Difference between revisions of "2007 AMC 10A Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | <cmath>\frac{6 @ 2}{6 \# 2} = \frac{(6)\times (2) - (2)^2}{(6) + (2) - (6) \cdot (2)^2} = \frac{8}{-16} = \frac{ | + | <cmath>\frac{6 @ 2}{6 \# 2} = \frac{(6)\times (2) - (2)^2}{(6) + (2) - (6) \cdot (2)^2} = \frac{8}{-16} = -\frac{1}{2} \Rightarrow \mathrm{(A)}</cmath> |
== See also == | == See also == | ||
Revision as of 17:43, 30 December 2019
Problem
Define 
and 
. What is 
?
Solution
![\[\frac{6 @ 2}{6 \# 2} = \frac{(6)\times (2) - (2)^2}{(6) + (2) - (6) \cdot (2)^2} = \frac{8}{-16} = -\frac{1}{2} \Rightarrow \mathrm{(A)}\]](http://latex.artofproblemsolving.com/7/0/2/7025af628f1ab6e6baf5bb0bf6419cde0e2a8315.png)
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. 
