Question Number 11351 by tawa last updated on 21/Mar/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{workdone}\:\mathrm{in}\:\mathrm{moving}\:\mathrm{a}\:\mathrm{paticle}\:\mathrm{once}\:\mathrm{around}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{C}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{xy}\:\mathrm{plane}.\:\mathrm{if}\:\mathrm{the}\:\mathrm{ellipse}\:\mathrm{has}\:\mathrm{centre}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{with}\:\mathrm{semi}\:\mathrm{major}\:\mathrm{and}\:\mathrm{semi}\:\mathrm{minor}\:\mathrm{axes}\: \\ $$$$\mathrm{4}\:\mathrm{and}\:\mathrm{3}\:\mathrm{respectively}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76855 by aliesam last updated on 31/Dec/19 Commented by aliesam last updated on 31/Dec/19 $${AB}={BC}=……={Fg}={GA} \\ $$$${prove}\:{that}\: \\ $$$${the}\:{area}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{7}{tan}\left(\frac{\pi}{\mathrm{14}}\right)\right) \\ $$ Commented…
Question Number 76556 by aliesam last updated on 28/Dec/19 Answered by benjo 1/2 santuyy last updated on 28/Dec/19 $${v}\:=\:\pi{R}^{\mathrm{2}} \:{h} \\ $$$${dv}/{dt}\:=\:{dv}/{dR}\:×\:{dR}/{dt} \\ $$$$=\:\mathrm{2}\pi{Rh}\:×\:{dR}/{dt} \\…
Question Number 76554 by aliesam last updated on 28/Dec/19 Answered by mr W last updated on 28/Dec/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76555 by aliesam last updated on 28/Dec/19 Answered by MJS last updated on 28/Dec/19 $$\mathrm{if}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{located}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same}\:\mathrm{lines} \\ $$$$\mathrm{blue}\:\mathrm{area}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$$$\mathrm{red}\:\mathrm{area}\:=\:\frac{\pi}{\mathrm{6}}{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{b}=\frac{\sqrt[{\mathrm{4}}]{\mathrm{27}}}{\:\sqrt{\mathrm{2}\pi}}{a} \\ $$$$\mathrm{s}=\frac{\pi}{\mathrm{3}}{b}=\frac{\sqrt[{\mathrm{4}}]{\mathrm{27}}\sqrt{\mathrm{2}\pi}}{\mathrm{6}}{a}…
Question Number 76524 by john santu last updated on 28/Dec/19 $${find}\:{vector}\:{unit}\:{perpendicular}\: \\ $$$${to}\:{vector}\:\overset{−} {{a}}=\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{and}\:\overset{−} {{b}}=\left(−\mathrm{1},\mathrm{0},\mathrm{2}\right) \\ $$ Answered by MJS last updated on 28/Dec/19 $$\begin{vmatrix}{\mathrm{1}}&{−\mathrm{1}}&{{u}_{{x}}…
Question Number 76475 by aliesam last updated on 27/Dec/19 Answered by mr W last updated on 27/Dec/19 $${y}=\mathrm{85}−\mathrm{85}\left(\frac{{x}−\mathrm{70}}{\mathrm{70}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${at}\:{x}=\mathrm{95}: \\ $$$${y}=\mathrm{85}−\mathrm{85}\left(\frac{\mathrm{95}−\mathrm{70}}{\mathrm{70}}\right)^{\mathrm{2}} =\mathrm{74}.\mathrm{16}\:{ft}…
Question Number 10862 by Saham last updated on 28/Feb/17 $$\mathrm{Given}\:\mathrm{that}:\:\:\hat {\mathrm{a}}\:=\:\mathrm{3i}\:+\:\mathrm{4j}\:+\:\mathrm{5k}\:\:\mathrm{and}\:\:\hat {\mathrm{b}}\:=\:\mathrm{2i}\:+\:\mathrm{2j}\:+\:\mathrm{3k}\:\:\mathrm{and}\:\:\:\hat {\mathrm{c}}\:=\:\mathrm{6i}\:−\:\mathrm{7j}\:−\:\mathrm{8k}. \\ $$$$\mathrm{find} \\ $$$$\mathrm{3}\hat {\mathrm{a}}\:+\:\mathrm{2}\hat {\mathrm{b}}\:−\:\mathrm{3}\hat {\mathrm{c}} \\ $$ Commented by Zainal…
Question Number 10597 by Saham last updated on 19/Feb/17 $$\mathrm{prove}\:\mathrm{that} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} −\:\mathrm{z}_{\mathrm{2}} \mid\:\leqslant\:\mid\mathrm{z}_{\mathrm{1}} \mid\:+\:\mid\mathrm{z}_{\mathrm{2}} \mid \\ $$ Answered by mrW1 last updated on 19/Feb/17…
Question Number 10562 by FilupS last updated on 18/Feb/17 $$\boldsymbol{{x}}=\begin{bmatrix}{{x}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} }\\{\:\vdots}\\{{x}_{{n}} }\end{bmatrix}\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\begin{bmatrix}{{y}_{\mathrm{1}} }\\{{y}_{\mathrm{2}} }\\{\:\vdots}\\{{y}_{{n}} }\end{bmatrix}\:\:\:\:\:\:\:\:\boldsymbol{{x}},\:\boldsymbol{{y}}\:\in\:\mathbb{R}^{{n}} \\ $$$$\: \\ $$$$\mathrm{1}.\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\:\boldsymbol{{x}},\:\mathrm{denoted}\:\mid\mid\boldsymbol{{x}}\mid\mid, \\ $$$$\:\:\:\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\sqrt{{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}}…