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anonymous

  • one year ago

The value of a particular item can be modeled by P(t) = P0(a)t where P is in dollars and t is the number of years since the item was purchased. Suppose the value of the item increases 5% each year and the item was purchased for $20. (a) Write a formula for P(t) according to the model. (b) How fast is the value of the item increasing when t = 5 years? Round your answer to two decimal places

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  1. Jhannybean
    • one year ago
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    a. you're given the initial price, \(P_0\) = $20 a = 0.05 Just plug in the values into your formula

  2. Jhannybean
    • one year ago
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    b. t = 5 years, a = 0.05, \(P_0\)= 20

  3. anonymous
    • one year ago
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    for a. i put P(t) = 20(.05)^t but i got it wrong

  4. Venomblast
    • one year ago
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    You are multiplying t

  5. anonymous
    • one year ago
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    sorry, its actually raised to the t power

  6. Venomblast
    • one year ago
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    So you would get, 20 * .05 * 5. To make this easier do 20*5 then multiply by .05 Hint: .05 = \[\frac{ 1 }{20 }\]

  7. Venomblast
    • one year ago
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    Oh. Then that changes the function completely!

  8. Venomblast
    • one year ago
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    So I assume your equation would be \[20 \times(.05)^{t}\]

  9. anonymous
    • one year ago
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    yeah, i typed that in but i got it wrong

  10. Venomblast
    • one year ago
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    This formula is similar to the compound formula. Put in your answer and tell me if it is correct?

  11. anonymous
    • one year ago
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    sorry, i am confused, for what letter a or b?

  12. Venomblast
    • one year ago
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    I gave you (a) already. Now plug in 5 for t

  13. Jhannybean
    • one year ago
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    Then you are multiplying by t and it is not raised to a power, perhaps?

  14. Jhannybean
    • one year ago
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    Fo a.

  15. anonymous
    • one year ago
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    wait sorry one second

  16. anonymous
    • one year ago
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    I'm trying to get a screenshot of it

  17. Venomblast
    • one year ago
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    substitute a with the appropriate variables.

  18. Venomblast
    • one year ago
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    Try this\[20(.05)^{t}\]

  19. anonymous
    • one year ago
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  20. anonymous
    • one year ago
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    yeah it gave that as wrong, not sure why

  21. Venomblast
    • one year ago
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    Try adding the times in. Could be a error. Maybe it s compounding formula. Try this \[20\times(1.05)^{t}\]

  22. anonymous
    • one year ago
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    ok that was right for the equation, but for the part b i plug in chug but it gives the wrong answer?

  23. anonymous
    • one year ago
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    i get 20(1.05)^5 = 25.52 rounding to two decimal places, says it is wrong tho

  24. Venomblast
    • one year ago
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    25.53

  25. anonymous
    • one year ago
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    Calculator gives me 25.52563 and i tried 25.52 and 25.53

  26. Venomblast
    • one year ago
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    when it says round to the nearest cent, It really telling you to round to the nears hundredth. Thus you look at the hundredth place which is a 2. look next to it to the right. if the number is 5 or above, you round up. The 2 becomes a 3.

  27. anonymous
    • one year ago
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    yea i tried 25.53 and 25.52 and both are wrong

  28. Venomblast
    • one year ago
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    That is very strange.

  29. anonymous
    • one year ago
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  30. anonymous
    • one year ago
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    would it maybe have to do with it asking for dollar/year?

  31. Venomblast
    • one year ago
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    dollar per year? That a strange way of saying it.

  32. anonymous
    • one year ago
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    wait wouldn't it be the derivative????

  33. Venomblast
    • one year ago
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    Why would it be the derivative?

  34. anonymous
    • one year ago
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    yeah nevermind, i don't know :(

  35. Venomblast
    • one year ago
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    If it's a physic problem, then yes I would take the derivative.

  36. Venomblast
    • one year ago
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    try 5.53

  37. Jhannybean
    • one year ago
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    Actually,you would have to take the derivative.

  38. anonymous
    • one year ago
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    yeah, my time limit, expired :( oh well got a 80% on it :)

  39. Jhannybean
    • one year ago
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    Do you still want to know how to figure it out? Or nah.

  40. anonymous
    • one year ago
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    yes please

  41. Venomblast
    • one year ago
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    Sorry man. It was confusing. It seems like an interest problem but then a physic problem?

  42. Venomblast
    • one year ago
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    If it is a physic problem, take the derivative (chain rule).

  43. anonymous
    • one year ago
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    no your fine man, i appreciate the help

  44. Jhannybean
    • one year ago
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    \[P(t) = 20\cdot 1.05^t\]Just to be clear... can you explain to me how you got \(a=1.05\) ? I know how this works, but I want you to explain it t me.

  45. Venomblast
    • one year ago
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    who me?

  46. anonymous
    • one year ago
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    to be honest i don't understand the 1.05 either

  47. Jhannybean
    • one year ago
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    @benitob

  48. Jhannybean
    • one year ago
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    Okay.

  49. Venomblast
    • one year ago
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    I can tell you why. When money is growing over certain amount of year (or days or month). You use that formula. P(1+i)^t. Where i is the interest rate, P is your initial amount and t is the amount of years/month/days

  50. Jhannybean
    • one year ago
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    So we have out function \(P(t) = P_0 \cdot a^t\). Were given that \(P_0 = 20\) an the growth factor = \(5\%\). Therefore for every year it is increasing... \(P(1) = \$20 + \left(\frac{0.05}{100}\right)\cdot \$20\)

  51. Venomblast
    • one year ago
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    Yes correct. You would get the constant rate of interest i if you are headed for the right direction

  52. Jhannybean
    • one year ago
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    ahh... im off on my decimal place.

  53. Venomblast
    • one year ago
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    yea that would be .0005 lol

  54. Jhannybean
    • one year ago
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    uld have said \[P(1) = \$20 + \left(\frac{5}{100}\right)\cdot \$20\]

  55. Venomblast
    • one year ago
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    I can prove it if you want me to. How i derive the formula.

  56. anonymous
    • one year ago
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    i understand the formula now, but what i don't understand is the part for b?

  57. Jhannybean
    • one year ago
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    That gives us, \(P(1) = \$20 + \$1 = \$21\) And now we want to find the growth factor, \(a\) to find the generalization of the growth function of \(t\) years.

  58. Jhannybean
    • one year ago
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    So we have \(P(1) = \$21,\qquad ~P_0 = \$20\qquad ~,~\qquad t=1\) \[P(t) = P_0 \cdot a^t\]\[P(1) = 20 \cdot a^1\]\[\$21= \$20 \cdot a^1\]\[a= \frac{21}{20} = 1.05\]\[\implies P(t) = \$20 \cdot 1.05^t\]

  59. Jhannybean
    • one year ago
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    Now moving on to part b.

  60. Jhannybean
    • one year ago
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    We want to find `how fast` the value is increasing with respect to time, therefore we're finding the rate of change of the growth function, \(\dfrac{dP}{dt}\)

  61. Jhannybean
    • one year ago
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    \[P(t)= 20 \cdot 1.05^t\]\[\frac{d}{dt} (P(t) = 20 \cdot 1.05^t) \]\[P'(t) = 20\frac{d}{dx}( 1.05^t)\qquad \qquad \frac{d}{dx}(a^x) = a^x \cdot \ln(a)\]\[P'(t) = 20[1.05^t \cdot \ln(1.05)]\]\[\color{red}{P'(t) \approx 0.98 \cdot 1.05^t}\]

  62. Jhannybean
    • one year ago
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    Sorry, I'm extremely lagged out, took a while to type up.

  63. anonymous
    • one year ago
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    no your fine, i see the answer now, i should've taken the derivative :/

  64. Jhannybean
    • one year ago
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    yeah, when a question asks you `how fast` something is changing within a given time, it's asking you basically for a certain value at a point. (your slope) :)

  65. Jhannybean
    • one year ago
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    I was working it out on a separate piece of paper while you guys were discussing the question earlier and i had to reread the question a few times to notice it was a rate of change question.

  66. anonymous
    • one year ago
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    thank you so much for explaining the question to me, i understand it well now, thank you!!!

  67. Jhannybean
    • one year ago
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    No problem :)

  68. Jhannybean
    • one year ago
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    Does it make sense to you too, @Venomblast ?

  69. Venomblast
    • one year ago
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    Haha yea. Sorry I went to sleep

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