1. **State the problem:** We are given a temperature model for coffee cooling: $$T(t) = 56e^{kt} + 22$$ where $T(t)$ is the temperature at time $t$ minutes, and we want to find the constant $k$ given that at $t=6$ minutes, $T=60^\circ C$. 2. **Use the given data to find $k$:** Substitute $t=6$ and $T=60$ into the model: $$60 = 56e^{6k} + 22$$ 3. **Isolate the exponential term:** $$60 - 22 = 56e^{6k}$$ $$38 = 56e^{6k}$$ 4. **Divide both sides by 56:** $$\frac{38}{56} = e^{6k}$$ Show cancellation: $$\frac{\cancel{38}}{\cancel{56}} = e^{6k}$$ (Here, 38 and 56 do not simplify nicely, so keep as fraction or decimal.) 5. **Take natural logarithm on both sides:** $$\ln\left(\frac{38}{56}\right) = 6k$$ 6. **Solve for $k$:** $$k = \frac{1}{6} \ln\left(\frac{38}{56}\right)$$ Calculate the fraction and logarithm: $$\frac{38}{56} \approx 0.67857$$ $$\ln(0.67857) \approx -0.388$$ So, $$k = \frac{-0.388}{6} = -0.0647$$ Rounded to two significant figures: $$k \approx -0.065$$ 7. **Interpretation:** The negative $k$ indicates the temperature decreases over time, as expected. --- 8. **Find time to reach $22^\circ C$:** Set $T(t) = 22$: $$22 = 56e^{kt} + 22$$ Subtract 22: $$0 = 56e^{kt}$$ Since $56e^{kt} = 0$ implies $e^{kt} = 0$, which is impossible (exponential never zero), the temperature approaches 22°C asymptotically but never actually reaches it. 9. **Conclusion:** The model predicts the coffee temperature will get closer and closer to 22°C but will never exactly reach it in finite time.