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sin-x-60-sin60-sin-x-60-sin30-sin-2x-30-sin60-sinx-sin60-1-x-0-60-x-

Question Number 154786 by Jamshidbek last updated on 21/Sep/21 $$\frac{\mathrm{sin}\left(\mathrm{x}+\mathrm{60}°\right)}{\mathrm{sin60}°}+\frac{\mathrm{sin}\left(\mathrm{x}+\mathrm{60}°\right)\centerdot\mathrm{sin30}°}{\mathrm{sin}\left(\mathrm{2x}+\mathrm{30}°\right)\centerdot\mathrm{sin60}°}=\frac{\mathrm{sinx}}{\mathrm{sin60}°}+\mathrm{1} \\ $$$$\mathrm{x}\in\left(\mathrm{0};\mathrm{60}°\right)\:\:\mathrm{x}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-154777

Question Number 154777 by Khalmohmmad last updated on 21/Sep/21 Answered by john_santu last updated on 21/Sep/21 $${it}\:{should}\:{be}\:\underset{{k}\rightarrow\mathrm{5}} {\mathrm{lim}}\left(\frac{{k}^{\mathrm{2}} −\mathrm{25}}{{k}−\mathrm{5}}\right)={k}+\mathrm{5} \\ $$$${or}\:{if}\:\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\left(\frac{{k}^{\mathrm{2}} −\mathrm{25}}{{k}−\mathrm{5}}\right)=\left(\frac{{k}^{\mathrm{2}} −\mathrm{25}}{{k}−\mathrm{5}}\right)×\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\left(\mathrm{1}\right)…

f-x-f-x-1-x-2-x-R-f-19-94-f-94-

Question Number 89188 by naka3546 last updated on 16/Apr/20 $${f}\left({x}\right)\:+\:{f}\left({x}−\mathrm{1}\right)\:\:=\:\:{x}^{\mathrm{2}} \:\:\:,\:\:\:{x}\:\in\:\mathbb{R} \\ $$$${f}\left(\mathrm{19}\right)\:\:=\:\:\mathrm{94} \\ $$$${f}\left(\mathrm{94}\right)\:\:=\:\:…\:\:? \\ $$ Answered by jagoll last updated on 16/Apr/20 $${i}\:{got}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{x}\left({x}+\mathrm{1}\right)…

G-46-27i-4-3i-G-2-

Question Number 154723 by liberty last updated on 21/Sep/21 $$\:{G}\:=\:\sqrt{\mathrm{46}−\mathrm{27}{i}}\:+\sqrt{−\mathrm{4}−\mathrm{3}{i}} \\ $$$$\:{G}^{\mathrm{2}} \:=? \\ $$ Commented by mathdanisur last updated on 21/Sep/21 $$\mathrm{G}\:=\:\mathrm{3}\:\sqrt{\mathrm{4}\:-\:\mathrm{3}\boldsymbol{\mathrm{i}}}\:+\:\sqrt{-\mathrm{4}\:-\:\mathrm{3}\boldsymbol{\mathrm{i}}}\:=\:\mathrm{5}\sqrt{\mathrm{2}}\:-\:\mathrm{3}\sqrt{\mathrm{2}}\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{G}^{\mathrm{2}}…

Question-154669

Question Number 154669 by SANOGO last updated on 20/Sep/21 Answered by ARUNG_Brandon_MBU last updated on 20/Sep/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right){dt}…