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Thread: The program says Im wrong but Im not so sure.... can someone verify?

  1. #1
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    The program says Im wrong but Im not so sure.... can someone verify?

    I just took an online quiz but I think some of the answers may be correct. can someone verify my answers truly are incorrect. Thanks you very very much!!

    The program says Im wrong but Im not so sure.... can someone verify?-wrong.jpg
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  2. #2
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    Re: The program says Im wrong but Im not so sure.... can someone verify?

    i) d is correct

    ii) d is correct

    iii) e is correct

    $y(x) = \tan\left( \dfrac{3x^2}{2} + c \right)$
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    Re: The program says Im wrong but Im not so sure.... can someone verify?

    Thank you very much! So I am just not understanding what I'm doing... Unfortunately I am in Houston and all the schools are closed down because of Harvey or I would go get tutor help. can you provide me steps on how you solved these? I would be very appreciative. Thanks for the support!
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    Re: The program says Im wrong but Im not so sure.... can someone verify?

    Quote Originally Posted by thatsmessedup View Post
    Thank you very much! So I am just not understanding what I'm doing... Unfortunately I am in Houston and all the schools are closed down because of Harvey or I would go get tutor help. can you provide me steps on how you solved these? I would be very appreciative. Thanks for the support!
    it's pretty much just plug and chug

    $y = e^{-4x}$

    $y' = -4 e^{-4x}$

    $y''=16 e^{-4x}$

    and just plug these into the possible answers listed. The $e^{-4x}$ will divide out and the algebra will either return 0 or not.

    In problem 2

    $y = C_1 e^{4x} + C_2 x e^{4x}$

    $y' = e^{4 x} \left(4 C_2 x+4 C_1+C_2\right)$

    $y'' = e^{4 x} \left(16 C_2 x+16 C_1+8 C_2\right)$

    and again plug all this in, divide out the $e^{4x}$ term, and the algebra will either be 0 or not.

    The program says Im wrong but Im not so sure.... can someone verify?-clipboard01.jpg

    In problem 3

    $y' = 3x(1+y^2)$

    $\dfrac{dy}{1+y^2} = 3x~dx$

    $\tan^{-1}(y) = \dfrac 3 2 x^2 + C$

    $y = \tan\left(\dfrac 3 2 x^2 + C\right)$
    Last edited by romsek; Sep 2nd 2017 at 03:54 PM.
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