How much water must be added to 12 quarts of a 10% solution of salt and water to reduce to a 6% solution?

Let x = quarts of water to be added

so then

12* 10% + x * 0% = (12+x) * 6% (this says 12 quarts of 10% solution + x quarts of 0%(water) equals the new total (12+x) of a 6% solution

1.2 = .72 + .06x

.48 = .06x

x = 8 (quarts of water to be added)

Step-by-step explanation:

Hope this helps

$15 each

step-by-step explanation:

$45/3 children= $15 each

< 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)