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:
2∙163+10∙162+12∙161+5∙160+2∙16-1 = 2∙4096+10∙256+12∙16+5∙1+2∙0.0625 = 8192+2560+192+5+0.125 = 10949.12510
got It: 2AC5.216 =10949.12510
Translate the number 10949.12510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
10949 | 8 | | | | |
-10944 | 1368 | 8 | | | |
5 | -1368 | 171 | 8 | | |
| 0 | -168 | 21 | 8 | |
| | 3 | -16 | 2 | |
| | | 5 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 125*8 |
1 | .0*8 |
the result of the conversion was:
10949.12510 = 25305.18
the Final answer: 2AC5.216 = 25305.18
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
2AC5.216 = 2 A C 5. 2 = 2(=0010) A(=1010) C(=1100) 5(=0101). 2(=0010) = 10101011000101.0012
the Final answer: 2AC5.216 = 10101011000101.0012
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
010101011000101.0012 = 010 101 011 000 101. 001 = 010(=2) 101(=5) 011(=3) 000(=0) 101(=5). 001(=1) = 25305.18
the Final answer: 2AC5.216 = 25305.18