Question Number 126549 by mohammad17 last updated on 21/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192080 by sciencestudentW last updated on 07/May/23 $${f}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}}\right)={x}^{\mathrm{3}} \:\: \\ $$$${f}^{−\mathrm{1}} \left({x}\right)+{f}\left(\mathrm{8}\right)=? \\ $$ Answered by AST last updated on 07/May/23 $${f}^{−\mathrm{1}}…
Question Number 192083 by sciencestudentW last updated on 07/May/23 $${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}\right)+{f}\left({x}+\mathrm{1}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=? \\ $$ Answered by AST last updated on 07/May/23…
Question Number 192077 by Mastermind last updated on 07/May/23 $$\mathrm{Let}\:\mathrm{H}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}−\mathrm{empty}\:\mathrm{subset}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{group}\:\mathrm{G},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{follow}− \\ $$$$\mathrm{ing}\:\mathrm{are}\:\mathrm{equivalent} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{H}\:\mathrm{is}\:\mathrm{a}\:\mathrm{subgroup}\:\mathrm{of}\:\mathrm{G} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\in\:\mathrm{H},\:\mathrm{ab}^{−\mathrm{1}} \:\in\:\mathrm{H} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\in\:\mathrm{ab}\:\in\:\mathrm{H} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{for}\:\mathrm{a}\:\in\:\mathrm{H},\:\mathrm{a}^{−\mathrm{1}} \:\in\:\mathrm{H} \\…
Question Number 126542 by MathSh last updated on 21/Dec/20 $$\mathrm{17}^{\mathrm{202}} \\ $$$${the}\:{end}\:\:\mathrm{3}\:{number}? \\ $$ Commented by AlagaIbile last updated on 21/Dec/20 Answered by Olaf last…
Question Number 192076 by sciencestudentW last updated on 07/May/23 $${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{6}{x}\:\:\:\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Answered by AST last updated on 07/May/23 $${y}={f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{6}{x}=\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9} \\…
Question Number 126543 by pticantor last updated on 21/Dec/20 $$\:\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{i}^{{k}} =??? \\ $$ Answered by MJS_new last updated on 21/Dec/20 $$=\underset{{k}=\mathrm{1}}…
Question Number 192073 by a.lgnaoui last updated on 07/May/23 $$\mathrm{Reponse}\:\mathrm{a}\:\:\mathrm{l}\:\mathrm{exercice}\:\mathrm{deja}\:\:\mathrm{pose} \\ $$ Answered by a.lgnaoui last updated on 07/May/23 Commented by a.lgnaoui last updated on…
Question Number 126538 by Ndala last updated on 21/Dec/20 $${if}\:\:{a}+{b}\geqslant{c}>\mathrm{0} \\ $$$${Prove}\:{that}\:\frac{{a}}{\mathrm{1}+{a}}+\frac{{b}}{\mathrm{1}+{b}}>\frac{{c}}{\mathrm{1}+{c}} \\ $$$$. \\ $$$$\mathrm{I}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help},\:\mathrm{please}! \\ $$ Commented by MJS_new last updated on 21/Dec/20…
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}}…