PLEASE HELP
I need this done is 10 minutes !!!

Triangle LMN has vertices at L(-1,5),M(-1,0),N(-2,5). Determine the vertices of image L’M’N if the pre image is rotated 90° counterclockwise about the origin.

L'(5, 1), M'(0, 1), N'(5,2)
L'(-1,-5), M'(-1,0), N'(-2,-5)
L'(-5, -1), M'(0, -1), N'(-5, -2).
L'(1, -5), M'(1, 0), N'(2,-5)

Answers

Answer 1

Triangle LMN has vertices at L(-1,5),M(-1,0),N(-2,5), then the vertices are L'(-5, -1), M'(0, -1), N'(-5, -2).

To rotate a point counterclockwise:

For a point (x, y):

x' = x*cos(theta) - y*sin(theta)

y' = x*sin(theta) + y*cos(theta)

In this situation, rotate the triangle 90° counterclockwise about the origin.

The transformation equations to each vertex of the triangle states that:

For vertex L(-1, 5):

x' = -1*cos(90°) - 5*sin(90°) = -1*0 - 5*1 = -5

y' = -1*sin(90°) + 5*cos(90°) = -1*1 + 5*0 = -1

Therefore, vertex L' is (-5, -1).

For vertex M(-1, 0):

x' = -1*cos(90°) - 0*sin(90°) = -1*0 - 0*1 = 0

y' = -1*sin(90°) + 0*cos(90°) = -1*1 + 0*0 = -1

Therefore, vertex M' is (0, -1).

For vertex N(-2, 5):

x' = -2*cos(90°) - 5*sin(90°) = -2*0 - 5*1 = -5

y' = -2*sin(90°) + 5*cos(90°) = -2*1 + 5*0 = -2

Therefore, vertex N' is (-5, -2).

Thus, the vertices of the image triangle L'M'N' after rotating the original triangle 90° counterclockwise about the origin are: L'(-5, -1), M'(0, -1), N'(-5, -2).

For more details regarding vertices, visit:

https://brainly.com/question/29154919

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Answer:

Distributive property says that:

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Here we can take the two terms inside the parentheses as A and B, and the term that multiplies them as C, then distributing we get:

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Step-by-step explanation:

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9514 1404 393

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01:11:1
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Step-by-step explanation:

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