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Tavrobel Tavrobel is online now
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Originally Bespoken by Jessweeee♪ View Post
and rubah's explanation is wrong (I think), the problem has three roots to it!
It's not. I got the same answer. If the intended answer is positive and the variable has a odd-numbered-exponent, then the answer has to be positive. (-4/3)^3 = -81/27. I'm just going to jump to the conclusion that you are overgeneralizing the need to add multiple answers (in the form of double and triple roots), which does not apply in this situation.

If you have a graphing calculator, you can confirm this by throwing in the function, Second-Trace, and going to the Zero option (set limits as close as possible to 0 for your min, and 10 for your max in ZStandard). It will give you 1.333.

EDIT: In ZStandard, it actually looks like a line. LAWLZ

THOSE WHO EDIT FURTHER: Maybe you should consolidate all of your math troubles into one thread, Britannia-man.

Last edited by Tavrobel; 05-06-2008 at 12:10 AM.

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05-06-2008, 12:05 AM
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Jessweeee♪ Jessweeee♪ is offline
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Okie dokie!


EDIT:

Is it weird that I find math fun now? I am genuinely having fun doing math homework. I wish I had more, it's the only homework I actually do ;_;

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05-06-2008, 12:09 AM
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Wait until you get to integrals. Mad amounts of fun to be had. Rubah can confirm.

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05-06-2008, 12:12 AM
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Or LaPlace transforms. You'll be disgusted that you've been lied to throughout calculus that there's not an easier way to do everything. There is.

{Proviso: I don't actually remember how to do LaPlace transforms. But they make everything easier.}

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05-06-2008, 12:18 AM
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jesse, with cubic functions, you can have either 1 real zero or 3. This is like some basic theorem of algebra or something.

for xn, you can have n, n-2, n-4 etc real zeros. (you can have 0 zeros for an even function but must have at least one real zero for an odd function, that is, n is even or n is odd).

since n=3 in this case, the function is odd and can have 3 or 1 real zeros.

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05-06-2008, 01:12 AM
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Originally Bespoken by Tavrobel View Post
Wait until you get to integrals. Mad amounts of fun to be had. Rubah can confirm.
No. Just no.


Integrals and fun dont mix. I found them quite hard at first. Now I find them easy. But fun..?




Naow.


(with the possible exception of integration by inspection, which makes you look like a genius.)



EDIT: Is it me or was that question piss easy? It just involves a bit of rearrangement and taking the cube root. Or am I missing something/cheating?

EDITGA: Yeah, Im not cheating, and yeah its 4/3. Ahh, I love fractions. Otherwise Id be writing 1.33333333333333333333333333333333333333 3333333333333333333333333333333333333333 3333333333333333333333333333333333333333 3...


(I think you get the point)

Last edited by Vivisteiner; 05-06-2008 at 10:38 PM.

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05-06-2008, 10:31 PM
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Originally Bespoken by Vivisteiner View Post
Originally Bespoken by Tavrobel View Post
Wait until you get to integrals. Mad amounts of fun to be had. Rubah can confirm.
No. Just no.

EDIT: Is it me or was that question piss easy? It just involves a bit of rearrangement and taking the cube root. Or am I missing something/cheating?

EDITGA: Yeah, Im not cheating, and yeah its 4/3. Ahh, I love fractions. Otherwise Id be writing 1.33333333333333333333333333333333333333 3333333333333333333333333333333333333333 3333333333333333333333333333333333333333 3...
Yes. Just yes.

Well, it's not exactly easy for someone who's never seen something like that before. It's like squares, but if you do that, then you're sure to get it wrong. Math causes psychological trauma, you know. People need examples first.

Or you could follow significant figures, and in this problem, it'd be two figures (81 has the least sig figs).

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05-06-2008, 10:58 PM
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oh god oh god oh god no not sigfigs what did we do to deserve this. . . NO

*melts*

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05-06-2008, 11:07 PM
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