Question Number 66285 by Rio Michael last updated on 12/Aug/19 $${What}\:{is}\:{the}\:{difference}\:{between} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{2}^{−} } {{lim}}\:\:{and} \\ $$$$\underset{{x}\rightarrow\mathrm{2}^{+} } {{lim}} \\ $$ Answered by MJS last…
Question Number 131795 by Dwaipayan Shikari last updated on 08/Feb/21 $$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{17}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{26}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{37}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{50}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{65}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{82}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{101}^{\mathrm{3}} }+… \\ $$ Answered…
Question Number 702 by 123456 last updated on 01/Mar/15 $$\mathrm{tan}\:{x}+\mathrm{tan}\:\mathrm{2}{x}+\mathrm{tan}\:\mathrm{4}{x}+\centerdot\centerdot\centerdot+\mathrm{tan}\:\mathrm{2}^{{n}} {x}=? \\ $$ Commented by prakash jain last updated on 01/Mar/15 $$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{expression}? \\ $$ Commented…
Question Number 66228 by Rio Michael last updated on 11/Aug/19 $${prove}\:{that}\: \\ $$$$\int_{\mathrm{2}} ^{\mathrm{4}} \frac{\mathrm{6}{x}\:+\mathrm{1}}{\left(\mathrm{2}{x}−\mathrm{3}\right)\left(\mathrm{3}{x}−\mathrm{2}\right)}{dx}\:=\:{ln}\:\mathrm{10} \\ $$ Commented by Prithwish sen last updated on 11/Aug/19…
Question Number 66227 by Rio Michael last updated on 11/Aug/19 $${Using}\:{a}\:{good}\:{counter}\:{procedure},\:{prove}\:{that}\: \\ $$$$\:\:\:\frac{\partial{y}}{\partial{x}}\:=\:\underset{\partial{x}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\left(\partial\:+\:{x}\right)\:−{f}\left({x}\right)}{\partial{x}} \\ $$$${for}\:{a}\:{given}\:{function}\:\:{f}\left({x}\right)\:{in}\:{x}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 66226 by Rio Michael last updated on 11/Aug/19 $${the}\:{equation}\:\:{f}\left({x}\right)=\mathrm{0}\:{has}\:{real}\:{roots}\:{in}\: \\ $$$${the}\:{interval}\:\left({a},\:{b}\right)\:{if} \\ $$$${A}\:\:\:\:−{f}\left({a}\right)>\mathrm{0}\:\:{and}\:{f}\left({b}\right)\:>\mathrm{0} \\ $$$${B}\:\:\:{f}\left({a}\right)\:<\mathrm{0}\:{and}\:{f}\left({b}\right)\:<\mathrm{0} \\ $$$${C}\:\:−{f}\left({a}\right)\:>\mathrm{0}\:\:{and}\:{f}\left({b}\right)\:=\mathrm{0} \\ $$$${D}\:\:{f}\left({a}\right)\:>\mathrm{0}\:\:{and}\:{f}\left({b}\right)\:<\:\mathrm{0} \\ $$ Commented by…
Question Number 66225 by Rio Michael last updated on 11/Aug/19 $${Given}\:{that}\:\:\:\:\:{f}\left({x}\right)=\begin{cases}{−{x}\:+\:\mathrm{1},\:\:{x}\leqslant\:\mathrm{3}_{} }\\{{kx}\:−\mathrm{8},\:\:\:\:{x}\:>\mathrm{3}}\end{cases} \\ $$$${is}\:{continuous}\:{then}\:\:{f}\left(\mathrm{5}\right)\:=\: \\ $$$${A}\:\:\:\mathrm{2} \\ $$$${B}\:\:\:\mathrm{0} \\ $$$${C}\:\:−\mathrm{2} \\ $$$${D}\:\:−\mathrm{1} \\ $$$$ \\…
Question Number 66216 by Rio Michael last updated on 11/Aug/19 $$\mid{a}\:\mid\:=\:\mathrm{3}\:,\mid{b}\mid=\:\mathrm{5}\:,\:{a}.{b}\:=−\mathrm{14} \\ $$$$\:\:\mid{a}\:−\:{b}\mid\:=\:? \\ $$ Commented by Rasheed.Sindhi last updated on 11/Aug/19 $$\mid{a}\:\mid\:=\:\mathrm{3}\:,\mid{b}\mid=\:\mathrm{5}\:\Rightarrow{a}.{b}\:\neq−\mathrm{14} \\ $$$${a}.{b}=−\mathrm{15}\:\:{or}\:\:{a}.{b}=\mathrm{15}…
Question Number 658 by 123456 last updated on 22/Feb/15 $${proof}\:{that}\:{n}!>\left(\frac{{n}}{\mathrm{3}}\right)^{{n}} ,{n}\in\mathbb{N}^{\ast} \\ $$ Commented by 123456 last updated on 20/Feb/15 $${n}=\mathrm{1}\Rightarrow\mathrm{1}!=\mathrm{1}>\frac{\mathrm{1}}{\mathrm{3}}=\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{1}} \\ $$$${n}=\mathrm{1}\Rightarrow\mathrm{0}!=\mathrm{1}>\frac{\mathrm{1}}{\mathrm{3}}\approx\mathrm{0}.\mathrm{33} \\ $$$${n}=\mathrm{2}\Rightarrow\mathrm{2}!=\mathrm{2}>\frac{\mathrm{4}}{\mathrm{9}}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}}…
Question Number 66160 by Tanmay chaudhury last updated on 09/Aug/19 Terms of Service Privacy Policy Contact: info@tinkutara.com