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patriot [66]
3 years ago
11

Which graph shows a set of ordered pairs that represent a function?

Mathematics
2 answers:
Yuri [45]3 years ago
4 0

Make XY tables for each option.

If any of the Tables have identical X numbers it is not a function.

The first option has two X's that re 2, so is not a function.

Second option has 2 x's that are 4's, so is not a function.

The third option has no repeating X values so is a function.

The fourth option has two -2's and two 0's so ids not a function.

The function is the third choice:

On a coordinate plane, solid circles appear at the following points: (negative 3, 2), (negative 2, 2), (0, 1), (1, 3), (2, negative 4), (4, negative 1).

KonstantinChe [14]3 years ago
4 0

Answer: On a coordinate plane, solid circles appear at the following points: (negative 3, 2), (negative 2, 2), (0, 1), (1, 3), (2, negative 4), (4, negative 1).

Step-by-step explanation:

A function is a relation in which only one output value is associated with each input variable .

In coordinate - plane , we denote x= independent variable or as input value.

y= dependent variable or output value.

The first option is not a function because there are 2 output values corresponding to one input value (2, negative 1) and (2, negative 3).

The second option is not a function because there are 2 output values corresponding to one input value (4, negative 3) ,(4, 4).

The fourth option is not a function because there are 2 output values corresponding to one input value  (negative 2, 1), (negative 2, negative 1) .

Only in third option , only one output value is associated with each input variable .

Therefore , the correct answer is "On a coordinate plane, solid circles appear at the following points: (negative 3, 2), (negative 2, 2), (0, 1), (1, 3), (2, negative 4), (4, negative 1)."

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Area of a triangle with points at (-9,5), (6,10), and (2,-10)
Ann [662]
First we are going to draw the triangle using the given coordinates. 
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Distance formula: d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Distance from point A to point B:
d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}
d_{AB}= \sqrt{(6+9)^2+(10-5)^2}
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Distance from point A to point C:
d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}
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Finally, to find the area of our triangle, we are going to use Heron's formula:
A= \sqrt{s(s-AB)(s-AC)(s-BC)}
A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}
A= \sqrt{27.41(11.6)(8.81)(7.01)}
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We can conclude that the perimeter of our triangle is 140.13 square units.

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We already have an equation with "y" on the left, above, so we just need to introduce the comparison symbol:

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Another way to write this is ...

  x + y ≥ 2

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