anonymous
  • anonymous
asdfghjkl
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
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anonymous
  • anonymous
What equations have you set up to model this problem?
anonymous
  • anonymous
hola
anonymous
  • anonymous
I'd first start by deciphering everything you have

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anonymous
  • anonymous
If you are unsure where to begin, review exponential functions (http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/ExpDecayL.htm) You should obtain \[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]
anonymous
  • anonymous
first lets make 2010 your P-not or \[P_0=P(0)\]
anonymous
  • anonymous
so at \[P(0)=50,000\]
anonymous
  • anonymous
for checkerville yes
anonymous
  • anonymous
lets rephrase what your equation is \[P=ce^{kt}\] where p is your population , c is a constant unknown atm and k is your rate. t is time. if we put in P(0) you get \[50,000=ce^{0}=c\] \[P=50,000e^{kt}\] since we know the rate at which it increases 7% \[P=50,000e^{.07t}\] next doing for checker ville you get \[P=70,000e^{-.04t}\]
anonymous
  • anonymous
At P1=P2 is when one will exceed the other and the other will be below soooo \[70,000e^{-.04t}=50,000e^{.07t}\] solve for t
anonymous
  • anonymous
i am going to make a guess that you are supposed to use \[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]
anonymous
  • anonymous
that is \[50000(1.07)^x=70000(.96)^x\]
anonymous
  • anonymous
essentially the same thing lol
anonymous
  • anonymous
divide by 50000 get \[ 91.07)^x=1.4(.96)^x\]divide by\[ (.96)^x\] get \[\left(\frac{1.07}{.96}\right)^x=1.4\]
anonymous
  • anonymous
@Outkast3r09 yeah it is the same except that i am thinking that because it say 7% per year it is using \((1.07)^x\) rather than \(e^{.07x}\) but i could be wrong
anonymous
  • anonymous
at any rate, solve via \[x=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}\]
anonymous
  • anonymous
say this exact same problem yesterday. they really reach for these word problems, don't they?
anonymous
  • anonymous
satellite, your equation is correct and your final solution is also correct
anonymous
  • anonymous
nah it's because i meant to put the rate as 1.07 and .96 so you'd get the same thing

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