Question Number 220958 by mr W last updated on 21/May/25 $${for}\:{x},\:{y},\:{z}\:>\mathrm{0}\:{find}\:{the}\:{maximum}\:{of} \\ $$$${x}^{{m}} {y}^{{n}} {z}^{{k}} \:{subject}\:{to}\:{ax}+{by}+{cz}={d}. \\ $$ Commented by mr W last updated on…
Question Number 220948 by fantastic last updated on 21/May/25 $$\int\:{x}^{\mathrm{2}} \sqrt{\mathrm{5}−{x}^{\mathrm{6}} }{dx} \\ $$ Answered by SdC355 last updated on 21/May/25 $${x}^{\mathrm{3}} ={u} \\ $$$$\frac{\mathrm{d}{u}}{\mathrm{d}{x}}=\mathrm{3}{x}^{\mathrm{2}}…
Question Number 220950 by Nicholas666 last updated on 21/May/25 $$ \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\pi} \int_{\:\mathrm{0}} ^{\:\mathrm{1}} \int_{\:\mathrm{0}} ^{\:\:\pi} \:\mathrm{sin}^{\:\mathrm{2}} \:{x}\:+\:{y}\:\mathrm{sin}\:{z}\:{dxdydz}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\pi\:\left(\mathrm{2}\:+\:\pi\right)\:\:\:\:\:\: \\ $$$$ \\ $$ Answered by…
Question Number 220947 by fantastic last updated on 21/May/25 $$\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{6}}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{{k}\pi}{\mathrm{6}}\right)} \\ $$ Answered by MrGaster last updated on 21/May/25 $$=\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\mathrm{2}\left[\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{6}}\right)−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{{k}\pi}{\mathrm{6}}\right)\right] \\…
Question Number 220995 by hardmath last updated on 21/May/25 Answered by Frix last updated on 22/May/25 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{red}\:\mathrm{line}\:=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{30}°}=\frac{{a}}{\mathrm{sin}\:\mathrm{15}°}\:\Rightarrow \\ $$$$\:\:\:\:\:{a}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{{a}}{\mathrm{sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{135}°−{x}\right)}=\frac{\sqrt{\mathrm{2}}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}\Rightarrow \\ $$$$\:\:\:\:\:{a}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}=\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}}…
Question Number 220863 by fantastic last updated on 20/May/25 $$\left(\mathrm{211}\right) \\ $$$$\:\: \\ $$$${Find}\:{the}\:{derivative}\:{of}\:\Delta{x},\:{where} \\ $$$$\Delta{x}=\begin{vmatrix}{{f}_{\mathrm{1}} \left({x}\right)}&{\phi_{\mathrm{1}} \left({x}\right)}&{\Psi_{\mathrm{1}} \left({x}\right)}\\{{f}_{\mathrm{2}} \left({x}\right)}&{\phi_{\mathrm{2}} \left({x}\right)}&{\Psi_{\mathrm{2}} \left({x}\right)}\\{{f}_{\mathrm{3}} \left({x}\right)}&{\phi_{\mathrm{3}} \left({x}\right)}&{\Psi_{\mathrm{3}} \left({x}\right)}\end{vmatrix}…
Question Number 220857 by fantastic last updated on 20/May/25 $${Prove}\:{that}\:\mathrm{tan}\:\mathrm{20}^{\mathrm{0}} \mathrm{tan40}^{\mathrm{0}} \:\mathrm{tan}\:\mathrm{80}^{\mathrm{0}} =\mathrm{tan}\:\mathrm{60}^{\mathrm{0}} \\ $$ Answered by fantastic last updated on 20/May/25 $$\mathrm{tan}\:\mathrm{20}^{\mathrm{0}} \mathrm{tan40}^{\mathrm{0}} \:\mathrm{tan}\:\mathrm{80}^{\mathrm{0}}…
Question Number 220858 by Rojarani last updated on 20/May/25 Commented by mr W last updated on 21/May/25 $$\mathrm{65}? \\ $$ Commented by Rojarani last updated…
Question Number 220852 by fantastic last updated on 20/May/25 $$\underset{{x}\rightarrow\mathrm{0}} {{Lim}}\left\{\frac{{xe}^{{x}} −{log}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }\right\} \\ $$ Answered by SdC355 last updated on 20/May/25 $$\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left({xe}^{{x}} −\mathrm{ln}\left({x}+\mathrm{1}\right)\right)}{\frac{\mathrm{d}\:}{\mathrm{d}{x}}\:{x}^{\mathrm{2}} }=\frac{\left({x}+\mathrm{1}\right){e}^{{x}}…
Question Number 220853 by fantastic last updated on 20/May/25 $${The}\:{two}\:{solutions}\:{of}\:{the}\:{equation}\:{are}\:{the}\:{same} \\ $$$${a}\left({b}−{c}\right){x}^{\mathrm{2}\:} +{b}\left({c}−{a}\right){x}+{c}\left({a}−{b}\right)=\mathrm{0} \\ $$$${Prove}\:{that}\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{2}}{{b}} \\ $$ Answered by fantastic last updated on 20/May/25 $${In}\:{any}\:{quadratic}\:{equation}\:\alpha{x}^{\mathrm{2}}…