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Answer :
Sure! Let's solve the problem step-by-step to find the number of classes, [tex]\( c \)[/tex], for which Class A and Class B cost the same.
## Fee Structure:
- Class A charges [tex]\( \$40 \)[/tex] per class without any registration fee.
- Class B charges a [tex]\( \$60 \)[/tex] registration fee plus [tex]\( \$25 \)[/tex] per class.
We want to determine for what number of classes, [tex]\( c \)[/tex], the total cost of both classes will be equal.
### Step-by-Step Solution:
1. Express the Cost for Each Class:
- For Class A, the cost after attending [tex]\( c \)[/tex] classes will be:
[tex]\[
\text{Cost of Class A} = 40c
\][/tex]
- For Class B, the cost will be:
[tex]\[
\text{Cost of Class B} = 60 + 25c
\][/tex]
2. Set the Costs Equal to Find [tex]\( c \)[/tex]:
To find the point where both classes cost the same, we set the expressions equal to each other:
[tex]\[
40c = 60 + 25c
\][/tex]
3. Solve for [tex]\( c \)[/tex]:
- First, we need to get all the terms involving [tex]\( c \)[/tex] on one side and the constants on the other:
[tex]\[
40c - 25c = 60
\][/tex]
- Simplify the left side:
[tex]\[
15c = 60
\][/tex]
- Finally, solve for [tex]\( c \)[/tex] by dividing both sides by 15:
[tex]\[
c = \frac{60}{15} = 4
\][/tex]
So, the equation from the step where we set the costs equal is:
[tex]\[
40c = 25c + 60
\][/tex]
And the number of classes, [tex]\( c \)[/tex], for which Class A and Class B cost the same is:
[tex]\[
c = 4
\][/tex]
## Fee Structure:
- Class A charges [tex]\( \$40 \)[/tex] per class without any registration fee.
- Class B charges a [tex]\( \$60 \)[/tex] registration fee plus [tex]\( \$25 \)[/tex] per class.
We want to determine for what number of classes, [tex]\( c \)[/tex], the total cost of both classes will be equal.
### Step-by-Step Solution:
1. Express the Cost for Each Class:
- For Class A, the cost after attending [tex]\( c \)[/tex] classes will be:
[tex]\[
\text{Cost of Class A} = 40c
\][/tex]
- For Class B, the cost will be:
[tex]\[
\text{Cost of Class B} = 60 + 25c
\][/tex]
2. Set the Costs Equal to Find [tex]\( c \)[/tex]:
To find the point where both classes cost the same, we set the expressions equal to each other:
[tex]\[
40c = 60 + 25c
\][/tex]
3. Solve for [tex]\( c \)[/tex]:
- First, we need to get all the terms involving [tex]\( c \)[/tex] on one side and the constants on the other:
[tex]\[
40c - 25c = 60
\][/tex]
- Simplify the left side:
[tex]\[
15c = 60
\][/tex]
- Finally, solve for [tex]\( c \)[/tex] by dividing both sides by 15:
[tex]\[
c = \frac{60}{15} = 4
\][/tex]
So, the equation from the step where we set the costs equal is:
[tex]\[
40c = 25c + 60
\][/tex]
And the number of classes, [tex]\( c \)[/tex], for which Class A and Class B cost the same is:
[tex]\[
c = 4
\][/tex]
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