# 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

-3900+Square root ()/-40.44

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