Here are the solutions to the mathematics problems: SECTION A (35 MARKS) 1. (6 Marks) a) Simplify: rac34 imes 3-232 b) Express 0.00032 in standard form. c) Rationalize and simplify: rac5sqrt(3)

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Asked on April 8, 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 · 2026-04-08T19:51:52.190Z

Here are the solutions to the mathematics problems: SECTION A (35 MARKS) 1. (6 Marks) a) Simplify: 3^4 × 3^-23^2 Step 1: Apply the exponent rule a^m × a^n = a^m+n to the numerator. 3^4 × 3^-2 = 3^4 + (-2) = 3^4-2 = 3^2 Step 2: Substitute this back into the expression. (3^2)/(3^2) Step 3: Apply the exponent rule (a^m)/(a^n) = a^m-n. 3^2-2 = 3^0 Step 4: Any non-zero number raised to the power of 0 is 1. 3^0 = 1 The simplified expression is 1. b) Express 0.00032 in standard form. Step 1: Move the decimal point to the right until there is one non-zero digit to its left. 0.00032 3.2 Step 2: Count how many places the decimal point was moved. It was moved 4 places to the right. Step 3: Since the decimal was moved to the right, the exponent of 10 will be negative. 3.2 × 10^-4 In standard form, it is 3.2 × 10^-4. c) Rationalize and simplify: (5)/(sqrt(3)) Step 1: To rationalize the denominator, multiply both the numerator and the denominator by sqrt(3). (5)/(sqrt(3)) × sqrt(3)sqrt(3) Step 2: Multiply the numerators and the denominators. 5sqrt(3)(sqrt(3))^2 Step 3: Simplify the denominator. 5sqrt(3)3 The rationalized and simplified expression is 5sqrt(3)3. d) Evaluate: sqrt(50) + 3sqrt(8) Step 1: Simplify sqrt(50) by finding the largest perfect square factor. sqrt(50) = sqrt(25 × 2) = sqrt(25) × sqrt(2) = 5sqrt(2) Step 2: Simplify sqrt(8) by finding the largest perfect square factor. sqrt(8) = sqrt(4 × 2) = sqrt(4) × sqrt(2) = 2sqrt(2) Step 3: Substitute the simplified radicals back into the expression. 5sqrt(2) + 3(2sqrt(2)) Step 4: Multiply the terms. 5sqrt(2) + 6sqrt(2) Step 5: Combine the like terms. (5+6)sqrt(2) = 11sqrt(2) The evaluated expression is 11sqrt(2). 2. (7 Marks) a) Expand and simplify: (2x-3)(x+5)-(x-4)^2 Step 1: Expand the first product (2x-3)(x+5) using FOIL. (2x)(x) + (2x)(5) + (-3)(x) + (-3)(5) 2x^2 + 10x - 3x - 15 2x^2 + 7x - 15 Step 2: Expand the second term (x-4)^2 using the formula (a-b)^2 = a^2 - 2ab + b^2. (x)^2 - 2(x)(4) + (4)^2 x^2 - 8x + 16 Step 3: Substitute the expanded terms back into the original expression. Remember to distribute the negative sign to all terms in the second expansion. (2x^2 + 7x - 15) - (x^2 - 8x + 16) 2x^2 + 7x - 15 - x^2 + 8x - 16 Step 4: Combine like terms. (2x^2 - x^2) + (7x + 8x) + (-15 - 16) x^2 + 15x - 31 The expanded and simplified expression is x^2 + 15x - 31. b) Factorize completely: 3x^2 - 11x - 4 Step 1: Find two numbers that multiply to 3 × -4 = -12 and add up to -11. These numbers are -12 and 1. Step 2: Rewrite the middle term using these two numbers. 3x^2 - 12x + x - 4 Step 3: Factor by grouping. 3x(x - 4) + 1(x - 4) Step 4: Factor out the common binomial factor (x-4). (x-4)(3x+1) The completely factorized expression is (x-4)(3x+1). c) Solve: 4x^2 - 9 = 0 Step 1: Add 9 to both sides of the equation. 4x^2 = 9 Step 2: Divide both sides by 4. x^2 = (9)/(4) Step 3: Take the square root of both sides, remembering both positive and negative roots. x = ±sqrt((9)/(4)) x = ±sqrt(9)sqrt(4) x = ±(3)/(2) The solutions are x = (3)/(2) or x = -(3)/(2). 3. (6 Marks) Solve: 3x + 2y = 17 (Equation 1) 2x - y = 4 (Equation 2) Step 1: Use the elimination method. Multiply Equation 2 by 2 to make the coefficients of y opposites. 2 × (2x - y = 4) 4x - 2y = 8 (Equation 3) Step 2: Add Equation 1 and Equation 3. (3x + 2y) + (4x - 2y) = 17 + 8 7x = 25 Step 3: Solve for x. x = (25)/(7) Step 4: Substitute the value of x into Equation 2 to solve for y. 2((25)/(7)) - y = 4 (50)/(7) - y = 4 Step 5: Isolate y. -y = 4 - (50)/(7) -y = (28)/(7) - (50)/(7) -y = -(22)/(7) y = (22)/(7) The solution is x = (25)/(7), y = (22)/(7). 4. (8 Marks) A projectile is modeled by h = -2t^2 + 12t + 5. a) Find the maximum height. The maximum height occurs at the vertex of the parabola. For a quadratic equation at^2 + bt + c, the t-coordinate of the vertex is given by t = (-b)/(2a). Step 1: Identify a and b from the equation h = -2t^2 + 12t + 5. a = -2, b = 12 Step 2: Calculate the time t at which the maximum height occurs. t = (-12)/(2(-2)) = (-12)/(-4) = 3 Step 3: Substitute t=3 into the height equation to find the maximum height h. h = -2(3)^2 + 12(3) + 5 h = -2(9) + 36 + 5 h = -18 + 36 + 5 h = 18 + 5 h = 23 The maximum height is 23 units. b) Determine the time when the projectile hits the ground. The projectile hits the ground when its height h is 0. Step 1: Set the equation to 0. -2t^2 + 12t + 5 = 0 Step 2: Use the quadratic formula t = -b ± sqrt(b^2 - 4ac)2a. Here a=-2, b=12, c=5. t = -12 ± sqrt(12^2 - 4(-2)(5))2(-2) t = -12 ± sqrt(144 + 40)-4 t = -12 ± sqrt(184)-4 Step 3: Calculate the square root of 184. sqrt(184) ≈ 13.56. t = (-12 ± 13.56)/(-4) Step 4: Calculate the two possible values for t. t_1 = (-12 + 13.56)/(-4) = (1.56)/(-4) = -0.39 t_2 = (-12 - 13.56)/(-4) = (-25.56)/(-4) = 6.39 Since time cannot be negative, we take the positive value. The projectile hits the ground at approximately 6.39 units of time. c) Find the average rate of change between t=1 and t=3. The average rate of change is given by (h(t_2) - h(t_1))/(t_2 - t_1). Step 1: Calculate h(1). h(1) = -2(1)^2 + 12(1) + 5 h(1) = -2 + 12 + 5 = 15 Step 2: Calculate h(3). h(3) = -2(3)^2 + 12(3) + 5 h(3) = -2(9) + 36 + 5 h(3) = -18 + 36 + 5 = 23 Step 3: Calculate the average rate of change. Average rate of change = (h(3) - h(1))/(3 - 1) = (23 - 15)/(2) = (8)/(2) = 4 The average rate of change between t=1 and t=3 is 4 units of height per unit of time.

Accepted Answer · 2026-04-08T19:51:52.190Z

Here are the solutions to the mathematics problems: SECTION A (35 MARKS) 1. (6 Marks) a) Simplify: 3^4 × 3^-23^2 Step 1: Apply the exponent rule a^m × a^n = a^m+n to the numerator. 3^4 × 3^-2 = 3^4 + (-2) = 3^4-2 = 3^2 Step 2: Substitute this back into the expression. (3^2)/(3^2) Step 3: Apply the exponent rule (a^m)/(a^n) = a^m-n. 3^2-2 = 3^0 Step 4: Any non-zero number raised to the power of 0 is 1. 3^0 = 1 The simplified expression is 1. b) Express 0.00032 in standard form. Step 1: Move the decimal point to the right until there is one non-zero digit to its left. 0.00032 3.2 Step 2: Count how many places the decimal point was moved. It was moved 4 places to the right. Step 3: Since the decimal was moved to the right, the exponent of 10 will be negative. 3.2 × 10^-4 In standard form, it is 3.2 × 10^-4. c) Rationalize and simplify: (5)/(sqrt(3)) Step 1: To rationalize the denominator, multiply both the numerator and the denominator by sqrt(3). (5)/(sqrt(3)) × sqrt(3)sqrt(3) Step 2: Multiply the numerators and the denominators. 5sqrt(3)(sqrt(3))^2 Step 3: Simplify the denominator. 5sqrt(3)3 The rationalized and simplified expression is 5sqrt(3)3. d) Evaluate: sqrt(50) + 3sqrt(8) Step 1: Simplify sqrt(50) by finding the largest perfect square factor. sqrt(50) = sqrt(25 × 2) = sqrt(25) × sqrt(2) = 5sqrt(2) Step 2: Simplify sqrt(8) by finding the largest perfect square factor. sqrt(8) = sqrt(4 × 2) = sqrt(4) × sqrt(2) = 2sqrt(2) Step 3: Substitute the simplified radicals back into the expression. 5sqrt(2) + 3(2sqrt(2)) Step 4: Multiply the terms. 5sqrt(2) + 6sqrt(2) Step 5: Combine the like terms. (5+6)sqrt(2) = 11sqrt(2) The evaluated expression is 11sqrt(2). 2. (7 Marks) a) Expand and simplify: (2x-3)(x+5)-(x-4)^2 Step 1: Expand the first product (2x-3)(x+5) using FOIL. (2x)(x) + (2x)(5) + (-3)(x) + (-3)(5) 2x^2 + 10x - 3x - 15 2x^2 + 7x - 15 Step 2: Expand the second term (x-4)^2 using the formula (a-b)^2 = a^2 - 2ab + b^2. (x)^2 - 2(x)(4) + (4)^2 x^2 - 8x + 16 Step 3: Substitute the expanded terms back into the original expression. Remember to distribute the negative sign to all terms in the second expansion. (2x^2 + 7x - 15) - (x^2 - 8x + 16) 2x^2 + 7x - 15 - x^2 + 8x - 16 Step 4: Combine like terms. (2x^2 - x^2) + (7x + 8x) + (-15 - 16) x^2 + 15x - 31 The expanded and simplified expression is x^2 + 15x - 31. b) Factorize completely: 3x^2 - 11x - 4 Step 1: Find two numbers that multiply to 3 × -4 = -12 and add up to -11. These numbers are -12 and 1. Step 2: Rewrite the middle term using these two numbers. 3x^2 - 12x + x - 4 Step 3: Factor by grouping. 3x(x - 4) + 1(x - 4) Step 4: Factor out the common binomial factor (x-4). (x-4)(3x+1) The completely factorized expression is (x-4)(3x+1). c) Solve: 4x^2 - 9 = 0 Step 1: Add 9 to both sides of the equation. 4x^2 = 9 Step 2: Divide both sides by 4. x^2 = (9)/(4) Step 3: Take the square root of both sides, remembering both positive and negative roots. x = ±sqrt((9)/(4)) x = ±sqrt(9)sqrt(4) x = ±(3)/(2) The solutions are x = (3)/(2) or x = -(3)/(2). 3. (6 Marks) Solve: 3x + 2y = 17 (Equation 1) 2x - y = 4 (Equation 2) Step 1: Use the elimination method. Multiply Equation 2 by 2 to make the coefficients of y opposites. 2 × (2x - y = 4) 4x - 2y = 8 (Equation 3) Step 2: Add Equation 1 and Equation 3. (3x + 2y) + (4x - 2y) = 17 + 8 7x = 25 Step 3: Solve for x. x = (25)/(7) Step 4: Substitute the value of x into Equation 2 to solve for y. 2((25)/(7)) - y = 4 (50)/(7) - y = 4 Step 5: Isolate y. -y = 4 - (50)/(7) -y = (28)/(7) - (50)/(7) -y = -(22)/(7) y = (22)/(7) The solution is x = (25)/(7), y = (22)/(7). 4. (8 Marks) A projectile is modeled by h = -2t^2 + 12t + 5. a) Find the maximum height. The maximum height occurs at the vertex of the parabola. For a quadratic equation at^2 + bt + c, the t-coordinate of the vertex is given by t = (-b)/(2a). Step 1: Identify a and b from the equation h = -2t^2 + 12t + 5. a = -2, b = 12 Step 2: Calculate the time t at which the maximum height occurs. t = (-12)/(2(-2)) = (-12)/(-4) = 3 Step 3: Substitute t=3 into the height equation to find the maximum height h. h = -2(3)^2 + 12(3) + 5 h = -2(9) + 36 + 5 h = -18 + 36 + 5 h = 18 + 5 h = 23 The maximum height is 23 units. b) Determine the time when the projectile hits the ground. The projectile hits the ground when its height h is 0. Step 1: Set the equation to 0. -2t^2 + 12t + 5 = 0 Step 2: Use the quadratic formula t = -b ± sqrt(b^2 - 4ac)2a. Here a=-2, b=12, c=5. t = -12 ± sqrt(12^2 - 4(-2)(5))2(-2) t = -12 ± sqrt(144 + 40)-4 t = -12 ± sqrt(184)-4 Step 3: Calculate the square root of 184. sqrt(184) ≈ 13.56. t = (-12 ± 13.56)/(-4) Step 4: Calculate the two possible values for t. t_1 = (-12 + 13.56)/(-4) = (1.56)/(-4) = -0.39 t_2 = (-12 - 13.56)/(-4) = (-25.56)/(-4) = 6.39 Since time cannot be negative, we take the positive value. The projectile hits the ground at approximately 6.39 units of time. c) Find the average rate of change between t=1 and t=3. The average rate of change is given by (h(t_2) - h(t_1))/(t_2 - t_1). Step 1: Calculate h(1). h(1) = -2(1)^2 + 12(1) + 5 h(1) = -2 + 12 + 5 = 15 Step 2: Calculate h(3). h(3) = -2(3)^2 + 12(3) + 5 h(3) = -2(9) + 36 + 5 h(3) = -18 + 36 + 5 = 23 Step 3: Calculate the average rate of change. Average rate of change = (h(3) - h(1))/(3 - 1) = (23 - 15)/(2) = (8)/(2) = 4 The average rate of change between t=1 and t=3 is 4 units of height per unit of time.

Here are the solutions to the mathematics problems: SECTION A (35 MARKS) 1. (6 Marks) a) Simplify: rac34 imes 3-232 b) Express 0.00032 in standard form. c) Rationalize and simplify: rac5sqrt(3)

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Here are the solutions to the mathematics problems: SECTION A (35 MARKS) 1. (6 Marks) a) Simplify: rac34 imes 3-232 b) Express 0.00032 in standard form. c) Rationalize and simplify: rac5sqrt(3)

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