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What is shown mendes phone number Find the point closest to the origin on the curve of intersection of plane 2y+4z=5 and cone z^2=4x^2+4y^2? Follow 2 answers Report Abuse Answers az-lender Best Answer: Consider the intersections of both these surfaces with the y-z plane. The intersection of the given plane with the y-z plane is a line whose slope in the y-z plane is -1/2. The intersection of the cone with the y-z plane is a pair of lines whose slopes are 2 and -2. Since the given plane is less steep than the sides of the cone, the curve of intersection is an ellipse rather than a hyperbola. Now consider the four vertices of the ellipse. Two of them are in the y-z plane, and the other two are either IN the x-z plane or quite close to it; I'll address that subject in a minute. The two vertices in the y-z plane are where 2y + 4 sqrt(4y^2) = 5, that is, where -6y = 5 OR 10y = 5: y = -5/6 and z = 5/3 OR y = 1/2 and z = 1. In 3-space, these points are P(0,1/2,1) and Q(0,-5/6,5/3). The distance from the origin to P is sqrt(1 The United Nations, the North Atlantic Treaty Organization (NATO), the Organization for Security and Cooperation in Europe, and the African Union are all examples of -. USFK Regulation on Marriage best body contouring treatment in dubai

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