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fruitbat Group Title

asdfghjkl

  • one year ago
  • one year ago

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  1. binarymimic Group Title
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    What equations have you set up to model this problem?

    • one year ago
  2. Outkast3r09 Group Title
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    hola

    • one year ago
  3. Outkast3r09 Group Title
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    I'd first start by deciphering everything you have

    • one year ago
  4. binarymimic Group Title
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    If you are unsure where to begin, review exponential functions (http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/ExpDecayL.htm) You should obtain \[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]

    • one year ago
  5. Outkast3r09 Group Title
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    first lets make 2010 your P-not or \[P_0=P(0)\]

    • one year ago
  6. Outkast3r09 Group Title
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    so at \[P(0)=50,000\]

    • one year ago
  7. Outkast3r09 Group Title
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    for checkerville yes

    • one year ago
  8. Outkast3r09 Group Title
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    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 \[P=70,000e^{-.04t}\]

    • one year ago
  9. Outkast3r09 Group Title
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    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

    • one year ago
  10. satellite73 Group Title
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    i am going to make a guess that you are supposed to use \[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]

    • one year ago
  11. satellite73 Group Title
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    that is \[50000(1.07)^x=70000(.96)^x\]

    • one year ago
  12. Outkast3r09 Group Title
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    essentially the same thing lol

    • one year ago
  13. satellite73 Group Title
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    divide by 50000 get \[ 91.07)^x=1.4(.96)^x\]divide by\[ (.96)^x\] get \[\left(\frac{1.07}{.96}\right)^x=1.4\]

    • one year ago
  14. satellite73 Group Title
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    @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

    • one year ago
  15. satellite73 Group Title
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    at any rate, solve via \[x=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}\]

    • one year ago
  16. satellite73 Group Title
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    say this exact same problem yesterday. they really reach for these word problems, don't they?

    • one year ago
  17. binarymimic Group Title
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    satellite, your equation is correct and your final solution is also correct

    • one year ago
  18. Outkast3r09 Group Title
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    nah it's because i meant to put the rate as 1.07 and .96 so you'd get the same thing

    • one year ago
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