1. Given that $ST^{1/3} = k$, where $k$ is a constant, determine the true statement about the variation of $S$ and $T$. 2. Given $w \propto \frac{x^m}{y^n}$ such that $w$ varies directly as the cube root of $x$ and inversely as the square of $y$, find $m$ and $n$. 3. Given $x$ varies directly as $y^2$, and $x = \frac{3}{5}$ when $y = \frac{1}{3}$, express $x$ in terms of $y$. 4. Given $P$ varies inversely as the square root of $Q$ and $P = 9$ when $Q = 16$, express $P$ in terms of $Q$. 5. Given values of $R$ and $T$: $R: 5, 6$ $T: 12, 10$ Find the variation equation involving $T$ and $R$. 6. Given $F \propto x$ for spring extension $x$ and load $F$, find $F$ when $x = 20$ cm, given $F = 4$ N when $x = 16$ cm. 7. Given $w$ varies inversely as $r^2$ and $w = 10$ when $r = 4$, find $w$ when $r = 8$. --- 1. From $ST^{1/3} = k$, rearranging for $S$: $$S = \frac{k}{T^{1/3}}$$ This means $S$ varies inversely as the cube root of $T$. 2. Given $w \propto \frac{x^m}{y^n}$, and $w$ varies directly as cube root of $x$ and inversely as square of $y$: $$w \propto \sqrt[3]{x} = x^{1/3}$$ $$w \propto \frac{1}{y^2} = y^{-2}$$ So $m = \frac{1}{3}$ and $n = 2$. 3. Since $x$ varies directly as $y^2$: $$x = k y^2$$ Given $x = \frac{3}{5}$ when $y = \frac{1}{3}$: $$\frac{3}{5} = k \left(\frac{1}{3}\right)^2 = k \frac{1}{9}$$ Multiply both sides by 9: $$9 \times \frac{3}{5} = \cancel{9} k \frac{\cancel{1}}{\cancel{9}} \Rightarrow k = \frac{27}{5}$$ Therefore: $$x = \frac{27}{5} y^2$$ 4. Since $P$ varies inversely as $\sqrt{Q}$: $$P = \frac{k}{\sqrt{Q}}$$ Given $P = 9$ when $Q = 16$: $$9 = \frac{k}{\sqrt{16}} = \frac{k}{4}$$ Multiply both sides by 4: $$k = 36$$ Therefore: $$P = \frac{36}{\sqrt{Q}}$$ 5. Check if $T$ varies inversely as $R$: $$T \times R = 5 \times 12 = 60$$ $$6 \times 10 = 60$$ Since $T R = 60$ constant, variation is: $$T = \frac{60}{R}$$ 6. Given $F \propto x$, so: $$F = k x$$ Given $F = 4$ N when $x = 16$ cm: $$4 = k \times 16 \Rightarrow k = \frac{4}{16} = \frac{1}{4}$$ Find $F$ when $x = 20$ cm: $$F = \frac{1}{4} \times 20 = 5$$ 7. Given $w$ varies inversely as $r^2$: $$w = \frac{k}{r^2}$$ Given $w = 10$ when $r = 4$: $$10 = \frac{k}{4^2} = \frac{k}{16}$$ Multiply both sides by 16: $$k = 160$$ Find $w$ when $r = 8$: $$w = \frac{160}{8^2} = \frac{160}{64} = \frac{5}{2}$$