Step-by-step explanation:
I assume that "ground" is at 0 ft height. which is in an actual scenario not airways the case.
y = -16x² + 64x + 89
shows us that the tower is 89 ft tall (the result for x = 0, at the start).
anyway, if the original assumption is correct, then we need to solve
0 = -16x² + 64x + 89
the general solution for such a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/)2a)
in our case
a = -16
b = 64
c = 89
x = (-64 ± sqrt(64² - 4×-16×89))/(2×-16) =
= (-64 ± sqrt(4096 + 5696))/-32 =
= (-64 ± sqrt(9792))/-32
x1 = (-64 + 98.95453501...)/-32 = -1.092329219... s
x2 = (-64 - 98.95453501...)/-32 = 5.092329219... s
the negative solution for time is but useful here (it would be the time calculated back to ground at the start).
so, x2 is our solution.
the rocket hits the ground after about 5.09 seconds.
Answer:
m = (p+7n)/3
Step-by-step explanation:
3m-7n = p
3m = p+7n
m = (p+7n)/3
Answer:
rgvrgvtgbrgvrgbthrfvrgvrhgbthgrvgvrgvthbtrfvrgvrgvhtbvrgvrgvhbt
Step-by-step explanation:
The addition of +7 influences the y values of the graph. Since it is positive, the graph of y=IxI would be moved 7 units upwards.
Hope I helped :)
