Chrisg094
  • Chrisg094
Simplify the result if possible
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Chrisg094
  • Chrisg094
|dw:1449785546677:dw|
Chrisg094
  • Chrisg094
\[f(x)=\frac{ x^2-49 }{ x^2+11x+24 }\]
calculusxy
  • calculusxy
is this two questions?

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Chrisg094
  • Chrisg094
Yes. Two separate
Chrisg094
  • Chrisg094
\[\frac{ (x-7)(x+7) }{ (x+3)(x+8) }\]
Chrisg094
  • Chrisg094
Thats what I got once I factored for the second problem. I don't even know where to start for the first problem.
Chrisg094
  • Chrisg094
And I don't know what to do next with those pair of binomials.
calculusxy
  • calculusxy
this is almost like adding regular fractions... you need to find the common denominator first to do so, you must multiply: \(x(x + 1)\) no need to multiply because if u were to simplify this you would get \(x(x+1)\) again. the denominator (common) will be \(x(x+1)\) and now you need to multiply the numerators and make sure that they are equivalent fractions
calculusxy
  • calculusxy
combine the numerators over the common denominator
Chrisg094
  • Chrisg094
so only multiply that x(x+1) to that first denominator?
calculusxy
  • calculusxy
\[\large \frac{ 3(x+1) + 8(x) }{ x(x+1) } = \frac{ 3x + 3 + 8x }{ x(x+1) } = \frac{ 11x + 3 }{ x(x+1) }\]
calculusxy
  • calculusxy
notice that we multiplied x with x+1 to get x(x+1) so we do 3(x+1) as well.
Chrisg094
  • Chrisg094
|dw:1449786325992:dw|
Chrisg094
  • Chrisg094
Oh I just saw what you wrote sorry.
Chrisg094
  • Chrisg094
Sow hat about the other problem?
calculusxy
  • calculusxy
i am looking that over now
Chrisg094
  • Chrisg094
oh, Okay
calculusxy
  • calculusxy
it seems correct :)
Chrisg094
  • Chrisg094
it can't be simplified any further?
calculusxy
  • calculusxy
there isn't any common factor between them so i don't think so
Chrisg094
  • Chrisg094
ok sound good. But know that I have you here I have this problem no one has been able to solve hopefully you can.
anonymous
  • anonymous
this would be cross multiply
calculusxy
  • calculusxy
woah... i didn't know about this. am i correct?
anonymous
  • anonymous
right about what?
Chrisg094
  • Chrisg094
|dw:1449786838729:dw|
anonymous
  • anonymous
I dont understand what you had put up. What grade are you in? @Chrisg094
Chrisg094
  • Chrisg094
it says write answer in interval notation. And btw, I don't think I remember anything about cross multiplication in my college algebra class.
calculusxy
  • calculusxy
u are in college?
Chrisg094
  • Chrisg094
Yes.
calculusxy
  • calculusxy
i am only in eighth grade!
anonymous
  • anonymous
ohh your in colledge. Sorry, I cant help you. I am in 8th grade
anonymous
  • anonymous
@calculusxy I had message you.
calculusxy
  • calculusxy
i know. i replied.
anonymous
  • anonymous
I got your message.
anonymous
  • anonymous
just tell me when you want me to stop messaging you.
whpalmer4
  • whpalmer4
\[\frac{3}{x} + \frac{8}{x+1}\]Here's a handy trick I saw here a while back: \[\frac{3}{x} + \frac{8}{x+1} = z\]\(z\) will be the value of the two fractions added together Let's get rid of the fractions by multiplying both sides of the equation by all of the denominators: \[x(x+1)*\frac{3}{x} + x(x+1)\frac{8}{x+1} = x(x+1)*z\] cancel out any matching terms in numerator and denominator: \[3(x+1) + 8x = xz(x+1)\]\[3x + 3 + 8x = xz(x+1)\]\[11x + 3 = xz(x+1)\]solve for \(z\), our answer:\[z = \frac{11x+3}{x(x+1)}\] The second problem is fully simplified. @Chrisg094
Chrisg094
  • Chrisg094
Thanks @whpalmer4 but what about the problem: \[f(x)=\sqrt{12-4x}\] a)? b)? c)? d)? and also, \[3+\sqrt{2x-3x}=x\]
Chrisg094
  • Chrisg094
sorry the 3 doesn't have an x i messed up its: \[3+\sqrt{2x-3}=x\]
whpalmer4
  • whpalmer4
I don't understand what a)? b)? c)? d)? are all about. Really much more convenient all around if you only post one problem in a thread...and include all of the information, such as instructions about what you're supposed to do!
whpalmer4
  • whpalmer4
If you can enlighten me on those topics, I'm happy to help...
Chrisg094
  • Chrisg094
sorry about that Im fairly new to this I will explain further.
Chrisg094
  • Chrisg094
so the first problem with the f(x) says find the domain of the square root function and writ in interval notation. Then the second problem says solve the radical equation.
Chrisg094
  • Chrisg094
as for the a) b) c) d) on that first f(x) problem I'm clueless. Maybe there supposed to represent each number @whpalmer4
whpalmer4
  • whpalmer4
\[f(x)=\sqrt{12−4x}\] Okay, so the problem wants you to find the domain of this function. The domain is the range of allowable values for the independent variable (\(x\) here). Can you think of any values of \(x\) that would not be allowable? \[3+\sqrt{2x-3}=x\]I would start by moving the \(3\) to the right side, then squaring both sides. You should be able to rearrange the equation and solve by factoring or the quadratic formula. Here's where the tricky part comes up: because we had a square root which we squared while solving, we may have introduced what is called an extraneous solution. You need to try all of the solutions you get in the original equation to make sure they actually work! One will, and one will not. Make sure you choose the right one as your answer!

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