Question Number 176610 by MASANJAJJ last updated on 23/Sep/22 Answered by BaliramKumar last updated on 23/Sep/22 $$\left({i}\right)\:\frac{\mathrm{3}×\mathrm{4}}{\mathrm{3}+\mathrm{4}}\:=\:\frac{\mathrm{12}}{\mathrm{7}}\:=\:\mathrm{1}\frac{\mathrm{5}}{\mathrm{7}}\:{hour} \\ $$$$\left({ii}\right)\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{6}}}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{4}+\mathrm{3}−\mathrm{2}}{\mathrm{12}}}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{5}}{\mathrm{12}}}\:=\:\frac{\mathrm{12}}{\mathrm{5}}\:=\:\mathrm{2}\frac{\mathrm{2}}{\mathrm{5}}\:=\:\mathrm{2}.\mathrm{4}\:{hour} \\ $$ Commented by Rasheed.Sindhi last…
Question Number 176607 by MASANJAJJ last updated on 23/Sep/22 Answered by FelipeLz last updated on 23/Sep/22 $${a}.\:{x}\left({y}+\mathrm{4}\right)\:=\:\left({x}+\mathrm{5}\right){y}\:\Rightarrow\:\cancel{{xy}}+\mathrm{4}{x}\:=\:\cancel{{xy}}+\mathrm{5}{y}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{5}{y}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\left({x}+\mathrm{5}\right){y}\:=\:\mathrm{400}\:\Rightarrow\:\frac{\mathrm{4}}{\mathrm{5}}×\left(\frac{\mathrm{5}}{\mathrm{4}}{y}^{\mathrm{2}} +\mathrm{5}{y}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}×\mathrm{400}\:\Rightarrow\:{y}^{\mathrm{2}} +\mathrm{4}{y}−\mathrm{320}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{y}\:=\:\frac{−\mathrm{4}\pm\sqrt{\mathrm{1296}}}{\mathrm{2}}\:=\:\begin{cases}{\mathrm{16}\:}\\{\cancel{−\mathrm{20}}}\end{cases} \\ $$$$\:\:\:\:\:\mathrm{16}\:{L}…
Question Number 110920 by Dwaipayan Shikari last updated on 31/Aug/20 Commented by Dwaipayan Shikari last updated on 31/Aug/20 $${I}\:{have}\:{found}\:{this}\:{while}\:{experimenting} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\mathrm{2}^{{n}} }=\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}.\mathrm{2}^{\mathrm{2}} }+….=−{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)={log}\left(\mathrm{2}\right)…
Question Number 110919 by Dwaipayan Shikari last updated on 31/Aug/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{r}} {r}!}\left(\underset{{k}=\mathrm{1}} {\overset{{r}} {\prod}}\left(\mathrm{2}{k}−\mathrm{1}\right)\right)\right) \\ $$ Answered by mnjuly1970 last updated on…
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