Question Number 204250 by Noorzai last updated on 10/Feb/24 Answered by AST last updated on 10/Feb/24 $${tan}\left(\mathrm{9}\right)+{tan}\left(\mathrm{81}\right)−\left({tan}\mathrm{27}+{tan}\mathrm{63}\right) \\ $$$$=\frac{{sin}\mathrm{9}}{{cos}\mathrm{9}}+\frac{{sin}\mathrm{81}}{{cos}\mathrm{81}}−\left(\frac{{sin}\mathrm{27}}{{cos}\mathrm{27}}+\frac{{sin}\mathrm{63}}{{cos}\mathrm{63}}\right) \\ $$$$=\frac{\mathrm{2}{sin}\left(\mathrm{9}+\mathrm{81}\right)=\mathrm{2}}{\mathrm{2}{cos}\mathrm{9}{cos}\mathrm{81}}−\frac{\mathrm{2}{sin}\left(\mathrm{27}+\mathrm{63}\right)=\mathrm{2}}{\mathrm{2}{cos}\left(\mathrm{27}\right){cos}\left(\mathrm{63}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{9}\right){cos}\left(\mathrm{9}\right)}−\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{27}\right){cos}\left(\mathrm{27}\right)}=\frac{\mathrm{2}}{{sin}\left(\mathrm{18}\right)}−\frac{\mathrm{2}}{{sin}\left(\mathrm{54}\right)} \\ $$$${Let}\:\theta=\mathrm{18}°\Rightarrow{sin}\left(\mathrm{5}\theta\right)=\mathrm{16}{sin}^{\mathrm{5}}…
Question Number 204262 by DEGWE last updated on 10/Feb/24 Answered by Frix last updated on 10/Feb/24 $$\mathrm{Charles}−\mathrm{Ange}\:\mathrm{LAISANT}\:\left(\mathrm{1905}\right): \\ $$$$\int{f}^{−\mathrm{1}} \left({x}\right){dx}={xf}^{−\mathrm{1}} \left({x}\right)−\left({F}\circ{f}^{−\mathrm{1}} \right)\left({x}\right)+{C} \\ $$$$\mathrm{with}\:{F}\left({x}\right)=\int{f}\left({x}\right){dx} \\…
Question Number 204273 by mustafazaheen last updated on 10/Feb/24 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{4}}\right)} } \\ $$$$\mathrm{Domain}\:\mathrm{f}\left(\mathrm{x}\right)\:=? \\ $$ Answered by Mathspace last updated on 11/Feb/24 $${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{−{ln}\mathrm{2}} }…
Question Number 204236 by Panav last updated on 09/Feb/24 $$\boldsymbol{{Helpful}}\:\boldsymbol{{questionfor}}\:\boldsymbol{{Olympiads}},\:\boldsymbol{{Find}}\:\boldsymbol{{Sol}}^{\boldsymbol{{n}}} . \\ $$ Commented by Panav last updated on 09/Feb/24 Answered by JDamian last updated…
Question Number 204233 by Perelman last updated on 09/Feb/24 Answered by Frix last updated on 09/Feb/24 $$\int\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}}{dx}\:\overset{{t}={x}^{\frac{\mathrm{1}}{\mathrm{4}}} } {=}\:\mathrm{4}\int\frac{{t}^{\mathrm{5}} }{{t}^{\mathrm{3}} +\mathrm{1}}{dt}= \\ $$$$=\mathrm{4}\int{t}^{\mathrm{2}}…
Question Number 204244 by pticantor last updated on 09/Feb/24 Answered by pticantor last updated on 09/Feb/24 $${please}\:{can}\:{some}\:{one}\:{help}\:{me}\:{to}\:{solve}\:{question}\:\mathrm{2}−{b}\: \\ $$$${for}\:{this}\:{exercise}? \\ $$ Answered by witcher3 last…
Question Number 204246 by Tutormalvis last updated on 09/Feb/24 $$\mathrm{l}\rfloor\infty \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 204230 by Davidtim last updated on 09/Feb/24 $${what}\:{does}\:{mean}\:{that}\:{we}\:{say}\:{C}^{°} ={F}^{°} \\ $$$${in}\:−\mathrm{40}? \\ $$ Commented by mr W last updated on 10/Feb/24 $${generally}\:{the}\:{temperature}\:{value}\:{in} \\…
Question Number 204226 by MathedUp last updated on 09/Feb/24 $$\mathrm{God}\:\mathrm{Damn}\:\mathrm{it}\:\mathrm{why}\:\mathrm{my}\:\mathrm{Integration}\:\mathrm{Test}\:\mathrm{is}\:\mathrm{Wrong}? \\ $$$$\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:{h}\centerdot{J}_{\nu} \left({h}\right)\:\mathrm{is}\:\mathrm{Conv}?? \\ $$$$\mathrm{Laplace}\:\mathrm{transform}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}} \centerdot \\ $$$$\mathrm{We}\:\mathrm{already}\:\mathrm{Know}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}} {J}_{\nu}…
Question Number 204227 by abdomath last updated on 09/Feb/24 Answered by AST last updated on 09/Feb/24 $${Let}\:{y}={f}\left({x}\right)\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{3}} =\left({x}+{y}\right)^{\mathrm{5}} \Rightarrow{x}^{\mathrm{2}} \left[{f}\left({x}\right)\right]^{\mathrm{3}} =\left({x}+{f}\left({x}\right)\right)^{\mathrm{5}} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \left[{f}\left({x}\right)\right]^{\mathrm{3}}…