Difference between revisions of "2006 AMC 10B Problems/Problem 2"
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Since <math> x \spadesuit y = (x+y)(x-y) </math>: | Since <math> x \spadesuit y = (x+y)(x-y) </math>: | ||
− | <math> 3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = -72 | + | <math> 3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{\textbf{(A)}-72}</math> |
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=1|num-a=3}} | |
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:33, 19 December 2021
Problem
For real numbers and
, define
. What is
?
Solution
Since :
See Also
2006 AMC 10B (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.