15x^4+6x^3-7x^2÷3x^2+1

Which correctly describes how the graph of the inequality 6y − 3x > 9 is shaded? -Above the solid line

baherus [9]

The statement which correctly describes the shaded region for the inequality is $\fbox{\begin\\\ Above the dashed line\\\end{minispace}}$

Further explanation:

In the question it is given that the inequality is $6y-3x>9$ .

The equation corresponding to the inequality $6y-3x>9$ is $6y-3x=9$ .

The equation $6y-3x=9$ represents a line and the inequality $6y-3x>9$ represents the region which lies either above or below the line $6y-3x=9$ .

Transform the equation $6y-3x=9$ in its slope intercept form as $y=mx+c$ , where $m$ represents the slope of the line and $c$ represents the $y$ -intercept.

$y$ -intercept is the point at which the line intersects the $y$ -axis.

In order to convert the equation $6y-3x=9$ in its slope intercept form add $3x$ to equation $6y-3x=9$ .

$6y-3x+3x=9+3x$

$6y=9+3x$

Now, divide the above equation by $6$ .

$\fbox{\begin\\\math{y=\dfrac{x}{2}+\dfrac{1}{2}}\\\end{minispace}}$

Compare the above final equation with the general form of the slope intercept form $\fbox{\begin\\\math{y=mx+c}\\\end{minispace}}$ .

It is observed that the value of $m$ is $\dfrac{1}{2}$ and the value of $c$ is $\dfrac{3}{2}$ .

This implies that the $y$ -intercept of the line is $\dfrac{3}{2}$ so, it can be said that the line passes through the point $\fbox{\begin\\\ \left(0,\dfrac{3}{2}\right)\\\end{minispace}}$ .

To draw a line we require at least two points through which the line passes so, in order to obtain the other point substitute $0$ for $y$ in $6y=9+3x$ .

$0=9+3x$

$3x=-9$

$\fbox{\begin\\\math{x=-3}\\\end{minispace}}$

This implies that the line passes through the point $\fbox{\begin\\\ (-3,0)\\\end{minispace}}$ .

Now plot the points $(-3,0)$ and $\left(0,\dfrac{3}{2}\right)$ in the Cartesian plane and join the points to obtain the graph of the line $6y-3x=9$ .

Figure $1$ shows the graph of the equation $6y-3x=9$ .

Now to obtain the region of the inequality $6y-3x>9$ consider any point which lies below the line $6y-3x=9$ .

Consider $(0,0)$ to check if it satisfies the inequality $6y-3x>9$ .

Substitute $x=0$ and $y=0$ in $6y-3x>9$ .

$(6\times0)-(3\times0)>9$

$0>9$

The above result obtain is not true as $0$ is not greater than $9$ so, the point $(0,0)$ does not satisfies the inequality $6y-3x>9$ .

Now consider $(-2,2)$ to check if it satisfies the inequality $6y-3x>9$ .

Substitute $x=-2$ and $y=2$ in the inequality $6y-3x>9$ .

$(6\times2)-(3\times(-2))>9$

$12+6>9$

$18>9$

The result obtain is true as $18$ is greater than $9$ so, the point $(-2,2)$ satisfies the inequality $6y-3x>9$ .

The point $(-2,2)$ lies above the line so, the region for the inequality $6y-3x>9$ is the region above the line $6y-3x=9$ .

The region the for the inequality $6y-3x>9$ does not include the points on the line $6y-3x=9$ because in the given inequality the inequality sign used is $>$ .

Figure $2$ shows the region for the inequality $\fbox{\begin\\\math{6y-3x>9}\\\end{minispace}}$ .

Therefore, the statement which correctly describes the shaded region for the inequality is $\fbox{\begin\\\ Above the dashed line\\\end{minispace}}$

Learn more:

A problem to determine the range of a function brainly.com/question/3852778
A problem to determine the vertex of a curve brainly.com/question/1286775
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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Linear inequality

Keywords: Linear, equality, inequality, linear inequality, region, shaded region, common region, above the dashed line, graph, graph of inequality, slope, intercepts, y-intercept, 6y-3x=9, 6y-3x>9, slope intercept form.