2000 AMC 10 Problems/Problem 9: Difference between revisions

Latest revision as of 22:37, 26 November 2011

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-==Problem==
+#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}}