A football was kicked through the air and it followed the path h\left(s\right)=-3s^2+75h(s)=−3s 2 +75 . Let h\left(s\right)h(s) represent the height of the soccer ball after, s, seconds.

Complete question is;

A soccer ball was kicked through the air and it followed the path. Let h(s) = −3s² +75 represent the height of the soccer ball after s seconds.

Find the time at which the soccer ball hits the ground.

s = 5 seconds.

Step-by-step explanation:

We are given the function;

h(s) = −3s² +75

Where;

h(s) is height at time s

s is the time taken

Now, the time at which the time at which the soccer ball hits the ground would be at h(s) = 0 since height is at 0 point.

Thus;

−3s² +75 = 0

−3s² = -75

s² = 75/3

s² = 25

s = √25

s = 5 seconds.

4+4+2x

or

2x+8

step-by-step explanation:

the answer is -1 1/5, also a

hope i

-lavira : )