Question Number 18461 by b.e.h.i.8.3.417@gmail.com last updated on 22/Jul/17 Answered by mrW1 last updated on 22/Jul/17 $$\left.\mathrm{1}\right) \\ $$$$\Delta_{\mathrm{AA}_{\mathrm{5B}} \mathrm{A}_{\mathrm{60C}} } =\frac{\mathrm{5}}{\mathrm{n}}×\frac{\mathrm{60}}{\mathrm{n}}×\Delta_{\mathrm{ABC}} \\ $$$$\Delta_{\mathrm{BCC}_{\mathrm{5A}} }…
Question Number 149510 by Tawa11 last updated on 05/Aug/21 Commented by Ar Brandon last updated on 06/Aug/21 $$\mathrm{36}−\mathrm{4}×\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{18cm}^{\mathrm{3}} \\ $$ Commented by Tawa11 last updated…
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Question Number 149068 by Tawa11 last updated on 02/Aug/21 Answered by mr W last updated on 03/Aug/21 Commented by mr W last updated on 02/Aug/21…
Question Number 149031 by puissant last updated on 02/Aug/21 Answered by EDWIN88 last updated on 02/Aug/21 $$\Rightarrow\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}+\mathrm{cos}\:\mathrm{4}{x}+\mathrm{1}+\mathrm{cos}\:\mathrm{6}{x}}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{6}{x}+\mathrm{cos}\:\mathrm{4}{x}+\mathrm{cos}\:\mathrm{2}{x}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{4}{x}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{4}{x}\left(\mathrm{2cos}\:\mathrm{2}{x}+\mathrm{1}\right)=−\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{2cos}\:^{\mathrm{2}}…