Q3 To solve the equation x^2 + 11x - 26 = 0, we can factor it as (x + 13)(x - 2) = 0. Setting each factor to zero gives x = -13 or x = 2.
Q4 To find when the ball hits the ground, we need to find the time when the height is 0. So we set the equation h = -16t^2 + 30t + 6 equal to 0: -16t^2 + 30t + 6 = 0. This can be simplified further to -8t^2 + 15t + 3 = 0. Using the quadratic formula, we get t ≈ 2.4 seconds.
Q5 The graph of y = -2x^2 + 5 will be shifted upwards by 5 units compared to the graph of y = -2x^2. This means that the vertex of the parabola is raised by 5 units on the y-axis.
Q3 Solve the equation: x2+11x-26=0
Q4 You throw a ball upward. Its height h, in feet, after t seconds can be modeled by the function h=-16t2+30t+6 After how many seconds will it hit the ground?
Q5 How is the graph of
y=-2x2+5 different from the graph of y=-2x2 ?
1 answer