Question 10 page 444
Two-Variable Inequalities
MAT 222 Week 2 Assignment
Instructor: Roosevelt Davis
1 October 2013
Two-Variable Inequalities
In this weeks’ assignment we were asked to solve problem 68 on page 539 (Dugopolski, 2012). After reading about inequalities I know now that they “consist of two simple inequalities joined with “and” or “or” “(Dugolpolski, 2012). When an ordered pair satisfies both of the inequalities then it is the “and” inequality. For the inequality to be an “or” it needs to satisfy one or both of the inequalities. Question 68 tells the number of refrigerators and TVs that will fit inside of an 18 wheeler truck. We are asked to write an inequality to describe the region of the graph that is shaded based upon the numbers provided. We are then asked to solve a multiple problems to determine how many refrigerators and TVs the 18 wheeler will fit at one time. First write the inequality to describe the shaded region of the graph. In the question the line is going from (0, 330) to (110, 0). The graph shows a solid line rather than a dashed line, meaning that the points on the line are part of the solution. I have Y as the TVs and X as the refrigerators. The formula used to write an inequality is: y=mx+b m=y2-y1/x2-x1 this tells me that my slope is: y=mx+b y=x+330 Slope intercept form
(1)y=(1)+330(1) multiply both sides by 1
Y=-3x+330
Y+3x=3x+3≤330 add both sides by 3
3x+y≤330 linear inequality for my line
The next question asked is will the truck hold 71 refrigerators and the 118 TVs. In order to answer the question I need to determine if points (71, 118) are within the shaded part of my graph.
X=71 y=118
3(71)+118≤330 multiply 3x71
213+118≤330 add 213+118
331≤330
The answer is no, the truck cannot hold 71 refrigerators or 118 TVs. The next question is asking if the 18 wheeler will hold 51 refrigerators and 176 TVs. In order to find the answer I need to determine if
MAT 222 Week 2 Assignment
Instructor: Roosevelt Davis
1 October 2013
Two-Variable Inequalities
In this weeks’ assignment we were asked to solve problem 68 on page 539 (Dugopolski, 2012). After reading about inequalities I know now that they “consist of two simple inequalities joined with “and” or “or” “(Dugolpolski, 2012). When an ordered pair satisfies both of the inequalities then it is the “and” inequality. For the inequality to be an “or” it needs to satisfy one or both of the inequalities. Question 68 tells the number of refrigerators and TVs that will fit inside of an 18 wheeler truck. We are asked to write an inequality to describe the region of the graph that is shaded based upon the numbers provided. We are then asked to solve a multiple problems to determine how many refrigerators and TVs the 18 wheeler will fit at one time. First write the inequality to describe the shaded region of the graph. In the question the line is going from (0, 330) to (110, 0). The graph shows a solid line rather than a dashed line, meaning that the points on the line are part of the solution. I have Y as the TVs and X as the refrigerators. The formula used to write an inequality is: y=mx+b m=y2-y1/x2-x1 this tells me that my slope is: y=mx+b y=x+330 Slope intercept form
(1)y=(1)+330(1) multiply both sides by 1
Y=-3x+330
Y+3x=3x+3≤330 add both sides by 3
3x+y≤330 linear inequality for my line
The next question asked is will the truck hold 71 refrigerators and the 118 TVs. In order to answer the question I need to determine if points (71, 118) are within the shaded part of my graph.
X=71 y=118
3(71)+118≤330 multiply 3x71
213+118≤330 add 213+118
331≤330
The answer is no, the truck cannot hold 71 refrigerators or 118 TVs. The next question is asking if the 18 wheeler will hold 51 refrigerators and 176 TVs. In order to find the answer I need to determine if