यदि \(s=\sqrt{242}+\sqrt{128}\) है तो \(\frac{s}{\sqrt{2}}\) का मान क्या है?
If \(s=\sqrt{242}+\sqrt{128}\), what is the value of \(\frac{s}{\sqrt{2}}\)?
Explanation opens after your attempt
B. (19)
Concept
\(\sqrt{242}=11\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so \(s=19\sqrt{2}\). Dividing gives (19).
Why this answer is correct
The correct answer is B. (19). \(\sqrt{242}=11\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so \(s=19\sqrt{2}\). Dividing gives (19).
Exam Tip
\(\sqrt{242}=11\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\), इसलिए \(s=19\sqrt{2}\) है। भाग देने पर (19) मिलता है।
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