Question Number 212877 by Spillover last updated on 25/Oct/24 Answered by Spillover last updated on 26/Oct/24 Answered by Spillover last updated on 26/Oct/24 Answered by…
Question Number 212878 by Spillover last updated on 25/Oct/24 Answered by som(math1967) last updated on 26/Oct/24 Commented by som(math1967) last updated on 26/Oct/24 $${let}\:{rad}\:{of}\:{green}\:{circle}={R} \\…
Question Number 212872 by Spillover last updated on 25/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212873 by Spillover last updated on 25/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212834 by efronzo1 last updated on 25/Oct/24 $$\:\:\:\:\underbrace{\:} \\ $$ Answered by golsendro last updated on 25/Oct/24 $$\:\:\mathrm{let}\:\begin{cases}{\mathrm{a}=\mathrm{sin}\:\mathrm{x}\:,\:\mathrm{b}=\mathrm{cos}\:\mathrm{x}}\\{\mathrm{c}=\mathrm{sin}\:\mathrm{y}\:,\:\mathrm{d}=\mathrm{cos}\:\mathrm{y}}\end{cases} \\ $$$$\:\:\mathrm{ac}\:+\:\mathrm{bd}\:=\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{y}\right) \\ $$$$\:\:\mid\mathrm{ac}+\mathrm{bd}\mid\:\leqslant\:\mathrm{1}\: \\…
Question Number 212867 by mr W last updated on 25/Oct/24 $${solution}\:{to}\:{Q}\mathrm{212752} \\ $$ Commented by mr W last updated on 26/Oct/24 Commented by mr W…
Question Number 212798 by MrGaster last updated on 24/Oct/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{{x}^{\mathrm{2}} }{{e}^{{x}} }} =? \\ $$ Answered by mehdee7396 last updated on 24/Oct/24 $${lim}_{{x}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}\:\:\:\:\&\:\:\:\:{lim}_{{x}\rightarrow\infty}…
Question Number 212827 by Spillover last updated on 25/Oct/24 Answered by A5T last updated on 25/Oct/24 Commented by A5T last updated on 25/Oct/24 $$\frac{{y}}{\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}}=\frac{{r}}{\frac{\mathrm{4}{r}}{\:\sqrt{\mathrm{3}}}}\Rightarrow{y}=\frac{\mathrm{2}{r}\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{4}{r}}=\frac{{r}}{\mathrm{2}} \\…
Question Number 212820 by Spillover last updated on 24/Oct/24 Answered by Spillover last updated on 25/Oct/24 Answered by Spillover last updated on 25/Oct/24 Terms of…
Question Number 212788 by MrGaster last updated on 24/Oct/24 $$ \\ $$$${Let}\:{f}\left({x}\right)\:\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{secondorder}\:\mathrm{continuoust} \\ $$$$\mathrm{derivaive},{t}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} },{g}\left({x},{y}\right)={f}\left(\frac{\mathrm{1}}{{r}}\right),\mathrm{ask}\:\frac{\partial^{\mathrm{2}} {g}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {g}}{\partial{y}^{\mathrm{2}} }. \\ $$ Terms of Service…