a ball is thrown vertically upward. after t seconds, its height h (in feet) is given by the function h(t) = 52t - 16t^2 . what is the maximum height that the ball will reach?
do not round your answer.

42.25 feet

Step-by-step explanation:

We know that after "t" seconds, its height "h" in feet is given by this function:

The maximum height is the y-coordinate of the vertex of the parabola. Then, we can use the following formula to find the corresponding value of "t" (which is the x-coordinate of the vertex):

In this case:

Substituting values, we get :

Substituting this value into the function to find the maximum height the ball will reach, we get:

42.25 feet

Step-by-step explanation:

The maximum of a quadratic can be found by finding the vertex of the parabola that the quadratic creates visually on a graph.

So first step to find the maximum height is to find the x-coordinate of the vertex.

After you find the x-coordinate of the vertex, you will want to find the y that corresponds by using the given equation, . The y-coordinate we will get will be the maximum height.

Let's start.

The x-coordinate of the vertex is .

compare to .

We have that .

Let's plug into   with those values.

with

.

The vertex's x-coordinate is 13/8.

Now to find the corresponding y-coordinate.

I'm going to just put this in the calculator:

So the maximum is 42.25 feet.

12/11c

or

12

c

11

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)