Difference between revisions of "1954 AHSME Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}\implies \frac{1}{16}+1-\frac{1}{8}-((-32)^4)^\frac{1}{5}\implies 1-\frac{1}{16}-\frac{1}{16}\implies1-\frac{1}{8}\implies\boxed{\textbf{(D) }\frac{7}{8}}</math>. | + | <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}\implies \frac{1}{16}+1-\frac{1}{8}-((-32)^4)^\frac{1}{5}\implies 1-\frac{1}{16}-\frac{1}{16}</math><math>\implies1-\frac{1}{8}\implies\boxed{\textbf{(D) }\frac{7}{8}}</math>. |
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| + | ==See Also== | ||
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| + | {{AHSME 50p box|year=1954|num-b=5|num-a=7}} | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 19:37, 17 February 2020
Problem 6
The value of 
is:
Solution
.
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
