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Exercise-ABC-is-a-triangle-AB-AC-2-and-BC-2-2-I-is-midle-of-BC-J-is-a-point-such-as-AJ-2-3-AI-J-is-the-center-of-gravity-of-the-triangle-1-a-we-define-the-set-T-of-point-M

Question Number 97174 by mathocean1 last updated on 06/Jun/20 $$\underset{−} {{Exercise}} \\ $$$${ABC}\:{is}\:{a}\:{triangle}.\:{AB}={AC}=\mathrm{2}\:{and} \\ $$$${BC}=\mathrm{2}\sqrt{\mathrm{2}}.\:{I}\:{is}\:{midle}\:{of}\:\left[{BC}\right]. \\ $$$${J}\:{is}\:{a}\:{point}\:{such}\:{as}\:\overset{\rightarrow} {{AJ}}=\frac{\mathrm{2}}{\mathrm{3}}\overset{\rightarrow} {{AI}}.\:{J}\:{is} \\ $$$${the}\:{center}\:{of}\:{gravity}\:{of}\:{the}\:{triangle}. \\ $$$$\left.\mathrm{1}\left.\right){a}\right)\:{we}\:{define}\:{the}\:{set}\left({T}\right)\:{of}\:\forall\:{point}\:{M} \\ $$$${of}\:{plane}:…

Question-162673

Question Number 162673 by mkam last updated on 31/Dec/21 Answered by mr W last updated on 31/Dec/21 $$\left({i}\right)×\mathrm{2}−\left({ii}\right): \\ $$$$−\mathrm{3}{x}=\mathrm{3}{t}+\mathrm{2}\:\Rightarrow{x}=−{t}−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$−\mathrm{1}+{t}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{{dy}}{{dt}}=\mathrm{2}{t}+\mathrm{1} \\ $$$$\frac{{dy}}{{dt}}={t}+\frac{\mathrm{4}}{\mathrm{3}} \\…

prove-that-cos-mx-cos-ny-cos-mx-ny-cos-mx-ny-2-help-me-sir-

Question Number 97142 by mhmd last updated on 06/Jun/20 $${prove}\:{that}\:{cos}\left({mx}\right){cos}\left({ny}\right)=\frac{{cos}\left({mx}+{ny}\right)+{cos}\left({mx}−{ny}\right)}{\mathrm{2}}\:\:? \\ $$$${help}\:{me}\:{sir}\:? \\ $$ Commented by mr W last updated on 06/Jun/20 $${just}\:{apply}\:{the}\:{formulae}\:{for} \\ $$$$\mathrm{cos}\:\left({a}+{b}\right)\:{and}\:\mathrm{cos}\:\left({a}−{b}\right).…

Question-31573

Question Number 31573 by mondodotto@gmail.com last updated on 10/Mar/18 Commented by MJS last updated on 10/Mar/18 $$\mathrm{messed}\:\mathrm{around}\:\mathrm{a}\:\mathrm{bit} \\ $$$$\mathrm{1}^{\mathrm{st}} \:\mathrm{thought}\:{y}={e}^{−{x}^{\mathrm{2}} } ,\:\mathrm{checking} \\ $$$$\mathrm{leads}\:\mathrm{to}\:−{x}×{e}^{−{x}^{\mathrm{2}} }…

Question-31572

Question Number 31572 by mondodotto@gmail.com last updated on 10/Mar/18 Answered by MJS last updated on 10/Mar/18 $${f}\left({x}\right)−{g}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} −\mathrm{8}{x} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{area}\:{A}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{zeros} \\ $$$$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{8}{x}=\mathrm{0} \\ $$$${x}_{\mathrm{1}}…