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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}}…

Question-31571

Question Number 31571 by mondodotto@gmail.com last updated on 10/Mar/18 Answered by Joel578 last updated on 10/Mar/18 $$\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{cosec}^{\mathrm{4}} \:\theta} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{4}} \:\theta}\:+\:\frac{\mathrm{1}}{\mathrm{cosec}^{\mathrm{4}} \:\theta} \\…

Question-31562

Question Number 31562 by mondodotto@gmail.com last updated on 10/Mar/18 Answered by Joel578 last updated on 10/Mar/18 $$\int\:{e}^{{ax}} \:+\:{bx}\:{dx}\:=\:\mathrm{4}{e}^{\mathrm{4}{x}} \:+\:\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{1}}{{a}}{e}^{{ax}} \:+\:\frac{{b}}{\mathrm{2}}{x}^{\mathrm{2}} \:=\:\mathrm{4}{e}^{\mathrm{4}{x}} \:+\:\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \\…