Difference between revisions of "2014 AMC 10B Problems/Problem 2"
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==Solution== | ==Solution== | ||
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}}=\frac{2^6+2^6}{1+1}={2^6} | + | <cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}=\frac{2^6+2^6}{1+1}={2^6}, resulting in 64. </cmath> |
| − | Therefore, the fraction equals to | + | Therefore, the fraction equals to $\boxed{{64 (\textbf{E})}}. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:21, 20 February 2014
Problem
What is 
?
Solution
We can synchronously multiply 
to the polynomials both above and below the fraction bar. Thus,
![\[\frac{2^3+2^3}{2^{-3}+2^{-3}}=\frac{2^6+2^6}{1+1}={2^6}, resulting in 64.\]](http://latex.artofproblemsolving.com/b/f/6/bf614ebb7956ead4b8acdfb55f3d4fd71c425ccd.png)
Therefore, the fraction equals to $\boxed{{64 (\textbf{E})}}.
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. 
