How do you simplify #-(5c^2d^5)/(8cd^5f^0)#?

Question

2889 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-05T12:23:35

See the entire simplification process below:

First, use this rule of exponents to eliminate the #f# term: #a^color(red)(0) = 1#

#-(5c^2d^5)/(8cd^5f^color(red)(0)) = -(5c^2d^5)/(8cd^(5)1) = -(5c^2d^5)/(8cd^(5))#

Next, cancel like terms:

#-(5c^2d^5)/(8cd^(5)) = -(5c^2color(red)(cancel(color(black)(d^5))))/(8c color(red)(cancel(color(black)(d^(5))))) = -(5c^2)/(8c)#

Now, use these two rules for exponents to complete the simplification:

#a = a^color(blue)(1)# or #a^color(blue)(1) = a# and #x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))#

#-(5c^2)/(8c) = -(5c^color(red)(2))/(8c^color(blue)(1)) = -(5c^(color(red)(2)-color(blue)(1)))/8 = -(5c^1)/8 = -(5c)/8#