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anonymous

  • one year ago

Can someone help me with this? √45n^5

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  1. SolomonZelman
    • one year ago
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    \(\large\sqrt{45n^5}\) like this?

  2. anonymous
    • one year ago
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    Yeah like that.

  3. SolomonZelman
    • one year ago
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    \(\color{black}{ \displaystyle \sqrt{45n^5}=\sqrt{9\cdot 5 \cdot n^4\cdot n} =\sqrt{9\cdot n^4}~\cdot ~\sqrt{5\cdot n}=\sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }\)

  4. SolomonZelman
    • one year ago
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    that is how i brake them down.

  5. SolomonZelman
    • one year ago
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    what is √9, can you tell me?

  6. anonymous
    • one year ago
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    3?

  7. SolomonZelman
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    How did we lose an n from n^5?

  9. SolomonZelman
    • one year ago
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    we didn't, all I did is used the fact that \(n^5=n^{4+1}=n^4\cdot n^1=n^4\cdot n\)

  10. SolomonZelman
    • one year ago
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    and of course, I rearranged them putting the terms that can be simplified in the beginning and then the terms that can't be simplified

  11. SolomonZelman
    • one year ago
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    \(\sqrt{n^4}\) is same as \(\sqrt[2]{n^4}\) and that is =\(n^{4/2}={\rm to~what?}\)

  12. anonymous
    • one year ago
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    Oh alright..

  13. SolomonZelman
    • one year ago
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    ok, did you get up to finding the \(\sqrt{n^4}\) ?

  14. anonymous
    • one year ago
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    Nope don't understand what's going on this time.

  15. SolomonZelman
    • one year ago
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    \(\sqrt{n^4}=\sqrt[2]{n^4}\) this is because whenever the root is not specificed, it is 2 (i.e. SQUARE root)

  16. SolomonZelman
    • one year ago
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    now, applying the rule: \(\sqrt[c]{a^b}=a^{b/c}\) we get that \(\sqrt[2]{n^4}=n^{4/2}=n^{2/1}=n^2\)

  17. anonymous
    • one year ago
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    Okay, didn't know that.

  18. SolomonZelman
    • one year ago
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    We had: \(\color{black}{ \displaystyle \sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }\) and now we had: \(\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }\)

  19. SolomonZelman
    • one year ago
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    oh, now you know.... :0

  20. anonymous
    • one year ago
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    Thanks.

  21. SolomonZelman
    • one year ago
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    \(\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }\) is just same as \(\color{black}{ \displaystyle3 n^2~\sqrt{5 n} }\) \(\bf ... \)yw

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