Question Number 208940 by efronzo1 last updated on 28/Jun/24 Answered by kapoorshah last updated on 28/Jun/24 $${r}\:=\:{OD}\:=\:\sqrt{\mathrm{10}} \\ $$$${BE}\:=\:\sqrt{\mathrm{15}} \\ $$$${BOE}\:\sim\:{BCA} \\ $$$$\frac{{BO}}{{BE}}\:=\:\frac{{BC}}{{AB}} \\ $$$$\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{15}}}\:=\:\frac{{BC}}{\mathrm{2}\sqrt{\mathrm{10}}}\:\:\Rightarrow\:{BC}\:=\:\frac{\mathrm{20}}{\:\sqrt{\mathrm{15}}}\:=\:\mathrm{5}.\mathrm{164}…
Question Number 208959 by hardmath last updated on 28/Jun/24 $$\mathrm{If}\:\:\:\boldsymbol{\mathrm{z}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{Find}\:\:\:\left(\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{2z}\right)\centerdot\left(\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\right)\:=\:? \\ $$ Answered by grigoriy last updated on 29/Jun/24 $$ \\…
Question Number 208943 by MWSuSon last updated on 28/Jun/24 Answered by A5T last updated on 28/Jun/24 $${a}.\:\frac{\left[{ADM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{DM}}{{AD}×\left({DM}×\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${b}.\:{Let}\:{AM}\:{and}\:{DN}\:{intersect}\:{at}\:{E} \\ $$$$\frac{\left[{DEM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\frac{{AD}}{\mathrm{2}}×{DM}}{{AD}×{DM}×\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${c}.\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$…
Question Number 208951 by hardmath last updated on 28/Jun/24 $$\mathrm{2}^{\mathrm{2024}} \::\:\mathrm{2024}\:=\:…\:\left(\mathrm{Remainder}\:=\:?\right) \\ $$ Answered by A5T last updated on 28/Jun/24 $$\left(\mathrm{2}^{\mathrm{11}} \right)^{\mathrm{184}} \:\:\overset{\mathrm{2024}} {\equiv}\:\mathrm{24}^{\mathrm{184}} \\…
Question Number 208945 by hardmath last updated on 28/Jun/24 $$\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}\:−\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:=\:\mathrm{0} \\ $$$$\left[\:\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\:\right] \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by Berbere last updated on…
Question Number 208912 by dimentri last updated on 27/Jun/24 $$\:\:\:\underbrace{\Subset} \underbrace{ \cancel{} }\pi \\ $$ Commented by Frix last updated on 27/Jun/24 $$\mathrm{33} \\ $$…
Question Number 208913 by efronzo1 last updated on 27/Jun/24 Answered by A5T last updated on 27/Jun/24 $$\frac{{sin}\beta}{\mathrm{6}}=\frac{\mathrm{1}}{{AC}};\frac{{sin}\left(\mathrm{60}−\beta\right)}{\mathrm{3}}=\frac{\mathrm{1}}{{AC}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{60}−\beta\right)}{{sin}\beta}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\beta\approx\mathrm{40}.\mathrm{8934} \\ $$$$\Rightarrow{sin}\beta\approx\mathrm{0}.\mathrm{654654};\:{we}\:{can}\:{then}\:{find}\:{AC}=\frac{\mathrm{6}}{{sin}\beta} \\ $$$${AC}\approx\mathrm{9}.\mathrm{165151} \\ $$…
Question Number 208931 by byaw last updated on 27/Jun/24 Answered by mr W last updated on 27/Jun/24 $$\left({a}\right) \\ $$$${R}={m}\left({g}+{a}\right)=\mathrm{0}.\mathrm{5}×\left(\mathrm{10}+\mathrm{2}\right)=\mathrm{6}\:{N} \\ $$$$ \\ $$$$\left({b}\right)\left({i}\right) \\…
Question Number 208915 by Tawa11 last updated on 27/Jun/24 Answered by mr W last updated on 27/Jun/24 Commented by mr W last updated on 27/Jun/24…
Question Number 208908 by alcohol last updated on 26/Jun/24 $${I}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{a}} \left(\mathrm{1}−{t}\right)^{{b}} {dt} \\ $$$${Note}\::\:{I}\left({a},{b}\right)\:=\:\frac{{a}}{{b}+\mathrm{1}}{I}\left({a}−\mathrm{1},{b}+\mathrm{1}\right) \\ $$$${show}\:{that} \\ $$$$\bullet\:{I}\left({a}+\mathrm{1},\:{b}\right)\:+\:{I}\left({a},{b}+\mathrm{1}\right)\:=\:{I}\left({a},\:{b}\right) \\ $$$$\bullet\:{find}\:{B}\left({a}+\mathrm{1},\:{b}+\mathrm{1}\right)\:{interms}\:{of}\:{B}\left({a},{b}\right) \\ $$$$\bullet\:{use}\:{I}\left({a},{b}\right)\:=\:{I}\left({b},{a}\right)\:{and}\:{deduce}\:{I}\left(\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\…