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Author: Tinku Tara

Question-200934

Question Number 200934 by sonukgindia last updated on 26/Nov/23 Answered by Frix last updated on 27/Nov/23 $$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\left(\mathrm{tan}\:\pi{x}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} {dx}\:\overset{{t}=\left(\mathrm{cot}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} } {=} \\ $$$$=−\frac{\mathrm{5}}{\pi}\underset{\infty} {\overset{\mathrm{0}}…

A-ball-lies-on-the-function-z-xy-at-the-point-1-2-2-Find-the-point-in-the-xy-plane-where-the-ball-will-touch-it-Calculus-2-problem-

Question Number 200929 by Akira181 last updated on 26/Nov/23 $$\mathrm{A}\:\mathrm{ball}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{function}\:{z}={xy}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{the}\:{xy}−\mathrm{plane}\:\mathrm{where}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{will} \\ $$$$\mathrm{touch}\:\mathrm{it}. \\ $$$$\mathrm{Calculus}\:\mathrm{2}\:\mathrm{problem}. \\ $$ Commented by Akira181 last updated…

1-9-1-18-1-30-1-45-1-63-1-84-

Question Number 200931 by Spillover last updated on 26/Nov/23 $$\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{45}}+\frac{\mathrm{1}}{\mathrm{63}}+\frac{\mathrm{1}}{\mathrm{84}}+…\infty=? \\ $$$$ \\ $$ Answered by MM42 last updated on 26/Nov/23 $${s}_{{n}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)\:\: \\ $$$$\Rightarrow{s}_{{n}}…

Question-200925

Question Number 200925 by akolade last updated on 26/Nov/23 Answered by mr W last updated on 26/Nov/23 $$\boldsymbol{{p}}=\overset{\rightarrow} {{AB}}=\left(\mathrm{6},\:−\mathrm{14}\right) \\ $$$${C}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{2}.\mathrm{5},\:\mathrm{1}.\mathrm{5}\right) \\ $$$${D}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{2}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{4},\:−\mathrm{2}\right) \\ $$$${E}=\left(\mathrm{1},\mathrm{5}\right)+\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{6},\:−\mathrm{14}\right)=\left(\mathrm{5}.\mathrm{5},\:−\mathrm{5}.\mathrm{5}\right)…

Question-200915

Question Number 200915 by Rupesh123 last updated on 26/Nov/23 Answered by Frix last updated on 26/Nov/23 $$\mathrm{Assume} \\ $$$$\int\frac{\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} }{{x}^{\frac{\mathrm{8}}{\mathrm{5}}} }{dx}=\frac{{p}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{{q}}{\mathrm{5}}} }{{x}^{\frac{{r}}{\mathrm{5}}} }…