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Math Help - Separating constants

  1. #1
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    Separating constants

    Hi guys, I need an idea of exactly what I'm supposed to do here. I'm working on a series of problems where the only instructions given are to 'separate the constants from the terms'

    The terms are: a. -2a^3/7

    b. (-3x)^2/8

    c. -1/5x

    Am I supposed to set them equal to 0 and get the variable on the left side?
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  2. #2
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    Quote Originally Posted by Lewong View Post
    Hi guys, I need an idea of exactly what I'm supposed to do here. I'm working on a series of problems where the only instructions given are to 'separate the constants from the terms'

    The terms are: a. -2a^3/7

    b. (-3x)^2/8

    c. -1/5x

    Am I supposed to set them equal to 0 and get the variable on the left side?
    No, you are not to "set them equal" to anything. You are only asked to "separate" the constant from the variable.

    However, I need to know if "-2a^3/7" is "-2a^(3/7)" or "-2a^(3)/7= (-2/7)a^3". In the first case, the "constant" is -2 while the "term" is a^(3/7). In the second, the "constant" is (-2/7) and the "term" is a^3.

    The same ambiguity applies to (b): is it (-3x)^(2/8) or ((-3x)^2)/8. Since the first results in (-3)^(1/4), which is not a real number, I suspect you meant ((-3a)^2)/8= ((-3)^2 a^2)/8= (9/8)a^2.

    Finally, in (c) do you mean -1/(5x) or (-1/5)x?
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  3. #3
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    I'll try to type these a little less crpytically here, heh

    a. -2a^3/7

    In this one 2a^3 is the numerator and 7 is the denominator, the negative sign is in the middle.



    b. (-3x)^2/8

    (-3x)^2 is the numerator, 8 is the denominator.



    c.-1/5x

    -1 is the numerator, 5x is the denominator.
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  4. #4
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    Quote Originally Posted by Lewong View Post
    I'll try to type these a little less crpytically here, heh

    a. -2a^3/7
    This is a weird kind of question, but I would do something like -\frac{2a^3}{7}=(-\frac{2}{7})(a^3)
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    Quote Originally Posted by Lewong View Post
    I'll try to type these a little less crpytically here, heh

    a. -2a^3/7

    In this one 2a^3 is the numerator and 7 is the denominator, the negative sign is in the middle.



    b. (-3x)^2/8

    (-3x)^2 is the numerator, 8 is the denominator.



    c.-1/5x

    -1 is the numerator, 5x is the denominator.
    Cover up the variable and that will leave a constant. Remember that \frac{ax}{b} = \frac{a}{b}x and the constant is \frac{a}{b}

    b. Note that (ab)^2 = a^2b^2

    c. Note that \frac{a}{bc} = \frac{a}{b} \times \frac{1}{c}
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  6. #6
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    Ok, so for (b) I would put: (-3^2/8)(x^2)

    Honestly, this is would I thought I was supposed to do originally, but it just seemed too simple in the context of a test containing much more advanced math.
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  7. #7
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    Quote Originally Posted by Lewong View Post
    Ok, so for (b) I would put: (-3^2/8)(x^2)

    Honestly, this is would I thought I was supposed to do originally, but it just seemed too simple in the context of a test containing much more advanced math.
    You could but it would be more proper to square the -3 to give \frac{9}{8}
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  8. #8
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    And be careful to distinguish between " -3^2", which is equal to -9, and (-3)^2, which is equal to 9. Use parentheses!
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