Difference between revisions of "2016 AMC 10B Problems/Problem 1"
(→Solution) |
(→Solution 2) |
||
Line 11: | Line 11: | ||
==Solution 2== | ==Solution 2== | ||
− | Substituting <math>\frac{1}{2}</math> for <math>a</math> in <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> goves us <math>\boxed{\textbf{(D) }10}</math>. | + | Substituting <math>\frac{1}{2}</math> for <math>a</math> in <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> goves us <math>\boxed{\textbf{(D) }10}</math>. ~peelybonehead |
==Video Solution== | ==Video Solution== |
Revision as of 23:59, 13 October 2022
Contents
[hide]Problem
What is the value of when ?
Solution
Factorizing the numerator, then becomes which is equal to which is .
Solution 2
Substituting for in goves us . ~peelybonehead
Video Solution
~savannahsolver
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by - |
Followed by Problem 2 | |
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.