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Bad White [126]
3 years ago
13

How many toys were stuffed animals

Mathematics
2 answers:
Cloud [144]3 years ago
7 0

Answer:

Model b 24 stuffed animals.

Step-by-step explanation:

Lets find what is 30% of 80 is:

30/100 x 80= 2400/100= 24 stuffed animals.

Model B shows right model and 24 stuffed animals.

Hope this helps, let me know if correct or incorrect.

Umnica [9.8K]3 years ago
4 0

Answer:

It would be the first one.

Step-by-step explanation:

The way you figure it out is that it said that there was a total of 80 stuffed animals, so model B is out. Then, you change the percent into a decimal by moving the decimal point over to the right 2 times and you get 0.30. Finally, you multiply 80 by 0.30 and you get 24 stuffed animals.

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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
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