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vesna_86 [32]
3 years ago
11

Helena has taken out a $9,300 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 6.4%, compounded monthly. By the time Helena graduates, how much greater will the amount of interest capitalized be than the minimum amount that she could pay to prevent interest capitalization? Round all dollar values to the nearest cent.
Mathematics
2 answers:
kherson [118]3 years ago
8 0
The effective annual interest rate is:
i = (1 + 0.064/12)^12 - 1 = 0.066
In year 1: the interest is $613.80 (multiple $9300 by 0.066)
In year 2: the interest is $654.31 (add interest from year 1 to $9300 and multiply by 0.066)
In year 3: the interest is $656.98 (do the same as year 2)
In year 4: the interest is $657.16

The total interest is: $2582.25
The present worth of this amount is:
P = 2582.23 / (1 + 0.066)^4 = $1999.72

The answer is $1999.72.
babymother [125]3 years ago
7 0

Answer:

$324.33

Step-by-step explanation:

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Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t
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Answer:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

Step-by-step explanation:

Given

\int\limits {\frac{x}{x^4 + 16}} \, dx

Required

Solve

Let

u = \frac{x^2}{4}

Differentiate

du = 2 * \frac{x^{2-1}}{4}\ dx

du = 2 * \frac{x}{4}\ dx

du = \frac{x}{2}\ dx

Make dx the subject

dx = \frac{2}{x}\ du

The given integral becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

Recall that: u = \frac{x^2}{4}

Make x^2 the subject

x^2= 4u

Square both sides

x^4= (4u)^2

x^4= 16u^2

Substitute 16u^2 for x^4 in \int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du

Simplify

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

In standard integration

\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)

So, the expression becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)

Recall that: u = \frac{x^2}{4}

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

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