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 Posted: Wed Sep 02, 2009 7:30 pm
I mean even when you get them they're SO MUCH WORK.
Stupid fractions...
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Posted: Wed Sep 02, 2009 7:34 pm
I knows. >.<" Kaddy's really unsure of what to do...
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Posted: Wed Sep 02, 2009 7:44 pm
3-(b²+1)
5+(b²+1)
...
-b² + 2
b² + 6
*thinks* I think that's it... unless you can factor that out and find a common factor.
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Posted: Wed Sep 02, 2009 7:53 pm
...what? That just made Kaddy confuzzled...
Kaddy's gotta go soon...-sighs- Very soon...
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Posted: Wed Sep 02, 2009 7:56 pm
I just simplified it in what I did.
If you can turn 2 - b² and 6 + b² into multiplication statements like (b+n)(b-n) or something like that and they have one that's the same, like they both have (b+n), then you can take that out of the equation.
... Aw, that sucks.
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Posted: Wed Sep 02, 2009 7:59 pm
It's already past Kaddy's bed time >.>"
I get it...sort of. Fractions always make Kaddy more confuzzled...
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Posted: Wed Sep 02, 2009 8:08 pm
Aw... *huggles*
They kind of do me too Kaddy.
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Posted: Wed Sep 02, 2009 8:10 pm
Hm...one more...no idea what to do for it...Oh well. -leaves it blank-
I'll be on after school tomorrow...^^" And early in the morning too. Like, 5 am early xD Too early for Brother and everyone else though.
-waves- Goodnight~ Thanks for all the help ^.^
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Posted: Wed Sep 02, 2009 8:16 pm
Goodnight Kaddy. You're welcome; any time. ^^
Okay see you then. x3
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Posted: Thu Sep 03, 2009 5:20 pm
x3 Opens!
Math times! xD
Okays...today we did adding, subtracting, multiplying, and dividing functions...x3 It's kinda confuzzling...I got the first one, but the second one confuzzles meh a bit with it's fraction-ness...
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Posted: Thu Sep 03, 2009 5:26 pm
Okay... For adding/subtracting fractions you find the Greatest Common Denominator. You multiply the two bottom parts together, then multiply the tops by the other bottom. If that makes sense...
Multiplying fractions You multiply each part. Top by Top, bottom by bottom. A lot easier.
Dividing fractions... Blegh. I know how but it's a pain.
Basically you put the two on top of each other. So it's (x/n)/(y/n). Which is like (x/n) times (n/y)... makes sense?
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Posted: Thu Sep 03, 2009 5:46 pm
Sort of...still confuzzling...xD Here's the problem:
f(x)
x
x+1
g(x)
x²-1
I haves to add, subract, multiply, and divide for that...
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Posted: Thu Sep 03, 2009 5:58 pm
So f(x)+g(x), f(x)-g(x), f(x)g(x), and f(x)/g(x)...
I'm going to do the fractions in sideways form 'cuz it's easier to type dear, mkay?
ADDING:
f(x) + g(x) = x/(x+1) + (x²-1)/1
=(x * 1)/(1(x+1)) + (x²-1)(x+1)/(1(x+1)
=x/(x+1) + (x^3 + x² - x - 1)/(x+1)
=(x^3 + x² -1)/(x+1)
... You might be able to factor that with one of the factors being "(x+1)" but I'm not sure. That should be enough. You get that? Then you do the same basic procedure for subtracting.
I'll help with dividing after this but multiplying just write it out and multiply horizontally. So it's the numerators multiplied over the denominators multiplied, mkay? Oh, if there's no fraction it's over 1 and what's there is the numerator. I hope you got that.
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Posted: Thu Sep 03, 2009 6:03 pm
...sort of...still confuzzling though xD Ugh, I hate math...
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Posted: Thu Sep 03, 2009 6:09 pm
Most of the world does dear... even if they all use parts of it.
Well if you don't get it here's another explanation... You multiply them by each other's bottoms over top of themselves. You do it so it's (x+1)/(x+1) because that equals one.
so if the equation's (f(x)/x) + g(x) then it's (f(x)/x)*(1/1) + g(x)*(x/x)
So it becomes (f(x)/x) + (g(x)/x). Then you just add the tops and keep the bottom the same.
Now, dividing's a real headache. Here's how I remember it.
f(x)/g(x)=f(x) * 1/g(x). You see how that works?
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