Chapter 5 - Exponents and Polynomials - 5.6 - Integral Exponents and Scientific Notation - Problem Set 5.6 - Page 223: 6

$\dfrac{16}{9}$

Using $a^{-x}=\dfrac{1}{a^x}$ or $\dfrac{1}{a^{-x}}=a^x$ and the laws of exponents, the given expression, $ \left( \dfrac{3}{4} \right)^{-2} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{3^{-2}}{4^{-2}} \\\\= \dfrac{4^2}{3^2} \\\\= \dfrac{16}{9} .\end{array}