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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Question Number 6464 by nburiburu last updated on 27/Jun/16 $$“{In}\:{a}\:{great}\:{reunion}\:{each}\:{person}\:{knows} \\ $$$${at}\:{least}\:{two}\:{different}\:{persons}\:{in}\:{there}''.\:{Show}\: \\ $$$${that}\:{this}\:{statement}\:{implies}\:{that}\:{exists} \\ $$$${one}\:{person}\:{who}\:{has}\:{to}\:{know}\:{everyone}. \\ $$ Commented by prakash jain last updated on…
Question Number 71849 by device4438043516@gmail.com last updated on 23/May/20 $$ \\ $$ Answered by MJS last updated on 21/Oct/19 $${P}\:\:\:{Q}\:\:{P}\Rightarrow{Q}\:\:{P}\wedge{Q}\:\:{P}\wedge{Q}={P} \\ $$$$\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{1}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0} \\…
Question Number 136697 by bramlexs22 last updated on 25/Mar/21 Commented by bramlexs22 last updated on 25/Mar/21 $${what}\:{the}\:{missing}\:{number} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 5450 by 3 last updated on 15/May/16 $${eqution} \\ $$ Answered by FilupSmith last updated on 15/May/16 $${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{with}\:\mathrm{solution}: \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}}…