write the conditions satisfied by 455,so that 64/455 has non-terminating repeating decimal expansion
Answers
Answered by
8
IF number is in denominator in 2^m × 5^n then it is terminating
Ex - 1/5= 1/5^1×2^0 (2^0=1)
In this case
64/455=2^6/5^1×7^1×13^1
Here it is nonterminating as 2 is not in denominator and when u solve 64/455=
0.1406593407 and hence it is non repeating
OR
(iii) 64/455
Factorize the denominator we get
455 =5 × 7 × 13
There are 7 and 13 also in denominator so denominator is not in form of 2m × 5n . so it is not terminating.
Ex - 1/5= 1/5^1×2^0 (2^0=1)
In this case
64/455=2^6/5^1×7^1×13^1
Here it is nonterminating as 2 is not in denominator and when u solve 64/455=
0.1406593407 and hence it is non repeating
OR
(iii) 64/455
Factorize the denominator we get
455 =5 × 7 × 13
There are 7 and 13 also in denominator so denominator is not in form of 2m × 5n . so it is not terminating.
Kush1601:
Bye
in which school you are studying
why do u want to know
First tell in which school u are studying
then I will tell
tell
bye
Kendriya Vidyalaya number 1 Arakkonam
bye
hi
Similar questions