Question Number 123776 by Dwaipayan Shikari last updated on 28/Nov/20 $${Another}\:{unanswered}\:{question} \\ $$$$\frac{{log}\mathrm{1}}{\:\sqrt{\mathrm{1}}}−\frac{{log}\mathrm{3}}{\:\sqrt{\mathrm{3}}}+\frac{{log}\mathrm{5}}{\:\sqrt{\mathrm{5}}}−\frac{{log}\mathrm{7}}{\:\sqrt{\mathrm{7}}}+..=\left(\frac{\pi}{\mathrm{4}}−\frac{\gamma}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\pi\right)\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}−.\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 189292 by Rupesh123 last updated on 14/Mar/23 Commented by talminator2856792 last updated on 14/Mar/23 $$\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{incomplete}. \\ $$$$\:\mathrm{it}\:\mathrm{doesnt}\:\mathrm{say}\:\mathrm{how}\:\mathrm{the}\:\mathrm{chords}\:\mathrm{are}\:\mathrm{chosen}\: \\ $$$$\:\mathrm{from}\:\mathrm{the}\:\mathrm{four}\:\mathrm{points}. \\ $$ Commented by…
Question Number 58195 by salaw2000 last updated on 19/Apr/19 $$\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}}=\mathrm{9} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$ Answered by Kunal12588 last updated on 19/Apr/19…
Question Number 58193 by PDF1104_ last updated on 19/Apr/19 $$\mathrm{If}\:{z}\:\in\:\mathbb{C}\:\:\mathrm{such}\:\mathrm{that}\:\mathfrak{R}\left({z}^{{n}} \right)>\mathrm{0}\:\mathrm{for}\:{n}\in\mathbb{N}^{+} . \\ $$$$\mathrm{Show}\:\mathrm{that}\:{z}\:\in\mathbb{R}^{+} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 58171 by pierre last updated on 19/Apr/19 $$\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{along} \\ $$$$\mathrm{a}\:\mathrm{smooth}\:\mathrm{wire}\:\mathrm{that}\:\mathrm{is}\:\mathrm{fixed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}. \\ $$$$\mathrm{The}\:\mathrm{polar}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire}\:\:\mathrm{is}\: \\ $$$$\mathrm{r}\:=\:\mathrm{ae}^{\mathrm{3}\theta} .\:\mathrm{The}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{with}\:\mathrm{a}\:\mathrm{cons} \\ $$$$\mathrm{tant}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{6}.\:\mathrm{At}\:\mathrm{time}\:\:\mathrm{t}\:=\:\mathrm{0}\:,\:\mathrm{the}\:\mathrm{par} \\ $$$$\mathrm{ticle}\:\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{with}\:\mathrm{polar}\:\mathrm{equation} \\ $$$$\left(\mathrm{a},\theta\right) \\ $$$$\left.\mathrm{a}\right)\mathrm{Find}\:\mathrm{the}\:\mathrm{transverse}\:\mathrm{and}\:\mathrm{radial}\:\mathrm{compo}…
Question Number 123694 by Dwaipayan Shikari last updated on 27/Nov/20 $${sinh}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\right)+{sinh}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)+{sinh}\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)+…. \\ $$ Commented by Dwaipayan Shikari last updated on 27/Nov/20 $${Another}\:{approximation}\:{of}\:{sin}\left(\frac{\pi}{\mathrm{9}}\right)…
Question Number 123653 by 777316 last updated on 27/Nov/20 Answered by Kunal12588 last updated on 27/Nov/20 $${d}=\mathrm{24}+{s} \\ $$$${d}−\mathrm{5}=\mathrm{5}\left({s}−\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{5}=\mathrm{24}−\mathrm{4}{s}+\mathrm{25} \\ $$$$\Rightarrow{s}=\frac{\mathrm{44}}{\mathrm{4}}=\mathrm{11} \\ $$$${d}=\mathrm{35}…
Question Number 123651 by 777316 last updated on 27/Nov/20 Answered by MJS_new last updated on 27/Nov/20 $${x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\left(\mathrm{14}+\mathrm{9}\right)^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }=\mathrm{17} \\ $$ Commented by malwan last…
Question Number 123650 by Dwaipayan Shikari last updated on 27/Nov/20 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{16}}+… \\ $$ Commented by MJS_new last updated on 27/Nov/20 $$\mathrm{sorry}\:\mathrm{I}\:\mathrm{misread}\:\mathrm{it}… \\ $$ Commented by…
Question Number 58091 by smiak8742 last updated on 17/Apr/19 $$\mathrm{27}{x}^{\mathrm{3}−\mathrm{1}=\mathrm{0}} \\ $$ Answered by MJS last updated on 17/Apr/19 $$\mathrm{27}{x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{27}} \\…