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tmybby

what is a simpler form for this radical expression

  • 10 months ago
  • 10 months ago

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  1. tmybby
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    • 10 months ago
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  2. satellite73
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    the square root of 16 is 4 is a start

    • 10 months ago
  3. tmybby
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    4g^5?

    • 10 months ago
  4. satellite73
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    and half of 10 is 5, giving you \[\sqrt{16g^{10}}=4g^5\]

    • 10 months ago
  5. satellite73
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    yes you got it

    • 10 months ago
  6. tmybby
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    @satellite73 ?

    • 10 months ago
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  7. satellite73
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    man they really reach for these word problems, don't they?

    • 10 months ago
  8. satellite73
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    perimeter is \[\sqrt{13}+\sqrt{13}+7\sqrt{13}+7\sqrt{13}\] add to get \(16\sqrt{13}\)

    • 10 months ago
  9. satellite73
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    \(\sqrt{20}+\sqrt{40}-\sqrt{5}\) you have to write the first two in simplest radical form do you know how to do that?

    • 10 months ago
  10. tmybby
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    no :/

    • 10 months ago
  11. satellite73
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    ok look to factor the number inside the radical as a perfect square, times something for example \(20=4\times 5\) and \(4\) is a square you get \[\sqrt{20}=\sqrt{4\times 5}=\sqrt{4}\sqrt{5}=2\sqrt{5}\] you don't need to write all those steps, that is only my explanation

    • 10 months ago
  12. satellite73
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    see if you can try that with \(\sqrt{45}\)

    • 10 months ago
  13. tmybby
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    ohhh okay!

    • 10 months ago
  14. satellite73
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    if you get stuck let me know, but it is not too bad for \(45\)

    • 10 months ago
  15. tmybby
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    \[\sqrt{45}=\sqrt{9x5}=\sqrt{9}\sqrt{5}\]?

    • 10 months ago
  16. satellite73
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    yes !

    • 10 months ago
  17. satellite73
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    so finish with \(\sqrt{9}\sqrt{5}=3\sqrt{5}\)

    • 10 months ago
  18. satellite73
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    now the whole problem is a matter of combining like terms for \[2\sqrt{5}+3\sqrt{5}-\sqrt{5}\] which is identical to \[2+3-1\] really, just stick a \(\sqrt{5}\) next to the answer

    • 10 months ago
  19. tmybby
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    \[4\sqrt{5}\]

    • 10 months ago
  20. satellite73
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    got it

    • 10 months ago
  21. satellite73
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    how about the last one \[9^{\frac{1}{3}}\times 81^{\frac{1}{3}}\]? you got that?

    • 10 months ago
  22. tmybby
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    I got 9 for the last one

    • 10 months ago
  23. satellite73
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    you win

    • 10 months ago
  24. tmybby
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    could you still help me?

    • 10 months ago
  25. satellite73
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    sure why not?

    • 10 months ago
  26. tmybby
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    • 10 months ago
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  27. satellite73
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    for a rational exponent, the power is the numerator and the root is the denominator

    • 10 months ago
  28. satellite73
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    your exponent will be negative, since the variable is in the denominator

    • 10 months ago
  29. tmybby
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    oh okay so it's C?

    • 10 months ago
  30. satellite73
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    should get \[\large 8x^{-\frac{15}{7}}\] yes, C

    • 10 months ago
  31. satellite73
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    for the second one i can't really read the exponents

    • 10 months ago
  32. tmybby
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    • 10 months ago
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  33. satellite73
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    is the exponent in the numerator a 3 or an 8 ?

    • 10 months ago
  34. tmybby
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    3

    • 10 months ago
  35. satellite73
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    oh, then \(\sqrt[3]{x^3}=x\) and you have \[\frac{x}{\sqrt[5]{x^2}}\]

    • 10 months ago
  36. satellite73
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    you can write this as \[\frac{x}{x^{\frac{2}{5}}}\] and then subtract the exponents

    • 10 months ago
  37. satellite73
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    \[\large x^{1-\frac{2}{5}}=x^{\frac{3}{5}}\]

    • 10 months ago
  38. tmybby
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    yay! :)

    • 10 months ago
  39. satellite73
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    that it?

    • 10 months ago
  40. tmybby
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    i have 3 more problems /:

    • 10 months ago
  41. satellite73
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    i am ready if you are

    • 10 months ago
  42. tmybby
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    okay hold on... :)

    • 10 months ago
  43. tmybby
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    • 10 months ago
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  44. satellite73
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    just like the previous one, multiply top and bottom by \(\sqrt{2}+\sqrt{5}\)

    • 10 months ago
  45. satellite73
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    \[\frac{\sqrt{2}+\sqrt{5}}{\sqrt2-\sqrt5}\times\frac{\sqrt{2}+\sqrt{5}}{\sqrt2+\sqrt5} \]

    • 10 months ago
  46. satellite73
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    \[\frac{2+2\sqrt{10}+5}{2-5}\] \[\frac{7+2\sqrt{10}}{-3}\]

    • 10 months ago
  47. tmybby
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  48. satellite73
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    any ideas for this one?

    • 10 months ago
  49. satellite73
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    one think should be ok, that \[\large (a^{-3})^{\frac{-2}{3}}=a^{-3\times \frac{-2}{3}}=a^2\]

    • 10 months ago
  50. satellite73
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    for \(8^{-\frac{2}{3}}\) you have an exponent with a a) three in the deminator b) 2 in the numerator and c) a minus sign they mean a) take the cube root b) square c) take the reciprocal

    • 10 months ago
  51. satellite73
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    therefore you get the cubed root of 8 is 2 2 squared is 4 the reciprocal of 4 is \(\frac{1}{4}\)

    • 10 months ago
  52. satellite73
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    final answer is therefor \[\frac{x^2}{4}\]

    • 10 months ago
  53. tmybby
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    okay! One more...

    • 10 months ago
  54. tmybby
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    • 10 months ago
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  55. satellite73
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    that is an easier one multiply top and bottom by \(\sqrt3\) to get \[\frac{5\sqrt{3}+\sqrt{6}}{3}\]

    • 10 months ago
  56. satellite73
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    typo there ! \[\frac{5\sqrt{3}-\sqrt{6}}{3}\]

    • 10 months ago
  57. tmybby
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    Thank you so much I really appreciate it!!

    • 10 months ago
  58. satellite73
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    yw

    • 10 months ago
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