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 Question 4. Question 4.1: The graph k is a parabola with the equation k(x) = ax^2. From the diagram, the parabola opens downwards, which means a < 0. The vertex of this parabola is at the origin (0,0), which is its maximum point. The range of k is all y-values less than or equal to the maximum value. Range of k: y 0 Question 4.2: The graph p is a straight line. The domain of any linear function is all real numbers. Domain of p: x R Question 4.3: Given that the gradient of AD (which is line p) is 1, and its equation is p(x) = mx + c. Step 1: Substitute the given gradient m=1 into the equation. p(x) = 1x + c p(x) = x + c Step 2: Use point C(0;-6), which lies on line p, to find the value of c. Substitute x=0 and p(x)=-6: -6 = 0 + c c = -6 Step 3: Write the equation of AD. p(x) = x - 6 Question 4.4: Determine the equation of k in the form of k(x) = ax^2 if B is equidistant from A to C. Step 1: Find the coordinates of point B. Since B is equidistant from A(6;0) and C(0;-6), B is the midpoint of AC. The midpoint formula is ((x_1+x_2)/(2), (y_1+y_2)/(2)). B = ((6+0)/(2), (0+(-6))/(2)) B = ((6)/(2), (-6)/(2)) B = (3, -3) Step 2: Use the coordinates of B to find the value of a in k(x) = ax^2. Point B(3;-3) lies on the parabola k(x) = ax^2. Substitute x=3 and k(x)=-3: -3 = a(3)^2 -3 = 9a a = (-3)/(9) a = -(1)/(3) Step 3: Write the equation of k. k(x) = -(1)/(3)x^2 Question 4.5: Calculate the co-ordinates of D. Step 1: Set the equations of line p and parabola k equal to each other to find their intersection points. We have p(x) = x - 6 and k(x) = -(1)/(3)x^2. x - 6 = -(1)/(3)x^2 Step 2: Solve the quadratic equation for x. Multiply by 3 to clear the fraction: 3(x - 6) = -x^2 3x - 18 = -x^2 Rearrange into standard form: x^2 + 3x - 18 = 0 Factor the quadratic equation: (x + 6)(x - 3) = 0 The solutions for x are x = -6 or x = 3. Step 3: Identify the coordinates of D. We know that x=3 corresponds to point B(3;-3). Therefore, x=-6 must correspond to point D. Substitute x=-6 into the equation of line p to find the y-coordinate of D: y = x - 6 y = -6 - 6 y = -12 The coordinates of D are (-6; -12). Question 4.6: What kind of a triangle is ACO according to the lengths of the sides and give a reason. Step 1: Identify the coordinates of the vertices. A(6;0), C(0;-6), O(0;0). Step 2: Calculate the lengths of the sides AO and CO. Length of AO: The distance from (0,0) to (6,0) is 6 units. Length of CO: The distance from (0,0) to (0,-6) is 6 units. Step 3: Classify the triangle based on side lengths. Since AO = CO = 6 units, two sides of the triangle are equal in length. Therefore, ACO is an isosceles triangle. Reason: Two sides are equal in length (AO = CO). Question 4.7: Without any calculations give the value of C. Step 1: Identify the vertices and their positions relative to the origin. A(6;0) is on the positive x-axis. C(0;-6) is on the negative y-axis. O(0;0) is the origin. Step 2: Determine the angle at O. Since AO lies on the x-axis and CO lies on the y-axis, the angle AOC is a right angle, i.e., 90^. Step 3: Use properties of ACO. From Question 4.6, we know that AO = CO = 6 units. Therefore, ACO is an isosceles right-angled triangle (right angle at O). In an isosceles right-angled triangle, the two non-right angles are equal and each measures 45^. Thus, C = ACO = 45^. C = 45^ That's 2 down. 3 left today — send the next one.