Difference between revisions of "2000 AMC 12 Problems/Problem 5"
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== Solution == | == Solution == | ||
− | When <math>\displaystyle x < 2,</math>, <math>x-2</math> is negative | + | When <math>\displaystyle x < 2,</math>, <math>x-2</math> is negative so <math>|x - 2| = 2-x</math> and <math>\displaystyle x - 2 = -p</math>. |
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Therefore: | Therefore: | ||
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<math>x=2-p</math> | <math>x=2-p</math> | ||
− | <math>\displaystyle x-p = (2-p)-p = 2-2p \ | + | <math>\displaystyle x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)} </math> |
== See also == | == See also == | ||
* [[2000 AMC 12 Problems]] | * [[2000 AMC 12 Problems]] | ||
− | *[[2000 AMC 12/Problem 4|Previous Problem]] | + | *[[2000 AMC 12 Problems/Problem 4|Previous Problem]] |
− | *[[2000 AMC 12/Problem 6|Next problem]] | + | *[[2000 AMC 12 Problems/Problem 6|Next problem]] |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 14:22, 16 November 2006
Problem
If where
then
Solution
When ,
is negative so
and
.
Therefore: