Question Number 63930 by Tawa1 last updated on 11/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63893 by mathmax by abdo last updated on 10/Jul/19 $$\left.\mathrm{1}\right)\:{simplify}\:{W}_{{n}} \left({z}\right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)….\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right)\:\left({z}\:{from}\:{C}\right) \\ $$$$\left.\mathrm{2}\right)\:{simplify}\:{P}_{{n}} \left(\theta\right)\:=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+{e}^{\mathrm{2}{i}\theta} \right)…..\left(\mathrm{1}+{e}^{{i}\mathrm{2}^{{n}} \theta} \right)\:{and}\:{sove} \\ $$$${P}_{{n}}…
Question Number 129394 by liberty last updated on 15/Jan/21 $$\begin{cases}{\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{17}=\mathrm{0}}\\{\mathrm{y}^{\mathrm{3}} −\mathrm{3y}^{\mathrm{2}} +\mathrm{5y}+\mathrm{11}=\mathrm{0}}\end{cases} \\ $$$$\:\mathrm{find}\:\mathrm{x}+\mathrm{y}\:. \\ $$ Answered by bemath last updated on 15/Jan/21…
Question Number 63858 by gunawan last updated on 10/Jul/19 $$\mathrm{if}\:\mathrm{a}_{\mathrm{1}} ,\:\mathrm{a}_{\mathrm{2}} ,\:\mathrm{a}_{\mathrm{3}} ,\:\mathrm{a}_{\mathrm{4}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{coefficient} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{four}\:\mathrm{four}\:\mathrm{consecutive} \\ $$$$\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} \\ $$$$\mathrm{then}\:\frac{\mathrm{a}_{\mathrm{1}} }{\mathrm{a}_{\mathrm{2}} +\mathrm{a}_{\mathrm{1}} }+\frac{\mathrm{a}_{\mathrm{3}} }{\mathrm{a}_{\mathrm{3}} +\mathrm{a}_{\mathrm{4}}…
Question Number 129364 by shaker last updated on 15/Jan/21 Answered by mindispower last updated on 15/Jan/21 $${let}\:\alpha={e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\alpha,\alpha^{\mathrm{3}} ,\alpha^{\mathrm{5}} ,\alpha^{\mathrm{7}} \:{are}\:{roots}\:{of}\:{x}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{2}}…
Question Number 129343 by Study last updated on 15/Jan/21 $${f}\left({x}+\mathrm{1}\right)={f}\left({x}−\mathrm{1}\right)={x}^{\mathrm{2}} \:\:\:\:\:\:\:\:{f}^{−\mathrm{1}} \left({x}\right)=?? \\ $$ Commented by bramlexs22 last updated on 15/Jan/21 $$\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{x}^{\mathrm{2}} \:? \\ $$…
Question Number 63803 by aliesam last updated on 09/Jul/19 Commented by Prithwish sen last updated on 09/Jul/19 $$\left(\mathrm{x}+\mathrm{iy}\right)^{\mathrm{3}} \:=\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{7}}\mathrm{i}\right)^{\mathrm{2}} }{\mathrm{1}−\sqrt{\mathrm{7}}\mathrm{i}}\:=\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{7}}\mathrm{i}\right)^{\mathrm{3}} }{\mathrm{8}} \\ $$$$\:\therefore\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{y}\:=\:\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$…
Question Number 63784 by MJS last updated on 09/Jul/19 $$\mathrm{question}\:\mathrm{63639}\:\mathrm{again} \\ $$$$\mathrm{prove}: \\ $$$$\forall{z}\in\mathbb{C}:\:\mid{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{2}} +{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{3}} +\mathrm{1}\mid\geqslant\mathrm{1} \\ $$ Commented by Prithwish sen last updated on…
Question Number 129320 by ajfour last updated on 14/Jan/21 Commented by gaferrafie last updated on 14/Jan/21 سلام چطور میشه تو خروجی آدرس سایت را حذف کرد Answered by ajfour last updated on 14/Jan/21 $${y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}}…
Question Number 129310 by I want to learn more last updated on 14/Jan/21 Commented by Tawa11 last updated on 23/Jul/21 $$\mathrm{like} \\ $$ Terms of…