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Math Help - Radical numbers homework due tomorrow

  1. #1
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    Radical numbers homework due tomorrow

    Simpify the following radicals.

    (9x)
    Would that be 9x

    -125c
    Would that be -5c

    Write ^7x by using rational exponents rather than radical notation.

    Use the laws of exponents to simplify. (x^-2/3 y)

    Find the domain of the given function. F(x)= 2x-3
    The 2x-3 is under the radical sign.

    Thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rowdy3 View Post
    (9x)
    Would that be 9x
    Actually \pm 9x.

    Quote Originally Posted by rowdy3 View Post
    -125c
    Would that be -5c
    Yes.
    Quote Originally Posted by rowdy3 View Post
    Write ^7x by using rational exponents rather than radical notation.
    \sqrt[n]{x} = x^{1/n}
    So what's the answer?
    Quote Originally Posted by rowdy3 View Post
    Use the laws of exponents to simplify. (x^-2/3 y)
    \left ( x^{-2/3}y \right )^3 = \left ( x^{1/3} \right )^3 y^3 = ?

    Quote Originally Posted by rowdy3 View Post
    Find the domain of the given function. F(x)= 2x-3
    The 2x-3 is under the radical sign.
    Can the number under the square root be negative?

    -Dan
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    Find the domain of the given function. F(x)= 2x-3
    The 2x-3 is under the radical sign.
    all real numbers where x is greater than or equal to 3/2

    Write ^7x by using rational exponents rather than radical notation.
    so would it be x^1/3

    (x^-2/3 y)
    Would the 3's cancel each other out? Would I get xy^3?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rowdy3 View Post
    Find the domain of the given function. F(x)= 2x-3
    The 2x-3 is under the radical sign.
    all real numbers where x is greater than or equal to 3/2
    (x^-2/3 y)
    Would the 3's cancel each other out? Would I get xy^3?
    Yes.

    Write ^7x by using rational exponents rather than radical notation.
    so would it be x^1/3
    Well, if I'm reading this correctly you have
    \sqrt[7]{x^3}
    so what does this become? What does the 7 do? And the 3 isn't a cube root so you don't do a 1/3....

    -Dan
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  5. #5
    Moo
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    Quote Originally Posted by topsquark View Post
    Actually \pm 9x.
    Hello,

    It depends on where the square is, (9x) or above the square root ?
    Actually, whatever... a square number is always positive and unless there is a - sign in front of a square root, it's positive...

    Isn't it ? ;s
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    It depends on where the square is, (9x) or above the square root ?
    Actually, whatever... a square number is always positive and unless there is a - sign in front of a square root, it's positive...

    Isn't it ? ;s
    \sqrt{9} = 3 by convention.

    \sqrt{9x^2} = \pm 3x because we don't know what x is.

    That's the way I learned it, anyway.

    But you do have a valid point: \left ( \sqrt{9x} \right ) ^2 = 9x, whereas \sqrt{(9x)^2} = \pm 9x, so we do need to know where that exponent is in the expression.

    -Dan
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  7. #7
    Moo
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    Yep, it has to be the absolute value of x actually, in the two configurations

    And this is if we work in \mathbb{R} of course
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  8. #8
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    1. 3 50 - 2 18 + 32

    2. 2 48x^5 + 4x 27x

    3. 7/5

    1. 350 - 218 + 32
    = 3(2*25) - 2(2*9) + (2*16)
    = 152 -62 + 42
    = 132

    2.2(48x^5) + 4x(27x)
    = 2(3*16x^4*x) + 4x(3*9x)
    = 8x(3x) + 12x(3x)
    = 20x(3x)
    Those boxes are square root signs.
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  9. #9
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    7/5
    My friend says its rad10 rad35/ 5.
    How did he get to that answer?

    Rationalize the denominator and simplify

    3/ 8+ 3
    He got (8rad3 + 3)/8
    How did he get that answer?
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