Question

Find the values of [6^-1+(3/2)^-1]^-2​

Answer 1

Answer:

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Answer 2

Answer:

$[6^{-1} + (\frac{3}{2})^{-1}]^{-2}$ = $\frac{36}{25}$

Step-by-step explanation:

here, $[6^{-1} + (\frac{3}{2})^{-1}]^{-2}$

      = $[\frac{1}{6} + \frac{2}{3}]^{-2}$                      [∵ $a^{-1} = \frac{1}{a}^{1}$]

      = $(\frac{5}{6})^{-2}$

      = $(\frac{6}{5})^{2}$

      = $\frac{36}{25}$

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