Complete the square to re-write the quadratic function in vertex form: y=x^2-x-4

b=-1
b/2=-0.5x-0.5
=0.25
y=x^2-x+0.25 -0.25-4
y=(x-0.5)^2-0.25-4
y=(x-0.5)^2-4.25
y=(x+1)^2+4

y=(x+1)^2+4

maybe - definition of supplementary angles. im not sure. sorry

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)

answer: it is a.,b.,d.

step-by-step explanation: