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Category: Mensuration

Question-50065

Question Number 50065 by peter frank last updated on 13/Dec/18 Answered by math1967 last updated on 13/Dec/18 $${y}^{\mathrm{2}} =\mathrm{4}{ax}\:\therefore{x}=\frac{{y}^{\mathrm{2}} }{\mathrm{4}{a}} \\ $$$${now}\:{lx}+{my}+{n}=\mathrm{0} \\ $$$$\Rightarrow\frac{{ly}^{\mathrm{2}} }{\mathrm{4}{a}}\:+{my}+{n}=\mathrm{0}…

Question-50047

Question Number 50047 by peter frank last updated on 13/Dec/18 Answered by peter frank last updated on 13/Dec/18 $$\mathrm{P}=\left(\mathrm{acos}\theta,\mathrm{bsin}\theta\right) \\ $$$$\mathrm{Q}=\left(-\mathrm{asin}\theta,\mathrm{bcos}\theta\right) \\ $$$$\mathrm{O}=\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\mathrm{L}^{\mathrm{2}}…

Question-49359

Question Number 49359 by peter frank last updated on 06/Dec/18 Answered by MJS last updated on 06/Dec/18 $$\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$$${y}=\frac{\mathrm{1}}{{x}}\:\Rightarrow\:\frac{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{{x}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\left({y}^{\mathrm{2}} −{y}+\mathrm{1}\right)^{\mathrm{2}} }{{y}\left({y}−\mathrm{1}\right)^{\mathrm{2}}…

Question-49358

Question Number 49358 by peter frank last updated on 06/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18 $$\left.\mathrm{1}\right)\mathrm{9}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\mathrm{6}\frac{{dx}}{{dt}}+{x}=−\mathrm{50}{sint} \\ $$$${step}\mathrm{1} \\ $$$$\mathrm{9}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}}…

Question-49354

Question Number 49354 by peter frank last updated on 06/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18 $$\left.\mathrm{2}\right){x}^{\mathrm{3}} −{qx}−{r}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{0}×{x}^{\mathrm{2}} +\left(−{q}\right){x}+\left(−{r}\right)=\mathrm{0} \\ $$$$\alpha+\beta+\gamma=\mathrm{0}…

Question-49296

Question Number 49296 by peter frank last updated on 05/Dec/18 Answered by kaivan.ahmadi last updated on 05/Dec/18 $$\frac{\mathrm{df}}{\mathrm{dx}}=\mathrm{2xysiny}^{\mathrm{2}} \mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{ycosy}^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{and} \\ $$$$\frac{\mathrm{df}}{\mathrm{dy}}=\mathrm{x}^{\mathrm{2}}…