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Look at the system of inequalities. $\newline$ $y \leq -x + 6$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 + 6

\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 with X-Axis: First, let's find the intersection of $y = -x + 6$ and the x-axis ( $y = 0$ ). $\newline$ Set $y$ to $0$ in the equation $y = -x + 6$ . $\newline$ $0 = -x + 6$ $\newline$ $x = 6$ $\newline$ So, the intersection with the x-axis is at $(6, 0)$ .
Find Intersection with Y-Axis: Next, find the intersection of $y = -x + 6$ and the y-axis ( $x = 0$ ). $\newline$ Set $x$ to $0$ in the equation $y = -x + 6$ . $\newline$ $y = -0 + 6$ $\newline$ $y = 6$ $\newline$ So, the intersection with the y-axis is at $(0, 6)$ .
Find Third Vertex: The third vertex is the intersection of the x-axis and y-axis, which is the origin $(0, 0)$ .
Check Vertices against Inequalities: Now, let's check if all the vertices satisfy the inequalities. $\newline$ For $(6, 0)$ : $y \leq -x + 6$ becomes $0 \leq -6 + 6$ , which is true. $\newline$ For $(0, 6)$ : $y \leq -x + 6$ becomes $6 \leq -0 + 6$ , which is true. $\newline$ For $(0, 0)$ : $y \leq -x + 6$ becomes $0 \leq -0 + 6$ , which is true. $\newline$ All vertices satisfy the inequalities.

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