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sana.almar
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with what
lala2
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this is it
lala2
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can anyone help?
tcarroll010
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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".
tcarroll010
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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
tcarroll010
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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.
lala2
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so you have to keep going?
tcarroll010
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Of course! In my first post, you should have read that there are 5 "+" signs, giving 6 total terms.
lala2
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oh!
lala2
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ok give me some time to do them.. could you check them when im done?
tcarroll010
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sure, np!
lala2
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sixth is
-1215y^5
tcarroll010
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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.
lala2
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I just punched it into the calc
tcarroll010
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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.
tcarroll010
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To start, what's 5C2 ?
tcarroll010
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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.
tcarroll010
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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
lala2
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oh
tcarroll010
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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
lala2
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so 720 (x^3)(y^2)
tcarroll010
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Yes, that is the correct 3rd term.
lala2
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ok!
lala2
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for the fourth what is the (5c3) and (5c5)
tcarroll010
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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.
lala2
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ok
lala2
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hello?
tcarroll010
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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
tcarroll010
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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)
lala2
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is 6th term -2430 (y^5)
tcarroll010
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For the 6th term:
(5C5)[(2x)^0][(-3y)^5] =
(1)(1)(-243y^5) =
-243y^5 =
-243(y^5)
lala2
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oh ok! I need some practice lol!
lala2
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thank you!
tcarroll010
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uw!
tcarroll010
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Good luck to you in all of your studies and thx for the recognition! @lala2
lala2
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thanks!