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Author: Tinku Tara

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Question Number 204510 by Ghisom last updated on 20/Feb/24 $$\mathrm{solve}\:\mathrm{for}\:{x}\neq{y}\wedge{y}\neq{z}\wedge{z}\neq{x} \\ $$$$\left(\mathrm{exact}\:\mathrm{solutions}\:\mathrm{required}\right) \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){x}}={y} \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){y}}={z} \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){z}}={x} \\ $$ Answered by Frix last updated…

Question-204517

Question Number 204517 by Lindemann last updated on 20/Feb/24 Answered by witcher3 last updated on 20/Feb/24 $$\left.=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} +\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right).\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}._{=\mathrm{0}}…

Question-204505

Question Number 204505 by cherokeesay last updated on 19/Feb/24 Answered by mr W last updated on 19/Feb/24 $$\left[\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1}\right]^{\mathrm{2}} +\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}\sqrt{\mathrm{1}−{r}}−\mathrm{1}\right]^{\mathrm{2}} =\mathrm{2}{r}−\mathrm{1}…