Find AC.

Assume all segments that appear to be tangent are tangent.

b. 2 and 4

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)

the equation of the parallel line would be y = 2x + 9

step-by-step explanation:

to start we need to find the slope of the original line. we can do that by looking at the coefficient of x, which is 2

since the slope of the original line is 2 and parallel lines have the same slope, we know the new line will also have 2 slope. now we can use that along with the point given in point-slope form to get the find equation.

y - y1 = m(x - x1)

y - 5 = 2(x + 2)

y - 5 = 2x + 4

y = 2x + 9

2/3=10/15

1/5=3/15

10+3=13/15

2 13/15 is the answer