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geniusboy [140]
2 years ago
11

You want to be able to withdraw $25,000 from your account each year for 20 years after you retire.

Mathematics
1 answer:
Nina [5.8K]2 years ago
3 0

Answer:

When using formulas in application, or memorizing them for tests, it is helpful to note the similarities and differences in the formulas so you don’t mix them up. Compare the formulas for savings annuities vs payout annuities.

Savings Annuity Payout Annuity

P

N

=

d

(

(

1

+

r

k

)

N

k

−

1

)

(

r

k

)

P

0

=

d

(

1

−

(

1

+

r

k

)

−

N

k

)

(

r

k

)

PAYOUT ANNUITY FORMULA

P

0

=

d

(

1

−

(

1

+

r

k

)

−

N

k

)

(

r

k

)

P0 is the balance in the account at the beginning (starting amount, or principal).

d is the regular withdrawal (the amount you take out each year, each month, etc.)

r is the annual interest rate (in decimal form. Example: 5% = 0.05)

k is the number of compounding periods in one year.

N is the number of years we plan to take withdrawals

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How can the bar symbol be used to show the quotient of 7÷11 if the division is carried on forever?
Thepotemich [5.8K]

\bf 7\div 11\implies 0.636363636363636........\implies 0.\overline{63}

4 0
3 years ago
(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

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3 years ago
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disa [49]

Answer:

The answer is 3,906.25

Step-by-step explanation:

All you have to do is divide 15,625 by 4! :)

6 0
2 years ago
Tom swims a 1/2 kilometers every 1/4 hour. How far will he swim in one hour.​
fgiga [73]

Answer:

2 kilometers

Step-by-step explanation:

every 1 kilometer is a 1/2 hour. double that and you get 2

8 0
3 years ago
F(x) = -2x - 1; Find f (x - 4)
qaws [65]

Answer:

B

if f(x-4) then we should put x-4 where the x is spouse to be so;

-2(x-4)-1 =

-2x + 8 - 1 =

-2x + 7

5 0
3 years ago
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