Question Number 163408 by mathlove last updated on 06/Jan/22 $$\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}=? \\ $$ Answered by mr W last updated on 06/Jan/22 $$\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}={x} \\ $$$$\left.\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}+\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}+\mathrm{3}{x}\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}\right.}\right)={x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}}…
Question Number 32328 by abdo imad last updated on 23/Mar/18 $${prove}\:{that}\:\mathrm{2}^{{n}+\mathrm{1}} \:{divide}\:\:{A}_{{n}} =\left[\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}{n}+\mathrm{1}} \right]. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 32312 by NECx last updated on 23/Mar/18 $${A}\:{small}\:{ball}\:{is}\:{dropped}\:{from}\:{a}\: \\ $$$${height}\:{of}\:\mathrm{1}{m}\:{into}\:{a}\:{horizontal} \\ $$$${floor}.{Each}\:{time}\:{it}\:{rebounces}\:{to} \\ $$$$\mathrm{3}/\mathrm{5}\:{of}\:{the}\:{height}\:{it}\:{has}\:{fallen}. \\ $$$$\left.{a}\right){show}\:{that}\:{when}\:{the}\:{ball}\:{strikes} \\ $$$${the}\:{ground}\:{for}\:{the}\:{third}\:{time}\:,{it} \\ $$$${has}\:{travelled}\:{a}\:{distance}\:{of}\:\mathrm{2}.\mathrm{92}{m} \\ $$$$\left.{b}\right){Show}\:{that}\:{the}\:{total}\:{distance} \\…
Question Number 163340 by mathlove last updated on 06/Jan/22 Answered by Rasheed.Sindhi last updated on 06/Jan/22 $$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}+{b}}={a}−{b} \\ $$$$=\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}}\right)\right] \\ $$$$=\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}}\right] \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}}\right)=\mathrm{2}×\frac{\mathrm{100}}{\mathrm{101}}=\frac{\mathrm{200}}{\mathrm{101}}\:\:\:\left({b}\right)\checkmark…
Question Number 162960 by mnjuly1970 last updated on 02/Jan/22 $$ \\ $$$$\:\:\:\:\mathrm{A}\:\in\:\mathrm{M}_{\:{n}×{n}} \:\:,\:\:\:\mathrm{A}^{\:\mathrm{3}} =\:\overset{−} {\mathrm{O}}\:\: \\ $$$$\:\:\:\:\:\mathrm{Find}\:,\:\:\:\:\:\:\:\:\left(\:\mathrm{A}−\mathrm{2I}\:\right)^{\:−\mathrm{1}} =? \\ $$$$ \\ $$ Answered by mathmax…
Question Number 162951 by Ar Brandon last updated on 02/Jan/22 $$\mathrm{4men}\:\mathrm{clear}\:\mathrm{a}\:\mathrm{farm}\:\mathrm{for}\:\mathrm{8}\:\mathrm{days}\:\mathrm{and}\:\mathrm{are}\:\mathrm{paid}\:\mathrm{24\$} \\ $$$$\mathrm{How}\:\mathrm{long}\:\mathrm{will}\:\mathrm{6}\:\mathrm{men}\:\mathrm{take}\:\mathrm{to}\:\mathrm{clear}\:\mathrm{the}\:\mathrm{same}\:\mathrm{farm} \\ $$$$\mathrm{if}\:\mathrm{they}\:\mathrm{are}\:\mathrm{paid}\:\mathrm{360\$}\:? \\ $$ Commented by mr W last updated on 03/Jan/22…
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 $${E}\left({x}\right)\:{denotes}\:{the}\:{integer}\:{part}\:{of}\:{x}\: \\ $$$$\left.{x}\in\right]\mathrm{0};\mathrm{1}\left[\:{determine}:\right. \\ $$$$\boldsymbol{{E}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}} \right)\:\boldsymbol{{and}}\:\boldsymbol{{E}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \right)\: \\ $$$$\boldsymbol{{calcul}}\:\boldsymbol{{li}}\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\boldsymbol{{E}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \right) \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{i}}\:\boldsymbol{{need}}\:\boldsymbol{{help}}\:\boldsymbol{{please}} \\…
Question Number 162390 by cortano last updated on 29/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162366 by cortano last updated on 29/Dec/21 $$\:{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…