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anonymous
 5 years ago
Add. Simplify by removing factors of 1
28zx + z7x
 
z^249x^2 z+7x
anonymous
 5 years ago
Add. Simplify by removing factors of 1 28zx + z7x   z^249x^2 z+7x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Add. simplify by removing factors of 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0step by step so i can make notes to do next one on my own plz

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1\[\frac{28zx}{(z7x)(z+7x)}+\frac{z7x}{z+7x}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i have to go but u need to find a common denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i really need help someone plz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i really need help someone plz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I see the exercise in mathematical notation now, so I know what to do. :) Here is how it goes... 1) 28zx/(z7x)(z+7x) (explanation: 28zx remains as is for now, but we have to focus on this part (z7x)(z+7x) so follow up close with what I'm doing here: the first part (z  7x)...z and 7x multiplies with the first and second of the second parenthesis like so: (z7x)(z+7x) = z^2 + 7zx  7zx  49x^2 and now you keep solving z^2  49x^2, and now you have 28zx/z^249x^2 + z7x/z+7x and now you kind of have to apply what you learned about adding fractions...) You multiply them in cross... 28zx * z + 7x and z^2  49x^2... and see what you get.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@tinaann: Did I help any? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but could you show ur work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Please forget everything I wrote earlier. :) I'm not so sure at the moment how I can do it, but I'll post as soon as I get it solved.
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