Question Number 97403 by M±th+et+s last updated on 07/Jun/20 Commented by M±th+et+s last updated on 07/Jun/20 $$\zeta:{zeta}\:{hurwtiz}\:{function} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 96930 by pticantor last updated on 21/Jun/20 $${E}\left({x}\right)\:{denotes}\:{the}\:{integer}\:{part}\:{of}\:{x}\: \\ $$$$\left.{x}\in\right]\mathrm{0};\mathrm{1}\left[\:{determine}:\right. \\ $$$$\boldsymbol{{E}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}} \right)\:\boldsymbol{{and}}\:\boldsymbol{{E}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \right)\: \\ $$$$\boldsymbol{{calcul}}\:\boldsymbol{{li}}\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\boldsymbol{{E}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \right) \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{i}}\:\boldsymbol{{need}}\:\boldsymbol{{help}}\:\boldsymbol{{please}} \\…
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Question Number 162366 by cortano last updated on 29/Dec/21 $$\:{Given}\:{that}\:{the}\:{solution}\:{set}\:{of}\:{the}\: \\ $$$$\:{quadratic}\:{inequality}\:{ax}^{\mathrm{2}} +{bx}+{c}\:>\mathrm{0} \\ $$$$\:{is}\:\left(\mathrm{2},\mathrm{3}\right).\:{Then}\:{the}\:{solution}\:{set}\: \\ $$$$\:{of}\:{the}\:{inequality}\:{cx}^{\mathrm{2}} +{bx}+{a}\:<\mathrm{0}\: \\ $$$$\:{will}\:{be}\: \\ $$ Answered by mr…
Question Number 96811 by Ar Brandon last updated on 05/Jun/20 $$\mathrm{Prove}\:\mathrm{that}; \\ $$$$\mathrm{a}\backslash\:\mathrm{A}+\mathrm{A}\centerdot\mathrm{B}=\mathrm{A}\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}\backslash\:\left(\mathrm{A}+\mathrm{B}^{} \right)\centerdot\left(\mathrm{A}+\overset{−} {\mathrm{B}}\right)=\mathrm{A} \\ $$$$\mathrm{b}\backslash\:\mathrm{A}\centerdot\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{A}\:\:\:\:\:\:\:\mathrm{d}\backslash\:\mathrm{A}+\overset{−} {\mathrm{A}B}=\mathrm{A}+\mathrm{B} \\ $$ Answered by 1549442205 last updated…
Question Number 31229 by abdo imad last updated on 03/Mar/18 $${simplify}\:{A}_{{n}} =\:{C}_{{n}} ^{\mathrm{1}} \:\:+\mathrm{2}\:{C}_{{n}} ^{\mathrm{2}} \:+\mathrm{3}\:{C}_{{n}} ^{\mathrm{3}} \:+…\:+\:{n}\:{C}_{{n}} ^{{n}} \:. \\ $$ Commented by abdo…
Question Number 31045 by abdo imad last updated on 02/Mar/18 $${simlify}\:{A}_{{n}} =\left({C}_{{n}} ^{\mathrm{1}} \right)^{\mathrm{2}} \:+\mathrm{2}\left({C}_{{n}} ^{\mathrm{2}} \right)^{\mathrm{2}} \:\:+{n}\:\left({C}_{{n}} ^{{n}} \right)^{\mathrm{2}} . \\ $$ Terms of…
Question Number 31044 by abdo imad last updated on 02/Mar/18 $${if}\:\mathrm{1}\:+\mathrm{2}^{{n}} \:+\mathrm{3}^{{n}} \:+\mathrm{4}^{{n}} =\mathrm{4}{q}\:+{r}\:\:{find}\:{r}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31043 by abdo imad last updated on 02/Mar/18 $${find}\:{n}\:{in}\:{ordre}\:{to}\:{have}\:\mathrm{7}\:{divide}\:{n}^{\mathrm{3}} \:+{n}−\mathrm{2} \\ $$ Commented by MJS last updated on 03/Mar/18 $$\mathrm{the}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{remains}\:\mathrm{of} \\ $$$${n}^{\mathrm{3}} /\mathrm{7}\:\mathrm{is}\:\langle\mathrm{1},\mathrm{1},\mathrm{6},\mathrm{1},\mathrm{6},\mathrm{6},\mathrm{0},…\rangle…
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