1. **State the problem:** (a) Write down the inequality shown on the graph. (b) Use the quadratic formula to solve the equation $$5x^2 - 13x - 12 = 0$$ and give answers correct to 2 decimal places. (c) Calculate the total surface area of the large solid given the volumes and surface area of the small solid. 2. **Part (a) - Inequality from graph:** The graph shows a line segment from $$x = -3$$ (open circle, not included) to $$x = 2$$ (closed circle, included). So the inequality is: $$-3 < x \leq 2$$ 3. **Part (b) - Quadratic formula:** The quadratic formula for solving $$ax^2 + bx + c = 0$$ is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a = 5$$, $$b = -13$$, $$c = -12$$. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-13)^2 - 4 \times 5 \times (-12) = 169 + 240 = 409$$ Calculate the square root: $$\sqrt{409} \approx 20.2237$$ Calculate the two roots: $$x_1 = \frac{13 + 20.2237}{10} = \frac{33.2237}{10} = 3.32$$ $$x_2 = \frac{13 - 20.2237}{10} = \frac{-7.2237}{10} = -0.72$$ 4. **Part (c) - Surface area of large solid:** Since the solids are mathematically similar, the ratio of volumes is related to the cube of the scale factor $$k$$: $$\frac{V_{large}}{V_{small}} = k^3$$ Given: $$V_{large} = 416$$ $$V_{small} = 52$$ Calculate $$k$$: $$k^3 = \frac{416}{52} = 8 \implies k = \sqrt[3]{8} = 2$$ The surface area scales as the square of $$k$$: $$\frac{A_{large}}{A_{small}} = k^2$$ Given $$A_{small} = 60$$, calculate $$A_{large}$$: $$A_{large} = 60 \times 2^2 = 60 \times 4 = 240$$ **Final answers:** (a) $$-3 < x \leq 2$$ (b) $$x = 3.32$$ or $$x = -0.72$$ (c) Total surface area of large solid = $$240$$ cm\textsuperscript{2}