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 h(t)= -16t^2+64t+80 How many seconds after being thrown will the ball hit the ground?

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The ball will hit the ground 5 seconds after being thrown.

Step-by-step explanation:

The correct function is: h(t)= -16t^2+64t+80

You can rewrite the Quadratic function given in the exercise with making h(t)=0. Then this is:

0= -16t^2+64t+80

Now you can simplify the equation dividing both sides by -16. So you get:

0= t^2-4t-5

To find the solution of the Quadratic equation, you can factor it. In order to do this, it is necessary to find two numbers whose sum is -4 and whose product is -5. These number would be -5 and 1.

Therefore, you get this result:

0= (t-5) (t+1)

t1=5

t2=-1

Since the time cannot be negative, you can conclude that the ball will hit the ground 5 seconds after being thrown.

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

the answer is b