Asked by Alice
Write the recurring decimal 0.abcabcabc… as a fraction.
a) (a+b+c)/999
b) (100a+10b+c)/999
c) (abc)/999 --> I did this but my teacher said I was wrong
d) (100c+10b+a)/999
a) (a+b+c)/999
b) (100a+10b+c)/999
c) (abc)/999 --> I did this but my teacher said I was wrong
d) (100c+10b+a)/999
Answers
R_scott
n = .abcabcabc…
multiplying ... 1000 n = abc.abcabcabc…
subtracting ... 999 n = abc
dividing ... n = abc / 999
curious as to which answer your teacher thought was correct
multiplying ... 1000 n = abc.abcabcabc…
subtracting ... 999 n = abc
dividing ... n = abc / 999
curious as to which answer your teacher thought was correct
Reiny
abc written as it, meant a*b*c, that is, they are multiplied
you will have to take their place-holder values
.abc = a/10 + b/100 + c/1000
= (100a + 10b + c)/1000
e.g. 0.234 = 2/10 + 3/100 + 4/1000
but since the abc repeats, we use 999 instead of 1000
so the correct answer is
(100a+10b+c)/999 , which would be b)
you will have to take their place-holder values
.abc = a/10 + b/100 + c/1000
= (100a + 10b + c)/1000
e.g. 0.234 = 2/10 + 3/100 + 4/1000
but since the abc repeats, we use 999 instead of 1000
so the correct answer is
(100a+10b+c)/999 , which would be b)
AJ L
abc would imply a*b*c!
100a+10b+1c would be more appropriate
100a+10b+1c would be more appropriate