Question

edfrgtrjupllkj,mnhbgf

Answer 1

@Kristen17

Answer 2

@ZachH54

Answer 3

@221emily

Answer 4

B.
f/g = (x^2+3x-4)/(x+4)
domain set the denom to zero.
x + 4 =0, x = -4 , all real numbers except at x= -4.

Factor out the top.
(x+4)(x-1)/(x+4)
cross out the x+4.
x-1 = f/g

A. is that a multiplication sign?..

Answer 5

Yes it's multiplication.

Answer 6

Thank you for B.

Answer 7

(f * g) = (x^2+3x-4) (x+4)
distribute it then.

Answer 8

Is that the final answer for A.?

Answer 9

No.
Multiply each term in the parantheses.
(x+4)(x^2+3x-4)
x^2(x)=
3x(x) =
-4(x) =
4(x^2) =
4(3x) =
4(-4) =
after finding that out combine the like terms to simplify.

Answer 10

I don't think that it is very nice of you to ask for answers. For one you should have at least provided your answers and your work for us to see that you attempted to solve the problem your self. I refuse to help you with this problem because I think that you only posted this question to get out of the work yourself. You were basically asking for the answer by saying "Is that the final answer for A.?" and "So what is the final answer sir?" and that is not what OS is about.

Answer 11

I don't think it's very nice of you when you tell me to tag and then you don't help at all. Yes I am looking for the answer. I am new on here, but I do know that Openstudy is about helping in studying. If it's wrong to ask for answer on here then I didn't know. This is not a test. If it was I would have seen your point. Also, when you said I could have at least provided my answers and work. If I knew how to do it, I wouldn't have asked for help. Before making this site, I've seen many post there questions not there work and answers, and they still got answers, and help too. Thank you for letting me know that, and I do not need you help I've gotten the help I needed. Thanks to Shamil98.

Answer 12

@shamil98 Like this?

x^2(x)= x^3
3x(x) =3x^2
-4(x) = -4x
4(x^2) = 4x^2
4(3x) = 12x
4(-4) =-16

Answer 13

@wowidontknow Welcome to OpenStudy.

Because you posted that you are new here, I invite you to take a look at the OpenStudy Code of Conduct. The users of this site are not to give answers with explanation or discussion.

I understand that you did not know that.

For your reference, here is a section of the OS Code of Conduct that speaks to doling out answers:

Code of conduct: OpenStudy values the Learning process - not the ‘Give you an answer’ process Don’t post only answers - guide the asker to a solution
http://openstudy.com/code-of-conduct

You may also want to read the "Terms and Conditions" section.
http://openstudy.com/terms-and-conditions

Much of that is legal "stuff," but do scroll to the very bottom of the document to read the Anti-Cheating policy.

Please understand that no one is accusing you of anything. And, I apologize that one of the Ambassadors did not greet you and explain all this.

Thanks for joining OpenStudy. We are glad you are here.

Answer 14

Thanks @Directrix for the explanation and welcome. I will take a good look at the links you've posted. That is very generous of you. Thanks again.

Answer 15

@wowidontknow You are welcome.