1640 - 64-12 - 3245 is equal to

The correct option is A 147/8 Given, (164)^0+64^ 1/2 32^4/5 ⇒ =1+18 16 =18 15 = 1198 ...

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Byju's Answer

Standard VII

Mathematics

Exponents with Like Bases

1640 + 64-12 ...

Question

$(\frac{1}{64})^{0} + (64)^{\frac{- 1}{2}} - (32)^{\frac{4}{5}}$ is equal to

- 15 \frac{7}{8}

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16 \frac{1}{8}

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- 14 \frac{7}{8}

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17 \frac{1}{8}

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Solution

The correct option is A $- 14 \frac{7}{8}$
Given,

${(\frac{1}{64})}^{0} + 64^{- \frac{1}{2}} - 32^{\frac{4}{5}}$

$\Rightarrow= 1 + \frac{1}{8} - 16$

$= \frac{1}{8} - 15$

$= \frac{- 119}{8}$

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{(\frac{1}{64})}^{0} + {(64)}^{\frac{- 1}{2}} + {(32)}^{\frac{4}{5}} - {(32)}^{\frac{- 4}{5}}

is equal to

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Q. Simplify :

\frac{4}{7} \sqrt{147} + \frac{3}{8} \sqrt{192} - \frac{1}{5} \sqrt{75}

Q. Find

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$\frac{1}{(\sqrt{9} - \sqrt{8})} - \frac{1}{(\sqrt{8} - \sqrt{7})} + \frac{1}{(\sqrt{7} - \sqrt{6})} - \frac{1}{(\sqrt{6} - \sqrt{5})} + \frac{1}{(\sqrt{5} - \sqrt{4})}$ is equal to:

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(164)0+(64)−12−(32)45 is equal to

The correct option is A −1478Given,(164)0+64−12−3245⇒=1+18−16=18−15=−1198

$(\frac{1}{64})^{0} + (64)^{\frac{- 1}{2}} - (32)^{\frac{4}{5}}$ is equal to

The correct option is A $- 14 \frac{7}{8}$
Given,

${(\frac{1}{64})}^{0} + 64^{- \frac{1}{2}} - 32^{\frac{4}{5}}$

$\Rightarrow= 1 + \frac{1}{8} - 16$

$= \frac{1}{8} - 15$

$= \frac{- 119}{8}$