Enter an inequality that represents the graph in the box.
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Answer:
Step-by-step explanation:
Easy. its [tex]x^{2} \int\limits^a_b {x} \, dx \\ \\ \left \{ {{y=2} \atop {x=2}} \right. \\ \neq \frac{x}{y} \beta \al\left \{ {{y=2} \atop {x=2}} \right. pha \neq \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \geq \\ \left \{ {{y=2} \atop {x=2}} \right. \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right][/tex]
The inequality that represents the graph is:
[tex]y\geq x+3[/tex]
By looking at the graph we observe that the graph is a solid line which passes through the point (-1,0) and (1,2).This means that the inequality is a inequality with a equality sign(i.e. not strict)
This means that the equation of line is:
[tex]y-2=\dfrac{2-0}{1-(-1)}\times (x-(-1))\\\\y-2=\dfrac{2}{2}\times (x+1)\\\\y-2=x+1\\\\y=x+1+2\\\\y=x+3[/tex]
The shaded region is away from the origin.
This means that the inequality does not pass the zero test.
Hence, the inequality is:
[tex]y\geq x+3[/tex]