check the picture below.
so, the rocket will come back to the ground when h(t) = 0, thus
![\bf h(t)=-3t^2+12t\implies \stackrel{h(t)}{0}=-3t^2+12t\implies 0=-3t(t-4)\\\\[-0.35em]~\dotfill\\\\0=-3t\implies 0=t\impliedby \textit{0 seconds when it took off from the ground}\\\\[-0.35em]~\dotfill\\\\0=t-4\implies 4=t\impliedby \textit{4 seconds later, it came back down}](https://tex.z-dn.net/?f=%5Cbf%20h%28t%29%3D-3t%5E2%2B12t%5Cimplies%20%5Cstackrel%7Bh%28t%29%7D%7B0%7D%3D-3t%5E2%2B12t%5Cimplies%200%3D-3t%28t-4%29%5C%5C%5C%5C%5B-0.35em%5D~%5Cdotfill%5C%5C%5C%5C0%3D-3t%5Cimplies%200%3Dt%5Cimpliedby%20%5Ctextit%7B0%20seconds%20when%20it%20took%20off%20from%20the%20ground%7D%5C%5C%5C%5C%5B-0.35em%5D~%5Cdotfill%5C%5C%5C%5C0%3Dt-4%5Cimplies%204%3Dt%5Cimpliedby%20%5Ctextit%7B4%20seconds%20later%2C%20it%20came%20back%20down%7D)
Answer:
he cut about 8 yards of fabric i think its right
Step-by-step explanation:
Answer: the answer should be B
9514 1404 393
Answer:
1. 5 seconds
2. 4.5 seconds
Step-by-step explanation:
Both of these questions can be solved the same way. Each quadratic gives height as a function of time. You're interested in the time at which the height is zero. Factor the quadratic and find the positive value of x that makes the relevant factor be zero.
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1) y = -15x(x -5)
x = 5 makes the second factor zero
The golf ball is in the air 5 seconds.
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2) y = -12x(x -4.5)
x = 4.5 makes the second factor zero
The football is in the air for 4.5 seconds.
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<em>Additional comment</em>
The "zero product rule" tells you a product will be zero if and only if one or more of the factors is zero. When a factored quadratic (or any other polynomial) is zero, one of the factors must be zero. If the factor is of the form (ax+b), then the corresponding solution is ...
ax+b = 0 ⇒ ax = -b ⇒ x = -b/a
That is, once you have the factored form, you can write the solutions "by inspection."
The first one is 8 and here's a graph. the second one is 5 again here's a graph
