Question

Evaluate:

Answer 1

\[(3^{2}/4^{2})^{-1/2}\]
\[ 3^{2}/4^{2}=(3/4)^{2}\]
then solve the rest

Answer 2

@jayanta that's not right.

Answer 3

Actually it's more like\[({9\over 16})^{-{1\over2}}=({16\over 9})^{1\over2}={16^{1\over2}\over 9^{1\over2}}={4\over3}\]

Answer 4

\[\LARGE \left(\frac{9}{16}\right)^{-\frac12}=\left(\frac{16}{9}\right)^{\frac12}\]
\[\LARGE \sqrt{\frac{16}{9}}=\frac{4}{3}\]