Question Number 36821 by Ajayraj1995 last updated on 06/Jun/18 $$>.\:\mathrm{2}{sin}\frac{\mathrm{5}\pi}{\mathrm{12}}{sin}\frac{\pi}{\mathrm{12}}\:{slove}\:{this}.? \\ $$ Commented by maxmathsup by imad last updated on 06/Jun/18 $$=\mathrm{2}\:{sin}\left(\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{12}}\right){sin}\left(\frac{\pi}{\mathrm{12}}\right)\:=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{12}}\right){sin}\left(\frac{\pi}{\mathrm{12}}\right)={sin}\left(\frac{\pi}{\mathrm{6}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$ Answered…
Question Number 102348 by mohammad17 last updated on 08/Jul/20 Answered by mr W last updated on 08/Jul/20 $${a}_{{n}} =\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}} \\…
Question Number 102347 by mohammad17 last updated on 08/Jul/20 Commented by bobhans last updated on 08/Jul/20 $$\mathrm{ln}\left(\mathrm{y}\right)\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}}\left(\mathrm{1}+\mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{ln}\left(\mathrm{y}\right)\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{2}}\right)}{{n}}\:=\:\mathrm{0} \\ $$$${y}\:=\:{e}^{\mathrm{0}} \:=\:\mathrm{1} \\…
Question Number 167872 by mkam last updated on 28/Mar/22 Commented by Tinku Tara last updated on 28/Mar/22 $${f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{continuous}\:\mathrm{at}\:\mathrm{all}\:{x}\in{Q} \\ $$ Commented by mokys last updated…
Question Number 167851 by otchereabdullai@gmail.com last updated on 27/Mar/22 $$\:\:\mathrm{A}\:\mathrm{particular}\:\mathrm{A}.\mathrm{P}\:\mathrm{has}\:\mathrm{a}\:\mathrm{positive}\: \\ $$$$\:\:\mathrm{common}\:\mathrm{difference}\:\mathrm{and}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\mathrm{for}\:\mathrm{any}\:\mathrm{three}\:\mathrm{adjacent}\:\mathrm{terms},\:\mathrm{three} \\ $$$$\:\mathrm{times}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{squares}\:\mathrm{exceed} \\ $$$$\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{by}\:\mathrm{37}.\mathrm{5}\:.\:\mathrm{Find} \\ $$$$\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}.\: \\ $$ Answered by mr…
Question Number 167853 by LEKOUMA last updated on 27/Mar/22 $${Par}\:{d}\acute {{e}velopement}\:{limit}\acute {{e}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)−\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{cos}\:{x}^{\mathrm{2}} } \\ $$ Answered by qaz last updated on…
Question Number 102301 by john santu last updated on 08/Jul/20 $${to}\:{all}\:{friends}\:{in}\:{this}\:{forum}. \\ $$$${i}'{ll}\:{be}\:{off}\:{for}\:{a}\:{while}\:.\:{thank}\:{you} \\ $$$${for}\:{the}\:{discussion}\:{in}\:{this}\:{forum} \\ $$$$.\:{see}\:{you}\:{another}\:{time}.\:\left({JS}\:\circledast\right) \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 167828 by otchereabdullai@gmail.com last updated on 26/Mar/22 $$\:\mathrm{Houses}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particular}\: \\ $$$$\:\mathrm{street}\:\mathrm{are}\:\mathrm{assigned}\:\mathrm{odd}\:\mathrm{numbers},\: \\ $$$$\:\mathrm{starting}\:\mathrm{from}\:\mathrm{11}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{551},\:\mathrm{how}\:\mathrm{many}\:\mathrm{houses} \\ $$$$\:\mathrm{are}\:\mathrm{there}? \\ $$ Answered by mr W last…
Question Number 102292 by Boykisss last updated on 08/Jul/20 Answered by Rio Michael last updated on 09/Jul/20 $$\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}\:+\:{kx}}}\:=\:\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}.}} \left(\mathrm{1}\:+\:{kx}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(−{x}\right)+\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}!}\left(−{x}\right)^{\mathrm{2}} \:+\:…\right]\left[\mathrm{1}\:+\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({kx}\right)+\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}}\left({kx}\right)^{\mathrm{2}} \right] \\…
Question Number 167820 by MWSuSon last updated on 26/Mar/22 $$\mathrm{given}\:\:\mathrm{x}^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}\Sigma\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}^{\mathrm{2}} },\:\mathrm{show}\:\mathrm{that}\:\Sigma\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$ Answered by Mathspace last updated on 26/Mar/22…