Answer:
y = -2x + 16
the y intercept is the y value at x = 0. On the graph there is a dot at 16 on x= 0, so the y intercept would be 16.
the only answer choice adding 16 sis y = -2x + 16
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:
AL=2πrh
Step - by - step explanation:
For this case we have the following equation:
s = root (S.A / 6)
Substituting values we have:
For S.A = 180:
s = root (180/6)
s = root (30)
For S.A = 120:
s = root (120/6)
s = root (20)
s = root (4 * 5)
s = 2 * root (5)
Subtracting both values we have:
root (30) - 2 * root (5)
Answer:
root (30) - 2 * root (5)
option 2
Answer:
Right Angles: C, F
Obtuse Angles: E, B
Acute Angles: A, D
Step-by-step explanation:
Right Angles are angles that are exactly 90 degrees.
Obtuse Angles are angles that are bigger than 90 degrees.
Acute Angles are angles that are smaller than 90 degrees.
