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Question-73722

Question Number 73722 by azizullah last updated on 15/Nov/19 Commented by TawaTawa last updated on 15/Nov/19 $$\mathrm{Kamal}\:\:=\:\:\mathrm{x} \\ $$$$\mathrm{Ali}\:\:=\:\:\mathrm{x}\:+\:\mathrm{50} \\ $$$$\mathrm{Total}\:\:=\:\:\mathrm{150} \\ $$$$\therefore\:\:\:\:\:\mathrm{x}\:+\:\mathrm{x}\:+\:\mathrm{50}\:\:=\:\:\mathrm{150} \\ $$$$\therefore\:\:\:\:\:\mathrm{2x}\:+\:\mathrm{50}\:\:=\:\:\mathrm{150}…

cos-x-7-1-3-sin-x-6-dx-

Question Number 139259 by bobhans last updated on 25/Apr/21 $$\:\int\:\frac{\mathrm{cos}\:\mathrm{x}+\sqrt[{\mathrm{3}}]{\mathrm{7}}}{\mathrm{sin}\:\mathrm{x}+\sqrt{\mathrm{6}}}\:\mathrm{dx}\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Apr/21 $$\int\frac{{cosx}+\sqrt[{\mathrm{3}}]{\mathrm{7}}}{{sinx}+\sqrt{\mathrm{6}}}{dx}={log}\left({sinx}+\sqrt{\mathrm{6}}\right)+\sqrt[{\mathrm{3}}]{\mathrm{7}}\:\int\frac{\mathrm{1}}{{sinx}+\sqrt{\mathrm{6}}}{dx} \\ $$$$={log}\left({sinx}+\sqrt{\mathrm{6}}\right)+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{7}}\:\int\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\sqrt{\mathrm{6}}}.\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:{tan}\frac{{x}}{\mathrm{2}}={t}…

Given-a-convex-hexagon-ABCDEG-satisfy-AB-BC-CD-DE-EF-FA-Suppose-ACE-is-a-right-triangle-BC-BE-DE-DA-FA-FC-min-

Question Number 139254 by mathdanisur last updated on 25/Apr/21 $${Given}\:{a}\:{convex}\:{hexagon}\:{ABCDEG} \\ $$$${satisfy}:\:{AB}={BC},\:{CD}={DE},\:{EF}={FA}. \\ $$$${Suppose}\:\bigtriangleup{ACE}\:{is}\:{a}\:{right}\:{triangle}. \\ $$$$\left(\frac{{BC}}{{BE}}+\frac{{DE}}{{DA}}+\frac{{FA}}{{FC}}\right)_{\boldsymbol{{min}}} =? \\ $$ Answered by mr W last updated…

Evaluate-the-integral-R-3x-2-14xy-8y-2-dxdy-for-the-region-R-in-the-1st-quadrant-bounded-by-the-lines-y-3-2-x-1-y-3-2-x-3-y-1-4-x-and-y-1-4-x-1-

Question Number 73715 by Learner-123 last updated on 15/Nov/19 $${Evaluate}\:{the}\:{integral}\:: \\ $$$$\underset{\:\mathbb{R}} {\int}\int\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{14}{xy}+\mathrm{8}{y}^{\mathrm{2}} \right){dxdy}\:{for}\:{the}\:{region} \\ $$$$\mathbb{R}\:\mathrm{in}\:{the}\:\mathrm{1}{st}\:{quadrant}\:{bounded}\:{by}\:{the} \\ $$$${lines}\:{y}=\frac{−\mathrm{3}}{\mathrm{2}}{x}+\mathrm{1},{y}=\frac{−\mathrm{3}}{\mathrm{2}}{x}+\mathrm{3},{y}=−\frac{\mathrm{1}}{\mathrm{4}}{x} \\ $$$${and}\:{y}=−\frac{\mathrm{1}}{\mathrm{4}}{x}+\mathrm{1}\:. \\ $$ Commented by…

Question-73712

Question Number 73712 by FCB last updated on 15/Nov/19 Answered by MJS last updated on 15/Nov/19 $$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\:=−\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}=+\infty \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}}…