Question

The decimal expansion of number 441/2²×5³×7 has​

Answer 1

Answer:

HEY MATE

Given :

To find the that the number \frac{441}{2^2 5^3 7}22537441 is rational & has terminating decimal expansion or not,.

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We know that,.

A rational number is in the form \frac{p}{q}qp (q≠0),.

The fraction \frac{441}{2^25^37}22537441 is in the form \frac{p}{q}qp (q≠0 As, 2²×5³×7 ≠ 0),.

Hence, it is rational number,.

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We know that,

A number with terminal decimal expansions have the denominator  in the form,

2^m 5^n2m5n where m & n ∈ W,.

The number \frac{441}{2^25^37} = \frac{3^27^2}{2^25^37} = \frac{3^27}{2^25^3} = \frac{63}{2^25^3}22537441=225373272=2253327=225363

Which the denominator is in the form,

2^m5^n = 2^25^32m5n=2253 with m = 2 , n = 3,.

Hence, it has terminal decimal expansion,.

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Answer 2

The decimal expansion of 441/2² × 5³ × 7 has terminating decimal expansion which terminates after 3 decimal places

Given : The fraction 441/2² × 5³ × 7

To find : Terminating / Non terminating decimal expansion

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

$\displaystyle \sf{ \frac{441}{ {2}^{2} \times {5}^{3} \times 7 } }$

We first simplify the given fraction

$\displaystyle \sf{ \frac{441}{ {2}^{2} \times {3}^{3} \times 7 } }$

$\displaystyle \sf{ = \frac{63}{ {2}^{2} \times {5}^{3} } }$

Since prime factorisation of the denominator contains only prime factors 2 and 5

So decimal expansion of the given fraction is terminating

Exponent of 2 = 2

Exponent of 5 = 3

Max { 2 , 3} = 3

So the fraction terminates after 3 decimal places

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