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4 years ago

What is a closed line of credit

2 answers:

7 0

You may start out with a "line of credit" agreement with a bank that allows you to create relatively small loans by the bank to yourself at a specified interest rate and under other specific terms. Of course you must repay these loans, with interest, and that's how the bank makes its money.

If you fail to repay the bank, the bank may close your account and demand that you immediately repay all the money you owe it. A "closed line of credit" is a pretty sad thing.

Vinil7 [7]4 years ago

4 0

A closed line of credit is a line of credit with a fixed total amount.

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24 = 3(n–5)<br><br>What is the answer

babunello [35]

Step-by-step explanation:

24=3n-15 or,25+15=3n or,39=3n or,39/3=n or,n=13 therefore the value of n is 13. I think this answer is helpful for you

8 0

3 years ago

Read 2 more answers

Find the length of the curve x=e^t e^{-t},\;\;y=5-2t,\;\;0 \le t \le 3.

Oksana_A [137]

$\begin{cases}x(t)=e^t+e^{-t}\\y(t)=5-2t\end{cases}\implies\begin{cases}x'(t)=e^t-e^{-t}\\y'(t)=-2\end{cases}$

The length of the curve is given by the integral

$\displaystyle\int_0^3\sqrt{(e^t-e^{-t})^2+(-2)^2}\,\mathrm dt$

Expand and rewrite the integrand:

$(e^t-e^{-t})^2+(-2)^2=e^{2t}+2+e^{-2t}$
$=e^{-2t}(e^{4t}+2e^{2t}+1)$
$=e^{-2t}(e^{2t}+1)^2$
$\implies\sqrt{(e^t-e^{-t})^2+(-2)^2}=\dfrac{e^{2t}+1}{e^t}$

Now the integral is

$\displaystyle\int_0^3\dfrac{e^{2t}+1}{e^t}\,\mathrm dt=\int_0^3(e^t+e^{-t})\,\mathrm dt$
$=2\displaystyle\int_0^3\cosh t\,\mathrm dt$
$=2\sinh t\bigg|_{t=0}^{t=3}$
$=2\sinh 3$

7 0

3 years ago

Help on this math stuff

liq [111]

Plan A would be better

6 0

4 years ago

Lesson 7: Comparing Relationships with Tables

makvit [3.9K]

Whats the proportional relationship??? I will edit my question when I get further information

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4 years ago

Use the Distributive Property to rewrite the algebraic expression. 9(3p + 6)

Tanya [424]

Answer:

27p+54

Step-by-step explanation:

Multiply 9 to all the numbers inside the parenthesis!

Hope this helps! :)

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3 years ago