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(B) To compare the numbers $\sqrt{3}, \sqrt[3]{4}, \sqrt[4]{10}$,we express them as powers with a common index.$1$. The indices are $2, 3, 4$. The lea

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

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(B) To compare the numbers $\sqrt{3}, \sqrt[3]{4}, \sqrt[4]{10}$,we express them as powers with a common index.$1$. The indices are $2, 3, 4$. The least common multiple $(LCM)$ of $2, 3, 4$ is $12$.$2$. Convert each number to the $12^{th}$ root:$\sqrt{3} = 3^{1/2} = 3^{6/12} = \sqrt[12]{3^6} = \sqrt[12]{729}$$\sqrt[3]{4} = 4^{1/3} = 4^{4/12} = \sqrt[12]{4^4} = \sqrt[12]{256}$$\sqrt[4]{10} = 10^{1/4} = 10^{3/12} = \sqrt[12]{10^3} = \sqrt[12]{1000}$$3$. Comparing the values inside the $12^{th}$ root: $256 $4$. Therefore,the ascending order is $\sqrt[12]{256} < \sqrt[12]{729} < \sqrt[12]{1000}$,which corresponds to $\sqrt[3]{4} < \sqrt{3} < \sqrt[4]{10}$.

Arrange the following numbers in ascending order: $\sqrt{3}, \sqrt[3]{4}, \sqrt[4]{10}$

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