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calculusxy

  • one year ago

In a certain coffee shop, a large latte costs 2 dollars more than a medium latte, and a medium latte costs 2 dollars more than a small latte. If 12 small lattes cost 2x dollars, how much will 5 large and 7 medium lattes cost? A. 2x + 24 B. 2x + 34 C. 2x + 48 D. 12x + 24 E. 24x + 3

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  1. calculusxy
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    what does x represent?

  3. calculusxy
    • one year ago
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    I don't know. That is confusing me.

  4. jim_thompson5910
    • one year ago
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    cost of large latte? medium? or small?

  5. calculusxy
    • one year ago
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    Can we assume that x=6?

  6. jim_thompson5910
    • one year ago
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    where are you getting that?

  7. calculusxy
    • one year ago
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    nvm

  8. jim_thompson5910
    • one year ago
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    S = cost of small M = cost of medium L = cost of large

  9. jim_thompson5910
    • one year ago
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    `a large latte costs 2 dollars more than a medium latte` so L = M+2 agreed?

  10. calculusxy
    • one year ago
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    yes

  11. jim_thompson5910
    • one year ago
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    ` medium latte costs 2 dollars more than a small latte` using the variables I set up, what equation do you get?

  12. calculusxy
    • one year ago
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    S = x M = x + 2 L = 2 + (x + 2) or 4 + x

  13. jim_thompson5910
    • one year ago
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    S = unknown (not x) M = S + 2 L = M + 2 = (S+2) + 2 = S + 4 `12 small lattes cost 2x dollars` 12S = 2x S = (2x)/(12) S = x/6 `how much will 5 large and 7 medium lattes cost?` 5L + 7M = 5(S+4) + 7(S+2)

  14. jim_thompson5910
    • one year ago
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    the next step is to plug S = x/6 into 5(S+4) + 7(S+2)

  15. calculusxy
    • one year ago
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    so we are learning about plugging in numbers. can we work around that level?

  16. jim_thompson5910
    • one year ago
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    you mean replacing S with some number like 7?

  17. calculusxy
    • one year ago
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    yeah

  18. jim_thompson5910
    • one year ago
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    how would you get to the answer? each answer has an x in it

  19. calculusxy
    • one year ago
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    so can we like plug in 7 for x?

  20. calculusxy
    • one year ago
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    I am really confused on this.

  21. jim_thompson5910
    • one year ago
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    Do you see how I got 5(S+4) + 7(S+2) ?

  22. calculusxy
    • one year ago
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    yes

  23. jim_thompson5910
    • one year ago
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    \[\Large 5(S+4) + 7(S+2) = 5(\frac{x}{6}+4) + 7(\frac{x}{6}+2) \] \[\Large 5(S+4) + 7(S+2) = 5(\frac{x}{6})+5(4) + 7(\frac{x}{6})+7(2) \] \[\Large 5(S+4) + 7(S+2) = \frac{5x}{6}+20 + \frac{7x}{6}+14 \] I'll let you simplify

  24. calculusxy
    • one year ago
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    12x/6 + 34

  25. jim_thompson5910
    • one year ago
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    12x/6 simplifies to ???

  26. calculusxy
    • one year ago
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    2x

  27. calculusxy
    • one year ago
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    THANK YOU!!!

  28. jim_thompson5910
    • one year ago
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    so 2x+34 is the answer

  29. jim_thompson5910
    • one year ago
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    np

  30. calculusxy
    • one year ago
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    I have another question

  31. calculusxy
    • one year ago
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    If \[a = 2x + 3 \] and \[b = 4d\] what is b in terms of a? A. \[(a - 3)^2\] B.\[2(a - 3)^2\] C. \[\frac{ (a-3)^2 }{ 4 }\] D. \[\frac{ (a + 3)^2 }{ 4 }\] E. \[a + 3\]

  32. jim_thompson5910
    • one year ago
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    b = 4d? is that a typo?

  33. calculusxy
    • one year ago
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    Yes sorry \[b = 4d^2\]

  34. jim_thompson5910
    • one year ago
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    and a = 2x + 3 is correct? or no?

  35. calculusxy
    • one year ago
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    that's correct

  36. jim_thompson5910
    • one year ago
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    I don't see how to connect a and b. They have no variables in common

  37. calculusxy
    • one year ago
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    i have checked with many numbers none of which seems to work

  38. jim_thompson5910
    • one year ago
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    something seems missing

  39. calculusxy
    • one year ago
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    i will check on that with my teacher. thank you :)

  40. jim_thompson5910
    • one year ago
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    np

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