Expert Mathematics Real Numbers Class 10 Level 19

\(\frac{1}{2^2\cdot 5^2\cdot 9}\) के बारे में सही कथन कौन-सा है?

Which statement is correct about \(\frac{1}{2^2\cdot 5^2\cdot 9}\)?

Explanation opens after your attempt
Correct Answer

B. असांत आवर्ती और (2) अनावर्ती आरंभिक अंकNon-terminating recurring with (2) initial non-repeating digits

Step 1

Concept

Since \(9=3^2\) remains, the decimal is non-terminating recurring. The larger exponent in \(2^2\cdot 5^2\) gives (2) initial non-repeating digits.

Step 2

Why this answer is correct

The correct answer is B. असांत आवर्ती और (2) अनावर्ती आरंभिक अंक / Non-terminating recurring with (2) initial non-repeating digits. Since \(9=3^2\) remains, the decimal is non-terminating recurring. The larger exponent in \(2^2\cdot 5^2\) gives (2) initial non-repeating digits.

Step 3

Exam Tip

\(9=3^2\) बचता है, इसलिए दशमलव असांत आवर्ती होगा। \(2^2\cdot 5^2\) की बड़ी घात (2) आरंभिक अनावर्ती भाग दिखाती है।

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FAQs

Mathematics Answer, Explanation and Revision Hints

\(\frac{1}{2^2\cdot 5^2\cdot 9}\) के बारे में सही कथन कौन-सा है? / Which statement is correct about \(\frac{1}{2^2\cdot 5^2\cdot 9}\)?

Correct Answer: B. असांत आवर्ती और (2) अनावर्ती आरंभिक अंक / Non-terminating recurring with (2) initial non-repeating digits. Explanation: \(9=3^2\) बचता है, इसलिए दशमलव असांत आवर्ती होगा। \(2^2\cdot 5^2\) की बड़ी घात (2) आरंभिक अनावर्ती भाग दिखाती है। / Since \(9=3^2\) remains, the decimal is non-terminating recurring. The larger exponent in \(2^2\cdot 5^2\) gives (2) initial non-repeating digits.

Which concept should I revise for this Mathematics MCQ?

Since \(9=3^2\) remains, the decimal is non-terminating recurring. The larger exponent in \(2^2\cdot 5^2\) gives (2) initial non-repeating digits.

What exam hint can help solve this Mathematics question?

\(9=3^2\) बचता है, इसलिए दशमलव असांत आवर्ती होगा। \(2^2\cdot 5^2\) की बड़ी घात (2) आरंभिक अनावर्ती भाग दिखाती है।

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