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ivettef365
 one year 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

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1For (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".

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1So, for the first term, "x^5" term, you have: (5C0)[(2x)^5][(3y)^0] (1)(32x^5)(1) 32x^5

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1For 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.

lala2
 one year ago
Best ResponseYou've already chosen the best response.0so you have to keep going?

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1Of course! In my first post, you should have read that there are 5 "+" signs, giving 6 total terms.

lala2
 one year ago
Best ResponseYou've already chosen the best response.0ok give me some time to do them.. could you check them when im done?

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1You have the fifth one right, but you better show me your work on the 3rd, 4th, and 6th because they are off.

lala2
 one year ago
Best ResponseYou've already chosen the best response.0I just punched it into the calc

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1yep. 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.

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1To start, what's 5C2 ?

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1Then 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.

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1There'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

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1And 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

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1Yes, that is the correct 3rd term.

lala2
 one year ago
Best ResponseYou've already chosen the best response.0for the fourth what is the (5c3) and (5c5)

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1Review what I did for 5C2. I went through a detailed derivation for how to calculate these. Use that as a guide.

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.15C3 = (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

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1This is why you have to write: (3y)^3 instead of 3y^3 Don't forget your signs. It's: 1080(x^2)(y^3)

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1For the 6th term: (5C5)[(2x)^0][(3y)^5] = (1)(1)(243y^5) = 243y^5 = 243(y^5)

lala2
 one year ago
Best ResponseYou've already chosen the best response.0oh ok! I need some practice lol!

tcarroll010
 one year ago
Best ResponseYou've already chosen the best response.1Good luck to you in all of your studies and thx for the recognition! @lala2
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