The problem you're facing asks to find $s(7)$, the position of a particle at time $t = 7$, given the velocity function $v(t) = -10t + 15t^2$ and the fact that $s(3) = 72$.

Steps to solve:

1. Understand the relationship: Velocity $v(t)$ is the derivative of the position $s(t)$. Therefore, to find $s(t)$, we need to integrate $v(t)$.

2. Integrate the velocity function: $v(t) = -10t + 15t^2$ To find $s(t)$, integrate $v(t)$ with respect to $t$: $s(t) = \int (-10t + 15t^2) \, dt = -5t^2 + 5t^3 + C$ where $C$ is the constant of integration.

3. Use the initial condition $s(3) = 72$ to find $C$: $s(3) = -5(3)^2 + 5(3)^3 + C = -5(9) + 5(27) + C = -45 + 135 + C = 90 + C$ Since $s(3) = 72$, we can solve for $C$: $90 + C = 72 \implies C = 72 - 90 = -18$ So, the position function is: $s(t) = -5t^2 + 5t^3 - 18$

4. Find $s(7)$: $s(7) = -5(7)^2 + 5(7)^3 - 18 = -5(49) + 5(343) - 18 = -245 + 1715 - 18 = 1450 - 18 = 1432$

So, $s(7) = 1432$, which matches the answer provided in the image. However, the error likely comes from rounding instructions or missing decimal precision. Since the value is exactly 1432.00, it's correctly rounded to two decimal places.

Would you like more details or have any further questions?

Here are five related questions to expand your understanding:

1. What is the significance of integrating a velocity function?
2. How would the answer change if the initial condition $s(3)$ were different?
3. What is the physical meaning of the constant of integration in this context?
4. How do you handle rounding errors in applied problems like this one?
5. Can you apply this method to non-polynomial velocity functions?

Tip: Always double-check rounding instructions in problems—small errors in rounding can affect whether an answer is accepted or rejected.