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Question-208540

Question Number 208540 by otchereabdullai@gmail.com last updated on 17/Jun/24 Answered by Sutrisno last updated on 18/Jun/24 $${misal}\::\:\mathrm{3}{x}=\mathrm{2}{tan}\theta \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta\rightarrow{dx}=\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\int\frac{\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta−\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{tan}\theta\right)^{\mathrm{2}} +\mathrm{4}}}.\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\…

Question-208502

Question Number 208502 by Kalebwizeman last updated on 17/Jun/24 Answered by A5T last updated on 17/Jun/24 $$=\sqrt{\mathrm{16}}+\sqrt{\mathrm{15}}−\sqrt{\mathrm{15}}−\sqrt{\mathrm{14}}+…−\sqrt{\mathrm{9}}−\sqrt{\mathrm{8}}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$ Commented by Kalebwizeman last updated on…

write-z-1-3-i-in-e-i-

Question Number 208398 by mokys last updated on 14/Jun/24 $${write}\:{z}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+{i}}\:{in}\:{e}^{{i}\theta} \\ $$ Answered by A5T last updated on 14/Jun/24 $${z}=\frac{\sqrt{\mathrm{3}}−{i}}{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left({i}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}−{i}}{\mathrm{4}},\:\mid{z}\mid=\sqrt{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\…