1110
10111
11101
111100
Correct answer is B
y = 23\(_{five}\) + 101\(_{three}\)
23\(_{five}\) = \(2 \times 5^1 + 3 \times 5^0\)
= 13\(_{ten}\)
101\(_{three}\) = \(1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0\)
= 10\(_{ten}\)
y\(_{ten}\) = 13\(_{ten}\) + 10\(_{ten}\)
= 23\(_{ten}\)
= 10111\(_{two}\)
Solve without using tables log5(62.5) - log5(\(\frac{1}{2}\))...
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