1. **Problem Statement:** Part A: The temperature $T(x)$ in °C decreases by 6.6 °C per km above sea level, starting from 16 °C at sea level ($x=0$ km). Part B: The air pressure $L(x)$ in hPa halves every 6 km increase in altitude, starting from 1013 hPa at sea level. --- 2. **Part A (Temperature function and value):** - The temperature decreases linearly with altitude. - The formula for a linear function is $T(x) = T_0 + mx$, where $T_0$ is the temperature at sea level and $m$ is the rate of change. - Here, $T_0 = 16$ and $m = -6.6$ (since temperature decreases). - So, $$T(x) = 16 - 6.6x$$ - To find temperature at $x=3.6$ km: $$T(3.6) = 16 - 6.6 \times 3.6 = 16 - 23.76 = -7.76$$ - Rounded to two decimals, $T(3.6) = -7.76$ °C. --- 3. **Part B (Air pressure function and calculations):** - Air pressure halves every 6 km, so after 6 km, pressure is $\frac{1}{2} \times 1013 = 506.5$ hPa. - This suggests an exponential decay model: $$L(x) = L_0 \cdot a^x$$ where $L_0 = 1013$ and $a$ is the base. - Since pressure halves every 6 km, $$L(6) = 1013 \cdot a^6 = \frac{1013}{2}$$ - Divide both sides by 1013: $$a^6 = \frac{1}{2}$$ - Taking the 6th root: $$a = \sqrt[6]{\frac{1}{2}} = (\frac{1}{2})^{\frac{1}{6}} \approx 0.8909$$ - The problem states $L(x) = 1013 \cdot 0.87^x$, which is an approximation. - So, the function is approximately: $$L(x) = 1013 \cdot 0.87^x$$ --- 4. **Calculate air pressure at Zugspitze (3 km) and Mount Everest (8 km):** - For Zugspitze ($x=3$): $$L(3) = 1013 \cdot 0.87^3 = 1013 \cdot 0.6585 = 667.6$$ - Rounded: 668 hPa. - For Mount Everest ($x=8$): $$L(8) = 1013 \cdot 0.87^8 = 1013 \cdot 0.343 = 347.5$$ - Rounded: 348 hPa. --- 5. **Calculate maximum height for minimum air pressure 400 hPa:** - Given $L(x) = 1013 \cdot 0.87^x$ and $L(x) = 400$: $$400 = 1013 \cdot 0.87^x$$ - Divide both sides by 1013: $$\frac{400}{1013} = 0.87^x$$ - Simplify fraction: $$\approx 0.3947 = 0.87^x$$ - Take natural logarithm on both sides: $$\ln(0.3947) = \ln(0.87^x) = x \ln(0.87)$$ - Solve for $x$: $$x = \frac{\ln(0.3947)}{\ln(0.87)}$$ - Calculate: $$\ln(0.3947) \approx -0.930$$ $$\ln(0.87) \approx -0.1398$$ - So, $$x = \frac{-0.930}{-0.1398} = 6.65$$ - Rounded to one decimal place: $6.6$ km. --- **Final answers:** - Part A: a) $T(x) = 16 - 6.6x$ b) $T(3.6) = -7.76$ °C - Part B: a) $L(x) = 1013 \cdot 0.87^x$ b) $L(3) = 668$ hPa, $L(8) = 348$ hPa c) Maximum height without mask: $6.6$ km