Question Number 10981 by ridwan balatif last updated on 05/Mar/17 Commented by prakash jain last updated on 06/Mar/17 $$\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+\mathrm{3}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\geqslant\mathrm{3}>\mathrm{0} \\ $$$$\mid\mathrm{3}{x}−\mathrm{6}\mid\geqslant\mathrm{0} \\…
Question Number 142049 by mohammad17 last updated on 25/May/21 $$\int_{\mathrm{1}} ^{\:\mathrm{3}} {e}^{{x}^{\mathrm{2}} } {dx} \\ $$$${help}\:{me}\:{sir}\: \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 76514 by mathmax by abdo last updated on 28/Dec/19 $${calculate}\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\alpha^{{k}\:} \:+\overset{−^{{k}} } {\alpha}\right)\:{with}\:\alpha\:{root}\:{of}\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\:=\mathrm{0} \\ $$ Commented by mathmax by abdo…
Question Number 142045 by Engr_Jidda last updated on 25/May/21 Answered by Dwaipayan Shikari last updated on 25/May/21 $${y}=\left(\frac{\mathrm{1}}{{x}}\right)^{{x}} \\ $$$${log}\left({y}\right)=−{xlog}\left({x}\right)\Rightarrow\frac{{y}'}{{y}}=−\mathrm{1}−{log}\left({x}\right)\Rightarrow{y}=−{x}^{−{x}} \left(\mathrm{1}+{log}\left({x}\right)\right) \\ $$ Commented by…
Question Number 10973 by ridwan balatif last updated on 05/Mar/17 $$\mid\mid{x}\mid+\mathrm{2x}\mid\leqslant\mathrm{3},\:\mathrm{interval}\:\mathrm{x}=…? \\ $$$$\mathrm{A}.−\mathrm{3}\leqslant{x}\leqslant\mathrm{3} \\ $$$$\mathrm{B}.\:{x}\geqslant\mathrm{0} \\ $$$$\mathrm{C}.\:{x}\leqslant\mathrm{0} \\ $$$$\mathrm{D}.\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\mathrm{E}.\:{x}\leqslant\mathrm{2} \\ $$ Commented by…
Question Number 142041 by nadovic last updated on 25/May/21 Answered by TheSupreme last updated on 25/May/21 $$\sqrt{−\mathrm{3}{a}+{b}}−\mathrm{2}=\mathrm{0} \\ $$$${b}−\mathrm{3}{a}=\mathrm{4} \\ $$$$\frac{−\mathrm{3}{a}}{\mathrm{2}\sqrt{−\mathrm{3}{a}+{b}}}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{6}{a}=\sqrt{−\mathrm{3}{a}+{b}} \\ $$$$\mathrm{36}{a}^{\mathrm{2}}…
Question Number 10970 by Joel576 last updated on 04/Mar/17 $${x}^{{x}^{{x}^{\iddots^{\mathrm{2}} } } } \:=\:\mathrm{2} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:? \\ $$ Answered by ridwan balatif last updated on…
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Question Number 142035 by Dwaipayan Shikari last updated on 25/May/21 $$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}/\mathrm{4}} }.\frac{\mathrm{3}^{\mathrm{1}/\mathrm{9}} }{\mathrm{4}^{\mathrm{1}/\mathrm{16}} }.\frac{\mathrm{5}^{\mathrm{1}/\mathrm{25}} }{\mathrm{6}^{\mathrm{1}/\mathrm{36}} }.\frac{\mathrm{7}^{\mathrm{1}/\mathrm{49}} }{\mathrm{8}^{\mathrm{1}/\mathrm{64}} }…={exp}\left(−\frac{\zeta'\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{log}\:\left(\mathrm{2}\right)\right) \\ $$ Answered by mindispower last…
Question Number 76497 by mathmax by abdo last updated on 27/Dec/19 $${calculate}\:{tan}\left(\mathrm{3}{x}\right)\:{interms}\:{of}\:{tanx} \\ $$ Answered by Rio Michael last updated on 27/Dec/19 $${tan}\mathrm{3}{x}\:=\:{tan}\left(\mathrm{2}{x}\:+\:{x}\right)\:=\:\frac{{tan}\mathrm{2}{x}\:+\:{tanx}}{\mathrm{1}−{tan}\mathrm{2}{xtanx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tan}^{\mathrm{2}}…