Motion and rate of change derivatives. attached is the question and where i am stuck, i would really appreciate the : )

A. 3 < t < 9

Step-by-step explanation:

When the particle is moving to the right, the velocity is positive.

To find the velocity, take the derivative of s(t) with respect to t.

s(t) = -t³/3 + 13t²/2 − 30t

s'(t) = -t² + 13t − 30

Find when the velocity is 0.

0 = -t² + 13t − 30

0 = t² − 13t + 30

0 = (t − 3) (t − 10)

t = 3, 10

Check the sign of s'(t) in each interval.

0 < t < 3, s'(t) < 0

3 < t < 9, s'(t) > 0

The particle moves to the right when 3 < t < 9.

96 39 cause 135 -96=39 im smart (i used a calculater)

answer: can you give the

step-by-step explanation: