algerba

How to figure out the temperature of a lake using x for the feet or the Depth and for the degrees Celsius use (3900x+1000)/(3x^2+100) with the answer of the degrees celsius being 8.44 looking like

8.44=(3900x+1000)/(3x^2+100)
i need the depth

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  1. Cross multiply to get:
    8.44(3x²+100)=3900x+1000
    Rearrange:
    25.32x²-3900x-156=0
    Solve for x using quadratic formula:
    x=-0.03999 or x=154.07 approx.

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  2. thank you ill try that

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  3. it worked i have been working for two days on that problem thank you so much

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  4. You're welcome!

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  5. I have another one instead of using the 8.44 I have to do the same question but now I have to put it with 6.74 like this

    6.74=(3900x+1000)/(3x^2+100)

    This very fustratimg I have been in school in so long after all I am 60yr old

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  6. I have another one instead of using the 8.44 I have to do the same question but now I have to put it with 6.74 like this

    6.74=(3900x+1000)/(3x^2+100)

    i have figure out that
    a=-20.22
    b= 3900
    c= 326

    On the quadritic formula table

    -3900+Square root ()/-40.44

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  7. Answered to your other post.

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