Question Number 183678 by ali009 last updated on 28/Dec/22 $${find}\:{the}\:{laplace}\:{invesrse}\:{for}\:{I}\left({s}\right) \\ $$$${I}\left({s}\right)=\frac{\mathrm{2000}{s}^{\mathrm{2}} }{\left({s}^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} \right)\left({s}^{\mathrm{2}} +\mathrm{400}{s}+\mathrm{2}×\mathrm{10}^{\mathrm{5}} \right)} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 117848 by john santu last updated on 14/Oct/20 $${Determine}\:{all}\:{functions}\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${such}\:{that}\:{the}\:{equality}\:{f}\left(\left[{x}\right]\:{y}\right)=\:{f}\left({x}\right)\:\left[{f}\left({y}\right)\:\right] \\ $$$${holds}\:{for}\:{all}\:{x},{y}\:\in\mathbb{R}\:.\:{Here}\:\:{by}\:\left[{x}\right]\:{we}\: \\ $$$${denote}\:{the}\:{greatest}\:{integer}\:{not}\:{exceeding}\:{x}. \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 117369 by bemath last updated on 11/Oct/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{units}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{2013}^{\mathrm{1}} +\mathrm{2013}^{\mathrm{2}} +\mathrm{2013}^{\mathrm{3}} +…+\mathrm{2013}^{\mathrm{2013}} \\ $$ Answered by JDamian last updated on 11/Oct/20 $${S}=\mathrm{2013}^{\mathrm{1}} +\mathrm{2013}^{\mathrm{2}} +\:\centerdot\centerdot\centerdot\:+\:\mathrm{2013}^{\mathrm{2013}}…
Question Number 117045 by bobhans last updated on 09/Oct/20 $$\mathrm{If}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{satisfying}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:=\:\mathrm{0}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}^{\mathrm{53}} +\mathrm{x}^{\mathrm{52}} +\mathrm{x}^{\mathrm{51}} +\mathrm{x}^{\mathrm{50}} +\mathrm{x}^{\mathrm{49}} \:? \\ $$ Answered by john…
Question Number 51507 by Rasheed.Sindhi last updated on 27/Dec/18 $$\:^{\bullet} \mathrm{739}\:{is}\:{a}\:\:{prime}\:{number}\:{and} \\ $$$${its}\:{reversed}\:{number}\:\mathrm{937}\:{is}\:{also} \\ $$$${prime}. \\ $$$${Determine}\:\mathrm{4}-{digit}\:{prime}\:{numbers} \\ $$$${whose}\:{reversed}\:{be}\:{also}\:{prime}. \\ $$$$\:^{\bullet} {If}\:{number}\:{of}\:{such}\:{primes}\:{is}\:{a} \\ $$$${function}\:{of}\:{number}\:{of}\:{digits}, \\…
Question Number 182374 by rs4089 last updated on 08/Dec/22 Commented by Rasheed.Sindhi last updated on 08/Dec/22 $${a}=\mathrm{10},{b}=\mathrm{13},{n}=\mathrm{3197} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$${a}=\mathrm{13},{b}=\mathrm{10},{n}=\mathrm{3197} \\ $$ Answered by…
Question Number 116725 by bemath last updated on 06/Oct/20 $$\mathrm{what}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{log}\:_{\mathrm{10}} \left(−\mathrm{1}\right)\:\mathrm{in}\: \\ $$$$\mathrm{complex}\:\mathrm{number} \\ $$ Commented by bemath last updated on 06/Oct/20 $$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$…
Question Number 116700 by bemath last updated on 06/Oct/20 $$\mathrm{How}\:\mathrm{many}\:\mathrm{positive}\:\mathrm{integral}\: \\ $$$$\mathrm{solutions}\:\mathrm{does}\:\mathrm{3x}+\mathrm{5y}=\mathrm{300}\:\mathrm{have}? \\ $$ Answered by mr W last updated on 06/Oct/20 $${x}=\mathrm{5}{p} \\ $$$${y}=\mathrm{3}{q}…
Question Number 182199 by depressiveshrek last updated on 05/Dec/22 $${Let}\:{S}=\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{7}\right\} \\ $$$${If}\:{we}\:{multiply}\:{atleast}\:\mathrm{2}\:{numbers} \\ $$$${from}\:{this}\:{set}\:{with}\:{each}\:{other},\:{what} \\ $$$${are}\:{the}\:{chances}\:{of}\:{the}\:{product}\:{to} \\ $$$${turn}\:{out}\:{to}\:{be}\:{divisible}\:{by}\:\mathrm{3}? \\ $$ Answered by Acem last updated…
Question Number 50960 by kaivan.ahmadi last updated on 22/Dec/18 $$\mathrm{4}\:\:\:\left(\mathrm{3}\right)\:\:\:\:\mathrm{2} \\ $$$$\mathrm{5}\:\:\:\:\mathrm{3}\:\:\:\left(?\right)\:\:\mathrm{1}\:\:\:\:\mathrm{1} \\ $$$$\mathrm{6}\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\left(\mathrm{8}\right)\:\:\:\:?\:\:\:\:\:?\:\:\:\:\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{each}\:\:\:? \\ $$ Terms of Service Privacy Policy…