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3 years ago

This is urgent, please answer!

Mathematics

1 answer:

In-s [12.5K]3 years ago

4 0

I think its C) bc you would take 3.14 x4= 12.56

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Suppose the desks are arranged in rows of 4. how many seating arrangements are there that put larry, moe, curly, and shemp in th

Paul [167]

$4!=24$

..............

7 0

3 years ago

What it the inequality for this interval? (-00, 2] u (8,00)

stepan [7]

Interval notation tells you what values where the function is defined. It’s defined all the way from negative infinity to positive 2, then there’s a gap all the way to positive 8, then we continue on to positive infinity. So x, would be literally every single number besides the number between 2 and 8. Oh and don’t forget that brackets mean included, so 2 is also a value that’s acceptable.

x is less than or equal to 2
and
x is greater than 8

5 0

3 years ago

Xy''+2y'-xy by frobenius method

aalyn [17]

First note that $x=0$ is a regular singular point; in particular $x=0$ is a pole of order 1 for $\dfrac2x$ .

We seek a solution of the form

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

where $r$ is to be determined. Differentiating, we have

$y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}$
$y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}$

and substituting into the ODE gives

$\displaystyle x\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-x\sum_{n\ge0}a_nx^{n+r}=0$
$\displaystyle \sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0$
$\displaystyle \sum_{n\ge0}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0$
$\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0$
$\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge2}a_{n-2}x^{n+r-1}=0$
$\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}\bigg((n+r)(n+r+1)a_n-a_{n-2}\bigg)x^{n+r-1}=0$

The indicial polynomial, $r(r+1)$ , has roots at $r=0$ and $r=-1$ . Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.

When $r=0$ , we have the recurrence

$a_n=\dfrac{a_{n-2}}{(n+1)(n)}$

valid for $n\ge2$ . When $n=2k$ , with $k\in\{0,1,2,3,\ldots\}$ , we find

$a_0=a_0$
$a_2=\dfrac{a_0}{3\cdot2}=\dfrac{a_0}{3!}$
$a_4=\dfrac{a_2}{5\cdot4}=\dfrac{a_0}{5!}$
$a_6=\dfrac{a_4}{7\cdot6}=\dfrac{a_0}{7!}$

and so on, with a general pattern of

$a_{n=2k}=\dfrac{a_0}{(2k+1)!}$

Similarly, when $n=2k+1$ for $k\in\{0,1,2,3,\ldots\}$ , we find

$a_1=a_1$
$a_3=\dfrac{a_1}{4\cdot3}=\dfrac{2a_1}{4!}$
$a_5=\dfrac{a_3}{6\cdot5}=\dfrac{2a_1}{6!}$
$a_7=\dfrac{a_5}{8\cdot7}=\dfrac{2a_1}{8!}$

and so on, with the general pattern

$a_{n=2k+1}=\dfrac{2a_1}{(2k+2)!}$

So the first indicial root admits the solution

$y=\displaystyle a_0\sum_{k\ge0}\frac{x^{2k}}{(2k+1)!}+a_1\sum_{k\ge0}\frac{x^{2k+1}}{(2k+2)!}$
$y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}$
$y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}$

which you can recognize as the power series for $\dfrac{\sinh x}x$ and $\dfrac{\cosh x}x$ .

To be more precise, the second series actually converges to $\dfrac{\cosh x-1}x$ , which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When $r=-1$ , we may seek a second solution of the form

$y=cy_1\ln x+x^{-1}\displaystyle\sum_{n\ge0}b_nx^n$

where $y_1=\dfrac{\sinh x+\cosh x-1}x$ . Substituting this into the ODE, you'll find that $c=0$ , and so we're left with

$y=x^{-1}\displaystyle\sum_{n\ge0}b_nx^n$
$y=\dfrac{b_0}x+b_1+b_2x+b_3x^2+\cdots$

Expanding $y_1$ , you'll see that all the terms $x^n$ with $n\ge0$ in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form $y_2=\dfrac1x$ . Adding this to $y_1$ , we end up with just $\dfrac{\sinh x+\cosh x}x$ .

This means the general solution for the ODE is

$y=C_1\dfrac{\sinh x}x+C_2\dfrac{\cosh x}x$

3 0

3 years ago

Can i have some help please uwu?

katovenus [111]

hope im right but i think it 1 sorry if im wrong

4 0

3 years ago

Read 2 more answers

A city's population is currently 500,000. If the population doubles every 65 years, what will the population be 130 years from n

kompoz [17]

Answer:

2,000,000

Step-by-step explanation:

Simple maths, really.

65 times 2 is 130, and the population is already 500,000 with no time passing. 65 years later, the population would be 1,000,000. 65 years after that, the population would double again, so the answer would be 2 million.

Hope this helped!

5 0

3 years ago