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Category: Coordinate Geometry

show-that-the-plane-x-2y-3z-d-0-is-perpendiculr-to-each-of-the-plane-is-2x-5y-4z-1-0-and-4x-7y-3z-2-0-

Question Number 8150 by trapti rathaur@ gmail.com last updated on 02/Oct/16 $${show}\:{that}\:{the}\:{plane}\:\:\:{x}+\mathrm{2}{y}−\mathrm{3}{z}+{d}=\mathrm{0}\:\:{is}\:{perpendiculr}\:{to}\:{each}\:{of} \\ $$$${the}\:{plane}\:{is}\:\:\:\mathrm{2}{x}+\mathrm{5}{y}+\mathrm{4}{z}+\mathrm{1}=\mathrm{0}\:\:{and}\:\:\:\mathrm{4}{x}+\mathrm{7}{y}+\mathrm{3}{z}+\mathrm{2}=\mathrm{0}\:.\: \\ $$ Commented by 123456 last updated on 03/Oct/16 $$\alpha:{x}+\mathrm{2}{y}−\mathrm{3}{z}+{d}=\mathrm{0} \\…

Question-139106

Question Number 139106 by mr W last updated on 23/Apr/21 Commented by mr W last updated on 25/Apr/21 $${an}\:{uniform}\:{dense}\:{rope}\:{with}\:{length}\:\boldsymbol{{L}}\: \\ $$$${and}\:{mass}\:\boldsymbol{{m}}\:{is}\:{pulled}\:{by}\:{a}\:{truck}\:{with} \\ $$$${constant}\:{speed}.\:{on}\:{the}\:{end}\:{of}\:{rope}\:{a} \\ $$$${mass}\:\boldsymbol{{M}}\:{is}\:{connected}.\:{if}\:{the}\:{friction}…

Given-that-the-function-f-x-x-3-is-differentiable-in-the-interval-2-2-Use-the-mean-value-theorem-to-find-the-value-of-x-for-which-the-tangent-to-the-curve-is-parallel-to-the-chord-through-the-p

Question Number 73542 by Rio Michael last updated on 13/Nov/19 $${Given}\:{that}\:{the}\:{function}\:{f}\left({x}\right)=\:{x}^{\mathrm{3}} \:{is}\:{differentiable} \\ $$$${in}\:{the}\:{interval}\:\left(−\mathrm{2},\mathrm{2}\right),\:{Use}\:{the}\:{mean}\:{value}\:{theorem} \\ $$$${to}\:{find}\:{the}\:{value}\:{of}\:{x}\:{for}\:{which}\:{the}\:{tangent}\:{to}\:{the}\:{curve} \\ $$$${is}\:{parallel}\:{to}\:{the}\:{chord}\:{through}\:{the}\:{point}\:\left(−\mathrm{2},\mathrm{8}\right)\:{and}\:\left(\mathrm{2},\mathrm{8}\right) \\ $$ Terms of Service Privacy Policy…

Question-138548

Question Number 138548 by I want to learn more last updated on 14/Apr/21 Answered by nimnim last updated on 14/Apr/21 $${Let}\:{the}\:{centre}\:{of}\:{the}\:{circle}\:{be}\:{C}\left({h},{k}\right) \\ $$$${Since},\:{it}\:{touches}\:{the}\:{x}−{axis}\:{at}\:{P}\left(−\mathrm{3},\mathrm{0}\right), \\ $$$$\:{CP}\bot\:{x}−{axis}\:{and}\:{h}=−\mathrm{3}…

Question-7258

Question Number 7258 by Tawakalitu. last updated on 19/Aug/16 Commented by Yozzia last updated on 19/Aug/16 $${y}'=\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}\Rightarrow{y}=\int\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}{dx}. \\ $$$${Let}\:{u}=\mathrm{5}−{x}^{\mathrm{2}} \Rightarrow{du}=−\mathrm{2}{xdx}\Rightarrow{xdx}=\frac{−\mathrm{1}}{\mathrm{2}}{du}. \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{du}}{\:\sqrt{{u}}}=\frac{−\mathrm{1}}{\mathrm{2}}\int{u}^{−\mathrm{1}/\mathrm{2}} {du}=\frac{−\mathrm{1}}{\mathrm{2}}×\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}}…