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n-360-1-or-n-360-for-plane-mirrors-inclined-an-angle-to-eachother-which-is-correct-

Question Number 132874 by aurpeyz last updated on 17/Feb/21 $${n}=\frac{\mathrm{360}}{\theta}−\mathrm{1}\:{or}\:{n}=\frac{\mathrm{360}}{\theta}\:{for}\:{plane} \\ $$$${mirrors}\:{inclined}\:{an}\:{angle}\:\theta\:{to} \\ $$$${eachother}.\:{which}\:{is}\:{correct}? \\ $$ Commented by Dwaipayan Shikari last updated on 17/Feb/21 $$\mu=\frac{\mathrm{360}°}{\theta°}−\mathrm{1}\:\:{or}\:\:\mu=\frac{\mathrm{2}\pi}{\theta}−\mathrm{1}\:\:\:\left({Radian}\right)…

dx-x-2-e-x-3-3-

Question Number 1799 by navajyoti.tamuli.tamuli@gmail. last updated on 30/Sep/15 $$\int\frac{{dx}}{{x}^{\mathrm{2}} .{e}^{−{x}^{\mathrm{3}} /\mathrm{3}} }=? \\ $$ Commented by 123456 last updated on 30/Sep/15 $$\mathrm{not}\:\mathrm{done}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{elementar}\:\mathrm{function} \\ $$…

Question-67307

Question Number 67307 by aliesam last updated on 25/Aug/19 Commented by mathmax by abdo last updated on 25/Aug/19 $${S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{cos}\left({n}\theta\right)}{\mathrm{2}^{{n}} }\:\Rightarrow\:{S}\:={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{in}\theta} }{\mathrm{2}^{{n}}…

7-couples-are-there-in-a-society-The-selection-of-pairs-are-to-be-done-for-a-mixed-double-badminton-tournament-In-how-many-ways-we-can-choose-these-pairs-

Question Number 1775 by lakshaysethi039 last updated on 26/Sep/15 $$\mathrm{7}\:{couples}\:{are}\:{there}\:{in}\:{a}\:{society}.\:{The}\:{selection}\:{of} \\ $$$$\:{pairs}\:{are}\:{to}\:{be}\:{done}\:{for}\:{a}\:{mixed}\:{double}\:{badminton} \\ $$$${tournament}.\:{In}\:{how}\:{many}\:{ways}\:{we}\:{can}\:{choose} \\ $$$$\:{these}\:{pairs}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Solve-e-2z-3-i-z-x-iy-

Question Number 132834 by mohammad17 last updated on 16/Feb/21 $${Solve}\:{e}^{\mathrm{2}{z}} =\sqrt{\mathrm{3}}−{i}\:\:\:,{z}={x}+{iy} \\ $$ Answered by mathmax by abdo last updated on 20/Feb/21 $$\mid\sqrt{\mathrm{3}}−\mathrm{i}\mid=\mathrm{2}\:\Rightarrow\sqrt{\mathrm{3}}−\mathrm{i}=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\right)\:=\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{6}}} =\mathrm{2e}^{\left(−\frac{\mathrm{i}\pi}{\mathrm{6}}+\mathrm{2ik}\pi\right)} \\…

Express-0-999-as-p-q-

Question Number 1736 by navajyoti.tamuli.tamuli@gmail. last updated on 04/Sep/15 $${Express}\:\mathrm{0}.\mathrm{999}…\:{as}\:\frac{{p}}{{q}} \\ $$ Answered by 123456 last updated on 04/Sep/15 $${x}=\mathrm{0}.\mathrm{999999}…. \\ $$$$\mathrm{10}{x}=\mathrm{9}.\mathrm{9999999}….. \\ $$$$\mathrm{9}{x}=\mathrm{9} \\…

dx-1-x-4-1-2-dx-1-ix-2-1-2-dx-1-ix-2-1-2-e-i-pi-4-d-e-i-pi-4-x-1-e-i-pi-4-x-2-1-2-e-i-pi-4-d-e-i-pi-4-x-1-e-i-pi-4-x-2-1-2-e-i-pi-4-ta

Question Number 132799 by Ñï= last updated on 16/Feb/21 $$\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}+{ix}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}−{ix}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \int\frac{{d}\left({e}^{{i}\frac{\pi}{\mathrm{4}}} {x}\right)}{\mathrm{1}+\left({e}^{{i}\frac{\pi}{\mathrm{4}}} {x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \int\frac{{d}\left({e}^{−{i}\frac{\pi}{\mathrm{4}}} \right){x}}{\mathrm{1}+\left({e}^{−{i}\frac{\pi}{\mathrm{4}}} {x}\right)^{\mathrm{2}} }…