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recurring decimals (1 Viewer)

x1

MÒderator
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ik that but how do u do it the fraction way?

like for 0.444444...
u do 4/10 + 4/100 + 4/1000 +...
then a=4/10 r=1/10

and then use sum to infinity equation.
 

enigma_1

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2.8888888

= 2+ (0.8+0.08+0.008+………)

R=0.1 a=0.8

:. = 2+ a/(1-r) = 2+ 0.8/(1-0.1) = 2+8/9 = 26/9
 

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