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anonymous

  • 5 years ago

The company discovered that it costs $45 to produce 4 camera cases, $257 to produce 7 camera cases, and $792 to produce 12 camera cases. Using quadratic function, find the cost of producing 1 camera case.

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  1. amistre64
    • 5 years ago
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    How we play in the matrix..... its the same as the other one I just did but without the letters cluttering it up

  2. amistre64
    • 5 years ago
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    (4,45) (7,257) (12,792) 16x^2 +4x +c = 45 49x^2 +7x +c = 257 144x^2 +12x +c= 792 | 16 4 1 || 45 | | 49 7 1 || 257 | |144 12 1 || 792 | is the setup

  3. amistre64
    • 5 years ago
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    | 16 4 1 || 45 | (1/16) | 1 1/4 1/16 || 45/16 | | 49 7 1 || 257 | |144 12 1 || 792 | | 1 1/4 1/16 || 45/16 | (-49) | 49 7 1 || 257 | |-49 -49/4 -49/16 || -2205/16 | | 49 28/4 16/16 || 4112/16 | ------------------------------- | 0 -21/4 -33/16 || 1907/16 | ------------------------------- | 1 1/4 1/16 || 45/16 | (-144) |144 12 1 || 792 | |-144 -144/4 -144/16 || -6480/16 | |144 48/4 16/16 || 12672/16 | ---------------------------------- | 0 -96/4 -128/16 || 6192/16 | ----------------------------------

  4. amistre64
    • 5 years ago
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    | -21/4 -33/16 || 1907/16 | (-4/21) | -96/4 -128/16 || 6192/16 | | 1 33/84 || -1907/84 | (96/4) | -96/4 -128/16 || 6192/16 | | 96/4 66/7 || -3814/7 | | -96/4 -56/7 || 2709/7 | ------------------------- | 0 10/7 || -1105/7 | ------------------------- | 10/7 || -1105/7 | (7/10) | 1 || -221/2 | c = 110.5

  5. amistre64
    • 5 years ago
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    c = -110.5 lol

  6. amistre64
    • 5 years ago
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    | -21/4 -33/16 || 1907/16 | | 0 1 || -1105/10 | (33/16) | -21/4 -33/16 || 3814/32 | | 0 33/16 || -7293/32 | --------------------------- | -21/4 0 || -3479/32 | (-4/21) | 1 0 || 3479/168 | b = 3479/168 c = -1105/10 its quicker on paper, and less mistakes... ill do it up and give you and answer ;)

  7. amistre64
    • 5 years ago
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    1 case costs $18.67 cents; if I did it right... and I have my doubts with it cause of all the fractions, but thats my best guess :)

  8. anonymous
    • 5 years ago
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    my computer just froze & shut everything down -_-

  9. anonymous
    • 5 years ago
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    Another approach to this problem is to use a curve fitting program. The general form of a quadratic equation is: y(x) = a*x^2 + b*x + c. Given a set of data, the curve fitting program will calculate the optimum values for the constants a, b and c using a "least squares" method. Refer to: http://mathworld.wolfram.com/LeastSquaresFitting.html Let y(x) be equal to the cost to produce a batch of x camera cases. From three direct measurements the company had determined that the cost to produce a batch of 4, 7 and 12 camera cases was 45, 257 and 792 dollars respectively. The company wants to know the approximate cost to produce 1 case based on their past production experience. The following Mathematica 8 statement will compute the values for a, b and c for this problem. Fit[{{4, 45}, {7, 257}, {12, 792}}, {1, x, x^2}, x] If the reader has no access to Mathematica 8, then copy and past the above into the "=" input box at www.WolframAlpha.com

  10. anonymous
    • 5 years ago
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    The result is 4.54167 x^2 + 20.7083 x -110.5 and evaluated at x=1 is -85.25 If the numbers above are rationalized,that is, made into fractions, the f[x] can be written as: \[ \frac{109}{24}x{}^{\wedge}2+\frac{497}{24}x-\frac{221}{2} \] Refer to the attachment, CameraCase.pdf, which is a plot of f(x). The plot shows that any batch less than 4 is a looser.

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