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    Trig help!!!!

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    Ahmed
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    Trig help!!!!

    Post by Ahmed on Wed Sep 09, 2009 6:09 pm

    PLEASE HELP! I am going crazy.

    If x+y=1, x^2+y^2=4, x^3+y^3=?

    ^x=exponent

    Lol, first day of Advanced Algebra and Trig and I feel like I'm lost -___-

    Please explain slowly.


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    Fangy
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    Re: Trig help!!!!

    Post by Fangy on Wed Sep 09, 2009 6:15 pm

    Lol, makes no sense to me. If you take the square roots of the second one, it would mean that x+y=2...
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    Re: Trig help!!!!

    Post by Ahmed on Wed Sep 09, 2009 6:18 pm

    This isn't simple Algebra that's why it doesn't sense to you (or me).


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    preid1220
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    Re: Trig help!!!!

    Post by preid1220 on Wed Sep 09, 2009 7:13 pm

    well if it follows a set pattern then the answer may be 6, it could also be 9. seeing as i failed trig it may not be that useful to listen to me tho


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    Re: Trig help!!!!

    Post by Fangy on Wed Sep 09, 2009 7:55 pm

    Is that actually trig? I think that's just algebra... I don't see anything related to trig in it (I myself am taking trig now)
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    Re: Trig help!!!!

    Post by preid1220 on Wed Sep 09, 2009 8:04 pm

    Fang wrote:Is that actually trig? I think that's just algebra... I don't see anything related to trig in it (I myself am taking trig now)
    trig and algebra go hand in hand


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    Re: Trig help!!!!

    Post by Ahmed on Wed Sep 09, 2009 9:48 pm

    Well it's "Advanced Algebra and Trig", this is also known as "Pre-Calculus"


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    Re: Trig help!!!!

    Post by Fangy on Thu Sep 10, 2009 12:14 am

    Not trig then.

    I'll post it on Sal's and see if they can help.
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    Re: Trig help!!!!

    Post by Fangy on Thu Sep 10, 2009 4:23 pm

    Here it is:

    How come you changed the result of the first equation into 2 rather than 1 ?

    I'd substitute for x or y from the first equation:

    x+y=1 --> y=1-x

    Then substitute y into the second equation and solve for x as a quadratic. Does that help? I get two possible values for each of x and y, but I'll leave you to work them out for yourself...
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    Re: Trig help!!!!

    Post by Ahmed on Thu Sep 10, 2009 7:06 pm

    Fang wrote:Here it is:

    How come you changed the result of the first equation into 2 rather than 1 ?

    I'd substitute for x or y from the first equation:

    x+y=1 --> y=1-x

    Then substitute y into the second equation and solve for x as a quadratic. Does that help? I get two possible values for each of x and y, but I'll leave you to work them out for yourself...

    No sorry. Thanks though Wink

    I finally understood it today Very Happy

    If x+y=1, x^2+y^2=4, x^3+y^3=?

    x^3+y^3=(x+y)(x^2-xy+y^2)
    x^3+y^3=(1)(4-xy)
    ----------
    (x+y)^2=x^2+2xy+y^2
    x^2+2xy+y^2=1
    (4)+2xy=1
    2xy=-3
    xy=-3/2
    ---------
    x^3+y^3=(1)(4-xy)
    x^3+y^3=(1)(4-(-3/2))
    x^3+y^3=(1)(4+3/2)
    x^3+y^3=(1)(11/2)
    x^3+y^3=11/2


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    Re: Trig help!!!!

    Post by Fangy on Thu Sep 10, 2009 9:24 pm

    Just btw that was from Slashing UK
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    Re: Trig help!!!!

    Post by Ahmed on Thu Sep 10, 2009 9:33 pm

    He probably saw it as a normal Algebra question...


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