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Question-226866

Question Number 226866 by Estevao last updated on 17/Dec/25 Commented by Jyrgen last updated on 17/Dec/25 $$\int\sqrt{\mathrm{1}+{x}^{{n}} }{dx}={x}\:_{\mathrm{1}} {F}_{\mathrm{2}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{{n}};\:\mathrm{1}+\frac{\mathrm{1}}{{n}};\:−{x}^{{n}} \right) \\ $$ Commented by…

Question-226855

Question Number 226855 by Kassista last updated on 17/Dec/25 Answered by mehdee7396 last updated on 17/Dec/25 $${x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}=\mathrm{4}\Rightarrow{x}=\mathrm{0},\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\Rightarrow{P}\left(\mathrm{0},,\mathrm{2}\right)\:\&\:\:{Q}\left(\frac{\mathrm{8}}{\mathrm{5}},−\frac{\mathrm{6}}{\mathrm{5}}\right)\Rightarrow{m}_{{PQ}} =−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{tan}\alpha=\frac{\mathrm{3}}{\mathrm{4}}\backsimeq\mathrm{36}^{\mathrm{0}} \Rightarrow\angle{POQ}\backsimeq\mathrm{126}^{\mathrm{0}}…

Question-226880

Question Number 226880 by Spillover last updated on 17/Dec/25 Answered by TonyCWX last updated on 18/Dec/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{2}^{{x}} \right]\mathrm{d}{x} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[{e}^{{x}\mathrm{ln}\left(\mathrm{2}\right)} \right]\mathrm{d}{x}…

Question-226882

Question Number 226882 by hardmath last updated on 17/Dec/25 Answered by breniam last updated on 17/Dec/25 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\:\frac{\sqrt[{{n}}]{\mathrm{2}^{{n}+\mathrm{1}} }−\mathrm{1}}{\:\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}\left(\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2}−\mathrm{1}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty}…

lim-x-0-lim-n-cos-x-2-cos-x-2-2-cos-x-2-n-

Question Number 226815 by CrispyXYZ last updated on 16/Dec/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}^{\mathrm{2}} }\:…\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}^{{n}} }\right)\right)\:=\:? \\ $$ Answered by AgniMath last updated on 16/Dec/25 $${sin}\mathrm{2}\theta\:=\:\mathrm{2}{sin}\theta{cos}\theta \\…