Consider the graph of the linear function h(x) = –6 + x. which quadrant will the graph not go through and why?

a. quadrant i, because the slope is negative and the y-intercept is positive
b. quadrant ii, because the slope is positive and the y-intercept is negative
c. quadrant iii, because the slope is negative and the y-intercept is positive
d. quadrant iv, because the slope is positive and the y-intercept is negative

well the answer to that is xoxoxo

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)