3 years ago

I need help asap math sucks and is hard 8th grade math

Mathematics

2 answers:

s2008m [1.1K]3 years ago

6 0

Answer:

Slope = $\frac{-1}{3}$

Step-by-step explanation:

(x₁ , y₁) = (-9 , -4) & (x₂ ,y₂) = (-3, -6)

Slope = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$= \frac{-6-[-4]}{-3-[-9]}\\\\= \frac{-6+4}{-3+9}\\\\= \frac{-2}{6}\\\\= \frac{-1}{3}$

amid [387]3 years ago

5 0

Answer:

Slope = 1/3

Step-by-step explanation:

You might be interested in

Solve the initial value problem <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D%20%2B%202%20%5C

pogonyaev

The characteristic equation for this ODE is

$r^2+2r+5=0$

which has roots at $r=-1\pm2i$ , so the general solution is

$y_c=(C_1\cos2x+C_2\sin2x)e^{-x}$

Given $y(0)=2$ , we have

$2=(C_1\cos0+C_2\sin0)e^{-0}\implies C_1=2$

and $y'(0)=2$ , we have (upon differentiating $y_c$ )

${y_c}'=((2C_2-C_1)\cos2x-(2C_1+C_2)\sin2x)e^{-x}$
$2=((2C_2-2)\cos0-(4+C_2)\sin0)e^{-0}$
$2=2C_2-2\implies C_2=2$

So the particular solution is

$y_c=(2\cos2x+2\sin2x)e^{-x}$