Difference between revisions of "2014 AMC 10B Problems/Problem 2"

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==Solution==
 
==Solution==
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We can synchronously multiply <math> \{2^3} </math> to the polynomials above and below  .nickels and 7 pennies. Therefore, Leah has <math>6\cdot5+7=\boxed{64 (\textbf{E})}</math>.
  
 
==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 11:55, 20 February 2014

Problem

What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$?

$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$

Solution

We can synchronously multiply $\{2^3}$ (Error compiling LaTeX. Unknown error_msg) to the polynomials above and below .nickels and 7 pennies. Therefore, Leah has $6\cdot5+7=\boxed{64 (\textbf{E})}$.

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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

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