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Author: Tinku Tara

Question-196714

Question Number 196714 by mr W last updated on 30/Aug/23 Answered by som(math1967) last updated on 30/Aug/23 $$\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}+{b}+{c}} \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−{abc}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)=\mathrm{0} \\ $$$$\:{a}+{b}=\mathrm{0}\Rightarrow{a}=−{b}\Rightarrow{c}=\mathrm{2022} \\…

Give-the-function-f-x-x-1-x-2-2-x-3-3-x-4-4-x-2022-2022-Find-extremes-of-f-x-

Question Number 196710 by tri26112004 last updated on 30/Aug/23 $${Give}\:{the}\:{function}: \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}−\mathrm{4}\right)^{\mathrm{4}} …\left({x}−\mathrm{2022}\right)^{\mathrm{2022}} \\ $$$${Find}\:{extremes}\:{of}\:{f}\left({x}\right)¿ \\ $$ Commented by mr W last updated…

Question-196738

Question Number 196738 by Tawa11 last updated on 30/Aug/23 Commented by Tawa11 last updated on 30/Aug/23 $$\mathrm{Please}\:\mathrm{is}\:\mathrm{this}\:\mathrm{solution}\:\mathrm{correct}???. \\ $$$$ \\ $$$$\mathrm{My}\:\mathrm{thinking}. \\ $$$$\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{force} \\ $$$$\mathrm{that}\:\mathrm{should}\:\mathrm{accelerate}\:\mathrm{the}\:\mathrm{body},\:\mathrm{not}\:\mathrm{resultant}\:\mathrm{force}.…

Question-196688

Question Number 196688 by cortano12 last updated on 29/Aug/23 Answered by Frix last updated on 29/Aug/23 $$\int\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:{dx}\:\overset{{t}=\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}} {=} \\ $$$$=\int\frac{{t}^{\mathrm{4}} −\mathrm{4}}{{t}^{\mathrm{2}} }{dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{4}}{{t}}=…

If-ax-loga-bx-logb-then-prove-that-x-1-ab-

Question Number 196690 by MATHEMATICSAM last updated on 29/Aug/23 $$\mathrm{If}\:\left({ax}\right)^{\mathrm{log}{a}} \:=\:\left({bx}\right)^{\mathrm{log}{b}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${x}\:=\:\frac{\mathrm{1}}{{ab}}\:. \\ $$ Answered by BaliramKumar last updated on 29/Aug/23 $$\mathrm{loga}\centerdot\mathrm{log}\left(\mathrm{ax}\right)\:=\:\mathrm{logb}\centerdot\mathrm{log}\left(\mathrm{bx}\right) \\…