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:
294
Step-by-step explanation:
Answer:
For number 1, -1.125 and -9/8. For number 2, -46.
Step-by-step explanation:
Answer: -$6,500
Step-by-step explanation:
Here we could , use the arithmetic progression where
T(2020 - 2010) = a + ( n - 1 )d
T10 = a + ( 10 - 1 )d --------------- 1
a = $25,000, n = 10 and d = 14% of $25,000 = $3,500 the common difference.
Note since it decreases the common difference d = -$3,500.
Now substitute for the values in the equation above.
T10 = 25,000 + 9 x -3,500
= $25,000 - $31,500
= -$6,500 (deficit )
Answer:
RT and KS
Step-by-step explanation:
isosceles triangle
