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Author: Tinku Tara

Hey-tinku-tara-I-couldn-t-graph-the-equation-

Question Number 213103 by MathematicalUser2357 last updated on 30/Oct/24 $$\mathrm{Hey}\:\mathrm{tinku}\:\mathrm{tara}, \\ $$$$\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{graph}\:\mathrm{the}\:\mathrm{equation}. \\ $$ Commented by Tinku Tara last updated on 30/Oct/24 $$\mathrm{We}\:\mathrm{are}\:\mathrm{aware}\:\mathrm{of}\:\mathrm{the}\:\mathrm{issue}.\:\mathrm{Will}\:\mathrm{be} \\ $$$$\mathrm{fixed}\:\mathrm{in}\:\mathrm{coming}\:\mathrm{days}.…

lim-x-0-2-1-e-1-x-

Question Number 213128 by klipto last updated on 30/Oct/24 $$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}^{+\:} } \frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{e}}^{−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} } \\ $$ Answered by a.lgnaoui last updated on 30/Oct/24 $$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}^{+\:} } \frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{e}}^{−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}}…

Hi-nikif90-can-you-please-look-at-q212921-and-provide-details-on-the-problem-that-are-facimg-

Question Number 213096 by Tinku Tara last updated on 30/Oct/24 $$\mathrm{Hi}\:\mathrm{nikif90} \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{look}\:\mathrm{at}\:\mathrm{q212921}\:\mathrm{and} \\ $$$$\mathrm{provide}\:\mathrm{details}\:\mathrm{on}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{that} \\ $$$$\mathrm{are}\:\mathrm{facimg} \\ $$ Answered by issac last updated on…

Uhhhh-can-you-guys-solve-Partial-differantial-equation-2-0-Cylinderical-Laplacian-case-2-1-1-2-2-2-2-z-2-Spherical-Laplacian-case-2-

Question Number 213097 by issac last updated on 30/Oct/24 $$\mathrm{Uhhhh}. \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{solve}\:\mathrm{Partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$$\bigtriangledown^{\mathrm{2}} \boldsymbol{\phi}=\mathrm{0} \\ $$$$\mathrm{Cylinderical}\:\mathrm{Laplacian}\:\mathrm{case} \\ $$$$\bigtriangledown^{\mathrm{2}} =\frac{\mathrm{1}}{\rho}\centerdot\frac{\partial\:\:}{\partial\rho}\left(\rho\frac{\partial\:\:}{\partial\rho}\right)+\left(\frac{\mathrm{1}}{\rho}\right)^{\mathrm{2}} \frac{\partial^{\mathrm{2}} \:}{\partial\phi^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} \:\:}{\partial{z}^{\mathrm{2}} }…

3x-2-5x-2-2x-3-dx-

Question Number 213098 by MathematicalUser2357 last updated on 30/Oct/24 $$\int\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\mathrm{d}{x}=? \\ $$ Answered by issac last updated on 30/Oct/24 $$\mathrm{Hmmmmm}….. \\ $$$$\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}=\frac{\mathrm{3}\left(\mathrm{10}{x}+\mathrm{2}\right)}{\mathrm{10}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}+\frac{\mathrm{7}}{\mathrm{5}\left(\mathrm{5}{x}^{\mathrm{2}}…

1-2-x-1-x-3-1-3-x-3-1-6-x-

Question Number 213099 by MathematicalUser2357 last updated on 30/Oct/24 $$\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}\right)−\left({x}−\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${x}=… \\ $$ Answered by MrGaster last updated on 30/Oct/24 $$\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}−{x}+\mathrm{3}=\frac{\mathrm{1}}{\mathrm{3}}{x}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{7}}{\mathrm{6}} \\…

Question-213054

Question Number 213054 by 281981 last updated on 29/Oct/24 Commented by Ghisom last updated on 30/Oct/24 $$\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{areas}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{and}\:\mathrm{its}\:\mathrm{circumcircle} \\ $$$$\mathrm{is} \\ $$$$\mathrm{1}:\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\:\mathrm{9}}\:\approx\:\mathrm{1}:\mathrm{2}.\mathrm{4} \\ $$$$\mathrm{here}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{is}…