Question Number 8300 by tawakalitu last updated on 06/Oct/16 Commented by ridwan balatif last updated on 07/Oct/16 $$\mathrm{solution} \\ $$$$\mathrm{1}.{v}=\sqrt{\mathrm{2gh}},\:\mathrm{where}\:\mathrm{h}\:\mathrm{is}\:\mathrm{height}\:\mathrm{and}\:\mathrm{g}\:\mathrm{is}\:\mathrm{gravitational}\:\mathrm{acceleration} \\ $$$$\:\:\:\:{v}=\sqrt{\mathrm{2}×\mathrm{9}.\mathrm{8}×\mathrm{5}} \\ $$$$\:\:\:\:{v}=\mathrm{9}.\mathrm{899}\:{m}/{s} \\…
Question Number 139371 by mohammad17 last updated on 26/Apr/21 $${by}\:{use}\:{Gamma}\:{function}\:{prove}\: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{8}}} {cos}^{\mathrm{3}} \mathrm{4}{xdx}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\:\pi} {sin}^{\mathrm{6}} \left(\frac{{x}}{\mathrm{2}}\right){cos}^{\mathrm{8}} \left(\frac{{x}}{\mathrm{2}}\right){dx}=\frac{\mathrm{5}\pi}{\mathrm{2}^{\mathrm{11}}…
Question Number 73832 by FCB last updated on 16/Nov/19 Commented by FCB last updated on 16/Nov/19 $$\boldsymbol{\mathrm{solve}} \\ $$ Commented by MJS last updated on…
Question Number 8297 by lepan last updated on 06/Oct/16 $$\underset{} {{B}y}\:{expessing}\:{each}\:{side}\:{of}\:{the} \\ $$$${equation}\:{in}\:{terms}\:{of}\:{tanA}\:,{or}\: \\ $$$${otherwise}\:{show}\:{that} \\ $$$$\frac{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}+\mathrm{1}}{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}−\mathrm{1}}=\frac{{tan}\left(\mathrm{45}°+{A}\right)}{{tanA}} \\ $$ Answered by Rasheed Soomro last updated…
Question Number 8296 by tawakalitu last updated on 06/Oct/16 $$\mathrm{Give}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{representation}\:\mathrm{of}\:\:\mathrm{2333}! \\ $$ Commented by 123456 last updated on 07/Oct/16 $${n}!=\underset{\mathrm{0}} {\overset{+\infty} {\int}}{t}^{{n}} {e}^{−{t}} {dt} \\…
Question Number 73828 by ajfour last updated on 16/Nov/19 Commented by ajfour last updated on 17/Nov/19 $${If}\:{on}\:{each}\:{face}\:{of}\:{the}\:{larger}\:{cube} \\ $$$${there}\:{is}\:\left({at}\:{least}\right)\:{one}\:{corner}\:{of}\: \\ $$$${the}\:{inner}\:{cube},\:{then} \\ $$$${find}\:{minimum}\:{value}\:{of}\:{ratio}\:{r}. \\ $$$$\:{r}=\frac{{s}}{{a}}\:=\:\frac{{edge}\:{length}\:{of}\:{inner}\:{cube}}{{edge}\:{length}\:{of}\:{outer}\:{cube}}\:.…
Question Number 8289 by rhm last updated on 07/Oct/16 $${what}\:{is}\:{value} \\ $$$${sin}\:\mathrm{36}° \\ $$$${plese}\:{give}\:{me}\:{answer} \\ $$ Commented by rhm last updated on 07/Oct/16 $${sir}\:{answer} \\…
Question Number 8287 by lepan last updated on 06/Oct/16 $${Show}\:{that}\:{tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}. \\ $$ Answered by ridwan balatif last updated on 06/Oct/16 $$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{sin}\left(\alpha+\beta\right)}{\mathrm{cos}\left(\alpha+\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha}{\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}{\left(\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}…
Question Number 139359 by Dwaipayan Shikari last updated on 26/Apr/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}\right)!} \\ $$ Commented by Dwaipayan Shikari last updated on 28/Apr/21 $$\frac{{cosh}\left(\mathrm{1}\right)+\mathrm{2}{cosh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\mathrm{3}} \\…
Question Number 139353 by mohammad17 last updated on 26/Apr/21 $$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{6}} \theta\:{cos}^{\mathrm{4}} \theta\:{d}\theta \\ $$ Answered by Dwaipayan Shikari last updated on 26/Apr/21 $$\int_{\mathrm{0}}…