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Question-199522

Question Number 199522 by cherokeesay last updated on 04/Nov/23 Answered by cortano12 last updated on 05/Nov/23 $$\left(\mathrm{g}_{\mathrm{1}} \right)\:\equiv\:\mathrm{y}=\mathrm{ax}+\mathrm{b}\:\left(\mathrm{tangent}\:\mathrm{line}\:\right) \\ $$$$\:\mathrm{where}\:\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\:\mathrm{1} \\ $$$$\:\mathrm{and}\:\mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\:\:\therefore\:\mathrm{y}=\mathrm{x}+\frac{\mathrm{1}}{\mathrm{8}} \\…

Question-199368

Question Number 199368 by sonukgindia last updated on 02/Nov/23 Answered by qaz last updated on 02/Nov/23 $$\int_{−\infty} ^{+\infty} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$$=\mathrm{2}\pi{i}\left(\left[\left({z}−{i}\right)^{−\mathrm{1}} \right]+\left[\left({z}−\mathrm{2}{i}\right)^{−\mathrm{1}}…

Question-199338

Question Number 199338 by sonukgindia last updated on 01/Nov/23 Answered by aleks041103 last updated on 01/Nov/23 $$\mathrm{5}^{\mathrm{2}^{{M}} } −\mathrm{1}=\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } \right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} =\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}} } +\mathrm{1}\right)\left(\mathrm{5}^{\mathrm{2}^{{M}−\mathrm{1}}…