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Author: Tinku Tara

Question-68965

Question Number 68965 by anaplak last updated on 17/Sep/19 Commented by anaplak last updated on 17/Sep/19 $${sir}\:{please}\:{give}\:{me}\:{the}\:{solution}\:{showing}\:{by}\:{taking}\:{right}\:{hand}\:{limit}\: \\ $$$${i}\:{mean}\:{to}\:{say}\:{in}\:{detailed} \\ $$ Commented by kaivan.ahmadi last…

Question-68962

Question Number 68962 by ahmadshah last updated on 17/Sep/19 Commented by mr W last updated on 17/Sep/19 $${assume}\:{that}\:{the}\:{volume}\:{of}\:{cone}\:{is}\:\mathrm{1}. \\ $$$${x}={height}\:{of}\:{cone} \\ $$$$\mathrm{1}−\left(\frac{\mathrm{8}}{{x}}\right)^{\mathrm{3}} =\left(\frac{{x}−\mathrm{2}}{{x}}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}}…

Question-68952

Question Number 68952 by aliesam last updated on 17/Sep/19 Answered by mind is power last updated on 17/Sep/19 $$\Rightarrow{x}−\mathrm{1}>\mathrm{0}\Rightarrow{x}>\mathrm{1} \\ $$$$\Leftrightarrow\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\leqslant\frac{{x}+\mathrm{1}+\mathrm{3}}{{x}−\mathrm{1}}\Rightarrow\left({x}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\left({x}+\mathrm{4}\right)\left({x}+\mathrm{1}\right) \\ $$$$\Leftrightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\leqslant{x}^{\mathrm{2}}…

Question-134484

Question Number 134484 by mohammad17 last updated on 04/Mar/21 Commented by Dwaipayan Shikari last updated on 04/Mar/21 $${General}\:{term}=\frac{{n}}{{n}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{{n}}{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\mathrm{2}{n}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\left({Sequence}\:{converges}\:{to}\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${And}\:{the}\:{series}\:{is}\:{Divergent}\:\underset{{n}=\mathrm{1}}…

Question-68946

Question Number 68946 by ramirez105 last updated on 17/Sep/19 Commented by kaivan.ahmadi last updated on 17/Sep/19 $$\frac{\partial{u}}{\partial{x}}=\left(\mathrm{2}{xy}−{tany}\right)\Rightarrow{u}\left({x},{y}\right)={x}^{\mathrm{2}} {y}−{xtany}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\…