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anonymous

  • one year ago

help with radicals

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  1. anonymous
    • one year ago
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    |@LeibyStrauss dw:1434330140647:dw|

  2. anonymous
    • one year ago
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    |dw:1434330297233:dw|

  3. anonymous
    • one year ago
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    @LegendarySadist

  4. UsukiDoll
    • one year ago
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    \[\sqrt{588v^3w^4}\]

  5. anonymous
    • one year ago
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    @UsukiDoll yea

  6. UsukiDoll
    • one year ago
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    shall we start on the exponents? it's easier

  7. anonymous
    • one year ago
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    @UsukiDoll ok

  8. UsukiDoll
    • one year ago
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    let's start at the fact that we are just dealing with square root so anything even will be square rooted quite nicely. let's start at the \[w^4\] now.. \[\sqrt{w^4}\] can also be written in exponential form, but since we are not dealing with cube root, square root is 2 and it's not written (that's normal) in the form of |dw:1434330705533:dw| \[w^{4/2} \] so what is 4/2 for w ?

  9. anonymous
    • one year ago
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    w^2

  10. anonymous
    • one year ago
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    @UsukiDoll w^2

  11. UsukiDoll
    • one year ago
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    yay! now let's deal with \[\sqrt{v^3}\] we can actually split the exponent up using one of the exponent laws \[v^{a}v^{b}\] for \[v^3 \rightarrow v^2v\] so \[\sqrt{v^2v}\]

  12. UsukiDoll
    • one year ago
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    so which one stays in the radical? anything not divisible by 2 stays in the radical . What is the square root of v^2?

  13. UsukiDoll
    • one year ago
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    what is \[\sqrt{v^2}\] focus on that part only... we can't take square root of v

  14. anonymous
    • one year ago
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    v^2

  15. anonymous
    • one year ago
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    is it just radv?

  16. UsukiDoll
    • one year ago
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    waitttttt.... \[\sqrt{v^3} \] you an split this up \[\sqrt{v^2v}\]

  17. UsukiDoll
    • one year ago
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    you can take the square root of the v^2 but not the v. so while the only v stays in the radical, you can take the square root of v^2

  18. UsukiDoll
    • one year ago
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    \[\sqrt{588v^2v(}w^2)\] this is what we have so far.

  19. UsukiDoll
    • one year ago
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    you can split any odd exponent number up in two.. like for example \[\sqrt{y^5} \rightarrow \sqrt{y^4y} \rightarrow y^2\sqrt{y}\]

  20. anonymous
    • one year ago
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    ohhh ok makes sense

  21. UsukiDoll
    • one year ago
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    for square roots we want even exponent numbers... for cube roots we want certain exponent numbers

  22. UsukiDoll
    • one year ago
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    anyway, now can you tell me what \[\sqrt{v^2v} \] is ?

  23. anonymous
    • one year ago
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    v^2 sq v

  24. UsukiDoll
    • one year ago
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    \[\sqrt{v^3} \rightarrow \sqrt{v^2}{v} -> ?????\] \[\sqrt{v^2} \rightarrow v^{\frac{2}{2}}\]

  25. UsukiDoll
    • one year ago
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    square root of v is fine, but that v^2 isn't... I have rewritten it in exponential form what is 2/2 ?

  26. anonymous
    • one year ago
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    just 1

  27. UsukiDoll
    • one year ago
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    yeah you will have just a v

  28. UsukiDoll
    • one year ago
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    \[\sqrt{588v(}w^2v)\] v to the first power or just v

  29. UsukiDoll
    • one year ago
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    now we have this nasty big number.. we need to not only split this in two but we need the largest perfect square number

  30. UsukiDoll
    • one year ago
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    otherwise you will be taking radicals over and over again and that takes a lot of time

  31. anonymous
    • one year ago
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    @UsukiDoll ok how do we do that

  32. UsukiDoll
    • one year ago
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    perfect square numbers are like 9 16 25 4 36

  33. anonymous
    • one year ago
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    49*12

  34. UsukiDoll
    • one year ago
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    let me check that

  35. UsukiDoll
    • one year ago
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    yeah that works... but we may need to take the radical again for 12 ... eh shouldn't be too bad

  36. UsukiDoll
    • one year ago
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    \[\sqrt{49 \times 12 [v]} (w^2v)\]

  37. UsukiDoll
    • one year ago
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    what is the square root of 49 ?

  38. anonymous
    • one year ago
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    7*7 49 12 4*3

  39. UsukiDoll
    • one year ago
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    one at a time...

  40. anonymous
    • one year ago
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    7 for 49

  41. UsukiDoll
    • one year ago
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    \[\sqrt{12 [v]} (7w^2v) \rightarrow \sqrt{(4)(3)[v]} (7vw^2)\]

  42. UsukiDoll
    • one year ago
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    mhm now we already have 12 = 4 x 3 so what's the square root of 4

  43. anonymous
    • one year ago
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    2

  44. UsukiDoll
    • one year ago
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    \[\sqrt{(3)[v]} (7(2)vw^2)\] we are almost done. we just need to combine 7 and 2.

  45. UsukiDoll
    • one year ago
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    what is 7 x 2 ?

  46. anonymous
    • one year ago
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    14

  47. UsukiDoll
    • one year ago
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    \[\sqrt{3v} (14vw^2)\] yup so this is your final answer

  48. anonymous
    • one year ago
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    ok thanks usuki

  49. anonymous
    • one year ago
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    wouldnt the 14vw^2 would go outside

  50. UsukiDoll
    • one year ago
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    :) I'm just rechecking with wolfram alpha , but this time I don't agree with it . the stupid calculator didn't simplify the w^4 in the square root... it is outside... the square root is just 3v. everything in the () represented outside.

  51. anonymous
    • one year ago
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    oh ok :)

  52. anonymous
    • one year ago
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    yay! thanks!

  53. UsukiDoll
    • one year ago
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    yes what we did manually is correct... wolfram alpha is making me mad on that v^3 part . the exponent is greater than 2 and can be split up. Wolfram says leave it alone (WRONG!) but purplemath says split it up. http://www.purplemath.com/modules/radicals2.htm see the example with the 12abc

  54. UsukiDoll
    • one year ago
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    the c^3 in that example has been split up to c^2 (c) then that c is left in the radical and the square root of c^2 is taken.

  55. anonymous
    • one year ago
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    ohh

  56. anonymous
    • one year ago
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    lol im confused now

  57. UsukiDoll
    • one year ago
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    wolfram can be sketchy at times... some users on here proven that ultimate calculator to be wrong several times. It shows that not everything is reliable... just focus on the purplemath site and what we did. splitting it up was the right thing to do because that's how I remembered it a long time ago.

  58. anonymous
    • one year ago
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    ok thanks

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