This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the problems from the image. 1. Curve and Line The curve is given by y = (6-x)(x+2). Expanding the curve equation: y = 6x + 12 - x^2 - 2x y = -x^2 + 4x + 12 The line is given by y = mx+c. a) Find the coordinates of the point A and B. From the graph, points A and B are the intersection points of the curve and the line. Point A is where x=0. Substitute x=0 into the curve equation: y = (6-0)(0+2) = 6 × 2 = 12 So, point A is (0, 12). Point B is where x=4. Substitute x=4 into the curve equation: y = (6-4)(4+2) = 2 × 6 = 12 So, point B is (4, 12). The coordinates are A(0, 12) and B(4, 12). b) Determine the equation of the straight line y = mx+c. The line passes through points A (0, 12) and B (4, 12). The slope m is given by: m = (y_2 - y_1)/(x_2 - x_1) = (12 - 12)/(4 - 0) = (0)/(4) = 0 The y-intercept c is the y-coordinate when x=0, which is 12 (from point A). So, the equation of the straight line is y = 0x + 12. The equation of the straight line is y = 12. c) Find the area between the curve y = (6-x)(x+2) and the line y = mx+c. The area between the curve y = -x^2 + 4x + 12 and the line y = 12 is given by the integral of the difference between the upper function and the lower function, from x=0 to x=4. In the interval [0, 4], the curve is above the line. Area = _0^4 ((-x^2 + 4x + 12) - 12) \,dx Area = _0^4 (-x^2 + 4x) \,dx Integrate the expression: (-x^2 + 4x) \,dx = -(x^3)/(3) + (4x^2)/(2) = -(x^3)/(3) + 2x^2 Now, evaluate the definite integral: Area = [ -(x^3)/(3) + 2x^2 ]_0^4 Area = ( -(4^3)/(3) + 2(4^2) ) - ( -(0^3)/(3) + 2(0^2) ) Area = ( -(64)/(3) + 2(16) ) - (0) Area = -(64)/(3) + 32 Area = (-64 + 96)/(3) Area = (32)/(3) The area between the curve and the line is (32)/(3) square units. 2. Square-based pyramid ABCD V Given: Square base ABCD with AD = DC = 6 cm. Height of pyramid OV = 10 cm, where O is the center of the base. a) State the projection of VA on the base ABCD. The projection of the vertex V onto the base is the point O. The point A is already on the base. Therefore, the projection of VA on the base ABCD is the line segment OA. b) Find: i) The length VA. First, find the length of the diagonal AC of the square base. Using the Pythagorean theorem in triangle ADC: AC^2 = AD^2 + DC^2 AC^2 = 6^2 + 6^2 = 36 + 36 = 72 AC = sqrt(72) = 6sqrt(2) cm The length OA is half of the diagonal AC: OA = (1)/(2) AC = (1)/(2) (6sqrt(2)) = 3sqrt(2) cm Now, consider the right-angled triangle VOA. VO is the height of the pyramid (10 cm) and OA is the distance from the center to a vertex of the base. Using the Pythagorean theorem in triangle VOA: VA^2 = VO^2 + OA^2 VA^2 = 10^2 + (3sqrt(2))^2 VA^2 = 100 + (9 × 2) VA^2 = 100 + 18 = 118 VA = sqrt(118) cm VA ≈ 1