Difference between revisions of "2004 AMC 10A Problems/Problem 2"

 
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== Problem ==
 
== Problem ==
For any three real numbers <math>a</math>, <math>b</math>, and <math>c</math>, with <math>b\neq c</math>, the operation <math>\otimes</math> is defined by:
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For any three [[real number]]s <math>a</math>, <math>b</math>, and <math>c</math>, with <math>b\neq c</math>, the [[operation]] <math>\otimes</math> is defined by:
 
<math>
 
<math>
 
\otimes(a,b,c)=\frac{a}{b-c}
 
\otimes(a,b,c)=\frac{a}{b-c}
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== Solution ==
 
== Solution ==
<math>\otimes</math><math>(\frac{1}{2-3}, \frac{2}{3-1}, \frac{3}{1-2})=</math><math>\otimes</math><math>(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\mathrm{(B)}</math>
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<math>\otimes</math><math> \left(\frac{1}{2-3}, \frac{2}{3-1}, \frac{3}{1-2}\right)=\displaystyle</math><math>\otimes</math><math>(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\mathrm{(B)}</math>
  
 
== See also ==
 
== See also ==

Revision as of 16:22, 11 September 2007

Problem

For any three real numbers $a$, $b$, and $c$, with $b\neq c$, the operation $\otimes$ is defined by: $\otimes(a,b,c)=\frac{a}{b-c}$ What is $\otimes$$( \otimes$$(1,2,3),$$\otimes$$(2,3,1),$$\otimes$$(3,1,2))$?

$\mathrm{(A) \ } -\frac{1}{2}\qquad \mathrm{(B) \ } -\frac{1}{4} \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac{1}{4} \qquad \mathrm{(E) \ } \frac{1}{2}$

Solution

$\otimes$$\left(\frac{1}{2-3}, \frac{2}{3-1}, \frac{3}{1-2}\right)=\displaystyle$$\otimes$$(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\mathrm{(B)}$

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions