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elena-s [515]
3 years ago
11

Cynthia wants to take out a $8500 loan with a 4.75% APR. She can afford to pay $245 per month for loan payments. How long should

he borrow the money so that she can afford the monthly payment?
Mathematics
1 answer:
coldgirl [10]3 years ago
5 0

Answer:

<em>She should borrow the money for 3.1161... years so that she can afford the monthly payment.</em>

Step-by-step explanation:

<u>Monthly payment formula</u> is:    M=P*\frac{r(1+r)^n}{(1+r)^n -1}  , where

M = Monthly payment amount, P = Loan amount, r = rate of interest per month and n = total number of months.

Given that, Cynthia wants to take out a $8500 loan with a 4.75% APR and she can afford to pay $245 per month.

That means,  P= 8500, M= 245 and r= \frac{0.0475}{12}= 0.0039583

Plugging these values into the above formula, we will get........

245=8500*\frac{0.0039583(1+0.0039583)^n}{(1+0.0039583)^n-1} \\ \\ 245=\frac{33.64555(1.0039583)^n}{(1.0039583)^n -1}\\ \\ 245(1.0039583)^n -245=33.64555(1.0039583)^n\\ \\ 211.35445(1.0039583)^n=245\\ \\ (1.0039583)^n=\frac{245}{211.35445}=1.15919\\ \\ n= log_{1.0039583}(1.15919)=37.39327

So, n= 37.39327 months =\frac{37.39327}{12} years = 3.1161... years

That means, she should borrow the money for 3.1161... years so that she can afford the monthly payment.

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If we substitute these values back into the expression...\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)

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The rest is just pure simplification:

3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}

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