Gcf of 3y^2,6y^3,10y^4
1 answer:
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Answer:
t ∈ {1, 3}
Step-by-step explanation:
You want to find t such that ...
h = 27
27 = -8t^2 +32t +3 . . . . . . substitute the expression for h
24 = -8t^2 +32t . . . . . . . . . subtract 3
-3 = t^2 -4t . . . . . . . . . . . . . divide by -8
1 = t^2 -4t +4 = (t -2)^2 . . . . add 4 to complete the square
±√1 = t -2 . . . . . . . . . . . . . . take the square root
t = 2 ± 1 . . . . . . . . . . . . . . . . add 2
t = 1 or 3
The object is 27 ft off the ground at t = 1 and again at t = 3.
