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anonymous

  • one year ago

AM I RIGHT?????

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  1. anonymous
    • one year ago
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    |dw:1433060237198:dw|

  2. anonymous
    • one year ago
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    c = 3A – (a – b) c = 3A – a – b<<< c = 3a + 3b + A c = 3Aab

  3. UsukiDoll
    • one year ago
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    What is this problem anyway?! are you solving for c?

  4. UsukiDoll
    • one year ago
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    and you forgot to distribute the negative sign -(a-b) = -a+b

  5. anonymous
    • one year ago
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    Yes sorry I forgot to put that there

  6. UsukiDoll
    • one year ago
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    |dw:1433061203222:dw|

  7. anonymous
    • one year ago
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    That's how I write my A's

  8. UsukiDoll
    • one year ago
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    \[A = \frac{1}{3}[(a-b)+c]\] so what is the question? solve for c?!

  9. UsukiDoll
    • one year ago
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    we don't even know what the original question is besides am I right? That doesn't give any information. What is the question and what is the goal of this?

  10. Michele_Laino
    • one year ago
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    step #1 in order to find c, we have to multiply both sides by 3, what do you get?

  11. UsukiDoll
    • one year ago
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    first of all, op is your question how do I solve in terms of c? is that why we have c = your answer

  12. anonymous
    • one year ago
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    Yes

  13. UsukiDoll
    • one year ago
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    ok then. so we have this equation \[A = \frac{1}{3}[(a-b)+c]\]

  14. anonymous
    • one year ago
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    Yes and we have to solve for C

  15. UsukiDoll
    • one year ago
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    ok so what do we do first?

  16. anonymous
    • one year ago
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    When I first did it I got c= -a+b+ 1/3A

  17. anonymous
    • one year ago
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    But it wasnt in the answer choices so I thought of the one that looked similar to it

  18. UsukiDoll
    • one year ago
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    if I was doing that problem I would distribute the 1/3 all around \[A = [(\frac{1}{3}a-\frac{1}{3}b)+\frac{1}{3}c]\]

  19. UsukiDoll
    • one year ago
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    then I would shift the 1/3a and 1/3b to the left

  20. UsukiDoll
    • one year ago
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    \[A -\frac{1}{3}a+\frac{1}{3}b=\frac{1}{3}c\]

  21. UsukiDoll
    • one year ago
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    and then multiply 3 on both sides \[3A-a+b=c\]

  22. UsukiDoll
    • one year ago
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    if I plug this guy back into the original equation I should have A = A or something\[A = \frac{1}{3}[(a-b)+3A-a+b]\]

  23. UsukiDoll
    • one year ago
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    \[A = \frac{1}{3}[(b-b)+3A-a+a]\] \[A = \frac{1}{3}[3A] \rightarrow A = \frac{3A}{3} \rightarrow A=A\]

  24. UsukiDoll
    • one year ago
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    a - a and -b+b cancel out I am left with 3A and I can just divide 3

  25. UsukiDoll
    • one year ago
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    what are your answer choices for this problem?

  26. anonymous
    • one year ago
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    c = 3A – (a – b) c = 3A – a – b c = 3a + 3b + A c = 3Aab

  27. anonymous
    • one year ago
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    I thought it was the second choice

  28. UsukiDoll
    • one year ago
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    \[3A-a+b=c\] it's the first choice if you distribute the negative

  29. UsukiDoll
    • one year ago
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    c = 3A – (a – b) c= 3A-a+b

  30. anonymous
    • one year ago
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    But why?

  31. anonymous
    • one year ago
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    No wait I figured out why and how you did it thank you !

  32. UsukiDoll
    • one year ago
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    medal? ^^

  33. anonymous
    • one year ago
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    and fan :)

  34. UsukiDoll
    • one year ago
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    yay ^^

  35. UsukiDoll
    • one year ago
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    whenever there is a negative sign on the right side of the equation you have to distribute it Ex 5x-(2x^2+4) 5x-2x^2-4

  36. anonymous
    • one year ago
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    Ohh I thought you had to do a whole different complicated equation

  37. anonymous
    • one year ago
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    you made it simple lol thanks for helping me again!

  38. UsukiDoll
    • one year ago
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    ^^ it's a very important rule

  39. anonymous
    • one year ago
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    I'm going to start doing problems like these so I wont forget the rule

  40. UsukiDoll
    • one year ago
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    ^_^ you can write a testimonial about me if you want xD

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