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3 years ago

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The amount of $15,000 is invested in two funds paying 2% and 5% simple interest. If the annual interest is $540, how much of the

$15,000 is invested in each interest rate?

1 answer:

Lostsunrise [7]3 years ago

8 0

Answer:

The amount $15000 is divided as

in 2% = $7000

in 5% = $8000

Step-by-step explanation:

It is given that the total amount is $15000 and rates of interests are 2% and 5%.

<em>Let the amount invested in 2% be "x", so the amount invested in 5% is (15000 - x).</em>

The annual interest on x part is given as,

= $(\frac{2}{100})(x)$

Similarly, The annual interest on (15000 - x) part is given as,

= $(\frac{5}{100})(15000 - x)$

the total annual interest is given as $540

Thus, <em>540 = $(\frac{5}{100})(15000 - x) + (\frac{2}{100})(x)$ </em>

<em>54000 = $2(x) + 75000 - 5(x)$ </em>

<em>21000 = 3(x)</em>

x = $\frac{21000}{3}$ = 7000

Thus the shares are $7000 and (15000 - 7000) $8000 .

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Which term describes that right of a lender to sell collateral to get back the principal if the borrower cannot repay the loan?

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Lien

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Answer:

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Step-by-step explanation:

we know that

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Let

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Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

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Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

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Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

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