Question Number 213097 by issac last updated on 30/Oct/24 $$\mathrm{Uhhhh}. \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{solve}\:\mathrm{Partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$$\bigtriangledown^{\mathrm{2}} \boldsymbol{\phi}=\mathrm{0} \\ $$$$\mathrm{Cylinderical}\:\mathrm{Laplacian}\:\mathrm{case} \\ $$$$\bigtriangledown^{\mathrm{2}} =\frac{\mathrm{1}}{\rho}\centerdot\frac{\partial\:\:}{\partial\rho}\left(\rho\frac{\partial\:\:}{\partial\rho}\right)+\left(\frac{\mathrm{1}}{\rho}\right)^{\mathrm{2}} \frac{\partial^{\mathrm{2}} \:}{\partial\phi^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} \:\:}{\partial{z}^{\mathrm{2}} }…
Question Number 213098 by MathematicalUser2357 last updated on 30/Oct/24 $$\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}….. \\ $$$$\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}=\frac{\mathrm{3}\left(\mathrm{10}{x}+\mathrm{2}\right)}{\mathrm{10}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}+\frac{\mathrm{7}}{\mathrm{5}\left(\mathrm{5}{x}^{\mathrm{2}}…
Question Number 213099 by MathematicalUser2357 last updated on 30/Oct/24 $$\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 $$\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}−{x}+\mathrm{3}=\frac{\mathrm{1}}{\mathrm{3}}{x}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{7}}{\mathrm{6}} \\…
Question Number 213123 by a.lgnaoui last updated on 30/Oct/24 $$\mathrm{factoriser} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1} \\ $$ Answered by MrGaster last updated on 30/Oct/24 $$\left({x}^{\mathrm{5}}…
Question Number 213054 by 281981 last updated on 29/Oct/24 Commented by Ghisom last updated on 30/Oct/24 $$\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{areas}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{and}\:\mathrm{its}\:\mathrm{circumcircle} \\ $$$$\mathrm{is} \\ $$$$\mathrm{1}:\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\:\mathrm{9}}\:\approx\:\mathrm{1}:\mathrm{2}.\mathrm{4} \\ $$$$\mathrm{here}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{is}…
Question Number 213055 by MrGaster last updated on 29/Oct/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{{x}} \mid\mathrm{sin}\:{t}\mid{dx}}{{x}} \\ $$$$ \\ $$ Answered by Ghisom last updated on…
Question Number 213045 by MrGaster last updated on 29/Oct/24 $$\overset{\rightharpoonup} {{a}},\overset{\rightharpoonup} {{b}ls}\:{the}\:{unit}\:{vector},\mid\overset{\rightharpoonup} {{a}}+\overset{\rightharpoonup} {{b}}\mid=\mathrm{1} \\ $$$${ask}\:\mid\overset{\rightharpoonup} {{a}}−\overset{\rightharpoonup} {{b}}\mid? \\ $$ Answered by mr W last…
Question Number 213047 by hardmath last updated on 29/Oct/24 $$\boldsymbol{\mathrm{n}}\:=\:\mathrm{15} \\ $$$$\mathrm{find}:\:\:\:\:\:\frac{\mathrm{4n}^{\mathrm{2}} \:\:+\:\:\mathrm{4n}\:\:+\:\:\mathrm{120}}{\:\sqrt{\mathrm{n}^{\mathrm{4}} \:\:+\:\:\mathrm{2n}^{\mathrm{3}} \:\:+\:\:\mathrm{n}^{\mathrm{2}} }}\:\:=\:\:? \\ $$ Answered by mehdee7396 last updated on 29/Oct/24…
Question Number 213073 by ajfour last updated on 29/Oct/24 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)\right\} \\ $$ Answered by universe last updated on 29/Oct/24 $${L}\:=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 213074 by Spillover last updated on 29/Oct/24 Answered by MrGaster last updated on 29/Oct/24 $$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \right)}{\mathrm{3}{z}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}}…