Answer
solution is shown in the table below:
\begin{equation}
\begin{array}{|c|c|c|}\hline \text { } & {\text { Binary } \text { }} & {\text { Equivalent base 10 form } \text { }} \\ \hline \mathrm{a} & {101010} & {42} \\ \hline \mathrm{b} & {100001} & {33} \\ \hline \mathrm{c} & {10111} & {23} \\ \hline \mathrm{d} & {0110} & {6} \\ \hline \mathrm{e} & {11111} & {31} \\ \hline\end{array}
\end{equation}
Work Step by Step
a.
Step $1 :$ To convert the binary representations to its equivalent base ten
representations, multiply every bit of the binary representation to
the power of two from right to left, start at $\left( 2^{0}\right)$
Step $2 :$ Represent the number in terms of its positions.
$\left(1 * 2^{5}\right)+ \left(0 * 2^{4}\right)+ \left(1 * 2^{3}\right)+\left(0^{*} 2^{2}\right)+\left(1 * 2^{1}\right) +\left(0 * 2^{0}\right)$
Step $3 :$ (0 * 8) + (1 * 4) + (0 * 2) + (1 * 1) = 32 + 0 + 8 + 0 + 2 + 0 = 42
Therefore, $(101010)_{2}=(42)_{10}$
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to solve [ b, c, d, e, f ] do the previous steps :
b. the solution is: 33
c. the solution is: 23
d. the solution is: 6
e. the solution is: 31
