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3 years ago

Find the amount in a account where $200 is invested at 3.5% compounded quarterly for a period of 3 years.

Mathematics

1 answer:

AlexFokin [52]3 years ago

3 0

The amount after 3 years would be $221.74

Explanation:

Given:

Principal, P = $200

Rate of interest, r = 3.5%

Time, t = 3 years

Amount, A = ?

We know:

$A = P(1+\frac{r}{100})^t$

Substituting the value we get:

$A = 200(1+\frac{3.5}{100} )^3\\\\A = 200 X 1.108\\\\A = 221.74$

Therefore, the amount after 3 years would be $221.74

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4 years ago

(1 point) a tank contains 1060 l of pure water. a solution that contains 0.06 kg of sugar per liter enters the tank at the rate

kirill115 [55]

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$S' = f(t,S) = \left(0.06 \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) - \left(\dfrac{S}{1060} \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) \ \Rightarrow \\ \\ S' = 0.54 \text{ kg}/\text{min} - \dfrac{9S}{1060}$

So yes, you enter S' = 0.54 - (9S/1060)

(c)

$\displaystyle\frac{dS}{dt} = 0.54 - \frac{9S}{1060} \ \Rightarrow\ \frac{dS}{dt} = \frac{572.4 - 9S}{1060}\ \Rightarrow\ \dfrac{dS}{572.4 - 9S} = \frac{1}{1060} dt\ \Rightarrow \\ \\
\int \dfrac{dS}{572.4 - 9S} = \int \frac{1}{1060} dt\ \Rightarrow\textstyle\ -\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t + C \\ \\
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$-\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t -\frac{1}{9} \ln 572.4\ \Rightarrow \\ \\
\ln|572.4 - 9S| = \ln 572.4 - \frac{9}{1060}t \ \Rightarrow \\ \\
|572.4 - 9S| = e^{\ln 572.4 - 9t/1060}\ \Rightarrow \\ \\
572.4 - 9S= \pm 572.4 e^{-9t/1060}\ \Rightarrow \\ \\
S = \frac{-1}{9}\left(-572.4 \pm 572.4 e^{-9t/1060}\right)$

But only (+) satisfies $S(0) = 0$

$S= -\frac{1}{9}\left(-572.4 + 572.4 e^{-9t/1060}\right) \\ \\
S= 63.6 - 63.6 e^{-9t/1060}\text{ kg}$

Enter in S = 63.6 - 63.6 * e^(-9t/1060)

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3 years ago