Question Number 60322 by Sardor2211 last updated on 19/May/19 Commented by Mr X pcx last updated on 19/May/19 $$\left({e}\right)\Leftrightarrow{xy}^{''} \:+{y}^{'} \:={x}^{\mathrm{3}} \:\:\:{let}\:{y}^{'} ={z}\:\Rightarrow \\ $$$${xz}^{'}…
Question Number 60321 by Sardor2211 last updated on 19/May/19 Commented by tanmay last updated on 19/May/19 $${not}\:{understood}\:{the}?{question}… \\ $$ Commented by maxmathsup by imad last…
Question Number 125760 by snipers237 last updated on 13/Dec/20 $${Let}\:{n}\geqslant\mathrm{1}\:{and}\:{integer},\:{P}_{{n}} \left({X}\right)=\left(\mathrm{1}+{X}\right)^{{n}} −\left(\mathrm{1}−{X}\right)^{{n}} \: \\ $$$$\left.\mathrm{1}\right)\:{Factorize}\:{P}_{{n}} \\ $$$$\left.\mathrm{2}\right){Deduce}\:\:{S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left[\mathrm{4}+{cotan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:\right] \\ $$ Answered by…
Question Number 125736 by snipers237 last updated on 13/Dec/20 $${P}_{{o}} =\mathrm{1}\:\:,{P}_{\mathrm{1}} =\mathrm{1}+{X}\:{and}\:{for}\:{all}\:{n}\geqslant\mathrm{1} \\ $$$${P}_{{n}+\mathrm{1}} ={P}_{{n}} +{XP}_{{n}−\mathrm{1}} \: \\ $$$${Explicit}\:\:{P}_{{n}} \:\:{and}\:{prove}\:{that}\:{its}\:{roots}\:{are}\:{all}\:{real}. \\ $$ Terms of Service…
Question Number 125737 by snipers237 last updated on 13/Dec/20 $$\:{Find}\:{C}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{e}^{−{u}} }\:{du} \\ $$$${Prove}\:{that}\:\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} \sqrt{\mathrm{1}+{x}^{{n}} }\:{dx}\:\underset{\infty} {\sim}\:\frac{{C}}{{n}}\: \\ $$ Answered by MJS_new last…
Question Number 125668 by mnjuly1970 last updated on 12/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:{evaluate}::: \\ $$$$\:\:\:\:\:\:\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left({ln}\left({cot}\left({x}\right)\right)\right){dx}=? \\ $$$$ \\ $$ Answered by mindispower last updated…
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Question Number 59975 by vajpaithegrate@gmail.com last updated on 16/May/19 Answered by tanmay last updated on 16/May/19 $${T}.{S}.{A}\:{of}\:{cone}=\pi{rl}+\pi{r}^{\mathrm{2}} \\ $$$$\left.{T}\left..{S}.{A}=\pi{r}\left({r}+\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }\:\right)\:\:\:\:\:\right]\right]{when}\:\left[{l}^{\mathrm{2}} ={r}^{\mathrm{2}} +{h}^{\mathrm{2}} \right] \\…
Question Number 59974 by vajpaithegrate@gmail.com last updated on 16/May/19 Answered by tanmay last updated on 16/May/19 $${lateral}\:{surface}\:{area}=\pi{rl} \\ $$$${given}\:{cos}\mathrm{45}^{{o}} =\frac{{h}}{{l}} \\ $$$${l}=\frac{{h}}{{cos}\mathrm{45}^{{o}} }={h}\sqrt{\mathrm{2}} \\ $$$${tan}\mathrm{45}^{{o}}…