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When the denominator of \dfrac{2}{\sqrt{3}} 3 ​ 2 ​ start fraction, 2, divided by, square root of, 3, end square root, end fraction is rationalized, it becomes \dfrac{2k}{3} 3 2k ​ start fraction, 2, k,
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When the denominator of \dfrac{2}{\sqrt{3}} 3 ​ 2 ​ start fraction, 2, divided by, square root of, 3, end square root, end fraction is rationalized, it becomes \dfrac{2k}{3} 3 2k ​ start fraction, 2, k, divided by, 3, end fraction. Find kkk

User Sreekar
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Answer:

Explanation:

k = 6

User Dan Mangiarelli
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