Question Number 35689 by prof Abdo imad last updated on 22/May/18 $${let}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} \sqrt{{n}^{\mathrm{2}} \:+{k}^{\mathrm{2}} }} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \\ $$ Commented…
Question Number 101172 by I want to learn more last updated on 01/Jul/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}: \\ $$$$\:\:\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{2}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}\left(\mathrm{x}\right),\:\:\:\:\:\:\mathrm{x}\:\geqslant\:\mathrm{0} \\ $$ Answered by 1549442205 last updated on…
Question Number 35624 by abdo mathsup 649 cc last updated on 21/May/18 $${let}\:{R}_{{n}} \:=\:\sum_{{k}={n}+\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{k}!} \\ $$$${find}\:{a}\:{equivalent}\:{of}\:{R}_{{n}} \:\:{when}\:{n}\rightarrow+\infty \\ $$ Terms of Service Privacy…
Question Number 35621 by abdo mathsup 649 cc last updated on 21/May/18 $${calculate}\:\:\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{3}{n}} }{\mathrm{3}{n}+\mathrm{1}}\:\:{after}\:{finding} \\ $$$${the}\:{radius}\:{of}\:{convergence}\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\mathrm{8}^{{n}} } \\ $$…
Question Number 35614 by abdo mathsup 649 cc last updated on 21/May/18 $${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\:\:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$ Terms of Service Privacy Policy…
Question Number 35579 by abdo mathsup 649 cc last updated on 20/May/18 $$\left({u}_{{n}} \right)\:{is}\:{a}\:{arithmetic}\:{sequence}\:{with}\:{u}_{\mathrm{0}} =\mathrm{1}\:{and} \\ $$$${u}_{\mathrm{5}} =\mathrm{11}\:\:\:\:\: \\ $$$$\left.\mathrm{1}\right)\:\:{find}\:{the}\:{value}\:{of}\:\:\:{S}_{{n}} \:\:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{u}_{{k}} ^{\mathrm{2}} }…
Question Number 101008 by Rio Michael last updated on 29/Jun/20 $$\mathrm{Given}\:{f}\left({x}\right)\:=\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}}{xe}^{\frac{\mathrm{1}}{{x}}} \:,\:{x}\:\neq\:\mathrm{0}}\\{\mathrm{0},\:{x}\:=\:\mathrm{0}\:}\end{cases} \\ $$$$\mathrm{find} \\ $$$$\left({i}\right)\:\mathrm{Thd}\:\mathrm{domain}\:\mathrm{of}\:{f} \\ $$$$\left({ii}\right)\:\mathrm{check}\:\mathrm{the}\:\mathrm{continuity}\:\mathrm{of}\:{f}\:\mathrm{at}\:{x}\:=\:\mathrm{0} \\ $$$$\left({iii}\right)\:\mathrm{check}\:\mathrm{its}\:\mathrm{differentiability}\:\mathrm{and}\:\mathrm{its}\:\mathrm{sign} \\ $$$$\left({i}\right)\:\mathrm{sketch}\:\mathrm{this}\:\mathrm{curve}\:\mathrm{and}\:\mathrm{find}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{f}\left({x}\right)\:\mathrm{and}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:{f}\left({x}\right) \\…
Question Number 100994 by mathmax by abdo last updated on 29/Jun/20 $$\mathrm{let}\:\:\mathrm{A}\:=\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{A}^{\mathrm{n}} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\mathrm{e}^{\mathrm{A}} \:,\mathrm{e}^{−\mathrm{A}} \\ $$$$\left.\mathrm{3}\right)\mathrm{determine}\:\mathrm{ch}\left(\mathrm{A}\right)\:\mathrm{and}\:\mathrm{sh}\left(\mathrm{A}\right)\:\:\mathrm{is}\:\mathrm{ch}^{\mathrm{2}} \mathrm{A}−\mathrm{sh}^{\mathrm{2}} \mathrm{A}\:=\mathrm{1}? \\ $$ Answered by…
Question Number 100951 by bemath last updated on 29/Jun/20 Commented by Rasheed.Sindhi last updated on 29/Jun/20 $${By}\:{john}\:{santu}\left({answer}\:{of}\right. \\ $$$$\left.{q}#\mathrm{97746}\right)\:{Same}\:{as}\:{above}. \\ $$$$\mathrm{2}^{{x}} \left(\mathrm{1}+\mathrm{2}^{{x}+\mathrm{1}} \right)\:=\:\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right) \\ $$$${show}\:{that}\:{the}\:{factors}\:{y}−\mathrm{1}\:{and}\:{y}+\mathrm{1}…
Question Number 35416 by abdo mathsup 649 cc last updated on 18/May/18 $${let}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}\:{simplify} \\ $$$$\left.{A}={sin}\left\{\:{arcsinx}\:\:+\mathrm{2}{arcsin}\right)\left(\mathrm{1}−{x}\right)\right\} \\ $$ Commented by abdo mathsup 649 cc last updated…