Question Number 60445 by ANTARES VY last updated on 21/May/19 Commented by bhanukumarb2@gmail.com last updated on 21/May/19 $${cauchy}\:{inequality}\: \\ $$ Answered by MJS last updated…
Question Number 191512 by mnjuly1970 last updated on 25/Apr/23 $$ \\ $$$$\:\:\:\:\:{Q}:\:\:\:\:\:\:\:{the}\:{equation}\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\lfloor\:\mathrm{cos}\left(\mathrm{4}{x}\:\right)\rfloor={m}.\mathrm{cos}\left(\mathrm{2}{x}\right) \\ $$$$\:\:\:{has}\:{no}\:\:{solution}\:.\:\:{x}\in\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:{find}\:{the}\:{acceptable} \\ $$$$\:\:\:\:\:\:\:{real}\:{values}\:{for}\:\:\:\:''{m}''. \\ $$ Answered…
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Question Number 125960 by AST last updated on 26/Sep/22 $$ \\ $$ Answered by mahdipoor last updated on 16/Dec/20 $$\begin{cases}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mid{a}\mid^{\mathrm{2}} +\mid{b}\mid^{\mathrm{2}} }\\{\mathrm{2}{ab}\leqslant\mathrm{2}\mid{a}\mid.\mid{b}\mid}\end{cases} \\…
Question Number 125950 by Mathgreat last updated on 15/Dec/20 Commented by Mathgreat last updated on 15/Dec/20 $${x},{y},{z}\:=\:? \\ $$ Answered by mahdipoor last updated on…
Question Number 60416 by Tawa1 last updated on 20/May/19 $$\mathrm{Sum}\:\mathrm{the}\:\mathrm{series}:\:\:\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{0}} \overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{1}} \:+\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{1}} \overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{2}} \:+\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{2}} \overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{3}} \:+\:…\:+\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}} \overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}\:+\:\mathrm{1}}…
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Question Number 191486 by MATHEMATICSAM last updated on 24/Apr/23 $${a}\sqrt{{a}}\:+\:{b}\sqrt{{b}}\:=\:\mathrm{183}\:\mathrm{and}\:{b}\sqrt{{a}}\:+\:{a}\sqrt{{b}}\:=\:\mathrm{182} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{9}}{\mathrm{5}}\:\left({a}\:+\:{b}\right)\:? \\ $$ Answered by mr W last updated on 24/Apr/23 $${let}\:{A}=\sqrt{{a}} \\ $$$${B}=\sqrt{{b}}…
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Question Number 60406 by Tawa1 last updated on 20/May/19 $$\mathrm{n}\:\in\:\mathbb{Z}^{+} ,\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\:\mathrm{x}^{−\mathrm{1}} \:\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\:\:\left(\mathrm{1}\:+\:\mathrm{x}\right)^{\mathrm{n}} \left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{n}} \\ $$ Answered by mr W last updated on 20/May/19 $$\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{n}}…