How do you simplify #(5 times q ^3) /sqrtq^11#?

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Answer 1 · 2016-02-15T20:01:46

#(5sqrt(q))/q^3" but "5q^(-5/2)" is better!"#

Assumption: You meant the question to look like

#(5xxq^3)/sqrt(q^11color(white)(.))#

Write as #" "5xx(q^3)/(q^(11/2))#

#=> 5xx q^(6/2-11/2)" "=" "5q^(-5/2)#

#=> 5/(sqrt(q^5))#

#=>5/(sqrt(q^2xxq^2xxqcolor(white)(.))#

#=> 5/(q^2sqrt(q))#

Not good practice to have a root as a denominator
So multiply by but in the form of #sqrt(q)/sqrt(q)#

Giving

#(5sqrt(q))/q^3" or just use the earlier value of " 5q^(-5/2)#