Question Number 176554 by mnjuly1970 last updated on 21/Sep/22 $$ \\ $$$$\:\:\:\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:\:=\varphi\:\:\left({Golden}\:{ratio}\right) \\ $$$$\:\:\:{then} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\:\mathrm{2000}} \:+\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2000}} }\:=\:? \\ $$$$ \\ $$ Answered by Peace…
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 111008 by Khanacademy last updated on 01/Sep/20 $$\frac{\sqrt{\boldsymbol{{x}}+\mathrm{1}}}{\boldsymbol{{y}}+\mathrm{2}}\:+\:\frac{\sqrt{\boldsymbol{{y}}+\mathrm{2}}}{\boldsymbol{{x}}+\mathrm{1}}\:=\mathrm{1}\:\:\:\:\:\:=>\:\:\boldsymbol{{x}}=? \\ $$ Commented by Rasheed.Sindhi last updated on 01/Sep/20 $${x}\:{in}\:{terms}\:{of}\:{y}? \\ $$ Commented by Khanacademy…
Question Number 45451 by Tinkutara last updated on 13/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18 $$\mid\left({x}+{iy}−\mathrm{3}−\mathrm{2}{i}\right)\mid=\mid\left({x}+{iy}\right){cos}\left(\frac{\pi}{\mathrm{4}}−{tan}^{−\mathrm{1}} \frac{{y}}{{x}}\right)\mid \\ $$$$\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }\:=\mid\left({x}+{iy}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}}…
Question Number 45450 by Tinkutara last updated on 13/Oct/18 Commented by Meritguide1234 last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18 $${a}_{\mathrm{1}} ^{\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 $$\mathrm{Posons}\:\:\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\frac{\mathrm{z}−\mathrm{3i}}{\mathrm{z}+\mathrm{i}}=\frac{\mathrm{x}+\left(\mathrm{y}−\mathrm{3}\right)\mathrm{i}}{\mathrm{x}+\left(\mathrm{y}+\mathrm{1}\right)\mathrm{i}}=\frac{\left[\mathrm{x}+\left(\mathrm{y}−\mathrm{3}\right)\mathrm{i}\right]\left[\mathrm{x}−\left(\mathrm{y}+\mathrm{1}\right)\mathrm{i}\right]}{\mathrm{x}^{\mathrm{2}}…
Question Number 176514 by CrispyXYZ last updated on 20/Sep/22 $$\bigtriangleup{ABC},\:\mathrm{sin}{A}=\mathrm{cos}{B}=\mathrm{tan}{C} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}^{\mathrm{3}} {A}+\mathrm{cos}^{\mathrm{2}} {A}−\mathrm{cos}{A}. \\ $$ Answered by mr W last updated on 21/Sep/22 $$\mathrm{sin}\:{A}=\mathrm{cos}\:{B}\:\Rightarrow{B}={A}−\frac{\pi}{\mathrm{2}}…
Question Number 45422 by Tawa1 last updated on 12/Oct/18 $$\mathrm{Three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}\:\mathrm{are}\:\mathrm{the}\:\mathrm{3rd},\:\mathrm{5th}\:\mathrm{and}\:\mathrm{8th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18 $${A}+\left(\mathrm{3}−\mathrm{1}\right){D}={a} \\ $$$${A}+\left(\mathrm{5}−\mathrm{1}\right){D}={ar} \\…
Question Number 176485 by Shrinava last updated on 20/Sep/22 Commented by mr W last updated on 20/Sep/22 $${x}=\mathrm{4}\:{is}\:{the}\:{solution}. \\ $$ Answered by a.lgnaoui last updated…
Question Number 176472 by mr W last updated on 19/Sep/22 $${solve}\:{for}\:{x},{y},{z}\:{with} \\ $$$${x}+{y}+{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{5} \\ $$ Answered by behi834171 last…