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Answer:
[tex]1001011_2[/tex] [tex]+[/tex] [tex]100011_2[/tex] [tex]=[/tex] [tex]6E_{hex}[/tex]
Explanation:
Required
[tex]1001011_2 + 100011_2 = []_{16}[/tex]
First, carry out the addition in binary
[tex]1001011_2[/tex] [tex]+[/tex] [tex]100011_2[/tex] [tex]=[/tex] [tex]1101110_2[/tex]
The step is as follows (start adding from right to left):
[tex]1 + 1 = 10[/tex] --- Write 0 carry 1
[tex]1 + 1 + 1(carry) = 11[/tex] ---- Write 1 carry 1
[tex]0 + 0 + 1(carry) = 1[/tex] ---- Write 1
[tex]1 + 0 = 1[/tex] --- Write 1
[tex]0 + 0 = 0[/tex] ---- Write 0
[tex]0 + 0 = 0[/tex] ---- Write 0
[tex]1 + 1 = 10[/tex] --- Write 0 carry 1
No other number to add ; So, write 1 (the last carry)
So, we have:
[tex]1001011_2[/tex] [tex]+[/tex] [tex]100011_2[/tex] [tex]=[/tex] [tex]1101110_2[/tex]
Next, convert [tex]1101110_2[/tex] to base 10 using product rule
[tex]1101110_2 = 1 * 2^6 +1 * 2^5 + 0 * 2^4 + 1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 0 * 2^0[/tex]
[tex]1101110_2 = 64 +32 + 0 + 8 + 4 + 2 + 0[/tex]
[tex]1101110_2 = 110_{10}[/tex]
Lastly, convert [tex]110_{10}[/tex] to hexadecimal using division and remainder rule
[tex]110/16 \to 6\ R\ 14[/tex]
[tex]6/16 \to 0\ R\ 6[/tex]
Write the remainder from bottom to top;
[tex]110_{10} = 6(14)_{hex}[/tex]
In hexadecimal
[tex]14 \to E[/tex]
So, we have:
[tex]110_{10} = 6E_{hex}[/tex]
Hence:
[tex]1001011_2[/tex] [tex]+[/tex] [tex]100011_2[/tex] [tex]=[/tex] [tex]6E_{hex}[/tex]