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Math Help - Help Fast!!!

  1. #1
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    Help Fast!!!

    I need help with few probs very fast plz help!!

    1)


    2)


    3)


    4)


    5)


    6)
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  2. #2
    Site Founder MathGuru's Avatar
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    heres the first one
    Attached Thumbnails Attached Thumbnails Help Fast!!!-math01.jpg  
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  3. #3
    Site Founder MathGuru's Avatar
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    For the second one solve the first part for y to get y = 2x-10 and then sub that into the second equation to get:

    x + 3(2x-10) = -9, now solve for x and then sub back in to either of the given equations and solve for y also.
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  4. #4
    Site Founder MathGuru's Avatar
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    Q#3

    9b^2c + 6c
    -----------
    ......3c

    Is the same as:

    9b^2c + 6c
    ------ -----
    ...
    3c.......3c


    or

    3b^2 + 2


    does that make sense?
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  5. #5
    Site Founder MathGuru's Avatar
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    #5

    notice that 27 is 3*(3^2)
    notice that 12 is 3*(2^2)

     -4\sqrt27 = -4\sqrt(3*3^2)=3*-4\sqrt3=-12\sqrt3
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  6. #6
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    Quote Originally Posted by mathelp
    I need help with few probs very fast plz help!!

    ...
    5)


    6)

    Hello,

    to Nr. 5:
    -4 \sqrt{27}+2\sqrt{12}=
    -4 \sqrt{9 \cdot 3}+2\sqrt{4 \cdot 3}=
    -4 \cdot 3 \sqrt{ 3}+2 \cdot 2 \sqrt{ 3}=
    -12 \sqrt{3}+4 \sqrt{3}= -8 \cdot \sqrt{3}

    to 6.:
    \left( \frac{5c^{-3} d^4}{p^{-2}} \right)^{-1}\ \cdot \ \left( \frac{p^7 d^{-1}}{c^2} \right)^2 =
    5^{-1} \cdot c^3 \cdot d^{-4} \cdot p^{-2} \cdot p^{14} \cdot d^{-2} \cdot c^{-2} =
    \frac{cp^{12}}{5d^6}

    Greetings

    EB
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  7. #7
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    Thnaks bro so much
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  8. #8
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    can you show me 2,4 how to do?
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  9. #9
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    #4

    Graph both equations on a graphing calculator and the point where they intersect is your answer of the form (x,y)
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  10. #10
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    Here is the graphs,
    Attached Thumbnails Attached Thumbnails Help Fast!!!-picture1.gif  
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