The value of -0-6-0 - -0-1-1-3-2-3-1 -3-2-3 - -1-3-1 is -3-2-3-22-3-1-2

The correct option is B 3/2On solving the given expression, (0.6)^0 (0.1)^ 1/(3/2^3)^ 1 (3/2)^3 + ( 1/3)^ 1 we get 1 1/(0.1)^1/(2^3/3)^1 (3/2)^3 + ( 3/1) ...

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Byju's Answer

Standard IX

Mathematics

Laws of Exponents for Real Numbers

The value of ...

Question

The value of $\frac{(0.6)^{0} - (0.1)^{- 1}}{(\frac{3}{2^{3}})^{- 1} (\frac{3}{2})^{3} + (- \frac{1}{3})^{- 1}}$ is _____________.

\frac{3}{2}

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\frac{- 3}{2}

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\frac{2}{3}

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\frac{- 1}{2}

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Solution

The correct option is B $\frac{- 3}{2}$
On solving the given expression,

$\frac{(0.6)^{0} - (0.1)^{- 1}}{(\frac{3}{2^{3}})^{- 1} (\frac{3}{2})^{3} + (- \frac{1}{3})^{- 1}}$

we get

$\frac{1 - \frac{1}{(0.1)^{1}}}{(\frac{2^{3}}{3})^{1} (\frac{3}{2})^{3} + (\frac{- 3}{1})^{1}}$

$[∵ x^{0} = 1 a n d (\frac{a}{b})^{- n} = (\frac{b}{a})^{n}]$

$= \frac{1 - 10}{\frac{2^{3}}{3} \times \frac{3^{3}}{2^{3}} + (- 3)} [(\frac{a}{b})^{n} = \frac{a^{n}}{b^{n}}]$

$= \frac{- 9}{3^{2} + (- 3)} = \frac{- 9}{9 - 3} = - \frac{9}{6} = - \frac{3}{2}$
$∴ Answer = \frac{- 3}{2}$

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The value of (0.6)0−(0.1)−1(323)−1(32)3+(−13)−1 is _____________.

The correct option is B −32On solving the given expression, (0.6)0−(0.1)−1(323)−1(32)3+(−13)−1 we get 1−1(0.1)1(233)1(32)3+(−31)1 [∵x0=1 and (ab)−n=(ba)n] =1−10233×3323+(−3)[(ab)n=anbn] =−932+(−3)=−99−3=−96=−32 ∴Answer=−32

The value of $\frac{(0.6)^{0} - (0.1)^{- 1}}{(\frac{3}{2^{3}})^{- 1} (\frac{3}{2})^{3} + (- \frac{1}{3})^{- 1}}$ is _____________.