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

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Hence, the fraction equals to <math>\boxed{{64 (\textbf{E})}}</math>.
 
Hence, the fraction equals to <math>\boxed{{64 (\textbf{E})}}</math>.
  
==Solution 2==
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==Video Solution==
We could also do quick math\, solving the exponents to get:  
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https://youtu.be/vLFULrT_7yk
<cmath>\frac{8+8}{</cmath>\frac{1}{8}+<cmath>\frac{1}{8}}\\=\frac{2^6+2^6}{1+1}\\={2^6}. </cmath>
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~savannahsolver
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:28, 12 September 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

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