Question Number 26133 by abdo imad last updated on 21/Dec/17 $${find}\:{the}\:{value}\:{of}\:\:\left({C}_{{n}} ^{\mathrm{0}\:\:} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{1}} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{2}} \right)^{\mathrm{2}} \:+…\left({C}_{{n}} ^{{n}} \right)^{\mathrm{2}} . \\ $$…
Question Number 157197 by MathSh last updated on 20/Oct/21 Answered by TheSupreme last updated on 20/Oct/21 $${I}=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{z}} {e}^{−{x}^{\mathrm{2}} } {dx}=\mathrm{0}.\mathrm{4} \\ $$$${I}^{\mathrm{2}} =\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{{z}}…
Question Number 157196 by MathSh last updated on 20/Oct/21 Answered by mindispower last updated on 21/Oct/21 $$\frac{\pi^{\mathrm{3}} }{\mathrm{64}}{ln}\left(\mathrm{2}\right)−\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\mathrm{tan}^{−\mathrm{1}} \left({x}\right)^{\mathrm{2}} {dx} \\ $$$$=−\mathrm{3}\int_{\mathrm{0}}…
Question Number 91660 by Zainal Arifin last updated on 02/May/20 Commented by Prithwish Sen 1 last updated on 02/May/20 $$\mathrm{LHS} \\ $$$$\sqrt{\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{21}}}{\mathrm{2}}}\:+\:\sqrt{\frac{\mathrm{16}+\mathrm{2}\sqrt{\mathrm{55}}}{\mathrm{2}}} \\ $$$$=\:\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}\:+\:\frac{\sqrt{\mathrm{5}}+\sqrt{\mathrm{11}}}{\:\sqrt{\mathrm{2}}}\:\:\:\:\boldsymbol{\mathrm{considering}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{value}} \\…
Question Number 26117 by ibraheem160 last updated on 20/Dec/17 $$\frac{\mathrm{1}}{\mathrm{6}}\sqrt{\frac{\mathrm{3}{log}\mathrm{1728}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{36}+\frac{\mathrm{1}}{\mathrm{3}}{log}\mathrm{8}}} \\ $$$${simplify}\:{the}\:{question}\:{above} \\ $$ Answered by ajfour last updated on 20/Dec/17 $$=\frac{\mathrm{1}}{\mathrm{6}}\sqrt{\frac{\mathrm{9log}\:\mathrm{12}}{\mathrm{1}+\mathrm{log}\:\mathrm{12}}}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{log}\:\mathrm{3}+\mathrm{2log}\:\mathrm{2}}{\mathrm{1}+\mathrm{log}\:\mathrm{3}+\mathrm{2log}\:\mathrm{2}}}\:. \\…
Question Number 157184 by cortano last updated on 20/Oct/21 $${p}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{ax}^{\mathrm{2}} +\left(\mathrm{3}{a}^{\mathrm{2}} +\mathrm{1}\right){x}−\left({a}^{\mathrm{3}} +{a}\right) \\ $$$${p}\left(\mathrm{2}\right)<\mathrm{0} \\ $$$${prove}\:{that}\:\mathrm{2}<{a}<\mathrm{3} \\ $$ Answered by ghimisi last updated…
Question Number 26115 by ktomboy1992 last updated on 21/Dec/17 $$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{3}\:\mathrm{thrn}\:\mathrm{find}\:\mathrm{the}\:\mathrm{valu}\:\mathrm{of}\left(\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \\ $$ Answered by $@ty@m last updated on 20/Dec/17 $$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{3}−\mathrm{2}=\mathrm{1}…
Question Number 91647 by mr W last updated on 02/May/20 $${solve} \\ $$$${x}\lfloor{x}\lfloor{x}\lfloor{x}\rfloor\rfloor\rfloor=\mathrm{2020} \\ $$ Answered by frc2crc last updated on 03/May/20 $${i}\:{used}\:{desmos}\:{x}\approx−\mathrm{6}.\mathrm{623} \\ $$…
Question Number 157163 by MathSh last updated on 20/Oct/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{solution} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{7}\centerdot\left(\mathrm{14xy}\:+\:\mathrm{3}\right) \\ $$ Commented by Rasheed.Sindhi last updated on 22/Oct/21…
Question Number 26067 by Joel578 last updated on 19/Dec/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{2}\:+\:\mathrm{3}^{\mathrm{2}} }{\mathrm{1}!\:+\:\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!}\:+\:\frac{\mathrm{3}\:+\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!\:+\:\mathrm{5}!}\:+\:…\:+\:\frac{\mathrm{2013}\:+\:\mathrm{2014}^{\mathrm{2}} }{\mathrm{2012}!\:+\:\mathrm{2013}!\:+\:\mathrm{2014}!\:+\:\mathrm{2015}!} \\ $$ Answered by iv@0uja last updated on 19/Dec/17 $$\:\:\:\:\:\frac{\mathrm{2}\:+\:\mathrm{3}^{\mathrm{2}}…