1. **State the problem:** Simplify the expression $9x^5 y^2 \times (-2x)^{-2} y^{-2}$. 2. **Recall the rules:** - When multiplying powers with the same base, add the exponents: $a^m \times a^n = a^{m+n}$. - A negative exponent means reciprocal: $a^{-n} = \frac{1}{a^n}$. 3. **Rewrite the expression:** $$9x^5 y^2 \times (-2x)^{-2} y^{-2} = 9x^5 y^2 \times \frac{1}{(-2x)^2} \times y^{-2}$$ 4. **Calculate $(-2x)^2$:** $$(-2x)^2 = (-2)^2 \times x^2 = 4x^2$$ 5. **Substitute back:** $$9x^5 y^2 \times \frac{1}{4x^2} \times y^{-2} = \frac{9x^5 y^2}{4x^2} \times y^{-2}$$ 6. **Simplify powers of $x$ and $y$:** - For $x$: $x^5 / x^2 = x^{5-2} = x^3$ - For $y$: $y^2 \times y^{-2} = y^{2-2} = y^0 = 1$ 7. **Final simplified expression:** $$\frac{9}{4} x^3$$ **Answer:** $\frac{9}{4} x^3$