Question Number 200860 by liuxinnan last updated on 25/Nov/23 $${when}\:{x}\in\left[\mathrm{0},\infty\right) \\ $$$$\left({x}+{m}\right){e}^{\mathrm{2}{x}} −{x}^{\mathrm{2}} \geqslant\left({x}+{m}\right){ln}\left({x}+{m}\right) \\ $$$${m}\in? \\ $$ Answered by witcher3 last updated on 25/Nov/23…
Question Number 200861 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $${J}=_{{x}={t}^{\frac{\mathrm{1}}{\mathrm{10}}} } \frac{\mathrm{1}}{\mathrm{10}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{5}}{\mathrm{1}+{t}}{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty}…
Question Number 200862 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $$\Phi \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{10}} }{dx}=_{{x}={t}^{\frac{\mathrm{1}}{\mathrm{10}}} } \frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}}…
Question Number 200863 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $${x}^{\mathrm{2}\pi} ={t}\:\Rightarrow{x}={t}^{\frac{\mathrm{1}}{\mathrm{2}\pi}} \:{and} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{\pi}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}\pi} }{dx}=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}}…
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Question Number 200859 by liuxinnan last updated on 25/Nov/23 $$\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…