Question

arrange in acsending order 4root3, 6root7, 12root48​

Answer 1

Step-by-step explanation:

arrange in acsending order 4root3, 6root7, 12root48

$\sqrt[4]{3} \: \sqrt[6]{7} \: \sqrt[12]{48}$

LCM OF 4,6,12 IS 12

CHANGE THE ORDER OF EACH SURD TI 12

$\sqrt[4]{3} = \sqrt[4 \times 3]{ {3}^{3} } = \sqrt[12]{27} - - - - > 1$

$\sqrt[6]{7} = \sqrt[6 \times2 ]{ {7}^{2} } = \sqrt[12]{49} \: \: \: \: - - - > 3$

$\sqrt[12]{48} - - - > 2$

s

$so \: ascending \: order \: \: is$

$\sqrt[4]{3} < \sqrt[12]{48} < \sqrt[6]{7}$