1. **State the problem:** Evaluate the expression $$\frac{2x^2}{y^{-1}}$$ when $$x=3$$ and $$y=-4$$. 2. **Recall the formula and rules:** The expression is a fraction with numerator $$2x^2$$ and denominator $$y^{-1}$$. Remember that $$y^{-1} = \frac{1}{y}$$, so dividing by $$y^{-1}$$ is the same as multiplying by $$y$$. 3. **Substitute the values:** $$\frac{2(3)^2}{(-4)^{-1}}$$ 4. **Calculate powers:** $$3^2 = 9$$ So the expression becomes: $$\frac{2 \times 9}{(-4)^{-1}} = \frac{18}{(-4)^{-1}}$$ 5. **Rewrite the denominator:** $$(-4)^{-1} = \frac{1}{-4}$$ 6. **Divide by a fraction:** Dividing by $$\frac{1}{-4}$$ is the same as multiplying by $$-4$$: $$\frac{18}{\frac{1}{-4}} = 18 \times (-4)$$ 7. **Multiply:** $$18 \times (-4) = -72$$ **Final answer:** $$-72$$