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Look at the system of inequalities. $\newline$ $y \leq -x + 5$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ The solution set is the triangular region where all the inequalities are true. $\newline$ What are the vertices of that triangular region? $\newline$ (,) $\newline$ (,) $\newline$ (,)

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Find the focus or directrix of a parabola

Full solution

Q. Look at the system of inequalities.

\newline

y \leq -x + 5

\newline

x \geq 0

\newline

y \geq 0

\newline

The solution set is the triangular region where all the inequalities are true.

\newline

What are the vertices of that triangular region?

\newline

(____,____)

\newline

(____,____)

\newline

(____,____)

Find Intersection of Inequalities: First, let's find the intersection of $y \leq -x + 5$ and $x \geq 0$ . $\newline$ Set $x = 0$ in the first inequality: $y \leq -0 + 5$ , so $y \leq 5$ . $\newline$ The point of intersection is $(0, 5)$ .
Intersection with x and y: Next, find the intersection of $y \leq -x + 5$ and $y \geq 0$ . $\newline$ Set $y = 0$ in the first inequality: $0 \leq -x + 5$ , so $x \leq 5$ . $\newline$ The point of intersection is $(5, 0)$ .
Origin as Vertex: Lastly, since $x \geq 0$ and $y \geq 0$ , the origin $(0, 0)$ is also a vertex of the triangular region.
Final Vertices: Now, we have all three vertices of the triangular region: $(0, 5)$ , $(5, 0)$ , and $(0, 0)$ .

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