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anonymous

  • 5 years ago

can someone help me with my problem?

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  1. anonymous
    • 5 years ago
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    and what is your problem jmfranco?

  2. anonymous
    • 5 years ago
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    2) We often use radio-isotopes to estimate the age of things (such as the Earth). Assuming the half-life of some isotope is 1000 years, and you found a rock of unknown age near the Mid Atlantic Ridge. You measured the amount of isotope in the rock and determined that 12.5% of the original amount when the rock formed was still present. How old is the rock

  3. anonymous
    • 5 years ago
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    Let's have a chat here franco. This is a growth and decay problem. Is that the topic you're studying?

  4. anonymous
    • 5 years ago
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    ok..yes it is

  5. anonymous
    • 5 years ago
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    great. now, sometimes growth and decay have been modeled by exponential functions. This is the model common model and the one you'll be using here. Do you know the form of the exponential function that is used to model growth and decay? is so, type it in here...

  6. anonymous
    • 5 years ago
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    I hope you don't mind that I'm not just giving you the answer, but you'll be better able to solve all the other problems you'll have this way. it will only take a few minutes.

  7. anonymous
    • 5 years ago
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    okay i don't really know exponential. i understand

  8. anonymous
    • 5 years ago
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    r us still there?

  9. anonymous
    • 5 years ago
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    yes, i'm here

  10. anonymous
    • 5 years ago
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    my connection just failed me for a bit, but i'm back

  11. anonymous
    • 5 years ago
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    this website is having a lot of traffic and seems slow, check out this site has a good explanation: http://serc.carleton.edu/quantskills/methods/quantlit/expGandD.html

  12. anonymous
    • 5 years ago
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    take a look and let me know if you have any questions

  13. anonymous
    • 5 years ago
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    please let me know once you have briefly looked at the page...

  14. anonymous
    • 5 years ago
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    then we can choose between those two equations and use the information you were given to solve the problem. it's not hard at all and you'll see.

  15. anonymous
    • 5 years ago
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    ok looked at it its sort of complicated can u go step by step

  16. anonymous
    • 5 years ago
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    sure

  17. anonymous
    • 5 years ago
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    so the two functions are n(t) = n0*e^(kt)

  18. anonymous
    • 5 years ago
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    now, the exponent of e is that decides whether we have a growth or decay. If it is negative, then the graph starts high and comes down (decay) and the opposite is true if the exponent is positive.

  19. anonymous
    • 5 years ago
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    with me so far?

  20. anonymous
    • 5 years ago
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    yes

  21. anonymous
    • 5 years ago
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    by the way, what grade you in? so I know what you sort of should know...

  22. anonymous
    • 5 years ago
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    college lol

  23. anonymous
    • 5 years ago
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    hehe, no worries. I just graduated a year ago myself. any how...i hope you're not rushed cuz I'm definitely not moving very quickly...let me know if you're rushed.

  24. anonymous
    • 5 years ago
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    so let's go over the equation in more detail...

  25. anonymous
    • 5 years ago
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    the site gives good description for each variable involved.

  26. anonymous
    • 5 years ago
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    lol cool. well i got this question and another, these questions r complicated and due at 4

  27. anonymous
    • 5 years ago
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    today sorry don't rush its okay.

  28. anonymous
    • 5 years ago
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    I'm gonna ask you to rephrase them for me here: what does N(t) mean? and the N_0(t) (that is the subscript zero), the k? the t?

  29. anonymous
    • 5 years ago
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    once you have those, we can replace the variables with the information in your problem. I just wanna make sure you're with me...

  30. anonymous
    • 5 years ago
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    Nt is quatity in time, t is time,n0 is initial quantity, k is a constant and eX is exponential function

  31. anonymous
    • 5 years ago
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    good

  32. anonymous
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    so, it's N(t) and it is a function of time. It's very important that you understand that. and eX is e^(x) which is the mathematically constant e (approx 2.71..) to power x.

  33. anonymous
    • 5 years ago
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    okay

  34. anonymous
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    so what the formulate says is N(t) [the amount as a function of time] = n0*e^(k*t) [the original amount times the exponential function to power k times t]

  35. anonymous
    • 5 years ago
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    k here is the rate of change. the larger is it, the faster the amount grows or, in this case, decays.

  36. anonymous
    • 5 years ago
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    What is the question asking of us? what do you think will be the variable that will be the final answer?

  37. anonymous
    • 5 years ago
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    stay with me franco, we're almost there

  38. anonymous
    • 5 years ago
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    not sure ive never done a prblem like this one

  39. anonymous
    • 5 years ago
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    the question asks you how old is the rock when only 12.5% of it is left..

  40. anonymous
    • 5 years ago
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    so we're looking for time, which will be contained in our variable t.

  41. anonymous
    • 5 years ago
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    a clue that is give to you without knowing it is the term "half-life"

  42. anonymous
    • 5 years ago
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    how long is the half-life of this rock?

  43. anonymous
    • 5 years ago
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    Franco?

  44. anonymous
    • 5 years ago
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    1000

  45. anonymous
    • 5 years ago
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    thank you.

  46. anonymous
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    so that means that in a 1000 years, we'll only have 50% of the original material left. that IS what half life means.

  47. anonymous
    • 5 years ago
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    okay so what do i do??

  48. anonymous
    • 5 years ago
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    so let's set up the equation

  49. anonymous
    • 5 years ago
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    actually, I'll let you do it: in 1000 years, we will only have 50% of the original material. can you set that up?

  50. anonymous
    • 5 years ago
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    i'll prompt with good questions: first: what is t?

  51. anonymous
    • 5 years ago
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    nooo :( i don't kno how this is complicateddd!

  52. anonymous
    • 5 years ago
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    okay: time in this case is 1000 years. second: notice how you can divide both sides by n0 (the original material) - once we do this, what do we have on the left side of the equation?

  53. anonymous
    • 5 years ago
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    this is only solving the first component of the problem.any way, when we divide both sides by n0, we get the equation: N(t)/n0 = e^(kt)

  54. anonymous
    • 5 years ago
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    are you with me?

  55. anonymous
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    in a thousand years, we know we will only have half left, so what will N(t)/n0 equal? think about this one cause it solves 50% of the problem!

  56. anonymous
    • 5 years ago
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    Franco? are you still with me?

  57. anonymous
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    is n0 1000

  58. anonymous
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    i didvide both sides by that?

  59. anonymous
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    n0 is the original amount. was that given in the probem? 1000 is the number of years...

  60. anonymous
    • 5 years ago
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    before i loose you here, we divide the variable n0 by both sides so that we can remove it from the right side and put it on the left. you understand what this results in?

  61. anonymous
    • 5 years ago
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    I miss half life problems, what is that calculus 1

  62. anonymous
    • 5 years ago
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    no what is the asnwer/

  63. anonymous
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    now the equation looks like this: \[n(t) \div n _{0} = e ^{k*t}\]

  64. anonymous
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    no, there is no calculus involved. this is still pre-calc

  65. anonymous
    • 5 years ago
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    do you understand how we get that equation?

  66. anonymous
    • 5 years ago
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    how do you know so much about math thierryu

  67. anonymous
    • 5 years ago
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    I studied engineering in college, i just graduated last year

  68. anonymous
    • 5 years ago
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    are you here to help us Amanda?

  69. anonymous
    • 5 years ago
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    Franco, do you see how we got that last equation I put in?

  70. anonymous
    • 5 years ago
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    yes

  71. anonymous
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    good

  72. anonymous
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    Now, if it has been a 1000 years and we only have 50% of the original materia.

  73. anonymous
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    so now if t=1000, we're gonna look at n0/N(1000). All we know about this fraction is that N(t) at t=1000 is 50% of n0, the original amount.

  74. anonymous
    • 5 years ago
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    digest that.

  75. anonymous
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    therefore, N(t)/n0 when t=1000 years should give us 1/2. do you agree?

  76. anonymous
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    I meant to say N(t)/n0 all along. sorry for the confusion.

  77. anonymous
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    bradley, you found me here!

  78. anonymous
    • 5 years ago
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    should just got to wolfram mathmatica, and type in halflife

  79. anonymous
    • 5 years ago
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    Franco, are you with me? do you see why N(t)/n0 at the half-time (t=1000) will give us 1/2 ?

  80. anonymous
    • 5 years ago
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    so wats the answer??? yea i see

  81. anonymous
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    i did but i will let you help him first

  82. anonymous
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    alright. so now we have 1/2 = e^(k*t)

  83. anonymous
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    im a girl

  84. anonymous
    • 5 years ago
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    and t= 1000 so we really have .5=e^(1000k)

  85. anonymous
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    franco plug in your time and your values for n and No

  86. anonymous
    • 5 years ago
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    ok

  87. anonymous
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    now we solve for k by first takiing the natural log of both sides: ln(.5) = ln(e^(1000k))

  88. anonymous
    • 5 years ago
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    we do this because we know that the ln(e^(x)) equals x.

  89. anonymous
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    so ln(e^(1000k)) reduces to simply 1000k!

  90. anonymous
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    now you can use your calculator to solve for k from teh equation: \[\ln(.5) = 1000*k\]

  91. anonymous
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    what is k?

  92. anonymous
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    once you find k, we go back to the board, but now we have everything we need expect time which is exactly what we're looking for!

  93. anonymous
    • 5 years ago
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    umm not sure..

  94. anonymous
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    why? because we only have 12.5% left. so the ratio of N(t)/n0 = .125

  95. anonymous
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    you're not sure about k?

  96. anonymous
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    there you'd just use a calculator; get the natural log of .5 and then divide it by 1000

  97. anonymous
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    are you with me now?

  98. anonymous
    • 5 years ago
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    i know this was elongated by if you go over this chat and solve the same or a similar problem, you'll get a hang of it. don't be discouraged.

  99. anonymous
    • 5 years ago
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    is it negative .0003

  100. anonymous
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    great. i didn't check but it looks about right

  101. anonymous
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    so on to solving the problem

  102. anonymous
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    i really havve to go can u just solve for me?

  103. anonymous
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    we only have 12.5% left; which means the ratio of N(t)/n0 = 12.5% = .125.

  104. anonymous
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    so since we now have k

  105. anonymous
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    we end up with \[.125 = e^(-.0003t)\]

  106. anonymous
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    remember how we solved for k? we do the same for t now

  107. anonymous
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    ln(.125) = ln(e^(-.0003*t)) = -.0003*t

  108. anonymous
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    so \[\ln(.125) = -.0003*t\]

  109. anonymous
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    can you solve for t? and that's gonna be the number of years that has passed for us to only have 12.5% of the rock left.

  110. anonymous
    • 5 years ago
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    save this chat and go over it when you get a chance and you should be able to do this types of problems on your own.

  111. anonymous
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    did you find t?

  112. anonymous
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    I, for one, would love to know what it is...

  113. anonymous
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    no

  114. anonymous
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    divde ln(.125) by .0003

  115. anonymous
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    yes, by negative .0003

  116. anonymous
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    so how long will have passed franco?

  117. anonymous
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    6900?

  118. anonymous
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    about that much

  119. anonymous
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    sounds about right. definitely way more than a thousand because it is way past the 50% mark. so that makes sense.

  120. anonymous
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    ok thanks

  121. anonymous
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    you're welcome.

  122. anonymous
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    bradley, did you solve yours?

  123. anonymous
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    nope, i am still lost

  124. anonymous
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    would you care to chat on the other thread

  125. anonymous
    • 5 years ago
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    since you're not rushed and i just got home, give me about half an hour to grab a snack and start dinner and I'll be back. I'm putting on a timer for 30 minutes sharp.

  126. anonymous
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    I may also have to review a little calculus ;)

  127. anonymous
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    haha okay i will be around

  128. anonymous
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    could you post the entire problem for me here again please?

  129. anonymous
    • 5 years ago
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    using lagrange multipliers find the maxium and minimum values of the function f(x,y,z)=xy^2z^3 subject to the contraints x^2+y^2+2z^2=25

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