I’m on a time limit please help!!!!!!!

Mathematics

2 answers:

Solnce55 [7]4 years ago

7 0

43,106 idk why they need so much

Sloan [31]4 years ago

3 0

Answer:

43,160

hope this helps :))

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A line contains the points (1, –6) and (–2, 6). What is the slope of a line that is perpendicular to this line?

OlgaM077 [116]

First, find the slope of the line contains the points (1,-6) and (-2,6) using slope formula
m = $\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

plug in the numbers
m = $\dfrac{6-(-6)}{-2-1}$
m = $\dfrac{6+6}{-3}$
m = $\dfrac{12}{-3}$
m = -4

Second, determine the slope of perpendicular line
The slope of the perpendicular line is the opposite and reciprocal from the other line. Thus, the slope is $\dfrac{1}{4}$

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3 years ago

Solve the given initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1

Yuri [45]

Let's check if the ODE is exact. To do that, we want to show that if

$\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0$

then $M_y=N_x$ . We have

$M_y=2(x+y)$
$N_x=2y+2x=2(x+y)$

so the equation is indeed exact. We're looking for a solution of the form $\Psi(x,y)=C$ . Computing the total differential yields the original ODE,

$\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$
$\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}$

Integrate both sides of the first PDE with respect to $x$ ; then

$\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)$

where $f(y)$ is a function of $y$ alone. Differentiate this with respect to $y$ so that

$\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)$
$\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)$
$f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C$

So the solution to this ODE is

$\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C$

i.e.

$\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C$