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 Posted: Wed Sep 02, 2009 6:11 pm
Yes, that's what I mean ^^"
I kind of get it...the only thing I'm confuzzled about really is about the exponents and the number besides the value that replaces x...
In that problem here's what I mean:
g(m+1)= 2(m + 1)² - 4(m + 1) + 2
The italicized part is what confuzzles me...because aren't we supposed to do the exponents first?
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Posted: Wed Sep 02, 2009 6:17 pm
You use "FOIL" to spread it out.
(m + 1)² = (m + 1)*(m + 1)
First you multiply the first two terms, m and m, to get m².
Then you do the outermost terms, 1 and m, to get m.
Then you multiply the inside terms, 1 and m as well, to get m... again.
Then you multiply the last two terms, 1 and 1, to get 1.
So you get
(m + 1)² = m² + m + m + 1
Which you simplify by making m + m = 2m.
Then multiply each of those trms by two.
First Oustide Inside Last. FOIL. ^^
Or you could learn... I think it was Pascal's Triangle? But that's Calculus and only works for exponents.
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Posted: Wed Sep 02, 2009 6:32 pm
Kaddy knows about foil and she was trying to use that...it's the 2 on the outside that kind of bothers me >.<" But I kind of get it nows...^^"
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Posted: Wed Sep 02, 2009 6:34 pm
Yep you just multiply each part by two. so 2(m²) + 2(2m) + 2(1). *nods*
Glad to be of help. ^^
Anyways... what's up? ^^
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Posted: Wed Sep 02, 2009 6:41 pm
Not much...nows...the second problem confuzzles Kaddy more...since it's cubed instead of squared...
...Kaddy is terrible at math...
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Posted: Wed Sep 02, 2009 6:44 pm
Yeah, but it's a lot simpler dear...
j(2a)= 1 - 4(2a)³
So it's just 2³ a³
Which is 1 - 8a³ . -nods-
Well you can always ask Eno if you need help. ^^
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Posted: Wed Sep 02, 2009 6:48 pm
...I get it! xD Thanks, Brother. -huggles- I'm am so glad...I've got you to help me >w< No one else understood what the Hell I was talking about...
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Posted: Wed Sep 02, 2009 6:59 pm
I'm glad I could help ^^
Eno was good at math. To a point.
He got a 4 on his AP Calc BC exam. =u=
And a 5 on the AB part.
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Posted: Wed Sep 02, 2009 7:05 pm
...then Brother is Kaddy's new math tutor. -nods- I've decided that xD
Okays, so...tell Kaddy if she's right for the third problem.
f(n-1) =2(n-1)²-(n-1)+9
=2(n-1)(n-1)-(n-1)+9
=2(n²-n-n+1)-(n-1)+9
=2n²+2n-2n+2-(n-1)+9
=2n²-5n+12
For some reason, though, I think I'm missing a step...
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Posted: Wed Sep 02, 2009 7:09 pm
Okay then. ^^
... Yeah, that's right. There's a typo in that third bit, a plus where it's supposed to be a minus, but your math's all right and you got the right answer. ^^
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Posted: Wed Sep 02, 2009 7:15 pm
Really? >w<
But don't two negatives make a positive when multiplied?
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Posted: Wed Sep 02, 2009 7:19 pm
No it's not there, it's on this one:
=2n² + 2n - 2n + 2 - (n-1) + 9
The bold one would be a minus.
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Posted: Wed Sep 02, 2009 7:24 pm
Ah. Okays <3 -fixes-
Ugh, fractions...
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Posted: Wed Sep 02, 2009 7:25 pm
Hehe... =u=
fractions are aaallways a pain. ;=.=
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Posted: Wed Sep 02, 2009 7:28 pm
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