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:
the correct answer is d
Step-by-step explanation:
Step-by-step explanation:
a) f(1) = -1 → x = -1
b) g(1) = Undefined
c) f(x) = 1 → x= Undefined
d) g(x) = 1 → x= 5
<h3>
Answer: Negative</h3>
Reason:
The template applies to any quadratic to graph out a parabola. The coefficient for the x^2 term is 'a', and it solely determines whether the parabola opens upward or downward.
If 'a' is negative, then the parabola opens downward. The way to remember this is that 'a' being negative forms a negative frown.
On the other hand if 'a' is positive, then it forms a positive smile, and the parabola opens upward.
In this case, the points are fairly close to a parabola opening downward. This means 'a' is negative and a < 0.