Question Number 79866 by 09264910412 last updated on 28/Jan/20 $${Chapter}\:\left(\mathrm{2}\right) \\ $$$${Remainder}\:{Theorem} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{and} \\ $$$${Factor}\:{Theorem} \\ $$$${Exercises}\left(\mathrm{2}.\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{Remainder}}\:\mathrm{T}\boldsymbol{\mathrm{hrorem}} \\ $$$$\mathrm{1}.\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\boldsymbol{\mathrm{x}}^{\mathrm{8}\:} +\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{5} \\ $$$$\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{1}\right).…
Question Number 145286 by physicstutes last updated on 03/Jul/21 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{function} \\ $$$$\:{f}\left({x}\right)\:=\sqrt[{\mathrm{3}}]{\mathrm{6}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} } \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{oblique}\:\mathrm{assymptote}\left(\mathrm{s}\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}. \\ $$ Answered by Olaf_Thorendsen last updated on 04/Jul/21…
Question Number 145273 by ArielVyny last updated on 03/Jul/21 $${show}\:{that}\:\forall{x}\in\mathbb{R}\:{x}−\mathrm{1}\leqslant{E}\left({x}\right)\leqslant{x} \\ $$ Answered by mathmax by abdo last updated on 04/Jul/21 $$\mathrm{we}\:\mathrm{have}\:\left[\mathrm{x}\right]\leqslant\mathrm{x}<\left[\mathrm{x}\right]+\mathrm{1}\:\Rightarrow\left[\mathrm{x}\right]\leqslant\mathrm{x}\:\mathrm{and}\:\mathrm{x}−\mathrm{1}<\left[\mathrm{x}\right]\:\Rightarrow \\ $$$$\mathrm{x}−\mathrm{1}<\left[\mathrm{x}\right]\leqslant\mathrm{x} \\…
Question Number 79719 by Ell Tee last updated on 27/Jan/20 $$\frac{\left(\mathrm{2}.\mathrm{39}\overset{\mathrm{2}} {\right)}−\left(\mathrm{1}.\mathrm{61}\overset{\mathrm{2}} {\right)}}{\mathrm{2}.\mathrm{39}−\mathrm{1}.\mathrm{61}}…? \\ $$$$ \\ $$ Commented by MJS last updated on 27/Jan/20 $$\frac{{a}^{\mathrm{2}}…
Question Number 145190 by qaz last updated on 03/Jul/21 $$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)=\mathrm{tan}^{−\mathrm{1}} \mathrm{x},\left(\mathrm{x}\neq\mathrm{0}\right) \\ $$$$\mathrm{Find}\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$ Answered by EDWIN88 last updated on 03/Jul/21 $$\left(\mathrm{1}\right){f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right) \\…
Question Number 79647 by john santu last updated on 27/Jan/20 $$\mathrm{discussion}\:\mathrm{back}\:\mathrm{with}\:\mathrm{mr}\:\mathrm{W}. \\ $$$$\mathrm{consider}\:\mathrm{this}\:\mathrm{equation}\: \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3}^{\mathrm{x}} −\mathrm{27}\right)=\mathrm{0} \\ $$$$\mathrm{does}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{have}\:\mathrm{two}\:\mathrm{roots} \\ $$$$\mathrm{or}\:\mathrm{three}\:\mathrm{roots}? \\ $$ Commented by…
Question Number 79625 by M±th+et£s last updated on 26/Jan/20 $${if}\:{n}>\mathrm{1}\:{prove}\:{that} \\ $$$$\mathrm{2}{ln}\left({n}\right)−{ln}\left({n}+\mathrm{1}\right)−{ln}\left({n}−\mathrm{1}\right)=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{6}} }+…= \\ $$ Answered by mind is power last updated on…
Question Number 14047 by FilupS last updated on 27/May/17 $${S}=\mathrm{1}+{i}−\mathrm{1}−{i}+\mathrm{1}+… \\ $$$$\frac{\mathrm{1}}{{i}}=−{i} \\ $$$${S}={i}\left(−{i}+\mathrm{1}+{i}−\mathrm{1}−{i}+\mathrm{1}+…\right) \\ $$$${S}={i}\left(−{i}+{S}\right) \\ $$$${S}=\mathrm{1}+{iS} \\ $$$${S}\left(\mathrm{1}−{i}\right)=\mathrm{1} \\ $$$$\therefore\:{S}=\frac{\mathrm{1}}{\mathrm{1}−{i}} \\ $$$$\: \\…
Question Number 79497 by john santu last updated on 25/Jan/20 $$\frac{\left(\mathrm{4}{x}−\mid{x}−\mathrm{6}\mid\right)\left(\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left({x}+\mathrm{4}\right)+\mathrm{1}\right)}{\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}^{\mid{x}\mid} } }\geqslant\mathrm{0} \\ $$ Answered by john santu last updated on 25/Jan/20…
Question Number 13934 by tawa tawa last updated on 25/May/17 Terms of Service Privacy Policy Contact: info@tinkutara.com