Question Number 188170 by mnjuly1970 last updated on 26/Feb/23 $$ \\ $$$$\:\:\:\:\:\:{solve} \\ $$$$ \\ $$$$\:\lfloor\:{cos}\:\left({x}\:\right)\rfloor\:+\:\lfloor\:{cos}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\rfloor+\:\lfloor\:−\mathrm{2}{cosx}\:\rfloor\:=\mathrm{0} \\ $$$$ \\ $$ Answered by Frix last updated…
Question Number 188156 by Shrinava last updated on 26/Feb/23 $$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{5a}\:=\:\mathrm{6b}\:=\:\mathrm{9c} \\ $$$$\left(\mathrm{abc}\right)_{\boldsymbol{\mathrm{min}}} \:=\:? \\ $$ Answered by mr W last updated on 26/Feb/23…
Question Number 188151 by Shrinava last updated on 26/Feb/23 $$\mathrm{Simplify}: \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}\:+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }}\:+…+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2022}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2023}^{\mathrm{2}} }} \\ $$ Answered by BaliramKumar last updated…
Question Number 188126 by Shrinava last updated on 25/Feb/23 $$\mathrm{In}\:\mathrm{convex}\:\mathrm{polygon}\:\:\mathrm{ABCD} \\ $$$$\mathrm{AB}\:=\:\mathrm{10}\:\sqrt{\mathrm{6}}\:\:,\:\:\mathrm{CD}\:=\:\mathrm{18} \\ $$$$\angle\:\mathrm{ABD}\:=\:\mathrm{60}°\:\:,\:\:\angle\:\mathrm{BDC}\:=\:\mathrm{45}° \\ $$$$\mathrm{and}\:\:\mathrm{BD}\:=\:\mathrm{13}\:\sqrt{\mathrm{6}}\:+\:\mathrm{9}\:\sqrt{\mathrm{2}} \\ $$$$\mathrm{find}\:\:\mathrm{AC}\:=\:? \\ $$ Terms of Service Privacy Policy…
Question Number 122580 by mathace last updated on 18/Nov/20 $${Solve} \\ $$$${x}^{{x}} =\mathrm{2}{x} \\ $$ Commented by MJS_new last updated on 18/Nov/20 $$\mathrm{obviously}\:{x}_{\mathrm{1}} =\mathrm{2} \\…
Question Number 188100 by Shrinava last updated on 25/Feb/23 Answered by a.lgnaoui last updated on 25/Feb/23 $$\bigtriangleup{ADC}\:\:\:\:\measuredangle{CDA}={x}+\mathrm{18};\:\:{AC}={b}\:{CD}= \\ $$$$\frac{\mathrm{sin}\:\mathrm{18}}{{b}}=\frac{\mathrm{sin}\:\left({x}+\mathrm{18}\right)}{{a}}\:\Rightarrow\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}=\frac{\mathrm{sin}\:\:\left(\boldsymbol{{x}}+\mathrm{18}\right)}{\mathrm{sin}\:\mathrm{18}}\:\left(\mathrm{1}\right)\: \\ $$$$\bigtriangleup{ABC}\:\:{BD}={AC}={a}\:\:{BC}={a}+{b} \\ $$$$\:\:\:\:\frac{\mathrm{sin}\left({x}+\mathrm{18}\right)\:}{{a}+{b}}=\frac{\mathrm{sin}\:\mathrm{18}}{{a}} \\ $$$${a}\left[\mathrm{sin}\:\left({x}+\mathrm{18}\right)−\mathrm{sin}\:\mathrm{18}\right]={b}\mathrm{sin}\:\mathrm{18}…
Question Number 57024 by pete last updated on 29/Mar/19 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{2x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{10}.\:\mathrm{Find}\:\mathrm{f}\left(\mathrm{3}\right). \\ $$ Commented by Joel578 last updated on 29/Mar/19 $${f}\left(\mathrm{3}\right)\:=\:{f}\left(\mathrm{4}\:−\:\mathrm{1}\right)\:=\:\mathrm{2}\left(\mathrm{4}^{\mathrm{3}} \right)\:−\:\mathrm{3}\left(\mathrm{4}^{\mathrm{2}} \right)\:+\:\mathrm{7}\left(\mathrm{4}\right)\:+\:\mathrm{10} \\…
Question Number 122551 by shaker last updated on 18/Nov/20 Answered by bramlexs22 last updated on 18/Nov/20 $$\:\int\:\frac{\mathrm{4sin}\:^{\mathrm{2}} \mathrm{2}{x}.\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{2cos}\:^{\mathrm{2}} {x}}\:{dx}\:=\:\mathrm{2}\int\:\frac{\mathrm{4sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}.\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx} \\…
Question Number 188082 by Shrinava last updated on 25/Feb/23 $$\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial} \\ $$$$\mathrm{If}\:\:\:\mathrm{P}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:=\:\mathrm{6x}^{\mathrm{4}} \:−\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{5} \\ $$$$\mathrm{Find}\:\:\:\mathrm{P}\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{1}\right)\:=\:? \\ $$ Answered by cortano12 last updated…
Question Number 188073 by Shrinava last updated on 25/Feb/23 $$\mathrm{If}\:\:\:\mathrm{x}\:=\:\sqrt{\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\:−\:\mathrm{1}}}\:\:\:\:\:\mathrm{find}\:\:\:\:\mathrm{5x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{1}=? \\ $$ Answered by som(math1967) last updated on 25/Feb/23 $${x}=\sqrt{\frac{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\…