Sarah kicked a ball in the air. The function
f
ff models the height of the ball (in meters) as a function of time (in seconds) after Sarah kicked it.

The ball hits the ground at 5 s

Step-by-step explanation:

The question seems incomplete and there is not enough data. However, we can work with the following function to understand this problem:

(1)

Where models the height of the ball in meters and the time.

Now, let's find the time when the ball Sara kicked hits the ground (this is when ):

(2)

Rearranging the equation:

(3)

Dividing both sides of the equation by :

(4)

This quadratic equation can be written in the form , and can be solved with the following formula:  

(5)  

Where:  

Substituting the known values:  

(6)  

Solving we have the following result:

 This means the ball hit the ground 5 seconds after it was kicked by Sara.

it is probably the graph on the bottom right.

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2