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Question Number 75186 by Hassen_Timol last updated on 08/Dec/19 $$\mathrm{7}{n}\:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{form}\:\mathrm{of}\:{n}\:? \\ $$ Commented by mr W last updated on 08/Dec/19 $$\mathrm{7}{n}=\mathrm{5}{m}+\mathrm{1} \\ $$$$\Rightarrow{n}=\mathrm{5}{k}+\mathrm{3}…
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Question Number 9648 by Chantria last updated on 22/Dec/16 Answered by sandy_suhendra last updated on 23/Dec/16 $$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{2}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\:>\:\mathrm{0} \\ $$$$\frac{\mathrm{x}^{\mathrm{6}} +\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}}…
Question Number 75180 by Rio Michael last updated on 08/Dec/19 $${There}\:{are}\:\mathrm{6}\:{girls}\:{and}\:{four}\:{boys}\:{in}\:{a}\:{class}. \\ $$$$\mathrm{3}\:{students}\:{are}\:{choosen}\:{at}\:{random}\:{so}\:{as}\:{to}\:{be} \\ $$$${awarded}\:{a}\:{scholaship}.{In}\:{how}\:{many}\:{ways}\:{can} \\ $$$${this}\:{be}\:{done}\:{if}\:{atlease}\:\mathrm{1}\:{boy}\:{and}\:\mathrm{1}\:{girl}\:{must} \\ $$$${in}\:{the}\:{selection} \\ $$$$ \\ $$ Commented by…
Question Number 75178 by Rio Michael last updated on 08/Dec/19 $${givn}\:{that}\:{z}\:=\:\mathrm{1}−{i}\sqrt{\mathrm{3}}\:{express}\:{z}\:{in}\:{the}\:{form}\: \\ $$$$\:{z}\:=\:{r}\left({cos}\theta\:+\:{isin}\theta\right),\:{hence}\:{express} \\ $$$${z}^{\mathrm{7}} \:{in}\:{the}\:{form}\:{re}^{{i}\theta} \\ $$ Answered by mr W last updated on…
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Question Number 140715 by mnjuly1970 last updated on 11/May/21 $$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:……{advanced}\:\:{calculus}…… \\ $$$$\:\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{2}} }{{cosh}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)}{dx}\overset{?} {=}\frac{\sqrt{\mathrm{2}}\:−\mathrm{2}}{\mathrm{4}}\:\sqrt{\pi}\:\zeta\:\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:………….. \\…
Question Number 75176 by 21042004 last updated on 08/Dec/19 $$\mathrm{P}=\sqrt{\mathrm{25}{x}−\mathrm{50}}−\mathrm{14}\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{4}}}+\sqrt{\mathrm{9}{x}−\mathrm{18}},\:\:{x}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Simplify}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{x}\:\mathrm{if}\:\mathrm{P}=\mathrm{3} \\ $$ Answered by Kunal12588 last updated on 08/Dec/19 $${P}=\sqrt{\mathrm{25}\left({x}−\mathrm{2}\right)}−\mathrm{14}\frac{\sqrt{{x}−\mathrm{2}}}{\:\sqrt{\mathrm{4}}}+\sqrt{\mathrm{9}\left({x}−\mathrm{2}\right)} \\…
Question Number 75177 by chess1 last updated on 08/Dec/19 Commented by chess1 last updated on 08/Dec/19 $$\mathrm{sir}\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power}\:\:\mathrm{please}\:\mathrm{solution} \\ $$ Commented by mind is power last…