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3 years ago

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Meg deposited a $3,000 bonus check in a new savings account. The account has an interest rate of 3% for 5 years. The interest is

compounded daily. How much money did Meg have at the end of the account term?

1 answer:

Korolek [52]3 years ago

3 0

Given that Meg deposited a $3,000 bonus check in a new savings account. The account has an interest rate of 3% for 5 years. The interest is compounded daily.

Now we need to find about how much money did Meg have at the end of the account term. So we can use compound interest formula which is given by:

$A=P\left(1+\frac{r}{n}\right)^{\left(nt\right)}$

P = initial deposit = 3000

r = rate of interest =3%=0.03

n= number of periods = 365 (for daily)

t= number of years = 5

Now plug these values into above formula ot get the final value that Meg would have at the end which is given by (A)

$A=3000\left(1+\frac{0.03}{365}\right)^{\left(365*5\right)}$

$A=3000\left(1+0.0000821917808219\right)^{\left(1825\right)}$

$A=3000\left(1.0000821917808219\right)^{\left(1825\right)}$

$A=3000(1.16182708115)$

A=3485.48124345

Hence final answer is approx $3485.48.

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so the flux over the closed region is

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The total flux over the closed surface is equal to the flux over its component surfaces, so we have

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$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

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with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

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$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

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