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lamthornhill Group Title

if the cost, C, for manufacturing x units of a certain product is given by C=x^2-5x+70, find the number of units manufactured at a cost of $8620

  • 2 years ago
  • 2 years ago

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  1. richyw Group Title
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    well \(C(x)=x^2-5x+70\) and you know the cost is $8620 so \[x^2-5x+70=8620\]now solve for x. Do you know how to do that?

    • 2 years ago
  2. lamthornhill Group Title
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    no, that is where I'm having the problem cause I don't understand algebra

    • 2 years ago
  3. Sheng Group Title
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    |dw:1352089616943:dw|

    • 2 years ago
  4. Sheng Group Title
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    quadratic formula, where a = 1, b= 5, c = -8550

    • 2 years ago
  5. Sheng Group Title
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    b = -5 sorry

    • 2 years ago
  6. richyw Group Title
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    alright well the first thing you usually want to do is write this in general form. To do this you subtract 8620 from both sides

    • 2 years ago
  7. lamthornhill Group Title
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    I have no idea what you are talking about

    • 2 years ago
  8. richyw Group Title
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    \[x^2-5x+70-8620=0\]\[x^2-5x-8850=0\]Now you want to find the roots to that. First you want to see if it can be factored, if it can then that's great and if not you need to use the quadratic formuala.

    • 2 years ago
  9. richyw Group Title
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    i'm getting there!

    • 2 years ago
  10. lamthornhill Group Title
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    ok

    • 2 years ago
  11. richyw Group Title
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    the quadratic formula finds the roots of quadratic polynomials. For a quadratic polynomial in general form, that is \(f(x)=ax^2+bx+c=0\), the quadratic formulas is \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • 2 years ago
  12. richyw Group Title
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    so looking at the equation we have \(x^2-5x-8850\), you can pick out the parameters a,b and c.

    • 2 years ago
  13. richyw Group Title
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    here the \(x^2\) term has nothing in front of it, so \(a=1\), the \(x\) term has a -5 in front of it so \(b=-5\), and \(c=8850\). Does that make sense?

    • 2 years ago
  14. richyw Group Title
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    sorry \(c=-8850\)

    • 2 years ago
  15. richyw Group Title
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    now plug those numbers into the the quadratic equation.

    • 2 years ago
  16. lamthornhill Group Title
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    sorry I'm not responding....... i'm writing and studying this so I can understand it

    • 2 years ago
  17. lamthornhill Group Title
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    I don't even know what the quadratic equation is

    • 2 years ago
  18. richyw Group Title
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    yup MEMORIZE the quadratic formula! also do you understand what the \(\pm\) sign means?

    • 2 years ago
  19. lamthornhill Group Title
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    no....... this is all new to me....... I didn't take algebra in high school

    • 2 years ago
  20. richyw Group Title
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    alright well a quadratic EQUATION is a polynomial equation of the second degree. The initial problem you have is a quadratic equation. The quadratic FORMULA is the formula I gave you. You use the quadratic formula to find the roots of quadratic equations.

    • 2 years ago
  21. richyw Group Title
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    This is the quadratic FORMULA.\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • 2 years ago
  22. lamthornhill Group Title
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    ok

    • 2 years ago
  23. richyw Group Title
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    the \(\pm\)is just used to save space, it means that this is actually two formulas...haha\[\frac{-b+\sqrt{b^2-4ac}}{2a}\]\[\frac{-b-\sqrt{b^2-4ac}}{2a}\]

    • 2 years ago
  24. lamthornhill Group Title
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    I have been on this type of problem now for 2 hours

    • 2 years ago
  25. richyw Group Title
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    but this equation is only useful if your equation is in general form. That was the first step that we did (when we subtracted 8620)

    • 2 years ago
  26. lamthornhill Group Title
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    ok

    • 2 years ago
  27. richyw Group Title
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    right so now just plug the numbers in \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-8550)}}{2(1)}\]

    • 2 years ago
  28. richyw Group Title
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    this gives \[x=\frac{5+\sqrt{25+4(8550)}}{2}\]\[x=\frac{5-\sqrt{25+4(8550)}}{2}\]

    • 2 years ago
  29. richyw Group Title
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    For the context of the question though you can't have negative sales, you you only need to plug in the top one...

    • 2 years ago
  30. richyw Group Title
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    sorry! how could I have forgotten this, if you were confused before, the quadratic equation gives you x... I'll put in an x= there for you :D. Anyways I don't have a calculator, so plug those two above in.\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • 2 years ago
  31. lamthornhill Group Title
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    for the top I came up 34225

    • 2 years ago
  32. richyw Group Title
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    nope that's not correct, you must have messed up on the calculator. Hand on i'll just check wolfram...

    • 2 years ago
  33. lamthornhill Group Title
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    ok... I'm not good with math.....

    • 2 years ago
  34. richyw Group Title
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    well we already finished the math, now it's just using the calculator :). But the two roots of this polynomial are x=-90 and x=95. Obviously the answer to the question is x=95, because you can't manufacture a negative number of products. Just you know you can always verify this on wolfram alpha http://www.wolframalpha.com/input/?i=x%5E2-5x%2B70%3D8620

    • 2 years ago
  35. richyw Group Title
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    It might seem compplicated but really all you did was two steps 1)Put equation in general form 2)Find the roots (in this case we found the roots using the quadratic equation).

    • 2 years ago
  36. lamthornhill Group Title
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    ty

    • 2 years ago
  37. richyw Group Title
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    No worries, and good luck with your math studies! I only learned algebra after high school as well. It just takes a LOT of work! It is sure worth it though.

    • 2 years ago
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