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3 years ago

10

Help due tonight plssss

1 answer:

postnew [5]3 years ago

6 0

Answer:

Perpendicular to base

Step-by-step explanation:

The cut creates two rectangular prisms (each angle must be 90 degrees)

The cut is perpendicular to the base

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4 years ago

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Find two linearly independent solutions to the equation y"-2xy'+2y=0 in the form of a power series.

ioda

We want a solution in the form

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

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting $y$ and its derivatives into the ODE,

$y''-2xy'+2y=0$

gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n-2\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}+2\sum_{n\ge0}a_nx^n=0$

Shift the index on the second sum to have it start at $n=1$ :

$\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}=\sum_{n\ge1}na_nx^n$

and take the first term out of the other two sums. Then we can consolidate the sums into one that starts at $n=1$ :

$\displaystyle(2a_2+2a_0)+\sum_{n\ge1}\bigg[(n+2)(n+1)a_{n+2}+(2-2n)a_n\bigg]x^n=0$

and so the coefficients in the series solution are given by the recurrence,

$\begin{cases}a_0=y(0)\\a_1=y'(0)\\(n+2)(n+1)a_{n+2}=2(n-1)a_n&\text{for }n\ge0\end{cases}$

or more simply, for $n\ge2$ ,

$a_n=\dfrac{2(n-3)}{n(n-1)}a_{n-2}$

Note the dependency between every other coefficient. Consider the two cases,

If $n=2k$ , where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

$k=1\implies n=2\implies a_2=-a_0=2^1\dfrac{(-1)}{2!}a_0$

$k=2\implies n=4\implies a_4=\dfrac{2\cdot1}{4\cdot3}a_2=2^2\dfrac{1\cdot(-1)}{4!}a_0$

$k=3\implies n=6\implies a_6=\dfrac{2\cdot3}{6\cdot5}a_4=2^3\dfrac{3\cdot1\cdot(-1)}{6!}a_0$

$k=4\implies n=8\implies a_8=\dfrac{2\cdot5}{8\cdot7}a_6=2^4\dfrac{5\cdot3\cdot1\cdot(-1)}{8!}a_0$

and so on, with the general pattern

$a_{2k}=\dfrac{2^ka_0}{(2k)!}\displaystyle\prod_{i=1}^k(2i-3)$

If $n=2k+1$ , then

$k=0\implies n=1\implies a_1=a_1$

$k=1\implies n=3\implies a_3=\dfrac{2\cdot0}{3\cdot2}a_1=0$

and we would see that $a_{2k+1}=0$ for all $k\ge1$ .

So we have

$y(x)=\displaystyle\sum_{k\ge0}\bigg[a_{2k}x^{2k}+a_{2k+1}x^{2k+1}\bigg]$

so that one solution is

$\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{2^k\prod\limits_{i=1}^k(2i-3)}{(2k)!}x^{2k}}$

and the other is

$\boxed{y_2(x)=a_1x}$

I've attached a plot of the exact and series solutions below with $a_0=y(0)=1$ , $a_1=y'(0)=1$ , and $0\le k\le5$ to demonstrate that the series solution converges to the exact one.

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3 years ago

Trigonometry Problem <br><br> Find the length of the missing side.

Vlad [161]

Answer:

x= 141.8179

Step-by-step explanation:

side a is 38, side b is 141.8179, side c is 146.8207

angle a is 15 angle b is 90 ancle c is 75

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4 years ago