Question Number 199391 by Calculusboy last updated on 03/Nov/23 Answered by AST last updated on 03/Nov/23 $$\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{10}\Rightarrow\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{10} \\ $$$$\Rightarrow\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}\right)=\mathrm{0}\Rightarrow{x}=\mathrm{4}\:{or}\:\mathrm{2}\Rightarrow\left({x},{y}\right)=\left(\mathrm{4},\mathrm{2}\right);\left(\mathrm{2},\mathrm{6}\right) \\…
Question Number 199385 by hardmath last updated on 02/Nov/23 $$\mathrm{Find}: \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\mathrm{15}} \:\sqrt{\mathrm{1}\:+\:\mathrm{3x}^{\mathrm{8}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by witcher3 last updated on 02/Nov/23…
Question Number 199380 by BHOOPENDRA last updated on 02/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199381 by Mingma last updated on 02/Nov/23 Answered by witcher3 last updated on 02/Nov/23 $$\left(\mathrm{202}\right)=\mathrm{2}.\mathrm{101} \\ $$$$\frac{\left(\mathrm{201}\right)!}{\mathrm{k}}\equiv\mathrm{0}\left[\mathrm{202}\right]\mathrm{0},\forall\mathrm{k}\in\left\{\mathrm{1},……\mathrm{201}\right)−\left\{\mathrm{2},\mathrm{101}\right) \\ $$$$\mathrm{J}\equiv\frac{\mathrm{201}!}{\mathrm{101}}+\frac{\mathrm{201}!}{\mathrm{2}}\left[\mathrm{202}\right] \\ $$$$\frac{\mathrm{201}!}{\mathrm{2}}=\mathrm{202}.\mathrm{3}.\mathrm{2}.\underset{\mathrm{k}=\mathrm{5},\mathrm{k}\neq\mathrm{101}} {\overset{\mathrm{201}} {\prod}}\mathrm{k}\equiv\mathrm{0}\left[\mathrm{202}\right]…
Question Number 199382 by mokys last updated on 02/Nov/23 $${if}\:{A}\:=\:\begin{bmatrix}{{cosh}\left({x}\right)\:\:\:\:\:\:{sinh}\left({x}\right)\:}\\{{sinh}\left({x}\right)\:\:\:\:\:\:{cosh}\left({x}\right)}\end{bmatrix}{find}\:{A}^{{k}} \:? \\ $$ Answered by manxsol last updated on 02/Nov/23 $${A}^{\mathrm{2}} ={cosh}^{\mathrm{2}} −{sinh}^{\mathrm{2}} =\mathrm{1} \\…
Question Number 199377 by universe last updated on 02/Nov/23 $$\:\:\int\underset{\mathrm{R}} {\int}\mathrm{cos}\:\left(\mathrm{max}\left\{\mathrm{x}^{\mathrm{3}} ,\:\mathrm{y}^{\mathrm{3}/\mathrm{2}} \right\}\right)\mathrm{dx}\:\mathrm{dy}\:,\:\mathrm{where}\:\mathrm{R}\:=\:\left[\mathrm{0},\mathrm{1}\right]×\left[\mathrm{0},\mathrm{1}\right] \\ $$ Answered by witcher3 last updated on 04/Nov/23 $$\mathrm{i}\:\mathrm{will}\:\mathrm{Try} \\ $$…
Question Number 199374 by cortano12 last updated on 02/Nov/23 Commented by cortano12 last updated on 02/Nov/23 $$\mathrm{prove}\:\mathrm{that}\:\:\cancel{\boldsymbol{{B}}} \\ $$ Commented by MathematicalUser2357 last updated on…
Question Number 199368 by sonukgindia last updated on 02/Nov/23 Answered by qaz last updated on 02/Nov/23 $$\int_{−\infty} ^{+\infty} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$$=\mathrm{2}\pi{i}\left(\left[\left({z}−{i}\right)^{−\mathrm{1}} \right]+\left[\left({z}−\mathrm{2}{i}\right)^{−\mathrm{1}}…
Question Number 199369 by universe last updated on 02/Nov/23 $$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \:\int_{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} ^{\mathrm{1}+\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{5}/\mathrm{2}} {dx}\:{dy}\:{dz}\:\:{is}…
Question Number 199370 by mathlove last updated on 02/Nov/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}{e}^{\mathrm{3}{x}} −\mathrm{9}{e}^{\mathrm{2}{x}} +\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{3}} }=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com