Question Number 157341 by mathocean1 last updated on 22/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157343 by mathocean1 last updated on 22/Oct/21 Answered by mindispower last updated on 23/Oct/21 $${X}^{\mathrm{2}} −{X}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{X}\in\left\{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$${u}_{{n}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{b}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\…
Question Number 91796 by zainal tanjung last updated on 03/May/20 $$\mathrm{Total}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{this}\:\mathrm{below}\:\mathrm{infinite}\:\mathrm{series}\:: \\ $$$$\left.\mathrm{1}\right).\:\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}+…\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=… \\ $$$$\left.\mathrm{2}\right).\:\:\mathrm{0}.\mathrm{6}+\mathrm{0}.\mathrm{06}+\mathrm{0}.\mathrm{006}+…\frac{\mathrm{6}}{\mathrm{10}^{\mathrm{n}} }=… \\ $$ Commented by mathmax by abdo last updated…
Question Number 157329 by MathsFan last updated on 22/Oct/21 $$\:\boldsymbol{{whats}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\: \\ $$$$\:\:\:\:!\mathrm{5} \\ $$ Answered by puissant last updated on 24/Oct/21 $$!\mathrm{5}=\mathrm{5}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{120}}\right)=\mathrm{5}!\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{120}}\right) \\ $$$$=\:\mathrm{5}!\left(\frac{\mathrm{60}}{\mathrm{120}}−\frac{\mathrm{20}}{\mathrm{120}}+\frac{\mathrm{5}}{\mathrm{120}}−\frac{\mathrm{1}}{\mathrm{120}}\right)=\:\mathrm{120}×\frac{\mathrm{44}}{\mathrm{120}}\:=\:\mathrm{44}.. \\…
Question Number 157272 by zakirullah last updated on 21/Oct/21 $${for}\:{solving}\:{equation}\:{which}\:{one}\:{we}\:{use} \\ $$$$\Rightarrow\:{and}\:=,\:{i}\:{mean}\:{where}\:{we}\:{use}\:\Rightarrow\:{and}\:{where}\:{we}\:{use},\:= \\ $$$${and}\:{where}\:{we}\:{use}\:{one}\:{of}\:{them}\:{to} \\ $$$${consider}\:{wrong}. \\ $$ Answered by TheHoneyCat last updated on 21/Oct/21…
Question Number 26200 by mubeen897@hotmail.com last updated on 22/Dec/17 Answered by ajfour last updated on 22/Dec/17 $$\left(\mathrm{0}.\mathrm{2}×\mathrm{10}\right)+\mathrm{10}+\left(\mathrm{0}.\mathrm{2}×\mathrm{20}\right)−\mathrm{30}+ \\ $$$$\:\:\:\:\:\:\:\mathrm{0}.\mathrm{5}{R}+\left(\mathrm{0}.\mathrm{5}×\mathrm{10}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}+\mathrm{10}+\mathrm{4}−\mathrm{30}+\mathrm{0}.\mathrm{5}{R}+\left[×\right]+\mathrm{5}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{0}.\mathrm{5}{R}\:=\:\mathrm{9}\:\:\:\Rightarrow\:\:\:{R}=\mathrm{18}\Omega\:. \\ $$…
Question Number 157230 by Jamshidbek last updated on 21/Oct/21 $$\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{100}} \:\: \\ $$$$\mathrm{Find}\:\mathrm{this}\:\mathrm{max}\:\mathrm{Koeffitcient} \\ $$ Answered by mr W last updated on 21/Oct/21 $${a}_{{n}} =\mathrm{3}^{{n}}…
Question Number 91690 by zainal tanjung last updated on 02/May/20 Commented by zainal tanjung last updated on 03/May/20 $$\mathrm{Hint}=\mathrm{Formula} \\ $$$$\bullet\:\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\bullet\:\:\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\bullet\:\left[\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}}…
Question Number 157217 by MathsFan last updated on 21/Oct/21 $${show}\:{that} \\ $$$$\:\frac{{cos}\mathrm{2}{x}+{cos}\mathrm{3}{x}+{cos}\mathrm{8}{x}}{{sin}\mathrm{2}{x}+{sin}\mathrm{3}{x}+{sin}\mathrm{8}{x}}={tanx} \\ $$ Commented by som(math1967) last updated on 21/Oct/21 $${please}\:{check}\:{your}\:{question} \\ $$ Commented…
Question Number 157216 by Khalmohmmad last updated on 21/Oct/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =? \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{tan}\:\mathrm{4}{x}}}{\mathrm{tan}\sqrt{\mathrm{4}{x}}}=? \\ $$ Commented by cortano last updated on 21/Oct/21…