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 out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+218x+75

6.81s

Step-by-step explanation:

First you must know that the velocity of the body is zero at the maximum height.

v =dy/dx

v= -32x+218

Since v = 0

-32x+218 =0

32x =218

x= 218/32

x = 6.81s

Hence the rocket reaches its maximum height after 6.81s

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2

step-by-step explanation:

is their an equation anywhere i can take a   look at it and see if i am right.