The main cables of a suspension bridge uniformly distribute the weight of the bridge when in the form of a parabola. The main cables of a particular bridge are attached to towers that are 600 ft apart. The cables are attached to towers at a height of 110 ft above the road. The cables are 10 ft above ground at their lowest points. If vertical support cables are at 50 ft intervals along the level roadway, what are the lengths of these vertical cables?a. if you knew the equation of tha parabola that the cable formed, what would you do with the information vertical support cables are at 50 ft intervals to find the length of these support cables?b. what are the coordinates for the vertex? What are the coordinates for the tops of the main towers?c. write the equation of the parabola going through these points in standard form.e. what are the x-coordinates of the vertical support cables?f. how long is each vertical support cable?please help!
First, let's see if we can come up with the equation of the parabola, with (x,y)(0,0) at the base of the first tower. Let the equation of the parabola be y a*x^2 +b*x +c We can determine these three constants by solving the following system of equations. y(0)110 c 110 y(600) a*600^2 +b*600+c 110 y(300) a*300^2 + b*300+c 10 The solution tells us that y(x) (1/900)*x^2 - (2/3)*x + 110 Part A: Given this equation, we can determine the length of the support cables by substituting x 50,100,150,200,250,300,350,400,450,500,5. into the equation since y is the height of the cable above the bridge. Part B. The vertex is at the minimum point: (300,10) Part C. normal form y(x) (1/900)*x^2 - (2/3)*x + 110 completed square form y(x) (1/900)*(x-300)^2 + 10 Part D. (I assume you missed D accidently) (0,110) (600,110) Part E. and F. (50,79.4) (100,54.44) (150,35.0) (200,21.11) (250,12.766) (300,10) (350,12.79) (400,21.11) (450,35) (500,54.4) (550,79.4)
There is a motorcycle junk yard in Yale,mi they might have it
Just guessing but it appears you ride to 'look sweet'. The bike will handle like sh!t with that sized tire on the front.
? equation of each parabola ya*x^2+b, where b10 ft; also know that y(50/2) a*(50/2)^2 +10 110 ft, hence a0.16; thus y(x) 0.16*x^2+10; ? elementary length dL√((dx)^2 +(dy)^2) dx*√(1 +(y’)^2) dx*√(1 +(2*0.16*x)^2) dx*√(1 +(0.32*x)^2); hence ? Lx*√(1 + 0.1024*x^2) + 3.125*ln(0.32*x+√(1 +(0.32*x)^2)); for limits x0 until 25; L210.233 ft; ? I hope u cope with rest ur self; click me if stuck though!
