\(\frac{\sqrt{72}+\sqrt{128}}{\sqrt{2}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{72}+\sqrt{128}}{\sqrt{2}}\)?
Explanation opens after your attempt
Correct Answer
B. (14)
Concept
\(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so the numerator is \(14\sqrt{2}\). Dividing gives (14).
Why this answer is correct
The correct answer is B. (14). \(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so the numerator is \(14\sqrt{2}\). Dividing gives (14).
Exam Tip
\(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\), इसलिए अंश \(14\sqrt{2}\) है। भाग देने पर (14) मिलता है।
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