(a) convert to hexadecimal: 1457.1110 Round to two digits past the hexadecimal point. (b) Convert your answer to binary, and then to octal Please show work and expain steps to solution. I amhaving some difficulty understanding and would appreciate someextra examples.

Finding out how many numbers you'll need is the first action you need to take. Hexadecimal is a base-16 coding system, hence we may ${\displaystyle {16}^0=1}$ and ${\displaystyle {16}^1=16}$ and ${\displaystyle {16}^2=256}$ and ${\displaystyle {16}^3=4096}$ since ${\displaystyle {16}^3}$ We know that the greatest digit is a multiple of 16 squared since is bigger than the number we require. First, we divide our decimal number by 256 while keeping out any decimals and simply utilizing the full amount. We receive 5 for our first digit because ${\displaystyle \frac{1457}{5}=5.69140625}$.
Working now with what is left over. Take the remaining to obtain ${\displaystyle 1457-256\times 5=177}$. Due to the fact that 177 is less than 256, this number is legitimate. Now that we've examined the second digit, which is represented by multiples of ${\displaystyle {16}^1.\frac{177}{16}=11.0625}$ So 11 or B is our second digit. Once more, we must locate the rest, thus ${\displaystyle 177-16\times 11=1}$ And once more, it is legitimate because one is less than 16. Our next digit in the ${\displaystyle {16}^0}$ power since our third digit is 3 and our remainder was 1. We currently have 5B1; however, we still need to consider the fractional portion of the number, which is the. or 11 in base 10 ${\displaystyle \frac{11}{100}}$.
In hex we our decimal point is ${\displaystyle \frac{1}{16}}$ and ${\displaystyle \frac{1}{{16}^2}}$ or ${\displaystyle {16}^{-1}}$ then ${\displaystyle {16}^{-2}}$ for two decimalplaces. So we need to know how many 16ths we havewith ${\displaystyle \frac{11}{100}}$. To do this we simple multiply .11 by 16 giving 1.76 so the first decimal point over is 1 and the remainderis ${\displaystyle \frac{11}{100}-\frac{1}{16}=.048}$ now with the now wee need to know what is howmany times ${\displaystyle {16}^{-2}}$ goes into .048 multiplying .048 by ${\displaystyle \frac{1}{\left({16}^{-2}\right)}=12}$.288 so the next digit is 12 or C if we wanted more numbers wewould continue by finding how many times ${\displaystyle {16}^{-3}}$ goes into theremainder of .${\displaystyle 048-12\times {16}^{-2}}$ However, because the issue only required two positions, we have arrived at the final solution of 5B1.1C.
Since binary is a multiple of hex, where base 2 goes into base 16, Part B is considerably simpler than Part A or ${\displaystyle 2^4=16}$ Therefore, all we need to do is convert each integer to binary immediately. The binary value of the first digit, "5", is changed to 5= 0101, followed by B=1011, 1= 0001 and 1= 0001, and C= 1100. then add the digits 5 and B and 1 together, yielding 010110110001.00011100 for base 2 or binary, followed by a decimal point, the numbers 1 and C. Now that Octal is base 8, we can accomplish this by directly translating from binary because $2^3=8$ Simply grouping the integers into groups of three and going rightward from the decimal point results in 000 and 111, which become 0 and 7, giving us everything we need to the right of the decimal point. 070 the last zero enters since there is nothing right of 111 now, going from right to left, we obtain 001=1, 110=6, 110=6, and 010=2 from the left of the decimal point. The conversion is reversed to provide 2661 for the portion on the left of the decimal point; combining the two parts yields 2661.070
in octal. Using the scientific function on the Windows calculator and turning on all the options in the view menue, then entering the number while decimal is chosen, then clicking on the base you want to go to, will allow you to check your results neatly. I hope you've found this useful.