\(\sqrt{2}\times\sqrt{3}\) किसके बराबर है?

What is \(\sqrt{2}\times\sqrt{3}\) equal to?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

B. \(\sqrt{6}\)

Step 1

Concept

\(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) and \(\sqrt{6}\) is irrational. In multiplication multiply the numbers inside roots.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{6}\). \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) and \(\sqrt{6}\) is irrational. In multiplication multiply the numbers inside roots.

Step 3

Exam Tip

\(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) होता है और \(\sqrt{6}\) अपरिमेय है। गुणन में अंदर की संख्याएँ गुणा करें।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

\(\sqrt{2}\times\sqrt{3}\) किसके बराबर है? / What is \(\sqrt{2}\times\sqrt{3}\) equal to?

Correct Answer: B. \(\sqrt{6}\). Explanation: \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) होता है और \(\sqrt{6}\) अपरिमेय है। गुणन में अंदर की संख्याएँ गुणा करें। / \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) and \(\sqrt{6}\) is irrational. In multiplication multiply the numbers inside roots.

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) and \(\sqrt{6}\) is irrational. In multiplication multiply the numbers inside roots.

What exam hint can help solve this Mathematics question?

\(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) होता है और \(\sqrt{6}\) अपरिमेय है। गुणन में अंदर की संख्याएँ गुणा करें।