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lala2 Group Title

help???!?

  • one year ago
  • one year ago

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  1. sana.almar Group Title
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    with what

    • one year ago
  2. lala2 Group Title
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    this is it

    • one year ago
  3. lala2 Group Title
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    can anyone help?

    • one year ago
  4. ivettef365 Group Title
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    Go to this website, enter the info and you get answer: http://calculator.tutorvista.com/math/597/binomial-expansion-calculator.html

    • one year ago
  5. tcarroll010 Group Title
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    For (a + b)^5, you would have: (5C0)(a^5)(b^0) + (5C1)(a^4)(b^1) + (5C2)(a^3)(b^2) + (5C3)(a^2)(b^3) + (5C4)(a^1)(b^4) + (5C5)(a^0)(b^5) Here, just substitute 2x for "a" and -3y for "b".

    • one year ago
  6. tcarroll010 Group Title
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    So, for the first term, "x^5" term, you have: (5C0)[(2x)^5][(-3y)^0] (1)(32x^5)(1) 32x^5

    • one year ago
  7. tcarroll010 Group Title
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    For the second term, the "x^4" term, you have: (5C1)[(2x)^4][(-3y)^1] (5)(16x^4)(-3y) -240(x^4)(y) Now, you can use these 2 examples and the general formula to derive the other 4 terms. Just use what I did as a guide.

    • one year ago
  8. lala2 Group Title
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    so you have to keep going?

    • one year ago
  9. tcarroll010 Group Title
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    Of course! In my first post, you should have read that there are 5 "+" signs, giving 6 total terms.

    • one year ago
  10. lala2 Group Title
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    oh!

    • one year ago
  11. lala2 Group Title
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    ok give me some time to do them.. could you check them when im done?

    • one year ago
  12. tcarroll010 Group Title
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    sure, np!

    • one year ago
  13. lala2 Group Title
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    sixth is -1215y^5

    • one year ago
  14. tcarroll010 Group Title
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    You have the fifth one right, but you better show me your work on the 3rd, 4th, and 6th because they are off.

    • one year ago
  15. lala2 Group Title
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    I just punched it into the calc

    • one year ago
  16. tcarroll010 Group Title
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    yep. I could just give you the answer, but that's not going to help you or anyone at all. Better if you show me what you did and I? can correct where you went wrong. You might want to start with the 3rd. All calculation, now.

    • one year ago
  17. tcarroll010 Group Title
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    To start, what's 5C2 ?

    • one year ago
  18. tcarroll010 Group Title
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    Then what's (2x)^3 ? Finally, what's (-3y)^2 ? You take the product of all 3 (that's including that 5C2) and that's your 3rd term.

    • one year ago
  19. tcarroll010 Group Title
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    There's your mistake: 5C2 is not 5, it's 10 5C2 = (5!) / [(3!)(2!)] = (5 x 4 x 3 x 2 x 1) / [(3 x 2 x 1)(2 x 1)] = (5 x 4) / (2 x 1) = 10

    • one year ago
  20. lala2 Group Title
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    oh

    • one year ago
  21. tcarroll010 Group Title
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    And you definitely should get into the extremely necessary habit of writing: (2x)^3 not 2x^3 The first is 8x^3 the second is 2x^3

    • one year ago
  22. lala2 Group Title
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    so 720 (x^3)(y^2)

    • one year ago
  23. tcarroll010 Group Title
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    Yes, that is the correct 3rd term.

    • one year ago
  24. lala2 Group Title
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    ok!

    • one year ago
  25. lala2 Group Title
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    for the fourth what is the (5c3) and (5c5)

    • one year ago
  26. tcarroll010 Group Title
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    Review what I did for 5C2. I went through a detailed derivation for how to calculate these. Use that as a guide.

    • one year ago
  27. lala2 Group Title
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    ok

    • one year ago
  28. lala2 Group Title
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    hello?

    • one year ago
  29. tcarroll010 Group Title
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    5C3 = (5!) / [(2!)(3!)] = (5 x 4 x 3 x 2 x 1) / [(2 x 1)(3 x 2 x 1)] = (5 x 4) / (2 x 1) = 10 It's the same as 5C2

    • one year ago
  30. tcarroll010 Group Title
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    This is why you have to write: (-3y)^3 instead of -3y^3 Don't forget your signs. It's: -1080(x^2)(y^3)

    • one year ago
  31. lala2 Group Title
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    is 6th term -2430 (y^5)

    • one year ago
  32. tcarroll010 Group Title
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    For the 6th term: (5C5)[(2x)^0][(-3y)^5] = (1)(1)(-243y^5) = -243y^5 = -243(y^5)

    • one year ago
  33. lala2 Group Title
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    oh ok! I need some practice lol!

    • one year ago
  34. lala2 Group Title
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    thank you!

    • one year ago
  35. tcarroll010 Group Title
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    uw!

    • one year ago
  36. tcarroll010 Group Title
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    Good luck to you in all of your studies and thx for the recognition! @lala2

    • one year ago
  37. lala2 Group Title
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    thanks!

    • one year ago
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