If sqrt{3} = 1.732,then the value of frac{9 + | vedclass

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(B) Given expression: $\frac{9 + 2\sqrt{3}}{\sqrt{3}}$Divide each term in the numerator by the denominator:$= \frac{9}{\sqrt{3}} + \frac{2\sqrt{3}}{\s

Vedclass · Accepted Answer

$7.196$

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(B) Given expression: $\frac{9 + 2\sqrt{3}}{\sqrt{3}}$Divide each term in the numerator by the denominator:$= \frac{9}{\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3}}$$= \frac{9}{\sqrt{3}} + 2$Rationalize the first term:$= \frac{9 	imes \sqrt{3}}{\sqrt{3} 	imes \sqrt{3}} + 2 = \frac{9\sqrt{3}}{3} + 2 = 3\sqrt{3} + 2$Substitute the value of $\sqrt{3} = 1.732$:$= 3(1.732) + 2$$= 5.196 + 2$$= 7.196$

If $\sqrt{3} = 1.732$,then the value of $\frac{9 + 2\sqrt{3}}{\sqrt{3}}$ is

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