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Alecsey [184]
3 years ago
14

Suppose that ​$17,183 is invested at an interest rate of 5.4​% per​ year, compounded continuously. ​a) Find the exponential func

tion that describes the amount in the account after time​ t, in years. ​b) What is the balance after 1​ year? 2​ years? 5​ years? 10​ years? ​c) What is the doubling​ time?

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
8 0

Answer:

Total = Principal * e ^ (rate * years)

Total = 17,183 * 2.718281828459 ^ (.054 * years)

After 1 year = 18,136.39

After 2 years = 19,142.68

After 5 years = 22,509.12

After 10 years = 29,486.15

We'll use this formula to find the doubling time:

Years = ln (total / principal) / rate

we'll use 200 for total and 100 for principal

Years = ln (200 / 100) / rate

Years = ln (2) / .054

Years = 0.69314718056 / .054

Years = 12.8360588993  Years Doubling Time

Step-by-step explanation:

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Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩
NISA [10]

Answer:

\theta = 108.29

Step-by-step explanation:

Given

u =

v =

Required:

Calculate the angle between u and v

The angle \theta is calculated as thus:

cos\theta = \frac{u.v}{|u|.|v|}

For a vector

A =

A = a * b

cos\theta = \frac{u.v}{|u|.|v|} becomes

cos\theta = \frac{.}{|u|.|v|}

cos\theta = \frac{2*6+4*-7}{|u|.|v|}

cos\theta = \frac{12-28}{|u|.|v|}

cos\theta = \frac{-16}{|u|.|v|}

For a vector

A =

|A| = \sqrt{a^2 + b^2}

So;

|u| = \sqrt{2^2 + 6^2}

|u| = \sqrt{4 + 36}

|u| = \sqrt{40}

|v| = \sqrt{4^2+(-7)^2}

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|v| = \sqrt{65}

So:

cos\theta = \frac{-16}{|u|.|v|}

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cos\theta = \frac{-16}{\sqrt{2600}}

cos\theta = \frac{-16}{\sqrt{100*26}}

cos\theta = \frac{-16}{10\sqrt{26}}

cos\theta = \frac{-8}{5\sqrt{26}}

Take arccos of both sides

\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})

\theta = cos^{-1}(\frac{-8}{5 * 5.0990})

\theta = cos^{-1}(\frac{-8}{25.495})

\theta = cos^{-1}(-0.31378701706)

\theta = 108.288386087

<em></em>\theta = 108.29<em> (approximated)</em>

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