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lala2

  • one year ago

help???!?

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  1. sana.almar
    • one year ago
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    with what

  2. lala2
    • one year ago
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    this is it

  3. lala2
    • one year ago
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    can anyone help?

  4. ivettef365
    • one year ago
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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

  5. tcarroll010
    • one year ago
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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".

  6. tcarroll010
    • one year ago
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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

  7. tcarroll010
    • one year ago
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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.

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

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

  10. lala2
    • one year ago
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    oh!

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

  12. tcarroll010
    • one year ago
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    sure, np!

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

  14. tcarroll010
    • one year ago
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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.

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

  16. tcarroll010
    • one year ago
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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.

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

  18. tcarroll010
    • one year ago
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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.

  19. tcarroll010
    • one year ago
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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

  20. lala2
    • one year ago
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    oh

  21. tcarroll010
    • one year ago
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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

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

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

  24. lala2
    • one year ago
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    ok!

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

  26. tcarroll010
    • one year ago
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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.

  27. lala2
    • one year ago
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    ok

  28. lala2
    • one year ago
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    hello?

  29. tcarroll010
    • one year ago
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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

  30. tcarroll010
    • one year ago
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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)

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

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

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

  34. lala2
    • one year ago
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    thank you!

  35. tcarroll010
    • one year ago
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    uw!

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

  37. lala2
    • one year ago
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    thanks!

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