Question

Which linear inequality represents the graph below help plss​

Answer 1

Answer:

 $y\ge -\frac{2}{3}x+1$ represents the graph. So, option A is true.

Step-by-step explanation:

Considering the linear inequality

$y\ge -\frac{2}{3}x+1$

as

$\mathrm{Domain\:of\:}\:-\frac{2}{3}x+1\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

and

$\mathrm{Range\:of\:}-\frac{2}{3}x+1:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

$x-\mathrm{axis\:interception\:points\:of\:}-\frac{2}{3}x+1:\quad \left(\frac{3}{2},\:0\right)$

$-\frac{2}{3}x+1=0:\quad x=\frac{3}{2}$

$\left(\frac{3}{2},\:0\right)$

$y-\mathrm{axis\:interception\:point\:of\:}-\frac{2}{3}x+1:\quad \left(0,\:1\right)$

$y=-\frac{2}{3}\cdot \:0+1$

$y=-0+1$

$y=1$

Thus,

$\mathrm{X\:Intercepts}:\:\left(\frac{3}{2},\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:1\right)$

$\mathrm{Slope\:of\:}-\frac{2}{3}x+1:\quad m=-\frac{2}{3}$

Also, (-3, 3) holds true.

as

$3\ge \:-\frac{2}{3}\cdot \left(-3\right)+1$

$3\ge \:\frac{2}{3}\cdot \:\:3+1$

$3\ge \:2+1$

$3\ge \:3$

WHICH IS TRUE!

Therefore,

                   $y\ge -\frac{2}{3}x+1$ represents the graph. So, option A is true.

Answer 2

Step-by-step explanation:

The boundary line goes through (-3,3) and (0,1)

The slope of this line is:

$m = \frac{1 - 3}{0 - - 3} = \frac{ - 2}{3} = - \frac{2}{3}$

The y-intercept is b=1.

The equation of the boundary line is given by

$y = mx + b$

Substitute and obtain:

$y = - \frac{2}{3} x + 1$

Since boundary line is shaded and the right half-plane of the boundary line is shaded, the corresponding inequality is

$y \geqslant - \frac{2}{3}x + 1$