Results for this submission Entered Answer Preview Result 1.9 dy (1.9/3.5)*sqrt(1+[(dy/dx)^2]) 1+ correct 3.5 dx 3.5 0 (35/38)*[(e^[(19*x)/35])+(e^[(-19*x)/35])]-(5/38) 3.5 0 correct correct 35 38 19z -19z e 35 + e 100 5 incorrect 38 At least one of the answers above is NOT correct. (1 point) A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the density p = 1.9 lb/ft. The coordinate system is chosen so that the y-axis passes through the lowest point P and the x-axis is a = 3.5 ft below P. If the absolute value of tension force in the cable at point P is T = 3.5, then the initial value problem that governs the shape of the cable is (Use the notation dy/dx for the first derivative) dy dr 9(0) == (0) = 1.9/3.5 sqrt(1+(dy/dx)^2) 3.5 0 The solution of this problem is y = y(x), where y(x) 35/38 (e^(19x/35) +e^(-19x/35))-5/38