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 $${C}_{\mathrm{0}} ^{{n}} +^{{n}}…
Question Number 200275 by sonukgindia last updated on 16/Nov/23 Answered by Mathspace last updated on 16/Nov/23 $${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{acosx}+\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{a}\frac{{e}^{{ix}}…
Question Number 200268 by cortano12 last updated on 16/Nov/23 Answered by ajfour last updated on 16/Nov/23 $${D}\:{origin}. \\ $$$${O}_{\mathrm{2}} \equiv\left({s}−{r},\:{s}−{r}\right) \\ $$$${O}_{\mathrm{3}} \equiv\left({r},\:{s}−{r}\right) \\ $$$${let}\:\:{eqn}\:{of}\:{line}\:{DF}\:{be}\:…
Question Number 200270 by ajfour last updated on 16/Nov/23 Commented by mr W last updated on 17/Nov/23 $${i}\:{think}\:{one}\:{vertex}\:{must}\:{lie}\:{at}\:{the} \\ $$$${center}\:{of}\:{a}\:{circle}\:{as}\:{in}\:{diagram}.\:{so} \\ $$$${s}_{{max}} ={radius}=\mathrm{1}. \\ $$…
Question Number 200265 by cherokeesay last updated on 16/Nov/23 Answered by witcher3 last updated on 16/Nov/23 $$\begin{cases}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}.\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}\right)=\mathrm{0}}\\{\left(\mathrm{2}\right)\Leftrightarrow\left(\mathrm{2}\right)}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0}}\\{\sqrt{\left(\mathrm{x}^{\mathrm{2}}…