Question 180iSimplify- 1-22 - 1-4 3 -1- 2-3

Given, [ ( 1/2 )^2 ( 1/4 )^3 ]^ 1× 2^ 3 = ( 1/4 1/64)^ 1× 2^ 3 = ( 16 1/64)^ 1× 2^ 3 = ( 15/64)^ 1× 2^ 3 =64/15×1/8 [ ∵ a^ m = 1/a^m] = 8/15 ...

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Standard IX

Mathematics

Product Law

Question 180i...

Question

Question 180(i)

Simplify

${[{(\frac{1}{2})}^{2} - {(\frac{1}{4})}^{3}]}^{- 1} \times 2^{- 3}$

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Solution

Given, ${[{(\frac{1}{2})}^{2} - {(\frac{1}{4})}^{3}]}^{- 1} \times 2^{- 3} = {(\frac{1}{4} - \frac{1}{64})}^{- 1} \times 2^{- 3} = {(\frac{16 - 1}{64})}^{- 1} \times 2^{- 3} = {(\frac{15}{64})}^{- 1} \times 2^{- 3} = \frac{64}{15} \times \frac{1}{8} [∵ a^{- m} = \frac{1}{a^{m}}] = \frac{8}{15}$

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Similar questions

Question 180(i)

Simplify

${[{(\frac{1}{2})}^{2} - {(\frac{1}{4})}^{3}]}^{- 1} \times 2^{- 3}$

Find $(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . . .$ to 10 terms.

Show that $\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . . + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

Q. Find the reciprocal of

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

Question 180(vi)

Simplify :
$\frac{(3^{- 2})^{2} \times (5^{2})^{- 3} \times (t^{- 3})^{2}}{(3^{- 2})^{5} \times (5^{3})^{(- 2)} \times (t^{- 4})^{3}}$

Question 180(vi)

Simplify :
$\frac{(3^{- 2})^{2} \times (5^{2})^{- 3} \times (t^{- 3})^{2}}{(3^{- 2})^{5} \times (5^{3})^{(- 2)} \times (t^{- 4})^{3}}$

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Question 180(i) Simplify [(12)2−(14)3]−1×2−3

Given, [(12)2−(14)3]−1×2−3=(14−164)−1×2−3=(16−164)−1×2−3=(1564)−1×2−3=6415×18 [∵a−m=1am]=815

Question 180(i)

Simplify

${[{(\frac{1}{2})}^{2} - {(\frac{1}{4})}^{3}]}^{- 1} \times 2^{- 3}$

Given, ${[{(\frac{1}{2})}^{2} - {(\frac{1}{4})}^{3}]}^{- 1} \times 2^{- 3} = {(\frac{1}{4} - \frac{1}{64})}^{- 1} \times 2^{- 3} = {(\frac{16 - 1}{64})}^{- 1} \times 2^{- 3} = {(\frac{15}{64})}^{- 1} \times 2^{- 3} = \frac{64}{15} \times \frac{1}{8} [∵ a^{- m} = \frac{1}{a^{m}}] = \frac{8}{15}$