Question Number 157258 by MathSh last updated on 21/Oct/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}\boldsymbol{\mathrm{k}}} \left(\mathrm{x}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}\boldsymbol{\mathrm{k}}} \left(\mathrm{x}\right)}\:=\:\mathrm{8}\:\:\:;\:\:\:\mathrm{k}\in\mathbb{Z} \\ $$ Commented by mr W last updated on 21/Oct/21 $$\left(\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}}…
Question Number 157231 by mnjuly1970 last updated on 21/Oct/21 $$ \\ $$$$\:\:\:\:\:\:\mathcal{SOLVE}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:+\:\lfloor\mathrm{2}{x}\:\rfloor\:+\lfloor\:\mathrm{3}{x}\:\rfloor=\:\mathrm{1} \\ $$$$−−−−−−−−−− \\ $$$$ \\ $$ Answered by Javokhir…
Question Number 91688 by s.ayeni14@yahoo.com last updated on 02/May/20 $${solve}\:{equations}\:{x}^{{x}\:} +\:{y}^{{y}} \:=\:\mathrm{31}\:{and}\:{x}\:+\:{y}\:=\:\mathrm{5} \\ $$ Commented by mr W last updated on 02/May/20 $${i}\:{think}\:{you}\:{know}\:{by}\:{yourself}\:{x}=\mathrm{2},\:{y}=\mathrm{3} \\ $$$${since}\:\mathrm{2}^{\mathrm{2}}…
Question Number 26153 by bbbbbb last updated on 21/Dec/17 $${ratio}\:{of}\:{income}\:{of}\:{two}\:{persons}\:{is} \\ $$$$\mathrm{9}\:{is}\:{to}\:\mathrm{7}.{ratio}\:{of}\:{their}\:{expenses} \\ $$$${is}\:\mathrm{4}\:{is}\:{to}\:\mathrm{3}\:.{every}\:{person}\:{saves}\: \\ $$$${rupees}\:\mathrm{200}.\:{find}\:{income}\:{of}\:{each}. \\ $$ Answered by ajfour last updated on 21/Dec/17…
Question Number 157222 by amin96 last updated on 21/Oct/21 Answered by Dimitri_01 last updated on 21/Oct/21 $$\mathrm{A}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{25}} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{100}\right)}=\frac{\mathrm{1}}{\mathrm{100}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{25}} {\sum}}\frac{\left({n}+\mathrm{100}\right)−{n}}{{n}\left({n}+\mathrm{100}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{100}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{25}} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{100}}\right)=\frac{\mathrm{1}}{\mathrm{100}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{101}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{102}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{103}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{25}}−\frac{\mathrm{1}}{\mathrm{125}}\right)…
Question Number 157223 by amin96 last updated on 21/Oct/21 Answered by Rasheed.Sindhi last updated on 21/Oct/21 $$\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)}…