Question Number 66906 by Kunal12588 last updated on 20/Aug/19 $${Intergrate}\:{I}=\int\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$ Commented by Prithwish sen last updated on 20/Aug/19 $$\frac{\mathrm{1}}{\mathrm{2i}}\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{it}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{it}^{\mathrm{2}} }\:\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left[−\mathrm{sin}^{−\mathrm{1}}…
Question Number 66909 by naka3546 last updated on 20/Aug/19 $$\int\:\:\frac{\sqrt{{x}^{\mathrm{2019}} +\mathrm{2019}}\:\:+\:\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2018}} +\mathrm{2018}}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{2017}} +\mathrm{2017}}\:\:+\:\:\sqrt[{\mathrm{5}}]{{x}^{\mathrm{2016}} +\mathrm{2016}}}\:\:{dx}\:\:=\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 66895 by Kunal12588 last updated on 20/Aug/19 $${x}+{y}+{z}=−{B} \\ $$$${xy}+{yz}+{zx}={C} \\ $$$${xyz}=−{D} \\ $$$${find}\:{x},{y}\:{and}\:{z}\:{in}\:{terms}\:{of}\:{B},{C}\:{and}\:{D} \\ $$ Answered by mr W last updated on…
Question Number 66890 by hmamarques1994@gmail.com last updated on 20/Aug/19 $$\: \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:+\:\boldsymbol{\mathrm{x}}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \:=\:\mathrm{20} \\ $$$$\: \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:=\:? \\ $$$$\: \\ $$ Answered by…
Question Number 132418 by MJS_new last updated on 14/Feb/21 $$\mathrm{to}\:\mathrm{Tinkutara} \\ $$$$\mathrm{strange}\:\mathrm{error}.\:\mathrm{try}\:\mathrm{to}\:\mathrm{edit}/\mathrm{comment}\:\mathrm{my} \\ $$$$\mathrm{answer}\:\mathrm{to}\:\mathrm{question}\:\mathrm{132402} \\ $$$$\mathrm{the}\:\mathrm{last}\:\mathrm{line}\:\mathrm{expands}\:\mathrm{in}\:\mathrm{height}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{on}\:\mathrm{my}\:\mathrm{device}… \\ $$ Commented by mr W last…
Question Number 132368 by otchereabdullai@gmail.com last updated on 13/Feb/21 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{log}_{\mathrm{2}} \mathrm{5}\:.\:\mathrm{log}_{\mathrm{4}} \mathrm{3}\:.\:\mathrm{log}_{\mathrm{25}} \mathrm{16}\:=\mathrm{log}_{\mathrm{2}} \mathrm{3} \\ $$ Answered by mr W last updated on 13/Feb/21 $${generally}\:\mathrm{log}_{{a}}…
Question Number 1255 by e.nolley@ieee.org last updated on 18/Jul/15 $$\left(\mathrm{X},\mathrm{F}\left(\mathrm{X}\right)\right),\:\mathrm{X}\in{C},\:\mathrm{F}\left(\mathrm{X}\right)\in{C},\:\:\mathrm{can}\:\mathrm{be}\:\: \\ $$$$\mathrm{plotted}\:\mathrm{in}\:\mathrm{R}^{\mathrm{2}} \:\left(\mathrm{Im}\:\mathrm{on}\:\mathrm{Y},\:\mathrm{Real}\:\mathrm{on}\:\mathrm{X}\right. \\ $$$$\left.\mathrm{axes}\right)\:\mathrm{as}\:\mathrm{a}\:\mathrm{directed}\:\mathrm{line}\:\mathrm{segment}, \\ $$$$\overset{} {\:}\overset{} {\left(\mathrm{X}\right)−−\gg\left(\mathrm{F}\left(\mathrm{X}\right)\right)}.\:\mathrm{This}\:\mathrm{has} \\ $$$$\mathrm{the}\:\mathrm{advantage}\:\mathrm{of}\:\mathrm{showing}\:\mathrm{vector}\: \\ $$$$\mathrm{ffields},\:\mathrm{fixed}\:\mathrm{points}\:\mathrm{and}\:\mathrm{bifurcations}. \\ $$$$…
Question Number 132303 by mohammad17 last updated on 13/Feb/21 Answered by EDWIN88 last updated on 13/Feb/21 $$\:\overset{\rightarrow} {\mathrm{a}}=\left(−\mathrm{1},\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:,\:\mathrm{1}\right)\: \\ $$$$\:\overset{\rightarrow} {\mathrm{b}}=\left(−\mathrm{3},\:−\mathrm{5ln}\:\mathrm{2},\:−\mathrm{4}\right) \\ $$$$\:\overset{\rightarrow} {\mathrm{c}}=\:\left(\mathrm{4},\:−\mathrm{5},\:\mathrm{5}\:\right)\: \\…
Question Number 132292 by Algoritm last updated on 13/Feb/21 Answered by mathmax by abdo last updated on 13/Feb/21 $$\frac{\mathrm{1}}{\left[\mathrm{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2x}\right]}=\left[\mathrm{x}\right]+\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{let}\:\left[\mathrm{x}\right]=\mathrm{n}\:\Rightarrow\mathrm{n}\leqslant\mathrm{x}<\mathrm{n}+\mathrm{1}\:\Rightarrow\mathrm{2n}\leqslant\mathrm{2x}<\mathrm{2n}+\mathrm{2} \\ $$$$\mathrm{if}\:\mathrm{2x}\in\left[\mathrm{2n},\mathrm{2n}+\mathrm{1}\left[\:\Rightarrow\mathrm{x}\in\left[\mathrm{n},\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\left[\:\Rightarrow\left[\mathrm{2x}\right]=\mathrm{2n}\right.\right.\right.\right. \\ $$$$\mathrm{e}\Rightarrow\frac{\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2n}}=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{3}}{\mathrm{2n}}=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{9}}{\mathrm{2n}}=\mathrm{3n}+\mathrm{1}\:\Rightarrow\mathrm{9}=\mathrm{2n}\left(\mathrm{3n}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{6n}^{\mathrm{2}}…
Question Number 132293 by Algoritm last updated on 13/Feb/21 Commented by mr W last updated on 13/Feb/21 $${x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$$${x}=\mathrm{0},\:{y}=\mathrm{1} \\ $$$${the}\:{proof}\:{you}\:{may}\:{ask}\:{for}\:{is}\:{thinking}. \\ $$ Answered…