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Firejay5

  • 2 years ago

Simplify. Show work and explain. 13. n^5 over n - 6 * n^2 - 6n over n^8

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  1. Mertsj
    • 2 years ago
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    Factor 3y^2 out of the numerator

  2. Mertsj
    • 2 years ago
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    Awesome!!

  3. Mertsj
    • 2 years ago
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    Remember to multiply by the reciprocal and factor n^2-6n by factoring out n

  4. Firejay5
    • 2 years ago
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    cancel out n - 6

  5. Mertsj
    • 2 years ago
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    yes

  6. Firejay5
    • 2 years ago
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    so now we have n^5 * n over n^8

  7. Mertsj
    • 2 years ago
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    no

  8. Firejay5
    • 2 years ago
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    what do you mean

  9. Mertsj
    • 2 years ago
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    \[\frac{n^5}{n-6}\times\frac{n(n-6)}{n^8}=\frac{1}{n^2}\]

  10. Firejay5
    • 2 years ago
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    @abb0t

  11. abb0t
    • 2 years ago
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    ?

  12. Firejay5
    • 2 years ago
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    I need someone to finish Mertsj's work

  13. zepdrix
    • 2 years ago
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    \[\large \frac{n^5}{n-6}\times \frac{n^2-6n}{n^8}\] Hmm do you understand what he did so far? That was a bunch of steps all done at once, I can understand if it was a little confusing.

  14. Firejay5
    • 2 years ago
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    I understand that, but I need the explanation of how that = 1 over n^2

  15. zepdrix
    • 2 years ago
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    \[\large \frac{n^5}{n-6}\times \frac{\color{orangered}{n^2-6n}}{n^8}\] For this orange part, factor an n out of each term.\[\large \frac{n^5}{n-6}\times \frac{\color{orangered}{n(n-6)}}{n^8}\]

  16. zepdrix
    • 2 years ago
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    From here we'll simply multiply across, Put brackets around the n-6 on the bottom, so the multiplication is a little clearer.\[\large \frac{n^5}{(n-6)}\times \frac{n(n-6)}{n^8}\] Then multiplying across gives us,\[\large \frac{n^5\times n(n-6)}{n^8(n-6)} \qquad = \qquad \frac{n^6(n-6)}{n^8(n-6)}\]Understand so far? :o

  17. Firejay5
    • 2 years ago
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    yes

  18. zepdrix
    • 2 years ago
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    The n-6's can divide out,\[\large \frac{n^6\cancel{(n-6)}}{n^8\cancel{(n-6)}} \qquad = \qquad \frac{n^6}{n^8}\]

  19. zepdrix
    • 2 years ago
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    From here, we want to remember our rules for exponents. When we divide terms of the same base, we `subtract` the exponents.\[\large \frac{n^6}{n^8} \qquad = \qquad n^{6-8}\]

  20. Firejay5
    • 2 years ago
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    then's it's 1 over n^2

  21. zepdrix
    • 2 years ago
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    \[\large n^{-2}\qquad = \qquad \frac{1}{n^2}\]Yes :)

  22. Firejay5
    • 2 years ago
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    you can do that OR the biggest exponent is on the bottom <--- 8 - 6 = 2

  23. Firejay5
    • 2 years ago
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    substract

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