Question Number 84293 by M±th+et£s last updated on 11/Mar/20 $${solve}\:{in}\:{R} \\ $$$${x}^{\left[{x}\right]} +{x}^{\mathrm{2}−\left[{x}\right]} ={x}^{\mathrm{2}} +\mathrm{1} \\ $$ Answered by TANMAY PANACEA last updated on 12/Mar/20…
Question Number 149817 by ajfour last updated on 07/Aug/21 $$\:\:\:{x}^{\mathrm{3}} −{x}={c} \\ $$$$\:\:{let}\:\:{x}={t}+{h} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{t}\left({t}^{\mathrm{2}} +\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={p} \\ $$$$\mathrm{3}{ht}^{\mathrm{2}}…
Question Number 149813 by mathdanisur last updated on 07/Aug/21 $$\int\left(−\mathrm{1}\right)^{\left[\boldsymbol{\mathrm{x}}\right]} \:\mathrm{dx}\:=\:? \\ $$ Commented by DonQuichote last updated on 07/Aug/21 $${deal}:\:{you}\:{show}\:{me}\:{your}\:{solution}\:{idea}\:{first} \\ $$$${then}\:{I}'{ll}\:{show}\:{you}\:{mine} \\ $$…
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