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Question-203457

Question Number 203457 by DEGWE last updated on 19/Jan/24 Answered by witcher3 last updated on 19/Jan/24 $$\mathrm{tanh}\left(\mathrm{x}\right)=\frac{\mathrm{e}^{\mathrm{2x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2x}} +\mathrm{1}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\frac{\mathrm{1}+\mathrm{3}^{\mathrm{x}} +\mathrm{x}^{\mathrm{3}} +\mathrm{sh}\left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{3}^{\mathrm{x}} −\mathrm{x}^{\mathrm{3}} −\mathrm{sh}\left(\mathrm{x}\right)}−\mathrm{1}}{\frac{\mathrm{1}+\mathrm{3}^{\mathrm{x}}…

Question-203392

Question Number 203392 by professorleiciano last updated on 18/Jan/24 Commented by a.lgnaoui last updated on 19/Jan/24 Commented by a.lgnaoui last updated on 20/Jan/24 $$\bigtriangleup\boldsymbol{\mathrm{OAB}}\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} =\mathrm{41}−\mathrm{40cos}\:\boldsymbol{\mathrm{x}}…

Question-203394

Question Number 203394 by professorleiciano last updated on 18/Jan/24 Answered by som(math1967) last updated on 18/Jan/24 $$\frac{\bigtriangleup{BPQ}}{\bigtriangleup{ABC}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{BQ}}{{BC}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{BQ}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\Rightarrow{BQ}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}\:{cm} \\ $$ Terms of…

Question-203374

Question Number 203374 by otchereabdullai@gmail.com last updated on 17/Jan/24 Answered by Calculusboy last updated on 18/Jan/24 $$\boldsymbol{{Solution}}:\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}\:\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{algebraic}}\:\boldsymbol{{methods}}\right) \\ $$$$\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}×\frac{\mathrm{5}\boldsymbol{{x}}}{\mathrm{5}\boldsymbol{{x}}}=\frac{\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}×\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{5}\boldsymbol{{x}}} \\ $$$$\boldsymbol{{NB}}:\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tanax}}}{\boldsymbol{{x}}}=\mathrm{1}\:\:\boldsymbol{{then}}\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}}…