Question Number 156448 by MathSh last updated on 11/Oct/21 $$\mathrm{If}\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b}<\pi\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{sin}\sqrt{\mathrm{ab}}}{\mathrm{sin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)}\:\geqslant\:\frac{\mathrm{32a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \sqrt{\mathrm{ab}}}{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{5}} } \\ $$ Answered by ghimisi last updated on 11/Oct/21 $${f}\left({t}\right)=\frac{{sint}}{{t}^{\mathrm{5}}…
Question Number 156447 by MathSh last updated on 11/Oct/21 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{functions}\:\:\mathrm{f}\::\:\mathbb{Z}\:\rightarrow\:\mathbb{R}\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{n}+\mathrm{m}\right)=\mathrm{nf}\left(\mathrm{n}\right)+\mathrm{mf}\left(\mathrm{m}\right)+\mathrm{nm}-\mathrm{n}-\mathrm{m} \\ $$$$\forall\mathrm{n};\mathrm{m}\in\mathbb{Z} \\ $$ Answered by ghimisi last updated on 11/Oct/21 $${m}=\mathrm{1};{n}=\mathrm{0}\Rightarrow{f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)−\mathrm{1}\Rightarrow\mathrm{0}=−\mathrm{1} \\…
Question Number 156423 by cortano last updated on 11/Oct/21 Commented by john_santu last updated on 11/Oct/21 $${answer}\:=\:\frac{\mathrm{81}\left(\sqrt{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}}\:{sq}\:{units} \\ $$ Commented by cortano last updated on…
Question Number 156404 by ajfour last updated on 10/Oct/21 $$\:\:\mathrm{x}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{cx}=\mathrm{m}+\mathrm{px}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{2x}^{\mathrm{4}} +\left(\mathrm{b}+\mathrm{p}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{m}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} =−\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)\pm\sqrt{\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{m}+\mathrm{d}}{\mathrm{2}}\right)} \\ $$$$\left(\mathrm{p}−\mathrm{b}\right)\mathrm{x}^{\mathrm{2}}…
Question Number 156400 by ajfour last updated on 10/Oct/21 Commented by ajfour last updated on 10/Oct/21 $$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\:\:\mathrm{f}\left(\mathrm{z}\right)=\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{2}} +\mathrm{b} \\ $$$$\mathrm{is}\:\mathrm{obtainable}\:\mathrm{by}\:\mathrm{the}\:\mathrm{welded} \\ $$$$\mathrm{parabola}\:\mathrm{wires}\:\left(\mathrm{with}\:\mathrm{mutually}\right. \\ $$$$\mathrm{perpendicular}\:\mathrm{planes};\:\:\mathrm{upper} \\…
Question Number 25317 by ibraheem160 last updated on 08/Dec/17 $${solvd}\:{for}\:{x}:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} +\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{{x}} =\mathrm{4} \\ $$ Answered by ajfour last updated on 08/Dec/17 $${let}\:\:\:{u}=\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\mathrm{ln}\:{u}={x}\mathrm{ln}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\…
Question Number 90842 by jagoll last updated on 26/Apr/20 $${x}^{\mathrm{2}} −\left({y}−{z}\right)^{\mathrm{2}} \:=\:\mathrm{3} \\ $$$${y}^{\mathrm{2}} \:−\:\left({z}−{x}\right)^{\mathrm{2}} \:=\:\mathrm{5} \\ $$$${z}^{\mathrm{2}} \:−\:\left({x}−{y}\right)^{\mathrm{2}} \:=\:\mathrm{12} \\ $$ Commented by john…
Question Number 156361 by mr W last updated on 10/Oct/21 $$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{1}}}\:=\mathrm{1}−\:\frac{\mathrm{2}}{\mathrm{x}} \\ $$ Commented by cortano last updated on 11/Oct/21 Commented by…
Question Number 156362 by MathSh last updated on 10/Oct/21 Answered by mindispower last updated on 10/Oct/21 $$\mid\Omega_{\mathrm{1}} +{i}\Omega_{\mathrm{2}} \mid^{\mathrm{2}} =\Omega_{\mathrm{1}} ^{\mathrm{2}} +\Omega_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{1}}…
Question Number 25283 by Tinkutara last updated on 07/Dec/17 Commented by Tinkutara last updated on 07/Dec/17 $${Solve}\:{for}\:{x}. \\ $$ Answered by naka3546 last updated on…