Question Number 203291 by zahaku last updated on 14/Jan/24 $${f}\left({x}\right)=\left\{\mathrm{1}+\frac{\sqrt{{x}^{\mathrm{2}} }}{{x}}\:\:\:{if}\:{x}#\mathrm{0}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:{if}\:\:{x}=\mathrm{0} \\ $$$${study}\:{the}\:{continuty}\:{of}\:{f}\:{in}\:\mathrm{0} \\ $$ Answered by esmaeil last updated on 15/Jan/24 $${f}\left({x}\right)=\begin{cases}{\mathrm{1}+\frac{\mid{x}\mid}{{x}}\:\:\:\:{if}\:\:{x}\neq\mathrm{0}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:{x}=\mathrm{0}}\end{cases}\:\:\:…
Question Number 203282 by mr W last updated on 14/Jan/24 Commented by cortano12 last updated on 14/Jan/24 $$\mathrm{AD}\parallel\mathrm{BC}\:? \\ $$ Commented by mr W last…
Question Number 203260 by a.lgnaoui last updated on 13/Jan/24 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2ln}\mid\mathrm{x}−\mathrm{1}\mid+\mathrm{ln}\left(\mathrm{2x}+\mathrm{1}\right) \\ $$$$\left.\bullet\mathrm{1}\right)−\mathrm{the}\:\mathrm{intersextion}\:\:\mathrm{Curve}\:\mathrm{with} \\ $$$$\mathrm{axe}\:\mathrm{XX}^{'} ? \\ $$$$\left.\bullet\mathrm{2}\right)−\mathrm{find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{Circle}\:\mathrm{tengent}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{Curve}\:\left(\mathrm{f}\right)\:\mathrm{and}\:\mathrm{axe}\:\left(\mathrm{O},\mathrm{Y}\right):\mathrm{Y}>\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{relative}\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\left.\bullet\mathrm{3}\right)−\mathrm{the}\:\mathrm{Area}\:\mathrm{delimited}\:\mathrm{by}\:\mathrm{x}=\mathrm{3} \\ $$$$\:\:\:\:\:\mathrm{max}\:\mathrm{relative}\:\mathrm{and}\:\mathrm{Circle}\:?…
Question Number 203244 by cortano12 last updated on 13/Jan/24 Answered by MathematicalUser2357 last updated on 14/Jan/24 $$\mathrm{If}\:\mathrm{you}\:\mathrm{took}\:\mathrm{that}\:\mathrm{image}\:\mathrm{on}\:\mathrm{FB}, \\ $$$$\mathrm{You}\:\mathrm{will}\:\mathrm{be}\:\mathrm{known}\:\mathrm{soon}. \\ $$$$\mathrm{Here}\:\mathrm{is}\:\mathrm{a}\:\mathrm{proof}: \\ $$ Commented by…
Question Number 203245 by hardmath last updated on 13/Jan/24 $$\mathrm{If}\:\:\:\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2019}}\:+\:\mathrm{1} \\ $$$$\mathrm{Find}: \\ $$$$\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{12x}\:−\:\mathrm{3} \\ $$ Answered by mr W last updated on…
Question Number 203247 by mathlove last updated on 13/Jan/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\:{tan}\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{x}\right)=? \\ $$ Answered by MM42 last updated on 13/Jan/24 $$={lim}_{{x}\rightarrow\mathrm{0}} \:−{xcot}\frac{\pi}{\mathrm{2}}{x}={lim}_{{x}\rightarrow\mathrm{0}} \:−\frac{\frac{\pi}{\mathrm{2}}{xcos}\frac{\pi}{\mathrm{2}}{x}}{{sin}\frac{\pi}{\mathrm{2}}{x}}×\frac{\mathrm{2}}{\pi} \\ $$$$=\:−\frac{\mathrm{2}}{\pi}\:\checkmark…
Question Number 203242 by sulaymonnorboyev140 last updated on 13/Jan/24 $${sin}\mathrm{6}°\centerdot{sin}\mathrm{12}°\centerdot{sin}\mathrm{24}°\centerdot{sin}\mathrm{28}° \\ $$ Answered by MathematicalUser2357 last updated on 15/Jan/24 $$\mathrm{4}.\mathrm{1498}×\mathrm{10}^{−\mathrm{3}} \\ $$ Terms of Service…
Question Number 203275 by Frix last updated on 13/Jan/24 $$\mathrm{Show}\:\mathrm{this}\:\mathrm{has}\:\mathrm{exactly}\:\mathrm{7}\:\mathrm{solutions}\:\mathrm{for}\:{x}\in\mathbb{C}: \\ $$$${x}^{\mathrm{ln}\:{x}} =\mathrm{1} \\ $$ Commented by aleks041103 last updated on 14/Jan/24 $${When}\:{you}\:{use}\:{ln}\left({x}\right)\:{do}\:{you}\:{imply}\:{the}\:{principle} \\ $$$${branch}\:{of}\:{this}\:{function}?…
Question Number 203236 by mustafazaheen last updated on 13/Jan/24 $$\mathrm{how}\:\mathrm{is}\:\mathrm{solution}\:\mathrm{Q203144} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203269 by ahmetgg last updated on 13/Jan/24 Answered by esmaeil last updated on 13/Jan/24 $${tan}\mathrm{10}=\frac{{AH}}{{BH}} \\ $$$${tan}\mathrm{20}=\frac{{HC}}{{BH}}\rightarrow\frac{{HC}}{{AH}}\approx\mathrm{2}.\mathrm{0642} \\ $$$${tan}\mathrm{50}=\frac{{OH}}{{AH}} \\ $$$${tanx}=\frac{{OH}}{{CH}}\rightarrow\frac{{tan}\mathrm{50}}{{tanx}}\approx\mathrm{2}.\mathrm{0642}\rightarrow \\ $$$${x}={tan}^{−\mathrm{1}}…