Menu Close

Category: Mensuration

Question-49298

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-49121

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}}…

Given-a-1-2-2-2-3-2-16-2-16-1-3-2-4-3-5-15-17-c-1-1-2-1-1-3-1-1-4-1-1-5-1-1-5-1-1-4-1-1-3-1-1-2-find-a-c-

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-48982

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-48920

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}}…