Four consecutive odd integers add up to -48. What is the least of these integers?

-9

Step-by-step explanation:

Four consecutive odd integers be (2x - 3),(2x -1),(2x + 1),(2x + 3)

2x-3 + 2x - 1 +2x+1 + 2x+3 = -48

2x + 2x + 2x +2x - 3 - 1 + 3 + 1 = -48

8x = -48

x = -48/8

x = -6

Least of these integer = 2x + 3 = 2*(-6) + 3 = -12 + 3 = -9

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)