Question Number 60628 by cesar.marval.larez@gmail.com last updated on 23/May/19 $${Who}\:\:{know}\:{dynamics}\:{about}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60576 by hovea cw last updated on 22/May/19 $$\mathrm{two}\:\mathrm{faire}\:\mathrm{dices}\:\mathrm{are}\:\mathrm{tossed}\:\mathrm{together} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\mathrm{total}\:\mathrm{score}\:\mathrm{is}\:\mathrm{atmost}\:\mathrm{4} \\ $$$$ \\ $$ Commented by hovea cw last updated…
Question Number 125662 by physicstutes last updated on 12/Dec/20 $$\:\mathcal{I}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:\mathrm{tanh}^{−\mathrm{1}} \:\mathrm{2}{x}\:{dx}\:=\:??? \\ $$ Answered by mathmax by abdo last updated on 12/Dec/20 $$\mathrm{th}\left(\mathrm{x}\right)=\mathrm{y}\:\Rightarrow\mathrm{y}=\frac{\mathrm{sh}\left(\mathrm{x}\right)}{\mathrm{ch}\left(\mathrm{x}\right)}=\frac{\mathrm{e}^{\mathrm{x}}…
Question Number 190702 by AROUNAMoussa last updated on 12/Apr/23 $$\boldsymbol{\mathrm{Calcule}}:\:\:\boldsymbol{\mathrm{I}}=\int_{\boldsymbol{\mathrm{o}}} ^{\frac{\boldsymbol{\pi}}{\mathrm{3}}} \boldsymbol{\mathrm{x}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{sinx}}\right)^{\mathrm{2}} \boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190283 by mnjuly1970 last updated on 31/Mar/23 $$ \\ $$$$\:\:{calculate} \\ $$$$ \\ $$$$\:\:\:\:{laplace}\:\:{transform}\:\: \\ $$$$\:\:\:\:\:\:\:\mathscr{L}\:\:\:\left\{\:\:\frac{\:{e}^{\:−\frac{\mathrm{1}}{{x}}} }{\:\sqrt{{x}}\:}\:\:\right\}\:=\:?\: \\ $$$$ \\ $$ Terms of…
Question Number 190186 by mnjuly1970 last updated on 29/Mar/23 $$ \\ $$$$\:\:\:\:\begin{array}{|c|}{\:\:\:\:\:\:\:\:\:\mathrm{If}\:\:,\:\Omega=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\:\mathrm{4cos}^{\:\mathrm{2}} \:\left(\mathrm{4}{x}\right)}{\mathrm{3}\left(\mathrm{1}+\mathrm{sin}^{\:\mathrm{2}} \left(\mathrm{2}{x}\:\right)\right)}\mathrm{d}{x}=\:{a}\sqrt{\mathrm{2}}\:\:+\:{b}\:\:\:\:\:\:\:\:\:\:}\\\hline\end{array}\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\mathrm{of}\:\:\:,\:\:{a}−\:{b}=? \\ $$$$ \\ $$ Commented…
Question Number 124131 by bramlexs22 last updated on 01/Dec/20 Answered by som(math1967) last updated on 01/Dec/20 $$\mathrm{Equation}\:\mathrm{of}\:\mathrm{plane} \\ $$$$\begin{vmatrix}{\mathrm{x}−\mathrm{1}}&{\mathrm{y}−\mathrm{1}}&{\mathrm{z}−\mathrm{1}}\\{\mathrm{1}−\mathrm{1}}&{−\mathrm{1}−\mathrm{1}}&{\mathrm{1}−\mathrm{1}}\\{−\mathrm{7}−\mathrm{1}}&{\mathrm{3}−\mathrm{1}}&{−\mathrm{5}−\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\begin{vmatrix}{\mathrm{x}−\mathrm{1}}&{\mathrm{y}−\mathrm{1}}&{\mathrm{z}−\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{0}}\\{−\mathrm{8}}&{\mathrm{2}}&{−\mathrm{6}}\end{vmatrix}=\mathrm{0} \\ $$$$\mathrm{12}\left(\mathrm{x}−\mathrm{1}\right)−\left(\mathrm{y}−\mathrm{1}\right)×\mathrm{0}+\left(\mathrm{z}−\mathrm{1}\right)×\left(−\mathrm{16}\right)=\mathrm{0} \\ $$$$\mathrm{12x}−\mathrm{16z}+\mathrm{4}=\mathrm{0}…
Question Number 58569 by peter frank last updated on 25/Apr/19 $${A}\:{vector}\:{has}\:{magnitude} \\ $$$$\mathrm{6}\:{and}\:{bearing}\:\mathrm{100}°.{write} \\ $$$${it}\:{in}\:{the}\:{form}\:{ai}+{bj} \\ $$ Answered by tanmay last updated on 25/Apr/19 $$\overset{\rightarrow}…
Question Number 123876 by john_santu last updated on 29/Nov/20 $${Find}\:{the}\:{distance}\:{from}\:{the}\: \\ $$$${point}\:{S}\left(\mathrm{1},\mathrm{1},\mathrm{5}\right)\:{to}\:{the}\:{line}\: \\ $$$${L}\::\:\begin{cases}{{x}=\mathrm{1}+{t}}\\{{y}=\mathrm{3}−{t}\:}\\{{z}=\mathrm{2}{t}}\end{cases}. \\ $$ Answered by mr W last updated on 29/Nov/20 $${Method}\:{I}…
Question Number 58312 by pete last updated on 21/Apr/19 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angle}\:\theta\:\mathrm{between}\:\mathrm{two}\:\mathrm{unit} \\ $$$$\mathrm{vectors}\:\underset{} {\hat {\mathrm{a}}}\:\mathrm{and}\:\underset{} {\hat {\mathrm{b}}}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{cos}\theta=\underset{} {\hat {\mathrm{a}}}\bullet\underset{} {\hat {\mathrm{b}}}. \\ $$$$\mathrm{Hence},\:\mathrm{given}\:\mathrm{that}\:\underset{} {\hat {\mathrm{a}}}=\underset{} {\mathrm{i}cosA}+\underset{}…