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calculusxy

  • one year ago

Help with exponents. MEDAL!! Question posted below ...

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  1. calculusxy
    • one year ago
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    \[7v^3 \times 10u^3v^5 \times 8uv^3\]

  2. calculusxy
    • one year ago
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    @jdoe0001

  3. Mehek14
    • one year ago
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    don't you multiply \(\bf{7*10*8}\) and add the exponents?

  4. calculusxy
    • one year ago
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    Okay so if we do that then we will have \[7 \times 10 \times 8 = 560\] But how do i add the exponents?

  5. Mehek14
    • one year ago
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    \(v^3+v^5+v^3=3+5+3=11\\v^11\\u^3+u^1\\3+1=4\\u^4\)

  6. Mehek14
    • one year ago
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    \(v^3+v^5+v^3=v^{11}\)

  7. Mehek14
    • one year ago
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    add numbers keep same variable

  8. calculusxy
    • one year ago
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    So would my answer be 560v^11v^4 ?

  9. calculusxy
    • one year ago
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    @Hero

  10. Hero
    • one year ago
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    You wrote two v's your u variable is missing.

  11. calculusxy
    • one year ago
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    Sorry i meant 560v^11u^4

  12. calculusxy
    • one year ago
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    Am I correct?

  13. calculusxy
    • one year ago
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    I really need help on this.

  14. calculusxy
    • one year ago
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    @zzr0ck3r

  15. Hero
    • one year ago
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    Correct, sorry.

  16. calculusxy
    • one year ago
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    Okay thank you. So I am going to do the next one: \[\large 9xy^2 \times 9x^5y^2\]

  17. calculusxy
    • one year ago
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    \[\large 9x^1y^2 \times 9x^5y^2 = 81x^6y^4\]

  18. calculusxy
    • one year ago
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    @Hero

  19. AbdullahM
    • one year ago
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    correct

  20. calculusxy
    • one year ago
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    Can i check more with you?

  21. AbdullahM
    • one year ago
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    absolutely :)

  22. calculusxy
    • one year ago
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    \[6m^3n^3 \times 8m^2n^3 = 48m^5n^6\]

  23. AbdullahM
    • one year ago
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    correct again :)

  24. calculusxy
    • one year ago
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    \[6x^2 \times 6x^3y^4 = 36x^5y^4\]

  25. AbdullahM
    • one year ago
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    correct :)

  26. calculusxy
    • one year ago
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    \[7u^2v^5 \times 9uv^3 = 63u^3v^8\]

  27. AbdullahM
    • one year ago
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    correct (:

  28. calculusxy
    • one year ago
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    \[uv \times 4uv^5 = 4u^2v^6\]

  29. AbdullahM
    • one year ago
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    and correct once again c:

  30. calculusxy
    • one year ago
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    \[10xy^3 \times 8x^5y^3 = 80x^6y^6\]

  31. AbdullahM
    • one year ago
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    correct

  32. calculusxy
    • one year ago
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    \[3u^4v^5 \times 7u^2v^3 = 21u^6v^8\]

  33. AbdullahM
    • one year ago
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    correct :)

  34. calculusxy
    • one year ago
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    \[(2x^2)^2 = 2x^4\]

  35. calculusxy
    • one year ago
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    \[(p^4)^4 = p^16\]

  36. calculusxy
    • one year ago
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    \[(2x^2)^2 = 2x^4\]

  37. AbdullahM
    • one year ago
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    uh-oh

  38. calculusxy
    • one year ago
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    just in case if you want to ask for (p^4)^4, my answer is p^16

  39. AbdullahM
    • one year ago
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    Use the formula \(\bf\Huge (ab)^m = a^mb^m\)

  40. AbdullahM
    • one year ago
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    That one is correct For that question, in latex you have to put p^{16}

  41. calculusxy
    • one year ago
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    \[p^{16}\]

  42. AbdullahM
    • one year ago
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    But \(\bf\Large (2x^2)^2 \neq 2x^4\)

  43. AbdullahM
    • one year ago
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    So use the formula I gave you :)

  44. calculusxy
    • one year ago
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    \[\large 2^2 \times (x^2)^2\]

  45. calculusxy
    • one year ago
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    \[4 \times x^4\] ?

  46. AbdullahM
    • one year ago
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    correct :D

  47. calculusxy
    • one year ago
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    @AbdullahM I juts want to make sure a couple more if u have no problem :)

  48. AbdullahM
    • one year ago
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    I would love to :)

  49. calculusxy
    • one year ago
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    thank you!

  50. calculusxy
    • one year ago
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    \[(k^3)^4\]

  51. calculusxy
    • one year ago
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    = \[k^12 \]

  52. AbdullahM
    • one year ago
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    correct :)

  53. calculusxy
    • one year ago
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    sorry i forgot the latex

  54. calculusxy
    • one year ago
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    \[k^{12}\]

  55. calculusxy
    • one year ago
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    OH wait..

  56. calculusxy
    • one year ago
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    \[7^2 \times k^2 = 12 \times k^2\]

  57. calculusxy
    • one year ago
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    i meant 14

  58. calculusxy
    • one year ago
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    \[\large 14 \times k^2\]

  59. AbdullahM
    • one year ago
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    So the question is \(\bf\Large 7^2 \times k^2 = 14 \times k^2\)

  60. calculusxy
    • one year ago
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    yes am i correct?

  61. AbdullahM
    • one year ago
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    No... \(\bf\Large 7^2 = 7\times 7\neq 2\times 7\)

  62. calculusxy
    • one year ago
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    oh my god.. silly mistake ! sorry i meant to say 49 x k^2

  63. AbdullahM
    • one year ago
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    good :)

  64. calculusxy
    • one year ago
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    \[\large (2b^2)^4 = 2^4 \times (b^2)^4 = 16b^8 \]

  65. calculusxy
    • one year ago
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    sorry but can u just say correct or incorrect really quickly? i need to go.

  66. AbdullahM
    • one year ago
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    correct

  67. AbdullahM
    • one year ago
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    sorry, I was looking at another tab :p

  68. calculusxy
    • one year ago
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    thank you so much for ur help! i truly appreciate it :)

  69. calculusxy
    • one year ago
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    sorry one quick question

  70. AbdullahM
    • one year ago
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    sure :D

  71. calculusxy
    • one year ago
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    \[2n^4 \times 5n^4 = 10n^8\]

  72. calculusxy
    • one year ago
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    and \[2n^4 \times 6n^4 = 12n^8 \]

  73. calculusxy
    • one year ago
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    those are basically the same

  74. AbdullahM
    • one year ago
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    correct for both :D

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