Determine the solution to 3-8-2x+1, and indicate the restrictions on the variable. Ruth solved the equation...

  

   

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Determine the solution to 3-8-2x+1, and indicate the restrictions on the variable. Ruth solved the equation 8=x-x-6 as follows. 8 = x-x-6 8-x = -x-6 x-8= x-6 x2-16x+64 = x-6 x-17x+70 = 0 (x-7)(x-10) = 0 x = 7,10 The extraneous root(s) introduced by Ruth is/are The time, t seconds, it takes a pendulum to complete one back and forth swing can be determined by the equation t = 0.32T, where is the length of the pendulum in inches. Determine the length of the pendulum if it takes 1.2 s to make a complete back and forth swing. The time, t seconds, it takes a pendulum to complete one back and forth swing can be determined by the equation t=0.32/1, where I is the length of the pendulum in inches. A clock manufacturer has an adjustable length pendulum with the shortest length being 7 in and the longest length being 11.5 in. Determine the range of swinging times for this pendulum. The function y = f(x) has a range of (-4,9]. The minimum value of y = T) is If 5f(1) = 20, then the point (1,y) on the graph of y = F) has a y-value of Answer: Determine the solution to 3-8-2x+1, and indicate the restrictions on the variable. Ruth solved the equation 8=x-x-6 as follows. 8 = x-x-6 8-x = -x-6 x-8= x-6 x2-16x+64 = x-6 x-17x+70 = 0 (x-7)(x-10) = 0 x = 7,10 The extraneous root(s) introduced by Ruth is/are The time, t seconds, it takes a pendulum to complete one back and forth swing can be determined by the equation t = 0.32T, where is the length of the pendulum in inches. Determine the length of the pendulum if it takes 1.2 s to make a complete back and forth swing. The time, t seconds, it takes a pendulum to complete one back and forth swing can be determined by the equation t=0.32/1, where I is the length of the pendulum in inches. A clock manufacturer has an adjustable length pendulum with the shortest length being 7 in and the longest length being 11.5 in. Determine the range of swinging times for this pendulum. The function y = f(x) has a range of (-4,9]. The minimum value of y = T) is If 5f(1) = 20, then the point (1,y) on the graph of y = F) has a y-value of Answer: