https://file.aiquickdraw.com/mathbot/file/1726695044448b5mm61xa.jpg

Let's first address the two math problems shown in the image.

**Problem 10:**

The equation given is:
\[
(x - 1)(x + 2) = 18
\]

To solve this:
1. Expand the left-hand side:
   \[
   (x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2
   \]
   
2. Set the equation equal to 18:
   \[
   x^2 + x - 2 = 18
   \]

3. Move all terms to one side to set the equation to 0:
   \[
   x^2 + x - 20 = 0
   \]

4. Factor the quadratic equation:
   \[
   (x - 4)(x + 5) = 0
   \]

5. Solve for \(x\):
   \[
   x - 4 = 0 \quad \text{or} \quad x + 5 = 0
   \]
   \[
   x = 4 \quad \text{or} \quad x = -5
   \]

So, the solutions to the equation are \(x = 4\) and \(x = -5\).

---

**Problem 11:**

The function provided is:
\[
h(t) = 5t - 10t^2 + 10, \quad t \geq 0
\]
where \(h(t)\) represents the height of the diver at time \(t\) seconds after launch.

This function is a quadratic equation, and we can analyze it further.

1. **Maximum Height:**

   Since the equation is quadratic and has a negative coefficient for \(t^2\), the graph is a parabola that opens downwards. The maximum height occurs at the vertex. The formula to find the time of the vertex is:
   \[
   t_{\text{max}} = \frac{-b}{2a}
   \]
   where \(a = -10\) and \(b = 5\). Substituting the values:
   \[
   t_{\text{max}} = \frac{-5}{2(-10)} = \frac{5}{20} = 0.25 \, \text{seconds}
   \]

2. **Height at Maximum:**
   Substitute \(t = 0.25\) into the height function to find the maximum height:
   \[
   h(0.25) = 5(0.25) - 10(0.25)^2 + 10
   \]
   Simplifying:
   \[
   h(0.25) = 1.25 - 10(0.0625) + 10 = 1.25 - 0.625 + 10 = 10.625 \, \text{metres}
   \]

Thus, the diver reaches a maximum height of **10.625 metres** at **0.25 seconds** after launch.

Would you like to explore any part of this further, or do you have specific questions?

### Follow-up questions:
1. What are the steps to solving any quadratic equation?
2. How can the vertex formula help in real-life problems involving parabolas?
3. How do you calculate the time it takes for the diver to reach the water?
4. How do you interpret the solutions to the quadratic equation in problem 10?
5. Can you graph both problems and discuss their geometric interpretations?

**Tip:** Always check if a quadratic equation can be factored easily before using the quadratic formula.

A diver launches herself off a springboard. The height of the diver, in metres, above the pool t seconds after launch can be modelled by the following function: h(t) = 5t - 10t^2 + 10, t ≥ 0.

This problem explores a diver's height over time using a quadratic function. Learn how to calculate the maximum height using the vertex formula and analyze the height function h(t) = 5t - 10t^2 + 10. Suitable for Grades 10-12.

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