Matemática, perguntado por duda595959, 10 meses atrás

sistema binário de 25 35 43 e 53

Soluções para a tarefa

Respondido por GeBEfte
3

25_{10}~=~1\times2^4+1\times2^3+0\times2^2+0\times2^1+1\times2^0\\\\\boxed{25_{10}~=~11001_2}\\\\\\35_{10}~=~1\times2^5+0\times2^4+0\times2^3+0\times2^2+1\times2^1+1\times2^0\\\\\boxed{35_{10}~=~100011_2}\\\\\\43_{10}~=~1\times2^5+0\times2^4+1\times2^3+0\times2^2+1\times2^1+1\times2^0\\\\\boxed{43_{10}~=~101011_2}\\\\\\53_{10}~=~1\times2^5+1\times2^4+0\times2^3+1\times2^2+0\times2^1+1\times2^0\\\\\boxed{53_{10}~=~110101_2}

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