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There are an infinite number of values between 0.01 and 0.027 for example .011 .0111 .01111 .01111 .011111 etc. , .012, .0121, .0122, etc.
In any base other than 2 (ie other than binary): 10101010 + 01111 = 10102121 In base 2 (binary): 10101010 + 01111 = 10111001
10101010 + 01111 = 10111001 (170+15 = 185)
with hexidecimal you neednt use as many chartictors to represent a number. in binary 15 would be 01111 where as in hex it would be E resulting in much quicker coding times
with hexidecimal you neednt use as many chartictors to represent a number. in binary 15 would be 01111 where as in hex it would be E resulting in much quicker coding times
Multiply by 16. ANSWER: The answer is bcd 11 = 3 111 = 7 1111 i= F 01111 =E and so foirth
3 bit stuffing is required
10,101,010 + 01,111 = 10,102,121Although you didn't say so, we suspect that you may have meantthe two numbers in the question to be binary (base-2) numbers.If so, then| 0 | 0 | 0 | 0= (170)100 | | | | = (15)10Their sum is (185)10
17
0 = 00000 1 = 00001 2 = 00010 3 = 00011 4 = 00100 5 = 00101 6 = 00110 7 = 00111 8 = 01000 9 = 01001 10 = 01010 11 = 01011 12 = 01100 13 = 01101 14 = 01110 15 = 01111 16 = 10000 17 = 10001 18 = 10010 19 = 10011 20 = 10100
3
01 = 00001 11 = 01011 21 = 10101 02 = 00010 12 = 01100 22 = 10110 03 = 00011 13 = 01101 23 = 10111 04 = 00100 14 = 01110 24 = 11000 05 = 00101 15 = 01111 25 = 11001 06 = 00110 16 = 10000 26 = 11010 07 = 00111 17 = 10001 27 = 11011 08 = 01000 18 = 10010 28 = 11110 09 = 01001 19 = 10011 29 = 11101 10 = 01010 20 = 10100 30 = 11111