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marta [7]
1 year ago
15

Olivia has taken out a $13,100 unsubsidized Stafford loan to pay for her college education. She plans to graduate in four years.

The loan has a duration of ten years and an interest rate of 7.6%, compounded monthly. By the time Olivia graduates, how much greater will the amount of interest capitalized be than the minimum amount that Olivia could pay to prevent interest capitalization? Round all dollar values to the nearest cent.
a. $654.45
b. $477.27
c. $995.60
d. $354.22
Mathematics
1 answer:
Mumz [18]1 year ago
6 0

If Olivia wants to prevent interest capitalization, she must pay the accrued interest each month. That amount is

... I = Prt = $13,100×0.076×(1/12) = $82.97

Over 4 years (48 months), these payments total $82.97×48 = $3982.56.

_____

If no payments are made, the loan balance grows by the multiplier (1 +r/12) each month. Then the amount of interest that will be capitalized at the end of 48 months is ...

... $13,100×((1 +0.076/12)^48 -1) = $4637.01

The difference in these amounts is ...

... $4637.01 -3982.56 = $654.45 . . . . . matches selection a.

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Notice that even tough our polynomial is written in descending form, x^3 is missing, so we are going to use 0x^3 to fill the gap:

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In our case the first term of the dividend is 3x^4 and the first term of the divisor is x, so \frac{3x^4}{x} =3x^3. Notice that 3x^3 is the first term of the quotient.

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The first term of our quotient (from the previous step) is 3x^3 and the divisor is (x-1), so 3x^3(x-2)=3x^4-6x^3. Now, we are going to subtract them from the respective term (the term with the same power) of the dividend. The respective terms of the dividend are 3x^4 and 0x^3, so 3x^4-3x^4=0 and 0x^3-(-6x^3)=0x^3+6x^3=6x^3

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The correct question is:

Consider the initial value problem

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Step-by-step explanation:

Given the differential equation

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Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

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The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

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