Question Number 195411 by sonukgindia last updated on 02/Aug/23 Answered by MM42 last updated on 02/Aug/23 $${lim}_{{x}\rightarrow\infty} \:\frac{\mathrm{2}×\mathrm{2}^{{x}} −\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{9}^{{x}} +\mathrm{4}^{{x}} }{\mathrm{8}×\mathrm{4}^{{x}} +\mathrm{3}^{{x}} −\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}^{{x}} } \\…
Question Number 195433 by sonukgindia last updated on 02/Aug/23 Answered by gatocomcirrose last updated on 02/Aug/23 $$\mathrm{2}\begin{vmatrix}{\mathrm{2}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}−\mathrm{3}\begin{vmatrix}{\mathrm{x}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{x}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+ \\ $$$$+\mathrm{i}\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{x}}&{−\mathrm{1}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{2}}\\{\mathrm{2}}&{\mathrm{x}}&{\mathrm{2}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left[−\mathrm{10}−\mathrm{5x}\right]−\mathrm{3}\left[−\mathrm{5x}−\mathrm{10}\right]+\mathrm{i}\left[−\mathrm{3x}^{\mathrm{2}} −\mathrm{x}+\mathrm{10}\right]+\left[\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}}…
Question Number 195385 by Rodier97 last updated on 01/Aug/23 $$ \\ $$$$ \\ $$$$\mathrm{lim}_{{x}\Rightarrow\mathrm{a}^{+} } \:\:\:\:\frac{\sqrt{{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{{x}−\mathrm{a}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:\:;\:\:\mathrm{a}\:>\:\mathrm{0} \\ $$ Answered by cortano12 last updated…
Question Number 195352 by sonukgindia last updated on 31/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195341 by MCH last updated on 31/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195342 by Matica last updated on 31/Jul/23 $$\:\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that}\:\:\forall{n}\:\in\:\mathbb{N}^{\ast} \:,\:\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{3}} \:<\:\left({n}+\mathrm{1}\right)^{\mathrm{3}{n}} \:. \\ $$$$\mathrm{2}.\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{in}\:\mathbb{Z}^{\mathrm{2}} \:: \\ $$$$\:\:\:\:\:{a}./\:\:\mathrm{2}{x}^{\mathrm{3}} +{xy}−\mathrm{7}=\mathrm{0}\:, \\ $$$$\:\:\:\:\:{b}./\:\:{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{7}\right)\left({x}+\mathrm{8}\right)={y}^{\mathrm{2}} . \\ $$…
Question Number 195343 by Matica last updated on 31/Jul/23 $$\:{hello}\:{everyone},\:{I}'{m}\:{a}\:{new}\:{student}. \\ $$$$\:{I}\:{beg}\:{your}\:{pardon}. \\ $$$$\:{Please}\:{tell}\:{me}\:{any}\:{good}\:{math}\:{book} \\ $$$$\:{you}\:{know}.\:{Thanks}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195364 by Rodier97 last updated on 01/Aug/23 $$ \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{natural}\:\mathrm{number}\:{n},\: \\ $$$$\mathrm{the}\:\mathrm{natural}\:\mathrm{number}\:\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} \:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:\mathrm{2}^{{n}} . \\ $$ Answered by Frix last…
Question Number 195290 by Matica last updated on 29/Jul/23 $$\:{A}\:{professor}\:{said}\:\:\mathrm{0}\mid\mathrm{0}\:{because}\:\mathrm{0}=\:\mathrm{0}×{a}+\mathrm{0}\:\:\:,\:{a}\in\:\mathbb{N}.\:{Can}\:{you}\:{prove}? \\ $$ Answered by Frix last updated on 29/Jul/23 $${a}\mid{b}\:\Leftrightarrow\:\frac{{b}}{{a}}\in\mathbb{Z}\:\mathrm{but}\:\frac{\mathrm{0}}{\mathrm{0}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\Rightarrow\:\mathrm{0}\mid\mathrm{0}\:\mathrm{is}\:\mathrm{meaningless} \\ $$ Terms of Service…
Question Number 195279 by edwinpollito last updated on 28/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com