Aparticle moves along the x–axis so that at time t its position is given by x(t) = (t+1)(t–3)3. for what values of t is the velocity of the particle increasing?

I'll assume that's an exponent at the end there:

The first derivative gives the velocity.  The second derivative gives the acceleration; increasing velocity is the same as positive acceleration.  So we want to find when the second derivative is positive.

Let's see if we can use to avoid multiplying this out.

That worked; let's do it again:

That's a nice parabola.  It's zero or negative for so positive everywhere else:

Increasing velocity when t < 1 or t > 3

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3