PLEASE HELP ASAP WILL GIVE BRAINLIEST

Mathematics

1 answer:

djverab [1.8K]3 years ago

4 0

It would be D
Rational functions are fractions where numerator and denominator are polynomials.

You might be interested in

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

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

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

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

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

2 years ago

When s is the open hemisphere x 2 + y 2 + z 2 = 1, z ≤ 0 , oriented by the inward normal pointing to the origin, then the bounda

Anit [1.1K]

The figure below shows a diagram of this problem. First of all we graph the hemisphere. This one has a radius equal to 1. Given that z ≤ 0 a sphere will be valid only in the negative z-axis, that is, we will get a half of a sphere that is the hemisphere shown in the figure. We know that this hemisphere is oriented by the inward normal pointing to the origin, then we have a Differential Surface Vector called N, using the Right-hand rule <span>the boundary orientation is </span>counterclockwise.

Therefore, the answer above False