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The formula P=0.64x^2-0.049+2 models the approx population P, in thousands, for a species of fish in a local pond, x yrs after 1997. During what yr will the population reach 42,568??? 2004,2005,2006, or 2007

Mathematics
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put P=42,568 and solve for x, and then add x to 1997 to find the answer
So solve for P first than add x to 42,568?
noo, the population is equalto 42,568 in a certain year, that's what we want to find... that year. to get that year, we know that x is the number of years after 1997 for which the population will be equal to 42,568... so we find x and add it to 1997 and voila, we know the year. :) is that making any sense so far?

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so 42,568=0.64x^2-0.049x+2 is the equation right? I think thats what I understood by what you just wrote
exactly... then we find x
I got x as 42,29,and -29
no that is not the right equation i don't think, because it gives the population in thousands
Well Mr. Satellite73 help plz [=
try this one \[ 42.568 = 0.64 x^2-0.049+2\]
http://www.wolframalpha.com/input/?i=42.568%3D0.64x%5E2-0.049%2B2
looks like the answer is closest to x = 8 so 8 years after 1997
aka 2005
Thank you [=
I think she missed the x after the 0.049 in her original question cos i noticed she wrote that in her reply.... so that makes it a quadratic, but i could be wrong
Yeah I missed the x after -0.049x
and btw why did u write 42.568 <<< and not 42,568 ?
because it says it gives the answer in thousands so if you get P = 4 that means 4,000
the answer appears to be 8 i redid it http://www.wolframalpha.com/input/?i=42.568%3D0.64x%5E2-0.049x%2B2
yeah it is, i solved it and it's approximately 8 so yeah :)
wolfram says 8 when i redid it
Gracias you guys I loved the help
anytime :)
yw

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