Question Number 222794 by MathematicalUser2357 last updated on 07/Jul/25 $$\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222787 by Nicholas666 last updated on 07/Jul/25 $$ \\ $$$$\:\:\:\:\:\int_{\mathrm{1}} ^{\:\pi/\mathrm{2}} \:\:\frac{\mathrm{4}^{−{x}} \:\centerdot\:{e}^{\mathrm{tan}\left({x}+{x}^{\mathrm{2}} \right)} \centerdot\:\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{3}} \right)}{\mathrm{1}\:+\:{x}}\:\:\mathrm{d}{x}\:\:\:\:\: \\ $$$$ \\ $$ Answered by gabthemathguy25…
Question Number 222744 by ErnestoOrlandoLisboa last updated on 06/Jul/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222747 by fantastic last updated on 06/Jul/25 Commented by fantastic last updated on 06/Jul/25 $${r}\:{in}\:{terms}\:{of}\:{R}\:{and}\:\alpha \\ $$ Answered by mr W last updated…
Question Number 222743 by hardmath last updated on 06/Jul/25 $$\mathrm{Compare}: \\ $$$$\boldsymbol{\mathrm{a}}\:=\:\mathrm{arcctg}\:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{b}}\:=\:\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{c}}\:=\:\mathrm{arctg}\:\sqrt{\mathrm{2}} \\ $$ Answered by mr W last updated on…
Question Number 222736 by Lekhraj last updated on 06/Jul/25 Answered by Raphael254 last updated on 08/Sep/25 $$\begin{array}{|c|c|}{\boldsymbol{{To}}\:\boldsymbol{{a}}\:\boldsymbol{{number}}\:\boldsymbol{{be}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{33},}\\{\boldsymbol{{it}}\:\boldsymbol{{needs}}\:\boldsymbol{{to}}\:\boldsymbol{{be}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{3}\:\boldsymbol{{and}}\:\mathrm{11}}\\\hline\end{array} \\ $$$$ \\ $$$${Why}? \\ $$$$ \\ $$$${ab}\mid{n}\:\Rightarrow\:{a}\mid{n}\:{and}\:{b}\mid{n}…
Question Number 222733 by gabthemathguy25 last updated on 06/Jul/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222730 by fantastic last updated on 06/Jul/25 Commented by fantastic last updated on 06/Jul/25 $${find}\:{x}\:{in}\:{terms}\:{of}\: \\ $$$$\alpha,\beta,\gamma, \\ $$ Answered by mr W…
Question Number 222756 by Osefavour last updated on 06/Jul/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}}{\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{xcos}}\left(\sqrt{\boldsymbol{\mathrm{x}}}\right)} \\ $$ Answered by gregori last updated on 07/Jul/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}\left(\mathrm{1}−\mathrm{cos}\:\sqrt{{x}}\:\right)\left(\mathrm{1}+\sqrt{\mathrm{cos}\:{x}}\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}}…
Question Number 222757 by chidera last updated on 06/Jul/25 Terms of Service Privacy Policy Contact: info@tinkutara.com