Question Number 31340 by NECx last updated on 06/Mar/18 $${Deduce}\:{the}\:{power}\:{series}\:{of}\:{sin}^{\mathrm{2}} {x}. \\ $$$${Hence}\:{show}\:{that}\:{if}\:{x}\:{is}\:{small}\:{then} \\ $$$$\left({sin}^{\mathrm{2}} {x}\:−\:{x}^{\mathrm{2}} {cosx}\right)/{x}^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{6}}\:−\:\frac{{x}^{\mathrm{2}} }{\mathrm{360}} \\ $$ Terms of Service Privacy…
Question Number 31314 by Abdullai otchere last updated on 08/Mar/18 $${let}\:{x}=\left\{\frac{\mathrm{1}}{{n}}\right\}_{{n}=\mathrm{1}} ^{\infty} {and}\:{y}=\left\{\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}_{{n}=\mathrm{1}} ^{\infty} {be}\: \\ $$$${a}\:{sequence}\:{of}\:{real}\:{numbers}\:{and} \\ $$$${l}_{\mathrm{2}\:} =\left\{{x}=\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{x}_{\mathrm{3}} ,…\right):\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{xi}\mid^{\mathrm{2}}…
Question Number 31306 by pieroo last updated on 05/Mar/18 $$\mathrm{Complete}\:\mathrm{the}\:\mathrm{square}\:\mathrm{in}\:\mathrm{y}^{\mathrm{2}} \:+\mathrm{8y}+\mathrm{9k}\:\mathrm{and}\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$ Answered by MJS last updated on 05/Mar/18 $$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}}…
Question Number 31281 by MURALI last updated on 05/Mar/18 $${find}\:{three}\:{nos}\:{in}\:{AP}\:{whose}\:{product} \\ $$$${is}\:{equal}\:{to}\:{the}\:{square}\:{of}\:{their}\:{sum}. \\ $$ Answered by Rasheed.Sindhi last updated on 07/Mar/18 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{number}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{is}\:\mathrm{d} \\…
Question Number 96764 by Rio Michael last updated on 04/Jun/20 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{P}\:\mathrm{of}\:\mathrm{mass}\:{m},\:\mathrm{is}\:\mathrm{projected}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A},\:\mathrm{on}\:\mathrm{horizontal}\:\mathrm{ground}.\:\mathrm{When}\:\mathrm{P}\:\mathrm{is}\:\mathrm{at}\:{x}\:\mathrm{above} \\ $$$$\mathrm{its}\:\mathrm{initial}\:\mathrm{position},\:\mathrm{its}\:\mathrm{speed}\:\mathrm{is}\:{v}.\:\mathrm{The}\:\mathrm{only}\:\mathrm{forces}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{P}\:\mathrm{is} \\ $$$$\mathrm{its}\:\mathrm{weight}\:\mathrm{and}\:\mathrm{resistance}\:{m}\mathrm{g}{kv}^{\mathrm{2}} .\:\mathrm{where}\:{k}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{constant}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{reached}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2g}{k}}\:\mathrm{ln}\left(\mathrm{1}\:+{ku}^{\mathrm{2}} \right). \\ $$$$\left(\mathrm{b}\right)\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{P}\:\mathrm{returns}\:\mathrm{to}\:\mathrm{A}\:\mathrm{is}\:\frac{{u}}{\:\sqrt{\mathrm{1}+\:{ku}^{\mathrm{2}} }}\:. \\…
Question Number 96761 by Rio Michael last updated on 04/Jun/20 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{P}\:\:\:\mathrm{moving}\:\mathrm{at}\:\mathrm{constant}\:\mathrm{angular}\:\mathrm{velocity} \\ $$$$\mathrm{describes}\:\mathrm{a}\:\mathrm{part}\:{y}\:=\:{f}\left(\theta\right).\:\mathrm{At}\:\mathrm{time}\:{t}\:=\:\mathrm{0},\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{with}\:\mathrm{coordinate}\:\left({a},\frac{\pi}{\mathrm{2}}\right)\:\mathrm{and}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\: \\ $$$$\mathrm{transverse}\:\mathrm{acceleration}\:\mathrm{of}\:−\mathrm{2}{a}\omega^{\mathrm{2}} \:\mathrm{sin}\theta.\:\mathrm{find}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{equation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{described}\:\mathrm{by}\:\mathrm{this}\:\mathrm{particle}.\mathrm{Show}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{radial}\:\mathrm{component}\:\mathrm{of}\:\mathrm{the}\:\:\mathrm{acceleration}\:\:\mathrm{of}\:\mathrm{P}\:\mathrm{is}\:−{a}\omega^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{cos}\:\theta\right). \\ $$…
Question Number 162229 by Tawa11 last updated on 27/Dec/21 Answered by mr W last updated on 28/Dec/21 $${u}_{{X}} ={speed}\:{of}\:{gear}\:{X} \\ $$$${t}_{{X}} ={torque}\:{of}\:{gear}\:{X} \\ $$$${u}_{{D}} =\mathrm{200}\:{rpm}\:\curvearrowright…
Question Number 96671 by 175 last updated on 03/Jun/20 Answered by maths mind last updated on 03/Jun/20 $${n}=\underset{{i}=\mathrm{1}} {\overset{{r}} {\prod}}{p}_{{i}} ^{{a}_{{i}} } \Rightarrow \\ $$$$\varphi\left({n}\right)=\underset{{i}=\mathrm{1}}…
Question Number 96667 by Ar Brandon last updated on 03/Jun/20 $$\mathcal{P}\mathrm{rove}\:\:\mathrm{that}\:\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$ Answered by Sourav mridha last updated on 03/Jun/20…
Question Number 162183 by Tawa11 last updated on 27/Dec/21 Answered by mr W last updated on 27/Dec/21 $${say}\:{point}\:{A}\:{is}\:\left({k},{h}\right),\:{then}\:{B}\:{is}\:\left({k}+{h},{h}\right). \\ $$$${h}\:{is}\:{the}\:{side}\:{length}\:{of}\:{the}\:{square}. \\ $$$${h}=\frac{{k}}{{e}^{{k}} } \\ $$$${h}=\frac{{k}+{h}}{{e}^{{k}+{h}}…