Difference between revisions of "2014 AMC 10B Problems/Problem 2"

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<cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}. </cmath>
 
<cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}. </cmath>
 
Hence, the fraction equals to <math>\boxed{{64 (\textbf{E})}}</math>.
 
Hence, the fraction equals to <math>\boxed{{64 (\textbf{E})}}</math>.
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==Video Solution==
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https://youtu.be/vLFULrT_7yk
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 09:00, 17 June 2020

Problem

What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$?

$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$

Solution

We can synchronously multiply ${2^3}$ to the polynomials both above and below the fraction bar. Thus, \[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\] Hence, the fraction equals to $\boxed{{64 (\textbf{E})}}$.

Video Solution

https://youtu.be/vLFULrT_7yk

~savannahsolver

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

We can factor the numerator and the denominator. The numerator becomes 2^3(1+1) and the denominator becomes 2^-3(1+1). The (1+1)'s cancel so we are left with 2^3 over 2^-3. This leaves us with 2^6 which equals to E) 64

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