Difference between revisions of "2000 AMC 10 Problems/Problem 9"

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==Problem==
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#REDIRECT [[2000 AMC 12 Problems/Problem 5]]
 
 
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==
 
 
 
<math>|x-2|=p</math>
 
 
 
<math>x<2</math>, so <math>2-x=p</math>.
 
 
 
<math>x+p=2</math>.
 
 
 
<math>x-p=2-2p</math>.
 
 
 
<math>\boxed{\text{C}}</math>
 
 
 
==See Also==
 
 
 
{{AMC10 box|year=2000|num-b=8|num-a=10}}
 

Latest revision as of 23:37, 26 November 2011