Asked by Pat
A student threw a ball directly upward from the balcony of a building. The height of the ball as measured from the ground "t" seconds after it was thrown, is given by the expression
h= -16t(2nd power) + 64t + 768
When does the ball reach the ground?
h= -16t(2nd power) + 64t + 768
When does the ball reach the ground?
Answers
Answered by
drwls
The ball reaches the ground when h = 0
That leaves you with the job of solving
-16t^2 + 64t + 768 = 0
That can also be rewritten (after dividing both sides by -16) as
t^2 -4t -48 = 0
That equation can be factored easily. There will be two possible answers; one is negative. Take the positive one.
That leaves you with the job of solving
-16t^2 + 64t + 768 = 0
That can also be rewritten (after dividing both sides by -16) as
t^2 -4t -48 = 0
That equation can be factored easily. There will be two possible answers; one is negative. Take the positive one.
Answered by
Anonymous
Then, 6x + 8y = 1 and 26x + 48y = 1 .
10 2
10 2
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