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anonymous

  • 5 years ago

The company discovered that it costs $28 to produce 2 camera cases, $50 to produce 4 camera cases, and $140 to produce 10 camera cases. Using the quadratic function, find the cost of producing 12 cameras.

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  1. anonymous
    • 5 years ago
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    P = ax^2+bx+c, P is total production cost, x is number of cases. solve for a,b,c using the three values you are given,

  2. anonymous
    • 5 years ago
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    so, 28 = 4a+2b+c 50 = 16a+4b+c 140 = 100a+10b+c

  3. anonymous
    • 5 years ago
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    solve for a,b,c

  4. anonymous
    • 5 years ago
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    how do i do that?

  5. anonymous
    • 5 years ago
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    do you know how to do matrices?

  6. anonymous
    • 5 years ago
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    no

  7. anonymous
    • 5 years ago
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    okay 28 = 4a+2b+c ---------------------------------------->Eqaution 1 50 = 16a+4b+c ---------------------------------------> Eqn 2 140 = 100a+10b+c----------------------------------> eqn 3 do eqn 3 - eqn 1 and do eqn 2 - eqn 1 and post what you got here

  8. anonymous
    • 5 years ago
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    still don't know how to figure it out

  9. anonymous
    • 5 years ago
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    Can you subtract equation 1 from equation 2 and post it here?

  10. anonymous
    • 5 years ago
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    38=12a+2b

  11. anonymous
    • 5 years ago
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    how did you get 38?

  12. anonymous
    • 5 years ago
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    32*

  13. anonymous
    • 5 years ago
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    i mean 22*

  14. anonymous
    • 5 years ago
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    ok so repost the correct equation again

  15. anonymous
    • 5 years ago
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    22=12a+2b

  16. anonymous
    • 5 years ago
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    okay good. Now, the same way, subtract equation 1 from equation 3

  17. anonymous
    • 5 years ago
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    112=96a+8b

  18. anonymous
    • 5 years ago
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    okay. Now, can you solve the set of equations 12a + 2b = 22 96a + 8b = 112 ?

  19. anonymous
    • 5 years ago
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    can you help me with it.

  20. anonymous
    • 5 years ago
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    okay, have you learned how to solve two equations for two unknown variables in school?

  21. anonymous
    • 5 years ago
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    not that i can remember.

  22. anonymous
    • 5 years ago
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    you haven't done basic algebra in school yet?

  23. anonymous
    • 5 years ago
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    alright anyway, 12a + 2b = 22 ------------>eqnA 96a + 8b = 112----------->eqnB

  24. anonymous
    • 5 years ago
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    do you know substitution to solve for a and b?

  25. anonymous
    • 5 years ago
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    okay just do 4 times eqn A and post it here.

  26. anonymous
    • 5 years ago
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    what do you mean

  27. anonymous
    • 5 years ago
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    If you multiply both sides of equation A by 4, what do you get?

  28. anonymous
    • 5 years ago
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    48a+2b=88

  29. anonymous
    • 5 years ago
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    try again

  30. anonymous
    • 5 years ago
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    that's multiplying both sides by 4.

  31. anonymous
    • 5 years ago
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    so 2b*4 = 2b?

  32. anonymous
    • 5 years ago
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    usually when i've done that, i've never done it to all. be more specific.

  33. anonymous
    • 5 years ago
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    48a+8b=88

  34. anonymous
    • 5 years ago
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    if a =4, then 4a =16 so we multiply both sides by 4. so if a+b = 4 then 4(a+b) =16

  35. anonymous
    • 5 years ago
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    okay good, so 48a + 8b = 88--------------->eqn. C Do you notice what we did there? we multiplied equation B by 4 so that both equation A and equation C have the term 8B in them. So now subtract equation C from equation A.

  36. anonymous
    • 5 years ago
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    i'm still completely confused

  37. anonymous
    • 5 years ago
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    what are you confused about?

  38. anonymous
    • 5 years ago
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    OK join me http://www.dabbleboard.com/draw?b=Guest654407&i=0&c=ce1b368e8fb4c87408d1e26d13c93f5bf92f9032 will explain in detail there

  39. anonymous
    • 5 years ago
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    okay i'm on there

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