Answer:
The value of the first "[tex]5[/tex]" in the number [tex]255,\!120[/tex] is ten times that of the second "[tex]5\![/tex]" in this number.
Step-by-step explanation:
What gives the number "[tex]255,\!120[/tex]" its value? Of course, each of its six digits has contributed. However, their significance are not exactly the same. For example, changing the first [tex]\verb!5![/tex] to [tex]\verb!6![/tex] would give [tex]2\mathbf{6}5,\!120[/tex] and increase the value of this number by [tex]10,\!000[/tex]. On the other hand, changing the second [tex]\verb!5!\![/tex] to [tex]\verb!6!\![/tex] would give [tex]25\mathbf{6},\!120[/tex], which is an increase of only [tex]1,\!000[/tex] compared to the original number.
The order of these two digits matter because the number "[tex]255,\!120[/tex]" is written using positional notation. In this notation, the position of each digits gives the digit a unique weight. For example, in [tex]255,\!120\![/tex]:
[tex]\begin{array}{|r||c|c|c|c|c|c|}\cline{1-7}\verb!Digit!& \verb!2! & \verb!5! & \verb!5! & \verb!1! & \verb!2! & \verb!0!\\\cline{1-7}\textsf{Index} & 5 & 4 & 3 & 2 & 1& 0 \\ \cline{1-7} \textsf{Weight} & 10^{5} & 10^{4} & 10^{3} & 10^{2} & 10^{1} & 10^{0}\\\cline{1-7}\end{array}[/tex].
(Note that the index starts at [tex]0[/tex] from the right-hand side.)
Using these weights, the value [tex]255,\!120[/tex] can be written as the sum:
[tex]\begin{aligned}& 255,\!120\\ &= 2 \times 10^{5} + 5 \times 10^{4} + 5 \times 10^{3} + 1 \times 10^{2} + 2 \times 10^{1} + 0 \times 10^{0} \\&=200,\!000 + 50,\!000 + 5,\!000 + 100 + 20 + 0 \end{aligned}[/tex].
As seen in this sum, the first "[tex]5[/tex]" contributed [tex]50,\!000[/tex] to the total value, while the second "[tex]5\![/tex]" contributed only [tex]5,\!000[/tex].
Hence: The value of the first "[tex]5[/tex]" in the number [tex]255,\!120[/tex] is ten times that of the second "[tex]5\![/tex]" in this number.
