Difference between revisions of "2006 AMC 12A Problems/Problem 2"
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<math>\textbf{(A)}\ -h\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3</math> | <math>\textbf{(A)}\ -h\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3</math> | ||
| − | == Solution == | + | == Solution 1 == |
By the definition of <math>\otimes</math>, we have <math>h\otimes h=h^{3}-h</math>. | By the definition of <math>\otimes</math>, we have <math>h\otimes h=h^{3}-h</math>. | ||
| − | Then, <math>h\otimes (h\otimes h)=h\otimes (h^{3}-h)=h^{3}-(h^{3}-h)= | + | Then, <math>h\otimes (h\otimes h)=h\otimes (h^{3}-h)=h^{3}-(h^{3}-h)=\boxed{\textbf{(C) }h}.</math> |
==Solution 2== | ==Solution 2== | ||
Revision as of 16:23, 16 December 2021
- The following problem is from both the 2006 AMC 12A #2 and 2006 AMC 10A #2, so both problems redirect to this page.
Contents
Problem
Define 
. What is 
?
Solution 1
By the definition of 
, we have 
.
Then, 
Solution 2
Substitute 
for 
. You get 
which is . That is , so the answer is 
.
~dragoon also aop is a god
See also
| 2006 AMC 12A (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 | |
| 2006 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. 
