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2 years ago

Which function forms a geometric sequence when x=1,2,3

Mathematics

1 answer:

taurus [48]2 years ago

4 0

Answer:

The correct answer is B.

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10. Which number sentence shows the<br> number of marbles that are either<br> red or blue?

nexus9112 [7]

Step-by-step explanation:

I think answer A should be the correct answer

7 0

2 years ago

I have $6.25 composed of quarters and dimes. there are 10 times more dimes than quarters. how many dimes do i have? the value of

ICE Princess25 [194]

1 quarter = 10 dimes - Total $1.25
2 quarters = 20 dimes - Total $2.50
4 quarters = 40 dimes - Total $5.00
5 Quarters = 50 dimes - Total $6.25

You have 50 dimes and 5 quarters

8 0

2 years ago

Read 2 more answers

Factor the expression using the GCF.<br> 22 + 11 =

Maurinko [17]

Answer should be 22= 11x2 11= 11x1, something like that :)

5 0

2 years ago

The sum of 10 and Hector's age is 69 . Use the variable H to represent Hector's age.

mamaluj [8]

Answer:

59? Is it H = 59?

6 0

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

2 years ago