college algebra worded problem

May 29th, 2015
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1.) The length of a rectangular field is 7/5 its width. If the perimeter of the field is 240 meters, what are the length and width of the field? Answer: Let L be the length and W be the width. L = (7/5)W Perimeter: 2L + 2W = 240, 2(7/5)W + 2W = 240

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1.) The length of a rectangular field is 7/5 its width. If the perimeter of the field is 240 meters, what are the length and width of the field?Answer:Let L be the length and W be the width. L = (7/5)WPerimeter: 2L + 2W = 240, 2(7/5)W + 2W = 240Solve the above equation to find: W = 50 m and L = 70 m.2.) The mass of the earth is about 230 000 000 000 kg. Express this mass in scientific notation.Answer: 2.3 x 10113.) For what value of the constant K does the equation 2x + Kx = 3 have one solution?Answer:Solve for x to obtain: x = 3/(2 + k)The given equation has one solution for all real values of k not equal to -2.4.) For what value of the constant K does the equation Kx2+ 2x = 1 have two real solutions?Answer:Rewrite the given quadratic equation in standard form: Kx2+ 2x - 1 = 0Discriminator = 4 - 4(K)(-1) = 4 + 4KFor the equation to have two real solutions, the discriminator has to be positive. Hence we need to solve the inequality 4 + 4K > 0.The solution set to the above inequality is given by: K > -1 for which the given equation has two real solutions.5.) For what value of the constant K does the system of equations 2x - y = 4 and 6x - 3y = 3K have an infinite number of solutions?Answer:For the system of equations 2x - y = 4 and 6x - 3y = 3K to have an infinite number of solutions, the two equations must be equivalent.If we multiply all terms of the first equation by 3, we obtain6x - 3y = 12for the two equat

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