Question Number 200251 by Calculusboy last updated on 16/Nov/23 Answered by Mathspace last updated on 16/Nov/23 $${f}\left({x},{y}\right)={sin}\left({e}^{{xy}} +{e}^{{x}} \right)\:\Rightarrow \\ $$$$\frac{\partial}{\partial{x}}{f}\left({x},{y}\right)=\frac{\partial}{\partial{x}}\left({e}^{{xy}} +{e}^{{x}} \right){cos}\left({e}^{{xy}} +{e}^{{x}} \right)…
Question Number 200304 by Calculusboy last updated on 16/Nov/23 Answered by Sutrisno last updated on 17/Nov/23 $${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty}…
Question Number 200240 by universe last updated on 16/Nov/23 $$\mathrm{the}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$${f}\left({x},{y}\right)\:=\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{3}{y}+\mathrm{11} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200242 by cherokeesay last updated on 16/Nov/23 Answered by cortano12 last updated on 17/Nov/23 $$\:\frac{\left({ab}+{c}\right){x}}{{b}−\frac{{c}}{{a}}\:}\:−\frac{\left({ab}−{c}\right){x}}{{b}+\frac{{c}}{{a}}}\:=\:\frac{{ab}−{c}}{{b}+\frac{{c}}{{a}}}\:−\frac{{ab}+{c}}{{b}−\frac{{c}}{{a}}} \\ $$$$\:\frac{{ax}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{ax}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{a}\left({ab}−{c}\right)}{{ab}+{c}}\:−\frac{{a}\left({ab}+{c}\right)}{{ab}−{c}} \\ $$$$\:\:\frac{{x}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{x}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{ab}−{c}}{{ab}+{c}}−\frac{{ab}+{c}}{{ab}−{c}} \\ $$$$\:\frac{\left({ab}+{c}\right)\left({x}+\mathrm{1}\right)}{{ab}−{c}}\:=\:\frac{\left({ab}−{c}\right)\left({x}+\mathrm{1}\right)}{{ab}+{c}} \\ $$$$\:\left({ab}+{c}\right)^{\mathrm{2}}…
Question Number 200300 by Calculusboy last updated on 16/Nov/23 Answered by MM42 last updated on 17/Nov/23 $${lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\left[{ln}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{1}} \right)+{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} +…+{ln}\left(\mathrm{1}+\left(\frac{{n}}{{n}}\right)^{\mathrm{2}} \right)\right.\right. \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\underset{{i}=\mathrm{1}} {\overset{{n}}…
Question Number 200236 by mr W last updated on 16/Nov/23 $${what}\:{is}\:{the}\:{smallest}\:{natural}\:{number} \\ $$$${which}\:{has}\:{at}\:{least}\:\mathrm{100}\:{divisors}? \\ $$ Answered by MM42 last updated on 16/Nov/23 $$\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{4}} ×\mathrm{5}×\mathrm{7}=\mathrm{45360}\:\checkmark…
Question Number 200302 by Calculusboy last updated on 16/Nov/23 Answered by Rasheed.Sindhi last updated on 18/Nov/23 $$\sqrt{{a}+{bx}}\:+\sqrt{{b}+{cx}}\:+\sqrt{{c}+{ax}}\:\:=\sqrt{{b}−{ax}}\:+\sqrt{{c}−{bx}}\:+\sqrt{{a}−{cx}}\: \\ $$$${a}+{bx}\geqslant\mathrm{0}\:\wedge\:{b}+{cx}\geqslant\mathrm{0}\:\wedge\:{c}+{ax}\geqslant\mathrm{0} \\ $$$$\left({a}+{b}+{c}\right)+\left({a}+{b}+{c}\right){x}\geqslant\mathrm{0}\Rightarrow{x}\geqslant−\frac{{a}+{b}+{c}}{{a}+{b}+{c}}=−\mathrm{1} \\ $$$$\begin{array}{|c|}{{x}\geqslant−\mathrm{1}}\\\hline\end{array} \\ $$$${b}−{ax}\geqslant\mathrm{0}\:\wedge\:{c}−{bx}\geqslant\mathrm{0}\:\wedge\:{a}−{cx}\geqslant\mathrm{0}…
Question Number 200298 by Calculusboy last updated on 16/Nov/23 Commented by Frix last updated on 17/Nov/23 $$−\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 200299 by Calculusboy last updated on 16/Nov/23 Answered by witcher3 last updated on 17/Nov/23 $$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\Omega=\int_{\infty} ^{\mathrm{0}} −\frac{\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{x}}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} }.\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{2}}…
Question Number 200284 by depressiveshrek last updated on 16/Nov/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{set}\:{A}\:\mathrm{containing}\:{n} \\ $$$$\mathrm{elements},\:\mid\mathcal{P}\left({A}\right)\mid=\mathrm{2}^{{n}} . \\ $$ Answered by AST last updated on 16/Nov/23 $${C}_{\mathrm{0}} ^{{n}} +^{{n}}…