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Question-197967

Question Number 197967 by Blackpanther last updated on 06/Oct/23 Answered by mr W last updated on 06/Oct/23 $${f}\left({x}\right)={a}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{10}\right) \\ $$$$\frac{\mathrm{20}}{\mathrm{3}}={a}\left(\mathrm{0}+\mathrm{1}\right)\left(\mathrm{0}−\mathrm{10}\right) \\ $$$$\Rightarrow{a}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${at}\:{x}=\frac{−\mathrm{1}+\mathrm{10}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}}: \\…

tan-pi-12-sin-sin-pi-12-cos-cos-pi-12-

Question Number 197944 by liuxinnan last updated on 05/Oct/23 $${tan}\frac{\pi}{\mathrm{12}}=\frac{{sin}\alpha−{sin}\frac{\pi}{\mathrm{12}}}{{cos}\alpha+{cos}\frac{\pi}{\mathrm{12}}} \\ $$$$\alpha=? \\ $$ Answered by MM42 last updated on 05/Oct/23 $${sin}\frac{\pi}{\mathrm{12}}×{cos}\alpha+{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}}={sin}\alpha×{cos}\frac{\pi}{\mathrm{12}}−{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}} \\ $$$$\Rightarrow{sin}\left(\alpha−\frac{\pi}{\mathrm{12}}\right)=\mathrm{2}{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}}={sin}\frac{\pi}{\mathrm{6}} \\…

Question-197917

Question Number 197917 by sonukgindia last updated on 04/Oct/23 Commented by Frix last updated on 04/Oct/23 $$\mathrm{I}\:\mathrm{have}\:\mathrm{more}\:\mathrm{joy}\:\mathrm{with}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one},\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{one} \\ $$$$\mathrm{is}\:\mathrm{boring}. \\ $$ Answered by…

Find-the-minimum-value-of-5t-2-8t-5-2-3-t-2-2t-2-3-where-2-3-lt-t-lt-2-3-

Question Number 197882 by CrispyXYZ last updated on 02/Oct/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{5}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}−\sqrt{\mathrm{3}}}\:\:\mathrm{where}\:\mathrm{2}−\sqrt{\mathrm{3}}<{t}<\mathrm{2}+\sqrt{\mathrm{3}}. \\ $$ Answered by mr W last updated on 03/Oct/23 $$\frac{\mathrm{5}{t}^{\mathrm{2}}…

Question-197874

Question Number 197874 by yaslm last updated on 01/Oct/23 Answered by aleks041103 last updated on 02/Oct/23 $${v}:\begin{cases}{\mathrm{0}\leqslant{z}\leqslant\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\\{\mathrm{0}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} }\end{cases} \\ $$$${in}\:{cylimdrical}\:{coordinates} \\…