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pogonyaev
2 years ago
7

Graph ∆KLM with vertices K(2,2), L(0,- 1), and M(-3,2). Graph ∆K'L'M' with a scale factor of 2. What are the coordinates of the

new vertices?
Mathematics
1 answer:
olasank [31]2 years ago
5 0

Answer:

The coordinates of the new vertices are:

K'(4,4), L'(0,- 2), and M'(-6,4).

Step-by-step explanation:

In order to dilate a shape by 2, you can simply multiply the coordinates of the orignal triangle each by 2 to find the new coordinates of the new vertices.

K(2,2), L(0,- 1), and M(-3,2)

K'(2*2,2*2), L'(0*2,- 1*2), and M'(-3*2,2*2).

Now become...

K'(4,4), L'(0,- 2), and M'(-6,4).

You can now graph ∆K'L'M' using these coordinates.

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Angles in a quadrilateral add to 360 degrees. So, x = 360-88-142-106=24 degrees

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1 year ago
Find the distance between each set of points. Round to the nearest tenth, when necessary. (-3, 2) and (1, -6)
vovangra [49]

Answer:

<h2>The answer is 8.9 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  } \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-3, 2) and (1, -6)

The distance between them is

d =   \sqrt{ ({ - 3 - 1})^{2}  +  ({2 + 6})^{2} }   \\  =  \sqrt{ ({ - 4})^{2} +  {8}^{2}  }  \\  =  \sqrt{16 + 64}  \\  =  \sqrt{80}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  = 4 \sqrt{5}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  = 8.9442719...

We have the final answer as

8.9 units to the nearest tenth

Hope this helps you

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