Whitney kicks a football off the ground. The height, h , in feet, of the football above the ground after t seconds is given by h ( t ) = − 8 t 2 + 32 t . How many seconds after the football is kicked does it hit the ground for the first time?

after 4seconds

Step-by-step explanation:

Given the height, h , in feet, of the football above the ground after t seconds expressed by h ( t ) = − 8 t^2 + 32 t, the height of the ball on the ground is 0feet.

Substitute h(t) = 0 into the expression and calculate t;

h ( t ) = − 8 t^2 + 32 t

0 = − 8 t^2 + 32 t

8t² = 32t

8t = 32

Divide both sides by 8

8t/8 = 32/8

t = 4s

Hence the football hits the ground after 4seconds

by definition, an arc subtended by a central angle of some given measure has the same measure. so .

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!