2000 AMC 10 Problems/Problem 9: Difference between revisions

Revision as of 21:52, 8 January 2009

If $|x-2|=p$ , where $x<2$ , then $x-p=$

$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 2-2p \qquad\mathrm{(D)}\ 2p-2 \qquad\mathrm{(E)}\ |2p-2|$

$|x-2|=p$

$x<2$ , so $2-x=p$ .

$x+p=2$ .

$x-p=2-2p$ .

$\boxed{\text{C}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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

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 ==Problem==
+If <math>|x-2|=p</math>, where <math>x<2</math>, then <math>x-p=</math>
+<math>\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 2-2p \qquad\mathrm{(D)}\ 2p-2 \qquad\mathrm{(E)}\ |2p-2|</math>
 ==Solution==