1. **State the problem:** We need to write a function $V(t)$ that represents the amount of coffee in milliliters (mL) at time $t$ seconds, where coffee is dropping into a 10-cup carafe at a rate of 1 mL/s. 2. **Understand the problem:** Each cup is typically 125 mL, so 10 cups equal $10 \times 125 = 1250$ mL. The carafe starts empty, and coffee is added at 1 mL per second. 3. **Write the function:** Since coffee is added at 1 mL/s, the amount of coffee after $t$ seconds is $t$ mL. If the carafe starts empty, the function is: $$V(t) = t$$ If the carafe already contains some coffee, say $V_0$ mL, then: $$V(t) = V_0 + t$$ 4. **Given the problem's expression $V(t) = \text{mL}(t) + 10$, it seems to add 10 mL initially. So the function is:** $$V(t) = t + 10$$ 5. **Domain:** Time $t$ cannot be negative, so: $$t \geq 0$$ 6. **Range:** The amount of coffee starts at 10 mL and increases by 1 mL per second, so: $$V(t) \geq 10$$ 7. **Summary:** - Function: $V(t) = t + 10$ - Domain: $t \geq 0$ - Range: $V(t) \geq 10$ This function models the amount of coffee in the carafe over time, starting with 10 mL and increasing by 1 mL every second.