Question Number 214264 by zhou0429 last updated on 03/Dec/24 Commented by Ghisom last updated on 05/Dec/24 $$\mathrm{it}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{find}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{solution}\:\mathrm{since}\:\left(\mathrm{1}\right)\:\sqrt{\mathrm{39}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{zero}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{polynomial}\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{the}\:\mathrm{real}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{polynomial}\:\mathrm{is}\:\mathrm{no}\:\mathrm{usable}\:\mathrm{expression} \\ $$$$\mathrm{just}\:\mathrm{use}\:\mathrm{a}\:\mathrm{good}\:\mathrm{calculator}…
Question Number 214281 by ajfour last updated on 03/Dec/24 Commented by ajfour last updated on 03/Dec/24 $${Q}.\:\mathrm{214132} \\ $$ Answered by ajfour last updated on…
Question Number 214279 by issac last updated on 03/Dec/24 $$\mathrm{evaluate} \\ $$$$\rho=\frac{\oint_{\:\mathcal{C}} \:\frac{{z}}{\mathrm{2sin}\left({z}\right)−\mathrm{2}{z}\mathrm{cos}\left({z}\right)−\pi}\:\mathrm{d}{z}}{\oint_{\:\mathcal{C}} \:\frac{\mathrm{2}}{\mathrm{2sin}\left({z}\right)−\mathrm{2}{z}\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}} \\ $$$$\mathcal{C}=\mid{z}−\frac{\mathrm{3}\pi}{\mathrm{4}}\mid=\frac{\pi}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 214258 by efronzo1 last updated on 03/Dec/24 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} −\mathrm{x}^{\mathrm{5}} }−\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} +\mathrm{5x}^{\mathrm{5}} }\:=? \\ $$ Answered by issac last updated on 03/Dec/24 $$−\mathrm{1}…
Question Number 214248 by mr W last updated on 03/Dec/24 Commented by mr W last updated on 02/Dec/24 $${a}\:{man}\:{with}\:{mass}\:{M}\:{is}\:{standing}\:{on} \\ $$$${the}\:{top}\:{of}\:{a}\:{twin}\:{step}\:{ladder}.\:{each} \\ $$$${ladder}\:{with}\:{length}\:{l}\:{has}\:{a}\:{mass}\:{m}. \\ $$$${both}\:{ladders}\:{form}\:{an}\:{angle}\:{of}\:\mathrm{45}°.…
Question Number 214251 by Hanuda354 last updated on 02/Dec/24 Commented by Hanuda354 last updated on 02/Dec/24 $$\mathrm{Determine}\:\:\mathrm{where}\:\:{f}\:\:\mathrm{is}\:\:\mathrm{continuous}\:\:\mathrm{algebraically}. \\ $$$$\mathrm{Write}\:\:\mathrm{in}\:\:\mathrm{interval}\:\:\mathrm{notation}. \\ $$ Answered by a.lgnaoui last…
Question Number 214244 by daniella last updated on 02/Dec/24 $$\mathrm{4}=\sqrt{{x}−\mathrm{7}} \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated on 02/Dec/24 $${x}−\mathrm{7}=\mathrm{16} \\ $$$${x}=\mathrm{23} \\…
Question Number 214228 by mr W last updated on 02/Dec/24 Commented by mr W last updated on 02/Dec/24 $${find}\:{area}\:{of}\:{the}\:{square} \\ $$ Answered by aleks041103 last…
Question Number 214247 by hardmath last updated on 02/Dec/24 $$\mathrm{If}\:\:\:\mathrm{2a}\:=\:\mathrm{1}\:−\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{2a}^{\mathrm{2}} \:+\:\sqrt{\mathrm{a}}}{\mathrm{2a}}\:=\:? \\ $$ Answered by A5T last updated on 02/Dec/24 $$\mathrm{2}{a}=\mathrm{1}−\mathrm{2}\sqrt{{a}}\Rightarrow\begin{cases}{\sqrt{{a}}=\frac{\mathrm{2}{a}−\mathrm{1}}{−\mathrm{2}}}\\{\left(\mathrm{2}{a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{a}\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}=\mathrm{8}{a}}\end{cases}…
Question Number 214256 by Einstein2006 last updated on 02/Dec/24 $$\boldsymbol{{lim}}_{\boldsymbol{{x}}\:\rightarrow\:\mathrm{1}} \propto.\boldsymbol{{arctan}}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\boldsymbol{{x}}}\:−\:\mathrm{1}\right) \\ $$$$\bullet\:\boldsymbol{{Calculons}}\:\boldsymbol{{la}}\:\boldsymbol{{limite}}\:\boldsymbol{{a}}\:\boldsymbol{{l}}'\boldsymbol{{intrieur}}: \\ $$$$\frac{\mathrm{2}}{\mathrm{1}\:+\:\boldsymbol{{x}}}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\:\rightarrow\:\mathrm{1}} {arctan}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\:{x}}\:−\:\mathrm{1}\right)=\:\boldsymbol{{arctan}}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{1}} \propto.{arctan}\left(\frac{\mathrm{2}}{\mathrm{1}+\:{x}}\:−\:\mathrm{1}\right)\:=\propto.\mathrm{0}\:…