Question

What is the decimal form of
129/2^2×5^7×7^5...

Answer 1

Answer:

here your answer

Step-by-step explanation:

129/2 25 77 5 will have a non-terminating repeating decimal expansion. = which is of the form 2 m· 5 n. ∴ 6/15 will have a terminating decimal expansion. ∵ 50 = 2 × 5 × 5 = 21 × 52, which is of the form 2 n · 5 m.

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Answer 2

The decimal form of $\displaystyle \sf{ \frac{129}{ {2}^{2} \times {5}^{7} \times {7}^{5} } }$ is non terminating

Given : $\displaystyle \sf{ \frac{129}{ {2}^{2} \times {5}^{7} \times {7}^{5} } }$

To find : The decimal form of the fraction

Tip :

$\displaystyle\sf{Fraction = \frac{Numerator}{Denominator} }$

A fraction is said to be terminating if prime factorisation of the denominator contains only prime factors 2 and 5

If the denominator is of the form

$\sf{Denominator = {2}^{m} \times {5}^{n} }$

Then the fraction terminates after N decimal places

Where N = max { m , n }

Solution :

Here the given fraction is

$\displaystyle \sf{ \frac{129}{ {2}^{2} \times {5}^{7} \times {7}^{5} } }$

We know that fraction is said to be terminating if prime factorisation of the denominator contains only prime factors 2 and 5

Since prime factorisation of denominator contains prime factor 7

So the decimal expansion of the given fraction is non terminating

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