Question Number 171145 by floor(10²Eta[1]) last updated on 08/Jun/22 $$\mathrm{let}\:\mathrm{p}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}\:\mathrm{constant}\:\mathrm{polynomial}\:\mathrm{with}\: \\ $$$$\mathrm{degree}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that}:\:\mathrm{p}\left(\mathrm{z}\right)=\mathrm{a}_{\mathrm{n}} \mathrm{z}^{\mathrm{n}} +…+\mathrm{a}_{\mathrm{0}} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{z}\in\mathbb{C}\:\mathrm{with}\mid\mathrm{z}\mid\geqslant\mathrm{max}\left\{\mathrm{1},\frac{\mathrm{2}}{\mid\mathrm{a}_{\mathrm{n}} \mid}\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mid\mathrm{a}_{\mathrm{i}} \mid\right\} \\ $$$$\mathrm{then}\:\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{a}_{\mathrm{n}} \mid\mid\mathrm{z}\mid^{\mathrm{n}} \leqslant\mid\mathrm{p}\left(\mathrm{z}\right)\mid\leqslant\frac{\mathrm{3}}{\mathrm{2}}\mid\mathrm{a}_{\mathrm{n}} \mid\mid\mathrm{z}\mid^{\mathrm{n}}…
Question Number 105605 by bobhans last updated on 30/Jul/20 $${prove}\:{by}\:{mathematical}\:{induction}\: \\ $$$$\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} +…+\left(\mathrm{2}{n}\right)^{\mathrm{3}} =\:\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$ Answered by bemath last updated…
Question Number 171116 by vonem1 last updated on 08/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 40042 by abdo mathsup 649 cc last updated on 15/Jul/18 $$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:\: \\ $$$${p}\left({x}\right)=\left(\mathrm{1}+{ix}\:+{x}^{\mathrm{2}} \right)^{{n}} −\left(\mathrm{1}−{ix}+{x}^{\mathrm{2}} \right)^{{n}} \:{with}\:{n}\:{integr} \\ $$$${natural} \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right)\:{inside}\:\:{C}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{give}\:{p}\left({x}\right)\:{at}\:{form}\:\:\Sigma\:{a}_{{p}}…
Question Number 171109 by akolade last updated on 08/Jun/22 Commented by Rasheed.Sindhi last updated on 09/Jun/22 $$\underset{\left(\mathrm{1}\right)} {\underbrace{{a}+{b}=\mathrm{2}}},\:\underset{\left({ii}\right)} {\underbrace{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{4}}},\:{a}^{\mathrm{2019}} +{b}^{\mathrm{2019}} =? \\ $$$$\left({i}\right)\Rightarrow\frac{{a}}{{b}}+\mathrm{1}=\frac{\mathrm{2}}{{b}}\Rightarrow\begin{array}{|c|}{\frac{{a}}{{b}}=\frac{\mathrm{2}−{b}}{{b}}….\left({A}\right)}\\\hline\end{array}…
Question Number 171085 by balirampatel last updated on 07/Jun/22 $$\mathrm{43}\:{devided}\:{by}\:{x}\:{remainder}\:{is}\:{x}−\mathrm{5}\:{how}\:{many}\:{value}\:{of}\:{x}? \\ $$ Answered by Rasheed.Sindhi last updated on 08/Jun/22 $$ \\ $$$${Assuming}: \\ $$$$\:\:\mathrm{1}<{divisor}<\mathrm{43}\:\wedge\:\mathrm{0}\leqslant{remainder}<\mathrm{43} \\…
Question Number 105534 by I want to learn more last updated on 29/Jul/20 Commented by I want to learn more last updated on 29/Jul/20 $$\mathrm{the}\:\mathrm{bracket}\:\mathrm{is}\:\mathrm{a}\:\mathrm{floor}\:\mathrm{function}\:\mathrm{sir}.\:\:\mathrm{error}\:\mathrm{in}\:\mathrm{bracket}.…
Question Number 105536 by I want to learn more last updated on 29/Jul/20 Answered by adhigenz last updated on 29/Jul/20 $$\alpha+\beta\:=\:\mathrm{5},\:\alpha\beta\:=\:−\mathrm{2} \\ $$$${x}_{{n}} \:=\:\alpha^{{n}} −\beta^{{n}}…
Question Number 39995 by behi83417@gmail.com last updated on 14/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18 $$\alpha+\beta={a}+{b}+\mathrm{1}\:\:\:\alpha\beta={a}+{b} \\ $$$$\alpha+\beta−\left({a}+{b}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\alpha+\beta−\alpha\beta−\mathrm{1}=\mathrm{0} \\ $$$$\alpha\left(\mathrm{1}−\beta\right)−\left(\mathrm{1}−\beta\right)=\mathrm{0} \\ $$$$\left(\alpha−\mathrm{1}\right)\left(\mathrm{1}−\beta\right)=\mathrm{0}…
Question Number 171046 by Beginner last updated on 06/Jun/22 Answered by thfchristopher last updated on 07/Jun/22 $$\mathrm{Let}\:{u}={a}+{bx} \\ $$$${du}={bdx},\:{x}=\frac{{u}−{a}}{{b}} \\ $$$$\therefore\int\frac{{x}^{\mathrm{2}} }{\left({a}+{bx}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{3}}…