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Math Help - Rational Problems

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    YOUR AWSOME!! but shouldnt it be 2x^2??
    correct, i'll change it
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  2. #17
    Ari
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    Ok this is the other problem i dont get(btw i love this forum now)


    simplify and state restrictions
    x^2+2x-63/7-x
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    Ok this is the other problem i dont get(btw i love this forum now)


    simplify and state restrictions
    x^2+2x-63/7-x
    do you mean \frac {x^2 + 2x - 63}{7 - x}? if so, you should type (x^2 + 2x - 63)/(7 - x)
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  4. #19
    Ari
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    Quote Originally Posted by Jhevon View Post
    do you mean \frac {x^2 + 2x - 63}{7 - x}? if so, you should type (x^2 + 2x - 63)/(7 - x)
    yeah thats what i meant, BTW how do you type the way you type? i cant get those fractions and stuff properly like you...
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    Ok this is the other problem i dont get(btw i love this forum now)


    simplify and state restrictions
    x^2+2x-63/7-x
    \frac {x^2 + 2x - 63}{7 - x} = \frac {(x + 9)(x - 7)}{-(x - 7)} = -(x + 9)

    the restriction is x \ne 7 since we have to consider the domain of the first function as well. in the original question, x = 7 makes the denominator zero

    Quote Originally Posted by Ari View Post
    yeah thats what i meant, BTW how do you type the way you type? i cant get those fractions and stuff properly like you...
    i use LaTex. see here
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  6. #21
    Ari
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    Oh OK so you factor the top and then just cancel them out i see.

    what about

    (5m^2+6m-8)/(5m^2-16m-16) * (20m^2+16m)/(4-m^2)


    <br />
\frac {5m^2+6m-8}{5m^2-16m-16} \cdot \frac {20m^2+16m}{4-m^2}<br />



    OMG I DID IT!
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  7. #22
    Ari
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    <br />
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}<br />


    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}<br />
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  8. #23
    is up to his old tricks again! Jhevon's Avatar
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    I hope you're not letting me do your homework for you... where's that suspicious smiley?
    Quote Originally Posted by Ari View Post
    Oh OK so you factor the top and then just cancel them out i see.

    what about

    (5m^2+6m-8)/(5m^2-16m-16) * (20m^2+16m)/(4-m^2)


    <br />
\frac {5m^2+6m-8}{5m^2-16m-16} \cdot \frac {20m^2+16m}{4-m^2}<br />
    same story here.

    \frac {5m^2 + 6m - 8}{5m^2 - 16m - 16} \cdot \frac {20m^2 + 16m}{4 - m^2} = \frac {(5m - 4) (m + 2)}{(m - 4) (5m + 4)} \cdot \frac {4m (5m + 4)}{(2 - m)(2 + m)}

    = \frac {4m(5m - 4)}{(m - 4)(2 - m)}

    what do you think the restrictions are? remember, consider the original function as well

    OMG I DID IT!
    haha, yes you did

    not so hard was it?
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  9. #24
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    <br />
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}<br />
    now that you can type with LaTex, try this on your own, and i will correct you if anything
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  10. #25
    Ari
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    Thats what i was planning on doing but i got an email saying you posted so im reading that ATM thanks!
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  11. #26
    Ari
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    m doesnt = 4 or -2 <---is that right?

    and im stuck here, do i cancel or recipricate and muiltyply?

    <br />
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}<br />


    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}<br />
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  12. #27
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    m doesnt = 4 or -2 <---is that right?
    what about the (2 - x) and the (5m + 4) in the denominators of the original fractions


    and im stuck here, do i cancel or recipricate and muiltyply?

    <br />
\frac {j^2+7j+12}{j^2+6j+9} \div \frac {j^2-16}{j^2+7j+12}<br />


    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \div \frac {(j-4)(j+4)}{(j+3)(j+4)}<br />
    multiply the first fraction by the reciprocal of the second fraction and see if anything cancels. if not, just multiply
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  13. #28
    Ari
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    so m doesnt = 2 but how do i do the (5m-4)?

    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j+4)}{(j-4)(j+4)}<br />

    <br />
\frac {(j+3)(j-3)}{(j+3)(j-4)}<br />

    did i cancel the right stuff?
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  14. #29
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    so m doesnt = 2 but how do i do the (5m-4)?
    set 5m + 4 = 0, solve for m


    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j+4)}{(j-4)(j+4)}<br />

    <br />
\frac {(j+3)(j-3)}{(j+3)(j-4)}<br />

    did i cancel the right stuff?
    no
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  15. #30
    Ari
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    m doesnt = -4/5

    then what am i supposed to cancel?
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