Question Number 2967 by Karting7442 last updated on 01/Dec/15 $${Powers}\:{of}\:{Monomials}\:\:\:\:\:\:{Alg}. \\ $$$$ \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{5}}{a}^{\mathrm{6}} {b}^{\mathrm{9}} \right)^{\mathrm{2}} \\ $$ Answered by Filup last updated on 02/Dec/15…
Question Number 134034 by mr W last updated on 27/Feb/21 $${how}\:{many}\:{zeros}\:{has}\:{the}\:{number} \\ $$$$\mathrm{1000}!\:{at}\:{the}\:{end}?\:{and}\:{what}\:{is}\:{the} \\ $$$${last}\:{digit}\:{before}\:{these}\:{zeros}? \\ $$ Answered by floor(10²Eta[1]) last updated on 27/Feb/21 $$\lfloor\frac{\mathrm{1000}}{\mathrm{5}}\rfloor+\lfloor\frac{\mathrm{1000}}{\mathrm{5}^{\mathrm{2}}…
Question Number 134005 by mr W last updated on 26/Feb/21 $${solve}\:{x}^{\mathrm{3}} −\mathrm{2}\lfloor{x}\rfloor=\mathrm{5} \\ $$ Answered by MJS_new last updated on 26/Feb/21 $${x}={i}\left[\mathrm{nteger}\:\mathrm{part}\right]+{f}\left[\mathrm{ractal}\:\mathrm{part}\right] \\ $$$$\left({i}+{f}\right)^{\mathrm{3}} −\mathrm{2}{i}=\mathrm{5}…
Question Number 2936 by Rasheed Soomro last updated on 30/Nov/15 $${Suppose}\:{I}\:{want}\:{to}\:{prove}\:{a}\:{statement},{say}\:{P}\left({n}\right),\: \\ $$$${for}\:{natural}\:{numbers}\:\boldsymbol{{mathematical}}\:\boldsymbol{{induction}} \\ $$$${is}\:{a}\:{proper}\:{tool}\:{for}\:{me}. \\ $$$$\mathcal{I}{f}\:{a}\:{P}\left({n}\right)\:{is}\:{required}\:{to}\:{prove}\:{for}\:{natural}\:{n}\geqslant{c}\: \\ $$$${mathematical}\:{induction}\:{is}\:{again}\:{a}\:{tool}\:{of}\:{proof}. \\ $$$$ \\ $$$$\mathcal{N}{ow}\:{suppose}\:{I}\:{have}\:{a}\:{statement}\:{which}\:{is}\:{true} \\ $$$${only}\:{for}\:\:\underset{−}…
Question Number 133996 by mr W last updated on 26/Feb/21 $${show}\:{that}\:\sqrt{\mathrm{2}}<\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\sqrt{\mathrm{3}} \\ $$ Answered by som(math1967) last updated on 26/Feb/21 $$\:\:\mathrm{2}^{\sqrt{\mathrm{2}}} <\mathrm{3}<\mathrm{2}^{\sqrt{\mathrm{3}}} \\ $$$${log}_{\mathrm{2}}…
Question Number 133991 by EDWIN88 last updated on 26/Feb/21 $$\:\mid\:\mathrm{3x}−\mid\:\mathrm{4x}+\mathrm{2}\:\mid\mid\:\geqslant\:\mathrm{4}\:>\:\mid\:\mathrm{5x}+\mathrm{8}\:\mid\: \\ $$ Answered by benjo_mathlover last updated on 26/Feb/21 $$\:\mid\mathrm{5x}+\mathrm{8}\mid\:<\:\mathrm{4}\:\leqslant\:\mid\:\mathrm{3x}−\mid\mathrm{4x}+\mathrm{2}\mid\mid \\ $$$$\left(\bullet\right)\:\mid\mathrm{5x}+\mathrm{8}\mid\:<\:\mathrm{4}\:\Rightarrow−\mathrm{4}<\mathrm{5x}+\mathrm{8}\:<\:\mathrm{4} \\ $$$$\Rightarrow−\mathrm{12}\:<\:\mathrm{5x}\:<\:−\mathrm{4}\:;\:−\frac{\mathrm{12}}{\mathrm{5}}\:<\mathrm{x}\:<−\frac{\mathrm{4}}{\mathrm{5}} \\…
Question Number 2917 by Rasheed Soomro last updated on 30/Nov/15 $$\mathcal{W}{hat}\:{is}\:{DMAS}\:\:{rule}?\:{Where}\:{is}\:{it}\:{followed}? \\ $$$${Do}\:{we}\:{follow}\:{this}\:{rule}?\: \\ $$ Answered by 123456 last updated on 30/Nov/15 $$\mathrm{if}\:\mathrm{i}\:\mathrm{understanded}\:\mathrm{it}\:\mathrm{about}\:\mathrm{order}\:\mathrm{of}\:\mathrm{operation} \\ $$$$\mathrm{like}…
Question Number 133981 by bemath last updated on 26/Feb/21 $$\mathrm{If}\:{x}\:=\:\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:\mathrm{then}\:\frac{{x}−\mathrm{1}}{\:\sqrt{{x}}}\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 26/Feb/21 $${x}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:=\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$…
Question Number 68433 by mind is power last updated on 10/Sep/19 $${hello} \\ $$$${i}\:{search}\:{som}\:{lectur}\:{about}\:{hypergeometric}\:{fonction}\mathrm{2}{F}_{\mathrm{1}} \left({a},{b},{c},{x}\right)=\frac{\Gamma\left({c}\right)}{\Gamma\left({a}\right)\Gamma\left({b}\right)}\sum_{{n}\geqslant\mathrm{0}} \frac{\Gamma\left({a}+{n}\right)\Gamma\left({b}+{n}\right)}{\Gamma\left({c}+{n}\right){n}!}{x}^{{n}} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 68422 by ajfour last updated on 10/Sep/19 Answered by ajfour last updated on 10/Sep/19 $${C}=\mathrm{10}{m}^{\mathrm{3}} {p}^{\mathrm{2}} +{b}\left({m}^{\mathrm{3}} +\mathrm{6}{m}^{\mathrm{2}} {p}+\mathrm{3}{mp}^{\mathrm{2}} \right) \\ $$$$\:+{c}\left(\mathrm{3}{m}^{\mathrm{2}} +\mathrm{6}{mp}+{p}^{\mathrm{2}}…