Question Number 155983 by MathsFan last updated on 06/Oct/21 $$\:{given}\:\:{y}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:,\:{then} \\ $$$$\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}+\boldsymbol{{k}}{y}=\mathrm{0}.\:{find}\:\boldsymbol{{k}} \\ $$ Commented by cortano last updated on 06/Oct/21 $$\mathrm{k}=\mathrm{1} \\ $$…
Question Number 155973 by Fareed last updated on 06/Oct/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{a}=? \\ $$$$\mathrm{whenever}\:\:\:\:\mathrm{f}\left(\mathrm{a}\right)=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{a}\right) \\ $$$$ \\ $$ Commented by MJS_new last…
Question Number 90435 by naka3546 last updated on 23/Apr/20 Commented by MJS last updated on 23/Apr/20 $$\mathrm{none}\:\mathrm{of}\:\mathrm{these}\:\mathrm{options} \\ $$$$\sqrt{\mathrm{2}\sqrt{\mathrm{3}\sqrt{\mathrm{4}\sqrt{\mathrm{5}\sqrt{…}}}}}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }} \mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }} \mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }} \mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}}…
Question Number 155959 by SANOGO last updated on 06/Oct/21 $${quel}\:{est}\:{le}\:{changement}\:{de}\:{variable}\:{qui}\:{permet}\:{de}\:{passer} \\ $$$${de}\:{l}'{equation}\:{differentielle}\:: \\ $$$${x}^{\mathrm{2}} {y}''−\mathrm{3}{xy}'+\mathrm{4}{y}=\mathrm{0} \\ $$$${a}\:{une}\:{equation}\:{lineaire}\:{d}'{ordre}\:\mathrm{2}\:\:{coefficient}\:{comstant}\:\:{en}\:{z} \\ $$ Answered by puissant last updated on…
Question Number 24865 by kosarrr last updated on 27/Nov/17 $$\frac{\mathrm{27}^{{x}−\mathrm{1}} +\mathrm{81}^{{x}} }{\mathrm{3}^{\mathrm{3}{x}} }=\frac{\mathrm{4}}{\mathrm{27}} \\ $$ Answered by iv@0uja last updated on 27/Nov/17 $$\frac{\mathrm{3}^{\mathrm{3}\left({x}−\mathrm{1}\right)} +\mathrm{3}^{\mathrm{4}{x}} }{\mathrm{3}^{\mathrm{3}{x}}…
Question Number 24866 by kosarrr last updated on 27/Nov/17 $$\mathrm{3}^{{x}} =\mathrm{2} \\ $$$$\mathrm{2}^{{y}} =\mathrm{4} \\ $$$$\frac{\left(\mathrm{3}^{{x}} \right)^{\mathrm{4}{y}−\mathrm{1}} ×\left(\mathrm{3}^{\mathrm{2}{x}} \right)^{{y}+\mathrm{1}} }{\left(\mathrm{3}^{\mathrm{3}{y}+\mathrm{2}} \right)^{{x}} }= \\ $$ Answered…
Question Number 90402 by bshahid010@gmail.com last updated on 23/Apr/20 Answered by MJS last updated on 23/Apr/20 $$\mathrm{for}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}:\:\frac{{d}}{{dx}}\left[\mathrm{sin}\:{x}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right]=\mathrm{cos}\:{x}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:{x}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:{x}=\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a}<\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$…
Question Number 155918 by Oberon last updated on 05/Oct/21 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{evaluate}\:\mathrm{this}\:\mathrm{sum}? \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{−\mathrm{n}} \mathrm{tan}\:\left(\mathrm{2}^{−\mathrm{n}} \right) \\ $$ Commented by horpy4 last updated on 05/Oct/21…
Question Number 24839 by kosarrr last updated on 27/Nov/17 $$\frac{\mathrm{5}^{\mathrm{2}{x}+\mathrm{1}} +\mathrm{5}^{\mathrm{2}{x}} }{\mathrm{5}^{\mathrm{2}{x}} +\mathrm{5}^{\mathrm{2}{x}−\mathrm{1}} }=\left(\mathrm{0}/\mathrm{04}\right)^{{x}−\mathrm{1}} \\ $$ Commented by behi.8.3.4.17@gmail.com last updated on 27/Nov/17 $$\boldsymbol{{LHS}}=\frac{\mathrm{5}^{\mathrm{2x}} \left(\mathrm{5}+\mathrm{1}\right)}{\mathrm{5}^{\mathrm{2x}}…
Question Number 24840 by kosarrr last updated on 27/Nov/17 Commented by prakash jain last updated on 27/Nov/17 $$\mathrm{Please}\:\mathrm{use}\:\mathrm{an}\:\mathrm{app}\:\mathrm{called}\:\mathrm{camscanner} \\ $$$$\mathrm{to}\:\mathrm{post}\:\mathrm{pictures}\:\mathrm{of}\:\mathrm{written}/\mathrm{printed} \\ $$$$\mathrm{material}. \\ $$$$\mathrm{Your}\:\mathrm{picture}\:\mathrm{is}\:\mathrm{very}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{read}. \\…