A ball is launched straight up in the air from a height of 6 feet. Its velocity (feet/second) t seconds after launch is given by f(t)=-32t+285 Between 2 seconds and 9 seconds the ball's height changed by feet. Round answer to nearest tenth.)

In between 2 second and 9 second , ball height is changed by 763 feet.

The velocity function is given by,

Velocity is defined as rate of change of displacement (or height) with respect to time.

Taking integration on both side.

Since, ball is launched straight up in the air from a height of 6 feet

So,      

We have to find ball height changed between 2 s and 9s.

Thus,  

Height change,

Therefore, In between 2 second and 9 second , ball height is changed by 763 feet.

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Step-by-step explanation:

A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :

f(t) = -32 t+285

Height of the ball is :

C is constant. Here the ball is launched from a height of 6 feet. So,

At t = 2 s,

At t = 9 s,

Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.