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anonymous

  • 5 years ago

hello, can you explain : 3 square root x^6 divided 3 square root 8x multiplied by 3 square root x^11. ???

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  1. shadowfiend
    • 5 years ago
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    Hm. Let me see if I have this right: \[\frac{3\sqrt{x^6}}{3\sqrt{8x}} \cdot 3\sqrt{x^{11}}\] Is that what the problem looks like?

  2. anonymous
    • 5 years ago
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    the 3 sqroot x^11 is under the radicand.

  3. shadowfiend
    • 5 years ago
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    Ok, gotcha. So we have: \[\frac{3\sqrt{x^6}}{3\sqrt{8x}\cdot 3\sqrt{x^{11}}}\]

  4. shadowfiend
    • 5 years ago
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    We can combine like terms by multiplying the 3s together and moving both of the bottom roots into the same one: \[\frac{3\sqrt{x^6}}{9\sqrt{8x\cdot x^{11}}}\] We see that we can combine the x terms in the denominator: \[\frac{3\sqrt{x^6}}{9\sqrt{8x^{12}}}\]

  5. anonymous
    • 5 years ago
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    thank you so very much!

  6. shadowfiend
    • 5 years ago
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    Next, we can actually take the roots. Since we have even powers, that's easy: \[\frac{3x^3}{9x^6\sqrt{8}}\] Then we can divide properly for the powers and the 3/9: \[\frac{1}{3x^3\sqrt{8}}\]

  7. shadowfiend
    • 5 years ago
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    Finally, we can simplify \(\sqrt{8}\) by remembering that 8 is 4 * 2 and that we can take the square root of 4: \[\frac{1}{3x^3(2\sqrt{2})}\] \[\frac{1}{6x^3\sqrt{2}}\]

  8. shadowfiend
    • 5 years ago
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    Most people also prefer keeping radicals out of the denominator; to do that, we can multiply by \(\sqrt{2}/\sqrt{2}\) and move the radical to the numerator: \[\frac{1}{6x^3\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\] \[\frac{\sqrt{2}}{6x^3\cdot 2}\] \[\frac{\sqrt{2}}{12x^3}\]

  9. shadowfiend
    • 5 years ago
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    No problem, glad to help :)

  10. anonymous
    • 5 years ago
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    thanks!

  11. shadowfiend
    • 5 years ago
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    Become a fan if you want! I'm trying to become a hero!

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