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Question-213216

Question Number 213216 by Spillover last updated on 01/Nov/24 Answered by MrGaster last updated on 01/Nov/24 $$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx}+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{log}^{\mathrm{2}}…

Question-213217

Question Number 213217 by Spillover last updated on 01/Nov/24 Answered by MrGaster last updated on 01/Nov/24 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}+{n}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{1}}…

Question-213250

Question Number 213250 by ajfour last updated on 01/Nov/24 Answered by a.lgnaoui last updated on 02/Nov/24 $$\mathrm{Calcul}\:\mathrm{de}\:\mathrm{l}\:\mathrm{aire} \\ $$$$ \\ $$$$\mathrm{BC}=\mathrm{a}\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\mathrm{DM}=\mathrm{a}\sqrt{\mathrm{2}}\:+\frac{\mathrm{a}}{\mathrm{2}}. \\ $$$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABDMC}}\right)=\mathrm{S}\left(\boldsymbol{\mathrm{ABC}}\right)+\mathrm{2}\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BDE}}\right)+\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCEF}}\right) \\ $$$$…

we-can-find-tan120-by-tan-180-60-but-can-not-find-by-tan-90-30-why-

Question Number 213175 by Davidtim last updated on 31/Oct/24 $${we}\:{can}\:{find}\:{tan}\mathrm{120}\:{by}\:{tan}\left(\mathrm{180}−\mathrm{60}\right) \\ $$$${but}\:{can}\:{not}\:{find}\:{by}\:{tan}\left(\mathrm{90}+\mathrm{30}\right)\:{why}? \\ $$ Answered by efronzo1 last updated on 31/Oct/24 $$\:\mathrm{tan}\:\mathrm{120}°=\:\mathrm{tan}\:\left(\mathrm{90}°+\mathrm{30}°\right)\:=−\:\mathrm{cot}\:\mathrm{30}°=−\sqrt{\mathrm{3}} \\ $$ Answered…

Question-213138

Question Number 213138 by Spillover last updated on 31/Oct/24 Commented by Frix last updated on 31/Oct/24 $$\mathrm{We}\:\mathrm{must}\:\mathrm{first}\:\mathrm{solve}\:\int\mathrm{sin}\:{t}^{\mathrm{3}} \:{dt}\:\mathrm{which}\:\mathrm{might} \\ $$$$\mathrm{be}\:\mathrm{possible}\:\mathrm{using} \\ $$$$\mathrm{sin}\:{t}^{\mathrm{3}} \:=\frac{\mathrm{e}^{\mathrm{i}{t}^{\mathrm{3}} } −\mathrm{e}^{−\mathrm{i}{t}^{\mathrm{3}}…

Question-213139

Question Number 213139 by Spillover last updated on 31/Oct/24 Answered by MrGaster last updated on 31/Oct/24 $${let}\:{u}=\sqrt{{x}}+\sqrt{{y}}+{z}\Rightarrow{dy}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}}}{dy}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{z}}}{dx} \\ $$$$\Rightarrow\mathrm{0}\leq{u}\leq\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}} \\ $$$$\Rightarrow{dxdydz}=\mathrm{8}{u}^{\mathrm{2}} {du} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}}…