What has to be added to the sum of 5/8 and 8/5 to get 3

31/40 has to be added to the sum of 5/8 and 8/5 to get 3.

Step-by-step explanation:

Solve this equation:  5/8 + 8/5 + f = 3, where f is the number that has to be added to the sum of 5/8 and 8/5.

The LCD of 5/8 and 8/5 is 40; thus, 5/8 + 8/5 = 25/40 + 64/40 = 89/40.

Then 89/40 + f = 120/40 (where 120/40 is equivalent to 3).

Subtracting 89/40 from both sides yields the desired result:

f = (120 - 89)/40 = 31/40

31/40 has to be added to the sum of 5/8 and 8/5 to get 3.

< 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)