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anonymous

  • one year ago

Mixture Problem We didn't go over them in class but they are on the online homework =(

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  1. anonymous
    • one year ago
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  2. zepdrix
    • one year ago
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    Let P=Pounds of Peanuts, C=Pounds of Cashews, R=Pounds of Raisins. Then we can establish an equation based on the first sentence they gave us,\[\large\rm P+C+R=9\]Some amount of Peanuts, Cashews and Raisins give us a total of 9 pounds, ya? :)

  3. anonymous
    • one year ago
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    Okay, that makes sense so far =)

  4. zepdrix
    • one year ago
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    We're going to set up another equation, this one is a little trickier. This one is based on prices.

  5. zepdrix
    • one year ago
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    The peanuts cost 1.50 per pound. So the total price we pay for ALL of our peanuts is \(\large\rm 1.50P\). Where P is the number of pounds of peanuts.

  6. zepdrix
    • one year ago
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    Similarly, our cashews will have a total price of \(\large\rm 2.00C\) While our raisins will total \(\large\rm 1.00r\)

  7. ♪Chibiterasu
    • one year ago
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    WUMBO

  8. zepdrix
    • one year ago
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    They tell us that the total costs, of all the nuts, is 13 dollars. So when we total up all of these totals:\[\large\rm 1.5P+2C+1R=13\]I dropped any unnecessary 0's, hopefully that wasn't confusing.

  9. anonymous
    • one year ago
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    All right, that makes sense =)

  10. zepdrix
    • one year ago
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    We can actually set up one more equation! :O

  11. zepdrix
    • one year ago
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    `twice as many peanuts` as `cashews`. So the pounds of peanuts, P, should be twice as large as the pounds of cashews, C.

  12. zepdrix
    • one year ago
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    We can write that relationship like this: \(\large\rm P=2C\) Maybe read that as, "peanuts equals twice the cashews"

  13. zepdrix
    • one year ago
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    We're going to go back and substitute this relationship into our other two equations.

  14. anonymous
    • one year ago
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    Okay!

  15. zepdrix
    • one year ago
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    Our weight equation: \[\large\rm \color{orangered}{P}+C+R=9\]Will become\[\large\rm \color{orangered}{2C}+C+R=9\] While our price equation:\[\large\rm 1.5\color{orangered}{P}+2C+1R=13\]will become\[\large\rm 1.5\color{orangered}{(2C)}+2C+1R=13\]

  16. anonymous
    • one year ago
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    Makes sense (Just as a side note, could this also be solved with a matrix?)

  17. zepdrix
    • one year ago
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    Mmm definitely. Lemme think..

  18. zepdrix
    • one year ago
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    \[\large\rm \left(\begin{array}{ccc|c}1 & 1 & 1 & 9\\ 1.5 & 2 & 1 & 13\end{array}\right)\]Ah sorry for slow :) I'm not really familiar with augmented matrices in latex lol Setting out our third column is a little tricky. Since \(\large\rm P=2C\), subtracting 2C from each side gives us: \(\large\rm 1P-2C+0R=0\) So our matrix would be:\[\large\rm \left(\begin{array}{ccc|c}1 & 1 & 1 & 9\\ 1.5 & 2 & 1 & 13\\ 1 &-2 & 0 & 0\end{array}\right)\]

  19. zepdrix
    • one year ago
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    Our third row* not third column, blah typo

  20. anonymous
    • one year ago
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    Okay! So then the answer would be (4,2,3)? And no worries! You saved my educational life! XP

  21. anonymous
    • one year ago
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    Nah wait, that's not one of the answers lol.

  22. zepdrix
    • one year ago
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    It's not? Hmm, that sounds right :o

  23. anonymous
    • one year ago
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    No wait, wrong problem! It is that. Lol sorry

  24. zepdrix
    • one year ago
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    you silly billy -_-

  25. zepdrix
    • one year ago
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    such a bum... just using a matrix calculator or something? lol

  26. anonymous
    • one year ago
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    Yeah, a TI84. Honestly the whole substitution thing loses me every time, not that your explanation was losing me! I was just curious about the matrix because I will probably use that on my test tomorrow.

  27. zepdrix
    • one year ago
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    matrices are extremely useful in real world problems :) So it's definitely not a waste of time to get comfortable with your calculator!

  28. anonymous
    • one year ago
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    Cool beans! Thanks so much =)

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