2014 AMC 10B Problems/Problem 2: Difference between revisions

Revision as of 12:18, 20 February 2014

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

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

We can synchronously multiply ${2^3}$ to the polynomials both above and below the fraction bar. Thus, $\frac{2^3+2^3}{2^{-3}+2^{-3}}$

\[implies \=\frac{2^6+2^6}{1+1}={2^6}\] (Error compiling LaTeX. Unknown error_msg)

, which can be calculated resulting in 64. Therefore, the fraction equals to $\boxed{{64 (\textbf{E})}}$ .

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 10 Problems and Solutions

@@ Line 5: / Line 5: @@
 ==Solution==
-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,
-Thus <cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}</cmath>
+<cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}</cmath>
-<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>.
+<cmath> implies \=\frac{2^6+2^6}{1+1}={2^6}</cmath>, which can be calculated resulting in 64. Therefore, the fraction equals to <math>\boxed{{64 (\textbf{E})}}</math>.
 ==See Also==
 {{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}}
 {{MAA Notice}}