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anonymous

  • one year ago

Simplify:

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  1. anonymous
    • one year ago
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    |dw:1440898252857:dw|

  2. LynFran
    • one year ago
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    not clear...

  3. anonymous
    • one year ago
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    \[(4-x^2)(1/3)(6x+1)^{-2/3}(6)-(6x+1)^{1/3}(-2x) / (4-x^2)^2\]

  4. LynFran
    • one year ago
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    6*1/3=2 ...\[\frac{[2( 4-x ^{2})(6x+1)^{\frac{ -2 }{ 3 }}]-[(6x+1)^{\frac{ 1 }{ 3 }}(2x)] }{ (4-x ^{2})^{2} }\]

  5. anonymous
    • one year ago
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    Do you start canceling?

  6. anonymous
    • one year ago
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    I have the answer, and that isn't right. I need to know how to work it.

  7. LynFran
    • one year ago
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    ok so we have ...

  8. LynFran
    • one year ago
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    \[\frac{ 2(6x+1)^{-1/3} [(6x+1)^{-1/3}(4-x ^{2})-(x)]}{ (4-x ^{2})^{2} }\]

  9. LynFran
    • one year ago
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    @Loser66 what do u think..

  10. Nnesha
    • one year ago
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    1/3 is an exponent at the numerator ?

  11. anonymous
    • one year ago
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    yes

  12. Nnesha
    • one year ago
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    \[\large\rm \frac{ 6 (4-x^2)^\frac{ 1 }{ 3 } (6x+1)^\frac{ -2 }{ 3}-(6x+1)^\frac{ 1 }{ 3 }(-2x) }{ (4-x^2)^2}\]hmm like this ?

  13. anonymous
    • one year ago
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    There is no 1/3 exponent for (4-x^2).

  14. anonymous
    • one year ago
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    \[(4-x^2)(1/3)(6x+1)^{-2/3}-(6x+1)^{1/3}(-2x) / (4-x^2)^2\] is the problem

  15. Nnesha
    • one year ago
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    ohh i asked before

  16. Nnesha
    • one year ago
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    \[\large\rm \frac{ 6(\frac{1}{3}) (4-x^2) (6x+1)^\frac{ -2 }{ 3}-(6x+1)^\frac{ 1 }{ 3 }(-2x) }{ (4-x^2)^2}\] looks right ?

  17. anonymous
    • one year ago
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    No

  18. Nnesha
    • one year ago
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    no hmm can you please take a screenshot

  19. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @ducksonquack \[(4-x^2)(1/3)(6x+1)^{-2/3}(6)-(6x+1)^{1/3}(-2x) / (4-x^2)^2\] \(\color{blue}{\text{End of Quote}}\) there is 6

  20. anonymous
    • one year ago
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  21. Nnesha
    • one year ago
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    alright thanks

  22. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @ducksonquack I have the answer, and that isn't right. I need to know how to work it. \(\color{blue}{\text{End of Quote}}\) what's the answer jsut want to check if i'm doing this right or not>.<

  23. anonymous
    • one year ago
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    \[2(5x^2+x+4) / (4-x^2)^2(6x+?)^{2/3}\]

  24. anonymous
    • one year ago
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    Didn't catch what the ? mark was when we were given the answers.

  25. Jhannybean
    • one year ago
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    Well that's in your problem, it states \((6x+1)\) in parenthesis.... so im sure the \(? = 1\)

  26. anonymous
    • one year ago
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    Okay. What's next?

  27. Jhannybean
    • one year ago
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    One second im solving it

  28. Jhannybean
    • one year ago
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    Alright got it, so...

  29. anonymous
    • one year ago
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    ?

  30. Jhannybean
    • one year ago
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    \[=\frac{\frac{1}{3}\cdot 6 \cdot (4-x^2)(6x+1)^{-2/3}-(6x+1)^{1/3}(-2x)}{(4-x^2)^2}\]\[=\frac{(6x+1)^{-2/3}\left(2\left[4-x^2-(6x+1)(-x)\right]\right)}{(4-x^2)^2}\]\[=\frac{2\left[4-x^2+6x+x\right]}{(4-x^2)^2(6x+1)^{2/3}}\]\[=\boxed{\frac{2(5x^2+x+4)}{(4-x^2)^2(6x+1)^{2/3}}}\]

  31. anonymous
    • one year ago
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    Can you explain it to me step by step?

  32. Jhannybean
    • one year ago
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    Okay so what you want to do first is look for a common like term. in this case it would be \((6x+1)^{-2/3}\). Notice how its a common multiplier in the numerator? We can pull it out. Now I pulled out \((6x+1)^{-2/3}\) instead of the \((6x+1)^{1/3}\) because outof the two, \((6x+1)^{-2/3}\) is the smaller power, therefore pulling it out as a common factor, I am left with \((6x+1)^1\) inside the brackets. (Remember: \((6x+1)^{-2/3} \cdot (6x+1)^1 = (6x+1)^{1/3}\)) Now I pulled out a 2 inside the brackets because that is another common like term between the entire function within the brackets. \([2(4-x^2)-(6x+1)(-2x)] \iff 2[(4-x^2-(6x+1)(-x)]\) Then I simplify my fraction even further by turning \((6x+1)^{-2/3}\) positive by putting it in the denominator. this turns it positive. Recall that: \(x^{-\#} \iff \dfrac{1}{x^{\#}}\) So far we have: \[\frac{2[4-x^2-(6x+1)(-x)]}{(4-x^2)^2(6x+1)^{2/3}}\] Now we simplify the stuff inside the brackets without multiplying in the 2. \[\frac{2[4-x^2 +6x+x]}{(4-x^2)^2(6x+1)^{2/3}}\] Simplify this a bit more and you'll have: \(\boxed{ \dfrac{2[5x^2+x+4]}{(4-x^2)^2(6x+1)^{2/3}}}\)

  33. Nnesha
    • one year ago
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    sorry i was afk great explanation jhanny!

  34. Jhannybean
    • one year ago
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    Thank you :)

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