Simplify to create an equivalent expression. 2 ( − 1 4 + r ) − ( − 3 r − 5 ) choose 1(choice a) a: 5r − 23 choice b) b: 5r − 33 (choice c) c: − 5r − 23 (choice d) d: 5r + 23

5r-23  Choice A

Step-by-step explanation:

2 ( − 1 4 + r ) − ( − 3 r − 5 )

distribute

2*-14 +2*r - 1*-3r -1*-5

-28 +2r +3r +5

combine like terms

2r+3r -28+5

5r-23

< 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)