How do you evaluate the expressions #(2y^2) /( -5y)# given x= 3 and y=5?

1 Answer
Jul 29, 2016

Answer:

There doesn't seem to be a variable #x# in the expression, so let's just ignore that for now. All you have to do is substitute the 2 #y#'s in the expression for #5#.

Explanation:

The expression now looks like this:

#(2(5)^2)/(-5(5))#

Next, solve:

#2(5)^2=50#

#-5*5=-25#

We now have #50/-25#, or #-2#.