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Lostsunrise [7]
3 years ago
13

Chan deposited money into his retirement account that is compounded annually at an interest rate of 7%. Chan thought the equival

ent quarterly interest rate would be 2%. Is Chan correct? If he is, explain why. If he is not correct, state what the equivalent quarterly interest rate is and show how you got your answer. (8 pts)
Mathematics
1 answer:
dedylja [7]3 years ago
4 0

Answer:

No, equivalent quarterly rate will be approx 1.75%

Step-by-step explanation:

Given that Chan deposited money into his retirement account that is compounded annually at an interest rate of 7%.

We know that there are 4 quarters in 1 year.

So to find that equivalent quarterly we will divide given yearly rate by number of quarters.

That means divide 7% by 4.

which gives 1.75%.

But that is different than Chan's though of 2% quarterly interest.

Hence Chan is wrong.

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iragen [17]
So all you do is you divide 2 1/2 by .25 and you get 10 then you know one fourth or 25% of that package is 10 berries so the rest or 30% left over is 30 berries. 
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I really need help with this question
Dominik [7]

Answer:

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5 0
3 years ago
Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
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5. In a dilation what mathematical operation is used with th scale factor?
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Step-by-step explanation:

multiplication is the mathematical operation used with scale factor

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Find the measure of m∠K
Jobisdone [24]

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Inscribed Angle = 1/2Intercepted Arc

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<k = 20

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