anonymous
  • anonymous
hello, can you explain : 3 square root x^6 divided 3 square root 8x multiplied by 3 square root x^11. ???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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shadowfiend
  • shadowfiend
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?
anonymous
  • anonymous
the 3 sqroot x^11 is under the radicand.
shadowfiend
  • shadowfiend
Ok, gotcha. So we have: \[\frac{3\sqrt{x^6}}{3\sqrt{8x}\cdot 3\sqrt{x^{11}}}\]

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shadowfiend
  • shadowfiend
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: \[\frac{3\sqrt{x^6}}{9\sqrt{8x^{12}}}\]
anonymous
  • anonymous
thank you so very much!
shadowfiend
  • shadowfiend
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: \[\frac{1}{3x^3\sqrt{8}}\]
shadowfiend
  • shadowfiend
Finally, we can simplify \(\sqrt{8}\) by remembering that 8 is 4 * 2 and that we can take the square root of 4: \[\frac{1}{3x^3(2\sqrt{2})}\] \[\frac{1}{6x^3\sqrt{2}}\]
shadowfiend
  • shadowfiend
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: \[\frac{1}{6x^3\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\] \[\frac{\sqrt{2}}{6x^3\cdot 2}\] \[\frac{\sqrt{2}}{12x^3}\]
shadowfiend
  • shadowfiend
No problem, glad to help :)
anonymous
  • anonymous
thanks!
shadowfiend
  • shadowfiend
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