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mestny [16]
3 years ago
10

• How do you use the distance formula and slope formula to prove properties of polygons?

Mathematics
1 answer:
patriot [66]3 years ago
6 0

Answer:

distance formula= \sqrt{x} (x_{2} -x_{1} )^{2} + (y_{2} -y_{1} )^{2} = find the length(units) using coordinates.

slope formula= m=\frac{rise}{run} =\frac{y_{2}-y_{1}  }{x_{2} -x_{1} }= determines slope of a line using two points on the line(coordinates)

When working with polygons the main properties which are important are:

The number of sides of the shape.

The angles between the sides of the shape.

The length of the sides of the shape.

length , width

straight lines , angles, vertices, vertex

A polygon is any 2-dimensional shape formed with straight lines.

a plane figure with at least three straight sides and angles, and typically five or more.

Step-by-step explanation:

<h2><u>answer= use distance and slope to determine that opposite side of a polygon are the same length and same slope</u></h2>
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Step-by-step explanation:

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The oil prices for 2014, rounded to the nearest dollar, were: 95, 101, 101, 102, 102, 106, 104, 97, 93, 84, 76, 59. what is the
kipiarov [429]
  • Interquartile Range (IQR) = Q_3-Q_1 , with Q_3 as the upper quartile and Q_1 as the lower quartile.

Firstly, rearrange the data so that it's in ascending order: \{59,76,84,93,95,97,101,101,102,102,104,106\}

Next, find the median:

\{59,76,84,93,95,\boxed{97,101,}101,102,102,104,106\}\\\\\frac{97+101}{2}\\\\\frac{198}{2}\\\\99

Now to find the lower quartile, find the "median" of the data set that's to the left of 99:

\{\overbrace{59,76,84,93,95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\{\overbrace{59,76,\boxed{84,93}95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\frac{84+93}{2}\\\\\frac{177}{2}\\\\88.5

Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:

\{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102

Now that we have the upper and lower quartile, subtract them:

102-88.5=13.5

<u>In short, the IQR of this data set is 13.5.</u>

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