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Question Number 211558 by mnjuly1970 last updated on 12/Sep/24 $$ \\ $$$$ \\ $$$$\:\:\:\:\:−−−−−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\Omega}=\:\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{n}}+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{n}}+\mathrm{1}}\:\right)=\:\boldsymbol{{a}\pi} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\boldsymbol{{a}}^{\mathrm{2}} =\:? \\ $$$$\:\:\:\:\:\:\:−−−−−−−−−−−− \\ $$…
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Question Number 210787 by mnjuly1970 last updated on 19/Aug/24 $$ \\ $$$$\:\begin{cases}{\:\:\mathrm{I}{f},\:\mathrm{D}\::\:{x}^{\mathrm{2}} \:+{y}^{\:\mathrm{2}} \:+\:{z}^{\:\mathrm{2}} \leqslant\mathrm{1}}\\{\:\Rightarrow\int\underset{\overset{} {\mathrm{D}}} {\int}\int\frac{\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{y}^{\:\mathrm{2}} }{{x}^{\mathrm{2}} \:+\:\mathrm{4}{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }\:{dxdydz}=?}\end{cases} \\ $$$$ \\…
Question Number 210566 by Erico last updated on 12/Aug/24 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{if}\:\left(\mathrm{x}\in\right]−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\left[\:\:\mathrm{y}\:=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{x}} \frac{\mathrm{dt}}{\mathrm{cos}\left(\mathrm{t}\right)}\:\right)\:\Rightarrow\:\:\left(\mathrm{y}\in\mathrm{IR}\:\:\:\mathrm{x}\:=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{y}} \frac{\mathrm{dt}}{\mathrm{cosh}\left(\mathrm{t}\right)}\:\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208931 by byaw last updated on 27/Jun/24 Answered by mr W last updated on 27/Jun/24 $$\left({a}\right) \\ $$$${R}={m}\left({g}+{a}\right)=\mathrm{0}.\mathrm{5}×\left(\mathrm{10}+\mathrm{2}\right)=\mathrm{6}\:{N} \\ $$$$ \\ $$$$\left({b}\right)\left({i}\right) \\…
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Question Number 208387 by Tawa11 last updated on 14/Jun/24 Find the value of the scalar for which the vector a = 3i + 2j is…
Question Number 207901 by Tawa11 last updated on 29/May/24 Answered by mr W last updated on 30/May/24 $$\boldsymbol{{v}}=\frac{{d}\boldsymbol{{r}}}{{dt}}={e}^{{t}} \left(\mathrm{cos}\:{t}−\mathrm{sin}\:{t}\right)\:\boldsymbol{{i}}+{e}^{{t}} \left(\mathrm{sin}\:{t}+\:\mathrm{cos}\:{t}\right)\:\boldsymbol{{j}} \\ $$$$\boldsymbol{{a}}=\frac{{d}\boldsymbol{{v}}}{{dt}}=−\mathrm{2}{e}^{{t}} \:\mathrm{sin}\:{t}\:\boldsymbol{{i}}+\mathrm{2}{e}^{{t}} \:\mathrm{cos}\:{t}\:\boldsymbol{{j}} \\…