It looks like you've uploaded an image showing a math problem about estimating instantaneous rates of change. Let me transcribe and solve one of the problems for you:

Problem 5:

"An object is sent through the air. Its height is modelled by the function $h(x) = -5x^2 + 3x + 65$, where $h(x)$ is the height of the object in meters, and $x$ is the time in seconds. Estimate the instantaneous rate of change in the object's height at $x = 3$ seconds."

To solve this, we'll compute the derivative of the height function, $h(x)$, to get the instantaneous rate of change, and then substitute $x = 3$.

Solution:

1. The height function is $h(x) = -5x^2 + 3x + 65$.
2. Differentiate $h(x)$ to find the rate of change function, $h'(x)$: $h'(x) = \frac{d}{dx}[-5x^2 + 3x + 65] = -10x + 3$
3. To estimate the instantaneous rate of change at $x = 3$ seconds, substitute $x = 3$ into $h'(x)$: $h'(3) = -10(3) + 3 = -30 + 3 = -27$

Thus, the instantaneous rate of change in the object's height at $x = 3$ seconds is $-27$ meters per second. This means the object is falling at a rate of 27 m/s at that moment.

Would you like me to solve the next problem (Problem 6) or provide further details on this solution?

Here are 5 related questions to expand your understanding:

1. What is the geometric interpretation of the derivative $h'(x)$ in the context of this problem?
2. How would the object's motion change if the coefficient of $x^2$ in $h(x)$ were positive?
3. What is the meaning of the constant term 65 in the function $h(x)$?
4. How does the sign of $h'(x)$ indicate whether the object is rising or falling at a given point?
5. How can you estimate instantaneous rates of change if a function is given in a table rather than an equation?

Tip:

When estimating instantaneous rates of change from a function, always remember that the derivative gives the slope of the tangent line at any point, which corresponds to the rate of change at that exact moment.