{c}{B=24/5*(2^{frac{9-4{3+1}})^{2}}}{C=frac{1+operatorname{Tan}(2pi/9)^2}{sqrt{(4^2+1)}}}{D=4frac{e^-5^frac{3{2}}}{sin(pi/4)}} ઉકેલો

$$\left. \begin{array} { c } { B = \frac { 24 } { 5 \times ( 2 ^ { \frac { 9 - 4 } { 3 + 1 } } ) ^ { 2 } } } \\ { C = \frac { 1 + \operatorname { Tan } ( 2 \frac { \pi } { 9 } ) ^ { 2 } } { \sqrt { ( 4 ^ { 2 } + 1 ) } } } \\ { D = 4 \frac { e ^ { - 5 ^ { \frac { 3 } { 2 } } } } { \sin ( \frac { \pi } { 4 } ) } } \end{array} \right.$$

B, C, D માટે ઉકેલો

$$B=\frac{3\sqrt{2}}{5}\approx 0.848528137$$

$$C=\frac{\sqrt{17}}{17\left(\cos(\frac{2\pi }{9})\right)^{2}}\approx 0.413302095$$

$$D=\frac{4\sqrt{2}}{e^{5\sqrt{5}}}\approx 0.000078889$$