Question Number 145664 by qaz last updated on 07/Jul/21 $$\mathrm{S}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{k}^{\mathrm{3}} =? \\ $$ Answered by mathmax by abdo last updated on 07/Jul/21…
Question Number 14594 by 1kanika# last updated on 02/Jun/17 Answered by Tinkutara last updated on 02/Jun/17 $${f}\left({x}\right)\:=\:{Q}\left({x}\:−\:\mathrm{2}\right)\left({x}\:−\:\mathrm{3}\right)\:+\:{ax}\:+\:{b} \\ $$$$\mathrm{3}{a}\:+\:{b}\:=\:\mathrm{2} \\ $$$$\mathrm{2}{a}\:+\:{b}\:=\:\mathrm{3} \\ $$$$\Rightarrow\:{a}\:=\:−\mathrm{1},\:{b}\:=\:\mathrm{5} \\ $$$$\mathrm{Remainder}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}…
Question Number 80131 by M±th+et£s last updated on 31/Jan/20 Commented by MJS last updated on 31/Jan/20 $$+\infty \\ $$$$\mathrm{e}^{\frac{\pi}{\mathrm{2}}} \approx\mathrm{4}.\mathrm{8} \\ $$$$\mathrm{4}.\mathrm{8}^{\mathrm{4}.\mathrm{8}} \approx\mathrm{1913} \\ $$$$……
Question Number 145666 by mathdanisur last updated on 07/Jul/21 Answered by Rasheed.Sindhi last updated on 07/Jul/21 $$\mid{x}+\mathrm{1}\mid+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}}+\mid{x}−{z}\mid+\mid{z}−\mathrm{3}\mid=\mathrm{4} \\ $$$$….. \\ $$$$…. \\ $$ Commented…
Question Number 145660 by imjagoll last updated on 07/Jul/21 Answered by Olaf_Thorendsen last updated on 08/Jul/21 $$!\mathrm{4}\:=\:\mathrm{4}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}\right) \\ $$$$!\mathrm{4}\:=\:\mathrm{9} \\ $$$$!\mathrm{5}\:=\:\mathrm{5}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}\right) \\ $$$$!\mathrm{5}\:=\:\mathrm{44} \\ $$$$!\mathrm{7}\:=\:\mathrm{7}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}+\frac{\mathrm{1}}{\mathrm{6}!}−\frac{\mathrm{1}}{\mathrm{7}!}\right)…
Question Number 80116 by mr W last updated on 31/Jan/20 $${Find} \\ $$$${S}_{{m}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\left({n}+{k}\right)}=? \\ $$$$\left({m}\geqslant\mathrm{2}\right) \\ $$ Commented by mr…
Question Number 80108 by jagoll last updated on 31/Jan/20 $${a},{b},{c}\:\in\mathbb{R} \\ $$$$\frac{{b}+{c}+{d}}{{a}}=\frac{{a}+{c}+{d}}{{b}}=\frac{{a}+{b}+{c}}{{d}}=\frac{{a}+{b}+{d}}{{c}}={r} \\ $$$${what}\:{is}\:{r}? \\ $$ Commented by john santu last updated on 31/Jan/20 $$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\:=\:\mathrm{x}…
Question Number 145641 by physicstutes last updated on 06/Jul/21 $$\mathrm{Let}\:\mathrm{g}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{given}\:\mathrm{by}\:\mathrm{g}\left({x}\right)\:=\:\mathrm{3}\:+\:\mathrm{4}{x}\:.\mathrm{Prove}\:\mathrm{by}\:\mathrm{induction} \\ $$$$\mathrm{that},\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:{n},\: \\ $$$$\mathrm{g}^{{n}} \left({x}\right)\:=\:\left(\mathrm{4}^{{n}} −\mathrm{1}\right)\:+\:\mathrm{4}^{{n}} \left({x}\right). \\ $$$$\mathrm{If}\:\mathrm{for}\:\mathrm{every}\:\mathrm{positive}\:\mathrm{integer}\:{k},\:\mathrm{we}\:\mathrm{inteprete}\:\mathrm{g}^{−{k}} \:\mathrm{as}\:\mathrm{the}\:\mathrm{inverse} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{g}^{{k}} .\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{above}\:\mathrm{formula}\:\mathrm{holds}\:\mathrm{alsl} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{negative}\:\mathrm{integers}\:{n}.…
Question Number 80093 by TawaTawa last updated on 30/Jan/20 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\:\:\:\:\:\mathrm{x}^{\sqrt{\mathrm{y}}} \:\:\:=\:\:\mathrm{64} \\ $$$$\:\:\:\:\:\mathrm{y}^{\sqrt{\mathrm{x}}} \:\:\:=\:\mathrm{81} \\ $$ Answered by MJS last updated on 30/Jan/20…
Question Number 14559 by tawa tawa last updated on 02/Jun/17 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x} \\ $$$$\frac{\mathrm{6x}\:+\:\mathrm{2a}\:+\:\mathrm{3b}\:+\:\mathrm{c}\:}{\mathrm{6x}\:+\:\mathrm{2a}\:−\:\mathrm{3b}\:−\:\mathrm{c}}\:=\:\frac{\mathrm{2x}\:+\:\mathrm{6a}\:+\:\mathrm{b}\:+\:\mathrm{3c}}{\mathrm{2x}\:+\:\mathrm{6a}\:−\:\mathrm{b}\:−\:\mathrm{3c}} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Jun/17 $$\frac{\left(\mathrm{6}{x}+\mathrm{2}{a}+\mathrm{3}{b}+{c}\right)+\left(\mathrm{6}{x}+\mathrm{2}{a}−\mathrm{3}{b}−{c}\right)}{\left(\mathrm{6}{x}+\mathrm{2}{a}+\mathrm{3}{b}+{c}\right)−\left(\mathrm{6}{x}+\mathrm{2}{a}−\mathrm{3}{b}−{c}\right)}= \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{6}{a}+{b}+\mathrm{3}{c}\right)+\left(\mathrm{2}{x}+\mathrm{6}{a}−{b}−\mathrm{3}{c}\right)}{\left(\mathrm{2}{x}+\mathrm{6}{a}+{b}+\mathrm{3}{c}\right)−\left(\mathrm{2}{x}+\mathrm{6}{a}−{b}−\mathrm{3}{c}\right)}\Rightarrow…