9:null 17:T40d,undefined We need to find the area enclosed by four boundary lines. Let's rewrite each line in slope-intercept form: x = 6 is a vertical line y = 0 is the x-axis 3x - 2y = 0 gives us y = 3x2 x - 4y + 10 = 0 gives us y = x + 104

Find where each pair of lines intersects: Intersection of y = 0 and y = 3x2: Setting equal: 0 = 3x2 gives us x = 0 Point: (0, 0) Intersection of y = 0 and x = 6: Point: (6, 0) Intersection of x = 6 and y = x + 104: Substituting: y = 6 + 104 = 164 = 4 Point: (6, 4) Intersection of y = 3x2 and y = x + 104: Setting equal: 3x2 = x + 104 Cross multiply: 12x = 2x + 20 Solving: 10x = 20, so x = 2 Then y = 3(2)2 = 3 Point: (2, 3)

Split the quadrilateral into two triang<=s: Triang<= 1: (0,0), (2,3), (6,0) Triang<= 2: (2,3), (6,4), (6,0) Using the triang<= area formula: Area = 12|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| Triang<= 1 area: 12|0(3-0) + 2(0-0) + 6(0-3)| = 12|0 + 0 - 18| = 9 Triang<= 2 area: 12|2(4-0) + 6(0-3) + 6(3-4)| = 12|8 - 18 - 6| = 12 16 = 8 Total Area = 9 + 8 = 17 square units18:T40d,undefined We need to find the area enclosed by four boundary lines. Let's rewrite each line in slope-intercept form: x = 6 is a vertical line y = 0 is the x-axis 3x - 2y = 0 gives us y = 3x2 x - 4y + 10 = 0 gives us y = x + 104

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