1) An equilateral triangle has side length 6. What is the area of the region outside the triangle and not more than 3 units from a point of the triangle. Think ! Please show all your work.
2)= sum of proper divisors (less than n) of n. What is <<<6>>>?
3) In a given sequence, for each positive integer n, the mean of the first n terms is n. What is the 2011th term?
4) Simplify... (3^2008)^2-(3^2006)^2/(3^2007)^2-(3^2005…
5) The positive integers A,B, A-B and A+B are all prime numbers. What is A+B+(A-B)+(A+B)
2)
3) In a given sequence, for each positive integer n, the mean of the first n terms is n. What is the 2011th term?
4) Simplify... (3^2008)^2-(3^2006)^2/(3^2007)^2-(3^2005…
5) The positive integers A,B, A-B and A+B are all prime numbers. What is A+B+(A-B)+(A+B)
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1) Three 3 by 6 rectangles for 3 * (3 * 6) = 54. One complete circle (three 1/3 circles, one at each vertex) for π*3^2 = 9π. Total is 54 + 9π.
2) Since <6> = 1 + 2 + 3 = 6, any number of <> around 6 is always 6.
3) The mean of the first n terms is their sum divided by n:
(sum of first n terms)/n = n
(sum of first n terms) = n^2
Well, the nth term is:
(sum of first n terms) - (sum of first n-1 terms)
n^2 - (n-1)^2
(n - (n-1))(n + n-1)
2n - 1
So the 2011th term is 2 * 2011 - 1 = 2022 - 1 = 2021.
4) I can't see the whole problem because you did not insert spaces. Y!A truncates long words. Also, I think you meant to have both the first two terms in the numerator, so you need parentheses around them. Likewise for the denominator. I'm going to assume you mean:
[(3^2008)^2 - (3^2006)^2] / [(3^2007)^2 - (3^2005)^2]
(3^2016 - 3^2012) / (3^2014 - 3^2010)
3^2012 * (3^4 - 1) / [3^2010 * (3^4 - 1)]
3^2
9
5) There is only one even prime, 2. If both A and B are odd, then both A-B and A+B are (different) even numbers, so both cannot be prime. Thus, B must be 2.
Now, A is some odd prime such that A-2, A and A+2 are all prime. Note that one of these 3 must be divisible by 3. The only number divisible by 3 that is prime is 3, so A - 2 = 3 meaning A = 5. The desired sum is:
2 + 5 + 3 + 7 = 17
2) Since <6> = 1 + 2 + 3 = 6, any number of <> around 6 is always 6.
3) The mean of the first n terms is their sum divided by n:
(sum of first n terms)/n = n
(sum of first n terms) = n^2
Well, the nth term is:
(sum of first n terms) - (sum of first n-1 terms)
n^2 - (n-1)^2
(n - (n-1))(n + n-1)
2n - 1
So the 2011th term is 2 * 2011 - 1 = 2022 - 1 = 2021.
4) I can't see the whole problem because you did not insert spaces. Y!A truncates long words. Also, I think you meant to have both the first two terms in the numerator, so you need parentheses around them. Likewise for the denominator. I'm going to assume you mean:
[(3^2008)^2 - (3^2006)^2] / [(3^2007)^2 - (3^2005)^2]
(3^2016 - 3^2012) / (3^2014 - 3^2010)
3^2012 * (3^4 - 1) / [3^2010 * (3^4 - 1)]
3^2
9
5) There is only one even prime, 2. If both A and B are odd, then both A-B and A+B are (different) even numbers, so both cannot be prime. Thus, B must be 2.
Now, A is some odd prime such that A-2, A and A+2 are all prime. Note that one of these 3 must be divisible by 3. The only number divisible by 3 that is prime is 3, so A - 2 = 3 meaning A = 5. The desired sum is:
2 + 5 + 3 + 7 = 17
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The area of a circle with radius 3 is pi * 3^2 = 9 pi.
The three corners of an equilateral triangle are each 60º. 60º is 1/6 of 360º, so the rest of the circle is (5/6) * 9pi = (15/2) * pi
Three of those is (45/2) * pi or 22.5 * pi
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1 + 2 + 3 = 6
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The mean of the first 2011 terms is 2011, so the total is 2011^2 = 4,044,121
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I don't have the rest of the equation, but (3^2006)^2 / (3^2007)^2 = (1/3)^2 = 1/9. I hope that helps.
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A + B + (A - B) + (A + B) = 3A + B
A = 5, B = 2, and 3A + B = 17
The three corners of an equilateral triangle are each 60º. 60º is 1/6 of 360º, so the rest of the circle is (5/6) * 9pi = (15/2) * pi
Three of those is (45/2) * pi or 22.5 * pi
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1 + 2 + 3 = 6
***
The mean of the first 2011 terms is 2011, so the total is 2011^2 = 4,044,121
***
I don't have the rest of the equation, but (3^2006)^2 / (3^2007)^2 = (1/3)^2 = 1/9. I hope that helps.
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A + B + (A - B) + (A + B) = 3A + B
A = 5, B = 2, and 3A + B = 17