How do you simplify ${(- (\frac{2}{3}))}^{- 2}$ $(-(2/3))^-2$ ?

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Answer 1 · 2016-09-11T13:42:39

The result is $\frac{9}{4}$ $9/4$ or $2 \frac{1}{4}$ $2 1/4$

${(- (\frac{2}{3}))}^{- 2} = {(- \frac{3}{2})}^{2} = \frac{9}{4}$ $(-(2/3))^(-2)=(-3/2)^2=9/4$

In the calculation I used the following property of powers:

$a^{- b} = {(\frac{1}{a})}^{b}$ $a^-b=(1/a)^b$

Answer 2 · 2016-09-11T13:43:22

$\frac{9}{4}$ $(9) / (4)$

We have: ${(- (\frac{2}{3}))}^{- 2}$ $(- ((2) / (3)))^(- 2)$

First, let's expand the parentheses:

$= {(- \frac{2}{3})}^{- 2}$ $= (- (2) / (3))^(- 2)$

Then, using the laws of exponents:

$= \frac{1}{{(- \frac{2}{3})}^{2}}$ $= (1) / ((- (2) / (3))^(2))$

$= \frac{1}{{(\frac{- 2}{3})}^{2}}$ $= (1) / (((- 2) / (3))^(2))$

$= \frac{1}{(\frac{{(- 2)}^{2}}{{(3)}^{2}})}$ $= (1) / (((- 2)^(2) / (3)^(2)))$

$= \frac{1}{(\frac{4}{9})}$ $= (1) / (((4) / (9)))$

$= 1 \cdot (\frac{9}{4})$ $= 1 cdot ((9) / (4))$

$= \frac{9}{4}$ $= (9) / (4)$

How do you simplify (−(23))−2(-(2/3))^-2?