How do you simplify #(20a^4b^-7c^6)/(-5ab^0c^2)#?

Question

564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-13T17:39:13

#=color(blue)(-4a^3b^-7c^4#

#(20a^4b^-7c^6)/(-5ab^0c^2)#

As per property:
#color(blue)(x^0 = 1#
So, #b^0 =1#

#=(20a^4b^-7c^6)/(-5a xx 1xxc^2)#

#=(20a^4b^-7c^6)/(-5ac^2)#

#=(20/-5) xx (a^4b^-7c^6)/(a^1c^2)#

As per property:
#color(blue)(x^m/x^n= x^(m-n)#

Applying the same to exponents of #a# and #c#:

#=(cancel20/-cancel5) xxcolor(blue)( (a^(4-1) xx b^-7 xx c^(6-2))#

#=(-4) xx (a^(3) xx b^-7 xx c^(4))#

#=color(blue)(-4a^3b^-7c^4#