Question Number 193543 by York12 last updated on 16/Jun/23 $${solve} \\ $$ Commented by Subhi last updated on 16/Jun/23 $${cd}!!! \\ $$ Commented by York12…
Question Number 193542 by SAMIRA last updated on 16/Jun/23 $$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calcul}\:\:\:\:\mathrm{h}\left(\mathrm{X}\right)\:=\:\mathrm{f}\left(\mathrm{a}+\mathrm{X}\right)\:−\mathrm{b} \\ $$2) determine a and b such that h is odd…
Question Number 193492 by Socracious last updated on 15/Jun/23 Answered by a.lgnaoui last updated on 15/Jun/23 $$\mathrm{DE}\mid\mid\mathrm{BC}\:\: \\ $$$$\mathrm{tan}\:\mathrm{30}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\boldsymbol{\mathrm{HB}}}{\boldsymbol{\mathrm{HA}}}=\frac{\boldsymbol{\mathrm{b}}}{\mathrm{2}\boldsymbol{\mathrm{x}}}\Rightarrow\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{b}}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{x}}.\frac{\boldsymbol{\mathrm{b}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Area}}=\frac{\mathrm{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$…
Question Number 193469 by Mingma last updated on 14/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193467 by aba last updated on 14/Jun/23 $$\mathrm{Proof}\:: \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\mathrm{sint}}+\sqrt{\mathrm{1}−\mathrm{sint}}}{\:\sqrt{\mathrm{1}+\mathrm{sint}}−\sqrt{\mathrm{1}−\mathrm{sint}}}\right)=\frac{\mathrm{t}}{\mathrm{2}}\: \\ $$ Commented by MM42 last updated on 14/Jun/23 $${attention} \\ $$$${in}\:{all}\:\:{three}\:{arguments}\:\:{we}\:{used}\:{equalities}…
Question Number 193484 by Frix last updated on 15/Jun/23 $$\mathrm{Nice}\:\mathrm{problem}: \\ $$$$\mathrm{Find}\:\mathrm{8}\:\mathrm{distinctive}\:\mathrm{numbers}\:\in\mathbb{N}\backslash\left\{\mathrm{0}\right\}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{these}\:\mathrm{are}\:\mathrm{simultaniously}\:\mathrm{true}: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:{a}+{b}+{c}+{d}\:=\:{e}+{f}+{g}+{h} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:=\:{e}^{\mathrm{2}} +{f}^{\mathrm{2}} +{g}^{\mathrm{2}} +{h}^{\mathrm{2}}…
Question Number 193411 by cortano12 last updated on 13/Jun/23 $$\:\:\:\:\: \\ $$$$ \\ $$ Commented by BaliramKumar last updated on 13/Jun/23 $$\mathrm{put}\:\:\:{x}\:=\:{cos}\mathrm{2}\theta \\ $$ Answered…
Question Number 193410 by cortano12 last updated on 13/Jun/23 $$\:\:\begin{cases}{\mathrm{x}=\sqrt{\mathrm{3}−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\\{\mathrm{y}=\sqrt{\mathrm{3}+\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\end{cases}\: \\ $$$$\:\:\:\:\underbrace{\boldsymbol{{x}}} \\ $$ Answered by aba last updated on 13/Jun/23 $$\mathrm{xy}=\sqrt{\mathrm{9}−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}\right)}=\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}}−\mathrm{1}\:\wedge\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{6} \\…
Question Number 193371 by gloriousman last updated on 11/Jun/23 $$\mathrm{Reduce}\:\mathrm{to}\:\mathrm{first}\:\mathrm{order}\:\mathrm{and}\:\mathrm{solve}\:, \\ $$$$\mathrm{showing}\:\mathrm{each}\:\mathrm{step}\:\mathrm{in}\:\mathrm{detail}. \\ $$$$\mathrm{1}.\:\mathrm{y}''\:+\left(\mathrm{y}'\right)^{\mathrm{3}} \mathrm{siny}=\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{y}''=\mathrm{1}+\left(\mathrm{y}'\right)^{\mathrm{2}} \\ $$$$ \\ $$ Answered by witcher3 last…
Question Number 193363 by MATHEMATICSAM last updated on 11/Jun/23 $${y}\:=\:\mathrm{4}\:×\:\mathrm{10}^{\mathrm{2}{x}} \\ $$$$\mathrm{Express}\:{x}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{y},\:\mathrm{giving}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{simplified}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{log}\:\mathrm{base}\:\mathrm{10}. \\ $$ Answered by aba last updated on 11/Jun/23 $$\mathrm{y}=\mathrm{4}×\mathrm{10}^{\mathrm{2x}} \:\Rightarrow\:\mathrm{10}^{\mathrm{2x}}…