Question Number 212762 by universe last updated on 23/Oct/24 Commented by MrGaster last updated on 23/Oct/24 $$\mathrm{Rewrite}\:\mathrm{the}\:\mathrm{integrand}\:\mathrm{function}\:\mathrm{as}: \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right) \\ $$$$\mathrm{Computational}\:\mathrm{integral}: \\ $$$$\int_{\mathrm{0}} ^{{n}}…
Question Number 212784 by Spillover last updated on 23/Oct/24 Answered by A5T last updated on 24/Oct/24 $${d}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}\left({r}−{a}\right){cos}\theta \\ $$$${c}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{r}\left({r}−{a}\right){cos}\theta…
Question Number 212783 by Spillover last updated on 23/Oct/24 Commented by York12 last updated on 25/Oct/24 $$ \\ $$$$ \\ $$$$ \\ $$ Answered by…
Question Number 212777 by Ghisom last updated on 23/Oct/24 $$\mathrm{question}\:\mathrm{212462} \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{160}}}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{95}{x}^{\mathrm{2}} +\mathrm{2560}}} \\ $$$$ \\…
Question Number 212779 by issac last updated on 23/Oct/24 $$\mathrm{excuse}\:\mathrm{me}???? \\ $$$$\mathrm{am}\:\mathrm{i}\:\mathrm{invisible}??\: \\ $$$$\mathrm{why}\:\mathrm{isn}'\mathrm{t}\:\mathrm{anyone}\:\mathrm{answering}\:\mathrm{my}\:\mathrm{qusestion} \\ $$$$ \\ $$ Commented by mr W last updated on…
Question Number 212775 by Davidtim last updated on 23/Oct/24 $${vector}×{scalar}=? \\ $$$${vector}\:{or}\:{scalar}? \\ $$ Commented by A5T last updated on 23/Oct/24 $${Q}\mathrm{207812};\:{similar}\:{idea}. \\ $$ Terms…
Question Number 212733 by cherokeesay last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $$\left({R}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 212752 by mr W last updated on 25/Oct/24 Commented by mr W last updated on 25/Oct/24 $${the}\:{uniform}\:{rod}\:{with}\:{length}\:\boldsymbol{{l}}\:{has}\: \\ $$$${mass}\:\boldsymbol{{m}}.\:{find}\:{the}\:{tensions}\:{in}\:{the} \\ $$$${strings}\:{a}\:{and}\:{b}. \\ $$…
Question Number 212720 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 $${AB}=\sqrt{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{6};\:{Let}\:\angle{OAB}=\theta \\ $$$$\frac{{sin}\theta}{{r}}=\frac{{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{{AB}}\Rightarrow{cos}\theta=\frac{\mathrm{3}}{{r}}\Rightarrow{sin}\mathrm{2}\theta=\frac{\mathrm{6}\sqrt{{r}^{\mathrm{2}} −\mathrm{9}}}{{r}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}}…
Question Number 212721 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $${In}\left[{ABC}\right];\:{tan}\theta=\frac{{R}}{{R}+\mathrm{2}{r}};{AC}=\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}}…