Answer :
Sure, let's solve each part of the problem step-by-step!
### (a) What is the tuition for 10 credits?
We have a piecewise function for the tuition, [tex]\( T(c) \)[/tex].
1. Identify which case to use:
- For the number of credits, [tex]\( c = 10 \)[/tex], we see that it falls in the range [tex]\( 8 < c \leq 18 \)[/tex].
2. Apply the correct formula:
- Since [tex]\( 10 \)[/tex] falls into the second case, we use the formula:
[tex]\[
T(c) = 800 + 190(c - 8)
\][/tex]
3. Substitute [tex]\( c = 10 \)[/tex] into the formula:
- Calculate the value:
[tex]\[
T(10) = 800 + 190(10 - 8)
\][/tex]
[tex]\[
T(10) = 800 + 190 \times 2
\][/tex]
[tex]\[
T(10) = 800 + 380
\][/tex]
[tex]\[
T(10) = 1180
\][/tex]
So, the tuition for 10 credits is [tex]$1180.
### (b) If the tuition was $[/tex]2510, how many credits were taken?
The goal is to find the number of credits [tex]\( c \)[/tex] that makes the tuition [tex]$2510.
1. Understand conditions for each case:
- Assume \( 0 \leq c \leq 8 \):
\[
T(c) = 100 + 190c
\]
If \( 100 + 190c = 2510 \), this simplifies to:
\[
190c = 2410
\]
\[
c = \frac{2410}{190} \approx 12.68
\]
Since \( c \) must be a whole number and within 0 to 8, this is not possible.
2. Case \( 8 < c \leq 18 \):
- Use the formula:
\[
T(c) = 800 + 190(c - 8)
\]
Set it equal to 2510:
\[
800 + 190(c - 8) = 2510
\]
Solve for \( c \):
\[
190(c - 8) = 2510 - 800
\]
\[
190(c - 8) = 1710
\]
\[
c - 8 = \frac{1710}{190}
\]
\[
c - 8 = 9
\]
\[
c = 17
\]
So, if the tuition was $[/tex]2510, the number of credits taken is 17 credits.
### (a) What is the tuition for 10 credits?
We have a piecewise function for the tuition, [tex]\( T(c) \)[/tex].
1. Identify which case to use:
- For the number of credits, [tex]\( c = 10 \)[/tex], we see that it falls in the range [tex]\( 8 < c \leq 18 \)[/tex].
2. Apply the correct formula:
- Since [tex]\( 10 \)[/tex] falls into the second case, we use the formula:
[tex]\[
T(c) = 800 + 190(c - 8)
\][/tex]
3. Substitute [tex]\( c = 10 \)[/tex] into the formula:
- Calculate the value:
[tex]\[
T(10) = 800 + 190(10 - 8)
\][/tex]
[tex]\[
T(10) = 800 + 190 \times 2
\][/tex]
[tex]\[
T(10) = 800 + 380
\][/tex]
[tex]\[
T(10) = 1180
\][/tex]
So, the tuition for 10 credits is [tex]$1180.
### (b) If the tuition was $[/tex]2510, how many credits were taken?
The goal is to find the number of credits [tex]\( c \)[/tex] that makes the tuition [tex]$2510.
1. Understand conditions for each case:
- Assume \( 0 \leq c \leq 8 \):
\[
T(c) = 100 + 190c
\]
If \( 100 + 190c = 2510 \), this simplifies to:
\[
190c = 2410
\]
\[
c = \frac{2410}{190} \approx 12.68
\]
Since \( c \) must be a whole number and within 0 to 8, this is not possible.
2. Case \( 8 < c \leq 18 \):
- Use the formula:
\[
T(c) = 800 + 190(c - 8)
\]
Set it equal to 2510:
\[
800 + 190(c - 8) = 2510
\]
Solve for \( c \):
\[
190(c - 8) = 2510 - 800
\]
\[
190(c - 8) = 1710
\]
\[
c - 8 = \frac{1710}{190}
\]
\[
c - 8 = 9
\]
\[
c = 17
\]
So, if the tuition was $[/tex]2510, the number of credits taken is 17 credits.
