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Question-193037

Question Number 193037 by DAVONG last updated on 02/Jun/23 Answered by Subhi last updated on 02/Jun/23 $${DC}^{\mathrm{2}} ={DB}.{AD} \\ $$$$\mathrm{20}^{\mathrm{2}} ={DB}.\left(\mathrm{9}+{DB}\right) \\ $$$${DB}^{\mathrm{2}} +\mathrm{9}{DB}−\mathrm{400}=\mathrm{0} \\…

Question-127490

Question Number 127490 by ZiYangLee last updated on 30/Dec/20 Commented by mr W last updated on 30/Dec/20 $${Q}\mathrm{45}. \\ $$$$\frac{{dx}}{{dt}}=\mathrm{3}{t}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dt}}=\mathrm{2}{t} \\ $$$$\frac{{dy}}{{dx}}=\frac{\frac{{dy}}{{dt}}}{\frac{{dx}}{{dt}}}=\frac{\mathrm{2}{t}}{\mathrm{3}{t}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}{t}}…