Answer:
Thus we find that velocity vector at time t is
(5t+15, 5t^2/2, 4t^2)
Step-by-step explanation:
given that acceleration vector is a funciton of time and at time t

v(t) can be obtained by integrating a(t)
v(t) = 
Thus we use the fact that acceleration is derivative of velocity and velocity is antiderivative of acceleration.
The arbitary constant normally used for integration C is here C vector = initial velocity (u0,v0,w0)
Position vector can be obtained by integrating v(t)
Thus we find that velocity vector at time t is
(5t+15, 5t^2/2, 4t^2)
Answer:
6
Step-by-step explanation:
to get cups to servings multiply by 3
if you need 18 servings then divide by 3
Ln(xy) - 2x =0
slope of the tangent line = derivative of the function
[ln(xy)]' = [2x]'
[1/(xy)] [y + xy'] = 2
y + xy' = 2(xy)
xy' = 2xy - y =y(2x-1)
y' = y(2x-1)/x
Now use x = -1 to find y and after to find y'
ln(xy) = 2x
x=-1
ln(-y) =-2
-y = e^-2
y = - e^-2
y' = [-e^-2][2(-1)-1]/(-1) = [e^-2](-2-1)= [e^-2](-3) = - 3e^-2
Answer: option 6. from the list
28-53= l -25l
Answer is 25
4x + 80 = 180
x = 25
2x + 80 = 130
2x = 50