Aball is thrown into the air from the top of a building. the height, h(t), of the ball above the ground t seconds after it is thrown can be modeled by h(t) = -16t2 + 64t + 80. how many seconds after being thrown will the ball hit the ground?

5 seconds

Step-by-step explanation:

Given : A ball is thrown into the air from the top of a building.

The height, h(t), of the ball above the ground t seconds after it is thrown can be modeled by .

When ball reaches the ground , then its height from ground become zero.

i.e.

Divide equation by 16 , we get

Since time cannot be negative , therefore t= 5

Hence, the ball will take 5 seconds ( after being thrown) to hit the ground.

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer