Question Number 5056 by Rasheed Soomro last updated on 06/Apr/16 $${x}\neq{y}\:\wedge\:{y}\neq\mathrm{0}\:\wedge\:\mathrm{log}_{\frac{\mathrm{x}}{\mathrm{y}}} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{3}} }\right)=? \\ $$ Commented by Yozzii last updated on 06/Apr/16 $${log}_{{x}/{y}} {x}^{\mathrm{2}}…
Question Number 70590 by behi83417@gmail.com last updated on 05/Oct/19 Commented by behi83417@gmail.com last updated on 05/Oct/19 $$\mathrm{convex}\:\mathrm{ABCD},\mathrm{is}\:\mathrm{given}\:\mathrm{with}: \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{ABCD}} =\boldsymbol{\mathrm{p}}\:\:\:,\:\:\:\:\:\mathrm{AB}+\mathrm{BD}+\mathrm{DC}=\boldsymbol{\mathrm{q}}\:\:\:. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{maximum}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{for}}:\:\boldsymbol{\mathrm{AC}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}} \\ $$$$\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{p}}\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{q}}. \\…
Question Number 136124 by bramlexs22 last updated on 18/Mar/21 $$ \\ $$Given a quadratic function f(x) =3-4k-(k+3) x-x^2, where k is a constant, is always…
Question Number 136121 by bramlexs22 last updated on 18/Mar/21 $$\mathrm{3}\:\mathrm{cos}\:{x}\:=\:\mathrm{13}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\:+\:\mathrm{17}\:\mathrm{cos}\:\left(\frac{{x}}{\mathrm{3}}\right) \\ $$ Answered by EDWIN88 last updated on 19/Mar/21 $$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{3}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{3cos}\:\mathrm{3t}\:=\:\mathrm{13sin}\:\mathrm{2t}+\mathrm{17cos}\:\mathrm{t} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{4cos}\:^{\mathrm{3}} \mathrm{t}−\mathrm{3cos}\:\mathrm{t}\right)=\mathrm{26sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:+\mathrm{17cos}\:\mathrm{t} \\ $$$$\mathrm{12cos}\:^{\mathrm{3}}…
Question Number 5051 by nhoduy last updated on 05/Apr/16 $$\mathrm{7}{x}−\mathrm{4}=\mathrm{0} \\ $$ Commented by nguyentai last updated on 05/Apr/16 $$\Leftrightarrow{x}=\mathrm{1} \\ $$ Answered by FilupSmith…
Question Number 70587 by behi83417@gmail.com last updated on 05/Oct/19 Commented by behi83417@gmail.com last updated on 05/Oct/19 $$\mathrm{ABCD},\mathrm{is}\:\mathrm{given}. \\ $$$$\mathrm{AB}=\boldsymbol{\mathrm{a}},\mathrm{BC}=\boldsymbol{\mathrm{b}},\mathrm{CD}=\boldsymbol{\mathrm{c}},\mathrm{DA}=\boldsymbol{\mathrm{d}},\mathrm{AC}=\boldsymbol{\mathrm{m}},\mathrm{BD}=\boldsymbol{\mathrm{n}}. \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}: \\ $$$$\:\:\:\:\:\left(\boldsymbol{\mathrm{mn}}\right)^{\mathrm{2}} =\left(\boldsymbol{\mathrm{ac}}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{bd}}\right)^{\mathrm{2}}…
Question Number 136123 by bramlexs22 last updated on 18/Mar/21 $$\frac{\left\{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{12}−{x}\right)^{\mathrm{2}} }\:+\sqrt[{\mathrm{3}}]{\left(\mathrm{12}−{x}\right)\left({x}−\mathrm{3}\right)}\:+\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:\right\}^{\mathrm{2}} }{\:\sqrt[{\mathrm{3}}]{\mathrm{12}−{x}}\:+\sqrt[{\mathrm{3}}]{{x}−\mathrm{3}}}\:=\:\frac{\mathrm{49}}{\mathrm{3}} \\ $$ Answered by EDWIN88 last updated on 19/Mar/21 $$\mathrm{let}\:\sqrt[{\mathrm{3}}]{\left(\mathrm{12}−\mathrm{x}\right)}\:=\:\mathrm{p}\:\&\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:=\:\mathrm{q} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{p}^{\mathrm{2}}…
Question Number 136122 by bramlexs22 last updated on 18/Mar/21 $$\mathrm{2}\left(\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}\right)\:+\mathrm{sin}\:\mathrm{2}{x}\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)=\mathrm{2sin}\:{x}\: \\ $$ Answered by EDWIN88 last updated on 18/Mar/21 $$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)+\mathrm{2sin}\:\mathrm{xcos}\:\mathrm{x}\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{x}\right)−\mathrm{2sin}\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)+\mathrm{2sin}\:\mathrm{x}\left[\:\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\:\right]=\mathrm{0}…
Question Number 5048 by LMTV last updated on 05/Apr/16 $$\int\left({x}\sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} {dx}=? \\ $$$$\int{y}\left(\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{2}} {dy}=? \\ $$$$\int\frac{\mathrm{1}}{\mathrm{4sec}\:{x}}{dx}=? \\ $$$${y}=\mathrm{sec}^{\mathrm{3}} \:\left(\mathrm{2}{x}+\mathrm{5}\right) \\ $$$${y}'=? \\ $$ Commented by…
Question Number 70582 by Raphael last updated on 05/Oct/19 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{y}=\frac{\mathrm{4}}{\:\sqrt{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)}}\:\mathrm{show}\:\mathrm{that}\:\mathrm{2}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{3x}^{\mathrm{2}} \mathrm{y}. \\ $$ Commented by kaivan.ahmadi last updated on 05/Oct/19 $$\mathrm{2}\left({x}^{\mathrm{3}} +\mathrm{1}\right)×\frac{−\frac{\mathrm{4}×\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}\sqrt{{x}^{\mathrm{3}}…