1. **State the problem:** We are given the function $H(t) = -4\sin\left(\frac{\pi}{3}(t - 1)\right) + 4$ describing Chloe's height in a wave pool over time $t$ seconds. 2. **Find the period (one full cycle) of the wave:** The general form for the period of a sine function $y = \sin(Bx)$ is $\text{Period} = \frac{2\pi}{B}$. Here, $B = \frac{\pi}{3}$, so $$\text{Period} = \frac{2\pi}{\frac{\pi}{3}} = 2\pi \times \frac{3}{\pi} = 6 \text{ seconds}.$$ This means the wave repeats every 6 seconds. 3. **Find Chloe's maximum height:** The amplitude $A$ is the coefficient of the sine function, here $-4$. The vertical shift is $+4$. Maximum height = vertical shift + amplitude magnitude = $4 + 4 = 8$ feet. 4. **Find Chloe's height at $t=6$ seconds:** $$H(6) = -4\sin\left(\frac{\pi}{3}(6 - 1)\right) + 4 = -4\sin\left(\frac{\pi}{3} \times 5\right) + 4 = -4\sin\left(\frac{5\pi}{3}\right) + 4.$$ Recall $\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}$, so $$H(6) = -4 \times \left(-\frac{\sqrt{3}}{2}\right) + 4 = 2\sqrt{3} + 4.$$ Thus, Chloe's height at 6 seconds is exactly $2\sqrt{3} + 4$ feet. 5. **Find when Chloe is closest to the bottom of the pool in the first 18 seconds:** The bottom corresponds to the minimum height, which occurs at the troughs of the sine wave. Since the period is 6 seconds, troughs occur every 6 seconds starting from the phase shift. Given the function, troughs are at approximately $t = 2.5, 8.5, 14.5$ seconds. 6. **Determine when Chloe is higher than 10 feet:** From the graph and function, the maximum height is 8 feet, so Chloe never exceeds 10 feet. **Final answers:** - a) One full cycle length: $6$ seconds - b) Maximum height: $8$ feet - c) Height at 6 seconds: $2\sqrt{3} + 4$ feet - d) Times closest to bottom: $2.5$, $8.5$, and $14.5$ seconds - e) Chloe is never higher than 10 feet.