Part A; There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve. An example of such system of equation is x < 0 y > 0 The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.
Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true. Substituting D(-4, 2) into the system we have: -4 < 0 2 > 0 as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities. Also, substituting E(-1, 5) into the system we have: -1 < 0 5 > 0 as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:Given that chicken can only be raised in the area defined by y > 3x - 4. To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area. For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true Therefore, the farms in which chicken can be raised are the farms at point A, D, E and F.
When converting 10cm to mm you get 100mm and when putting 100mm to 103mm its visible that the 103mm is larger, but by how much? The difference is 0.3cm and/or 3mm.