Question Number 20739 by Tinkutara last updated on 02/Sep/17 $$\mathrm{If}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mathrm{2},\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{3},\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{4}\:\mathrm{and} \\ $$$$\mid\mathrm{2}{z}_{\mathrm{1}} \:+\:\mathrm{3}{z}_{\mathrm{2}} \:+\:\mathrm{4}{z}_{\mathrm{3}} \mid\:=\:\mathrm{4},\:\mathrm{then}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mid\mathrm{8}{z}_{\mathrm{2}} {z}_{\mathrm{3}} \:+\:\mathrm{27}{z}_{\mathrm{3}} {z}_{\mathrm{1}} \:+\:\mathrm{64}{z}_{\mathrm{1}} {z}_{\mathrm{2}}…
Question Number 151806 by liberty last updated on 23/Aug/21 Commented by ghimisi last updated on 23/Aug/21 $${x}={y}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by ghimisi last…
Question Number 20726 by Tinkutara last updated on 01/Sep/17 $$\mathrm{Five}\:\mathrm{distinct}\:\mathrm{2}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{geometric}\:\mathrm{progression}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{middle} \\ $$$$\mathrm{term}. \\ $$ Answered by dioph last updated on 02/Sep/17 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{GP}\:\mathrm{be}\:\left\{{a},{qa},{q}^{\mathrm{2}} {a},{q}^{\mathrm{3}}…
Question Number 20723 by rajpurohithakshay999@gmail.com last updated on 01/Sep/17 $$\int\sqrt{{x}}\:.{sinx}.{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151789 by mathdanisur last updated on 23/Aug/21 Answered by ArielVyny last updated on 23/Aug/21 $${we}\:{have}\:\mathrm{0}\leqslant\lambda\leqslant\mathrm{1}\rightarrow\mathrm{0}\leqslant\lambda+{a}+{b}\leqslant\mathrm{1}+{a}+{b} \\ $$$${and}\:{we}\:{admit}\:{that}\:{a}+{b}\geqslant\lambda+\mathrm{1};{b}+{c}\geqslant\lambda+\mathrm{1} \\ $$$${c}+{a}\geqslant\lambda+\mathrm{1}\:\:\left({note}\:{that}\:{abc}=\mathrm{1}\right) \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}\leqslant\frac{\mathrm{1}}{\lambda+{a}+{b}}\:\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}}…
Question Number 151791 by mathdanisur last updated on 23/Aug/21 Answered by ghimisi last updated on 23/Aug/21 $${x}_{{i}} ^{\mathrm{4}} +\mathrm{1}\geqslant\mathrm{2}{x}_{{i}} ^{\mathrm{2}} \Rightarrow{x}_{{i}} ^{\mathrm{4}} −{x}_{{i}} ^{\mathrm{2}} +\mathrm{1}\geqslant{x}_{{i}}…
Question Number 151790 by mathdanisur last updated on 23/Aug/21 Answered by Olaf_Thorendsen last updated on 24/Aug/21 $${u}_{\mathrm{1}} \:=\:\sqrt{\mathrm{99}} \\ $$$${u}_{{n}} \:=\:\sqrt{\mathrm{102}−\mathrm{3}{n}+{u}_{{n}−\mathrm{1}} } \\ $$$${u}_{\mathrm{33}} \:=\:\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+…+\sqrt{\mathrm{96}+\sqrt{\mathrm{99}}}}}}…
Question Number 151782 by gloriousman last updated on 23/Aug/21 $$ \\ $$$$\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c}\geqslant\mathrm{0}\:\mathrm{and}\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{y}−\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{z}+\mathrm{2}}{\mathrm{3}}, \\ $$$$\mathrm{min}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)=? \\ $$$$ \\ $$ Answered by liberty last…
Question Number 151768 by mathdanisur last updated on 22/Aug/21 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}{\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Aug/21…
Question Number 86230 by M±th+et£s last updated on 27/Mar/20 $${is}\:\:\left(−\mathrm{1}\right)^{\frac{{m}}{{n}}} \:=\left(\sqrt[{{n}}]{\left(−\mathrm{1}\:\right)^{{m}} }\right)\:{or}\:=\left(\sqrt[{{n}}]{−\mathrm{1}}\right)^{{m}} \\ $$$${or}\:{both}\:{of}\:{them}\:{are}\:{fault}\:{and}\:{why}\:? \\ $$ Answered by MJS last updated on 27/Mar/20 $$\left(−\mathrm{1}\right)^{\frac{{m}}{{n}}} =\left(\mathrm{e}^{\mathrm{i}\pi}…