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:
answer of this question is option A x-9 this represent solution
1. 3p: Add 6 to negative 3, and carry over the "p"
3. 6x: Subtract 1 from 7 and carry over the "x"
5. -4v: Add 6 to negative 10 and carry over the "v"
7. -4r + 9: Add 5 to -9 and carry over the "r". Then put "+9" after the variable, since the other 9 was by itself.
9. 14n: Add 5 to 9 and carry over the "n".
I hope this helps!
Answer: >72
Step-by-step explanation:
x*1/3-6=<18
1/3*x=<24
x=<72
Christine is more than 72 years old
Answer:
Step-by-step explanation:
as given in question that a > 0 so
if we put a=1
we get g(x) = f(x)
now put a =2
we get
g(x) = 2 f(x)
here we can see that g(x) would always be greater than or equals to f(x)
so we can say that the graph of g(x) will never be narrower than graph of g(x)
