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tatuchka [14]
3 years ago
15

How much money should be invested today in an account that earns 3.5%, compound daily, in order to accumulate $75000 in 10 years

(assume n=365)
Mathematics
1 answer:
dangina [55]3 years ago
4 0

\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill &\$75000\\ P=\textit{original amount deposited}\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\dotfill &10 \end{cases} \\\\\\ 75000=Pe^{0.035\cdot 10}\implies 75000=Pe^{0.35}\implies \cfrac{75000}{e^{0.35}}=P \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 52851.61\approx P~\hfill

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Step-by-step explanation:

a) To find this, we use the midpoint formula

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(x1,y1) = (3,8)

(x2,y2) = (-5,1)

(x,y) = (3 -5)/2 , (8 + 1)/2 = (-1, 4.5)

b) To calculate the distance D between two points, we use the formula;

D = √(x2-x1)^2 + (y2-y1)^2

D = √(-5-3)^2 + (1-8)^2

D = √(-8)^2 + (-7)^2

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D = √113

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C) We use the slope formula here

m = (y2-y1)/(x2-x1)

m = (1-8)/(-5-3) = -7/-8 = 7/8

6 0
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Svetlanka [38]
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