Difference between revisions of "2006 AMC 12B Problems/Problem 2"
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<math>4\spadesuit 5=-9</math> | <math>4\spadesuit 5=-9</math> | ||
| − | <math>3\spadesuit -9=-72 \Rightarrow \text{( | + | <math>3\spadesuit -9=-72 \Rightarrow \text{(A)}</math> |
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== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2006|ab=B|num-b=1|num-a=3}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 09:44, 4 July 2013
Problem
For real numbers 
and 
, define 
. What is 
?
Solution
See also
| 2006 AMC 12B (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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
