One small piece of ash was ejected from the volcano with an initial velocity of 336ft/sec. the height h, in feet, of the ash projectile is given by the equation: h= -16t^2 + 336t where t is the time in sec. the graph of this equation will be a parabola. we will assume that the volcano has no height. h= 0 at t= 0.
1. when does the ash projectile reach its maximum height?
2.what is its maximum height?
3.when does the ash projectile return to the ground?

(a)t=10.5 seconds

(b)1764 feet

(c)t=21 seconds

Step-by-step explanation:

The equation of the projectile is given by:

(1)The ash projectile reaches its maximum height when the slope of the line is equal to zero.

First, we find the derivative of H.

When

Divide both sides by -32

t= 10.5 seconds

(2) The maximum height occurs at the point where t= 10.5

(3)The ash projectile returns back to the ground when its Height, H =0

The ash projectile returns to the ground after 21 seconds.

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation:

x-150=475   x=625

step-by-step explanation:

if the bike after the sale ($150) is $475 then just simply add 150 to 475.