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Question-198186

Question Number 198186 by Tawa11 last updated on 13/Oct/23 Answered by mr W last updated on 13/Oct/23 $${y}\left({t}\right)=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi{t}}{{T}}\right) \\ $$$${y}\left(\mathrm{0}.\mathrm{4}\right)=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi×\mathrm{0}.\mathrm{4}}{\mathrm{1}.\mathrm{0}}\right)=\mathrm{0}.\mathrm{09}\:{cm} \\ $$ Commented by Tawa11…

Question-198052

Question Number 198052 by Mastermind last updated on 09/Oct/23 Answered by Sutrisno last updated on 09/Oct/23 $${x}={asin}\theta\:\Rightarrow\:{dx}={acos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} {a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\left({a}^{\mathrm{2}} −{a}^{\mathrm{2}} {sin}^{\mathrm{2}}…

Question-198018

Question Number 198018 by obia last updated on 07/Oct/23 Answered by a.lgnaoui last updated on 08/Oct/23 $$\boldsymbol{\mathrm{Z}}=\sqrt{\mathrm{3}}\:+\mathrm{1}+\:\boldsymbol{\mathrm{i}}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right) \\ $$$$ \\ $$$$\left.\bullet\mathrm{1}\right):\:\:\boldsymbol{\mathrm{Z}}^{\mathrm{2}} =\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\:−\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\:\right)+\mathrm{2}\boldsymbol{\mathrm{i}}\left(\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{3}}\:−\:\mathrm{1}\right) \\ $$$$\:\:\:=\mathrm{4}\sqrt{\mathrm{3}}\:\:+\mathrm{4}\boldsymbol{\mathrm{i}}\:\:=\mathrm{4}\left(\sqrt{\mathrm{3}}\:+\boldsymbol{\mathrm{i}}\right) \\…

If-z-1-2-and-z-2-2-Show-that-z-1-z-2-2-z-1-z-2-2-z-2-z-2-2-and-deduce-that-2-2-2-2-Thank-you-in-advan

Question Number 197851 by Mastermind last updated on 01/Oct/23 $$\mathrm{If}\:\mathrm{z}_{\mathrm{1}} \:=\:\sqrt{\frac{\alpha−\beta}{\mathrm{2}}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} \:=\:\sqrt{\frac{\alpha+\beta}{\mathrm{2}}}\:.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\:=\:\mathrm{2}\mid\mathrm{z}\mid^{\mathrm{2}} +\mid\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{deduce}\:\mathrm{that}\: \\…