Question

simplify
$(2 {}^{ - 1} \div 3 {}^{ - 1} ) {}^{3} \times ( \frac{ - 9}{4} ) {}^{ - 2}$
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Answer 1

Answer:

Here's your solution in above attachments.

Hope it's helpful..

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Answer 2

Answer:

$\dfrac{2}{3}$

Step-by-step explanation:

To Solve:

$= (2 {}^{ - 1} \div 3 {}^{ - 1} ){}^{3} \times ( \dfrac{-9}{4} ){}^{-2}$

Solution:

$= (2 {}^{ - 1} \div 3 {}^{ - 1} ){}^{3} \times ( \frac{-9}{4} ){}^{-2} \\ \\ = {( \frac{1}{2} \div \frac{1}{3} )}^{3} \times {( - \frac{4}{9})}^{2} \\ \\ = {( \frac{1}{2} \times 3) }^{3} \times {( \frac{4}{9})}^{2} \\ \\ = {( \frac{3}{2})}^{3} \times \{{ { (\frac{2}{3})}^{2} \}}^{2} \\ \\ ={( \frac{3}{2})}^{3} \times {( \frac{2}{3}) }^{4} \\ \\ = {( \frac{2}{3})}^{ - 3} \times {( \frac{2}{3}) }^{4} \\ \\ = {( \frac{2}{3}) }^{(4 - 3)} \\ \\ = {( \frac{2}{3}) }^{1} \\ \\ = \frac{2}{3}$

How you can solve ?

This question is related to Laws of Indices.

1. Laws of Indices definitely used.

2. First break the exponent in smallest form i.e 4 = 2² , 9 = 3²

3. Use BODMAS

4. If there is a negative sign in exponent but the the power is even (2,4,8...) So negative sing will no longer help.

[As (-)×(-) = + ]

5. Simplify till Answer

Simple isn't it ?

${\boxed{\red{\textsf{Some Important Laws of Indices}}}}$

${a}^{n}.{a}^{m}={a}^{(n + m)}$

${a}^{-1}=\dfrac{1}{a}$

$\dfrac{{a}^{n}}{ {a}^{m}}={a}^{(n-m)}$

${({a}^{c})}^{b}={a}^{b\times c}={a}^{bc}$

${a}^{\frac{1}{x}}=\sqrt[x]{a} \\\\ a^0 = 1$

$[\text{where all variables are real and greater than 0}]$