Question Number 143979 by mathdanisur last updated on 20/Jun/21 Answered by mitica last updated on 20/Jun/21 $${p}={x}+{y}+{z};{q}={xy}+{yz}+{xz};{r}={xyz}\Rightarrow{p}^{\mathrm{2}} \geqslant\mathrm{3}{q} \\ $$$$\mathrm{3}\left({p}+\frac{\mathrm{3}{r}}{{q}}\right)^{\mathrm{4}} =\mathrm{3}\left(\frac{{p}}{\mathrm{3}}+\frac{{p}}{\mathrm{3}}+\frac{{p}}{\mathrm{3}}+\frac{\mathrm{3}{r}}{{q}}\right)^{\mathrm{4}} \overset{{am}−{gm}} {\geqslant} \\ $$$$\geqslant\mathrm{3}\left(\mathrm{4}\centerdot\sqrt[{\mathrm{4}}]{\frac{{p}}{\mathrm{3}}\centerdot\frac{{p}}{\mathrm{3}}\centerdot\frac{{p}}{\mathrm{3}}\centerdot\frac{\mathrm{3}{r}}{{q}}}\right)^{\mathrm{4}}…
Question Number 78440 by ajfour last updated on 17/Jan/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}\:=\:{p} \\ $$$$\:\:\left({a}−\mathrm{2}{b}\right)\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)=−\mathrm{24}{q} \\ $$$${Find}\:\:\:\frac{{a}+{b}}{\mathrm{3}}\:\:\:\:{in}\:{terms}\:{of}\:{p},\:{q}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12905 by 433 last updated on 06/May/17 $$\mathrm{sin}\:\left({x}\right)=\mathrm{3cos}\:\left({x}\right)\:\Leftrightarrow\mathrm{tan}\:\left({x}\right)=\mathrm{3} \\ $$$${True}\:{or}\:{false}? \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 $$\left.\mathrm{1}\right){false}\:{if}:\:{cosx}=\mathrm{0}\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){true}\:{if}\:{x}\neq\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}},{k}\in\boldsymbol{{Z}}\:. \\…
Question Number 12903 by tawa last updated on 06/May/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 $$\boldsymbol{{y}}^{'} =\mathrm{6}\boldsymbol{{x}}−\mathrm{1}\Rightarrow\boldsymbol{{m}}_{\boldsymbol{{t}}} =\mathrm{6}×\mathrm{1}−\mathrm{1}=\mathrm{5} \\ $$$$\boldsymbol{{y}}−\boldsymbol{{y}}_{\boldsymbol{{t}}} =\boldsymbol{{m}}_{\boldsymbol{{t}}} \left(\boldsymbol{{x}}−\boldsymbol{{x}}_{\boldsymbol{{t}}} \right)\Rightarrow\boldsymbol{{y}}−\mathrm{2}=\mathrm{5}\left(\boldsymbol{{x}}−\mathrm{1}\right) \\…
Question Number 12902 by tawa last updated on 06/May/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 $$\left.\mathrm{1}\right)\:\boldsymbol{{c}}=\mathrm{0}\:\:\:\:\:\:\:\left(\boldsymbol{{x}}=\mathrm{0}\Rightarrow\boldsymbol{{y}}=\mathrm{0}\right) \\ $$$$\boldsymbol{{y}}^{'} =\mathrm{2}\boldsymbol{{ax}}+\boldsymbol{{b}}\:\Downarrow \\ $$$$\left.\mathrm{2}\right)\begin{cases}{\mathrm{2}\boldsymbol{{a}}+\boldsymbol{{b}}=\mathrm{4}}\\{\mathrm{4}\boldsymbol{{a}}+\boldsymbol{{b}}=\mathrm{5}}\end{cases}\Rightarrow\mathrm{2}\boldsymbol{{a}}=\mathrm{1}\Rightarrow\boldsymbol{{a}}=\frac{\mathrm{1}}{\mathrm{2}},\boldsymbol{{b}}=\mathrm{3} \\ $$$$\boldsymbol{{y}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}\:\:\:.\:\blacksquare…
Question Number 143970 by ZiYangLee last updated on 20/Jun/21 $$\mathrm{Given}\:\mathrm{that}\:\omega\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}, \\ $$$$\omega^{\mathrm{7}} =\mathrm{1},\:\omega\neq\mathrm{1},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\omega^{\mathrm{1}} +\omega^{\mathrm{2}} +\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{5}} +\omega^{\mathrm{6}} . \\ $$ Answered by…
Question Number 143965 by Khalmohmmad last updated on 20/Jun/21 Commented by Canebulok last updated on 20/Jun/21 $$\boldsymbol{{Solution}}: \\ $$$$\Rightarrow\:\frac{{log}\left({x}\right)}{{log}\left(\mathrm{3}\right)}\:+\:\frac{{log}\left(\mathrm{5}\right)}{{log}\left({x}\right)}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{log}\left({x}\right)^{\mathrm{2}} \:+\:{log}\left(\mathrm{5}\right){log}\left(\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\: \\…
Question Number 12895 by tawa last updated on 06/May/17 Commented by tawa last updated on 06/May/17 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$ Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on…
Question Number 143964 by naka3546 last updated on 20/Jun/21 $${x}_{\mathrm{1}} \:{and}\:\:{x}_{\mathrm{2}} \:\:{are}\:\:{solutions}\:\:{of}\:\:{equality}\:: \\ $$$$\:\:\mathrm{cos}\:\left(\frac{\pi{x}+\pi}{\mathrm{6}}\right)\:−\:\mathrm{sin}\:\left(\frac{\pi{x}−\pi}{\mathrm{6}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\mathrm{3}}\:\:\:,\:\:\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{12} \\ $$$${Find}\:\:{the}\:\:{value}\:\:{of}\:\:{x}_{\mathrm{1}} +\:{x}_{\mathrm{2}} \:. \\ $$ Commented by Canebulok last updated…
Question Number 12894 by tawa last updated on 06/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 $${S}=\pi\frac{{D}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}×\pi\frac{{D}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}}=\frac{\pi}{\mathrm{4}}\left(\mathrm{84}^{\mathrm{2}} −\mathrm{3}×\mathrm{42}^{\mathrm{2}} \right)=\mathrm{441}\pi\:.\blacksquare \\ $$ Commented…