anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
Which one is the question?
anonymous
  • anonymous
the one that says "ok" ^_^

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

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