Question Number 197792 by Mastermind last updated on 28/Sep/23 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{y}''\:+\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{x}^{\mathrm{3}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{y}''\:+\:\mathrm{y}^{'} \:−\:\mathrm{2y}\:=\:\mathrm{x}\:+\:\mathrm{sin2x},\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{y}''\:−\:\mathrm{y}'\:=\:\mathrm{xe}^{\mathrm{x}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\:\mathrm{1} \\ $$$$ \\ $$$$ \\…
Question Number 197640 by Mastermind last updated on 25/Sep/23 $${Water}\:{is}\:{leaking}\:{from}\:{a}\:{hemispheric} \\ $$$${bowl}\:{of}\:{radius}\:\mathrm{20}{cm}\:{at}\:{the}\:{rate}\:{of}\: \\ $$$$\mathrm{0}.\mathrm{5}{cm}^{\mathrm{3}} /{s}.\:{Find}\:{the}\:{rate}\:{at}\:{which}\:{surface} \\ $$$${area}\:{of}\:{the}\:{water}\:{decreasing}\:{when}\:{the} \\ $$$${water}\:{level}\:{is}\:{halfway}\:{from}\:{the}\:{top}. \\ $$$$ \\ $$$${Thank}\:{you} \\ $$…
Question Number 197470 by Mastermind last updated on 18/Sep/23 $${Solve}\:{the}\:{following}\:{equation} \\ $$$${x}\:+\:\mathrm{2}{y}\:+\:\mathrm{2}{z}\:=\:\mathrm{0} \\ $$$$\mathrm{2}{x}\:+\:{y}\:−\:\mathrm{2}{z}\:=\mathrm{0} \\ $$$$\mathrm{3}{x}\:+\:\mathrm{4}{y}\:−\:\mathrm{6}{z}\:=\mathrm{0} \\ $$$$\mathrm{3}{x}\:−\:\mathrm{11}{y}\:+\:\mathrm{12}{z}\:=\:\mathrm{0} \\ $$ Answered by MathedUp last updated…
Question Number 197459 by MathematicalUser2357 last updated on 18/Sep/23 $$\mathrm{Is}\:\mathrm{complex}\:\mathrm{infinity}\:\mathrm{big}? \\ $$$$\overset{\sim} {\infty}=\infty\centerdot\left(\mathrm{1}+{i}\right) \\ $$$$\mathrm{Their}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{is}\:\mathrm{big} \\ $$$$\mid\overset{\sim} {\infty}\mid>\mid\infty\mid \\ $$ Commented by TheHoneyCat last updated…
Question Number 197449 by mohaa last updated on 17/Sep/23 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197371 by Mastermind last updated on 15/Sep/23 Answered by Rasheed.Sindhi last updated on 15/Sep/23 $${d}:{dog},\:{c}:{cat},\:{r}:{rat} \\ $$$${d}+{r}=\mathrm{20}…\left({i}\right) \\ $$$${c}+{r}=\mathrm{10}…\left({ii}\right) \\ $$$${d}+{c}=\mathrm{24}…\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right):…
Question Number 197184 by pete last updated on 10/Sep/23 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{log}\left(−\mathrm{log}{i}\right)=\mathrm{log}\left(\frac{\pi}{\mathrm{2}}\right)−{i}\frac{\pi}{\mathrm{2}} \\ $$ Answered by Frix last updated on 10/Sep/23 $$\mathrm{ln}\:\left(−\mathrm{ln}\:\mathrm{i}\right)\:=\mathrm{ln}\:\frac{\pi}{\mathrm{2}}\:−\mathrm{i}\frac{\pi}{\mathrm{2}} \\ $$$$−\mathrm{ln}\:\mathrm{i}\:=\mathrm{e}^{\mathrm{ln}\:\frac{\pi}{\mathrm{2}}\:−\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$−\mathrm{ln}\:\mathrm{i}\:=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}} \\…
Question Number 197185 by MathematicalUser2357 last updated on 10/Sep/23 $$\mathrm{Simplify} \\ $$$$\sqrt[{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}]{\frac{\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}} \centerdot\left(\sqrt{\mathrm{3}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{12}}} }{\left(\sqrt{\mathrm{6}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} }} \\ $$ Answered by som(math1967) last updated on…
Question Number 197052 by pete last updated on 07/Sep/23 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{2}} \\ $$ Answered by Frix last updated on 07/Sep/23 $$\mathrm{sin}\:{x}\:=\sqrt{\mathrm{2}} \\ $$$${x}=\left(\mathrm{2}{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi\pm\mathrm{i}\:\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{You}\:\mathrm{get}\:\mathrm{this}\:\mathrm{using}…
Question Number 197089 by pete last updated on 07/Sep/23 $$\mathrm{Simplify}\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}}\right)^{\mathrm{10}} \\ $$ Answered by JDamian last updated on 07/Sep/23 $$\left(\frac{{z}}{{z}^{\ast} }\right)^{{n}} =\left(\frac{\cancel{\mid{z}\mid}\centerdot{e}^{{i}\varphi} }{\cancel{\mid{z}\mid}\centerdot{e}^{−{i}\varphi} }\right)^{{n}} ={e}^{{i}\mathrm{2}{n}\varphi}…