Question Number 122605 by bramlexs22 last updated on 18/Nov/20 Answered by Dwaipayan Shikari last updated on 18/Nov/20 $${T}_{{n}} =\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${S}_{{n}} +\mathrm{1}=\overset{{n}} {\sum}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${S}_{{n}}…
Question Number 122484 by JBocanegra last updated on 17/Nov/20 $$\:\frac{\mathrm{4}{a}}{\mathrm{3}{b}\centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \centerdot{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \centerdot{b}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}\centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }\right)\left(\frac{{a}}{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \centerdot{b}^{\frac{\mathrm{4}}{\mathrm{3}}} }\right) \\ $$$$\left(\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}\centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }\right)\left({a}^{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \centerdot{b}^{−\frac{\mathrm{4}}{\mathrm{3}}} \right)=\left(\mathrm{2}^{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}} \right)\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 188012 by Shlock last updated on 24/Feb/23 Commented by Rasheed.Sindhi last updated on 25/Feb/23 $${Perhaps}\:{there}'{s}\:{an}\:{error}\:{in}\:{some} \\ $$$${figures}\:{of}\:{the}\:{question}.{Have}\:{you} \\ $$$${an}\:{answer}\:{of}\:{the}\:{question}? \\ $$ Terms of…
Question Number 56857 by azam2412 last updated on 25/Mar/19 $${romi}−{romo}=\mathrm{0} \\ $$$$\mathcal{L}.{x}_{\mathrm{0}} +\mathcal{G}.{y}_{{n}} =\mathcal{L}.{x}_{{n}} +\mathcal{G}.{y}_{\mathrm{0}} \\ $$$$\mathcal{L}\left({x}_{\mathrm{0}} −{x}_{{n}} \right)=\mathcal{G}\left({y}_{\mathrm{0}} −{y}_{{n}} \right) \\ $$$${y}_{\mathrm{0}} =\frac{\mathcal{L}}{\mathcal{G}}.\left({x}_{\mathrm{0}} −{x}_{{n}}…
Question Number 56852 by peter frank last updated on 25/Mar/19 Answered by MJS last updated on 25/Mar/19 $$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:{f}\circ{g}\:\mathrm{means}\:{f}\left({g}\right)\:\mathrm{or}\:{g}\left({f}\right) \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{only}\:\mathrm{a}\:\mathrm{matter}\:\mathrm{of}\:\mathrm{thinking}\:\mathrm{logically} \\ $$$${f}\left({g}\left({x}\right)\right)=\begin{cases}{\mathrm{1};\:{x}<−\mathrm{1}}\\{\mathrm{3};\:−\mathrm{1}\leqslant{x}<−\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{4}{x}^{\mathrm{2}} ;\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{2};\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}}\\{\mathrm{1};\:{x}>\mathrm{1}}\end{cases} \\ $$$${g}\left({f}\left({x}\right)\right)=\begin{cases}{\mathrm{1};\:{x}<−\mathrm{1}}\\{\mathrm{2}{x}^{\mathrm{2}}…
Question Number 56845 by Tawa1 last updated on 25/Mar/19 Answered by math1967 last updated on 25/Mar/19 $$\left(\mathrm{11}…\mathrm{108}{times}\right)×\mathrm{10}+\mathrm{1}+\left(\mathrm{22}..\mathrm{108}{times}\right)×\mathrm{10}+\mathrm{2} \\ $$$$……\left(\mathrm{77}..\mathrm{108}{times}\right)×\mathrm{10}+\mathrm{7} \\ $$$$\frac{\mathrm{11}….\mathrm{108}{times}+…..\mathrm{77}…\mathrm{108}{times}}{\mathrm{37}}+\frac{\mathrm{1}+..\mathrm{7}}{\mathrm{37}} \\ $$$${remainder}\:\mathrm{0}\:+{remainder}\mathrm{28} \\ $$$${so}\:\mathrm{28}\:{ans}…
Question Number 187869 by BaliramKumar last updated on 23/Feb/23 $$\frac{\mathrm{41}!}{\mathrm{47}}\:{find}\:{remaider} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Feb/23 $${According}\:{to}\:{Wilson}'{s}\:{theorem}: \\ $$$$\left(\mathrm{47}−\mathrm{1}\right)!\equiv−\mathrm{1}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{46}!\equiv−\mathrm{1}+\mathrm{47}\left({mod}\:\mathrm{47}\right) \\…
Question Number 56785 by Runzzy last updated on 23/Mar/19 Commented by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19 $${rechek}\:{question} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 56763 by Tawa1 last updated on 23/Mar/19 $$\left(\mathrm{a}\right)\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{rectangle}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circle}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\:=\:\:\mathrm{a}^{\mathrm{2}} \:. \\ $$$$ \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Name}\:\mathrm{the}\:\mathrm{rectangle}\:\mathrm{so}\:\mathrm{formed} \\ $$ Answered by kaivan.ahmadi last…
Question Number 56711 by Tawa1 last updated on 22/Mar/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shotest}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{line} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{x}\:−\:\mathrm{8}}{\mathrm{3}}\:=\:\frac{\mathrm{y}\:−\:\mathrm{2}}{\mathrm{4}}\:=\:\frac{\mathrm{z}\:+\:\mathrm{1}}{\mathrm{1}}\:,\:\:\:\:\:\:\:\frac{\mathrm{x}\:−\:\mathrm{3}}{\mathrm{3}}\:=\:\frac{\mathrm{y}\:+\:\mathrm{4}}{\mathrm{5}}\:=\:\frac{\mathrm{z}\:−\mathrm{2}}{\mathrm{2}} \\ $$ Answered by mr W last updated on 23/Mar/19 $${using}\:{vector}\:{method}: \\ $$$$…