Me out guys. the photo will show you what to do. 50 points cause i need it done fast

a) starting height: 5.5 ft

b) hang time: 5.562 seconds

c) maximum height: 126.5 ft

d) time to maximum height: 2.75 seconds

Step-by-step explanation:

a) The starting height is the height at t=0.

h(0) = -16·0 +88·0 +5.5

h(0) = 5.5

The starting height is 5.5 feet.

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b) The ball is in the air between t=0 and the non-zero time when h(t) = 0. We can find the latter by solving ...

-16t^2 + 8t +5.5 = 0

t^2 -(11/2)t = 5.5/16 . . . . . subtract 5.5, then divide by -16

t^2 -(11/2)t +(11/4)^2 = (5.5/16) +(11/4)^2 . . . . complete the square

(t -11/4)^2 = 126.5/16 . . . . . . . . . . . . . . . . . . . . call this [eq1] for later use

t -11/4 = √7.90625

t = 2.75 +√7.90625 ≈ 5.562

The ball will be in the air about 5.562 seconds.

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c) If we multiply [eq1] above by -16 and add the constant on the right, we get the vertex form of the height equation:

h(t) = -16(t -11/4) +126.5

The vertex at (2.75, 126.5) tells us ...

The maximum height of the ball is 126.5 feet.

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d) That same vertex point tells us ...

The maximum height will be reached at t = 2.75 seconds.

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If you really need answers fast, a graphing calculator can give them to you in very short order (less than a minute).

step-by-step explanation:

3024 rhe base is what up in worry about

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x