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!

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

20%

step-by-step explanation: