Can someone tell me where I went wrong? Livestock feed. Soybean meal is 16% protein and corn meal is 9% protein. How many pounds of each should be mixed to get a 350-lb mixture that is 12% protein? Here is what I came up with. Let s = pounds of soybean Let c = pounds of corn meal s + c = 350 16s = 9c = 12 ( s+c) = 12(350)2 -9s-9c +165 + 9c = 3150 + 4200 7s =1050àdivide by 7 S= 150 150+ c = 350 150 + 200 150 + 200=350 C= 200
My teacher said this was wrong 16s = 9c = 12 ( s+c) = 12(350)2 ; incorrect and then gave me this hint, but i am confused Let s be the Soybean meal, and let c be the corn meal Write an equation so that the sum of s and c is 350 lbs Then write a second equation that the sum of 16% s and 9% c is 12% of 350. Then solve for s and c.
@andjie your first equation s + c = 350 is correct! For your second equation, percents should be expressed as decimals. It should be 0.16s + 0.09c = 0.12(350). Do you understand?
you can multiply everything in equation 2 by 100 to get 16s+9c = 12*350
easier to go right to \[16x+9(350-x)=12\times 350\]
if you put \(x\) as the amount of soy, the amount of corn must be \(350-x\)
@calculusfunctions How was mine wrong if you can explain that and was that the only thing wrong with my whole solution
See that is the probelm after that happens I get confused on my whole set up
@andjie the second equation is in terms of the amount of protein in the mixture. A percent is a measurement out of 100. Thus 36% for example, would be 36/100 or 0.36. Now if there are s pounds of soybean and soybean has 16% pounds of protein, then the part that is 16% is 0.16s. Similarly for cornmeal it is 0.09c. Now there is 12% protein in the mixture of the two, and there is 350 lb of the mixture., hence the total amount of protein is 0.12(350). Thus the the second equation should be 0.16s + 0.09c = 0.12(350). Understand?
Wow I understand it now
Thank you so much for helping me understand I was going crazy
Great! Do you want me to stick around to check your solution and final answer?
I would appreciate that I have problems setting everything up
let me see if I can do this
No problem. I'll stay for as long as you need me to.
Nope I still don't have it hold on
That's OK! Don't rush, take your time.
0.16s + 0.9c = 0.12(350) -0.9s-0.9c +165 + 9c = 3150 + 4200 7s =1050divide by 7 S= 150 150+ c = 350 150 + 200 150 + 200=350 C= 200
It is like I keep losing what I am suppose to do next it is irritating
No @andjie it's 0.09c, not 0.9c.
s + c = 350 0.16s + 0.09c = 0.12(350) -0.09s-0.09c +165 + 0.09c = 3150 + 4200 7s =1050divide by 7 S= 150 150+ c = 350 150 + 200 150 + 200=350 C= 200
YES! c = 200 and s = 150. Perfect!
So that is what I was missing that one tiny aspect although in math terms simple is a big mistake
Yes, do you understand how to deal with percents now?
yes multiply the number by 100 and you will get the decimal form?
You mean divide by 100 to change percents to decimals.
oh divide wow I will get it
That's OK, you did just fine!
Anything else?
So my setup is complete now
I do have one more question actually I just want you to let me know if I did it correctly if you have time
Sure. Go ahead.
a = 1 + b, b = 5 − 2a b = 5-2(1=b) b = 5-2-2b b + 2b = 5-2 3b = 3 b = 1 a = 1 + b a = 1 + 1 a = 2 b = 5 – 2a 1 = 5 – 2(2) 1 = 5 – 4 a=2, b=1 The lines intersect at (2, 1), so they are consistent and are dependant. a= (0, -1), (3, 2) b= (0, 5), (3, -1)
Solve each system graphically. Be sure to check your solution. If a system has an infinite number of solutions, use set−builder notation to write the solution set. If a system has no solution, state this.
@andjie Perfect!!!!
Really!
:)
Yes, really!!!
Awesome thank you for all your help it means the world to me!
Welcome anytime! Anything else right now?
No not right now!
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