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AngelaB97

  • one year ago

what is 6 --- ^3√4

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  1. AngelaB97
    • one year ago
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    what is |dw:1441484588912:dw|

  2. AngelaB97
    • one year ago
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    how do you rationalize it?

  3. triciaal
    • one year ago
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    |dw:1441486622928:dw|

  4. triciaal
    • one year ago
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    to rationalize the denominator means to get rid of the radical. multiply by 1 expressed as a fraction

  5. AngelaB97
    • one year ago
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    before you continue can i just ask is there some sort of rule for what you did when you crossed out those fractions. Meaning, could i be able to do that with any sort of radical that i want to rationalize?

  6. triciaal
    • one year ago
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    |dw:1441486781234:dw|

  7. triciaal
    • one year ago
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    rules for exponents the nth root is the same as a fractional exponent.

  8. AngelaB97
    • one year ago
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    the answer is supposed to be |dw:1441486876177:dw|

  9. Excalibur0126
    • one year ago
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    Or in decimal form, 3.7797631...

  10. triciaal
    • one year ago
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    |dw:1441486898347:dw|

  11. triciaal
    • one year ago
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    give me a minute

  12. AngelaB97
    • one year ago
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    sure

  13. triciaal
    • one year ago
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    |dw:1441487116337:dw||dw:1441487355866:dw|

  14. triciaal
    • one year ago
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    I think I messed up on something @zepdrix please correct

  15. zepdrix
    • one year ago
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    Hey Angela :) In order to rationalize this thing, you'd like the power on the 4 to be a 1. I like the first step that Tricia applied, writing the 4 with a rational exponent.\[\large\rm \frac{6}{\sqrt[3]{4}}=\frac{6}{4^{1/3}}\]So we don't want a 1/3 power, we want a 1 power down there. What do we have to add to 1/3 to get 1?

  16. AngelaB97
    • one year ago
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    hey :)) we have to add 2/3

  17. zepdrix
    • one year ago
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    Good good good. We would like 2/3 exponent, we'll leave the base the same. So we actually want to multiply top and bottom by \(\large\rm 4^{2/3}\) Our rules of exponents will give us \(\large\rm 4^{1/3}\cdot4^{2/3}=4^{1/3+2/3}\) in the denominator!

  18. triciaal
    • one year ago
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    @zepdrix thanks

  19. zepdrix
    • one year ago
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    \[\large\rm \frac{6}{4^{1/3}}\left(\frac{4^{2/3}}{4^{2/3}}\right)=\frac{6\cdot4^{2/3}}{4^{3/3}}\]Ok with that step Angela? :o

  20. triciaal
    • one year ago
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    |dw:1441488139781:dw|

  21. AngelaB97
    • one year ago
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    6 * 42/3 divided by 4

  22. zepdrix
    • one year ago
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    \[\large\rm \frac{\color{orangered}{6}\cdot 4^{2/3}}{\color{orangered}{4}}\]Mmmm k good. This can be simplified a little bit further since 6 and 4 share a common factor.

  23. AngelaB97
    • one year ago
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    so on the bottom we get 2 and the 6 turns into a 3?

  24. zepdrix
    • one year ago
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    \[\large\rm \frac{3\cdot4^{2/3}}{2}\]Yayyyy good job \c:/

  25. AngelaB97
    • one year ago
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    so we don't need to do any further simplifying? because the answer in my book shows |dw:1441488618998:dw| @zepdrix

  26. zepdrix
    • one year ago
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    OH interesting :O Ok sec I think about it

  27. zepdrix
    • one year ago
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    Ok maybe this is a better route to take then..\[\large\rm \frac{6}{4^{1/3}}=\frac{3\cdot2}{4^{1/3}}=\frac{3\cdot2}{(2^2)^{1/3}}=\frac{3\cdot2}{2^{2/3}}\]

  28. zepdrix
    • one year ago
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    I'm rewriting the 6 as 3*2. I'm rewriting 4 as 2 squared. and then applying exponent rule, when we have a power and a power like that, we multiply. So 2 times 1/3 gave me 2/3 for the power on the 2.

  29. zepdrix
    • one year ago
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    \[\large\rm =\frac{3\cdot2^1}{2^{2/3}}\]And now we can apply one of our exponent rules from here: \(\large\rm \frac{x^{a}}{x^b}=x^{a-b}\) Do you see how we can apply that to the 2's?

  30. AngelaB97
    • one year ago
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    yes thank you very much

  31. triciaal
    • one year ago
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    @AngelaB97 sorry if we confuse you but hopefully you understand more about the rules of exponents and how to rationalize it is very easy to make mistakes if not careful

  32. AngelaB97
    • one year ago
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    @zepdrix @triciaal thank you both for your help!

  33. zepdrix
    • one year ago
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    yay team \c:/

  34. triciaal
    • one year ago
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    welcome anytime

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