Difference between revisions of "2006 AMC 10A Problems/Problem 2"

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<math> \mathrm{(A) \ } -h\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } h\qquad \mathrm{(D) \ } 2h\qquad \mathrm{(E) \ } h^3 </math>
 
<math> \mathrm{(A) \ } -h\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } h\qquad \mathrm{(D) \ } 2h\qquad \mathrm{(E) \ } h^3 </math>
 
== Solution ==
 
== Solution ==
Plugging our values in the function, we have  
+
Plugging our values into the defined [[operation]], we have  
  
<center><math>\displaystyle h^3-h</math></center>
+
<center><math>\displaystyle h\otimes h = h^3-h</math></center>
  
Plugging in the function once more, we have
+
Plugging in the [[function]] once more, we have
  
<center><math>\displaystyle h^3-(h^3-h)=h,
+
<center><math>\displaystyle h\otimes(h\otimes h) = h\otimes (h^3 - h) = h^3-(h^3-h)=h</math></center>
(C)</math></center>
+
 
 +
which is choice <math>(C)</math>.
  
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]

Revision as of 08:14, 28 July 2006

Problem

Define $x\otimes y=x^3-y$. What is $h\otimes (h\otimes h)$?

$\mathrm{(A) \ } -h\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } h\qquad \mathrm{(D) \ } 2h\qquad \mathrm{(E) \ } h^3$

Solution

Plugging our values into the defined operation, we have

$\displaystyle h\otimes h = h^3-h$

Plugging in the function once more, we have

$\displaystyle h\otimes(h\otimes h) = h\otimes (h^3 - h) = h^3-(h^3-h)=h$

which is choice $(C)$.

See Also