How do you find the quotient of #(3r^4)/k^2div(18r^3)/k#?

2 Answers
Feb 10, 2017

See the entire solution process below:

Explanation:

First, we can rewrite this expression as:

#((3r^4)/k^2)/((18r^3)/k)#

We can now use this rule for dividing fractions:

#(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))#

#(color(red)(3r^4)/color(blue)(k^2))/(color(green)(18r^3)/color(purple)(k)) = (color(red)(3r^4) xx color(purple)(k))/(color(blue)(k^2) xx color(green)(18r^3)) = (3kr^4)/(18k^2r^3) = (kr^4)/(6k^2r^3)#

We can now use these rules for exponents to further simplify the expression:

#a = a^color(red)(1)#

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

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

#(kr^4)/(6k^2r^3) = (k^1r^4)/(6k^2r^3) = r^(4-3)/(6k^(2 - 1)) = r^1/(6k^1) = r/(6k)#

Feb 10, 2017

#r/(6k)#

Explanation:

To divide by a decimal. multiply by its reciprocal.

#(3r^4)/(k^2) color(blue)(div (18r^3)/k)#

#=(3r^4)/(k^2) color(blue)(xx k/(18r^3))#

#=(cancel3kr^4)/(cancel18_6 k^2r^3)" "larr# subtract indices of like bases

#=r/(6k)#