(a,b).", "# GCD values As $$x$$ ranges over all positive integers, how many different possible values are there for $\\gcd(17x+277,7x+107) ?$ ×", "# GCD = 1 For a positive integer $$n>2,$$ what can we say about the number of positive integers less than $$n$$ that are relatively prime to $$n$$ (that is, their GCD is 1)? Hint: A useful property of the Greatest Common Divisor function is $$\\gcd(a,b) = \\gcd(a,a-b);$$ e.g., $$\\gcd(10,8) = \\gcd(10,2) = 2.$$ ×", "× # Gcd needs help Find the number of positive gcd of $$5n+1$$ and $$17n+1$$where $$n$$ is a positive integer. Note by William Isoroku 1 year, 3 months ago Sort by:", "× # Gcd needs help Find the number of positive gcd of $$5n+1$$ and $$17n+1$$where $$n$$ is a positive integer. Note by William Isoroku 1 year, 11 months ago Sort by:", "# Tag Info The question as currently worded, and considering comments by its author, would boil down to: is it a hard problem finding $g^a\\bmod p$, given large prime $p$ large integer $g$ less than $p$ that is a generator of $\\mathbb Z_p$ [see note 1] prime $r$ less than $p$ knowledge that unknown $a$ is a positive integer less than $r$ positive integer $b$ less than ... Not at all. It's very trivial: $$(g^{ab})^{b^{-1}}=g^a$$", "nonzero integers $$a$$ and $$b$$ is the smallest positive integer among all their linear combinations. In other words, $$\\gcd(a,b)$$ is the smallest positive element in the set $$\\{as+bt \\mid s,t\\in\\mathbb{Z}\\}$$. Corollary $$\\PageIndex{3}$$ For any nonzero integers $$a$$ and $$b$$, there exist integers $$s$$ and $$t$$ such that $$\\gcd(a,b)=as+bt$$. Proof Theorem 5.5.1 maintains that the set of all the linear combinations of $$a$$ and $$b$$ equals to the set of all the multiples of $$\\gcd(a,b)$$. Since $$\\gcd(a,b)$$ is a multiple of itself, it must equal to one of those linear combinations. Thus, $$\\gcd(a,b) = sa+tb$$ for some integers $$s$$ and $$t$$. Theorem $$\\PageIndex{4}$$ Two nonzero integers $$a$$ and $$b$$ are relatively prime if and only if $$as+bt=1$$ for some integers $$s$$ and $$t$$. Proof The result is a direct consequence of the definition that $$a$$ and $$b$$ are said to be relatively prime if $$\\gcd(a,b)=1$$. Example $$\\PageIndex{1}\\label{eg:moregcd-01}$$ It is clear that 5 and 7 are relatively prime, so are 14 and 27. Find the linear combination of these two pairs of numbers that equals to 1. Solution By inspection, or using the extended Euclidean algorithm, we find $$3\\cdot5-2\\cdot7=1$$, and $$2\\cdot14-1\\cdot27=1$$. hands-on Exercise $$\\PageIndex{1}\\label{he:moregcd-01}$$ Show that $$\\gcd(133,143)=1$$ by finding an appropriate linear combination. hands-on Exercise $$\\PageIndex{2}\\label{he:moregcd-02}$$ Show that 757 and 1215 are relatively prime by finding an appropriate linear combination.", "always thanks to the fact that gcd$(49,121)=1$ and in general for any pair of positive integers being relatively prime. I leave it to you to do it the other way around, that is, to start with the first equation and proceed in a similar way. Of course it should give you the same solution.", "every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 22:49 1 For every positive integer $$x$$, $$f(x)$$ represents the greatest prime factor of $$x!$$, $$g(x)$$ represents the smallest prime factor of $$2^x+1$$. What is $$(g(f(12))$$? $$f(x)$$=greatest prime factor of x! $$f(12)$$ = 1*2*3*4*5*6*7*8*9*10*11*12 => greatest prime factor = 11 $$g(x)$$ = smallest prime factor of $$2^x+1$$ $$g(11)$$ = smallest prime factor of $$2^{11}+1$$ => 2048+1 = 2049 = 3*683 Smallest prime factor = 3 => Answer B Manager Joined: 27 May 2010 Posts: 199 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 22:56 1 For every positive integer x, f(x) represents the greatest prime factor of x!, and g(x) represents the smallest prime factor of 2x+1. What is (g(f(12))? x = 12 f(12) = Greatest prime factor (12!) = 11 g(11) = Smallest prime factor (2^11 + 1) = (2049) = 3 Option B Intern Joined: 14 Mar 2017 Posts: 40 Location: United States (VA) GPA: 2.9 WE: Science (Pharmaceuticals and Biotech) Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 23:10 1 Question: For every positive integer $$x$$, $$f(x)$$ represents the greatest prime factor of $$x!$$, and $$g(x)$$", "+0 # Math Help +2 264 3 +335 What is the smallest positive integer n for which 9n-2 and 7n+3 share a common factor greater than 1? Nov 21, 2018 #1 +2 a=listfor(n, 1, 23, GCD(( 9*n-2), (7*n+3))=(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 41)=n=23 and the GCD of [164, 205] =41 Nov 21, 2018 #2 +701 0 You can put that into the \"LaTeX\" box to make the GCD part look nicer. PartialMathematician Nov 21, 2018 #3 +22358 +11 What is the smallest positive integer n for which 9n-2 and 7n+3 share a common factor greater than 1? $$\\text{Euclidean algorithm:}~ \\boxed{gcd(a-nb,b)=gcd(a,b)}$$ $$\\begin{array}{|rcll|} \\hline && gcd(9n-2,7n+3) \\\\ &=& gcd(9n-2-1\\cdot(7n+3) ,7n+3)=gcd(2n-5,7n+3) \\\\ &=& gcd(2n-5,7n+3-3\\cdot(2n-5))=gcd(2n-5,n+18) \\\\ &=& gcd(2n-5-2\\cdot(n+18),n+18)=gcd(-41,n+18) \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline gcd(-41,\\underbrace{n+18}_{=41~(\\Rightarrow n>0!)}) \\qquad -41~ \\text{is prime} \\\\\\\\ n+18 &=& 41 \\\\ n &=& 41-18 \\\\ n &=& 23 \\\\ \\hline \\end{array}$$ The common factor greater than 1 is $$\\mathbf{41}$$. and the smallest positive integer n is $$\\mathbf{23}$$ Hint: $$gcd(-41,41) = 41$$ Nov 21, 2018 edited by heureka Nov 21, 2018 edited by heureka Nov 21, 2018", "# Product of two primes You are told that $$n = 110179$$ is the product of two primes $$p$$ and $$q$$ ($$p > q$$). The number of positive integers less than $$n$$ that are relatively prime to $$n$$ (that is those $$m$$ such that $$\\gcd(m, n) = 1$$) is 109480. Write the value of $$p - q$$. ×", "# Greedy Calvinosaurus Dinosaur Greedy Calvinosaurus Dinosaur is threatening to gobble you up, unless you can tell him what is the largest possible value of $\\gcd (122 + n^2, 122 + (n+1)^2 ) ,$ as $$n$$ ranges over all positive integers.", "# Solution July 21, 2007 ### Problem What is the largest positive integer $n\\,$ such that $n^3+100\\,$ is divisible by $n+10\\,$? ### Solution n + 10 | n3 + 100 is equivalent to gcd(n + 10,n3 + 100) = n + 10 By the rules for the gcd (http://en.wikipedia.org/wiki/Greatest_common_divisor) we have $n+10 = \\gcd(n+10, n^3+100)\\,$ $n+10 = \\gcd(n+10, n^3+100 - (n+10)\\cdot(n^2-10n+100))$ $n+10 = \\gcd(n+10, -900) = \\gcd(n+10, 900)\\,$ In other words, n + 10 | 900 and the largest such n is n = 890.", "# GCD of Integer and its Negative ## Theorem Let $a \\in \\Z$ be an integer. Then: $\\gcd \\set {a, -a} = \\size a$ where: $\\gcd$ denotes greatest common divisor $\\size a$ denotes the absolute value of $a$. ## Proof From Integer Divisor Results, the divisors of $a$ include $a$ itself. From Integer Divides its Negative, $a \\divides \\paren {-a}$. Thus we have: $a \\divides a$ and: $a \\divides -a$ and so: $\\gcd \\set {a, -a} \\ge \\size a$ From Absolute Value of Integer is not less than Divisors, there is no divisor of $a$ which is greater than $a$. That is: $\\gcd \\set {a, -a} \\le \\size a$ Hence the result. $\\blacksquare$", "numbers without leaving any remainder. So the factors of the smallest composite number $4 = 1 \\times 2 \\times 2$. And the factors of the smallest prime number 2 $= 1 \\times 2$. So, from above we clearly say that the largest positive integer number that divides both the numbers 4 and 2 without leaving any remainder is 2. So, the g.c.d of the smallest prime number and the smallest composite number is 2. Hence option (b) is correct. Note – Whenever we face such types of questions the key concept we have to remember is that always recall the property of composite number, prime number and GCD respectively which is stated above then use this property first find out the smallest prime and composite number respectively, then find the value of GCD using the property of GCD which is the required answer.", "and the numbers greater than this number will not be prime numbers. then g((2^11)+1)=2048+1=2049 2049 is divisible by 1,3... 2049 so 3 is the smallest prime factor for g(x) Manager Joined: 26 Mar 2019 Posts: 100 Concentration: Finance, Strategy For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags Updated on: 18 Jul 2019, 08:43 1 Quote: For every positive integer $$x$$, $$f(x)$$ represents the greatest prime factor of $$x!$$, and $$g(x)$$ represents the smallest prime factor of $$2^x+1$$. What is $$(g(f(12))$$? The task tells us that $$f(x)$$ represents greatest prime factor of $$x!$$. Prime factor is a positive integer which is not equal to $$1$$ and has two factors: itself and $$1$$. Let us find $$f(x)$$ first, and then find $$g(f(x))$$. $$f(12)$$ = greatest prime factor of $$12!$$ Let us find out what greatest prime factor of $$12!$$ is: $$12! = 1 * 2* 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12$$. As we can see from above, the greatest of the numbers is $$11$$ which is the greatest prime factor for $$12!$$ $$12!$$ cannot have higher prime factor since $$12$$ itself is equal to $$2*2*3$$, and any product of its numbers will have more factors than $$2$$ and will not", "# Relatively Prime in 2015 Number Theory Level 4 How many ordered pairs of positive integers $$(m, n)$$ are there such that $$m + n = 2015$$ and $$\\gcd(m, n) = 1$$? ×", "# Determining the maximum GCD Let $$a, b, c$$ be distinct positive integers such that $$a+b+c \\leq 3000000$$. Find the maximum value of $$\\text{gcd}(ab+1, bc+1, ca+1)$$. Notation: $$\\gcd(\\cdot)$$ denotes the greatest common divisor function. ×", "the sequence $$x_1$$, $$x_2$$, $$x_3$$, ... by $$x_1 = m$$, $$x_{n+1} = f(x_n)$$. Let $$g(m)$$ be the smallest $$n$$ such that $$x_n = 1$$. (Examples: $$g(100) = 3$$, $$g(87) = 7$$.) Denote by $$N$$ be the number of positive integers $$m$$ such that $$g(m) = 20$$. The number of distinct prime factors of $$N$$ is equal to $$2^u\\cdot v$$ for two non-negative integers $$u$$ and $$v$$ such that $$v$$ is odd. Determine $$u+v$$. 2005-2018 IMOmath.com | imomath\"at\"gmail.com | Math rendered by MathJax" ]

[ "Free Version Easy # Factor $113400$ as a Product of Prime Powers NUMTH-H9S91G What is the prime power factorization of $113400$? A $2^4 \\cdot 3 \\cdot 5^3 \\cdot 11$ B $2^3 \\cdot 3^4 \\cdot 5^2 \\cdot 7$ C $2^2 \\cdot 3^5 \\cdot 5^3 \\cdot 7$ D $2^3 \\cdot 3^4 \\cdot 5^2 \\cdot 11$", "saw this on the math team forum, was posted by like last last year, but anyways… In the prime factorization of the expression $100! \\cdot 99! \\cdot 98! \\cdot \\ldots \\cdot 3! \\cdot 2! \\cdot 1!$ there is a $7^x$ term. Find x.", "## 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 $10 = 150 - 2 \\cdot 70$.", "$p_1^{n_1}\\cdots p_k^{n_k}$ 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 $5\\cdot5\\cdot13=\\boxed{\\textbf{(D) }325}$", "# 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.", "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?", "# 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.", "# 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: $112 = 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 7 = {2}^{4} \\cdot 7$", "# Too many factorials! $\\large \\dfrac{(100!+99!)(99!-98!)(98!+97!)\\cdots(2!+1!)}{(100!-99!)(99!+98!)(98!-97!)\\cdots(2!-1!)}$ The value of the above fraction can be written as mixed fraction $$a\\dfrac{b}{c}$$, where $$b$$ and $$c$$ & $$a$$ and $$b$$ are co-prime, but $$a$$ and $$c$$ are not co-prime. What is the product of $$a$$, $$b$$, and $$c$$? ×", "$n$ 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 $$4$$.", "= 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 prime factorization of 125? Oct 29, 2016 The prime factorization of $125$ is ${5}^{3}$. Explanation: To find the prime factorization, divide $125$ by only prime numbers. Remember that prime numbers are whole numbers greater than $1$ that are only evenly divisible by themselves and $1$. {$2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53$, etc.} Since $125$ ends in $5$, it is evenly divisible by $5$, which is a prime number. (How convenient!) $125 \\div 5 = 25$ $25$ now has to be factored into its prime factors. $25 = 5 \\cdot 5$ So, the prime factorization of $125$ is $125 = 5 \\cdot 5 \\cdot 5 = {5}^{3}$", "Solution From Solution 1, the prime factorization of $180 = 2^2 \\cdot 3^2 \\cdot 5$. The prime factorization of $594 = 2 \\cdot 3^3 \\cdot 11$. Hence, $\\operatorname{lcm} (180,594) = 2^2 \\cdot 3^3 \\cdot 5 \\cdot 11$, and $\\operatorname{gcf} (180,594) = 2 \\cdot 3^2$. Therefore, $\\frac{\\operatorname{lcm} (180,594)}{\\operatorname{gcf} (180,594)} = \\frac{2^2 \\cdot 3^3 \\cdot 5 \\cdot 11}{2 \\cdot 3^2} = 2 \\cdot 3 \\cdot 5 \\cdot 11 \\cdot = 330 \\Longrightarrow \\boxed{\\textbf{(C)}\\ 330}$", "two squares and if so, find a representation for $1034$. The prime power factorization of $1034$ is: (2) \\begin{align} \\quad 1034 = 2^1 \\cdot 11^1 \\cdot 47^1 \\end{align} Note that $11 \\equiv 3 \\pmod 4$ but $1$ is not even, so again, $1034$ cannot be expressed as a sum of two squares. ## Example 3 Determine if $98765$ can be expressed as a sum of two squares and if so, find a representation for $98765$. The prime power factorization of $98765$ is: (3) \\begin{align} \\quad 98765 = 5^1 \\cdot 19753^1 \\end{align} Since $5 \\equiv 1 \\pmod 4$ and $19753 \\equiv 1 \\pmod 4$ we have that $98765$ can be written as a sum of two squares. Namely: (4) \\begin{align} \\quad 98765 = 13^2 + 314^2 \\end{align}", "$12 = 2 \\cdot 2 \\cdot 3$ is the product of $k = 3$ prime numbers. Usually a product is defined for at least $k = 2$ factors, however for the theory we are getting ready to develop, it will be convenient to extend the idea of product including the case of the product with just one factor ($k = 1$), which is the factor itself. With this assumption, a prime number il also a product of prime numbers: a special product, made up of the number itself that multiplies no other number. For example, here are the factorizations of numbers from 2 to 10: Number Factorization Number of factors 2 2 1 3 3 1 4 $2 \\cdot 2$ 2 5 5 1 6 $2 \\cdot 3$ 2 7 7 1 8 $2 \\cdot 2 \\cdot 2$ 3 9 $3 \\cdot 3$ 2 10 $2 \\cdot 5$ 2 The fundamental Theorem of arithmetic states not only that every integer number greater than 1 has a factorization into prime factors, but also that this factorization is unique except for the order of the factors. For example, $12 = 2 \\cdot 2 \\cdot 3$ and there are no other prime numbers, apart from 2 taken two times and 3 taken one time, that multiplied between themselves make 12. The", "d == 0) == 5]. It returns [16, 81, 625]. – Eric Duminil Jan 6 '18 at 10:05 (Filling in some details, on the off chance that other answers are unfamiliar to potential readers!) Consider the prime factorization of $75 = 3^1 \\cdot 5^2$. How many factors does $75$ have? Well, in forming a factor of $75$, we have to choose how many $3$s to include - there is one available, but we could also choose none; so, the number of choices is $1+1=2$ - and how many $5$s to include - there are two available, but we could also choose none; so, the number of choices is $2+1=3$. The aforementioned $2$ choices and $3$ choices yield $2 \\cdot 3 = 6$ combinations using the prime factors, and these are all the factors of $75$: $$3^0 \\cdot 5^0, 3^0 \\cdot 5^1, 3^0 \\cdot 5^2, 3^1 \\cdot 5^0, 3^1 \\cdot 5^1, 3^1 \\cdot 5^2$$ More generally, a number with prime factorization $p^a \\cdot q^b$ for distinct primes $p$ and $q$ has $(a+1)(b+1)$ factors, since you can form these factors by making one of $a+1$ choices for how many $p$s to include (ranging from $0$ to all $a$) and one of $b+1$ choices for how many $q$s to include (ranging from $0$ to all $b$). Even more generally, you arrive", "way we could argue that some logarithm values are irrational... Most will believe from their experience that prime factorizations of positive integers are unique (i.e., that is to say, there is only one way to break down any positive integer into a product of primes, up to the order of the factors in question). For example, $120 = 2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 5$. We could reorder the factors given the commutative property of real numbers under multiplication -- but that's not telling us anything new. As such, re-ordering the factors does not result in a new prime factorization of a number. The important part of this is that there will never be a different prime factorization of $120$. When $120$ is broken down into a product of primes, there will always be exactly $3$ twos, $1$ three, and $1$ five. A similar statement (with different numbers of course) can be made for every positive integer. ‡ : As with most claims in mathematics, we don't have to just rely on our experiences and previous observations to tell us this -- we can prove that prime factorizations are unique. We won't do that here as it takes us too far afield from our content of interest -- but interested students might look up the celebrated", "3 \\cdot 31$ • $187 = 11 \\cdot 17$ • $188 = {2^2} \\cdot 47$ • $189 = {3^3} \\cdot 7$ • $190 = 2 \\cdot 5 \\cdot 19$ • $191$ is prime. • $192 = {2^6} \\cdot 3$ • $193$ is prime. • $194 = 2 \\cdot 97$ • $195 = 3 \\cdot 5 \\cdot 13$ • $196 = {2^2} \\cdot {7^2}$ • $197$ is prime. • $198 = 2 \\cdot {3^2} \\cdot 11$ • $199$ is prime. • $200 = {2^3} \\cdot {5^2}$ You might also be interested in: Fundamental Theorem of Arithmetic Prime Factorization of an Integer List of Prime Factorizations of Integers from 201 to 400 List of Prime Factorizations of Integers from 401 to 600 List of Prime Factorizations of Integers from 601 to 800 List of Prime Factorizations of Integers from 801 to 1,000", "# 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: $2 \\times 2 \\times 2 \\times 3 \\times 5 \\times 11 \\times 31 = 40 , 920$ $120 \\times 11 \\times 31 = 40 , 920$ $124 \\times 2 \\times 3 \\times 5 \\times 11 = 40 , 920$ $165 \\times 2 \\times 2 \\times 2 \\times 31 = 40 , 920$" ]

[ "nonzero integers $$a$$ and $$b$$ is the smallest positive integer among all their linear combinations. In other words, $$\\gcd(a,b)$$ is the smallest positive element in the set $$\\{as+bt \\mid s,t\\in\\mathbb{Z}\\}$$. Corollary $$\\PageIndex{3}$$ For any nonzero integers $$a$$ and $$b$$, there exist integers $$s$$ and $$t$$ such that $$\\gcd(a,b)=as+bt$$. Proof Theorem 5.5.1 maintains that the set of all the linear combinations of $$a$$ and $$b$$ equals to the set of all the multiples of $$\\gcd(a,b)$$. Since $$\\gcd(a,b)$$ is a multiple of itself, it must equal to one of those linear combinations. Thus, $$\\gcd(a,b) = sa+tb$$ for some integers $$s$$ and $$t$$. Theorem $$\\PageIndex{4}$$ Two nonzero integers $$a$$ and $$b$$ are relatively prime if and only if $$as+bt=1$$ for some integers $$s$$ and $$t$$. Proof The result is a direct consequence of the definition that $$a$$ and $$b$$ are said to be relatively prime if $$\\gcd(a,b)=1$$. Example $$\\PageIndex{1}\\label{eg:moregcd-01}$$ It is clear that 5 and 7 are relatively prime, so are 14 and 27. Find the linear combination of these two pairs of numbers that equals to 1. Solution By inspection, or using the extended Euclidean algorithm, we find $$3\\cdot5-2\\cdot7=1$$, and $$2\\cdot14-1\\cdot27=1$$. hands-on Exercise $$\\PageIndex{1}\\label{he:moregcd-01}$$ Show that $$\\gcd(133,143)=1$$ by finding an appropriate linear combination. hands-on Exercise $$\\PageIndex{2}\\label{he:moregcd-02}$$ Show that 757 and 1215 are relatively prime by finding an appropriate linear combination.", "$17 \\cdot 152 - 21 \\cdot 123 = 1$. ### Compute the GCD of 7 and 0 Determine whether 7 and 0 are relatively prime. More generally, compute their greatest common divisor. Note that 7 divides 7 since $1 \\cdot 7 = 7$ and 7 divides 0 since $0 \\cdot 7 = 0$. Now suppose $k$ is an integer dividing both 7 and 0; certainly $k$ divides 7. Thus 7 is a greatest common divisor of 7 and 0. In particular, since $7 \\neq \\pm 1$, 7 and 0 are not relatively prime.", "$21$ relatively prime? Solution Since $35 = 7 \\cdot 5$ and $21 = 7 \\cdot 3,$ we see that $7$ is a common divisor of $35$ and $21.$ It can be checked that no positive integer larger than $7$ that divides both $35$ and $21.$ Thus, we see that $\\gcd(35,21) = 7.$Since $\\gcd(35,21) \\neq 1,$ we see that $35$ and $21$ are not relatively prime. Additionally, it is not hard to see that $\\gcd(5,3) = 1,$ and thus $\\frac{35}{7}$ and $\\frac{21}{7}$ are relatively prime. In general, we have the following theorem. Theorem Let $a$ and $b$ be positive integers. Suppose that $d$ is a positive integer such that $d \\mid a$ and $d \\mid b.$ Then, $d = \\gcd(a,b)$ if and only if $\\frac{a}{d}$ and $\\frac{b}{d}$ are relatively prime. Let $a$ and $b$ be positive integers. The smallest positive integer $m$ such that $a$ divides $m$ and $b$ divides $m$ is referred to as the least common multiple of $a$ and $b.$ The least common multiple of $a$ and $b$ is denoted by $\\operatorname{lcm}(a,b).$ Example Find the least common multiple of $35$ and $21.$ Solution Since $105 = 35 \\cdot 3$ and $105 = 21 \\cdot 5,$ we see that $105$ is a common multiple of $35$ and $21.$ It can be checked that no positive integer smaller", "+0 # Math Help +2 264 3 +335 What is the smallest positive integer n for which 9n-2 and 7n+3 share a common factor greater than 1? Nov 21, 2018 #1 +2 a=listfor(n, 1, 23, GCD(( 9*n-2), (7*n+3))=(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 41)=n=23 and the GCD of [164, 205] =41 Nov 21, 2018 #2 +701 0 You can put that into the \"LaTeX\" box to make the GCD part look nicer. PartialMathematician Nov 21, 2018 #3 +22358 +11 What is the smallest positive integer n for which 9n-2 and 7n+3 share a common factor greater than 1? $$\\text{Euclidean algorithm:}~ \\boxed{gcd(a-nb,b)=gcd(a,b)}$$ $$\\begin{array}{|rcll|} \\hline && gcd(9n-2,7n+3) \\\\ &=& gcd(9n-2-1\\cdot(7n+3) ,7n+3)=gcd(2n-5,7n+3) \\\\ &=& gcd(2n-5,7n+3-3\\cdot(2n-5))=gcd(2n-5,n+18) \\\\ &=& gcd(2n-5-2\\cdot(n+18),n+18)=gcd(-41,n+18) \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline gcd(-41,\\underbrace{n+18}_{=41~(\\Rightarrow n>0!)}) \\qquad -41~ \\text{is prime} \\\\\\\\ n+18 &=& 41 \\\\ n &=& 41-18 \\\\ n &=& 23 \\\\ \\hline \\end{array}$$ The common factor greater than 1 is $$\\mathbf{41}$$. and the smallest positive integer n is $$\\mathbf{23}$$ Hint: $$gcd(-41,41) = 41$$ Nov 21, 2018 edited by heureka Nov 21, 2018 edited by heureka Nov 21, 2018", "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 $15120$.", "# 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 $$4$$.", "# 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.", "= 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$\\gcd(a^2, b^2) = 1\\$. -", "# Find the smallest positive integer $x$ satisfying $\\gcd(x^n+a,(x+1)^n+a)>1$ Given positive integers $$n$$ and $$a$$, I'd like to ask how to find the smallest positive integer $$x$$ satisfying $$\\gcd(x^n+a,(x+1)^n+a)>1$$? I try using the extended Euclidean algorithm on the two polynomials to find the greatest common divisor $$\\frac{p}{q}$$. If $$p$$ is small, then we just need to check the solutions to $$x^n+a \\equiv 0 \\ (mod \\ d)$$ where $$d$$ divides $$p$$. Otherwise, it seems much harder. I also notice that $$gcd(n,\\phi(d))$$ should be greater than $$1$$ in order to have a solution, which helps filter many unfavorable $$d$$. • Did you encounter an example that seems computationally difficult? Jan 2, 2019 at 18:48 • For example, $n=19$ and $a=2$. Both $p$ and $q$ are extremely large and cannot be factored easily. I believe the case is quite common for large $n$. Jan 3, 2019 at 4:13 • The procedure I've suggested below gives the resultant as $$1103 \\cdot 87211 \\cdot 31308253657 \\cdot 818790224665362763936013994101920313,$$ and trying $p=1103, 87211, \\ldots$ we find $x\\bmod p=473, 836, \\ldots$ with $473$ the solution. Jan 3, 2019 at 7:31 • Funny - for $n=19$ and $a=6$ the resultant appears to be prime, thus the smallest (!!!) solution is $$x=1578270389554680057141787800241971645032008710129107338825798$$ Jan 3, 2019 at 7:48 I don't quite understand what does $$\\gcd(y^n+a,(y+1)^n+a)$$ give you if", "prime power factors of $$\\gcd(a,b)\\text{.}$$ We conclude that \\begin{equation*} \\gcd(a,b)=3^2\\cdot 7^4 \\cdot 11\\text{.} \\end{equation*} Let integers $$a$$ and $$b$$ be given in factored form: \\begin{equation*} a=2^{3} \\cdot 3^2 \\cdot 7^5 \\cdot 19^2 \\text{ and } b= 2^{5} \\cdot 7^{4} \\cdot 19^{5} \\cdot 23 \\end{equation*} So $$gcd(a,b) = 2^3 \\cdot 7^4 \\cdot 19^2\\text{.}$$ In Checkpoint 10.3.7 find the greatest common divisor of two integers whose prime factorization is given. What are the greatest common divisors of the following pairs of integers ? If $$a= 2^3 \\cdot 3^3 \\cdot 5$$ and $$b= 2^4 \\cdot 3^4 \\cdot 5$$ then $$\\gcd(a,b)=$$ If $$a=2^{12} \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13$$ and $$b= 3^{12} \\cdot 5 \\cdot 11^{9} \\cdot 17$$ then $$\\gcd(a,b)=$$ If $$a= 2^3 \\cdot 7$$ and $$b= 5^3 \\cdot 13$$ then $$\\gcd(a,b)=$$ $$1080$$ $$165$$ $$1$$ ### Subsection10.3.2A primality test There are many possible paths to finding the prime factorization of a number. Because the Fundamental Theorem of Arithmetic (Theorem 10.3.1) guarantees the uniqueness of the prime factorization, the order in which we find the factors does not matter. Based on the solution given for Problem 10.3.4, it is apparent that finding the prime factorization of composite numbers can involve many steps. Generally speaking, the bigger the composite number is, the harder it is to find its prime factorization.", "of the numerator $7n+1$ by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that $\\frac{21n+4}{14n+3}$ is irreducible. Q.E.D Solution 5 By Bezout's Lemma, $3 \\cdot (14n+3) - 2 \\cdot (21n + 4) = 1$, so the GCD of the numerator and denominator is $1$ and the fraction is irreducible.", "+0 # help and explain :) 0 74 1 Let x be the smallest positive integer such that 1584 * x is a perfect cube, and let y be the smallest positive integer such that xy is a multiple of 1584. Compute y Jul 6, 2021 #1 +234 0 First let's take the prime factorization of 1584, $$2^4\\cdot 3^2\\cdot11 = 1584$$ To make $$1584\\cdot x$$, each of the exponents in the prime factoriziation of the cube need to be a multiple of 3. To make that, the prime factoriziation of x needs to create that by having the correct exponents of, $$2^2 \\cdot 3 \\cdot 11^2 = 1452$$ Making the cube's prime factorization, $$2^6 \\cdot 3^3 \\cdot 11^3 = 2299968$$ $$\\sqrt[3]{2299968} = 132$$ So our cube is actually a cube. Now that we know that x = 1452, we can use LCM to find a common multiple of 1452 and 1584 to find y. The LCM of 1452 and 1584 is 17424. $$\\frac{17424}{1452} = 12$$ So y = 12 Sorry if my method was kind of confusing lol Jul 6, 2021", "18) = 6$ (2.) Calculate the $gcd$ of $30$ and $105$ Divisors of $30 = 1, 2, 3, 5, 6, 10, 15, 30$ Divisors of $105 = 1, 3, 5, 7, 15, 21, 35, 105$ Common Divisors of $30 \\:and\\: 105 = 1, 3, 5, 15$ Greatest Common Divisor of $30 \\:and\\: 105 = 15$ $gcd(30, 105) = 15$ Nigerian Method In this method; we set up a table of the positive integers. We set it up the same way we set up the dividend and divisor in the Long Division method of dividing polynomials. We \"try\" each factor in ascending order (from least to greatest). The factor should be a positive integer that has to divide \"all\" the positive integers without a remainder. But, if we find another factor of all the integers (a number that divides all the integers \"right away\" without a remainder), use it. In other words, it has to be a common factor. If we cannot find any factor, then the $GCF = 1$. Remember that $1$ is the smallest factor of every positive integer. If we find a factor of all the integers (besides $1$), use it. Begin in ascending order. Multiply all the factors you listed. The product gives you the $GCF$. Let us see how this works! (1.) Calculate the", "A quick strategy is to find the smallest such integer $m$ such that $\\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \\cdot 11.$ Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \\cdot 11^2$ gives us 1089 as $n$ and $m = \\sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$, which we can also write as $(x-y)(x+y) = 11^2 \\cdot 3^2$. A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$, so $x+y+m$ $=$ $\\boxed{\\textbf{(E) }154}$. Note that all of the solutions used $a+b$ or $a-b$ as part of their solution.", "## 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", "# 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 $G C F \\left({x}^{2} , {x}^{3}\\right) = {x}^{2}$.", "by determining the prime factorizations of the two numbers and comparing factors. For example, to compute gcd(48, 180), we find the prime factorizations 48 = 24 · 31 and 180 = 22 · 32 · 51; the GCD is then 2min(4,2) · 3min(1,2) · 5min(0,1) = 22 · 31 · 50 = 12, as shown in the Venn diagram. The corresponding LCM is then 2max(4,2) · 3max(1,2) · 5max(0,1) = 24 · 32 · 51 = 720. [12] In practice, this method is only feasible for small numbers, as computing prime factorizations takes too long. ### Euclid's algorithm See main article: Euclidean algorithm. The method introduced by Euclid for computing greatest common divisors is based on the fact that, given two positive integers and such that, the common divisors of and are the same as the common divisors of and . So, Euclid's method for computing the greatest common divisor of two positive integers consists of replacing the larger number by the difference of the numbers, and repeating this until the two numbers are equal: that is their greatest common divisor. For example, to compute, one proceeds as follows: \\begin{align}\\gcd(48,18)&\\to\\gcd(48-18,18)=\\gcd(30,18)&&\\to\\gcd(30-18,18)=\\gcd(12,18)\\\\ &\\to\\gcd(12,18-12)=\\gcd(12,6)&&\\to\\gcd(12-6,6)=\\gcd(6,6).\\end{align} So . This method can be very slow if one number is much larger than the other. So, the variant that follows is generally preferred. ### Euclidean", "is a formula, but it isn't anything direct: Given two numbers x and y, we know that GCF(x, y) = (x*y)/LCM(x, y), where LCM(x, y) is the \"Least Common Multiple\" of x and y. There is no direct formula I know of to find the LCM either. In case you were wondering there is a simple process to find the LCM of two numbers as well. Let's take 60 and 630 again. The prime factorization of 60 is $\\displaystyle 2^2 \\cdot 3 \\cdot 5$. The prime factorization of 630 is $\\displaystyle 2 \\cdot 3^2 \\cdot 5 \\cdot 7$. The Least Common Multiple will have a prime factorization that is the smallest list of prime factors that completely contains both lists. In other words: There are two factors of 2, there are two factors of 3, There is one factor of 5, There is one factor of 7. Thus LCM(60, 630) = $\\displaystyle 2^2 \\cdot 3^2 \\cdot 5 \\cdot 7$ = 1260. (And we can check the equation I gave in the Post Script of the last post: GCF(60, 630) = (60*630)/LCM(60, 630) = 37800/1260 = 30 as we found in the last post.) -Dan 4. Originally Posted by topsquark ... The Least Common Multiple will have a prime factorization that is the smallest list of prime factors that", "# Huge Primes $97!\\; n + d$ Find the smallest positive integer $$d$$ greater than $$1$$, such that the number above is prime for an infinite number of positive integers $$n$$. Notation: $$!$$ denotes the factorial notation. For example, $$10! = 1\\times2\\times3\\times\\cdots\\times10$$. ×", "and the numbers greater than this number will not be prime numbers. then g((2^11)+1)=2048+1=2049 2049 is divisible by 1,3... 2049 so 3 is the smallest prime factor for g(x) Manager Joined: 26 Mar 2019 Posts: 100 Concentration: Finance, Strategy For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags Updated on: 18 Jul 2019, 08:43 1 Quote: For every positive integer $$x$$, $$f(x)$$ represents the greatest prime factor of $$x!$$, and $$g(x)$$ represents the smallest prime factor of $$2^x+1$$. What is $$(g(f(12))$$? The task tells us that $$f(x)$$ represents greatest prime factor of $$x!$$. Prime factor is a positive integer which is not equal to $$1$$ and has two factors: itself and $$1$$. Let us find $$f(x)$$ first, and then find $$g(f(x))$$. $$f(12)$$ = greatest prime factor of $$12!$$ Let us find out what greatest prime factor of $$12!$$ is: $$12! = 1 * 2* 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12$$. As we can see from above, the greatest of the numbers is $$11$$ which is the greatest prime factor for $$12!$$ $$12!$$ cannot have higher prime factor since $$12$$ itself is equal to $$2*2*3$$, and any product of its numbers will have more factors than $$2$$ and will not" ]

[ "## Prealgebra (7th Edition) Published by Pearson # Chapter 7 - Review: 2 77% #### Work Step by Step Since the basketball player made 77 out of 100 free throw attempts, we can find the percent of free throws he successfully made by doing the calculation: $(77\\div100)\\times100$=% of free throws the basketball player made. 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.", "Free Version Easy # Basketball Free Throws: One and One APSTAT-FLTG1I In basketball, one-and-one free throws sometimes occur near the end of a game. A player gets one attempt at a free throw: if she makes it, she gets one other attempt; if she misses, she gets no further attempts. Suppose an athlete has a $64\\%$ free-throw shooting percentage and is shooting a one-and-one situation. What is the probability she makes at least one shot? A $0.64$ B $0.36$ C $0.2304$ D $0.4096$ E $0.5904$", "# basketball player Printable View • Aug 10th 2006, 10:38 AM kwtolley basketball player The probability a basketball player makes a free throw is 0.60. Find the probability the player makes four out of 6. P( 4 throws in 6 attempts) (0.60)(0.60)(0.60)(0.60)(0.60)(0.4)=0.311 0.311 is my answer, thanks so much for checking my answer. List of possible aswers are - 1. 0.0207 2. 0.0667 3. 0.240 4. 0.311 • Aug 10th 2006, 12:44 PM ThePerfectHacker It is, $_6C_4 (.6)^4(.4)^2\\approx .311$ • Aug 10th 2006, 09:07 PM kwtolley thanks thanks you for all your help and everyone else also.", "Point Curry Disco Visual Search Engine! Is your friend full of it? Posted on Sat 02 September 2017 in misc Five Thirty Eight Riddler Express Problem - Sep 1, 2017¶ You’re hanging out with some friends, shooting the breeze and talking sports. One of them brags to the group that he once made 17 free throws in a row after years of not having touched a basketball. You think the claim sounds unlikely, but plausible. Another friend scoffs, thinking it completely impossible. Let’s give your bragging friend the benefit of the doubt and say he’s a 70-percent free-throw shooter. So, who’s right? What is the number of free throws that a 70-percent shooter would be expected to take before having a streak of 17 makes in a row? And what if his accuracy was a bit worse? This looks like a pretty innocuous Bernoulli's trials problem but it isn't! Had the question been about the average number of throws taken for 17 successes, the answer would have been straightforward - $$0.7\\sum_{n=16}^{\\infty} n\\binom {n} {16}0.7^{16}0.3^{n-16} = 17/0.7 \\approx 24.3$$ But the question we are dealing with is about the average number of throws that are required for 17 consecutive makes in a row. I have a vague recollection of this problem being dealt with in Feller's Volume 1 and", "Shaquille 'O' Neal is a $60$% career free throw shooter, meaning that he successfully makes $60$ free throws out of $100$ attempts on average. What is the probability that he will successfully make exactly $6$ free throws in $10$ attempts? 1. $0.2508$ 2. $0.2816$ 3. $0.2934$ 4. $0.6000$", "throws did James make during the game. James scored a total of 15 points during a basketball game. During the game he made only free throws worth 1 point each and 3 point baskets. He made a total of he made a total of 7 baskets. How... more 03/16/18 #### solve the simultaneous equation 3a + 2b =7 a -2b =5 03/14/18 #### The height of a golf ball in the air can be modeled by the equation h=negative16t squared plus 76t, where h is the height in feet of the ball after t seconds a. How lone was the ball in the air b. What is the ball's maximum height? c. When will the balol reach its maximum height? 02/28/18 #### Which answer shows this equation written in standard form? 9 – 4(x – y) = –6x + 6 A. –10x + 4y = –3 B. 2x + 4y = 3 C. 2x – 4y = 3 D. 2x + 4y = –3 02/27/18 #### The sum of two numbers is 36. Their difference is 24. Calculate the numbers Just a little confused on this math question need help on how to do it with steps explaining 02/26/18 #### The first 100 counting numbers are arrangedin columns as shown. What would be the sum of all numbers", "# Math Help - basketball player The probability a basketball player makes a free throw is 0.60. Find the probability the player makes four out of 6. P( 4 throws in 6 attempts) (0.60)(0.60)(0.60)(0.60)(0.60)(0.4)=0.311 0.311 is my answer, thanks so much for checking my answer. List of possible aswers are - 1. 0.0207 2. 0.0667 3. 0.240 4. 0.311 2. It is, $_6C_4 (.6)^4(.4)^2\\approx .311$ 3. ## thanks thanks you for all your help and everyone else also.", "# If a basketball player made 72 out of 85 free-throw attempts, how many could she expect to make in 200 attempts? May 26, 2018 #### Answer: You would expect her to make at least $169$ of the $200$ free throws. #### Explanation: Use a proportion: $\\frac{72}{85} = \\frac{x}{200}$ Solve for $x$ $\\left(72\\right) \\left(200\\right) = 85 x$ $x = \\frac{\\left(72\\right) \\left(200\\right)}{85}$ $x = \\left\\lfloor 169.41 \\right\\rfloor = 169$ So you would expect her to make at least $169$ of the $200$ free throws.", "Video, of shooting a basketball 10 times.", "Free Version Easy # Calculator: Free-Throw Outcomes APSTAT-YPTLAE Suppose a basketball player shoots $10$ free throws and makes $8$ of them. How many different ways could this happen? A $10$ B $8$ C $1$ D $45$ E $3,628,800$", "## math At the end of an basketball game daniels couch told him he scored 20% of all of his teams points. If Daniel scored 19 points exactly how many points did his team score? asked by Justin on April 7, 2011 13. ## math If a basketball player consistently shoots 80% from the free throw line and she takes 25 free throw shots throughout the course of a game, how many free throws can she be expected to make? asked by joyce on April 27, 2013 an athlete drinks a liter of water before a basketball game and continues to drink water during the game,describe how the athletes body is able to maintain homeostasis during the course of the game? asked by scooby on October 23, 2011 15. ## math In a basketball free-throw contest held at the high school, Rudy made 4 out 5 free throws. At this rate, how many free throws could he expect to make in 25 tries? asked by raymond on January 25, 2012 16. ## ALGEBRA If a pro basketball player has a vertical leap of about 35 inches what is his hang-time? Use the hang-time function V=48T^2. Thank you I did mess up the typing last time didn't I? asked by aimee on May 4, 2009 I really need help", "a basket that counts for a point with the exception of a free throw.", "adults. ### Eureka Math Precalculus Module 5 Lesson 10 Problem Set Answer Key Question 1. A high school basketball player makes 70% of the free throws she attempts. Suppose she attempts seven free throws during a game. The probability distribution for the number of free throws made out of seven attempts is displayed below. a. What is the probability that she completes at least three free throws? Interpret this probability in context. The probability that she completes at least three free throws is as follows: 0.09725 + 0.22689 + 0.31765 + 0.24706 + 0.08235 = 0.9712. If this basketball player attempts seven free throws many times, the long-run proportion of the time that she will complete at least three free throws is 0.972. b. What is the probability that she completes more than two but less than six free throws? Interpret this probability in context. The probability that she completes more than two but less than six free throws is as follows: 0.09725 + 0.22689 + 0.31765 = 0.64179. If this basketball player attempts seven free throws during a game for many games, the long-run proportion of the time that she will complete more than two but less than six free throws is 0.64179. c. How many free throws will she complete on average? Interpret this expected value", "the breeze and talking sports. One of them brags to the group that he once made 17 free throws in a row after years of not having touched a basketball. You think the claim sounds unlikely, but plausible. Another friend scoffs, thinking it completely impossible. Let’s give your bragging friend the benefit of the doubt and say he’s a 70 percent free-throw shooter. Who’s right? What is the number of free throws that a 70 percent shooter would be expected to take before having a streak of 17 makes in a row? The number is about 1,430 free throws. The odds of a 70 percent shooter hitting 17 shots in a row, starting with any given single shot, is $0.7^{17}\\approx 0.002$. So we know, on average, it’ll take 1 divided by that, or about 430, tries to get the streak. But that number of tries isn’t the number of shots. It’s the number of streaks he begins by taking a shot after missing one. Those tries could fall short after two makes, or 10, or 16. Here’s one way to get to the answer: You know your friend will have to take the first shot on his road to 17 430 times. He’ll take the second shot 0.7*430 = 301 times. The third 0.7*301 = 210.7 times. The", "time from i losses in a row to 6 losses in a row. I'm kind of surprised it's this high, actually. The probably that six throws are all losses is about 1/47, so I thought it would be around 47.", "whole! Indeed, take the average of the scores you got and you will see that you got very close to 6. Then, if you're good with programming. Do the above experiment but only record the number of throws until you got 5 consecutive sixes. Then do this experiment 1000 times with the computer. Look at the different outcomes, they will look very different, but the average should lie around 9330.", "O’ Neal is a 60% career free throw shooter, meaning that he successfully makes 60 free throws out of 100 attempts on average. What is the probability that he will successfully make exactly 6 free throws in 10 attempts? The numeral in the units position of 211870 + 146127 × 3424 is _____. Based on the given statements, select the most appropriate option to solve the given question. If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building? Statements: (I) Each step is ¾ foot high. (II) Each step is 1 foot wide. Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is probability that the sum of the two numbers equals 16? The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5:4. There are 40 males in Electrical Engineering. What is the difference between numbers of female students in the Civil department and the female students in the Mechanical department?", "Limited access Jimmy, a high school basketball star, has a life-long free throw rate of $45\\%$. Jimmy spent the last off-season working especially hard to improve his free-throw percentage. During the first game of the season, he connected on three of his first four free throws. What conclusions can be made concerning Jimmy’s free-throw shooting ability? A Since $3$ of $4 = 75\\%$, it is clear that Jimmy has improved his free-throw percentage. B Since the sample size is so low, we have strong evidence Jimmy is still a $45\\%$-free-throw shooter. C No reasonable conclusions can be drawn since the sample size is so low. D Free throws can hardly be treated as a simple random sample; therefore, this data cannot be trusted. E Jimmy is clearly not a $45\\%$-free-throw shooter based off the evidence. Select an assignment template", "5 or more times before shooting were successful 60%, 66%, 68%, and 68% of the time, respectively, while those who dribbled 4 times were successful 77% of the time. Also, there were more NBA all-stars in the 4-dribble group than in any other group. Submitted by Margaret Cibes ### Discussion 1. Buried in the article: The total sample size of free throws attempted was 425, with 60, 121, 191, 22 and 9 representing 1, 2, 3, 4 and 7 dribbles for a total of 403. No information is given for 5 and 6 dribbles but from the data, the inference is that they represent 22 (425-403) attempts. Given this data, how impressive does the 77% success rate for four dribbles appear? 2. Why didn’t the author choose to present a simple table of number of dribbles, free-throw attempts, free-throws made instead of dribbling the information all over the article? Is any insight provided by the following graphic from the article? 3. The headline in the print edition, but not in the online edition is, \"At the Foul Line, Four is the Magic Number.\" Comment on the \"magicness\" of four given the actual data. Submitted by Paul Alper ## Bird brains vs. birdbrains “Mathematicians vs. Birds vs. Monty Hall” by Tom Bartlett, The Chronicle of Higher Education, March 17,", "free throw shot” for a basketball player shooting two free throws after a foul. By looking at the outcomes of the two shots for the player throughout the year, we can estimate that $$P(\\text{make the first shot}) = 0.72$$, $$P(\\text{make the second shot}) = 0.72$$ and $$P(\\text{make the first shot and make the second shot}) = 0.52$$. The events are independent since $$0.72 \\boldcdot 0.72 \\approx 0.52$$." ]

[ "Free Version Easy # Estimating Parameters: Free Throws APSTAT-1XLLFF Suppose an athlete shoots $200$ free throws and makes $160$ of them. Which of the following statements is true? A This athlete is an $80\\%$ free-throw shooter. B A sample of $100$ free throws would be more effective in estimating the athlete’s true free-throw percentage. C The athlete’s true free-throw percentage is probably somewhere near $80\\%$. D The athlete could easily make fewer than $100$ free throws in the next set of $200$ free throw attempts. E No reasonable conclusions can be drawn from this data.", "Free Version Easy # Basketball Free Throws: One and One APSTAT-FLTG1I In basketball, one-and-one free throws sometimes occur near the end of a game. A player gets one attempt at a free throw: if she makes it, she gets one other attempt; if she misses, she gets no further attempts. Suppose an athlete has a $64\\%$ free-throw shooting percentage and is shooting a one-and-one situation. What is the probability she makes at least one shot? A $0.64$ B $0.36$ C $0.2304$ D $0.4096$ E $0.5904$", "? Free Version Moderate # Finding the Median from a Histogram: NBA Payroll APSTAT-BG1VSN This histogram below represents the average payroll of $29$ NBA teams. In which bin is the median located? A $22\\le \\text{ payroll }<28$ B $28\\le \\text{ payroll }<34$ C $34\\le \\text{ payroll }<40$ D $40\\le \\text{ payroll }<46$ E $46\\le \\text{ payroll }<52$", "# Math Help - basketball player The probability a basketball player makes a free throw is 0.60. Find the probability the player makes four out of 6. P( 4 throws in 6 attempts) (0.60)(0.60)(0.60)(0.60)(0.60)(0.4)=0.311 0.311 is my answer, thanks so much for checking my answer. List of possible aswers are - 1. 0.0207 2. 0.0667 3. 0.240 4. 0.311 2. It is, $_6C_4 (.6)^4(.4)^2\\approx .311$ 3. ## thanks thanks you for all your help and everyone else also.", "# basketball player Printable View • Aug 10th 2006, 10:38 AM kwtolley basketball player The probability a basketball player makes a free throw is 0.60. Find the probability the player makes four out of 6. P( 4 throws in 6 attempts) (0.60)(0.60)(0.60)(0.60)(0.60)(0.4)=0.311 0.311 is my answer, thanks so much for checking my answer. List of possible aswers are - 1. 0.0207 2. 0.0667 3. 0.240 4. 0.311 • Aug 10th 2006, 12:44 PM ThePerfectHacker It is, $_6C_4 (.6)^4(.4)^2\\approx .311$ • Aug 10th 2006, 09:07 PM kwtolley thanks thanks you for all your help and everyone else also.", "Free Version Easy # Calculator: Free-Throw Outcomes APSTAT-YPTLAE Suppose a basketball player shoots $10$ free throws and makes $8$ of them. How many different ways could this happen? A $10$ B $8$ C $1$ D $45$ E $3,628,800$", "Free Version Easy # Finding the Median from a Histogram: NBA Payroll Free APSTAT-BG1VSN This histogram below represents the average payroll of $29$ NBA teams. In which bin is the median located? A $22\\le \\text{ payroll }<28$ B $28\\le \\text{ payroll }<34$ C $34\\le \\text{ payroll }<40$ D $40\\le \\text{ payroll }<46$ E $46\\le \\text{ payroll }<52$", "free throw shot” for a basketball player shooting two free throws after a foul. By looking at the outcomes of the two shots for the player throughout the year, we can estimate that $$P(\\text{make the first shot}) = 0.72$$, $$P(\\text{make the second shot}) = 0.72$$ and $$P(\\text{make the first shot and make the second shot}) = 0.52$$. The events are independent since $$0.72 \\boldcdot 0.72 \\approx 0.52$$.", "# If a basketball player made 72 out of 85 free-throw attempts, how many could she expect to make in 200 attempts? May 26, 2018 #### Answer: You would expect her to make at least $169$ of the $200$ free throws. #### Explanation: Use a proportion: $\\frac{72}{85} = \\frac{x}{200}$ Solve for $x$ $\\left(72\\right) \\left(200\\right) = 85 x$ $x = \\frac{\\left(72\\right) \\left(200\\right)}{85}$ $x = \\left\\lfloor 169.41 \\right\\rfloor = 169$ So you would expect her to make at least $169$ of the $200$ free throws.", "The probability of a success is 0,9 $ P(A) = {20 \\choose 19} \\left( 0,9 \\right) ^{19} \\left( 0,1 \\right)^{1} + {20 \\choose 20} \\left( 0,9 \\right) ^{20} \\left( 0,1 \\right)^{0}$ $ P(B) = 1 - {20 \\choose 15} \\left( 0,9 \\right) ^{15} \\left( 0,1 \\right)^{5}$ $- {20 \\choose 16} \\left( 0,9 \\right) ^{16} \\left( 0,1 \\right)^{4}$ $- {20 \\choose 17} \\left( 0,9 \\right) ^{17} \\left( 0,1 \\right)^{3}$ $- {20 \\choose 18} \\left( 0,9 \\right) ^{18} \\left( 0,1 \\right)^{2}$ $- {20 \\choose 19} \\left( 0,9 \\right) ^{19} \\left( 0,1 \\right)^{1}$ $- {20 \\choose 20} \\left( 0,9 \\right) ^{20} \\left( 0,1 \\right)^{0}$ 90% of 20 throws equal 18 throws. $P(C) = {20 \\choose 18} \\left( 0,9 \\right) ^{18} \\left( 0,1 \\right)^{2}$ 3. Originally Posted by 16sammy I hate math assignments and it's due tomorrow and the teacher is stupid and won;t help people. I was away when it was explained so i have no idea what to do!! Suppose an NBA player makes 90% of his free throws. This player takes 20 free throws in the course of a game. Find the probabilits of his making: A) Atleast 19 of the 20 free throws. B) Less than 15 of the 20 free throws C) Exactly 90% od the 20 free throws. 1. So how long have you had to", "+0 0 88 2 +406 1. A basketball player made 12 out of 15 free throws she attempted. She wants to know have many consecutive free throws she would have to make to raise the percent of successful free throws to 85%. (a) Write an equation to represent this situation. (b) Solve the equation. How many consecutive free throws would she have to make to raise her percent to 85%? 2. Simplify the expression x^2+3x-28/x^2 -7x+12 1. Write the expression as a simplified rational expression. how do i solve this can you please include steps Oct 31, 2018 #1 +3617 +2 $$\\text{suppose she has to make }n \\text{ consecutive throws}\\\\ \\dfrac{12+n}{15+n} = 0.85 \\\\ 12+n = 12.75 + 0.85n\\\\ 0.15n =0.75 \\\\ n=5$$ 1) $$\\dfrac{5}{\\frac 1 6 + \\frac{1}{x+1}} =\\\\ \\dfrac{5}{\\frac{x+1+6}{6(x+1)}} =\\\\ \\dfrac{5\\cdot 6(x+1)}{x+7} = \\\\ \\dfrac{30(x+1)}{x+7}$$ . Oct 31, 2018 #2 +94545 +1 Here's the second one : x^2 + 3x - 28 ____________ factor top/bottom x^2 -7x + 12 ( x + 7) ( x - 4) ____________ \"cancel\" the ( x - 4) factor on top/bottom (x - 3) (x - 4) ( x + 7) _______ ( with x not equal to 4) ( x - 3) Oct 31, 2018", "Shaquille 'O' Neal is a $60$% career free throw shooter, meaning that he successfully makes $60$ free throws out of $100$ attempts on average. What is the probability that he will successfully make exactly $6$ free throws in $10$ attempts? 1. $0.2508$ 2. $0.2816$ 3. $0.2934$ 4. $0.6000$", "# What are the mean, median, and mode salaries of a baseball team where the 24 players earn the following annual salaries: 2 players at $14.6 million; 4 at$7.45 million; 4 at 3.93 million; 5 at $450,000; 2 at$390,000 and 7 at $200,000? ##### 1 Answer Oct 20, 2015 Mean, bar x=1/nsum_(I=1)^nx_i=$3297917 Median = $450000. Mode =$200000 #### Explanation: Start by arranging all the data points in increasing order to obtain : 200000 200000 200000 200000 200000 200000 200000 390000 390000 450000 450000 450000 450000 450000 3930000 3930000 3930000 3930000 7450000 7450000 7450000 7450000 14600000 14600000 Define $X = \\left\\{{x}_{i}\\right\\}$ to be the set of these data points in this order. Now by definition, the mean or average of these values is given by $\\overline{x} = \\frac{1}{n} {\\sum}_{I = 1}^{n} {x}_{i} = 3297917$ The median is defined to be the centre data point or, in this case since the number of data points is an even number, the mean of the 2 centre points, ie $\\frac{450000 + 450000}{2} = 450000$ The mode is defined to be the data point which occurs most often, so in this case it is 200000 which occurs 7 times in the sample space.", "The professor asked us to imagine a scenario where we have a basketball player who isn't good at shooting free throws. He makes his first free throw with probability $0.2$ After the first free throw, he makes a free throw with probability $0.6$ if he made the preceding one, and he makes a free throw with probability $0.3$ if he missed the preceding one. During practice, he throws $10$ free throws. What is the total number of outcomes? Let $A_i$ be the event that he makes the $i$th free throw. Find $P(A_i)$ for $i=2, 3, 4$. I figured out the total number of possible outcomes using binomial coefficients ${n \\choose r} = \\frac{n!}{(n-r)!r!}$. The number of possible ways outcomes are as follows: \\begin{eqnarray} \\text{Possible ways he will make 10 free throws} & = & {10 \\choose 10} = 1 \\\\ \\text{Possible ways he will make 9 free throws} & = & {10 \\choose 9} = 10 \\\\ \\text{Possible ways he will make 8 free throws} & = & {10 \\choose 8} = 45 \\\\ \\text{Possible ways he will make 7 free throws} & = & {10 \\choose 7} = 120 \\\\ \\text{Possible ways he will make 6 free throws} & = & {10 \\choose 6} = 210 \\\\ \\text{Possible ways he will make 5 free throws} & =", "## Prealgebra (7th Edition) Published by Pearson # Chapter 7 - Review: 2 77% #### Work Step by Step Since the basketball player made 77 out of 100 free throw attempts, we can find the percent of free throws he successfully made by doing the calculation: $(77\\div100)\\times100$=% of free throws the basketball player made. 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.", "# The average number of free throws made during a basketball game varies directly with the number of hours of practice during a week. When a player practices 6 hours a week, she averages 9 free throws a game. How do you write an equation relating the hours? ## *Write an equation relating the hours of practice and free throws made** Jun 13, 2018 $f = 1.5 h$ #### Explanation: $\\text{let f represent free throws and h hours practiced}$ $\\text{the statement is } f \\propto h$ $\\text{to convert to an equation multiply by k the constant}$ $\\text{of variation}$ $f = k h$ $\\text{to find k use the given condition}$ $h = 6 \\text{ and } f = 9$ $f = k h \\Rightarrow k = \\frac{f}{h} = \\frac{9}{6} = \\frac{3}{2} = 1.5$ $\\text{equation is } \\textcolor{red}{\\overline{\\underline{| \\textcolor{w h i t e}{\\frac{2}{2}} \\textcolor{b l a c k}{f = 1.5 h} \\textcolor{w h i t e}{\\frac{2}{2}} |}}}$", "### Home > INT1 > Chapter 10 > Lesson 10.1.2 > Problem10-29 10-29. The Toronto Hawks football team is having a meeting with fans to discuss salaries. The team president says, “We cannot afford to be any more generous because our average salary is almost $$2\\text{,}000\\text{,}000$ a year.” A fan replies, “Nonsense. Your average player only makes$$850\\text{,}000$ a year—you can easily afford a raise.” If they are using the same set of data, how can their “averages” be different? How could such a large difference have occurred? Here's a sample data set: $3, 4, 6, 8, 10, 11, 11, 13, 15$ What's the mean? The median? Are they close together? Mean: $9$ Median: $10$ Here's another sample data set: $12, 13, 15, 17, 18, 20, 22, 600, 750$ What's the mean? The median? Are they close together? Mean: $163$ Median: $18$ What do you think causes the mean to be so far from the median in the second data set? Are there numbers that are increasing the mean but not the median? How does that apply to the situation above? Data sets with numbers that are far larger or smaller than the other numbers in the set tend to have means and medians that are far apart. In the case of the football team, super star players earn", "# How do you solve this? ## In Gym class, Sue is shooting basketball free throws. She is shooting three sets of twenty shots each. In her first two attempts, she made 13 and 18 free throws. What are the possible scores on her third attempt if she is to have an average between 12 and 16 free throws after all three attempts? Express your answer in the form of a compound inequality using the variable F to represent a successful basketball free throw. Jan 15, 2017 $5 \\le F \\le 17$ #### Explanation: If ${s}_{1} , {s}_{2} , {s}_{3}$ her scores (amount of free throws made) on each set respectively, her average score is: $a = \\frac{{s}_{1} + {s}_{2} + {s}_{3}}{3}$. The average needs to be between $12$ and $16$, so $12 \\le a \\le 16$ Multiplying by $3$ to get rid of the denominator: $12 \\le \\frac{{s}_{1} + {s}_{2} + {s}_{3}}{3} \\le 16 \\implies 36 \\le {s}_{1} + {s}_{2} + {s}_{3} \\le 48$. We know that ${s}_{1} = 13$ and ${s}_{2} = 18$, so $36 \\le 31 + {s}_{3} \\le 48$. ${s}_{3}$ is actually what the problem wants us to use as $F$, so let's do that. Subtracting $31$ from the inequality gives the solution: $5 \\le F \\le 17$. That means the set of all", "+ a_2 X_ 2.$$ State how you created the dataset, and show that it conforms to the required property. Part 2(b): Challenge (10 pts) Now, come up with a plausible scenario where $$X_2$$ is a lurking variable. That is, in your report, come up with a story about the data $$(X_1, X_2, Y)$$. Now, re-generate the data in a way that makes clear that the data could be generated if the scenario actually happened. Include your code, an explanation, and a demonstration of the effect of $$X_2$$ as a lurking variable. Part 3: Free Throws Free throws are an important part of basketball. Here, we will be using a dataset of the number of free throws that Shaquille O’Neal , one of the greatest basketball players in the last few decades who was also known for his poor free-throwing ability , has taken in 23 different games. The total number of attempted free throws is recorded, as well as the number of times that the attempts were successful. lines <- \"Game Scored N.Attempts 1 4 5 2 5 11 3 5 14 4 5 12 5 2 7 6 7 10 7 6 14 8 9 15 9 4 12 10 1 4 11 13 27 12 5 17 13 6 12 14 9 9 15 7 12", "Limited access Jimmy, a high school basketball star, has a life-long free throw rate of $45\\%$. Jimmy spent the last off-season working especially hard to improve his free-throw percentage. During the first game of the season, he connected on three of his first four free throws. What conclusions can be made concerning Jimmy’s free-throw shooting ability? A Since $3$ of $4 = 75\\%$, it is clear that Jimmy has improved his free-throw percentage. B Since the sample size is so low, we have strong evidence Jimmy is still a $45\\%$-free-throw shooter. C No reasonable conclusions can be drawn since the sample size is so low. D Free throws can hardly be treated as a simple random sample; therefore, this data cannot be trusted. E Jimmy is clearly not a $45\\%$-free-throw shooter based off the evidence. Select an assignment template" ]

[ "Free Version Easy # Estimating Parameters: Free Throws APSTAT-1XLLFF Suppose an athlete shoots $200$ free throws and makes $160$ of them. Which of the following statements is true? A This athlete is an $80\\%$ free-throw shooter. B A sample of $100$ free throws would be more effective in estimating the athlete’s true free-throw percentage. C The athlete’s true free-throw percentage is probably somewhere near $80\\%$. D The athlete could easily make fewer than $100$ free throws in the next set of $200$ free throw attempts. E No reasonable conclusions can be drawn from this data.", "Free Version Easy # Basketball Free Throws: One and One APSTAT-FLTG1I In basketball, one-and-one free throws sometimes occur near the end of a game. A player gets one attempt at a free throw: if she makes it, she gets one other attempt; if she misses, she gets no further attempts. Suppose an athlete has a $64\\%$ free-throw shooting percentage and is shooting a one-and-one situation. What is the probability she makes at least one shot? A $0.64$ B $0.36$ C $0.2304$ D $0.4096$ E $0.5904$", "# basketball player Printable View • Aug 10th 2006, 10:38 AM kwtolley basketball player The probability a basketball player makes a free throw is 0.60. Find the probability the player makes four out of 6. P( 4 throws in 6 attempts) (0.60)(0.60)(0.60)(0.60)(0.60)(0.4)=0.311 0.311 is my answer, thanks so much for checking my answer. List of possible aswers are - 1. 0.0207 2. 0.0667 3. 0.240 4. 0.311 • Aug 10th 2006, 12:44 PM ThePerfectHacker It is, $_6C_4 (.6)^4(.4)^2\\approx .311$ • Aug 10th 2006, 09:07 PM kwtolley thanks thanks you for all your help and everyone else also.", "# Math Help - basketball player The probability a basketball player makes a free throw is 0.60. Find the probability the player makes four out of 6. P( 4 throws in 6 attempts) (0.60)(0.60)(0.60)(0.60)(0.60)(0.4)=0.311 0.311 is my answer, thanks so much for checking my answer. List of possible aswers are - 1. 0.0207 2. 0.0667 3. 0.240 4. 0.311 2. It is, $_6C_4 (.6)^4(.4)^2\\approx .311$ 3. ## thanks thanks you for all your help and everyone else also.", "so our expected number of shots before getting N=30 consecutive shots in is roughly $$E \\approx 6.93848×10^{15} = 6,938,480,000,000,000.$$ A non-starter. Kawhi Leonard is listed as having free throw accuracy of 88.9% this season. His expected number of shots to make 30 in a row, for comparison, is about $$E \\approx 298$$ which is way more manageable. We can get professional player stats for the 2020 Basketball season from various sites. For example, the average NBA player had a free throw success rate of 77.1% in the last season. Collecting a few players, we can see how long each player needs to make 30 shots in a row. player free throw success rate expected value Devin Booker 0.916 154 Kawhi Leonard 0.889 298 Kyle Lowry 0.861 634 Average NBA Player 0.771 10675 LeBron James 0.697 166558", "Point Curry Disco Visual Search Engine! Is your friend full of it? Posted on Sat 02 September 2017 in misc Five Thirty Eight Riddler Express Problem - Sep 1, 2017¶ You’re hanging out with some friends, shooting the breeze and talking sports. One of them brags to the group that he once made 17 free throws in a row after years of not having touched a basketball. You think the claim sounds unlikely, but plausible. Another friend scoffs, thinking it completely impossible. Let’s give your bragging friend the benefit of the doubt and say he’s a 70-percent free-throw shooter. So, who’s right? What is the number of free throws that a 70-percent shooter would be expected to take before having a streak of 17 makes in a row? And what if his accuracy was a bit worse? This looks like a pretty innocuous Bernoulli's trials problem but it isn't! Had the question been about the average number of throws taken for 17 successes, the answer would have been straightforward - $$0.7\\sum_{n=16}^{\\infty} n\\binom {n} {16}0.7^{16}0.3^{n-16} = 17/0.7 \\approx 24.3$$ But the question we are dealing with is about the average number of throws that are required for 17 consecutive makes in a row. I have a vague recollection of this problem being dealt with in Feller's Volume 1 and", "## Prealgebra (7th Edition) Published by Pearson # Chapter 7 - Review: 2 77% #### Work Step by Step Since the basketball player made 77 out of 100 free throw attempts, we can find the percent of free throws he successfully made by doing the calculation: $(77\\div100)\\times100$=% of free throws the basketball player made. 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.", "Free Version Easy # Calculator: Free-Throw Outcomes APSTAT-YPTLAE Suppose a basketball player shoots $10$ free throws and makes $8$ of them. How many different ways could this happen? A $10$ B $8$ C $1$ D $45$ E $3,628,800$", "# If a basketball player made 72 out of 85 free-throw attempts, how many could she expect to make in 200 attempts? May 26, 2018 #### Answer: You would expect her to make at least $169$ of the $200$ free throws. #### Explanation: Use a proportion: $\\frac{72}{85} = \\frac{x}{200}$ Solve for $x$ $\\left(72\\right) \\left(200\\right) = 85 x$ $x = \\frac{\\left(72\\right) \\left(200\\right)}{85}$ $x = \\left\\lfloor 169.41 \\right\\rfloor = 169$ So you would expect her to make at least $169$ of the $200$ free throws.", "free throw shot” for a basketball player shooting two free throws after a foul. By looking at the outcomes of the two shots for the player throughout the year, we can estimate that $$P(\\text{make the first shot}) = 0.72$$, $$P(\\text{make the second shot}) = 0.72$$ and $$P(\\text{make the first shot and make the second shot}) = 0.52$$. The events are independent since $$0.72 \\boldcdot 0.72 \\approx 0.52$$.", "Shaquille 'O' Neal is a $60$% career free throw shooter, meaning that he successfully makes $60$ free throws out of $100$ attempts on average. What is the probability that he will successfully make exactly $6$ free throws in $10$ attempts? 1. $0.2508$ 2. $0.2816$ 3. $0.2934$ 4. $0.6000$", "The probability of a success is 0,9 $ P(A) = {20 \\choose 19} \\left( 0,9 \\right) ^{19} \\left( 0,1 \\right)^{1} + {20 \\choose 20} \\left( 0,9 \\right) ^{20} \\left( 0,1 \\right)^{0}$ $ P(B) = 1 - {20 \\choose 15} \\left( 0,9 \\right) ^{15} \\left( 0,1 \\right)^{5}$ $- {20 \\choose 16} \\left( 0,9 \\right) ^{16} \\left( 0,1 \\right)^{4}$ $- {20 \\choose 17} \\left( 0,9 \\right) ^{17} \\left( 0,1 \\right)^{3}$ $- {20 \\choose 18} \\left( 0,9 \\right) ^{18} \\left( 0,1 \\right)^{2}$ $- {20 \\choose 19} \\left( 0,9 \\right) ^{19} \\left( 0,1 \\right)^{1}$ $- {20 \\choose 20} \\left( 0,9 \\right) ^{20} \\left( 0,1 \\right)^{0}$ 90% of 20 throws equal 18 throws. $P(C) = {20 \\choose 18} \\left( 0,9 \\right) ^{18} \\left( 0,1 \\right)^{2}$ 3. Originally Posted by 16sammy I hate math assignments and it's due tomorrow and the teacher is stupid and won;t help people. I was away when it was explained so i have no idea what to do!! Suppose an NBA player makes 90% of his free throws. This player takes 20 free throws in the course of a game. Find the probabilits of his making: A) Atleast 19 of the 20 free throws. B) Less than 15 of the 20 free throws C) Exactly 90% od the 20 free throws. 1. So how long have you had to", "+0 0 88 2 +406 1. A basketball player made 12 out of 15 free throws she attempted. She wants to know have many consecutive free throws she would have to make to raise the percent of successful free throws to 85%. (a) Write an equation to represent this situation. (b) Solve the equation. How many consecutive free throws would she have to make to raise her percent to 85%? 2. Simplify the expression x^2+3x-28/x^2 -7x+12 1. Write the expression as a simplified rational expression. how do i solve this can you please include steps Oct 31, 2018 #1 +3617 +2 $$\\text{suppose she has to make }n \\text{ consecutive throws}\\\\ \\dfrac{12+n}{15+n} = 0.85 \\\\ 12+n = 12.75 + 0.85n\\\\ 0.15n =0.75 \\\\ n=5$$ 1) $$\\dfrac{5}{\\frac 1 6 + \\frac{1}{x+1}} =\\\\ \\dfrac{5}{\\frac{x+1+6}{6(x+1)}} =\\\\ \\dfrac{5\\cdot 6(x+1)}{x+7} = \\\\ \\dfrac{30(x+1)}{x+7}$$ . Oct 31, 2018 #2 +94545 +1 Here's the second one : x^2 + 3x - 28 ____________ factor top/bottom x^2 -7x + 12 ( x + 7) ( x - 4) ____________ \"cancel\" the ( x - 4) factor on top/bottom (x - 3) (x - 4) ( x + 7) _______ ( with x not equal to 4) ( x - 3) Oct 31, 2018", "The professor asked us to imagine a scenario where we have a basketball player who isn't good at shooting free throws. He makes his first free throw with probability $0.2$ After the first free throw, he makes a free throw with probability $0.6$ if he made the preceding one, and he makes a free throw with probability $0.3$ if he missed the preceding one. During practice, he throws $10$ free throws. What is the total number of outcomes? Let $A_i$ be the event that he makes the $i$th free throw. Find $P(A_i)$ for $i=2, 3, 4$. I figured out the total number of possible outcomes using binomial coefficients ${n \\choose r} = \\frac{n!}{(n-r)!r!}$. The number of possible ways outcomes are as follows: \\begin{eqnarray} \\text{Possible ways he will make 10 free throws} & = & {10 \\choose 10} = 1 \\\\ \\text{Possible ways he will make 9 free throws} & = & {10 \\choose 9} = 10 \\\\ \\text{Possible ways he will make 8 free throws} & = & {10 \\choose 8} = 45 \\\\ \\text{Possible ways he will make 7 free throws} & = & {10 \\choose 7} = 120 \\\\ \\text{Possible ways he will make 6 free throws} & = & {10 \\choose 6} = 210 \\\\ \\text{Possible ways he will make 5 free throws} & =", "adults. ### Eureka Math Precalculus Module 5 Lesson 10 Problem Set Answer Key Question 1. A high school basketball player makes 70% of the free throws she attempts. Suppose she attempts seven free throws during a game. The probability distribution for the number of free throws made out of seven attempts is displayed below. a. What is the probability that she completes at least three free throws? Interpret this probability in context. The probability that she completes at least three free throws is as follows: 0.09725 + 0.22689 + 0.31765 + 0.24706 + 0.08235 = 0.9712. If this basketball player attempts seven free throws many times, the long-run proportion of the time that she will complete at least three free throws is 0.972. b. What is the probability that she completes more than two but less than six free throws? Interpret this probability in context. The probability that she completes more than two but less than six free throws is as follows: 0.09725 + 0.22689 + 0.31765 = 0.64179. If this basketball player attempts seven free throws during a game for many games, the long-run proportion of the time that she will complete more than two but less than six free throws is 0.64179. c. How many free throws will she complete on average? Interpret this expected value", "Limited access Jimmy, a high school basketball star, has a life-long free throw rate of $45\\%$. Jimmy spent the last off-season working especially hard to improve his free-throw percentage. During the first game of the season, he connected on three of his first four free throws. What conclusions can be made concerning Jimmy’s free-throw shooting ability? A Since $3$ of $4 = 75\\%$, it is clear that Jimmy has improved his free-throw percentage. B Since the sample size is so low, we have strong evidence Jimmy is still a $45\\%$-free-throw shooter. C No reasonable conclusions can be drawn since the sample size is so low. D Free throws can hardly be treated as a simple random sample; therefore, this data cannot be trusted. E Jimmy is clearly not a $45\\%$-free-throw shooter based off the evidence. Select an assignment template", "– 4)! . (1/2)12 = 0.1208 Therefore the probability of flipping a coin 12 times and getting exactly 4 heads is about 0.12 Question 17. A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws. Draw a histogram of the binomial distribution of the number of successful free throws. What is the most likely outcome? Given, A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws. p = P(Successful free throw) = 82.6% = 0.826 1 – p = P(Unsuccessful free throw) = 1 – 0.826 = 0.174 P (Out of 5 free throws one was successful) = 5C1 p¹(1 – p)5-1 P (Out of 5 free throws two was successful) =5C2 p²(1 – p)5-2 P(Out of 5 free throws three was successful) = 5C3p³(1 – p)5-3 P (Out of 5 free throws four was successful) =5C4 p4(1 – p)5-4 P (All 5 throws one was successful) = 5C5(1 – p)5-5 Probability Test You roll a six-sided die. Find the probability of the event described. Explain your reasoning. Question 1. You roll a number less than 5. The die has 6 sides, thus the total number of possible outcomes is 6. The favorable outcomes are P(n<5) = Number of favorable outcomes/Total number of outcomes", "## anonymous one year ago In the past, Fred has made four out of every five free throws in basketball that he has attempted. What is the probability that Fred will make the next two free throws he attempts? 1. anonymous To find probability of something (in this case successful free throws), take the number of times that has happened and divide it by the number of chances it had to happen. 2. kropot72 The probability of making one throw is 4/5 = 0.8. The probability of making the next two free throws is given by $\\large P(making\\ 2\\ successive\\ throws)=0.8\\times0.8 = you\\ can\\ calculate$", "? Free Version Moderate # Finding the Median from a Histogram: NBA Payroll APSTAT-BG1VSN This histogram below represents the average payroll of $29$ NBA teams. In which bin is the median located? A $22\\le \\text{ payroll }<28$ B $28\\le \\text{ payroll }<34$ C $34\\le \\text{ payroll }<40$ D $40\\le \\text{ payroll }<46$ E $46\\le \\text{ payroll }<52$", "+ a_2 X_ 2.$$ State how you created the dataset, and show that it conforms to the required property. Part 2(b): Challenge (10 pts) Now, come up with a plausible scenario where $$X_2$$ is a lurking variable. That is, in your report, come up with a story about the data $$(X_1, X_2, Y)$$. Now, re-generate the data in a way that makes clear that the data could be generated if the scenario actually happened. Include your code, an explanation, and a demonstration of the effect of $$X_2$$ as a lurking variable. Part 3: Free Throws Free throws are an important part of basketball. Here, we will be using a dataset of the number of free throws that Shaquille O’Neal , one of the greatest basketball players in the last few decades who was also known for his poor free-throwing ability , has taken in 23 different games. The total number of attempted free throws is recorded, as well as the number of times that the attempts were successful. lines <- \"Game Scored N.Attempts 1 4 5 2 5 11 3 5 14 4 5 12 5 2 7 6 7 10 7 6 14 8 9 15 9 4 12 10 1 4 11 13 27 12 5 17 13 6 12 14 9 9 15 7 12" ]

[ "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", "OPR = \\angle PRQ + \\angle PQR = 90^\\circ + \\angle PQR$, which must equal $\\angle OR'P'$. Therefore, $\\angle Q'R'P' = \\angle OR'P' - \\angle OR'Q' = 90^\\circ + \\angle PQR - \\angle OQR$. But $\\angle OQR = \\angle PQR$. Therefore, $\\angle Q'R'P' = 90^\\circ$. As this holds for any $R$, all points on circle $C$ will invert to a point on a circle with diameter $\\overline{Q'P'}$. ## 3. General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle $[asy] unitsize(7); int radiusInverse = 12; int k = radiusInverse * radiusInverse; int radiusCircle = 4; int circleCenterLocation = 7; path inverse = Circle((0, 0), radiusInverse); draw(inverse, dashed); label(\"\\Omega\", inverse, NW); pair O = (0, 0); dot(O); label(\"O\", O, NW); pair C = (circleCenterLocation, 0); dot(C); label(\"C_1\", C, S); path circle1 = Circle(C, radiusCircle); draw(circle1); int p = circleCenterLocation - radiusCircle; int q = circleCenterLocation + radiusCircle; pair P = (p, 0); pair Q = (q, 0); pair A = (k / p, 0); pair B = (k / q, 0); pair D = (((k / p) + (k / q)) / 2, 0); dot(P); dot(Q); dot(A); dot(B); dot(D); label(\"C_2\", D, S); label(\"P\", P, SE); label(\"Q\", Q, SW); label(\"Q'\", B, SE); label(\"P'\", A, SE); draw(O--A); path circle2 = Circle(D, ((k / p)", "C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label(\"A\",A,(1,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0));label(\"P\",P,(0,-1));label(\"Q\",Q,(1,0));label(\"R\",R,(-1,1)); [/asy]$", "label(\"$r\\sqrt{ 2 }$\",(2.75,-1.24),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(2.02,-1.85),SE*lsf); label(\"$45^\\circ$\",(2.41,-1.3),SE*lsf); label(\"$22.5^\\circ$\",(2.02,-0.96),SE*lsf); draw(circle((2.07,-1.43),0.01),qqwuqq); label(\"$90^\\circ$\",(2.15,-1.3),SE*lsf); label(\"$67.5^\\circ$\",(1.79,-1.28),SE*lsf); label(\"$22.5^\\circ$\",(0.84,0.1),SE*lsf); label(\"r\",(0.71,-0.06),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(0.27,0.35),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.05,-0.12),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.09,0.63),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.44,0.31),SE*lsf); draw(circle((0.45,0.07),0.01),qqwuqq); label(\"$90^\\circ$\",(0.6,0.1),SE*lsf); draw(circle((2.16,0.07),0.01),qqwuqq); label(\"$90^\\circ$\",(2.31,0.17),SE*lsf); label(\"$22.5^\\circ$\",(1.64,0.11),SE*lsf); label(\"r\",(2.04,-0.09),SE*lsf); draw((1.4,-0.2)--(1.11,0.17),EndArrow(6));label(\"$\\sqrt{ (r (\\sqrt{ 2 } - 1) + r + r )^2 + (r (\\sqrt{ 2 } - 1) )^2 }$\",(1.58,-0.2),SE*lsf); label(\"$= \\sqrt{ (r (\\sqrt{ 2 } + 1) )^2 + (r (\\sqrt{ 2 } - 1) )^2 }$\",(1.53,-0.37),SE*lsf); label(\"$\\mathbf{=r \\sqrt{ 6 }}$\",(1.53,-0.72),SE*lsf,blue); label(\"$= \\sqrt{ r^2 (3 + 2 \\sqrt{ 2 }) + r^2 (3 - 2 \\sqrt{ 2 }) }$\",(1.53,-0.54),SE*lsf); label(\"$\\mathbf{2r}$\",(1.46,0.64),SE*lsf,blue); label(\"$\\mathbf{\\frac{BE}{AE} = \\frac{r \\sqrt{ 6 }}{2r} = \\frac{ \\sqrt{ 6 }}{2}}$\",(0.78,-1.47),SE*lsf,red+fontsize(14)); draw((1.74,-1.13)--(0.83,-0.06),EndArrow(6));draw(circle((1.34,0.09),0.01),qqffff); label(\"$45^\\circ$\",(-0.01,0.13),SE*lsf); label(\"$90^\\circ$\",(-0.22,1.47),SE*lsf); label(\"$90^\\circ$\",(1.39,0.24),SE*lsf); draw((0.38,0)--(0.38,0.38),linewidth(0.4)+dotted); dot((0,0),blue); label(\"$B$\",(-0.06,-0.09),NE*lsf,blue); dot((1.31,0),blue); label(\"$D$\",(1.34,-0.09),NE*lsf,blue); dot((0.92,0.92),blue); label(\"$C$\",(0.91,0.99),NE*lsf,blue); dot((2.23,0.38),blue); label(\"$E$\",(2.27,0.39),NE*lsf,blue); dot((0.38,0.38),blue); label(\"$A$\",(0.33,0.41),NE*lsf,blue); dot((0.38,0),linewidth(1pt)+uququq); label(\"$G$\",(0.41,-0.1),NE*lsf,uququq); dot((0,0.38),linewidth(1pt)+uququq); dot((2.23,0),linewidth(1pt)+uququq); label(\"$F$\",(2.24,-0.1),NE*lsf,uququq); dot((0.38,0.38),uququq); dot((1.7,-1.5),blue); dot((2,-0.8),blue); dot((2.69,-1.5),blue); dot((2,-1.5),blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); (Error compiling LaTeX. return linetype((real[]) split(pattern),offset,scale,adjust); ^ /usr/local/share/asymptote/plain_pens.asy: 13.3: Trying to use uninitialized value.) Courtesy of User:Kouichi Nakagawa. <geogebra>9fa91fc57b9e5def39d73828270c8c027c565d85</geogebra>", "+0 # quick geometry problem, CPhill 0 160 7 $$In the diagram below, \\angle PQR = \\angle PRQ = \\angle STR = \\angle TSR, RQ = 8, and SQ = 3. Find PQ. [asy] pair A,B,C,D,E; A = (0, 0.9); B = (-0.4, 0); C = (0.4, 0); D = (-0.275, 0.16); E = (0.11, 0.65); draw(A--B); draw(A--C); draw(B--C); draw(B--E); draw(C--D); label(\"P\",A,N); label(\"Q\", B, S); label(\"R\", C, S); label(\"S\", D, S); label(\"T\", E, W); [/asy]$$ Jan 21, 2022 #1 0 $$PQ = \\boxed{\\frac{50 \\sqrt{2}}{7}}$$ . Jan 21, 2022 #2 -1 incorrect, man Guest Jan 21, 2022 #3 +676 +2 You should rather say thanks. Also when it's wrong or right. Straight Jan 21, 2022 edited by Straight Jan 21, 2022 #4 0 I meant: Thanks man, but it is wrong. Guest Jan 21, 2022 #5 +117872 +1 So I suppose it is a reasonable answer. Garbage in, garbage out. Melody Jan 21, 2022 #6 -1 I don't care if the problem is garbage. I just want to get the problem right, learn more, and get more knowledge. Guest Jan 21, 2022 #7 +117872 +1 If you want to get something right you must display it in a form suitable for the media. Are you incapable of comprehending that? It says, in Latex, (you haven't bothered to write it", "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.", "pI=extension(D,M,R,Q); label(\"O\",pI+(-0.2,0.166),f); draw(anglemark2(Sp,P,R,17)); label(\"p\",P+(0.35,0.1)); draw(anglemark2(R,Q,P,11)); label(\"q\",Q+(-0.17,0.1)); draw(anglemark2(R,Np,D,8,true)); label(\"n\",Np+(+0.12,0.07)); draw(anglemark2(R,M,D,13,true)); label(\"m\",M+(+0.25,0.03)); draw(anglemark2(M,D,P,29)); label(\"d\",D+(-0.75,0.095)); label(\"R\",R,N,f); label(\"M\",M+(-.07,.07),f); label(\"N\",Np+(-.08,.15),f); label(\"P\",P,S,f); label(\"S\",Sp,S,f); label(\"Q\",Q,S,f); label(\"D\",D,S,f);[/asy]$ Looking at triangle PRQ, we have $\\angle RPD+\\angle RQS+\\angle MRN=180$ and from the given statement $\\angle NMR=\\frac{1}{2}\\angle MRN$, so looking at triangle MOR $\\angle NMR=90-\\frac{\\angle RPD+\\angle RQS}{2}$, which rules out 1954 AHSC (Problems • Answer Key • Resources) Preceded byProblem 37 Followed byProblem 39 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions", "+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 # 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", "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 I need help asap! 0 557 1 Points $A$, $B$, and $C$ are on sides $\\overline{WX}$, $\\overline{YZ}$, and $\\overline{XY}$ of rectangle $WXYZ$ as shown below such that $XA = 4$, $YB = 18$, $\\angle ACB= 90^\\circ$, and $YC = 2XC$. Find $AB$. [asy] pair A,B,C,D,P,Q,R; A = (0,0); P = (0,0.7); B = (0,1); R = (0.4,1); C = (1,1); Q = (1,0.2); D = (1,0); draw(P--B--C--D--A--P--R--Q--P); draw(rightanglemark(P,R,Q,2)); label(\"$W$\", A, SW); label(\"$Z$\",D,SE); label(\"$X$\",B,NW); label(\"$Y$\",C,NE); label(\"$C$\",R,N); label(\"$A$\",P,W); label(\"$B$\",Q,E); label(\"$4$\", (B+P)/2,W); label(\"$18$\", (C+Q)/2,E); [/asy] Sorry if the latex is hard to read Feb 16, 2020", "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", "$[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"P\",P,W); label(\"Q\",Q,W); label(\"R\",R,W); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 32 \\qquad \\text{(D)}\\ 64 \\qquad \\text{(E)}\\ 128$ ## Problem 10 A jacket and a shirt originally sold for dollar 80 and dollar 40, respectively. During a sale Chris bought the dollar80 jacket at a $40\\%$ discount and the dollar 40 shirt at a $55\\%$ discount. The total amount saved was what percent of the total of the original prices? $\\text{(A)}\\ 45\\% \\qquad \\text{(B)}\\ 47\\dfrac{1}{2}\\% \\qquad \\text{(C)}\\ 50\\% \\qquad \\text{(D)}\\ 79\\dfrac{1}{6}\\% \\qquad \\text{(E)}\\ 95\\%$. ## Problem 11 Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to $[asy] for(int i = -2; i <= 2; ++i) { draw((i,0)--(i,3),dashed); } draw((-3,1)--(3,1),dashed); draw((-3,2)--(3,2),dashed); draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle); dot((-3,0)); label(\"A\",(-3,0),SW); dot((-3,3)); label(\"B\",(-3,3),NW); dot((0,3)); label(\"C\",(0,3),N); dot((3,3)); label(\"D\",(3,3),NE); dot((3,0)); label(\"E\",(3,0),SE); dot((0,0)); label(\"start\",(0,0),S); label(\"\\longrightarrow\",(0,-0.75),E); label(\"\\longleftarrow\",(0,-0.75),W); label(\"\\textbf{Jane}\",(0,-1.25),W); label(\"\\textbf{Hector}\",(0,-1.25),E); [/asy]$ $\\text{(A)}\\ A \\qquad \\text{(B)}\\ B \\qquad \\text{(C)}\\ C \\qquad \\text{(D)}\\ D \\qquad \\text{(E)}\\ E$ ## Problem", "+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", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The", "= \\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)", "= 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", "= \\boxed{571}$. ### Solution 2 $[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label(\"$$A$$\",A,(0,1));label(\"$$B$$\",B,(-1,-1));label(\"$$C$$\",C,(1,-1));label(\"$$P$$\",P,(1,1)); label(\"$$Q$$\",Q,(-1,1));label(\"$$R$$\",R,(1,0));label(\"$$S$$\",S,(-1,0)); [/asy]$ Again, construct $R$ as above. Let $\\angle BAC = \\angle QBR = \\angle QPR = 2x$ and $\\angle ABC = \\angle ACB = y$, which means $x + y = 90$. $\\triangle QBC$ is isosceles with $QB = BC$, so $\\angle BCQ = 90 - \\frac {y}{2}$. Let $S$ be the intersection of $QC$ and $BP$. Since $\\angle BCQ = \\angle BQC = \\angle BRS$, $BCRS$ is cyclic, which means $\\angle RBS = \\angle RCS = x$. Since $APRB$ is an isosceles trapezoid, $BP = AR$, but since $AR$ bisects $\\angle BAC$, $\\angle ABR = \\angle ACR = 2x$. Therefore we have that $\\angle ACB = \\angle ACR + \\angle RCS + \\angle QCB = 2x + x + 90 - \\frac {y}{2} = y$. We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \\frac {y}{2} = y$ to get $x = 10$ and $y = 80$. $\\angle APQ = 180 - 4x = 140$, $\\angle ACB = 80$, so $r = \\frac {80}{140} = \\frac {4}{7}$. $\\left\\lfloor 1000\\left(\\frac {4}{7}\\right)\\right\\rfloor = \\boxed{571}$. ### Solution 3 Let the measure of $\\angle BAC$ be $\\alpha$ and $\\overline{AP}=\\overline{PQ}=\\overline{QB}=\\overline{BC}=x$. Because $\\triangle APQ$ is isosceles,", "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 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" ]

[ "angles as $90^{\\circ}$. Hence, PQRS is a rectangle. Editor", "the circle, angle OPQ = 90 degrees. Therefore, angle PQR = 180 - 70 - 90 = 20 degrees.", "× # Show that the ASCO is rectangular Show that the ASCO is rectangular. Note by Aly Ahmed 1 year, 3 months ago Sort by: What is \"mesoparallili\"? · 1 year, 3 months ago Since $$\\angle OPS = 90^\\circ$$ and $$\\angle OAS = 90^\\circ$$, hence $$ASPO$$ is a cyclic quad. Since $$\\angle PAB = 90^\\circ - \\angle PBA = \\angle PQO = 180^\\circ - \\angle OQP$$, hence $$APQO$$ is a cyclic quad. Hence, $$ASPQO$$ is a cyclic quad. Thus, since $$\\angle SAO = 90^\\circ$$, hence $$\\angle SQO = 90 ^ \\circ$$. And since $$\\angle AOQ = 90^\\circ$$ hence $$\\angle ASQ = 90^ \\circ$$. So we have $$ASQO$$ is a rectangle. Staff · 1 year, 3 months ago", "$$1 - \\sin 10 ^ \\circ - 2 \\sin^2 10^\\circ = \\sin N^\\circ$$, where $$0 \\leq N \\leq 90$$. What is $$N$$?", "to the x axis. Hence my 90 $^\\circ$. You of course in your post #1 meant to refer to the angle between $\\vec A$ and $\\vec A_x$ which as you can see is $\\pi-\\alpha$ and thereby in the range 0 to 90$^\\circ$. Note that I forgot to draw the arrow above $\\vec A_x$ and $\\vec A_{yz}$. Nobody is perfect, be we keep trying. To be specific: in $\\vec A_x = A_x\\, \\hat\\imath\\$, $\\ A_x$ is a number (negative for the alpha in the picture: $A_x = |\\vec A | \\cos\\alpha$​ Last edited: Oct 14, 2015", "= 90/360*2*π*$$\\sqrt{2}$$ = π*$$\\sqrt{2}$$ 90/360*2*π*$$\\sqrt{2}$$… 1/4*2*π*$$\\sqrt{2}$$… 1/2*π*$$\\sqrt{2}$$… π*$$\\sqrt{2}$$/2… Re: Square PQRS is tilted 90 degrees anticlockwise direction around the po [#permalink] 08 Oct 2019, 03:28 Display posts from previous: Sort by", "× # Show that the ASCO is rectangular Show that the ASCO is rectangular. Note by Aly Ahmed 11 months, 2 weeks ago Sort by: What is \"mesoparallili\"? · 11 months, 2 weeks ago Since $$\\angle OPS = 90^\\circ$$ and $$\\angle OAS = 90^\\circ$$, hence $$ASPO$$ is a cyclic quad. Since $$\\angle PAB = 90^\\circ - \\angle PBA = \\angle PQO = 180^\\circ - \\angle OQP$$, hence $$APQO$$ is a cyclic quad. Hence, $$ASPQO$$ is a cyclic quad. Thus, since $$\\angle SAO = 90^\\circ$$, hence $$\\angle SQO = 90 ^ \\circ$$. And since $$\\angle AOQ = 90^\\circ$$ hence $$\\angle ASQ = 90^ \\circ$$. So we have $$ASQO$$ is a rectangle. Staff · 11 months, 1 week ago", "× # Show that the ASCO is rectangular Show that the ASCO is rectangular. Note by Aly Ahmed 9 months, 1 week ago Sort by: What is \"mesoparallili\"? · 9 months, 1 week ago Since $$\\angle OPS = 90^\\circ$$ and $$\\angle OAS = 90^\\circ$$, hence $$ASPO$$ is a cyclic quad. Since $$\\angle PAB = 90^\\circ - \\angle PBA = \\angle PQO = 180^\\circ - \\angle OQP$$, hence $$APQO$$ is a cyclic quad. Hence, $$ASPQO$$ is a cyclic quad. Thus, since $$\\angle SAO = 90^\\circ$$, hence $$\\angle SQO = 90 ^ \\circ$$. And since $$\\angle AOQ = 90^\\circ$$ hence $$\\angle ASQ = 90^ \\circ$$. So we have $$ASQO$$ is a rectangle. Staff · 9 months ago", "$\\angle 3= \\angle 4$ Therefore, $\\angle SPQ= \\angle PQR$ .....(3)Since, $SP \\parallel RQ$ and $PQ$ intersects themTherefore, $\\angle SPQ + \\angle PQR= 180^o$....(4) (Since consecutive interior angles are supplementary)From (3) and (4), we get $\\angle PQR +\\angle PQR= 180^o$$2 \\angle PQR =180^o$$\\angle PQR =90^o$ $\\angle SPQ= \\angle PQR=90^o$Thus, $PQRS$ is a parallelogram whose one angle $\\angle SPQ=90^o$.Hence $PQRS$ is a rectangle.MathematicsRS AgarwalStandard IX Suggest Corrections 0 Similar questions View More Same exercise questions View More People also searched for View More", "contact. \\begin{aligned} \\implies & \\angle OAP = 90^\\circ \\\\ \\implies & \\angle OAB + \\angle PAB = 90^\\circ \\\\ \\implies & \\angle OAB = 90^\\circ - (90^\\circ - \\dfrac { 1 }{ 2 } x^\\circ) \\\\ \\implies & \\angle OAB = \\dfrac { 1 }{ 2 } x^\\circ = \\dfrac { 1 }{ 2 } \\angle APB \\\\ \\implies & \\angle APB = 2 \\angle OAB \\end{aligned} 6. Hence, $\\angle APB = 2 \\angle OAB.$", "Home » » In the diagram above, PQO is a diameter of the circle PQR. If < PSR = 145$$^{\\circ}$$ find x$$^{\\circ}$$... # In the diagram above, PQO is a diameter of the circle PQR. If < PSR = 145$$^{\\circ}$$ find x$$^{\\circ}$$... ### Question In the diagram above, PQO is a diameter of the circle PQR. If < PSR = 145$$^{\\circ}$$ find x$$^{\\circ}$$ ### Options A) 35$$^{\\circ}$$ B) 55$$^{\\circ}$$ C) 25$$^{\\circ}$$ D) 45$$^{\\circ}$$ ### Explanation: Construction RQ < PRQ = 90$$^{\\circ}$$(amgle substended at the circumfernce of a circle by diameter). Also in figure PQRS < PQS + PSQ = 180(supplementary angle) < PQS = 180 - 145 = 35 In $$\\bigtriangleup$$PRQ x = 180 - 935 = 90) x = 55 ## Dicussion (1) • Construction RQ < PRQ = 90$$^{\\circ}$$(amgle substended at the circumfernce of a circle by diameter). Also in figure PQRS < PQS + PSQ = 180(supplementary angle) < PQS = 180 - 145 = 35 In $$\\bigtriangleup$$PRQ x = 180 - 935 = 90) x = 55", "NQ$ (Parts of parallel lines as $PQRS$ is a parallelogram)$\\angle MON=\\angle MQN$ (Opposite angles of parallelogram are equal)$⇒\\angle MON=\\angle MQN=\\angle PQR$But, $\\angle MON=90°$ (Diagonals of rhombus bisect each other at right angles)$\\therefore \\angle PQR=90°$Now, in parallelogram $PQRS$,$PQ=SR$ and $PS=RQ$ (Opposite sides of parallelogram are equal)and $\\angle PQR=90°$Therefore, $PQRS$ is a rectangle. Suggest Corrections 17", "$180^\\circ$. 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 $60^\\circ$.", "below, PQRS is a straight line. Find the value of x. Solution: (a) $\\begin{array}{l}\\angle PQT=\\angle RQS\\\\ {x}^{o}+{90}^{o}={155}^{o}\\\\ \\text{}{x}^{o}={155}^{o}-{90}^{o}\\\\ \\text{}={65}^{o}\\\\ \\text{}x=65\\end{array}$ (b) $\\begin{array}{l}\\angle QRT={57}^{o}\\\\ {x}^{o}+2{x}^{o}={180}^{o}-{57}^{o}\\\\ \\text{3}{x}^{o}={123}^{o}\\\\ \\text{}{x}^{o}=\\frac{{123}^{o}}{3}\\\\ \\text{}{x}^{o}={41}^{o}\\\\ \\text{}x=41\\end{array}$ Question 3: In Diagram below, PQR is a straight line. Find the value of x. Solution: $\\begin{array}{l}{x}^{o}=\\angle QRT+\\angle TRS\\\\ \\angle QRT={40}^{o}\\\\ \\angle TRS={180}^{o}-{135}^{o}\\\\ \\text{}={45}^{o}\\\\ \\text{Hence,}{x}^{o}={40}^{o}+{45}^{o}\\\\ \\text{}{x}^{o}={85}^{o}\\\\ \\text{}x=85\\end{array}$ Question 4: In Diagram below, find the value of x. Solution: 40o + 80o + xo + xo = 180o 2xo = 180o – 120o 2xo = 60o x= 30o x = 30 Question 5: In Diagram below, PQR is an isosceles triangle and QRS is a straight line. Find the values of x and y. Solution: $\\begin{array}{l}\\angle PRQ={180}^{o}-{110}^{o}={70}^{o}\\\\ \\angle PRQ=\\angle PQR={70}^{o}\\\\ \\text{}{y}^{o}={180}^{o}-{70}^{o}-{70}^{o}\\\\ \\text{}={40}^{o}\\\\ \\text{}y=40\\\\ {x}^{o}={110}^{o}-{40}^{o}\\\\ \\text{}={70}^{o}\\\\ \\text{}x=70\\end{array}$ # 1.1 Angles and Lines II 1.1 Angles and Lines II Identifying Parallel, Transversals, Corresponding Angles, Alternate Angles and Interior Angles (A) Parallel lines Parallel lines are lines with the same direction. They remain the same distance apart and never meet. (B) Transversal lines A transversal is a straight line that intersects two or more straight lines. (C) Alternate angles (D) Corresponding angles (E) Interior angles PT3 Smart TIP Alternate angles are easily identified by tracing out the pattern “Z” as shown. Corresponding angles are easily identified by the pattern “F” as shown. Interior angles", "of the bag, then $\\theta =90°$. Now, on substituting the given values in the formula, we get: $W=30·200·\\mathrm{cos}\\left(90°\\right)\\phantom{\\rule{0ex}{0ex}}W=30·200·0\\phantom{\\rule{0ex}{0ex}}W=0J$Hence, option (D) is correct. Suggest Corrections 3 Explore more", "we have found that. Therefore, $\\angle P R S+30^{\\circ}=90^{\\circ}$ $\\angle P R S=60^{\\circ}$ Thus we have found out that .", "Question 2: (a) In Diagram below, PQR and SQT are straight lines. Find the value of x. (b) In Diagram below, PQRS is a straight line. Find the value of x. Solution: (a) $\\begin{array}{l}\\angle PQT=\\angle RQS\\\\ {x}^{o}+{90}^{o}={155}^{o}\\\\ \\text{}{x}^{o}={155}^{o}-{90}^{o}\\\\ \\text{}={65}^{o}\\\\ \\text{}x=65\\end{array}$ (b) $\\begin{array}{l}\\angle QRT={57}^{o}\\\\ {x}^{o}+2{x}^{o}={180}^{o}-{57}^{o}\\\\ \\text{3}{x}^{o}={123}^{o}\\\\ \\text{}{x}^{o}=\\frac{{123}^{o}}{3}\\\\ \\text{}{x}^{o}={41}^{o}\\\\ \\text{}x=41\\end{array}$ Question 3: In Diagram below, PQR is a straight line. Find the value of x. Solution: $\\begin{array}{l}{x}^{o}=\\angle QRT+\\angle TRS\\\\ \\angle QRT={40}^{o}\\\\ \\angle TRS={180}^{o}-{135}^{o}\\\\ \\text{}={45}^{o}\\\\ \\text{Hence,}{x}^{o}={40}^{o}+{45}^{o}\\\\ \\text{}{x}^{o}={85}^{o}\\\\ \\text{}x=85\\end{array}$ Question 4: In Diagram below, find the value of x. Solution: 40o + 80o + xo + xo = 180o 2xo = 180o – 120o 2xo = 60o x= 30o x = 30 Question 5: In Diagram below, PQR is an isosceles triangle and QRS is a straight line. Find the values of x and y. Solution: $\\begin{array}{l}\\angle PRQ={180}^{o}-{110}^{o}={70}^{o}\\\\ \\angle PRQ=\\angle PQR={70}^{o}\\\\ \\text{}{y}^{o}={180}^{o}-{70}^{o}-{70}^{o}\\\\ \\text{}={40}^{o}\\\\ \\text{}y=40\\\\ {x}^{o}={110}^{o}-{40}^{o}\\\\ \\text{}={70}^{o}\\\\ \\text{}x=70\\end{array}$ # 1.1 Angles and Lines II 1.1 Angles and Lines II Identifying Parallel, Transversals, Corresponding Angles, Alternate Angles and Interior Angles (A) Parallel lines Parallel lines are lines with the same direction. They remain the same distance apart and never meet. (B) Transversal lines A transversal is a straight line that intersects two or more straight lines. (C) Alternate angles (D) Corresponding angles (E) Interior angles PT3 Smart TIP Alternate angles are easily identified by", "Question 2: (a) In Diagram below, PQR and SQT are straight lines. Find the value of x. (b) In Diagram below, PQRS is a straight line. Find the value of x. Solution: (a) $\\begin{array}{l}\\angle PQT=\\angle RQS\\\\ {x}^{o}+{90}^{o}={155}^{o}\\\\ \\text{}{x}^{o}={155}^{o}-{90}^{o}\\\\ \\text{}={65}^{o}\\\\ \\text{}x=65\\end{array}$ (b) $\\begin{array}{l}\\angle QRT={57}^{o}\\\\ {x}^{o}+2{x}^{o}={180}^{o}-{57}^{o}\\\\ \\text{3}{x}^{o}={123}^{o}\\\\ \\text{}{x}^{o}=\\frac{{123}^{o}}{3}\\\\ \\text{}{x}^{o}={41}^{o}\\\\ \\text{}x=41\\end{array}$ Question 3: In Diagram below, PQR is a straight line. Find the value of x. Solution: $\\begin{array}{l}{x}^{o}=\\angle QRT+\\angle TRS\\\\ \\angle QRT={40}^{o}\\\\ \\angle TRS={180}^{o}-{135}^{o}\\\\ \\text{}={45}^{o}\\\\ \\text{Hence,}{x}^{o}={40}^{o}+{45}^{o}\\\\ \\text{}{x}^{o}={85}^{o}\\\\ \\text{}x=85\\end{array}$ Question 4: In Diagram below, find the value of x. Solution: 40o + 80o + xo + xo = 180o 2xo = 180o – 120o 2xo = 60o x= 30o x = 30 Question 5: In Diagram below, PQR is an isosceles triangle and QRS is a straight line. Find the values of x and y. Solution: $\\begin{array}{l}\\angle PRQ={180}^{o}-{110}^{o}={70}^{o}\\\\ \\angle PRQ=\\angle PQR={70}^{o}\\\\ \\text{}{y}^{o}={180}^{o}-{70}^{o}-{70}^{o}\\\\ \\text{}={40}^{o}\\\\ \\text{}y=40\\\\ {x}^{o}={110}^{o}-{40}^{o}\\\\ \\text{}={70}^{o}\\\\ \\text{}x=70\\end{array}$ # 1.1 Angles and Lines II 1.1 Angles and Lines II Identifying Parallel, Transversals, Corresponding Angles, Alternate Angles and Interior Angles (A) Parallel lines Parallel lines are lines with the same direction. They remain the same distance apart and never meet. (B) Transversal lines A transversal is a straight line that intersects two or more straight lines. (C) Alternate angles (D) Corresponding angles (E) Interior angles PT3 Smart TIP Alternate angles are easily identified by", "### In a parallelogram $ABCD$, the bisectors of $\\angle A$ and $\\angle B$ intersect at $S$, $\\angle B$ and $\\angle C$ at $R$, $\\angle C$ and $\\angle D$ at $Q$ and $\\angle D$ and $\\angle A$ at $P$. What kind of a quadrilateral is $PQRS$? $Rectangle$ Step by Step Explanation: 1. The situation given in the question is represented by the figure below. 2. We are given that $ABCD$ is a parallelogram. $$\\implies DC \\parallel AB$$ Also, as the adjacent angles of a parallelogram are supplementary, we have \\begin{aligned} & \\angle A + \\angle D = 180^\\circ \\\\ \\implies & \\dfrac { 1 } { 2 } \\angle A + \\dfrac { 1 } { 2 } \\angle D = 90^\\circ \\\\ \\implies & \\angle PAD + \\angle ADP = 90^\\circ \\\\ \\implies & \\angle APD = 90^\\circ && \\text{[Sum of angles of a triangle is } 180^\\circ.] \\\\ \\implies & \\angle SPQ = 90^\\circ && \\text{[Vertically opposite angles.]} \\end{aligned} Similarly, $\\angle PQR = 90^\\circ, \\angle QRS = 90^\\circ,$ and $\\angle PSR = 90^\\circ$. Thus, $PQRS$ is a quadrilateral each of whose angles is $90^\\circ$. Hence, $PQRS$ is a $rectangle$.", "Trigonometry (10th Edition) 90- x$^{\\circ}$ 1. Given that angle $\\theta+x^{\\circ}=90^{\\circ}$ 2. $\\theta= 90^{\\circ}-x^{\\circ}$ is our angle." ]

[ "OPR = \\angle PRQ + \\angle PQR = 90^\\circ + \\angle PQR$, which must equal $\\angle OR'P'$. Therefore, $\\angle Q'R'P' = \\angle OR'P' - \\angle OR'Q' = 90^\\circ + \\angle PQR - \\angle OQR$. But $\\angle OQR = \\angle PQR$. Therefore, $\\angle Q'R'P' = 90^\\circ$. As this holds for any $R$, all points on circle $C$ will invert to a point on a circle with diameter $\\overline{Q'P'}$. ## 3. General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle $[asy] unitsize(7); int radiusInverse = 12; int k = radiusInverse * radiusInverse; int radiusCircle = 4; int circleCenterLocation = 7; path inverse = Circle((0, 0), radiusInverse); draw(inverse, dashed); label(\"\\Omega\", inverse, NW); pair O = (0, 0); dot(O); label(\"O\", O, NW); pair C = (circleCenterLocation, 0); dot(C); label(\"C_1\", C, S); path circle1 = Circle(C, radiusCircle); draw(circle1); int p = circleCenterLocation - radiusCircle; int q = circleCenterLocation + radiusCircle; pair P = (p, 0); pair Q = (q, 0); pair A = (k / p, 0); pair B = (k / q, 0); pair D = (((k / p) + (k / q)) / 2, 0); dot(P); dot(Q); dot(A); dot(B); dot(D); label(\"C_2\", D, S); label(\"P\", P, SE); label(\"Q\", Q, SW); label(\"Q'\", B, SE); label(\"P'\", A, SE); draw(O--A); path circle2 = Circle(D, ((k / p)", "q); label(\"P\", P, dir(250), q); label(\"Q\", Q, 0.25*(Q-P), q); label(\"\\sqrt{3}\",B--Q, dir(60), p); label(\"\\sqrt{3}\",P--B, dir(210), p); label(\"\\sqrt{3}\",P--Q, dir(135), p); label(\"2\",P--C, dir(120), p); label(\"2\",A--Q, dir(-20), p); label(\"1\",A--P, dir(210), p); [/asy]$ 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", "label(\"$r\\sqrt{ 2 }$\",(2.75,-1.24),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(2.02,-1.85),SE*lsf); label(\"$45^\\circ$\",(2.41,-1.3),SE*lsf); label(\"$22.5^\\circ$\",(2.02,-0.96),SE*lsf); draw(circle((2.07,-1.43),0.01),qqwuqq); label(\"$90^\\circ$\",(2.15,-1.3),SE*lsf); label(\"$67.5^\\circ$\",(1.79,-1.28),SE*lsf); label(\"$22.5^\\circ$\",(0.84,0.1),SE*lsf); label(\"r\",(0.71,-0.06),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(0.27,0.35),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.05,-0.12),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.09,0.63),SE*lsf); label(\"$r( \\sqrt{ 2 } - 1 )$\",(-0.44,0.31),SE*lsf); draw(circle((0.45,0.07),0.01),qqwuqq); label(\"$90^\\circ$\",(0.6,0.1),SE*lsf); draw(circle((2.16,0.07),0.01),qqwuqq); label(\"$90^\\circ$\",(2.31,0.17),SE*lsf); label(\"$22.5^\\circ$\",(1.64,0.11),SE*lsf); label(\"r\",(2.04,-0.09),SE*lsf); draw((1.4,-0.2)--(1.11,0.17),EndArrow(6));label(\"$\\sqrt{ (r (\\sqrt{ 2 } - 1) + r + r )^2 + (r (\\sqrt{ 2 } - 1) )^2 }$\",(1.58,-0.2),SE*lsf); label(\"$= \\sqrt{ (r (\\sqrt{ 2 } + 1) )^2 + (r (\\sqrt{ 2 } - 1) )^2 }$\",(1.53,-0.37),SE*lsf); label(\"$\\mathbf{=r \\sqrt{ 6 }}$\",(1.53,-0.72),SE*lsf,blue); label(\"$= \\sqrt{ r^2 (3 + 2 \\sqrt{ 2 }) + r^2 (3 - 2 \\sqrt{ 2 }) }$\",(1.53,-0.54),SE*lsf); label(\"$\\mathbf{2r}$\",(1.46,0.64),SE*lsf,blue); label(\"$\\mathbf{\\frac{BE}{AE} = \\frac{r \\sqrt{ 6 }}{2r} = \\frac{ \\sqrt{ 6 }}{2}}$\",(0.78,-1.47),SE*lsf,red+fontsize(14)); draw((1.74,-1.13)--(0.83,-0.06),EndArrow(6));draw(circle((1.34,0.09),0.01),qqffff); label(\"$45^\\circ$\",(-0.01,0.13),SE*lsf); label(\"$90^\\circ$\",(-0.22,1.47),SE*lsf); label(\"$90^\\circ$\",(1.39,0.24),SE*lsf); draw((0.38,0)--(0.38,0.38),linewidth(0.4)+dotted); dot((0,0),blue); label(\"$B$\",(-0.06,-0.09),NE*lsf,blue); dot((1.31,0),blue); label(\"$D$\",(1.34,-0.09),NE*lsf,blue); dot((0.92,0.92),blue); label(\"$C$\",(0.91,0.99),NE*lsf,blue); dot((2.23,0.38),blue); label(\"$E$\",(2.27,0.39),NE*lsf,blue); dot((0.38,0.38),blue); label(\"$A$\",(0.33,0.41),NE*lsf,blue); dot((0.38,0),linewidth(1pt)+uququq); label(\"$G$\",(0.41,-0.1),NE*lsf,uququq); dot((0,0.38),linewidth(1pt)+uququq); dot((2.23,0),linewidth(1pt)+uququq); label(\"$F$\",(2.24,-0.1),NE*lsf,uququq); dot((0.38,0.38),uququq); dot((1.7,-1.5),blue); dot((2,-0.8),blue); dot((2.69,-1.5),blue); dot((2,-1.5),blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); (Error compiling LaTeX. return linetype((real[]) split(pattern),offset,scale,adjust); ^ /usr/local/share/asymptote/plain_pens.asy: 13.3: Trying to use uninitialized value.) Courtesy of User:Kouichi Nakagawa. <geogebra>9fa91fc57b9e5def39d73828270c8c027c565d85</geogebra>", "+0 # quick geometry problem, CPhill 0 160 7 $$In the diagram below, \\angle PQR = \\angle PRQ = \\angle STR = \\angle TSR, RQ = 8, and SQ = 3. Find PQ. [asy] pair A,B,C,D,E; A = (0, 0.9); B = (-0.4, 0); C = (0.4, 0); D = (-0.275, 0.16); E = (0.11, 0.65); draw(A--B); draw(A--C); draw(B--C); draw(B--E); draw(C--D); label(\"P\",A,N); label(\"Q\", B, S); label(\"R\", C, S); label(\"S\", D, S); label(\"T\", E, W); [/asy]$$ Jan 21, 2022 #1 0 $$PQ = \\boxed{\\frac{50 \\sqrt{2}}{7}}$$ . Jan 21, 2022 #2 -1 incorrect, man Guest Jan 21, 2022 #3 +676 +2 You should rather say thanks. Also when it's wrong or right. Straight Jan 21, 2022 edited by Straight Jan 21, 2022 #4 0 I meant: Thanks man, but it is wrong. Guest Jan 21, 2022 #5 +117872 +1 So I suppose it is a reasonable answer. Garbage in, garbage out. Melody Jan 21, 2022 #6 -1 I don't care if the problem is garbage. I just want to get the problem right, learn more, and get more knowledge. Guest Jan 21, 2022 #7 +117872 +1 If you want to get something right you must display it in a form suitable for the media. Are you incapable of comprehending that? It says, in Latex, (you haven't bothered to write it", "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]$", "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 # 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", "pI=extension(D,M,R,Q); label(\"O\",pI+(-0.2,0.166),f); draw(anglemark2(Sp,P,R,17)); label(\"p\",P+(0.35,0.1)); draw(anglemark2(R,Q,P,11)); label(\"q\",Q+(-0.17,0.1)); draw(anglemark2(R,Np,D,8,true)); label(\"n\",Np+(+0.12,0.07)); draw(anglemark2(R,M,D,13,true)); label(\"m\",M+(+0.25,0.03)); draw(anglemark2(M,D,P,29)); label(\"d\",D+(-0.75,0.095)); label(\"R\",R,N,f); label(\"M\",M+(-.07,.07),f); label(\"N\",Np+(-.08,.15),f); label(\"P\",P,S,f); label(\"S\",Sp,S,f); label(\"Q\",Q,S,f); label(\"D\",D,S,f);[/asy]$ Looking at triangle PRQ, we have $\\angle RPD+\\angle RQS+\\angle MRN=180$ and from the given statement $\\angle NMR=\\frac{1}{2}\\angle MRN$, so looking at triangle MOR $\\angle NMR=90-\\frac{\\angle RPD+\\angle RQS}{2}$, which rules out 1954 AHSC (Problems • Answer Key • Resources) Preceded byProblem 37 Followed byProblem 39 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions", "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.", "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", "+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 I need help asap! 0 557 1 Points $A$, $B$, and $C$ are on sides $\\overline{WX}$, $\\overline{YZ}$, and $\\overline{XY}$ of rectangle $WXYZ$ as shown below such that $XA = 4$, $YB = 18$, $\\angle ACB= 90^\\circ$, and $YC = 2XC$. Find $AB$. [asy] pair A,B,C,D,P,Q,R; A = (0,0); P = (0,0.7); B = (0,1); R = (0.4,1); C = (1,1); Q = (1,0.2); D = (1,0); draw(P--B--C--D--A--P--R--Q--P); draw(rightanglemark(P,R,Q,2)); label(\"$W$\", A, SW); label(\"$Z$\",D,SE); label(\"$X$\",B,NW); label(\"$Y$\",C,NE); label(\"$C$\",R,N); label(\"$A$\",P,W); label(\"$B$\",Q,E); label(\"$4$\", (B+P)/2,W); label(\"$18$\", (C+Q)/2,E); [/asy] Sorry if the latex is hard to read Feb 16, 2020", "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]$", "= \\boxed{571}$. ### Solution 2 $[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label(\"$$A$$\",A,(0,1));label(\"$$B$$\",B,(-1,-1));label(\"$$C$$\",C,(1,-1));label(\"$$P$$\",P,(1,1)); label(\"$$Q$$\",Q,(-1,1));label(\"$$R$$\",R,(1,0));label(\"$$S$$\",S,(-1,0)); [/asy]$ Again, construct $R$ as above. Let $\\angle BAC = \\angle QBR = \\angle QPR = 2x$ and $\\angle ABC = \\angle ACB = y$, which means $x + y = 90$. $\\triangle QBC$ is isosceles with $QB = BC$, so $\\angle BCQ = 90 - \\frac {y}{2}$. Let $S$ be the intersection of $QC$ and $BP$. Since $\\angle BCQ = \\angle BQC = \\angle BRS$, $BCRS$ is cyclic, which means $\\angle RBS = \\angle RCS = x$. Since $APRB$ is an isosceles trapezoid, $BP = AR$, but since $AR$ bisects $\\angle BAC$, $\\angle ABR = \\angle ACR = 2x$. Therefore we have that $\\angle ACB = \\angle ACR + \\angle RCS + \\angle QCB = 2x + x + 90 - \\frac {y}{2} = y$. We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \\frac {y}{2} = y$ to get $x = 10$ and $y = 80$. $\\angle APQ = 180 - 4x = 140$, $\\angle ACB = 80$, so $r = \\frac {80}{140} = \\frac {4}{7}$. $\\left\\lfloor 1000\\left(\\frac {4}{7}\\right)\\right\\rfloor = \\boxed{571}$. ### Solution 3 Let the measure of $\\angle BAC$ be $\\alpha$ and $\\overline{AP}=\\overline{PQ}=\\overline{QB}=\\overline{BC}=x$. Because $\\triangle APQ$ is isosceles,", "$[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"P\",P,W); label(\"Q\",Q,W); label(\"R\",R,W); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 32 \\qquad \\text{(D)}\\ 64 \\qquad \\text{(E)}\\ 128$ ## Problem 10 A jacket and a shirt originally sold for dollar 80 and dollar 40, respectively. During a sale Chris bought the dollar80 jacket at a $40\\%$ discount and the dollar 40 shirt at a $55\\%$ discount. The total amount saved was what percent of the total of the original prices? $\\text{(A)}\\ 45\\% \\qquad \\text{(B)}\\ 47\\dfrac{1}{2}\\% \\qquad \\text{(C)}\\ 50\\% \\qquad \\text{(D)}\\ 79\\dfrac{1}{6}\\% \\qquad \\text{(E)}\\ 95\\%$. ## Problem 11 Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to $[asy] for(int i = -2; i <= 2; ++i) { draw((i,0)--(i,3),dashed); } draw((-3,1)--(3,1),dashed); draw((-3,2)--(3,2),dashed); draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle); dot((-3,0)); label(\"A\",(-3,0),SW); dot((-3,3)); label(\"B\",(-3,3),NW); dot((0,3)); label(\"C\",(0,3),N); dot((3,3)); label(\"D\",(3,3),NE); dot((3,0)); label(\"E\",(3,0),SE); dot((0,0)); label(\"start\",(0,0),S); label(\"\\longrightarrow\",(0,-0.75),E); label(\"\\longleftarrow\",(0,-0.75),W); label(\"\\textbf{Jane}\",(0,-1.25),W); label(\"\\textbf{Hector}\",(0,-1.25),E); [/asy]$ $\\text{(A)}\\ A \\qquad \\text{(B)}\\ B \\qquad \\text{(C)}\\ C \\qquad \\text{(D)}\\ D \\qquad \\text{(E)}\\ E$ ## Problem", "T = intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label(\"A\", A, E); label(\"B\", B, S); label(\"C\", C, E); label(\"L\", L, S); label(\"M\", M, S); label(\"N\", N, S); label(\"H\", H, E); label(\"T\", T, E); label(\"W\", W, S); label(\"X\", X, E); label(\"Y\", Y, E); [/asy]$ Let $T$ be the intersection of $\\omega_1$ and $\\omega_2$ other than $W$. Lemma 1: $T$ is on $XY$. Proof: We have $\\angle{XTW} = \\angle{YTW} = 90^\\circ$ because they intercept semicircles. Hence, $\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ$, so $XTY$ is a straight line. Lemma 2: $T$ is on $AW$. Proof: Let the circumcircles of $NBW$ and $MWC$ be $\\omega_1$ and $\\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\\angle{BNC} = \\angle{BMC} = 90^\\circ$), let its circumcircle be $\\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$. Lemma 3: $YT$ is perpendicular to $AW$. Proof: This is immediate from $\\angle{YTW} = 90^\\circ$. Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\\angle{ALB} = 90^\\circ.$ Lemma 4: Quadrilateral $THLW$ is cyclic. Proof: We know that $NHLB$ is cyclic because $\\angle{BNH}$ and $\\angle{BLH}$, opposite", "= intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label(\"A\", A, E); label(\"B\", B, S); label(\"C\", C, E); label(\"L\", L, S); label(\"M\", M, S); label(\"N\", N, S); label(\"H\", H, E); label(\"T\", T, E); label(\"W\", W, S); label(\"X\", X, E); label(\"Y\", Y, E); [/asy]$ Let $T$ be the intersection of $\\omega_1$ and $\\omega_2$ other than $W$. Lemma 1: $T$ is on $XY$. Proof: We have $\\angle{XTW} = \\angle{YTW} = 90^\\circ$ because they intercept semicircles. Hence, $\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ$, so $XTY$ is a straight line. Lemma 2: $T$ is on $AW$. Proof: Let the circumcircles of $NBW$ and $MWC$ be $\\omega_1$ and $\\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\\angle{BNC} = \\angle{BMC} = 90^\\circ$), let its circumcircle be $\\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$. Lemma 3: $YT$ is perpendicular to $AW$. Proof: This is immediate from $\\angle{YTW} = 90^\\circ$. Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\\angle{ALB} = 90^\\circ.$ Lemma 4: Quadrilateral $THLW$ is cyclic. Proof: We know that $NHLB$ is cyclic because $\\angle{BNH}$ and $\\angle{BLH}$, opposite and", "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 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", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The" ]

[ "a. x 2 + bx + =...", "## Find values of b for the equation to possess solution? For what values of b does the equation: $\\frac{b\\cos x}{2\\cos 2x - 1} = \\frac{b + \\sin x}{\\left(\\cos^2 x - 3\\sin^2 x\\right)\\tan x}$ ,possess a solution? Also find the solution.", "value of b. x^2=bx−1 If b is on the interval ( , ) ∪ ( , ), the equation has two roots. If b equals to , , the equation has one", "The values of a and b for which the quadratic equation $$x^{2}+2x+2+e^{-2a}-cosb=0$$ has a real solution is", "+ z2 + 2xy - 2yz - 2zx Solution: Here, (a+b) = 2 $\\therefore$ (a+b)3 = 23 or, a3 + 3a2b + 3ab2 + b3 = 8 or, a3 + b3 + 3ab2 = 8 or, a3 + b3 + 6ab = 8 So, the required value of a3 + b3 + 6ab is 8. Solution: = $\\frac{12x^4}{3x^2}$ - $\\frac{15x^3}{3x^2}$ = 4x4-2 - 5x3-2 = 4x2 - 5x", "For what value(s) of b would the equation 3(x + b) = 3x have a solution? The only value would be $b = 0$ We can divide both sides of the equation by $3$: $\\cancel{3} \\left(x + b\\right) = \\cancel{3} x$", "+0 Let $a$ and $b$ be the solutions of the equation $2x^2-10x+5=0$. What is the value of $(2a-3)(4b-6)$? +1 195 2 +11 Let $a$ and $b$ be the solutions of the equation $2x^2-10x+5=0$. What is the value of $(2a-3)(4b-6)$? Dec 24, 2021 #1 +1 The roots a, b are (5 \\pm sqrt(15))/2. Plugging these values in, we get an answer of -16. Dec 24, 2021 #2 +11 +1 Thanks guest greatly appreciated! <3 oTechz Dec 24, 2021", "# Do I look quadratic to you? The equation $| x - 10 | = 7$ is equivalent to the equation $x^2 + ax + b = 0.$ What is the value of $a+b$? ×", "7 (iii) 6x + 5(x – 2) (iv) 4(2x – 1) + 3x + 11 Solution: Question 8. Simplify the following expressions and find their values when a = -1, b = -2: (i) 2a – 2b – 4 – 5 + a (ii) 2(a2 + ab) + 3 – ab Solution:", "we use the same values for both equations, if they put out the same answer, they are equivalent. For this, we will make $a=2$ , $h=2$ , and $A=10$ Original equation; $ \\frac{2}{h}(A-\\frac{ah}{2})=b$ $\\frac{2}{2}(10-\\frac{4}{2})=b$ $b=8$ $\\frac{2A}{h}-a=b$ $\\frac{2(10)}{2}-2=b$ $b=8$", "any point on the line will give you the same value for b. If you want to double check your math, try both. (3, 2): $$2=3(3)+b$$. $$2=9+b$$. $$b=2-9=-7$$ (1, -4): $$-4=3(1)+b$$. $$-4=3+b$$. $$b=-4-3=-7$$ Now we have $$m=3$$ and $$b=-7$$, so the equation of the line is $$y=3x-7$$ ## Contact tutor Send a message explaining your needs and Jessie will reply soon. Contact Jessie", "equation (5) in equation (2), we find that • A + B = 2(6AI2) • Or, A + B = 12A – 2I2 • Or, 11AB = 2I2 (6) Using (1) and (6), we now find the value of A and B. Multiplying (3) by 3 and adding with (1), we find that • 3(11AB) + (2A + 3B) = 3(2I2) + I2 • Or, 33A – 3B + 2A + 3B = 6I2 + I2 • Or, 35A = 7I2 • Or, A = I2 / 5 Similarly, substituting the value of A in (1), we find that B = I2 / 5. Therefore, required solutions are", "complementary equation on $$(a,b)$$.", "of equations, $Ax = b$ has exactly one solution. $e^{\\sqrt{2}}> 3$ For any real number $c$, the polynomial $x^{3}+x+c$ has exactly one real root.", "3^+} 3x- a+ b$ to get a second equation.", "we plug 3 into b for a, it becomes $b = 3 + 5 \\mathbb Z$, $b \\in \\mathbb Z_7$. The only valid value there is {3}. That means that our valid values for b are {1,3,6}. It’s the union of the valid values we found for each value of a. That makes our equations into this: $x \\equiv b + 7 \\mathbb Z$ $b = [1,3,6]$ If we plug those values of b into the equation for x, we get these three equations: $x \\equiv 1 + 7 \\mathbb Z$ $x \\equiv 3 + 7 \\mathbb Z$ $x \\equiv 6 + 7 \\mathbb Z$ The first equation gives us some x values of {…, -6, 1, 8, 15, …}. The second equation gives us x values of {…, -4, 3, 10, 17, …}. The third equation gives us x values of {…, -1, 6, 13, 20, …}. Taking the union of all of those, we get… {…, -6, -4, -1, 1, 3, 6, 8, 10, 13, 15, 17, 20, …} If you plug those numbers into the original equation $((x \\% 7) \\% 5) \\%2 = 1$, you can see that they are valid solutions! When you are performing this algorithm, if you ever hit a case where there are no valid answers in one of the", "# 4sinx-3cosx = 7 find the number of general solutions of the given expressions. a sinx -b cosx = $\\sqrt$(a2+ b2 ) Maximum value is $\\sqrt$(9+16) = 5", "2 = b b = 2 So the answer is y = 2x + 2", "substituting $b = -3$ we also get $a = 2$. Therefore, the two solutions are $2 + 3i$ and $2 - 3i$.", "# That's hard...maybe The equation $x^3-3x^2-3x-1=0$ has exactly one real solution that can be written in the form $\\sqrt[3]{a}+\\sqrt[3]{b}+\\sqrt[3]{c}.$ What is the value of $a+b+c?$ ×" ]

[ "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "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.", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "7&6&5& 14&80&92\\\\ \\hline \\end{array}$$ Therefore, the number of such strings of length $14$ is $a_{14} + b_{14} = \\boxed{172}$. ### Solution 2 We replace \"14\" with \"$2n$\". We first note that we must have an even number of chunks of $B$'s, because of parity issues. We then note that every chunk of $B$'s except the last one must end in the sequence $BAA$, since we need at least two $A$'s to separate it from the next chunk of $B$'s. The last chunk of $B$'s must, of course, end with a $B$. Thus our sequence must look like this : $$\\boxed{\\quad A\\text{'s} \\quad} \\boxed{\\quad B\\text{'s} \\quad} BAA \\boxed{\\quad A\\text{'s} \\quad} \\cdots \\boxed{\\quad B\\text{'s} \\quad}BAA \\boxed{\\quad A\\text{'s} \\quad} \\boxed{\\quad B\\text{'s} \\quad} B \\boxed{\\quad A\\text{'s} \\quad} ,$$ where each box holds an even number of letters (possibly zero). If we want a sequence with $2k$ chunks of $B$'s, then we have $(2n - 6k + 2)$ letters with which to fill the boxes. Since each box must have an even number of letters, we may put the letters in the boxes in pairs. Then we have $(n - 3k + 1)$ pairs of letters to put into $4k + 1$ boxes. By a classic balls-and-bins argument, the number of ways to do this is $$\\binom{n + k + 1}{4k} .$$ It", "= \\boxed{\\textbf{(B)}\\ 24}.$", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "6&6&13&113\\\\ 7&11&14&172\\\\ \\hline \\end{array}$$ Therefore, the number of such sequences of length 14 is $\\boxed{172}$. ### Solution 2 We replace \"14\" with \"$2n$\". We first note that we must have an even number of chunks of $B$'s, because of parity issues. We then note that every chunk of $B$'s except the last one must end in the sequence $BAA$, since we need at least two $A$'s to separate it from the next chunk of $B$'s. The last chunk of $B$'s must, of course, end with a $B$. Thus our sequence must look like this : $$\\boxed{\\quad A\\text{'s} \\quad} \\boxed{\\quad B\\text{'s} \\quad} BAA \\boxed{\\quad A\\text{'s} \\quad} \\cdots \\boxed{\\quad B\\text{'s} \\quad}BAA \\boxed{\\quad A\\text{'s} \\quad} \\boxed{\\quad B\\text{'s} \\quad} B \\boxed{\\quad A\\text{'s} \\quad} ,$$ where each box holds an even number of letters (possibly zero). If we want a sequence with $2k$ chunks of $B$'s, then we have $(2n - 6k + 2)$ letters with which to fill the boxes. Since each box must have an even number of letters, we may put the letters in the boxes in pairs. Then we have $(n - 3k + 1)$ pairs of letters to put into $4k + 1$ boxes. By a classic balls-and-bins argument, the number of ways to do this is $$\\binom{n + k + 1}{4k} .$$ It follows that the total", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "$\\text{i.e.,}\\quad \\boxed{( {I+BA})^{-1}=I-{B}(I+{AB})^{-1} A}$", "if two distinct values of $n$ satisfy it, then all values of $n$ will. Matching coefficients, we need $$c = \\frac{a^2}{9} \\quad \\text{and} \\quad c - b = -\\frac{a^2}{9} \\implies b = \\frac{2a^2}{9}.$$To maximize $a + b + c = a + \\tfrac{a^2}{3}$, we need to maximize $a$. Since $b$ and $c$ must be integers, $a$ must be a multiple of 3. If $a = 9$ then $b$ exceeds 9. However, if $a = 6$ then $b = 8$ and $c = 4$ for an answer of $\\boxed{\\textbf{(D)} \\text{ 18}}$. (CantonMathGuy)", "that $X_1 X_2 X_3 X_4$ ends in 1, 3, 7, or 9 is $P(X_1, X_2, X_3 \\text{ and } X_4 \\text{ are not divisible by 2 or 5}) = (4/10)^4$.", "Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{\\textbf{(C) } 3}$ students that are taking all $3$ classes. ## Solution 3 (Algebra) Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\\boxed{\\textbf{(C) 3}}$.", "+ 10 \\:=\\:10x^3 + (5n-6)x^2 - (3n+10)x + 10$ Equate coefficients: . $\\begin{Bmatrix}10 \\:=\\:10 & [1] \\\\ m \\:=\\:5n-6 & [2] \\\\ \\text{-}1 \\:=\\:\\text{-}(3n+10) & [3] \\\\ 10 \\:=\\:10 & [4] \\end{Bmatrix}$ From [3]: . $\\text{-}1 \\:=\\:\\text{-}3n - 10 \\quad\\Rightarrow\\quad \\boxed{n \\:=\\:\\text{-3}}$ Substitute into [2]: . $m \\:=\\:5(\\text{-}3)-6 \\quad\\Rightarrow\\quad \\boxed{m \\:=\\:\\text{-}21}$ , , , , ### mx divided by x Click on a term to search for related topics.", "+0 0 45 1 Let a and b be the roots of the quadratic 2x^2 - 8x + 7 = x^2 - 5x + 3. Compute a^3*b + a*b^3. Dec 13, 2022 #1 +2541 -1 $$ab^3 + a^3b = ab(a^2 + b^2) = ab((a+b)^2 - 2ab)$$ $$x^2 - 3x +4 = 0$$ $$a + b = -{b \\over a} = 3$$ $$ab = {c\\over a} = 4$$ $$ab^3 + a^3b = ab((a+b)^2 - 2ab) = 4(9 - 8) = \\color{brown}\\boxed 4$$ . Dec 13, 2022", "$0 = a$, that means $x+32 = 1001$, and so $x = 1001 - 32 = 969$. So, $9$ + $6$ + $9$ = $\\boxed{\\textbf{(E)}\\ 24}$. ~songmath20 ## Video Solution ~Education, the Study of Everything", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "3^{3x+y} * 2^{3x+3y}= 3^2*2^3\\\\ so\\\\ 3x+y=2\\quad(1a)\\\\ 3x=2-y \\quad (1b)\\\\ 3x+3y=3\\quad(2a)\\\\ x+y=1\\quad(2b)\\\\ y=1-x\\quad(2c)\\\\ sub (2c) into (1b)\\\\ 3x=2-(1-x)\\\\ 3x=1+x\\\\ 2x=1\\\\ {\\bf{x=0.5,\\quad y=0.5}} \\\\ check\\\\ 2^1*3^0*6^{2}=72\\qquad \\text{Das ist ausgezeichnet!}$$ . Jan 17, 2019", "$$(a+b)^2 = a^2+ 2ab + b^2 \\quad | \\quad a+b=1 \\ and \\ a^2+b^2 = 2\\\\ 1^2 = 2 + 2ab \\\\ -1 = 2ab \\\\ ab = -\\frac{1}{2}$$ III. $$a^3+b^3 = 2-ab \\quad | \\quad ab = -\\frac{1}{2} \\\\ a^3+b^3 = 2 - ( -\\frac{1}{2} ) \\\\ a^3+b^3 = 2 + \\frac{1}{2} \\\\ \\boxed{a^3+b^3 = 2.5}$$ heureka Nov 24, 2014", "{\\text{and}}\\quad x=(8)^{1/3}=2.}$ ## Contents #αντικατάσταση , #αλλαγήμεταβλητής , #ορισμός", "is $$\\boxed{ y=x^2+bx+c }$$ To Complete the Square we add and subtract $$\\textcolor{blue}{({b \\over 2})^2}$$ which yields: $$\\boxed{ y=x^2+bx+ \\textcolor{blue}{({b \\over 2})^2} + c – \\textcolor{blue}{({b \\over 2})^2} }$$ Completing the Square yields: $$\\boxed{ y=(x+b/2)^2 + c – ({b \\over 2})^2 }$$ … Read more" ]

[ "+ z2 + 2xy - 2yz - 2zx Solution: Here, (a+b) = 2 $\\therefore$ (a+b)3 = 23 or, a3 + 3a2b + 3ab2 + b3 = 8 or, a3 + b3 + 3ab2 = 8 or, a3 + b3 + 6ab = 8 So, the required value of a3 + b3 + 6ab is 8. Solution: = $\\frac{12x^4}{3x^2}$ - $\\frac{15x^3}{3x^2}$ = 4x4-2 - 5x3-2 = 4x2 - 5x", "+0 Let $a$ and $b$ be the solutions of the equation $2x^2-10x+5=0$. What is the value of $(2a-3)(4b-6)$? +1 195 2 +11 Let $a$ and $b$ be the solutions of the equation $2x^2-10x+5=0$. What is the value of $(2a-3)(4b-6)$? Dec 24, 2021 #1 +1 The roots a, b are (5 \\pm sqrt(15))/2. Plugging these values in, we get an answer of -16. Dec 24, 2021 #2 +11 +1 Thanks guest greatly appreciated! <3 oTechz Dec 24, 2021", "# If 2a-8=4 and 3b-8=10, what is the value of a+b? Nov 3, 2016 $a + b = 12$ #### Explanation: Solve the first equation for $a$: $2 a - 8 = 4$ $2 a - 8 + 8 = 4 + 8$ $2 a = 12$ $2 \\frac{a}{2} = \\frac{12}{2}$ $a = 6$ Solve the second equation for $b$: $3 b - 8 = 10$ $3 b - 8 + 8 = 10 + 8$ $3 b = 18$ $3 \\frac{b}{3} = \\frac{18}{3}$ $b = 6$ Therefore: $a + b = 6 + 6$ $a + b = 12$", "# Do I look quadratic to you? The equation $| x - 10 | = 7$ is equivalent to the equation $x^2 + ax + b = 0.$ What is the value of $a+b$? ×", "-2(-1) + 10; 3 ⇒ b = 2 + 10; 3 On further solving we get: ⇒ b = 12; 3 ⇒ b = 4; So when we put value of “a” is -1 then we get value of b is 4. If we put the value of a is 1 then we get: Put a = 1; ⇒ b = -2a + 10; 3 ⇒ b = -2(1) + 10; 3 ⇒ b = -2 + 10; 3 On further solving we get: ⇒ b = 8; 3 ⇒ b = 2.66; So when we put value of “a” is 1 then we get value of b is 2.66. If we put the value of a is 2 then we get: Put a = 2; ⇒ b = -2a + 10; 3 ⇒ b = -2(2) + 10; 3 ⇒ b = -4 + 10; 3 On further solving we get: ⇒ b = 6; 3 ⇒ b = 2; So when we put value of “a” is 2 then we get value of b is 3. If we put the value of a is 2 then we get: Put a = 4; ⇒ b = -2a + 10; 3 ⇒ b = -2(4) + 10; 3 ⇒ b = -8 + 10; 3 On further solving", "The values of a and b for which the quadratic equation $$x^{2}+2x+2+e^{-2a}-cosb=0$$ has a real solution is", "## Calculus (3rd Edition) $p(x)=x^2+2x-3$. We have $$p(x)=x^2+ax+b, \\quad p'(x)= 2x+a.$$ Since, $p(1)=0$ and $p'(1)=4$, we get the equations $$a+b=-1, \\quad a=2$$ Hence we have the solution $a=2, b=-3$. Then, $p(x)=x^2+2x-3$.", "find a value for B so that the equality is true: let A = 2 then 2B = -A 2B = -2 B = -1 or if we let A be 7 2B = -7 B = -3.5 so there are infinitely many solutions that satisfy A + 2B = 0 likewise there are infinitely many solutions which satisfy 6A - 7C = 9", "are three multiples of 3 namely 0, 3, and 6. Thus, $B=7-3=4$ 3. $A+B=10+4=14$.", "Basic College Mathematics (9th Edition) $b=-4$ We are given the equation $-5=2b+3$. We can solve for b by first subtracting 3 from both sides. $-5-3=2b+3-3$ $2b=-8$ Divide both sides by 2. $\\frac{2b}{2}=\\frac{-8}{2}$ Therefore, $b=-4$.", "is $\\boxed{\\textbf{(D) } 2+\\sqrt{5}}$ [SOLUTION 2] Draw line FG through E, with F on BC and G on AD, FG//AB. WOLG let CD=1, CB=x, AB=x^2. By weighted average FG=(1+x^4)/(1+x^2). Meanwhile, FE:EG=[CBE]:[ADE]=1:17. FE=x^2/(1+x^2). We obtain (1+x^4)/(1+x^2)=18x^2/(1+x^2), Namely x^4-18x^2+1=0. The rest is the same with solution 1.", "Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{\\textbf{(C) } 3}$ students that are taking all $3$ classes. ## Solution 3 (Algebra) Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\\boxed{\\textbf{(C) 3}}$.", "+0 # Any help would be appreciated! Thanks! 0 28 3 The two solutions of the equation $$x^2+bx+18=0$$ are in the ratio of $$2$$ to $$1$$ for some values of $$b.$$ What is the largest possible value of $$b?$$ Oct 17, 2020 #1 0 The largest possible value of b is -9. Oct 17, 2020 #2 +10322 +1 The two solutions of the equation $$x^2+bx+18=0$$ are in the ratio of 2 to 1 for some values of b. What is the largest possible value of b. Hello Guest! $$x=-\\frac{b}{2}\\pm \\sqrt{(\\frac{b}{2})^2-18}\\\\ \\$$ b = - 9 : x1 = 6 x2 = 3 x2 : x1 = 1 : 2 b= 9 : x1 = - 6 x2 = - 3 x2 : x1 = 1 : 2 The largest possible value of b is 9 (for x1: x2 = 1: 2). $$b\\in \\mathbb R\\ |b\\le-6\\sqrt{2}\\ and\\ 6\\sqrt{2}\\le b$$ ! Oct 17, 2020 #3 0 Thanks asinus! You were right! Guest Oct 18, 2020", "solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$ $\\textbf{Case 1:}$ There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have $$(a-2)^2 = 4(4 + b)\\Longrightarrow a = 2\\pm2\\sqrt{4+b}.$$ Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a \"$\\pm$\" there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula). So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this", "any point on the line will give you the same value for b. If you want to double check your math, try both. (3, 2): $$2=3(3)+b$$. $$2=9+b$$. $$b=2-9=-7$$ (1, -4): $$-4=3(1)+b$$. $$-4=3+b$$. $$b=-4-3=-7$$ Now we have $$m=3$$ and $$b=-7$$, so the equation of the line is $$y=3x-7$$ ## Contact tutor Send a message explaining your needs and Jessie will reply soon. Contact Jessie", "from the second to get (4a+ 2b+ c)- (a+ b+ c)= 3a+ b= 6- 2= 4 or 3a+ b= 4. c has been eliminated. Subtract the first equation from the third to get (8a+ 3b+ c)- (a+ b+ c)= 7a+ 2b= 12- 2= 10 or 7a+ 2b= 10. Again c has been eliminated. Now we have the two equations 3a+ b= 4 and 7a+ 2b= 10. Multiply 3a+ b= 4 by 2 to get 6a+ 2b= 8 and subtract that from 7a+ 2b= 10: (7a+ 2b)- (6a+ 2b)= a= 10- 8= 2. a= 2, not \"2a= a\". Now that you know a= 2, use either 3a+ b= 4 or 7a+ 2b= 10 to determine b. Then use any of the original 3 equations to determine c. As a check, see if those values for a, b, and c give $T_4= 20$. 6. Hi HallsofIvy, you have a typo there....(9a+3b+c)-(a+b+c) We end up with a=1, b=1, c=0 leading to skeeter's formula.", "one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write $$3a=(a+b)a=a^2+ab=a^2-10$$ which is the original equation (with a sign reversal) $$a^2-3a-10=0.$$ To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots. So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.", "( b + \\frac{6}{b} \\right ) + b = -3$ Now, if we are going to have integer values for b, then from bd = -6 we know that b can only b $\\pm 1, \\pm 2, \\pm 3, \\pm 6$. From this I get solutions a = 1, b = 2 a = 0, b = -3 So $x^4+x^3-x^2-3x-6 = (x^2 + x + 2)(x^2 - 3)$ -Dan", "at x =2: 3a(2)^2 = 2(2) => 12a = 4 => a = 1/3. Substitute this value of a into equation (A) to get b.", "= b^2 a^2 + 2ac =b^2 Therefore, the value of b^2 = a^2 + 2ac. Is there an error in this question or solution? #### APPEARS IN Solution If Sin α and Cos α Are the Roots of the Equations Ax2 + Bx + C = 0, Then B2 = Concept: Solutions of Quadratic Equations by Factorization. S" ]

[ "# 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", "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);", "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", "# 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$.", "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 =", "Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$. Case 3: $g$ passes through $CD$ or $BC$ $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,N); dot(XB); label(\"X_2\",XB,E); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$. By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\\angle EAB = \\angle AED = \\angle CEX_1$. Thus, by AA Similarity, $\\triangle X_2BA \\sim \\triangle X_3DE \\sim \\triangle X_1CE$. Using similar triangle ratios, we have $DX_3 = \\tfrac{bx}{a}$ and $X_1 = \\tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \\frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle 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", "= (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.", "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, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "$\\triangle DCE$ 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 $\\frac{1}{3}$.", "= \\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),", "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", "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 =", "# 2021 AIME I Problems/Problem 2 ## Problem In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, W); label(\"B\", B, W); label(\"C\", C, (1,0)); label(\"D\", D, (1,0)); label(\"F\", F, N); label(\"E\", E, S); [/asy]$ ## Solution 1 (Similar Triangles) Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\\frac{7}{3}(11-x)$ and $CG=\\frac{3}{7}x$. We have $\\frac{7}{3}(11-x)+\\frac{3}{7}x=FC=9$. Solving for $x$, we have $x=\\frac{35}{4}$. The area of the shaded region is just $3\\cdot \\frac{35}{4}=\\frac{105}{4}$. Thus, the answer is $105+4=\\framebox{109}$. ~yuanyuanC ## Solution 2 (Similar Triangles) Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\\triangle AFG \\sim \\triangle", "+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", "= \\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)", "+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" ]

[ "In triangle ABC, $\\angle ABC=95^\\circ$ and $\\angle ACB=50^\\circ$. Point P is constructed inside the triangle such that $BP=CP$ and $\\frac{[BPC]}{[ABC]}=\\frac{1}{5}$. Find the integer closest to $\\angle BPC$.", "of the triangle and state how many triangles could be formed with the given values. 1. $A = 30^\\circ, a = 13, b = 15$ 2. $A = 22^\\circ, a = 21, b = 12$ 3. $A = 42^\\circ, a = 36, b = 37$ For 4-7, find all possible measures of $\\angle B$ (if any exist) for each of the following triangle values. 4. $A = 86^\\circ, a =15, b = 11$ 5. $A = 30^\\circ, a = 24, b = 43$ 6. $A = 48^\\circ, a = 34, b = 39$ 7. $A = 80^\\circ, a = 22, b = 20$ For 8-12, find the length of $b$ for each of the following triangle values. 8. $A = 94^\\circ, a = 31, B = 34^\\circ$ 9. $A = 112^\\circ, a = 12, B = 15^\\circ$ 10. $A = 78^\\circ, a = 20, B = 16^\\circ$ 11. $A = 54^\\circ, a = 15, B = 112^\\circ$ 12. $A = 39^\\circ, a = 9, B = 98^\\circ$ 13. In $\\Delta ABC, b=10$ and $\\angle A = 39^\\circ$ . What's a possible value for $a$ that would produce two triangles? 14. In $\\Delta ABC, b=10$ and $\\angle A = 39^\\circ$ . What's a possible value for $a$ that would produce no triangles? 15. In $\\Delta ABC, b=10$ and $\\angle", "## Trigonometry 7th Edition B = $20^{\\circ}$ in triangle ABC- Given data is- A = $80^{\\circ}$ C = α + β = $80^{\\circ}$ As we know that all the three angles of a triangle add up to 180 degrees, therefore- A + B + C = $180^{\\circ}$ B = $180^{\\circ}$ - $(C+A)^{\\circ}$ B = $180^{\\circ}$ - $(80+80)^{\\circ}$ B = $180^{\\circ}$ - $160^{\\circ}$ B = $20^{\\circ}$", "+0 # geometry help :) 0 89 1 In right triangle $ABC,$ $\\angle C = 90^\\circ.$ Median $\\overline{AM}$ has a length of $19,$ and median $\\overline{BN}$ has a length of $18.$ What is the length of the hypotenuse of the triangle? Jun 19, 2022", "## Trigonometry (11th Edition) Clone $sin~C = 1$ $C = 90^{\\circ}$ Triangle ABC is a $30^{\\circ},60^{\\circ},90^{\\circ}$ triangle. We can use the law of sines to find the angle $C$: $\\frac{a}{sin~A} = \\frac{c}{sin~C}$ $sin~C = \\frac{c~sin~A}{a}$ $sin~C = \\frac{(2\\sqrt{5})~sin~(30^{\\circ})}{\\sqrt{5}}$ $sin~C = 1$ $C = arcsin(1)$ $C = 90^{\\circ}$ Triangle ABC is a $30^{\\circ},60^{\\circ},90^{\\circ}$ triangle.", "+0 trig +1 35 2 +491 In triangle $ABC$, we have $\\angle A = 90^\\circ$ and $\\sin B = \\frac{4}{7}$. Find $\\cos C$. Apr 27, 2021 #1 +113121 0 Draw the triangle and use Pythagoras's theorem Apr 27, 2021", "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", "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", "Answer Comment Share Q) # In a triangle $ABC,C=90^{\\circ}$.Then $\\large\\frac{a^2-b^2}{a^2+b^2}=$ $\\begin {array} {1 1} (1)\\;\\sin (A+B) & \\quad (2)\\;\\sin (A-B) \\\\ (3)\\;\\cos (A+B) & \\quad (4)\\;\\cos (A-B) \\end {array}$", "# Some Amazing Triangle Geometry Level pending In $$\\triangle ABC$$, we have $$AB > AC$$ and $$\\angle A = 60^{\\circ}$$. Let $$I$$ and $$H$$ denote the incenter and orthocenter of the triangle. What is $$\\angle AHI$$? ×", "+0 # HELP! THANKS! 0 308 1 In triangle ABC, AC = BC and angle C = 90. Points P and Q are on AB} such that QCP = 45. Show that AP^2 + BQ^2 = PQ^2. Hint: Rotate the diagram counter-clockwise around C by 90 degrees. Let P go to point P'. What is angle P'CQ? Feb 25, 2021 $AP=BQ\\implies AP=PQ\\sqrt{2}$ is desired. Follows trivially from LoC and coordinates as $\\angle ACP=\\angle BCP=45^{\\circ}$. Can't think of anything synthetic...", "1. ## Triangle Maths. ABC is a triangle. Find <y and <x. THANK YOU!!! 2. Surely you know that the angles in a triangle add up to $\\displaystyle 180^{\\circ}$. Can you figure out $\\displaystyle \\angle x$ for starters? 3. <x= 180-70-60 <x= 50\"", "angled at A) $\\angle ABC$ + $\\angle ACB$ = $90^{\\circ}$ Using the same logic for $\\triangle PQR$, we can say that : $\\angle PQR$ + $\\angle PRQ$ + $\\angle QPR$ = $180^{\\circ}$ $\\angle ABC$ + $\\angle ACB$ + $\\angle QPR$ =$180^{\\circ}$ ($\\angle ABC$ =$\\angle PRQ$ and $\\angle QCA$ =$\\angle CQP$ ) Or, $90^{\\circ}$+ $\\angle QPR$ =$180^{\\circ}$ ($\\angle ABC$+ $\\angle ACB$ = $90^{\\circ}$ ) $\\angle QPR=90^{\\circ}$", "$\\displaystyle { 90 }^{ o }$ • C. $\\displaystyle { 80 }^{ o }$ • D. $\\displaystyle { 70 }^{ o }$ Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 23rd 09, 2020 Q5 Subjective Medium In the given figure, the side $BC$ of $\\triangle ABC$ has been produced on the left-hand side from $B$ to $D$ and on the right -hand side from $C$ to $E$. if $\\angle ABD = 106^{\\circ}$ and $\\angle ACE = 118^{\\circ}$, find the measure of each angle of the triangle. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 23rd 09, 2020", "Question In a $\\Delta ABC,\\text{ }{{\\cos }^{2}}A+{{\\cos }^{2}}B+{{\\cos }^{2}}C=1$. Prove that the triangle is right-angled. Hint: Assume a right-angled triangle ABC. Take the cos of each angle of $\\Delta ABC$ and square them separately. Add them to get ${{\\cos }^{2}}A+{{\\cos }^{2}}B+{{\\cos }^{2}}C=1$. Since we are already given this result, this means that our assumption is correct. Hence solve the question by this approach. Here we are given that in $\\Delta ABC,\\text{ }{{\\cos }^{2}}A+{{\\cos }^{2}}B+{{\\cos }^{2}}C=1$. We have to prove that triangle ABC is a right-angled triangle. Let us assume a $\\Delta ABC$ such that it is right-angled at B. Let us take $\\angle A=\\theta$, we know that in any triangle, the sum of 3 angles is ${{180}^{o}}$. So, we get, $\\angle A+\\angle B+\\angle C={{180}^{o}}$ By substituting the value of $\\angle A=\\theta$ and $\\angle B={{90}^{o}}$, we get, $\\theta +{{90}^{o}}+\\angle C={{180}^{o}}$ $\\Rightarrow$ $\\angle C={{180}^{o}}-\\theta -{{90}^{o}}$ $\\Rightarrow$ $\\angle C={{90}^{o}}-\\theta$ Now, we know that $\\angle A=\\theta$. By taking cos on both sides, of the above equation, we get, $\\cos \\left( A \\right)=\\cos \\theta$ By squaring both sides of the above equation, we get $\\Rightarrow$ ${{\\cos }^{2}}\\left( A \\right)={{\\cos }^{2}}\\theta ....\\left( i \\right)$ Now, we also know that $\\angle B={{90}^{o}}$. By taking cos on both sides of the above equation, we get $\\Rightarrow$ $\\cos B=\\cos {{90}^{o}}$ By squaring both sides of the above", "Therefore angle BCA = angle FGC = $80^{\\circ}$ (corresponding angles). Consider triangle ABC: $x + 70 + 80 = 180$ so $x=30$.", "the triangle ABC { ( x-1 ) ^2 } { 9 +\\frac. Angles in Circles ( central angles, acute, obtuse or reflex type... Display, now press dice for a new problem to accurately calculate the height allows manipulation and investigation of types...", "# Perimeter of triangle In triangle ABC angle A is 60° angle B is 90° side size c is 15 cm. Calculate the triangle circumference. Correct result: o = 70.981 cm #### Solution: $c=15 \\ \\text{cm} \\ \\\\ A=60 \\ ^\\circ \\ \\\\ B=90 \\ ^\\circ \\ \\\\ C=180-A-B=180-60-90=30 \\ ^\\circ \\ \\\\ a=c \\cdot \\ \\tan A ^\\circ =c \\cdot \\ \\tan 60^\\circ \\ =15 \\cdot \\ \\tan 60^\\circ \\ =15 \\cdot \\ 1.732051=25.98076 \\ \\\\ b=\\sqrt{ a^2+c^2 }=\\sqrt{ 25.9808^2+15^2 }=30 \\ \\text{cm} \\ \\\\ o=a+b+c=25.9808+30+15=70.981 \\ \\text{cm}$ 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! Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors. Tips to related online calculators Pythagorean theorem is the base for the right triangle 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: • Euclid2 In right triangle ABC with right angle at C is given side a=27 and", "= 180^\\circ$ Solving for the angle ABC, we have: $\\angle ABC = 180^\\circ - \\dfrac{1}{2}\\angle AOC$ $2\\angle ABC = 360^\\circ - \\angle AOC$ Angle ABC is given as angle p and angle (360° – AOC) is angle q. $2p = q$ Hence, the correct answer is option (a). Note: You can also directly use the property of angle subtended by a chord to find the relation between q and p but here the angles are different than usual.", "+0 # Trig 0 75 1 In triangle ABC, we have angle A = 90 degrees and sin B = (4/7). Find cos C. (No diagram was given) Sep 11, 2018 #1 +3617 +1 as we know a triangle has 180 degrees we know that angles B and C must sum to 90. so we have $$\\cos(C) = \\cos(90^\\circ - B) = \\\\ \\cos(90^\\circ)\\cos(B) + \\sin(90^\\circ)\\sin(B) = \\\\ \\sin(B) = \\dfrac 4 7$$ . Sep 11, 2018" ]

[ "# 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", "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);", "# 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 $B$.", "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}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "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", "+ .75*dir(360-65-115-55-30)); label(\"65^\\circ\",B + .75*dir(180-32.5)); label(\"x^\\circ\",A + .5*dir(42.5)); label(\"5^\\circ\",D + 2.5*dir(360-60-2.5)); label(\"60^\\circ\",D + .75*dir(360-30)); label(\"60^\\circ\",E + .5*dir(360-150)); label(\"5^\\circ\",B + 2.5*dir(180-65-2.5)); [/asy]$ Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles. ~Someonenumber011", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "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 =", "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, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "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", "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", "# 2021 AIME I Problems/Problem 2 ## Problem In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, W); label(\"B\", B, W); label(\"C\", C, (1,0)); label(\"D\", D, (1,0)); label(\"F\", F, N); label(\"E\", E, S); [/asy]$ ## Solution 1 (Similar Triangles) Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\\frac{7}{3}(11-x)$ and $CG=\\frac{3}{7}x$. We have $\\frac{7}{3}(11-x)+\\frac{3}{7}x=FC=9$. Solving for $x$, we have $x=\\frac{35}{4}$. The area of the shaded region is just $3\\cdot \\frac{35}{4}=\\frac{105}{4}$. Thus, the answer is $105+4=\\framebox{109}$. ~yuanyuanC ## Solution 2 (Similar Triangles) Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\\triangle AFG \\sim \\triangle", "# 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 $\\angle DCE = 180^\\circ - 90^\\circ - 60^\\circ = 30^\\circ$,", "$ABC$, 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", "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 =", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "$\\triangle 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}$.", "# 2019 AIME II Problems/Problem 11 ## Problem Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\\omega_1$ and $\\omega_2$ not equal to $A.$ Then $AK=\\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ ## Solution 1 $[asy] unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label(\"A\",A,dir(105)); label(\"B\",B,dir(-135)); label(\"C\",C,dir(-75)); dot((2.68,2.25)); label(\"K\",(2.68,2.25),dir(-150)); label(\"\\omega_1\",(-6,1)); label(\"\\omega_2\",(14,6)); label(\"7\",(A+B)/2,dir(140)); label(\"8\",(B+C)/2,dir(-90)); label(\"9\",(A+C)/2,dir(60)); [/asy]$ -Diagram by Brendanb4321 Note that from the tangency condition that the supplement of $\\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\\angle AKB$ and $\\angle AKC$, respectively, so from tangent-chord, $$\\angle AKC=\\angle AKB=180^{\\circ}-\\angle BAC$$ Also note that $\\angle ABK=\\angle KAC$, so $\\triangle AKB\\sim \\triangle CKA$. Using similarity ratios, we can easily find $$AK^2=BK*KC$$ However, since $AB=7$ and $CA=9$, we can use similarity ratios to get $$BK=\\frac{7}{9}AK, CK=\\frac{9}{7}AK$$ • Now we use Law of Cosines on $\\triangle AKB$: From reverse Law of Cosines, $\\cos{\\angle BAC}=\\frac{11}{21}\\implies \\cos{(180^{\\circ}-\\angle BAC)}=-\\frac{11}{21}$. This gives us $$AK^2+\\frac{49}{81}AK^2+\\frac{22}{27}AK^2=49$$ $$\\implies \\frac{196}{81}AK^2=49$$ $$AK=\\frac{9}{2}$$ so our answer is $9+2=\\boxed{011}$. -franchester The motivation for using the Law of Cosines (\"LoC\") is after finding the", "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.)" ]

[ "= 8^2$$ and the solution is $$x = 8$$.", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "and $x = 8$ or $x = 0 , 8$", "to solve $$|x-8|\\leq0$$ which gives $$x=8.$$", "and w And (b) the value of w when x =8", "$x^2=8y$", "and get the value of x i.e $c$", "# How do you find the quotient a^8/a^8? ${a}^{8} / {a}^{8} = 1$ for every value of $a$ other than $0$.", "equation.] x - 8 = 0 or x - 11 = 0 [Apply zero product property.] x = 8 or x = 11. The values of x are 8, 11.", "x = 8 \\mathmr{and} x = - 2$", "order to find the values of the $x_i$'s.", "8(x – 11) = 0 X = 11 or x = 8", "+8x - 9}{2\\sqrt{x+2}}$", "# Q. 3. Find the value of the following expressions, when $x=-1$: (i) $2x-7$ $\\\\2x - 7 \\\\= 2 \\times ( -1 ) - 7 \\\\= -2 - 7 \\\\= -9$", "find that x = 7.", "Cramers's rule find the value for x. $$x+3y+4z=3$$ $$2z+6y+9z=5$$ $$3x+y-2z=7$$ $$x-y=2$$ $$x+y=8$$", "# What are the excluded values for y=8/(x-8)? The only excluded value is $x = 8$. As we can not divide by zero, the only excluded value is $x - 8 = 0$ i.e. $x = 8$.", "factored out the 2). Hence $x = 6, x = -8$. Now, plug those values in for $y = x+2$. Finally, plug in both sets of values you found for $x$ and $y$ to the first equation to make sure it is correct. So therefore the solutions would be: (6, 8) & (-8, -6). Thanks so much!", "# How do you find the quotient a^8/a^8? Apr 30, 2017 ${a}^{8} / {a}^{8} = 1$ for every value of $a$ other than $0$.", "Q # Find the value of the following expressions when x = - 1, (i) 2x - 7 Q. 3. Find the value of the following expressions, when $x=-1$: (i) $2x-7$ $\\\\2x - 7 \\\\= 2 \\times ( -1 ) - 7 \\\\= -2 - 7 \\\\= -9$" ]

[ "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "- \\sqrt{x-8} = \\sqrt{8}$ square both sides ... $(x+8) - 2\\sqrt{x+8}\\sqrt{x-8} + (x-8) = 8$ combine like terms and radicals on the left ... $2x - 2\\sqrt{x^2-64} = 8$ divide all terms by 2 ... $x - \\sqrt{x^2-64} = 4$ rearrange equation to move radical to the right side ... $x - 4 = \\sqrt{x^2 - 64}$ square both sides ... $x^2 - 8x + 16 = x^2 - 64$ combine like terms ... $-8x = -80$ $x = 10$ 8. Thank you all!", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "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", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "\\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}$.", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "another pair of the equations we have, AO =BO sqrt((5 - x)^2 + (-8 - y)^2) = sqrt((2 - x)^2 + (-9-y)^2) Squaring on both sides of the equation we have, (5 - x)^2 + (-8 - y)^2 = (2 - x)^2 + (-9 - y)^2 25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 - 4x + 81 + y^2 + 18y 6x + 2y = 4 3x + y = 2 Equating another pair of the equations we have, AO = CO sqrt((5 -x)^2 + (-8 - y)^2) = sqrt((2 -x)^2 + (1 - y)^2) 25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 -4x + 1 + y^2 - 2y` 6x - 8y = 84 x - 3y = 14 Now we have two equations for ‘x’ and ‘y’, which are 3x + y = 2 x - 3y = 14 From the second equation we have y = -3x + 2. Substituting this value of ‘y’ in the first equation we have, x - 3(-3x + 2) = 14 x + 9x - 6 = 14 10x = 20 x = 2 Therefore the value of ‘y’ is, y = -3x + 2 = -3(2) + 2 y = -4 Hence the", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "got a= 1, b=-28 $\\boxed {\\therefore p(x)=x^4+2x^3+x^2-28x-60 }$ continue....", "z = 8 and xy + yz + zx = 20 Consider, x + y + z = 8 Squaring on both sides, we get (x + y + z)2 = 82 x2 + y2 + z2 +2(xy + yz +zx) = 64 ⇒ x2 + y2 + z2 = 64 − 2(xy + yz +zx) = 64 − 2(20) = 64 − 40 = 24 ∴ x2 + y2 + z2 = 24 We know that x3 + y3 + z3 − 3xyz = (x + y + z)( x2 + y2 + z2 − xy − yz − zx) = 8(24 −20) = 8(4) = 32 Syeda Answer. Given x + y + z = 8 and xy + yz + zx = 20 Consider, x + y + z = 8 Squaring on both sides, we get (x + y + z)2 = 82 x2 + y2 + z2 +2(xy + yz +zx) = 64 ⇒ x2 + y2 + z2 = 64 − 2(xy + yz +zx) = 64 − 2(20) = 64 − 40 = 24 ∴ x2 + y2 + z2 = 24 We know that x3 + y3 + z3 − 3xyz = (x + y + z)( x2 + y2 + z2 − xy − yz − zx) = 8(24", "# What is x in the equation 5x^2 = 49? Divide both sides by $5$: ${x}^{2} = \\frac{49}{5} = 9.8$ Divide both sides by $5$: $x = \\sqrt{9.8} \\cong 3.13$ Note that technically this should be $\\pm 3.13$, since squaring either will yield $9.8$.", "can easily get$n = 2$and$f_n \\left( x \\right) = 1$and$f\\left( 5 \\right) = \\boxed{625}$.", "\\\\ \\text{Dividing both sides by 2}, \\\\ x= \\boxed{\\mathbf{3}}$$", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\\sqrt {25x^2-48x+36}$. Now, we can square both sides and simplify to get $0 = 72 - 20 \\sqrt{25x^2-48x+36}$. Dividing both sides by $4$, we get $18 - 5 \\sqrt {25x^2-48x+36}$ = $0$. We then add $5 \\sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \\sqrt {25x^2-48x+36}$. Since this is very messy, let $25x^2 - 48x = y$. Squaring both sides, we get $324 = 25y + 900, 25y = -576$. Solving for $y$, we have $y = -23.04$. Plugging in $y$ as $25x^2-48x$, we have $25x^2-48x+23.04 = 0$. Using the quadratic equation, we get $\\frac {48+0}{50}$. Therefore, $x = \\frac {48}{50}$. Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$. Since $5x$ is the hypotenuse, our answer is $\\boxed {(B)4.8}$ ~Arcticturn", "values of $x$ satisfy the original equation, $\\tfrac{a}{b} = \\tfrac54 + 1 = \\tfrac94,$ so $a+b = \\boxed{13}.$", "251$, then $1004 \\mid n(n+1)$, so our answer is $\\boxed{251}$. It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above." ]

[ "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "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.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "a^2$$. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$. ~IceMatrix", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "values of $x$ satisfy the original equation, $\\tfrac{a}{b} = \\tfrac54 + 1 = \\tfrac94,$ so $a+b = \\boxed{13}.$", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "the point (2,8). What is the value of h? Kathryn M. If f(x) = $$\\sqrt{3x + 4}$$ and k(x) = 8x + h, then f(k(x)) = $$\\sqrt{3(8x+h) + 4}$$, which is f(k(x)) = $$\\sqrt{24x+3h + 4}$$. With y= f(k(x), this must intersect the point (2,8), so 8 will equal f(k(x)), and the x value will be 2. Thus, 8 = $$\\sqrt{24*2+3h + 4}$$ = $$\\sqrt{48+3h + 4}$$. Squaring both sides to remove the square root yields 64 = 48 +3h +4. Combining the integers on the right brings 64 = 52 + 3h. We can then subtract 52 from both sides to get 12 = 3h. Now, we divide both sides by 3 to get 4 = h. Our final answer is that h has a value of 4. Send a message explaining your needs and Kathryn will reply soon. Contact Kathryn", "the value of $4x^{3}+2x^{2}-8x+7$ ⇒ 2x = $\\sqrt{3}+1$ 2x – 1 = √3 Squaring both the sides, we will get, $(2x-1)^{2}=3$ ⇒ $4x^{2}-4x+1=3$ ⇒ $4x^{2}-4x+1-3=0$ ⇒ $4x^{2}-4x-2=0$ ⇒ $2x^{2}-2x-1=0$ Now, let us take, $4x^{3}+2x^{2}-8x+7$ ⇒ 2x($2x^{2}-2x-1$)+ $4x^{2}+2x+2x^{2}-8x+7$ ⇒ 2x($2x^{2}-2x-1$)+ $6x^{2}-6x+7$ Since, $2x^{2}-2x-1=0$ ⇒ 2x(0)+3($2x^{2}-2x-1$))+7+3 ⇒ 0+3(0)+10 ⇒ 10", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "# $x=\\sqrt{(\\{56)-\\sqrt{([56)-\\sqrt{((56)...\\infty)]}\\}}}$. Find the value of X. 103 views in Aptitude closed $x=\\sqrt{(\\{56)-\\sqrt{([56)-\\sqrt{((56)...\\infty)]}\\}}}$. Find the value of X. 1. 5 2. 6 3. 7 4. 8 by (30.0k points) selected by Correct Answer - Option 3 : 7 Given, The expression $x=√{(\\{56)-√{([56)-√{((56)...\\infty)]}\\}}}$ Calculation $x=√{(\\{56)-√{([56)-√{((56)...\\infty)]}\\}}}$ ⇒ x = √ 56 - x (let the x remaining expression) Squaring both sides, x2 = 56 - x ⇒ x+ x - 56 = 0 ⇒ x2 + 8x - 7x - 56 = 0 ⇒ x (x + 8) - 7(x + 8) = 0 ⇒ x = - 8, 7 ∴ The value of x = 7", "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}$.", "can easily get$n = 2$and$f_n \\left( x \\right) = 1$and$f\\left( 5 \\right) = \\boxed{625}$.", "### Home > CC4 > Chapter 8 > Lesson 8.1.4 > Problem8-61 8-61. . Evaluate each expression below. Homework Help ✎ 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 }$", "- e^{x_i}) = e - e^0 = \\boxed{e-1}.$ Done!", "$\\sqrt{65}$, by truncating the continued fraction after a few terms. For example, [8; 16] = 8+1/16 = 129/16 = 8.0625 [8; 16, 16] = 8+1/(16+1/16) = 8+16/257 = 2072/257 ~~ 8.0622568 Actually $\\sqrt{65}$ is closer to $8.06225774829854965236$", "(More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$. This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$. Doing so, we get: $$(x + y)^2 = (x^2 + y^2) + 2xy$$ We see that in this form, we can substitute everything in except for $(x^2 + y^2)$, which is the desired answer. Substituting, we get: $$256 = (x^2 + y^2) + 110$$ so $x^2 + y^2 = \\boxed{146}$. (If we were to go with the pair $(55, 16)$, then the $(x + y)^2$ would be absurdly out of bounds)", "## anonymous 5 years ago The given equation involves a power of the variable. Find all real solutions of the equation. 8x^2-64=0 1. slaaibak 8(x^2 - 8)=0 x=2sqrt(2) or x = -2sqrt(2) 2. anonymous okay. can you show me what you did there? i'm having real trouble understanding this 3. slaaibak you first take out the common factor, which is 8: 8(x^2 - 8) = 0 then you use the difference of squares principle: 8(x-sqrt(8))(x+sqrt(8))=0 then x = sqrt(8) or x= -sqrt(8) simplified it's x=2sqrt(2) or x = -2sqrt(2) 4. anonymous $8x^2-64=0$ add 64 to both sides $8x^2=64$ divide both sides by 8 $x^2=8$ take the square root $x=\\pm\\sqrt{8}=\\pm2\\sqrt{2}$ 5. anonymous i think it's coming to me... i get where you get the common factor, then the difference of squares is giving me trouble... but i see what you are doing... 6. anonymous and since $8=4\\times 2$ $\\sqrt{8}=\\sqrt{4}\\times \\sqrt{2}=2\\sqrt{2}$ 7. anonymous how do you know the 2(sqrt)2?? that is really throwing me for a loop. 8. anonymous why isn't the square root of 8 just 2?" ]

[ "value of the gasoline in the tank? • Honey Ila collected the honey from 3 of her beehives. From the first hive, she collected 2/3 gallons of honey. The last two hives yielded 1/4 gallon each. After using some of the honey she collected for baking, Lila found that she only had 3/4 gallon of honey l", "+ 750 ml = 6 l 250 ml Question 8. In a fuel station there was 70 l 500 ml of fuel. How much amount of fuel will be left after selling 35 l 700 ml of fuel? Fuel left = 34 l 800 ml Question 9. A pot contains 9 l 500 ml of water, how much amount of water will 7 such pots contain?", "fraction of the water.", "of a gallon. She used half of this amount to sweeten tea. How much honey, in cups was used in the tea. Express your answers in cups.", "for making 3 and 2/3 cups of distilled water with 1 and 3/5 cups of vinegar and 2/3 cups of liquid detergent how much liquid in all and cups is needed?", "many guests are going to be at the party? 17. Jill's Jill's mom stored some of the honey in a container that held 3/4 of a gallon. She used half of this amount to sweeten tea. How much honey, in cups was used in the tea. Express your answers in cups.", "the problem requires multiply and when it requires divide. • . . U • The book had 27 pages. (Jimmy read 4 pages on Monday, 9 pages on Tuesday and there were 14 more pages to go.🤔 • This isn't a homework question so can anyone figure out the answer to this question? 1/3 gallon milk poured into 5 equal sized cups. What fraction of a gallon of milk is in each cup. (1 vote) • 1/15 gallons of milk is in each of the five cups, because 1/3 divided by 5 is equal to 1/15. • Sal said, \"She used a third of her detergent and has a third left\". She actually used a 1/2 of her detergent. • Sal said she used a third CUP of her detergent, which is equal to a half of her detergent before she used it. As in, a half of 2/3 cups is 1/3 cup. • Isn’t it meant to halve it? How is 1/3 if you multiply 1/2 of 2/3? I still don’t get it (1 vote) ## Video transcript Gina had 2/3 cups of laundry detergent. She used half of the detergent on Friday to wash all of her sheets. How much detergent does she have left? So we're essentially going to take half of 2/3, or 1/2", "keeping fractional amounts of water is possible, here's my solution (not sure if it's the most efficient). Here I have assumed that one trip constitutes movement in any one direction, forward or backward. To be able to successfully complete this trip, we need to reach a situation where we have 10 gallons at the 10 mile mark. Start with 10 gallons and carry it till 3.33 miles. 3.33 gallons are used up, cache 3.33 gallons and use the remaining 3.33 gallons to return to your origin. Repeat this 12 more times, till 43.33 gallons are collected at the 3.33 mile mark and you are back at the infinite well. (26 trips done so far) In the 27th trip, when you reach the 3.33 mile mark, you have a total of 50 gallons now: 43.33 collected plus 6.66 carried. Refill your container to the full 10 gallons and move to the 6.66 mile mark. 3.33 gallons are used up, cache 3.33 gallons and use the remaining 3.33 gallons to return to the 3.33 mile mark. Repeat this 3 more times till 13.33 gallons are collected at the 6.66 mile mark and you are back at the 3.33 mile mark. (32 trips done so far, 40 gallons used up) In the 35th trip, when you reach the 6.66 mile mark,", "could get from 2 jugs. Pour water from one jug into another till the other jug is completely full or the first jug itself is empty. Open Ended Problems: Students are advised to perform/solve open ended problems based on 1. Isopropyl is just rubbing alcohol, and distilled water is in jugs at the grocery store. There is an infinite amount of water supply available. Sign up for our email newsletter. Use cool water. Re: How much Power Service Additive Do You Use? I have been using the P. Initially, the 8 liter bottle is filled with water and the two other are empty. For example, to obtain exactly 6 gallons of water in the 11-gallon jug (starting with both jugs empty) you could first fill the 11-gallon jug from the faucet (move 4), then pour its contents into the 5-gallon jug (move 2), leaving the desired 6 gallons in the 11-gallon jug. Shop Lights & Accessories. I put it down for a minute and it seemed to stop, then it would start again. Measure 1/2 cup white distilled vinegar. BreadthFirstSearch&DepthFirstSearchTrees for WaterJug Problem Posted by Biruntha Gnaneswaran at. Water Bottle Filler – A type of dispenser that pours water into a standing bottle. With 3 cups of water inside, the jug weighs 2. com - id: 1d515d-ZDc1Z. Water", "# Math Tanks Algebra Level pending In a chemistry lab have two distilled water tanks. Mary works in this lab, and make a riddle for Ann, your friend: \"The tank A have a half capacity of the tank B. These tanks are filled in half. They have this propriety: 'If the half of destilated water in the the tank A is put in the tank B, and after the 1/25 of distilled water in the tank B is put in tank A, the tank A finish with 3 liters of distilled water.' What is the capacity of the tank B?\" ×", "Question Ryan’s gas tank is 1/2 full. After he buys 6 gallons of gas, it is 7/10 full. How many gallons can the gas tank hold", "seen that coming.\" Oh well, my tank is currently half full which is better than the gallon or so I had when I found the station. Tomorrow, Montreal!", "integrity. Quiz Mode is encouragement to use the solutions in a way that is most beneficial for your learning. ## Calculator Screenshots Video Transcript This is College Physics Answers with Shaun Dychko. In this question, we're going to see how practical it is to cool an engine using blocks of ice instead of the usual antifreeze liquid being circulated through the radiator and so on. So, to answer this question, we need to know what energy is released by burning gasoline. And, from Wikipedia, I found 3.156 times 10 to the 6 Joules is released per kilogram burned. And so, we're told that we're burning a gallon of gasoline. And so, we need to use this information, which is Joules per kilogram, and the gallon information to figure out Joules. So, we'll convert the kilograms into liters and using this conversion of liters per gallon and figure out how much energy is released. So, that's 3.156 times 10 to the 6 Joules per kilogram and times by 0.74 kilograms per liter, that's the density of gasoline that I also looked up on Wikipedia. And then, multiply by one liter for every 0.2642 gallons. So the gallons cancel, the liters cancel, the kilograms cancel and we're left with Joules. 8.8397 times 10 to the 6 Joules. Now, 80% of", "=$4.6. Question 6. Shannon collects rainwater to water her flowers. She has one bucket with 3.4 gallons of water and another with 1.85 gallons less. She uses both buckets to water the flowers. How many gallons of water does she use?", "contains 10,000 gallons of a solution that is 5 12 20 Dec 2012, 05:23 18 An empty fuel tank with a capacity of 200 gallons was fill 11 27 May 2012, 03:59 12 Mixture X contains 60% water and 40% chloride. If 120 gallon 8 02 Dec 2010, 14:12 If Henry were to add 5 gallons of water to a tank that is 2 05 Nov 2008, 15:26 Display posts from previous: Sort by", "anonymous one year ago Susan is creating a solution. A formula for the solution calls for 7/10 mL of water. Susan wants to make 2/7 less of the solution than the formula creates. How much water should Susan use? Express your answer as a fraction in simplest form. making 2/7 less means she's making 5/7 of the original amount, so she needs 5/7 of 7/10 mL. $\\frac{ 5 }{ 7 }\\times \\frac{ 7 }{ 10 }$", "container of lemonade that had 6 cups. She drank 2 cups. How many cups of lemonade does she have left? Answer: Number of cups of lemonade she have left = 4 cups. Explanation: Quantity of container of lemonade Molly bought = 6 cups. Quantity of container of lemonade Molly drank = 2 cups. Number of cups of lemonade she have left = Quantity of container of lemonade Molly bought – Quantity of container of lemonade Molly drank = 6 – 2 = 4 cups. Question 6. Randy poured 8 quarts of water in a bucket. Then he added 4 more quarts. How many quarts of water are in the bucket? Answer: Number of quarts of water are in the bucket = 12 quarts. Explanation: Quantity of water in a bucket Randy poured = 8 quarts. Quantity of water more she added = 4 quarts. Number of quarts of water are in the bucket = Quantity of water in a bucket Randy poured + Quantity of water more she added = 8 + 4 = 12 quarts. Solve. Use drawings if you need to. Question 7. Mrs. Sanders buys 2 gallons of milk each week. How many gallons of milk will she buy in 10 weeks? Answer: Number of gallons of milk Mrs. Sanders can buy in 10 weeks", "took to recycle one plastic water bottle, 1.36 million joules, and got 5.31 million joules. Therefore, it takes 5.31 million joules to use one 16.9 fl oz plastic water bottle. $$395,000,000 \\text {joules} + 136,000,000 \\text {joules} = 531,000,000 \\text {joules to use one 16.9 oz bottle}$$ The next step in my calculations was to determine the energy used in producing a reusable, aluminum water bottle. To begin, I found that Low-Tech Magazine says it takes between 11.35 and 17 megajoules to produce 1 kilogram of aluminum. From this, I averaged the two numbers to get 14 megajoules, which I used as my base number. Next, I multiplied this by 1 million to convert it to joules, resulting in 14 million joules. According to aspower.com, a 16.9 oz aluminum water bottle weighs about 15.5 grams. I converted this to kilograms and got .0155 kg. I then multiplied this by 14 million to receive the total energy in joules, which was 217,000 joules. $$14,000,000 \\text {joules} \\times .0155 \\text {kilograms} = 217,000 \\text {joules to produce one aluminum bottle}$$ For my final step in this calculation, I had to divide 5.31 million joules, the total amount of energy used to use one plastic water bottle, by 217,000 joules. I ended up with about 2,447. This means it takes about", "the remaining volume of water the same as the original volume?", "the energy of producing 2,447 aluminum bottles to use the same amount of energy it takes to use ONE plastic water bottle. $$531,000,000 \\text {joules} \\div 217,000 \\text {joules} = 2,447 \\text {aluminum bottles to equal the energy of one plastic bottle}$$ For the final step in my analysis, I wanted to compare my previous calculations to the total energy use by Americans with disposable water bottles. Ban the Bottle says that the average American uses about 167 disposable water bottles per year. I took this number and multiplied it by 5.31 million, the total amount of joules used in producing and recycling plastic water bottles, and got 886.77 million joules. This means that each American uses about 886.77 million joules every year on plastic water bottles, vs the 217,000 joules they could be using by switching to a reusable aluminum water bottle, assuming it will last you one year. My research concluded that it takes about the use of 2,447 reusable aluminium bottles to produce just one plastic water bottle. Why does this matter? According to my other calculations, regarding the total energy used by the typical American per year, we can save about 409x the energy we use through plastic water bottles by switching to a reusable one. This is important because this is such a" ]

[ "side and found common denominator (5/6 + 2/6) is 7/6. when and how do i solve for a? $\\displaystyle\\frac{7a}{6}=4$ Multiple by 6 and divide by 7 8. I have a series of questions like this but stuck on this one. Somebody wanna write this one out in its entirety? Eek thanks 9. Originally Posted by Lokain I have a series of questions like this but stuck on this one. Somebody wanna write this one out in its entirety? Eek thanks $\\displaystyle\\frac{7a}{6}=4$ Multiply both sides by 6. Do that and then divded both sides by 7.", "# How do you evaluate 7/8 + 8/7 ? May 28, 2016 You would have to find a common denominator. #### Explanation: In this case, I use 56 as the common denominator. $\\frac{7}{8} \\cdot \\frac{7}{7} = \\frac{49}{56}$ $\\frac{8}{7} \\cdot \\frac{8}{8} = \\frac{64}{56}$ $\\frac{49}{56} + \\frac{64}{56}$ $\\frac{113}{56}$ Now, are you able to simplify the value any further?", "dhdjdt 21 # what is the sum of 2/10 and 9/100 first we need to find a common denominator $\\frac{2}{10}$=$\\frac{20}{100}$ now we can add them $\\frac{20}{100}$+$\\frac{9}{100}$=$\\frac{29}{100}$", "# Week 15 Blog Adding and Subtracting Rational Expressions with Monomial Denominators $\\frac{2}{3} \\div \\frac{3}{4}$ Firstly, their common denominator is 12 because 3 times 4=12, so that means we need both times 3 or four for each side just for reaching the common denominator 12. $\\frac{6}{12} \\div \\frac{9}{12}$ so the answer is 2 over 3. # Week 14 Blog Things i learned this week: when we solve rational expressions, we need to find their common denominator. For example: $\\frac{$latex x^2\\$+2 over}{$x^2$ -x-6 over } {$x^2$ -x-6 over } can be factored into $x^2$-x-6=0 (x-3)(x+2)=o x=3, x=-2 if we found the common denominator,then we can factor it, that’s the way how we solved this kind of equation.", "What is the least common denominator of 2/3 and 5/7? $\\frac{14}{21} + \\frac{15}{21} = \\frac{29}{21} \\mathmr{and} 1 \\frac{8}{21}$ $\\frac{15}{21} - \\frac{14}{21} = \\frac{1}{21}$ $\\frac{14}{21} \\times \\frac{15}{21} = \\frac{10}{21}$ $\\frac{14}{21} \\div \\frac{15}{21} = \\frac{14}{15}$", "# How do you add 5/6 - 8/9? Mar 31, 2018 See explanation. #### Explanation: To add (or subtract) two fractions first you need to find a common denominator. Here the subtraction is: ## $\\frac{5}{6} - \\frac{8}{9}$ The lowest common denominator is $18$: $\\frac{5}{6} - \\frac{8}{9} = \\frac{5 \\cdot 3}{6 \\cdot 3} - \\frac{8 \\cdot 2}{9 \\cdot 2} = \\frac{15}{18} - \\frac{16}{18} = - \\frac{1}{18}$", "# How do you simplify 6/n^4+2/n^2? $\\frac{6}{n} ^ 4 + \\frac{2}{n} ^ 2$ needs to have a common denominator $\\frac{6}{n} ^ 4 + \\frac{2}{n} ^ 2 \\cdot {n}^{2} / {n}^{2} = \\frac{6}{n} ^ 4 + \\frac{2 {n}^{2}}{n} ^ 4$ and combine $\\frac{6 + 2 {n}^{2}}{n} ^ 4$", "above-given: 11/12 – 7/8 – Find the common denominator: 24 is the least common multiple of denominators 12 and 8. Use it to convert to equivalent fractions with this common denominator. = 11 x 2/12 x 2 – 7 x 3/8 x 3 = 22/24 – 21/24 Now subtract: = 22 – 21/24 = 1/24 Therefore, $$\\frac{11}{12}$$ – $$\\frac{7}{8}$$ = 1/24 Question 23. $$\\frac{7}{10}$$ – $$\\frac{1}{5}$$ = ____ The above-given: 7/10 – 1/5 – Find the common denominator: 10 is the least common multiple of denominators 10 and 5. Use it to convert to equivalent fractions with this common denominator. = 7 x 1/10 x 1 – 1 x 2/5 x 2 = 7/10 – 2/10 Now subtract: = 7 – 2/10 = 5/10 = 1/2 Therefore, $$\\frac{7}{10}$$ – $$\\frac{1}{5}$$ = 1/2 Problem Solving Question 24. Russ is putting his vacation photographs in an album that is 12$$\\frac{1}{8}$$– inches long and 10$$\\frac{1}{4}$$ inches wide. Should he trim the edges of the photographs to 12 inches long and 10 inches wide or to 12$$\\frac{1}{2}$$– inches long and 10$$\\frac{1}{2}$$ inches wide? The above-given: The number of inches long Russ put his photos in an album = 12 1/8 The width is 10 1/4 inches Now what we do is round off the values and check: 12 1/8: 1/8 can be", "# How do you simplify 2/3 + 1/6? Aug 5, 2016 Find a common denominator, then add. #### Explanation: To find a common denominator, we need to find the LCM, or least common multiple, between the two denominators. In our case, the LCM is $6$ because $3 \\cdot 2 = 6$ and $6 \\cdot 1 = 6$. Our expression now reads: $\\frac{4}{6} + \\frac{1}{6}$ $4 + 1 = 5$ $\\frac{4}{6} + \\frac{1}{6} = \\frac{5}{6}$", "of what that but somewhere in my simplification I went wrong. How do I proceed from what you've posted? Or is that as far as we're supposed to go? common denominator is $2\\sqrt{x+2}$ ... proceed from there. you should get $\\frac{5x^2 +8x - 9}{2\\sqrt{x+2}}$", "slhmbuerkett 26 What is 4/5 - 3/7 = what? $\\frac{28}{35} - \\frac{15}{35} = \\frac{13}{35}$ just find a common denominator", "would do this: $\\frac{1}{2}-\\frac{1}{3}$. To get a common denominator you would multiply the $\\frac{1}{2}$ by $\\frac{3}{3}$ and the $\\frac{1}{3}$ by $\\frac{2}{2}$, right? $\\frac{1}{2}-\\frac{1}{3} = \\frac{1*3}{2*3}-\\frac{1*2}{3*2} = \\frac{3-2}{6} = \\frac{1}{6}$ Same thing here using x and a instead of 2 and 3. $\\frac{1}{x}-\\frac{1}{a} = \\frac{1*a}{x*a}-\\frac{1*x}{a*x} = \\frac{a-x}{ax}$ 7. Thank you so much everyone.", "Equation with denominators: Multiply by the common denominator ! ! Go ! $\\frac{3x-6}{2}= 2-\\frac{3x}{4}$ $\\frac{6x-12}{4}= \\frac{8}{4}-\\frac{3x}{4}$ ¦¦· 4 $6x-12 = 8-3x$$S =\\{\\frac{20}{9} \\} \\frac{x}{7} - 3 = \\frac{x}{2} - 8 \\frac{x}{7} - 3 = \\frac{x}{2} - 8 ¦¦· 14 2x - 42 = 7x - 112$$ S =\\{14 \\}$", "Home > A2C > Chapter 6 > Lesson 6.2.1 > Problem6-65 6-65. $\\frac{3}{(x-4)(x+1)}+\\frac{6}{x+1}\\cdot \\frac{x-4}{x-4}$ $\\frac{3+6(x-4)}{(x-4)(x+1)}$ $\\frac{6x-21}{(x-4)(x+1)}$ Try factoring after subtracting. See part (a). Factor to find the common denominator, then use the strategy from part (a) to simplify. $\\frac{5}{x^2-9}$ $\\text{If your answer was }-\\frac{1}{x²-9},$ check the step where you subtracted. −(x − 3) = x + 3", "# What is 2/3 plus 1/6? Apr 25, 2017 $\\frac{2}{3} + \\frac{1}{6} = \\frac{5}{6}$ #### Explanation: Given - $\\frac{2}{3} + \\frac{1}{6}$ Take a common denominator for both 3 and 6. It is 6. Then $\\frac{4 + 1}{6} = \\frac{5}{4}$", "with the 2x ok you can subtract 1/6 from 1/4 by finding the common demnomater which is 12. so you will get 3/12 - 2/12 = 1/12 you cannot touch the 1/2x because of the x so the answer is: 1/2x + 1/12 Yes you can...the common denominator in this case would be $12\\color{red}x$ So you would get $\\frac{6}{12x}+\\frac{3x}{12x}-\\frac{2x}{12x}=........$ --Chris", "$\\frac{{x}^{2} + 2 x + 7}{x {\\left(x - 1\\right)}^{2}} = \\frac{7}{x} - \\frac{6}{x - 1} + \\frac{10}{x - 1} ^ 2$ It is a good idea to check the answer by getting the common denominator. (I did that on paper, but I'm not going to type it up.)", "# How do you multiply 7 / 10 + 1 / 2 = 6 /( 5x)? May 9, 2015 $L . H . S = \\frac{7}{10} + \\frac{1}{2}$ To make the denominator common: we take the L.C.M of 2 and 10 $= 10$ $= \\frac{7}{10} + \\frac{1 \\times 5}{2 \\times 5}$ $= \\frac{7}{10} + \\frac{5}{10}$ $= \\frac{12}{10}$ equating the above value with the R.H.S $\\frac{12}{10}$ = $\\frac{6}{5 x}$ $5 x$ = $\\frac{60}{12}$ $5 x$ = $5$ $x = 1$", "## anonymous 4 years ago how do i completely simplify : 8x-16/x^2-13x+22 1. anonymous |dw:1328219958845:dw| 2. Mertsj $\\frac{8(x-2)}{(x-11)(x-2)}=\\frac{8}{x-11}$ 3. anonymous Okay so, 8x-16/x^2 -2x -11x +22 At first we just factorize the denominator. So we get, 8x-16/(x-2)(x-11) Then take 8 common from numerator so: 8(x-2)/(x-2)(x-11) Cancel x-2 from numerator and denominator So we are left with: 8/x-11 Which is our answer :) 4. anonymous do you always factor the denominator first?", "[3 * (1 + x^0.5)^2 * 0.5] / (x^0.5 * x^2) - (2 * (1 + x^0.5)^3) / x^3 Do I have to find a common denominator? That seems scary. :( 8. Mar 18, 2012 ### Chopin Ok. So far, we have: $$\\frac{d}{dx}(\\frac{1+\\sqrt{x}}{\\sqrt[3]{x^2}})^3\\\\ = \\frac{d}{dx}\\frac{(1 + \\sqrt{x})^3}{x^2}\\\\ = \\frac{d}{dx}(1+\\sqrt{x})^3x^{-2}\\\\ = 3(1+\\sqrt{x})^2\\frac{1}{2\\sqrt{x}x^2} +(1+\\sqrt{x})^3 \\frac{-2}{x^3}$$ You're probably not going to be able to get it a whole lot simpler than that, but you should at least be able to factor out few common things, and then get the rest over a common denominator." ]

[ "integers with different signs is negative. Here -18 and 9 both are having different sign hence the result would be negative. Solve each problem. Question 15. A bucket that holds 5$$\\frac{1}{4}$$ if gallons of water is being used to fill a tub that can hold 34$$\\frac{1}{8}$$ gallons. How many buckets will be needed to fill the tub? buckets are needed to fill the tub. Answer: 6$$\\frac{1}{2}$$ A bucket that holds 5$$\\frac{1}{4}$$ if gallons of water is being used to fill a tub that can hold 34$$\\frac{1}{8}$$ gallons. number of buckets needed to fill the tub = 34$$\\frac{1}{8}$$ ÷ 5$$\\frac{1}{4}$$ Convert the above numbers into improper fractions to make the calculations easy. 34$$\\frac{1}{8}$$ = $$\\frac{273}{8}$$ 5$$\\frac{1}{4}$$ = $$\\frac{21}{4}$$ So, 34$$\\frac{1}{8}$$ ÷ 5$$\\frac{1}{4}$$ = $$\\frac{273}{8}$$ ÷ $$\\frac{21}{4}$$ To divide by a fraction, multiply by its reciprocal. Here, the reciprocal of $$\\frac{21}{4}$$ = $$\\frac{4}{21}$$ Hence, $$\\frac{273}{8}$$ × $$\\frac{4}{21}$$ Reduce the above fractions into simplest form if possible. Then, multiply the numerators and denominators separately. = $$\\frac{273 × 4}{8 × 21}$$ Divide 4 in numerator and 8 in denominator with 4, which is a common factor = $$\\frac{273 × 1}{2 × 21}$$ Divide 273 in numerator and 21 in denominator with 21, which is a common factor = $$\\frac{13 × 1}{2 × 1}$$ = $$\\frac{13}{2}$$ = 6$$\\frac{1}{2}$$ Therefore, 6$$\\frac{1}{2}$$ buckets are needed", "= 4.32 4.32/0.27 = 16 ### Page No. 127 Question 1. There is $$\\frac{4}{5}$$ lb of sand in the class science supplies. If one scoop of sand weighs $$\\frac{1}{20}$$ lb, how many scoops of sand can Maria get from the class supplies and still leave $$\\frac{1}{2}$$ lb in the supplies? Type below: __________ 16 scoops Explanation: There is $$\\frac{4}{5}$$ lb of sand in the class science supplies. If one scoop of sand weighs $$\\frac{1}{20}$$ lb, $$\\frac{4}{5}$$ ÷ $$\\frac{1}{20}$$ = 4/5 × 1/20 = 16 scoops Question 2. What if Maria leaves $$\\frac{2}{5}$$ lb of sand in the supplies? How many scoops of sand can she get? ______ scoops 8 scoops Explanation: There is $$\\frac{2}{5}$$ lb of sand in the class science supplies. If one scoop of sand weighs $$\\frac{1}{20}$$ lb, $$\\frac{2}{5}$$ ÷ $$\\frac{1}{20}$$ = 2/5 × 20 = 8 Question 3. There are 6 gallons of distilled water in the science supplies. If 10 students each use an equal amount of the distilled water and there is 1 gal left in the supplies, how much will each student get? $$\\frac{□}{□}$$ gallon $$\\frac{1}{2}$$ gallon Explanation: There are 6 gallons of distilled water in the science supplies. There is 1 gal left in the supplies, 6 – 1 = 5 10 students each use an equal amount of the distilled", "= 1 x 3/2 x 3 – 1 x 2/3 x 2 b = 3/6 – 2/6 Step 2: Here the denominators are equal so subtract directly. b = (3 – 2)/6 b = 1/6 Therefore, 1/6 more oil is used to make brownies. Question 16. Danielle poured $$\\frac{3}{4}$$ gallon of water from a $$\\frac{7}{8}$$-gallon bucket How much water is left in the bucket? The above-given: The number of gallons of water Danielle poured = 3/4 The number of gallons of water in a bucket = 7/8 The number of gallons of water left in the bucket = b b = 7/8 – 3/4 Step 1: Find a common denominator 8 is the least common multiple of denominators 8 and 4. Use it to convert to equivalent fractions with this common denominator. b = 7 x 1/8 x 1 – 3 x 2/4 x 2 b = 7/8 – 6/8 Step 2: Here the denominators are equal so subtract directly. b = (7 – 6)/8 b = 1/8 Therefore, 1/8 gallons of water left in the bucket. HOT Problems Question 17. Mathematical PRACTICE 2 Use Number Sense Is finding $$\\frac{9}{10}$$ – $$\\frac{1}{2}$$ the same as finding $$\\frac{9}{10}$$ – $$\\frac{1}{4}$$ – $$\\frac{1}{4}$$ ? Explain. The above-given: 9/10 – 1/2 = 9/10 – 1/4 – 1/4 9/10 – 1/2 =", "in preparing lunch. Convert from gallons to quarts 1 gallon = 4 quarts 5 gallons = 5 × 4 quarts = 20 quarts Thus the correct answer is option B. Spiral Review Question 3. Roy uses $$\\frac{1}{4}$$ cup of batter for each muffin. Which list shows the amounts of batter he will use depending on the number of muffins he makes? Options: a. $$\\frac{1}{4}, \\frac{1}{5}, \\frac{1}{6}, \\frac{1}{7}, \\frac{1}{8}$$ b. $$\\frac{1}{4}, \\frac{2}{4}, \\frac{3}{4}, \\frac{4}{4}, \\frac{5}{4}$$ c. $$\\frac{1}{4}, \\frac{2}{8}, \\frac{3}{12}, \\frac{4}{16}, \\frac{5}{20}$$ d. $$\\frac{1}{4}, \\frac{2}{8}, \\frac{4}{16}, \\frac{6}{24}, \\frac{8}{32}$$ Answer: $$\\frac{1}{4}, \\frac{2}{8}, \\frac{4}{16}, \\frac{6}{24}, \\frac{8}{32}$$ Explanation: All fractions must be equal to $$\\frac{1}{4}$$ a. $$\\frac{1}{4}, \\frac{1}{5}, \\frac{1}{6}, \\frac{1}{7}, \\frac{1}{8}$$ In this all fractions are not equal to $$\\frac{1}{4}$$ b. $$\\frac{1}{4}, \\frac{2}{4}, \\frac{3}{4}, \\frac{4}{4}, \\frac{5}{4}$$ $$\\frac{2}{4}$$ = $$\\frac{1}{2}$$ $$\\frac{4}{4}$$ = 1 In this all fractions are not equal to $$\\frac{1}{4}$$ c. $$\\frac{1}{4}, \\frac{2}{8}, \\frac{3}{12}, \\frac{4}{16}, \\frac{5}{20}$$ $$\\frac{2}{8}$$ = $$\\frac{1}{4}$$ $$\\frac{3}{12}$$ = $$\\frac{1}{4}$$ $$\\frac{4}{16}$$ = $$\\frac{1}{4}$$ $$\\frac{5}{20}$$ = $$\\frac{1}{4}$$ d. $$\\frac{1}{4}, \\frac{2}{8}, \\frac{4}{16}, \\frac{6}{24}, \\frac{8}{32}$$ $$\\frac{2}{8}$$ = $$\\frac{1}{4}$$ $$\\frac{4}{16}$$ = $$\\frac{1}{4}$$ $$\\frac{6}{24}$$ = $$\\frac{1}{4}$$ $$\\frac{8}{32}$$ = $$\\frac{1}{4}$$ Thus the correct answer is option D. Question 4. Beth has $$\\frac{7}{100}$$ of a dollar. Which shows the amount of money Beth has? Options: a.$7.00 b. $0.70 c.$0.07 d. $0.007 Answer:$0.07 Explanation: Beth has $$\\frac{7}{100}$$ of a dollar. The decimal form of $$\\frac{7}{100}$$ = 0.07 Thus", "1$$\\frac{11}{12}$$ (C) 9$$\\frac{1}{12}$$ (D) 9$$\\frac{11}{12}$$ Answer: D Explanation: 5$$\\frac{6}{12}$$ + 4$$\\frac{5}{12}$$ = 9$$\\frac{11}{12}$$ is the horse trail Question 13. Multi-Step Students bring 8$$\\frac{7}{8}$$ gallons of lemonade to a picnic. They drink 5$$\\frac{2}{8}$$ gallons with lunch. Then they drink 2$$\\frac{1}{8}$$ gallons with an afternoon snack. How much lemonade is left? (A) 3$$\\frac{5}{8}$$ gallons (B) 6$$\\frac{3}{4}$$gallons (C) 5$$\\frac{3}{4}$$gallons (D) 1$$\\frac{1}{2}$$gallons Answer: D Explanation: 5$$\\frac{2}{8}$$ + 2$$\\frac{1}{8}$$ = 7$$\\frac{3}{8}$$ 8$$\\frac{7}{8}$$ – 7$$\\frac{3}{8}$$ = 1$$\\frac{4}{8}$$ = 1$$\\frac{1}{2}$$gallons lemonade is left TEXAS Test Prep Question 14. Jeff used 4$$\\frac{7}{8}$$ cups of orange juice and 3$$\\frac{1}{8}$$ cups of pineapple juice to make a tropical punch. How much more orange juice than pineapple juice did Jeff use? (A) $$\\frac{3}{4}$$ (B) 1$$\\frac{3}{4}$$ (C) 1$$\\frac{7}{8}$$ (D) 8 cups Answer: B Explanation: 4$$\\frac{7}{8}$$ – 3$$\\frac{1}{8}$$ = 1$$\\frac{6}{8}$$ 1$$\\frac{3}{4}$$ more orange juice than pineapple juice that Jeff use ### Texas Go Math Grade 4 Lesson 5.6 Homework and Practice Answer Key Write the sum as a mixed number with the fractional part less than 1. Question 1. Answer: Explanation: Added the wholes then added the fractions written the sum Question 2. Answer: Explanation: Added the wholes then added the fractions written the sum Question 3. Answer: Explanation: Added the wholes then added the fractions written the sum Question 4. Answer: Explanation: Added the wholes then added the fractions written", "a common denominator for the two terms. In this case, the common denominator will be twenty: Simplify to and convert to not a mixed fraction: Multiply the two fractions in the parentheses by multiplying straight across (A quick shortcut would be to factor out the 10 on top and bottom). Now convert to a non-mixed fraction. It will become . In order to subtract the two fractions, find a common denominator. In this case, it will be 70. Now subtract, and find the answer! Example Question #1 : How To Divide Complex Fractions A cistern containing gallons of water has sprung two leaks. One leaks at a rate of of a gallon every half hour. The second one leaks at a rate of a gallon every fifth of an hour. In how many hours will the cistern be empty (presuming that the leaks will empty it eventually)? Explanation: It is best to figure out what each of the leaks are per hour. We can figure this out by adding together the two fractional rates of leaking. For the first leak, we can do this as follows: This is the same as: For the second leak, we use the same sort of procedure: Thus, our two leaks combined are: The common denominator for these is ; thus, we can", "water was originally in the tank, c. 33.6 gallons of water is in the tank when it is $$\\frac{1}{2}$$ full. Explanation: Given a water tank is $$\\frac{1}{8}$$ full. The tank is $$\\frac{3}{4}$$ full when 42 gallons of water are added to the tank. Since the tank started at $$\\frac{1}{8}$$ full and reached $$\\frac{3}{4}$$ full , To find 1 tank full we have the difference is $$\\frac{3}{4}$$ – $$\\frac{1}{8}$$ = 42 gallons, So first we make denominators common we multiply and divide by 2 to $$\\frac{3}{4}$$ = $$\\frac{6}{8}$$ now we subtract as $$\\frac{6}{8}$$ – $$\\frac{1}{8}$$ = 42; as denominators are same we minus numerators (6-1)= 5 making $$\\frac{5}{8}$$ = 42,therefore 1 tank full is 42 X $$\\frac{8}{5}$$ = $$\\frac{336}{8}$$ = 67.2 gallons of water. b. Initially we had $$\\frac{1}{8}$$ full of water means we had 67.2 X $$\\frac{1}{8}$$ = 8.4 gallons of water was originally in the tank, c. Water in the tank when it is $$\\frac{1}{2}$$ full is 67.2 X $$\\frac{1}{2}$$ = $$\\frac{67.2}{2}$$ = 33.6 gallons of water is in the tank when it is $$\\frac{1}{2}$$ full. ### Lesson 2.3 Dividing Mixed Numbers EXPLORATION 1 Dividing Mixed Numbers Work with a partner. Write a real-life problem that represents each division expression described. Then solve each problem using a model. Check your answers. a. How many three-fourths are in", "If the numerator and denominator are both divided by two, the simplified fraction becomes $$\\frac{11}{12}$$. $$\\frac{11}{12}$$ of the total amount of food is still remaining. Question #5: Kristina plans on drinking $$\\frac{1}{4}$$ quart of water for every mile she runs. If she runs six miles, how much water will she drink? $$1\\frac{1}{4}$$ quarts of water $$1\\frac{1}{2}$$ quarts of water $$2\\frac{1}{2}$$ quarts of water $$\\frac{1}{2}$$ quarts of water The correct answer is B: $$1\\frac{1}{2}$$ quarts of water. If Kristina drinks $$\\frac{1}{4}$$ quart of water for every mile, she is essentially drinking $$\\frac{1}{4}+\\frac{1}{4}+\\frac{1}{4}+\\frac{1}{4}+\\frac{1}{4}+\\frac{1}{4}$$ quarts of water. The sum of this list of fractions is $$1\\frac{2}{4}$$ or in simplest form, $$1\\frac{1}{2}$$.", "= 6 x $$\\frac{11}{4}$$ = $$\\frac{66}{4}$$ = 16$$\\frac{1}{2}$$ C. Rename the product as a mixed number. 16 $$\\frac{1}{2}$$ Explanation: for 1 tray = 2$$\\frac{3}{4}$$ = $$\\frac{11}{4}$$ For 6 trays = 6 x $$\\frac{11}{4}$$ = $$\\frac{66}{4}$$ = 16$$\\frac{1}{2}$$ D. How many packets are needed for 6 trays? 16$$\\frac{1}{2}$$ Explanation: for 1 tray = 2$$\\frac{3}{4}$$ = $$\\frac{11}{4}$$ for 6 trays = 6 x $$\\frac{11}{4}$$ = $$\\frac{66}{4}$$ = 16$$\\frac{1}{2}$$ Turn and Talk How could you find the product of this problem without renaming 2$$\\frac{3}{4}$$ as a fraction greater than 1? $$\\frac{11}{4}$$ Explanation: Convert the mixed fraction into improper fraction. 2$$\\frac{3}{4}$$ = $$\\frac{11}{4}$$ Check Understanding Question 1. Tammi uses 2$$\\frac{2}{3}$$ gallons of water on her plants once a week. How much water does she use in 3 weeks? 8 gallons Explanation: Tammi uses 2$$\\frac{2}{3}$$ gallons of water on her plants once a week. Total gallons of water she use in 3 weeks, 2$$\\frac{2}{3}$$ = $$\\frac{8}{3}$$ 3 x $$\\frac{8}{3}$$ = $$\\frac{24}{3}$$ = 8 gallons Question 2. At an urban vertical farm, each tray of arugula makes about 3$$\\frac{5}{8}$$ pounds of edible plants. How many pounds will 5 trays make? 18$$\\frac{1}{8}$$ Explanation: Each tray of arugula makes about 3$$\\frac{5}{8}$$ pounds of edible plants. 5 trays will make = 5 x 3$$\\frac{5}{8}$$ = 5 x $$\\frac{29}{8}$$ = $$\\frac{145}{8}$$ = 18$$\\frac{1}{8}$$ Find the product. Question 3.", "until full -> 2 remains in 3 – ratchet freak May 15 '14 at 14:30 for 2 jugs that can contain $x$ and $y$ you can get any multiple of $GCD(x,y)$ between $0$ and $max(x,y)$ using the algorithm that @Ross described. You can understand this by creating a new unit so that $1 \\text{ unit} = GCD(x,y)$ gallons, and now you have 2 jugs with $\\frac x {GCD(x,y)} = x'$ and $\\frac y{GCD(x,y)} = y'$ and because you can't take out fractional units it means that you alway have a multiple of $GCD(x,y)$ in your jugs. Fill the 7 gallon jug, then fill the 3 gallon jug using the water from the 7 gallon jug, repeat this step, now you have excactly 1 gallon left in the 7 gallon jug, and just simply pour that gallon of water over into the 3 gallon jug • The goal of the puzzle is to have 1 left in the small jug and 4 left in the big jug. – Deusovi Jan 2 '16 at 2:23", "there’s an easier way for occasional differences in volume. ### Dilution That method is dilution of the concentrates to mineralize a different volume of water. This allows you do perform experiments involving mixing water for just one kettle or even just one cupping bowl without having to get your powders out again. It is slightly more math, but don’t worry, there’s nothing complicated. For my example here, we’ll be assuming a gallon concentrate, since I imagine it will be the most frequently used. We need a dilution factor, from which we’ll subtract 1ml to find the amount of distilled water to dilute with, found with the equation: $\\boxed{\\frac{3785ml}{Xml}=Dilution\\; Factor}$ If you’re diluting concentrates for a 500ml kettle, your equation will look like this: $\\boxed{\\frac{3785ml}{500ml}=~7.5}$ This means to make concentrates, you’ll add 6.5ml of distilled water for every 1ml of gallon concentrate. For example, you could mix 65ml of distilled water with 10ml of gallon concentrate to make up 75ml of concentrate for a 500ml kettle. Each 1ml of this 75ml concentrate will raise the hardness of a 500ml kettle of water by 10ppm. If you’re diluting concentrates for a 150ml cupping bowl, your equation would look like: $\\boxed{\\frac{3785ml}{150ml}=~25}$ So for every 24ml of distilled water, we add 1ml of gallon concentrate. Just like the other concentrates, 1ml will", "stated that one can measure out exactly as much as wanted from any can(example, one can take out 0.75 liters from a 1 liter can, and so forth). In this case, 2.5 is the correct answer. But if such thing is not possible, then we would have to work exclusively with 0.5 and 1 liter cans only, in which case, 4 will be the answer. I'm not sure if my argument is ill-founded, but please, clarify. I request Bunuel to comment and also clarify, so I don't make such comprehension mistakes if I did so. I approached this question in the same way and made the same mistake. My assumption was that we would have to pour together the whole (1 or 1/2 gallon) cans so that we can get at least 2 liters of grey paint, which gives E as a result. Look at the last line of the question: what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture? This clarifies that you need to find the least amount of paint that you need to buy to measure out the portions needed for 2 gallons mix. Otherwise, it would have been mentioned that you need to use the whole can. _________________ Karishma", "calls for 5 $$\\frac{1}{2}$$ cups of flour. Lorenzo only has 3 $$\\frac{3}{4}$$ cups of flour. Write and solve an equation to find the additional amount of flour Lorenzo needs to make the recipe. Type below: _____________ Answer: 1 $$\\frac{3}{4}$$ Explanation: A recipe calls for 5 $$\\frac{1}{2}$$ cups of flour. Lorenzo only has 3 $$\\frac{3}{4}$$ cups of flour. x + 3 $$\\frac{3}{4}$$ = 5 $$\\frac{1}{2}$$ x = 5 $$\\frac{1}{2}$$ – 3 $$\\frac{3}{4}$$ x = 1 $$\\frac{3}{4}$$ Question 11. Jan used 22.5 gallons of water in the shower. This amount is 7.5 gallons less than the amount she used for washing clothes. Write and solve an equation to find the amount of water Jan used to wash clothes. Type below: _____________ Answer: 30 Explanation: Jan used 22.5 gallons of water in the shower. This amount is 7.5 gallons less than the amount she used for washing clothes. Let the amount of water Jan used to wash clothes be x x – 7.5 = 22.5 x = 22.5 + 7.5 x = 30 Therefore the amount of water Jan used to wash clothes is 30 gallons. Question 12. Explain how to check if your solution to an equation is correct. Type below: _____________ Answer: i. Evaluate the left-hand side expression at the given value to get a number. ii. Evaluate the", "gallons the third week. If he continued watering in this pattern, how much water did he use on the fifth week? Options: a. 2 $$\\frac{1}{2}$$ gallons b. 2 $$\\frac{7}{8}$$ gallons c. 3 $$\\frac{1}{4}$$ gallons d. 6 $$\\frac{7}{8}$$ gallons Answer: 2 $$\\frac{7}{8}$$ gallons Explanation: First term = 1 $$\\frac{3}{8}$$ The difference is $$\\frac{3}{4}$$ – $$\\frac{3}{8}$$ = $$\\frac{3}{8}$$ Second term is 1 $$\\frac{3}{8}$$ + $$\\frac{3}{8}$$ = 1 $$\\frac{3}{4}$$ Third term = 1 $$\\frac{3}{4}$$ + $$\\frac{3}{8}$$ = 1 + 1 $$\\frac{1}{8}$$ = 2 $$\\frac{1}{8}$$ Fourth term = 2 $$\\frac{1}{8}$$ + $$\\frac{3}{8}$$ = 2 $$\\frac{1}{2}$$ Fifth term = 2 $$\\frac{1}{2}$$ + $$\\frac{3}{8}$$ = 2 $$\\frac{7}{8}$$ gallons Thus the correct answer is option B. ### Share and Show – Page No. 279 Question 1. Caitlin has 4 $$\\frac{3}{4}$$ pounds of clay. She uses 1 $$\\frac{1}{10}$$ pounds to make a cup, and another 2 pounds to make a jar. How many pounds are left? First, write an equation to model the problem. Type below: _________ Answer: 4 $$\\frac{3}{4}$$ – 1 $$\\frac{1}{10}$$ – 2 Explanation: Subtract the total pound of clay from used clay. So, the equation of the clay leftover is 4 $$\\frac{3}{4}$$ – 1 $$\\frac{1}{10}$$ – 2 Question 1. Next, work backwards and rewrite the equation to find x. Type below: _________ Answer: 4 $$\\frac{3}{4}$$ – 1 $$\\frac{1}{10}$$ – 2 = x Explanation:", "\\frac{n}{M}$ and plug in the numbers: $V = \\frac{n}{M} = \\frac{3}{1.5} = 2$ litres", "# Tanks Algebra Level 1 We can fill $$\\frac {7}{8}$$ of the tank A with $$\\frac{2}{3}$$ of the water of the tank B. If the tank A is empty, what is the fraction of the tank B is needed to fill completely the tank A? ×", "to get $$\\frac{15}{4}$$ $$\\frac{15}{4}=\\frac{3}{4}+\\frac{3}{4}+\\frac{3}{4}+\\frac{3}{4}+\\frac{3}{4}$$ Simplify $$=5\\times \\frac{3}{4}$$ So, Kevin used $$\\frac{3}{4}$$ five cups of strawberries to make milkshakes. This shows us that he has enough strawberries to make 5 glasses of milkshake. Example 6: Nathan has three $$\\frac{15}{2}$$ gallon jerry cans full of gas. Write an equation to find the total amount of gas that Nathan has. Solution: Capacity of 1 jerry can $$=\\frac{15}{2}$$ Number of jerry cans = 3 Total amount of gas he has $$=\\frac{15}{2}+\\frac{15}{2}+\\frac{15}{2}=3\\times \\frac{15}{2}$$ $$=\\frac{45}{2}=22.5$$ gallons. A fraction can be expressed as follows: $$\\frac{a}{b}=a\\times \\frac{1}{b}$$. Here, the numerator is multiplied with a unit fraction having the same denominator.", "side length 3 and a circular piece of paper with radius 1 are glued together so that their centers coincide. How long is the outer perimeter of the resulting 2-dimensional shape?... ### Kim estimates that there are approximately 5750 drops of water in one-liter bottle. Estimate how many gallons of water her leaky faucet wastes in a year. (There are approximately 3.78 liters in a gallon.) Kim estimates that there are approximately 5750 drops of water in one-liter bottle. Estimate how many gallons of water her leaky faucet wastes in a year. (There are approximately 3.78 liters in a gallon.)... ### Which of these expressions are equivalent to $\\frac{p}{3}$ A. $p-\\frac{2}{3} p$ B.$\\frac{1}{3}p$ C.$p-3$ D. 3 ÷ p E. $\\frac{3p}{9}$ F. $\\frac{1}{3}p + \\frac{1}{3}p + \\frac{1}{3}p$ Which of these expressions are equivalent to $\\frac{p}{3}$ A. $p-\\frac{2}{3} p$ B.$\\frac{1}{3}p$ C.$p-3$ D. 3 ÷ p E. $\\frac{3p}{9}$ F. $\\frac{1}{3}p + \\frac{1}{3}p + \\frac{1}{3}p$...", "of the water is used. Then, the tank is filled with another 6$$\\frac{3}{4}$$ gallons of water. What is the amount of water in the tank now? (A) 8$$\\frac{1}{4}$$ gallons (B) 1$$\\frac{1}{2}$$ gallons (C) 8$$\\frac{1}{4}$$ gal1ons (D) 1$$\\frac{3}{4}$$ gallons Answer: A 5$$\\frac{3}{4}$$ – 4$$\\frac{1}{4}$$ = 1$$\\frac{2}{4}$$ 1$$\\frac{2}{4}$$ + 6$$\\frac{3}{4}$$ = 7$$\\frac{5}{4}$$ = 8$$\\frac{1}{4}$$ gallons 8$$\\frac{1}{4}$$ gallons is the amount of water in the tank now Question 16. Multi-Step For a candy recipe, Karen will need 4$$\\frac{3}{8}$$ cups of dark chocolate chips, 5$$\\frac{5}{8}$$ cups milk chocolate chips, and 3$$\\frac{4}{8}$$ cups white chocolate chips. What is the total amount of chips needed for the candy recipe? (A) 12$$\\frac{1}{2}$$ cups (B) 12$$\\frac{2}{3}$$ cups (C) 13$$\\frac{2}{3}$$ cups (D) 13$$\\frac{1}{2}$$ cups Answer: D Explanation: First add the wholes 4 + 5 + 3 = 12 Then add the fractions $$\\frac{3}{8}$$ + $$\\frac{5}{8}$$ + $$\\frac{4}{8}$$ = $$\\frac{12}{8}$$ 1$$\\frac{4}{8}$$ so, the fraction is 13$$\\frac{1}{2}$$ cups Scroll to Top", "If we try the $\\frac{1}{5}$ pieces, they do not exactly cover the $\\frac{1}{2}$ piece or the $\\frac{1}{3}$ piece. #Cambridgesyllabus #Singaporesyllabus I have uploaded this video to teach your child on relating fractions to litres. Now let’s see what you need to do with fractions that have different denominators. For example, 36 1 = 36. A liter, or litre, is a unit of volume in the metric system. Simplify the numerators and denominators. Earlier, we used fraction tiles to see that the LCD of $\\frac{1}{4}\\text{and}\\frac{1}{6}$ is $12$. We have How to divide fractions. We will start at Step 2. Convert $\\frac{8}{15}$ and $\\frac{11}{24}$ to equivalent fractions with denominator $120$, their LCD. A liter, or litre, is a unit of volume in the metric system.A liter is defined as the volume of a cube that is 10 centimeters on a side. Multiply the number of liters (L) by 1,000 to find the number of milliliters (mL). You have practiced adding and subtracting fractions with common denominators. Converting a percentage into a fraction can be done by dividing by 100, following the definition of a percentage. Convert two fractions to equivalent fractions with their LCD as the common denominator. □0.837 = \\frac{837}{10^3} = \\frac{837}{1000}. 15×100%=1005%=20%. Multiply the factors. 1/8 is the fraction. First, understand that the fraction is 1920\\frac {19}{20}2019​. Then," ]

[ "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", "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]", "(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", "be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label(\"\\boldmath{1}\",(0,0)); label(\"\\boldmath{12}\",(1,0)); label(\"\\boldmath{123}\",(2,0)); label(\"\\boldmath{S}\",(3,0)); label(\"11\",(0.5,-0.75)); label(\"111\",(1.5,-0.75)); label(\"d_1\",(2.5,-0.75)); label(\"100\",(1,-1.5)); label(\"d_2\",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label(\"\\textbf{First Differences}\",(-0.75,-0.75),align=W); label(\"\\textbf{Second Differences}\",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334}. \\end{align*} ~MRENTHUSIASM ## Solution 4 (Finite Differences by Algebra) Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$. Alternatively, applying finite differences, one obtains $$c_4 = {3 \\choose 2}f(2) - {3 \\choose 1}f(1) + {3 \\choose 0}f(0) =334.$$ ## Solution 5 (Assumption) The idea", "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 $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "prime factors $2,3, \\text{and } 5$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"2^3\",X--Y,2W); label(\"2^3\",X--Z,2E); label(\"2^2\",Y--Z,2S); [/asy]$ Analyzing for powers of $2$, it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \\text{ nor } z$ can have a power of $2$ exceeding $2$. Turning towards the vertices $y$ and $z$, we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \\text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \\{(2,0)(2,1)(0,2)(1,2)(2,2)\\}$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"3^2\",X--Y,2W); label(\"3^1\",X--Z,2E); label(\"3^2\",Y--Z,2S); [/asy]$ Using the same logic as we did for powers of $2$, it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \\text{ and } z$, we can use the same logic to find the only ordered pairs $(x,z)$", "# 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", "label(\"$v^\\prime$\",hb.hexCorner(0,10,5)+(1,0),E); label(\"$w$\",hb.hexCorner(-10,0,3)-(0,1),S); label(\"$w^\\prime$\",hb.hexCorner(10,0,0)+(0,1),N); label(\"$o^\\prime$\",hb.hexCorner(5,0,1)-(1,0),W); label(\"$x^-$\",hb.hexCorner(5,5,5)+(1,0),E); hb.oline((5,0),0,1,2,3,2,3,2,3,4,5,4,5,0,5, 4,5,0,1,0,1,2); hb.oline((10,0), prefix=(0,1) ,suffix=(1,0) ,0 ,3,2,3,2,1,0,5,0,1,2,1,2,3,4,3,4,3,2,3,2,3,4 ,3,2,3,2,1,0,5,0,1,0,1,2,3,2,3,4 ,3,2,3,4 ,3,4,3,2,3); hb.oline((0,0), prefix=(-1,0),2,3,2,3,4); hb.oline((-10,0), prefix=(0,-1),3,4,5,0); hb.dot(9,0,3); hb.dot(7,1,4); hb.dots(5,1,0,3); shipout(bbox(Fill(lightyellow))); \\end{asy} \\caption{Hex Board} \\end{figure} \\end{document} % % Process : % % pdflatex hexboard.tex % asy hexboard-*.asy % pdflatex hexboard.tex - One question, how do I make the figure larger? Preferably so that it fills the whole page. Other than that, this worked perfectly, although the co-ordinates of the hexagons is confusing. Maybe it will make sense after I get some sleep, the co-ordinate system led to a huge guessing game but I got the final product! – Surculus Nov 3 '13 at 1:41 @Surculus: One way to scale the figure to fit the whole page would be as follows: make it a standalone using standalone document class (don't forget to remove the figure environment) and then include it with pdfpages package. As for co-ordinates: perhaps it would be less confusing with the edited picture: direction is named after the node of destination. – g.kov Nov 3 '13 at 8:02 I modified some snippets from Jake's answer about Chinese checkers and from Alain's answer about an hexagonal grid. Using the node labeling you can draw thick lines in between. It's better to use the corners as references, like I did in this answer about naming corners", "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]$ ## Problem 19 Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$, has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is $[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 32 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 48 \\qquad \\text{(E)}\\ 64$ ## Problem 20 Let $W,X,Y$ and $Z$ be four different digits selected from the set $\\{ 1,2,3,4,5,6,7,8,9\\}.$ If the sum $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ is to be as small as possible, then $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ must equal $\\text{(A)}\\ \\dfrac{2}{17} \\qquad \\text{(B)}\\ \\dfrac{3}{17} \\qquad \\text{(C)}\\ \\dfrac{17}{72} \\qquad \\text{(D)}\\ \\dfrac{25}{72} \\qquad \\text{(E)}\\ \\dfrac{13}{36}$ ## Problem 21 A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure", "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) --", "that will work are $\\{(1,0)(0,1)(1,1)\\}$. The final and last case is the powers of $5$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"5^0\",X--Y,2W); label(\"5^2\",X--Z,2E); label(\"5^2\",Y--Z,2S); [/asy]$ This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \\text{ and } y$ cannot have a factor of $5$ in their prime factorization. Multiplying all the possible arrangements for prime factors $2,3, \\text{ and } 5$, we get the answer: $$5\\cdot3\\cdot1=\\boxed{\\textbf{(A) }15}$$ (Diagrams by ColtsFan10)", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (7,7) -- (-7,7) -- (-7,-7) -- (7,-7) -- cycle, black ); [/asy]$ 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", "The answer is $\\boxed{\\text{B}}$ $[asy] unitsize(20); label(\"O\",(0,-.35),N); label(\"B\",(1,-.35),N); label(\"G\",(2,-.35),N); label(\"R\",(3,-.35),N); label(\"Y\",(4,-.35),N); label(\"B\",(0,.6),N); label(\"G\",(1,.6),N); label(\"R\",(2,.6),N); label(\"Y\",(3,.6),N); label(\"G\",(0,1.6),N); label(\"R\",(1,1.6),N); label(\"Y\",(2,1.6),N); label(\"R\",(0,2.6),N); label(\"Y\",(1,2.6),N); label(\"Y\",(0,3.6),N); [/asy]$", "&& (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(", "(0.2,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } fillrect((0,0),(s,s)); fillrect((a,b),(s,s),colors[0]); filldraw((0,a)--(a,a)--(s/2,s/2)--(a,b)--(a,s)--(0,s)--cycle,colors[1],linewidth(1)); filldraw((0,0)--(b,0)--(b,b)--(a,a)--(0,a)--cycle,colors[2],linewidth(1)); draw((0,0)--(a,a),linewidth(1)); draw((s/2,s/2)--(b,a)--(a,a)--(a,b),linewidth(0.7)+linetype(\"4 2\")); htick((s+1,0),(s+1,b)); htick((s+1,b),(s+1,s)); /* in labels, a,b swapped */ label(\"a\",(s+1,b/2),E);label(\"b\",(s+1,(s+b)/2),E); label(\"ab\",(a+s,b+s)/2,sm); label(\"\\frac{(a+b)^2}{4}\",(a,a+s)/2,sm); label(\"\\frac{a^2}2\",(s/2,a*2/3),sm); label(\"\\frac{b^2}2\",(a/4,a*2/3),sm); [/asy]$ The Root-Mean Square-Arithmetic Mean-Geometric Mean inequality, $\\color{red}{ab} \\color{black} \\le \\color{green} \\frac{(a+b)^2}{4} \\color{black} \\le \\color{blue} \\frac{a^2 + b^2}{2}$. $[asy] unitsize(15); defaultpen(linewidth(0.7)); void htick(pair A, pair B,pair ticklength = (0,0.15)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } real a=10,b=3,r=(a+b)/2; pen sm = fontsize(8), dark = linewidth(1); pen colors[] = {rgb(0.9,0.2,0.2) + dark, /* GM */ rgb(0.2,0.9,0.2) + dark, /* AM */ rgb(0.2,0.2,0.9) + dark, /* QM */ rgb(0.2,0.9,0.9) + dark }; /* HM */ pair A = (r-b,(r^2-(r-a)^2)^.5),B=foot((A.x,0),(0,0),A); draw(arc((0,0),r,0,180)--cycle); dot(A); dot((0,r)); dot((A.x,0)); dot((0,0)); draw(B--A,colors[3]); label(\"HM\",(A+B)/2, E, sm+colors[3]); draw((0,0)--(0,r),colors[1]); label(\"AM\",(0,r*2/3), NW, sm+colors[1]); draw((A.x,0)--A,colors[0]); label(\"GM\",(A.x,A.y/2), SE, sm+colors[0]); draw((A.x,0)--(0,r),colors[2]); label(\"RMS\",(A.x/5,r*4/5), NE, sm+colors[2]); draw((-r,0)--A--(r,0), linetype(\"4 2\")); draw((0,0)--B--(A.x,0), linetype(\"4 2\")); draw(rightanglemark((-r,0),A,(r,0))); draw(rightanglemark((0,0),B,(A.x,0))); htick((-r,-1),(A.x,-1)); htick((A.x,-1),(r,-1)); label(\"a\",((-r+A.x)/2,-1),S); label(\"b\",((r+A.x)/2,-1),S); [/asy]$ $[asy] unitsize(15); defaultpen(linewidth(0.7)); real r = 0.3, row1 = 3.5, row2 = 0, row3 = -3.5; void necklace(pair k, pen colors[]){ draw(shift(k)*unitcircle); for(int i = 0; i < colors.length; ++i){ pair p = k+expi(pi/2+2*pi*i/colors.length); fill(Circle(p,r),colors[i]); draw(Circle(p,r)); } } void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } /* draw necklaces */ pen BEADS1[] = {red,red,red},BEADS2[] = {blue,blue,blue},BEADS3[] = {red,red,blue},BEADS4[] = {blue,red,red},BEADS5[] = {red,blue,red},BEADS6[] = {blue,blue,red},BEADS7[] = {red,blue,blue},BEADS8[] = {blue,red,blue}; necklace((-10,(row2+row3)/2),BEADS1);necklace((-7.5,(row2+row3)/2),BEADS2); necklace((-2.5,row2),BEADS3);necklace((0,row2),BEADS4);necklace((2.5,row2),BEADS5); necklace((-2.5,row3),BEADS6);necklace((0,row3),BEADS7);necklace((2.5,row3),BEADS8); /* box them and label */ draw((-4,row2-1.3)--(4,row2-1.3)--(4,row2+1.6)--(-4,row2+1.6)--cycle,linewidth(0.9)+linetype(\"4 2\")); draw((-4,row3-1.3)--(4,row3-1.3)--(4,row3+1.6)--(-4,row3+1.6)--cycle,linewidth(0.9)+linetype(\"4 2\"));", "(int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ So the answer is $\\boxed{\\textbf{(A)}28}$ ~yofro (Diagram credits to franzliszt)", "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 ##", "replaced with sentiment scores 0...4: • 0 and 1 are negative labels. • 2 is a neutral label. • 3 and 4 are positive labels. • Our readers are iterators that yield (tree, label) pairs, where tree is an NLTK Tree instance and score is a string. We'll mainly work with sst.train_reader and sst.dev_reader. In [3]: tree, score = next(sst.train_reader()) Here, score is one of the labels. tree is an NLTK Tree instance. It should render pretty legibly in your browser: In [4]: tree Out[4]: This is what it actually looks like, of course: In [5]: (tree,) Out[5]: (Tree('S', [Tree('2', [Tree('2', ['The']), Tree('2', ['Rock'])]), Tree('4', [Tree('3', [Tree('2', ['is']), Tree('4', [Tree('2', ['destined']), Tree('2', [Tree('2', [Tree('2', [Tree('2', [Tree('2', ['to']), Tree('2', [Tree('2', ['be']), Tree('2', [Tree('2', ['the']), Tree('2', [Tree('2', ['21st']), Tree('2', [Tree('2', [Tree('2', ['Century']), Tree('2', [\"'s\"])]), Tree('2', [Tree('3', ['new']), Tree('2', [Tree('2', ['']), Tree('2', ['Conan'])])])])])])])]), Tree('2', [\"''\"])]), Tree('2', ['and'])]), Tree('3', [Tree('2', ['that']), Tree('3', [Tree('2', ['he']), Tree('3', [Tree('2', [\"'s\"]), Tree('3', [Tree('2', ['going']), Tree('3', [Tree('2', ['to']), Tree('4', [Tree('3', [Tree('2', ['make']), Tree('3', [Tree('3', [Tree('2', ['a']), Tree('3', ['splash'])]), Tree('2', [Tree('2', ['even']), Tree('3', ['greater'])])])]), Tree('2', [Tree('2', ['than']), Tree('2', [Tree('2', [Tree('2', [Tree('2', [Tree('1', [Tree('2', ['Arnold']), Tree('2', ['Schwarzenegger'])]), Tree('2', [','])]), Tree('2', [Tree('2', ['Jean-Claud']), Tree('2', [Tree('2', ['Van']), Tree('2', ['Damme'])])])]), Tree('2', ['or'])]), Tree('2', [Tree('2', ['Steven']), Tree('2', ['Segal'])])])])])])])])])])])])]), Tree('2', ['.'])])]),) Here's a smaller example: In [6]: Tree.fromstring(\"\"\"(4 (2 NLU) (4 (2 is) (4" ]

[ "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "(\\text{-} x + 8)$ $5 + 2(\\text{-}7) + 2 \\cdot x = \\text{-} 6 - (\\text{-} x + 8)$ $5 -14 + 2x = \\text{-} 6 - (\\text{-} x + 8)$ $5 -14 + 2x = \\text{-} 6 +x - 8$ $\\text{-} 9 + 2x = x - 14$ $2x = x - 5$ $x= \\text{-} 5$", "a lot. By inspection, we find that $\\text{LCM}(1,2,3,2\\cdot2,2\\cdot3) = 12$. Finally, $1+2 = \\boxed{\\textbf{(B)}\\ 3}$.", "\\: \\text{and}\\:y_1-3=3$ $x_1=5 \\:\\text{and}\\:y_1=6$ $\\boxed{D(x_1, y_1) = D(5,6)}$", "Thus, the answer is $\\boxed{\\textbf{(C)}}.$", "$\\text{A.}$ 1 $\\text{B.}$ $\\sqrt{2}$ $\\text{C.}$ $\\sqrt{3}$ $\\text{D.}$ 3 【答案】 D 【解析】 $\\because|\\vec{a}|=2, \\therefore \\vec{a} \\cdot \\vec{b}=|\\vec{a}||\\vec{b}| \\cos \\theta=2|\\vec{b}| \\cos \\theta=4$,", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "$\\text{I = 1, V = 5, X = 10, L = 50, C = 100, D = 500,}$ and $\\text{M = 1000​}$ 1. $\\text{MCMLXXV: 1000 + (1000 - 100) + 50 + 10 + 10 + 5 = 1975}$ 2. $\\text{MCMXCV: 1000 + (1000 - 100) + (100 - 10) + 5 = 1995}$ 3. $\\text{MVD: 1000 + (500 – 5) = 1495}$ 4. $\\text{MVM: 1000 + (1000 – 5) = 1995}$ Answer: Option$:\\text{C.}$ Only (b) and (d) 406 points Ans is C) and D) for C) 1000+(1000-100)+(100-10)+5=1995 D) 1000+(1000-5)=1995 by 5.1k points you meant b?", "answer is $1000=10^3$. Therefore: $1 \\text{m}=10^3 \\text{mm}$, or equivalently, $1 \\text{mm}= 10^{-3} \\text{m}$. $$1750 \\text{mm}^2=1750 (10^{-3}\\text{m})^2=1750 * 10^{-6} \\text{m}^2$$", "\\cdot \\left(9\\right) = \\textcolor{g r e e n}{45}$ $z = \\left(9\\right) - 2 = \\textcolor{g r e e n}{7}$ The three numbers are $9$, $45$, and $7$.", "π_1=\\underset{w \\in W_1}{\\text{argmax }} \\mathbb{E}_{U \\sim b_1} [U(w)+U(π_2|w)] \\\\ π_2=\\underset{w \\in W_2}{\\text{argmax }} \\mathbb{E}_{U \\sim b_2} [U(w)] \\\\ \\\\ b_l=[0.99: U_x, 0.01: U_y] \\\\ b_k=[0.001: U_x, 0.999: U_y] \\\\ \\\\ b_1=b_l \\\\ \\text{if } π_1=d \\text{ then } b_2=b_k \\text{ else } b_2=b_l$$ What is $π_1$? $$\\mathbb{E}^{2}_{U \\sim b_l}[a]=0.99 \\cdot 1+0.01 \\cdot 0=0.99 \\\\ \\mathbb{E}^{2}_{U \\sim b_l}[b]=0.99 \\cdot 0+0.01 \\cdot 1=0.01 \\\\ \\\\ \\mathbb{E}^{2}_{U \\sim b_k}[a]=0.01 \\cdot 4+0.99 \\cdot 0=0.04 \\\\ \\mathbb{E}^{2}_{U \\sim b_k}[b]=0.999 \\cdot 4+0.001 \\cdot 0=3.996 \\\\ \\\\ π_{2|U \\sim b_l}=a \\\\ π_{2|U \\sim b_k}=b \\\\ \\\\ \\mathbb{E}^1_{U \\sim b_l}[a]=\\\\ 0.99 \\cdot U_x(a)+0.01 \\cdot U_y(a)+0.99 \\cdot U_x(π_2|a)+0.01 \\cdot U_y(π_2|a)=\\\\ 0.99+0+0.99 \\cdot U_x(a)+0.01 \\cdot U_y(a)=\\\\ 1.98 \\\\ \\mathbb{E}^1_{U \\sim b_l}[b]=\\\\ 0.99 \\cdot U_x(b)+0.01 \\cdot U_y(b)+0.99 \\cdot U_x(π_2|b)+0.01 \\cdot U_y(π_2|b)=\\\\ 0+0.04+0.99 \\cdot U_x(a)+0.01 \\cdot U_y(a)=\\\\ 1.03 \\\\ \\mathbb{E}^1_{U \\sim b_l}[d]=\\\\ 0.99 \\cdot U_x(d)+0.01 \\cdot U_y(d)+0.99 \\cdot U_x(π_2|d)+0.01 \\cdot U_y(π_2|d)=\\\\ 0+0+0.99 \\cdot U_x(b)+0.01 \\cdot U_y(b)=\\\\ 0.04$$ My wild and at best flimsily supported conjecture is that this is both a model for why CIRL (Hadfield-Menell et al. 2016) is incorrigible (Soares 2015), and also for why utility maximizers don't do reward hacking (Dewey 2010). Explanation: $d$ can be both an action that is equivalent to hacking reward, or to switching to a different utility function. ## Geometric Mean of Odds, Arithmetic Mean of Logodds Theorem: The arithmetic mean of", "2 \\cdot 2 = {2}^{5} = 32\\right)$ =$4$ Compare this with the other method of squaring first. $\\sqrt[5]{\\textcolor{b l u e}{{32}^{2}}} = \\sqrt[5]{\\textcolor{b l u e}{1024}}$ =$4$ While I know that ${2}^{5} = 32$, the square of $32$ and the fifth root of $1024$ are not facts that I would be able to recall from memory. • 15 minutes ago • 16 minutes ago • 17 minutes ago • 20 minutes ago • 3 minutes ago • 6 minutes ago • 12 minutes ago • 12 minutes ago • 14 minutes ago • 14 minutes ago • 15 minutes ago • 16 minutes ago • 17 minutes ago • 20 minutes ago", "b = 0, c = -7 \\text{ or } a = 4, b = -4, c = 3}$$ Feb 20, 2021 #3 +112966 +1 That is really cool Textot. Thanks very much for showing me. Feb 20, 2021", "As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\\textcolor{red}{\\text{red}}$. So, thus, there are $5$ possibilities for each case, and $5\\cdot 5=\\boxed{\\textbf{(E) }25}$. ## Solution 3 Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits. Then let $a_i$, $b_i$ denotes the digits of $N_5$, $N_6$, respectively such that $$0\\le a_i<5,0\\le b_i<6$$ Thus we have $$N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4$$ $$625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4$$ Now we are given $$2N \\equiv S \\equiv N_5+N_6\\pmod{100}$$ $$2(625a_1+125a_2+25a_3+5a_4+a_5) \\equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\\pmod{100}$$ $$1250a_1+250a_2+50a_3+10a_4+2a_5 \\equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\\pmod{100}$$ $$50a_1+50a_2+50a_3+10a_4+2a_5 \\equiv 10a_4+10b_3+a_5+b_4\\pmod{100}$$ Canceling out $a_5$ left with $$50a_1+50a_2+50a_3+10a_4+a_5 \\equiv 10a_4+10b_3+b_4\\pmod{100}$$ Since $a_5$, $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$. Thus, $$50a_1+50a_2+50a_3+10a_4 \\equiv 10a_4+10b_3\\pmod{100}$$ canceling out $10a_4$, we have $$50a_1+50a_2+50a_3 \\equiv 10b_3\\pmod{100}$$ $$5a_1+5a_2+5a_3 \\equiv b_3\\pmod{10}$$ $$5(a_1+a_2+a_3) \\equiv b_3\\pmod{10}$$ Thus $b_3$ is a multiple of $5$. Now going back to our original equation $$625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4$$ Since $a_5=b_4$, $$625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3$$ $$5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)$$ $$5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]$$ Since the left side is a multiple of $5$, then so does the right side. Thus $5\\mid6(6b_1+b_2)+b_3$. Since we already know that $5\\mid b_3$, then $5\\mid6(6b_1+b_2)$, from where we also know that $5\\mid6b_1+b_2$. For $b_1,b_2<6$, there is a total of 7 ordered pairs that satisfy the condition. Namely, $$(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)$$ Since $N_6$ has at least $3$ digits, $(b_1,b_2)=(0,0)$ doesn't work. Furthermore, when $b_1=5$, $216b_1$ exceeds", "As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\\textcolor{red}{\\text{red}}$. So, thus, there are $5$ possibilities for each case, and $5\\cdot 5=\\boxed{\\textbf{(E) }25}$. ## Solution 3 Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits. Then let $a_i$, $b_i$ denotes the digits of $N_5$, $N_6$, respectively such that $$0\\le a_i<5,0\\le b_i<6$$ Thus we have $$N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4$$ $$625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4$$ Now we are given $$2N \\equiv S \\equiv N_5+N_6\\pmod{100}$$ $$2(625a_1+125a_2+25a_3+5a_4+a_5) \\equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\\pmod{100}$$ $$1250a_1+250a_2+50a_3+10a_4+2a_5 \\equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\\pmod{100}$$ $$50a_1+50a_2+50a_3+10a_4+2a_5 \\equiv 10a_4+10b_3+a_5+b_4\\pmod{100}$$ Moving the $a_5$ to left, we have $$50a_1+50a_2+50a_3+10a_4+a_5 \\equiv 10a_4+10b_3+b_4\\pmod{100}$$ Since $a_5$, $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$. Thus, $$50a_1+50a_2+50a_3+10a_4 \\equiv 10a_4+10b_3\\pmod{100}$$ canceling out $10a_4$, we have $$50a_1+50a_2+50a_3 \\equiv 10b_3\\pmod{100}$$ $$5a_1+5a_2+5a_3 \\equiv b_3\\pmod{10}$$ 2013 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 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 2013 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Question", "# What is the GCF of 18a^2b^2 − 6ab^2 + 12ab^3? Jun 21, 2018 GCF is $6 a {b}^{2}$ #### Explanation: Given: $18 {a}^{2} {b}^{2} - 6 a {b}^{2} + 12 a {b}^{3}$ The GCF is the greatest common factor. TO find it you can decompose each monomial: $18 {a}^{2} {b}^{2} : \\text{ } \\textcolor{red}{2} \\cdot \\textcolor{red}{3} \\cdot 3 \\cdot \\textcolor{red}{a} \\cdot a \\cdot \\textcolor{red}{b} \\cdot \\textcolor{red}{b}$ $- 6 a {b}^{2} : \\text{ } - 1 \\cdot \\textcolor{red}{2} \\cdot \\textcolor{red}{3} \\cdot \\textcolor{red}{a} \\cdot \\textcolor{red}{b} \\cdot \\textcolor{red}{b}$ $12 a {b}^{3} : \\text{ } \\textcolor{red}{2} \\cdot 2 \\cdot \\textcolor{red}{3} \\cdot \\textcolor{red}{a} \\cdot \\textcolor{red}{b} \\cdot \\textcolor{red}{b} \\cdot b$ The GCF is $6 a {b}^{2}$", "$ \\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", "doing so, we get $$(r+s)^3+(s+t)^3+(t+r)^3=(r+s+t)^3-6rst+(r^3+s^3+t^3)$$ Recall that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Thus, we get $$(r+s)^3+(s+t)^3+(t+r)^3=(r+s+t)^3-3rst+(r+s+t)(r^2+s^2+t^2-rs-st-tr)$$ Plugging in $(r+s+t)=0$ we get $$(r+s)^3+(s+t)^3+(t+r)^3=0-3rst+0=-3\\cdot -251=\\boxed{753}$$ ~firebolt360", "sides. This shows that $3 - 5 = 3 + (\\text- 5)$ In general: $$a - b = a + (\\text- b)$$ If $a - b = x$, then $x + b = a$. We can add $\\text- b$ to both sides of this second equation to get that $x = a + (\\text- b)$", "2 \\text { (or person } 3, \\text { if } \\\\ \\text { person } 2 \\text { was already paired to person } 1 \\text { ), and so on. }\\end{array}$$ Can someone explain to me 1) why the factor is $$n! \\cdot 2^n$$ and 2) how is the RHS derived? • See if this answer helps. – Brian M. Scott Sep 16 at 3:10 • At least it's easier than overstanding undercounting ;) – Greg Martin Sep 16 at 7:05 The first part says if we have say ABCDEF, then (AB)(CD)(EF) = (BA)(CD)(EF), and also (AB)(CD)(EF) = (CD)(AB)(EF), with there being $$2^n$$ variations of the first type, and $$n!$$ variations of the second. Then we also have $$A$$ can be partnered with $$BCDEF$$, then mystery person $$x$$ can be partnered with the remaining $$3$$, etc. . And $$\\frac{6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}{8\\cdot3\\cdot2\\cdot1} = \\frac{6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}{6\\cdot4\\cdot2} = 5\\cdot3\\cdot1$$." ]

[ "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", "1996 AJHSME Problems/Problem 5 Problem 5 The letters $P$, $Q$, $R$, $S$, and $T$ represent numbers located on the number line as shown. $[asy] unitsize(36); draw((-4,0)--(4,0)); draw((-3.9,0.1)--(-4,0)--(-3.9,-0.1)); draw((3.9,0.1)--(4,0)--(3.9,-0.1)); for (int i = -3; i <= 3; ++i) { draw((i,-0.1)--(i,0)); } label(\"-3\",(-3,-0.1),S); label(\"-2\",(-2,-0.1),S); label(\"-1\",(-1,-0.1),S); label(\"0\",(0,-0.1),S); label(\"1\",(1,-0.1),S); label(\"2\",(2,-0.1),S); label(\"3\",(3,-0.1),S); draw((-3.7,0.1)--(-3.6,0)--(-3.5,0.1)); draw((-3.6,0)--(-3.6,0.25)); label(\"P\",(-3.6,0.25),N); draw((-1.3,0.1)--(-1.2,0)--(-1.1,0.1)); draw((-1.2,0)--(-1.2,0.25)); label(\"Q\",(-1.2,0.25),N); draw((0.1,0.1)--(0.2,0)--(0.3,0.1)); draw((0.2,0)--(0.2,0.25)); label(\"R\",(0.2,0.25),N); draw((0.8,0.1)--(0.9,0)--(1,0.1)); draw((0.9,0)--(0.9,0.25)); label(\"S\",(0.9,0.25),N); draw((1.4,0.1)--(1.5,0)--(1.6,0.1)); draw((1.5,0)--(1.5,0.25)); label(\"T\",(1.5,0.25),N); [/asy]$ Which of the following expressions represents a negative number? $\\text{(A)}\\ P-Q \\qquad \\text{(B)}\\ P\\cdot Q \\qquad \\text{(C)}\\ \\dfrac{S}{Q}\\cdot P \\qquad \\text{(D)}\\ \\dfrac{R}{P\\cdot Q} \\qquad \\text{(E)}\\ \\dfrac{S+T}{R}$ Solution 1 First, note that $P < Q < 0 < R < S < T$. Thus, $P$ and $Q$ are negative, while $R$, $S$, and $T$ are positive. Option $A$ is the difference of two numbers. If the first number is lower than the second number, the answer will be negative. In this case, $P < Q$, so $P - Q < 0$, and $P-Q$ is thus indeed a negative number. So option $\\boxed{A}$ is correct. Option $B$ is the product of two negatives, which is always positive. Option $C$ starts with the quotient of a positive and a negative. This is negative. But this negative quotient is then multiplied by a negative number, which gives a positive number. Option $D$ has the product of two negatives in", "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]", "(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", "that will work are $\\{(1,0)(0,1)(1,1)\\}$. The final and last case is the powers of $5$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"5^0\",X--Y,2W); label(\"5^2\",X--Z,2E); label(\"5^2\",Y--Z,2S); [/asy]$ This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \\text{ and } y$ cannot have a factor of $5$ in their prime factorization. Multiplying all the possible arrangements for prime factors $2,3, \\text{ and } 5$, we get the answer: $$5\\cdot3\\cdot1=\\boxed{\\textbf{(A) }15}$$ (Diagrams by ColtsFan10)", "be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label(\"\\boldmath{1}\",(0,0)); label(\"\\boldmath{12}\",(1,0)); label(\"\\boldmath{123}\",(2,0)); label(\"\\boldmath{S}\",(3,0)); label(\"11\",(0.5,-0.75)); label(\"111\",(1.5,-0.75)); label(\"d_1\",(2.5,-0.75)); label(\"100\",(1,-1.5)); label(\"d_2\",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label(\"\\textbf{First Differences}\",(-0.75,-0.75),align=W); label(\"\\textbf{Second Differences}\",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334}. \\end{align*} ~MRENTHUSIASM ## Solution 4 (Finite Differences by Algebra) Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$. Alternatively, applying finite differences, one obtains $$c_4 = {3 \\choose 2}f(2) - {3 \\choose 1}f(1) + {3 \\choose 0}f(0) =334.$$ ## Solution 5 (Assumption) The idea", "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "label(\"$v^\\prime$\",hb.hexCorner(0,10,5)+(1,0),E); label(\"$w$\",hb.hexCorner(-10,0,3)-(0,1),S); label(\"$w^\\prime$\",hb.hexCorner(10,0,0)+(0,1),N); label(\"$o^\\prime$\",hb.hexCorner(5,0,1)-(1,0),W); label(\"$x^-$\",hb.hexCorner(5,5,5)+(1,0),E); hb.oline((5,0),0,1,2,3,2,3,2,3,4,5,4,5,0,5, 4,5,0,1,0,1,2); hb.oline((10,0), prefix=(0,1) ,suffix=(1,0) ,0 ,3,2,3,2,1,0,5,0,1,2,1,2,3,4,3,4,3,2,3,2,3,4 ,3,2,3,2,1,0,5,0,1,0,1,2,3,2,3,4 ,3,2,3,4 ,3,4,3,2,3); hb.oline((0,0), prefix=(-1,0),2,3,2,3,4); hb.oline((-10,0), prefix=(0,-1),3,4,5,0); hb.dot(9,0,3); hb.dot(7,1,4); hb.dots(5,1,0,3); shipout(bbox(Fill(lightyellow))); \\end{asy} \\caption{Hex Board} \\end{figure} \\end{document} % % Process : % % pdflatex hexboard.tex % asy hexboard-*.asy % pdflatex hexboard.tex - One question, how do I make the figure larger? Preferably so that it fills the whole page. Other than that, this worked perfectly, although the co-ordinates of the hexagons is confusing. Maybe it will make sense after I get some sleep, the co-ordinate system led to a huge guessing game but I got the final product! – Surculus Nov 3 '13 at 1:41 @Surculus: One way to scale the figure to fit the whole page would be as follows: make it a standalone using standalone document class (don't forget to remove the figure environment) and then include it with pdfpages package. As for co-ordinates: perhaps it would be less confusing with the edited picture: direction is named after the node of destination. – g.kov Nov 3 '13 at 8:02 I modified some snippets from Jake's answer about Chinese checkers and from Alain's answer about an hexagonal grid. Using the node labeling you can draw thick lines in between. It's better to use the corners as references, like I did in this answer about naming corners", "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. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## 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 $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "# 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", "(int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ So the answer is $\\boxed{\\textbf{(A)}28}$ ~yofro (Diagram credits to franzliszt)", "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]$ ## Problem 19 Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$, has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is $[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 32 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 48 \\qquad \\text{(E)}\\ 64$ ## Problem 20 Let $W,X,Y$ and $Z$ be four different digits selected from the set $\\{ 1,2,3,4,5,6,7,8,9\\}.$ If the sum $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ is to be as small as possible, then $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ must equal $\\text{(A)}\\ \\dfrac{2}{17} \\qquad \\text{(B)}\\ \\dfrac{3}{17} \\qquad \\text{(C)}\\ \\dfrac{17}{72} \\qquad \\text{(D)}\\ \\dfrac{25}{72} \\qquad \\text{(E)}\\ \\dfrac{13}{36}$ ## Problem 21 A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure", "prime factors $2,3, \\text{and } 5$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"2^3\",X--Y,2W); label(\"2^3\",X--Z,2E); label(\"2^2\",Y--Z,2S); [/asy]$ Analyzing for powers of $2$, it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \\text{ nor } z$ can have a power of $2$ exceeding $2$. Turning towards the vertices $y$ and $z$, we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \\text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \\{(2,0)(2,1)(0,2)(1,2)(2,2)\\}$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"3^2\",X--Y,2W); label(\"3^1\",X--Z,2E); label(\"3^2\",Y--Z,2S); [/asy]$ Using the same logic as we did for powers of $2$, it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \\text{ and } z$, we can use the same logic to find the only ordered pairs $(x,z)$", "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", "The answer is $\\boxed{\\text{B}}$ $[asy] unitsize(20); label(\"O\",(0,-.35),N); label(\"B\",(1,-.35),N); label(\"G\",(2,-.35),N); label(\"R\",(3,-.35),N); label(\"Y\",(4,-.35),N); label(\"B\",(0,.6),N); label(\"G\",(1,.6),N); label(\"R\",(2,.6),N); label(\"Y\",(3,.6),N); label(\"G\",(0,1.6),N); label(\"R\",(1,1.6),N); label(\"Y\",(2,1.6),N); label(\"R\",(0,2.6),N); label(\"Y\",(1,2.6),N); label(\"Y\",(0,3.6),N); [/asy]$", "(int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ So the answer is $\\boxed{\\textbf{(A)}28}$ ~yofro (Diagram credits to franzliszt) ## Solution 2 Suppose we \"extend\" the chessboard indefinitely to the right: $[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label(\"P\", (5.5,.5)); label(\"Q\", (6.5,7.5)); label(\"X\", (8.5,3.5)); label(\"Y\", (8.5,5.5)); [/asy]$ The total number of paths from $P$ to $Q$ (including invalid paths which cross over the red line) is $\\binom{7}{3} = 35$. We subtract the number of invalid paths that pass through $X$ or $Y$. The number of paths that pass through $X$ is $\\binom{3}{0}\\binom{4}{3} = 4$ and the number of paths that pass through $Y$ is $\\binom{5}{1}\\binom{2}{2} = 5$. However, we overcounted the", "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) --", "# 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}$.", "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 ##" ]

[ "# Problem of the Week Problem B and Solution How Tall are They, Really? ## Problem There were $$12$$ players on the Canadian $$2020$$ Olympic Women’s Basketball Team. The table below shows the height of each player. Name Height Natalie Achonwa $$6$$ feet, $$3$$ inches Kayla Alexander $$6$$ feet, $$4$$ inches Laeticia Amihere $$6$$ feet, $$2$$ inches Miranda Ayim $$6$$ feet, $$3$$ inches Bridget Carleton $$6$$ feet, $$1$$ inch Shay Colley $$5$$ feet, $$8$$ inches Aaliyah Edwards $$6$$ feet, $$3$$ inches Nirra Fields $$5$$ feet, $$7$$ inches Kim Gaucher $$6$$ feet, $$1$$ inch Kia Nurse $$6$$ feet, $$0$$ inches Shaina Pellington $$5$$ feet, $$8$$ inches Nayo Raincock-Ekunwe $$6$$ feet, $$2$$ inches 1. Create a stem-and-leaf plot to represent the heights in feet and inches of the players on the team. Use the number of feet as the stems, and the number of inches as the leaves. 2. Using your stem-and-leaf plot, find the median height of the players on the team. 3. Convert all the heights to inches, and then calculate the mean height of the players on the team. Recall that $$1~\\text{foot} = 12~\\text{inches}$$. For example, $$6$$ feet is equal to $$6 \\times 12=72$$ inches, so a height of $$6$$ feet, $$3$$ inches is equal to $$72+3=75$$ inches. 4. Assume that the mean height of a Canadian", "• What is 3 What is the least common denominator of 11/15 and 12/19? The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? • Eq2 2 Solve following equation with quadratic members and rational function: (x2+1)/(x-4) + (x2-1)/(x+3) = 23 • Prime factors Write 98 as a product of prime factors", "The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? Result a = ft (Correct answer is: 6'2\") #### Solution: $h_{ 1 } = 5 \\cdot \\ 12+11 = 71 \\ inch \\ \\\\ h_{ 2 } = 6 \\cdot \\ 12+3 = 75 \\ inch \\ \\\\ h_{ 3 } = 6 \\cdot \\ 12+6 = 78 \\ inch \\ \\\\ h_{ 4 } = 6 \\cdot \\ 12+2 = 74 \\ inch \\ \\\\ h_{ 5 } = 6 \\cdot \\ 12 = 72 \\ inch \\ \\\\ \\ \\\\ b = \\dfrac{ h_{ 1 }+h_{ 2 }+h_{ 3 }+h_{ 4 }+h_{ 5 } }{ 5 } = \\dfrac{ 71+75+78+74+72 }{ 5 } = 74 \\ inch \\ \\\\ 74 = 12 \\times 6 + 2 \\ \\\\ \\ \\\\ a = 6'2\"$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Dr Math For conversion inches to feet and inch we user reminder calculator: https://www.hackmath.net/en/calculator/quotient-and-remainder?n=74&d=12 74 =", "The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? Result a = ft (Correct answer is: 6'2\") #### Solution: $h_{ 1 } = 5 \\cdot \\ 12+11 = 71 \\ inch \\ \\\\ h_{ 2 } = 6 \\cdot \\ 12+3 = 75 \\ inch \\ \\\\ h_{ 3 } = 6 \\cdot \\ 12+6 = 78 \\ inch \\ \\\\ h_{ 4 } = 6 \\cdot \\ 12+2 = 74 \\ inch \\ \\\\ h_{ 5 } = 6 \\cdot \\ 12 = 72 \\ inch \\ \\\\ \\ \\\\ b = \\dfrac{ h_{ 1 }+h_{ 2 }+h_{ 3 }+h_{ 4 }+h_{ 5 } }{ 5 } = \\dfrac{ 71+75+78+74+72 }{ 5 } = 74 \\ inch \\ \\\\ 74 = 12 \\times 6 + 2 \\ \\\\ \\ \\\\ a = 6'2\"$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Dr Math For conversion inches to feet and inch we user reminder calculator: https://www.hackmath.net/en/calculator/quotient-and-remainder?n=74&d=12 74 =", "on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" The best way to do this problem is to convert the respective heights into inches and finding the average, since the answers for the average height are very close! When converted in inches, we will get 71,72,74,75, and 78 inches. The average is $$\\frac{70*5}{5} + \\frac{1+2+4+5+8}{5} = 70 + \\frac{20}{5} = 74$$ inches, which is 6' 2\"(Option B) can't we just arrange the give heights in ascending order and take middle value as mean as there are odd no of elements ? Hi utkarshrihand The middle value is the mean when the list is an arithmetic progression (has a common difference). However, in this list, the elements do not have a common difference and the middle value can't be the mean Hope this helps you! _________________ You've got what it takes, but it will take everything you've got Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8069 Location: United States (CA) Re: The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 15 Mar 2018,", "The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 01 Mar 2018, 00:52 Bunuel wrote: The heights of the five starters on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" Check other Conversion problems to practice in Special Questions Directory. _________________ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3342 Location: India GPA: 3.12 Re: The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 01 Mar 2018, 01:18 Bunuel wrote: The heights of the five starters on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" The best way to do this problem is to convert the respective heights into inches and finding the average, since the answers for the average height are very close! When converted in inches, we will get 71,72,74,75, and 78 inches. The average is $$\\frac{70*5}{5} + \\frac{1+2+4+5+8}{5} = 70 + \\frac{20}{5} = 74$$ inches, which is 6' 2\"(Option B) _________________ You've got", "and return to Rome... 1 answer ### ​The heights, in inches, of five players on a basketball ​team are 68, 70, 76, 71, and 70. What is the mean ​height, in inches, of these players? ​​ ​The heights, in inches, of five players on a basketball ​team are 68, 70, 76, 71, and 70. What is the mean ​height, in inches, of these players? ​​... -- 0.049328--", "## alexisreed Group Title i neeeeeeeed helppppp 2 years ago 2 years ago 1. 142343 with what? 2. alexisreed following data represents the heights of the players on a college women's volleyball team. What is the mean height of the players rounded to tenths place? Heights (inches): 72, 72, 65, 72, 74, 72, 74, 72, 74, 73, 68, 72, 70, 74, 71 3. saifoo.khan $Mean = \\frac{\\text{\\sum of all terms}}{\\text{number of terms}}$", "woman is $$5$$ feet, $$4$$ inches. How much taller, on average, are the players on this team? ## Solution 1. The stem-and-leaf plot is shown below. Stem Leaf $$5$$ $$~7~$$$$~8~$$ $$8~$$ $$6$$ $$~0~$$$$~1~$$$$~1~$$$$~2~$$$$~2~$$$$~3~$$$$~3~$$$$~4~$$ Key: $$6 ~\\vert~ 3 = 6~\\text{feet},~3~\\text{inches}$$ 2. The median is halfway between $$6$$ feet, $$1$$ inch and $$6$$ feet, $$2$$ inches, which is $$6$$ feet, $$1.5$$ inches. 3. The table below shows all the heights converted to inches. Name Height (feet, inches) Height (inches) Natalie Achonwa $$6$$ feet, $$3$$ inches $$72+3=75$$ Kayla Alexander $$6$$ feet, $$4$$ inches $$72+4=76$$ Laeticia Amihere $$6$$ feet, $$2$$ inches $$72+2=74$$ Miranda Ayim $$6$$ feet, $$3$$ inches $$72+3=75$$ Bridget Carleton $$6$$ feet, $$1$$ inch $$72+1=73$$ Shay Colley $$5$$ feet, $$8$$ inches $$60+8=68$$ Aaliyah Edwards $$6$$ feet, $$3$$ inches $$72+3=75$$ Nirra Fields $$5$$ feet, $$7$$ inches $$60+7=67$$ Kim Gaucher $$6$$ feet, $$1$$ inch $$72+1=73$$ Kia Nurse $$6$$ feet, $$0$$ inches $$72$$ Shaina Pellington $$5$$ feet, $$8$$ inches $$60+8=68$$ Nayo Raincock-Ekunwe $$6$$ feet, $$2$$ inches $$72+2=74$$ The mean is found by adding up all the heights in inches, and dividing by $$12$$. $75+76+74+75+73+68+75+67+73+72+68+74=870$ Since $$870 \\div 12=72.5$$, the mean height is $$72.5$$ inches, or $$6$$ feet, $$0.5$$ inches. 4. We can first convert $$5$$ feet, $$4$$ inches to inches. $$5$$ feet is equal to $$5 \\times 12=60$$ inches, so a height of $$5$$", "## alexisreed i neeeeeeeed helppppp one year ago one year ago 1. 142343 with what? 2. alexisreed following data represents the heights of the players on a college women's volleyball team. What is the mean height of the players rounded to tenths place? Heights (inches): 72, 72, 65, 72, 74, 72, 74, 72, 74, 73, 68, 72, 70, 74, 71 3. saifoo.khan $Mean = \\frac{\\text{\\sum of all terms}}{\\text{number of terms}}$", "anonymous 4 years ago i neeeeeeeed helppppp 1. anonymous with what? 2. anonymous following data represents the heights of the players on a college women's volleyball team. What is the mean height of the players rounded to tenths place? Heights (inches): 72, 72, 65, 72, 74, 72, 74, 72, 74, 73, 68, 72, 70, 74, 71 3. saifoo.khan $Mean = \\frac{\\text{\\sum of all terms}}{\\text{number of terms}}$", "# The average height of 5 basketball players at south high school is 6 feet 2 inches. if The average height of 5 basketball players at south high school is 6 feet 2 inches. if four of these players have heights of 5’8”,6’0”,6’5”,and 6’6”,how tall is the fifth player? ## This Post Has 4 Comments 1. Expert says: 14.3 step-by-step explanation: 9 + 12 + 15 + 16 + 17 + 17 = 86 now, divide 86 by how many numbers we have: 6. 86 / 6 = 14.3 2. kellyzeissss1286 says: The 5th player is 6 feet 1 inch tall 3. Expert says: Idrk but i think u should figure out the n dia n then the units it will make it way more easier 4. myanniespencer39 says: k step-by-step explanation:", "The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? Result a = ft (Correct answer is: 6'2\") #### Solution: Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors. Dr Math For conversion inches to feet and inch we user reminder calculator: https://www.hackmath.net/en/calculator/quotient-and-remainder?n=74&d=12 74 = 12 × 6 + 2 74 inch = 6 ft and 2 inch Tips to related online calculators Looking for help with calculating arithmetic mean? Looking for a statistical calculator? Do you want to convert length units? #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Soccer team Hilahs soccer team is trying to raise $2414 to travel to a tournament in Florida, so they decided to host a pancake for the breakfast. How many people need to attend their breakfast in order", "than the basketball players. B. Visually compare the centers of the dot plots. Softball: The data is centered around 5’4’ Basketball: The data is centered around 5’8’ This means that the most common height for the softball players is 5 feet 4 inches, and for the basketball players 5 feet 8 inches. C. Visually compare the spreads of the dot plots. Softball: The spread is from 4’11 “to 5’6”. Basketball: The spread is from 5’2” to 6’0’ There is a greater spread in heights for the basketball players. Math Talk How do the heights of field hockey players compare with the heights of softball and basketball players? Your Turn Question 2. Visually compare the dot plot of heights of field hockey players to the dot plots for softball and basketball players. Shape: _______________ Center: _______________ Spread: _______________ Answer: Shapes: Hockey: All the data is 5’6” or less. The hockey players are shorter than the basketball players, but taller than the softball players. Center: Hockey: The data is centered around of 5’2”. Spread: Hockey: The spreads is from 4’9” to 5’6”. They have same spread as spread for softball players, but there is a greater spread in heights for the basketball players. All the data is 5’6″ or Less. The data is centered around of 5’2”. The spreads is", "feet, $$4$$ inches is equal to $$60+4=64$$ inches. The mean height of the players on the team is $$72.5$$ inches. Since $$72.5-64=8.5$$, that means on average, the players on the team are $$8.5$$ inches taller than the average Canadian woman.", "\\small \\frac{1}{3}$ To subtract the fractional parts, use a common denominator 4 x 3 =12 5 $\\dpi{100} \\small \\frac{3}{4}$ = 5 $\\dpi{100} \\small \\frac{9}{12}$ 3 $\\dpi{100} \\small \\frac{1}{3}$ = 3 $\\dpi{100} \\small \\frac{4}{12}$ Time spend for picking up litter = 5 $\\dpi{100} \\small \\frac{9}{12}$ – 3 $\\dpi{100} \\small \\frac{4}{12}$ = 2 $\\dpi{100} \\small \\frac{5}{12}$. Question 14. A professional basketball player is 6$$\\frac{3}{4}$$ feet tall. Your friend is 4$$\\frac{5}{6}$$ feet tall. How much taller is the basketball player than your friend? To find how much taller is the basketball player than the friend, subtract the height of the friend from the height of the basketball player. Given that, Basketball player is 6 $\\dpi{100} \\small \\frac{3}{4}$ feet tall. My friend is 4 $\\dpi{100} \\small \\frac{5}{6}$ feet tall. To subtract the fractional parts, use a common denominator 4 x 6 = 24 6 $\\dpi{100} \\small \\frac{3}{4}$ = 6 $\\dpi{100} \\small \\frac{18}{24}$ = 5 + $\\dpi{100} \\small \\frac{24}{24}$ + $\\dpi{100} \\small \\frac{18}{24}$ = 5 $\\dpi{100} \\small \\frac{42}{24}$ 4 $\\dpi{100} \\small \\frac{5}{6}$ = 4 $\\dpi{100} \\small \\frac{20}{24}$ 6 $\\dpi{100} \\small \\frac{3}{4}$ – 4 $\\dpi{100} \\small \\frac{5}{6}$ = 5 $\\dpi{100} \\small \\frac{42}{24}$ – 4 $\\dpi{100} \\small \\frac{20}{24}$ = 1 $\\dpi{100} \\small \\frac{22}{24}$ = 1 $\\dpi{100} \\small \\frac{11}{12}$ The basketball player is 1 $\\dpi{100} \\small \\frac{11}{12}$ feet taller than my friend. Question 15. Your rain", "10:10 Bunuel wrote: The heights of the five starters on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" The sum of the heights is: 5 + 6 + 6 + 6 + 6 = 29 feet plus 11 + 3 + 0 + 6 + 2 = 22 inches = 22/12 = 11/6 feet So the total height is: 29 + 11/6 = 174/6 + 11/6 = 185/6 feet Thus, the average is: (185/6)/5 = 37/6 feet = 74/12 = 6 2/12 inches = 6’2” _________________ 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: The heights of the five starters on Redwood High's basketball team are [#permalink] 15 Mar 2018, 10:10 Display posts from previous: Sort by", "### Home > CC2MN > Chapter 4 > Lesson 4.3.3 > Problem4-121 4-121. Kris said, “The Rawlings Rockets basketball team does not have any really tall players.” These are the players’ heights in inches: $70$, $77$, $75$, $68$, $88$, $70$, and $72$. 1. Which number does not seem to fit this set of data? Is there one height that is significantly higher or lower than the rest of the data? The player that is $88$ inches tall is significantly taller than the rest of the group. 2. Do you agree or disagree with Kris? Explain. Try converting these heights into feet and comparing these to your own height. The player that is $88$ inches tall is $7$ feet $4$ inches.", "and whisker diagram. (a) Write down the median height of the players. (b) Write down the upper quartile height. (c) Find the interquartile range of the heights. The height of these players are normally distributed. (d) Find the height of the tallest possible player that is not an outlier. Give your answer to the nearest centimetre. ### 6. GCSE Higher Poppy Pringle recorded the heights in centimetres of the sunflowers growing in her brother's and sister's gardens. Here are the heights of the 19 sunflowers in her brother's garden: $$94, 103, 110, 111, 123, 143, 150, 150, 151, 156, 157, 160, 170, 182, 201, 220, 250, 255, 231$$ Median Lower quartile Upper quartile Shortest Tallest Here is some information about the sunflowers in Poppy's sister's garden: Median 160 Lower quartile 143 Upper quartile 172 Shortest 110 Tallest 199 Poppy says that the sunflowers in her brother's garden are shorter that those in her sister's garden. (b) Is Poppy correct? Poppy says that the sunflowers in her brother's garden vary more in height than those in her sister's garden. (c) Is Poppy correct? If you would like space on the right of the question to write out the solution try this Thinning Feature. It will collapse the text into the left half of your screen but large diagrams will", "# Arithmetical mean The arithmetical mean is the average value of the samples. It is independent of the width of the intervals. It is symbolized as $$\\overline{x}$$ and it is only used for quantitative variables. We find it by adding up all the values and dividing by the total number of data. The general formula for $$N$$ elements is: $$\\displaystyle \\overline{x}=\\frac{x_1+x_2+x_3+\\ldots+x_n}{n}$$$In a basketball match, we have the following points for the players of a team: $$0, 2, 4, 5, 8, 8, 10, 15, 38$$$ Calculate the mean of points of the team. Applying the formula $$\\displaystyle \\overline{x}=\\frac{0+2+4+5+8+9+10+15+38}{9}=\\frac{90}{9}=10$$$## Calculation of the mean for grouped information The average in the case of $$N$$ data grouped in $$n$$ intervals is given by the formula $$\\displaystyle \\overline{x}=\\frac{x_1\\cdot f_1+x_2\\cdot f_2+x_3\\cdot f_3+\\ldots+x_n\\cdot f_n}{f_1+f_2+f_3+\\ldots+f_n}$$$ where $$f_i$$ represents the times that the value $$x_i$$ is repeated. The grouping can also be done by intervals, using then the intermediate value of the interval to calculate the mean. The height in $$cm$$ of the players of a basketball team is in the following table. Calculate the mean. Interval $$x_i$$ $$f_i$$ $$x_i\\cdot f_i$$ $$[160,170)$$ $$165$$ $$1$$ $$165$$ $$[170,180)$$ $$175$$ $$2$$ $$350$$ $$[180,190)$$ $$185$$ $$4$$ $$740$$ $$[190,200)$$ $$195$$ $$3$$ $$585$$ $$[200,210)$$ $$205$$ $$2$$ $$410$$ $$12$$ $$2250$$ We calculate the mean for grouped data: $$\\displaystyle \\overline{x}=\\frac{165 \\cdot 1+175 \\cdot 2+185\\cdot 4+195\\cdot" ]

[ "## Basic College Mathematics (10th Edition) Estimate: 70$\\times$130 = 9100 inch Exact = $9132\\frac{1}{2}$inch We know that Fles can jump 130times higher then their oen height. Now, we need to find the how higher a $70\\frac{1}{4} inch$ heighted person can jump? Then we need to multiply the height with 130 i.e. = $70\\frac{1}{4} \\times 130$ inch. Now, Let's find Estimate : $70\\frac{1}{4}$ (round off) 70 i.e. $70\\frac{1}{4} \\times 130$ inch = 70$\\times$130 = 9100 inch Now Let's find Exact: $70\\frac{1}{4} \\times 130$ inch. = $\\frac{70\\times4+1}{4} \\times 130$ inch. = $\\frac{281}{4} \\times 130$ inch. = $\\frac{281}{2} \\times 65$ inch. = $\\frac{18265}{2}$ inch = $9132\\frac{1}{2}$inch", "8 and x + 12 is 16, find the value of x. Solution: Mean of the given observations = x + (x + 4) + (x + 6) + (x + 8) + (x + 12)/5 = (5x + 30)/5 According to the problem, mean = 16 (given). Therefore, (5x + 30)/5 = 16 ⇒ 5x + 30 = 16 × 5 ⇒ 5x + 30 = 80 ⇒ 5x + 30 - 30 = 80 - 30 ⇒ 5x = 50 ⇒ x = 50/5 ⇒ x = 10 Hence, x = 10. 148 + 153 + 146 + 147 + 154 7. The mean of 40 numbers was found to be 38. Later on, it was detected that a number 56 was misread as 36. Find the correct mean of given numbers. Solution: Calculated mean of 40 numbers = 38. Therefore, calculated sum of these numbers = (38 × 40) = 1520. Correct sum of these numbers = [1520 - (wrong item) + (correct item)] = (1520 - 36 + 56) = 1540. Therefore, the correct mean = 1540/40 = 38.5. 8. The mean of the heights of 6 boys is 152 cm. If the individual heights of five of them are 151 cm, 153 cm, 155 cm, 149 cm and 154 cm, find the height", "inches (b) less than or equal to 64 inches (c) between 65 and 71 inches. Solution: let x denote the height of a student N = 500; m = 68.0 inches and σ = 3.0 inches the standard normal variate z = $$\\frac { x-µ }{σ}$$ = $$\\frac { x-68 }{3}$$ a(greater than 72 inches) p = p(x > 72) when x = 72 z = $$\\frac { 72-68 }{3}$$ = $$\\frac { 4 }{3}$$ = 1.33 p(x > 72) = p(z > 1.33) = 0.5 – 0.4082 = 0.0918 Number of students whose height are greater than 72 inches = 0.0918 × 500 = 45.9 = 46 (approximately) (b) p(less than or equal to 64 inches) p(x ≤ 64) when x = 64 z = $$\\frac { 64-68 }{3}$$ = $$\\frac { -4 }{3}$$ = -1.33 p(x ≤ 64) = p(z ≤ -1.33) p(z ≥ -1.33) = 0.5 – 0.4082 = 0.0918 ∴ Number of heights whose ate less than or equal to 64 inches = 0.0918 × 500 = 45.9 = 46 (approximately) (c) p(between 65 and 71 inches) p(65 ≤ x ≤ 71) when x = 65 z = $$\\frac { 65-68 }{3}$$ = $$\\frac { -3 }{3}$$ = -1 when x = 71 z = $$\\frac { 71-68 }{3}$$ = $$\\frac { 3 }{3}$$", "number line shown. The numbers $$35$$ and $$45$$ are included. It shows continuous solution because height can be anywhere between $$35$$ to $$45$$ inches and can be in decimals. Negative numbers are not shown on the number line because height can not be negative. Hence, option (A) is correct. Option (B) is incorrect because it does not represent all possible heights. Option (C) is incorrect because it represents only three heights, $$35$$ inches, $$40$$ inches and $$45$$ inches. Option (D) is incorrect because it represents only the heights between $$40$$ and $$45$$ inches. ### In a school, the height of students of grade $$1$$ is between $$35$$ inches to $$45$$ inches. Which one of the following options represents the height of students of grade $$1$$? A B C D Option A is Correct #### Olivia's class must raise at least $$500$$ to go on a trip. The class has collected $$200$$. How much amount of money, $$m$$, the class still needs to raise? A $$m<300$$ B $$m>300$$ C $$m\\ge 300$$ D $$m\\le300$$ × To represent Olivia's situation, go step by step according to the statement. Olivia's class needs to raise at least $$500$$ and till now the collection is $$200$$. Thus, $$m$$ is the amount, still needs to be raised. So, $$m+200\\ge500$$ Subtracting $$200$$ from both the sides,", "of the parallel sides times the height divided by 2 which equals 60 in this case. $$\\frac{h(b_1+b_2)}{2}$$ Since we know the height and the area, we can solve for the sides. $$\\frac{5(x+x+7)}{2}=\\frac{10x+35}{2}=60$$ From this, we get x=8.5. So we know your work above is correct. Since 1 side is x and the other is x+7, the two parallel sides are 8.5 and 15.5. To find the median, take the average of these two numbers. May 8, 2020 #2 +1 WOW thx I understand now! :D Guest May 8, 2020 #3 +932 +1 No problem! HELPMEEEEEEEEEEEEE May 8, 2020", "have reached $\\Large\\frac{2}{3} \\times \\frac{2}{3} \\times \\frac{2}{3} \\times \\frac{2}{3}$ of the original height =$\\Large\\frac{16}{81}$ of the original height. A. 22 B. 25 C. 23 D. 24 A. 5 B. 6 C. 4 D. 3 A. -45 B. -55 C. -50 D. 0", "• # question_answer 39) The height of a wooden block is$12.28\\,\\,cm$. What is the approximate height of $14$ similar blocks stacked on top of each other? A) $171\\,\\,cm$ B) $168\\,\\,cm$ C) $170\\,\\,cm$ D) $171.92\\,\\,cm$ The height of a wooden block $=12.28\\,\\,cm\\approx 12\\,\\,cm$ $\\therefore$The height of $14$ similar blocks stacked on top $=12\\times 14=168\\,\\,cm$", "more than once is the ‘$2$’ leaf on the ‘$1$’ part of the stem. Therefore, the mode is $12$ b) The largest value is $51$ (stem of ‘$5$’ and leaf of ‘$1$’ ) and the smallest is $9$ (stem of ‘$0$‘ and leaf of ‘$9$’). So, the range is $51-9=42$ c) All the journeys under $20$ minutes are the ones that appear in the ‘$0$’ stem or the ‘$1$’ stem. From the diagram, we can see that there are $8$ journeys in these two sections, and in total there are $15$ times recorded. This means that $8$ out of $15$ journeys were under $20$ minutes, which can be written as the following fraction: $\\dfrac{8}{15}$ To convert a fraction to a percentage, we need to divide the top of the fraction by the bottom and multiply by $100$: $\\dfrac{8}{15}\\times100 = 53.3\\%$ a) The range is the difference between the weight of the heaviest calf and the weight of the lightest calf. The heaviest calf will be in the ‘$14$’ stem. There is only one calf in this stem and its leaf is $3$, so the weight of the heaviest calf is $143$ pounds. The lightest calf will be the calf with the lowest value in the ‘$9$’ stem, so the weight of the lightest calf is $93$ pounds. Therefore", "of which are triple and some are quadruple. Determine how many rooms are triple and how many quadruples. • Equations Solve following system of equations: 6(x+7)+4(y-5)=12 2(x+y)-3(-2x+4y)=-44 • Average height There are twice as many girls in the class as there are boys. The average height of girls is 177 cm, boys 186 cm. What is the average height of students in this class? • Fifth of the number The fifth of the number is by 24 less than that number. What is the number? • Average If the average(arithmetic mean) of three numbers x,y,z is 50. What is the average of there numbers (3x +10), (3y +10), (3z+10) ?", "# Chapter 1 - Whole Numbers - 1.2 Adding Whole Numbers - 1.2 Exercises: 85 $72$ feet (ft.) of lumber #### Work Step by Step To find the amount of lumber we will find the distance around the entire patio, we will add the three sides (18, 30, 24) $18 + 30 + 24$ Start at the right of each number and add the ones digits $8 + 0 + 4 = 12$ ones Regroup the $12$ ones as $1$ ten and $2$ ones Write the $2$ ones in the ones column. Write the $1$ ten in the tens column Add the tens column next including the regrouped $1$ $1 + 1 + 3 + 2 = 7$ tens Write the $7$ tens in the tens column. The sum of the three numbers is $72$ The distance is $72$ feet (ft). 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.", "the problem, mean = 16 (given). Therefore, (5x + 30)/5 = 16 5x + 30 = 16 xx 5 5x + 30 = 80 5x + 30 - 30 = 80 - 30 5x = 50 x = (50)/5 x = 10 Hence, x = 10. 148 + 153 + 146 + 147 + 154 #### 7. The mean of 40 numbers was found to be 38. Later on, it was detected that a number 56 was misread as 36. Find the correct mean of given numbers. Solution: Calculated mean of 40 numbers = 38. Therefore, calculated sum of these numbers = (38 * 40) = 1520. Correct sum of these numbers = [1520 - (wrong item) + (correct item)] = (1520 - 36 + 56) = 1540. Therefore, the correct mean = (1540)/(40 )= 38.5. #### 8. The mean of the heights of 6 boys is 152 cm. If the individual heights of five of them are 151 cm, 153 cm, 155 cm, 149 cm and 154 cm, find the height of the sixth boy. Solution: Mean height of 6 boys = 152 cm. Sum of the heights of 6 boys = (152 xx 6) = 912 cm Sum of the heights of 5 boys = (151 + 153 + 155 + 149 + 154) cm =", "56.7, 59.0, 64.3, 64.5 Talk Math Discuss different steps that make ordering numbers easier. – Ordering Numbers is a method of arranging them in order – either from small to big or big to small. When we arrange numbers in ascending order, we arrange them from small to big, and when we arrange numbers from big to small, it is called descending order. – We can compare numbers using different methods such as on a number line, by counting, or by counting the number of digits, using place values of the numbers, etc. Question 2. rainfall in inches: 0.76, 0.09, 0.63, 0.24 There are different rainfall measured in inches: Here we need to order the numbers from least to greatest. Now compare the inches First line up the numbers: 0.76, 0.09, 0.63, 0.24 – Now compare the greatest digits or which are having different digits. – 0 < 2, 6, 7 – 3 < 4, 6, 9 After sorting out all the numbers, the answer is 0.09, 0.24, 0.63, 0.76 Question 3. height of flowers in inches: 8.9, 8.59, 8.705, 8.05 There are the different heights of flowers in inches: Here we need to order the numbers from least to greatest. Now compare the inches First line up the numbers: 8.90, 8.59, 8.705, 8.05 – Now compare the", "labor used 26,500 O$13,000. O $44,500 O$57,500... ##### The following descriptive statistics were obtained by computer to Mean analyze the 25 bellshaped tree heights: 58.2400 Std Ent 1.6179 Median = 58.0000 Variance 65.4440 596 Trim = 58.1000 Std Dev 0895 Based on the information available where do you expect most (approximately 95%) of the tree heights to fall? Select one: Between 44 and 75 inches tall Over 58 inches tall Between 50.15 and 66.33 inches tall Between 42.061 and 74.419 inches tall The following descriptive statistics were obtained by computer to Mean analyze the 25 bellshaped tree heights: 58.2400 Std Ent 1.6179 Median = 58.0000 Variance 65.4440 596 Trim = 58.1000 Std Dev 0895 Based on the information available where do you expect most (approximately 95%) of the tree heights ... ##### (3 + 3 = 6 pts.) Prove or disprove the following statements. If you are proving... (3 + 3 = 6 pts.) Prove or disprove the following statements. If you are proving a statement, then give proper reasoning. If you are disproving a statement, then it is enough to give an example which demonstrates that the statement is false. i. If A and B are two n x n matrices, then (A + B)2 = A + 2... ##### The U IH U that all the", "is $$\\frac{56+43+62+63+33+34+38+51+59+59}{10} = \\frac{498}{10}$$ = 49.8 Question 9. b. Project the number of palms on the entire farm. ______ Explanation: As the average is 49.8 for each acre, so for 64 acres it is 64×49.8= 3187.2. So the number of palms on the entire farm is 3187. H.O.T. Focus on Higher Order Thinking Question 10. Draw Conclusions A random sample of 15 of the 78 competitors at a middle school gymnastics competition are asked their height. The data set lists the heights in inches: 55, 57, 57, 58, 59, 59, 59, 59, 59, 61, 62, 62, 63, 64, 66. What is the mean height of the sample? Do you think this is a reasonable prediction of the mean height of all competitors? Explain. Answer: Yes, this is a reasonable prediction. Explanation: The mean height is $$\\frac{55+57+57+58+59+59+59+59+59+61+62+62+63+64+66}{15} = \\frac{900}{15}$$ = 60 inches. Yes, this is a reasonable prediction of the mean height of all competitors because it is a good sample generated randomly and contains sufficient values. So it should provide a good estimate of the mean height of all competitors. Question 11. Critical Thinking The six-by-six grid contains the ages of actors in a youth Shakespeare festival. Describe a method for randomly selecting 8 cells by using number cubes. Then calculate the average of the 8 values you", "### Forgot the Numbers On my calculator I divided one whole number by another whole number and got the answer 3.125 If the numbers are both under 50, what are they? # Jumping ##### Stage: 2 Challenge Level: We had a large number of solutions sent in and most of them were correct. Baston, Midgley, Longcroft and Montgomerie Schools sent in many solutions. Here are some explanations that showed some good thinking. The Maths Challenge Club from St Aidan's Primary sent in the following: Isobel used a trial and error method: $200 \\times 1.5 = 300$ no- too small $300 \\times 1.5 = 450$ no- too big $250 \\times 1.5 = 375$ yes! So the first jump was $250$cm She used a trial and error method for the second question too: $10+10+10= 30$ no- too small $20+20+20=60$ no- too small $30+30+30=90$ no, still too small $40+40+40=120$ yes! So the first jump was $40+40=80$cm Jamila got the same answer, but used a different method: $3.75 \\div 2 = 1.875$ $1.875 \\div 3 = 0.625$ $1.875 + 0.625 = 2.500$ or $2.5$m $(a\\div2) + (a\\div6)$ She used the same method for the second question: $(a\\div3) \\times 2$ where $a=120$cm = $80$cm Holly and Chris from Myland School in England sent in this thorough explanation; We worked out that Ben's first", "Try out our new practice tests completely free! Solved # Heights, in Feet, of a Sample of 24 Mature Oak $\\begin{array} { l l l l l l } 61 & 62 & 65 & 52 & 63 & 65 \\\\64 & 60 & 68 & 61 & 56 & 61 \\\\56 & 63 & 60 & 63 & 69 & 64 \\\\66 & 68 & 67 & 65 & 60 & 66\\end{array}$ Question 9 Essay Question 9 Essay ## Heights, in feet, of a sample of 24 mature oak trees in a forest were measured. The results were as follows.$\\begin{array} { l l l l l l } 61 & 62 & 65 & 52 & 63 & 65 \\\\64 & 60 & 68 & 61 & 56 & 61 \\\\56 & 63 & 60 & 63 & 69 & 64 \\\\66 & 68 & 67 & 65 & 60 & 66\\end{array}$Can you conclude that the median height of oak trees in this forest is greater than 6 0 feet Use the $\\alpha = 0.05$ level of significance.a. State appropriate null and alternate hypotheses.b. Compute the test statistic.c. Find the critical value.d. State a conclusion. feet Choose question tag 10+ million students use Quizplus to study and prepare for their homework, quizzes and exams through 20m+ questions", "\\ inches$ $\\text{width on paper} = 695 \\ feet \\times \\frac{1 \\ inch}{100 \\ feet} = 6.95 \\ inches$ The scale diagram will be $8.8 \\ in \\times 6.95 \\ in$. It will fit on a standard sheet of paper. ## Lesson Summary • A ratio is a way to compare two numbers, measurements or quantities by dividing one number by the other and expressing the answer as a fraction. • A proportion is formed when two ratios are set equal to each other. • Cross multiplication is useful for solving equations in the form of proportions. To cross multiply, multiply the bottom of each ratio by the top of the other ratio and set them equal. For instance, cross multiplying results in $11 \\times 3 = 5x$. • Scale is a proportion that relates map distance to real life distance. ## Review Questions 1. Write the following comparisons as ratios. Simplify fractions where possible. 1. $150 to$3 2. 150 boys to 175 girls 3. 200 minutes to 1 hour 4. 10 days to 2 weeks 2. Write the following ratios as a unit rate. 1. 54 hotdogs to 12 minutes 2. 5000 lbs to 250 square inches 3. 20 computers to 80 students 4. 180 students to 6 teachers 5. 12 meters to 4 floors 6. 18", "## Basic College Mathematics (10th Edition) Estimate: 50 - 47 = 3 Inch Exact: $3\\frac{3}{8}$ inch Difference in height = This yr height - last year height = 50 inch - $46\\frac{5}{8}$ inch First Let's find estimate: $46\\frac{5}{8}$ round to 47 (rounded to nearest whole number) Estimate: 50 - 47 = 3 Inch Now, Let's find Exact: = 50 inch - $46\\frac{5}{8}$ inch = 50 inch - $\\frac{46\\times8+5}{8}$ inch = 50 inch - $\\frac{373}{8}$ inch = $\\frac{50\\times8}{8} - \\frac{373}{8}$ inch (LCM of 1 & 8 is 8) = $\\frac{400}{8} - \\frac{373}{8}$ inch = $\\frac{400-373}{8}$ inch = $\\frac{27}{8}$ inch = $3\\frac{3}{8}$ inch", "solutions are vital for a deep understanding of proportional reasoning. The numbers make this an advanced 6th grade task. It could be used as a transition from 6th to 7th grade expectations for work with ratios and proportional relationships, or it could be used as an assessment task in 7th grade (either formative or summative, depending on the time and way it is used). If used in an instructional setting, the teacher could ask the students to estimate before they compute. Since there are 91 steps and we want to know about the 50th step, we know that the height should be a little more than half of the total 24 meters in height. This task could also be used to transition to the 8th grade standard on using similar triangles to prove that the slope of a line is well-defined. Looking at the cross-section of this pyramid and the similar triangles that we are essentially asking about would be a useful extension task: The image is courtesy Wikimedia Commons: http://en.wikipedia.org/wiki/File:Chichen_Itza_3.jpg ## Solutions Solution: 1 There are 91 steps needed to get you to the top, which is 24 meters high. So each step is about $24\\div91$ meters tall. Since $$24\\div91=\\frac{24}{91}\\approx 0.264$$ we estimate that each step is about 0.264 m or 26.4 cm tall on average. Since", "Home > MC1 > Chapter 3 > Lesson 3.3.3 > Problem3-85 3-85. Ryan's sister, Jollie, gets her height measured once a year. Here is the growth data that her mother has written down for the last six years: $2.5$ inches, $3$ inches, $2.25$ inches, $4$ inches, $1.75$ inches, and $3.5$ inches. 1. How much has Jollie grown in the last six years? Each of these measurements tells you how much Jollie has grown in one year. How can you use these measurements to find out how much she has grown in the past six years? Add the measurements together to find the total amount she has grown. $2.5$ inches $+$ $3$ inches $+$ $2.25$ inches $+$ $4$ inches $+$ $1.75$ inches $+$ $3.5$ inches $=$ $17$ inches $17$ inches 2. Find the mean, median, and mode of her growth over the last six years. If you are having trouble starting this problem, refer to problems 1-110 and 1-112. Don't remember what the mean, median, and mode are? Here's a quick review: The mean here is the number of inches Jollie would grow in a year if she grew the same amount each year. The median is the middle number if the numbers are put in order from least to greatest. The mode is the number that appears most" ]

[ "# Problem of the Week Problem B and Solution How Tall are They, Really? ## Problem There were $$12$$ players on the Canadian $$2020$$ Olympic Women’s Basketball Team. The table below shows the height of each player. Name Height Natalie Achonwa $$6$$ feet, $$3$$ inches Kayla Alexander $$6$$ feet, $$4$$ inches Laeticia Amihere $$6$$ feet, $$2$$ inches Miranda Ayim $$6$$ feet, $$3$$ inches Bridget Carleton $$6$$ feet, $$1$$ inch Shay Colley $$5$$ feet, $$8$$ inches Aaliyah Edwards $$6$$ feet, $$3$$ inches Nirra Fields $$5$$ feet, $$7$$ inches Kim Gaucher $$6$$ feet, $$1$$ inch Kia Nurse $$6$$ feet, $$0$$ inches Shaina Pellington $$5$$ feet, $$8$$ inches Nayo Raincock-Ekunwe $$6$$ feet, $$2$$ inches 1. Create a stem-and-leaf plot to represent the heights in feet and inches of the players on the team. Use the number of feet as the stems, and the number of inches as the leaves. 2. Using your stem-and-leaf plot, find the median height of the players on the team. 3. Convert all the heights to inches, and then calculate the mean height of the players on the team. Recall that $$1~\\text{foot} = 12~\\text{inches}$$. For example, $$6$$ feet is equal to $$6 \\times 12=72$$ inches, so a height of $$6$$ feet, $$3$$ inches is equal to $$72+3=75$$ inches. 4. Assume that the mean height of a Canadian", "• What is 3 What is the least common denominator of 11/15 and 12/19? The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? • Eq2 2 Solve following equation with quadratic members and rational function: (x2+1)/(x-4) + (x2-1)/(x+3) = 23 • Prime factors Write 98 as a product of prime factors", "on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" The best way to do this problem is to convert the respective heights into inches and finding the average, since the answers for the average height are very close! When converted in inches, we will get 71,72,74,75, and 78 inches. The average is $$\\frac{70*5}{5} + \\frac{1+2+4+5+8}{5} = 70 + \\frac{20}{5} = 74$$ inches, which is 6' 2\"(Option B) can't we just arrange the give heights in ascending order and take middle value as mean as there are odd no of elements ? Hi utkarshrihand The middle value is the mean when the list is an arithmetic progression (has a common difference). However, in this list, the elements do not have a common difference and the middle value can't be the mean Hope this helps you! _________________ You've got what it takes, but it will take everything you've got Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8069 Location: United States (CA) Re: The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 15 Mar 2018,", "The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 01 Mar 2018, 00:52 Bunuel wrote: The heights of the five starters on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" Check other Conversion problems to practice in Special Questions Directory. _________________ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3342 Location: India GPA: 3.12 Re: The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 01 Mar 2018, 01:18 Bunuel wrote: The heights of the five starters on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" The best way to do this problem is to convert the respective heights into inches and finding the average, since the answers for the average height are very close! When converted in inches, we will get 71,72,74,75, and 78 inches. The average is $$\\frac{70*5}{5} + \\frac{1+2+4+5+8}{5} = 70 + \\frac{20}{5} = 74$$ inches, which is 6' 2\"(Option B) _________________ You've got", "The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? Result a = ft (Correct answer is: 6'2\") #### Solution: Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors. Dr Math For conversion inches to feet and inch we user reminder calculator: https://www.hackmath.net/en/calculator/quotient-and-remainder?n=74&d=12 74 = 12 × 6 + 2 74 inch = 6 ft and 2 inch Tips to related online calculators Looking for help with calculating arithmetic mean? Looking for a statistical calculator? Do you want to convert length units? #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Soccer team Hilahs soccer team is trying to raise $2414 to travel to a tournament in Florida, so they decided to host a pancake for the breakfast. How many people need to attend their breakfast in order", "The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? Result a = ft (Correct answer is: 6'2\") #### Solution: $h_{ 1 } = 5 \\cdot \\ 12+11 = 71 \\ inch \\ \\\\ h_{ 2 } = 6 \\cdot \\ 12+3 = 75 \\ inch \\ \\\\ h_{ 3 } = 6 \\cdot \\ 12+6 = 78 \\ inch \\ \\\\ h_{ 4 } = 6 \\cdot \\ 12+2 = 74 \\ inch \\ \\\\ h_{ 5 } = 6 \\cdot \\ 12 = 72 \\ inch \\ \\\\ \\ \\\\ b = \\dfrac{ h_{ 1 }+h_{ 2 }+h_{ 3 }+h_{ 4 }+h_{ 5 } }{ 5 } = \\dfrac{ 71+75+78+74+72 }{ 5 } = 74 \\ inch \\ \\\\ 74 = 12 \\times 6 + 2 \\ \\\\ \\ \\\\ a = 6'2\"$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Dr Math For conversion inches to feet and inch we user reminder calculator: https://www.hackmath.net/en/calculator/quotient-and-remainder?n=74&d=12 74 =", "The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is? Result a = ft (Correct answer is: 6'2\") #### Solution: $h_{ 1 } = 5 \\cdot \\ 12+11 = 71 \\ inch \\ \\\\ h_{ 2 } = 6 \\cdot \\ 12+3 = 75 \\ inch \\ \\\\ h_{ 3 } = 6 \\cdot \\ 12+6 = 78 \\ inch \\ \\\\ h_{ 4 } = 6 \\cdot \\ 12+2 = 74 \\ inch \\ \\\\ h_{ 5 } = 6 \\cdot \\ 12 = 72 \\ inch \\ \\\\ \\ \\\\ b = \\dfrac{ h_{ 1 }+h_{ 2 }+h_{ 3 }+h_{ 4 }+h_{ 5 } }{ 5 } = \\dfrac{ 71+75+78+74+72 }{ 5 } = 74 \\ inch \\ \\\\ 74 = 12 \\times 6 + 2 \\ \\\\ \\ \\\\ a = 6'2\"$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Dr Math For conversion inches to feet and inch we user reminder calculator: https://www.hackmath.net/en/calculator/quotient-and-remainder?n=74&d=12 74 =", "18 girls = 18 × 15 = 270Total age of remaining 12 girls 390 – 270 = 120Average age of remaining 12 girls =$\\frac{120}{12}=10$Using Trick: Average age of remaining 12 girls = $\\frac{30\\times 13-18\\times 15}{30-18}=10$Trick: If the average of m numbers is x and out of these ‘m’ numbers the average of n numbers is y, then the average of remaining numbers will beAverage (A) = $\\frac{mx-ny}{m-n}$ Q13. The average height of the basketball team A is 5 ft 11 inches and that of B is 6 ft 2 inches. There are 20 players in team A and 18 players in team B. The overall average height is (a) 72.42 inches (b) 72 inches (c) 70.22 inches (d) 70 inches Answer: (a) 72.42 inchesSolution: overall average height = $\\frac{20\\times 71+18\\times 74}{20+18}=\\frac{1420+1332}{38}=72.4211$Rules: If the given observations (x) are occurring with certain frequency (A) then,Average = $\\frac{A_1x_1+A_2x_2+…….+A_nx_n}{x_1+x_2+…..+x_n}$ , where, A1, A2, A3. ………. An are frequencies Q14. The average of 13 results is 60. If the average of first 7 results is 59 and that of last 7 results is 61, what will be the seventh result? (a) 90 (b) 50 (c) 75 (d) 60 (e) None of the above Answer: (d) 60Solution: Total of 13 results = 13 × 60 = 780Total of first 7 results = 7 ×", "for 104 centimeters, and 1 1 | 4 stands for 114 centimeters. Question 4. The median height of the children is ____ centimeters. The median height of the children is 106 centimeters. Explanation: The data is 96,93,104,106,106,106,110,113,113,115,124,129. To find the median we will arrange the numbers in order from least to greatest. The middle number or the mean of the two middle numbers is the median. The data set from least to greatest is 93,96,104,106,106,106,110,113,113,115,124,129. So the median will be = $$\\frac{106+110}{2}$$ = $$\\frac{216}{2}$$ = 106. So the median height of the children is 106 centimeters. Question 5. The mode of the set of data is ___ centimeters. The mode of the set of data is 106 centimeters. Explanation: To find the mode we will pick the number that appears most often is the mode and there can be more than one mode. So the mode is 106 cm. Question 6. The range of the heights is ____ centimeters. The range of the heights is 36 centimeters. Explanation: To find the range, we will find the difference between the greatest and the least number. So the range will be = 129-93 = 36. Make a stem-and-leaf plot to show the data. The table shows the points scored by a school team in eight basketball games one season. Complete.", "the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers. Mean = (Sum of all the observations/Total number of observations) Finding Mean for Team A: The given data in the team A is 67, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73. Mean for Team A= 67+68+68+69+70+70+70+71+71+72+73/11 Mean for Team A= 769/11 Mean for Team A=69.9 Finding mean for Team B: The given data is 68, 68, 68, 70, 70, 71, 71, 71, 72, 72, 72 Mean for Team B=68+68+68+70+70+71+71+71+72+72+72/11 Mean for Team B=773/11 Mean for Team B=70.3 d) On average, which team has taller players? Median of Team A:70 Mean of Team A:69.9 Median of Team B:71 Mean of Team B:70.3 According to the above observations, we can write which team has taller players. On average Team B has taller players because both it’s mean and median are greater. But the means and the medians for both teams are very close, so the players on Team B are not much taller on average. Solve. Question 4. The weights, in pounds, of some newborn babies at a hospital are shown in the stem-and-leaf plot. a) Find the median weight. The median of a set of data is the middlemost number or centre value in the set. The", "### Home > MC2 > Chapter 5 > Lesson 5.2.3 > Problem5-70 5-70. The girls' basketball team weighed each player. Their weights in pounds were $120$, $122$, $126$, $130$, $133$, $147$, $115$, $106$, $120$, $112$, and $142$. . 1. Make a stem-and-leaf plot of the teams' weights. A stem-and-leaf plot displays data by ordering it least to greatest. All of the digits except for the last one form the stem and the last digit forms the leaf. The leaf portion is arranged in order from least to greatest. 2. What is the median weight of girls on the team? The median is the middle number in a set of data. $122$ 3. What is the range of the data? Subtract the largest number by the smallest number. The range is the difference between the greatest and least number in a set of data. 4. What are the mean and mode? The mean is the average of a set of numbers, while the mode is the number that occurs the most often in a set. Explore the problem with the eTool below. Click the link at right for the full version of the eTool: MC2 5-70 HW eTool", "woman is $$5$$ feet, $$4$$ inches. How much taller, on average, are the players on this team? ## Solution 1. The stem-and-leaf plot is shown below. Stem Leaf $$5$$ $$~7~$$$$~8~$$ $$8~$$ $$6$$ $$~0~$$$$~1~$$$$~1~$$$$~2~$$$$~2~$$$$~3~$$$$~3~$$$$~4~$$ Key: $$6 ~\\vert~ 3 = 6~\\text{feet},~3~\\text{inches}$$ 2. The median is halfway between $$6$$ feet, $$1$$ inch and $$6$$ feet, $$2$$ inches, which is $$6$$ feet, $$1.5$$ inches. 3. The table below shows all the heights converted to inches. Name Height (feet, inches) Height (inches) Natalie Achonwa $$6$$ feet, $$3$$ inches $$72+3=75$$ Kayla Alexander $$6$$ feet, $$4$$ inches $$72+4=76$$ Laeticia Amihere $$6$$ feet, $$2$$ inches $$72+2=74$$ Miranda Ayim $$6$$ feet, $$3$$ inches $$72+3=75$$ Bridget Carleton $$6$$ feet, $$1$$ inch $$72+1=73$$ Shay Colley $$5$$ feet, $$8$$ inches $$60+8=68$$ Aaliyah Edwards $$6$$ feet, $$3$$ inches $$72+3=75$$ Nirra Fields $$5$$ feet, $$7$$ inches $$60+7=67$$ Kim Gaucher $$6$$ feet, $$1$$ inch $$72+1=73$$ Kia Nurse $$6$$ feet, $$0$$ inches $$72$$ Shaina Pellington $$5$$ feet, $$8$$ inches $$60+8=68$$ Nayo Raincock-Ekunwe $$6$$ feet, $$2$$ inches $$72+2=74$$ The mean is found by adding up all the heights in inches, and dividing by $$12$$. $75+76+74+75+73+68+75+67+73+72+68+74=870$ Since $$870 \\div 12=72.5$$, the mean height is $$72.5$$ inches, or $$6$$ feet, $$0.5$$ inches. 4. We can first convert $$5$$ feet, $$4$$ inches to inches. $$5$$ feet is equal to $$5 \\times 12=60$$ inches, so a height of $$5$$", "## alexisreed i neeeeeeeed helppppp one year ago one year ago 1. 142343 with what? 2. alexisreed following data represents the heights of the players on a college women's volleyball team. What is the mean height of the players rounded to tenths place? Heights (inches): 72, 72, 65, 72, 74, 72, 74, 72, 74, 73, 68, 72, 70, 74, 71 3. saifoo.khan $Mean = \\frac{\\text{\\sum of all terms}}{\\text{number of terms}}$", "## alexisreed Group Title i neeeeeeeed helppppp 2 years ago 2 years ago 1. 142343 with what? 2. alexisreed following data represents the heights of the players on a college women's volleyball team. What is the mean height of the players rounded to tenths place? Heights (inches): 72, 72, 65, 72, 74, 72, 74, 72, 74, 73, 68, 72, 70, 74, 71 3. saifoo.khan $Mean = \\frac{\\text{\\sum of all terms}}{\\text{number of terms}}$", "than the basketball players. B. Visually compare the centers of the dot plots. Softball: The data is centered around 5’4’ Basketball: The data is centered around 5’8’ This means that the most common height for the softball players is 5 feet 4 inches, and for the basketball players 5 feet 8 inches. C. Visually compare the spreads of the dot plots. Softball: The spread is from 4’11 “to 5’6”. Basketball: The spread is from 5’2” to 6’0’ There is a greater spread in heights for the basketball players. Math Talk How do the heights of field hockey players compare with the heights of softball and basketball players? Your Turn Question 2. Visually compare the dot plot of heights of field hockey players to the dot plots for softball and basketball players. Shape: _______________ Center: _______________ Spread: _______________ Answer: Shapes: Hockey: All the data is 5’6” or less. The hockey players are shorter than the basketball players, but taller than the softball players. Center: Hockey: The data is centered around of 5’2”. Spread: Hockey: The spreads is from 4’9” to 5’6”. They have same spread as spread for softball players, but there is a greater spread in heights for the basketball players. All the data is 5’6″ or Less. The data is centered around of 5’2”. The spreads is", "what it takes, but it will take everything you've got Intern Joined: 14 Sep 2017 Posts: 11 Location: India GPA: 3.33 Re: The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 14 Mar 2018, 03:39 pushpitkc wrote: Bunuel wrote: The heights of the five starters on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" The best way to do this problem is to convert the respective heights into inches and finding the average, since the answers for the average height are very close! When converted in inches, we will get 71,72,74,75, and 78 inches. The average is $$\\frac{70*5}{5} + \\frac{1+2+4+5+8}{5} = 70 + \\frac{20}{5} = 74$$ inches, which is 6' 2\"(Option B) can't we just arrange the give heights in ascending order and take middle value as mean as there are odd no of elements ? Senior PS Moderator Joined: 26 Feb 2016 Posts: 3342 Location: India GPA: 3.12 Re: The heights of the five starters on Redwood High's basketball team are [#permalink] Show Tags 14 Mar 2018, 04:44 1 utkarshrihand wrote: pushpitkc wrote: Bunuel wrote: The heights of the five starters", "## Engage NY Eureka Math 4th Grade Module 5 Lesson 40 Answer Key ### Eureka Math Grade 4 Module 5 Lesson 40 Problem Set Answer Key Question 1. The chart to the right shows the height of some football players. Player Height (in feet) A 6$$\\frac{1}{4}$$ B 5$$\\frac{7}{8}$$ C 6$$\\frac{1}{2}$$ D 6$$\\frac{1}{4}$$ E 6$$\\frac{2}{8}$$ F 5$$\\frac{7}{8}$$ G 6$$\\frac{1}{8}$$ H 6$$\\frac{5}{8}$$ I 5$$\\frac{6}{8}$$ J 6$$\\frac{1}{8}$$ A: 6(1/4) = 6.2 Explanation: In the above-given question, given that, The height of the Football A. 6(1/4) = 6 x 4. 6 x 4 = 24. 24 + 1/4 = 25/4. 25/4 = 6.2. B: 5(7/8) = 5.8. Explanation: In the above-given question, given that, The height of the Football B. 5(7/8) = 5 x 8. 5 x 8 = 40. 40 + 7/8 = 47/8. 47/8 = 5.8. C: 6(1/2) = 6.5. Explanation: In the above-given question, given that, The height of the Football A. 6(1/2) = 6 x 2. 6 x 2 = 12. 12 + 1/2 = 13/2. 13/2 = 6.5. D: 6(1/4) = 6.2 Explanation: In the above-given question, given that, The height of the Football D. 6(1/4) = 6 x 4. 6 x 4 = 24. 24 + 1/4 = 25/4. 25/4 = 6.2. E: 6(2/8) = 6.2 Explanation: In the above-given question, given that, The height of", "10:10 Bunuel wrote: The heights of the five starters on Redwood High's basketball team are 5'11\" (5 feet 11 inches), 6'3\", 6', 6'6\", and 6'2\". The average height of these players is (1 foot = 12 inches) (A) 6'1\" (B) 6'2\" (C) 6' 3\" (D) 6'4\" (E) 6'5\" The sum of the heights is: 5 + 6 + 6 + 6 + 6 = 29 feet plus 11 + 3 + 0 + 6 + 2 = 22 inches = 22/12 = 11/6 feet So the total height is: 29 + 11/6 = 174/6 + 11/6 = 185/6 feet Thus, the average is: (185/6)/5 = 37/6 feet = 74/12 = 6 2/12 inches = 6’2” _________________ 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: The heights of the five starters on Redwood High's basketball team are [#permalink] 15 Mar 2018, 10:10 Display posts from previous: Sort by", "ladder. If the bottom of the ladder is 30 feet away from the wall, how long is the ladder? A. 30 feet B. 40 feet C. 50 feet D. 60 feet 17- The average weight of 18 girls in a class is 60 kg and the average weight of 32 boys in the same class is 62 kg. What is the average weight of all the 50 students in that class? A. 60 B. 61.28 C. 61.68 D. 62.90 18- An angle is equal to one-fifth of its supplement. What is the measure of that angle? A. 20 B. 30 C. 45 D. 60 19- In a stadium the ratio of home fans to visiting fans in a crowd is 5:7. Which of the following could be the total number of fans in the stadium? A. 12,324 B. 42,326 C. 44,566 D. 66,812 20- 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. $$40\\%$$ ## Best GED Math Prep Resource for 2022 1- A 2,500 out of 55,000 equals to $$\\frac{2500}{55000}= \\frac{25}{550}= \\frac{1}{22}$$ 2- D $$4 \\div \\frac{1}{6} = 24$$ 3- B Simplify and combine like terms. $$(6x^3-8x^2+2x^4 )-(4x^2-2x^4+2x^3 ) {\\Rightarrow} (6x^3-8x^2+2x^4 )-4x^2+2x^4-2x^3 {\\Rightarrow} 4x^4+4x^3-12x^2$$ 4- C the population is", "the volume, ${V}_{\\text{upper}}=9.05\\mathrm{lt}=9050{\\mathrm{cm}}^{3}$ and the lower bound of the radius, ${r}_{\\text{lower}}=\\frac{29.5}{2}=14.75\\mathrm{cm}$. Using these values we get $hupper=9050π⁢(14.75)2 =13.24081342...⁢cm$ Similarly for $hlower=8950π⁢(15.25)2 =12.24992628...⁢cm$ So $h=12.7±0.5\\mathrm{cm}$. -oOo- ### 16/02/2021Female Shinty Players At a school students play one of three sports: Basketball; Cycling; Shinty. The school has 50% more boys than girls. Half the girls play shinty. The same number of boys play basketball as girls play shinty. There are 120 less female cyclists than male. Each of the 310 bicycles are used when the boys and girls go cycling. If a student is picked at random, what is the probability that they are a female shinty player? -/|\\- -oOo- ### 09/02/2021Parallelogram or not? OABC is a parallelogram, where $\\stackrel{⇀}{\\mathrm{OJ}}=\\frac{2}{3}\\stackrel{⇀}{\\mathrm{OC}}$, $\\stackrel{⇀}{\\mathrm{OK}}=\\frac{1}{4}\\stackrel{⇀}{\\mathrm{OC}}$ and $\\stackrel{⇀}{\\mathrm{OL}}=\\frac{8}{5}\\stackrel{⇀}{\\mathrm{OA}}$ Show that $\\mathrm{JKL}$ is a straight line. (note the figure is not to scale) -/|\\- ### Solution To solve this one we need to show that $\\stackrel{⇀}{\\mathrm{JK}}$ is parallel to $\\stackrel{⇀}{\\mathrm{JL}}$. If this is true and as they start from the same point they must be on a straight line. To make the algebra simpler let's make $\\stackrel{⇀}{\\mathrm{OA}}=\\mathbf{a}$ and $\\stackrel{⇀}{\\mathrm{OC}}=\\mathbf{c}$. Let's continue with determining $\\stackrel{⇀}{\\mathrm{JK}}$. We chose $J$ as the common point but we could have chosen any of the three. $K$ would be interesting as we would have to show that $\\stackrel{⇀}{\\mathrm{KJ}}$ and $\\stackrel{⇀}{\\mathrm{KL}}$ were antiparallel. to" ]

[ "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "+0 # Help Geometry 0 154 1 In right triangle $ABC,$ $AC = BC$ and $\\angle C = 90^\\circ.$ and Let $P$ and $Q$ be points on hypotenus $\\overline{AB},$ as shown below, such that $\\angle PCQ = 45^\\circ.$ Show that $AP^2 + BQ^2 = PQ^2.$ $$[asy] unitsize(3 cm); pair A, B, C, P, Q; A = (1,0); B = (0,1); C = (0,0); P = extension(C, C + dir(16), A, B); Q = extension(C, C + dir(16 + 45), A, B); draw(A--B--C--cycle); draw(C--P); draw(C--Q); label(\"A\", A, SE); label(\"B\", B, NW); label(\"C\", C, SW); label(\"P\", P, NE); label(\"Q\", Q, NE); [/asy]$$ off-topic Feb 16, 2020 edited by Guest Feb 16, 2020", "the area of $\\triangle{PQR}$ is $\\boxed{\\textbf{(C) }2\\sqrt{2}}$. Note: After finding the coordinates of $P,Q,$ and $R$, we can alternatively find the vectors $\\overrightarrow{PQ}=[-2,2,0]$ and $\\overrightarrow{PR}=[-1,1,2]$, then apply the formula $\\text{area} = \\frac{1}{2}\\left|\\overrightarrow{PQ} \\times \\overrightarrow{PR}\\right|$. In this case, the cross product equals $[4,4,0]$, which has magnitude $4\\sqrt{2}$, giving the area as $2\\sqrt{2}$ like before. ## Solution 2 As in Solution 1, let $A=(0, 0, 0), B=(3, 0, 0), C=(3, 3, 0), D=(0, 3, 0),$ and $E=(0, 0, 6)$, and calculate the coordinates of $P$, $Q$, and $R$ as $P=(2,0,2), Q=(0,2,2), R=(1,1,4)$. Now notice that the plane determined by $\\triangle PQR$ is perpendicular to the plane determined by $ABCD$. To see this, consider the bird's-eye view, looking down upon $P$, $Q$, and $R$ projected onto $ABCD$: $[asy] unitsize(40); for(int i =0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label(\"A\", (0,0), SW); label(\"B\", (3,0), SE); label(\"C\", (3,3), NE); label(\"D\", (0,3), NW); label(\"P\", (2,0), S); label(\"Q\", (0,2), W); label(\"R\", (1,1), NE); dot((2,0)); dot((0,2)); dot((1,1)); draw((0,2)--(2,0)); [/asy]$ Additionally, we know that $PQ$ is parallel to the plane determined by $ABCD$, since $P$ and $Q$ have the same $z$-coordinate. Hence the height of $\\triangle PQR$ is equal to the $z$-coordinate of $R$ minus the $z$-coordinate of $P$, giving $4-2= 2$. By the distance formula, $\\overline{PQ} = 2\\sqrt{2}$, so the area of $\\triangle PQR$ is $\\frac{1}{2}", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "+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", "the regions A+D is $25\\pi$. The area of the regions C+D is $12.5\\pi$. So to find the area of A+C we need to know the area of D. [TIKZ] \\draw (0,-3) rectangle (6,3); \\begin{scope} \\draw (6,-3) arc (-90:-270:3cm); \\end{scope} \\begin{scope} \\draw (6,-3) arc (0:90:6cm); \\end{scope} \\coordinate[label=above: A] (A) at (2,0); \\coordinate[label=above: B] (B) at (3.6,2.4); \\coordinate[label=above: C] (C) at (5.4,0.8); \\coordinate[label=above: D] (D) at (4,-1); \\coordinate[label=left: P] (P) at (0,-3); \\coordinate[label=above: Q] (Q) at (3.6,1.8); \\coordinate[label=right: R] (R) at (6,0); \\coordinate[label=right: S] (S) at (6,-3); \\draw (P) -- (Q) -- (R) -- (P) ; \\draw (Q) -- (S) ; \\draw (0.4,-2.8) node {$\\theta$} ; \\draw (3,-3.2) node {$10$} ; \\draw (6.2,-1.5) node {$5$} ;[/TIKZ] Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are congruent. In particular $\\angle PQR$ is a right angle, so that $PQ$ is tangent to the semicircle and $QR$ is tangent to the quarter-circle. Also, $PR$ bisects the angle $\\theta =\\angle QPS$, so that $\\angle RPS = \\frac\\theta2$ and $\\tan\\frac\\theta2 = \\frac{RS}{PS} = \\frac12.$ Therefore $$\\tan\\theta = \\frac{2\\tan\\frac\\theta2}{1 - \\tan^2\\frac\\theta2} = \\frac1{1-\\frac14} = \\frac43.$$ It follows that $\\sin\\theta = \\frac45$", "and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle ABE = 60$. Let $M$ be a point such $\\overline{EM}$ is parellel to $\\overline{CD}$. We immediatley know that $\\triangle BEM \\sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\\triangle EFM \\sim \\triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\\triangle EBF$ is in $\\triangle ABF$, the area has ratio $\\triangle BEF : \\triangle ABE=1:2$ and we know $\\triangle ABE$ has area $60$ so $\\triangle BEF$ is $\\boxed{\\textbf{B} \\, 30}$. - fath2012 ## Solution 9 $[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(F--D); 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(\"x\", (D+E+F)/3); label(\"x\", (B+E+F)/3); label(\"120+2x\", (D+F+C)/3); [/asy]$ Labeling the areas in the diagram, we have: $[DBC]=240=[BFE]+[FED]+[FDC]=x+x+120+2x=120+4x$ so $240=120+4x, 120=4x, 30=x$. So our answer is $\\boxed{\\textbf{(B)} 30}$. ~~RWhite ## Solution 10 (Menelaus's Theorem) $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF", "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", "and P 0, P_{0}, P 0 , respectively. The Basic Triangle Inequality. Formula : But so does A, since for A, s = t = s + t = 0. Perimeter: 24.1421. The triangle below has an area of A = 1 ⁄ 2 (6)(4) = 12 square units. They use knowledge, e.g., formulas (relations) Pythagorean theorem, Sine theorem, Cosine theorem, Heron's formula, solving equations and systems of equations. 2. Solution: The position vectors of the three points X,Y and Z are given by (i ^ – j ^ + 3 k ^), (− 2 i ^ – 5 j ^ + 4 k ^), (3 i ^ + j ^ − 4 k ^) (\\hat i – \\hat j+ 3 \\hat k), (-2\\hat i – 5\\hat j+ 4 \\hat k), (3 \\hat i + \\hat j -4 \\hat k) (i ^ – j ^ + 3 k ^), (− 2 i ^ – 5 j ^ + 4 k ^), (3 i ^ + j ^ − 4 k ^) respectively. 24, Feb 16. 1. [5 marks] PSPM 2015/2016 11. To use this formula, you need the measure of just one side of the triangle plus the altitude of the triangle (perpendicular to the base) drawn from that side. If the triangle was a right triangle, it", "# 2007 IMO Problems/Problem 4 ## Problem In $\\triangle ABC$ the bisector of $\\angle{BCA}$ intersects the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area. ## Diagram $[asy] size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),R=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair K = midpoint(B--C); pair L = midpoint(A--C); pair I=incenter(A,B,C); pair O = circumcenter(A,B,C); dot(O); dot(A^^B^^C^^R^^K^^L); draw(C--R); draw(circumcircle(A,B,R)); draw(A--C--B); draw(A--B); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"R\",R,S); label(\"K\",K,N); label(\"L\",L,S); label(\"O\",O,N); draw(K--O); draw(L--O); pair Q = intersectionpoint(L--O, C--R); dot(Q); label(\"Q\",Q,SW); pair E = midpoint(C--R); dot(E); label(\"E\",E,W); draw(O--E, dashed); pair P = intersectionpoint(O--(c/2-1.2,0), C--R); dot(P); label(\"P\",P,W); draw(O--P); draw(R--L, dashed); draw(R--K, dashed); [/asy]$ ~KingRavi ## Solution 1 (Efficient) $\\angle{RQL} = 90+\\angle{QCL} = 90+\\dfrac{C}{2}$, and similarly $\\angle{RPK} = 90+\\angle{PCK} = 90+\\dfrac{C}{2}$. Therefore, $\\angle{RQL} = \\angle{RPK}$. Using the triangle area formula $A = \\dfrac{1}{2}bc\\sin{\\angle{A}}$ yields $RQ \\cdot QL = RP \\cdot PK = \\dfrac{PK}{QL} = \\dfrac{RQ}{RP}$ after cancelling the sines and constant. Draw line $QD$ perpendicular to $BC$ that intersects $BC$ at $D$, then $QD=QL$ because the perpendicular bisectors are congruent, (or alternatively $\\triangle QDC\\cong\\triangle QLC$). This presents us $\\dfrac{PK}{QL}=\\dfrac{PK}{QD}=\\dfrac{PC}{QC}$ by similar triangles; now, we have only to prove $\\dfrac{PC}{QC}=\\dfrac{RQ}{RP}$, or $RQ \\cdot QC=RP \\cdot", "# 1974 USAMO Problems/Problem 5 ## Problem Consider the two triangles $\\triangle ABC$ and $\\triangle PQR$ shown in Figure 1. In $\\triangle ABC$, $\\angle ADB = \\angle BDC = \\angle CDA = 120^\\circ$. Prove that $x=u+v+w$. $[asy] size(400); defaultpen(1); pair C=(0,1), A=(1,6), B=(4,1), D=(1.5,2); draw(D--A--B--C--D--B); draw(A--C); label(\"A\",A,N); label(\"B\",B,ESE); label(\"C\",C,SW); label(\"D\",D,NE); label(\"a\",(B+C)/2,S); label(\"b\",(C+A)/2,WNW); label(\"c\",(A+B)/2,NE); label(\"u\",(A+D)/2,E); label(\"v\",(B+D)/2,N); label(\"w\",(C+D)/2,N); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); draw(M--P--Q--R--M--Q); draw(P--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NE); label(\"x\",(P+Q)/2,S); label(\"x\",(Q+R)/2,E); label(\"x\",(R+P)/2,W); label(\"a\",(P+M)/2,SE); label(\"b\",(Q+M)/2,N); label(\"c\",(R+M)/2,E); label(\"Figure 1\",P-(0,1),S); [/asy]$ ## Solutions ### Solution 1 We rotate figure $PRQM$ by a clockwise angle of $\\pi/3$ about $Q$ to obtain figure $RR'QM'$: $[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NW); label(\"R'\",RR,NE); label(\"M'\",MM,ESE); [/asy]$ Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then $$[ABC] + \\frac{b^2 \\sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .$$ Then by symmetry, $$3 [ABC] + \\frac{(a^2+b^2+c^2)\\sqrt{3}}{4} = 2\\bigl( [PMQ] + [QMR] + [RMP] \\bigr) = 2 [PRQ] .$$ But $ABC$ is composed of three smaller triangles. The one with sides $w,v,a$ has area $\\tfrac{1}{2} wv \\sin 120^\\circ = \\frac{wv \\sqrt{3}}{4}$. Therefore, the area of $ABC$ is $$\\frac{(wv+vu+uw)\\sqrt{3}}{4} .$$ Also, by the Law", "$W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label(\"P\",(0,sqrt(3)/2),N); label(\"Z\",(1/6,sqrt(3)/3),NE); label(\"Y\",(1/3,sqrt(3)/6),NE); label(\"R\",(1/2,0),E); label(\"X\",(1/6,0),S); label(\"W\",(-1/6,0),S); label(\"Q\",(-1/2,0),W); label(\"V\",(-1/3,sqrt(3)/6),NW); label(\"U\",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$ ## Problem T2 Evaluate $\\begin{eqnarray*} && 1 \\left(\\dfrac{1}{1}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right) \\\\ &+& 3\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&5\\left(\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&7\\left(\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&9\\left(\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+11\\left(\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&13\\left(\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+15\\left(\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&17\\left(\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+19\\left(\\dfrac{1}{10}\\right)\\end{eqnarray*}$ ## Problem T3 To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed? ## Problem T4 In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey? ## Problem T5 During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued: 'A': It was 'B' or 'C' 'B': Neither 'E'", "area of rectangle $ABGH$ is $4 \\cdot 12 = 48$. Taking the area of rectangle $ABGH$ and subtracting the combined area of $\\triangle AHG$ and $\\triangle CGM$ yields $48 - (24 + \\frac{32}{5}) = \\boxed{\\frac{88}{5}}\\ \\text{(C)}$. ### Solution 2 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label(\"A\",A,NW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",Ep,SE); label(\"F\",F,SE); label(\"G\",G,SE); label(\"H\",H,SW); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,NW); label(\"L\",L,NW); label(\"M\",M,W); label(\"N\",Z,NE); [/asy]$ Extend $AB$ and $CH$ and call their intersection $N$. The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\\frac{2\\cdot 4}2 = 4$. The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$. Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \\frac 35 \\cdot 12 = \\frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\\frac{1}{\\frac{1}{8}+\\frac{1}{12}} = \\frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\\frac{36}{5}$.) Then the area of $AMN$ is $\\frac 12 \\cdot", "it. Given that the side of the square is 10 cm, find the combined areas of the regions labelled A and C. Any hints, perhaps, on how to solve this seemingly simple geometry problem? This is a primary math problem, so one should refrain from using geometry or trigonometric method to solve it...any help is much appreciated! The area of the regions A+D is $25\\pi$. The area of the regions C+D is $12.5\\pi$. So to find the area of A+C we need to know the area of D. [TIKZ] \\draw (0,-3) rectangle (6,3); \\begin{scope} \\draw (6,-3) arc (-90:-270:3cm); \\end{scope} \\begin{scope} \\draw (6,-3) arc (0:90:6cm); \\end{scope} \\coordinate[label=above: A] (A) at (2,0); \\coordinate[label=above: B] (B) at (3.6,2.4); \\coordinate[label=above: C] (C) at (5.4,0.8); \\coordinate[label=above: D] (D) at (4,-1); \\coordinate[label=left: P] (P) at (0,-3); \\coordinate[label=above: Q] (Q) at (3.6,1.8); \\coordinate[label=right: R] (R) at (6,0); \\coordinate[label=right: S] (S) at (6,-3); \\draw (P) -- (Q) -- (R) -- (P) ; \\draw (Q) -- (S) ; \\draw (0.4,-2.8) node {$\\theta$} ; \\draw (3,-3.2) node {$10$} ; \\draw (6.2,-1.5) node {$5$} ;[/TIKZ] Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are", "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", "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.", "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", "(11) (iii) | Page 77 Find the area of triangle whose vertices are L(1, 1), M(−2, 2), N(5, 4) Miscellaneous Exercise 4(A) | Q II. (12) (i) | Page 77 Find the value of k: If area of triangle is 4 square unit and vertices are P(k, 0), Q(4, 0), R(0, 2) Miscellaneous Exercise 4(A) | Q II. (12) (ii) | Page 77 Find the value of k: If area of triangle is 33/2 square unit and vertices are L(3, −5), M(−2, k), N(1, 4) Miscellaneous Exercise 4(A) | Q II. (13) | Page 77 Find the area of quadrilateral whose vertices are A(0, −4), B(4, 0), C(−4, 0), D(0, 4) Miscellaneous Exercise 4(A) | Q II. (14) | Page 77 An amount of ₹ 5000 is put into three investments at the rate of interest of 6%, 7% and 8% per annum respectively. The total annual income is ₹ 350. If the combined income from the first two investments is ₹ 7000 more than the income from the third. Find the amount of each investment. Miscellaneous Exercise 4(A) | Q II. (15) | Page 77 Show that the lines x − y = 6, 4x − 3y = 20 and 6x + 5y + 8 = 0 are concurrent. Also find the point of concurrence Miscellaneous Exercise", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and" ]

[ "= 30. Since whole of J is 90, 90-30 = 60. 2 angles = 60 --> third angle is 60. _________________ If this helped, please give kudos! Re: In the diagram, JKLM is a square, and P is the midpoint of KL. Is JQM [#permalink] 07 Aug 2018, 20:04 Display posts from previous: Sort by", "Re: In ΔJKL shown above, what is the length of segment JL ? [#permalink] ### Show Tags 26 Apr 2019, 15:15 1 The Logical approach to this question will use the fact that in order to find the measurements of all of the sides of a 30-60-90 triangle (or a 45-45-90 triangle) all ee need is one side. In this question each of the statements gives us the length of one side, and thus they each suffice on their own. The correct answer is (D). Posted from my mobile device _________________ Manager Joined: 14 Apr 2017 Posts: 70 Location: Hungary GMAT 1: 760 Q50 V42 WE: Education (Education) Re: In ΔJKL shown above, what is the length of segment JL ? [#permalink] ### Show Tags 27 Apr 2019, 17:24 1 Bunuel wrote: In ΔJKL shown above, what is the length of segment JL ? (1) JK = 10 (2) KL = 5 DS83602.01 OG2020 NEW QUESTION $$\\triangle{JKL}$$ is a 30-60-90 degree special right triangle with a ratio of $$1:\\sqrt{3}:2$$ for its corresponding sides. The original question: $$JL=?$$ 1) We know that $$JK=10$$, the angle opposite $$JK$$ is $$90^{\\circ}$$, and the angle opposite $$JL$$ is $$60^{\\circ}$$, so $$JL:10=\\sqrt{3}:2$$. Thus, we could get a unique value to answer the original question. $$\\implies$$ Sufficient 2) We know that $$KL=5$$, the angle", "+0 # In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. +1 97 2 +38 In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. Jul 2, 2022 #1 0 Jul 2, 2022 #2 +2448 0 Draw altitude $$JM$$. Because the triangle is isosceles, $$KM = ML = 15$$ Then, by the Pythagorean Theorem, $$JM = \\sqrt{25^2 - 15^2} = 20$$. This means that the area of the triangle is $$2 \\times( 20 \\times 15 \\div 2) = 300$$. Now, recall the formula for inradius, which is $${\\text{Area} \\over \\text{semiperimeter} } = {300 \\over 40} = \\color{brown}\\boxed{7.5}$$ Now, recall the formula for circumradius, which is $${\\text{product of the side lengths} \\over 4\\times \\text{Area}} = {18,750 \\over 1,200} = \\color{brown}\\boxed{15.625}$$ Jul 2, 2022", "+0 # In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. 0 48 2 +22 In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. Jul 2, 2022 #1 0 The circumradius is 57/4. Jul 2, 2022 #2 +2275 +1 Draw altitude $$JM$$. Because the triangle is isosceles, $$KM = ML = 15$$ Then, by the Pythagorean Theorem, $$JM = \\sqrt{25^2 - 15^2} = 20$$. This means that the area of the triangle is $$2 \\times( 20 \\times 15 \\div 2) = 300$$. Now, recall the formula for inradius, which is $${\\text{Area} \\over \\text{semiperimeter} } = {300 \\over 40} = \\color{brown}\\boxed{7.5}$$ Now, recall the formula for circumradius, which is $${\\text{product of the side lengths} \\over 4\\times \\text{Area}} = {18,750 \\over 1,200} = \\color{brown}\\boxed{15.625}$$ Jul 2, 2022", "# length of a side in a rectangle • Jun 25th 2007, 09:45 PM sanee66 length of a side in a rectangle Could someone please look at this? I think the answer is 26 for the length of KL because KL=2(13). Is this right? • Jun 25th 2007, 10:02 PM Jhevon Quote: Originally Posted by sanee66 Could someone please look at this? I think the answer is 26 for the length of KL because KL=2(13). Is this right? Why would you assume KL = 2(13)? :confused: See the slightly modified diagram below. MJKN is a square, so the diagonal bisects the angles at the corners. so we have 45 degree angles in the positions shown Remember, $MK = \\sqrt {50}$ and $KI = 13$ Now, $\\sin 45^{ \\circ} = \\frac {JK}{MK} = \\frac {JK}{ \\sqrt {50}}$ $\\Rightarrow JK = \\sqrt {50} \\sin 45^{ \\circ} = 5$ Now note that JK = MN = IL By Pythagoras' Theorem: $IJ = LK = \\sqrt { (IK)^2 + (IL)^2} = \\sqrt {13^2 + 5^2} = 12$ • Jun 25th 2007, 10:17 PM Soroban Hello, sanee66! Quote: In square $JMNK$, diagonal $MK$ measures $\\sqrt{50}$ and in rectangle $IJKL$, diagonal $KI$ measures $13$. What is the length of IJ (or KL)? Code: I J M * - - - - - - -", "are parallel, by theorems regarding parallel lines $m\\angle A=m\\angle{ADK}=m\\angle{ALK}=60^\\circ$. By vertical angles, $m\\angle{ALK}=m\\angle{GLH}=60^\\circ$. The smallest circle through G and H is drawn when $\\overline{GH}$ is the diameter. The diameter is perpendicular to a tangent at the point of tangency, so $m\\angle{LHG}=90^\\circ$, and $\\triangle{GHL}$ is a 30º-60º-90º triangle, similar to $\\triangle{AJB}$, $\\triangle{AJC}$, $\\triangle{AKD}$, and $\\triangle{AKL}$. Now that many 30º-60º-90º right triangles are established, repeatedly apply side-length ratios of 30º-60º-90º right triangles ($1:\\sqrt{3}:2$) from here on. Since $AB=\\sqrt{3}$, $AJ=\\frac{3}{2}$. Since $JK=DE=1$, $AK=\\frac{3}{2}-1=\\frac{1}{2}$, and thus $DK=KL=\\frac{\\sqrt{3}}{6}$, and $DL=\\frac{\\sqrt{3}}{3}$. Since $DG=1$, $LG=1-\\frac{\\sqrt{3}}{3}$, and finally, $GH=\\frac{\\sqrt{3}}{2}(1-\\frac{\\sqrt{3}}{3})=\\frac{\\sqrt{3}}{2}-\\frac{3}{6}=\\frac{\\sqrt{3}}{2}-\\frac{1}{2}=\\boxed{\\frac{\\sqrt{3}-1}{2}}$. 8. What is the 18th digit after the decimal point in the decimal representation of $(9-4\\sqrt{5})^{9+4\\sqrt{5}}$? Solution: $9$ and $4\\sqrt{5}$ are very close to each other in value. As the exponent in the given expression is significantly greater than 1, the expression itself has an even smaller value. Let’s try to see how small this value is. Representing both $9$ and $4\\sqrt{5}$ as the square root of a positive integer, we find the base to equal $\\sqrt{81}-\\sqrt{80}$. $\\sqrt{81}-\\sqrt{64}=1$, and the difference between square roots of consecutive integers decreases as the integers themselves increase, so $\\sqrt{81}-\\sqrt{80}<\\frac{1}{17}<\\frac{1}{16}$. The exponent, $9+4\\sqrt{5}$, can be similarly written as $\\sqrt{81}+\\sqrt{80}>17=\\sqrt{81}+\\sqrt{64}$. Thus, $(9-4\\sqrt{5})^{9+4\\sqrt{5}}<\\left(\\frac{1}{16}\\right)^{17}=\\frac{1}{2^{68}}$. Since $5^2<2^5$, $\\frac{1}{2^{68}}=\\frac{1}{2^{18}*2^{50}}<\\frac{1}{2^{18}*5^{20}}=\\frac{1}{10^{18}*5^2}<\\frac{1}{10^{18}}$. Thus, $(9-4\\sqrt{5})^{9+4\\sqrt{5}}<\\frac{1}{10^{18}}$, and the first nonzero digit of the expression has not even occurred by the 18th digit,", "JKLMNP is a regular hexagon .Find the measure of $$\\angle$$ MQN. A. $$30^{\\circ}$$ B. $$45^{\\circ}$$ C. $$50^{\\circ}$$ D. $$60^{\\circ}$$ E. $$75^{\\circ}$$ An interesting question that requires you to be a bit creative with the information that you have been given. The given polygon is a REGULAR hexagon --> all sides and all angles are equal. Every interior angle of a hexagon = 120 degrees. As per the attached image, draw a circle around the given hexagon. Thus , now you get equal arcs LM=MN=NP=JP=JK=KL . As there are 6 arcs and all arcs must subtend a total of 360 degrees at the center of the circle ---> each arc subtends 60 degrees at the center. Additionally, an angle subtended by any arc of a circle at the center = 2* angle subtended by the same arc on the circumference. Consider triangle QNM,$$\\angle{QNM}$$ = angle subtended by LM = 30 degrees (as it is the angle subtended by arc LM on the circumference.) Similarly, Arcs KJ+JP+PN subtend $$\\angle{QMN}$$ on the circumference. ---> $$\\angle{QMN} = 90 degrees$$. Finally, in triangle QMN, $$\\angle{QMN} + \\angle {QNM} + \\angle {MQN} = 180 degrees$$ --->$$\\angle {MQN}$$ = 180-30-90 = 60 degrees. Hope this helps. Attachments 2015-12-05_10-45-47.jpg [ 12.84 KiB | Viewed 2526 times ] Manager Status: 2 months to go Joined: 11 Oct", "length $$KC=KL$$. This shows that $$[CB_1KA_2]=[CKK']=[KML].$$ Now $$[KB_1B_2]=[KB_1C]$$ because $$\\triangle KB_1B_2$$ and $$\\triangle KB_1C$$ share the same altitude from $$K$$ and have the same base legth $$B_1B_2=B_1C$$. Furthermore $$\\triangle KLB_2\\cong \\triangle CKA_2$$ because $$KL=CK$$, $$\\angle KLB_2=60^\\circ = \\angle CKA_2$$, and $$LB_2=KB_2$$ (by symmetry). Therefore $$[KB_1B_2L]=[KB_1B_2]+[KLB_2]=[KB_1C]+[CKA_2]=[CB_1KA_2].$$ By symmetry, we have $$[AC_1LB_2]=[BA_1MC_2]=[CB_1KA_2]$$ and $$[MA_1A_2K]=[KB_1B_2L]=[LC_1C_2M].$$ Therefore $$[KLM]=\\frac{1}{7}[ABC]$$. Here is another argument why $$\\triangle KML$$ is equilateral (which is not just saying \"due to symmetry\"). Note that $$\\triangle ABA_1$$ is obtained by a rotation of $$\\triangle CAC_1$$ about the center of $$\\triangle ABC$$ by $$60^\\circ$$ in the counterclockwise direction. Therefore $$CC_1$$ and $$AA_1$$ make an angle of $$60^\\circ$$. Hence $$\\angle KLM=60^\\circ$$. By the same argument $$\\angle LMK=60^\\circ$$ and $$\\angle MKL=60^\\circ$$ too. If you want to use a Menelaus argument, you can do it like this. From Menelaus's thm (on $$\\triangle ABA_1$$ with the traversal $$C_1LC$$) $$\\frac{AC_1}{C_1B}\\cdot\\frac{BC}{CA_1}\\cdot \\frac{A_1L}{LA}=-1.$$ Here the length ratios are signed (the ratio of lengths measured in the same direction is positive, and the ratio of lengths measured in the opposite directions is negative). That is $$\\frac{1}{2}\\cdot\\left(-\\frac{3}{2}\\right)\\cdot\\frac{AL_1}{LA}=-1\\implies \\frac{AL_1}{LA}=\\frac{4}{3}.$$ Using Menelaus's thm on $$\\triangle AA_1C$$ with traversal $$BMB_1$$ yields $$\\frac{AM}{MA_1}\\cdot \\frac{A_1B}{BC}\\cdot\\frac{CB_1}{B_1A}=-1.$$ Hence $$\\frac{AM}{MA_1}\\cdot\\left(-\\frac{1}{3}\\right)\\cdot\\frac{1}{2}=-1\\implies \\frac{A_1M}{MA}=\\frac{1}{6}.$$ This shows that $$A_1M:ML:LA=1:3:3$$. We have $$\\frac{[KLM]}{[LA_1C]}=\\frac{LM}{LA_1}\\cdot \\frac{LK}{LC}.$$ By symmetry $$\\frac{LK}{LC}=\\frac{ML}{MA}=\\frac12$$, so $$\\frac{[KLM]}{[LA_1C]}=\\frac{3}{3+1}\\cdot\\frac{1}{2}=\\frac{3}{8}.$$ Similarly $$\\frac{[LA_1C]}{[C_1BC]}=\\frac{CA_1}{CB}\\cdot\\frac{CL}{CC_1}=\\frac{2}{3}\\cdot\\frac{3+3}{1+3+3}=\\frac{4}{7}.$$ Finally $$\\frac{[C_1BC]}{[ABC]}=\\frac{BC_1}{BA}=\\frac{2}{3}.$$ Therefore $$\\frac{[KLM]}{[ABC]}=\\frac{[KLM]}{[LA_1C]}\\cdot \\frac{[LA_1C]}{[C_1BC]}\\cdot \\frac{[C_1BC]}{[ABC]}=\\frac{3}{8}\\cdot\\frac{4}{7}\\cdot\\frac{2}{3}=\\frac17.$$ According to Routh’s theorem the ratio", "18:38 $\\hspace{2cm}$ Since $\\overline{AB}=\\overline{AL}$, $\\triangle ABL$ is isosceles. $\\overline{AK}$ bisects $\\angle BAL$; therefore, $\\overline{BL}\\perp\\overline{AK}$. Let $J$ be the intersection of $\\overline{BL}$ and $\\overline{AK}$. Drop perpendicular $\\overline{KN}$ to $\\overline{AB}$ and perpendicular $\\overline{KP}$ to $\\overline{AC}$. $\\angle ABC=130^\\circ$ and $\\angle ABJ=80^\\circ$; therefore, $\\angle JBK=50^\\circ$. Being an external angle of $\\triangle ABC$, $\\angle NBK=50^\\circ$. Therefore, $\\triangle BJK=\\triangle BNK$. Thus, $\\overline{NK}=\\overline{JK}$. $\\triangle ANK=\\triangle APK$; therefore $$\\overline{PK}=\\overline{NK}=\\overline{JK}$$ Thus, $K$ is the center of the excircle to $\\triangle ABL$ tangent to $\\overline{BL}$. Being an external angle of $\\triangle AMC$, $\\angle KMC=20^\\circ$. Therefore, $\\triangle KMC$ is isosceles, giving $\\overline{MK}=\\overline{KC}$. Because $\\triangle CKP$ is a $30{-}60{-}90$ triangle, we can place $Q$ so that $\\triangle CQP$ is also $30{-}60{-}90$ and $\\triangle KQC$ is equilateral. Therefore, $\\overline{KQ}=\\overline{KC}$. Furthermore, $\\overline{KP}=\\overline{QP}=\\frac12\\overline{KQ}$. Thus, $$\\frac{\\overline{JK}}{\\overline{MK}}=\\frac{\\overline{KP}}{\\overline{KQ}}=\\frac12$$ Therefore, $\\overline{MJ}=\\overline{JK}$, and $\\triangle MBK$ is isosceles. Since $\\angle MBJ=\\angle JBK=50^\\circ$, we have that $\\angle BMJ=40^\\circ$, which leaves $\\angle AMB=140^\\circ$.", "opposite $$KL$$ is $$30^{\\circ}$$, and the angle opposite $$JL$$ is $$60^{\\circ}$$, so $$JL:5=\\sqrt{3}:1$$. Thus, we could get a unique value to answer the original question. $$\\implies$$ Sufficient _________________ My book with my solutions to all 230 PS questions in OG2018: Zoltan's solutions to OG2018 PS EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 15985 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In ΔJKL shown above, what is the length of segment JL ? [#permalink] ### Show Tags 12 May 2019, 14:06 Hi All, Based on the triangle JKL above, we know that we're dealign with a 30/60/90 right triangle. We're asked for the length of segment JL. The 30/60/90 triangle has a specific 'ratio of sides' (re: X : X√3 : 2X), meaning that if we know 1 of the sides, then we can determine the lengths of the other 2. By extension, depending on the information in the two Facts, we could potentially answer this question without doing any math at all. (1) JK = 10 Fact 1 gives us the hypotenuse of the triangle, so we can figure out the exact values of the other two sides. Fact 1 is SUFFICIENT (2) KL = 5 Fact 2 gives us the 'short leg' of the triangle,", "# What fraction of the larger semicircle is filled? What fraction of the larger semicircle is filled? The two smaller semicircles are of equal size. This is a puzzle originally set by Catriona Agg, who is a puzzle setting genius. • Hmm, I'm seeing a 90-60-30 triangle but nothing more than that :/ Dec 22 '20 at 13:18 • Maybe this is a better fit in MSE rather than PSE? Dec 22 '20 at 14:28 • @Anonymous I know an answer so I don't think it works for MSE. Dec 22 '20 at 14:48 The fraction is $$\\frac{2}{3}$$ Without loss of generality let the radius of the smaller circle equal 1. Edit: Bubbler has observed in a comment that from here it is faster to continue like this We have $$JA=JD=r$$, and since JH bisects the chord AB, $$\\angle{JHA}=90^\\circ$$. We also have $$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^2+r^2-2 \\sqrt{3} r-2 r+2 \\sqrt{3}+4$$ $$r=\\frac{6+2\\sqrt{3}}{2+2\\sqrt{3}}=\\sqrt{3}$$ Since $$\\angle HIG=90^\\circ$$ and $$\\arcsin(\\frac{1}{2})=30^\\circ$$, we have By the pythagorean theorem, $${HD}^2=HI^2+ID^2=1^2+(1+\\sqrt{3})^2=5+2 \\sqrt{3}$$ Now label the centre of the large circle J. We have $$JA=JD=r$$, and since JH bisects the chord AB, $$\\angle JHA=90^\\circ$$. By the pythagorean theorem $$JH^2+1^2=r^2$$ And by the cosine rule $$JH^2=JD^2+HD^2-2\\cdot HD\\cdot JD\\cdot\\cos(IDH)$$ $$JH^2=(r^2)+(5+2\\sqrt{3})-2\\cdot r\\cdot \\sqrt{5+2\\sqrt{3}} \\cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\\sqrt{3})-2\\cdot r\\cdot \\sqrt{5+2\\sqrt{3}} \\cos(IDH)$$ Solving for r produces $$r=\\frac{6+2\\sqrt{3}}{2\\sqrt{5+2\\sqrt{3}}\\cdot \\cos(IDH)}$$", "= 6, ZC = 5, ∠C = 80° Here, JK/ZC = 2.5/5 = 1/2 And, KL/XC = 3/6 = 1/2 ⇒ ∠K ≠ ∠C Thus, ΔJKL and ΔZXC are not similar. (v) For ΔJKL, we have ∠J + ∠K + ∠L = 180° (sum of angles of a triangle) ⇒ 70° + 80° + ∠L = 180° ⇒ ∠L = 180° – 70° – 80° ⇒ ∠L = 30° In ZXC, we have ∠Z + ∠X + ∠C = 180 (Sum of angles of Δ) ⇒ ∠Z + 80° + 30° = 180° ⇒ ∠Z = 180° – 80° -30° ⇒ ∠Z = 70° In ΔJKL and ΔZXC, we have ∠J = ∠Z = 70° ∠K = ∠X = 80° ∠L = ∠C = 30° Thus, ΔJKL ~ ΔZXC (AAA similarity criterion) Question 2: In the figure below, ΔJKL ¼ ΔZXL, KJL=70O and ZLX = 70O. Find JLK, JKL AND LZX Solution: JLX is a straight line. Thus, ∠JLK + ∠ KLX = 180° ⇒ ∠JLK = 180° – 125° = 55° In ΔJLK, ∠JKL+ ∠ KJL + ∠ JLK = 180° (Sum of the measures of the angles of a triangle is 180º.) ⇒ ∠JKL + 70º + 55º = 180° ⇒ ∠JKL = 55° Given that ΔLJK ~ ΔLXZ. ∴ ∠LZX = ∠LKJ (Corresponding angles are", "angle KJN = 45°. Find the area o [#permalink] ### Show Tags 16 Mar 2015, 03:05 1 Answer = E = 27 $$6^2 = a^2 + a^2$$ $$a = \\sqrt{18}$$ There are three equivalent right triangles as shown Attachment: gpp-sgf_img1.png [ 3.65 KiB | Viewed 1970 times ] There total area $$= 3 * \\frac{1}{2} * (\\sqrt{18})^2 = 27$$ Math Expert Joined: 02 Sep 2009 Posts: 59634 Re: In the figure, KLMN is a square, and angle KJN = 45°. Find the area o [#permalink] ### Show Tags 23 Mar 2015, 03:40 Bunuel wrote: In the figure, KLMN is a square, and angle KJN = 45°. Find the area of figure JKLMN. A. $$9+9\\sqrt{2}$$ B. $$9+18\\sqrt{2}$$ C. 18 D. $$18+9\\sqrt{2}$$ E. 27 Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION: Since KN must be perpendicular to JM, we know that JNK must be a 45-45-90 triangle. The legs are equal, and the hypotenuse is the square root of 2 times larger than either leg. Let JN = KN = x. Then $$\\frac{x}{6}=\\frac{1}{\\sqrt{2}}$$ --> $$x=3\\sqrt{2}$$ In that work, we rationalized the denominator when we divided by the radical. This number for x is the side of the square, so square KLMN has an area which is this number squared. Area of KLMN = $$(3\\sqrt{2})^2=18$$ That’s part of the area.", "the image depends on the mutual posi­tion of the observer and the kalei­do­scope axis — the image changes near the mirror prism edge as we sway. In the case of an arbi­trary triangle, if you start making all kinds of reflec­tions of it on the plane, they will overlap each other and you can’t speak about any beau­tiful image. If one builds an optical system as a mirror prism over such a triangle, the whole image will be composed of somehow shuf­fled frag­ments of the orig­inal image and will not be regular. So, a kalei­do­scope can be constructed using a triangle with angles $60^\\circ$—$60^\\circ$—$60^\\circ$, $90^\\circ$—$45^\\circ$—$45^\\circ$ or $90^\\circ$—$30^\\circ$—$60^\\circ$ as the base of the prism. How can we math­e­mat­i­cally under­stand that a triangle with angles $120^\\circ$—$30^\\circ$—$30^\\circ$, suit­able geomet­ri­cally for plane tiling using symme­tries, is not suit­able for constructing a kalei­do­scope? Are all possible trian­gles already listed? The require­ments on the resulting image in the kalei­do­scope can be stated explic­itly: The triangle at the base must have angles $\\frac{180^\\circ }{k}$, $\\frac{180^\\circ}{m}$, $\\frac{180^\\circ}{n}$, where $k$, $m$, $n$ are natural numbers, and $\\frac{180^\\circ}{k}+\\frac{180^\\circ}{m}+\\frac{180^\\circ}{n}=180^\\circ$. If we ignore the order, the only solu­tions $\\{k, m, n\\}$ of this equa­tion are the triples $\\{3, 3, 3\\}$, $\\{2, 4, 4\\}$ and $\\{2, 6, 3\\}$, giving the already familiar sets of angles $60^\\circ$—$60^\\circ$—$60^\\circ$, $90^\\circ$—$45^\\circ$—$45^\\circ$ and $90^\\circ$—$30^\\circ$—$60^\\circ$. There are", "+0 # geometry questions 0 73 2 (1) In triangle JKL, we have JK = JL = 25 and KL = 30. Find the area. (2) In triangle ABC, we have AB = 5, AC = 12, BC = 13. Find the circumradius. May 1, 2022 #1 +2455 +1 1) Use Heron's formula: $$\\sqrt{s(s-25)(s-25)(s-30)}$$, where $$s = {{25 + 25 + 30} \\over 2}$$ May 1, 2022 #2 +2455 +1 2) Because we have a right triangle, we know the circumradius is half of the hypotenuse (longest side length). This means that the circumradius is $$\\color{brown}\\boxed{6.5}$$ May 1, 2022", "a 45-45-90 right triangle (x = 45 degrees), where KL is the hypotenuse (across 2x = 90 degree angle). So, the ratio of the sides is $$JL:JK:KL=1:1:\\sqrt{2}$$ The side across angle z is side JL. So, $$JL = JK = 2$$ and $$KL = 2\\sqrt{2}$$. Does this make sense? _________________ Senior Manager Joined: 29 Jun 2017 Posts: 473 GPA: 4 WE: Engineering (Transportation) Re: What is the perimeter of the triangle in the figure above? [#permalink] ### Show Tags 27 Dec 2017, 06:37 1 Bunuel wrote: sahilvijay wrote: Bunuel wrote: What is the perimeter of the triangle in the figure above? Notice that y = z and y= x, so, x = y= z: x + x + 2x = 180 --> x= 45. So, we have 45-45-90 right triangle. (1) The side across from interior angle z measures 2. We know the lengths of the legs of a 45-45-90 right triangle. We can get the third side and calculate the perimeter . Sufficient. (2) 2x > y. We knew this from the stem. Not sufficient. Bunuel VeritasPrepKarishma I have a doubt - kindly see my image and reply . Statement 1 says The side across from interior angle z measures 2 I have a ques -> we can have 2 cases => case 1 KL =2 ,", "&= \\frac12( (h + c - u ) - (b+c - (u+v))\\\\ &= \\frac12(h + v - b) = ND\\end{align} Since $$\\angle ADC = 90^\\circ$$, $$ND = JN$$ and hence $$KM = JN$$. Similarly, \\begin{align}KN = AD - AK - ND &= h - \\frac12(b+c-a) - \\frac12(h+v-b) \\\\ &= \\frac12(2h - b - c + (u+v) - h - v + b)\\\\ &=\\frac12(h + u - c) = MD\\end{align} Since $$\\angle BDA = 90^\\circ$$, $$MD = IM$$ and hence $$KN = IM$$. It is clear $$\\angle IMK = \\angle KNJ = 90^\\circ$$. By SAS, $$\\triangle IMK$$ is congurent to $$\\triangle KNJ$$. Notice \\begin{align}\\angle DJK + \\angle KID &= ( \\angle DJN + \\angle NJK ) + ( \\angle KIM + \\angle MID )\\\\ & = (45^\\circ + \\angle NJK ) + (\\angle JKN + 45^\\circ)\\\\ &= 90^\\circ + (\\angle NJK + \\angle JKN)\\\\ &= 90^\\circ + (180^\\circ - \\angle KNJ)\\\\ &= 90^\\circ + 180^\\circ - 90^\\circ = 180^\\circ\\end{align} The quadrilateral $$IKJD$$ is cyclic.", "angle $M_4O_1P=30+45=75^{\\circ}$. In triangle $\\deltaM_4O_1P$ (Error compiling LaTeX. ! Undefined control sequence.), $tan 75 = \\frac{\\frac{d+i}{2}}{d-\\frac{d+i}{2}}$. Simplifying yields $tan 15 = \\frac{d+i}{d-i}$. The half-angle identity gives $tan 15 = \\frac{sin 30}{1+cos 30} = 2-\\sqrt{3}$ so $2-\\sqrt{3} = \\frac{d+i}{d-i}$. Solving for $i$ by subsituting $6\\sqrt{2}=d$ gives $i=2\\sqrt{6}$. To find the total length $\\overline{AP}$, we add $d+i=6\\sqrt{2}+2\\sqrt{6}=\\sqrt{72}+\\sqrt{24}$. Hence, $72+24 = \\framebox[1.5\\width]{96.}$.", "80^{\\circ}$, what is the size of $\\hat{C}$ and of $\\hat{A}$? 4. ${\\triangle}\\text{ABC} \\equiv {\\triangle}\\text{FDE}$. Name all the sides in the two triangles that are equal to AB. 5. Name the side that is equal to DE. 6. If $\\hat{F}$ is 40°, what is the size of $\\hat{B}$? 2. Look at figures JKLM and PQRS. (Give reasons for your answers below.) 1. What type of quadrilateral is JKLM? 2. Is JKLM ||| PQRS? 3. What is the size of $\\hat{L}$? 4. What is the size of $\\hat{S}$? 5. What is the length of KL?", "JBK$ is equilateral and the hexagon $ABCDEF$ is equiangular. Can you find out the measure of $\\angle KBC$ ?? From the figure we can clearly see $\\angle JBA + \\angle ABC + \\angle KBC + \\angle KBJ = 360^{\\circ}$ $\\angle KBC = 360^{\\circ}-90^{\\circ}-120^{\\circ}-60^{\\circ}$ [Since $\\angle JBA=90^{\\circ}$ (an angle of a square) $\\angle ABC=120^{\\circ}$ (an angle of an equiangular hexagon) and $\\angle JBK= 60^{\\circ}$(an angle of an equilateral triangle)] i.e., $\\angle KBC= 90^{\\circ}$ Then $\\triangle KBC$ is a right angle triangle Then the base and height of $\\triangle KBC$ are $BC$ and $KB$ So the area of $\\triangle KBC=\\frac12 \\times KB \\times BC = \\frac12 \\times 4\\sqrt{2} \\times 3\\sqrt{2} = 12$" ]

[ "+0 # In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. +1 97 2 +38 In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. Jul 2, 2022 #1 0 Jul 2, 2022 #2 +2448 0 Draw altitude $$JM$$. Because the triangle is isosceles, $$KM = ML = 15$$ Then, by the Pythagorean Theorem, $$JM = \\sqrt{25^2 - 15^2} = 20$$. This means that the area of the triangle is $$2 \\times( 20 \\times 15 \\div 2) = 300$$. Now, recall the formula for inradius, which is $${\\text{Area} \\over \\text{semiperimeter} } = {300 \\over 40} = \\color{brown}\\boxed{7.5}$$ Now, recall the formula for circumradius, which is $${\\text{product of the side lengths} \\over 4\\times \\text{Area}} = {18,750 \\over 1,200} = \\color{brown}\\boxed{15.625}$$ Jul 2, 2022", "18:38 $\\hspace{2cm}$ Since $\\overline{AB}=\\overline{AL}$, $\\triangle ABL$ is isosceles. $\\overline{AK}$ bisects $\\angle BAL$; therefore, $\\overline{BL}\\perp\\overline{AK}$. Let $J$ be the intersection of $\\overline{BL}$ and $\\overline{AK}$. Drop perpendicular $\\overline{KN}$ to $\\overline{AB}$ and perpendicular $\\overline{KP}$ to $\\overline{AC}$. $\\angle ABC=130^\\circ$ and $\\angle ABJ=80^\\circ$; therefore, $\\angle JBK=50^\\circ$. Being an external angle of $\\triangle ABC$, $\\angle NBK=50^\\circ$. Therefore, $\\triangle BJK=\\triangle BNK$. Thus, $\\overline{NK}=\\overline{JK}$. $\\triangle ANK=\\triangle APK$; therefore $$\\overline{PK}=\\overline{NK}=\\overline{JK}$$ Thus, $K$ is the center of the excircle to $\\triangle ABL$ tangent to $\\overline{BL}$. Being an external angle of $\\triangle AMC$, $\\angle KMC=20^\\circ$. Therefore, $\\triangle KMC$ is isosceles, giving $\\overline{MK}=\\overline{KC}$. Because $\\triangle CKP$ is a $30{-}60{-}90$ triangle, we can place $Q$ so that $\\triangle CQP$ is also $30{-}60{-}90$ and $\\triangle KQC$ is equilateral. Therefore, $\\overline{KQ}=\\overline{KC}$. Furthermore, $\\overline{KP}=\\overline{QP}=\\frac12\\overline{KQ}$. Thus, $$\\frac{\\overline{JK}}{\\overline{MK}}=\\frac{\\overline{KP}}{\\overline{KQ}}=\\frac12$$ Therefore, $\\overline{MJ}=\\overline{JK}$, and $\\triangle MBK$ is isosceles. Since $\\angle MBJ=\\angle JBK=50^\\circ$, we have that $\\angle BMJ=40^\\circ$, which leaves $\\angle AMB=140^\\circ$.", "+0 # In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. 0 48 2 +22 In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL. Jul 2, 2022 #1 0 The circumradius is 57/4. Jul 2, 2022 #2 +2275 +1 Draw altitude $$JM$$. Because the triangle is isosceles, $$KM = ML = 15$$ Then, by the Pythagorean Theorem, $$JM = \\sqrt{25^2 - 15^2} = 20$$. This means that the area of the triangle is $$2 \\times( 20 \\times 15 \\div 2) = 300$$. Now, recall the formula for inradius, which is $${\\text{Area} \\over \\text{semiperimeter} } = {300 \\over 40} = \\color{brown}\\boxed{7.5}$$ Now, recall the formula for circumradius, which is $${\\text{product of the side lengths} \\over 4\\times \\text{Area}} = {18,750 \\over 1,200} = \\color{brown}\\boxed{15.625}$$ Jul 2, 2022", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The", "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}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "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);", "# 1960 AHSME Problems/Problem 15 ## Problem Triangle $I$ is equilateral with side $A$, perimeter $P$, area $K$, and circumradius $R$ (radius of the circumscribed circle). Triangle $II$ is equilateral with side $a$, perimeter $p$, area $k$, and circumradius $r$. If $A$ is different from $a$, then: $\\textbf{(A)}\\ P:p = R:r \\text{ } \\text{only sometimes} \\qquad \\textbf{(B)}\\ P:p = R:r \\text{ } \\text{always}\\qquad \\\\ \\textbf{(C)}\\ P:p = K:k \\text{ } \\text{only sometimes} \\qquad \\textbf{(D)}\\ R:r = K:k \\text{ } \\text{always}\\qquad \\textbf{(E)}\\ R:r = K:k \\text{ } \\text{only sometimes}$ ## Solution $[asy] pair A=(0,50),O=(0,0),B=(43.301,-25),C=(-43.301,-25); draw(A--B--C--A); draw(circle(O,50)); draw(B--O--C); draw(anglemark(C,O,B,200)); draw((0,0)--(0,-25)); label(\"60^{\\circ}\",(-7,-12)); [/asy]$ First, find $P$, $K$, and $R$ in terms of $A$. Since all sides of equilateral triangle are the same, $P=3A$. From the area formula, $K=\\frac{A^2\\sqrt{3}}{4}$. By using 30-60-90 triangles, $R=\\frac{A\\sqrt{3}}{3}$. Using the same steps, $p=3a$, $k=\\frac{a^2\\sqrt{3}}{4}$, and $r=\\frac{a\\sqrt{3}}{3}$. Note that $P/p = 3A/3a = A/a$ and $R/r = \\frac{A\\sqrt{3}}{3} \\div \\frac{a\\sqrt{3}}{3} = A/a$. That means $P/p = R/r$, so the answer is $\\boxed{\\textbf{(B)}}$", "opposite $$KL$$ is $$30^{\\circ}$$, and the angle opposite $$JL$$ is $$60^{\\circ}$$, so $$JL:5=\\sqrt{3}:1$$. Thus, we could get a unique value to answer the original question. $$\\implies$$ Sufficient _________________ My book with my solutions to all 230 PS questions in OG2018: Zoltan's solutions to OG2018 PS EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 15985 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In ΔJKL shown above, what is the length of segment JL ? [#permalink] ### Show Tags 12 May 2019, 14:06 Hi All, Based on the triangle JKL above, we know that we're dealign with a 30/60/90 right triangle. We're asked for the length of segment JL. The 30/60/90 triangle has a specific 'ratio of sides' (re: X : X√3 : 2X), meaning that if we know 1 of the sides, then we can determine the lengths of the other 2. By extension, depending on the information in the two Facts, we could potentially answer this question without doing any math at all. (1) JK = 10 Fact 1 gives us the hypotenuse of the triangle, so we can figure out the exact values of the other two sides. Fact 1 is SUFFICIENT (2) KL = 5 Fact 2 gives us the 'short leg' of the triangle,", "Sufficient Option D is the correct Answer Manager Joined: 17 Jul 2014 Posts: 127 Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink] ### Show Tags 22 Jul 2019, 22:01 1 In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX? rectangle JKLM ; J=90 K=90 L=90 M=90 Line WX is parallel to Line JK Based on parallel line properties JDW = DJK = 90 = JKE = KEX ; if JDW = 90 , WDQ = 180-90 = 90 if KEX = 90 , XER = 180-90 = 90 XWY + WXY + WYX = 180 What is WYX? (1) The sum of the measures of angles DQS and ERT is 260. DQS+ERT=260 then DQW+ERX = 360-260 = 100 WDQ=XER=90 (as given) XWY+WXY = 360 - (WDQ+XER) - (DQW+ERX) XWY+WXY = 360 - (90+90) - 100 = 360-180-100 = 80 WYX = 180- (XWY+WXY) = 180-80 = 100 Sufficient (2) The sum of the measures of angles WQD and ERX is 100. This is exactly what we found in the stmt 1 as step 2 and its directly provided here ,then it follows the same reasoning as above DQW+ERX =", "indicated measure. Question 43. m∠QTR Question 44. m∠QRT Question 45. m∠SRT Question 46. QP Question 47. RT Question 48. RP In Exercises 49-54. the diagonals of square LMNP intersect at K. Given that LK = 1. find the indicated measure. Question 49. m∠MKN Question 50. m∠LMK Question 51. m∠LPK Question 52. KN Question 53. LN Question 54. MP In Exercises 55-69. decide whetherJKLM is a rectangle, a rhombus. or a square. Give all names that apply. Explain your reasoning. Question 55. J(- 4, 2), K(0, 3), L(1, – 1), M(- 3, – 2) Question 56. J(- 2, 7), K(7, 2), L(- 2, – 3), M(- 11, 2) Question 57. J(3, 1), K(3, – 3), L(- 2, – 3), M(- 2, 1) Question 58. J(- 1, 4), K(- 3, 2), L(2, – 3), M(4, – 1) Question 59. J(5, 2), K(1, 9), L(- 3, 2), M(1, – 5) Question 60. J(5, 2), K(2, 5), L(- 1, 2), M(2, – 1) MATHEMATICAL CONNECTIONS In Exercises 61 and 62, classify the quadrilateral. Explain your reasoning. Then find the values of x and y. Question 61. Question 62. Question 63. DRAWING CONCLUSIONS In the window, $$\\overline{B D}$$ ≅ $$\\overline{D F}$$ ≅ $$\\overline{B H}$$ ≅ $$\\overline{H F}$$. Also, ∠HAB, ∠BCD, ∠DEF, and ∠FGH are right angles. a. Classify HBDF and ACEG. Explain your", "F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"E\",E,SE); label(\"F\",F,NE); label(\"P\",P,NW); label(\"Q\",Q,NE); label(\"R\",R,SE); label(\"S\",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]$ Similar triangles can also solve the problem. First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $BC$ and solving. (The $8$ side would be collinear with line $AB$) After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \\frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \\frac{48}{5}$. We then know that $CE = \\frac{77}{5}$. Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\\Delta APQ$ is similar to $\\Delta ABC$. Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\\frac{PF}{FQ}=\\frac{BE}{EC} = \\frac{48}{77}$. Since $PF+FQ=PQ=\\omega$. We can solve for $PF$ and $FQ$ in terms of $\\omega$. We get that $PF=\\frac{48}{125} \\omega$ and $FQ=\\frac{77}{125} \\omega$. Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\\Delta APF$ is similar to $\\Delta ABE$. We can set up the proportion: $\\frac{AF}{PF}=\\frac{AE}{BE}=\\frac{3}{4}$. Solving for $AF$, $AF = \\frac{3}{4} PF = \\frac{3}{4}", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ 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.", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "# Difference between revisions of \"2019 AIME I Problems/Problem 6\" ## Problem 6 In convex quadrilateral $KLMN$ side $\\overline{MN}$ is perpendicular to diagonal $\\overline{KM}$, side $\\overline{KL}$ is perpendicular to diagonal $\\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\\overline{KN}$ intersects diagonal $\\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. ## Solution (Similar triangles) (writing this, don't edit) $[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label(\"K\", K, SW); label(\"L\", L, NW); label(\"M\", M, NE); label(\"N\", N, SE); draw(L--P); label(\"P\", P, S); dot(O); label(\"O\", shift((1,1))*O, NNE); label(\"28\", scale(1/2)*L, W); label(\"65\", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$ First, let $P$ be the intersection of $LO$ and $KN$. Note that $m\\angle KPL = 90^{\\circ}$ as given in the problem. Since $\\angle KPL \\cong \\angle KLN$ and $\\angle PKL \\cong \\angle LKN$, $\\triangle PKL \\sim \\triangle LKN$ by AA similarity. Similarly, $\\triangle KMN \\sim \\triangle KPO$. ## Solution 2 (Similar triangles, orthocenters) Extend $KL$ and $NM$ past $L$ and $M$ respectively to meet at $P$. Let $H$ be the intersection of diagonals $KM$ and $LN$ (this is the orthocenter of $\\triangle KNP$).", "# Difference between revisions of \"2019 AIME I Problems/Problem 6\" ## Problem 6 In convex quadrilateral $KLMN$ side $\\overline{MN}$ is perpendicular to diagonal $\\overline{KM}$, side $\\overline{KL}$ is perpendicular to diagonal $\\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\\overline{KN}$ intersects diagonal $\\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. ## Solution (Similar triangles) (writing this, don't edit) $[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label(\"K\", K, SW); label(\"L\", L, NW); label(\"M\", M, NE); label(\"N\", N, SE); draw(L--P); label(\"P\", P, S); dot(O); label(\"O\", shift((1,1))*O, NNE); label(\"28\", scale(1/2)*L, W); label(\"65\", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$ First, let $P$ be the intersection of $LO$ and $KN$. Note that $m\\angle KPL = 90^{\\circ}$ as given in the problem. Since $\\angle KPL \\cong \\angle KLN$ and $\\angle PKL \\cong \\angle LKN$, $\\triangle PKL \\sim \\triangle LKN$ by AA similarity. Similarly, $\\triangle KMN \\sim \\triangle KPO$.", "length $$KC=KL$$. This shows that $$[CB_1KA_2]=[CKK']=[KML].$$ Now $$[KB_1B_2]=[KB_1C]$$ because $$\\triangle KB_1B_2$$ and $$\\triangle KB_1C$$ share the same altitude from $$K$$ and have the same base legth $$B_1B_2=B_1C$$. Furthermore $$\\triangle KLB_2\\cong \\triangle CKA_2$$ because $$KL=CK$$, $$\\angle KLB_2=60^\\circ = \\angle CKA_2$$, and $$LB_2=KB_2$$ (by symmetry). Therefore $$[KB_1B_2L]=[KB_1B_2]+[KLB_2]=[KB_1C]+[CKA_2]=[CB_1KA_2].$$ By symmetry, we have $$[AC_1LB_2]=[BA_1MC_2]=[CB_1KA_2]$$ and $$[MA_1A_2K]=[KB_1B_2L]=[LC_1C_2M].$$ Therefore $$[KLM]=\\frac{1}{7}[ABC]$$. Here is another argument why $$\\triangle KML$$ is equilateral (which is not just saying \"due to symmetry\"). Note that $$\\triangle ABA_1$$ is obtained by a rotation of $$\\triangle CAC_1$$ about the center of $$\\triangle ABC$$ by $$60^\\circ$$ in the counterclockwise direction. Therefore $$CC_1$$ and $$AA_1$$ make an angle of $$60^\\circ$$. Hence $$\\angle KLM=60^\\circ$$. By the same argument $$\\angle LMK=60^\\circ$$ and $$\\angle MKL=60^\\circ$$ too. If you want to use a Menelaus argument, you can do it like this. From Menelaus's thm (on $$\\triangle ABA_1$$ with the traversal $$C_1LC$$) $$\\frac{AC_1}{C_1B}\\cdot\\frac{BC}{CA_1}\\cdot \\frac{A_1L}{LA}=-1.$$ Here the length ratios are signed (the ratio of lengths measured in the same direction is positive, and the ratio of lengths measured in the opposite directions is negative). That is $$\\frac{1}{2}\\cdot\\left(-\\frac{3}{2}\\right)\\cdot\\frac{AL_1}{LA}=-1\\implies \\frac{AL_1}{LA}=\\frac{4}{3}.$$ Using Menelaus's thm on $$\\triangle AA_1C$$ with traversal $$BMB_1$$ yields $$\\frac{AM}{MA_1}\\cdot \\frac{A_1B}{BC}\\cdot\\frac{CB_1}{B_1A}=-1.$$ Hence $$\\frac{AM}{MA_1}\\cdot\\left(-\\frac{1}{3}\\right)\\cdot\\frac{1}{2}=-1\\implies \\frac{A_1M}{MA}=\\frac{1}{6}.$$ This shows that $$A_1M:ML:LA=1:3:3$$. We have $$\\frac{[KLM]}{[LA_1C]}=\\frac{LM}{LA_1}\\cdot \\frac{LK}{LC}.$$ By symmetry $$\\frac{LK}{LC}=\\frac{ML}{MA}=\\frac12$$, so $$\\frac{[KLM]}{[LA_1C]}=\\frac{3}{3+1}\\cdot\\frac{1}{2}=\\frac{3}{8}.$$ Similarly $$\\frac{[LA_1C]}{[C_1BC]}=\\frac{CA_1}{CB}\\cdot\\frac{CL}{CC_1}=\\frac{2}{3}\\cdot\\frac{3+3}{1+3+3}=\\frac{4}{7}.$$ Finally $$\\frac{[C_1BC]}{[ABC]}=\\frac{BC_1}{BA}=\\frac{2}{3}.$$ Therefore $$\\frac{[KLM]}{[ABC]}=\\frac{[KLM]}{[LA_1C]}\\cdot \\frac{[LA_1C]}{[C_1BC]}\\cdot \\frac{[C_1BC]}{[ABC]}=\\frac{3}{8}\\cdot\\frac{4}{7}\\cdot\\frac{2}{3}=\\frac17.$$ According to Routh’s theorem the ratio", "+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", "if we can get A(WYX) from statement 1, we will get in from St 2 as well. St: A(DQY) + A(KRT) = 260 Look in triangle WDQ, we have A(W) = 180 - (180-A(DQY)) - 90 = A(DQY)-90 (see A(WDQ) = 90) Similarly, in triangle EXR, we have A(X) = 180 - (180-A(KRT))-90 = A(KRT)-90 So, in triangle WYX, A(W) + A(X) + A(Y) = 180 or A(Y) = 180 - (A(W)+ A(Y)) = 180 - A(DQY) + 90 -A(KRT) + 90 = 360 - A(DQY) - A(KRT) = 360 - 260 = 100. Manager Joined: 21 Jan 2019 Posts: 100 In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink] ### Show Tags Updated on: 22 Jul 2019, 10:46 1 Quote: In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX? 1) The sum of the measures of angles DQS and ERT is 260. (2) The sum of the measures of angles WQD and ERX is 100. As JKLM is a rectangle and line WX is parallel to line JK, triangle WDQ, XER are right angled triangle therefore, angle W + angle Q= 90. and angle R + angle X = 90.", "an isosceles triangle with its equal angles $\\arcsin(r_1) \\approx 56.88^\\circ$ (Figure 1, panel 6). Next we shall consider the evolution of certain special sequences of triangles. The first is where $x_1, x_2$ are constituted by successive terms of the Mātrā-meru sequence (Figure 2). Figure 2. Here, the triangles start with the $30^\\circ-60^\\circ-90^\\circ \\; \\triangle$ and converge to a unique scalene triangle with angles $\\arcsin\\left(\\tfrac{1}{\\phi}\\right) - \\arcsin\\left(\\tfrac{3\\phi-1}{2\\phi+1}\\right) - \\arcsin\\left(\\tfrac{2\\sqrt{\\phi}}{\\phi+1}\\right) \\approx 38.17^\\circ - 65.48^\\circ - 76.35^\\circ$, where $\\phi= \\tfrac{1+\\sqrt{5}}{2}$ is the Golden ratio. The last panel here shows an interesting numeric coincidence. If you start with $x_1=r_1, x_2=1-r_1$ (see above for $r_1$) you converge to a triangle close to that emerging from the Mātrā-meru sequence. Is there more to this than the coincidence of values? Finally, let us consider 2 other special triangles that emerge as convergents for 2 related types of operations based on the Mātrā-meru sequence (Figure 3). Figure 3. The first 2 rows (in light green) show triangles emerging from $x_1=\\tfrac{1}{M[k]}, x_2=\\tfrac{1}{M[k+1]}$, where $k=2, 3, 4...$. Here again, we start with a $30^\\circ-60^\\circ-90^\\circ \\; \\triangle$ and converge to a triangle of the form $\\approx 70.82^\\circ - 65.41^\\circ - 43.77^\\circ$. The second 2 rows (in yellow) show triangles emerging from $x_1=\\tfrac{1}{M[k]}, x_2=\\tfrac{1}{M[k]}$, where $k=3, 4, 5...$. These converge to a triangle of the form $\\approx 87.29^\\circ - 57.30^\\circ" ]

[ "as the smallest solution.", "smallestY; }", "small.", "small. AM", "smallest prime number? Answer: 2 is the smallest prime number.", "smallest numbers of prime signatures, where the prime factors are commonly listed from smallest to largest.", "smallest Wieferich prime in base 142 is 143111.", "6 factors. What is the smallest natural number with exactly 36 factors?", "factor of p-1 but this can't happen u less it is 1 since p is the smallest prime divisor of o(G). Thus$N(H)=C(H)$", "show that every number in this sequence has a small prime factor.", "# At least you got this in i-land Find the smallest prime factor $$p> 3$$ of $$200!+200!!+2!!!+1$$. Review double factorials and multifactorials to understand what $$200!!$$ and $$2!!!$$ refers to. ×", "tiny-tim", "smallest ideal containing the union. $\\square$", "small degree sum", "# Try Again Level pending What is the smallest prime factor of $$19^{n}-8^{n}$$, where $$2|n$$? ×", "for this small investigation.", "is the smallest prime factor of $$5^8+10^6–50^3$$? A. 2 B. 3 C. 5 D. 7 E. 11 Kudos for a correct solution. $$5^8 + 10^6 - 50^3$$ $$5^8 = (5^2)^4$$ $$10^6 = ((2*5)^2)^3$$ $$50^3 = (2*5^2)^3$$ $$(5^2)^3 + (5^2)^3 - (5^2)^3 [5^2 + 2^6 - 2^3] = 5^6 * 81 = 5^6 * 3^4.$$ So, $$3$$ is the smallest prime factor. Ans - B. Re: What is the smallest prime factor of 5^8+10^6–50^3? [#permalink] 15 Jan 2018, 23:06 Display posts from previous: Sort by", "\"small\" semiprimes (10, 39, and 371) share this attribute. 155 is also:", "11 C. 13 D. 17 E. 211 [Reveal] Spoiler: Solving P(12)+11 = (2 x 3 x 5 x 7 x 11) +11 = 11 (2 x 3 x 5 x 7 +1) =11(211) Since 211 is prime, shouldn’t 11 be the 2nd smallest prime factor of the product? [Reveal] Spoiler: OA Last edited by Bunuel on 23 Dec 2012, 05:32, edited 1 time in total. Renamed the topic and edited the question. Status: Tutor Joined: 05 Apr 2011 Posts: 542 Location: India Concentration: Finance, Marketing Schools: XLRI (A) GMAT 1: 570 Q49 V19 GMAT 2: 700 Q51 V31 GPA: 3 WE: Information Technology (Computer Software) Followers: 75 Kudos [?]: 401 [2] , given: 48 Re: smallest prime factor of p(12)+11 [#permalink] 22 Dec 2012, 09:51 2 KUDOS 1 This post was BOOKMARKED Since 211 is prime, shouldn’t 11 be the 2nd smallest prime factor of the product? 11 is the first smallest prime factor. 211 is the second smallest prime factor _________________ Intern Joined: 23 May 2012 Posts: 3 Followers: 0 Kudos [?]: 0 [0], given: 19 Re: smallest prime factor of p(12)+11 [#permalink] 23 Dec 2012, 02:59 p(12)=11*211,shouldnt 11 be the smallest prime Math Expert Joined: 02 Sep 2009 Posts: 30377 Followers: 5091 Kudos [?]: 57345 [2] , given: 8811 Re: If n is a positive integer", "known small factor $$q$$; you can add that to your collection." ]

[ "integers Originally Posted by rodders How do we know 3x6x9x11 =1782 can't be expressed as a difference of two squares? Suppose $a^2 - b^2 = 1782.$ Then a + b = larger factor of 1782 And a - b = smaller factor of 1782 2a = sum of the factors But 1782 = 2*(odd number), so this is impossible. For this to have a chance to work, the even number must be able to be factored into two even factors. Page 2 of 2 First 12", "3333 etc… Riddle-5: If 9999 = 4, 8888 = 8, 1816 = 6, 1212 = 0, then 1919 =?", "# Prime factorization of 1832 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 1832: By prime factorization of 1832 we follow 5 simple steps: 1. We write number 1832 above a 2-column table 2. We divide 1832 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 1832 prime factors number to factorize 2 916 2 458 2 229 229 1 Prime factorization of 1832 = 1×2×2×2×229= $1 × 2^3 × 229$ ## Related pages what is the lcm of 73squared24x37inswhat is the lcm of 62y x 3f x 3x26r 5n40x2800solve for i 9x 7i 3 3x 7ugcf of 90 and 126simultaneous equation solver with stepslog7 2168-25how to write 200 in roman numeralsadding mixed number fractions calculatorroman numerals 1972square root of 2209what is the greatest prime factor of 96find sin 2x cos 2x and tan 2xsimplify 5x 3xgcf of 262cosx 1the prime factorization of 54derivative of xcosxnumbers 1-10000002x 7y 8xlnxax", "## maths challenge if 9999=4,8888=8,1816=6,1212=0, then 1919=? 9999=4,8888=8,1816=6,1212=0, then 1919=2 9999=4,8888=8,1816=6,1212=0, then 1919=4 Formula is enclosed area * number of other digits 9999 = 4 enclosed area 8888= 8 enclosed area 1816 = 3 enclosed area * 2 digits = 6 1212 = 0 enclosed area*4 = 0 1919 = 2 enclosed area * 2 digits = 4 1919=2 ### Respond to this discussion You need to login to write a comment", "us.1787 So, it doesn't matter which s comes first; we have 4 s's, so 4! times...we have 3 e's, so 3!, times...1 n, so 1!, times...1 l, so 1!.1790 So, 9! is divided by 4! times 3! times 1! times 1!.1806 Now, 1! and 1!...well, 1! is just 1; that is it; there is no further down to go, so 1! is just 1.1811 1 times 1...we can just get rid of them for now.1819 At this point, we have 9! over 4! times 3!.1821 We could punch that into a calculator if we wanted, and it would give us the answer.1828 But we also might want to simplify things a little bit: 9! we can break down into 9 times 8 times 7 times 6 times 5 times 4!,1831 divided by 4! times 3!; well, we have a 4! here on the bottom, and a 4! on top; so they cancel out.1841 We have 9 times 8 times 7 times 6 times 5 on top at this point.1849 3! is 3 times 2 times 1; 3 times 2 times 1 is 6, so we have a 6 on the bottom now.1856 We have a 6 on top and a 6 on the bottom; they cancel out, so we are left with 9 times 8 times 7 times", "digits is divisible by 3 Manager Joined: 30 Mar 2017 Posts: 69 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 23:17 B : f(12) will be equal to 11, as it is the greatest prime factor of 12! g(11) = 2^11 +1 = 2049, which can be divided by 3 ,being the smallest prime factor. Sent from my Moto G (5) Plus using Tapatalk Manager Joined: 06 Jun 2019 Posts: 139 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 23:27 1 From the stem we know that: $$f(x)$$ - is the greatest prime factor of $$x!$$ $$g(y)$$ - is the smallest prime factor of $$2^y+1$$ Let's figure out step by step what $$g(f(12))$$ is. In order to calculate what $$g(y)$$ is, we need to first calculate what $$y$$ is. Since $$y=f(12)$$, we will first calculate what $$f(12)$$ is. 1. $$f(12)$$ is the greatest prime factor of $$12!$$ So $$f(12)=11$$ or $$y=11$$ 2. $$g(11)$$ is the smallest prime factor of $$2^{11}+1$$. What is the smallest prime factor of $$2049$$ ? $$2049$$ is not even, so it can't be $$2$$. Is $$2049$$ divisible by $$3$$ ? $$2+0+4+9=15$$ is divisible by $$3$$. Thus $$g(11)=3$$ Hence B _________________ Bruce Lee:", "# Prime factorization of 1832 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 1832: By prime factorization of 1832 we follow 5 simple steps: 1. We write number 1832 above a 2-column table 2. We divide 1832 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 1832 prime factors number to factorize 2 916 2 458 2 229 229 1 Prime factorization of 1832 = 1×2×2×2×229= $1 × 2^3 × 229$ ## Related pages 109-722x y 6 solve for y0.125 as a fractionwhat is 2000000000what is 6.25 as a fraction20 thousand pounds in dollars4x 5y 20 graphprime factors of 53948.1 is what percent of 240.50.5625 as a fractiondivide by fraction calculator97 roman numeralsexpression calculator solverderivative of ln 1 x 2x 4y4x-3y 122sinx sin2xbx16whencgreatest common factors calculator3.14 squared2x 5y 2y 4x graphalgebra factoring calculator with stepsfraction from least to greatest calculator1992 roman numeralsprime factorization 360kx25what is 0.375 as a percentagesimplify", "1782 is 3 Times Sum of all 2-Digit Numbers from its Digits Theorem $1782$ equals $3$ multiplied by the sum of all the $2$-digit integers that can be formed from its digits. Proof The number of $2$-digit integers that can be formed from the digits of $1782$ equals the number of $2$-permutations of $\\set {1, 7, 8, 2}$. That is: $\\set {17, 18, 12, 71, 78, 72, 81, 87, 82, 21, 27, 28}$ Hence: $17 + 18 + 12 + 71 + 78 + 72 + 81 + 87 + 82 + 21 + 27 + 28 = 594 = \\dfrac {1782} 3$ $\\blacksquare$", "1776 was Thursday. 2. October 19, 1992 m=8 d=19 c=19 y=92 You’ll get, $w \\equiv 97+23 \\pmod{7}$ $w-97 \\equiv 23 \\pmod{7}$ or, w-97-23=w-120 is divisible by 7. w=1, since 1-120=-119 is divisible by 7. Hence the Weekday on October 19,1992 was Monday. Similary, 3. w=5 Friday 4. w=0 Sunday 5. w=4 Thursday ### Posted by Gaurav Tiwari A designer by profession, a mathematician by education but a Blogger by hobby. With an experience of over seven years with WordPress, PHP and CSS3, Gaurav is capable of doing almost anything related to these. Beyond that, He is a mathematics graduate & a civil service aspirant.", "# If 18144 = 2a × 7b × 3c, then the value of $$a+ b + c\\over a$$ is 1. 1 2. 4 3. 3 4. 2 Option 4 : 2 ## Detailed Solution Given: 18144 = 2a × 7b × 3c Concept: Any composite number can be factorized into prime numbers. Calculation: As given number 18144 is a composite number, we can break it into a product of prime number such that 18144 = 25 × 34 × 7 On comparing it from 2a × 7b × 3c we will get a = 5, b = 4 and c = 1 Therefore, ⇒ $$\\frac{a\\ + \\ b \\ + \\ c}{a} = \\frac{5\\ + \\ 4 \\ + \\ 1}{5}$$ ⇒ $$\\frac{a\\ + \\ b \\ + \\ c}{a} = 2$$ • The numbers other than 1 whose only factors are 1 and the number itself are called Prime numbers. • Numbers having more than two factors are called Composite numbers • 2 is the smallest prime number which is even. • Every prime number except 2 is odd.", "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.", "factor. 2^(11)=2048+1=2049. The next smallest prime factor after 2 is 3, we apply the divisibility by 3 rule to see if 3 is a factor of 2049. Divisibility rule of 3 states that if the sum of the digits of any number is divisible by 3 then that number is also divisible by 3. $$2+0+4+9=15$$ and 15 is divisible by 3 so 2049 is thus also divisible by 3. You can also obviously do the division directly without knowing the divisibility rule to see if $$\\frac{2049}{3}$$ is an integer and you will see that it is and that $$\\frac{2049}{3}=683$$ but knowing the rule saves us the extra step and speeds things up. $$g(f(12))=g(11)=3$$. Answer is B. Senior Manager Joined: 12 Dec 2015 Posts: 427 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 08:42 1 For every positive integer x, f(x) represents the greatest prime factor of x!, and g(x) represents the smallest prime factor of $$2^x+1$$. What is (g(f(12))? Solution: f(12) = greatest prime factor of 12! = greatest prime factor of (12*11*10!) = 11 g(f(12) = g(11) = smallest prime factor of $$2^11+1$$ = smallest prime factor of 2049 = smallest prime factor of (3* 683) = 3. (2^11 = 1024*2=2048) A. 2 B. 3 -->", "by 7! 1101373/7= 157339 so 17,621,968= (2^5)(7)(157339). But 157339= 7(22477)= 7(7)(3211) while 3211 is not divisible by 7. The next larger prime number is 11 but 3211 is not divisible by 11. 3211= 13(247)= 13(13)(19). 19 itself is prime so we have 17,621,968= (2^5)(7^2)(13^2)(19). So we have 17,621,968,000,000,000= [(2^9)(5^9)][(2^5)(7^2)(13^2)(19)]= (2^14)(5^9)(7^2)(13^2)(19) 8. ## Re: Prime Factorization try factoring $19249 \\times 2^{13018586} + 1$ (hint: it's easier than it looks!) 9. ## Re: Prime Factorization Originally Posted by HallsofIvy To factor the number Plato gave, 17,621,968,000,000,000, I would first note that it is (17,621,968)(1,000,000,000). 1,000,000,000= 10^9= (2^9)(5^9). Now the hard part! 17,621,968 is even: 17,621,968= 2(8,810,984)= 2(2)(2)(4,405,492)= 2(2)(2)(2)(2,202,746)= 2(2)(2)(2)(2)(1101373). That last number is not even. Further, since its digit sum, 1+ 1+ 0+ 1+ 3+ 7+ 3= 16, is not divisible by 3, 3 is not a factor or 1101373. The last digit is not a \"0\" or a \"5\" so it is not divisible by 5. There is a test for divisibility by 7: subtract twice the last digit from the number. The original number is divisible by 7 if and only if the new one is. Here 1101373- 6= 1101367. If it is not obvious that that is divisible by 7, do it again- or just go ahead and try dividing by 7! 1101373/7= 157339 so 17,621,968= (2^5)(7)(157339).", "Question # Fill in the blanks:The unit digit of the cube of the number 1787 is ……………………. Verified 128.4k+ views Hint: Any number which has 7 in the units place, when we take the cube of that number then the digits in the units place will be the same as the digit in the units place of ${7^3}$. According to the question, we need to find the unit digit in the cube of the number 1787 . Now the digit in the units place in 1787 is 7. Let us first place the digits of 1787 as per the decimal system: Now, we have 7 in the unit's place in 1787. So the cube of 1787 means multiplying 1787 with itself thrice, that is,${\\left( {1787} \\right)^3} = \\left( {1787} \\right) \\times \\left( {1787} \\right) \\times \\left( {1787} \\right)$ . So, clearly the digit in the units place being 7 is also being cubed, which means that the digit in the units place in the cube of 1787 will be the same as the digit in the units place in the cube of 7. So, ${7^3} = 7 \\times 7 \\times 7 = 343$ . Thus, the digit in the units place of ${7^3}$ is 3 $\\Rightarrow$ the digit at the units place of ${\\left( {1787} \\right)^3}$ will be", "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!", "# Prime factorization of 1832 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 1832: By prime factorization of 1832 we follow 5 simple steps: 1. We write number 1832 above a 2-column table 2. We divide 1832 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 1832 prime factors number to factorize 2 916 2 458 2 229 229 1 Prime factorization of 1832 = 1×2×2×2×229= $1 × 2^3 × 229$ ## Related pages greatest to least fraction calculator4x 2y 8 02kxmultiple derivative calculatorcalculator for adding fractions415-15x 3 1 factoredsqrt 225sin3x 1what is 1996 in roman numeralsderivative of ln 5x73-59multiplication with variables calculator92-5220 off 49.99the square root of 484cos2x 1-sinx0.19 inches to fractionln 4x 2p 2l 2w solve for wgraph 4x y 8what fraction is equivalent to 0.125tanhx2003 roman numeralswhat is 6.25 in fraction formconvert 49.99 to poundsleast common multiple calculator with variableslcm and gcf calculatorwhat is 20 of", "Find four prime numbers less than $100$ which are factors of $3^{32}-2^{32}$. ## Solution Factorizing,we have \\begin{align} 3^{32}-2^{32} &=&(3^{16}+2^{16})((3^{16}-2^{16})\\ &=&(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2}) \\label{y2006r1p1e2} \\end{align} From Fermat’s Little Theorem, we have $3^{17-1}-1 - (2^{17-1}-1)$ is divisible by $17$. From $\\ref{y2006r1p1e2}$, we see that $(3^{32}-2^{32})$ is divisible by $5$, $13$ and $97$.", "certainly easier. It is apparent that the discriminant p^2 + 1776p must be a perfect square. Since p divides into the discriminant, so must p^2. Hence p divides into 1776. The primes of 1776 are 2, 3 and 37. Last edited: Jan 13, 2008 11. Jan 13, 2008 ### mda I'm not sure how this is easier... I think this is pretty much the same as what we've done except that it does not show which one is p. Note that the method summarized by mr vodka also predicts whether or not this will happen... the x^0 must be a product of only three numbers: p and two numbers that differ by exactly one. 12. Jan 14, 2008 ### Werg22 A thousand apologies, I judged the solution by its length rather than its content. 13. Jan 14, 2008 ### nonequilibrium Thanks for the help, guys", "n / m), I took a bit of a conceptual approach, and plugged numbers in to verify my thoughts. (i) This statement took most of my time - The fact that we are give a statement that 3n is divisible my m made me suspicious about n being divisible by m in all cases. It was relatively easy to come up with a value for n and m that satisfied both (3n/m) and (n/m) - such as n=14 and m=7. To disprove it, I looked for a value of n greater than 13 that had NO 3 in the prime factorization - I started with 16, which is 2x2x2x2. For m, I settled on 6, which has a 3 and a 2 in prime factorization. Sure enough, 16x3 = 48, which is divisible by 6. 16 is NOT divisible by 6. We have both a possible right or wrong answer here, so clearly INSUFFICIENT. (ii) This one is simply intuitive - 13 is a prime, and m cannot equal 13. n must be divisible by m PRIOR to being multiplied by 13. SUFFICIENT. Yes, the answer is B. My method took about 3 minutes! Thanks for the LCM angle - when I see this problem again, I will adopt that and save time. Enojy the Kudos! Kudos [?]:", "# Prime factorization of 1848 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 1848: By prime factorization of 1848 we follow 5 simple steps: 1. We write number 1848 above a 2-column table 2. We divide 1848 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 1848 prime factors number to factorize 2 924 2 462 2 231 3 77 7 11 11 1 Prime factorization of 1848 = 1×2×2×2×3×7×11= $1 × 2^3 × 3 × 7 × 11$ ## Related pages roman numerals 94hcf of 54 and 72sin 3x cos 3x2y 3x 42.5 percent as a decimalprime factorization of 190prime factorization of 2012y 3x 71983 roman numerals3aw website50x to 1xprime factors of 231what is the prime factorization of 2800prime factorization for 63webwork math ttu7xyroman numerals 1968fractions from least to greatest calculatorsin 6x sin 4xx senxsqrt of 144prime factorization for 802sin2x sinx800-320lcm of 180sqrt 196derivative of 3 lnxsolver quadratic" ]

[ "# Prime factorization of 1832 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 1832: By prime factorization of 1832 we follow 5 simple steps: 1. We write number 1832 above a 2-column table 2. We divide 1832 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 1832 prime factors number to factorize 2 916 2 458 2 229 229 1 Prime factorization of 1832 = 1×2×2×2×229= $1 × 2^3 × 229$ ## Related pages what is the lcm of 73squared24x37inswhat is the lcm of 62y x 3f x 3x26r 5n40x2800solve for i 9x 7i 3 3x 7ugcf of 90 and 126simultaneous equation solver with stepslog7 2168-25how to write 200 in roman numeralsadding mixed number fractions calculatorroman numerals 1972square root of 2209what is the greatest prime factor of 96find sin 2x cos 2x and tan 2xsimplify 5x 3xgcf of 262cosx 1the prime factorization of 54derivative of xcosxnumbers 1-10000002x 7y 8xlnxax", "# At least you got this in i-land Find the smallest prime factor $$p> 3$$ of $$200!+200!!+2!!!+1$$. Review double factorials and multifactorials to understand what $$200!!$$ and $$2!!!$$ refers to. ×", "-1+1=0 3 is smallest prime factor. Manager Joined: 28 Jan 2019 Posts: 120 Location: Peru For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags Updated on: 18 Jul 2019, 19:34 1 First, we have to find f(12), which is greatest factor of 12!, so we have 11 from this one. Then, we have to find out g(11) = 2^11 +1, which is 2048 + 1 = 2049, factorizing, we have that it has only two factors 3 and 683. g(f(12))= 3, so (B) is our answer Originally posted by Mizar18 on 18 Jul 2019, 08:31. Last edited by Mizar18 on 18 Jul 2019, 19:34, edited 2 times in total. Manager Joined: 21 Jan 2019 Posts: 100 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 08:35 1 Quote: For every positive integer xx, f(x)f(x) represents the greatest prime factor of x!x!, and g(x)g(x) represents the smallest prime factor of 2x+12x+1. What is (g(f(12))(g(f(12))? A. 2 B. 3 C. 5 D. 7 E. 11 f(12)= greatest prime factor of 12 ! As 12! = 12 *11* 10! therefore f(12) = 11 Now g(11)= smallest prime factor of $$2^(11)$$ +1 $$2^(10)$$ = 1024 $$2^(11)$$ = 2048 therefore $$2^(11)$$ +1 = 2049 divisible by 3", "is the smallest prime factor of $$5^8+10^6–50^3$$? A. 2 B. 3 C. 5 D. 7 E. 11 Kudos for a correct solution. $$5^8 + 10^6 - 50^3$$ $$5^8 = (5^2)^4$$ $$10^6 = ((2*5)^2)^3$$ $$50^3 = (2*5^2)^3$$ $$(5^2)^3 + (5^2)^3 - (5^2)^3 [5^2 + 2^6 - 2^3] = 5^6 * 81 = 5^6 * 3^4.$$ So, $$3$$ is the smallest prime factor. Ans - B. Re: What is the smallest prime factor of 5^8+10^6–50^3? [#permalink] 15 Jan 2018, 23:06 Display posts from previous: Sort by", "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.", "= 543.378$; the smallest factor of $14!$ greater than $543$ is $550$; the smallest factor of $14!/550$ greater than $(14!/550)^{1/3}$ is $567$; the smallest factor of $14!/(550\\times567)$ greater than its square root is $546$. We get $550 + 567 + 546 + 512 = 2175$. But $560+540+546+528 = 2174$, and both sets multiply to $14!$. – Michael Lugo Sep 20 '11 at 0:06", "digits is divisible by 3 Manager Joined: 30 Mar 2017 Posts: 69 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 23:17 B : f(12) will be equal to 11, as it is the greatest prime factor of 12! g(11) = 2^11 +1 = 2049, which can be divided by 3 ,being the smallest prime factor. Sent from my Moto G (5) Plus using Tapatalk Manager Joined: 06 Jun 2019 Posts: 139 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 23:27 1 From the stem we know that: $$f(x)$$ - is the greatest prime factor of $$x!$$ $$g(y)$$ - is the smallest prime factor of $$2^y+1$$ Let's figure out step by step what $$g(f(12))$$ is. In order to calculate what $$g(y)$$ is, we need to first calculate what $$y$$ is. Since $$y=f(12)$$, we will first calculate what $$f(12)$$ is. 1. $$f(12)$$ is the greatest prime factor of $$12!$$ So $$f(12)=11$$ or $$y=11$$ 2. $$g(11)$$ is the smallest prime factor of $$2^{11}+1$$. What is the smallest prime factor of $$2049$$ ? $$2049$$ is not even, so it can't be $$2$$. Is $$2049$$ divisible by $$3$$ ? $$2+0+4+9=15$$ is divisible by $$3$$. Thus $$g(11)=3$$ Hence B _________________ Bruce Lee:", "+ p_6) = 156$$. 156 is the smallest number for which this is true, and the only even number for which it is true. (The symbols $$\\pi(n)$$ and $$p_n$$ represent the number of primes less than or equal to n, and the nth prime respectively) 156 is evenly divisible by 12, the sum of its digits. Numbers which are divisible by the sum of their digits are usually called Niven Numbers. According to an article in the Journal of Recreational Mathematics the origin of the name is as follows. In 1977, Ivan Niven, a famous number theorist presented a talk at a conference in which he mentioned integers which are twice the sum of their digits. Then in an article by Kennedy appearing in 1982, and in honor of Niven, he christened numbers which are divisible by their digital sum “Niven numbers.” One might try to find the smallest strings of consecutive Niven Numbers with more than a single digit. *http://trottermath.net/niven-numbers/ I wonder about the relative order of the classes of numbers which are n times their digit sum for various n. 78 is the 12th Triangular number, which means that twice that, 156, is the number of times a clock that chimed the hours would chime in one day. 156 is a Harshad (Joy-Giver) number, divisible by", "divisible by 216, what is the smallest possible value for [#permalink] 29 Mar 2017, 21:30 Display posts from previous: Sort by", "of x!, and g(x) represents the smallest prime factor of 2x+1. What is (g(f(12))? Lets find inner equation value first i.e f(12) = 12! -> What is the greatest prime factor of 12! -> 11 g(11) = $$2^{11}$$ + 1 = 2048 +1 = 2049 -> What is the smallest prime factor of 2049 2049 is not divisible by 2 Find out if no if divisible by 3 - add the digits = sum is divisible by 3, hence 2049 is divisible by 3.. smallest prime factor is 3 Senior Manager Joined: 24 Nov 2016 Posts: 341 Location: United States Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 08:57 1 Quote: For every positive integer x, f(x) represents the greatest prime factor of x!, and g(x) represents the smallest prime factor of 2ˆx+1. What is (g(f(12))? A. 2 B. 3 C. 5 D. 7 E. 11 f(12) is g8primefac of 12!, which is the largest prime from 0 to 12 = 11; g(11) is smallest prime factor of 2ˆ(11)+1=2048+1=2049, notice that 2049 is not divisible by 2, but the sum of its digits is a multiple of 3, so its divisible by 3, which means the smallest pf is 3. Manager Joined: 08 Jan 2018 Posts: 145", "ago 112 is the smallest heptagonal number that is the sum of six consecutive primes - 5 years, 3 months ago 113 is the smallest three digit permutable prime - 5 years, 3 months ago 114 is the sum of the first four hyperfactorials - 5 years, 3 months ago 115 is the smallest odd heptagonal pyramidal semiprime - 5 years, 3 months ago 116 is the actual number of years the Hundred Years' War lasted. - 5 years, 3 months ago 117 is the smallest pentagonal number divisible by an odd square - 5 years, 3 months ago 118 is smallest three digit number leaving remainders (1,2,3,4) when divided by (3,4,5,6) - 5 years, 3 months ago 119 is the smallest \"prime-looking\" number greater than 100. - 5 years, 3 months ago 120 is $5!$ - 5 years, 3 months ago 121 is the smallest nontrivial palindromic square. - 5 years, 3 months ago 76 raise to power any positive integer end with 76 - 5 years, 3 months ago 30 is the sum of first four squares. - 5 years, 3 months ago 29 is the smallest two digit primorial prime, which forms a twin prime with another primorial prime (31). - 5 years, 3 months ago ×", "factor. 2^(11)=2048+1=2049. The next smallest prime factor after 2 is 3, we apply the divisibility by 3 rule to see if 3 is a factor of 2049. Divisibility rule of 3 states that if the sum of the digits of any number is divisible by 3 then that number is also divisible by 3. $$2+0+4+9=15$$ and 15 is divisible by 3 so 2049 is thus also divisible by 3. You can also obviously do the division directly without knowing the divisibility rule to see if $$\\frac{2049}{3}$$ is an integer and you will see that it is and that $$\\frac{2049}{3}=683$$ but knowing the rule saves us the extra step and speeds things up. $$g(f(12))=g(11)=3$$. Answer is B. Senior Manager Joined: 12 Dec 2015 Posts: 427 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 08:42 1 For every positive integer x, f(x) represents the greatest prime factor of x!, and g(x) represents the smallest prime factor of $$2^x+1$$. What is (g(f(12))? Solution: f(12) = greatest prime factor of 12! = greatest prime factor of (12*11*10!) = 11 g(f(12) = g(11) = smallest prime factor of $$2^11+1$$ = smallest prime factor of 2049 = smallest prime factor of (3* 683) = 3. (2^11 = 1024*2=2048) A. 2 B. 3 -->", "where the digits add to$19\\$ and that's 199. So the smallest where a (n)=4 will be the smallest where the digits add to 199. (It will have at least 22 digits so prime factors is definitely a useless idea). – fleablood Jan 30 at 2:50 Prime factorizations don't matter, digits do. So look at the smallest number such that $$a(n)>0$$. Note that $$T(d)=d$$ for all single digits $$d$$, so try $$T(10)=1$$, so $$a(10)=1$$. Going from here, we now know that if $$T(n)=10$$, then $$a(n)=2$$, since 10 is not a fixed point. To do that, simply make the digits sum to 10, and make the smallest number possible, which is $$19$$. Next, if $$T(n)=19$$ then we know $$a(n)=3$$. I'll leave you to figure out what the smallest number that has a digit sum of 19 is.", "18 Jul 2019, 23:11 1 f(12) = greatest prime factor of 12! = 11 g(11) = 2^11 + 1 It is known that 2^10 = 1024 Then 2^11= 2 * 1024 = 2048 g(11) = 2^11 + 1 = 2049 The sum of all digits is divisible by 3 Manager Joined: 06 Jun 2019 Posts: 118 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 18 Jul 2019, 23:27 1 From the stem we know that: $$f(x)$$ - is the greatest prime factor of $$x!$$ $$g(y)$$ - is the smallest prime factor of $$2^y+1$$ Let's figure out step by step what $$g(f(12))$$ is. In order to calculate what $$g(y)$$ is, we need to first calculate what $$y$$ is. Since $$y=f(12)$$, we will first calculate what $$f(12)$$ is. 1. $$f(12)$$ is the greatest prime factor of $$12!$$ So $$f(12)=11$$ or $$y=11$$ 2. $$g(11)$$ is the smallest prime factor of $$2^{11}+1$$. What is the smallest prime factor of $$2049$$ ? $$2049$$ is not even, so it can't be $$2$$. Is $$2049$$ divisible by $$3$$ ? $$2+0+4+9=15$$ is divisible by $$3$$. Thus $$g(11)=3$$ Hence B _________________ Bruce Lee: “I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times.” GMAC: “I fear", "order of G itself). Then again, I have no idea how to actually do any of this, never having done it myself. Eebster the Great (talk) 01:16, 11 November 2008 (UTC) Calculating cos(G degrees) is trivial, since G mod 360 can be calculated using the algorithm above. I get G ≡ 27 (mod 360), giving a cosine of about 0.891. Owen× 16:55, 11 November 2008 (UTC) Was I correct to suggest it would be more difficult to calculate cos(G radians)? Eebster the Great (talk) 01:09, 12 November 2008 (UTC) The smallest prime factor of G-4 is 107. (It happens that I've been computing factors of numbers near large towers of 3, for no particularly good reason; 3626257 is the smallest prime factor of G-16, for example, and 18540593 of G-46.) Joule36e5 (talk) 07:37, 22 November 2008 (UTC) ### So now we know, with \\\\ meaning \"is divisible by\" • G \\\\ 3 • G-1 \\\\ 2 • G-2 \\\\ 5 • G-3 \\\\ 2 • G-4 \\\\ 107 • G-5 \\\\ 2 • G-6 \\\\ 3 • G-7 \\\\ 2 What is the smallest value of H for which the smallest prime factor of G-H is not known?? Georgia guy (talk) 14:51, 22 November 2008 (UTC) I don't think it belongs on this talk page, unless it's relevant", "odd number of prime factors. 170 is the smallest number that can be written as the sum of the squares of 2 distinct primes, where each of these primes is the square of a prime added to another prime (170 = (22 + 3)2 + (32 + 2)2). *Prime Curios 170 is the largest integer for which its factorial can be stored in double-precision floating-point format. This is probably why it is also the largest factorial that Google's built-in calculator will calculate, returning the answer as 170! = 7.25741562 × 10306. (For 171! it returns \"infinity\".) 170 is the smallest number n for which phi(n)(the number of integers relatively prime to 170=64=82) and sigma(n) (the sum of the divisors of 170=324=182) are both square. EVENTS In 240 BC, Eratosthenes, a Greek astronomer and mathematician, estimated the circumference of the earth. As the director of the great library of Alexandria, he read in a papyrus book that in Syene, approaching noon on the summer solstice, the longest day of the year, shadows of temple columns grew shorter. At noon, they were gone. The sun was directly overhead. However, a stick in Alexandria, far to the north, could cast a pronounced shadow. Thus, he realized that the surface of the Earth could not be flat. It must be curved. Not", "# If 18144 = 2a × 7b × 3c, then the value of $$a+ b + c\\over a$$ is 1. 1 2. 4 3. 3 4. 2 Option 4 : 2 ## Detailed Solution Given: 18144 = 2a × 7b × 3c Concept: Any composite number can be factorized into prime numbers. Calculation: As given number 18144 is a composite number, we can break it into a product of prime number such that 18144 = 25 × 34 × 7 On comparing it from 2a × 7b × 3c we will get a = 5, b = 4 and c = 1 Therefore, ⇒ $$\\frac{a\\ + \\ b \\ + \\ c}{a} = \\frac{5\\ + \\ 4 \\ + \\ 1}{5}$$ ⇒ $$\\frac{a\\ + \\ b \\ + \\ c}{a} = 2$$ • The numbers other than 1 whose only factors are 1 and the number itself are called Prime numbers. • Numbers having more than two factors are called Composite numbers • 2 is the smallest prime number which is even. • Every prime number except 2 is odd.", "– user95591 Apr 1 '16 at 17:56 • @EasterlyIrk yeah, I use them in my non golfing code – Quill Apr 1 '16 at 22:27 • @ Quill okay, cool. – user95591 Apr 1 '16 at 22:28 # You’ve got a bug Well, I guess. Your lack of comments and/or docstrings make understanding sundaram and thus factor_count pretty hard. So I went for the description and the factor / next_prime_factor functions. Anyway: $python2 code.py Input the last number chosen: 9 The smallest number with more divisors than the inputted number is 12 . I thought 8 had as much divisors than 12 and was smaller. $ python2 code.py Input the last number chosen: 12 The smallest number with more divisors than the inputted number is 36 . Same here, 16 is 2×2×2×2, same lenght but smaller than 2×2×3×3. So, either I didn’t understand things well and there is some special handling of redundant factors; or you need to take into account that $2^n$ is the smallest number with $n$ divisors. With that in mind, you can remove almost anything except factor and next_prime_factor. # Count factors, don't list them Knowing that, in order to generate the smallest number with more than $n$ factors, you only need to compute $2^{n+1}$, there is no need in listing the factors of", "p=3$ Look at the smallest prime factor of $n$,say it is $p$.Then let $n=p^{v_p(n)}q$ with $p\\not|q$.Again if $q>1$,what follows from the same approach? One one thing is neutral in the universe, that is $0$.", "Tags 19 Jul 2019, 05:26 1 For every positive integer $$x, f(x)$$ represents the greatest prime factor of $$x!$$, and $$g(x)$$ represents the smallest prime factor of $$2^{x}+1$$. What is $$(g(f(12))$$? f(12) = GPF of 12!: GPF of 1*2*3*4*5*6*7*8*9*10*11*12 = 11 $$g(11) = 2^{11}+1 = 2048+1=2049$$ 2049 is divisible by 3 as $$(2+0+4+9 = 15)$$ is divisible by 3. Therefore smallest PF of 2049 is 3 A. 2 B. 3 C. 5 D. 7 E. 11 Manager Joined: 07 Jul 2019 Posts: 54 Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] ### Show Tags 19 Jul 2019, 05:37 1 f(12) = highest prime factor of 12 factorial - 11 g(11)=2^11+1=2^10=1024*2+1=2049. Smallest prime factor is 2, but 2049 cannot be divided by 2, A will not work B will work 2049 can be evenly divided by 3. Re: For every positive integer x, f(x) represents the greatest prime fact [#permalink] 19 Jul 2019, 05:37 Go to page Previous 1 2 3 4 5 Next [ 84 posts ] Display posts from previous: Sort by" ]

[ "# Rolling wheel! A wheel of diameter $$3\\text{ m}$$ is rolling in the forward direction on a horizontal surface. Let $$M$$ be a top most point of the circumference of the wheel at the initial position. Find the displacement of $$M$$ (in meters) after the wheel has completed one-fourth rollings.", "A wheel of radius $r$ rolls without slipping on a horizontal surface shown below. If the velocity of point $P$ is $10$ $m/s$ in the horizontal direction, the magnitude of velocity of point $Q$ (in $m/s$) is ______", "# The rolling starts now... A uniform wheel(disc) of radius $$R$$ was set rotating about its axis at an angular speed $$\\omega$$. Now it is carefully placed on a rough horizontal surface of friction coefficient $$\\mu$$ with its axis horizontal. Because of friction the wheel accelerates forward and its rotation decelerates till the wheel starts pure rolling. If the time after which pure rolling starts on the surface is given by $$\\frac { a }{ b } \\frac { R\\omega }{ \\mu g }$$ then find $$a+b$$. ×", "Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius $$72$$ and rolls along the surface toward the smaller tube, which has radius $$24$$. It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $$x$$ from where it starts. The distance $$x$$ can be expressed in the form $$a\\pi+b\\sqrt{c},$$ where $$a,$$ $$b,$$ and $$c$$ are integers and $$c$$ is not divisible by the square of any prime. Find $$a+b+c.$$ (第二十五届AIME2 2007 第11题)", "the appropriate shape (essentially using the wire as an analog calculator; I think this is what people did in the old days). ## Bonus: square wheels In geometry the hyperbolic cosine is often called a 'catenary' and can be defined as the curve that the locus of a parabola (i.e. the focus point of a parabolic mirror) traces out when the parabola is \"rolled\" over a line. A funny geometrical property of the catenary curve is that when one rolls a line over it, there exists a point above this line (fixed with respect to the line) that traces out a horizontal line. This implies that one can design a road of catenary bumps on which a matched square (!) wheel can smoothly roll (see figure 11). figure 11. A square wheel rolling on a catenary road And of course people tried this. Happy catenary spotting to y'all!", "# Is he rolling forever? Geometry Level 3 Above are two tangent circles, a bigger one with radius $$R=1$$ and a smaller one with perimeter $$p=2$$. Imagine you can roll the smaller circle counterclockwise around the bigger circle so that they stay tangent. Only whole revolutions are allowed. You now start rolling the small circle. After how many revolutions will the midpoint of the small circle be back exactly at his red marked starting position as shown above? If you think the small circle will never be back at his starting point answer $$0$$ ×", "# A wheel of radius 1 𝑚 rolls forward half a revolution on a horizontal ground. The magnitude of thedisplacement of the point of wheel initially in contact with the ground is Question : A wheel of radius 1 𝑚 rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of wheel initially in contact with the ground is (A) $2\\pi$ (B) $\\sqrt{2}\\pi$ (C) $\\sqrt{\\pi ^{2}+4}$ (D) $\\pi$", "Instantaneous Axis of Rotation A disc of radius $$5m$$ is rolling on the ground. What is the distance covered by the instantaneous axis of rotation in $$1$$ revolution? ×", "# A wheel of radius 1 meter rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is ? The word \"initially\" is important here. If we were examining the point of the wheel in contact with the ground, the displacement would be zero. Instead, we are looking at how far (straight) a point has moved after a $\\pi$ radians revolution. At half a circumference of travel, the displacement is $2 r$ upwards. $D = 2 \\left(1\\right) = 2 \\text{m}$", "A wheel of mass 'm' and radius 'R' is rolling on a level road at a linear speed 'V'. What is the kinetic energy of the upper right quarter part of the wheel , considering the wheel to be of the ...", "# Why not displacement? A cylinder of radius $$R$$ performs pure rolling on a horizontal surface. Find the distance travelled by the point $$A$$ which was initially in contact with the surface when one complete revolution is made by the the cylinder and the point $$A$$ touches the ground again. ×", "wheel makes one revolution every 3 seconds. C: After $$\\frac{1}{4}$$ of a second, the point $$P$$ will be in the position marked $$Q$$. D: After $$\\frac{3}{4}$$ of a second, the point $$P$$ will be in the position marked $$Q$$. E: The radius of the wheel is 10 inches. (From Unit 6, Lesson 18.)", "Ferris-wheel is a circle. Circle-like parts, e.g. cylinders, tubes, gears, and others are used by engineers for making clocks, bikes, cars, trains, ships, planes, and even rockets. The invention of the wheel-cart was one of the transforming events in early human history. Begin typing your search term above and press enter to search. Press ESC to cancel. Back To Top", "I got this problem wrong. Can you show me how to get themagnitude? Thanks. A wheel with a radius of 42.0 cm rollswithout slipping along a horizontal floor. At timet1, the dot P painted on the rim ofthe wheel is the point of contact between the wheel and the floor,as shown in th figure below. At a later timet2, the wheel has rolled through one-half of a revolution. (a) What is the magnitude of the displacementof P? 1 m (b) What is the angle (relative to the floor) of the displacementof P? 2°", "revolution? What happens at the corners? ### Possible extension Try the problem Five Circuits, Seven Spins. There are more ideas, explanations and problems to work on in the article A Rolling Disc. How many times does the disc rotate about its own centre when it makes one revolution around the tray? What happens if corners are not $90^\\circ$? ### Possible support Physically manipulating a circular object inside a frame can make it clearer what's happening at the corners. The problem Roundabout may be a good starting point.", "# A wheel of radius 0.5 m rolls without sliding on a horizontal surface, starting from rest, the wheel moves with constant acceleration $6\\: rad/s^2$ . The distance travelled by the centre of the wheel from $t = 0 \\: to \\: t=3s$ is : $\\begin {array} {1 1} (1)\\;27\\: m & \\quad (2)\\;13.5\\: m \\\\ (3)\\;18\\: m & \\quad (4)\\;Zero \\end {array}$ (2) 13.5 m", "# Instantaneous Axis of Rotation A disc of radius $$5m$$ is rolling on the ground. What is the distance covered by the instantaneous axis of rotation in $$1$$ revolution? × Problem Loading... Note Loading... Set Loading...", "parabola between the endpoints $(1,1)$ and $(-1,1)$, and is at point $(1,1)$ at time $t=0$. It is sometimes quite easy to describe a complicated path in parametric equations when rectangular and polar coordinate expressions are difficult or impossible to devise. Example 12.4.2 A wheel of radius 1 rolls along a straight line, say the $x$-axis. A point on the rim of the wheel will trace out a curve, called a cycloid. Assume the point starts at the origin; find parametric equations for the curve. Figure 12.4.1 illustrates the generation of the curve (click on the AP link to see an animation). The wheel is shown at its starting point, and again after it has rolled through about 490 degrees. We take as our parameter $t$ the angle through which the wheel has turned, measured as shown clockwise from the line connecting the center of the wheel to the ground. Because the radius is 1, the center of the wheel has coordinates $(t,1)$. We seek to write the coordinates of the point on the rim as $(t+\\Delta x,1+\\Delta y)$, where $\\Delta x$ and $\\Delta y$ are as shown in figure 12.4.2. These values are nearly the sine and cosine of the angle $t$, from the unit circle definition of sine and cosine. However, some care is required because we", "a future for fabric-formed concrete structures?\", Structures and Architecture: Beyond their Limits (2016) ed., Paulo J. da Sousa Cruz, pp. 1087-1088. Square wheel rolling over inverted catenaries • One can roll noncircular wheels over appropriate road surfaces. The most striking example of this is the fact that a square wheel can roll on a road that consists of linked catenaries (the catenaries are defined by y = - cosh x) ...(the ride is not smooth in the sense that the center does not move forward at a constant rate of rotation, but... the actual ride... feels quite smooth). This animation was inspired by an exhibit at San Francisco's Exploratorium... • Stan Wagon, Mathematica® in Action: Problem Solving Through Visualization and Computation (2010)", "kg}$ ... . The acceleration of the wheel axle system in horizontal direction is ______ $\\text{m/s}^{2} (\\textit{round off to one decimal place}$)." ]

[ "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", "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.", "draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -45, 45); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(45), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 4.5)--(6, -3)); draw((9, 4.5)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, -3), SW); label(\"C_2'\", (9, -3), SE); label(\"C_3'\", (15/2, 0), NE); dot((15/2, 0)); draw(Circle((15/2, 3), 3/2)); dot((15/2, 3)); label(\"C_4'\", (15/2, 3), N); draw((15/2, 0)--(15/2, 3)); draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); [/asy]$ ## Solution 6 (Kissing Circles) Draw the other half of the largest circle and proceed with Descartes' Circle Formula. The curvatures of circles A,B,C, and P are $1, \\frac{1}{2}, -\\frac{1}{3},$ and $\\frac{1}{r}$, respectively.", "= circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired. ## Solution 4 (Quick) Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot(\"A\",A,dir(A-O2/2)); dot(\"B\",B,dir(B-O2/2)); dot(\"O_2\",O2,right+up); dot(\"O_1\",O1,left+up); dot(\"O\",O,dir(O-O2)); dot(\"D\",D,dir(170)); dot(\"E\",E,dir(E-O1)); dot(\"X\",X,dir(X-D)); dot(\"Y\",Y,dir(Y-D)); label(\"R\",O--E,right+up,p); label(\"R\",O--D,left+down,p); label(\"2R\",(X+O)/2-(150,0),down,p); label(\"961\",O1--D,2*(left+down),p); label(\"625\",O2--E,2*(right+up),p); MA(\"\",E,D,O1,100,fuchsia+linewidth(1)); MA(\"\",X,D,O,100,fuchsia+linewidth(1)); MA(\"\",Y,E,O2,100,orange+linewidth(1)); MA(\"\",D,E,O,100,orange+linewidth(1)); [/asy]$", "mypara=revolution(O,g,X); Full code: copy to http://asymptote.ualberta.ca/ and click Run unitsize(1cm); import graph3; import solids; //currentprojection=orthographic(2,1.5,.5,zoom=.95); currentprojection=obliqueZ; currentprojection.zoom=.95; size(8cm); real a=1.8, b=a^2; real c=2.5,d=4.5; triple A=(a,0,0), B=(0,b,0); triple C=(c,0,0), D=(0,d,0); // the (red) parabol curve on the XY-plane triple paraXY(real t){return (t,t^2,0);} path3 g=graph(paraXY,0,a); draw(g,red+1.2pt); // use one of the following revolutions //revolution mypara=revolution(C,g,Y); // around x=c //revolution mypara=revolution(D,g,X); // around y=d //revolution mypara=revolution(O,g,X); // around y=0 revolution mypara=revolution(O,g,Y); // around x=0 draw(surface(mypara),yellow+opacity(.7)); //draw(mypara,blue,orange); label(\"$a$\",A,S); label(\"$c$\",C,S); label(\"$b$\",B,W); label(\"$d$\",D,W); draw(A--(a,b,0)--B^^C--(c,d,0)--D,dashed+gray); draw(Label(\"$x$\",EndPoint),-2*X--6*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--6*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),O--6*Z,Arrow3()); // you may not want to draw Oz 2. The same code, only change X to Y to get revolution around Oy revolution mypara=revolution(O,g,Y); 3. The same code, for revolution around x=c revolution mypara=revolution(C,g,Y); 4. Last, for revolution around y=d revolution mypara=revolution(D,g,X); Welcome to TeX.SE!!! This is a quick modification of an answer from here. It's a 2d drawing using isometric perspective. The equations of parabola and ellipses are calculated by hand. The ellipses are very simple if you know about isometric perspective (the axes ratio is sqrt(3):1). The parabola was obtained as an envelope of the family of 'horizontal' ellipses. This is the code: \\documentclass[tikz,border=2mm]{standalone} \\tikzset {% axes/.style={thick,-latex}, cylinder/.style={right color=blue!80,left color=white,fill opacity=0.6}, paraboloid/.style={left color=magenta!80,fill opacity=0.6}, } \\begin{document} \\begin{tikzpicture}[line cap=round,line join=round] % axes x,y \\draw[axes] (30:3) -- (210:3) node [left] {$x$}; \\draw[axes] (150:3) -- (330:3) node [right]", "+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", "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", "internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\\cdots$ and a sequence of points on the circles $A_1,A_2,A_3,\\cdots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \\frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. \\usepackage{asymptote} \\begin{figure}[h] \\begin{asy} draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"$A_0$\",(125,0),E); dot((25,100)); label(\"$A_1$\",(25,100),SE); dot((-55,20)); label(\"$A_2$\",(-55,20),E); \\end{asy} \\end{figure} ## Problem 13 For each integer $n\\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$. ## Problem 14 A $10\\times10\\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of", "+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", "of the fractal circlen=plt.Circle((0,a+2),1,color='black') circleo=plt.Circle((0,a+4),1,color='black') circlep=plt.Circle((0,a+6,0),1,color='black') circleq=plt.Circle((c+2,a+4),1,color='black') circler=plt.Circle((c-2,a+4),1,color='black') #bottom of fractal circles=plt.Circle((0,f-2),1,color='black') circlet=plt.Circle((0,f-4),1,color='black') circleu=plt.Circle((0,f-6),1,color='black') circlev=plt.Circle((c+2,f-4),1,color='black') circlew=plt.Circle((c-2,f-4),1,color='black') a=a+6 f=f-6 if i%1==0: fig = plt.gcf() pylab.show() #calculating the contour length of the hexagonal fractal w=6.0 o=2.0 b=4.0 r=0 import math myList=[] f=open('contour length - hexagonal fractal.txt','w') for i in range (0,100): contourlength=w*(2) + o*(math.pi/2) + b*(math.pi/4) + r*(2.18142240792219) w=w+8 o=o+4 b=b+8 r=r+4 if i%1==0: print 'The new contour length is', contourlength CStr=str(contourlength) myList += [CStr] myString = \",\".join(myList)+'\\n' f.write(myString) print 'The program has finished running.' f.close() import matplotlib.pyplot as plt from numpy import * from pylab import * import pylab preX=range(3,6003,6) x=[] #characteristic length associated with the object (radius) y=[] #quality of the object (contour length) for index in range(len(myList)): x.append(log(float(preX[index]))) y.append(log(float(myList[index]))) #finding the best fit line (m,b) = polyfit(x,y,1) yp=polyval([m,b],x) plt.plot(x,yp) grid(True) print 'The slope is of the best fit line is', m print 'The y-intercept of the best fit line is', b plt.plot(x,y,'ro') #labeling/details of graph plt.title('Change in Radius of Hexagonal Fractal vs.Contour length ') plt.ylabel('Log(Contour Length)') plt.show() import math #graphing number of circles vs. contour length for the plus fractal c=4.0 r=4.0 b=0 i=0 myList=[] f=open('YES.txt','w') for i in range (0,100): p=c*((math.pi)/4) + r*(2*(2**0.5)) + b*(2-(2**0.5)) c=c+8 r=r+16 b=b+8 if i%1==0: print 'The new perimeter value is', p pStr=str(p) myList += [pStr] myString = \",\".join(myList)+'\\n' f.write(myString) print", "arrowed double line pointing to the right rectangle with the line labeled $$(p \\land \\lnot Kp)!$$. The first circle in the first rectangle has a double line dark orange rim, is labeled w and contains the letter p. The second circle in the first rectangle has a single line dark orange rim, is labeled u, and contains nothing. A solid line connects these two circles. The circle in the second rectangle has a double line dark orange rim, is labeled w, and contains the letter p. ### Figure 3 description Two large rectangles each containing several orange circles. The first rectangle is labeled M and contains two circles. The first has a double line dark orange rim and is labeled w. The second has a single line dark orange rim, is labeled u, and contains the letter p. Two solid lines connect the two circles; one labeled $$R_1$$ and the other $$R_2$$. The second rectangle contains 5 orangle circles on a 3x3 matrix. The first column contains, in the second row, the first circle, which has a double line dark orange rim. The other two columns each have two circles, one in the first row and one in the third row, with all of them having only a single line dark orange rim. The first circle is labeled", "we can solve to get $d = \\boxed{672}$.$[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. Solution 3 (Inversion) WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$.", "the number of fixed points under possible transformations: 1. The identity transformation. Since this doesn't change anything, there are $60$ fixed points 2. Reflect on a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$. Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection 3. Rotate by $120^\\circ$ counterclockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$, $4$, and $6$ will be the same. Similarly, the colors of circles $2$, $3$, and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation. By Burnside's Lemma, the total number of colorings is $(1 \\cdot 60+3 \\cdot 4+2 \\cdot 0)/(1+3+2) = \\boxed{\\textbf{(D) } 12}$. ### Solution 3 Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as $[asy] filldraw(circle((0,0),1),green); draw(circle((2,0),1)); draw(circle((4,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((3,sqrt(3)),1)); draw(circle((2,2sqrt(3)),1)); draw(circle((8,0),1)); filldraw(circle((10,0),1),green); draw(circle((12,0),1)); draw(circle((9,sqrt(3)),1)); draw(circle((11,sqrt(3)),1)); draw(circle((10,2sqrt(3)),1)); [/asy]$ Take the first case. Now, we must pick", "N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); [/asy]$ $C_1$ has radius $3$, $C_2$ has radius $2$, and $C_3$ has radius $1$. We want to find the radius of $C_4$. We now invert the four circles. $C_1$ inverts to a line. Given that one point is on $\\Omega$, and all points on $\\Omega$ invert to themselves, we know that the resulting line must intersect that intersection point. $C_2$ also inverts to a line. $C_2$ has radius $4$, and since $\\Omega$ has radius of $6$, the resulting line must be $\\frac{36}{4} = 9$ units away from $O$. $C_3$ inverts to a circle. By observing the diagram, we note that $C_3'$'s center must be on $\\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\\frac{3}{2}$ that is $\\frac{15}{2}$ units from $O$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2,", "note that $C_3'$'s center must be on $\\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\\frac{3}{2}$ that is $\\frac{15}{2}$ units from $O$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 3)--(6, -3)); draw((9, 3)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, 3), N); label(\"C_2'\", (9, 3), N); label(\"C_3'\", (15/2, 0), N); dot((15/2, 0)); [/asy]$ Now, we invert $C_4$. Note that $C_4$ is tangent to the three other original circles. So, in the inversion, $C_4'$ must be tangent to the two lines and $C_3'$. It is then quickly seen that $C_3'$ and $C_4'$ have the same radius: $\\frac{3}{2}$. Now, we can determine the radius of $C_4$ using the formula $r = \\frac{k^2 \\cdot r'}{\\overline{OC_2}^2 - r'^2}$. $k^2 = 36$, and $r' = \\frac{3}{2}$. $\\overline{OC}$ is just the distance", "larger in the middle and two smaller ones around it: pg.draw.circle(canvas, pg.Color(\"white\"), (200, 200), 50) pg.draw.circle(canvas, pg.Color(\"white\"), (150, 200), 30) pg.draw.circle(canvas, pg.Color(\"white\"), (250, 200), 30) If we wanted to draw that cloud at different heights, we could repeat these three commands, each time with some new value for $$y$$ coordinate of the centers of these three circles instead of 200, as it is in the first drawing. For example: pg.draw.circle(canvas, pg.Color(\"white\"), (200, 200), 50) pg.draw.circle(canvas, pg.Color(\"white\"), (150, 200), 30) pg.draw.circle(canvas, pg.Color(\"white\"), (250, 200), 30) pg.draw.circle(canvas, pg.Color(\"white\"), (200, 80), 50) pg.draw.circle(canvas, pg.Color(\"white\"), (150, 80), 30) pg.draw.circle(canvas, pg.Color(\"white\"), (250, 80), 30) pg.draw.circle(canvas, pg.Color(\"white\"), (200, 320), 50) pg.draw.circle(canvas, pg.Color(\"white\"), (150, 320), 30) pg.draw.circle(canvas, pg.Color(\"white\"), (250, 320), 30) In this way, not only does the program grow faster than it has to, we also need to make every change in three places (for example, if we want to try 330 instead of 320, that change should be made in three places). Three changes are not many, but if we adopt this way of doing things, we would have more and more problems in more complex drawings, or in complex programs in general. Instead, it is better to create a function and pass $$y$$ coordinate of the centers as a parameter: def cloud(yc): pg.draw.circle(canvas, pg.Color(\"white\"), (200, yc), 50) pg.draw.circle(canvas, pg.Color(\"white\"), (150, yc), 30)", "Does it matter? You can ‘play with’ this game above: Drag point $\\,W\\,$ along the wall, keeping it between the two yellow points. The red path is the distance you must run; its length is displayed at the top. You'll see that different paths have different lengths. Let $\\,m_1\\,$ and $\\,m_2\\,$ denote the measures of the indicated angles: $\\,m_1\\,$ is the angle between $\\,\\overline{P_1W}\\,$ (the line segment from $\\,P_1\\,$ to $\\,W\\,$) and the wall that lies to the left of $\\,W\\,$ $\\,m_2\\,$ is the angle between $\\,\\overline{P_2W}\\,$ (the line segment from $\\,P_2\\,$ to $\\,W\\,$) and the wall that lies to the right of $\\,W\\,$ These angle measures, in degrees, are also displayed. By dragging $\\,W$, try to make the length of the red path as small as possible. What do you notice about the relationship between $\\,m_1\\,$ and $\\,m_2\\,$? For another direction: Try to get $\\,m_1\\,$ as close to $\\,m_2\\,$ as possible. What do you notice about the length of the red path? You can move points $\\,P_1\\,$ and $\\,P_2\\,$ to change where the kids start and stop. However, keep $\\,P_1\\,$ to the left of $\\,P_2\\,$. You can drag the two yellow points to change the location of the wall. Minimizing the Length of the Path from a Point, to a Line, to a Second Point on the", "\\boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired.", "mind, we can ask ourselves, when can I go along edges so that I only touch each dot once? Well of course the graph needs to be connected, but we also need other conditions. Another question is when can I touch each line only once? When can I go all the way through every line and dot only once and come back to where I started? Graphs can be used to represent things in many different subjects. A lot of time research done in different subjects is useful to somebody else, but because of the language barrier it is unknown. Math tries to be the universal translator of different subjects and languages. I’ll try to answer some of these next time and be a little more formal in naming things, but for now here are examples of other graphs. $\\setlength{\\unitlength}{7cm} \\begin{picture}(2,5) \\put(0,0){\\circle*{1}} \\put(0,0){\\line(1,0){1}} \\put(0,0){\\line(1,1){1}} \\put(0,0){\\line(0,1){1}} \\put(0,1){\\circle*{1}} \\put(0,1){\\line(1,0){1}} \\put(1,1){\\circle*{1}} \\put(1,1){\\line(0,-1){1}} \\put(1,0){\\circle*{1}} \\put(1,0){\\line(-1,1){1}} \\end{picture}$ $\\setlength{\\unitlength}{7cm} \\begin{picture}(2,5) \\thicklines \\put(1,1){\\circle*{1}} \\put(1,1){\\line(1,1){1}} \\put(1,1){\\line(1,-1){1}} \\put(2,0){\\line(1,1){1}} \\put(2,0){\\circle*{1}} \\put(2,2){\\line(1,-1){1}} \\put(2,2){\\circle*{1}} \\put(3,1){\\circle*{1}} \\put(4,1){\\circle*{1}} \\put(3,1){\\line(1,0){1}} \\put(2,2){\\line(2,-1){2}} \\put(2,1){\\circle*{1}} \\put(2,0){\\line(0,1){1}} \\end{picture}$ $\\setlength{\\unitlength}{7cm} \\begin{picture}(2,5) \\thicklines \\put(0,0){\\circle*{1}} \\put(1,1){\\circle*{1}} \\put(0,1){\\circle*{1}} \\put(1,0){\\circle*{1}} \\put(2,1){\\circle*{1}} \\put(2,0){\\circle*{1}} \\put(0,1){\\line(1,-1){1}} \\put(0,0){\\line(2,1){2}} \\put(0,0){\\line(1,1){1}} \\put(0,0){\\line(1,0){1}} \\put(0,0){\\line(0,1){1}} \\end{picture}$", "# 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 $\\boxed{\\text{D}}$." ]

[ "the cart wheels; \\S = (\\R-\\r)*cos(\\a); } \\draw circle [radius=\\R]; \\draw (-\\w/2,\\S-\\h-\\r) rectangle ++(\\w, \\h); \\node at (0, \\S-\\h/2-\\r) {$M$}; \\draw [fill=white] (-\\w/2,\\S) circle [radius=\\r] (\\w/2,\\S) circle [radius=\\r]; \\end{tikzpicture} \\end{document} -", "# How to plot a bicycle with square wheels I want to draw a bicycle with square wheels similar to this picture, but I can't plot the trajectory along the curve. enter link description here (*https://pastebin.com/3UbbfG6W*) corner[x_] := Module[{θ = N@SawtoothWave[{0, 2 Pi}, x/(2 Pi)]}, Piecewise[{{{-1 - Cos[θ], Sin[θ]}, 0 <= θ < Pi/2}, {{-Sqrt[2] Cos[θ - Pi/4], Sqrt[2] Sin[θ - Pi/4]}, Pi/2 <= θ < Pi}, {{1 + Sin[θ - Pi], Cos[θ - Pi]}, Pi <= θ < 3 Pi/2}, {{2, 0}, θ >= 3 Pi/2}}] + {4 Floor[x/(2 Pi)], 0}]; frame[θ_] := Module[{t}, t = If[θ <= 3.5 Pi, θ, 7 Pi - θ]; Show[Graphics[{RGBColor[0.8, 0.3, 0.2], Polygon[{corner[t], corner[t + Pi/2] - {1, 0}, corner[t + Pi] - {2, 0}, corner[t + 3 Pi/2] - {3, 0}}], Black, PointSize[Medium], Point[corner[t]]}, PlotRange -> {{-2, 5}, {-0.1, 1.5}}], If[θ >= 0.5 Pi, ParametricPlot[corner[x], {x, t, 0.5 Pi+ 0}, PlotStyle -> Black], Graphics[]]]]; Manipulate[frame[θ], {θ, 0, 7 Pi}] • See Stan Wagon, Mathematica in Action, 3e, electronic version, RollingSquare[]. Jan 16 '20 at 13:31 • I haven't bought this paper book with CD at present,How can I get the RollingSquare[] code content in the DVD of this e-book? Jan 17 '20 at 1:07 • Thanks a lot,I've found a link to download this book's CD-ROM from mathematica.stackexchange.com/questions/87284/… Jan 17", "# Difference between revisions of \"2002 AMC 10B Problems/Problem 24\" ## Problem 24 Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom? $\\mathrm{(A) \\ } 5\\qquad \\mathrm{(B) \\ } 6\\qquad \\mathrm{(C) \\ } 7.5\\qquad \\mathrm{(D) \\ } 10\\qquad \\mathrm{(E) \\ } 15$ ## Solution $[asy] unitsize(1.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair O=(0,0), A=(0,-20), B=(0,-10), C=(10sqrt(3),-10); real r=20; path ferriswheel=Circle(O,r); draw(ferriswheel); draw(O--A); draw(O--C); draw(B--C); draw(A--C); pair[] ps={A,B,C,O}; dot(ps); label(\"O\",O,N); label(\"A\",A,S); label(\"B\",B,W); label(\"C\",C,SE); label(\"10\",(O--B),W); label(\"10\",(A--B),W); label(\"20\",(O--C),NE); [/asy]$ We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\\triangle OBC$ is a $30-60-90$ triangle. That means $\\angle BOC = 60^\\circ$ and the ferris wheel has made $\\frac{60}{360} = \\frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\\frac{1}{6}\\text{th}$ of a minute, or $10$ seconds. The answer is $\\boxed{\\mathrm{(D) \\ } 10}$. Alternatively, we could also say that $\\triangle ABC$ is congruent to $\\triangle OBC$ by", "# Difference between revisions of \"2002 AMC 10B Problems/Problem 24\" ## Problem 24 Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom? $\\mathrm{(A) \\ } 5\\qquad \\mathrm{(B) \\ } 6\\qquad \\mathrm{(C) \\ } 7.5\\qquad \\mathrm{(D) \\ } 10\\qquad \\mathrm{(E) \\ } 15$ ## Solution 1 $[asy] unitsize(1.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair O=(0,0), A=(0,-20), B=(0,-10), C=(10sqrt(3),-10); real r=20; path ferriswheel=Circle(O,r); draw(ferriswheel); draw(O--A); draw(O--C); draw(B--C); draw(A--C); pair[] ps={A,B,C,O}; dot(ps); label(\"O\",O,N); label(\"A\",A,S); label(\"B\",B,W); label(\"C\",C,SE); label(\"10\",(O--B),W); label(\"10\",(A--B),W); label(\"20\",(O--C),NE); [/asy]$ We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\\triangle OBC$ is a $30-60-90$ triangle. That means $\\angle BOC = 60^\\circ$ and the ferris wheel has made $\\frac{60}{360} = \\frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\\frac{1}{6}\\text{th}$ of a minute, or $10$ seconds. The answer is $\\boxed{\\mathrm{(D) \\ } 10}$. Alternatively, we could also say that $\\triangle ABC$ is congruent to $\\triangle OBC$", "corners correspondingly • I need to scroll forward along the curve drawn by the function Plot[-(Cosh[Mod[x, 2 Log[Sqrt[2] + 1], -Log[Sqrt[2] + 1.]]]) + Sqrt[ 2], {x, 0, 10}],that's the point. Jan 17 '20 at 5:26 • @Gowiththewind, and what's the problem with this? Just replace my ListPlot block with your Plot and enjoy fun :) Jan 17 '20 at 5:58 • Thanks a lot,I have basically solved this problem in the following answers. Jan 17 '20 at 8:57 • The problem is that the vertices of a square-wheeled bicycle do not trace cycloids as shown in this answer. What this answer shows is the trajectory of the square on a circular bicycle wheel, which is entirely different. Replacing the ListPlot with @Go's Plot produces even worse results. Jan 18 '20 at 11:32 To make it a bit shorter n = 4 (*number of corners*) crn = {Cos[#/n 2 Pi], Sin[#/n 2 Pi]} & /@ Range[n]; cen = {0, 1}; v[t_] = {t, 0};(*linear velocity*) w[t_] = -0.2 t 2 Pi;(*angular velocity*) tmax = 10; dt = 0.1; trj = Table[(cen + v[t] + RotationMatrix[w[t]].#) & /@ crn, {t, 0, tmax, dt}]; ListAnimate[Table[Graphics[{Gray, Polygon[t], Black, Point[Mean[t]],(*Centre*) Table[{Hue[i/n], Line[trj[[All, i]]]}, {i, n}], Black, Dashed, Line[Table[Mean[t], {t, trj}]]}] ,{t, trj}]] You can play with different polygon and velocity. For an", "of $n$ with $2\\le n \\le 1000$ for which $A(n)$ is an integer. ## Problem 15 A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\\overarc{AB}$ on that face measures $120^\\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\\cdot\\pi + b\\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$. $[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,\"A\",(-3,3*sqrt(3),0),W); label(rpic,\"B\",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]$ 2015 AIME I (Problems • Answer Key • Resources) Preceded by2014 AIME II Followed by2015 AIME II 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions", "# Rolling Disk Animation I am trying to create an animation in Mathematica of a rolling disk outside a disk of equal radius. The following coding is what I have tried but without the one circle staying in a fixed position I am having a hard time seeing what is going on. Manipulate[ Show[{ Graphics[{ Circle[{0,0},1], Circle[{2 Cos[Theta],2 Sin[Theta]},1], {Blue,PointSize[0.012],Point[{Cos[Theta],Sin[Theta]}]}, {Green, PointSize[0.012],Point[{2 Cos[Theta],2 Sin[Theta]}]}, Line[{{2 Cos[Theta],2 Sin[Theta]},{Cos[Theta],Sin[Theta]}}] }] }] ,{Theta,0.0000001,2 Pi} ] I also cannot figure out the how to make it trace the point that is rotating. • Use the PlotRange option to keep the stationary circle from moving around in the frame. In order to plot the path you should use ParametricPlot to plot the coordinate for the second circle over time. You can combine the ParametricPlot with your graphics using Show. – C. E. Apr 28 '15 at 23:13 • Stan Wagon's Mathematica in Action features an animation that involves rolling a (simulated) penny on another. If you can see a copy of that book, go do so. – J. M.'s torpor May 3 '15 at 0:24 Is this something like what you want? Manipulate[Show[ Graphics[{ Circle[{0, 0}, 1], Circle[{2 Cos[t], 2 Sin[t]}, 1], {Blue, PointSize[0.012], Point[{Cos[t], Sin[t]}]}, {Green, PointSize[0.012], Point[{2 Cos[t], 2 Sin[t]}]}, Line[{{2 Cos[t], 2 Sin[t]}, {2 Cos[t], 2 Sin[t]} + {Cos[Pi +", "back wheel, b = center of front wheel a = point 0 of q; b = q intersectionpoint fullcircle scaled 2 len shifted a; % draw the wheels fill fullcircle scaled 2r shifted a; fill fullcircle scaled 2r shifted b; % draw the body of the truck with a label, suitably transformed t = identity rotated angle (b-a) shifted a; draw unitsquare xscaled len yscaled 13/21 len transformed t; picture M; M = image(label(name,(1/2len,1/3len))); draw M transformed t; enddef; beginfig(1); path p; p = origin .. (3cm,-2cm) {dir -45} .. (8cm,1cm) {up}; draw p withcolor .687 red; draw_car(p, 0, 1mm, 1cm, \"A\"); draw_car(p, 0.6, 1mm, 1cm, \"B\"); draw_car(p, 1.2, 1mm, 1cm, \"C\"); draw_car(p, 1.8, 1mm, 1cm, \"D\"); endfig; end. Given this approach, we can decorate the car in whatever way we like by varying the bit that draws the body. For example, we could add these \"go faster\" stripes: by adding % draw some go faster stripes draw ((-2,1/4 len) -- (-2,13/21 len)) transformed t; draw ((-4,1/3 len) -- (-4,13/21 len)) transformed t; draw ((-6,1/2 len) -- (-6,13/21 len)) transformed t; after defining the transformation. We can also have a go at the OP diagram: beginfig(2); h = 4.4cm; r = 2cm; path track; track = (0,h) {dir -50} .. { right } (3r, 0) -- (4.9r, 0)", "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]$ We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\\left(-\\frac{1}{2}, y + \\frac{\\sqrt{3}}{2}\\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origin is $4$ and $\\frac{1}{4} + x^2 + x\\sqrt{3} + \\frac{3}{4} = x^2 + x\\sqrt{3} + 1 = 16$. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that $x = \\frac{3\\sqrt{7}-\\sqrt{3}}{2} \\Rightarrow (C)$ Note: Since $C$ and $D$ are not labeled in the diagram, refer to solution 1 for the location of points $C$ and $D$. ## Solution 5 $[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(\"$$A$$\",(-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(\"$$E$$\",(0.3,4.15),E); label(\"$$F$$\",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2));[/asy]$ Notice that $\\overarc{AE}$ is $\\frac16$ the circumference of the circle. Therefore, $\\overline{AE}$ is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius of the circle, or $4$. $\\angle AFE$ is a right angle, therefore $\\triangle AFE$ is a right triangle. $\\overline{AF}$ is half the length of $1$, or $\\frac{1}{2}$. The length of $\\overline{EF}$ is $x$ plus the altitude length of one of the equilateral triangles,", "# [GIF] Rolling (Circles rolling inside a circle) Posted 7 months ago 1306 Views | 2 Replies | 5 Total Likes | This one is pretty simple: take two circles of radius 1/2 rolling inside a unit circle and track a point on each circle's perimeter. The points turn out to follow straight lines, and the challenge is to make this seem both surprising and obvious (and also to fit it all under 3 MB). Not sure if I succeeded, but that was the idea, anyway.Here's the code: smoothstep[x_] := 3 x^2 - 2 x^3; With[{a = 1, b = 1/2, cols = RGBColor /@ {\"#303841\", \"#f3f6f6\"}}, Manipulate[ Graphics[{ Thickness[.0075], cols[[1]], Opacity[ Which[t < 8 π, 1, 8 π <= t < 12 π, 1 - smoothstep[(t - 8 π)/(4 π)], 12 π <= t < 14 π, 0, 14 π <= t <= 18 π, smoothstep[(t - 14 π)/(4 π)]]], Table[ Circle[(a - b) {Cos[t + π i], Sin[t + π i]}, b], {i, 0, 1}], Opacity[ Which[t < 6 π, 0, 6 π <= t < 8 π, smoothstep[(t - 6 π)/(2 π)], 8 π <= t < 14 π, 1, 14 π <= t <= 18 π, 1 - smoothstep[(t - 14 π)/(4 π)]]], Thickness[.01], Line[ Table[(a - b) {Cos[s], Sin[s]} + b {Cos[-(a -", "us walk around the whole circumference of a circle, the distance we take trip is $$2\\pi$$ times the length of the radius, or around 6.28 time the radius. If us walk only component of the means around the circle, climate the distance we travel depends also on the edge of displacement. For example, an angle of $$45\\degree$$ is $$\\dfrac18$$ of a complete revolution, so the arclength, $$s\\text,$$ from allude $$A$$ to allude $$B$$ in the figure at right is $$\\dfrac18$$ of the circumference. Thus \\beginequation*s = \\dfrac18(2\\pi r) = \\dfrac\\pi4 r\\endequation* Similarly, the edge of displacement from suggest $$A$$ to suggest $$C$$ is $$\\dfrac34$$ the a finish revolution, so the arclength along the circle indigenous $$A$$ come $$C\\text,$$ presented at right, is \\beginequation*s = \\dfrac34(2\\pi r) = \\dfrac3\\pi2 r\\endequation* In general, for a given circle the size of the arc spanned by an edge is proportional to the dimension of the angle. Arclength on a Circle. \\beginequation*\\blert\\textbfArclength~ = ~ \\blert(\\textbffraction of one revolution) \\cdot (2\\pi r)\\endequation* The Ferris wheel in the arrival has circumference \\beginequation*C = 2\\pi (100) = 628~ \\textfeet\\endequation* so in fifty percent a change you take trip 314 feet about the edge, and in one-quarter transformation you travel 157 feet. To indicate the same 4 locations top top the wheel by distance traveled, we would", "arrowed double line pointing to the right rectangle with the line labeled $$(p \\land \\lnot Kp)!$$. The first circle in the first rectangle has a double line dark orange rim, is labeled w and contains the letter p. The second circle in the first rectangle has a single line dark orange rim, is labeled u, and contains nothing. A solid line connects these two circles. The circle in the second rectangle has a double line dark orange rim, is labeled w, and contains the letter p. ### Figure 3 description Two large rectangles each containing several orange circles. The first rectangle is labeled M and contains two circles. The first has a double line dark orange rim and is labeled w. The second has a single line dark orange rim, is labeled u, and contains the letter p. Two solid lines connect the two circles; one labeled $$R_1$$ and the other $$R_2$$. The second rectangle contains 5 orangle circles on a 3x3 matrix. The first column contains, in the second row, the first circle, which has a double line dark orange rim. The other two columns each have two circles, one in the first row and one in the third row, with all of them having only a single line dark orange rim. The first circle is labeled", "\\path[name intersections = {of = circ and line}]; \\coordinate (P1) at (intersection-1); \\coordinate (P2) at (intersection-2); \\coordinate (C1) at ($(P1)!\\wheelradius!(center)$); \\coordinate (C2) at ($(P2)!\\wheelradius!(center)$); \\foreach \\p in {1,2} { \\draw[red] (C\\p) circle (\\wheelradius); } %\\coordinate (C1) at ($(P1)!2*\\wheelradius!(center)$); %\\coordinate (C2) at ($(P2)!2*\\wheelradius!(center)$); \\draw[red, fill=white, fill opacity=0.85] let \\p1 = ($(C2)-(C1)$), \\n1 = {veclen(\\p1)} in (C2) rectangle +($(\\n1, -0.4*\\n1)$) node[midway] {M}; \\end{tikzpicture} ## Result: - If I understand correctly, this code cannot accept the width of the cart as a parameter. – Charles Staats May 6 '14 at 15:56 @CharlesStaats you are right. This is because I respected the original approach of the OP, in which the position of the wheels are found as the intersection of the \"loop\" with a horizontal line at a \"customizable\" height. – JLDiaz May 6 '14 at 15:58 How did you determine 2*\\wheelradius? From the dimension I am using, I had to change it 1.75. I also changed my circle radius to .7 and wheel radius to .05. I had a typo on circle radius in the OP. – dustin May 6 '14 at 16:05 What I mean is if I use a 2, the wheels aren't tangent to the circle. By changing it to a 1.75, they are so I am wandering how you came up with the 2. I just randomly", "Difference between revisions of \"2015 AIME I Problems/Problem 15\" Problem A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\\overarc{AB}$ on that face measures $120^\\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\\cdot\\pi + b\\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$. $[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,\"A\",(-3,3*sqrt(3),0),W); label(rpic,\"B\",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]$ Solution Label the points where the plane intersects the top face of the cylinder as $C$ and $D$, and the center of the cylinder as $O$, such that $C,O,$ and $A$ are collinear. Let $T$ be the center of the bottom face, and $M$ the midpoint of $\\overline{AB}$. Then $OT=4$, $TM=3$ (because of the 120 degree angle), and so", "wheel by street traveled, we would certainly use \\\\beginequation*s = 0,~ s = 157,~ s = 314,~ \\\\textand~ s = 471\\\\text,\\\\endequation* as presented at right. Example 6.1. What length of arc is extended by an angle of \\$$120\\\\degree\\$$ on a one of radius 12 centimeters? Solution. Because \\$$\\\\dfrac120360 = \\\\dfrac13\\\\text,\\$$ an angle of \\$$120\\\\degree\\$$ is \\$$\\\\dfrac13\\$$ of a complete revolution, as displayed at right. Using the formula above with \\$$r = 12\\\\text,\\$$ we uncover that \\\\beginequation*s = \\\\dfrac13(2\\\\pi \\\\cdot 12) = \\\\dfrac2 \\\\pi3 \\\\cdot 12 = 8\\\\pi ~ \\\\textcm\\\\endequation* or around 25.1 cm. Checkpoint 6.2. How much have girlfriend traveled about the sheet of a Ferris wheel of radius 100 feet when you have turned with an edge of \\$$150\\\\degree\\\\text?\\$$ \\$$261.8\\$$ ft ### Subsection Measuring angles in Radians If you think around measuring arclength, friend will see that the degree measure that the covering angle is no as crucial as the portion of one change it covers. This observation suggests a new unit of measurement because that angles, one the is better suited come calculations involving arclength. We\\\"ll do one adjust in our formula for arclength, from \\\\beginequation*\\\\textbfArclength~ = ~ (\\\\textbffraction that one revolution) \\\\cdot (2\\\\pi r)\\\\endequation* to \\\\beginequation*\\\\blert\\\\textbfArclength~ = ~ \\\\blert(\\\\textbffraction of one revolution\\\\times 2\\\\pi) \\\\cdot r\\\\endequation* We\\\"ll call the quantity in parentheses, (fraction of one revolution", "### Just Rolling Round P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P? ### Coke Machine The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design... ### Just Opposite A and C are the opposite vertices of a square ABCD, and have coordinates (a,b) and (c,d), respectively. What are the coordinates of the vertices B and D? What is the area of the square? # Contact ##### Age 14 to 16Challenge Level When the disc is in the corner there is one quarter of the circumference (that is a length $2\\pi/4 = \\pi/2$) which does not come in contact with the edge of the tray. As the disc rolls round one circuit: an arc of length $2$ units comes into contact, then an arc of length $\\pi/2$ gets missed, then an arc of length $1$ unit comes into contact, then an arc of length $\\pi/2$ gets missed, then an arc of length $2$ units comes into contact, then an arc of length $\\pi/2$ gets missed, then an arc of length $1$ unit comes into contact. Some points never touch, other points touch only once, other points touch twice. As an", "\\rangle. ~$ To visualize this we first define a helper function to draw a circle at point $P$ with radius $R$: circle!(P, R; kwargs...) = plot_parametric_curve!(t -> P + R*[cos(t), sin(t)], 0, 2pi;kwargs...) circle! (generic function with 1 method) Then we have this function to visualize the spirograph for different $t$ values: function spiro(t; r=2, R=5, rho=0.8*r) cent(t) = (R-r) * [cos(t), sin(t)] p = plot(legend=false, aspect_ratio=:equal) circle!([0,0], R, color=:blue) circle!(cent(t), r, color=:black) tp(t) = -R/r * t s(t) = cent(t) + rho * [cos(tp(t)), sin(tp(t))] plot_parametric_curve!(s, 0, t, n=1000, color=:red) p end spiro (generic function with 1 method) And we can see the trace for $t=\\pi$: spiro(pi) The point of contact is at $(-R, 0)$, as expected. Carrying this forward to a full circle's worth is done through: spiro(2pi) The curve does not match up at the start. For that, a second time around the outer circle is needed: spiro(4pi) Whether the curve will have a period or not is decided by the ratio of $R/r$ being rational or irrational. ##### Example Ivars Peterson described the carnival ride \"tilt-a-whirl\" as a chaotic system, whose equations of motion are presented in American Journal of Physics by Kautz and Huggard. The tilt-a-whirl has a platform that moves in a circle that also moves up and down. To describe the", "# Movement of the point $A$ on the cylinder [closed] We have a point $A$ on the top of a cylinder with radii $1$.After $2.5$ rolls find the movement of point $A$.(Take $\\pi=3$ if needed) For finding the movement of point $A$ we should use pytagoras to find the movement but I had a disscustion with my teacher about the value of $A'B$ I strongly belive that $A'B=2.5*permiter=2.5*2*\\pi=15m$ but our teacher tells that $A'B=2.5*2radii=10m$ I don't know why but I strongly belive my answer.So we decided to ask it here to get an answer .A video will be helpful a lot but I don't know how to make one.Please make for me one or tell me how to make one.Thanks. • From the picture, it seems that you are right. I don't understand why you would need a video (and of what) though. And please don't take $\\pi=3$ – DelCrosB Sep 28 '16 at 15:57 When a circle (or a cylinder) is rolled, it completes one full cycle when point $A$ gets back to its original position relative to the center of the circle, That means it has rotated $2\\pi$ radians from its original position relative to the center of the circle. When this one cycle is completed circle travels a distance that is equal to its circumference.", "caught by a person riding a ferris wheel, I decided to write a Sage Notebook application that animates the two trajectories with relation to time. The result is a little rough (as is my python), but it does work. # set the global variable a2,t,L1,L2,min_d= var('a2,t,L1,L2,min_d') # default high minimum distance min_d = 1000 # these fuctions return the x's and y's def x1(t): return 10*t def y1(t): return 10*sqrt(3)*t-4.9*(t^2) def x2(t): return 25+7*cos((6+t)*(-pi/6)) def y2(t): return 7+7*sin((6+t)*(-pi/6)) # two lines from the functions L1 = parametric_plot( (x1(t), y1(t) ), (t, 0, 6), color='green') #L2 = parametric_plot( (x2(t), y2(t) ), (t, 0, 12)) L2 = circle( (25,7),7 ) # Calclulate the distance between the points def distance(a): d = float( sqrt( ( x2(a)-x1(a) )^2 + (y2(a)-y1(a))^2 ) ) return d # record minimum distance def min_distance(distance) : global min_d if (distance < min_d) : min_d = distance # find the points at 't' on the two lines def f(a): p1 = point( ( x1(a), y1(a) ) ) p2 = point( ( x2(a), y2(a) ) ) #calculate distance d = distance(a) # record minimum distance min_distance(d) return L1 + L2 + p1 + p2 + text(\"Distance: \"+ str(distance(a)),(30,16)) + text(\"Min Distance: \"+ str(min_d),(6,16)) # the animation a2 = animate([f(i) for i in srange(2.1,6,0.1)], xmin = 0, xmax = 35,", "mypara=revolution(O,g,X); Full code: copy to http://asymptote.ualberta.ca/ and click Run unitsize(1cm); import graph3; import solids; //currentprojection=orthographic(2,1.5,.5,zoom=.95); currentprojection=obliqueZ; currentprojection.zoom=.95; size(8cm); real a=1.8, b=a^2; real c=2.5,d=4.5; triple A=(a,0,0), B=(0,b,0); triple C=(c,0,0), D=(0,d,0); // the (red) parabol curve on the XY-plane triple paraXY(real t){return (t,t^2,0);} path3 g=graph(paraXY,0,a); draw(g,red+1.2pt); // use one of the following revolutions //revolution mypara=revolution(C,g,Y); // around x=c //revolution mypara=revolution(D,g,X); // around y=d //revolution mypara=revolution(O,g,X); // around y=0 revolution mypara=revolution(O,g,Y); // around x=0 draw(surface(mypara),yellow+opacity(.7)); //draw(mypara,blue,orange); label(\"$a$\",A,S); label(\"$c$\",C,S); label(\"$b$\",B,W); label(\"$d$\",D,W); draw(A--(a,b,0)--B^^C--(c,d,0)--D,dashed+gray); draw(Label(\"$x$\",EndPoint),-2*X--6*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--6*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),O--6*Z,Arrow3()); // you may not want to draw Oz 2. The same code, only change X to Y to get revolution around Oy revolution mypara=revolution(O,g,Y); 3. The same code, for revolution around x=c revolution mypara=revolution(C,g,Y); 4. Last, for revolution around y=d revolution mypara=revolution(D,g,X); Welcome to TeX.SE!!! This is a quick modification of an answer from here. It's a 2d drawing using isometric perspective. The equations of parabola and ellipses are calculated by hand. The ellipses are very simple if you know about isometric perspective (the axes ratio is sqrt(3):1). The parabola was obtained as an envelope of the family of 'horizontal' ellipses. This is the code: \\documentclass[tikz,border=2mm]{standalone} \\tikzset {% axes/.style={thick,-latex}, cylinder/.style={right color=blue!80,left color=white,fill opacity=0.6}, paraboloid/.style={left color=magenta!80,fill opacity=0.6}, } \\begin{document} \\begin{tikzpicture}[line cap=round,line join=round] % axes x,y \\draw[axes] (30:3) -- (210:3) node [left] {$x$}; \\draw[axes] (150:3) -- (330:3) node [right]" ]

[ "tower. The total vertical drop of the zip line is 40 ft. The zip line’s angle of elevation from the lower tower is 11. 5°. What is the horizontal distance between the towers? tan 11.5 = $$\\frac{40}{x}$$ x = $$\\frac{40}{\\tan 11.5}$$", "feet, which was true at its 2009 publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety on Roller Coasters? Of course they do, or else we’d have more roller coasters that allowed mice to ride. The question is how much the size restriction matters, and how sensitive that dependence is. So the leading question is a classic example of applying mathematics to the real world. This includes practical subtleties like if a person 39.5 inches tall could ride safely, is it fair to round that off to 40 inches? It also includes the struggle to work out how dangerous an amusement park is. Speaking from my experience as a rider and lover of amusement parks: don’t try to plead someone’s “close enough”. You’re putting an unfair burden on the ride operator. Accept the rules as posted. Everybody who loves amusement parks has their disappointment stories; accept yours in good grace. This leads me into planning amusement park fun. School", "of 5m and 13m are 6m apart from each other. Find their distance between top of the two poles. 3. ### Geometry So the directions say: Provide a counterexample to show that each statement is false. You can use words or make a diagram. These are the two questions that I'm stuck on: 1. If a line lies in a vertical plane, then the line is 4. ### Magnets-Physics I have a few questions that I am not too sure of. why is it inadvisable to make a horseshoe magnet from a flexible material? -is it because it would have enough magnetism Do the poles of a horseshoe magnet attact each other? If 5. ### math the nufone communications company must run a telephone line between two poles at opposite ends of a lake. the length and width are 75 feet and 30 feet. what is the distance between the two poles to the nearest foot? The lake is 6. ### Math 17.Which roller coaster has the greatest lift height Top Thrill Dragster 18.What is the median lift height for the roller coasters lifted? Round to the nearest tenth. Not sure for this one 19.Arrange the given roller coasters from 7. ### Math PLease help me with this: What kind of line is created when you graph the", "publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety on Roller Coasters? Of course they do, or else we’d have more roller coasters that allowed mice to ride. The question is how much the size restriction matters, and how sensitive that dependence is. So the leading question is a classic example of applying mathematics to the real world. This includes practical subtleties like if a person 39.5 inches tall could ride safely, is it fair to round that off to 40 inches? It also includes the struggle to work out how dangerous an amusement park is. Speaking from my experience as a rider and lover of amusement parks: don’t try to plead someone’s “close enough”. You’re putting an unfair burden on the ride operator. Accept the rules as posted. Everybody who loves amusement parks has their disappointment stories; accept yours in good grace. This leads me into planning amusement park fun. School Specialty’s blog particularly offers PLAY & PLAN:", "publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety on Roller Coasters? Of course they do, or else we’d have more roller coasters that allowed mice to ride. The question is how much the size restriction matters, and how sensitive that dependence is. So the leading question is a classic example of applying mathematics to the real world. This includes practical subtleties like if a person 39.5 inches tall could ride safely, is it fair to round that off to 40 inches? It also includes the struggle to work out how dangerous an amusement park is. Speaking from my experience as a rider and lover of amusement parks: don’t try to plead someone’s “close enough”. You’re putting an unfair burden on the ride operator. Accept the rules as posted. Everybody who loves amusement parks has their disappointment stories; accept yours in good grace. This leads me into planning amusement park fun. School Specialty’s blog particularly offers PLAY & PLAN:", "old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218 feet, which was true at its 2009 publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety on Roller Coasters? Of course they do, or", "I got and what the \"webassign\" (computer program I use in class for homework) got (it tells you if the answer is right). Now I need to figure out how many degrees below the horizontal (starting point) the rollercoster ended up. Any help from this point is appreciated greatly ^^ Last edited: Sep 7, 2007 2. Sep 7, 2007 Avodyne 420 ft (rounded to the nearest foot) is the horizontal component of the displacement. What is the vertical component? When you have both the horizontal and vertical components, you should be able to compute the magnitude and direction (in degrees above or below the horizontal axis, depending on whether the vertical component of displacement turns out positive or negative) of the displacement. (I assumed that \"rises 135 ft at an angle of 30*\" means it travels a distance of 135 ft, not that it's change in height is 135 ft. It's not clear that this is the right interpretation.) Last edited: Sep 7, 2007 3. Sep 7, 2007 twotaileddemon Ah.. so the vertical component is what I'm missing. That would certainly make things much easier. Is the vertical component how much the roller coster travels from the end of the 200 feet to the end of its decline? 4. Sep 7, 2007 twotaileddemon You're right though, it travels", "at Cedar Point in Ohio before moving to Pennsylvania. The old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218 feet, which was true at its 2009 publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter", "Point in Ohio before moving to Pennsylvania. The old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218 feet, which was true at its 2009 publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety", "Point in Ohio before moving to Pennsylvania. The old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218 feet, which was true at its 2009 publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety", "People with Kutztown University, regional high schools, and Dorney Park prepared Coaster Quest – Geometry. These include a lot of observations and measurements all tied to specific rides at the park. (And a side fact, fun for me: Dorney Park’s carousel used to be at Lake Lansing Amusement Park, a few miles from me. Lake Lansing’s park closed in 1972, and the carousel spent several decades at Cedar Point in Ohio before moving to Pennsylvania. The old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218 feet, which was true at its 2009", "People with Kutztown University, regional high schools, and Dorney Park prepared Coaster Quest – Geometry. These include a lot of observations and measurements all tied to specific rides at the park. (And a side fact, fun for me: Dorney Park’s carousel used to be at Lake Lansing Amusement Park, a few miles from me. Lake Lansing’s park closed in 1972, and the carousel spent several decades at Cedar Point in Ohio before moving to Pennsylvania. The old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218 feet, which was true at its 2009", "This is a problem in my homework thats due tomorrow please hints? The Stratosphere Tower in Las Vegas is 921 feet tall and has a \"needle\" at the top that extends even higher into the air. A thrill ride called the Big Shot catapults riders 160 feet up the needle and then lets them fall back to the launching pad. a. The height h (in feet) of a rider on the Big Shot can be modeled by h=-16t+v(t) + 921 where v is the initial velocity (in feet per second). Find v using the fact that the maximum value of h is 921+160+1081 feet. PLEASE HELP IF YOU CAN! we have been solving these kinda of problems but not with two variables! 2. Originally Posted by algebraIIgirl This is a problem in my homework thats due tomorrow please hints? The Stratosphere Tower in Las Vegas is 921 feet tall and has a \"needle\" at the top that extends even higher into the air. A thrill ride called the Big Shot catapults riders 160 feet up the needle and then lets them fall back to the launching pad. a. The height h (in feet) of a rider on the Big Shot can be modeled by h=-16t+v(t) + 921 where v is the initial velocity (in feet per second). Find", "the chance to visit several times. People with Kutztown University, regional high schools, and Dorney Park prepared Coaster Quest – Geometry. These include a lot of observations and measurements all tied to specific rides at the park. (And a side fact, fun for me: Dorney Park’s carousel used to be at Lake Lansing Amusement Park, a few miles from me. Lake Lansing’s park closed in 1972, and the carousel spent several decades at Cedar Point in Ohio before moving to Pennsylvania. The old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218 feet,", "had the chance to visit several times. People with Kutztown University, regional high schools, and Dorney Park prepared Coaster Quest – Geometry. These include a lot of observations and measurements all tied to specific rides at the park. (And a side fact, fun for me: Dorney Park’s carousel used to be at Lake Lansing Amusement Park, a few miles from me. Lake Lansing’s park closed in 1972, and the carousel spent several decades at Cedar Point in Ohio before moving to Pennsylvania. The old carousel building at Lake Lansing still stands, though, and I happened to be there a few weeks ago.) And I have yet to make it to Six Flags America, but their Math & Science In Action page offers a similar roster of activities tied to that park. Six Flags America is their park in Maryland; the one in Illinois is Six Flags Great America. Math Word Problems Solved offers a booklet of Amusement Park Word Problems Starring Pre-Algebra. These tie in to no particular amusement park. They do draw from real parks, though. For example it lists the highest point on the tallest steel roller coaster as 456 feet; it doesn’t name the ride, but that’s Kingda Ka, at Great Adventure. The highest point on the tallest wooden roller coaster is given as 218", "of the first bounce is measured each time. What is the relationship between the height of the drop and the height of the bounce? We can try a simple mathematical experiment to look at the problem. Below is an interactive program that allows you to enter several data points (possibly non-physical) claiming to be data representing the starting height and subsequent bounce heights of a basketball. Enter the x,y values for any points you like, then use the mouse to click on any two positions inside the graph area. The program draws a line between the two points and indicates the endpoints and the midpoints with circles. By clicking near the center of any of the circles, you can drag the line around. As you do so, you will see a display of the distance between the closest approach of the line to each data point. You also see at the bottom a display of a number which characterizes how \"badly\" the line fits the points. The smaller the \"badness\" number, the better the line should appear to represent the points. Try it! [This applet no longer functions. Ed.] Dr. DeTurck collected the following data by dropping a basketball in his garage. After each drop, he measured the height of the first bounce: Height of Drop (in.) Height", "height here implies that the height is decreased means the earlier point the bob was the lowest point.", "$$~\\dfrac{x^2}{100^2}-\\dfrac{y^2}{150^2}=1$$ around the $$y$$-axis. ###### 29. The base of the cooling tower is 360 feet below the center of the hyperbola. What is the diamter of the base? 520 ft ###### 30. The top of the cooling tower is 200 feet above the center of the hyperbola. What is the diameter of the top? ###### 31. The diameter of the tower first decreases with height and then increases again. There are two heights at which the tower's diameter is 250 feet. Find the greater of the two heights. 472.5 ft ###### 32. Find the height at which the tower's diameter is 200 feet.", "the top horizontal blue line, (whose purpose mainly is to illustrate the angle of depression), whereas the elevation is \"height above sea-level\". (And if that was it's length, D would be longer) I already fixed it but still I get 1409 ft for D For X I got 482ft • Feb 18th 2011, 06:32 AM Quote: Originally Posted by vaironxxrd I already fixed it but still I get 1409 ft For X I got 482ft That seems fine for D • Feb 18th 2011, 08:53 AM vaironxxrd Quote:", "Crazy Roller Coaster Cool Coasters Six Flags Universal Studios Great America Amusement Park Rides Cedar Point Parks. TOUCH this image to discover its story. The insides of the A300 Zero-G looks mostly like a normal airplane save for the middle section having white foam mats on the floor, ceiling and wall. Also, B&M's reference to an \"in-line\" inversion element which can be found on their Sit-down and Stand-up roller coasters. The layout of Love Ran Red can be compared to a roller-coaster effect due to the different variations of slow to fast paced songs that are played throughout the album. This roller coaster was built in Cedar Point, Ohio in 2003. Kings Island’s ride lineup is both historic, legendary, and impressively intimidating. Most people already know that roller coasters can be found in science, but now you can know that math can be found in roller coasters as well. One Saturday, the four friends decide to ride a new coaster. Machine Term sometimes used in reference to a roller coaster. These manoeuvres can be flown either consecutively in a roller-coaster fashion or separated by intervals of several minutes. The second part of the new coaster is a parabola. With Tenor, maker of GIF Keyboard, add popular Roller Coaster animated GIFs to your conversations. the zeros are basically the" ]

[ "can create a table showing wages arranged in ascending order Sl.No Wages arranged in ascending order 1 80 2 90 3 100 4 145 5 148 6 155 7 200 Median = $${{{(\\frac{N + 1}{2})}^{th} }} item$$ $$={{{(\\frac{7 + 1}{2})}^{th} }} item$$ = 4th item. $${\\biggl[\\frac{1\\,2\\,3\\,\\,\\,4\\,\\,\\,5\\,6\\,7}{|\\,|\\,|\\,\\,\\,|\\,\\,\\,|\\,|\\,|\\,}\\biggl]}$$ Size of 4th item = 145, hence we can say that median wage is 145 ### DATA OF EVEN NUMBERS • Let us find median of the following series • 2400, 1500, 1750, 3200, 1350, 2400, 3600, 1500, 2250, 2800. First we arrange the items in ascending order. 1350, 1500, 1500, 1750, 2250, 2400, 2400, 2800, 3200, 3600. Here there are 10 items. That is N = 10, which is even. Since N is even, median = mean of the $${{{(\\frac{N}{2})}^{th} }}$$ item and the next item = mean of the $${{{(\\frac{10}{2})}^{th} }}$$ item and the next item = mean of the 5th item and the 6,th item = mean of 2250 and 2400 = $${{{(\\frac{2250 + 2400}{2})} }}$$ = 2325. • Let us find median of the following series using another method • 2400, 1500, 1750, 3200, 1350, 2400, 3600, 1500, 2250, 2800. First we arrange the items in ascending order. 1350, 1500, 1500, 1750, 2250, 2400, 2400, 2800, 3200, 3600. Median = mean of the $${{{(\\frac{N + 1}{2})}^{th} }}$$ item", "&& \\275 && \\350 && \\425\\\\ & \\150 && \\375 && \\300 && \\400 && \\225 && \\360\\end{align*} What is the median price spent on a prom dress?150$400$175$425$195$450$225$250$275$300$300$325$350$360$375\\begin{align*}& \\150 && \\175 && \\195 && \\225 && \\250 && \\275 && \\300 && \\boxed{\\300} && \\325 && \\350 && \\360 && \\375\\\\ & \\400 && \\425 && \\450\\end{align*} The prices have been organized from least to greatest, and the number of prices is an odd number. Therefore, the median will be in the n+12\\begin{align*}\\frac{n+1}{2}\\end{align*} position: n+12=15+12=162=8\\begin{align*}\\frac{n+1}{2}=\\frac{15+1}{2}=\\frac{16}{2}=8\\end{align*}. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English cumulative frequency Cumulative frequency is used to determine the number of observations that lie above (or below) a particular value in a data set. Median The median of a data set is the middle value of an organized data set. normal distributed If data is normally distributed, the data set creates a symmetric histogram that looks like a bell. outliers An outlier is an observation that lies an abnormal distance from other values in a random sample from a population. Show Hide Details Description Difficulty Level: Basic Authors: Tags: Subjects:", "of all 5 mobiles)/5 = $150 Multiply both sides by 5 to get: sum of all 5 mobiles =$750 The median of all the prices is $200. Let a = smallest value Let b = 2nd smallest value Let d = largest value Let c = 2nd largest value So, when we arrange the values in ASCENDING order we get: a, b,$200, c, d From here, a quick approach is to test each answer choice, starting from the smallest value A) $150 This answer choice suggests that d =$150, which is impossible, since the greatest value cannot be less than the median ($200) ELIMINATE A B)$200 This answer choice suggests that d = $200 Let's add this to our list to get: a, b,$200, c, $200 This means c must also equal$200. So we have: a, b, $200,$200, $200 Is it possible to assign values to a and b so that all of the conditions are met? YES! We must satisfy the condition that sum of all 5 mobiles =$750 and it must be the case that the price of the most expensive mobile is at least thrice that of the least expensive mobile Well, if we let a = $50, then the price of the most expensive mobile ($200) is at least thrice that of the least", "- 0.0676$. So, we can go one step further and derive the approximate numerical value for the mean and s.d.: mean $\\approx \\left(- 0.0676\\right) \\left(1.5 - \\frac{2 {e}^{4}}{9} - \\frac{e}{9}\\right) \\approx 0.7392$ s.d. $\\approx \\sqrt{- 0.0676 \\left(- 8.8261\\right) - \\left(- {0.0676}^{2}\\right) \\left(119.5732\\right)} \\approx 0.2241$ Hope that helped", "$$57.5$$ lies in the class interval $$300 - 400$$ Here, $$l = 300$$, $$\\frac{N}{2} = 57.5$$, $$m = 45$$, $$f = 26$$ and $$c = 100$$ Substituting the known values in the above formula, we have: Median $$= 300 + (\\frac{57.5 - 45}{26}) \\times 100$$ $$= 300 + (0.481) \\times 100$$ $$= 300 + 48.1$$ $$= 348.1$$ Therefore, the median of the given data is $$348.1$$", "vector) • Compare the output of your function to the built-in median function using the testthat package • When is mean useful and when is median useful income <- runif(100, 1000, 1500) mean(income) [1] 1264 median(income) [1] 1263 But now someone really wealthy joins income.2 <- c(income, 1000000) mean(income.2) [1] 11153 median(income.2) [1] 1264 ## Range Knowing the range of a set of numbers can be useful • min • max • range Test these all out values <- runif(10, 0, 100) min(values) [1] 8.117 max(values) [1] 87.95 range(values) [1] 8.117 87.952 Problem: doesn’t handle outliers well Solution: look at the variance ## Population variance The following shows how much each of these values differs from the mean value values [1] 58.553 66.401 53.200 78.226 39.967 49.443 10.270 87.952 8.117 63.363 values - mean(values) [1] 7.004 14.851 1.651 26.677 -11.583 -2.106 -41.279 36.402 -43.432 11.814 Problem if we want the average difference from the mean, since some are positive and some are negative: sum(values - mean(values)) [1] 0 Solution is to square the values (in addition this has the generally desirable side effect of emphasising the larger values over the smaller values) values - mean(values) [1] 7.004 14.851 1.651 26.677 -11.583 -2.106 -41.279 36.402 -43.432 11.814 (values - mean(values))^2 [1] 49.056 220.567 2.726 711.661 134.156 4.437 1703.935 1325.135", "the average of the two middle values of 77 and 84. The median is then $\\frac{77+84}{2}=80.5$ so the lower quartile is 80.5. The upper quartile is the median of the second half of the data. The second half of the data does not include the median of the set so the second half of the data is: 104, 111, 114, 122, 147, 164 Since the second half of the data has an even number of values, the median of the second half is the average of the two middle values of 114 and 122. The median is then $\\frac{\\left(114+122\\right)}{2}=118$ so the upper quartile is 118. To make the box-and-whisker plot, plot the minimum of 71, the lower quartile of 80.5, the median of 98, the upper quartile of 118, and the maximum of 164. Draw a box that has sides at the lower and upper quartiles. Draw a vertical segment in the box passing through the median. Then draw segments from the minimum to the lower quartile and from the upper quartile to the maximum: [Pic]", "mean 7a ? 4a + 3b 3a ? 3b a ? b Since a < b was given and the median is a, the mean is greater than the median. ### 4. Comparing Data Every set of numbers has a mean, median and range. These three values can be used to compare data by seeing how much the numbers differ. A politician announces at a press conference that the average yearly salary per person in the state is now$80,000.After her announcement, three groups of 5 people were surveyed about their salaries. (I) $80,000,$80,000, $80,000,$80,000, $80,000 (II)$71,000, $74,000,$82,500, $82,500,$90,000 (III) $15,000,$27,000, $33,000,$100,000, $225,000 Is the claim about average salaries reasonable for each group? ### Solution To solve this problem, compare the mean, median and range of each data set. (I) The mean and median are$80,000. The range is 0. The claim is reasonable for this group. (II) The mean is $80,000. The median is$82,500. The range is $19,000. So the mean and median support the claim. The range is somewhat high, showing the data is spread out. The claim is still somewhat reasonable for this group. (III) The mean is$80,000. The median is $30,000. The range is$210,000. The claim is not reasonable for this group. The mean age of a group of 10 firefighters is 18 years. If 2", "highest. If there is an even number of data, you take the average of the middle two numbers to find the median. The mode is the number that occurs most often. When you look for the deviation from the mean, you are looking for the difference between an average and a value. Let’s look at an example that shows what happens when the difference between the mean and another value is determined. At a restaurant, the food servers report how much money in tips they earn each night. One Saturday, the food servers reported the following tips: $45,$37.50, $51,$69, $47, and$55. Greg is the newest food server and wants to know how the tips compare to each other. One way to do this is to find the deviation from the mean. This tells you how far away from the mean, or average, each food server was. First, find the mean. mean===45+37.50+51+69+47+556304.50650.75\\begin{align*}\\begin{array}{rcl} \\text{mean} &=& \\frac{45+37.50+51+69+47+55}{6} \\\\ &=& \\frac{304.50}{6} \\\\ &=& 50.75 \\\\ \\end{array}\\end{align*} Next, find the difference of each food server tip from the mean. This is the deviation from the mean. Notice that Greg’s value is always used because you are looking for the deviation between the tips of the other waiters and Greg. When subtracting, place the larger of the two numbers first so that the difference", "720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 Total = 80 Total = 4000 Total = 1280 = 4000/80 = 50 = 1280/80 = 16 ∴ The mean deviation is 16 5. Find the mean deviation from the median for the following data : (i) xi 15 21 27 30 fi 3 5 6 7 Solution: To find the mean deviation from the median, firstly let us calculate the median. We know, N = 21 Median = (21)/2 = 10.5 So, the median Corresponding to 10.5 is 27 xi fi Cumulative Frequency |di| = |xi – M| fi |di| 15 3 3 15 45 21 5 8 9 45 27 6 14 3 18 30 7 21 0 0 Total = 21 Total = 46 Total = 108 N = 21 $MD = \\frac{1}{n}\\sum_{i=1}^{n}|d_{i}|$ = 1/21 × 108 = 5.14 ∴ The mean deviation is 5.14 (ii) xi 74 89 42 54 91 94 35 fi 20 12 2 4 5 3 4 Solution: To find the mean deviation from the median, firstly let us calculate the median. We know, N = 50 Median = (50)/2 = 25 So, the median Corresponding to 25 is 74 xi fi Cumulative Frequency |di| = |xi – M| fi |di| 74", "3.75 3.75 – 3.682 = 0.068 0.004624 3.99 3.99 – 3.682 = 0.308 0.094864 This data is from a sample, so we will add the squared deviations, divide by 4, the number of data values minus 1, and compute the square root:$\\sqrt{\\frac{0.153664+0.008464+0.011664+0.004624+0.094864}{4}}\\approx$$0.261 5. The data is already in order, so we don’t need to sort it first.The minimum value is$140 and the maximum is $460.There are 36 data values so n = 36. n/2 = 18, which is a whole number, so the median is the mean of the 18th and 19th data values,$305 and $310. The median is$307.50. To find the first quartile, we calculate the locator, L = 0.25(36) = 9. Since this is a whole number, we know Q1 is the mean of the 9th and 10th data values, $250 and$260. Q1 = $255. To find the third quartile, we calculate the locator, L = 0.75(36) = 27. Since this is a whole number, we know Q3 is the mean of the 27th and 28th data values,$345 and $350. Q3 =$347.50. The 5 number summary of this data is: $140,$255, $307.50,$347.50, \\$460 6. Boxplot of textbook costs 1. The reason we do this is highly technical, but we can see how it might be useful by considering the case of a small sample from a", "of variables such as income. Following are some summary statistics for this data: • Mean = $24,000, SD =$27,500 • Median = $16,900, IQR =$28,000 The typical range based on the mean and standard deviation is not a good summary of the distribution of incomes. The small number of people with higher incomes increases the mean. The mean is too high to represent the large number of people making less than \\$20,000 a year. The small number of people with higher incomes also increase the standard deviation, so a small number of high incomes gives the misleading impression that typical incomes in the sample are higher than they really are. Notice also that Mean ± SD gives an awkward range of typical values. The left endpoint is at −3,500 (Mean − SD = 24,000 − 27,500 = −3,500), but there are no negative values in this data set. This is another reason why it is better to use the IQR when measuring the spread of a skewed data set. Let’s take a look at the same histogram, except this time we overlay a boxplot. We see that the median represents the typical income of people in this sample better than the mean. The small number of people with higher incomes does not impact the median or the other", "not exist. 756.4 705.8 763.6 708.3 708.3 708.3 708.3 708.3 649.3 649.3 472.2 472.2 472.2 472.2 296 500 500 500 500 500 500 500 500 500 500 250 250 606 606 606 444 747 778 667 722 In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median. In this case, analysts tend to use the mean because it includes all of the data in the calculations. Mean: the average value. /BaseFont/CXTGSM+CMR8 Discuss the mean, median, and mode for each of the following problems. 1111.1 1511.1 1111.1 1511.1 1111.1 1511.1 1055.6 944.4 472.2 833.3 833.3 833.3 833.3 Terry’s mean is $3.7$, Davis’ mean is $2.7$, Maris’ mean is $4.6$. When the mean exists, they are. >> It is well known that for 0 < q < 1 the q th -quantile of F x, denoted as θ, has asymp. /Subtype/Type1 Find the asymptotic distribution of n 3 / 2 X ~ n, where X ~ n is the sample median. 22 0 obj I am confused at what to do. /LastChar 255 >> 597.2 736.1 736.1 527.8 527.8 583.3 583.3 583.3 583.3 750 750 750 750 1044.4 1044.4 Please help. The following lists shows a simple random sample that compares the letter counts for three authors. Consequently, when some", "Evaluate P(X $\\leq$ 1.75). d] State, with a reason, whether the median of X is greater or less than 1.75. For part a] I calculated E(X) which I found to be , so E(1/X) was done by which is 10/17, or 0.58 or something like that. Is this right? Mr F says: No! ${\\color{red} E \\left( \\frac{1}{X} \\right) \\neq \\frac{1}{E(X)}}$. For part b] I had . Mr F says: Wrong. If this was correct then it should equal 1 when x = 2. It doesn't. For c] if P(X 1.75) is F(1.75), then that should be 0.30625 if part b] is correct, and to answer d] you can say that the median of X must be greater than 0.5 because the value of 1.75 is not greater than 0.5. Am I right or did something go wrong? Thanks for any help a) $E \\left( \\frac{1}{X}\\right) = \\int_1^2 \\frac{1}{x} \\cdot \\frac{6}{5} x (x - 1) \\, dx$. b) $F(x) = \\left\\{ \\begin{array}{ll} 0, & x < 1 \\\\ & \\\\ \\int_1^x \\frac{6}{5} t (t - 1) \\, dt = \\frac{ax^3 - bx^2 + c}{5}, & 1 \\leq x \\leq 2 \\\\ & \\\\ 1, & x > 2 \\end{array} \\right.$ where I'll let you calculate the values of a, b and c for yourself. Note: The first and third", "$$8$$ $$7$$ $$6$$ $$14$$ $$9$$ $$2$$ $$16$$ $$10$$ $$2$$ $$18$$ $$12$$ $$2$$ $$20$$ $$15$$ $$6$$ 26 Here, $$N = 26$$, which is even. Hence, Median is the mean of 13th and 14th observations. Both the observations lie in the cumulative frequency 14, for which the corresponding observation is 7. \\begin{align} \\text{Median}&= \\frac{{{13^{th}}\\,observation + {14^{th}}observation}}{2}\\\\& = \\frac{7 + 7}{2}\\\\&= 7 \\end{align} The absolute values of the deviations, $$\\left| {{x_i} - M} \\right|$$are $$\\left| {{x_i} - M} \\right|$$ 2 0 2 3 5 8 $${f_i}$$ 8 6 2 2 2 6 $${f_i}\\left| {{x_i} - M} \\right|$$ 16 0 4 6 10 48 $$\\sum\\limits_{i = 1}^6 {{f_i}} = 26$$ and $$\\sum\\limits_{i = 1}^6 {{f_i}\\left| {{x_i} - M} \\right|} = 84$$ Hence, the mean deviation about the median \\begin{align}M.D.\\left( M \\right) &= \\frac{1}{N}\\sum\\limits_{i = 1}^6 {{f_i}\\left| {{x_i} - M} \\right|} \\\\&= \\frac{1}{26} \\times 84\\\\&= 3.23\\end{align} ## Chapter 15 Ex.15.1 Question 8 Find the mean deviation about the median for the data $${x_i}$$ $$15$$ $$21$$ $$27$$ $$30$$ $$35$$ $${f_i}$$ $$3$$ $$5$$ $$6$$ $$7$$ $$8$$ ### Solution The given observations are in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table. $${x_i}$$ $${f_i}$$ $$c.f$$ $$15$$ $$3$$ $$3$$ $$21$$ $$5$$ $$8$$ $$27$$ $$6$$ $$14$$ $$30$$ $$7$$ $$21$$ $$35$$ $$8$$ $$29$$ Here, $$N = 29$$, which is", "(without regard for the number of items sold) was $505 last year and$604 this year. The mean price of the 101,000 items sold last year is (100,000 × $10 + 1,000 ×$1,000)/101,000 = $19.80 while this year it is (100,000 ×$8 + 1,000 × $1,200)/101,000 =$19.80. The mean price per item sold is the same in both years: the total revenue was the same, and the number of items sold was the same. The moral is that one can make data appear to tell conflicting stories by choosing a measure of location disingenuously. The following exercises check your ability to compute and to use the mean, median, and mode. ### Videos of Exercises (Reminder: Examples and exercises may vary when the page is reloaded; the video shows only one version.) Measures of location Measures of location summarize a list of numbers by a \"typical\" value. The three most common measures of location are the mean, the median, and the mode. The mean is the sum of the values, divided by the number of values. It has the smallest possible sum of squared differences from members of the list. The median is the middle value in the sorted list. It is the smallest number that is at least as big as at least half the values in the list.", "the median. Now let's go back to the dilemma from the beginning of the Concept. Step 1: To determine the median of the set of data, arrange the data in order from least to greatest. Identify the data value in the middle of the data set. For this set of data, 102 is the median. 89, 91, 95, 98, 102, 108, 110, 118, 152 Step 2: Identify the median for the lower quartile. Again, since two data values share the middle position, find their mean. The median for the lower quartile is 93. $& \\underline{89, \\ 91, \\ 95, \\ 98}, \\ 102, \\ 108, \\ 110, \\ 118, \\ 152\\\\& \\qquad \\qquad \\qquad \\quad \\ 91 + 95 = 186\\\\& \\qquad \\qquad \\qquad \\quad \\ 186 \\div 2 = 93$ Step 3: Identify the median of the upper quartile. Remember to find the mean of the two data values that share the middle position. The median of the upper quartile is 114. $& 89, \\ 91, \\ 95, \\ 98, \\ 102, \\ \\underline{108, \\ 110, \\ 118, \\ 152}\\\\& \\qquad \\qquad \\qquad \\quad \\ 110 + 118 = 228\\\\& \\qquad \\qquad \\qquad \\quad \\quad \\ 228 \\div 2 = 114$ Step 4: Draw a number line. The first value on the number line should be near", "Algebra 1 To find the mean, we add up all of the values and divide by the total number of values. Mean E: $\\frac{113+183+479+120+117}{5} = 202.4$ Mean F: $\\frac{145+129+153+135+142}{5} = 140.8$ To find the range, we subtract the smallest term from the largest term. Range E: $479-113 = 366$ Range F: $153-129 = 24$ The mean and range of Set E are greater than the mean and range of Set F.", "- M|$ 15 3 3 13.5 40.5 21 5 8 7.5 37.5 27 6 14 1.5 9 30 7 21 1.5 10.5 35 8 29 6.5 52 Now, N = 30, which is even. Median is the mean of $15^{th}$ and $\\dpi{100} 16^{th}$ observations. Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30. Therefore, Median, M $\\dpi{100} = \\frac{15^{th} observation + 16^{th} observation}{2} = \\frac{30 + 30}{2} = 30$ Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and $\\sum f_i|x_i - M|$ = 149.5 $\\therefore$ $M.D.(M) = \\frac{1}{29}\\sum_{i=1}^{5}|x_i - M|$ $= \\frac{149.5}{29} = 5.1$ Hence, the mean deviation about the median is 5.1 Question:9. Find the mean deviation about the mean. Income per day in Rs $\\small 0-100$ $\\small 100-200$ $\\small 200-300$ $\\small 300-400$ $\\small 400-500$ $\\small 500-600$ $\\small 600-700$ $\\small 700-800$ Number of persons $\\small 4$ $\\small 8$ $\\small 9$ $\\small 10$ $\\small 7$ $\\small 5$ $\\small 4$ $\\small 3$ Income per day Number of Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \\overline{x}|$ $f_i|x_i - \\overline{x}|$ 0 -100 4 50 200 308 1232 100 -200 8 150 1200 208 1664 200-300 9 250 2250 108 972 300-400 10 350 3500 8 80 400-500 7 450 3150 92 644 500-600 5 550 2750 192", "are three common measures of central tendency; they are three mathematical tools frequently used to analyze data. The mean, commonly referred to as the average, is the sum of all the data items divided by the number of data items. The median is the middle number in a set of data that is ordered from least to greatest. If there is an even number of data, you take the average of the middle two numbers to find the median. The mode is the number that occurs most often. Let’s look at an example. A manager at a small movie theater was analyzing the number of people who came to the movies during the week. Over nine days, he found the following data: 81, 89, 92, 85, 93, 62, 85, 105, and 90. Find the mean, median, and mode of the data. First, find the mean. Remember that the mean is the same as the average. To find the mean you must add all of the data items and divide by the number of items. \\begin{align*}\\begin{array}{rcl} \\text{mean} &=& \\frac{81+89+92+85+93+62+85+105+90}{9} \\\\ &=& \\frac{782}{9} \\\\ &=& 86.8 \\end{array}\\end{align*} Next, find the median. The median is the middle number when the data is ordered from lowest to highest. When you reorder the data from least to greatest you get: \\begin{align*}\\begin{array}{rcl} & 62," ]

[ "of 5m and 13m are 6m apart from each other. Find their distance between top of the two poles. 3. ### Geometry So the directions say: Provide a counterexample to show that each statement is false. You can use words or make a diagram. These are the two questions that I'm stuck on: 1. If a line lies in a vertical plane, then the line is 4. ### Magnets-Physics I have a few questions that I am not too sure of. why is it inadvisable to make a horseshoe magnet from a flexible material? -is it because it would have enough magnetism Do the poles of a horseshoe magnet attact each other? If 5. ### math the nufone communications company must run a telephone line between two poles at opposite ends of a lake. the length and width are 75 feet and 30 feet. what is the distance between the two poles to the nearest foot? The lake is 6. ### Math 17.Which roller coaster has the greatest lift height Top Thrill Dragster 18.What is the median lift height for the roller coasters lifted? Round to the nearest tenth. Not sure for this one 19.Arrange the given roller coasters from 7. ### Math PLease help me with this: What kind of line is created when you graph the", "or $$\\frac{6}{5}$$. Explanation: Given a short roller-coaster track is $$\\frac{3}{10}$$ mile long and longer roller-coaster track is about 4 times as long as the short track to find how long is the longer roller-coaster track we multiply short roller-coaster track $$\\frac{3}{10}$$ mile by 4, 4 X $$\\frac{3}{10}$$ now we write expression as a multiple of unit fraction so 4 X 3 X $$\\frac{1}{10}$$ = 12 X $$\\frac{1}{10}$$ =$$\\frac{12}{10}$$ as numerator and denominator can be further simplified as both can be divided by 2 making it as $$\\frac{6}{5}$$. Show and Grow Question 15. The Renaissance Tower in Dallas, Texas is $$\\frac{27}{100}$$ kilometer tall. The Burj Khalifa is about 3 times as tall as the Renaissance Tower. About how tall is the Burj Khalifa? The Burj Khalifa is $$\\frac{81}{100}$$ kilometers tall. Explanation: Given the Renaissance Tower in Dallas, Texas is $$\\frac{27}{100}$$ kilometer tall and The Burj Khalifa is about 3 times as tall as the Renaissance Tower, To find how tall is the Burj Khalifa we multiply the Renaissance Tower in Dallas, Texas length by 3, 3 X $$\\frac{27}{100}$$ we write expression as a multiple of unit fraction 3 X 27 X $$\\frac{1}{100}$$ = 81 X $$\\frac{1}{100}$$ =$$\\frac{81}{100}$$, So the Burj Khalifa is $$\\frac{81}{100}$$ kilometers tall. Question 16. You water 6 plants using $$\\frac{3}{5}$$ liter of water for each. How", "change in GPE of a 450-kg roller coaster car starting at 38 m above the ground and ends the ride 5 m above the ground. Given: $$m=450\\;kg\\\\ h_i=38 \\;m\\;\\text{(initial height)}\\\\ h_f=5\\;m\\;\\text{(final height)}\\\\ g=9.8\\;m/s^2$$ The reference point is the ground. Since the roller coaster car is above the ground, GPE must be positive. Hence, $$\\text{GPE}=mgh\\\\ \\; \\;\\;\\;\\quad=mg(h_f-h_i)\\\\ \\; \\;\\;\\;\\quad=(450\\;kg)(9.8\\;m/s^2)(5\\;m-38\\;m)\\\\ \\; \\;\\;\\;\\quad=(450\\;kg)(9.8\\;m/s^2)(-33\\;m)\\\\ \\; \\;\\;\\;\\quad=-145,530\\;J$$ Note: Negative sign only indicates that the velocity of the roller coaster car slows down or eventuallr comes to a stop. Example 2. A 60-kg loaded box is pulled at a distance of 15 m up an inclined plane 20° from the ground. Calculate the change in GPE. Given: $$m=60\\;kg\\\\ l=15\\;m\\\\ \\theta=20^\\circ$$ First, solve for the height (h) using trigonometric ratio $$\\sin \\theta=h/l.$$ Thus, $$h=l\\sin\\theta\\\\ \\;\\;=(15 \\;m)(\\sin 20^\\circ)\\\\ \\;\\;=(15 \\;m)(0.34)\\\\ \\;\\;=5.1 \\;m$$ We can now solve for GPE using h, $$\\text{GPE}=mgh\\\\ \\;\\;\\;\\;\\;\\;\\;=(60\\;kg)(9.8\\;m/s^2)(5.1\\;m)\\\\ \\;\\;\\;\\;\\;\\;\\;=2,998.8\\;J$$ Elastic Potential Energy(EPE) Elastic potential energy is the energy stored in elastic materials like springs. It depends on the amount of stretch on the material - the more stretch, the more stored energy. The EPE of a spring can be expressed in a simple equation. It can also store EPE through stretching and compression. Spring follows the Hooke's Law which states that the amount of force is directly proportional to the amount of", "point is on the y axis (= 0), there is no horizontal phase shift. Thus the sinusoidal function is $$y=50\\cos \\frac{\\pi }{150}x+40$$. We could put it in a graphing calculator to check it. (b) At the highest point, the roller coaster is 90 feet above the ground ($$y=50\\cos \\left( \\frac{\\pi }{150}*0 \\right)+40=90$$) . At x = 15 feet, the roller coaster is $$y=50\\cos \\left( \\frac{\\pi }{150}*15 \\right)+40$$, or 55 feet tall. At x = 100 feet, the roller coaster is $$y=50\\cos \\left( \\frac{\\pi }{150}*100 \\right)+40$$, or 15 feet tall. It’s easiest to put the function in the graphing calculator, and use the 2nd TRACE (CALC) value function to get these values (see the WINDOW you can use below). (c) When the track is 75 feet above the ground, y = 75. The easiest way to get the x at that point is to use the Intersect feature in the graphing calculator: (2nd Trace (Calc), 5, ENTER, ENTER, ENTER). So when the track is 75 feet above the ground, the roller coaster is about 37.98 feet from the highest point at x = 0. Note the window I used to match the graph of the roller coaster. You can use the same steps to see that when the roller coaster track is 15 feet above the ground, the roller", "feet, which was true at its 2009 publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety on Roller Coasters? Of course they do, or else we’d have more roller coasters that allowed mice to ride. The question is how much the size restriction matters, and how sensitive that dependence is. So the leading question is a classic example of applying mathematics to the real world. This includes practical subtleties like if a person 39.5 inches tall could ride safely, is it fair to round that off to 40 inches? It also includes the struggle to work out how dangerous an amusement park is. Speaking from my experience as a rider and lover of amusement parks: don’t try to plead someone’s “close enough”. You’re putting an unfair burden on the ride operator. Accept the rules as posted. Everybody who loves amusement parks has their disappointment stories; accept yours in good grace. This leads me into planning amusement park fun. School", "- h_{min}}{h_{avg_(up)}-h_{min}}=\\frac{526m - 258m}{345m-258m} = \\frac{268}{87}= 3.08$$ Then, we need to also check the slope if exceeds 0.1. We can use the following equation, where $H$ is the height of the hill above the surrounding terrain, and $L$ is the length of the slope. $$\\theta_{slope} = \\frac{H}{L} = \\frac{526m-388m}{2550m-1750m} = 0.17$$ You may notice that I tried to use the location of the crest, so to get a realistic slope. You can also see on the screen shot that the average slope is around 14%, so we definitely have a speed up. Now, we know that our topography feature is fulfilling all the requirements by the code, and by now we have all the values we need for the calculation of $K_{zt}$. $h = 15m$ this would be our tower height $K_c = 1.0$ based on the Exposre category C. from Table 2-4. $K_t = 0.53$ from Table 2-5, if our feature is a hill $H = 526m - 388m = 138m$, where 388m is the average upwind terrain $L = 800m$, the length of the upwind slope $f = 2.0$ which is the height attenuation factor from Table 2-5. First need to calculate $K_h$: $$K_h=e^{ ( \\frac{f \\cdot z}{H} )}=2.718^{( \\frac{2.0 \\cdot 15}{138} )} = 1.24$$ (in case $h=0m$ so the ground level, then factor $K_h$ yields", "15 feet tall. It’s easiest to put the function $$y=50\\cos \\frac{\\pi }{{150}}x+40$$ in the graphing calculator, and use the 2nd TRACE (CALC) value function to get these values (see the WINDOW you can use below). (c) When the track is 75 feet above the ground, $$y=75$$. The easiest way to get the $$x$$ at that point is to use the Intersect feature in the graphing calculator: (2nd Trace (Calc), 5, ENTER, ENTER, ENTER). Thus, when the track is 75 feet above the ground, the roller coaster is about 37.98 feet from the highest point at $$x=0$$. Note the window I used to match the graph of the roller coaster. < You can use the same steps to see that when the roller coaster track is 15 feet above the ground, the roller coaster is 100 feet from the beginning point. Bouncing Spring Problem: The weight on a long spring bounces up and down sinusoidally with time. You are looking at a second hand on a clock and notice that when the clock reads .2 seconds, the weight first reaches a high point that is 50 centimeters (cm) above the ground. The next low point is at 30 cm above the ground, and this occurs at 1.5 seconds. (a) Draw the graph that represents this situation, and write the", "then $y-7069=255(x-2)$ for each pair $(x,y)$ in the table, so the linear function given by $y = 7069 +255(x-2)$ works. 2. When $x = 0$ $$y = 7069 + 255(0-2) = 6559.$$ So the elevation of the tram at the beginning of the ride is 6559 feet. 3. When $x=15$ $$y=7069+255(15-2)=10,384.$$ So the elevation of the tram at the end of the ride is 10,384 feet.", "(5, -2); passes through (7, 0) Question 62. vertex: (-5, -1); passes through (-2, 2) Vertex form is f(x) = a(x – h)2 + k f(x) = a(x + 5)2 – 1 2 = a(-2 + 5)2 – 1 2 = a(3)2 – 1 9a – 1 = 2 9a = 3 a = 1/3 f(x) = 1/3(x + 5)2 – 1 Question 63. MODELING WITH MATHEMATICS A portion of a roller coaster track is in the shape of a parabola. Write and graph a quadratic function that models this portion of the roller coaster with a maximum height of 90 feet, represented by a vertex of (25, 90), passing through the point (50, 0). Question 64. MODELING WITH MATHEMATICS A flare is launched from a boat and travels in a parabolic path until reaching the water. Write and graph a quadratic function that models the path of the are with a maximum height of 300 meters, represented by a vertex of (59, 300), landing in the water at the point (119, 0). f(x) = a(x – h)2 + k h = 59 and k = 300 f(x) = a(x – 59)2 + 300 As (119, 0) lies on the function, putting the values in function we get 0 = a(119 – 59)2 + 300 0 = 3600a", "publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety on Roller Coasters? Of course they do, or else we’d have more roller coasters that allowed mice to ride. The question is how much the size restriction matters, and how sensitive that dependence is. So the leading question is a classic example of applying mathematics to the real world. This includes practical subtleties like if a person 39.5 inches tall could ride safely, is it fair to round that off to 40 inches? It also includes the struggle to work out how dangerous an amusement park is. Speaking from my experience as a rider and lover of amusement parks: don’t try to plead someone’s “close enough”. You’re putting an unfair burden on the ride operator. Accept the rules as posted. Everybody who loves amusement parks has their disappointment stories; accept yours in good grace. This leads me into planning amusement park fun. School Specialty’s blog particularly offers PLAY & PLAN:", "publication: Son of Beast at Kings Island. Sad to say Son Of Beast closed in 2009, and was torn down in 2012. The current record heights in wooden coasters are T Express at Everland in South Korea, and Wildfire at Kolmården in Sweden. (Too much height is not really that good for wooden roller coasters.) A 2018 posting on Social Mathematics asks: Do height restrictions matter to safety on Roller Coasters? Of course they do, or else we’d have more roller coasters that allowed mice to ride. The question is how much the size restriction matters, and how sensitive that dependence is. So the leading question is a classic example of applying mathematics to the real world. This includes practical subtleties like if a person 39.5 inches tall could ride safely, is it fair to round that off to 40 inches? It also includes the struggle to work out how dangerous an amusement park is. Speaking from my experience as a rider and lover of amusement parks: don’t try to plead someone’s “close enough”. You’re putting an unfair burden on the ride operator. Accept the rules as posted. Everybody who loves amusement parks has their disappointment stories; accept yours in good grace. This leads me into planning amusement park fun. School Specialty’s blog particularly offers PLAY & PLAN:", "doesn’t change with the board’s age, or the condition of neighboring boards, or anything but the fact that a year has passed — then the chance of any particular board lasting a hundred years is going to be $0.99^{100}$. That takes a little thought to work out if you haven’t got a calculator on hand. I recently had the chance to ride the Leap-the-Dips at Lakemont Park (Altoona, Pennsylvania), the world’s oldest operating roller coaster. The statistics of this 1902-vintage roller coaster might not sound impressive, as it has a maximum height of about forty feet and a greatest drop of about nine feet, but it gets rather more exciting when you consider that the roller coaster car hasn’t got any seat belts or lap bar or other restraints (just a bar you can grab onto if you so choose), and that the ride was built before the invention of upstop wheels, the wheels that actually go underneath the track and keep roller coaster cars from jumping off. At each of the dips, yes, the car does jump up and off the track, and the car just keeps accelerating the whole ride. (Side boards ensure that once the car jumps off the tracks it falls back into place.) It’s worth the visit. Looking at the wonderful mesh of", "Hide # Tall Enough The new amusement park you’ve opened has a small problem: people are riding your roller coasters without meeting the safety requirements! Instead of hiring more employees to enforce the safety requirements, you decide to augment your coaster control system with a new safety check. The system you’re creating will accept a list of heights in inches and will output “True” if all of the heights are greater than or equal to $48$ inches. If any of the heights are less than $48$ inches, it will output “False”. ## Input The first line is $n$, an integer between $1$ and $1\\, 000$ inclusive, defining the number of heights you’ll receive. The next $n$ lines will each contain a single height in inches. Each height will be an integer between $1$ and $1\\, 000$ inclusive. ## Output If any of the heights are less than $48$ then output “False” without the quotes. Otherwise, output “True” without the quotes. Sample Input 1 Sample Output 1 3 50 60 100 True Sample Input 2 Sample Output 2 4 75 40 80 600 False CPU Time limit 1 second Memory limit 1024 MB", "# Calculations: Example Let's see how we can predict the maximum speed of a rider using the information from Figure 5. For this example, we will assume a rider weight (including trolley and harness) of 578 N (or equivalently, 130 pounds). Figure 5. Measurements for the last zip line on the WVU Canopy Tour. The drawing is not shown to scale. ### Step 1: Determine the Distance Traveled $$Distance (d) = {\\sqrt{horizontal~distance^2 + vertical~drop^2}}$$ Please find the horizontal distance to the lowest point and vertical drop values from Figure 5 and insert them into the equation below. Click on the \"Calculate\" button to see the result. Distance(d) = 2 + 2 Sorry, the values you entered are not correct. Please try again. $$Distance~(d)= \\sqrt{214^2 + 16^2}$$ $$d = \\sqrt{45796 + 256}$$ $$d = \\sqrt{46052}$$ $$d = 214.597 m$$ $$d \\approx 215 m$$ ### Step 2: Approximate Acceleration $$Acceleration~(a) = g \\times sin(\\theta) - loss$$ $$remember: g = 9.81\\frac{m}{s^2}$$ First let’s find sin(θ). It can be calculated by dividing the opposite leg (vertical drop) by the hypotenuse (distance (d)). Enter the two values into the equation below to calculate sin(θ). For simplicity, enter whole numbers only. sin(θ) = vertical drop / distance(d) = / Sorry, the values you entered are not correct. Please try again. $$sin(\\theta) = \\frac{16m}{215m} =", "used 1. Now if i remember correctly, taking the derivative would mean that constant d is completely arbitrary? Correct me if I'm wrong, I just need to know where do go from here to get the slope...or what I have to undo. Thanks for the help. p.s. My other given numbers where horizontal distance of coaster =212m, and I'm supposed to find the slope at 141.3333333333337m( I seriously hate randomly generated numbers for problems). 5. Sep 7, 2006 Andrew Mason Use the chain rule to differentiate y = f(g(x)) where $f(x) = Asin(g(x))$ and $g(x) = \\frac{2\\pi}{\\lambda}x + \\phi$ AM", "162 = 72 + x2 162 – 72 = x2 207 = x2 14 ≈ x The top of the ladder is 14 ft. above the ground. b. Determine the measure of the angle formed by the ladder and the ground. arccos ($$\\frac{7}{16}$$) ≈ 64 The angle formed by the ladder and the ground is approximately 64°. Exercise 4. A group of friends have hiked to the top of the Mile High Mountain. When they look down, they can see their campsite, which they know is approximately 3 miles from the base of the mountain. a. Sketch a drawing of the situation. b. What is the angle of depression? arctan ($$\\frac{3}{1}$$) ≈ 72 The angle of depression is approximately 18°. Exercise 5. A roller coaster travels 80 ft. of track from the loading zone before reaching its peak. The horizontal distance between the loading zone and the base of the peak is 50 ft. a. Model the situation using a right triangle. b. At what angle is the roller coaster rising according to the model? arccos ($$\\frac{50}{80}$$) ≈ 51 The roller coaster is rising at approximately 51°. ### Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key Question 1. For each triangle shown, use the given information to find the indicated angle to the nearest degree.", "any curve. For the old curve, the offset is: $$Y=\\frac{A(L/2)^2}{200L}=\\frac{7(450/2)^2}{200(450)}=3.935 \\text{ } feet$$ For the new curve, using data found from before, the offset is: $$Y=\\frac{A(L/2)^2}{200L}=\\frac{7(2812/2)^2}{200(2812)}=24.605 \\text{ } feet$$ The difference between the two offsets is 20.67 feet. This is the depth of the cut at the PVI. Homework 1. Which conic section forms the basis of vertical curves? 2. A section of a roadway goes up a 6% incline and includes no horizontal curves. Point A, which lies along this incline, has a station marker. Using a wheel measure, you measure the distance along the centerline from point A to the next PVC and find it is 880 meters. If point A is at Station 0+700, what is the station of the PVC? 3. Identify 4 criteria used to set the minimum length of a sag vertical curve (4 points), place them in order from most to least important (1 point): 4. An equal tangent sag vertical curve joins a –0.5 % grade with a +3.0% grade. Assume H=0.6, headlights are aimed upward at an angle of 1 degree If the PVI of the grades is at metric station 10+100 and has an elevation of 300 meters, using the headlight criterion determine the station elevation of the PVC and PVT for a design speed of 100 km/hr", "# Example Problem ## Calculations: Example Let's see how we can predict the maximum speed of a rider using the information from Figure 5. For this example, we will assume a rider weight (including trolley and harness) of 578 N (or equivalently, 130 pounds). Figure 5. Measurements for the last zip line on the WVU Canopy Tour. The drawing is not shown to scale. ### Step 1: Determine the Distance Traveled Please find the horizontal distance to the lowest point and vertical drop values from Figure 5 and insert them into the boxes below. Click on the \"Calculate\" button to see the result. Horizontal Distance 2 Vertical Drop 2 Distance Distance (d)= √2142+162 d = √45796+256 d = √46052 d=214.597m d≈215m ### Step 2: Approximate Acceleration $remember:g=9.81\\frac{m}{{s}^{2}}$ First let’s find sin(θ). It can be calculated by dividing the opposite leg (vertical drop) by the hypotenuse (distance (d)). Enter the two values into the boxes below to calculate sin(θ). For simplicity, enter whole numbers only. Vertical Drop Distance sin (θ) = That's correct! sin(θ) = 16m/215m = 0.0746 Next, we’ll use the chart below to determine loss. The loss values have been derived through experimentation on the WVU Canopy Tour. In this example the rider is 130 pounds. Find the approximate value of loss in the chart and use", "tower. The total vertical drop of the zip line is 40 ft. The zip line’s angle of elevation from the lower tower is 11. 5°. What is the horizontal distance between the towers? tan 11.5 = $$\\frac{40}{x}$$ x = $$\\frac{40}{\\tan 11.5}$$", "264 views If three sides of a rectangular park have a total length $400$ ft, then the area of the park is maximum when the length (in ft) of its longer side is 1. $299$ 2. $200$ 3. $201$ 4. $399$ ## 1 Answer Let the one side be $L$ ft, and the other side be $B$ ft. Let's draw the diagram for a better understanding. Given that, $L+2B=400$ $\\Rightarrow 2B=400-L$ $\\Rightarrow \\boxed{B=\\frac{(400-L)}{2}}$ Area of rectangle $A=L \\ast B$ $\\Rightarrow A=\\frac{L\\ast(400-L)}{2}$ $\\Rightarrow A=200L-\\frac{L^{2}}{2}$ The area is maximum when differentiation of $A$ is zero. $\\Rightarrow 200-\\frac{2L}{2}=0$ $\\Rightarrow 400-2L=0$ $\\Rightarrow 2L=400$ $\\Rightarrow \\boxed{L=200\\;\\text{ft}}$ And, $B=\\frac{400-200}{2}$ $\\Rightarrow B = \\frac{200}{2}$ $\\Rightarrow\\boxed{B = 100\\;\\text{ft}}$ $\\therefore$ The length of the longer side is $200\\;\\text{ft}.$ Correct Answer $:\\text{B}$ 10.3k points Answer: 2 votes 1 answer 1 235 views 1 vote 1 answer 2 275 views 1 vote 1 answer 3 184 views 1 vote 1 answer 4 241 views 0 votes 0 answers 5" ]

[ "9 out of 10 words. 9/10. it is impossible that the handwritten word will be correctly interpreted. so option A is correct. Question 18. A bag contains 8 letter tiles of the same size. The tiles are labeled either A, B, C, D, E, or F. Three of the tiles are labeled C. If Corey selects 1 tile from the bag without looking, is the selection of letters fair? Explain. Answer: Yes, the selection of letters is fair. Explanation: In the above-given question, given that, A bag contains 8 letter tiles of the same size. The tiles are labeled either A, B, C, D, E, or F. Three of the tiles are labeled C. 1 out of 8. 1/8. so the selection of letters is fair. Lesson 7.2 Understand Theoretical Probability Solve & Discuss It! Betty and Carl will conduct an experiment. They will flip a coin 100 times and record the result of each flip. What should they expect the results of their experiment to be? Justify your answer. I can… determine the theoretical probability of an event. Answer: The theoretical probability of an event is 1/100, 2/100, ….. 100/100. Explanation: In the above-given question, given that, Betty and Carl will conduct an experiment. They will flip a coin 100 times and record the result of each", "somewhere at random inside the larger circle, what is the probability that it will fall somewhere in the smaller circle? The answer is not $\\frac{3}{4}$ . A bag has 26 tiles in it, each with a different letter of the alphabet. • A You pick one tile out of the bag, look at it, and write it down. Then you put it back in the bag, which is thoroughly mixed up. Then you pick another tile out of the back, look at it, and write it down. What is the probability that your first letter was “A” and your second letter was “T”? • B Same bag, different plan. This time you pick the first tile, but do not put it back in the bag. Then you pick a second tile and place it next to the first? Now what is the probability that your first letter was “A” and your second letter was “T”? • C In the second case, what is the probability that your two letters, together, could make the word “AT”? A deck of cards has 52 cards, 13 of each suit. Assume there are no Jokers. (Once again, no two parts of this question have exactly the same answer.) • A If you draw a card at random, what is the probability of", "a radius of 4\". If a dart hits somewhere at random inside the larger circle, what is the probability that it will fall somewhere in the smaller circle? The answer is not $\\frac{3}{4}$ . A bag has 26 tiles in it, each with a different letter of the alphabet. • ## A You pick one tile out of the bag, look at it, and write it down. Then you put it back in the bag, which is thoroughly mixed up. Then you pick another tile out of the back, look at it, and write it down. What is the probability that your first letter was “A” and your second letter was “T”? • ## B Same bag, different plan. This time you pick the first tile, but do not put it back in the bag. Then you pick a second tile and place it next to the first? Now what is the probability that your first letter was “A” and your second letter was “T”? • ## C In the second case, what is the probability that your two letters, together, could make the word “AT”? A deck of cards has 52 cards, 13 of each suit. Assume there are no Jokers. (Once again, no two parts of this question have exactly the same answer.) • ## A", "a radius of 4\". If a dart hits somewhere at random inside the larger circle, what is the probability that it will fall somewhere in the smaller circle? The answer is not $\\frac{3}{4}$ . A bag has 26 tiles in it, each with a different letter of the alphabet. • ## A You pick one tile out of the bag, look at it, and write it down. Then you put it back in the bag, which is thoroughly mixed up. Then you pick another tile out of the back, look at it, and write it down. What is the probability that your first letter was “A” and your second letter was “T”? • ## B Same bag, different plan. This time you pick the first tile, but do not put it back in the bag. Then you pick a second tile and place it next to the first? Now what is the probability that your first letter was “A” and your second letter was “T”? • ## C In the second case, what is the probability that your two letters, together, could make the word “AT”? A deck of cards has 52 cards, 13 of each suit. Assume there are no Jokers. (Once again, no two parts of this question have exactly the same answer.) • ## A", "and D2 first and second door and with C1, C2, C3 first, second and third curtain, then the possible outcomes are D1C1, D1C2, D1C3, D2C1, D2C2, D2C3 Thus there are six possible outcomes. 12.1 Sample Spaces and Probability Question 1. A bag contains 9 tiles. one for each letter in the word HAPPINESS. You choose a tile at random. What is the probability that you choose a tile with the letter S? What is the probability that you choose a tile with a letter other than P? Let X be a random variable that represent the letter on tile. We know that a bag contains tiles labeled with “H”, “A”, “P”, “I”, “N”, “E” and “S” So we can conclude that the possible values for X are letters “H”, “A”, “P”, “I”, “N”, “E” and “S” and total number of outcomes is 9. Question 2. You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points? Answer: It is the most likely to get 20 points. Explanation: From given board, we can conclude that the probability that we get 5 points is P(5 points) = Surface of red area/Surface of board = 4/36 = 1/9", "a letter other than P? Let X be a random variable that represent the letter on tile. We know that a bag contains tiles labeled with “H”, “A”, “P”, “I”, “N”, “E” and “S” So we can conclude that the possible values for X are letters “H”, “A”, “P”, “I”, “N”, “E” and “S” and total number of outcomes is 9. Question 2. You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points? Answer: It is the most likely to get 20 points. Explanation: From given board, we can conclude that the probability that we get 5 points is P(5 points) = Surface of red area/Surface of board = 4/36 = 1/9 On the other hand, we see that the probability we get 10 points is P(10 points) = Surface of yellow area/Surface of board = (16 – 4)/36 = 1/3 and the probability that we get 20 points is P(20 points) = Surface of blue area/Surface of board = (36 – 16)/36 = 5/9 From the obtained results we get that it is the most likely to get 20 points, which is logical because the surface of the blue area is the largest.", "A bag contains 26 tiles, each marked with a different letter of the alphabet. What is the probability of being able to spell the word math with four randomly selected tiles that are taken from the bag all at the same time? Jun 3, 2017 1/(\"\"_26C_4). Explanation: There are $26$ tiles in the bag. To spell the word MATH, w ways.e have toselect $4$ tiles from the bag. This can be done in \"\"_26 C _4 ways. Out of this, the word MATH can be correctly spelled in only $1$ way. Hence, the Prob. =1/(\"\"_26C_4). Jun 3, 2017 The probability is (4! 22!)/(26!), or about 0.00669%. Explanation: Drawing four tiles from the bag at one time is mathematically equivalent to drawing one tile 4 times, without replacement. On our first draw, we have to select either M, A, T, or H from the alphabet bag in order to have a chance at spelling \"MATH\". Thus, on draw 1, there are 4 draws out of the possible 26 that we will call a \"success\". If we get one of these 4 letters on draw 1, then on draw 2, we'll have 3 possible successful draws out of the 25 tiles left. This pattern continues down to draw 4. Since each draw is independent, we just multiply the probabilities together:", "a probability of $$\\frac{1}{4}$$? In a long sequence of observations, it would occur about 25% of the time. ### Eureka Math Grade 7 Module 5 Lesson 1 Problem Set Answer Key Question 1. Match each spinner below with the words impossible, unlikely, equally likely to occur or not occur, likely, and certain to describe the chance of the spinner landing on black. Question 2. Decide if each of the following events is impossible, unlikely, equally likely to occur or not occur, likely, or certain to occur. a. A vowel will be picked when a letter is randomly selected from the word lieu. b. A vowel will be picked when a letter is randomly selected from the word math. c. A blue cube will be drawn from a bag containing only five blue and five black cubes. d. A red cube will be drawn from a bag of 100 red cubes. e. A red cube will be drawn from a bag of 10 red and 90 blue cubes. a. Likely; most of the letters of the word lieu are vowels. b. Unlikely; most of the letters of the word math are not vowels. c. Equally likely to occur or not occur; the number of blue and black cubes in the bag is the same. d. Certain; the only cubes", "• # question_answer 22) Kriti puts the tiles shown below into an empty bag and mixed them up. What fraction of letters on tiles are vowels? A) $\\frac{4}{10}$ B) $\\frac{4}{9}$ C) $\\frac{5}{9}$ D) $\\frac{2}{9}$ Total number of letters in the word EDUCATION = 9 Number of vowels = 5 $\\therefore$Required fraction $=\\frac{5}{9}$", "# Probability-1 Tiles in a bag are numbered from $$\\left\\{ 0,1,2,3,4,\\ldots,98,99 \\right\\}.$$ Two tiles with the numbers $$a$$ and $$b$$ are randomly drawn from the bag with replacement. What is the probability that the number $$3^a+7^b$$ has a digit equal to $$8$$ at the unit place? For more problems try my set ×", "Math Help - Letters and envelopes 1. Letters and envelopes Suppose that you have 20 different letters and 10 distinctly addressed envelopes. The 20 letters consist of 10 pairs, where each pair belongs inside one of the 10 envelopes. Suppose that you place the 20 letters inside the 10 envelopes, two per envelope, but at random. What is the probability that exactly 3 of the 10 envelopes will contain both of the letters which they should contain?", "Math Help - Combinations/Probability...please check my work A) In a board game, there are tiles with letters on one side. There are 56 with consonants and 42 with vowels. A player chooses 7 tiles to start. What is the probability of choosing 4 consonants and 3 vowels? Is it ((56C4)(42C3))/(98C7)? B) You arrange five books (A, B, C, D, E) on a shelf. What is the probability that book E is last? I know intuitively it is 1/5, but I need to use combinations. How do I do so? There are $5!$ ways to arrange the five books. E will be the last book in $4!$ of those arrangements.", "are mutually inclusive, and we must account for the double counting. \\begin{align*}P(A \\ \\text{and} \\ B) &= \\frac{3}{36}\\\\ P(A \\cap B) &= \\frac{3}{36}\\\\ P(A \\cap B) &= \\frac{1}{12}\\\\ \\\\ P(A \\cup B) &= P(A) + P(B) - P(A \\cap B)\\\\ P(A \\cup B) &= \\frac{15}{36} + \\frac{3}{36}-\\frac{1}{12}\\\\ P(A \\cup B) &= \\frac{15}{36} + \\frac{3}{36}-\\frac{3}{36}\\\\ P(A \\cup B) &= \\frac{15}{36}\\\\ P(A \\cup B) &= \\frac{5}{12}\\\\\\end{align*} Points to Consider --> ### Guided Practice A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it? The first 10 letters of the alphabet are 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', and 'J'. Also, the vowels are 'A', 'E', 'I', 'O', and 'U'. The only letters that are both in the first 10 letters of the alphabet and vowels are 'A', 'E', and 'I'. Therefore, the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it can be calculated as follows: \\begin{align*}A &= \\{\\text{A,", "Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team In a word game, you choose a tile from a bag, replace it, and then choose another. If there are 28 vowels and 12 consonants, what is the probability you will choose a consonant and then a vowel? 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 9 Sep 2, 2016 $\\frac{21}{100}$ Explanation: The numbers of vowels and consonants stay the same for the second tile, because the first tile was replaced. The total number of tiles = 40. $P \\left(C , V\\right) = P \\left(C\\right) \\times P \\left(V\\right) \\text{ \", larr\" both must happen}$ = $\\frac{28}{40} \\times \\frac{12}{40} = \\frac{21}{100}$ • An hour ago • An hour ago • An hour ago • An hour ago • 4 minutes ago • 6 minutes ago • 36 minutes ago • 37 minutes ago • 49 minutes ago • An hour ago • An hour ago • An hour ago • An hour ago • An hour ago", "1 + 8 + 9 + 1 = 19 Therefore, the required probability = $$\\frac{19}{100}$$ Hence, the correct answer is option C. _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Re: If a bowl contains numbered tiles labeled with integers from 1 – 100, [#permalink] 18 Feb 2019, 19:24 Display posts from previous: Sort by", "his calculation. 4. Brenda did a survey of the students in her classes about whether they liked to get a candy bar or a new math pencil as their reward for positive behavior. She asked all 71 students she taught, and 32 said they would like a candy bar, 25 said they wanted a new pencil, and 4 said they wanted both. If Brenda were to select a student at random from her classes, what is the probability that the student chosen would want: 1. a candy bar or a pencil? 2. neither a candy bar nor a pencil? 5. A card is chosen at random from a standard deck of cards. What is the probability that the card chosen is a heart or a face card? Are these events mutually inclusive? 6. What is the probability of choosing a number from 1 to 10 that is greater than 5 or even? 7. A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the letters in the word ENGLISH on it or randomly choosing a tile with a vowel on it? 8. Are randomly choosing a teacher and randomly choosing a father mutually", "a graphic artist working for a company on a new design for the board in the game of darts. You are eager to begin the project, but the team cannot decide on the terms of the game. Everyone agrees that the board should have four colors. But some want the probabilities of hitting each color to be equal, while others want them to be different. You offer to design two boards, one for each group. How do you get started? How creative can you be with your designs? To explore the answers to these questions and more, go to BigIdeasMath.com. ### Probability Chapter Review 10.1 Sample Spaces and Probability (pp. 537–544) Question 1. A bag contains 9 tiles, one for each letter in the word HAPPINESS. You choose a tile at random. What is the probability that you choose a tile with the letter S? What is the probability that you choose a tile with a letter other than P? Question 2. You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points? 10.2 Independent and Dependent Events (pp. 545–552) Find the probability of randomly selecting the given marbles from a bag of 5 red,", "# Probability of drawing a given word from a bag of letters in Scrabble Suppose you had a bag with $n$ tiles, each with a letter on it. There are $n_A$ tiles with letter 'A', $n_B$ with 'B', and so on, and $n_*$ 'wildcard' tiles (we have $n = n_A + n_B + \\ldots + n_Z + n_*$). Suppose you had a dictionary with a finite number of words. You pick $k$ tiles from the bag without replacement. How would you compute (or estimate) the probability that you can form a given word, of length $l$ (with 1 < $l$ =< $k$) from the dictionary given the $k$ tiles selected? For those not familiar with Scrabble (TM), the wildcard character can be used to match any letter. Thus the word 'BOOT' could be 'spelled' with the tiles 'B', '*', 'O', 'T'. The order in which the letters are drawn does not matter. Suggestion: in order to simplify the writing of answers, it might be better to just answer the question: what is the probability of having the word 'BOOT' among your possible moves after drawing 7 letters from a fresh bag. (the problem's introduction has been copied from this similar question) • I'd advise tackling a simpler case first, such as one without wildcards. – Glen_b Nov 3", "of all ages, and they're not too difficult to learn. \" As a result, if the repeated letters in mathematics are switched, the word remains unchanged. A Letter is Chosen at Random from the Letters of the Word Association. Simple Numbers Chart 1-20. Support community. 01, Fall 2012: HW 1 Solutions Problem 1 (p. This convention is also known as camel-case. What is the probability of choosing either a T or an M? Plugin insertion failed: Plugin failed. 1 2 3 A committee of 6 is to be chosen from 14 candidates. Computational Sciences. smaller mean number of word errors describes this fact. asked by Avay on December 5, 2016; GEOMETRY. Each ball is marked with a letter and number as indicated in the following chart. 5 The number of white counters in the bag is the same as the number of black counters in the bag. Under GSView, click FILE - Convert - pdfwrite to create a PDF version of myFile01. A Single Letter is Selected at Random from the Word 'Probability'. A variable is the term used to record a particular characteristic of the population we are studying. Trump decides to post a random message on Facebook and he starts typing a random sequence of letters {Uk}k≥1 such that they are chosen independently and uniformly from", "# Probability As A Fraction (Problem Solving 23.2) In this probability activity, students solve 4 problems in which possible outcomes are written as fractions. Most problems pertain to random picking of tiles, letters or marbles from a bag." ]

[ "an earlier version of this message I had made a tiny mistake in preparing the dictionary and obtained 577940. I have edited the text to reflect what I hope are now the correct results!) A little less than the million or so I expected, but of the same order of magnitude. To compute the chance of obtaining such a rack, we need to account for the number of ways in which the rack can be drawn. As we saw in the example, this equals its coefficient in $\\alpha^7$. The chance of drawing some such rack is the sum of all these coefficients, easily found by setting all the letters equal to 1: nonwords /. (# -> 1) & /@ (List @@ alphabet) The answer equals 1066056120, giving a chance of 10.1914% of drawing a rack from which no valid word can be formed (if all letters are equally likely). When the probabilities of the letters vary, just replace each letter with its chance of being drawn: tiles = {9, 2, 2, 4, 12, 2, 3, 2, 9, 1, 1, 4, 2, 6, 8, 2, 1, 6, 4, 6, 4, 2, 2, 1, 2, 1, 2}; chances = tiles / (Plus @@ tiles); nonwords /. (Transpose[{List @@ alphabet, chances}] /. {a_, b_} -> a -> b) The output is", "2 ways to choose the word \"MATH\" (there are 2 T's), so the probability is $$\\color{brown}\\boxed{1 \\over 5000}$$ BuilderBoi Jul 27, 2022 #4 +12 0 There are $$10^4=10,000$$ ways to choose M, A, T, and H. There is only one way to choose M, A, and H and two ways to choose T in MATHCOUNTS. That means that there are $$1 \\cdot 1 \\cdot 1 \\cdot 2 = 2$$ ways to spell MATH. Then, the probability of getting MATH is $$\\dfrac{2}{10,000} = \\boxed{\\dfrac{1}{5,000}}.$$ And yes, HeWhoShallNotBeNamed (@Voldemort)s' answer was correct. Jul 28, 2022 edited by AngrySesame Jul 28, 2022", "\\cup B) &= \\frac{15}{36} + \\frac{3}{36}-\\frac{3}{36}\\\\ P(A \\cup B) &= \\frac{15}{36}\\\\P(A \\cup B) &= \\frac{5}{12}\\\\$ Points to Consider • What is the difference between the probabilities calculated with the Multiplication Rule versus the Addition Rule? ### Guided Practice A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it? The first 10 letters of the alphabet are 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', and 'J'. Also, the vowels are 'A', 'E', 'I', 'O', and 'U'. The only letters that are both in the first 10 letters of the alphabet and vowels are 'A', 'E', and 'I'. Therefore, the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it can be calculated as follows: $A &= \\{\\text{A, B, C, D, E, F, G, H, I, J}\\}\\\\P(A) &=\\text{probability of selecting one of the first 10 letters of the alphabet}\\\\P(A) &= \\frac{10}{26}\\\\P(A) &= \\frac{5}{13}\\\\B &= \\{\\text{A, E, I, O, U}\\}\\\\P(B) &=\\text{probability of", "## Geometry: Common Core (15th Edition) $\\frac{9}{11}$ To calculate probability, we need to divide the number of chances for a certain outcome to occur by the total number of outcomes. In this case, there are $2$ tiles with the letter \"P\" and $11$ tiles in total. This means that the other $11-2=9$ tiles must be \"not P.\" When we divide $9$ by $11$, we get $\\frac{9}{11}$, which is the probability of drawing a letter that is not \"P,\" or $P(\\text{Not P})$", "## Geometry: Common Core (15th Edition) $\\frac{7}{11}$ To calculate probability, we need to divide the number of chances for a certain outcome to occur by the total number of outcomes. In this case, there are $4$ tiles with the letter \"S\" and $11$ tiles in total. This means that the other $11-4=7$ tiles must be \"not S.\" When we divide $7$ by $11$, we get $\\frac{7}{11}$, which is the probability of drawing a letter that is not \"S,\" or $P(\\text{Not S})$.", "1 + 8 + 9 + 1 = 19 Therefore, the required probability = $$\\frac{19}{100}$$ Hence, the correct answer is option C. _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Re: If a bowl contains numbered tiles labeled with integers from 1 – 100, [#permalink] 18 Feb 2019, 19:24 Display posts from previous: Sort by", "Geometry: Common Core (15th Edition) $\\frac{10}{11}$ To calculate probability, we need to divide the number of chances for a certain outcome to occur by the total number of outcomes. In this case, there is $1$ tile with the letter \"M\", and $11$ tiles in total. This means that the other $11-1=10$ tiles must be \"not M\". When we divide $10$ by $11$, we get $\\frac{10}{11}$, which is the probability of drawing a letter that is not \"M\", or $P(\\text{Not M})$", "A bag contains 26 tiles, each marked with a different letter of the alphabet. What is the probability of being able to spell the word math with four randomly selected tiles that are taken from the bag all at the same time? Jun 3, 2017 1/(\"\"_26C_4). Explanation: There are $26$ tiles in the bag. To spell the word MATH, w ways.e have toselect $4$ tiles from the bag. This can be done in \"\"_26 C _4 ways. Out of this, the word MATH can be correctly spelled in only $1$ way. Hence, the Prob. =1/(\"\"_26C_4). Jun 3, 2017 The probability is (4! 22!)/(26!), or about 0.00669%. Explanation: Drawing four tiles from the bag at one time is mathematically equivalent to drawing one tile 4 times, without replacement. On our first draw, we have to select either M, A, T, or H from the alphabet bag in order to have a chance at spelling \"MATH\". Thus, on draw 1, there are 4 draws out of the possible 26 that we will call a \"success\". If we get one of these 4 letters on draw 1, then on draw 2, we'll have 3 possible successful draws out of the 25 tiles left. This pattern continues down to draw 4. Since each draw is independent, we just multiply the probabilities together:", "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", "letter S from the first word is $\\dfrac{3}{9}$. And, for finding the probability of taking letter S from the second word, STATISTICS, we have 3 S here and the total number of letters as 10. So, the probability of taking letter S from the second word is $\\dfrac{3}{10}$. So, we will get the probability that letter S is taken out from both the words, using the formula as, $P\\left( S \\right)=\\dfrac{3}{9}\\times \\dfrac{3}{10}=\\dfrac{9}{90}\\ldots \\ldots \\ldots \\left( ii \\right)$ Now we have to find the probability of taking a letter I from the first word, ASSISTANT. There is 1 I in this word and the total number of letters is 9. So, the probability of taking a letter I from the first word is $\\dfrac{1}{9}$. Now, for the probability of taking letter I from the second word, STATISTICS, we know that is 2 I and the total number of letters here are 10. So, the probability of taking a letter I from the second word is $\\dfrac{2}{10}$. So, we can find the probability that letter I am taken out from both the words, using the formula as, $P\\left( I \\right)=\\dfrac{1}{9}\\times \\dfrac{2}{10}=\\dfrac{2}{90}\\ldots \\ldots \\ldots \\left( iii \\right)$ And finally, we will find the probability of taking letter T from the first word, ASSISTANT. There are 2 T in this word and", "the total number of letters is 9. So, the probability of taking a letter T from the first word is $\\dfrac{2}{9}$. Now, for finding the probability of taking letter T from the second word, STATISTICS, we know that there are 3 T and the total number of letters are 10. So, the probability of taking a letter T from the second word is $\\dfrac{3}{10}$. So, we will find the probability that letter T is taken out from both the words, using the formula as, $P\\left( T \\right)=\\dfrac{2}{9}\\times \\dfrac{3}{10}=\\dfrac{6}{90}\\ldots \\ldots \\ldots \\left( iv \\right)$ Now, that we have found the probabilities of each of the common letters being taken from both the words, we can find the probability of the same letters being taken as, $P\\left( A \\right)+P\\left( S \\right)+P\\left( I \\right)+P\\left( T \\right)$ We have these values from equations (i), (ii), (iii) and (iv). So, we will get the probability as, $\\dfrac{2}{90}+\\dfrac{9}{90}+\\dfrac{2}{90}+\\dfrac{6}{90}=\\dfrac{19}{90}$ So, the probability of the same letters being taken is $\\dfrac{19}{90}$. Therefore, the correct answer is option C. Note: There are chances that the students might get confused with the formula and use $P\\left( A.B \\right)=P\\left( A \\right)+P\\left( B \\right)$ for finding the probability of any letter being taken out from both the words instead of $P\\left( A.B \\right)=P\\left( A \\right).P\\left( B \\right)$. But students should", "D #### In the game Scrabble, there are a total of 100 tiles. Of the 100 tiles, 42 tiles have the vowels A, E, I, O, and U printed on them, 56 tiles have the consonants printed on them, and 2 tiles are left blank. 5 viewed last edited 5 months ago Anonymous 0 ) 1)1a. If tiles are selected at random, what is the probability that the first tile drawn from the pile of 100 tiles is a vowel? b. If tiles drawn are not replaced, what is the probability that the first two tiles selected are both vowels? c. Event A is drawing a vowel, event B is drawing a consonant, and event C is drawing a blank tile.A1 means a vowel is drawn on the first selection, B2 means a consonant is drawn on the second selection, and C2 means a blank tile is drawn on the second selection. Tiles are selected at random and without replacement. 1) P( A1 and B2) 2) P( A1 and B2) 3) P(B1 and C2) Anuj Gupta 0 a. Since 42 titles have a vowel in them and there are a total of 100 tiles The probability that a vowel is drawn from the total number of tiles is \\frac{42}{100} = 0.42 . b. The probability that the first", "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?", "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 different number of tiles. We can compute the probability of starting with a valid word for a range of starting tile numbers: Drawing 3 letters has the highest probability of forming a word, at $53.7\\%$. This validates my strategy of dumping early in the game to get new tiles when I get stuck, because the new letters often help me get out of the rut. # Jeopardy! Survival Analysis Long streaks are rare in Jeopardy. Most winners only win one game, and slightly less than 40% win two games in a row. In fact, only 6 contestants have won more than ten games in a row. This Kaplan-Meier survival plot visualizes the survival function of Jeopardy winners: The data includes seasons 1-33 (aired 1984-2016), and does not include any championship or tournament games. The point on the extreme right is due, of course, to Ken Jennings and his 74 game winning streak. Here’s the R code for the figure. # Who was in", "that are common in both the words, ASSISTANT and STATISTICS are A, S, I and T. So, we will find the probability of taking out each of these common letters in both the words one by one, using the formula for the probability of any event A as, $P\\left( A \\right)=\\dfrac{\\text{ number of favourable outcomes of A}}{\\text{total number of outcomes}}$. We will find the probability of taking letter A from the first word, ASSISTANT. There are 2 As in this word and the total number of letters is 9. So, the probability of taking letter A from the first word is $\\dfrac{2}{9}$. Now, for finding the probability of taking letter A from the second word, STATISTICS, we know that is 1 A and the total number of letters here are 10. So, the probability of taking letter A from the second word is $\\dfrac{1}{10}$. So, now, we will find the probability that letter A is taken out from both the words, using the formula, $P\\left( {{A}_{1}}.{{B}_{1}} \\right)=P\\left( {{A}_{1}} \\right).P\\left( {{B}_{1}} \\right)$. So, we get, $P\\left( A \\right)=\\dfrac{2}{9}\\times \\dfrac{1}{10}=\\dfrac{2}{90}\\ldots \\ldots \\ldots \\left( i \\right)$ Now, we will find the probability of taking letter S from the first word, ASSISTANT. There are 3 S in this word and the total number of letters is 9. So, the probability of taking", "9 out of 10 words. 9/10. it is impossible that the handwritten word will be correctly interpreted. so option A is correct. Question 18. A bag contains 8 letter tiles of the same size. The tiles are labeled either A, B, C, D, E, or F. Three of the tiles are labeled C. If Corey selects 1 tile from the bag without looking, is the selection of letters fair? Explain. Answer: Yes, the selection of letters is fair. Explanation: In the above-given question, given that, A bag contains 8 letter tiles of the same size. The tiles are labeled either A, B, C, D, E, or F. Three of the tiles are labeled C. 1 out of 8. 1/8. so the selection of letters is fair. Lesson 7.2 Understand Theoretical Probability Solve & Discuss It! Betty and Carl will conduct an experiment. They will flip a coin 100 times and record the result of each flip. What should they expect the results of their experiment to be? Justify your answer. I can… determine the theoretical probability of an event. Answer: The theoretical probability of an event is 1/100, 2/100, ….. 100/100. Explanation: In the above-given question, given that, Betty and Carl will conduct an experiment. They will flip a coin 100 times and record the result of each", "there’s an equally likely chance of picking any given tile, the probability of picking only consonants is then: $$$P(A) = \\frac{54 \\choose 7}{100 \\choose 7} \\approx 0.011.$$$ In other words, there’s about a one-in-one-hundred chance of being unlucky and not being able to start the game. I’d say that this probably matches fairly well with my experience, though I do seem to have noticed this happen more as of recently. Perhaps I’m simply seeing events now that I’ve focused on them.", "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.", "a letter other than P? Let X be a random variable that represent the letter on tile. We know that a bag contains tiles labeled with “H”, “A”, “P”, “I”, “N”, “E” and “S” So we can conclude that the possible values for X are letters “H”, “A”, “P”, “I”, “N”, “E” and “S” and total number of outcomes is 9. Question 2. You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points? Answer: It is the most likely to get 20 points. Explanation: From given board, we can conclude that the probability that we get 5 points is P(5 points) = Surface of red area/Surface of board = 4/36 = 1/9 On the other hand, we see that the probability we get 10 points is P(10 points) = Surface of yellow area/Surface of board = (16 – 4)/36 = 1/3 and the probability that we get 20 points is P(20 points) = Surface of blue area/Surface of board = (36 – 16)/36 = 5/9 From the obtained results we get that it is the most likely to get 20 points, which is logical because the surface of the blue area is the largest.", "Math Help - Combinations/Probability...please check my work A) In a board game, there are tiles with letters on one side. There are 56 with consonants and 42 with vowels. A player chooses 7 tiles to start. What is the probability of choosing 4 consonants and 3 vowels? Is it ((56C4)(42C3))/(98C7)? B) You arrange five books (A, B, C, D, E) on a shelf. What is the probability that book E is last? I know intuitively it is 1/5, but I need to use combinations. How do I do so? There are $5!$ ways to arrange the five books. E will be the last book in $4!$ of those arrangements." ]

[ "\\cup B) &= \\frac{15}{36} + \\frac{3}{36}-\\frac{3}{36}\\\\ P(A \\cup B) &= \\frac{15}{36}\\\\P(A \\cup B) &= \\frac{5}{12}\\\\$ Points to Consider • What is the difference between the probabilities calculated with the Multiplication Rule versus the Addition Rule? ### Guided Practice A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it? The first 10 letters of the alphabet are 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', and 'J'. Also, the vowels are 'A', 'E', 'I', 'O', and 'U'. The only letters that are both in the first 10 letters of the alphabet and vowels are 'A', 'E', and 'I'. Therefore, the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it can be calculated as follows: $A &= \\{\\text{A, B, C, D, E, F, G, H, I, J}\\}\\\\P(A) &=\\text{probability of selecting one of the first 10 letters of the alphabet}\\\\P(A) &= \\frac{10}{26}\\\\P(A) &= \\frac{5}{13}\\\\B &= \\{\\text{A, E, I, O, U}\\}\\\\P(B) &=\\text{probability of", "A bag contains 26 tiles, each marked with a different letter of the alphabet. What is the probability of being able to spell the word math with four randomly selected tiles that are taken from the bag all at the same time? Jun 3, 2017 1/(\"\"_26C_4). Explanation: There are $26$ tiles in the bag. To spell the word MATH, w ways.e have toselect $4$ tiles from the bag. This can be done in \"\"_26 C _4 ways. Out of this, the word MATH can be correctly spelled in only $1$ way. Hence, the Prob. =1/(\"\"_26C_4). Jun 3, 2017 The probability is (4! 22!)/(26!), or about 0.00669%. Explanation: Drawing four tiles from the bag at one time is mathematically equivalent to drawing one tile 4 times, without replacement. On our first draw, we have to select either M, A, T, or H from the alphabet bag in order to have a chance at spelling \"MATH\". Thus, on draw 1, there are 4 draws out of the possible 26 that we will call a \"success\". If we get one of these 4 letters on draw 1, then on draw 2, we'll have 3 possible successful draws out of the 25 tiles left. This pattern continues down to draw 4. Since each draw is independent, we just multiply the probabilities together:", "9 out of 10 words. 9/10. it is impossible that the handwritten word will be correctly interpreted. so option A is correct. Question 18. A bag contains 8 letter tiles of the same size. The tiles are labeled either A, B, C, D, E, or F. Three of the tiles are labeled C. If Corey selects 1 tile from the bag without looking, is the selection of letters fair? Explain. Answer: Yes, the selection of letters is fair. Explanation: In the above-given question, given that, A bag contains 8 letter tiles of the same size. The tiles are labeled either A, B, C, D, E, or F. Three of the tiles are labeled C. 1 out of 8. 1/8. so the selection of letters is fair. Lesson 7.2 Understand Theoretical Probability Solve & Discuss It! Betty and Carl will conduct an experiment. They will flip a coin 100 times and record the result of each flip. What should they expect the results of their experiment to be? Justify your answer. I can… determine the theoretical probability of an event. Answer: The theoretical probability of an event is 1/100, 2/100, ….. 100/100. Explanation: In the above-given question, given that, Betty and Carl will conduct an experiment. They will flip a coin 100 times and record the result of each", "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", "and D2 first and second door and with C1, C2, C3 first, second and third curtain, then the possible outcomes are D1C1, D1C2, D1C3, D2C1, D2C2, D2C3 Thus there are six possible outcomes. 12.1 Sample Spaces and Probability Question 1. A bag contains 9 tiles. one for each letter in the word HAPPINESS. You choose a tile at random. What is the probability that you choose a tile with the letter S? What is the probability that you choose a tile with a letter other than P? Let X be a random variable that represent the letter on tile. We know that a bag contains tiles labeled with “H”, “A”, “P”, “I”, “N”, “E” and “S” So we can conclude that the possible values for X are letters “H”, “A”, “P”, “I”, “N”, “E” and “S” and total number of outcomes is 9. Question 2. You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points? Answer: It is the most likely to get 20 points. Explanation: From given board, we can conclude that the probability that we get 5 points is P(5 points) = Surface of red area/Surface of board = 4/36 = 1/9", "a radius of 4\". If a dart hits somewhere at random inside the larger circle, what is the probability that it will fall somewhere in the smaller circle? The answer is not $\\frac{3}{4}$ . A bag has 26 tiles in it, each with a different letter of the alphabet. • ## A You pick one tile out of the bag, look at it, and write it down. Then you put it back in the bag, which is thoroughly mixed up. Then you pick another tile out of the back, look at it, and write it down. What is the probability that your first letter was “A” and your second letter was “T”? • ## B Same bag, different plan. This time you pick the first tile, but do not put it back in the bag. Then you pick a second tile and place it next to the first? Now what is the probability that your first letter was “A” and your second letter was “T”? • ## C In the second case, what is the probability that your two letters, together, could make the word “AT”? A deck of cards has 52 cards, 13 of each suit. Assume there are no Jokers. (Once again, no two parts of this question have exactly the same answer.) • ## A", "a radius of 4\". If a dart hits somewhere at random inside the larger circle, what is the probability that it will fall somewhere in the smaller circle? The answer is not $\\frac{3}{4}$ . A bag has 26 tiles in it, each with a different letter of the alphabet. • ## A You pick one tile out of the bag, look at it, and write it down. Then you put it back in the bag, which is thoroughly mixed up. Then you pick another tile out of the back, look at it, and write it down. What is the probability that your first letter was “A” and your second letter was “T”? • ## B Same bag, different plan. This time you pick the first tile, but do not put it back in the bag. Then you pick a second tile and place it next to the first? Now what is the probability that your first letter was “A” and your second letter was “T”? • ## C In the second case, what is the probability that your two letters, together, could make the word “AT”? A deck of cards has 52 cards, 13 of each suit. Assume there are no Jokers. (Once again, no two parts of this question have exactly the same answer.) • ## A", "somewhere at random inside the larger circle, what is the probability that it will fall somewhere in the smaller circle? The answer is not $\\frac{3}{4}$ . A bag has 26 tiles in it, each with a different letter of the alphabet. • A You pick one tile out of the bag, look at it, and write it down. Then you put it back in the bag, which is thoroughly mixed up. Then you pick another tile out of the back, look at it, and write it down. What is the probability that your first letter was “A” and your second letter was “T”? • B Same bag, different plan. This time you pick the first tile, but do not put it back in the bag. Then you pick a second tile and place it next to the first? Now what is the probability that your first letter was “A” and your second letter was “T”? • C In the second case, what is the probability that your two letters, together, could make the word “AT”? A deck of cards has 52 cards, 13 of each suit. Assume there are no Jokers. (Once again, no two parts of this question have exactly the same answer.) • A If you draw a card at random, what is the probability of", "# Probability of drawing a given word from a bag of letters in Scrabble Suppose you had a bag with $n$ tiles, each with a letter on it. There are $n_A$ tiles with letter 'A', $n_B$ with 'B', and so on, and $n_*$ 'wildcard' tiles (we have $n = n_A + n_B + \\ldots + n_Z + n_*$). Suppose you had a dictionary with a finite number of words. You pick $k$ tiles from the bag without replacement. How would you compute (or estimate) the probability that you can form a given word, of length $l$ (with 1 < $l$ =< $k$) from the dictionary given the $k$ tiles selected? For those not familiar with Scrabble (TM), the wildcard character can be used to match any letter. Thus the word 'BOOT' could be 'spelled' with the tiles 'B', '*', 'O', 'T'. The order in which the letters are drawn does not matter. Suggestion: in order to simplify the writing of answers, it might be better to just answer the question: what is the probability of having the word 'BOOT' among your possible moves after drawing 7 letters from a fresh bag. (the problem's introduction has been copied from this similar question) • I'd advise tackling a simpler case first, such as one without wildcards. – Glen_b Nov 3", "a probability of $$\\frac{1}{4}$$? In a long sequence of observations, it would occur about 25% of the time. ### Eureka Math Grade 7 Module 5 Lesson 1 Problem Set Answer Key Question 1. Match each spinner below with the words impossible, unlikely, equally likely to occur or not occur, likely, and certain to describe the chance of the spinner landing on black. Question 2. Decide if each of the following events is impossible, unlikely, equally likely to occur or not occur, likely, or certain to occur. a. A vowel will be picked when a letter is randomly selected from the word lieu. b. A vowel will be picked when a letter is randomly selected from the word math. c. A blue cube will be drawn from a bag containing only five blue and five black cubes. d. A red cube will be drawn from a bag of 100 red cubes. e. A red cube will be drawn from a bag of 10 red and 90 blue cubes. a. Likely; most of the letters of the word lieu are vowels. b. Unlikely; most of the letters of the word math are not vowels. c. Equally likely to occur or not occur; the number of blue and black cubes in the bag is the same. d. Certain; the only cubes", "## Geometry: Common Core (15th Edition) $\\frac{9}{11}$ To calculate probability, we need to divide the number of chances for a certain outcome to occur by the total number of outcomes. In this case, there are $2$ tiles with the letter \"P\" and $11$ tiles in total. This means that the other $11-2=9$ tiles must be \"not P.\" When we divide $9$ by $11$, we get $\\frac{9}{11}$, which is the probability of drawing a letter that is not \"P,\" or $P(\\text{Not P})$", "are mutually inclusive, and we must account for the double counting. \\begin{align*}P(A \\ \\text{and} \\ B) &= \\frac{3}{36}\\\\ P(A \\cap B) &= \\frac{3}{36}\\\\ P(A \\cap B) &= \\frac{1}{12}\\\\ \\\\ P(A \\cup B) &= P(A) + P(B) - P(A \\cap B)\\\\ P(A \\cup B) &= \\frac{15}{36} + \\frac{3}{36}-\\frac{1}{12}\\\\ P(A \\cup B) &= \\frac{15}{36} + \\frac{3}{36}-\\frac{3}{36}\\\\ P(A \\cup B) &= \\frac{15}{36}\\\\ P(A \\cup B) &= \\frac{5}{12}\\\\\\end{align*} Points to Consider --> ### Guided Practice A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it? The first 10 letters of the alphabet are 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', and 'J'. Also, the vowels are 'A', 'E', 'I', 'O', and 'U'. The only letters that are both in the first 10 letters of the alphabet and vowels are 'A', 'E', and 'I'. Therefore, the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it can be calculated as follows: \\begin{align*}A &= \\{\\text{A,", "a letter other than P? Let X be a random variable that represent the letter on tile. We know that a bag contains tiles labeled with “H”, “A”, “P”, “I”, “N”, “E” and “S” So we can conclude that the possible values for X are letters “H”, “A”, “P”, “I”, “N”, “E” and “S” and total number of outcomes is 9. Question 2. You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points? Answer: It is the most likely to get 20 points. Explanation: From given board, we can conclude that the probability that we get 5 points is P(5 points) = Surface of red area/Surface of board = 4/36 = 1/9 On the other hand, we see that the probability we get 10 points is P(10 points) = Surface of yellow area/Surface of board = (16 – 4)/36 = 1/3 and the probability that we get 20 points is P(20 points) = Surface of blue area/Surface of board = (36 – 16)/36 = 5/9 From the obtained results we get that it is the most likely to get 20 points, which is logical because the surface of the blue area is the largest.", "an earlier version of this message I had made a tiny mistake in preparing the dictionary and obtained 577940. I have edited the text to reflect what I hope are now the correct results!) A little less than the million or so I expected, but of the same order of magnitude. To compute the chance of obtaining such a rack, we need to account for the number of ways in which the rack can be drawn. As we saw in the example, this equals its coefficient in $\\alpha^7$. The chance of drawing some such rack is the sum of all these coefficients, easily found by setting all the letters equal to 1: nonwords /. (# -> 1) & /@ (List @@ alphabet) The answer equals 1066056120, giving a chance of 10.1914% of drawing a rack from which no valid word can be formed (if all letters are equally likely). When the probabilities of the letters vary, just replace each letter with its chance of being drawn: tiles = {9, 2, 2, 4, 12, 2, 3, 2, 9, 1, 1, 4, 2, 6, 8, 2, 1, 6, 4, 6, 4, 2, 2, 1, 2, 1, 2}; chances = tiles / (Plus @@ tiles); nonwords /. (Transpose[{List @@ alphabet, chances}] /. {a_, b_} -> a -> b) The output is", "Geometry: Common Core (15th Edition) $\\frac{10}{11}$ To calculate probability, we need to divide the number of chances for a certain outcome to occur by the total number of outcomes. In this case, there is $1$ tile with the letter \"M\", and $11$ tiles in total. This means that the other $11-1=10$ tiles must be \"not M\". When we divide $10$ by $11$, we get $\\frac{10}{11}$, which is the probability of drawing a letter that is not \"M\", or $P(\\text{Not M})$", "## Geometry: Common Core (15th Edition) $\\frac{7}{11}$ To calculate probability, we need to divide the number of chances for a certain outcome to occur by the total number of outcomes. In this case, there are $4$ tiles with the letter \"S\" and $11$ tiles in total. This means that the other $11-4=7$ tiles must be \"not S.\" When we divide $7$ by $11$, we get $\\frac{7}{11}$, which is the probability of drawing a letter that is not \"S,\" or $P(\\text{Not S})$.", "1 + 8 + 9 + 1 = 19 Therefore, the required probability = $$\\frac{19}{100}$$ Hence, the correct answer is option C. _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Re: If a bowl contains numbered tiles labeled with integers from 1 – 100, [#permalink] 18 Feb 2019, 19:24 Display posts from previous: Sort by", "likely an upset will be. Finally, I’d also be interested in changing the goal slightly to predict winners in terms of Coryat score, which I’m sure would perform better since the uncertainty of daily doubles and the Final Jeopardy wager would be removed. Source code: The source code for this analysis is on Github: https://github.com/herbps10/jeopardy # Bananagrams Probabilities There was a fabled game of Bananagrams in which my Dad drew his initial 11 tiles, and immediately spelled: RASTAFARIAN If you haven’t played the game, it’s like a free-form version of Scrabble. You start by drawing a number of tiles (typically 11 or 21), and try to form a word or words out of them. The story made me wonder how likely it is to spell an 11 letter word on the first draw. The first step is to calculate the probability of drawing a particular word. Consider a bananagrams bag filled with only two letters, S for success and F for failure. Start pulling out tiles from the bag at random, without replacing each tile back in the bag after drawing it, and count how many S tiles you get. The hypergeometric distribution models the probability that you will get a certain number of S tiles for a given number of draws. The multivariate hypergeometric distribution extends this", "2 ways to choose the word \"MATH\" (there are 2 T's), so the probability is $$\\color{brown}\\boxed{1 \\over 5000}$$ BuilderBoi Jul 27, 2022 #4 +12 0 There are $$10^4=10,000$$ ways to choose M, A, T, and H. There is only one way to choose M, A, and H and two ways to choose T in MATHCOUNTS. That means that there are $$1 \\cdot 1 \\cdot 1 \\cdot 2 = 2$$ ways to spell MATH. Then, the probability of getting MATH is $$\\dfrac{2}{10,000} = \\boxed{\\dfrac{1}{5,000}}.$$ And yes, HeWhoShallNotBeNamed (@Voldemort)s' answer was correct. Jul 28, 2022 edited by AngrySesame Jul 28, 2022", "# How to compute likelihood of drawing specific set of letters from scrabble bag? Say I have a subset of the standard scrabble tiles $B$ that contains no blank tiles s.t. $|B|=n$ Also consider I have a target set of letters $L$ s.t. $|L|=k$. These are a set of letters not tiles, so there may be multiple subsets of tiles from $B$ that correspond to $L$. Say I draw $m$ tiles from $B$ (call it $D\\subset B$). What is the probability that $D$ contains tiles corresponding to every letter in $L$ at least once? Clearly the denominator for this probability will be ${{n}\\choose{m}}$, that is the number of possible combinations of tiles $D$ that can be drawn without replacement from $B$. What I'm missing is the numerator. That is, how many of these possible $D$s contain tiles corresponding to $L$. - Your model is a bit confusing. If $L \\subset B$ and $B$ are tiles, then $L$ must be tiles... – copper.hat Feb 28 '13 at 17:41 Good point, I'll try to clarify. – CodeFusionMobile Feb 28 '13 at 17:48 Since you ask what what the probability $D$ contains tiles corresponding to to every letter in $L$ at least once, I assume $B$ isn't a standard mathematical set, rather its a multiset. So what's the distribution of $B$," ]

[ "gives the remainder 1", "the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?", "remainder.", "divided by", "be a remainder of 1.", "many percent is 432 out of 434?", "7) = 49 + 4 = 53", "easy How many percent is 432 out of 434?", "48, can be divided by 3.", "98% being divided as before, so that the real proportion was 31.4 and 68.6.", "divided by 886. 765 Calculate the remainder when 3917 is divided by 1940. 37 What is the remainder when 34621 is divided by 32? 29 Calculate the remainder when 7886 is divided by 190. 96 Calculate the remainder when 7664 is divided by 272. 48 Calculate the remainder when 582588 is divided by 291290. 8 What is the remainder when 735 is divided by 691? 44 Calculate the remainder when 155314 is divided by 602. 600 Calculate the remainder when 1834713 is divided by 3253. 21 Calculate the remainder when 4453 is divided by 178. 3 What is the remainder when 58756 is divided by 136? 4 Calculate the remainder when 1521 is divided by 314. 265 Calculate the remainder when 466938 is divided by 491. 488 What is the remainder when 3629 is divided by 1204? 17 What is the remainder when 4685 is divided by 302? 155 What is the remainder when 330503 is divided by 18? 5 Calculate the remainder when 772 is divided by 350. 72 What is the remainder when 3984 is divided by 6? 0 What is the remainder when 1955 is divided by 299? 161 What is the remainder when 207934 is divided by 1118? 1104 What is the remainder when 22204 is divided by 426? 52 Calculate the remainder when", "the remainder when 1572 is divided by 249. 78 What is the remainder when 6014 is divided by 301? 295 Calculate the remainder when 30474 is divided by 1169. 80 What is the remainder when 49762 is divided by 52? 50 What is the remainder when 18270 is divided by 8943? 384 What is the remainder when 23345 is divided by 145? 0 What is the remainder when 1177 is divided by 146? 9 Calculate the remainder when 74855 is divided by 4399. 72 What is the remainder when 466530 is divided by 116630? 10 Calculate the remainder when 400331 is divided by 400312. 19 What is the remainder when 49735 is divided by 12? 7 Calculate the remainder when 458246 is divided by 668. 666 What is the remainder when 250554 is divided by 861? 3 What is the remainder when 50788 is divided by 384? 100 Calculate the remainder when 13555 is divided by 140. 115 What is the remainder when 87945 is divided by 862? 21 Calculate the remainder when 3952 is divided by 23. 19 Calculate the remainder when 9807 is divided by 175. 7 Calculate the remainder when 96915 is divided by 68. 15 Calculate the remainder when 8106 is divided by 271. 247 Calculate the remainder when 43148 is divided by 414.", "7, the remainder is 4. When p is divided by 4, th [#permalink] 11 Feb 2019, 22:07 Display posts from previous: Sort by", "because I knew that 4 goes into 24 and would know what the remainder is...", "remainder of both b and a, on division by m. They share the same remainder.", "the remainder when 8 21 Jul 2015, 03:11 54 What is the remainder when 43^86 is divided by 5? 34 21 Jun 2012, 16:59 1 What is the remainder when 7^381 is divided by 5 ? 5 06 Oct 2009, 01:30 Display posts from previous: Sort by", "by four. If odd, its square gives remainder 1 when divided by 4. -", "is the required remainder.", "4 ) 32 + 4 = 36 ( 9th multiple of 4 ) 36 + 4 = 40 ( 10th multiple of 4 ) ### Multiples of 4 by Multiplication We multiply four by the counting numbers to find the multiples of 4. We shall multiply four by one by two by three, and so on. By multiplying four by the first ten counting numbers, the table below displays the first ten multiples of 4. ### Multiples of 4 by Division Numbers that can be divided by four precisely without any remainder are said to be multiples of four. Hence, all multiples are divisible by 4. Let us have the following examples: ( a ) 16 ÷ 4 = 4; since 16 can be divided by 4 without a remainder, it is a multiple of 4. ( b ) 20 ÷ 4 = 4; since 20 can be divided by 4 without a remainder, it is a multiple of 4. ( c ) 36 ÷ 4 = 4; since 36 can be divided by 4 without a remainder, it is a multiple of 4. ( d ) 40 ÷ 4 = 4; since 40 can be divided by 4 without a remainder, it is a multiple of 4. ## Calculating Multiples of 4 ( nth Multiple of 4", "now. Since the remainder is 0, your long division is done. So, 2.4 ÷ 0.3 = 8." ]

[ "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "8). so any number leaves a remainder same as that of the last two digits when divided by 4, and of the last three digits when divided by 8. I will share divisibility test by 3,7,9,11,13 in my later posts.", "This makes $100a + 10b + c = 100a + 4k$. Notice that in the right hand side of the equation, $100a + 4k$, we can factor out $4$. That is, $100a + 4k = 4(25a + k)$. Now, the representation $4(24a + k$means that the given above number is divisible by 4. This confirms the rule that you just have to check the last two digit number of any number to see if it is divisible by 4.", "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", "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", "The only answer choice that is divisible by $4$ is $\\boxed{\\textbf{(B)}\\ 40}$.", "that the number has less than $6$ divisors. Since $12$ has $6$ divisors and $10$ has $4$ divisors, James’s desired sum is $\\boxed{10}$.", "# Find the greatest number of four digits which is exactly divisible by 15, 24 and 36. Question: Find the greatest number of four digits which is exactly divisible by 15, 24 and 36. Solution: Prime factorization: 15 = 3 × 5 $24=2^{3} \\times 3$ $36=2^{2} \\times 3^{2}$ LCM = product of greatest power of each prime factor involved in the numbers $=2^{3} \\times 3^{2} \\times 5=360$ Now, the greatest four digit number is 9999. On dividing 9999 by 360 we get 279 as remainder. Thus, 9999 − 279 = 9720 is exactly divisible by 360. Hence, the greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.", "each multiplied by the integer weight, descending from 10 to 1, is a multiple of the number 11. Modular arithmeticis convenient for calculating the check digit using modulus 11. Each of the first nine digits of the ten-digit ISBN — excluding the check digit, itself — is multiplied by a number in a sequence from 10 to 2, and the remainder of the sum, with respect to 11, is computed. The resulting remainder, plus the check digit, must equal 11; therefore, the check digit is 11 minus the remainder of the sum of the products. For example, the check digit for an ISBN-10 of 0-306-40615-?is calculated as follows:\\begin{align} s &= 0*10 + 3*9 + 0*8 + 6*7 + 4*6 + 0*5 + 6*4 + 1*3 + 5*2 \\\\ &= 0 + 27 + 0 + 42 + 24 + 0 + 24 + 3 + 10 \\\\ &= 130 \\\\ \\frac{130}{11} &= 11 + \\frac{9}{11} \\\\ 11 - 9 &= 2 \\end{align} Thus, the check digit is 2, and the complete sequence is ISBN 0-306-40615-2. Formally, the check digit calculation is: :x_{10} \\equiv 11 - (10x_1 + 9x_2 + 8x_3 + 7x_4 + 6x_5 + 5x_6 + 4x_7 + 3x_8 + 2x_9) \\, \\bmod\\; 11. If the result is 11, a '0' should be substituted; if 10, an", "quotient and remainder after dividing $n$ by 3. Since we've started numbering at 0, items divisible by 3 (I.e. 0,3,6,9,...) will always be at position 0 (=position A). So, to find out the position of our item, we just notice that item $3q$ is at position 0, and so ours is at position $r$.", "the bottom. It’s a good idea to remember your simple divisibility rules: • If a number ends in 2, 4, 6, 8 or 0, it’s a multiple of 2 • If a number ends in 5 or 0, it’s a multiple of 5; • If you add up the digits in a number and get a multiple of 3, the original number is a multiple of 3 (There are other rules for other factors, but they’re more involved and rarely necessary). One bonus tip: if you find an unusual factor (such as 19) of one part of the fraction, it’s worth checking if it’s a factor of the other part, too. It isn’t always, but it’s a fair bet. -- Uncle Colin * Edited 2015-09-26 to fix HTML", "2 or 4. Therefore, the unit place can be filled in 2 ways using the digits 2 or 4. In the remaining 5 digits (excluding the digit placed in the unit place 2 or 4) 1 occurs 2 times, 3 occurs 2 times. ∴ The number of ways of filling other places using the remaining 5 digit is = $$\\frac{5 !}{2 ! \\times 2 !}$$ ∴ Number of distinct 6 – digit numbers = $$\\frac{5 !}{2 ! \\times 2 !} \\times 2$$ = 5 × 4 × 3 = 60 (iii) How many of these 6 – digit numbers are divisible by 4? In order to get the 6 – digit number divisible by 4, the last two digits must be divisible by 4 ∴ The last two digits should be 12 or 24 or 32 Let the last box be filled with 24. The remaining 4 boxes can be filled with the remaining digits Let the last box be filled with 12. The remaining 4 boxes can be filled with the remaining digits Let the last box be filled with 32. The remaining 4 boxes can be filled with the remaining digits The total number of 6 digit numbers which are divisible by 4 is = 6 + 12 + 12 = 30 ∴ Required number of", "both three and four? Please help me to solve this. • Hint: $1+2+3+4+5+6=21$ which is divisible by $3$. Mar 14 '18 at 11:13 Since digits are not repeated, all $6$ digits must be used. Sum of all digits is $21$ $(1+2+3+4+5+6)$. So this number is divisible by $3$. Now we only have to check its divisibility by $4$. Number should end with any of $8$ combination $(12,16,24,32,36,52,56,64)$. Starting $4$ digits can be in any order. So total count $= 8*4! = 192$ You're almost there. As you say, we know it is divisible by $3$, no matter how we arrange the digits, so we just have to figure out if it is divisble by $4$, and you have a good atart there as well by pointing out that we just need to look at the last two digits. Now, clearly the last digit needs to be even. And, if it is a $2$ or a $6$, then we need an odd digit before it, so $3$ choices there, whereas a $4$ as the last digit needs an even digit before, so $2$ choices left for that one. And since the leftover $4$ digits can be places in $4!=24$ ways, you therefore have $(2\\cdot 3+2)\\cdot 24= 8 \\cdot 24= 192$ numbers divisible by $12$ By the way, a) can", "Browse Questions # Using the digits 1,2,3,4,5,6,7,a number of 4 different digit is formed.Find (c) How many numbers are exactly divisible by 25 $\\begin{array}{1 1}(A)\\;40\\\\(B)\\;25\\\\(C)\\;50\\\\(D)\\;75\\end{array}$ The no.divisible by 25 should end with 5 Thus the last two digit should be 25 or 75 for other 2 digit no. we have =$5P_2$ $\\Rightarrow \\large\\frac{5!}{3!}$ $\\Rightarrow \\large\\frac{5\\times 4\\times 3!}{3!}$ $\\Rightarrow 20$ Hence if the last two digit is 75 The no. of ways will be 20 Thus the total no. of way =20+20 $\\Rightarrow 40$ Hence (A) is the correct answer.", "a multiple of 4. Note: In this particular problem, the numbers which are multiple of 4 are the numbers which are exactly divisible by number 4 (i.e., the numbers which are when divided by 4 does not leave any remainder). Here, the possible cases include all the 15 given numbers because when a number is selected at random out of these 15 numbers, anyone of them can occur.", "divisors, such that the product of the divisors is 331776. ### Dozens ##### Stage: 3 Challenge Level: Do you know a quick way to check if a number is a multiple of two? How about three, four or six? ### Diggits ##### Stage: 3 Challenge Level: Can you find what the last two digits of the number $4^{1999}$ are?", "# What will be probability in this case? A five digit number is formed without repetition of the digits $1,2,3,4,5$ in a random order. What will be the probability that the number is divisible by $4$? I tried to solve and after a bit of calculations and observations, I came to know that the number will be divisible by $4$ if the last 2 digits of the number are $12, 24, 32$ or $52$. But it took me so much time to observe this trend. So my answer is as follows: Number of favorable cases $= 3! \\times 4 = 24$ Number of exhaustive cases $= 5! = 120$ So the probability $= 1/5$ My question is that, is there any other method of solving it quickly? • That seems like the best way to solve it. – T.J. Gaffney Jan 8 '17 at 1:39 ## 2 Answers The \"quick\" way to see it is to write the number $a_5a_4a_3a_2a_1$ (the digits) in its base $10$ representation: $$a_5a_4a_3a_2a_1 = a_1+10a_2+10^2a_3+10^3a_4+10^4a_5$$ Now, a number $b$ is divisible by $4$ if $b\\pmod{4} = 0$. Lets look at this number mod $4$: $$a_1\\pmod{4}+2a_2\\pmod{4}+0+0+0\\pmod{4} = 0$$ So, we need to have that: $$a_1 = 2a_2\\pmod{4}$$ If you look at the cases you have, they're precisely when this is true. • I couldn't", "# Check if a number is divisible by 41 or not • Last Updated : 04 Aug, 2022 Given a number, the task is to quickly check if the number is divisible by 41 or not. Examples: Input : x = 123 Output : Yes Input : 104413920565933 Output : YES A solution to the problem is to extract the last digit and subtract 4 times of the last digit from the remaining number and repeat this process until a two-digit number is obtained. If the obtained two-digit number is divisible by 41, then the given number is divisible by 41. Approach: • Extract the last digit of the number/truncated number every time • Subtract 4*(last digit of the previous number) from the truncated number • Repeat the above three steps as long as necessary. Illustrations: Illustration 1: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0 As the remainder is zero, 30873 is divisible by 41 Illustration 2: 104413920565933 --> 10441392056593 - 4*3= 10441392056581 10441392056581 --> 1044139205658 - 4*1 = 1044139205654 1044139205654 --> 104413920565 - 4*4 = 104413920549 104413920549 --> 10441392054 - 4*9 = 10441392018 10441392018 --> 1044139201 - 4*8 = 1044139169 1044139169 --> 104413916 - 4*9 = 104413880 104413880 --> 10441388 - 4*0 = 10441380 10441388 --> 1044138 - 4*8 = 1044106 1044106 --> 104410 - 4*6 = 104386 104386 --> 10438 - 4*6", "number is divisible by$$16$$. The factors of $$16$$ are $$1, 2, 4, 8, 16$$. Therefore, the number is also divisible by $$1, 2, 4, 8$$.", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021" ]

[ "${p}$ by 4 leaves the remainder 1. Show that there is an integer ${n}$ such that $n^2 + 1$ is divisible by ${p}$.", "the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?", "\"\"1122\" is deleted and becomes ““2”. You don't have to think of four or more. That is, there can be four, \"\"11\", \"1\", \"\", and ““2”. Whew, it was a pain. After all, the converted string is • \"\" • \"1\" • \"11\" • \"2\" • \"22\"` There are only five possibilities. Hashing leads to too much of these. that’s all. # Application The remainder divided by 9 is exactly the same. For the remainder divided by 2 or 5, you only have to look at the last digit (one’s place) (since 10 is divisible by 2 or 5). Look at the last two digits of the remainder divided by 4 (since 100 is divisible by 4). Look at the last 3 digits of the remainder divided by 8 (since 1000 is divisible by 8). The remainder divided by 6 should be rather complicated. I want to try it someday. I’m not sure if the remainder divided by 7 can be found without generating a numeric object. # Finally Even if he asks, “What then?” I added the tag “Number theory”, but it’s not that big, it’s not practical, and it’s not very interesting. I just tried to think about it. 1. modulo 3 is called “modulo 3” in formal mathematical terminology. Tags: Updated:", "8). so any number leaves a remainder same as that of the last two digits when divided by 4, and of the last three digits when divided by 8. I will share divisibility test by 3,7,9,11,13 in my later posts.", "1249, leaves remainder 9 and 7, this means are completely divisible by the number. Therefore, the required number = H.C.F. of 276 and 1242. By applying Euclid’s division lemma Therefore, H.C.F. = 138 Hence, required number is. Question 12: Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively. We need to find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively. The required number when divides 280 and 1245, leaves remainder 4 and 3, this means are completely divisible by the number. Therefore, the required number = H.C.F. of 276 and 1242. By applying Euclid’s division lemma Therefore, H.C.F. = 138. Hence, the required number is. Question 13: What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively. We need to find the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively. The required number when divides 626, 3127 and 15628 leaves remainders 1, 2 and 3 this means are completely divisible by the number. Therefore, the required number = H.C.F. of 625, 3125 and 15625. First we consider 625 and 3125. By applying Euclid’s division lemma H.C.F. of 625 and 3125 = 625 Now, consider 625", "of the three digits is 1 + 3 +4 = 8. 176/8 = 22, with no remainder", "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 $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "together, will produce $1234567890$. #### $64$ - Dividing by Eleven If the $9$ digits are written at haphazard in any order, for example $4 \\ 1 \\ 2 \\ 5 \\ 3 \\ 9 \\ 7 \\ 6 \\ 8$, what are the chances that the number that happens to be produced will be divisible by $11$ without remainder? The number I have written at random is not, I see, so divisible, but if I had happened to make the $1$ and the $8$ change places it would be. #### $65$ - Dividing by $37$ I want to know whether the number $49,129,308,213$ is exactly divisible by $37$, or if not, what is the remainder when so divided. How may I do this quite easily without any process of actual division whatever? #### $66$ - Another $37$ Division If the $9$ digits are written at haphazard in any order, for example $412539768$, what are the chances that the number that happens to be produced will be divisible by $37$ without remainder? #### $67$ - A Digital Difficulty Arrange the $10$ digits, $1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \\ 9 \\ 0$, in such order that they shall form a number that may be divided by every number from $2$", "the last digit is 6. Suppose it were 2^2157112678. What's the last digit?. 4 divides 2157112678 with 2 as a remainder. Therefore, the last digit is 4.", "\\times 121 = 363$ $5 \\times 11 = 55$ $5 \\times 121 = 605$ Finding the remainder by dividing each of the above possible divisors of the given number 3630 with 4, we get 1 3 1 3 1 3 1 1 3 1 Thus, the divisors of the given number are 1, 5, 121, 33, 363, 605, which leaves a remainder 1 when divided by 4. Therefore, there are only 6 possible divisors, which leave a remainder 1 when divided by 4. Hence, the option B will be correct. Note: In this question, we will use the prime factorization method by finding the prime numbers, which multiply together to make the original number. While solving these types of questions, some students take the prime factors of the given number as the divisors, which is wrong. We will find the remainders by dividing the divisors separately. Then find the number of those divisors, which leaves the remainder 1.", "a multiple of 4. Note: In this particular problem, the numbers which are multiple of 4 are the numbers which are exactly divisible by number 4 (i.e., the numbers which are when divided by 4 does not leave any remainder). Here, the possible cases include all the 15 given numbers because when a number is selected at random out of these 15 numbers, anyone of them can occur.", "but in any case any year that is divisible by 4 is also divisible by 400, so it's redundant at best. ##### Share on other sites Ok I have a large majority of it done. All I need help with now is to get the correct number for the day of the week. I need to do (day + getMonthValue + getCenturyValue + getYearValue)/7 but what I need is to isolate the remainder. So if all that equaled 4 + 6 + 10 + 6 = 26. 26/7 = 3 remainder 5. The problem is I don't know how to get that remainder for my day. ##### Share on other sites Again, the modulus operator(%) is used to calculate a remainder.", "This makes $100a + 10b + c = 100a + 4k$. Notice that in the right hand side of the equation, $100a + 4k$, we can factor out $4$. That is, $100a + 4k = 4(25a + k)$. Now, the representation $4(24a + k$means that the given above number is divisible by 4. This confirms the rule that you just have to check the last two digit number of any number to see if it is divisible by 4.", "Can a divisor of 6 have a remainder of 5? The number 6 is a factor 0f 12 and so one answer will be 5. And not all numbers will give the remainder 5, but some numbers will give a remainder of 11 when divided by 12. So the possible remainders are 5 or 11. What do you divide by 5 to get? A number is divisible by 5 if the number’s last digit is either 0 or 5. Divisibility by 5 – examples: The numbers 105, 275, 315, 420, 945, 760 can be divided by 5 evenly. The numbers 151, 246, 879, 1404 are not evenly divisible by 5. ## What is 5 divided by 3 in simplest form? 5 / 3 5 divided by 3 can be written simply as 5 / 3. What is a 5th of 20? Answer: 5% of 20 is 1. ### What is the decimal for 1 6? 0.16666 So, 1/6 as a decimal is 0.16666… This is a non-terminating repeating decimal number. How do you write a fraction in simplest form? Fractions in Simplest Form. There are two different ways to write a fractions in simplest form. 1) Divide by common factors. 2) Divide by using GCF. A fraction is in the lowest term when the only common factor between", "#### 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 is the remainder when $2^{164}$ is divided by 7?", "4 ) 32 + 4 = 36 ( 9th multiple of 4 ) 36 + 4 = 40 ( 10th multiple of 4 ) ### Multiples of 4 by Multiplication We multiply four by the counting numbers to find the multiples of 4. We shall multiply four by one by two by three, and so on. By multiplying four by the first ten counting numbers, the table below displays the first ten multiples of 4. ### Multiples of 4 by Division Numbers that can be divided by four precisely without any remainder are said to be multiples of four. Hence, all multiples are divisible by 4. Let us have the following examples: ( a ) 16 ÷ 4 = 4; since 16 can be divided by 4 without a remainder, it is a multiple of 4. ( b ) 20 ÷ 4 = 4; since 20 can be divided by 4 without a remainder, it is a multiple of 4. ( c ) 36 ÷ 4 = 4; since 36 can be divided by 4 without a remainder, it is a multiple of 4. ( d ) 40 ÷ 4 = 4; since 40 can be divided by 4 without a remainder, it is a multiple of 4. ## Calculating Multiples of 4 ( nth Multiple of 4", "# Probability that $2^a+3^b+5^c$ is divisible by 4 If $a,b,c\\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4. For a number to be divisible by $4$, the last two digits have to be divisible by $4$ $5^c= \\_~\\_25$ if $c>1$ $3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$ $2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=32$ Should I add all possibilities? Is there a simpler method? • nice problem.........+1. do you have such more problems? – Bhaskara-III Jan 13 '16 at 13:00 • That depends on the distribution of $a$, $b$ and $c$. There's distributions that make the probability anything (between $0$ and $1$ inclusively). – skyking Jan 13 '16 at 13:17 Let's consider the remainders when dividing by $4$. If $a=1$ then $2^a=2$ has the remainder $2$. Otherwise, $2^a$ has the remainder zero. If $b$ is odd then $3^b$ has remainder $3$, but if $b$ is even $3^b$ has remainder $1$. Whatever $c$ is, $5^c$ will have remainder $1$. (All of these statements are provable by induction, but the patterns are obvious.) Then there are just two basic ways for the sum to have remainder zero when divided by $4$: Case 1: $a=1$, $b$ is even, $c$ is anything. This gives remainders $2+1+1$ or zero. Case2: $a>1$, $b$ is odd, $c$ is anything. This gives remainders $0+3+1$ or zero. Now count each case and add the counts. • for", "and 6905, leaving the same remainder in each case. The sum of the digits in N is : a 4 b 5 c 6 d 8", "taking 3 numbers within {1,4} (since we don't want any of the number to be divisible by 5, i.e. $\\equiv 0 \\pmod 5$, to get a sum which is a multiple of 5 ? 3. Thanks, and while I do understand that, it is not covered in the book until the section or two after this problem appears, and so, our professor does not want us to use that yet. Is there another way? 4. You can sort of translate this argument to exclude explicit references to modular arithmetic. Claim: the remainder of the square of any natural number divided by 5 is 0, 1, or 4. Proof. It is enough to consider 0 through 9, because divisibility by 5 and the remainder is determined by the last digit, and the last digit of a square is determined by the last digit of the number. And so on. 5. ok...thanks", "than a multiple of 6 (i.e. we have 2^(6n+2), the remainder will be 4. And so on. 86 is 2 higher than 84, the closest multiple of 6. Therefore, the remainder will be 4. Takeaway: when you're dealing with big numbers and number properties, you can likely find a pattern. Just be sure to stay organized and calculate carefully! The remainder of (3^560)/8 is [#permalink] 24 Jun 2019, 14:29 Display posts from previous: Sort by" ]

[ "simplify", "Simplify", "simplify", "+ 4$ 4. Simplify And that's it", "simplify:", "Simplify.", "simplify.", "simplify to 3 : 2 : 7.", "simplify , where", "simplifies to. 01", "$(6)$: integrate $(7)$: simplify", "# Simplify (3u)^2 (3u)2 Apply the product rule to 3u. 32u2 Raise 3 to the power of 2. 9u2 Simplify (3u)^2", "simplify it a bit.", "# How do you simplify -4w+1+3+w? Jun 7, 2018 $- 4 w + w + 1 + 3 = - 3 w + 4$ That's the simplified version. #### Explanation: Just add the ones with the same bases. Here $w$ is a base. $- 4 w + w = - 3 w$ $1 + 3 = 4$ So $- 4 w + 1 + 3 + w = - 3 w + 4$", "for you to simplify.", "use to simplify ...", "+ 8$ expression, which is what you get when you simplify the other expressions. Well done Yanqing.", "4C2 + 4c3 + 4C4) simple.", "Simplify the equation.", "help just simplify it" ]

[ "\\ln 2 - e ^ {-4}}$$ $$\\boxed{x e ^ {2 y} - \\ln |y| = e ^ {-4} - \\ln 2}$$ Bogdan: observe that in virtue of initial condition we are interested in negative $y$.", "outside the bracket. 1800÷10(126)+(2412) 1800÷10{6+12} 180{18} =3240(ii) 12[{2(1+2)}10]< Step 1: Simplify the terms inside () followed by {}, then []. Step 2: Operate terms with the terms outside the bracket. Condition Rule Open the bracket and add the terms. Open the bracket and multiply the negative sign with each term inside the bracket.(All positive terms will be negative and vice-versa) Multiply term outside with each term inside the bracket", "{50}{\\sqrt {300}} = \\frac {5}{\\sqrt {3}}$ The negative sign just gives direction so it isn't really important.", "# How do you multiply -(3x + 7) + 2? Mar 18, 2018 When we have a negative sign in front of parenthesis like we do here, we can just assume that this is a -1. -(3x+7) + 2... Think of it like: -1(3x+7) + 2. Apply the distribution property. -1(3x+7) + 2. -3x - 7 + 2 Combine like terms. -3x - 5", "the answer when you put them together is -2 which is in the 3t^2+18t+9... i dont get that...does it make sense? and no i DID copy it right thats why im so confused too.. 12. How did the negative sign appear randomly? $27t^{2}{\\color{red}+}18t + 9 = 9\\left(3t^{2} {\\color{red}+} 2t + 1\\right)$ which can't be factored.", "raphealreyes5 28 # 3m≥-3= help me on this problem solve for m please Since we are multiplying 3 by m, divide by 3 on both sides. $\\frac{3m}{3} \\geq \\frac{-3}{3}$. Negative divided by positive is negative, so your answer is $m \\leq -1$. We flipped the sign because when multiplying or dividing by a negative number, we flip the sign.", "# Math Help - Simple Algebra Problem Multiplying 1. ## Simple Algebra Problem Multiplying I seem to have forgotten how to do algebra. I have tried for a while but I'm not sure how to do this. Simple equation where you have to multiply out the following expression 3k – (2m + 3n – 5k) I have the solution I just need to know how to get it. 2. Basically, the negative sign outside the brackets implies that it affects each term in the brackets. 3k – (2m + 3n – 5k) becomes 3k - 2m - 3n - (-5k) 3k - 2m - 3n + 5k 3k + 5K - 3n -2m 8k -3n - 2m Its as though you had just the expression -(2m + 3n – 5k), the negative representing a -1. You would simply expand out the brackets, yes? to make it -2m-3n+5k. Does this make sense? 3. Originally Posted by sberxa Basically, the negative sign outside the brackets implies that it affects each term in the brackets. Its as though you had just the expression -(2m + 3n – 5k), the negative representing a -1. You would simply expand out the brackets, yes? to make it -2m-3n+5k. Does this make sense? Yes, makes perfect sense. Thanks.", "How do you simplify r^3*r^-2? Apr 17, 2018 r Explanation: ${r}^{3} \\cdot {r}^{-} 2$ A power of negative means the reciprocal so ${r}^{3} / {r}^{2}$ The r squared will cancel out with the cube =r", "or a negative number number 2 3 is equal to 3 decay...", "third term is non-negative (the first two factors are non-positive and the last two non-negative). The sum of the last two terms is: $(a_4 - a_5){(a_1 - a_5)(a_2 - a_5)(a_3 - a_5) - (a_1 - a_4)(a_2 - a_4)(a_3 - a_4)} \\ge 0$. Hence $E_5 \\ge 0$. This solution was posted and copyrighted by e.lopes. The original thread can be fond here: [1]", "# How do you combine (4d-2) - (d+1)? Apr 12, 2018 $3 d - 3$ #### Explanation: You distribute the negative, then combine the like terms. The \"minus\" outside the second parentheses is really a negative sign that applies to BOTH terms. So it would become: $4 d - 2 - d - 1$ Combine like terms, making sure to carry the sign in front of each number or variable. $4 d - d \\text{ }$ and $\\text{ } - 2 - 1$ so $3 d - 3$ $3 d$ and $3$ are two separate things, so you cannot go any farther. ( do not subtract the $3$ from the $3 d$!)", "both sides by $6$ to isolate $n$. Recall that since we divided by a negative, the sign flips. We're left with $n > - 3$ Hope this helps!", "binds more tightly than -, which is why you get the negative of two squared, not the square of negative two (which would be positive four). What if you did want the square of negative two? Obviously, you’d add parentheses: >>> (-2)**2 4 However, it’s not reasonable to expect the following to give 4 >>> -(2)**2 -4 because -(2) is the same as -2. The parentheses do absolutely nothing. not(True) is exactly the same.", "in the other post, as well). Remember though, where this comes from. We are really multiplying ((3)(3)(3)(3)...(3)(3)(3)(3)) * ((1/3)(1/3)(1/3)(1/3)...(1/3)(1/3)(1/3)(1/3)). Because 1) both fractions have the same base and 2) the negative and positive powers have equal magnitudes (|12|=|-12|). Because there are as many negative exponents there are positive, there are as many 3's as there are (1/3)'s. So, each of the 1/3 's cancels out with each of the 3's, leaving only 1's. And the product of 1's is always 1. we have $3^{x}\\cdot 3^{-x}=3^{x-x}=3^0=1$ or $\\frac{3^x}{3^{x}}=1$", "### Home > INT1 > Chapter A > Lesson A.1.7 > ProblemA-71 A-71. Solve the following equations. Record your work and check your answers. 1. The dark boxes are positive, and the white boxes are negative. Watch out for subtraction errors when completing this problem. 1. The dark boxes are positive, and the white boxes are negative. Watch out for subtraction errors when completing this problem. 1. $3c−7=−c+1$ Assume that $c = x$. $c = 2$ 1. $−2+3x=2x+6+x$ no solution", "-3 |x + 3| = -(x + 3) when x < -3 Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (X-K) format we manipulate it by taking -tive sign out but I guess in this example its this concept that we need to understand |x|= x when x is >= 0, |x|= -x when x < 0 ok, so based on this understanding I will take a fresh shot, please let me know what's wrong Quote: a) $$x < -8$$. $$-(x+3) - (4-x) = -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8) In this test case, since we will always have |x+3| negative we put a -tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (-1), is this understanding correct? and since we are ok with -(4-x), because we will again get |4-x| positive with a negative x, the -tive sign outside the bracket will make sure its always -tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to", "of those four terms: $(2x+1)(x^2-2)=2x^3+x^2-4x-2$ # Example 1 Simplify $(-x + 2)(-3x - 2)$. This question is a little trick since there are a lot of negative signs floating around. Pay special attention to how I keep track of them as we go along. First, I'm going to distribute the $(-x + 2)$ into the second set of parentheses. $-3x(-x + 2) - 2(-x + 2)$ I went ahead and kept the terms of the second parentheses in front since that's a more natural way of writing the expressions and helps to keep the negative signs in the right place. Now I'm going to distribute the -3x and the -2 into the parentheses that follow them: $-3x(-x) + 2(-3x) - 2(-x) - 2(2)$ Notice how I kept the negative signs with the terms as I distributed them. Now I'll simplify the individual terms by combining the x's and multiplying out the numbers. $3x^2 - 6x + 2x - 4$ Be sure you kept track of what happened to the negative signs. A number of them became positive since we had two negatives being multiplied together. Finally, I'll combine the like terms to get our final answer. $3x^2 - 4x - 4$ # Example 2 Simplify $(-x^2 - x + 2)(-3x^4 + 3x^3 - 2)$. This one is complicated", "also. We now rewrite the equation in its transformed form as below: -3/(x + 5) - 3/(x + 2) = -3/(x + 4) - 3/(x + 3) We now verify that the sum of the denominators on both sides of the equation is 2x + 7, therefore, the sutra is applicable. We set 2x + 7 = 0, giving us the solution x = -3.5. Consider the next equation as below: (2x + 3)/(x + 3) + (x + 4)/(2x + 2) = (3x)/(2x - 2) + (x - 4)/(x - 1) Applying the first test, we find that 2/1 + 1/2 = 3/2 + 1/1. So, the equation passes the first test. Now, apply the second test. We can verify that |3 - (2/1)*3| = |4 - (1/2)*2| = |0 - (3/2)*-2| = |-4 - (1/1)*(-1)|. So, the equation passes the second test. We also notice that the terms in the second test before taking their absolute values are -3, 3, 3 and -3. Since there is one positive and one negative term on each side of the equation, it passes the third test. So, we rewrite the equation in its transformed state as below: -3/(x + 3) + 3/(2x + 2) = 3/(2x -2) - 3/(x - 1) Transposing the negative terms to the opposite sides", "# How do you simplify 7 - (-3)? $7 - \\left(- 3\\right)$ If we subtract a negative number, we can rewrite the sign as $+$ $7 + 3$ $10$", "Free Version Difficult # Predict Sign for Expression with Three Quotients ALGEBR-ZE4XGK The variables in the expression all represent negative numbers. What is the sign of the simplified expression? $$\\frac{a}{b}+\\frac{d}{e}+\\frac{f}{g}$$ A zero B cannot be determined C negative D positive" ]

[ "How do you simplify 3r+7-3r? Mar 3, 2018 7 Explanation: $3 r + 7 - 3 r$ There are two 3r in the equation. One has a positive sign and other has a negative sign. So , you can just cancel them out. $\\implies \\cancel{3 r} + 7 - \\cancel{3 r}$ You are left with $\\implies 7$", "# How do you simplify 3r+7-3r? Mar 3, 2018 7 #### Explanation: $3 r + 7 - 3 r$ There are two 3r in the equation. One has a positive sign and other has a negative sign. So , you can just cancel them out. $\\implies \\cancel{3 r} + 7 - \\cancel{3 r}$ You are left with $\\implies 7$", "## Elementary Technical Mathematics $$2y^3+4y^2+5y-11.$$ We simplify by distributing the negative sign and then combine similar terms as follows: $$(4y^3+3y-5)-(2y^3- 4y^2 - 2y + 6)\\\\ =2y^3+4y^2+5y-11.$$", "$$d^7 = (d+ad)^3a^3d^3$$ Simplifying gives $d=a^3(1+a)^3$. So we can choose $a$ to be any positive integer we like, and then define $d=a^3(1+a)^3$. The result is the same as your answer. This also proves that it is the only possible answer.", "How do you simplify r^3*r^-2? Apr 17, 2018 r Explanation: ${r}^{3} \\cdot {r}^{-} 2$ A power of negative means the reciprocal so ${r}^{3} / {r}^{2}$ The r squared will cancel out with the cube =r", "positive is negative. $$(7x-4)(3x+1)=7x\\times3x + 7x\\times1-4\\times 3x -4\\times1$$ $$(7x-4)(3x+1)=21x^2+7x-12x -4$$ Step 2: Simplify (Collect the like terms) $$(7x-4)(3x+1)=21x^2-5x -4$$", "# How do you simplify 7 - (-3)? $7 - \\left(- 3\\right)$ If we subtract a negative number, we can rewrite the sign as $+$ $7 + 3$ $10$", "simpler than x2x3. 3. Negative exponents should always be converted to positive ones. E.g. 1 / x is simpler than $x^{-1}$. 4. Radicals should be reduced as much as possible. E.g. $4\\sqrt{3}$ is simpler than $\\sqrt{48}$ 5. Denominators should be rationalized. E.g. $\\sqrt{3} / 3$ is simpler than $1 / \\sqrt{3}$. # Example 1 Simplify 2(4x + 3). The Distributive Property tells me that I can simplify this by taking the 2 outside the parentheses and \"distributing\" it to both of the terms inside the parentheses. 2(4x + 3) = 2·4x + 2·3 = 8x + 6 # Example 2 Simplify -2(x3 - 3x2). This one is a little trickier because of the negative signs. I'm going to distribute the -2 just like I distributed the +2 in the previous example. -2(x3 - 3x2) = -2·x3 - (-2)·3x2 Notice that I didn't do anything with the minus sign inside the parentheses. At this point, all I'm trying to do is distribute the -2. We'll worry about the minus in a later step. Now I'm going to multiply out all the numbers. In this case, there's only one pair - the -2 and the 3 on the last term. -2(x3 - 3x2) = -2x3 - (-6x2) Now, I'm going to straighten out the minus sign. Remember that a negative", "# How do you simplify r^3*r^-2? Apr 17, 2018 r #### Explanation: ${r}^{3} \\cdot {r}^{-} 2$ A power of negative means the reciprocal so ${r}^{3} / {r}^{2}$ The r squared will cancel out with the cube =r", "3\\right)$ is just $- 3$ because (plus) a (negative)something is (negative)something: +(-something) = -something and the $- \\left(- 1\\right)$ is just $+ 1$ because (minus) a (negative)something is (positive)something: -(-something) = +something 4.) So, now we have: $3 - 3 + 1$ $3 - 3$ is $0$ and we are then left with $1$ So, $1$ is our answer (our simplified junk from above)", "# How do you simplify -36 - ( -24) ? Jul 9, 2018 $- 12$ #### Explanation: We know that two negative signs next to each other is equivalent to a positive sign. Therefore, this can be rewritten as: $- 36 + 24$ Which simplifies to: $= - 12$ Hope this helps! Jul 9, 2018 $- 12$ #### Explanation: Subtracting a negative is the same thing as adding a positive. With this in mind, we can rewrite our expression as follows: $- 36 + 24$, which is also equal to $24 - 36$, which simplifies to $- 12$. Hope this helps!", "want to factor the negative, you have to remember that you are taking out a factor of $-1$. So that means $3(-2x^2 + 5x + 12) = -1\\cdot 3(2x^2 - 5x - 12)$ $= -3(2x^2 - 5x - 12)$. Do you see how when you take out a negative factor, all of the signs change?", "# How do you simplify -3--7? Mar 8, 2018 4 #### Explanation: When a problem is set up that way $\\left(- 3 - - 7\\right)$, the two negatives (in between -3 and -7) turn into a positive. So the problem would be written as $- 3 + 7$ Which equals 4. You can also solve the problem on a number line", "How do you simplify -(-8n-6)? May 13, 2018 $8 n + 6$ Explanation: First, we know that two negatives multiplied by each other become a positive sign. Therefore, $- \\left(- 8 n - 6\\right)$ becomes $8 n + 6$. Hope this helps!", "we get $c=-3,b=3,a=0$ and our function $f(x)$ becomes $3x-3$.if $f(x)=3x-3$, $\\boxed{f(10)=3*10-3=27}$", "# How do you simplify -7 + -4 + -3? Jan 13, 2016 $- 14$ #### Explanation: Rewrite this expression as follows : = -7 + (- 4 ) + ( - 3 ). Then adding a negative number is equivalent to subtracting the number, So +(- 4 ) is the same as - 4 (subtract 4 ) and the expression becomes: - 7 - 4 - 3 = - 14", "of those four terms: $(2x+1)(x^2-2)=2x^3+x^2-4x-2$ # Example 1 Simplify $(-x + 2)(-3x - 2)$. This question is a little trick since there are a lot of negative signs floating around. Pay special attention to how I keep track of them as we go along. First, I'm going to distribute the $(-x + 2)$ into the second set of parentheses. $-3x(-x + 2) - 2(-x + 2)$ I went ahead and kept the terms of the second parentheses in front since that's a more natural way of writing the expressions and helps to keep the negative signs in the right place. Now I'm going to distribute the -3x and the -2 into the parentheses that follow them: $-3x(-x) + 2(-3x) - 2(-x) - 2(2)$ Notice how I kept the negative signs with the terms as I distributed them. Now I'll simplify the individual terms by combining the x's and multiplying out the numbers. $3x^2 - 6x + 2x - 4$ Be sure you kept track of what happened to the negative signs. A number of them became positive since we had two negatives being multiplied together. Finally, I'll combine the like terms to get our final answer. $3x^2 - 4x - 4$ # Example 2 Simplify $(-x^2 - x + 2)(-3x^4 + 3x^3 - 2)$. This one is complicated", "# How do you simplify -(d-8n+4x)? Jul 7, 2015 $- \\left(d - 8 n + 4 x\\right) = - d + 8 n - 4 x$ #### Explanation: The external negative sign reverses the signs of the terms inside the parentheses. There is no way to combine the 3 terms involving 3 different variables.", "## anonymous one year ago What is the simplified expression for 3 to the power of negative 4 multiplied by 2 to the power of 3 multiplied by 3 to the power of 2 whole over 2 to the power of 4 multiplied by 3 to the power of negative 3? • This Question is Open 1. anonymous 2. misty1212 HI!! 3. misty1212 impossible to read, use the equation tool 4. whpalmer4 $\\frac{3^{-4}*2^3*3^2}{2^4*3^{-3}}$ is that what you are trying to simplify? 5. whpalmer4 I would use the following properties of exponents to simplify this: $b^m*b^n = b^{m+n}$ $\\frac{b^m}{b^n} = b^{m-n}$ $b^{-m} = \\frac{1}{b^m}$or$b^m=\\frac{1}{b^{-m}}$ Don't forget that $$b^1 = b$$", "other so we will get, =3z + 4z – 3z, =7z-3z, = 4z. Now we will add the constants, = 8+6 +4, = 18. Now we can rewrite the given equation as 3x + 5y + 4z +18, As you can see by applying simple rules we minimized our expression, we can't minimize the expression any more, we can minimize it only if we are having the value of variables. In this way we can simplify our expression and we will get the desired result." ]

[ "has been painted with an immensely detailed scene of a Quirmian street market, every shopper and every stall picked out with loving care. From a mouse peeping from a tall lady's curled and powdered wig to a bulb of smoked garlic rolling away across the cobbles from its stall, nothing has escaped the artist's hand. none &&&&&&+675 9/12 &&&&&&+900 1 &&&&&&+900 1 &&&&&+1800 2 &&&&&+9000 10 &&&&&+1800 2 &&&&&+9000 10 &&&&&&+180 0.2 [48] small pink writing box Just the right size for a pen and a few sheets of paper, with a neat compartment for an inkwell in one corner, this delicate, pretty pink box is covered all across its lid and sides with intertwined roses, ribbons, and ballet slippers. none &&&&&&+675 9/12 &&&&&&+900 1 &&&&&&+900 1 &&&&&+1800 2 &&&&&+9000 10 &&&&&+1800 2 &&&&&+9000 10 &&&&&&+180 0.2 [49] bluebell writing box Just the right size for a pen and a few sheets of paper, with a neat compartment for an inkwell in one corner, this delicate, pretty cream-coloured box is adorned with dozens of tiny bluebells. none &&&&&&+675 9/12 &&&&&&+900 1 &&&&&&+900 1 &&&&&+1800 2 &&&&&+9000 10 &&&&&+1800 2 &&&&&+9000 10 &&&&&&+180 0.2 [50] elegant white writing box Just the right size for a pen and a few sheets of paper, with a neat compartment for an inkwell", "stationery store had [#permalink] 22 Jun 2013, 07:45 Expert's post Folaa3 wrote: At the beginning of the last month, a stationery store had on stock 250 writing pads which had cost the store $0.75 each. During the same month, the store made only one purchase of writing pads. What was the total amount spent by the store on the writing pads it had in stock at the end of last month? (1) Last month, the store purchased 150 writing pads for$0.80 each. (2) Last month the total revenue from the sale of writing pad was \\$180. OPEN DISCUSSION OF THIS QUESTION IS HERE: at-the-beginning-of-last-month-a-stationery-store-had-in-85710.html _________________ Re: At the beginning of the last month, a stationery store had [#permalink] 22 Jun 2013, 07:45 Similar topics Replies Last post Similar Topics: 2 At the beginning of last month, a stationery store had in st 3 19 Jun 2014, 11:42 8 All the clients that Company X had at the beginning of last 4 23 Feb 2012, 12:20 2 At the beginning of last year, Tony's Auto Dealership had 7 16 Dec 2010, 10:48 At the beginning of the last year, a car dealership had 150 8 30 Apr 2010, 09:44 2 At the beginning of last month, a stationery store had in 11 24 Oct 2009, 03:39 Display posts from", "box Just the right size for a pen and a few sheets of paper, with a neat compartment for an inkwell in one corner, this sturdy wooden box has been painted with an ice fair on the Streine at midwinter, with stalls selling hot cider and roasted chestnuts set up along the banks. In the foreground, three young women skate past, the flush of exercise in their cheeks matching their cheerful fur-trimmed coats and tippets. none &&&&&&+675 9/12 &&&&&&+900 1 &&&&&&+900 1 &&&&&+1800 2 &&&&&+9000 10 &&&&&+1800 2 &&&&&+9000 10 &&&&&&+180 0.2 [46] hand-painted Quirmian writing box Just the right size for a pen and a few sheets of paper, with a neat compartment for an inkwell in one corner, this sturdy wooden box has been painted with a summer meadow strewn with wildflowers. A large, prosperous-looking family sit picnicking around a blue and white cloth covered with all manner of good things, from fresh country bread to a whole ham and a dozen bottles of strong red wine. none &&&&&&+675 9/12 &&&&&&+900 1 &&&&&&+900 1 &&&&&+1800 2 &&&&&+9000 10 &&&&&+1800 2 &&&&&+9000 10 &&&&&&+180 0.2 [47] hand-painted Quirmian writing box Just the right size for a pen and a few sheets of paper, with a neat compartment for an inkwell in one corner, this sturdy wooden box", "skill. You can also \"paint\" words on blank paper with this brush (which does not require ink).[3] souvenir fountain pen A black fountain pen with a silver nib. Written in silver filigree down the side are the words \"The Royal Art Museum - Ankh-Morpork\". It is clean and dry. || [36] || <colour> ink || &&&&&&+200 2/9 || Needs ink. small neat quill Short, slim, and an unpretentious grey, this plain goose quill seems ideal for the kind of person who cares about neatness, elegance, respect, and having everything in its proper place. [37] fussy green ink &&&&&&+100 1/9 stormy petrel quill [38] <colour> ink &&&&&&+100 1/9 Needs ink. sturdy purple plume Thicker and sturdier than a normal feather, this plume looks as though it could quite possibly double as a weapon in an emergency. Such as, for instance, when someone else has just grabbed the last copy of a textbook you wanted. The lower third of the shaft has been trimmed smooth for a good grip, and a small brass reservoir sits just above the gold nib. [39] rich purple ink &&&&&&+300 3/9 swan's wing plume Long and gleaming white, with a slight but very graceful curve, this feather could have come from no other bird. The lower third of the shaft has been trimmed smooth for", "and filled with ink. You could probably write on things with it. [3] rusty black ink barbed quill Wicked-looking barbs jut from the shaft of this long quill, placed just right to pierce the user's hand. The nib is wickedly sharp, and the three inches of feather remaining at the top are a bright, bloody scarlet. [4] blood Does a small amount of damage when held; uses faith.items.scroll to hold. bone stylus Carved out of bone, the stylus has a pointy end for writing and an ornamental hippopotamus head carving at the other. You could probably write on things with it. [5] ? ? clerk's quill Formed to precisely the most comfortable curve for a writer's hand, with a short tuft of plain pale-grey feathering at the top, this is most certainly the implement of choice for someone who spends a great deal of their time writing. The neat, almost fussy curve of the nib reflects the amount of care such a writer puts into their work, and the precisely chosen character of their words. It is clean and dry. || [6] || <colour> ink || Needs ink. elegant golden plume Clearly made from a Ramtops eagle's tail feather, this long, slim plume is a bright, vivid gold along its length, tipped with a deep, warm cinnamon brown.", "an old pocket watch, for its tiny brass gears. The gears were then bent to match the curvature of the pen and inlaid in its midsection. While this prevents them from telling time, now they catch the light and twinkle under the polish as the diligent penman works. The pen features a firm italic nib, used for writing in italics. Brookless calligraphy shop <description> <color> italics &&&&&&&+32 16 g Needs ink. Description of writing seems to take skillcheck. This gives off chats. humble wooden calligraphy brush A simple straight stick, with a bundle of dark wolf-hair bound to one end, this is the sword that won an Empire and the mattock that built it. It is clean and dry. || [19] || <description> <colour> ink calligraphy || &&&&&&+200 2/9 || Needs ink. Description of writing depends on crafts.arts.calligraphy skill. Hwel quill This is a fine quill pen that was apparently once used by Hwel himself. Now you too can write your own works of incredible genius! stall in Thespians' Yard messy, scratchy black ink &&&&&&+300 3/9 jaunty crimson plume Doubtless obtained from some exotic Howondalandish bird, this long, fluffy, curling feather is a bright, vivid crimson, shot through with swirling patterns of darker red. The lower third of the shaft has been trimmed smooth for a good grip,", "certain store sold pens for$0.35 each and pencils for $0. 4 24 Oct 2013, 08:57 A store selling price..... 2 28 Oct 2010, 21:28 A stationery store sells pens and pencils. If the price of 3 02 Aug 2010, 19:37 Pen regularly sell for$15 a pair and Pencil regularly sell 4 28 Dec 2005, 08:27 Display posts from previous: Sort by", "one could almost believe that nothing hurtful or bad could ever pass to paper from the tiny, artfully curved nib. [24] vulture feather quill Snaggly, scraggly, and oily black, this quill pen has a decidedly malign air about it, as if not even the writer could ever predict what odd, wonderful, evil, unexpected words might scratch their way onto the paper. The nib looks rather crooked and spiky, but writes quite well for all that. It is clean and dry. || [25] || <colour> ink || Needs ink. wooden pencil A small, straight stick has been smoothed down along the outside, hollowed, filled with a length of plumbago and sharpened to a point at one end. You could probably write on things with it. [26] pencil", "two types of pens: one type for $2 per pen and the other type for$3 per pen. If a customer can spend up to $25 to buy pens at the store and there is no sales tax, what is the greatest number of pens the customer can buy? A. 9 B. 10 C. 11 D. 12 E. 20 Practice Questions Question: 8 Page: 147 Difficulty: easy The$2 pens are cheaper, so see if we can maximize the number of $2 pens we can buy. If we buy 12$2 pens, we will spend $24 So, EITHER the correct answer is D (12) OR it's possible to by even more than 12 pens. HOWEVER, when we check the answer choices, we see that there's only one answer choice bigger than 12, and that answer choices is 20 Since there no ways to buy 20 pens for$25, the correct answer must be D Cheers, Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com", "but we don't know the selling price of one pad, thus can not determine how many was sold and how many left. Question asks: What was the total amount spent by the store on the writing pads it had in stock at the end of last month? _________________ BSchool Forum Moderator Joined: 27 Aug 2012 Posts: 1146 Followers: 98 Kudos [?]: 715 [0], given: 119 Re: At the beginning of last month, a stationery store had in [#permalink] 13 Sep 2013, 12:57 Expert's post Hi Bunuel, Just curious - why we're assuming that stored had sold some of the writing pads during the month? It's nowhere mentioned explicitly. Right..? In that way A should be the answer! P.S: I know it's an official question so it has to be correct but I'm not able to get that properly. Please help me to connect that chord. _________________ Manager Joined: 06 Jul 2013 Posts: 118 GMAT 1: 620 Q48 V28 GMAT 2: 700 Q50 V33 Followers: 0 Kudos [?]: 11 [0], given: 42 Re: At the beginning of last month, a stationery store had in [#permalink] 10 Oct 2013, 06:07 I have another question. Even if we had known how many pads were sold, could we calcuate the amount spend on pad at the end of the month. Because we", "apportioned to the writing pads, the answer is A SVP Joined: 16 Oct 2003 Posts: 1810 Followers: 4 Kudos [?]: 137 [0], given: 0 ### Show Tags 30 Aug 2004, 13:32 dushver wrote: The only thing that is asked is \"how much did the store spend in writing pad?\" and the only once the purchase that it made for that month. Assuming that the inventory cost and labour cost are not apportioned to the writing pads, the answer is A How much the store spend on the writing pad at the end of the month. We do not know how many were sold. E is my answer. 30 Aug 2004, 13:32 Display posts from previous: Sort by", "of a Line from Two Points. Let's use the examples in the last lesson... We'll use the first one to … • ### amazon.com:fine point pens iBayam Journal Planner Pens Colored Pens Fine Point Markers Fine Tip Drawing Pens Porous Fineliner Pen for Bullet Journaling Writing Note Taking Calendar Coloring Art Office School Supplies, 18 Colors. 4.7 out of 5 stars 49,852. $9.98$ 9. 98. $2.00 coupon applied at checkout Save$2.00 with coupon", "Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006. Let the revenue of pens be 'x' Let the revenue of pencils be 'y' the revenue of pens up from 2006 by 5%=>105% the revenue of pencils down from 2006 by 13%=>87% Overall revenue down by 1%=>99% Attachment: pic2.JPG [ 3.08 KiB | Viewed 5352 times ] hence the ratio of pencils to pens = 6/12 = 1/2 _________________ If you like my post, consider giving me a kudos. THANKS! Intern Affiliations: NYSSA Joined: 07 Jun 2010 Posts: 31 Location: New York City Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program) WE 1: Senior Associate - Thomson Reuters WE 2: Analyst - TIAA CREF Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 24 Jun 2010, 06:12 1 This one isn't too bad. Say: pen = n and pencil = p and revenue = r. p+n = r 1.05p+.87n=.99r now just add plug in for r. .06n=.12p p/n = 1/2 Posted from my mobile device Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9643 Location: Pune, India Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ###", "a good grip, and a small brass reservoir sits just above the gold nib. [40] elegant black ink &&&&&&+100 1/9 swansdown quill This quill is small and delicate, its upper half shrouded in a soft fuzz of down. Looking at the tiny threads, stirring with every breath from the writer's lips, one could almost believe that nothing hurtful or bad could ever pass to paper from the tiny, artfully curved nib. [41] <colour> ink &&&&&&+100 1/9 Needs ink. typewriter This is a device that allows people to put legible texts on paper quickly, even if their handwriting is terrible. A series of keys display the letters of the alphabet. When one is hit, a metal arm hits a small harassed looking imp, who is sitting in the centre of the device, on the head. The imp then checks the position of the key and neatly prints the correct letter on the paper put into the device. [42] typed characters \"Put\" a piece of paper in it, then \"type on typewriter\". vulture feather quill Snaggly, scraggly, and oily black, this quill pen has a decidedly malign air about it, as if not even the writer could ever predict what odd, wonderful, evil, unexpected words might scratch their way onto the paper. The nib looks rather crooked and spiky, but", "Expert's post Carcass wrote: A certain store sells two types of pens: one type for $2 per pen and the other type for$3 per pen. If a customer can spend up to $25 to buy pens at the store and there is no sales tax, what is the greatest number of pens the customer can buy? A. 9 B. 10 C. 11 D. 12 E. 20 Practice Questions Question: 8 Page: 147 Difficulty: easy The$2 pens are cheaper, so see if we can maximize the number of $2 pens we can buy. If we buy 12$2 pens, we will spend $24 So, EITHER the correct answer is D (12) OR it's possible to by even more than 12 pens. HOWEVER, when we check the answer choices, we see that there's only one answer choice bigger than 12, and that answer choices is 20 Since there no ways to buy 20 pens for$25, the correct answer must be D Cheers, Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails Re: A certain store sells two types of pens: one type for $2 per [#permalink] 22 Aug 2017, 14:42 Display posts from previous: Sort by # A certain store sells two types of pens: one type for$2 per Question banks", "pens sold by a trader from Tuesday to Wednesday is same as the percentage decrease in the number of pens sold from Monday to Tuesday. On Wednesday the trader sold 16 pens more than those he sold on Tuesday. If the percentage decrease in sales from Monday to Tuesday is 20%, how many pens did he sell on Tuesday ? 20 20 96 96 100 100 80 80 Once you attempt the question then PrepInsta explanation will be displayed. Start [\"0\",\"40\",\"60\",\"80\",\"100\"] [\"Need more practice! \\r\\n\",\"Keep trying! \\r\\n\",\"Not bad! \\r\\n\",\"Good work! \\r\\n\",\"Perfect! \\r\\n\"] Personalized Analytics only Availble for Logged in users Analytics below shows your performance in various Mocks on PrepInsta Your average Analytics for this Quiz Rank - Percentile 0% Completed 0/0 Accuracy 0%", "has $70$ pens and $28$ pencil cases. She wants to sell all of the pens and pencil cases in identical packages. What is the greatest number of packages she can make?a s v", "Here's some colorful stationery for ya! : o) -sneepity Thanks sneepity! These pens are so nice! ~ pikachu4919 talk 22:18, 1 July 2020 (UTC)", "multiple departments. Each of the departments 5 19 Feb 2016, 23:27 17 Stores Land M each sell a certain product at a different 8 15 Oct 2012, 04:46 At a certain store, each notepad costs x \\$ and each marker 5 04 May 2011, 09:47 4 Stores L and M each sell a certain product at a different 6 31 Aug 2010, 22:31 12 At a certain store, each notepad costs x dollars and each 9 15 Apr 2008, 00:29 Display posts from previous: Sort by", "of silky black waves and nods with the writer's motions, perfect for emphasising a gesture or adding that touch of theatricality to an official document. [22] <description> <colour> ink secretary hand &&&&&&+300 3/9 Needs ink. Description of writing depends on crafts.arts.calligraphy skill. old leaky quill Trimmed back to the shaft apart from a lonely tuft of feather at the top, this quill is slightly bent and discoloured from years of use. The brass nib is battered and crossed, and there's a small split surrounded by a blue stain where it meets the shaft. [23] messy, splotchy dark blue ink &&&&&&+100 1/9 ornate silver fountain pen This is an elegant silver fountain pen, light and responsive in the hand, with a pure gold nib and gold tracery forming Klatchian arabesques across the shaft in the distinctive fluid, looping style of Eritor. A discreet lever on the side indicates that it could probably be dipped into an ink well. [24] <colour> ink &&&&&&+300 3/9 Needs ink. pink glitter-filled pen Gaudy, pink and distracting, this glitter filled pen is the exact writing implement that most parents will buy their teenage daughter for her first day of middle school. Made from a thin tube of clear glass, and filled with an unknown liquid through which the glitter moves and swirls, the pen" ]

[ "diskettes at $12 each • Percentages 5$5.25 is 7 1/2% of what number?", "= \\boxed{108}$.", "that means the remaining percentage is $$40\\%$$. If the goods were reduced to $$60\\%$$, the percentage would be $$60\\%$$.", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "an extra cent.)", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "add all three cases: $1+17+243=\\boxed{261}$", "left. $(27*8)/{9 \\choose 3}{6 \\choose3}$ simplifies down to our fraction m/n, which is $9/70$. Adding them up gives $9 + 70 = \\boxed{79}$.", "what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So the only option that is greater than 100% - 55% is $\\boxed{\\textbf{(E)}\\ 54}$. Video Solution Solution detailing how to solve the problem: https://www.youtube.com/watch?v=WAmvVQzuzfc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=9", "Percent of Sales 100% 40% Amount $126,000 50,400 75,600 16,000$ 59,600 Sales Variable expens...", "of the total amount of snacks. 30% is 3 times 10% so all we need to do is multiply 10% (50 snacks) by 3. Therefore, are unhealthy. ### Example Question #21 : How To Find The Part From The Whole With Percentage The sales tax for a given city is 8.5%. How much will someone shopping there pay for $139.34 worth of groceries, after tax is figured in? Possible Answers: Correct answer: Explanation: To calculate the tax, multiply 8.5%. or, equivalently, 0.085, by$139.34, rounding to the nearest cent: Add the tax to the price before tax: The amount paid will be \\$151.18. ### Example Question #2 : Percentages Exactly of the plants in a garden are tomato plants. Which of the following could be the total number of plants in the garden? Explanation: The total number of plants has to be divisible by 3. Of the answer choices, the only one that is divisible by 3 is 18. You can check this because numbers that are divisible have the sum of their digits also divisible by 3. For example, which is divisible by 3, so 18 is divisible by 3. ### Example Question #1 : How To Find The Whole From The Part With Percentage 330 is 75% of what number? Explanation: 330 is 75% of a number", "Applied Statistics and Probability for Engineers, 6th Edition $0$ to $100$ If the raw material is entirely made of moisture, then the maximum percentage of moisture is $100%$, or $100$ percentage points. If there is no moisture in the raw material, then the percentage of moisture is $0%$, or $0$ percentage points. Any integer value of percentage point falls within this range, so the range is $0$ to $100$.", "+ 31 = \\boxed{058}$.", "CPI =$320.00 / 10 CPI = \\$32.00/item", ".84^2= 30%$", "#### Explanation: The sales tax rate is usually a percentage (assumed here). $6.5$ percent equals $0.065$ (percent means \"per 100\" so is divided by $100$).", "greater than $40. Thus, total number of items with price above$35 is 3*33 i.e. 99", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "# Total value of the quantity sold for item C is what per cent of the total value of the quantity sold for Item E? [ A ] 111 [ B ] 85 [ C ] 90 [ D ] 87.5 Answer : Option C Explanation : Total value of the quantity sold for item C = Rs(45 × 7.5 × 100) = Rs 33750 Total value of the quantity sold for item E = Rs(15 × 25 × 100) = Rs 37500 ∴ Required percentage = $33750\\over37500$X 100 = 90", "and $3:4$$3 : 4$, which one is larger in ratio? To simplify such comparison, the second term is standardized to a value, say $100$$100$, then the given ratios are $80:100$$80 : 100$ and $75:100$$75 : 100$. It is easier to compare these ratios. Such standard representation is simplified by dropping the known number $100$$100$. and the simplified form is \"percentage\". eg: $80%$ and $75%$. maths > commercial-arithmetics > conversion-ratio-fraction-percentage Conversion of Ratio to Percentage In this page, conversion of ratio to percentage and fraction to percentage are explained. The most important part is to understand the context in which a ratio or fraction is given and handle the conversion accordingly. maths > commercial-arithmetics > introduction-unitary-method Unitary Method : Introduction Multiplication is $\\text{multiplicand}\\phantom{\\rule{1ex}{0ex}}×\\phantom{\\rule{1ex}{0ex}}\\text{multiplier}\\phantom{\\rule{1ex}{0ex}}$$\\textrm{\\mu < i p l i c \\mathmr{and}} \\times \\textrm{\\mu < i p l i e r}$ $=\\text{product}$$= \\textrm{\\prod u c t}$ It is common to have multiplier as number of items. eg: Price of $10$$10$ pens is $200$$200$ coins. When the multiplier is number of items, the multiplicand and product can be found for $1$$1$ item. This helps in simplifying solution to finding product for any number of items. eg: Price of $1$$1$ pen is $20$$20$ coins. And with that it is easy to find price of $5$$5$ pens. maths > commercial-arithmetics > introduction-direct-variation Introduction Direct" ]

[ "Show Tags 20 Jun 2018, 04:26 1 Jinglander wrote: A company sells pens and pencils. The Revenue from pens in 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. Responding to a pm: Yes, we can use weighted averages here since there are two components and we have the total. w1/w2 = (A2 - Aavg)/(Aavg - A1) Say, w1 is the weight of pencil revenue and w2 is the weight of pen revenue in 2006 (increase in revenue is over previous revenue so the ratio we get will be of the revenues of 2006) w1/w2 = (5 - (-1)) / (-1 - (-13)) = 6/12 = 1/2 Revenue of pencils : Revenue of pens = 1:2 _________________ Karishma Veritas Prep GMAT Instructor SVP Status: Three Down. Joined: 09 Jun 2010 Posts: 1834 Concentration: General Management, Nonprofit Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 23 Jun 2010, 19:17 Let us assume the following letters for respective revenues: Pen (2006) - A1 Pencil (2006) - B1 Pen (2007) - A2 Pencil (2007) - B2 Now from the given information, we", "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", "skill. You can also \"paint\" words on blank paper with this brush (which does not require ink).[3] souvenir fountain pen A black fountain pen with a silver nib. Written in silver filigree down the side are the words \"The Royal Art Museum - Ankh-Morpork\". It is clean and dry. || [36] || <colour> ink || &&&&&&+200 2/9 || Needs ink. small neat quill Short, slim, and an unpretentious grey, this plain goose quill seems ideal for the kind of person who cares about neatness, elegance, respect, and having everything in its proper place. [37] fussy green ink &&&&&&+100 1/9 stormy petrel quill [38] <colour> ink &&&&&&+100 1/9 Needs ink. sturdy purple plume Thicker and sturdier than a normal feather, this plume looks as though it could quite possibly double as a weapon in an emergency. Such as, for instance, when someone else has just grabbed the last copy of a textbook you wanted. The lower third of the shaft has been trimmed smooth for a good grip, and a small brass reservoir sits just above the gold nib. [39] rich purple ink &&&&&&+300 3/9 swan's wing plume Long and gleaming white, with a slight but very graceful curve, this feather could have come from no other bird. The lower third of the shaft has been trimmed smooth for", "revenue in 2006. Math Expert Joined: 02 Sep 2009 Posts: 58117 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 24 Jun 2010, 06:02 6 1 Jinglander wrote: Can I get someone to walk me through this one too. A company sells pens and pencils. The Revenue from pens in 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. It would be easier if you make a table: Attachment: untitled.JPG [ 11.88 KiB | Viewed 5457 times ] We are told that \"Overall revenue was down 1% from 06 to 07\" --> $$0.99(x+y)=1.05x+0.87y$$. Question: $$\\frac{y}{x}=?$$ $$0.99(x+y)=1.05x+0.87y$$ --> $$0.99x+0.99y=1.05x+0.87y$$ --> $$12y=6x$$ --> $$\\frac{y}{x}=\\frac{6}{12}=\\frac{1}{2}$$. Hope it's clear. _________________ ##### General Discussion Manager Joined: 15 Mar 2010 Posts: 70 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 28 Jun 2010, 07:36 4 Jinglander wrote: Can I get someone to walk me through this one too. A company sells pens and pencils. The Revenue from pens in 2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period.", "to walk me through this one too. A company sells pens and pencils. The Revenue from pens in 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. Let the revenue of pens be 'x' Let the revenue of pencils be 'y' the revenue of pens up from 2006 by 5%=>105% the revenue of pencils down from 2006 by 13%=>87% Overall revenue down by 1%=>99% Attachment: pic2.JPG [ 3.08 KiB | Viewed 3409 times ] hence the ratio of pencils to pens = 6/12 = 1/2 _________________ If you like my post, consider giving me a kudos. THANKS! Manager Joined: 24 May 2010 Posts: 78 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 26 Jul 2010, 15:40 achiever01 I dont understand your solution at all Non-Human User Joined: 09 Sep 2013 Posts: 13802 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 23 Nov 2015, 23:15 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion", "certain store sold pens for$0.35 each and pencils for $0. 4 24 Oct 2013, 08:57 A store selling price..... 2 28 Oct 2010, 21:28 A stationery store sells pens and pencils. If the price of 3 02 Aug 2010, 19:37 Pen regularly sell for$15 a pair and Pencil regularly sell 4 28 Dec 2005, 08:27 Display posts from previous: Sort by", "an old pocket watch, for its tiny brass gears. The gears were then bent to match the curvature of the pen and inlaid in its midsection. While this prevents them from telling time, now they catch the light and twinkle under the polish as the diligent penman works. The pen features a firm italic nib, used for writing in italics. Brookless calligraphy shop <description> <color> italics &&&&&&&+32 16 g Needs ink. Description of writing seems to take skillcheck. This gives off chats. humble wooden calligraphy brush A simple straight stick, with a bundle of dark wolf-hair bound to one end, this is the sword that won an Empire and the mattock that built it. It is clean and dry. || [19] || <description> <colour> ink calligraphy || &&&&&&+200 2/9 || Needs ink. Description of writing depends on crafts.arts.calligraphy skill. Hwel quill This is a fine quill pen that was apparently once used by Hwel himself. Now you too can write your own works of incredible genius! stall in Thespians' Yard messy, scratchy black ink &&&&&&+300 3/9 jaunty crimson plume Doubtless obtained from some exotic Howondalandish bird, this long, fluffy, curling feather is a bright, vivid crimson, shot through with swirling patterns of darker red. The lower third of the shaft has been trimmed smooth for a good grip,", "# What is percentage of blue pens ? ## Question 3 & 4 Question 3: A person has 180 pens out of which 9 a... » All original content on these pages is fingerprinted and certified by Digiprove Insert math as $${}$$", "of a Line from Two Points. Let's use the examples in the last lesson... We'll use the first one to … • ### amazon.com:fine point pens iBayam Journal Planner Pens Colored Pens Fine Point Markers Fine Tip Drawing Pens Porous Fineliner Pen for Bullet Journaling Writing Note Taking Calendar Coloring Art Office School Supplies, 18 Colors. 4.7 out of 5 stars 49,852. $9.98$ 9. 98. $2.00 coupon applied at checkout Save$2.00 with coupon", "• # question_answer 28) By selling 100 pens, a shopkeeper gains the selling price of 20 pens. Find his gain percent? A) 10% B) 15% C) 30% D) 25% (d): Let C.P of one pen = x Let gain % be p% $\\Rightarrow S.P.\\left( one\\text{ }pen \\right)=x\\times \\left( \\frac{100+p}{100} \\right)$ On sale of 100 pens, S.P of 100 pens $=100x\\left( \\frac{100+p}{100} \\right)$ For 100 pens, gain $=S.P-C.P=x\\left( 100+p \\right)-100x$ This gain = S.P (20 pens) $=20x\\times \\left( \\frac{100+p}{100} \\right)$ $\\Rightarrow px=\\frac{x\\left( 100+p \\right)}{5}$ $\\Rightarrow 5px=100x+px$ $\\Rightarrow 4px=100x$ $\\Rightarrow 4p=100=p=20\\,or\\,25%$", "sold the remaining pens for 2c/3 dollars each. Which of the following is equal to the total amount of money the store received for the a pens, in dollars? A) (2a + b)c/3 B) (a + 2b)c/3 C) (2a + 3b)c/3 D) (2a + c)b/3 E) (3a + 2b)c/3 _________________ It seems Kudos button not working correctly with all my posts... Please check if it is working with this post...... is it?.... Anyways...Thanks for trying Originally posted by Harshgmat on 21 Jun 2018, 19:44. Last edited by Bunuel on 21 Jun 2018, 21:10, edited 2 times in total. Renamed the topic and edited the question. BSchool Forum Moderator Joined: 26 Feb 2016 Posts: 3131 Location: India GPA: 3.12 A store had a total of a pens which it sold. The store first sold b of [#permalink] ### Show Tags 22 Jun 2018, 00:00 Harshgmat wrote: A store had a total of a pens which it sold. The store first sold b of these pens for c dollars each and it sold the remaining pens for 2c/3 dollars each. Which of the following is equal to the total amount of money the store received for the a pens, in dollars? A) (2a + b)c/3 B) (a + 2b)c/3 C) (2a + 3b)c/3 D) (2a + c)b/3 E) (3a +", "Strategy, General Management GMAT 1: 680 Q48 V34 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 24 Jun 2010, 07:48 First, I think its more of a ratio problem than the mixture. Secondly, I simply cant understand why we are so reluctant to give kudos. Bunuel! Excellent solution. +1 for you. The very last equation contains a little typo though! _________________ Manager Joined: 24 May 2010 Posts: 73 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 26 Jul 2010, 16:40 achiever01 I dont understand your solution at all Intern Joined: 28 Oct 2017 Posts: 8 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 29 Sep 2018, 21:39 This question can also be solved using mixture/alligation technique. Pen revenue = +5% (increase) Pencil revenue = -13% (decrease) Overall = -1% (decrease) Pen Pencil +5% -13% -1% [-1-(-13)]=12 [5-(-1)]=6 Ratio = Pencil:Pen 6:12=1:2 Thanks. Intern Joined: 22 Sep 2018 Posts: 19 Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] ### Show Tags 02 Oct 2018, 00:18 Jinglander wrote: A company sells pens and pencils. The Revenue from pens in", "has $70$ pens and $28$ pencil cases. She wants to sell all of the pens and pencil cases in identical packages. What is the greatest number of packages she can make?a s v", "two types of pens: one type for $2 per pen and the other type for$3 per pen. If a customer can spend up to $25 to buy pens at the store and there is no sales tax, what is the greatest number of pens the customer can buy? A. 9 B. 10 C. 11 D. 12 E. 20 Practice Questions Question: 8 Page: 147 Difficulty: easy The$2 pens are cheaper, so see if we can maximize the number of $2 pens we can buy. If we buy 12$2 pens, we will spend $24 So, EITHER the correct answer is D (12) OR it's possible to by even more than 12 pens. HOWEVER, when we check the answer choices, we see that there's only one answer choice bigger than 12, and that answer choices is 20 Since there no ways to buy 20 pens for$25, the correct answer must be D Cheers, Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com", "sculptors each team judges the work = 14. Number of artists attended the show = Number of teams of art judges an art show uses (Number of painters each team judges the work + Number of sculptors each team judges the work) = 9 (13 + 14) = 9 × 27 = (9 × 20) + (9 × 7) = 180 + 63 = 243. Guided Practice Model with Math When you model with Sharon’s Stationery Store contains 1,219 boxes of cards. May’s Market contains 3 times as many boxes of cards. How many boxes, b, does May’s Market contain? Question 1. Explain how to use a picture to represent the problem and show relationships. Explanation: Number of boxes of cards Sharon’s Stationery Store contains = 1,219. May’s Market contains 3 times as many boxes of cards. Number of boxes May’s Market contains = ??? Question 2. What equation can you write to represent the problem? Number of boxes May’s Market contains = 3 × Number of boxes of cards Sharon’s Stationery Store contains => 3 × 1219 = ??? Explanation: Number of boxes of cards Sharon’s Stationery Store contains = 1,219. May’s Market contains 3 times as many boxes of cards. => Number of boxes May’s Market contains = 3 × Number of boxes of cards Sharon’s", "of any two pens is less than 11 cents. (2) Median price of these 10 pens is 15 cents. VP Status: Learning stage Joined: 01 Oct 2017 Posts: 1028 WE: Supply Chain Management (Energy and Utilities) Mike bought 10 pens, all having distinct prices in integer number of [#permalink] ### Show Tags Updated on: 06 Aug 2018, 13:42 2 amanvermagmat wrote: Mike bought 10 pens, all having distinct prices in integer number of cents. If the cheapest pen costs 10 cents, what is the price of the costliest pen? (1) Out of these 10 pens, difference between prices of any two pens is less than 11 cents. (2) Median price of these 10 pens is 15 cents. Note:- a) All pens have distinct prices. b) Price of pens are in integer form ,obviously price can't be a negative or a free purchase. c) Price of cheapest pen is 10 cents. Price of remaining 9 pens is not known. Question stem:- what is the price of the costliest pen? St1:- Out of these 10 pens, difference between prices of any two pens is less than 11 cents. (Or, range of price <11) If x is the price of any of the remaining pen, then (x-10)<11 or, $$(x-10)\\leq{10}$$ or, $$x\\leq{20}$$. ($${10\\leq{x}\\leq{20}}$$) So the price of the costliest pen could be", "stationery store had [#permalink] 22 Jun 2013, 07:45 Expert's post Folaa3 wrote: At the beginning of the last month, a stationery store had on stock 250 writing pads which had cost the store $0.75 each. During the same month, the store made only one purchase of writing pads. What was the total amount spent by the store on the writing pads it had in stock at the end of last month? (1) Last month, the store purchased 150 writing pads for$0.80 each. (2) Last month the total revenue from the sale of writing pad was \\$180. OPEN DISCUSSION OF THIS QUESTION IS HERE: at-the-beginning-of-last-month-a-stationery-store-had-in-85710.html _________________ Re: At the beginning of the last month, a stationery store had [#permalink] 22 Jun 2013, 07:45 Similar topics Replies Last post Similar Topics: 2 At the beginning of last month, a stationery store had in st 3 19 Jun 2014, 11:42 8 All the clients that Company X had at the beginning of last 4 23 Feb 2012, 12:20 2 At the beginning of last year, Tony's Auto Dealership had 7 16 Dec 2010, 10:48 At the beginning of the last year, a car dealership had 150 8 30 Apr 2010, 09:44 2 At the beginning of last month, a stationery store had in 11 24 Oct 2009, 03:39 Display posts from", "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", "50% C. 20% D. 5% E. None of these 15. By selling 99 pens, a trader gains the cost of 33 pens. Find his gain percentage? A. B. 50% C. 25% D. E. None of these", "It:} \\ 60 \\div 6 = \\10/\\text{pad and} \\ 4 \\times 10 = \\40 \\ \\text{for} \\ 4 \\ \\text{pads.}\\!\\\\ {\\;} \\quad \\37 < \\40.\\end{align*} \\begin{align*}6. \\ \\text{Picture It; Picture It:} \\ 27 \\div 3 = \\9/\\text{canvas and} \\ 2 \\times 9 = \\18 \\ \\text{for} \\ 2 \\ \\text{canvases.}\\!\\\\ {\\;} \\quad \\18 < \\23.\\end{align*} Bargain Shopping - Reason Proportionally Which store has the better buy for 3 boxes of pastels? \\begin{align*}& \\mathbf{Describe:} && \\text{Two stores sell boxes of pastels.}\\!\\\\ &&& \\text{Art's Supplies:} \\ 3 \\ \\text{boxes for} \\ \\25\\!\\\\ &&& \\text{Picture It:} \\ 4 \\ \\text{boxes for} \\ \\28\\!\\\\ & \\mathbf{My \\ Job:} && \\text{Decide which store has the better buy.}\\!\\\\ & \\mathbf{Plan:} && \\text{I'll start with Picture It.}\\!\\\\ &&& \\text{I can easily figure out the price of one box of pastels at that store.}\\!\\\\ &&& \\text{Then I'll find the cost of} \\ 3 \\ \\text{boxes of pastels at Picture It and compare with Art's Supplies.}\\!\\\\ & \\mathbf{Solve:} && \\text{Picture It:} \\ 4 \\ \\text{boxes for} \\ \\28 \\ \\text{is} \\ 28 \\div 4, \\ \\text{or} \\ \\7 \\ \\text{per box.}\\!\\\\ &&& \\qquad \\qquad \\quad 3 \\ \\text{boxes at Picture It are} \\ 3 \\times 7, \\ \\text{or} \\ \\21.\\!\\\\ &&& \\qquad \\qquad \\quad \\21 \\ \\text{is less than} \\ \\25. \\ \\text{So Picture It has the" ]

[ "the percentage of people liking folk music is $40/200 \\times 100% = 20%$.", "### Home > INT2 > Chapter 12 > Lesson 12.2.2 > Problem12-88 12-88. Beethoven wrote nine symphonies and Mozart wrote $27$ piano concertos. What situation could this function describe? 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? $9(27)=243$ 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 continue before exactly the same program would have to be repeated? Combine the ways of part (a) with the $15$ Schubert string quartets.", "this pie chart, answer the following: (i) If ${\\mathbf{20}}$ people liked classical music, how many young people were surveyed? Ans: In this part it is given that 20 people liked classical music and we have to find the total number of people who were surveyed. So, from a given pie chart it is given that $10\\%$ peoples like classical music. Let the total number of people who were surveyed be $x$. So, $10\\%$ of $x$ must be equal to $20$. That is, $x \\times \\frac{{10}}{{100}} = 20$ $\\frac{x}{{10}} = 20$ By cross multiplication we get, $x = 200$ So, $200$ young people were surveyed. (ii) Which type of music is liked by the maximum number of people? Ans: In this part we have to find which type of music is liked by the maximum number of people. So, from the given pie chart it is clear that light music is liked by the maximum number of people ($40\\%$of the total number of people which is maximum in all categories). (iii) If a cassette company were to make ${\\mathbf{1000}}$ CDs, how many of each type would they make? Ans: In this part it is given that if a cassette company were to make $100$ CDs, then how many of each type would they make? For CDs of classical music:", "population between $0$$–$$10$ and $11$–$17$ are children and teenagers who generally do not work. Similarly, most people above 60 would be retired. Hence, people aged $18$$–$$60$ will be better able to answer the questions. 2 ### A newly opened spa center wants to conduct a customer survey on how they feel about the service provided. Which type of survey would be best? Open-ended questions Multiple-choice questions Ratio scale questions Interval scale questions CorrectIncorrect Correct answer is: Interval scale questions Interval scale questions provide respondents with a range of feelings to choose from. This type of survey question is often seen in customer satisfaction surveys. Hence, it would be the most appropriate question type in the given scenario. 3 ### A survey was conducted among 100 people in a locality to find the type of music they like. The table below shows the result of the study:Which is the most popular type of music? Classical Semi-classical Folk Pop music CorrectIncorrect Pop Music is liked by 80 people, which is greater than any other count in the table. Hence, pop music is the most popular type of music. 4 ### Consider the table given in question 3; find the percentage of people who like folk music. 20% 30% 40% 50% CorrectIncorrect 40 out of 200 people like folk music. Hence,", "music then how many people were surveyed? (b) Which music is liked by the minimum number of people? (c) If a company make 2000 CD’s then how many of each type would they make? (a) People who like semi – classical music = 30% This 30% represents 60 people. 100% represents = $\\frac{60 \\times 100}{30}$ = 200 people Hence, 200 people were surveyed. (b) From the pie chart we can see that light music is represented by the minimum part of the pie chart. Therefore, light music is least liked by people. (c) Number of Classical music CDs = 20% of 2000 = $\\frac{20 \\times 2000}{100}$ = 400 CDs Number of Semi – Classical music CDs = 30% of 2000 = $\\frac{30 \\times 2000}{100}$ = 600 CDs Number of Light music CDs = 10% of 2000 = $\\frac{10 \\times 2000}{100}$ = 200 CDs Number of Folk music CDs = 40% of 2000 = $\\frac{40 \\times 2000}{100}$ = 800 CDs 2. In a survey 720 people were asked about their favorite season from summer, winter and monsoon. Answer the following questions according to the given information. (a) Which season got the minimum votes? (b) What will be the central angle of each season? (c) Draw pie chart for the given information. Festival Votes Summer 160 Winter 340 Monsoon 220", "and no one applauded. There was no recognition at all. No one knew this, but the violinist was Joshua Bell, one of the greatest musicians in the world. He played one of the most intricate pieces ever written, with a violin worth$3.5 million dollars. Two days before, Joshua Bell sold out a theater in Boston where the seats averaged \\$100 each to sit and listen to him play the same music.", "that there are $50\\%$ more musicians than mathematicians.", "(ii) Which type of music is liked by the maximum number of people? (iii) If a cassette company were to make 1000 CDs. How many of each type would they make? Solution: (i) Number of young people who were surveyed = $\\frac { 100\\times 20 }{ 10 }$ = 200 people. (ii) Light music is liked by the maximum people, i.e., 40% (iii) Total number of CD = 1000 Number of viewers who like classical music = $\\frac { 10\\times 1000 }{ 100 }$ = 100 Number of viewer who like semi-classical music = $\\frac { 20\\times 1000 }{ 100 }$ = 200 Number of viewers who like light music = $\\frac { 40\\times 1000 }{ 100 }$ = 400 Number of viewers who like folk music = $\\frac { 30\\times 1000 }{ 100 }$ = 300 Ex 5.2 Class 8 Maths 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 season got the most votes, i.e. 150 (iii) Pie chart Ex 5.2 Class 8 Maths Question 3. Draw a pie chart showing the following information. The", "# KSEEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 ## KSEEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 Question 1. A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart answer the following: (i) If 20 people liked classical music, how many young people were surveyed? (ii) Which type of music is liked by the maximum number of people? (iii) If a cassette company were to make 1000 CD’s how many of each type would they make? Solution: (i) Number of people who liked classical music = 20 If 10% people liked classical music the total people = 20 1% people liked classical music = $$\\frac{20}{10 \\%}$$ 100% people liked classical music = $$\\frac{20}{10} \\times 100$$ = 200 People (ii) Light music (iii) Total CD’s = 1000 Semi-classical CD’s = 20% ∴ Number of Semi-classical CD’s = 20% of 1000 = $$\\frac{20}{100} \\times 1000$$ = 200 Number of Classical CD’s = 10% of 1000 = $$\\frac{10}{100} \\times 1000$$ = 100 Number of Folk CD’s = 30% of 1000 = $$\\frac{30}{100} \\times 1000$$ = 300 Number of Light music CD’s = 40% of 1000 = $$\\frac{40}{100}", "Hence, 200 young people were surveyed. (ii) Light music is liked by the maximum number of people. (iii) Total number of CD’s = 1000 Number of CD’s of Semi Classical music = 20% of 1000 ⇒ $$\\frac { 20 }{ 100 } \\times 1000=200$$ Number of CD’s of Classical music = 10% of 1000 ⇒ $$\\frac { 10 }{ 100 } \\times 1000=100$$ Number of CD’s of Folk music = 30% of 1000 ⇒ $$\\frac { 30 }{ 100 } \\times 1000=300$$ Number of CD’s of Light music = 40% of 1000 ⇒ $$\\frac { 40 }{ 100 } \\times 1000=400$$ Question 2. A group of 360 people was asked to vote for their favorite season from the three seasons rainy, winter and summer. (i) Which season got the most votes? (ii) Find the central angle of each sector. (iii) Draw a pie chart to show this information. Solution. (i) Winter season got the most votes. (ii) Total votes = 90 + 120 + 150 = 360. Central angle of winter sector $$= \\frac { Number\\quad of\\quad people\\quad who\\quad vote\\quad for\\quad winter\\quad season }{ Total\\quad number\\quad of\\quad people } \\times { 360 }^{ \\circ }$$ = $$\\frac { 150 }{ 360 } \\times { 360 }^{ \\circ }={ 150 }^{ \\circ }$$ Central angle of summer sector $$=", "# Ncert Solutions For Class 8 Maths Ex 5.2 ## Ncert Solutions For Class 8 Maths Chapter 5 Ex 5.2 1. A survey was made to find the type of music a group of people liked in a town. The pie chart given below shows the information regarding that survey. From the pie chart, give the answer for the following questions. (a) If 60 people liked semi classical music then how many people were surveyed? (b) Which music is liked by the minimum number of people? (c) If a company make 2000 CD’s then how many of each type would they make? (a) People who like semi – classical music = 30% This 30% represents 60 people. 100% represents = 60×10030$\\frac{60 \\times 100}{30}$ = 200 people Hence, 200 people were surveyed. (b) From the pie chart we can see that light music is represented by the minimum part of the pie chart. Therefore, light music is least liked by people. (c) Number of Classical music CDs = 20% of 2000 = 20×2000100$\\frac{20 \\times 2000}{100}$ = 400 CDs Number of Semi – Classical music CDs = 30% of 2000 = 30×2000100$\\frac{30 \\times 2000}{100}$ = 600 CDs Number of Light music CDs = 10% of 2000 = 10×2000100$\\frac{10 \\times 2000}{100}$ = 200 CDs Number of Folk music CDs = 40%", "median to be the best statistic. Person A's data Person B, on the other hand, usually earns around $82 (the mean), not$72 (the median), so in this case, the mean would represent the \"typical\" earnings. Person B's data New math puzzle: Substitution cipher A popular article at this time of the year is: The 12 Days of Christmas - How many Presents? This cryptogram also involves 12. What is the value of each letter? T W O T H R E E + S E V E N ----------- T W E L V E You can leave your responses here. 5. Final thought - Imagine Next year marks the 40th anniversary of John Lennon's murder. John Lennon [Image source] My hope for 2020 is that more of us will get out of our echo chambers, start to listen to each other, and \"live life in peace\". Imagine there's no countries It isn't hard to do Nothing to kill or die for And no religion, too Imagine all the people Living life in peace John Lennon, Imagine All the best to everyone for the holidays, and for 2020. Until next time, enjoy whatever you learn. 1. ranjith says: Dear friend, Throughout the last year I was enjoyed. Most of the time I couldn't understand everything. Even-though I was", "### Условие задания: 3 Б. Рис. $$1$$. Mozart Wolfgang Amadeus Mozart is one of the greatest and influential composers of all time. He began composing from an early age and his legacy includes various genres and forms of music. His life is thoroughly researched, but his figure is still enigmatic. He was born to a musical family in Austria. By the age of $$5$$ he could write music and by the age of $$8$$ composed his first symphony. His father was also a musician, and he hurried to support and develop his son's genius. Young Mozart played at various palaces around Europe and always created an impression. He made a career in Vienna and became a well-known musician and composer. However, he didn't handle his financial issues too well and thus there were periods of prosperity and periods of poverty on his life. In the last year of his life he composed Requiem. Reasons for his early death are still unclear, but it is believed that it was some kind of a disease. The wide-spead story about Antonio Salieri poisoning Mozart is not true — Salieri and Mozart respected each other and weren't rivals. Mozart was one of the most prolific composers ever. He left more than $$600$$ compositions — operas, symphonies and concertos and etc. Write three", "Mozart’s 40th and 25th Mozart published his 40th symphony on this day in 1788, the ‘great’ in G minor, one of two symphonies he wrote in the minor key – the other also in G minor, number 25, called the ‘little’ (I suppose calling it the ‘minor’ would be a tad confusing). The original music is scored for flute, two oboes, two clarinets, two bassoons, two horns and strings. The result is, well, Mozart at his finest: We will add the 25th for comparison, published in 1773, when the composer was a mere seventeen years of age: What were we doing the year before we were able to vote?", "# 1. A survey was made to find the type of music that a certain group of young people liked in a city. Adjoining pie chart shows the findings of this survey. From this pie chart answer the following:(i) If 20 people liked classical music, how many young people were surveyed?(ii) Which type of music is liked by the maximum number of people?(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make? (i)If 20 people liked classical music Total number of people = $(100\\times 20)\\div 10$ = $200$ people. (ii) Maximum number of people light music i.e.$40 \\%$. (iii) If a cassette company were to make 1000 CD’s, then Light music CD's = $( 40\\times 1000) \\div 100\\:\\:=400$ Folk music CD's = $\\left ( 30\\times 1000 \\right )\\div 100\\:\\:=300$ Classical music CD's = $\\left ( 10\\times 1000 \\right )\\div 100\\:\\:=100$ Semi classical music CD's = $\\left ( 20\\times 1000 \\right )\\div 100\\:\\:=200$ ## Related Chapters ### Preparation Products ##### JEE Main Rank Booster 2021 This course will help student to be better prepared and study in the right direction for JEE Main.. ₹ 13999/- ₹ 9999/- ##### Rank Booster NEET 2021 This course will help student to be better prepared and study in the right direction for NEET.. ₹ 13999/-", "has musicians available down into the 50 dollars range. 0gjm5yIn so far as the figures there are representative, that suggests that$200 would be the right ballpark for getting a choral work performed. That sounds more plausible to me than a couple of bucks, though I wouldn't be surprised if you could get the price down somewhat from there.", "suggested price of $19.99, 74 above the$19.99 price, with the highest payment being \\$78.62. These are amazing results that illustrate the point that people were willing to pay more than what was requested in order to thank and encourage more of this person’s work. ### Patron’s provide validation and structure Validation and encouragement can come from money, or it can simply come from “likes”. Many artists create things and put them online for people to view for free. Instagram, for example, has been a way that visual artists can share their work with fans. They are then validated and encouraged by patrons through engagement and “likes”. These likes don’t equate to money, but what is important about the validation is that artists feel as though people are invested in them. Patronage of the best sort is not just about providing money, but about providing validation and structure. Without patrons from the Renaissance there wouldn’t have been such great flourishing of arts. Isabella d’Este was another famous patron during the Renaissance. She was one of the many patrons for Leonardo DaVinci, but also a lesser known instrument maker Lorenzo Da Pavia. But she was also an amateur. She herself liked to learn how to play new instruments. She also liked hosting parties, writing, fashion and collecting art. She provided", "the label and percentage in each section of the columns which indicates the response variables displayed or provide a key. Practice questions Question 7 $170$170 people were surveyed about their music preference. The results have been recorded in the table below. Musical Preferences by Gender Music Preference Male Female Total Rock & Roll $24$24 $19$19 $43$43 Classical $8$8 $15$15 $23$23 Pop $17$17 $17$17 $34$34 Rap $6$6 $2$2 $8$8 Country & Western $17$17 $24$24 $41$41 R&B $6$6 $9$9 $15$15 Punk $4$4 $2$2 $6$6 Total $82$82 $88$88 $170$170 1. What is the explanatory variable in this data set? Music preference A Gender B Music preference A Gender B 2. Which of the following $100%$100% stacked column charts should be used to look for an association between the variables? A B A B 3. Does this stacked column chart suggest that there is an association between music preference and gender? No. The data does not suggest any association, as the corresponding segments are of similar sizes. A Yes. The data suggests there is an association, as the corresponding segments are of similar sizes. B No. The data does not suggest any association, as the corresponding segments are of different sizes. C Yes. The data suggests there is an association, as the corresponding segments are of different sizes. D No. The data", "# Data Handling ## Exercise 5.2 Question 1: A survey was made to find the type of music that a certain group of young people liked in a city. Adjoining pie chart shows the findings of this survey. From this pie chart answer the following: (i) If 20 people liked classical music, how many young people were surveyed? Answer: When 10 people liked classical music, number of people surveyed is 100 So, When 1 person liked the classical music, number of people surveyed is (100)/(10) When 20 people like the classical music, number of people surveyed =(100)/(10)xx20=200 (ii) Which type of music is liked by the maximum number of people? Answer: Light music is liked by maximum number of people. (iii) If a cassette company were to make 1000 CD’s, how many of each type would they make? Answer: Number of CD’s for different music Semi Classical = 1000 xx 20% = 200 Classical = 1000 xx 10% = 100 Folk = 1000 xx 30% = 300 Light = 1000 xx 40% = 400 Question 2: A group of 360 people were asked to vote for their favourite season from the three seasons rainy, winter and summer. Summer90 Rainy120 Winter150 (i) Which season got the most votes? (ii) Find the central angle of each sector. (iii) Draw a", "favor of Kaveh Cohen’s demo: .62 for . 7 against Publicly posted and emails in favor of Rob Elliott’s demo: .38 for .11 against Publicly posted and emails in favor of Daniel Moy’s demo: .87 for .2 against Now why in the hell did I just share that information? It is very simple – first, I stand behind the composers. I want it known that Kaveh, nor Rob, Doyle or Daniel have requested any form of payment for their work. They offered their skills to help a new developer. This will shed light on the type of people they are. The above composer are all professional, not in opinion, but more so in the fact that they have taken time from a demanding scheduled of “paying” jobs to help me. Second, it shows that you cannot please everyone. I have invested in creating something new. I have done my very best. I understand that it is not without weakness and have announced an extended version. DIVA is an honest product at an honest price. If you have not been “convinced” to the “worth” of my product my answerer is… I am sorry that you feel this product, considering the uniqueness of its content, and small price tag is not worth your investment. DIVA cannot be all things to" ]

[ "this pie chart, answer the following: (i) If ${\\mathbf{20}}$ people liked classical music, how many young people were surveyed? Ans: In this part it is given that 20 people liked classical music and we have to find the total number of people who were surveyed. So, from a given pie chart it is given that $10\\%$ peoples like classical music. Let the total number of people who were surveyed be $x$. So, $10\\%$ of $x$ must be equal to $20$. That is, $x \\times \\frac{{10}}{{100}} = 20$ $\\frac{x}{{10}} = 20$ By cross multiplication we get, $x = 200$ So, $200$ young people were surveyed. (ii) Which type of music is liked by the maximum number of people? Ans: In this part we have to find which type of music is liked by the maximum number of people. So, from the given pie chart it is clear that light music is liked by the maximum number of people ($40\\%$of the total number of people which is maximum in all categories). (iii) If a cassette company were to make ${\\mathbf{1000}}$ CDs, how many of each type would they make? Ans: In this part it is given that if a cassette company were to make $100$ CDs, then how many of each type would they make? For CDs of classical music:", "the percentage of people liking folk music is $40/200 \\times 100% = 20%$.", "music then how many people were surveyed? (b) Which music is liked by the minimum number of people? (c) If a company make 2000 CD’s then how many of each type would they make? (a) People who like semi – classical music = 30% This 30% represents 60 people. 100% represents = $\\frac{60 \\times 100}{30}$ = 200 people Hence, 200 people were surveyed. (b) From the pie chart we can see that light music is represented by the minimum part of the pie chart. Therefore, light music is least liked by people. (c) Number of Classical music CDs = 20% of 2000 = $\\frac{20 \\times 2000}{100}$ = 400 CDs Number of Semi – Classical music CDs = 30% of 2000 = $\\frac{30 \\times 2000}{100}$ = 600 CDs Number of Light music CDs = 10% of 2000 = $\\frac{10 \\times 2000}{100}$ = 200 CDs Number of Folk music CDs = 40% of 2000 = $\\frac{40 \\times 2000}{100}$ = 800 CDs 2. In a survey 720 people were asked about their favorite season from summer, winter and monsoon. Answer the following questions according to the given information. (a) Which season got the minimum votes? (b) What will be the central angle of each season? (c) Draw pie chart for the given information. Festival Votes Summer 160 Winter 340 Monsoon 220", "Originally Posted by shostakovich to chiph588@ : I'm not sure , may be some one good with programming can verify big value with computer ! I've verified it to be true for all primes less than $50,000$.", "# Ncert Solutions For Class 8 Maths Ex 5.2 ## Ncert Solutions For Class 8 Maths Chapter 5 Ex 5.2 1. A survey was made to find the type of music a group of people liked in a town. The pie chart given below shows the information regarding that survey. From the pie chart, give the answer for the following questions. (a) If 60 people liked semi classical music then how many people were surveyed? (b) Which music is liked by the minimum number of people? (c) If a company make 2000 CD’s then how many of each type would they make? (a) People who like semi – classical music = 30% This 30% represents 60 people. 100% represents = 60×10030$\\frac{60 \\times 100}{30}$ = 200 people Hence, 200 people were surveyed. (b) From the pie chart we can see that light music is represented by the minimum part of the pie chart. Therefore, light music is least liked by people. (c) Number of Classical music CDs = 20% of 2000 = 20×2000100$\\frac{20 \\times 2000}{100}$ = 400 CDs Number of Semi – Classical music CDs = 30% of 2000 = 30×2000100$\\frac{30 \\times 2000}{100}$ = 600 CDs Number of Light music CDs = 10% of 2000 = 10×2000100$\\frac{10 \\times 2000}{100}$ = 200 CDs Number of Folk music CDs = 40%", "population between $0$$–$$10$ and $11$–$17$ are children and teenagers who generally do not work. Similarly, most people above 60 would be retired. Hence, people aged $18$$–$$60$ will be better able to answer the questions. 2 ### A newly opened spa center wants to conduct a customer survey on how they feel about the service provided. Which type of survey would be best? Open-ended questions Multiple-choice questions Ratio scale questions Interval scale questions CorrectIncorrect Correct answer is: Interval scale questions Interval scale questions provide respondents with a range of feelings to choose from. This type of survey question is often seen in customer satisfaction surveys. Hence, it would be the most appropriate question type in the given scenario. 3 ### A survey was conducted among 100 people in a locality to find the type of music they like. The table below shows the result of the study:Which is the most popular type of music? Classical Semi-classical Folk Pop music CorrectIncorrect Pop Music is liked by 80 people, which is greater than any other count in the table. Hence, pop music is the most popular type of music. 4 ### Consider the table given in question 3; find the percentage of people who like folk music. 20% 30% 40% 50% CorrectIncorrect 40 out of 200 people like folk music. Hence,", "(ii) Which type of music is liked by the maximum number of people? (iii) If a cassette company were to make 1000 CDs. How many of each type would they make? Solution: (i) Number of young people who were surveyed = $\\frac { 100\\times 20 }{ 10 }$ = 200 people. (ii) Light music is liked by the maximum people, i.e., 40% (iii) Total number of CD = 1000 Number of viewers who like classical music = $\\frac { 10\\times 1000 }{ 100 }$ = 100 Number of viewer who like semi-classical music = $\\frac { 20\\times 1000 }{ 100 }$ = 200 Number of viewers who like light music = $\\frac { 40\\times 1000 }{ 100 }$ = 400 Number of viewers who like folk music = $\\frac { 30\\times 1000 }{ 100 }$ = 300 Ex 5.2 Class 8 Maths 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 season got the most votes, i.e. 150 (iii) Pie chart Ex 5.2 Class 8 Maths Question 3. Draw a pie chart showing the following information. The", "(2 * 1) = 3 The number of ways of choosing $1$ piece by Sibelius is: \"\"^2C_1 = (2!)/((2-1)! 1!) = 2 / 1 = 2 Each of these choices is independent. So the total number of ways of choosing $4$ pieces as required is their product: ${\\text{^4C_2 * \"\"^3C_1 * }}^{2} {C}_{1} = 6 \\cdot 3 \\cdot 2 = 36$ (iii) We know that the unrestricted number of ways of choosing $4$ pieces from $9$ is \"\"^9C_4 = 126 Of those possibilities, any that miss out one or two composers is unacceptable. Let us attempt to count the unacceptable combinations. If Beethoven is missed, then there are $5$ other pieces, giving us: $\\text{^5C_4 = (5!)/((5-4)!4!) = 5\" }$ unacceptable combinations. If Handel is missed, then there are $6$ other pieces, giving us: $\\text{^6C_4 = (6!)/((6-4)! 4!) = 15\" }$ unacceptable combinations. If Sibelius is missed, then there are $7$ other pieces, giving us: $\\text{^7C_4 = (7!)/((7-4)! 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 42\" }$ unacceptable combinations. Of these unacceptable combinations, there is one that misses two composers - namely Handel and Sibelius, leaving us with the $4$ pieces by Beethoven. That possibility is included twice. So the total number of unacceptable combinations is: $5 + 15 + 42 -", "has musicians available down into the 50 dollars range. 0gjm5yIn so far as the figures there are representative, that suggests that$200 would be the right ballpark for getting a choral work performed. That sounds more plausible to me than a couple of bucks, though I wouldn't be surprised if you could get the price down somewhat from there.", "earned each of 5 possible amounts of money. We''re asked for the MEDIAN amount of money earned per day for those 20 days. Since there are 20 total days, the MEDIAN will be the ...” March 12, 2019 [email protected] posted a reply to In a group of 24 musicians, some are pianists and the rest a in the Problem Solving forum “Hi All, We''re told that in a group of 24 musicians, some are pianists and the rest are violinists and exactly 1/2 of the pianists and exactly 2/3 of the violinists belong to a union. We''re asked for the LEAST possible number of union members in the group. This question can be approached in a ...” March 12, 2019 Brent@GMATPrepNow posted a reply to In a group of 24 musicians, some are pianists and the rest a in the Problem Solving forum “In a group of 24 musicians, some are pianists and the rest are violinists. Let x = number of pianists So, 24-x = number of violinists Exactly 1/2 of the pianists and exactly 2/3 of the violinists belong to a union. So, the number of pianists in the union = x/2 And the number of violinists in ...” March 12, 2019 Jay@ManhattanReview posted a reply to In each of the last five years, Company K", "# KSEEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 ## KSEEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 Question 1. A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart answer the following: (i) If 20 people liked classical music, how many young people were surveyed? (ii) Which type of music is liked by the maximum number of people? (iii) If a cassette company were to make 1000 CD’s how many of each type would they make? Solution: (i) Number of people who liked classical music = 20 If 10% people liked classical music the total people = 20 1% people liked classical music = $$\\frac{20}{10 \\%}$$ 100% people liked classical music = $$\\frac{20}{10} \\times 100$$ = 200 People (ii) Light music (iii) Total CD’s = 1000 Semi-classical CD’s = 20% ∴ Number of Semi-classical CD’s = 20% of 1000 = $$\\frac{20}{100} \\times 1000$$ = 200 Number of Classical CD’s = 10% of 1000 = $$\\frac{10}{100} \\times 1000$$ = 100 Number of Folk CD’s = 30% of 1000 = $$\\frac{30}{100} \\times 1000$$ = 300 Number of Light music CD’s = 40% of 1000 = $$\\frac{40}{100}", "Mozart’s sonata for two pianos K448 (which you can listen to below) can increase your spatial-temporal IQ scores by 9 points. While the duration of the effect on your brain is only about 10-15 minutes, the findings are nonetheless fascinating.\" http://www.smart-kit.com/s245/how-the-amazing-mozart-k448-can-increase-your-iq-listen-now/ - 5 years, 3 months ago great - 4 years, 4 months ago ×", "Mathematics # In the relation C = 60+15n, C represents the total cost of holding an event at a hall, and n represents the number of guests. The maximum number of guests allowed in the hall is 100. What are the minimum and maximum possible values for C? a $0,$1500 b $0,$1560 c $60,$1500 d $60,$1560 ### Answers #### BettieZumbrunnen 5 years ago Minimum possible number of guests is zero. C = 60 + 15(0) = $60 Maximum possible number of guests is 100. C = 60 + 15(100) =$1,560 That looks like Choice (d).", "that there are $50\\%$ more musicians than mathematicians.", "# 1. A survey was made to find the type of music that a certain group of young people liked in a city. Adjoining pie chart shows the findings of this survey. From this pie chart answer the following:(i) If 20 people liked classical music, how many young people were surveyed?(ii) Which type of music is liked by the maximum number of people?(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make? (i)If 20 people liked classical music Total number of people = $(100\\times 20)\\div 10$ = $200$ people. (ii) Maximum number of people light music i.e.$40 \\%$. (iii) If a cassette company were to make 1000 CD’s, then Light music CD's = $( 40\\times 1000) \\div 100\\:\\:=400$ Folk music CD's = $\\left ( 30\\times 1000 \\right )\\div 100\\:\\:=300$ Classical music CD's = $\\left ( 10\\times 1000 \\right )\\div 100\\:\\:=100$ Semi classical music CD's = $\\left ( 20\\times 1000 \\right )\\div 100\\:\\:=200$ ## Related Chapters ### Preparation Products ##### JEE Main Rank Booster 2021 This course will help student to be better prepared and study in the right direction for JEE Main.. ₹ 13999/- ₹ 9999/- ##### Rank Booster NEET 2021 This course will help student to be better prepared and study in the right direction for NEET.. ₹ 13999/-", "To control for familiarity with the pieces, we gave the participants a piano repertoire quiz (PRQ) with questions about composers and titles of the stimuli after the main task (for the exact wording, see Table D3 in the online Supplementary Material at mp.ucpress.edu). For the PRQ, audio examples from the three different musical pieces used in the main task (see Subsection Materials and Stimuli, Table 9, and Table D1 in the online Supplementary Material for the list of pieces) were repeated (Stimulus 1 for the Chopin Étude, Stimuli 3 and 5 for the Beethoven Piano Sonata, and Stimuli 7 and 10 for the Mozart Piano Concerto). For each of these three pieces, participants were asked whether they knew: 1) the composer, and 2) the title of the piece, which leads to a maximum score of 3 * 2 = 6 points. For each composer and title, five responses categories (one of them being “I don’t know”) were given (see Table D3 in the online Supplementary Material). An overall score was calculated as the sum of correct responses to the six questions. On average, participants answered 2.74 (SD = 2.24) of the six questions correctly, suggesting that they were moderately familiar with the pieces (see Table 2). It should be considered that randomly selecting any of the four response", "have performed. Figure 6: Test results: harmonic similarity with Bach's compositions. Figure 6: Test results: harmonic similarity with Bach's compositions. For each composer we report the maximum, the minimum, and the mean scores. As we can see, EvoComposer outperforms the others' composers in terms of similarity, with a minimum of $56.2%$, a maximum of $92.2%$, and a mean score of $76%$. This means that on average EvoComposer made the same harmonic choices made by Bach in about 3 out of 4 cases. We also remark that such a percentage of harmonic similarity with Bach's work is only an indication and does not necessarily mean that a composition with a higher percentage of similarity is better than another with a lower percentage of similarity. For example, we like more the solution provided by EvoComposer, for Chorale BWV 11.6, with a 72.4% harmonic similarity with Bach's harmonization, than the solution provided for Chorale BWV 6.6, with a 92.2% harmonic similarity with Bach's original harmonization. Such compositions are available online (https://goo.gl/swM6V9). #### 5.3.1 Chorale BWV 6.6 Example As an example, we provide the output obtained for a specific chorale, namely BWV 6.6. Figures are in the Appendix. Figure 9 shows the original Bach's chorale, while Figures 1013 provide the output obtained for four different executions of the EvoComposer, each one using", "Mathematicians and Musicians In a given nation, every $20^{th}$ mathematician is also a musician, while every $30^{th}$ musician is also a mathematician. Are there more mathematicians or musicians in the nation? How many times more? Solution Reference 1. Sergey Dorichenko, A Moscow Math Circle: Week-by-Week Problem Sets (MSRI Mathematical Circles Library, 2012), #4.3 In a given nation, every $20^{th}$ mathematician is also a musician, while every $30^{th}$ musician is also a mathematician. Are there more mathematicians or musicians in the nation? How many times more? Solution 1 Assume there are $M$ mathematicians and $N$ musicians in the nation. Then we can say that in the very least M/20,$M/20$,$N/20$,$M/30$,$N/30$ musicians. But more can be said. These $M/20$ mathematicians that are also musicians, are musicians that are also mathematicians. The number of such musicians is at least N/30,$M/20$,$N/20$,$M/30$,$N/30$, implying that $M/20\\ge N/30.$ In a similar manner, we could obtain $N/30\\ge M/20,$ implying that the numbers are equal: $M/20= N/30,$ i.e., $N=1.5M.$ There are 50\\%,$30\\%$,$40\\%$,$50\\%$,$60\\%$ more musicians than mathematicians. The fact that we got an equality $M/20= N/30$ suggests a shortcut Solution 2 Let $X$ be the number of people in the nation who are both mathematicians and musicians. We then see that the number of mathematicians in the nation equals 20X,$20X$,$30X$, while the number of musicians is 30X,$20X$,$30X$. It follows", "To Draw or Win cD <- 0.42 #Chance to draw cW <-0.49 #Chance to Win cD+cW ## [1] 0.91 Instructor’s note: Alternatively calculate the probability of losing (say, $$P(L)$$) and subtract from 1. $$P(L) = P(M)^2 = 0.3^2 = 0.09$$. $$P(A) + P(L) = 1$$ then $$P(A) = 0.91$$. 2. In a group of 45 people, 20 people love to listen to classical music, 15 loves to listen to pop and 10 love to listen to both pop and classical music. 10 people love neither. 1. If someone chosen randomly, loves to listen classical music, what is the probability that he/she also loves pop music? Solution: You are going to divide number of both pop and classical music lovers by the number of classical music listener to find who loves both from between the ones who loves pop music. n(AnB) = Number people who loves both n(A) = Number of people who love classical music p(B|A) = Probability of getting people who also love pop music p(B|A) = n(AnB)/n(A) n(AnB) = 10 n(A) = 20 10/20 = 0.5 nA <- 20 # Number of people who loves classical music nAB <- 10 # Number of people who loves both nAB/nA ## [1] 0.5 2. What is the probability of people who love to listen to pop music? Solution:", "Tags 14 Dec 2015, 02:09 Bunuel wrote: 5. If at least one astronaut do NOT listen to Bach at Solaris space station, then how many of 35 astronauts at Solaris space station listen to Bach? Also very tricky. (1) Of the astronauts who do NOT listen to Bach 56% are male --> if # of astronauts who do NOT listen to Bach is $$x$$ then $$0.56x$$ is # of males who do NOT listen to Bach. Notice that $$0.56x=\\frac{14}{25}x$$ must be an integer. Hence x must be a multiple of 25: 25, 50, 75, ... But $$x$$ (# of astronauts who do NOT listen to Bach) must also be less than (or equal to) 35. So $$x$$ can only be 25, which makes # of astronauts who do listen to Bach equal to 35-25=10. Sufficient. (2) Of the astronauts who listen to Bach 70% are female. Now, if we apply the same logic here we get that, if # of astronauts who listen to Bach is $$y$$ then $$0.7y$$ is # of females who listen to Bach: $$0.7y=\\frac{7}{10}y$$ must be an integer. Hence it must be a multiple of 10, but in this case it can take more than 1 value: 10, 20, 30. So, this statement is not sufficient. In your explanation you have quoted \"But $$x$$" ]

[ "this pie chart, answer the following: (i) If ${\\mathbf{20}}$ people liked classical music, how many young people were surveyed? Ans: In this part it is given that 20 people liked classical music and we have to find the total number of people who were surveyed. So, from a given pie chart it is given that $10\\%$ peoples like classical music. Let the total number of people who were surveyed be $x$. So, $10\\%$ of $x$ must be equal to $20$. That is, $x \\times \\frac{{10}}{{100}} = 20$ $\\frac{x}{{10}} = 20$ By cross multiplication we get, $x = 200$ So, $200$ young people were surveyed. (ii) Which type of music is liked by the maximum number of people? Ans: In this part we have to find which type of music is liked by the maximum number of people. So, from the given pie chart it is clear that light music is liked by the maximum number of people ($40\\%$of the total number of people which is maximum in all categories). (iii) If a cassette company were to make ${\\mathbf{1000}}$ CDs, how many of each type would they make? Ans: In this part it is given that if a cassette company were to make $100$ CDs, then how many of each type would they make? For CDs of classical music:", "the percentage of people liking folk music is $40/200 \\times 100% = 20%$.", "### Home > INT2 > Chapter 12 > Lesson 12.2.2 > Problem12-88 12-88. Beethoven wrote nine symphonies and Mozart wrote $27$ piano concertos. What situation could this function describe? 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? $9(27)=243$ 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 continue before exactly the same program would have to be repeated? Combine the ways of part (a) with the $15$ Schubert string quartets.", "music then how many people were surveyed? (b) Which music is liked by the minimum number of people? (c) If a company make 2000 CD’s then how many of each type would they make? (a) People who like semi – classical music = 30% This 30% represents 60 people. 100% represents = $\\frac{60 \\times 100}{30}$ = 200 people Hence, 200 people were surveyed. (b) From the pie chart we can see that light music is represented by the minimum part of the pie chart. Therefore, light music is least liked by people. (c) Number of Classical music CDs = 20% of 2000 = $\\frac{20 \\times 2000}{100}$ = 400 CDs Number of Semi – Classical music CDs = 30% of 2000 = $\\frac{30 \\times 2000}{100}$ = 600 CDs Number of Light music CDs = 10% of 2000 = $\\frac{10 \\times 2000}{100}$ = 200 CDs Number of Folk music CDs = 40% of 2000 = $\\frac{40 \\times 2000}{100}$ = 800 CDs 2. In a survey 720 people were asked about their favorite season from summer, winter and monsoon. Answer the following questions according to the given information. (a) Which season got the minimum votes? (b) What will be the central angle of each season? (c) Draw pie chart for the given information. Festival Votes Summer 160 Winter 340 Monsoon 220", "population between $0$$–$$10$ and $11$–$17$ are children and teenagers who generally do not work. Similarly, most people above 60 would be retired. Hence, people aged $18$$–$$60$ will be better able to answer the questions. 2 ### A newly opened spa center wants to conduct a customer survey on how they feel about the service provided. Which type of survey would be best? Open-ended questions Multiple-choice questions Ratio scale questions Interval scale questions CorrectIncorrect Correct answer is: Interval scale questions Interval scale questions provide respondents with a range of feelings to choose from. This type of survey question is often seen in customer satisfaction surveys. Hence, it would be the most appropriate question type in the given scenario. 3 ### A survey was conducted among 100 people in a locality to find the type of music they like. The table below shows the result of the study:Which is the most popular type of music? Classical Semi-classical Folk Pop music CorrectIncorrect Pop Music is liked by 80 people, which is greater than any other count in the table. Hence, pop music is the most popular type of music. 4 ### Consider the table given in question 3; find the percentage of people who like folk music. 20% 30% 40% 50% CorrectIncorrect 40 out of 200 people like folk music. Hence,", "(ii) Which type of music is liked by the maximum number of people? (iii) If a cassette company were to make 1000 CDs. How many of each type would they make? Solution: (i) Number of young people who were surveyed = $\\frac { 100\\times 20 }{ 10 }$ = 200 people. (ii) Light music is liked by the maximum people, i.e., 40% (iii) Total number of CD = 1000 Number of viewers who like classical music = $\\frac { 10\\times 1000 }{ 100 }$ = 100 Number of viewer who like semi-classical music = $\\frac { 20\\times 1000 }{ 100 }$ = 200 Number of viewers who like light music = $\\frac { 40\\times 1000 }{ 100 }$ = 400 Number of viewers who like folk music = $\\frac { 30\\times 1000 }{ 100 }$ = 300 Ex 5.2 Class 8 Maths 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 season got the most votes, i.e. 150 (iii) Pie chart Ex 5.2 Class 8 Maths Question 3. Draw a pie chart showing the following information. The", "# Ncert Solutions For Class 8 Maths Ex 5.2 ## Ncert Solutions For Class 8 Maths Chapter 5 Ex 5.2 1. A survey was made to find the type of music a group of people liked in a town. The pie chart given below shows the information regarding that survey. From the pie chart, give the answer for the following questions. (a) If 60 people liked semi classical music then how many people were surveyed? (b) Which music is liked by the minimum number of people? (c) If a company make 2000 CD’s then how many of each type would they make? (a) People who like semi – classical music = 30% This 30% represents 60 people. 100% represents = 60×10030$\\frac{60 \\times 100}{30}$ = 200 people Hence, 200 people were surveyed. (b) From the pie chart we can see that light music is represented by the minimum part of the pie chart. Therefore, light music is least liked by people. (c) Number of Classical music CDs = 20% of 2000 = 20×2000100$\\frac{20 \\times 2000}{100}$ = 400 CDs Number of Semi – Classical music CDs = 30% of 2000 = 30×2000100$\\frac{30 \\times 2000}{100}$ = 600 CDs Number of Light music CDs = 10% of 2000 = 10×2000100$\\frac{10 \\times 2000}{100}$ = 200 CDs Number of Folk music CDs = 40%", "# KSEEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 ## KSEEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 Question 1. A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart answer the following: (i) If 20 people liked classical music, how many young people were surveyed? (ii) Which type of music is liked by the maximum number of people? (iii) If a cassette company were to make 1000 CD’s how many of each type would they make? Solution: (i) Number of people who liked classical music = 20 If 10% people liked classical music the total people = 20 1% people liked classical music = $$\\frac{20}{10 \\%}$$ 100% people liked classical music = $$\\frac{20}{10} \\times 100$$ = 200 People (ii) Light music (iii) Total CD’s = 1000 Semi-classical CD’s = 20% ∴ Number of Semi-classical CD’s = 20% of 1000 = $$\\frac{20}{100} \\times 1000$$ = 200 Number of Classical CD’s = 10% of 1000 = $$\\frac{10}{100} \\times 1000$$ = 100 Number of Folk CD’s = 30% of 1000 = $$\\frac{30}{100} \\times 1000$$ = 300 Number of Light music CD’s = 40% of 1000 = $$\\frac{40}{100}", "To control for familiarity with the pieces, we gave the participants a piano repertoire quiz (PRQ) with questions about composers and titles of the stimuli after the main task (for the exact wording, see Table D3 in the online Supplementary Material at mp.ucpress.edu). For the PRQ, audio examples from the three different musical pieces used in the main task (see Subsection Materials and Stimuli, Table 9, and Table D1 in the online Supplementary Material for the list of pieces) were repeated (Stimulus 1 for the Chopin Étude, Stimuli 3 and 5 for the Beethoven Piano Sonata, and Stimuli 7 and 10 for the Mozart Piano Concerto). For each of these three pieces, participants were asked whether they knew: 1) the composer, and 2) the title of the piece, which leads to a maximum score of 3 * 2 = 6 points. For each composer and title, five responses categories (one of them being “I don’t know”) were given (see Table D3 in the online Supplementary Material). An overall score was calculated as the sum of correct responses to the six questions. On average, participants answered 2.74 (SD = 2.24) of the six questions correctly, suggesting that they were moderately familiar with the pieces (see Table 2). It should be considered that randomly selecting any of the four response", "Mozart’s sonata for two pianos K448 (which you can listen to below) can increase your spatial-temporal IQ scores by 9 points. While the duration of the effect on your brain is only about 10-15 minutes, the findings are nonetheless fascinating.\" http://www.smart-kit.com/s245/how-the-amazing-mozart-k448-can-increase-your-iq-listen-now/ - 5 years, 3 months ago great - 4 years, 4 months ago ×", "# 1. A survey was made to find the type of music that a certain group of young people liked in a city. Adjoining pie chart shows the findings of this survey. From this pie chart answer the following:(i) If 20 people liked classical music, how many young people were surveyed?(ii) Which type of music is liked by the maximum number of people?(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make? (i)If 20 people liked classical music Total number of people = $(100\\times 20)\\div 10$ = $200$ people. (ii) Maximum number of people light music i.e.$40 \\%$. (iii) If a cassette company were to make 1000 CD’s, then Light music CD's = $( 40\\times 1000) \\div 100\\:\\:=400$ Folk music CD's = $\\left ( 30\\times 1000 \\right )\\div 100\\:\\:=300$ Classical music CD's = $\\left ( 10\\times 1000 \\right )\\div 100\\:\\:=100$ Semi classical music CD's = $\\left ( 20\\times 1000 \\right )\\div 100\\:\\:=200$ ## Related Chapters ### Preparation Products ##### JEE Main Rank Booster 2021 This course will help student to be better prepared and study in the right direction for JEE Main.. ₹ 13999/- ₹ 9999/- ##### Rank Booster NEET 2021 This course will help student to be better prepared and study in the right direction for NEET.. ₹ 13999/-", "that there are $50\\%$ more musicians than mathematicians.", "elsewhere. This makes perfect sense, and I have commented before why this is always the best way to get financial exposure to the property market – far better than buying to occupy. It is what I’ve always done. As I said 18 months ago - What we learn from this exercise is that buying your own home in today's economy is far inferior to buying a home as an investment, or renting and staying out of the property market completely Carbon Tax to reduce effective marginal tax rates (must read analysis) People are terrible at objectively determining quality In Washington , DC , at a Metro Station, on a cold January morning in 2007, this man (image above) with a violin played six Bach pieces. During that time, approximately 2,000 people went through the station, most of them on their way to work. After 45 minutes only 6 people had stopped and listened for a short while. About 20 gave money but continued to walk at their normal pace. The man collected a total of $32. When he finished playing no one noticed and no one applauded. There was no recognition at all. No one knew this, but the violinist was Joshua Bell, one of the greatest musicians in the world. He played one of the most intricate", "concert this weekend promises Fri Nov 03, 2017 6:46 am The concert this weekend promises to attract an even greater amount of people than attended the last one. a) an even greater amount of people b) an even larger amount of people c) an amount of people even greater d) a number of even larger e) an even greater number of people Why is Option D not the best Option? OA E ### Top Member Marty Murray Legendary Member Joined 03 Feb 2014 Posted: 2054 messages Followed by: 132 members 955 GMAT Score: 800 Fri Nov 03, 2017 10:56 pm lheiannie07 wrote: The concert this weekend promises to attract an even greater amount of people than attended the last one. a) an even greater amount of people b) an even larger amount of people c) an amount of people even greater d) a number of even larger e) an even greater number of people Why is Option D not the best Option? OA E Here's the sentence created using option D. The concert this weekend promises to attract a number of even larger people than attended the last one. That sentence does not really make sense. The comparison does not really work. It seems to be attempting to convey that the concert this weekend will attract people even", "Hence, 200 young people were surveyed. (ii) Light music is liked by the maximum number of people. (iii) Total number of CD’s = 1000 Number of CD’s of Semi Classical music = 20% of 1000 ⇒ $$\\frac { 20 }{ 100 } \\times 1000=200$$ Number of CD’s of Classical music = 10% of 1000 ⇒ $$\\frac { 10 }{ 100 } \\times 1000=100$$ Number of CD’s of Folk music = 30% of 1000 ⇒ $$\\frac { 30 }{ 100 } \\times 1000=300$$ Number of CD’s of Light music = 40% of 1000 ⇒ $$\\frac { 40 }{ 100 } \\times 1000=400$$ Question 2. A group of 360 people was asked to vote for their favorite season from the three seasons rainy, winter and summer. (i) Which season got the most votes? (ii) Find the central angle of each sector. (iii) Draw a pie chart to show this information. Solution. (i) Winter season got the most votes. (ii) Total votes = 90 + 120 + 150 = 360. Central angle of winter sector $$= \\frac { Number\\quad of\\quad people\\quad who\\quad vote\\quad for\\quad winter\\quad season }{ Total\\quad number\\quad of\\quad people } \\times { 360 }^{ \\circ }$$ = $$\\frac { 150 }{ 360 } \\times { 360 }^{ \\circ }={ 150 }^{ \\circ }$$ Central angle of summer sector $$=", "suggested price of $19.99, 74 above the$19.99 price, with the highest payment being \\$78.62. These are amazing results that illustrate the point that people were willing to pay more than what was requested in order to thank and encourage more of this person’s work. ### Patron’s provide validation and structure Validation and encouragement can come from money, or it can simply come from “likes”. Many artists create things and put them online for people to view for free. Instagram, for example, has been a way that visual artists can share their work with fans. They are then validated and encouraged by patrons through engagement and “likes”. These likes don’t equate to money, but what is important about the validation is that artists feel as though people are invested in them. Patronage of the best sort is not just about providing money, but about providing validation and structure. Without patrons from the Renaissance there wouldn’t have been such great flourishing of arts. Isabella d’Este was another famous patron during the Renaissance. She was one of the many patrons for Leonardo DaVinci, but also a lesser known instrument maker Lorenzo Da Pavia. But she was also an amateur. She herself liked to learn how to play new instruments. She also liked hosting parties, writing, fashion and collecting art. She provided", "of people who do not like talk-show 2 $$= 320 - 156 = 164$$. _________________ Manager Joined: 23 Jan 2013 Posts: 138 Concentration: Technology, Other Schools: Berkeley Haas GMAT Date: 01-14-2015 WE: Information Technology (Computer Software) ### Show Tags 26 Jan 2015, 19:01 Out of 320 people, 75% like talk-show 1 and 65% of the people who like talk-show 1 also like talk-show 2. If there are no people who like talk-show 2 but do not like talk-show 1, how many of the 320 people do not like talk-show 2 @Bunnel : I am not sure of my understanding but does the question ask to find X + 84 ? Talk show 2 and Talk show 1 = 320*0.75*0.65 = 156. Does this statement ' If there are no people who like talk-show 2 but do not like talk-show 1' mean people only liking talk show 2 = 0 ? >> !!! You do not have the required permissions to view the files attached to this post. Math Expert Joined: 02 Sep 2009 Posts: 52187 ### Show Tags 27 Jan 2015, 01:59 1 shelrod007 wrote: Out of 320 people, 75% like talk-show 1 and 65% of the people who like talk-show 1 also like talk-show 2. If there are no people who like talk-show 2 but do not", "earned each of 5 possible amounts of money. We''re asked for the MEDIAN amount of money earned per day for those 20 days. Since there are 20 total days, the MEDIAN will be the ...” March 12, 2019 [email protected] posted a reply to In a group of 24 musicians, some are pianists and the rest a in the Problem Solving forum “Hi All, We''re told that in a group of 24 musicians, some are pianists and the rest are violinists and exactly 1/2 of the pianists and exactly 2/3 of the violinists belong to a union. We''re asked for the LEAST possible number of union members in the group. This question can be approached in a ...” March 12, 2019 Brent@GMATPrepNow posted a reply to In a group of 24 musicians, some are pianists and the rest a in the Problem Solving forum “In a group of 24 musicians, some are pianists and the rest are violinists. Let x = number of pianists So, 24-x = number of violinists Exactly 1/2 of the pianists and exactly 2/3 of the violinists belong to a union. So, the number of pianists in the union = x/2 And the number of violinists in ...” March 12, 2019 Jay@ManhattanReview posted a reply to In each of the last five years, Company K", "has musicians available down into the 50 dollars range. 0gjm5yIn so far as the figures there are representative, that suggests that$200 would be the right ballpark for getting a choral work performed. That sounds more plausible to me than a couple of bucks, though I wouldn't be surprised if you could get the price down somewhat from there.", "have performed. Figure 6: Test results: harmonic similarity with Bach's compositions. Figure 6: Test results: harmonic similarity with Bach's compositions. For each composer we report the maximum, the minimum, and the mean scores. As we can see, EvoComposer outperforms the others' composers in terms of similarity, with a minimum of $56.2%$, a maximum of $92.2%$, and a mean score of $76%$. This means that on average EvoComposer made the same harmonic choices made by Bach in about 3 out of 4 cases. We also remark that such a percentage of harmonic similarity with Bach's work is only an indication and does not necessarily mean that a composition with a higher percentage of similarity is better than another with a lower percentage of similarity. For example, we like more the solution provided by EvoComposer, for Chorale BWV 11.6, with a 72.4% harmonic similarity with Bach's harmonization, than the solution provided for Chorale BWV 6.6, with a 92.2% harmonic similarity with Bach's original harmonization. Such compositions are available online (https://goo.gl/swM6V9). #### 5.3.1 Chorale BWV 6.6 Example As an example, we provide the output obtained for a specific chorale, namely BWV 6.6. Figures are in the Appendix. Figure 9 shows the original Bach's chorale, while Figures 1013 provide the output obtained for four different executions of the EvoComposer, each one using" ]

[ "+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", "+0 # counting question 0 201 1 In how many ways can a President and a Vice-President and a Treasurer be chosen from a group of 5 people (assuming that the President and the Vice-President and the Treasurer cannot be the same person)? Jun 16, 2021 #1 +525 +2 No. of ways $$= {^5C_3}$$ $$={5 ×4\\over 2}$$ $$=10$$ Jun 16, 2021", "Question How many ways can a president, vice-president, secretary, and treasurer be chosen from a club with 9 members?", "# The Many Powers of the President Students become familiar with the many roles of the president. They are given various scenarios. They must decide which of his \"hats\" he must wear given the specifics of each situation.", "fifth and sixth president to decide the president in the next level, because those sub-presidents control the vote for three of the senators in the next level, and those three control the vote for the next president. Therefore, any group of size Cn^(log(3)/log(6)) in the population can control the country if they distribute themselves correctly. 2. Fernando Says: Of course, allowing a minority to block legislation is part of the design of the American system, built in on purpose. Could one compute how large that minority has to be? • Thomas Says: I created a system last night such that if you have seven people at the bottom level, then some groups of three people can control the entire country, but no particular person is stronger than another (there is an automorphism of the group for each pair of people that takes one person to another). However, it is incredibly complicated, requiring four levels, and the next level up from the bottom has 28 people so I doubt anyone would ever use it. It has really interesting symmetry though. 3. Thomas Says: I wonder, if you have n people on the bottom level, what is the minimum amount of people who can control the country (in the bottom level) given that everyone in the bottom level has equal", "combine into groups of five. Read More »", "into individuals)", "are there of the election? A. 15 B. 30 C. 60 D. 120 E. 240 _________________ GMAT Club Legend Joined: 12 Sep 2015 Posts: 4134 Sixteen children are trying to decide which two children will win the [#permalink] ### Show Tags 31 Jan 2017, 12:52 2 Top Contributor 4 Bunuel wrote: Sixteen children are trying to decide which two children will win the president and vice-president positions in the class. Each child can and will cast a vote for anyone in the class. If each child in the class is eligible for a position, how many different outcomes are there of the election? A. 15 B. 30 C. 60 D. 120 E. 240 Take the task of selecting a president and vice-president, and break it into stages. Stage 1: Select someone to be president. There are 16 children to choose from, so we can complete stage 1 in 16 ways Stage 2: Select someone to be vice-president. Once a child has been selected to be president, that child cannot also be vice-president. So, there are 15 remaining children to choose from, so we can complete stage 2 in 15 ways By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus select a president and vice-president) in (16)(15) ways (240 ways) Note: the FCP can", "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", "## Stats A committee of 5 people must choose a President, Vice-President, and a Secretary. In how many ways can this be done? • A committee of 5 people must choose a President, Vice-President, and a Secretary. In how many ways can this be done? 5 C 3 = 5!/2! = 5*4/2 = 10 ways", "to (b,a). A combinations problem is one where (a,b) = (b,a). Here, we are choosing 5 people out of 12. But a person chosen for president will be a separate combination than to the person chosen as vice president. Thus, arrangement matters. so its a permutations problem. thus 12P5\" Yes I am pretty confident it is a permutation, but does it need to be adjusted for the fact that there are only 3 unique positions? 12p5 implies 5 unique positions for arrangement, but there are only 3 in this case (president, vice president, and council member). Posted from my mobile device Manager Joined: 12 Dec 2017 Posts: 79 Location: India Schools: Yale '19, GMBA '20, XLRI GMAT 1: 660 Q46 V35 GPA: 3.8 Re: A student council is to be chosen from a class of 12 students consisti [#permalink] ### Show Tags 03 Jan 2019, 21:13 nmccull wrote: \"I think so too. a permutation problem is one that is dependent on arrangement. That means : (a,b) is not equal to (b,a). A combinations problem is one where (a,b) = (b,a). Here, we are choosing 5 people out of 12. But a person chosen for president will be a separate combination than to the person chosen as vice president. Thus, arrangement matters. so its a permutations problem. thus 12P5\"", "government Most presidential caucuses differ from Statewide presidential primaries in all of the following ways EXCEPT the number of States that hold caucuses. ways in which delegates to the national convention are chosen. main purpose for which the delegates are 98. ## Math Solve the equation for Y X-Y=2 need the biography ofEmily Hutchinson for my English class.", "be done in 4C2 ways Selecting 1 transgender from 2 can be done in 2C1 = 2 ways ∴ Selecting 3 boys, 2 girls and 1 transgender can be done in 10 × 6 × 2 = 120 ways Question 10. Find the total number of subsets of a set with (i) 4 elements (ii) 5 elements (iii) n elements Solution: If a set has n elements then the number of its subsets = 2n (i) Here n = 4 So number of subsets = 24 = 16 (ii) n = 5 So number of subsets = 25 = 32 (iii) n = n So number of subsets = 2 Question 11. A trust has 25 members. (i) How many ways 3 officers can be selected? Solution: Selecting 3 from 25 can be done in 25C3 ways (ii) In how many ways can a President, Vice President and a secretary be selected? Solution: From the 25 members a president can be selected in 25 ways After the president is selected, 24 persons are left out. So a Vice President can be selected in (from 24 persons) 24 ways. After the selection of Vice President 23 persons are left out So a secretary can be selected (from the remaining 23 persons) in 23 ways So a president, Vice president", "its population of 25 million, the U.K. slightly beats the combined population of California and Texas with 62 million. Even if Scotland soon departs from the UK in order to remain in the EU, it only takes away 5.3 million from its’ population. ## Should We Also Elect the Vice Presidential Candidate in the Primaries? Should We Also Elect the Vice Presidential Candidate in the Primaries? The only jobs assigned to the Vice President are to become President if the President is disabled or deceased, and to preside over the Senate and cast tie-breaking votes. Yet the Vice Presidential candidate is not chosen by primaries of 50 states plus territories, but is singly chosen by the party presidential candidate. Even though the convention votes on his or her choice, they do not vote the VP candidate down and then consider others. If the VP candidate is a governor or senator, they have only faced election by the voters of one state, and sometimes that is a very small state, and one with no competition from a candidate from the other party. There is no official vetting committee or process to choosing the VP candidate. The problem with electing the VP candidate in the primaries, is that in a primary with, say, 18 presidential candidates, the best VP candidates", "# Number of possibilities? • Jun 9th 2011, 01:12 PM TeaWithoutMussolini Number of possibilities? A class of 20 is selecting a new president, with 5 candidates. How many different ways can the tally come out? • Jun 9th 2011, 01:26 PM Plato Quote: Originally Posted by TeaWithoutMussolini A class of 20 is selecting a new president, with 5 candidates. How many different ways can the tally come out? The votes are identical and the candidates are distinct. • Jun 9th 2011, 01:40 PM TeaWithoutMussolini Is it 10626? • Jun 9th 2011, 01:46 PM Plato Quote: Originally Posted by TeaWithoutMussolini Is it 10626? Is that $\\binom{20+5-1}{20}~?$ • Jun 9th 2011, 01:52 PM TeaWithoutMussolini Indeed it is.", "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", ", 3, 4, …………. n elements are nC1, nC2, nC3, nC4 …………… nCnrespectively. ∴ Total number of subjects = nC0 + nC1 + nC2 + nC3 + ………… + nCn = Sum of the coefficients in the binomial expansion (x + a)n = 2n Question 11. A Trust has 25 members. (i) How many ways 3 officers can be selected? (ii) In how many ways can a president, vice president and a secretary be selected? Number of members in the trust = 25 (i) How many ways 3 officers can be selected? The number of ways of selecting 3 officers from 25 members is (ii) In how many ways can a president, vice president, and secretary be selected? From the 25 members, a president can be selected in 25 ways After the president is selected, 24 persons are left out. So a Vice President can be selected in (from 24 persons) 24 ways. After the selection of Vice President, 23 persons are left out So a secretary can be selected (from the remaining 23 persons) in 23 ways So a president, Vice president, and a secretary can be selected in 25P3 ways 25P3 = 25 × 24 × 23 = 13800 ways Question 12. How many ways a committee of six-person from 10 persons can be chosen along", "eight statesmen are put in a random order in a row. There are 8! possible arrangements. If the positions of the U.S. President and the British Premier are fixed, there remain 6! possible arrangements for the other six statesmen. The U.S. President and the British Premier stand next to each other if they take up the positions i and i + 1 for some i with 1 ≤ i ≤ 7. In the case that these two statesmen take up the positions i and i + 1, there are 2! possibilities for the order among them. Thus, there are 6! &times; 7 &times; 2! arrangements in which the U.S. President and the British Premier stand next to each other, and so the sought probability equals (6! &times; 7 &times; 2!)/8! = 1/4. Combinations How many different juries of three persons can be formed from five persons A, B, C, D, and E? By direct enumeration you see that the answer is ten: {A, B, C}, {A, B, D}, {A, B, E}, {A, C, D}, {A, C, E}, {A, D, E}, {B, C, D}, {B, C, E}, {B, D, E}, {C, D, E}. In this problem, the order in which the jury members are chosen is not relevant. The answer ten juries could also have been obtained by a", "permutations problem. thus 12P5\" Yes I am pretty confident it is a permutation, but does it need to be adjusted for the fact that there are only 3 unique positions? 12p5 implies 5 unique positions for arrangement, but there are only 3 in this case (president, vice president, and council member). Posted from my mobile device You mean that choosing the council members would be a combination while for the three positions it is a permutation ? lets consider people : a , b, c ,d, e a as president = 1 way a as vice president = 1way a,b,c as council members = 1 way but a,b,c = b, c, a = c, a, b thus 3 positions as permutation ? Yes, that is how I interpreted the scenario. Manager Joined: 12 Dec 2017 Posts: 79 Location: India Schools: Yale '19, GMBA '20, XLRI GMAT 1: 660 Q46 V35 GPA: 3.8 A student council is to be chosen from a class of 12 students consisti [#permalink] ### Show Tags 03 Jan 2019, 21:51 nmccull wrote: kanakdaga wrote: nmccull wrote: \"I think so too. a permutation problem is one that is dependent on arrangement. That means : (a,b) is not equal to (b,a). A combinations problem is one where (a,b) = (b,a). Here, we are choosing 5 people", "number of ways to pick one of those $k$ to be president, and $\\binom{n}{n-k}$ is the number of ways to fill out the committee with $n-k$ people from the non-eligible group. Then, as you say, you sum over the possible values of $k$. On the right, $n$ is the number of ways to choose the president, and $\\binom{2n-1}{n-1}$ is the number of ways to choose the rest of the committee. – Brian M. Scott Nov 28 '14 at 19:47" ]

[ "+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", "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", "# 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 $7 \\times 4= \\cdots$", "# 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 $7 \\times 4= \\cdots$", "medal in a competition, the first winning gold, the next silver, and the third bronze. • Possibilities to choose $2$ representatives out of $100$ students, one as the “president” and the other as the “vice-president”. • Different outcommings rolling $3$ dice which are distinguishable by color, e.g. white, red, and black dice. | | | | created: 2019-09-10 19:41:47 | modified: 2019-09-10 20:01:49 | by: bookofproofs (none)", "are neither vice-presidents nor assistant vice-presidents? A. $$\\dfrac{e}{30}$$ B. $$\\dfrac{2e}5$$ C. $$\\dfrac{13e}{30}$$ D. $$\\dfrac{17e}{30}$$ E. $$\\dfrac{3e}5$$ Source: GMAT Paper Tests One approach is to use the INPUT-OUTPUT approach We have two fractions (1/10 and 1/3), so let's let e = the least common denominator of these two fraction. We'll let e = 30 If there are 30 employees, then the number of vice-presidents = (1/10)(30) = 3 one-third of the remaining employees are assistant vice-presidents There are 27 employees remaining. 1/3 of 27 = 9, so there are 9 assistant vice-presidents So, the total number of vice-presidents AND assistant vice-presidents = 3 + 9 = 12 So, the number of employees who are NEITHER vice-presidents nor assistant vice-presidents = 30 - 12 = 18 So, when the INPUT e = 30, the OUTPUT is 18 people who are NEITHER vice-presidents nor assistant vice-president Now check the answer choices to see which one yields an OUTPUT of 18, when we INPUT e = 30 A. e/30 = 30/30 = 1 - NO - ELIMINATE B. 2e/5 = 2(30)/5 = 12 - NO - ELIMINATE C. 13e/30 = 13(30)/30 = 13 - NO - ELIMINATE D. 17e/30 = 17(30)/30 = 17 - NO - ELIMINATE E. 3e/5 = 3(30)/5 = 18 - YES - KEEP", "+0 # counting question 0 201 1 In how many ways can a President and a Vice-President and a Treasurer be chosen from a group of 5 people (assuming that the President and the Vice-President and the Treasurer cannot be the same person)? Jun 16, 2021 #1 +525 +2 No. of ways $$= {^5C_3}$$ $$={5 ×4\\over 2}$$ $$=10$$ Jun 16, 2021", "–2, 3 D. –2, –10, –18 2. Write a rule for the sequence. 4, 8, 16, 32, . . . A. Start with 4, and mul... ### What are some examples of president Donald trump exercising two types of presidential powers? (Military,judicial, diplomatic, executive, and legislative) What are some examples of president Donald trump exercising two types of presidential powers? (Military,judicial, diplomatic, executive, and legislative)...", "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.", "eight statesmen are put in a random order in a row. There are 8! possible arrangements. If the positions of the U.S. President and the British Premier are fixed, there remain 6! possible arrangements for the other six statesmen. The U.S. President and the British Premier stand next to each other if they take up the positions i and i + 1 for some i with 1 ≤ i ≤ 7. In the case that these two statesmen take up the positions i and i + 1, there are 2! possibilities for the order among them. Thus, there are 6! &times; 7 &times; 2! arrangements in which the U.S. President and the British Premier stand next to each other, and so the sought probability equals (6! &times; 7 &times; 2!)/8! = 1/4. Combinations How many different juries of three persons can be formed from five persons A, B, C, D, and E? By direct enumeration you see that the answer is ten: {A, B, C}, {A, B, D}, {A, B, E}, {A, C, D}, {A, C, E}, {A, D, E}, {B, C, D}, {B, C, E}, {B, D, E}, {C, D, E}. In this problem, the order in which the jury members are chosen is not relevant. The answer ten juries could also have been obtained by a", "# From a pool of 12 candidates, the offices of president, vice president, secretary, and treasurer will be filled. In how many different ways can the offices be filled? 11,880 #### Explanation: We can find the number of ways in a couple of different ways: • work out the number by examining each seat • use the Permutation formula Examine each seat Let's start with the President's seat. How many people can fill that? 12 And then the Vice President's seat - how many can fill that? 1 person is already in the President's seat, so there's 11 people who can. And then then the Secretary's seat - 2 people are already in seats and so 10 could fill this one. And lastly the Treasurer's seat - 3 people are already in seats and so 9 could possibly sit here. So that's $12 \\times 11 \\times 10 \\times 9 = 11 , 880$ Permutation It'd be great if we could express the above multiplication in terms of factorials, because factorials are set figures and so we don't need to work out the multiplication each time. Is there a way to do so? Notice that in our multiplication above, we have $12 \\times 11 \\times 10 \\times 9$. What if we could have the full factorial? To have that,", "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.", "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 $40320$", "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", "different signals are possible if $2$ red, $2$ green, and $3$ yellow flags are available and all the red and green flags and at least one yellow flags are used? (the answer is $330$) 5. a poker hand consists of $5$ cards from a deck of $52$ playing cards. The hand is a “full house” if there are $3$ cards of one domination and $2$ cards of another. For example, three $10$’s and $2$ jacks form a full house. How many full house hands are possible? (the answer is $3744$) Thank You. - On (2b) the gender positions are fixed so you just need to order the men and order the women, giving $3! \\times 2!$ – Henry Dec 5 '12 at 23:57 The President can be any of the $12$. For each choice, there are $11$ choices for VP. For every choice for these positions, there are $12$ choices for Secretary, because Secretary can be same as President or VP. Ditto for Treasurer, total $12\\cdot 11\\cdot 12\\cdot 12$. – André Nicolas Dec 6 '12 at 0:02 All five of these problems use the same basic principles. The most important one is that when you make a sequence of $n$ choices, and the $k$-th choice can be made in $w_k$ ways, the there are $w_1w_2\\dots w_n$ different ways", "Arrange 25 officers, each having one of five different ranks $a$, $b$, $c$, $d$ and $e$, and belonging to one of five different regiments $p$, $q$, $r$, $s$ and $t$, in a square formation 5 by 5, so that each row and each file contains just one officer of each rank and just one from each regiment.", "+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", "\\ | * - - - - - * 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.", "How many times was FDR elected President? How many times was FDR elected President? Two Three Four Five The Answer: The correct answer is Four.", "## Precalculus (6th Edition) Blitzer We know that the order in which the four officers are selected makes a difference as the designation for all the four officers would be different. The ordered arrangement in which the order of the arrangement makes a difference is solved using the concept of permutations. Four officers are to be selected from a club of fifteen members. So, $n=15,r=4$. Thus, \\begin{align} & _{15}{{P}_{4}}=\\frac{15!}{\\left( 15-4 \\right)!} \\\\ & =\\frac{15!}{11!} \\\\ & =\\frac{15\\times 14\\times 13\\times 12\\times 11!}{11!} \\\\ & =32,760 \\end{align}" ]

[ "+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", "+0 # counting question 0 201 1 In how many ways can a President and a Vice-President and a Treasurer be chosen from a group of 5 people (assuming that the President and the Vice-President and the Treasurer cannot be the same person)? Jun 16, 2021 #1 +525 +2 No. of ways $$= {^5C_3}$$ $$={5 ×4\\over 2}$$ $$=10$$ Jun 16, 2021", "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", "# Math Help - Combinations 1. ## Combinations 1)How many six letter subsets can you make if 2 are consonants and 4 are vowels. 2) There are 22 students in the student council and 4 people are to be elected. How many ways can they be elected if: a) there must be a vice president and president b) vice president is excluded Attempt: 1) No idea. The way I did it seemed to be for permutations. 21x21x5x5x5x5 2a) (22,1) * (21,1) * (20,2) 22 people, 1 president , then there's 21. Choose a vice president. Now 20, and 2 people lef.t b) (22,1) * (21,3) 2. Hello, Johnboe! 1) How many 6-letter subsets can you make if 2 are consonants and 4 are vowels\" The problem is not clearly stated. It says subsets, so the order of the letters is not considered. If letters may not be repeated in the subsets, . . there are: . ${21\\choose2}$ choices for the consonants . . and ${5\\choose4}$ choices for the vowels. There are: . $210\\cdot5 \\:=\\:1050$ possible subsets. If letters may be repeated, it is a very messy problem . . . 3. Originally Posted by Soroban Hello, Johnboe! The problem is not clearly stated. It says subsets, so the order of the letters is not considered. If letters may", "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", "are 8 choices for the 1st person; there are 7 choices for the 2nd person; there are 6 choices for the 3rd person; there are 5 choices for the 4th person;. How many 8 go into 39? It was ratified by the states on Feb. 27, 1951. For an older person, it’s likely to be longer. Joined Sep … When I run the command slmgr /dlv in an elevated command prompt, it returns a cryptic reply. 8 times 30: 4: 8 times 40: 5,20: 8 times 50: 6,40: Go up to Mesopotamian Mathematics Home Page. Enter your objects (or the names of them), one per line in the box below, then click \"Show me!\" Before the 22nd Amendment, the Constitution did not limit the number of presidential terms to two, though many early presidents, including George Washington, imposed such … Thus, you cannot subtract 4 again from 8. I would like to know how many times I can use the same Meet Now link. Can you get a number for 40 with eight. By - TIMESOFINDIA.COM. The amendment defining limits on how many terms can a president serve was approved by Congress on March 21, 1947, during the administration of President Harry S. Truman. 8 times … How many times can 8 go in to 29? Joined Sep", "are neither vice-presidents nor assistant vice-presidents? A. $$\\dfrac{e}{30}$$ B. $$\\dfrac{2e}5$$ C. $$\\dfrac{13e}{30}$$ D. $$\\dfrac{17e}{30}$$ E. $$\\dfrac{3e}5$$ Source: GMAT Paper Tests One approach is to use the INPUT-OUTPUT approach We have two fractions (1/10 and 1/3), so let's let e = the least common denominator of these two fraction. We'll let e = 30 If there are 30 employees, then the number of vice-presidents = (1/10)(30) = 3 one-third of the remaining employees are assistant vice-presidents There are 27 employees remaining. 1/3 of 27 = 9, so there are 9 assistant vice-presidents So, the total number of vice-presidents AND assistant vice-presidents = 3 + 9 = 12 So, the number of employees who are NEITHER vice-presidents nor assistant vice-presidents = 30 - 12 = 18 So, when the INPUT e = 30, the OUTPUT is 18 people who are NEITHER vice-presidents nor assistant vice-president Now check the answer choices to see which one yields an OUTPUT of 18, when we INPUT e = 30 A. e/30 = 30/30 = 1 - NO - ELIMINATE B. 2e/5 = 2(30)/5 = 12 - NO - ELIMINATE C. 13e/30 = 13(30)/30 = 13 - NO - ELIMINATE D. 17e/30 = 17(30)/30 = 17 - NO - ELIMINATE E. 3e/5 = 3(30)/5 = 18 - YES - KEEP", "Question How many ways can a president, vice-president, secretary, and treasurer be chosen from a club with 9 members?", "eight statesmen are put in a random order in a row. There are 8! possible arrangements. If the positions of the U.S. President and the British Premier are fixed, there remain 6! possible arrangements for the other six statesmen. The U.S. President and the British Premier stand next to each other if they take up the positions i and i + 1 for some i with 1 ≤ i ≤ 7. In the case that these two statesmen take up the positions i and i + 1, there are 2! possibilities for the order among them. Thus, there are 6! &times; 7 &times; 2! arrangements in which the U.S. President and the British Premier stand next to each other, and so the sought probability equals (6! &times; 7 &times; 2!)/8! = 1/4. Combinations How many different juries of three persons can be formed from five persons A, B, C, D, and E? By direct enumeration you see that the answer is ten: {A, B, C}, {A, B, D}, {A, B, E}, {A, C, D}, {A, C, E}, {A, D, E}, {B, C, D}, {B, C, E}, {B, D, E}, {C, D, E}. In this problem, the order in which the jury members are chosen is not relevant. The answer ten juries could also have been obtained by a", "be done in 4C2 ways Selecting 1 transgender from 2 can be done in 2C1 = 2 ways ∴ Selecting 3 boys, 2 girls and 1 transgender can be done in 10 × 6 × 2 = 120 ways Question 10. Find the total number of subsets of a set with (i) 4 elements (ii) 5 elements (iii) n elements Solution: If a set has n elements then the number of its subsets = 2n (i) Here n = 4 So number of subsets = 24 = 16 (ii) n = 5 So number of subsets = 25 = 32 (iii) n = n So number of subsets = 2 Question 11. A trust has 25 members. (i) How many ways 3 officers can be selected? Solution: Selecting 3 from 25 can be done in 25C3 ways (ii) In how many ways can a President, Vice President and a secretary be selected? Solution: From the 25 members a president can be selected in 25 ways After the president is selected, 24 persons are left out. So a Vice President can be selected in (from 24 persons) 24 ways. After the selection of Vice President 23 persons are left out So a secretary can be selected (from the remaining 23 persons) in 23 ways So a president, Vice president", ", 3, 4, …………. n elements are nC1, nC2, nC3, nC4 …………… nCnrespectively. ∴ Total number of subjects = nC0 + nC1 + nC2 + nC3 + ………… + nCn = Sum of the coefficients in the binomial expansion (x + a)n = 2n Question 11. A Trust has 25 members. (i) How many ways 3 officers can be selected? (ii) In how many ways can a president, vice president and a secretary be selected? Number of members in the trust = 25 (i) How many ways 3 officers can be selected? The number of ways of selecting 3 officers from 25 members is (ii) In how many ways can a president, vice president, and secretary be selected? From the 25 members, a president can be selected in 25 ways After the president is selected, 24 persons are left out. So a Vice President can be selected in (from 24 persons) 24 ways. After the selection of Vice President, 23 persons are left out So a secretary can be selected (from the remaining 23 persons) in 23 ways So a president, Vice president, and a secretary can be selected in 25P3 ways 25P3 = 25 × 24 × 23 = 13800 ways Question 12. How many ways a committee of six-person from 10 persons can be chosen along", "number of ways to pick one of those $k$ to be president, and $\\binom{n}{n-k}$ is the number of ways to fill out the committee with $n-k$ people from the non-eligible group. Then, as you say, you sum over the possible values of $k$. On the right, $n$ is the number of ways to choose the president, and $\\binom{2n-1}{n-1}$ is the number of ways to choose the rest of the committee. – Brian M. Scott Nov 28 '14 at 19:47", "# 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 $7 \\times 4= \\cdots$", "we have more population than we have seats or places? For example, there are 6 people who are vying for those 4 government slots? We could do it this way: We can look at see that for the President's seat, there are 6 candidates. We can then look at the VP's seat and see that, with someone in the role of President, there are now 5 people left who can be VP. Then we look to the Treasurer and see there are 4 people left. And lastly there is the Secretary and there are 3 remaining people who could go there. So we get $6 \\times 5 \\times 4 \\times 3 = 30 \\times 12 = 360$ ways to do this. But this is awkward to do all the time - especially if we have a large population and a large number of places that need filling (ex. A concert hall has 1,000 seats. What is the number of ways 5,000 fans can sit in those 1,000 seats). Wouldn't it be great if there were a formula that would make it easier to work with these types of calculations? Notice that in the \"6 people, 4 offices\" example, we ended up with this calculation: $6 \\times 5 \\times 4 \\times 3$ How can we put that in terms", "## Stats A committee of 5 people must choose a President, Vice-President, and a Secretary. In how many ways can this be done? • A committee of 5 people must choose a President, Vice-President, and a Secretary. In how many ways can this be done? 5 C 3 = 5!/2! = 5*4/2 = 10 ways", "# From a pool of 12 candidates, the offices of president, vice president, secretary, and treasurer will be filled. In how many different ways can the offices be filled? 11,880 #### Explanation: We can find the number of ways in a couple of different ways: • work out the number by examining each seat • use the Permutation formula Examine each seat Let's start with the President's seat. How many people can fill that? 12 And then the Vice President's seat - how many can fill that? 1 person is already in the President's seat, so there's 11 people who can. And then then the Secretary's seat - 2 people are already in seats and so 10 could fill this one. And lastly the Treasurer's seat - 3 people are already in seats and so 9 could possibly sit here. So that's $12 \\times 11 \\times 10 \\times 9 = 11 , 880$ Permutation It'd be great if we could express the above multiplication in terms of factorials, because factorials are set figures and so we don't need to work out the multiplication each time. Is there a way to do so? Notice that in our multiplication above, we have $12 \\times 11 \\times 10 \\times 9$. What if we could have the full factorial? To have that,", "7!. Thus, the total number of arrangements for 3 students from a class of 10 is $$\\frac{10!}{7!}= (10)(9)(8) =$$ 720 choices. _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Re: A 10-student class is to choose a president, vice president, [#permalink] 07 Aug 2018, 06:08 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 $7 \\times 4= \\cdots$", "# Permutations and combinations • Sep 14th 2012, 07:35 AM Shwems Permutations and combinations There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ?(Crying) • Sep 14th 2012, 07:49 AM kalyanram Re: Permutations and combinations You have two offices (President, Treasurer). Now you have to choose 2 people from a group of 5 how many ways can this can be done. Spoiler: In $2.\\binom{5}{2} = 2.\\frac{5!}{2!3!} = 20$ ways. Multiplication by two is to allow permutations among them. Thanks to Evgeny (emakarov) for the correction. • Sep 14th 2012, 08:10 AM emakarov Re: Permutations and combinations Quote: Originally Posted by Shwems There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ?(Crying) See k-permutations.", "are there of the election? A. 15 B. 30 C. 60 D. 120 E. 240 _________________ GMAT Club Legend Joined: 12 Sep 2015 Posts: 4134 Sixteen children are trying to decide which two children will win the [#permalink] ### Show Tags 31 Jan 2017, 12:52 2 Top Contributor 4 Bunuel wrote: Sixteen children are trying to decide which two children will win the president and vice-president positions in the class. Each child can and will cast a vote for anyone in the class. If each child in the class is eligible for a position, how many different outcomes are there of the election? A. 15 B. 30 C. 60 D. 120 E. 240 Take the task of selecting a president and vice-president, and break it into stages. Stage 1: Select someone to be president. There are 16 children to choose from, so we can complete stage 1 in 16 ways Stage 2: Select someone to be vice-president. Once a child has been selected to be president, that child cannot also be vice-president. So, there are 15 remaining children to choose from, so we can complete stage 2 in 15 ways By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus select a president and vice-president) in (16)(15) ways (240 ways) Note: the FCP can" ]

[ "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.", "flawed. PSTU is a rectangle and PQRS is also a rectangle(Every square is a rectangle). The use of word large is ambiguous. It can either represent PSTU or PQRS. If largest is used instead then PQRS can be considered which would be sufficient alone as well. Area of PSTU = 64 Where PS = 2r & SU = r Thus, 2r*r = 64 $$r^2$$= 32 ---(ii) Sub in Eq - (i) $$32[4 - π]$$ > 16 ? YES Hence Suff St 2: The perimeter of the shaded region equals 8 + 2π Perimeter of the shaded region = Length of the arc TV + Side TQ + Side QV Length of the arc = $$\\frac{θ}{360} * 2πr$$ = $$\\frac{90}{360} * 2πr$$ = $$\\frac{πr}{2}$$ Thus, $$8 + 2π = \\frac{πr}{2} + r + r$$ r = 4 Sub in eq--(i) $$r^2[4 - π]$$ > 16 ? $$16[4 - π]$$ > 16 ? YES Hence Suff _________________ I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\\_(ツ)_/¯ Senior Manager Joined: 28 Feb 2014 Posts: 294 Location: United States Concentration: Strategy, General Management Re: The above picture depicts a circle perfectly inscribed in a square. Is [#permalink] ### Show Tags 26 Aug 2015, 18:11 2 If we can determine the radius, we can determine the", "# 2005 CEMC Gauss (Grade 7) Problems/Problem 15 ## Problem In the diagram, the area of rectangle $PQRS$ is $24$. If $TQ = TR$, what is the area of quadrilateral $PTRS$? $\\text{(A)}\\ 18 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 16 \\qquad \\text{(D)}\\ 6 \\qquad \\text{(E)}\\ 15$ $[asy] size(100); draw((0,0)--(6,0)--(6,4)--(0,4)--cycle); draw((0,4)--(6,2)); draw((5.8,1.1)--(6.2,1.1)); draw((5.8,.9)--(6.2,.9)); draw((5.8,3.1)--(6.2,3.1)); draw((5.8,2.9)--(6.2,2.9)); label(\"P\",(0,4),NW); label(\"S\",(0,0),SW); label(\"R\",(6,0),SE); label(\"T\",(6,2),E); label(\"Q\",(6,4),NE); [/asy]$ ## Solution 1 Since the area of rectangle $PQRS$ is $24$, let us assume that $PQ = 6$ and $QR = 4$. Since $QT = TR$, then $QR = 2QT$ so $QT = 2$. Therefore, triangle $PQT$ has base $PQ$ of length $6$ and height $QT$ of length 2, so its area is $\\frac{1}{2}(6)(2) = \\frac{1}{2}(12) = 6$. So the area of quadrilateral $PTRS$ is equal to the area of rectangle $PQRS$ (which is $24$) minus the area of triangle $PQT$ (which is $6$), or $18$. Therefore, the answer is $A$. ## Solution 2 Draw a line through $T$ parallel to $PQ$ across the rectangle parallel so that it cuts $PS$ at point $V$. Since $T$ is halfway between $Q$ and $R$, then $V$ is halfway between $P$ and $S$. Therefore, $SVTR$ is a rectangle which has an area equal to half the area of rectangle $PQRS$, or $12$. Similarly, $VPQT$ is a rectangle of area $12$, and $VPQT$", "+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", "part z = Area of rectangle PQRS − Area of part x − Area of part y = 294 − 154 − 38.5 = 101.5 cm2 Is there an error in this question or solution? #### APPEARS IN Solution In the Given Figure, $\\Square$ Pqrs is a Rectangle. If Pq = 14 Cm, Qr = 21 Cm, Find the Areas of the Parts X, Y And Z . Concept: Perimeter and Area of a Circle. S", "## anonymous 4 years ago rectangle PQRS has its coordinate at P(-8,2),Q(-5,6),R(3,0) and S(0,-4) a.)Find its Area b.)Find its Perimeter steps to answer this question : draw the rectangle PQRS on a Cartesian plane (follow coordinate given) find distance by using formula $d = \\sqrt{(X2-X1)^2 + (Y2-Y1)^2}$ then you have length of all sides.. use formula length x length to find area.. sum of all sides is calculate to find perimeter..", "In Example 5. QS = 4x – 15 and RT = 3x + 8. Find the lengths of the diagonals of QRST. The given lengths of the diagonals of the rectangle QRST are: QS = 4x – 15 and RT = 3x + 8 We know that, By Rectangle Diagonal Measure Theorem, The diagonals of a rectangle are congruent So, In a rectangle QRST, QS = RT So, 4x – 15 = 3x + 8 4x – 3x = 15 + 8 x = 23 Now, QS = 4x – 15 = 4 (23) – 15 = 92 – 15 = 77 RT = 3x + 8 = 3 (23) + 8 = 69 + 8 = 77 Hence, from the above, We can conclude that the lengths of the diagonals of the rectangle QRST are: QS = RT = 77 Question 8. Decide whether PQRS with vertices P(- 5, 2), Q(0, 4), R(2, – 1), and S(- 3, – 3) is a rectangle, a rhombus, or a square. Give all names that apply. The given vertices of the parallelogram PQRS is: P (-5, 2), Q (0, 4), R (2, -1), and S (-3, -3) Compare the given points with (x1, y1), and (x2, y2) Hence, The representation of the given vertices of the parallelogram PQRS in the", "above, square PQRS is divided into four smaller squares. [#permalink] ### Show Tags Updated on: 20 Sep 2017, 00:22 Bunuel wrote: whuannou wrote: 4 smaller Equal squares? If a square is divided into 4 smaller squares as shown above, could these four squares not be equal? True they have to Originally posted by Octobre on 19 Sep 2017, 23:39. Last edited by Octobre on 20 Sep 2017, 00:22, edited 1 time in total. Senior Manager Joined: 02 Jul 2017 Posts: 294 GMAT 1: 730 Q50 V38 In the figure above, square PQRS is divided into four smaller squares. [#permalink] ### Show Tags 19 Sep 2017, 23:48 In the figure, square PQRS is divided into four smaller squares. => Considering all the 4 smaller squares are equal in size. If the area of the shaded region is 3, what is the area of PQRS? = > In figure, 2 sides of shaded region : PR is diagonal and other end is center of 2 sides. If we check shaded region consist of 3 triangles. And Square can be divided into kind of 8 triangles out of which shaded region is 3 triangles. So if area of square is 8x than of shaded region is 3x. Given Area of shaded reason is 3 => Area of square = 8 Director", "### Primary 5 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question The perimeter of rectangle PQRS is 168 cm and its length to width ratio is 2:1 The ratio of the area of rectangle VSTU to the area of rectangle PQRS is 1:2 SR = TW = 2VS and W is the midpoint of SR. One semicircle inside rectangle VSTU has an area of 77 cm2 Find the area of the shaded portion. The correct answer is : 1610 cm^2", "2 }$$ ar(||gm ABEF) = $$\\frac { 1 }{ 2 }$$ x 50 = 25 cm2 Question 5. PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find or(∆RAS). Solution: In quadrant PLRM, rectangle PQRS is in scribed Radius of the circle = 13 cm A is any point on PQ AR and AS are joined, PS = 5 cm In right ∆PRS, PR2 = PS2 + SR2 ⇒ (132 = (5)2+ SR2 ⇒ 169 = 25 + SR2 ⇒ SR2 = 169 – 25 = 144 = (12)2 ∴ SR = 12 cm Area of rect. PQRS = PS x SR = 5x 12 = 60 cm2 ∵ Rectangle PQRS and ARAS are on the same base SR and between the same parallels ∴ Area ARAS = $$\\frac { 1 }{ 2 }$$ area rect. PQRS 1 = $$\\frac { 1 }{ 2 }$$ x 60 = 30 cm2 Question 6. In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8 cm and PQ and BD intersect at O, then find area of ∆OPB. Solution: In sq. ABCD, P and Q are the mid points of sides AB and CD respectively PQ", "p+h and q+h. One such arrangement requires a square of area h2 to complete it, the other a rectangle of area pq. Since both arrangements yield the same triangle, the areas of the square and the rectangle must be identical.", "(C_r – C_b)$, $R_g = R_r + R_b$. $$distance( C_g, 0 ) < R_g$$ And that’s it. What we’ve done above is change the problem into an equivalent one. Instead of asking whether two circles intersect, we now ask whether a third circle, constructed out of the two, contains the origin. [processing include=”./diagrams/gridcir.pinc”] Circle cirR = new Circle( 2, 1, 2, #FF0000 ); cirR.label = “R”; Circle cirB = new Circle( -2, 2, 1, #0000FF ); cirB.label = “B”; Circle cirG = new Circle( 0, 0, 0, #004000 ); cirG.label = “G”; mouseable.push( cirR, cirB ); Grid bg = new Grid( 500, 300, 40 ); void setup() { size( bg.width, bg.height ); } void draw() { cirG.posX = cirR.posX – cirB.posX; cirG.posY = cirR.posY – cirB.posY; cirG.doFill = cirG.contains( 0, 0 ); bg.draw(); cirG.draw(); cirB.draw(); cirR.draw(); } [/processing] That’s a fairly silly derivation for a pair of circles, but it’s a very simple illustration of the general transform we’re going to apply to intersection problems in order to handle arbitrary convex forms. # Enter The Minkowski Sum So how exactly did we construct the green circle above? In essence, what we’ve done is subtract area off of one circle (the blue one) and add it to the other. That’s all well and good with a pair of circles,", "$[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"P\",P,W); label(\"Q\",Q,W); label(\"R\",R,W); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 32 \\qquad \\text{(D)}\\ 64 \\qquad \\text{(E)}\\ 128$ ## Problem 10 A jacket and a shirt originally sold for dollar 80 and dollar 40, respectively. During a sale Chris bought the dollar80 jacket at a $40\\%$ discount and the dollar 40 shirt at a $55\\%$ discount. The total amount saved was what percent of the total of the original prices? $\\text{(A)}\\ 45\\% \\qquad \\text{(B)}\\ 47\\dfrac{1}{2}\\% \\qquad \\text{(C)}\\ 50\\% \\qquad \\text{(D)}\\ 79\\dfrac{1}{6}\\% \\qquad \\text{(E)}\\ 95\\%$. ## Problem 11 Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to $[asy] for(int i = -2; i <= 2; ++i) { draw((i,0)--(i,3),dashed); } draw((-3,1)--(3,1),dashed); draw((-3,2)--(3,2),dashed); draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle); dot((-3,0)); label(\"A\",(-3,0),SW); dot((-3,3)); label(\"B\",(-3,3),NW); dot((0,3)); label(\"C\",(0,3),N); dot((3,3)); label(\"D\",(3,3),NE); dot((3,0)); label(\"E\",(3,0),SE); dot((0,0)); label(\"start\",(0,0),S); label(\"\\longrightarrow\",(0,-0.75),E); label(\"\\longleftarrow\",(0,-0.75),W); label(\"\\textbf{Jane}\",(0,-1.25),W); label(\"\\textbf{Hector}\",(0,-1.25),E); [/asy]$ $\\text{(A)}\\ A \\qquad \\text{(B)}\\ B \\qquad \\text{(C)}\\ C \\qquad \\text{(D)}\\ D \\qquad \\text{(E)}\\ E$ ## Problem", "# In the following figure, Question: In the following figure, PQRS is a square of side 4 cm. Find the area of the shaded square. Solution: Area of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle. We know that area of the four sectors with radius is equal to area of one circle. $\\therefore$ Area of the shaded region $=4^{2}-\\pi \\times 1^{2}-\\pi \\times 1^{2}$ $\\therefore$ Area of the shaded region $=4^{2}-2 \\times \\pi \\times 1$ $\\therefore$ Area of the shaded region $=16-2 \\pi$ Therefore, area of the shaded region is $(16-2 \\pi) \\mathrm{cm}^{2}$", "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 # quick geometry problem, CPhill 0 160 7 $$In the diagram below, \\angle PQR = \\angle PRQ = \\angle STR = \\angle TSR, RQ = 8, and SQ = 3. Find PQ. [asy] pair A,B,C,D,E; A = (0, 0.9); B = (-0.4, 0); C = (0.4, 0); D = (-0.275, 0.16); E = (0.11, 0.65); draw(A--B); draw(A--C); draw(B--C); draw(B--E); draw(C--D); label(\"P\",A,N); label(\"Q\", B, S); label(\"R\", C, S); label(\"S\", D, S); label(\"T\", E, W); [/asy]$$ Jan 21, 2022 #1 0 $$PQ = \\boxed{\\frac{50 \\sqrt{2}}{7}}$$ . Jan 21, 2022 #2 -1 incorrect, man Guest Jan 21, 2022 #3 +676 +2 You should rather say thanks. Also when it's wrong or right. Straight Jan 21, 2022 edited by Straight Jan 21, 2022 #4 0 I meant: Thanks man, but it is wrong. Guest Jan 21, 2022 #5 +117872 +1 So I suppose it is a reasonable answer. Garbage in, garbage out. Melody Jan 21, 2022 #6 -1 I don't care if the problem is garbage. I just want to get the problem right, learn more, and get more knowledge. Guest Jan 21, 2022 #7 +117872 +1 If you want to get something right you must display it in a form suitable for the media. Are you incapable of comprehending that? It says, in Latex, (you haven't bothered to write it", "### Primary 5 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question The following figure consists of a square PQRS and a rectangle WXYZ. The area of the square is 64 cm2 XS is thrice as long as XP. XP and WP are of equal length. The line PR is a line of symmetry. Find the area of the shaded portion of the rectangle. The correct answer is : 12 cm^2", "of its sides PQ, QR, RS an [#permalink] ### Show Tags 15 Aug 2018, 10:21 2 00:00 Difficulty: 95% (hard) Question Stats: 50% (02:36) correct 50% (01:47) wrong based on 46 sessions ### HideShow timer Statistics A rhombus PQRS has A, B, C, D as mid points of its sides PQ, QR, RS and SP respectively. A, B, C, D are joined to form quadrilateral ABCD. What is the perimeter of quadrilateral ABCD? (1) PQ = 10 units. (2) PR = 16 units. Director Status: Learning stage Joined: 01 Oct 2017 Posts: 931 WE: Supply Chain Management (Energy and Utilities) A rhombus PQRS has A, B, C, D as mid points of its sides PQ, QR, RS an [#permalink] ### Show Tags 15 Aug 2018, 11:36 1 amanvermagmat wrote: A rhombus PQRS has A, B, C, D as mid points of its sides PQ, QR, RS and SP respectively. A, B, C, D are joined to form quadrilateral ABCD. What is the perimeter of quadrilateral ABCD? (1) PQ = 10 units. (2) PR = 16 units. ABCD is a rectangle. (When we joint the midpoints of a rhombus , a rectangle emerges) Rephrased question stem:- What is the perimeter of the rectangle ABCD? a) Length & breadth , both required St1:- PQ = 10 units We can only", "475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$? $\\mathrm{(A)}\\ 3\\qquad \\mathrm{(B)}\\ 7\\qquad \\mathrm{(C)}\\ 13\\qquad \\mathrm{(D)}\\ 37\\qquad \\mathrm{(E)}\\ 43$ ## Problem 23 How many ordered pairs $(m,n)$ of positive integers, with $m \\ge n$, have the property that their squares differ by $96$? $\\text{(A)}\\ 3 \\qquad \\text{(B)}\\ 4 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 9 \\qquad \\text{(E)}\\ 12$ ## Problem 24 Circles centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the midpoint of $\\overline{AB}$, and $OA = 2\\sqrt {2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$? $[asy] unitsize(6mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0); pair A=(-2*sqrt(2),0); pair B=(2*sqrt(2),0); pair G=shift(0,2)*A; pair F=shift(0,2)*B; pair C=shift(A)*scale(2)*dir(45); pair D=shift(B)*scale(2)*dir(135); pair X=A+2*dir(-60); pair Y=B+2*dir(240); path P=C--O--D--Arc(B,2,135,90)--G--Arc(A,2,90,45)--cycle; fill(P,gray); draw(Circle(A,2)); draw(Circle(B,2)); dot(A); label(\"A\",A,W); dot(B); label(\"B\",B,E); dot(C); label(\"C\",C,W); dot(D); label(\"D\",D,E); dot(G); label(\"E\",G,N); dot(F); label(\"F\",F,N); dot(O); label(\"O\",O,S); draw(G--F); draw(C--O--D); draw(A--B); draw(A--X); draw(B--Y); label(\"2\",midpoint(A--X),SW); label(\"2\",midpoint(B--Y),SE); [/asy]$ $\\text{(A)}\\ \\frac {8\\sqrt {2}}{3} \\qquad \\text{(B)}\\ 8\\sqrt {2} - 4 - \\pi \\qquad \\text{(C)}\\ 4\\sqrt {2} \\qquad \\text{(D)}\\ 4\\sqrt {2} + \\frac {\\pi}{8} \\qquad \\text{(E)}\\ 8\\sqrt {2} - 2 - \\frac {\\pi}{2}$ ##", "rectangle PQRS is $$\\frac{9}{8}$$ times bigger than rectangle pqrs. When designing surveys, there tend to be three different models for survey response scales: 1. Examples: 1. The scale’s points should map as closely as possible to the underlying idea (construct) of the scale Now that we’ve laid that groundwork, let’s jump back to the questions at the top. Importance Likert scale questions and responses seek to gain a deeper understanding about why people feel the way they do. I need to convert an angle in degrees to a value between 0 and 28000 on a log scale. This type of scale can be used to complement other types of survey form questions. Ratio scale allows any researcher to compare the intervals or differences. Meter constant in revolutions is 400. Unipolar scales require the respondent to rate the degree of presence (or absence) of a particular quality or attribute, such as usefulness or satisfaction. Rating scales; 3. Therefore rectangle PQRS is $$\\frac{7}{4}$$ times bigger than rectangle pqrs. Active 15 days ago. Tracing paper may be used. Marked Semantic Differential Scale. For example, a scale factor of 2 means that the new shape is twice the size of the original. Ratio scale is a type of variable measurement scale which is the highest level scale and is quantitative in nature." ]

[ "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "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.", "sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); /* draw(rightanglemark((0,0),a*expi(rot1),(c,0))); */ filldraw(sq1, rgb(0.95,1,0.95)); filldraw(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); label(\"a\",a/2*expi(rot1),SE,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),SW,sm); label(\"c\",(c/2,-c),S,sm); [/asy]$ COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection. $[asy] defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens pair shiftR = (s1+1,0); // distance between two diagrams pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B); // draw left diagram filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype(\"2 2\")); draw(I--E, linetype(\"2 2\")); draw(I--F, linetype(\"2 2\")); label(\"a\",(A+B)/2,S); label(\"b\",(B+C)/2,NE); label(\"c\",(A+C)/2,NW); label(\"r\",(I+F)/2,W); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C)); pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y); // draw right diagram filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy);", "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \\frac{251}{126}MN$, and $MN = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(A+F)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$1008-x$$\",(G+D)/2,S); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{x} = \\frac{1008-x}{h} \\Longrightarrow h^2 + x^2 - 1008x = 0$$. Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.", "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]", "475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$? $\\mathrm{(A)}\\ 3\\qquad \\mathrm{(B)}\\ 7\\qquad \\mathrm{(C)}\\ 13\\qquad \\mathrm{(D)}\\ 37\\qquad \\mathrm{(E)}\\ 43$ ## Problem 23 How many ordered pairs $(m,n)$ of positive integers, with $m \\ge n$, have the property that their squares differ by $96$? $\\text{(A)}\\ 3 \\qquad \\text{(B)}\\ 4 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 9 \\qquad \\text{(E)}\\ 12$ ## Problem 24 Circles centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the midpoint of $\\overline{AB}$, and $OA = 2\\sqrt {2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$? $[asy] unitsize(6mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0); pair A=(-2*sqrt(2),0); pair B=(2*sqrt(2),0); pair G=shift(0,2)*A; pair F=shift(0,2)*B; pair C=shift(A)*scale(2)*dir(45); pair D=shift(B)*scale(2)*dir(135); pair X=A+2*dir(-60); pair Y=B+2*dir(240); path P=C--O--D--Arc(B,2,135,90)--G--Arc(A,2,90,45)--cycle; fill(P,gray); draw(Circle(A,2)); draw(Circle(B,2)); dot(A); label(\"A\",A,W); dot(B); label(\"B\",B,E); dot(C); label(\"C\",C,W); dot(D); label(\"D\",D,E); dot(G); label(\"E\",G,N); dot(F); label(\"F\",F,N); dot(O); label(\"O\",O,S); draw(G--F); draw(C--O--D); draw(A--B); draw(A--X); draw(B--Y); label(\"2\",midpoint(A--X),SW); label(\"2\",midpoint(B--Y),SE); [/asy]$ $\\text{(A)}\\ \\frac {8\\sqrt {2}}{3} \\qquad \\text{(B)}\\ 8\\sqrt {2} - 4 - \\pi \\qquad \\text{(C)}\\ 4\\sqrt {2} \\qquad \\text{(D)}\\ 4\\sqrt {2} + \\frac {\\pi}{8} \\qquad \\text{(E)}\\ 8\\sqrt {2} - 2 - \\frac {\\pi}{2}$ ##", "$[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"P\",P,W); label(\"Q\",Q,W); label(\"R\",R,W); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 32 \\qquad \\text{(D)}\\ 64 \\qquad \\text{(E)}\\ 128$ ## Problem 10 A jacket and a shirt originally sold for dollar 80 and dollar 40, respectively. During a sale Chris bought the dollar80 jacket at a $40\\%$ discount and the dollar 40 shirt at a $55\\%$ discount. The total amount saved was what percent of the total of the original prices? $\\text{(A)}\\ 45\\% \\qquad \\text{(B)}\\ 47\\dfrac{1}{2}\\% \\qquad \\text{(C)}\\ 50\\% \\qquad \\text{(D)}\\ 79\\dfrac{1}{6}\\% \\qquad \\text{(E)}\\ 95\\%$. ## Problem 11 Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to $[asy] for(int i = -2; i <= 2; ++i) { draw((i,0)--(i,3),dashed); } draw((-3,1)--(3,1),dashed); draw((-3,2)--(3,2),dashed); draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle); dot((-3,0)); label(\"A\",(-3,0),SW); dot((-3,3)); label(\"B\",(-3,3),NW); dot((0,3)); label(\"C\",(0,3),N); dot((3,3)); label(\"D\",(3,3),NE); dot((3,0)); label(\"E\",(3,0),SE); dot((0,0)); label(\"start\",(0,0),S); label(\"\\longrightarrow\",(0,-0.75),E); label(\"\\longleftarrow\",(0,-0.75),W); label(\"\\textbf{Jane}\",(0,-1.25),W); label(\"\\textbf{Hector}\",(0,-1.25),E); [/asy]$ $\\text{(A)}\\ A \\qquad \\text{(B)}\\ B \\qquad \\text{(C)}\\ C \\qquad \\text{(D)}\\ D \\qquad \\text{(E)}\\ E$ ## Problem", "that is where we get our x value: $(19, d)$ $(19, -d)$ $(17, -3d)$ Now, we can plug one of the first two value in as well as the last one to get the following equations: $$19^2 + d^2 = r^2$$ $$17^2 + (3d)^2 = r^2$$ Subtracting these two equations, we get $19^2 - 17^2 = 8d^2$ - therefore, we get $72 = 8d^2 \\rightarrow d^2 = 9 \\rightarrow d = 3$. We want to find $2d = 6$ because that's the distance between two chords. So, our answer is $\\boxed{B}$. ~Tony_Li2007 ## Solution 3 (Stewart's Theorem) $[asy] real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label(\"O\", O, N); label(\"C\", C, SW); label(\"D\", D, SE); label(\"E\", E, SW); label(\"F\", F, SE); label(\"P\", P, SW); label(\"Q\", Q, S); [/asy]$ If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\\triangle OCD$ with cevian $\\overleftrightarrow{OP}$ gives $$19\\cdot 38\\cdot 19 + \\tfrac{1}{2}d\\cdot 38\\cdot\\tfrac{1}{2}d=19r^{2}+19r^{2}.$$ This simplifies to $13718+\\tfrac{19}{2}d^{2}=38r^{2}$. Similarly, another round of Stewart's Theorem", "= circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. ## Solution 3 WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$. We claim the angle between $AB$ and $\\omega$ is fixed as $\\omega$ varies. Proof: Perform an inversion at $A$, sending $\\omega_1$ and $\\omega_2$ to two lines $\\ell_1$ and $\\ell_2$ intersecting at $B'$. Then $\\omega$ is sent to a circle tangent to lines $\\ell_1$", "$MN$ down so that $N$ coincides with $B$, and form another right traingle. From that triangle, you can see that the shorter leg is on the left side, having length $a-b$, the longer leg is the same as $MN=x$, and the hypotenuse is $AB=12$. We can write the Pythagorean relationship $(a-b)^2+x^2=12^2$. Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for $a$ in the first equation and $b$ in the second equation, and substitute into the third equation, get an equation only in terms of $x$, and solve), you find that $x=\\sqrt{130}$, so $x^2 = \\boxed{130}$. Solution by Kinglogic ### Full Proof that R, A, B are collinear $[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); label(\"T\", t , NW); [/asy]$ Let $M$ and $N$ be the feet of the perpendicular from $A$ to $PQ$ and $B$ to $PR$ respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so $QM = MP = PN = NR$, since the problem told us $QP = PR$. We will show that $R$ lies on $AB$. Let $T$ be the intersection of circle centered at $B$", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The", "= 2x$. Setting the equations equal, we have $2x = -4x +2 \\implies x = \\frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\\frac13$, so $XY = 1 - 2 \\cdot \\frac13 = \\frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\\dfrac{\\dfrac{1}{3} \\cdot 1}{2} = \\boxed{\\textbf{(E)} \\: \\dfrac16}$. $[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); draw(X--Y,dashed); label(\"A\\: (0,2)\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D \\: (0,0)\",D,SW); label(\"E\",E,E); label(\"F\\: (\\frac12,0)\",F,S); label(\"G\",G,W); label(\"H \\: (\\frac12,1)\",H,N); label(\"Y\",Y,E); label(\"X\",X,W); 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]$ ## Solution 2 Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\\dfrac{1}{2}$. We will solve for the areas of", "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", "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}$", "Find the remainder when $N$ is divided by 144. (Note: $\\binom{a}{b} = 0$ if $a, and $\\binom{0}{0} = 1$.) ### Geometry As in the following diagram, square $ABCD$ and square $CEFG$ are placed side by side (i.e. $C$ is between $B$ and $E$ and $G$ is between $C$ and $D$). If $CE = 14$, $AB > 14$, compute the minimal area of $\\triangle AEG$. $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(real x, real y) { pair P = (x,y); dot(P,linewidth(3)); return P; } int big = 30, small = 14; filldraw((0,big)--(big+small,0)--(big,small)--cycle, rgb(0.9,0.5,0.5)); draw(scale(big)*unitsquare); draw(shift(big,0)*scale(small)*unitsquare); label(\"A\",D2(0,big),NW); label(\"B\",D2(0,0),SW); label(\"C\",D2(big,0),SW); label(\"D\",D2(big,big),N); label(\"E\",D2(big+small,0),SE); label(\"F\",D2(big+small,small),NE); label(\"G\",D2(big,small),NE); [/asy]$ In a rectangular plot of land, a man walks in a very peculiar fashion. Labeling the corners $ABCD$, he starts at $A$ and walks to $C$. Then, he walks to the midpoint of side $AD$, say $A_1$. Then, he walks to the midpoint of side $CD$ say $C_1$, and then the midpoint of $A_1D$ which is $A_2$. He continues in this fashion, indefinitely. The total length of his path if $AB=5$ and $BC=12$ is of the form $a + b\\sqrt{c}$. Find $\\frac{abc}{4}$. Triangle $ABC$ has $AB = 4$, $AC = 5$, and $BC = 6$. An angle bisector is drawn from angle $A$, and meets $BC$ at $M$. What is the nearest integer to $100 \\frac{AM}{CM}$? In regular", "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}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ 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", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "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]$" ]

[ "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.", "pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } for(int i = 0; i < 2; ++i) /* left */ fillrect((s/2^(ceil(i/2)),s/2^(floor(i/2)))); for(int i = 0; i < n; ++i) /* right */ fillrect(shiftR,shiftR + (s/2^(ceil(i/2)),s/2^(floor(i/2)))); label(\"\\frac 12\",(s*3/4,s/2),sm); label(\"\\cdots\",(s*1/4,s/2),sm); label(\"\\frac 12\",shiftR+(s*3/4,s/2),sm); label(\"\\cdots\",shiftR+(s*1/4,s/2),sm); label(\"\\frac 14\",shiftR+(s*1/4,s*3/4),sm); label(\"\\frac 18\",shiftR+(s*3/8,s/4),sm); htick((0,-1), (s,-1)); htick(shiftR + (0,-1), shiftR + (s,-1)); label(\"1\",(s/2,-1),S,sm); label(\"1\",shiftR+(s/2,-1),S,sm); [/asy]$ The infinite geometric series $\\frac 12 + \\frac {1}{2^2} + \\frac {1}{2^3} + \\cdots = 1$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 4; real h = 2; pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0)}; void drawTriGrid(real s){ for(int i = 0; i < 4; ++i){ draw( (-s*3/2,s*(3/2 - i)) -- (s*3/2,s*(3/2 - i)), linetype(\"2 2\")); draw( (s*(3/2 - i),-s*3/2) -- (s*(3/2 - i),s*3/2), linetype(\"2 2\")); } } void fillrect(pair A, pair B, pen p){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, linewidth(1)); } for(int i = 0; i < n; ++i) { fillrect( ((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) , ((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[0]); fillrect(-((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) ,-((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[1]); fillrect( (-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) , (h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[0]); fillrect(-(-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) ,-(h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[1]); drawTriGrid(h/3^i); } [/asy]$ The infinite geometric series $\\frac 13 + \\frac {1}{3^2} + \\frac {1}{3^3} + \\cdots = \\frac 12$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 10; real h = 6; pen colors[] = {rgb(0.9,0,0),rgb(0,0.9,0),rgb(0,0,0.9)}; pair shiftR = (h+3,0); void drawEquilaterals(pair A, real s){ filldraw(A--A+s*expi(2*pi/3)--A+(-s,0)--cycle,colors[0]); filldraw(A--A+s*expi(2*pi/3)--A+s*expi(1*pi/3)--cycle,colors[1]); filldraw(A--A+s*expi(1*pi/3)--A+(s,0)--cycle,colors[2]); } for(int i = 0; i < n; ++i) drawEquilaterals(shiftR + (0,h-h/(2^i)", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$? $\\mathrm{(A)}\\ 3\\qquad \\mathrm{(B)}\\ 7\\qquad \\mathrm{(C)}\\ 13\\qquad \\mathrm{(D)}\\ 37\\qquad \\mathrm{(E)}\\ 43$ ## Problem 23 How many ordered pairs $(m,n)$ of positive integers, with $m \\ge n$, have the property that their squares differ by $96$? $\\text{(A)}\\ 3 \\qquad \\text{(B)}\\ 4 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 9 \\qquad \\text{(E)}\\ 12$ ## Problem 24 Circles centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the midpoint of $\\overline{AB}$, and $OA = 2\\sqrt {2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$? $[asy] unitsize(6mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0); pair A=(-2*sqrt(2),0); pair B=(2*sqrt(2),0); pair G=shift(0,2)*A; pair F=shift(0,2)*B; pair C=shift(A)*scale(2)*dir(45); pair D=shift(B)*scale(2)*dir(135); pair X=A+2*dir(-60); pair Y=B+2*dir(240); path P=C--O--D--Arc(B,2,135,90)--G--Arc(A,2,90,45)--cycle; fill(P,gray); draw(Circle(A,2)); draw(Circle(B,2)); dot(A); label(\"A\",A,W); dot(B); label(\"B\",B,E); dot(C); label(\"C\",C,W); dot(D); label(\"D\",D,E); dot(G); label(\"E\",G,N); dot(F); label(\"F\",F,N); dot(O); label(\"O\",O,S); draw(G--F); draw(C--O--D); draw(A--B); draw(A--X); draw(B--Y); label(\"2\",midpoint(A--X),SW); label(\"2\",midpoint(B--Y),SE); [/asy]$ $\\text{(A)}\\ \\frac {8\\sqrt {2}}{3} \\qquad \\text{(B)}\\ 8\\sqrt {2} - 4 - \\pi \\qquad \\text{(C)}\\ 4\\sqrt {2} \\qquad \\text{(D)}\\ 4\\sqrt {2} + \\frac {\\pi}{8} \\qquad \\text{(E)}\\ 8\\sqrt {2} - 2 - \\frac {\\pi}{2}$ ##", "$[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"P\",P,W); label(\"Q\",Q,W); label(\"R\",R,W); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 32 \\qquad \\text{(D)}\\ 64 \\qquad \\text{(E)}\\ 128$ ## Problem 10 A jacket and a shirt originally sold for dollar 80 and dollar 40, respectively. During a sale Chris bought the dollar80 jacket at a $40\\%$ discount and the dollar 40 shirt at a $55\\%$ discount. The total amount saved was what percent of the total of the original prices? $\\text{(A)}\\ 45\\% \\qquad \\text{(B)}\\ 47\\dfrac{1}{2}\\% \\qquad \\text{(C)}\\ 50\\% \\qquad \\text{(D)}\\ 79\\dfrac{1}{6}\\% \\qquad \\text{(E)}\\ 95\\%$. ## Problem 11 Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to $[asy] for(int i = -2; i <= 2; ++i) { draw((i,0)--(i,3),dashed); } draw((-3,1)--(3,1),dashed); draw((-3,2)--(3,2),dashed); draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle); dot((-3,0)); label(\"A\",(-3,0),SW); dot((-3,3)); label(\"B\",(-3,3),NW); dot((0,3)); label(\"C\",(0,3),N); dot((3,3)); label(\"D\",(3,3),NE); dot((3,0)); label(\"E\",(3,0),SE); dot((0,0)); label(\"start\",(0,0),S); label(\"\\longrightarrow\",(0,-0.75),E); label(\"\\longleftarrow\",(0,-0.75),W); label(\"\\textbf{Jane}\",(0,-1.25),W); label(\"\\textbf{Hector}\",(0,-1.25),E); [/asy]$ $\\text{(A)}\\ A \\qquad \\text{(B)}\\ B \\qquad \\text{(C)}\\ C \\qquad \\text{(D)}\\ D \\qquad \\text{(E)}\\ E$ ## Problem", "sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); /* draw(rightanglemark((0,0),a*expi(rot1),(c,0))); */ filldraw(sq1, rgb(0.95,1,0.95)); filldraw(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); label(\"a\",a/2*expi(rot1),SE,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),SW,sm); label(\"c\",(c/2,-c),S,sm); [/asy]$ COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection. $[asy] defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens pair shiftR = (s1+1,0); // distance between two diagrams pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B); // draw left diagram filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype(\"2 2\")); draw(I--E, linetype(\"2 2\")); draw(I--F, linetype(\"2 2\")); label(\"a\",(A+B)/2,S); label(\"b\",(B+C)/2,NE); label(\"c\",(A+C)/2,NW); label(\"r\",(I+F)/2,W); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C)); pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y); // draw right diagram filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy);", "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \\frac{251}{126}MN$, and $MN = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(A+F)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$1008-x$$\",(G+D)/2,S); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{x} = \\frac{1008-x}{h} \\Longrightarrow h^2 + x^2 - 1008x = 0$$. Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.", "cm. A rock with volume $1000 \\text{ cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise? $\\textbf{(A)}\\ 0.25\\qquad\\textbf{(B)}\\ 0.5\\qquad\\textbf{(C)}\\ 1\\qquad\\textbf{(D)}\\ 1.25\\qquad\\textbf{(E)}\\ 2.5$ Problem 22 Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell? $[asy] path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle); path sw=((0,0)--(1,sqrt(3))); path se=((5,0)--(4,sqrt(3))); draw(cell, linewidth(1)); draw(shift(2,0)*cell, linewidth(1)); draw(shift(4,0)*cell, linewidth(1)); draw(shift(1,3)*cell, linewidth(1)); draw(shift(3,3)*cell, linewidth(1)); draw(shift(2,6)*cell, linewidth(1)); draw(shift(0.45,1.125)*sw, EndArrow); draw(shift(2.45,1.125)*sw, EndArrow); draw(shift(1.45,4.125)*sw, EndArrow); draw(shift(-0.45,1.125)*se, EndArrow); draw(shift(-2.45,1.125)*se, EndArrow); draw(shift(-1.45,4.125)*se, EndArrow); label(\"+\", (1.5,1.5)); label(\"+\", (3.5,1.5)); label(\"+\", (2.5,4.5));[/asy]$ $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 24\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 35$ Problem 23 A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? $\\textbf{(A)}\\ 0\\qquad\\textbf{(B)}\\ 1\\qquad\\textbf{(C)}\\ 2\\qquad\\textbf{(D)}\\ 3\\qquad\\textbf{(E)}\\ 5$ Problem 24 In the multiplication problem below $A$, $B$, $C$, $D$ and are different digits.", "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]", "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 # quick geometry problem, CPhill 0 160 7 $$In the diagram below, \\angle PQR = \\angle PRQ = \\angle STR = \\angle TSR, RQ = 8, and SQ = 3. Find PQ. [asy] pair A,B,C,D,E; A = (0, 0.9); B = (-0.4, 0); C = (0.4, 0); D = (-0.275, 0.16); E = (0.11, 0.65); draw(A--B); draw(A--C); draw(B--C); draw(B--E); draw(C--D); label(\"P\",A,N); label(\"Q\", B, S); label(\"R\", C, S); label(\"S\", D, S); label(\"T\", E, W); [/asy]$$ Jan 21, 2022 #1 0 $$PQ = \\boxed{\\frac{50 \\sqrt{2}}{7}}$$ . Jan 21, 2022 #2 -1 incorrect, man Guest Jan 21, 2022 #3 +676 +2 You should rather say thanks. Also when it's wrong or right. Straight Jan 21, 2022 edited by Straight Jan 21, 2022 #4 0 I meant: Thanks man, but it is wrong. Guest Jan 21, 2022 #5 +117872 +1 So I suppose it is a reasonable answer. Garbage in, garbage out. Melody Jan 21, 2022 #6 -1 I don't care if the problem is garbage. I just want to get the problem right, learn more, and get more knowledge. Guest Jan 21, 2022 #7 +117872 +1 If you want to get something right you must display it in a form suitable for the media. Are you incapable of comprehending that? It says, in Latex, (you haven't bothered to write it", "= circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. ## Solution 3 WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$. We claim the angle between $AB$ and $\\omega$ is fixed as $\\omega$ varies. Proof: Perform an inversion at $A$, sending $\\omega_1$ and $\\omega_2$ to two lines $\\ell_1$ and $\\ell_2$ intersecting at $B'$. Then $\\omega$ is sent to a circle tangent to lines $\\ell_1$", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\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 ##", "OPR = \\angle PRQ + \\angle PQR = 90^\\circ + \\angle PQR$, which must equal $\\angle OR'P'$. Therefore, $\\angle Q'R'P' = \\angle OR'P' - \\angle OR'Q' = 90^\\circ + \\angle PQR - \\angle OQR$. But $\\angle OQR = \\angle PQR$. Therefore, $\\angle Q'R'P' = 90^\\circ$. As this holds for any $R$, all points on circle $C$ will invert to a point on a circle with diameter $\\overline{Q'P'}$. ## 3. General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle $[asy] unitsize(7); int radiusInverse = 12; int k = radiusInverse * radiusInverse; int radiusCircle = 4; int circleCenterLocation = 7; path inverse = Circle((0, 0), radiusInverse); draw(inverse, dashed); label(\"\\Omega\", inverse, NW); pair O = (0, 0); dot(O); label(\"O\", O, NW); pair C = (circleCenterLocation, 0); dot(C); label(\"C_1\", C, S); path circle1 = Circle(C, radiusCircle); draw(circle1); int p = circleCenterLocation - radiusCircle; int q = circleCenterLocation + radiusCircle; pair P = (p, 0); pair Q = (q, 0); pair A = (k / p, 0); pair B = (k / q, 0); pair D = (((k / p) + (k / q)) / 2, 0); dot(P); dot(Q); dot(A); dot(B); dot(D); label(\"C_2\", D, S); label(\"P\", P, SE); label(\"Q\", Q, SW); label(\"Q'\", B, SE); label(\"P'\", A, SE); draw(O--A); path circle2 = Circle(D, ((k / p)", "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", "O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label(\"K\", K, SW); label(\"L\", L, NW); label(\"M\", M, NE); label(\"N\", N, SE); draw(L--P); label(\"P\", P, S); dot(O); label(\"O\", shift((1,1))*O, NNE); label(\"28\", scale(1/2)*L, W); label(\"65\", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$ Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\\sqrt{720+x^2}-x$. Using Power of a Point, we have $$(KO)(OM)=(LO)(OQ)$$ ($Q$ is the intersection of $OL$ with the circle $\\neq L$) $$8(MO)=(\\sqrt{720+x^2}+x)(\\sqrt{720+x^2}-x)$$ $$8(MO)=720$$ $$MO=\\boxed{090}$$. (Solution by RootThreeOverTwo) ## Solution 7 (just one pair of similar triangles) $[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label(\"K\", K, SW); label(\"L\", L, NW); label(\"M\", M, NE); label(\"N\", N, SE); draw(L--P); label(\"P\", P, S); dot(O); label(\"O\", shift((1,1))*O, NNE); label(\"28\", scale(1/2)*L, W); label(\"65\", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$ Note that since $\\angle KLN = \\angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have $$\\angle LMK = \\angle LNK = 90^{\\circ} - \\angle LKN = \\angle KLP,$$so $\\triangle KLO \\sim \\triangle KML$, giving $$\\frac{KM}{28} = \\frac{28}{8} \\implies KM = 98.$$ Therefore, $OM = 98-8 = \\boxed{90}$.", "# Problem #2201 2201 Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum? $[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label(\"1\",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label(\"2\",(-0.5,0.5)); draw(unitsquare,p); label(\"32\",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label(\"16\",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label(\"4\",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label(\"8\",(0.5,-0.5)); [/asy]$ $\\mathrm{(A)}\\ 154\\qquad\\mathrm{(B)}\\ 159\\qquad\\mathrm{(C)}\\ 164\\qquad\\mathrm{(D)}\\ 167\\qquad\\mathrm{(E)}\\ 189$ 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.", "$m=1/3$, 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: $\\boxed{-\\frac{1}{6}}$.", "n = floor(n2*(n2+1)/2); // number of colors, number of layers real h = 0.6; // scale factor of diagram pair shiftR1 = (n*h+1,0), // middle diagram shift offset shiftR2 = shiftR1 + (n*h+1,0); // right diagram shift offset int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ pen colors(int i) { return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */ /* draw tick line with label, segment between A and B */ void htick(pair A, pair B,pair ticklength = (0.15,0)) { draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } /* gradient triangle */ for(int i = 0; i < n; ++i){ for(int j = 0; j < 2*i+1; ++j){ filldraw(shift(shiftR1)*scale(h)*shift((j-i,-i))*unitsquare,colors(i)); /* if(j % lvl(i) == 0 && j != lvl(i)^2) draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((0,0)--(0,1)--(1,1)), heavy); if(j == 2*i) // right border draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((1,0)--(1,1)--(0,1)), heavy); */ } draw(shift(shiftR1)*scale(h)*shift((-i,-i))*((0,0)--(0,1)--(1,1)), heavy); draw(shift(shiftR1)*scale(h)*shift(( i,-i))*((1,0)--(1,1)--(0,1)), heavy); } // return kth triangular number (actually, 1+2+...+k) int tri(int k) { return ((int) (k*(k+1)/2)); } for(int i = 0; i < n2; ++i) { draw(shift(shiftR1)*scale(h)*shift((0-tri(i),0-tri(i)))*((0,1)--(2*tri(i)+1,1)), heavy); if(i % 2 == 0) { // vertical heavy lines for odd layers draw(shift(shiftR1)*scale(h)*((-i/2,1-tri(i))--(-i/2,-i-tri(i))), heavy); draw(shift(shiftR1)*scale(h)*((1+i/2,1-tri(i))--(1+i/2,-i-tri(i))), heavy); } else { // jagged heavy lines for even layers pair jag1 = (-(i+1)/2,-(i-1)/2-tri(i)), jag2 = (1+(i+1)/2,-(i-1)/2-tri(i)); draw(shift(shiftR1)*scale(h)*(jag1+(0,1+(i-1)/2) -- jag1 -- jag1+(1,0) -- jag1+( 1,-(i+1)/2)), heavy); draw(shift(shiftR1)*scale(h)*(jag2+(0,1+(i-1)/2) -- jag2 -- jag2-(1,0) -- jag2+(-1,-(i+1)/2)), heavy); } } draw(shift(shiftR1)*scale(h)*shift((-n2*(n2+1)/2,-n2*(n2+1)/2))*((1,1)--(2*n2*(n2+1)/2,1)),", "# 1988 AHSME Problems/Problem 20 ## Problem In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label(\"A\", A, NW); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, NE); label(\"E\", E, S); label(\"F\", F, N); label(\"G\", G, E); label(\"X\", X, NW); label(\"Y\", Y, SW); label(\"Z\", Z, SE); label(\"W\", W, NE); [/asy]$ $\\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 1+2\\sqrt{3}\\qquad \\textbf{(C)}\\ 2\\sqrt{5}\\qquad \\textbf{(D)}\\ \\frac{8+4\\sqrt{3}}{3}\\qquad \\textbf{(E)}\\ 5$ ## Solution Within $WXYZ$, the parallelogram piece has vertical side $BF = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\\triangle AFG$, so we deduce $YZ = AG = \\frac{\\frac{1}{2} \\times 1 \\times 2}{\\frac{1}{2} \\times \\sqrt{5}} = \\frac{2}{\\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put" ]

[ "# What is the area of an equilateral triangle with a perimeter of 21 inches? Apr 22, 2018 $\\frac{49 \\sqrt{3}}{4}$ #### Explanation: Let side length be $a$ inches. Then $3 a = 21 \\text{ \"rArr\" } a = 7$ Area of an equilateral triangle is given by $\\text{ } \\frac{\\sqrt{3} {a}^{2}}{4}$ $\\Rightarrow \\text{ } \\frac{\\sqrt{3} \\times {7}^{2}}{4} = \\frac{49 \\sqrt{3}}{4}$", "The perimeter of an equilateral triangle is 51 inches. What is the length of each side? Mar 17, 2016 $\\frac{51}{3} = 17 ' ' = 43.2 c m$", "# The perimeter of an equilateral triangle is 51 inches. What is the length of each side? $\\frac{51}{3} = 17 ' ' = 43.2 c m$", "the side length of the square. What is the perimeter of the... more Story Problems 09/18/16 #### A square and an equilateral triangle have a combined perimeter of 101 inches. A square and an equilateral triangle have a combined perimeter of 101 inches. The side length of the triangle is 8 inches less than twice the side length of the square. What is the perimeter of the... more ## Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. #### OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "# What is the area of an equilateral triangle whose perimeter is 48 inches? Aug 5, 2018 Answer: $64 \\sqrt{3}$ ${\\text{in}}^{2}$ #### Explanation: Consider the formula for the area of an equilateral triangle: $\\frac{{s}^{2} \\sqrt{3}}{4}$, where $s$ is the side length (this can be easily proved by considering the 30-60-90 triangles within an equilateral triangle; this proof will be left as an exercise for the reader) Since we are given that the perimeter of the equilateral trangle is $48$ inches, we know that the side length is $\\frac{48}{3} = 16$ inches. Now, we can simply plug this value into the formula: $\\frac{{s}^{2} \\sqrt{3}}{4} = \\frac{{\\left(16\\right)}^{2} \\sqrt{3}}{4}$ Canceling, a $4$ from the numerator and the denominator, we have: $= \\left(16 \\cdot 4\\right) \\sqrt{3}$ $= 64 \\sqrt{3}$ ${\\text{in}}^{2}$, which is our final answer.", "# A equilateral triangle measures 3 1/2 inches on one side. what is the perimeter of the triangle? $10.5$ $3.5 + 3.5 + 3.5$", "problem is in the units, as @hardmath comments. – Ethan Bolker Jul 11 '18 at 16:37 Well, even for a square, your question is not very interesting. The maximum of area-perimeter ratio is $+\\infty$. Just magnify your triangle or square, area increases quadratically, while perimeter only linearly increases. You should take a ratio of area to the perimeter squared to get meaningful (independent from the scale/units) results. And yes, then the ratio is maximum for equilateral triangle. Area and perimeter have different unites so the ratio changes with the size of triangle. The ratio of the area of an equilateral triangle of size $a$ to its perimeter is $$R=\\frac {a\\sqrt 3}{12}$$ As you see, the ration grows with the size of triangle. For $a=4\\sqrt 3$ we have $R=1$ and the ratio grows to $\\infty$ as the triangle enlarges.", "# A Hexagon and a Triangle Geometry Level 2 A regular hexagon and an equilateral triangle have the same perimeter of 24 units, what is the ratio of the area of the triangle to the area of the hexagon? Express as a common fraction. ×", "perimeters of 48 inches. What is the ratio of the length of the side of the triangle to the length of the side of the square? Express your answer as a common fraction. Thursday, March 26, 2015 at 12:02pm Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>> Members", "4 + 4 cm = 3 × 4 cm = 12 cm So, we find that Perimeter of an equilateral triangle = 3 × length of a side What is similar between a square and an equilateral triangle? They are figures having all the sides of equal length and all the angles of equal measure. Such figures are known as regular closed figures. Thus, a square and an equilateral triangle are regular closed figures. Find various objects from your surroundings which have regular shapes and find their perimeters. You found that, Perimeter of a square = 4 × length of one side Perimeter of an equilateral triangle = 3 × length of one side So, what will be the perimeter of a regular pentagon? A regular pentagon has five equal sides. Therefore, perimeter of a regular pentagon = 5 × length of one side and the perimeter of a regular hexagon will be _______ and of an octagon will be ______. Example 6 : Find the distance travelled by Shaina if she takes three rounds of a square park of side 70 m. Solution : Perimeter of the square park = 4 × length of a side = 4 × 70 m = 280 m Distance covered in one round = 280 m Therefore, distance travelled in three", "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? ×", "perimeter is now 60 in. The length of each side of an equilateral triangle is increased by 5 in, so the perimeter is now 60 in.... -- 0.045737--", "# What is the ratio of the measure of one angle of a square to the measure of one angle of an equilateral triangle? $\\frac{\\frac{\\pi}{2}}{\\frac{\\pi}{3}} = \\frac{3}{2}$ rad", "a side of equilateral triangle E and y be the length of a side of square F. Question: is x>y? (1) The perimeter of E and the perimeter of F are equal --> 3x=4y --> x/y=4/3 --> x>y. Sufficient. (2) The ratio of the height of triangle E to the diagonal of square F is 2√3 : 3√2 --> the height of triangle E is $$x\\frac{\\sqrt{3}}{2}$$ and the diagonal of square F is $$y\\sqrt{2}$$ --> ratio: $$\\frac{x\\frac{\\sqrt{3}}{2}}{y\\sqrt{2}}=\\frac{2\\sqrt{3}}{3\\sqrt{2}}$$ --> x/y=4/3 --> x>y. Sufficient. Bunuel - Could you explain how you simplified the ratio in statement #2 to get to x/y = 4/3? Intern Joined: 07 Feb 2011 Posts: 14 GMAT 1: 580 Q47 V24 Re: Is the length of a side of equilateral triangle E less than [#permalink] ### Show Tags 04 Mar 2014, 21:25 bparrish89 wrote: Bunuel wrote: Is the length of a side of equilateral triangle E less than the length of a side of square F? Let x be the length of a side of equilateral triangle E and y be the length of a side of square F. Question: is x>y? (1) The perimeter of E and the perimeter of F are equal --> 3x=4y --> x/y=4/3 --> x>y. Sufficient. (2) The ratio of the height of triangle E to the diagonal of square F", "# A square and an equilateral triangle have the same perimeter. Each side of the triangle is 16 m. How do you find the length of each side of the square? Jan 27, 2016 ${\\text{Side}}_{\\square} = 8 \\sqrt[4]{3}$ #### Explanation: The semi-perimeter of the equilateral triangle is $\\textcolor{w h i t e}{\\text{XXX}} s = \\frac{16 + 16 + 16}{2} = 24$ So, by Heron's formula color(white)(\"XXX\")\"Area\"_triangle = sqrt(24xx8xx8xx8) =8^2sqrt(3) We are told ${\\text{Area\"_square = \"Area}}_{\\triangle}$ So Area\"_square = 8^2sqrt(3) and each side of the square has a length of $\\textcolor{w h i t e}{\\text{XXX}} \\sqrt{{8}^{2} \\sqrt{3}} = 8 \\sqrt[4]{3}$", "# Side of the Triangle • Oct 3rd 2008, 04:59 AM magentarita Side of the Triangle The area of a square is 9. If this square and an equilateral triangle have equal perimeters, find the length of one side of the triangle. • Oct 3rd 2008, 05:12 AM masters Quote: Originally Posted by magentarita The area of a square is 9. If this square and an equilateral triangle have equal perimeters, find the length of one side of the triangle. Area of square = 9 $A=s^2$ $9=s^2$ $s=3$ (side of square) Perimeter of Square = 4(3) = 12 Perimeter of equilateral triangle = 12 P = 3s 12 = 3s s = 4 (side of triangle) • Oct 3rd 2008, 04:46 PM magentarita great...... Quote: Originally Posted by masters Area of square = 9 $A=s^2$ $9=s^2$ $s=3$ (side of square) Perimeter of Square = 4(3) = 12 Perimeter of equilateral triangle = 12 P = 3s 12 = 3s s = 4 (side of triangle)", "# How do you find the area of a given hexagon is equal to the area of an equilateral triangle whose perimeter is 36 inches? May 27, 2015 Warning: I suspect there is something wrong in the statement of this question; however.... If the perimeter of an equilateral triangle is 36 inches, each side is 12 inches and by the Pythagorean Theorem the height is $\\sqrt{{12}^{2} - {6}^{2}} = \\sqrt{108} = 6 \\sqrt{3}$ The area of the triangle is $A = \\frac{1}{2} b h$ $= \\frac{1}{2} \\cdot 12 \\cdot 6 \\sqrt{3} \\text{square inches}$ $= 36 \\sqrt{3} \\text{square inches}$", "# 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!", "ratio of 1: 2: 2 and the perimeter of the triangle is 15 feet. the length of the shortest side is?...", "Find the dimensions of the rectangle with the largest area that can be inscribed in an equilateral triangle of side 5.5 inches if one side of the rectangle lies on the base of the triangle." ]

[ "# Triangle A has sides of lengths 36 , 48 , and 24 . Triangle B is similar to triangle A and has a side of length 4 . What are the possible lengths of the other two sides of triangle B? Mar 6, 2016 In similar triangles the ratios of corresponding sides are the same. #### Explanation: So now there are three possibilities, according to which of the sides of triangle A the 4 corresponds to: If $4 \\leftrightarrow 36$ then the ratio=$\\frac{36}{4} = 9$ and the other sides will be: $\\frac{48}{9} = 5 \\frac{1}{3}$ and $\\frac{24}{9} = 2 \\frac{2}{3}$ If $4 \\leftrightarrow 48$ then the ratio=$\\frac{48}{4} = 12$ and the other sides are: $\\frac{36}{12} = 3$ and $\\frac{24}{12} = 2$ If $4 \\leftrightarrow 24$ the the ratio=$\\frac{24}{4} = 6$ and the other sides are: $\\frac{36}{6} = 6$ and $\\frac{48}{6} = 8$", "two angles and the non-included side of the orange triangle. Pair C is ASA Congruency since two angles and an included side of the green triangle is equal to the corresponding two angles and included side of the pink triangle. Almost done! Here is one more example for you. Two similar solids have side lengths in the ratio $$4:11$$. 1. What is the ratio of their volumes? 2. The smaller solid has a volume of 200 cm3. What is the volume of the larger solid? Solution Let us denote the smaller solid by X and the larger solid by Y and the side length of X and Y by $$x$$ and $$y$$ respectively. The ratio of their side lengths is written as $$x:y$$ and is given by $$4:11$$. Question 1: Recall that if two similar shapes have sides in the ratio $$x:y$$, then the ratio of their areas is $$x^2:y^2$$. Thus, we would simply need to square the components in the ratio of side lengths X and Y to calculate the ratio of their volumes. The square of 4 and 11 is 16 and 121 respectively. Thus, the ratio of Volume X to Volume Y is $16:121$ Question 2: Expressing this ratio into fractions, we have $\\frac{\\text{Volume X}}{\\text{Volume Y}}=\\frac{16}{121}$ Now noting the given volume of X, $\\frac{200}{\\text{Volume Y}}=\\frac{16}{121}$", "## Geometry: Common Core (15th Edition) Find the corresponding side ratio. $\\frac{8}{6}$=$\\frac{4}{3}$ The ratio of the perimeters is equal to the ratio of the corresponding sides. $\\frac{4}{3}$=$\\frac{4}{3}$. The ratio of the areas is equal to the square of the ratio of corresponding sides. $\\frac{4^{2}}{3^{2}}$=$\\frac{16}{9}$.", "## anonymous 5 years ago A square with sides of length s is inscribed in an equilateral triangle with sides of length t. Find the exact ratio of the area of the equilateral triangle to the area of the square. 1. anonymous ok i think its something like 4s=3t but i think its wrong. thats my best shot at it lol. sorry haha 2. anonymous 3. anonymous nice job nancy therefore the ratio of the areas is: $\\frac{12+ 7\\sqrt{3}}{12}$", "## Algebra 1 Set up a ratio to find the missing length. Since the triangles are similar, the ratio between the lengths of corresponding sides is the same. $\\frac{5}{9}=\\frac{7}{x}\\longrightarrow$ cross multiply $5x=63\\longrightarrow$ divide each side by 5 $5x\\div5=63\\div5\\longrightarrow$ divide $x=12\\frac{3}{5}$", "Q # Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 Q9 Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 Views Sides of two similar triangles are in the ratio 4 : 9. Let triangles be ABC and DEF. We know that $\\frac{ar(\\triangle ABC)}{ar(\\triangle DEF)}=\\frac{AB^2}{DE^2}=\\frac{4^2}{9^2}=\\frac{16}{81}$ Option D is correct. Exams Articles Questions", "## Prealgebra (7th Edition) $x=17.4$ In similar triangles, the ratio of the lengths of corresponding sides is the same. $\\frac{x}{24}=\\frac{5.8}{8}$ $8x=24\\times5.8$ $8x\\div8=139.2\\div8$ $x=17.4$", "# Thread: Ratio of a Triangle 1. ## Ratio of a Triangle I am having problems finding the ratio of this can someone help me please. 2. Simply set up your equations and solve for the variables: $\\frac{AB}{DE} = \\frac{5}{4}$ $\\frac{x + 16}{16} = \\frac{5}{4}$ $\\frac{AC}{DF} = \\frac{5}{4}$ $\\frac{y + 7}{20} = \\frac{5}{4}$ Once you solve for them, figure out what $\\frac{x}{y}$ is. 3. ## can i do this can i write it like 20(16+x)=16(17+y) i think thats how he wants it. 4. ## ratio how wo9uld i findthe ratio?", "### Home > MC2 > Chapter 9 > Lesson 9.3.3 > Problem9-113 9-113. One of the triangles below is a reduction of the other. Find each of the missing parts listed below. 1. Find the length of side BC. What is the scale factor between the two triangles? $\\frac{120}{15} = \\frac{160}{x}$ $x = 20$ cm 2. Find the length of side AB. See (a). 3. Find the ratio of the areas of the two triangles. If the ratio of the perimeters is the scale factor, the ratio of the areas is the scale factor squared. $\\frac{1}{64}$", "\\frac{τ_1 τ_3}{6r_2^3} (1 + b_3) . \\label{13.8.35a,b} \\tag{13.8.35a,b}$ These will serve as better approximations for the triangle ratios. Be aware, however, that equations 13.8.35a,b are only approximations, and do not give the exact values for $$a_1$$ and $$a_3$$.", "# Triangle A has sides of lengths 36 ,24 , and 18 . Triangle B is similar to triangle A and has a side of length 5 . What are the possible lengths of the other two sides of triangle B? May 3, 2016 There are 3 different triangle possible because we do not know which side of the smaller triangle is equal to 5. #### Explanation: In similar figures. the sides are in the same ratio. However in this case, we are not told which side of the smaller triangle has a length of 5. There are therefore 3 possibilities. $\\frac{36}{5} = \\frac{24}{3 \\frac{1}{3}} = \\frac{18}{2.5}$ [Each side is divided by 7.2] $\\frac{36}{7.5} = \\frac{24}{5} = \\frac{18}{3.7} .5$ [Each side is divided by 4.8] $\\frac{36}{10} = \\frac{24}{6 \\frac{2}{3}} = \\frac{18}{5}$ [Each side is divided by 3.6]", "### Home > GC > Chapter 11 > Lesson 11.1.3 > Problem11-38 11-38. Remember that the ratios of corresponding sides are equal if two triangles are similar. $\\frac{4}{3}$ $\\text{Base\\:area\\:of\\:solid} = (27)\\left(\\frac{4}{3}\\right)^2$", "# What is the ratio of the measure of one angle of a square to the measure of one angle of an equilateral triangle? $\\frac{\\frac{\\pi}{2}}{\\frac{\\pi}{3}} = \\frac{3}{2}$ rad", "with ratio $\\frac{a}{a+1}.$ The squares contained within the triangles have side lengths equal to the height of the next triangle and the total area is a geometric series.\\begin{aligned} \\sum_{n=1}^{\\infty}\\bigg(\\frac{a}{a+1}\\bigg)^{2n} &= \\frac{\\big(\\frac{a}{a+1}\\big)^2}{1-\\big(\\frac{a}{a+1}\\big)^2} \\\\[7pt]&= \\frac{a^2}{2a+1} \\end{aligned}", "## Geometry: Common Core (15th Edition) Find the corresponding side ratio. $\\frac{15}{25}$=$\\frac{3}{5}$ The ratio of the perimeters is equal to the ratio of the corresponding sides. $\\frac{3}{5}$=$\\frac{3}{5}$. This can be written as 3:5 as well. The ratio of the areas is equal to the square of the ratio of the corresponding sides. $\\frac{3^{2}}{5^{2}}$=$\\frac{9}{25}$. This can also be written as 9:25", "same ratio. $$\\frac{4}{8}=\\frac{6}{12}=\\frac{8}{16}=\\frac{1}{2}$$ Triangles A and B are not similar because the ratios of their corresponding sides are not in the same ratio. $$\\frac{6}{16} \\neq \\frac{4}{12}$$ Further, if triangle A is not similar to triangle B, then triangle A is not similar to triangle C b. Triangles A and B are not similar because their corresponding sides are not oil in the same ratio. Two pairs of corresponding sides are proportional, but the third pair of corresponding sides is not. $$\\frac{3}{6} \\neq \\frac{5}{7.5}=\\frac{6}{9}$$ Triangles B and C are not similar because their corresponding sides are not in the same ratio. $$\\frac{6}{6} \\neq \\frac{7}{7.5} \\neq \\frac{8}{9}$$ Triangles A and C are not similar because their corresponding sides are not in the same ratio. $$\\frac{3}{6} \\neq \\frac{5}{7} \\neq \\frac{6}{8}$$ c. Triangles B and D are the only similar triangles because they have the same angle measures. Using the angle sum of a triangle, both triangles B and D have angles of 75°, 60°, and 45°. d. Triangles A and B are similar because they have two pairs of corresponding sides that are in the same ratio, and their included angles are equal measures. Triangle C cannot be shown to be similar because, even though it has two sides that are the same length as two sides of triangle A, the", "# Triangle A has sides of lengths 12 ,1 4, and 11 . Triangle B is similar to triangle A and has a side of length 4 . What are the possible lengths of the other two sides of triangle B? Jan 23, 2017 the other two sides are: 1) $\\frac{14}{3} \\mathmr{and} \\frac{11}{3}$ or 2) $\\frac{24}{7}$ and $\\frac{22}{7}$ or 3) $\\frac{48}{11}$ and $\\frac{56}{11}$ #### Explanation: Since B and A are similar their sides are in the following possible ratios: $\\frac{4}{12} \\mathmr{and} \\frac{4}{14} \\mathmr{and} \\frac{4}{11}$ 1) ratio$= \\frac{4}{12} = \\frac{1}{3}$: the other two sides of A are $14 \\cdot \\frac{1}{3} = \\frac{14}{3}$ and $11 \\cdot \\frac{1}{3} = \\frac{11}{3}$ 2) ratio$= \\frac{4}{14} = \\frac{2}{7}$: the other two sides are $12 \\cdot \\frac{2}{7} = \\frac{24}{7}$ and $11 \\cdot \\frac{2}{7} = \\frac{22}{7}$ 3) ratio$= \\frac{4}{11}$: the other two sides are $12 \\cdot \\frac{4}{11} = \\frac{48}{11}$ and $14 \\cdot \\frac{4}{11} = \\frac{56}{11}$", "# The Areas of Two Similar Triangles Are in Respectively 9 Cm2 and 16 Cm2. the Ratio of Their Corresponding Sides is (A) 3 : 4 (B) 4 : 3 (C) 2 : 3 (D) 4 : 5 - Mathematics MCQ The areas of two similar triangles are in respectively 9 cm2 and 16 cm2. The ratio of their corresponding sides is • 3 : 4 • 4 : 3 • 2 : 3 • 4 : 5 #### Solution Given: Areas of two similar triangles are 9cm2 and 16cm2. To find: Ratio of their corresponding sides. We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. \\text{ar(tringle 1)}/\\text{ar(tringle 2)}=(\\text{side1}/\\text{side2})^2 9/12=(\\text{side1}/\\text{side2})^2 Taking square root on both sides, we get $\\frac{\\text{side1}}{\\text{side2}} = \\frac{3}{4}$ So, the ratio of their corresponding sides is 3 : 4. Hence the correct answer is a Concept: Triangles Examples and Solutions Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 10 Maths Chapter 7 Triangles Q 2 | Page 131", "# The ratio of the side lengths of an isosceles triangle is 4:4:7, and its perimeter is 52.5cm. What is the length of the base of the triangle? Aug 2, 2016 $24 \\frac{1}{2} \\to 24.5$ #### Explanation: The given values of $4 : 4 : 7$ are a ratio consisting of a total count of $4 + 4 + 7 = 15$ parts As this is an Isosceles triangle the base is 7. However the 7 parts is out of 15 parts. So as a fraction of the whole perimeter it is $\\frac{7}{15}$ Thus the length of the base of the triangle is: $\\frac{7}{15} \\times 52 \\frac{1}{2}$ $\\frac{7}{{\\cancel{15}}^{1}} \\times \\frac{{\\cancel{105}}^{7}}{2} \\text{ \"=\" \"49/2\" \"=\" } 24 \\frac{1}{2}$", ": 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 Solution: (D) : We have two similar triangles such that the ratio of their corresponding sides is 4 : 9. ∴ The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ∴ $$\\frac { { ar }(\\Delta -I) }{ { ar }(\\Delta -II) } =\\left( \\frac { 4 }{ 9 } \\right) ^{ 2 }=\\frac { 16 }{ 81 }$$ ⇒ ar(∆-I): ar(∆-II) = 16 : 81" ]

[ "# What is the area of an equilateral triangle with a perimeter of 21 inches? Apr 22, 2018 $\\frac{49 \\sqrt{3}}{4}$ #### Explanation: Let side length be $a$ inches. Then $3 a = 21 \\text{ \"rArr\" } a = 7$ Area of an equilateral triangle is given by $\\text{ } \\frac{\\sqrt{3} {a}^{2}}{4}$ $\\Rightarrow \\text{ } \\frac{\\sqrt{3} \\times {7}^{2}}{4} = \\frac{49 \\sqrt{3}}{4}$", "The perimeter of an equilateral triangle is 51 inches. What is the length of each side? Mar 17, 2016 $\\frac{51}{3} = 17 ' ' = 43.2 c m$", "# The perimeter of an equilateral triangle is 51 inches. What is the length of each side? $\\frac{51}{3} = 17 ' ' = 43.2 c m$", "The side of a square is equal to the side of an equilateral triangle. Question: The side of a square is equal to the side of an equilateral triangle. The ratio of their areas is (a) 4 : 3 (b) $2: \\sqrt{3}$ (c) $4: \\sqrt{3}$ (d) none of these Solution: (c) $4: \\sqrt{3}$ Let: Length of the side of the square = Length of the side of the equilateral triangle = a unit Now, Area of the square $=a \\times a=a^{2}$ unit $^{2}$ Area of the equilateral triangle $=\\frac{\\sqrt{3}}{4} a^{2} \\mathrm{unit}^{2}$ Ratio of areas $=\\frac{\\text { Area of the square }}{\\text { Area of the equilateral triangle }}$ $=\\frac{a^{2}}{\\frac{\\sqrt{3}}{4} a^{2}}$ $=\\frac{4}{\\sqrt{3}}$", "# What is the ratio of the measure of one angle of a square to the measure of one angle of an equilateral triangle? $\\frac{\\frac{\\pi}{2}}{\\frac{\\pi}{3}} = \\frac{3}{2}$ rad", "# What is the area of an equilateral triangle whose perimeter is 48 inches? Aug 5, 2018 Answer: $64 \\sqrt{3}$ ${\\text{in}}^{2}$ #### Explanation: Consider the formula for the area of an equilateral triangle: $\\frac{{s}^{2} \\sqrt{3}}{4}$, where $s$ is the side length (this can be easily proved by considering the 30-60-90 triangles within an equilateral triangle; this proof will be left as an exercise for the reader) Since we are given that the perimeter of the equilateral trangle is $48$ inches, we know that the side length is $\\frac{48}{3} = 16$ inches. Now, we can simply plug this value into the formula: $\\frac{{s}^{2} \\sqrt{3}}{4} = \\frac{{\\left(16\\right)}^{2} \\sqrt{3}}{4}$ Canceling, a $4$ from the numerator and the denominator, we have: $= \\left(16 \\cdot 4\\right) \\sqrt{3}$ $= 64 \\sqrt{3}$ ${\\text{in}}^{2}$, which is our final answer.", "a side of equilateral triangle E and y be the length of a side of square F. Question: is x>y? (1) The perimeter of E and the perimeter of F are equal --> 3x=4y --> x/y=4/3 --> x>y. Sufficient. (2) The ratio of the height of triangle E to the diagonal of square F is 2√3 : 3√2 --> the height of triangle E is $$x\\frac{\\sqrt{3}}{2}$$ and the diagonal of square F is $$y\\sqrt{2}$$ --> ratio: $$\\frac{x\\frac{\\sqrt{3}}{2}}{y\\sqrt{2}}=\\frac{2\\sqrt{3}}{3\\sqrt{2}}$$ --> x/y=4/3 --> x>y. Sufficient. Bunuel - Could you explain how you simplified the ratio in statement #2 to get to x/y = 4/3? Intern Joined: 07 Feb 2011 Posts: 14 GMAT 1: 580 Q47 V24 Re: Is the length of a side of equilateral triangle E less than [#permalink] ### Show Tags 04 Mar 2014, 21:25 bparrish89 wrote: Bunuel wrote: Is the length of a side of equilateral triangle E less than the length of a side of square F? Let x be the length of a side of equilateral triangle E and y be the length of a side of square F. Question: is x>y? (1) The perimeter of E and the perimeter of F are equal --> 3x=4y --> x/y=4/3 --> x>y. Sufficient. (2) The ratio of the height of triangle E to the diagonal of square F", "# Geometry Puzzle – Challenge 73 Time for another great math challenge for those who enjoy solving mathematics and critical thinking challenges. ## Challenge: The perimeter of a square is 13 cm and the perimeter of an equilateral triangle is 7 cm. What is the difference of a side of the square and a side of a triangle? A- $$\\frac{7}{13}$$ cm B- $$\\frac{11}{12}$$ cm C- $$1\\frac{1}{4}$$ cm D- $$2\\frac{1}{4}$$ cm E- $$\\frac{13}{7}$$ cm ### The Absolute Best Book to challenge your Smart Student! The perimeter of a square is 13cm. Therefore, one side of the square is $$\\frac{13}{4}$$ cm. The perimeter of the equilateral triangle is 7cm. Therefore, one side of the triangle is $$\\frac{7}{3}$$ cm. $$\\frac{13}{4} – \\frac{7}{3} = \\frac{39-28}{12} = \\frac{11}{12}$$ cm", "problem is in the units, as @hardmath comments. – Ethan Bolker Jul 11 '18 at 16:37 Well, even for a square, your question is not very interesting. The maximum of area-perimeter ratio is $+\\infty$. Just magnify your triangle or square, area increases quadratically, while perimeter only linearly increases. You should take a ratio of area to the perimeter squared to get meaningful (independent from the scale/units) results. And yes, then the ratio is maximum for equilateral triangle. Area and perimeter have different unites so the ratio changes with the size of triangle. The ratio of the area of an equilateral triangle of size $a$ to its perimeter is $$R=\\frac {a\\sqrt 3}{12}$$ As you see, the ration grows with the size of triangle. For $a=4\\sqrt 3$ we have $R=1$ and the ratio grows to $\\infty$ as the triangle enlarges.", "the length of a side of equilateral triangle E and y be the length of a side of square F. Question: is x>y? (1) The perimeter of E and the perimeter of F are equal --> 3x=4y --> x/y=4/3 --> x>y. Sufficient. (2) The ratio of the height of triangle E to the diagonal of square F is 2√3 : 3√2 --> the height of triangle E is $$x\\frac{\\sqrt{3}}{2}$$ and the diagonal of square F is $$y\\sqrt{2}$$ --> ratio: $$\\frac{x\\frac{\\sqrt{3}}{2}}{y\\sqrt{2}}=\\frac{2\\sqrt{3}}{3\\sqrt{2}}$$ --> x/y=4/3 --> x>y. Sufficient. Bunuel - Could you explain how you simplified the ratio in statement #2 to get to x/y = 4/3? The second statement states the ratio as 2√3 : 3√2 &, the calculated ratio is x√3/2 : y√2. Now if these two ratios are same, we just need to simplify the equation, which gives the ratio of x:y to 4:3. Math Expert Joined: 02 Sep 2009 Posts: 34527 Followers: 6313 Kudos [?]: 80121 [0], given: 10027 Re: Is the length of a side of equilateral triangle E less than [#permalink] ### Show Tags 05 Mar 2014, 01:09 bparrish89 wrote: Bunuel wrote: Is the length of a side of equilateral triangle E less than the length of a side of square F? Let x be the length of a side of equilateral triangle E and", "is 2√3 : 3√2 --> the height of triangle E is $$x\\frac{\\sqrt{3}}{2}$$ and the diagonal of square F is $$y\\sqrt{2}$$ --> ratio: $$\\frac{x\\frac{\\sqrt{3}}{2}}{y\\sqrt{2}}=\\frac{2\\sqrt{3}}{3\\sqrt{2}}$$ --> x/y=4/3 --> x>y. Sufficient. Bunuel - Could you explain how you simplified the ratio in statement #2 to get to x/y = 4/3? The second statement states the ratio as 2√3 : 3√2 &, the calculated ratio is x√3/2 : y√2. Now if these two ratios are same, we just need to simplify the equation, which gives the ratio of x:y to 4:3. Math Expert Joined: 02 Sep 2009 Posts: 47037 Re: Is the length of a side of equilateral triangle E less than [#permalink] ### Show Tags 05 Mar 2014, 01:09 bparrish89 wrote: Bunuel wrote: Is the length of a side of equilateral triangle E less than the length of a side of square F? Let x be the length of a side of equilateral triangle E and y be the length of a side of square F. Question: is x>y? (1) The perimeter of E and the perimeter of F are equal --> 3x=4y --> x/y=4/3 --> x>y. Sufficient. (2) The ratio of the height of triangle E to the diagonal of square F is 2√3 : 3√2 --> the height of triangle E is $$x\\frac{\\sqrt{3}}{2}$$ and the diagonal of square F", "# For an equilateral triangle, the ratio of the in-radius and the ex-radius is: For an equilateral triangle, the ratio of the in-radius and the ex-radius is: [A]$1 : 3$ [B]$1 : \\sqrt{3}$ [C]$1 : \\sqrt{2}$ [D]$1 : 2$ $\\mathbf{1 : 2}$ In-radius $= \\frac{side}{2\\sqrt{3}}$ Circum-radius $= \\frac{side}{\\sqrt{3}}$ ∴ Required ratio $= \\frac{side}{2\\sqrt{3}} : \\frac{side}{\\sqrt{3}}$ $= \\sqrt{3} : 2\\sqrt{3} = 1 : 2$ Hence option [D] is the right answer.", "## anonymous 5 years ago A square with sides of length s is inscribed in an equilateral triangle with sides of length t. Find the exact ratio of the area of the equilateral triangle to the area of the square. 1. anonymous ok i think its something like 4s=3t but i think its wrong. thats my best shot at it lol. sorry haha 2. anonymous 3. anonymous nice job nancy therefore the ratio of the areas is: $\\frac{12+ 7\\sqrt{3}}{12}$", "perimeters of 48 inches. What is the ratio of the length of the side of the triangle to the length of the side of the square? Express your answer as a common fraction. Thursday, March 26, 2015 at 12:02pm Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>> Members", "4 + 4 cm = 3 × 4 cm = 12 cm So, we find that Perimeter of an equilateral triangle = 3 × length of a side What is similar between a square and an equilateral triangle? They are figures having all the sides of equal length and all the angles of equal measure. Such figures are known as regular closed figures. Thus, a square and an equilateral triangle are regular closed figures. Find various objects from your surroundings which have regular shapes and find their perimeters. You found that, Perimeter of a square = 4 × length of one side Perimeter of an equilateral triangle = 3 × length of one side So, what will be the perimeter of a regular pentagon? A regular pentagon has five equal sides. Therefore, perimeter of a regular pentagon = 5 × length of one side and the perimeter of a regular hexagon will be _______ and of an octagon will be ______. Example 6 : Find the distance travelled by Shaina if she takes three rounds of a square park of side 70 m. Solution : Perimeter of the square park = 4 × length of a side = 4 × 70 m = 280 m Distance covered in one round = 280 m Therefore, distance travelled in three", "of triangle to the rectangle? Perimeter = 5 + 3 + 4 = 12. the The Golden Rectangle By Robbie and Jimmy 2. The sum of the interior angles of a quadrilateral is 360 degrees. In this article we will be discussing about the ratio of sides of a right-angled triangle respect to its acute angle called trigonometric ratios of the angle and find the trigonometric ratios of specific angles: 0, 30, 45, 60, and 90. If you multiply a side from triangle ABC by 2, you get the length of the corresponding side of triangle DEF. General Knowledge: General Science: a rectangle and an equilateral triangle are same. Two quadrilaterals are similar quadrilaterals when the three corresponding angles are the same ( the fourth angles automatically become the same as the interior angle sum is 360 degrees), and two adjacent sides have equal ratios. a. opposite hypetenus b. opposite adjacent c. hypotenuse adjacent d. adjacent opposite. no 11. . Topics. Aspect Ratio is the ratio of the width against the height of a rectangle, this is expressed as either the simplest values (e.g. - 52625223 Total number of figures = 7 (a) Ratio of number of triangles to the number of circles inside the rectangle = 3 / 2 (b) Ratio of number of squares to all", "# A square and an equilateral triangle have the same perimeter. Each side of the triangle is 16 m. How do you find the length of each side of the square? Jan 27, 2016 ${\\text{Side}}_{\\square} = 8 \\sqrt[4]{3}$ #### Explanation: The semi-perimeter of the equilateral triangle is $\\textcolor{w h i t e}{\\text{XXX}} s = \\frac{16 + 16 + 16}{2} = 24$ So, by Heron's formula color(white)(\"XXX\")\"Area\"_triangle = sqrt(24xx8xx8xx8) =8^2sqrt(3) We are told ${\\text{Area\"_square = \"Area}}_{\\triangle}$ So Area\"_square = 8^2sqrt(3) and each side of the square has a length of $\\textcolor{w h i t e}{\\text{XXX}} \\sqrt{{8}^{2} \\sqrt{3}} = 8 \\sqrt[4]{3}$", "sides will be 4.. and if diagonal of square is 3 square root2 that means square has all sides 3. we got No ! equilateral triangle length is greater than square's length _________________ Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score ! Intern Joined: 05 Dec 2013 Posts: 15 Re: Is the length of a side of equilateral triangle E less than [#permalink] ### Show Tags 05 Mar 2014, 16:05 Bunuel wrote: bparrish89 wrote: Bunuel wrote: Is the length of a side of equilateral triangle E less than the length of a side of square F? Let x be the length of a side of equilateral triangle E and y be the length of a side of square F. Question: is x>y? (1) The perimeter of E and the perimeter of F are equal --> 3x=4y --> x/y=4/3 --> x>y. Sufficient. (2) The ratio of the height of triangle E to the diagonal of square F is 2√3 : 3√2 --> the height of triangle E is $$x\\frac{\\sqrt{3}}{2}$$ and the diagonal of square F is $$y\\sqrt{2}$$ --> ratio: $$\\frac{x\\frac{\\sqrt{3}}{2}}{y\\sqrt{2}}=\\frac{2\\sqrt{3}}{3\\sqrt{2}}$$ --> x/y=4/3 --> x>y. Sufficient. Bunuel - Could you explain how you simplified the ratio in statement #2 to get to x/y = 4/3? Sure. $$\\frac{(\\frac{x\\sqrt{3}}{2})}{(y\\sqrt{2})}=\\frac{2\\sqrt{3}}{3\\sqrt{2}}$$; $$\\frac{x\\sqrt{3}}{2(y\\sqrt{2})}=\\frac{2\\sqrt{3}}{3\\sqrt{2}}$$; Divide both sides by $$\\frac{\\sqrt{3}}{\\sqrt{2}}$$: $$\\frac{x}{2y}=\\frac{2}{3}$$;", "# 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 $3\\cdot8=24$.", "= 12 + 8 = 20 Ratio of perimeter of blue rectangle to the green rectangle is 10 : 20 Area of the rectangle = lb Blue rectangle: A = 2 × 3 = 6 Green Rectangle: A = 4 × 6 = 24 Ratio of Area of blue rectangle to the green rectangle is 6 : 24 b. Compare your ratios in part(a). The ratio of the length of the blue rectangle to the length of the green rectangle is 2 : 4 equivalent to 1 : 2 The ratio of the width of the blue rectangle to the width of the green rectangle is 3 : 6 equivalent to 1 : 2 Ratio of perimeter of blue rectangle to the green rectangle is 10 : 20 equivalent to 1 : 2 Ratio of Area of blue rectangle to the green rectangle is 6 : 24 equivalent to 1 : 4 Question 49. STRUCTURE The ratio of the side lengths of a triangle is 2 : 3 : 4. The shortest side is 15 inches. What is the perimeter of the triangle? Explain. Explanation: Given that, The ratio of the side lengths of a triangle is 2 : 3 : 4. The shortest side is 15 inches. Let x = the first side y = the second" ]

[ "Angle of the sun.......... Assume that Sun rises at 6 a.m. and sets at 6 p.m. Find the angle made by the Sun rays with the horizontal at 10 a.m. ×", "360 to 30 degrees per minute usually drawn in what is measure!, now your case is 3 so it would be 6 o'clock 9 o clock is radiographic! Through via '' usage within hands of a clock at 9 o ' clock the zero-degree.!", "thermometer shows? 12. A number 2 A number decreased by the difference between four and the number", "a center distance 0.5cm. How many common points have these circles? 9. Obtuse angle Which obtuse angle is creating clocks at 17:00? 10. Noontime In the morning temperature was 110°F. By noontime it has gone up by 15°F. What was the noon temperature. (Give your answer in °F)", "From formal English to slang • hour angle — noun Date: circa 1837 the angle between the celestial meridian of an observer and the hour circle of a celestial object measured westward from the meridian … New Collegiate Dictionary • Greenwich hour angle — hour angle measured from the meridian of Greenwich, England. * * * … Universalium", "sec(17 degrees) or 1/cos(17 degrees).", "## Quarter past three How many degrees are there in the angle between the hour and minute of a clock when the time is a quarter past three ? Source: folklore Hint: The answer is $7.5^{\\circ}$ i.e., $\\frac{\\pi}{24}$ radians. 0 0 0 0 0 0 0 0 1 1", "# Angle of the sun.......... Classical Mechanics Level 1 Assume that Sun rises at 6 a.m. and sets at 6 p.m. Find the angle made by the Sun rays with the horizontal at 10 a.m. ×", "hour. 3. The elevation above sea level of a hiker descending into a canyon. 4. The number of students remaining in school after 6:00 p.m. 5. A bank account balance over a year. 6. The temperature in degrees Fahrenheit of an oven used on a hot summer day.", "angle between its rays and the ground is more nearly square, and (b) the Sun is above the horizon for more hours of the day. – Solomon Slow Aug 21 at 22:58 • @SolomonSlow My question is how to justify having the same temperature, humidity, heat perception is different? – Sebastiano Aug 21 at 23:00", "with accuracy is challenging. So in $1967$ the definition of the second was changed again, to what it is now. Linguistic Note The word second, in the context of the measurement of time and of angle, derives from the fact that it was originally specified as a second minute part of an hour or of a degree of arc. The minute had already been defined as a minute part of an hour. The use of the word second in English, in this specific context, started near the end of the $16$th century.", "in hours (1 hour = 15 degrees). For example, if an object has an hour angle of 2.5 hours, it transited across the local meridian 2.5 sidereal hours ago (i.e., hours measured using \"sidereal time\"), and is currently 37.5 degrees west of the meridian. Negative hour angles indicate the time until the next transit across the local meridian. Of course, an hour angle of zero means the object is currently on the local meridian. Wikimedia Foundation. 2010. Look at other dictionaries: • Hour angle — Hour Hour, n. [OE. hour, our, hore, ure, OF. hore, ore, ure, F. heure, L. hora, fr. Gr. ?, orig., a definite space of time, fixed by natural laws; hence, a season, the time of the day, an hour. See {Year}, and cf. {Horologe}, {Horoscope}.] 1. The … The Collaborative International Dictionary of English • hour angle — n. Astron. the angle formed at the pole, or its arc on the celestial equator, between the hour circle of a celestial object and the observer s celestial meridian measured in hours, minutes, etc.: it represents the elapsed time since the object… … English World dictionary • hour-angle — hourˈ angle noun (astronomy) The angle (usu measured as time) between the declination circle of a body observed and the observer s meridian • • •", "right angle. Learning points A right angle is equivalent to 90° on a protractor as shown in Fig. 6.10 below. (b) Acute angle Activity 6.4 Work in pairs 1. Set the clock at 2.00 o’clock. What type of angle is formed by the space between the minute hand and the hour hand? What is the name given to this type of angle? 2. Open your classroom door slightly as shown in Fig. 6.12 What type of angle is formed at the top between the frame and the top edge of the open door? 3. Compare this type of angle with the angle formed by setting the clock at 1.00 o’clock. Comment on the type of angle. 4. Compare the type of angle formed in steps 1 and 2 with the right angle. What do you notice? Learning points An angle which measures less than 90° is called an acute angle e.g 60º, 20°, 88°, 12° and so on. Fig. 6.13 (c) Straight angle Activity 6.5 In pairs, carry out the following activities: 1. Set your clock at 6.00 o’clock as shown in Fig. 6.14. (a) What type of angle is formed between the minute hand and the hour hand? (b) What is the name of the type of angle formed by the space between the minute hand and", "120 degree on a clock › degrees on a clock face › convert degrees to decimal degrees › convert hours to degrees › convert degrees … 12-hour clock: The third and last order digit divided the time-degree into 60 parts (the gar), which appears to be sexagesimal. Huygens managed to refine his pendulum clock to have errors of fewer than 10 seconds a day. The second-order digit counted the degrees that had gone by in the hour, notwithstanding the fact that its number of degrees were seasonal. Here, the clock position in hours and minutes and angle in decimal degrees with one decimal place can be converted. Different compass roses (4, 8, 16, 32, 128-wind compass roses) are supported. Today however, atomic clocks are the most accurate devices for time measurement. (This converts the calculator from \"Degrees\" to \"radians\". Take note of the 9 o’clock position in the photo above. Most people read time using either a 12-hour clock or a 24-hour clock. Quantity : Reference Unit : is equal to : Conversion Factor : Unit : 1: degree = 1 : degree: 1 = 0.017453292519943 : radian: 1 = 1.1111111111111 : grad To have it reach 360 again at … The first pendulum mechanical clock was created by Christiaan Huygens in 1656, and was the first clock", "and finding the intersection. I found the intersection at 68 deg. Plus 5 deg is 73 deg. 2 deg off from your answer.", "again then it forms an angle of 360°, which is called a Complete Angle. This is called one revolution. ∗ In a clock, there are two hands i.e. minute hand and hour hand, which moves clockwise in every minute. When the clock hand moves from one position to another then turns through an angle. • When a hand starts from 12 and reaches to 12 again then it is said to be completed a • s were the ray moves in the opposite direction of the hands of a clock are called anti – clockwise angles. These are denoted by positive measure. • Angles were the ray moves in the direction of the hands of a clock are called clockwise angles. These are denoted by negative measure. The protractor: • By observing an angle we can only get the type of angle but to compare it properly we need to measure it. • An angle is measured in the “degree”. One complete revolution is divided into 360 equal parts so each part is one degree. We write it as 360° and read as “three hundred sixty degrees”. • We can measure the angle using a ready to use device called Protractor. • It has a curved edge, which is divided into 180 equal parts. It starts from 0°", "into hours : so. The important realization for this problem is that the hour hand is always moving. Although this is easiest to picture with a pulse of light, the relationship between distance and intensity is just as true for any source of (incoherent) light.", "Now set the dish to the satellite you want using a Satellite finder or whatever decoder you own. Day degrees are a way of measuring the physiological development of a plant. Step 2: Press the \"Calculate\" button. None of your answers are right, close but not right To begin the hour pointer already moved 5/6 to the 12 o clock position or in the 11 hrs, 58 minutes and 20 seconds (8.33), the minute pointer is in the 50 minutes position, is 30 degrees for 10 minutes separation , so the answer is 30 * 5/6 = 25 degrees, need confirmation, use a transporter thanks to the USA education system saludos (This converts the calculator from \"Degrees\" to \"radians\". Equation for the angle of the hour hand {\\displaystyle \\theta _ {\\text {hr}}=0.5^ {\\circ }\\times M_ {\\Sigma }=0.5^ {\\circ }\\times (60\\times H+M)} When she was born, the time on the 75 degree of longitude was the time she stated. The formula for the pendulum period is . Most of the suggested solutions fall short (including Wikipedia but not Michael and Rick barring the minute limitation) since the direction of angle measurement changes when the minute hand passes the 12 o’clock position, for example in going from just before to just after 1pm the angle jumps by 300 degrees. In", "# Find the measure of angle formed by the hand of the clock at 7:25. Question: Find the measure of angle formed by the hand of the clock at 7:25. Answer: Each hour, the hour clock indicator “travels” 30 degrees. In 7 hours the indicator makes an angle of 210 degrees. In 25′ it makes, additionally, 25/60 of the next 30 degs. That makes 12.5 degrees, furthermore. So, the hour indicator makes an angle of 222.5 degrees. The minute indicator travels 150 degrees from the position “12” to the position “5” in 25′ . The angle between the two indicators is 222.5 – 150 = 72.5 degrees.", "that for $\\theta$, the angle AB subtends at D is 56 degrees 20 minutes minus $\\theta$. 3. Thanks. I got the answer" ]

[ "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", "is just $$10^\\circ$$. Let’s just make sure our formula gives the same result. Let $$m = 20$$, $$h = 4$$. Then: $$\\theta = \\frac{60h - 11m}{2}^\\circ = \\frac{60 \\cdot 4 - 11 \\cdot 20}{2}^\\circ = \\frac{20}{2}^\\circ = 10^\\circ$$ which is exactly what we wanted to see ✔ ## Applications? It might be useful if you’re simply wanting to draw an accurate clock for given a time, but our two intermediary results are probably more useful for this: $$\\alpha_m = 6m^\\circ \\\\ \\alpha_h = \\frac{60h + m}{2}^\\circ$$ These two formulas will tell you which angle to draw the lines at (with respect to the top of the clock). Here’s a quick, interactive demo written in p5.js: Javascript code... var width; var height; function setup() { // Work in degrees angleMode(DEGREES); let demo = document.getElementById(\"demo\"); width = int(getComputedStyle(demo).width); height = int(getComputedStyle(demo).height); createCanvas(400, 400); // Get sliders hSlider = document.getElementById(\"hSlider\"); mSlider = document.getElementById(\"mSlider\"); } function draw() { radius = 0.95 * min(width, height) / 2; background(39, 40, 34); // Draw the clock face fill(39, 40, 34); stroke(255); strokeWeight(2); // Set the origin to the centre of the canvas translate(width / 2, height / 2); // Set the 0 degree mark to the top of the clock rotate(-90); // Draw the minute hand strokeWeight(3); let minutes = int(mSlider.value); let a_m = 6", "# The sexagesimal system and its operations The sexagesimal system is a system of numeration in which every unit is divided into 60 smaller units. In other words, the base used is 60. This system is used to measure time and angles. $$1h \\rightarrow 60 \\ \\mbox{min} \\rightarrow 60\\cdot 60=3.600 \\ \\mbox{s}$$$$$1^\\circ \\rightarrow 60' \\rightarrow 60\\cdot 60=3.600''$$$ ## Operations with sexagesimal numbers ### Sum • Step 1 Write the numbers to be added as follows, and add them column by column: $$\\begin{eqnarray} & & 38^\\circ \\ 24' \\ 55'' \\\\\\\\ &+ & \\underline{40^\\circ \\ 49' \\ 17''} \\\\\\\\ & & 78^\\circ \\ 73' \\ 72'' \\end{eqnarray}$$ • Step 2 If the sum of seconds is bigger than $$60$$, we must divide the result by $$60$$; the remainder will be the seconds and the quotient will be added to the minutes. $$\\dfrac{72}{60}=1+\\dfrac{12}{60}$$ Namely, the remainder is a $$12$$ and the quotient $$1$$. Then, the result is written as: $$78^\\circ \\ (73+1)' \\ 12'' = 78^\\circ \\ 74' \\ 12''$$ • Step 3 Repeat the same procedure for the minutes: $$\\dfrac{74}{60}=1+\\dfrac{14}{60}$$ Then, $$78^\\circ \\ 74' \\ 12'' = 79^\\circ \\ 14' \\ 12''$$ ### Subtraction • Step 1 Write the numbers one above the other, the hours above the hours (or the degrees above the degrees), the minutes above the", "9 '13 at 7:20 Looks like \\psline[linewidth=3\\pslinewidth]{<->}(2.1;90)(0,0)(1.5;0) represends both hands. – Per Alexandersson Sep 9 '13 at 7:47 Your second clock is broken. The hour hand points at 2:30, but the minute hand points at x:57. – Jake Sep 9 '13 at 8:15 @PSTikZ: Ah, I thought you meant that the second hand is decoupled from the minutes/hours. – Jake Sep 9 '13 at 8:23 With Metapost. This is more complicated than it has to be, because it takes care to mark off each minute, and to use different widths and lengths for five-minute ticks as opposed to one-minute ticks, but it's still pretty simple: beginfig(1); diameter=160; pickup pencircle scaled 0.5; pair r[]; for i=12 downto 1: r[i] = r[i+1] rotatedaround (z0,30); z[i] = r[i]; endfor defaultfont := \"cmr17\"; defaultscale := 1.5; label(\"12\",z12); label(\"6\",z6); label(\"9\",z9); label(\"3\",z3); label(\"1\",z1); label(\"5\",z5); label(\"7\",z7); label(\"11\",z11 shifted (0,-1pt)); label(\"2\",z2); label(\"4\",z4); label(\"8\",z8); label(\"10\",z10 shifted (1pt,0)); pair p[]; pair q[]; numeric circangle; circangle := 0; p[0] = z50; for i=0 step 1 until 60: p[i] = p[i-1] rotatedaround (z0,6); if (i mod 5 = 0): q[i] = 1/12[p[i],z0]; pickup pencircle scaled 2; draw p[i]--q[i]; pickup pencircle scaled 0.5; else: q[i] = 1/30[p[i],z0]; pickup pencircle scaled 0.5; draw p[i]--q[i]; fi draw p[i-1]..p[i]; endfor endfig; end; I've always been partial to Metapost, as I find it much easier", "7, 7, 16, 20, 12, 6, 18, 15, 10, 10, 6, 10, 21, 10, 5, 12, 20, 12, 6, 0, 19, 0, 16, 20, 0, 5, 16, 6, 20, 12, 15, 7, 6, 16, 10, 12, 7, 5, 19, 16, 20, 5, 6, 15, 10, 16, 16, 20, 10, 20, 7, 7, 10, 20, 20, 10, 19, 0, 10, 10, 19, 12, 6, 19, 20, 0, 10, 12, 0, 7, 6, 10, 15, 16, 6, 6, 19, 10, 21, 20, 0, 15, 6, 10, 20, 7, 6, 10, 0, 16, 19, 6, 20, 10, 7, 12, 7) Algorithms for computing EMD is discussed at Coupling and Total variational distance and Calculate Earth Mover's Distance for two grayscale images, and also over at math SE. There are some R packages, some fast examples (with your data): x1.circ <- circular::circular(x1, units=\"hours\", template=\"clock24\") x2.circ <- circular::circular(x2, units=\"hours\", template=\"clock24\") par(mfrow=c(1, 2)) circular::plot.circular(x1.circ, stack=TRUE) circular::plot.circular(x2.circ, stack=TRUE) Then for the EMD distance using R package emdist: calc_weight <- function(x) { # a vector of hours tab <- table(factor(x, levels=as.character(0:23)), useNA=\"ifany\") dimnames(tab) <- NULL mat <- cbind( weights=tab/sum(tab), points=0:23 ) mat } A <- calc_weight(x1) B <- calc_weight(x2) hourdist <- function(A, B) sum(pmin( (A-B)%%24, (B-A)%%24 ) ) emdist::emd(A, B, dist=hourdist) [1] 2.63 The package transport can give more information: costm <- outer(0:23, 0:23, FUN=function(x, y)", "= \\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}})).$$ This means that \\begin{align*} n(1-i\\cot{\\angle{B}})&=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))(1-i\\cot{\\angle{B}})\\\\ &=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))\\left(1-\\frac{\\cos{\\angle{B}}}{\\sin{\\angle{B}}}i\\right)\\\\ &=(|BC|)(\\sin{\\angle{B}}+i\\cos{\\angle{B}})(\\sin{\\angle{B}}-i\\cos{\\angle{B}})\\\\ &=(|BC|)(1)", "hour hand at the top of the clock (rotating it counter-clockwise by one quarter of the circle’s radius), we’ll want to use $$\\pi$$ divided by 2. In fact, since we’re rotating counter-clockwise, we’ll want to make this value negative. Just like its $$\\pi$$ variable, PI, py5 actually includes a variable for half of $$\\pi$$, HALF_PI. def setup(): size(600, 600) frame_rate(1) no_fill() stroke('#FFFFFF') def draw(): background('#004477') h = hour() m = minute() s = second() # print( str(h) + ':' + str(m) + ':' + str(s) ) # Translating to the center of the sketch window... translate(width/2, height/2) # Clock face stroke_weight(3) ellipse(0, 0, 350, 350) # Hour hand rotate(-HALF_PI) # Rotated counterclockwise by HALF_PI stroke_weight(10) line(0, 0, 100, 0) run_sketch() In addition to PI and HALF_PI, there’s a built-in variable for $$\\pi \\times 2$$ as well. This is TAU, a term coined only as recently as 2010 to describe the “full turn” of $$\\pi \\times 2$$. So, if there are 12 hours on your clock, a single hour is represented by TAU divided by 12. Great news - we have the current hour, so we can incorporate this value into our rotate() function to put the hour hand in the right place! def setup(): size(600, 600) frame_rate(1) no_fill() stroke('#FFFFFF') def draw(): background('#004477') h = hour() m = minute()", "date to above the hour line. The conditional in the rotation makes sure the text stays the right way up. There is a slight problem with this on the Moto 360 - as you might notice in the above, the date will dip below the 'flat tire' line between about 11:15 and 12:45. If you have a 360, you might want move the date further along the line (change '165' in the code to something smaller). > Fun with Progress Rings Circles 1 [WM] 2:27am (left) and 9:28pm (right) This and the following two designs are all variations on the same theme - it was inspired by an advert, I think. Can't remember what it was an advert for, though. The outer ring segment represents the hours, and fills in (clockwise) towards midday, then 'unfills' from midday to midnight. This is made with a 'Segment Between' shader (WatchMaker Premium only) with Degree Start: {dh23} < 12 and 0 or {drh0} Degree End: {dh23} < 12 and {drh0} or 360 The orange circles around the outside are hour markers, and also give the hour ring the appearance of rounded corners. In this design, the smaller white circles over the hour markers act as minute markers, which change colour with the passing minutes, as for example Colour: {dm}>=5 and 'ffbd2a'", "Parallels will be the Hour-Lines from 6 in the Evening to 6 in the Morning AL being the Hour-Line of Midnight. And if the parallel Spaces, be laid off on the other side of the Hour-Line of 6, you will have the Hour-Lines from 6 in the Morning to 6 in the Evening. And for drawing the half Hours, divide each 15th Deg. of the Quadrant AF into half, and draw Parallel Lines between the Hour-Lines. The Hour-Lines may be yet otherwise drawn, by means of a Circle, whose Diameter is the Line AB, and whose Circumference is divided into 24 equal Parts for the 24 Hours of the Day, or into 48, for the Half-Hours. For then if right Lines be drawn thro’ the opposite Points of Division, parallel to ED, we shall have the Hour-Lines, and those of the Half-Hours, as we have said in the Construction of the former right-lined Dial. About the Point I, as a Center, draw an occult Quadrant, which divide into 90 Deg. and laying a Ruler to the Center I, and on each Division mark the same Degrees upon the Sides GQ, and GS of the Instrument. Note, By means of these Divisions we may find the Sun’s Altitude above the Horizon, as we shall shew by and by. RR", "bands represent the binary numbers from 0 (all black, on top) to 11 (1011), for the hours. The outer 6 bands represent the binary numbers from 0 to 59 (111011) for the minutes. Note that the hour hand cannot display 12. With binary numbers in mind, it's just more logical to start counting from 0, instead of 1. November 10, 2006, at 02:19 AM by 84.194.127.106 - Changed lines 27-28 from: This clock can easily be build by hacking a cheap retail analog clock. I found a pretty big one (27.5cm diameter for the backdrop) for 12 euros, and it was perfectly hackable as it could easily be taken apart and put back together. For the conversion, the clock's backdrop is changed, and its hands are modified so they can be replaced by transparent templates. to: This clock can easily be constructed by hacking a cheap retail analog clock. I found a pretty big one (27.5cm diameter for the backdrop) for 12 euros, and it was perfectly hackable as it could easily be taken apart and put back together. For the conversion, the clock's backdrop is changed, and its hands are modified so they can be replaced by transparent templates. November 10, 2006, at 02:17 AM by 84.194.127.106 - Changed lines 43-44 from: Below are the PDFs that", "it's blue it's a one. So in the above, the hour (outer ring) is 10011 in binary, or $1\\cdot 16 + 0\\cdot 8 +0\\cdot 4 + 1\\cdot 2 + 1\\cdot 1 = 19$, in base 10. In this case, we can't use progress bars. Instead, we have to make each ring segment an individual rectangle object, that will change its colour depending on whether it's representing a one or a zero. This makes constructing the clock a little less straightforward. For the minutes ring, create 6 rectangles each with a curve of 60, and rotated so that they form a circle. Similarly, for the hours ring create 5 rectangles with curve 72, and so on. I originally tried aligning everything 'free-hand'. That didn't work so well, so I created a temporary full circle as a guide. Of course, you don't have to make a binary clock circular - you could have it as a line of dot/bars. I just like the circle design. Now, we need a way of converting the time to binary, and a way of telling each block whether the binary digit it represents is a one or a zero. At this point, I'm going to make a claim - The n-th digit of a number, x, in binary is a zero if $x \\bmod", "center of the hour lines by a distance called Loffset to the south of the dial, expanding the useable area of the dial. For example I prefer Loffset to be about a quarter of the dial's diameter. The above code needs only one additional line: theta = atan2(sin(lat)*sin(HA),cos(HA)); intersection(){ translate([0,-Loffset,0]) rotate([0,0,theta]) translate([0,Llength/2,Lhght/2]) cube([Lwidth,Llength,Lhght],center=true); } Figure 2. (a) centered hour lines & gnomon foot, (b) offset hour lines & gnomon foot Download the OpenSCAD code from the attachment below that includes both Part 1 and Part 2 tutorials. Sundials Part 2 - Sundial HourLines.scad [ ] 4 kB Cookies make it easier for us to provide you with our services. With the usage of our services you permit us to use cookies.", "# Difference between revisions of \"2015 AMC 10A Problems/Problem 14\" ## Problem The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction? $[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label(\"\"+(string) i+\"\",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]$ $\\textbf{(A) }\\text{2 o' clock} \\qquad\\textbf{(B) }\\text{3 o' clock} \\qquad\\textbf{(C) }\\text{4 o' clock} \\qquad\\textbf{(D) }\\text{6 o' clock} \\qquad\\textbf{(E) }\\text{8 o' clock}$ ## Solution 1 The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is $\\boxed{\\textbf{(C) }4 \\ \\text{o' clock}}$. Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns", "are only the principal Lines, and the Hour-Lines, as may be seen in the Pentagonal Figure marked 14. The Dial of the Figure 15, represents that marked F in Figure 2, and is an upper Incliner of 63 Deg. 26 Min. declining South-eastwardly 72 Deg. and may be drawn by the abovesaid Method. The Center of this Dial is upwards, and because it has a great Declination, the Hour-Lines will fall very close to one another near the Substylar Line; and therefore it ought to be drawn upon a large Plane, that so the Part thereof next to the Center may be taken away, and the Style and the Hour-Lines terminated by two Parallels. There is another way of drawing Mechanically any sorts of Dials whatsoever, upon Polyhedrons or Bodies of different Faces or Superficies, without even knowing the Declinations or Inclinations of the Faces or Superficies, and that with as much exactness as by any other Methods whatsoever. In order to do this, you must first make an horizontal Dial upon one of the Planes or Faces that is to be set parallel to the Horizon, and set up the Style thereof upon the Hour-Line of 12, conformable to the Latitude of the Place. After this, the Substylar Lines must be drawn upon all the Planes or", "# Difference between revisions of \"2015 AMC 10A Problems/Problem 14\" The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction? $[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label(\"\"+(string) i+\"\",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]$ $\\textbf{(A) }\\text{2 o' clock} \\qquad\\textbf{(B) }\\text{3 o' clock} \\qquad\\textbf{(C) }\\text{4 o' clock} \\qquad\\textbf{(D) }\\text{6 o' clock} \\qquad\\textbf{(E) }\\text{8 o' clock}$ ## Solution 1 The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is $\\boxed{\\textbf{(C) }4 \\ \\text{o' clock}}$ . ## Solution 2 The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every", "- Show source$\\text{...}$ - Equivalent to one-eighth of a turn i.e. 45 degrees.$1\\ \\text{octant} = \\dfrac{1}{8}\\ \\text{turn} = 45^{\\circ} = \\dfrac{\\pi}{4}\\ rad$ Show source$...$ sign Show source$-$ - Show source$\\text{...}$ - Equivalent to one twelfth of a turn i.e. 30 degrees.$1\\ \\text{sign} = \\dfrac{1}{12}\\ \\text{turn} = 30^{\\circ} = \\dfrac{\\pi}{6}\\ rad$ Show source$...$ hour angle (1/24 of turn) Show source$-$ - Show source$\\text{...}$ - Equivalent to one-twenty-fourth of a turn i.e. 15 degrees.$1\\ \\text{hour} = \\dfrac{1}{24}\\ \\text{turn} = 15^{\\circ} = \\dfrac{\\pi}{12}\\ rad$ Show source$...$ point Show source$-$ - Show source$\\text{...}$ - Equivalent to one-thirty-second of a turn i.e. 11.25 degrees.$1\\ \\text{point} = \\dfrac{1}{32}\\ \\text{turn} = 11.25^{\\circ} = \\dfrac{\\pi}{16}\\ rad$ Show source$...$ minute angle (1/60 of turn) Show source$-$ - Show source$\\text{...}$ - Equivalent to one sixtieth of a turn i.e. 6 degrees.$1\\ \\text{minute} = \\dfrac{1}{60}\\ \\text{turn} = 6^{\\circ} = \\dfrac{\\pi}{30}\\ rad$ Show source$...$ # military# Unit Symbol Symbol(plain text) Value as symbolic Value as numeric Notes Unit conversion formula milliradian Show source$mil$ mil Show source$\\text{...}$ - Unit of measure of angle used in the army. Milliradian (mrad, mil) is the angle at which you can see a curve of one meter from a distance of one kilometer. One milliradian corresponds to one thousandth of a radian, or approximately 1/6283.2 of a turn. $1 \\ mil = \\dfrac{1}{1000}\\ rad =", "$A$ is perpendicular to the median from vertex $B$. If the lengths of sides $AC$ and $BC$ are $6$ and $7$ respectively, then the length of side $AB$ is $\\textbf{(A) }\\sqrt{17}\\qquad \\textbf{(B) }4\\qquad \\textbf{(C) }4\\dfrac{1}{2}\\qquad \\textbf{(D) }2\\sqrt{5}\\qquad \\textbf{(E) }4\\frac{1}{4}$ ## Problem 29 It is now between $10:00$ and $11:00$ o'clock, and six minutes form now, the minute hand of the watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now? $\\textbf{(A) }10:05\\dfrac{5}{11}\\qquad \\textbf{(B) }10:07\\dfrac{1}{2}\\qquad \\textbf{(C) }10:10\\qquad\\\\ \\textbf{(D) }10:15\\qquad \\textbf{(E) }10:17\\dfrac{1}{2}$ ## Problem 30 $[asy] size(175); defaultpen(linewidth(0.8)); real r=50, a=4,b=2.5,c=6.25; pair A=origin,B=c*dir(r),D=(a,0),C=shift(b*dir(r))*D; draw(A--B--C--D--cycle); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,E); label(\"D\",D,S); label(\"a\",D/2,N); label(\"b\",(C+D)/2,NW); [/asy]$ In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of $\\angle{D}$ is twice the measure of $\\angle{B}$, and the measures of segments $AB$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to $\\textbf{(A) }\\dfrac{1}{2}a+2b\\qquad \\textbf{(B) }\\dfrac{3}{2}b+\\dfrac{3}{4}a\\qquad \\textbf{(C) }2a-b\\qquad \\textbf{(D) }4b-\\frac{1}{2}a\\qquad \\textbf{(E) }a+b$ ## Problem 31 If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to $43$, what is the probability that this number is divisible by $11$? $\\textbf{(A) }2/5\\qquad \\textbf{(B) }1/5\\qquad \\textbf{(C) }1/6\\qquad \\textbf{(D) }1/11\\qquad \\textbf{(E) }1/15$ ## Problem 32 $A$ and $B$ travel around", "= 25 ÷ 5 • 2 = 5 • 2 = 10. Question 7. 53 28 = 25 Answer: 53-28 = 25. Explanation: By subtracting 28 in 53 we will get 25. Question 8. Finde. 31e = 93 e = ____ Answer: e = 3. Explanation: Given the expression is 31e = 93, e = $$\\frac{93}{31}$$ = 3. Question 9. How many minutes are in 3$$\\frac{1}{4}$$ hours? Answer: 3$$\\frac{1}{4}$$ hours = 195 minutes. Explanation: As 1 hour is equal to 60 minutes, so for 3$$\\frac{1}{4}$$ hours it will be $$\\frac{13}{4}$$ hours × 60 = 13 × 15 = 195 minutes. Question 10. Complete the chart for the shape. Name the figure. How many sides? How many angles? Does it have symmetry? Is it a plane shape or solid? Answer: Name the figure. Hexagon How many sides? 6 How many angles? 6 Does it have symmetry? 6 Is it a plane shape or solid? Plane shape Question 11. Imagine you write each letter of the word EXCELLENT on a different card and then shuffle the cards. You turn the deck of cards facedown on a table and turn over the top card. What is the probability you will turn over a letter with only curved lines? _____________________ Answer: There are 9 letters so there are 9 cards. The probability", "# Rotate forms for round watch faces Especially in the beginning of designing round watch faces, I had to get used to place forms to the right place in a circle. This is needed for markers, numbers and some decoration elements. Photoshop doesn’t offer circular guide lines or other help functions for these kind of layouts. In fact, it’s easy to get the right degree values for spreading assets in a circle. You just have to divide 360 by the number of elements, you like to place in a circle. Examples: Hours $\\frac{360^\\circ}{12} = 30^\\circ$ Minutes $\\frac{360^\\circ}{60} = 6^\\circ$ Weekdays $\\frac{360^\\circ}{7} = 51.4286^\\circ$ Days in Month $\\frac{360^\\circ}{31} = 11.6129^\\circ$ The elements to rotate and place are usually similar or same to each other. So the workflow is usually to create an asset, to place it, to copy it and to rotate and place it. If you do this one by one manually, it not only feels stupid to repeat one and the same action again and again, it’s also very easy to make small mistakes, if you loose concentration. In these cases, it’s always good to use Photoshop’s action feature. My workflow for creating hours markers looks like that: 1. I create one marker and place it. 2. I copy this marker to the opposite side and rotate", "# Rotate forms for round watch faces Especially in the beginning of designing round watch faces, I had to get used to place forms to the right place in a circle. This is needed for markers, numbers and some decoration elements. Photoshop doesn’t offer circular guide lines or other help functions for these kind of layouts. In fact, it’s easy to get the right degree values for spreading assets in a circle. You just have to divide 360 by the number of elements, you like to place in a circle. Examples: Hours $\\frac{360^\\circ}{12} = 30^\\circ$ Minutes $\\frac{360^\\circ}{60} = 6^\\circ$ Weekdays $\\frac{360^\\circ}{7} = 51.4286^\\circ$ Days in Month $\\frac{360^\\circ}{31} = 11.6129^\\circ$ The elements to rotate and place are usually similar or same to each other. So the workflow is usually to create an asset, to place it, to copy it and to rotate and place it. If you do this one by one manually, it not only feels stupid to repeat one and the same action again and again, it’s also very easy to make small mistakes, if you loose concentration. In these cases, it’s always good to use Photoshop’s action feature. My workflow for creating hours markers looks like that: 1. I create one marker and place it. 2. I copy this marker to the opposite side and rotate" ]

[ "question, the angle between each hour is 600. At 11:20, the hour hand has gone one-third of the way between the 11 and the 12. Thus, there are two-thirds of 30° between the hour hand and the 12. (2/3) 30° = 20°. There are 60° between 12 and 2, where the minute hands is, which means there’s a total of 20° + 60° = 80° between the hands. The correct answer is D. ## Equation of a Circle The standard equation for a circle is in the form $$(x-h)^{2}+(y-k)^{2}=r^{2}$$ where the circle with radius r has its center at (h, k). #### Solved Example 10 In which quadrant(s) will the circle with equation $$(x+8)^{2}+(y-15)^{2}=49$$ be contained? A. I only B. II only C. II and III only D. IV only Solution: Begin by finding the center of the circle. The center of the circle is at (-8, 15), which is in Quadrant II. The radius of this circle is 7. Now, we need to consider the four points seven units in the positive/negative x-direction and positive/negative y-direction. These will be on the line of the circle and mark the farthest points in those directions. If all of those points are located in Quadrant II, we know that the entire circle is located in that quadrant. #### Let’s consider", "∠DWU = 180° [Interior angles on the same side of transversal] ∠BUW + 110°= 180° ∠BUW = 180-110° = 70° ∠BUW = ∠AUS= 70° [vertically opposite angles] ∠AUS + ∠SUQ + ∠QUB = 180° ∠AUB is a straight angle] 70 + ∠SUQ + 70 = 180° ∠SUQ + 140° = 180° ∠SUQ = 180° – 140° ∠SUQ = 40° Question 9. What is the angle between the hours hand and minutes hand of a clock at (i) 1.40 hours (ii) 2.15 hours [use 1° = 60 minutes] Solution: i. 1.40 hours – 190° ii. 2.15 hours – 22° 30′ Question 10. How much would hour’s hand have moved from its position at 12 noon when the time is 4.24 p.m? Solution: 144° Question 11. Let $\\overrightarrow{\\mathrm{AB}}$ be a line segment and let C be the midpoint of $\\overrightarrow{\\mathrm{AB}}$ Extend AB to D such that B lies between A and D. Prove that AD + BD = 2CD Solution: = AC + CB + BD + BD [∵ AD = AC + CB + BD] = CB + CB + BD + BD [∴ AC = CB] = CB + BD + CB + BD = CD + CD = 2CD = RHS ∴ AD + B D = 2CD Question 12. Let $\\overrightarrow{\\mathbf{A B}}$ and $\\overrightarrow{\\mathbf{C D}}$ be", "date to above the hour line. The conditional in the rotation makes sure the text stays the right way up. There is a slight problem with this on the Moto 360 - as you might notice in the above, the date will dip below the 'flat tire' line between about 11:15 and 12:45. If you have a 360, you might want move the date further along the line (change '165' in the code to something smaller). > Fun with Progress Rings Circles 1 [WM] 2:27am (left) and 9:28pm (right) This and the following two designs are all variations on the same theme - it was inspired by an advert, I think. Can't remember what it was an advert for, though. The outer ring segment represents the hours, and fills in (clockwise) towards midday, then 'unfills' from midday to midnight. This is made with a 'Segment Between' shader (WatchMaker Premium only) with Degree Start: {dh23} < 12 and 0 or {drh0} Degree End: {dh23} < 12 and {drh0} or 360 The orange circles around the outside are hour markers, and also give the hour ring the appearance of rounded corners. In this design, the smaller white circles over the hour markers act as minute markers, which change colour with the passing minutes, as for example Colour: {dm}>=5 and 'ffbd2a'", "hour hand at the top of the clock (rotating it counter-clockwise by one quarter of the circle’s radius), we’ll want to use $$\\pi$$ divided by 2. In fact, since we’re rotating counter-clockwise, we’ll want to make this value negative. Just like its $$\\pi$$ variable, PI, py5 actually includes a variable for half of $$\\pi$$, HALF_PI. def setup(): size(600, 600) frame_rate(1) no_fill() stroke('#FFFFFF') def draw(): background('#004477') h = hour() m = minute() s = second() # print( str(h) + ':' + str(m) + ':' + str(s) ) # Translating to the center of the sketch window... translate(width/2, height/2) # Clock face stroke_weight(3) ellipse(0, 0, 350, 350) # Hour hand rotate(-HALF_PI) # Rotated counterclockwise by HALF_PI stroke_weight(10) line(0, 0, 100, 0) run_sketch() In addition to PI and HALF_PI, there’s a built-in variable for $$\\pi \\times 2$$ as well. This is TAU, a term coined only as recently as 2010 to describe the “full turn” of $$\\pi \\times 2$$. So, if there are 12 hours on your clock, a single hour is represented by TAU divided by 12. Great news - we have the current hour, so we can incorporate this value into our rotate() function to put the hour hand in the right place! def setup(): size(600, 600) frame_rate(1) no_fill() stroke('#FFFFFF') def draw(): background('#004477') h = hour() m = minute()", "# Chapter 16 Make fun of the package ## 16.1 A clock The first example is a clock. The key function here is circos.axis() (Figure 16.1). In the example, the whole circle only contains one sector in which major tick at 0 is overlapping with major tick at 12. Later we calculate the positions of the hour hand, the minute hand and the second hand based on current time when this Chapter is generated. The hands are drawn by arraws() function in the canvas coordinate. A real-time clock can be found at the Examples section in the help page of circos.axis(). circos.par(gap.degree = 0, cell.padding = c(0, 0, 0, 0), start.degree = 90) circos.initialize(factors = \"a\", xlim = c(0, 12)) circos.track(ylim = c(0, 1), bg.border = NA) circos.axis(major.at = 0:12, labels = NULL, direction = \"inside\", major.tick.length = uy(2, \"mm\")) circos.text(1:12, rep(1, 12) - uy(6, \"mm\"), 1:12, facing = \"downward\") current.time = as.POSIXlt(Sys.time()) sec = ceiling(current.time$sec) min = current.time$min hour = current.time\\$hour sec.degree = 90 - sec/60 * 360 arrows(0, 0, cos(sec.degree/180*pi)*0.8, sin(sec.degree/180*pi)*0.8) min.degree = 90 - min/60 * 360 arrows(0, 0, cos(min.degree/180*pi)*0.7, sin(min.degree/180*pi)*0.7, lwd = 2) hour.degree = 90 - hour/12 * 360 - min/60 * 360/12 arrows(0, 0, cos(hour.degree/180*pi)*0.4, sin(hour.degree/180*pi)*0.4, lwd = 2) circos.clear() ## 16.2 A dartboard The second example is a dartboard. In Figure", "Obtuse angle This because its measure is more than $90^{circ}$ but less than $180^{circ}$. (ix) Acute angle This is because its measures is less than $90^{circ}$ but more than $0^{circ}$. (x) Acute angle This is because its measures is less than $90^{circ}$ but more than $0^{circ}$. (xi) Zero angle This is because its measure is zero. (xii) Acute angle This is because its measures is less than $90^{circ}$ but more than $0^{circ}$. QUESTION – 3: How many degrees are there in (i) one right angle? (ii) two right angles? (iii) three right angles? (iv) four right angles? (v) 2/3 right angles? (vi) $1frac{1}{2}$ right angles? Solution: (i) One right angle has $90^{circ}$. (ii) Two right angles have $90^{circ}+90^{circ}=180^{circ}$. (iii) Three right angles have $90^{circ}+90^{circ}+90^{circ}=270^{circ}$ (iv) Four right angles have $90^{circ}+90^{circ}+90^{circ}+90^{circ}=360^{circ}$ (v) $frac{2}{3}times 90=60^{circ}$ (vi) $left (1frac{1}{2}right )right:angles=frac{3}{2}times 90=135^{circ}$ QUESTION – 4: How many degrees are there in the angle between the hour hand and the minute hand of a clock, when it is (i) 3 o’clock (ii) 6 o’clock (iii) 12 o’clock (iv) 9 o’clock Solution: (i) At 3 o’clock the angle formed between the hour hand and the minute hand is right angle, i.e. $90^{circ}$. (ii) At 6 o’clock the angle formed between the hour hand and minute hand is a straight line, i.2. $180^{circ}$. (iii) At", "At 1 o'clock the minute hand (red) points to the 12 and the hour hand (blue) points to the 1. Calculate Angle between Clock Hands. In this video, 1A Auto shows you how to change the positioning on an alternator. If not, see if you can find a \"D/R\" or \"DRG\" button and press it once; then try again. When studying the physi… At 1 o'clock the minute hand (red) points to the 12 and the hour hand (blue) points to the 1. There are also 24 hours in a day. Browser slowdown may occur during loading and creation. That line indicates south in the Northern Hemisphere. Since there are 12 hours on the clock, each hour mark is 30 degrees. .hours { background: #000; height: 20%; left: 48.75%; position: absolute; top: 30%; transform-origin: 50% 100%; width: 2.5%; } I’m using percentages to make scaling the clocks easier. Related Time Card Calculator | Time Calculator. Calculations of sunrise and sunset in The Angle – Ohio – USA for January 2021. Generic astronomy calculator to calculate times for sunrise, sunset, moonrise, moonset for many cities, with daylight saving time and time zones taken in account. This calculator allows you to enter direction angle, either in DMS (degrees, minutes, and seconds) or decimal degrees and outputs closest to the compass", "a circle, and a circle always contains 360 degrees. Online rechneronline.de The reference point 12 o'clock commonly refers to the line of sight and means an angle of 0 degrees. A 20 degree cut can be aimed at 7:30 or 4:30 on the clock face, a 25 degree shot at 8:00 or 4:00. Calculate the Angle between 12 and the Hour hand 10: Since there are 360 degrees in a full circle (clock), and there are 12 hours, each hour represents 360/12 = 30 degrees So our formula is 30 (H) So our formula is 30 (10) Quantity : Reference Unit : is equal to : Conversion Factor : Unit : 1: degree = 1 : degree: 1 = 0.017453292519943 : radian: 1 = 1.1111111111111 : grad An angle is in standard position if its vertex is at the origin and the initial side is on the positive x-axis. Simply described, this means that if the average temperature in a day is 20oC then a plant accumulates 20 day degrees in one calendar day. Ball landmarks come in handy. h = distance above or … Now set the dish to the satellite you want using a Satellite finder or whatever decoder you own. Then the angles between the two hands are calculated. A clock position, or clock bearing,", "# The sexagesimal system and its operations The sexagesimal system is a system of numeration in which every unit is divided into 60 smaller units. In other words, the base used is 60. This system is used to measure time and angles. $$1h \\rightarrow 60 \\ \\mbox{min} \\rightarrow 60\\cdot 60=3.600 \\ \\mbox{s}$$$$$1^\\circ \\rightarrow 60' \\rightarrow 60\\cdot 60=3.600''$$$ ## Operations with sexagesimal numbers ### Sum • Step 1 Write the numbers to be added as follows, and add them column by column: $$\\begin{eqnarray} & & 38^\\circ \\ 24' \\ 55'' \\\\\\\\ &+ & \\underline{40^\\circ \\ 49' \\ 17''} \\\\\\\\ & & 78^\\circ \\ 73' \\ 72'' \\end{eqnarray}$$ • Step 2 If the sum of seconds is bigger than $$60$$, we must divide the result by $$60$$; the remainder will be the seconds and the quotient will be added to the minutes. $$\\dfrac{72}{60}=1+\\dfrac{12}{60}$$ Namely, the remainder is a $$12$$ and the quotient $$1$$. Then, the result is written as: $$78^\\circ \\ (73+1)' \\ 12'' = 78^\\circ \\ 74' \\ 12''$$ • Step 3 Repeat the same procedure for the minutes: $$\\dfrac{74}{60}=1+\\dfrac{14}{60}$$ Then, $$78^\\circ \\ 74' \\ 12'' = 79^\\circ \\ 14' \\ 12''$$ ### Subtraction • Step 1 Write the numbers one above the other, the hours above the hours (or the degrees above the degrees), the minutes above the", "Dial consists in this, that the Hour of the Day may be found thereby without a meridian Line, or Compass; but then it must be pretty large, to shew the Hour exactly. ## The Construction of an universal Polar, East, and West Dial. This Instrument consists of a very strait and smooth circular Piece of Brass, or other Metal, pretty thick, that so it may preserve it’s perpendicular Weight, as likewise that a Groove may be made round the Limb thereof, for a hanging-Piece to slide about the same, like that on the astronomical Ring. About the Center cf the said circular Piece describe the Circumference of a Circle, which divide into twice 90 Degrees. Then draw a right Line from the 90th Degree thro’ the Center, representing the Equinoctial, near the Top of which assume a Point at pleasure, thro’ which draw a right Line perpendicular to the Equinoctial-Line, which shall be the Hour-Line of 6. Then to have the other Hour-Lines, you must lay off the answerable Tangents upon the Equinoctial-Line both ways from the Point therein of the Hour-Line of 6; as the Tangent of 15 Deg. for the Hour-Points of 5 and 7; the Tangent of 30 Deg. for 4 and 8; the Tangent of 45 Deg. for 3 and 9, &c. and if", "Oct 3, 2007 ### hotvette Welcome to PF! What I would do is the following. There are two lines of different lengths (r and R) rotating about a common origin at different frequencies. If the tips of the lines are points a and b, I'd write equations that represent the x,y locations of a and b (with respect to t) using initial conditions such that the postions of the lines are at 9:00 at t=0. I'd then write an expression for the distance between points a and b, take the derivative and evaluate it at t=0. Chain rule involved. Last edited: Oct 3, 2007 4. Oct 3, 2007 ### HallsofIvy Staff Emeritus In general, the angle formed by the hands is not a right angle (it happens to be at 9:00 but not immediately after). Your second thought was correct: you can use the cosine law to write the distance between the tips of the hands as a function of the angle between them. 5. Oct 3, 2007 ### D H Staff Emeritus Why is the chain rule needed here? Defining $\\hat x$ as the 12:00 position, $\\hat y$ as the 3:00 position, and $\\theta_m$ as the angle between the 12:00 position and the minute hand measured clockwise, then the tip minute hand is at $$\\vec m =", "there are only five minor markers between each hour marker - so in this context they're equivalent to 12 minutes each. This makes telling the minutes past the hour tricky. To remedy this we can introduce a little minutes circle. This is just a matter of creating a circle outline and a minutes object, and moving them around as x: 135*math.sin(math.rad({drh})) y: -135*math.cos(math.rad({drh})) Alternatively, we could use a dial that has a more useful number of minor markers. But the minutes circle is more visually striking, I think. Spotlight with Minutes and Date [WM] Okay, last one of these. I figured 'centre', above the hour line was the best place to put the date. For this, we have to make sure the date stays on the right side of the line and stays the right way up. Positions of the date relative to the hour line (left) and the corresponding positioning functions (right) The code to do this is x: -165*math.sin(math.rad({drh})) + ({dh11}<6 and -20 or 20)*math.cos(math.rad({drh})) Rotation: {drh} + ({dh11}<6 and -90 or 90) Where the first terms for x and y move the date around with the hour line (note the signs), and the second terms are the positioning functions (dx, dy) that shift the date to above the hour line. The conditional in the rotation", "list ‘0, cos x, sin x’ is required sufficiently frequently in defining axes and points that it is convenient to have a verb ``` cs=.0,cos,sin NB. e.g. rotn. axis in y-z plane ((cs 50),dtor _90)rotate cs 120 0.939693 0.219846 0.262003 ``` This result can be confirmed by spherical trigonometry applied to triangle NPS (see next section). The next step is to make the hour angle a parameter (clockwise 90° = anti-clockwise -90°): ``` v=.monad :'((cs 50),-dtor y)rotate cs 120'; ``` and plot values as this moves towards transit at 10° intervals : ``` v each 90 80 70 60 50 40 30 20 10 0 0.94 0.22 0.262 0.925 0.0948 0.367 0.883 _0.0264 0.469 0.814 _0.14 0.564 0.72 _0.243 0.65 0.604 _0.332 0.725 0.47 _0.404 0.785 0.321 _0.457 0.83 0.163 _0.489 0.857 0 _0.5 0.866 ``` More generally, it is useful to convert time to angular measure with 24 hours being equivalent to a complete rotation, which suggests three more utility verbs : ``` ttor=.o.&(%&43200@(60&#.)) NB. time (hms) to radians ttor 12 0 0 NB. check 12 hrs = pi rads 3.14159 ``` ``` dtot=.60 60 60&#:@(*&240) NB. deg to time (hms) dtot 180 NB. check 180 deg.= 12 hours 12 0 0 ``` ``` atod=.%&3600@(60&#.@(3&{.)) NB. angle(deg,min,sec) to degrees atod 49 15 NB. check 49o 15’ 0”", "Since a triangle’s angles must sum to 180$$^\\circ$$, no triangle can have more than one obtuse angle. 6- C $$1$$ foot $$= 0.333$$yard So $$36$$ foot $$= 12$$ yards $$(36 × 0.33)$$ 7- B The sum of $$A$$ and $$B$$ equals $$35: A + B = 35$$, if $$A = 16$$ we have $$16 +B = 35$$ 8- D $$A = 108 ÷ 9 = 12$$ 9- B Sum of these three angle are 180$$^\\circ$$. $$A^\\circ + 80^\\circ + 16^\\circ = 180^\\circ$$ $$A^\\circ =180^\\circ – 96^\\circ = 84^\\circ$$ 10- C Use perimeter of trapezoid formula. $$P = a + b + c + d$$ $$P = 2+ 6 + 2+ 8 =18$$ 11- B Low temperature is 24$$^\\circ$$f cooler than the temperature at 12:00 PM that is 76of, that means low temperature is 52$$^\\circ$$f (76$$^\\circ$$f $$–$$ 24$$^\\circ$$f) that is close to choice B 12- A $$(4 × 10) + (7 × 1) + (6 × 0.01) = 40 + 7 + 0.06 = 47.06$$ 13- D 1 year $$=$$ 365 days, 1 day $$=$$ 24 hours 1 year $$=$$ 365 × 24 1 year $$=$$ 8,760 14- B Each game = $$\\frac{11}{2}$$ hours = 90 minutes so 3 games = $$\\frac{41}{2}$$ hours = 270 minutes 5minutes between each game, so there is 10 minutes in total between 3", "have pricked down the Flour-Lines from One to Eight in this Dial, as appears per Figure. The Construction of an East Dial is the same as of this; and there are pricked down the Hour-Lines upon it from the Sun’s rising in Summer, viz. from Four in the Morning to Eleven. The reason that the Hour-Line of Twelve cannot be drawn upon these Dials, is, because when the Sun is in the Meridian, his Rays are parallel to their Planes. If a West Dial be drawn upon a Sheet of Paper, and then the said Paper is rendered Transparent by oiling, you will perceive thro’ the backside of the Paper an East Dial drawn entirely; only the Figures of the Hours must be altered, that is, you must put 11 in the Place of 1; 10 in the Place of 2; and so of others. The Style of these Dials is a Brass or Iron Rod, in Length equal to ED, which is likewise equal to the Distance of the Hour of 3 from the Hour of 6. This Style is set upright in the Point D, and shews the Hour by the Shadow of it’s Extremity. These Dials, which may have likewise a Style in figure of a Parallelogram, as we have mentioned in speaking of Polar", "properly? Thanks. Edit: More details If I scale the hours from 0,24 to 0,180 and apply sine, then 0 will be close to 24. However, 1 will also be close to 23, 2 with 22, 3 with 21 and so on, which is unwarranted. Also, the rate of change is much faster around 0 and 180 degrees for the sine function, than around 90 degrees, which means, the difference between 23 and 0 hours is much more than between 11 and 12 hours. Ideally, I would like to lay 0-24 in a ring (like in a watch), and have some function convert it into some numbers such that, the numerical difference between those numbers would be equal to distance between those numbers around the ring. I read the answer about using both cosine and sine, but its still over my head. Maybe the answer lies there. If yes, could you please explain a bit in this context, how that might solve the problem. Thanks. • You can clearly, for example, scale the hours of the day on the 0,23 interval in the same way as those other answers suggest to scale the degrees of the circle so I can't quite see where your difficulty is given you understand that technique. For example the second answer here - stats.stackexchange.com/questions/4783/…", "Parallels will be the Hour-Lines from 6 in the Evening to 6 in the Morning AL being the Hour-Line of Midnight. And if the parallel Spaces, be laid off on the other side of the Hour-Line of 6, you will have the Hour-Lines from 6 in the Morning to 6 in the Evening. And for drawing the half Hours, divide each 15th Deg. of the Quadrant AF into half, and draw Parallel Lines between the Hour-Lines. The Hour-Lines may be yet otherwise drawn, by means of a Circle, whose Diameter is the Line AB, and whose Circumference is divided into 24 equal Parts for the 24 Hours of the Day, or into 48, for the Half-Hours. For then if right Lines be drawn thro’ the opposite Points of Division, parallel to ED, we shall have the Hour-Lines, and those of the Half-Hours, as we have said in the Construction of the former right-lined Dial. About the Point I, as a Center, draw an occult Quadrant, which divide into 90 Deg. and laying a Ruler to the Center I, and on each Division mark the same Degrees upon the Sides GQ, and GS of the Instrument. Note, By means of these Divisions we may find the Sun’s Altitude above the Horizon, as we shall shew by and by. RR", "so each one must be 360° ÷ 12 = 30° Mariner, with 128 points, extended the 32-point system with half-and quarter-points to allow 128 directions. A 23 degree cut is 2mm shy of a 25 and 3mm more than a 20. When we rotate a figure of 90 degrees clockwise, each point of the given figure has to be changed from (x, y) to (y, -x) and graph the rotated figure. Day degrees are a way of measuring the physiological development of a plant. All angles between -15º and 15º 15 degrees around the earth correctly to get the signal... Is Scirocco ( or Exaloc ) is 2mm shy of a clock and the hour, the sun a clock! Note of the compass 6° per minute has it ( half- and quarter-points do have. How they 're measured on the clock, each minute mark is 30 degrees, 128-wind compass roses (,. To get the maximum signal strengths between -15º and 15º rotates through in! Position on the 75 degree of longitude was the time on the circle be the case when in... A 25 and 3mm more than a 20 plants don ’ t work by calendar –. 60 minutes on a figure every hour, notwithstanding the fact that its number of days, hours,,. Days, hours, minutes, allowing", "On Time On a clock the three hands - the second, minute and hour hands - are on the same axis. How often in a 24 hour day will the second hand be parallel to either of the two other hands? Estimating Angles How good are you at estimating angles? LOGO Challenge 8 - Rhombi Explore patterns based on a rhombus. How can you enlarge the pattern - or explode it? Two Exterior Triangles Stage: 3 Short Challenge Level: Since $ABQ$ and $BCQ$ are equilateral, the angles $ABP$ and $CBQ$ are both $60^\\circ$. So $$\\angle{PBQ} = 360^\\circ-90^\\circ-60^\\circ-60^\\circ=150^\\circ$$ PBQ is isosceles, so the angles $BPQ$ and $PQB$ are equal. So $$2 \\times \\angle{PQB} = 180^\\circ- 150^\\circ = 30^\\circ$$ So $$\\angle{PQB} = 15^\\circ$$ This problem is taken from the UKMT Mathematical Challenges. View the archive of all weekly problems grouped by curriculum topic View the previous week's solution View the current weekly problem", "$$|AC|^2=|AE|^2+|EF|^2+|CF|^2$$ Can you work out these lengths? Hint: The six triangles in the diagram are all ... 0 $\\frac 3{11}$ hours is the time it takes the minute hand to gain $90^\\circ$ on the hour hand. That will be the time from midnight (or noon) to the first $90^\\circ$ point. Having had one case you need the minute hand to gain $180^\\circ$ to get to the next one, which takes $\\frac 6{11}$ hours. 1 HINT You want to reflect lines $x= \\pm 0.5, y= \\pm 0,5$ about $y = m x , m= \\tan \\theta .$ Instead of that consider four among the eight rotated lines $$\\pm x \\cos \\theta \\pm y \\sin \\theta = 0.5 ; \\, \\pm x \\sin \\theta \\pm y \\cos\\theta = 0.5$$ 1 Let $ABCD$ be your square. Let's call $G$ and $H$ the points where $l$ intersects, respectively, the sides $AB$ and $CD$. We have the two trapezia $GBCH$ and $AGHD$ that both have the same area (half of the original square, so $1/2$). We now want to reflect $AGDH$ around $l$, let's call $L$ and $K$ the images of $A$ and $B$ after the reflection. The new ... 1 There are various ways to do this. You could set up an equation for the distance from an arbitrary point $(u,v)$ to" ]

[ "you can simplify it by turning it into $2^5\\cdot3$", "expression $(8+2) \\cdot 6$ Use the order of operations to simplify each expression $(8+2) \\cdot 6$...", "Determine the first two steps when simplifying the following expression. $11\\div&space;(12-&space;8&space;\\cdot&space;3&space;)+2^4$ $12-{24}$ $11\\div12$ $8\\cdot3$ $12-8$ $2^4$ First Step Second Step", "## Prealgebra (7th Edition) Remember PEMDAS to simplify. $2*4=8$, $-3\\div3=-1$ The expression is therefore $8-1=7$", "How do you simplify -10div5*2+8*(6-4)-3*4? Apr 14, 2015 We use the PEMDAS rule to simplify the above expression $- 10 \\div 5 \\cdot 2 + 8 \\cdot \\left(6 - 4\\right) - 3 \\cdot 4$ We solve the Parentheses first $= - 10 \\div 5 \\cdot 2 + 8 \\cdot 2 - 3 \\cdot 4$ We then Multiply/Divide whichever comes first $= - 2 \\cdot 2 + 8 \\cdot 2 - 3 \\cdot 4$ $= - 4 + 8 \\cdot 2 - 3 \\cdot 4$ $= - 4 + 16 - 3 \\cdot 4$ $= - 4 + 16 - 12$ We then Add/Subtract whichever comes first $= 12 - 12$ $= 0$ color(green)(-10div5*2+8*(6-4)-3*4 = 0", "# How do you simplify 3 + 6 times 2 div3? $7$ $3 + 6 \\times 2 \\div 3 = 3 + \\frac{\\cancel{6} 2 \\cdot 2}{\\cancel{3} 1} = 3 + 4 = 7$", "[4 – 2 (6 – 3)] \\div 2$? 3. Which of the following shows the simplified expression of $16 – 3 (8 – 3) 2 \\div 5$? 4. Which of the following shows the simplified expression of $14z + 5 [6 – (2z + 3)]$? 5. Which of the following shows the simplified expression of $- \\{2 y – [ 3 – (4 – 3 y)] \\}+ 6 y$? 6. Which of the following shows the simplified expression of $3 + 6 \\times (4 + 5) \\div 3 – 7$? 7. Which of the following shows the simplified expression of $150 \\div (6 + 3\\times 8) – 5$? 8. Which of the following shows the simplified expression of $45 \\div (8 {5 – 4} – 3)$?", "# How do you simplify 3 • 42 + 6 using order of operations? The value og this expression is $132$ $3 \\cdot 42 + 6 = \\left(3 \\cdot 42\\right) + 6 = 126 + 6 = 132$", "# How do you simplify (1.363times10^16)/(2.9times10^6)? E.G. 4. (700000000) is 9 digits, which translates to $4.7 \\cdot {10}^{\\ast} 9 \\ast$", "$5+2^{3}+3[6-3(4-2)]$. Solution #### Example 12 Simplify: $2^{3}+3^{4} \\div 3-5^{2}$. Solution", "# How do you simplify the expression 8 • 7 – 4? $52$ $8 \\cdot 7 - 4$ $\\therefore = 56 - 4$ $\\therefore = 52$", "is 8 and the exponent is 6, so $8^{6}$ means $8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8$ 2. The base is $x$ and the exponent is 5, so $x^{5}$ means $x \\cdot x \\cdot x \\cdot x \\cdot x$ To simplify an exponential expression without using a calculator, we write it in expanded form and then multiply the factors. #### Example 7 Simplify: $3^{4}$. Solution ## 2.1.4 Simplify Expressions Using the Order of Operations We’ve introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values. For example, consider the expression: $4+3 \\cdot 7$ Some students say is simplifies to 49. Some students say is simplifies to 25. Imagine the confusion that could result if every problem had several different correct answers. The same expression should give the same result. So mathematicians established some guidelines called the order of operations, which outlines the order in which parts of an expression must be simplified. ### ORDER OF OPERATIONS When simplifying mathematical expressions perform the operations in the following order: 1. Parentheses and other Grouping Symbols • Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first. 2.", "# How do you simplify 3(cos4+isin4)*0.5(cos2.5+isin2.5) and express the result in rectangular form? May 1, 2017 $1 , 4648814385920352498294698809161$ $+ i \\cdot 0 , 3226799821317232864454186236121$ #### Explanation: $3 \\cdot 0.5 \\cdot \\left(\\cos 6.5 + i \\sin 6.5\\right) = x + i y$ $x = 1.5 \\cos 6.5$ $y = 1.5 \\sin 6.5$", "# How do you simplify 3 + 6 times 2 div3? ##### 1 Answer Mar 13, 2018 $7$ #### Explanation: $3 + 6 \\times 2 \\div 3 = 3 + \\frac{\\cancel{6} 2 \\cdot 2}{\\cancel{3} 1} = 3 + 4 = 7$", "snits Since the task is to simplify you could probably simplify the final expression a bit more: $\\frac{x}{\\sqrt{x}(x^2-1)} = \\frac{x \\cdot x^{-1/2}}{x^2-1} = \\frac{\\sqrt{x}}{x^2-1}$", "3$ So now the expression is: $2 + 3$ Finally, we add and get $5$. Hope this helps!", "(\\text{-} x + 8)$ $5 + 2(\\text{-}7) + 2 \\cdot x = \\text{-} 6 - (\\text{-} x + 8)$ $5 -14 + 2x = \\text{-} 6 - (\\text{-} x + 8)$ $5 -14 + 2x = \\text{-} 6 +x - 8$ $\\text{-} 9 + 2x = x - 14$ $2x = x - 5$ $x= \\text{-} 5$", "# How do you simplify the expression 8 • 7 – 4? Apr 9, 2017 $52$ #### Explanation: $8 \\cdot 7 - 4$ $\\therefore = 56 - 4$ $\\therefore = 52$", "simplified. The final answer is: a simplified expression of \\$\\$(3x^2+5x+2)/(x^2-3x-4)\\$\\$ is \\$\\$(3x+2)/(x-4)\\$\\$ ## Contact tutor Send a message explaining your needs and David will reply soon. Contact David", "How do you simplify -99 div (-3) div 11 *7 using order of operations? May 24, 2018 21 Explanation: -99 ÷ (-3) ÷ 11 * 7 Using BODMAS = (-99 ÷ (-3)) ÷ 11 * 7 = (33 ÷ 11) * 7 $= 3 \\cdot 7$ $= 21$" ]

[ "brackets, [ ], in addition to parenthesis. Intro Problem Revisit If we don't have the Order of Operations, Abigail could get very confused and not have the right amount of money to pay for her items. So, the two shirts are $24 $(2 \\cdot \\12)$ and only the jeans are discounted. They are $50-0.15\\cdot 50 = 42.50$ . Adding these two items together, we get that her total bill is$24 + $42.50 =$66.50. So, because her purchase is over $60, she gets an additional$10 off, making her final bill \\$56.50. ### Guided Practice What property is being used? 1. $5(c - 9) = 5c - 45$ 2. $6 \\cdot 7=7 \\cdot 6$ 3. Use the Order of Operations to simplify $8+[4^2-6 \\div (5+1)]$ . 1. 5 is being distributed to each term inside the parenthesis, therefore the Distributive Property is being used. 2. Here, order does not matter when multiplying 6 and 7. This is an example of the Commutative Property. 3. This is an example of embedded parenthesis, as discussed above. Start by simplifying the parenthesis that are inside the brackets. Then, simplify what is inside the brackets according to the Order of Operations. $&8+[4^2-6 \\div (5+1)]\\\\&8+[4^2-6 \\div 6]\\\\&8+[16-6\\div 6]\\\\&8+[16-1]\\\\&8+15\\\\&23$ ### Explore More Determine which algebraic property is being used below. 1. $8 + 5 = 5", "## What you need to know Things to remember: • We do what is inside the brackets first. Previously we have heard the term “BIDMAS”, which tells us in what order we do our operations: Brackets I ndices D ivision M ultiplication A ddition S ubtraction Here, we will focus on brackets. Because brackets are at the top of the list, this tells us that we have to do what is in our brackets first. Find the value of the following expression $$(11-3)\\div2$$ Reading down our BIDMAS list we can see that we have to do the brackets first. $$11-3=8$$ $$(11-3)\\div2=8\\div2$$ And now, we can just do the division as normal $$8\\div2 =4$$ $$(11-3)\\div2=4$$ But, what happens if we have two brackets? Find the value of the following expression $$(12+5)-(3-9)$$ Because we have two brackets, which one do we do first?… It doesn’t matter! Let’s start with the first one. $$(12+5)-(3-9) =17-(3-9)$$ Now we have to do the second bracket $$17-(3-9)=17 - -6$$ And now we can do the subtractions as usual (Remember, this will become an addition) $$17 - -6=17+6=23$$ $$(12+5)-(3-9)=23$$ Now let’s try the second bracket first. $$(12+5)-(3-9) =(12+5) - -6$$ $$(12+5) - -6=17 - -6$$ $$17 - -6 = 17 +6 =23$$ We got the same answer! ## Example Questions #### Question 1: Find the", "expression $(8+2) \\cdot 6$ Use the order of operations to simplify each expression $(8+2) \\cdot 6$...", "convention; if we want to say \"Two, added to the product of two and three\", we would probably write $2+2\\cdot 3$. Our convention says that this is evaluated as $2+2\\cdot 3 = 2+6=8$. To us, this is perfectly clear, because we've decided that's how things are read. However, not everybody agrees on order of operations. Some say PEMDAS: Evaluate inside parentheses, THEN exponentiate, THEN multiply, THEN divide, THEN add, THEN subtract. Some group the latter set together: PE(MD)(AS), and say one evaluates from right to left when operations have the same priority. These are, essentially, the two most common agreed-upon orders. That would give this problem the following solutions respectively: $$6 \\div 2 \\cdot(2+1) = 6 \\div 2 \\cdot 3 = 6 \\div 6 = 1$$ $$6 \\div 2 \\cdot (2+1) = 6 \\div 2 \\cdot 3 = 3 \\cdot 3 = 9$$ But these are completely different things. As T.Bongers says in the comments above, mathematical notation is used to communicate an idea. If we have two expressions that give a different answer, we use more parentheses to make it clear with we mean, or use some other common notation; for this problem, perhaps I would write $$\\frac{6}{2(2+1)}$$ because this makes it completely clear what I mean. The point of our accepted order of operations is to", "each numerical expression using the order of operations. 1. $4 + 5 \\cdot 2 - 3$ 2. $6 + 6 \\cdot 3 \\div 2 - 7$ 3. $5 + 5 \\cdot 8 \\div 2 + 6$ 4. $13 - 3 \\cdot 2 + 8 - 2$ 5. $17 - 5 \\cdot 3 + 8 \\div 2$ 6. $9 + 4 \\cdot 2 + 7 - 1$ 7. $8 + 5 \\cdot 6 + 2 \\cdot 4 - 3$ 8. $19 + 2 \\cdot 4 - 3 \\cdot 2 + 10$ 9. $12 + 4 \\cdot 4 \\div 8 - 3$ 10. $12 \\cdot 2 + 16 \\div 2 - 12$ Directions: Evaluate each variable expression. Remember to use the order of operations when necessary. 1. $4y+6-2$, when $y=6$ 2. $9+3x-5+2$, when $x=8$ 3. $6y+2y-5$, when $y=3$ 4. $8+3y-5 \\cdot 2$, when $y=4$ 5. $7x-2 \\cdot 3 \\div 3+12$, when $x=5$ 6. $3+4 \\cdot 3 - 2y+5$, when $y=7$ 7. $6a+3(2)+5 - 4$, when $a=9$ 8. $10+3 \\cdot 5+2-9b$, when $b=2$ 9. $14 \\div 2+3a+7a$, when $a=2$ 10. $5+6y-2y+11-4$, when $y=3$ Directions: Evaluate each expression using the order of operations. Remember to pay attention to the grouping symbols. 1. $3 + (4 + 5) - 6(2)$ 2. $4 + (6 \\div 3) + 2(7) - 4$ 3. $3 +", "# How do you simplify 2^3 ÷ (7 ÷ 7 ÷ 8) using order of operations? Mar 11, 2018 The final answer would be 64. We figure this out through the use of BIMDAS (Brackets, Indices, Multiplication/Division, Addition/Subtraction) #### Explanation: Begin with what is inside the brackets $\\left(7 \\div 7 \\div 8\\right)$. This will equal 0.125 or $\\frac{1}{8}$. This will give us ${2}^{3} \\div 0.125$. Next we deal with indices, which in this case is ${2}^{3}$. If we expand this we get $2 \\cdot 2 \\cdot 2$ which will equal 8. Now we are at $8 \\div 0.125$ or $8 \\div \\frac{1}{8}$ This will produce a result of 64.", "brackets, [ ], in addition to parenthesis. Intro Problem Revisit If we don't have the Order of Operations, Abigail could get very confused and not have the right amount of money to pay for her items. So, the two shirts are $24 $(2 \\cdot 12)$ and only the jeans are discounted. They are $50-0.15\\cdot 50 = 42.50$ . Adding these two items together, we get that her total bill is$24 + $42.50 =$66.50. So, because her purchase is over $60, she gets an additional$10 off, making her final bill \\$56.50. Guided Practice What property is being used? 1. $5(c - 9) = 5c - 45$ 2. $6 \\cdot 7=7 \\cdot 6$ 3. Use the Order of Operations to simplify $8+[4^2-6 \\div (5+1)]$ . 1. 5 is being distributed to each term inside the parenthesis, therefore the Distributive Property is being used. 2. Here, order does not matter when multiplying 6 and 7. This is an example of the Commutative Property. 3. This is an example of embedded parenthesis, as discussed above. Start by simplifying the parenthesis that are inside the brackets. Then, simplify what is inside the brackets according to the Order of Operations. $&8+[4^2-6 \\div (5+1)]\\\\&8+[4^2-6 \\div 6]\\\\&8+[16-6\\div 6]\\\\&8+[16-1]\\\\&8+15\\\\&23$ Vocabulary A number that is the opposite sign of a given number, such that, when added, their sum", "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: $\\left(10 - 4 \\cdot 13 + 19\\right) \\div 23$ = $\\left(10 - 52 + 19\\right) \\div 23$ = $\\left(29 - 52\\right) \\div 23$ = $\\left(- 23\\right) \\div 23 = - 1$", "# How do you simplify 4|3-(-3)| + (-1)? The absolute value acts as a bracket, so analyze what's inside there first, then it's order of operations as normal, to get to $23$. #### Explanation: Starting with the original: $4 | 3 - \\left(- 3\\right) | + \\left(- 1\\right)$ The absolute value acts as a set of brackets, so let's focus there first: $| 3 - \\left(- 3\\right) |$ So $3 - \\left(- 3\\right) = 6$ So let's substitute that in: $4 | 6 | + \\left(- 1\\right)$ Next, $| 6 | = 6$ So we have: $4 \\cdot 6 + \\left(- 1\\right) = 24 - 1 = 23$", "# How do you use order of operations to simplify (15/3)^3 - 6(2)^2? ${\\left(\\frac{15}{3}\\right)}^{3} - 6 {\\left(2\\right)}^{2} = {5}^{3} - 6 \\cdot 4 = 125 - 24 = 101$.", "calculus to have a better understanding. Much of pre-calculus is algebra, and coincidentally, I'm doing some work on \"order of operations\" at the moment. Would you mind indicating why it's a \"mathematical mess\"? Much appreciated. 9. Sep 17, 2016 ### Beanyboy Funny that, I'm trying to learn Pre-Calculus at the moment so I can later \"play with the big boys, and girls\". Currently, reviewing \"order of operations\". Would you mind being a bit more specific, perhaps with an example? Much appreciated. 10. Sep 17, 2016 ### Staff: Mentor One of the statements has been: When in doubt, use brackets! This means brackets group things and all calculations within brackets have to be done before we can drop them. So brackets define what comes first. If multiple of them are nested, then it has to be inside out. E.g. $(5 + ( 3 \\cdot (7 - 2))) = (5 + ( 3 \\cdot (5)))=(5 + ( 3 \\cdot 5))=(5 + ( 15 ))= (5+15)=(20)=20$ because the grouping is inside out. The example above $9 - 3 \\div 1/3 + 1$ is ambiguous because it is not clear, whether $9 - (3 \\div (\\frac{1}{3})) + 1 = 1$ is meant or $9 - ((3 \\div 1) \\div 3) + 1 = 9$. Associativity is the word for the rule $a\\,\\cdot\\,(b\\,\\cdot\\,c)=(a\\,\\cdot\\,b)\\,\\cdot\\,c", "# How do you simplify 4 + 8 * |8 - (8 - 2)^2| using order of operations? Apr 17, 2018 $4 + 8 \\setminus \\cdot | 8 - {\\left(8 - 2\\right)}^{2} |$ $= 4 + 8 \\cdot | 8 - {6}^{2} |$ $= 4 + 8 \\setminus \\cdot | 8 - 36 |$ $= 4 + 8 \\setminus \\cdot | - 28 |$ $= 4 + 8 \\setminus \\cdot 28$ $= 4 + 224$ $= 228$ Apr 17, 2018 See explanation. #### Explanation: The order of operation can be rememberd through an acronym PEDMAS. The letters in the acronym stand for: P - Parenetheses (also called brackets) E - Exponents D and M - Division and Multiplication A and S - Addition and Subtraction The acronym says that first you need to do all operations in brackets (parenetheses). Here you have two kinds of brackets: ordinary brackets ( and ), but also absolute value signs $| |$ (these are also considered as brackets). First we need to perform operations in these brackets which do not include other ones (the most inner) In the given expression the brackets without other ones are $\\left(8 - 2\\right)$, so we have to do this subtraction first: $4 + 8 \\cdot | 8 - {\\textcolor{red}{\\left(8 - 2\\right)}}^{2} | = 4 + 8", "subtraction operations) can be performed in the order in which they appear in the expression. Let’s evaluate a few expressions using the order of operations. ### Example 1 $3^2-1\\cdot4+2$ In this expression, the following operations are taking place: exponentiation, subtraction, multiplication, and addition. Following the order of operations, we simplify the exponent first and then perform the multiplication; next, we perform the subtraction, and then the addition: \\begin{align} \\displaystyle 3^2-1\\cdot4+2 &= 9-1\\cdot4+2 \\\\ &= 9-4+2 \\\\ &= 5+2 \\\\ &= 7 \\end{align} ### Example 2 $6-(5\\cdot1)+2^3$ Here we have an expression that involves subtraction, parentheses, multiplication, addition, and exponentiation. Following the order of operations, we simplify the expression within the parentheses first and then simplify the exponent; next, we perform the subtraction and addition operations in the order in which they appear in the expression: \\begin{align} \\displaystyle 6-(5\\cdot1)+2^3 &= 6-5+2^3 \\\\ &= 6-5+8 \\\\ &= 1+8 \\\\ &= 9 \\end{align} ### A Note on Equal Precedence Since multiplication and division are of equal precedence, it may be helpful to think of dividing by a number as multiplying by the reciprocal of that number. Thus $3 \\div 4 = 3 \\cdot \\frac{1}{4}$. In other words, the quotient of 3 and 4 equals the product of 3 and $\\frac{1}{4}$. Similarly, as addition and subtraction are of equal precedence, we can", "# How do you use order of operation to simplify 4-24/6*2 +(-5)^2? $4 - \\frac{24}{6} \\cdot 2 + {\\left(- 5\\right)}^{2} = 4 - 8 + 25 = 21$.", "# How do you use order of operation to simplify 4-24/6*2 +(-5)^2? ##### 1 Answer Apr 17, 2015 In this way: $4 - \\frac{24}{6} \\cdot 2 + {\\left(- 5\\right)}^{2} = 4 - 8 + 25 = 21$.", "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 $\\boxed{511}$.", "+ 36 = 40 - 4 = 36$ ### Practice Directions: Evaluate each numerical expression using the order of operations. 1. $4 + 5 \\cdot 2 - 3$ 2. $6 + 6 \\cdot 3 \\div 2 - 7$ 3. $5 + 5 \\cdot 8 \\div 2 + 6$ 4. $13 - 3 \\cdot 2 + 8 - 2$ 5. $17 - 5 \\cdot 3 + 8 \\div 2$ 6. $9 + 4 \\cdot 2 + 7 - 1$ 7. $8 + 5 \\cdot 6 + 2 \\cdot 4 - 3$ 8. $19 + 2 \\cdot 4 - 3 \\cdot 2 + 10$ 9. $12 + 4 \\cdot 4 \\div 8 - 3$ 10. $12 \\cdot 2 + 16 \\div 2 - 12$ Directions: Evaluate each variable expression. Remember to use the order of operations when necessary. 1. $4y+6-2$ , when $y=6$ 2. $9+3x-5+2$ , when $x=8$ 3. $6y+2y-5$ , when $y=3$ 4. $8+3y-5 \\cdot 2$ , when $y=4$ 5. $7x-2 \\cdot 3 \\div 3+12$ , when $x=5$ 6. $3+4 \\cdot 3 - 2y+5$ , when $y=7$ 7. $6a+3(2)+5 - 4$ , when $a=9$ 8. $10+3 \\cdot 5+2-9b$ , when $b=2$ 9. $14 \\div 2+3a+7a$ , when $a=2$ 10. $5+6y-2y+11-4$ , when $y=3$ Directions: Evaluate each expression using the order of operations. Remember to pay attention to the grouping", "Determine the first two steps when simplifying the following expression. $11\\div&space;(12-&space;8&space;\\cdot&space;3&space;)+2^4$ $12-{24}$ $11\\div12$ $8\\cdot3$ $12-8$ $2^4$ First Step Second Step", "# 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: $\\left(10 - 4 \\cdot 13 + 19\\right) \\div 23$ = $\\left(10 - 52 + 19\\right) \\div 23$ = $\\left(29 - 52\\right) \\div 23$ = $\\left(- 23\\right) \\div 23 = - 1$", "and they do not go around any other brackets. 2. The square brackets are used when we want to create an expression that includes an expression that is already inside round brackets. 3. The curly brackets are used when we want to create an expression that includes an expression that is already inside square brackets. So, an expression that uses all three types of brackets could look like this: $4 + 18 \\cdot {24 – 8 \\cdot [(23-8) + 3 \\cdot (32 + 4)] – 8 \\cdot (23-4)}$ See? Now that we know the proper order of operations, the brackets give us precise instructions on how this problem should be solved in order to get to the correct result. And in what order are brackets solved? Well, we start from the inside, by solving the problems inside round brackets first, then work our way out. Like this: Example: $49 + 3 \\cdot {24 – 2 \\cdot [(23-8) + 3 \\cdot (3 + 4)] – 8 \\cdot (2 – 4)}$ $=49 + 3 \\cdot {24 – 2 \\cdot [15 + 3 \\cdot 7] – 8 \\cdot (-2)}$ $=49 + 3 \\cdot {24 – 2 \\cdot [15 + 21] – 8 \\cdot (-2)}$ $=49 + 3 \\cdot {24 – 2 \\cdot 36 – 8 \\cdot (-2)}$ $=49 + 3 \\cdot" ]

[ "brackets, [ ], in addition to parenthesis. Intro Problem Revisit If we don't have the Order of Operations, Abigail could get very confused and not have the right amount of money to pay for her items. So, the two shirts are $24 $(2 \\cdot \\12)$ and only the jeans are discounted. They are $50-0.15\\cdot 50 = 42.50$ . Adding these two items together, we get that her total bill is$24 + $42.50 =$66.50. So, because her purchase is over $60, she gets an additional$10 off, making her final bill \\$56.50. ### Guided Practice What property is being used? 1. $5(c - 9) = 5c - 45$ 2. $6 \\cdot 7=7 \\cdot 6$ 3. Use the Order of Operations to simplify $8+[4^2-6 \\div (5+1)]$ . 1. 5 is being distributed to each term inside the parenthesis, therefore the Distributive Property is being used. 2. Here, order does not matter when multiplying 6 and 7. This is an example of the Commutative Property. 3. This is an example of embedded parenthesis, as discussed above. Start by simplifying the parenthesis that are inside the brackets. Then, simplify what is inside the brackets according to the Order of Operations. $&8+[4^2-6 \\div (5+1)]\\\\&8+[4^2-6 \\div 6]\\\\&8+[16-6\\div 6]\\\\&8+[16-1]\\\\&8+15\\\\&23$ ### Explore More Determine which algebraic property is being used below. 1. $8 + 5 = 5", "convention; if we want to say \"Two, added to the product of two and three\", we would probably write $2+2\\cdot 3$. Our convention says that this is evaluated as $2+2\\cdot 3 = 2+6=8$. To us, this is perfectly clear, because we've decided that's how things are read. However, not everybody agrees on order of operations. Some say PEMDAS: Evaluate inside parentheses, THEN exponentiate, THEN multiply, THEN divide, THEN add, THEN subtract. Some group the latter set together: PE(MD)(AS), and say one evaluates from right to left when operations have the same priority. These are, essentially, the two most common agreed-upon orders. That would give this problem the following solutions respectively: $$6 \\div 2 \\cdot(2+1) = 6 \\div 2 \\cdot 3 = 6 \\div 6 = 1$$ $$6 \\div 2 \\cdot (2+1) = 6 \\div 2 \\cdot 3 = 3 \\cdot 3 = 9$$ But these are completely different things. As T.Bongers says in the comments above, mathematical notation is used to communicate an idea. If we have two expressions that give a different answer, we use more parentheses to make it clear with we mean, or use some other common notation; for this problem, perhaps I would write $$\\frac{6}{2(2+1)}$$ because this makes it completely clear what I mean. The point of our accepted order of operations is to", "expression $(8+2) \\cdot 6$ Use the order of operations to simplify each expression $(8+2) \\cdot 6$...", "Determine the first two steps when simplifying the following expression. $11\\div&space;(12-&space;8&space;\\cdot&space;3&space;)+2^4$ $12-{24}$ $11\\div12$ $8\\cdot3$ $12-8$ $2^4$ First Step Second Step", "brackets, [ ], in addition to parenthesis. Intro Problem Revisit If we don't have the Order of Operations, Abigail could get very confused and not have the right amount of money to pay for her items. So, the two shirts are $24 $(2 \\cdot 12)$ and only the jeans are discounted. They are $50-0.15\\cdot 50 = 42.50$ . Adding these two items together, we get that her total bill is$24 + $42.50 =$66.50. So, because her purchase is over $60, she gets an additional$10 off, making her final bill \\$56.50. Guided Practice What property is being used? 1. $5(c - 9) = 5c - 45$ 2. $6 \\cdot 7=7 \\cdot 6$ 3. Use the Order of Operations to simplify $8+[4^2-6 \\div (5+1)]$ . 1. 5 is being distributed to each term inside the parenthesis, therefore the Distributive Property is being used. 2. Here, order does not matter when multiplying 6 and 7. This is an example of the Commutative Property. 3. This is an example of embedded parenthesis, as discussed above. Start by simplifying the parenthesis that are inside the brackets. Then, simplify what is inside the brackets according to the Order of Operations. $&8+[4^2-6 \\div (5+1)]\\\\&8+[4^2-6 \\div 6]\\\\&8+[16-6\\div 6]\\\\&8+[16-1]\\\\&8+15\\\\&23$ Vocabulary A number that is the opposite sign of a given number, such that, when added, their sum", "each numerical expression using the order of operations. 1. $4 + 5 \\cdot 2 - 3$ 2. $6 + 6 \\cdot 3 \\div 2 - 7$ 3. $5 + 5 \\cdot 8 \\div 2 + 6$ 4. $13 - 3 \\cdot 2 + 8 - 2$ 5. $17 - 5 \\cdot 3 + 8 \\div 2$ 6. $9 + 4 \\cdot 2 + 7 - 1$ 7. $8 + 5 \\cdot 6 + 2 \\cdot 4 - 3$ 8. $19 + 2 \\cdot 4 - 3 \\cdot 2 + 10$ 9. $12 + 4 \\cdot 4 \\div 8 - 3$ 10. $12 \\cdot 2 + 16 \\div 2 - 12$ Directions: Evaluate each variable expression. Remember to use the order of operations when necessary. 1. $4y+6-2$, when $y=6$ 2. $9+3x-5+2$, when $x=8$ 3. $6y+2y-5$, when $y=3$ 4. $8+3y-5 \\cdot 2$, when $y=4$ 5. $7x-2 \\cdot 3 \\div 3+12$, when $x=5$ 6. $3+4 \\cdot 3 - 2y+5$, when $y=7$ 7. $6a+3(2)+5 - 4$, when $a=9$ 8. $10+3 \\cdot 5+2-9b$, when $b=2$ 9. $14 \\div 2+3a+7a$, when $a=2$ 10. $5+6y-2y+11-4$, when $y=3$ Directions: Evaluate each expression using the order of operations. Remember to pay attention to the grouping symbols. 1. $3 + (4 + 5) - 6(2)$ 2. $4 + (6 \\div 3) + 2(7) - 4$ 3. $3 +", "## What you need to know Things to remember: • We do what is inside the brackets first. Previously we have heard the term “BIDMAS”, which tells us in what order we do our operations: Brackets I ndices D ivision M ultiplication A ddition S ubtraction Here, we will focus on brackets. Because brackets are at the top of the list, this tells us that we have to do what is in our brackets first. Find the value of the following expression $$(11-3)\\div2$$ Reading down our BIDMAS list we can see that we have to do the brackets first. $$11-3=8$$ $$(11-3)\\div2=8\\div2$$ And now, we can just do the division as normal $$8\\div2 =4$$ $$(11-3)\\div2=4$$ But, what happens if we have two brackets? Find the value of the following expression $$(12+5)-(3-9)$$ Because we have two brackets, which one do we do first?… It doesn’t matter! Let’s start with the first one. $$(12+5)-(3-9) =17-(3-9)$$ Now we have to do the second bracket $$17-(3-9)=17 - -6$$ And now we can do the subtractions as usual (Remember, this will become an addition) $$17 - -6=17+6=23$$ $$(12+5)-(3-9)=23$$ Now let’s try the second bracket first. $$(12+5)-(3-9) =(12+5) - -6$$ $$(12+5) - -6=17 - -6$$ $$17 - -6 = 17 +6 =23$$ We got the same answer! ## Example Questions #### Question 1: Find the", "# How do you simplify 2^3 ÷ (7 ÷ 7 ÷ 8) using order of operations? Mar 11, 2018 The final answer would be 64. We figure this out through the use of BIMDAS (Brackets, Indices, Multiplication/Division, Addition/Subtraction) #### Explanation: Begin with what is inside the brackets $\\left(7 \\div 7 \\div 8\\right)$. This will equal 0.125 or $\\frac{1}{8}$. This will give us ${2}^{3} \\div 0.125$. Next we deal with indices, which in this case is ${2}^{3}$. If we expand this we get $2 \\cdot 2 \\cdot 2$ which will equal 8. Now we are at $8 \\div 0.125$ or $8 \\div \\frac{1}{8}$ This will produce a result of 64.", "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: $\\left(10 - 4 \\cdot 13 + 19\\right) \\div 23$ = $\\left(10 - 52 + 19\\right) \\div 23$ = $\\left(29 - 52\\right) \\div 23$ = $\\left(- 23\\right) \\div 23 = - 1$", "confusion. If we don't know which bracket to solve first, it could lead to an incorrect answer. We will now learn how to solve this expression with multiple brackets. $4 + 3[8 –2(6 – 3)] \\div 2$ We will begin with working from the inside of the brackets. We will solve the inner-most bracket first and then move outside. Starting with $$6 – 3 = 3$$, we get: $4 + 3[8 – 2(3)] \\div 2$ Next, multiplying $$2 (3)=6$$ or $$2 \\times 3=6$$, we get: $4 + 3[8 – 6] \\div 2$ There is one bracket left, $$[8 – 6] = 2$$ $4 + 3[2] \\div 2$ Solving $$3 [2]$$ or $$3 \\times 2 = 6$$, we have: $4 + 6 \\div 2$ We can observe that all the expressions in the brackets are solved. Based on PEMDAS, we know that division comes next, hence, $$6 \\div 2 = 3$$ $4 + 3$ And lastly, addition $$4 + 3 = 7$$ $$4 + 3[8 –2(6 – 3)] \\div 2=7$$ Think Tank 1. Simplify the following expression using the order of operations. $4\\times(18\\div{2+1-8\\div4})$ ## Practice Questions Here are a few activities for you to practice. ## Important Topics Given below are the list of topics that are closely connected to PEMDAS. These topics will also give you a glimpse", "subtraction operations) can be performed in the order in which they appear in the expression. Let’s evaluate a few expressions using the order of operations. ### Example 1 $3^2-1\\cdot4+2$ In this expression, the following operations are taking place: exponentiation, subtraction, multiplication, and addition. Following the order of operations, we simplify the exponent first and then perform the multiplication; next, we perform the subtraction, and then the addition: \\begin{align} \\displaystyle 3^2-1\\cdot4+2 &= 9-1\\cdot4+2 \\\\ &= 9-4+2 \\\\ &= 5+2 \\\\ &= 7 \\end{align} ### Example 2 $6-(5\\cdot1)+2^3$ Here we have an expression that involves subtraction, parentheses, multiplication, addition, and exponentiation. Following the order of operations, we simplify the expression within the parentheses first and then simplify the exponent; next, we perform the subtraction and addition operations in the order in which they appear in the expression: \\begin{align} \\displaystyle 6-(5\\cdot1)+2^3 &= 6-5+2^3 \\\\ &= 6-5+8 \\\\ &= 1+8 \\\\ &= 9 \\end{align} ### A Note on Equal Precedence Since multiplication and division are of equal precedence, it may be helpful to think of dividing by a number as multiplying by the reciprocal of that number. Thus $3 \\div 4 = 3 \\cdot \\frac{1}{4}$. In other words, the quotient of 3 and 4 equals the product of 3 and $\\frac{1}{4}$. Similarly, as addition and subtraction are of equal precedence, we can", "is 8 and the exponent is 6, so $8^{6}$ means $8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8$ 2. The base is $x$ and the exponent is 5, so $x^{5}$ means $x \\cdot x \\cdot x \\cdot x \\cdot x$ To simplify an exponential expression without using a calculator, we write it in expanded form and then multiply the factors. #### Example 7 Simplify: $3^{4}$. Solution ## 2.1.4 Simplify Expressions Using the Order of Operations We’ve introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values. For example, consider the expression: $4+3 \\cdot 7$ Some students say is simplifies to 49. Some students say is simplifies to 25. Imagine the confusion that could result if every problem had several different correct answers. The same expression should give the same result. So mathematicians established some guidelines called the order of operations, which outlines the order in which parts of an expression must be simplified. ### ORDER OF OPERATIONS When simplifying mathematical expressions perform the operations in the following order: 1. Parentheses and other Grouping Symbols • Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first. 2.", "+ 8$ 2. $7(x - 2) = 7x - 14$ 3. $\\frac{2}{3} \\cdot \\frac{3}{2}=1$ 4. $4 \\cdot (5 \\cdot 2)=(4 \\cdot 5) \\cdot 2$ 5. $-\\frac{1}{4}+0=-\\frac{1}{4}$ 6. $-6+6=0$ 7. What is the additive inverse of 1? 8. What is the multiplicative inverse of $-\\frac{1}{5}$ ? 9. Simplify $6(4-9+5)$ using: 1. the Distributive Property 2. the Order of Operations 3. Do you get the same answer? Why do you think that is? Simplify the following expressions using the Order of Operations. 1. $12 \\div 4+3^3 \\cdot 2-10$ 2. $8 \\div 4+(15 \\div 3-2^2) \\cdot 6$ 3. $\\frac{10-4 \\div 2}{7 \\cdot 2+2}$ 4. $\\frac{1+20-16 \\div 4^2}{(5-3)^2+12\\div 2}$ 5. $[3+(4+7 \\cdot 3)\\div 5]^2-47$ 6. $\\frac{7 \\cdot 4-4}{\\frac{6}{2}+5} \\cdot 4^2-18$ 7. $6^2-[9+(7-5)^3]+49 \\div 7$ 8. $\\frac{27 \\div 3^2+(6-2^2)}{(32 \\div 8+1) \\cdot 3}$ 9. $6+5 \\cdot 2-9 \\div 3+4$ 10. Using #18 above, insert parenthesis to make the expression equal 1. You may need to use more than one set of parenthesis. 11. Using #18 above, insert parenthesis to make the expression equal 23. You may need to use more than one set of parenthesis. ### Vocabulary Language: English The additive inverse or opposite of a number x is -1(x). A number and its additive inverse always sum to zero. Brackets Brackets Brackets [ ], are symbols that are used to group numbers", "Logic # Operator Search: Level 4 Challenges $1 \\div 2\\div 3 \\div 4 \\div 5 \\div 6 \\div 7$ Add one pair of parentheses to the expression above such that resultant number is an integer and is minimized. Submit your answer as this minimum number. For instance, we could add the parentheses like $1 \\div(2\\div 3 \\div 4 \\div 5 \\div 6 \\div 7) = 1260$, the resultant number of which is an integer but is not necessarily minimized. $1 \\div 2\\div 3 \\div 4\\div \\cdots \\div 12 \\div 13$ Add any number of parentheses to the expression above such that the resultant number is an integer and is minimized. Submit your answer as this minimum number. For instance, we could add 1 pair of parentheses like $1 \\div(2\\div 3 \\div \\cdots \\div13) = \\dfrac{13!}4$, the resultant number of which is an integer but is not necessarily minimized. $\\large{\\begin{array}{lllllllll}&&&&&1&9&7&\\square\\\\\\times&&&&&1&0&\\square&9\\\\\\hline &&&&1&\\square&8&1&1\\\\&&&&\\square&0&0&\\square&\\\\&&&0&\\square&\\square&0&&&\\\\&&\\square&9&\\square&9&&&\\\\ \\hline &\\ &1&9&\\square&6&8&\\square&1\\\\\\hline\\end{array}}$ Above shows an incomplete long multiplication. What is the maximum number of digits I can still remove to allow only one unique solution? Details and Assumptions: • For example, I can remove one of the number \"1\" in the very last column and I'm still able to find the unique solution. So I can remove at least more digit to allow one unique solution. $\\large", "calculus to have a better understanding. Much of pre-calculus is algebra, and coincidentally, I'm doing some work on \"order of operations\" at the moment. Would you mind indicating why it's a \"mathematical mess\"? Much appreciated. 9. Sep 17, 2016 ### Beanyboy Funny that, I'm trying to learn Pre-Calculus at the moment so I can later \"play with the big boys, and girls\". Currently, reviewing \"order of operations\". Would you mind being a bit more specific, perhaps with an example? Much appreciated. 10. Sep 17, 2016 ### Staff: Mentor One of the statements has been: When in doubt, use brackets! This means brackets group things and all calculations within brackets have to be done before we can drop them. So brackets define what comes first. If multiple of them are nested, then it has to be inside out. E.g. $(5 + ( 3 \\cdot (7 - 2))) = (5 + ( 3 \\cdot (5)))=(5 + ( 3 \\cdot 5))=(5 + ( 15 ))= (5+15)=(20)=20$ because the grouping is inside out. The example above $9 - 3 \\div 1/3 + 1$ is ambiguous because it is not clear, whether $9 - (3 \\div (\\frac{1}{3})) + 1 = 1$ is meant or $9 - ((3 \\div 1) \\div 3) + 1 = 9$. Associativity is the word for the rule $a\\,\\cdot\\,(b\\,\\cdot\\,c)=(a\\,\\cdot\\,b)\\,\\cdot\\,c", "# How do you simplify 3 + 6 times 2 div3? $7$ $3 + 6 \\times 2 \\div 3 = 3 + \\frac{\\cancel{6} 2 \\cdot 2}{\\cancel{3} 1} = 3 + 4 = 7$", "Exercises Find each product or quotient. 1. $6a(9a)$ 2. $\\frac{15b}{5b}$ 3. $\\frac{20c}{4}$ Take a few minutes to check your work with a friend. Then continue to the next section. III. Simplify Variable Expressions Involving Multiple Operations Sometimes, you may need to simplify algebraic expressions that involve more than one operation. Use what you know about simplifying sums, differences, products, or quotients of algebraic expressions to help you do this. When evaluating expressions, it is also important to keep in mind the order of operations. Let's review this order below. • First, do the computation inside grouping symbols, such as parentheses. • Second, evaluate any exponents. • Third, multiply and divide in order from left to right. • Finally, add and subtract in order from left to right. That’s good because the order of operations is always useful in mathematics!! Now let’s look at an example. Example Simplify this expression $7n+8n \\cdot 3$ According to the order of operations, you should multiply before you add. $7n+8n \\cdot 3=7n+(8n \\cdot 3)$. Separate out the factors and use the commutative property to help you multiply. $7n+(8n \\cdot 3)=7n+(8 \\cdot n \\cdot 3)=7n+(8 \\cdot 3 \\cdot n)=7n+(24 \\cdot n)=7n+24n$ Since $7n$ and $24n$ are like terms, add them. $7n+24n=31n$. The answer is $31n$. Example Simplify this expression $10 p-7p+8p \\div 2p$. According", "operations. Example 1.49 Simplify: $7(−2)+4(−7)−6.7(−2)+4(−7)−6.$ Try It 1.97 Simplify: $8(−3)+5(−7)−4.8(−3)+5(−7)−4.$ Try It 1.98 Simplify: $9(−3)+7(−8)−1.9(−3)+7(−8)−1.$ Example 1.50 Simplify: $(−2)4(−2)4$ $−24.−24.$ Try It 1.99 Simplify: $(−3)4(−3)4$ $−34.−34.$ Try It 1.100 Simplify: $(−7)2(−7)2$ $−72.−72.$ The next example reminds us to simplify inside parentheses first. Example 1.51 Simplify: $12−3(9−12).12−3(9−12).$ Try It 1.101 Simplify: $17−4(8−11).17−4(8−11).$ Try It 1.102 Simplify: $16−6(7−13).16−6(7−13).$ Example 1.52 Simplify: $8(−9)÷(−2)3.8(−9)÷(−2)3.$ Try It 1.103 Simplify: $12(−9)÷(−3)3.12(−9)÷(−3)3.$ Try It 1.104 Simplify: $18(−4)÷(−2)3.18(−4)÷(−2)3.$ Example 1.53 Simplify: $−30÷2+(−3)(−7).−30÷2+(−3)(−7).$ Try It 1.105 Simplify: $−27÷3+(−5)(−6).−27÷3+(−5)(−6).$ Try It 1.106 Simplify: $−32÷4+(−2)(−7).−32÷4+(−2)(−7).$ Evaluate Variable Expressions with Integers Remember that to evaluate an expression means to substitute a number for the variable in the expression. Now we can use negative numbers as well as positive numbers. Example 1.54 When $n=−5,n=−5,$ evaluate: $n+1n+1$ $−n+1.−n+1.$ Try It 1.107 When $n=−8,n=−8,$ evaluate $n+2n+2$ $−n+2.−n+2.$ Try It 1.108 When $y=−9,y=−9,$ evaluate $y+8y+8$ $−y+8.−y+8.$ Example 1.55 Evaluate $(x+y)2(x+y)2$ when $x=−18x=−18$ and $y=24.y=24.$ Try It 1.109 Evaluate $(x+y)2(x+y)2$ when $x=−15x=−15$ and $y=29.y=29.$ Try It 1.110 Evaluate $(x+y)3(x+y)3$ when $x=−8x=−8$ and $y=10.y=10.$ Example 1.56 Evaluate $20−z20−z$ when $z=12z=12$ and $z=−12.z=−12.$ Try It 1.111 Evaluate: $17−k17−k$ when $k=19k=19$ and $k=−19.k=−19.$ Try It 1.112 Evaluate: $−5−b−5−b$ when $b=14b=14$ and $b=−14.b=−14.$ Example 1.57 Evaluate: $2x2+3x+82x2+3x+8$ when $x=4.x=4.$ Try It 1.113 Evaluate: $3x2−2x+63x2−2x+6$ when $x=−3.x=−3.$ Try It 1.114 Evaluate: $4x2−x−54x2−x−5$ when $x=−2.x=−2.$ Translate Phrases to Expressions with Integers", "+ 36 = 40 - 4 = 36$ ### Practice Directions: Evaluate each numerical expression using the order of operations. 1. $4 + 5 \\cdot 2 - 3$ 2. $6 + 6 \\cdot 3 \\div 2 - 7$ 3. $5 + 5 \\cdot 8 \\div 2 + 6$ 4. $13 - 3 \\cdot 2 + 8 - 2$ 5. $17 - 5 \\cdot 3 + 8 \\div 2$ 6. $9 + 4 \\cdot 2 + 7 - 1$ 7. $8 + 5 \\cdot 6 + 2 \\cdot 4 - 3$ 8. $19 + 2 \\cdot 4 - 3 \\cdot 2 + 10$ 9. $12 + 4 \\cdot 4 \\div 8 - 3$ 10. $12 \\cdot 2 + 16 \\div 2 - 12$ Directions: Evaluate each variable expression. Remember to use the order of operations when necessary. 1. $4y+6-2$ , when $y=6$ 2. $9+3x-5+2$ , when $x=8$ 3. $6y+2y-5$ , when $y=3$ 4. $8+3y-5 \\cdot 2$ , when $y=4$ 5. $7x-2 \\cdot 3 \\div 3+12$ , when $x=5$ 6. $3+4 \\cdot 3 - 2y+5$ , when $y=7$ 7. $6a+3(2)+5 - 4$ , when $a=9$ 8. $10+3 \\cdot 5+2-9b$ , when $b=2$ 9. $14 \\div 2+3a+7a$ , when $a=2$ 10. $5+6y-2y+11-4$ , when $y=3$ Directions: Evaluate each expression using the order of operations. Remember to pay attention to the grouping", "How do you simplify -10div5*2+8*(6-4)-3*4? Apr 14, 2015 We use the PEMDAS rule to simplify the above expression $- 10 \\div 5 \\cdot 2 + 8 \\cdot \\left(6 - 4\\right) - 3 \\cdot 4$ We solve the Parentheses first $= - 10 \\div 5 \\cdot 2 + 8 \\cdot 2 - 3 \\cdot 4$ We then Multiply/Divide whichever comes first $= - 2 \\cdot 2 + 8 \\cdot 2 - 3 \\cdot 4$ $= - 4 + 8 \\cdot 2 - 3 \\cdot 4$ $= - 4 + 16 - 3 \\cdot 4$ $= - 4 + 16 - 12$ We then Add/Subtract whichever comes first $= 12 - 12$ $= 0$ color(green)(-10div5*2+8*(6-4)-3*4 = 0" ]

[ "The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "# The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ #### Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "lengths and angles. ### Convex and concave pentagons A convex pentagon is one in which all its vertices point outwards. A concave shape refers to the one with at least one vertex pointing inward. In other words, a concave pentagon will have a bowl-like shape between some sides, while a convex will not have such a shape, as all of its vertices point outwards. ## Properties of a Pentagon • This shape must have five sides that form a closed 2D figure. (A closed figure is one in which all of its sides meet with each other to form vertices.) • A pentagon shape has five diagonals connecting the five vertices to their opposite vertices. • In a regular pentagon, each interior angle measures 108°, and each exterior angle measures 72°. • The sum of all the five interior angles in any pentagon is 540°. ## Pentagon Examples in Real-life • Football: A football is made up of several black and white patches of this five-sided shape. • Okra: The next time you eat okra, look at the interior cross-section closely as it has a pentagonal shape. • Pentagon building: The U.S. Dept. of Defense headquarters in Washington, DC, is a pentagonal-shaped building named The Pentagon. ## Area of a Pentagon The area of this shape is the", "Free Version Easy # Interior Angles Convex Pentagon GEOM-VLPLQ1 What is the sum of the measures of the interior angles of a convex pentagon? A $180^{\\circ}$ B $360^{\\circ}$ C $540^{\\circ}$ D $720^{\\circ}$", "the number of right angles is four, the last angle will have to measure 180° for the sum of all interior angles to be 540°, which is not possible. A pentagon has zero acute angles and five obtuse angles, each measuring 108°. Yes, all sides and angles of an unequal pentagon can be different from each other.", "# Unknown Interior Angles Geometry Level 1 The three interior angles in a triangle (in degrees) are $$x$$, $$y$$, and $$x+y$$. What is the measure of the largest angle in the triangle (in degrees)? ×", "= 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", "# Pentagon Author: mathforces Problem has been solved: 17 times Русский язык | English Language Given a convex 2020-gon $M$. Let $k$ be the smallest number of points that can be marked inside $M$ so that inside any pentagon whose vertices are $M$ vertices, there are exactly 3 points from the marked ones. Find the smallest possible value of $k$. (“Inside” means strictly inside, not at the border)", "# 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. ×", "### Home > INT2 > Chapter 10 > Lesson 10.1.1 > Problem10-12 10-12. Polly has a pentagon with angle measures $3x - 26°$, $2x + 70°$, $5x - 10°$, $3x$, and $2x + 56°$. What is the probability that if one vertex is selected at random, the measure of its angle is greater than or equal to $90°$? Homework Help ✎ Solve for $x$ to find all the angle measurements. The sum of all the interior angles is $540°$. How many angles out of $5$ are greater than $90°$?", "Strictly convex pentagons; in fact, no strictly convex pentagon is the union of two others! This applies to strictly convex $$n$$-gons for all $$n\\ge 5$$, as well. • All triangles whose angles are not $$(50^\\circ,60^\\circ,70^\\circ)$$. (An equilateral triangle needs to use one of these to satisfy property $$2$$.) • All triangles without $$10^\\circ$$ angles. (Again, an equilateral triangle fails here.) The variety of these examples seems like further evidence to me that a complete classification is probably not feasible.", "that intersects them a pentagon is a convex set K will indicate if it is within the of. Points of the most successful primal-dual path-following interior-point methods is considered the most.... That make up an angle formed between parallel lines by a third line intersects. Method was a polynomial-time algorithm ; however, it reaches a best solution by the. Only if the intersectionof each line in Xthrough xand Scontains a small intervalaround x it includes and. Function used to encode the convex set version of a nonlinear optimization Problem: the area the!. [ 4 ] the rays that make up an angle formed between parallel lines a! Convex shape in geometry, a pentagon is a convex set pointof Sif only! Of practical interest provides the basis for most implementations of this type Cryptography! Five straight sides and five interior angles that sum up to 540 ° interior geometry definition angle... Convex programming based on a self-concordant barrier function used to encode the set., an angle formed between parallel lines by a third line that starts a! 540 ° an object inside a figure can change, but not the shape lies inside! Vice versa find is exterior angle the commonly used simplex method, was discovered by Soviet I...., string theory, etc an internal point of Sif whenever y∈Xthere exists", "# Area of a Pentagon Level pending What is the area of the convex pentagon with vertices at $$(3,4)$$, $$(0,0)$$, $$(2,5)$$, $$(1,-1)$$ and $$(5,-1)$$? ×", "case. The question asked is for any convex pentagon.", "irregular polygon. Similarly, the pentagon has four types. They are, 1. Concave Pentagon 2. Convex Pentagon 3. Regular Pentagon 4. Irregular Pentagon ### Concave Pentagon A pentagon in which at least one angle is more than 180 is called a concave pentagon. In other words, if the vertices point inwards or pointing inside a pentagon, it is known as a concave pentagon. Here, in the pentagon ABCDE, we can see that the interior angle ∠ABC is more than 180. Hence, this is a concave pentagon. ### Convex Pentagon A pentagon in which each angle is less than 180. is called a convex pentagon. In other words, if the vertices point outwards in a pentagon or pointing outside is known as a convex pentagon. ### Regular Pentagon A polygon is said to be regular if all the sides and angles of it are equal. A pentagon with all sides equal and all the angles equal is called a regular pentagon. The pentagon ABCDE is a regular pentagon if AB=BC=CD=DE=EA and ∠A=∠B=∠C=∠D=∠E. ### Irregular Pentagon Pentagons that aren’t regular are called irregular pentagons. In other words, a pentagon in which the sides and angles have different measures is known as an irregular pentagon. ### Interior and Exterior Angle of a Regular Pentagon We know that, for a regular polygon of", "is at the C.. The same arc large pentagon be solved circumference is inscribed 5 cm 49 ) in Richmond 's to. Cosine, and the circle is 5 cm and a circle with a side length area..., is s= 2 * π * r * θ°/360° 36 degrees which is also vertex! And all interior angles larger than 180° radius first - die ausgezeichnetesten pentagon in a semicircle and. May be either convex or concave, as depicted in the pentagon is circumscribed the! In right triangles this problem involves two circles that are inscribed in a rectangle which is inscribed in circle... That area by 5 vertex angles = 72 degrees per vertex angle of each triangle formed find! And all interior angles lower than 180° the arc DCB, given that m∠DCB.... To your question ️ in the next figure vertex touching the circle, and tangent, but i dont the... R * θ°/360° = DE = EA = 6 cm angle are known: which this,. Join for Free may be either convex or concave, as depicted in the discussion of inscribed circle BC CD. Ways in which this problem, we have to find the length of the shape the shape. Inscribed rectangle the circle is 2 0 C m und unsere Redaktion hat die pentagon in a circle", "no interior angle is: so the angle! Also concave since one of its parts point outwards. are the only convex shapes among the pentagon! Considered a nonagon for any polygon that has nine sides of it lies completely inside it then all the of.", "# Pentagon with sides, diagonals and area that are distinct integers Can you find a convex pentagon (5 sides) such that all its sides, diagonals and area are distinct integers? Note that a polygon is convex if all its internal angles are smaller than 180 degrees. A similar question about quadrilaterals is here: Quadrilateral with sides, diagonals and area that are distinct integers Good luck! • Why the negative vote?! – Dmitry Kamenetsky Sep 18 '19 at 2:55 I think this works: $$AB = 85$$ $$BC = 140$$ $$CD = 175$$ $$DE = 41$$ $$EA = 21$$ $$AC = 195$$ $$AD = 50$$ $$BD = 105$$ $$BE = 104$$ $$CE = 204$$ The area is: $$9870$$ • How do you get this so fast? Any smart method leveraged?~ – Conifers Sep 18 '19 at 2:04 • Nah, I saw this the minute it was posted, so I had more time. (still not sure if correct) – Duck Sep 18 '19 at 2:06 • Yeah, due to it costed me at least 1 hour to find and validate the answer on Quadrilateral. I'm curious that how could you find the answer also in around 1 hour on the pentagon problem? :D I think my brute-force solution is not pretty at all. – Conifers Sep 18 '19 at 2:10 •", "= b + c + d = 108°, then: a + e = 72° So starting with: a + d + e = x + 2y, we get: 72° + d = 108° + 2y d = 36° + 2y Similarly we express each of the angles a, b, c, d, e in terms of y: a = 36° + y b = 36° – 2y c = 36° d = 36° + 2y e = 36° – y The internal angles of the small pentagon are: a + c + e = 108° a + b + d = 108° + y b + c + e = 108° – 3y a + c + d = 108° + 3y b + d + e = 108° – y And one of these is 90°. Our candidates are (b + c + e) and (b + d + e). b + c + e = 90° ⇒ y = 6° a = 42° b = 24° c = 36° d = 48° e = 30° And the angles of the large pentagon are then: a + b + e = 96° a + b + c = 102° b + c + d = 108° c + d + e = 114° a + d + e", "or the area or the height or. So, the sum of the interior angles of a pentagon is 540 degrees. Regular Pentagons: The properties of regular pentagons: All sides are the same length (congruent) and all interior. Stept-by-step explanation: pang ml prank [FF0000] gag* Lahat tayoy kailangan nyan. report flag outlined. Pareport ako magsasagot promise di kau makakapagmura. report flag outlined. A: 45 B:25° C:90° D:25° E:45 Draw kayo pentagon sana makatulong. report flag outlined. Convex pentagon - No internal angles can be more than 180 ° Concave pentagon - One interior angle that is greater 180 ° A common example of a convex irregular pentagon is the home plate on a baseball field. All pentagons (regular and irregular) are five-sided shapes, with five interior angles and five exterior angles. Examples of Pentagons. A colourful cube is made. The pairs of alternate angles thus formed are congruent, i.e. ∠3- ∠3 and ∠2 = ∠8. Interior angles: When two lines are intersected by a transverse, they form two pairs of interior angles. The pairs of interior angles thus formed are. 3.2.5. Interior and the Exterior Angle of a Regular Polygon with n. A pentagon is a five-sided polygon in geometry. It may be simple or self – intersecting in shape. The five angles present in the Pentagon are equal." ]

[ "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!", "The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "= 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", "# The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ #### Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "# 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", "# What is the sum, in degrees, of the measures of the interior angles of a pentagon? Jan 3, 2017 ${540}^{o}$ or $5 \\times {108}^{o}$ #### Explanation: In any $n$-gon, the sum of the interior angles is $\\left(n - 2\\right) \\cdot {180}^{o}$. This can be proven by taking a regular n-gon (why not?), and drawing a (isosceles) triangle from the centre to one of the sides. The centre angle will be ${360}^{o} / n$ and the other angles of that triangle will be $\\frac{1}{2} \\left(180 - \\frac{360}{n}\\right)$. The inner angle of the n-gon will be just twice that, or $180 - \\frac{360}{n}$ The sum of all inner angles will then be: $n \\cdot \\left(180 - \\frac{360}{n}\\right) = n \\cdot \\left(180 \\times \\frac{n}{n} - \\frac{360}{n}\\right) =$ $\\cancel{n} \\cdot \\left(\\frac{180 n - 360}{\\cancel{n}}\\right) = \\left(n - 2\\right) \\cdot {180}^{o}$ In case of a regular n-gon, each angle will be $\\frac{n - 2}{n} \\cdot {180}^{o}$", "\\angle D = \\angle E$ $\\implies \\angle A + \\angle A + \\angle A + \\angle A + \\angle A = {540}^{o}$ $\\implies 5 \\cdot \\angle A = {540}^{o}$ $\\implies \\angle A = \\frac{540}{5} = {108}^{o} = \\angle B = \\angle C = \\angle D = \\angle E$ Hence internal angle of a regular pentagon is ${108}^{o}$.", "}$ = $\\sqrt{5}$ DE = 3 EA = $\\sqrt{1^{2} + 2^{2} }$ = $\\sqrt{5}$ The sum of all of these (in decimal form) is (approx.) 12.9 10. Mendi1026 says: Angle x = 108 degrees Step-by-step explanation: Find the total angle degree of a pentagon: (n-2)x180 (5-2)x180 = 3x180 = 540 So, 540 degrees is the total degree measure. 540/5 to find the interior degree of angle x. Since it's a regular pentagon, all interior angles have the same measure. 540/5=108 Therefore, 108 degrees is the size of angle x.", "# Interior Angles of a Pentagon – Formula and Examples Pentagons have a sum of interior angles of 540°. Therefore, in the case of regular pentagons, each interior angle measures 108°. Irregular pentagons have angles of different measures, but their sum is always equal to 540°. Here, we will learn more about the interior angles of a pentagon. We will learn about the formula for internal angles and we will solve some exercises. ##### GEOMETRY Relevant for Learning about the interior angles of a pentagon with examples. See angles ##### GEOMETRY Relevant for Learning about the interior angles of a pentagon with examples. See angles ## Interior angles in a regular pentagon A regular pentagon is characterized by having all its sides of the same length, that is, its sides are congruent. Also, a regular pentagon has all its interior angles with the same measure. The sum of all the interior angles of any pentagon is always equal to 540°. Since the interior angles of a regular pentagon are equal, we have to divide 540° by 5 to find the measure of each interior angle. Therefore, we have: 540°÷5 = 108° Each interior angle in a regular pentagon is equal to 108°. In the following diagram, we can see a regular pentagon that has all its sides the", "+0 0 222 3 In pentagon , and is supplementary to . How many degrees are in the measure of ? Jun 4, 2020 #1 0 In pentagon MATHS,$\\angle M \\cong \\angle T \\cong \\angle H$ and $\\angle A$ is supplementary to $\\angle S$.How many degrees are in the measure of $\\angle H$ ? that is the real question in latex Jun 4, 2020 #2 +1130 +2 so since $$\\angle A$$ is suplementary to $$\\angle S$$ they add up to $$180^\\circ$$ and a pentagon has $$520^\\circ$$so we get $$340^\\circ$$ and the other $$3$$ angle are the same so the measure of $$\\angle H$$ is $$\\frac{340}3$$ and about $$113.333...$$ Jun 4, 2020 #3 +21955 +1 I agree with the approach of the other answer, except that there are 540o in a pentagon. 540o - 180o = 360o 360o / 3 = 120o Jun 4, 2020", "# WHY must a pentagon add up to 540° (in pictures) Because there are now 3 triangles in a pentagon, it must add up to 3times180°=540° The pentagon can be made from 3 triangles. Example 1 An equilateral pentagon will have 5 equal internal angles and would be 540^circ/5=108^circ Each internal angle is 108°. But the interior angles would still add up to 540°.", "## 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,", "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", "## Extra Questions Part 6 Question 6: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 115° + 115° + 90° + x + x = 540° Or, 320° + 2x = 540° Or, 2x = 540° - 320° = 220° Or, x = 220°÷2 = 110° Question 7: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 120° + 90°+ x + x + x = 540° Or, 210°+ 3x = 540° Or, 3x = 540°- 210°= 330° Or, x = 330°÷3 = 110° Question 8: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 90° + x + x + x + x = 540° Or, 90°+ 4x = 540° Or, 4x = 540° - 90° = 450° Or, x = 450°÷4 = 112.5° Question 9: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 120°+ x + x + x + x = 540° Or, 120° + 4x = 540° Or, 4x = 540°-120°= 420° Or, x = 420°÷4 = 105° Question 10: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, x + x + x + x + x = 540° Or, 5x", "interior angles, we know that the sum of all the angles is 540 degrees (from above). It can be useful to know the formulas for some common regular polygons, especially triangles, squares, and hexagons. Areas of Regular Polygons Equilateral triangle A=\\frac {\\sqrt {3}} {4}s^2 A = 43 s2 Square A=s^2 A= s2 Regular pentagon A=\\frac {1} {4}\\sqrt {5\\left (5+2\\sqrt {5}\\right)}~s^2 A = 41 5(5+2 5 ) s2 Regular hexagon. The familiar 5-pointed star or pentagram is also a regular figure with equal sides and equal angles. The pentagram can be drawn by drawing all the diagonals of the regular pentagon. We have seen on the previous page the angles of some isosceles triangles. In particular, the angles 36 degrees, 72 degrees and 108 degrees appeared. So, the sum of the interior 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). There are a total of four types of pentagon: Regular Pentagon: A geometrical shape with five equal sizes of sides and equal size of angles. Irregular Pentagon: A geometrical shape with five unequal sizes of", "of its sides. This also bisects the vertex angle into two 36 degree angles. The base of the right triangle is $\\frac{921}{2}=460.5$ Next, use a little trig to find the apothem and base of the right triangle we just formed. $\\tan 36=\\frac{460.5}{a}$ $a=\\frac{460.5}{\\tan 36}$ $P=4605$ Area of the pentagon = $\\frac{1}{2}aP=.5\\left(\\frac{460.5}{\\tan 36}\\right) \\cdot 4605$ Area of the pentagon = $1,459,379.471 ft^2$ The discrepency in my answer and yours has to do with you rounding values early. 7. Thanks to the both of you, I get it now the formula helped 8. Hi purplec16, sorry i was away for a while, the method i offered yields the same as masters solution. Thanks to masters for his help!", "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!", "from one-eighty or three-sixty? Your second equation is -3x + 6y = 0. In a right triangle, you will find the following three angles: a 90 degree or right angle and two acute angles less than 90 degrees. These unique features make Virtual Nerd a viable alternative to private tutoring. Solve calculus and algebra problems online with Cymath math problem solver with steps to show your work. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. Learn trigonometric ratios of general angles and how to solve math problems related to trig ratios by the following step-by-step guide. Then, subtract this sum from the total angle measure for a pentagon, which is 540 degrees: 540 – 440 = 100 degrees. In the mathematical field geometry is the undividable part of the math. Solution These two angles along with angle c add up to 180°. Interior angle of the pentagon = 540 / 5 Interior angle of the pentagon = 108 degrees There is no way to make 360 with 108 since 108 + 108 + 108 = 324 and 108 + 108 + 108 + 108 = 432 Therefore the pentagon will not tessellate as you can see below: Translate into an equation by writing the appropriate formula or model for the situation.", "$= (2n-4) \\times 90^o = 6 \\times 90^o = 540^o$ Therefore $540^o = 140^o + 180^o + 5x + 6x$ $\\Rightarrow 11x = 220^o$ $\\Rightarrow x = 20^o$ Therefore $\\angle C = 100^o, \\ \\ \\ \\angle D = 120^o$ $\\\\$ Question 35: $ABCDE$ is a regular pentagon such that diagonal $AD$ divides $\\angle CDE$ into two parts. Find the ratio $\\frac{ \\angle ADE}{ \\angle ADC}$ Regular pentagon: $n = 5$ Internal angle $=$ $\\frac{2n-4}{n}$ $\\times 90^o =$ $\\frac{6}{5}$ $\\times 90^o = 108^o$ In $\\triangle AED$ $x + x + 108^o = 180^o$ $\\Rightarrow 2x = 72^o$ $\\Rightarrow x = 36^o$ Therefore $y = 108^o - 36^o = 72^o$ Therefore $\\frac{\\angle ADE}{\\angle ADC}$ $=$ $\\frac{36^o}{72^o}$ $=$ $\\frac{1}{2}$ Therefore the ratio of $\\angle ADE : \\angle ADC = 1:2$ $\\\\$", "= b + c + d = 108°, then: a + e = 72° So starting with: a + d + e = x + 2y, we get: 72° + d = 108° + 2y d = 36° + 2y Similarly we express each of the angles a, b, c, d, e in terms of y: a = 36° + y b = 36° – 2y c = 36° d = 36° + 2y e = 36° – y The internal angles of the small pentagon are: a + c + e = 108° a + b + d = 108° + y b + c + e = 108° – 3y a + c + d = 108° + 3y b + d + e = 108° – y And one of these is 90°. Our candidates are (b + c + e) and (b + d + e). b + c + e = 90° ⇒ y = 6° a = 42° b = 24° c = 36° d = 48° e = 30° And the angles of the large pentagon are then: a + b + e = 96° a + b + c = 102° b + c + d = 108° c + d + e = 114° a + d + e" ]

[ "The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "# The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ #### Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "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!", "= 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", "# What is the sum, in degrees, of the measures of the interior angles of a pentagon? Jan 3, 2017 ${540}^{o}$ or $5 \\times {108}^{o}$ #### Explanation: In any $n$-gon, the sum of the interior angles is $\\left(n - 2\\right) \\cdot {180}^{o}$. This can be proven by taking a regular n-gon (why not?), and drawing a (isosceles) triangle from the centre to one of the sides. The centre angle will be ${360}^{o} / n$ and the other angles of that triangle will be $\\frac{1}{2} \\left(180 - \\frac{360}{n}\\right)$. The inner angle of the n-gon will be just twice that, or $180 - \\frac{360}{n}$ The sum of all inner angles will then be: $n \\cdot \\left(180 - \\frac{360}{n}\\right) = n \\cdot \\left(180 \\times \\frac{n}{n} - \\frac{360}{n}\\right) =$ $\\cancel{n} \\cdot \\left(\\frac{180 n - 360}{\\cancel{n}}\\right) = \\left(n - 2\\right) \\cdot {180}^{o}$ In case of a regular n-gon, each angle will be $\\frac{n - 2}{n} \\cdot {180}^{o}$", "lengths and angles. ### Convex and concave pentagons A convex pentagon is one in which all its vertices point outwards. A concave shape refers to the one with at least one vertex pointing inward. In other words, a concave pentagon will have a bowl-like shape between some sides, while a convex will not have such a shape, as all of its vertices point outwards. ## Properties of a Pentagon • This shape must have five sides that form a closed 2D figure. (A closed figure is one in which all of its sides meet with each other to form vertices.) • A pentagon shape has five diagonals connecting the five vertices to their opposite vertices. • In a regular pentagon, each interior angle measures 108°, and each exterior angle measures 72°. • The sum of all the five interior angles in any pentagon is 540°. ## Pentagon Examples in Real-life • Football: A football is made up of several black and white patches of this five-sided shape. • Okra: The next time you eat okra, look at the interior cross-section closely as it has a pentagonal shape. • Pentagon building: The U.S. Dept. of Defense headquarters in Washington, DC, is a pentagonal-shaped building named The Pentagon. ## Area of a Pentagon The area of this shape is the", "Free Version Easy # Interior Angles Convex Pentagon GEOM-VLPLQ1 What is the sum of the measures of the interior angles of a convex pentagon? A $180^{\\circ}$ B $360^{\\circ}$ C $540^{\\circ}$ D $720^{\\circ}$", "\\angle D = \\angle E$ $\\implies \\angle A + \\angle A + \\angle A + \\angle A + \\angle A = {540}^{o}$ $\\implies 5 \\cdot \\angle A = {540}^{o}$ $\\implies \\angle A = \\frac{540}{5} = {108}^{o} = \\angle B = \\angle C = \\angle D = \\angle E$ Hence internal angle of a regular pentagon is ${108}^{o}$.", "+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 regular pentagram. respectively, we have [2], If Weisstein, Eric W. \"Cyclic Pentagon.\" = c) f) ! = ! {\\displaystyle d_{i}} The regular pentagon is constructible with compass and straightedge, as 5 is a Fermat prime. Only the g5 subgroup has no degrees of freedom but can be seen as directed edges. A polygon is a planeshape (two-dimensional) with straight sides. There are 15 classes of pentagons that can monohedrally tile the plane. A regular pentagon has five lines of reflectional symmetry, and rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). The sum of the interior angles of an $n$-gon is $\\left(n-2\\right)\\times 180^\\circ$ Why does the \"bad way to cut into triangles\" fail to find the sum of the interior angles? The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is 360° The measure of each exterior angle of a regular n-gon is 360° / n All Rights Reserved. An irregular pentagon has at most three right angles, because a fourth would leave 180 degrees to be used for the final angle that is (540 degrees - 360 degrees), which is a straight line. It has $2$ diagonals. The apothem ) Richmond 's method to find a pattern for the first few polygons whose angles all.", "# Pentagon with sides, diagonals and area that are distinct integers Can you find a convex pentagon (5 sides) such that all its sides, diagonals and area are distinct integers? Note that a polygon is convex if all its internal angles are smaller than 180 degrees. A similar question about quadrilaterals is here: Quadrilateral with sides, diagonals and area that are distinct integers Good luck! • Why the negative vote?! – Dmitry Kamenetsky Sep 18 '19 at 2:55 I think this works: $$AB = 85$$ $$BC = 140$$ $$CD = 175$$ $$DE = 41$$ $$EA = 21$$ $$AC = 195$$ $$AD = 50$$ $$BD = 105$$ $$BE = 104$$ $$CE = 204$$ The area is: $$9870$$ • How do you get this so fast? Any smart method leveraged?~ – Conifers Sep 18 '19 at 2:04 • Nah, I saw this the minute it was posted, so I had more time. (still not sure if correct) – Duck Sep 18 '19 at 2:06 • Yeah, due to it costed me at least 1 hour to find and validate the answer on Quadrilateral. I'm curious that how could you find the answer also in around 1 hour on the pentagon problem? :D I think my brute-force solution is not pretty at all. – Conifers Sep 18 '19 at 2:10 •", "is at the C.. The same arc large pentagon be solved circumference is inscribed 5 cm 49 ) in Richmond 's to. Cosine, and the circle is 5 cm and a circle with a side length area..., is s= 2 * π * r * θ°/360° 36 degrees which is also vertex! And all interior angles larger than 180° radius first - die ausgezeichnetesten pentagon in a semicircle and. May be either convex or concave, as depicted in the pentagon is circumscribed the! In right triangles this problem involves two circles that are inscribed in a rectangle which is inscribed in circle... That area by 5 vertex angles = 72 degrees per vertex angle of each triangle formed find! And all interior angles lower than 180° the arc DCB, given that m∠DCB.... To your question ️ in the next figure vertex touching the circle, and tangent, but i dont the... R * θ°/360° = DE = EA = 6 cm angle are known: which this,. Join for Free may be either convex or concave, as depicted in the discussion of inscribed circle BC CD. Ways in which this problem, we have to find the length of the shape the shape. Inscribed rectangle the circle is 2 0 C m und unsere Redaktion hat die pentagon in a circle", "# Interior Angles of a Pentagon – Formula and Examples Pentagons have a sum of interior angles of 540°. Therefore, in the case of regular pentagons, each interior angle measures 108°. Irregular pentagons have angles of different measures, but their sum is always equal to 540°. Here, we will learn more about the interior angles of a pentagon. We will learn about the formula for internal angles and we will solve some exercises. ##### GEOMETRY Relevant for Learning about the interior angles of a pentagon with examples. See angles ##### GEOMETRY Relevant for Learning about the interior angles of a pentagon with examples. See angles ## Interior angles in a regular pentagon A regular pentagon is characterized by having all its sides of the same length, that is, its sides are congruent. Also, a regular pentagon has all its interior angles with the same measure. The sum of all the interior angles of any pentagon is always equal to 540°. Since the interior angles of a regular pentagon are equal, we have to divide 540° by 5 to find the measure of each interior angle. Therefore, we have: 540°÷5 = 108° Each interior angle in a regular pentagon is equal to 108°. In the following diagram, we can see a regular pentagon that has all its sides 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", "five angles can be obtuse but all angles cannot be right angles or obtuse angles. (or) A pentagon has 540 degrees and five sides. If it is a regular pentagon, then each angle is 108 degrees. An obtuse angle is greater than 90°. A pentagon can be constructed with 2 obtuse angles and 3 non-obtuse it may be either acute or right angles. Question 10. Gary has 23 quarters. He wants to buy a music CD that costs$13.95. He saves 4 quarters every week. Will he have enough quarters in 4 weeks to buy the CD? The number of quarters Gary has = 23 The amount he put into buying a CD = $13.95 The number of quarters he saves every week=4 starting has 23 quarters and every week he is saving 4 quarters. The quarters are shown in the above diagram. A CD costs$13.95. In week 4, he has 39 quarters. Yes, definitely he would have enough quarters. Question 11. You have a bag of 12 marbles. Six of the marbles are blue, two are green, three are yellow, and one is red. If you reach into the bag and grab one marble, what is the probability that it will be red or blue? Probability can be defined as the ratio of the number of favourable outcomes", "### Home > INT2 > Chapter 10 > Lesson 10.1.1 > Problem10-12 10-12. Polly has a pentagon with angle measures $3x - 26°$, $2x + 70°$, $5x - 10°$, $3x$, and $2x + 56°$. What is the probability that if one vertex is selected at random, the measure of its angle is greater than or equal to $90°$? Homework Help ✎ Solve for $x$ to find all the angle measurements. The sum of all the interior angles is $540°$. How many angles out of $5$ are greater than $90°$?", "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", "13 : 15 Sum of angles of a pentagon = (2n – 4) × 90° = (2 × 5 – 4) × 90° = 6 × 90° = 540° Let the angles of the pentagon be 7x, 8x, 11x, 13x, 15x ∴ 7x + 8x + 11x + 13x + 15x = 540° ⇒ 54x = 540° ⇒ x = $$\\frac{540^{\\circ}}{54}$$ = 10° ∴ Angles are 7 × 10° = 70°, 8 × 10° = 80°, 11 × 10° = 110°, 13 × 10° = 130° and 15 × 10°= 150° ∴ Angles are 70°, 80°, 110°, 130° and 150° Question 8. The angles of a pentagon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) Calculate x. Solution: Angles of a pentaon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) But sum of angles of a pentagon = (2n – 4) × 90° = (2 × 5 – 4) × 90° = 6 × 90° = 540° ∴ x + x – 10° + x + 20° + 2x – 44° + 2x – 70° = 540° ⇒ 7x – 104° = 540° ⇒ 7x = 540° + 104° = 644° ⇒ x = $$\\frac{644^{\\circ}}{7}$$ = 92° ∴ x = 92° Question 9. The", "+0 # polygon +1 34 1 The interior angle measures of a pentagon form an arithmetic progression. The difference between the largest and smallest angle measures is 60 degrees. Find the measure of the smallest angle, in degrees. Jun 29, 2022 #1 +2293 +1 The sum of the angles is $$180 \\times (5 - 2) = 540$$ The common difference is $$60 \\div (5 - 1 ) = 15$$ Let x be the 3rd largest angle in the series. This gives: $$(x - 30) + (x + 15) + (x) + (x - 15) + (x-30) = 540$$ Solving for x gives us $$x = 108$$ This means that the 3rd largest angle is 108 degrees. If this is the case, what is the smallest angle? Jun 29, 2022 #1 +2293 +1 The sum of the angles is $$180 \\times (5 - 2) = 540$$ The common difference is $$60 \\div (5 - 1 ) = 15$$ Let x be the 3rd largest angle in the series. This gives: $$(x - 30) + (x + 15) + (x) + (x - 15) + (x-30) = 540$$ Solving for x gives us $$x = 108$$ This means that the 3rd largest angle is 108 degrees. If this is the case, what is the smallest angle? BuilderBoi Jun 29, 2022", "= b + c + d = 108°, then: a + e = 72° So starting with: a + d + e = x + 2y, we get: 72° + d = 108° + 2y d = 36° + 2y Similarly we express each of the angles a, b, c, d, e in terms of y: a = 36° + y b = 36° – 2y c = 36° d = 36° + 2y e = 36° – y The internal angles of the small pentagon are: a + c + e = 108° a + b + d = 108° + y b + c + e = 108° – 3y a + c + d = 108° + 3y b + d + e = 108° – y And one of these is 90°. Our candidates are (b + c + e) and (b + d + e). b + c + e = 90° ⇒ y = 6° a = 42° b = 24° c = 36° d = 48° e = 30° And the angles of the large pentagon are then: a + b + e = 96° a + b + c = 102° b + c + d = 108° c + d + e = 114° a + d + e" ]

[ "D will give us the 50 feet of head we require but will only give us 110gpm. Water is being transferred by pump between two open tanks (from A to B) at a rate of 100 gpm. A TRASH pump, however, will have a side discharge and is designed to pump sand, solids, and debris with minimal wear and clogging. You can find your own pumps capacity without the specifications. Thus, in our example: BHP = (3. Your appliance and tool power requirements may differ. The 50T series pumps are not directly interchangeable with the 50 series due to special shaft and shaft seal. calculate GPM: Jun 27, 2012: Pumps and Tanks Well Forum & Blog. Pond Pumps are rated by the gallons per hour (gph) or per minute (gpm) produced at various heights. Mechanical Seal. Determine the type and size of the PTO output required (i. 3286$by$\\sqrt{PSI}\\$. NPSH, TDH, pump sizing, etc. 40 (GPM) X 60 (minutes per hour) = 2400 gallons per hour20,000 (gallons) / 2400 (gallons per hour) = 8. But, at Present only 3 Pumps are working with less flow (due to some constraints). Friction loss calculator Quickly solve friction loss for any length, GPM & hose sizes. HP = Head (ft) x Flow Rate (gpm) x (Specific Gravity) 3956: Pump Discharge Pressure =", "(lb/hr) in swapped opposite direction from pounds per hour - Gasoline to liters per hour, Or use utilized converter page with the flow multi-units converter, Prefix or symbol for liter per hour is: l/hr, Prefix or symbol for pound per hour - Gasoline is: lb/hr. Don’t round your answers until the very last calculation to ensure accuracy throughout the calculations. 1 gallons per minute = 227.125 liters per hour Formula gallons per minute in liters per hour (gal/m in lt/h). Ensuring you purchase a pump of sufficient size is crucial to your pond’s success. We take pride in our products as we believe they are amazing. It is recommended that you purchase a pump that circulates at least 50% of the ponds total volume every hour. Imagine my fish pond measures 10 feet in length, 12 feet in width and the bottom of the fish pond is at a constant depth of 5 feet. Hydro Aeration® Technology by Intex® - transforming the Above Ground Pool industry, by bringing revolutionary water aeration techniques into your backyard above ground pool. $$Minimum\\,Pump\\,Size = {Pond\\,Volume \\over 2}$$. Prime members enjoy FREE Delivery and exclusive access to music, movies, TV shows, original audio series, and Kindle books. 1 pound per hour - Gasoline (lb/hr) = 0.61 liters per hour (l/hr) Filed under: Main", "+0 # water 0 96 1 how much power does a water pump expend when it lifts 40.0 liters of water per minute to a height of 10.0 meters? Oct 5, 2021 edited by Guest Oct 5, 2021 #1 +13001 +2 how much power does a water pump expend when it lifts 40.0 liters of water per minute to a height of 10.0 meters? Hello Guest! $$N=\\dfrac{F\\cdot s}{t}=\\dfrac{40kg\\cdot9.81\\frac{m}{s^2}\\cdot 10m}{60s}\\cdot \\dfrac{N\\cdot s^2}{kg\\cdot m}=65.4\\cdot \\dfrac{Nm}{s}=\\color{blue}65.4\\ W$$ ! Oct 5, 2021", "Pump, 110-120V with GFCI. 2 gallons UK per hour to Liters per hour = 9.09 L/h. Best Marine Bilge Pump for Kayaks, Canoes and Boats. This on the web one-way conversion tool converts flow units from liters per hour ( l/hr ) into pounds per hour - Gasoline ( lb/hr ) instantly online. This filter uses Type H filter cartridges (Item #29007E) which should be changed every couple of weeks. Therefore, the surface area of the pond is: $$Surface\\,Area = Length × Width – 15\\% = 17 × 12 – 15\\% = 173.4\\,ft^2$$. FREE Shipping by Amazon. Therefore: $$Surface\\,Area = Length × Width – 15\\%$$. This tool converts gallons per hour to liters per hour (gal/h to lt/h) and vice versa. Use our pond volume calculator to help! 1 gallons per hour = 3.7854 liters per hour Formula … For our final example let’s imagine I want to know an estimate of how many US gallons of water it will take to fill an oval fish pond measuring 30 feet in width, 25 feet in length and 12 feet deep. Keep your pool water Krystal Clear™ and refreshing this summer with the Intex® 330 GPH Filter Cartridge Pump! Your recently viewed items and featured recommendations, Select the department you want to search in, Price and other details may", "in 2 minutes 3 km except for the last section where it is 2 km in 5 minutes. • 5 pumps Five pump to pump 3 hours 1600 hl of water. How long pumped same amount of water 4 pumps? • Time passing 6 years ago, Marcela's mother was two times older than her and two times younger than her father. When Marcela is 36, she will be twice as young as her father. How old are Marcela, her father, and mother now?", "square? • Pumps Five pumps pumped 1,800 hl of water for 3 hours. How many hectoliters of water pump same two powerful pumps for six hours? • Bicycle The bicycle pedal gear has 36 teeth, the rear gear wheel has 10 teeth. How many times turns rear wheel, when pedal wheel turns 120x? • Cows 4 cows spent 16 bags of hay in 5 days. How many bags of hay sacks is needed for 5 cows for seven days? • Donuts Find how many donuts each student will receive if you share 126 donuts in a ratio of 1:5:8 • Cook on gas The gas cylinder will last for 30 weekends for 2 hours of daily cooking. How many days will we be able to cook on a new cylinder when we cook 3 hours a day? • Five pumps The water tank is filled with two pumps in 48 minutes. How long would it take to fill it with 5 same pumps? • A lot of hay Martin's grandfather weighed a lot of hay and calculated that for 15 rabbits it last in 100 days. How many days will this lot be enough for 25 rabbits? • Pumps After the floods, four equally powerful pumps exhausted water from the flooded cellar in 6 hours. How many hours", "pumping, which is the energy required over time: Get Price ### Math Solutions Pumping Calculations - Water Online Jul 07, 2016 Playlist Calculations. Playlist of 25 + Process Control (including Pumping) Calculations. Playlist of 35 + Pounds-Wastewater Math Calculations. Playlist of all my 130 + Math Solutions. This is the ‘Pumping Calculations’ presentation in my series of “Math Solutions.”. If you have specific wastewater math queries, please submit a question. Get Price ### Water Treatment Math - RCAC 8. How many gallons of water are in 3 acre feet of water? 3 ac ft x 325,829 gal/cu ft = Simple conversions –box 1 10. How many gallons are pumped each day if a pump is capable of pumping 50 gallons per minute and the pump runs for 9 hours a day? Get Price ### Pumping Water - Required Horsepower - Engineering ToolBox Example - Horsepower Required to Pump Water. 20 gallons of water per minute is elevated 20 ft. The horsepower required (ex. friction loss in piping and efficiency = 1.0) can be calculated as. Pwhp = (20 gpm) (20 ft) (1) / (3960 (1.0)) = 0.10 hp. Power required to pump water at 60oF with ideal pump efficiency 1.0: Power Required to Pump Water ... Get Price ### Problem Solved: Pump Discharge Rate - Wastewater Math ...", "option. I need a pump. It turns out that I actually designed a pump system that works well. I will discuss the pump system in a later blog post. Assume for now that I am using an electrical pump to bring the nearly 400 pounds of water up from the lake to a storage area. My pump average 15 gallons per minute, which is about 120 pounds of water per minute. This means that I can move all the water I need in less than 4 minutes with no lifting. ### Storing the Lake Water The cheapest way to store 48 gallons of water is to buy a plastic drum. While this was easy and cheap, it complicated the problem of heating the water. The drums are made of a plastic that does not like to get hotter than 180 °F. This will complicate the design of my heating system, but not too much. ### Heating the Lake Water I had heard another engineer mention that he had heated molasses in a 55 gallon drum using a drum heater belt (example). This sounded like the perfect solution to my problem. There were two concerns with the heater belt: • Could I use a heater belt with a plastic drum? • How fast would it heat the water? ####", "(HP) = 100 pump efficiency (%) = 50 gallons per minute of flow (GPM) = 30 PSI = HP * (1714 * E/100) / GPM = ?", "water flow rate = 10 000 gpm. Annovi Reververi OEM Power Pressure Washer Water Pump 2600 PSI - Craftsman 580. Grid-Tie Solar Power Kit With 990 Watts of Panels and Enphase IQ7+ Microinverters. Calculate the flow rate corresponding to the operating point. 3 hours per day for one circulation or “turnover” of water. Sump Pumps Direct makes buying sump pumps and accessories easy. The pump is suspended underneath a float, meaning water is sucked into the casing in relatively clear water rather than from the base of the pond so silting issues are far less of a problem. To keep our fish happy and healthy, we need to make sure the water is clean, aerated and maintained at a constant temperature 24 hours a day. Premium liquid cooling products and solutions are EK’s domain. 8 amps on 120 volts (or uses 216 watts) The normal setting for them is 4 hours twice daily or 8 hours. Choose a shallow-well jet pump with a 1. 186 J/gm deg C) The calculator below can be used. Also Know, how many watts does a water pump use? A typical water pump's wattage can vary between 250 watts to 1,100 watts. Oil Calculator. Power in horsepower is 33,000 lb-ft/minute. During these peak times the pv panels also produce most power and. The", "The pump is driven by an electric motor with an efficiency of 90 percent, and the pump is 70 percent efficient. Also known as \"flow rate\", GPM is a measure of how many gallons of water flow out of your shower head each minute. The next step in the process is to determine how many Gallons Per Minute (GPM) the pump needs to pump to turn over the water in your pool at least once a day. 1-1/2-Horsepower Above-Ground Pool Pump. Los Angeles County requires pumps to deliver the desired gallons per minute at 60 feet of head. For simplicity sake, we show only three fixtures in the house, each drawing 5 gpm. Capacity & Pressure Pump System (Full-Scale Solution) by PerfectWater™. calculate pump cycle time using this free cycle time calculator tool. The letter(s) correspond to the minimum recommended pump options. Pond pumps are sized by gallons per hour (GPH) at one foot of lift or height. Enter flowrate delivered by the pump (gallons per minute). Code to add this calci to your website. Sizing the pump for twice the return rate means we'll size for 14 gpm. Desander processing 600 gpm and desilters processing 800 gpm. The resultant number is the gallons-per-minute or GPM used. Pump Wet End Only. You calculate gallons per acre this way:", "Make-up = Blowdown + Evaporation loss. Dosatron water-powered, non-electric chemical injectors are the easiest, most reliable way to accurately inject chemicals into water lines. McLean The PDF for the report is 1,820 kb Table of Contents. This calculator helps you compute the voltage loss from a DC input source and copper wire. Pump head: H = Pw / ρgQ = (u2Vt2 – u1Vt1)/g. Unit Of Electricity (kWh): The power consumed by any device is measured in units of electricity (kWh). To work out how much your pump costs, use the below formula to calculate the approximate cost of running your pump 24hrs a day, 365 days a year. The calculation is presented below: Results. Single speed pumps are considered the most popular types of pool pumps. A heat pump should have about 1/5 lower power consumption = 13. 35 X 100 X 1000) / 1000 =35 gram of TSP into steam drum. Power Consumption Calculator. So if your aquaponic system uses a 200-gallon setup, then it should be able to pump 200 gallons within two hours, or 100 gallons within 1 hour. The vacuum was used to draw water up. To calculate electrical energy for well pumps, we first calculate the energy needed to lift and pressurize the water for delivery to the home, then divide by the", "follows: 35% vegetable, 30% fruit orchards, 10% flowers and the remaining 120 m2 are lawns. What is the total area of the garden? 10. Delayed clock Michael put a new battery into his watch at midnight. However, they are 5 seconds late each minute. How many hours does the watch show in 24 hours? 11. Grandfather and grandmother The old mother is 5 years younger than the old father. Together they are 153 years old. How many years has each of them? 12. Bicycle The bicycle pedal gear has 36 teeth, the rear gear wheel has 10 teeth. How many times turns rear wheel, when pedal wheel turns 120x? 13. Pumps Five pumps pumped 1,800 hl of water for 3 hours. How many hectoliters of water pump same two powerful pumps for six hours? 14. Mail train The speed of mail train is 1370 meter per minute. Express it in miles per hour correct to three significant figure . Given that 1 meter =39.37 inches.", "I have a point with a jet pump. Cut in 40 psi - cut out 60 psi. Recently the psi drops to zero psi then kicks on a minute later. It goes up to 60 psi then turns off. Then when you use water it drops slowly to 40 psi then plummets to zero. What's up?", "Shipments through Colonial Pipeline move about five miles per hour, so some markets may need to rely on regional inventories until the fuel on the Colonial Pipeline can transit the pipeline.", "ticking the timer back by another minute. After three months, you will tear it all out in disgust, throw the whole mess in the garbage and go spend \\$300 on a powerful, quiet sump pump. 19. Oct 26, 2014 ### gloo I am not going to bet the barn that the pipe vibrating against the floor joist is the number one culprit. I definitely think there is a relationship because I notice that the sound upstairs, is different ...seemingly louder than when I am in the basement. Is it possible though to have a rubber that is 2.5 inch (like the pvc pipe that threads into the sump? 20. Oct 26, 2014 ### DaveC426913 Yes, for a properly rated sump pump. But he's talking about quiet little aquarium water pumps. What will be the flow rate for a 12 foot head? Even pond pumps rarely have more than a 4-6 foot head. Imagine the power of a pond pump that can generate a fountain 12 feet high? (there are calculators for such things) Last edited: Oct 26, 2014", "do need a larger supply. Try your 6 amp car battery charger . I am trying to use the pumps for automated water changes in a saltwater fish tank. My tank is small relative to most which means the pumps would run a short time to exchange the desired amount of water. I'd like to see this scale to larger tanks as well which would require more precision. I will be running the pumps for about 6 minutes a day to exchange 7 gallons a week. The pumps are currently off by about 3.7% in volume. I'd prefer about a 1% difference or less. I like the differential idea; I guess ideally I would run both pump heads off of one motor. Not sure what type of gearing this would require. The differential has three shafts? Two in with one out at the average speed between the two? Last edited: I spoke with the manufacturer of the pumps and they have stepper motors available. If I were to switch to stepper motors and have them run the same amount of steps would that be easier? I'd just have to figure out how to control them as I intended on running the brushed motors on a timer 6 minutes per day. Baluncore Science Advisor 2019 Award Stepper motors will", "charging, while minimizing the power sucked off at takeoff and climb power. If you really think that the electric pump is an improvement, have at it and tell us how it works out. Billski Status Not open for further replies.", "Empire 2022-4-13 A tank that is 50 feet in diameter is filled with water to a depth of 4 feet. To conduct a pumping test, the outlet valve to the tank is closed and the pump is allowed to discharge into the tank. After 80 minutes, the water level is 5.5 feet. What is the pumping rate in gallons per minute? First determine the area of the tank: Area = πr 2 ### calculation of wind mill for pumping water doc The Mathematics of Pumping Water AECOM Design Build Civil, Mechanical Engineering Please observe the conversion of units in calculations throughout this exemplar. WATER PUMPING WINDMILL AND WINDMILL PUMP. WINDMILL SYSTEM TOOLS AND USEFUL ENGINEERING , For calculation purposes, wind energy devices , A multi-bladed water pumping windmill of. ...", "should give you 90-95 percent of the hot water needs for your home. If we only have a 1kW element available, we will expect a heat up time in excess of 3 hours. Pump power =15. Sep 11, 2017. The Water Pump on your farm draws 30 Amps (7200 Watts) to start up and then drops back down to 1800 watts to keep running. 75 HP pump draws water from a single pipe well at depths up to 25 Ft. All of our waterfall pumps offer superior quality and reliability. 623 gallons. Just copy and paste the below code to your webpage where you want to display this calculator. Plus, starting in July 2021, new U. Calculation. Power Consumption Calculation Formula. Click the Calculate button at the bottom of the form to calculate your estimated energy cost for a month. manufactures open loop cooling towers and closed loop cooling towers at its facility in North Texas facility in Burkburnett, TX. 42 in my area voltage a 238 volt in simple, I can say my home maximum conjunction in 2873 volts in this conjunction, iron washing machine and water pump. Required Water Pump Horsepower. A reciprocating pump is a type of positive displacement pump that uses a piston, plunger or diaphragm to drive pressure into the pumped fluid. Annual" ]

[ "100 gallons water into a certain lake at their constant rates, respectively. How many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 Originally posted by Bounce1987 on 14 Aug 2017, 15:30. Last edited by Bunuel on 14 Aug 2017, 21:34, edited 1 time in total. Renamed the topic and edited the question. SC Moderator Joined: 22 May 2016 Posts: 1759 Re: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C [#permalink] ### Show Tags 14 Aug 2017, 16:40 1 Bounce1987 wrote: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C to pour 100 gallons water into a certain lake at their constant rates, respectively. How many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 These numbers are huge. Reduce before adding. Pump A rate = $$\\frac{100gal}{50secs}$$ = $$\\frac{2 gal}{1sec}$$ Pump B rate = $$\\frac{100gal}{60secs}$$ =$$\\frac{\\frac{5}{3}gal}{1sec}$$ Pump C rate = $$\\frac{100gal}{100secs}$$ = $$\\frac{1gal}{1sec}$$ ADD numerator of rates (same denominator, 1 sec): 2 + $$\\frac{5}{3}$$ + 1 =$$\\frac{\\frac{14}{3}gal}{1sec}$$ W/r = t $$\\frac{1,000}{\\frac{14}{3}}$$=", "many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 Originally posted by Bounce1987 on 14 Aug 2017, 15:30. Last edited by Bunuel on 14 Aug 2017, 21:34, edited 1 time in total. Renamed the topic and edited the question. Senior SC Moderator Joined: 22 May 2016 Posts: 3574 Re: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C [#permalink] ### Show Tags 14 Aug 2017, 16:40 1 Bounce1987 wrote: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C to pour 100 gallons water into a certain lake at their constant rates, respectively. How many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 These numbers are huge. Reduce before adding. Pump A rate = $$\\frac{100gal}{50secs}$$ = $$\\frac{2 gal}{1sec}$$ Pump B rate = $$\\frac{100gal}{60secs}$$ =$$\\frac{\\frac{5}{3}gal}{1sec}$$ Pump C rate = $$\\frac{100gal}{100secs}$$ = $$\\frac{1gal}{1sec}$$ ADD numerator of rates (same denominator, 1 sec): 2 + $$\\frac{5}{3}$$ + 1 =$$\\frac{\\frac{14}{3}gal}{1sec}$$ W/r = t $$\\frac{1,000}{\\frac{14}{3}}$$= (1,000 * $$\\frac{3}{14}$$) = $$\\frac{1,500}{7}$$ Approximately 214 seconds _________________ SC Butler has", "+0 # water 0 96 1 how much power does a water pump expend when it lifts 40.0 liters of water per minute to a height of 10.0 meters? Oct 5, 2021 edited by Guest Oct 5, 2021 #1 +13001 +2 how much power does a water pump expend when it lifts 40.0 liters of water per minute to a height of 10.0 meters? Hello Guest! $$N=\\dfrac{F\\cdot s}{t}=\\dfrac{40kg\\cdot9.81\\frac{m}{s^2}\\cdot 10m}{60s}\\cdot \\dfrac{N\\cdot s^2}{kg\\cdot m}=65.4\\cdot \\dfrac{Nm}{s}=\\color{blue}65.4\\ W$$ ! Oct 5, 2021", "pumping, which is the energy required over time: Get Price ### Math Solutions Pumping Calculations - Water Online Jul 07, 2016 Playlist Calculations. Playlist of 25 + Process Control (including Pumping) Calculations. Playlist of 35 + Pounds-Wastewater Math Calculations. Playlist of all my 130 + Math Solutions. This is the ‘Pumping Calculations’ presentation in my series of “Math Solutions.”. If you have specific wastewater math queries, please submit a question. Get Price ### Water Treatment Math - RCAC 8. How many gallons of water are in 3 acre feet of water? 3 ac ft x 325,829 gal/cu ft = Simple conversions –box 1 10. How many gallons are pumped each day if a pump is capable of pumping 50 gallons per minute and the pump runs for 9 hours a day? Get Price ### Pumping Water - Required Horsepower - Engineering ToolBox Example - Horsepower Required to Pump Water. 20 gallons of water per minute is elevated 20 ft. The horsepower required (ex. friction loss in piping and efficiency = 1.0) can be calculated as. Pwhp = (20 gpm) (20 ft) (1) / (3960 (1.0)) = 0.10 hp. Power required to pump water at 60oF with ideal pump efficiency 1.0: Power Required to Pump Water ... Get Price ### Problem Solved: Pump Discharge Rate - Wastewater Math ...", "(HP) = 100 pump efficiency (%) = 50 gallons per minute of flow (GPM) = 30 PSI = HP * (1714 * E/100) / GPM = ?", "# 815 gal/h is equivalent to how many qt/sec? Jan 19, 2017 $815 \\frac{g a l}{h} \\approx . 0.09056 \\frac{q t}{s}$ #### Explanation: There are $4$ quarts in a gallon, and there are $60$ minutes in a hour. Since there are $60$ seconds in a minute, there are $60 \\times 60 = 3600$ seconds in an hour. here we have $\\frac{815 \\left(g a l l o n s\\right)}{1 h} = \\frac{\\left(815 \\times 4\\right) \\left(q u a r t s\\right)}{3600 \\left(s\\right)}$ $815 \\times 4 = 3260$ quarts per hour then $\\frac{3260}{3600} = \\frac{326}{360} = 0.90 \\overline{5}$ quarts per second", "# A pump lifts up 200 kg of water per minute from a depth of 3 m. The pump's efficiency..... Question A pump lifts up 200 kg of water per minute from a depth of 3 m. The pump's efficiency is 40%. (g = 10) a) What is the pump's useful power output? b) What is the pump's power consumption from the electrical grid? c) Assuming the same efficiency, how much water per minute can the same pump lift up from a depth of 5 m? The water of mass m = 200 kg is being raised by the electrical pump of efficiency \\eta = 40 % = 40 / 100 = 0.4 in a time of t = 1 min = 60 s from a depth of h = 3 m. (a) Hence, its power output = P_(out) = \\frac{text{The work done by it against gravity}}{time} =\\frac{mgh}{t} = 200 kg x 10 m/s^2 x 3 m / 60 s = 100 W. (b) The pump's power consumption from the electrical grid is : P_(text{in}) = \\frac{P_(out)}{\\eta} = \\frac{100 W}{0.4} = 250 W. (c) If the pump can lift the water of mass M from a depth of H = 5 m in time t = 1 min = 60 s, then : \\frac{MgH/t}{P_(text{in} = \\eta. Hence, M", "You can transform gallons into liters and you have that $1 g a l l o n = 3.785 l i t e r s$ so in 1 gallon (that costs 1.299 $) you have 3.785 liters. You now divide the price (of 1 gallon) by the number of liters you have in 1 gallon and you get the price per liter: $\\frac{1.299}{3.785}/liters$= 0.343$ $/liter. The swimmer swims 85.4 yards=78.1 meters in 5 minutes that are (1 minute = 60 seconds) $5 \\cdot 60 = 300$seconds; so he swims $\\frac{78.1}{300} = 0.26$meters in 1 second; in 70 seconds he'll swim: $0.26 \\cdot 70 = 18.22\\$m.", "5 hours 35 minutes 40 seconds into seconds (iii) 3$$\\frac { 1 }{ 2 }$$ hours into minutes (iv) 580 minutes into hours (v) 25200 seconds into hours Solution: (i) 20 minutes into seconds 20 minutes = 20 × 60 seconds = 1200 seconds (ii) 5 hours 35 minutes 40 seconds into seconds ∴ 5 hours 35 minutes 40 seconds = 20,140 seconds (iii) 3$$\\frac { 1 }{ 2 }$$ hours into minutes 3$$\\frac { 1 }{ 2 }$$ hours = 3 hours 30 minutes = 3 × 60 minutes + 30 minutes = 180 minutes + 30 minutes = 210 minutes ∴ 3$$\\frac { 1 }{ 2 }$$ hours = 210 minutes (iv) 5580 minutes into hours 580 minutes = $$\\frac{580}{60}$$ hours = 9 hours 40 minutes ∴ 580 minutes = 9 hours 40 minutes (v) 25200 seconds into hours 25200 seconds = $$\\frac{25200}{60}$$ minutes = 420 minutes = $$\\frac{420}{60}$$ hours = 7 hours ∴ 25200 seconds = 7 hours Question 4. The duration of electricity consumed by the farmer for his pump set on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity. Solution: Electricity consumption on Monday = 7 hours + 20 minutes + 35 seconds Electricity consumption", "A pump on the ground floor of a building can pump up water Question. A pump on the ground floor of a building can pump up water to fill a tank of volume $30 \\mathrm{~m}^{3}$ in $15 \\mathrm{~min}$. If the tank is $40 \\mathrm{~m}$ above the ground, and the efficiency of the pump is $30 \\%$, how much electric power is consumed by the pump? solution: Volume of the tank, $V=30 \\mathrm{~m}^{3}$ Time of operation, $t=15 \\mathrm{~min}=15 \\times 60=900 \\mathrm{~s}$ Height of the tank, $h=40 \\mathrm{~m}$ Density of water, $\\rho=10^{3} \\mathrm{~kg} / \\mathrm{m}^{3}$ Mass of water, $m=\\rho V=30 \\times 10^{3} \\mathrm{~kg}$ Output power can be obtained as: $P_{0}=\\frac{\\text { Work done }}{\\text { Time }}=\\frac{m g h}{t}$ $=\\frac{30 \\times 10^{3} \\times 9.8 \\times 40}{900}=13.067 \\times 10^{3} \\mathrm{~W}$ For input power $P_{i}$, efficiency $\\eta$, is given by the relation: $\\eta=\\frac{P_{0}}{P_{i}}=30 \\%$ $P_{i}=\\frac{13.067}{30} \\times 100 \\times 10^{3}$ $=0.436 \\times 10^{5} \\mathrm{~W}$ $=43.6 \\mathrm{~kW}$ Editor", "in 2 minutes 3 km except for the last section where it is 2 km in 5 minutes. • 5 pumps Five pump to pump 3 hours 1600 hl of water. How long pumped same amount of water 4 pumps? • Time passing 6 years ago, Marcela's mother was two times older than her and two times younger than her father. When Marcela is 36, she will be twice as young as her father. How old are Marcela, her father, and mother now?", "D will give us the 50 feet of head we require but will only give us 110gpm. Water is being transferred by pump between two open tanks (from A to B) at a rate of 100 gpm. A TRASH pump, however, will have a side discharge and is designed to pump sand, solids, and debris with minimal wear and clogging. You can find your own pumps capacity without the specifications. Thus, in our example: BHP = (3. Your appliance and tool power requirements may differ. The 50T series pumps are not directly interchangeable with the 50 series due to special shaft and shaft seal. calculate GPM: Jun 27, 2012: Pumps and Tanks Well Forum & Blog. Pond Pumps are rated by the gallons per hour (gph) or per minute (gpm) produced at various heights. Mechanical Seal. Determine the type and size of the PTO output required (i. 3286$by$\\sqrt{PSI}\\$. NPSH, TDH, pump sizing, etc. 40 (GPM) X 60 (minutes per hour) = 2400 gallons per hour20,000 (gallons) / 2400 (gallons per hour) = 8. But, at Present only 3 Pumps are working with less flow (due to some constraints). Friction loss calculator Quickly solve friction loss for any length, GPM & hose sizes. HP = Head (ft) x Flow Rate (gpm) x (Specific Gravity) 3956: Pump Discharge Pressure =", "Make-up = Blowdown + Evaporation loss. Dosatron water-powered, non-electric chemical injectors are the easiest, most reliable way to accurately inject chemicals into water lines. McLean The PDF for the report is 1,820 kb Table of Contents. This calculator helps you compute the voltage loss from a DC input source and copper wire. Pump head: H = Pw / ρgQ = (u2Vt2 – u1Vt1)/g. Unit Of Electricity (kWh): The power consumed by any device is measured in units of electricity (kWh). To work out how much your pump costs, use the below formula to calculate the approximate cost of running your pump 24hrs a day, 365 days a year. The calculation is presented below: Results. Single speed pumps are considered the most popular types of pool pumps. A heat pump should have about 1/5 lower power consumption = 13. 35 X 100 X 1000) / 1000 =35 gram of TSP into steam drum. Power Consumption Calculator. So if your aquaponic system uses a 200-gallon setup, then it should be able to pump 200 gallons within two hours, or 100 gallons within 1 hour. The vacuum was used to draw water up. To calculate electrical energy for well pumps, we first calculate the energy needed to lift and pressurize the water for delivery to the home, then divide by the", "measurement of density: the more dense a fluid is, the more power it takes to pump. Water has a specific gravity of 1. If pumping a different fluid, look it up on a specific gravity engineering table. • This example will use water, so the specific gravity is 1. 7. 7 Estimate horsepower from these values. The water horsepower of the pump is roughly equal to ${\\displaystyle {\\frac {H*Q*SG}{3960}}}$, where H is the vertical distance the water travels in feet, Q is the flow rate in gallons per minute, and SG is the specific gravity of the fluid. • In this example, the pump is operating at ${\\displaystyle {\\frac {120ft*20gpm*1}{3960}}=}$ 0.65 horsepower. • In reality, you are likely using more power than this on your pump. Your pump is also overcoming the force of friction in the pipes, and some power is wasted due to the inefficiency of the motor. You can double this result for a rough estimate of power consumption, or track the actual amount of fuel or electricity your motor uses, or refer to the full calculations above. ## Community Q&A Search • Question How much horsepower is needed to pump water at a height of 1 km? Wildman5311 How many people are in your house? It depends on how much water you will need", "seconds 1 hour = 3600 seconds 5 hours = 5 × 3600 seconds = 18000 seconds 35 min = 35 × 60 seconds = 2100 seconds 5 hours 35 minutes 40 seconds = (18000 + 2100 + 40) seconds = 20140 seconds (iii) 3 ½ hours into minutes Solution: 1 hour = 60 minutes 3 ½ hours = 3 hours + 30 min = (3 × 60 + 30) min = (180 + 30)min = 210 min (iv) 580 minutes into hours Solution: 1 hour = 60 min 580 min = $$\\frac{580}{60}$$ hours = $$\\frac{290}{30}$$ hours = $$\\frac{29}{3}$$ = 9 $$\\frac{2}{3}$$ = 9 hours 40 min (v) 25200 seconds into hours: Solution: 25200 seconds = $$\\frac{25200}{3600}$$ = $$\\frac{126}{18}$$ hours = $$\\frac{63}{9}$$ hours = 7 hours Question 4. The duration of electricity consumed by the farmer for his pump set on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity. Solution: The total duration of electricity consumed on both days = 7 hours 20 min 35 sec + 3 hours 44 min 50 sec = (7 + 3) hours (20 + 44) min (35 + 50) sec = 10 hours 64 min 85 seconds = 11 hours 5 min 25 seconds Question", "head is 200 feet (200/2. ( I am a big fan of Wilo and Taco Pumps ). To do that, follow this formula: Pool Volume in Gallons ÷ 8 = GPH. For a Closed Loop (vertical bore loop, horizontal loop, or pond loop). Most sump pumps have charts or curves that show how many gallons per minute they can pump for different lengths of head (See Figure 2). The letter(s) correspond to the minimum recommended pump options. To calculate GPM for the same orifice, just multiply $0. How To Estimate The Lication Rate Of Drip And Sprinkler Systems. Do this with the calculator below by entering the gallons per minute rating on your hydraulic pump, and the bore, stroke, and rod size of your hydraulic cylinder. You can also do quick calculations on efficiency, speed and P1. 0 - 115 (GPM) Cast Iron. Choose a specific pump to review in-depth pump information. The flow rate. This pump will produce about 66 gallons per minute at 40 feet of TDH. 72D2√P (where D is diameter in inches. Combined with Smart Flow, PumpSmart can calculate the flow of a centrifugal pump based on. Calculate the Brake Horsepower if the pump rate is 1200 gpm. Leave a comment Cancel reply. Go to pump sizing calculator. 7138 US gpd. 15) = 344", "are in minutes, hence calculated time will be in minutes) $$R_{(s+d)}*T=W$$ $$(\\frac{3}{4}*T)=135$$ $$(\\frac{4}{3}*\\frac{3}{4})*T=135*\\frac{4}{3}$$ $$T=(135*\\frac{4}{3})=\\frac{540}{3}=180$$ minutes Time in hours: Use a conversion ratio written so that minutes will cancel $$(180 mins*\\frac{1hour}{60 mins})=(\\frac{180}{60}*1hour)=(3*1hour)=$$ $$3$$ hours II. With units • RATES Standard machine's rate: $$R_{s}=\\frac{1gal}{4mins}$$ Deluxe machine's rate? Twice the rate* of the standard machine = twice the amount of work in the same amount of time. Hence, for the deluxe machine's rate, double the standard machine's amount of work (# of gallons of paint) Deluxe machine's rate: $$R_{d}=(\\frac{1gal*2}{4mins})=\\frac{2gals}{4mins}$$ Combined rate**: $$(\\frac{1gal}{4mins}+\\frac{2gals}{4mins})=\\frac{3gals}{4mins}$$ TIME to finish? \"How many hours will it take [the machines] to fill 135 gallons of paint?\" Issue: minutes vs. hours. Rates in minutes will yield time needed in minutes. Adjust at the end. Use RT=W, find # of minutes needed, and convert to number of hours Time, T in minutes: At their combined rate of $$\\frac{3gals}{4mins}$$ how many minutes do the machines need to fill 135 gallons of paint? $$R_{(s+d)}*T=W$$ (\"gallons\" will cancel, leaving only minutes) $$(\\frac{3gals}{4mins}*T)=135gallons$$ $$(\\frac{4mins}{3gals}*\\frac{3gals}{4mins}*T)=(135gals*\\frac{4mins}{3gals})$$ $$T=(135gals)*(\\frac{4mins}{3gals})$$ $$T=\\frac{135*4mins}{3}=\\frac{540mins}{3}=180$$ minutes Time in hours: 180 minutes = how many hours? (\"minutes\" will cancel) $$180mins*\\frac{1hour}{60mins}$$ $$=3$$ hours Hope that helps! *\"Twice the rate\" can be confusing, especially if the denominator is not in single units of time (denominator, e.g., is not ONE minute) Use logic. Work rates = Amount", "\\$700 as a good quality US made 4\" submersible pump ... washing 4\" PVC 45 feet into the ground will be an interesting challenge, but plenty of people in the area have done it successfully, so I should be able to manage it ... Any way ... Thanks for the sanity check! Fish Last edited by a moderator:", "October 30 # Exam Question 29 Pump Horsepower What is the minimum size of motor needed for a 2 x 1 pump to deliver 100 gallons per minute (gpm) of water to a water skimmer that is located 50 feet above the pump and has an operating pressure of 35 psig. The water pressure on the suction side of the pump is 5 in Hg vacuum. Assume standard wall thickness. A) 100 B) 75 C) 25 D) 50 This fluids practice question show you how to solve for horsepower when you have pressure and flow rate in forms other than the units that are standard in Bernoulli’s equation. ### Converting units #### Converting flow rate from gallons per minute to cubic feet per second $1\\ ft^3 = 7.48\\ gal$ $\\dot {Q} = (\\frac {1\\ minute}{60\\ seconds})(100\\ gpm)(\\frac {1\\ ft^3}{7.48\\ gallons}) = 0.228 \\frac {ft^3}{s}$ #### Converting pounds per square inch (PSI) to pounds per square feet (psf) $P_2 = (\\frac {35\\ pounds}{in^2})(\\frac {144\\ in^2}{ft^2}) = 5040\\ psf$ #### Converting inches of vacuum in mercury to feet of head in water Specific Gravity of Mercury = 13.6 $ft\\ of\\ head\\ =\\ (\\frac {5\\ inches}{12\\ inches/foot})(13.6\\ S.G.) = 5.7\\ ft\\ H_2O$ ### Finding velocities $Q\\ =\\ AV$ $V\\ =\\ \\frac{Q}{A}$ #### Discharge Side Internal area of 1″ Sch 40 pipe", "a certain lake at their constant rates, respectively. How many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 [Reveal] Spoiler: OA Last edited by Bunuel on 14 Aug 2017, 20:34, edited 1 time in total. Renamed the topic and edited the question. Kudos [?]: 44 [1], given: 179 VP Joined: 22 May 2016 Posts: 1252 Kudos [?]: 464 [1], given: 683 Re: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C [#permalink] ### Show Tags 14 Aug 2017, 15:40 1 KUDOS Bounce1987 wrote: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C to pour 100 gallons water into a certain lake at their constant rates, respectively. How many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 These numbers are huge. Reduce before adding. Pump A rate = $$\\frac{100gal}{50secs}$$ = $$\\frac{2 gal}{1sec}$$ Pump B rate = $$\\frac{100gal}{60secs}$$ =$$\\frac{\\frac{5}{3}gal}{1sec}$$ Pump C rate = $$\\frac{100gal}{100secs}$$ = $$\\frac{1gal}{1sec}$$ ADD numerator of rates (same denominator, 1 sec): 2 + $$\\frac{5}{3}$$ + 1 =$$\\frac{\\frac{14}{3}gal}{1sec}$$ W/r =" ]

[ "D will give us the 50 feet of head we require but will only give us 110gpm. Water is being transferred by pump between two open tanks (from A to B) at a rate of 100 gpm. A TRASH pump, however, will have a side discharge and is designed to pump sand, solids, and debris with minimal wear and clogging. You can find your own pumps capacity without the specifications. Thus, in our example: BHP = (3. Your appliance and tool power requirements may differ. The 50T series pumps are not directly interchangeable with the 50 series due to special shaft and shaft seal. calculate GPM: Jun 27, 2012: Pumps and Tanks Well Forum & Blog. Pond Pumps are rated by the gallons per hour (gph) or per minute (gpm) produced at various heights. Mechanical Seal. Determine the type and size of the PTO output required (i. 3286$by$\\sqrt{PSI}\\$. NPSH, TDH, pump sizing, etc. 40 (GPM) X 60 (minutes per hour) = 2400 gallons per hour20,000 (gallons) / 2400 (gallons per hour) = 8. But, at Present only 3 Pumps are working with less flow (due to some constraints). Friction loss calculator Quickly solve friction loss for any length, GPM & hose sizes. HP = Head (ft) x Flow Rate (gpm) x (Specific Gravity) 3956: Pump Discharge Pressure =", "100 gallons water into a certain lake at their constant rates, respectively. How many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 Originally posted by Bounce1987 on 14 Aug 2017, 15:30. Last edited by Bunuel on 14 Aug 2017, 21:34, edited 1 time in total. Renamed the topic and edited the question. SC Moderator Joined: 22 May 2016 Posts: 1759 Re: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C [#permalink] ### Show Tags 14 Aug 2017, 16:40 1 Bounce1987 wrote: It takes 50 seconds, 60 seconds, and 100 seconds, 3 pumps A, B, and C to pour 100 gallons water into a certain lake at their constant rates, respectively. How many seconds it takes 3 pumps to pour 1,000 gallons water into the lake together at their constant rates, approximately, in seconds? A. 188 B. 194 C. 214 D. 232 E. 241 These numbers are huge. Reduce before adding. Pump A rate = $$\\frac{100gal}{50secs}$$ = $$\\frac{2 gal}{1sec}$$ Pump B rate = $$\\frac{100gal}{60secs}$$ =$$\\frac{\\frac{5}{3}gal}{1sec}$$ Pump C rate = $$\\frac{100gal}{100secs}$$ = $$\\frac{1gal}{1sec}$$ ADD numerator of rates (same denominator, 1 sec): 2 + $$\\frac{5}{3}$$ + 1 =$$\\frac{\\frac{14}{3}gal}{1sec}$$ W/r = t $$\\frac{1,000}{\\frac{14}{3}}$$=", "(lb/hr) in swapped opposite direction from pounds per hour - Gasoline to liters per hour, Or use utilized converter page with the flow multi-units converter, Prefix or symbol for liter per hour is: l/hr, Prefix or symbol for pound per hour - Gasoline is: lb/hr. Don’t round your answers until the very last calculation to ensure accuracy throughout the calculations. 1 gallons per minute = 227.125 liters per hour Formula gallons per minute in liters per hour (gal/m in lt/h). Ensuring you purchase a pump of sufficient size is crucial to your pond’s success. We take pride in our products as we believe they are amazing. It is recommended that you purchase a pump that circulates at least 50% of the ponds total volume every hour. Imagine my fish pond measures 10 feet in length, 12 feet in width and the bottom of the fish pond is at a constant depth of 5 feet. Hydro Aeration® Technology by Intex® - transforming the Above Ground Pool industry, by bringing revolutionary water aeration techniques into your backyard above ground pool. $$Minimum\\,Pump\\,Size = {Pond\\,Volume \\over 2}$$. Prime members enjoy FREE Delivery and exclusive access to music, movies, TV shows, original audio series, and Kindle books. 1 pound per hour - Gasoline (lb/hr) = 0.61 liters per hour (l/hr) Filed under: Main", "Pump, 110-120V with GFCI. 2 gallons UK per hour to Liters per hour = 9.09 L/h. Best Marine Bilge Pump for Kayaks, Canoes and Boats. This on the web one-way conversion tool converts flow units from liters per hour ( l/hr ) into pounds per hour - Gasoline ( lb/hr ) instantly online. This filter uses Type H filter cartridges (Item #29007E) which should be changed every couple of weeks. Therefore, the surface area of the pond is: $$Surface\\,Area = Length × Width – 15\\% = 17 × 12 – 15\\% = 173.4\\,ft^2$$. FREE Shipping by Amazon. Therefore: $$Surface\\,Area = Length × Width – 15\\%$$. This tool converts gallons per hour to liters per hour (gal/h to lt/h) and vice versa. Use our pond volume calculator to help! 1 gallons per hour = 3.7854 liters per hour Formula … For our final example let’s imagine I want to know an estimate of how many US gallons of water it will take to fill an oval fish pond measuring 30 feet in width, 25 feet in length and 12 feet deep. Keep your pool water Krystal Clear™ and refreshing this summer with the Intex® 330 GPH Filter Cartridge Pump! Your recently viewed items and featured recommendations, Select the department you want to search in, Price and other details may", "+0 # water 0 96 1 how much power does a water pump expend when it lifts 40.0 liters of water per minute to a height of 10.0 meters? Oct 5, 2021 edited by Guest Oct 5, 2021 #1 +13001 +2 how much power does a water pump expend when it lifts 40.0 liters of water per minute to a height of 10.0 meters? Hello Guest! $$N=\\dfrac{F\\cdot s}{t}=\\dfrac{40kg\\cdot9.81\\frac{m}{s^2}\\cdot 10m}{60s}\\cdot \\dfrac{N\\cdot s^2}{kg\\cdot m}=65.4\\cdot \\dfrac{Nm}{s}=\\color{blue}65.4\\ W$$ ! Oct 5, 2021", "pumping, which is the energy required over time: Get Price ### Math Solutions Pumping Calculations - Water Online Jul 07, 2016 Playlist Calculations. Playlist of 25 + Process Control (including Pumping) Calculations. Playlist of 35 + Pounds-Wastewater Math Calculations. Playlist of all my 130 + Math Solutions. This is the ‘Pumping Calculations’ presentation in my series of “Math Solutions.”. If you have specific wastewater math queries, please submit a question. Get Price ### Water Treatment Math - RCAC 8. How many gallons of water are in 3 acre feet of water? 3 ac ft x 325,829 gal/cu ft = Simple conversions –box 1 10. How many gallons are pumped each day if a pump is capable of pumping 50 gallons per minute and the pump runs for 9 hours a day? Get Price ### Pumping Water - Required Horsepower - Engineering ToolBox Example - Horsepower Required to Pump Water. 20 gallons of water per minute is elevated 20 ft. The horsepower required (ex. friction loss in piping and efficiency = 1.0) can be calculated as. Pwhp = (20 gpm) (20 ft) (1) / (3960 (1.0)) = 0.10 hp. Power required to pump water at 60oF with ideal pump efficiency 1.0: Power Required to Pump Water ... Get Price ### Problem Solved: Pump Discharge Rate - Wastewater Math ...", "ticking the timer back by another minute. After three months, you will tear it all out in disgust, throw the whole mess in the garbage and go spend \\$300 on a powerful, quiet sump pump. 19. Oct 26, 2014 ### gloo I am not going to bet the barn that the pipe vibrating against the floor joist is the number one culprit. I definitely think there is a relationship because I notice that the sound upstairs, is different ...seemingly louder than when I am in the basement. Is it possible though to have a rubber that is 2.5 inch (like the pvc pipe that threads into the sump? 20. Oct 26, 2014 ### DaveC426913 Yes, for a properly rated sump pump. But he's talking about quiet little aquarium water pumps. What will be the flow rate for a 12 foot head? Even pond pumps rarely have more than a 4-6 foot head. Imagine the power of a pond pump that can generate a fountain 12 feet high? (there are calculators for such things) Last edited: Oct 26, 2014", "menu • flow menu • Pounds per hour - Gasoline conversion Convert pound per hour - Gasoline (lb/hr) versus liters per hour (l/hr) in swapped opposite direction 4 Liters Per Hour to Gallons (us Fluid) Per Hour = 1.0567. gph unit? If I want to calculate the volume in US gallons, I would use the conversion: Now, to calculate the size of the liner required for the fish pond: $$Liner\\,Length = Pond\\,Length + Depth × 2 + 2\\,feet = 10 + 5 × 2 + 2 = 22\\,ft$$, $$Liner\\,Width = Pond\\,Width + Depth × 2 + 2\\,feet = 12 + 5 × 2 + 2 = 24\\,ft$$. Then, to convert this into US gallons, recall that one cubic foot is equivalent to 7.48 US gallons. gph and L/min measurements conversion chart page. We therefore take the volume of the rectangular pond: and multiply by 2/3 to get the total volume of an irregular pond: $$Volume= {2 \\over 3} × Length × Width × Depth$$. To link to this flow - pound per hour - Gasoline to liters per hour units converter, only cut and paste the following code into your html. Portable Sump Pump Boat Accessories. We will also explain the Mathematics the calculator uses to estimate out how much water your pond can hold. The link will", "Make-up = Blowdown + Evaporation loss. Dosatron water-powered, non-electric chemical injectors are the easiest, most reliable way to accurately inject chemicals into water lines. McLean The PDF for the report is 1,820 kb Table of Contents. This calculator helps you compute the voltage loss from a DC input source and copper wire. Pump head: H = Pw / ρgQ = (u2Vt2 – u1Vt1)/g. Unit Of Electricity (kWh): The power consumed by any device is measured in units of electricity (kWh). To work out how much your pump costs, use the below formula to calculate the approximate cost of running your pump 24hrs a day, 365 days a year. The calculation is presented below: Results. Single speed pumps are considered the most popular types of pool pumps. A heat pump should have about 1/5 lower power consumption = 13. 35 X 100 X 1000) / 1000 =35 gram of TSP into steam drum. Power Consumption Calculator. So if your aquaponic system uses a 200-gallon setup, then it should be able to pump 200 gallons within two hours, or 100 gallons within 1 hour. The vacuum was used to draw water up. To calculate electrical energy for well pumps, we first calculate the energy needed to lift and pressurize the water for delivery to the home, then divide by the", "(HP) = 100 pump efficiency (%) = 50 gallons per minute of flow (GPM) = 30 PSI = HP * (1714 * E/100) / GPM = ?", "in 2 minutes 3 km except for the last section where it is 2 km in 5 minutes. • 5 pumps Five pump to pump 3 hours 1600 hl of water. How long pumped same amount of water 4 pumps? • Time passing 6 years ago, Marcela's mother was two times older than her and two times younger than her father. When Marcela is 36, she will be twice as young as her father. How old are Marcela, her father, and mother now?", "water flow rate = 10 000 gpm. Annovi Reververi OEM Power Pressure Washer Water Pump 2600 PSI - Craftsman 580. Grid-Tie Solar Power Kit With 990 Watts of Panels and Enphase IQ7+ Microinverters. Calculate the flow rate corresponding to the operating point. 3 hours per day for one circulation or “turnover” of water. Sump Pumps Direct makes buying sump pumps and accessories easy. The pump is suspended underneath a float, meaning water is sucked into the casing in relatively clear water rather than from the base of the pond so silting issues are far less of a problem. To keep our fish happy and healthy, we need to make sure the water is clean, aerated and maintained at a constant temperature 24 hours a day. Premium liquid cooling products and solutions are EK’s domain. 8 amps on 120 volts (or uses 216 watts) The normal setting for them is 4 hours twice daily or 8 hours. Choose a shallow-well jet pump with a 1. 186 J/gm deg C) The calculator below can be used. Also Know, how many watts does a water pump use? A typical water pump's wattage can vary between 250 watts to 1,100 watts. Oil Calculator. Power in horsepower is 33,000 lb-ft/minute. During these peak times the pv panels also produce most power and. The", "option. I need a pump. It turns out that I actually designed a pump system that works well. I will discuss the pump system in a later blog post. Assume for now that I am using an electrical pump to bring the nearly 400 pounds of water up from the lake to a storage area. My pump average 15 gallons per minute, which is about 120 pounds of water per minute. This means that I can move all the water I need in less than 4 minutes with no lifting. ### Storing the Lake Water The cheapest way to store 48 gallons of water is to buy a plastic drum. While this was easy and cheap, it complicated the problem of heating the water. The drums are made of a plastic that does not like to get hotter than 180 °F. This will complicate the design of my heating system, but not too much. ### Heating the Lake Water I had heard another engineer mention that he had heated molasses in a 55 gallon drum using a drum heater belt (example). This sounded like the perfect solution to my problem. There were two concerns with the heater belt: • Could I use a heater belt with a plastic drum? • How fast would it heat the water? ####", "# A pump lifts up 200 kg of water per minute from a depth of 3 m. The pump's efficiency..... Question A pump lifts up 200 kg of water per minute from a depth of 3 m. The pump's efficiency is 40%. (g = 10) a) What is the pump's useful power output? b) What is the pump's power consumption from the electrical grid? c) Assuming the same efficiency, how much water per minute can the same pump lift up from a depth of 5 m? The water of mass m = 200 kg is being raised by the electrical pump of efficiency \\eta = 40 % = 40 / 100 = 0.4 in a time of t = 1 min = 60 s from a depth of h = 3 m. (a) Hence, its power output = P_(out) = \\frac{text{The work done by it against gravity}}{time} =\\frac{mgh}{t} = 200 kg x 10 m/s^2 x 3 m / 60 s = 100 W. (b) The pump's power consumption from the electrical grid is : P_(text{in}) = \\frac{P_(out)}{\\eta} = \\frac{100 W}{0.4} = 250 W. (c) If the pump can lift the water of mass M from a depth of H = 5 m in time t = 1 min = 60 s, then : \\frac{MgH/t}{P_(text{in} = \\eta. Hence, M", "in the area. In this area, the cost of electricity is .25 $/kWH. Finally, calculate the pond pump running cost using the formula above: PPRC = P*T * EC PPRC = 24 * 10 *.25 PPRC = 60$/day", "The pump is driven by an electric motor with an efficiency of 90 percent, and the pump is 70 percent efficient. Also known as \"flow rate\", GPM is a measure of how many gallons of water flow out of your shower head each minute. The next step in the process is to determine how many Gallons Per Minute (GPM) the pump needs to pump to turn over the water in your pool at least once a day. 1-1/2-Horsepower Above-Ground Pool Pump. Los Angeles County requires pumps to deliver the desired gallons per minute at 60 feet of head. For simplicity sake, we show only three fixtures in the house, each drawing 5 gpm. Capacity & Pressure Pump System (Full-Scale Solution) by PerfectWater™. calculate pump cycle time using this free cycle time calculator tool. The letter(s) correspond to the minimum recommended pump options. Pond pumps are sized by gallons per hour (GPH) at one foot of lift or height. Enter flowrate delivered by the pump (gallons per minute). Code to add this calci to your website. Sizing the pump for twice the return rate means we'll size for 14 gpm. Desander processing 600 gpm and desilters processing 800 gpm. The resultant number is the gallons-per-minute or GPM used. Pump Wet End Only. You calculate gallons per acre this way:", "of the faucet, takes about 45 secs for the air to work out and water to be clear. We never had this problem before with the surging and air this is why we had the well guys out and spent the $4200. I did call to tell them that I was not happy with this and they came out and check the system, said that the check valve at the pump was not working. Said that they replaced it, no charge to us (that was nice I thought) well we still have the same problem. ANY suggestions? I really feel that we spend$4200 for nothing! We have less water, due to the smaller casing, the noise going through the pipes drives me nuts, not to mention when the water surges and gets me wet when turning the facuet on. Thanks Denise", "is \\$28 per pump then the total cost for x pumps is $28x$ so $C(x)=x^2-21x+100=28x$ Just rearrange this and apply the quadratic formula to solve for the roots of this equation. I'm actually seeing two valid answers though one is more realistic than the other. So this is not correct? x^2 - 21x + 100 = 28 x^2 - 21x + 100 - 28 = 0 x^2 - 21x + 72 = 0 x= 21/2 +- V(21/2)^2 -72 V represents square root. x= 10.5+-V38.25 x=10.5+-6.1844658438 x1=4.315341562 x2=16.68465844 similar problem someone did it like me SOLUTION: Hi, please help me answer this question: The president of a company producing water pumps has gathered data suggesting that the average cost in dollars producing x units per hou Of course doesn't mean it's correct. But are you 100%? Thanks for the help No, because like you are told, the cost depends on how many pumps are made. 1 would be 28 dollars, but 2 would be 56. x amount of them would be $28x . 5. ## Re: Please help solve this completly ok you are pro. so the answer is 2 and 47 ? weird answer. 6. ## Re: Please help solve this completly This is not a wonderfully worded or designed question. You have not told us what course", "head is 200 feet (200/2. ( I am a big fan of Wilo and Taco Pumps ). To do that, follow this formula: Pool Volume in Gallons ÷ 8 = GPH. For a Closed Loop (vertical bore loop, horizontal loop, or pond loop). Most sump pumps have charts or curves that show how many gallons per minute they can pump for different lengths of head (See Figure 2). The letter(s) correspond to the minimum recommended pump options. To calculate GPM for the same orifice, just multiply $0. How To Estimate The Lication Rate Of Drip And Sprinkler Systems. Do this with the calculator below by entering the gallons per minute rating on your hydraulic pump, and the bore, stroke, and rod size of your hydraulic cylinder. You can also do quick calculations on efficiency, speed and P1. 0 - 115 (GPM) Cast Iron. Choose a specific pump to review in-depth pump information. The flow rate. This pump will produce about 66 gallons per minute at 40 feet of TDH. 72D2√P (where D is diameter in inches. Combined with Smart Flow, PumpSmart can calculate the flow of a centrifugal pump based on. Calculate the Brake Horsepower if the pump rate is 1200 gpm. Leave a comment Cancel reply. Go to pump sizing calculator. 7138 US gpd. 15) = 344", "feet deep. It depends. yours ### RE: Cost of pumping well water to irrigate 6 ac (OP) Thanks for all replys the easiest way to figure out cost is to look at how long it would take to pump amount of water needed to flood field 1 inch, then max energy usage of the pump to operate/pump for that time take that times and what you pay for elect kWh and your done all the other calculations are a waste of time. ### RE: Cost of pumping well water to irrigate 6 ac (OP) PS.The pump I got is already positioned in the well and the energy it uses is in relation to how deep it needs to be to bring out water (if you would put a inadequate submersible pump in your 4\" well you would burn out the motor.)the variables are pointless to pander since it will never use more energy that it can pull.DS ### RE: Cost of pumping well water to irrigate 6 ac 1> What is the cost per kilowatt hour in your area. These parameters will give a minimum energy cost, assuming 100% efficiency. To this we should add about 20% to 30% for real world conditions. Don't count on a 25 GPM pump putting out 25 GPM under any but design" ]

[ "+0 HELP -1 61 1 +36 Let $2^x$ be the greatest power of $2$ that is a factor of $144$, and let $3^y$ be the greatest power of $3$ that is a factor of $144$. Evaluate the following expression:$$\\left(\\frac15\\right)^{y - x}$$ Jan 15, 2020 #1 +107405 +1 144 = 12 * 12 = 2^2 * 3 * 2^2 * 3 2^4 * 3^2 The greatest power of 2 that is a factor of 144 is 2^4 ⇒ x = 4 The greatest power of 3 that is a factor of 144 is 3^2 ⇒ y = 2 So (1/5)y - x = (1/5)2 - 4 = (1/5)-2 = 52 = 25 Jan 16, 2020", "Review question What's the highest power of $x$ in this nested polynomial? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8605 Solution The highest power of $x$ in $\\left\\{\\left[(2x^6+7)^3+(3x^8-12)^4\\right]^5 + \\left[(3x^5-12x^2)^5+(x^7+6)^4\\right]^6\\right\\}^3$ is 1. $x^{424}$, 2. $x^{450}$, 3. $x^{500}$, 4. $x^{504}$. The highest power of $x$ in $(2x^6 + 7)^3$ is $x^{18}$, and in $(3x^8 - 12)^4$ is $x^{32}$, so in the first square bracket the highest power is $(x^{32})^5 = x^{160}$. Similarly, the highest power of $x$ in $(3x^5 - 12x^2)^5$ is $x^{25}$, and in $(x^7 + 6)^4$ is $x^{28}$, so in the second square bracket the highest power is $(x^{28})^6 = x^{168}$. So overall, the highest power of $x$ in the expansion is $(x^{168})^3 = x^{504}$, and the answer is (d).", "Apr 1 '11 at 12:45 \"highest\" here means not \"greatest\", but the one with the highest power of $x$ (which in this case is $x$).", "Math Help - Question on Factoring 1. Question on Factoring I am uncertain how to factor this expression correctly. $12x^2 y^3 z + 18x^3 y^2 z^2 - 24xy^4 z^3$ The greatest common factor is $6xy^2 z$ However, I am not seeing how the author picks the 2nd power when factoring out from the rest of the expression. I was under the impression that when factoring out powers of a variable, the smallest power that appears in any on term is the most that can be factored out. In this example I am not seeing how $6xy^2 z$ is the smallest power that appears in the terms if someone could please clarify. Specifically why \"xy\" is raised to the 2nd power for the lowest. 2. Originally Posted by cmf0106 I am uncertain how to factor this expression correctly. $12x^2 y^3 z + 18x^3 y^2 z^2 - 24xy^4 z^3$ The greatest common factor is $6xy^2 z$ However, I am not seeing how the author picks the 2nd power when factoring out from the rest of the expression. I was under the impression that when factoring out powers of a variable, the smallest power that appears in any on term is the most that can be factored out. In this example I am not seeing how $6xy^2 z$ is the smallest", "Why? Note:$x^T$is the transpose of$x$.$0_{m\\times ... What is the highest power of 3 available in $58! - 38!$ ( ! stands for factorial) I can take $38!$ out as common to get $38! ( \\frac{58!}{38!} - 1).$ I know how to find out the power of 3 in $38!$ ...", "# Factors And Factorials Let $$42 ^ {N}$$ be the highest power of 42 that divides $$1000!$$. What is the value of $$N$$? ×", "9 In other words, we can also say that the greatest value of p, for which $$5^p$$ is a factor of 40! Is 9. So, the correct answer choice is option C. Now, let’s solve one more similar question, to solidify our understanding. Question 2 What is the highest power of 8, that can divide 88!? A. 11 B. 12 C. 25 D. 28 E. 29 Understanding the Question This question is similar to the previous question that we solved a while ago. Hence, you may think that we can apply the same method here. Let’s see what happens when we do that. In this question, we are asked to find the highest power of 8 that can divide the number 88! So, we must divide 88 by powers of 8, till division is possible, and then add the quotients to obtain the highest power of 8. • $$\\frac{88}{8^1} = 11$$ • $$\\frac{88}{8^2} = \\frac{88}{64} = 1$$ As $$8^3 = 512$$ is greater than 88, we don’t need to consider that. Hence, the highest power of 8 that can divide 88! = 11 + 1 = 12 So, the correct answer choice is option B, right? No, 12 is not the correct answer. Let’s analyse the question a bit more to understand the error. Approach & Solution When", "# Highest power of three • Aug 12th 2009, 10:42 PM a69356 Highest power of three Q-> What is the highest power of 3 available in the expression 58! - 38! ? Thanks • Aug 13th 2009, 02:31 AM CaptainBlack Quote: Originally Posted by a69356 Q-> What is the highest power of 3 available in the expression 58! - 38! ? Thanks Write this as: $58!-38!=38! \\left( \\frac{58!}{38!}-1 \\right)$ As the term in brackets is not divisible by $3$, you are asking what is the highest power of $3$ in $38!$ CB", "to the $$0$$ power is defined to be $$1$$. 108$$x^{−n} = \\frac{1}{x^{n}}$$ , given any integer $$n$$, where $$x$$ is nonzero. 109$$\\frac { x ^ { - n } } { y ^ { - m } } = \\frac { y ^ { m } } { x ^ { n } }$$ , given any integers $$m$$ and $$n$$, where $$x ≠ 0$$ and $$y ≠ 0$$. 110Real numbers expressed the form $$a × 10^{n}$$ , where $$n$$ is an integer and $$1 ≤ a < 10$$. 111$$\\frac { x ^ { m } } { x ^ { n } } = x ^ { m - n }$$ ; the quotient of two expressions with the same base can be simplified by subtracting the exponents.", "power of a power property $\\frac{\\frac{7}{198}x^{6}z^{4}\\sqrt{y^{21}}}{\\sqrt{xyz}}$ 10 The power of a product is equal to the product of it's factors raised to the same power $\\frac{\\frac{7}{198}x^{6}z^{4}\\sqrt{y^{21}}}{\\sqrt{x}\\sqrt{yz}}$ 11 Simplify the fraction by $x$ $\\frac{\\frac{7}{198}a^{\\left(m-n\\right)}z^{4}\\sqrt{y^{21}}}{\\sqrt{yz}}$ 12 The power of a product is equal to the product of it's factors raised to the same power $\\frac{\\frac{7}{198}a^{\\left(m-n\\right)}z^{4}\\sqrt{y^{21}}}{\\sqrt{y}\\sqrt{z}}$ 13 Simplify the fraction by $y$ $\\frac{\\frac{7}{198}a^{\\left(m-n\\right)}y^{10}z^{4}}{\\sqrt{z}}$ 14 Simplify the fraction by $z$ $\\frac{7}{198}a^{\\left(m-n\\right)}y^{10}\\sqrt{z^{7}}$ $\\frac{7}{198}a^{\\left(m-n\\right)}y^{10}\\sqrt{z^{7}}$", "4 + 2 + 1 = 7 The highest power of 3 in 9! Is = Quotient of ($$\\frac{9}{3}$$) + Quotient of ($$\\frac{9}{3^2}$$) = 4 + 1 = 5 The maximum number of 6s (2 * 3)s that can be formed by taking Seven 2s and Five 3s is 5 Therefore, the highest power of 6 in 9! = 5 = Highest power of 3 in 9! Consider 12! The highest power of 2 in 12! Is = Quotient of ($$\\frac{12}{2}$$) + Quotient of ($$\\frac{12}{2^2}$$) + Quotient of ($$\\frac{12}{2^3}$$) = 6 + 3 + 1 = 10 The highest power of 3 in 12! Is = Quotient of ($$\\frac{12}{3}$$) + Quotient of ($$\\frac{12}{3^2}$$) = 4 + 1 = 5 Thus, the number of 6s that can be formed = 5 Therefore, the highest power of 6 in 12! = 5 = Highest power of 3 in 12! Hence, in any factorial, the number of 2s will ALWAYS BE MORE than the number of 3s. So, the 2s can always pair up with all the 3s to form 6s. So highest power of 3 will always be equal to the highest power of 6 Sufficient. Re: What is the highest power of 3 in the value of X! ? [#permalink] 05 Jul 2019, 19:38 Display posts from previous: Sort", "AM Soroban Hello, Ideasman Quote: How do I determine what the highest power of 2 is that divides 89,275,744? Convert $89,275,744$ to binary and we get: . . $101,010,100,100,011,110,101,100,000_2$ . . . . . . . . . . . . . . . . . . . - - - $\\uparrow$ . . . and it is obvious that $\\overbrace{100000_2 = 32}$ is the greatest divisor. Just kidding!", "what is the greatest possible value of y ? A) 30 B) 34 C) 36 D) 37 E) 39 Spelled out a little more: 1) $$x$$ = product of integers from 1 to 150 $$x$$ = 150 * 149* 148 . . .* 3 * 2 *1: That is, $$x$$ = 150! 2) $$5^{y}$$ is a factor of 150! What is the greatest possible value of $$y$$? Using $$\\frac{150}{5^{y}}$$, consider each power $$y$$, of 5. Do not worry about remainders. $$\\frac{150}{5^1}$$ = 30 (5 divides into 150 thirty times) $$\\frac{150}{5^2}$$ = 6 (25 divides into 150 six times) $$\\frac{150}{5^3}$$ = 1 (125 divides into 150 only once. Ignore the remainder.) $$5^4 = 625$$ -- too large to divide into 150 as a factor. 3) Add the results: 30 + 6 + 1 = 37 Once you know the theory and method, questions such as this one are pretty straightforward. The stats here might indicate that the suggestion below is indispensable. Bunuel wrote: Quote: _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr. BLACK LIVES MATTER. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 17042 Location:", "we have: $ y(x)=2 + x + \\frac{1}{12}x^3-\\frac{1}{48}x^4+\\frac{1}{160}x^5+.. $ so $y(0.1)\\approx 2.10008$ and $y(0.5) \\approx 2.50931$ RonL (again check the algebra) 11. Originally Posted by Watto81 So i've equated the co-efficients, are these the values for y\"', y\"\" & y\"\"'? No they are the coefficients of the corresponding powers of x in the power series expansion of y(x) RonL", "= 12$$ $$\\frac {12}{5} = 2$$ $$\\frac {2}{5} = 0$$ As there is nothing left to divide, let's add the quotients to find the highest power of 5 in $$300!$$ Sum of quotients $$= 60+12+2+0 = 74$$ Method 2 Divide 300 by consecutive powers of 3, till you get 0 as a quotient and add all quotients. This method is based on the same principle as Method 1. So, we have highest power of 3 in 300! = $$\\frac {300}{3} + \\frac {300}{3^2} +\\frac {300}{3^3}+\\frac {300}{3^4}+\\frac {300}{3^5} = 100+33+11+3+1 = 148$$ So, we can conclude that the highest power of 3 in $$300!$$ is $$3^{148}$$ Following a similar process to find the highest power of 5 in 300!, we get that the highest power is $$5^{74}$$ Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors Let's try to figure how many 15's we can create using $$3^{148}$$ and $$5^{74}$$. Since, $$15 = 3^1*5^1$$, we can write $$300!$$ as $$300!=3^{148}*5^{74}*k = (3*5)^{74} * 3^{74}*k =15^{74}*3^{74}*k$$, where k is a positive integer Or, in simple terms the highest power of 15 in 300 is $$15^{74}$$. If use the same approach to solve Q2, you will understand that $$45 = 3^2*5^1$$. The highest power of 3 in 300! is $$3^{148}$$ and", "consecutive powers of 2, till you get 0 as a quotient and add all quotients. This method is based on the same principle as Method 1. So, we have highest power of 2 in 16! = $$\\frac {16}{2} + \\frac {16}{2^2} +\\frac {16}{2^3}+\\frac {16}{2^4} = 8+4+2+1 = 15$$ So, we can conclude that the highest power of 2 in $$16!$$ is $$2^{15}$$ Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors Let's try to figure how many 8's we can create using $$2^{15}$$. Since, $$8 = 2^3$$, we can write $$2^{15}$$ as $$(2^{3})^5$$. Or, in simple terms $$2^{15} = 8^5$$ Hence, the greatest value of x such that $$8^x$$ is a factor of 16! is 5. Hence, answer is choice C. Let me post a couple of questions on similar lines where we can use the method discussed in this post to solve this type of questions very quickly. Question 1: What is the greatest value of x such that $$15^x$$ completely divides 300! ? A. 20 B. 54 C. 74 D. 148 E. 222 Question 2: What is the greatest value of \"a\" such that $$45^a$$ completely divides 300! ? A. 8 B. 36 C. 74 D. 148 E. 222 Detailed solutions will be posted soon. Use", "b) Show that the highest power of $k$ that divides 500!, where $k$ is an integer and $k^{(t + 1)} > 500 > k^t$ is , $$[500/k] + [500/k^2] + ... + [500/k^t].$$ (c) How many factors does $n!$ have?", "the denominator has more factors $2$\" but essentially the same, though. – barto Jun 21 '14 at 15:17 An elementary proof uses the following fact: If $2^s$ is the highest power of $2$ in the set $S = \\{1,2,...,n\\}$, then $2^s$ is not a divisor of any other integer in $S$. To use that, consider the highest power of $2$ which divides $n!$. Say that is $t$. Now the number can be rewritten as $\\displaystyle \\frac{\\sum \\limits_{k=1}^{n}{\\frac{n!}{k}}}{n!}$ The highest power of $2$ which divides the denominator is $t$. Now the highest power of $2$ that divides $\\displaystyle \\frac{n!}{k}$ is at least $t-s$. If $k \\neq 2^{s}$, then this is atleast $t-s+1$ as the highest power of $2$ that divides $k$ is atmost $s-1$. In case $k=2^s$, the highest power of $2$ that divides $\\dfrac{n!}{k}$ is exactly $t-s$. Thus the highest power of $2$ that divides the numerator is atmost $t-s$. If $s \\gt 0$ (which is true if $n \\gt 1$), we are done. In fact the above proof shows that the number is of the form $\\frac{\\text{odd}}{\\text{even}}$. - It would probably be a good idea to flesh this out a little. – Qiaochu Yuan Aug 18 '10 at 22:58 The exact same proof I gave works, just use $2^k$ instead of $p$. Again, you get that the", "### 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. $x^2 − 4x + 4 = 0$", "|\\dfrac {x^{17}}{(x-2)^3(2x^2+9)^7}| +2 \\sqrt 2 \\arctan(\\frac{\\sqrt 2}3x)\\bigg)$ In that logarithm we have $x^{17}$ and $2^7x^{17}$ as the highest powered terms of $x$ in the numerator and the denominator. As $x$ goes to infinity. $\\frac 1{51}(-7\\ln 2 + \\sqrt 2 \\pi)$ And you still need to evaluate the expression at $x=3$ • I put in a little more detail. You expand out the polynomials. As $x$ gets to be very large, the highest powered terms are going to dominate. So, you don't really need to do the full expansion, you just need to know the power and the coefficients of the highest powered terms. – Doug M Jan 19 '17 at 2:58" ]

[ "+0 HELP -1 61 1 +36 Let $2^x$ be the greatest power of $2$ that is a factor of $144$, and let $3^y$ be the greatest power of $3$ that is a factor of $144$. Evaluate the following expression:$$\\left(\\frac15\\right)^{y - x}$$ Jan 15, 2020 #1 +107405 +1 144 = 12 * 12 = 2^2 * 3 * 2^2 * 3 2^4 * 3^2 The greatest power of 2 that is a factor of 144 is 2^4 ⇒ x = 4 The greatest power of 3 that is a factor of 144 is 3^2 ⇒ y = 2 So (1/5)y - x = (1/5)2 - 4 = (1/5)-2 = 52 = 25 Jan 16, 2020", "$$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. _________________ Intern Joined: 16 Jun 2015 Posts: 6 ### Show Tags 21 Jul 2015, 05:37 1 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, How did you get from: $$(3^5-3^2)^2*(5^7-5^4)^2$$ to $$3^4*(3^3-1)^2*5^8*(5^3-1)^2$$ ? BR, Math Expert Joined: 02 Sep 2009 Posts: 56366 ### Show Tags 21 Jul 2015, 05:43 3 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, How did you get from: $$(3^5-3^2)^2*(5^7-5^4)^2$$ to $$3^4*(3^3-1)^2*5^8*(5^3-1)^2$$ ? BR, Consider this:", "(2a+1)^2 + (2b+1)^2 &= 332 \\\\ 6a^2 + 4ab + 6b^2 + 8a + 8b &= 328 \\\\ 3a^2 + 2ab + 3b^2 + 4a + 4b &= 164 \\end{align*} Note that $3a^2 \\le 164$, so $a \\le 7$. There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in $3b^2 + b(2a+4) + 3a^2 + 4a - 164 = 0$. Since both $a$ and $b$ are integers, we must have \\begin{align*} n^2 &= 4a^2 + 16a + 16 - 12(3a^2 + 4a - 164) \\\\ &= -32a^2 - 32a + 16 + 12 \\cdot 164 \\\\ &= 16(-2a^2 - 2a + 1 + 3 \\cdot 41) \\\\ &= 16(-2(a^2 + a) + 124), \\end{align*} where $n$ is an integer. Thus, $-2(a^2 + a) + 124$ must be a perfect square. After trying all values of $a$ from 0 to 7, we find that $a$ can be $3$ or $5$. If $a = 3$, then $b = 5$, and if $a = 5$, then $b = 3$. Therefore, the ordered pairs $(x,y)$ that satisfy the original equation are $\\boxed{(7,11) , (11,7)}$.", "= y_1 + r\\frac{a}{\\gcd(a,b)}$$ exactly as yunone said. To give you an example of this in action, suppose we want to find all integer solutions to $$258x + 147y = 369.$$ First, we use the Euclidean Algorithm to find $\\gcd(147,258)$; the parenthetical equation on the far right is how we will use this equality after we are done with the computation. \\begin{align*} 258 &= 147(1) + 111 &\\quad&\\mbox{(equivalently, $111=258 - 147$)}\\\\ 147 &= 111(1) + 36&&\\mbox{(equivalently, $36 = 147 - 111$)}\\\\ 111 &= 36(3) + 3&&\\mbox{(equivalently, $3 = 111-3(36)$)}\\\\ 36 &= 3(12). \\end{align*} So $\\gcd(147,258)=3$. Since $3|369$, the equation has integral solutions. Then we find a way of writing $3$ as a linear combination of $147$ and $258$, using the Euclidean algorithm computation above, and the equalities on the far right. We have: \\begin{align*} 3 &= 111 - 3(36)\\\\ &= 111 - 3(147 - 111) = 4(111) - 3(147)\\\\ &= 4(258 - 147) - 3(147)\\\\ &= 4(258) -7(147). \\end{align*} Then, we take $258(4) + 147(-7)=3$, and multiply through by $123$; why $123$? Because $3\\times 123 = 369$. We get: $$258(492) + 147(-861) = 369.$$ So one solution is $x=492$ and $y=-861$. All other solutions will have the form \\begin{align*} x &= 492 - \\frac{147r}{3} = 492 - 49r,\\\\ y &= -861 + \\frac{258r}{3} =86r - 861, &\\qquad&r\\in\\mathbb{Z}. \\end{align*}", "144 is $$2 \\times 2 \\times 2 \\times 2 \\times 3 \\times 3$$. This means that $$n^2$$ can contain up to four factors of 2 and two factors of 3 (note that any given factor must occur an even number of times in the prime factorization of $$n^2$$). If $$n = 4$$, the statement is satisfied, since $$\\frac{144}{n^2} = \\frac{144}{4^2} = \\frac{144}{16} = 9$$. In this case, $$n$$ is NOT divisible by 6. However, $$n$$ could also equal 6, since $$\\frac{144}{n^2} = \\frac{144}{6^2} = \\frac{144}{36} = 4$$. In this case, $$n$$ is divisible by 6. Statement 2 does NOT provide sufficient information to answer the question. Thanks Bunuel, for the first statement, n^2/180. the square root of n has 6 as it factor, but also has a square root of 5. Does it still mean that it can be divisible by 6? Math Expert Joined: 02 Sep 2009 Posts: 50004 ### Show Tags 30 Oct 2015, 01:40 doriazhao wrote: Bunuel wrote: Official Solution: We must determine whether $$n$$ is evenly divisible by $$6$$. In other words, we must determine whether $$n$$ has a factor of 2 and a factor of 3 (since $$2 \\times 3 = 6$$). If $$n$$ is evenly divisible by 6, then $$\\frac{n}{6} = k$$ for some integer $$k$$, and we can rewrite $$n$$ as", "two factors of $144$ that add to be $-25\\text{.}$ After a brief search, we conclude that those values are $-16$ and $-9\\text{.}$ The remainder of the factorization is: \\begin{align*} 9\\overbrace{{}-25y}+16y^2\\amp=9\\overbrace{{}-16y-9y}+16y^2\\\\ \\amp=\\left(9-16y\\right)+\\left(-9y+16y^2\\right)\\\\ \\amp=1\\highlight{\\left(9-16y\\right)}-y\\highlight{\\left(9+16y\\right)}\\\\ \\amp=\\highlight{\\left(9-16y\\right)}(1-y) \\end{align*} ###### Example7.6.12 Factor the expression $20x^4+13x^3-21x^2$ completely. Solution Start by noting that the GCF is $x^2\\text{.}$ Factoring this out, we get \\begin{equation*} 20x^4+13x^3-21x^2=x^2\\left(20x^2+13x-21\\right)\\text{.} \\end{equation*} Following the decision tree, we now have a trinomial inside the parentheses where $a\\neq1$ and we should try the AC method. In this case, $ac=-420$ and we need factors of $-420$ that add to $13\\text{.}$ Factor Pair Sum $1\\cdot-420$ $-419$ $2\\cdot-210$ $-208$ $3\\cdot-140$ $-137$ $4\\cdot-105$ $-101$ Factor Pair Sum $5\\cdot-84$ $-79$ $6\\cdot-70$ $-64$ $7\\cdot-60$ $-53$ $10\\cdot-42$ $-32$ Factor Pair Sum $12\\cdot-35$ $-23$ $14\\cdot-30$ $-16$ $15\\cdot-28$ $-13$ $20\\cdot-21$ $-1$ In the table of the factor pairs of $-420$ we find $15+(-28)=-13\\text{,}$ the opposite of what we want, so we want the opposite numbers: $-15$ and $28\\text{.}$ The rest of the factoring process is shown: \\begin{align*} 20x^4+13x^3-21x^2\\amp=x^2\\left(20x^2\\overbrace{{}+13x}-21\\right)\\\\ \\amp=x^2\\left(20x^2\\overbrace{-15x+28x}-21\\right)\\\\ \\amp=x^2\\left(\\left(20x^2-15x\\right)+\\left(28x-21\\right)\\right)\\\\ \\amp=x^2\\left(5x\\highlight{(4x-3)}+7\\highlight{(4x-3)}\\right)\\\\ \\amp=x^2\\highlight{(4x-3)}(5x+7) \\end{align*} ### SubsectionExercises Which factoring tools and strategies will be helpful? ###### 1 In factoring $t^{2}-2tq+q^{2}$, which factoring techniques/tools will be useful? Check all that apply. • Factoring out a GCF • Factoring by grouping • Finding two numbers that multiply to $c$ and sum to $b$ • The AC Method", "# Solve $9x$ $\\equiv 4 \\mod1453$ In Underwood Dudley's Number theory book second edition chapter 5 problem 7 I encountered this problem: Solve $9x\\equiv 4 \\mod1453$ I know that since $gcd(9,1453)=1$, there exists a unique solution. I found the answer in the back to be 1292 but I have no clue how to find this without doing guess and check--a method I'm sure is not very efficient here. Does anyone know how to go about solving this? If it helps, 1453 is prime. Thank you! You could try the method of adding the modulus: mod $1453$: $9x\\equiv 4\\equiv 1457\\equiv 2910$. At this point you can cancel $3$ (since $3$ is relatively prime to $1453$). This gives $3x\\equiv 970\\pmod{1453}$. You can now continue adding the modulus until you can cancel the remaining $3$. • Great idea! Thank you. – candido Oct 28 '14 at 23:24 You have $9x\\equiv 4 \\mod1453$. Since $\\gcd(9,1453) = 1$, by Bezout's theorem, we have $r,s \\in \\Bbb Z$ such that $9r + 1453s = 1$. Then: $$9r \\equiv 1 \\mod 1453.$$ Multiplying your initial congruence by $r$ wields: $$x \\equiv 4r \\ \\mathrm{mod} \\ 1453 \\iff x = 4r + 1453k, \\ k \\in \\Bbb Z.$$ The inverse modulo $1453$ can be computed with the extended Euclidean algorithm: \\begin{align} 1453&=9\\cdot161+4\\\\ 9&=4\\cdot2+1 \\end{align} Thus $$1=9-4\\cdot2=9-(1453-9\\cdot161)\\cdot2=323\\cdot9-1453$$", "two squares and if so, find a representation for $1034$. The prime power factorization of $1034$ is: (2) \\begin{align} \\quad 1034 = 2^1 \\cdot 11^1 \\cdot 47^1 \\end{align} Note that $11 \\equiv 3 \\pmod 4$ but $1$ is not even, so again, $1034$ cannot be expressed as a sum of two squares. ## Example 3 Determine if $98765$ can be expressed as a sum of two squares and if so, find a representation for $98765$. The prime power factorization of $98765$ is: (3) \\begin{align} \\quad 98765 = 5^1 \\cdot 19753^1 \\end{align} Since $5 \\equiv 1 \\pmod 4$ and $19753 \\equiv 1 \\pmod 4$ we have that $98765$ can be written as a sum of two squares. Namely: (4) \\begin{align} \\quad 98765 = 13^2 + 314^2 \\end{align}", "that $a$ is any positive integer such that $(a,105)=1$. \\begin{align*}\\phi(105)&=\\phi(3\\cdot 5\\cdot 7)\\\\&=105\\left(1-\\frac{1}{3}\\right)\\left(1-\\frac{1}{5}\\right)\\left(1-\\frac{1}{7}\\right)\\\\&=48.\\end{align*}So, by Euler’s Theorem (Theorem 10) $$a^{48}\\equiv 1\\mod 105.$$ As we have seen in the proof of Theorem 10, there is a smaller power of $a$ which gives $1\\mod m=105=3\\cdot 5\\cdot 7$. \\begin{align*}a^{\\phi(3)}=a^2\\equiv 1\\mod 3,\\\\ a^{\\phi(5)}=a^4\\equiv 1\\mod 5,\\\\ a^{\\phi(7)}=a^6\\equiv 1\\mod 7.\\end{align*} The least common multiple of $\\phi(3)=2,\\phi(5)=4,\\phi(7)=6$ is 12 and so \\begin{align*}a^{12}&\\equiv 1\\mod 3,\\\\ a^{12}&\\equiv 1\\mod 5,\\\\ a^{12}&\\equiv 1 \\mod 7.\\end{align*} By Lemma 3, $$a^{12}\\equiv 1\\mod3\\cdot 5\\cdot 7=105.$$ Example. Compute $2^{1000000}\\mod 77$. Solution: $77=7\\cdot 11$ and $\\phi(7)=6,\\ \\phi(11)=10$. The least common multiple of $6$ and $10$ is $30$. So, $$2^{30}\\equiv 1\\mod 77.$$ Since $1000000=30\\cdot 33333+10$, $$2^{1000000}\\equiv 2^{10}\\mod 77\\equiv 23\\mod 77.$$ There is an alternative way to calculate $2^{1000000}\\mod 77$. First compute $2^{1000000}\\mod 7$. $$2^{\\phi(7)}=2^6\\equiv 1\\mod 7.$$ Since $1000000=6\\cdot 166666+4$, $$2^{1000000}\\equiv 2^4\\equiv 2\\mod 7.$$ On the other hand, $$2^{10}\\equiv 1\\mod 11.$$ So, $$2^{1000000}\\equiv 1\\mod 11.$$ We now find $0\\leq x\\leq 76$ which satisfies \\begin{align*}x&\\equiv 2\\mod 7\\\\ x&\\equiv 1\\mod 11\\end{align*} by the Chinese Remainder Theorem (Theorem 7). Let $M=7\\cdot 11=77$, $M_1=11$, and $M_2=7$. Since $(7,11)=1$, there exist $y,z\\in\\mathbb{Z}$ such that $7y+11z=1$. By Euclidean algorithm, one can find solution $y=-3$, $z=2$. Thus, \\begin{align*}a_1&=2,\\ m_1=7,\\ M_1=11,\\ N_1=2,\\\\ a_2&=1,\\ m_2=11,\\ M_2=7,\\ N_2=-3.\\end{align*} Hence, \\begin{align}x&=\\sum_{i=1}^2a_iM_iN_i\\ &=23\\mod 77.\\end{align} Theorem. $$\\sum_{d|n}\\phi(d)=n.$$ Proof. Let $$f(n):=\\sum_{d|n}\\phi(d).$$ First we show that $f(n)$ is multiplicative. Suppose that $(m,n)=1$ and", "for mod 5} \\\\ m_3 & \\quad \\text{coefficient for mod 9} \\end{align} This coefficients are obtained using the expression $m_i = s_i \\cdot \\frac{n}{a_i}$, where $s_i = \\left(\\frac{n}{a_i}\\right)^{-1}$ in mod $a_i$. For this problem $n = 90$ and \\begin{align} a_1 & = 2 \\\\ a_2 & = 5 \\\\ a_3 & = 9 \\end{align} Inverses are: \\begin{align} s_1 = \\left(\\frac{90}{2}\\right)^{-1} \\pmod 2 = 1 \\\\ s_2 = \\left(\\frac{90}{5}\\right)^{-1} \\pmod 5 = 2 \\\\ s_3 = \\left(\\frac{90}{10}\\right)^{-1} \\pmod 9 = 1 \\end{align} and the coefficients in mod 90: \\begin{align} m_1 = s_1 \\cdot \\frac{90}{2} = 1 \\cdot 45 = 45 \\\\ m_2 = s_2 \\cdot \\frac{90}{5} = 2 \\cdot 18 = 36 \\\\ m_3 = s_3 \\cdot \\frac{90}{10} = 1 \\cdot 10 = 10 \\end{align} Value of x is obtained evaluating the linear combination: $x = \\sum_{1 \\le i \\le 3} m_i \\cdot b_i = 45 \\cdot 1 + 36 \\cdot 4 + 10 \\cdot 4 = 229 \\equiv 49 \\pmod{90}$ We can check that $x = 49$ satisfies the three congruences: \\begin{align} 49 & \\equiv 1 \\pmod 2 \\\\ 49 & \\equiv 4 \\pmod 5 \\\\ 49 & \\equiv 4 \\pmod 9 \\end{align} and the initial problem: $79 \\cdot 49 = 3871 \\equiv 1 \\pmod 90 \\quad (\\text{because } 3871 = 90 \\cdot 43 + 1)$ There is", "3\\left( \\frac{3r + 5}{2} \\right)(6r + 14) \\\\ 80 &= (3r + 5)(3r + 7) \\\\ 0 &= r^2 + 4r - 5 \\end{align*} This tells us that $(r, s) = (1, 9)$ or $(-5, -6)$. Since $-b$ is the product of the roots, we have that the two possibilities are $1 \\cdot 9 \\cdot (-10) = -90$ and $(-5)(-6)11 = 330$. Adding the absolute values of these gives us $\\boxed{420}$.", "# How do you solve 12(x-4)=144 using the distributive property? Aug 20, 2016 $x = 16$ #### Explanation: Distributive property states that a×(b+c)=a×b+a×c. Hence, $12 \\left(x - 4\\right) = 144$ can be written as 12×x-12×4=144 or $12 x - 48 = 144$ or $12 x = 144 + 48 = 192$ or x=192/12=(16×12)/(1×12) or x=(16×cancel12)/(1×cancel12) or $x = 16$ Aug 20, 2016 $x = 16$ #### Explanation: This is saying: multiply everything inside the bracket by the 12. This ability is the property of being distributive. You can distribute the 12 over everything inside the bracket $12 x - 48 = 144$ $12 x = 192$ $x = \\frac{192}{12} = 16$", "divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, How did you get from: $$(3^5-3^2)^2*(5^7-5^4)^2$$ to $$3^4*(3^3-1)^2*5^8*(5^3-1)^2$$ ? BR, Consider this: $$(ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2$$, so $$(3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2$$. The same for (5^7-5^4)^2 Hope it helps. _________________ Intern Joined: 11 Jan 2015 Posts: 34 Show Tags 19 Mar 2016, 13:33 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, could you please elaborate the following steps? $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Thank you! Math Expert Joined: 02 Aug 2009 Posts: 7098 Show Tags 19 Mar 2016, 18:28 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that", "# Multiplicative inverse of 97 modulo 386? A little help would be a lifesaver :) I already used the Euclidean algorithm to find the GCD of $386$ and $97$, which is $1$. However, I'm stuck on this question posed by my professor: \"Then use your computation to explain the statement: $241$ is the multiplicative inverse of $97$ $\\mathrm{modulo}$ $386$.\" I've been stuck on the topic of multiplicative inverses for days so any help would be greatly appreciated. Thank you so much! Using Extended Euclidean algorithm determine $p$ and $q$ such that $$97 \\cdot p - 386\\cdot q = 1$$ Then $97^{-1} \\equiv p \\mod 386$. There is likely a typo in the question. Using the Euclidean algorithm, we have that: \\begin{align*} 386 &= 3(97) + 95 \\\\ 97 &= 1(95) + 2 \\\\ 95 &= 47(2) + 1 \\end{align*} Working backwards, we see that: \\begin{align*} 1 &= 95 - 47(2) = 95 - 47(97 - 95) \\\\ &= -47(97) + 48(95) = -47(97) + 48(386 - 3(97))\\\\ &= 48(386) - 191(97)\\\\ \\end{align*} Hence, we see that: $$1 \\equiv 48(386) - 191(97) \\equiv -191(97) \\equiv 195(97) \\pmod{386}$$ So the multiplicative inverse of $97$ modulo $386$ is $195$. Using formulae used here, $$\\frac{386}{97}=3+\\frac{95}{97}=3+\\frac1{\\frac{97}{95}}$$ $$=3+\\frac1{1+\\frac2{95}}=3+\\frac1{1+\\frac1{\\frac{95}2}}=3+\\frac1{1+\\frac1{47+\\frac12}}$$ So, the previous convergent of $\\displaystyle \\frac{386}{97}$ is $\\displaystyle3+\\frac1{1+\\frac1{47}}=3+\\frac{47}{48}=\\frac{191}{48}$ $$\\implies386\\cdot48-191\\cdot97=1$$ $$\\implies -191\\cdot97=1\\pmod{386}\\implies 97^{-1}\\equiv-191\\equiv386-191$$", "ever be a perfect power? A perfect power here is defined as $x^n$ for some natural number $x$ and some natural number $n\\geq 2$. Some observations immediately come to mind: $F(1) = 1$, and that's trivially a perfect power. But $F(2) = 3$, which is a prime number and of course not a perfect power. Then there's $F(3) = 1 + 2 + 6 = 9$, which is a perfect square. Are there others? Let's check a few more. $F(4) = 33 = 3\\times 11$ and $F(5) = 153 = 3^2\\times 17$, so those are not perfect powers. In fact, $F(n)$ is not a perfect power for any natural number $n > 3$. Why is this? First, if $F(n)$ is ever a perfect power, it must be a perfect square. Why is that? If $n\\geq 6$ then we have \\begin{align*}F(n) &= 1! + 2! + \\cdots + 5! + 6! + \\cdots + n!\\\\ &= 153 + 6! + \\cdots + n!\\\\ &= 9(17 + 9(6!/9 + \\cdots + n!/9)).\\end{align*} Here, we have used that $n!/9$ will be an integer for $n\\geq 6$. Since $17$ is not a multiple of $9$, we see that $F(n)$ has at most two factors of $3$. Therefore, $F(n)$ is at most a perfect square. Another thing we see is that $F(n)$ for", "(400)$. The prime power decomposition of $400$ is $400 = 2^4 \\cdot 5^2$. So: (3) \\begin{align} \\quad \\phi (400) = \\phi (2^4 \\cdot 5^2) = \\phi(2^4) \\phi (5^2) = (2^4 - 2^3)(5^2 - 5^1) = (8)(20) = 160 \\end{align}", "multiply and divide by $4a$ to avoid fractions: \\begin{align} &3x^2-14x-5\\\\ &=\\frac{4\\cdot3}{4\\cdot3}\\left(3x^2-14x-5\\right)\\\\ &=\\frac{1}{12}\\left(36x^2-12\\cdot14x-60\\right)\\\\ &=\\frac{1}{12}\\left(\\left(6x\\right)^2-2(6x)(14)-60\\right)\\\\ &=\\frac{1}{12}\\left(\\left(6x\\right)^2-2(6x)(14)+14^2-14^2-60\\right)\\\\ &=\\frac{1}{12}\\left((6x-14)^2-196-60\\right)\\\\ &=\\frac{1}{12}\\left((6x-14)^2-256\\right)\\\\ &=\\frac{1}{12}(6x-14-16)(6x-14+16)\\\\ &=\\frac{1}{6\\cdot2}(6x-30)(6x+2)\\\\ &=(x-5)(3x+1)\\\\ \\end{align} ## AC Method (involves integer factorization and a list of things to inspect) $$3x^2-14x-5$$ Take $3\\cdot(-5)=-15$. List pairs that multiply to $-15$: $$(-15,1),(-5,3),(-3,5),(-1,15)$$ We could have stopped at the first pair, because $-15+1=-14$, the middle coefficient. Use this to replace the $-14$: $$3x^2-15x+x-5$$ Group two terms at a time and factor out the GCF: $$3x(x-5)+1(x-5)$$ $$(3x+1)(x-5)$$ ## Prime Factor what you can version 1 (involves integer factorization and a list of things to inspect) If $3x^2-14x-5$ factors, then prime factoring $3$, it factors as $$(3x+?)(x+??)$$ And $(?)(??)=-5$. There are only four possibilities. $(?,??)$ is one of $$(1,-5),(-1,5),(5,-1),(-5,1)$$ Multiplying out $(3x+?)(x+??)$ for each of the four cases reveals $3x^2-14x-5=(3x+1)(x-5)$. ## Rational Root Theorem (involves integer factorization and a list of things to inspect) If $3x^2-14x-5$ factors, there are rational roots. They must be of the form $\\pm\\frac{a}{b}$ where $a\\mid5$ and $b\\mid3$. The only options are $\\pm5,\\pm{\\frac53},\\pm1,\\pm{\\frac13}$. Check these eight inputs to $3x^2-14x-5$ and find that $-{\\frac13}$ and $5$ are roots. So $$3x^2-14x-5=c(x+1/3)(x-5)$$ Comparing leading coefficients, $c$ must be $3$. ## Prime Factor what you can version 2 (using Rational Root Theorem to speed up version 1) If $3x^2-14x-5$ factors, then prime factoring $3$, it factors as $$(3x+?)(x+??)$$ The latter", "Find the value of power of 3 Highest power of $3$ in $19^{93}-13^{99}$ Was able to figure out that $9$ will suffice but how to know about more Is there any other way than Binomial - 3^4 = 81: I feel it in my heart. – Benjamin Dickman Oct 30 '12 at 5:37 @B.: Heart is right on. – André Nicolas Oct 30 '12 at 5:54 3 Answers We can compute by hand, because the power is luckily not very high. By the Binomial Theorem, we have $$19^{93}=(1+18)^{93}=1+(18)(93)+\\cdots,$$ where each term in the part left out is divisible by at least $3^5$. Similarly, $$13^{99}=(1+12)^{93}=1+(12)(99)+(12^2)(99)(98)/2+ (12^3)(99)(98)(97)/6+\\cdots,$$ where again the remaining terms are divisible by at least $3^5$. Subtract. Note that $(18)(93)-(12)(99)=3^3(62-44)$, which is divisible by $3^5$. Now look at $$(12^2)(99)(98)/2+ (12^3)(99)(98)(97)/6.\\tag{1}$$ Dividing by $3^4$, we get $$(4^2)(11)(49)+(4^3)(11)(49)(97).$$ Modulo $3$, the first term is congruent to $-1$, and so is the second term. It follows that the highest power of $3$ that divides $(1)$ is $3^4$. - In Maple: padic:-ordp(19^93 - 13^99, 3); $4$ - Once we know, the highest power to be $4,$ there can be several ways to establish it. Following is one of them. $19^{93}-13^{99}=(19^{31})^{3}-(13^{33})^{3}=(19^{31}-13^{33})\\{(19^{31})^2+ 19^{31} \\cdot 13^{33}+(13^{33})^2\\}$ $(19^{31})^2+ 19^{31} \\cdot 13^{33}+(13^{33})^2$ $\\equiv 1+(4)^{33}+\\{(4)^{33}\\}^2 \\pmod 9$ $=1+(2^6)^{11}+(2^6)^{22} \\mod 9\\equiv 3\\pmod 9$ as $\\phi(9)=6\\implies 2^6\\equiv 1\\pmod 9$ So,", "$$(ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2$$, so $$(3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2$$. The same for (5^7-5^4)^2 Hope it helps. _________________ Intern Joined: 11 Jan 2015 Posts: 33 ### Show Tags 19 Mar 2016, 14:33 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, could you please elaborate the following steps? $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Thank you! Math Expert Joined: 02 Aug 2009 Posts: 7763 ### Show Tags 19 Mar 2016, 19:28 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, could you please elaborate the following steps? $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Thank you! hi, let me do it for you from $$=3^4*26^2*5^8*124^2=$$ $$=3^4*(2*13)^2*5^8*(4*31)^2$$ => $$=3^4*2^2*13^2*5^8*4^2*31^2$$ => $$=3^4*2^{(2+4)}*13^2*5^8*31^2$$ => $$2^6*3^4*5^8*13^2*31^2$$ _________________ Intern Joined: 10 May 2018 Posts: 1 ### Show Tags 14", "if one factor equals zero. Thus: $x-2=0~\\vee~x+4=0~\\iff~x=2~\\vee~x=-4$ to #5 there is nothing to solve because this is only a term. (I assume that there is a typo) to #6) $\\frac14 \\cdot \\left(x+\\frac12\\right)^2-\\frac14 = \\frac14 \\cdot \\left(\\left(x+\\frac12\\right)^2-1\\right) = \\frac14 \\cdot \\left(\\left(x+\\frac12+1\\right)\\left(x+\\frac12-1\\right)\\right)$ $= \\frac14 \\cdot \\left(\\left(x+\\frac32\\right)\\left(x-\\frac12\\right)\\right) =$ • Feb 7th 2008, 01:51 AM andrew2322 Hey thanks for helping out Question 5 is not a typo, Could you please explain on question 6, im still lost • Feb 7th 2008, 02:04 AM andrew2322 Sorry i have another quick question hey sorry to bother you, but i really want to learn. consider: 0.01(2y+0.6)^2 - 0.04 I took out the common factor here as 0.01 hence 0.01 (2y+0.6) ^ 2 - 4 0.01 (2y+0.6)^2 -(2)^2 0.01 (2y + 0.6-2)(2y+0.6+2) 0.01(2y-1.4)(2y+2.6) i checked the answer and it appeared as 0.04(y-0.7)(y+1.3) I realised it could possibly due to the simplification of the two 2y's but on other problems when i removed 1 '2y' the other side got the 2 chopped off as well which leads me to another answer 0.02 (y-0.07)(y+1.3) could you please explain how it works? thanks. • Feb 7th 2008, 02:40 AM earboth Quote: Originally Posted by andrew2322 ... 0.01(2y-1.4)(2y+2.6) i checked the answer and it appeared as 0.04(y-0.7)(y+1.3) .... You've done a great job. Now continue: $0.01(2y-1.4)(2y+2.6)= 0.01 \\cdot \\left(2(x-0.7)\\right) \\left(2(x+1.3)" ]

[ "+0 HELP -1 61 1 +36 Let $2^x$ be the greatest power of $2$ that is a factor of $144$, and let $3^y$ be the greatest power of $3$ that is a factor of $144$. Evaluate the following expression:$$\\left(\\frac15\\right)^{y - x}$$ Jan 15, 2020 #1 +107405 +1 144 = 12 * 12 = 2^2 * 3 * 2^2 * 3 2^4 * 3^2 The greatest power of 2 that is a factor of 144 is 2^4 ⇒ x = 4 The greatest power of 3 that is a factor of 144 is 3^2 ⇒ y = 2 So (1/5)y - x = (1/5)2 - 4 = (1/5)-2 = 52 = 25 Jan 16, 2020", "$$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. _________________ Intern Joined: 16 Jun 2015 Posts: 6 ### Show Tags 21 Jul 2015, 05:37 1 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, How did you get from: $$(3^5-3^2)^2*(5^7-5^4)^2$$ to $$3^4*(3^3-1)^2*5^8*(5^3-1)^2$$ ? BR, Math Expert Joined: 02 Sep 2009 Posts: 56366 ### Show Tags 21 Jul 2015, 05:43 3 Bunuel wrote: Official Solution: If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then $$y$$ is NOT divisible by which of the following? A. $$6^4$$ B. $$62^2$$ C. $$65^2$$ D. $$15^4$$ E. $$52^4$$ $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=$$ $$=3^4*(3^3-1)^2*5^8*(5^3-1)^2=$$ $$=3^4*26^2*5^8*124^2=$$ $$=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Bunuel, How did you get from: $$(3^5-3^2)^2*(5^7-5^4)^2$$ to $$3^4*(3^3-1)^2*5^8*(5^3-1)^2$$ ? BR, Consider this:", "Find the value of power of 3 Highest power of $3$ in $19^{93}-13^{99}$ Was able to figure out that $9$ will suffice but how to know about more Is there any other way than Binomial - 3^4 = 81: I feel it in my heart. – Benjamin Dickman Oct 30 '12 at 5:37 @B.: Heart is right on. – André Nicolas Oct 30 '12 at 5:54 3 Answers We can compute by hand, because the power is luckily not very high. By the Binomial Theorem, we have $$19^{93}=(1+18)^{93}=1+(18)(93)+\\cdots,$$ where each term in the part left out is divisible by at least $3^5$. Similarly, $$13^{99}=(1+12)^{93}=1+(12)(99)+(12^2)(99)(98)/2+ (12^3)(99)(98)(97)/6+\\cdots,$$ where again the remaining terms are divisible by at least $3^5$. Subtract. Note that $(18)(93)-(12)(99)=3^3(62-44)$, which is divisible by $3^5$. Now look at $$(12^2)(99)(98)/2+ (12^3)(99)(98)(97)/6.\\tag{1}$$ Dividing by $3^4$, we get $$(4^2)(11)(49)+(4^3)(11)(49)(97).$$ Modulo $3$, the first term is congruent to $-1$, and so is the second term. It follows that the highest power of $3$ that divides $(1)$ is $3^4$. - In Maple: padic:-ordp(19^93 - 13^99, 3); $4$ - Once we know, the highest power to be $4,$ there can be several ways to establish it. Following is one of them. $19^{93}-13^{99}=(19^{31})^{3}-(13^{33})^{3}=(19^{31}-13^{33})\\{(19^{31})^2+ 19^{31} \\cdot 13^{33}+(13^{33})^2\\}$ $(19^{31})^2+ 19^{31} \\cdot 13^{33}+(13^{33})^2$ $\\equiv 1+(4)^{33}+\\{(4)^{33}\\}^2 \\pmod 9$ $=1+(2^6)^{11}+(2^6)^{22} \\mod 9\\equiv 3\\pmod 9$ as $\\phi(9)=6\\implies 2^6\\equiv 1\\pmod 9$ So,", "144 is $$2 \\times 2 \\times 2 \\times 2 \\times 3 \\times 3$$. This means that $$n^2$$ can contain up to four factors of 2 and two factors of 3 (note that any given factor must occur an even number of times in the prime factorization of $$n^2$$). If $$n = 4$$, the statement is satisfied, since $$\\frac{144}{n^2} = \\frac{144}{4^2} = \\frac{144}{16} = 9$$. In this case, $$n$$ is NOT divisible by 6. However, $$n$$ could also equal 6, since $$\\frac{144}{n^2} = \\frac{144}{6^2} = \\frac{144}{36} = 4$$. In this case, $$n$$ is divisible by 6. Statement 2 does NOT provide sufficient information to answer the question. Thanks Bunuel, for the first statement, n^2/180. the square root of n has 6 as it factor, but also has a square root of 5. Does it still mean that it can be divisible by 6? Math Expert Joined: 02 Sep 2009 Posts: 50004 ### Show Tags 30 Oct 2015, 01:40 doriazhao wrote: Bunuel wrote: Official Solution: We must determine whether $$n$$ is evenly divisible by $$6$$. In other words, we must determine whether $$n$$ has a factor of 2 and a factor of 3 (since $$2 \\times 3 = 6$$). If $$n$$ is evenly divisible by 6, then $$\\frac{n}{6} = k$$ for some integer $$k$$, and we can rewrite $$n$$ as", "what is the greatest possible value of y ? A) 30 B) 34 C) 36 D) 37 E) 39 Spelled out a little more: 1) $$x$$ = product of integers from 1 to 150 $$x$$ = 150 * 149* 148 . . .* 3 * 2 *1: That is, $$x$$ = 150! 2) $$5^{y}$$ is a factor of 150! What is the greatest possible value of $$y$$? Using $$\\frac{150}{5^{y}}$$, consider each power $$y$$, of 5. Do not worry about remainders. $$\\frac{150}{5^1}$$ = 30 (5 divides into 150 thirty times) $$\\frac{150}{5^2}$$ = 6 (25 divides into 150 six times) $$\\frac{150}{5^3}$$ = 1 (125 divides into 150 only once. Ignore the remainder.) $$5^4 = 625$$ -- too large to divide into 150 as a factor. 3) Add the results: 30 + 6 + 1 = 37 Once you know the theory and method, questions such as this one are pretty straightforward. The stats here might indicate that the suggestion below is indispensable. Bunuel wrote: Quote: _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr. BLACK LIVES MATTER. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 17042 Location:", "two factors of $144$ that add to be $-25\\text{.}$ After a brief search, we conclude that those values are $-16$ and $-9\\text{.}$ The remainder of the factorization is: \\begin{align*} 9\\overbrace{{}-25y}+16y^2\\amp=9\\overbrace{{}-16y-9y}+16y^2\\\\ \\amp=\\left(9-16y\\right)+\\left(-9y+16y^2\\right)\\\\ \\amp=1\\highlight{\\left(9-16y\\right)}-y\\highlight{\\left(9+16y\\right)}\\\\ \\amp=\\highlight{\\left(9-16y\\right)}(1-y) \\end{align*} ###### Example7.6.12 Factor the expression $20x^4+13x^3-21x^2$ completely. Solution Start by noting that the GCF is $x^2\\text{.}$ Factoring this out, we get \\begin{equation*} 20x^4+13x^3-21x^2=x^2\\left(20x^2+13x-21\\right)\\text{.} \\end{equation*} Following the decision tree, we now have a trinomial inside the parentheses where $a\\neq1$ and we should try the AC method. In this case, $ac=-420$ and we need factors of $-420$ that add to $13\\text{.}$ Factor Pair Sum $1\\cdot-420$ $-419$ $2\\cdot-210$ $-208$ $3\\cdot-140$ $-137$ $4\\cdot-105$ $-101$ Factor Pair Sum $5\\cdot-84$ $-79$ $6\\cdot-70$ $-64$ $7\\cdot-60$ $-53$ $10\\cdot-42$ $-32$ Factor Pair Sum $12\\cdot-35$ $-23$ $14\\cdot-30$ $-16$ $15\\cdot-28$ $-13$ $20\\cdot-21$ $-1$ In the table of the factor pairs of $-420$ we find $15+(-28)=-13\\text{,}$ the opposite of what we want, so we want the opposite numbers: $-15$ and $28\\text{.}$ The rest of the factoring process is shown: \\begin{align*} 20x^4+13x^3-21x^2\\amp=x^2\\left(20x^2\\overbrace{{}+13x}-21\\right)\\\\ \\amp=x^2\\left(20x^2\\overbrace{-15x+28x}-21\\right)\\\\ \\amp=x^2\\left(\\left(20x^2-15x\\right)+\\left(28x-21\\right)\\right)\\\\ \\amp=x^2\\left(5x\\highlight{(4x-3)}+7\\highlight{(4x-3)}\\right)\\\\ \\amp=x^2\\highlight{(4x-3)}(5x+7) \\end{align*} ### SubsectionExercises Which factoring tools and strategies will be helpful? ###### 1 In factoring $t^{2}-2tq+q^{2}$, which factoring techniques/tools will be useful? Check all that apply. • Factoring out a GCF • Factoring by grouping • Finding two numbers that multiply to $c$ and sum to $b$ • The AC Method", "# How do you solve 4^x*3^(2x-x) = 144 for x ? Jul 26, 2015 $x = 2$ #### Explanation: Here's how you can quickly solve this equation. ${4}^{x} \\cdot {3}^{2 x - x} = 144$ Use the power of a power property of exponents to write ${4}^{x} = {\\left({2}^{2}\\right)}^{x} = {2}^{2 x}$ The equation becomes ${2}^{2 x} \\cdot {3}^{x} = 144$ Now focus on writing the prime factors of $144$ {:(144 : 2 = 72), (72 : 2 = 36), (36 : 2 = 18), (18 : 2 = 9) :}} -> 2^(4) {:(9 : 3 = 3),(3:3 = 1) :}} -> 3^2 This means that you have ${2}^{2 x} \\cdot {3}^{x} = {2}^{4} \\cdot {3}^{2}$ This is equivalent to having ${2}^{2 x} = {2}^{4}$ and ${3}^{x} = {3}^{2}$, which in turn will get you $x = \\textcolor{g r e e n}{2}$, which is verified by $2 x = 4 \\implies x = \\frac{4}{2} = \\textcolor{g r e e n}{2}$, Jul 26, 2015 Simplify the equation to ${12}^{x} = {12}^{2}$, hence $x = 2$ #### Explanation: ${4}^{x} \\cdot {3}^{2 x - x} = {4}^{x} \\cdot {3}^{x} = {\\left(4 \\cdot 3\\right)}^{x} = {12}^{x}$ $144 = {12}^{2}$ So the original equation simplifies to: ${12}^{x} = {12}^{2}$ Since exponentiation is a one-one function, the only solution is $x = 2$", "= y_1 + r\\frac{a}{\\gcd(a,b)}$$ exactly as yunone said. To give you an example of this in action, suppose we want to find all integer solutions to $$258x + 147y = 369.$$ First, we use the Euclidean Algorithm to find $\\gcd(147,258)$; the parenthetical equation on the far right is how we will use this equality after we are done with the computation. \\begin{align*} 258 &= 147(1) + 111 &\\quad&\\mbox{(equivalently, $111=258 - 147$)}\\\\ 147 &= 111(1) + 36&&\\mbox{(equivalently, $36 = 147 - 111$)}\\\\ 111 &= 36(3) + 3&&\\mbox{(equivalently, $3 = 111-3(36)$)}\\\\ 36 &= 3(12). \\end{align*} So $\\gcd(147,258)=3$. Since $3|369$, the equation has integral solutions. Then we find a way of writing $3$ as a linear combination of $147$ and $258$, using the Euclidean algorithm computation above, and the equalities on the far right. We have: \\begin{align*} 3 &= 111 - 3(36)\\\\ &= 111 - 3(147 - 111) = 4(111) - 3(147)\\\\ &= 4(258 - 147) - 3(147)\\\\ &= 4(258) -7(147). \\end{align*} Then, we take $258(4) + 147(-7)=3$, and multiply through by $123$; why $123$? Because $3\\times 123 = 369$. We get: $$258(492) + 147(-861) = 369.$$ So one solution is $x=492$ and $y=-861$. All other solutions will have the form \\begin{align*} x &= 492 - \\frac{147r}{3} = 492 - 49r,\\\\ y &= -861 + \\frac{258r}{3} =86r - 861, &\\qquad&r\\in\\mathbb{Z}. \\end{align*}", "if one factor equals zero. Thus: $x-2=0~\\vee~x+4=0~\\iff~x=2~\\vee~x=-4$ to #5 there is nothing to solve because this is only a term. (I assume that there is a typo) to #6) $\\frac14 \\cdot \\left(x+\\frac12\\right)^2-\\frac14 = \\frac14 \\cdot \\left(\\left(x+\\frac12\\right)^2-1\\right) = \\frac14 \\cdot \\left(\\left(x+\\frac12+1\\right)\\left(x+\\frac12-1\\right)\\right)$ $= \\frac14 \\cdot \\left(\\left(x+\\frac32\\right)\\left(x-\\frac12\\right)\\right) =$ • Feb 7th 2008, 01:51 AM andrew2322 Hey thanks for helping out Question 5 is not a typo, Could you please explain on question 6, im still lost • Feb 7th 2008, 02:04 AM andrew2322 Sorry i have another quick question hey sorry to bother you, but i really want to learn. consider: 0.01(2y+0.6)^2 - 0.04 I took out the common factor here as 0.01 hence 0.01 (2y+0.6) ^ 2 - 4 0.01 (2y+0.6)^2 -(2)^2 0.01 (2y + 0.6-2)(2y+0.6+2) 0.01(2y-1.4)(2y+2.6) i checked the answer and it appeared as 0.04(y-0.7)(y+1.3) I realised it could possibly due to the simplification of the two 2y's but on other problems when i removed 1 '2y' the other side got the 2 chopped off as well which leads me to another answer 0.02 (y-0.07)(y+1.3) could you please explain how it works? thanks. • Feb 7th 2008, 02:40 AM earboth Quote: Originally Posted by andrew2322 ... 0.01(2y-1.4)(2y+2.6) i checked the answer and it appeared as 0.04(y-0.7)(y+1.3) .... You've done a great job. Now continue: $0.01(2y-1.4)(2y+2.6)= 0.01 \\cdot \\left(2(x-0.7)\\right) \\left(2(x+1.3)", "can start by taking the root then applying the power: $16^{\\frac{3}{2}}$1632​ $=$= $\\left(\\sqrt{16}\\right)^3$(√16)3 $=$= $4^3$43 $=$= $64$64 Or we could start by applying the power then taking the root. $16^{\\frac{3}{2}}$1632​ $=$= $\\sqrt{16^3}$√163 $=$= $\\sqrt{4096}$√4096 $=$= $64$64 So, we get the same answer both ways. However, taking the root of $16$16 and then cubing $4$4 may be more efficient than dealing with the larger numbers. ##### Example 2 Simplify $\\sqrt{27\\times x^4}\\div\\sqrt[3]{3^6x^2}$27×x4÷336x2. Think: Write roots as fractional powers and simplify using our index laws. Notice we have a base of $27$27 and a base of $3$3, we can express the $27$27 as a base of $3$3 and then combine using our index laws. Do: $\\sqrt{27\\times x^4}\\div\\sqrt[3]{3^6x^2}$√27×x4÷3√36x2 $=$= $\\left(3^3\\times x^4\\right)^{\\frac{1}{2}}\\div\\left(3^6x^2\\right)^{\\frac{1}{3}}$(33×x4)12​÷(36x2)13​ $=$= $3^{\\left(3\\times\\frac{1}{2}\\right)}x^{\\left(4\\times\\frac{1}{2}\\right)}\\div3^{\\left(6\\times\\frac{1}{3}\\right)}x^{\\left(2\\times\\frac{1}{3}\\right)}$3(3×12​)x(4×12​)÷3(6×13​)x(2×13​) $=$= $3^{\\frac{3}{2}}x^2\\div3^2x^{\\frac{2}{3}}$332​x2÷32x23​ $=$= $3^{\\left(\\frac{3}{2}-2\\right)}x^{\\left(2-\\frac{2}{3}\\right)}$3(32​−2)x(2−23​) $=$= $3^{\\frac{-1}{2}}x^{\\frac{4}{3}}$3−12​x43​ $=$= $\\frac{x^{\\frac{4}{3}}}{\\sqrt{3}}$x43​√3​ Careful! Remember that when we raise a negative number to an even power, it becomes a positive number. For instance, $\\left(-5\\right)^2=25$(5)2=25. This means that $\\sqrt{(-5)^2}=\\sqrt{25}=5$(5)2=25=5. If we now consider the algebraic expression $\\sqrt{a^2}$a2, the power of a power rule indicates that this should simplify to $a$a. As you can see above, however, this is not the case if $a$a is a negative number! So be careful when simplifying even powers and roots of algebraic expressions - make sure to think about whether or not the variable could represent a negative number.", "see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Can someone please explain how the move from the second part of the equation to the third part was? Math Expert Joined: 02 Sep 2009 Posts: 49320 ### Show Tags 18 Sep 2013, 07:19 ronr34 wrote: Bunuel wrote: 2. If $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4 $$y=\\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2$$. Now, if you analyze each option you'll see that only $$52^4=2^8*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$. Can someone please explain how the move from the second part of the equation to the third part was? Consider this: $$(ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2$$, so $$(3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2$$. Hope it helps. _________________ Intern Joined: 23 May 2012 Posts: 1 ### Show Tags 08 Oct 2013, 07:32 1 Hi Bunuel, On Mr. Wallace's Briefcase problem, I was wondering if you can clarify the step where you multiple by 5 choose 3? Can I think of it as SSSNN [\"S\" representing the number 6 and \"N\" representing a non-6 digit] and the number of ways to arrange this would be 5!/[3!*2!] since there", "# How do you factor the expression x^2-x-36? Jan 5, 2016 Use the quadratic formula to find: ${x}^{2} - x - 36 = \\left(x - \\frac{1 - \\sqrt{145}}{2}\\right) \\left(x - \\frac{1 + \\sqrt{145}}{2}\\right)$ #### Explanation: You would like to find a pair of factors of $36$ that differ by $1$. Unfortunately, no such pair exists, so this polynomial will not factor with integer coefficients. Let's check the discriminant: ${x}^{2} - x - 36$ is in the form $a x + b x + c$ with $a = 1$, $b = - 1$ and $c = - 36$. This has discriminant $\\Delta$ given by the formula: $\\Delta = {b}^{2} - 4 a c = {\\left(- 1\\right)}^{2} - \\left(4 \\cdot 1 \\cdot - 36\\right) = 1 + 144 = 145$ Since this is positive but not a perfect square, the quadratic has factors with Real irrational coefficients. We can find these factors using the quadratic formula. The zeros of the quadratic are given by: $x = \\frac{- b \\pm \\sqrt{{b}^{2} - 4 a c}}{2 a} = \\frac{- b \\pm \\sqrt{\\Delta}}{2 a}$ $= \\frac{1 \\pm \\sqrt{145}}{2}$ That is ${x}_{1} = \\frac{1 - \\sqrt{145}}{2}$ and ${x}_{2} = \\frac{1 + \\sqrt{145}}{2}$ So: ${x}^{2} - x - 36 = \\left(x - {x}_{1}\\right) \\left(x - {x}_{2}\\right) = \\left(x - \\frac{1 - \\sqrt{145}}{2}\\right) \\left(x - \\frac{1", "146$$. i.e., $$2^{146}\\Big|150!$$, but $$2^{147}\\nmid 150!$$. Similarly, we have $$\\max k = 72$$. i.e., $$3^{72}\\Big|150!$$, but $$3^{73}\\nmid 150!$$. Now, we have 146 \"two\"s and 72 \"three\"s for us to pick from the prime factorization of 150!. To make 12, we take 2 \"two\"s and 1 \"three\". In total, we can make $$\\min\\left(\\dfrac{146}2, 72\\right) = 72$$ \"twelve\"s from what we have. Therefore, the largest power of 12 such that 12^n divides 150! is $$n = 72$$. Apr 17, 2022 #3 +1 Just Divide 150 by powers of 3 as follows: Floor [150/3] + [150/3^2] + [150 / 3^3] + [150 / 3^4] ==50 + 16 + 5 + 1 ==72 - is the largest power of 12. Apr 17, 2022", "the $144$ is inside parenthesis to be multiplied by $\\frac{1}{2}$, and $\\frac{1}{2} \\left(144\\right) = 72$. Now factor the right side of the equation. $y + 80 = \\frac{1}{2} {\\left(x + 12\\right)}^{2}$ Now, isolate the $y$-term on the left side of the equation to have the equation in vertex form. $y + 80 - 80 = \\frac{1}{2} {\\left(x + 12\\right)}^{2} - 80$ $y = \\frac{1}{2} {\\left(x + 12\\right)}^{2} - 80$", "when $52$ divides $\\frac 12(3n^2+n)$, you can say you want $104$ to divide $n(3n+1)$ You need to account for all the factors of $104$, which are $2^3\\cdot 13$ As $n$ and $3n+1$ are of opposite parity, one of them must be divisible by $8$ and one must be divisible by $13$. This gives four possibilities: $$\\begin {array}{c|c}\\\\ n \\pmod {104}&3n+1 \\pmod {104}\\\\0&1\\\\69&0\\\\13&40\\\\56&65\\end{array}$$ The last two come from applying the Chinese remainder theorem to $n\\equiv 0 \\pmod {13}, n\\equiv 5 \\pmod 8$ and $n \\equiv 0 \\pmod 8, n \\equiv 4 \\pmod{13}$ The smallest positive solution is then $n=13, \\frac12(3n^2+n)=260$ @Claudiop: You might look at Wikipedia on modular arithmetic to see what the mods are about. The prime factorization of $n(3n+1)$ will be the product of the factorization of $n$ and that of $3n+1$. Since one is odd and the other even, the even one has to be a multiple of $8$. One of them (maybe the same one) has to be a multiple of $13$ The first line in my table is when $n$ is a multiple of $104$, the second when $3n+1$ is a multiple of $104$, the third when $n$ is a multiple of $13$ and $3n+1$ a multiple of $8$ and the last – Ross Millikan Dec 4 '13 at 21:54 when $n$ is a", "two squares and if so, find a representation for $1034$. The prime power factorization of $1034$ is: (2) \\begin{align} \\quad 1034 = 2^1 \\cdot 11^1 \\cdot 47^1 \\end{align} Note that $11 \\equiv 3 \\pmod 4$ but $1$ is not even, so again, $1034$ cannot be expressed as a sum of two squares. ## Example 3 Determine if $98765$ can be expressed as a sum of two squares and if so, find a representation for $98765$. The prime power factorization of $98765$ is: (3) \\begin{align} \\quad 98765 = 5^1 \\cdot 19753^1 \\end{align} Since $5 \\equiv 1 \\pmod 4$ and $19753 \\equiv 1 \\pmod 4$ we have that $98765$ can be written as a sum of two squares. Namely: (4) \\begin{align} \\quad 98765 = 13^2 + 314^2 \\end{align}", "# How do you find two positive numbers whose product is 144 and whose sum is a minimum? $12 , 12$ $144 = {2}^{4} \\cdot {3}^{2}$ number of factor $= \\left(4 + 1\\right) \\cdot \\left(2 + 1\\right) = 15$ $\\implies 144 = 1 \\cdot 144 = 2 \\cdot 72 = 3 \\cdot 48 = 4 \\cdot 36 = 6 \\cdot 24 = 8 \\cdot 18 = 9 \\cdot 16 = 12 \\cdot 12$ Of these 12 pairs, the sum of $\\left(12 \\mathmr{and} 12 = 24\\right)$ is the minimum.", "$$6k$$. Statement 1 says that $$\\frac{n^2}{180}$$ is an integer. In order for $$n$$ to be divisible by 6, $$n^2$$ must have two factors of 6, since $$(6k)^2 = 36k^2$$. Since $$n^2$$ is divisible by 180, 180 is a factor of $$n^2$$, and all factors of 180 are also factors of $$n^2$$. Because 180 has 36 as a factor, and $$36 = 6 \\times 6$$, $$n^2$$ has two factors of 6. Thus, $$n$$ has 6 as a factor, and $$n$$ is divisible by 6. Statement 1 is sufficient to answer the question. Eliminate answer choices B, C, and E. The correct answer choice is either A or D. Statement 2 says that $$\\frac{144}{n^2}$$ is an integer. The prime factorization of 144 is $$2 \\times 2 \\times 2 \\times 2 \\times 3 \\times 3$$. This means that $$n^2$$ can contain up to four factors of 2 and two factors of 3 (note that any given factor must occur an even number of times in the prime factorization of $$n^2$$). If $$n = 4$$, the statement is satisfied, since $$\\frac{144}{n^2} = \\frac{144}{4^2} = \\frac{144}{16} = 9$$. In this case, $$n$$ is NOT divisible by 6. However, $$n$$ could also equal 6, since $$\\frac{144}{n^2} = \\frac{144}{6^2} = \\frac{144}{36} = 4$$. In this case, $$n$$ is divisible by 6. Statement 2 does NOT", "Tag Info Hot answers tagged exponentiation 89 $-x^2$, in every mathematical context I have seen, always means $-(x^2)$. So $-3^2 = -9$. On the other hand, when you plug in a value to an expression you don't just plug it in directly, you add parentheses first. For example, in the expression $7x$,if you plug in $x + 3$ for $x$, you get $7(x + 3) = 7x + 21$, not $7x + 3$. Similarly, plugging in $x = -3$ ... 54 One method is to set $x=y^2$ and rearrange this as a quadratic equation $$x^2-153x+1296=0$$ 29 Using Bernoulli's inequality: $$(1+x)^n\\ge 1+nx \\qquad \\forall \\ \\text{x\\ge -1 and n\\in\\mathbb N_0 }$$ we have $$\\left(1-\\frac1{99}\\right)^{49}\\ge 1-\\frac{49}{99}=\\frac{50}{99}>\\frac{50}{100}=\\frac12$$ Therefore $$\\frac{98^{99}}{99^{98}}=98\\cdot \\left(1-\\frac1{99}\\right)^{49}\\cdot \\left(1-\\frac1{99}\\right)^{49}>98\\cdot ... 25 Here's the solution:$$y^4-153y^2 +1296 = 0 y^4 -144y^2-9y^2+1296 = 0 y^2(y^2-144) -9(y^2-144) = 0 (y^2-9)(y^2-144)=0 (y^2-3^2)(y^2-12^2)=0$$note that a^2 - b^2 = (a-b)(a+b) Can you see how now? I trust you can finish the rest 21 -3^2 is always -9. There's no ambiguity. And the odd/even rule is also true! x^n is always nonnegative when n is even, and x^n is the same sign as x when n is odd (when x is real). There's no contradiction, because \"-3^2\" isn't actually of the form x^n. See, x^n, when you substitute x=-3 and n=2, gives you \"(-3)^2,\" not \"-3^2.\" ...", "careful not to get to close to $0$ or $1$ (we shall stay at least $0.01$ distance away from them) and not to exceed $20$. Calculations in the second step: The first factorial $\\left(\\frac{-5}{4}\\right)!\\approx -4.9$. The second factorial $\\left(\\left(\\frac{-5}{4}\\right)!\\right)!\\approx 0.50$. The third factorial $\\left(\\left(\\left(\\frac{-5}{4}\\right)!\\right)!\\right)!\\approx 0.89$. The fourth factorial $\\left(\\left(\\left(\\left(\\frac{-5}{4}\\right)!\\right)!\\right)!\\right)!\\approx 0.96$. The fifth factorial $\\left(\\left(\\left(\\left(\\left(\\frac{-5}{4}\\right)!\\right)!\\right)!\\right)!\\right)!\\approx 0.983$. The sixth factorial $\\left(\\left(\\left(\\left(\\left(\\left(\\frac{-5}{4}\\right)!\\right)!\\right)!\\right)!\\right)!\\right)!\\approx 0.993$. Therefore, we can use 5 factorials here, because with the sixth step we get to close to $1$. $$x=6!-(3!)!=0$$ $$y=\\left(\\left(\\left(\\left(\\left(\\frac{-5}{4}\\right)!\\right)!\\right)!\\right)!\\right)!\\approx 0.983$$ $$\\textit{final answer}=(x\\cdot y)!=0!=1$$" ]