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Angelina_Jolie [31]
3 years ago
7

Removing which point from the coordinate plane would make the graph a function of x? On a coordinate plane, points are at (negat

ive 2, negative 3), (negative 2, 1), (negative 4, 3), (0, 4), (1, 1), and (2, 3). (–4, 3) (–2, 1) (0, 4) (1, 1)
Mathematics
2 answers:
salantis [7]3 years ago
7 0

Answer:

  (-2, 1)

Step-by-step explanation:

For a relation consisting of (x, y) pairs to be a function, all of the x-values must be unique. In the given relation, points (-2, -3) and (-2, 1) have the same x-value. Removing either point will make the relation a function.

Of these, the only one listed among answer choices is (-2, 1).

kogti [31]3 years ago
5 0

Answer:

-2 , 1

Step-by-step explanation:

good luck love

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Answer:

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Step-by-step explanation:

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Given segments AB and CD intersect at E.
nata0808 [166]

The length of a segment is the distance between its endpoints.

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<u>(a) Length of AB</u>

We have:

\mathbf{A = (1,2)}

\mathbf{B = (4,5)}

The length of AB is calculated using the following distance formula

\mathbf{AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}

So, we have:

\mathbf{AB = \sqrt{(1 - 4)^2 + (2 - 5)^2}}

\mathbf{AB = \sqrt{18}}

Simplify

\mathbf{AB = 3\sqrt{2}}

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First, we calculate the length of CD using:

\mathbf{CD = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}

Where:

\mathbf{C = (2, 4)}

\mathbf{D = (2, 1)}

So, we have:

\mathbf{CD = \sqrt{(2 -2)^2 + (4 - 1)^2}}

\mathbf{CD = \sqrt{9}}

\mathbf{CD = 3}

By comparison

\mathbf{CD \ne AB}

Hence, AB and CD are not congruent

<u>(c) AB bisects CD or not?</u>

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\mathbf{AB = \frac 12 \times CD}

The above equation is not true, because:

\mathbf{3\sqrt 2 \ne \frac 12 \times 3}

Hence, AB does not bisect CD

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If CD bisects AB, then:

\mathbf{CD = \frac 12 \times AB}

The above equation is not true, because:

\mathbf{3 \ne \frac 12 \times 3\sqrt 2}

Hence, CD does not bisect AB

Read more about lengths and bisections at:

brainly.com/question/20837270

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Answer:

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