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Author: Tinku Tara

Does-anyone-know-how-this-works-I-have-d-x-2-cy-2-dy-And-my-physics-teacher-says-it-is-or-can-be-a-harmonic-function-0-Can-anyone-explain-

Question Number 206273 by EmGent last updated on 10/Apr/24 $$\mathrm{Does}\:\mathrm{anyone}\:\mathrm{know}\:\mathrm{how}\:\mathrm{this}\:\mathrm{works}\:? \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:{d}\psi\:=\:\left({x}^{\mathrm{2}} -{cy}^{\mathrm{2}} \right){dy} \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{my}\:\mathrm{physics}\:\mathrm{teacher}\:\mathrm{says}\:\mathrm{it}\:\mathrm{is}\:\left(\mathrm{or}\:\mathrm{can}\right. \\ $$$$\left.\mathrm{be}\right)\:\mathrm{a}\:\mathrm{harmonic}\:\mathrm{function}\:\left(\Delta\psi\:=\:\mathrm{0}\right) \\ $$$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{explain}\:? \\…

Question-206216

Question Number 206216 by cortano21 last updated on 09/Apr/24 Answered by A5T last updated on 09/Apr/24 $${HC}=\sqrt{\mathrm{108}+\mathrm{4}}=\sqrt{\mathrm{112}}=\mathrm{4}\sqrt{\mathrm{7}} \\ $$$${Let}\:{HC}\:{meet}\:{the}\:{circumcircle}\:{of}\:{the}\:{hexagon} \\ $$$${at}\:{L}\:{then},\:{HC}×{HL}={FH}×{HE}=\mathrm{8}\Rightarrow{HL}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}} \\ $$$${sin}\angle{EHC}=\frac{\sqrt{\mathrm{108}}}{\:\sqrt{\mathrm{112}}}=\frac{\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}\Rightarrow{cos}\angle{EHC}=\frac{\sqrt{\mathrm{7}}}{\mathrm{14}} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}}…

xsinx-1-cosx-dx-

Question Number 206200 by Shrodinger last updated on 09/Apr/24 $$\int\frac{{xsinx}}{\mathrm{1}−{cosx}}{dx} \\ $$ Answered by Frix last updated on 09/Apr/24 $$\int\frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}{dx}=\mathrm{i}\int\frac{{x}\left(\mathrm{e}^{\mathrm{i}{x}} +\mathrm{1}\right)}{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}}{dx}= \\ $$$$=\mathrm{i}\int{xdx}+\mathrm{2i}\int\frac{{x}}{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}}{dx}…

Expand-x-2-2x-3-respect-to-x-2-a-x-2-2-2-x-2-3-b-x-2-2-2-x-2-3-c-x-2-2-2-x-2-3-d-x-2-2-2-x-2-3-is-it-taylors-theorem-

Question Number 206230 by BaliramKumar last updated on 09/Apr/24 $$\mathrm{Expand}\:\:\:\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{3}\:\:\:{respect}\:{to}\:{x}\:=\:−\mathrm{2}. \\ $$$$\left(\mathrm{a}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{b}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{c}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} +\:\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{d}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:−\:\mathrm{2}\right)\:−\:\mathrm{3} \\ $$$$…

Question-206198

Question Number 206198 by lmcp1203 last updated on 09/Apr/24 Answered by A5T last updated on 09/Apr/24 $$\frac{{sin}\mathrm{4}\theta}{{AD}}=\frac{{sin}\left(\mathrm{90}−\mathrm{3}\theta\right)}{{AB}};\frac{{sin}\left(\theta\right)}{{AD}}=\frac{{sin}\left(\mathrm{90}−\mathrm{4}\theta\right)}{{CD}={AB}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{4}\theta\right)}{{sin}\left(\mathrm{90}−\mathrm{3}\theta\right)}=\frac{{sin}\left(\theta\right)}{{sin}\left(\mathrm{90}−\mathrm{4}\theta\right)}\Rightarrow{x}=\mathrm{20}° \\ $$ Answered by lmcp1203 last…

OA-4-x-OB-7-5-and-AB-5-units-

Question Number 206227 by bett last updated on 09/Apr/24 $${OA}=\left(\overset{{x}} {\mathrm{4}}\right)\:{OB}=_{\mathrm{7}} ^{\mathrm{5}} \:{and}\:{AB}=\mathrm{5}\:{units} \\ $$ Answered by A5T last updated on 09/Apr/24 $$\sqrt{\left(\mathrm{5}−{x}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{5}\Rightarrow\mathrm{25}−\mathrm{16}=\mathrm{3}^{\mathrm{2}}…