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.
I believe the answer would be D.
Answer:
2794000000
Step-by-step explanation:
2.794 x 10^9
= 2794000000
First you need to figure out the slope:
y2-y1/x2-x1=(1+5)/(-5-7)=-1/2
y+5=-1/2(x-7)
y=-1/2x-3/2
Answer:
The answer is c.
Step-by-step explanation:
12*12 = 144 so It would be 12
7*7= 49 therefore, it would be 7
