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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | We have | ||
+ | <math></math>\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = <math>\boxed{{64 (\textbf{E})}}</math>.<math></math> | ||
==Video Solution== | ==Video Solution== |
Revision as of 15:17, 24 April 2022
Problem
What is ?
Solution
We can synchronously multiply to the polynomials both above and below the fraction bar. Thus, Hence, the fraction equals to .
Solution 2
We have $$ (Error compiling LaTeX. Unknown error_msg)\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = .$$ (Error compiling LaTeX. Unknown error_msg)
Video Solution
~savannahsolver
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.