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Look at the system of inequalities. $\newline$ $y \leq -\frac{1}{2}x + 5$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 -\frac{1}{2}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 with $x=0$ : First, let's find the intersection of $y = -\frac{1}{2}x + 5$ and $x = 0$ . $\newline$ Substitute $x = 0$ into $y = -\frac{1}{2}x + 5$ . $\newline$ $y = -\frac{1}{2}(0) + 5$ $\newline$ $y = 5$ $\newline$ So, one vertex is $(0, 5)$ .
Find Intersection with $y=0$ : Next, find the intersection of $y = -\frac{1}{2}x + 5$ and $y = 0$ . $\newline$ Set $y = 0$ in the equation $y = -\frac{1}{2}x + 5$ . $\newline$ $0 = -\frac{1}{2}x + 5$ $\newline$ $\frac{1}{2}x = 5$ $\newline$ $x = 10$ $\newline$ So, another vertex is $(10, 0)$ .
Find Origin Intersection: Lastly, find the intersection of $x = 0$ and $y = 0$ . This is simply the origin, so the third vertex is $(0, 0)$ .

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