Chapter 5 - Exponents and Polynomials - Chapter 5 Review Problem Set - Page 231: 55

$\dfrac{16}{9}$

Using $a^{-x}=\dfrac{1}{a^x}$, the given expression, $ \left( \dfrac{3}{4} \right)^{-2} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{1}{\left( \dfrac{3}{4} \right)^2} \\\\= \dfrac{1}{\dfrac{9}{16}} \\\\= 1\div\dfrac{9}{16} \\\\= 1\cdot\dfrac{16}{9} \\\\= \dfrac{16}{9} .\end{array}