Plot ΔABC on graph paper with points A(3,3), B(1,1), and C(6,1). Reflect ΔABC by multiplying the y-coordinates of the vertices by −1. Then use the function (x, y)→(x−6,y−3) to translate the resulting triangle. Name the coordinates of the vertices of the result. Question 3 options:

A'(3,-3), B'(1,-1), C'(6,-1)

A'(-6,-3), B'(-4,-5), C'(-4,0)

A'(-3,3), B'(-1,1), C'(-1,6)

A'(-3,-6), B'(-5,-4), C'(0,-4)

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< 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)

answer: the angel abc is 30°,and the abc is a triangle where the angel bac is 90°

so ac is the half of 24=12ft,and from the pythagorean theorem we find ba

ba=√bc^2-ac^2=√576-144=√432=12√3ft

so bac is 12+12√3=12(1+√3)ft