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Author: Tinku Tara

Question-198184

Question Number 198184 by Blackpanther last updated on 13/Oct/23 Answered by Rasheed.Sindhi last updated on 13/Oct/23 $$\mathrm{Sector}\:\mathrm{DFG}=\mathrm{quarter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{3} \\ $$$$=\pi\left(\mathrm{3}\right)^{\mathrm{2}} /\mathrm{4}=\frac{\mathrm{9}\pi}{\mathrm{4}} \\ $$$${Shaded}\:{area}\:{in}\:{one}\:{circle}={Sector}\:\mathrm{DFG}−\blacktriangle\mathrm{DFG} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{9}\pi}{\mathrm{4}}−\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}\pi−\mathrm{18}}{\mathrm{4}} \\…

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…