Question Number 122518 by Dwaipayan Shikari last updated on 17/Nov/20 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{11}}+\frac{\mathrm{1}}{\mathrm{13}}−\frac{\mathrm{1}}{\mathrm{17}}+… \\ $$ Commented by Dwaipayan Shikari last updated on 17/Nov/20 $$\frac{\mathrm{1}}{\mathrm{2}}{tanx}=\frac{\mathrm{1}}{\pi−\mathrm{2}{x}}−\frac{\mathrm{1}}{\pi+\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{3}\pi−\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{3}\pi+\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{5}\pi−\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{5}{x}+\mathrm{2}\pi}+… \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{1}}{\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}}−\frac{\mathrm{1}}{\pi+\frac{\mathrm{2}\pi}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{3}\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{5}\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}}−.. \\…
Question Number 122514 by hedaru last updated on 17/Nov/20 Commented by liberty last updated on 18/Nov/20 $$\mathrm{A}=\mathrm{P}\left(\mathrm{1}+\mathrm{3}.\frac{\mathrm{8}}{\mathrm{100}}\right)\:=\:\mathrm{2000}\left(\mathrm{1}.\mathrm{24}\right)=\mathrm{2480} \\ $$$$\mathrm{CI}\:=\:\mathrm{2480}−\mathrm{2000}=\mathrm{480} \\ $$$$\mathrm{2000}×\mathrm{1}.\mathrm{24} \\ $$$$\mathrm{2480}.\mathrm{0} \\ $$…
Question Number 56971 by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19 Commented by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19 $${source}\:..{book} \\ $$ Answered by mr W last updated…
Question Number 56949 by rajesh4661kumar@gamil.com last updated on 27/Mar/19 Commented by MJS last updated on 27/Mar/19 $$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{read}\:\mathrm{the}\:\mathrm{number}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{circle} \\ $$ Terms of Service Privacy Policy…
Question Number 56932 by turbo msup by abdo last updated on 27/Mar/19 $${let}\:{f}_{{n}} \left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{{n}} } \\ $$$${with}\:{n}\:{from}\:{N}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\mathrm{1}.\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}_{{n}} \left({t}\right) \\…
Question Number 56900 by bshahid010@gmail.com last updated on 26/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19 $${f}\left({x}\right)=\mid\left({x}−{a}\right)^{\mathrm{3}} \mid+\mid\left({x}−{b}\right)^{\mathrm{3}} \mid \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left({x}−{a}\right)^{\mathrm{3}} +\left({x}−{b}\right)^{\mathrm{3}} \:\:\:\:\:{when}\:{x}>{b} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left({b}−{a}\right)^{\mathrm{3}}…
Question Number 56890 by Ismael002 last updated on 25/Mar/19 $${d}^{\mathrm{2}} {y}/{dx}^{\mathrm{2}} ={x}^{\mathrm{2}} {y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 56854 by Tinku Tara last updated on 25/Mar/19 $$\mathrm{There}\:\mathrm{was}\:\mathrm{a}\:\mathrm{post}\:\mathrm{sime}\:\mathrm{time}\:\mathrm{back} \\ $$$$\mathrm{about}\:\mathrm{not}\:\mathrm{being}\:\mathrm{able}\:\mathrm{to}\:\mathrm{backup} \\ $$$$\mathrm{or}\:\mathrm{restore}.\:\mathrm{Can}\:\mathrm{anyone}\:\mathrm{send}\:\mathrm{the}\:\mathrm{requirdd} \\ $$$$\mathrm{information}\:\mathrm{if}\:\mathrm{you}\:\mathrm{faced}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{problem}? \\ $$ Commented by Hassen_Timol last…
Question Number 56847 by Ismael002 last updated on 25/Mar/19 $${d}\mathrm{2}{y}/{dx}\mathrm{2}={x}\mathrm{2}{y}=\mathrm{0} \\ $$ Commented by ajfour last updated on 25/Mar/19 $$\mathrm{do}\:\mathrm{you}\:\mathrm{mean} \\ $$$$\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }−\mathrm{x}^{\mathrm{2}} \mathrm{y}=\mathrm{0}\:\:?…
Question Number 56851 by behi83417@gmail.com last updated on 25/Mar/19 $${let}:\:\left[\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{n}}} =\sqrt{\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{n}}−\mathrm{1}} }+\sqrt{\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{n}}−\mathrm{2}} },\boldsymbol{\mathrm{u}}_{\mathrm{0}} =\mathrm{1},\boldsymbol{\mathrm{u}}_{\mathrm{1}} =\mathrm{1}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{n}}} }\right)=? \\ $$ Terms of Service…