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Question Number 156834 by MathSh last updated on 16/Oct/21 Answered by MJS_new last updated on 16/Oct/21 $${x}=\mathrm{2} \\ $$ Commented by MathSh last updated on…
Question Number 91277 by john santu last updated on 29/Apr/20 $${p}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2003}}−\frac{\mathrm{1}}{\mathrm{2004}} \\ $$$${q}=\frac{\mathrm{1}}{\mathrm{1003}}+\frac{\mathrm{1}}{\mathrm{1004}}+…+\frac{\mathrm{1}}{\mathrm{2004}} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} \:=\: \\ $$ Answered by naka3546 last updated on…
Question Number 156809 by MathSh last updated on 15/Oct/21 $$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:=\:\sqrt{\mathrm{40x}^{\mathrm{2}} \:+\:\mathrm{32x}\:-\:\mathrm{16}} \\ $$ Commented by MathSh last updated on 15/Oct/21 $$\mathrm{Yes}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\…
Question Number 156808 by MathSh last updated on 15/Oct/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\int\:\frac{\mathrm{x}^{\mathrm{7}} \:-\:\mathrm{x}^{\mathrm{5}} \:+\:\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{x}}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{10}} }\:\mathrm{dx}\:\:;\:\:\mathrm{x}\in\mathbb{R} \\ $$ Commented by tabata last updated on 15/Oct/21…
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Question Number 156806 by MathSh last updated on 15/Oct/21 Answered by mr W last updated on 16/Oct/21 $$\frac{{AB}}{{AC}}=\frac{\mathrm{sin}\:\mathrm{60}}{\mathrm{sin}\:\mathrm{20}} \\ $$$$\frac{{BC}}{{AC}}=\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\mathrm{20}} \\ $$$$\frac{{DC}}{{AC}}=\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\mathrm{30}} \\ $$$$\frac{{AB}×{DC}}{{BC}×{AC}}=\frac{{AB}}{{AC}}×\frac{{AC}}{{BC}}×\frac{{DC}}{{AC}} \\…
Question Number 91258 by M±th+et+s last updated on 28/Apr/20 $${prove}\:{that} \\ $$$$\:\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\alpha,\beta,\beta−{a}+\mathrm{1},−\mathrm{1}\right)=\frac{\Gamma\left(\beta−{a}+\mathrm{1}\right)\Gamma\left(\frac{\beta}{\mathrm{2}}+\mathrm{1}\right)}{\Gamma\left(\beta+\mathrm{1}\right)\Gamma\left(\frac{\beta}{\mathrm{2}}−\alpha+\mathrm{1}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 25709 by ajfour last updated on 13/Dec/17 $${Q}.\:\mathrm{25444}\:\:\left({another}\:{solution}\right) \\ $$$$\:\:\bar {{z}}+\mathrm{1}\:={iz}^{\mathrm{2}} +\mid{z}\mid^{\mathrm{2}} \\ $$$${let}\:\:{z}={x}+{iy}\:,\:\Rightarrow \\ $$$${x}−{iy}+\mathrm{1}={i}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)−\mathrm{2}{xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${or}\:\:\left({x}−{y}\right)^{\mathrm{2}} ={x}+\mathrm{1}\:,{and}\: \\…