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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}\: \\…

Solve-the-following-differential-equation-1-y-y-e-x-x-3-y-0-2-y-0-0-2-y-y-2y-x-sin2x-y-0-1-y-0-0-3-y-y-xe-x-y-0-2-y-0-1-Thank-

Question Number 197792 by Mastermind last updated on 28/Sep/23 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{y}''\:+\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{x}^{\mathrm{3}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{y}''\:+\:\mathrm{y}^{'} \:−\:\mathrm{2y}\:=\:\mathrm{x}\:+\:\mathrm{sin2x},\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{y}''\:−\:\mathrm{y}'\:=\:\mathrm{xe}^{\mathrm{x}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\:\mathrm{1} \\ $$$$ \\ $$$$ \\…

Water-is-leaking-from-a-hemispheric-bowl-of-radius-20cm-at-the-rate-of-0-5cm-3-s-Find-the-rate-at-which-surface-area-of-the-water-decreasing-when-the-water-level-is-halfway-from-the-top-Thank-you

Question Number 197640 by Mastermind last updated on 25/Sep/23 $${Water}\:{is}\:{leaking}\:{from}\:{a}\:{hemispheric} \\ $$$${bowl}\:{of}\:{radius}\:\mathrm{20}{cm}\:{at}\:{the}\:{rate}\:{of}\: \\ $$$$\mathrm{0}.\mathrm{5}{cm}^{\mathrm{3}} /{s}.\:{Find}\:{the}\:{rate}\:{at}\:{which}\:{surface} \\ $$$${area}\:{of}\:{the}\:{water}\:{decreasing}\:{when}\:{the} \\ $$$${water}\:{level}\:{is}\:{halfway}\:{from}\:{the}\:{top}. \\ $$$$ \\ $$$${Thank}\:{you} \\ $$…

Solve-the-following-equation-x-2y-2z-0-2x-y-2z-0-3x-4y-6z-0-3x-11y-12z-0-

Question Number 197470 by Mastermind last updated on 18/Sep/23 $${Solve}\:{the}\:{following}\:{equation} \\ $$$${x}\:+\:\mathrm{2}{y}\:+\:\mathrm{2}{z}\:=\:\mathrm{0} \\ $$$$\mathrm{2}{x}\:+\:{y}\:−\:\mathrm{2}{z}\:=\mathrm{0} \\ $$$$\mathrm{3}{x}\:+\:\mathrm{4}{y}\:−\:\mathrm{6}{z}\:=\mathrm{0} \\ $$$$\mathrm{3}{x}\:−\:\mathrm{11}{y}\:+\:\mathrm{12}{z}\:=\:\mathrm{0} \\ $$ Answered by MathedUp last updated…