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gayaneshka [121]

3 years ago

5

Hunter took out a $3500 loan with interest to pay for some business supplies. He will repay The loan and three years with monthl

y payments of $109.68. What is the total amount of interest hundred will pay? ( there's no answer choices)

1 answer:

almond37 [142]3 years ago

7 0

Answer: $448.48

Step-by-step explanation:

The amount Hunter will pay in total is;

= 109.68 * 12 months * 3 years

= $‭3,948.48‬

Interest he will pay = Amount paid - Loan amount

= ‭3,948.48‬ - 3,500

= $448.48

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-2/3f=9.2<br> -1/8n = -2/7 <br> 12/20n = -36/44<br> -64.5g = -25.8 <br> -15/28 = -9/10

user100 [1]

Step-by-step explanation:

-2/3f =9.2

cross multiply

27,6= -2f

divide both sides by 2

f = - 13.8

-1/8n = -2/7

cross multiply

-16 = -7n

divide

n = 2.29

12/20n = -36/44

cross multiply

528 = - 720n

divide

n= -0.733

-64.5g = -25.8

divide both sides by -64.5

g = -0.4

the last one has no variable

5 0

3 years ago

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

Anika [276]

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

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

4 0

3 years ago

Answer these please tell me <br> 1. What the error was<br> 2. How can U check ur answer

emmainna [20.7K]

Answer:

The first one: the error was that the 3 shouldn’t have been moved but the 5 should have. The correct answer is k=-8

The second one: the error was that they added the 6 and 4, you would only get -10 if the 4 was - the correct answer is -2

You can check your answer by plugging in your answer to the variable and making sure both side equal each other

Step-by-step explanation:

The first one: the 5 should be - from both side so you get k=-8

The second one: the 4 should be + to both sides. And the -6 and +4 should be added together to get n=-2

Remember you always want to get the variables by themselves.

Sorry that was a lot. Hope it helps!

8 0

3 years ago

Plase dont awnser just for points

egoroff_w [7]

So 3 2/3 is about 3.67 so it would go in between 3.65 and 3.7. So it would be the second answer.

5 0

4 years ago

Read 2 more answers

Hey can some one help me out

Archy [21]

I believe the correct situation is C

3 0

3 years ago