Question Number 191595 by MATHEMATICSAM last updated on 26/Apr/23 $${a}\:=\:\frac{{xy}}{{x}\:+\:{y}}\:,\:{b}\:=\:\frac{{xz}}{{x}\:+\:{z}}\:\mathrm{and}\:{c}\:=\:\frac{{yz}}{{y}\:+\:{z}}\:. \\ $$$$\mathrm{Represent}\:{x}\:\mathrm{in}\:{a},\:{b},\:{c}\:\mathrm{form}.\:\left[{x},\:{y},\:{z}\:\neq\:\mathrm{0}\right] \\ $$ Answered by mehdee42 last updated on 26/Apr/23 $$\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:\:\:\left({i}\right)\:,\:\:\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}\:\:\left({ii}\right)\:\:,\:\:\:\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\:\:\left({iii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\overset{\left({iii}\right)} {\Rightarrow}\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{c}}…
Question Number 191589 by MATHEMATICSAM last updated on 28/Apr/23 $${a}^{{x}} \:=\:{bc},\:{b}^{{y}} \:=\:{ca},\:{c}^{{z}} \:=\:{ab}. \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\:=\:\mathrm{2}. \\ $$$$\left(\mathrm{Without}\:\mathrm{using}\:\mathrm{log}\right) \\ $$$${a}\:\neq\:{b}\:\neq\:{c} \\ $$ Commented by mr W…
Question Number 60500 by prof Abdo imad last updated on 21/May/19 $${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{A}^{−\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{e}^{{A}} \:\:\:{and}\:{e}^{−\mathrm{2}{A}} \:. \\ $$ Commented by maxmathsup…
Question Number 60501 by prof Abdo imad last updated on 21/May/19 $${let}\:{A}\:=\begin{pmatrix}{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right){calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{e}^{{A}} \:\:\:{and}\:{e}^{−{A}} \:. \\ $$$$ \\ $$ Commented by maxmathsup…
Question Number 191569 by Shrinava last updated on 26/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126030 by Fareed last updated on 16/Dec/20 $$\mathrm{2}^{\mathrm{x}} =\mathrm{4x} \\ $$$$\mathrm{solve}\:\mathrm{it}\:\mathrm{please} \\ $$ Answered by Dwaipayan Shikari last updated on 16/Dec/20 $$\mathrm{2}^{{x}} =\mathrm{4}{x}…
Question Number 191553 by MATHEMATICSAM last updated on 25/Apr/23 $$\mathrm{If}\:{a}^{\mathrm{2}} \:+\:{a}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find}\:{a}^{\mathrm{5}} \:+\:{a}^{\mathrm{4}} \:+\:\mathrm{1}. \\ $$ Commented by Tinku Tara last updated on 25/Apr/23 $${a}=\omega,\:\omega^{\mathrm{2}} ,\:\mathrm{where}\:\omega\:\mathrm{is}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}.…
Question Number 191555 by MATHEMATICSAM last updated on 25/Apr/23 $$\mathrm{Solve} \\ $$$$\sqrt{{x}}\:+\:{y}\:=\:\mathrm{11} \\ $$$${x}\:+\:\sqrt{{y}}\:=\:\mathrm{7} \\ $$ Commented by Frix last updated on 26/Apr/23 $$\mathrm{Use}\:\mathrm{your}\:\mathrm{common}\:\mathrm{sense} \\…
Question Number 191552 by MATHEMATICSAM last updated on 25/Apr/23 $$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:\mathrm{3}{x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\left({x}^{\mathrm{2}} \:+\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$ Answered by mr W last updated on 25/Apr/23…
Question Number 60475 by Tawa1 last updated on 21/May/19 Answered by tanmay last updated on 21/May/19 $$\frac{{dp}}{{dx}}=\frac{{f}\left({x}\right)\frac{{dg}\left({x}\right)}{{dx}}−\left[{g}\left({x}\right)−\mathrm{1}\right]\frac{{df}\left({x}\right)}{{dx}}}{\left\{{f}\left({x}\right)\right\}^{\mathrm{2}} } \\ $$$${p}'\left(\mathrm{2}\right)=\frac{{f}\left(\mathrm{2}\right){g}'\left(\mathrm{2}\right)−\left[{g}\left(\mathrm{2}\right)−\mathrm{1}\right]{f}'\left(\mathrm{2}\right)}{{f}^{\mathrm{2}} \left(\mathrm{2}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)\left(−{sin}\mathrm{2}\right)−\left[{cos}\mathrm{2}−\mathrm{1}\right]\mathrm{5}}{\left(−\mathrm{1}\right)^{\mathrm{2}} } \\…