Question Number 95846 by naka3546 last updated on 28/May/20 $${A}\:=\:\left\{\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…\:,\:\mathrm{2020}\:\right\} \\ $$$${how}\:\:{many}\:\:{zero}\:\:{on}\:\:{A}\:? \\ $$ Answered by Rasheed.Sindhi last updated on 28/May/20 $$'\_'\:\mathrm{is}\:\mathrm{nonzero}\:\mathrm{digit}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}. \\ $$$$\mathrm{1}-\mathrm{digit}\:\mathrm{numbers}: \\…
Question Number 161377 by stelor last updated on 17/Dec/21 $${please}\:{calculate}\:{A}\:{and}\:\:{B}. \\ $$$$ \\ $$$${A}\:=\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{9}}\right)\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{16}}\right)…\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{4}\:\mathrm{084}\:\mathrm{441}}\right)\:\:\left({and}\:\mathrm{2021}^{\mathrm{2}\:} =\:\mathrm{4084441}\:\right) \\ $$$${B}\:=\:\left(\mathrm{1}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \:−\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \right)\:+\:\left(\:\mathrm{5}^{\mathrm{2}} \:−\:\mathrm{6}^{\mathrm{2}} \:−\mathrm{7}^{\mathrm{2}} \:+\mathrm{8}^{\mathrm{2}} \right)\:+\left(\mathrm{9}^{\mathrm{2}}…
Question Number 161378 by KONE last updated on 22/Dec/21 $${determine},\:{pour}\:{tout}\:\alpha\in\mathbb{R},\:\:{la}\:{nature} \\ $$$$\:{de}\:{la}\:{serie}\:{de}\:{terme}\:{general} \\ $$$${U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left({k}^{\mathrm{2}} +\left({n}−{k}\right)^{\mathrm{2}} \right)^{\alpha} } \\ $$$${besoin}\:{d}'{aide}\:{svp} \\ $$$${please}\:{help}\:{me} \\…
Question Number 161356 by naka3546 last updated on 17/Dec/21 $${a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0} \\ $$$$ \\ $$$${Find}\:\:{the}\:\:{value}\:\:{of} \\ $$$$\:\:\:\left(\frac{{b}−{c}}{{a}}\:+\:\frac{{c}−{a}}{{b}}\:+\:\frac{{a}−{b}}{{c}}\right)\left(\frac{{a}}{{b}−{c}}\:+\:\frac{{b}}{{c}−{a}}\:+\:\frac{{c}}{{a}−{b}}\right)\:. \\ $$ Commented by MJS_new last updated on 17/Dec/21…
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Question Number 161338 by akolade last updated on 16/Dec/21 Commented by cortano last updated on 16/Dec/21 $$\:\begin{cases}{\mathrm{16}−{x}^{\mathrm{2}} >\mathrm{0}\Rightarrow\left({x}−\mathrm{4}\right)\left({x}+\mathrm{4}\right)<\mathrm{0}\:;−\mathrm{4}<{x}<\mathrm{4}}\\{\mathrm{3}{x}−\mathrm{4}>\mathrm{0}\Rightarrow{x}>\frac{\mathrm{4}}{\mathrm{3}}}\\{\mathrm{3}{x}−\mathrm{4}\neq\mathrm{1}\:\Rightarrow{x}\neq\frac{\mathrm{5}}{\mathrm{3}}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{16}−{x}^{\mathrm{2}} =\left(\mathrm{3}{x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{16}−{x}^{\mathrm{2}} =\mathrm{9}{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{16}…
Question Number 161335 by kapoorshah last updated on 16/Dec/21 Commented by mr W last updated on 17/Dec/21 $${see}\:{Q}\mathrm{161366} \\ $$ Terms of Service Privacy Policy…
Question Number 30259 by mondodotto@gmail.com last updated on 19/Feb/18 Commented by prof Abdo imad last updated on 20/Feb/18 $${we}\:{have}\:\:\frac{{x}}{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{{lnx}}=\:\frac{{xlnx}\:−{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right){lnx}}=\frac{{u}\left({x}\right)}{{v}\left({x}\right)} \\ $$$${we}\:{hsve}\:{u}\left(\mathrm{1}\right)={v}\left(\mathrm{1}\right)=\mathrm{0}\:{let}\:{use}\:{hospital}\:{theorem} \\ $$$${u}^{'} \left({x}\right)=\mathrm{1}+{lnx}\:−\mathrm{1}\Rightarrow{u}^{''} \left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:{anf}\:{u}^{''}…
Question Number 30257 by mondodotto@gmail.com last updated on 19/Feb/18 Commented by rahul 19 last updated on 19/Feb/18 $$\mathrm{Sandwhich}\:\mathrm{theorem}\:! \\ $$ Commented by abdo imad last…
Question Number 30258 by mondodotto@gmail.com last updated on 19/Feb/18 Answered by $@ty@m last updated on 19/Feb/18 $${Let}\:\mathrm{sec}\:{x}−\mathrm{1}={z} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {xdx}={dz} \\ $$$${dx}=\frac{{dz}}{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}}=\frac{{dz}}{{z}\left({z}+\mathrm{2}\right)} \\ $$$${I}=\int\frac{{dx}}{{z}^{\mathrm{2}}…