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 $${C}=\mathrm{10}{m}^{\mathrm{3}} {p}^{\mathrm{2}} +{b}\left({m}^{\mathrm{3}} +\mathrm{6}{m}^{\mathrm{2}} {p}+\mathrm{3}{mp}^{\mathrm{2}} \right) \\ $$$$\:+{c}\left(\mathrm{3}{m}^{\mathrm{2}} +\mathrm{6}{mp}+{p}^{\mathrm{2}}…
Question Number 68414 by behi83417@gmail.com last updated on 10/Sep/19 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{any}\:\mathrm{value}\:\mathrm{for} \\ $$$$\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:\mathrm{from}\:\mathrm{below}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equetions}? \\ $$$$\begin{cases}{\boldsymbol{\mathrm{sina}}+\boldsymbol{\mathrm{sinb}}=\boldsymbol{\mathrm{sinc}}}\\{\boldsymbol{\mathrm{cosa}}+\boldsymbol{\mathrm{cosb}}=\boldsymbol{\mathrm{cosc}}}\end{cases} \\ $$ Commented by kaivan.ahmadi last updated on 10/Sep/19 $$\begin{cases}{{sinacosa}+{sinbcosa}={sinccosa}}\\{−{sinacosa}−{sinacosb}=−{sinacosc}}\end{cases}\Rightarrow \\…
Question Number 2874 by RasheedAhmad last updated on 29/Nov/15 $${What}\:{is}\:{the}\:{meaning}\:{of} \\ $$$$\left({i}\right)\:{z}\rightarrow\mathrm{0}\:\:\:\left({ii}\right)\:{z}\rightarrow{z}_{\mathrm{0}} \:\:\:{z},{z}_{\mathrm{0}} \in\mathbb{C} \\ $$ Answered by Filup last updated on 29/Nov/15 $$\left(\mathrm{1}\right) \\…
Question Number 2845 by Rasheed Soomro last updated on 28/Nov/15 $$\mathcal{W}{hile}\:{you}\:{are}\:{in}\:{between}\:{the}\:{project} \\ $$$$\mathcal{I}\:{am}\:{trying}\:{to}\:{improve}\:{my}\:{digestiblity}\:{to} \\ $$$${digest}\:{the}\:{concept}\:{of}\:'{analytical}\:{continuation}'. \\ $$$$ \\ $$$${First}\:{we}\:{make}\:{aformula}\:{to}\:{sum}\:{n}\:{terms}\:{of}\:{a}\:{powe}\:{series}: \\ $$$$\frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} \\ $$$${latter}\:{we}\:{change}\:{it}\:{for}\:\mid{x}\mid<\mathrm{1}\:{and}\:{n}\rightarrow\infty\:\left[{x}^{{n}}…
Question Number 2843 by prakash jain last updated on 28/Nov/15 $$\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$${n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left({n}+{r}−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divible}\:\mathrm{by}\:{r}! \\ $$$$ \\ $$$${n},{r}\in\mathbb{N}. \\ $$ Commented by 123456 last updated on…