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  1. #1
    Lucianberg's Avatar
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    I need a test checker

    So, I did a maths test today. The professor allowed us to take the tasks home so I decided to upload them here since we're not getting the test untill Tuesday.If you could calulate these and give me the results so I can know what I'm on...

    Task #1: Calculate:

    a) (-2mbn)4

    b) a-2+a-3 ; if a=2

    Task #2: Divide on factors:

    a) a4b2-c2

    Task #3: Shorten the fraction:

    a3 - b3___
    a2+ab+b2

    Task #4: Calculate:
    (the underline is the main fraction since it has double fractions)

    x/a - x/2a____
    x2/a2 - x2/2a2

    @Alen
    @iverson954360
    Last edited by Lucianberg; 12-08-2011 at 07:27 AM.
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  2. #2
    Chester Bennington's Avatar
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    Work it out for yourself and you'll feel much better for it.

  3. #3
    Alen's Avatar
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    (-2^m b^n )^4=((-1)⋅2^m⋅b^n )^4=(-1)^4 (2^m )^4 (b^n )^4=2^4m b^4n=16^m b^4n
    a^(-2)+a^(-3)=a⋅a^(-3)+1⋅a^(-3)=(a+1) a^(-3)=(2+1)/2^3 =3/8
    a^4 b^2-c^2=(a^2 b)^2-c^2=(a^2 b-c)(a^2 b+c)
    (a^3-b^3)/(a^2+ab+b^2 )=((a-b)(a^2+ab+b^2))/(a^2+ab+b^2 )=a-b
    (x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(x/2a)/(x^2/(2a^2 ))=(2a^2 x)/(2ax^2 )=a/x
    Or with le substitution (most likely what the prof wanted):
    (x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(l-l/2)/(l^2-l^2/2),l=x/a
    ((l-l/2)⋅2)/((l^2-l^2/2)⋅2)=(2l-l)/(2l^2-l^2 )=l/l^2 =1/l=a/x

  4. #4
    Lucianberg's Avatar
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    Quote Originally Posted by Alen View Post


    (-2^m b^n )^4=((-1)⋅2^m⋅b^n )^4=(-1)^4 (2^m )^4 (b^n )^4=2^4m b^4n=16^m b^4n
    a^(-2)+a^(-3)=a⋅a^(-3)+1⋅a^(-3)=(a+1) a^(-3)=(2+1)/2^3 =3/8
    a^4 b^2-c^2=(a^2 b)^2-c^2=(a^2 b-c)(a^2 b+c)
    (a^3-b^3)/(a^2+ab+b^2 )=((a-b)(a^2+ab+b^2))/(a^2+ab+b^2 )=a-b
    (x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(x/2a)/(x^2/(2a^2 ))=(2a^2 x)/(2ax^2 )=a/x
    Or with le substitution (most likely what the prof wanted):
    (x/a-x/2a)/(x^2/a^2 -x^2/(2a^2 ))=(l-l/2)/(l^2-l^2/2),l=x/a
    ((l-l/2)⋅2)/((l^2-l^2/2)⋅2)=(2l-l)/(2l^2-l^2 )=l/l^2 =1/l=a/x
    OMFG I'm getting an F...
    I only solved 2/5

    Now my GPA is 4.2 or something like that
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  5. #5
    iverson954360's Avatar
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    This exam is not hard at all, you just need to practice your laws of exponents as well as a few rules pertaining to polynomial functions. Alen made solving the second one look harder than it needed to be. When you see something raised to a negative power you must put it on the opposite side of the fraction so in this case if you have all whole numbers over an a 1 ie. 2/1, 3/1, etc. then when a negative exponent comes into play you just flip the fraction ie (2^-2)/1 goes to 1/(2^2) which you then just solve simply as 1/4.

  6. #6
    Alen's Avatar
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    Quote Originally Posted by iverson954360 View Post
    This exam is not hard at all, you just need to practice your laws of exponents as well as a few rules pertaining to polynomial functions. Alen made solving the second one look harder than it needed to be. When you see something raised to a negative power you must put it on the opposite side of the fraction so in this case if you have all whole numbers over an a 1 ie. 2/1, 3/1, etc. then when a negative exponent comes into play you just flip the fraction ie (2^-2)/1 goes to 1/(2^2) which you then just solve simply as 1/4.
    I broke it up in a manner that I though would make it easiest to understand what I did Easiest way to "solve" it would be to simply solve

  7. #7
    Lucianberg's Avatar
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    @Alen the second one where your result was 3/8, couldn't I just went with:

    a-2+a-3 ; if a=2

    2-2 + 2-3=
    4 + (-8)= -4?
    Last edited by Lucianberg; 12-08-2011 at 11:52 AM.
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  8. #8
    Alen's Avatar
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    Quote Originally Posted by Chuck View Post
    @Alen the second one where your result was 3/8, couldn't I just went with:

    a-2+a-3 ; if a=2

    2-2 + 2-3=
    -4 + 8 = 4?
    Nope.


  9. #9
    Lucianberg's Avatar
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    @Alen
    So dead...

    Another question...

    Shouldn't the tast go like this?

    x/a - x/2a
    x2/a2 - x2/2a2=

    2x-x/2a
    2x2-x2/2a2=

    x/2a
    x2/2a2=

    2a2x/2ax2=

    2a/x
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  10. #10
    Alen's Avatar
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    Quote Originally Posted by Chuck View Post
    @Alen2a2x/2ax2=

    2a/x
    Nope.


  11. #11
    Lucianberg's Avatar
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    OMFG!!!!!!
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