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barxatty [35]
2 years ago
7

Mr. and Mrs. Williams hope to send their daughter to college in twelve years. How much money should they invest now at an intere

st rate of 8.5% per year,
compounded continuously, in order to be able to contribute $9000 to her education?
Mathematics
1 answer:
Marta_Voda [28]2 years ago
5 0

~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$9000\\ P=\textit{original amount deposited}\\ r=rate\to 8.5\%\to \frac{8.5}{100}\dotfill &0.085\\ t=years\dotfill &12 \end{cases} \\\\\\ 9000=Pe^{0.085\cdot 12}\implies 9000Pe^{1.02}\implies \cfrac{9000}{e^{1.02}}=P\implies 3245.35\approx P

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3 years ago
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Line l1 passes through (-2, 5), and (-1, -10)
zalisa [80]

The equations of the lines l1 with points (-2, 5), and (-1, -10) and the line l2 with points (5, 15), and (3, 8), gives the coordinate of the intersection between the lines as the point; (-45/37, -250/37)

<h3>Which method can be used to describe the lines to find the intersection point?</h3>

The slope, m1, of line 1 l1 is found as follows;

  • m1 = (5 - (-10))/(-2- (-1)) = -15

The equation of line l1 in point and slope form is therefore;

y1 - 5 = -15•(x - (-2))

Which gives;

y1 = -15•(x - (-2)) + 5 = -15•x - 25

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The slope, m2, of line 2 l2 is found as follows;

m2 = (15 - 8)/(5 - 3) = 3.5

Equation of line l2 is therefore;

y2 - 15 = 3.5•(x - 5)

Which gives;

y2 = 3.5•(x - 5) + 15 = 3.5•x - 2.5

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At the intersection point, we have;

y1 = y2

Therefore;

-15•x - 25 = 3.5•x - 2.5

18.5•x = -22.5

x = -22.5/18.5 = -45/37

y2 = 3.5•x - 2.5

At the intersection point, we have;

y = y2 = 3.5×(-22.5/18.5) - 2.5 = -250/37

y = -250/37

The coordinates of the intersection between the lines is therefore;

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Answer:

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