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Author: Tinku Tara

Find-the-cardano-s-solution-of-the-equation-28x-3-9x-2-1-0-

Question Number 200460 by faysal last updated on 19/Nov/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cardano}'\mathrm{s}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{equation}\:\mathrm{28x}^{\mathrm{3}} −\mathrm{9x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$ Answered by Frix last updated on 19/Nov/23 $$\mathrm{28}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}}…

Let-u-n-k-1-n-n-n-2-k-2-for-n-N-gt-0-Show-that-u-n-n-gt-0-is-increasing-

Question Number 200521 by brahim_mekkaoui last updated on 19/Nov/23 $$\mathrm{Let}\:{u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{n}}{{n}^{\mathrm{2}} +{k}^{\mathrm{2}} }\:\:\mathrm{for}\:{n}\in\mathbb{N}_{>\mathrm{0}} \:\:. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\left({u}_{{n}} \right)_{{n}>\mathrm{0}} \:\mathrm{is}\:\mathrm{increasing}. \\ $$ Terms of Service…

Question-200444

Question Number 200444 by Calculusboy last updated on 18/Nov/23 Answered by Frix last updated on 19/Nov/23 $$\int\mathrm{e}^{−\mathrm{i}{x}^{\mathrm{2}} } {dx}\:\overset{{t}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} {x}} {=} \\ $$$$=\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{2}}\int\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{4}}\int\frac{\mathrm{2e}^{{t}^{\mathrm{3}}…

Question-200379

Question Number 200379 by cortano12 last updated on 18/Nov/23 Commented by Frix last updated on 18/Nov/23 $$\mathrm{No}\:\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}.\:\mathrm{Transform}\:\mathrm{to} \\ $$$$\begin{cases}{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}{y}+\frac{{x}^{\mathrm{2}} +{x}−\mathrm{20}}{\mathrm{4}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\\{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}{y}+\frac{\mathrm{3}{x}−\mathrm{20}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\end{cases} \\…