This transfer is possible in two ways: direct transfer and using the decimal system.
First we will perform the translation through the decimal system
let\'s translate to decimal like this:
11∙162+9∙161+15∙160+1∙16-1+10∙16-2+12∙16-3+11∙16-4 = 11∙256+9∙16+15∙1+1∙0.0625+10∙0.00390625+12∙0.000244140625+11∙1.52587890625E-5 = 2816+144+15+0.0625+0.0390625+0.0029296875+0.0001678466796875 = 2975.1046600341796910
got It: B9f.1acb16 =2975.1046600341796910
Translate the number 2975.1046600341796910 в octal like this:
the Integer part of the number is divided by the base of the new number system:
2975 | 8 | | | |
-2968 | 371 | 8 | | |
7 | -368 | 46 | 8 | |
| 3 | -40 | 5 | |
| | 6 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 10466003417969*8 |
0 | .83728*8 |
6 | .69824*8 |
5 | .58594*8 |
4 | .6875*8 |
5 | .5*8 |
4 | .0*8 |
5 | .0*8 |
the result of the conversion was:
2975.1046600341796910 = 5637.06545458
the Final answer: B9f.1acb16 = 5637.06545458
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
B9f.1acb16 = B 9 f. 1 a c b = B(=1011) 9(=1001) f(=1111). 1(=0001) a(=1010) c(=1100) b(=1011) = 101110011111.00011010110010112
the Final answer: B9f.1acb16 = 101110011111.00011010110010112
Fill in the number with missing zeros on the right
let\'s make a direct translation from binary to post-binary like this:
101110011111.0001101011001011002 = 101 110 011 111. 000 110 101 100 101 100 = 101(=5) 110(=6) 011(=3) 111(=7). 000(=0) 110(=6) 101(=5) 100(=4) 101(=5) 100(=4) = 5637.0654548
the Final answer: B9f.1acb16 = 5637.0654548