Question Number 197848 by cortano12 last updated on 01/Oct/23 Answered by mr W last updated on 01/Oct/23 $${f}\left({x}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}}+\frac{{b}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \:{x}}\geqslant\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{sin}^{\mathrm{2}} \:{x}}=\left({a}+{b}\right)^{\mathrm{2}}…
Question Number 197865 by Blackpanther last updated on 01/Oct/23 Answered by mr W last updated on 02/Oct/23 Commented by mr W last updated on 02/Oct/23…
Question Number 197850 by tri26112004 last updated on 01/Oct/23 $${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{2}{sin}\:{x}\:−\:\mathrm{1}\right)\left(\mathrm{2}{cos}\:\mathrm{2}{x}\:+\:\mathrm{2}{sin}\:{x}\:+\mathrm{1}\right)\:=\:\mathrm{3}\left(\mathrm{1}−\mathrm{2}{cos}\:\mathrm{2}{x}\right) \\ $$ Answered by Frix last updated on 01/Oct/23 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$−\mathrm{2sin}^{\mathrm{3}} \:{x}\:−\mathrm{sin}^{\mathrm{2}}…
Question Number 197851 by Mastermind last updated on 01/Oct/23 $$\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} \\ $$$$\mathrm{that}\:\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\:=\:\mathrm{2}\mid\mathrm{z}\mid^{\mathrm{2}} +\mid\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{deduce}\:\mathrm{that}\: \\…
Question Number 197860 by cortano12 last updated on 01/Oct/23 $$\:\:\:\:\frac{\mathrm{2}^{\mathrm{17}} +\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{15}} +\ldots+\mathrm{1}}{\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{7}} +\mathrm{2}^{\mathrm{6}} +\ldots+\mathrm{1}}\:=\:?\: \\ $$ Answered by mr W last updated on…
Question Number 197847 by CrispyXYZ last updated on 01/Oct/23 $$\mathrm{If}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} <\mathrm{1},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of} \\ $$$$\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}. \\ $$ Answered by Frix last updated on 01/Oct/23…
Question Number 197874 by yaslm last updated on 01/Oct/23 Answered by aleks041103 last updated on 02/Oct/23 $${v}:\begin{cases}{\mathrm{0}\leqslant{z}\leqslant\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\\{\mathrm{0}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} }\end{cases} \\ $$$${in}\:{cylimdrical}\:{coordinates} \\…
Question Number 197820 by lemonstre last updated on 30/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197821 by mnjuly1970 last updated on 30/Sep/23 $$ \\ $$$$\:\:\:\:{find}\:{the}\:{value}\:\:{of}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:\mathrm{ln}\:\left(\:\mathrm{1}+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }\:\right)}{\mathrm{2}\:+\:{x}^{\:\mathrm{2}} }\:{dx}\:=\:? \\ $$$$ \\ $$ Answered…
Question Number 197822 by mokys last updated on 30/Sep/23 $${find}\:{maximum}\:{of}\:\mid{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{3}\mid\:? \\ $$ Commented by AST last updated on 30/Sep/23 $$\infty \\ $$ Commented by…