Question Number 28166 by abdo imad last updated on 21/Jan/18 $${let}\:{give}\:{w}=\:{e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:\:{and}\:\:{Z}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{w}^{{k}^{\mathrm{2}} } \:\:\:{find}\:\mid{Z}\mid^{\mathrm{2}} \:{in} \\ $$$${form}\:{of}\:{double}\:{sum}. \\ $$ Terms of Service Privacy…
Question Number 28165 by abdo imad last updated on 21/Jan/18 $${let}\:{give}\:{w}=\:{e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:{calculate}\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{1}+{w}^{{k}} \right)^{{n}} \:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28143 by ktomboy1992 last updated on 21/Jan/18 $$\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{3} \\ $$$$\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=? \\ $$ Answered by mrW2 last updated on 21/Jan/18 $$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{9}…
Question Number 28139 by math solver last updated on 21/Jan/18 Answered by mrW2 last updated on 21/Jan/18 $${A}=\frac{\pi}{\mathrm{2}}−{B} \\ $$$$\frac{{A}−{B}}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}−{B} \\ $$$$\mathrm{tan}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−{B}\right)=\frac{\mathrm{1}−\mathrm{tan}\:{B}}{\mathrm{1}+\mathrm{tan}\:{B}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{4}−\mathrm{4}\:\mathrm{tan}\:{B}=\mathrm{1}+\mathrm{tan}\:{B} \\…
Question Number 28124 by ajfour last updated on 20/Jan/18 $${f}\left({R}^{+} \rightarrow{R}\right)\:{is}\:{a}\:{differentiable} \\ $$$${function}\:{obeying} \\ $$$$\mathrm{2}{f}\left({x}\right)={f}\left({xy}\right)+{f}\left(\frac{{x}}{{y}}\right) \\ $$$${for}\:{all}\:{x},{y}\:\in\:{R}^{+} \:{and}\: \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0},\:{f}\:'\left(\mathrm{1}\right)=\mathrm{1}\:. \\ $$$${Find}\:{f}\left({x}\right).\:{More}\:{questions}\:{may} \\ $$$${follow}.. \\…
Question Number 159188 by cortano last updated on 14/Nov/21 $$\:{If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots}\:{of} \\ $$$${equation}\:\sqrt{\frac{{t}}{\mathrm{1}−{t}}}\:+\:\sqrt{\frac{\mathrm{1}−{t}}{{t}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}\:. \\ $$$${Find}\:\mathrm{2}\alpha+\mathrm{3}\beta\:. \\ $$ Commented by Rasheed.Sindhi last updated on 14/Nov/21 $${Do}\:{you}\:{mean}\:\:\sqrt{\frac{{t}}{\mathrm{1}−{t}}}\:+\:\sqrt{\frac{\mathrm{1}−{t}}{{t}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\…
Question Number 159179 by cortano last updated on 14/Nov/21 $$\:\:\mathcal{X}^{\mathrm{3}} −\mathrm{4}\mathcal{X}^{\mathrm{2}} −\mathrm{6}\mathcal{X}−\mathrm{24}\:=\:\mathrm{0} \\ $$ Answered by mr W last updated on 14/Nov/21 $${let}\:\mathcal{X}={x}+\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}^{\mathrm{3}}…
Question Number 159173 by HongKing last updated on 13/Nov/21 Answered by qaz last updated on 14/Nov/21 $$\Omega=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)}\right) \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\frac{\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)−\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)}{\mathrm{1}+\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}}…
Question Number 159170 by mnjuly1970 last updated on 13/Nov/21 Commented by cortano last updated on 13/Nov/21 $$\Rightarrow\sqrt[{\mathrm{3}}]{{x}}\:+\sqrt[{\mathrm{3}}]{{a}−{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{2}{a}−{x}}\:=\:\mathrm{0} \\ $$$$\Rightarrow{x}+{a}−{x}−\mathrm{2}{a}+{x}\:=\:−\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)} \\ $$$$\Rightarrow{x}−{a}\:=\:−\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}}…
Question Number 159152 by HongKing last updated on 13/Nov/21 Answered by mr W last updated on 13/Nov/21 $${let}\:{x}={n}+{f}\:{with}\:{n}\in\mathbb{Z},\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$${RHS}={n}+\mathrm{2022}={LHS}\geqslant\mathrm{2021}{n}\: \\ $$$$\Rightarrow{n}\leqslant\frac{\mathrm{2022}}{\mathrm{2020}}\:\Rightarrow{n}\leqslant\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${RHS}={n}+\mathrm{2022}={LHS}<\mathrm{2021}\left({n}+\mathrm{1}\right)\: \\…