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anonymous
 5 years ago
Would the following fraction be considered to be fully simplified?
x^42x^35x^218x36

x^55x^4+6x^3+8x^240x+48
anonymous
 5 years ago
Would the following fraction be considered to be fully simplified? x^42x^35x^218x36  x^55x^4+6x^3+8x^240x+48

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This was the result I got when multiplying two fractions. I know my multiplication was correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm guessing you'd have to solve the top and the bottom separately for it to be simplified.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, there is no solving. My instructions are to simplify completely. This is where I am in the problem. Does that mean I need to combine all of the x's and the whole numbers in front of them, and then subtract the smaller total from the larger total, and get an x with a single variable on whatever side of the equation that happens to be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or just the 36/48 part...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe its just the 36/48 part.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay yea, so then that would make the final answer: \[x^42x^35x^218x3 \]  \[x^55x^4+6x^3+8x^240x+12\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lost me their, man. I'm only in 8th grade.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Eh... Does that look completely simplified to you then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. Doesn't look simplified. I would just try out all the options and see which one does look like its in its final "simplified" form...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yo no entiendo las cosas que este chico esta preguntando me...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, then how about the original question... See what you can come up with... \[(x^22x+4)/(x^25x+6)\div(x^29)/(x^3+8)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Darn it. I messed it up. I can't solve that right now. I'm studying at the same time I do this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry man. You're on your own until someone with the proper knowledge can help you. :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Si no sabes cómo hacerlo, ¿por qué estás tratando de ayudar?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this the original expression \[{x^22x+4 \over x^25x+6}\div {x^29 \over x^3+8}?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At the very top is what I got from long multiplication. I just need to know if that form is correctly simplified since the exponents on the variables aren't all the same, or if I have to simplify those variables with the same exponents if possible.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It does not look difficult at all. Just a minute.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0such as 2x^3/6x^3, would you simplify that to x^3/3x^3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I forgot the negative in that answer, sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure it's division, not multiplication?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I flipped the fraction on the right side to turn the division sign into a multiplication sign.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No problem. \[={x^22x+4 \over (x3)(x2)}.{(x+2)(x^22x+4) \over (x3)(x+3)}\] \[{(x^22x+4)^2(x+2) \over (x3)^2(x+3)(x2)}\] For me, this is the best simplification.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ahhh, I got on here yesterday to ask about factoring it and someone told me not to do that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you know this type of question usually involve some cancellation, which would be the case if it was a multiplication.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But I get why you did it. It IS more simple in factored form.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From what I remember of my first attempt at factoring it, all of the factored trinomials did not cancel.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have the work around here, just one secon.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, nothing can be cancelled. Is this a question from your teacher or from the book?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is a study guide for my final exam for college algebra. There are no answers given, so I want to make sure I have the idea down.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see. Good luck. I'll be around of you have any questions that I have answers for.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But in that form you gave, couldn't you factor it further?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but x^22x+4 would be (x2)(x2) wouldn't it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait no, it won't work that way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But I get what you are saying. Factoring would be the simplest way to completely simplify a problem with two fractions and any kind of operation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well there is one thing you could do to the denominator. The denominator can be written as (x^29)(x3)(x2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, (x^22x+4)/1 would cross cancel with 1/(x+2)(x^22x+4) which is the result of x^3+8, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But what I wrote before is more simplified than this, I would say.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, and I appreciate the help very much, I will let you move on to other things, but you helped me figure something out :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome, but you can't cancel them. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if there was multiplication in the center you could though, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could cancel many things actually.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Like this: \[(x^22x+4) * (1)\] \[(1) (x+2)(x^22x+4)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0eh, everything after the 1 is on the right side of that problem. Cant figure out how to like up the equation things

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It would seem like, in that scenario, you would have 1/1 * 1/(x+2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, try it this way: 1 5  *  5 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am applying that principle to the fractions. You helped me figure it out though :D Thank you again!
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