Question Number 16719 by chux last updated on 25/Jun/17 $$\mathrm{what}\:\mathrm{are}\:\mathrm{best}\:\mathrm{apps}\:\mathrm{to}\:\mathrm{use}\:\mathrm{on} \\ $$$$\mathrm{android}\:\mathrm{phones}\:\mathrm{for}\:\mathrm{architectural} \\ $$$$\mathrm{works}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147784 by Khalmohmmad last updated on 23/Jul/21 Answered by Olaf_Thorendsen last updated on 23/Jul/21 $${F}\left({x}\right)\:=\:\int\frac{{dx}}{\:\sqrt{\mathrm{7}{x}^{\mathrm{2}} −\mathrm{8}}} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{7}}}} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\mathrm{argch}\left(\frac{{x}}{\:\sqrt{\frac{\mathrm{8}}{\:\mathrm{7}}}}\right)+\mathrm{C} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\mathrm{argch}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}{x}\right)+\mathrm{C}…
Question Number 147768 by aliibrahim1 last updated on 23/Jul/21 Answered by liberty last updated on 23/Jul/21 $${Let}\:{BC}\:=\:{x}\:\Rightarrow\mathrm{tan}\:\mathrm{44}°=\frac{{CD}}{{x}} \\ $$$$\Rightarrow{CD}={x}\:\mathrm{tan}\:\mathrm{44}° \\ $$$$\mathrm{L}{et}\:{AC}\:={y}\Rightarrow\mathrm{tan}\:\mathrm{48}°=\frac{{CD}}{{y}} \\ $$$$\Rightarrow{CD}={y}\:\mathrm{tan}\:\mathrm{48}° \\ $$$${where}\:{y}=\sqrt{\mathrm{30}^{\mathrm{2}}…
Question Number 147756 by Sozan last updated on 23/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147711 by lapache last updated on 22/Jul/21 $${Determiner}\:{l}'{original}\:{de}\:{laplace} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Olaf_Thorendsen last updated on 23/Jul/21 $$\mathrm{G}\left({x}\right)\:=\:{F}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}}…
Question Number 147706 by aliibrahim1 last updated on 22/Jul/21 Answered by mr W last updated on 22/Jul/21 Commented by aliibrahim1 last updated on 22/Jul/21 $${sorry}\:{sir}\:{i}\:{tried}\:{to}\:{build}\:{on}\:{that}\:{before}\:{sending}\:{it}\:{didnt}\:{work}\:{with}\:{me}…
Question Number 147686 by 0731619 last updated on 22/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147666 by Fabricio last updated on 22/Jul/21 $$\mathrm{6bhh} \\ $$$$\mathrm{bnn77} \\ $$$$\mathrm{chb65b} \\ $$$$\mathrm{fbb7}\sqrt{} \\ $$$$\mathrm{m977} \\ $$$$\mathrm{tgvn} \\ $$ Terms of Service…
Question Number 82125 by liki last updated on 18/Feb/20 Commented by john santu last updated on 18/Feb/20 $$\mathrm{log}_{{y}} \:\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{log}_{{x}} \:\left({y}\right)} \\ $$$$\Rightarrow\mathrm{log}_{{y}} \:\left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({x}\right)}\:=\:\mathrm{64} \\…
Question Number 82115 by Khyati last updated on 18/Feb/20 $${Q}.\:{Find}\:{the}\:{number}\:{of}\:{solution}\:{of}\:{thd} \\ $$$${equation}\:{tanx}\:+\:{secx}\:=\:\mathrm{2}\:{cosx}\:{lying}\:{in} \\ $$$${the}\:{interval}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]\:?? \\ $$ Answered by MJS last updated on 18/Feb/20 $$\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}\:=\mathrm{2cos}\:{x} \\…