Answer
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Hint: First see carefully that the given rational number is in fraction form such that the denominator is not any multiple or power of 10. So first we will try to convert it to a power of 10 and then will go for the decimal system. We can see in the denominator the numbers are 2 and 5. Their product gives 10 but unless their powers are the same. So first make the power the same and then follow the process above.
Complete step-by-step answer:
Given the rational number is,
\[\dfrac{{229}}{{{2^2} \times {5^7}}}\]
Now we will multiply the numerator and denominator by \[{2^5}\],
\[ = \dfrac{{229 \times {2^5}}}{{{2^2} \times {5^7} \times {2^5}}}\]
Taking the base of 2 nearer,
\[ = \dfrac{{229 \times {2^5}}}{{{2^2} \times {2^5} \times {5^7}}}\]
We know that, \[{a^m}{a^n} = {a^{m + n}}\] so applying this we get,
\[ = \dfrac{{229 \times {2^5}}}{{{2^7} \times {5^7}}}\]
Now we know \[{a^m}{b^m} = {\left( {ab} \right)^m}\]and taking the fifth power of 2 so we can write,
\[ = \dfrac{{229 \times 32}}{{{{\left( {2 \times 5} \right)}^7}}}\]
Taking the product in both numerator and denominator,
\[ = \dfrac{{7328}}{{{{10}^7}}}\]
Taking the power of 10,
\[ = \dfrac{{7328}}{{1,00,00,000}}\]
Now on dividing we get,
\[ = 0.0007328\]
This states that the decimal representation of the rational number will terminate after 7 decimal places.
So, the correct answer is “0.0007328”.
Note: Note that, decimal is related to 10 or powers of 10 generally. So we tried to get the denominator as that. Also note that termination is nothing but removal. So there are 7 digits after decimal. So the termination will be after 7 decimal places. Remember the rules of indices very well!
Also note that, since the power of 2 is smaller than the power of 5 we raised the power of 2; not decreased the power of 5.
Complete step-by-step answer:
Given the rational number is,
\[\dfrac{{229}}{{{2^2} \times {5^7}}}\]
Now we will multiply the numerator and denominator by \[{2^5}\],
\[ = \dfrac{{229 \times {2^5}}}{{{2^2} \times {5^7} \times {2^5}}}\]
Taking the base of 2 nearer,
\[ = \dfrac{{229 \times {2^5}}}{{{2^2} \times {2^5} \times {5^7}}}\]
We know that, \[{a^m}{a^n} = {a^{m + n}}\] so applying this we get,
\[ = \dfrac{{229 \times {2^5}}}{{{2^7} \times {5^7}}}\]
Now we know \[{a^m}{b^m} = {\left( {ab} \right)^m}\]and taking the fifth power of 2 so we can write,
\[ = \dfrac{{229 \times 32}}{{{{\left( {2 \times 5} \right)}^7}}}\]
Taking the product in both numerator and denominator,
\[ = \dfrac{{7328}}{{{{10}^7}}}\]
Taking the power of 10,
\[ = \dfrac{{7328}}{{1,00,00,000}}\]
Now on dividing we get,
\[ = 0.0007328\]
This states that the decimal representation of the rational number will terminate after 7 decimal places.
So, the correct answer is “0.0007328”.
Note: Note that, decimal is related to 10 or powers of 10 generally. So we tried to get the denominator as that. Also note that termination is nothing but removal. So there are 7 digits after decimal. So the termination will be after 7 decimal places. Remember the rules of indices very well!
Also note that, since the power of 2 is smaller than the power of 5 we raised the power of 2; not decreased the power of 5.
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