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Look at the system of inequalities. $\newline$ $y \leq -x + 8$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 + 8

\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 $y = -x + 8$ and the x-axis ( $y = 0$ ). $\newline$ Set $y$ to $0$ in the equation $y = -x + 8$ and solve for $x$ . $\newline$ $0 = -x + 8$ $\newline$ $x = 8$ $\newline$ So, one vertex is at $(8, 0)$ .
Find Intersection with Y-Axis: Next, find the intersection of $y = -x + 8$ and the y-axis ( $x = 0$ ). $\newline$ Set $x$ to $0$ in the equation $y = -x + 8$ and solve for $y$ . $\newline$ $y = -0 + 8$ $\newline$ $y = 8$ $\newline$ So, another vertex is at $(0, 8)$ .
Find Third Vertex: The third vertex is the intersection of the x-axis and y-axis, which is the origin. $\newline$ So, the third vertex is at $(0, 0)$ .

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Question

The equation of a circle is

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\newline

1

\newline

(A) The center is

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and the radius is

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\newline

(B) The center is

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y=(x-3)(x+9)

\newline

The given equation is graphed in the

x y

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x

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\newline

1

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-3

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-3

9

\newline

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\newline

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Question

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is graphed in the

x y

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\newline

1

\newline

x

\newline

y

\newline

x

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\newline

y

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Algebra - Order of Operations Algebra - Distributive Property `X` and `Y` Axes Geometry - Scalene Triangle Common Multiple Geometry - Quadrant