The decimal expansion of frac{15}{2^2} termin | vedclass

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(B) To find the number of decimal places for a rational number $\frac{p}{q}$ where $q = 2^n \cdot 5^m$,the number of decimal places is equal to $\max(

Vedclass · Accepted Answer

$2$

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(B) To find the number of decimal places for a rational number $\frac{p}{q}$ where $q = 2^n \cdot 5^m$,the number of decimal places is equal to $\max(n, m)$.Given the fraction is $\frac{15}{2^2}$.Here,the denominator is $2^2 \cdot 5^0$.Comparing this with $2^n \cdot 5^m$,we get $n = 2$ and $m = 0$.The maximum value of $n$ and $m$ is $\max(2, 0) = 2$.Therefore,the decimal expansion terminates after $2$ decimal places.Verification: $\frac{15}{2^2} = \frac{15}{4} = 3.75$,which has $2$ decimal places.

The decimal expansion of $\frac{15}{2^2}$ terminates after how many decimal places?

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