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Look at the system of inequalities. $\newline$ $y \leq -x + 10$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 + 10

\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 y-intercept: First, let's find the intersection of $y = -x + 10$ and $x = 0$ , which is the y-axis. $\newline$ Substitute $x = 0$ into $y = -x + 10$ . $\newline$ $y = -0 + 10$ $\newline$ $y = 10$ $\newline$ So, one vertex is $(0, 10)$ .
Find x-intercept: Next, find the intersection of $y = -x + 10$ and $y = 0$ , which is the x-axis. $\newline$ Substitute $y = 0$ into $y = -x + 10$ . $\newline$ $0 = -x + 10$ $\newline$ $x = 10$ $\newline$ So, another vertex is $(10, 0)$ .
Find third vertex: Lastly, since $x \geq 0$ and $y \geq 0$ , the third vertex is where both $x$ and $y$ are zero. $\newline$ So, the third vertex is $(0, 0)$ .

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