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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{Choose the incorrect relation(s) from the following}\hspace{.33em}\\~\\ & i.)\ \sqrt{6}+\sqrt{2}=\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & ii.)\ \sqrt{6}+\sqrt{2}<\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & iii.)\ \sqrt{6}+\sqrt{2}>\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & a.)\ i.) \ \text{and}\ iii.)\hspace{.33em}\\~\\ & b.)\ ii.) \ \text{and}\ iii.)\hspace{.33em}\\~\\ & c.)\ i.) \hspace{.33em}\\~\\ & d.)\ ii.) \hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i have to solve this with in a minute

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0If a calculator is allowed you could just calculate the decimal values of both sides and check if they are equal.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1lol no calculator is allowed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2In view of arriving at a contradiction, lets suppose that \(\sqrt{6}+\sqrt{2}=\sqrt{5}+\sqrt{3}\). Square both sides and get \(6+2+2\sqrt{12}=5+3+2\sqrt{15}\). \(\implies \sqrt{12}=\sqrt{15}\) which is wrong. So the initial assumption is wrong.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Alternatively you may work it like this : \[\large{\begin{align} \sqrt{6}+\sqrt{2}~~&\stackrel{?}{}~~\sqrt{5}+\sqrt{3}\\~\\ (\sqrt{6}+\sqrt{2})^2~~&\stackrel{?}{}~~(\sqrt{5}+\sqrt{3})^2\\~\\ 6+2+2\sqrt{12}~~&\stackrel{?}{}~~ 5+3+2\sqrt{15}\\~\\ \sqrt{12}~~&\stackrel{?}{}~~ \sqrt{15}\\~\\ \sqrt{12}~~&\lt~~ \sqrt{15} \end{align}}\] since the numbers are positive, the implications will work in the reverse direction also. read it from bottom to top.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1option 1 is incorrect
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