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

(Solution)
(Solution)
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We can synchronously multiply <math> {2^3} </math> to the polynomials both above and below the fraction bar.  
 
We can synchronously multiply <math> {2^3} </math> to the polynomials both above and below the fraction bar.  
 
Thus <cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}</cmath>
 
Thus <cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}</cmath>
<math></math>implies \=\frac{2^6+2^6}{1+1}={2^6}<math>, which can be calculated resulting in 64. Therefore, the fraction equals to </math>\boxed{{64 (\textbf{E})}}$.
+
<math></math> implies \=\frac{2^6+2^6}{1+1}={2^6}<math>, which can be calculated resulting in 64. Therefore, the fraction equals to </math>\boxed{{64 (\textbf{E})}}$.
  
 
==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 12:16, 20 February 2014

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 \[\frac{2^3+2^3}{2^{-3}+2^{-3}}\] $$ (Error compiling LaTeX. ! Missing $ inserted.) implies \=\frac{2^6+2^6}{1+1}={2^6}$, which can be calculated resulting in 64. Therefore, the fraction equals to$\boxed{{64 (\textbf{E})}}$.

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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

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