Question Number 126632 by BHOOPENDRA last updated on 22/Dec/20 Commented by BHOOPENDRA last updated on 22/Dec/20 $${thanku}\:{sir} \\ $$ Answered by Ar Brandon last updated…
Question Number 192160 by universe last updated on 10/May/23 $$\mathrm{if}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{are}\:\mathrm{three}\:\mathrm{distinct}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:\frac{\mathrm{x}}{\mathrm{y}−{z}}+\frac{\mathrm{y}}{\mathrm{z}−\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{x}−\mathrm{y}}\:=\:\mathrm{0}\:\mathrm{then}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\Sigma\:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{y}−\mathrm{z}\right)^{\mathrm{2}} } \\ $$ Commented by mehdee42 last updated on 09/May/23…
Question Number 192138 by Mastermind last updated on 09/May/23 $$\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\left\{_{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{0},\mathrm{0}\right)} ^{\frac{\mathrm{x}^{\mathrm{2}} \mathrm{y}}{\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{2y}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x},\mathrm{y}\right)\neq\:\left(\mathrm{0},\mathrm{0}\right)} \right. \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{directional}\:\mathrm{derivative}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{direction}\:\mathrm{of}\:\mathrm{an}\:\mathrm{arbitrary}\:\mathrm{unit}\:\mathrm{vector} \\ $$$$\phi\:\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right),\:\mathrm{but}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{not}\:\mathrm{continous}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right)\: \\ $$…
Question Number 192129 by universe last updated on 08/May/23 $$\:{prove}\:{that} \\ $$$$\:\mid{a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\mid\:+\:\mid{a}\:−\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\mid\:=\:\mid{a}+{b}\mid\:+\mid{a}−{b}\mid \\ $$$${a},{b}\:\in\:\mathbb{C} \\ $$ Answered by AST last updated…
Question Number 192126 by universe last updated on 08/May/23 $$\:\:\:\:\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\:\:\:\:\:\:\mid\boldsymbol{{z}}\mid\:>\:\frac{\mid\boldsymbol{{Re}}\left(\boldsymbol{{z}}\right)\mid\:+\mid\boldsymbol{{Im}}\left(\boldsymbol{{z}}\right)\mid}{\mathrm{2}}\:\:,\:\:\:\forall\boldsymbol{{z}}\in\mathbb{C} \\ $$ Commented by York12 last updated on 09/May/23 $${sir}\:{how}\:{can}\:{I}\:{reach}\:{you}\:{out}\:,\:{I}\:{need}\:{to}\:{ask}\:{several}\:{questions} \\ $$ Answered…
Question Number 192094 by Mastermind last updated on 08/May/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{a}\:\mathrm{subgroup} \\ $$$$\mathrm{S}\:\mathrm{of}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{group}\:\mathrm{G},\:\mathrm{always}\:\mathrm{divide} \\ $$$$\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{group}\:\mathrm{G}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192095 by Mastermind last updated on 07/May/23 $$\mathrm{Prove}\:\mathrm{a}\:\mathrm{non}−\mathrm{empty}\:\mathrm{set}\:\mathrm{S}\:\mathrm{of}\:\mathrm{a}\:\mathrm{group} \\ $$$$\mathrm{G}\:\mathrm{wrt}\:\mathrm{binary}\:\mathrm{operation}\:\ast\:\mathrm{is}\:\mathrm{a}\:\mathrm{sub}− \\ $$$$\mathrm{group}\:\mathrm{of}\:\mathrm{G}.\:\mathrm{Iff}\: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{a},\mathrm{b}\:\in\:\mathrm{S}\:\Rightarrow\:\mathrm{a}\ast\mathrm{b}\in\mathrm{S} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{a}\:\in\:\mathrm{S}\:\Rightarrow\:\mathrm{a}^{−\mathrm{1}} \:\in\:\mathrm{S}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Hello}…
Question Number 192077 by Mastermind last updated on 07/May/23 $$\mathrm{Let}\:\mathrm{H}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}−\mathrm{empty}\:\mathrm{subset}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{group}\:\mathrm{G},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{follow}− \\ $$$$\mathrm{ing}\:\mathrm{are}\:\mathrm{equivalent} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{H}\:\mathrm{is}\:\mathrm{a}\:\mathrm{subgroup}\:\mathrm{of}\:\mathrm{G} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\in\:\mathrm{H},\:\mathrm{ab}^{−\mathrm{1}} \:\in\:\mathrm{H} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\in\:\mathrm{ab}\:\in\:\mathrm{H} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{for}\:\mathrm{a}\:\in\:\mathrm{H},\:\mathrm{a}^{−\mathrm{1}} \:\in\:\mathrm{H} \\…
Question Number 61003 by Tawa1 last updated on 28/May/19 Answered by tanmay last updated on 28/May/19 $${PQ}={diameter}=\sqrt{\left(\mathrm{4}−\mathrm{0}\right)^{\mathrm{2}} +\left(\mathrm{0}−\mathrm{2}\right)^{\mathrm{2}} }\:=\mathrm{2}\sqrt{\mathrm{5}}\: \\ $$$${radius}=\sqrt{\mathrm{5}}\: \\ $$$${centre}=\left(\frac{\mathrm{0}+\mathrm{4}}{\mathrm{2}},\frac{\mathrm{2}+\mathrm{0}}{\mathrm{2}}\right)\rightarrow\left(\mathrm{2},\mathrm{1}\right) \\ $$$${eqn}\:{circle}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}}…
Question Number 60984 by naka3546 last updated on 28/May/19 $$\frac{{a}}{{a}−{b}}\:\:+\:\:\frac{{b}}{{b}−{c}}\:\:+\:\:\frac{{c}}{{c}−{a}}\:\:=\:\:\mathrm{4} \\ $$$${ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{abc}\:+\:{ca}^{\mathrm{2}} \:\:=\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a} \\ $$$$\left(\frac{{a}}{{a}−{b}}\right)^{\mathrm{3}} \:\:+\:\:\left(\frac{{b}}{{b}−{c}}\right)^{\mathrm{3}} \:\:+\:\:\left(\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:\:=\:\:? \\ $$$$…