Question Number 125526 by ka7th last updated on 11/Dec/20 Answered by Dwaipayan Shikari last updated on 11/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{n}^{{m}} =\frac{\mathrm{1}}{{m}+\mathrm{1}}\left({n}^{{m}+\mathrm{1}} \begin{pmatrix}{{m}+\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\beta_{\mathrm{0}} +{n}^{{m}} \begin{pmatrix}{{m}+\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\beta_{\mathrm{1}} +{n}^{{m}−\mathrm{1}}…
Question Number 59991 by selimatas01 last updated on 16/May/19 Commented by Mr X pcx last updated on 16/May/19 $${let}\:\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cosx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int_{−\infty}…
Question Number 191060 by Mingma last updated on 17/Apr/23 Answered by witcher3 last updated on 18/Apr/23 $$\mathrm{BS}=\mathrm{a}\frac{\mathrm{d}}{\mathrm{b}},\mathrm{BR}=\frac{\mathrm{d}}{\mathrm{b}}\mathrm{c} \\ $$$$\mathrm{AP}=\mathrm{c}\frac{\mathrm{d}}{\mathrm{a}},\mathrm{AQ}=\mathrm{b}\frac{\mathrm{d}}{\mathrm{a}} \\ $$$$\mathrm{CM}=\mathrm{a}.\frac{\mathrm{d}}{\mathrm{c}},\mathrm{NC}=\frac{\mathrm{d}}{\mathrm{c}}\mathrm{b} \\ $$$$\mathrm{c}\left(\mathrm{1}−\frac{\mathrm{d}}{\mathrm{a}}\right)=\mathrm{BP},\mathrm{c}\left(\mathrm{1}−\frac{\mathrm{d}}{\mathrm{b}}\right)=\mathrm{RA} \\ $$$$\mathrm{PR}=\mathrm{c}\left(\frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{b}}−\mathrm{1}\right)…
Question Number 59990 by selimatas01 last updated on 16/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191063 by Mingma last updated on 17/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125524 by Algoritm last updated on 11/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191062 by Mingma last updated on 17/Apr/23 Answered by mr W last updated on 17/Apr/23 Commented by mr W last updated on 17/Apr/23…
Question Number 191057 by 073 last updated on 17/Apr/23 Commented by 073 last updated on 19/Apr/23 $$\mathrm{solution}\:\mathrm{please}?? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 59987 by Tawa1 last updated on 16/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125523 by Algoritm last updated on 11/Dec/20 Answered by Dwaipayan Shikari last updated on 11/Dec/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}{x}+\mathrm{3}\right){log}\left(\frac{{x}+\mathrm{2}}{{x}}\right) \\ $$$$=\left(\mathrm{2}{x}+\mathrm{3}\right)\left(\frac{\mathrm{2}}{{x}}\right)=\mathrm{4}+\frac{\mathrm{6}}{{x}}=\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{log}\left(\mathrm{1}+{z}\right)={z} \\ $$ Terms…