1. **Problem statement:** A metal cools at a constant rate. After 19 minutes, its temperature is 401 °C, and after 42 minutes, it is 148 °C. We need to find how time and temperature are related and the initial temperature. 2. **Formula used:** Since temperature changes at a constant rate, we use the linear equation for temperature $T$ as a function of time $t$: $$T = mt + b$$ where $m$ is the rate of change (slope) and $b$ is the initial temperature (intercept). 3. **Calculate the rate of change $m$:** $$m = \frac{T_2 - T_1}{t_2 - t_1} = \frac{148 - 401}{42 - 19} = \frac{-253}{23} = -11$$ This means the temperature decreases by 11 °C per minute. 4. **Write the linear equation:** $$T = -11t + b$$ 5. **Find the initial temperature $b$:** Use one point, for example at $t=19$, $T=401$: $$401 = -11 \times 19 + b$$ $$401 = \cancel{-209} + b$$ $$b = 401 + 209 = 610$$ 6. **Interpretation:** - As time increases, temperature decreases. - Temperature decreases at 11 °C per minute. - Initial temperature when cooling started ($t=0$) was 610 °C. **Final answers:** (a) As time increases, the temperature of the metal decreases. The temperature of the metal decreases at a rate of 11 °C per minute. (b) The initial temperature was 610 °C.