Question Number 160416 by alephzero last updated on 29/Nov/21 $${guys}\:{help} \\ $$$${what}\:{does}\:\coprod\:{mean}? \\ $$ Commented by yeti123 last updated on 29/Nov/21 $$\mathrm{A}\:\cup\:\mathrm{B}\:=\:\mathrm{A}\:\sqcup\:\mathrm{B}\:=\:\mathrm{A}\:\coprod\:\mathrm{B} \\ $$$$\mathrm{maybe}…\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure} \\…
Question Number 94877 by mathocean1 last updated on 21/May/20 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\right]: \\ $$$$\mathrm{sin}\left(\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by mathocean1 last updated on 21/May/20 $$\mathrm{Please}\:\mathrm{sir},\:\mathrm{is}\:\mathrm{there}\:\mathrm{only}\:\mathrm{2}\:\mathrm{solutions}\: \\ $$$$\in\:\left[\mathrm{0};\mathrm{2}\pi\right]\:? \\…
Question Number 94862 by mathocean1 last updated on 21/May/20 $$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\mathrm{cos}^{\mathrm{6}} {x}+{sin}^{\mathrm{6}} {x}=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{5}+\mathrm{3}{cos}\left(\mathrm{4}{x}\right)\right) \\ $$$$\mathrm{By}\:\mathrm{using}\:\mathrm{the}\:\mathrm{formula}: \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \right) \\ $$ Commented…
Question Number 29324 by mondodotto@gmail.com last updated on 07/Feb/18 Answered by ajfour last updated on 07/Feb/18 $${e}^{{y}} \left(\frac{{dy}}{{dx}}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{tan}^{−\mathrm{1}} {y}+\left(\frac{{x}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{tan}^{−\mathrm{1}} {y}}{\left[{e}^{{y}}…
Question Number 29320 by Glorious Man last updated on 07/Feb/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{argument}\:\mathrm{of}\:\frac{−\mathrm{1}+\mathrm{2i}}{\mathrm{1}−\mathrm{3i}}\:. \\ $$ Commented by abdo imad last updated on 08/Feb/18 $$\mid−\mathrm{1}+\mathrm{2}{i}\mid=\sqrt{\mathrm{5}}\:{and}\:−\mathrm{1}+\mathrm{2}{i}=\sqrt{\mathrm{5}}\left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:+{i}\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)={r}\:{e}^{{i}\theta} \Rightarrow \\ $$$${r}=\sqrt{\mathrm{5}}\:{and}\:{cos}\theta=\:\frac{−\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:{and}\:{sin}\theta\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow{tan}\theta\:=−\mathrm{2}…
Question Number 94855 by mathocean1 last updated on 21/May/20 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{in}\:\mathbb{R} \\ $$$${x}\sqrt{{x}+\mathrm{3}}−\mathrm{4}\sqrt{{x}+\mathrm{3}}+\mathrm{2}{x}−\mathrm{8}=\mathrm{0} \\ $$ Answered by ElOuafi last updated on 21/May/20 $${let}\:;\:{x}\geqslant−\mathrm{3}\:{then}\: \\ $$$${x}\sqrt{{x}+\mathrm{3}}−\mathrm{4}\sqrt{{x}+\mathrm{3}}+\mathrm{2}{x}−\mathrm{8}=\mathrm{0} \\…
Question Number 94849 by mathocean1 last updated on 21/May/20 $$\mathrm{Calculate}\:\mathrm{limits}\:\mathrm{of}\:{f}\:\mathrm{at}\:+\infty;\:−\mathrm{1}\:\mathrm{and}\:\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}+\mathrm{3}−\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$ Answered by mathmax by abdo last updated on 21/May/20 $$\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{2x}+\mathrm{3}+\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}}…
Question Number 160371 by stelor last updated on 28/Nov/21 $${li}\underset{{t}\rightarrow{a}} {{m}}\frac{\left({alnt}−{tlna}\right)^{\mathrm{2}} \left({tlnt}−{alna}\right)}{\mathrm{2}\left({t}−{a}\right)\left({t}−{a}−{aln}\frac{{t}}{{a}}\right)} \\ $$$${please}\:{help}\:{me}. \\ $$ Answered by mathmax by abdo last updated on 28/Nov/21…
Question Number 29286 by mrW2 last updated on 06/Feb/18 $${To}\:{Developer}\:{Tinku}\:{Tara}: \\ $$$${Since}\:{the}\:{recent}\:{updates},\:{it}\:{is}\:{not}\:{more} \\ $$$${possible}\:{to}\:{change}\:{an}\:{existing}\:{image}\:{via} \\ $$$${Edit}\:{Post}''.\:{This}\:{was}\:{possible}\:{in}\:{earlier} \\ $$$${versions}\:{of}\:{the}\:{app}.\:{Can}\:{you}\:{please} \\ $$$${fix}\:{this}\:{problem}?\:{Thank}\:{you}! \\ $$ Commented by Rasheed.Sindhi…
Question Number 160344 by krupa last updated on 28/Nov/21 $${compare}\:{the}\:{following}\:{numbers}\:\frac{\mathrm{12}}{\mathrm{15}}\:,\frac{\mathrm{3}}{\mathrm{5}} \\ $$ Answered by FelipeLz last updated on 28/Nov/21 $$\frac{\mathrm{12}}{\mathrm{15}}\:=\:\frac{\mathrm{4}}{\mathrm{5}}\:>\:\frac{\mathrm{3}}{\mathrm{5}}\:\Rightarrow\:\frac{\mathrm{12}}{\mathrm{15}}\:>\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$ Answered by Rasheed.Sindhi…