Difference between revisions of "2000 AMC 12 Problems/Problem 5"
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If <math>|x - 2| = p</math>, where <math>x < 2</math>, then <math>x - p =</math> | If <math>|x - 2| = p</math>, where <math>x < 2</math>, then <math>x - p =</math> | ||
− | <math> \ | + | <math> \textbf{(A) \ -2 } \qquad \textbf{(B) \ 2 } \qquad \textbf{(C) \ 2-2p } \qquad \textbf{(D) \ 2p-2 } \qquad \textbf{(E) \ |2p-2| } </math> |
== Solution == | == Solution == |
Revision as of 09:45, 8 November 2021
- The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.
Problem
If , where
, then
Solution
When
is negative so
and
.
Thus .
See also
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 |
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.