1. **Problem statement:** We need to find a rule for the volume $V$ (in litres) of water in a flask as a function of time $t$ (in minutes), given that the flask is filled at a constant rate. 2. **General formula:** Since the filling rate is constant, the volume $V$ changes linearly with time $t$. The general linear formula is: $$V = mt + b$$ where $m$ is the rate of change (litres per minute) and $b$ is the initial volume at $t=0$. 3. **Part i:** Initially, $V=0$ at $t=0$, and after 1 minute, $V=4$ litres. Using $V=mt+b$, substitute $t=0$, $V=0$: $$0 = m \times 0 + b \Rightarrow b=0$$ Substitute $t=1$, $V=4$: $$4 = m \times 1 + 0 \Rightarrow m=4$$ **Rule:** $$V = 4t$$ 4. **Part ii:** Initially, $V=0$ at $t=0$, and after 3 minutes, $V=9$ litres. At $t=0$, $V=0$ gives $b=0$. At $t=3$, $V=9$: $$9 = 3m + 0 \Rightarrow m=3$$ **Rule:** $$V = 3t$$ 5. **Part iii:** After 1 minute, $V=2$ litres; after 2 minutes, $V=3$ litres. We have two points: $(1,2)$ and $(2,3)$. Calculate slope $m$: $$m = \frac{3 - 2}{2 - 1} = 1$$ Use point-slope form with point $(1,2)$: $$2 = 1 \times 1 + b \Rightarrow b = 2 - 1 = 1$$ **Rule:** $$V = t + 1$$ 6. **Part iv:** After 1 minute, $V=3.5$ litres; after 2 minutes, $V=5$ litres. Points: $(1,3.5)$ and $(2,5)$. Slope $m$: $$m = \frac{5 - 3.5}{2 - 1} = 1.5$$ Using point $(1,3.5)$: $$3.5 = 1.5 \times 1 + b \Rightarrow b = 3.5 - 1.5 = 2$$ **Rule:** $$V = 1.5t + 2$$ 7. **Part b:** Find initial volume $b$ for parts iii and iv. - Part iii: $b=1$ litre - Part iv: $b=2$ litres 8. **Part c:** Write your own information that gives the rule $V = -t + b$. For example, if initially the flask has 5 litres ($b=5$), and water is being removed at 1 litre per minute ($m=-1$), then: $$V = -t + 5$$ This means at $t=0$, $V=5$ litres, at $t=1$, $V=4$ litres, etc. **Final answers:** - i) $V=4t$ - ii) $V=3t$ - iii) $V=t+1$ - iv) $V=1.5t+2$ - Initial volumes: iii) 1 litre, iv) 2 litres - Example for $V=-t+b$: $V=-t+5$ with initial volume 5 litres and decreasing volume over time.