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ivettef365Best ResponseYou've already chosen the best response.0
Go to this website, enter the info and you get answer: http://calculator.tutorvista.com/math/597/binomialexpansioncalculator.html
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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".
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
So, for the first term, "x^5" term, you have: (5C0)[(2x)^5][(3y)^0] (1)(32x^5)(1) 32x^5
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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.
 10 months ago

lala2Best ResponseYou've already chosen the best response.0
so you have to keep going?
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Of course! In my first post, you should have read that there are 5 "+" signs, giving 6 total terms.
 10 months ago

lala2Best ResponseYou've already chosen the best response.0
ok give me some time to do them.. could you check them when im done?
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
You have the fifth one right, but you better show me your work on the 3rd, 4th, and 6th because they are off.
 10 months ago

lala2Best ResponseYou've already chosen the best response.0
I just punched it into the calc
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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.
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
To start, what's 5C2 ?
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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.
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Yes, that is the correct 3rd term.
 10 months ago

lala2Best ResponseYou've already chosen the best response.0
for the fourth what is the (5c3) and (5c5)
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Review what I did for 5C2. I went through a detailed derivation for how to calculate these. Use that as a guide.
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
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)
 10 months ago

lala2Best ResponseYou've already chosen the best response.0
is 6th term 2430 (y^5)
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
For the 6th term: (5C5)[(2x)^0][(3y)^5] = (1)(1)(243y^5) = 243y^5 = 243(y^5)
 10 months ago

lala2Best ResponseYou've already chosen the best response.0
oh ok! I need some practice lol!
 10 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Good luck to you in all of your studies and thx for the recognition! @lala2
 10 months ago
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