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calculusxy

  • one year ago

Help with exponents! @hartnn

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  1. calculusxy
    • one year ago
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    \[\huge (2v)^2 \times 2v^2\]

  2. calculusxy
    • one year ago
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    \[\large (2v)^2 = 2^2 \times v^2 = 4v^2\] \[\large 4v^2 \times 2v^2 = 8v^2 \] or does it equal to\[\large 8v^4\]

  3. Nnesha
    • one year ago
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    \[\huge\rm (ab)^m =a^m b^m\] apply this exponent first both number in the parentheses are raising to the m power

  4. Nnesha
    • one year ago
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    looks good!

  5. calculusxy
    • one year ago
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    So which one does it equal to?

  6. calculusxy
    • one year ago
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    \[\large 8v^2 \] or \[\large 8v^4\]

  7. Nnesha
    • one year ago
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    ohh i see well when we multiply same bases we should `ADD` their exponents

  8. calculusxy
    • one year ago
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    oh okay so it is \[8v^4 \] right?

  9. Nnesha
    • one year ago
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    remember it's not `combine like terms` \[2x+3x=(2+3)x\] when we add/subtract like terms variable stay the same but when we multiply them we should add their exponents

  10. Nnesha
    • one year ago
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    yes right

  11. calculusxy
    • one year ago
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    I have another question @Nnesha

  12. Nnesha
    • one year ago
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    okay :=)

  13. Nnesha
    • one year ago
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    when you `ADD or subtract ` like terms u just have to deal with the coefficients like \[2x+3x=(2+3)x\] but when we multiply same bases we should `add` their exponents and multiply the coefficient\[\huge\rm \color{Red}{1}x^m · \color{blue}{1}x^n=(\color{red}{1} ·\color{blue}{1})x^{m+n}\]

  14. calculusxy
    • one year ago
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    \[\huge \frac{ 2x^2y^4 \times 4x^2y^4 \times 3x }{ 3x^{-3}y^2 }\]

  15. Nnesha
    • one year ago
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    ayoooXD

  16. anonymous
    • one year ago
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    I guess you forget what he/she did there lol

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  17. Nnesha
    • one year ago
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    multiply the coefficients and add the exponent of the same base

  18. Nnesha
    • one year ago
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    \[\huge \frac{ 2x^2y^4 \times 4x^2y^4 \times 3x }{ 3x^{-3}y^2 }\] can be written as \[\frac{ (2·4·3)(x^2·x^2·x)(y^4·y^4) }{3x^{-3}y^2 }\]

  19. Nnesha
    • one year ago
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    and x is same as x^1

  20. calculusxy
    • one year ago
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    \[\large \frac{ 16x^5y^8 }{3x^{-3}y^2 } = \frac{ 16x^8y^6 }{ 3 }\]

  21. Nnesha
    • one year ago
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    (2 times 3 times 4) isn't equal to 16 :=) x^8 and y^6 is correct

  22. calculusxy
    • one year ago
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    sorry 24

  23. Nnesha
    • one year ago
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    yes 24/3 simplify done!

  24. calculusxy
    • one year ago
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    8

  25. Nnesha
    • one year ago
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    that's it great job!

  26. calculusxy
    • one year ago
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    Thank you! If i need more help can i mention you...?

  27. Nnesha
    • one year ago
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    sure!

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