129 by ^ 2 into 5 power 7 into 7 ^ 5
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kundan1419
Primary School Math 5+3 pts
Solve: 129/2^2×5^7×7^5...
Report by Shafra22 06.03.2019
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mridulaa28
Mridulaa28Ambitious
To have an terminating decimal expansion the denominator must have only 2 and 5 as its factors in the simlified form of fraction for any degree or power of 2 and 5
denominator = 2^2 × 5^7 × 7^5
Since it have 7 also in its prime factors. Thus it doesn't have an terminating decimal expansion.
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