Question 667884
I try doing but I got stuck on the last part. How do you solve ((3x^2)(6x^-1)^2/(16y^3)(5y^-1)). I got 108x^0/80y^2 but you have to simplify it which is 27/20y^2 but I do not understand how they got the 27 from.


You're correct up to this point: {{{108x^0/80y^2}}}


{{{108x^0}}} = {{{108 * x^0}}} = 108 * 1 ------ Anything to the "0" power = 1


We now have: {{{108/80y^2}}}, which when simplified results in: {{{highlight_green(27/20y^2)}}} --- Dividing numerator and denominator by GCF, 4


Send comments and “thank-yous” to “D” at MathMadEzy@aol.com