Question Number 157223 by amin96 last updated on 21/Oct/21 Answered by Rasheed.Sindhi last updated on 21/Oct/21 $$\underset{−} {\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }+\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 157220 by MathSh last updated on 21/Oct/21 $$\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{find}\:\:\sqrt{\mathrm{x}\:+\:\mathrm{1}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 21/Oct/21 $$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}};\:\:\sqrt{\mathrm{x}\:+\:\mathrm{1}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}}}\:=\:? \\…
Question Number 26147 by pieroo last updated on 21/Dec/17 $$\mathrm{There}\:\mathrm{are}\:\mathrm{5}\:\mathrm{more}\:\mathrm{girls}\:\mathrm{than}\:\mathrm{boys}\:\mathrm{in}\:\mathrm{a}\:\mathrm{class}.\:\mathrm{If}\:\mathrm{2}\:\mathrm{boys}\:\mathrm{join} \\ $$$$\mathrm{the}\:\mathrm{class},\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{girls}\:\mathrm{to}\:\mathrm{boys}\:\mathrm{will}\:\mathrm{be}\:\mathrm{5}:\mathrm{4}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{of}\:\mathrm{girls}\:\mathrm{in}\:\mathrm{the}\:\mathrm{class}. \\ $$ Answered by ajfour last updated on 21/Dec/17 $${g}={b}+\mathrm{5}\:={b}+\mathrm{2}+\mathrm{3}\:\:\:\:\:\:\:….\left({i}\right) \\…
Question Number 26143 by chantriachheang last updated on 21/Dec/17 $$\mathrm{2000}^{\mathrm{3000}} \:\:\boldsymbol{{vs}}\:\mathrm{3000}^{\mathrm{2000}} \\ $$$$ \\ $$$$\:\boldsymbol{{who}}\:\boldsymbol{{is}}\:\boldsymbol{{stronger}}\:? \\ $$ Commented by Tinkutara last updated on 22/Dec/17 You can look at my question number 21781. Also this link is helpful: https://artofproblemsolving.com/community/c4h1522793…
Question Number 91671 by M±th+et+s last updated on 02/May/20 $${show}\:{that} \\ $$$$\sqrt[{{ln}\left({x}\right)}]{{x}}={e} \\ $$ Commented by mr W last updated on 02/May/20 $$\sqrt[{{ln}\left({x}\right)}]{{x}}={x}^{\frac{\mathrm{1}}{\mathrm{ln}\:{x}}} ={e}^{\mathrm{ln}\:\left({x}^{\frac{\mathrm{1}}{\mathrm{ln}\:{x}}} \right)}…
Question Number 26133 by abdo imad last updated on 21/Dec/17 $${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}}}\:. \\…