How do you simplify # (5sqrt20 )/ sqrt5#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Jun 27, 2016 #(5sqrt20)/sqrt5=10# Explanation: #(5sqrt20)/sqrt5# = #(5sqrt(2xx2xx5))/sqrt5# = #(5xx2xxsqrt5)/sqrt5# = #(5xx2xxcancelsqrt5)/(1cancel(sqrt5)# = #5xx2=10# Answer link Related questions How do you simplify #(2sqrt2 + 2sqrt24) * sqrt3#? How do you simplify #sqrt735/sqrt5#? How do you rationalize the denominator and simplify #1/sqrt11#? How do you multiply #sqrt[27b] * sqrt[3b^2L]#? How do you simplify #7sqrt3 + 8sqrt3 - 2sqrt2#? How do you simplify #sqrt468 #? How do you simplify #sqrt(48x^3) / sqrt(3xy^2)#? How do you simplify # sqrt ((4a^3 )/( 27b^3))#? How do you simplify #sqrt140#? How do you simplify #sqrt216#? See all questions in Square Root Impact of this question 2089 views around the world You can reuse this answer Creative Commons License