The curve is given by the equation y = 2x(3-x). a) State the coordinates of the point B. b) Show that the value of k is (5)/(2). c) Find an equation of the straight line passing through D and E.

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Asked on April 12, 2026Mathematics

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.

Accepted Answer · 2026-04-12T07:59:21.351Z

Morning Tangunu — let's get this done. The curve is given by the equation y = 2x(3-x). a) State the coordinates of the point B. Point B is where the curve crosses the x-axis. This means y=0. Step 1: Set y=0 in the curve equation. 0 = 2x(3-x) Step 2: Solve for x. This equation gives two solutions: 2x=0 x=0 (which corresponds to point O) or 3-x=0 x=3. So, point B has an x-coordinate of 3. The coordinates of point B are (3, 0). b) Show that the value of k is (5)/(2). The straight line crosses the y-axis at point E with coordinates (0, 5). It meets the curve at points C and D. The coordinates of point D are (k, k). Since point D (k, k) lies on the curve, its coordinates must satisfy the curve's equation y = 2x(3-x). Step 1: Substitute x=k and y=k into the curve equation. k = 2k(3-k) Step 2: Solve for k. k = 6k - 2k^2 2k^2 - 5k = 0 Factor out k: k(2k - 5) = 0 This gives two possible values for k: k=0 or 2k-5=0. If k=0, then point D would be (0,0), which is point O. From the diagram, D is clearly not O. Therefore, 2k-5=0 2k=5 k=(5)/(2). Thus, the value of k is (5)/(2). c) Find an equation of the straight line passing through D and E. Point E is (0, 5). Point D is (k, k) = ((5)/(2), (5)/(2)). Step 1: Calculate the gradient m of the line using points E and D. m = (y_2 - y_1)/(x_2 - x_1) = (5)/(2) - 5(5)/(2) - 0 m = (5)/(2) - (10)/(2)(5)/(2) = (-5)/(2)(5)/(2) = -1 Step 2: Use the point-slope form y - y_1 = m(x - x_1) or the slope-intercept form y = mx + c. Since point E is (0, 5), the y-intercept c is 5. Using m=-1 and c=5: y = -1x + 5 The equation of the straight line is y = -x + 5. d) Show that the x-coordinates of the point C is 1. Point C is an intersection of the curve y = 2x(3-x) and the straight line y = -x + 5. Step 1: Set the equations equal to each other to find the intersection points. 2x(3-x) = -x + 5 6x - 2x^2 = -x + 5 Step 2: Rearrange into a quadratic equation. 2x^2 - 7x + 5 = 0 Step 3: Solve the quadratic equation for x. We can factorize the quadratic equation: 2x^2 - 2x - 5x + 5 = 0 2x(x-1) - 5(x-1) = 0 (2x-5)(x-1) = 0 This gives two solutions for x: 2x-5=0 x=(5)/(2) or x-1=0 x=1. The x-coordinate x=(5)/(2) corresponds to point D. Therefore, the x-coordinate of point C is 1. e) Calculate the area of the region R bounded by the curve. The region R is bounded by the curve y = 2x(3-x) and the straight line y = -x + 5. The region is between the x-coordinates of C and D. The x-coordinate of C is 1. The x-coordinate of D is (5)/(2). The area R is given by the integral of the upper function minus the lower function, from x=1 to x=(5)/(2). The upper function is the curve y_c = 2x(3-x) = 6x - 2x^2. The lower function is the line y_l = -x + 5. Step 1: Set up the definite integral for the area. Area R = _1^(5)/(2) [(6x - 2x^2) - (-x + 5)]\,dx Area R = _1^(5)/(2) (6x - 2x^2 + x - 5)\,dx Area R = _1^(5)/(2) (-2x^2 + 7x - 5)\,dx Step 2: Integrate the expression. Area R = [ -(2x^3)/(3) + (7x^2)/(2) - 5x ]_1^(5)/(2) Step 3: Evaluate the definite integral. First, evaluate at x=(5)/(2): F((5)/(2)) = -(2)/(3)((5)/(2))^3 + (7)/(2)((5)/(2))^2 - 5((5)/(2)) F((5)/(2)) = -(2)/(3)((125)/(8)) + (7)/(2)((25)/(4)) - (25)/(2) F((5)/(2)) = -(125)/(12) + (175)/(8) - (25)/(2) Find a common denominator, which is 24: F((5)/(2)) = -(125 × 2)/(24) + (175 × 3)/(24) - (25 × 12)/(24) F((5)/(2)) = -(250)/(24) + (525)/(24) - (300)/(24) F((5)/(2)) = (-250 + 525 - 300)/(24) = (275 - 300)/(24) = -(25)/(24) Next, evaluate at x=1: F(1) = -(2(1)^3)/(3) + (7(1)^2)/(2) - 5(1) F(1) = -(2)/(3) + (7)/(2) - 5 Find a common denominator, which is 6: F(1) = -(2 × 2)/(6) + (7 × 3)/(6) - (5 × 6)/(6) F(1) = -(4)/(6) + (21)/(6) - (30)/(6) F(1) = (-4 + 21 - 30)/(6) = (17 - 30)/(6) = -(13)/(6) Finally, calculate the area: Area R = F((5)/(2)) - F(1) Area R = -(25)/(24) - (-(13)/(6)) Area R = -(25)/(24) + (13)/(6) Convert (13)/(6) to have a denominator of 24: Area R = -(25)/(24) + (13 × 4)/(6 × 4) Area R = -(25)/(24) + (52)/(24) Area R = (52 - 25)/(24) = (27)/(24) Simplify the fraction: Area R = (9)/(8) a) The coordinates of point B are (3, 0). b) The value of k is (5)/(2). c) The equation of the straight line is y = -x + 5. d) The x-coordinate of point C is 1. e) The area of the region R is (9)/(8) units^2. Got more? Send 'em.

Accepted Answer · 2026-04-12T07:59:21.351Z

Morning Tangunu — let's get this done. The curve is given by the equation y = 2x(3-x). a) State the coordinates of the point B. Point B is where the curve crosses the x-axis. This means y=0. Step 1: Set y=0 in the curve equation. 0 = 2x(3-x) Step 2: Solve for x. This equation gives two solutions: 2x=0 x=0 (which corresponds to point O) or 3-x=0 x=3. So, point B has an x-coordinate of 3. The coordinates of point B are (3, 0). b) Show that the value of k is (5)/(2). The straight line crosses the y-axis at point E with coordinates (0, 5). It meets the curve at points C and D. The coordinates of point D are (k, k). Since point D (k, k) lies on the curve, its coordinates must satisfy the curve's equation y = 2x(3-x). Step 1: Substitute x=k and y=k into the curve equation. k = 2k(3-k) Step 2: Solve for k. k = 6k - 2k^2 2k^2 - 5k = 0 Factor out k: k(2k - 5) = 0 This gives two possible values for k: k=0 or 2k-5=0. If k=0, then point D would be (0,0), which is point O. From the diagram, D is clearly not O. Therefore, 2k-5=0 2k=5 k=(5)/(2). Thus, the value of k is (5)/(2). c) Find an equation of the straight line passing through D and E. Point E is (0, 5). Point D is (k, k) = ((5)/(2), (5)/(2)). Step 1: Calculate the gradient m of the line using points E and D. m = (y_2 - y_1)/(x_2 - x_1) = (5)/(2) - 5(5)/(2) - 0 m = (5)/(2) - (10)/(2)(5)/(2) = (-5)/(2)(5)/(2) = -1 Step 2: Use the point-slope form y - y_1 = m(x - x_1) or the slope-intercept form y = mx + c. Since point E is (0, 5), the y-intercept c is 5. Using m=-1 and c=5: y = -1x + 5 The equation of the straight line is y = -x + 5. d) Show that the x-coordinates of the point C is 1. Point C is an intersection of the curve y = 2x(3-x) and the straight line y = -x + 5. Step 1: Set the equations equal to each other to find the intersection points. 2x(3-x) = -x + 5 6x - 2x^2 = -x + 5 Step 2: Rearrange into a quadratic equation. 2x^2 - 7x + 5 = 0 Step 3: Solve the quadratic equation for x. We can factorize the quadratic equation: 2x^2 - 2x - 5x + 5 = 0 2x(x-1) - 5(x-1) = 0 (2x-5)(x-1) = 0 This gives two solutions for x: 2x-5=0 x=(5)/(2) or x-1=0 x=1. The x-coordinate x=(5)/(2) corresponds to point D. Therefore, the x-coordinate of point C is 1. e) Calculate the area of the region R bounded by the curve. The region R is bounded by the curve y = 2x(3-x) and the straight line y = -x + 5. The region is between the x-coordinates of C and D. The x-coordinate of C is 1. The x-coordinate of D is (5)/(2). The area R is given by the integral of the upper function minus the lower function, from x=1 to x=(5)/(2). The upper function is the curve y_c = 2x(3-x) = 6x - 2x^2. The lower function is the line y_l = -x + 5. Step 1: Set up the definite integral for the area. Area R = _1^(5)/(2) [(6x - 2x^2) - (-x + 5)]\,dx Area R = _1^(5)/(2) (6x - 2x^2 + x - 5)\,dx Area R = _1^(5)/(2) (-2x^2 + 7x - 5)\,dx Step 2: Integrate the expression. Area R = [ -(2x^3)/(3) + (7x^2)/(2) - 5x ]_1^(5)/(2) Step 3: Evaluate the definite integral. First, evaluate at x=(5)/(2): F((5)/(2)) = -(2)/(3)((5)/(2))^3 + (7)/(2)((5)/(2))^2 - 5((5)/(2)) F((5)/(2)) = -(2)/(3)((125)/(8)) + (7)/(2)((25)/(4)) - (25)/(2) F((5)/(2)) = -(125)/(12) + (175)/(8) - (25)/(2) Find a common denominator, which is 24: F((5)/(2)) = -(125 × 2)/(24) + (175 × 3)/(24) - (25 × 12)/(24) F((5)/(2)) = -(250)/(24) + (525)/(24) - (300)/(24) F((5)/(2)) = (-250 + 525 - 300)/(24) = (275 - 300)/(24) = -(25)/(24) Next, evaluate at x=1: F(1) = -(2(1)^3)/(3) + (7(1)^2)/(2) - 5(1) F(1) = -(2)/(3) + (7)/(2) - 5 Find a common denominator, which is 6: F(1) = -(2 × 2)/(6) + (7 × 3)/(6) - (5 × 6)/(6) F(1) = -(4)/(6) + (21)/(6) - (30)/(6) F(1) = (-4 + 21 - 30)/(6) = (17 - 30)/(6) = -(13)/(6) Finally, calculate the area: Area R = F((5)/(2)) - F(1) Area R = -(25)/(24) - (-(13)/(6)) Area R = -(25)/(24) + (13)/(6) Convert (13)/(6) to have a denominator of 24: Area R = -(25)/(24) + (13 × 4)/(6 × 4) Area R = -(25)/(24) + (52)/(24) Area R = (52 - 25)/(24) = (27)/(24) Simplify the fraction: Area R = (9)/(8) a) The coordinates of point B are (3, 0). b) The value of k is (5)/(2). c) The equation of the straight line is y = -x + 5. d) The x-coordinate of point C is 1. e) The area of the region R is (9)/(8) units^2. Got more? Send 'em.

The curve is given by the equation y = 2x(3-x). a) State the coordinates of the point B. b) Show that the value of k is (5)/(2). c) Find an equation of the straight line passing through D and E.

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The curve is given by the equation y = 2x(3-x). a) State the coordinates of the point B. b) Show that the value of k is (5)/(2). c) Find an equation of the straight line passing through D and E.

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