Question Number 196097 by leicianocosta last updated on 18/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196099 by leicianocosta last updated on 18/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196098 by sonukgindia last updated on 18/Aug/23 Answered by mr W last updated on 18/Aug/23 $${x}^{\frac{\mathrm{1}}{\mathrm{2}^{{x}^{\mathrm{4}} } }} ={x}^{\frac{\mathrm{1}}{\mathrm{4}^{{x}^{\mathrm{2}} } }} \\ $$$$\Rightarrow{x}=\mathrm{0}…
Question Number 196063 by Rodier97 last updated on 17/Aug/23 $$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{calcul}\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\sqrt{\mathrm{u}}\:.\mathrm{arctan}\left(\mathrm{u}\right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$ \\ $$$$…
Question Number 196062 by qaz last updated on 17/Aug/23 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left[{gcd}\left({i},{j}\right)=\mathrm{1}\right]=?\:\:\:\:\:\:\:\:\:,\left[{D}\right]=\begin{cases}{\mathrm{1},\:\:\:{D}\:{is}\:{ture}.}\\{\mathrm{0},\:\:\:\:{D}\:{is}\:{false}.\:\:\:}\end{cases} \\ $$ Commented by mr W last updated on 17/Aug/23 $${i}\:{don}'{t}\:{think}\:{we}\:{can}\:{get}\:{it}\:{in}\:{a}\:{closed}…
Question Number 196085 by sonukgindia last updated on 17/Aug/23 Answered by MM42 last updated on 17/Aug/23 $$\left(\mathrm{1}^{\mathrm{17}} +\mathrm{2023}^{\mathrm{17}} \right)+\left(\mathrm{2}^{\mathrm{17}} +\mathrm{2022}^{\mathrm{17}} \right)+…+\mathrm{1012}^{\mathrm{17}} = \\ $$$$\left(\mathrm{1}+\mathrm{2023}\right){k}_{\mathrm{1}} +\left(\mathrm{2}+\mathrm{2022}\right){k}_{\mathrm{2}}…
Question Number 196081 by sonukgindia last updated on 17/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196049 by CrispyXYZ last updated on 17/Aug/23 $${a},\:{b},\:{c}\:>\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of} \\ $$$$\underset{\mathrm{cyc}} {\sum}\:\sqrt{\frac{{a}+{b}}{{a}+{b}+{c}}}\:. \\ $$ Commented by Frix last updated on 17/Aug/23 $${b}={pa}\wedge{c}={qa} \\ $$$$\Sigma=\frac{\sqrt{{p}+\mathrm{1}}+\sqrt{{p}+{q}}+\sqrt{{q}+\mathrm{1}}}{\:\sqrt{{p}+{q}+\mathrm{1}}}…
Question Number 196035 by sonukgindia last updated on 16/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195996 by sonukgindia last updated on 15/Aug/23 Answered by Rasheed.Sindhi last updated on 15/Aug/23 $${Let}\:{radius}\:{of}\:{blue}\:{quarter}={r} \\ $$$${and}\:{of}\:{green}={R} \\ $$$${R}={r}+\mathrm{1} \\ $$$${Diagonal}={R}+{r} \\ $$$$\left({R}+{r}\right)^{\mathrm{2}}…