Question Number 208023 by efronzo1 last updated on 02/Jun/24 Answered by mr W last updated on 02/Jun/24 $${x}={a}+\mid{r}\mid\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{y}={b}+\mid{r}\mid\:\mathrm{sin}\:\theta \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\…
Question Number 207991 by efronzo1 last updated on 02/Jun/24 $$\:\:\:\:\:\mathrm{y}\:\underline{\underbrace{\mathcal{W}}} \\ $$ Answered by mr W last updated on 02/Jun/24 $${particular}\:{solution}: \\ $$$${y}={A}\:\mathrm{sin}\:{x}+{B}\:\mathrm{cos}\:{x} \\ $$$${y}'={A}\:\mathrm{cos}\:{x}−{B}\:\mathrm{sin}\:{x}…
Question Number 207984 by Thomaseinstein last updated on 02/Jun/24 Answered by Frix last updated on 02/Jun/24 $${p},\:{q}\:>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−\frac{\mathrm{1}}{{q}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{3}{p}^{\mathrm{2}} }{\:\mathrm{3}−\mathrm{4}{p}^{\mathrm{2}} } \\…
Question Number 207985 by efronzo1 last updated on 02/Jun/24 Answered by mr W last updated on 02/Jun/24 $${say}\:{AC}={s}={AB}={CB} \\ $$$${AP}^{\mathrm{2}} ={s}×{AM}\:\Rightarrow{AP}=\sqrt{{s}×{AM}} \\ $$$${CP}^{\mathrm{2}} ={s}×{CN}\:\Rightarrow{CP}=\sqrt{{s}×{CN}} \\…
Question Number 207986 by Thomaseinstein last updated on 02/Jun/24 $$\left({x}−\mathrm{3}\right)^{\sqrt{{x}−\mathrm{3}\:}} \:\:=\:\mathrm{3} \\ $$ Answered by Frix last updated on 02/Jun/24 $${x}−\mathrm{3}>\mathrm{0} \\ $$$$\sqrt{{x}−\mathrm{3}}\:\mathrm{ln}\:\left({x}−\mathrm{3}\right)\:=\mathrm{3} \\ $$$${x}=\mathrm{3}+\mathrm{e}^{\mathrm{2}{t}}…
Question Number 208011 by necx122 last updated on 02/Jun/24 Commented by necx122 last updated on 02/Jun/24 $${If}\:{the}\:{ratio}\:{of}\:<{ARS}\::\:<{BRT}\:=\:\mathrm{9}:\mathrm{4}, \\ $$$${find}\:{the}\:{ratio}\:{of}\:{area}\:{ABC}\::\:{area}\:{ARS} \\ $$ Commented by A5T last…
Question Number 208037 by hardmath last updated on 02/Jun/24 $$\mathrm{Find}:\:\:\:\mathrm{2}\:\mathrm{log}_{\sqrt{\mathrm{5}}} \:\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:\centerdot\:\mathrm{log}_{\sqrt{\boldsymbol{\mathrm{sin}}\:\frac{\boldsymbol{\pi}}{\mathrm{7}}}} \:\:\mathrm{5}\:\:=\:\:? \\ $$ Answered by mr W last updated on 02/Jun/24 $$=\mathrm{2}×\frac{\mathrm{log}\:\left(\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)}{\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{2}}}×\frac{\mathrm{log}\:\mathrm{5}}{\frac{\mathrm{log}\:\left(\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)}{\mathrm{2}}} \\ $$$$=\mathrm{8}×\frac{\cancel{\mathrm{log}\:\left(\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)}}{\cancel{\mathrm{log}\:\mathrm{5}}}×\frac{\cancel{\mathrm{log}\:\mathrm{5}}}{\cancel{\mathrm{log}\:\left(\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right)}}…
Question Number 208034 by ali009 last updated on 02/Jun/24 Commented by ali009 last updated on 02/Jun/24 $${how}\:{that}\:{can}\:{be}\:{proved}? \\ $$ Answered by Frix last updated on…
Question Number 208003 by efronzo1 last updated on 02/Jun/24 Answered by A5T last updated on 02/Jun/24 Commented by A5T last updated on 02/Jun/24 $${PM}=\frac{\mathrm{18}}{{s}};{Let}\:{line}\:{through}\:{Q}\:{parallel}\:{to}\:{KN}\:{meet} \\…
Question Number 207980 by mnjuly1970 last updated on 01/Jun/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{In}\:\:\:{A}\overset{\Delta} {{B}C}\::\:\:{B}=\:\mathrm{90}^{\:\mathrm{o}} \: \\ $$$$\:\mathrm{BB}'\:\:\bot\:\mathrm{CC}'\:\left(\:\mathrm{BB}'\:{and}\:\mathrm{CC}'\:{are}\:{medians}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\frac{\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\boldsymbol{{m}}_{\boldsymbol{{a}}} \:}\:=\:? \\ $$$$\boldsymbol{{note}}:\:\:\mid\:\mathrm{CC}'\:\mid\:=\:\boldsymbol{{m}}_{\boldsymbol{{c}}} \:\:,\:\mid\mathrm{AA}'\mid=\:\boldsymbol{{m}}_{\boldsymbol{{a}}}…