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anonymous

  • 5 years ago

ok, here's another exponent problem! If you guys don't mind...cube root of s over s^2 = ?

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  1. anonymous
    • 5 years ago
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    \[\sqrt[3]{s}\div s ^{2}\]

  2. mathteacher1729
    • 5 years ago
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    I believed this was asked just a little while ago today...

  3. anonymous
    • 5 years ago
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    why is to the -1/3 and not positive 1/3?

  4. anonymous
    • 5 years ago
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    sorry, my bad!

  5. anonymous
    • 5 years ago
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    \[s ^{1/3} . s ^{-2}\] = s\[s ^{-5/6}\]

  6. amistre64
    • 5 years ago
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    \[{s^{1/3} \over s^2} = {1 \over s^{2-\frac{1}{3}}}\]

  7. amistre64
    • 5 years ago
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    1/s^(5/3)

  8. anonymous
    • 5 years ago
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    thank you, jgeorge! Thanks so much!

  9. anonymous
    • 5 years ago
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    any time :)

  10. anonymous
    • 5 years ago
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    Amistre64, where in the heck have you been? I have calculus problems to solve, you know!!! ; )

  11. amistre64
    • 5 years ago
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    i had to do laundry; and air up my tire; put oil in the car... you know, nonmathical stuff lol

  12. anonymous
    • 5 years ago
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    Well, next time do those things when I DON'T need your help! Seriously, how have you been? Can you help me now with the antiderivative of that exponent you gave me? And wait a minute, why is it to the -5//6 and not -5/3?

  13. amistre64
    • 5 years ago
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    he confused 3*2

  14. amistre64
    • 5 years ago
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    2-(1/3) = (3*2 -1)/3

  15. anonymous
    • 5 years ago
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    so it is s^-5/3, then? right?

  16. amistre64
    • 5 years ago
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    yes

  17. amistre64
    • 5 years ago
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    and you want to derive that?

  18. anonymous
    • 5 years ago
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    Need help finding the antiderivative of this if you don't mind then. Let me try to attach an equation just to look professional, like i know what I'm doing! ; )

  19. amistre64
    • 5 years ago
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    no keeping secrets now lol

  20. anonymous
    • 5 years ago
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    \[1+s ^{-5/3}\]

  21. anonymous
    • 5 years ago
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    Need to find the antid of that! I know it's s-something. but it's the "something" i get stuck on, sort of!

  22. amistre64
    • 5 years ago
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    \[(-5/3)s^{(-5/3 - 3/3)}\]

  23. amistre64
    • 5 years ago
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    \[\frac{-5}{3 s^2 \sqrt[3]{s^2}}\]

  24. amistre64
    • 5 years ago
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    \(*\sqrt[3]{s}\) top and bottom to rationalise the denm

  25. amistre64
    • 5 years ago
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    \[\frac{-5 \sqrt[3]{s}}{3s^3}\]

  26. amistre64
    • 5 years ago
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    does that help?

  27. anonymous
    • 5 years ago
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    Well I am trying to figure out the first step where had -5 over all that gobbledy-gook. I don't get that part at all!

  28. amistre64
    • 5 years ago
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    its the same steps as derive \(x^4\)

  29. amistre64
    • 5 years ago
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    \(\ 'exp' * x^{(exp-1)}\)

  30. anonymous
    • 5 years ago
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    Where did the s^2 come fronm in the denominator? I see about the cube root, but not that.

  31. amistre64
    • 5 years ago
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    lets take this one step at a time.... \[s^{(a)} \iff a*s^{a-1}\] \[\frac{-5}{3}s^{(\frac{-5}{3}-\frac{3}{3})} \iff \frac{-5 s^{-8/3}}{3}\]

  32. amistre64
    • 5 years ago
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    \[\frac{-5}{3s^{8/3}} \iff \frac{-5}{3s^{6/3}s^{2/3}}\]

  33. amistre64
    • 5 years ago
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    \[\frac{-5}{3s^2 s^{2/3}}*\frac{s^{1/3}}{s^{1/3}}=\frac{-5s^(1/3)}{3s^2 s}\]

  34. anonymous
    • 5 years ago
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    Ok, I see where you are getting those numbers, but why do you have to do the 6/3 2/3 thing in the denominator in the first place? Why do you have to break it up like that? You can't just leave it? Why 2 "s" 's?

  35. amistre64
    • 5 years ago
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    \[\frac{-5 \sqrt[3]{s}}{3s^3}\]

  36. amistre64
    • 5 years ago
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    for the same reason why we dont leave \(\sqrt{36}\) as it is :)

  37. anonymous
    • 5 years ago
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    ok, but why the s^3? I though it was s^2 in the denominator! aaarrrgh!

  38. amistre64
    • 5 years ago
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    \[\sqrt[3]{s.s.s.s.s.s.s.s} \iff s^2 \sqrt[3]{s^2}\]

  39. amistre64
    • 5 years ago
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    \[\sqrt[3]{s^3 s^3 s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2}\]

  40. anonymous
    • 5 years ago
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    ok, go back up if you would to where I said,"ok, I see where you get those numbers, but why do you have to do the 6/3 2/3 thing in the denominator." See that post of mine? the one right before it is yours. THATS what I don't understand!

  41. anonymous
    • 5 years ago
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    The s^1/3 part.

  42. amistre64
    • 5 years ago
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    and the posts i just did explain that :)

  43. anonymous
    • 5 years ago
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    ok, then I am just an idiot, because I don't get it at all. What to do?

  44. amistre64
    • 5 years ago
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    \(\frac{-5}{3s^{8/3}}\) is what we get in the denominator right?

  45. amistre64
    • 5 years ago
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    your question is about the s^(8/3) and how we play with it

  46. anonymous
    • 5 years ago
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    Is that 3s^8/3? It's quite small! But yes that is my question. How to play with it. But i have other things I would rather play with. Exponents AINT one of em!

  47. amistre64
    • 5 years ago
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    Does \(s^{8/3}\) equal \(\sqrt[3]{s^8}\)?

  48. anonymous
    • 5 years ago
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    yes, i would have to say it does. ok, go on... please!

  49. amistre64
    • 5 years ago
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    Does \(s^8\) mean: \(s.s.s.s.s.s.s.s\) ?

  50. anonymous
    • 5 years ago
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    yes.

  51. amistre64
    • 5 years ago
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    Does \(s.s.s.s.s.s.s.s\) equal \((s.s.s).(s.s.s).(s.s)\) ? Does \((s.s.s).(s.s.s).(s.s) \iff s^3 s^3 s^2\) ?

  52. anonymous
    • 5 years ago
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    yes, but is there a reason you split it up like that and not like (s.s.s.s).(s.s).(s.s)?

  53. amistre64
    • 5 years ago
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    yes theres; our radical is a \(\sqrt[3]{...}\) which means we want to group these into \(s^3\) .

  54. anonymous
    • 5 years ago
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    OH!!!!

  55. amistre64
    • 5 years ago
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    and heres why: \(\sqrt[3]{s^8} \iff \sqrt[3]{s^3s^3s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\)

  56. amistre64
    • 5 years ago
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    \[\sqrt[3]{s^3} = s\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2 }\]

  57. anonymous
    • 5 years ago
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    OMGosh! Ok, so back to the problem...I don't remember where we were even!

  58. amistre64
    • 5 years ago
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    these are the basic steps that you tend to do in your head to avoid all the work of reproving them time and again lol

  59. anonymous
    • 5 years ago
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    Maybe in YOUR head, but not MINE!

  60. amistre64
    • 5 years ago
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    yes, it does help to actually step thru them to gain insight ;)

  61. amistre64
    • 5 years ago
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    ok so we are left with: \[\frac{-5}{3s^2\sqrt[3]{s^2}}\] and thats fine but, we maybe want to get rid of the radical in the bottom part

  62. amistre64
    • 5 years ago
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    we 'know' \(\sqrt[3]{s^3}=s\) so we want to multily top and bttom by \(\sqrt[3]{s}\) to get rid of it in the bottom right?

  63. anonymous
    • 5 years ago
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    right.

  64. amistre64
    • 5 years ago
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    \[\frac{-5 \sqrt[3]{s}}{3s^3}\]

  65. anonymous
    • 5 years ago
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    wait, if you multiply the numerator by that radical and that radical =s, why do you still have the radical sign there?

  66. amistre64
    • 5 years ago
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    \[-5 * \sqrt[3]{s} = -5\sqrt[3]{s}\]

  67. anonymous
    • 5 years ago
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    yeah, but it's the cube root of s^3. That just equals s I thought?

  68. amistre64
    • 5 years ago
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    \[3s^2\sqrt[3]{s^2}*\sqrt[3]{s}\iff3s^2\sqrt[3]{s^2.s}\iff3s^2\sqrt[3]{s^3}\iff 3s^2.s\] \[\implies 3s^3 \]

  69. anonymous
    • 5 years ago
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    Are they all this hard? How in the heck do they expect newcomers here to calculus to know all this crap? I really don't understand! It's so hard!!!!! : (

  70. amistre64
    • 5 years ago
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    they expect you to have learned this in college algebra; since thats all it is

  71. amistre64
    • 5 years ago
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    if they have to reteach you what you should have already learned... then it gets rather ...... misses the point of dividing mathinto teachable sections lol

  72. anonymous
    • 5 years ago
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    BTW your answers are always the best and easiest to work with! As are you! Thank you so much! I am going to go and try towrok this problem...will you be around? Oh, and the college algebra thing? I got a LOW C! I couldn't do it then; I aced calc in college 20 years ago, and now don't know a thing, but could help my daughter with high school algebra II like a pro (sort of).

  73. amistre64
    • 5 years ago
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    :) youre doing fine :)

  74. anonymous
    • 5 years ago
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    I get the point! Do you know of a good website with the exponent rules on it?

  75. anonymous
    • 5 years ago
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    Actually, somehow in calc I'm getting a 97%!

  76. amistre64
    • 5 years ago
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    my sirttes are all books; but i know a good practie site for math: interactmath.com

  77. amistre64
    • 5 years ago
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    sirttes eh.... my uiniversal translator is on the fritz i think lol

  78. anonymous
    • 5 years ago
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    Yeah, what is a sirttes anyways?

  79. amistre64
    • 5 years ago
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    it my brain cells jockeying for position lol

  80. anonymous
    • 5 years ago
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    Will you be around for some of today? Can I visit you here again if I need to?

  81. amistre64
    • 5 years ago
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    ill be here as long as the weather holds up; im under the veranda since the libraries are closed

  82. anonymous
    • 5 years ago
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    Ok, then again...thanks from the bottom of my heart for all your wonderful help! You're the best! Giving me all that time and patience...you're great, thanks

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