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Math Help - Need Help!

  1. #1
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    Exclamation Need Help!

    I have few probs i need to help with here are cupple i did them on photoshop heres the link to it CLICK HERE TO SEE THE PROBS
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    Problem 3]
    \frac{x-3}{4}-\frac{x}{7}=7
    Multiply by Common Denominator to get,
    28\cdot\frac{x-3}{4}-28\cdot\frac{x}{7}=28\cdot 7, thus,
    7(x-3)-4x=196
    Thus,
    7x-21-4x=196
    Thus,
    3x-21=196
    Thus,
    3x=217
    Thus,
    x=\frac{217}{3}=72.\bar 3
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  3. #3
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    Problem 4]
    \frac{m^4(n^{-3})^2}{(m^{-2})^{-3}n^0} \left(\frac{m^{-4}n^8}{n^8m^{-2}} \right)^2
    Open parantheses,
    \frac{m^4n^{-6}}{m^6n^0}\cdot \frac{m^{-8}n^{16}}{n^{16}m^{-4}} \right)
    Use rule of exponents,
    m^{-2}n^{-6}m^{-4}n^0
    Use rule of exponents again ( n^0=1),
    m^{-6}n^{-6}
    Q.E.D.
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    Problem 5]
    f(-\frac{2}{3})=\frac{-\frac{2}{3}}{2}=-\frac{1}{3}
    Q.E.D.
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  5. #5
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    Quote Originally Posted by autopimp
    I have few probs i need to help with here are cupple i did them on photoshop heres the link to it CLICK HERE TO SEE THE PROBS
    Problem 2.

    Solve by substitution:
    2x+3y=7
    x-2y=7

    In the substitution method we solve one of the equations for one unknown and plug it into the other...this creates an equation with only one unknown in it. I tend to pick the simplest variable to solve for. In this case that looks to be the x in the second equation.

    So solving the second equation for x I get:
    x=2y+7

    Plugging this value of x into the first equation I get:
    2(2y+7)+3y=7.

    I'll let you take it from here. You may plug this value of y into any of the original equations, or more simply, into your x equation. I get (x,y) = (5,-1).

    -Dan
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  6. #6
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    Quote Originally Posted by autopimp
    I have few probs i need to help with here are cupple i did them on photoshop heres the link to it CLICK HERE TO SEE THE PROBS
    Hello,

    to problem nr. 6: Transform the quotient into a product:
    \frac{2x-10}{x^2-25}\ \cdot \ \frac{x^2-4x-5}{x^2-2x-15}
    factorize all numerators and denominators:
    \frac{2(x-5)}{(x+5)(x-5)}\ \cdot \ \frac{(x+1)(x-5)}{(x+3)(x-5)}
    simplify and you'll get:
    \frac{2(x+1)}{(x+5)(x+3)}

    Greetings

    EB
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  7. #7
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    Quote Originally Posted by autopimp
    I have few probs i need to help with here are cupple i did them on photoshop heres the link to it CLICK HERE TO SEE THE PROBS
    Hello,

    to problem nr. 8:
    Calculate the roots partially:
    3\cdot \sqrt{32}-5\cdot \sqrt{50}-8 \cdot \sqrt{45} + 7\cdot \sqrt{20}=
    3\cdot  4\sqrt{2}-5\cdot  5\sqrt{2}-8 \cdot  3\sqrt{5} + 7\cdot  2  \sqrt{5}=
    -13 \sqrt{2}-10 \sqrt{5}

    Greetings

    EB
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