SOLUTION: i hope you can solve the following for me: 1. x2m+n (2m+n being the exponent) ------ xm+3n (m+3n being the exponent) 2. 22m+2 (2m+2 being the exponent)

Question 441213: i hope you can solve the following for me:
1. x2m+n (2m+n being the exponent)
------
xm+3n (m+3n being the exponent)

2. 22m+2 (2m+2 being the exponent)
-----
2(2m-1)2 ( (2m-1)2 being the exponent)

3. (x+1)3n-2 (3n-2 being the exponent)
---------
(x+1)2-3n (2-3n being the exponent)

THANK YOU
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

 

Hi, Note the  use of ^ (uppercase 6)

Refer to the Laws governing exponents:

**



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1. 

2. 

3. 

 

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