Question Number 80206 by peter frank last updated on 01/Feb/20 Commented by mr W last updated on 01/Feb/20 $${the}\:{language}\:{of}\:{the}\:{question}\:{is}\:{not} \\ $$$${to}\:{understand}. \\ $$$${please}\:{make}\:{the}\:{question}\:{clear}!\:{what} \\ $$$${exact}\:{is}\:{to}\:{prove}?\:…
Question Number 14658 by tawa tawa last updated on 03/Jun/17 $$\mathrm{Given}\:\mathrm{that}: \\ $$$$\mathrm{x}\:=\:\left(\sqrt{\mathrm{2}}\:+\:\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \:−\:\left(\sqrt{\mathrm{2}}\:−\:\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{Show}\:\mathrm{that}\:,\:\:\:\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{3x}\:=\:\mathrm{2} \\ $$ Answered by Tinkutara last updated on…
Question Number 14564 by tawa tawa last updated on 02/Jun/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{2}\:\mathrm{digits}\:\mathrm{of}\:\:\:\:\:\:\mathrm{2}^{\mathrm{613}} \\ $$ Commented by tawa tawa last updated on 02/Jun/17 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$ Commented…
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Question Number 80036 by jagoll last updated on 30/Jan/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}=\: \\ $$ Commented by john santu last updated on 30/Jan/20 $$\underset{\mathrm{p}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{p}}…
Question Number 14486 by FilupS last updated on 01/Jun/17 $${S}=\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+… \\ $$$$\therefore{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n} \\ $$$$\: \\ $$$${S}=\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n}^{\mathrm{1}−{s}} \right) \\…
Question Number 145474 by Ppmaurya last updated on 05/Jul/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}\:\mathrm{C}\:\mathrm{Classes}\: \\ $$$$\:\:\mathrm{M}.\mathrm{M}.−\mathrm{30}\:\:\:\:\mathrm{10th}−\mathrm{Maths}\:\mathrm{Test}\:\:\:\mathrm{Time}\:\mathrm{Duration}:\mathrm{40min}. \\ $$$$\:\:\mathrm{If}\:\:\mathrm{the}\:\:\mathrm{given}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{then} \\ $$$$\:\mathrm{write}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{and}\:\mathrm{write}\:\:\:\mathrm{three}\:\mathrm{more}\:\mathrm{terms} \\ $$$$\left.\:\:\:\mathrm{Q}.\mathrm{1}\right)\:\:\mathrm{3},\:\mathrm{3}+\sqrt{\mathrm{2},}\:\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}},\:\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}…….. \\ $$$$\: \\ $$$$\left.\:\:\mathrm{Q}.\mathrm{2}\right)\:\:\sqrt{\mathrm{2}},\:\sqrt{\mathrm{8}}\:,\sqrt{\mathrm{18}},\sqrt{\mathrm{32}},…… \\ $$$$\left.\:\:\mathrm{Q}.\mathrm{3}\right)\:\:\mathrm{Derive}\:\mathrm{the}\:\mathrm{formula}\:\:\mathrm{a}_{\mathrm{n}} =\:\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}…
Question Number 79866 by 09264910412 last updated on 28/Jan/20 $${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} \\…