Difference between revisions of "2000 AMC 12 Problems/Problem 5"
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+ | {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #5]] and [[2000 AMC 10 Problems|2000 AMC 10 #9]]}} | ||
== Problem == | == Problem == | ||
− | If <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 == | ||
− | When <math> | + | When <math>x < 2,</math> <math>x-2</math> is negative so <math>|x - 2| = 2-x = p</math> and <math>x = 2-p</math>. |
− | + | Thus <math>x-p = (2-p)-p = 2-2p</math>. | |
+ | <math>\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}</math> | ||
− | + | ==Video Solution by Daily Dose of Math== | |
− | + | https://youtu.be/albUhCOwv3Y?si=4XcusOEp70EA6XKr | |
+ | |||
+ | ~Thesmartgreekmathdude | ||
== See also == | == See also == | ||
− | + | {{AMC10 box|year=2000|num-b=8|num-a=10}} | |
− | + | {{AMC12 box|year=2000|num-b=4|num-a=6}} | |
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− | |||
− | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:42, 15 July 2024
- 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 .
Video Solution by Daily Dose of Math
https://youtu.be/albUhCOwv3Y?si=4XcusOEp70EA6XKr
~Thesmartgreekmathdude
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.