Question Number 212033 by RojaTaniya last updated on 27/Sep/24 $$\:{Find}\:{last}\:{two}\:{digit}\:{of}\:\mid\mathrm{33}^{\mathrm{22}} −\mathrm{22}^{\mathrm{33}} \mid \\ $$ Answered by Frix last updated on 27/Sep/24 $$\mathrm{22}^{\mathrm{33}} −\mathrm{33}^{\mathrm{22}} = \\…
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Question Number 211992 by ajfour last updated on 26/Sep/24 Commented by ajfour last updated on 26/Sep/24 $${Find}\:{r}\:{in}\:{terms}\:{of}\:{R}\:{and}\:{a}. \\ $$ Commented by ajfour last updated on…
Question Number 211995 by SVMEHTA last updated on 26/Sep/24 $$\int\sqrt{}{tanx}\:{dx} \\ $$ Commented by Frix last updated on 26/Sep/24 $$\mathrm{Use}\:{t}=\sqrt{\mathrm{tan}\:{x}} \\ $$ Commented by BHOOPENDRA…
Question Number 211989 by Spillover last updated on 26/Sep/24 Answered by som(math1967) last updated on 26/Sep/24 $$\:{cos}\alpha{cos}\beta+{sin}\alpha{sin}\beta+{cos}\beta{cos}\gamma \\ $$$$+{sin}\beta{sin}\gamma+{cos}\gamma{cos}\alpha+{sin}\gamma{sin}\alpha \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}\left({cos}\alpha{cos}\beta+{cos}\beta{cos}\gamma+{cos}\gamma{cos}\alpha\right) \\ $$$$+\mathrm{2}\left({sin}\alpha{sin}\beta+{sin}\beta{sin}\gamma+{sin}\alpha{sin}\gamma\right)…
Question Number 211986 by Spillover last updated on 26/Sep/24 Answered by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Commented by…
Question Number 211987 by Spillover last updated on 26/Sep/24 Commented by MathematicalUser2357 last updated on 26/Sep/24 $$\mathrm{No}\:\mathrm{closed}\:\mathrm{forms}\:\mathrm{for}\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:{x}}}−\sqrt{\mathrm{tan}\:{x}}\right){dx}. \\ $$$$\mathrm{But}\:\mathrm{you}\:\mathrm{can}\:\mathrm{just}\:\mathrm{approximate}. \\ $$ Commented…
Question Number 212011 by a.lgnaoui last updated on 28/Sep/24 $$\boldsymbol{\mathrm{D}}\mathrm{eterminer}:\:\:\boldsymbol{\mathrm{R}}\mathrm{1}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{2}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{3} \\ $$$$\boldsymbol{\mathrm{pour}}\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{12}\boldsymbol{\mathrm{cm}}\:\:\:\: \\ $$$$\boldsymbol{\mathrm{EF}}\://\:\boldsymbol{\mathrm{MN}}\:;\:\:\boldsymbol{\mathrm{EF}}\:\boldsymbol{\mathrm{Tangent}}\:\boldsymbol{\mathrm{aux}}\:\boldsymbol{\mathrm{cercles}}:\: \\ $$$$\:\:\boldsymbol{\mathrm{C}}\mathrm{1}\left(\boldsymbol{\mathrm{R}}\mathrm{1}\right)\:\:\:\boldsymbol{\mathrm{C}}\mathrm{2}\left(\boldsymbol{\mathrm{R}}\mathrm{2}\right)\:\:;\:\:\:\boldsymbol{\mathrm{EF}}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\boldsymbol{\mathrm{MN}}=\boldsymbol{\mathrm{b}} \\ $$$$\boldsymbol{\mathrm{MN}}:\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{au}}\:\boldsymbol{\mathrm{cercle}}\:\boldsymbol{\mathrm{C}}\mathrm{2} \\ $$$$\boldsymbol{\mathrm{OM}}=\boldsymbol{\mathrm{ON}}=\frac{\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\measuredangle\mathrm{MON}=\mathrm{2}\boldsymbol{\mathrm{x}}\:\:\: \\ $$$$ \\ $$ Commented…
Question Number 212006 by RojaTaniya last updated on 26/Sep/24 Answered by a.lgnaoui last updated on 26/Sep/24 $$\:\:\mathrm{posons}\:\:\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{x}}+\mathrm{100} \\ $$$$\:\:\frac{\left(\boldsymbol{\mathrm{z}}−\mathrm{2}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{2}\right)^{\mathrm{5}} }{\left(\boldsymbol{\mathrm{z}}−\mathrm{1}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}} +\mathrm{80}\boldsymbol{\mathrm{z}}^{\mathrm{3}} +\mathrm{160}\boldsymbol{\mathrm{z}}}{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}}…