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lala2

  • 2 years ago

help???!?

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  1. sana.almar
    • 2 years ago
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    with what

  2. lala2
    • 2 years ago
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    this is it

  3. lala2
    • 2 years ago
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    can anyone help?

  4. ivettef365
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    so you have to keep going?

  9. tcarroll010
    • 2 years 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
    • 2 years ago
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    oh!

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

  12. tcarroll010
    • 2 years ago
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    sure, np!

  13. lala2
    • 2 years ago
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    sixth is -1215y^5

  14. tcarroll010
    • 2 years 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
    • 2 years ago
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    I just punched it into the calc

  16. tcarroll010
    • 2 years 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
    • 2 years ago
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    To start, what's 5C2 ?

  18. tcarroll010
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    oh

  21. tcarroll010
    • 2 years 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
    • 2 years ago
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    so 720 (x^3)(y^2)

  23. tcarroll010
    • 2 years ago
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    Yes, that is the correct 3rd term.

  24. lala2
    • 2 years ago
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    ok!

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

  26. tcarroll010
    • 2 years 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
    • 2 years ago
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    ok

  28. lala2
    • 2 years ago
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    hello?

  29. tcarroll010
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    is 6th term -2430 (y^5)

  32. tcarroll010
    • 2 years 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
    • 2 years ago
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    oh ok! I need some practice lol!

  34. lala2
    • 2 years ago
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    thank you!

  35. tcarroll010
    • 2 years ago
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    uw!

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

  37. lala2
    • 2 years ago
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    thanks!

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