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Author: Tinku Tara

A-triangle-is-formed-by-the-three-straight-line-y-m-1-x-a-m-1-y-m-2-x-a-m-2-y-m-3-x-a-m-3-prove-that-its-orthocenter-always-lies-on-the-line-x-a-0-

Question Number 76717 by peter frank last updated on 29/Dec/19 $${A}\:{triangle}\:{is}\:{formed}\:{by} \\ $$$${the}\:{three}\:{straight}\:{line} \\ $$$${y}={m}_{\mathrm{1}} {x}+\frac{{a}}{{m}_{\mathrm{1}} } \\ $$$${y}={m}_{\mathrm{2}} {x}+\frac{{a}}{{m}_{\mathrm{2}} } \\ $$$${y}={m}_{\mathrm{3}} {x}+\frac{{a}}{{m}_{\mathrm{3}} }…

Question-142248

Question Number 142248 by gsk2684 last updated on 28/May/21 Commented by gsk2684 last updated on 31/May/21 $$\mathrm{let}\:\mathrm{a}>\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{D}=\left\{\left(\mathrm{x},\mathrm{y}\right)/\mathrm{0}\leq\mathrm{x}\leq\mathrm{1},\mathrm{0}\leq\mathrm{y}\leq\mathrm{3x}\right\} \\ $$$$\mathrm{consider}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integral} \\ $$$$\int\underset{\mathrm{D}} {\int}\left(\mathrm{3x}+\mathrm{y}\right)\left(\mathrm{3x}−\mathrm{y}\right)^{\mathrm{a}} \mathrm{dxdy}…

If-u-arcsin-x-y-arctan-y-x-show-that-x-u-dx-y-u-dy-0-

Question Number 76715 by peter frank last updated on 29/Dec/19 $${If}\:{u}={arc}\mathrm{sin}\:\frac{{x}}{{y}}+{arc}\mathrm{tan}\:\frac{{y}}{{x}} \\ $$$${show}\:{that}\: \\ $$$${x}\frac{\partial{u}}{{dx}}+{y}\frac{\partial{u}}{{dy}}=\mathrm{0} \\ $$ Commented by abdomathmax last updated on 30/Dec/19 $$\frac{\partial{u}}{\partial{x}}\:=\frac{\mathrm{1}}{{y}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}}…

Question-76711

Question Number 76711 by aliesam last updated on 29/Dec/19 Answered by MJS last updated on 30/Dec/19 $${x}>\mathrm{0}\wedge{y}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} >\mathrm{0}\wedge{x}^{\mathrm{2}} +{y}^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{0}<{x}^{\mathrm{3}} +{y}^{\mathrm{3}} \leqslant{x}−{y}\:\Rightarrow\:\mathrm{0}<{x}−{y}\:\Leftrightarrow\:{y}<{x}…

Question-142246

Question Number 142246 by mathdanisur last updated on 28/May/21 Answered by mr W last updated on 28/May/21 $${V}=\frac{\mathrm{1}}{\mathrm{6}}\begin{pmatrix}{{a}}&{{a}}&{{a}}\\{{b}}&{\mathrm{1}}&{{b}}\\{{a}}&{{b}}&{{b}}\end{pmatrix} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[{a}\left({b}−{b}^{\mathrm{2}} \right)−{a}\left({b}^{\mathrm{2}} −{ab}\right)+{a}\left({b}^{\mathrm{2}} −{a}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}{a}\left({b}−{a}\right)\left(\mathrm{1}−{b}\right)…