संख्या 1980 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 1980?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3^2\times5\times11\)

Step 1

Concept

Write \(1980=198\times10\).

Step 2

Why this answer is correct

\(198=2\times3^2\times11\) and \(10=2\times5\), so \(1980=2^2\times3^2\times5\times11\).

Step 3

Exam Tip

Since 2 appears twice, write \(2^2\). चरण 1: \(1980=198\times10\) लिखें। चरण 2: \(198=2\times3^2\times11\) और \(10=2\times5\), इसलिए \(1980=2^2\times3^2\times5\times11\)। चरण 3: 2 दो बार आता है, इसलिए \(2^2\) लिखें।

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संख्या 1980 का सही अभाज्य गुणनखंडन क्या है? / What is the correct prime factorisation of 1980?

Correct Answer: A. \(2^2\times3^2\times5\times11\). Explanation: चरण 1: \(1980=198\times10\) लिखें। चरण 2: \(198=2\times3^2\times11\) और \(10=2\times5\), इसलिए \(1980=2^2\times3^2\times5\times11\)। चरण 3: 2 दो बार आता है, इसलिए \(2^2\) लिखें। / Step 1: Write \(1980=198\times10\). Step 2: \(198=2\times3^2\times11\) and \(10=2\times5\), so \(1980=2^2\times3^2\times5\times11\). Step 3: Since 2 appears twice, write \(2^2\).

Which concept should I revise for this Mathematics MCQ?

Write \(1980=198\times10\).

What exam hint can help solve this Mathematics question?

Since 2 appears twice, write \(2^2\). चरण 1: \(1980=198\times10\) लिखें। चरण 2: \(198=2\times3^2\times11\) और \(10=2\times5\), इसलिए \(1980=2^2\times3^2\times5\times11\)। चरण 3: 2 दो बार आता है, इसलिए \(2^2\) लिखें।