1. **State the problem:** Given that $P$ varies directly as $Q$ and the cube of $R$, find the values of $x$ and $y$ from the table where $P=0.012$, $Q=0.5$, $R=0.2$ for the first set, and $P=0.0081$, $Q=y$, $R=0.3$ for the second set, and $x$ corresponds to $P$ when $Q=4$ and $R=2$. 2. **Formula and explanation:** Since $P$ varies directly as $Q$ and $R^3$, we write: $$P = k Q R^3$$ where $k$ is the constant of proportionality. 3. **Find $k$ using the first set of values:** $$0.012 = k \times 0.5 \times (0.2)^3$$ Calculate $0.2^3$: $$0.2^3 = 0.008$$ Substitute: $$0.012 = k \times 0.5 \times 0.008$$ Simplify: $$0.012 = k \times 0.004$$ Divide both sides by $0.004$: $$k = \frac{0.012}{0.004}$$ Show cancellation: $$k = \frac{\cancel{0.012}}{\cancel{0.004}} = 3$$ 4. **Calculate $x$ when $Q=4$ and $R=2$:** $$x = k \times 4 \times (2)^3$$ Calculate $2^3$: $$2^3 = 8$$ Substitute: $$x = 3 \times 4 \times 8$$ Multiply: $$x = 96$$ 5. **Calculate $y$ when $P=0.0081$ and $R=0.3$:** $$0.0081 = 3 \times y \times (0.3)^3$$ Calculate $0.3^3$: $$0.3^3 = 0.027$$ Substitute: $$0.0081 = 3 \times y \times 0.027$$ Simplify: $$0.0081 = 0.081 y$$ Divide both sides by $0.081$: $$y = \frac{0.0081}{0.081}$$ Show cancellation: $$y = \frac{\cancel{0.0081}}{\cancel{0.081}} = 0.1$$ **Final answers:** $$x = 96$$ $$y = 0.1$$