Menu Close

Category: None

Version-2-084-has-fixes-for-all-crashes-which-were-reported-on-playstore-in-last-week-If-anyone-is-facing-crashes-please-update-to-v2-084-and-see-if-the-problem-is-solved-If-the-problem-is-still-pr

Question Number 98723 by Tinku Tara last updated on 15/Jun/20 $$\mathrm{Version}\:\mathrm{2}.\mathrm{084}\:\mathrm{has}\:\mathrm{fixes}\:\mathrm{for}\:\mathrm{all} \\ $$$$\mathrm{crashes}\:\mathrm{which}\:\mathrm{were}\:\mathrm{reported}\:\mathrm{on} \\ $$$$\mathrm{playstore}\:\mathrm{in}\:\mathrm{last}\:\mathrm{week}. \\ $$$$\mathrm{If}\:\mathrm{anyone}\:\mathrm{is}\:\mathrm{facing}\:\mathrm{crashes}\:\mathrm{please} \\ $$$$\mathrm{update}\:\mathrm{to}\:\mathrm{v2}.\mathrm{084}\:\mathrm{and}\:\mathrm{see}\:\mathrm{if}\:\mathrm{the}\: \\ $$$$\mathrm{problem}\:\mathrm{is}\:\mathrm{solved}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{still}\:\mathrm{present},\:\mathrm{please} \\ $$$$\mathrm{send}\:\mathrm{us}\:\mathrm{an}\:\mathrm{email}\:\mathrm{as}\:\mathrm{the}\:\mathrm{problem}…

Question-164237

Question Number 164237 by DAVONG last updated on 15/Jan/22 Answered by mathmax by abdo last updated on 15/Jan/22 $$=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{1}} \:\:\:\:\frac{\mathrm{nx}^{\mathrm{n}−\mathrm{1}} −\mathrm{n}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}\:=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{1}} \frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}−\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\left(\mathrm{hospital}\:\mathrm{2times}\right)…

Question-164227

Question Number 164227 by SANOGO last updated on 15/Jan/22 Answered by TheSupreme last updated on 15/Jan/22 $$\begin{vmatrix}{\mathrm{4}}&{\mathrm{6}}&{\mathrm{0}}\\{−\mathrm{3}}&{−\mathrm{5}}&{\mathrm{0}}\\{−\mathrm{3}}&{−\mathrm{6}}&{−\mathrm{5}}\end{vmatrix}−\lambda\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\left\{\begin{matrix}{{x}}\\{{y}}\\{{z}}\end{matrix}\right\}=\left\{\begin{matrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{matrix}\right\} \\ $$$${det}\left({A}−\lambda{I}\right)=\mathrm{0} \\ $$$$\left[\left(\mathrm{4}−\lambda\right)\left(−\mathrm{5}−\lambda\right)+\mathrm{18}\right]\left(−\mathrm{5}−\lambda\right)=\mathrm{0} \\ $$$$\left[−\mathrm{20}−\mathrm{4}\lambda+\mathrm{5}\lambda+\lambda^{\mathrm{2}} +\mathrm{18}\right]\left(−\mathrm{5}−\lambda\right)=\mathrm{0} \\…

0-e-ax-sin-mx-x-dx-tan-1-m-a-a-gt-0-

Question Number 98678 by PRITHWISH SEN 2 last updated on 15/Jun/20 $$\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ax}}} \frac{\mathrm{sin}\:\boldsymbol{\mathrm{mx}}}{\boldsymbol{\mathrm{x}}}\:\boldsymbol{\mathrm{dx}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{a}}}\right),\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$ Commented by Tinku Tara last updated on…

prove-0-a-ln-1-ax-1-x-2-dx-1-2-ln-1-a-2-tan-1-a-a-gt-0-

Question Number 98677 by PRITHWISH SEN 2 last updated on 15/Jun/20 $$\mathrm{prove} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{ax}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$ Answered by maths mind…