Given that \(\sqrt 6= 2.55,\) then the value of \(\sqrt\frac23+\sqr

1.	Given that \(\sqrt 6= 2.55,\) then the value of \(\sqrt\frac23+\sqrt[3]\frac32\) is(a) 4.48 (b) 4.49 (c) 4.5 (d) 4.675
Answer» (d) 4.492 \(\sqrt\frac23+3\times\sqrt\frac32 = \frac{\sqrt2}{\sqrt3}+3\times\frac{\sqrt3}{\sqrt2}\) \(\frac{\sqrt2\times\sqrt3}{\sqrt3\times\sqrt3} + 3\times\frac{\sqrt3\times\sqrt2}{\sqrt2\times\sqrt2}\) (Rationalising the denominator) = \(\frac{\sqrt6}{3}+\frac{3\times\sqrt6}{2}=\frac{2\sqrt6+9\sqrt6}{6}\) = \(\frac{11\sqrt6}{6}=\frac{11\times2.45}{6}=\frac{26.95}{6}\) = 4.492

Given that \(\sqrt 6= 2.55,\) then the value of \(\sqrt\frac23+\sqrt[3]\frac32\) is(a) 4.48 (b) 4.49 (c) 4.5 (d) 4.675

Answer»

(d) 4.492

\(\sqrt\frac23+3\times\sqrt\frac32 = \frac{\sqrt2}{\sqrt3}+3\times\frac{\sqrt3}{\sqrt2}\)

\(\frac{\sqrt2\times\sqrt3}{\sqrt3\times\sqrt3} + 3\times\frac{\sqrt3\times\sqrt2}{\sqrt2\times\sqrt2}\)

(Rationalising the denominator)

= \(\frac{\sqrt6}{3}+\frac{3\times\sqrt6}{2}=\frac{2\sqrt6+9\sqrt6}{6}\)

= \(\frac{11\sqrt6}{6}=\frac{11\times2.45}{6}=\frac{26.95}{6}\) = 4.492