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 208909 by lepuissantcedricjunior last updated on 26/Jun/24 $$\:\:\:\:\:\boldsymbol{{soit}}\:\boldsymbol{{la}}\:\boldsymbol{{fonction}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{x}}\:\:\boldsymbol{{definie}} \\ $$$$\boldsymbol{{sur}}\:\mathbb{R}\:\boldsymbol{{on}}\:\boldsymbol{{note}}\:\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{f}}^{−\mathrm{1}} \left(\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{alors}}\:\boldsymbol{{que}}\:\:\boldsymbol{{la}}\:\boldsymbol{{primitive}}\:\boldsymbol{{G}}\left(\boldsymbol{{x}}\right)=\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{g}}\left(\boldsymbol{{t}}\right)\boldsymbol{{dt}} \\ $$ Commented by Ghisom last updated…
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 208849 by SANOGO last updated on 24/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208823 by Ismoiljon_008 last updated on 23/Jun/24 $$ \\ $$$$\:\:\:{If}\:{a}+{b}+{c}=\mathrm{15},\:{then}\:{find}\:{the}\:{smallest}\:{value}\: \\ $$$$\:\:\:{of}\:{the}\:{expression}\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{b}^{\mathrm{2}} +\mathrm{9}}+\sqrt{{c}^{\mathrm{2}} +\mathrm{16}}. \\ $$$$\:\:\:\:\:{Help}\:{please} \\ $$ Answered by mr W…
Question Number 208819 by Ismoiljon_008 last updated on 23/Jun/24 Answered by Ismoiljon_008 last updated on 23/Jun/24 $${help}\:{please} \\ $$ Answered by A5T last updated on…
Question Number 208814 by Ismoiljon_008 last updated on 23/Jun/24 Commented by Ismoiljon_008 last updated on 23/Jun/24 $${Help}\:{me}\:{please} \\ $$ Commented by Ismoiljon_008 last updated on…
Question Number 208809 by kgmxdd last updated on 23/Jun/24 Commented by Frix last updated on 23/Jun/24 $$\mathrm{As}\:\mathrm{always}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 208676 by RoseAli last updated on 20/Jun/24 Answered by Berbere last updated on 20/Jun/24 $$\sqrt[{\mathrm{4}}]{{a}}+\sqrt[{\mathrm{4}}]{{b}}=\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}.\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{2}{a}}+\sqrt[{\mathrm{4}}]{\mathrm{2}{b}}=\sqrt{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}+\sqrt[{\mathrm{4}}]{\frac{{b}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${b}=\mathrm{0}\:\Rightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1}\:{no}\:{solution}\:\mathbb{N} \\…
Question Number 208629 by RoseAli last updated on 19/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com