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Question-153082

Question Number 153082 by liberty last updated on 04/Sep/21 Answered by Rasheed.Sindhi last updated on 04/Sep/21 $$\frac{{L}}{{W}}=\frac{{W}}{{L}−{W}}\Rightarrow{L}^{\mathrm{2}} −{LW}={W}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{L}}{{W}}\right)^{\mathrm{2}} −\frac{{L}}{{W}}−\mathrm{1}=\mathrm{0} \\ $$$$\frac{{L}}{{W}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\left(−\mathrm{1}\right)}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${Negative}\:{value}\:{rejected}…

Question-153054

Question Number 153054 by daus last updated on 04/Sep/21 Answered by bobhans last updated on 04/Sep/21 $$\begin{cases}{{g}\left({x}\right)=\mathrm{1}−\mathrm{2}{x}}\\{{f}\left({x}\right)={kx}^{\mathrm{2}} +{m}}\end{cases}\:\Rightarrow\left({f}\bullet{g}\right)\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5} \\ $$$$\Rightarrow{k}\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{2}} +{m}\:=\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5} \\ $$$$\Rightarrow{k}\left(\mathrm{4}{x}^{\mathrm{2}}…

lim-x-0-x-e-x-1-2-e-x-1-x-3-

Question Number 153039 by ZiYangLee last updated on 04/Sep/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({e}^{{x}} +\mathrm{1}\right)−\mathrm{2}\left({e}^{{x}} −\mathrm{1}\right)}{{x}^{\mathrm{3}} }\:=\:? \\ $$ Answered by puissant last updated on 04/Sep/21 $$={lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}}…

Question-153038

Question Number 153038 by 7770 last updated on 04/Sep/21 Answered by mr W last updated on 04/Sep/21 $${say}\:{BC}={DC}={AD}=\mathrm{1} \\ $$$${BD}=\mathrm{2}×\mathrm{1}×\mathrm{sin}\:\mathrm{24}=\mathrm{2}\:\mathrm{sin}\:\mathrm{24} \\ $$$$\mathrm{96}−\frac{\mathrm{180}−\mathrm{48}}{\mathrm{2}}=\mathrm{30} \\ $$$$\angle{A}=\alpha \\…

y-log-1-cos-x-dy-dx-

Question Number 153033 by ZiYangLee last updated on 04/Sep/21 $${y}=\mathrm{log}\:\left(\mathrm{1}+\mathrm{cos}\:{x}\right) \\ $$$$\frac{{dy}}{{dx}}= \\ $$ Answered by puissant last updated on 04/Sep/21 $${y}={ln}\left(\mathrm{1}+{cosx}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{−{sinx}}{\mathrm{1}+{cosx}} \\…