Graph a triangle (LMN) and rotate it 180° around the origin to create triangle L′M′N′. Describe the transformation using words. Make sure you refer to the characteristics and the coordinates. Draw a line through points L and L′ and through M and M′. What do you notice about the lines you drew? Do you think you would notice the same characteristics if you drew another line through points N and N′? How do you know?

the shape would stay the same but the triangle will be in the opposite quadrant of the graph. the lines would line up with each other.

hope this helps

There isn’t much to describe when there aren’t points assigned to lmn, but being that it is rotated around the origin 180 degrees, two of the points of l'm'n' would be further form their original position, and the other would be relatively closer. : ))

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)