Question Number 68732 by ravi Raj last updated on 15/Sep/19 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 134270 by abdurehime last updated on 01/Mar/21 $$\mathrm{proof}\:\mathrm{that} \\ $$$$\left.\begin{cases}{\mathrm{k}}\\{\mathrm{n}}\end{cases}\right\}=\frac{\mathrm{k}!}{\mathrm{n}!×\left(\mathrm{k}−\mathrm{n}\right)!} \\ $$ Commented by mr W last updated on 02/Mar/21 $${you}\:{mean}\:\left(_{{n}} ^{{k}} \right)=\frac{{k}!}{{n}!×\left({k}−{n}\right)!}.…
Question Number 134267 by bramlexs22 last updated on 01/Mar/21 $$\begin{cases}{\mathrm{x}\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{x}\equiv\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{4}\:\right)}\end{cases} \\ $$ Answered by EDWIN88 last updated on 01/Mar/21 $$\begin{cases}{\mathrm{x}=\mathrm{5k}+\mathrm{4}}\\{\mathrm{x}=\mathrm{4}\ell+\mathrm{3}}\end{cases}\:\Leftrightarrow\:\mathrm{5k}+\mathrm{4}\:=\:\mathrm{4}\ell+\mathrm{3} \\ $$$$\ell\:=\:\frac{\mathrm{5k}+\mathrm{1}}{\mathrm{4}}\:;\:\left(\mathrm{k},\ell\right)\:=\:\left(\mathrm{3},\mathrm{4}\right),\:\left(\mathrm{7},\mathrm{9}\right),\left(\mathrm{11},\mathrm{14}\right),…, \\ $$$$\left(\mathrm{4}\lambda−\mathrm{1},\:\mathrm{5}\lambda−\mathrm{1}\right)\:\Rightarrow\mathrm{then}\:\mathrm{x}=\mathrm{5}\left(\mathrm{4}\lambda−\mathrm{1}\right)+\mathrm{4} \\…
Question Number 68728 by Rio Michael last updated on 15/Sep/19 $${dear}\:{scientist}. \\ $$$${i}\:{did}\:{some}\:{research}\:{on}\:{the}\:{energy}\:{obtained}\:{from}\:{the}\:{sun} \\ $$$${any}\:{other}\:{source}\:\:{by}\:{a}\:{liquid},\:{of}\:{density}\:\rho\:,\:{velocity}\:\:{v},\:\:{viscosity}\:\eta\:{and}\:{distance}\: \\ $$$${travelled}\:\:\:{d}. \\ $$$$ \\ $$$${i}\:{came}\:{out}\:{with}\:{the}\:{equation}\: \\ $$$$\:\:\:{E}\:=\:{k}\:\left(\:{v}\rho\:\eta^{\mathrm{3}} \:{d}^{\mathrm{2}} \right)…
Question Number 3193 by Filup last updated on 07/Dec/15 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}: \\ $$$$\underset{{i}={a}} {\overset{{n}} {\sum}}{i}=\underset{{i}={a}} {\overset{{n}} {\prod}}{i}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({n},\:{a}\right)\in\mathbb{Z} \\ $$ Commented by Filup last updated on 07/Dec/15…
Question Number 3190 by Filup last updated on 07/Dec/15 $$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{curve}\:\mathrm{between}\:{f}\left({a}\right)\:\mathrm{and}\:{f}\left({b}\right)? \\ $$ Commented by 123456 last updated on 07/Dec/15 $$\mathrm{wath}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{by}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{curve}? \\ $$$$\mathrm{if}\:\mathrm{i}\:\mathrm{undestand}\:\mathrm{right},\:\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{continuous} \\…
Question Number 134259 by Dwaipayan Shikari last updated on 01/Mar/21 $$\frac{\mathrm{1}}{\mathrm{998}}+\frac{\mathrm{1}}{\mathrm{998}.\mathrm{1995}}+\frac{\mathrm{1}}{\mathrm{998}.\mathrm{1995}.\mathrm{2992}}+\frac{\mathrm{1}}{\mathrm{998}.\mathrm{1995}.\mathrm{2992}.\mathrm{3989}}+…=\frac{\mathrm{1}}{\mathrm{997}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 3185 by Yozzi last updated on 06/Dec/15 $${I}'{ve}\:{got}\:{a}\:{Cambridge}\:{interview}\:{for} \\ $$$${mathematics}\:{coming}\:{up}\:{and}\:{I}'{m} \\ $$$${wondering}\:{what}\:{pieces}\:{of}\:{advice}\:{I} \\ $$$${could}\:{possibly}\:{get}\:{from}\:{members}\:{of} \\ $$$${this}\:{forum}.\:{I}'{ve}\:{been}\:{doing}\:{some}\:{work}\: \\ $$$${in}\:{an}\:{attempt}\:{to}\:{prepare}\:{so}\:{that}'{s}\:{why} \\ $$$${I}\:{haven}'{t}\:{been}\:{posting}\:{questions}\:{in} \\ $$$${relation}\:{to}\:{series}\:{and}\:{sequences}\: \\…
Question Number 68721 by ahmadshah last updated on 15/Sep/19 Commented by Prithwish sen last updated on 15/Sep/19 $$\mathrm{3} \\ $$ Answered by $@ty@m123 last updated…
Question Number 134252 by bramlexs22 last updated on 01/Mar/21 $$\:\mathrm{solve}\:\begin{cases}{\mathrm{x}\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right)}\\{\mathrm{x}\equiv\:\mathrm{5}\:\left(\mathrm{mod}\:\mathrm{7}\right)}\end{cases} \\ $$ Answered by EDWIN88 last updated on 01/Mar/21 $$\begin{cases}{\mathrm{x}\:=\:\mathrm{3k}+\mathrm{2}\:;\mathrm{k}\in\mathbb{Z}}\\{\mathrm{x}=\mathrm{7m}+\mathrm{5};\:\mathrm{m}\in\mathbb{Z}}\end{cases}\:\Rightarrow\:\mathrm{3k}+\mathrm{2}\:=\:\mathrm{7m}+\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{m}=\frac{\mathrm{3k}−\mathrm{3}}{\mathrm{7}}\:;\:\left(\mathrm{k},\mathrm{m}\right)=\left(\mathrm{8},\mathrm{3}\right),\left(\mathrm{15},\mathrm{6}\right),\left(\mathrm{22},\mathrm{9}\right) \\ $$$$…,\:\left(\mathrm{7}\lambda+\mathrm{1},\:\mathrm{3}\lambda\right) \\…