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:
6 units
Step-by-step explanation:
HOPE THIS HELPED
Answer:
y=2/7x-4
Step-by-step explanation:
Write in Slope intercept form
y=mx+b
m = slope
b = y intercept
x = x intercept
y = (0,-4)
So,
y = 2/7x+(-4) substitute
y = 2/7x-4
Answer:
47.5 and 2853.56 m
Step-by-step explanation:
Graph the equation on desmos and use the graph to find the maxima and the roots
