1. **Problem 7a:** Calculate the cost to park for 10 days given the rate is $20 per day for the first 7 days and $17 per day thereafter. 2. **Formula:** Total cost = (cost per day for first 7 days) \times 7 + (cost per day after 7 days) \times (number of days - 7) 3. **Calculation:** - Cost for first 7 days = $20 \times 7 = 140$ - Days after 7 = $10 - 7 = 3$ - Cost for next 3 days = $17 \times 3 = 51$ - Total cost = $140 + 51 = 191$ 4. **Answer:** The cost to park for 10 days is $191. 5. **Problem 7b:** Draw the graph of the cost function $C(d)$ where $d$ is the number of days. 6. **Function definition:** $$ C(d) = \begin{cases} 20d & \text{if } 0 < d \leq 7 \\ 140 + 17(d - 7) & \text{if } d > 7 \end{cases} $$ 7. This is a piecewise linear function with a slope of 20 for $d \leq 7$ and slope 17 for $d > 7$. 8. **Problem 8:** Evaluate the piecewise function $$ f(x) = \begin{cases} 1 - x & \text{if } x \leq -1 \\ x^2 & \text{if } x > -1 \end{cases} $$ 9. Calculate: - $f(-2) = 1 - (-2) = 1 + 2 = 3$ - $f(-1) = 1 - (-1) = 1 + 1 = 2$ - $f(0) = 0^2 = 0$ 10. The graph consists of a line $y = 1 - x$ for $x \leq -1$ and a parabola $y = x^2$ for $x > -1$. 11. **Problem 9:** Write the function represented by the graph described. 12. From the description: - For $x$ from $-4$ to $-1$, the function is linear decreasing from $y=4$ to $y=1$. - At $x=-1$, there is a jump discontinuity: $f(-1) = 1$ (filled point), but the limit from the right is 2 (open circle). - For $x > -1$, the function is quadratic increasing from $y=0$ at $x=0$ upwards. 13. Define the function: $$ f(x) = \begin{cases} -x & \text{if } x \leq -1 \\ x^2 & \text{if } x > -1 \end{cases} $$ 14. This matches the linear segment (since $-(-4) = 4$, $-(-1) = 1$) and the quadratic segment for $x > -1$. **Final answers:** - 7a: $191$ - 7b: $C(d) = \begin{cases} 20d & d \leq 7 \\ 140 + 17(d-7) & d > 7 \end{cases}$ - 8: $f(-2) = 3$, $f(-1) = 2$, $f(0) = 0$ - 9: $f(x) = \begin{cases} -x & x \leq -1 \\ x^2 & x > -1 \end{cases}$