Question Number 123362 by bemath last updated on 25/Nov/20 Answered by liberty last updated on 25/Nov/20 Commented by mnjuly1970 last updated on 25/Nov/20 $${nice}\:{very}\:{nice}\:{sir}\:{liberty}\:… \\…
Question Number 188898 by Shrinava last updated on 08/Mar/23 $$\mathrm{In}\:\bigtriangleup\mathrm{ABC}\:\mathrm{holds}:\:\:\:\Sigma\:\frac{\mathrm{2}\:+\:\sqrt{\mathrm{3}}\:\mathrm{tan}\:\frac{\mathrm{B}}{\mathrm{2}}}{\mathrm{1}\:+\:\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\mathrm{A}}{\mathrm{2}}}\:\geqslant\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 57818 by Tawa1 last updated on 12/Apr/19 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19 $$\frac{{a}^{} +{b}+{c}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{27}}\geqslant{abc}\:\:\:{let}\:{abc}=\frac{\mathrm{1}}{\mathrm{27}}−\delta \\ $$$${LHS} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{a}^{\mathrm{2}} }{{b}}+\frac{{b}^{\mathrm{2}}…
Question Number 188879 by Nimnim111118 last updated on 08/Mar/23 $${Find}\:{the}\:{sum}\:{of}\:{all}\:{three}\:{digit}\:{numbers} \\ $$$${started}\:{with}\:{odd}\:{number}\:{when}\:{each}\:{digit} \\ $$$${are}\:{different}. \\ $$$$ \\ $$$${Please}\:{help}… \\ $$ Commented by mr W last…
Question Number 188864 by Mingma last updated on 08/Mar/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 57791 by Tawa1 last updated on 12/Apr/19 $$\:\mathrm{If}\:\:\:\:\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\:=\:\:\mathrm{1}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \:\:=\:\:\mathrm{3}\:\: \\ $$$$\mathrm{then}\:\:\:\:\:\:\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \:+\:\mathrm{c}^{\mathrm{5}\:\:} =\:\:?…
Question Number 123265 by Study last updated on 24/Nov/20 Commented by Study last updated on 24/Nov/20 $${which}\:{is}\:{a}\:{polynomial}?\:{why}? \\ $$ Commented by mathmax by abdo last…
Question Number 188786 by Rupesh123 last updated on 07/Mar/23 Commented by Rupesh123 last updated on 07/Mar/23 Prove that: Answered by mr W last updated on 07/Mar/23…
Question Number 188776 by depressiveshrek last updated on 06/Mar/23 $$\mathrm{Prove}\:\mathrm{that}\:{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2} \\ $$$$\mathrm{for}\:\mathrm{any}\:{n}\in\mathbb{Z} \\ $$ Commented by Frix last updated on 08/Mar/23 $$\mathrm{2}\mid{n}\:\Rightarrow\:\mathrm{2}\mid{n}^{\mathrm{2}} \wedge\mathrm{2}\mid\mathrm{3}{n} \\…
Question Number 188774 by Shrinava last updated on 06/Mar/23 Answered by aleks041103 last updated on 09/Mar/23 $${Diagonlize}\:{A},\:{i}.{e}.\:{A}={P}^{−\mathrm{1}} {DP},\:{where}\:{D} \\ $$$${is}\:{diagonal}. \\ $$$${Now}\:{you}\:{can}\:{chose}\:{as}\:{many}\:{as}\:{you}\:{want} \\ $$$${two}\:{more}\:{diagonal}\:{matrices}\:{D}_{\mathrm{1}} \:{and}\:{D}_{\mathrm{2}}…