Question Number 222523 by hu last updated on 29/Jun/25 $${x}+{V}−{J}\left({x}\right)\frac{{ustx}}{{zac}^{\mathrm{2}} {x}}={x}−\frac{{ustx}_{\mathrm{0}} }{\:\sqrt{{x}}\psi+\zeta\left(\frac{−\mathrm{1}+{v\%}}{\pi+\mathrm{2}}+\mathscr{L}_{{l}\left({x}\rightarrow\mathrm{0}\right)} ^{\:\:{v\%}} {X}_{\mathrm{2}{x}^{\mathrm{2}} } ^{\:\mathrm{1}} \right)} \\ $$ Answered by wewji12 last updated on…
Question Number 222516 by MrGaster last updated on 29/Jun/25 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{integer}\:{k},\mathrm{how}\:\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{incomplete}\:\mathrm{general} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\mathrm{the}\:\mathrm{non}-\mathrm{trivial}\:\mathrm{integer} \\ $$$$\mathrm{solutions}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{Diophantine}\:\mathrm{equation}: \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{ka}^{\mathrm{2}} {b}^{\mathrm{2}} ={c}^{\mathrm{4}} +{d}^{\mathrm{4}} +{kc}^{\mathrm{2}} {d}^{\mathrm{2}}…
Question Number 222519 by MrGaster last updated on 29/Jun/25 $$\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{+\infty} \sqrt{\mathrm{cosh}\:{x}}−\sqrt{\mathrm{sinh}\:{x}}{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\:\sqrt{\pi}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222546 by Tawa11 last updated on 29/Jun/25 Commented by Tawa11 last updated on 29/Jun/25 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{AB}. \\ $$ Commented by fantastic last updated on…
Question Number 222541 by Nicholas666 last updated on 29/Jun/25 $$ \\ $$$$\:\:\:\mathrm{very}\:\mathrm{very}\:\mathrm{crazy}\:\mathrm{problem},\:\:\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{found} \\ $$$$\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{result}\:\mathrm{of}\:\mathrm{this}\:\mathrm{integral}; \\ $$$$\:\:\:\:\:\:\:\:\int\frac{\mathrm{1}}{\:\sqrt{{z}}\:+\:\sqrt{{z}−{h}}\:+\:\sqrt{{z}−\mathrm{2}{h}}}\:{dz} \\ $$$$ \\ $$ Commented by Ghisom last updated…
Question Number 222532 by MrGaster last updated on 29/Jun/25 $$\mathrm{Prove}: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}{n}} }\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \frac{\mathscr{K}_{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{192}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{4}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)+\frac{\mathrm{16}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{15}\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)^{\mathrm{3}} −\frac{\mathrm{148}}{\pi}\beta\left(\mathrm{4}\right),\mathscr{K}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\ldots+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$…
Question Number 222533 by Tawa11 last updated on 29/Jun/25 Answered by fantastic last updated on 29/Jun/25 Commented by fantastic last updated on 29/Jun/25 $${Total}\:{area}\:{of}\:{the}\:{triangle}=\mathrm{54}+\mathrm{42}=\mathrm{96}{cm}^{\mathrm{2}} \\…
Question Number 222529 by Nicholas666 last updated on 29/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−{x}+\frac{\mathrm{1}}{{x}}\:} \:{dx}\:=\:?? \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 222531 by BHOOPENDRA last updated on 29/Jun/25 $${The}\:{foot}\:{of}\:{the}\:{perpendicular}\:{from}\: \\ $$$${a}\:{point}\:{of}\:{the}\:{circle}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1},{z}=\mathrm{0} \\ $$$${to}\:{the}\:{plan}\:\mathrm{2}{x}+\mathrm{3}{y}+{z}=\mathrm{6}\:{lie}\:{on}\:{curve}−−−−−− \\ $$ Answered by mr W last updated on…
Question Number 222487 by MathematicalUser2357 last updated on 28/Jun/25 $$'{Delete}\:{all}\:{lines}'\:{function}\:{deletes}\:{all}\:{lines}\:{without}\:{asking}\:{after}\:{deleting}\:{all}\:{lines}\:{for}\:{the}\:{first}\:{time}\:{in}\:{the}\:{equation}\:{editor} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com