1. **State the problem:** a. Write down the inequality shown on the number line with an open circle at -3 and a closed circle at 2 connected by a solid line. b. Solve the quadratic equation $5x^2 - 13x - 12 = 0$ using the quadratic formula and give answers correct to 2 decimal places. c. Given two mathematically similar solids with volumes 416 cm$^3$ (large) and 52 cm$^3$ (small), and the total surface area of the small solid is 60 cm$^2$, find the total surface area of the large solid. 2. **Part a: Write the inequality** The open circle at -3 means $x > -3$ (not including -3). The closed circle at 2 means $x \leq 2$ (including 2). The solid line connects these points, so the inequality is: $$-3 < x \leq 2$$ 3. **Part b: Solve $5x^2 - 13x - 12 = 0$ using the quadratic formula** The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=-13$, and $c=-12$. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-13)^2 - 4 \times 5 \times (-12) = 169 + 240 = 409$$ Calculate the square root: $$\sqrt{409} \approx 20.2237$$ Find the two solutions: $$x_1 = \frac{13 + 20.2237}{2 \times 5} = \frac{33.2237}{10} = 3.32$$ $$x_2 = \frac{13 - 20.2237}{10} = \frac{-7.2237}{10} = -0.72$$ Rounded to 2 decimal places: $$x = 3.32 \quad \text{or} \quad x = -0.72$$ 4. **Part c: Calculate the total surface area of the large solid** Since the solids are mathematically similar, the ratio of volumes is related to the scale factor $k$ by: $$\frac{V_{large}}{V_{small}} = k^3$$ Calculate $k$: $$k^3 = \frac{416}{52} = 8 \implies k = \sqrt[3]{8} = 2$$ The surface area scales as the square of the scale factor: $$\frac{A_{large}}{A_{small}} = k^2 = 2^2 = 4$$ Given $A_{small} = 60$ cm$^2$, find $A_{large}$: $$A_{large} = 4 \times 60 = 240 \text{ cm}^2$$ **Final answers:** - a. $$-3 < x \leq 2$$ - b. $$x = 3.32 \text{ or } x = -0.72$$ - c. $$240 \text{ cm}^2$$