Question Number 202447 by maths_plus last updated on 26/Dec/23 $$\mathrm{rationnalise}\:\mathrm{le}\:\mathrm{denominateur}\:\mathrm{de}\: \\ $$$$\mathrm{x}\:=\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$ Answered by cortano12 last updated on 27/Dec/23 $$\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}}…
Question Number 202415 by MathematicalUser2357 last updated on 26/Dec/23 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\int{g}'\left({x}\right){f}'\left({g}\left({x}\right)\right){dx}\:\mathrm{is}… \\ $$ Answered by cortano12 last updated on 26/Dec/23 $$\:\mathrm{let}\:\mathrm{u}=\mathrm{g}\left(\mathrm{x}\right)\Rightarrow\mathrm{du}=\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$\:\mathrm{I}=\:\int\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{f}\:'\left(\mathrm{g}\left(\mathrm{x}\right)\right)\:\mathrm{dx}\: \\ $$$$\:\:\:=\:\int\:\mathrm{f}\:'\left(\mathrm{u}\right)\:\mathrm{du}=\:\int\:\frac{\mathrm{df}\left(\mathrm{u}\right)}{\mathrm{du}}.\:\mathrm{du} \\…
Question Number 202408 by Tawa11 last updated on 26/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202436 by MATHEMATICSAM last updated on 26/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{3}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{frame}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} ,\:\gamma^{\mathrm{2}} \:.\: \\ $$ Answered by aleks041103 last updated…
Question Number 202406 by mou0113 last updated on 26/Dec/23 Answered by witcher3 last updated on 26/Dec/23 $$\mathrm{f}\left(\mathrm{s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{s}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dt}\Rightarrow\mathrm{f}'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}}…
Question Number 202400 by MATHEMATICSAM last updated on 26/Dec/23 $$\mathrm{Solve}\:\mathrm{for}\:{a},\:{b}\:\mathrm{and}\:{c} \\ $$$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}\:+\:{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$$\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by mr W last updated on…
Question Number 202388 by Calculusboy last updated on 25/Dec/23 $$\:\:\boldsymbol{{P}}\:\boldsymbol{{rove}}\:\boldsymbol{{that}}:\:\:\:\:\int\:\frac{\boldsymbol{{dx}}}{\boldsymbol{{b}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{c}}}=\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}\right)}{\:\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}+\boldsymbol{{C}} \\ $$$$\boldsymbol{{if}}\:\:\boldsymbol{{a}}\centerdot\left(\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} \right)>\mathrm{0} \\ $$$$ \\ $$ Answered by witcher3…
Question Number 202356 by hardmath last updated on 25/Dec/23 $$\mathrm{Find}: \\ $$$$\mathrm{1}−\left(\mathrm{sin30}°\right)^{\mathrm{2}} \:+\:\left(\mathrm{sin30}°\right)^{\mathrm{4}} \:−\:\left(\mathrm{sin30}°\right)^{\mathrm{6}} \:+\:… \\ $$ Answered by Rasheed.Sindhi last updated on 25/Dec/23 $$\mathrm{1}−\left(\mathrm{sin30}°\right)^{\mathrm{2}}…
Question Number 202390 by Calculusboy last updated on 25/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202359 by hardmath last updated on 25/Dec/23 $$\mathrm{2}\:,\:\mathrm{8}\:,\:\mathrm{32}\:,\:…\:\mathrm{geometfic}\:\mathrm{serie} \\ $$$$\mathrm{for}\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{m}}} \:>\:\mathrm{1024}\:\:\:\mathrm{find}\:\:\:\mathrm{min}\left(\mathrm{m}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 25/Dec/23 $$\mathrm{b}_{\mathrm{n}} =\mathrm{ar}^{\mathrm{n}−\mathrm{1}} \:;\:\mathrm{a}=\mathrm{2}\:,\mathrm{r}=\mathrm{8}/\mathrm{2}=\mathrm{4}…