Using 1, 3, 5, 9, and 9 create a problem in which the answer equals 5. you may use operations more than once.

you add 1+9 which would be 10 then minus 5 which would be 5

1 plus 5 equals 6 plus 9 equals 15 minis 9 equals 6 minis 1 equals 5

< 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)

ιт ιѕ 6/23

step-by-step explanation:

нσρє ι ¢συℓ∂ нєℓρ!