Please refer to a similar question for details.
As #5/12# is already in smallest form and factors of denominators are #2xx2xx3#. As prime factors are other than #2# and #5#, we have to long divide #5# by #12# and answer will have repeating numbers after decimal point.
#color(white)(xxxxxx)0.4 color(white)(x)1color(white)(x)6color(white)(x)6#
#color(white)(xx) 12| bar(5color(white)(x).0color(white)(x) 0)color(white)(x) 0#
#color(white)(xxxxx)ul(4color(white)("x)8)color(red)(darr#
#color(white)(xxxxxxx)2 color(white)(x) "0"#
#color(white)(xxxxxxx)ul(1 color(white)(x) 2 )"#
#color(white)(xxxxxxxxx) 8 color(white)(x) "0"#
#color(white)(xxxxxxxxx)ul(7 color(white)(x) 2 )"#
#color(white)(xxxxxxxxxxx) 8 color(white)(x) "0"#
#color(white)(xxxxxxxxxxx)ul(7 color(white)(x) 2 )"#
#color(white)(xxxxxxxxxxxxx) 8 #
As the remainder continues to be #8#, the digit #8# will repeat endlessly and hence
#5/12=0.41666666666..# or #0.41bar6#
