Using the monthly payments formula, it is found that a car with a value of at most $25,293.
It is given by:
[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]
In which:
In this problem, we have that the parameters are given as follows:
A = 400, n = 70, r = 0.035.
Hence:
r/12 = 0.035/12 = 0.002917.
Then we have to solve for P to find the maximum value of the car.
[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]
[tex]400 = P\frac{0.002917(1.002917)^{70}}{(1.002917)^{70}-1}[/tex]
[tex]P = \frac{400[(1.002917)^{70}-1]}{0.002917(1.002917)^{70}}[/tex]
P = $25,293.
More can be learned about the monthly payments formula at https://brainly.com/question/26267630
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