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anonymous
 3 years ago
help???!?
anonymous
 3 years ago
help???!?

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ivettef365
 3 years ago
Best ResponseYou've already chosen the best response.0Go to this website, enter the info and you get answer: http://calculator.tutorvista.com/math/597/binomialexpansioncalculator.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For (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".

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, for the first term, "x^5" term, you have: (5C0)[(2x)^5][(3y)^0] (1)(32x^5)(1) 32x^5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you have to keep going?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Of course! In my first post, you should have read that there are 5 "+" signs, giving 6 total terms.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok give me some time to do them.. could you check them when im done?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You have the fifth one right, but you better show me your work on the 3rd, 4th, and 6th because they are off.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I just punched it into the calc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep. 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To start, what's 5C2 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There'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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, that is the correct 3rd term.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for the fourth what is the (5c3) and (5c5)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Review what I did for 5C2. I went through a detailed derivation for how to calculate these. Use that as a guide.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05C3 = (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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is why you have to write: (3y)^3 instead of 3y^3 Don't forget your signs. It's: 1080(x^2)(y^3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is 6th term 2430 (y^5)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the 6th term: (5C5)[(2x)^0][(3y)^5] = (1)(1)(243y^5) = 243y^5 = 243(y^5)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok! I need some practice lol!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Good luck to you in all of your studies and thx for the recognition! @lala2
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