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}} \\ $$$$\mathrm{1}−\mathrm{64}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} =\left(\mathrm{1}−\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} \\ $$$$\mathrm{1}−\mathrm{64}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} =\mathrm{1}−\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)+\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{2}}…
Question Number 49111 by Pk1167156@gmail.com last updated on 03/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 03/Dec/18 Commented by Pk1167156@gmail.com last updated on 03/Dec/18 Very nice sir Commented…
Question Number 114638 by bobhans last updated on 20/Sep/20 $${Given}\:{a}\:=\:\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+\mathrm{16}^{\mathrm{2}} −\mathrm{16}}{\mathrm{1}.\mathrm{3}+\mathrm{2}.\mathrm{4}+\mathrm{3}.\mathrm{5}+…+\mathrm{15}.\mathrm{17}} \\ $$$$\:\:\:{c}\:=\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right).\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right).\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right).\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right). \\ $$$${find}\:{a}×{c}\:=\: \\ $$ Answered by bemath…
Question Number 48984 by peter frank last updated on 30/Nov/18 Commented by mr W last updated on 01/Dec/18 $${director}\:{circle}\:{of}\:{a}\:{circle}\:{is} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}} \\ $$$${or}…
Question Number 48982 by peter frank last updated on 30/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 01/Dec/18 $$\left.\mathrm{2}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}={sin}\left(\frac{\pi}{\mathrm{6}}\right)\:\:\:\frac{\mathrm{1}}{\mathrm{2}}={cos}\left(\frac{\pi}{\mathrm{6}}\right) \\ $$$${e}^{{i}\theta} ={cos}\theta+{isin}\theta\:\leftarrow{demoivers}\:{theorem} \\ $$$${e}^{−{i}\theta} ={cos}\theta−{isin}\theta \\…
Question Number 48920 by peter frank last updated on 30/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18 $${y}={e}^{−\mathrm{2}{mx}} {sin}\mathrm{4}{mx} \\ $$$$\frac{{dy}}{{dx}}={e}^{−\mathrm{2}{mx}} ×\mathrm{4}{mcos}\mathrm{4}{mx}−\mathrm{2}{me}^{−\mathrm{2}{mx}} {sin}\mathrm{4}{mx} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{me}^{−\mathrm{2}{mx}}…
Question Number 114447 by bemath last updated on 19/Sep/20 Commented by bemath last updated on 19/Sep/20 $${gave}\:{kudos}\:{all}\:{master}\: \\ $$ Answered by abdullahquwatan last updated on…
Question Number 114071 by bobhans last updated on 17/Sep/20 $${Find}\:{the}\:{solution}\:{set}\: \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}}\:\geqslant\:{x} \\ $$ Commented by prakash jain last updated on 17/Sep/20 $$ \\…