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Question Number 151164 by puissant last updated on 18/Aug/21 Commented by puissant last updated on 18/Aug/21 $${solve}\:{in}\:\mathbb{R} \\ $$ Answered by MJS_new last updated on…
Question Number 150977 by aupo14 last updated on 17/Aug/21 Answered by talminator2856791 last updated on 17/Aug/21 $$\:\sqrt{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 19884 by ajfour last updated on 17/Aug/17 Commented by ajfour last updated on 17/Aug/17 $$\:\:{A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:;\:\:{B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)\:\:;\:{C}\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} \right)\:\:; \\ $$$$\:\:{D}\left({x}_{\mathrm{4}}…
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Question Number 150864 by puissant last updated on 16/Aug/21 Answered by ajfour last updated on 16/Aug/21 $$\mathrm{70}°=\theta,\:\mathrm{30}°=\alpha,\:\mathrm{40}°=\beta,\:\mathrm{6}{cm}={r} \\ $$$${A}_{\mathrm{1}} =\left({r}−\mathrm{2}{r}\mathrm{cos}\:\theta\right){r}\mathrm{sin}\:\beta \\ $$$${A}_{\mathrm{2}} =\frac{{r}\left(\mathrm{1}−\mathrm{2cos}\:\theta\right)\left\{{r}\mathrm{sin}\:\theta−{r}\mathrm{sin}\:\beta\right\}}{\mathrm{2}} \\ $$$${A}_{\mathrm{3}}…
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Question Number 150801 by puissant last updated on 15/Aug/21 Commented by puissant last updated on 15/Aug/21 $${find}\:{tan}\theta\:{please}.. \\ $$ Commented by mr W last updated…
Question Number 150793 by cherokeesay last updated on 15/Aug/21 $$\left({R}−{r}\right)^{\mathrm{2}} \:+\:{R}^{\mathrm{2}} \:=\:\left({R}\:+\:{r}\right)^{\mathrm{2}} \:\Leftrightarrow \\ $$$${R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} −\mathrm{2}{rR}\:+\:{R}^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} \:+\mathrm{2}{rR}\:\Rightarrow \\ $$$${R}^{\mathrm{2}} \:=\:\mathrm{4}{rR}\:\Rightarrow\:{r}\:=\:\frac{{R}}{\mathrm{4}}\:=\:\mathrm{1}{cm} \\ $$$$\mathscr{A}_{{S}}…
Question Number 150794 by cherokeesay last updated on 15/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com