Find the number set which satisfies each of the problems. If 7 is subtracted from the absolute value of the sum of a number and 6, the result is 3.

x=4      or    x= - 16

Step-by-step explanation:

|x+6|

Now subtract 7 which equals to 3.

|x+6|-7=3

|x+6|=10

Now remove the mode by adding plus minus sign in the front of 10.

x+6=±10

x+6=10 or x+6=-10

x=4      or x=-16

1/2*qr*rs

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)