The height of a rock t seconds after it is dropped off a cliff is given by the equation h= -16t^2 +256. how many seconds after the rock is dropped will it hit the ground?

4 seconds

Step-by-step explanation:

To hit the ground, the height here would be zero

Hence We equate the h here to zero.

This means 0 = -16t^2 + 256

16t^2 = 256

t^2 = 256/16

t*2 = 16

t = 4s

The rock will hit the ground after

it will take 4 seconds after the rock is dropped to hit the ground.

Step-by-step explanation:

The equation representing the height of the rock, h dropped off a cliff after it has fallen for t seconds is

h= -16t^2 +256

By the time the rock hits the ground, its height above the ground would be zero. Thus, h = 0. Therefore

0 = -16t^2 +256

16t^2 = 256

t^2 = 256/16 = 16

Taking square root of both sides of the equation, it becomes

t = √16 = 4

idk

step-by-step explanation:

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing im pretty sure that is the right answer

step-by-step explanation: