Question

are the term in ascending order √2,√3,√3,√2,-√3,2+√3​

Answer 1

Answer:

no, they are not

Step-by-step explanation:

-$\sqrt{3}$,$\sqrt{2}, \sqrt{2}, \sqrt{3}, \sqrt{3}, 2+\sqrt{3}$ is the valid ascending order

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Answer 2

Answer:

Step-by-step explanation:

Given:\sqrt[4]{3},\sqrt[3]{2},\sqrt[3]{4}

(i)\sqrt[4]{3}=3^{\frac{1}{4}}

(ii)\sqrt[3]{2}=2^{\frac{1}{3}}

(iii)\sqrt[3]{4}=4^{\frac{1}{3}}

LCM of 4,3,3 = 12.

So, LCM of denominator of exponents is 12.

(i)\sqrt[4]{3}=3^{\frac{1}{4}} = 3^{\frac{1}{4} * \frac{3}{3}} = 3^{\frac{3}{12}} = \sqrt[12]{3^3} = \sqrt[12]{27}

(ii)\sqrt[3]{2} = 2^{\frac{1}{3}} = 2^{\frac{1}{3} * \frac{4}{4}} = 2^{\frac{4}{12}} = \sqrt[12]{2^4} = \sqrt[12]{16}

(iii)\sqrt[3]{4} = 4^{\frac{1}{3}} = 4^{\frac{1}{3} * \frac{4}{4}} = 4^{\frac{4}{12}} = \sqrt[12]{4^4} = \sqrt[12]{256}

Ascending order of the numbers is:

\boxed{\therefore\sqrt[3]{2}, \sqrt[4]{3}, \sqrt[3]{4}}