Aball is thrown upward from ground level. its height h, in feet, above the ground after t seconds is h-48t -16t^2.
find the maximum height of the ball.

36 feet.

Step-by-step explanation:

We have been given that a ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is . We are asked to find the maximum height of the ball.

We can see that our given equation is a downward opening parabola, so its maximum height will be the vertex of the parabola.

To find the maximum height of the ball, we need to find y-coordinate of vertex of parabola.

Let us find x-coordinate of parabola using formula .

So, the x-coordinate of the parabola is . Now, we will substitute in our given equation to find y-coordinate of parabola.

Therefore, the maximum height of the ball is 36 feet.

sadsdda

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°