Difference between revisions of "1956 AHSME Problems/Problem 1"
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Therefore, the answer is <math>\fbox{(A) 10}</math> | Therefore, the answer is <math>\fbox{(A) 10}</math> | ||
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{{AHSME box|year=1956|before=1956 AHSME|num-a=2}} | {{AHSME box|year=1956|before=1956 AHSME|num-a=2}} | ||
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Revision as of 23:40, 30 December 2015
1956 AHSME Problem #1
The value of 
when 
is:
Solution
Simple substitution yields
![\[2 + 2(2^2) = 2 + 2(4) = 10\]](http://latex.artofproblemsolving.com/8/3/f/83f3051ce6110cecd8c706dbc393ac683b6f13f7.png)
Therefore, the answer is 
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by 1956 AHSME |
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. 
