Question Number 126895 by mohammad17 last updated on 25/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126886 by SOMEDAVONG last updated on 25/Dec/20 $$\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}\:+….+\:\frac{\mathrm{1}}{\mathrm{2n}}\right) \\ $$ Answered by Dwaipayan Shikari last updated on 25/Dec/20 $$\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}}{{n}}}+….+\frac{\mathrm{1}}{\mathrm{1}+\frac{{n}}{{n}}}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\frac{{r}}{{n}}}…
Question Number 126869 by Mustafa2020 last updated on 25/Dec/20 $$\int_{\mathrm{0}} ^{\infty} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192406 by josemate19 last updated on 17/May/23 $$\int\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{{x}}\:{dx}??? \\ $$ Answered by mehdee42 last updated on 17/May/23 $${x}=\mathrm{2}{chu}\Rightarrow{dx}=\mathrm{2}{shudu} \\ $$$$\Rightarrow{I}=\mathrm{2}\int\:\frac{{sh}^{\mathrm{2}} {u}}{{chu}}\:{du}\:=\mathrm{2}\int\:\left({chu}−\frac{\mathrm{1}}{{chu}}\:\right){du} \\…
Give-the-function-x-2-7x-8-0-have-two-roots-x-1-and-x-2-No-solving-the-function-Find-x-1-3-x-2-2023-
Question Number 192390 by TUN last updated on 16/May/23 $${Give}\:{the}\:{function}: \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}−\mathrm{8}=\mathrm{0} \\ $$$${have}\:{two}\:{roots}\:{x}_{\mathrm{1}\:} {and}\:{x}_{\mathrm{2}} \\ $$$${No}\:{solving}\:{the}\:{function}\: \\ $$$${Find}:\:{x}_{\mathrm{1}} ^{\mathrm{3}} +{x}_{\mathrm{2}} +\mathrm{2023} \\ $$…
Question Number 126848 by mohammad17 last updated on 24/Dec/20 Answered by mathmax by abdo last updated on 24/Dec/20 $$\mid\mathrm{z}−\mathrm{1}+\mathrm{i}\mid=\mathrm{1}\:\:\:\mathrm{put}\:\mathrm{z}=\mathrm{x}+\mathrm{iy}\:\Rightarrow \\ $$$$\mid\mathrm{z}−\mathrm{1}+\mathrm{i}\mid=\mid\mathrm{x}+\mathrm{iy}−\mathrm{1}+\mathrm{i}\mid=\mid\mathrm{x}−\mathrm{1}+\mathrm{i}\left(\mathrm{y}+\mathrm{1}\right)\mid=\sqrt{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \:+\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mid\mathrm{z}−\mathrm{1}+\mathrm{i}\mid=\mathrm{1}\:\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 126836 by mohammad17 last updated on 24/Dec/20 Commented by mohammad17 last updated on 24/Dec/20 $${help}\:{me}\:{sir} \\ $$ Commented by mohammad17 last updated on…
Question Number 126837 by mohammad17 last updated on 24/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126835 by mohammad17 last updated on 24/Dec/20 Answered by liberty last updated on 25/Dec/20 $$\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:\:−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}}\\{\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\begin{pmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\begin{pmatrix}{\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{pmatrix}\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\:{x}'+{y}'\right)}\\{\frac{\mathrm{1}}{\mathrm{2}}\left(−{x}'+\sqrt{\mathrm{3}}\:{y}'\right.}\end{pmatrix}…
Question Number 126827 by harckinwunmy last updated on 24/Dec/20 Answered by AST last updated on 24/Dec/20 $$\mathrm{Q33}. \\ $$$$\frac{\mathrm{4}+\mathrm{7}{i}}{\mathrm{1}−\mathrm{3}{i}} \\ $$$$\mathrm{Rationalising} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{4}+\mathrm{7}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}{\left(\mathrm{1}−\mathrm{3}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}=\frac{\mathrm{4}+\mathrm{12}{i}+\mathrm{7}{i}+\left(\mathrm{21}{i}^{\mathrm{2}} \right)}{\mathrm{1}^{\mathrm{2}} −\left(\mathrm{3}{i}\right)^{\mathrm{2}}…