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>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/vLFULrT_7yk | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Revision as of 10:00, 17 June 2020
Contents
Problem
What is 
?
Solution
We can synchronously multiply 
to the polynomials both above and below the fraction bar. Thus,
![\[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\]](http://latex.artofproblemsolving.com/5/f/6/5f61712212b2338ec47df60099f80f0e4a58c157.png)
Hence, the fraction equals to 
.
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. 
We can factor the numerator and the denominator. The numerator becomes 2^3(1+1) and the denominator becomes 2^-3(1+1). The (1+1)'s cancel so we are left with 2^3 over 2^-3. This leaves us with 2^6 which equals to E) 64
