Question Number 93933 by oustmuchiya@gmail.com last updated on 16/May/20 $${Let}\:\ast'\:{be}\:{the}\:{binary}\:{operation}\:{on}\:{the}\:{set}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\}\:{defined}\:{by}\:\mathrm{a}\ast'\mathrm{b}=\mathrm{H}.\mathrm{C}.\mathrm{F}{of}\:{a}\:{and}\:{b}.\: \\ $$$${Is}\:{the}\:{operation}\:\ast'\:{same}\:{as}\:{the}\:{operation}\:\ast\:{defined}\:{above}?\:{justify}\:{your}\:{answer}. \\ $$ Commented by Rasheed.Sindhi last updated on 16/May/20 $${Operation}\:\ast'\:{is}\:{defined}\:{in}\:{the} \\ $$$${question}.{What}\:{do}\:{you}\:{mean}\:{by} \\…
Question Number 93567 by oustmuchiya@gmail.com last updated on 13/May/20 $${Given}\:{that}\:{A}=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5}\right\}\:{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{7}\right\}\:{and}\:{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\}\:{prove}\:{that} \\ $$$$\left(\mathrm{i}\right)\:\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{C}\:=\:\mathrm{A}\cap\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{ii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C}\:=\:\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right) \\ $$$$\left(\mathrm{iii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\:=\:\left(\mathrm{A}\cup\mathrm{C}\right)\cup\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{iv}\right)\:\left(\mathrm{A}\cap\mathrm{C}\right)\cup\mathrm{B}\:=\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{C}\cup\mathrm{B}\right) \\ $$ Commented by Ritu Jain last…
Question Number 158522 by amin96 last updated on 05/Nov/21 $$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(\zeta\left({n}\right)−{n}\right)}{\mathrm{2}^{{n}} }=? \\ $$ Answered by qaz last updated on 06/Nov/21 $$\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{n}\right)−\mathrm{n}}{\mathrm{2}^{\mathrm{n}}…
Question Number 25723 by Joel578 last updated on 13/Dec/17 $$\mathrm{Given}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{n}} \:\mathrm{are}\:\mathrm{non}−\mathrm{negative} \\ $$$$\mathrm{integers}\:\mathrm{and}\:\mathrm{satisfy} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{a}_{\mathrm{1}} } }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{{a}_{\mathrm{2}} } }\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{2}^{{a}_{{n}} } }\:=\:\frac{\mathrm{1}}{\mathrm{3}^{{a}_{\mathrm{1}} } }\:+\:\frac{\mathrm{2}}{\mathrm{3}^{{a}_{\mathrm{2}}…
Question Number 24930 by 8006539252 last updated on 29/Nov/17 $${a}+{a}= \\ $$ Answered by 8006539252 last updated on 29/Nov/17 $$\mathrm{2}{a} \\ $$ Terms of Service…
Question Number 24366 by Tinkutara last updated on 16/Nov/17 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{7}-\mathrm{element}\:\mathrm{set}\:{A}\:=\:\left\{{a},\:{b},\:{c},\right. \\ $$$$\left.{d},\:{e},\:{f},\:{g}\right\},\:\mathrm{find}\:\mathrm{a}\:\mathrm{collection}\:{T}\:\mathrm{of}\:\mathrm{3}- \\ $$$$\mathrm{element}\:\mathrm{subsets}\:\mathrm{of}\:{A}\:\mathrm{such}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{pair}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{from}\:{A}\:\mathrm{occurs}\:\mathrm{exactly} \\ $$$$\mathrm{in}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{subsets}\:\mathrm{of}\:{T}. \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 155277 by amin96 last updated on 28/Sep/21 Answered by Rasheed.Sindhi last updated on 28/Sep/21 $$\left({big}-{square}-{area}\right)−\left({non}-{shadow}-{area}\right) \\ $$$${big}-{square}-{area}=\mathrm{8}^{\mathrm{2}} =\mathrm{64}\:{sq}-{units} \\ $$$$\:\:\:\:\:\:\underset{−} {{Non}-{shadow}-{area}} \\ $$$$\mathrm{A}:{two}-{triangles}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{1}.\mathrm{2}\right)=\mathrm{2}\:{unit}^{\mathrm{2}}…
Question Number 154544 by amin96 last updated on 19/Sep/21 $$\begin{array}{|c|c|}{\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}}\\{\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} ^{\left(\mathrm{2}\right)} }{\boldsymbol{\mathrm{n}}^{\mathrm{3}} }+\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} ^{\left(\mathrm{3}\right)} }{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }=\frac{\mathrm{21}}{\mathrm{8}}\boldsymbol{\zeta}\left(\mathrm{6}\right)+\boldsymbol{\zeta}^{\mathrm{2}} \left(\mathrm{3}\right)}\\\hline\end{array} \\ $$$${by}\:{Math}.{Amin}\:\:\mathrm{11}.{fb}.\mathrm{96}…
Question Number 88761 by M±th+et£s last updated on 12/Apr/20 $${if}\:{a}_{{n}} =\frac{{n}!}{{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}} \\ $$$${and}\:{b}_{{n}} =\frac{\left(\mathrm{2}{n}\right)!\sqrt{{n}}}{\mathrm{4}^{{n}} \:\left({n}!\right)^{\mathrm{2}} } \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\mathrm{1} \\ $$$${find}\:\underset{{n}\rightarrow\infty} {{lim}b}_{{n}}…
Question Number 88473 by M±th+et£s last updated on 10/Apr/20 Commented by kaivan.ahmadi last updated on 11/Apr/20 $${av}_{\mathrm{1}} +{b}\left({v}_{\mathrm{1}} +{v}_{\mathrm{2}} \right)+{c}\left({v}_{\mathrm{1}} +{v}_{\mathrm{2}} +{v}_{\mathrm{3}} \right)=\mathrm{0}\Rightarrow \\ $$$$\left({a}+{b}+{c}\right){v}_{\mathrm{1}}…