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> | ||
== Solution == | == Solution == | ||
| − | 6@2 must be equal to 6*2-2^2 which is 8. 6#2 is equal to 6+2-6*2^2 which is 8-24 = -16. Therefore {frac}(6@2)/(6#2) must be equal to 8/-16 = -1/2. Therefore the solution is A. | + | <math>6@2</math> must be equal to <math>6*2-2^2</math> which is 8. <math>6#</math>2 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>8/-16 = -1/2</math>. Therefore the solution is <math>A</math>. |
== See also == | == See also == | ||
Revision as of 14:05, 27 January 2021
Problem
Define 
and 
. What is 
?
Solution
must be equal to 
which is 8. $6#$ (Error compiling LaTeX. Unknown error_msg)2 is equal to 
which is 
. Therefore ${frac}(6@2)/(6#2)$ (Error compiling LaTeX. Unknown error_msg) 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. 
