Question

Evaluate 15 by root 10 + root 20 + root 40 -root 5 - root 80 , it being given that root 5 is = 2 .236 and root 10 is= 3 . 162

Answer 1

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Answer 2

Answer:

$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}=5.398$

Step-by-step explanation:

$Given \\\sqrt{5}=2.236,\:\sqrt{10}=3.162$

$i)\sqrt{10}--(1)$

$ii)\sqrt{20}=\sqrt{2^{2}\times 5}=2\sqrt{5}--(2)$

$iii)\sqrt{5}--(3)$

$iv)\sqrt{40}=\sqrt{2^{2}\times 10}=2\sqrt{10}--(4)$

$v)\sqrt{80}=\sqrt{4^{2}\times 5}=4\times 5--(5)$

$Now,\\\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}\\=\frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}\\=\frac{15}{3\sqrt{10}-3\sqrt{5}}\\=\frac{15}{3(\sqrt{10}-\sqrt{5})}$

/* Rationalising the denominator, we get

$=\frac{15(\sqrt{10}+\sqrt{5}}{3(\sqrt{10}-\sqrt{5})(\sqrt{10}+\sqrt{5})}\\=\frac{15(\sqrt{10}+\sqrt{5}}{3[(\sqrt{10})^{2}-(\sqrt{5})^{2}]}\\=\frac{15(\sqrt{10}+\sqrt{5}}{3(10-5)}\\=\frac{15(\sqrt{10}+\sqrt{5})}{3\times 5 }\\=\frac{15(\sqrt{10}+\sqrt{5})}{15}\\=\sqrt{10}+\sqrt{5}\\=3.162+2.236\\=5.398$

Therefore,

$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}=5.398$

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