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Question Number 141087 by greg_ed last updated on 15/May/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{masters}}\:! \\ $$$$\boldsymbol{\mathrm{look}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{thing}}\:\boldsymbol{\mathrm{carefully}}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\mathrm{3}\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\mathrm{6}\:\mathrm{7}\:\mathrm{8}\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10}\:\mathrm{11}\:\mathrm{12}\:\mathrm{13}\:\mathrm{14}\:\mathrm{15}\:\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{17}\:\mathrm{18}\:\mathrm{19}\:\mathrm{20}\:…………….. \\ $$$$\:\:\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{line}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{column}}\:\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{number}}\:\mathrm{795471}\:\boldsymbol{\mathrm{will}}\:\boldsymbol{\mathrm{appear}}\:! \\…
Question Number 75257 by must1987 last updated on 09/Dec/19 $${Let}\:{A}=\left\{\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$${B}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{10}\right\}\:{and} \\ $$$${C}=\left\{{x}:\mathrm{3}{x}+\mathrm{6}=\mathrm{0}\:{or}\:\mathrm{2}{x}+\mathrm{6}=\mathrm{0}\right\}.{Find} \\ $$$${a}.\:{A}\cup{B}. \\ $$$${b}.\:{is}\left({A}\cup{B}\right)\cup{C}={A}\cup\left({B}\cup{C}\right)? \\ $$ Answered by 21042004 last updated…
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Question Number 75254 by must1987 last updated on 09/Dec/19 Answered by 21042004 last updated on 09/Dec/19 $$\mathrm{1}.\:{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{6}\right\} \\ $$$$\left.\mathrm{2}.\:{a}\right)\:{A}\cup{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\} \\ $$$$\left.\:\:\:\:\:{b}\right)\:{Yes} \\ $$$$\left.\mathrm{3}.\:{a}\right)\:{A}'=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9}\right\} \\ $$$$\left.\:\:\:\:\:{b}\right)\:{A}\cap{A}'=\varnothing…
Question Number 9487 by Joel575 last updated on 12/Dec/16 $$\mathrm{A}\:\mathrm{ticket}\:\mathrm{contains}\:\mathrm{9}\:\mathrm{digits}\:\mathrm{serial}\:\mathrm{number}, \\ $$$$\mathrm{which}\:\mathrm{only}\:\mathrm{composed}\:\mathrm{from}\:\mathrm{number}\:\mathrm{1},\:\mathrm{2},\:\mathrm{or}\:\mathrm{3} \\ $$$$\mathrm{Ticket}\:\mathrm{can}\:\mathrm{be}\:\mathrm{colored}\:\mathrm{red},\:\mathrm{blue},\:\mathrm{or}\:\mathrm{green}. \\ $$$$\mathrm{If}\:\mathrm{2}\:\mathrm{different}\:\mathrm{tickets}\:\mathrm{have}\:\mathrm{different}\:\mathrm{serial}\:\mathrm{numbers}, \\ $$$$\mathrm{they}\:\mathrm{have}\:\mathrm{different}\:\mathrm{color}\:\mathrm{too} \\ $$$$ \\ $$$$\mathrm{John}\:\mathrm{has}\:\mathrm{a}\:\mathrm{blue}\:\mathrm{ticket}\:\mathrm{with}\:\mathrm{serial}\:\mathrm{number}\:\mathrm{311111111} \\ $$$$\mathrm{Ann}\:\mathrm{has}\:\mathrm{a}\:\mathrm{red}\:\mathrm{ticket}\:\mathrm{with}\:\mathrm{serial}\:\mathrm{number}\:\mathrm{111111111} \\…
Question Number 74246 by TawaTawa last updated on 20/Nov/19 Answered by MJS last updated on 20/Nov/19 $$\mathrm{1}\:\mathrm{2}\:\mathrm{49} \\ $$$$… \\ $$$$\mathrm{1}\:\mathrm{25}\:\mathrm{26} \\ $$$$\mathrm{24}\:\mathrm{triples} \\ $$$$\mathrm{2}\:\mathrm{3}\:\mathrm{47}…
Question Number 8545 by Rishabh#1 last updated on 15/Oct/16 $$\mathrm{18}:\mathrm{48}\:::\:\mathrm{180}:? \\ $$$$\mathrm{Answer}\:\mathrm{optioms} \\ $$$$\mathrm{a}.\:\mathrm{392}\:\mathrm{b}.\:\mathrm{294}\:\mathrm{c}.\mathrm{230} \\ $$ Commented by Rishabh#1 last updated on 16/Oct/16 $$\mathrm{Answer}\:\mathrm{given}\:\mathrm{is}\:\mathrm{294}.\:\mathrm{I}\:\mathrm{am}\:\mathrm{not} \\…
Question Number 8164 by 666225 last updated on 02/Oct/16 $${if}\:\boldsymbol{{p}}\:{is}\:{prima}\:{show}\:{that}\:\sqrt{\boldsymbol{{p}}}\:{irasional} \\ $$ Commented by 123456 last updated on 02/Oct/16 $$\mathrm{1}\mid{p},{p}\mid{p},{p}>\mathrm{1} \\ $$$${p}\mid{a}^{\mathrm{2}} \Leftrightarrow{p}\mid{a} \\ $$…
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