anonymous
  • anonymous
ok, here's another exponent problem! If you guys don't mind...cube root of s over s^2 = ?
Mathematics
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anonymous
  • anonymous
ok, here's another exponent problem! If you guys don't mind...cube root of s over s^2 = ?
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
\[\sqrt[3]{s}\div s ^{2}\]
mathteacher1729
  • mathteacher1729
I believed this was asked just a little while ago today...
anonymous
  • anonymous
why is to the -1/3 and not positive 1/3?

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anonymous
  • anonymous
sorry, my bad!
anonymous
  • anonymous
\[s ^{1/3} . s ^{-2}\] = s\[s ^{-5/6}\]
amistre64
  • amistre64
\[{s^{1/3} \over s^2} = {1 \over s^{2-\frac{1}{3}}}\]
amistre64
  • amistre64
1/s^(5/3)
anonymous
  • anonymous
thank you, jgeorge! Thanks so much!
anonymous
  • anonymous
any time :)
anonymous
  • anonymous
Amistre64, where in the heck have you been? I have calculus problems to solve, you know!!! ; )
amistre64
  • amistre64
i had to do laundry; and air up my tire; put oil in the car... you know, nonmathical stuff lol
anonymous
  • anonymous
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?
amistre64
  • amistre64
he confused 3*2
amistre64
  • amistre64
2-(1/3) = (3*2 -1)/3
anonymous
  • anonymous
so it is s^-5/3, then? right?
amistre64
  • amistre64
yes
amistre64
  • amistre64
and you want to derive that?
anonymous
  • anonymous
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! ; )
amistre64
  • amistre64
no keeping secrets now lol
anonymous
  • anonymous
\[1+s ^{-5/3}\]
anonymous
  • anonymous
Need to find the antid of that! I know it's s-something. but it's the "something" i get stuck on, sort of!
amistre64
  • amistre64
\[(-5/3)s^{(-5/3 - 3/3)}\]
amistre64
  • amistre64
\[\frac{-5}{3 s^2 \sqrt[3]{s^2}}\]
amistre64
  • amistre64
\(*\sqrt[3]{s}\) top and bottom to rationalise the denm
amistre64
  • amistre64
\[\frac{-5 \sqrt[3]{s}}{3s^3}\]
amistre64
  • amistre64
does that help?
anonymous
  • anonymous
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!
amistre64
  • amistre64
its the same steps as derive \(x^4\)
amistre64
  • amistre64
\(\ 'exp' * x^{(exp-1)}\)
anonymous
  • anonymous
Where did the s^2 come fronm in the denominator? I see about the cube root, but not that.
amistre64
  • amistre64
lets take this one step at a time.... \[s^{(a)} \iff a*s^{a-1}\] \[\frac{-5}{3}s^{(\frac{-5}{3}-\frac{3}{3})} \iff \frac{-5 s^{-8/3}}{3}\]
amistre64
  • amistre64
\[\frac{-5}{3s^{8/3}} \iff \frac{-5}{3s^{6/3}s^{2/3}}\]
amistre64
  • amistre64
\[\frac{-5}{3s^2 s^{2/3}}*\frac{s^{1/3}}{s^{1/3}}=\frac{-5s^(1/3)}{3s^2 s}\]
anonymous
  • anonymous
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?
amistre64
  • amistre64
\[\frac{-5 \sqrt[3]{s}}{3s^3}\]
amistre64
  • amistre64
for the same reason why we dont leave \(\sqrt{36}\) as it is :)
anonymous
  • anonymous
ok, but why the s^3? I though it was s^2 in the denominator! aaarrrgh!
amistre64
  • amistre64
\[\sqrt[3]{s.s.s.s.s.s.s.s} \iff s^2 \sqrt[3]{s^2}\]
amistre64
  • amistre64
\[\sqrt[3]{s^3 s^3 s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2}\]
anonymous
  • anonymous
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!
anonymous
  • anonymous
The s^1/3 part.
amistre64
  • amistre64
and the posts i just did explain that :)
anonymous
  • anonymous
ok, then I am just an idiot, because I don't get it at all. What to do?
amistre64
  • amistre64
\(\frac{-5}{3s^{8/3}}\) is what we get in the denominator right?
amistre64
  • amistre64
your question is about the s^(8/3) and how we play with it
anonymous
  • anonymous
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!
amistre64
  • amistre64
Does \(s^{8/3}\) equal \(\sqrt[3]{s^8}\)?
anonymous
  • anonymous
yes, i would have to say it does. ok, go on... please!
amistre64
  • amistre64
Does \(s^8\) mean: \(s.s.s.s.s.s.s.s\) ?
anonymous
  • anonymous
yes.
amistre64
  • amistre64
Does \(s.s.s.s.s.s.s.s\) equal \((s.s.s).(s.s.s).(s.s)\) ? Does \((s.s.s).(s.s.s).(s.s) \iff s^3 s^3 s^2\) ?
anonymous
  • anonymous
yes, but is there a reason you split it up like that and not like (s.s.s.s).(s.s).(s.s)?
amistre64
  • amistre64
yes theres; our radical is a \(\sqrt[3]{...}\) which means we want to group these into \(s^3\) .
anonymous
  • anonymous
OH!!!!
amistre64
  • amistre64
and heres why: \(\sqrt[3]{s^8} \iff \sqrt[3]{s^3s^3s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\)
amistre64
  • amistre64
\[\sqrt[3]{s^3} = s\] \[s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2 }\]
anonymous
  • anonymous
OMGosh! Ok, so back to the problem...I don't remember where we were even!
amistre64
  • amistre64
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
anonymous
  • anonymous
Maybe in YOUR head, but not MINE!
amistre64
  • amistre64
yes, it does help to actually step thru them to gain insight ;)
amistre64
  • amistre64
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
amistre64
  • amistre64
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?
anonymous
  • anonymous
right.
amistre64
  • amistre64
\[\frac{-5 \sqrt[3]{s}}{3s^3}\]
anonymous
  • anonymous
wait, if you multiply the numerator by that radical and that radical =s, why do you still have the radical sign there?
amistre64
  • amistre64
\[-5 * \sqrt[3]{s} = -5\sqrt[3]{s}\]
anonymous
  • anonymous
yeah, but it's the cube root of s^3. That just equals s I thought?
amistre64
  • amistre64
\[3s^2\sqrt[3]{s^2}*\sqrt[3]{s}\iff3s^2\sqrt[3]{s^2.s}\iff3s^2\sqrt[3]{s^3}\iff 3s^2.s\] \[\implies 3s^3 \]
anonymous
  • anonymous
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!!!!! : (
amistre64
  • amistre64
they expect you to have learned this in college algebra; since thats all it is
amistre64
  • amistre64
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
anonymous
  • anonymous
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).
amistre64
  • amistre64
:) youre doing fine :)
anonymous
  • anonymous
I get the point! Do you know of a good website with the exponent rules on it?
anonymous
  • anonymous
Actually, somehow in calc I'm getting a 97%!
amistre64
  • amistre64
my sirttes are all books; but i know a good practie site for math: interactmath.com
amistre64
  • amistre64
sirttes eh.... my uiniversal translator is on the fritz i think lol
anonymous
  • anonymous
Yeah, what is a sirttes anyways?
amistre64
  • amistre64
it my brain cells jockeying for position lol
anonymous
  • anonymous
Will you be around for some of today? Can I visit you here again if I need to?
amistre64
  • amistre64
ill be here as long as the weather holds up; im under the veranda since the libraries are closed
anonymous
  • anonymous
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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