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    (Original post by Zacken)
    The 'd' isn't a variable, in my opinion, so should be textified.

    oh ok

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    (Original post by EricPiphany)
    \displaystyle \int_0^4 \sqrt{4-x^2} \ \mathrm{d}x =\pi
    is it possible to change \sqrt{x^2} into a single term of  x?
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    (Original post by Zacken)
    You might want to use \mathrm{d}x (\mathrm{d}x) instead of dx (dx).
    You should edit everyone's \LaTeX until they learn to use the proper  \mathrm{d}
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    (Original post by notnek)
    You should edit everyone's \LaTeX until they learn to use the proper  \mathrm{d}
    your latex word has downs

    what happened to the midgey a?
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    (Original post by thefatone)
    better not use  tex\ then\ right?
    Uh I think it is better, you just might not wanna type all this


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    (Original post by thefatone)
    is it possible to change \sqrt{x^2} into a single term of  x?
    What? I don't think what I wrote is actually correct. One tip: It suck writing latex on a crap phone
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    (Original post by Student403)
    Uh I think it is better, you just might not wanna type all this


    lol i just put a ton of enters in because i though it'd make a difference
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    Now solve that cubic
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    (Original post by EricPiphany)
    What? I don't think what I wrote is actually correct. One tip: It suck writing latex on a crap phone
    This should be better
    \displaystyle \int_0^2 \sqrt{2^2-x^2} \ \mathrm{d}x=\pi
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    (Original post by 13 1 20 8 42)
    Now solve that cubic
    I have a smart calculator
    -0.3380272249, -0.1643197209 \pm 2.803909762i
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    (Original post by EricPiphany)
    I have a smart calculator
    -0.3380272249, -0.1643197209 \pm 2.803909762i
    I prefer the exact (real) answer, beautiful...

    Name:  0000lol.png
Views: 88
Size:  1.9 KB
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    (Original post by 13 1 20 8 42)
    I prefer the exact (real) answer, beautiful...

    Name:  0000lol.png
Views: 88
Size:  1.9 KB
    We all recognise wolfram
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     PoOp
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    Did not come here for maths....
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    (Original post by EricPiphany)
    We all recognise wolfram
    What's wrong with using Wolfram..
    Though I assume they just use the ugly general cubic formula which anyone can I guess
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    \displaystyle \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} \ \mathrm{d}x=\pi

    \displaystyle \int_{-\infty}^{+\infty} \frac{1}{1+x^2} \ \mathrm{d}x=\pi
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    (Original post by 13 1 20 8 42)
    What's wrong with using Wolfram..
    Though I assume they just use the ugly general cubic formula which anyone can I guess
    nothing wrong
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    (Original post by 13 1 20 8 42)
    I prefer the exact (real) answer, beautiful...

    Name:  0000lol.png
Views: 88
Size:  1.9 KB
    what s that horrendous thing? i'll try and write that when i come back from my driving lesson
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    (Original post by thefatone)
    what s that horrendous thing? i'll try and write that when i come back from my driving lesson
    root of your equation...sounds like good practice
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    (Original post by 13 1 20 8 42)
    root of your equation...sounds like good practice
    oh god oh lord pls help me now xD
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    (Original post by 13 1 20 8 42)
    root of your equation...sounds like good practice
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    x = \dfrac{1}{9} \left( -2 -106 \sqrt[3] {\dfrac{2}{9\times \sqrt{7437} -83} + 2^{2/3} \times \sqrt[3] 9 \times \sqrt {7437} -83 \right)



    i have failed help
 
 
 
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