64-125 -2-3-1- 256-625 1-4- -25-64 0 -

The correct option is B 9/4 (64125)^ 2/3÷1(256/625)^1/4+((√(25))√(64))^0 ⇒ #160; (45)^ 2÷145+1 ⇒ #160; (45)^ 2×(45)^1+1 ...

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Byju's Answer

Standard VIII

Mathematics

Adding Consecutive Odd Numbers

64/125 -2/3÷1...

Question

${(\frac{64}{125})}^{- 2 / 3} \div \frac{1}{{(256 / 625)}^{1 / 4}} + {(\frac{(\sqrt{25})}{\sqrt[3]{64}})}^{0} =$

\frac{9}{2}

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\frac{9}{4}

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\frac{8}{2}

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\frac{8}{4}

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Solution

The correct option is B $\frac{9}{4}$
${(\frac{64}{125})}^{- 2 / 3} \div \frac{1}{(256 / 625)^{1 / 4}} + {(\frac{(\sqrt{25})}{\sqrt[3]{64}})}^{0}$

$\Rightarrow$ ${(\frac{4}{5})}^{- 2} \div \frac{1}{\frac{4}{5}} + 1$

$\Rightarrow$ ${(\frac{4}{5})}^{- 2} \times {(\frac{4}{5})}^{1} + 1$

$\Rightarrow$ ${(\frac{4}{5})}^{- 2 + 1} + 1$

$\Rightarrow$ ${(\frac{4}{5})}^{- 1} + 1$

$\Rightarrow$ $\frac{5}{4} + 1$

$\Rightarrow$ $\frac{9}{4}$

$∴$ ${(\frac{64}{125})}^{- 2 / 3} \div \frac{1}{(256 / 625)^{1 / 4}} + {(\frac{(\sqrt{25})}{\sqrt[3]{64}})}^{0} = \frac{9}{4}$

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Similar questions

Q. Solve :

{(\frac{64}{125})}^{- 2 / 3} + \frac{1}{{(\frac{256}{625})}^{1 / 4}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0} = \frac{61}{16}

Evaluate: ${(\begin{matrix} \frac{64}{125} \end{matrix})}^{\frac{- 2}{3}} \div \frac{1}{(\frac{256}{625})^{\frac{1}{4}}} + (\begin{matrix} \frac{(\sqrt{25})}{\sqrt[3]{64}} \end{matrix})$

Q. Simplify:
(i)

(27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{- 1}

(ii)

{(\frac{64}{125})}^{\frac{- 2}{3}} \div \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} \times \frac{\sqrt{25}}{\sqrt[3]{64}}

(iii)

{⎡ ⎢ ⎢ ⎣ 5 {⎛ ⎜ ⎝ 8^{\frac{1}{3}} + 27^{\frac{1}{3}} ⎞ ⎟ ⎠}^{3} ⎤ ⎥ ⎥ ⎦}^{\frac{1}{4}}

Prove that :

$(i) (\sqrt{3 \times 5^{- 3}} \div \sqrt[3]{3^{- 1}} \sqrt{5}) \times \sqrt[6]{3 \times 5^{6}} = \frac{3}{5} (i i) 9^{\frac{3}{2}} - 3 \times 5^{0} - {(\frac{1}{81})}^{\frac{- 1}{2}} = 15 (i i i) {(\frac{1}{4})}^{- 2} - 3 \times 8^{\frac{2}{3}} \times 4^{0} + {(\frac{9}{16})}^{\frac{- 1}{2}} = \frac{16}{3} (i v) \frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{\frac{- 1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{- 7}{5}}}{4^{\frac{- 3}{5}} \times 6} = 10 (v) \sqrt{\frac{1}{4}} + (0.01)^{\frac{- 1}{2}} - (27)^{\frac{2}{3}} = \frac{3}{2} (v i) \frac{2^{n} + 2^{n - 1}}{2^{n + 1} - 2^{n}} = \frac{3}{2} (v i i) {(\frac{64}{125})}^{\frac{- 2}{3}} + \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0} = \frac{61}{16} (v i i i) \frac{3^{- 3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{\frac{1}{25}} \times (15)^{\frac{- 4}{3}} \times 3^{\frac{1}{3}}} = 28 \sqrt{2} (i x) \frac{(0.6)^{0} - (0.1)^{- 1}}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(\frac{1}{3})}^{- 1}} = - \frac{3}{2}$

Q. (1) 5, 3, 2, 9 (2) 2, 4, 3, 9
(3) 1, 4, 3, 8 (4) 3, 2, 3, 8

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(64125)−2/3÷1(256/625)1/4+((√25)3√64)0=

The correct option is B 94(64125)−2/3÷1(256/625)1/4+((√25)3√64)0⇒ (45)−2÷145+1⇒ (45)−2×(45)1+1⇒ (45)−2+1+1⇒ (45)−1+1⇒ 54+1⇒ 94∴ (64125)−2/3÷1(256/625)1/4+((√25)3√64)0=94

${(\frac{64}{125})}^{- 2 / 3} \div \frac{1}{{(256 / 625)}^{1 / 4}} + {(\frac{(\sqrt{25})}{\sqrt[3]{64}})}^{0} =$