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Math Help - Still lost part 11

  1. #1
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    Still lost part 11

    Any help would be greatly appreciated.

    Thanks once again.
    Luis
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  2. #2
    dan
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    ok I read the attached...but i could not understand what you want to know...
    what do you need help with???

    or did you just want someone to check your work??

    dan
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  3. #3
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    Dan,

    This is an online class I am taking; (which I would not recommend to anyone, take the class in a traditional setting) check my work, give me the anwers, it doesn't really matter. These are the only nine questions I can't answer. I just want to get them done. Call me lazy, whatever...some things just weren't meant for people to understand and I think algebra is it for me.

    Thanks,
    Luis
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FLO21I99 View Post
    Dan,

    This is an online class I am taking; (which I would not recommend to anyone, take the class in a traditional setting) check my work, give me the anwers, it doesn't really matter. These are the only nine questions I can't answer. I just want to get them done. Call me lazy, whatever...some things just weren't meant for people to understand and I think algebra is it for me.

    Thanks,
    Luis
    I know exactly what you mean, but NEVER admit that when teachers are around.

    -Dan
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  5. #5
    dan
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    Quote Originally Posted by topsquark View Post
    I know exactly what you mean, but NEVER admit that when teachers are around.

    -Dan

    haha... we should start a club....

    dan
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  6. #6
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    Quote Originally Posted by FLO21I99 View Post
    Dan,

    This is an online class I am taking; (which I would not recommend to anyone, take the class in a traditional setting) check my work, give me the anwers, it doesn't really matter. These are the only nine questions I can't answer. I just want to get them done. Call me lazy, whatever...some things just weren't meant for people to understand and I think algebra is it for me.

    Thanks,
    Luis
    In the first question shown below in the attachment, we see that g(x)=a x^3
    grows more quickly as x -> +inf, than does f(x)=x^3.

    So choose lets say x=2, then we see that:

    8 a>8,

    so a>1.

    RonL
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    Last edited by CaptainBlack; September 21st 2006 at 09:57 PM.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by FLO21I99 View Post
    Any help would be greatly appreciated.

    Thanks once again.
    Luis
    Second question (shown in attachment).

    If you multiply out the factors the leading lerm is x^3. This dominates
    when |x| is large. So as x-> -inf f(x) ->-inf, and as x->inf f(x)->inf.

    Also f(x)=0 has three roots: 4, 0, -1. So

    f(x)<0, when x<-1,
    f(x)>0, when -1<x<0,
    f(x)<0, when 0<x<4,
    f(x)>0, when x>4

    So f(x)>0 on (-1,0)U(4,inf)

    RonL
    Attached Thumbnails Attached Thumbnails Still lost part 11-gash.jpg  
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