Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

fruitbat

  • 3 years ago

asdfghjkl

  • This Question is Closed
  1. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What equations have you set up to model this problem?

  2. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hola

  3. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'd first start by deciphering everything you have

  4. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    first lets make 2010 your P-not or \[P_0=P(0)\]

  6. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so at \[P(0)=50,000\]

  7. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for checkerville yes

  8. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \[P=70,000e^{-.04t}\]

  9. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  10. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i am going to make a guess that you are supposed to use \[50000(1 + 0.07)^{x} = 70000(1 - 0.04)^{x}\]

  11. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is \[50000(1.07)^x=70000(.96)^x\]

  12. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    essentially the same thing lol

  13. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    divide by 50000 get \[ 91.07)^x=1.4(.96)^x\]divide by\[ (.96)^x\] get \[\left(\frac{1.07}{.96}\right)^x=1.4\]

  14. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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

  15. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    at any rate, solve via \[x=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}\]

  16. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    say this exact same problem yesterday. they really reach for these word problems, don't they?

  17. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    satellite, your equation is correct and your final solution is also correct

  18. Outkast3r09
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nah it's because i meant to put the rate as 1.07 and .96 so you'd get the same thing

  19. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy