Question Number 5254 by Rasheed Soomro last updated on 02/May/16 $$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which} \\ $$$$\left(\boldsymbol{\mathrm{a}}\right)\:\mathrm{5}\:\mathrm{children}\:\mathrm{can}\:\mathrm{be}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{groups} \\ $$$$\mathrm{of}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:, \\ $$$$\left(\boldsymbol{\mathrm{b}}\right)\:\mathrm{9}\:\mathrm{children}\:\mathrm{can}\:\mathrm{be}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{groups} \\ $$$$\mathrm{of}\:\mathrm{5}\:\mathrm{and}\:\mathrm{4}, \\ $$$$\mathrm{Hence}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in} \\ $$$$\mathrm{which}\:\mathrm{9}\:\mathrm{children}\:\mathrm{can}\:\mathrm{be}\:\mathrm{divided}\:\mathrm{into} \\ $$$$\mathrm{groups}\:\mathrm{of}\:\mathrm{2},\mathrm{3}\:\mathrm{and}\:\mathrm{4}.…
Question Number 136320 by radians last updated on 20/Mar/21 Answered by Olaf last updated on 20/Mar/21 $$\mathrm{coefficient}\:{x}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{same}\:\mathrm{as}\:{x}^{\mathrm{3}} \:: \\ $$$$\Rightarrow\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2020}} {a}^{\mathrm{2}} {b}^{\mathrm{2018}} \:=\:\mathrm{C}_{\mathrm{3}}…
Question Number 70784 by rajesh4661kumar@gmail.com last updated on 08/Oct/19 Commented by MJS last updated on 08/Oct/19 $$\mathrm{really}\:\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{3}\sqrt{{x}+\mathrm{1}}}\:\mathrm{not}\:\frac{{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{3}\right)\sqrt{{x}+\mathrm{1}}}? \\ $$ Commented by rajesh4661kumar@gmail.com last…
Question Number 70782 by oyemi kemewari last updated on 08/Oct/19 $$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{get}\:\mathrm{my}\:\mathrm{password}\:?\:\mathrm{i}\:\:\mathrm{can}'\mathrm{t}\:\mathrm{remeber} \\ $$ Commented by Tinku Tara last updated on 08/Oct/19 There is no option to retrieve forgotten password. Commented by oyemi…
Question Number 70783 by naka3546 last updated on 08/Oct/19 $$\frac{\mathrm{5}}{\mathrm{4}\centerdot\mathrm{9}}\:+\:\mathrm{2}\left(\frac{\mathrm{9}}{\mathrm{16}\centerdot\mathrm{25}}\right)\:+\:\mathrm{3}\left(\frac{\mathrm{13}}{\mathrm{36}\centerdot\mathrm{49}}\right)\:+\:\mathrm{4}\left(\frac{\mathrm{17}}{\mathrm{64}\centerdot\mathrm{81}}\right)\:+\:\mathrm{5}\left(\frac{\mathrm{21}}{\mathrm{100}\centerdot\mathrm{121}}\right)\:+\:\ldots\:=\:\:? \\ $$ Commented by tw000001 last updated on 08/Oct/19 $$\mathrm{Well},\mathrm{this}\:\mathrm{one}\:\mathrm{is}\:\mathrm{unconverge}\:\mathrm{series}\:\mathrm{because} \\ $$$$\mathrm{difference}\:\mathrm{and}\:\mathrm{ratio}\:\mathrm{are}\:\mathrm{different}. \\ $$$$\mathrm{But}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{approximately} \\…
Question Number 70780 by MJS last updated on 08/Oct/19 $$\frac{\mathrm{1}}{{t}+\sqrt{{u}}+\sqrt{{v}}}= \\ $$$$=\frac{\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}{\left({t}+\sqrt{{u}}+\sqrt{{v}}\right)\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}= \\ $$$$=\frac{{t}^{\mathrm{3}} −\left(\sqrt{{u}}+\sqrt{{v}}\right){t}^{\mathrm{2}} −\left(\sqrt{{u}}−\sqrt{{v}}\right)^{\mathrm{2}} {t}+\left({u}−{v}\right)\left(\sqrt{{u}}−\sqrt{{v}}\right)}{{t}^{\mathrm{4}} −\mathrm{2}\left({u}+{v}\right){t}^{\mathrm{2}} +\left({u}−{v}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{t}+\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}}= \\…
Question Number 136313 by aurpeyz last updated on 20/Mar/21 $$\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{dx} \\ $$ Answered by liberty last updated on 20/Mar/21 $$\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:\sqrt{\mathrm{1}−\mathrm{9}{x}^{−\mathrm{2}} }}\:=\:\int\:\frac{{x}^{−\mathrm{3}} }{\:\sqrt{\mathrm{1}−\mathrm{9}{x}^{−\mathrm{2}}…
Question Number 5237 by sanusihammed last updated on 02/May/16 $${find}\:{the}\:{domain}\:{and}\:{range}\:{of}\: \\ $$$${y}\:=\:\left[_{\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:<\mathrm{2}} ^{\:\mathrm{3}{x}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\geqslant\:\mathrm{2}} \right. \\ $$ Answered by FilupSmith last updated on 02/May/16 $${Domain}\:=\:\left\{{x}:\:−\infty\leqslant{x}\leqslant\infty\right\}…
Question Number 5236 by sanusihammed last updated on 02/May/16 $${find}\:{the}\:{range}\:{and}\:{domain}\:{of}\: \\ $$$$\left({i}\right)\:\:{y}\:=\:\mid−{x}\mid \\ $$$$\left({ii}\right)\:{y}\:=\:\mid{x}−\mathrm{5}\mid \\ $$ Answered by FilupSmith last updated on 02/May/16 $$\left({i}\right) \\…
Question Number 5235 by sanusihammed last updated on 02/May/16 $${Find}\:{the}\:{range}\:{and}\:{domain}\:{of}\: \\ $$$${y}\:=\:\frac{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{10}\right)}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}} \\ $$ Answered by Yozzii last updated on 02/May/16 $${x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}=\left({x}+\mathrm{3}\right)^{\mathrm{2}}…