Difference between revisions of "2014 AMC 10B Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
We have | We have | ||
| − | <cmath>\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 = \boxed | + | <cmath>\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 = 64,</cmath> so our answer is <math>\textbf{\boxed{(E)}}</math>. |
==Video Solution== | ==Video Solution== | ||
Revision as of 16:18, 24 April 2022
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 
.
Solution 2
We have
![\[\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64,\]](http://latex.artofproblemsolving.com/2/f/7/2f746975ff2b7b39e12fbe1865ec81670932b221.png)
so our answer is 
.
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. 
