Question Number 61495 by bhanukumarb2@gmail.com last updated on 03/Jun/19 Commented by bhanukumarb2@gmail.com last updated on 03/Jun/19 $${prove}\:{second}\:{in}\:{which}\:{book}\:{i}\:{can}\:{get}\: \\ $$$${these}\:{type}\:{approximation} \\ $$ Commented by bhanukumarb2@gmail.com last…
Question Number 61490 by MJS last updated on 03/Jun/19 $$\sqrt{{a}−\sqrt{{a}+{x}}}+\sqrt{{a}+\sqrt{{a}−{x}}}=\mathrm{2}{x} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{Sir}\:\mathrm{Aifour}\:\mathrm{and}\:\mathrm{me}\:\mathrm{found} \\ $$$$ \\ $$$$\mathrm{trivial}\:\mathrm{solution}\:{a}={x}=\mathrm{0} \\ $$$$ \\ $$$${a},\:{x}\:\in\mathbb{R} \\ $$$$ \\ $$$${x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left({r}+\sqrt{{r}^{\mathrm{2}} +\mathrm{4}}\right)\sqrt{\mathrm{2}\left(\mathrm{4}{a}−\mathrm{1}\right)−{r}^{\mathrm{2}}…
Question Number 127006 by liberty last updated on 26/Dec/20 $${Two}\:{pipes}\:{A}\:{and}\:{B}\:{together}\:{can}\:{fill}\: \\ $$$${a}\:{cistern}\:{in}\:\mathrm{5}\:{hours}.\:{Had}\:{they}\:{been}\:{opened} \\ $$$${separately},\:{then}\:{B}\:{would}\:{have}\:{taken}\: \\ $$$$\mathrm{6}\:{hours}\:{more}\:{than}\:{A}\:{to}\:{fill}\:{the}\:{cistern}. \\ $$$${How}\:{much}\:{time}\:{will}\:{be}\:{taken}\:{by}\:{A}\:{to}\:{fill} \\ $$$${the}\:{cistern}\:{separately}?\: \\ $$ Answered by bramlexs22…
Question Number 61470 by Rasheed.Sindhi last updated on 03/Jun/19 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{besides} \\ $$$$\left\{\mathrm{x}=\mathrm{a},\mathrm{y}=\mathrm{b}\right\}\:\mathrm{or}\:\left\{\mathrm{x}=\mathrm{b},\mathrm{y}=\mathrm{a}\right\}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\:\:\:\:\mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b}\:\:\wedge\:\mathrm{x}^{\mathrm{7}} +\mathrm{y}^{\mathrm{7}} =\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} \:\:? \\ $$$$ \\ $$ Answered…
Question Number 192542 by peter frank last updated on 20/May/23 Answered by Frix last updated on 20/May/23 $$\sqrt{\mathrm{i}}=\sqrt{\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} }=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$$$\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}} =\:\:\:\:\:\:\:\:\:\:\left[\left({a}^{{b}} \right)^{{c}}…
Question Number 126990 by physicstutes last updated on 26/Dec/20 $$\mathrm{Obtain}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\: \\ $$$$\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{{n}} {\int}}\left[{x}\right]\:{dx}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\:\mathrm{where}\:\left[{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}\:\mathrm{of}\:{x}. \\ $$ Answered by mr W last updated…
Question Number 126989 by arash sharifi last updated on 25/Dec/20 $$\int{e}^{\sqrt{{x}}} {dx} \\ $$ Answered by bramlexs22 last updated on 26/Dec/20 $${let}\:\sqrt{{x}}\:=\:{u}\:\Rightarrow{x}\:=\:{u}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{dx}\:=\:\mathrm{2}{u}\:{du}\: \\…
Question Number 192508 by Spillover last updated on 19/May/23 Commented by Frix last updated on 19/May/23 $$\mathrm{They}\:\mathrm{solved}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{question}\:\mathrm{for}\:\mathrm{you}. \\ $$$$\mathrm{Are}\:\mathrm{you}\:\mathrm{too}\:\mathrm{stupid}\:\mathrm{or}\:\mathrm{simply}\:\mathrm{too}\:\mathrm{lazy}? \\ $$ Answered by Spillover last…
Question Number 126966 by ajfour last updated on 25/Dec/20 Commented by ajfour last updated on 25/Dec/20 $${With}\:{the}\:{help}\:{of}\:{the}\:{intersection} \\ $$$${of}\:{the}\:{inclined}\:{parabola},\:{find}\: \\ $$$${a}\:{root}\:{of}\:{the}\:{cubic}:\:\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}−\boldsymbol{{c}}. \\ $$$${Assume}\:{shape}\:{of}\:{parabola}\:{the} \\…
Question Number 126950 by AST last updated on 05/Dec/22 Answered by floor(10²Eta[1]) last updated on 25/Dec/20 $$\mathrm{A}.\mathrm{1}+\mathrm{2}^{\mathrm{2}^{\mathrm{n}} } +\mathrm{2}^{\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \equiv\mathrm{1}+\left(−\mathrm{1}\right)^{\mathrm{2}^{\mathrm{n}} } +\left(−\mathrm{1}\right)^{\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \equiv\mathrm{1}+\mathrm{1}+\mathrm{1}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{3}\right)…