The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is
at (-3, -4). Point B is at (-4, 3). Point C is at (2, 2). Point D is at (1, -2). Point E is at (5, -4). Point F is at (3, 4). Part A: Using the graph above, create a system of inequalities that only contain points C and F in the overlapping shaded regions. Explain how the lines will be graphed and shaded on the coordinate grid above. Part B: Explain how to verify that the points C and F are solutions to the system of inequalities created in Part A. Part C: Natalie can only attend a school in her designated zone. Natalie's zone is defined by y < −2x + 2. Explain how you can identify the schools that Natalie is allowed to attend.
Part A; There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by <span>(-3, -4), (-4, 3), (1, -2) and (5, -4) 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 first quadrant of the coordinate system. The area above the x-axis and to the right of the y-axis is shaded.
Part 2: It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(2, 2) into the system we have: 2 > 0 2 > 0 as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.
Part C: Given that </span><span>Natalie
can only attend a school in her designated zone and that Natalie's zone is
defined by y < −2x + 2.
To identify the schools that
Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.
For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true
For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true
For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false
For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true
For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false
For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D. </span>
The approximate number of pink chips is 40. Here's why: what you need to do is set up the problem as shown below. 12 goes above 30 because you have to know what the total number of chips chosen is in order to relate it to 100 chips. From there, all you have to do is cross-multiply and divide (or my middle school teacher used to call it fish.... I'll explain why). First you are going to draw a line from the 100 to the 12 and multiply them. then you are going to curve the line that you will draw from the 12 to the 30 and divide the product of 12 and 100 by 30. Then you will draw a line from the 30 to the X meaning that the quotient that you find from the product of 120 and 100 divided by 30 equals X. Hope that helps.