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If x is equal to the sum of the even integers from m to n, inclusive,
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06 Aug 2021, 09:08
06 Aug 2021, 09:08
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Question Stats:
68% (01:57) correct
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If x is equal to the sum of the even integers from m to n, inclusive, where m and n are positive even integers, which of the following represents the value of x in terms of m and n?
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2-m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2-m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
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If x is equal to the sum of the even integers from m to n, inclusive,
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07 Aug 2021, 08:04
07 Aug 2021, 08:04
1
Carcass wrote:
If x is equal to the sum of the even integers from m to n, inclusive, where m and n are positive even integers, which of the following represents the value of x in terms of m and n?
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2-m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2-m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
Let us assume values for \(m\) and \(n\)
\(m = 2\)
\(n = 8\)
Sum = \(2 + 4 + 6 + 8 = 20\)
Check the option choices;
A. \((\frac{2+8}{2})(\frac{8-2}{2} + 1)\)
\((\frac{10}{2})(\frac{6}{2} + 1)\)
\((5)(3 + 1) = 20\)
B. \(3(m+n)\)
\(3(2 + 8) = 30\)
C. \(\frac{n^2-m^2}{2}\)
\(\frac{8^2-2^2}{2}\)
\(\frac{64-4}{2} = 30\)
D. \(6(m+n)\)
\(6(2 + 8) = 60\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
\((\frac{2+8}{2})(\frac{8-2}{2})\)
\((\frac{10}{2})(\frac{6}{2})\)
\((5)(3) = 15\)
Hence, option A
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Re: If x is equal to the sum of the even integers from m to n, inclusive,
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15 Jan 2022, 12:27
15 Jan 2022, 12:27
1
Carcass wrote:
If x is equal to the sum of the even integers from m to n, inclusive, where m and n are positive even integers, which of the following represents the value of x in terms of m and n?
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2-m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
A. \((\frac{m+n}{2})(\frac{n-m}{2} + 1)\)
B. \(3(m+n)\)
C. \(\frac{n^2-m^2}{2}\)
D. \(6(m+n)\)
E. \((\frac{m+n}{2})(\frac{n-m}{2})\)
For this question, it's pretty easy to test values.
Let's say m = 2 and n = 6
So, x = 2 + 4 + 6 = 12
So, we'll plug m = 2 and n = 6 to be chances choice to see which one yields a value of 12.
A. \((\frac{2+6}{2})(\frac{6-2}{2} + 1)=(4)(3)=12\). PERFECT!
B. \(3(2+6)=24\). No good. We need an output of 12. Eliminate.
C. \(\frac{6^2 - 2^2}{2}=16\). No good. We need an output of 12. Eliminate.
D. \(6(2+6)=48\). No good. We need an output of 12. Eliminate.
E. \((\frac{2+6}{2})(\frac{6-2}{2})=(4)(2)=8\). No good. We need an output of 12. Eliminate.
