Difference between revisions of "2007 AMC 10A Problems/Problem 2"
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<math>\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}</math> | <math>\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}</math> | ||
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== Solution == | == Solution == | ||
| − | < | + | <math>6@2</math> must be equal to <math>6*2-2^2</math> which is 8. <math>6\# 2</math> is equal to <math>6+2-6*2^2</math> which is <math>8-24 = -16</math>. Therefore <math>\frac{6@2}{6\# 2}</math> must be equal to <math>\frac{8}{-16} = -\frac{1}{2}</math>. Therefore the solution is <math>A</math>. |
== See also == | == See also == | ||
Latest revision as of 14:07, 27 January 2021
Problem
Define 
and 
. What is 
?
Solution
must be equal to 
which is 8. 
is equal to 
which is 
. Therefore 
must be equal to 
. Therefore the solution is 
.
See also
| 2007 AMC 10A (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. 
