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Author: Tinku Tara

Question-67775

Question Number 67775 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{side}\:{s}=\sqrt{\mathrm{196}}=\mathrm{14} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{sector}'\mathrm{s}\:\mathrm{area}\:=\frac{\pi}{\mathrm{8}}{s}^{\mathrm{2}} =\frac{\mathrm{49}\pi}{\mathrm{2}}\:\mathrm{minus} \\ $$$$\mathrm{the}\:\mathrm{white}\:\mathrm{segment}\:\mathrm{which}\:\mathrm{intersects}\:\mathrm{the} \\ $$$$\mathrm{diagonal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square} \\…

0-2pi-dx-5-3sin-2x-

Question Number 133305 by liberty last updated on 21/Feb/21 $$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{5}+\mathrm{3sin}\:\mathrm{2x}}\:=? \\ $$ Answered by physicstutes last updated on 21/Feb/21 $$\mathrm{set}\:{t}\:=\:\mathrm{tan}\:{x} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:=\:\frac{\mathrm{2}\:\mathrm{tan}\:{x}}{\mathrm{1}+\:\mathrm{tan}^{\mathrm{2}} {x}}\:\:=\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…

Question-67770

Question Number 67770 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${AD}+{DE}={AE}=\mathrm{6} \\ $$$$\Rightarrow\:{AD}={q};\:{DE}=\mathrm{6}−{q}\:\:\left[\Rightarrow\:{q}<\mathrm{6}\right] \\ $$$${AB}={DE}−{AD}=\mathrm{6}−\mathrm{2}{q}\:\:\left[\Rightarrow\:{q}<\mathrm{3}\right] \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{{q}}\\{\mathrm{0}}\end{pmatrix}\:\:{E}=\begin{pmatrix}{\mathrm{6}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{line}\:{AB}:\:{y}={x}\mathrm{tan}\:\mathrm{60}°\:\Leftrightarrow\:{y}=\sqrt{\mathrm{3}}{x}\:\:\left[\mathrm{1}\right]…

in-quantum-physics-is-this-a-true-statement-h-h-bar-d-dt-t-t-0-d-dt-t-t-t-t-dx-d-dt-dx-H-dx-1-ih-H-dx-1-ih-

Question Number 2234 by madscientist last updated on 10/Nov/15 $${in}\:{quantum}\:{physics}\:{is}\:{this}\:{a}\:{true}\: \\ $$$${statement}?\:{h}={h}\:{bar} \\ $$$$\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)\rangle=\mathrm{0} \\ $$$$\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)=\int\psi^{\ast} \left({t}\right)\psi\left({t}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int\frac{{d}\psi^{\ast} }{{dt}}\psi{dx}+\int\psi^{\ast} {H}\psi{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{{ih}}\int\left({H}\psi\right)^{\ast} \psi{dx}+\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx}…

f-x-x-2-3x-10-f-x-2x-3-and-f-1-5-what-indicate-these-vslue-f-x-andf-1-in-geometrical-meaning-

Question Number 2229 by tabrez8590@gmail last updated on 09/Nov/15 $${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{10} \\ $$$$\Rightarrow{f}^{'} \left({x}\right)=\mathrm{2}{x}+\mathrm{3}\:{and} \\ $$$${f}^{'} \left(\mathrm{1}\right)=\mathrm{5}\:{what}\:{indicate}\:{these}\:{vslue}\: \\ $$$${f}'\left({x}\right){andf}'\left(\mathrm{1}\right)\:{in}\:{geometrical}\:{meaning} \\ $$ Answered by prakash jain…

Question-133296

Question Number 133296 by rs4089 last updated on 21/Feb/21 Answered by SEKRET last updated on 21/Feb/21 $$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\mathrm{3}\boldsymbol{\mathrm{xy}}'+\boldsymbol{\mathrm{y}}=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\mathrm{3}\boldsymbol{\mathrm{xy}}'+\boldsymbol{\mathrm{y}}=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{y}}=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} \:\:\:\:\boldsymbol{\mathrm{y}}'=\boldsymbol{\mathrm{mx}}^{\boldsymbol{\mathrm{m}}−\mathrm{1}} \:\:\:\:\:\boldsymbol{\mathrm{y}}''=\boldsymbol{\mathrm{m}}\left(\boldsymbol{\mathrm{m}}−\mathrm{1}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}−\mathrm{2}} \\…