Question Number 151311 by Ahmed1hamouda last updated on 20/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20204 by mondodotto@gmail.com last updated on 24/Aug/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20203 by mondodotto@gmail.com last updated on 24/Aug/17 Commented by Einstein Newton last updated on 24/Aug/17 $$\left(\mathrm{a}\right)\:\mathrm{20}\:+\:\mathrm{37}\:−\:\mathrm{12}\:=\:\mathrm{45} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{20}\:+\:\mathrm{37}\:=\:\mathrm{57} \\ $$ Terms of Service…
Question Number 20201 by khamizan833@gmail.con last updated on 24/Aug/17 $$\:^{\mathrm{4}} \sqrt{\mathrm{49}\:−\:\mathrm{20}\sqrt{\mathrm{6}}}\:= \\ $$ Answered by Einstein Newton last updated on 24/Aug/17 $$\sqrt{\mathrm{49}\:−\:\mathrm{20}\sqrt{\mathrm{6}}}\:=\:\sqrt{\mathrm{5}^{\mathrm{2}} \:+\:\left(\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{5}\right)\left(\mathrm{2}\sqrt{\mathrm{6}}\right)} \\…
Question Number 151269 by tabata last updated on 19/Aug/21 Commented by tabata last updated on 19/Aug/21 $$?????? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 20195 by mondodotto@gmail.com last updated on 23/Aug/17 Answered by Tinkutara last updated on 23/Aug/17 $$\mathrm{Let}\:{x}\:=\:\mathrm{2}^{{y}} \\ $$$$\mathrm{2}^{{y}\left({y}+\mathrm{4}\right)} \:=\:\mathrm{2}^{\mathrm{5}} \\ $$$${y}^{\mathrm{2}} \:+\:\mathrm{4}{y}\:−\:\mathrm{5}\:=\:\mathrm{0} \\ $$$$\left({y}\:+\:\mathrm{5}\right)\left({y}\:−\:\mathrm{1}\right)\:=\:\mathrm{0}…
Question Number 85708 by Roland Mbunwe last updated on 24/Mar/20 $${li}\underset{{n}−\infty} {{m}}\:\:/\frac{{U}_{{n}} +\mathrm{1}}{{Un}}/\:\:\:>\mathrm{0} \\ $$$${Test}\:{for}\:{convergence} \\ $$ Answered by Rio Michael last updated on 24/Mar/20…
Question Number 85709 by otchereabdullai@gmail.com last updated on 24/Mar/20 Answered by MJS last updated on 24/Mar/20 $$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{many}\:\mathrm{times}\:\mathrm{before} \\ $$$$\mathrm{the}\:\mathrm{posts}\:\mathrm{are}\:\mathrm{50},\:\mathrm{the}\:\mathrm{cable}\:\mathrm{is}\:\mathrm{80},\:\mathrm{the}\:\mathrm{space} \\ $$$$\mathrm{under}\:\mathrm{the}\:\mathrm{cable}\:\mathrm{is}\:\mathrm{10} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{effective}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{is}\:\mathrm{40}\:\Rightarrow…
Question Number 85706 by naka3546 last updated on 24/Mar/20 $${Find}\:\:{the}\:\:{general}\:\:{solution}\:\:{of} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} \:\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}\:\:{dx}\:+\:\mathrm{5}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}\:\:{y}\:{dy}\:\:=\:\:\mathrm{0} \\ $$ Answered by mind is power last updated on…
Question Number 20132 by mondodotto@gmail.com last updated on 22/Aug/17 Answered by mrW1 last updated on 22/Aug/17 $$\mathrm{log}\:\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{x}}{\mathrm{25}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{10}^{\frac{\mathrm{x}}{\mathrm{25}}} \\ $$$$\mathrm{x}=\pm\mathrm{10}^{\frac{\mathrm{x}}{\mathrm{50}}} \\ $$$$\mathrm{50}×\frac{\mathrm{x}}{\mathrm{50}}=\pm\mathrm{10}^{\frac{\mathrm{x}}{\mathrm{50}}}…