How do you write #-0.5555...#, the #5# repeating, as a decimal?

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Answer 1 · 2017-08-29T14:52:42

See a solution process below:

First, equate: #-0.bar5# to #x# giving:

#x = -0.bar5#

Next, multiply each side of the equation by #10# giving:

#10x = -5.bar5#

Then subtract each side of the first equation from each side of the second equation giving:

#10x - x = -5.bar5 - (-0.bar5)#

Now, solve this equation for #x#:

#10x - 1x = -(5 + 0.bar5) - (-0.bar5)#

#(10 - 1)x = -5 - 0.bar5 + 0.bar5#

#9x = -5 - 0#

#9x = -5#

#9x/color(red)(9) = -5/color(red)(9)#

#color(red)(cancel(color(black)(9)))x/cancel(color(red)(9)) = -5/9#

#x = -5/9#

#-0.bar5 = -5/9#