1. **Problem c:** A solid metal prism with volume 500 cm^3 is melted and made into 6 identical spheres. Calculate the radius of each sphere correct to the nearest cm. Step 1: Find the volume of one sphere by dividing the total volume by 6. $$V_{sphere} = \frac{500}{6} \approx 83.33 \text{ cm}^3$$ Step 2: Use the formula for the volume of a sphere: $$V = \frac{4}{3} \pi r^3$$ Step 3: Solve for radius $r$: $$r^3 = \frac{3V}{4\pi} = \frac{3 \times 83.33}{4\pi} \approx \frac{250}{12.566} \approx 19.89$$ Step 4: Calculate $r$: $$r = \sqrt[3]{19.89} \approx 2.71 \text{ cm}$$ Step 5: Round to nearest cm: $$r \approx 3 \text{ cm}$$ --- 2. **Problem 8:** A bag contains 6 red, 4 blue, and 5 green counters. i) Find the probability that a counter chosen at random is not green. Step 1: Total counters: $$6 + 4 + 5 = 15$$ Step 2: Counters not green: $$6 + 4 = 10$$ Step 3: Probability: $$P(\text{not green}) = \frac{10}{15} = \frac{2}{3}$$ ii) Two counters are drawn without replacement. a) Probability both are blue. Step 1: Probability first blue: $$\frac{4}{15}$$ Step 2: Probability second blue after first blue taken: $$\frac{3}{14}$$ Step 3: Combined probability: $$\frac{4}{15} \times \frac{3}{14} = \frac{12}{210} = \frac{2}{35}$$ b) Probability first is green and second is red. Step 1: Probability first green: $$\frac{5}{15} = \frac{1}{3}$$ Step 2: Probability second red after first green taken: $$\frac{6}{14} = \frac{3}{7}$$ Step 3: Combined probability: $$\frac{1}{3} \times \frac{3}{7} = \frac{1}{7}$$ --- 3. **Problem 9:** i) The 3rd term of an arithmetic sequence is 12 and the 8th term is 27. Find the first term and common difference. Step 1: Use formula for nth term: $$a_n = a + (n-1)d$$ Step 2: Write equations: $$a + 2d = 12$$ $$a + 7d = 27$$ Step 3: Subtract first from second: $$5d = 15 \Rightarrow d = 3$$ Step 4: Substitute $d$ back: $$a + 2 \times 3 = 12 \Rightarrow a = 12 - 6 = 6$$ ii) The first four terms of a sequence are 5, 12, 23, 38, ... a) Find an expression for the nth term. Step 1: Find differences: $$12 - 5 = 7, \quad 23 - 12 = 11, \quad 38 - 23 = 15$$ Step 2: Second differences: $$11 - 7 = 4, \quad 15 - 11 = 4$$ Step 3: Since second difference is constant, sequence is quadratic: $$a_n = An^2 + Bn + C$$ Step 4: Use first three terms to form equations: $$A(1)^2 + B(1) + C = 5$$ $$A(2)^2 + B(2) + C = 12$$ $$A(3)^2 + B(3) + C = 23$$ Step 5: Simplify: $$A + B + C = 5$$ $$4A + 2B + C = 12$$ $$9A + 3B + C = 23$$ Step 6: Subtract first from second: $$3A + B = 7$$ Step 7: Subtract second from third: $$5A + B = 11$$ Step 8: Subtract these two: $$(5A + B) - (3A + B) = 11 - 7 \Rightarrow 2A = 4 \Rightarrow A = 2$$ Step 9: Substitute $A=2$ into $3A + B = 7$: $$3(2) + B = 7 \Rightarrow B = 7 - 6 = 1$$ Step 10: Substitute $A=2$, $B=1$ into $A + B + C = 5$: $$2 + 1 + C = 5 \Rightarrow C = 2$$ Step 11: Final formula: $$a_n = 2n^2 + n + 2$$ b) Find the 10th term: Step 1: Substitute $n=10$: $$a_{10} = 2(10)^2 + 10 + 2 = 2(100) + 10 + 2 = 200 + 12 = 212$$ --- 4. **Problem 10:** Calculate the perimeter of the grass lawn which is the shaded region between two sectors. Given: - Larger sector radius $R = 10$ m - Smaller sector radius $r = 2$ m - Sector angle $\theta = 40^\circ$ Step 1: Convert angle to radians: $$\theta = 40^\circ = \frac{40 \pi}{180} = \frac{2\pi}{9} \text{ radians}$$ Step 2: Calculate arc length of larger sector: $$L_{large} = R \times \theta = 10 \times \frac{2\pi}{9} = \frac{20\pi}{9} \text{ m}$$ Step 3: Calculate arc length of smaller sector: $$L_{small} = r \times \theta = 2 \times \frac{2\pi}{9} = \frac{4\pi}{9} \text{ m}$$ Step 4: Calculate perimeter of grass lawn: It consists of two straight lines (DE and FG) which are the difference of radii, and two arcs (EF and DG). Straight lines DE and FG each equal: $$R - r = 10 - 2 = 8 \text{ m}$$ Step 5: Total perimeter: $$P = L_{large} + L_{small} + 2(R - r) = \frac{20\pi}{9} + \frac{4\pi}{9} + 2 \times 8 = \frac{24\pi}{9} + 16 = \frac{8\pi}{3} + 16$$ Step 6: Approximate numeric value: $$\frac{8\pi}{3} \approx 8.38$$ $$P \approx 8.38 + 16 = 24.38 \text{ m}$$ Final answer rounded to two decimal places: $$\boxed{24.38 \text{ m}}$$