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the-sequence-is-partial-from-sequence-1-n-1-n-N-1-1-n-n-2-1-2n-1-3-1-2n-chouse-the-right-answer-

Question Number 113957 by mohammad17 last updated on 16/Sep/20 $${the}\:{sequence}\:………..\:{is}\:{partial}\:{from}\:{sequence}\:\langle\frac{\mathrm{1}}{{n}+\mathrm{1}}\rangle_{{n}\in{N}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\langle\frac{\mathrm{1}−{n}}{{n}}\rangle\:\:\:\:\:\:\:\left(\mathrm{2}\right)\langle\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\rangle\:\:\:\:\:\:\left(\mathrm{3}\right)\langle\frac{\mathrm{1}}{\mathrm{2}{n}}\rangle\: \\ $$$$ \\ $$$${chouse}\:{the}\:{right}\:{answer} \\ $$ Terms of Service Privacy Policy…

Question-48396

Question Number 48396 by behi83417@gmail.com last updated on 23/Nov/18 Commented by MJS last updated on 23/Nov/18 $$\frac{\mathrm{2}{a}\pm\sqrt{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}} \\…

to-tinku-tara-sir-for-some-reasons-i-don-t-get-any-notifications-about-updates-to-my-posts-can-you-please-check-sir-thank-you-

Question Number 179429 by mr W last updated on 29/Oct/22 $${to}\:{tinku}\:{tara}\:{sir}: \\ $$$${for}\:{some}\:{reasons}\:{i}\:{don}'{t}\:{get}\:{any} \\ $$$${notifications}\:{about}\:{updates}\:{to}\:{my} \\ $$$${posts}.\:{can}\:{you}\:{please}\:{check}\:{sir}?\: \\ $$$${thank}\:{you}! \\ $$ Commented by ARUNG_Brandon_MBU last…

Question-48283

Question Number 48283 by naka3546 last updated on 21/Nov/18 Commented by maxmathsup by imad last updated on 22/Nov/18 $${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{\mathrm{2}{k}−\mathrm{1}}\:\:{changement}\:{of}\:{indice}\:{k}−\mathrm{1}={j}\:{give} \\ $$$${S}_{{n}}…

lim-x-0-xlnx-

Question Number 113812 by Khalmohmmad last updated on 15/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}xlnx}=? \\ $$ Answered by bemath last updated on 15/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:{x}}{\frac{\mathrm{1}}{{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{{x}}}{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:=\:\mathrm{0} \\…

find-the-greatest-coeeficient-in-the-expansion-6-4x-3-

Question Number 113796 by aurpeyz last updated on 15/Sep/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coeeficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\: \\ $$$$\left(\mathrm{6}−\mathrm{4x}\right)^{−\mathrm{3}} \\ $$ Answered by mr W last updated on 15/Sep/20 $$\left(\mathrm{6}−\mathrm{4}{x}\right)^{−\mathrm{3}} =\mathrm{6}^{−\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}{x}\right)^{−\mathrm{3}}…