Question Number 200859 by liuxinnan last updated on 25/Nov/23 $$\sqrt[{\mathrm{0}}]{{x}}=? \\ $$ Commented by mr W last updated on 25/Nov/23 $${not}\:{defined},\:{just}\:{like}\:\frac{\mathrm{1}}{\mathrm{0}}. \\ $$ Terms of…
Question Number 200882 by Rupesh123 last updated on 25/Nov/23 Commented by AST last updated on 26/Nov/23 $${Do}\:{you}\:{know}\:{the}\:{answer}? \\ $$ Commented by mr W last updated…
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Question Number 200876 by mrityun05867 last updated on 25/Nov/23 $${x}^{\mathrm{3cosec}\:^{\mathrm{2}} \left(\right.} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200877 by mathlove last updated on 25/Nov/23 Commented by mr W last updated on 25/Nov/23 $${we}\:{know}\:{a}^{\mathrm{log}_{{a}} \:{b}} ={b},\:{so}\:{we}\:{can}\:“{see}'' \\ $$$${that}\:{x}=\mathrm{3}\:{is}\:{a}\:{root}. \\ $$ Commented…
Question Number 200873 by cortano12 last updated on 25/Nov/23 Commented by witcher3 last updated on 26/Nov/23 $$\mathrm{tan}\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\in\left[−\sqrt{\mathrm{3}},\sqrt{\mathrm{3}}\right] \\ $$$$\mathrm{tg}\left(\mathrm{3}.\frac{\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{3tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{tg}^{\mathrm{3}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{3tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3x}}{\mathrm{2}}\right)}=\mathrm{2}+\mathrm{cos}\left(\mathrm{x}\right) \\…
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Question Number 200864 by Rydel last updated on 25/Nov/23 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{xE}\left({x}\right)+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{sin}\:{x}}} \\ $$ Answered by Mathspace last updated on 25/Nov/23 $$={lim}_{{x}\rightarrow+\infty} \frac{{xE}\left({x}\right)+\mathrm{3}}{{x}\sqrt{\mathrm{1}+\frac{{sinx}}{{x}^{\mathrm{2}} }}} \\…
Question Number 200844 by darklord last updated on 24/Nov/23 Commented by som(math1967) last updated on 24/Nov/23 $$?? \\ $$ Commented by darklord last updated on…