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: 6. b) Find the value of p from 24 = 6 + 2p Step 1: Subtract 6 from both sides of the equation. 24 - 6 = 2p 18 = 2p Step 2: Divide both sides by 2. (18)/(2) = p 9 = p The value of p is 9. c) Find the value of y from equation (y^2)/(8) = 8 Step 1: Multiply both sides of the equation by 8. y^2 = 8 × 8 y^2 = 64 Step 2: Take the square root of both sides. Remember to include both positive and negative roots. y = ±sqrt(64) y = ± 8 The value of y is ± 8. SECTION C: 10 MARKS 7. a) Describe the shape that is formed by the coordinates A(2,4), B(4,1), C(2,-2), D(-4,1). Step 1: Calculate the lengths of the sides. Length AB: sqrt((4-2)^2 + (1-4)^2) = sqrt(2^2 + (-3)^2) = sqrt(4+9) = sqrt(13) Length BC: sqrt((2-4)^2 + (-2-1)^2) = sqrt((-2)^2 + (-3)^2) = sqrt(4+9) = sqrt(13) Length CD: sqrt((-4-2)^2 + (1-(-2))^2) = sqrt((-6)^2 + (3)^2) = sqrt(36+9) = sqrt(45) Length DA: sqrt((2-(-4))^2 + (4-1)^2) = sqrt(6^2 + 3^2) = sqrt(36+9) = sqrt(45) Step 2: Observe the side lengths. We have two pairs of equal-length adjacent sides: AB = BC and CD = DA. This is the definition of a kite. We can also check the diagonals: Diagonal AC connects (2,4) and (2,-2). This is a vertical line segment. Diagonal BD connects (4,1) and (-4,1). This is a horizontal line segment. Vertical and horizontal lines are perpendicular. In a kite, the diagonals are perpendicular. The shape formed by the coordinates is a kite. b) Find the average of 20 kg, 35 kg, 15 kg, 10 kg, and 20 kg. Step 1: Sum all the values. Sum = 20 + 35 + 15 + 10 + 20 = 100 kg Step 2: Count the number of values. There are 5 values. Step 3: Divide the sum by the count. Average = 100 kg5 = 20 kg The average is 20 kg. 8. a) Find the size of angle EBD. Step 1: Recognize that angles on a straight line sum to 180^. The angles CBE, EBD, and DBA lie on the straight line CA. So, CBE + EBD + DBA = 180^. Step 2: Substitute the given angle measures into the equation. 100^ + 2m + (4m + 30^) = 180^ Step 3: Combine like terms. 100 + 30 + 2m + 4m = 180 130 + 6m = 180 Step 4: Solve for m. 6m = 180 - 130 6m = 50 m = (50)/(6) = (25)/(3) Step 5: Calculate the size of angle EBD, which is 2m. EBD = 2m = 2 × (25)/(3) = (50)/(3)^ The size of angle EBD is (50)/(3)^ or approximately 16.67^. b) Find the area of the shaded part. The shaded part is a right-angled triangle. The sides are (6x-2) m, (20-5x) m, and the hypotenuse is 26 m. Step 1: Use the Pythagorean theorem to find the value of x. (6x-2)^2 + (20-5x)^2 = 26^2 (36x^2 - 24x + 4) + (400 - 200x + 25x^2) = 676 61x^2 - 224x + 404 = 676 61x^2 - 224x + 404 - 676 = 0 61x^2 - 224x - 272 = 0 This quadratic equation is difficult to factor directly. Let's check for integer solutions by testing small integer values for x. If x=4: 61(4^2) - 224(4) - 272 = 61(16) - 896 - 272 = 976 - 896 - 272 = 80 - 272 = -192 ≠ 0. If x=5: 61(5^2) - 224(5) - 272 = 61(25) - 1120 - 272 = 1525 - 1120 - 272 = 405 - 272 = 133 ≠ 0. Let's re-examine the problem. It's possible there's a simpler way or an integer value for x that makes the sides work out. If the sides are integers, we could look for Pythagorean triples. 26 is 2 × 13. Common triples are (5,12,13), so (10,24,26) is a possibility. If the legs are 10 and 24: Case 1: 6x-2 = 10 6x = 12 x=2. Then 20-5x = 20-5(2) = 20-10 = 10. This doesn't match 24. Case 2: $6x-2 = 24