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Thread: Please help with these three questions they the only three I couldn't do from the ex

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    Please help with these three questions they the only three I couldn't do from the ex


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    Re: Please help with these three questions they the only three I couldn't do from the

    Quote Originally Posted by kpad5991 View Post
    You have already tried finding the indefinite integral for all three of them?

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    Re: Please help with these three questions they the only three I couldn't do from the

    Try differentiating the denominator of the first one and tell me whst you get

    For the second one think of what you can do to a fraction when the numerator is made up of two different variables that are being added together

    For the third one how can you turn the fraction into a reciprocal
    Last edited by ichila101; 14 Jan 2018 at 6:56 PM.
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    Re: Please help with these three questions they the only three I couldn't do from the

    Is it y=ln (x^2+5x+4) + C

    And then you substitute (1,1) to find C. But the solution is y=log (x^2+5x+4/10) +1

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    Re: Please help with these three questions they the only three I couldn't do from the

    I figured out 9d
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    Re: Please help with these three questions they the only three I couldn't do from the

    Quote Originally Posted by kpad5991 View Post
    Is it y=ln (x^2+5x+4) + C

    And then you substitute (1,1) to find C. But the solution is y=log (x^2+5x+4/10) +1
    Yes thats right and what did you get when you substituted (1,1)

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    Re: Please help with these three questions they the only three I couldn't do from the

    C=0

    So y = log (x^2 +5x+4)

    But the solution is as mentioned above y= log (x^2+5x+4/10) +1

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    Re: Please help with these three questions they the only three I couldn't do from the

    Quote Originally Posted by kpad5991 View Post
    C=0

    So y = log (x^2 +5x+4)

    But the solution is as mentioned above y= log (x^2+5x+4/10) +1
    You subbed the x=1 in log((1)^2 + 5(1) + 4)?
    C shouldnt be equating to 0
    Last edited by ichila101; 14 Jan 2018 at 7:12 PM.
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    Re: Please help with these three questions they the only three I couldn't do from the

    AAah okay I get what I did wrong thanks now I need 9e

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    Re: Please help with these three questions they the only three I couldn't do from the

    Quote Originally Posted by kpad5991 View Post
    AAah okay I get what I did wrong thanks now I need 9e
    Did you try turning it into reciprocal and then integrating it?

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    Re: Please help with these three questions they the only three I couldn't do from the

    Okay I got 9e finally but how is y(e) = e+1!??

    I got y= 2+x -log x

    So y(e) = 2+e-log e
    Last edited by kpad5991; 14 Jan 2018 at 7:41 PM.

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    Re: Please help with these three questions they the only three I couldn't do from the

    Quote Originally Posted by kpad5991 View Post
    Okay I got 9e finally but how is y(e) = e+1!??

    I got y= 2+x -log x

    So y(e) = 2+e-log e
    Evaluate (keeping in mind that logarithms are implicitly in base ).
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    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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    Re: Please help with these three questions they the only three I couldn't do from the

    Quote Originally Posted by kpad5991 View Post
    Okay I got 9e finally but how is y(e) = e+1!??

    I got y= 2+x -log x

    So y(e) = 2+e-log e
    What does loge (e) equate to

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