Menu Close

Category: Others

Question-216621

Question Number 216621 by Tawa11 last updated on 12/Feb/25 Answered by A5T last updated on 12/Feb/25 Commented by A5T last updated on 12/Feb/25 $$\left[\mathrm{OA}_{\mathrm{1}} \mathrm{DT}\right]=\left[\mathrm{ABCT}\right]\:\mathrm{when}\:\mathrm{both}\:\mathrm{cars}\:\mathrm{meet}\:\mathrm{at}\:\mathrm{time}\:\mathrm{t}…

Question-216593

Question Number 216593 by Tawa11 last updated on 11/Feb/25 Answered by A5T last updated on 11/Feb/25 $$\mathrm{At}\:\mathrm{vertex},\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2ax}+\mathrm{b}=\mathrm{0}\Rightarrow\mathrm{18a}+\mathrm{b}=\mathrm{0}\Rightarrow\mathrm{b}=−\mathrm{18a} \\ $$$$−\mathrm{14}=\mathrm{81a}+\mathrm{9b}+\mathrm{c} \\ $$$$\Rightarrow\mathrm{c}=−\mathrm{14}+\mathrm{81a} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{a}−\mathrm{18a}−\mathrm{14}+\mathrm{81a}=\mathrm{64a}−\mathrm{14} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}}…

Question-216477

Question Number 216477 by Jubr last updated on 08/Feb/25 Answered by A5T last updated on 08/Feb/25 $$\left.\mathrm{2}\right)\:\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}=\mathrm{u}\:;\:\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}=\mathrm{v} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{2}\:;\:\mathrm{uv}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{−\mathrm{3}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}}=\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{2}}…