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i have forgotten my password. how may i retrieve it? please help me or forward me to one of the developers please

$$\mathrm{i}\:\mathrm{have}\:\mathrm{forgotten}\:\mathrm{my}\:\mathrm{password}. \\ $$$$\mathrm{how}\:\mathrm{may}\:\mathrm{i}\:\mathrm{retrieve}\:\mathrm{it}? \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{or}\:\mathrm{forward}\:\mathrm{me}\:\mathrm{to} \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{developers}\:\mathrm{please} \\ $$

Question Number 119564 Answers: 0 Comments: 0

((3/x) − ((15)/(2y)) ) ÷ (6/(xy)) = ((6y − 15x)/(2xy)) × ((xy)/6).

$$\left(\frac{\mathrm{3}}{{x}}\:−\:\frac{\mathrm{15}}{\mathrm{2}{y}}\:\right)\:\boldsymbol{\div}\:\frac{\mathrm{6}}{{xy}}\:=\:\frac{\mathrm{6}{y}\:−\:\mathrm{15}{x}}{\mathrm{2}{xy}}\:×\:\frac{{xy}}{\mathrm{6}}. \\ $$

Question Number 119529 Answers: 3 Comments: 7

Question Number 119463 Answers: 1 Comments: 2

one cube of side length 3cm is painted by three colour say (red blue and black) opposte face same colour Now cube is cut into 27 small cube i)How many cube has three face coloured ii)How many cube two face coloured iii)How many cube has no colour in its face

$${one}\:{cube}\:{of}\:{side}\:{length}\:\mathrm{3}{cm}\:{is}\:{painted}\:{by}\:{three}\:{colour} \\ $$$${say}\:\left({red}\:{blue}\:{and}\:{black}\right)\:{opposte}\:{face}\:{same}\:{colour} \\ $$$${Now}\:{cube}\:{is}\:{cut}\:{into}\:\mathrm{27}\:{small}\:{cube} \\ $$$$\left.{i}\right){How}\:{many}\:{cube}\:{has}\:{three}\:{face}\:{coloured} \\ $$$$\left.{ii}\right){How}\:{many}\:{cube}\:{two}\:{face}\:{coloured} \\ $$$$\left.{iii}\right){How}\:{many}\:{cube}\:\:{has}\:{no}\:{colour}\:{in}\:{its}\:{face} \\ $$

Question Number 119376 Answers: 3 Comments: 2

Solve for x in the equation below ax^2 +bx + c = 0.

$${Solve}\:{for}\:\boldsymbol{{x}}\:{in}\:{the}\:{equation}\:{below} \\ $$$${ax}^{\mathrm{2}} \:+{bx}\:+\:{c}\:=\:\mathrm{0}. \\ $$

Question Number 119316 Answers: 0 Comments: 0

((Σ_(n=1) ^∞ (1/n^n ))/(Σ_(n=1) ^∞ (−1)^(n+1) (1/n^n )))

$$\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{n}} }}{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}^{{n}} }} \\ $$

Question Number 119112 Answers: 0 Comments: 2

Seems my recent post was removed by Tinkutara.

$$\boldsymbol{\mathrm{Seems}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{recent}}\:\boldsymbol{\mathrm{post}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{removed}}\:\boldsymbol{\mathrm{by}} \\ $$$$\boldsymbol{\mathrm{Tinkutara}}.\: \\ $$

Question Number 118952 Answers: 2 Comments: 0

Salut Pouvez vous m′aider avec ce devoir? Examiner si les series suivantes sont absolument convergentes ou semi−convergentes Σ_(k=0) ^n (((-1)^(k-1) )/((k+1)^2 )) Σ_(k=0) ^n (((-1)^(k-1) )/k) Etudier la convergence des series suivantes Σ_(k=1) ^n (((2k)!)/2^k ) Σ_(k=1) ^n ((1/2))^k

$$\mathrm{Salut} \\ $$$$\mathrm{Pouvez}\:\mathrm{vous}\:\mathrm{m}'\mathrm{aider}\:\mathrm{avec}\:\mathrm{ce}\:\mathrm{devoir}? \\ $$$$ \\ $$$$\mathrm{Examiner}\:\mathrm{si}\:\mathrm{les}\:\mathrm{series}\:\mathrm{suivantes}\:\mathrm{sont}\: \\ $$$$\mathrm{absolument}\:\mathrm{convergentes}\:\mathrm{ou}\: \\ $$$$\mathrm{semi}−\mathrm{convergentes} \\ $$$$ \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\left(-\mathrm{1}\right)^{\mathrm{k}-\mathrm{1}} }{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\left(-\mathrm{1}\right)^{\mathrm{k}-\mathrm{1}} }{\mathrm{k}} \\ $$$$ \\ $$$$\mathrm{Etudier}\:\mathrm{la}\:\mathrm{convergence}\:\mathrm{des}\:\mathrm{series}\:\mathrm{suivantes} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\left(\mathrm{2k}\right)!}{\mathrm{2}^{\mathrm{k}} } \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \\ $$$$ \\ $$

Question Number 118685 Answers: 0 Comments: 1

Converting decimal numbers to base 10

$$\boldsymbol{{Converting}}\:\boldsymbol{{decimal}}\:\boldsymbol{{numbers}}\:\boldsymbol{{to}}\:\boldsymbol{{base}}\:\mathrm{10} \\ $$

Question Number 118594 Answers: 0 Comments: 0