1954 AHSME Problems/Problem 6: Difference between revisions
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}\implies1-\frac{1}{8}\implies\boxed{\textbf{(D) }\frac{7}{8}}</math>. | ||
Revision as of 14:59, 23 June 2018
Problem 6
The value of 
is:
Solution
.
