Question Number 122668 by mathocean1 last updated on 18/Nov/20 $${Esta}\:{has}\:{a} \\ $$$$\:{directory}.\:{it}\:{contains}\: \\ $$$${n}\:\left({n}<\mathrm{1000}\right)\:{pages}\:{where}\: \\ $$$$\mathrm{999991}\:{subscribers}\:{are}\:{registered} \\ $$$${and}\:{each}\:{page}\:{contains}\:{the}\: \\ $$$${same}\:{numbers}\:{of}\:{subscribers}. \\ $$$${How}\:{many}\:{pages}\:{does}\:{this} \\ $$$${directory}\:{has}? \\…
Question Number 188203 by moh777 last updated on 26/Feb/23 Answered by mr W last updated on 26/Feb/23 $${g}=\mathrm{32}.\mathrm{2}\:{ft}/{s}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right) \\ $$$${h}_{\mathrm{1}} ={v}_{\mathrm{1}} \left({t}+\mathrm{3}\right)−\frac{{g}\left({t}+\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{2}}…
Question Number 122638 by Backer last updated on 18/Nov/20 Commented by Backer last updated on 18/Nov/20 $$\mathrm{Hi} \\ $$$$\mathrm{Can}\:\mathrm{somebody}\:\mathrm{do}\:\mathrm{me}\:\mathrm{a}\:\mathrm{favor}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{problem}? \\ $$ Terms of…
Question Number 122641 by Dwaipayan Shikari last updated on 18/Nov/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 57074 by Tawa1 last updated on 30/Mar/19 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}. \\ $$$$\:\:\:\:\:\:\overset{\mathrm{x}} {\:}\mathrm{C}_{\mathrm{y}\:+\:\mathrm{1}} \:\:=\:\:\mathrm{20},\:\:\:\:\:\:\:\:\:\:\:\overset{\mathrm{x}\:−\:\mathrm{1}} {\:}\mathrm{C}_{\mathrm{y}} \:\:=\:\:\mathrm{10} \\ $$ Answered by mr W last updated on…
Question Number 57075 by Tawa1 last updated on 30/Mar/19 Commented by Tawa1 last updated on 30/Mar/19 $$\mathrm{Please}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{the}\:−\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{becomes}\:−\:\mathrm{13} \\ $$$$\mathrm{and}\:\mathrm{the}\:+\:\mathrm{60}\:\mathrm{outside}.\:\mathrm{and}\:\mathrm{how}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\:\mathrm{8}.\:\:\mathrm{Using}\:\mathrm{the}\:\mathrm{same}\:\mathrm{method}. \\ $$ Commented by 121194 last…
Question Number 57062 by olalekan2 last updated on 29/Mar/19 $${find}\:{the}\:{sum}\:{of}\:{all} \\ $$$${three}\:{digital}\:{natural} \\ $$$${numbers}\:{that}\:{are}\: \\ $$$${divisible}\:{by}\:\mathrm{7} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 29/Mar/19…
Question Number 188107 by pascal889 last updated on 25/Feb/23 Answered by CElcedricjunior last updated on 25/Feb/23 $$\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}+\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\: \\ $$$$\exists\boldsymbol{{ssi}}\:\boldsymbol{{x}}\neq−\mathrm{2}\:\boldsymbol{{et}}\:\boldsymbol{{x}}\neq\mathrm{2} \\ $$$$=>\boldsymbol{{x}}^{\mathrm{2}} \left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}+\mathrm{4}\right)+\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 122518 by Dwaipayan Shikari last updated on 17/Nov/20 $$\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} \\ $$$$\mathrm{CI}\:=\:\mathrm{2480}−\mathrm{2000}=\mathrm{480} \\ $$$$\mathrm{2000}×\mathrm{1}.\mathrm{24} \\ $$$$\mathrm{2480}.\mathrm{0} \\ $$…