Category: Geometry

Question Number 227495 by Ari last updated on 04/Feb/26 Commented by Ari last updated on 04/Feb/26 $${thequestionisbelow} \\ $$ Terms of Service Privacy Policy Contact:…

Question Number 227357 by mr W last updated on 18/Jan/26 Answered by A5T last updated on 18/Jan/26 $$\mathrm{Let}\:\parallel\:\mathrm{be}\:\mathrm{x}\:,\:\mathrm{tangent}\:\mathrm{be}\:\mathrm{t}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{be}\:\mathrm{r}. \\ $$$$\mathrm{x}\left(\mathrm{2x}\right)=\mathrm{t}^{\mathrm{2}} \Rightarrow\mathrm{t}=\mathrm{x}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\left(\mathrm{2r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }=\mathrm{t}−\mathrm{r}\Rightarrow\mathrm{t}=\mathrm{r}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)…

Question Number 227155 by zahraa last updated on 03/Jan/26 Answered by Ghisom_ last updated on 03/Jan/26 $${x}=\mathrm{2}{y}'+\mathrm{ln}\:{y}' \\ $$$$\mathrm{e}^{{x}} =\mathrm{e}^{\mathrm{2}{y}'} {y}' \\ $$$$\mathrm{2e}^{{x}} =\mathrm{e}^{\mathrm{2}{y}'} \mathrm{2}{y}'…

Question Number 227096 by Jubr last updated on 30/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 226910 by Estevao last updated on 18/Dec/25 Answered by Estevao last updated on 18/Dec/25 $$>{Ashed}\:{Area} \\ $$ Answered by fantastic2 last updated on…