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2 years ago

Linear relationships are (sometimes, always, or never) examples of directly proportional relationships

Mathematics

1 answer:

lys-0071 [83]2 years ago

5 0

Sometimes, it has to cross the orgin or (0,0)

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Two marbles are drawn without replacement from a box with 3 white, 2 green, 2 red, and 1 blue marble. Find the probability that

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White 3/8
Green 2/8
Red 2/8
Blue 1/8

Blue 1/7

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2 years ago

The sum of two consecutive numbers is 71. Find the smallest number in the list

bazaltina [42]

35+36=71
35 is the answer

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2 years ago

Maggie purchased a 2-pound roast to fix for dinner. How many ounces does the roast weigh?

Misha Larkins [42]

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2 years ago

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What is the gradient of a line perpendicular to a line with a gradient of -6

alexira [117]

Answer:

undefined -6 = 6

Step-by-step explanation:

im guessing but thinking

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2 years ago

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 $a_4=\dfrac1{728}$ .

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2 years ago