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anonymous

  • one year ago

HELP PLEASE!!!!!! I know the answer I just need to know about the exponents...

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  1. anonymous
    • one year ago
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    \[\sqrt{72x ^{3}y ^{16}}\]

  2. Nnesha
    • one year ago
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    you can convert square root of something to 1/2 exponents right \[\huge\rm \sqrt{x} =x^\frac{ 1 }{ 2 }\] what are the factors of 75 ?

  3. Nnesha
    • one year ago
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    x^3 can be written as x^2 times x \[\huge\rm \sqrt{72 · x^2 · x· y^{16}} = \sqrt{72x} · x^\frac{ 2 }{ 2 } y^\frac{ 16 }{ 2 }\]

  4. Nnesha
    • one year ago
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    let me know if you have question about that^^^^

  5. anonymous
    • one year ago
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    why do you separate the x^2 and x

  6. Nnesha
    • one year ago
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    so i can take square root of x^2 we have to take out the variables from the square root or another way of taking square root of x^3 would |dw:1441409742550:dw|

  7. anonymous
    • one year ago
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    and the Y too can you explain the Y why it goes outside

  8. Nnesha
    • one year ago
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    how many times does 2 goes into 3 ? one time right and remainder would be one |dw:1441409878485:dw|

  9. Nnesha
    • one year ago
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    like i said we can convert root to exponent form \[\huge\rm \sqrt[n]{x^m} = x^\frac{ m }{ n }\]

  10. anonymous
    • one year ago
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    yea

  11. anonymous
    • one year ago
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    what about the y

  12. Nnesha
    • one year ago
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    16 is divisible by 2 so i just convert square root to 1/2 exponent form |dw:1441410065097:dw|

  13. anonymous
    • one year ago
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    so if it can be divided it's automatically on the outside of the root?

  14. phi
    • one year ago
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    the easiest way to do these is as Nnesha showed. change square root to "exponent form" \[ \sqrt{y^{16} }= \left(y^{16}\right)^\frac{1}{2}\] now use the rule \( (a^b)^c= a^{bc} \) to write that as \(y^{16\cdot \frac{1}{2}}= y^8\) notice there is nothing strange (no square roots) it is just y^8

  15. anonymous
    • one year ago
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    yea

  16. Nnesha
    • one year ago
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    yea here are some example \[\huge\rm \sqrt{a^8} = a^\frac{ 8 }{ 2 }=y^{4} \] \[\huge\rm \sqrt[3]{y^9} =y^\frac{ 9 }{ 3 }=y^{3}\] of the exponent of the variable not divisible by the root like for x^3 you can still write it as \[\huge\rm \sqrt{x^3} =x^\frac{ 3 }{ 2 }\] but the remainder is one so that's mean one x would remain under the root hope it makes sense

  17. Nnesha
    • one year ago
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    if*

  18. phi
    • one year ago
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    the other way to think of it is \[ \sqrt{x \cdot x} = x\] if instead of x we had y^8 \[ \sqrt{y^8 \cdot y^8} = y^8\] that is the same pattern (and of course you know inside the square root y^8 * y^8 is y^16 so \[ \sqrt{y^{16}}=\sqrt{y^8 \cdot y^8} = y^8\]

  19. anonymous
    • one year ago
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    okay so i guess i was confused about the 16 cuz i thought it would be a y^4 because 4^2=16 but I think i got it now

  20. phi
    • one year ago
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    yes, it can get confusing . the 16 is an exponent. that is different from an ordinary number 16. if you get confused , remember y^16 is short for y*y*y*y..*y (y times itself 16 times) the square root of that is 1/2 of those y's or y*y*y*y*y*y*y*y (y times itself 8 times)

  21. anonymous
    • one year ago
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    got it now thanks phi!!

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