Question Number 162866 by HongKing last updated on 01/Jan/22 Answered by aleks041103 last updated on 02/Jan/22 $$\begin{vmatrix}{\mathrm{1}}&{{k}}&{{k}}\\{\mathrm{2}{n}}&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}&{{k}^{\mathrm{2}} +{k}}\\{\mathrm{2}{n}−\mathrm{1}}&{{k}^{\mathrm{2}} }&{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\end{vmatrix}= \\ $$$$=\mathrm{1}.\begin{vmatrix}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}&{{k}^{\mathrm{2}} +{k}}\\{{k}^{\mathrm{2}}…
Question Number 162859 by HongKing last updated on 01/Jan/22 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\int}}\:\frac{\mathrm{e}^{\boldsymbol{\mathrm{cos}}\:\mathrm{2}\boldsymbol{\mathrm{x}}} \:\centerdot\:\mathrm{sin}\left(\mathrm{x}\:+\:\mathrm{sin}\:\mathrm{2x}\right)}{\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:=\:\frac{\pi{e}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162856 by HongKing last updated on 01/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162854 by HongKing last updated on 01/Jan/22 $$\mathrm{let}\:\:\mathrm{a};\mathrm{b};\mathrm{c}\geqslant\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}\:-\:\mathrm{1}}{\:\sqrt{\mathrm{b}\:+\:\mathrm{3}}}\:+\:\frac{\mathrm{b}\:-\:\mathrm{1}}{\:\sqrt{\mathrm{c}\:+\:\mathrm{3}}}\:+\:\frac{\mathrm{c}\:-\:\mathrm{1}}{\:\sqrt{\mathrm{a}\:+\:\mathrm{3}}}\:\geqslant\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31771 by rahul 19 last updated on 14/Mar/18 $${Consider}\:{a}\:{sequence}\:{in}\:{the}\:{form}\:{of} \\ $$$${groups}\:\left(\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{4},\mathrm{4},\mathrm{4}\right), \\ $$$$\left(\mathrm{5},\mathrm{5},\mathrm{5},\mathrm{5},\mathrm{5}\right),………… \\ $$$${then}\:{the}\:\mathrm{2000}{th}\:{term}\:{of}\:{the}\:{above}\: \\ $$$${sequence}\:{is}\::\:? \\ $$ Commented by Joel578 last…
Question Number 97306 by Abdulrahman last updated on 07/Jun/20 $$\mathrm{if}\:\:\:\:\:\:\:\:\:\mathrm{sin14}=\mathrm{x} \\ $$$$\mathrm{then} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{22}−\mathrm{cos}^{\mathrm{2}} \mathrm{8}=? \\ $$ Commented by john santu last updated on…
Question Number 97303 by bobhans last updated on 07/Jun/20 $$\boldsymbol{\mathrm{G}}\mathrm{iven}\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} \:=\:\mathrm{0}\:,\:\mathrm{y}_{\mathrm{1}} \:+\:\mathrm{y}_{\mathrm{2}} +\mathrm{y}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{x}_{\mathrm{1}} \mathrm{y}_{\mathrm{1}} +\:\mathrm{x}_{\mathrm{2}} \mathrm{y}_{\mathrm{2}} \:+\:\mathrm{x}_{\mathrm{3}} \mathrm{y}_{\mathrm{3}} \:=\:\mathrm{0}\:.\:\mathrm{The}\:\mathrm{value} \\…
Question Number 31763 by RAMANUJAN last updated on 14/Mar/18 $${please}\:{find}\:{the}\:{integral}\:{solutions}\:\left({x}\:{and}\:{y}\right)\: \\ $$$$\left({xy}−\mathrm{7}\right)^{\mathrm{2}} \:={x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \\ $$ Answered by MJS last updated on 14/Mar/18 $$\left(−\mathrm{7},\mathrm{0}\right) \\…
Question Number 162819 by mnjuly1970 last updated on 01/Jan/22 $$ \\ $$$$\:\:\:\:\:#{combinatorial}\:{mathematics}# \\ $$$$\:\:\:\:\:\:\mathrm{I}{n}\:{how}\:{many}\:\:{subsets}\:{of}\:\mathrm{10}\: \\ $$$${members}\:{of}\:\:{the}\:{set}\:,\:\left\{\:\mathrm{1},\:\mathrm{2},\:…,\:\mathrm{20}\:\right\} \\ $$$$\:\:\:{is}\:{there}\:{no}\:{difference}\:{between} \\ $$$$\:{two}\:{members}\:{of}\:\:\:\:\mathrm{5}\:?\: \\ $$$${a}:\:\:\mathrm{2}^{\:\mathrm{10}} \:\:\:\:\:\:{b}\::\:\:\mathrm{3}^{\:\mathrm{5}} \:\:\:\:\:\:\:{c}\::\:\:\:\mathrm{2}^{\:\mathrm{8}} \:\:\:\:\:\:\:\:{d}:\:\:\:\mathrm{10}^{\:\mathrm{4}}…
Question Number 162814 by amin96 last updated on 01/Jan/22 Answered by MathsFan last updated on 01/Jan/22 $$\:{let}\:\:{log}_{\mathrm{3}} {x}={u}\:\rightarrow\:{x}=\mathrm{3}^{{u}} \\ $$$$\:\mathrm{2}^{{u}} =\sqrt{\mathrm{3}^{{u}} }+\mathrm{1} \\ $$$$\:\mathrm{2}^{{u}} −\sqrt{\mathrm{3}^{{u}}…