1. **Stating the problem:** We have two quantities whose sum is $y$. One varies directly as $x^2$ and the other varies inversely as $x$. Given $y=32$ when $x=2$ and $y=86$ when $x=4$, we need to find: (i) an equation for $y$ in terms of $x$. (ii) the value of $y$ when $x=3$. 2. **Setting up the equation:** Let the first quantity be $A$ and the second be $B$. Since $A$ varies directly as $x^2$, we write $A = kx^2$ for some constant $k$. Since $B$ varies inversely as $x$, we write $B = \frac{m}{x}$ for some constant $m$. The sum is $y = A + B = kx^2 + \frac{m}{x}$. 3. **Using given values to find constants:** When $x=2$, $y=32$: $$32 = k(2)^2 + \frac{m}{2} = 4k + \frac{m}{2}$$ When $x=4$, $y=86$: $$86 = k(4)^2 + \frac{m}{4} = 16k + \frac{m}{4}$$ 4. **Solving the system of equations:** Multiply the first equation by 4 to clear denominators: $$4 \times 32 = 4 \times 4k + 4 \times \frac{m}{2} \Rightarrow 128 = 16k + 2m$$ The second equation is: $$86 = 16k + \frac{m}{4}$$ 5. **Subtract the second equation from the first:** $$128 - 86 = (16k + 2m) - (16k + \frac{m}{4})$$ $$42 = 2m - \frac{m}{4} = \frac{8m}{4} - \frac{m}{4} = \frac{7m}{4}$$ Multiply both sides by $\frac{4}{7}$: $$m = \frac{42 \times 4}{7} = 24$$ 6. **Find $k$ using $m=24$ in one equation:** From $86 = 16k + \frac{24}{4}$: $$86 = 16k + 6$$ Subtract 6: $$80 = 16k$$ Divide both sides by 16: $$k = \frac{80}{16} = 5$$ 7. **Write the equation for $y$:** $$y = 5x^2 + \frac{24}{x}$$ 8. **Find $y$ when $x=3$:** $$y = 5(3)^2 + \frac{24}{3} = 5 \times 9 + 8 = 45 + 8 = 53$$ --- 9. **Part (b): Express 0.0098 in standard form:** Standard form means writing the number as $a \times 10^n$ where $1 \leq a < 10$ and $n$ is an integer. Move the decimal point 3 places to the right: $$0.0098 = 9.8 \times 10^{-3}$$ **Final answers:** (i) $y = 5x^2 + \frac{24}{x}$ (ii) $y = 53$ when $x=3$ (b) $0.0098 = 9.8 \times 10^{-3}$