Answer:
B) x^2+6x+8
Explanation:
x-4 | x^3+2x^2-16x-32
- x^3-4x^2 <-- (x-4)(x^2)
_________________
6x^2-16x-32
- 6x^2-24x <-- (x-4)(6x)
_________________
8x-32
- 8x-32 <- (x-4)(8)
___________________________
0 | x^2+6x+8
This means the answer is B) x^2+6x+8
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
I believe it is acceleration!
2.27 mps repeating.
This is the last question ill ever answer here. Thanks for being the last.
