How do you write #0.86# repeating as a fraction?

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Answer 1 · 2017-04-11T21:14:23

#86/99 = 0.bar(86)#

Let #x = 0.86868686....#

There are #2# digits recurring, multiply by #100#

#100 xx x = 86.868686.....#

If you subtract: #100x -x#, we get:

#100x = 86.86868686....#
#ul(" -x = -0.8686868686...)#
#99x = 86.000000000000....#

Now find a value for #x#

#x = 86/99#

Check with a calculator:

#86 div 99 = 0.86868686.. = 0.bar(86)#