By substituting the value of x in the following, compute y and z Â
(x)10 + (35) 8 = ( y ) 2 = ( z ) 16
ii (x)10 × (1010110) 2 = ( y ) 4 = ( z ) 10 Â
value of x is 29
Part i
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 290_{10} \\\\\n + 280_{10}\n\\end{array}}" "\\underline{\\begin{array}{cc}\n X \\\\\n + Y\n\\end{array}}"
Z
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 290_{10} \\\\\n + 720_{10}\n\\end{array}}"
101010
Hence discarding the leading 1 the result is 10.
Part ii
(x)10 × (1010110) 2 = ( y ) 4 = ( z ) 10 Â
(1010110)2=8610
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 10_{10} \\\\\n + 86_{10}\n\\end{array}}" "\\underline{\\begin{array}{cc}\n X \\\\\n + Y\n\\end{array}}"
Z
"\\underline{\\begin{array}{cc}\n \\space \\space \\space \\space 490_{10} \\\\\n + 086_{10}\n\\end{array}}"
57610
Hence the result is 576
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