If you mean: (0, 11) then y = x+11
011 & 009 Since no decimal point is given , then drop the prefix zeroes. Hence 11 & 9 . It follows that 11 > 9
Base2 011 = 11 Base3 011 = 10 Any base above that: Base2(11) equals 3
Divide the binary number into 3 bit segments starting from the right, then convert each 3 bit section into its decimal equivalent. 111 011 110 101 = 7 3 6 5 (111 = 7, 011 = 3, 110 = 6, and 101 = 5).
Convert each "digit" of the octal into a triplet of binary digits, according to the following rule: Octal Binary 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111 So, for example, octal 53 = binary 101 011 [= decimal 43]
If you mean 011 then it is simply 11.0
.011
0.5 is greater than 0.11
.011
Eight thousand and eleven
It is 0.098 814 229 249 011 857 707 5 where the underlined 22-digit string repeats.
50,000 plus 1,000 plus 10 plus 1.
011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg
If you mean: (0, 11) then y = x+11
011 = (0 x 100) + (1 x 10) + (1 x 1)
011 & 009 Since no decimal point is given , then drop the prefix zeroes. Hence 11 & 9 . It follows that 11 > 9
Base2 011 = 11 Base3 011 = 10 Any base above that: Base2(11) equals 3