1. Make r the subject in the equation $x = \frac{3r}{r+2}$. Step 1: Multiply both sides by $(r+2)$ to eliminate the denominator: $$x(r+2) = 3r$$ Step 2: Distribute $x$: $$xr + 2x = 3r$$ Step 3: Get all terms involving $r$ on one side: $$xr - 3r = -2x$$ Step 4: Factor out $r$: $$r(x - 3) = -2x$$ Step 5: Divide both sides by $(x - 3)$: $$r = \frac{-2x}{x - 3}$$ 2. Find the value of $x$ when $\cos x = -0.866$ for $0^\circ < x < 180^\circ$. Step 1: Recall that $\cos 150^\circ = -\frac{\sqrt{3}}{2} \approx -0.866$. Step 2: Since cosine is negative in the second quadrant and $0^\circ < x < 180^\circ$, the solution is: $$x = 150^\circ$$ 3. Eugene bought a car for 55000 at the end of 2024. The car depreciates by 10% every year. Find the value at the end of 2026. Step 1: Depreciation rate per year is 10%, so the value after each year is multiplied by $0.9$. Step 2: Number of years from end 2024 to end 2026 is 2. Step 3: Calculate the value: $$\text{Value} = 55000 \times 0.9^2 = 55000 \times 0.81 = 44550$$ Final answers: 1. $r = \frac{-2x}{x - 3}$ 2. $x = 150^\circ$ 3. Value at end of 2026 is 44550