Question Number 49253 by cesar.marval.larez@gmail.com last updated on 04/Dec/18 Answered by afachri last updated on 05/Dec/18 $$\boldsymbol{\mathrm{no}}.\:\mathrm{2} \\ $$$$\boldsymbol{{f}}'\left({x}\right)\:=\:\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\frac{\:\:\:\:\:\left(\frac{\mathrm{3}\left({x}\:+\:{h}\right)\:+\:\mathrm{2}}{\mathrm{5}\left({x}\:+\:{h}\right)\:−\:\mathrm{2}}\:\:−\:\:\frac{\mathrm{3}{x}\:+\:\mathrm{2}}{\mathrm{5}{x}\:−\:\mathrm{2}\:}\right)\:\:\:\:}{{h}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\mathrm{1}}{{h}}\left(\:\frac{\:\mathrm{3}{x}\:+\:\mathrm{3}{h}\:+\:\mathrm{2}\:}{\mathrm{5}{x}\:+\:\mathrm{5}{h}\:−\:\mathrm{2}}\:\:−\:\:\frac{\:\mathrm{3}{x}\:+\:\mathrm{2}}{\mathrm{5}{x}\:−\:\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}}…
Question Number 49251 by cesar.marval.larez@gmail.com last updated on 04/Dec/18 Commented by afachri last updated on 05/Dec/18 $$\mathrm{unfortunately}\:\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{learning}\:\mathrm{integral} \\ $$$$\mathrm{yet} \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
Question Number 49249 by maxmathsup by imad last updated on 04/Dec/18 $${smplify}\:{A}_{{np}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{cos}\left({pk}\right)\:\:{and}\:{B}_{{np}} \:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left({pk}\right)\:{with}\:{p}\:{fromN} \\ $$ Terms of Service Privacy Policy…
Question Number 49248 by maxmathsup by imad last updated on 04/Dec/18 $$\left.\mathrm{1}\right)\:{solve}\:{z}^{\mathrm{4}} =\mathrm{1}+{i}\sqrt{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{1}−{i}\sqrt{\mathrm{3}}{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){factorze}\:{inside}\:{R}\left[{x}\right]\:{the}\:{polynom}\:{p}\left({x}\right). \\ $$ Answered by Smail last updated…
Question Number 49244 by maxmathsup by imad last updated on 04/Dec/18 $${let}\:{w}\:{from}\:{C}\:{and}\:{w}^{{n}} \:=\mathrm{1}\:{find}\:{the}\:{value}\:{of}\: \\ $$$${S}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}} ^{{k}} \:{w}^{{k}} \:. \\ $$ Answered by Smail…
Question Number 49245 by maxmathsup by imad last updated on 04/Dec/18 $${solve}\:{inside}\:{C}:\:\mathrm{1}+\left({z}−\mathrm{1}\right)^{\mathrm{3}} \:+\left({z}−\mathrm{1}\right)^{\mathrm{6}} =\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 49246 by maxmathsup by imad last updated on 04/Dec/18 $${simplify}\:\:\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({e}^{{i}\frac{\mathrm{4}{k}\pi}{{n}}} \:−\mathrm{2}{cos}\theta\:{e}^{\frac{{i}\mathrm{2}\pi}{{n}}} \:+\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 49242 by maxmathsup by imad last updated on 04/Dec/18 $${let}\:{z}\:{from}\:{C}\:{and}\:\theta\:{from}\:{R}\:{and}\:{z}^{\mathrm{2}} \:+\mathrm{2}{zcos}\theta\:+\mathrm{1}\:=\mathrm{0}\:{find}\:{the}\:{value}\:{of} \\ $$$${z}^{\mathrm{2}{n}} \:+\mathrm{2}{zcos}\left({n}\theta\right)+\mathrm{1}\:. \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 49241 by maxmathsup by imad last updated on 04/Dec/18 $${let}\:{z}\:={r}\:{e}^{{i}\theta} \:\:\:{find}\:{the}\:{value}\:{of}\: \\ $$$${P}_{{n}} =\left({z}+\overset{−} {{z}}\right)\left({z}^{\mathrm{2}} \:+\overset{−^{\mathrm{2}} } {{z}}\right)…..\left({z}^{{n}} \:+\overset{−^{{n}} } {{z}}\right)\:. \\ $$…
Question Number 49237 by cesar.marval.larez@gmail.com last updated on 04/Dec/18 $${Find}\:{a}\:{polynomial}\:{f}\left({x}\right)\in{Q}\left[{x}\right]\:{such} \\ $$$${that}\:{its}\:{main}\:{coefficient}\:{is}\:\mathrm{1}\:{and}\: \\ $$$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\in{V}\left({f}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com