Could someone tell me if my answer was correct? I will attach the equation and available answers in a comment! Its just a algebra question but im evidently unable to focus rite now >.<

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Could someone tell me if my answer was correct? I will attach the equation and available answers in a comment! Its just a algebra question but im evidently unable to focus rite now >.<

Mathematics
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Which one is the question?
the one that says "ok" ^_^

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Other answers:

I see.
The process is confusing me im pretty sure this is one of the lessons that i missed :P
So the question is to simplify: \[{4 \over m+2}+{2 \over m-2}?\]
The m + 2 and m - 2 is confusing me :P And it says to Add and Simplify.
Good. First, what is the LCM of (m+2) and (m-2).
Thats the part that was confusing me ^_^ lol
Haha I see. Do (m+2) and (m-2) have any common factors?
Yep lol >.< i think urghh i should of paid attention :P and i just didnt know how to write it bc of the - and + signs >.< lol
Well they don't have a common factor, and because of that the LCM of them is their multiplication.
Ahhhhhh alright i get it so far. ^_^
so (m+2)(m-2)=m^2-4. That's would be our LCM.
That's good. Now, we have to apply this to our equation.
expression*
ohhhh haha wow ya bc since they dont have a common factors u'd multiply them together. alright.
Exactly.
That means, we will multiply the first term (top and bottom) by (m-2). and multiply the second term (top and bottom by (m+2).
Does that make sense?
Yes so far so good so then the answer would be the "ya" one? lol like my labels? :D
>.< i think... :P
\[{4 \over m+2}+{2 \over m-2}={4(m-2) \over m^2-4}+{2(m+2) \over m^2-4}={4m-8+2m+4 \over m^2-4}={6m-4 \over m^2-4}\]
OR as in the choices: \[{6m-4 \over (m+2)(m-2)}\]
WOOOOOOOOOOOOO!!!! so i was rite huh?!?! YAY YAY YAY!!
lol ^_^
Haha Yeah you were :)
:) Thanks, helped my bunches! :D
np
Oh! and do you think you could help me on part of this other question? :) plesss? lol
This is it. do u think u will help meh? :)
so ok is the original expression?
You make strange names? :P
My question is about the denominators again so would there be common factors? >.< i got confused again :P lol and ya i am making strange names ^^
The same first question again, do they have common factors?
And yea this is the original expression .. pretty sure lol
and i thought they did >.< like since 3x can go into 6x.. but then i got lost arghh
Okay can you factor any of the two denominators?
No i dont think so
Wait which version of factoring? >.<
3x+5 can't be factored.. (6x+10)=2(3x+5).. right?
Right.
Okay now your LCD should be each one of the factor to be taken only one. the factors we have are (3x+5) and 2. so our LCD would 2(3x+5).
I don't think that was clear enough, was it?
Not really im sitting here repeating it over in my head going wth.... ^_^
LOL
Well, you can see that (3x+5) occur in both denominators. right?
after factorization*
Yep i see that.
\[{2 \over 3x+5}-{1 \over 6x+10}={2 \over 3x+5}-{1 \over 2(3x+10)}\] Right?
Now list all factors in both denominators.
I mean 2(3x+5) :)
Ahhhh so then i see why you would do that. But couldnt you just multiply the 3x + 5 expression by 2 and then subtract?
the two equations with the same denominator i mean ^^
Yes you could, but then you have to simplify after that.
that wouldn't be the simplest form.
Ahhh rite, ^_^ ok then go on lol
Well you're saying that you multiply the (3x+5) by 2, that's right. And that's what I was trying to say when I said that 2(3x+5) is the LCD. So then we keep the 6x+10 and multiply the (3x+5) term (top and bottom) by 2.
:)
ohh so the answer would be B haha wow i rly need to stop doubting myself argh :P
LOL thanks for the help! AGAIN ^_^
But you know why we can do that. That's because the denominator of the second term is nothing but 2 times the denominator of the first term.
AnwarA, how did you make the rational expression look like that ?
It's 2(3x+5).. just typo :P
all I can do is\[{a+bi}/{10a-c}\]
how did you make it go under a long bar ?
lol think person means like how do u ya long bar ^^
Oh.. {a+bi over 10a=c}
And hello yuki.
:)
>.< i will never understand computer tech/ math peoples smartness :P
haha. I think I am the only one who does it this way :P
\[\sqrt(b^2-4ac) \over 2a\]
that is sooo cooooool !! I am goint to use it from now on, thanks!
LOL obviously not anymore!! :D and i see someone ik viewing the question but he isnt in our meeting place!! >:P lol (inside joke cant rly explain it ^^)
\[{\sqrt{b^2-4ac} \over 2a} \] :P
can you do "plus or minus" as well ?
Haha
>.< smart people arg lol
Yep anything
\[{-b \pm \sqrt{b^2-4ac} \over 2a}\]
how did you make the sqrt top part long ?! I have so much to learn lol
Haha try that yourself :P
is there a place that have all of that written down ?
*sigh* so much to learn to little time lmao im jk :)
just use {}, with anything that you want to make long :)
\[\sqrt{10x+b}\]
Hah, i think its funnay how i have no idea how the heck u guys are doing that ^^
\[e^{ANWAR}={1 \over \sqrt{yuki}^{dolly}}\] :P that does not mean anything by the way
\[\int\limits _{a,b}_{b,a}\]
lol im above yuki hee hee!!!!
I am in the top, that's what matters :D
\[\ln(\ln (x)) = 0\]
:P looks more like im the glue that holds everything together mwuahahaha!!!
take ln of both sides to bring me down :(
\[+,- \pm \]
Ha ! found it lo
You're learning :P
\[\cos(a \pm b) = \cos(a)\cos(b) \mp \sin(a)\sin(b)\]
im not ^_^ LOL
You want to?
But you can't use your weird names there :P
yuki, whenever you use these things, rmr to thank me first
awww yep sure the person i am waiting for isnt talking so sure :) lol
\[\int\limits_{a,b \int\limits_{1}^{2}}\]
Haha
whom are you waiting for? xD
I definitely will AnwarA
his name starts with an I and he is rlyyyy sweet and hot and all tht jaz oh and did i mention it was u ? xD lmao
haha anwar is all like wat the hell.... LOL
Thanks buddy!! I was just kidding. I did nothing, anyone who knows that would help do that. :)
@dolly: LOL
\[{a+bi \over a-bi} *{a+bi \over a-bi}\]
^_^
why do you think I was "all like what the hell"? :)
haha to what i said lol jeezus anwar pay attention!! >:D im jk :)
Haha, I read that and I wasn't "all like what the hell" at all :P
yuki, what grade are you at?
LOL then i was wrong i thought u would be ^_^ lol and prolly college or high school senior!! lol
I think wherever I need info about anyone in this site, I should just come straight to you. You got it all :O
Do you work for the CIA?
haha its easy i just notice the kind of math they know how to do, the way they type.. etc :) nm too it :D
so you are not here to gather informations about us for some agencies?
Uhh, no. lol That would only be extremely creepy lol
:) Anyway, I am off to bed. Yuki would be more of help than I am, specially that he now knows how to do the equations thing :D.
Bye all!!
LOL okie dokie nite buddaaayyyy!!! :)
sigma

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