A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ x+2 }{ (x4)^{2} } \frac{ x }{x4 }\] this is my first one

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0hey @Hero can you help me out?

Hero
 2 years ago
Best ResponseYou've already chosen the best response.1Hint: Multiply the second fraction by (x4)/(x4)

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ x^24 }{ x^216 }?\]

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0@Hero is this right?

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0how would i go about doing this?

Hero
 2 years ago
Best ResponseYou've already chosen the best response.1I'm only going to show you this once.

Hero
 2 years ago
Best ResponseYou've already chosen the best response.1\[\space\space\space\space\frac{x + 2}{(x  4)^2}  \frac{x}{x4}\]\[=\frac{x + 2}{(x  4)^2}  \frac{x}{x4} \times\frac{x4}{x4}\]\[=\frac{x + 2}{(x  4)^2}  \frac{x(x4)}{x4(x4)}\]\[=\frac{x + 2}{(x  4)^2}  \frac{x^24x}{(x4)^2}\]\[=\frac{(x + 2)  (x^2  4x)}{(x4)^2}\]\[=\frac{x + 2  x^2 + 4x}{(x4)^2}\]\[=\frac{x^2 + 4x + x + 2}{(x4)^2}\]\[=\frac{x^2 + 5x + 2}{(x4)^2}\]\[=\frac{(x^2  5x  2)}{(x4)^2}\]\[=\frac{x^2  5x  2}{(x4)^2}\]

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0thanks allot for the help!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.