A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. y=−16x^2+
180x+63

To find the maximum of a function in ax^2 + bx + c form, you can use  the formula: maximum = c - (b^2 / 4a).

In this question, a = -16, b = 180, and c = 63.

Maximum = 63 - (180^2/4(-16))

Maximum  = 569.25

Finally, rounding this to the nearest tenth of a foot, the final answer is 569.3 ft.

Let me know if this helps!

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

if i put something wrong your not gonna like it so rather not say nothing.