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  1. #1
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    Help Help!

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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by autopimp
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    $\displaystyle \frac{y^0x^{-8}}{x^{-7}z} \left ( \frac{(y^5)^{-2}}{(z^0)^9} - \frac{x^{11}}{y^4} \right )$

    First, anything to the zeroth power is 1, so:
    $\displaystyle \frac{1*x^{-8}}{x^{-7}z} \left ( \frac{(y^5)^{-2}}{(1)^9} - \frac{x^{11}}{y^4} \right )$

    Now, $\displaystyle x^{-a} \equiv \frac{1}{x^a}$ and $\displaystyle \frac{1}{x^{-a}} \equiv x^a$, so:
    $\displaystyle \frac{x^7}{x^8z} \left ( \frac{1}{(y^5)^2} - \frac{x^{11}}{y^4} \right )$

    Cancelling the x's in the first fraction:
    $\displaystyle \frac{1}{xz} \left ( \frac{1}{(y^5)^2} - \frac{x^{11}}{y^4} \right )$

    Now, $\displaystyle (x^a)^b = x^{ab}$, so:
    $\displaystyle \frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}}{y^4} \right )$

    Now we add the fractions by getting a common denominator, which is $\displaystyle y^{10}$:
    $\displaystyle \frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}}{y^4} * \frac{y^6}{y^6} \right )$
    $\displaystyle \frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}y^6}{y^4y^6} \right )$
    $\displaystyle \frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}y^6}{y^{10}} \right )$
    $\displaystyle \frac{1}{xz} \left ( \frac{1-x^{11}y^6}{y^{10}} \right )$

    I am leaving the answer in this form because nothing will cancel upon multiplying the $\displaystyle \frac{1}{xz}$.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by autopimp
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    2. $\displaystyle \frac{5-\sqrt2}{\sqrt2}$

    We need to remove the $\displaystyle \sqrt2$ from the denominator. So I will multiply the numerator and denominator by $\displaystyle \sqrt2$:
    $\displaystyle \frac{5-\sqrt2}{\sqrt2}*\frac{\sqrt2}{\sqrt2}$
    $\displaystyle \frac{5 \sqrt2-\sqrt2 \sqrt2}{\sqrt2 \sqrt2}$
    $\displaystyle \frac{5 \sqrt2-2}{2}$.

    3. $\displaystyle \frac{x-2}{5}-\frac{x-7}{3}=-1$
    We need to get a common denominator, which in this case is 15:
    $\displaystyle \frac{x-2}{5}*\frac{3}{3}-\frac{x-7}{3}*\frac{5}{5}=-1$

    $\displaystyle \frac{3x-6}{15}-\frac{5x-35}{15}=-1$

    $\displaystyle \frac{3x-6-(5x-35)}{15}=-1$

    $\displaystyle \frac{3x-6-5x+35}{15}=-1$

    $\displaystyle \frac{-2x+29}{15}=-1$

    Now we multiply both sides of the equation by 15:
    $\displaystyle -2x+29=-15$

    $\displaystyle -2x+29-29=-15-29$

    $\displaystyle -2x=-44$

    $\displaystyle \frac{-2x}{-2}=\frac{-44}{-2}$

    $\displaystyle x=22$.

    4. $\displaystyle 3\sqrt{32}-5\sqrt{50}-8\sqrt{45}+7\sqrt{20}$

    Let's simplify some of these radicals:
    $\displaystyle \sqrt{32}=\sqrt{2*16}=4\sqrt2$
    $\displaystyle \sqrt{50}=\sqrt{2*25}=5\sqrt2$
    etc.

    Once you get those, plug them into your original expression:
    $\displaystyle 3(4\sqrt2)-5(5\sqrt2)+... $
    $\displaystyle 12\sqrt2-25\sqrt2 + ...$

    Then just collect like terms. My final answer was $\displaystyle -13\sqrt2-10\sqrt5$.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by autopimp
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    5. $\displaystyle \frac{5x-2}{x^2-5x-14}-\frac{3}{x^2+2x}$

    First, factor the denominators:
    $\displaystyle \frac{5x-2}{(x-7)(x+2)}-\frac{3}{x(x+2)}$

    A good common denominator for this would be: $\displaystyle x(x-7)(x+2)$. So:
    $\displaystyle \frac{5x-2}{(x-7)(x+2)}*\frac{x}{x}-\frac{3}{x(x+2)}*\frac{(x-7)}{(x-7)}$

    $\displaystyle \frac{x(5x-2)}{x(x-7)(x+2)}-\frac{3(x-7)}{x(x-7)(x+2)}$

    $\displaystyle \frac{x(5x-2)-3(x-7)}{x(x-7)(x+2)}$

    By the way, you rarely want to multiply out the denominator. Just leave it like it is.

    $\displaystyle \frac{(5x^2-2x)-(3x-21)}{x(x-7)(x+2)}$

    $\displaystyle \frac{5x^2-2x-3x+21}{x(x-7)(x+2)}$

    $\displaystyle \frac{5x^2-5x+21}{x(x-7)(x+2)}$

    At this stage ALWAYS look to see if the numerator factors. In this case, it doesn't so we are done.

    7. $\displaystyle \frac{x}{x^2-16}$ divided by $\displaystyle \frac{x^3}{x^2-14x+40}$

    When you divide by a fraction, you are multiplying by the inverse of the fraction:
    $\displaystyle \frac{x}{x^2-16}*\frac{x^2-14x+40}{x^3}$

    At this point, factor what you can, then terms will cancel.

    -Dan
    Last edited by topsquark; Mar 7th 2006 at 04:40 PM. Reason: Hit the wrong button!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by autopimp
    I puted all in img Click here
    6. This is similar to what I did in 1. Group the powers of 10 together and apply the exponent rules I listed in 1. I get 250000.

    8.
    $\displaystyle 3x+5y=5$
    $\displaystyle x-2y=9$

    For the substitution method we pick one equation to solve for one variable and put it into the remaining equation. I usually pick the easiest one to solve, so I'll solve the second equation for x:
    $\displaystyle x=2y+9$

    Now we put this in as the value for x in the first equation:
    $\displaystyle 3(2y+9)+5y=5$
    This is now one equation in one unknown. Solve this for y and use your x equation to find x. I get (x,y)=(5,-2).

    -Dan
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