Find the value of 0-6 0 - 0-1 -1 3 8 -1 3 2 3 - -1 3 -1-

The correct option is C 3 2 Now ( 0.6 ) #160; ^ 0 0.1 ^ 1 ( 3 8 #160; ) #160; ^ 1 ( 3 2 #160; ) #160; ^ 3 +( 1 3 #160; ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

Find the valu...

Question

Find the value of $\frac{{(0.6)}^{0} - {(0.1)}^{- 1}}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(\frac{- 1}{3})}^{- 1}}$ .

\frac{- 2}{3}

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\frac{3}{2}

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- \frac{3}{2}

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\frac{2}{3}

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Solution

The correct option is C $- \frac{3}{2}$
$N o w \frac{{(0.6)}^{0} - {0.1}^{- 1}}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(\frac{- 1}{3})}^{- 1}} \to \frac{1 - \frac{1}{0.1}}{\frac{1}{\frac{3}{8}} \times \frac{3^{3}}{2^{3}} + ⎛ ⎜ ⎜ ⎜ ⎝ \frac{1}{\frac{- 1}{3}} ⎞ ⎟ ⎟ ⎟ ⎠} = \frac{1 - 10}{\frac{8}{3} \times \frac{3^{3}}{2^{3}} - (\frac{3}{1})} \to \frac{- 9}{\frac{2^{3}}{2^{3}} \times 3^{3 - 1} - 3} = \frac{- 9}{3^{2} - 3} = \frac{- 9}{9 - 3} = \frac{- 9}{6} = - \frac{3}{2}$

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$\frac{3}{4} \div 2 \frac{1}{4}$ of $\frac{2}{3} - \frac{(\frac{1}{2}) - (\frac{1}{3})}{(\frac{1}{2}) + (\frac{1}{3})} \times 3 \frac{1}{3} + \frac{5}{6}$ is

A. $\frac{29}{12}$

\frac{49}{54}

\frac{2}{3}

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Q. Find the value of:
(i)

(3^{0} + 4^{- 1}) \times 2^{2}

(ii)

(2^{- 1} \times 4^{- 1}) \div 2^{- 2}

(iii)

{(\frac{1}{2})}^{- 2} + {(\frac{1}{3})}^{- 2} + {(\frac{1}{4})}^{- 2}

(iv)

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(v)

{({(\frac{- 2}{3})}^{- 2})}^{2}

Q. Find the value of

{{(\frac{1}{3})}^{- 2} - {(\frac{1}{2})}^{- 3}} \div {(\frac{1}{4})}^{- 2}

Q. Simplify:

[{(\frac{1}{2})}^{- 3} + {(\frac{1}{3})}^{- 3} + {(\frac{1}{4})}^{- 3}] \div {(3)}^{2}

Q. Find the value of

∣ ∣ ∣ \frac{1}{2} - 1 ∣ ∣ ∣ - ∣ ∣ ∣ \frac{2}{3} - 1 ∣ ∣ ∣ - ∣ ∣ ∣ \frac{3}{4} - 1 ∣ ∣ ∣ - ∣ ∣ ∣ \frac{4}{5} - 1 ∣ ∣ ∣

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Find the value of (0.6)0−(0.1)−1(38)−1(32)3+(−13)−1.

The correct option is C −32Now(0.6)0−0.1−1(38)−1(32)3+(−13)−1→1−10.1138×3323+⎛⎜ ⎜ ⎜⎝1−13⎞⎟ ⎟ ⎟⎠=1−1083×3323−(31)→−92323×33−1−3=−932−3=−99−3=−96=−32

Find the value of $\frac{{(0.6)}^{0} - {(0.1)}^{- 1}}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(\frac{- 1}{3})}^{- 1}}$ .