To determine the correct slope that models the situation described, we need to calculate the slope based on the change in amount remaining after each monthly payment.
Barton started with a loan balance of $2,625 and makes monthly payments of $175. The slope \( m \) of a line can be represented by the formula:
\[ m = \frac{\text{change in amount}}{\text{change in months}} \]
In this case, the change in amount is the payment amount (which is decreasing the balance), and the change in months is the amount of time that has passed (1 month per payment).
If he pays $175 every month, the balance after 1 month is:
\[ 2625 - 175 = 2450 \]
To find the slope after the first payment:
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Change in amount: From $2,625 to $2,450, the change is: \[ 2450 - 2625 = -175 \]
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Change in months: After 1 month, the change is \( 1 - 0 = 1 \).
Therefore, the slope after one month could be modeled as:
\[ m = \frac{-175}{1} = -175 \]
Now, we can compare it to the options provided:
- \( m = \frac{9 - 8}{1,050 - 1,225} \) — Not relevant; values don't fit financial model.
- \( m = \frac{525 - 700}{11 - 12} \) — Not relevant; values don't fit financial model.
- \( m = \frac{1,050 - 1,225}{9 - 8} \) — Not relevant; values don't fit.
- \( m = \frac{2,450 - 2,625}{0 - 1} \) — This correctly represents the change from the starting balance.
Calculating \( m \) for the last option:
\[ m = \frac{2450 - 2625}{0 - 1} = \frac{-175}{-1} = 175 \]
Thus, the proper slope indicative of the situation is represented by:
\[ m = \frac{2,450 - 2,625}{0 - 1} \]
So, the last option \( m=2,450-2,6250-1 \) is the one that models the correct slope.