Question

rewrite the following irrational numbers in descending order: 3rdroot of 10, 3rd root of 36, root of 3, 6th root of 5, 10th root of 60​

Answer 1

Answer:

$\sqrt[3]{36} > \sqrt[3]{10} > \sqrt{3} > \sqrt[10]{60} > \sqrt[6]{5}$
Solution:

3rdroot of 10=>
$\sqrt[3]{10} = 2.154$
3rd root of 36=>
$\sqrt[3]{36} \\ \\ = 3.30$
root of 3=>
$\sqrt{3} \\ \\ = 1.73$
6th root of 5=>
$\sqrt[6]{5} \\ \\ = 1.30$
10th root of 60=>
$\sqrt[10]{60} \\ \\ = 1.50$
So descending arrangement of numbers are

3.30,2.154,1.732,1.50,1.30

$\sqrt[3]{36} > \sqrt[3]{10} > \sqrt{3} > \sqrt[10]{60} > \sqrt[6]{5}$