How do you simplify #(6qr^3)/(-2q^5r)#? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 1 Answer MeneerNask May 22, 2016 You can write numbers, #q#'s and #r#''s apart Explanation: #=-6/2xxq/q^5xxr^3/r# Then split up the powers of #q# and #r# #=-3xx(cancelq)/(cancelqxxq^4)xx(cancelrxxr^2)/cancelr# #=-(3r^2)/q^4# Answer link Related questions What is the quotient of powers property? How do you simplify expressions using the quotient rule? What is the power of a quotient property? How do you evaluate the expression #(2^2/3^3)^3#? How do you simplify the expression #\frac{a^5b^4}{a^3b^2}#? How do you simplify #((a^3b^4)/(a^2b))^3# using the exponential properties? How do you simplify #\frac{(3ab)^2(4a^3b^4)^3}{(6a^2b)^4}#? Which exponential property do you use first to simplify #\frac{(2a^2bc^2)(6abc^3)}{4ab^2c}#? How do you simplify #(x^5y^8)/(x^4y^2)#? How do you simplify #[(2^3 *-3^2) / (2^4 * 3^-2)]^2#? See all questions in Exponential Properties Involving Quotients Impact of this question 1129 views around the world You can reuse this answer Creative Commons License