Category: Algebra

Question Number 45512 by Tawa1 last updated on 13/Oct/18 $$\mathrm{Calculate}:\:\:\:\frac{\left(\mathrm{2}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{4}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{6}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{8}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{10}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{12}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{1}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{3}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{5}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{7}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{9}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{11}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$ Commented…

Question Number 111029 by Khanacademy last updated on 01/Sep/20 Answered by mindispower last updated on 02/Sep/20 $$=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right)^{\mathrm{2}} {sin}\left(\mathrm{2}\pi−\mathrm{2}{x}\right){sin}\left(\frac{\pi}{\mathrm{2}}{cos}\left(\pi−{x}\right)\right)}{\pi−\mathrm{2}{x}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi^{\mathrm{2}} −\mathrm{2}{x}\pi+{x}^{\mathrm{2}}…

Question Number 176555 by Shrinava last updated on 21/Sep/22 $$\mathrm{ABCD}−\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{M}\in\mathrm{Int}\left(\mathrm{ABCD}\right)\:,\:\mathrm{F}−\mathrm{area}\:,\:\mathrm{s}−\mathrm{semiperimetr} \\ $$$$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}−\mathrm{sides}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{MA}^{\mathrm{4}} }{\mathrm{b}}\:+\:\frac{\mathrm{MB}^{\mathrm{4}} }{\mathrm{c}^{\mathrm{4}} }\:+\:\frac{\mathrm{MC}^{\mathrm{4}} }{\mathrm{d}}\:+\:\frac{\mathrm{MD}^{\mathrm{4}} }{\mathrm{a}}\:\geqslant\:\frac{\mathrm{2F}^{\mathrm{2}} }{\mathrm{s}} \\ $$ Commented…

Question Number 45451 by Tinkutara last updated on 13/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18 $$\mid\left({x}+{iy}−\mathrm{3}−\mathrm{2}{i}\right)\mid=\mid\left({x}+{iy}\right){cos}\left(\frac{\pi}{\mathrm{4}}−{tan}^{−\mathrm{1}} \frac{{y}}{{x}}\right)\mid \\ $$$$\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }\:=\mid\left({x}+{iy}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}}…

Question Number 176513 by CrispyXYZ last updated on 20/Sep/22 $${z}\in\mathbb{C},\:\frac{{z}−\mathrm{3i}}{{z}+\mathrm{i}}\in\mathbb{R}^{−} \:,\:\:\frac{{z}−\mathrm{3}}{{z}+\mathrm{1}}\in\mathbb{I} \\ $$$$\mathrm{find}\:{z}. \\ $$ Answered by a.lgnaoui last updated on 20/Sep/22 $$\mathrm{Posons}\:\:\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\frac{\mathrm{z}−\mathrm{3i}}{\mathrm{z}+\mathrm{i}}=\frac{\mathrm{x}+\left(\mathrm{y}−\mathrm{3}\right)\mathrm{i}}{\mathrm{x}+\left(\mathrm{y}+\mathrm{1}\right)\mathrm{i}}=\frac{\left[\mathrm{x}+\left(\mathrm{y}−\mathrm{3}\right)\mathrm{i}\right]\left[\mathrm{x}−\left(\mathrm{y}+\mathrm{1}\right)\mathrm{i}\right]}{\mathrm{x}^{\mathrm{2}}…