Difference between revisions of "2014 AMC 10B Problems/Problem 2"
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We can synchronously multiply <math> {2^3} </math> to the polynomials both above and below the fraction bar. Thus, | 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> | + | <cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}</cmath>\ |
<cmath>\=\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>. | <cmath>\=\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>. | ||
Revision as of 12:19, 20 February 2014
Problem
What is ?
Solution
We can synchronously multiply to the polynomials both above and below the fraction bar. Thus, \
\[\=\frac{2^6+2^6}{1+1}={2^6}\] (Error compiling LaTeX. Unknown error_msg)
, which can be calculated resulting in 64. Therefore, the fraction equals to .
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.