Question Number 4844 by 123456 last updated on 17/Mar/16 $${y}\left({x}\right)={f}\left({x}\right)+{g}\left({x}\right)+{h}\left({x}\right) \\ $$$${y}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{sin}\:{x}+{x}\left(\mathrm{1}−{x}\right) \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{even} \\ $$$${g}\left({x}\right)\:\mathrm{is}\:\mathrm{odd} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)−{h}\left(\mathrm{0}\right)=? \\ $$ Commented by Rasheed Soomro…
Question Number 4841 by awan last updated on 17/Mar/16 $$ \\ $$$${what}\:{mean}\:{for}\:{symbol}\:\Pi? \\ $$ Commented by 123456 last updated on 17/Mar/16 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{x}_{{i}} ={x}_{\mathrm{1}}…
Question Number 4839 by FilupSmith last updated on 17/Mar/16 $$\left(\mathrm{1}\right)\:\:\:{S}_{\mathrm{1}} =\sqrt{\mathrm{1}−\mathrm{4}{x}} \\ $$$$\left(\mathrm{2}\right)\:\:\:{S}_{\mathrm{2}} =\sqrt{\mathrm{1}+\mathrm{4}{x}} \\ $$$$\mathrm{For}\:{S}_{\mathrm{1}} ,{S}_{\mathrm{2}} \in\mathbb{Z},\:{x}=? \\ $$ Commented by Yozzii last updated…
Question Number 135910 by liberty last updated on 17/Mar/21 $${Algebra}\: \\ $$$$\:\:\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:{p}+{q}\sqrt{\mathrm{2}} \\ $$$${then}\:\frac{{q}^{\mathrm{2}} }{{p}\sqrt{\mathrm{2}}}\:+\:{p}^{\mathrm{2}} {q}\:=? \\ $$ Answered by som(math1967) last updated on 17/Mar/21…
Question Number 70370 by Rasheed.Sindhi last updated on 04/Oct/19 $${If}\:\:{gcd}\left({p}\:,\:{q}\right)=\mathrm{1},{prove}\:{that} \\ $$$$\:\:\:\:\:{gcd}\left({p}\left({p}+{q}\right)\:,\:{q}\left({p}+{q}\right)\:,\:{pq}\right)=\mathrm{1} \\ $$$$\mathrm{R}\boldsymbol{\mathrm{elated}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{Q}}#\mathrm{69939} \\ $$ Commented by mind is power last updated on 03/Oct/19…
Question Number 135906 by SWPlaysMC last updated on 16/Dec/21 $${Edit}:\:{this}\:{is}\:{already}\:{fixed}.\:{Therefore}\:{this}\:{thread}\:{is}\:{closed}… \\ $$$$ \\ $$$$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{glitch}! \\ $$$$\mathrm{Whenever}\:\mathrm{you}\:\mathrm{backspace}\:\mathrm{from}\:\mathrm{a}\:\mathrm{line}\:\mathrm{to}\:\mathrm{delete}\:\mathrm{it},\:\mathrm{you}'\mathrm{ll}\:\mathrm{end}\:\mathrm{up}\:\mathrm{directly}\:\mathrm{in}\:\mathrm{front}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{thing}\:\mathrm{you}\:\mathrm{typed}, \\ $$$$\mathrm{whatever}\:\mathrm{it}\:\mathrm{is}. \\ $$$$ \\ $$$$\mathrm{Also},\:\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{for}\:\mathrm{this}\:\mathrm{app}\:\mathrm{to}\:\mathrm{run}\:\mathrm{in}\:\mathrm{portrait}\:\mathrm{orrentation}\:\mathrm{on}\:\mathrm{tablets}?\:\mathrm{Because}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{get}\:\mathrm{it}\:\mathrm{to}\:\mathrm{rotate} \\ $$$$\mathrm{to}\:\mathrm{that}\:\mathrm{mode}.\:\mathrm{I}\:\mathrm{saw}\:\mathrm{in}\:\mathrm{your}\:\mathrm{screenshots}\:\mathrm{on}\:\mathrm{the}\:\mathrm{Google}\:\mathrm{Play}\:\mathrm{page}\:\mathrm{that}\:\mathrm{it}\:\mathrm{was}\:\mathrm{possible},\:\mathrm{but}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{get}\:\mathrm{it}\:\mathrm{to}\:\mathrm{work}. \\…
Question Number 70369 by ahmadshahhimat775@gmail.com last updated on 03/Oct/19 Commented by MJS last updated on 03/Oct/19 $$\mathrm{see}\:\mathrm{question}\:\mathrm{69588} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 70364 by 20190927 last updated on 03/Oct/19 $$\mathrm{solve}\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} −\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} } \\ $$ Commented by kaivan.ahmadi last updated on 03/Oct/19 $${lim}_{{x}\rightarrow\mathrm{0}} \frac{\left(\mathrm{1}−{x}\right){e}^{{x}} −\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 70360 by Hassen_Timol last updated on 03/Oct/19 $$\underset{{n}\rightarrow+\infty} {\mathrm{l}im}\:\:\:\:\frac{\sqrt{{n}\:+\:\mathrm{1}\:}−\:{n}}{\:\sqrt{{n}\:+\:\mathrm{1}}\:+\:{n}}\:\:=\:\:? \\ $$ Answered by mind is power last updated on 03/Oct/19 $$\frac{\sqrt{{n}+\mathrm{1}}−{n}}{\:\sqrt{{n}+\mathrm{1}}+{n}}=\frac{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}−\mathrm{1}\right)}{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}+\mathrm{1}\right)}=\frac{\sqrt{\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}}…
Question Number 4825 by sanusihammed last updated on 16/Mar/16 $${Find}\:{the}\:{value}\:{of}\: \\ $$$$ \\ $$$$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$ Commented by prakash jain last updated on 16/Mar/16 $$\mathrm{Assuming}\:\mathrm{the}\:\mathrm{series}\:\mathrm{goes}\:\mathrm{infinitely}…