Question

Root 3,cube root 2, fourth root 7 form an asending order

Answer 1

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Step-by-step explanation:

$\sqrt{3} \: \: \: \: \sqrt[3]{2} \: \: \: \: \: \sqrt[4]{7} \\ {3}^{ \frac{1}{2} } \: \: \: \: {2}^{ \frac{1}{3} } \: \: \: \: {7}^{ \frac{1}{4} } \\ lcm \: of \: 2 \: \: \: \: 3 \: \: \: \: 4 = 12 \\ {3}^{ \frac{1}{2} \times \frac{6}{6} } \: \: \: \: \: {2}^{ \frac{1}{3} \times \frac{4}{4} } \: \: \: \: \: {7}^{ \frac{1}{4} \times \frac{3}{3} } \\ {3}^{ \frac{6}{12} } \: \: \: \: \: {2}^{ \frac{4}{12} } \: \: \: \: {7}^{ \frac{3}{12} } \: \: \\ ascending \: order \\ {3}^{ \frac{6}{12} } \: \: \: \: {2}^{ \frac{4}{12} } \: \: \: \: \: {7}^{ \frac{3}{12} } \\ ansssssssssssss$