The value of frac{sqrt{32}+sqrt{48}}{sqrt{8}+ | vedclass

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Question

(D) Given expression: $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$Simplify the square roots in the numerator and denominator:$\sqrt{32} = \sqrt{16

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given expression: $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$Simplify the square roots in the numerator and denominator:$\sqrt{32} = \sqrt{16 	imes 2} = 4\sqrt{2}$$\sqrt{48} = \sqrt{16 	imes 3} = 4\sqrt{3}$$\sqrt{8} = \sqrt{4 	imes 2} = 2\sqrt{2}$$\sqrt{12} = \sqrt{4 	imes 3} = 2\sqrt{3}$Substitute these values back into the expression:$\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$Factor out the common terms:$\frac{4(\sqrt{2} + \sqrt{3})}{2(\sqrt{2} + \sqrt{3})}$Cancel the common term $(\sqrt{2} + \sqrt{3})$:$\frac{4}{2} = 2$Hence,the correct option is $(d)$.

The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to

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