Difference between revisions of "1992 AHSME Problems/Problem 1"
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<math>6(4x+5\pi) = 2 \cdot 3(4x+5\pi) = 2 \cdot P</math> | <math>6(4x+5\pi) = 2 \cdot 3(4x+5\pi) = 2 \cdot P</math> | ||
| − | <math>6(8x+10\pi) = 2 \cdot 6(4x+5\pi) = 2 \cdot 2P</math> | + | <math>6(8x+10\pi) = 2 \cdot 6(4x+5\pi) = 2 \cdot 2P = \fbox{4P}</math> |
== See also == | == See also == | ||
Revision as of 19:38, 2 March 2017
Problem
If 
then 
Solution
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
