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anonymous

  • one year ago

Which of the following expressions are equivalent? Justify your reasoning. Please help Thank you!! A. \[^{4\sqrt{x ^{3}}}\] B. \[\frac{ 1 }{ x ^{-1} }\] C.\[^{10}\sqrt{x ^{5}}\times x ^{4}\times x ^{2}\] D. \[x ^{\frac{ 1 }{ 3 }}\times x ^{\frac{ 1 }{ 3 }}\times x ^{\frac{ 1 }{ 3 }}\]

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  1. anonymous
    • one year ago
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    @Nnesha hey you think you can help me real quick please?

  2. Nnesha
    • one year ago
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    first one is \[\huge\rm \sqrt[4]{x^3}\] ?

  3. anonymous
    • one year ago
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    Yeah

  4. Nnesha
    • one year ago
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    alright you should know some exponent rules for that question when we multiply same bases we should `add` exponents \[\huge\rm x^m \times x^n=x^{m+n}\] \[\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }\] you can convert root to an exponent form

  5. Nnesha
    • one year ago
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    exponent if there is a negative exponent `flip` the fraction \[\huge\rm x^{-m}=\frac{ 1 }{ x^m }\]when you flip the fraction sign of the exponent would change

  6. Nnesha
    • one year ago
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    so let's do D first \[\huge\rm x^\frac{ 1 }{ 3} \times x^\frac{ 1 }{ 3 } \times x^\frac{ 1 }{ 3 }\]

  7. anonymous
    • one year ago
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    for the first one it would be changed to |dw:1442352168592:dw|

  8. Nnesha
    • one year ago
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    yes right!

  9. Nnesha
    • one year ago
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    what about B ?

  10. anonymous
    • one year ago
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    would it be |dw:1442352301079:dw|

  11. Nnesha
    • one year ago
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    perfecT!

  12. Nnesha
    • one year ago
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    can you try c ?

  13. Nnesha
    • one year ago
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    first rewrite sqrt{x^5} in exponent form

  14. Nnesha
    • one year ago
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    is it \[\huge\rm \sqrt[10]{x^5}\] ?

  15. anonymous
    • one year ago
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    yeah

  16. Nnesha
    • one year ago
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    ok

  17. anonymous
    • one year ago
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    would C. be |dw:1442352798917:dw|

  18. Nnesha
    • one year ago
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    hmm not write \[\huge\rm x^{\frac{ 5 }{ 10 }+4+2}\] when we multiply same bases we should add their exponents so 5/10+4+2 = ?

  19. Nnesha
    • one year ago
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    right* not write

  20. anonymous
    • one year ago
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    wouldn't it be 11/10

  21. Nnesha
    • one year ago
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    hmm no there is denominator so we should find common denomiantor

  22. Nnesha
    • one year ago
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    |dw:1442353149290:dw| 4+2 = 6 so 5/10+6/1

  23. Nnesha
    • one year ago
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    |dw:1442353207919:dw| it's like cross multiplication but don't forget the positive sign when we find common denominator we should multiply the `numerator` of 1st fraction by the denominator of 2nd fraction and multiply the numerator of *2nd *fraction by the denominator of first fraction

  24. Nnesha
    • one year ago
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    no common denominator is 10 so that would stay the same |dw:1442353470101:dw| like this

  25. Nnesha
    • one year ago
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    let me know if you didn't understand that :=)

  26. anonymous
    • one year ago
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    would we simplify it or that would be the end

  27. Nnesha
    • one year ago
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    simplify that

  28. anonymous
    • one year ago
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    65/10

  29. Nnesha
    • one year ago
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    reduce the fraction

  30. anonymous
    • one year ago
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    13/2

  31. anonymous
    • one year ago
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    13/2

  32. Nnesha
    • one year ago
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    yes right so x^{13}/2 \[\huge\rm x^\frac{ 1 }{ 3} \times x^\frac{ 1 }{ 3 } \times x^\frac{ 1 }{ 3 }\]

  33. Nnesha
    • one year ago
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    what about D ??^

  34. anonymous
    • one year ago
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    would it be x^1

  35. Nnesha
    • one year ago
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    yes right

  36. anonymous
    • one year ago
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    so the two equivalent ones are D. and B.

  37. Nnesha
    • one year ago
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    yes right

  38. anonymous
    • one year ago
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    thank you again so much

  39. Nnesha
    • one year ago
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    yw :=)

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