The question is in the image

h(t) = -5*t^2 + 20*t  + 2

Step-by-step explanation:

First, we know that the motion equation of the ball will be quadratic, so we write the equation:

h(t) = a*t^2 + b*t + c

Now we need to work with the data in the table.

h(1) = 17

h(3) = 17

h(1) = h(2) = 17

Then we have a symmetry around:

(3 - 1)/2 + 1 = 2

Remember that the symmetry is around the vertex of the parabola, then we can conclude that the vertex of the parabola is the point:

(2, h(2)) = (2, 22)

Remember that for a quadratic equation:

y = a*x^2 + b*x + c

with a vertex (h, k)

we can rewrite our function as:

y = a*(x - h)^2 + k

So in this case, we can rewrite our function as:

h(t) = a*(t - 2)^2 + 22

To find the value of a, notice the first point in the table:

(0, 2)

then we have:

h(0) = 2 = a*(0 - 2)^2 + 22

         = 2 = a*(-2)^2 + 22

            2 = a*(4) + 22

           2 - 22 = a*(4)

            -20/4 = -5 = a

Then our function is:

h(t) = -5*(t - 2)^2 + 22

Now we just expand it:

h(t) = -5*(t^2 - 4*t + 4) + 22

h(t) = -5*t^2 + 20*t  + 2

The correct option is the first one.

the answer is c hope it

step-by-step explanation:

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.