Question Number 161899 by HongKing last updated on 23/Dec/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:-\infty} {\overset{\:\infty} {\int}}\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:\:;\:\:\mathrm{n}\in\mathbb{Z} \\ $$ Answered by Ar Brandon last updated on…
Question Number 161884 by HongKing last updated on 23/Dec/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt[{\mathrm{7}}]{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{-\mathrm{1}} }\:=\:\mathrm{1}\:+\:\sqrt[{\mathrm{7}}]{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:-\:\mathrm{2}^{-\mathrm{1}} } \\ $$ Commented by mr W last updated on…
Question Number 96340 by Rio Michael last updated on 31/May/20 $$\mathrm{The}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{two}\:\mathrm{circles}\:{S}_{\mathrm{1}} \:\mathrm{and}\:{S}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:{S}_{\mathrm{1}} :\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}{y}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:{S}_{\mathrm{2}} :\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\mathrm{4}{x}\:+\:\mathrm{2}{y}\:+\mathrm{1}\:=\:\mathrm{0}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{S}_{\mathrm{1}}…
Question Number 161868 by cortano last updated on 23/Dec/21 $$\:\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} \:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} }\:=\:\mathrm{2}^{{x}} \\ $$$$\:\:{x}=? \\ $$ Answered by Rasheed.Sindhi last updated on 23/Dec/21 $${a}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:,\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}×\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}} \\…
Question Number 161866 by Tawa11 last updated on 23/Dec/21 Commented by mr W last updated on 23/Dec/21 $$\frac{{BC}}{\mathrm{10}}=\frac{\mathrm{6}}{\mathrm{6}+\mathrm{9}}\:\Rightarrow{BC}=\mathrm{4} \\ $$$$\frac{{AC}}{\mathrm{6}}=\frac{\mathrm{7}.\mathrm{5}}{\mathrm{9}}\:\Rightarrow{AC}=\mathrm{5} \\ $$$$\frac{{AX}}{\mathrm{6}}=\frac{\mathrm{5}−{AX}}{\mathrm{4}}\:\Rightarrow{AX}=\mathrm{3}\:\Rightarrow{XC}=\mathrm{5}−\mathrm{3}=\mathrm{2} \\ $$ Commented…
Question Number 161861 by Rasheed.Sindhi last updated on 23/Dec/21 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={n}^{\mathrm{2}} +{n}−\mathrm{2} \\ $$ Commented by mr W…
Question Number 96329 by M±th+et+s last updated on 31/May/20 $${x}\lfloor{x}\lfloor{x}\lfloor{x}\rfloor\rfloor\rfloor=\mathrm{88} \\ $$$${x}>\mathrm{0} \\ $$ Commented by prakash jain last updated on 31/May/20 $$\mathrm{88}=\mathrm{2}^{\mathrm{3}} ×\mathrm{11} \\…
Question Number 96321 by Mathudent last updated on 31/May/20 $${It}\:{is}\:{given}\:{that}\:{x}^{\mathrm{2}} =\mathrm{2}^{{x}} .\:{Find}\:{x}. \\ $$ Commented by bemath last updated on 31/May/20 $$\mathrm{look}\:\mathrm{qn}\:\mathrm{96241} \\ $$ Answered…
Question Number 161854 by mnjuly1970 last updated on 23/Dec/21 Commented by cortano last updated on 23/Dec/21 $$\:{yx}^{\mathrm{2}} −{yx}−\mathrm{6}{y}={x}^{\mathrm{2}} −{ax}+{a} \\ $$$$\:\left({y}−\mathrm{1}\right){x}^{\mathrm{2}} +\left({a}−{y}\right){x}−\left(\mathrm{6}{y}+{a}\right)=\mathrm{0} \\ $$$$\:\Delta\geqslant\mathrm{0} \\…
Question Number 161848 by mnjuly1970 last updated on 23/Dec/21 $$ \\ $$$$\:\:\:\:\:\:\mathrm{I}{f}\:\:\:\:{tan}\:\left(\alpha\:\right)=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{K}=\frac{\:\:\mathrm{1}+{sin}\left(\:\mathrm{8}\:\alpha\right)−{cos}\:\left(\mathrm{8}\alpha\:\right)}{\mathrm{1}+{sin}\left(\:\mathrm{8}\alpha\:\right)\:+\:{cos}\:\left(\mathrm{8}\:\alpha\:\right)}\:=? \\ $$$$ \\ $$$$ \\ $$ Commented…