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.

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.