Question Number 218738 by Spillover last updated on 14/Apr/25 Answered by som(math1967) last updated on 15/Apr/25 Commented by som(math1967) last updated on 15/Apr/25 $${area}\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{a}×{b}×{sinA} \\…
Question Number 218674 by Spillover last updated on 14/Apr/25 Answered by Nicholas666 last updated on 14/Apr/25 $$\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{4}} }{\:\sqrt{\mathrm{8}+\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{4}} \:}}\right)\sqrt{\frac{\mathrm{2}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}…
Question Number 218739 by Spillover last updated on 14/Apr/25 Answered by A5T last updated on 15/Apr/25 $$\mathrm{AB}=\mathrm{c}\:;\:\mathrm{BC}=\mathrm{a}\:;\:\mathrm{CA}\:=\mathrm{b}\:;\:\mathrm{AF}=\mathrm{h}_{\mathrm{a}} \\ $$$$\mathrm{AF}\bot\mathrm{BC}\:;\:\mathrm{BG}\bot\mathrm{DC}\:;\:\mathrm{DH}\bot\mathrm{BC}\: \\ $$$$\mathrm{S}_{\bigtriangleup\mathrm{ABC}} =\frac{\mathrm{ah}_{\mathrm{a}} }{\mathrm{2}}\Rightarrow\mathrm{h}_{\mathrm{a}} =\frac{\mathrm{2S}_{\bigtriangleup\mathrm{ABC}} }{\mathrm{a}}…
Question Number 218675 by Spillover last updated on 14/Apr/25 Answered by Nicholas666 last updated on 14/Apr/25 $$ \\ $$$$\:\:\:\:\frac{\mathrm{2020}^{\mathrm{2}} +\mathrm{1}}{\mathrm{0}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\mathrm{4080400}}{\mathrm{1}}\:=\mathrm{4080401}\:\checkmark \\ $$$$ \\ $$…
Question Number 218733 by Spillover last updated on 14/Apr/25 Commented by A5T last updated on 15/Apr/25 $$\mathrm{This}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{contradiction}\left(\mathrm{if}\:\mathrm{the}\:\mathrm{other}\:\mathrm{chord}\right. \\ $$$$\left.\mathrm{were}\:\mathrm{also}\:\mathrm{a}\:\mathrm{diameter}\right). \\ $$ Answered by mr W…
Question Number 218734 by Spillover last updated on 14/Apr/25 Answered by mr W last updated on 15/Apr/25 Commented by mr W last updated on 15/Apr/25…
Question Number 218735 by Spillover last updated on 14/Apr/25 Answered by som(math1967) last updated on 15/Apr/25 $${let}\:{rad}\:{of}\:{large}\:{circle}\:={R} \\ $$$${rad}.{of}\:{small}\:{circle}={r} \\ $$$${AD}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}×{R}×\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{2} \\ $$$$\:\Rightarrow{R}=\mathrm{2}−\sqrt{\mathrm{2}}…
Question Number 218662 by Nicholas666 last updated on 14/Apr/25 $$\: \\ $$$$\:\:\:\:{Prove}:\:\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{sin}\left({x}\right)}{{x}}\:{dx}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$ Answered by SdC355 last updated on 14/Apr/25…
Question Number 218656 by Mamadi last updated on 14/Apr/25 $${resolve}\:{the}\:{equation}\:{with}\:{unknow}\:{p} \\ $$$${P}\:\:{is}\:{polynom}\: \\ $$$$\left.\mathrm{1}\right)\:{P}\left({X}^{\mathrm{2}} \right)=\left({X}^{\mathrm{2}} +\mathrm{1}\right){P}\left({X}\right) \\ $$$$\left.\mathrm{2}\right)\:{P}\:\mathrm{0}{P}\:={P} \\ $$ Answered by MrGaster last updated…
Question Number 218657 by SdC355 last updated on 14/Apr/25 $$……. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com