Question Number 211276 by mnjuly1970 last updated on 04/Sep/24 $$ \\ $$$$\:\:{Prove}\:,\:{in}\:{A}\overset{\Delta} {{B}C}\:\::\: \\ $$$$ \\ $$$$\:\:\:\:\:\frac{{cosA}}{{sin}^{\mathrm{2}} {A}}\:+\:\frac{{cosB}}{{sin}^{\mathrm{2}} {B}}\:\:+\frac{{cosC}}{{sin}^{\mathrm{2}} {C}}\:\geqslant\:\frac{{r}}{{R}} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{r}\::\:{incircle}\:\:{radius} \\…
Question Number 211277 by mr W last updated on 04/Sep/24 Commented by mr W last updated on 04/Sep/24 $${find}\:{the}\:{volume}\:{of}\:{the}\:{tetrahedron} \\ $$$${in}\:{terms}\:{of}\:{a},{b},{c},\:\theta\:{with}\:\mathrm{0}<\theta\leqslant\mathrm{120}°. \\ $$ Commented by…
Question Number 211279 by Davidtim last updated on 04/Sep/24 Commented by Davidtim last updated on 04/Sep/24 $${T}_{\mathrm{1}} =? \\ $$$${T}_{\mathrm{2}} =? \\ $$$${T}_{\mathrm{3}} =? \\…
Question Number 211262 by mathlove last updated on 03/Sep/24 Answered by mahdipoor last updated on 03/Sep/24 $${f}\left({m}\right)+{f}\left(\mathrm{1}−{m}\right)=\frac{{a}^{{m}−\mathrm{1}} \:}{{a}^{{m}} +{b}}+\frac{{a}^{−{m}} }{{a}^{\mathrm{1}−{m}} +{b}}= \\ $$$$\frac{{a}^{{m}−\mathrm{1}} \left({a}^{\mathrm{1}−{m}} +{b}\right)+{a}^{−{m}}…
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Question Number 211258 by otchereabdullai@gmail.com last updated on 02/Sep/24 Answered by A5T last updated on 02/Sep/24 $$\Rightarrow{tan}\mathrm{2}\theta=\frac{\mathrm{28}}{\mathrm{9}}\Rightarrow\theta=\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{28}}{\mathrm{9}}\right)}{\mathrm{2}}\approx\mathrm{36}.\mathrm{091}° \\ $$ Terms of Service Privacy Policy…
Question Number 211252 by RojaTaniya last updated on 02/Sep/24 Answered by Frix last updated on 02/Sep/24 $$\left[\mathrm{1}\right]×\left({x}−{y}\right)\:\Rightarrow\:{x}^{\mathrm{3}} −{y}^{\mathrm{3}} =\mathrm{39}\left({x}−{y}\right)\:\:\left[\mathrm{1}{a}\right] \\ $$$$\left[\mathrm{2}\right]×\left({y}−{z}\right)\:\Rightarrow\:{y}^{\mathrm{3}} −{z}^{\mathrm{3}} =\mathrm{49}\left({y}−{z}\right)\:\:\left[\mathrm{2}{a}\right] \\ $$$$\left[\mathrm{3}\right]×\left({z}−{x}\right)\:\Rightarrow\:{z}^{\mathrm{3}}…
Question Number 211255 by ajfour last updated on 02/Sep/24 $$\sqrt{{a}+\sqrt{{b}−{x}}+\sqrt{{b}−\sqrt{{a}+{x}}}}=\mathrm{2}{x} \\ $$$${solve}\:{for}\:{x}.\:\:\:\: \\ $$ Commented by Ghisom last updated on 03/Sep/24 $$\sqrt{{a}+\sqrt{{b}−{x}}+\sqrt{{b}−\sqrt{{a}+{x}}}}=\mathrm{2}{x} \\ $$$${x}=\left({b}−\left(\mathrm{4}{x}^{\mathrm{2}} −\sqrt{{b}−{x}}−{a}\right)^{\mathrm{2}}…
Question Number 211245 by Ghisom last updated on 01/Sep/24 $$\mathrm{prove}: \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}^{\alpha−\mathrm{1}} }{{t}^{\pi} +\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{sin}\:\alpha} \\ $$ Answered by Frix last updated on 02/Sep/24…
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