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anonymous

  • one year ago

can someone help me with this? [(16/m-3)-(4/m-4)] / [(16/m^2)-(m-4/m-3)}

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  1. freckles
    • one year ago
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    \[[\frac{16}{m-3}-\frac{4}{m-4}] \div [\frac{16}{m^2}-\frac{m-4}{m-3}] ?\]

  2. anonymous
    • one year ago
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    yes its a complex fraction

  3. freckles
    • one year ago
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    do you know how to combine the first set of fractions that do you know how to write \[\frac{16}{m-3}-\frac{4}{m-4} \text{ as one fraction }?\]

  4. anonymous
    • one year ago
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    [16(m-4)-4(m-3)/(m-3)(m-4)] ? i havent figured out how to use the equation writer yet sorry! so it becomes 16m-64-4m+12, 12m-52/(m-3)(m-4) ?????

  5. freckles
    • one year ago
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    \[\frac{16(m-4)-4(m-3)}{(m-3)(m-4)}=\frac{16m-64-4m+12}{(m-3)(m-4)} \\ =\frac{12m-52}{(m-3)(m-4)}\] that is awesome now we have to work with the second pair of fractions that follow the division sign

  6. freckles
    • one year ago
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    \[\frac{16}{m^2}-\frac{m-4}{m-3}=\frac{16(m-3)-m^2(m-4)}{m^2(m-3)} \\ =\frac{16m-48-m^3+4m^2}{m^2(m-3)}=\frac{-m^3+4m^2+16m-48}{m^2(m-3)}\] so you have: \[\frac{12m-52}{(m-3)(m-4)} \div \frac{-m^3+4m^2+16m-48}{m^2(m-3)} \\ \text{ which we can change \to multiplication } \\ \text{ so we have} \\ \frac{12m-52}{(m-3)(m-4)} \times \frac{m^2(m-3)}{-m^3+4m^2+16m-48}\] notice I just flipped the second fraction

  7. freckles
    • one year ago
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    you should see something already that cancels

  8. anonymous
    • one year ago
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    m-3

  9. anonymous
    • one year ago
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    hello?

  10. freckles
    • one year ago
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    yes the (m-3) cancels

  11. anonymous
    • one year ago
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    now whhat

  12. anonymous
    • one year ago
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    @freckles

  13. freckles
    • one year ago
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    multiply top and bottom

  14. freckles
    • one year ago
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    we know to this because there is a operation of multiplication between the fractions

  15. freckles
    • one year ago
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    \[\frac{(12m-52)(m^2)}{(m-4)(-m^3+4m^2+16m-48)}\]

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