+0  
 
0
228
2
avatar

Let . When  is written out as a repeating decimal, what is the sum of the digits in a single repeating period? (In other words, what is the sum of the digits covered by the repeat bar?)

Guest May 24, 2018
 #1
avatar+94181 
+1

Let       \(y=(0.\overline{12})+(0.\overline{345})\).

 

When y is written out as a repeating decimal, what is the sum of the digits in a single repeating period? (In other words, what is the sum of the digits covered by the repeat bar?)

 

Let t =  0.1212121212........

then 100t= 12.12121212.....

100t - t = 12

99t=12

t= 12/99

 

0.1212121212... = 12/99

 

Now do the same thing with 0.345345345.....

only since there are 3 digits repeating you will have to multiply by 1000.

 

Now if you understand what i have written you can finish it yourself.

Melody  May 24, 2018
 #2
avatar+92855 
+1

0.12 (repetaing)  = 12/99 =   4/33

 

0.345 (repeating)  = 345/999  = 115/333

 

So 

 

4/33  + 115/333  =

 

[4* 333 + 115 * 33] / [ 33 *333] =    1709 / 3663  =   .466557   (repeating)

 

So...the sum is  33

 

 

 

cool cool cool

CPhill  May 25, 2018

12 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.