\(3\sqrt{5}+\sqrt{125} = 17.88\)
\(\Rightarrow 3\sqrt{5}+\sqrt{25\times5} = 17.88\)
\(\Rightarrow 3\sqrt{5}+5\sqrt{5} = 17.88\)
\(\Rightarrow8\sqrt{5} = 17.88\)
\(\Rightarrow \sqrt{5}=2.235\)
\(\therefore \sqrt{80}+6\sqrt{5} =\sqrt{16\times5}+6\sqrt{5}\)
\(= 4\sqrt{5}+6\sqrt{5}\)
\(= 10\sqrt{5} = (10\times2.235)=22.35\)
\( \sqrt{4a^{2}-4a+1}+3a = \sqrt{(a^{2})+(2a)^{2}-2\times1\times2a}+3a\)
= \( \sqrt{(1-2a)^{2}}+3a\)
= (1 - 2a) + 3a
= (1 + a)
= (1 + 0.1039)
= 1.1039
\(x =\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}\times\frac{(\sqrt{3}+1)}{(\sqrt{3}+1)} = \frac{(\sqrt{3}+1)}{(3-1)}^{2} = \frac{3+1+2\sqrt{3}}{2} = 2+\sqrt{3}.\)
\(y =\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}\times\frac{(\sqrt{3}-1)}{(\sqrt{3}-1)} = \frac{(\sqrt{3}-1)}{(3-1)}^{2} = \frac{3+1-2\sqrt{3}}{2} = 2-\sqrt{3}\)
\(\therefore a^{b}+a^{b} = (2+\sqrt{3})^{2}+(2-\sqrt{3})^{2}\)
= 2(4 + 3)
= 14
Money collected = (59.29 x 100) paise = 5929 paise.
\(\therefore Number of members = \sqrt{5929} = 77.\)
\( \sqrt{(7+3\sqrt{5)}(7-3\sqrt{5)}}=\sqrt{(7)^{2}-(3\sqrt{5})^{2}} = \sqrt{49-45} = \sqrt{4} = 2\)
\(\frac{\sqrt{5}}{2} - \frac{10}{\sqrt{5}}+\sqrt{125} = \frac{(\sqrt{5})^{2}-20+2\sqrt{5}\times5\sqrt{5}}{2\sqrt{5}}\)
= \(\frac{5-20+50}{2\sqrt{5}}\)
= \(\frac{35}{2\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}\)
= \(\frac{35\sqrt{5}}{10}\)
= \(\frac{7\times2.236}{2}\)
= 7 x 1.118
= 7.826
Given Exspression = \( \frac{25}{11}\times\frac{14}{5}\times\frac{11}{14} = 5\)
Let \(\sqrt{0.0169\times x} = 1.3\)
Then, 0.0169x = (1.3)2 = 1.69
\(\Rightarrow x= \frac{1.69}{0.0169} = 100\)
\((\sqrt{3}-\frac{1}{\sqrt{3}})^{2} = (\sqrt{3})^{2}+(\frac{1}{\sqrt{3}})^{2}-2\times\sqrt{3}\times\frac{1}{\sqrt{3}}\)
\(= 3+\frac{1}{3}-2\)
\(= 1+\frac{1}{3}\)
\(= \frac{4}{3}\)
A number ending in 8 can never be a perfect square.
