The operation # is defined for all nonzero f and y by f#y = f + f/y. If a>0, then 1#(1#a) = Answers:	A) a	B) a+1	C) a/(a+1)	D) (a+2)/(a+1)	E) (2a+1)/(a+1)
f#y = f + f/y. If a&gt;0, then 1#(1#a) =   You can see that a stands for y and 1 stands for f. I like to work with numbers so I did f=1 a=2 and a=y so y=2   1#a= 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 a, which is 2:   A. a --&gt; 2   B. a+1 --&gt; 2+1=3   C. a/(a+1)--&gt;2/(2+1)=2/3   D. (a+2)/(a+1)--&gt;(2+2)/(2+1)=4/3   E. (2a+1)/(a+1)--&gt;(2*2+1)/(2+1)=5/3   So (2a+1)/(a+1) is the answer.
<result>E</result>
The operation # is defined for all nonzero f and y by f#y = f + f/y. If a>0, then 1#(1#a) =