Arrange the following in ascending order of magnitude: $\sqrt[4]{90}, \sqrt[3]{10}, \sqrt{6}$

\sqrt{3} < \sqrt[4]{10} < \sqrt[3]{6}

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\sqrt{3} > \sqrt[4]{10} > \sqrt[3]{6}

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\sqrt{3} > \sqrt[4]{10} < \sqrt[3]{6}

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\sqrt{3} < \sqrt[4]{10} > \sqrt[3]{6}

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Solution

The correct option is A $\sqrt{3} < \sqrt[4]{10} < \sqrt[3]{6}$
$\sqrt[4]{10} = \sqrt[12]{1000}$
$\sqrt[3]{6} = \sqrt[12]{1296}$
$\sqrt{3^{6}} = \sqrt[12]{729}$
$\sqrt[12]{1296} > \sqrt[12]{1000} > \sqrt[12]{729}$

$\sqrt[3]{6} > \sqrt[4]{10} > \sqrt{3}$
or $\sqrt{3} < \sqrt[4]{10} < \sqrt[3]{6}$

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Similar questions

(1) \frac{- 3}{5}, \frac{7}{- 10}, \frac{- 5}{6}

(2) \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{- 7}{12}

Arrange the following in ascending order: $\frac{3}{12}, \frac{3}{7}, \frac{3}{10}, \frac{3}{4}$

4^{3} + 5^{3} + 6^{3} + \dots . . + 10^{3} =

Simplify:

$\frac{3^{- 4} \times 10^{- 5} \times (625)}{5^{- 3} \times 6^{- 4}}$

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