The operation # is defined for all nonzero x and y by x#y = x + x/y. If S>0, then 1#(1#S) =  Pick:
A) S B) S+1 C) S/(S+1) D) (S+2)/(S+1) E) (2S+1)/(S+1)
x#y = x + x/y. If S&gt;0, then 1#(1#S) =   You can see that a stands for y and 1 stands for x. I like to work with numbers so I did x=1 S=2 and S=y so y=2   1#S= 1+(1/2)= 
<gadget id="calculator">1+(1/2)</gadget>
<output>3/2 = around 1.5</output>
1.5 so now a is 1.5 --&gt; 1+(1/1.5)=1 2/3 = 5/3   So now you can check by filling in S, which is 2:   A. S --&gt; 2   B. S+1 --&gt; 2+1=3   C. S/(S+1)--&gt;2/(2+1)=2/3   D. (S+2)/(S+1)--&gt;(2+2)/(2+1)=4/3   E. (2S+1)/(S+1)--&gt;(2*2+1)/(2+1)=5/3   So (2S+1)/(S+1) is the answer.
<result>E</result>
The operation # is defined for all nonzero x and y by x#y = x + x/y. If S>0, then 1#(1#S) =