Question Number 195898 by Calculusboy last updated on 12/Aug/23 Answered by qaz last updated on 13/Aug/23 $${a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} −{a}_{{n}+\mathrm{1}} {a}_{{n}} =\mathrm{2}\:\:\:\:\Rightarrow{a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} =\mathrm{2}{n}+{a}_{\mathrm{1}} {a}_{\mathrm{2}} =\mathrm{2}\left({n}+\mathrm{1}\right)…
Question Number 195892 by AROUNAMoussa last updated on 12/Aug/23 Answered by mr W last updated on 13/Aug/23 $$\mathrm{1}+\sqrt{\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{4}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} }+\mathrm{2} \\ $$$$=\mathrm{7}+\mathrm{4}\sqrt{\mathrm{2}} \\…
Question Number 195895 by Erico last updated on 12/Aug/23 $$\mathrm{Calcul}\:\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \mathrm{t}\sqrt{\mathrm{tan}\left(\mathrm{t}\right)}\:\mathrm{dt} \\ $$ Answered by witcher3 last updated on 13/Aug/23 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}\right).\sqrt{\mathrm{t}}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}}…
Question Number 195885 by cortano12 last updated on 12/Aug/23 $$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\sqrt{\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:=\:\mathrm{0}\: \\ $$ Answered by mokys last updated on 12/Aug/23 $$\frac{{dy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\:+\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:{d}\left(\mathrm{0}\right) \\…
Question Number 195884 by Humble last updated on 12/Aug/23 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{by}\:\mathrm{the}\:\mathrm{method}\: \\ $$$$\mathrm{of}\:\mathrm{Differentiation} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$ Terms of Service Privacy Policy…
Question Number 195880 by aawb_247 last updated on 12/Aug/23 Commented by aawb_247 last updated on 12/Aug/23 $${anyone}\:{can}\:{help}..\:{what}\:{a}\:{value}\:{of}\:{x}? \\ $$ Answered by mr W last updated…
Question Number 195883 by Humble last updated on 12/Aug/23 $$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{a}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by},\:\mathrm{V}=\mathrm{k}\left(\frac{\mathrm{l}}{\mathrm{t}}−\mathrm{at}\right), \\ $$$$\mathrm{where}\:\mathrm{k}\:\mathrm{is}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{and}\:\mathrm{l}\:\mathrm{is}\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{travel}\:\mathrm{by}\:\mathrm{boat}\:\mathrm{in}\:\mathrm{time}\:\mathrm{t}\:\mathrm{and}\:\mathrm{a}\:\mathrm{is}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{water}.\: \\ $$$$\mathrm{If}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{change}\:\mathrm{in}\:\mathrm{l}\:\mathrm{from}\:\mathrm{2cm}\:\mathrm{to}\:\mathrm{1cm}\:\mathrm{in}\:\mathrm{time}\:\mathrm{2sec}.\:\mathrm{to}\:\mathrm{1sec}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{water}\:\mathrm{changes}\:\mathrm{from}\:\mathrm{0}.\mathrm{95m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{2m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{Find}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{boat}. \\ $$ Terms of Service…
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Question Number 195872 by CrispyXYZ last updated on 12/Aug/23 $${z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \in\mathbb{C}.\mid{z}_{\mathrm{1}} \mid=\mid{z}_{\mathrm{2}} \mid=\mid{z}_{\mathrm{3}} \mid=\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)\left({z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{3}} +{z}_{\mathrm{1}} \right)}{{z}_{\mathrm{1}} {z}_{\mathrm{2}}…
Question Number 195874 by Humble last updated on 12/Aug/23 Answered by mr W last updated on 12/Aug/23 $${x}={v}_{\mathrm{0}} \mathrm{sin}\:\theta\:{t}+\frac{{g}\:\mathrm{cos}\:\alpha}{\mathrm{2}}\:{t}^{\mathrm{2}} \\ $$$${z}={v}_{\mathrm{0}} \mathrm{cos}\:\theta{t}−\frac{{g}\:\mathrm{sin}\:\alpha}{\mathrm{2}}\:{t}^{\mathrm{2}} \\ $$$$\left.{a}\right) \\…