Question Number 208872 by Ismoiljon_008 last updated on 26/Jun/24 $$ \\ $$$$\:\:\:{Find}\:{the}\:{side}\:{of}\:{a}\:{triangle}\:{if}\:{the}\:{distances} \\ $$$$\:\:\:{from}\:{an}\:{arbitrary}\:{point}\:{inside}\:{a}\:{regular}\:{triangle}\: \\ $$$$\:\:\:{to}\:{its}\:{vertices}\:{are}\:{m},\:{n}\:{and}\:{k}. \\ $$$$\:\:{Help}\:{please} \\ $$ Answered by mr W last…
Question Number 208891 by efronzo1 last updated on 26/Jun/24 $$\:\:\:\underline{\kappa} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208900 by KradecaNaLukcheta last updated on 26/Jun/24 $${Does}\:{anyone}\:{know}\:{of}\:{an}\:{intuition} \\ $$$${behind}\:{the}\:{integral}\:{form}\:{of}\:{the} \\ $$$${remainder}\:{in}\:{Taylor}'{s}\:{theorem}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208871 by Shrodinger last updated on 26/Jun/24 $${L}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}{dx} \\ $$ Answered by Sutrisno last updated on 26/Jun/24 $${misal} \\ $$$$\sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}={p}\rightarrow{x}=\frac{−\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}}{\mathrm{5}{p}^{\mathrm{2}}…
Question Number 208896 by efronzo1 last updated on 26/Jun/24 Answered by MM42 last updated on 27/Jun/24 $${s}_{\mathrm{1}} =\mathrm{32}−\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:;\:\:{x}=\mathrm{8}{sin}\theta \\ $$$$\Rightarrow=\mathrm{32}−\mathrm{64}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{cos}^{\mathrm{2}}…
Question Number 208880 by efronzo1 last updated on 26/Jun/24 Answered by Frix last updated on 26/Jun/24 $${a},\:{b},\:{c}\:>\mathrm{0}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{{a}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}>\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{question}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$ Terms of Service Privacy…
Question Number 208861 by mokys last updated on 25/Jun/24 Commented by mokys last updated on 25/Jun/24 $${solve}\:{this} \\ $$ Commented by mokys last updated on…
Question Number 208862 by THORR last updated on 25/Jun/24 $${x}=\:{b}^{\mathrm{2}} \: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208852 by Tawa11 last updated on 25/Jun/24 Answered by Frix last updated on 25/Jun/24 $$\mathrm{3}^{−\mathrm{3}\left({x}^{\mathrm{2}} −{x}\right)} ={x}^{\mathrm{2}} −{x}\:\Rightarrow\:{x}^{\mathrm{2}} −{x}>\mathrm{0}\:\Leftrightarrow\:{x}<\mathrm{0}\vee{x}>\mathrm{1} \\ $$$${t}={x}^{\mathrm{2}} −{x}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{t}+\mathrm{1}}}{\mathrm{2}} \\…
Question Number 208855 by efronzo1 last updated on 25/Jun/24 Answered by mr W last updated on 25/Jun/24 $${let}\:{t}=\mathrm{1}+\mid{x}\mid\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{2024}^{−{t}} −\lambda}{\mathrm{2024}^{−{t}} −\lambda^{−\mathrm{1}} }=\lambda\:\mathrm{2024}^{{t}} \\ $$$$\mathrm{2024}^{−{t}}…