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Look at the system of inequalities. $\newline$ $y \leq -x + 3$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 + 3

\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 Vertex $(0, 3)$ : First, let's find the intersection of $y = -x + 3$ and $x = 0$ . Substitute $x = 0$ into $y = -x + 3$ . $y = -0 + 3$ $y = 3$ So, one vertex is $(0, 3)$ .
Find Vertex $(3, 0)$ : Next, find the intersection of $y = -x + 3$ and $y = 0$ . Substitute $y = 0$ into $y = -x + 3$ . $0 = -x + 3$ $x = 3$ So, another vertex is $(3, 0)$ .
Find Origin $0, 0$ : Finally, find the intersection of $x = 0$ and $y = 0$ . This is simply the origin, so the last vertex is \(0, 0\).

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Question

The equation of a circle is

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. What are the center and radius of the circle?

\newline

1

\newline

(A) The center is

(-4,0)

and the radius is

1

\newline

(B) The center is

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and the radius is

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(C) The center is

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(D) The center is

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Question

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

\newline

The given equation is graphed in the

x y

-plane. Which of the following are

x

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\newline

1

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-3

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\newline

-3

9

\newline

3

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\newline

3

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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