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A creature on the moon throws a rock off a cliff. the height of the rock above the surface of the moon, in meters, t seconds after it is thrown is given by h(t) = -t^2 +8t +10. when does the rock first reach a height of 12 meters?

 Jul 19, 2022
 #1
avatar+99 
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t^2-8t-10=0

t^2-8t=10

t^2-8t+16=10+16

(t-4)^2=26

t-4=+(26^(1/2))

t=4+(26^(1/2))

t=9.1s

 Jul 19, 2022
 #2
avatar+2448 
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\(-t^2 + 8t + 10 = 12\)

\(-t^2 + 8t - 2 = 0\)

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(t = {-8 \pm \sqrt{-8^2-(4 \times -1 \times -2)} \over 2 \times -1}\)

\(t= {-8 \pm \sqrt{56} \over -2}\)

\(t = {-8 \pm 2\sqrt{14} \over -2}\)

\(t = {4 \pm \sqrt{14}}\)

\(t = \color{brown}\boxed{4 - \sqrt{14}}\)

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 Jul 19, 2022

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