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Math Help - Beginning Algebra Final Exam review pt. 2

  1. #1
    Junior Member Cait's Avatar
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    Red face Beginning Algebra Final Exam review pt. 2

    oK i had this all typed out and my computer died >.<

    I have my final exam review here..


    Here's the probs:

    http://aycu10.webshots.com/image/158...9289183_rs.jpg



    Lets start out with 11, 15, 16...



    #15.. i did..

    x^2-12x=-36

    devide by -12..

    x^2+x=3

    =

    2x^2=3

    devide by 2

    x^2= 3/2

    and.. Yeah thats not right. Its supposed to be x=6




    Thanks everyone ^^;;


    edit: errr.. i guess the attach image thing does work o.0 i never saw it till now..
    Attached Thumbnails Attached Thumbnails Beginning Algebra Final Exam review pt. 2-mathsux.jpg  
    Last edited by Cait; May 6th 2007 at 05:35 PM.
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  2. #2
    Junior Member Cait's Avatar
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    Angry

    o.-..

    Okay now...

    11, 16, 18, 20, 21, 22, and 24 are now (er... for the most part) solved.


    This leaves

    15,23,25,26 that still arent done.


    for 11.. do i gotta factor a x-3 and a x+2 out somehow?????
    Last edited by Cait; May 6th 2007 at 05:36 PM.
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  3. #3
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    ok, I have worked out 11 for you (will do the rest later, I'm tried zzzzz)
    So anyway....

    lets factorise the bottom parts
    x^2 - x - 6 = (x + 2) (x - 3)
    x^2 - 8x + 15 = (x - 5) (x - 3)

    So they cancel the top so we are left with:

    1/ (x - 3) + 1/ (x - 5)
    get a common denominator
    (times (x-3) to the right side, times (x-5) to the left side)
    So we get:

    (x - 5)/ (x^2 - 8x + 15) + (x - 3)/ (x^2 - 8x + 15)

    it becomes:

    (2x - 8)/ (x^2 - 8x +15)

    and as we know...
    (x^2 - 8x +15) = (x - 3) (x - 5)
    so answer =
    (2x - 8)/ (x - 3) (x - 5)

    Happy to help :P I'll be back later, but in the mean-time some other helpful member may have helped you out Goodnight... zzzzzzzzzzz
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  4. #4
    Junior Member Cait's Avatar
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    Unhappy

    Ok thanks got that one done.



    Just 5 left now..


    erm.

    ON this page..



    ...with #16.. would you do... y^3 - y^2 to equal just y??


    and on 23... because theres an = sign there and not a plus would you still times the left side by x-3 and the right by x+1??

    I tried that and it didnt work for me..

    Thanks
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  5. #5
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    so I'll do 16 today :P lol

    y^3 - y^2 - 6y = 0

    so lets take out the y so we get:

    y(y^2 - y - 6) = 0

    y^2 - y - 6 FACTORISE to!

    (y + 2) (y - 3)

    so we now have:

    y(y + 2) (y - 3) = 0

    this correct? lol




    btw on question 11
    you can factorise it further to:

    2(x -4)/(x - 3) (x - 5)
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  6. #6
    Junior Member Cait's Avatar
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    Talking

    Ah okay..


    I dont need the rest now.. (exam in an hour >.<)

    But i suppose you could do them for fun if you'd like! lol!


    But thanks for replying on those 2 !!!!!!!!

    and..

    *sniff* because my exam is today, no more math for me... and more than likely, I wont be back (XD sorry, no offense but i dispise math >.<) so i'd like to give a special thanks to everyone who has dealed with my stupidity over the past ~2 month, and helped me through this, I OWE YOU GUYS!!

    But definatelly thanks to Jhevon, helping in the middle of the night, and answering MOST of my questions.. so i could understand it.. you are awesome!

    So, I'll be seein ya, math help forum!!

    (Ill definately recommend you to anyone i find that needs help )

    ~Cait
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