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anonymous

  • 4 years ago

Square root of 8y^5 X the square root of 40y^2

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  1. anonymous
    • 4 years ago
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    to multiply variables with exponents, you add the exponents together

  2. anonymous
    • 4 years ago
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    oh there's a square root

  3. anonymous
    • 4 years ago
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    \[8y^5(\sqrt{40y^2})?\]

  4. anonymous
    • 4 years ago
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    you can pull a y out of the square root because \[\sqrt{y^2}=y\]

  5. anonymous
    • 4 years ago
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    you can also pull a 2 out of the square root because of \[\sqrt{2^2*10}\]

  6. anonymous
    • 4 years ago
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    \[\sqrt{8y^{5}} * \sqrt{40y ^{2}}\]no its

  7. anonymous
    • 4 years ago
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    ok

  8. anonymous
    • 4 years ago
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    you can put it all under one square root

  9. anonymous
    • 4 years ago
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    such as this \[\sqrt{(8y^5)(40y^2)}\]

  10. anonymous
    • 4 years ago
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    then i got \[\sqrt{320y ^{7}}\]

  11. anonymous
    • 4 years ago
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    yep

  12. anonymous
    • 4 years ago
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    now you have to pull numbers and variables out

  13. anonymous
    • 4 years ago
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    as a hint, 64*5=320

  14. anonymous
    • 4 years ago
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    then change 320 to \[8\sqrt{5} \] then \[8\sqrt{5y ^{7}}\] but i dont know what to do next

  15. anonymous
    • 4 years ago
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    yes \[8\sqrt{y^7}\]

  16. anonymous
    • 4 years ago
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    im sorry \[8\sqrt{5y^7}\]

  17. anonymous
    • 4 years ago
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    now you can factor y

  18. anonymous
    • 4 years ago
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    if there was a y^2 in there, you would just take out the y, correct?

  19. anonymous
    • 4 years ago
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    into 4 and 3

  20. anonymous
    • 4 years ago
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    yeah that too

  21. anonymous
    • 4 years ago
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    so what if you factored \[y^7\] as \[y^6y\]

  22. anonymous
    • 4 years ago
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    i thought i have to factor it into a perfect square?

  23. anonymous
    • 4 years ago
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    \[y^6=(y^3)^2\], if you remember the rules of exponents

  24. anonymous
    • 4 years ago
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    so can you pull y^6 out now?

  25. anonymous
    • 4 years ago
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    \[8\sqrt{5y ^{6}}y\]

  26. anonymous
    • 4 years ago
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    yes, but now \[y^6=(y^3)^2\], which is a perfect square

  27. anonymous
    • 4 years ago
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    \[8\sqrt{5y^6y}\]

  28. anonymous
    • 4 years ago
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    \[=8\sqrt{5(y^3)^2y}\]

  29. anonymous
    • 4 years ago
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    if i said y^3=x, then you would see an x^2 inside of the square root

  30. anonymous
    • 4 years ago
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    and you'd pull the x out of \[\sqrt{x^2}\]

  31. anonymous
    • 4 years ago
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    see what i'm talking about?

  32. anonymous
    • 4 years ago
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    but then i have 8y^3 square root of 5y^3y

  33. anonymous
    • 4 years ago
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    nope, you'd pull the y^3 out of the root, leaving you with only the other y

  34. anonymous
    • 4 years ago
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    \[8\sqrt{5(y^3)^2y} = 8y^3\sqrt{5y} = \]

  35. anonymous
    • 4 years ago
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    forget that last = sign, that is it. it's done now

  36. anonymous
    • 4 years ago
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    there's no way to make perfect squares from factors of 5 or y

  37. anonymous
    • 4 years ago
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    understand that when you raise a power to a power, you multiply the exponents

  38. anonymous
    • 4 years ago
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    i kind of get it i think..i just have to have more practice because it's still kind of confusing but ill get it

  39. anonymous
    • 4 years ago
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    yeah that's all it takes is practice

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