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anonymous

  • 5 years ago

Help with rationalizing denominators equation is 1 over the square root 96 not sure what to do

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  1. anonymous
    • 5 years ago
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    \[1 FRACTION \sqrt{96}\]

  2. anonymous
    • 5 years ago
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    \[1/\sqrt{96} \times \sqrt{96}/\sqrt{96} = \sqrt{96}/9\]

  3. anonymous
    • 5 years ago
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    \[1/\sqrt{96}\]\[\sqrt{96}/(\sqrt{96}*\sqrt{96})\]\[\sqrt{96}/96\]

  4. anonymous
    • 5 years ago
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    I understand I have to write the smallest number of a perfect square, but unsure of how to water that down. haha

  5. anonymous
    • 5 years ago
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    1 is the numerator. 96 is the denominator

  6. anonymous
    • 5 years ago
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    96|2 48|2 24|2 12|2 06|2 03|3 01|- \[\sqrt{2^{5}*3}=4\sqrt{2*3}=4\sqrt{?}\]

  7. anonymous
    • 5 years ago
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    2?

  8. anonymous
    • 5 years ago
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    \[4\sqrt{6}/96 = \sqrt{6}/\]

  9. anonymous
    • 5 years ago
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    \[\sqrt{6}/24 \]

  10. anonymous
    • 5 years ago
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    would you mind explaining it?

  11. anonymous
    • 5 years ago
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    ?

  12. anonymous
    • 5 years ago
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    in english haha

  13. anonymous
    • 5 years ago
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    to simplify a sqrt, you devide the number by primary numbers: 2,3,5,7,11,13,... until you get 1. The amount of primary is the amount of times you are multiplying it so as i did: 96|2 96/2 = 48 48|2 48/2 = 24 24|2 24/2 = 12 12|2 12/2 = 06 06|2 06/2 = 03 03|3 i can't devide 3 by 2 to get a non-decimal number so, i devide by the next primary which is 3 -> 03/3 = 1 01|- I have five 2's and one 3 therefore, 2^5*3 = 96

  14. anonymous
    • 5 years ago
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    \[2^{5}*3^{1} = 2^{2}*2^{2}*2^{1}*3^{1} = 96\]

  15. anonymous
    • 5 years ago
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    I somewhat get it a bit better.

  16. anonymous
    • 5 years ago
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    Times both the numerator and denomiator by \[\sqrt{96}\] to get \[1*\sqrt{96}/\sqrt{96}*\sqrt{96}\] Then the denominator becomes 96 and the numerator becomes \[\sqrt{96}\] You then can simplify the numerator; \[\sqrt{96} = \sqrt{16*64} =4\sqrt{6}\] As now both the numerator and denominator are divisible by 4 we can divide them both to make \[\sqrt{6}/24\]

  17. anonymous
    • 5 years ago
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    what should the answer look like?

  18. anonymous
    • 5 years ago
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    ah, I got it. I am just hazy on the whole find 1 thing that you mentioned earlier. how did you determine 16 and 64 for the square roots of 96??

  19. anonymous
    • 5 years ago
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    The answer is the smallest surd you can get ; \[\sqrt{6}\] over the smallest integer you can get; 24 which equates to \[\sqrt{6}/24\] ^^^ that's your answer..

  20. anonymous
    • 5 years ago
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    got that. so I have 10 over square root of 45 first step, mult 45 by 45

  21. anonymous
    • 5 years ago
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    OH DEAR>> I meant 16*6 not *64.. my computer is very slow so i can't see my errors in typing

  22. anonymous
    • 5 years ago
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    so is it that 6 is the smallest square root, and 16 is because it goes into 96 6 times--leaving a relation between 6 and 16?

  23. anonymous
    • 5 years ago
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    yeah :)

  24. anonymous
    • 5 years ago
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    ohhhkay cool. thanks dude!

  25. anonymous
    • 5 years ago
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    what if it has a factored numerator?such as \[\sqrt{6}/\sqrt{28}\]

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