1. **Question 1: Rectangular Plot** a) Calculate the area of the rectangular plot. The formula for area of a rectangle is: $$\text{Area} = \text{length} \times \text{width}$$ Given length = 24 m, width = 15 m, $$\text{Area} = 24 \times 15 = 360 \text{ square meters}$$ b) Find the dimensions on the scale drawing with scale 1 cm : 3 m. Length on drawing: $$\frac{24}{3} = 8 \text{ cm}$$ Width on drawing: $$\frac{15}{3} = 5 \text{ cm}$$ c) Calculate the total cost of fencing around the perimeter. Perimeter of rectangle: $$P = 2(\text{length} + \text{width}) = 2(24 + 15) = 2 \times 39 = 78 \text{ m}$$ Cost of fencing: $$78 \times 12 = 936$$ 2. **Question 2: Superannuation Contributions** Boaz's annual salary = 112000 a) Employer's superannuation (15%): $$112000 \times 0.15 = 16800$$ b) Boaz's superannuation (5%): $$112000 \times 0.05 = 5600$$ c) Income less Boaz's contribution: $$112000 - 5600 = 106400$$ d) Total superannuation paid: $$16800 + 5600 = 22400$$ 3. **Question 3: Angles and Distances** a) Angle of depression from building to car is given as 30°. b) Angle of elevation from building to airplane is given as 13°. c) Angle of elevation from car to top of building equals the angle of depression from building to car, so: $$30^\circ$$ d) Distance of car from building base: Using angle of depression 30° and building height 45 m, $$\tan 30^\circ = \frac{\text{height}}{\text{distance}} \Rightarrow \text{distance} = \frac{45}{\tan 30^\circ}$$ $$\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$$ $$\text{distance} = \frac{45}{0.577} \approx 78 \text{ meters}$$ 4. **Question 4: Construction Site Areas and Perimeter** a) Area A1: Given dimensions 12 m x 4 m, $$\text{Area } A1 = 12 \times 4 = 48 \text{ m}^2$$ b) Area A2: (likely rectangle 8 m x 4 m), $$A2 = 8 \times 4 = 32 \text{ m}^2$$ c) Area A3: (likely 2 m x 15 m), $$A3 = 2 \times 15 = 30 \text{ m}^2$$ d) Total area: $$48 + 32 + 30 = 110 \text{ m}^2$$ e) Perimeter of composite object (sum of all outer sides): Sum lengths: top 12, right side 4 + 4 + 15 = 23, bottom 2 + 8 = 10, left side 15 Perimeter: $$12 + 23 + 10 + 15 = 60 \text{ m}$$ 5. **Question 5: Quadratic Function Analysis** Given $$y = -\frac{3}{2}(x+2)^2 - 4$$ a) Axis of symmetry is vertical line at $$x = -2$$ b) Vertex is $$(-2, -4)$$ c) y-intercept: put $$x = 0$$ $$y= -\frac{3}{2}(0+2)^2 -4 = -\frac{3}{2} \times 4 - 4 = -6 -4 = -10$$ So y-intercept is $$-10$$ d) Expand function: $$(x+2)^2 = x^2 + 4x + 4$$ $$y = -\frac{3}{2}x^2 - 6x - 6 - 4 = -\frac{3}{2}x^2 - 6x - 10$$ e) Discriminant $$\Delta = b^2 - 4ac$$ where $$a = -\frac{3}{2}$$, $$b = -6$$, $$c = -10$$ $$\Delta = (-6)^2 - 4 \times -\frac{3}{2} \times -10 = 36 - 4 \times \frac{3}{2} \times 10 = 36 - 60 = -24$$ Negative discriminant means no real roots. 6. **Question 6: Rainfall Data Analysis** a) Range = max - min rainfall = 280 - 0 = 280 mm b) Days with rainfall <161 mm: Sum days in 0-40,41-80,81-120,121-160: $$10 + 16 + 25 + 30 = 81 \text{ days}$$ c) Days with rainfall between 45 and 200 mm: 41-80 (16 days), 81-120 (25), 121-160 (30), 161-200 (22) Total: $$16 + 25 + 30 + 22 = 93$$ days. But total given is 90 days, so possibly a data misinterpretation. Considering that 45 to 200 mm covers 41-80 to 161-200 bands, Total days: $$16 + 25 + 30 + 22 = 93$$ (cross-check with total of 90 days; so likely question expects sum of 16 + 25 + 30 + 22 = 93 or assume typo, accept 93.) d) Most frequent rainfall is the class with maximum days: 121-160 mm with 30 days e) Days with rainfall > 200 mm: 201-240 (12) + 241-280 (5) = 17 days 7. **Question 7: Angles T, W, X, Y, Z in Diagram** Given SS = 65° and one angle 55°, assume these are alternate or corresponding angles. Since no numeric values for T, W, X, Y, Z besides what is given and only SS=65°, T = 65° (as SS likely stands for supplementary angles). If needed, assuming angles T + 65° = 180° (linear pair), then: $$T = 180° - 65° = 115°$$ Without diagram values for others, cannot calculate precisely. 8. **Question 8: Regression Equation S = 2.2A + 1400** a) Rate of increase in sales per K1 advertising spent is 2.2 b) Predicted sales for K10,000 advertising: $$S = 2.2 \times 10000 + 1400 = 22000 + 1400 = 23400$$ c) Advertising for sales of K16,000: $$16000 = 2.2A + 1400 \Rightarrow 2.2A = 16000 -1400 = 14600$$ $$A = \frac{14600}{2.2} = 6636.36$$ d) Profit from K30,000 advertising (Assuming profit equals sales): $$S = 2.2 \times 30000 + 1400 = 66000 + 1400 = 67400$$ 9. **Question 9: Solve System and Quadratic Equation** a) Solve system: $$2x - y = 5$$ $$x + 4y = 7$$ From first equation: $$y = 2x - 5$$ Substitute into second: $$x + 4(2x - 5) = 7$$ $$x + 8x - 20 = 7$$ $$9x = 27$$ $$x = 3$$ Then: $$y = 2(3) - 5 = 6 - 5 = 1$$ Solution: $$(3, 1)$$ b) Solve quadratic using formula for $$x^2 - 2x - 15 = 0$$ Here $$a=1, b=-2, c=-15$$ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2} = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2}$$ Roots: $$x = \frac{2 + 8}{2} = 5$$ $$x = \frac{2 - 8}{2} = -3$$ 10. **Question 10: Graph Questions** a) Slope of straight line passing through (-5, 0) and (10,5): $$m = \frac{5 - 0}{10 - (-5)} = \frac{5}{15} = \frac{1}{3}$$ b) Parabola's y-intercept at $$x=0$$, given near (0,0), so approximately 0 c) Equation of straight line using point-slope: $$y - 0 = \frac{1}{3}(x + 5) \Rightarrow y = \frac{1}{3}x + \frac{5}{3}$$ d) Equation of parabola given vertex (-2.5, -15), opening upwards. General form: $$y = a(x + 2.5)^2 - 15$$ Using y-intercept at (0,0): $$0 = a (0 + 2.5)^2 - 15 = a(6.25) - 15 \Rightarrow a = \frac{15}{6.25} = 2.4$$ So: $$y = 2.4 (x + 2.5)^2 - 15$$ e) X-intercepts: solve $$y=0$$ $$0 = 2.4 (x + 2.5)^2 - 15 \Rightarrow (x + 2.5)^2 = \frac{15}{2.4} = 6.25$$ $$x + 2.5 = \pm 2.5$$ Roots: $$x = -2.5 + 2.5 = 0$$ $$x = -2.5 - 2.5 = -5$$