Question Number 50047 by peter frank last updated on 13/Dec/18 Answered by peter frank last updated on 13/Dec/18 $$\mathrm{P}=\left(\mathrm{acos}\theta,\mathrm{bsin}\theta\right) \\ $$$$\mathrm{Q}=\left(-\mathrm{asin}\theta,\mathrm{bcos}\theta\right) \\ $$$$\mathrm{O}=\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\mathrm{L}^{\mathrm{2}}…
Question Number 49651 by ajfour last updated on 08/Dec/18 Commented by ajfour last updated on 08/Dec/18 $${If}\:\:{AP}\:=\:{a}\:,\:{BP}\:=\:{b}\:,\:{CP}\:=\:{c}\:,\:{DP}\:=\:{d}\:, \\ $$$${find}\:{sides}\:\boldsymbol{{p}},\:\boldsymbol{{q}}\:\:{of}\:{rectangle}. \\ $$ Commented by mr W…
Question Number 180599 by Crabby last updated on 14/Nov/22 $${Find}\:{the}\:{length}\:{PQ}. \\ $$ Answered by MJS_new last updated on 14/Nov/22 $${P},\:{Q}\:\in\mathbb{R}^{{n}} \\ $$$${P}=\left({p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} ,\:…\:{p}_{{n}} \right)\wedge{Q}=\left({q}_{\mathrm{1}}…
Question Number 49359 by peter frank last updated on 06/Dec/18 Answered by MJS last updated on 06/Dec/18 $$\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$$${y}=\frac{\mathrm{1}}{{x}}\:\Rightarrow\:\frac{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{{x}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\left({y}^{\mathrm{2}} −{y}+\mathrm{1}\right)^{\mathrm{2}} }{{y}\left({y}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 49358 by peter frank last updated on 06/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18 $$\left.\mathrm{1}\right)\mathrm{9}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\mathrm{6}\frac{{dx}}{{dt}}+{x}=−\mathrm{50}{sint} \\ $$$${step}\mathrm{1} \\ $$$$\mathrm{9}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}}…
Question Number 49356 by peter frank last updated on 06/Dec/18 Commented by peter frank last updated on 07/Dec/18 $$\mathrm{am}\:\mathrm{getting}\:\mathrm{greatest}\:\mathrm{area}\:\mathrm{is}\:\:\mathrm{4}\pi\mathrm{a}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
Question Number 49354 by peter frank last updated on 06/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18 $$\left.\mathrm{2}\right){x}^{\mathrm{3}} −{qx}−{r}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{0}×{x}^{\mathrm{2}} +\left(−{q}\right){x}+\left(−{r}\right)=\mathrm{0} \\ $$$$\alpha+\beta+\gamma=\mathrm{0}…
Question Number 49296 by peter frank last updated on 05/Dec/18 Answered by kaivan.ahmadi last updated on 05/Dec/18 $$\frac{\mathrm{df}}{\mathrm{dx}}=\mathrm{2xysiny}^{\mathrm{2}} \mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{ycosy}^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{and} \\ $$$$\frac{\mathrm{df}}{\mathrm{dy}}=\mathrm{x}^{\mathrm{2}}…
Question Number 49298 by peter frank last updated on 05/Dec/18 Answered by Kunal12588 last updated on 05/Dec/18 $$\left.{b}\right)\:\mathrm{3}{ac}\left({a}+{c}\right)+{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \\ $$$$=\mathrm{3}{ac}\left({a}+{c}\right)+\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)+\mathrm{3}{abc}…
Question Number 49121 by Pk1167156@gmail.com last updated on 03/Dec/18 Answered by mr W last updated on 03/Dec/18 $$\mathrm{1}−\left(\frac{\mathrm{8}}{{h}}\right)^{\mathrm{3}} =\left(\frac{{h}−\mathrm{2}}{{h}}\right)^{\mathrm{3}} \\ $$$$\mathrm{1}−\mathrm{64}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} =\left(\mathrm{1}−\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} \\ $$$$\mathrm{1}−\mathrm{64}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} =\mathrm{1}−\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)+\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{2}}…