1. **Problem Statement:** We have two functions describing water flow in a tank over time $t$ from 0 to 7. - Water pumped in: $N(t) = 3(1.23)^t$ - Water drained: $C(t) = -9(0.74)^t + 10$ 2. **Understanding the functions:** - $N(t)$ is an increasing exponential function because the base $1.23 > 1$. - $C(t)$ is a decreasing exponential function shifted by 10 because $0.74 < 1$ and it has a negative coefficient. 3. **Evaluate at $t=0$:** - $N(0) = 3(1.23)^0 = 3 \times 1 = 3$ - $C(0) = -9(0.74)^0 + 10 = -9 \times 1 + 10 = 1$ 4. **Evaluate at $t=7$:** - $N(7) = 3(1.23)^7$ Calculate $1.23^7$: $$1.23^7 \approx 1.23^3 \times 1.23^4 = (1.8609) \times (2.289) \approx 4.26$$ So, $$N(7) \approx 3 \times 4.26 = 12.78$$ - $C(7) = -9(0.74)^7 + 10$ Calculate $0.74^7$: $$0.74^7 \approx 0.74^3 \times 0.74^4 = (0.405) \times (0.299) \approx 0.121$$ So, $$C(7) \approx -9 \times 0.121 + 10 = -1.089 + 10 = 8.911$$ 5. **Interpretation:** - At $t=0$, water pumped in is 3 units, drained is 1 unit. - At $t=7$, water pumped in increases to about 12.78 units, drained decreases to about 8.91 units. 6. **Graph shapes:** - $N(t)$ grows exponentially from 3 upwards. - $C(t)$ starts near 1 and approaches 10 from below as $t$ increases. This analysis helps understand the water volume changes in the tank over time.