- Fri Jul 20, 2012 11:06 pm
- Forum: Intermediate Algebra
- Topic: Identifying Square Root Functions
- Replies:
**1** - Views:
**2122**

Hi, I understand the basic concept of a square root function. I just don't quite get how to identify them. To be more specific, 1) is y=sqrroot of x a square root function? I'm assuming yes. 2) What about y=x^2? is this a square root function? I understand it is the inverse of y=sqrroot of x. But, w...

- Thu Aug 04, 2011 12:19 am
- Forum: Intermediate Algebra
- Topic: Finding domain and range from equation of inverse function
- Replies:
**1** - Views:
**1182**

Can someone please tell me the domain and range of this equation: http://www.mathway.com/math_image.aspx?p=fSMB02ESMB03-1SMB02eSMB03%28x%29SMB01SMB02RSMB03x+2+1SMB02rSMB03?p=122?p=22 I know the answer. It is: The domain is all values of x that are greater than or equal to -2. The range (y) is all re...

- Fri May 13, 2011 9:54 pm
- Forum: Intermediate Algebra
- Topic: Adding/Subtracting Rational Expressions w Unlike Denominator
- Replies:
**1** - Views:
**1970**

Hi, Can someone please show me how this problem is solved: http://www.mathway.com/math_image.aspx?p=SMB02FSMB037mSMB10mSMB02ESMB032SMB02eSMB03-36SMB02fSMB03-SMB02FSMB033SMB10m-6SMB02fSMB03?p=98?p=42 Here is one of my attempts: First, I found LCD, so I can subtract the two expressions. I did so by mu...

- Sat May 07, 2011 8:50 pm
- Forum: Intermediate Algebra
- Topic: Dividing Rational Expressions
- Replies:
**1** - Views:
**1007**

Hi, can someone please help me understand how this problem is solved? This is the problem: http://www.mathway.com/math_image.aspx?p=SMB02FSMB03xSMB02ESMB032SMB02eSMB03-6x+15SMB10ySMB02fSMB03SMB11SMB02FSMB03x-5SMB105zSMB02fSMB03?p=122?p=42 This is as far as I get (multiply first fraction by reciproca...

- Wed Mar 30, 2011 12:18 am
- Forum: Beginning Algebra
- Topic: converting equation to quadratic equation form
- Replies:
**8** - Views:
**4891**

stapel_eliz wrote:I haven't checked your math, but since 343 is not a multiple of 5, clearly this fraction cannot be reduced.

makes sense

Thanks for bearing with my questions

- Tue Mar 29, 2011 5:35 pm
- Forum: Beginning Algebra
- Topic: converting equation to quadratic equation form
- Replies:
**8** - Views:
**4891**

I kind of just forgot how to do deal with the left-hand side. Specifically how to multiply it. To review how to multiply out the square, try this lesson . :wink: Yes, thank you very much. I ended up getting this: y=5x^2/343 - 90x/49 + 402/7 Can I simplify this even more? I know I can divide 402 by ...

- Tue Mar 29, 2011 12:55 am
- Forum: Beginning Algebra
- Topic: converting equation to quadratic equation form
- Replies:
**8** - Views:
**4891**

5\left(\frac{x}{7}-9\right)^2=7y+3 Multiply out the left-hand side. Subtract the 3 over to the left-hand side, and divide through by 7. Simplify to get "y=" the quadratic form. :wink: I understand that part. I kind of just forgot how to do deal with the left-hand side. Specificall...

- Tue Mar 29, 2011 12:19 am
- Forum: Beginning Algebra
- Topic: converting equation to quadratic equation form
- Replies:
**8** - Views:
**4891**

Hi, can someone please help me help convert this equation into a quadratic equation (i.e. of the form y=ax^2+bx+c) 5(x/7-9)^2=7y+3 How would I go about simplifying the left side, I don't really know how to deal with x/7-9. How do I subtract x/7 and 9. I forgot how to do this :( Just quickly tell me...

- Mon Mar 28, 2011 8:47 pm
- Forum: Beginning Algebra
- Topic: converting equation to quadratic equation form
- Replies:
**8** - Views:
**4891**

Hi, can someone please help me help convert this equation into a quadratic equation (i.e. of the form y=ax^2+bx+c) 5(x/7-9)^2=7y+3 How would I go about simplifying the left side, I don't really know how to deal with x/7-9. How do I subtract x/7 and 9. I forgot how to do this :( Just quickly tell me ...

- Fri Mar 18, 2011 8:30 pm
- Forum: Beginning Algebra
- Topic: Factoring polynomials using the distributive property
- Replies:
**2** - Views:
**2864**

stapel_eliz wrote:onmyown wrote:4) Used distributive property:

4x(x-2) + 4y(2-x)

5) Therefore, the solution to the problem is the binomial:

(4x+4y)(x-2)

How did the "2 - x" turn into "x - 2"? Is 5 - 3 the same value as 3 - 5?

I got the right answer now. Thanks