Question

Which linear inequality represents the graph below help plss​

Answer 1

Explanation:

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

The slope of this line is:

`m = (1 - 3)/(0 - - 3) = ( - 2)/(3) = - (2)/(3)`

The y-intercept is b=1.

The equation of the boundary line is given by

`y = mx + b`

Substitute and obtain:

`y = - (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 - (2)/(3)x + 1`

Answer 2

`y\ge -(2)/(3)x+1` represents the graph. So, option A is true.

Explanation:

Considering the linear inequality

`y\ge -(2)/(3)x+1`

as

`\mathrm{Domain\:of\:}\:-(2)/(3)x+1\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}`

and

`\mathrm{Range\:of\:}-(2)/(3)x+1:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<f\left(x\right)<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}`

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

`-(2)/(3)x+1=0:\quad x=(3)/(2)`

`\left((3)/(2),\:0\right)`

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

`y=-(2)/(3)\cdot \:0+1`

`y=-0+1`

`y=1`

Thus,

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

`\mathrm{Slope\:of\:}-(2)/(3)x+1:\quad m=-(2)/(3)`

Also, (-3, 3) holds true.

as

`3\ge \:-(2)/(3)\cdot \left(-3\right)+1`

`3\ge \:(2)/(3)\cdot \:\:3+1`

`3\ge \:2+1`

`3\ge \:3`

WHICH IS TRUE!

Therefore,

`y\ge -(2)/(3)x+1` represents the graph. So, option A is true.