Answer:
1.33×10^-6
Step-by-step explanation:
5.32 x 10^-7= 5.32×10^(-5-2)= 5.32×10^-2×10^(-5)
=0.0532×10^-5
(0.0532×10^-5) (2.5 x 10^-5)
0.0532×2.5 × 10^-5
0.133×10^-5= 1.33×10^-1 ×10^-5
1.33×10^-6
Answer:
x<-11/26
Step-by-step explanation:
1/5(2/3x-1/2)>x+1/3
2/15x-1/10>x+1/3
2/15x-x>1/3+1/10
-13/15x>11/30
x<(11/30)(-15/13)
x<-11/26
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)