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anonymous

  • 5 years ago

3nth sqrt 50x^2z^5 X 3nth sqrt 15y^3

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  1. anonymous
    • 5 years ago
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    In this case, it's not so nice because you don't have things you can easily divide. But fractional exponents are valid notationally

  2. anonymous
    • 5 years ago
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    So \[\sqrt[3]{a^2} = a^{\frac{2}{3}}\]

  3. anonymous
    • 5 years ago
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    well idk if this is right but i got 5sqrt 6yz^2

  4. anonymous
    • 5 years ago
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    Wait, the original expression you have is \[\sqrt[3]{50x^2z^5} \times \sqrt[3]{15y^3}\] ?

  5. anonymous
    • 5 years ago
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    yeah. because i broke 50 into radical 25 and and 2

  6. anonymous
    • 5 years ago
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    Ok, so yes. If you bring the insides of both together you can factor out a 5 and a z.

  7. anonymous
    • 5 years ago
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    So the only problem is that you should have a y and a z on the outside as well.

  8. anonymous
    • 5 years ago
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    \[\sqrt[3]{50*15x^2y^3z^5} = \sqrt[3]{6*5^3x^2y^3z^3z^2}\] And we take out everything with a power of 3. \[= 5yz*\sqrt[3]{6x^2z^2}\]

  9. anonymous
    • 5 years ago
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    ohh. okaiee. that makes sense. thanks once again

  10. anonymous
    • 5 years ago
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    All of this is just remembering your rules for exponents when you multiply or raise to a power. a \[a^b*a^c = a^{b+c}\] \[(a^b)^c = a^{b*c}\]

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