3900 workers in 3 buildings. twice as many work in largest building as the smallest of the 3 . there are 500 more workers in the second largest building than the smallest. how many workers are in each building

There are 3900 workers in the three main buildings downtown. Twice as many people work in the largest building as in the smallest of the three. There are 500 more workers in the second-largest building than in the smallest building. How many workers are in each building?

There are 850 workers in smallest building and 1700 workers in largest building and 1350 workers in second largest building

Let "b" be the number of workers in smallest building

Given that Twice as many people work in the largest building as in the smallest of the three

number of workers in largest building = 2b

Given that There are 500 more workers in the second-largest building than in the smallest building

Number of workers in second largest building = 500 + b

Given that there are 3900 workers in 3 buildings

b + 2b + 500 + b = 3900

4b + 500 = 3900

4b = 3900 - 500

4b = 3400

b = 850

Thus there are 850 workers in smallest building

workers in largest building = 2b = 2(850) = 1700

workers in second largest building = 500 + b = 500 + 850 = 1350

Thus there are 850 workers in smallest building and 1700 workers in largest building and 1350 workers in second largest building

its electric no smoke

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)