anonymous
  • anonymous
help???!?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
with what
anonymous
  • anonymous
this is it
anonymous
  • anonymous
can anyone help?

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ivettef365
  • ivettef365
Go to this website, enter the info and you get answer: http://calculator.tutorvista.com/math/597/binomial-expansion-calculator.html
anonymous
  • anonymous
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".
anonymous
  • anonymous
So, for the first term, "x^5" term, you have: (5C0)[(2x)^5][(-3y)^0] (1)(32x^5)(1) 32x^5
anonymous
  • anonymous
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.
anonymous
  • anonymous
so you have to keep going?
anonymous
  • anonymous
Of course! In my first post, you should have read that there are 5 "+" signs, giving 6 total terms.
anonymous
  • anonymous
oh!
anonymous
  • anonymous
ok give me some time to do them.. could you check them when im done?
anonymous
  • anonymous
sure, np!
anonymous
  • anonymous
sixth is -1215y^5
anonymous
  • anonymous
You have the fifth one right, but you better show me your work on the 3rd, 4th, and 6th because they are off.
anonymous
  • anonymous
I just punched it into the calc
anonymous
  • anonymous
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.
anonymous
  • anonymous
To start, what's 5C2 ?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
oh
anonymous
  • anonymous
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
anonymous
  • anonymous
so 720 (x^3)(y^2)
anonymous
  • anonymous
Yes, that is the correct 3rd term.
anonymous
  • anonymous
ok!
anonymous
  • anonymous
for the fourth what is the (5c3) and (5c5)
anonymous
  • anonymous
Review what I did for 5C2. I went through a detailed derivation for how to calculate these. Use that as a guide.
anonymous
  • anonymous
ok
anonymous
  • anonymous
hello?
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
is 6th term -2430 (y^5)
anonymous
  • anonymous
For the 6th term: (5C5)[(2x)^0][(-3y)^5] = (1)(1)(-243y^5) = -243y^5 = -243(y^5)
anonymous
  • anonymous
oh ok! I need some practice lol!
anonymous
  • anonymous
thank you!
anonymous
  • anonymous
uw!
anonymous
  • anonymous
Good luck to you in all of your studies and thx for the recognition! @lala2
anonymous
  • anonymous
thanks!

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