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2 years ago

Graph the solution to the following system of inequalities. y>−3x8 y>_(−3x-4)

Mathematics

1 answer:

ikadub [295]2 years ago

7 0

The graph of the system of inequalities y > 3x + 8 and y ≤ -4 - 3x is shown in the picture.

<h3>What is inequality?</h3>

It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

It is given that:

The system of inequalities:

y > 3x + 8

y ≤ -4 - 3x

The above inequalities represent the linear equation:

The graph is shown in the picture.

Thus, the graph of the system of inequalities y > 3x + 8 and y ≤ -4 - 3x is shown in the picture.

Learn more about the inequality here:

brainly.com/question/19491153

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Find the slope of each side of quadrilateral ABCD Then determine whether the quadrilateral is a parallelogram.

lara31 [8.8K]

AB=4
BC=1/4
CD=4
DA=1/4
Yes, quadrilateral ABCD is a parallelogram.

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3 years ago

What is the answer to 38-19+12

Naily [24]

The answer is 31.

38 - 19 = 19

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4 years ago

Using power series, solve the LDE: (2x^2 + 1) y" + 2xy' - 4x² y = 0 --- - -- -

sattari [20]

We're looking for a solution of 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 these into the ODE gives

$\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0$

Shifting indices to get each term in the summand to start at the same power of $x$ and pulling the first few terms of the resulting shifted series as needed gives

$2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0$

Then the coefficients in the series solution are given according to the recurrence

$\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}$

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of $y(0)=y'(0)=1$ , a degree-8 approximation of

$y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}$

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).