A storage tank. I hope this helps ^^
Answer:the car was traveling at a speed of 80 ft/s when the brakes were first applied.
Step-by-step explanation:
The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,
a = - 16ft/s^2
While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.
This means that it travelled a distance,
s = 200 feet
We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.
Since the car came to a stop, its final velocity, v = 0
Applying Newton's equation of motion,
v^2 = u^2 + 2as
0 = u^2 - 2 × 16 × 200
u^2 = 6400
u = √6400
u = 80 ft/s
Answer:
4
Step-by-step explanation:
ΔABC is congruent with ΔDEF. That means the corresponding sides are congruent.
Look at the order of the letters. EF are the last two letters of ΔDEF. So this corresponds with the last two letters of ΔABC. Therefore, EF ≅ BC.
EF ≅ BC
x² + 8x = 48
x² + 8x − 48 = 0
(x + 12) (x − 4) = 0
x = -12 or 4
Since x can't be negative, x = 4.
Hello,
f(x)=2x^2-3x+4
f(3a)=2(3a)^2-3*(3a)+4=2*9a²-9a+4=18a²-9a+4
We are given with
P(pop quiz) = 60%
P(not do homework) = 85%
And the condition that
P (pop quiz and do homework) > 5%
So,
P (pop quiz) x P (do homework)
P (pop quiz) x ( 1 - P (not do homework) )
60% x ( 1 - 85%)
The result is greater than five percent so he will not do his homework.
