Question Number 37122 by JOHNMASANJA last updated on 09/Jun/18 $${draw}\:\Delta{A}\overset{} {{B}C}\:{and}\:{its}\:{image}\:\:\Delta{A}'{B}^{'} {C}^{'} \: \\ $$$${after}\:{a}\:{reflection}\:{in}\:{line}\:{y}={x} \\ $$$${if}\:{A}\left(\mathrm{0},\mathrm{3}\right),{B}\left(\mathrm{3},\mathrm{0}\right),{C}\left(\mathrm{3},\mathrm{2}\right).{what}\:{is}\:{the} \\ $$$${line}\:{of}\:{symmetry}\:\:{of}\:{the}\:{two}\:{figures}? \\ $$ Terms of Service Privacy…
Question Number 102651 by Dwaipayan Shikari last updated on 10/Jul/20 $$\mathrm{12}+\mathrm{14}+\mathrm{24}+\mathrm{58}+\mathrm{164}….\mathrm{upto}\:\mathrm{nth}\:\mathrm{terms} \\ $$ Commented by prakash jain last updated on 10/Jul/20 $$\mathrm{see}\:\mathrm{arguments}\:\mathrm{in}\:\mathrm{q102652} \\ $$ Terms…
Question Number 168185 by Mastermind last updated on 05/Apr/22 $${Separate}\:{cos}^{−\mathrm{1}} {e}^{{i}\theta} \:{into}\: \\ $$$${real}\:{and}\:{imaginary}\:{parts}. \\ $$$$ \\ $$$${Mastermind} \\ $$ Terms of Service Privacy Policy…
Question Number 168187 by Mastermind last updated on 05/Apr/22 $${Prove}\:{that}\::\: \\ $$$${sinh}^{−\mathrm{1}} {tan}\theta\:=\:{log}\:{tan}\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by peter frank last updated…
Question Number 102635 by ketto255 last updated on 10/Jul/20 Answered by bobhans last updated on 10/Jul/20 $$\mathrm{sin}\:\left(\mathrm{90}^{{o}} −\theta\right)\:=\:\mathrm{cos}\:\theta\:=\pm\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}\:= \\ $$$$\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{25}}}\:=\:\pm\:\frac{\sqrt{\mathrm{23}}}{\mathrm{5}} \\ $$ Answered by…
Question Number 37089 by behi83417@gmail.com last updated on 08/Jun/18 Commented by math khazana by abdo last updated on 09/Jun/18 $$\left.\mathrm{1}\right)\:{f}\:{id}\:{defined}\:{on}\:\left[\mathrm{0},+\infty\left[\right.\right. \\ $$$${f}^{'} \left({x}\right)=\:\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{x}}}\left(\mathrm{1}+\sqrt{{x}}\right)\:−\sqrt{\mathrm{1}+{x}}\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} } \\…
Question Number 102627 by Rio Michael last updated on 10/Jul/20 $$\mathrm{show}\:\mathrm{that}:\:\mathrm{cos}\theta\:+\:\mathrm{cos2}\theta\:+\:….\mathrm{cos}\:{n}\theta=\:\frac{\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}\:+\mathrm{1}\right)\theta\:\mathrm{sin}\frac{\mathrm{1}}{\mathrm{2}}{n}\theta}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{n}\theta} \\ $$$$\mathrm{Show}\:\mathrm{that}:\:\mathrm{sin}\:\theta\:+\:\mathrm{sin}\:\mathrm{2}\theta\:+\:….+\:\mathrm{sin}\:{n}\theta\:=\:\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}\:+\:\mathrm{1}\right)\theta\:\mathrm{sin}\frac{\mathrm{1}}{\mathrm{2}}{n}\theta}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{n}\theta} \\ $$$$\mathrm{where}\:\theta\:\in\:\mathbb{R}\:\mathrm{and}\:\theta\:\neq\mathrm{2}\pi{k}\:,\:{k}\:\in\mathbb{Z} \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 37084 by Rio Mike last updated on 08/Jun/18 $$\:\:{solve}\:{using}\:{matrix}\:{method} \\ $$$$\:\:\:\:\:\:{x}\:−\:{y}=\:\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{2}{x}\:−\:\mathrm{3}{y}=\:\mathrm{5} \\ $$ Commented by math khazana by abdo last updated…
Question Number 37081 by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18 Commented by candre last updated on 09/Jun/18 $$\mathrm{techinaly}\:\mathrm{newton}\:\mathrm{Segond}\:\mathrm{law}\:\mathrm{is}\:\mathrm{write}\:\mathrm{as} \\ $$$${F}=\frac{\partial{p}}{\partial{t}} \\ $$$$\mathrm{but}\:\mathrm{since}\:\mathrm{most}\:\mathrm{case}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{you}\:\mathrm{can}\:\mathrm{simplify}\:\mathrm{to} \\ $$$${F}=\frac{\partial}{\partial{t}}\left({mv}\right)={m}\frac{\partial{v}}{\partial{t}}={ma} \\ $$…
Question Number 102598 by dw last updated on 10/Jul/20 Answered by mr W last updated on 10/Jul/20 $${z}=\mathrm{2}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$$\mid{z}−{i}\mid=\sqrt{\mathrm{4cos}^{\mathrm{2}} \:\theta+\left(\mathrm{2}\:\mathrm{sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{sin}\:\theta} \\ $$$$\mid{z}+{i}\mid=\sqrt{\mathrm{4cos}^{\mathrm{2}} \:\theta+\left(\mathrm{2}\:\mathrm{sin}\:\theta+\mathrm{1}\right)^{\mathrm{2}}…