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3 years ago

Answer please lot's of points.

Mathematics

2 answers:

Svetlanka [38]3 years ago

3 0

Answer:

t≈8.0927

Step-by-step explanation:

h(t) = -16t^2 + 128t +12

We want to find when h(t) is zero ( or when it hits the ground)

0 = -16t^2 + 128t +12

Completing the square

Subtract 12 from each side

-12 = -16t^2 + 128t

Divide each side by -16

-12/-16 = -16/-16t^2 + 128/-16t

3/4 = t^2 -8t

Take the coefficient of t and divide it by 8

-8/2 = -4

Then square it

(-4) ^2 = 16

Add 16 to each side

16+3/4 = t^2 -8t+16

64/4 + 3/4= (t-4)^2

67/4 = (t-4)^2

Take the square root of each side

±sqrt(67/4) =sqrt( (t-4)^2)

±1/2sqrt(67) = (t-4)

Add 4 to each side

4 ±1/2sqrt(67) = t

The approximate values for t are

t≈-0.092676

t≈8.0927

The first is before the rocket is launched so the only valid answer is the second one

prisoha [69]3 years ago

3 0

Answer:

your correct answer is t

Step-by-step explanation:

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