Soybean meal is 16% protein and corn meal is 9% protein. How many pounds of each should be mixed together to get a 350-lb mixture that is 12% protein.
i get stuck after i make a chart for it :(
2007-11-17 05:09:32 · 5 answers · asked by thiswasjustforphysics 3 in Education & Reference ➔ Homework Help
the amount of protein in S lbs of soybean meal is .16S
the amount of protein in C lbs of corn meal is .09C
We know that S + C = 350 (lbs of soybean meal + lbs of corn meal = 350 lbs mixture)
We also know that .16S + .09C = .12(350)
So we have 2 variables and 2 equations, which means we can solve it! Let's write our equations:
S + C = 350
.16S + .09C = 42
Let's solve for S in the first equation and substitute into the second:
S + C = 350
S = 350 - C
.16S + .09C = 42
.16(350-C) + .09C = 42
56 - .16C + .09C = 42
56 - .07C = 42
-.07C = -14
C = -14/-.07
C = 200
So S = 150
200 lbs of cornmeal and 150 lbs of soybean meal
Hope this helps! :)
2007-11-17 05:19:06 · answer #1 · answered by disposable_hero_too 6 · 1⤊ 0⤋
Value Amount Total
Soybean .16 X .16(X)
Meat .09 350-x .09(350-X)
Mixture .12 350 .12(350)
So then you just solve:
.16(X) +.09 (350-X) = .12(350)
edit: Whoa this looked different when i typed it then when i submitted it but basically it is just
.16(X) + .09(350-X) = .12(350)
2007-11-17 05:15:03 · answer #2 · answered by Brock 3 · 0⤊ 0⤋
Soybean = S
Corn = C
16% = 0.16
9% =0.09
12% =0.12
S +C =350
0.16S +0.09C = 350(0.12)
S = 350 -C
0.16(350 -C) +0.09C = 350(0.12) =42
56 -0.16C +0.09C = 42
-0.07C = 42 -56 = -14
C = -14/(-0.07) =200
C = 200 lbs
S = 150 lbs
2007-11-17 05:16:39 · answer #3 · answered by Any day 6 · 1⤊ 0⤋
a equals X B equals Y Cequals M simple algebra
2007-11-17 05:14:39 · answer #4 · answered by vanessa 6 · 0⤊ 0⤋
Chuck it all away and get a Big Mac with fries.
2007-11-17 05:12:35 · answer #5 · answered by Anonymous · 1⤊ 1⤋