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Math Help - lots of math problems, and fast????

  1. #1
    Member cassiopeia1289's Avatar
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    lots of math problems, and fast????

    ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

    1: (simplify with only positive exponents)
    sqrt(4x-16) / 4thsqrt (x-4)^3

    and I can only make it 2(square root of (x-4))times(4th square root of (x-4) / x-4
    is that right?

    2: [(1/x^-2) + (4/x^-1*y^-1) + (1/y^-2)]^(-1/2)
    I got square root of (x^2 + 4xy + y^2)/x^2 + 4xy + y^2
    is that right?

    3: find domain: y=log(2x-12)
    is it all real numbers?
    4: same deal, y=square root of tanx
    all real numbers?
    5: y=sqrt (square root) of (x-3) - sqrt(x+3)
    all real numbers?

    6: (solve albsolute value inequalities)
    abs(x+1) < & = abs(x-3)
    is is 2 <&= x <&= -4?

    7: solve and sign chart:
    2x^2 + 4x <&= 3
    is it x= .581, 2.581?

    8: use synthetic division to factor P(x) then solve P(x)=0
    I've tried all numbers and can't get it to work
    P(x) = x^3 - 6x^2 + 3x - 10

    9: find vertical and horizontal asymptotes
    y= (x+4)/(x^2 - 1)
    and
    y= (x^2 - x - 6) / (x^3 - x^2 + x - 6)


    please please please help! I need it by tomorrow morning, so as soon as possible would be unbelievably helpful
    THANK YOU!
    -meg-

    (you don't have to know all of them either, just please I'll take anything, I'm kinda desperare at this point)
    also, I work backwards, giving me the answer will do, i can figure it out myself, in fact, I'd probably learn it better if i figured it out by myself, I just need input
    Last edited by cassiopeia1289; August 12th 2007 at 03:20 PM.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    First read the LaTeX Tutorial

    3) To get the domain of y=\log(2x-12) you gotta set two things: 2x-12 can't be zero nor negative, then you have to set 2x-12>0

    The rest are similar.
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  3. #3
    Eater of Worlds
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    8: use synthetic division to factor P(x) then solve P(x)=0
    I've tried all numbers and can't get it to work
    P(x) = x^{3} - 6x^{2} + 3x - 10
    Are you sure that isn't x^{3}+6x^{2}+3x-10?.

    The other one doesn't factor very nicely and has two non real solutions.



    9: find vertical and horizontal asymptotes
    y= \frac{(x+4)}{(x^2 - 1)}
    and
    y= \frac{(x^2 - x - 6)}{(x^3 - x^2 + x - 6)}
    Asymptotes are easy if you just know some rules. To find the vertical asymptotes, find what x value(s) makes the denominator equal 0.

    To find the horizontal asymptotes, if the power of the numerator is less than the power of the denominator the x-axis is the horizontal asymptote.
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  4. #4
    Member cassiopeia1289's Avatar
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    ^
    That's exactly what I thought, it might be a mistake ... erk
    and thanks for the asymptotes
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  5. #5
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    Quote Originally Posted by cassiopeia1289 View Post
    ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

    1: (simplify with only positive exponents)
    sqrt(4x-16) / 4thsqrt (x-4)^3

    and I can only make it 2(square root of (x-4))times(4th square root of (x-4) / x-4
    is that right?

    ...
    Hello,

    \frac{\sqrt{4x-16}}{\sqrt[4]{(x-4)^3}} = 2 \cdot (x-4)^{\frac{1}{2}} \cdot (x-4)^{-\frac{3}{4}} = 2 \cdot (x-4)^{-\frac{1}{4}}=\frac{2}{(x-4)^{\frac{1}{4}}}
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  6. #6
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    Quote Originally Posted by cassiopeia1289 View Post
    ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

    ...

    2: [(1/x^-2) + (4/x^-1*y^-1) + (1/y^-2)]^(-1/2)
    I got square root of (x^2 + 4xy + y^2)/x^2 + 4xy + y^2
    is that right?

    ...
    I assume that you mean:

    \left(\frac{1}{x^{-2}} + \frac{4}{x^{-1} \cdot y^{-1}} + \frac{1}{y^{-2}}\right)^{-\frac{1}{2}}

    If so your answer is right but I would add that x and y must be unequal zero and the complete term in bracket must be greater than zero.
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  7. #7
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    Quote Originally Posted by cassiopeia1289 View Post
    ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

    ...
    4: same deal, y=square root of tanx
    all real numbers?
    5: y=sqrt (square root) of (x-3) - sqrt(x+3)
    all real numbers?

    ...
    to #4:

    y = \sqrt{\tan(x)}. That means \tan(x) \geq 0 which is only possible if D=\{x|x \in [n \cdot \pi, n \cdot \pi + \frac{\pi}{2}]\} with n \in \mathbb{Z}

    to #5:

    I assume that you mean:

    y=\sqrt{\sqrt{x-3} - \sqrt{x+3}}

    The domain of \sqrt{x-3} is D_1=\{x|x\geq 3\}

    The domain of \sqrt{x+3} is D_2=\{x|x\geq -3\}

    But because \sqrt{x-3} < \sqrt{x+3} the radicand is allways negative that means the domain is the empty set.

    = + = + = + = + = + = + = + = + = + = + = + = + = + = + = + = + = + = +

    If you mean:

    y=\sqrt{x-3} - \sqrt{x+3} then the domain is: D=\{x|x\geq 3\}
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  8. #8
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    Quote Originally Posted by cassiopeia1289 View Post
    ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

    ...

    6: (solve albsolute value inequalities)
    abs(x+1) < & = abs(x-3)
    is is 2 <&= x <&= -4?

    ...
    Translate your inequality: "The graph of the left function should be below the graph of the right function." (see attachment)

    |x+1| \leq |x+3|~\Longleftrightarrow~|x+1| - |x+3|\leq 0

    Since
    |x+1|=\left\{\begin{array}{lr}-(x+1) & x < -1 \\x+1 & x\geq -1 \end{array}\right.
    and
    |x-3|=\left\{\begin{array}{lr}x-3 & x \geq 3 \\-(x-3) & x< 3 \end{array}\right.

    your inequality becomes:
    \left\{\begin{array}{lr}-(x+1)-(-(x-3))\leq 0 & x <-1 \\ x+1-(-(x-3))\leq 0 & -1\leq x < 3 \\ x+1 - (x-3) \leq 0 &x \geq 3\end{array}\right.

    Solve for x and you'll get:

    x < -1 ~\wedge -1 \leq x < 1. That means the solution is the set of x with x < 1.

    I've attached a sketch of the 2 functions. On the x-axis I've marked the x values which make the inequality true.
    Attached Thumbnails Attached Thumbnails lots of math problems, and fast????-betrg_unglg.gif  
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