Question Number 126765 by mnjuly1970 last updated on 24/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{elementary}\:\:{mathematics}… \\ $$$$\:\:\:{if}\:\:\:\:\:\mathrm{13}\:\mid\mathrm{9}^{\mathrm{51}} +{k}+\mathrm{1}\:\:\:,\:{k}\in\mathbb{N} \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:\:{k}_{\left({min}\right)} \:=? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by floor(10²Eta[1])…
Question Number 191731 by Spillover last updated on 29/Apr/23 $$\mathrm{Use}\:\mathrm{laws}\:\mathrm{of}\:\mathrm{algebra}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following} \\ $$$$\left(\mathrm{a}\right)\left[\left(\mathrm{B}−\mathrm{A}\right)\mathrm{u}\left(\mathrm{A}−\mathrm{B}\right)\right]=\left[\left(\mathrm{AuB}\right)−\left(\mathrm{AnB}\right)\right] \\ $$$$\left(\mathrm{b}\right)\mathrm{A}\bigtriangledown\left(\mathrm{AnB}\right)=\mathrm{A}−\mathrm{B} \\ $$ Answered by AST last updated on 29/Apr/23 $$\left({a}\right)\:{B}−{A}={B}\cap{A}^{'} \:\:\:{and}\:\:{A}−{B}={A}\cap{B}^{'}…
Question Number 191501 by normans last updated on 24/Apr/23 $$ \\ $$$$\:\:\:\:{find}\:{a}\:{solution}; \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{e}}^{\boldsymbol{{x}}} \:=\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right) \\ $$$$ \\ $$ Commented by mehdee42 last…
Question Number 190573 by sami123 last updated on 06/Apr/23 $$\hat {{a}}\mathrm{2}+\mathrm{2}{ab}+\hat {{b}}\mathrm{2} \\ $$ Commented by a.lgnaoui last updated on 06/Apr/23 $$\mathrm{1}\bullet{pour}\:{editer}\:\:{exposant}\left({a}^{\mathrm{2}} \right) \\ $$$${dans}\:{clavier}\:{ecrire}\:\:{a}\:\:{puis}\:\:\:\:…
Question Number 124899 by Don08q last updated on 06/Dec/20 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124409 by oustmuchiya@gmail.com last updated on 03/Dec/20 $${write}\:\mathrm{A}'\cup\:{B}'\:{in}\:{a}\:{disjoint}\:{set} \\ $$ Answered by Ar Brandon last updated on 03/Dec/20 $$\left(\mathrm{A}\cup\mathrm{B}\right)−\left(\mathrm{A}\cap\mathrm{B}\right) \\ $$ Terms of…
Question Number 189672 by Bidhan12 last updated on 20/Mar/23 Commented by Rasheed.Sindhi last updated on 20/Mar/23 $${B}\supseteq{A}\:{or}\:{A}\subseteq{B} \\ $$ Commented by som(math1967) last updated on…
Question Number 189091 by mnjuly1970 last updated on 12/Mar/23 $$ \\ $$$$\:\:\:\:\mathrm{1}\::\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\:\mathrm{1}\:\right)^{\:{n}} \mathrm{H}_{\:{n}} }{{n}^{\:\mathrm{2}} }\:=\:? \\ $$$$\:\:\:\:\mathrm{2}\::\:\:\:\:\:\eta\:\left(−\mathrm{1}\:\right)=\:? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$…
Question Number 123287 by mnjuly1970 last updated on 24/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}\:… \\ $$$$\:\:\:\:\:\:\:{number}\:{theory} \\ $$$$\:\:\:\:\:\:\:\:\:\:{prove}\:{thar}\:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{32}} +\mathrm{1}\overset{\mathrm{641}} {\equiv}\mathrm{0}\:\checkmark \\ $$$$\:\:\:{notice}:\:{without}\:{calculator}\:{and}\:{only} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{with}\:{the}\:{use}\:{of}\:{congruence}\:{properties}.. \\ $$ Answered…
Question Number 122436 by sahiljakhar04 last updated on 17/Nov/20 $$\boldsymbol{\mathrm{Mean}}\::\:\mathrm{It}\:\mathrm{is}\:\mathrm{found}\:\mathrm{by}\:\mathrm{adding}\:\mathrm{all}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{observation}\:\mathrm{and}\:\mathrm{dividing}\:\mathrm{it}\:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{observations}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{denoted}\:\mathrm{by}\:\bar {{x}}. \\ $$$$\mathrm{So},\:\bar {{x}}\:=\:\frac{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} }{{n}}.\:\mathrm{For}\:\mathrm{an}\:\boldsymbol{\mathrm{ungrouped}}\:\boldsymbol{\mathrm{frequency}}\:\boldsymbol{\mathrm{distribution}},\:\mathrm{it}\:\mathrm{is}\:\bar {\boldsymbol{{x}}}\:=\:\frac{\underset{\boldsymbol{{i}}\:=\:\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{f}}_{\boldsymbol{{i}}} \boldsymbol{{x}}_{\boldsymbol{{i}}} }{\underset{\boldsymbol{{i}}\:=\:\mathrm{1}} {\overset{\boldsymbol{{n}}}…