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abbie1 Group Title

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  • 2 years ago
  • 2 years ago

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  1. freewilly922 Group Title
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    Convert them to exponents and then divide. \[\sqrt[n]{a} = a^{1/n}\]

    • 2 years ago
  2. freewilly922 Group Title
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    So for example \[\sqrt[4]{x^3}\text{ divided by }\sqrt[3]{x^2} \] would be \[\frac{x^{\frac{3}{4}}}{x^{\frac{2}{3}}}\] or \[x^{\frac{3}{4}-\frac{2}{3}}\]

    • 2 years ago
  3. freewilly922 Group Title
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    Which would then be \[x^{\frac{1}{12}}=\sqrt[12]{x}\]

    • 2 years ago
  4. freewilly922 Group Title
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    Sometimes the test wants you to give you answer in terms of a radical, then you might have to convert it back. If they will accept the exponent version, then that is the way to go.

    • 2 years ago
  5. freewilly922 Group Title
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    yes, what is \[\frac{x^{5/12}}{x^{1/6}}\] Just because they want it in radical form as the answer doesn't mean that you can't do all the work in exponent form and then reconvert. \[\frac{x^a}{x^b}=x^{a-b}\] This seems easier to me than trying to figure out the radical forms.

    • 2 years ago
  6. freewilly922 Group Title
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    Um. No. What is \[\frac{5}{12}-\frac{1}{6}\]?

    • 2 years ago
  7. freewilly922 Group Title
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    You are forgetting to put them into common denoms. 5\12 - 1\6 = 5\12 -2\12 = 3\12 = 1\4

    • 2 years ago
  8. freewilly922 Group Title
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    When you divide same base different exponents you subtract the exponents from one another. \[\frac{aaaaa}{aa}=\frac{a^5}{a^2}\] \[\frac{aaa}{1}\frac{aa}{aa}=aaa = a^3= a^{5-2}=a^3\]

    • 2 years ago
  9. freewilly922 Group Title
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    No, x^2/3 is wrong. Sorry. If you want to do it in radical form you can. \[\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}\] can also be done realizing that \[\sqrt[5]{a} = \sqrt[10]{a^2}\] so you can do it that way. You need to convert the index 6 to a twelve

    • 2 years ago
  10. freewilly922 Group Title
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    Sure.

    • 2 years ago
  11. gohangoku58 Group Title
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    index 12√x^5 x index 6√x doesn't it mean, \(\huge \sqrt[12]{x^5} \times\sqrt[6]x\)

    • 2 years ago
  12. freewilly922 Group Title
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    It does say to divide the expressions though. \[\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}\to\frac{\sqrt[12]{x^5}}{\sqrt[12]{x^2}}\to \] \[\sqrt[12]{\frac{x^5}{x^2}}\to\sqrt[12]{x^{5-2}}\to\sqrt[12]{x^3}\to \sqrt[4]{\sqrt[3]{x^3}}\to\sqrt[4]{x}\]

    • 2 years ago
  13. freewilly922 Group Title
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    The main methods you need to know is that 1) \[\sqrt[5]{x} = \sqrt[10]{x^2}\] 2) \[\frac{\sqrt[12]{a}}{\sqrt[12]{b}}=\sqrt[12]{\frac{a}{b}}\] Yes the fourth root of x is the answer in radical form.

    • 2 years ago
  14. freewilly922 Group Title
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    The reason why I suggested immediately to use your exponent conversion is that, if this is algII, you probably learned the exponent rules prior to this and can then use them rather than memorize another set of rules for radicals.

    • 2 years ago
  15. freewilly922 Group Title
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    for questions like these, you need to be able to write down the 4 or so rules of exponents and the radical methods as well. Good luck and don't give up. exponents give a lot of people confusion.

    • 2 years ago
  16. gohangoku58 Group Title
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    i hope to question was avtually to divide... its bit unclear....

    • 2 years ago
  17. gohangoku58 Group Title
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    *actually

    • 2 years ago
  18. freewilly922 Group Title
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    yes it was. I am working on the fact that it said to divide the two expressions.

    • 2 years ago
  19. gohangoku58 Group Title
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    the 'x' in middle is throwing me off...

    • 2 years ago
  20. gohangoku58 Group Title
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    can u confirm the question abbie ?

    • 2 years ago
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