8^3=512
5*12=60
512+60= 572
Given :
A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .
To Find :
A. The vector position of the particle at any time t .
B. The acceleration of the particle at any time t .
Solution :
A )
Position of vector v is given by :

B )
Acceleration a is given by :

Hence , this is the required solution .
Answer:
B ($9.78)
Step-by-step explanation:
Answer:
37.3032cm^2
Step-by-step explanation:
Given data
Diameter= 0.6cm
Radius= 0.3cm
Lenght = 19.5cm
Surface area (A) =2πrh+2πr^2
Substitute
A=2*3.14*0.3*19.5+2*3.14*0.3^2
A=36.738+0.5652
A=37.3032
Hence the surface area is
37.3032cm^2
Answer:
x+((4 + 1/3))=(-(2 + 5/6))
6x = -43
x≈ -7.166667