Question Number 110367 by mathdave last updated on 28/Aug/20 Answered by Dwaipayan Shikari last updated on 28/Aug/20 $$\left.\mathrm{1}−\left({log}\left(−\mathrm{1}\right)\right)\right)^{\mathrm{2}} −\mathrm{4}{ilog}\left({i}\right) \\ $$$$\mathrm{1}−\left(\pi{i}\right)^{\mathrm{2}} −\mathrm{4}{i}\left(\frac{\pi{i}}{\mathrm{2}}\right) \\ $$$$\left(\pi+\mathrm{1}\right)^{\mathrm{2}} \\…
Question Number 110365 by mathdave last updated on 28/Aug/20 Answered by Rasheed.Sindhi last updated on 29/Aug/20 $${wxyz}=\mathrm{120}\:,\:{wxz}−{y}=\mathrm{26}, \\ $$$${xyz}−{w}=\mathrm{58}\:\:,\:\:{wz}+{xy}=\mathrm{22} \\ $$$$\frac{{wxyz}}{{y}}−{y}=\mathrm{26}\Rightarrow\frac{\mathrm{120}}{{y}}−{y}=\mathrm{26}….\left({ii}\right) \\ $$$$\frac{{wxyz}}{{w}}−{w}=\mathrm{58}\Rightarrow\frac{\mathrm{120}}{{w}}−{w}=\mathrm{58}…\left({iii}\right) \\ $$$$\frac{{wxyz}}{{xy}}+{xy}=\mathrm{22}\Rightarrow\frac{\mathrm{120}}{{xy}}+{xy}=\mathrm{22}…\left({iv}\right)…
Question Number 110350 by bobhans last updated on 28/Aug/20 $$\:\:\:\frac{{bob}}{{hans}} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}−\frac{\mathrm{9}}{{x}}\right)}{\mathrm{tan}\:\left({x}−\mathrm{3}\right)\mathrm{cos}\:\left(\frac{\mathrm{9}}{{x}}−{x}\right)}= \\ $$$$\left(\mathrm{2}\right)\left({x}\:\mathrm{tan}^{−\mathrm{1}} \left({y}\right)\right){dx}\:+\left(\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\right).\:{dy}\:=\mathrm{0}\: \\ $$ Answered by john santu last…
Question Number 44811 by jasno91 last updated on 05/Oct/18 Commented by math khazana by abdo last updated on 05/Oct/18 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{4}\:\Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{16}\:\Rightarrow{x}^{\mathrm{2}} \:+\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\mathrm{16}\:\Rightarrow \\ $$$${x}^{\mathrm{2}}…
Question Number 44812 by jasno91 last updated on 05/Oct/18 Commented by math khazana by abdo last updated on 05/Oct/18 $${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}\:=\mathrm{12}^{\mathrm{2}} \:−\mathrm{28}\:=\mathrm{144}−\mathrm{28} \\…
Question Number 44809 by jasno91 last updated on 05/Oct/18 Commented by Joel578 last updated on 05/Oct/18 $$\mathrm{Just}\:\mathrm{substitute}\:{x}\:\mathrm{with}\:\mathrm{12} \\ $$$$\mathrm{9}\left(\mathrm{12}^{\mathrm{2}} \right)\:+\:\mathrm{24}\left(\mathrm{12}\right)\:+\:\mathrm{16} \\ $$ Terms of Service…
Question Number 44810 by jasno91 last updated on 05/Oct/18 Answered by Kunal12588 last updated on 05/Oct/18 $$\left(\mathrm{6}{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{6}{x}\right)\left(\mathrm{5}{y}\right)+\left(\mathrm{5}{y}\right)^{\mathrm{2}} =\left(\mathrm{6}{x}−\mathrm{5}{y}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9} \\ $$ Terms…
Question Number 44807 by jasno91 last updated on 05/Oct/18 Answered by Kunal12588 last updated on 05/Oct/18 $$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$$$\mathrm{1}.\:\left(\mathrm{50}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{2500}+\mathrm{16}+\mathrm{400}=\mathrm{2916} \\ $$$$\mathrm{2}.\:\left(\mathrm{80}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{6400}+\mathrm{4}+\mathrm{320}=\mathrm{6734}…
Question Number 44808 by jasno91 last updated on 05/Oct/18 Commented by Kunal12588 last updated on 05/Oct/18 $${read}\:{some}\:{books} \\ $$ Answered by Kunal12588 last updated on…
Question Number 44806 by jasno91 last updated on 05/Oct/18 Answered by Kunal12588 last updated on 05/Oct/18 $${let}\:{the}\:{numbers}\:{be}\:{x}\:{and}\:{y} \\ $$$${x}:{y}=\mathrm{3}:\mathrm{5} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow{x}=\frac{\mathrm{3}{y}}{\mathrm{5}}\:\:\:….\left(\mathrm{1}\right) \\ $$$$\left({x}+\mathrm{10}\right):\left({y}+\mathrm{10}\right)=\mathrm{5}:\mathrm{7} \\ $$$$\Rightarrow{x}=\frac{\mathrm{5}}{\mathrm{7}}\left({y}+\mathrm{10}\right)−\mathrm{10}\:\left(\mathrm{2}\right)…