Question Number 149793 by mathdanisur last updated on 07/Aug/21 $$\mathrm{Compare}: \\ $$$$\mathrm{tan}\left(\mathrm{11}°\right)\:\:\:\mathrm{and}\:\:\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$ Commented by DonQuichote last updated on 07/Aug/21 $$\frac{\mathrm{1}}{\mathrm{5}}\in\mathbb{Q}\:{but}\:{tan}\:\mathrm{11}°\:\notin\mathbb{Q} \\ $$$${what}\:{else}\:{can}\:{I}\:{tell}\:{you}? \\…
Question Number 149795 by mathdanisur last updated on 07/Aug/21 $$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}\centerdot\mathrm{3}\centerdot\mathrm{5}\centerdot\mathrm{7}\centerdot\:…\:\centerdot\left(\mathrm{2n}-\mathrm{1}\right)}{\mathrm{2}\centerdot\mathrm{4}\centerdot\mathrm{6}\centerdot\:…\:\centerdot\mathrm{2n}}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 07/Aug/21 $$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}…..\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2}.\mathrm{4}.\mathrm{6}….\left(\mathrm{2n}\right)}\:=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2}^{\mathrm{n}} \:\mathrm{n}!}…
Question Number 149781 by mathdanisur last updated on 07/Aug/21 $$\mathrm{a}\:\:−\:\:\sqrt{\frac{\mathrm{20}}{\mathrm{a}}}\:=\:\mathrm{7}\:\:\Rightarrow\:\:\sqrt{\mathrm{5a}}\:−\:\mathrm{a}\:=\:? \\ $$ Commented by amin96 last updated on 07/Aug/21 $$\sqrt{\frac{\mathrm{20}}{{a}}}={a}−\mathrm{7}\:\:\:\Rightarrow\:\:\sqrt{\mathrm{5}{a}}=\frac{{a}^{\mathrm{2}} −\mathrm{7}{a}}{\mathrm{2}}\:\: \\ $$$$\sqrt{\mathrm{5}{a}}−{a}=\frac{\mathrm{5}{a}−{a}^{\mathrm{2}} }{\:\sqrt{\mathrm{5}{a}}+{a}}=\frac{\mathrm{5}{a}−{a}^{\mathrm{2}} }{\frac{{a}^{\mathrm{2}}…
Question Number 18704 by gourav~ last updated on 28/Jul/17 Answered by Principal last updated on 28/Jul/17 $$\mathrm{Circumference}\:\mathrm{of}\:\mathrm{circular}\:\mathrm{wire} \\ $$$$=\:\mathrm{2}\pi×\mathrm{7}.\mathrm{5}\:=\:\mathrm{15}\pi\:\mathrm{cm} \\ $$$$\theta\:=\:\frac{{l}}{{r}}\:=\:\frac{\mathrm{15}\pi\:\mathrm{cm}}{\mathrm{120}\:\mathrm{cm}}\:=\:\frac{\pi}{\mathrm{8}}\:\mathrm{rad}\:=\:\mathrm{22}.\mathrm{5}° \\ $$ Terms of…
Question Number 149778 by mathdanisur last updated on 07/Aug/21 $$\mathrm{Simplify}: \\ $$$$\sqrt{\mathrm{8}\sqrt{\mathrm{24}\sqrt{\mathrm{8}\sqrt{\mathrm{24}…}}}}\:\:=\:? \\ $$ Commented by amin96 last updated on 07/Aug/21 $$\sqrt{\mathrm{8}\sqrt{\mathrm{24}{x}}}={x}\:\:\Rightarrow\:\:\:{x}^{\mathrm{2}} =\mathrm{8}\sqrt{\mathrm{24}{x}}\Rightarrow\:\:{x}^{\mathrm{4}} =\mathrm{64}\centerdot\mathrm{24}{x} \\…
Question Number 149775 by mathdanisur last updated on 07/Aug/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:=\:\left(\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \right)^{\mathrm{2}} \\ $$ Commented by amin96 last updated on 07/Aug/21 $$\left(\mathrm{9}−\mathrm{4}\right)^{{x}}…
Question Number 149768 by mathdanisur last updated on 07/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149770 by mathdanisur last updated on 07/Aug/21 $$\mathrm{6}\boldsymbol{{z}}^{\mathrm{4}} \:=\:\boldsymbol{{z}}^{\boldsymbol{{z}}^{\mathrm{2}} } \:\:\Rightarrow\:\:\mathrm{1}\:+\:\boldsymbol{{z}}^{\mathrm{2}} \:+\:\boldsymbol{{z}}^{\mathrm{4}} \:=\:? \\ $$ Commented by amin96 last updated on 07/Aug/21 $${z}^{\mathrm{2}}…
Question Number 149766 by mathdanisur last updated on 07/Aug/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}}\:=\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$ Commented by amin96 last updated on 07/Aug/21 $${x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{y}\:\:\:\:\Rightarrow\:\:{y}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{10}}{\mathrm{3}}…
Question Number 84188 by jagoll last updated on 10/Mar/20 $$\mathrm{if}\:\mathrm{2x}+\mathrm{3y}\:=\:\mathrm{2020}? \\ $$$$\mathrm{find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{3x}+\mathrm{2y}\:\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{natural} \\ $$$$\mathrm{number} \\ $$ Commented by jagoll last updated on 10/Mar/20 $$\mathrm{this}\:\mathrm{diopthantine}\:\mathrm{equation}\:? \\…