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3 years ago

9

Avery gets newsletters by e-mail. He gets one for sports every 5 days, one for model railroads every 10 days, and one for music

every 8 days. If he got all three today, how many more days will it be until he gets all three newsletters on the same day again answers

1 answer:

pav-90 [236]3 years ago

7 0

5, 10 and 8 all need to equal the same number for it to be the exact same day. 5, 10, and 8 go into 40

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For the differential equation 3x^2y''+2xy'+x^2y=0 show that the point x = 0 is a regular singular point (either by using the lim

Svetlanka [38]

Given an ODE of the form

$y''(x)+p(x)y'(x)+q(x)y(x)=f(x)$

a regular singular point $x=c$ is one such that $p(x)$ or $q(x)$ diverge as $x\to c$ , but the limits of $(x-c)p(x)$ and $(x-c)^2q(x)$ as $x\to c$ exist.

We have for $x\neq0$ ,

$3x^2y''+2xy'+x^2y=0\implies y''+\dfrac2{3x}y'+\dfrac13y=0$

and as $x\to0$ , we have $x\cdot\dfrac2{3x}\to\dfrac23$ and $x^2\cdot\dfrac13\to0$ , so indeed $x=0$ is a regular singular point.

We then look for a series solution about the regular singular point $x=0$ of the form

$y=\displaystyle\sum_{n\ge0}a_nx^{n+k}$

Substituting into the ODE gives

$\displaystyle3x^2\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}+2x\sum_{n\ge0}a_n(n+k)x^{n+k-1}+x^2\sum_{n\ge0}a_nx^{n+k}=0$

$\displaystyle3\sum_{n\ge2}a_n(n+k)(n+k-1)x^{n+k}+3a_1k(k+1)x^{k+1}+3a_0k(k-1)x^k$
$\displaystyle+2\sum_{n\ge2}a_n(n+k)x^{n+k}+2a_1(k+1)x^{k+1}+2a_0kx^k$
$\displaystyle+\sum_{n\ge2}a_{n-2}x^{n+k}=0$

From this we find the indicial equation to be

$(3(k-1)+2)ka_0=0\implies k=0,\,k=\dfrac13$

Taking $k=\dfrac13$ , and in the $x^{k+1}$ term above we find $a_1=0$ . So we have

$\begin{cases}a_0=1\\a_1=0\\\\a_n=-\dfrac{a_{n-2}}{n(3n+1)}&\text{for }n\ge2\end{cases}$

Since $a_1=0$ , all coefficients with an odd index will also vanish.

So the first three terms of the series expansion of this solution are

$\displaystyle\sum_{n\ge0}a_nx^{n+1/3}=a_0x^{1/3}+a_2x^{7/3}+a_4x^{13/3}$

with $a_0=1$ , $a_2=-\dfrac1{14}$ , and $a_4=\dfrac1{728}$ .

6 0

4 years ago

You have 9 chairs arranged in a circle, and wish to seat 9 people (one person per seat). The one constraint is that person A can

stiv31 [10]

Answer:

30240 number of ways are there to seat them

Step-by-step explanation:

Total number of ways of arranging 9 people on 9 chairs in circular manners = (9-1)! = 8! =

number of ways A sit always sit next to B = AB together makes a single and

therefore total number of arrangements for this = 7+(AB) = 8 that is 8 persons sitting in circular manner

number of ways = (8-1)! =7! = 5040

likewise number of arrangements for A and C will be = 5040

Total number of ways such that A cannot sit next to B or C = total ways of 9 persons - total number of A always sitting next to B - total number of ways always sitting next to C = 8! - 7!-7!

= 40320- 5040-5040

= 30240

=

5 0

3 years ago

Joshua wrote 13 articles for the school newspaper this year. Paulette wrote 7 more articles than Joshua. Jeff wrote as many arti

Keith_Richards [23]

53 articles? Because josh wrote 13 plus 20 that Paulette wrote then Jeff wrote as many as Paulette so another 20.

6 0

4 years ago

-9 squared minus -6 squared plus 3 times negative 20

blagie [28]

Wow thats a complex question

6 0

3 years ago

Read 2 more answers

In one week, a campground rented 18 cabins at $225 each. About how much did they collect in rent altogether?

zvonat [6]

You would have to multiply $225• 18 which would equal to $4050 in all.

4 0

3 years ago