1. **Problem statement:** We analyze the water volume function $f(t)$ over time $t$ hours, given a graph with volume in cubic meters. 2. **Volume at 5 hours:** From the graph, find $f(5)$, the volume at $t=5$ hours. 3. **Time interval with volume at least 350 m³:** Find $t$ such that $f(t) \geq 350$. 4. **Momentary rate of change at 2 hours:** Calculate $f'(2)$, the derivative of $f$ at $t=2$ hours, representing the instantaneous change rate. 5. **Time when the basin is empty given constant rate after 15 hours:** Given $f'(15)$ remains constant until $f(t)=0$, find $t$ where $f(t)=0$. 6. **Interpretation of $f(t+6) = f(t) - 350$:** This means the volume decreases by 350 m³ every 6 hours. **Step 1: Volume at 5 hours** From the graph, $f(5) \approx 500$ m³ (estimate from the curve). **Step 2: Time interval with $f(t) \geq 350$** Find $t$ where the curve is above or equal to 350 m³. From the graph, this is approximately from $t=2$ hours to $t=16$ hours. **Step 3: Momentary rate of change at 2 hours** The slope of the tangent at $t=2$ is $f'(2)$. Estimate slope from graph: if volume changes from about 600 m³ at $t=0$ to about 500 m³ at $t=5$, slope is $$f'(2) \approx \frac{f(5)-f(0)}{5-0} = \frac{500-600}{5} = \frac{-100}{5} = -20 \text{ m}^3/\text{h}.$$ **Step 4: Time when basin is empty with constant rate after 15 hours** Given $f'(15) \approx$ slope of tangent near $t=15$ to $20$ is about $-70$ m³/h (steeper slope). Volume at $t=15$ is about 350 m³. Using linear decrease: $$f(t) = f(15) + f'(15)(t-15) = 350 - 70(t-15).$$ Set $f(t)=0$: $$0 = 350 - 70(t-15)$$ $$70(t-15) = 350$$ $$t-15 = \frac{350}{70} = 5$$ $$t = 20 \text{ hours}.$$ **Step 5: Interpretation of $f(t+6) = f(t) - 350$** This means every 6 hours, the volume decreases by 350 m³. A solution is any $t$ where this holds, e.g., $t=0$: $$f(6) = f(0) - 350.$$