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Question-106572

Question Number 106572 by DeepakMahato last updated on 06/Aug/20 Commented by Rasheed.Sindhi last updated on 06/Aug/20 $$\:\:\:\:\underset{−} {\:\:\:\:\:\:\:{Still}\:{another}\:{way}\:\:\:\:\:\:\:\:} \\ $$$${Roots}\left({zeros}\right)\:{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${are}: \\ $$$$\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}}…

sen-7-

Question Number 106559 by MessiasAntonii last updated on 05/Aug/20 $${sen}\left(\mathrm{7}\right)= \\ $$ Commented by bobhans last updated on 06/Aug/20 $$\mathrm{cos}\:\left(\mathrm{14}°\right)=\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{7}°\right) \\ $$$$\mathrm{sin}\:\left(\mathrm{7}°\right)=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{14}°\right)}{\mathrm{2}}}\:\smile\overset{\bullet} {\bullet}\smile \\…

The-LCM-and-GCF-of-x-18-and-60-are-360-and-6-respectively-find-the-value-of-x-

Question Number 41003 by mondodotto@gmail.com last updated on 30/Jul/18 $$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{LCM}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{GCF}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{{x}},\mathrm{18}\:\boldsymbol{\mathrm{and}}\:\mathrm{60} \\ $$$$\boldsymbol{\mathrm{are}}\:\mathrm{360}\:\boldsymbol{\mathrm{and}}\:\mathrm{6}\:\boldsymbol{\mathrm{respectively}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{{x}} \\ $$ Answered by $@ty@m last updated on 31/Jul/18 $${Let}\:{x}=\mathrm{6}{k} \\…

Solve-1-cos-70-2-cos-10-

Question Number 106493 by mohammad17 last updated on 05/Aug/20 $${Solve}? \\ $$$$\left(\mathrm{1}\right){cos}\left(\mathrm{70}\right) \\ $$$$\left(\mathrm{2}\right){cos}\left(\mathrm{10}\right) \\ $$ Commented by malwaan last updated on 05/Aug/20 $${cos}\:\mathrm{10}\:=\:{cos}\:\mathrm{70}\:+\:{cos}\:\mathrm{50} \\…

Question-106488

Question Number 106488 by Algoritm last updated on 05/Aug/20 Commented by Dwaipayan Shikari last updated on 05/Aug/20 $$\:{Q}\mathrm{106379} \\ $$$$\frac{\mathrm{1}.\mathrm{2}^{\mathrm{2}} }{\mathrm{10}}+\frac{\mathrm{2}.\mathrm{3}^{\mathrm{2}} }{\mathrm{10}}+…..=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{10}^{{n}}…