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 new operations (Posted on 2004-03-31)
Mathematicians have just created three new operation symbols, @, \$,and _, each doing a different thing to a pair of numbers (All of these symbols can be derived from already-known operations and notations). The following equations using the new symbols are correct:

5@5= 2/5
2@2= 1
2@3= 5/9
3@4= 7/16
2@4= 3/8

1\$0= undefined
1\$4= 1/10
4\$1= 1/10
10\$10= 1/50
(1@1) \$ 1= (√2)/10

2_2= 6
2_3= 9
6_2= 42
3_5= 30
5_6= 90

If the previous equations are correct, compute the following equation.

1 @ (1 \$ (1_1))=

 Submitted by Victor Zapana Rating: 3.3000 (10 votes) Solution: (Hide) The answer is 30. Just to let you know... x @ y = (x + y) / y^2 x \$ y = (√(xy)) / (5xy) x _ y = [x(x + 1)y] / 2 = y[x + (x-1) + (x-2) + ... + 2 + 1]

 Subject Author Date Answer K Sengupta 2008-03-03 10:53:02 re: Very poor Jack McBarn 2004-04-05 14:35:31 re: Very poor Federico Kereki 2004-04-01 12:29:02 re(3): And \$ Makes 30 ... ? Tristan 2004-04-01 04:05:23 Thank you Robert Mendez 2004-04-01 03:29:50 re(2): And \$ Makes 30 ... ? Victor Zapana 2004-03-31 22:06:58 re(2): And \$ Makes 30 ...er...6 I mean Richard 2004-03-31 20:09:52 re: And \$ Makes 30 ... ? Tristan 2004-03-31 18:33:18 And \$ Makes 30 Richard 2004-03-31 16:40:37 re: ERRing Thoughts Ady TZIDON 2004-03-31 15:57:04 The _ operator Brian Smith 2004-03-31 14:52:43 Solution for the @ operator Larry 2004-03-31 14:39:51 Beginning Thoughts TomM 2004-03-31 14:36:59 Very poor e.g. 2004-03-31 13:57:37

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