Answer:
It goes to zero three times
Step-by-step explanation:
s(t) = e^ cos(x)
To find the velocity, we have to take the derivative of the position
ds/dt = -sin x e^ cos x dx/dt
Now we need to find when this is equal to 0
0 = -sin x e^ cos x
Using the zero product property
-sin x=0 e^cos x= 0
sin x = 0
Taking the arcsin of each side
arcsin sinx= arcsin 0
x = 0 ,pi, 2 pi
e^cos x= 0
Never goes to zero
The velocity is equal to 0 for 3 times.
Given position function s = ecos(x)
Its velocity function, s' = ds/dt = e(-sinx)dx/dt
Between [0,2π], s'=0, -e(sinx)dx/dt=0
sinx=0
x=0, π, 2π
Each roll is $1.43.
20.02 ÷ 14 = 1.43.
1.43 × 14 = 20.02.
5
You first start by finding the acceleration of the problem.
a=f/m
Which this case is 45.
The question is asking " How many times greater must it be than A?"
You then divide
45/9
Your final answer is 5.
