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In-an-arithmetic-progression-the-ninth-term-is-greater-than-the-second-term-and-the-sum-of-the-first-term-with-the-fifth-term-is-20-What-is-the-fifth-term-

Question Number 72835 by Maclaurin Stickker last updated on 03/Nov/19 $${In}\:{an}\:{arithmetic}\:{progression}\:{the} \\ $$$${ninth}\:{term}\:{is}\:{greater}\:{than}\:{the}\:{second} \\ $$$${term}\:{and}\:{the}\:{sum}\:{of}\:{the}\:{first}\:{term} \\ $$$${with}\:{the}\:{fifth}\:{term}\:{is}\:\mathrm{20}.\:{What}\:{is} \\ $$$${the}\:{fifth}\:{term}? \\ $$ Answered by $@ty@m123 last…

if-x-2-x-2-2-2-2-x-16-x-16-

Question Number 138367 by KwesiDerek last updated on 12/Apr/21 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{−\mathrm{2}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{16}} +\boldsymbol{\mathrm{x}}^{−\mathrm{16}} =? \\ $$ Commented by Kamel last updated on 12/Apr/21…

2-8-2-8-

Question Number 138350 by abenarhodym last updated on 12/Apr/21 $$\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{8}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{8}}}\right) \\ $$ Answered by Ñï= last updated on 12/Apr/21 $$\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{8}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{8}}}\right)=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{2}^{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}} \right)=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{2}^{\mathrm{2}}…

If-z-1-6-cos-pi-4-sin-pi-4-and-z-2-2-cos-pi-5-i-sin-pi-5-calculate-z-1-z-2-

Question Number 72787 by Maclaurin Stickker last updated on 02/Nov/19 $${If}\:{z}_{\mathrm{1}} =\mathrm{6}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)\:{and} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{5}}+\mathrm{i}×\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\right)\:{calculate}\:\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }. \\ $$ Commented by MJS last updated on…