If the displacement of a particle is given by the formula S = 3t^3 - 20t^2 + 40t. Find (i) the displacement after 3 seconds (ii) the formula for the velocity at any time t (iii) the value of t when the particle is not moving (iv) the initial velocity of the particle (v) the formula for the acceleration at any time t (vi) the initial acceleration of the particle.​

I) displacement is s(3) = 81 - 180 +120

II) below. Iii) v= 0 when. 9t^2 -40t +40 = 0

IV) below. V) when t = 0 a(0) = -40

Step-by-step explanation: speed is first derivative of distance. V = s' (t)

V(t) = 9t^2 -40t +40

Acceleration is second derivative of distance. A= v'(t) = 18t -40

-13

step-by-step explanation:

f(x)=2x-5 for f(-4)

f(-4)   means let x=-4

f(-4) = 2(-4) -5

f(-4) =-8-5

      = -13

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b