Question Number 207218 by hardmath last updated on 09/May/24 $$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:\:+\:\:\frac{\mathrm{6}\:\centerdot\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{10}}{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{2}}\:\:=\:\:\mathrm{4}\:\centerdot\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{2}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{x}\:=\:? \\ $$ Answered by Frix last updated on…
Question Number 207207 by hardmath last updated on 09/May/24 $$\mathrm{100}\:\mathrm{x}^{\boldsymbol{\mathrm{lg}}\:\boldsymbol{\mathrm{x}}} \:\:=\:\:\mathrm{x}^{\mathrm{3}} \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by mr W last updated on 09/May/24 $$\mathrm{100}={x}^{\mathrm{3}−\mathrm{log}\:{x}} \\…
Question Number 207219 by hardmath last updated on 09/May/24 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{tg}\:\mathrm{20}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{20}}\:+\:\frac{\mathrm{tg}\:\mathrm{21}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{21}}\:+…+\:\frac{\mathrm{tg}\:\mathrm{70}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{70}} \\ $$ Answered by Berbere last updated on 10/May/24 $$\frac{{tg}\left({a}\right)}{\mathrm{1}+{tg}^{\mathrm{2}}…
Question Number 207169 by hardmath last updated on 08/May/24 $$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{3}} \:\sqrt{\mathrm{9}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by lepuissantcedricjunior last updated on 08/May/24 $$\int_{\mathrm{0}} ^{\mathrm{3}} \sqrt{\mathrm{9}−\boldsymbol{\mathrm{x}}^{\mathrm{2}}…
Question Number 207170 by hardmath last updated on 08/May/24 $$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \:\sqrt{\mathrm{16}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Commented by MM42 last updated on 08/May/24 $${x}=\mathrm{4}{sin}\theta\Rightarrow{dx}=\mathrm{4}{cos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}}…
Question Number 207128 by hardmath last updated on 07/May/24 $$\mathrm{log}_{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\mathrm{x}}} \:\:\mathrm{sinx}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{log}_{\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}} \:\:\mathrm{tanx}}\:\:=\:\:? \\ $$ Answered by Frix last updated on 07/May/24 $$=\frac{\mathrm{ln}\:{s}}{\mathrm{ln}\:{t}}−\frac{\mathrm{ln}\:{c}}{\mathrm{ln}\:{t}}=\frac{\mathrm{ln}\:{s}\:−\mathrm{ln}\:{c}}{\mathrm{ln}\:\frac{{s}}{{c}}} \\ $$$$\mathrm{If}\:{c}>\mathrm{0}\wedge{s}>\mathrm{0}\:\mathrm{this}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1},\:\mathrm{else}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\…
Question Number 207129 by hardmath last updated on 07/May/24 $$\mathrm{Find}:\:\:\:\:\:\int\:\frac{\mathrm{x}\:+\:\mathrm{4}}{\mathrm{x}\:+\:\mathrm{3}}\:\mathrm{dx}\:=\:? \\ $$ Answered by mathzup last updated on 08/May/24 $$\int\frac{{x}+\mathrm{4}}{{x}+\mathrm{3}}{dx}=\int\frac{{x}+\mathrm{3}+\mathrm{1}}{{x}+\mathrm{3}}{dx}\:=\int{dx}\:+\int\frac{{dx}}{{x}+\mathrm{3}} \\ $$$$={x}+{ln}\mid{x}+\mathrm{3}\mid\:+{c} \\ $$ Terms…
Question Number 207156 by hardmath last updated on 07/May/24 $$\mathrm{Find}:\:\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\mathrm{10}°}\:\:=\:\:? \\ $$ Answered by A5T last updated on 07/May/24 $$=\frac{\mathrm{2}{sin}\mathrm{30}}{{sin}\mathrm{10}}−\frac{\mathrm{2}{cos}\mathrm{30}}{{cos}\mathrm{10}}=\frac{\mathrm{2}\left[{sin}\mathrm{30}{cos}\mathrm{10}−{cos}\mathrm{30}{sin}\mathrm{10}\right]}{{sin}\mathrm{10}{cos}\mathrm{10}} \\ $$$$=\frac{\mathrm{2}\left[{sin}\left(\mathrm{30}−\mathrm{10}\right)\right]}{{sin}\mathrm{10}{cos}\mathrm{10}}=\frac{\mathrm{4}{sin}\mathrm{20}}{\mathrm{2}{sin}\mathrm{10}{cos}\mathrm{10}={sin}\mathrm{20}}=\mathrm{4} \\ $$ Terms…
Question Number 207130 by hardmath last updated on 07/May/24 $$\mathrm{Find}:\:\:\:\int\:\frac{\mathrm{x}\:−\:\mathrm{1}}{\mathrm{x}\:−\:\mathrm{2}}\:\mathrm{dx}\:\:=\:\:? \\ $$ Answered by Frix last updated on 07/May/24 $$\int\frac{{x}\pm{a}}{{x}\pm{b}}{dx}=\int{dx}+\left(\pm{a}\mp{b}\right)\int\frac{{dx}}{{x}\pm{b}}= \\ $$$$={x}+\left(\pm{a}\mp{b}\right)\mathrm{ln}\:\mid{x}\pm{b}\mid\:+{C} \\ $$ Commented…
Question Number 207157 by hardmath last updated on 07/May/24 $$\mathrm{Find}:\:\:\:\frac{\mathrm{5}\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}}{\frac{\mathrm{1}}{\mathrm{tg}\:\mathrm{75}°}\:\:−\:\:\mathrm{tg}\:\mathrm{75}°}\:\:=\:\:? \\ $$ Answered by A5T last updated on 07/May/24 $$\frac{\mathrm{1}}{{tan}\mathrm{75}}−{tan}\mathrm{75}=\frac{{cos}\mathrm{75}}{{sin}\mathrm{75}}−\frac{{sin}\mathrm{75}}{{cos}\mathrm{75}}=\frac{{cos}^{\mathrm{2}} \mathrm{75}−{sin}^{\mathrm{2}} \mathrm{75}}{{sin}\mathrm{75}{cos}\mathrm{75}} \\ $$$$=\frac{\mathrm{2}{cos}\mathrm{150}}{{sin}\mathrm{150}}=\frac{\mathrm{2}\left(\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}}=−\mathrm{2}\sqrt{\mathrm{3}} \\…