This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s translate to decimal like this:
1∙82+5∙81+7∙80+6∙8-1+7∙8-2 = 1∙64+5∙8+7∙1+6∙0.125+7∙0.015625 = 64+40+7+0.75+0.109375 = 111.85937510
got It: 157.678 =111.85937510
Translate the number 111.85937510 в hexadecimal like this:
the Integer part of the number is divided by the base of the new number system:
| 111 | 16 | |
| -96 | 6 | |
| F | | |
 |
the Fractional part of the number is multiplied by the base of the new number system:
 |
| 0. | 859375*16 |
| D | .75*16 |
| C | .0*16 |
the result of the conversion was:
111.85937510 = 6F.DC16
the Final answer: 157.678 = 6F.DC16
now let\'s make the transfer using the decimal system.
let\'s do a direct translation from octal to binary like this:
157.678 = 1 5 7. 6 7 = 1(=001) 5(=101) 7(=111). 6(=110) 7(=111) = 001101111.1101112
the Final answer: 157.678 = 1101111.1101112
Fill in the number with missing zeros on the left
Fill in the number with missing zeros on the right
let\'s do a direct translation from binary to hexadecimal like this:
01101111.110111002 = 0110 1111. 1101 1100 = 0110(=6) 1111(=F). 1101(=D) 1100(=C) = 6F.DC16
the Final answer: 01101111.110111008 = 6F.DC16