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Mashutka [201]
3 years ago
5

Please help 100 points

Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0

Answer:

Choices A, C, E

Step-by-step explanation:

The prices are proportional, so divide any price by the corresponding number of pounds to find the unit cost.

$1.47/(3 lb) = $0.49/lb

The unit cost is $0.49 per lb.

Now we look in the choices to see which choice has a unit price of $0.49/lb.

We divide each price by its number of pounds to fund each unit cost. Every choice with a unit cost of $0.49/lb is an answer.

A $0.98/(2 lb) = $0.49/lb     Choice A works

B $4.45/(7 lb) = $0.64/lb      Choice B does not work

C $2.94/(6 lb) = $0.49/lb     Choice C works

D $0.54/(1 lb) = $0.54/lb     Choice D does not work

E $3.92/(8 lb) = $0.49/lb     Choice E works

Answer: Choices A, C, E

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36 minutes

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2 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.
5 0
3 years ago
Explain how to use 8×5=40 to find 8×6.
Zina [86]

Answer:

Well there are many different ways to find 8x5=40 to 8x6=__

You could add 8 to 40 and get 48.

or you could draw pictures to represent the 48.

8x6=48

7 0
3 years ago
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