Difference between revisions of "1954 AHSME Problems/Problem 6"
Katzrockso (talk | contribs) (Created page with "== Problem 6== The value of <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}</math> is: <math>\textbf{(A)}\ 1 ...") |
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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)^ | + | <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 | |
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| All AHSME Problems and Solutions | ||
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